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Netanyahu 'caught on tape negotiating secret deal with press baron'
The political scandal swirling around Benjamin Netanyahu intensified on Sunday after it was reported he had been caught on tape offering financial incentives to a newspaper publisher in return for better coverage.
The Israeli prime minister spent eight hours last week being questioned by police about two separate allegations of criminal wrongdoing.
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Haaretz reported that Mr Netanyahu was caught on tape in a secret conversation with Mr Mozes last year.
The prime minister allegedly offered to scale back the circulation of a free newspaper owned by one of his political allies, which had been undercutting the sales of Mr Mozes’s Yedioth Ahronoth.
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Yedioth Ahronoth has been deeply critical of Mr Netanyahu in the past. The newspaper did not respond to the allegations that its publisher was involved in talks with the prime minister.
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David Bitan, an Israeli MP and close associate of Mr Netanyahu, said the Israeli leader would not resign even if was charged with a crime.
Mr Netanyahu’s predecessor, Ehud Olmert, is in prison after being convicted of corruption. | 0.003742 |
\begin{document}
\date{}
\def\a{\alpha}
\def\la{\lambda}
\def\si{\sigma}
\def\z{\zeta}
\def\phi{\varphi}
\def\ka{\kappa}
\def\pd{\partial}
\def\d{{\rm d}}
\def\~#1{\widetilde #1}
\def\.#1{\dot #1}
\def\^#1{\widehat #1}
\def \q{\quad}
\def\={\, =\, }
\def\beq{\begin{equation}}
\def\eeq{\end{equation}}
\def \sy {symmetry}
\def \sys {symmetries}
\def \eq {equation}
\def\ni{\noindent}
\newtheorem{theorem}{Theorem}
\newtheorem{Proposition}{Proposition}
\title{On the connections between symmetries and conservation rules
of dynamical systems}
\author{
Giampaolo Cicogna\thanks{Email: [email protected]}
\\~\\
Dipartimento di Fisica ``E.Fermi'' dell'Universit\`a di Pisa\\
and Istituto Nazionale di Fisica Nucleare, Sez. di Pisa \\~\\
Largo B. Pontecorvo 3, Ed. B-C, I-56127, Pisa, Italy }
\maketitle
\begin{abstract}
The strict connection between Lie point-symmetries of a
dynamical system and its constants of motion is discussed and emphasized, through
old and new results. It is shown in particular how the knowledge of a
symmetry of a dynamical system can allow to obtain
conserved quantities which are invariant under the symmetry.
In the case of Hamiltonian dynamical systems it is shown that, if the
system admits a symmetry of ``weaker''
type (specifically, a $\lambda$ or a $\Lambda$-symmetry), then the generating function
of the symmetry is not a conserved quantity, but the deviation from the exact
conservation is ``controlled'' in a well defined way.
Several examples illustrate the various aspects.
\end{abstract}
\bigskip \ni
{\it PACS}: 02.20.Sv; 02.30.Hq, {\it MOS}: {34A05; 37C80}
\medskip
\ni
{\it Keywords}: {Dynamical Systems; Lie point-symmetries;
$\lambda$-symmetries; \\ constants of motion; Hamiltonian dynamical systems;
conservation rules}
\bigskip\ni
{\bf Talk given at the ICNAAM Conference, Halkidiki (Greece), \\ September 2011}
\smallskip
\section{Introduction}
The role and the relevance of methods based on the analysis of symmetry
properties of differentialequations (both ordinary and partial) are well
known, for what concerns not only the problem of finding explicit solutions, but
also of examining ``structural'' properties (a typical and relevant
feature is, e.g., the presence of conservation rules).
There is an enormous literature on this subject: see e.g.
\cite{Ovs,Olv,CRC,BA} for some classical texts where general procedures and
standard applications can be found. However, there is a particular
context where \sy\ methods meet some intrinsic difficulty: this is the
case of dynamical systems (DS), i.e. systems of first-order
time-evolution differential \eq s of the form
\beq\label{DS1}\.u_a\=f_a(u,t)\q\q\ u_a=u_a(t)\q\ (a=1,\ldots, n) \eeq
with $\.u=\d u/\d t$ and (sufficiently smooth) given functions
$f_a=f_a(u,t)$.
The present paper is devoted to investigate precisely this case.
I am referring more specifically to Lie point-\sys
, i.e. to continuous transformations generated by infinitesimal vector fields $X$
which can be written in the form
\beq\label{Xt} X\=\phi_a(u,t)\frac{\pd}{\pd u_a}+\tau(u,t)\frac{\pd}{\pd t}
\equiv \phi\cdot\nabla_u+\tau\pd_t\ .\eeq
The problem of finding all \sys\ admitted by a DS is
quite difficult. On the other hand, expectedly, the presence of some \sy\
is strictly related to the determination of first integrals of the
differential problem, also called, in this context, constants of motion of the
dynamical flow. The determination of constants of motion for a time-evolution process
is clearly a basic result, not only in view of obtaining full solutions of
the problem, but also for their physical interpretation as ``conserved
quantities'' along the time evolution.
Sect.2 is devoted to present the \eq s providing the conditions for the
existence of Lie point-\sys\ of a given DS, and to discuss the close
connection existing between the problem of solving these \eq s and of
detecting constants of motion. It can be noticed that this connection, in the
present context,
is not {\it directly} related to the celebrated Noether theorem. It will be
useful, instead, in view of our discussion, to recall two classical, perhaps
less known,
old results. In few words, these results are very interesting as they
illustrate the strict relationship existing, also in the context of DS,
between the
notions of \sys\ and of constants of motion, but on the other hand they are not very
useful in practice because they require the knowledge of ``many'' constants of motion in order
to find (possibly all) \sys\ of the DS.
In Sect.3, the point of view is partly
reversed: I will assume that {\it just only one} \sy\ is known.
Indeed, although finding all \sys\ can be a difficult problem,
it often happens that one is able to detect a single \sy , and
often this \sy\ has a rather simple expression. Sect.4 is devoted to show
two quite general new classes of examples where this situation occurs. The
first one includes Lorenz-like DS, the second one deals with systems
related to higher order ODE's.
I will then show that this \sy\ can be used to introduce some suitable
``\sy -adapted coordinates'' where the problem of determining the
constants of motion becomes easier and provides conserved quantities which are also invariant under
the \sy . This is completely confirmed by the examples
considered in Sect.4.
In addition, it is shown that a ``weaker'' notion of \sy\ can be
introduced to the same purpose. I am referring to the notion of
$\la$-\sy\ \cite{MR1,MR2}, which will be briefly recalled especially for what
concerns its application to DS \cite{MRV,PLA}. In particular, the presence
of a $\la$-\sy\ allows the introduction of suitable coordinates exactly
as standard \sys .
In Sect.5, the particular case of {\it Hamiltonian} DS will be considered.
A classical result ensures that if such a DS exhibits a \sy\ admitting a generating
function $G$, this
function is automatically a conserved quantity invariant under the \sy .
If instead the DS admits a $\la$-\sy , the generating function is no
longer a conserved quantity, but the ``breaking'' of the conservation is
``controlled'' in a well defined way. In this context, also a
generalization of the notion of $\la$-\sy\ will be usefully introduced \cite{PLA,JN}.
Two examples illustrate the results, with the explicit calculation of the
``deviation'' from the exact conservation rule.
This is a full paper presented within ICNAAM 2011; a very short and preliminary
sketch of part of these results can be found in the enlarged
Abstracts of the Conference Proceedings \cite{Proc}.
\section{Symmetries and constants of motion: some classical facts}
Following the standard procedure, see e.g. \cite{Olv,BA}, a vector field $X$
(\ref{Xt}) is a Lie point-\sy\ for the DS (\ref{DS1}) (according to an
usually accepted abuse of language, I will denote by $X$ both the \sy\ and
its infinitesimal generator) if the following condition is satisfied
(sum over repeated indices unless otherwise stated)
\beq\label{syt}[\,f, \phi\,]_a\=-\frac{\pd}{\pd
t}(\phi_a-\tau\,f_a)+\frac{\pd\tau}{\pd
u_b}f_af_b\q\q\q (a,b=1,\ldots,n)\eeq
where $[\,f, \phi\,]_a$ is defined by
\[ [\,f, \phi\,]_a\= f_b\frac{\pd}{\pd u_b}\phi_a-\phi_b\frac{\pd}{\pd
u_b}f_a \equiv (f\cdot\nabla)\phi_a-(\phi\cdot\nabla) f_a\ .\]
It is not restrictive to put $\tau=0$, possibly introducing
``evolutionary'' vector field
\[X_{\rm ev}\! := (\phi-\tau\.u)\cdot\nabla\=(\phi-\tau
f)\cdot\nabla\,\equiv\~\phi\cdot\nabla\]
so the \sy\ condition becomes
\beq\label{sy}[\,f,\phi\,]\,_a+\frac{\pd}{\pd t}\phi_a\=0\ .\eeq
Despite this apparently simple form, it is in general very difficult to
obtain a complete solution to this set of determining \eq s.
{\it In principle } there are $n$ (functionally independent, locally
defined) solutions $\phi^{(a)}$; denoting by $\ka=\ka(u,t)$ any
constant of motion of the DS, i.e. any function such that
\[D_t\ka\equiv
\pd_t\ka+f\cdot\nabla\ka\=0 \]
the most general \sy\ of the DS can be written as
\[X\=\sum_{a=1}^n\ka^{(a)}\phi^{(a)}\cdot\nabla\equiv \ka^{(a)}X^{(a)}\ .\]
Apart from this very general result, the relationship between
\sys\ and constants of motion is actually much closer. To illustrate this point, and
also in view of our discussion, let me recall the
two following, perhaps less known, classical results.
\bigskip\noindent
{\bf a}. This result is due to Ovsjannikov \cite{OvNov} and shows
how \sys\ of a DS can be deduced from the knowledge of its constants of motion.
\begin{Proposition} Assume that $n$ functionally independent constants of motion $\ka^{(a)}$
of the given DS are known; then the linear system of $n^2$ \eq s
\[\sum_{a=1}^n p_{ab}\frac{\pd \ka^{(a)}}{\pd u_c}\=\delta_{bc}\]
can be solved for the $n^2$ quantities $p_{ab}$. Then
\[X^{(a)}\=\sum_{b=1}^n p_{ab}\frac{\pd}{\pd u_b}\]
are $n$ independent \sys\ for the DS.
\end{Proposition}
\bigskip\noindent
{\bf b}. The following result, based on the notion of Liouville vector field, has been
restated by G. \"Unal \cite{Un}, and also used by J. Zhang and Y. Li \cite{ZL}.
Let me recall the main fact in the following form.
A DS is said to admit a {\it Liouville vector field} $Y=\psi(u,t)\cdot\nabla$
if
\[\pd_t\psi_a+[f,\psi]_a+({\tt Div}\,f)\psi_a\=0\]
Clearly, if ${\tt Div}\,f=0$ then $Y$ is a standard \sy ;
if instead ${\tt Div}\,f\not=0$, putting
$$Y=q\,X$$
then $X$ is a standard \sy\ for the DS if $q$ is a scalar function solving
\[\pd_tq+f\cdot\nabla q+({\tt Div}\,f)q=0\ .\]
Then one has:
\begin{Proposition}
If the Liouville vector field $Y$ satisfies
${\tt Div}\,\psi=0$, then there are $n-1$ constants of motion $\^\ka^{(a)}$ such that
\beq\label{ep} Y\= \psi_a\frac{\pd}{\pd
u_a}\=\varepsilon_{abc\ldots
l}\^\ka^{(1)}_{,b}\,\^\ka^{(2)}_{,c}\ldots\^\ka^{(n-1)}_{,l}\frac {\pd}{\pd
u_a}
\eeq
where $\^\ka^{(a)}_{,b}=\pd\^\ka^{(a)}/\pd u_b$.
In addition,
the above constants of motion $\^\ka^{(a)}$ are invariant
under both the vector fields $Y$ and $X$:
\[Y\^\ka^{(a)}=X\^\ka^{(a)}\=0\ .\]
\end{Proposition}
\medskip
The last sentence says that the quantities $\^\ka^{(a)}$ are
simultaneously
invariant under the dynamical flow and under the \sy
\footnote{
In \cite{Un} this result is stated saying that the quantities
$\^\ka^{(a)}$ appearing in (\ref{ep}) are `the' first integrals of the DS.
Clearly, not all the first integrals satisfy (\ref{ep}) nor are \sy
-invariant.}.
\bigskip
Both the above results are conceptually greatly relevant, but clearly
of little practical use if one wants to explicitly find \sys\ (or constants of motion
as well) of a given DS. In the following, I try to partly reverse the
approach:
I will assume that just {\it only one} \sy\ is known, and then try to
deduce any possible information from it.
\section{Symmetry adapted coordinates}
Often, given a DS, one \sy\ of it is easily seen, either by direct
inspection or by simple calculations, as we shall see in the following
section. Then (remembering also Proposition 2) the idea is to
use invariants under this \sy\ to construct (one or more) constants of motion.
To this purpose, the presence of a $\la$-\sy\ (instead of a standard one)
may equally well help in the
calculations. Let me briefly recall the basic definitions of $\la$-\sy\
for what concerns the application in this context.
The notion of $\la$-\sy\ has been
originally introduced in 2001 by C. Muriel \& J.L.Romero in the context
of ODE's \cite{MR1,MR2}. Since then, this notion has received many very
important applications and extensions, which
cannot be recalled here (for a fairly complete list of references,
see e.g. \cite{Gtw,GaC}). In our case, a DS admits a
$\la$-\sy\ $X=\phi\cdot\nabla$ if there is a $C^\infty$ function
$\la=\la(u,\.u,t)$
such that the following condition holds
\beq\label{ll} [\,f,\phi\,]\,_a+\frac{\pd}{\pd t}\phi_a\=-\la\,\phi_a \eeq
to be compared with the standard condition (\ref{sy}).
$\la$-\sys\ are not properly \sys , indeed, e.g., they do not transform
solutions into other solutions,
nevertheless they share with standard \sys\ many useful properties;
in particular they indicate, as well as standard \sys ,
a convenient choice of variables in view of our procedure.
Let us assume then that the given DS admits either a standard or a
$\la$-\sy\ $X=\phi\cdot\nabla$. Introduce then $n$ functionally independent
quantities which are left fixed by this \sy : choose the time $t$ as
one of these, and the remaining $n-1$, denoted by $w_j=w_j(u)$, independent
of $t$:
\[ X\,w_j\=X\,t\=0 \q\q (j=1,\ldots,n-1)\ .\]
Let $\z$ be the ``rectifying'' coordinate along the action of $X$, i.e.
\[X\,\z\=1\q\q {\rm or}\q\q X\=\frac{\pd}{\pd \z}\ .\]
Notice that, even in the case of $\la$-\sy , all these coordinates depend
only on $X$ and not on $\la$. Choose now $w_j,\z$ as new dependent
variables (with $t$ still as independent one), and rewrite the DS
in terms of these, i.e. in the form $\.w_j=W_j,\,\.\z=Z$: one immediately has that
the r.h.s. $W_j,Z$ of the new DS turn out to be independent of $\z$ if the
\sy\ is
standard \cite{Olv}, and that only $Z$ may depend on $\z$ if the \sy\ is
a $\la$-\sy\ \cite{MRV,PLA}: this can be summarized writing
\beq\label{wz} \.w_j \=W_j(w,t) \q , \q \.\z \=Z(w,[\z],t)\ .\eeq
If now we look for constants of motion of the DS {\it expressed as functions of}
$w_j,\z,t$,
i.e. $\ka=\ka(w,\z,t)$, we can conclude with the following
\begin{Proposition}
The constants of motion $\ka^{(a)}(w,\z,t)$ solve the characteristic \eq
\[\frac{\d w_1}{W_1}\=\ldots\= \frac{\d w_{n-1}}{W_{n-1}}\=\frac{\d\z}{Z}\=\d t\]
where $W_j,Z$ are defined in (\ref{wz}).
\end{Proposition}
The advantage of this procedure is clear: we have a {\it reduction} of
the initial problem to a system of $n-1$ \eq s involving $n-1$ variables
$w_j$; for the same reason, also the search for the constants of motion through the above
characteristic \eq\ is easier. One obtains in this way, by
construction, conserved quantities which are also \sy -invariant; this
agrees of course with the classical Frobenious theorem \cite{Olv}.
Notice that this is a special case of a
more general problem of finding suitable reduction procedures of DS; this
and other related aspects will be discussed in a paper by G. Gaeta, S.
Walcher and the present author (in preparation).
\section{Two classes of examples}
The two following propositions provide two quite general classes of DS
where the presence of one \sy\ is guaranteed and constants of motion can be successfully
deduced.
\begin{Proposition}
A DS of the form
$$ \.u_a\=\si_{(a)}u_a+g_a(u) \q\q (a=1,\ldots,n; \
{\rm no\ sum\ over}\ a)$$
where $\si_{(a)}$=const ($\not= 0$) admits the \sy
$$X\=\exp(\la t)\,g\cdot\nabla$$
$\left(\la=const\ (possibly\ zero)\right)$ if $g_a(u)$ have the form
\[ g_a(u)\=u_a^{ \left(1-(\la/\si_{(a)})\right)} P_a(u)
\q\q (a,b,c=1,\ldots,n)\]
where $P_a$ are any smooth functions of the ratios
$u_b^{\si_{(c)}}/{u_c^{\si_{(b)}}}$ with
``exchanged'' exponents.
\end{Proposition}
Several examples of this situation are known: it includes e.g. generalized
Lorenz systems \cite{ZL}, etc.
\medskip\noindent
{\it Remark.} If $\la\not=0$, one can equivalently say that
$X':=g\cdot\nabla$ is a {\it $\la$-\sy }
with $\la(u,\.u,t)=\la=$ const.
This follows from the general property that
if $X_\la$ is a $\la$-\sy\ with some $\la(u,\.u, t)$ then
\[X:=\exp\left(\int\la(u,\.u,t)\,\d t\right)X_\la\]
is a (possibly {\it nonlocal}) standard \sy
\footnote{According to this remark, all examples given in \cite{ZL}
are actually equivalent to standard (not properly $\la$) \sys .}.
\medskip
Here an explicit example for Proposition 4.
\medskip
\noindent
{\it Example 1. }
With $n=3$ and $u\equiv(x,y,z)$ consider the DS:
\[ \.x=-x+x^2P_1 \ ,\ \.y=-y+y^2P_2 \ ,\
\.z=-2z+z^{(3/2)}P_3\]
where $P_a$ are functions of $x/y,x^2/z$, and which can be cast in analytic form
\[\.x\!=\!-x+Q_1(xy,y^2,z)\ ,\ \.y\!=\!-y+Q_2(x^2,xy,z)\ ,\ \.z\!=
\!-2z+Q_3(x^3,yz,xz)\ .\]
This DS admits the \sy
\[X\=e^t\left(Q_1\frac{\pd}{\pd x}+Q_2\frac{\pd}{\pd y}+
Q_3\frac{\pd}{\pd z}\right)\ .\]
Choose e.g. (this is a variant of an example given in \cite{Un})
$$Q_1=z-2y^2\q ,\q Q_2=2xy \q ,\q Q_3=4xz$$
then, with the notations of the above section,
\[w_1=z/y^2\q , \q w_2\=x^2+y^2-z/2\ ,\]
\[\z\=\frac{e^{-t}}{4v}\log\Big|\frac{x-v}{x+v}\Big| \q\q \q
({\rm here} \q v=|w_2|^{1/2})\ .\]
Following the above procedure, the DS becomes
\[\.w_1=0\q , \q\.w_2=-2w_2\q , \q\.\z=e^{-t}\]
and exactly {\it three} functionally independent constants of motion can be found
\[\ka_1=w_1=z/y^2\q , \q \ka_2=e^{2t}(x^2+y^2-z/2)\]
\[\ka_3=e^{-t} \left(1+\frac{1}{4v}\log\Big|\frac{x-v}{x+v}\Big|\right)\ .\]
\bigskip
\begin{Proposition}
Let $n=2,\,u\equiv(x,y)$; the DS
\[\.x\=y\q\q\.y\=y^2\gamma^{-1}\gamma_x+\gamma\,F(\gamma^{-1}y)\]
where $\gamma=\gamma(x)\not=0$ and $F$ are any given smooth
functions, admits the \sy
\[X\=\gamma\frac{\pd}{\pd x}+y\gamma_x\frac{\pd}{\pd y}\ .\]
\end{Proposition}
Here one has
\[w\=\gamma^{-1}y\q\q , \q\q \z\=\int\gamma^{-1}\d x\]
and the DS becomes
\[\.w\=F(w)\q\q ,\q\q \.\z\=w\ .\]
Two constants of motion are easily obtained:
\[\ka_1\=\z-\int w\,F^{-1}(w)\,\d w \q\q ,\q\q \ka_2\=t-\int F^{-1}(w)\,\d w\ .\]
This type of DS is specially interesting because the DS is equivalent to
the ODE
\[\ddot x\=\.x^2\gamma^{-1}\gamma_x+\gamma\,F(\gamma^{-1}\.x)\]
and the above \sy\ of the DS is in this case automatically extended to
become a \sy\ for the ODE:
\[ X_{ODE}\=\gamma\frac{\pd}{\pd x}\=\frac{\pd}{\pd\z }\ .\]
Notice that the ODE becomes just $\ddot\z=F(\dot\z)=F(w)$.
Similarly, constants of motion for the DS become first integrals for the ODE simply
replacing $y$ with $\.x$.
It is clearly possible to extend in a suitable way this example to DS and to the
corresponding ODE to the case $n>2$.
An explicit example follows.
\medskip\noindent
{\it Example 2. }
Choosing $\gamma=e^x,\,F=-y^2\,e^{-2x}=-w^2$, the DS is
\[\.x\=y\q\ , \q\ \.y\=y^2\left(1-e^{-x}\right)\]
with \sy\ $X=\exp(x)(\pd_x+y\pd_y)$, and the ODE is
\[\ddot x\=\.x^2-e^{-x}\,\.x^2\ .\]
Thanks to the new variables, the general solution is easily get
and two constants of motion for the DS (and for the ODE, replacing $y$ with $\.x$) are
\[ \ka_1\=\log|y|-x-e^{-x}\q\ , \q\ \ka_2\=t-e^x/y\ .\]
\section{Hamiltonian DS, $\Lambda$-\sys\ and ``controlled failure'' of
conservation rules}
Let me now consider the specially interesting case in which the DS is a
{\it Hamiltonian} DS, i.e. the DS is obtained from a given Hamiltonian
function $H$.
Changing accordingly the notations, with $n=2m$, the $n$ variables
$u=u_a(t)$ are replaced by the canonical variables $q_\a(t),p_\a(t)
\,(\a=1,\ldots,m)$:
\[u\equiv(q_1,\ldots,q_n,p_1,\ldots,p_n)\equiv(q,p)\in{\bf R}^{2m}\]
and the DS is the system of the Hamilton \eq s of motion for the given Hamiltonian
$H=H(q,p,t)$:
\[\.u\=J\nabla H\=F(u,t) \q\ ; \q\ \nabla\equiv(\nabla_q,\nabla_p)\]
where
\[J\=\pmatrix {0 & I_m \cr -I_m & 0} \]
is the standard symplectic matrix and $I_m$ the $m\times m$ identity
matrix. In the same way, vector fields $X$ will be written
\[X\=\phi_\a(q,p,t)\frac{\pd}{\pd q_\a}+\psi_\a(q,p,t)\frac{\pd}{\pd
p_\a}\equiv\Phi\cdot\nabla \q\q ;\q\q \Phi\equiv(\phi_\a,\psi_\a).\]
I will now restrict the attention on vector fields admitting a {\it generating
function} $G=G(q,p,t)$, i.e. vector fields $X$ satisfying
\[\Phi\=J\nabla G \q {\rm or} \q \phi=\nabla_p G,\,\psi=-\nabla_q G\ . \]
The connection between \sys\ and constants of motion in the Hamiltonian context is even
more stringent. It is well known indeed, since the end of XIX century \cite{LCiv}
(see also, e.g., \cite{Olv}), that if $X$ is a Lie
point-\sy\ for a Hamiltonian DS, i.e. $[F,\Phi]+\pd_t\Phi=0$, then
\beq\label{DTG} \nabla (D_tG)\=0 \q\q {\rm or}\q\q D_tG\=g(t)\ .\eeq
This follows from the identity
\beq\label{LP} \nabla_a (D_tG)=\nabla_a(\{G,H\}+\pd_tG)=-J_{ab}([F,\Phi ]_b
+\pd_t\Phi_b) \q\q (a,b=1,\ldots,n)\eeq
where $\{\cdot,\cdot\}$ is the standard Poisson bracket.
Then $G$ is a constant of motion apart from an additional time dependent function $g$.
Let me now consider for simplicity generating functions $G(q,p)$ not depending
explicitly on $t$; combining this with the obvious property $X(G)=~0$ if
$X=J\nabla G\cdot\nabla$, the following standard result can be stated for
convenience
\begin{Proposition}
Let the Hamiltonian DS $\.u=J\nabla H$ admit a \sy\ $X=\Phi\cdot\nabla$ where
$\Phi=J\nabla G$. Then the generating function $G(q,p)$ is a conserved
quantity invariant under $X$.
\end{Proposition}
In terms of our previous arguments, $G$ can then be chosen as one of \sy
-invariant variables $w$, and in this case it is automatically (and
trivially) also a constant of motion.
A less trivial and more interesting situation occurs if the
Hamiltonian DS does admit some $X$ as a $\la$-\sy : in this case the
identity (\ref{LP}), thanks to (\ref{ll}) becomes (let me now write $\.G$
instead of $D_tG$)
\beq\label{Gll}\nabla_a(\.G)\=\la(J\Phi)_a\=-\la\,\nabla_a\,G\ .\eeq
In addition, in this context, it may be useful to introduce an extension
of the notion of $\la$-\sy , replacing the scalar function $\la$ with a
$n\times n$ matrix $\Lambda$ (depending in general on $q,p,\.q,\.p,t$):
the $\la$-\sy\ condition
(\ref{ll}) for the DS $\.u=F(u)=J\nabla H$ is replaced by (see \cite{PLA,JN})
\beq\label{LL} [F,\Phi]_a\=-(\Lambda\Phi)_a\eeq
and the above identity (\ref{LP}) must be modified accordingly, giving
the following
\begin{Proposition}
If a DS admits $X=J\nabla G\cdot\nabla$ as a $\Lambda$-\sy , then $\.G$ obeys the
\eq
\beq\label{LA}\nabla_a(\.G)\=(J\,\Lambda\, \Phi)_a\=(J\,\Lambda\,J\,\nabla)_a\,G \ .\eeq
\end{Proposition}
The two \eq s (\ref{Gll},\ref{LA}) clearly point out an interesting
property of $\la$ (or $\Lambda$) \sys : they can be viewed as
``perturbations'' of the ``exact'' \sy . Equations (\ref{Gll},\ref{LA})
indeed express the ``deviation'' from the exact conservation rule $\.G=0$
produced by the presence of a nonzero $\la$ (or $\Lambda$).
Some examples will clarify this point. The first one deals with the case
of a $\la$-\sy\ (i.e. with a scalar function $\la$); it is quite simple and can
be useful to illustrate the idea. The second one deals with a
Toda-like Hamiltonian and a $\Lambda$-\sy\ with the introduction of a $\Lambda$
matrix. In both cases, the deviation from the exact conservation of $G$
will be explicitly evaluated and the ``controlled'' failure of the
conservation rule clearly described.
\medskip\noindent
{\it Example 3.}
Consider the Hamiltonian in $m=2$ degrees of freedom
\[H\=\frac 1 2 p_1^2+\frac 1 2 p_2^2+\frac 1 2 q_1p_1^3+\frac 1 2
q_2^2p_1^2\ .\]
The vector field $X=\pd/\pd q_1$ is a $\la$-\sy\ for the Hamilton \eq s
of motion $\.u=F(u)=J\nabla H$
(which can be easily written), with $\la$ given by the scalar function
$\la=3p_1^2/2$, namely
\[ [F,\Phi ]\=-\frac 3 2 p_1^2\Phi\q\q , \q\q \Phi=(1,0,0,0)^t \ .\]
Expectedly, the generating function $G=p_1$ is not conserved, indeed one
has
\[\.G\=-\frac 1 2 G^3\=-\frac 1 3 \la\, G\ .\]
Elementary integration gives $G(t)=G_0\big(1+t\,G_0^2\big)^{-1/2}$ with
$G_0=G(0)$, and
\[\.G\=-\frac 1 2 G_0^3 \big(1+t\,G_0^2\big)^{-3/2}\]
which precisely expresses ``how much'' $G$ is not conserved and indicates
in particular that $G$ is ``almost conserved" for great values of $t$.
\medskip\noindent
{\it Example 4.}
Consider now the following $2$ degrees of freedom Toda Hamiltonian
\[ H\=\frac 1 2 p_1^2+\frac 1 2 p_2^2+e^{q_1+q_2}+e^{q_1-q_2}\ .\]
It is easy to write the corresponding Hamilton \eq s of motion and to verify
that the vector field
\[ X\=\frac{\pd}{\pd q_1}+\frac{\pd}{\pd q_2}\]
with generating function $G=p_1+p_2$ is a $\Lambda$-\sy\ for this system with
$\Lambda$ given by the $4\times 4$ matrix
\[ \Lambda\=-2\,e^{q_1+q_2} \pmatrix {0 & 0 \cr I_2 & 0} \]
namely
\[ [F,\Phi]\=-\Lambda\Phi \q\q ,\q\q \Phi=(1,1,0,0)^t\ .\]
In agreement with the above discussion and Proposition 7, one obtains
\[ \.G=\.p_1+\.p_2=-2\,e^{q_1+q_2}\q\q ,\q\q
\nabla(\.G)=-2\,e^{q_1+q_2}\Phi\ .\]
Introducing the variables
\[w_1=q_1-q_2 \q ,\q w_2=p_1-p_2 \q ,\q w_3=G=p_1+p_2 \q ,\q \z=q_1+q_2\]
the DS becomes
\[ \.w_1=w_2 \q ,\q \.w_2=-2e^{w_1} \q ,\q \.w_3=-2e^\z \q ,\q \.\z=w_3\ .\]
It can be noted that this DS has not the ``reduced" form as said in
Proposition 3 and eq. (\ref{wz}), where $\z$ is present only in the r.h.s. of $\.\z$:
indeed the reduced form (\ref{wz})
is granted only if $\Lambda$ is a scalar, $\Lambda=\la I$. Anyway, the
system is easily solvable; in particular one has
\[\z\=2\log\Big(\frac {|c_1|}{\cosh (c_1t+c_2)}\Big) \q , \q \.G\=-2e^\z \]
\[ G\=-2c_1\tanh(c_1t+c_2) \q ,\q |\.G(t)|\le 2 |c_1|\]
where $c_1,c_2$ are arbitrary constants, which shows that $G$ is not a
conserved quantity, as expected, however -- for any choice of
$c_1,c_2$ -- both $\.G(t)$ and $G(t)$ are determined and bound quantities.
\section{Conclusion}
Finding Lie point-\sys\ of a dynamical system is in general a quite
difficult task, and it is strictly connected with the searching for its constants of motion.
In this paper I have discussed and emphasized this close connection, using old and new
results. I have shown in particular that the knowledge of a \sy\ of the
DS can allow to directly obtain quantities which are conserved and invariant
under the \sy .
The particular case of Hamiltonian DS is specially
interesting: a classical result ensures that if a \sy\ of the DS admits a
generating function $G$, then $G$ is automatically a constant of motion and a
\sy-invariant. If instead the Hamiltonian DS admits a \sy\ of ``weaker''
type (specifically, a $\la$ or a $\Lambda$-\sy ) then the generating function
is no longer a conserved quantity, but we have seen that the deviation from the exact
conservation rule $\.G=0$ is ``controlled'' in a well defined way. | 0.005621 |
TITLE: Prove the inequality $4(a+b+c)^3\ge 27(a^2b+b^2c+c^2a)$
QUESTION [1 upvotes]: $$4(a+b+c)^3\ge 27(a^2b+b^2c+c^2a)$$, $a,b,c\ge0$
I tried to work out this inequality the way I did
So we have to show $$ 4(a + b + c) ^ 3 \ge 27 (a ^ 2b + b ^ 2c + c ^ 2a + abc) $$
One way is to use cyclic symmetry, and WLOG assumes $ a $ is $ a, b, c $ minute. Then we can write $ b = a + x, c = a + y $, where $ x, y \ge 0 $. Now the inequality reduces to 9 $$ (x ^ 2-xy + y ^ 2) + (x-2y) ^ 2 (4x + y) \ge 0 $$ which is obvious. Also from the above, we get then equality is possible iff $ x = y = 0 $ or when $ a = 0, x = 2y $, i.e. when $ (a, b, c) = (1, 1, 1) $ or permutation $ (0, 2, 1) $.
I think it's a mistake please help
REPLY [2 votes]: The last inequality should be
$$9(x^2-xy+y^2)a+4x^3-15x^2y+12xy^2+4y^3 \geqslant 0,$$
or
$$9(x^2-xy+y^2)a+(4x+y)(x-2y)^2 \geqslant 0.$$
It's called BW method. | 0.016039 |
TITLE: Proof that an embedding into $\ell^1$ is compact
QUESTION [3 upvotes]: Prove that any sequence $(x^{(n)})_{n\in\mathbb{N}}\subseteq\ell^1$ such that $\sum_{k=1}^\infty k\lvert x_k^{(n)}\lvert\leq1$ for all $n\in\mathbb{N}$ has a convergent subsequence.
My thoughts on this: Clearly $\lvert x^{(n)}_k\lvert\leq\frac{1}{k}$ uniform in $n$. Therefore the sequence $x_1^{(n)}$ has a convergent subsequence $x_1^{(\tilde{n}_k)}$. Further extracting subsequences for any fixed $N\in\mathbb{N}$ I find $({n_l})_{l\in\mathbb{N}}\subseteq\mathbb{N}$ such that $x_k^{(n_l)}$ converges to some $x_k$ for all $1\leq k\leq N$ as $n\rightarrow\infty$.
If for every $\epsilon>0$ I could find some $N\in\mathbb{N}$ such that $\sum_{k=N+1}^\infty \lvert x_k^{(n)}\lvert<\epsilon$ uniform in $n$ the proof would be complete. But I don't see why this should be true.
Can you give me some hint?
REPLY [0 votes]: I think you may want a proof like this:
By the given condition we know $|x_k^{(n)}|\le1$ for all $n$ and $k$. So for $k=1$, we may pick up a convergent subsequence $\{x_1^{n^1_j}\}_{j\in\mathbb{N}}$. Then based on this subsequence we pick up for $k=2$ a convergent subsequence $\{x_2^{n^2_j}\}_{j\in\mathbb{N}}$. Keep doing in this way and get a sequence for each $k$, and then take the diagonal elements $n^j_j$. It can be verified that $\{x^{n^j_j}\}_{j\in\mathbb{N}}$ converge pointwisely to some $\tilde{x}$. For simplicity we denote this new sequence $\{x^{(m)}\}_{m\in\mathbb{N}}$. It can be shown that $\sum_{k=1}^{\infty}k|\tilde{x}_k|\le1$, hence $\tilde{x}\in\mathcal{l}^1$. In fact, since $\{x^{(m)}\}_{m\in\mathbb{N}}$ converge pointwisely to $\tilde{x}$, for finite $N$ we have $\sum_{k=1}^{N}k|\tilde{x}_k|\le1$, then passing $N$ to infinity we know $\sum_{k=1}^{\infty}k|\tilde{x}_k|\le1$.
Then we show that for every $\epsilon>0$ we can find some $N\in\mathbb{N}$ such that $\underset{m}{\sup}\sum_{k=N+1}^{\infty}|x_k^{(m)}|<\epsilon$. In fact, only need to observe that
$$\sum_{k=N+1}^{\infty}|x_k^{(m)}|\le\frac{1}{N+1}\sum_{k=N+1}^{\infty}k|x_k^{(m)}|\le\frac{1}{N+1},\qquad\forall m.$$
Now we're already to prove your claim. For any $\epsilon>0$ take $N_1$ such that $\underset{m}{\sup}\sum_{k=N_1+1}^{\infty}|x_k^{(m)}|<\epsilon/4$. Note that $\tilde{x}\in\mathcal{l}^1$, so we can find $N_2$ such that $\sum_{k=N_2+1}^{\infty}|\tilde{x}_k|<\epsilon/4$. Therefore, if taking $N_3=\max\{N_1,N_2\}$, then
$$\underset{m}{\sup}\sum_{k=N_3+1}^{\infty}|x_k^{(m)}-\tilde{x}_k|\le\underset{m}{\sup}\sum_{k=N_3+1}^{\infty}|x_k^{(m)}|+\sum_{k=N_3+1}^{\infty}|\tilde{x}_k|<\epsilon/2.$$
On the other hand, since $\{x^{m}\}_{m\in\mathbb{N}}$ converge pointwisely to $\tilde{x}$, we can find $N_4$ such that
$$\sum_{k=1}^{N_3}|x_k^{(m)}-\tilde{x}_k|<\epsilon/2,\qquad\forall m>N_4.$$
Combining the above two we know
$$\|x^{(m)}-\tilde{x}\|_{\mathcal{l}^1}\le\sum_{k=1}^{N_3}|x_k^{(m)}-\tilde{x}_k|+\underset{m}{\sup}\sum_{k=N_3+1}^{\infty}|x_k^{(m)}-\tilde{x}_k|<\epsilon,\qquad\forall m>N_4.$$
This means $\{x^{(m)}\}$ converge to $\tilde{x}$ in $\mathcal{l}^1$. Proof is finished. | 0.03257 |
TITLE: Derivative of continuous function exists if limit of derivative exists
QUESTION [11 upvotes]: I'm stuck on this old qualifier problem. I suppose one could do it using the basic definitions of continuity and differentiability, but is there a simpler way? (For example, using DCT, FTC, Lebesgue differentiation theorem, etc.)
Let $f:\mathbb{R} \mapsto \mathbb{R}$ be continuous. Suppose $f$ is differentiable away from $0$ and lim$_{x \to 0} f^\prime(x)$ exists. Show $f^\prime(0)$ exists.
REPLY [0 votes]: By L'Hôpital's rule, we know that for any function $f$, defined in a neighbourhood of $0$,
$$
\lim_{x\to0}\frac{f(x)-f(0)}{x}=\lim_{x\to0}f'(x)
$$
provided that $f'$ is defined in a punctured neighbourhood of $0$, and the limit on the RHS exists. Since we are given these hypotheses in the question, it follows that $f'$ is both defined and continuous at $0$.
Although this is not directly relevant to the question at hand, it's worth noting that it is possible for $f'(0)$ to exist even if $\lim_{x\to 0}f'(x)$ does not. Consider, for instance, the function
$$
f(x)=\begin{cases}
x^2\sin(1/x) & x\neq0 \, , \\
0 & x = 0 \, .
\end{cases}
$$
Note also that this answer is essentially equivalent to Daniel Fischer's, as L'Hôpital's rule is derived from the mean value theorem. | 0.005316 |
\begin{document}
\begin{center}
{\Large\bf
Coxeter Decompositions of Bounded Hyperbolic Pyramids and Triangular
Prisms.
}
\medskip
{\Large A.~Felikson}
\vspace{35pt}
\parbox{10.5cm}
{\scriptsize
{\bf Abstract.}
Coxeter decompositions of hyperbolic simplices where studied in
math.MG/0212010 and math.MG/0210067.
In this paper we use the methods of these works
to classify Coxeter decompositions of
bounded convex pyramids and triangular prisms in the hyperbolic space
$\H^3$.
}
\end{center}
\section*{Introduction}
Let $P$ be a convex polyhedron in the hyperbolic space $H^3$.
\begin{define}
A polyhedron $P$ is called a {\bf Coxeter polyhedron} if all dihedral
angles of $P$ are integer parts of $\pi$.
\end{define}
\begin{define}\label{def1}
A polyhedron $P$ admits a {\bf Coxeter decomposition}
if $P$ can be tiled by finitely many Coxeter polyhedra
such that any two tiles
having a common face are symmetric with respect to this face.
\end{define}
In this paper we classify Coxeter decompositions of
bounded convex pyramids and triangular prisms in the hyperbolic space
$\H^3$.
\subsection*{Basic definitions}
If $P$ admits a Coxeter decomposition
we also say that $P$ is a {\bf quasi-Coxeter} polyhedron.
The tiles in Definition~\ref{def1} are called
{\bf fundamental polyhedra} and denoted by $F$.
Clearly, any two fundamental polyhedra are congruent to each other.
A plane $\alpha$ containing a face of a fundamental polyhedron
is called a {\bf mirror} if $\alpha$ contains no face of $P$.
\medskip
In this paper any polyhedron is either a quasi-Coxeter
polyhedron or a polyhedron bounded by the mirrors of some
Coxeter decomposition.
By the "decomposition" we always mean a Coxeter decomposition.
\begin{define}
Given a Coxeter decomposition of a polyhedron $P$,
a {\bf dihedral angle} of $P$ formed up by facets
$\alpha$ and $\beta$ is called {\bf fundamental}
if no mirror contains $\alpha \cap \beta$.
In this case the {\bf edge} $\alpha \cap \beta$ of $P$ is called
{\bf fundamental} too. \\
A {\bf vertex} $A$ is called {\bf fundamental}
if no mirror contains $A$.
\end{define}
\section{Fundamental polyhedron}
Let $P$ be a bounded quasi-Coxeter polyhedron.
From now on $P$ is either
a bounded pyramid or a bounded triangular prism in $\H^3$.
\vspace{6pt}
\noindent
{\bf Notation.}\\
Denote by $\alpha \cap \beta$
the intersection of
the sets $\alpha$ and $\beta$ in the {\it inner}
part of $\H^3$.\\
Let $\alpha$ be a face of a polyhedron, denote by $\overline
\alpha$ a plane containing $ \alpha$.
\begin{lemma}\label{inter}
Let $F$ be a Coxeter polyhedron.\\
Then either $F$ is a tetrahedron or $F$ has two faces $\alpha$
and $\beta$ such that
$\overline \alpha \cap \overline \beta =\emptyset$.
\end{lemma}
\begin{proof}
For any polyhedron without obtuse
angles we have
$$ \alpha \cap \beta =\emptyset
\quad \Rightarrow \quad
\overline \alpha \cap \overline \beta =\emptyset$$
(see~\cite{A2}).
Any Coxeter polyhedron has no obtuse angles.
Thus, it is enough
to prove that $F$ has faces
$\alpha$ and $\beta$ such that
$\alpha \cap \beta =\emptyset.$
Suppose that $\alpha \cap \beta \ne \emptyset$
for any faces $\alpha$ and $\beta$ of $F$.
Let $A$ be an arbitrary vertex of $F$. Let
$\alpha_1$,...,$\alpha_k$ be all the faces of $F$
containing $A$.
Let $\beta$ be any face of $F$ such that $A \notin \beta$.
By the assumption,
$\alpha_i \cap \beta \ne \emptyset \quad \forall i=1,...,k$.
Therefore, $F$ has no faces except $\alpha_i$ and $\beta$,
i.e. $F$ is a pyramid.
Suppose that $A$ is not an ideal vertex of $F$.
Then $A$ is incident exactly to three faces of $F$
and $F$ is a tetrahedron.
Suppose that $A$ is an ideal vertex of $F$ and
the pyramid $F$ is not a tetrahedron.
Consider two faces (say, $\alpha_1$ and $\alpha_3$) having no
common edges.
Since $A\notin \H^3$, we have
$\alpha_1 \cap \alpha_3=\emptyset$.
\end{proof}
\begin{lemma}
\label{l.3.1}
If $P$ is a bounded pyramid, then $F$ is a tetrahedron.
\end{lemma}
\begin{proof}
Suppose that $F$ is not a tetrahedron.
By Lemma~\ref{inter}, $F$ has two faces $\alpha$ and $ \beta$
such that $\overline \alpha \cap \overline \beta =\emptyset$.
Let $F_0$ be a fundamental polyhedron in $P$;
let $\alpha_0$ and $\beta_0$ be its disjoint faces.
Consider a sequence of fundamental polyhedra $F_i\in P$, $i\in
\mathbf Z$, such that
$\alpha_i=\alpha_{i+1}$, if $i$ is odd, and
$\beta_i=\beta_{i+1}$, if $i$ is even
(see Fig.~\ref{seq1}).
The sequence is finite,
since $P$ contains finitely many fundamental polyhedra.
The polyhedra $F_i$ cannot make a cycle,
since $\alpha_i\cap \beta_i =\emptyset$.
Let $F_k$ and $F_s$ be the endpoints
of the sequence. Then $\alpha_k$ or $\beta_k$ belongs to
some face $\gamma$ of $P$.
Similarly, $\alpha_s$ or $\beta_s$ belongs to
some face $\delta$ of $P$.
Obviously, $\overline \alpha_i \cap \overline \beta_j=\emptyset$
for any
$i,j$. But $\gamma$ intersects $\delta$, since $P$ is a bounded
pyramid. The contradiction shows that $F$ is a tetrahedron.
\begin{figure}[!h]
\begin{center}
\input pic/seq1.tex
\end{center}
\caption{A sequence of the fundamental polyhedra.}
\label{seq1}
\end{figure}
\end{proof}
\noindent
{\bf Remark.} The idea of the proof of Lemma~\ref{l.3.1}
belongs to O.~V.~Schwarzman.
\vspace{7pt}
Let $\alpha$ be a mirror in a Coxeter decomposition of the
triangular prism $P$. Then
$\alpha$ intersects $P$ as shown in one
of Fig.~\ref{mirrors}.1--\ref{mirrors}.16.
\begin{define}\label{mirr}
We say that a mirror is {\bf pentagonal} if it intersects the prism as
shown in Fig.~\ref{mirrors}.16.
\end{define}
\begin{define}
We say that a triangular quasi-Coxeter prism $P$ is {\bf
minimal} if any prism $P'$ inside $P$
is fundamental.
\end{define}
\begin{figure}[!h]
\begin{center}
\input pic/mirrors.tex
\end{center}
\caption{Mirrors in the prism.}
\label{mirrors}
\end{figure}
\begin{lemma}
\label{l5}
Let $P$ be a minimal prism and $F$ be a fundamental polyhedron.
Suppose that $F$ is neither a tetrahedron nor a triangular prism.
Then any mirror is pentagonal and any dihedral angle of $P$ is
fundamental.
\end{lemma}
\begin{proof}
Suppose that the decomposition contains a tetrahedron or
a pyramid. Then by Lemma~\ref{l.3.1} $F$ is a tetrahedron.
Hence, $P$ contains neither pyramid nor tetrahedron.
Since $P$ is minimal, $P$ contains no smaller triangular prisms.
Therefore, any mirror is pentagonal
(see Fig.~\ref{mirrors}).
Clearly, all dihedral angles of $P$ are fundamental.
\end{proof}
\begin{lemma}
Let $P$ be a triangular prism. Then $F$ is either a tetrahedron or
a triangular prism.
\end{lemma}
\begin{proof}
It is sufficient to prove the lemma for a minimal prism $P$.
Suppose that $F$ is neither a tetrahedron nor a triangular prism.
Then by Lemma~\ref{l5} any mirror is pentagonal and any dihedral
angle of $P$ is fundamental.
Let $ABCDE$ be any pentagonal mirror (see Fig.~\ref{5}).
This mirror separates the points
$M_1$, $N_1$, $N_3$ from the points $M_2$, $M_3$, $N_2$.
\begin{figure}[!h]
\begin{center}
{\scriptsize
\input pic/5.tex
}
\end{center}
\caption{A pentagonal mirror.}
\label{5}
\end{figure}
We say that a polyhedron $W$ is {\bf good} if
it contains the points $M_2$, $M_3$, $N_2$
and if it can be cut out of $P$ by a single pentagonal
mirror.
Let $W$ be a minimal good polyhedron
(i.e. no polyhedron inside $W$ is good).
The minimal polyhedron exists, since the decomposition contains
finitely many mirrors. Suppose that $W$ has vertices
$M_2M_3N_2ABCDE$ (see Fig.~\ref{5}).
Suppose that $W$ has a non-fundamental dihedral angle $\alpha$.
By Lemma~\ref{l5} $\alpha$ is an angle formed up by $ABCDE$ and some
face of $P$. Let $\Pi$ be a mirror decomposing the angle
$\alpha$. Clearly, $\Pi$ does not separate the points $M_1$,
$N_1$ and $N_3$ one from another. Since $\Pi$ is pentagonal,
it separates all the points above from the points
$M_2$, $M_3$, $N_2$. Therefore, $\Pi$ cuts out of $P$ a good
polyhedron contained in $W$. This contradicts to the minimal
property of $W$.
Thus, any dihedral angle of $W$ is fundamental,
and $W$ is a Coxeter polyhedron.
The
prolongations of two disjoint faces cannot intersect each other
(see~\cite{A2}).
This property is broken in the polyhedron $W$
(consider the faces $AEN_2$ and $CDM_3$ which prolongations have
a common line $N_1N_3$). The contradiction shows that $F$ is
either a tetrahedron or a triangular prism.
\end{proof}
\section{Decompositions of pyramids.}
This section describes the way to classify all Coxeter
decompositions of bounded hyperbolic pyramids.
Suppose that a pyramid $P$ has only five vertices
$OA_1A_2A_3A_4$ (where $A_1A_2A_3A_4$ is a base of $P$ and $O$ is an
apex). By Lemma~\ref{l.3.1} $F$ is a tetrahedron. Suppose that any edge
$OA_i$ is fundamental. Consider a small sphere $s$ centered in $O$. The
intersection $s\bigcap P$ is a spherical quadrilateral $q$. Any angle of
$q$ is fundamental. This contradicts to the fact that the sum of the
angles of a spherical quadrilateral should be greater than $2\pi$.
Hence, we can assume that there is a mirror $m$ through $OA_1$.
This mirror decomposes the pyramid into two smaller pyramids.
One of the smaller pyramids is a tetrahedron and another is
either a tetrahedron or a small quadrilateral pyramid.
Thus, any minimal quadrilateral pyramid
consists of two tetrahedra (not necessary fundamental). The
Coxeter decompositions of hyperbolic tetrahedra
are classified in~\cite{arxiv-tetr}.
Therefore, we can find all
decompositions of minimal quadrilateral pyramids.
Using the decompositions of minimal pyramids we
can find the decompositions of greater pyramids too.
So, using this algorithm, it is possible to find all
decompositions of quadrilateral pyramids.
Analogously, it is possible to classify the decompositions of
pyramids with bigger number of vertices.
Any pyramid $OA_1...A_n$ has a decomposed edge $OA_i$.
So, the pyramid is decomposed into two smaller pyramids which
possible decompositions we should enumerate at the
previous steps.
This procedure realized by a computer leads to a large list of
quadrilateral quasi-Coxeter pyramids, several
pentagonal quasi-Coxeter pyramids, and exactly one hexagonal
quasi-Coxeter pyramid.
See Table~2 for the result.
\section{ Decompositions of prisms into prisms}
In this section both $P$ and $F$ are triangular prisms.
For any prism we say that a triangular face is a
{\bf base} and a quadrilateral face is a {\bf side}.
\begin{lemma}\label{bok}
No base of $F$ belongs to a side of $P$.
\end{lemma}
\begin{proof}
Suppose that a side $s$ of $P$ contains a base of a fundamental
prism $F_1$. Consider a sequence of fundamental prisms
$F_1$, $F_2$,...,$F_n$,...,
such that $F_i\in P$, and $F_i$ and $F_{i+1}$ have a common
base. This chain is finite, since $P$ contains finitely many
fundamental prisms. The prolongations of the bases of $F_1$
have no common points, since $F_1$ is a Coxeter prism.
So, the prolongations of the bases of two different prisms $F_i$
have no common points. Therefore, the prisms $F_1$,
$F_2$,...,$F_n$, cannot make a cycle and the sequence has an
endpoint $F_n$. One of the bases of $F_n$ belongs to some face
$b$ of $P$ (otherwise the sequence has a prolongation). The
face $s$ cannot intersect the face $b$, since these faces are
prolongations of the bases of $F_i$. This contradicts to the
fact that the side $s$ intersects each face of $P$.
\end{proof}
\begin{lemma} \label{osn}
No side of $F$ belongs to a base of $P$.
\end{lemma}
\begin{proof}
Denote by $b$ and $s$ the bases of $P$.
Suppose that a side $a_0$ of a fundamental prism $F_0$
belongs to $b$. Consider
a sequence of fundamental prisms
$F_{-k}$,...,$F_0$,...$F_l$, $F_i\in P$,
such that $F_i$ and $F_{i+1}$ have a common base.
The sequence is finite, and the prisms $F_i$ cannot make a cycle.
Let $a_{-k}$ and $b_l$ be the bases of the endpoints
$F_{-k}$ and $F_l$. By Lemma~\ref{bok} the bases
$a_{-k}$ and $b_l$ belong to two different bases of $P$.
If $a_{-k}$ belongs to the base $s$ then the prolongations
of $a_{-k}$ and $a_0$ have a common point
(see Fig.~\ref{chain}). This is impossible,
since the prolongations of the bases have no common points.
\begin{figure}[!h]
\begin{center}
{\scriptsize
\input pic/chain.tex
}
\end{center}
\caption{No side of $F$ belongs to a base of $P$.}
\label{chain}
\end{figure}
\end{proof}
\begin{lemma}\label{krishki}
Let $b$ be a base of $P$, $b_1$ and $b_2$ be the bases of
$F$. Suppose that $b$ is non-fundamental. Then the tiling of
$b$ by the faces of $F$ is a Coxeter decomposition. A
fundamental polyhedron of this decomposition is a triangle
$b_1$ or $b_2$.
\end{lemma}
\begin{proof}
By Lemma~\ref{osn} any tile is a triangle.
The adjacent tiles of the tiling are the faces of $F$
having a common edge. Thus, if one of the tiles is
a base $b_1$ then the other tiles are the copies of this
base. Clearly, this tiling is a Coxeter decomposition.
\end{proof}
By Lemma~\ref{bok} any side $s$ of $P$ is tiled by the sides of $F$.
This tiling is not necessary a Coxeter decomposition,
but Lemma~\ref{lat}~and~\ref{l_boka} show that this tiling
is very simple.
\begin{lemma}
\label{lat}
Let $s$ be a side of $P$.
Then the tiling of $s$ is a "lattice"
(see Fig.~\ref{lattice}a), i.e. the tiling
has the following properties:\\
\phantom{iii}
(1) any vertex of the face $s$ belongs to a unique tile;\\
\phantom{iii}
(2) any point of any edge of the quadrilateral $s$
belongs to either one or two tiles.\\
\phantom{iii}
(3) any point in the inner part of $s$ belongs to one, two or four tiles.
\end{lemma}
\begin{figure}[!h]
\begin{center}
{\scriptsize
\input pic/latticed.tex
}
\end{center}
\caption{A tiling of a side looks like a lattice.}
\label{lattice}
\end{figure}
\begin{proof}
We say that an edge $AB$ of a tile is {\bf horizontal }
if $AB$ belongs to a base of a fundamental prism;
otherwise we say that $AB$ is {\bf vertical}.
Let $ABCD$ be any tile which vertical edges are $AB$ and $CD$.
Consider a sequence of tiles such that
this sequence contains $ABCD$, and
any two neighboring tiles have a common horizontal edge.
We say that this sequence is {\bf vertical}.
By the same way construct a {\bf horizontal} sequence
where the neighboring tiles have a common vertical edge.
Evidently, the vertical sequence ends at two different horizontal
edges of $s$ and the horizontal sequence ends at two different
vertical edges. Let us show that this condition is broken if
the lemma is false.
1). Let $A$ be a vertex of the face $s$. Suppose that $A$
belongs to more then one tile (see Fig.~\ref{lattice}b).
Consider a vertical sequence through the tile $ABCD$.
No pair of tiles from this sequence can be separated
from each other by the line $AB$.
Therefore, this sequence cannot reach the horizontal edge $AE$,
and (1) is proved.
2). Let $A$ be an inner point of a horizontal edge $MN$.
Suppose that $A$ belongs to more then two tiles.
Then one of these tiles has no edges in $MN$
(see Fig.~\ref{lattice}c).
At least two lines through $A$ are vertical edges of some tiles.
Consider a vertical sequence through the tile $ABCD$.
It is evident that the angle $\angle BAE$ contains any tile
from this sequence. Thus, the sequence cannot reach the horizontal
edge $MN$.
The same reasoning works for the points on the
vertical edges, and (2) is proved.
3). Let $A$ be an inner point of the tile $s$. Clearly, $A$
cannot be incident to exactly three tiles. Suppose that $A$ is a vertex of
five or more tiles.
Let $a_1$ be any tile and $a_1$,...,$a_k$
be an upper part of the vertical sequence containing $a_1$
(i.e. $a_k$ is the upper endpoint
of the sequence). We say that $a_1$ is a tile of the level $k$.
We say that a point $A$ is a point of the level $k$, if $A$ belongs
to a tile of the level $k$. If $A$ belongs to several tiles,
we define the level of $A$ as a maximum.
The part 2) of this proof establishes the lemma for the points
of level 0.
The same reasoning
shows that if the lemma is true for the tiles of
the level $k$ then it is true for the tiles of the level $k+1$.
Therefore, the lemma is true for a tile of any level.
\end{proof}
\begin{lemma} \label{l_boka}
Any side of $P$ is tiled as shown in
Fig.~\ref{boka}.
\end{lemma}
\begin{figure}[!h]
\begin{center}
{\scriptsize
\input pic/boka.tex
}
\end{center}
\caption{Tilings of sides.}
\label{boka}
\end{figure}
\begin{proof}
Let $s$ be a side of $P$ and $b_1$, $b_2$, $b_3$ be the sides
of $F$.
Since $F$ is a Coxeter polyhedron, no angle of $b_i$ is obtuse.
Therefore, any horizontal line is perpendicular to any vertical line
(where the word "line" means an intersection of $s$ with
some mirror). By the same reason the lines are perpendicular to
the edges of $s$.
At most one line may be perpendicular to
the pair of opposite edges of a quadrilateral
in $\H^2$. Thus, there is at most one vertical line and at most
one horizontal line.
\end{proof}
\begin{lemma}\label{l_base}
The bases of $P$ are tiled as shown in Fig.~\ref{base}.
\end{lemma}
\begin{figure}[!h]
\begin{center}
{\scriptsize
\input pic/base.tex
}
\caption{Decompositions of bases}
\vspace{1pt}
(see the angles of fundamental triangles under
decompositions).
\end{center}
\label{base}
\end{figure}
\begin{proof}
Let $s$ be a base of $P$.
By Lemma~\ref{krishki} $s$ is either fundamental or a
quasi-Coxeter. The Coxeter decompositions of triangles are
classified in~\cite{Pink}.
By Lemma~\ref{l_boka} no edge of the triangle $s$ may be
decomposed into more than two parts. The decompositions
satisfying this condition are listed in Fig.~\ref{base}.
\end{proof}
\begin{theorem}
Let $P$ be a quasi-Coxeter triangular prism such that $F$ is a
triangular prism. Then the decomposition is one of the
decompositions listed in Table~\ref{prism}.
\end{theorem}
\begin{proof}
It follows from Lemma~\ref{lat}
that two bases of $P$ are decomposed in the same way.
By Lemma~\ref{l_base} there are five possibilities for
decompositions of bases.
Consider two cases.
1). Suppose that the bases are fundamental.
Then the sides are either fundamental or decomposed
like in Fig.~\ref{boka}(c).
Obviously all the sides of $P$ are decomposed in the
same way. This leads to the decomposition shown in the upper part
of Table~\ref{prism}
(since any mirror intersects a boundary of $P$,
there cannot be another decomposition).
2). Suppose that the bases are decomposed. Each decomposition of
the base corresponds to two possible decompositions of the
sides (with the horizontal line or without).
Thus, each decomposition of the base corresponds to
two decompositions of $P$. Some of the dihedral angles of $F$ are
uniquely determined by the combinatorial structure of the
decomposition. We need only to check that the rest dihedral angles
of $F$ can be prescribed in such a way that the dihedral
angles satisfies the condition of Andreev's theorem(see~\cite{A1}).
If the bases are decomposed as in
Fig.~\ref{base}(a) or \ref{base}(c), then it is possible to satisfy
the Andreev's theorem. Otherwise, it is impossible.
\end{proof}
\section{ Decompositions of prisms into tetrahedra}
In this section $P$ is a triangular prism and $F$ is a
tetrahedron.
Any face of $P$ is obviously tiled by triangles,
but in most cases this tiling is not a Coxeter
decomposition.
At first, we derive some information about tilings of bases.
\begin{define}
Let $ABCD$ be a fundamental tetrahedron. An edge $AB$ is called
{\bf $k$-edge} if the dihedral angle formed by $ABC$ and $ABD$
equals $\frac{\pi}{k}$.
\end{define}
\begin{lemma}\label{tile}
Let $p$ be a base of $P$. If any flat angle of $p$ is
fundamental then the tiling of $p$ consists of a unique
triangle. If $p$ has a flat angle decomposed into three parts
then $p$ is bounded by 2-edge, 3-edge and 5-edge.
\end{lemma}
\begin{proof}
Let $f$ be a face of $F$ and $\bar f $ be a plane containing
$f$. Obviously, $\bar f $ is tiled by triangles congruent
to the faces of $F$. To find this tiling consider a face $f$
and the adjacent to $f$ triangles $f_1$, $f_2$ and $f_3$ at the
plane $\bar f$.
These triangles $f_i$ are congruent either to
$f$ or to the other faces of $F$ (we obtain the same face if the
dihedral angle is equal to $\frac{\pi}{k}$, where $k$ is even,
otherwise we obtain another face). Adding face by face we can
prolong the tiling. The only problem is how many triangles are
incident to a fixed vertex.
To solve this problem suppose that $A$ is a vertex of $F$
incident to $k$-edge, $l$-edge and $m$-edge. Suppose that
$\bar f$ contains $A$. Then a decomposition of a small
three-dimensional spherical neighborhood of $A$ is similar to
a Coxeter decomposition of a sphere with fundamental triangle
($\frac{\pi}{k},\frac{\pi}{l},\frac{\pi}{m}$).
An intersection of the neighborhood with $\bar f$ corresponds
to a spherical line in the decomposition of the sphere.
So, to find how many rays starting from $A$ belong to $\bar
f$, it is sufficient to count a number of vertices incident to a
corresponding spherical line.
Thus, for any face of any Coxeter tetrahedron we can find a
tiling of the corresponding plane. In these tilings any triangle
with fundamental angles is a face of $F$. This proves the first
part of the lemma. A triangle with an angle decomposed into
three parts was found in one of these tilings only.
This triangle is bounded by
2-edge, 3-edge and 5-edge and tiled as shown in
Fig.~\ref{treug}.
\begin{figure}[!h]
\begin{center}
{\scriptsize
\input pic/treug.tex
}
\end{center}
\caption{The edge labeled by $k$ is a $k$-edge ($k=2,3,5$). }
\label{treug}
\end{figure}
\end{proof}
\begin{lemma}\label{l_imp}
Let $A$ be a non-fundamental vertex of $P$.
Then there exists a non-fundamental edge incident to $A$.
\end{lemma}
\begin{proof}
Suppose that any edge incident to $A$ is fundamental
(i.e. three dihedral angles incident to $A$ are fundamental).
Consider a small sphere $s$ centered in $A$. The decomposition
of $P$ restricted to $s$ is a Coxeter decomposition of a
spherical triangle $p=s\cap P$. Evidently, any angle of $p$
is fundamental.
It is easy to check that in this condition
$p$ is decomposed as shown in Fig.~\ref{fund_s}(a).
Any angle of $p$ is a right angle.
\begin{figure}[!h]
\begin{center}
{\scriptsize
\input pic/imp.tex
}
\caption{If $A$ is a non-fundamental vertex
but the edges ended in $A$ are fundamental,
then $AB$ and $AC$ are $2$-edges and the tiling of $\triangle ABC$
is impossible. }
\label{fund_s}
\end{center}
\end{figure}
Thus the neighborhood of $A$ is decomposed as shown in
Fig.~\ref{fund_s}(b). Any edge of $P$ incident to $A$ is a
2-edge. Consider a base $ABC$ of the prism $P$. The angle $A$ of
this triangle is decomposed into three parts by the lines $AN$
and $AM$. It follows from Lemma~\ref{tile} that the sides
of $ABC$ should be 2-edge, 3-edge and 5-edge. This contradicts
to the fact that $AB$ and $AC$ are 2-edges.
\end{proof}
\begin{lemma}
Let $P$ be a triangular prism admitting a Coxeter decomposition
into tetrahedra. Then $P$ has a non-fundamental dihedral angle.
\end{lemma}
\begin{proof}
Suppose that any dihedral angle of $P$ is fundamental.
Then by Lemma~\ref{l_imp}
any
vertex of $P$ is fundamental. By Lemma~\ref{tile} any base of $P$ is
congruent to a single face of $F$. Consider a fundamental tetrahedron
$F_0$ containing a base $\alpha$. Since any dihedral angle of $P$ is
fundamental, the sides of $P$ are the faces of $F_0$. This is
impossible.
\end{proof}
\subsection*{How to find all decompositions of prisms into tetrahedra }
We have already proved that any prism with tetrahedral
fundamental polyhedron has a non-fundamental dihedral angle.
It follows from Fig.~\ref{mirrors}(1)--(4) that any prism
consists of the tetrahedra, the smaller prisms and
the quadrilateral pyramid. If we know the decompositions of the smaller
parts we can find the decomposition of the whole prism.
Recall that a prism $P$ is called {\bf minimal} if $P$
is non-fundamental and any prism inside $P$ is fundamental.
\begin{define}
We say that a minimal prism is a {\bf prism of level 0}.
A non-fundamental prism $P$ is a {\bf prism of level $k+1$}
if $P$ contains a prism of level $k$ but any prism
inside $P$ contains no prism of level $k$.
\end{define}
At first, we classify decompositions of minimal
prisms. Evidently, a minimal prism contains no mirrors shown in
Fig.~\ref{mirrors}(2)--\ref{mirrors}(4).
Thus, it contains a mirror shown in Fig.~\ref{mirrors}(1).
Let this mirror be $A_1B_2B_3$ in Fig.~\ref{3tetr}.
Then $A_1$ is a non-fundamental vertex, and
by Lemma~\ref{l_imp} there is a mirror containing an edge ended in $A_1$.
This edge cannot be $A_1B_1$, since the prism is minimal.
Hence, one of the edges $A_1A_2$ and $A_1A_3$ (say, $A_1A_2$)
is non-fundamental.
Therefore, the prism is decomposed into three tetrahedra.
Decompositions of tetrahedra are classified in~\cite{arxiv-tetr}.
Thus, we can classify decompositions of minimal prisms.
\begin{figure}[!h]
\begin{center}
\input pic/3tetr.tex
\caption{A minimal prism consists of three (possibly
non-fundamental) tetrahedra.}
\label{3tetr}
\end{center}
\end{figure}
Suppose that we know the classification of decompositions for
the prisms of the levels smaller than $k+1$. Now we can show how to
find a classification for the prisms of the level $k+1$. Suppose
that $P$ has a dihedral angle decomposed as shown in
Fig.~\ref{mirrors}(2)--\ref{mirrors}(4). By the definition,
any prismatic part is a prism of the level smaller than $k+1$.
Thus, we know the list of possible decompositions of
prismatic parts. Decompositions of tetrahedral parts
are known too. Therefore, we can classify all
possible decompositions of $P$. Suppose now that $P$ has no
mirrors shown in Fig.~\ref{mirrors}(2)--\ref{mirrors}(4).
Then $P$ has a mirror shown in Fig.~\ref{mirrors}(1).
Moreover, there are two mirrors of this type decomposing
$P$ into three tetrahedra
(we use Lemma~\ref{l_imp} again). Combining the decompositions
of these tetrahedra we can find all possible decompositions of
$P$.
Thus, we have an algorithm leading to the classification of
decompositions for the prisms which level is smaller than any
fixed number. In fact, there is no prism of level eight;
hence, there cannot be prism of greater level. Therefore,
this algorithm classifies all decompositions of bounded
triangular prisms. See Table~2 for the classification.
\clearpage
\section*{Tables}
Table~1 contains all decompositions of bounded triangular
prisms which fundamental polyhedron is a triangular prism.
Table~2 contains the list of decompositions of convex bounded
pyramids and triangular prisms which fundamental polyhedron is a
tetrahedron. Here we describe the structure of Table~2.
\begin{itemize}
\item
Horizontal lines separate polyhedra with different
fundamental tetrahedra. For any fundamental tetrahedron the
table contains three columns. The left column contains the list of
decompositions of bounded tetrahedra
(the tetrahedron number 0 is the fundamental one).
The right column lists decompositions of bounded
triangular prisms. The column at the middle contains bounded
pyramids. At first we list quadrilateral pyramids which can
be combined from two (possibly non-fundamental) tetrahedra.
Then, after a dotted line, we list the rest quadrilateral
pyramids. After a new dotted line we list pentagonal
pyramids and at last we list hexagonal pyramids.
(There is no Coxeter decomposition of heptagonal pyramid).
\end{itemize}
The polyhedra and their decompositions are represented in
Table~2 as it is described below.
\begin{itemize}
\item
A tetrahedron with dihedral angles
$\frac{k_i\pi}{q_i}$ $i=1$,...,6 is represented by the
following diagram:
the nodes of the diagram correspond to the faces of the tetrahedron,
two nodes are connected by a $q$-fold edge decomposed into $k$ parts
if the corresponding faces form up a dihedral angle $\frac{k\pi}{q}$.\\
The numerator $k_i$ is a number of parts in the dihedral angle
(for example, a fraction $\frac{2}{4}$ denotes a right angle
decomposed into two parts).
The faces of the tetrahedron are numbered by the following way:
the nodes of the diagram have the numbers 0,1,2,3 from the left to the
right. (The numeration will be used below).
\item
A quadrilateral pyramid $OA_1A_2A_3A_4$ with the base $A_1A_2A_3A_4$
is represented by eight dihedral angles
$$(\widehat{A_1A_2},\widehat{A_2A_3},\widehat{A_3A_4},\widehat{A_4A_1};
\widehat{OA_1},\widehat{OA_2},\widehat{OA_3},\widehat{OA_4}).$$
The dihedral angles are the rational fractions,
where the numerator $q$ means that the corresponding dihedral angle
is decomposed into $q$ parts (we always omit a multiple $\pi$).
A triangular face $OA_iA_{i+1}$
has a number $i$.
\item
By the same way we represent pentagonal and hexagonal
pyramids (we use ten and twelve angles correspondingly).
\item
A triangular prism $A_1A_2A_3B_1B_2B_3$ (where
$A_1A_2A_3$ and $B_1B_2B_3$ are the bases) is represented by nine
dihedral angles
$$(\widehat {A_3B_3},\widehat {A_1B_1},\widehat {A_2B_2};
\widehat {A_1A_2}; \widehat {A_2A_3},\widehat {A_3A_1};
\widehat {B_1B_2},\widehat {B_2B_3},\widehat {B_3B_1}).$$
The dihedral angles are the rational fractions, where
the numerator is a number of parts in the dihedral
angle and the multiple $\pi$ is omitted. The faces are numbered
in the following way: $A_1A_2A_3$ has number 0,
$B_1B_2B_3$ has number 1, nd $A_iA_{i+1}B_{i+1}B_i$ has number
$i+1$.
\item
Any decomposition which is a superposition of some other decompositions
is labeled by a star.
\end{itemize}
\Remark
In this notation each pyramid or triangular prism can be written
by several ways (for instance, the order of angles changes
while we rotate the polyhedron). To unify the notation we
choose a record which stands first in the alphabetical order.
\begin{itemize}
\item
For any polyhedron except fundamental tetrahedra we
need some way to reconstruct the decomposition.
For this aim we put some numbers under the diagram of the
polyhedron. These seven numbers (denoted by
$t$,$k$,$l$ and $m$,$n$,$p$,$q$) show the following:\\
$t$ is a number of the polyhedron (for any fundamental
tetrahedron we start a new numeration),\\
$k$ is a number of fundamental tetrahedra in the decomposition,\\
$l$ is a number of gluings used to obtain this decomposition
(if the decomposition was obtained by a gluing of two polyhedra
with $l=l_1$ and $l=l_2$ then
$l=1+\max\{l_1,l_2\}$),\\
$m$ and $n$ are the numbers of the polyhedra which should be
glued together to obtain the decomposition.\\
$p$ and $q$ are the numbers of the glued faces of the polyhedra
$m$ and $n$ respectively,\\
The numbers $m$ and $n$ are accompanied by the labels:
we write "tet", "pyr" or "pri"
to show that a corresponding polyhedron
is a tetrahedron, a quadrilateral pyramid or a triangular prism.
(Any tetrahedron admitting a Coxeter decomposition consists of
two smaller tetrahedra,
so we omit the label "tet" in the left column.)
\end{itemize}
\begin{table}[!h]
\caption{Coxeter decompositions of prisms into prisms.}
\begin{center}
\input pic/prism_e.tex
\end{center}
\label{prism}
\end{table}
\clearpage
\begin{center}
{\normalsize
Table~2. Coxeter decompositions of pyramids and prisms\\
which fundamental polyhedron is a tetrahedron}
\vspace{15pt}
\input pic/tabpr4e.tex
\end{center}
\pagebreak
\clearpage | 0.031788 |
\begin{document}
\begin{frontmatter}
\newtheorem{theo}{Theorem}
\newtheorem{lemma}[theo]{Lemma}
\newtheorem{remark}[theo]{Remark}
\newtheorem{cor}[theo]{Corollary}
\newtheorem{proof}[theo]{Proof}
\newtheorem{definition}[theo]{Definition}
\title{Interpolated variational iteration method for initial value problems}
\author{Davod Khojasteh Salkuyeh$^\dag$\footnote{Corresponding author} and Ali Tavakoli$^\S$ }
\address{$^\dag$Faculty of Mathematical Sciences, University of Guilan, Rasht, Iran\\
e-mail: [email protected], [email protected]\\[2mm]
$^\S$Mathematics Department, Vali-e-Asr University of Rafsanjan, Iran.\\
e-mail: [email protected]}
\begin{abstract}
\indent In order to solve an initial value problem by the variational iteration method, a sequence of functions is produced which converges to the solution under some suitable conditions. In the nonlinear case, after a few iterations the terms of the sequence become complicated, and therefore, computing a highly accurate solution would be difficult
or even impossible. In this paper, for one-dimensional initial value problems, we propose a new
approach which is based on approximating each term of the sequence by a piecewise linear function.
Moreover, the convergence of the method is proved. Three illustrative examples are given to show the superiority
of the proposed method over the classical variational iteration method.
\end{abstract}
\begin{keyword}
Variational iteration method \sep piecewise linear function \sep interpolated \sep convergence.
\MSC 34A34, 65L05.
\end{keyword}
\end{frontmatter}
\section{Introduction} \label{SEC1}
The He's variational iteration method (VIM) \cite{He1,He4} is a powerful mathematical technique to solve linear and nonlinear problems that can be implemented easily in practice. It has been successfully applied for solving various ODEs and PDEs \cite{He1,HeIJMPB,He4,He7,He8,He9}. Convergence of the VIM has been investigated in some papers \cite{SalkuyehCAMWA,Tatari}. In \cite{Zhao}, Zhao and Xiao applied the VIM for solving singular perturbation initial value problems and investigated its convergence. Recently, Wu and Baleanu in \cite{WuBaleanu2} have introduced a new way to define the Lagrange multipliers in the VIM for solving fractional differential equations with the Caputo derivatives. They also developed the VIM to $q$-fractional difference equations in \cite{WuBaleanu1}. The review article \cite{GCWu} concerns some new applications of the VIM to numerical simulations of differential equations and fractional differential equations.
Consider the following differential equation
\begin{equation}\label{Eq001}
\mathcal{L}u(t)+\mathcal{N}u(t)=g(t),
\end{equation}
\noindent where $\mathcal{L}$ and $\mathcal{N}$ are, respectively, linear and nonlinear operators and $g$ is an inhomogeneous term.
Given an initial guess $u_0(t)$, the VIM to solve (\ref{Eq001}) takes the following form
\begin{equation}\label{Eq002}
u_{m+1}(t)= u_m(t)+\int_{t_0}^{t} \lambda(s,t) (\mathcal{L}u_n(s)+
\mathcal{N}u_m(s)-g(s))ds,\quad m=0,1,2,\ldots.
\end{equation}
where $\lambda$ is a general Lagrange multiplier (for more details see \cite{He1,He4,He7}).
As we see the VIM generates a sequence of functions that can converge to the solution of the problem under some suitable conditions. After a few iterations each term of this sequence often involves a definite integral whose integrand contains several nonlinear terms. In this case, computing a high accuracy solution would be difficult or even impossible by the common softwares such as MAPLE, MATHEMATICA or MATLAB. See \cite{Abbasbandy} for an application of the VIM to quadratic Riccati differential equations that illustrates this. To overcome this problem, Geng et al. in \cite{Geng} have introduced a piecewise variational iteration method for the quadratic Riccati differential equation. Their main idea is that for solving the problem in a large interval, the interval is split into several subintervals and the problem is solved in each subinterval by the VIM in a progressive manner. In this way, the accuracy of the computed solution can be considerably improved. However, the main mentioned problem is still unsolved, since in each subinterval the VIM is directly implemented.
In this paper, to solve one-dimensional initial value problems, we modify the VIM in such a way that each term of the sequence is interpolated by a piecewise linear function. Hereafter, the proposed method is called the IVIM (for interpolated VIM). In spite of the VIM, the IVIM does not need any symbolic computation and all of the computations are done numerically. Therefore, we can compute hundreds of the sequence terms in a small amount of time.
The rest of the paper is organized as follows. The IVIM is introduced in Section \ref{SEC3}. Convergence of the proposed method is discussed in Section \ref{SEC3}. Section \ref{SEC4} is devoted to presenting three numerical examples. Some concluding remarks are given in Section \ref{SEC5}.
\section{The IVIM } \label{SEC2}
Consider the one-dimensional initial value problem
\begin{equation}\label{Eq003}
\left\{
\begin{array}{ll}
u'(t)=f(t,u(t)), & t\in [a,T], \\
u(a)=u_a.
\end{array}
\right.
\end{equation}
For simplicity, we assume that $u_a=0$, otherwise one can use a simple change of variable $\tilde{u}=u-u_a$ to obtain $\tilde{u}(a)=0$. In this case, Eq. (\ref{Eq002}) becomes
\begin{equation}\label{Eq004}
u_{m+1}(t)= u_m(t)+\int_{a}^{t} \lambda(s,t) \left( u_m'(s)-f(s,u_m(s))\right)ds,\quad m=0,1,2,\ldots,
\end{equation}
where $u_0(t)$ satisfies the initial condition $u_0(a)=0$. By integration by parts, Eq. (\ref{Eq004}) can be rewritten as
\begin{equation}\label{Eq005}
u_{m+1}(t)= G_m(t)-\int_{a}^{t}H_m(s,t)ds,
\end{equation}
where
\begin{eqnarray*}
G_m(t) \hspace{-0.2cm}&=&\hspace{-0.2cm} (1+\lambda(t,t))u_m(t)-\lambda(a,t)u_m(a),
\end{eqnarray*}
and
\begin{eqnarray*}
H_m(s,t) \hspace{-0.2cm}&=&\hspace{-0.2cm} \frac{\partial \lambda}{\partial s}(s,t)~u_m(s)+\lambda(s,t)f(s,u_m(s)).
\end{eqnarray*}
In order to present the IVIM, we take a natural number $n$ and discretize the interval $[a,T]$ into $n-1$ subintervals with step size $h=(T-a)/(n-1)$ and grid points
\[
t_i=a+(i-1)h,\quad i=1,2,\ldots,n.
\]
Now, we define the well-known B-spline basis functions (see \cite{Cheney}) of first-order on the nodal points $t_i$, i.e.
\begin{eqnarray*}
\varphi_i(t) &=& \left\{
\begin{array}{ll}
\displaystyle\frac{t-t_{i-1}}{h}, & t_{i-1} \leq t < t_i, \\[4mm]
\displaystyle\frac{t_{i+1}-t}{h}, & t_{i} \leq t \leq t_{i+1}, \hspace{2cm} i=2,3,\ldots,n-1,\\[4mm]
\quad 0 , & t<t_{i-1}~\textrm{or}~t>t_{i+1},\\[4mm]
\end{array}
\right. \\
\varphi_n(t) &=&\left\{
\begin{array}{ll}
\displaystyle\frac{t-t_{n-1}}{h}, & t_{n-1} \leq t \leq t_n, \\[4mm]
\quad 0 , & \quad t<t_{n-1},
\end{array}
\right. \\
\end{eqnarray*}
on $[a,T]$. Each function $\varphi_i$ is equal to $1$ at grid point $t_i$, and equal to zero at other grid points. Let
\[
X_h=\textrm{span}\{\varphi_{i}:i=2,3,\ldots,n\}.
\]
Every $v^{(h)}$ in $X_h$ is a piecewise linear function of the form
\[
v^{(h)}(t)=\sum_{i=2}^n \alpha_i \varphi_i (t),\quad t\in [a,T].
\]
Obviously, $\alpha_i$ is the value of $v^{(h)}$ at the grid point $x_i$, i.e., $\alpha_i=v^{(h)}(t_i)$. Furthermore, $v^{(h)}(a)=0$, since $\varphi_i (a)=0$, $i=2,3,\ldots,n$. See \cite{Reddy} for more details about the properties of this kind of functions.
We now return to Eq. (\ref{Eq005}). The aim is to recursively compute $u_{i}$'s. Let $u_m$ has been computed and we intend to compute $u_{m+1}$.
The main idea of our method is to compute a piecewise linear interpolation $u_{m+1}^{(h)}$ of $u_{m+1}$ in $X_h$, instead of computing $u_{m+1}(t)$.
In fact, instead of $u_{m+1}$, we consider its projection onto $X_h$. To do so, we have
\begin{equation}\label{Eq006}
u_{m+1}^{(h)}(t)=\displaystyle \sum_{r=2}^n u_{m+1}(t_r)\varphi_{r}(t) ,\quad t\in [a,T].
\end{equation}
All we need here is to compute $u_{m+1}(t_i)$, $i=2,3,\ldots,n$. By using Eq. (\ref{Eq005}), we have
\begin{equation}\label{Eq007}
u_{m+1}(t_i)= G_m(t_i)-\int_{a}^{t_i}H_m(s,t_i)ds.
\end{equation}
Now, we can approximate the integral term in the latter equation by different methods of numerical integration. But, we again compute a piecewise linear interpolation of $H_m(s,t_i)$ in $X_h$ as
\[
H_m^{(h)}(s,t_i)=\sum_{r=2}^n H_m(t_r,t_i)\varphi_r(s),\quad s\in [a,t_i].
\]
Replacing $H_m(s,t_i)$ by $H_m^{(h)}(s,t_i)$ in Eq. (\ref{Eq004}) yields
\begin{equation}\label{Eq008}
u_{m+1}(t_i) \approx G_m(t_i)-\sum_{r=2}^n H_m(t_r,t_i)\int_{a}^{t_i} \varphi_r(t)dt =G_m(t_i)-\sum_{r=2}^n H_m(t_r,t_i)\mu_{ri},
\end{equation}
where $\mu_{ri}=\displaystyle\int_{a}^{t_i} \varphi_r(t)dt$, $r=2,3,\ldots,n$. It is easy to see that
\[
\mu_{ri}=\left\{
\begin{array}{ll}
0, & i\leq r-1, \\[2mm]
\displaystyle \frac{h}{2}, & i=r, \\ [4mm]
\displaystyle h , & i\geq r+1.
\end{array}
\right.
\]
Now, from this and Eq. (\ref{Eq008}) we get
\begin{equation}\label{Eq009}
u_{m+1}(t_i)\approx \hat{u}_{m+1}(t_i)=G_m(t_i)-h\sum_{r=2}^{i-1} H_{m}(t_r,t_i)-\frac{h}{2}H_m(t_i,t_i),\quad i=2,3,\ldots,n.
\end{equation}
Then, from (\ref{Eq006}) we set
\begin{equation}\label{Eq010}
u_{m+1}^{(h)}(t)\approx \displaystyle \sum_{r=2}^n \hat{u}_{m+1}(t_r)\varphi_{r}(t).
\end{equation}
\noindent {\bf Remark 1.} Consider the special case that $\mathcal{L}u\equiv u'+\alpha u$, for some $\alpha \in \mathbb{R}$. This means that $f(t,u(t))=-\alpha u(t)-\mathcal{N}u(t)+g(t)$, where $\mathcal{N}u(t)$ is the nonlinear term appeared in Eq. (\ref{Eq001}). In this case, we have $\lambda(s,t)=-e^{\alpha(s-t)}$, and therefore $1+\lambda(t,t)=0$. Obviously, from Eq. (\ref{Eq004}) we see that $u_m(a)=0$, for $m=0,1,\ldots$. Hence, $G_m(t)=0$, for $m=0,1,\ldots$. On the other hand, we have
\begin{eqnarray}
\nonumber H_m(s,t) \hspace{-0.2cm}&=&\hspace{-0.2cm} e^{\alpha(s-t)}(-\alpha u_m(s)+f(s,u_m(s))) \\
\nonumber \hspace{-0.2cm}&=&\hspace{-0.2cm} e^{\alpha(s-t)}(-\alpha u_m(s)+\alpha u_m(s)+\mathcal{N}u_m(s)-g(s))\\
\hspace{-0.2cm}&=&\hspace{-0.2cm} e^{\alpha(s-t)}(\mathcal{N}u_m(s)-g(s)). \label{Eq011}
\end{eqnarray}
\section{Convergence of the method}\label{SEC3}
In this section, we present the convergence of the VIM and the IVIM. We first recall a lemma which gives conditions for the existence and uniqueness of a solution to the problem (\ref{Eq003}).
\begin{lemma}
Consider the initial-value problem (\ref{Eq003}), where $f:[a,T]\times \Bbb R\rightarrow\Bbb R$ is continuous with respect to the first variable and satisfies a Lipschitz condition with respect to the second variable, i.e.,
\begin{equation}\label{conv00}
|f(t,u(t))-f(t,v(t))|\leq \gamma(t) |u(t)-v(t)|,\quad t\in [a,T],
\end{equation}
where $\gamma(t)$ is a nonnegative continuous function. Then, there exists a unique solution to the problem.
\end{lemma}
\noindent{\bf Proof}. See \cite{buc,tav}.\\
To establish convergence of the VIM, we state the following theorem.
\begin{theo}\label{theo1}
Let $f:[a,T]\times C^1([a,T])\rightarrow\Bbb R$ be a continuous function with respect to its first variable which satisfies the Lipschitz condition (\ref{conv00}).
If $u_m(t)\in C^1([a,T])$ for $m=0,1,2,\ldots$, then the sequence defined by (\ref{Eq004}) converges to the solution of (\ref{Eq003}).
\end{theo}
\noindent {\bf Proof}.
By Eq. (\ref{Eq004}), we have
\begin{equation}\label{conv01}
u_{m+1}(t)= \int_{a}^{t} ((1+\lambda(s,t)) u_m'(s)-\lambda(s,t)f(s,u_m(s)))ds,\quad m=0,1,2,\ldots,
\end{equation}
since $u_m(a)=0$.
Let $u(t)$ be the exact solution of Eq. (\ref{Eq003}). By Eq. (\ref{conv01}), it is readily seen that
\begin{eqnarray}\label{conv02}
u(t)= \int_{a}^{t} ((1+\lambda(s,t)) u'(s)-\lambda(s,t)f(s,u(s)))ds.
\end{eqnarray}
Let $E_m(t)=u_m(t)-u(t)$ be the error produced by the VIM at $m$th iteration.
Since the first derivative of $E_0(t)$ is continuous on the closed interval $[a,T]$, there exists a constant $M$ such that
\begin{eqnarray}\label{conv04}
|E_0'(t)|\leq M|E_0(t)|,\qquad \forall t\in[a,T].
\end{eqnarray}
From Eqs. (\ref{conv01}), (\ref{conv02}) and (\ref{conv00}), it follows that
\begin{eqnarray}
\nonumber |E_{1}(t)| \hspace{-0.2cm}&=&\hspace{-0.2cm} |\int_{a}^{t} ((1+\lambda(s,t)) E_0'(s)-\lambda(s,t)(f(s,u_0(s))-f(s,u(s))))ds| \\
\hspace{-0.2cm}& \leq &\hspace{-0.2cm} \int_{a}^{t} |1+\lambda(s,t)||E_0'(s)| ds+\int_{a}^{t}|\gamma(s)\lambda(s,t)||E_0(s)| ds. \label{conv05}
\end{eqnarray}
Then, it turns out that by Eqs. (\ref{conv04}) and (\ref{conv05}),
\begin{equation}\label{conv06}
|E_{1}(t)|\leq M_a \int_{a}^{t} |E_0(s) | ds,
\end{equation}
in which
\[
M_a=\max\large\{M \max_{s,t\in[a,T]} |1+\lambda(s,t)|, \max_{s,t \in[a,T]}|\gamma(s)\lambda(s,t)| \large\}.
\]
Let $\|E_i\|=\displaystyle\max_{t\in [a,T]} |E_i(t)|,$ $ i=0,1,\ldots$. Hence,
\begin{equation}\label{conv07}
|E_{1}(t)|\leq \|E_{0}\| M_a(t-a).
\end{equation}
Therefore, similar to Eq. (\ref{conv06}) and using Eq. (\ref{conv07}), we deduce
\begin{eqnarray*}
|E_{2}(t)| \hspace{-0.2cm}& \leq &\hspace{-0.2cm} M_a\int_{a}^{t}|E_1(s)| ds \leq M_a^2 \|E_{0}\|\int_{a}^{t}(s-a) ds = \|E_{0}\|\frac{M_a^2(t-a)^2}{2!},\\
|E_{3}(t)| \hspace{-0.2cm}& \leq &\hspace{-0.2cm} M_a\int_{a}^{t}|E_2(s)| ds \leq M_a^3 \|E_{0}\|\int_{a}^{t}(s-a)^2 ds = \|E_{0}\|\frac{M_a^3(t-a)^3}{3!},
\end{eqnarray*}
and in general,
\begin{eqnarray*}
|E_{m}(t)|\leq \|E_{0}\| \frac{M_a^m(t-a)^m}{m!}\leq \|E_{0}\|\frac{M_a^m(T-a)^m}{m!}.
\end{eqnarray*}
Then $\|E_{m}\|\rightarrow 0$ as $m$ tends to $\infty$ and the convergence of the method is established.\qquad $\Box$
\bigskip
Now, we are ready to prove the convergence of the IVIM.
\bigskip
\begin{theo}
Let $f:[a,T]\times C^1([a,T])\rightarrow\Bbb R$ be two times continuously differentiable function with respect to the first variable and satisfies the Lipschitz condition (\ref{conv00}). If the VIM of (\ref{Eq004}) is convergent, then so is the IVIM of (\ref{Eq010}).
\end{theo}
{\bf Proof.}
Define $\bar{u}_{m}(t):=\sum\limits_{i=2}^n \hat{u}_{m}(t_i)\varphi_{i}(t)$. For each grid point $t_r$, we have $\bar{u}_{m}(t_r)=\hat{u}_{m}(t_r)$. Also, by Eq. (\ref{Eq009}), we have
\begin{eqnarray}
\nonumber \hat{u}_{m+1}(t_i) \hspace{-0.2cm}&= &\hspace{-0.2cm} (1+\lambda(t_i,t_i))u_m(t_i)-h\sum\limits_{r=2}^{i-1} (\frac{\partial\lambda}{\partial s}(s,t_i)\mid_{s=t_r}u_m(t_r)+\lambda(t_r,t_i)f(t_r,u_m(t_r)))\\
\hspace{-0.2cm}& &\hspace{-0.2cm} -\frac{h}{2}((\frac{\partial\lambda}{\partial s}(s,t_i)\mid_{s=t_i}u_m(t_i)+\lambda(t_i,t_i)f(t_i,u_m(t_i))). \label{conv08}
\end{eqnarray}
On the other hand, by Eq. (\ref{Eq005}) it is clear that the exact solution $u(t)$ at $t=t_i$ satisfies the following relation
\begin{eqnarray}\label{conv09}
u(t_i)=(1+\lambda(t_i,t_i))u(t_i)-\int\limits_a^{t_i}(\frac{\partial \lambda}{\partial s}(s,t_i)~u(s)+\lambda(s,t_i)f(s,u(s)))ds.
\end{eqnarray}
Now, by the trapezoidal rule of integration, we have\\[4mm]
$\displaystyle \int_a^{t_i}(\frac{\partial \lambda}{\partial s}(s,t_i)~u(s)+\lambda(s,t_i)f(s,u(s)))ds$
\begin{eqnarray}
\nonumber \qquad \hspace{-0.2cm}& = &\hspace{-0.2cm} \frac{h}{2}(\frac{\partial\lambda}{\partial s}(s,t_i)\mid_{s=t_i}u(t_i)+\lambda(t_i,t_i)f(t_i,u(t_i))+\lambda(a,t_i)f(a,0)) \\
\nonumber \hspace{-0.2cm}& &\hspace{-0.2cm} +h\sum_{r=2}^{i-1} (\frac{\partial\lambda}{\partial s}(s,t_i)\mid_{s=t_r}u(t_r) +\lambda(t_r,t_i)f(t_r,u(t_r)))\\
\hspace{-0.2cm}& &\hspace{-0.2cm} -\frac{t_i-a}{12}h^2 \frac{d^2}{ds^2}f(s,u(s))|_{s=\xi}, \label{conv10}
\end{eqnarray}
where $\xi\in(a,t_i)$. Let $\bar{E}_m(t_i)=\bar{u}_{m}(t_i)-u(t_i)=\hat{u}_{m}(t_i)-u(t_i)$ and $E_m(t_i)=u_m(t_i)-u(t_i)$. By Eqs. (\ref{conv08}), (\ref{conv09}) and (\ref{conv10}), we get
\begin{eqnarray*}
\nonumber |\bar{E}_{m+1}(t_i)| \hspace{-0.2cm}& = &\hspace{-0.2cm} |\hat{u}_{m+1}(t_i)-u(t_i)| \\
\nonumber \hspace{-0.2cm}& = &\hspace{-0.2cm} |(1+\lambda(t_i,t_i))E_m(t_i)-h\sum\limits_{r=2}^{i-1} (\frac{\partial\lambda}{\partial s}(s,t_i)\mid_{s=t_r}E_m(t_r)\vspace{.3cm}\\
\nonumber \hspace{-0.2cm}& & \hspace{-0.2cm} +\lambda(t_r,t_i)(f(t_r,u_m(t_r))-f(t_r,u(t_r))))
-\frac{h}{2}(\frac{\partial\lambda}{\partial s}(s,t_i)\mid_{s=t_i}E_m(t_i)\\
\nonumber \hspace{-0.2cm}& &\hspace{-0.2cm}+\lambda(t_i,t_i)(f(t_i,u_m(t_i))-f(t_i,u(t_i))))-\frac{h}{2}\lambda(a,t_i)f(a,0)\\
\hspace{-0.2cm}& & \hspace{-0.2cm} +\frac{t_i-a}{12}h^2 \frac{d^2}{ds^2}f(s,u(s))|_{s=\xi}|.
\end{eqnarray*}
Now, by the Lipschitz condition, it follows that
\begin{eqnarray*}
\nonumber|\bar{E}_{m+1}(t_i)| \hspace{-0.2cm}& \leq &\hspace{-0.2cm} |1+\lambda(t_i,t_i)|~|E_m(t_i)|+h\sum_{r=2}^{i-1} (|\frac{\partial\lambda}{\partial s}(s,t_i)\mid_{s=t_r}|~ |E_m(t_r)|\\
\nonumber\hspace{-0.2cm}&&\hspace{-0.2cm} +|\gamma(t_r)\lambda(t_r,t_i)|~|E_m(t_r)|)
+\frac{h}{2}(|\frac{\partial\lambda}{\partial s}(s,t_i)\mid_{s=t_i}|~|E_m(t_i)|\\
\nonumber\hspace{-0.2cm}&&\hspace{-0.2cm}+|\gamma(t_i)\lambda(t_i,t_i)|~|E_m(t_i)|+\frac{h}{2}|\lambda(a,t_i)f(a,0)|
+\frac{t_i-a}{12}h^2 | \frac{d^2}{ds^2}f(s,u(s))|_{s=\xi}|\\
\nonumber\hspace{-0.2cm}&\leq&\hspace{-0.2cm} K_1|E_m(t_i)|+hK_2\sum_{r=2}^{i-1} |E_m(t_r)|+\frac{h}{2}K_2 |E_m(t_i)|\\
\hspace{-0.2cm}&&\hspace{-0.2cm} +
\frac{h}{2}|\lambda(a,t_i)f(a,0)|+\frac{t_i-a}{12}h^2K_3,
\end{eqnarray*}
where $|1+\lambda(t,t)|\leq K_1$, $|\frac{\partial\lambda}{\partial s}(s,t)|+|\gamma(t)\lambda(s,t)|\leq K_2$
and $|\frac{d^2}{ds^2}f(s,u(s))|\leq K_3$ for some constants $K_1,K_2$ and $K_3$. Now, if the VIM is convergent, then $|E_m(t_i)|\rightarrow 0$ as $m\rightarrow\infty$ and therefore, as a result when $m\rightarrow \infty$ and $h\rightarrow 0$, we have $|\bar{E}_{m}(t_i)|\rightarrow 0$ for each grid point $t_i$.
It is necessary to mention that
\[
\lim_{m\rightarrow\infty}\lim_{h\rightarrow 0} h\sum_{r=2}^{i-1}|E_{m}(t_r)|=\lim_{m\rightarrow\infty}\int_{x_1}^{x_{i-1}} |E_{m}(t)|dt=0.
\]
Therefore the proof is completed.
\qquad $\Box$
\section{Illustrative examples}\label{SEC4}
In this section, we present three numerical examples to show the effectiveness of our method. In order to do so, we compare the results of the IVIM to those of the VIM.
\bigskip
\noindent{\bf Example 1.} We consider the quadratic Riccati differential equation of the form
\begin{equation}\label{Eq012}
\left\{
\begin{array}{ll}
u'(t)=2u(t)-u^2(t)+1, & t\in [0,1], \\
u(0)=0.
\end{array}
\right.
\end{equation}
The exact solution of (\ref{Eq012}) is given by (see for example \cite{Abbasbandy})
\[
u(t)=1+\sqrt{2} \tanh \left(\sqrt{2}t+\frac{1}{2} \log \frac{\sqrt{2}-1}{\sqrt{2}+1} \right).
\]
We assume that $\mathcal{L}u=u'-2u$, $\mathcal{N}u= u^2$ and $g\equiv 1$. By setting $\alpha=-2$, according to Remark 1 we get $G_m\equiv 0$ and
\[
H_m(s,t)=e^{2(t-s)}(u^2_m(s)-1),
\]
for $(s,t)\in [0,1]\times [0,1]$. Therefore, Eq. (\ref{Eq005}) takes the following form
\[
u_{m+1}(t)=\int_{0}^t e^{2(t-s)}(1-u^2_m(s))ds,
\]
which is equivalent to the iteration produced by the VIM. In Figure \ref{Fig1-Ex1}, the approximate solutions computed by the VIM and the IVIM for different values of $m$ together with the exact solution are depicted. {In the IVIM}, we have considered $n=41$. The figure shows that the solution computed by the IVIM is in good agreement with that of the VIM and both of them converge to the exact solution. Table \ref{Tbl1} shows a comparison between the CPU times (in seconds) for computing the approximate solutions computed by these two methods. As we see, the CPU times for computing the approximated solution obtained by the VIM increase drastically as $m$ increases, whereas those of the IVIM are negligible.
For more investigation we set $m=10$ and $n=33,65,129,257$. In Figure \ref{Fig2-Ex1} the $\log_{10}$ of the absolute error of the computed solutions by the IVIM, denoted by $\log_{10} {\rm Err}$, are displayed. The CPU times for all of the four values of $n$ are $0.000$. Similarly, for $m=40$ and $n=1000,2000,3000,4000$, the results are shown in Figure \ref{Fig3-Ex1}. Here the CPU times for computing the approximate solutions by the IVIM are $0.265,0.859,1.891,3.203$, respectively, for $n=1000,2000,3000,4000$. These figures show that the computed solutions obtained by the IVIM are in good agreement with the exact solution. Moreover, the results show that the CPU times for setting up the IVIM is very small in comparison to the VIM.
\begin{table}
\caption{CPU times (in seconds) for computing the approximate solutions by the VIM and the IVIM for Example 1.}
\vspace{-0.2cm}
\begin{center}
\begin{tabular}{lcccccccccc}\hline
$m$ \qquad\qquad & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline
VIM & 0.015 & 0.093 & 0.109 & 0.204 & 0.672 & 2.485 & 20.034 & 383.469\\
IVIM & 0.000 & 0.000 & 0.000 & 0.000 & 0.000 & 0.000 & 0.000 & 0.000\\ \hline
\end{tabular}
\end{center}
\label{Tbl1}
\end{table}
\begin{figure}
\centering
\includegraphics[height=6cm,width=8cm]{Fig1-Ex1.eps}\includegraphics[height=6cm,width=8cm]{Fig2-Ex1.eps}
\includegraphics[height=6cm,width=8cm]{Fig3-Ex1.eps}\includegraphics[height=6cm,width=8cm]{Fig4-Ex1.eps}
\includegraphics[height=6cm,width=8cm]{Fig5-Ex1.eps}\includegraphics[height=6cm,width=8cm]{Fig6-Ex1.eps}
\caption{A comparison between the exact and the approximate solutions computed by the VIM and the IVIM for the Riccati differential equation for Example 1.}\label{Fig1-Ex1}
\end{figure}
\begin{figure}
\centering
\includegraphics[height=9cm,width=12cm]{FigB-Ex1.eps}
\caption{$\log_{10}$ of the absolute error of the solutions provided by the IVIM for $m=10$ and $n=33,65,129,257$ in Example 1.}\label{Fig2-Ex1}
\end{figure}
\begin{figure}
\centering
\includegraphics[height=9cm,width=12cm]{FigC-Ex1.eps}
\caption{$\log_{10}$ of the absolute error of the solutions provided by the IVIM for $m=10$ and $n=1000,2000,3000,4000$ in Example 1.}\label{Fig3-Ex1}
\end{figure}
\bigskip
\noindent{\bf Example 2.} We consider the initial value problem
\begin{equation}\label{Eq013}
\left\{
\begin{array}{ll}
u'(t)=\frac{5}{3} \sqrt[5]{u^2(t)} \cos t, & t\in [0,3], \\
u(0)=0.
\end{array}
\right.
\end{equation}
The exact solution of the problem is $u(t)= t\sqrt[3]{t^2} \sin t$. Let $\mathcal{L}u= u'$, $\mathcal{N}u(t)=-\frac{5}{3} \sqrt[5]{u^2(t)} \cos t$ and $g\equiv 0$. Setting $\alpha=0$, from Remark 1 we get $G_m\equiv 0$ and
\[
H_m(s,t)=-\frac{5}{3} \sqrt[5]{u_m^2(s)} \cos s,
\]
for $(s,t)\in [0,3]\times [0,3]$. Therefore, from Eq. (\ref{Eq005}) the VIM takes the following form
\[
u_{m+1}(t)=\int_{0}^t \frac{5}{3} \sqrt[5]{u_m^2(s)} \cos s~ds.
\]
We have provided a MAPLE code for computing $u_m$'s. In Figure \ref{Fig1-Ex2} we have depicted the exact solution together with the computed solutions by one iteration of the VIM and IVIM. As is seen, there is a good agreement between the computed solutions by the VIM and IVIM. For $m\geq 2$ the VIM could not provide an explicit expression for the $u_m$'s, while the IVIM is well suited for sufficiently large values of $m$ and $n$ (Figures 5 and 6).
\begin{figure}
\centering
\includegraphics[height=9cm,width=12cm]{Fig1-Ex2.eps}
\caption{Exact solution together with the approximate solutions computed by one iteration of the VIM and the IVIM for Example 2.}\label{Fig1-Ex2}
\end{figure}
To see the efficiency of the IVIM, we consider the solutions provided by the IVIM with $m=10$ and different values of $n$ ($=33,65,129$ and $257$). Figure \ref{Fig2-Ex2} shows the $\log_{10}$ of the absolute error for the given solutions by IVIM. As we can see, the CPU times for computing these solutions are $0.000, 0.016,0.031$ and $0.047$, respectively. Similar results have been shown in Figure \ref{Fig3-Ex2} for $m=40$ and $n=1000,2000,3000$ and $4000$. The CPU times for computing the approximate solutions are $0.281,0.891,2.234$ and $3.484$, respectively.
\begin{figure}
\centering
\includegraphics[height=9cm,width=12cm]{Fig2-Ex2.eps}\caption{$\log_{10}$ of the absolute error of the computed solutions by the IVIM for $m=10$ and $n=33,65,129,257$ for Example 2.}\label{Fig2-Ex2}
\end{figure}
\begin{figure}
\centering
\includegraphics[height=9cm,width=12cm]{Fig3-Ex2.eps}\caption{$\log_{10}$ of the absolute error of the computed solutions by the IVIM for $m=40$ and $n=1000,2000,3000,4000$ for Example 2.}\label{Fig3-Ex2}
\end{figure}
\bigskip
\noindent{\bf Example 3.} Although we have presented the IVIM for the first-order initial value problems, by this example we illustrate that the method can be implemented for the higher-order initial value problems. To do so, consider the second-order initial value problem
\begin{equation}\label{Ex3Eq1}
\left\{
\begin{array}{ll}
u''(t)-2u'(t)^2+u'(t)+u(t)=g(t), & t\in [0,1.5], \\
u(0)=u'(0)=0, &
\end{array}
\right.
\end{equation}
where $g(t)=t+2\sin^2 \frac{t}{2}-8\sin^4 \frac{t}{2}$. The exact solution of this problem is $u(t)=x-\sin t$. Letting $v(t)=u'(t)$ for $t\in [0,1.5]$, Eq. (\ref{Ex3Eq1}) can be written as a
system of two simultaneous first-order differential equations
\begin{equation}\label{Ex3Eq2}
\left\{
\begin{array}{ll}
u'(t)=v(t),\\
v'(t)-2v(t)^2+v(t)+u(t)=g(t),
\end{array}
\right.
\end{equation}
with the initial condition $u(0)=v(0)=0$. Having in mind the exact solution of (\ref{Ex3Eq1}), we see that the exact solution of the problem (\ref{Ex3Eq2}) is given by $(u(t),v(t))=(t-\sin t, 1-\cos t)$.
To implement the VIM for solving (\ref{Ex3Eq2}), in the first equation of (\ref{Ex3Eq2}) the linear and nonlinear terms are chosen as $u'$ and $-v$, respectively. In the same way, for the second equation in (\ref{Ex3Eq2}), $v'+v$ and $-2v^2+u$ are considered as the linear and nonlinear terms, respectively.
Similar to the previous examples the VIM iteration takes the following form (for more details the reader can refer to \cite{Zhao}):
\begin{equation}\label{Ex3Eq3}
\left\{
\begin{array}{ll}
u_{m+1}(t)=u_m(t)-\displaystyle\int_{0}^{t} \left( u_m'(s)-v_m(s) \right) ds,\\[4mm]
v_{m+1}(t)=v_m(t)-\displaystyle\int_{0}^{t} e^{s-t}\left( v_m'(s)-2v_m(s)^2+v_m(s)+u_m(s)-g(s)\right)ds,
\end{array}
\right.
\end{equation}
where $u_0$ and $v_0$ are two given functions satisfying $u_0(0)=v_0(0)=0$. Obviously, from (\ref{Ex3Eq3}) it follows that $u_m(0)=v_m(0)=0$. Therefore, by straightforward integration by parts Eq. (\ref{Ex3Eq3}) can be written in the simpler form
\begin{equation}\label{Ex3Eq4}
\left\{
\begin{array}{ll}
u_{m+1}(t)=\displaystyle\int_{0}^{t} v_m(s)ds,\\[4mm]
v_{m+1}(t)=\displaystyle\int_{0}^{t} e^{s-t}\left(2v_m(s)^2-u_m(s)+g(s)\right)ds\\[4mm]
\hspace{1.45cm}=\displaystyle\int_{0}^{t} e^{s-t}\left(2v_m(s)^2+g(s)\right)ds
-\displaystyle\int_{0}^{t} e^{s-t}u_m(s)ds .
\end{array}
\right.
\end{equation}
To implement the IVIM, we set
\begin{eqnarray*}
\bar{H}_m(s,t) &=& -v_m(s), \\
\hat{H}_m(s,t) &=& -e^{s-t}\left(2v_m(s)^2+g(s)\right), \\
\tilde{H}_m(s,t) &=& e^{s-t}u_m(s).
\end{eqnarray*}
In this case, Eq. (\ref{Ex3Eq4}) can be rewritten as
\begin{equation}\label{Ex3Eq5}
\left\{
\begin{array}{ll}
u_{m+1}(t)=-\displaystyle\int_{0}^{t} \bar{H}_m(s,t)ds,\\[4mm]
v_{m+1}(t)=-\displaystyle\int_{0}^{t} \hat{H}_m(s,t) ds-\displaystyle\int_{0}^{t} \tilde{H}_m(s,t) ds .
\end{array}
\right.
\end{equation}
Taking Eq. (\ref{Eq005}) into account, one can approximate $u_{m+1}$ and $v_{m+1}$ by the idea described in Section \ref{SEC3}. Since the implementation of the method is straightforward, the details of the method are omitted here. Since, $\hat{H}_m$ involves the nonlinear term $2v_m^2$, the main difficulties mentioned in Example 1 arise here, too.
For the numerical results we use $u_0(t)=v_0(t)=0$ for $t\in [0,1.5]$ as the starting point. We now present the numerical results of using the IVIM for solving the problem. To this end, we first take $n=20$. In Figure \ref{Ex3Fig1} the graphs of $u_{m}$ and $v_{m}$ for $m=1,2,3,4,5$, computed by the VIM and the IVIM, together with $u(t)=t-\sin t$ and $v(t)=1-\cos t$ are depicted. This figure illustrates that, in the IVIM, $u_m(t)\rightarrow u(t)$ and $v_m(t)\rightarrow v(t)$ for $t\in [0,1.5]$ as $m$ tends to infinity. Moreover, it shows that there is a good agreement between the solution provided by the IVIM and the VIM when the VIM gives a suitable solution. In addition, the VIM can not provides any suitable expression for $v_4$, $u_m$ and $v_m$ for $m\geq 5$. This is due to the nonlinear term involving $v_{m+1}$. For more investigation we report the CPU times (in seconds) for computing the solutions by the VIM and the IVIM methods in Table \ref{Tbl2}. As we can observe, the CPU times for the VIM are very small, whereas those of the VIM drastically grows as $m$ increases.
\begin{figure}
\centering
\includegraphics[height=4.5cm,width=7cm]{Ut1.eps}\includegraphics[height=4.5cm,width=7cm]{Vt1.eps}
\includegraphics[height=4.5cm,width=7cm]{Ut2.eps}\includegraphics[height=4.5cm,width=7cm]{Vt2.eps}
\includegraphics[height=4.5cm,width=7cm]{Ut3.eps}\includegraphics[height=4.5cm,width=7cm]{Vt3.eps}
\includegraphics[height=4.5cm,width=7cm]{Ut4.eps}\includegraphics[height=4.5cm,width=7cm]{Vt4.eps}
\includegraphics[height=4.5cm,width=7cm]{Ut5.eps}\includegraphics[height=4.5cm,width=7cm]{Vt5.eps}
\caption{A comparison between the graphs of $u(t)=t-\sin t$ and $v(t)=u'(t)=1-\cos(t)$ , respectively, with $u_m(t)$ and $v_m(t)$, for $m=1,2,3,4,5$.}\label{Ex3Fig1}
\end{figure}
\begin{table}
\caption{CPU times (in seconds) for computing the approximate solutions by the VIM and the IVIM for Example 3.}
\vspace{-0.2cm}
\begin{center}
\begin{tabular}{lcccccccccc}\hline
$m$ \qquad\qquad & 1 & 2 & 3 & 4 & 5 \\ \hline
VIM & 0.046 & 0.313 & 1.531 & 16.750 & 576.4 (Fail) \\
IVIM & 0.010 & 0.012 & 0.012 & 0.014 & 0.015 \\ \hline
\end{tabular}
\end{center}
\label{Tbl2}
\end{table}
Finally, we set $n=1000$ and compute the approximate solutions of the problem by the IVIM for $m=5$ and $m=15$. In Figure \ref{Fig2Ex3}, $\log_{10}$ of the absolute error of the computed solutions is displayed. It is noted that the VIM gives the approximate solutions in 0.082 and 0.201 seconds for $m=5$ and $m=15$, respectively. As we see, the IVIM provides quite suitable solutions for the problem in a reasonable amount of time.
\begin{figure}
\centering
\includegraphics[height=6.cm,width=8cm]{ErrUEx3.eps}\includegraphics[height=6.cm,width=8cm]{ErrVEx3.eps}
\caption{ $\log_{10}$ of the absolute error of the solution computed by the IVIM for Example 3 for $n=1000$ and $m=5,15$.}\label{Fig2Ex3}
\end{figure}
\section{Conclusion}\label{SEC5}
We have proposed the interpolated VIM (IVIM) for solving one-dimensional initial value problems. The convergence of the method together with some numerical examples have been investigated. The numerical results show that the IVIM is more effective than the VIM. We have also shown the method can be easily implemented to the systems of ODEs.
The following aspects can be considered in future works:
\begin{enumerate}
\item Generalization of the method to two- or three-dimensional initial value problems;
\item Implementation of the idea for the other methods such as the homotopy perturbation method;
\item Using another type of basis functions such as the cubic spline functions and the Chebyshev polynomials.
\end{enumerate}
\section*{Acknowledgments}
The authors would like to thank the anonymous referees and the editors of the journal for their valuable comments and suggestions which substantially improved the quality of the paper. The work of the first author is partially supported by University of Guilan. | 0.005701 |
Fishing Report 19 November 2018
Monday
Currently we have a gentle NE blowing ideal for fishing at Port Dunford. Cloudy with a max of 24. Wave height 1.4 period 9 barometer 1017-1010.
High Tide 01-09 Low Tide 07.15
Tuesday
NE gusting 22 at fist peaking at 27 late afternoon. Cloudy with a max of 36. Wave height 2 –a.8 period 7 barometer 1004-996
High Tide 01.47 Low Tide 07.51
Wednesday
S wind gusting 19-24 later in the day. Cloudy with a max of 22. Wave height 2.2 -3 period 7-9 barometer 1010-1011.
High Tide 02.22 Low Tide 08.25
Thursday
SW gusting 14 swinging to a SE gusting 6 by midday. Partly cloudy with a max of 24. Wave height2.5 period 11 barometer 1015-1012.
High Tide 06.56 Low Tide 08.58 heavies that water motor bikes the bikes left early and wecod. There was some bird activity as in them flying along and boing some hovering but no diving.
From the shop.
We are expecting our summer order in tomorrow, we have bought in some new lines. Kingfisher have bought out their new catalogue with some very exciting new ranges and changes to some of the old, pop in and let us know what you want us to stock. | 0.004754 |
.
Make-A-Wish needed to keep costs low while upgrading their communications system.
Food for the Poor depended on contact center technology for fundraising, and they turned to Mitel for help with efficiency and connectivity. | 0.638849 |
\begin{document}
\author{Cl\'ement Canc\`es\thanks{UPMC Univ Paris 06, UMR 7598, Laboratoire Jacques-Louis Lions, F-75005, Paris, France (\href{mailto:[email protected]}{\tt [email protected]})}
\thanks{The author is partially supported by GNR MoMaS}}
\title{Asymptotic behavior of two-phase flows in heterogeneous porous media for capillarity depending only on space.\\ {II}. Non-classical shocks to model oil-trapping}
\date{\today}
\maketitle
\begin{abstract}
We consider a one-dimensional problem modeling two-phase flow in heterogeneous porous media made of two homogeneous subdomains, with discontinuous capillarity at the interface between them.
We suppose that the capillary forces vanish inside the domains, but not on the interface.
Under the assumption that the gravity forces and the capillary forces are oriented in opposite directions,
we show that the limit, for vanishing diffusion, is not in general the optimal entropy solution of the hyperbolic scalar conservation law as in the first paper of the series~\cite{NPCX}. A non-classical shock can occur at the interface, modeling oil-trapping.
\end{abstract}
\begin{keywords} scalar conservation laws with discontinuous flux, non-classical shock, two-phase flow, porous media, discontinuous capillarity
\end{keywords}
\begin{AMS} 35L65, 35L67, 76S05
\end{AMS}
\section{Introduction}
The models of two-phase flows provide good first approximations to predict the motions of oil in the subsoil. Although the theoretical knowledge concerning the question of the existence and the uniqueness of the solution to such models for homogeneous porous media
\cite{AKM90,CJ86} and for media with regular enough variations \cite{Chen01} is quite complete, few results are available for discontinuous media, as for example media made of several rock types \cite{ABE96, BPvD03, BLS09, FVbarriere, CGP09, EEM06}.
\vskip 10pt
One says that oil-trapping occurs when some oil can not pass through interfaces between
different rocks. Such a phenomenon plays an important role in the basin modeling, to predict the position of eventual reservoirs where oil could be collected.
As already explained in \cite{BPvD03,vDMdN95}, discontinuities of the capillary pressure field can induce the so-called oil-trapping phenomenon.
\vskip 10pt
The effects of capillarity, which play a crucial role in oil-trapping, seem to play a less important role concerning the motion of oil in homogeneous porous media, and can sometime be neglected to provide the so-called Buckley-Leverett equation.
\vskip 10pt
In this paper, we show that even if the dependence of the capillary pressure with respect to the oil-saturation of the fluid vanishes, the capillary pressure field still plays a crucial role to determine the saturation profile. In order to carry out this study, we restrict our frame to the one-dimensional case.
We will strongly use some recent results \cite{BLS09, FVbarriere,CGP09} obtained on flows in heterogeneous media with discontinuous capillary forces.
\vskip 10pt
We consider a one-dimensional porous medium, made of two different rocks, represented by $\O_1=\R_-^\star$ and $\O_2 = \R_+^\star$.
Let $\pi(u,x)$ be the capillary pressure, then it it is well known (see e.g. the introduction of the associate paper~\cite{NPCX})
that, if both phases have different densities, the equation governing the two phase flow can be written
\be\label{dep_NC}
\partial_t u + \partial_x \Big( q c(u,x) + g(u,x) \left(1 - C\partial_x \pi(u,x)\right) \Big) = 0,
\ee
where $u$ is the \emph{oil saturation} of the fluid, $q$ is the total flow rate, supposed to be a nonnegative constant, $C$ is a constant depending on the buoyancy forces and
$$c(u,x) = c_i(u), \quad g(u,x)=g_i(u), \quad \textrm{ and } \quad
\pi(u,x)=\pi_i(u) \quad \textrm{ if }x\in \O_i.$$
The functions $c_i$ are supposed to be increasing and Lipschitz continuous with $c_i(0)=0$ and $c_i(1) = 1$, while $g_i$ are supposed to be Lipschitz continuous, strictly positive in $(0,1)$ satisfying $g_i(0)=g_i(1) = 0$ and $\pi_i$ are increasing Lipschitz continuous functions.
\vskip 10pt
Physical experiments suggest that the dependence
of $\pi_i$ with respect to $u$ can be weak, at least for $u$ far from $0$ and $1$.
So we want to choose $\pi_1(u)=P_1$, and $\pi_2(u)=P_2$. The equation \refe{dep_NC}
turns formally to the scalar conservation law with discontinuous flux function
\begin{equation}\label{eq:base}
\partial_t u + \partial_x f(u,x) =0,
\end{equation}
where $f(u,x)$ (resp. $f_i(u)$) is equal to $q c(u,x) - g(u,x)$ (resp. $q c_i(u) - g_i(u)$).
\vskip 10pt
Such conservation laws have been widely studied in the last years. For a large overview on this topic, we refer to the introduction
of \cite{BKT09}, or in a lesser extent to the associated paper \cite{NPCX}. In particular, it has been proven by Adimurthi, Mishra and Veerappa Gowda \cite{AMV05} that there might exist an infinite number of $L^1$-contraction semi-groups corresponding to the equation~\eqref{eq:base}.
Among them, in the case where the functions $f_i$ have at most a single extremum in (0,1), we mention the so-called \emph{optimal entropy solution} which corresponds to the unique entropy solution in the case of a continuous flux function $f_1=f_2=f$. We refer to \cite{AMV05} and to the first part of this communication~\cite{NPCX} for a discussion on the so-called \emph{optimal entropy condition}.
\vskip 10pt
In the sequel of this paper, we suppose that
\vskip 10pt
\begin{itemize}
\item[{\bf (H1)}] for $i\in\{1,2\}$, there exists a value $u_i^\star\in [0,1)$ such that $f_i(u_i^\star) = q$, $f_i$ is increasing on $[0,u_i^\star]$ and $f_i(s) >q $ for all $s\in (u_i,1)$.
\end{itemize}
\vskip 10pt
We refer to Figure~\ref{fig:f_i} for an illustration of the previous assumption.
\begin{figure}[htb]
\begin{center}
\resizebox{8cm}{!}{
\input{g_i.pstex_t}
}
\caption{example of functions $f_i$ satisfying Assumption~{\bf (H1)}. Note the we have note supposed, as it is done in~\cite{AJV04,BKT09}, that $f_i$ has a single local extremum in $(0,1)$, but all the extrema have to be strictly greater than $q$.}\label{fig:f_i}
\end{center}
\end{figure}
We denote by $$\phii(u)= C \int_0^u g_i(s)\, ds.$$
For technical reasons, we have to assume that
\vskip 10pt
\begin{itemize}
\item[{\bf (H2)}] there exist $R>0$, $\alpha>0$ and $m\in(0,1)$ such that
\be f_1\circ\varphi_1^{-1}(s)\ge q+ R (\varphi_1(1)-s)^m \quad \textrm{ if } s\in [\varphi_1(1)-\a,\varphi_1(1)]. \label{hyp:Holder}
\ee
\end{itemize}
\vskip 10pt
These assumptions are fulfilled by models widely used by the engineers,
for which a classical choice of $c_i, g_i$ is
$$
c_i(u) = \frac{u^{a_i}}{u^{\a_i} + \frac{a}{b} (1-u)^{\b_i}}, \qquad g_i(u)=K_i \frac{u^{\a_i} (1-u)^{\b_i}}{b u^{a_i} + a (1-u)^{\beta_i}},
$$
where $\a_i, \beta_i \ge 1$ and $a,b$ are given constants.
\vskip 10pt
The goal of this paper is to show that if the capillary forces at the level of the interface $\{x=0\}$ are
oriented in the opposite sense with respect to the gravity forces (in our case $P_1<P_2$), then
a \emph{non classical} stationary shock can occur at the interface. It was shown by Kaasschieter~\cite{Kaa99}
that if the capillary pressure field is continuous at the interface (corresponding to the case $P_1 = P_2$), then
the good notion of solution is the one of \emph{optimal entropy solution}, computed by Adimurthi, Jaffr\'e and Veerappa Gowda
using a Godunov-type scheme~\cite{AJV04}.
We have pointed out in \cite{NPCX} that if the capillary forces and the gravity forces are oriented in the same sense, the
good notion of solution is also the one of \emph{optimal entropy solution}. If the assumptions stated above are fulfilled, if $P_1<P_2$ and if the initial data $u_0$ is large enough to ensure that both phases move in opposite directions, i.e.
\be\label{hyp:large_init}
u_i^\star \le u_0(x) \le 1 \quad \textrm{ a.e. in }\O_i,
\ee
we will show that the limit is not the optimal entropy solution, but the entropy solution to the problem
\be\label{Plim}\tag{$\mathcal{P}_{\lim_{}}$}
\left\{
\begin{array}{l}
\partial_t u + \partial_x f_i(u)=0, \\
u(x=0^-)=1 \textrm{ and } u(x=0^+)=u_2^\star, \\
u(t=0)=u_0.
\end{array}
\right.
\ee
In the sequel, we denote by $a^+ $ (resp. $a^-$) the positive (resp. negative)
part of $a$, i.e. $\max(0,a)$ (resp. $\max(0,-a)$), and for $i=1,2$, for $u,\k\in[0,1]$, one denotes
by
$$
\Phi_{i+}(u,\k)= \left\{\begin{array}{ll}
f_i(u)-f_i(\k) &\textrm{ if } u\ge \k,\\
0 &\textrm{ otherwise, }
\end{array}\right.
$$
$$
\Phi_{i-}(u,\k)= \left\{\begin{array}{ll}
f_i(\k)-f_i(u) &\textrm{ if } u\le \k,\\
0 &\textrm{ otherwise, }
\end{array}\right.
$$
and
$$
\Phi_i(u,\k)= \Phi_{i+}(u,\k)+\Phi_{i-}(u,\k)=f_i(\max(u,\k))-f_i(\min(u,\k)).
$$
We can now define the notion of solution to \refe{Plim}, which is in fact an entropy
in each subdomain $\O_i$, with an internal boundary condition at the level of
the interface.
\vskip 10pt
\begin{definition}[solution to \refe{Plim}]\label{Sol_NC_def}
Let $u_0\in L^\infty(\R)$, $u_i^\star \le u_0(x) \le 1$ a.e. in $\O_i$,
A function $u$ is said to be a solution of \refe{Plim} if it belongs to $L^\infty(\R\times\R_+)$,
$u_i^\star \le u \le 1$ a.e. in $\OiT$, and for $i=1,2$, for all $\psi\in \Dd^+(\overline\O_i\times\R_+)$, for all $\k\in[0,1]$,
\beqn
\lefteqn{ \int_{\R_+} \int_{\O_i}(u(x,t)-\k)^\pm \partial_t \psi dxdt + \int_{\O_i}(u_0(x)-\k)^\pm \psi(x,0) dx} \nn \\
&&\ds \hspace{10pt}+ \int_{\R_+} \int_{\O_i} \Phi_{i\pm}(u(x,t),\k) \partial_x \psi(x,t) dxdt
+ M_{f_i} \int_{\R_+} (\overline u_i -\k)^\pm \psi(0,t) dt \ge 0, \label{sol_NC_for}
\eeqn
where $M_{f_i}$ is a Lipschitz constant of $f_i$, and $\overline u_1=1$, $\overline u_2=u_2^\star$.
\end{definition}
\vskip 10pt
For a given $u_0$ in $L^\infty(\R)$, there exists a unique solution $u$ to \refe{Plim}
in the sense of Definition \ref{Sol_NC_def}, which is in fact made on an apposition of two
entropy solutions in $\R_\pm\times\R_+$. We refer to \cite{MNRR96,Otto96_CRAS} and \cite{Vov02} for proofs of existence and uniqueness to solutions to the problem~\eqref{Plim}. Moreover, thanks to \cite{Cont_L1}, one can suppose that $u$ belongs to $\Cc(\R_+;L^1_{loc}(\R))$.
\vskip 10pt
\begin{theorem}\label{thm:well-posed_Plim}
Let $u_0\in L^\infty(\R)$ with $u_i^\star \le u_0 \le 1$ a.e. in $\O_i$, then there exists a unique solution to \eqref{Plim} in the sense of Definition~\ref{Sol_NC_def}. Furthermore, if $v$ is another solution to \eqref{Plim} corresponding to $v_0\in L^\infty(\R)$ with $u_i^\star \le v_0 \le 1$ a.e. in $\O_i$, then for all $\R>0$, for all $t\in \R_+$
\be\label{eq:L1_contract_Plim}
\int_{-R}^R (u(x,t) - v(x,t))^\pm dx \le \int_{-R - M_f t}^{R + M_f t}(u_0(x) - v_0(x))^\pm dx
\ee
where $M_f$ is a Lipschitz constant of both $f_i$.
\end{theorem}
\vskip 10pt
Assume now that both phases move in the same direction:
\be\label{hyp:small_init}
0 \le u_0(x) \le u_i^\star \quad \textrm{ a.e. in }Ê \O_i,
\ee
then it will be shown that the relevant solution $u$ to the problem is the unique entropy solution defined below.
\vskip 10pt
\begin{definition}\label{Def:entro}
A function $u$ is said to be an entropy solution if it belongs to $L^\infty(\R\times\R_+)$,
$0\le u \le u_i^\star$ a.e. in $\OiT$, and for $i=1,2$, for all $\psi\in \Dd^+(\R\times\R_+)$, for all $\k\in[0,1]$,
\beqn
\lefteqn{ \int_{\R_+} \int_{\R}\left| u(x,t)-\k \right| \partial_t \psi dxdt + \int_{\R} \left|u_0(x)-\k\right| \psi(x,0) dx} \nn \\
&&\ds + \int_{\R_+}\hspace{-3pt}\sum_{i\in\{1,2\}} \hspace{-3pt}\int_{\O_i}\hspace{-3pt} \Phi_{i}(u(x,t),\k) \partial_x \psi(x,t) dxdt
+ |f_2(\k) - f_1(\k)|\hspace{-3pt}\int_{\R_+}\hspace{-3pt} \psi(0,t) dt \ge 0. \label{eq:entro_for_Tow}
\eeqn
\end{definition}
Thanks to Assumption~{\bf (H1)}, there exist no $\chi\in[0, \max u_i^\star]$ such that $f_1(\chi) = f_2(\chi)$, $f_1$ is decreasing and $f_2$ is increasing on $(\chi-\delta, \chi+\delta)$ for some $\delta>0$. Then the notion of entropy solution described by \eqref{eq:entro_for_Tow} introduced by Towers \cite{Tow00,Tow01} is equivalent to the notion of \emph{optimal entropy solution} introduced in \cite{AMV05} (see also \cite{BKT09}). We take advantage of this by using the very simple algebraic relation~\eqref{eq:entro_for_Tow}.
\vskip 10pt
It has been proven that the entropy solution $u$ exists and is unique for general flux functions $f$ \cite[Chapters~4 and 5]{Bachmann_These}. In particular, the following comparison and $L^1$-contraction principle holds.
\vskip 10pt
\begin{theorem}\label{thm:entro}
Let $u_0\in L^\infty(\R)$ with $0 \le u_0 \le u_i^\star$ a.e. in $\O_i$, then there exists a unique entropy solution in the sense of Definition~\ref{Def:entro}. Furthermore, if $v$ is another entropy solution corresponding to $v_0\in L^\infty(\R)$ with $0 \le v_0 \le u_i^\star$ a.e. in $\O_i$, then for all $\R>0$, for all $t\in \R_+$
\be\label{eq:L1_contract_entro}
\int_{-R}^R (u(x,t) - v(x,t))^\pm dx \le \int_{-R - M_f t}^{R + M_f t}(u_0(x) - v_0(x))^\pm dx
\ee
where $M_f$ is a Lipschitz constant of both $f_i$.
\end{theorem}
\subsection{non classical shock at the interface}
As already mentioned, the optimal entropy solution can be seen as a extension to the case of discontinuous flux functions of the
usual entropy solution \cite{K70} obtained for a regular flux function. We will now illustrate that it is not the case with the solution to~\eqref{Plim}.
Assume for the moment (it will be proved later) that in the case where $u_0(x) \in (u_i^\star,1)$ a.e. in $\O_i$, the corresponding solution $u$ to \eqref{Plim} admits $\overline u_i$ as strong trace on the interface. One has the following \emph{Rankine-Hugoniot} relation
$$
f_1(\overline u_1) = f_2(\overline u_2) = q,
$$
then $u$ is a weak solution to~\eqref{eq:base}, i.e. it satisfies for all $\psi \in \Dd(\R\times\R_+)$:
\be\label{weak_form}
\int_{\R_+}\int_{\R} u \partial_t \psi\, dxdt+ \int_\R u_0 \psi(\cdot, 0) \, dx + \int_{\R_+}\int_{\R} f(u,\cdot) \partial_x \psi\, dxdt = 0.
\ee
\vskip 10pt
Firstly, suppose for the sake of simplicity that $f_1(u) = f_2(u) = f(u)$, and that $q=0$, then $u_i^\star = 0$ for $i\in\{1,2\}$.
The function
$$
u(x) = \left\{ \begin{array}{ll}
1 & \textrm{ if }x<0,\\
0 & \textrm{ if }x>0
\end{array}\right.
$$
is then a steady solution to~\eqref{Plim} satisfying~\eqref{sol_NC_for}. However, since
$$
\frac{f(1) - f(s)}{1-s} <0 \quad \textrm{ for all } s \in (0,1),
$$
the discontinuity at $\{x=0\}$ does not fulfill the usual \emph{Oleinik entropy condition} (see e.g.~\cite{Smo94}). This discontinuity is thus said to be a
\emph{non-classical shock}.
\vskip 10pt
Suppose now that $f_1'(1) <0$ and that $f_2'(u_2^\star) >0$, then the pair $(1,u_2^\star)$ is a stationary \emph{undercompressible shock-wave}, that are prohibited for optimal entropy solutions \cite{AMV05} as for classical entropy solutions in the case of regular flux functions.
\vskip 10pt
\begin{remark}\label{rmk:AB}
It has been pointed out in \cite{AMV05} that allowing a \emph{connection} $(A,B)$, i.e. a stationary undercompressible wave between the left state $A$ and the right state $B$ at the interface lead to another \mbox{$L^1$-contraction} semi-group (see \cite{AMV05,BKT09,GNPT07}), which is so-called entropy solution of type $(A,B)$. However, we rather use the denomination \emph{non-classical shock} for the connection between $A$ and $B$ since, as stressed above, the corresponding solution violates some fundamental properties of the classical entropy solutions.
\end{remark}
\subsection{oil-trapping modeled by the non-classical shock}\label{subsec:oil-trapping}
In this section, we assume that $q=0$.
Let $u$ be the solution of the problem \eqref{Plim} corresponding to the initial data $u_0$. Assume that $u$ admits strong traces on the interface. The flow-rate of oil going from $\O_1$ to $\O_2$ through the interface is given by
$$f_1(\overline u_1) = f_2(\overline u_2) =0.$$
Thus the oil cannot overcome the interface from $\O_1$ to $\O_2$, thus if one supposes that $u_0$ belongs to $L^\infty(\R)$, with $0\le u_0 \le 1$ a.e., then the quantity of oil standing between $x=-R$ ($R$ is an arbitrary positive number) and $x=0$ can only grow.
\vskip 10pt
Indeed, let $t_2>t_1\ge 0$, let $\zeta_n(x) = \min(1, n(x+R)^+,n x^-)$ and
$\theta_m(t) = \min(1, m(t-t_1), m(t_2-t))$. Choosing $\psi(x,t)=\zeta_n(x) \theta_m(t)$ in \eqref{weak_form} for $m,n\in \N$ yields, using the positivity of $f_1$
$$
\int_{t_1}^{t_2}\left( \int_{-R}^0u(x,t) \zeta_n(x)dx\right) \partial_t\theta_m(t) dt + \int_{t_1}^{t_2}\theta_m(t)
\left( \frac1n \int_{-1/n}^0 f_1(u(x,t)) dx \right)dt \le 0.
$$
Since $u$ admits a strong trace on the interface,
$$
\lim_{n\to\infty} \frac1n \int_{-1/n}^0 f_1(u(x,t)) dx = f_1(\overline u_1) = 0.
$$
Then we obtain
\begin{equation}\label{ot_1}
\int_{t_1}^{t_2}\left( \int_{-R}^0u(x,t) dx\right) \partial_t\theta_m(t) dt \le 0.
\end{equation}
The solution $u$ belong to $\Cc(\R_+;L^1(\R))$ thanks to \cite{Cont_L1}, thus taking the limit as $m\to \infty$ in \refe{ot_1} provides
$$
\int_{-R}^0 u(x,t_1)\, dx \le \int_{-R}^0 u(x, t_2)\, dx.
$$
\vskip 10pt
Suppose now that $q\ge0$. Thanks to what follows, we are able to solve the Riemann problem at the interface for any initial data
$$
u_0(x) = \left\{ \begin{array}{lll}
u_\ell & \textrm{ if }Ê& x<0,\\
u_r & \textrm{ if }Ê& x>0.
\end{array}\right.
$$
The study of the Riemann problem is carried out in Section~\ref{sec:Riemann}, leading to the following result.
\begin{itemize}
\item If $u_\ell> u_1^\star$, then $u_1 = 1$ and $u_2 = u_2^\star$. We obtain the expected non-classical shock at the interface.
\item If $u_\ell \le u_1^\star$, then $u_1 = u_\ell$ and $u_2$ is the unique value of $[0,u_2^\star]$ such that $f_2(u_2) = f_1(u_\ell)$.
\end{itemize}
Using Assumption~{\bf (H1)}, this particularly implies that in both cases, the flux at the interface is given by
\be\label{eq:trace_1}
f_1(u_1) = f_2(u_2) = G_1(u_\ell, 1)
\ee
where $G_1$ is the Godunov solver corresponding to the flux function $f_1$:
$$
G_1(a,b) = \left\{\begin{array}{lll}
\ds \min_{s\in[a,b]} f_1(s) & \textrm{ if } & a \le b,\\
\ds \max_{s\in[b,a]} f_1(s) & \textrm{ if } & a > b.
\end{array}\right.
$$
This particularly yields that for any initial data $u_0\in L^\infty(\R)$ with $0\le u_0 \le 1$, the restriction $u_{|_{\O_1}}$ of the solution $u$ to $\O_1$ is the unique entropy solution to
\be\label{syst:u_1}
\left\{\begin{array}{lll}
\partial_t u + \partial_x f_1(u) = 0 & \textrm{ in } \O_1\times\R_+,\\
u(\cdot, 0) = u_0 & \textrm{ in } \O_1,\\
u(0,\cdot) = \gamma & \textrm{ in }Ê\R_+
\end{array}\right.
\ee
for $\gamma = 1$. Since the solution $u$ to the problem~\eqref{syst:u_1} is a non-decreasing function of the prescribed trace $\gamma$ on $\{x=0\}$, we can claim as in \cite{NPCX} that
$$
u_{|_{\O_1}} = \sup_{\begin{subarray}{c}\gamma\in L^\infty(\R_+) \\
0 \le \gamma \le 1
\end{subarray}}
\left\{\ v \textrm{ solution to }\eqref{syst:u_1}\ \right\}.
$$
In particular, $u$ is the unique weak solution (i.e. satisfying \eqref{weak_form}) that is entropic in each subdomain and that minimizes the flux through the interface.
\subsection{organization of the paper}
We will introduce a family of approximate problems in Section
\ref{approx_sect_NC}, which takes into account the capillarity, with small
dependance $\eps$ of the capillary pressure with respect to the
saturation. We use the transmission conditions introduced in \cite{BLS09,FVbarriere,CGP09,Sch08} to connect the capillary pressure at the interface.
For $\eps>0$, the problem~\refe{Pe_NC} admits a unique solution $\ue$ thanks to~\cite{FVbarriere} and it is recalled that a comparison principle holds for the solutions of the approximate problem~\eqref{Pe_NC}. Particular sub- and super-solution are derived in order to show that if $u_0(x) \ge u_I^\star$ a.e. in $\O_i$, then the limit $u$ of the approximate solutions $\left(\ue \right)_{\eps>0}$ as $\eps$ tends to $0$. An energy estimate is also derived.
\vskip 10pt
In Section~\ref{conv_NC_sec}, letting $\eps$ tend to $0$, since no strong pre-compactness can be derived on $\left( \ue \right)_\eps>0$ in $L^1_{loc}(\R\times\R_+)$ from the available estimates, we use the notion of process solution \cite{EGH00}, which is equivalent to the notion of measure valued solution introduced by DiPerna \cite{DP85} (see also \cite{MNRR96,Sze91}). The uniqueness of such a process solution allows us to claim that $(\ue)$ converges strongly in $L^1_{loc}((\R\times\R_+)$ towards the unique solution to \eqref{Plim}.
\vskip 10pt
In Section~\ref{entro_small}, it is shown that if both phases move in the same direction, that is if $0 \le u_0 \le u_i^{\star}$ a.e. in $\O_i$, then $\left(\ue\right)$ converges towards the unique entropy solution to the problem in the sense of Definition~\ref{Def:entro}.
\vskip 10pt
In Section~\ref{sec:Riemann}, we complete the study of the Riemann problem at the interface.
\section{The approximate problem}\label{approx_sect_NC}
In this section, we take into account the effects of the capillarity, supposing that
they are small. We will so build an approximate problem \refe{Pe_NC}, whose unknown
$\ue$ will depend on a small parameter $\eps$ representing the dependance
of the capillary pressure with respect to the saturation.
We assume for the sake of simplicity that the capillary pressure in $\O_i$ is
given by:
\be\label{pi_NC}
\pi_i^\eps(\ue)=P_i + \eps \ue.
\ee
It has been shown simultaneously in \cite{BLS09} and in \cite{CGP09} that a good way to connect the capillary pressures at the interface is to require
\be\label{connect_p_NC}
\t \pi_1^\eps(\ue_1)\cap \t \pi_2^\eps(\ue_2) \neq \emptyset,
\ee
where $\ue_1$ and $\ue_2$ are the traces of $\ue$ on the interface, and
where $\t \pi_i^\eps$ is the monotonous graph given by
$$
\t \pi_i^\eps(s)=\left\{
\begin{array}{ll}
\pi_i^\eps(s) & \textrm{ if } s\in (0,1),\\
(-\infty, P_i] & \textrm{ if } s=0,\\
{[} P_i+\eps,\infty) & \textrm{ if } s=1.
\end{array}
\right.
$$
We suppose that the capillary force is oriented in the sense of decreasing $x$, i.e.
$P_1<P_2$ (the capillary force goes from the high capillary pressure to the low capillary pressure).
Since $\eps$ is assumed to be a small parameter, we can suppose that
$0<\eps<P_2-P_1$, so that the relation \refe{connect_p_NC} turns to
\be\label{connect_p_NC2}
\ue_1=1 \textrm{ or }Ê\ue_2=0.
\ee
The flux function in $\O_i$ is then given by:
$$
F_i^\eps(x,t)= f_i(\ue)(x,t)-\eps \partial_x \phii(\ue)(x,t).
$$
Because of the conservation of mass, we require the continuity
of the flux functions at the interface. Thus the approximate problem becomes
\be\label{Pe_NC}\tag{$\mathcal{P}^\eps$}
\left\{
\begin{array}{l}
\partial_t \ue + \partial_x F_i^\eps=0, \\
\ue(x=0^-)=1 \textrm{ or } \ue(x=0^+)=0, \\
F_1^\eps (0^-) = F_2^\eps(0^+), \\
u(t=0)=u_0.
\end{array}
\right.
\ee
We are not able to prove the uniqueness of a weak solution of \refe{Plim} if
the flux $F_i^\eps$ "only" belongs to $L^2(\overline\O_i\times\R_+)$, and we will
define the notion of prepared initial data, so that the flux belongs to $L^\infty(\O_i\times\R_+)$.
In this latter case, the uniqueness holds.
\subsection{bounded flux solutions}
We define now the notion of bounded flux solution, that was introduced in this framework in
\cite{FVbarriere,CGP09}.
\vskip 5pt
\begin{definition}[bounded flux solution to \refe{Pe_NC}]\label{bounded_Def_NC}
Let $u_0\in L^\infty(\R)$, $0\le u_0 \le 1$, a function $\ue$
is said to be a bounded flux solution if
\begin{enumerate}
\item $\ue \in L^\infty(\R\times\R_+)$, $0\le u \le 1$;
\item $\partial_x \phii(\ue) \in L^\infty(\O_i\times\R_+)\cap L^2_{loc}(\R_+;L^2(\O_i))$;
\item $u_1^\eps(t)\left(1- u_2^\eps(t)\right) = 0$ for almost all $t\ge 0$, where $u_i^\eps$ denotes the trace of $u^\eps_{|_{\O_i}}$ on $\{x=0\}$.
\item $\forall \psi\in \Dd(\R\times\R_+)$,
\beqn
\lefteqn{ \int_{\R_+}\int_\R \ue(x,t)\partial_t \psi(x,t) dxdt + \int_\R u_0(x)\psi(x,0)dx }\nn \\
&& \ds \hspace{20pt} + \int_{\R_+}\sum_{i\in\{1,2\}}\int_{\O_i} \left[ f_i(\ue) -\eps \partial_x \phii(\ue) \right]
\partial_x \psi(x,t) dxdt = 0. \label{bounded_for_NC}
\eeqn
\end{enumerate}
\end{definition}
\vskip 10pt
\begin{remark}\label{continuite_temporelle}
Such a bounded-flux $\ue$ solution belongs to $\Cc(\R_+;L^1_{\rm loc}(\R))$, in the sense that there exists
$\tilde\ue$ in $\Cc(\R_+;L^1_{\rm loc}(\R))$ such that $\ue(t)=\tilde\ue(t)$ for almost all $t\ge 0$ (see \cite{Cont_L1}). More precisely,
all $t\ge 0$ is a Lebesgue point for $\ue$. So, the slight abuse of notation consisting in considering $\ue(t)$ for all $t\ge 0$ will not lead to any confusion.
\end{remark}
\vskip 10pt
\begin{proposition}\label{comp_prop}
Let $u$ and $v$ be two bounded-flux solutions associated to initial data $u_0,v_0$,
then for all $\psi\in\Dd^+(\R\times\R_+)$,
\beqn
&\ds \int_{\R_+} \int_\R \left( u - v \right)^\pm \partial_t \psi dxdt + \int_\R \left( u_0 - v_0 \right)^\pm \psi(\cdot,0) dx&\nn\\
&\ds + \int_{\R_+} \sum_i \int_{\O_i} \left( \Phi_{i\pm}(u, v) - \eps \partial_x \left( \phii(u) - \phii(v)\right)^\pm \right) \partial_x\psi dxdt \ge 0. &\label{comp_prop_eq}
\eeqn
\end{proposition}
\vskip 10pt
We state now a theorem which is a generalization in the case of unbounded domains of Theorem~3.1 and Theorem~4.1 stated in \cite{FVbarriere}.
\vskip 10pt
\begin{theorem}[existence--uniqueness for bounded flux solutions]\label{comp_bounded_NC}
Let $u_0\in L^\infty(\R)$ with $0 \le u_0 \le 1$ such that:
\begin{itemize}
\item there exists a function $\hat u\in L^\infty(\R)$, with $0 \le \hat u \le 1$ a.e. in $\R$, satisfying $\partial_x \hat u \in L^1(\R)\cap L^\infty(\R)$ and such that $(u_0-\hat u)\in L^1(\R)$
\item $\partial_x \phii(u_0)\in L^\infty(\O_i)$
\item $\lim_{x\nearrow0} u_0(x) = 1$ or $\lim_{x\searrow0} u_0(x) = 0$.
\end{itemize}
Then there exists a unique bounded flux
solution $\ue$ to the problem \refe{Pe_NC} in the sense of Definition \ref{bounded_Def_NC} satisfying $(\ue - \hat u) \in L^1(\R)$.
Furthermore, if $\ue, v^\eps$ are two bounded flux solutions associated to initial data
$u_0,v_0$ then
\be\label{eq:comp_bounded}
u_0(x) \ge v_0(x) \textrm{ a.e. in } \R \quad \Rightarrow \ue(x,t) \ge v^\eps(x,t) \textrm{ a.e. in }\R \textrm{ for all }t\ge 0.
\ee
\end{theorem}
\vskip 10pt
Obviously, the existence of a bounded flux solution can not be extended to any initial data
in $L^1(\R)$. Indeed, the initial data $u_0$ has at least to involve bounded initial flux, i.e.
$\partial_x \phii(u_0) \in L^\infty(\R)$. An additional natural assumption is needed to
ensure the existence of such a bounded flux solution : the connection in the graphical
sense of the capillary pressures at the interface.
\vskip 10pt
If $(u_0 - \hat u)$ and $(v_0 - \hat u)$ belong to $L^1$ for the same $\hat u$, then \eqref{comp_prop_eq} yields that the bounded flux solutions
$\ue$ and $v^\eps$ corresponding to $u_0$ and $v_0$ satisfy the following contraction principle: $\forall t\in \R_+$,
$$
\int_\R (\ue(x,t) - v^\eps(x,t))^\pm dx \le \int_\R (u_0(x) - v_0(x))^\pm dx,
$$
providing the uniqueness result stated in Theorem~\ref{comp_bounded_NC}.
\subsection{particular sub- and super-solutions}\label{sub-super_sec}
We will study particular steady states of the approximate problem~\eqref{Pe_NC}. We will consider
steady bounded flux solutions $s^\eps$ corresponding to a zero water flow rate, i.e.
\be\label{eq:steady}
f_i(s^\eps) - \eps \frac{{\rm d} }{{\rm d}x} \phii(s^\eps)= q\quad \textrm{ in } \O_i.
\ee
For $\eps>0$, there are infinitely many solutions $s^\eps$ of the equation~\eqref{eq:steady}. We will
construct some particular solutions, that will permit us to show that the limit $u$ as $\eps$ tends to $0$
of bounded flux solutions $\ue$ corresponding to large initial data admits the expected strong traces on
the interface $\{x=0\}$.
\vskip 10pt
We will introduce now particular solutions of the ordinary differential equation
\be\label{EDO_NC}
y'=f_i\circ\phii^{-1}(y) - q.
\ee
\begin{lemma}\label{EDO_lem}
Let $\eta>0$, there exists a solution $y^\eta$ to \eqref{EDO_NC} for $i=1$ which is nondecreasing on $(-\infty,-0]$
equal to $\varphi_1(1)$ on $[-\eta,0)$, satisfying $y^\eta(x) < \varphi_1(1)$ if $x<-\eta$ and $\lim_{x\to-\infty} y^\eta(x) = u_1^\star$.
\end{lemma}
\vskip 10pt
\begin{proof}
Consider the problem
\be\label{eq:EDO_1}
\left\{\begin{array}{rclll}
w'(x) &=&R(\phii(1)-w(x))^m & \textrm{ if } & x < -\eta, \\
w(-\eta) &=& \varphi_1(1),
\end{array}\right.
\ee
where $R$ and $m$ are constants given by Assumption~{\bf (H2)}. The function
$$
w^\eta(x) = \phii(1)-\left( R(1-m)(-x-\eta) \right)^{\frac1{1-m}}
$$
is a solution of~\eqref{eq:EDO_1}. Because of {\bf (H2)}, there exists a neighborhood $(-\eta - \delta, -\eta]$ of $\eta$ such that
$w^\eta$ is a super-solution of the problem
\be\label{eq:EDO_2}
\left\{\begin{array}{rclll}
y'(x) &=&f_1\circ\varphi_1^{-1}(y) - q & \textrm{ if } & x < -\eta, \\
y(-\eta) &=& \varphi_1(1).
\end{array}\right.
\ee
Then there exists $y^\eta$ solution to \eqref{eq:EDO_2} such that $y^\eta(x) = \varphi_1(1)$ if $x\in (-\eta,0)$ and
$$y^\eta(x) \le w^\eta(x) \quad \textrm{ on } (-\eta - \delta, -\eta].
$$
In particular, $y^\eta$ is not constant equal to 1.
Thanks to {\bf (H1)}, the function $y^\eta$ is increasing on the set $\{\ x\in \O_1\ | \ y^\eta(x) \in ( \varphi_1(u_1^\star), \varphi_1(1) )\ \}$.
Assume that there exists $x_\star<-{\eta}$ such that $y^\eta(x_\star) = \varphi_1(u_1^\star)$, then one sets
$y^\eta(x) = \varphi_1(u_1^\star)$ for all $x\in(-\infty, x_\star]$. If $y^\eta(x) > \varphi_1(u_1^\star)$ for all $x<0$, then
$y^\eta$ is increasing on $(-\infty, -\eta)$. Thus it admits a limit as $x$ tends to $-\infty$, and it is clear that the
only possible limit is $u_1^\star$.
\end{proof}
\vskip 10pt
\begin{lemma}\label{lem:EDO_2}
Let $\eta>0$, then there exists a solution $z^\eta$ to \eqref{EDO_NC} for $i=2$ which is nondecreasing on $\R$ satisfying
$z^{\eta}(x) \le \varphi_2\left(\frac{1+u_2^\star}{2}\right)$ if $x\le \eta$, $z^{\eta}(x) \ge \varphi_2\left(\frac{1+u_2^\star}{2}\right)$ if $x\ge \eta$ and
$\lim_{x\to\infty} z^\eta(x) = \varphi_2(1)$, $\lim_{x\to -\infty} z(x)= u_2^\star$.
\end{lemma}
\vskip 10pt
\begin{proof}
The problem
$$
\left\{\begin{array}{rclll}
z'(x) &=& f_2\circ\varphi_2^{-1}(z(x)) - q& \textrm{ if } & x\in \R,\\[5pt]
z(\eta) &=& \varphi_2\left(\frac{1+u_2^\star}{2}\right).
\end{array}\right.
$$
admits a (unique) solution in $\Cc^1(\R,[0,1])$. Since $u_2^\star$ is a constant solution of~\eqref{EDO_NC} for $i=2$, then one has $z(x) \ge u_2^\star$ in
$\R$. Thanks to {\bf (H1)}, the function $z$ is nondecreasing. This implies that it admits limits respectively in $-\infty$ and in $+\infty$. The only possible values for this limits are respectively $u_2^\star$ and $1$.
\end{proof}
\vskip 10pt
\begin{proposition}\label{prop:sub-super}
Let $\eta>0$, then there exists two families of steady bounded flux solutions $\left(\underline s^{\eps,\eta}\right)_{\eps>0}$ and $\left(\overline s^{\eps,\eta}\right)_{\eps>0}$ tending in $L^1_{loc}(\R)$ as $\eps\to0$ respectively towards
$$
\underline s^{\eta} : x\mapsto \left\{
\begin{array}{lll}
u_1^\star & \textrm{ if } & x< -\eta,\\
1 & \textrm{ if } & x\in (-\eta ,0),\\
u_2^\star & \textrm{ if }& x >0,
\end{array}\right.
$$
and
$$
\overline s^{\eta} : x\mapsto \left\{
\begin{array}{lll}
1 & \textrm{ if } & x< 0,\\
u_2^\star & \textrm{ if } & x\in (0,\eta),\\
1 & \textrm{ if }& x >\eta.
\end{array}\right.
$$
\end{proposition}
\vskip 10pt
\begin{proof}
We set
\be\label{eq:us}
\underline{s}^{\eps,\eta}(x) = \left\{\begin{array}{lll}
y^\eta\left(\frac{x+\eta}{\eps} -\eta \right) & \textrm{ if } & x< 0,\\
u_2^\star & \textrm{ if }& x >0,
\end{array}\right.
\ee
and
\be\label{eq:os}
\overline{s}^{\eps,\eta}(x) = \left\{\begin{array}{lll}
1 & \textrm{ if }Ê& x<0, \\
z^\eta\left(\frac{x-\eta}{\eps} + \eta \right) & \textrm{ if } & x> 0,
\end{array}\right.
\ee
where the functions $y^\eta$ and $z^\eta$ have been defined in Lemmas~\ref{EDO_lem} and \ref{lem:EDO_2}. Since the functions $\phii(\underline{s}^{\eps,\eta})$ and $\phii(\overline{s}^{\eps,\eta})$ are monotone in $\O_i$, there derivatives $\frac{\rm d}{{\rm d}x} \phii(\underline{s}^{\eps,\eta})$ and $\frac{\rm d}{{\rm d}x} \phii(\overline{s}^{\eps,\eta})$ belong to $L^1(\R)$, and also to $L^\infty(\R)$ because $\underline{s}^{\eps,\eta}$ and $\overline{s}^{\eps,\eta}$ are solutions to~\eqref{eq:steady}. Thus they belong to $L^2(\R)$. Hence, for fixed $\eps$,
$\underline{s}^{\eps,\eta}$ and $\overline{s}^{\eps,\eta}$ are bounded flux solutions to the problem~\eqref{Pe_NC}. The convergence
as $\eps\to0$ towards the functions $\overline s^\eta$ and $\underline s^\eta$ is a direct consequence of Lemmas~\ref{EDO_lem} and \ref{lem:EDO_2}.
\end{proof}
\subsection{a ${ L^2((0,T);H^1(\O_i))}$ estimate}
Our goal is now to derive an estimate which ensures that the effects of capillarity
vanish almost everywhere in $\O_i\times\R_+$ as $\eps$ tends to $0$.
\vskip 10pt
\begin{proposition}\label{L2H1_NC}Let $u_0\in L^\infty(\R)$ with $0 \le u_0 \le 1$ a.e. satisfying the assumptions of Theorem~\ref{comp_bounded_NC} and let $\ue$ be the corresponding bounded flux solution. Then for all $\eps\in(0,1)$, for all $T>0$,
there exists $C$ depending only on $u_0,g_i,\phii,T$ such that
\be\label{L2H1_NC_2}
\sqrt{\eps} \| \partial_x \phii(\ue) \|_{L^2(\OiT)} \le C.\ee
This particularly ensures that
\be\label{L2H1_NC_3}
\eps \| \partial_x\phii(\ue)\|_{L^2(\OiT)} \to 0 \quad \textrm{ as } \eps\to0.
\ee
\end{proposition}
\vskip 10pt
The idea of the proof of Proposition~\ref{L2H1_NC} is formally to choose
$(\ue-\hat u)\psi$ as test function in \refe{bounded_for_NC} for a function $x\mapsto\psi(x)$
compactly supported in $\O_i$. Using the fact that the flux $F_i^\eps$ is uniformly
bounded in $L^\infty(\OiT)$, we can let $\psi$ tend towards $\chi_{\O_i}$, with
$\chi_{\O_i}(x)=1$ if $x\in \O_i$ and $0$ otherwise, and the estimate \refe{L2H1_NC_2}
follows. To obtain \refe{L2H1_NC_3}, it suffices to multiply \refe{L2H1_NC_2} by
$\sqrt\eps$. We refer to \cite[Proposition~2.3]{NPCX} for a more details on the proof of
Proposition~\ref{L2H1_NC}.
\subsection{approximation of the initial data}
In order to ensure that the limit $u$ of the approximate solutions $\ue$ as $\eps\to 0$ admits the expected strong traces on the interface $\{x=0\}$,
we will perturb the initial data $u_0$.
\vskip 10pt
\begin{lemma}\label{lem:init_eta_eps}
Let $u_0\in L^\infty(\R)$ satisfying~\eqref{hyp:large_init}, then there exists $\left(u_0^{\eps,\eta}\right)_{\eps,\eta}$ such that
\begin{enumerate}
\item[(a).] $\underline s^{\eps,\eta} (x) \le u_0^{\eps,\eta}(x) \le \overline s^{\eps,\eta}(x)$ a.e. in $\R$, where the functions $\underline s^{\eps,\eta}$ and $\overline s^{\eps,\eta}$ are defined in~\eqref{eq:us}-\eqref{eq:os},
\vskip 5pt
\item[(b).] $\eps\left\|\partial_x \varphi_i(u_0^{\eps,\eta})\right\|_{L^\infty(\O_i)} \le C$ where $C$ depends neither on $\eps$ nor on $\eta$,
\vskip 5pt
\item[(c).] $u_0^{\eps,\eta} \to u_0$ in $L^1_{loc}(\R)$ as $(\eps,\eta)\to (0,0)$.
\end{enumerate}
\end{lemma}
\vskip 10pt
\begin{proof}
Let $\left(\rho_n\right)_{n\in \N^\star}$ be a sequence of mollifiers, then $\rho_n\ast u_0$ is a smooth function tending $u_0$ as $n\to\infty$.
Then, for $\eps>0$, we choose $n\in \N^\star$ such that
\be\label{eq:init_45}
\max\left\{ n, \left\| \partial_x \phii(u_0\ast \rho_n) \right\|_{L^\infty(\O_i)} \right\} \ge \frac1\eps,
\ee
and we define
\be\label{eq:inti_eps_eta}
u_0^{\eps,\eta}(x) = \max\left\{ \underline s^{\eps,\eta}(x), \min\left\{ \overline{s}^{\eps,\eta}(x), u_0\ast\rho_n(x)\right\}\right\}.
\ee
The point (a) is a direct consequence of~\eqref{eq:inti_eps_eta}. Letting $(\eps,\eta)\to (0,0)$ in~\eqref{eq:inti_eps_eta} yields
$$
\lim_{(\eps,\eta)\to(0,0)} u_0^{\eps,\eta}(x) = \max\left\{ u_i^\star, \min\left\{1, u_0(x)\right\}\right\}.
$$
Since $u_0$ is supposed to satisfy~\eqref{hyp:large_init}, this provides
$$
\lim_{(\eps,\eta)\to(0,0)} u_0^{\eps,\eta}(x) = u_0(x) \quad \textrm{ a.e. in } \R.
$$
The point (c) follows. In order to establish (b), it suffices to note that there exist an open subset $\omega$ of $\R$ such that
$u_0^{\eps,\eta}(x)$ is equal to $u_0\ast \rho_n(x)$ for $x\in\omega$, and such that
$u_0^{\eps,\eta}(x)$ is either equal to $\underline s^{\eps,\eta}(x)$ or to $\overline s^{\eps,\eta}(x)$ on $\omega^c = \R\setminus\omega$.
It follows from~\eqref{eq:init_45} that
$$\eps \left\|\partial_x \phii(u_0^{\eps,\eta}) \right\|_{L^\infty(\O_i\cap\omega)} \le 1.$$
One has
$$
f_i(u_0^{\eps,\eta})(x) - \eps \partial_x \phii(u_0^{\eps,\eta})(x) = q\quad \textrm{ a.e. in } \O_i\cap\omega^c,
$$
thus
$$\eps \left\|\partial_x \phii(u_0^{\eps,\eta}) \right\|_{L^\infty(\O_i\cap\omega^c)} \le \| q - f_i \|_{L^\infty(u_i^\star,1)} .$$
This concludes the proof of Lemma~\ref{lem:init_eta_eps}.
\end{proof}
\vskip 10pt
\begin{definition}\label{Def:prepared}
A function $u_0$ is said to be a prepared initial data if it satisfies $(1-u_0)\in L^1(\R)$, $\partial_x \phii(u_0) \in L^\infty(\O_i)$ and
\be\label{eq:prepared}
\underline s^{\eps,\eta} (x) \le u_0(x) \le \overline s^{\eps,\eta}(x)\textrm{ a.e. in } \R
\ee
for some $\eps>0,\eta>0$.
\end{definition}
\vskip 10pt
Since the function $(\eps,\eta) \mapsto \underline s^{\eps,\eta}$ is decreasing with respect to both arguments and since the
function $(\eps,\eta) \mapsto \overline s^{\eps,\eta}$ is increasing with respect to both arguments, if $u_0$
satisfies~\eqref{eq:prepared} for $\eps=\eps_0$ and $\eta=\eta_0$, then $u_0$ satisfies~\eqref{eq:prepared} for all
$(\eps,\eta)$ such that $\eps\le \eps_0$ and $\eta\le\eta_0$. So the following Proposition is a direct consequence from~\eqref{eq:comp_bounded}.
\vskip 10pt
\begin{proposition}\label{prop:comp_prepared}
Let $u_0$ be a prepared initial data satisfying~\eqref{eq:prepared} for $\eps=\eps_0$ and $\eta = \eta_0$, then for all $\eps\le \eps_0$,
for all $\eta\le\eta_0$, the solution $\ue$ to \eqref{Pe_NC} satisfies
$$
\underline{s}^{\eps,\eta}(x) \le \ue(x,t) \le \overline{s}^{\eps,\eta}(x) \quad\textrm{ for a.e. } (x,t) \in \R\times\R_+.
$$
\end{proposition}
\section{Convergence}\label{conv_NC_sec}
\subsection{a compactness result}
Since ${(\ue)}_\eps$ is uniformly bounded between 0 and 1, there exists $u\in L^\infty(\R\times(0,T))$
such that $\ue \to u$ is the $L^\infty$ weak-star sense. This is of course insufficient to pass
in the limit in the nonlinear terms. Either greater estimates are needed, like for example a
$BV$-estimate introduced in the work of Vol$'$pert \cite{Vol67} and in \cite{NPCX}, or
we have to use a weaker compactness result. This idea motivates the introduction
of Young measures as in the papers of DiPerna \cite{DP85} and Szepessy \cite{Sze91},
or equivalently the notion of nonlinear weak star convergence, introduced in \cite{EGGH98}
and \cite{EGH00}, which leads to the notion of process solution given in
Definition~\ref{process_def}.
\vskip 10pt
\begin{theorem}[Nonlinear weak star convergence]\label{NLW*}
Let $\Qq$ be a Borelian subset of $\R^k$, and $(u_n)$ be a bounded sequence in $L^\infty(\Qq)$.
Then there exists $u\in L^\infty(\Qq\times(0,1))$, such that up to a subsequence, $u_n$ tends to $u$
"in the non linear weak star sense" as $n\to \infty$, i.e.: $\forall g\in \Cc(\R,\R), $
$$
g(u_n)\to \int_0^1 g(u(\cdot,\a))d\a \textrm{ for the weak star topology of }
L^\infty(\Qq) \textrm{ as }n\to\infty.
$$
\end{theorem}
\vskip 10pt
We refer to \cite{DP85} and \cite{EGH00} for the proof of Theorem~\ref{NLW*}.
\subsection{convergence towards a process solution}
Because of the lack of compactness, we have to introduce the notion of process solution, inspired from the notion of {\em measure valued solution} introduced by DiPerna \cite{DP85}.
\vskip 10pt
\begin{definition}[process solution to \refe{Plim}]\label{process_def}
A function $u\in L^\infty(\R\times\R_+\times(0,1))$ is said to be a process solution to
\refe{Plim} if $0\le u \le 1$ and for $i=1,2$, $\forall \psi\in \Dd^+(\overline\O_i\times\R_+)$, $\forall\k\in[0,1]$,
\beqn
\lefteqn{\ds\int_{\R_+}\int_{\O_i}\int_0^1 (u(x,t,\a)-\k)^\pm \partial_t\psi(x,t)d\a dxdt +
\int_{\O_i} (u_0(x)-\k)^\pm \psi(x,0)dx }\nn\\
&&\ds + \int_{\R_+}\int_{\O_i}\int_0^1 \Phi_{i\pm}(u(x,t,\a),\k) \partial_x\psi(x,t) d\a dxdt
+ M_{f_i}\int_{\R_+} (\overline u_i-\k)^\pm \psi(0,t)dt \ge 0,\nn
\eeqn
where $M_{f_i}$ is any Lipschitz constant of $f_i$, $\overline u_1=1$ and $\overline u_2=u_2^\star$.
\end{definition}
\vskip 10pt
\begin{lemma}\label{lem:strong_traces}
Let $u_0$ be a $\eta$-prepared initial data in the sense of Definition~\ref{Def:prepared} for some $\eta>0$, and let ${(\ue)}_\eps$ be the corresponding
family of approximate solutions. Then
\be\label{eq:strong_trace_1}
\ue(x,t) \to 1\quad \textrm{ for a.e. }(x,t)\in (-\eta,0)\times\R_+,
\ee
\be\label{eq:strong_trace_2}
\ue(x,t) \to u_2^\star \quad\textrm{ for a.e. }(x,t)\in (0,\eta)\times\R_+.
\ee
\end{lemma}
\vskip 10pt
\begin{proof}
Firstly, since $u_0$ is a $\eta$-prepared initial data, there exists $\eps_0>0$ such that
$$
\underline{s}^{\eps_0,\eta} \le u_0 \le \overline{s}^{\eps_0,\eta}.
$$
Then it follows from Proposition~\ref{prop:comp_prepared} that for all $\eps\in(0,\eps_0)$, for a.e. $(x,t)\in \R\times\R_+$
\be\label{eq:comp_47}
\underline{s}^{\eps,\eta}(x) \le \ue(x,t) \le \overline{s}^{\eps,\eta}(x).
\ee
This particularly shows that for all $\eps\in(0,\eps_0)$, for a.e. $(x,t)\in (-\eta,0)\times\R_+$,
$$
\ue(x,t) = 1,
$$
thus \eqref{eq:strong_trace_1} holds. The assertion~\eqref{eq:strong_trace_2} can be obtained by using Proposition~\ref{prop:sub-super} and the dominated convergence theorem.
\end{proof}
\vskip 10pt
\begin{proposition}[convergence towards a process solution]\label{conv_proc}
Let $u_0$ be a prepared initial data in the sense of Definition~\ref{Def:prepared}, and let ${(\ue)}_\eps$ be the corresponding
family of approximate solutions. Then, up to an extraction, $u^\eps$ converges in the nonlinear
weak-star sense towards a process solution $u$ to the problem \refe{Plim}.
\end{proposition}
\vskip 10pt
\begin{proof}
Since $\ue$ is a weak solution of \refe{Pe_NC}, which is a non-fully degenerate parabolic problem, i.e.
$\phii^{-1}$ is continuous, it follows from the work of Carrillo \cite{Car99} that $\ue$ is an entropy
weak solution, i.e.: $\forall \psi\in \Dd^+(\O_i\times\R_+)$, $\forall\k\in[0,1]$,
\beqn
\lefteqn{ \int_{\R_+}\int_{\O_i} (\ue(x,t)-\k)^\pm \partial_t \psi(x,t) dxdt
+ \int_{\O_i} (u_0(x)-\k)^\pm \psi(x,0) dx}\nn\\
&&\ds\hspace{10pt} + \int_{\R_+} \int_{\O_i} \left[\Phi_{i\pm}(\ue(x,t),\k)- \eps\partial_x(\phii(\ue)(x,t)-\phii(\k))^\pm \right]
\partial_x \psi(x,t)dxdt\ge 0.\nn\label{entro_for_NC}
\eeqn
This family of inequalities is only available for non-negative functions $\psi$ compactly
supported in $\O_i$, and so vanishing on the interface $\{x=0\}$. To overpass this difficulty, we
use cut-off functions $\chi_{i,\delta}$. \label{cut-off}
Let $\delta>0$, we denote by $\chi_{i,\delta}$ a smooth non-negative function,
with $\chi_{i,\delta}(x)=0$ if $x\notin \O_i$, and
$\chi_{i,\delta}(x)=1$ if $x\in\O_i$, $|x|\ge \delta$.
Let $\psi\in \Dd^+(\overline\O\times\R_+)$, then $\psi\chi_{i,\delta}\in \Dd^+(\O_i\times\R_+)$ can be used
as test function in \refe{entro_for_NC}. This yields
\beqn
\lefteqn{\ds \int_{\R_+}\int_{\O_i} (\ue-\k)^\pm \partial_t \psi \chi_{i,\delta}dxdt
+ \int_{\O_i} (u_0-\k)^\pm \psi(\cdot,0)\chi_{i,\delta} dx} \nn\\
&&\ds \hspace{20pt}+ \int_{\R_+} \int_{\O_i} \left[\Phi_{i\pm}(\ue,\k)- \eps\partial_x(\phii(\ue)-\phii(\k))^\pm \right]
\partial_x \psi \chi_{i,\delta}dxdt \nn \\
&&\ds \hspace{20pt} +\!\!\int_{\R_+} \int_{\O_i}\left[\Phi_{i\pm}(\ue,\k)- \eps\partial_x(\phii(\ue)-\phii(\k))^\pm \right]
\psi\partial_x \chi_{i,\delta}dxdt\ge 0.\label{entro_for_NC2}
\eeqn
We can now let $\eps$ tend to $0$. Thanks to Theorem~\ref{NLW*}, there exists $u\in L^\infty(\R\times\R_+\times(0,1))$ such that
\begin{eqnarray}
\lefteqn{ \lim_{\eps\to0} \int_{\R_+}\int_{\O_i} (\ue(x,t)-\k)^\pm \partial_t \psi(x,t) \chi_{i,\delta}(x)dxdt = } \nn \\
&&\hspace{50pt} \int_{\R_+}\int_{\O_i} \int_0^1(u(x,t,\a)-\k)^\pm \partial_t \psi(x,t) \chi_{i,\delta}(x)d\a dxdt, \label{eq:NLW*_1}
\end{eqnarray}
\begin{eqnarray}
\lefteqn{ \lim_{\eps\to0} \int_{\R_+}\int_{\O_i} \Phi_{i\pm}(\ue(x,t),\k) \partial_x \psi(x,t) \chi_{i,\delta}(x)dxdt = } \nn \\
&&\hspace{50pt} \int_{\R_+}\int_{\O_i} \int_0^1 \Phi_{i\pm}(u(x,t,\a),\k) \partial_x \psi(x,t) \chi_{i,\delta}(x)d\a dxdt . \label{eq:NLW*_2}
\end{eqnarray}
Thanks to Proposition~\ref{L2H1_NC}, one has
$$
\eps\partial_x(\phii(\ue)-\phii(\k))^\pm \textrm{ tends to } 0 \quad\textrm{ a.e. in } \OiT \textrm{ as }\eps\to 0,
$$
then
\be\label{lim_phi(u)}
\lim_{\eps\to0} \int_{\R_+}\int_{\O_i} \eps\partial_x(\phii(\ue)(x,t) - \phii(\k))^\pm \partial_x \left( \psi(x,t) \chi_{i,\delta}(x)\right) dxdt = 0.
\ee
Since $u_0$ is supposed to be a $\eta$-prepared initial data for some $\eta>0$, we can claim thanks to
Lemma~\ref{lem:strong_traces} that $\ue(x,t)$ converges almost everywhere on $(-\eta,\eta)\times\R_+$ towards $\overline u_i$ if
$x\in\O_i$. Since for $\delta<\eta$ small enough, the support of $\partial_x \chi_{1,\delta}$ is
included in the set where $\ue$ converges strongly, one has
\begin{eqnarray}
\lefteqn{
\lim_{\eps\to0} \int_{\R_+}\int_{\O_i} \Phi_{i\pm}(\ue(x,t),\k) \psi(x,t) \partial_x\chi_{i,\delta}(x)dxdt = } \nn \\
&&\hspace{100pt}
\int_{\R_+}\int_{\O_i} \Phi_{i\pm}(\overline u_i,\k) \psi(x,t) \partial_x \chi_{i,\delta}(x)dxdt . \label{eq:NLW*_3}
\end{eqnarray}
We let now $\delta$ tend to $0$. Since $\chi_{i,\delta}(x)$ tends to $1$ a.e. in $\O_i$, \eqref{eq:NLW*_1} and \eqref{eq:NLW*_2} respectively provide
\begin{eqnarray}
\lefteqn{ \lim_{\delta\to0} \lim_{\eps\to0} \int_{\R_+}\int_{\O_i} (\ue(x,t)-\k)^\pm \partial_t \psi(x,t) \chi_{i,\delta}(x)dxdt = } \nn \\
&&\hspace{100pt} \int_{\R_+}\int_{\O_i} \int_0^1(u(x,t,\a)-\k)^\pm \partial_t \psi(x,t) d\a dxdt, \label{eq:NLW*_4}
\end{eqnarray}
\begin{eqnarray}
\lefteqn{ \lim_{\delta\to 0}\lim_{\eps\to0} \int_{\R_+}\int_{\O_i} \Phi_{i\pm}(\ue(x,t),\k) \partial_x \psi(x,t) \chi_{i,\delta}(x)dxdt = } \nn \\
&&\hspace{100pt} \int_{\R_+}\int_{\O_i} \int_0^1 \Phi_{i\pm}(u(x,t,\a),\k) \partial_x \psi(x,t) d\a dxdt . \label{eq:NLW*_5}
\end{eqnarray}
One has also
\be\label{eq:NLW*_6}
\lim_{\delta\to 0} \int_{\O_i} (u_0(x)-\k)^\pm \psi(x,0)\chi_{i,\delta}(x) dx = \int_{\O_i} (u_0(x)-\k)^\pm \psi(x,0)dx.
\ee
One has $$\left| \Phi_{i\pm}(\overline u_i,\k) \right| \le M_{f_i} \left( \overline u_i - \k \right)^\pm$$
then
\begin{eqnarray*}
\lefteqn{\left|\int_{\R_+}\int_{\O_i} \Phi_{i\pm}(\overline u_i,\k) \psi(x,t) \partial_x \chi_{i,\delta}(x)dxdt\right|}\\
&&\hspace{100pt}\le M_{f_i} \left( \overline u_i - \k \right)^\pm \int_{\R_+}\int_{\O_i} \psi(x,t)\left| \partial_x \chi_{i,\delta}(x) \right| dxdt.
\end{eqnarray*}
Since $\left| \partial_x \chi_{i,\delta} \right|$ tends to $\delta_{x=0}$ in the $\Mm(\R)$-weak star sense where $$\Big<\delta_{x=0},\zeta\Big>_{\Mm(\R),\Cc_0(\R)} = \zeta(0),$$
we obtain that
\begin{eqnarray}
\lefteqn{
\liminf_{\delta\to0}\lim_{\eps\to0} \int_{\R_+}\int_{\O_i} \Phi_{i\pm}(\ue(x,t),\k) \psi(x,t) \partial_x\chi_{i,\delta}(x)dxdt \ge
} \nn \\
&&\hspace{180pt}
M_{f_i} \left( \overline u_i - \k \right)^\pm\int_{\R_+}\psi(0,t)dt . \label{eq:NLW*_7}
\end{eqnarray}
Using~\eqref{lim_phi(u)},\eqref{eq:NLW*_4},\eqref{eq:NLW*_5},\eqref{eq:NLW*_6},\eqref{eq:NLW*_7} in \eqref{entro_for_NC2} shows that $u$ is a process solution in the sense of Definition~\ref{process_def}.
\end{proof}
\subsection{uniqueness of the (process) solution}
It is clear that the notion of process solution is weaker than the one of solution
given in Definition~\ref{Sol_NC_def}.
We state here a theorem which claims the equivalence of the two notions,
i.e. any process solution is a solution in the sense of Definition~\ref{Sol_NC_def}.
Furthermore, such a solution is unique, and a $L^1$-contraction principle can
be proven.
\vskip 10pt
\begin{theorem}[uniqueness of the (process) solutions]\label{uni_NC}
There exists a unique process solution $u$ to the problem~\refe{Plim}, and
furthermore this solution does not depend on $\a$, i.e. $u$ is a solution
to the problem \refe{Plim} in the sense of definition~\ref{Sol_NC_def}.
Furthermore, if $u_0$, $v_0$ are two initial data in $L^\infty(\R)$ satisfying~\eqref{hyp:large_init} and let
$u$ and $v$ be two solutions associated to those initial data, then
for all $t\in[0,T)$,
\be\label{contract_NC}
\int_{-R}^{R} (u(x,t)-v(x,t))^\pm dx \le \int_{-R - M_f t}^{R+M_f t} (u_0(x)-v_0(x))^\pm dx.
\ee
\end{theorem}
\vskip 10pt
This theorem is a consequence of \cite[Theorem 2]{Vov02}. Let $u(x,t,\a)$ and $v(x,t,\b)$ be two process solutions corresponding to initial data $u_0$ and $v_0$. Classical Kato inequalities can be derived in each $\O_i \times\R_+$ by using the doubling variable technique: $\forall \psi\in \Dd^+(\O_i\times\R_+)$,
\begin{eqnarray*}
\lefteqn{\int_{\R_+}\int_{\O_i}\int_0^1\int_0^1 (u(x,t,\a) - v(x,t,\b))^\pm\partial_t \psi(x,t) d\a d\b dxdt }\\
&&\hspace{50pt}+ \int_{\O_i} (u_0(x) - v_0(x))^\pm \psi(x,0) dx \\
&&\hspace{50pt}+ \int_{\R_+}\int_{\O_i}\int_0^1\int_0^1\Phi_{i\pm}(u(x,t,\a),v(x,t,\b)) \partial_x \psi(x,t) d\a d\b dxdt \ge 0.
\end{eqnarray*}
The treatment of the boundary condition at the interface is an adaptation to the case of process solution
to the work of Otto summarized in~\cite{Otto96_CRAS} and detailed in~\cite{MNRR96} leading to (see
\cite[Lemma 2]{Vov02}): $\forall \psi\in \Dd^+(\overline\O_i\times\R_+)$,
\begin{eqnarray}
\lefteqn{\int_{\R_+}\int_{\O_i}\int_0^1\int_0^1 (u(x,t,\a) - v(x,t,\b))^\pm\partial_t \psi(x,t) d\a d\b dxdt } \nn \\
&&\hspace{30pt}+ \int_{\O_i} (u_0(x) - v_0(x))^\pm \psi(x,0) dx \nn\\
&&\hspace{30pt}+ \int_{\R_+}\int_{\O_i}\int_0^1\int_0^1\Phi_{i\pm}(u(x,t,\a),v(x,t,\b)) \partial_x \psi(x,t) d\a d\b dxdt \ge 0. \label{eq:Kato_2}
\end{eqnarray}
Choosing
$$\psi_\eps(x,s) = \left\{\begin{array}{lll}1&\textrm{ if } &|x|\le R+ M_f s ,\\
\ds \frac{R+ M_f s + \eps - |x|}{\eps}&\textrm{ if } & R+ M_f t \le |x| \le R+ M_f s + \eps \\
0 &\textrm{ if } &|x|\ge R+ M_f s + \eps
\end{array}\right.$$
if $s \le t$ and $\psi_{\eps}(x,s)=0$ if $s > t$
as test function in~\eqref{eq:Kato_2} and letting $\eps$ tend to $0$ provide the expected $L^1$-contraction principle~\eqref{contract_NC}.
\vskip 10pt
Finally, if $u$ and $\t u$ are two process solutions associated
to the same initial data $u_0$, we obtain a $L^1$-contraction principle of the
following form: for a.e. $t\in\R_+$,
$$
\int_\R \int_0^1 \int_0^1 (u(x,t,\a)-\t u(x,t,\beta))^\pm d\a d\beta dx \le 0,
$$
thus $u(x,t,\a) = \t u(x,t,\b)$ a.e. in $\R\times\R_+\times(0,1)\times(0,1)$. Hence $u$ does not depend on the process variable $\a$.
\vskip 10pt
\begin{theorem}\label{conv_en_NC}
Let $u_0$ be a prepared initial data in the sense of Definition~\ref{Def:prepared}, and let $\ue$ be the corresponding solution to
the approximate problem \refe{Pe_NC}. Then $\ue$ converges to the unique
solution $u$ to \eqref{Plim} associated to initial data $u_0$ in the $L^p((0,T);L^q(\R))$-sense, for all
$p,q\in [1,\infty)$.
\end{theorem}
\vskip 10pt
\begin{proof}
We have seen in Proposition~\ref{conv_proc} that $\ue$ converges up to an extraction
towards a process solution. The family ${(\ue)}_\eps$ admits so a unique
adherence value, which is a solution thanks to Theorem~\ref{uni_NC},
thus the whole family converges towards this unique limit $u$.
\vskip 10pt
Let $K$ denotes a compact subset of $\R\times[0,T]$, then one has
$$
\iint_K (\ue - u)^2 dxdt = \iint_K \left(\ue\right)^2 dx -2 \iint_K \ue u dx + \iint_K u^2 dx.
$$
Since $\ue$ converges in the nonlinear weak star sense towards $u$,
$$
\lim_{\eps\to 0} \iint_K \left(\ue\right)^2 dx = \iint_K u^2 dx.
$$
Moreover, $\ue$ converges in the $L^\infty$ weak star topology towards $u$,
then
$$
\lim_{\eps\to 0} \iint_K \ue u dx = \iint_K u^2 dx.
$$
Thus we obtain
$$
\lim_{\eps\to 0}\iint_K (\ue - u)^2 dxdt =0.
$$\label{ae_conv}
One concludes using the fact the $|\ue - u| \le 1$ for all $\eps>0$.
\end{proof}
\subsection{initial data in $L^\infty(\R)$}
In this section, we extend the result of Theorem~\ref{conv_en_NC} to any initial data in $L^\infty(\R)$ satisfying~\eqref{hyp:large_init} thanks to density argument.
\vskip 10pt
\begin{theorem}
Let $u_0\in L^\infty(\R)$ satisfying \eqref{hyp:large_init}, and let $\left(u_{0,n}\right)_{n\in \N^\star}$ be a sequence of prepared initial data tending to $u_0$ in $L^1_{loc}(\R)$. Then the sequence $\left(u_n\right)_n$ of solutions to~\eqref{Plim} corresponding to the sequence $\left(u_{0,n}\right)$ of initial data converges in $\Cc(\R_+;L^1_{loc}(\R))$ towards the unique solution to~\eqref{Plim} corresponding to solution the initial data $u_0$.
\end{theorem}
\vskip 10pt
\begin{proof}
First, note that for all $u_0\in L^\infty(\R)$ satisfying \eqref{hyp:large_init}, there exists a sequence $\left(u_{0,n}\right)_{n\in \N^\star}$ of prepared initial data tending to $u_0$ in $L^1_{loc}(\R)$ thanks to Lemma~\ref{lem:init_eta_eps}.
\vskip 10pt
Thanks to~\eqref{contract_NC}, one has for $n,m\in \N^\star$, for all $t\in\R_+$
$$
\int_{-R}^{R} (u_n(x,t)-u_m(x,t))^\pm dx \le \int_{-R - M_f t}^{R+M_f t} (u_{0,n}(x)-u_{0,m}(x))^\pm dx,
$$
then $\left(u_n\right)_n$ is a Cauchy sequence in $\Cc(\R_+;L^1_{loc}(\R))$. In particular, there exists $u$ such that
$$
\lim_{n\to\infty} u_n = u \quad \textrm{ in } \Cc(\R_+; L^1_{loc}(\R)).
$$
It is then easy to check that $u$ is the unique solution to~\eqref{Plim}.
\end{proof}
\section{Entropy solution for small initial data}\label{entro_small}
In this section, we suppose that the initial data $u_0$ belongs to $L^1(\R)$, and that
\be\label{H_u0_entro}
0 \le u_0 \le u_i^\star\quad \textrm{ a.e. in }\O_i.
\ee
This initial data can be smoothed using following lemma whose proof is almost the same as the proof of Lemma~\ref{lem:init_eta_eps}.
\vskip 10pt
\begin{lemma}\label{lem:prepared_2}
There exists $\left(u_0^\eps\right)_{\eps>0}\subset L^1(\R)$ such that
\begin{itemize}
\item $\partial_x \phii(u_0^\eps)\in L^\infty(\O_i)$,
\item ${\rm ess}\lim_{x\nearrow 0} u_0^\eps(x) = 1$,
\item $\lim_{\eps\to0} u_0^\eps = u_0$ in $L^1_{loc}(\R)$.
\end{itemize}
\end{lemma}
\vskip 10pt
For all $\eps>0$, there exists a unique bounded flux solution $\ue$ to~\eqref{Pe_NC} corresponding to $u_0^\eps$ thanks to Theorem~\ref{comp_bounded_NC}. The following theorem claims that as $\eps$ tends to $0$, $\ue$ tends to the unique entropy solution in the sense of Definition~\ref{Def:entro}.
\vskip 10pt
\begin{theorem}[convergence towards the entropy solution]
Let $u_0\in L^\infty(\R)$ satisfying~\eqref{H_u0_entro} and let $\left(u_0^\eps\right)_\eps$ be a family of approximate initial data built in Lemma~\ref{lem:prepared_2}. Let $\ue$ be the bounded flux solution to~\eqref{Pe_NC} corresponding to $u_0^\eps$, then $\ue$ converges to $u$ in $L^1_{loc}(\R\times\R_+)$ as $\eps$ tends to $0$ where $u$ is the unique entropy solution in the sense of Definition~\ref{Def:entro}.
\end{theorem}
\vskip 10pt
\begin{proof}
Using the technics introduced in \cite[Proposition 2.8]{NPCX}, we can show that for all $\lambda\in[0,q]$ there exists a steady solution $\k_\lambda^\eps$ to the problem \eqref{Pe_NC}, corresponding to a constant flux
$$f_i(\k^\eps_\lambda)-\eps \partial \phii(\k^\eps_\lambda) = \lambda,$$
and such that this solution converges uniformly on each compact subset of $\R^\star$ as $\eps$ tends to $0$ towards
$$\k_\lambda(x)= \min_\k\left\{f(\k,x) = \lambda\right\}.$$
Following the idea of Audusse and Perthame~\cite{AP05}, we will now compare the limit $u$ of $\ue$ as $\eps$ to $0$ with the steady state $\k_\lambda$. Let $\l\in [0,q]$. Since $\ue$ and $\k^\eps_\l$ are both bounded flux solutions, it follows from Proposition~\ref{comp_prop} that for all $\psi\in \Dd^+(\R\times{\R_+})$,
\beqn
\lefteqn{\ds \int_{\R_+} \int_\R \left( \ue - \k^\eps_\l\right)^\pm \partial_t \psi dxdt + \int_\R \left( u_0^\eps - \k^\eps_\l\right)^\pm \psi(\cdot,0) dx}\nn\\
&&\ds\hspace{20pt} + \int_{\R_+} \sum_i \int_{\O_i} \left( \Phi_{i\pm}(\ue, \k_\l^\eps) - \eps \partial_x \left( \phii(\ue) - \phii(\k_\l^\eps)\right)^+ \right) \partial_x\psi dxdt \ge 0. \label{comp_kle}
\eeqn
Choosing $\l=q$ and $\psi(x,t)=(T-t)^+ \xi(x)$ for some arbitrary $T>0$ and some $\xi\in\Dd^+(\R)$ yields
\be\int_0^T \int_\O (\ue-\k^\eps_q)^+ \xi dxdt \le \int_0^T (T-t) \sum_{i=1,2} \int_{\O_i}
\eps \partial_x \left(\phii(\ue) - \phii(\k^\eps_q)\right)^+ \partial_x \xi dxdt. \label{comp_ke}
\ee
Since $\ue$ is bounded between $0$ and $1$, it converges in the nonlinear weak star sense, thanks to Theorem~\ref{NLW*} towards a function $u\in L^\infty(\R\times{\R_+}\times(0,1))$, with
$0 \le u \le 1$ a.e.. Then \eqref{comp_ke} provides
\be
u \le \k_q = u_i^\star \quad \textrm{ a.e. in }\O_i\times{\R_+}\times(0,1). \label{max_entro}
\ee
Let $\l\in [0,q]$, then taking the limit for $\eps\to 0$ in \eqref{comp_kle} yields
\beqn
\lefteqn{\ds \int_{\R_+} \int_\R\int_0^1 \left| u - \k_\l\right| \partial_t \psi d\a dxdt + \int_\R \left| u_0 - \k_\l\right| \psi(\cdot,0) dx}\nn\\
&&\ds\hspace{100pt} + \int_{\R_+} \sum_i \int_{\O_i} \int_0^1 \Phi_{i}(u, \k_\l) \partial_x\psi d\a dxdt \ge 0. \label{comp_kl}
\eeqn
Suppose that $u_2^\star\ge u_1^\star$. Let $\k\in [0,u_2^\star]$, we denote by $\tilde\k = f_1^{-1}(f_2(\k))\cap [0,u_1^\star]$. Then choosing $\l = f_2(\k)$ in \eqref{comp_kl}, and letting $\eps$ tend to $0$ gives: $\forall\k\in [0,u_2^\star]$, $\forall \psi\in\Dd^+(\R\times{\R_+})$,
\begin{eqnarray}
\lefteqn{\int_0^T \int_{\O_1} \int_0^1 | u - \t \k| \partial_t \psi d\a dxdt+ \int_{\O_1} |u_0- \t \k | \psi(\cdot,0) dx}\nn\\
&&\hspace{50pt}+ \ds \int_{\R_+} \int_{\O_2} \int_0^1 | u - \k| \partial_t \psi d\a dxdt+ \int_{\O_2} |u_0-\k | \psi(\cdot,0) dx \nn\\
&&\hspace{50pt}+ \int_{\R_+} \int_0^1 \left( \int_{\O_1} \Phi_1(u, \t \k) \partial_x \psi dx + \int_{\O_2} \Phi_2(u,\k) \partial_x \psi dx\right)d\a dt \ge 0. \label{comp2_k}
\end{eqnarray}
It follows from the work of Jose Carrillo \cite{Car99} that the following entropy inequalities hold for test functions compactly supported in $\O_1$: $\forall\k\in [0,1]$, $\forall \psi\in\Dd^+(\O_1\times{\R_+})$,
\begin{eqnarray}
\lefteqn{ \ds \int_{\R_+} \int_{\O_1} | \ue - \k| \partial_t \psi dxdt+ \int_{\O_1} |u_0^\eps-\k | \psi(\cdot,0) dx }\nn\\
&&\hspace{50pt}+ \int_{\R_+} \int_{\O_1} \left( \Phi_1(\ue,\k) -\eps \partial_x \left| \varphi_1(\ue) - \varphi_1(\k) \right| \right) \partial_x \psi dxdt \ge 0. \label{eq:entro_car}
\end{eqnarray}
Thus letting $\eps$ tend to $0$ in~\eqref{eq:entro_car} provides: $\forall \psi\in\Dd^+(\O_1\times{\R_+})$, $\forall\k\in[0,1]$,
\begin{eqnarray}
\lefteqn{\ds \int_{\R_+} \int_{\O_1} \int_0^1 | u - \k| \partial_t \psi d\a dxdt+ \int_{\O_1} |u_0-\k | \psi(\cdot,0) dx }\nn\\
&&\hspace{100pt} + \int_{\R_+} \int_{\O_1} \int_0^1 \Phi_1(u,\k) \partial_x \psi d\a dxdt \ge 0. \label{comp1_k}
\end{eqnarray}
Let $\delta>0$, and let $\psi\in \Dd^+(\R\times{\R_+})$, we define
$$\psi_{1,\delta}(x,t) = \psi (x,t) \chi_{1,\delta}(x),\qquad \psi_{2,\delta}=\psi-\psi_{1,\delta},$$ where $\chi_{1,\delta}$ is the cut-off function introduced in section~\ref{cut-off}. Then using $\psi_{1,\delta}$ as test function in \eqref{comp1_k} and $\psi_{2,\delta}$ in \eqref{comp2_k} leads to:
\beqn
\lefteqn{ \ds \int_{\R_+}\int_\R\int_0^1 | u - \k | \partial_x \psi d\a dxdt + \int_{\R} |u_0-\k | \psi(\cdot,0) dx }\nn \\
&&\ds\hspace{20pt} + \int_{\R_+}\sum_i\int_{\O_i}\int_0^1 \Phi_i(u,\k) \partial_x \psi d\a dxdt \nn \\
&&\ds\hspace{20pt} + \int_{\R_+}\int_{\O_1}\int_0^1 \left( \Phi_1(u,\k) - \Phi_1(u,\t\k) \right) \psi \partial_x \chi_{1,\delta} d\a dxdt\ge \Rr(\k,\psi,\delta), \label{comp3_k}
\eeqn
where $\lim_{\delta\to0} \Rr(\k,\psi,\delta) = 0.$ Since $f_1$ is increasing on $[0,u_1^\star]$ and $f_1([u_1^\star,1)) \subset [q,\infty)$, either
$\k\le u_1^\star$, or $f_1(\k) \ge f_1(u_1^\star).$ This ensures that
$$\Phi_1(u,\k) = | f_1(u) - f_1(\k) |, \qquad \forall u \in [0,u_1^\star],\quad \forall k\in [0,u_2^\star].$$
This yields
\begin{eqnarray}
\left|\Phi_1(u,\k) - \Phi_1(u,\t\k) \right|&=& \big|| f_1(u) - f_1(\k) | - | f_1(u)- f_1(\t\k) | \big| \nn\\
&\le& | f_1(\k) - f_1(\t\k) | = | f_1(\k) - f_2(\k) |. \label{triangle_k}
\end{eqnarray}
Taking the inequality \eqref{triangle_k} into account in \eqref{comp3_k}, and letting $\delta\to0$ provides: \\
$\forall \k\in [0,\| u \|_\infty]$, $\forall \psi\in \Dd^+(\R\times{\R_+})$,
\begin{eqnarray*}
\lefteqn{ \int_{\R_+}\int_\R\int_0^1 | u - \k | \partial_x \psi d\a dxdt + \int_{\R} |u_0-\k | \psi(\cdot,0) dx } \nn \\
&& \ds\hspace{20pt} + \int_{\R_+}\sum_i\int_{\O_i}\int_0^1 \Phi_i(u,\k) \partial_x \psi d\a dxdt
+ |f_1(\k) - f_2(\k)| \int_0^T \psi (0,\cdot) dt \ge 0.
\end{eqnarray*}
Using the work of Florence Bachmann~\cite[Theorem 4.3]{Bachmann_These}, we can claim that $u$ is the unique entropy solution to the problem. Particularly, $u$ does not depend on $\a$ (introduced for the nonlinear weak star convergence). As proven in the proof of Theorem~\ref{conv_en_NC}, this implies that $\ue$ converges in $L^1_{loc}(\R\times\R_+)$ towards $u$.
\end{proof}
\section{Resolution of the Riemann problem}\label{sec:Riemann}
In this section, we complete the resolution of the Riemann problem at the interface $\{x=0\}$, whose result has been given in section~\ref{subsec:oil-trapping}. Consider the initial data
$$u_0(x) = \left\{Ê\begin{array}{lll}
u_\ell & \textrm{ if } & x<0,\\
u_r & \textrm{ if }Ê& x>0.
\end{array}\right.$$
We aim to determine the traces $(u_1,u_2)$ at the interface of the solution $u(x,t)$ corresponding to $u_0$. This resolution has already been performed in the following cases.
\begin{itemize}
\item[(a).] $u_1^\star < u_\ell \le 1$ and $u_2^\star \le u_r < 1$: it has been seen that $u_1 = 1$ and $u_2 = u_2^\star$. \vskip 5pt
\item[(b).] $0 \le u_\ell \le u_1^\star$ and $0 \le u_r \le u_2^\star$: Since $u$ is the unique optimal entropy solution studied in \cite{AMV05,Kaa99}, then $u_1 = u_\ell$ and $u_2$ is the unique value in $[0,u_2^\star]$ such that $f_1(u_\ell) = f_2(u_2)$.
\end{itemize}
\vskip 10pt
\noindent
In the cases
\begin{itemize}
\item[(c).] $u_1^\star < u_\ell \le 1$ and $u_r = 1$,\vskip 5pt
\item[(d).] $u_\ell = u_1^\star$ and $u_r = 1$,
\end{itemize}
it is possible to approach the solution $u$ by bounded flux solutions $\ue$ that are constant equal to $1$ in $\O_2\times\R_+$. Then one obtains $u_1 = u_2 = 1$ for the case (c) and $u_1= u_1^\star$ and $u_2 = 1$ for the case (d).
\vskip 10pt
\noindent The last points we have to consider are
\begin{itemize}
\item[(e).] $u_1^\star < u_\ell \le 1$ and $0 \le u_r < u_2^\star$,
\vskip 5pt
\item[(f).] $0 \le u_\ell \le u_1^\star$ and $u_2^\star < u_r \le 1$.
\end{itemize}
To perform the study of the two last cases (e) and (f), we need the following lemmas that can be proved using similar arguments than those used in \cite{FVbarriere}, particularly concerning the treatment of the boundary condition imposed on $\{x=0\}$.
\vskip 10pt
\begin{lemma}\label{lem:e}
Let $u_r\in [0,u_2^\star)$. For all $\eps>0$, there exists a function $v^\eps$ solution to the problem
\be\label{syst:Pe_2}
\left\{\begin{array}{lll}
\partial_t v^{\eps} + \partial_x \big( f_2(v^{\eps}) - \eps \partial_x \varphi_2(v^{\eps}) \big) = 0 & \textrm{ if } & x>0,\ t>0, \\[5pt]
f_2(v^{\eps}) - \eps \partial_x \varphi_2(v^{\eps}) = f_2(u_2^\star) & \textrm{ if } & x=0,\ t>0, \\[5pt]
v^{\eps} = u_r & \textrm{ if } & x>0 , \ t=0,
\end{array}
\right.
\ee
satisfying furthermore $u_r\le v^{\eps} \le u_2^\star$ and $\partial_x \varphi_2(v^{\eps}) \in L^\infty(\R_+\times\R_+)$.
\end{lemma}
\vskip 10pt
\begin{lemma}\label{lem:f}
Let $u_\ell \in [0,u_1^\star]$, $u_r\in (u_2^\star,1]$ and let $u_2$ be the unique value of $[0,u_2^\star]$ such that $f_2(u_2) = f_1(u_\ell)$.
For all $\eps>0$ there exists a function $w^{\eps}$ solution to the problem
\be\label{syst:Pe_22}
\left\{\begin{array}{lll}
\partial_t w^{\eps} + \partial_x \big( f_2(w^{\eps}) - \eps \partial_x \varphi_2(w^{\eps}) \big) = 0 & \textrm{ if } & x>0,\ t>0, \\[5pt]
f_2(w^{\eps}) - \eps \partial_x \varphi_2(w^{\eps}) = f_2(u_2) = f_1(u_\ell) & \textrm{ if } & x=0,\ t>0, \\[5pt]
w^{\eps} = u_r & \textrm{ if } & x>0 , \ t=0,
\end{array}
\right.
\ee
satisfying furthermore $u_2 \le w^{\eps} \le u_r$ and $\partial_x \varphi_2(w^{\eps}) \in L^\infty(\R_+\times\R_+)$.
\end{lemma}
\vskip 10pt
\textbf{The case (e).} Assume that $u_\ell> u_1^\star$ and $u_r< u_2^\star$.
Let $\left(u_0^\eta\right)_\eta$ be a family of initial data such that $\partial_x \phii(u_0^\eta) \in L^\infty(\O_i)$, $u_0^\eta(x) = 1$ for $x \in (-\eta,0)$, $u_0^\eta(x) \in [u_\ell, 1]$ for a.e. $x\in \O_1$, $u_0^\eta(x) = u_r$ a.e. in $\O_2$ and such that
$$
\left\| u_0^\eta - u_\ell \right\|_{L^1(\O_1)} \le 2\eta.
$$
Then thanks to Theorem~\ref{comp_bounded_NC}, there exists a unique bounded flux solution $u^{\eps,\eta}$ to the problem~\eqref{Pe_NC} corresponding to the initial data $u_0^\eta$.
It is easy to check that the solution defined in $\Omega_2\times\R_+$ by the function $v^\eps$ introduced in Lemma~\ref{lem:e} and coinciding in $\O_1\times\R_+$ with the unique bounded flux solution corresponding to the initial data
$$
\tilde u_0^\eta(x) = \left\{Ê\begin{array}{lll}
u_0^\eta(x) & \textrm{ if }Ê& x<0,\\[5pt]
1 & \textrm{ if }Ê& x>0.
\end{array}\right.
$$
In particular, as $\eps$ tends to $0$, it follows from arguments similar to those developed in the previous sections that $u^{\eps,\eta}$ converges in $L^1_{loc}(\overline\O_i\times\R_+)$ towards the unique entropy solution to the problem
problem
\be\label{eq:Riemann_e1}
\left\{ \begin{array}{lll}
\partial_t u^\eta + \partial_x f_1(u^\eta) = 0 & \textrm{ if } x<0,\ t>0,\\[5pt]
u^\eta = 1 & \textrm{ if } x=0,\ t>0,\\[5pt]
u^\eta = u_0^\eta & \textrm{ if } x<0,\ t=0.
\end{array}\right.
\ee
\be\label{eq:Riemann_e2}
\left\{ \begin{array}{lll}
\partial_t u + \partial_x f_2(u) = 0 & \textrm{ if } x>0,\ t>0,\\[5pt]
u = u_2^\star & \textrm{ if } x=0,\ t>0,\\[5pt]
u = u_r & \textrm{ if } x>0,\ t=0.
\end{array}\right.
\ee
Note that the trace condition on the interface $\{x=0\}$ in \eqref{eq:Riemann_e2} is fulfilled in a strong sense since $u_r \le u(x,t) \le u_2^\star$ a.e. in $\O_2\times\R_+$ and $f_2$ is increasing on $[u_r, u_2^\star]$.
\vskip 10pt
The solution to~\eqref{eq:Riemann_e1} depends continuously on the initial data in $L^1_{loc}$. Hence, letting $\eta$ tend to $0$ in~\eqref{eq:Riemann_e1} provides that the limit $u$ of $u^\eta$ is the unique entropy solution to the problem
$$
\left\{ \begin{array}{lll}
\partial_t u+ \partial_x f_1(u) = 0 & \textrm{ if } x<0,\ t>0,\\[5pt]
u = 1 & \textrm{ if } x=0,\ t>0,\\[5pt]
u = u_\ell & \textrm{ if } x<0,\ t=0.
\end{array}\right.
$$
Note that since $u_1^\star < u_\ell \le u \le 1$ and $\min_{s\in[u,1]} f_1(s) = f_1(1) = q$, the trace prescribed on the interface $\{x= 0\}$ is fulfilled in a strong sense. This particularly yields that in the case (e), the solution to the Riemann problem is given by
$$
u_1 = 1, \qquad u_2 =u_2^\star.
$$
\vskip 5pt
\textbf{The case (f).} Following the technique used in \cite{NPCX} and in Section~\ref{entro_small}, there exists a unique function
$u_\ell^\eps$ solution to the problem:
$$
\left\{
\begin{array}{lll}
\ds f_1(u^\eps_\ell)- \eps\frac{\rm d}{{\rm d}x} \varphi_1(u_\ell^\eps) = f_1(u_\ell) & \textrm{ if } & x<0,\\[5pt]
u_{\ell}^{\eps}(0) = 1 & \textrm{ if } & x=0.
\end{array}\right.
$$
Let $\ue$ be the function defined by
$$
\ue(x,t) = \left\{ \begin{array}{lll}
u_\ell^\eps(x) & \textrm{Êif }Ê& x<0,\ t\ge0,\\[5pt]
w^\eps(x,t) & \textrm{ if } & x>0,\ t\ge 0,
\end{array}\right.
$$
where $w^\eps$ is the function introduced in Lemma~\ref{lem:f}.
Then $\ue$ is a bounded flux solution to the problem~\eqref{Pe_NC} in the sense of Definition~\ref{bounded_Def_NC}.
\vskip 10pt
One has
$$u_\ell^\eps \to u_\ell\quad \textrm{Êin }ÊL^1_{loc}(\O_1)\quad \textrm{Êas } \eps \to 0,$$
and
$$
w^\eps \to w\quad \textrm{Êin }ÊL^1_{loc}(\O_2\times\R_+)\quad \textrm{Êas } \eps \to 0
$$
where $w$ is the unique solution to
$$
\left\{
\begin{array}{lll}
\partial_t w + \partial_x f_2(w) = 0 & \textrm{ if } x >0,\ t>0, \\[5pt]
w = u_2 = f_2^{-1}\circ f_1(u_\ell) & \textrm{ if } x=0, \ t>0, \\[5pt]
w= u_r & \textrm{ if } x>0, \ t=0.
\end{array}\right.
$$
Since $w(x,t) \in [u_2,u_r]$ a.e. in $\O_2 \times \R_+$ and since $\min_{s\in [u_2,w]} f_2(s) = f_2(u_2) = f_1(u_\ell)$, the trace $w=u_2$ is satisfied in
a strong sense on $\{x=0\}$.
This yields that the solution to the Riemann problem in the case (f) is given by
$$
u_1 = u_\ell, \qquad u_2 = f_2^{-1}\circ f_1(u_\ell).
$$
\section{Conclusion}
The model presented here shows that for two-phase flows in heterogeneous porous media with negligible dependance of the capillary pressure with respect to the saturation, the good notion of solution is not always the entropy solution presented for example in \cite{AJV04,Bachmann_These}, and particular care as to be taken with respect to the orientation of the gravity forces. Indeed, some non classical shock can appear at the discontinuities of the capillary pressure field, leading to the phenomenon of oil trapping. We stress the fact that the non classical shocks appearing in our case have a different origin, and a different behavior of those suggested in the recent paper \cite{vDPP07} (see also \cite{LeFloch02}). Indeed, in this latter paper, this lack of entropy was caused by the introduction of the dynamical capillary pressure \cite{HG90,HG93,Pav89}, i.e. the capillary pressure is supposed to depend also on $\partial_t u$. In our problem, the lack of entropy comes only from the discontinuity of the porous medium.
\vskip 5pt
In order to conclude this paper, we just want to stress that this model of piecewise constant capillary pressure curves can not lead to some interesting phenomenon. Indeed, if the capillary pressure functions $\pi_i$ are such that $\pi_1((0,1))\cap \pi_2((0,1))\neq\emptyset$, it appears in \cite[Section 6]{FVbarriere} (see also \cite{BPvD03}) that some oil can overpass the boundary, and that only a finite quantity of oil can be definitely trapped. Moreover, this quantity is determined only by the capillary pressure curves and the difference between the volume mass of both phases, and does not depend on $u_0$. The model presented here, with total flow-rate $q$ equal to zero, do not allow this phenomenon, and all the oil present in $\O_1$ at the initial time remains trapped in $\O_1$ for all $t\ge 0$.
\bibliographystyle{plain}
\bibliography{ccances}
\end{document} | 0.004643 |
Democratic Gov. Jerry Brown will seek an unprecedented fourth term in today’s election, facing Republican Neel Kashkari, who has focused his campaign on creating jobs and improving education.
Brown has done little campaigning on behalf of his re-election, concentrating instead on trying to win passage of Propositions 1 and 2, the water bond and “rainy day” measures on today’s ballot.
Brown has is a former Treasury Department official making his first run for office. He has recently focused on has.
Following his government service, Kashkari worked for the Newport Beach-based global investment management firm PIMCO, resigning last year to explore running for governor. | 0.286468 |
I around the globe (she is taking small groups to Hawaii and Greece in 2018). Jane’s herb school, Dandelion Herbal Center, is nestled in the redwoods in beautiful and remote Humboldt County, California. An ideal location to connect with the spirit of plants! Through her Festival of Herbs series, she invites herbalists like Rosemary Gladstar, Pam Montgomery, Christopher Hobbs and others from across the U.S. to visit and share their wisdom to the herbal community. She is a gift to us all and I am so happy to share a little of her story with you!
Enjoy this episode of The Herb Walk Podcast and Subscribe today to catch up on all of Season 1 and find out when Season 2 is released next year! | 0.250424 |
Germany: travelogue February 2010
Every year we take our granddaughter Esmee for a holiday at least twice a year for a week, In 2009, 2010 and 2011 we spent a week in February in the same park in the Eifel mountain area at Gunderath. We have made a travelogue from our trip in 2010 but used also photographs from 2009 and 2011 since we did a lot of the same things in those years.
To Monreal and a vulcano museum
To Eifel-Zoo at Lünebach-Pronsfeld
Playing in the snow and we visit Bitburg
Exploring the Mosel valley and castles
To Idar-Oberstein to buy stones
To the ruin of Kronenburg and the kasteel at Vianden
Visiting Emsflower on our way home | 0.135498 |
We have had an exciting week in year one......the children all received letters from their penpals at Our Lady Star of the Sea and they have written some lovely letters back. We have continued to learn about weight and length in maths, we had lots of fun flying paper planes and measuring how far they traveled with our learning partners.
Next Wednesday we go to Skipton Castle, please make sure your child comes in full PE kit. We will not be going to the shop so they will not need any money. Please let me know if your child is travel sick so we can keep an extra eye on them.
Learning logs have gone home tonight and they are due back on Wednesday 6th April. Please practice the spellings with your children.
Have a lovely weekend and enjoy the gorgeous sunshine whilst you can!
Miss Dixon and Mrs Veasey | 0.14815 |
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Other Top Job Vacancies In Singapore: | 0.01637 |
Sommelier Nick Demos
Old Edwards Inn and Spa will once again host its Release the Rosé Wine Dinner at The Farm at Old Edwards on Friday, April 14. It’s an evening of food, live music, wine, and camaraderie.
Rosé is one of the most fashionable swirls to top a stem in today’s wine world, even for men. Rosés have sashayed up to the proverbial plate, not only in popularity but in quality as well. Shedding the misperception that all rosés are sweet, and sporting a palate-pleasing acidity, today’s French-style versions are flavorful, complex, even crisp and dry; and their sales are growing 10 times faster than overall table wines.
In April, just in time for the spring releases, Old Edwards is pairing these stylish sips with the culinary imaginings of Madison’s chefs. Live music, rustic ambiance, and fun camaraderie will make for another magical evening at The Farm. It is the perfect time to get your #rosévibes on in time for the spring season.
Sommelier Nick Demos will be present to discuss the different pairings and educate diners on the joys of rosés. Nick, a native of Western North Carolina, has been immersed in wine culture since an early age. His grandfather opened the very first wine shop in downtown Asheville, North Carolina, just after Prohibition.
In July of 2016, Mr. Demos passed the ever-challenging Advanced Sommelier Exam through the Court of Master Sommeliers. He is currently pursuing his studies as a candidate for the Master Sommelier Diploma. His natural inclination for hospitality and his passion for the art of wine guarantee a sparkling evening.
Let’s not forget about the culinary part of the evening. Executive Chef Chris Huerta and his team have created five fresh and innovative dishes, each course paired perfectly with a different rosé. The cost of the bubbly evening is $135 per person. | 0.129064 |
Last Updated on January 4, 2016 by bvnadmin
[vc_row full_width=”” parallax=”” parallax_image=””][vc_column width=”1/1″][vc_column_text]No matter if you use clippers or a tried-and-true razor to shave, dealing with bumps over time can be annoying.Depending on your skin’s sensitivity, and your DNA, you might regularly get pseudofolliculitis barbae, commonly called razor bumps, which are caused by hair that curls back into the skin producing an inflammatory reaction. Lessening any chance of developing bumps also involves making sure to stick to a solid grooming game plan while being patient with skin while you shave.
We’ve put together a few hacks to help reduce bumps on both face and head — and show you how to get closer to having shaving Jedi status:[/vc_column_text][vc_column_text]
MAINTAIN YOUR CLIPPERS
[/vc_column_text][vc_column_text css_animation=””]In the same way you keep a car maintained, it’s best to also keep your grooming tools clean and ready to use. To start, after each shave, clean and oil your clippers. With regular use [6-12 months], make sure the blade of your clippers hasn’t become dull. Remember, if you have sensitive skin, a fresh blade might be too sharp, meaning it can also pull at hair and skin instead of clipping it, so be gentle when using a new blade.[/vc_column_text][vc_single_image image=”4186″ border_color=”grey” img_link_large=”” img_link_target=”_self” img_size=”full”][vc_separator color=”grey” align=”align_center” style=”” border_width=”” el_width=””][vc_column_text]
MAKE TIME FOR GROOMING
[/vc_column_text][vc_column_text]Grooming on the fly is just as bad as zooming in your car at a high speed through traffic. The chances for an accident are greatly increased. The same goes with doing a rush job on your face or even your head. If you’re in a hurry to get to work or running late to meet a date, then ditch the idea of a quick shave. Being too hasty can lead to unwanted nicks and unsightly bumps. Instead of rushing, plan your grooming schedule a week in advance and make shaving your face or head a nighttime practice. This is a clear win-win: A pre-bedtime shave will give your skin a chance to heal while you sleep.[/vc_column_text][vc_column_text]
GIVE YOURSELF A BETTER SHAVE
[/vc_column_text][vc_column_text]You too can become a Jedi, but much like Luke Skywalker, becoming a Jedi requires brushing up on your technique and skills. To better learn the art of shaving, start here. Pay attention to which direction your facial hair grows, avoid shaving against the grain and don’t stretch skin while you shave or repeatedly shave in one area. Make sure to open your pores with a warm towel before shaving and close them by using cold water. If you’re using clippers, don’t place clippers directly on the skin but maintain a close shave.[/vc_column_text][vc_separator color=”grey” align=”align_center” style=”” border_width=”” el_width=””][vc_column_text]
CREATE A BETTER SKIN-CARE ROUTINE
[/vc_column_text][vc_column_text]One of the best ways to lessen your chances for bumps is to regularly exfoliate the skin on both face and head in order to remove dead layers of skin. Using a mild exfoliating scrub can also help to unclog pores and lessen any chance of getting acne, all while helping to achieve a more healthy, smooth skin. Try using an exfoliating scrub once or twice a week (figure out how sensitive your skin is first) especially during the winter months when your face and head are more exposed to colder temperatures.
To keep your skin in good shape and prepped for a shave or the winter’s elements, you have to maintain an arsenal of grooming goodies such as Kyoku for Men Exfoliating Facial Scrub, $18; Ursa Major Super Natural Skin Care 4-in-1 Essential Face Tonic, $26; Morihata Binchotan Facial Soap, various prices; and Fig+Yarrow Yarrow Buttercream, $48.
Here’s an added tip: go to your medicine cabinet or kitchen for home hacks such as diluted apple cider vinegar, witch hazel, or even tree tea oil which helps lessen inflammation.[/vc_column_text][vc_single_image image=”4187″ border_color=”grey” img_link_large=”” img_link_target=”_self” img_size=”full”][vc_separator color=”grey” align=”align_center” style=”” border_width=”” el_width=””][vc_column_text]
GIVE YOUR SKIN A BREAK
[/vc_column_text][vc_column_text]If you’ve noticed more bumps after shaving…Read more at bevelcode.com[/vc_column_text][vc_separator color=”grey” align=”align_center” style=”” border_width=”” el_width=””][vc_column_text]Words by By Marques Harper[/vc_column_text][/vc_column][/vc_row] | 0.918916 |
While listening to Popshop Radio 003 by The Knocks for the um-teenth time there’s always one track that gets me bobbing my head, tapping my feet and belting the lyrics. It’s from Swedish Electronic Pop duo Icona Pop. These two ladies are electric-sounding like Ladytron, vocally strong like Robyn and just as edgy as The Sounds.
Their sound is uplifting, empowering and highly danceable. Their EP ‘Night Like This’ was just released this month on iTunes. For now, here are some of their tracks:
Icona Pop – “Top Rated” – Download (Thanks The Chuckness)
Icona Pop – “Nights Like This”
Icona Pop – “Lovers to Friends”
Icona Pop – “Manners”
SoundCloud:
Twitter:. | 0.304538 |
\begin{document}
\title{An outline of obstruction theories of extensions via track categories}
\author{MAriam Pirashvili}
\maketitle
\begin{abstract}
Abelian track categories can be classified via the third Baues-Wirsching cohomology of small categories. This approach is used in this paper to compare and classify different generalisations of the obstruction theory of non-abelian group extensions, due to Cegarra, Garz\'on and Grandjean, Cegarra, Garz\'on and Ortega, and Chen, Du and Wang.
\end{abstract}
\section{Introduction}
The $3$-dimensional cohomology groups appear in the study of extensions of non-abelian groups. The general problem is that of constructing all short exact sequences given end groups $G$ and $\Pi$ and given an abstract kernel $\eta: \Pi\to G$. In \cite{e-m}, Eilenberg and MacLane showed that the existence of such extensions depends on an element, referred to as the obstruction, in the third cohomology $H^3(\Pi,Z(G))$.
There have been numerous generalisations of this obstruction theory. In \cite{cegarra,ortega}, Cegarra et al. defined graded extensions of categories by a group, with and without a monoidal structure on the category. More recently, in \cite{chen}, the authors defined extensions of groupoids by another groupoid. These theories are also related by the fact that they are all special cases of Grothendieck cofibrations.
In fact, we show in this paper that using the Baues-Wirsching cohomology for small categories, we can give a unified viewpoint for all these theories. We construct a class in the third Baues-Wirsching cohomology, which in different settings connects all of these classes in a non-trivial way. We exploit the crucial observation of Baues-Jibladze that abelian track categories are classified by the third Baues-Wirsching cohomology. Any such track category defines a class in the third cohomology. We show that all the above-mentioned classes can be obtained by choosing the appropriate track category.
This allows for a unified view of these related construction, as well as allowing for comparisons between them.
In particular, to obtain all Eilenberg-Maclane classes, we construct an abelian track category which has not been considered before. The objects of said track category are groups with surjective homomorphisms as $1$-morphisms and $2$-morphisms given by conjugation. This gives a class in $H^3({\mathcal G}_\sim, D^{\mathcal G})$, where ${\mathcal G}_\sim$ is the homotopy category of said track category, and the associated natural system $D^{\mathcal G}$ is the functor which assigns to each group its centre. Then for each group, we can consider the restriction to the one-object subcategory, with $1$-morphisms the outer automorphisms of the group and the centre of the group as the associated natural system, and thus recover the associated Eilenberg-Maclane class.
\section{Third cohomology and abelian track categories}
Track categories are groupoid enriched categories. A track category is \emph{abelian} if the automorphism group of any 1-arrow is abelian. By a fundamental observation of Baues and Jibladze \cite{BJ}, any abelian track category $\mathcal T$ defines an element in third cohomology, called the \emph{global Toda class} of $\mathcal T$.
In this section we recall the basic definitions and facts needed to explain this result.
\subsection{Category of factorizations and natural systems} For a category $\bI$, one denotes by $\bF\bI$ the
\emph{category of factorizations of} $\bI$ \cite{BW}. Let us recall
that objects of the category $\bF\bI$ are morphisms $\alpha:i\to j$
of $\bI$. A morphism from $\alpha$ to $\beta:k\to l$ in $\bF\bI$ is a
pair $(\xi,\eta)$, where $\xi:k\to i$ and $\eta:j\to l$ are
morphisms in $\bI$ such that
$$
\beta=\eta\circ\alpha\circ\xi.
$$
In other words, the following diagram
$$
\xymatrix{
j\ar[r]^\eta&l\\
i\ar[u]^\alpha&k\ar[l]^\xi\ar[u]_\beta }
$$
commutes. If $(\xi',\eta')$ is also a morphism in $\bF\bI$ from $\beta:k\to l$ to $\beta':k'\to l'$:
$$\xymatrix{
j\ar[r]^\eta&l\ar[r]^{\eta'}&l'\\
i\ar[u]^\alpha&k\ar[l]^\xi\ar[u]_\beta&k'\ar[l]^{\xi'}\ar[u]_{\beta'}
}
$$
then the composite morphism $\alpha\to\beta'$ in $\bF\bI$ is defined by
$$
(\xi',\eta')(\xi,\eta) = (\xi\xi',\eta'\eta).
$$
Clearly $(\xi,\eta)$ is an isomorphism in $\bF\bI$ iff both $\xi$ and $\eta$ are isomorphisms in $\bI$.
Let $\bI$ be a small category. A \emph{natural system of abelian groups} on $\bI$
is a functor $D$ from the category $\bF\bI$ to the category $\ab$ of abelian
groups. For a natural system $D$ we usually denote the value of $D$
on $\alpha:i\to j$ by $D_\alpha$ as well as $D(\alpha)$. If $\alpha$
is the identity $\id_i:i\to i$ we write $D_i$ instead of
$D_{\id_i}$. For morphisms $\xi:k\to i$ and $\eta:j\to l$ we also
denote actions of $D(\xi,\id_j):D_\alpha\to D_{\alpha\xi}$, resp.
$D(\id_i,\eta):D_\alpha\to D_{\eta\alpha}$ on an element $a\in
D_\alpha$ by $\xi^*(a)$, resp. $\eta_*(a)$ or as well by $a\xi$,
resp. $\eta a$. It is clear that $$(\xi ,\eta
)=(\xi,\id_j)(\id_i,\eta ) =(\id_i,\eta)(\xi,\id_j),$$ i.~e. in
$D_{\eta\alpha\xi}$ one has $\xi^*\eta_*(a)=\eta_*\xi^*(a)$ for any $a\in D_\alpha$.
\subsection{Baues-Wirsching cohomology}
Let $D$ be a natural system of abelian groups on a small category $\bI$. Following Baues
and Wirsching \cite{BW} we define the cochain complex
$C^*(\bI;D)$ by
$$
C^0(\bI;D)=\prod_{i\in\ob(\bI)}D_i
$$
and
$$
C^n(\bI;D)=\prod_{i_0\xot{\i_1}\cdots\xot{\i_n}i_n}D_{\i_1\cdots
\i_n}, \quad n>0$$
The boundary operator is defined by $d=\sum_{i=0}^n (-1)^i\dd^i$, where the coface operators $\dd^m:C^n(\bI;D)\to C^{n+1}(\bI;D)$ are
defined as follows. If $m=0$ one puts:
$$(\dd^0(f))(i_0\xot{\i_1}\cdots\xot{\i_{n+1}}i_{n+1})
={\i_1}_*f(i_1\xot{\i_2}\cdots\xot{\i_{n+1}}i_{n+1}),$$
For $0<m<n$ one puts
$$(\dd^m(f))(i_0\xot{\i_1}\cdots\xot{\i_{n+1}}i_{n+1})
=f(i_0\xot{\i_1}\cdots\xot \i_{m-1}\xot{\i_m
\i_{m+1}}i_{\nu+1}\xot{}\cdots
\xot{\i_{n+1}}i_{n+1})$$
and finally for $m=n$ one puts
$$(\dd^n(f))(i_0\xot{\i_1}\cdots\xot{\i_{n+1}}i_{n+1})
=(\i_{n+1})^*f(i_0\xot{\i_1}\cdots\xot{\i_n}i_n)$$
for $n>0$.
\begin{De} Let $D$ be a natural system of abelian groups on a small category $\bI$. The
\emph{cohomology $H^*(\bI;D)$ of $\bI$ with coefficients in} $D$ is
defined as the cohomotopy of the cochain complex $C^*(\bI;D)$.
\end{De}
We will now discuss functorial properties with respect of the first variable. Let $q:{\bf C}\to \bI$ be a functor and $D$ be a natural system of abelian groups on $\bI$. Then we have a natural system $D_q$ on $\bf C$, given by $\alpha\mapsto D_{q(\alpha)}$, where $\alpha$ is a morphism of $\bf C$. In this notation there is a cochain map $q^*:C^*(\bI,D)\to C^*({\bf C},D_q)$ given by
$$q^*(f)(c_0\xot{\alpha_1}\cdots\xot{\alpha_{n+1}}c_{n+1})=f(q(c_0)\xot{q(\alpha_1)}\cdots\xot{q(\alpha_{n+1})}q(c_{n+1}))$$
In particular we have induced map in cohomology
$$q^*:H^n(\bI,D)\to H^n({\bf C},D_q),\ n\geq0.$$
{\bf Of the special interests is the case, when $\bf C$ is a subcategory of $\bI$ and $q$ is the inclusion. In this case insteate $D_q$ we write simply $D$ and the above map in cohomology is called the \emph{restriction homomorphism} and is denoted by $Res:H^*(\bI,D)\to H^*({\bf C},D)$.}
\subsection{Functors and bifunctors as natural systems}. There are functors
$$\bF\bI\xto{q} \bI^{op}\times \bI$$
and $$p_1: \bI^{op}\times \bI\to \bI^{op}, \quad p_2: \bI^{op}\times \bI\to \bI$$
given respectively by
$$q(\alpha:i\to j)=(i,j), \ p_1(i,j)=i, \ p_2(i,j)=j.$$
Thus any bifunctor $B: \bI^{op}\times \bI\to\ab$, resp. any covariant functor $F:\bI\to \ab$ or any contravariant functor $G:\bI^{op}\to \ab$, gives rise to natural systems on $\bI$ given by $D(\alpha:i\to j)=B(i,j)$ respectively $D(\alpha:i\to j)=F(j)$ or $D(\alpha:i\to j)=G(i)$. In what
follows we will consider the functors and bifunctors as natural systems in this way. Thus there are well-defined cochain complexes $C^*(\bI,B)$, $C^*(\bI,F)$, $C^*(\bI,G)$ and cohomologies $H^*(\bI,B)$, $H^*(\bI,F)$, $H^*(\bI,G)$.
The groups $H^*(\bI,F)$ are quite classical and coincide with right derived functors of the limit, studied for example in \cite{laudal}.
\subsection{The case of groupoids} Let $\bG$ be a small groupoid (for example, it can be a group, considered as a one object category). There is a functor
$\kappa:\bG\to \bF\bG$ which is given on objects by $\kappa(g)=(g\xto{\id_g} g)$. The functor $\kappa$ takes a morphism $\eta:g\to h$ of $\bG$ to the morphism $(\eta^{-1},\eta):\id_g\to \id_h$. In other words, one has a commutative diagram
$$
\xymatrix{
g\ar[r]^\eta&h\\
g\ar[u]^{\id}&h\ar[l]^{\eta^{-1}}\ar[u]_\id }
$$
\begin{Le}\label{nsgpd} The functor $\kappa$ is an equivalence of categories.
\end{Le}
\begin{proof} Define the functor $\iota: \bF\bG\to \bG$ on objects by $$\iota(x\xto{\alpha} y)=y=codomain (\alpha)$$
and on morphisms by $\iota(\xi,\eta)=\eta$. Here $(\xi,\eta)$ is a morphism $\alpha\to \beta$ in the category $\bF\bG$, thus one has a commutative diagram
$$
\xymatrix{
y\ar[r]^\eta&v\\
x\ar[u]^\alpha&u\ar[l]^\xi\ar[u]_\beta }
$$
Then obviously $\iota\circ \kappa=\id_{\bG}$. Moreover there is a natural isomorphism of functors $\theta:\kappa\circ \iota\to \id_{\bF\bG}$. Here $\theta(x\xto{\alpha} y)=(\alpha,\id_y)$. This follows from the following commutative diagram
$$
\xymatrix{
y\ar[r]^\id&y\\
y\ar[u]^\id&x\ar[l]^\alpha\ar[u]_\alpha }
$$
\end{proof}
Let $\bG$ be a small groupoid. Then there is an isomorphism of categories $\bG^{op} \to \bG$ which is the identity on objects and takes a morphism to its inverse. Hence the category of contravariant functors from $\bG$ to $\ab$ is isomorphic to the category of covariant functors from $\bG$ to $\ab$ and both categories are equivelent to the category of natural systems on $\bG$ thanks to Lemma \ref{nsgpd}.
Especially nice is the case when $\bG$ is a one object category corresponding to a group $G$. In this case, a covariant functor $\bG\to\ab$ is nothing but a left $G$-module. Any such $A$ gives rise to a natural system $D$ on $\bG$, where $D_x=A$ for all $x\in G$. Moreover, for any $a\in A$, $x,y,z\in G$, the maps
$$y_*:D_{x}=A\to A=D_{xy}, \quad z^*:D_x=A\to A=D_{zx}$$
are given by $y_*(a)=ya$ and $z^*(a)=a$ respectively.
Conversely, if $D$ is a natural system on $\bG$ then one obtains a $G$ module $A$ as follows: $A=D_1$ (here $1$ is the unit of $G$ considered as the identity morphism of $\bG$). The left action of $G$ on $A$ is given by
$$xa=x_*{(x^{-1})}^*a$$
for $a\in A$ and $x\in G$. Thus, the category of natural systems on $\bG$ is equivalent to the category of left $G$-modules. By comparing the Baues-Wirsching cochain complex with the classical complex used in group cohomology, we see that $H^*(\bG,D)=H^*(G,A)$, where $A=D_1$ with the above actions.
Actually this isomorphism can be generalised to groupoids. This follows from the fact that any groupoid is equivalent to a groupoid which is a disjoint union of one object groupoids (i.e. groups considered a one object categories). Denote by $\Lambda$ the set of connected components of $\bG$ and for each $\lambda\in \Lambda$ choose an object $x_\lambda$ in the connected component corresponding to $\lambda$. Denote by $G_\lambda$ the group of automorphisms of $x_\lambda$ then \begin{equation}\label{grcoh}H^*(\bG,D)\cong \prod_\lambda H^*(G_\lambda, A_\lambda)\end{equation}
where $A_\lambda$ is the evaluation of $D$ on the morphism ${\id}_{x_\lambda}:x_\lambda \to x_\lambda$.
\subsection{Track categories}
Recall that a track category (known also as a groupoid enriched category) ${\bf T}$ has objects $A,B,C, \cdots $ and for any two objects $A$ and $B$ a small groupoid ${\bf T}(A,B)$ is given, called the hom-{\it groupoid} of ${\bf T}$.
Moreover, for any triple of objects $A,B,C$ we have a composition functor
$${\bf T}(B,C)\times {\bf T}(A,B) \to {\bf T}(A,C).$$
For any object $A$ the identity $1_A$ is given which is an object of ${\bf T}(A,A)$. These data must satisfy the usual equations of associativity and identity. Objects of the
category ${\bf T}(A,B)$ are called $1$-{\it morphisms} in ${\bf T}$, while morphisms
from ${\bf T}(A,B)$ are called $2$-{\it morphisms} or \emph{tracks}. $1$-morphisms are denoted by $f,g$
etc, while $2$-morphisms are denoted by $\alpha,\bb$ etc. If $f$ is a $1$-morphism
from $A$ to $B$ we write $f:A\to B$, while for a $2$-morphism $\alpha$ from
$f$ to $g$ we write $\alpha:f\then g$. For $1$-morphisms we use multiplicative notation, while for $2$-morphisms we use additive
notation. So we write $gf:A\to C$ for the composite of $1$-morphisms $f:A\to B$ and $g:B\to C$. We write $\bb+\alpha:f\then h$ for
the composite of $2$-morphisms $\alpha:f\then g$ and $\bb:g\then h$.
Similarly, by $0_f$ or simply by $0$ we denote the identity morphism of the object $f$ in the category ${\bf T}(A,B)$.
A $1$-morphism $g:B\to C$ induces the functors
$$g_*:{\bf T} (A,B)\to {\bf T}(A,C), \ \ f\mapsto gf, \ \ \alpha\mapsto g_*\alpha,$$
$$g^*:{\bf T} (C,D)\to {\bf T}(B,D), \ \ h\mapsto hg, \ \ \bb\mapsto g^*\bb.$$
These functors are restrictions of the composition functors. It follows from the definition that the following relations hold:
$$(\alpha +\bb)+\gamma=\alpha +(\bb+\gamma),\leqno {\rm TR \ 1}$$
$$\alpha +0=\alpha=0+\alpha, \leqno {\rm TR \ 2}$$
$$f^*(\alpha +\bb)=f^*(\alpha) +f^*(\bb),\leqno {\rm TR \ 3} $$
$$g_*(\alpha +\bb)=g_*(\alpha) +g_*(\bb),\leqno {\rm TR \ 4}$$
$$f^*(0)=0=g_*(0),\leqno {\rm TR \ 5}$$
$$(ff_1)^*=f_1^*f^*, \ 1^*=1,\leqno {\rm TR \ 6}$$
$$(gg_1)_*=g_*g_{1*}, \ 1_*=1,\leqno {\rm TR \ 7}$$
$$g_*f^*=f^*g_*,\leqno {\rm TR \ 8}$$
$$f_1^*(\beta)+g_*(\alpha )=g_{1*}(\alpha)+f^*(\beta). \leqno {\rm TR \ 9}$$
The following diagram explains the 1-morphisms and $2$-morphisms in TR 9:
$$\xymatrix{A\ar@/^/[r]^{f} \ar@/_/[r]_{f_1}&B\ar@/^/[r]^{g}\ar@/_/[r]_{g_1} & C}, \ \ \alpha:f\then f_1, \ \ \beta:g \then g_1. $$
The equality TR 9 holds in ${\bf T}(gf,g_1f_1)$. The common value in TR 9 is denoted by $\beta * \alpha$ and is called the \emph{Godement product}.
A basic example of a track category is $\bf Cat$, the $2$-category of small categories. The objects of $\bf Cat$ are small categories, $1$-morphisms are functors and $2$-morphisms are natural isomorphisms. It has several interesting $2$-subcategories, for example the track category $\bf Gpd$ of groupoids, functors and their natural transformations.
Let us return to general track categories.
There are several categories associated to a track category ${\bf T}$.
The most important for us is the category ${\bf T} _0$, which has the same objects as ${\bf T}$, and the morphisms in ${\bf T}_0$ are the $1$-morphisms of ${\bf T}$.
Another category which can be associated to ${\bf T}$ is the category ${\bf T}_1$. It has the same objects as ${\bf T}$. The morphisms $A\to B$ in ${\bf T}_1$ are triples $(f,f_1,\alpha)$ where $f,f_1:A\to B$ are $1$-morphisms in ${\bf T}$ and $\alpha:f\then f_1$ is a $2$-morphism in $\bf T$. The composition in ${\bf T}_1$ is defined by $$(g,g_1,\beta)\circ (f,f_1,\alpha)=(gf,g_1f_1,\beta*\alpha).$$
One then has the source and target functions
$${\bf T}_0\buildrel s\over \longleftarrow {\bf T}_1 \buildrel t\over
\longrightarrow {\bf T}_0,$$ where both functors are identity on objects and on morphisms are given by $s(f,f_1,\alpha)=f$ and
$t(f,f_1,\alpha)=f_1$. Sometimes we also write
$${\bf T}_1 \rightrightarrows {\bf T}_0$$ instead of a track category ${\bf T}$.
The \emph{homotopy category} ${\bf T}_\sim$ of a track category $\bf T$ is the category whose objects are the same as for $\bf T$, while morphisms are homotopy classes of $1$-arrows of $\bf T$. Recall that two $1$-arrows $f,g:A\to B$ are \emph{homotopic} if there is a track $\alpha:f\then g$.
A \emph{strict $2$-functor} $F:{\bf T}\to{\bf T}'$ from a track category ${\bf T}$ to a
track category ${\bf T}'$ assigns to each $A\in\ob{\bf T}$ an object
$F(A)\in\ob({\bf T} ')$, to each 1-morphism $f:A\to B$ in ${\bf T}$ -- a 1-morphism
$F(f):F(A)\to F(B)$ in ${\bf T}'$, and to each 2-morphism $\alpha:f\then g$ for
$f,g:A\to B$, a 2-morphism $F(\alpha):F(f)\then F(g)$ in a functorial
way, i.~e. so that one gets functors
$$F_{A,B}:{\bf T}(A,B)\to {\bf T} '\left(F(A),F(B)\right).$$ Moreover
these assignments are compatible with identities and composition, or
equivalently induce a functor ${\bf T}_1\to{{\bf T}'}_1$, that is,
$F(1_A)=1_{F(A)}$ for $A\in\ob({\bf T})$, $F(fg)=F(f)F(g)$,
$F(\alpha*\beta)=F(\alpha)*F(\beta)$ and $F(0)=0$.
We also need a weaker version which is called a \emph{$2$-functor}. We will need it only in the case when the source category is an ordinary category, considered as a track category (where the only $2$-morphisms are $0:f\then f$).
\begin{De} \label{2fun} Let ${\bf T}$ be a $2$-category and let $\bI$ be a category. A $2$-functor $F:\bI\dashrightarrow {\bf T}$ consists of the following data:
\smallskip
-- an object $F(i)$ for each object $i\in\bI$,
\smallskip
-- a $1$-morphism $F(\alpha):F(i)\to F(j)$ for each morphism $\alpha:i\to j$ in $\bI$,
\smallskip
-- a track $F(\alpha,\beta):F(\alpha) F(\beta)\then F(\alpha\beta)$ for all composable arrows $$\xymatrix{i\ar[r]^{\beta}&j\ar[r]^{\alpha}&k}$$ of the category $\bI$.
\smallskip
\noindent One requires that the following conditions hold:
{\rm (i)} $F(\alpha,\id)=0$ and $F(\id,\alpha)=0$,
{\rm (ii)} for all composable arrows
$$\xymatrix{i\ar[r]^{\gamma}&j\ar[r]^{\beta}&k\ar[r]^{\alpha}&l}$$
of the category $\bI$ one has the following equality in ${\bf T}(F(\alpha\beta\gamma), F(\alpha) F(\beta) F(\gamma))$:
\begin{equation}
\label{pseudofunctor}
F(\alpha,\beta\gamma)+F(\alpha)_*f(\beta,\gamma)=F(\alpha\beta,\gamma)+F(\gamma)^*F(\alpha,\beta).
\end{equation}
\end{De}
\subsection{Track categories and natural systems}
Recall that ${\bf T}_0$ denotes the underlying category of a track category ${\bf T}$. For any morphism $f:A\to B$ of ${\bf T}_0$ we let $\Aut(f)$ be the collection of all automorphisms of $f$ in the category ${\bf T}(A,B)$. Thus, this is the collection of all $2$-morphisms $\alpha:f\then f$. It follows from TR 1 and TR 2 that $\Aut(f)$ is a group. Moreover, for any morphism $g:B\to C$ of the category ${\bf T}_0$, we have maps $g_*:\Aut(f)\to \Aut(gf)$ and $f^*:\Aut(g)\to \Aut(gf)$, which are group homomorphisms thanks to TR 3 -- TR 5. Moreover, in this way one obtains a natural system $\Aut$ of groups on ${\bf T}_0$. This follows from the identities TR 6 -- TR 8. The natural system $\Aut$ is sometimes denoted by $\Aut^{\bf T}$ in order to explicitly show the dependence on the track category $\bf T$.
\subsection{Abelian track categories, linear track extensions and the global Toda class} Recall that a groupoid $\bG$ is \emph{abelian} if for any object $x$ of $\bG$ the group of authomorphisms of $x$ is abelian.
A track category ${\bf T}$ is an {\it abelian track category} if $\bf T(A,B)$ is an abelian groupoid for all $A,B\in\ob \bf T$. In this case $\Aut^{\bf T}$ is a natural system of abelian groups.
\begin{De} \cite{BJ} Let $D$ be a natural system of abelian groups on a category ${\bf C}$. A \emph{linear track extension} of $\bf C$ by $D$, denoted by
$$D + \xto{\sigma} {\bf T}_1 \rightrightarrows {\bf T}_0\xto{q} {\bf C},$$
is a track category ${\bf T}$ equipped with a functor $q:{\bf T}_0\to {\bf C}$ such that $q$ induces an isomorphism of categories ${\bf T}_\sim\to {\bf C}$, and an isomorphism $\sigma: D_q\to \Aut^{\bf T}$, where $D_q$ is a natural system of abelian groups on ${\bf T}_0$ given by $f\mapsto D_{q(f)}$. Thus, such a linear extension consists of a collection of isomorphisms of groups $\sigma_f:D_{q(f)}\to Aut(f)$ for each $1$-arrow $f:A\to B$ of ${\bf T}$, which have the following properties.
\begin{enumerate}[(i)]
\item The functor $q$ is full and the identity on objects. In addition, for $f,g:A\to B$ in ${\bf T}_0$ we have $q(f)=q(g)$ iff there exists a track $f \then g$. In other words, the functor $q$ identifies $\bf C$ with ${\bf T}_\sim$.
\item For $\alpha:f\then g$ and $\xi:D_{q(f)}=D_{q(g)}$ we have $\sigma_f(\xi)=-\alpha +\sigma_g(\xi)+\alpha$.
\item For any 1-arrows $A\xto{f} B\xto{g} C\xto{h} D$ in ${\bf T}_0$ and any $\xi\in D_{q(g)}$ one has
$$h_*\sigma_g(\xi)=\sigma_{hg}(q(h)_*\xi), \quad f^*\sigma_g(\xi)=\sigma_{gf}( q(f)^*\xi).$$
\end{enumerate}
\end{De}
Given a $2$-functor $F: {\bf B}\to {\bf I}$, we can compose with $q$ to get a functor ${\bar F}: {\bf I}\to {\bf C}$. This is called a \emph{collective character}, following the classic terminology by Teichm\"uller \cite{teich}. Conversely, we can ask whether for a given collective character ${\bar F}: {\bf I}\to {\bf C}$ there exists a lifting as a 2-functor $F:{\bf I}\to {\bf T}$ and if it exists, how to classify all such liftings.
As we will see, the answer depends on an obstruction class in third cohomology. This construction is analogous to the one by Teichm\"uller for a similar situation with linear algebras, and to that by Eilenberg-MacLane, for the study of obstructions to group extensions with non-abelian kernels.
To state the corresponding results, we will fix some notation. If $F: {\bf I}\to {\bf T}$ is a 2-functor and ${\bar F}: {\bf I}\to {\bf C}$ is the corresponding collective character, then for any object $i$ of ${\bf I}$ we have ${\bar F}(i)=F(i)$. Thus if a given collective character ${\bar F}: {\bf I}\to {\bf C}$ has a lifting $F$, then $F(i)$ must be ${\bar F}(i)$. Thus we are searching to find a pair $(F_1,F_2)$ of functions known as a factor set or $2$-cocycle. The function $F_1$ assigns a $1$-morphism $F_1(\alpha)$ to each morphism $\alpha:i\to j$ in $\bf I$ with the property $qF_1(\alpha)={\bar F}(\alpha)$, and a track $F_2(\alpha,\beta):F(\alpha)F(\beta)\then F(\alpha\beta)$ for all composable arrows $i\xto{\beta}j\xto{\alpha} k$ such that the conditions (i) and (ii) of Definition \ref{2fun} hold.
Two such $2$-cocycles $(F_2,F_1)$ and $(F_2',F_1')$ are equivalent if there exists a function $\varphi$, which assigns to any arrow $\alpha:i\to j$ a track $\varphi(\alpha):F_1(\alpha)\then F_1'(\alpha)$ such that for all composable arrows
$$\xymatrix{i\ar[r]^{\beta}&j\ar[r]^{\alpha}&k}$$ of the category $\bI$ one has the following commutative diagram of tracks
$$\xymatrix{F_1(\alpha)\circ F_1(\beta)\ar[rr]^{F_2(\alpha,\beta)}\ar[dr]_{F_1(\alpha)_*\varphi(\beta)}&&F_1(\alpha\beta)\ar[r]^{\varphi(\alpha\beta)}&F_1'(\alpha\beta)\\
&F_1(\alpha)\circ F_1'(\beta)\ar[rr]_{F_1'(\beta)^*\varphi(\alpha)}&&F_1'(\alpha)\circ F_1'(\beta)\ar[u]^{F_2'(\alpha,\beta)}
}.$$
In other words, one has the following equality in ${\bf T}(F_1(\alpha)\circ F_1(\beta), F_1'(\alpha)\circ F_1'(\beta))$
$$F_2'(\alpha,\beta) = \varphi(\alpha\beta)+F_2(\alpha,\beta)-F(\alpha)_*\varphi(\beta)-F_1'(\beta)^*\varphi(\alpha).$$
The set of all such factor sets up to equivalence with fixed ${\bar F}$ is denoted by $H^2({\bf I}, {\bf T})$.
Let $D + \xto{\sigma} {\bf T}_1 \rightrightarrows {\bf T}_0\xto{q} {\bf C}$ be a linear track extension of a category $\bf C$ by a natural system $D$. Then \cite{BJ} the \emph{universal Toda bracket} $\grupo{\bf T}$ is the element of $H^3({\bf C},D)$ represented by the following cocycle: choose for each morphism $ f$ of ${\bf T}_\sim$ a representative 1-arrow $s(f)$, that is $q(s(f))=f$. Furthermore, choose a track $\varphi(f,g):s(f)s(g)\then s(fg)$ for any composable pair $A\xto{g}B\xto{f} C$ of the category $\bf C$. Such pair $(s, \varphi)$ is called a \emph{section} of $q$.
It follows that $h^*\varphi_{f,g}+\varphi_{fg,h}$ and $f_*\varphi_{g,h}+\varphi_{f,gh}$ both define tracks $s(fgh)\then s(f)s(g)s(h)$ and hence
\begin{equation}\label{deftd}
f_*\varphi(g,h)+ \varphi(f,gh)-\varphi(fg,h)-h^*\varphi(f,g)
\end{equation}
determines an element in $\Aut(s(f)s(g)s(h))$. Going back via $\sigma$, this determines an element $c(f,g,h)\in D_{fgh}$. Varying $f,g,h$ one obtains a 3-dimensional cocycle, whose class in cohomology is $\grupo{{\bf T}}\in H^3({\bf C},D)$.
\begin{Le}\label{1.5.m}
\label{cohomology}
\begin{enumerate}[(i)]
\item Let $q:{\bf T}\to {\bf C}$ be the canonical strict $2$-functor, which assigns the identity to all tracks (here and elsewhere, categories are considered as track categories with trivial tracks). Then $q$ has a section as a 2-functor iff $\grupo{\bf T}=0$.
\item More generally, given a category ${\bf I}$ and a functor $\bar F:{\bf I}\to {\bf C}$, we can lift the functor $\bar F$ to ${\bf T}$ as a $2$-functor if and only if the pull-back $Obs_{\bar F} = {\bar F}^*(\grupo{{\bf I}})\in H^3({\bf I}, D_{\bar F} )$ is zero.
\item Suppose we are given a collective character $\bar F:{\bf I}\to {\bf C}$ which has a lift as a $2$-functor $F: {\bf I}\to {\bf T}$, so the obstruction $Obs_{\bar F}$ is zero. Then the set of all such liftings of $\bar F$ up to equivalence is in one-to-one correspondence with elements in $H^2({\bf I}, D_{\bar F})$. In fact, the set of such linear track extensions up to equivalence is a torsor over $H^2({\bf I}, D_{\bar F})$.
\end{enumerate}
\end{Le}
\begin{proof}
\begin{enumerate}[(i)]
\item Let $(s,\varphi)$ be a section of $q$. For it to be a $2$-functor, the equation (\ref{pseudofunctor}) has to hold. This is equivalent to the vanishing of the $3$-cocycle $c$ defined in the equation (\ref{deftd}). Conversely, let $c$ defined in the equation (\ref{deftd}) be a coboundary, i.e. $c(f,g,h)=f_*\rho(g,h)+ \rho(f,gh)-\rho(fg,h)-h^*\rho(f,g)$ for a function $\rho$ which assigns to each composable pair of morphisms $A\xto{g}B\xto{f} C$ an element in $D_{fg}$. We have a corresponding track $\bar \rho (f,g): = \sigma_{s(f)s(g)}(\rho(f,g))\in \Aut(s(f)s(g))$.
We define $\tau = \varphi - \bar\rho.$ It is clear that $\tau(f,g)$ is a track $s(f)s(g)s(h)\then s(fgh)$. So, the pair $(s,\tau)$ is also a section, for which the corresponding $3$-cocycle $c_\tau(f,g,h)$ vanishes and hence $(s,\tau)$ defines a section of $q$ which is a $2$-functor.
\item This is a formal consequence of part (i) and the pull-back construction.
\item
We describe an action
$$H^2({\bf I}, D_{\bar F})\times H^2({\bf I}, {\bf T})\to H^2({\bf I}, {\bf T}).$$
Let $h$ be a normalised $2$-cocycle representing the class of $[h]\in H^2({\bf I}, D_{\bar F})$, and let $(F_2,F_1)$ be a factor set representing the class $[F_2,F_1]\in H^2({\bf I}, {\bf T})$. Then the result of the action is the class represented by the factor set $(F_2',F_1)$, where
$$F_2'(\alpha,\beta)= \sigma_{F_1(\alpha\beta)}(h(\alpha,\beta)) +F(\alpha,\beta).$$
To see that this is well-defined, and the action is transitive follows closely to the proof of Theorem $9$ in \cite{cegarra}.
\end{enumerate}
\end{proof}
The most important example of linear track extensions arises from abelian track categories \cite{BJ}. Namely, if $\bf T$ is an abelian track category, then obviously $\Aut^{\bf T}$ is a natural system of abelian groups on ${\bf T}_0$. It was proved by Baues and Jiblazde that there exists a unique (up to isomorphism) natural system of abelian groups $D^{\bf T}$ on ${\bf T}_\sim$ and an isomorphism of natural systems $\sigma:D^{\bf T}_q\to \Aut^{\bf T}$. Hence any such track category defines an element $\grupo{{\bf T}}\in H^3({\bf T}_\sim, D^{\bf T})$.
\section{Applications to Crossed modules, the Eilenberg-MacLane obstruction class and the Teichm\"uller class}
\subsection{Crossed modules and track categories}
Recall that a crossed module is a group homomorphism $\delta : T\to R$ together with an action of $R$ on $T$ satisfying:
$$\delta (^rt)=r\delta (t)r^{-1} \ {\rm and} \
^{\delta t}s=tst^{-1}, \ r\in R, t,s\in T.$$
It follows from the definition that the image $Im (\delta)$ is a normal subgroup of $R$, and the kernel $Ker (\delta)$ is in the centre of $T$. Moreover, the action of $R$ on $T$ induces an
action of $G$ on $M=Ker (\delta)$, where $G=Coker(\delta)$. So we have an exact sequence
$$0\to M\to T\xto{\delta} R\xto{p} G\to 0,$$
called a \emph{crossed extension} of a group $G$ by a $G$-module $M$. Such an extension defines an element in $H^3(G,M)$, see for example \cite[Ch. IV. Section 5] {kenn}. We recall the construction of this class.
Choose a pair of maps $(s:G\to R$, $\ss:G\times G\to T)$ for which the following hold:
$$ps(x)=x, \quad s(x)s(y)=\delta(\ss(x,y))s(xy), \ x,y\in G.$$
Then the map
\begin{equation}\label{3cycle}
f(x,y,z)=\,^x \sigma(y,z)\ss(x,yz)\ss(xy,z)^{-1}\ss(x,y)^{-1}\end{equation}
defines a $3$-cocycle $f\in Z^3(G,M)$. The class of this cocycle in $H^3(G,M)$ is denoted by $\grupo{\delta}$.
We will show that this classical construction is a special case of the global Toda bracket construction. To this end, for a group $L$ we let $\underline{L}$ denote the one object category whose morphisms are the elements of $L$. Denote by ${\bf T}^\delta$ the following track category. It has only one object, 1-arrows are elements of $R$ and the composition is induced by the product rule in $R$. A track from $r$ to $r'$ in ${\bf T}^{\delta}$ is an element $t\in T$ such that $r'=\delta(t)r$ (observe that now we use multiplicative notation for tracks, which should not cause any complications) and we write $t:r\then r'$. Moreover, for a track $t:r\then r'$ and $x,y\in R$, considered as 1-arrows, we set $$x_*(t)=\, ^xt:xr\then xr' \quad {\rm and} \quad y^*(t)=t:ry\then r'y.$$
One easily sees that we indeed obtain a track category, with
$$({\bf T}^\delta)_0=\underline{R}, \quad ({\bf T}^\delta)_\sim=\underline{G}.$$
Furthermore, the natural system $D^{\bf T}$ is nothing but the $G$-module $M$. This easily follows from the fact that $\Aut(r)=M$ for all $r\in R$.
\begin{Le}
\label{Toda}
One has the equality
$$\grupo{{\bf T}^{\delta}}=\grupo{\delta}.$$
\end{Le}
\begin{proof} By the isomorphism (\ref{grcoh}) the cohomology groups in questions are the same. The rest follows by comparing the equations (\ref{deftd}) and (\ref{3cycle}) and the fact that $y^*(t)=t$ for all $y\in R$, $t\in T$.
\end{proof}
\subsection{The track category related to the category of groups and the Eilenberg-MacLane obstruction class}
Recall that to given groups $\Pi$, $G$ and a group homomorphism $\eta:\Pi\to \Out(G)$, Eilenberg and MacLane associated an obstruction class ${\mathcal Obs}_{\eta}\in H^3(\Pi,{\sf Z}(G))$, where ${\sf Z}(G)$ denotes the centre of $G$. In this section we construct a class, which is independent of the groups $\Pi$ and $G$, and which restricts to all these classes, see Proposition \ref{u_cl} below. Before we state this result explicitly, let us recall some basic facts related to the class ${\mathcal Obs}_{\eta}$, following \cite{homology}.
If
$$E: \quad \quad 1\to G\xrightarrow{\varkappa} B \xrightarrow{\sigma} \Pi \to 1$$
is a group extension, the epimorphism $\sigma$ induces a homomorphism $\eta: \Pi \to \Out(G)$, where $\Out(G)$ is the group of outer automorphisms of $G$. This happens in the following way: We have an action of $B$ on $G$ via conjugation, which gives us the homomorphism $\theta: B\to \Aut(G)$. Because $\theta(\varkappa (G))\subset {\sf In}(G)$, where ${\sf In}(G)$ is the group of inner automorphisms, we have the induced homomorphism $\eta: \Pi \to \Aut(G) / {\sf In}(G) = \Out(G)$.
We say that the extension $E$ has \emph{conjugation class $\eta$}: thus $\eta$ records in which way $G$ appears as a normal subgroup of $B$. Conversely, call a pair of groups $\Pi$, $G$ together with a homomorphism $\eta: \Pi\to \Out(G)$ an \emph{abstract kernel}.
The general problem of group extensions is constructing all extensions $E$ to a given abstract kernel $(\Pi, G, \eta)$. Given such an abstract kernel however, there might not necessarily exist an extension of $\Pi$ by $G$ that induces $\eta$.
The obstruction to the existence of an extension $$1\to G\to B \to \Pi \to 1$$ that induces $\eta$ is given by a certain class ${\mathcal Obs}_\eta$ in the third cohomology group $H^3(\Pi, {\sf Z}(G))$, where ${\sf Z}(G)$ is the centre of $G$.
To each group $G$ we associate the canonical crossed module $\mu: G\to \Aut(G)$. The kernel and cokernel of $\mu$ make up the crossed extension
$$0\to {\sf Z}(G)\to G\to \Aut(G)\to \Out(G)\to 1.$$
Here, ${\sf Z}(G)$ is the kernel of $\mu$, while $\Out(G)$ is the cokernel. As outlined in the previous section, this crossed extension leads to an element $Cl_G$ in the third cohomology group $H^3(\Out(G), {\sf Z}(G))$. Now we use the pull-back construction of the homomorpism $\eta$ to create another crossed extension of $\Pi$ via ${\sf Z}(G)$:
$$
\xymatrix{
0\ar[r]&{\sf Z}(G)\ar[r]\ar[d]_{id}&G\ar[r]\ar[d]^{id}&X\ar[r]\ar[d]&\Pi\ar[d]^\eta\ar[r]&1\\
0\ar[r]&{\sf Z}(G)\ar[r]&G\ar[r]&\Aut(G)\ar[r]&\Out(G)\ar[r]&1.}
$$
The class of this crossed extension in $H^3(\Pi,{\sf Z}(G))$ is denoted by ${\mathcal Obs}_{\eta}$. Thus we have the basic formula
\begin{equation}\label{obs=cl} {\mathcal Obs}_{\eta}=\eta^*(Cl_G),\end{equation}
where $\eta^*$ is the induced homomorphism $H^3(\Out(G),{\sf Z}(G))\to H^3(\Pi,{\sf Z}(G))$.
Now we are in the position to formulate the following classical result of Eilenberg and MacLane, see \cite[Ch.IV. Section 6]{kenn} or \cite[Ch. IV] {homology}.
\begin{Th}(Eilenberg-MacLane)
The abstract kernel $(\Pi,G,\eta)$ has an extension if and only if ${\mathcal Obs}_{\eta}=0$.
\end{Th}
Next, we construct a universal class $\grupo{\mathcal G}$, such that all obstruction classes ${\mathcal Obs}_{\eta}$ are restrictions of $\grupo{\mathcal G}$. To this end, let us consider the following track category $\mathcal G$. The objects of $\mathcal G$ are groups, the $1$-morphisms are surjective homomorphisms, and the set of $2$-morphisms between two $1$-morphisms $f,g:G\to H$ is given by $\{b\in H | f(a) = b+g(a)-b\}\subseteq H$.
Let us denote the underlying category by ${\mathcal G}_0$, and the homotopy category by ${\mathcal G}_\sim$. Thus ${\mathcal G}_0$ is the category whose objects are groups and whose morphisms are surjective group homomorphisms, while the category ${\mathcal G}_\sim$ has as objects groups, while the morphisms are conjugacy classes of surjective homomorphisms. By our description of 2-morphisms, for each morphism $f:G\to H$ the group $\Aut(f)$ is the group $${\sf Z}_f = \{b\in H | f(a)+b=b+f(a) \}.$$ Since $f$ is surjective, we see that ${\sf Z}_f$ is the centre of $H$ and hence is abelian. Therefore, we have an abelian track extension
$$0\to D^{\mathcal G}\to {\mathcal G}_1 \rightrightarrows {\mathcal G}_0\to {\mathcal G}_\sim\to 1,$$
with the natural system $D$ being given by abelian groups
$$D^{\mathcal G}_{f:G\to H}={\sf Z}(H).$$
It follows that $D^{\mathcal G}$ in our case is a natural system induced by the functor $${\mathcal G}_\sim\to \ab, \quad H\mapsto {\sf Z}(H).$$
Thus we obtain the class $\grupo{{\mathcal G}}\in H^3({\mathcal G}_\sim, D^{\mathcal G})$, where the last group is the cohomology of the category ${\mathcal G}_\sim$ with coefficients in a functor $H\mapsto {\sf Z}(H)$.
{\bf Remark}. Since the track category $\mathcal G$ is not small, we have to be more careful in this place. To avoid set theoretical problems, we have to fix two universes ${\mathfrak U}_1\subset {\mathfrak U}_2$. The elements of ${\mathfrak U}_1$ are called sets, while elements of ${\mathfrak U}_2$ are called classes. The objects of our categories are classes, while
morphisms between two fixed objects form a set. In this framework the definition of a cochain complex of a category still makes sense, but as a cochain complex of classes, and hence, cohomologies are not sets in general, but classes. Regarding our primary interest $H^*({\mathcal G}_\sim, D^{\mathcal G})$, we conjecture that nevertheless they are sets. The same reasoning is applied to other non-small categories considered below.
Let $\eta:G\to \Out(\Pi)$ be an abstract kernel. Then $\eta$ can be considered as a functor $\underline{\eta}:\underline{G}\to {\mathcal G}_\sim$, which sends the unique object of $\underline{G}$ to $\Pi\in {\mathcal G}_\sim$.
\begin{Pro}\label{u_cl}
For every abstract kernel $\eta:G\to \Pi$, we have the induced homomorphism
$$\underline{\eta}^*:H^3({\mathcal G}_\sim, D^{\mathcal G})\to H^3(\Out(G),{\sf Z}(G))$$
and the following equality
\begin{equation}\label{obs=eta} {\mathcal Obs}_{\eta}=\underline{\eta}^*(
\grupo{\mathcal G}).\end{equation}
Moreover, if we identify the one object category $\underline{\Out(\Pi)}$ with a subcategory of ${\mathcal G}_\sim$ which has a single object $\Pi$ and morphisms are isomorphisms in ${\mathcal G}_\sim$, then we have
\begin{equation}\label{cl=res} Cl_\Pi=Res (\grupo{{\mathcal G}})). \end{equation}
\end{Pro}
\begin{proof} Thanks to the equality (\ref{obs=cl}), the equality (\ref{cl=res}) implies (\ref{obs=eta}). To show the equality (\ref{cl=res}), we consider the crossed module $\delta:\Pi\to \Aut(\Pi)$, where $\delta$ sends element of $\Pi$ to the inner automorphisms of $\Pi$. The rest follows from the fact that the track category ${\bf T}^\delta$ (see the previous section) is isomorphic to the one object track subcategory of $\mathcal G$ with object the group $\Pi$, where $1$-morphisms are isomorphism of $\Pi$, while 2-morphisms are the same as in ${\mathcal G}$.
\end{proof}
\subsection{The track category related to the category of rings and the Teichm\"uller class}
Let $A$ be a ring. Denote by $\Aut(A)$ the group of ring automorphisms of $A$. Denote by ${\sf U}(A)$ the group of invertible elements of $A$. There is an obvious homomorphism of groups $\partial:{\sf U}(A)\to \Aut(A)$, which sends an invertible element $a$ to the inner automorphism $\partial_a:A\to A$, where $\partial_a(x)=axa^{-1}$. Then ${\sf U}(A)\xto{\partial} \Aut(A)$ is a crossed module and
$$e_A:\quad 0\to {\sf U}({\sf Z}(A))\to {\sf U}(A)\xto{\partial} \Aut(A)\to \Out(A)\to 1$$
is a crossed extension, where ${\sf Z}(A)$ is the centre of $A$ and $\Out(A)={\sf Coker}(\partial)$.
This crossed module and the corresponding element $h_{A}$ in $H^3(\Out(A),{\sf U}({\sf Z}(A)))$ play an important role in the work of Huebschmann \cite{hueb} on the Teichm\"uller class in the Galois theory of rings.
If $A$ is varied, one obtains different classes and the relationship between these classes in unclear. We will now prove that these classes are in fact a restriction of a unique class, which is independent of the chosen ring $A$.
Let us consider the following track category $\mathcal R$. The objects of $\mathcal R$ are rings, the $1$-morphisms are surjective homomorphisms of rings, and the set of $2$-morphisms between two $1$-morphisms $f,g:A\to B$ is given by
$$\{b\in {\sf U}(B) | f(a) = bg(a)b^{-1}\}\subseteq B.$$
Therefore, for each morphism $f:A\to B$, $\Aut(f)$ is the abelian group ${\sf Z}_f = \{b\in {\sf U}(B) | f(a)b=bf(a) \}$, which is the centre of ${\sf U}(B)$. Let us denote the underlying category by ${\mathcal R}_0$, and the homotopy category by ${\mathcal R}_\sim$. Hence we have an abelian track extension
$$0\to D^{\mathcal R}\to {\mathcal R}_1 \rightrightarrows {\mathcal R}_0\to {\mathcal R}_\sim\to 1,$$
where the natural system $D^{\mathcal R}$ assigns the abelian group ${\sf Z}(U(B))$ to a morphism $A\to B$ in ${\mathcal R}_\sim.$ In fact, this natural system is induced by the functor
$${\mathcal R}_\sim\to\ab, \quad B\mapsto {\sf Z}(B).$$
In this way one obtains the element in $H^3({\mathcal R}_\sim, D^{\mathcal R})$, which is denoted by $\grupo{\mathcal R}$ and is called the \emph{global Teichm\"uller class}.
For a fixed ring $A$ we can consider a track subcategory of $\mathcal R$, which has just one single object $A$ and the 1-morphisms of which are automorphisms. The tracks $f\Longrightarrow g$ are the same as in $\mathcal R$. Then this track subcategory is the track category corresponding to the crossed extension $e_A$. Hence we obtain
$$h_A=Res(\grupo{\mathcal R}).$$
\section{Applications to graded extensions of (monoidal) categories}
\subsection{Recollection on Grothendieck cofibrations}
Let $P:{\mathcal E}\to {\mathcal B}$ be a functor and $B$ be an object of ${\mathcal B}$. The \emph{fibre} ${\mathcal E}_B$ of $P$ over $B$ is the subcategory of ${\mathcal E}$ consisting of all morphisms $f$ such that $P(f)=id_B$. In particular, the objects of ${\mathcal E}_B$ are such objects $E$ of ${\mathcal E}$ that $P(E)=B$. Denote by $i_B:{\mathcal E}_B \hookrightarrow {\mathcal E}$ the inclusion functor. We say that a morphism $f:E\to F$ in the category ${\mathcal E}$ is \emph{over} $a:A\to B$ if $P(f)=a$. This of course implies that $E\in {\mathcal E}_A$ and $F\in {\mathcal E}_B$.
A morphism $f:E\to F$ in the category ${\mathcal E}$ is called \emph{cocartesian} over $a=P(f):A\to B$, if each morphism $g:E\to G$ in $\mathcal E$ and each decomposition of $b=P(g)=c\circ a$ in $\mathcal B$ uniquely determines a morphism $h:F\to G$ in $\mathcal E$ over $c$ with $g=h\circ f$:
$$\xymatrix{ &&&&&&&\\
{\mathcal E}\ar[dddd]^P && &&&& G\\
&&&& E\ar[r]_f\ar[rru]^g&F\ar[ru]_h&\\
&&&&&&&\\
& && &&& C\\
{\mathcal B} &&&& A\ar[r]_a \ar[urr]^b &B\ar[ur]_c\\
&&&&&&&\\
}
$$
The functor $P:{\mathcal E}\to {\mathcal B}$ is a \emph{cofibration} if above each morphism $A=P(E)\to B$ in $\mathcal B$ there is a cocartesian morphsim $E\to F$.
A \emph{cleavage} in a cofibration $f:E\to F$ is a choice, for each object $E$ of $\mathcal E$ and morphism $a:A=P(E)\to B$ in $\mathcal B$, of a cocartesian morphism $a_*:E\to F$ above $a$. If $P$ is equipped with a cleavage, it is said to be \emph{cloven}. The cleavage defines the functor $a_*:{\mathcal E}_A\to {\mathcal E}_B$, which sends an object $E$ over $A$ to the codomain of the cocartesian morphism $a_*$. In fact,
$$A\mapsto {\mathcal E}_A,\ a\mapsto a_*$$ yields a $2$-functor ${\mathcal B}\to {\mathcal Cat}$. Conversely, having a $2$-functor ${\mathcal B}\xto{\Psi} {\mathcal Cat}$, one can construct the category ${\mathcal E}={\mathcal B}\int \Phi$, known as the Grothendieck construction. The objects of ${\mathcal B}\int \Phi$ are pairs $(A,x)$, where $A$ is an object of $\mathcal B$ and $x$ is an object of $\Psi(A)$. A morphism $(A,x)\to (B,y)$ in ${\mathcal B}\int \Phi$ is a pair $(a,\alpha)$, where $a:A\to B$ is a morphism in $\mathcal B$, while $\alpha:\Phi(a)(x)\to y$ is a morphism in $\Psi(B)$. The composite of morphisms $(A,x)\xto{(a,\alpha)} (B,y)$ and $(B,y)\xto{(b,\beta)}$ is $(A,x)\xto{(c,\gamma)} (C,z)$, where $c=ba$, while $\gamma$ is the composite
$$c_*(x)=\Phi(ba)(x)\to \Phi(b)(\Phi(a)(x))\xto{\Phi(\alpha)(x)} \Phi(b)(y)\xto{\beta} z.$$
Consider the functor $P:{\mathcal B}\int \Phi\to {\mathcal B}$ , where
$$P(A,x)=A \quad {\rm and} \quad P(a,\alpha)=a.$$
One easily checks that the morphisms of the form $(a,id)$ are cocartesian and hence $P$ is a cofibration. Grothendieck proved that in this way one obtains a one-to -one correspondence between cofibarions over ${\mathcal B}$ (up to equivalence) and $2$-functors from ${\mathcal B}\to {\mathcal Cat}$.
\begin{Le} If ${\mathcal E}$ and ${\mathcal B}$ are groupoids and $P$ is full, then
$P$ is a cofibration.
\end{Le}
\begin{proof} The result follows from the fact that any isomorphism is cocartesian.
\end{proof}
\subsection{The class of Cegarra-Garz\'on-Grandjean}
In \cite{cegarra}, the authors define an extension of a category $\mathcal{C}$ by a group $G$. In the case when $\mathcal C$ is the category $\underline{\Pi}$ associated to a group $\Pi$ one recovers the classical theory of group extensions.
A stable $G$-grading on a category $\mathcal{D}$ is a functor $g:\mathcal{D}\to \underline{G}$ such that for every object $A\in \mathcal{D}$ and $x\in G$, there is an isomorphism $\kappa$ in $\mathcal{D}$ with source $A$ and such that $g(\kappa) = x$. We refer to $g(\kappa)$ as the grade of $\kappa$. Then we can define the category $Ker(\mathcal{D})$ as the subcategory consisting of all morphisms of grade $1$.
A \emph{$G$-graded extension} \cite{cegarra} of a category $\mathcal{C}$ is a stably $G$-graded category whose kernel is isomorphic to $\mathcal{C}$.
In this setting, by a factor set, or $2$-cocycle on $G$ with coefficients in $\mathcal{C}$, we shall mean a $2$-functor $\underline{G} \to \emph{Cat}$, the category of small categories, that associates the category $\mathcal{C}$ to the unique object of $\underline{G}$. This is a special case of the Grothendieck construction where $\underline{G}$ is the one-object category associated to the group $G$, rather than a general (small) category.
The centre of $\mathcal{C}$, ${\sf Z}(\mathcal{C})$, is defined as the set of all natural transformations $u:id_{\mathcal{C}}\to id_{\mathcal{C}}$, where $id_{\mathcal{C}}$ is the identity functor, and ${\sf Z}(\mathcal{C})^*$ denotes the abelian group of the units in ${\sf Z}(\mathcal{C})$, that is, the abelian group of all natural isomorphisms of $id_\mathcal{C}$ with itself. The group of outer autoequivalences of $\mathcal{C}$, $\Out(\mathcal{C})$, is the set of isomorphism classes of autoequivalences of $\mathcal{C}$ with the multiplication induced by the composition of autoequivalences.
In this setting, we also have a theory of obstructions, and a collective character $\Phi:G\to \Out(\mathcal{C})$ for such an extension plays a similar role to an abstract kernel. To it, we can associate an element $k(\Phi)\in H^3_\Phi(G, {\sf Z}(\mathcal{C})^*)$, called the \emph{Teichm{\"u}ller class}, which is constructed analogously to a classic construction by Teichm{\"u}ller for a similar situation with linear algebras, and to that by Eilenberg--Mac Lane, for the study of obstructions to group extensions with non-abelian kernels considered above.
\begin{Th}
\cite{cegarra} A collective character $\Phi:G\to \Out(\mathcal{C})$ is realizable if and only if its
Teichm{\"u}ller obstruction class $k(\Phi)\in H^3_\Phi(G, {\sf Z}(\mathcal{C})^*)$ vanishes.
\end{Th}
Let us consider the track category ${\mathcal Cat}$, the objects of which are small categories, and the $1$- and $2$-morphisms are functors and natural transformations, respectively.
\begin{Le}\label{niso_abe}
If we restrict the $1$-morphisms in the track category ${\mathcal Cat}$ to equivalences and the $2$-morphisms to natural isomorphisms, we will have defined an abelian track category.
\end{Le}
\begin{proof}
We need to show that for each equivalence $f:C\to D$ in ${\mathcal Cat}$, the group $\Aut(f)$ is abelian. In order to do so, we show that for any two automorphisms $\alpha$, $\beta$ of $f$, there exists a unique isomorphism $u\in \Aut(Id_D)$ such that $\beta = f^*u+\alpha=\alpha+f^*u$. In fact, $\alpha+\beta=\alpha+(f^*u+\alpha)=(\alpha+f^*u)+\alpha=\beta+\alpha.$
For any object $Y\in D$, since $f$ is essentially surjective, we can select an object $X\in C$ and an isomorphism of $D$, $\eta_{X,Y}: f(X)\to Y$. We then define $u_Y:Y\then Y$, $Y\in D$, by $u_Y = \eta_{X,Y} -\alpha_X + \beta_X -\eta_{X,Y}$. First, we observe that this morphism does not depend on the choice of $\eta$. Indeed, for another family of isomorphisms $\mu_{X',Y}: f(X')\to Y$, since $f$ is fully faithful, there will exist a unique isomorphism $\phi_{X,X'}:X\to X'$ in $C$, such that $f(\phi_{X,X'}) = -\mu_{X',Y}+\eta_{X,Y}$. Hence, by the naturality of $\alpha$ and $\beta$, we have that $u_Y = \eta_{X,Y} -\alpha_X + \beta_X -\eta_{X,Y} = \mu_{X',Y} +f(\phi_{X,X'}) -\alpha_X+\beta_X -f(\phi_{X,X'}) - \mu_{X',Y} = \mu_{X',Y} -\alpha_{X'}+\beta_{X'} - \mu_{X',Y}$.
Next, we check that $u$ is a natural transformation. Let $\psi:Y\to Y'$ be a morphism in $D$. Since $f$ is an equivalence, there exists a morphism $\phi:X\to X'$ in $C$ such that $f(\phi_{X,X'})=-\eta_{X',Y'}+\psi_{Y,Y'}+\eta_{X,Y}$. Then the naturality of $\alpha$ and $\beta$ implies that
\begin{align*}
\psi_{Y,Y'}+u_Y+\eta_{X,Y}&=\psi_{Y,Y'}+\eta_{X,Y} -\alpha_X + \beta_X= \eta_{X',Y'}+f(\phi_{X,X'})-\alpha_X + \beta_X \\
&=\eta_{X',Y'}-\alpha_X' + \beta_X'+f(\phi_{X,X'})=u_{Y'}+\psi_{Y,Y'}+\eta_{X,Y},
\end{align*}
and so $\psi_{Y,Y'}+u_Y=u_{Y'}+\psi_{Y,Y'}$.
For all $X\in C$, we have $u_{f(X)}=-\alpha_X + \beta_X$, implying that $\beta = \alpha+f^*u$. The equation TR 9 provides the second part of the equation.
As for the uniqueness of $u$, let us suppose that $\alpha+f^*u= \alpha+f^*v$ for $u,v\in \Aut(Id_D)$. Because $\alpha$ is an isomorphism, this implies that $f^*u=f^*v$, and so $u=v$ because $f$ is an equivalence.
\end{proof}
The abelian track category $\mathcal Cat$ defined above defines a class $\grupo{{\mathcal Cat}}$ in the third cohomology $H^3({\mathcal Cat}_\sim, D^{{\mathcal Cat}})$.
\begin{Le}
For every small category ${\mathcal Cat}$, we have the restriction
$$r:H^3({\mathcal Cat}_\sim, D^{{\mathcal Cat}})\to H^3(\Out({\mathcal Cat}),{\sf Z}({\mathcal Cat})^*),$$
with $k(\Phi) = \Phi^*(r(\grupo{{\mathcal Cat}}))$.
\end{Le}
\subsection{The class of Cegarra-Garz\'on-Ortega}
We would like to apply the Toda class to monoidal categories. To obtain an appropriate abelian track category we consider the following $2$-category ${\mathcal Mcat}$. Objects of ${\mathcal Mcat}$ are small monoidal categories, $1$-morphisms are monoidal equivalences and $2$-morphisms are natural isomorphisms. It follows from Lemma \ref{niso_abe} that ${\mathcal Mcat}$ is an abelian track category. The corresponding homotopy category ${\mathcal Mcat}_\sim$ is a groupoid. As any abelian track category, it defines a class
$$\grupo{{\mathcal Mcat}}\in H^3({\mathcal Mcat}_\sim, D^{\mathcal Mcat}).$$
Here $D^{\mathcal Mcat}$ is a natural system such that for any monoidal equivalence $f:({\mathcal C},\otimes)\to {\mathcal C'},\otimes')$ the abelian group $D^{\mathcal Mcat}_{q(f)}$ is isomorphic to the following group
$${\sf Z}_f=\{\alpha:f\then f| \ \alpha \ {\rm is \ a} \ {\rm monoidal} \ {\rm isomorphism}\}.$$
Here as usual $q$ denotes the natural functor ${\mathcal Mcat}_0 \to {\mathcal Mcat}_\sim$.
For $f=\id_{\mathcal C}$, the group ${\sf Z}_{\id_{\mathcal C}}$ is known as the \emph{centre} of the monoidal category $({\mathcal C},\otimes)$, and is denoted by ${\sf Z}({\mathcal C},\otimes)^*$, see \cite[p.631]{ortega}.
For a monoidal category $({\mathcal C}, \otimes)$ the group of automorphisms of $({\mathcal C}, \otimes)$ in ${\mathcal Mcat}_\sim$ is known as the Picard group of $({\mathcal C}, \otimes)$ \cite{ortega} and is denoted by ${\sf Pic}({\mathcal C}, \otimes)$, compare with \cite[p.633]{ortega}. Thus ${\sf Pic}({\mathcal C}, \otimes)$ is the set of isomorphic classes of monoidal autoequivalences of $({\mathcal C}, \otimes)$, with the multiplication induced by the composition of monoidal autoequivalences.
Let $G$ be a group. A group homomorphism $\varrho\to {\sf Pic}({\mathcal C}, \otimes)$ can be seen as a functor $\underline{\varrho}:\underline{G}\to {\mathcal Mcat}_\sim$, which sends the unique object of $\underline{G}$ to $({\mathcal C},\otimes)$. Hence we obtain the class
$${\mathcal T}(\varrho):=\varrho^*(\grupo{{\mathcal Mcat}})
\in H^3(G, {\sf Z}({\mathcal C},\otimes)^*)$$
which was considered in \cite[637]{ortega}. The map given by $\varrho\mapsto {\mathcal T}(\varrho)$ is known as the \emph{Teichm\"uller obstruction map}, \cite[637]{ortega}. The name comes from the problem to lift the homomorphism $\varrho$, equivalently the functor $\underline{\varrho}:\underline{G}\to {\mathcal Mcat}_\sim$ to a $2$-functor $\underline{G}\to {\mathcal Mcat}$. Thanks to Lemma \ref{1.5.m} such a lifting exists iff ${\mathcal T}(\varrho)=0$ and if this is the case. then set of equivalence classes of such liftings is a torsor over $H^2(G, {\sf Z}({\mathcal C},\otimes)^*)$. So, we obtain the main results of \cite{ortega}, see Proposition 4.1. Theorem 5.1 and Theorem 5.2 of \cite{ortega}.
Moreover, a variant of this theory is also considered in \cite{ortega}, where instead of monoidal categories , so called $k$-linear monoidal categories are considered, where $k$ is a fixed commutative ring. By considering abelian track category of all small $k$-linear monoidal categories, $k$-linear monoidal equivalences and $k$-linear monoidal natural isomorphisms, one obtains similar results for $k$-linear monoidal categories. Details left to an interested reader.
\subsection{Obstruction class of Chen-Du-Wang}
In the recent preprint \cite{chen}, the authors extended the theory of group extensions to groupoids. They, in fact, considered extensions of Lie groupoids. We will show how this theory (in the discrete case) can be obtained as a particular case of our approach.
Let $\sf K$ be a groupoid. All groupoids are assumed to be nonempty. The set of objects of ${\sf K}$ will be denoted by ${\sf K}_0$, while ${\sf K}_1$ denotes the set of morphisms of ${\sf K}$. For an arrow $\alpha\in {\sf K}_1$, we let $s_{\sf K}(\alpha)$ be the source of $\alpha$, while $t_{\sf K}(\alpha)$ denotes the target of $\alpha$.
A \emph{Chen-Du-Wang extension} (compare with \cite[Definition 3.1] {chen}) of a groupoid $\sf K$ by a groupoid $\sf A$ is the following data:
\begin{enumerate}[i)]
\item A groupid $\sf G$ and a morphism of groupoids $\varPhi=(\varPhi_0,\varPhi_1): {\sf G}\to {\sf K}$.
\item The identification ${\sf G}_i={\sf A}_i\times {\sf K}_i$, such that $\varPhi_i$ is the projection to the second factor, $i=1,2$.
\item The source map satisfies $s_{\sf G} =s_{\sf A}\times s_{\sf K}$.
\item For objects $k\in {\sf K}_0$, $a\in {\sf A}_0$, one has
$$ \id_{(a,k)}=(\id_{a},\id_{k}).$$
\item The target map satisfies the equation
$$t_{\sf G}(\alpha, \id_k)=(t_{\sf A}(\alpha), k).$$
Here $k\in {\sf K}_0$ and $\alpha$ is a morphism of $\sf A$.
\item For any composable pair of morphisms $\alpha$ and $\beta$ of $\sf A$, one has
$$(\alpha,\id_k)\circ (\beta,\id_k)=(\alpha\beta,\id_k).$$
\end{enumerate}
If ${\sf G}$ is a Chen-Du-Wang extension of a groupoid ${\sf K}$ by ${\sf A}$, we write
$$1\to {\sf A}\to {\sf G}\xto{\varPhi} {\sf K}\to 1.$$
\begin{Le} If $1\to {\sf A}\to {\sf G}\xto{\varPhi} {\sf K}\to 1$ is a Chen-Du-Wang extension, then $\varPhi$ is a Grothendieck cofibration and for any object $k\in {\sf K}_0$, the maps given by
$$a\mapsto (a,k), \ \ \alpha\mapsto (\alpha,\id_k), \ a\in {\sf A}_0, \alpha\in {\sf A}_1$$
yield an isomorphism of groupoids
$${\sf A}\to \Phi^{-1}(k).$$
\end{Le}
\begin{proof} According to the Lemma \ref{gr-ful=cof}, we only need to check that $\varPhi$ is full, but this follows from the condition ii). According to conditions iv) and vi) these maps define indeed a functor. Objects of $\Phi^{-1}(k)$ have the form $(a,k)$, where $a\in {\sf A}_0$. Moreover, morphisms in $\Phi^{-1}(k)$ have the form $(\alpha,\id_k)$ and hence the above functor is an isomorphism.
\end{proof}
As in any Grothendieck cofibration, the assignment $k\mapsto \Phi^{-1}(k)$ defines a $2$-functor. Hence we immediately obtain the following fact (compare with \cite[Section 3.3]{chen}).
\begin{Co} If $1\to {\sf A}\to {\sf G}\xto{\varPhi} {\sf K}\to 1$ is a Chen-Du-Wang extension, then there is a $2$-functor from the groupoid $\sf K$ to the 2-category of groupoids, which has the same value ${\sf A}$ on each object of ${\sf K}$ and for the morphism $\kappa:k_1\to k_2$ of the groupoid $\sf K$, the induced functor $$\kappa_*:{\sf A}=\Phi^{-1}(k_1)\to \varPhi^{-1}(k_2)={\sf A}$$
is given on objects by
$$\kappa_*(a)=b$$
where the $(b,k_2)$ is the target of $(\id_a,\kappa)$.
\end{Co}
Conversely, having such a $2$-functor $\varPsi$, the Grothendieck construction ${\sf K}\int \varPsi$ gives a Chen-Du-Wang extension, which explains \cite[Theorem 4.1]{chen}.
Hence the classification of Chen-Du-Wang extensions completely reduces to the study of appropriate $2$-functors, which can be done based on properties of the Toda class of the abelian track category ${\sf SGpd}$. Objects of ${\sf SGpd}$ are small groupoids, $1$-morphisms are isomorphisms of groupoids and $2$-morphisms are natural isomorphisms of isomorphisms of groupoids. We will see that the corresponding Toda class $\grupo{{\sf SGpd}}\in H3({\sf SGpd}_\sim,D^{\sf SGpd})$ explains several results obtained in \cite{chen}. First of all, observe that for any small groupoid $\sf A$, we have
$$D^{\sf SGpd}_{id_{\sf A}}=\{\xi:\id_{\sf A}\then \id_{\sf A}\}.$$ Thus
$D^{\sf SGpd}_{id_{\sf A}}={\sf Z}({\sf A})$ is the centre of $\sf A$. The second observation is the fact that the full subcategory of ${\sf SGpd}_\sim$ corresponding to the object $\sf A$ is the one object category corresponding to the group (in the notations \cite{chen}) $\overline{{\sf SAut(A)}}$, which is the group of isomorphism classes of automorphisms of ${\sf A}$. Hence by restricting the class $\grupo{{\sf SGpd}}$ to this subcategory one obtains the class
$$Res_{\sf A}{\grupo{{\sf SGpd}}} \in H^3(\overline{{\sf SAut(A)}}, {\sf Z}({\sf A})).$$
Let us take a small groupoid ${\sf K}$ and let $\bar{\Lambda}$ be a functor from ${\sf K}$ to the category ${\sf SGpd}_\sim$ such that all objects of $\sf K$ map to ${\sf A}$. Such a functor is called a \emph{band} in \cite[Definition 3.4]{chen}. The problem still to be answered is under what conditions $\bar{\Lambda}$ can be lifted to a $2$-functor $\Lambda:{\sf K}\to {\sf SGpd}$. By Lemma \ref{1.5.m} this happens if and only if the class
$$\bar{\Lambda}^*(Res_{\sf A}{\grupo{{\sf SGpd}}}\in H^3({\sf K}, {\sf Z}({\sf A}))$$
is zero. Moreover, if this happens then isomorphism classes of such liftings form a torsor over $H^2({\sf K}, {\sf Z}({\sf A}))$. These results reprove \cite[Theorem 4.4, Theorem 4.5 and Theorem 5.3]{chen}. | 0.004 |
TITLE: How many $n$-pointed stars are there?
QUESTION [7 upvotes]: Say we have $n$ distinct points spaced evenly in a circle. Define a star as a connected graph with these points as vertices and with $n$ edges, no two having the same endpoints. We think of two stars as being equivalent if they differ only by a rotation (if they differ by a reflection I consider them as different objects). I'd like to know how many different stars there are on $n$ points.
For example there are only two 4-pointed stars: a square and a bowtie. There are four 5-pointed stars:
And I think there are twelve 6-pointed stars:
(I'm not completely sure that I have exhausted the possibilities for 6 points).
A quick search of the online encyclopedia of integer sequences hasn't revealed any obvious candidates.
Another way to phrase the question:
Let $\rho = (1 2 \ldots n)\in S_n$. Say that two permutations $\sigma, \tau\in S_n$ are equivalent if $\sigma = \rho^{-k}\tau\rho^k$ for some $k$. How many equivalence classes contain order $n$ permutations?
REPLY [4 votes]: Deriving the general formula for this sequence (and its reflection-insensitive analogue) is precisely the content of:
Golomb, S.W.; Welch, L.R., "On the enumeration of polygons", Amer. Math. Monthly, 67 (1960), 349-353.
The authors derive the result as a clever application of Burnside's Lemma, giving for the count of $n$-pointed stars
$$
E(n)
=
\left\{
\begin{array}{ll}
F(n) , & \textrm{$n$ odd}\\
F(n) + \displaystyle{\frac{1}{2 n^2} \cdot 2^{n / 2} {n \choose 2} {n \choose 2}!}, & \textrm{$n$ even}
\end{array}
\right. ,
$$
where
$$F(n) := \frac{1}{2 n^2} \sum_{d \mid n} \phi^2\left(\frac{n}{d}\right) d! \left(\frac{n}{d}\right)^d .$$
Here, $\phi$ is the Euler totient function.
As Dan Uznanski's answer has pointed out, this is OEIS A000939, Number of inequivalent n-gons.
Remark As Empy2's answer observes, the procedure for computing should be easier for prime $n$. Indeed, for odd primes $p$ the above formula simplifies to
$$E(p) = \frac{1}{2 p} [(p - 1)^2 + (p - 1)!] .$$ It follows from the fact that $E(p)$ is an integer that $p \mid [(p - 1)^2 + (p - 1)!]$, and rearranging gives $$(p - 1)! \equiv -1 \pmod p ,$$ which recovers one direction of Wilson's Theorem. | 0.057259 |
Big Walls to Fill
Art's Cutting -- and Moneyed -- Edge Makes a Splash at Art Basel Miami Beach
Monday, December 10, 2007; Page. | 0.013286 |
TITLE: If $i^2=-1$, then what about $(-i)^2=-1$?
QUESTION [6 upvotes]: By definition, $i^2=-1$, right?
But one can then clearly deduce that $(-i)^2=-1$. The only difference I see is that one is $-1$ times the second.
So what allows us to differentiate between $i$ and $-i$? Can they be used synonymous? That is, does nothing happen if all of a sudden we were to switch $i$ and $-i$ in math, so long as we are consistent?
REPLY [9 votes]: There is some interesting mathematics bound up in your question. There is an automorphism of $\mathbb C$ as a field extension of $\mathbb R$ which is defined by sending $i$ to $-i$. Provided you are consistent (including very great care with signs) everything algebraic works nicely. This reflects the fact that in building $\mathbb C$ from $\mathbb R$ we make an arbitrary choice of a square root of $-1$ to call $i$.
For amusement there is an automorphism of $\mathbb Q(\sqrt 2)$ as an extension of $\mathbb Q$ which sends $\sqrt 2$ to $-\sqrt 2$ - for much the same reason. However, interesting things happen to the metric (distance between points) - so though the algebraic properties are retained, it is not true that "everything" remains the same. | 0.04095 |
REMINDER – Due today!
Thank you to the students who have contacted us to confirm their graduation by filling out the questionnaire! We’d like to make one final push for graduate information before we publish our special graduate e-newsletter.
If you are a graduate who hasn’t submitted your information to us yet, please do so here,! We still want to hear from you! Our final deadline will be this Friday, May 17th. Thank you | 0.296534 |
\section{Preliminaries}
In this section, we introduce some definitions that are useful in our analysis later.
Consider a proper\footnote{An extended real-valued function $h$ is proper if its domain $\dom h := \{ x: h(x) < \infty \}$ is nonempty.} and lower-semicontinuous function $h:\mathbb{R}^d \to \mathbb{R}$ which is {\em not} necessarily smooth nor convex. We introduce the following generalized notion of derivative for the function $h$.
\begin{definition}(Subdifferential and critical point, \cite{vari_ana})\label{def:sub}
The Frech\'et subdifferential $\widehat\partial h$ of function $h$ at $x\in \dom h$ is the set of $u\in \mathbb{R}^d$ defined as
\begin{align*}
\widehat\partial h(x) := \bigg\{u: \liminf_{z\neq x, z\to x} \frac{h(z) - h(x) - u^\intercal(z-x)}{\|z-x\|} \ge 0 \bigg\},
\end{align*}
and the limiting subdifferential $\partial h$ at $x\in\dom h$ is the graphical closure of $\widehat\partial h$ defined as:
\begin{align*}
\partial h(x) := \{ u: \exists x_k \to x, h(x_k) \to h(x), u_k \in \widehat{\partial} h(x_k) \to u \}.
\end{align*}
The set of critical points of $h$ is defined as $\mathbf{\crit}\!~h := \{ x: \zero\in\partial h(x) \}$.
\end{definition}
Note that when the function $h$ is continuously differentiable, the limiting sub-differential $\partial h$ reduces to the usual notion of gradient $\nabla h$.
Next, we introduce the Kurdyka-{\L}ojasiewicz (K{\L}) property of a function $h$. Throughout, we define the distance between a point $x\in \mathbb{R}^d$ and a set $\Omega \subseteq \mathbb{R}^d$ as $\dist_\Omega(x) := \inf_{w\in \Omega} \|x - w\|$.
\begin{definition}(\KL property, \cite{Bolte2014})\label{def: KL}
A proper and lower-semicontinuous function $h$ is said to satisfy the \KL property if for every compact set $\Omega\subset \dom h$ on which $h$ takes a constant value $h_\Omega \in \mathbb{R}$, there exist $\varepsilon, \lambda >0$ such that for all $x \in \{z\in \mathbb{R}^d : \dist_\Omega(z)<\varepsilon, h_\Omega < h(z) <h_\Omega + \lambda\}$, the following inequality is satisfied
\begin{align}\label{eq: KL}
\varphi' \left(h(x) - h_\Omega\right) \dist_{\partial h(x)}(\zero) \ge 1,
\end{align}
where $\varphi'$ is the derivative of function $\varphi: [0,\lambda) \to \mathbb{R}_+$, which takes the form $\varphi(t) = \frac{c}{\theta} t^\theta$ for some $c>0, \theta\in (0,1]$.
\end{definition}
To elaborate, consider the case where $h$ is differentiable. Then, the \KL property in \cref{eq: KL} can be rewritten as
\begin{align}
h(x) - h_\Omega \le C \|\nabla h(x)\|^{p} \label{eq: KLsimple}
\end{align}
for some constant $C>0$ and $p\in (1, +\infty)$. In fact, \Cref{eq: KLsimple} can be viewed as a generalization of the gradient dominance condition
that corresponds to the special case of $p = 2$. A large class of functions have been shown to satisfy the \KL property, e.g., sub-analytic functions, logarithm and exponential functions, etc \cite{Bolte2007}. These function classes cover most of nonconvex objective functions encountered in practical applications, e.g., logistic loss, vector and matrix norms, rank, and polynomial functions, etc. Please refer to \cite[Section 5]{Bolte2014} and \cite[Section 4]{Attouch2010} for more example functions.
To handle non-smooth objective functions, we introduce the following notion of proximal mapping.
\begin{definition}(Proximal mapping)\label{def:prox}
For a proper and lower-semicontinuous function $h$, its proximal mapping at $x\in \mathbb{R}^d$ with parameter $\eta > 0$ is defined as:
\begin{align}
\prox{\eta h}(x) := \argmin_{z\in \mathbb{R}^d} \bigg\{h(z) + \frac{1}{2\eta}\|z - x\|^2\bigg\}.
\end{align}
\end{definition} | 0.009385 |
\begin{document}
\title{Refined analytic torsion for twisted de Rham complexes}
\author{Rung-Tzung Huang}
\address{Institute of Mathematics, Academia Sinica, Nankang 11529, Taipei, Taiwan}
\email{[email protected]}
\subjclass[2010]{Primary: 58J52}
\keywords{determinant, analytic torsion, twisted de Rham complex}
\begin{abstract}
Let $E$ be a flat complex vector bundle over a closed oriented odd dimensional manifold $M$ endowed with a flat connection $\nabla$. The refined analytic torsion for $(M,E)$ was defined and studied by Braverman and Kappeler. Recently Mathai and Wu defined and studied the analytic torsion for the twisted de Rham complex with an odd degree closed differential form $H$, other than one form, as a flux and with coefficients in $E$. In this paper we generalize the construction of the refined analytic torsion to the twisted de Rham complex. We show that the refined analytic torsion of the twisted de Rham complex is independent of the choice of the Riemannian metric on $M$ and the Hermitian metric on $E$. We also show that the twisted refined analytic torsion is invariant (under a natural identification) if $H$ is deformed within its cohomology class. We prove a duality theorem, establishing a relationship between the twisted refined analytic torsion corresponding to a flat connection and its dual. We also define the twisted analogue of the Ray-Singer metric and calculate the twisted Ray-Singer metric of the twisted refined analytic torsion. In particular we show that in case that the Hermtitian connection is flat, the twisted refined analytic torsion is an element with the twisted Ray-Singer norm one.
\end{abstract}
\maketitle
\section{Introduction}
Let $E$ be a flat complex vector bundle over a closed oriented odd dimensional manifold $M$ endowed with a flat connection $\nabla$. Braverman and Kappeler \cite{BK1,BK3,BK4,BK5,BK6} defined and studied the refined analytic torsion for $(M,E)$, which can be viewed as a refinement of the Ray-Singer torsion \cite{RS} and an analytic analogue of the Farber-Turaev torsion, \cite{FT1,FT2,Tu1,Tu2}. It was shown that the refined analytic torsion is closely related with the Farber-Turaev torsion, \cite{BK1,BK3,BK6,H}.
In \cite{MW,MW1} Mathai and Wu generalize the classical construction of the Ray-Singer torsion to the twisted de Rham complex with an odd degree closed differential form $H$, other than one form, as a flux and with coefficients in $E$. The twisted de Rham complex is the $\mathbb{Z}_2$-graded complex $(\Omega^\bullet(M,E),\nabla^H)$, where $(\Omega^\bullet(M,E)$ is the space of differential forms with coefficients in $E$ and $\nabla^H:=\nabla + H\wedge \cdot$. Its cohomology $H^{\bullet}(M,E,H)$ is called the twisted de Rham cohomology. Mathai and Wu \cite{MW} defined the analytic torsion of the twisted de Rham complex $\tau(M,E,H) \in \Det\big(H^{\bullet}(M,E,H)\big)$ as a ratio of $\zeta$-regularized determinants of partial Laplacians, multiplied by the ratio of volume elements of the cohomology groups. They showed that when $\dim M$ is odd, $\tau(M,E,H)$ is independent of the choice of the Riemannian metric on $M$ and the Hermitian metric on $E$. They also showed that the torsion $\tau(M,E,H)$ is invariant (under a natural identification) if $H$ is deformed within its cohomology class and discussed its connection with the generalized geometry \cite{Gu2}.
In this paper we define the refined analytic torsion for the twisted de Rham complex $\rho_{\operatorname{an}}(\nabla^H)\in \Det \big( H^{\bullet} (M,E,H)\big)$. We show that the twisted refined analytic torsion $\rho_{\operatorname{an}}(\nabla^H)$ is independent of the choice of the Riemannian metric on $M$ and the Hermitian metric on $E$. We then show that the torsion $\rho_{\operatorname{an}}(\nabla^H)$ is invariant (under a natural identification) if $H$ is deformed within its cohomology class. We also establish a duality theorem, establishing a relationship between the twisted refined analytic torsion corresponding to a flat connection and its dual, which is a twisted analogue of Theorem 10.3 of \cite{BK3}. In the end we define the twisted analogue of the Ray-Singer metric and then calculate the twisted Ray-Singer norm of the twisted refined analytic torsion. In particular we show that in case of flat Hermtitian metric, the twisted refined analytic torsion is a canonical choice of an element with the twisted Ray-Singer norm one.
The paper is organized as follows. In Section \ref{S:rtrev}, we review some standard materials about determinant lines of a $\mathbb{Z}_2$-graded finite dimensional complex. Then we define and calculate the refined torsion of the $\mathbb{Z}_2$-graded finite dimensional complex with a chirality operator. In Section \ref{S:gdtoso}, we define the graded determinant of the twisted version of the odd signature operator of a flat vector bundle $E$ over a closed oriented odd dimensional manifold $M$. We use this graded determinant to define a canonical element $\rho_H$ of the determinant line of the twisted de Rham cohomology of the vector bundle $E$. We study the relationship between this graded determinant and the $\eta$-invariant of the twisted odd signature operator. In Section \ref{S:madrat}, we first study the metric dependence of the canonical element $\rho_H$ and then use this element to construct the refined analytic torsion twisted by the flux form $H$. We show that the twisted refined analytic torsion is independent of the metric $g^M$ and the representative $H$ in the cohomology class $[H]$. In Section \ref{S:dttrat}, we first review the concept of the dual of a complex and construct a natural isomorphism between the determinant lines of a $\mathbb{Z}_2$-graded complex and its dual. We then establish a relationship between the twisted refined analytic torsion corresponding to a flat connection and that of its dual. In Section \ref{S:cwtat}, we first define the twisted Ray-Singer metric and then calculate the twisted Ray-Singer norm of the twisted refined analytic torsion.
Throughout this paper, the bar over an integer means taking the value modulo $2$.
\subsection*{Acknowledgement}
The author would like to thank Maxim Braverman for suggesting this problem.
\section{The refined torsion of a $\mathbb{Z}_2$-graded finite dimensional complex with a chirality operator}\label{S:rtrev}
In this section we first review some standard materials about determinant lines of a $\mathbb{Z}_2$-graded finite dimensional complex. Then we define and calculate the refined torsion of the $\mathbb{Z}_2$-graded finite dimensional complex with a chirality operator. The contents are $\mathbb{Z}_2$-graded analogues of Section 2, Section 4 and Section 5 of \cite{BK3}.
Throughout this section $\bf{k}$ is a field of characteristic zero.
\subsection{The determinant line of a $\mathbb{Z}_2$-graded finite dimensional complex}
Given a $\bf{k}$-vector space $V$ of dimension $n$, the determinant line of $V$ is the line $\Det(V):=\wedge^nV$, where $\wedge^nV$ denotes the $n$-th exterior power of $V$. By definition, we set $\operatorname{Det}(0):=\bf{k}$. Further, we denote by $\Det(V)^{-1}$ the dual line of $\Det(V)$.
Let
\begin{equation}\label{E:utdetline}
0 \stackrel{}{\longrightarrow} \ C^0 \ \stackrel{\partial_0}{\longrightarrow}\ C^1\ \stackrel{\partial_1}{\longrightarrow} \ \cdots \stackrel{\partial_{m-1}}{\longrightarrow}\ \ C^m \ \stackrel{\partial_m}{\longrightarrow} 0
\end{equation}
be an odd length, i.e. $m=2r-1$ being a positive odd integer, cochain complex of finite dimensional $\bf{k}$-vector spaces. Set
$$C^{\bar{0}}=C^{\operatorname{even}}=\bigoplus_{i=0}^{r-1} C^{2i}, \qquad C^{\bar{1}}=C^{\operatorname{odd}}=\bigoplus_{i=0}^{r-1} C^{2i+1}.$$
Let
\begin{equation}\label{E:detline}
(C^\bullet,d) \ : \ \cdots \stackrel{d_{\bar{1}}}{\longrightarrow} \ C^{\bar{0}} \ \stackrel{d_{\bar{0}}}{\longrightarrow}\ C^{\bar{1}}\ \stackrel{d_{\bar{1}}}{\longrightarrow} \ C^{\bar{0}} \ \stackrel{d_{\bar{0}}}{\longrightarrow} \cdots
\end{equation}
be a $\mathbb{Z}_2$-graded cochain complex of finite dimensional $\bf{k}$-vector spaces. For example, we can choose $d_{\bar{k}}=\sum_{i, i=k\, \text{mod}\, 2}\partial_i$. Denote by $H^{\bar{k}}(d_{\bar{k}}), (k=0,1)$ its cohomology. Set
\begin{equation}
\Det(C^\bullet)\,:=\,\Det \big( C^{\bar{0}} \big) \otimes \Det \big( C^{\bar{1}} \big)^{-1}, \qquad \Det(H^\bullet(d))\,:=\, \Det\big( H^{\bar{0}}(d_{\bar{0}})\big) \otimes \Det\big( H^{\bar{1}}(d_{\bar{1}})\big)^{-1}.
\end{equation}
\subsection{The fusion isomorphisms}(cf. \cite[Subsection 2.3]{BK3})
For two finite dimensional $\bf{k}$-vector spaces $V$ and $W$, we denote by $\mu_{V,W}$ the canonical fusion isomorphism,
\begin{equation}\label{E:fusio}
\mu_{V,W} : \Det(V) \otimes \Det(W) \to \Det(V \oplus W).
\end{equation}
For $v \in \Det(V)$, $w \in \Det(W)$, we have
\begin{equation}\label{E:fusio1}
\mu_{V,W}(v \otimes w)= (-1)^{\dim V \cdot \dim W} \mu_{W,V}(w \otimes v).
\end{equation}
By a slight abuse of notation, denote by $\mu_{V,W}^{-1}$ the transpose of the inverse of $\mu_{V,W}$.
Similarly, if $V_1, \cdots, V_r$ are finite dimensional $\bf{k}$-vector spaces, we define an isomorphism
\begin{equation}\label{E:mulfus}
\mu_{V_1,\cdots, V_r}: \Det(V_1) \otimes \cdots \otimes \Det(V_r) \to \Det(V_1 \oplus \cdots \oplus V_r).
\end{equation}
\subsection{The isomorphism between the determinant lines of a $\mathbb{Z}_2$-graded complex and its cohomology }\label{SS:isomo}
For $k=0,1$, fix a direct sum decomposition
\begin{equation}\label{E:decompbha}
C^{\bar{k}}=B^{\bar{k}} \oplus H^{\bar{k}} \oplus A^{\bar{k}},
\end{equation}
such that $B^{\bar{k}}\oplus H^{\bar{k}}=(\Ker d_{\bar{k}}) \cap C^{\bar{k}}$ and $B^{\bar{k}}= d_{\overline{k+1}}\big(C^{\overline{k+1}}\big)=d_{\overline{k+1}}\big(A^{\overline{k+1}}\big)$. Then $H^{\bar{k}}$ is naturally isomorphic to the cohomology $H^{\bar{k}}(d_{\bar k})$ and $d_{\bar{k}}$ defines an isomorphism $d_{\bar{k}}: A^{\bar{k}} \to B^{\overline{k+1}}$.
Fix $c_{\bar{k}} \in \Det(C^{\bar{k}})$ and $a_{\bar{k}} \in \Det(A^{\bar{k}})$. Let $d_{\bar{k}}(a_{\bar{k}}) \in \Det(B^{\overline{k+1}})$ denote the image of $a_{\bar{k}}$ under the map $\Det(A^{\bar{k}}) \to \Det(B^{\overline{k+1}})$ induced by the isomorphism $d_{\bar{k}}:A^{\bar{k}} \to B^{\overline{k+1}}$. Then there is a unique element $h_{\bar{k}} \in \Det(H^{\bar{k}})$ such that
\begin{equation}\label{E:cbahce}
c_{\bar{k}}= \mu_{B^{\bar{k}},H^{\bar{k}},A^{\bar{k}}}\big(d_{\overline{k+1}}(a_{\overline{k+1}})\otimes h_{\bar{k}} \otimes a_{\bar{k}}\big),
\end{equation}
where $\mu_{B^{\bar{k}},H^{\bar{k}},A^{\bar{k}}}$ is the fusion isomorphism, cf. \eqref{E:mulfus}, see also \cite[Subsection 2.3]{BK3}.
Define the canonical isomorphism
\begin{equation}\label{E:isomorphism}
\phi_{C^\bullet}=\phi_{(C^{\bullet},d)} \,:\,\Det(C^{\bullet} ) \longrightarrow \Det(H^{\bullet}(d)),
\end{equation}
by the formula
\begin{equation}\label{E:phic}
\phi_{C^\bullet}: c_{\bar{0}} \otimes c_{\bar{1}}^{-1} \mapsto (-1)^{\mathcal{N}(C^{\bullet})} h_{\bar{0}} \otimes h_{\bar{1}}^{-1},
\end{equation}
where
\begin{equation}\label{E:ncbullet}
\mathcal{N}(C^\bullet) \, := \, \frac{1}{2} \sum_{k=0,1} \dim A^{\bar k} \cdot \big(\dim A^{\bar k}+(-1)^{k+1}\big).
\end{equation}
\subsection{The fusion isomorphism for $\mathbb{Z}_2$-graded complexes}
Let $C^\bullet=C^{\bar 0} \oplus C^{\bar 1}$ and $\widetilde{C}^\bullet=\widetilde{C}^{\bar 0} \oplus \widetilde{C}^{\bar 1}$ be finite dimensional $\mathbb{Z}_2$-graded $\bf{k}$-vector spaces. The fusion isomorphism
\[
\mu_{C^\bullet ,\widetilde{C}^{\bullet}} : \Det(C^{\bullet}) \otimes \Det(\widetilde{C}^{\bullet}) \to \Det(C^{\bullet} \oplus \widetilde{C}^{\bullet}),
\]
is defined by the formula
\begin{equation}\label{E:grafus}
\mu_{C^{\bullet},\widetilde{C}^{\bullet}} := (-1)^{\mathcal{M}(C^{\bullet},\widetilde{C}^{\bullet})} \mu_{C^{\bar 0},\widetilde{C}^{\bar 0}} \otimes \mu_{C^{\bar 1},\widetilde{C}^{\bar 1}}^{-1},
\end{equation}
where
\begin{equation}\label{E:mvw123}
\mathcal{M}(C^{ \bullet},\widetilde{C}^{ \bullet}) := \dim C^{\bar 1} \cdot \dim \widetilde{C}^{\bar 0}.
\end{equation}
The following lemma is a $\mathbb{Z}_2$-graded analogue of \cite[Lemma 2.7]{BK3} and \cite[Lemma 2.4]{FT2}. The proof is a slight modification of the proof of \cite[Lemma 2.7]{BK3}.
\begin{lemma}\label{L:commu}
Let $(\,C^\bullet, d\,)$ and $(\,\widetilde{C}^\bullet, \widetilde{d}\,)$ be $\mathbb{Z}_2$-graded complexes with finite dimensional $\bf{k}$-vector spaces. Further, assume that the Euler characteristics $\chi(C^\bullet)=\chi(\widetilde{C}^\bullet)=0$. Then the following diagram commutes:
\begin{equation}\label{E:comm}
\begin{CD}
\Det(C^\bullet) \otimes \Det(\widetilde{C}^\bullet) & @>\phi_{C^\bullet}\otimes \phi_{\widetilde{C}^\bullet}>> & \Det\big(H^\bullet(d)\big) \otimes \Det\big(\, H^\bullet(\widetilde{d})\,\big) \\
@V \mu_{C^\bullet,\widetilde{C}^\bullet} VV & & @VV \mu_{H^\bullet(d),H^\bullet(\widetilde{d})} V \\
\Det(C^\bullet \oplus \widetilde{C}^\bullet) & @> \phi_{C^\bullet \oplus \widetilde{C}^\bullet} >> & \Det\big(\, H^\bullet(d \oplus \widetilde{d})\, \big) \cong \Det\big( \, H^\bullet(d) \oplus H^\bullet(\widetilde{d})\, \big)
\end{CD}
\end{equation}
\end{lemma}
\begin{proof}
Proceed similar procedures as the proof of Lemma 2.7 of \cite{BK3}, cf. \cite[P. 152-153]{BK3}, we conclude that to prove the commutativity of the diagram \eqref{E:comm} it remains to show that, mod $2$,
\begin{equation}\label{E:eukccnmm}
\mathcal{N}(C^\bullet \oplus \widetilde{C}^\bullet)+\mathcal{N}(C^\bullet)+\mathcal{N}( \widetilde{C}^\bullet)+\mathcal{M}(C^\bullet, \widetilde{C}^\bullet) +\mathcal{M}(H^\bullet, \widetilde{H}^\bullet)
\end{equation}
\[
\equiv \sum_{k=0,1}\big( \dim A^{\bar k} \cdot \dim \widetilde{A}^{\overline{k+1}}+ \dim H^{\bar k} \cdot \dim \widetilde{A}^{\overline{k+1}}+\dim A^{\bar k} \cdot \dim \widetilde{H}^{\bar k}\big).
\]
Using the identity
\begin{equation}\label{E:usid1}
\frac{(x+y)(x+y+(-1)^j)}{2}-\frac{x(x+(-1)^j)}{2}-\frac{y(y+(-1)^j)}{2}=xy,
\end{equation}
where $x,y \in \mathbb{C}, j \in \mathbb{Z}_{\ge 0}$, we have
\begin{equation}\label{E:nnn}
\mathcal{N}(C^\bullet \oplus \widetilde{C}^\bullet) - \mathcal{N}(C^\bullet)- \mathcal{N}( \widetilde{C}^\bullet) = \sum_{k=0,1} \dim A^{\bar k} \cdot \dim \widetilde{A}^{\bar k}.
\end{equation}
By \eqref{E:decompbha} and the equalities $\dim A^{\overline{k+1}}= \dim B^{\bar k}$, $\dim \widetilde{A}^{\overline{k+1}}= \dim \widetilde{B}^{\bar k}$, we have
\begin{equation}\label{E:cbha1}
\dim C^{\bar k} = \dim A^{\bar k} + \dim A^{\overline{k+1}} + \dim H^{\bar k}, \quad \dim \widetilde{C}^{\bar k} = \dim \widetilde{A}^{\bar k} + \dim \widetilde{A}^{\overline{k+1}} + \dim \widetilde{H}^{\bar k}.
\end{equation}
By \eqref{E:mvw123}, \eqref{E:cbha1} and a straightforward computation, we obtain, modulo $2$,
\begin{equation}\label{E:lonequ}
\begin{array}{l}
\mathcal{M}(C^\bullet, \widetilde{C}^\bullet) +\mathcal{M}(H^\bullet, \widetilde{H}^\bullet) + \sum_{k=0,1}\big( \dim A^{\bar k} \cdot \dim \widetilde{A}^{\overline{k+1}}+ \dim H^{\bar k} \cdot \dim \widetilde{A}^{\overline{k+1}}+\dim A^{\bar k} \cdot \dim \widetilde{H}^{\bar k}\big)\\
\\
= \sum_{k=0,1}\dim A^{\bar k} \cdot \dim \widetilde{A}^{\bar k} + \dim A^{\bar 1}\cdot (\dim \widetilde{H}^{\bar 0}+ \dim \widetilde{H}^{\bar 1}) + (\dim H^{\bar 0}+ \dim H^{\bar 1})\cdot \dim \widetilde{A}^{\bar 1}.
\end{array}
\end{equation}
By \eqref{E:nnn}, \eqref{E:lonequ} and the assumption that the Euler characteristic of the complex $(C^\bullet,d)$ (resp. $(\widetilde{C}^\bullet,\widetilde{d})$) is zero, i.e. $\sum_{k=0,1} \dim H^{\bar k} \equiv 0 \, (\operatorname{mod}\, 2)$ (resp. $\sum_{k=0,1}\dim \widetilde{H}^{\bar k} \equiv 0 (\operatorname{mod}\, 2) \,$), we obtain the equality \eqref{E:eukccnmm}.
\end{proof}
\subsection{The refined torsion of a finite dimensional $\mathbb{Z}_2$-graded complex with a chirality operator}
Let $(C^\bullet,d)$ be a $\mathbb{Z}_2$-graded complex defined as \eqref{E:detline}. A {\em chirality operator} is an involution $\Gamma : C^\bullet \to C^\bullet$ such that $\Gamma(C^{\bar{k}})=C^{\overline{k+1}}, k=0,1$. For $c_{\bar{k}} \in \Det(C^{\bar{k}})$, we denote by $\Gamma c_{\bar{k}} \in \Det(C^{\overline{k+1}})$ the image of $c_{\bar{k}}$ under the isomorphism $\Det(C^{\bar{k}}) \to \Det(C^{\overline{k+1}})$ induced by $\Gamma$.
Fix a nonzero element $c_{\bar{0}} \in \Det(C^{\bar{0}})$ and consider the element
\begin{equation}\label{E:elec}
c_{\Gamma}:=(-1)^{\mathcal{R}(C^\bullet)} \cdot c_{\bar{0}} \otimes (\Gamma c_{\bar{0}})^{-1} \in \Det(C^\bullet),
\end{equation}
where
\begin{equation}\label{E:rcbullet}
\mathcal{R}(C^\bullet ) \, := \, \frac{1}{2} \dim C^{\bar 0} \cdot (\dim C^{\bar 0}+1).
\end{equation}
The element defined in \eqref{E:elec} is a $\mathbb{Z}_2$-graded analogue of the $\mathbb{Z}$-graded one as defined in \cite[(4-1)]{BK3}, by Braverman-Kappeler, and is chosen to fit the $\mathbb{Z}_2$-graded setting.
\begin{definition}\label{D:elerho1}
The {\em refined torsion} of the pair $(C^\bullet,\Gamma)$ is the element
\begin{equation}\label{E:elerho}
\rho_{\Gamma}=\rho_{C^{\bullet},\Gamma}:= \Phi_{C^{\bullet}}(c_{\Gamma}),
\end{equation}
where $\Phi_{C^{\bullet}}$ is the canonical map defined by \eqref{E:isomorphism}.
\end{definition}
The following is the $\mathbb{Z}_2$-graded analogue of Lemma 4.7 of \cite{BK3}.
\begin{lemma}
Let $(C^\bullet,d)$ and $(\widetilde{C}^\bullet,\widetilde{d})$ be $\mathbb{Z}_2$-graded complexes defined as \eqref{E:detline} and let $\Gamma : C^\bullet \to C^\bullet$, \, $\widetilde{\Gamma} : \widetilde{C}^\bullet \to \widetilde{C}^\bullet$ be chirality operators. Then $\widehat{\Gamma}:=\Gamma \oplus \widetilde{\Gamma} : C^\bullet \oplus \widetilde{C}^\bullet \to C^\bullet \oplus \widetilde{C}^\bullet$ is a chirality operator on the direct sum complex $(C^\bullet \oplus \widetilde{C}^\bullet ,d \oplus \widetilde{d})$ and
\begin{equation}\label{E:disuto}
\rho_{\widehat{\Gamma}}=\mu_{H^\bullet(d),H^\bullet(\widetilde{d})}\big( \rho_\Gamma \otimes \rho_{\widetilde{\Gamma}} \big).
\end{equation}
\end{lemma}
\begin{proof}
Clearly, $\widehat{\Gamma}^2=1$ and $\widehat{\Gamma}(C^{\bar k} \oplus \widetilde{C}^{\bar k})=C^{\overline{k+1}}\oplus \widetilde{C}^{\overline{k+1}}$. Hence, $\widehat{\Gamma}$ is a chirality operator. By Lemma \ref{L:commu}, to prove \eqref{E:disuto} it is enough to show that
\begin{equation}\label{E:maidp}
c_{\widehat{\Gamma}}=\mu_{C^\bullet,\widehat{C}^\bullet}\big( c_\Gamma \otimes c_{\widetilde{\Gamma}} \big).
\end{equation}
Fix nonzero elements $c_{\bar 0} \in \Det(C^{\bar 0}), \widetilde{c}_{\bar 0} \in \Det(\widetilde{C}^{\bar 0})$ and set $\widehat{c}_{\bar 0} = \mu_{C^{\bar 0},\widetilde{C}_{\bar 0}}(c_{\bar 0} \otimes \widetilde{c}_{\bar 0})$. We denote the operators induced by $\Gamma$ and $\widetilde{\Gamma}$ on $\Det(C^\bullet)$ and $\Det(\widetilde{C}^\bullet)$ by the same letters. Thus,
\[
\widehat{\Gamma}\, \widehat{c}_{\bar 0} = (\Gamma \oplus \widetilde{\Gamma}) \circ \mu_{C^{\bar 0}, \widetilde{C}_{\bar 0}}(c_{\bar 0} \otimes \widetilde{c}_{\bar 0}) = \mu_{C^{\bar 1},\widetilde{C}^{\bar 1}}(\Gamma c_{\bar 0} \otimes \widetilde{\Gamma} \widetilde{c}_{\bar 0}).
\]
Hence, it follows from \eqref{E:grafus} and \eqref{E:elec} that
\begin{align}\label{E:watd}
\mu_{C^\bullet,\widetilde{C}^\bullet}(c_\Gamma \otimes c_{\widetilde{\Gamma}}) & = (-1)^{\mathcal{M}(C^\bullet, \widetilde{C}^\bullet)+\mathcal{R}(C^\bullet)+\mathcal{R}(\widetilde{C}^\bullet)} \cdot \widehat{c}_{\bar 0} \otimes (\widehat{\Gamma} \, \widehat{c}_{\bar 0})^{-1} \nonumber \\
& = (-1)^{\mathcal{M}(C^\bullet, \widetilde{C}^\bullet)+\mathcal{R}(C^\bullet)+\mathcal{R}(\widetilde{C}^\bullet)-\mathcal{R}(C^\bullet \oplus \widetilde{C}^\bullet)} \cdot c_{\widehat{\Gamma}}.
\end{align}
Using the identity \eqref{E:usid1}, we obtain from \eqref{E:rcbullet}
\begin{equation}\label{E:rrmr}
\mathcal{R}(C^\bullet)+\mathcal{R}(\widetilde{C}^\bullet)-\mathcal{R}(C^\bullet \oplus \widetilde{C}^\bullet) = \dim C^{\bar 0} \cdot \dim \widetilde{C}^{\bar 0}.
\end{equation}
Using the isomorphism $\Gamma : C^{\bar 0} \to C^{\bar 1}$, one sees that $\dim C^{\bar 0} = \dim C^{\bar 1}$. Combining this fact with \eqref{E:mvw123} and \eqref{E:rrmr}, we conclude that
\begin{equation}\label{E:id13}
\mathcal{M}(C^\bullet, \widetilde{C}^\bullet)+\mathcal{R}(C^\bullet)+\mathcal{R}(\widetilde{C}^\bullet)-\mathcal{R}(C^\bullet \oplus \widetilde{C}^\bullet) \equiv 0 \quad \text{mod} \ 2.
\end{equation}
The identity \eqref{E:maidp} follows from \eqref{E:watd} and \eqref{E:id13}.
\end{proof}
\subsection{Dependence of the $\mathbb{Z}_2$-graded refined torsion on the chirality operator}
Suppose that $\Gamma_t,t \in \mathbb{R}$, is a smooth family of chirality operators on the $\mathbb{Z}_2$-graded complex $(C^{ \bullet},d)$. Let $\dot{\Gamma}_t:C^{\bar k} \to C^{\overline{k+1}},k=0,1$, denote the derivative of $\Gamma_t$ with respect to $t$. Then, for $k=0,1$, the composition $\dot{\Gamma_t}\circ \Gamma_t$ maps $C^{\bar k}$ into itself. Define the {\em supertrace} $\operatorname{Tr}_s(\dot{\Gamma_t}\circ \Gamma_t)$ of $\dot{\Gamma_t}\circ \Gamma_t$ by the formula
\begin{equation}\label{E:supertrace}
\operatorname{Tr}_s(\dot{\Gamma_t}\circ \Gamma_t):= \operatorname{Tr}(\dot{\Gamma_t}\circ \Gamma_t|_{C^{\bar 0}})- \operatorname{Tr}(\dot{\Gamma_t}\circ \Gamma_t|_{C^{\bar 1}}).
\end{equation}
The following proposition is the $\mathbb{Z}_2$-graded analogue of Proposition 4.9 of \cite{BK3}. We modify the proof of Proposition 4.9 of \cite{BK3} slightly to fit our setting.
\begin{proposition}
Let $(C^{ \bullet},d)$ be a $\mathbb{Z}_2$-graded complex of finite dimensional $\bf{k}$-vector spaces and let $\Gamma_t : C^{\bar{k}} \to C^{\overline{k+1}}, t \in \mathbb{R}$, be a smooth family of chirality operators on $C^{ \bullet}$. Then the following equality holds
\begin{equation}\label{E:varfor}
\frac{d}{dt}\rho_{\Gamma_t}= \frac{1}{2} \operatorname{Tr}_s(\dot{\Gamma_t}\circ \Gamma_t) \cdot \rho_{\Gamma_t}.
\end{equation}
\end{proposition}
\begin{proof}
Let $\Gamma_{t,\bar{0}}$ denote the restriction of $\Gamma_t$ to $C^{\bar 0}$. We denoted the map $\Det(C^{\bar 0}) \to \Det(C^{\bar{1}})$ induced by $\Gamma_t$ by the same symbol $\Gamma_t$ above. To avoid confusion we denote this map by $\Gamma^{\Det}_{t,\bar{0}}$ in the proof.
For $ t_0 \in \mathbb{R}$, we have $\Gamma_{t,\bar{0}}=\Gamma_{t,\bar{0}} \circ \Gamma_{t_0,\bar{1}}\Gamma_{t_0,\bar{0}}$.
Hence,
\[
\frac{d}{dt}\Big|_{t=t_0} \Gamma_{t,\bar{0}}^{\Det} = \frac{d}{dt} \Big|_{t=t_0} \Big[ \Det(\Gamma_{t,\bar{0}} \circ \Gamma_{t_0,\bar{1}}) \Gamma_{t_0,\bar{0}}^{\Det} \Big] = \operatorname{Tr}( \dot{\Gamma}_{t_0, \bar{0} } \circ \Gamma_{t_0,\bar{1}})\Gamma^{\Det}_{t_0,\bar{0}},
\]
where for the latter equality we used the fact that for any smooth family of operators $A_t:C^{\bar{1}} \to C^{\bar{1}}$, one has $\frac{d}{dt} \Det(A_t)=\operatorname{Tr}(\dot{A_t}A_t^{-1})\cdot \Det(A_t)$ and that $\Gamma_{t_0,\bar{0}}^{-1}=\Gamma_{t_0,\bar{1}}$. Hence, for any nonzero element $c_{\bar 0} \in \Det(C^{\bar 0})$, we have
\begin{equation}\label{E:varif}
\frac{d}{dt}(\Gamma_{t,\bar{0}}^{\Det}(c_{\bar 0}))^{\pm}= \pm \operatorname{Tr}(\dot{\Gamma}_{t,\bar{0}} \circ \Gamma_{t,\bar{1}})\cdot (\Gamma^{\Det}_{t,\bar{0}})^{\pm}.
\end{equation}
By \eqref{E:varif} and the definition \eqref{E:elec} of $c_\Gamma$, we obtain
\begin{equation}\label{E:varform}
\frac{d}{dt}c_{\Gamma_t}=- \operatorname{Tr}(\dot{\Gamma}_{t,\bar{0}}\circ \Gamma_{t,\bar{1}}) \cdot c_{\Gamma_t}.
\end{equation}
Since $\Gamma_{t,\bar{0}} \circ \Gamma_{t,\bar{1}}=1$, we have
\[
0=\frac{d}{dt} \operatorname{Tr}(\Gamma_{t,\bar{0}} \circ \Gamma_{t,\bar{1}})= \operatorname{Tr}(\dot{\Gamma}_{t,\bar{0}} \circ \Gamma_{t,\bar{1}})+\operatorname{Tr}(\Gamma_{t,\bar{0}} \circ \dot{\Gamma}_{t,\bar{1}}).
\]
Hence,
\begin{equation}\label{E:trpm}
\operatorname{Tr}(\dot{\Gamma}_{t,\bar{0}} \circ \Gamma_{t,\bar{1}}) = - \operatorname{Tr}(\dot{\Gamma}_{t,\bar{1}} \circ \Gamma_{t,\bar{0}}).
\end{equation}
Combining \eqref{E:varform} with \eqref{E:trpm}, we obtain \eqref{E:varfor}.
\end{proof}
\subsection{The signature operator}
We now introduce the $\mathbb{Z}_2$-graded analogue of the $\mathbb{Z}$-graded finite dimensional odd signature operator of \cite[Section 5]{BK3}.
The {\em signature operators} $\B_{\overline{k}},k=0,1$ are defined by the formula
\begin{equation}\label{E:sigop}
\B_{\bar{k}} := \Gamma d_{\bar{k}} + d_{\overline{k+1}} \Gamma.
\end{equation}
Define
\begin{equation}\label{E:cjpm}
C^{\bar{k}}_+ := \Ker(d_{\overline{k+1}} \circ \Gamma) \cap C^{\bar{k}} = \Gamma(\Ker d_{\overline{k+1}} \cap C^{\overline{k+1}}), \qquad
C^{\bar{k}}_- := \Ker d_{\bar{k}} \cap C^{\bar{k}}.
\end{equation}
Let $\B^{\pm}_{\bar{k}}$ denote the restriction of $\B_{\bar{k}}$ to $C^{\bar{k}}_{\pm}$. Then one has
\begin{equation}\label{E:imbp}
\im \B^+_{\bar{k}} \subseteq \im (\Gamma \circ d_{\bar{k}}|_{C^{\bar{k}}}) \subseteq \Gamma(\Ker d_{\overline{k+1}}|_{C^{\overline{k+1}}}) \subseteq C^{\bar{k}}_+ ;
\end{equation}
\begin{equation}\label{E:imbm}
\im \B^-_{\bar{k}} \subseteq \im ( d_{\overline{k+1}} \circ\Gamma|_{C^{\bar{k}}}) \subseteq \im (d_{\overline{k+1}}|_{C^{\overline{k+1}}}) \subseteq C^{\bar{k}}_-.
\end{equation}
Hence,
$$
\B^+_{\bar{k}}=\Gamma \circ d_{\bar{k}} : C^{\bar{k}}_+ \to C^{\bar{k}}_+, \quad \B^-_{\bar{k}}=d_{\overline{k+1}} \circ \Gamma : C^{\bar{k}}_- \to C^{\bar{k}}_-.
$$
Note thar $\B_{\bar{k}}=\Gamma \circ \B_{\overline{k+1}} \circ \Gamma$.
The following lemma is the $\mathbb{Z}_2$-graded analogue of \cite[Lemma 5.2]{BK3}. The proof is a verbatim repetition of the proof of \cite[Lemma 5.2]{BK3}, we skip the proof.
\begin{lemma}\label{L:ac}
Suppose that the signature operators $\B_{\bar{k}},k=0,1$ are bijective. Then the complex $(C^{\bullet},d)$ is acyclic and
\begin{equation}\label{E:cbij}
C^{\bar{k}}=C^{\bar{k}}_+ \oplus C^{\bar{k}}_-.
\end{equation}
\end{lemma}
\subsection{Calculation of the refined torsion in case $\B$ is bijective}
In this subsection we compute the $\mathbb{Z}_2$-graded refined torsion in the case that $\B_{\bar{k}},k=0,1$ are bijective. Assume that the signature operators $\B_{\bar{k}},k=0,1$ are bijective. Then, by Lemma~\ref{L:ac}, the complex $(C^{\bullet},d)$ is acyclic. Note that $\Gamma \B^-_{\bar 0}\Gamma=\B^+_{\bar 1}$. Hence $\Det(\B^-_{\bar 0})=\Det(\B^+_{\bar 1})$. Then we have the following definition.
\begin{definition}
The {\em graded determinant} of the signature operator $\B_{\bar{0}}$ is defined by the formula
\begin{equation}\label{E:figraddet}
\Det_{\operatorname{gr}}(\B_{\bar{0}}):=\Det(\B^+_{\bar{0}})/\Det(-\B^-_{\bar{0}})=\Det(\B^+_{\bar{0}})/\Det(-\B^+_{\bar{1}}).
\end{equation}
\end{definition}
The following proposition is the $\mathbb{Z}_2$-graded analogue of \cite[Proposition 5.6]{BK3}. We modify the proof of \cite[Proposition 5.6]{BK3} slightly to fit our setting.
\begin{proposition}\label{P:rhogddet1}
Suppose that the signature operators $\B_{\bar k},k=0,1$ are invertible and, hence, the complex $(C^{\bullet},d)$ is acyclic. Then
\begin{equation}\label{E:rhogddet}
\rho_{\Gamma}=\Det_{\operatorname{gr}}(\B_{\bar{0}}).
\end{equation}
\end{proposition}
\begin{proof}
We choose the decomposition \eqref{E:decompbha} to be $C^{\bar{k}}=C^{\bar{k}}_- \oplus C^{\bar{k}}_+$ and define elements $c_{\bar{k}}$ as follows. Fix a nonzero element $a_{\bar{k}} \in \Det(C^{\bar{k}}_+)$ and set
\[
c_{\bar{k}}= \mu_{C^{\bar{k}}_-,C^{\bar{k}}_+}(\Gamma a_{\overline{k+1}} \otimes a_{\bar{k}}),
\]
where $\mu_{C^{\bar{k}}_-,C^{\bar{k}}_+}$ is the fusion isomorphism, cf. \eqref{E:fusio}, see also \cite[(2-5)]{BK3}.
Note that, by \eqref{E:fusio1},
\begin{align}
\Gamma c_{\bar{k}} & = \mu_{C^{\overline{k+1}}_+,C^{\overline{k+1}}_-}(a_{\overline{k+1}} \otimes \Gamma a_{\bar{k}})\nonumber \\
&= (-1)^{\dim C^{\overline{k+1}}_+ \cdot \dim C^{\overline{k+1}}_-} \cdot \mu_{C^{\overline{k+1}}_-,C^{\overline{k+1}}_+}(\Gamma a_{\bar{k}} \otimes a_{\overline{k+1}}) \nonumber \\
&= (-1)^{\dim C^{\overline{k+1}}_+ \cdot \dim C^{\overline{k+1}}_-} \cdot c_{\overline{k+1}}.
\end{align}
Thus, from \eqref{E:elec}, we obtain
\begin{align}\label{E:cgs}
c_{\Gamma} & = (-1)^{\mathcal{R}(C^{\bullet})} \cdot c_{\bar{0}} \otimes (\Gamma c_{\bar{0}})^{-1} \nonumber \\
& = (-1)^{\mathcal{R}(C^{\bullet})+ \dim C^{\bar{1}}_+ \cdot \dim C^{\bar{1}}_-} \cdot c_{\bar{0}} \otimes c_{\bar{1}}^{-1}.
\end{align}
Hence, by \eqref{E:elerho} and \eqref{E:phic}, to compute $\rho_{\Gamma}$ we need to compute the elements $h_{\bar{k}} \in \Det(H^{\bar{k}}) \cong \bf{k}$.
If $L$ is a complex line and $x,y \in L$ with $y \not=0$, we denote by $[x:y] \in \bf{k}$ the unique number such that $x=[x:y]y$. Then
\begin{align}\label{E:hrel}
h_{\bar{k}} & = [c_{\bar{k}}: \mu_{C^{\bar{k}}_-,C^{\bar{k}}_+}(d_{\overline{k+1}}a_{\overline{k+1}} \otimes a_{\bar{k}})] \nonumber \\
&= [\mu_{C^{\bar{k}}_-,C^{\bar{k}}_+}(\Gamma a_{\overline{k+1}} \otimes a_{\bar{k}}):\mu_{C^{\bar{k}}_-,C^{\bar{k}}_+}(d_{\overline{k+1}}a_{\overline{k+1}} \otimes a_{\bar{k}})] \nonumber \\
& = [\Gamma a_{\overline{k+1}} : d_{\overline{k+1}}a_{\overline{k+1}}] \nonumber \\
&= [a_{\overline{k+1}}: \Gamma d_{\overline{k+1}} a_{\overline{k+1}}]\nonumber \\
& = \Det (\Gamma d_{\overline{k+1}})^{-1}.
\end{align}
Combining \eqref{E:phic} and \eqref{E:hrel}, we obtain
\begin{align}\label{E:phccc}
\Phi_{C^{\bullet}} ( c_{\bar{0}} \otimes c_{\bar{1}}^{-1} ) & = (-1)^{\mathcal{N}(C^{\bullet})} \cdot \Det(\Gamma d_{\overline{0}})\cdot \Det (\Gamma d_{\overline{1}})^{-1} \nonumber \\
& = (-1)^{\mathcal{N}(C^{\bullet}) + \dim C^{\bar{1}}_+} \cdot \Det (\Gamma d_{\overline{0}})\cdot \Det (- \Gamma d_{\overline{1}})^{-1}.
\end{align}
Combining \eqref{E:elerho}, \eqref{E:figraddet}, \eqref{E:cgs} and \eqref{E:phccc}, we have
\[
\rho_{\Gamma}= \Phi_{C^{\bullet}}(c_{\Gamma}) = (-1)^{\mathcal{R}(C^{\bullet})+ \dim C^{\bar{1}}_+ \cdot \dim C^{\bar{1}}_- + \mathcal{N}(C^{\bullet}) + \dim C^{\bar{1}}_+} \cdot \Det_{\operatorname{gr}}(\B_{\bar{0}}).
\]
Hence, we are remaining to show that
\begin{equation}\label{E:fcbullet}
\mathcal{F}(C^{ \bullet}):=\mathcal{R}(C^{\bullet})+ \dim C^{\bar{1}}_+ \cdot \dim C^{\bar{0}}_+ + \mathcal{N}(C^{ {\bullet}}) + \dim C^{\bar{1}}_+ \equiv 0 \quad \text{mod} \ 2,
\end{equation}
here we use the fact that $\Gamma \B^+_{\bar{0}} \Gamma = \B^-_{\bar{1}}$.
By the fact that $\Gamma \B^+_{\bar{1}} \Gamma = \B^-_{\bar{0}}$ and \eqref{E:rcbullet}, we have
\begin{align}\label{E:rcbullet1}
\mathcal{R}(C^{\bullet}) &= \frac{1}{2}(\dim C^{\bar 0}_+ + \dim C^{\bar 0}_-)\cdot(\dim C^{\bar 0}_+ + \dim C^{\bar 0}_- +1)\nonumber \\
& = \frac{1}{2}(\dim C^{\bar 0}_+ + \dim C^{\bar 1}_+)\cdot (\dim C^{\bar 0}_+ + \dim C^{\bar 1}_+ +1) \nonumber \\
& = \frac{1}{2}(\dim C^{\bar 0}_+)^2 + \frac{1}{2}(\dim C^{\bar 1}_+)^2 + \dim C^{\bar 0}_+ \cdot \dim C^{\bar 1}_+ + \frac{1}{2} \dim C^{\bar 0}_+ + \frac{1}{2} \dim C^{\bar 1}_+.
\end{align}
Recall that, \eqref{E:ncbullet},
\begin{align}\label{E:ncbullet1}
\mathcal{N}(C^{\bullet}) & = \frac{1}{2}\dim C^{\bar 0}_+(C^{\bar 0}_+ +1) + \frac{1}{2}\dim C^{\bar 1}_+(C^{\bar 1}_+ -1) \nonumber \\ & = \frac{1}{2}(\dim C^{\bar 0}_+)^2 + \frac{1}{2} \dim C^{\bar 0}_+ + \frac{1}{2}(\dim C^{\bar 1}_+)^2 - \frac{1}{2} \dim C^{\bar 1}_+.
\end{align}
Combining \eqref{E:fcbullet}, \eqref{E:rcbullet1} and \eqref{E:ncbullet1}, we have
\begin{equation}\label{E:fcbullet1}
\mathcal{F}(C^{ \bullet})=(\dim C^{\bar 0}_+)^2 + (\dim C^{\bar 1}_+)^2 + 2\dim C^{\bar 0}_+ \cdot \dim C^{\bar 1}_+ + \dim C^{\bar 0}_+ + \dim C^{\bar 1}_+.
\end{equation}
By \eqref{E:fcbullet1} and the fact that for any $x \in \mathbb{Z}, x(x+1) \equiv 0\, (\text{mod}\, 2)$, we obtain
\[
\mathcal{F}(C^{ \bullet})=0.
\]
\end{proof}
\subsection{Calculation of the refined torsion in case $\B$ is not bijective}
In this subsection we compute the $\mathbb{Z}_2$-graded refined torsion in the case that $\B_{\bar{k}},k=0,1$ are not bijective. Note that the operator $\B^2_{\bar{k}}$ maps $C^{\bar{k}}$ into itself. For an arbitrary interval $\mathcal{I}$, denote by $C^{\bar{k}}_{\mathcal{I}} \subset C^{\bar{k}}$ the linear span of the generalized eigenvectors of the restriction of $\B^2_{\bar{k}}$ to $C^{\bar{k}}$, corresponding to eigenvalue $\lambda$ with $\lambda \in \mathcal{I}$. Since both $\Gamma$ and $d_{\bar{k}}$ commute with $\B_{\bar{k}}$ (and, hence, with $\B^2_{\bar{k}}$), $\Gamma (C^{\bar{k}}_{\mathcal{I}}) \subset C^{\overline{k+1}}_{\mathcal{I}}$ and $d_{\bar{k}}(C^{\bar{k}}_{\mathcal{I}}) \subset C^{\overline{k+1}}_{\mathcal{I}}$. Hence, we obtain a subcomplex $C^{\bullet}_{\mathcal{I}}$ of $C^{\bullet}$ and the restriction $\Gamma_{\mathcal{I}}$ of $\Gamma$ to $C^{\bullet}_{\mathcal{I}}$ is a chirality operator for $C^{\bullet}_{\mathcal{I}}$. We denote by $H^{\bullet}_{\mathcal{I}}(d)$ the cohomology of the complex $(C^\bullet_{\mathcal{I}},d_{\mathcal{I}})$. Denote by $d_{\bar{k},\mathcal{I}}$ and $\B_{\bar{k},\mathcal{I}}$ the restrictions of $d_{\bar{k}}$ and $\B_{\bar{k}}$ to $C^{\bar{k}}_{\mathcal{I}}$. Then $\B_{\bar{k},\mathcal{I}} = \Gamma_{\mathcal{I}} d_{\bar{k},\mathcal{I}}+d_{\overline{k+1},\mathcal{I}}\Gamma_{\mathcal{I}}$.
\begin{lemma}
If $0 \notin \mathcal{I}$, then the complex $(C^{\bullet}_{\mathcal{I}},d_{\mathcal{I}})$ is acyclic.
\end{lemma}
\begin{proof}
If, for $k=0,1$, $x \in \Ker d_{\bar{k},\mathcal{I}}$, then $\B^2_{\bar{k},\mathcal{I}}x = (d_{\overline{k+1}}\Gamma)^2x \in \im d_{\overline{k+1},\mathcal{I}} \subset \Ker d_{\overline{k},\mathcal{I}}$. Since the operators $\B^2_{\bar{k},\mathcal{I}}:C^{\bar{k}}_{\mathcal{I}} \to C^{\bar{k}}_{\mathcal{I}},k=0,1$ are invertible, we conclude that $\Ker d_{\overline{k},\mathcal{I}} = \im d_{\overline{k+1},\mathcal{I}}$.
\end{proof}
For each $\lambda \ge 0$, $C^{\bullet}=C^{\bullet}_{[0,\lambda]} \oplus C^{\bullet}_{(\lambda,\infty)}$ and $H^{\bullet}(d)=0$ whereas $H^{\bullet}_{[0,\lambda]}(d) \cong H^{\bullet}(d)$. Hence there are canonical isomorphisms
\[
\Phi_\lambda : \Det(H^{\bullet}_{(\lambda,\infty)}(d)) \to \mathbb{C}, \qquad \Psi_{\lambda}: \Det(H^{\bullet}_{[0,\lambda]}(d)) \to \Det(H^{\bullet}(d)).
\]
In the sequel, we will write $t$ for $\Phi_\lambda(t) \in \mathbb{C}$.
The following proposition is $\mathbb{Z}_2$-graded analogue of \cite[Proposition 5.10]{BK3}.
\begin{proposition}\label{P:indedec}
Let $(C^{\bullet},d)$ be a $\mathbb{Z}_2$-graded complex of finite dimensional $\bf{k}$-vector spaces and let $\Gamma$ be a chirality operator on $C^{\bullet}$. Then, for each $\lambda \ge 0$,
\[
\rho_{\Gamma}= \Det_{\operatorname{gr}}(\B_{(\lambda, \infty)}^{\bar{0}}) \cdot \rho_{\Gamma_{[0,\lambda]}},
\]
where we view $\rho_{\Gamma_{[0,\lambda]}}$ as an element of $\Det(H^{\bullet}(d))$ via the canonical isomorphism $\Psi_{\lambda}: \Det(H^{\bullet}_{[0,\lambda]}(d)) \to \Det(H^{\bullet}(d))$.
\end{proposition}
\begin{proof}
Recall the natural isomorphism
\begin{equation}\label{E:isofu}
\Det(H^{\bar{k}}_{[0,\lambda]}(d) \otimes H^{\bar{k}}_{(\lambda,\infty)}(d)) \cong \Det(H^{\bar{k}}_{[0,\lambda]}(d) \oplus H^{\bar{k}}_{(\lambda,\infty)}(d))=\Det(H^{\bar{k}}(d))
\end{equation}
From Definition~\ref{D:elerho1}, Proposition~\ref{P:rhogddet1} and \eqref{E:isofu}, we obtain the result.
\end{proof}
\section{Graded determinant of the twisted odd signature operator}\label{S:gdtoso}
In this section we define the graded determinant of the odd signature operator, cf. \cite{APS2,Gi}, twisted by a flux form $H$, of a flat vector bundle $E$ over a closed oriented odd dimensional manifold $M$. We use this graded determinant to define an element of the determinant line of the twisted de Rham cohomology of the vector bundle $E$. We also study the relationship between this graded determinant and the $\eta$-invariant of the twisted odd signature operator.
\subsection{The twisted odd signature operator}
Let $M$ be a closed oriented smooth manifold of odd dimension $m=2r-1$ and let $E$ be a complex vector bundle over $M$ endowed with a flat connection $\nabla$. We denote by $\Omega^p(M,E)$ the space of $p$-forms with values in the flat bundle $E$, i.e., $\Omega^p(M,E)=\Gamma(\wedge^p(T^*M)_{\mathbb{R}}\otimes E)$ and by
\[
\nabla : \Omega^\bullet(M,E) \to \Omega^{\bullet+1}(M,E)
\]
the covariant differential induced by the flat connection on $E$. Fix a Riemannian metric $g^M$ on $M$ and let $\star : \Omega^\bullet(M,E) \to \Omega^{m-\bullet}(M,E)$ denote the Hodge $\star$-operator. We choose a Hermitian metric $h^E$ so that together with the Riemannian metric $g^M$ we can define a scalar product $<\cdot,\cdot>_M$ on $\Omega^{ \bullet}(M,E)$. Define the chirality operator $\Gamma = \Gamma(g^M):\Omega^\bullet(M,E) \to \Omega^\bullet(M,E)$ by the formula, cf. \cite[(7-1)]{BK3},
\begin{equation}\label{E:chirality}
\Gamma \omega := i^r(-1)^{\frac{q(q+1)}{2}} \star \omega, \quad \omega \in \Omega^q(M,E),
\end{equation}
where $r$ given as above by $r=\frac{m+1}{2}$. The numerical factor in \eqref{E:chirality} has been chosen so that $\Gamma^2=\operatorname{Id}$, cf. Proposition 3.58 of \cite{BGV}.
Assume that $H$ is an odd degree closed differential form on $M$. Let $\Omega^{\bar{0}}(M,E):=\Omega^{\even}(M,E)$, $\Omega^{\bar{1}}(M,E):=\Omega^{\odd}(M,E)$ and $\nabla^H:=\nabla+H \wedge \cdot$. We assume that $H$ does not contain a 1-form component, which can be absorbed in the flat connection $\nabla$.
\begin{definition}
The twisted odd signature operator is the operator
\begin{equation}\label{E:toso}
\B^H=\B(\nabla^H,g^M) := \Gamma \nabla^H + \nabla^H \Gamma : \Omega^{ \bullet}(M,E) \to \Omega^{ \bullet}(M,E).
\end{equation}
We denote by $\B^H_{\bar k}$ the restriction of $\B^H$ to the space $\Omega^{\bar k}(M,E),k=0,1$.
\end{definition}
\subsection{$\zeta$-function and $\zeta$-regularized determinant}
In this subsection we briefly recall some definitions of $\zeta$-regularized determinants of non self-adjoint elliptic operators. See \cite[Section 6]{BK3} for more details. Let $D:C^\infty(M,E) \to C^\infty(M,E)$ be an elliptic differential operator of order $n \ge 1$. Assume that $\theta$ is an Agmon angle, cf. for example, Definition 6.3 of \cite{BK3}. Let $\Pi: L^2(M,E) \to L^2(M,E)$ denote the spectral projection of $D$ corresponding to all nonzero eigenvalues of $D$. The $\zeta$-function $\zeta_\theta(s,D)$ of $D$ is defined as follows
\begin{equation}\label{E:zetdef}
\zeta_\theta(s,D)= \operatorname{Tr} \Pi D_\theta^{-s}, \quad \operatorname{Re} s > \frac{\dim M}{n}.
\end{equation} It was shown by Seeley \cite{Se} (See also \cite{Sh}) that $\zeta_\theta(s,D)$ has a meromorphic extension to the whole complex plane and that $0$ is a regular value of $\zeta_\theta(s,D)$.
\begin{definition}
The $\zeta$-regularized determinant of $D$ is defined by the formula
\[
\Det'_\theta(D):= \exp \Big(\, -\frac{d}{ds}\Big|_{s=0}\zeta_\theta(s,D)\,\Big).
\]
\end{definition}
We denote by
\[
\LDet'_\theta(D)= - \frac{d}{ds}\Big|_{s=0}\zeta_\theta(s,D).
\]
Let $Q$ be a $0$-th order pseudo-differential projection, ie. a $0$-th order pseudo-differential operator satisfying $Q^2=Q$. We set
\begin{equation}\label{E:zetadefpro}
\zeta_\theta(s,Q,D)= \operatorname{Tr} Q \Pi D_\theta^{-s}, \quad \operatorname{Re} s > \frac{\dim M}{n}.
\end{equation}
The function $\zeta_\theta(s,Q,D)$ also has a meromorphic extension to the whole complex plane and, by Wodzicki, \cite[Section 7]{Wo1}, it is regular at $0$.
\begin{definition}
Suppose that $Q$ is a $0$-th order pseudo-differential projection commuting with $D$. Then $V:=\operatorname{Im} Q$ is $D$ invariant subspace of $C^\infty(M,E)$. The $\zeta$-regularized determinant of the restriction $D|_V$ of $D$ to $V$ is defined by the formula
\[
\Det'_\theta(D|_V):= e^{\LDet'_\theta(D|_V)},
\]
where
\begin{equation}\label{E:ldetv}
\LDet'_\theta(D|_V)=-\frac{d}{ds}\Big|_{s=0} \zeta_\theta(s,Q,D).
\end{equation}
\end{definition}
\begin{remark}
The prime in $\Det'_\theta$ and $\LDet'_\theta$ indicates that we ignore the zero eigenvalues of the operator in the definition of the regularized determinant. If the operator is invertible we usually omit the prime and write $\Det_\theta$ and $\LDet_\theta$ instead.
\end{remark}
\subsection{The graded determinant of the twisted odd signature operator}\label{SS:gdto}
Note that for each $k=0,1$, the operator $(\B^H)^2$ maps $\Omega^{\bar k}(M,E)$ to itself. Suppose that $\mathcal{I}$ is an interval of the form $[0,\lambda],(\lambda, \mu],$ or $(\lambda,\infty)$ $(\mu > \lambda \ge 0)$. Denote by $\Pi_{(\B^H)^2,\mathcal{I}}$ the spectral projection of $(\B^H)^2$ corresponding to the set of eigenvalues, whose absolute values lie in $\mathcal{I}$. set
\[
\Omega^{ \bullet}_{\mathcal{I}}(M,E) := \Pi_{(\B^H)^2,\mathcal{I}}(\Omega^{ \bullet}(M,E)) \subset \Omega^{ \bullet}(M,E).
\]
If the interval $\mathcal{I}$ is bounded, then, cf. Section 6.10 of \cite{BK3}, the space $ \Omega^{ \bullet}_{\mathcal{I}}(M,E)$ is finite dimensional.
For each $k=0,1$, set
\begin{equation}\label{E:split}
\begin{array}{l}
\Omega^{\bar k}_{+,\mathcal{I}}(M,E) := \Ker(\nabla^H \Gamma) \cap \Omega^{\bar k}_{\mathcal{I}}(M,E) = \big(\Gamma(\Ker \nabla^H) \big) \cap \Omega^{\bar k}_{\mathcal{I}}(M,E);\\
\\
\Omega^{\bar k}_{-,\mathcal{I}}(M,E) := \Ker(\Gamma \nabla^H ) \cap \Omega^{\bar k}_{\mathcal{I}}(M,E) = \Ker \nabla^H \cap \Omega^{\bar k}_{\mathcal{I}}(M,E).
\end{array}
\end{equation}
Then
\begin{equation}\label{E:splitt1}
\Omega^{\bar k}_{\mathcal{I}}(M,E) = \Omega^{\bar k}_{+,\mathcal{I}}(M,E) \oplus \Omega^{\bar k}_{-,\mathcal{I}}(M,E) \quad \text{if} \quad 0 \notin \mathcal{I}.
\end{equation}
We consider the decomposition \eqref{E:splitt1} as a grading of the space $\Omega^{\bar k}_{\mathcal{I}}(M,E)$, and refer to $\Omega^{\bar k}_{+,\mathcal{I}}(M,E)$ and $\Omega^{\bar k}_{-,\mathcal{I}}(M,E)$ as the positive and negative subspaces of $\Omega^{\bar k}_{\mathcal{I}}(M,E)$. Denote by $\B^{H}_\mathcal{I}$ and $\B^{H}_{\bar{k},\mathcal{I}}$ the restrictions of $\B^H$ to the subspaces $\Omega^{ \bullet}_{\mathcal{I}}(M,E)$ and $\Omega^{\bar k}_{\mathcal{I}}(M,E)$ respectively. Then $\B^{H}_{\bar{k},\mathcal{I}}$ maps $\Omega^{\bar k}_{\pm,\mathcal{I}}(M,E)$ to itself. Let $\B^{H,\pm}_{\bar{k},\mathcal{I}}$ denote the restriction of $\B^{H}_{\bar{k},\mathcal{I}}$ to the subspace $\Omega^{\bar k}_{\pm,\mathcal{I}}(M,E)$. Clearly, the operator $\B^{H,\pm}_{\bar{k},\mathcal{I}}$ are bijective whenever $0 \notin \mathcal{I}$. Note that $\Gamma \B^{H,-}_{\bar{0},\mathcal{I}}\Gamma=\B^{H,+}_{\bar{1},\mathcal{I}}$. Hence $\Det_\theta(\B^{H,-}_{\bar{0},\mathcal{I}})=\Det_\theta(\B^{H,+}_{\bar{1},\mathcal{I}})$. Then we have the following definition.
\begin{definition}\label{D:graddet01}
Suppose that $0 \notin \mathcal{I}$. The graded determinant of the operator $\B^{H,\mathcal{I}}_{\bar 0}$ is defined by
\begin{equation}\label{E:graddet}
\Det_{\gr,\theta}(\B^{H}_{\bar{0},\mathcal{I}}) \, := \, \frac{\Det_\theta(\B^{H,+}_{\bar{0},\mathcal{I}})}{\Det_\theta(-\B^{H,-}_{\bar{0},\mathcal{I}})} \, = \, \frac{\Det_\theta(\B^{H,+}_{\bar{0},\mathcal{I}})}{\Det_\theta(-\B^{H,+}_{\bar{1},\mathcal{I}})} \in \mathbb{C} \backslash \{ 0 \},
\end{equation}
where $\Det_\theta$ denotes the $\zeta$-regularized determinant associated to the Agmon angle $\theta \in (-\pi,0)$, cf. for example, Section 6 of \cite{BK3}.
\end{definition}
We define, cf. \eqref{E:ldetv},
\begin{equation}\label{E:ldetgr}
\LDet_{\gr,\theta}(\B^{H}_{\bar{0},\mathcal{I}})=\LDet_\theta(\B^{H,+}_{\bar{0},\mathcal{I}})-\LDet_\theta(-\B^{H,-}_{\bar{0},\mathcal{I}})=\LDet_\theta(\B^{H,+}_{\bar{0},\mathcal{I}})-\LDet_\theta(-\B^{H,+}_{\bar{1},\mathcal{I}}).
\end{equation}
It follows from formula (6-17) of \cite{BK3} that \eqref{E:graddet} is independent of the choice of $\theta \in (-\pi, 0)$.
\subsection{The canonical element of the determinant line} It is not difficult to check that $(\nabla^H)^2=0$.
Clearly, $\nabla^H : \Omega^{\bar k}(M,E) \to \Omega^{\overline{k+1}}(M,E)$ and $\Gamma : \Omega^{\bar k}(M,E) \to \Omega^{\overline{k+1}}(M,E)$. Hence we can consider the following twisted de Rham complex with chirality operator $\Gamma$:
\begin{equation}\label{E:ztcomplex}
\big(\, \Omega^{ \bullet}(M,E), \nabla^H \, \big): \cdots \stackrel{\nabla^H}{\longrightarrow} \Omega^{\bar 0}(M,E ) \stackrel{\nabla^H}{\longrightarrow} \Omega^{\bar 1}(M,E ) \stackrel{\nabla^H}{\longrightarrow} \Omega^{\bar 0}(M,E )\stackrel{\nabla^H}{\longrightarrow} \cdots.
\end{equation}
We define the {\em twisted de Rham cohomology groups of $(\Omega^{\bullet}(M,E),\nabla^H)$} as
\[
H^{\bar{k}}(M,E,H) \, \equiv \, H^{\bar{k}}(\nabla^H) \, := \, \frac{\Ker(\nabla^H : \Omega^{\bar{k}}(M,E) \to \Omega^{\overline{k+1}}(M,E))}{\im (\nabla^H : \Omega^{\overline{k+1}}(M,E) \to \Omega^{\bar{k}}(M,E))}, \quad k=0,1.
\]
The groups $H^{\bar{k}}(M,E,H),k=0,1$ are independent of the choice of the Riemannian metric on $M$ or the Hermitian metric on $E$. Suppose that $H$ is repalced by $H'=H-dB$ for some $B \in \Omega^{\bar{0}}(M)$, there is an isomorphism $\varepsilon_B:=e^B \wedge \cdot : \Omega^{ \bullet}(M,E) \to \Omega^{ \bullet}(M,E)$ satisfying
\[
\varepsilon_B \circ \nabla^H = \nabla^{H'} \circ \varepsilon_B.
\]
Therefore $\varepsilon_B$ induces an isomorphism on the twisted de Rham cohomology, also denote by $\varepsilon_B$,
\begin{equation}\label{E:epsiso}
\varepsilon_B: H^\bullet(M,E,H) \to H^\bullet(M,E,H').
\end{equation}
Denote by $(\nabla^{H}_{\bar k})^*$ the adjoint of $\nabla^H_{\bar k}$ with respect to the scalar product $<\cdot,\cdot>_M$. Then the Laplacians
\[
\Delta_{\bar k}=\Delta^H_{\bar k}:=(\nabla_{\bar k}^{H})^*\nabla_{\bar k}^H + \nabla_{\overline{k+1}}^H(\nabla_{\overline{k+1}}^{H})^*, \quad k=0,1
\]
are elliptic operators and therefore the complex \eqref{E:ztcomplex} is elliptic. By Hodge theory, we have the isomorphism $\Ker \Delta_{\bar k} \cong H^{\bar{k}}(M,E,H),k=0,1$. For more details of the twisted de Rham cohomology, cf. for example \cite{MW}.
Since $\nabla^H$ commutes with $\B^H$, the subspace $\Omega^{ \bullet}_{\mathcal{I}}(M,E)$ is a subcomplex of the twisted de Rham complex $(\Omega^{\bullet}(M,E), \nabla^H)$. Clearly, for each $\lambda \ge 0$, the complex $\Omega^{ \bullet}_{(\lambda, \infty)}(M,E)$ is acyclic. Since
\begin{equation}\label{E:omespl}
\Omega^{ \bullet}(M,E) = \Omega^{ \bullet}_{[0,\lambda]}(M,E) \oplus \Omega^{ \bullet}_{(\lambda,\infty)}(M,E),
\end{equation}
the cohomology $H^{ \bullet}_{[0,\lambda]}(\nabla^H) \equiv H^{ \bullet}_{[0,\lambda]}(M,E,H)$ of the complex $\big( \Omega^{ \bullet}(M,E), \nabla^H \big)$ is naturally isomorphic to the cohomology $H^{\bullet}(M,E,H)$. Let $\Gamma_{\mathcal{I}}$ denote the restriction of $\Gamma$ to $\Omega^{ \bullet}_{\mathcal{I}}(M,E)$. For each $\lambda \ge 0$, let
\begin{equation}\label{E:canonele}
\rho_{\Gamma_{[0,\lambda]}} = \rho_{\Gamma_{[0,\lambda]}}(\nabla^H,g^M) \in \Det\big(H^{ \bullet}_{[0,\lambda]}(M,E,H)\big)
\end{equation}
denote the refined torsion of the twisted finite dimensional complex $\big( \Omega^{ \bullet}_{[0,\lambda]}(M,E), \nabla^H \big)$ corresponding to the chirality operator $\Gamma_{[0,\lambda]}$, cf. Definition~\ref{D:elerho1}.
We view $\rho_{\Gamma_{[0,\lambda]}}$ as an element of $\Det\big(H^\bullet(M,E,H)\big)$ via the canonical isomorphism between $H^\bullet(M,E,H)$ and $H^{\bullet}_{[0,\lambda]}(M,E,H)$.
\begin{proposition}
Assume that $\theta \in (-\pi,0)$ is an Agmon angle for the operator $\B^H_{\bar 0}$. Then the element
\begin{equation}\label{E:rhoh}
\rho_{H}=\rho(\nabla^H,g^M):= \Det_{\gr,\theta}(\B^{H}_{\bar{0},(\lambda,\infty)}) \cdot \rho_{\Gamma_{[0,\lambda]}} \in \Det\big(H^\bullet(M,E,H)\big)
\end{equation}
is independent of the choice $\lambda \ge 0$. Further, $\rho_H$ is independent of the choice of the Agmon angle $\theta \in (-\pi,0)$ of $\B^H_{\bar 0}$.
\end{proposition}
\begin{proof}
Clearly, for $0 \le \lambda \le \mu$, we have
\begin{equation}\label{E:indlam}
\Det_{\operatorname{gr}}(\B_{\bar{0},(\lambda, \infty)}^{H})=\Det_{\operatorname{gr}}(\B_{\bar{0},(\lambda, \mu]}^{H}) \cdot \Det_{\operatorname{gr}}(\B_{\bar{0},(\mu, \infty)}^{H}).
\end{equation}
From Proposition~\ref{P:indedec}, \eqref{E:indlam} and (6-17) of \cite{BK3}, we obtain the result.
\end{proof}
\subsection{The $\eta$-invariant}
In this subsection we recall the definition of the $\eta$-invariant of a non-self-adjoint elliptic operator $D$, cf. \cite{Gi}, \cite[Subsection 6.15]{BK3}.
\begin{definition}
Let $D:C^\infty(M,E) \to C^\infty(M,E)$ be an elliptic differential operator of order $n \ge 1$ whose leading symbol is self-adjoint with respect to some given Hermitian metric on $E$. Assume that $\theta$ is an Agmon angle for $D$, cf. Definition 6.3 of \cite{BK3}. Let $\Pi_>$ (resp. $\Pi<$) be a pseudo-differential projection whose image contains the span of all generalized eigenvectors of $D$ corresponding to eigenvalues $\lambda$ with $\operatorname{Re} >0$ (resp. with $\operatorname{Re} <0$) and whose kernel contains the span of all generalized eigenvectors of $D$ corresponding to eigenvalues $\lambda$ with $\operatorname{Re} \le 0$ (resp. with $\operatorname{Re} \ge 0$). We define the $\eta$-function of $D$ by the formula
\[
\eta_\theta(s,D)=\zeta_\theta(s, \Pi_>,D)- \zeta_\theta(s, \Pi_<,-D).
\]
\end{definition} Note that, by the above definition, the purely imaginary eigenvalues of $D$ do not contribute to $\eta_\theta(s,D)$.
It was shown by Gilkey, \cite{Gi}, that $\eta_\theta(s,D)$ has a meromorphic extension to the whole complex plane $\mathbb{C}$ with isolated simple poles, and that it is regular at $0$. Moreover, the number $\eta_\theta(0,D)$ is independent of the Agmon angle $\theta$.
Since the leading symbol of $D$ is self-adjoint, the angles $\pm \pi/2$ are principal angles for $D$ cf. \cite[Definition 6.2]{BK3}. Hence, there are at most finitely many eigenvalues of $D$ on the imaginary axis. Let $m_+(D)$ (resp. $m_-(D)$) denote the number of eigenvalues of $D$, counted with their algebraic multiplicities, on the positive (resp. negative) part of the imaginary axis. Let $m_0(D)$ denote the algebraic multiplicity of $0$ as an eigenvalue of $D$.
\begin{definition}\label{D:etainv}
The $\eta$-invariant $\eta(D)$ of $D$ is defined by the formula
\begin{equation}\label{E:etainv11}
\eta(D)=\frac{\eta_\theta(0,D)+m_+(D)-m_-(D)+m_0(D)}{2}.
\end{equation}
\end{definition}
Since $\eta_\theta(0,D)$ is indenpendent of the choice of the Agmon angle $\theta$ for $D$, cf. \cite{Gi}, so is $\eta(D)$. Note that the defintion of $\eta(D)$ is slightly different from the one proposed by Gilkey in \cite{Gi}. See \cite[Remark 2.5]{BK3}.
Denote by $\eta(\nabla^H)=\eta(\B^H_{\bar 0})$ the $\eta$-invariant of the restriction $\B^H_{\bar 0}$ of the twisted odd signature operator $\B^H$ to $\Omega^{\bar 0}(M,E)$.
\subsection{Relationship with the $\eta$-invariant}
In this subsection we study the relationship between \eqref{E:graddet} and the $\eta$-invariant of $\B_{\bar{0},(\lambda, \infty)}^{H}$.
To simplify the notation set
\begin{equation}\label{E:etano}
\eta_\lambda(\nabla^H):=\eta(\B_{\bar{0},(\lambda, \infty)}^{H})
\end{equation}
and
\begin{equation}\label{E:xino}
\xi_\lambda = \xi_\lambda(\nabla^H,g^M,\theta) = \frac{1}{2}\Big(\, \LDet_{2\theta}\big((\B_{\bar{0},(\lambda, \infty)}^{H,+})^2\big) - \LDet_{2\theta}\big((\B_{\bar{1},(\lambda, \infty)}^{H,+})^2\big) \, \Big).
\end{equation}
Let $P^\pm_{\bar{k},\mathcal{I}}, k=0,1$ be the orthogonal projection onto the closure of the subspace $\Omega^{\bar k}_{\pm,\mathcal{I}}(M,E)$. Set
\begin{equation}\label{E:dno}
d_{\bar{k},\lambda}^{\mp}:= \operatorname{rank}(\operatorname{Id}-P^{\pm}_{\bar{k},[0,\lambda]}) = \dim \Omega^{\bar k}_{\mp,\mathcal{I}}(M,E), \quad k=0,1.
\end{equation}
If $\mathcal{I} \subset \mathbb{R}$ we denote by $L_{\mathcal{I}}$ the solid angle
\[
L_{\mathcal{I}}=\{\rho e^{i \theta}: 0 < \rho < \infty, \theta \in \mathcal{I} \}.
\]
\begin{proposition}
Let $\nabla$ be a flat connection on a vector bundle $E$ over a closed Riemannian manifold $(M,g^M)$ of odd dimension $m=2r-1$ and $H$ is an odd-degree closed differential form, other than one form, on $M$. Assume $\theta \in (-\pi/2,0)$ is an Agmon angle for the twisted odd signature operator $\B_{\bar{0},(\lambda, \infty)}^{H}$ such that there are no eigenvalues of $\B^H$ in the solid angles $L_{(-\pi/2,\theta]}$ and $L_{(\pi/2,\theta + \pi]}$. Then, for every $\lambda \ge 0$, cf. \eqref{E:ldetgr},
\begin{equation}\label{E:ldetetarell}
\LDet_{\gr,\theta}(\B^{H}_{\bar{0},(\lambda,\infty)})=\xi_\lambda - i \pi \eta_\lambda(\nabla^H) -\frac{i \pi}{2}\sum_{k=0,1}(-1)^k d^-_{\bar{k},\lambda}.
\end{equation}
\end{proposition}
\begin{proof}
From Definition \ref{D:etainv} of the $\eta$-invariant it follows that
\begin{equation}\label{E:eteq1}
\eta(-\B_{\bar{1},(\lambda, \infty)}^{H,+}) = - \eta(\B_{\bar{1},(\lambda, \infty)}^{H,+}).
\end{equation}
By the fact that $\Gamma \B_{\bar{1},(\lambda, \infty)}^{H,+} \Gamma = \B_{\bar{0},(\lambda, \infty)}^{H,-}$,
we have
\begin{equation}\label{E:eteq2}
\eta(\B_{\bar{1},(\lambda, \infty)}^{H,+}) = \eta(\B_{\bar{0},(\lambda, \infty)}^{H,-}).
\end{equation}
Combining \eqref{E:etano}, \eqref{E:eteq1} with \eqref{E:eteq2}, we have
\begin{equation}\label{E:etarelat}
\eta(\B_{\bar{0},(\lambda, \infty)}^{H,+})-\eta(-\B_{\bar{1},(\lambda, \infty)}^{H,+})=\eta(\B_{\bar{0},(\lambda, \infty)}^{H,+})+\eta(\B_{\bar{0},(\lambda, \infty)}^{H,-})=\eta_\lambda(\nabla^H).
\end{equation}
By \cite[(4.34)]{BK1}, for $k=0,1$, we have
\begin{equation}\label{E:ldetpmrel}
\LDet_\theta(\pm \B^{H,+}_{\bar{k},\mathcal{I}}) \, = \, \frac{1}{2}\LDet_{2\theta}\big((\B_{\bar{k},(\lambda, \infty)}^{H,+})^2\big)- i \pi \Big(\, \eta(\pm \B^{H,+}_{\bar{k},\mathcal{I}})-\frac{\zeta_{2\theta}(0, \big(\B_{\bar{k},(\lambda, \infty)}^{H,+}\big)^2)}{2} \,\Big)
\end{equation}
and, by \cite[(6-6)]{BK3} and \eqref{E:dno}, we have
\begin{equation}\label{E:zettw}
\zeta_{2\theta}(0, \big(\B_{\bar{0},(\lambda, \infty)}^{H,+}\big)^2)-\zeta_{2\theta}(0, \big(\B_{\bar{1},(\lambda, \infty)}^{H,+}\big)^2)= -\sum_{k=0,1}(-1)^k d^-_{\bar{k},\lambda}.
\end{equation}
Combining \eqref{E:xino}, \eqref{E:etarelat}, \eqref{E:ldetpmrel} with \eqref{E:zettw}, we obtain the result.
\end{proof}
\section{Metric anomaly and the definition of the refined analytic torsion twisted by a flux form}\label{S:madrat}
In this section we study the metric dependence of the element $\rho_H=\rho(\nabla^H,g^M)$ defined in \eqref{E:rhoh}. We then use this element to construct the {\em twisted refined analytic torsion}, which is a canonical element of the determinant line $\Det(H^\bullet(M,E,H))$. We also show that the twisted refined analytic torsion is independent of the metric $g^M$ and the representative $H$ in the cohomology class $[H]$.
\subsection{Relationship between $\rho_H(t)$ and the $\eta$-invariant}
Suppose that $g^M_t, t \in \mathbb{R}$, is a smooth family of Riemannian metrics on $M$. Let
\[
\rho_H(t)=\rho(\nabla^H,g^M_t) \in \Det\big( H^\bullet(M,E,H) \big)
\]
be the canonical element defined in \eqref{E:rhoh}.
Let $\Gamma_t$ denote the chirality operator corresponding to the metric $g^M_t$, cf. \eqref{E:chirality}, and let $\B^H(t)=\B(\nabla^H,g_t^M)$ denote the twisted odd signature operator corresponding to the Riemannian metric $g^M_t$ and let $\B^{H,+}(t)$ denote the restriction of $\B^H(t)=\B(\nabla^H,g_t^M)$ to $\Omega^{\bullet}_+(M,E)$.
Fix $t_0 \in \mathbb{R}$ and choose $\lambda \ge 0$ such that there are no eigenvalues of $(\B^{H}(t_0))^2$ of absolute value $\lambda$. Further, assume that $\lambda$ is big enough so that the real parts of eigenvalues of $(\B^{H}_{(\lambda,\infty)}(t_0))^2$ are all greater than $0$. Then there exists $\delta > 0$ small enough such that the same holds for the spectrum of $(\B^{H}(t))^2$ for $t \in (t_0-\delta,t_0+\delta)$. In particular, $d^-_{\bar{k},\lambda}$, cf. \eqref{E:dno}, is independent of $t \in (t_0-\delta,t_0+\delta)$. Set
\[
\eta_\lambda(\nabla^H,t):=\eta(\B_{\bar{0},(\lambda, \infty)}^{H}(t)), \quad
\xi_\lambda(t,\theta) = \xi_\lambda(\nabla^H,g_t^M,\theta).
\]
By definition \eqref{E:rhoh},
\[
\rho_H(t)= \Det_{\gr,\theta}(\B^{H}_{\bar{0},(\lambda,\infty)}(t)) \cdot \rho_{\Gamma_{t,[0,\lambda]}}.
\]
Assume that $\theta_0 \in (\pi/2,0)$ is an Agmon angle for $\B^{H}(t_0)$ such that there are no eigenvalues of $\B^{H}(t_0)$ in $L_{(-\pi/2,\theta_0]}$ and $L_{(-\pi/2,\theta_0 + \pi)}$. Choose $\delta > 0$ so that for every $t \in (t_0-\delta,t_0+\delta)$ both $\theta_0$ and $\theta_0+\pi$ are Agmon angles of $\B^{H}(t_0)$. For $t \not= t_0$, it might happen that there are eigenvalues of $\B_{\bar{k},(\lambda, \infty)}^{H}(t)$ in $L_{(-\pi/2,\theta_0]}$ and $L_{(-\pi/2,\theta_0 + \pi)}$. Hence, \eqref{E:ldetetarell} is not necessarily true. However, from the independence of the Agmon angle of the $\zeta$-function, cf. \cite[(6-16)]{BK3}, and \eqref{E:xino}, we conclude that for every angle $\theta \in (-\pi/2,0)$, so that $\theta$ and $\theta + \pi$ are Agmon angles for $\B_{\bar{k},(\lambda, \infty)}^{H}(t)$,
\[
\xi_{\lambda}(t,\theta) \equiv \xi_{\lambda}(t,\theta_0) \quad \text{mod} \, \pi i.
\]
Hence, from \eqref{E:ldetetarell}, we obtain
\begin{equation}\label{E:pmfor}
\rho_H(t)= \pm e^{\xi_{\lambda}(t,\theta_0)} \cdot e^{-i \pi \eta_\lambda(\nabla^H,t)} \cdot e^{\frac{i \pi}{2}\sum_{k=0,1}(-1)^k d^-_{\bar{k},\lambda}} \cdot \rho_{\Gamma_{t,[0,\lambda]}}.
\end{equation}
\begin{lemma}\label{L:meind}
Under the above assumptions, the product
\[
e^{\xi_{\lambda}(t,\theta_0)} \cdot \rho_{\Gamma_{t,[0,\lambda]}} \in \Det\big( H^{\bullet}\big(M,E,H\big) \big)
\]
is independent of $t \in (t_0-\delta, t_0+\delta)$.
\end{lemma}
\begin{proof}
Let $\Gamma_t$ denote the chirality operator corresponding to the metric $g_t^M$. Following the $\mathbb{Z}$-graded case, cf. \cite{RS}, we set
\begin{align}\label{E:plva3}
f(s,t)& = \sum_{k=0,1}(-1)^k \int_0^\infty u^{s-1} \Tr \Big[\exp \big(-u(\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big]du \nonumber \\
& = \Gamma(s) \sum_{k=0,1}(-1)^k \zeta \Big(s, (\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \Big)
\end{align}
Using the fact that
$$
\Gamma_t (\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \Gamma_t = (\nabla^H \Gamma_t )^2 \big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)},
$$ we also have
\begin{align}\label{E:plva4}
f(s,t)& = -\sum_{k=0,1}(-1)^k \int_0^\infty u^{s-1} \Tr \Big[\exp \big(-u(\nabla^H\Gamma_t )^2 \big|_{\Omega^{\bar k}_{-,(\lambda,\infty)}(M,E)}\big) \Big]du \nonumber \\
& = -\Gamma(s) \sum_{k=0,1}(-1)^k \zeta \Big(s, (\nabla^H\Gamma_t )^2 \big|_{\Omega^{\bar k}_{-,(\lambda,\infty)}(M,E)} \Big)
\end{align}
We denote by $\dot{\Gamma}_t$ with respect to the parameter $t$. Then
\begin{align}\label{E:plva1}
\frac{d}{dt} (\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} & = \dot{\Gamma}_t \Gamma_t (\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \nonumber \\
& + (\Gamma_t \nabla^H) \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \dot{\Gamma}_t \Gamma_t (\Gamma_t \nabla^H) \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}
\end{align}
where we used that $\Gamma^2_t=1$.
Similarly, we have
\begin{equation}\label{E:plva2}
\frac{d}{dt} ( \nabla^H \Gamma_t)^2 \big|_{\Omega^{\bar k}_{-,(\lambda,\infty)}(M,E)} = ( \nabla^H \Gamma_t)^2\Gamma_t \dot{\Gamma}_t \big|_{\Omega^{\bar k}_{-,(\lambda,\infty)}(M,E)}+ ( \nabla^H \Gamma_t)\Gamma_t \dot{\Gamma}_t( \nabla^H \Gamma_t)\big|_{\Omega^{\bar k}_{-,(\lambda,\infty)}(M,E)}.
\end{equation}
If $A$ is of trace class and $B$ is a bounded operator, it is well known that $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$. Hence, by this fact and the semi-group property of the heat operator, we have
\begin{align}\label{E:trequality1}
& \Tr \Big[ (\Gamma_t \nabla^H) \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \dot{\Gamma}_t \Gamma_t (\Gamma_t \nabla^H) \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-u(\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big] \nonumber \\
= & \Tr \Big[ \exp \big(-\frac{u}{2}(\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) (\Gamma_t \nabla^H) \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \nonumber \\
& \cdot \dot{\Gamma}_t \Gamma_t (\Gamma_t \nabla^H) \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-\frac{u}{2}(\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big]\nonumber \\
= & \Tr \Big[\dot{\Gamma}_t \Gamma_t (\Gamma_t \nabla^H) \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-\frac{u}{2}(\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \nonumber \\
& \cdot \exp \big(-\frac{u}{2}(\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) (\Gamma_t \nabla^H) \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \Big] \nonumber \\
= & \Tr \Big[ \dot{\Gamma}_t \Gamma_t (\Gamma_t \nabla^H) \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-u(\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) (\Gamma_t \nabla^H) \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \Big] \nonumber \\
= & \Tr \Big[ \dot{\Gamma}_t \Gamma_t (\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\exp \big(-u(\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big],
\end{align}
here in the last equality we used the fact that
\begin{align}
& (\Gamma_t \nabla^H) \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-u(\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) (\Gamma_t \nabla^H) \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \nonumber \\
= &(\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\exp \big(-u(\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big). \nonumber
\end{align}
By \eqref{E:plva1}, \eqref{E:plva3} and \eqref{E:trequality1}, we have
\begin{equation}\label{E:plva5}
\frac{d}{dt}f(s,t) = \sum_{k=0,1}(-1)^k \int_0^\infty u^{s-1} \Tr \Big[- 2u \dot{\Gamma}_t \Gamma_t (\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\exp \big(-u(\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big]du.
\end{equation}
Simlarly, we have
\begin{equation}\label{E:plva6}
\frac{d}{dt}f(s,t) = -\sum_{k=0,1}(-1)^k \int_0^\infty u^{s-1} \Tr \Big[- 2u \Gamma_t \dot{\Gamma}_t (\nabla^H \Gamma_t )^2 \big|_{\Omega^{\bar k}_{-,(\lambda,\infty)}(M,E)}\exp \big(-u(\nabla^H \Gamma_t )^2 \big|_{\Omega^{\bar k}_{-,(\lambda,\infty)}(M,E)}\big) \Big]du.
\end{equation}By \eqref{E:plva5}, \eqref{E:plva6} and the fact that $\dot{\Gamma}_t \Gamma_t = - \Gamma_t \dot{\Gamma}_t$, we conclude that
\begin{align}\label{E:plva7}
\frac{d}{dt}f(s,t) & = - \sum_{k=0,1}(-1)^k \int_0^\infty u^{s} \Tr \Big[ \dot{\Gamma}_t \Gamma_t (\B_{\bar{k},(\lambda,\infty)}^{H}(t))^2 \exp \big(-u(\B_{\bar{k},(\lambda,\infty)}^{H}(t))^2 \big) \Big]du \nonumber \\
& = \sum_{k=0,1}(-1)^k \int_0^\infty u^{s} \frac{d}{du} \Tr \Big[ \dot{\Gamma}_t \Gamma_t \exp \big(-u(\B_{\bar{k},(\lambda,\infty)}^{H}(t))^2 \big) \Big]du \nonumber \\
& = - s\sum_{k=0,1}(-1)^k \int_0^\infty u^{s-1} \Tr \Big[ \dot{\Gamma}_t \Gamma_t \exp \big(-u(\B_{\bar{k},(\lambda,\infty)}^{H}(t))^2 \big) \Big]du,
\end{align}
where we used the integration by parts for the last equality.
Since $(\B^H(t))^2$ is an elliptic differential operator, the dimension of $\Omega^\bullet_{[0,\lambda]}(M,E)$ is finite. Let $\varepsilon \not=0$ be a small enough real number so that $(\B^H(t))^2 + \varepsilon$ is bijective and $2\theta_0$ is an Agmon angle for $(\B^H(t))^2 + \varepsilon$. Then we can rewrite \eqref{E:plva7} as
\begin{align}\label{E:plva8}
\frac{d}{dt}f(s,t) & = - s\sum_{k=0,1}(-1)^k \int_0^1 u^{s-1} \Tr \Big[ \dot{\Gamma}_t \Gamma_t \big(\exp \big(-u(\B_{\bar{k}}^{H}(t))^2+\varepsilon \big) \Big]du \nonumber \\
& - s\sum_{k=0,1}(-1)^k \int_1^\infty u^{s-1} \Tr \Big[ \dot{\Gamma}_t \Gamma_t \big(\exp \big(-u(\B_{\bar k}^{H}(t))^2+\varepsilon \big) \Big]du \nonumber \\
& + s\sum_{k=0,1}(-1)^k \int_0^1 u^{s-1} \Tr \Big[ \dot{\Gamma}_t \Gamma_t \big(\exp \big(- u (\B_{\bar{k},[0,\lambda]}^{H}(t))^2 \big) \Big]du \nonumber \\
& + s \sum_{k=0,1}(-1)^k \int_1^\infty u^{s-1} \Tr \Big[ \dot{\Gamma}_t \Gamma_t \big(\exp \big(- u (\B_{\bar{k},[0,\lambda]}^{H}(t))^2 \big) \Big]du.
\end{align}
Since $\dot{\Gamma}_t \Gamma_t$ is a local quantitity and the dimension of the manifold $M$ is odd, the asymptotic expansion as $u \downarrow 0$ for $\Tr \Big[ \dot{\Gamma}_t \Gamma_t \big(\exp \big(-u(\B^{H}(t))^2+\varepsilon \big) \Big]$ does not contain a constant term. Therefore the integrals of the first term on the right hand side of \eqref{E:plva8} do not have poles at $s=0$. On the other hand, because of exponential decay of $\Tr \Big[ \dot{\Gamma}_t \Gamma_t \big(\exp \big(-u(\B^{H}(t))^2+\varepsilon \big) \Big]$ and $\Tr \Big[ \dot{\Gamma}_t \Gamma_t \big(\exp \big(- u (\B_{[0,\lambda]}^{H}(t))^2 \big) \Big]$ for large $u$, the integrals of the second term and the fourth term on the right hand side of \eqref{E:plva8} are entire functions in $s$. Hence we have
\begin{align}\label{E:plva9}
\frac{d}{dt}\Big|_{s=0}f(s,t) & = \Big(s \sum_{k=0,1}(-1)^k \int_0^1 u^{s-1} \Tr \Big[ \dot{\Gamma}_t \Gamma_t \Big|_{\Omega^{\bar k}_{[0,\lambda]}(M,E)} \Big] du \Big) \Big|_{s=0} \nonumber \\
& = \sum_{k=0,1}(-1)^k \Tr \Big[ \dot{\Gamma}_t \Gamma_t \Big|_{\Omega^{\bar k}_{[0,\lambda]}(M,E)} \Big]
\end{align}
and, by \eqref{E:plva3},
\begin{equation}\label{E:plva10}
\frac{d}{dt}\Big|_{s=0}sf(s,t) = \frac{d}{dt}\Big|_{s=0} \sum_{k=0,1}(-1)^k \zeta \Big(s, (\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \Big) =0.
\end{equation}
By \eqref{E:xino} and \eqref{E:plva3}, we know that
\begin{equation}\label{E:plva11}
\xi_{\lambda}(t,\theta_0)= -\frac{1}{2}\operatorname{lim}_{s \to 0}\Big[ f(s,t)-\frac{1}{s} \sum_{k=0,1}(-1) \zeta(0, (\Gamma_t \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}) \Big].
\end{equation}
Combining \eqref{E:plva9}, \eqref{E:plva10} with \eqref{E:plva11}, we obtain
\begin{equation}\label{E:xivarfor}
\frac{d}{dt}\Big|_{t=0}\xi_{\lambda}(t,\theta_0)= -\frac{1}{2}\sum_{k=0,1}(-1)^k \Tr \Big[ \dot{\Gamma}_t \Gamma_t \Big|_{\Omega^{\bar k}_{[0,\lambda]}(M,E)} \Big]
\end{equation}
Combining \eqref{E:varfor} with \eqref{E:xivarfor}, we obtain
\[
\frac{d}{dt}(e^{\xi_{\lambda}(t,\theta_0)} \cdot \rho_{\Gamma_{t,[0,\lambda]}})=0.
\]
\end{proof}
We need the following lemma, which is a slight modification of the result in Subsection 9.3 of \cite{BK3}.
\begin{lemma}\label{L:etaf}
For any $t_1,t_2 \in (t_0-\delta, t_0 + \delta)$, we have
\[
\eta_\lambda(\nabla^H,t_1)-\eta_\lambda(\nabla^H,t_2) \equiv \eta(\B^{H}_{\bar{0}}(t_1)) - \eta(\B^{H}_{\bar{0}}(t_2)), \quad \operatorname{mod} \ \mathbb{Z}.
\]
\end{lemma}
Let $\B^H_{\trivial}=\B_{\bar 0}(\nabla^H_{\trivial},g^M):\Omega^{\bar 0}(M,E) \to \Omega^{\bar 0}(M,E)$ denote the even part of twisted odd signature operator corresponding to the metric $g^M$ and the trivial line bundle over $M$ endowed with the trivial connection $\nabla_{\trivial}$. Put
\[
\eta_{\trivial}:=\frac{1}{2} \eta(0,\B_{\trivial}^{H}).
\]
We now need to study the dependence of $\eta(\B^{H}_{\bar{0}})$ on the Riemannian metric $g^M$. This was essentially done in \cite{APS2} and \cite{Gi}.
\begin{lemma}\label{L:rhopm}
The function $\eta(\B^{H}_{\bar{0}}(t))-\operatorname{rank}(E)\eta_{\trivial}(t)$ is, modulo $\mathbb{Z}$, independent of $t \in (t_0-\delta, t_0+\delta)$.
\end{lemma}
\subsection{Removing the metric anomaly and the definition of the twisted refined analytic torsion}
The following theorem is the main theorem of this subsection.
\begin{theorem}
Let $M$ be an odd dimensional oriented closed Riemannian manifold. Let $(E,\nabla,h^E)$ be a flat complex vector bundle over $M$ and $H$ is a closed differential form on $M$ of odd degree, other than one form. Then the element
\begin{equation}\label{E:reffor}
\rho(\nabla^H,g^M) \cdot e^{i \pi (\operatorname{rank} (E))\eta_{\trivial}} \in \Det\big(H^\bullet(M,E,H)\big),
\end{equation}
where $\nabla^H:=\nabla + H \wedge \cdot$ and $\rho(\nabla^H,g^M) \in \Det\big(H^\bullet(M,E,H)\big)$ is defined in \eqref{E:rhoh}, is independent of $g^M$.
\end{theorem}
\begin{proof}
Consider a smooth family $g^M_t, t \in \mathbb{R}$ of Riemannian metrics on $M$. From \eqref{E:pmfor}, we obtain for $t \in (t_0-\delta, t_0+\delta)$
\begin{equation}\label{E:meana}
\rho^H(t) \cdot e^{i \pi (\operatorname{rank} (E))\eta_{\trivial}}=\pm e^{\xi_{\lambda}(t,\theta_0)} \cdot e^{-i \pi \eta_\lambda(\nabla^H,t)} \cdot e^{\frac{i \pi}{2}\sum_{k=0,1}(-1)^kd_{\bar{k},\lambda}} \cdot \rho_{\Gamma_{t,[0,\lambda]}} \cdot e^{i \pi (\operatorname{rank} (E))\eta_{\trivial}}.
\end{equation}
Combining \eqref{E:xivarfor}, \eqref{E:meana} with Lemma \ref{L:rhopm} we conclude that for any $t_1,t_2 \in (t_0-\delta,t_0+\delta)$
\[
\rho^H(t_1) \cdot e^{i \pi (\operatorname{rank} (E))\eta_{\trivial}}= \pm \rho^H(t_2) \cdot e^{i \pi (\operatorname{rank} (E))\eta_{\trivial}}.
\]
Since the function $\rho^H(t) \cdot e^{i \pi (\operatorname{rank} (E))\eta_{\trivial}}$ is continuous and nonzero, the sign in the right hand side of the equality must be positive. This proves the statement.
\end{proof}
\begin{definition}\label{D:maindef}
Let $M$ be an odd dimensional oriented closed Riemannian manifold. Let $(E,\nabla,h^E)$ be a flat complex vector bundle over $M$ and $H$ is a closed differential form on $M$ of odd degree, other than one form. The {\em twisted refined analytic torsion} $\rho_{\operatorname{an}}(\nabla^H)$ is the element of $\Det\big(H^\bullet(M,E,H)\big)$ defined by \eqref{E:reffor}.
\end{definition}
\subsection{Variation of refined analytic torsion with respect to the flux in a cohomology class}
Suppose that the (real) flux form $H$ is deformed smoothly along a one-parameter family with parameter $v \in \mathbb{R}$ in such a way that the cohomology class $[H] \in H^{\bar 1}(M, \mathbb{R})$ is fixed. Then $\frac{d}{dv}H=-dB$ for some form $B \in \Omega^{\bar 0}(M)$ that depends smoothly on $v$. Let $\beta=B \wedge \cdot$.
Fix $v_0 \in \mathbb{R}$ and choose $\lambda > 0$ such that there are no eigenvalues of $(\B^H)^2(v_0)$ of absolute value $\lambda$. Further, assume that $\lambda$ is big enough so that the real parts of eigenvalues of $(\B^{H}_{(\lambda,\infty)}(v_0))^2$ are all greater than $0$. Then there exists $\delta > 0$ small enough such that the same holds for the spectrum of $(\B^{H}(v))^2$ for $v \in (v_0-\delta,v_0+\delta)$. For simplicity, we often omit the parameter $v$ in the notations of operators in the following discussion.
We have the following two lemmas, see also Lemma 3.5 and lemma 3.7 of \cite{MW}.
\begin{lemma}\label{L:flva1}
Under the above assumptions, we have
\[
\frac{d}{dv} \xi_\lambda = \sum_{k=0,1}(-1)^k \Tr (\beta \big|_{\Omega^{\bar k}_{[0,\lambda]}(M,E)}).
\]
\end{lemma}
\begin{proof}
As in the proof of Lemma~\ref{L:meind}, we set
\begin{equation}\label{E:fvh1}
f(s,v) = \sum_{k=0,1}(-1)^k \int_0^\infty u^{s-1} \Tr \Big[\exp \big(-u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big]du.\\
\end{equation}
We note that $B$, hence $\beta$, is real. By \eqref{E:fvh1} and the fact that
\[
\frac{d}{dv} \nabla^H = [\beta, \nabla^H],
\]
we have
\begin{align}\label{E:fvh2}
& \frac{d}{dv}f(s,v) \nonumber \\ & = \sum_{k=0,1}(-1)^k \int_0^\infty u^{s-1} \Tr \Big[-u \big(\Gamma [\beta, \nabla^H] \Gamma \nabla^H \big)\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\exp \big(-u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big]du \nonumber \\
&+ \sum_{k=0,1}(-1)^k \int_0^\infty u^{s-1} \Tr \Big[-u \big(\Gamma \nabla^H \Gamma [\beta, \nabla^H]\big)\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big]du.
\end{align}
Using the fact that $\Gamma^2=1$, we have
\begin{align}\label{E:fvh3}
& \Tr \Big[ \big(\Gamma \beta \nabla^H \Gamma \nabla^H \big)\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big] \nonumber \\
& = \Tr \Big[ \big(\Gamma \beta \Gamma \big)\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \cdot \big( \Gamma \nabla^H \big)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big].
\end{align}
By using the trace property, $\Gamma^2=1$, and the semi-group property of the heat operator, we have
\begin{align}\label{E:fvh4}
& \Tr \Big[ \big(\Gamma \nabla^H \beta \Gamma \nabla^H \big)\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big( -u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \big) \Big] \nonumber \\
& = \Tr \Big[ \big(\Gamma \nabla^H \Gamma \cdot \Gamma \beta \Gamma \nabla^H \big)\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-\frac{1}{2}u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \nonumber\\
& \qquad \cdot \exp \big(-\frac{1}{2}u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big] \nonumber \\
&= \Tr \Big[ \exp \big(-\frac{1}{2}u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \cdot \big(\Gamma \nabla^H \Gamma \cdot \Gamma \beta \Gamma \nabla^H \big)\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \nonumber \\ & \qquad \cdot \exp \big(-\frac{1}{2}u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big] \nonumber \\
& = \Tr \Big[ \big( \Gamma \beta \Gamma \nabla^H \big)\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-\frac{1}{2}u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big)\nonumber \\ & \qquad \cdot \exp \big(-\frac{1}{2}u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \cdot \big(\Gamma \nabla^H \Gamma \big)\big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)} \Big] \nonumber \\
& = \Tr \Big[ \big( \Gamma \beta \Gamma \big)\big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)} \cdot \big( \nabla^H \Gamma \big)^2\big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)} \exp \big(-u( \nabla^H \Gamma)^2 \big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)}\big) \Big].
\end{align}
For the last equality of \eqref{E:fvh4}, we used the fact that
\begin{align*}
& \nabla^H\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big)\big(\Gamma \nabla^H \Gamma \big)\big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)}\\
& = \big( \nabla^H \Gamma \big)^2\big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)} \exp \big(-u( \nabla^H \Gamma)^2 \big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)}\big).
\end{align*}
By combining \eqref{E:fvh3} with \eqref{E:fvh4}, we obtain
\begin{align}\label{E:fvha1}
& \sum_{k=0,1}(-1)^k \int_0^\infty u^{s} \Tr \Big[ \big(\Gamma [\beta, \nabla^H] \Gamma \nabla^H \big)\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big] du \nonumber \\
& = \sum_{k=0,1}(-1)^k \int_0^\infty u^{s} \Big( \, \Tr \Big[ \big(\Gamma \beta \Gamma \big)\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \cdot \big( \Gamma \nabla^H \big)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big] \nonumber \\
& \qquad - \Tr \Big[ \big( \Gamma \beta \Gamma \big)\big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)} \cdot \big( \nabla^H \Gamma \big)^2\big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)} \exp \big(-u( \nabla^H \Gamma)^2 \big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)}\big) \Big] \, \Big) du \nonumber \\
& = \sum_{k=0,1}(-1)^k \int_0^\infty u^{s} \Tr \Big[ \big(\Gamma \beta \Gamma \big)\big|_{\Omega^{\bar k}_{(\lambda,\infty)}(M,E)}\cdot (\B^H_{(\lambda,\infty)})^2\exp \big(-u(\B^H_{(\lambda,\infty)})^2 \Big]du.
\end{align}
Similarly, we have
\begin{align}\label{E:fvh5}
& \Tr \Big[ \big(\Gamma \nabla^H \Gamma \nabla^H \beta \big)\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big] \nonumber \\
& = \Tr \Big[ \beta\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \cdot \big( \Gamma \nabla^H \big)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big]
\end{align}
and
\begin{align}\label{E:fvh6}
& \Tr \Big[ \big(\Gamma \nabla^H \Gamma \beta \nabla^H \big)\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big( -u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \big) \Big] \nonumber \\
& = \Tr \Big[ \beta\big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)} \cdot \big( \nabla^H \Gamma \big)^2\big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)} \exp \big(-u( \nabla^H \Gamma)^2 \big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)}\big) \Big].
\end{align}
By combining \eqref{E:fvh5} with \eqref{E:fvh6}, we have
\begin{align}\label{E:fvha2}
& \sum_{k=0,1}(-1)^k \int_0^\infty u^{s} \Tr \Big[ \big(\Gamma \nabla^H \Gamma [\beta, \nabla^H] \big)\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big] du \nonumber \\
& = \sum_{k=0,1}(-1)^k \int_0^\infty u^{s} \Big( \, \Tr \Big[ \beta\big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \cdot \big( \Gamma \nabla^H \big)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)} \exp \big(-u(\Gamma \nabla^H)^2 \big|_{\Omega^{\bar k}_{+,(\lambda,\infty)}(M,E)}\big) \Big] \nonumber \\
& \qquad - \Tr \Big[ \beta\big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)} \cdot \big( \nabla^H \Gamma \big)^2\big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)} \exp \big(-u( \nabla^H \Gamma)^2 \big|_{\Omega^{\overline{k+1}}_{-,(\lambda,\infty)}(M,E)}\big) \Big] \, \Big) du \nonumber \\
& = \sum_{k=0,1}(-1)^k \int_0^\infty u^{s} \Tr \Big[ \beta\big|_{\Omega^{\bar k}_{(\lambda,\infty)}(M,E)} \cdot (\B^H_{(\lambda,\infty)})^2\exp \big(-u(\B^H_{(\lambda,\infty)})^2 \Big]du.
\end{align}
Combining \eqref{E:fvha1}, \eqref{E:fvha2}, with \eqref{E:fvh2}, we obtain
\begin{align*}
\frac{d}{dv}f(s,v) & = -\sum_{k=0,1}(-1)^k \int_0^\infty u^{s} \Tr \Big[ \big(\Gamma \beta \Gamma \big)\big|_{\Omega^{\bar k}_{(\lambda,\infty)}(M,E)}\cdot (\B^H_{(\lambda,\infty)})^2\exp \big(-u(\B^H_{(\lambda,\infty)})^2 \Big]du \nonumber \\
& - \sum_{k=0,1}(-1)^k \int_0^\infty u^{s} \Tr \Big[ \beta\big|_{\Omega^{\bar k}_{(\lambda,\infty)}(M,E)} \cdot (\B^H_{(\lambda,\infty)})^2 \exp \big(-u(\B^H_{(\lambda,\infty)})^2 \big) \Big]du \\
&= - 2\sum_{k=0,1}(-1)^k \int_0^\infty u^{s} \Tr \Big[ \beta\big|_{\Omega^{\bar k}_{(\lambda,\infty)}(M,E)} \cdot (\B^H_{(\lambda,\infty)})^2 \exp \big(-u(\B^H_{(\lambda,\infty)})^2 \big) \Big]du,
\end{align*}
where for the latter equality we used the fact that $\Tr (\beta \big|_{\Omega^{\bar k}_{[0,\lambda]}(M,E)})=\Tr (\Gamma \beta \Gamma \big|_{\Omega^{\bar k}_{[0,\lambda]}(M,E)})$.
The rest is similar to the proof of Lemma \ref{L:meind}.
\end{proof}
\begin{lemma}\label{L:flva2}
Under the same assumptions, along any one parameter deformation of $H$ that fixes the cohomology class $[H]$, the element can be chosen so that
\[
\frac{d}{dv} \rho_{\Gamma_{[0,\lambda]}} = - \sum_{k=0,1}(-1)^k \Tr (\beta \big|_{\Omega^{\bar k}_{[0,\lambda]}(M,E)}) \rho_{\Gamma_{[0,\lambda]}},
\]
where we identify $\Det\big(H^\bullet(M,E,H)\big)$ along the deformation using \eqref{E:epsiso}.
\end{lemma}
\begin{proof}
In order to compare the elements $\rho_{\Gamma_{[0,\lambda]}} \in \Det\big(H^\bullet(M,E,H)\big)$ at different values of $v$. We choose a reference point, say $v=0$, and let $H^{(0)}$, $\rho^{(0)}_{\Gamma_{[0,\lambda]}}$ be the values of $H$, $\rho_{\Gamma_{[0,\lambda]}}$, respectively, at $v=0$. By \eqref{E:epsiso}, we have the isomorphism
\[
\Det(\varepsilon_B): \Det\big(H^\bullet(M,E,H^{(0)})\big) \to \Det\big(H^\bullet(M,E,H)\big).
\]
Since $\varepsilon_B=e^\beta$ on $\Omega^\bullet(M,E)$, we have, for $k=0,1$,
\[
\frac{d}{dv}(\Det(\varepsilon_B))^{-1}\rho_{\Gamma_{[0,\lambda]}}\, = \, - \sum_{k=0,1}(-1)^k \Tr (\beta \big|_{\Omega^{\bar k}_{[0,\lambda]}(M,E)})(\Det(\varepsilon_B))^{-1} \rho_{\Gamma_{[0,\lambda]}}.
\]
The result follows.
\end{proof}
The argument of the following lemma is similar to the argument of Lemma 9.4 of \cite{BK3}.
\begin{lemma}\label{L:flva0}
For any $v_1,v_2 \in (v_0-\delta, v_0 + \delta)$, we have
\[
\eta_\lambda(\nabla^H,v_1)-\eta_\lambda(\nabla^H,v_2) \equiv \eta(\B^{H}_{\bar{0}}(v_1)) - \eta(\B^{H}_{\bar{0}}(v_2)), \quad \operatorname{mod} \ \mathbb{Z}.
\]
\end{lemma}
Again we need to study the dependence of $\eta(\B^{H}_{\bar{0}})$ on the parameter $v$. As Lemma \ref{L:rhopm}, this was essentially done in \cite{APS2} and \cite[P. 52]{Gi}.
\begin{lemma}\label{L:flva3}
The function $\eta(\B^{H}_{\bar{0}}(v))-\operatorname{rank}(E)\eta_{\trivial}(v)$ is, modulo $\mathbb{Z}$, independent of $v \in \mathbb{R}$.
\end{lemma}
Now we have the main theorem of this subsection.
\begin{theorem}
Let $M$ be an odd dimensional oriented closed Riemannian manifold. Let $(E,\nabla,h^E)$ be a flat complex vector bundle over $M$. Suppose that $H$ and $H'$ are closed differential forms on $M$ of odd degrees representing the same deRham cohomology class, and let $B$ be an even form so that $H'=H-dB$. Then the refined analytic torsion $\rho_{\operatorname{an}}(\nabla^{H'})=\Det (\varepsilon_B)(\rho_{\operatorname{an}}(\nabla^{H}))$.
\end{theorem}
\begin{proof}
Again we choose a reference point, say $v=0$, and let $H^{(0)}$, $\rho^{(0)}_{\Gamma_{[0,\lambda]}}$ be the values of $H$, $\rho_{\Gamma_{[0,\lambda]}}$, respectively, at $v=0$. By \eqref{E:epsiso}, we have the isomorphism
\[
\Det(\varepsilon_B): \Det\big(H^\bullet(M,E,H^{(0)})\big) \to \Det\big(H^\bullet(M,E,H)\big).
\]
Recall that $\varepsilon_B=e^\beta$ on $\Omega^\bullet(M,E)$. By combining Lemma \ref{L:flva1}, Lemma \ref{L:flva2} with Lemma \ref{L:flva3}, we conclude that $(\Det(\varepsilon_B))^{-1}\rho_{\operatorname{an}}(\nabla^{H})$ is, up to sign, invariant along the deformation. Since the function $(\Det(\varepsilon_B))^{-1}\rho_{\operatorname{an}}(\nabla^{H})$ is continuous and nonzero, the sign in the right hand side of the equality must be positive. This proves the statement.
\end{proof}
As pointed out by Mathai and Wu, \cite{MW}, that when $H$ is a $3$-form on $M$, the deformation of the Riemannian metric $g^M$ and that of the flux $H$ within its cohomology class can be interpreted as a deformation of generalized metrics on $M$ and the analytic torsion should be defined for generalized metrics so that the deformations of $g^M$ and $H$ are unified. Similarly, the the refined analytic torsion should be also defined for generalized metrics so that the deformations of $g^M$ and $H$ are unified.
\section{A duality theorem for the twisted refined analytic torsion}\label{S:dttrat}
In this section we first review the concept of the dual of a complex and construct a natural isomorphism between the determinant lines of a $\mathbb{Z}_2$-graded complex and its dual. We then show that this isomorphism is compatible with the canonical isomorphism \eqref{E:isomorphism}. Finally we establish a relationship between the twisted refined analytic torsion corresponding to a flat connection and that of its dual. The contents of this section are $\mathbb{Z}_2$-graded analogues of Sections 3 and 10 of \cite{BK3}. Throughout this section, $\bf{k}$ is a field of characteristic zero endowed with an involutive automorphism $\tau:\bf{k} \to \bf{k}.$
The main examples are $\bf{k}=\mathbb{C}$ with $\tau$ being the complex conjugation and $\bf{k}=\mathbb{R}$ with $\tau$ being the identity map.
\subsection{The $\mathbb{Z}_2$-graded $\tau$-dual space}\label{SS:duakcplx}
If $V,W$ are $\bf{k}$-vector spaces, a map $f:V \to W$ is said to be $\tau-linear$ if
\[
f(x_1v_1+x_2v_2)=\tau(x_1)v_1+\tau(x_2)v_2, \, \text{for any} \, v_1,v_2 \in V, x_1,x_2 \in \bf{k}.
\]
The linear space $V^*=V^{*_\tau}$ of all $\tau$-linear maps $V \to \bf{k}$ is called the $\tau-dual \ space \ to \ V$. There are natural $\tau$-linear isomorphisms, cf. \cite[Subsection 3.1]{BK3},
\begin{equation}\label{E:albev}
\alpha_V:\Det(V^*) \rightarrow \Det(V)^{-1}, \quad \beta_V:\Det(V) \rightarrow \Det(V^*)^{-1}.
\end{equation}
Then for any $v \in \Det(V)$, we have, cf. \cite[(3-8)]{BK3},
\begin{equation}\label{E:alinvb}
\big( \alpha_V^{-1}(v^{-1}) \big)^{-1} = (-1)^{\dim V} \beta_V(v),
\end{equation}
Let $V$ and $W$ be $\bf{k}$-vector spaces, then, for any $v \in \Det(V),w \in \Det(W)$, we have, cf. \cite[(3-9)]{BK3},
\begin{equation}\label{E:dualmu}
\big( \mu_{V,W}(v \otimes w)\big)^{-1}= \alpha_{V \oplus W} \circ \mu_{V^*,W^*}\big( \alpha_V^{-1}(v^{-1}) \otimes \alpha_W^{-1}(w^{-1}) \big).
\end{equation}
Let $T:V \to W$ be a $\bf{k}$-linear map. The $\tau-adjoint \ of \ T$ is the linear map
\[
T^*:W^* \rightarrow V^*
\]
such that
\[
(T^*w^*)(v) = w^*(Tv), \ \text{for all} \ v \in V, w^* \in W^*.
\]
If $T$ is bijective, then, for any nonzero $v \in \Det(V)$, we have, cf. \cite[(3-11)]{BK3},
\begin{equation}\label{E:tastal}
T^*\alpha_W^{-1}\big( (Tv)^{-1} \big) = \alpha_V^{-1}(v^{-1}).
\end{equation}
Let $V^0, V^1, \cdots, V^m$ be finite dimensional $\bf{k}$-vector space, where $m=2r-1$ is an odd integer. Denote by $V^{\bar 0}=\bigoplus_{i=0}^{r-1} V^{2i}$ and $V^{\bar 1}=\bigoplus_{i=0}^{r-1} V^{2i+1}$. Let $V^{\bullet} = V^{\bar 0} \oplus V^{\bar 1}$ be a finite dimensional $\mathbb{Z}_2$-graded $\bf{k}$-vector space. We define the $\mathbb{Z}_2-graded \ (\tau-) \ dual \ space\ \widehat{V} = \widehat{V}^{\bar 0} \oplus \widehat{V}^{\bar 1}$ by
\[
\widehat{V}^{\bar k} := (V^{\overline{k+1}})^*, \quad k=0,1.
\]
Then \eqref{E:albev} induces a $\tau$-linear isomorphism
\begin{equation}\label{E:alvhatv}
\alpha_{V^\bullet}:\Det(V^\bullet) \rightarrow \Det(\widehat{V}^\bullet),
\end{equation}
defined by
\begin{equation}\label{E:allinve1}
\alpha_{V^\bullet}(v_{\bar 0} \otimes (v_{\bar 1})^{-1}) = (-1)^{\mathcal{M}(V^\bullet)} \cdot \alpha^{-1}_{V^{\bar 1}}(v_{\bar 1}^{-1}) \otimes \alpha_{V^{\bar 0}}(v_{\bar 0}),
\end{equation}
where $v_{\bar k} \in \Det(V^{\bar k}), k=0,1$ and, cf. \eqref{E:mvw123},
\[
\mathcal{M}(V^\bullet) = \mathcal{M}(V^\bullet, V^\bullet)= \dim V^{\bar 0} \cdot \dim V^{\bar 1}.
\]
\subsection{The dual complex of a $\mathbb{Z}_2$-graded complex}
Consider the $\mathbb{Z}_2$-graded complex \eqref{E:detline} of finite dimensional $\bf{k}$-vector spaces. The dual complex of the $\mathbb{Z}_2$-graded complex \eqref{E:detline} is the complex
\begin{equation}\label{E:detline1}
(\widehat{C}^\bullet,d^*) \ : \ \cdots \stackrel{d^*}{\longrightarrow} \ \widehat{C}^{\bar{0}} \ \stackrel{d^*}{\longrightarrow}\ \widehat{C}^{\bar{1}}\ \stackrel{d^*}{\longrightarrow} \ \widehat{C}^{\bar{0}} \ \stackrel{d^*}{\longrightarrow} \cdots,
\end{equation}
where $\widehat{C}^{\bar k}=(C^{\overline{k+1}})^*$ and $d^*$ is the $\tau$-adjoint of $d$. Then the cohomology $H^{\bar k}(d^*)$ of $\widehat{C}^\bullet$ is natural isomorphic to the $\tau$-dual space to $H^{\overline{k+1}}(d) \, (k=0,1)$. Hence by \eqref{E:alvhatv}, we obtain $\tau$-linear isomorphisms
\begin{equation}\label{E:alphah0}
\begin{array}{l}
\alpha_{C^\bullet}: \Det(C^\bullet) \rightarrow \Det(\widehat{C}^\bullet),
\\
\alpha_{H^\bullet(d)}: \Det(H^\bullet(d)) \rightarrow \Det(H^\bullet(d^*)).
\end{array}
\end{equation}
The following lemma is the $\mathbb{Z}_2$-graded analogue of Lemma 3.6 of \cite{BK3}. The proof is similar to the proof of Lemma 3.6 of \cite{BK3}. We skip the proof.
\begin{lemma}\label{L:comm22}
Let $(C^\bullet,d)$ be a $\mathbb{Z}_2$-graded complex of finite dimensional $\bf{k}$-vector spaces, defined as \eqref{E:detline}. Further, assume that the Euler characteristics $\chi(C^\bullet)=\chi(\widehat{C}^\bullet)=0$. Then the following diagramm commutes:
\begin{equation}\label{E:comm2}
\begin{CD}
\Det(C^\bullet) & @>\phi_{C^\bullet} >> & \Det(H^\bullet(d)) \\
@V \alpha_{C^\bullet} VV & & @VV \alpha_{H^\bullet(d)} V \\
\Det(\widehat{C}^\bullet) & @> \phi_{\widehat{C}^\bullet} >> & \Det\big( H^\bullet(d^*) \big)
\end{CD}
\end{equation}
where $\phi_{C^\bullet}$ and $\phi_{\widehat{C}^\bullet}$ are defined as in \eqref{E:phic}.
\end{lemma}
\subsection{The refined torsion of the $\mathbb{Z}_2$-graded dual complex}
Suppose now that $\bf{k}$ is endowed with an involutive endomorphism $\tau$. Let $\widehat{C}^\bullet$ be the $\tau$-dual complex of $C$ and let $\alpha_{C^\bullet}: \Det(C^\bullet) \to \Det(\widehat{C}^\bullet)$ denote the $\tau$-isomorphism defined in \eqref{E:alphah0}. Let $\Gamma^*$ be the $\tau$-adjoint of $\Gamma$. Then $\Gamma^*$ is a chirality operator for the complex $\widehat{C}^\bullet$.
The following lemma is the $\mathbb{Z}_2$-graded analogue of Lemma 4.11 of \cite{BK3}.
\begin{lemma}\label{L:rhostarrho}
In the situation described above,
\begin{equation}\label{E:lega00}
\rho_{\Gamma^*} = \alpha_{H^\bullet(d)}(\rho_\Gamma).
\end{equation}
\end{lemma}
\begin{proof}
Fix $c_{\bar 0} \in \Det(C^{\bar 0})$ and set
\begin{equation}\label{E:lega1}
\widehat{c}_{\bar 0} = \alpha^{-1}_{C^{\bar 0}}\big( (\Gamma c_{\bar 0})^{-1} \big) \in \Det(\widehat{C}^{\bar 0}).
\end{equation}
Then, by \eqref{E:tastal},
\begin{equation}\label{E:lega2}
\Gamma^* \widehat{c}_{\bar 0} = \alpha^{-1}_{C^{\bar 0}}(c_{\bar 0}^{-1}) \in \Det(\widehat{C}^{\bar 0}).
\end{equation}
Using \eqref{E:alinvb}, we obtain from \eqref{E:lega1} and \eqref{E:lega2}, that
\begin{equation}\label{E:lega3}
\beta_{C^{\bar 0}}(c_{\bar 0}) = (-1)^{\dim C^{\bar 0}} \cdot (\Gamma^* \widehat{c}_{\bar 0})^{-1}, \quad \beta_{C^{\bar 1}}(\Gamma c_{\bar 0}) = (-1)^{\dim C^{\bar 0}} \cdot \widehat{c}_{\bar 0}^{-1}.
\end{equation}
Combining \eqref{E:allinve1}, \eqref{E:elec}, \eqref{E:lega1}, \eqref{E:lega2} and \eqref{E:lega3}, we have
\begin{equation}\label{E:lega4}
\alpha_{C^\bullet}(c_\Gamma) = (-1)^{\mathcal{M}(C^\bullet)+\dim C^{\bar 0}} \cdot c_{\Gamma^*}.
\end{equation}
By \eqref{E:elerho}, $\rho_\Gamma= \phi_{C^\bullet}(c_\Gamma)$. Therefore, from Lemma \ref{L:comm22}, we obtain
\begin{equation}\label{E:lega5}
\alpha_{H^\bullet(d)}(\rho_\Gamma) = \phi_{\widehat{C}^\bullet} \circ \alpha_{C^\bullet}(c_\Gamma) = (-1)^{\mathcal{M}(C^\bullet)+\dim C^{\bar 0}} \cdot \rho_{\Gamma^*}.
\end{equation}
By \eqref{E:mvw123} and the fact $\dim C^{\bar 0} = \dim C^{\bar 1}$, we get
\begin{equation}\label{E:lega6}
\mathcal{M}(C^\bullet)+\dim C^{\bar 0} = \dim C^{\bar 1} \cdot \dim C^{\bar 0} + \dim C^{\bar 0} = \dim C^{\bar 0} \cdot (\dim C^{\bar 0} + 1) \equiv 0, \quad \text{mod} \ 2.
\end{equation}
Combining \eqref{E:lega5} with \eqref{E:lega6}, we obtain \eqref{E:lega00}.
\end{proof}
\subsection{The duality theorem}\label{SS:duaconne1}
Suppose that $M$ is a closed oriented manifold of odd dimension $m=2r-1$. Let $E \to M$ be a complex vector bundle over $M$ and let $\nabla$ be a flat connection on $E$. Fix a Hermitian metric $h^E$ on $E$. Denote by $\nabla'$ the connection on $E$ dual to the connection $\nabla$, cf. \cite[Subsection 10.1]{BK3}. We denote by $E'$ the flat bundle $(E,\nabla')$, referring to $E'$ as the dual of the flat vector bundle $E$. Using the construction of Section \ref{SS:duakcplx}, with $\tau: \mathbb{C} \to \mathbb{C}$ be the complex conjugation, we have the canonical anti-linear isomorphism
\begin{equation}\label{E:antiiso}
\alpha : \Det \big( H^\bullet(M,E,H) \big) \rightarrow \Det \big( H^\bullet(M,E',H) \big).
\end{equation}
The following theorem is the main result of this section and is the twisted analogue of Theorem 10.3 of \cite{BK3}.
\begin{theorem}\label{T:duality0}
Let $E \to M$ be a complex vector bundle over a closed oriented odd dimensional manifold $M$ and let $\nabla$ be a flat connection on $E$. Denote by $\nabla'$ the connection dual to $\nabla$ with respect to a Hermitian metric $h^E$ on $E$ and let $H$ be a odd-degree cloed form, other than one form, on $M$. Then
\begin{equation}\label{E:aloh15}
\alpha(\rho_{\an}(\nabla^H)) = \rho_{\an}(\nabla'^H)\cdot e^{2 \pi i \big( \bar{\eta}(\nabla^H,g^M)-(\operatorname{rank}E)\eta_{\trivial}(g^M) \big)},
\end{equation}
where $\alpha$ is the anti-linear isomorphism \eqref{E:antiiso} and $g^M$ is any Riemannian metric on $M$.
\end{theorem}
The rest of this section is concerned about the proof of Thoerem \ref{T:duality0}.
\subsection{A choice of $\lambda$}\label{SS:anglecho}
Assume that no eigenvalue of $\B^H_{(\lambda,\infty)}$ lies in the solid angles $L_{[-\theta-\pi,\theta]}$ and $L_{[-\theta,\theta+\pi]}$, cf. \cite[Subsection 10.4 and 10.5]{BK3}, then it follows that no eigenvalue of $(\B^H_{(\lambda,\infty)})^2$ lies in the solid angles $L_{[-2\theta,2\theta+2\pi]}$.
Let $\B'^H$ denote the twisted odd signature operator associated to the connection $\nabla'$, the odd-degree form $H$ and the Riemannian metric $g^M$. One can check that
\begin{equation}\label{E:nahdu1}
(\nabla^H)^*=\Gamma \nabla'^H \Gamma \ \text{and} \ (\nabla'^H)^*=\Gamma \nabla^H \Gamma.
\end{equation}
Using \eqref{E:toso}, \eqref{E:nahdu1} and the equality $\Gamma^*=\Gamma$, one can see that the adjoint $\B'^H$ of $\B^H$ satisfies
\begin{equation}\label{E:nahdu2}
(\B^H)^*=\B'^H.
\end{equation}
The choice of the angle $\theta$ guarantees that $\pm 2 \theta$ are Agmon angles for the operator $(\Gamma \nabla'^H)^2=\big((\Gamma \nabla^H)^2\big)^*$. In particular, for each $\lambda \ge 0$, the number $\xi_\lambda(\nabla'^H,g^M,\theta)$ can be defined by the formula \eqref{E:xino}, with the same angle $\theta$ and with $\nabla^H$ replaced by $\nabla'^H$ everywhere.
The following lemma is twisted analogue of Lemma 10.6 of \cite{BK3} and the proof is similar to the proof of Lemma 10.6 of \cite{BK3}. We skip the proof.
\begin{lemma}\label{L:xiducong}
Let $\theta$ be as above and let $\lambda \ge 0$ be big enough so that the operator $\B^H_{(\lambda,\infty)}$ does not have purely imaginary eigenvalues, cf. \cite[Subsection 10.5]{BK3}. Then
\[
\xi_\lambda(\nabla'^H,g^M,\theta) = \bar{\xi}_\lambda(\nabla^H,g^M,\theta),
\]
and
\begin{equation}\label{E:etanooo}
\eta_\lambda(\nabla'^H)=\bar{\eta}_\lambda(\nabla^H),
\end{equation}
where $\bar{z}$ denotes the complex conjugate of the number $z \in \mathbb{C}$.
\end{lemma}
\subsection{Small eigenvalues of $\B^H$ and $\B'^H$}
We define $\Omega^{\bar k}_{\pm,\mathcal{I}}(M,E'), \Omega^{\bar k}_{\pm}(M,E')$, and $\Omega^{\bar k}(M,E')$ in similar ways as $\Omega^{\bar k}_{\pm,\mathcal{I}}(M,E), \Omega^{\bar k}_{\pm}(M,E)$ and $\Omega^{\bar k}(M,E)$, respectively. As \eqref{E:dno}, for $k=0,1$, set
\begin{equation}\label{E:dno1}
d_{\bar{k},\lambda}^{\pm} = \dim \Omega^{\bar k}_{\pm,[0,\lambda]}(M,E), \quad {d'}_{\bar{k},\lambda}^{\pm} = \dim \Omega^{\bar k}_{\pm,[0,\lambda]}(M,E').
\end{equation}
From the fact that
$\Gamma \B_{\bar{k},(\lambda, \infty)}^{H,\pm} \Gamma = \B_{\overline{k+1},(\lambda, \infty)}^{H,\mp}$, we conclude that
\begin{equation}\label{E:aaa}
\Gamma \Big( \Omega^{\bar k}_{\pm,[0,\lambda]}(M,E) \Big) = \Big( \Omega^{\overline{k+1}}_{\mp,[0,\lambda]}(M,E) \Big), \quad k=0,1.
\end{equation}
Therefore
\[
d_{\bar{k},\lambda}^{\pm} = d_{\overline{k+1},\lambda}^{\mp}, \quad k=0,1.
\]
Hence
\begin{equation}\label{E:ddequ1}
\sum_{k=0,1}(-1)^k d_{\bar{k},\lambda}^- = d_{\bar{0},\lambda}^- - d_{\bar{1},\lambda}^- \equiv d_{\bar{0},\lambda}^- + d_{\bar{0},\lambda}^+ = \dim \Omega^{\bar 0}_{[0,\lambda]}(M,E), \quad \text{mod} \ 2\mathbb{Z}.
\end{equation}
From \eqref{E:nahdu2} and \eqref{E:aaa}, we obtain
\begin{equation}\label{E:ddequ2}
\dim \Omega^{\bar 0}_{[0,\lambda]}(M,E)=\dim \Omega^{\bar 1}_{[0,\lambda]}(M,E)=\dim \Omega^{\bar 0}_{[0,\lambda]}(M,E')=\dim \Omega^{\bar 1}_{[0,\lambda]}(M,E').
\end{equation}
Hence by \eqref{E:ddequ1} and \eqref{E:ddequ2}, we have
\begin{equation}\label{E:ddequ3}
\sum_{k=0,1}(-1)^k {d'}_{\bar{k},\lambda}^- = \dim \Omega^{\bar 0}_{[0,\lambda]}(M,E), \quad \text{mod} \ 2\mathbb{Z}.
\end{equation}
By the definition \eqref{E:etainv11}, we obtain
\begin{equation}\label{E:ddequ4}
2\eta(\B^H_{\bar{0},[0,\lambda]}) \equiv \dim \Omega^{\bar 0}_{[0,\lambda]}(M,E), \quad \text{mod} \ 2\mathbb{Z}.
\end{equation}
From \eqref{E:etano}, \eqref{E:ddequ3} and \eqref{E:ddequ4}, we obtain, modulo $2\mathbb{Z}$,
\begin{align}\label{E:ddequ5}
2\eta(\B^H_{\bar{0}}(\nabla^H)) & = 2 \eta(\B^H_{\bar{0},(\lambda, \infty)}(\nabla^H)) + 2\eta(\B^H_{\bar{0},[0,\lambda]}) \nonumber \\
& \equiv 2 \eta_\lambda(\nabla^H) + \sum_{k=0,1}d^-_{\bar{k},\lambda}.
\end{align}
Similarly,
\begin{equation}\label{E:ddequ6}
2\eta(\B^H_{\bar{0}}(\nabla'^H)) \equiv 2 \eta_\lambda(\nabla'^H) + \sum_{k=0,1}d^-_{\bar{k},\lambda}, \quad \text{mod} \ 2\mathbb{Z}.
\end{equation}
\subsection{Proof of Theorem \ref{T:duality0}}
Let $\rho'_{\Gamma_{[0,\lambda]}}$ be the twisted refined torsion of the complex $\Omega_{[0,\lambda]}^\bullet(M,E')$ associated to the restriction of $\Gamma$ to $\Omega_{[0,\lambda]}^\bullet(M,E')$.
By Lemma \ref{L:rhostarrho}, \eqref{E:antiiso} and the fact that $\Gamma^*=\Gamma$, cf. \cite[Proposition 3.58]{BGV},
we obtain
\begin{equation}\label{E:rhpri5}
\rho'_{\Gamma_{[0,\lambda]}} = \alpha(\rho_{\Gamma_{[0,\lambda]}})
\end{equation}
From \eqref{E:rhoh}, \eqref{E:ldetetarell} and Definition \ref{D:maindef}, we obtain
\begin{equation}\label{E:ldetetarell123}
\rho_{\operatorname{an}}(\nabla^H) = \rho_{\Gamma_{[0,\lambda]}} \cdot \exp\Big(\xi_\lambda(\nabla^H,g^M,\theta) - i \pi \eta_\lambda(\nabla^H) -\frac{i \pi}{2}\sum_{k=0,1}(-1)^k d^-_{\bar{k},\lambda} + i \pi (\operatorname{rank} (E))\eta_{\trivial} \Big).
\end{equation}
Since $\alpha$ is an anti-linear isomorphism, $\alpha(\rho_{\an} \cdot z)=\alpha(\rho_{\an}) \cdot \bar{z}$ for any $z \in \mathbb{C}$. Hence, from \eqref{E:rhpri5} and \eqref{E:ldetetarell123}, we get
\begin{equation}\label{E:rhpri3}
\alpha \big( \rho_{\operatorname{an}}(\nabla^H) \big) = \rho'_{\Gamma_{[0,\lambda]}} \cdot \exp\Big(\bar{\xi}_\lambda(\nabla^H,g^M,\theta) + i \pi \bar{\eta}_\lambda(\nabla^H) + \frac{i \pi}{2}\sum_{k=0,1}(-1)^k d^-_{\bar{k},\lambda} - i \pi (\operatorname{rank} (E))\eta_{\trivial} \Big).
\end{equation}
Using Lemma \ref{L:xiducong} and the analogue of \eqref{E:ldetetarell123} for $\rho_{\an}(\nabla'^H)$, we obtain from \eqref{E:rhpri3}
\begin{equation}\label{E:rhpri30}
\alpha \big( \rho_{\operatorname{an}}(\nabla^H) \big) = \rho_{\operatorname{an}}(\nabla'^H) \cdot \exp\Big(2 i \pi \bar{\eta}_\lambda(\nabla^H) + i \pi \sum_{k=0,1}(-1)^k d^-_{\bar{k},\lambda} - i \pi (\operatorname{rank} (E))\eta_{\trivial} \Big).
\end{equation}
From \eqref{E:rhpri30} and \eqref{E:ddequ6}, we obtain \eqref{E:aloh15}.
\section{Comparison with the twisted analytic torsion}\label{S:cwtat}
In this section we first define the twisted Ray-Singer metric $\| \cdot \|^{\RS}_{\Det ( H^\bullet(M,E,H) )}$ and then calculate the twisted Ray-Singer norm $\| \rho_{\an}(\nabla^H) \|^{\RS}_{\Det ( H^\bullet(M,E,H) )}$ of the twisted refined analytic torsion. In particular, we show that, if $\nabla$ is a Hermitian metric, then $\| \rho_{\an}(\nabla^H) \|^{\RS}_{\Det ( H^\bullet(M,E,H) )}=1$.
\subsection{The twisted analytic torsion}\label{SS:twrst1}
Let $E \to M$ be a complex vector bundle over a cloed oriented manifold $M$ of odd dimension $m=2r-1$. Let $\nabla$ be a flat connection on $E$ and $H$ be a odd degree closed form, other than one form, on $M$. Fix a Riemannian metric $g^M$ on $M$ and a Hermitian metric $h^E$ on $E$. Let ${\nabla^H}^*$ denote the adjoint of $\nabla^H:=\nabla +H \wedge \cdot$ with respect to the scalar product $<\cdot, \cdot>_M$ on $\Omega^\bullet(M,E)$ defined by $h^E$ and the Riemannian metric $g^M$.
Now let
\[
\Delta^H = {\nabla^H}^* \nabla^H + \nabla^H {\nabla^H}^*
\]
be the Laplacian twisted by the form $H$. We denote by $\Delta^H_{\bar k}$ the restriction of $\Delta^H$ to $\Omega^{\bar k}(M,E), k=0,1$. Assume that $\mathcal{I}$ is an interval of the form $[0,\lambda], (\lambda, \mu], (\lambda, \infty) (\mu \ge \lambda \ge 0)$ and let $\Pi_{\Delta^H_{\bar{k}},\mathcal{I}}$ be the spectral projection of $\Delta^H_{\bar k}$ corresponding to $\mathcal{I}$, cf. Subsection~\ref{SS:gdto}. Set
\[
\check{\Omega}^{\bar k}_{\mathcal{I}}(M,E):= \Pi_{\Delta^H_{\bar k},\mathcal{I}} \big( \Omega^\bullet(M,E) \big) \subset \Omega^{\bullet}(M,E).
\]
Let $\Delta^{H,\mathcal{I}}_k$ denote the restriction of $\Delta^H_{\bar k}$ to $\check{\Omega}^{\bar k}_{\mathcal{I}}(M,E)$ and define
\begin{equation}\label{E:trs11}
\trs_{\mathcal{I}} = \trs_{\mathcal{I}}(\nabla^H):=\exp \Big( \frac{1}{2} \sum_{k=0,1}(-1)^{k+1} \LDet'_{-\pi}\big( {\nabla^H}^* \nabla^H \big)\big|_{\Omega^{\bar k}_{\mathcal{I}}(M,E)} \Big).
\end{equation}
It is not difficult to check that, for any non-negative, real numbers $\mu \ge \lambda \ge 0$,
\begin{equation}\label{E:spli2}
\trs_{(\lambda,\infty)} = \trs_{(\lambda,\mu]}\cdot \trs_{[\mu,\infty)}.
\end{equation}
Note that if $\eta_{\bar k}$ is the unit volume element of $H^{\bar k}(M,E,H), k=0,1$, then
\begin{equation}\label{E:tatmw}
\tau(M,E,H):= \big(\trs_{(0,\infty)}\big)^{-1} \cdot \eta_{\bar 0} \otimes \eta_{\bar 1}^{-1} \in \Det \big( H^\bullet(M,E,H) \big)
\end{equation}
is the {\em twisted analytic torsion}, introduced by V. Mathai and S. Wu in \cite{MW}.
For each $\lambda >0$, the cohomology of the finite dimensional complex $\big( \check{\Omega}^\bullet_{[0,\lambda]}(M,E),\nabla^H \big)$ is naturally isomorphic to $H^\bullet(M,E,H)$. Identifying these two cohomology spaces, we then obtain from \eqref{E:phic} an isomorphism
\begin{equation}\label{E:philam}
\phi_\lambda = \phi_{\check{\Omega}^\bullet_{[0,\lambda]}(M,E)} : \Det \big( \check{\Omega}^\bullet_{[0,\lambda]}(M,E) \big) \rightarrow \Det \big( H^\bullet(M,E,H) \big).
\end{equation}
The scalar product $<\cdot,\cdot>$ on $\check{\Omega}^\bullet_{[0,\lambda]}(M,E) \subset \Omega^\bullet(M,E)$ defined by $g^M$ and $h^E$ induces a metric ${\| \cdot \|}_{\Det \big( \check{\Omega}^\bullet_{[0,\lambda]}(M,E)\big)}$ on the determinant line $\Det \big( \check{\Omega}^\bullet_{[0,\lambda]}(M,E)\big)$. Let ${\|\cdot \|_{\lambda}}$ denote the metric on the determinant line $\Det\big(H^\bullet(M,E,H) \big)$ such that the isomorphism \eqref{E:philam} is an isometry. Then, for $c \in \Det \big( \check{\Omega}^\bullet_{[0,\lambda]}(M,E) \big)$, we have
\begin{equation}\label{E:isoequiv}
{\| c \|}_{\Det \big( \check{\Omega}^\bullet_{[0,\lambda]}(M,E)\big)} = {\|\phi_\lambda(c) \|_{\lambda}}.
\end{equation}
Using the Hodge theory, we have the canonical identification
\[
H^{\bar k}(M,E,H) \cong \Ker \Delta^H_{\bar k}, \quad k=0,1.
\]
By their inclusion in $\Omega^{\bar k}(M,E)$, the space of twisted harmonic forms $\Ker \Delta^H_{\bar k}$ inherits a metric. We denote by $|\cdot|_{\Det\big( H^\bullet(M,E,H) \big)}$ the corresponding metric on $\Det\big( H^\bullet(M,E,H) \big)$. By definition
\begin{equation}\label{E:metricequ}
{\| \cdot \|}_{\Det \big( \check{\Omega}^\bullet_{\{0\}}(M,E)\big)} = |\cdot|_{\Det\big( H^\bullet(M,E,H) \big)}.
\end{equation}
The following is twisted analogue of \cite[Proposition 1.5]{BGS}. See also \cite{RuSe} for the contact version.
\begin{proposition}
\begin{equation}\label{E:metrican}
\|\cdot \|_\lambda = |\cdot|_{\Det\big( H^\bullet(M,E,H) \big)} \cdot \trs_{(0,\lambda]}.
\end{equation}
\end{proposition}
\begin{proof}
For $k=0,1$, fix $c'_{\bar k} \in \Det \big( \check{\Omega}_{\{0\}}^{\bar k}(M,E)\big) \cong \Det \big( H^{\bar k}(M,E,H) \big)$ and $c''_{\bar k} \in \Det \big( \check{\Omega}_{(0,\lambda]}^{\bar k}(M,E)\big)$. Then, using the natural isomorphism
\begin{align*}
& \Det \big( \check{\Omega}_{\{0\}}^{\bar k}(M,E)\big) \otimes \Det \big( \check{\Omega}_{(0,\lambda]}^{\bar k}(M,E)\big) \nonumber \\
& \cong \Det \big(\check{\Omega}_{\{0\}}^{\bar k}(M,E) \oplus \check{\Omega}_{(0,\lambda]}^{\bar k}(M,E) \big) \nonumber \\
& = \Det \big( \check{\Omega}_{[0,\lambda]}^{\bar k}(M,E) \big),
\end{align*}
we can regard the tensor product $c_{\bar k}:= c'_{\bar k} \otimes c''_{\bar k}$ as an element of $\Det \big( \check{\Omega}_{[0,\lambda]}^{\bar k}(M,E) \big)$. Denote by
\[
\check{\Omega}^{\bar{k},-}_{\mathcal{I}}(M,E) = \Ker \nabla^H \cap \check{\Omega}^{\bar{k}}_{\mathcal{I}}(M,E), \, \quad
\check{\Omega}^{\bar{k},+}_{\mathcal{I}}(M,E) = \Ker {\nabla^H}^* \cap \check{\Omega}^{\bar{k}}_{\mathcal{I}}(M,E).
\]
Since the complexes $\check{\Omega}^{\bar k}_{(0,\lambda]}(M,E), k=0,1$, are acyclic, it is not difficult to see that each $\check{\Omega}^{\bar k}_{(0,\lambda]}(M,E)$ splits orthogonally into
\begin{equation}\label{E:splittt}
\check{\Omega}^{\bar k}_{(0,\lambda]}(M,E) = \check{\Omega}^{\bar{k},-}_{(0,\lambda]}(M,E) \oplus \check{\Omega}^{\bar{k},+}_{(0,\lambda]}(M,E).
\end{equation}
Take $a_{\bar k} \in \Det \big(\check{\Omega}^{\bar{k},+}_{(0,\lambda]}(M,E)\big)$ so that $c''_{\bar k}=\nabla^H (a_{\overline{k+1}}) \wedge a^{\bar k}$. Denote by $c=c'\otimes c''$, where $c'=c'_{\bar 0} \otimes {c'_{\bar 1}}^{-1}$ and $c''=c''_{\bar 0} \otimes {c''}^{-1}_{\bar 1}$. Then
\begin{align}\label{E:split1}
\|c \|_{\Det \big( \check{\Omega}^{\bullet}_{[0,\lambda]}(M,E) \big)} & = \|c'\|_{\Det\big( \check{\Omega}_{\{0\}}^\bullet(M,E) \big)} \times \| c'' \|_{\Det \big( \check{\Omega}_{(0,\lambda]}^{\bar k}(M,E)\big)}\nonumber \\
& = \|c'\|_{\Det\big( \check{\Omega}_{\{0\}}^\bullet(M,E) \big)} \times \|a_{\bar 0}\|_{\Det \big( \check{\Omega}^{\bar{0},+}_{(0,\lambda]}(M,E) \big)}\times \|\nabla^H(a_{\bar 1})\|_{\Det \big( \check{\Omega}^{\bar{0},-}_{(0,\lambda]}(M,E) \big)}
\nonumber \\
& \times \Big(\|a_{\bar 1}\|_{\Det \big( \check{\Omega}^{\bar{1},+}_{(0,\lambda]}(M,E) \big)} \Big)^{-1}
\times \Big(\|\nabla^H(a_{\bar 0})\|_{\Det \big( \check{\Omega}^{\bar{1},-}_{(0,\lambda]}(M,E) \big)} \Big)^{-1}.
\end{align}
where $\| \cdot \|_{V}$ denotes the naturally induced norm on the subspace $V$. The space $\check{\Omega}^{\bar k}_{(0,\lambda]}(M,E)$ splits orthogonally into eigenspaces.
\[
\check{\Omega}^{\bar k}_{(0,\lambda]}(M,E)= \oplus_{\nu \le \lambda}\check{\Omega}^{\bar k}_{\{\nu\}}(M,E)
\]
Given $\nu \in (0,\lambda]$, we choose an orthogonal basis $(v_1, \cdots, v_{n_k})$ of each eigenspace $\check{\Omega}^{\bar{k},+}_{\{ \nu \}}(M,E)$ and choose the element $a_{\bar{k},\nu}=v_1 \wedge \cdots \wedge v_{n_k} \in \Det \big( H^{\bar{k},+}_{\{\nu \}}(M,E,H) \big)$, where $n_k=\dim \check{\Omega}^{\bar{k},+}_{\{ \nu \}}(M,E)$. Then
\begin{align}\label{E:split2}
\| \nabla^H (a_{\bar{k},\nu}) \|_{\Det\big( \check{\Omega}^{\overline{k+1},-}_{\{ \nu \}}(M,E) \big)} & = \| \nabla^H v_1 \wedge \cdots \wedge \nabla^H v_{n_k}\|_{\Det\big( \check{\Omega}^{\overline{k+1},-}_{\{ \nu \}}(M,E) \big)} \nonumber \\
& = \| \nabla^H v_1\|_{\Det\big( \check{\Omega}^{\overline{k+1},-}_{\{ \nu \}}(M,E) \big)} \times \cdots \times \| \nabla^H v_{n_k}\|_{\Det\big( \check{\Omega}^{\overline{k+1},-}_{\{ \nu \}}(M,E) \big)} \nonumber \\
& = \nu^{\frac{n_k}{2}} \|v_1\|_{\Det\big( \check{\Omega}^{\bar{k},+}_{\{ \nu \}}(M,E) \big)} \times \cdots \times \|v_{n_k}\|_{\Det\big( \check{\Omega}^{\bar{k},+}_{\{ \nu \}}(M,E) \big)} \nonumber \\
& = \nu^{\frac{n_k}{2}}\| a_{\bar{k},\nu}\|_{\Det\big( \check{\Omega}^{\bar{k},+}_{\{ \nu \}}(M,E) \big)}.
\end{align}
By combining \eqref{E:split1} with \eqref{E:split2}, we obtain
\begin{equation}\label{E:split3}
{\|c \|}_{\Det\big( \check{\Omega}_{[0,\lambda]}^\bullet(M,E) \big)} = \|c'\|_{\Det\big( \check{\Omega}_{\{0\}}^\bullet(M,E) \big)} \times \prod_{k=0,1} \big(\Det_{-\pi}\big( \nabla^H {\nabla^H}^* \big) \big|_{ \Omega^{\bar k}_{(0,\lambda]}(M,E)} \big)^{(-1)^{k+1}/2}.
\end{equation}
By \eqref{E:trs11}, \eqref{E:isoequiv}, \eqref{E:metricequ} and \eqref{E:split3}, we obtain the result.
\end{proof}
By \eqref{E:metrican}, we have, for $0 \le \lambda \le \mu$,
\begin{equation}\label{E:spli1}
\|\cdot \|_\mu = \| \cdot \|_\lambda \cdot \trs_{(\lambda,\mu]}.
\end{equation}
The {\em twisted Ray-Singer metric} on $\Det \big( H^\bullet(M,E,H) \big)$ is defined by the formula
\begin{equation}\label{E:metric}
\| \cdot \|^{\RS}_{\Det \big( H^\bullet(M,E,H) \big)}:= \| \cdot \|_\lambda \cdot \trs_{(\lambda,\infty)}, \quad \lambda \ge 0.
\end{equation}
It follows immediately from \eqref{E:spli2} and \eqref{E:spli1} that $\| \cdot \|_{\Det \big( H^\bullet(M,E,H) \big)}$ is independent of the choice of $\lambda \ge 0$. Note that for $\lambda =0$, by \eqref{E:tatmw} and \eqref{E:metric}, we have
\[
\| \tau(M,E,H) \|^{\RS}_{\Det \big( H^\bullet(M,E,H) \big)}=1.
\]
\begin{theorem}\label{T:comthm1}
Let $E$ be a complex vector bundle over a closed oriented odd dimensional manifold $M$ and let $\nabla$ be a flat connection on $E$. Further, let $H$ be a odd-degree closed form on $M$ and denote by $\nabla^H:=\nabla+H \wedge \cdot$. Then
\begin{equation}\label{E:comthm2}
\| \rho_{\an}(\nabla^H) \|^{\RS}_{\Det\big( H^\bullet(M,E,H) \big)}= e^{\pi \operatorname{Im} \eta(\nabla^H,g^M)},
\end{equation}
where
\[
\eta(\nabla^H,g^M) = \eta\big( \B_{\bar 0}(\nabla^H,g^M) \big).
\]
In particular, if $\nabla$ is a Hermitian connection, then $\| \rho_{\an}(\nabla^H) \|^{\RS}_{\Det\big( H^\bullet(M,E,H) \big)}=1$.
\end{theorem}
The rest of this section is concerned with the proof of Theorem \ref{T:comthm1}.
\subsection{Comparison between the twisted Ray-Singer metrics associated to a connection and to its dual}
We assume that $\theta \in (-\pi/2,0)$ is any Agmon angle for the twisted odd signature operator $\B^H$ such that no eigenvalue of $\B^H_{(\lambda,\infty)}$ lies in the solid angles $L_{[-\theta-\pi,-\pi/2]}$, $L_{(-\pi/2,\theta]}$, $L_{[-\theta,\pi/2)}$ and $L_{(\pi/2,\theta+\pi]}$, cf. \cite[Subsection 11.4]{BK3}
As in Subsection \ref{SS:duaconne1}, let $\nabla'$ be the connection dual to $\nabla$ with respect to the Hermitian metric $h^E$ and let $E'$ denote the flat bundle $(E,\nabla')$. Let $H$ be an odd degree closed form, other than one form, on $M$ and denote by $\nabla^H:=\nabla + H \wedge \cdot$. Let
\[
{\Delta'}^H=({\nabla'^H})^* \nabla'^H + \nabla'^H (\nabla'^H)^*,
\]
denote the twisted Laplacian of the connection $\nabla'$ twisted by the form $H$. For any $\lambda \ge 0$, we denote by
\[
\check{\Omega}^\bullet_{[0,\lambda]}(M,E') \subset \Omega^\bullet(M,E')
\]
the image of the spectral projection $\Pi_{\Delta'^H,[0,\lambda]}$, cf. Subsection \ref{SS:gdto}. As in Subsection \ref{SS:twrst1}, we use the scalar product induced by $g^M$ and $h^E$ on $\check{\Omega}^\bullet_{[0,\lambda]}(M,E')$ to construct a metric $\| \cdot \|'_\lambda$ on $\Det \big( H^\bullet(M,E',H) \big)$ and we define the twisted Ray-Singer metric on $\Det \big( H^\bullet(M,E',H) \big)$ by the formula
\begin{equation}\label{E:metricdual}
\| \cdot \|^{\RS}_{\Det \big( H^\bullet(M,E',H) \big)}:= \| \cdot \|'_\lambda \cdot \trs_{(\lambda,\infty)}(\nabla'^H), \quad \lambda \ge 0.
\end{equation}
As the untwisted case, cf. \cite[Subsection 11.6]{BK3}, we have the following identification,
\[
\check{\Omega}^\bullet_{[0,\lambda]}(M,E') \cong \check{\Omega}^{m-\bullet}_{[0,\lambda]}(M,E)^*,
\]
which preserves the scalar products induced by $g^M$ and $h^E$ on $\check{\Omega}^\bullet_{[0,\lambda]}(M,E')$ and $\check{\Omega}^{m-\bullet}_{[0,\lambda]}(M,E)^*$. Hence, the anti-linear isomorphism $\alpha$, cf. \eqref{E:antiiso}, is an isometry with respect to the metrics $\|\cdot \|_\lambda$ and $\|\cdot \|'_\lambda$. In particular,
\[
\|\rho_{\an}(\nabla^H) \|_\lambda = \|\alpha(\rho_{\an}(\nabla^H)) \|'_\lambda.
\]
It follows from \eqref{E:aloh15} that
\begin{equation}\label{E:rholamim}
\|\rho_{\an}(\nabla^H) \|_\lambda = \| \rho_{\an}({\nabla}'^H) \|'_\lambda \cdot e^{2\pi \operatorname{Im} \eta(\nabla^H,g^M)}.
\end{equation}
We need the following lemma. For untwisted case, cf. for example \cite[Lemma 8.8]{BK1}.
\begin{lemma}
\begin{equation}\label{E:trl32}
\trs_{(\lambda,\infty)}(\nabla'^H)=\trs_{(\lambda,\infty)}(\nabla^H).
\end{equation}
\end{lemma}
\begin{proof}
From \eqref{E:nahdu1}, we have
\begin{equation}\label{E:nahnainterw}
(\nabla^H)^*\nabla^H = \Gamma \nabla'^H \Gamma \nabla^H = \Gamma (\nabla'^H \Gamma \nabla^H \Gamma) \Gamma = \Gamma \nabla'^H (\nabla'^H)^* \Gamma.
\end{equation}
As in \eqref{E:splittt}, we have
\begin{equation}\label{E:splittt0}
\check{\Omega}^{\bar k}_{(\lambda,\infty)}(M,E) = \check{\Omega}^{\bar{k},-}_{(\lambda,\infty)}(M,E) \oplus \check{\Omega}^{\bar{k},+}_{(\lambda,\infty)}(M,E).
\end{equation}
The operator $\nabla^H$ maps $\check{\Omega}^{\bar{k},+}_{(\lambda,\infty)}(M,E)$ isomorphically onto $\check{\Omega}^{\overline{k+1},-}_{(\lambda,\infty)}(M,E)$ and
\begin{equation}\label{E:intewin2}
\nabla^H\big|_{\check{\Omega}^{\bar{k},+}_{(\lambda,\infty)}(M,E)} \big((\nabla^H)^* \nabla^H\big) \big|_{\check{\Omega}^{\bar{k},+}_{(\lambda,\infty)}(M,E)} = \big(\nabla^H (\nabla^H)^* \big|_{\check{\Omega}^{\overline{k+1},-}_{(\lambda,\infty)}(M,E)}\big) \cdot \nabla^H\big|_{\check{\Omega}^{\bar{k},+}_{(\lambda,\infty)}(M,E)}.
\end{equation}
Hence, by \eqref{E:intewin2}, we have
\begin{equation}\label{E:interwi1}
\LDet'_{-\pi}\big( (\nabla'^H)^* \nabla'^H \big)\big|_{\Omega^{\bar{k},+}_{(\lambda,\infty)}(M,E)} = \LDet'_{-\pi}\big( \nabla'^H (\nabla'^H)^* \big)\big|_{\Omega^{\overline{k+1},-}_{(\lambda,\infty)}(M,E)}.
\end{equation}
From \eqref{E:nahnainterw}, we obtain
\begin{align*}
&\log \trs_{(\lambda,\infty)}(\nabla^H)\\
&= \frac{1}{2} \sum_{k=0,1}(-1)^{k+1} \LDet'_{-\pi}\big( (\nabla^H)^* \nabla^H \big)\big|_{\Omega^{\bar{k},+}_{(\lambda,\infty)}(M,E)} \\
& = \frac{1}{2} \sum_{k=0,1}(-1)^{k+1} \LDet'_{-\pi}\big( \Gamma \nabla'^H (\nabla'^H)^* \Gamma \big)\big|_{\Omega^{\bar{k},+}_{(\lambda,\infty)}(M,E)} \\
& = \frac{1}{2} \sum_{k=0,1}(-1)^{k+1} \LDet'_{-\pi}\big( \nabla'^H (\nabla'^H)^* \big)\big|_{\Omega^{\overline{m-k},-}_{(\lambda,\infty)}(M,E)} \\
&= \frac{1}{2} \sum_{k=0,1}(-1)^k \LDet'_{-\pi}\big( \nabla'^H (\nabla'^H)^* \big)\big|_{\Omega^{\bar{k},-}_{(\lambda,\infty)}(M,E)} \\
&= \frac{1}{2} \sum_{k=0,1}(-1)^{k+1} \LDet'_{-\pi}\big( (\nabla'^H)^* \nabla'^H \big)\big|_{\Omega^{\bar{k},+}_{(\lambda,\infty)}(M,E)},
\end{align*}
where we use \eqref{E:interwi1} for the last equality. This proves the lemma.
\end{proof}
Hence, from \eqref{E:metric}, \eqref{E:metricdual}, \eqref{E:rholamim} and \eqref{E:trl32}, we conclude that
\begin{equation}\label{E:rhoimet}
\|\rho_{\an}(\nabla^H) \|^{\RS}_{\Det \big( H^\bullet(M,E,H) \big)} = \| \rho_{\an}({\nabla}'^H) \|^{\RS}_{\Det \big( H^\bullet(M,E',H) \big)} \cdot e^{2\pi \operatorname{Im} \eta(\nabla^H,g^M)}.
\end{equation}
\subsection{Direct sum of a connection and its dual}
Let
\[
\widetilde{\nabla} = \left(
\begin{array}{cc}
\nabla & 0 \\
0 & \nabla'
\end{array}
\right)
\]
denote the flat connection on $E \oplus E$ obtained as a direct sum of the connections $\nabla$ and $\nabla'$. Denote by $$\widetilde{\nabla}^H:= \left(
\begin{array}{cc}
\nabla^H & 0 \\
0 & \nabla'^H
\end{array}
\right).$$
Then a discussion similar to \cite[Subsection 11.7]{BK3}, where the untwisted case was treated, one easily obtains that,
\[
\rho_{\an}(\widetilde{\nabla}^H)= \mu_{H^\bullet(M,E,H),H^\bullet(M,E',H)}\big( \rho_{\an}(\nabla^H) \otimes \rho_{\an}(\nabla'^H) \big)
\]
and
\[
\| \rho_{\an}(\widetilde{\nabla}^H) \|^{\RS}_{\Det \big( H^\bullet(M,E \oplus E',H \big)}=\| \rho_{\an}(\nabla^H) \|^{\RS}_{\Det \big( H^\bullet(M,E,H \big)} \cdot \| \rho_{\an}(\nabla'^H) \|^{\RS}_{\Det \big( H^\bullet(M, E',H \big)}.
\]
Combining this later equality with \eqref{E:rhoimet}, we get
\[
\| \rho_{\an}(\widetilde{\nabla}^H) \|^{\RS}_{\Det \big( H^\bullet(M,E \oplus E',H \big)} = \big( \| \rho_{\an}(\nabla^H) \|^{\RS}_{\Det \big( H^\bullet(M,E,H \big)} \big)^2 \cdot e^{-2\pi \operatorname{Im} \eta(\nabla^H,g^M)}.
\]
Hence, \eqref{E:comthm2} is equivalent to the equality
\begin{equation}\label{E:comrart}
\| \rho_{\an}(\widetilde{\nabla}^H) \|^{\RS}_{\Det \big( H^\bullet(M,E \oplus E',H \big)} = 1.
\end{equation}
By a slight modification of the deformation argument in \cite[Section 11, P.205-211]{BK3}, where the untwisted case was treated, we can obtain \eqref{E:comrart}. Hence, we finish the proof of Theorem \ref{T:comthm1}. | 0.001497 |
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- Affection – The first step is to show your children tons of affection. Knowing they are loved and cared for is a big part of helping kids deal with adversity. Failures in life are easier to take when they know they are loved.
- Praise – Always praise your child’s accomplishments and be specific when you do. Give them feedback that links their achievements to what they did to accomplish it. An example would be to state how all the extra studying paid off with a good grade on a test.
- Monitor – Pay attention to what your child is watching on television, the games they play and the books they read. You should monitor their activities to make sure they’re not exposed to too many bad influences. Also, make sure they’re not attempting to accomplish too much, as taking on more than they can reasonably handle can set them up for failure.
- Be an example – Children will often mirror their parent’s attitudes, so be a good example. Kids will only learn to be optimistic if you show them how. Parents who are continually complaining or are pessimistic will pass those traits on to their children.
- Reinforcement – When your child expresses optimism, be sure to comment on their great attitude. This positive reinforcement will put the emphasis on the desired behavior instead of focusing on the negative.
- Accentuate the positive – When something bad happens, always try to find the bright side. If bad weather cancels an event, find something fun to do instead. Accentuate the positive by commenting how you wouldn’t have had so much fun if the picnic wasn’t cancelled.
- Thought catching – Teach children how to do thought catching to prevent negative behavior. When something bad happens ask them what thoughts they had so they’re aware of them and can capture the negativity. Kids often have negative thoughts without ever realizing it.
-.
- Promote success – Encourage kids with age appropriate activities they are able to excel at to promote success. Having unrealistic expectations only sets children up for failure, so make sure they’re not trying to do too much.
- Laugh – The best way to encourage an optimistic attitude is with lots of laughter. Teach children how to laugh at themselves and not take everything so seriously. Laughing is the best way to diffuse a troubling situation.... and totally agree with all 10 points!
Thanks Isabel.
What's funny is so many people say our mind is our most valuable tool, but we have gyms all over the place to work out our bodies.
In addition, these points are so simple to do at home, while a child is growing up, and yet many people don't do it ;(
It saddens me.
I guess that's why it's a large part of my book.
This is great advice for those of us who don't have kids and just want to be more positive. Especially the thought catching part - as I think we adults aren't aware how negative we are at times.
Well put Michelle.
#1 on my list is model it! You've got it there as "Be an example..."
Thanks another really good one!
Thanks for sharing that Bruce.
This is such an important post. I don't know that I'm a pessimist, but I certainly consider myself a realist. I need to make sure that I model positive behavior for my children. I agree completely with what you wrote about optimism helping people bounce back from negative or trying happenings. I want that for my children, so it's important for me to model this behavior for them.
From the perspective of a child who grew up in a mostly negative or neutral family, it's vitally important! It's most of what my book is about.
Great list. I've never heard of stopping the negative thinking as "thought catching," but I really like that. I'm going to remember that!
It's funny how some phrases are so obvious once we've made sense of them... | 0.996143 |
TITLE: Numerical complexity of different quantum chemistry approaches
QUESTION [6 upvotes]: For the introductory course for the students I am trying to collect the brief overview of different quantum chemistry methods and their numerical complexity. The second point is surprisingly poorly explained in all textbooks I was able to find. In the best case, it is a small note like
The most computationally expensive step in the Hartree–Fock procedure is the formation of the two-electron part, $G$, of the Fock matrix; the computation of G requires $O(n^4)$ steps, where $n$ is the number of basis functions (with integral screening this cost asymptotically approaches $O(n^2)$)
[source is "Parallel computing in quantum chemistry" by C.L. Janssen, 2008], but there are no estimates for all methods discussed in the book and I have problems to find this kind of information elsewhere.
Probably, there exists a more extended discussion of this point with more details (which part of the algorithm limits the complexity, what can be done to improve the scalability, etc.)?
REPLY [1 votes]: For your basic single-reference quantum chemistry ground state methods:
HF: ${\cal O}(N^4)$
DFT: ${\cal O}(N^3)$ or ${\cal O}(N^4)$ with HF-exchange
MP2: ${\cal O}(N^5)$
CCSD: ${\cal O}(N^6)$
and excited state methods:
TDHF: ${\cal O}(N^4)$
TDDFT: ${\cal O}(N^3)$ or ${\cal O}(N^4)$
$\mathrm{G}_0\mathrm{W}_0$: ${\cal O}(N^4)$
GW: ${\cal O}(N^5)$
EOMCCSD: ${\cal O}(N^6)$
It's probably more useful to know why these methods come with the aforementioned cost scalings. It's all about identifying the rate-limiting step in the calculation, which is typically a type of tensor contraction. Let's take Hartree-Fock for example. The most costly step comes from the construction of the effective one-body interaction, which for a conventional closed-shell calculation, looks like this
$$f_{\mu \nu} = 2h_{\mu \nu} + g_{\mu \nu}$$
$$g_{\mu \nu} = \sum_{\rho \sigma} \langle \mu \rho | \nu \sigma \rangle P_{\rho \sigma} -
\frac{1}{2}\langle \mu \rho | \sigma \nu \rangle P_{\rho \sigma}$$
We can identify the first term in $g_{\mu\nu}$ as the Coulomb interaction and the second term as exchange. If one were to unravel the above tensor contraction to execute using some nested do/for loops, one would need something like this (using Fortran 90 as an example syntax):
do p = 1,N
do q = 1,N
K = 0.0
J = 0.0
do r = 1,N
do s = 1,N
K = K + V(p,q,r,s) * P(q,s)
J = J + V(p,q,s,r) * P(q,s)
end do
end do
g(p,q) = 2*K + J
end do
end do
Clearly, the above loop has a computational expense that scales as $\mathcal{O}(N^4)$ because we have 4 nested loops, each running from $1$ to $N$. Another example we can do is CCSD. If you derive the CCSD equations (in spinorbital form for simplicity), one finds that the most expensive term is the so-called ladder diagram contribution $
\frac{1}{4}\sum_{mnef} v_{mn}^{ef} t_{ef}^{ij} t_{ab}^{mn}$ where $v_{pq}^{rs}$ are the antisymmetrized electron repulsion integrals and $t_{ab}^{ij}$ are the 2-body cluster amplitudes. If we were to repeat the above procedure of naively unravelling into nested "do" loops, one find that the ladder diagram has a cost that scales as $\mathcal{O}(N^8)$. Why then is CCSD listed as $\mathcal{O}(N^6)$? The answer is that the tensor contraction can be broken up into two steps. If one first computes $\chi_{mn}^{ij} = \sum_{ef} v_{mn}^{ef} t_{ef}^{ij}$, this operation scales as $\mathcal{O}(N^6)$. Then, one computes $\sum_{mn} \chi_{mn}^{ij} t_{ab}^{mn}$ which also scales as $\mathcal{O}(N^6)$. Thus, you can compute the $\mathcal{O}(N^8)$ contraction at the cost of no more than multiple $\mathcal{O}(N^6)$ operations, hence CCSD scales as $\mathcal{O}(N^6)$! These types of time-saving factorizations are easily deduced by inspection of the diagrammatic form of these many-body equations (or, one can just look at the tensor equations directly, however, it's a bit harder that way). Optimal factorization is of utmost importance in any quantum chemistry code.
And you can also reduce the scaling of MP2, TDHF, and HF using density-fitting and Cholesky decompositions by an order of magnitude to ${\cal O}(N^4)$ and ${\cal O}(N^3)$, respectively. There are also density-fitting type speedups (called least-squares tensor hypercontraction, or LS-THC) that can be applied to coupled-cluster that reduce the scaling of CCSD to ${\cal O}(N^4)$ which is probably the best you can get. Certainly, every method can have its scaling reduced dramatically by introducing some kind of orbital localization because the steep scaling of quantum chemistry stems from the fact that even atoms far away from each other, say on opposite ends of a molecule, end up interacting because the standard MO basis used is delocalized. However, localization comes with a heavy cost of losing orthonormality, so it's generally tricky implement. Actually, a very promising class of methods are the rank-compression methods like Cholesky decomposition because you can very robustly reduce the time needed for rate-limiting tensor contractions while having a very controllable and systematic error - something most other "fast scaling" approaches do not have. | 0.283606 |
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Content Writer | Research and Writing
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Scheme
Rec.INN
CAS registry number (Chemical Abstracts Service)
0001405-52-3
Chemical Formula
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Molecular Weight
1673
Therapeutic Category
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Chemical Name
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Foreign Names
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- Sulfomixina (Spanish)
Generic Name
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\begin{document}
\RRNo{7548}
\makeRR
\allowdisplaybreaks
\section{Introduction}\label{sec:intro}
Data approximation using sparse linear expansions from overcomplete dictionaries has become a central theme in signal and image processing with applications ranging from data acquisition (compressed sensing) to denoising and compression. Dictionaries can be seen as collections of vectors $\{\phi_j\}$ from a Banach space $X$ equipped with a norm $\norm{\cdot}{X}$, and one wishes to approximate data vectors $f$ using $k$-term expansions $\sum_{j \in I} c_j \phi_j$ where $I$ is an index set of size $k$. Formally, using the matrix notation $\dict c = \sum_j c_j \phi_j$ and denoting $\norm{c}{0} = \sharp \{j, c_j \neq 0\}$ the number of nonzero components in the vector $c$ , we can define the (nonlinear) set of all such $k$-term expansions
\[
\Sigma_k(\dict) := \left\{\dict c,\quad \norm{c}{0} \leq k\right\}.
\]
\subsection{Best $k$-term approximation}
A first question we may want to answer, for each data vector $f$, is: {\em how well can we approximate it using elements of $\Sigma_k(\dict)$}? The error of best $k$-term approximation is a quantitative answer for a fixed $k$:
\[
\sigma_k(f,\dict) := \inf_{y \in \Sigma_k(\dict)} \norm{f-y}{X}.
\]
A more global view is given by the largest approximation rate $s>0$ such that\footnote{
The notation $a \lesssim b$ indicates the existence of a finite constant $C$ such that $a \leq C \cdot b$. The notation $a \asymp b$ means that we have both $a \lesssim b$ and $b \lesssim a$. As usual, $C$ will denote a generic finite constant, independent from the other quantities of interest. Different occurences of this notation in the paper may correspond to different values of the constant.}
\[
\sigma_k(f,\dict) \lesssim k^{-s},\quad \forall k \geq 1.
\]
To measure more finely the rate of approximation, one defines for $0<q<\infty$ \cite[Chapter 7, Section 9]{DeVore:1996aa}
\begin{equation}
\label{eq:DefApproxNorm}
\seminorm{f}{\cA_q^s(\dict)} := \left(\sum_{k \geq 1} \left[k^s \sigma_k(f,\dict)\right]^q k^{-1} \right)^{1/q}
\asymp \left(\sum_{j \geq 0} \left[2^{js} \sigma_{2^j}(f,\dict)\right]^q\right)^{1/q}.
\end{equation}
and the associated {\em approximation spaces}
\begin{equation}
\label{eq:DefApproxSpace}
\cA_q^s(\dict) := \{f \in \cH, \seminorm{f}{\cA_q^s(\dict)}<\infty\}\quad
\end{equation}
\subsection{Sparse or compressible representations}
Alternatively, we may be interested in {\em sparse / compressible representations} of $f$ in the dictionary. Suppose the vectors forming $\dict$ are quasi-normalized in $X$: for all $j$, $0< c \leq \norm{\phi_j}{X} \leq C <
\infty$.
Then using $\ell^\tau$ (quasi)-norms (in particular, $0 < \tau \leq 1$) one defines\footnote{It has been shown in~\cite{Gribonval:2004ab} that under mild assumptions on the dictionary, such as Eq.~\eqref{eq:DefUpperFrameBound}, the definition~\eqref{eq:DefSparseNorm} is fully equivalent to the more general topological definition of $\norm{f}{\ell^\tau(\dict)}$ introduced in~\cite{DeVore:1996aa}.}
\begin{equation}
\label{eq:DefSparseNorm}
\norm{f}{\ell^\tau(\dict)} := \inf_{c | \dict c = f} \norm{c}{\tau}
\end{equation}
and the associated {\em sparsity spaces} (also called {\em smoothness spaces}, for when $\dict$ is, e.g., a wavelet frame, they indeed characterize smoothness on the Besov scale)
\[
\ell^\tau(\dict) := \{f, \norm{f}{\ell^\tau(\dict)}<\infty\}.
\]
\subsection{Direct and inverse estimates}
Interestingly, the above defined concepts are related. In a Hilbert space $X = \cH$, when $\dict$ satisfies the upper bound
\begin{equation}
\label{eq:DefUpperFrameBound}
\norm{\dict c}{\cH}^2 \leq \ufbound \cdot \norm{c}{2}^2,\quad \forall c \in \ell^2,
\end{equation}
the sparsity spaces for $0<\tau<2$ are characterized as
\[
\ell^\tau(\dict) = \{f,\ \exists c,\ \norm{c}{\ell^\tau}<\infty, f = \dict c\}= \dict \ell^\tau,
\]
and for any $s>0$ we have the so-called {\em Jackson inequality}
\begin{equation}
\label{eq:DefJineq}
\sigma_k(f,\dict) \leq C_\tau(\ufbound) \cdot \norm{f}{\ell^\tau(\dict)} \cdot k^{-s},\quad s = \frac1\tau-\frac12,\quad \forall f\in \ell^\tau(\dict),\, \forall k\in\bN
\end{equation}
where, as indicated by the notation, the constant $C_\tau(\ufbound)$ only depends on $\tau$ and the upper bound $\ufbound$ in~\eqref{eq:DefUpperFrameBound}.
Note that the upper bound~\eqref{eq:DefUpperFrameBound} holds true whenever the dictionary is a {\em frame}: $\ufbound$ is then called the {\em upper frame bound}, and we will use this terminology.
When $\dict$ is an orthogonal basis, a converse result is true: if $\sigma_k(f,\dict)$ decays as $k^{-s}$ then $\norm{f}{\ell^\tau_w(\dict)} < \infty$, where $\ell^\tau_w$ is a weak $\ell^\tau$ space \cite{DeVore:1996aa} and $s = 1/\tau-1/2$. More generally, inverse estimates are related to a {\em Bernstein inequality} \cite{DeVore:1988aa,DeVore:1993aa}.
\begin{equation}
\label{eq:DefBineq}
\norm{f_k}{\ell^\tau(\dict)} \leq C \cdot k^{r} \cdot \norm{f_k}{\cH},\quad \forall f_k \in \Sigma_k(\dict), \forall k.
\end{equation}
The inequality \eqref{eq:DefBineq} is related to the so-called Bernstein-Nikolsky inequality, we refer the reader to \cite{Bang:1991aa,DeVore:1993aa} for more information.
\subsection{When approximation spaces are sparsity spaces}
When a Jackson inequality holds together with a Bernstein inequality with matching exponent $r = 1/\tau-1/2$, it is possible to characterize (with equivalent (quasi)-norms) the approximation spaces $\cA_q^s(\dict)$ as real interpolation spaces \cite[Chapter 7]{DeVore:1993aa} between $\cH$, denoted $(\cH,\ell^\tau(\dict))_{\theta,q}$, where $s = \theta r$, $0<\theta<1$.
The definition of real interpolation spaces will be recalled in Section~\ref{sec:interpolation}. Let us just mention here that it is based on decay properties of the {\em $K$-functional}
\begin{equation}
\label{eq:DefKfunctional}
K(f,t;\cH,\ell^p(\dict))
= \inf_{c\in \ell^p} \left\{\norm{f-\dict c}{\cH} + t \norm{c}{p}\right\}.
\end{equation}
A priori, without a more explicit description of real interpolation spaces, the characterization of approximation spaces as interpolation spaces may seem just a sterile pedantic rewriting.
Fortunately, we show in Section~\ref{sec:interpolation} ({\em Theorem~\ref{th:Interpolation}}) that the Bernstein inequality~\eqref{eq:DefBineq}, together with the upper frame bound~\eqref{eq:DefUpperFrameBound}, allows to directly identify approximation spaces with sparsity spaces, with equivalent (quasi)-norms, for certain ranges of parameters. In particular, the following result can be obtained as a consequence of Theorem~\ref{th:Interpolation}.
\begin{theorem}
\label{th:FullSpaceEquality}
Suppose that $\dict$ satisfies the upper frame bound~\eqref{eq:DefUpperFrameBound} with constant $B$ as well as the Bernstein inequality~\eqref{eq:DefBineq} with some $0<\tau \leq 1$, with exponent $r=1/\tau-1/2$ and constant $C$. Then we have
\begin{equation}
\label{eq:FullSpaceEquality}
\cA_\tau^r(\dict) = \ell^\tau(\dict)
\end{equation}
with equivalent norms, i.e.
\[
c_1(B,C) \cdot \norm{f}{\ell^\tau(\dict)}
\leq \norm{f}{\cA_\tau^r(\dict)} := \norm{f}{\cH}+\seminorm{f}{\cA_\tau^r(\dict)}
\leq c_2(B,C) \cdot \norm{f}{\ell^\tau(\dict)}
\]
where the constants only depend on $B$ and $C$.
\end{theorem}
In other words, under the assumptions of Theorem~\ref{th:FullSpaceEquality}, a data vector $f \in \cH$ can be approximated by $k$-term expansions at rate $k^{-r}$ (in the sense $f \in \cA_\tau^r(\dict)$, where $r = 1/\tau-1/2$) if, {\em and only if}, $f$ admits a sparse {\em representation} $f = \sum_j c_j \phi_j$ with $\sum_j |c_j|^\tau < \infty$.
\subsection{Ideal {\em vs} practical approximation algorithms}
Consider a function $f$ that can be approximated at rate $k^{-r}$ using $k$-term expansions from $\dict$: $\sigma_k(f,\dict) \lesssim k^{-r}, \forall k \geq 1$. Under the assumptions of the above Theorem, we can conclude that the function $f$ indeed admits a {\em representation} $f = \sum_j c_j \phi_j$ with $\sum_j |c_j|^\tau < \infty$. Suppose that we know how to compute such a representation (e.g., that we can solve the optimization problem $\min \|c\|_\tau\ \mbox{subject to}\ f = \dict c$). Then, sorting the coefficients in decreasing order of magnitude $|c_{j_m}| \geq |c_{j_{m+1}}|$, one can build a simple sequence of $k$-term approximants $f_m := \sum_{m=1}^k c_{j_m} \phi_{j_m}$ which converge to $f$ at the rate $r$: $\|f-f_k\|_\cH \lesssim k^{-r}$. Note that one may not however be able to guarantee that $\|f-f_k\|_\cH \leq C \sigma_k(f,\dict)$ for a fixed constant $C<\infty$.
A special case of interest is $\tau=1$, where the optimization problem
\[
\min \|c\|_1 \ \mbox{subject to}\ f = \dict c
\]
is convex, and the unit ball in $\ell^1(\dict)$ is simply the convex hull of the symmetrized dictionary $\{\pm \phi_j\}_j$
with $\phi_j$ the atoms of the dictionary $\dict$. Therefore, under the assumptions of the above Theorem for $\tau = 1$, if a function can be approximated at rate $k^{-1/2}$ then, after proper rescaling, it belongs to the convex hull of the symmetrized dictionary, and there exists constructive algorithms such as Orthonormal Matching Pursuit ~\cite{MZ93,PRK93} which are guaranteed to provide the rate of approximation $k^{-1/2}$ \cite[Theorem 3.7]{DeVore:1996aa}.
\input{intro15}
\subsection{Incoherence and the Restricted Isometry Property}
The above examples illustrate that the Bernstein inequality (and its nice consequences such as Theorem~\ref{th:FullSpaceEquality}) can be fairly fragile. However, this could be misleading, and we will now show that in a certain sense "most" dictionaries satisfy the inequality in a robust manner.
In a previous work we showed that {\em incoherent} frames \cite{Gribonval:2006aa} satisfy a "robust" Bernstein inequality, although with an exponent $r = 2(1/\tau-1/2)$ instead of the exponent $s = 1/\tau-1/2$ that would match the Jackson inequality. This inequality is then robust, because small enough perturbations of incoherent dictionaries remain incoherent.
In the last decade, a very intense activity related to Compressed Sensing \cite{Donoho:2006aa} has lead to the emergence of the concept of Restricted Isometry Property (RIP) \cite{Candes:2005aa,Candes:2006aa}, which generalizes the notion of coherence. A dictionary $\dict$ is said to satisfy the RIP of order $k$ with constant $\ripc$ if, for any coefficient sequence $c$ satisfying $\norm{c}{0}\leq k$, we have
\begin{equation}
\label{eq:DefRIP}
(1-\ripc) \cdot \norm{c}{2}^2 \leq \norm{\dict c}{\cH}^2 \leq (1+\ripc) \cdot \norm{c}{2}^2.
\end{equation}
The RIP has been widely studied for random dictionaries, and used to relate the minimum $\ell^1$ norm solution $c^\star$ of an inverse linear problem $f=\dict c$ to a "ground truth" solution $c_0$ which is assumed to be ideally sparse (or approximately sparse).
In this paper, we are a priori not interested in "recovering" a coefficient vector $c_0$ from the observation $f = \dict c_0$. Instead, we wish to understand how the rate of ideal (but NP-hard) $k$-term approximation of $f$ using $\dict$ is related to the existence of a representation with small $\ell^\tau$ norm.
In Section~\ref{sec:finitedim}, we study finite-dimensional dictionaries, where it turns out that the lower bound in the RIP~\eqref{eq:DefRIP} provides an appropriate tool to obtain Bernstein inequalities with controlled constant\footnote{The control of constants is the crucial part, since in finite dimension all norms are equivalent, which implies that the Bernstein inequality is always trivially satisfied.}.
Namely, we say that the dictionary $\dict$ satisfies $\LRIP(k,\ripc)$ with a constant $\ripc<1$ provided that
\begin{equation}\label{eq:RIP21}
\norm{\dict c}{\cH}^2 \geq (1-\ripc) \cdot
\norm{c}{2}^2,
\end{equation}
for any sequence $c$ satisfying $\norm{c}{0}\leq k$.
We prove ({\em Lemma~\ref{le:RIPBineq}}) that in $\cH = \bR^N$ the lower frame bound $\lfbound>0$ and the $\LRIP(\slevel N,\ripc)$, imply a Bernstein inequality for $0<\tau \leq 2$ with exponent $r = 1/\tau-1/2$. As a result we have:
\begin{theorem}
\label{th:RIPimpliesCaract}
Let $\dict$ be an $m\times N$ frame with frame bounds $0<\lfbound \leq \ufbound < \infty$. Assume that $\dict$ satisfies $\LRIP(\slevel N,\ripc)$, where $\ripc<1$ and $0<\slevel <1$. Then
\begin{itemize}
\item for $0<\tau\leq 2$, the Bernstein inequality~\eqref{eq:DefBineq} holds with exponent $r = 1/\tau-1/2$ and a constant $
C_\tau(\lfbound,\ripc,\slevel) < \infty$ ({\em cf} Eq.~\eqref{eq:ConstantBernstein})
\item for $0<\tau \leq 1$, $0<\theta<1$, we have, with equivalent norms,
\[
\cA_\tau^r(\dict) = \ell^\tau(\dict) = (\cH,\ell^p(\dict))_{\theta,\tau}, \quad \frac1\tau = \frac{\theta}{p} + \frac{1-\theta}{2}.
\]
\end{itemize}
The constant $C_\tau(\lfbound,\ripc,\slevel)$ and the constants in norm equivalences may depend on $\lfbound, \ufbound, \ripc$, and $\slevel$, {\em but they do not depend on the dimension} $N$.
\end{theorem}
For random Gaussian dictionaries, the typical order of magnitude of $\lfbound,\ripc(\slevel)$ is known and governed by the aspect ratio $\redund := N/m$ of the dictionary, provided that it is sufficiently high dimensional (its number of rows should be above a threshold $m(\redund)$ implicitly defined in Section~\ref{sec:finitedim}). We obtain the following theorem.
\begin{theorem}
\label{th:GaussianFullSpaceEquality}
Let $\dict$ be an $m \times N$ matrix with i.i.d. Gaussian entries $\mathcal{N}(0,1/m)$. Let $\redund:=N/m$ be the redundancy of the dictionary. If $m \geq m(\redund)$ then, except with probability at most $10\redund^2 \cdot \exp(-\gamma(\redund) m)$, we have
for all $0<\tau\leq 1$ the equality
\begin{equation}
\label{eq:GaussianFullSpaceEquality}
\cA_\tau^r(\dict) = \ell^\tau(\dict) = (\cH,\ell^p(\dict))_{\theta,\tau},\quad r=1/\tau-1/2= \theta(1/p-1/2).
\end{equation}
with equivalent norms.
The constants driving the equivalence of the norms are universal: they only depend on $\tau$ and the redundancy factor $\redund$ but {\em not} on the individual dimensions $m$ and $N$. Similarly $\gamma(\redund)$ and $m(\redund)$ only depend on $\redund$.
For $\redund \geq 1.27$ we have $\gamma(\redund) > 7 \cdot 10^{-6}$. For large $\redund$ we have $\gamma(\redund)\approx 0.002$.
\end{theorem}
Indeed, for random Gaussian dictionaries in high-dimension, with high probability, the Bernstein inequality holds for all $0< \tau \leq 2$ with constants driven by the aspect ratio $\redund := N/m$ but otherwise {\em independent of the dimension $N$}.Using the notion of decomposable dictionary \cite[Theorem 3.3]{Gribonval:2006aa}, this finite dimensional result can be easily adapted to build arbitrarily overcomplete dictionaries in infinite dimension that satisfy the equality~\eqref{eq:FullSpaceEquality}.
The result of Theorem~\ref{th:GaussianFullSpaceEquality} should be compared to our earlier result for incoherent frames obtained in \cite{Gribonval:2006aa}. In \cite{Gribonval:2006aa} we found an incoherent dictionary with aspect ratio (approximately) 2 for which the Bernstein inequality \eqref{eq:DefBineq} can be shown to hold only for the exponent $r=2(1/\tau-1/2)$, i.e., for $r$ twice as large as the Jackson exponent $s= 1/\tau-1/2$.
Theorem~ \ref{th:GaussianFullSpaceEquality} illustrates that the result in \cite{Gribonval:2006aa} really corresponds to a ``worst case'' behaviour and there are indeed many dictionaries (according to the Gaussian measure: the overwhelming majority of dictionaries) with a much better behaviour with respect to Bernstein estimates. This holds true even for aspect ratios $\redund$ that can be arbitrarily large.
\subsection{Conclusion and discussion}
The restricted isometry property is a concept that has been essentially motivated by the understanding of sparse regularization for linear inverse problems such as compressed sensing.
Beyond this traditional use of the concept, we have shown new connections between the RIP and nonlinear approximation.
The main result we obtained is that, from the point of view of nonlinear approximation, a frame which satisfies a nontrivial restricted property $\delta_k < 1$ (i.e., in the regime $k \propto N$) behaves like an orthogonal basis: the optimal rate of $m$-term approximation can be achieved with an approach that does not involve solving a (potentially) NP-hard problem to compute the best $m$-term approximation for each $m$. In such nice dictionaries, near optimal $k$-term approximation can be achieved in two steps, like in an orthonormal basis:
\begin{itemize}
\item decompose the data vector $f = \sum_j c_j \phi_j$, with coefficients as sparse as possible in the sense of minimum $\ell^\tau$ norm;
\item keep the $m$ largest coefficients to build the approximant $f_m := \sum_{j \in I_m} c_j \phi_j$.
\end{itemize}
The second main result is that redundant dictionaries with the above property are not the exception, but rather the rule. While it is possible to build nasty overcomplete dictionaries either directly or by arbitrarily small perturbations of some "nice" dictionaries", in a certain sense the vast majority of overcomplete dictionaries are nice.
One should note that several results of this paper are expressed in finite dimension, where all norms are equivalent. The strength of the results is therefore not the mere existence of inequalities between norms, but in the fact that the involved constants do not depend on the dimension. From a numerical perspective, the control of these constants has essentially an impact in (very) large dimension, and it is not clear whether the constants numerically computed for random dictionaries are useful for dimensions less than a few millions.
A few key questions remains open. For a given data vector $f$, it is generally not known in advance to which $\ell^\tau(\dict)$ space $f$ belongs:
under which conditions is it possible to efficiently compute a sparse decomposition $f = \sum_j c_j \phi_j$ which is guaranteed to be near optimal in the sense that $\|c\|_\tau$ is almost minimum whenever $f \in \ell^\tau(\dict)$? Can $\ell^1$ minimization (which is convex) be used and provide near best performance under certain conditions ? This is left to future work.
\input{interpolation}
\input{fragile}
\input{random}
\appendix
\input{AppIncreasingFunction}
\bibliographystyle{abbrv}
\bibliography{bernstein}
\end{document} | 0.035591 |
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TITLE: If every proper subring is noetherian then the ring is noetherian?
QUESTION [3 upvotes]: If every proper subring is noetherian then the ring is noetherian ?
For example every proper subring of $\mathbb{Z} $ is $m\mathbb{Z}$ and we know that $m\mathbb{Z} $ is noetherian.
REPLY [3 votes]: If you require that the ring $R$ has a unit but do not require your subrings to contain the unit, then this is trivial. Indeed, if $R$ is any non-Noetherian unital ring, let $I\subset R$ be a non-finitely generated proper ideal (a non-finitely generated ideal in $R$ must be proper since $R$ has a unit). Then $I$ is a non-Noetherian subrng of $R$ (since $I$ itself is a non-finitely generated ideal of $I$).
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\begin{document}
\begin{abstract}
We show the existence of generalized clusters of a finite or even infinite number of sets, with minimal total perimeter and given total masses, in metric measure spaces homogeneous with respect to a group acting by measure preserving homeomorphisms,
for a quite wide range of perimeter functionals.
Such generalized clusters are a natural ``relaxed'' version of a cluster and can be thought of as ``albums'' with possibly infinite pages, having a minimal cluster drawn on each page, the total perimeter and the vector of masses being calculated by summation over all pages, the total perimeter being minimal among all generalized clusters with the same masses. The examples include any anisotropic perimeter in a Euclidean space, as well as a hyperbolic plane with the Riemannian perimeter and Heisenberg groups with a canonical left invariant perimeter or its equivalent versions.
\end{abstract}
\maketitle
\section{Introduction}
In a finite-dimensional Euclidean space $\R^n$ we call
$\vec E = (E_1,\dots, E_N)$ with $N\in \N\cup\{\infty\}$ an \textit{$N$-cluster} of sets
if each $E_j\subset \R^n$ is Borel (possibly empty), and
$\mathcal{L}^n(E_i\cap E_j)=0$ for all $i\neq j$, $\mathcal{L}^n$ standing for the Lebesgue measure.
If $P(M)$ stands for the classical (Euclidean) perimeter of the set $M\subset \R^n$, we set
\begin{align*}
\vec m(\vec E) &:= \enclose{\mathcal{L}^n(E_1),\dots,\mathcal{L}^n(E_N)} \\
P(\vec E) &:=
\frac 1 2 \sum_{j\geq 1} P(E_j)+
\frac 1 2 P\left(\bigcup_{j\geq 1} E_j\right)
\end{align*}
An $N$-cluster $\vec E$ is usually called
\emph{minimal}, or \textit{isoperimetric}, if
\[
P(\vec E) = \inf\ENCLOSE{P(\vec F)\colon \vec m(\vec F)=\vec m(\vec E)}.
\]
Existence of \textit{finite} (i.e. with $N <+\infty$) isoperimetric clusters for the classical (Euclidean) perimeter in a Euclidean space
has been proven in~\cite{Maggi12-GMTbook} for each given vector of masses $\vec m$. The technique of the proof has been
further extended to several other situations, for various different notions of perimeter. More has been done for the problem of existence of \textit{isoperimetric sets}, i.e.\ in our terminology, minimal $1$-clusters. It has been accomplished
for different versions of perimeter in finite-dimensional spaces such as Riemannian manifolds
(see the list of relevant results in the Introduction to~\cite{MondinoNard16-isoperim} as well as the recent papers~\cite{CeNo18,FloreNard20-isoperimRiemann,AntFogPoz21}) and even in more general spaces with Ricci curvature bounds~\cite{AnBrFoPo21,AntPasPoz21}. However these techniques are based on regularity of isoperimetric
sets, and are quite difficult even in the Euclidean case. Moreover, such results are only valid for \textit{finite} clusters.
The difficulty to prove the existence by purely variational techniques, even
in the Euclidean space and just for isoperimetric sets (i.e. just $1$-clusters) is only due to noncompactness of the ambient space
(in compact spaces it is trivial), and comes from the fact
that minimizing sequences of sets may ``escape to infinity'' and lose part of their volumes. It is however well-known that
the limits of the pieces going to infinity are still isoperimetric sets for their own volumes (which may be lower than the original requirement), see e.g.~\cite{DePhilFranzPrat17,PratSar18-isoperim} where the problem of existence of isoperimetric sets has been studied for
quite a general family of weighted perimeters and volume measures. The same is known to happen also with \textit{finite} isoperimetric clusters, see~\cite{Scattaglia20-clust} where a formula for the perimeter of a minimal finite cluster is proven (again, for possibly weighted perimeters and volumes) suggesting that a minimal cluster is given by the union of the limit cluster (with possibly lower volumes) for a minimizing sequence
and a ``cluster at infinity'' which has precisely the missing volumes.
Therefore to prove the existence of an isoperimetric set or cluster one needs to appropriately adjust the sets by adding necessary volume which requires a great deal of heavy regularity techniques. However it is important to emphasize that the problem is not merely technical. In fact, even for $N=1$ (isoperimetric sets) it might happen that in some Riemannian manifolds there are no isoperimetric sets of some or even every volume, see~\cite{Ritore01-isoperim-symm} as well as~\cite{Nardulli14_isoperimgener,MondinoNard16-isoperim}.
Here we consider a more general situation of a metric measure space $(X,d,\mu)$ (with distance $d$ and measure $\mu$) instead of an Euclidean space with Lebesgue measure, which is \textit{homogeneous} with respect to some group acting by measure preserving homeomorphisms.
We show that in such a situation one can easily prove by means of concentration compactness-like technique (i.e. similar to~\cite{Lions84-conccomp}), the existence of ``generalized'' minimal clusters (finite or possibly infinite) for each given vector of masses. Such generalized clusters are in a sense a natural ``relaxed'' version of a cluster and can be thought of as ``albums'' with possibly infinite pages, having a minimal $N$-cluster drawn on each page, the total perimeter and the vector of masses being calculated by summation over all pages, the total perimeter being minimal among all generalized clusters with the same vector of masses. Such a generalized cluster in fact keeps track of all the parts of minimal sequences ``escaping at infinity'' by placing them on possibly different pages.
The toy examples we provide include finite or infinite minimal clusters for any anisotropic perimeter in $\R^n$, as well as in a hyperbolic plane with the Riemannian perimeter and in Heisenberg groups with a canonical left invariant perimeter or its equivalent versions. In all these examples there is some natural discrete group acting on the respective space properly discontinuously and cocompactly by measure preserving isometries (of course, the existence of isoperimetric sets is also known in such situations).
Note that the technique developed here is purely variational and does not involve any regularity-type arguments, thus allowing to obtain existence of generalzied minimal clusters for the whole range of perimeter functionals. The cost paid for such a simplicity is of course that one can say nothing a priori about the regularity of generalized clusters; and it has to be noted that even very weak regularity properties of a minimal generalized cluster would allow to conclude
the existence of a solution to the original problem (i.e.\ the existence of minimal clusters in the original space).
The paper is structured as follows. After introducing the general axiomatic notion of a perimieter functional by a set of quite weak requirements
in Section~\ref{sec:notation}, we provide in Section~\ref{sec:compsets} the general semicontinuity and compactness result for this functional
for sequences of sets and in Section~\ref{sec:compclusters} for sequences of clusters. This allows to prove the existence of generalized
optimal clusters in Section~\ref{sec:existgencluster}. Finally, in Section~\ref{sec_ClusterExamples} we give some basic applications of such existence results for various types of perimeter functional in different environments, in partcicular, in a finite-dimensional normed space (i.e.\ with either classical or anisotropic perimeter), in a hyperbolic space and in te Heisenberg group.
\section{Notation and prelimiaries}
\label{sec:notation}
In the sequel we suppose $(X,d,\mu)$ to be a metric measure space with distance $d$ and nonnegative $\sigma$-finite Radon measure
$\mu$ with $\mu(X)\neq 0$, and $G$ be a discrete, countable, topological group acting properly discontinuously on $X$
(i.e.\ $\{g\in G\colon gK\cap K \neq \emptyset\}$ is finite for all compact $K\subset X$) by homeomorphisms preserving the measure $\mu$.
We denote by $\mathcal{B}(X)$ the Borel $\sigma$-algebra of $X$, and by $\mathcal{A}(X)$ the class of open subsets of $X$.
The metric space $(X,d)$ is said to have Heine-Borel property, if every closed ball is compact.
By $L^1(\mu)$ (resp.\ $L^1_\loc(\mu)$) we denote the usual Lebesgue space of $\mu$-integrable (resp.\ $\mu$-integrable over compact sets) functions over $X$, and if $X=\R^n$, we denote by $C_0^\infty(\R^n)$ the set of infinitely differentiable functions with compact support. As usual, for an $E\subset X$ we let $\bar E$ stand for the closure of $E$ and $\mathbf{1}_E$ for the characteristic function of $E$, while $B_r(x)$ denotes the open ball of radius $r>0$ centered at $x\in X$.
We write $g_k\to\infty$ for a sequence of $g_k\in G$, if $\lim_k g_k=\infty$ in the one-point compactifictaion
$G\cup\{\infty\}$ of $G$, that is, for every finite $F\subset G$ the set of $\{k\in \N\colon g_k\in F\}$ is finite.
In what follows we let $P\colon \mathcal{B}(X)\times \mathcal{A}(X)\to \R^+\cup\{\infty\}$ be a ``relative perimeter'' functional
for which we customarily abbreviate $P(E):=P(E,X)$,
satisfying
\begin{itemize}
\item[(semicontinuity)] $P(D, U) \le \liminf_k P(D_k,U)$ whenever $D_k \stackrel{L^1_\loc(\mu)}\longrightarrow D$,
\item[(Beppo Levi)] If $U_k \nearrow X$, then $P(D, U_k)\nearrow P(D)$ as $k\to\infty$,
\item[(monotonicity)] $P(B, U)\leq P(B, V)$ as $U\subset V$,
\item[(superadditivity)] $P(B,U)\geq \sum_{k=1}^M P(B, U_k)$ whenever $U=\bigsqcup_{i=1}^M U_k$,
\item[($G$-invariance)] $P(gB,gU)= P(B,U)$ for every $g\in G$,
\item[(compactness)]
if $E_k\subset X$ satisfy $\displaystyle\sup_k P(E_k,U)<+\infty$
for some precompact set $U$, then
there exists an $E\subset X$ such that, up to a subsequence,
$E_k \cap U\stackrel{L^1(\mu)}\longrightarrow E$.
\end{itemize}
\section{Compactness and semicontinuity for sequences of sets}
\label{sec:compsets}
\subsection{Semicontinuity}
We start with the following semicontinuity statement which will be further applied both to perimeters and measures.
\begin{theorem}[semicontinuity]
\label{th:comp_semicont}
Assume that $(X,d)$ has Heine-Borel property.
Let $E_k\subset X$ be a sequence of Borel sets, and
$E^i\subset X$, $i\in I$, $I$ at most countable, $g_k^i\in G$ such that:
\begin{gather}
\label{eq_diffg1}
\lim_k (g_k^{i'})^{-1}g_k^i = +\infty \qquad \mbox{for all $i\neq i'$}\\
\label{eq_convE1}
(g_k^{i})^{-1} E_k \stackrel{L^1_\loc(\mu)}\longrightarrow E^i \qquad \text{as $k\to+\infty$}.
\end{gather}
Let also $F\colon \mathcal{B}(X)\times \mathcal{A}(X)\to \R$ be a functional satisfying
\begin{itemize}
\item[(semicontinuity)] $F(B, U) \le \liminf_k F(B_k,U)$ whenever $B_k \stackrel{L^1_\loc(\mu)}\longrightarrow B$,
\item[(Beppo Levi)] If $U_k \nearrow X$, then $F(D, U_k)\nearrow F(D, X)$ as $k\to\infty$,
\item[(monotonicity)] $F(D, U)\leq F(D, V)$ as $U\subset V$,
\item[(superadditivity)] $F(B,U)\geq \sum_{k=1}^M F(B, U_k)$ whenever $U=\bigsqcup_{i=1}^M U_k$,
\item[($G$-invariance)] $F(gB,gU)= F(B,U)$ for every $g\in G$.
\end{itemize}
If $G$ acts on $X$ properly discontinuously, then
\begin{gather*}
\sum_i F(E^i,X) \le \liminf_k F(E_k,X).
\end{gather*}
\end{theorem}
\begin{proof}
Up to a subsequence (not relabeled) we may
suppose that \[\liminf_k F(E_k,X)=\lim_k F(E_k,X).\]
Fix arbitrary $M\in \N$, $x_0\in X$ and $R>0$ and denote for brevity $U:=B_R(x_0)$.
Since $U$ is precompact by Heine-Borel property, and the action of $G$ on $X$ is properly discontinuous, then
$gU\cap U\neq \emptyset$ only for a finite number of $g\in G$. Thus
from~\eqref{eq_diffg1} we get that the sets
$g_k^1 U, \dots, g_k^M U$ are pairwise disjoint for all sufficiently large $k$.
Hence we get
\begin{equation}\label{eq_semicontF1}
\begin{aligned}
\lim_k F(E_k,X) & \geq \limsup_k F\left( E_k,\bigsqcup_{i=1}^M g_k^i U \right) \quad \mbox{by monotonicity of $F$}\\
& = \limsup_k \sum_{i=1}^M F(E_k, g_k^i U) \quad \mbox{by superadditivity of $F$}\\
& \geq \sum_{i=1}^M \liminf_k F(E_k, g_k^i U).
\end{aligned}
\end{equation} But $F(E_k,g_k^i U)) = F((g_k^i)^{-1} E_k,U)$ and from~\eqref{eq_convE1}
using semicontinuity of $F$ we obtain
\begin{equation}\label{eq_semicontF2}
\liminf_k F(E_k,g_k^i U) = \liminf_k F((g_k^i)^{-1} E_k,U)\ge F(E^i,U).
\end{equation}
The inequalities~\eqref{eq_semicontF1} and\eqref{eq_semicontF2} together give
\[
\lim_k F(E_k, X)\ge \sum_{i=1}^M F(E^i,U)
=\sum_{i=1}^M F(E^i,B_R(x_0)).
\]
Letting now $R\to+\infty$, we obtain
\[
\lim_k F(E_k,X)\ge \sum_{i=1}^M F(E^i,X)
\]
and finally letting $M\to +\infty$ we get
\[
\lim_k F(E_k,X) \ge \sum_i F(E^i, X).
\]
as desired.
\end{proof}
\begin{corollary}\label{co_comp_semicont1}
Let $(X,d,\mu)$, $E_k$,
$E^i$, $I$, $g_k^i$ be as in Theorem~\ref{th:comp_semicont}.
If $G$ acts on $X$ properly discontinuously, then
\begin{align*}
\sum_i P(E^i) &\le \liminf_k P(E_k),\\
\sum_i \mu(E^i) &\le \liminf_k \mu(E_k).
\end{align*}
\end{corollary}
\begin{proof}
Apply Theorem~\ref{th:comp_semicont} with $F(B,U):=P(E,U)$, and then with $F(B,U):=\mu(E\cap U)$.
\end{proof}
\subsection{Compactness}
The following theorem is our main technical tool to prove the existence of generalized isoperimetric clusters.
\begin{theorem}[concentration compactness]
\label{th:comp_comp}
Assume that $(X,d)$ has Heine-Borel property.
Let
\[E_k\in \mathcal{B}(X), \quad \mu(E_k)=m,\quad \sup_k P(E_k) = P < +\infty.\]
Suppose that
\begin{itemize}
\item[(i)] $G$ act on $X$ properly discontinuously,
\item[(ii)] there is a precompact $B\in \mathcal{B}(X)$ with $\mu(\partial B)=0$,
such that
$GB=X$ and $\mu(g B \cap B)=0$ for all $g\in G$ except $g=1$.
\end{itemize}
Assume further that
\begin{itemize}
\item[(iii)] there is an $\varepsilon>0$ and a precompact open $V\subset X$ with
$B\subset V$ such that
the local isoperimetric inequality
\begin{equation}
\label{eq_relisoperim1a}
P(E,V)\geq f(\mu(E\cap V))
\end{equation}
holds for all Borel $E\subset X$
with $\mu(E\cap V)\leq \varepsilon$ with some nondecreasing function $f\colon\R^+\to \R^+$ such that
$f(0)=0$ and $f'(0)=+\infty$.
\end{itemize}
Then there are: a subsequence $E_k$, some Borel sets
$E^i\subset X$,
and $g_k^i\in G$, $i\in I$, $I$ at most countable,
such that
\begin{gather}
\label{eq_diffg2}
\lim_k (g_k^{i'})^{-1}g_k^i = +\infty \qquad \mbox{for all $i\neq i'$}\\
\label{eq_convE2}
(g_k^i)^{-1} E_k \stackrel{L^1_\loc(\mu)}\longrightarrow E^i,
\qquad k\to +\infty\\
\label{eq_convE3}
\sum_i \mu(E^i) = m.
\end{gather}
\end{theorem}
\begin{remark}\label{rm_isoperim1}
In many examples condition~(iii) of the above Theorem~\ref{th:comp_comp} is verified with $f(t):=Ct^\alpha$ for some $\alpha \in (0,1)$ and $C>0$, namely,~\eqref{eq_relisoperim1a} reads as
the local isoperimetric inequality
\begin{equation}
\label{eq_relisoperim1}
P(E,V)\geq C\mu(E\cap V)^\alpha.
\end{equation}
\end{remark}
\begin{proof}
For each $k$ let $h_k^j$ be an enumeration of $G$
such that
the sequence
$j\mapsto \mu(E_k \cap h_k^j V)$ is not-increasing. Let $U$ stand for the interior of $B$.
One observes that $U\neq \emptyset$, because $\mu(B)>0$ (otherwise from $GB=X$ one would have $\mu(X)=0$), and $\mu(\partial B)=0$ hence $\mu(U)= \mu(B\setminus \partial B)>0$.
Note that
$Q_k^j:= h_k^j U$ are open sets since $G$ acts by homeomorphisms.
If there are infinitely many sets $h_k^j V$ in which $E_k$
has positive measure we avoid to enumerate those
with zero measure.
By the compactness property of the perimeter functional,
for each $j$ we may suppose that up to a subsequence
$((h_k^j)^{-1}E_k) \cap U$ converges in $L^1(\mu)$ as $k\to +\infty$
and hence there exists $F^j\subset U$ such that
\begin{equation}\label{eq_gkjEk1}
(h_k^j)^{-1} (E_k \cap Q_k^j) = ((h_k^j)^{-1}E_k)\cap U \stackrel{L^1(\mu)}\longrightarrow F^j,
\qquad \text{as $k\to \infty$}.
\end{equation}
By a diagonal argument we may choose a single subsequence of $k$ (not relabeled) such that~\eqref{eq_gkjEk1} holds for all $j\in \N$.
The rest of the proof is divided in several steps.
{\sc Step 1}. We claim that $\sum_j \mu(F^j) = m$.
To show the claim, we first note that
since the action of $G$ on $X$ is properly discontinuous, we have
\[
\nu:=\#\{g\in G\colon g\bar V\cap \bar V\neq \emptyset\} <\infty.
\]
Applying Lemma~\ref{lm_groupPart1} with $K:=\bar V$
one has a partition of $G$ is at most $\nu$ disjoint subsets $\mathcal{F}_j$ such that the sets
$h\bar V$ are mutually disjoint for all $h\in \mathcal{F}_j$.
Thus for each fixed $k\in \N$
one can find a finite partition $\N = \bigsqcup_{\lambda=1}^\nu J_{k,\lambda}$ of all indices $j$
such that for $j \in J_{k,\lambda}$ the sets
$h_k^j \bar V$ are pairwise disjoint, by setting
$J_{k,\lambda} := \{j\in \N\colon h_k^j \in \mathcal{F}_\lambda\}$.
Since $j\mapsto \mu(E_k\cap h_k^j(V))$ is non-increasing, we have
\begin{equation}\label{eq_bdmu1}
\begin{aligned}
\mu(E_k \cap h_k^n V)
&\leq \frac 1 n \sum_{j=1}^n \mu(E_k \cap h_k^j V)
\leq \frac 1 n \sum_{j=1}^{+\infty} \mu(E_k \cap h_k^j V) \\
&=
\frac 1 n \sum_{\lambda =1}^\nu \sum_{j\in J_{k,\lambda}} \mu(E_k \cap h_k^j V)
\leq \frac \nu n \mu(E_k) = \frac{\nu m}{n}.
\end{aligned}
\end{equation}
In particular, by~\eqref{eq_bdmu1} for every $\delta>0$
one can choose an
$N=N(\varepsilon,\delta )\in \N$ such that for all $j\ge N$ we have
\begin{eqnarray}
\label{eq_mukeps1}
\mu(E_k \cap h_k^j V) \leq \varepsilon,&\quad\text{hence also}\\
\label{eq_mukeps2}
\mu(E_k \cap Q_k^j) \leq \varepsilon,
\end{eqnarray}
and, moreover,
\begin{equation}\label{eq_mukeps3}
\frac{\mu(E_k \cap Q_k^j)}{f(\mu(E_k \cap Q_k^j))}\leq \delta.
\end{equation}
Thus for such $j\in \N$, using~\eqref{eq_mukeps2},~(iii) and the invariance of perimeter,
we get
\begin{equation}\label{eq_bdper1}
\begin{aligned}
f(\mu(E_k \cap Q_k^j))
&\le f(\mu((h_k^j)^{-1}E_k \cap V))\\
&\leq P((h_k^j)^{-1}E_k ,V)
=
P(E_k ,h_k^j V)
.
\end{aligned}
\end{equation}
Therefore, for $n\geq N$ we have
\begin{equation*}\label{eq_summu1}
\begin{aligned}
\sum_{j=n}^{+\infty} \mu(E_k \cap Q_k^j)
&\le \sum_{j=n}^{+\infty} \frac{\mu(E_k \cap Q_k^j)}{f(\mu(E_k \cap Q_k^j))}
f(\mu(E_k \cap Q_k^j)) \\
&\le \sum_{j=n}^{+\infty} \frac{\mu(E_k \cap Q_k^j)}{f(\mu(E_k \cap Q_k^j))}
P(E_k,h_k^jV) \quad\mbox{[by~\eqref{eq_bdper1}]}\\
&\le \delta
\sum_{j=n}^{+\infty} P(E_k,h_k^j V) \quad\mbox{[by~\eqref{eq_mukeps3}]}\\
&\le \delta
\sum_{\lambda=1}^\nu \sum_{j\in J_{k,\lambda}} P(E_k,h_k^j V) \\
& \leq \delta
\sum_{\lambda=1}^\nu P(E_k,\bigsqcup_{j\in J_{k,\lambda}} h_k^j V) \quad\mbox{[by superadditivity of perimeter]}\\
& \leq \delta
\sum_{\lambda=1}^\nu P(E_k)\quad\mbox{[by monotonicity of perimeter]}\\
& = \nu \delta
P(E_k) \leq \nu\delta P.
\end{aligned}
\end{equation*}
Since $\delta>0$ is arbitrary,
we get
\[
\lim_{n\to +\infty}
\sup_k \sum_{j=n}^{+\infty} \mu(E_k \cap Q_k^j) = 0.
\]
But we also have
\[\mu(F^j) = \lim_k \mu(E_k \cap Q_k^j).\]
Apply Lemma~\ref{lm:equisummability} with
$m_j:=\mu(F^j)$, $m_{k,j}:=\mu(E_k \cap Q_k^j)$,
recalling that
\begin{align*}
\sum_i m_{k,j}
&= \mu\big( E_k \cap \bigcup_j h_k^j U \big)
\qquad \text{[since $\mu(hU\cap U)=0$ for all $h\in G$]}
\\
&= \mu\big( E_k \cap \bigcup_j h_k^j B \big)
\quad\text{[since $\mu(h(B\setminus U))=\mu(B\setminus U)=0$ for all $h\in G$]}\\
&=\mu(E_k)=m\qquad \text{[since $GB=X$]},
\end{align*}
to obtain $\sum_j \mu(F^j) = m$ as claimed.
{\sc Step 2: construction of $g_k^i$ and $E^i$}.
Define on $\N$ an equivalence relation by letting
$j\sim j'$ whenever the set $\{(h_k^{j'})^{-1}h_k^j\colon k\in \N\}\subset G$ is finite
and let $I:=\N/{\sim}$ be the quotient set.
For each $i\in I$ let $\underline{i}:= \min i$.
Passing to a subsequence in $k$ we might and shall suppose that for all $i\in I$ and all $j\in i$
the sequence $\{(h_k^{\underline i})^{-1} h_k^j\}_k$ (which by assumption assumes a finite number of values) is actually constant, which will be denoted $h^j\in G$.
Clearly for $j\in i$ all $h^j$ are distinct
because so are $h_k^j\in G$.
Define $g_k^i := h_k^{\underline{i}}$
and suppose that, up to a subsequence,
the sets $(g_k^i)^{-1}E_k$ converge in $L^1_\loc(\mu)$
to some Borel set $E^i$ as $k\to +\infty$.
Hence~\eqref{eq_diffg2} and~\eqref{eq_convE2} hold by construction.
{\sc Step 3}. It remains to prove~\eqref{eq_convE3}. To this aim observe that from~\eqref{eq_gkjEk1} with $j\in i$ we get
\begin{align*}
\mu(F^j) &\leftarrow
\mu(E_k \cap Q_k^j)= \mu\left((g_k^i)^{-1} (E_k \cap Q_k^j) \right) \\
&= \mu\left((g_k^i)^{-1} E_k \cap (g_k^i)^{-1} h_k^j U\right)\\
&= \mu\left((g_k^i)^{-1} E_k \cap h^j U\right)\\
&\to \mu\left(E^i\cap h^j U\right)
\end{align*}
as $k\to\infty$, so that $\mu(F^j) =\mu(E^i\cap h^j U)$ whenever $j\in i$.
Therefore
\begin{align*}
\lim_k \mu(E_k) &= m = \sum_j \mu(F^j) \quad\text{[by Step~1]}\\
& = \sum_i \sum_{j\in i} \mu(E^i\cap h^j U) = \sum_i \mu\big(E^i\cap \bigcup_{j\in i}h^jU\big) \leq \sum_i \mu\left(E^i\right),
\end{align*}
the reverse inequality coming from Theorem~\ref{th:comp_semicont}.
\end{proof}
\section{Compactness and semicontinuity for sequences of clusters}
\label{sec:compclusters}
The following definition of an isoperimetric cluster is an obvious extension
of the classical one provided in the Introduction from a Euclidean to a general metric measure space.
\begin{definition}[isoperimetric clusters]
We say that $\vec E = (E_j)_{j=1}^N$ with $N\in \N\cup\{\infty\}$, is an $N$-cluster in $X$
if each $E_j\subset X$ is a Borel set and
$\mu(E_i\cap E_j)=0$ for all $i\neq j$.
If $N$ is finite we have $\vec E = (E_1,\dots, E_N)$, if $N=\infty$
we have $\vec E = (E_1,\dots,E_n,\dots)$.
We set
\begin{align*}
\vec m(\vec E) &:= \enclose{\mu(E_j)}_{j=1}^N \\
P(\vec E) &:=
\frac 1 2 \sum_{j\geq 1} P(E_j)+
\frac 1 2 P\left(\bigcup_{j\geq 1} E_j\right)
\end{align*}
An $N$-cluster $\vec E$
will be called
\emph{minimal}, or \textit{isoperimetric}, if
\[
P(\vec E) = \inf\ENCLOSE{P(\vec F)\colon \vec m(\vec F)=\vec m(\vec E)}.
\]
\end{definition}
\begin{theorem}[concentration compactness for clusters]\label{th_comp_clust}
Assume that $(X,d)$ has Heine-Borel property.
Let $\vec E_k$ be a sequence of $N$-clusters in $X$,
$N\in \N \cup \{\infty\}$,
with
\[\vec m(\vec E_k) = \vec m\in \R^N\quad\mbox{and }
P(\vec E_k)\le P <+\infty.
\]
Then under conditions~(i),~(ii) and~(iii) of Theorem~\ref{th:comp_comp}
there exist $N$-clusters $\vec E^i$
with $i\in I$, $I$ at most countable, such that
\begin{gather*}
\sum_i \vec m(\vec E^i) = \vec m\\
\sum_i P(\vec E^i) \le \liminf_k P(\vec E_k)
\end{gather*}
and there exist $g_k^i\in G$ such that up to a subsequence
\[
(g_k^i)^{-1}(E_k)_j \to (E^i)_j\qquad \text{as $k\to +\infty$ in $L^1_\loc(\mu)$}
\]
for all $j=1,\ldots, N$ and $i\in I$.
\end{theorem}
\begin{proof}
Up to a subsequence suppose that $\liminf_k P(\vec E_k) = \lim_k P(\vec E_k)$.
Let
\[
F_k := \bigsqcup_{j=1}^N (E_k)_j.
\]
Clearly $P(F_k)\le 2 P(\vec E_k)\leq 2P$, and hence we can apply Theorem~\ref{th:comp_comp}
to find $F^i\subset X$, $g_k^i\in G$, $\lim_k (g_k^{i^\prime})^{-1}g_k^i= \infty$ for $i\neq i^\prime$
such that, up to a subsequence,
$(g_k^i)^{-1}F_k\to F^i$ in $L^1_\loc(\mu)$ as $k\to\infty$ with
$\displaystyle \sum_i \mu(F^i)=\mu(F_k) = \sum_{j=1}^N m_j$.
Since $P((E_k)_j) \le P(\vec E_k) \le P$,
by the compactness assumption on the perimeter functional, up to subsequences we can
define the sets
\[E^i_j := L^1_\loc(\mu)\text{-}\lim_k (g_k^i)^{-1}(E_k)_j.\]
By Lemma~\ref{lm:union} then
\[
F^i = \bigsqcup_{j=1}^N E^i_j.
\]
By Theorem~\ref{th:comp_semicont} we have $\displaystyle \sum_i \mu(E^i_j)\le \mu((E_k)_j)$.
Therefore, since
\[
\sum_j\sum_i \mu(E^i_j)=\sum_i\sum_j \mu(E^i_j)=\sum_i\mu(F^i)=\mu(F_k)=\sum_j \mu((E_k)_j),
\]
one gets $\displaystyle \sum_i \mu(E^i_j)=\mu((E_k)_j)$.
Hence $\vec E^i = (E^i_1,\dots,E^i_N)$ is an $M$-cluster, $M\le N$, with
\begin{align*}
\sum_i \vec m(\vec E^i) &= \vec m,\\
P(\vec E^i) &=\frac 1 2 \sum_{j=1}^N P(E^i_j) + \frac 1 2 P(E^i).
\end{align*}
Again by Theorem~\ref{th:comp_semicont}
we have
\begin{align*}
\sum_i P(F^i) &\le \liminf_k P(F_k), \\
\sum_i P(E^i_j) &\le \liminf_k\sum_j P((E_k)_j),\qquad j=1,\dots,N
\end{align*}
Summing up these relationships we get
\begin{align*}
\sum_i P(\vec E^i) &\le \lim_k P(\vec E_k)
\end{align*}
as claimed.
\end{proof}
\section{Existence of generalized isoperimeric clusters}
\label{sec:existgencluster}
We introduce now the notion of a \textit{generalized isoperimeric cluster}.
\begin{definition}[generalized isoperimetric clusters]
Let
$X^\infty:= \Z\times X$
and let $\phi_i\colon X\to X^\infty$ be the inclusion $\phi_i(x) = (i,x)$.
We call
$\vec E = (E_i)_{i=1}^N\subset X^\infty$ a generalized $N$-cluster in $X$
with $N\in \N\cup\{\infty\}$, if for each $j\in \Z$ the vector $\vec E^j$ with components
$E^j_i:=\phi_j^{-1}(E_i) \subset X$, $i=1,\ldots, N$
is a $N$-cluster in $X$.
We set
\begin{align*}
\vec m^\infty(\vec E) &:= \sum_j \vec m(\vec E^j) = \enclose{\sum_j\mu(E_1^j),\dots,\sum_j \mu(E_N^j)} \\
P^\infty(\vec E) &:= \sum_j P(\vec E^j).
\end{align*}
A generalized $N$-cluster $\vec E\in X^\infty$ will be called
\emph{minimal}, or \textit{isoperimetric}, if
\[
P^\infty(\vec E) = \inf\ENCLOSE{P^\infty(\vec F)\colon \vec m(\vec F)=\vec m(\vec E)}.
\]
\end{definition}
It is convenient to think of a generalized cluster as a sequence of $N$-clusters $\vec E^i$
drawn each on a page $X_i:=\{i\}\times X$ of the ``album'' $X^\infty$
and $\phi_i^{-1}$ as an extraction of the $i$-th page from this album.
The following immediate observation is worth mentioning.
\begin{proposition}\label{prop_genclustMinpage1}
Every page of a generalzied minimal cluster is a minimal cluster, namely,
for every $j\in \N$ one has that $\phi_j^{-1}(\vec E)$ is a minimal cluster.
\end{proposition}
\begin{proof}
If not, there is an $N$-cluster $\vec F^j$ such that
\[
P(\vec F^j) < P(\phi_j^{-1}(\vec E))
\quad\mbox{and } \vec m(\vec F^j)=\vec m(\phi_j^{-1}(\vec E)).
\]
Then for the generalized cluster $\vec F$ defined by
\begin{align*}
\phi_k^{-1}(\vec F) & = \phi_k^{-1}(\vec E),\qquad k\neq j,\\
\phi_j^{-1}(\vec F) & = \vec F^j,
\end{align*}
one has $P^\infty(\vec F)<P^\infty(\vec E)$ contradicting the minimality of $\vec E$.
\end{proof}
We show now the existence of minimal generalized clusters.
\begin{theorem}[existence of generalized minimal clusters]\label{th_exist_generclust1}
Assume that $(X,d)$ has Heine-Borel property.
If conditions~(i),~(ii) and~(iii) of Theorem~\ref{th:comp_comp} hold,
condition~(iii) being satisfied with $f\colon \R^+\to \R^+$ subadditive
(this is true in particular when $f(t):= Ct^\alpha$, with $C>0$, $\alpha\in (0,1)$),
then for an arbitrary $\vec m \in \R^N_+$ there exists
a minimal generalized $N$-cluster $\vec E$ with $\vec m^\infty(\vec E)=\vec m$.
\end{theorem}
\begin{proof}
Let us define over $X^\infty$ the distance
\[
d^\infty(x,y)
:= \begin{cases}
\arctan d(x',y'), &\text{if $x=(i,x')$ and $y=(i,y')$ for some $i\in \N$},\\
\frac{\pi}{2}, & \text{otherwise},
\end{cases}
\]
the measure $\mu^\infty$ on $X^\infty$ by
\[
\mu^\infty(E) := \sum_j \mu(\phi_j^{-1}(E)), \qquad E \subset X^\infty
\]
and define $G^\infty = \Z\times G$ the product group acting over $X^\infty$ by
\[
(n,g)(m,x) = (n+m, g(x))\qquad g\in G, x\in X, n,m\in \Z.
\]
Notice that $P^\infty$ and $\mu^\infty$ satisfy monotonicity, superadditivity, $G$-invariance
and compactness properties of section~\ref{sec:notation}.
The semicontinuity property is simply an application of Fatou Lemma and
Beppo Levi property is an application of Beppo Levi theorem.
We claim that (i) of Theorem~\ref{th:comp_comp} holds.
In fact if $K^\infty$ is a compact set in $X^\infty$ then the set $J:=\{j\in \Z\colon (j,x)\in K^\infty\ \text{for some $x\in X$}\}$
is finite.
Let $K:= \{x\in X\colon (j,x)\in K^\infty\text{ for some $j\in J$}\}$.
Since $J$ is finite $K$ is compact in $X$.
Then the set
\[
\{(n,g)\in G^\infty\colon (n,g)K^\infty \cap K^\infty \neq \emptyset\}
\subset (J-J)\times \{g\in G\colon gK\cap K\neq \emptyset\}
\]
is finite because if $(j,x)\in(n,g)K^\infty\cap K^\infty$ then $x\in K$, $g(x)\in K$, $j\in J$ and $n+j\in J$.
Also the condition (ii) of Theorem~\ref{th:comp_comp} is satisfied for the metric measure space $X^\infty$ with
this group action.
In fact the set $U^\infty:=\{0\}\times U$ satisfies
$G^\infty \overline {U^\infty} = X^\infty$
and $\mu^\infty(g\overline {U^\infty} \cap \overline{U^\infty})=0$ for all $g\in G^\infty$ except when $g$
is the neutral element in $G^\infty$.
For condition (iii) of Theorem~\ref{th:comp_comp} just notice that if $\mu^\infty(E)<\eps$ implies that
$\mu(E^j)< \eps$ (where $E^j=\phi_j^{-1}(E)$)
hence $P(E^j,V)\ge f(\mu(E_j\cap V))$ and summing up, using the subadditivity
of $f$,
\begin{align*}
P^\infty(E,\{0\}\times V) &
\ge \sum_j f(\mu(E_j\cap V)) \\
&\ge f\left(\sum_j \mu(E_j\cap V)\right)
= f(\mu^\infty(E\cap \{0\}\times V)).
\end{align*}
Let $\vec E_k$ be a sequence of generalized $N$-clusters in $X$ satisfying $\vec m^\infty(\vec E_k) = \vec m\in \R^N$.
By Theorem~\ref{th_comp_clust} applied to $\vec E_k$, $X^\infty$, $\mu^\infty$, $P^\infty$
in place of $X$, $\mu$, $P$ respectively,
we have the existence $N$-clusters $\vec E^i\in X^\infty$ for $i\in I$ with $I$ at most countable,
such that
\begin{equation}
\label{eq:min_gen_clust1}
\sum_i \vec m^\infty(\vec E^i) = \vec m,
\qquad
\sum_i P^\infty(\vec E^i) \le \liminf_k P^\infty(\vec E_k).
\end{equation}
Given an injective map $f\colon I\times \Z\to \Z$ we can define the
generalized $N$-cluster $\vec E=(E_1,\dots, E_N)$ in $X$ such that
\[
\phi_{f(i,j)}^{-1}(E_n) = \phi_j^{-1}(E^i_n), \qquad
\text{for every $n=1,\dots N$, $j\in \Z$, $i\in I$.}
\]
Hence \eqref{eq:min_gen_clust1} reads now
\[
\vec m^\infty(\vec E) = \vec m,
\qquad
P^\infty(\vec E) \le \liminf_k P^\infty(\vec E_k)
\]
which means that $\vec E$ is a generalized minimal $N$-cluster in $X$.
\end{proof}
\begin{remark}
Note that a minimal \textit{generalized} cluster can be thought in many particular cases as a kind of natural relaxation of
the notion of a minimal cluster. In fact, if it is possible, say, to cut away from each page everything outside of a sufficiently large ball, without changing too much the perimeter and the volumes, so that what remains on each a page is a bouunded cluster, then put by the group action all the bounded clusters from each page to just one page, and finally adjust the volumes, say, by adding small isoperimetric sets or at least sets with sufficiently small perimeters, we get that for every $\varepsilon>0$ and for each generalized c $N$-cluster $\vec E$ there is an $N$-cluster $\vec E'$ with the same volumes, i.e. $\vec m^\infty (\vec E')= \vec m (\vec E)$ and \[P(\vec E')\leq P^\infty(\vec E) +\varepsilon.\] This can be done for instance in a Euclidean space (the cutting of a large bounded set from each page without changing too much the volumes and the perimeter can be done in view of the coarea formula; in more general spaces even a coarea type inequality would suffice).
\end{remark}
\section{Basic examples}\label{sec_ClusterExamples}
For merely illustrative purposes we provide here several examples of existence of finite or infinite generalized isoperimetric clusters
in a finite-dimensional vector space (for any anisotropic perimeter related to some norm), in the hyperbolic plane and in the Heisenberg groups.
Note that even such simple examples in fact provide immediately the existence resluts for the whole range of equivalent perimeters
in the mentioned spaces, without requiring the study of regularity properties of such clsuters and/or just isoperimetric sets for each particular perimeter. It is also worth mentioning that in the same way one can formulate similar existence results in many more interestng geometries (in particular, in higher dimensional hyperbolic spaces, or more general Carnot groups).
\subsection{Finite-dimensional space}
Let $X$ be an $n$-dimensional finite dimensional space equipped with any norm $\|\cdot\|$, $\mu$ be the Lebesgue measure on $X$,
$G:=\Z^n$ acting by translations, and $P$ be the relative perimeter functional
in $X$ corresponding to the chosen norm, i.e.\
\[
P(E,U) := \sup
\left\{
\int_U \mathbf{1}_E \mathrm{div} v\, d\mu \colon
\|v\|_\infty\leq 1
\right\},
\]
the $\sup$ being taken over smooth vector fields $v$ with compact support on $H$, with
$\|v\|_\infty:= \sup_{p\in X} \|v(p)\|$.
The following corollary is a direct consequence of Theorem~\ref{th_exist_generclust1}.
\begin{corollary}\label{co_existclust_findim1}
For every $\vec m \in \R^N_+$ there exists
a minimal generalized $N$-cluster $\vec E$ in $X$ with $\vec m^\infty(\vec E)=\vec m$.
\end{corollary}
It is worth mentioning that there is nothing obligatory in the choice of the group $G=\Z^n$; instead, another some other crystallographic group could have been chosen.
\subsection{Hyperbolic space}
Taking $H$ to be the hyperbolic plane, $\mu$ be its canonical volume measure,
$G$ be any countable Fuchsian group (i.e.\ a discrete subgroup of isometries of $H$) acting properly discontinuously and cocompactly on $H$
(e.g. one may take $G$ to be the classical Fucsian group providing the tiling of $H$ into isometric Schwartz triangles)
and $P$ be the classical Riemannian relative perimeter functional
in $H$, i.e.\
\[
P(E,U) := \sup
\left\{
\int_U \mathbf{1}_E \mathrm{div}_H v\, d\mu \colon
\|v\|_\infty\leq 1
\right\},
\]
the $\sup$ being taken over smooth vector fields $v$ with compact support on $H$, with
$\|v\|_\infty:= \sup_{p\in H} |v_p|_p$, $|\cdot|_p$ standing for the Riemannian norm of a vector in $T_p H$.
\begin{corollary}\label{co_existclust_hyperb1}
For every $\vec m \in \R^N_+$ there exists
a minimal generalized $N$-cluster $\vec E$ in $H$ with $\vec m^\infty(\vec E)=\vec m$.
\end{corollary}
\begin{proof}
Note that the perimeter functional $P$ clearly satisfies the semicontinuity, monotonicity, superadditivity, compactness, $G$-invariance and Beppo Levi properties,
and
conditions~(i) and~(ii) of Theorem~\ref{th:comp_comp} are satisfied as well
with $B$ given by Lemma~\ref{lm_Voronoi1}.
Condition~(iii) of Theorem~\ref{th:comp_comp} is satisfied as in Remark~\ref{rm_isoperim1} with $V\supset B$ being a sufficiently large ball containing $B$,
and $\alpha := 1/2$, $\varepsilon < \mu(V)/2$ due to the relative isoperimetric inequality over a compact Riemannian manifold
$\bar V$~\cite{keselman05-isopermhyperb1}. The claim is now a direct application of Theorem~\ref{th_exist_generclust1}.
\end{proof}
\begin{remark}\label{rm_existclust_hyperb1}
If, chosen a Fuchsian group $G$, there is an equivalent Finslerian structure on $H$ (with the norm $\|\cdot\|_p$ in each $T_p H$ equivalent to the Riemannian one $|\cdot|_p$), invariant under the action of $G$, then the same argument gives the existence for every $\vec m \in \R^N_+$
of a minimal generalized $N$-cluster $\vec E$ in $H$ with $\vec m^\infty(\vec E)=\vec m$, for the perimeter $P_G$ relative to this Finslerian structure instead of $P$, i.e.\ for
\[
P_G(E,U) := \sup
\left\{
\int_U \mathbf{1}_E \mathrm{div}_H v\, d\mu \colon
\|v\|_{G,\infty}\leq 1
\right\},
\]
the $\sup$ being taken over smooth vector fields $v$ with compact support on $H$, with
$\|v\|_{G,\infty}:= \sup_{p\in H} \|v_p\|_p$. To see this it is enough to note that $P_G$ and $P$ are equivalent.
\end{remark}
\subsection{Heisenberg groups}
Taking $H^n$ to be the Heisenberg group of topological dimension $2n+1$,
$\mu$ be its Haar measure,
$G$ be the respective discrete Heisenberg group (i.e. once $H^{n}$ is canonically associated with a group of matrices with real entries,
$G$ is associated with the subgroup of such matrices with integer entries), $\{X_i\}_{i=1}^{2n}$ be left-invariant vector fields
satisfying the H\"{o}rmander condition,
and $P$ be the sub-Riemannian relative perimeter functional
corresponding to the choice of $X_i$,
defined as the total variation measure of the vector ($\R^{2n}$-valued) measure
\[
D_X \mathbf{1}_1:= (D_{X_1} \mathbf{1}_E, \ldots, D_{X_{2n}} \mathbf{1}_E),
\]
where the distribution $D_{X_i} f$ is defined
by its action $\langle \varphi, D_{X_i} f\rangle$
on every test function $\varphi_\in C_0^\infty(\R^{2n})$
by the formula
\[
\langle \varphi, D_{X_i} f\rangle := - \int_{\R^{2n}} f X_i\cdot \nabla \varphi\, dx - \int_{\R^{2n}} f \varphi \mathrm{div}\,X_i\, dx.
\]
Note that there are several equivalent definitions of
this perimeter given by theorem~3.1 of~\cite{AmbrosioGhezziMagn15-BVsubriem}. We also observe that one of the equivalent definitions
is given in~[section 5.3]\cite{Miranda03-BVgoodMMS} and used in particular in~\cite{DanGarofNhieu98-CarnotTraceIneq}.
\begin{corollary}\label{co_existclust_heisenb1}
For every $\vec m \in \R^N_+$ there exists
a minimal generalized $N$-cluster $\vec E$ in $H^n$ with $\vec m^\infty(\vec E)=\vec m$.
\end{corollary}
\begin{proof}
We equip $H^n$
with either the Carnot-Caratheodory distance
or any equivalent left-invariant distance $d$, so that now $G$ acts by isometries.
The claim follows from Theorem~\ref{th_exist_generclust1} since he perimeter functional $P$ satisfies the semicontinuity, monotonicity, superadditivity, Beppo Levi and $G$-invariance properties, as well as
conditions~(i) and~(ii) of Theorem~\ref{th:comp_comp} (with $B$ given by Lemma~\ref{lm_Voronoi1}),
while
\begin{itemize}
\item condition~(iii) of Theorem~\ref{th:comp_comp} is satisfied as in Remark~\ref{rm_isoperim1} with $V\supset B$ sufficiently large ball containing $U$,
and $\alpha := (Q-1)/Q$, $Q:=2n+2$ standing for the homogeneous dimension of $H^n$, $\varepsilon < \mu(V)/2$ due to the relative isoperimetric inequality in a ball of a Carnot-Caratheodory space (theorem~1.6 from~\cite{DanGarofNhieu98-CarnotTraceIneq}),
\item compactness property is given by theorem~3.7 from~\cite{Miranda03-BVgoodMMS}.
\end{itemize}
\end{proof}
\begin{remark}\label{rm_existclust_heisenb1}
Similarly to Remark~\ref{rm_existclust_hyperb1} if one takes in $H^n$ any perimeter
$P$ defined as $P(E,U):=|D\mathbf{1}_E| (U)$ with the metric total variation $u\mapsto |Du|$ defined with respect to any chosen left $G$-invariant distance in $H^n$, then word-to-word repetition of the proof of the above Corollary~\ref{co_existclust_heisenb1} gives
the existence for every $\vec m \in \R^N_+$
of a minimal generalized $N$-cluster $\vec E$ in $H$ with $\vec m^\infty(\vec E)=\vec m$ for such a perimeter.
\end{remark}
\appendix
\section{Useful facts about group actions}
We collect here the following lemma, useful in applications of our results to verify condition~(ii) of Theorem~\ref{th:comp_comp}, and some easy remarks related to it.
\begin{lemma}\label{lm_Voronoi1}
Let $(X,d,\mu)$ be a complete locally compact metric space with nonnegative $\sigma$-finite Radon measure.
If $G$ is a
topological group acting
on $X$ by homeomorphisms
preserving $\mu$-nullsets, and
\begin{itemize}
\item[(i)] the action of $G$ is proper discontinuous,
\item[(ii)] the set of $x\in X$ having nontrivial stabilizer subgroup
$G_x:=\{g\in G\colon gx=x\}$, i.e.\ with $G_x\neq \{1\}$, is $\mu$-negligible, i.e.\
$\mu(S)=0$,
\item[(iii)] and $X$ is compactly generated, i.e. there exists a compact $K\subset X$ such that $GK=X$.
\end{itemize}
Then there
is a precompact Borel set $B\subset X$ (a fundamental domain for the action of $G$)
such that $G\bar B=X$, $\mu(\partial B)=0$,
and
$\mu(gB \cap B)=0$. Moreover, if the action of $G$ on $X$ is free, then one can choose $B$ so as to have $G B=X$ and
$gB \cap B=\emptyset$ for all $g\in G$, $g\neq 1$.
\end{lemma}
\begin{remark}\label{rem_cocomp1}
It is a more or less folkloric fact that the condition~(iii) of Lemma~\ref{lm_Voronoi1} for locally compact metric space $X$
is equivalent to \textit{cocompactness} of the action of $G$, that is, to compactness of the quotient space $X/G$.
In fact, since the natural projection map $\pi \colon X\to X/G$ is continuous, then condition~(iii) implies that $X/G=\pi(K)$ is compact. Vice versa, note that $\pi$ is also an open map (as a projection map under group action, namely, because for every open
$U\subset X$ one has $\pi^{-1}(\pi(U)))=\cup_{g\in G} gU$ is open since so is each $gU$). Therefore, taking a cover $\{U_\lambda\}$ of $X$ by precompact open sets, we have that $\{\pi(U_\lambda)\}$ is an open cover of $X/G$, and if $X/G$ is compact, extracting a
finite subcover $\{\pi(U_{\lambda_j})\}_{j=1}^m$, $m\in \N$ of $\{\pi(U_\lambda)\}$, we get that
\[
K:=\bigcup_{j=1}^m \bar U_{\lambda_j}
\]
is a compact set satisfying $GK=X$.
\end{remark}
\begin{remark}\label{rem_cocomp2}
Another immediate observation worth mentioning is that for the condition~(iii) of Lemma~\ref{lm_Voronoi1}
to hold, it is necessary that $G$ be infinite, unless, of course, $X$ is compact.
\end{remark}
\begin{proof}
For the readers' convenience we provide a detailed proof divinding it in two steps.
{\sc Step 1}. We show the existence for each $x\in X\setminus S$ of an open precompact ball $U_x\subset X$ with
$x\in U_x$ such that
\[
gU_x\cap U_x=\emptyset\quad\mbox{for all $g\in G$, $g\neq 1$}
\]
and $\mu(\partial U_x)=0$.
Let now
\[
D_x:=\{ y\in X\colon d(y,x)< d(gy, x)\quad\mbox{for all $g\in G$, $g\neq 1$}\}
\}
\]
stand for the Dirichlet-Voronoi fundamental domain for the action of $G$, so that
\[
g D_x\cap D_x=\emptyset\quad\mbox{for all } g\in G, g\neq 1.
\]
In fact, if $y\in g D_x$, that is, $gy\in D_x$, then $y:=g^{-1}gy \not\in D_x$.
Since $X$ is locally compact, then there is an $r>0$ such that the balls $B_\varepsilon(x)$ with $\varepsilon<r$ are precompact.
We claim that $D_x$ contains a small ball $B_\rho(x)$, $\rho<r$. In fact, otherwise there is a sequence of points $y_k\in X$ and a sequence of
$g_k\in G$ such that
$\lim _k y_k = x$ and
\[
d(y_k,x)\geq d(g_k y_k, x)
\]
so that
also $\lim_k g_k y_k = x$.
This is only possible when the set
$\{g_k\}$ is finite, because when $\varepsilon <r$ one has that
$g \bar B_\varepsilon (x)\cap \bar B_\varepsilon (x)\neq \emptyset$ for only a finite number of $g\in G$, again by proper discontinuous action.
Therefore, there is a $g\in G$, $g\neq 1$, such that up to a subsequence (not relabeled) we have
$\lim _k g y_k = x$, which imples that $gx=x$, that is a contradiction to $x\not\in S$.
It remains now to observe that since $B_\rho(x)\subset D_x$, we may find a small ball $U_x:= B_s(x)$ such that
$U_x\subset B_\rho(x)$ (hence is precompact) and $\mu(\partial U_x)=0$ (all but a countable number of $s<\rho$ will suffice).
{\sc Step 2}. If $S\neq\emptyset$, we let for every $k\in \N$ let $S_k\supset S$ be an open set such that $\mu(\bar S_k)\leq 1/k$
(otherwise, if $S=\emptyset$, we just let $S_k:=\emptyset$).
Consider a finite cover $\{S_k, U_1,\ldots, U_{n_k}\}$ of $K$ in $X$, where
each $U_i=U_{x_i}$ is a precompact open
sets containing an $x_i\in K\setminus S$ such that
\[
gU_i\cap U_i=\emptyset\quad\mbox{for all $g\in G$, $g\not\in G_x$}
\]
and $\mu(\partial U_i)=0$ for all $i$.
Without loss of generality we may assume
that
\begin{equation}\label{eq_Ujpart1}
U_j\subset \left(\bigcup_{i=1}^{n_k} U_i\right)\cup S_k
\end{equation}
for all $j>n_k$.
We define inductively disjoint sets
\begin{align*}
V_1 &:= U_1,\\
V_{i+1} &:= U_{i+1}\setminus \bigcup_{j=1}^i G U_j.
\end{align*}
Note that since $\bar U_j$ are all compact and $G$ acts properly discontinuously, then
each $G U_j$ in the definition of $V_i$ can be substituted by a finite union of $g U_j$ over a finite subset
of $g\in G$ (depending of course on $i$ and $j$): in fact,
\[
g U_j\cap U_{i+1} \subset g \overline{(U_j\cup U_{i+1})}\cap \overline{(U_j\cup U_{i+1})}\neq \emptyset
\]
for an at most finite set of $g\in G$. Therefore, recalling also that
$\partial (g U_j)= g\partial U_j$ (since the action of $g$ is a
homeomorphism), and
hence $\mu (\partial (g U_j))= \mu (g\partial U_j)=0 $ (since the action preserves $\mu$-nullsets), we get
$\mu(\partial V_i)=0$ for all $i$.
Seting now
\[
\tilde B_k:=\bigcup_{i=1}^{n_k} V_i,
\]
we get that $\tilde B_k$ is a Borel set satisfying
\begin{equation}\label{eq_deftildeB1}
g\tilde B_k \cap \tilde B_k=\emptyset\quad\mbox{for all $g\in G$, $g\neq 1$}
\end{equation}
because so are all $V_i$, $\mu(\partial \tilde B_k)=0$.
We now define
\[
B_k:= \tilde B_k \cup S_k \quad\text{and } B:=\cap_k B_k.
\]
Note that the sequence of sets $B_k$ is nonincreasing in view of~\eqref{eq_Ujpart1}.
By~\eqref{eq_deftildeB1} one has $gB_k \cap B_k\subset S_k\cup g S_k$.
Thus
\[
gB\cap B \subset \bigcap_k (S_k\cup gS_k)
\]
for all $g\in G$, $g\neq 1$.
Since $S_k\cup gS_k$ is a decreasing sequence of sets and
\[
\mu(S_k\cup gS_k)\leq \mu(S_k) + \mu(gS_k)\to \mu(S) + \mu(gS)=0
\]
as $k\to\infty$, since the action of $G$ preserves $\mu$-nullsets, then
$\mu(gB\cap B) =0$
for all $g\in G$, $g\neq 1$.
Further,
\begin{align*}
\mu (\partial B)\leq \mu((\partial B)\setminus \bar S_k) + \mu(\bar S_k) \leq \mu (\partial \tilde B_k) + \mu(\bar S_k) = \mu(\bar S_k) \leq 1/k,
\end{align*}
hence $\mu(\partial B)=0$.
Finally,
$GB_k=X$ because $GB_k \supset GK=X$.
This means that for each $x\in X$ there is an $y_k \in B_k$ and $g_k\in G$ such that
$g_ky_k=x$. Since $B_k$ is a decreasing sequence of precompact sets, then there is a compact $\tilde K$ such that $\{y_k\}\subset \tilde K$. Thus $g_k\tilde K$ have a common point which can only happen for a finite number of $g_k$ by discontiuity of action. Thus, up to a subsequence we have $g_k=g\in G$, and $gy_k=x$. Extracting a further convergent subsequence of
$y_k$, we have $\lim_k y_k=y\in \bar B$ and $gy=x$, hence $G\bar B=X$.
Note that
when $S=\emptyset$, then all $S_k=\emptyset$ and hence $B=B_k=\tilde B_k$ for some finite $k$, concluding the proof.
\end{proof}
The following easy technical lemma is also used.
\begin{lemma}\label{lm_groupPart1}
Let $G$ be a group acting on a set $X$ and $K\subset X$.
If
\[
\nu:=\#\{g\in G\colon gK\cap K\neq \emptyset\} <\infty,
\]
then there is a partition of $G$ into at most $\nu$ disjoint subsets $\mathcal{F}_j$, such that
$hK\cap gK =\emptyset$
for every couple
$\{h, g\}\subset \mathcal{F}_j$ for each
$j=1,\ldots,\nu$, and each $\mathcal{F}_j$ is a maximal subset of $G$ with this property, i.e.\ for every
$\tilde h\not \in \mathcal{F}_j$ there is a $\tilde g\in \mathcal{F}_j$ such that
$\tilde h K\cap \tilde g K \neq\emptyset$.
\end{lemma}
\begin{proof}
We construct the sets $\mathcal{F}_j$ inductively.
Namely, by Zorn's lemma there is a set $\mathcal{F}_1\subset G$ maximal (with respect to inclusion) containing $1$ such that
$hK\cap gK =\emptyset$
for every couple
$\{h, g\}\in \mathcal{F}_1$.
Once $\mathcal{F}_j$ are constructed for $j=1,\ldots, k$, if there an $h\not\in \cup_{j=1}^k \mathcal{F}_j$, we take
$\mathcal{F}_{k+1}$ to be a maximal (with respect to inclusion) subset of $G$ containing $h$ such that
$hK\cap gK =\emptyset$
for every couple
$\{h, g\}\in \mathcal{F}_{k+1}$ (the existence is again guaranteed by Zorn's lemma).
It is easy to see that the induction finishes after at most $\nu$ steps, i.e.\ we may construct at most $\nu$ nonempty sets
$\mathcal{F}_j$.
In fact, if we managed to construct nonempty sets $\mathcal{F}_1,\ldots, \mathcal{F}_{\nu+1}$,
then by maximality of each of them, for an $h\in \mathcal{F}_{\nu+1}$ there are $h_k\in \mathcal{F}_k$, such that
$h_k K \cap h K \neq \emptyset$, or, equivalently, $h^{-1}h_k K \cap K\neq \emptyset$,
where $k=1,\ldots,\nu$. In other words,
$\{h^{-1}h_k\}_{k=1}^{\nu} \subset \{g\in G\colon gK\cap K\neq \emptyset\}$,
and since all $h^{-1}h_k$ are distinct, these two sets must coincide. Hence, for some $k$ one has $h^{-1}h_k=1$, that is,
$h=h_k$, which is impossible, and thus only sets $\mathcal{F}_1,\ldots, \mathcal{F}_\nu$ are nonempty.
\end{proof}
\section{Auxiliary lemmata}
We collect here, mainly for the sake of completeness and readers' convenience, some auxiliary lemmata of more or less folkloric nature.
\begin{lemma}[equisummability]\label{lm:equisummability}
Suppose that $m_{k,i}\geq 0$, and:
\begin{gather*}
\lim_{k\to +\infty} \sum_i m_{k,i} = m, \\
\lim_k m_{k,i} = m_i\\
\lim_n \enclose{\sup_k \sum_{i=n}^{+\infty} m_{k,i}} = 0.
\end{gather*}
Then
\[
\sum_i m_i = m.
\]
\end{lemma}
\begin{proof} For every $\eps>0$ there is an $n_\eps\in \N$ such that
for all $k\in \N$ and $n\geq n_\eps$ one has
\[
\sum_{i=n}^{+\infty} m_{k,i} < \eps , \quad\text{and}\quad \sum_{i<n} m_{k,i}\le \sum_i m_{k,i} \le \sum_{i<n} m_{k,i}+\eps.
\]
Hence either $m=+\infty$ and $m_i=+\infty$ for some $i<n_\eps$, or $m\in\R$.
In the first case the thesis follows. In the second letting $k\to +\infty$ we obtain
for $n\geq n_\eps$ the estimate
\[ m-\eps \le \sum_{i=1}^n m_i \le m.
\]
Letting now $n\to +\infty$, one gets
\[
m-\eps \le \sum_{i} m_i \le m
\]
for all $\eps>0$ implying the thesis.
\end{proof}
\begin{lemma}\label{lm:union}
Suppose that for all $k\in \N$, $j=1,\ldots, N$, where $N\in \N\cup\{\infty\}$ the sets
$A_k\subset X$, $A_k^j\subset X$ be $\mu$-measurable and
\[
\mu\left(A_k \triangle \bigcup_{j=1}^N A_k^j\right)=0, \qquad \mu\left(A_k^j \cap A_k^i\right)=0 \quad \mbox{whenever } i\neq j.
\]
If $A_k\to A$ and $A_k^j\to A^j$ in $L^1_\loc(\mu)$ as $k\to\infty$, then
\[
\mu\left(A\triangle \bigcup_{j=1}^N A^j\right)=0, \qquad \mu\left(A^j \cap A^i\right)=0 \quad \mbox{whenever } i\neq j.
\]
\end{lemma}
\begin{proof}
It suffices to write for every compact $K\subset X$ the relationships
\begin{align*}
\mu\left(\left( A_k \triangle \bigcup_{j=1}^N A_k^j\right) \cap K\right) &=\int_K \left|\mathbf{1}_{A_k}(x)-\left( \sum_{j=1}^N \mathbf{1}_{A_k^j}(x)\right)\right| \,d\mu(x),\\
\mu\left(\left( A_k^j \cap A_k^i\right) \cap K\right) &=\int_K \mathbf{1}_{A_k^j}(x)\mathbf{1}_{A_k^i}(x) \,d\mu(x),
\end{align*}
and pass to the limit as $k\to \infty$.
\end{proof}
\bibliographystyle{plain} | 0.015991 |
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TITLE: Reference for a proof: $\mathsf{Hom}(V, W)\longrightarrow \mathsf{Hom}(W^*, V^*), f\longmapsto f^*$, is an isomorsphim.
QUESTION [2 upvotes]: Let $V$ and $W$ be two vector spaces over a field $\mathbb K$. The transpose of a linear map $f:V\longrightarrow W$ is the linear map $f^*:W^*\longrightarrow V^*$ given by $$f^*(u):=u\circ f.$$ I read in this page that whenever $V$ or $W$ are finite dimensional then the map $$\mathsf{Hom}(V, W)\longrightarrow \mathsf{Hom}(W^*, V^*), f\longmapsto f^*,$$ is an isomorphism. I looked for the proof in some standard books and couldn't find it, can anyone recomend me a reference where I can find the proof?
Thanks.
REPLY [1 votes]: The map is injective. If $f\ne0$, take $v\in V$ with $w=f(v)\ne0$. Then there exists $u\in W^*$ with $u(w)\ne0$ (complete $w$ to a basis of $W$). Therefore $f^*(u)\ne0$, because $u\circ f(v)=u(w)\ne0$.
Suppose $V$ and $W$ are finite dimensional. Then the existence of an injective map $\operatorname{Hom}(V,W)\to\operatorname{Hom}(W^*,V^*)$ implies
$$
\dim\operatorname{Hom}(V,W)\le\dim\operatorname{Hom}(W^*,V^*)
$$
(it's easy to see $\operatorname{Hom}(V,W)$, $V^*$ and $W^*$ are finite dimensional as well).
By the same reason,
$$
\dim\operatorname{Hom}(W^*,V^*)\le\dim\operatorname{Hom}(V^{**},W^{**})
$$
Now we can easily build an isomorphism
$$
\operatorname{Hom}(V^{**},W^{**})\to\operatorname{Hom}(V,W)
$$
by using the fact that the canonical maps $V\to V^{**}$ and $W\to W^{**}$ are isomorphisms.
An alternative proof for surjectivity in the finite dimensional case.
Fix a basis of $W$, say $\{w_1,\dots,w_n\}$ and consider that a map $g\colon W^*\to V^*$ is determined by its action on the dual basis $\{w_1^*,\dots,w_n^*\}$. This means that, given arbitrary elements $u_1,\dots,u_n\in V^*$, we need to find $f\colon V\to W$ such that
$$
f^*(w_i^*)=u_i,\qquad i=1,\dots,n
$$
This means $w_i^*\circ f=u_i$. Let $\{v_1,\dots,v_m\}$ be a basis of $V$. We must have $w_i^*(f(v_j))=u_i(v_j)$.
Writing $f(v_j)=\sum_{k=1}^n \alpha_{kj}w_k$ this amounts to
$$
\alpha_{ij}=u_i(v_j)
$$
and we can so define $f$ with the required property. | 0.006963 |
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Learning Curves of the Airtraq and the Macintosh Laryngoscopes for Tracheal Intubation by Novice Laryngoscopists: A Clinical StudyDi Marco, Pierangelo; Scattoni, Lorena; Spinoglio, Annamaria; Luzi, Marta; Canneti, Alessandra; Pietropaoli, Paolo; Reale, CarloSurvey of Anesthesiology: December 2011 - Volume 55 - Issue 6 - p 311–312 doi: 10.1097/01.SA.0000407043.33009.11 Techniques and Monitoring Buy Author InformationAuthors Article MetricsMetrics Department of Anesthesia and Intensive Care Medicine, Sapienza University of Rome, Rome, Italy. © 2011 Lippincott Williams & Wilkins, Inc. | 0.303058 |
\section*{Appendix}
\section{Lower bounds for sample complexity of \erspuddc algorithm in the Bernoulli-Rademacher model}\SectionName{bernoulli-rademacher-lower-bound}
In this section we prove that the modification introduced in this paper to \erspuddc algorithm is necessary in order to guarantee the correctness for arbitrary Bernoulli-subgaussian $X$ with strictly subquadratic number of samples $p$. More concretely, we prove that if $X$ follows the Bernoulli-Rademacher model, and $p = \Oh(n^{2 - \varepsilon}\log n)$, then the \erspud algorithm actually fails to recover $A$ and $X$ with probability at least $1 - \tilde{\Oh}(\frac{1}{n^{\varepsilon}})$. The proof of this theorem relies on few technical lemmas, they are presented later in this section.
\begin{theorem}\TheoremName{main-lb}
For every constant $C$ and $\varepsilon \leq 1$, there exist $C'$, such that for sufficiently large $n$ if $2 n \leq p \leq C n^{2 - \varepsilon} \log n$ and $X \in \R^{n\times p}$ follows the Bernoulli-Rademacher model with sparsity parameter $\theta := C' \frac{\log n}{n}$, then the \erspuddc algorithm fails to recover $X$ with probability at least $1 - \Oh(\frac{\log^5 n}{n^\varepsilon})$.
\end{theorem}
\begin{proof}
We shall first prove that once following events happens simultaneously, the \erspuddc algorithm fails to recover $X$. Later on we will prove that each of those events fails with probability at most $\Oh(\frac{\log^5 n}{n^\varepsilon})$ --- that will be enough to conclude the statement of the theorem.
In what follows, let $X_i$ be the $i$-th column of $X$, $j_* \in [n]$ be the index of the row of $X$ with largest number of non-zero entries (for concreteness, the smallest such index), and $K$ be some universal constant the same for fourth and fifth event, and will be specified later. Consider the following events
\begin{enumerate}
\item Matrix $X$ is of full rank.
\item For every $i \in [p-1]$ it holds that $|\supp(X_i) \cap \supp(X_{i+1})| < 2$.
\item For every $i \in [p-1]$ it holds that $j_* \not\in \left(\supp(X_i) \cap \supp(X_{i+1})\right)$.
\item Every column of $X$ has at least $K \log n$ nonzero entries.
\item The number of rows of $X$ with largest support size is smaller than $K \log n$.
\end{enumerate}
Let us call those events $\mathcal{E}_1, \ldots, \mathcal{E}_5$ respectively. We claim that under $\mathcal{E}_1, \ldots, \mathcal{E}_5$, the $j_*$-th row of $X$ would not be recovered by the \erspuddc algorithm. Indeed, assume for the proof by contradiction, that solving the optimization problem $\min_{w} \|w^T Y\|_1$ subject to $r_*^T w = 1$, yields a solution such that $w^T Y$ is proportional to the $j_*$-th row of $X$, for some $r_*$ which is sum of two consecutive columns of $Y$. By condition $\mathcal{E}_1$, it means that the solution to the equivalent problem of $\min_{z} \|z^T X\|_1$ subject to $b_*^T z = 1$ is $z = \pm e_{j_*}$, where $b_*$ is a sum of corresponding columns of $X$.
Observe that, because the matrix $X$ has entries in $\{-1, 0, 1\}$, the $\ell_1$ norm and sparsity of each row is equal, that is for every $k$ we have $\| e_k^T X \|_1 = \|e_k^T X\|_0$.
Now, by condition $\mathcal{E}_2$, at most one coordinate of $b_*$ is of absolute value $2$, and all other are either $\pm 1$ or $0$. Moreover, if the condition $\mathcal{E}_3$ holds, the entry with absolute value $2$ is not $j_*$-th. If for some $k$, we had $(b_*)_k = 2$, then taking $z := \frac{1}{2} e_k$, would yield a feasible solution to the optimization problem mentioned above, and one with smaller value of objective function --- $\frac{1}{2}\|e_k^T X\|_0$ as opposed to $\|e_{j_*}^T X\|_0$; similarly for $(b_*)_k = - 2$. Therefore, all non-zero entries of $b_*$ must have absolute value $1$.
Moreover $b_*$ has support of size at least $K \log n$ (as $\mathcal{E}_4$ holds and $b_*$ is a sum of two columns of $X$ with almost disjoint support by $\mathcal{E}_2$), hence $|\supp b_*|$ is strictly larger than the number of rows of $X$ with largest support --- this number of rows is less than $K \log n$ by $\mathcal{E}_5$. In particular, there is some $k \in \supp(b_*)$, such that the $k$-th row of $X$ has strictly smaller support size than $j_*$-th row. Again, if this is the case $z = e_{j_*}$ is not a solution to the optimization problem $\min_{z} \|z^T X\|_1$ subject to $b_*^T z = 1$ --- as $z := e_k$ (or $-e_k$) is feasible and with strictly smaller objective value.
From this contradiction we conclude, that once all the events in preceding list hold simultaneously, the \erspuddc algorithm fails in recovering $X$, and therefore fails to recover the hidden decomposition.
It is now enough to show that each of those events $\mathcal{E}_1 \ldots \mathcal{E}_5$ fails with probability at most $\Oh(\frac{\log^5 n}{n^\varepsilon})$.
Event $\mathcal{E}_1$ fails with probability at most $n(1 - \frac{C'\log n}{n})^n \lesssim \frac{1}{n^{C' - 1}}$ by \Lemma{X-full-rank} and assumption that $p > 2n$. For $C' \geq 1$ and large enough $n$ this quantity is smaller than $\frac{\log^5 n}{n^\varepsilon}$.
For event $\mathcal{E}_2$, it holds with probability $1 - \Oh(\frac{\log^5}{n^\varepsilon})$ simply by union bound --- for every fixed $i$, we have $\P(\supp(X_i) \cap \supp(X_{i+1}) \geq 2) \leq \binom{n}{2} \theta^4 \lesssim \frac{\log^4 n}{n^2}$ --- and we need a union bound over $p \leq C n^{2 - \varepsilon} \log n$ such events.
To bound the probability of event $\mathcal{E}_3$, let random set $S \subset [p]$ be the support of $j_*$-th row of $X$. In what follows we will condition implicitly on $S \not= \emptyset$ as $S$ is empty with exponentially small probability.
Expected support size of any single row is $p\theta > 2 \log n$, therefore by Chernoff and union bound, we deduce that except with probability smaller than $\frac{1}{n}$ all rows of $X$ has support size smaller than $C_2 p\theta$ for some universal constant $C_2$. Conditioned on $|S| < C_2 p \theta$, the distribution of $S$ is invariant under permutations of $[n]$, in a sense that for fixed set $S_0\subset [n]$ probability $\P(S = S_0 | |S| < C_2 p \theta)$ depends only on the size of $S_0$. In such a case, and because of additional conditioning on $S$ being nonempty, by \Lemma{S-negative-correlation} for every $i \not= j$ we have
\begin{equation*}
\P(i \in S | j \in S \land |S| < C_2 p \theta) \leq \P(i \in S | |S| < C_2 p \theta)
\end{equation*}
In particular, for fixed $i \in [p-1]$, we have
\begin{align*}
\P\left(i \in S \, \land\, (i+1) \in S \,|\, |S| < C_2 p \theta \right) & \leq \P\left(i \in S | |S| < C_2 p \theta\right) \P\left(i+1 \in S | |S| < C_2 p \theta \right) \\
& \leq \frac{C_2^2 p^2 \theta^2}{p^2} \\
& \lesssim \frac{\log^2 n}{n^2}
\end{align*}
Hence, by union bound over all $i \in [p-1]$, it follows that
\begin{equation*}
\P(\lnot \mathcal{E}_3) \leq \P(|S| \geq C_2 p \theta) + \P(\lnot \mathcal{E}_3 | |S| < C_2 p \theta) \lesssim \frac{\log^3 n}{n^{\varepsilon}}
\end{equation*}
For $\mathcal{E}_4$, we know that the expected number of non-zero entries in a column of $X$ is $\theta n = C' \log n$ --- by the Chernoff bound, probability that any such column has sparsity smaller than $K \log n$ is much smaller than $\frac{1}{n^4}$ if we set $C'$ large enough depending on $K$. Therefore by union bound, they all have sparsity at least $K \log n$ simultaneously with probability at least $1 - \frac{1}{n^2}$.
In order to bound probability of failure for the event $\mathcal{E}_5$, let $s_i\in \N$ for $i \in [n]$ be the size of the support of the $i$-th row of $X$. Clearly $s_i$ are Binomial random variable with parameters $(p, \theta)$. Take $\gamma := \frac{K_1 \log n}{n}$ (with some constant $K_1$ that will be specified later) and let $T_0 \in [p]$ be the largest number such that $\P(s_i \geq T_0) \geq \gamma$. We want to apply \Lemma{no-large-gaps} for all random variables $s_i$. Observe that in this setting $\E s_i = p\theta \ll \frac{p}{8}$, and on the other hand $\E s_i \geq 2 \log n$.
Moreover, observe that $T_0 \leq 4 \E s_i$ --- it is enough to show that $\P(s_i \geq 4 \E s_i) \leq \gamma$, and this fact follows from Chernoff bound if $K_1$ is large enough constant. Therefore, we can apply \Lemma{no-large-gaps} to conclude that
\begin{equation*}
\P(S_i \geq T_0) \leq K_2 \gamma \max(1, \frac{T_0}{\E S_i}) \leq K_3 \gamma
\label{}
\end{equation*}
where $K_2$ and $K_3$ are some universal constants.
Now we want to show that with probability at least $1 - \frac{1}{n}$, number of $s_i$ that are not smaller than $T_0$, is between $1$ and $4 K_3 \gamma n = 4 K_3 K_1 \log n =: K_4 \log n$. If we consider indicator random variables $M_i \in \{0, 1\}$, such that $M_i = 1$ if and only if $S_i \geq T_0$, by previous discussion we know that $\gamma \leq \P(M_i = 0) \leq K_3 \gamma$, and all $M_i$ are independent. We can now apply the Chernoff bound to bound the probability of $\P(\sum M_i < 1)$ and $\P(\sum M_i \geq 4 K_3 \gamma n)$ --- again, if $K_1$ is large enough, each of those is much smaller than $\frac{1}{2n}$ --- we can now fix constant $K_1$ large enough so that all three Chernoff bounds yield desired inequalities.
Finally, if the number of rows with support larger than $T_0$ is between one and $K_4 \log n$, then clearly at most $K_4 \log n$ has the largest support --- and therefore the event $\mathcal{E}_5$ holds with $K := K_4$.
\end{proof}
We now prove certain technical lemmas that were used in the proof of \Theorem{main-lb}.
\begin{lemma}
\LemmaName{no-large-gaps}
For $\theta \in (0,1)$ and $p\in \N$, let $Q$ be a Binomial random variable with parameters $(p, \theta)$ and let $\gamma \in (0,1)$ be some fixed threshold. Moreover, let $T_0$ be the largest natural number such that
\begin{equation*}
\P(Q \geq T_0) \geq \gamma
\end{equation*}
and assume that $T_0 \leq \frac{p}{2}$. Then
\begin{equation*}
\P(Q \geq T_0) \leq K \gamma \max(1, \frac{T_0}{\E Q})
\end{equation*}
for some universal constant $K$.
\end{lemma}
\begin{proof}
We start with bounding the ratio
\begin{align*}
\frac{\P(Q = T_0)}{\P(Q = T_0 + 1)} & = \frac{ \binom{p}{T_0} \theta^{T_0} (1-\theta)^{p - T_0}}{\binom{p}{T_0 + 1} \theta^{T_0 + 1} (1-\theta)^{p - T_0 - 1}} \\
& = \frac{T_0 + 1}{p - T_0} \cdot \frac{1 - \theta}{\theta} \\
& \leq K_1 \frac{T_0}{p \theta} \\
& = K_1 \frac{T_0}{\E Q}
\end{align*}
We can rephrase it as $\P(Q = T_0) \leq K_1 \frac{T_0}{\E Q} \P(Q = T_0 + 1)$, and clearly $\P(Q = T_0 + 1) \leq \P(Q \geq T_0 + 1)$. Therefore
\begin{equation}
\P(Q = T_0) \leq K_1 \frac{T_0}{\E Q} \P(Q \geq T_0 + 1)
\label{}
\end{equation}
We can now directly bound the desired probability as follows
\begin{align*}
\P(Q \geq T_0) & = \P(Q = T_0) + \P(Q \geq T_0 + 1) \\
& \leq \left(K_1 \frac{T_0}{\E Q} + 1 \right) \P( Q \geq T_0 + 1) \\
& \leq \left(K_1 \frac{T_0}{\E Q} + 1 \right) \gamma
\label{}
\end{align*}
Where the last inequality follows from the assumption that $T_0$ were largest such that $\P(Q \geq T_0) \geq \gamma$.
\end{proof}
\begin{lemma}
\LemmaName{X-full-rank}
Let $X\in \R^{n\times p}$ follow the Bernoulli-Rademacher model with sparsity parameter $\theta$ and $p > n$. Then matrix $X$ is of rank $n$ with probability at least $1 - n\left(1 - \theta\right)^{p-n}$.
\end{lemma}
\begin{proof}
Let $W_i \subset \mathbb{R}^p$ be a subspace of $\mathbb{R}^p$ spanned by first $i-1$ rows of $X$. We wish to prove that $i$-th row of $X$ lies in $W_i$ with probability at most $\left(1 - \theta\right)^{p - n}$ --- if we do this, the claim will follow by the union bound. Fix some $i$, and let $v$ be the $i$-th row of $X$; moreover, let $W^{\bot}$ be the orthogonal complement of $W_i$. Clearly $\dim W^{\bot} \geq p - n$. We will show that for any fixed $W^{\bot}$
\begin{equation*}
\P(v \bot W^{\bot}) \leq (1-\theta)^{\dim W^{\bot}}
\end{equation*}
Indeed, let $q := \dim W^{\bot}$, and consider sequence of indices $i_1, \ldots i_q$ together with a basis $u_1, \ldots u_q$ of $W^{\bot}$ such that for every $r$ we have $(u_r)_{i_r} \not= 0$, and for every pair $s < r$ we have $(u_s)_{i_r} = 0$ --- such a basis and sequence of indices exists by Gaussian elimination. Now, by the chain rule, we have
\begin{align}
\P\left(v \bot W^{\bot}\right) & = \prod_{r=1}^q \P\left(\langle v, u_r \rangle = 0 | \forall_{s < r} \langle v, u_s \rangle = 0\right) \EquationName{v-on-W}
\end{align}
Let us fix some $r$ now. We want to show that $\P\left(\langle v, u_r \rangle = 0 | \forall_{s < r} \langle v, u_r \rangle = 0\right) < (1-\theta)$. Observe that the event $\forall_{s < r} \langle v, u_s \rangle = 0$ is independent of $v_{i_r}$; moreover, if we fix all values of $v_j$ for $j\not= i_r$, probability of $\langle v, u_r \rangle = 0$ is at most $(1-\theta)$ --- there is at most one value of $v_{i_r}$ that would make this inner product equal zero, and $v_{i_r}$ assumes every value with probability at most $(1-\theta)$. Therefore
\begin{align*}
\P\left(\langle v, u_r \rangle = 0 | \forall_{s < r} \langle v, u_r \rangle = 0\right) & = \E\left(\P\left(\langle v, u_r \rangle = 0 | v_1, \ldots \hat{v_{i_r}} \ldots v_p\right) | \forall_{s < r} \langle v, u_r\rangle = 0\right) \\
& \leq \E \left( 1-\theta | \forall_{s < r} \langle v, u_r\rangle = 0\right) \\
& = 1 - \theta
\end{align*}
Where $v_1, \ldots \hat{v_{i_r}} \ldots v_d$ denotes omitting the $i_r$-th index in this sequence.
We can plug this back to \Equation{v-on-W} to conclude that $\P\left(v \bot W^{\bot}\right) \leq (1-\theta)^{\dim W^{\bot}}$ and the statement of the lemma follows.
\end{proof}
\begin{lemma}
Let $S \subset [n]$ be a random set, with permutationally invariant distribution, i.e. such that for fixed $S_0$, $\P(S = S_0)$ depends only on the size of $S_0$. Assume moreover, that $S$ is nonempty almost surely. Then for any $i \neq j \in [n]$, we have $\P(i \in S | j \in S) \leq \P(i \in S)$. \LemmaName{S-negative-correlation}
\end{lemma}
\begin{proof}
For $k \in \{0, \ldots n\}$, let $p_k := \P(|S| = k)$. Observe that for fixed $i\in S$
\begin{equation}
\P(i \in S) = \sum_{k=1}^n p_k \P(i \in S | |S| = k) = \sum_{k=1}^n p_k \frac{k}{n}
\label{ }
\end{equation}
On the other hand
\begin{align*}
\P(i \in S | j \in S) & = \frac{1}{1 - p_0} \left( \sum_{k=1}^n p_k \P(i\in S | j\in S, |S| = k) \right) \\
& = \frac{1}{1 - p_0} \left( \sum_{k=1}^n p_k \frac{k-1}{n-1} \right) \\
& = \sum_{k=1}^n p_k \frac{k-1}{n-1}
\label{}
\end{align*}
where the last equality follows from the assumption $\P(S = \emptyset) = 0$.
Then the statement of the lemma follows by explicitly comparing two expressions for $\P(i \in S)$ and $\P(i \in S | j \in S)$, and using inequality $\frac{k-1}{n-1} \leq \frac{k}{n}$.
\end{proof} | 0.011551 |
It's shortsighted leadership to attempt to permanently squeeze extra time from workers so they can't catch their breath and recharge.
Are there people in your company who are working inflamed? Angry, hot around the collar, on guard, suspicious,
untrusting, disbelieving, on edge,
ready to react? All are potentially dangerous
to their company's smooth functioning — and to
their personal safety.
Cutting-edge medical research is replete with references
to inflammation. Brent Bauer, M.D., wrote in
the Mayo Clinic Health Letter, "It seems as though
everyone is talking about inflammation, especially
the fact that it appears to play a role in many chronic
diseases." These illnesses run the gamut, from heart/
cardiovascular problems to Alzheimer's, acne, arthritis,
psoriasis, cancer, asthma, lupus, and many more.
Acute (short-term) inflammation isn't necessarily
bad. In a healthy person, it arises from the immune
system's mobilizing in response to illness or injury,
doing its job and then standing down. But many
people are chronically inflamed. Their innate defense
system rampages out of control, fighting on against
an enemy that's not there (or never was), sometimes
attacking the very body it's supposed to protect. This
can lead to autoimmune disease, disintegration, and
loss of all kinds of function.
I've seen similar processes occurring within organizations.
People working at a feverish pace leading to burnout.
Trust in leaders crumbling. Anger leading first to disaffection,
then to barely being present, and ultimately even not showing up
at all. And accidents or illness stemming from misdirected attention,
where workers were "seeing red" rather than noting changing
hazards and calmly making needed head-it-off-at-the-pass adaptations.
The good news is that leaders can harness energy toward productivity
and safety; the bad part is ineff ective leaders can push
people over the edge. Some leaders overly rely on charging up
workers, lighting a fire underneath them, pushing them to perform,
squeezing as much out of them as possible (while sometimes
taking back pay or benefits). It is important to keep an organization
moving, creative, warm — but in a controlled manner. There's a
marked distinction between flames in a fireplace that heat a home
vs. a raging fire that consumes it.
Think of leadership as setting up a thermostatic climate where
work flows at the optimal temperature. We've seen heating and
cooling systems actually working against each other.
Experience shows leaders in highest-performing companies create
and monitor strong balance, make sure situations don't stay too
hot, promote healing aft er angry negotiations or disaff ected takeaways,
and help people align toward safely accomplishing critical
tasks. Here are some methods leaders can use to heal inflammation
before it results in chronic dysfunction or spiraling breakdown:
Boost healthy intake. The wrong foods (fats, sugar, gluten for
some, etc.) can elevate whole-body inflammation. Similarly, kneejerk
communications may raise a company's temperature
during times of tension. Always assume
that whatever you write or say, even if "in private,"
may be overheard or read by someone, fueling a destructive
rumor-mill cycle. Temper communications
others will ingest.
Exercise. Cardiovascular conditioning helps reduce
inflammation; similarly, encouraging people,
committees, and groups to move forward can head
off a destructive frustration-anger-inflammation
cycle. So make sure Safety committees have real
training, tangible objectives, as well as adequate
budget and power to make concrete and visible improvements.
Build in recovery time. Align your expectations
(and staffing) to encourage all workers to actually
take replenishing breaks, lunch periods, and vacations.
It's shortsighted leadership to attempt to permanently
squeeze extra time from workers so they
can't catch their breath and recharge.
Reduce overreaction. Everyone watches how
leaders respond to unforeseen circumstances. The
most potent executives control themselves first. They
don't run amok or frantic; rather, they first take time to gather their
thoughts, make peace with their emotions, and then communicate
cogently and reassuringly. Similarly, don't allow yourself to speak
harshly or "shoot the messenger."
Minimize over-stress. Stress, like inflammation, is mostly a
problem when it doesn't abate. Th ree of the most potent organizational
stress reducers are: 1. allowing people to take as much control
of their own work as possible (within guidelines), 2. employing
appropriate humor (that doesn't make anyone the butt of a joke),
and 3. encouraging social support (i.e., opportunities for people to
get together, make contact, safely vent, and feel part of a team).
Best leaders watch and then regulate inflammation before it
gets to a fevered pitch, heightening their workers' and company's
health, productivity, and safety.: | 0.995076 |
Summary
At first blush, Earth appears to have power to spare. The total power from sunlight striking the ground is a whopping 101,000 terawatts, and experts estimate that we could capture enough of that to exceed by a wide margin the 15 terawatts of power that the world's population now consumes. But sunshine and other forms of renewable energy such as wind deliver energy in a far less dense form than coal, oil, or natural gas. Another issue: The sun doesn't necessarily shine the brightest and the wind doesn't blow the fiercest where most people live. And technologies have yet to emerge to store and transport vast amounts of energy generated from sunshine or wind. The energy problem is also a water problem. Electricity from solar thermal technologies uses 68% more water than electricity from coal. Use power from biomass crops and you'll also use hundreds of liters of water to grow the fuel. | 0.985097 |
Dear Jacob,
First off, I would like to tell you how interesting of a character you are. I love the fact that you put lots of feelings into everything. You're alos a one of a kind guy. I can't tell you how amazed I am sometimes the way you handle things. With this whole August and Marlena thing, i feel you've handled it very weel. I really want you to get Marlena because i know you feel shes just the greatest thing. I also don't like August because of how he treats Marlena, is mean to ytou now and then, and has that problem of his where he can be cool one minute and scary the next. I think you should just ditch the circuz for a number of reasons. 1. You have never been payed yet because Uncle Al is cheating you out of your money. 2. It's a dirty place, you sleep on a crappy cot in the room where the horses are kept. 3. You could still be a vet without being the circus and be with the animals you love. 4. I feel like your wasting your life away dreaming about Marlena and just working your butt off for nothing. Anyway, I hope you get a pay check soon and i hope you work things out with August and Marlena.
-Colin Moynihan | 0.011443 |
This lovely fresh autumn salad is crunchy, colourful and satisfying. Smoked Kelp is used here instead of salt or any other seasoning; it is a lovely vegetarian alternative to bacon for those who like the smoked flavour in food.
Continue reading "Beetroot, Fennel & Feta Salad flavoured with Smoked Kelp" » | 0.039563 |
No.1 Melvin Fields, Kinlough
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& dishwasher 3.7 x 8.4
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with fully tiled ensuite 3.7 x 4.6
Bedroom 2 3.7 x 3.7
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Bathroom – fully tiled 1.8 x 3.6
Outside
Spacious Garage with roller doors
Potting Area beside garage
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Good on-site parking bays
Features
Oil fired central heating
uPVC double glazing
Site Area c. 600 sq m
BER: B3
BER No. 101511939
EPI: 142kWh/m²/yr | 0.000947 |
TITLE: Deceptively simple inequality involving expectations of products of functions of just one variable
QUESTION [7 upvotes]: For a proof to go through in a paper I am writing, I need to prove, as an auxiliary step, the following deceptively simple inequality:
$$E(X^a) E(X^{a+1} \ln X) > E(X^{a+1})E(X^a \ln X) $$
where $X>e$ has a continuous distribution and $0<a<1$. The intuition, in one sentence, is that if you start from
$$E(X^a) E(X^a \ln X) = E(X^a)E(X^a \ln X) $$
it "pays more" (in terms of expected values) to place the added $X$ multiplying larger quantities $(X^a \ln X)$ than smaller quantities $(X^a)$. Simulations have confirmed the intuition, at least up to now. However, although I have tried to prove this inequality for days, using other well-known inequalities as well as relationships between expectations of products, products of expectations, and covariances, I have not been successful so far.
Something that seems related is that we know that
$$E\left(\prod_i^n f_i(X)\right)>\prod_i^nE(f_i(X)) $$
as long as the functions $f_1\ldots f_n$ are continuous monotonic functions of $X$, and are all, for instance, increasing and satisfy $f_i(X)>0$ (e.g, John Gurland's "Inequalities of Expectations of Random Variables Derived by Monotonicity or Convexity", The American Statistician, April 1968). The inequality I am trying to prove is, in a sense, "in between" the two sides in the inequality above.
Any suggestion would be very greatly appreciated.
REPLY [3 votes]: Thanks for the insightful and fun problem. Here is a proof (I think) via the Cauchy-Schwarz inequality. Consider the function
$$
f(t) \equiv \frac{ \mathbb E[X^{a+t} \ln X] } { \mathbb E[X^{a+t}] }.
$$
So the target inequality is $f(1) > f(0)$. We can show this by proving $f(t)$ is increasing, or $f'(t) \ge 0$.
But this is easy, because
$$
\begin{aligned}
f'(t)
&=
\frac{d}{dt} \left(
\frac{ \mathbb E[e^{(a+t)\ln X} \ln X] } { \mathbb E[e^{(a+t) \ln X}] }
\right)
\\
&=
\frac{ \mathbb E\left[ \frac{d}{dt} e^{(a+t)\ln X} \ln X \right] }
{ \mathbb E\left[e^{(a+t) \ln X} \right] }
-
\mathbb E[ e^{(a+t)\ln X} \ln X ]
\frac{ \mathbb E\left[ \frac{d}{dt} e^{(a+t) \ln X} \right] }
{ \mathbb E[e^{(a+t) \ln X}]^2 }
\\
%&=
%\frac{ \mathbb E\left[ e^{(a+t)\ln X} (\ln X)^2 \right] }
%{ \mathbb E\left[e^{(a+t) \ln X} \right] }
%-
%\mathbb E\left[ e^{(a+t) \ln X} \ln X \right]
%\frac{ \mathbb E\left[ e^{(a+t) \ln X} \ln X \right] }
%{ \mathbb E\left[e^{(a+t) \ln X}\right]^2 }
\\
&=\frac{
\mathbb E[X^{a+t} (\ln X)^2] \, \mathbb E[X^{a+t}]
-
\mathbb E[X^{a+t} (\ln X)]^2
} {
\mathbb E\left[X^{a+t}\right]^2
} \ge 0.
\qquad (1)
\end{aligned}
$$
The numerator of (1) is nonnegative by the
Cauchy-Schwarz inequality.
That is, with $U = X^{\frac{a+t}{2}} \ln X, V = X^{\frac{a+t}{2}}$, we have
$$
\mathbb E\left[U^2 \right] \mathbb E\left[V^2\right] \ge \mathbb E[U \, V]^2.
\qquad (2)
$$
It remains to argue that the equality cannot hold for all $t \in [0,1]$, which is easy.
Alternative to the Cauchy-Schwarz inequality (2)
Alternatively, we can show (1) directly by observing that
$$
\mathbb E\left[X^{a+t}(y - \ln X)^2 \right] \ge 0,
$$
holds for all $y$ (for the quantity of averaging is nonnegative), i.e., the quadratic polynomial
$$
\begin{aligned}
p(y)
&=
\mathbb E\left[X^{a+t}\right] y^2
- 2 \, \mathbb E\left[X^{a+t} \ln X\right] y
+ \mathbb E\left[X^{a+t} (\ln X)^2\right]
\\
&\equiv A \,y^2 - 2 \, B \, y + C,
\end{aligned}
$$
has no zero.
Thus the discriminant of $p(y)$, which is $4B^2 - 4AC$, must be non-positive. This means $AC \ge B^2$, or
$$
\mathbb E\left[X^{a+t}\right] \,
\mathbb E\left[X^{a+t} (\ln X)^2\right]
\ge
\mathbb E\left[X^{a+t} \ln X\right]^2.
$$
Further discussion
There is a more intuitive interpretation of (1). We define the characteristic function of $\ln X$ as
$$
F(t) \equiv \log \left\{ \mathbb E\left[ X^{a+t} \right] \right\}.
$$
We find $f(t) = F'(t)$, and $f'(t) = F''(t) \ge 0$. In other words, (1) is a generalized statement of that the second cumulant of $\ln X$ is non-negative at nonzero $a+t$. | 0.025269 |
\begin{document}
\title{Factorizable enriched categories and applications}
\author{Aura B\^{a}rde\c{s}}
\author{Drago\c{s} \c{S}tefan}
\address{University of Bucharest, Faculty of Mathematics and Informatics,
Bucharest, 14 Academiei Street, Ro-010014, Romania.}
\email{aura\[email protected]}
\address{University of Bucharest, Faculty of Mathematics and Informatics,
Bucharest, 14 Academiei Street, Ro-010014, Romania.}
\email{[email protected]}
\subjclass[2000]{Primary 18D20; Secondary 18D10; 16Sxx}
\date{}
\begin{abstract}
We define the twisted tensor product of two enriched categories, which
generalizes various sorts of `products' of algebraic structures, including
the bicrossed product of groups, the twisted tensor product of (co)algebras
and the double cross product of bialgebras. The key ingredient in the
definition is the notion of simple twisting systems between two enriched
categories. To give examples of simple twisted tensor products we introduce
matched pairs of enriched categories. Several other examples related to
ordinary categories, posets and groupoids are also discussed.
\end{abstract}
\keywords{Enriched category, twisting system, twisted tensor product,
matched pair, bicrossed product.}
\maketitle
\tableofcontents
\section*{Introduction}
The most convenient way to explain what we mean by the factorization problem
of an algebraic structure is to consider a concrete example. Chronologically
speaking, the first problem of this type was studied for groups, see for
instance \cite{Mai,Ore,Za,Sze,Tak}. Let $G$ be a group. Let $H$ and $K$
denote two subgroups of $G.$ One says that $G$ factorizes through $H$ and $K$
if $G=HK$ and $H\cap K=1.$ Therefore, the factorization problem for groups
means to find necessary and sufficient conditions which ensure that $G$
factorizes through the given subgroups $H$ and $K$. Note that, if $G$
factorizes through $H$ and $K$ then the multiplication induces a canonical
bijective map $\varphi :H\times K\rightarrow G,$ which can be used to
transport the group structure of $G$ on the Cartesian product of $H$ and $K.$
We shall call the resulting group structure the bicrossed product of $H$ and
$K$, and we shall denote it by $H\Join K$. The identity element of $H\Join K$
is $(1,1)$, and its group law is uniquely determined by the `twisting' map
\begin{equation*}
R:K\times H\rightarrow H\times K,\quad R(k,h):=\varphi ^{-1}(kh).
\end{equation*}
Obviously, $R$ is induced by a couple of functions $\triangleright :K\times
H\rightarrow H$ and $\triangleleft :K\times H\rightarrow K$ such that $
R(k,h)=(k\triangleright h,k\triangleleft h).$ Using this notation the
multiplication on $H\Join K$ can be written as
\begin{equation*}
(h,k)\cdot (h^{\prime },k^{\prime })=\left( h(k\triangleright h^{\prime
}),(k\triangleleft h^{\prime })k^{\prime }\right) .
\end{equation*}
The group axioms easily imply that $(H,K,\triangleright ,\triangleleft )$ is
a matched pair of groups, in the sense of \cite{Tak}. Conversely, any
bicrossed product $H\Join K$ factorizes through $H$ and $K.$ In conclusion,
a group $G$ factorizes through $H$ and $K$ if and only if it is isomorphic
to the bicrossed product $H\Join K$ associated to a certain matched pair $
(H,K,\triangleright ,\triangleleft ).$
Similar `products' are known in the literature for many other algebraic
structures. In \cite{Be}, for a distributive law $\lambda:G\circ
F\rightarrow F\circ G$ between two monads, Jon Beck defined a monad
structure on $F\circ G,$ which can be regarded as a sort of bicrossed
product of $F$ and $G$ with respect to the twisting natural transformation $
\lambda.$
The twisted tensor product of two $\mathbb{K}$-algebras $A$ and $B$ with
respect to a $\mathbb{K}$-linear twisting map $R:B\otimes _{\mathbb{K}
}A\rightarrow A\otimes _{\mathbb{K}}B$ was investigated for instance in \cite
{Ma1}, \cite{Tam}, \cite{CSV}, \cite{CIMZ}, \cite{LPvO} and \cite{JLPvO}. It
is the analogous in the category of associative and unital algebras of the
bicrossed product of groups. The classical tensor product of two algebras,
the graded tensor product of two graded algebras, skew algebras, smash
products, Ore extensions, generalized quaternion algebras, quantum affine
spaces and quantum tori are all examples of twisted tensor products.
Another class of examples, including the Drinfeld double and the double
crossed product of a matched pair of bialgebras, comes from the theory of
Hopf algebras, see \cite{Ma2}. Some of these constructions have been
generalized for bialgebras in monoidal categories \cite{BD} and bimonads
\cite{BV}.
Enriched categories have been playing an increasingly important role not
only in Algebra, but also in Algebraic Topology and Mathematical Physics,
for instance. They generalize usual categories, linear categories, Hopf
module categories and Hopf comodule categories. Monoids, algebras,
coalgebras and bialgebras may be regarded as enriched categories with one
object.
Our aim in this paper is to `categorify' the factorization problem, i.e. to
answer the question when an enriched category factorizes through a couple of
enriched subcategories. Finding a solution at this level of generality would
allow us to approach in an unifying way all factorization problems that we
have already mentioned. Moreover, it would also provide a general method for
producing new non-trivial examples of enriched categories.
In order to define factorizable enriched categories, we need some notation.
Let $\boldsymbol{C}$ be a small enriched category over a monoidal category $(
\boldsymbol{M},\otimes ,\mathbf{1}).$ Let $S$ denote the set of objects in $
\boldsymbol{C}$. For the hom-objects in $\boldsymbol{C}$ we use the notation
$_{x}C_{y}.$ The composition of morphisms and the identity morphisms in $
\boldsymbol{C}$ are defined by the maps $_{x}c_{z}^{y}:{}_{x}C_{y}\otimes
{}_{y}C_{z}\rightarrow {}_{x}C_{z}$ and $1_{x}:\mathbf{1}\rightarrow
{}_{x}C_{x},$ respectively. For details, the reader is referred to the next
section. We assume that $\boldsymbol{A}$ and $\boldsymbol{B}$ are two
enriched subcategories of $\boldsymbol{C}$. The inclusion functor $
\boldsymbol{\alpha} :\boldsymbol{A}\rightarrow \boldsymbol{C}$ is given by a
family $\{_{x}\alpha _{y}\}_{x,y\in S}$ of morphisms in $\boldsymbol{M},$
where $_{x}\alpha _{y}:{}_{x}{A}_{y}\rightarrow {}${}$_{x}C_{y}.$ If $
\boldsymbol{\beta }$ is the inclusion of $\boldsymbol{B}$ in $\boldsymbol{C}$
, then for $x,$ $y$ and $u$ in $S$ we define
\begin{equation*}
_{x}\varphi _{y}^{u}:{}_{x}A_{u}\otimes {}_{u}B{_{y}\rightarrow {}_{x}{C}
_{y},\quad }_{x}\varphi _{z}^{y}={}_{x}c_{y}^{u}\circ (_{x}{\alpha }
_{u}\otimes {}_{u}{\beta }_{y}).
\end{equation*}
{Assuming that all }$S$-indexed families of objects in $\boldsymbol{M}$ have
a coproduct it follows that the maps $\{_{x}\varphi _{y}^{u}\}_{u\in S}$
yield a unique morphism
\begin{equation*}
_{x}\varphi _{y}:{}\textstyle\bigoplus\limits_{u\in S}{}_{x}A_{u}\otimes
{}_{u}B{_{y}\rightarrow {}_{x}{C}_{y}.}
\end{equation*}
We say that $\boldsymbol{C}$ {factorizes} {through} $\boldsymbol{A} $ {and} $
\boldsymbol{B}$ if and only if all $_{x}\varphi _{y}$ are invertible. An
enriched category $\boldsymbol{C}$ is called {factorizable} if it factorizes
through $\boldsymbol{A}$ and $\boldsymbol{B},$ for some $\boldsymbol{A}$ and
$\boldsymbol{B}$.
In Theorem \ref{teo1}, our first main result, under the additional
assumption that the tensor product on $\boldsymbol{M}$ is distributive over
the direct sum, we show that to every $\boldsymbol{M}$-category $\boldsymbol{
C}$ that factorizes through $\boldsymbol{A} $ and $\boldsymbol{B}$
corresponds a twisting system between $\boldsymbol{B}$ and $\boldsymbol{A},$
that is a family ${R}:=\{_{x}R_{z}^{y}\}_{x,y,z\in S}$ of morphisms
\begin{equation*}
_{x}R_{z}^{y}:{}_{x}B_{y}\otimes {}_{y}A{_{z}\rightarrow } \textstyle
\bigoplus\limits_{u\in S}{}_{x}A_{u}\otimes {}_{u}B{_{z}}
\end{equation*}
which are compatible with the composition and identity maps in $\boldsymbol{A
} $ and $\boldsymbol{B}$ in a certain sense.
Trying to associate to a twisting system ${R}:=\{_{x}R_{z}^{y}\}_{x,y,z\in
S} $ an $\boldsymbol{M}$-category we encountered some difficulties due to
the fact that, in general, the image of $_{x}R_{z}^{y}$ is too big.
Consequently, in this paper we focus on the particular class of twisting
systems for which there is a function $\left\vert \cdots \right\vert
:S\times S\times S\rightarrow S$ such that the image of $_{x}R_{z}^{y}$ is
included into $_{x}A_{\left\vert xyz\right\vert }\otimes {}_{|xyz|}B{_{z},}$
for every $x,y,z\in S$. These twisting systems are characterized in
Proposition \ref{thm R simplu}. A more precise description of them is given
in Corollary \ref{cor: R simplu1}, provided that $\boldsymbol{M}$ satisfies
an additional condition (\dag ), see \S \ref{fa:dag}. A similar result is
obtained in Corollary \ref{cor R simplu2} for a linear monoidal category.
In this way we are led in \S \ref{STS} to the definition of simple twisting
systems. For such a twisting system ${R}$ between $\boldsymbol{B}$ and $
\boldsymbol{A},$ in Theorem \ref{thm:TTP} we construct an $\boldsymbol{M}$
-category $\boldsymbol{A}\otimes _{{R}}\boldsymbol{B}$ which factorizes
through $\boldsymbol{A}$ and $\boldsymbol{B}.$ Since it generalizes the
twisted tensor product of algebras, $\boldsymbol{A}\otimes _{{R}}\boldsymbol{
B}$ will be called the twisted tensor product of $\boldsymbol{A}$ and $
\boldsymbol{B}.$
In the third section we consider the case when $\boldsymbol{M}$ is the
monoidal category of coalgebras in a braided category $\boldsymbol{M}
^{\prime }$. In this setting, we prove that there is an one-to-one
correspondence between simple twisting systems and matched pair of enriched
categories, see \S \ref{fa:MP} for the definition of the latter notion. We
shall refer to the twisted tensor product of a matched pair as the bicrossed
product. By construction, the bicrossed product is a category enriched over $
\boldsymbol{M^{\prime }}$, but we prove that it is enriched over $
\boldsymbol{M}$ as well.
More examples of twisted tensor products of enriched categories are given in
the last part of the paper. By definition, usual categories are enriched
over $\boldsymbol{Set},$ the category of sets. Actually, they are enriched
over the monoidal category of coalgebras in $\boldsymbol{Set}$. Hence,
simple twisting systems and matched pairs are equivalent notions for usual
categories. Moreover, if $\boldsymbol{A}$ and $\boldsymbol{B}$ are thin
categories (that is their hom-sets contain at most one morphism), then we
show that any twisting system between $\boldsymbol{B}$ and $\boldsymbol{A}$
is simple, so it corresponds to a uniquely determined matched pair of
categories. We use this result to investigate the twisting systems between
two posets.
Our results may be applied to algebras in a monoidal category $\boldsymbol{M}
$, which are enriched categories with one object. Therefore, we are also
able to recover all bicrossed product constructions that we discussed at the
beginning of this introduction.
Finally, we prove that the bicrossed product of two groupoids is also a
groupoid, and we give an example of factorizable groupoid with two objects.
\section{Preliminaries and notation.}
Mainly for fixing the notation and the terminology, in this section we
recall the definition of enriched categories, and then we give some example
that are useful for our work.
\begin{fact}[Monoidal categories.]
Throughout this paper $(\boldsymbol{M},\otimes ,\mathbf{1},a,l,r)$ will
denote a monoidal category with associativity constraints $a_{X,Y,Z}:\left(
X\otimes Y\right) \otimes Z\rightarrow X\otimes \left( Y\otimes Z\right) $
and unit constraints $l_{X}:\mathbf{1}\otimes X\rightarrow X$ and $
r_{X}:X\otimes \mathbf{1}\rightarrow X.$ The class of objects of $
\boldsymbol{M}$ will be denoted by $M_{0}$. Mac Lane's Coherence Theorem
states that given two parenthesized tensor products of some objects $
X_{1},\dots ,X_{n}$ in $\boldsymbol{M}$ (with possible arbitrary insertions
of the unit object $\boldsymbol{1}$) there is a unique morphism between them
that can be written as a composition of associativity and unit constraints,
and their inverses. Consequently, all these parenthesized tensor products
can be identified coherently, and the parenthesis, associativity constraints
and unit constraints may be omitted in computations. Henceforth, we shall
always ignore them. The identity morphism of an object $X$ in $\boldsymbol{M}
$ will be denoted by the same symbol $X.$
By definition, the tensor product is a functor. In particular, for any
morphisms $f^{\prime }:X^{\prime }\rightarrow Y^{\prime \prime }$ and $
f^{\prime \prime }:X^{\prime \prime }\rightarrow Y^{\prime \prime }$ in $
\boldsymbol{M}$ the following equations hold
\begin{equation} \label{ec:TP=functor}
\left( f^{\prime }\otimes Y^{\prime \prime }\right) \circ \left( X^{\prime
}\otimes f^{\prime \prime }\right) =f^{\prime }\otimes f^{\prime \prime
}=\left( Y^{\prime }\otimes f^{\prime \prime }\right) \circ \left( f^{\prime
}\otimes X^{\prime \prime }\right) .
\end{equation}
If the coproduct of a family $\{X_{i}\}_{i\in I}$ of objects in $\boldsymbol{
M}$ exists, then it will be denoted as a pair $\left( \textstyle
\bigoplus\limits {}_{i\in I}X_{i},\left\{ \sigma _{i}\right\} _{i\in
I}\right) ,$ where the maps $\sigma _{i}:X_{i}\rightarrow \textstyle
\bigoplus\limits {}_{i\in I}X_{i}$ are the canonical inclusions.
\end{fact}
\begin{fact}[The opposite monoidal category.]
\label{CatMonOpusa} If $(\boldsymbol{M},\otimes ,\mathbf{1,}a,l,r)$ is a
monoidal category, then one constructs the monoidal category $(\boldsymbol{M}
^{o},\otimes ^{o},\mathbf{1}^{o},a^{o},l^{o},r^{o})$ as follows. By
definition, $\boldsymbol{M}^{o}$ and $\boldsymbol{M}$ share the same objects
and identity morphisms. On the other hand, for two objects $X,Y$ in $
\boldsymbol{M}$, one takes $\mathrm{Hom}_{\boldsymbol{M}^{o}}(X,Y):={}
\mathrm{Hom}_{\boldsymbol{M}}(Y,X)$. The composition of morphisms in $
\boldsymbol{M}^{o}$
\begin{equation*}
\bullet :\mathrm{Hom}_{\boldsymbol{M}^{o}}(Y,Z)\times \mathrm{Hom}_{
\boldsymbol{M}^{o}}(X,Y){}\rightarrow \mathrm{Hom}_{\boldsymbol{M}
^{o}}(X,Z){}
\end{equation*}
is defined by the formula $f\bullet g:=g\circ f,$ for any $f:Z\rightarrow Y$
and $g:Y\rightarrow X$ in $\boldsymbol{M}.$ The monoidal structure is
defined by $X\otimes ^{o}Y=X\otimes Y$ and $\boldsymbol{1}^{o}=\boldsymbol{1}
$. The associativity and unit constraints in $\boldsymbol{M}^{o}$ are given
by $a^{o}_{X,Y,Z}=a_{X,Y,Z}^{-1},$ $l^{o}=l_{X}^{-1}$ and $r^{o}=r_{X}^{-1}.$
If, in addition $\boldsymbol{M}$ is braided monoidal, with braiding $\chi
_{X,Y}:X\otimes Y\rightarrow Y\otimes X$ then $\boldsymbol{M}^{o}$ is also
braided, with respect to the braiding $\chi ^{o}$ defined by $\chi
^{o}_{X,Y}:=( \chi _{X,Y})^{-1}.$
\end{fact}
\begin{definition}
Let $S$ be a set. We say that a monoidal category $\boldsymbol{M}$ is $S$
\emph{-distributive} if every $S$-indexed family of objects in $\boldsymbol{
M\ }$has a coproduct, and the tensor product is distributive to the left and
to the right over any such coproduct. More precisely, $\boldsymbol{M}$ is $S$
-distributive if for any family $\left\{ X_{i}\right\} _{i\in S}$ the
coproduct $(\textstyle\bigoplus\nolimits_{i\in S}X_{i},\{\sigma _{i}\}_{i\in
S})$ exists and, for an arbitrary object $X$,
\begin{equation*}
(X\otimes (\textstyle\bigoplus\limits_{i\in S}X_{i}),\{X\otimes \sigma
_{i}\}_{i\in S})\text{\qquad and\qquad }((\textstyle\bigoplus\limits_{i\in
S}X_{i})\otimes X,\{\sigma _{i}\otimes X\}_{i\in S})
\end{equation*}
are the coproducts of $\{X\otimes X_{i}\}_{i\in S}$ and $\{X_{i}\otimes
X\}_{i\in S},$ respectively. Note that all monoidal categories are $S$
-distributive, provided that $S$ is a singleton (i.e. the cardinal of $S$ is
1).
\end{definition}
\begin{fact}[Enriched categories.]
An \emph{enriched category} $\boldsymbol{C}$ over $(\boldsymbol{M},\otimes ,
\mathbf{1}),$ or an $\boldsymbol{M}$-category for short, consists of:
\begin{enumerate}
\item A class of objects, that we denote by ${C}_{0}.$ If ${C}_{0}$ is a set
we say that $\boldsymbol{C}$ is \emph{small}.
\item A hom-object $_{x}{C}_{y}$ in $\boldsymbol{M}$, for each $x$ and $y$
in ${C}_{0}.$ It plays the same role as $\mathrm{Hom}_{\boldsymbol{C}}(y,x)$
, the set of morphisms from $y$ to $x$ in an ordinary category $\boldsymbol{C
}$.
\item A morphism $_{x}c_{z}^{y}:{}_{x}{C}_{y}\otimes {}_{y}{C}
_{z}\rightarrow {}_{x}{C}_{z},$ for all $x,$ $y$ and $z$ in ${C}_{0}.$
\item A morphism $1_{x}:\mathbf{1\rightarrow {}}_{x}{C}_{x},$ for all $x$ in
${C}_{0}.$
\end{enumerate}
\noindent By definition one assumes that the diagrams in Figure~\ref{ec:
M-categorie} are commutative, for all $x,y,z$ and $t$ in ${C}_{0}.$ The
commutativity of the square means that the composition of morphisms in $
\boldsymbol{C}$, defined by $\left\{ _{x}c_{z}^{y}\right\} _{z,y,z\in C_{0}}$
, is \emph{associative}. We shall say that $1_{x}$ is the \emph{identity
morphism} of $x\in C_{0}$.
\begin{figure}[h]
\begin{equation*}
\xymatrix{ {}\sv{_x{C}_y\otimes{}_y{C}_z\otimes{}_z{C}_t}
\ar[rr]^-{_x{c}^y_z\otimes{}_z{C}_t}\ar[d]_-{_x{C}_y\otimes{}_y{c}^z_t} &
&\sv{_x{C}_z\otimes{}_z{C}_t}\ar[d]^-{_x{c}^z_t}\\
\sv{_x{C}_y\otimes{}_y{C}_t}\ar[rr]_-{_x{c}^y_t} & &\sv{_x{C}_t}}\qquad
\xymatrix{ \sv{_x{C}_y\otimes{}_y{C}_y} \ar[rd]_-{_xc^y_y} &\sv{_x{C}_y}
\ar[l]_-{_xC_y\otimes{}1_y}\ar[r]^-{1_x\otimes{}_xC_y}
&\sv{_x{C}_x\otimes{}_x{C}_y}\ar[ld]^-{_xc^x_y}\\ & \sv{_x{C}_y}\ar@{=}[u] &
}
\end{equation*}
\caption{The definition of enriched categories.}
\label{ec: M-categorie}
\end{figure}
An $\boldsymbol{M}$-functor $\boldsymbol{\alpha }:\boldsymbol{C}\rightarrow
\boldsymbol{C}^{\prime }$ is a couple $(\alpha _{0},\{_{x}\alpha
_{y}\}_{x,y\in C_{0}})$, where $\alpha _{0}:C_{0}\rightarrow C_{0}^{\prime }$
is a function and $_{x}\alpha _{y}:{}_{x}C_{y}\rightarrow {}_{x^{\prime
}}C^{\prime }_{y^{\prime }}$ is a morphism in $\boldsymbol{M}$ for any $
x,y\in C_{0},$ where for simplicity we denoted $\alpha _{0}(u)$ by $
u^{\prime }$, for any $u\in C_{0}.$ By definition, $\alpha _{0}$ and $
_{x}\alpha _{y}$ must satisfy the following conditions
\begin{equation*}
_{x}\alpha _{x}\circ 1_{x}^{C}=1_{x^{\prime }}^{D}\qquad \text{and\qquad }
_{x^{\prime }}d_{z^{\prime }}^{y^{\prime }}\circ (_{x}\alpha _{y}\otimes
{}_{y}\alpha _{z})={}_{x}\alpha _{z}\circ {}_{x}c_{z}^{y}.
\end{equation*}
\end{fact}
\begin{fact}
\label{fa:not} To work easier with tensor products of hom-objects in $
\boldsymbol{M}$-categories we introduce some new notation. Let $S$ be a set
and for every $i=1,\dots ,n+1$ we pick up a family $\left\{
_{x}X_{y}^{i}\right\} _{x,y\in S}$ of objects in $\boldsymbol{M}$. If $
x_{1},\dots ,x_{n+1}\in S$ then the tensor product $_{x_{0}}X_{x_{1}}^{1}
\otimes {}_{x_{1}}X_{x_{2}}^{2}\otimes \cdots \otimes
{}_{x_{n-1}}X_{x_{n}}^{n}\otimes {}_{x_{n}}X_{x_{n+1}}^{n+1}$ will be
denoted by $_{x_{0}}X_{x_{1}}^{1}X_{x_{2}}^{2}\cdots
{}{}_{x_{n}}X_{x_{n+1}}^{n+1}.$ Assuming that $\boldsymbol{M}$ is $S$
-distributive and fixing $x_{0}$ and $x_{n+1},$ one can construct
inductively the iterated coproduct
\begin{equation}
_{x_{0}}X_{\overline{x}_{1}}^{{1}}\cdots {}_{\overline{x}_{n-1}}X_{\overline{
x}_{n}}^{n}X_{x_{n+1}}^{n+1}:=\textstyle\bigoplus_{x_{1}\in S}\cdots
\textstyle\bigoplus_{x_{n}\in S}{}_{x_{0}}X_{x_{1}}^{1}\cdots
{}_{x_{n-1}}X_{x_{n}}^{n}X_{x_{n+1}}^{n+1}. \label{cop}
\end{equation}
It is not difficult to see that this object is a coproduct of $
\{_{x_{0}}X_{x_{1}}^{1}\cdots {}_{x_{n}}X_{x_{n+1}}^{n+1}\}_{\left(
x_{1},\dots ,x_{n}\right) \in S^{n}}.$ Moreover, as a consequence of the
fact that the tensor product is distributive over the direct sum, we have
\begin{equation}
_{x_{0}}X_{\overline{x}_{1}}^{1}\cdots {}_{\overline{x}_{n-1}}X_{\overline{x}
_{n}}^{n}X_{x_{n+1}}^{n+1}\cong \textstyle\bigoplus_{x_{1}\in S}\cdots
\textstyle\bigoplus_{x_{n}\in S}{}_{x_{0}}X_{x_{\pi (1)}}^{1}X_{x_{\pi
(2)}}^{2}\cdots {}_{x_{\pi (n)}}X_{x_{n+1}}^{n+1} \label{ec:suma-tensor}
\end{equation}
for any permutation $\pi $ of the set $\{1,2,\dots ,n\}.$ The inclusion of $
_{x_{0}}X_{x_{1}}^{1}\cdots {}_{x_{n}}X_{x_{n+1}}^{n+1}$ into the coproduct
defined in \eqref{cop} is also inductively constructed as the composition of
the following two arrows
\begin{equation*}
_{x_{0}}X_{x_{1}}^{1}\otimes {}_{x_{1}}X_{x_{2}}^{1}\cdots
{}_{x_{n}}X_{x_{n+1}}^{n+1}\longrightarrow {}_{x{}_{0}}X_{x_{1}}^{1}\otimes
{}_{x_{1}}X_{\overline{x}_{2}}^{1}\cdots {}_{\overline{x}
_{n}}X_{x_{n+1}}^{n+1}\hookrightarrow \textstyle\bigoplus_{x_{1}\in
S}{}_{x{}_{0}}X_{x_{1}}^{1}\otimes {}_{x_{1}}X_{\overline{x}_{2}}^{1}\cdots
{}_{\overline{x}_{n}}X_{x_{n+1}}^{n+1}{},
\end{equation*}
where the first morphism is the tensor product between the identity of $
_{x_{0}}X_{x_{1}}^{1}$ and the inclusion of $_{x_{1}}X_{x_{2}}^{1}\cdots
{}_{x_{n}}X_{x_{n+1}}^{n+1}$ into $_{x_{1}}X_{\overline{x}_{2}}^{1}\cdots
{}_{\overline{x}_{n}}X_{x_{n+1}}^{n+1}.$ Clearly, for every $x_{n+1}\in S,$
\begin{equation*}
_{\overline{x}_{0}}X_{\overline{x}_{1}}^{1}\cdots {}_{\overline{x}_{n-1}}X_{
\overline{x}_{n}}^{n}X_{x_{n+1}}^{n+1}:=\textstyle\bigoplus_{x_{0}\in
S}{}_{x_{0}}X_{\overline{x}_{1}}^{1}\cdots {}_{\overline{x}_{n-1}}X_{
\overline{x}_{n}}^{n}X_{x_{n+1}}^{n+1}
\end{equation*}
is the coproduct of $\{_{x_{0}}X_{x_{1}}^{1}\cdots
{}_{x_{n}}X_{x_{n+1}}^{n+1}\}_{\left( x_{0},x_{1},\dots ,x_{n}\right) \in
S^{n}}.$ The objects $_{x_{0}}X_{\overline{x}_{1}}^{1}\cdots {}_{\overline{x}
_{n-1}}X_{\overline{x}_{n}}^{n}X_{\overline{x}_{n+1}}^{n+1}$ and $_{
\overline{x}_{0}}X_{\overline{x}_{1}}^{1}\cdots {}_{\overline{x}_{n-1}}X_{
\overline{x}_{n}}^{n}X_{\overline{x}_{n+1}}^{n+1}$ are analogously defined.
A similar notation will be used for morphisms. Let us suppose that $
_{x}\alpha _{y}^{i}{}{}$ is a morphism in $\boldsymbol{M}$ with source $
{}_{x}X_{y}^{i}$ and target $_{x}Y_{y}^{i},$ where $x,y\in S$ and $i\in
\{1,\dots ,n+1\}.$ We set
\begin{equation*}
_{x_{1}}\alpha _{x_{2}}^{1}\alpha _{x_{3}}^{2}\cdots {}_{x_{n}}\alpha
_{x_{n+1}}^{n+1}:={}_{x_{0}}\alpha _{x_{1}}^{1}\otimes \dots \otimes
{}_{x_{n}}\alpha _{x_{n+1}}^{n+1}.
\end{equation*}
By the universal property of coproducts, $\{_{x_{0}}\alpha
_{x_{1}}^{1}\cdots {}_{x_{n}}\alpha _{x_{n+1}}^{n+1}\}_{\left( x_{1}\cdots
x_{n}\right) \in S^{n}}$ induces a unique map $_{x_{0}}\alpha _{\overline{x}
_{1}}^{1}\cdots {}_{\overline{x}_{n-1}}\alpha _{\overline{x}
_{n}}^{n-1}\alpha _{x_{n+1}}^{n}\ $that commutes with the inclusions. In a
similar way one constructs
\begin{equation*}
_{\overline{x}_{0}}\alpha _{\overline{x}_{1}}^{1}\cdots {}_{\overline{x}
_{n-1}}\alpha _{\overline{x}_{n}}^{n}\alpha _{x_{n+1}}^{n+1},\qquad
{}_{x_{0}}\alpha _{\overline{x}_{1}}^{1}\cdots {}_{\overline{x}_{n-1}}\alpha
_{\overline{x}_{n}}^{n}\alpha _{\overline{x}_{n+1}}^{n+1}\qquad \text{and}
\qquad {}_{\overline{x}_{0}}\alpha _{\overline{x}_{1}}^{1}\cdots {}_{
\overline{x}_{n-1}}\alpha _{\overline{x}_{n}}^{n}\alpha _{\overline{x}
_{n+1}}^{n+1}.
\end{equation*}
To make the above notation clearer, let us have a look at some examples. Let
$\boldsymbol{A}$ and $\boldsymbol{B}$ be two $\boldsymbol{M}$-categories
such that $A_{0}=B_{0}=S$. Recall that the hom-objects in $\boldsymbol{A}$
and $\boldsymbol{B}$ are denoted by $_{x}A_{y}$ and $_{x}B_{y}$. Hence, $_{
\overline{x}}{A}_{y}=\textstyle\bigoplus\limits_{x\in S}{}{}{_{x}A_{y}.}$ We
also have ${_{x}A_{y}B_{z}A_{t}}={}_{x}{A}_{y}\otimes {}_{y}{B}_{z}\otimes
{}_{z}{A}_{t}$ and
\begin{equation*}
{_{x}A_{\overline{y}}B_{\overline{z}}A_{t}=}\textstyle\bigoplus\limits_{y\in
S}\textstyle\bigoplus\limits_{z\in S}{_{x}A_{y}B_{z}A_{t}\cong }\textstyle
\bigoplus\limits_{z\in S}\textstyle\bigoplus\limits_{y\in S}{
_{x}A_{y}B_{z}A_{t}\cong }\textstyle\bigoplus\limits_{y,z\in S}{
_{x}A_{y}B_{z}A_{t}.}
\end{equation*}
\noindent Since we have agreed to use the same notation for an object and
its identity map, we can write $_{x}B_{y}\alpha _{z}A_{t}\beta _{u}$ instead
of $Id_{_{x}B_{y}}\otimes {}_{y}\alpha _{z}\otimes Id_{{}_{z}A_{t}}\otimes
{}_{t}\beta _{u}$, for any morphisms $_{y}\alpha _{z}$ and $_{t}\beta _{u}$
in $\boldsymbol{M}.$ The maps $_{\overline{x}}a_{z}^{y}:{}_{\overline{x}
}A_{y}A_{z}\longrightarrow {}_{\overline{x}}A_{z}$ and $_{x}a_{\overline{z}
}^{y}:{}_{x}A_{y}A_{\overline{z}}\longrightarrow {}_{x}A_{\overline{z}}$ are
induced by the composition in $\boldsymbol{A}$, that is by the set $
\{_{x}a_{z}^{y}\}_{z\in S}$. For example, the former map is uniquely defined
such that its restriction to $_{x}A_{y}A_{z}$ and $\sigma _{x,z}\circ
\,_{x}a_{z}^{y}$ coincide for all $x\in S,$ where $\sigma _{x,z}$ is the
inclusion of $_{x}A_{z}$ into $_{\overline{x}}A_{z}.$ Similarly, $_{x}a_{z}^{
\overline{y}}:$ ${}_{x}A_{\overline{y}}A_{z}\longrightarrow {}_{x}A_{z}$ is
the unique map whose restriction to $_{x}A_{y}A_{z}$ is $_{x}a_{z}^{y},$ for
all $y\in S.$
For more details on enriched categories the reader is referred to \cite{Ke}.
We end this section giving some examples of enriched categories.
\end{fact}
\begin{fact}[The category $\boldsymbol{Set}$.]
The category of sets is monoidal with respect to the Cartesian product. The
unit object is a fixed singleton set, say $\{\emptyset \}.$ The coproduct in
$\boldsymbol{Set}$ is the disjoint union. Since the disjoint union and the
Cartesian product commute, $\boldsymbol{Set}$ is $S$-distributive for any
set $S$. Clearly, a $\boldsymbol{Set}$-category is an ordinary category. If $
\boldsymbol{C}$ is such a category, then an element $f\in {}_{x}{C}_{y}$
will be thought of as a morphism from $y$ to $x,$ and it will be denoted by $
f:y\rightarrow x,$ as usual. In this case we shall say that $y$
(respectively $x)$ is the domain or the source (respectively the codomain or
the target) of $f.$ The same notation and terminology will be used for
arbitrary $\boldsymbol{M}$-categories, whose objects are sets.
\end{fact}
\begin{fact}[The category $\mathbb{K}$-$\boldsymbol{Mod}$.]
Let $\mathbb{K}$ be a commutative ring. The category of $\mathbb{K}$-modules
is monoidal with respect to the tensor product of $\mathbb{K}$-modules. The
unit object is $\mathbb{K},$ regarded as a $\mathbb{K}$-module. This
monoidal category is $S$-distributive for any $S$. By definition, a $\mathbb{
K}$-\emph{linear category} is an enriched category over $\mathbb{K}$-$
\boldsymbol{Mod}$.
\end{fact}
\begin{fact}[The category $\Lambda $-$\boldsymbol{Mod}$-$\Lambda .$]
Let $\Lambda $ be a $\mathbb{K}$-algebra and let $\Lambda $-$\boldsymbol{Mod}
$-$\Lambda $ denote the category of left (or right) modules over $\Lambda
\otimes _{\mathbb{K}}\Lambda ^{o},$ where $\Lambda ^{o}$ is the opposite
algebra of $\Lambda .$ Thus, $M$ is an object in $\Lambda $-$\boldsymbol{Mod}
$-$\Lambda $ if, and only if, it is a left and a right $\Lambda $-module and
these structures are compatible in the sense that
\begin{equation*}
a\cdot m=m\cdot a\text{\qquad and\qquad }(x\cdot m)\cdot y=x\cdot (m\cdot y)
\end{equation*}
for all $a\in \mathbb{K},$ $x,y\in \Lambda $ and $m\in M.$ A morphism in $
\Lambda $-$\boldsymbol{Mod}$-$\Lambda $ is a map of left and right $\Lambda $
-modules. The category of $\Lambda $-bimodules is monoidal with respect to $
(-)\otimes _{\Lambda }(-).$ The unit object in $\Lambda $-$\boldsymbol{Mod}$-
$\Lambda $ is $\Lambda ,$ regarded as a $\Lambda $-bimodule. This monoidal
category also is $S$-distributive for any $S$.
\end{fact}
\begin{fact}[The category $H$-$\boldsymbol{Mod}.$]
Let $H$ be a bialgebra over a commutative ring $\mathbb{K}$. The category of
left $H$-modules is monoidal with respect to $(-)\otimes _{\mathbb{K}}(-)$.
If $M$ and $N$ are $H$-modules, then the $H$-action on $M\otimes N$ is given
by
\begin{equation*}
h\cdot m\otimes n=\textstyle\sum {}h_{\left( 1\right) }\cdot m\otimes
h_{\left( 2\right) }\cdot n.
\end{equation*}
In the above equation we used the $\Sigma $-notation $\Delta \left( h\right)
=\textstyle\sum {}$ $h_{\left( 1\right) }\otimes h_{\left( 2\right) }$. The
unit object is $\mathbb{K}$, which is an $H$-module with the trivial action,
induced by the counit of $H$. This category is $S$-distributive, for any $S$
. An enriched category over $H$-$\boldsymbol{Mod}$ is called $H$-\emph{
module category}.
\end{fact}
\begin{fact}[The category $\boldsymbol{Comod}$-$H$.]
Dually, the category of right $H$-comodules is monoidal with respect to $
(-)\otimes _{\mathbb{K}}(-)$. The coaction on and $M\otimes _{\mathbb{K}}N$
is defined by
\begin{equation*}
\rho (m\otimes n)=\textstyle\sum {}m_{\langle 0\rangle }\otimes n_{\langle
0\rangle }\otimes m_{\langle 1\rangle }n_{\langle 1\rangle },
\end{equation*}
where $\rho (m)=\textstyle\sum {}m_{\langle 0\rangle }\otimes n_{\langle
0\rangle }$, and a similar $\Sigma $-notation was used for $\rho (n).$ This
category is $S$-distributive, for any set $S$. By definition, an $H$-\emph{
comodule category} is an enriched category over $\boldsymbol{Comod}$-$H$.
\end{fact}
\begin{fact}[{The category $[\boldsymbol{A},\boldsymbol{A}].$}]
\label{Cat[A,A]}Let $\boldsymbol{A}$ be a small category, and let $[
\boldsymbol{A},\boldsymbol{A}]$ denote the category of all endofunctors of $
\boldsymbol{A}$. Therefore, the objects in $[\boldsymbol{A},\boldsymbol{A}]$
are functors $F:\boldsymbol{A}\rightarrow \boldsymbol{A},$ while the set $
_{F}[\boldsymbol{A},\boldsymbol{A}]_{G}$ contains all natural
transformations $\mu :G\rightarrow F$. The composition in this category is
the composition of natural transformations. The category $[\boldsymbol{A},
\boldsymbol{A}]$ is monoidal with respect to the composition of functors. If
$\mu :F\rightarrow G$ and $\mu ^{\prime }:F^{\prime }\rightarrow G^{\prime }$
are natural transformations, then the natural transformations $\mu F^{\prime
}$ and $G\mu ^{\prime }$ are given by
\begin{align*}
\mu F^{\prime }:F\circ F^{\prime }\rightarrow G\circ G^{\prime }& ,\qquad
\left( \mu F^{\prime }\right) _{x}:=\mu _{F^{\prime }(x)}, \\
G\mu ^{\prime }:G\circ F^{\prime }\rightarrow G\circ G^{\prime }& ,\qquad
\left( G\mu ^{\prime }\right) _{x}:=G(\mu _{x}^{\prime }).
\end{align*}
We can now define the tensor product of $\mu $ and $\mu ^{\prime }$ by
\begin{equation*}
\mu \otimes \mu ^{\prime }:=G\mu ^{\prime }\circ \mu F^{\prime }=\mu
G^{\prime }\circ F\mu ^{\prime }.
\end{equation*}
Even if $\boldsymbol{A}$ is $S$-distributive, $[\boldsymbol{A},\boldsymbol{A}
]$ may not have this property. In spite of the fact that, by assumption, any
$S$-indexed family in $[\boldsymbol{A},\boldsymbol{A}]$ has a coproduct, in
general this does not commute with the composition of functors.
Nevertheless, as we have already noticed, $[\boldsymbol{A},\boldsymbol{A}]$
is $S$-distributive if $\left\vert S\right\vert =1$.
This remark will allow us to apply our main results to an $[\boldsymbol{A},
\boldsymbol{A}]$-category $\boldsymbol{C}$ with one object $x$. Hence $
F:={}_{x}{C}_{x}$ is an endofunctor of $\boldsymbol{A}$, and the composition
and the identity morphisms in $\boldsymbol{C}$ are uniquely defined by
natural transformations
\begin{equation*}
\mu :F\circ F\rightarrow F\qquad \text{and}\qquad \iota :\text{Id}_{
\boldsymbol{A}}\rightarrow F.
\end{equation*}
The commutativity of the diagrams in Figure~\ref{ec: M-categorie} is
equivalent in this case with the fact that $\left( F,\mu ,\iota \right) $ is
a \emph{monad}, see \cite{Be} for the definition of monads. In conclusion,
monads are in one-to-one correspondence to $[\boldsymbol{A},\boldsymbol{A}]$
-categories with one object.
\end{fact}
\begin{fact}[The category $\boldsymbol{Opmon(M).}$]
Let $\left( \boldsymbol{M},\otimes ,1\right) $ be a monoidal category. An
opmonoidal functor is a triple $\left( F,\delta ,\varepsilon \right) $ that
consists of
\begin{enumerate}
\item A functor $F:\boldsymbol{M}\rightarrow \boldsymbol{M}$.
\item A natural transformation $\delta :=\left\{ \delta _{x,y}\right\}
_{(x,y)\in {M}_{0}\times {M}_{0}}$, with $\delta _{x,y}:F\left( x\otimes
y\right) \rightarrow F\left( x)\otimes F(y\right) .$
\item A map $\varepsilon :F(\mathbf{1})\rightarrow \mathbf{1}$ in $
\boldsymbol{M}.$
\end{enumerate}
\noindent In addition, the transformations $\delta $ and $\varepsilon $ are
assumed to render commutative the diagrams in Figure~\ref{fig:opmon}. An
opmonoidal transformation $\alpha :(F,\delta ,\varepsilon )\rightarrow
(F^{\prime },\delta ^{\prime },\varepsilon ^{\prime })$ is a natural map $
\alpha :F\rightarrow F^{\prime }$ such that, for arbitrary objects $x$ and $
y $ in $\boldsymbol{M},$
\begin{equation*}
(\alpha _{x}\otimes \alpha _{y})\circ \delta _{x,y}=\delta _{x,y}^{\prime
}\circ \alpha _{x\otimes y}\qquad \text{and\qquad }\varepsilon ^{\prime
}\circ \alpha _{\mathbf{1}}=\varepsilon .
\end{equation*}
Obviously the composition of two opmonoidal transformations is opmonoidal,
and the identity of an opmonoidal functor is an opmonoidal transformation.
The resulting category will be denoted by $\boldsymbol{Opmon(M)}.$ For two
opmonoidal functors $(F,\delta ,\varepsilon )$ and $(F^{\prime },\delta
^{\prime },\varepsilon ^{\prime })$ one defines
\begin{equation*}
(F,\delta ,\varepsilon )\otimes (F^{\prime },\delta ^{\prime },\varepsilon
^{\prime }):=\left( F\circ F^{\prime },\delta _{F^{\prime },F^{\prime
}}\circ F(\delta ^{\prime }),\varepsilon \circ F(\varepsilon ^{\prime
})\right) ,
\end{equation*}
where $\delta _{F^{\prime },F^{\prime }}=\left\{ \delta _{F^{\prime
}(x),F^{\prime }(y)}\right\} _{x,y\in M_{0}}$. On the other hand, if $\mu
:F\rightarrow G$ and $\mu ^{\prime \prime }\rightarrow G^{\prime }$ are
opmonoidal transformations, then $\mu \otimes \mu ^{\prime }:=\mu G^{\prime
}\circ F\mu ^{\prime }$ is opmonoidal too. One can see easily that $\otimes $
defines a monoidal structure on $\boldsymbol{Opmon(M)}$ with unit object $(
\text{Id}_{\boldsymbol{M}},\left\{ \text{Id}_{x\otimes y}\right\} _{x,y\in
M_{0}},\text{Id}_{\mathbf{1}}).$
\begin{figure}[tbh]
\begin{equation*}
\xymatrix{ \sv{F(x\otimes y\otimes z) \ar[rr]^-{\delta_{x\otimes
y,z}}}\ar[d]|-{\sa{\delta_{x, y\otimes z}}} & & \sv{F(x\otimes y)\otimes
F(z)} \ar[d]|-{\sa{\delta_{x, y}\otimes F(z)}}\\ \sv{F(x)\otimes F(y\otimes
z)}\ar[rr]_-{\sa{F(x)\otimes \delta_{y,z}}} & & \sv{F(x)\otimes F(y)\otimes
F(z) }}\qquad \xymatrix{ \sv{F(x)\otimes F(\mathbf{1})}
\ar[r]^-{\sa{F(x)\otimes\varepsilon_{\mathbf{1}}}} & \sv{F(x)} &
\sv{F(\mathbf{1})\otimes F(x)}\ar[l]_-{\sa{\varepsilon_{\mathbf{1}}\otimes
F(x)}}\\ & \sv{F(x)}\ar[ul]^-{\delta_{x,\mathbf{1}}}
\ar[ur]_-{\sa{\delta_{\mathbf{1},x}}}\ar@{=}[u] & }
\end{equation*}
\caption{The definition of opmonoidal functors.}
\label{fig:opmon}
\end{figure}
\end{fact}
\begin{fact}[The categories $\boldsymbol{Alg(M)}$ and $\boldsymbol{Coalg(M)}
. $]
Let $\left( \boldsymbol{M},\otimes ,\mathbf{1},\chi\right) $ be a braided
monoidal category with braiding $\chi:=\left\{ \chi _{x,y}\right\} _{\left(
x,y\right) \in {M}_{0}\times {M}_{0}}$, where $\chi_{x,y}:x\otimes
y\rightarrow y\otimes x.$ The category $\boldsymbol{Alg(M)}$ of all algebras
in $\boldsymbol{M}$ is monoidal too. Recall that an algebra in $\boldsymbol{M
}$ is an $\boldsymbol{M}$-category with one object. As in \S \ref{Cat[A,A]},
such a category is uniquely determined by an object $X$ in $\boldsymbol{M}$
and two morphisms $m:X\otimes X\rightarrow X$ (the multiplication) and $u:
\mathbf{1}\rightarrow X$ (the unit). The commutativity of the diagrams in
Figure~\ref{ec: M-categorie} means that the algebra is associative and
unital. If $\left( X,m,u\right) $ and $\left( X^{\prime },m^{\prime
},u^{\prime }\right) $ are algebras in $\boldsymbol{M},$ then $X\otimes
X^{\prime }$ is an algebra in $\boldsymbol{M}$ with multiplication
\begin{equation*}
\left( m\otimes m^{\prime }\right) \circ \left( X\otimes \chi_{X^{\prime
},X}\otimes X^{\prime }\right) :\left( X\otimes X^{\prime }\right) \otimes
\left( X\otimes X^{\prime }\right) \rightarrow X\otimes X^{\prime }
\end{equation*}
and unit $u\otimes u^{\prime }:\mathbf{1}\rightarrow X\otimes X^{\prime }$.
The monoidal category $\boldsymbol{Coalg(M)}$ of coalgebras in $\boldsymbol{M
}$ can be defined in a similar way. Alternatively, one may take $\boldsymbol{
Coalg(M)}:=\boldsymbol{Alg(M^{o})^{o}}.$ Note that the monoidal category of
coalgebras in $\boldsymbol{M}$ and the monoidal category of algebras in $
\boldsymbol{M}^{o}$ are opposite each other.
It is not hard to see that $\boldsymbol{Coalg(M)}$ is $S$-distributive,
provided that $\boldsymbol{M}$ is so.
\end{fact}
\section{Factorizable $\boldsymbol{M}$-categories and
twisting systems.}
In this section we define factorizable $\boldsymbol{M}$-categories and
twisting systems. We shall prove that to every factorizable system
corresponds a certain twisting system. Under a mild extra assumption on the
monoidal category $\boldsymbol{M}$, we shall also produce enriched
categories using a special class of twisting systems that we call simple.
Throughout this section $S$ denotes a fixed set. We assume that all $
\boldsymbol{M}$-categories that we work with are small, and that their set
of objects is $S$.
\begin{fact}[Factorizable $\boldsymbol{M}$-categories.]
Let $\boldsymbol{C}$ be a small enriched category over $(\boldsymbol{M}
,\otimes ,\mathbf{1})$. We assume that $\boldsymbol{M}$ is $S$-distributive.
Suppose that $\boldsymbol{A}$ and $\boldsymbol{B}$ are $\boldsymbol{M}$
-subcategories of $\boldsymbol{C}.$ Note that, by assumption, $
A_{0}=B_{0}=C_{0}=S.$ For $x,$ $y$ and $u$ in $S$ we define
\begin{equation}
_{x}\varphi _{y}^{u}:{}_{x}{A}{}_{u}{B}_{y}\rightarrow {}_{x}{C}_{y},\qquad
{}_{x}\varphi _{y}^{u}:={}_{x}c_{y}^{u}\circ {}_{x}\alpha _{u}\beta _{y},
\label{ec:phi^u}
\end{equation}
where $\boldsymbol{\alpha }:\boldsymbol{A}\rightarrow \boldsymbol{C}$ and $
\boldsymbol{\beta }:\boldsymbol{B}\rightarrow \boldsymbol{C}$ denote the
corresponding inclusion $\boldsymbol{M}$-functors. By the universal property
of coproducts, for every $x$ and $y$ in $S,$ there is $_{x}\varphi
_{y}:{}_{x}A_{\overline{u}}B{_{y}\rightarrow {}_{x}{C}_{y}}$ such that
\begin{equation}
_{x}\varphi _{y}\circ {}_{x}\sigma _{y}^{u}={}_{x}\varphi _{y}^{u},
\label{ec:phi}
\end{equation}
where $_{x}\sigma _{y}^{u}$ is the canonical inclusion of $_{x}A_{u}B_{y}$
into ${}_{x}A_{\overline{u}}B_{y}.$ Note that by the universal property of
coproducts $_{x}\varphi _{y}={}_{x}c_{y}^{\overline{u}}\circ {}_{x}\alpha _{
\overline{u}}\beta _{y},$ as we have $_{x}\alpha _{\overline{u}}\beta
_{y}\circ {}_{x}\sigma _{y}^{u}={}_{x}\tau _{y}^{u}\circ {}_{x}\alpha
_{u}\beta _{y}$ and $_{x}c_{y}^{\overline{u}}\circ {}_{x}\tau
_{y}^{u}={}_{x}c_{y}^{u},$ where $_{x}\tau _{y}^{u}$ denotes the inclusion
of $_{x}C_{u}C_{y}$ into ${}_{x}C_{\overline{u}}C_{y}.$
We shall say that $\boldsymbol{C}$ \emph{factorizes} through $\boldsymbol{A}$
and $\boldsymbol{B}$ if $_{x}\varphi _{y}$ is an isomorphism, for all $x$
and $y$ in $S$. By definition, an $\boldsymbol{M}$-category $\boldsymbol{C}$
is \emph{factorizable }if it factorizes through $\boldsymbol{A} $ and $
\boldsymbol{B},$ where $\boldsymbol{A}$ and $\boldsymbol{B}$ are certain $
\boldsymbol{M}$-subcategories of $\boldsymbol{C}.$
\end{fact}
\begin{fact}[The twisting system associated to a factorizable $\boldsymbol{M}
$-category.]
Let $\boldsymbol{C}$ be an enriched category over a monoidal category $(
\boldsymbol{M},\otimes ,\boldsymbol{1})$. We assume that $\boldsymbol{M}$ is
$S$-distributive. The family ${R}:=\{_{x}R_{z}^{y}\}_{x,y,z\in S}$ of
morphisms $_{x}R_{z}^{y}:{}_{x}{B}_{y}A_{z}\rightarrow {}_{x}A_{\overline{u}
}B_{y}$ is called a \emph{twisting system} if the four diagrams in Figure
\ref{fig:R_1} are commutative for all $x,$ $y,$ $z$ and $t$ in $S$.
\begin{figure}[tbh]
\centering
$
\xymatrix{
\sv{_xB_yB_zA_t}
\ar[d]|{\sa{RI\circ IR}}
\ar[r]^-{\sa{bI}}
& \sv{_xB_zA_t}
\ar[d]|{R}\\
\sv{_xA_{\overline v}B_{\overline u}B_t}
\ar@/_0px/[r]_-{\sa{Ib}}
& \sv{_xA_{\overline v}B_t} }\quad
\xymatrix{
\sv{_xB_yA_zA_t}
\ar[d]|{\sa{IR\circ RI}}
\ar[r]^-{\sa{Ia}}
& \sv{_xB_yA_t}
\ar[d]|{R}\\
\sv{_xA_{\overline v}A_{\overline u}B_t}
\ar@/_0px/[r]_-{\sa{aI}}
& \sv{_xA_{\overline u}B_t} } \quad
\xymatrix{
\sv{{}_{x}A_{y}}
\ar[r]^-{\sa{1^BI}}
\ar[d]|{\sa{I 1^B}}
&\sv{_xB_{x}A_{y}}
\ar[d]|{\sa{R}}\\
\sv{_xA_{y}B_{y}}
\ar[r]_-{\sa{\sigma}}
& \sv{_xA_{\overline u}B_y}}\quad
\xymatrix{
\sv{{}_{x}B_{y}}
\ar[r]^-{\sa{I 1^A}}
\ar[d]|{\sa{1^AI}}
&\sv{_xB_yA_y}\ar[d]|{\sa{R}}\\
\sv{_xA_xB_y}\ar[r]_-{\sa{\sigma}}
&\sv{_xA_{\overline u}B_y}} $
\caption{The definition of twisting systems.}
\label{fig:R_1}
\end{figure}
\newline
Let us briefly explain the notation that we used in these diagrams. As a
general rule, we omit all subscripts and superscripts denoting elements in $
S $, and which are attached to a morphism. The symbol $\otimes $ is also
omitted. For example, $a$ and $1^{A}$ (respectively $b$ and $1^{B}$) stand
for the suitable composition maps and identity morphisms in $\boldsymbol{A}$
(respectively $\boldsymbol{B}$). The identity morphism of an object in $
\boldsymbol{M}$ is denoted by $I$. Thus, by $Ia:{}_{x}B_{y}A_{z}A_{t}
\rightarrow {}_{x}B_{y}A_{t}$ we mean $_{x}B_{y}\otimes {}_{y}a_{t}^{z}$. On
the other hand, $aI:{}_{x}A_{\overline{v}}A_{\overline{u}}B_{t}\rightarrow
{}_{x}A_{\overline{u}}B_{t}$ is a shorthand notation for $_{x}a_{\overline{u}
}^{\overline{v}}B_{t}$, which in turn is the unique map induced by $\left\{
_{x}\sigma _{t}^{u}\circ {}_{x}a_{u}^{v}B_{t}\right\} _{u,v\in S}.$ We shall
keep the foregoing notation in all diagrams that we shall work with.
We claim that to every factorizable $\boldsymbol{M}$-category $\boldsymbol{C}
$ corresponds a certain twisting system. By definition, the map $_{x}\varphi
_{y}$ constructed in (\ref{ec:phi}) is invertible for all $x$ and $y$ in $S.$
Let $_{x}\psi _{y}$ denote the inverse of $_{x}\varphi _{y}$. For $x$, $y$
and $z$ in $S$, we can now define
\begin{equation}
_{x}R_{z}^{y}:{}_{x}B_{y}A_{z}\rightarrow {}_{x}A_{\overline{u}}B_{z},\quad
_{x}R_{z}^{y}:={}_{x}\psi _{z}\circ {}_{x}c_{z}^{y}\circ {}_{x}\beta
_{y}\alpha _{z}. \label{ec:R}
\end{equation}
\end{fact}
\begin{theorem}
\label{teo1} If $\boldsymbol{C}$ is a factorizable enriched category over an
$S$-distributive monoidal category $\boldsymbol{M}$, then the maps in (\ref
{ec:R}) define a twisting system.
\end{theorem}
\begin{proof}
Let us first prove that the first diagram in Figure \ref{fig:R_1} is
commutative. We fix $x,$ $y,$ $z$ and $t$ in $S,$ and we consider the
following diagram.
\begin{equation*}
\xymatrix{\ar@{}[drr] |*+[o][F-]{A} \sv{_xB_yA_{\overline u}B_t}
\ar[rr]^{\sa{\beta{}\alpha{}\beta{}}} \ar[d]|{\sa{\beta{}\alpha{}I}} &
\ar@{}[d]{} & \sv{_xC_yC_{\overline u}C_t} \ar@{}[ddddr]|-*+[o][F-]{F}
\ar[r]^{\sa{Ic}} \ar@{=}[d]{} & \sv{_xC_yC_t} \ar[dddd]|{c}\\
\sv{_xC_yC_{\overline u}B_t} \ar[rr]|{\sa{II\beta{}}} \ar@{}[drr]
|*+[o][F-]{B} \ar[d]|{\sa{cI}} & \ar@{}[d]{} & \sv{_xC_yC_{\overline u}C_t}
\ar[d]|{\sa{cI}} & \\ \sv{_xC_{\overline u}B_t} \ar@{}[drr]|*+[o][F-]{C}
\ar[rr]|{\sa{I\beta{}}} \ar[d]|<(.25){\sa{\psi I}} & \ar@{}[d]{} &
\sv{_xC_{\overline u}C_t} \ar@{=}[d]{} & \\ \sv{_xA_{\overline
v}B_{\overline u}B_t} \ar@{}[dr] |*+[o][F-]{D}
\ar@/^0px/[r]^-{\sa{\alpha{}\beta{}\beta{}}} \ar[d]|{\sa{Ib}} &
\sv{_xC_{\overline v}C_{\overline u}C_t} \ar@{}[dr] |*+[o][F-]{E}
\ar@/^0px/[r]|-{\sa{cI}} \ar[d]|{\sa{Ic}} & \sv{_xC_{\overline u}C_t}
\ar[d]|{\sa{c}} & \\ \sv{_xA_{\overline v}B_t} \ar[r]_{\sa{\alpha{}\beta{}}}
& \sv{_xC_{\overline v}C_t} \ar[r]_{\sa{c}} & \sv{_xC_t} \ar@{=}[r]{} &
\sv{_xC_t} }
\end{equation*}
Since the tensor product in a monoidal category is a functor, that is in
view of \eqref{ec:TP=functor}, we have
\begin{equation}
_{x}C_{y}C_{u}\beta _{t}\circ {}_{x}\beta _{y}\alpha _{u}B_{t}={}_{x}\beta
_{y}\alpha _{u}\beta _{t}, \label{eq:ab}
\end{equation}
for any $u$ in $S.$ Hence by the universal property of the coproduct and the
construction of the maps $_{x}C_{y}C_{\overline{u}}\beta _{t},{}$ $_{x}\beta
_{y}\alpha _{\overline{u}}B_{t}$ and ${}_{x}\beta _{y}\alpha _{\overline{u}
}\beta _{t}$ we deduce that the relation which is obtained by replacing $u$
with $\overline{u}$ in (\ref{eq:ab}) holds true. This means that the square
(A) is commutative. Proceeding similarly one shows that (B) is commutative
as well. Furthermore, $_{x}c_{\overline{u}}^{\overline{v}}C_{t},$ $
_{x}\alpha _{\overline{v}}\beta _{\overline{u}}\beta _{t}$ and $_{x}\psi _{
\overline{u}}B_{t}$ are induced by $\left\{ _{x}c_{u}^{\overline{v}}\otimes
{}_{u}C_{t}\right\} _{u\in S},$ $\left\{ _{x}\alpha _{\overline{v}}\beta
_{u}\otimes {}_{u}\beta _{t}\right\} _{u\in S}$ and $\left\{ _{x}\psi
_{u}\otimes {}_{u}B_{t}\right\} _{u\in S},$ respectively. Hence their
composite $\lambda :={}_{x}c_{\overline{u}}^{\overline{v}}C_{t}\circ
{}_{x}\alpha _{\overline{v}}\beta _{\overline{u}}\beta _{t}\circ {}_{x}\psi
_{\overline{u}}B_{t}$ is induced by $\left\{ \lambda _{u}\right\} _{u\in S},$
where
\begin{equation*}
\lambda _{u}=\left( _{x}c_{u}^{\overline{v}}\otimes {}_{u}C_{t}\right) \circ
\left( _{x}\alpha _{\overline{v}}\beta _{u}\otimes {}_{u}\beta _{t}\right)
\circ \left( _{x}\psi _{u}\otimes {}_{u}B_{t}\right) =\left( _{x}c_{u}^{
\overline{v}}\circ {}_{x}\alpha _{\overline{v}}\beta _{u}\circ {}_{x}\psi
_{u}\right) \otimes {}_{u}\beta _{t}=\left( _{x}\varphi _{u}\circ {}_{x}\psi
_{u}\right) \otimes {}_{u}\beta _{t}.
\end{equation*}
Since $_{x}\psi _{u}$ is the inverse of $_{x}\varphi _{u}$ it follows that $
\lambda _{u}={}_{x}C_{u}\beta _{t},$ for every $u\in S.$ In conclusion
\begin{equation*}
_{x}c_{\overline{u}}^{\overline{v}}C_{t}\circ {}_{x}\alpha _{\overline{v}
}\beta _{\overline{u}}\beta _{t}\circ {}_{x}\psi _{\overline{u}
}B_{t}={}_{x}C_{\overline{u}}\beta _{t},
\end{equation*}
so (C) is a commutative square. Since $\boldsymbol{\beta }$ is an $
\boldsymbol{M}$-functor it follows that $\left\{ _{x}C_{v}c_{t}^{u}\circ
{}_{x}\alpha {}_{v}\beta _{u}\beta _{t}{}\right\} _{u,v\in S}$ and $\left\{
_{x}\alpha {}_{v}\beta _{t}\circ {}{}_{x}A_{v}b_{t}^{u}\right\} _{u,v\in S}$
are equal. Therefore these families induce the same morphism, that is
\begin{equation*}
_{x}C_{\overline{v}}c_{t}^{\overline{u}}\circ {}_{x}\alpha {}_{\overline{v}
}\beta _{\overline{u}}\beta _{t}{}={}_{x}\alpha {}_{\overline{v}}\beta
_{t}\circ {}{}_{x}A_{\overline{v}}b_{t}^{\overline{u}}.
\end{equation*}
Hence (D) is commutative too. Since the composition of morphisms in $
\boldsymbol{C}$ is associative, we have
\begin{equation*}
_{x}c_{t}^{\overline{v}}\circ {}_{x}C_{\overline{v}}c_{t}^{\overline{u}
}={}_{x}c_{t}^{\overline{u}}\circ {}_{x}c_{\overline{u}}^{\overline{v}
}C_{t}\qquad \text{and\qquad }_{x}c_{t}^{y}\circ {}_{x}C_{y}c_{t}^{\overline{
u}}={}_{x}c_{t}^{\overline{u}}\circ {}_{x}c_{\overline{u}}^{y}C_{t}.
\end{equation*}
These equations imply that (E) and (F)\ are commutative. Summarizing, we
have just proved that all diagrams (A)-(F) are commutative. By diagram
chasing it results that the outer square is commutative as well, that is
\begin{equation*}
{}_{x}\varphi _{t}\circ {}_{x}A_{\overline{v}}{b}_{t}^{\overline{u}}\circ
{}_{x}R_{\overline{u}}^{y}B_{t}={}_{x}c_{t}^{y}\circ {}_{x}\beta _{y}\varphi
_{t}.
\end{equation*}
Left composing and right composing both sides of this equation by $_{x}\psi
_{t}$ and ${}_{x}B_{y}R_{t}^{z},$ respectively, yield
\begin{align*}
{}_{x}A_{\overline{v}}b_{t}^{\overline{u}}\circ {}_{x}R_{\overline{u}
}^{y}b_{t}\circ {}_{x}B_{y}R_{t}^{z}& ={}_{x}\psi _{t}\circ
{}_{x}c_{t}^{y}\circ {}_{x}\beta _{y}\varphi _{t}\circ {}_{x}B_{y}R_{t}^{z}
\\
& ={}_{x}\psi _{t}\circ {}_{x}c_{t}^{y}\circ {}_{x}\beta _{y}\varphi
_{t}\circ {}_{x}B_{y}\psi _{t}\circ {}_{x}B_{y}c_{t}^{z}\circ
{}_{x}B_{y}\beta _{z}\alpha _{t} \\
& ={}_{x}\psi _{t}\circ {}_{x}c_{t}^{y}\circ {}_{x}C_{y}{}c_{t}^{z}\circ
{}_{x}\beta _{y}\beta _{z}\alpha _{t},
\end{align*}
where for the second and third relations we used the definition of $
_{y}R_{t}^{z}$ and that $_{y}\varphi _{t}$ and $_{y}\psi _{t}$ are inverses
each other. On the other hand, the definition of ${}_{x}R_{t}^{z},$ the fact
that $\boldsymbol{\beta }$ is a functor and associativity of the composition
in $\boldsymbol{C\ }$imply the following sequence of identities
\begin{align*}
{}_{x}R_{t}^{z}\circ {}{}_{x}b_{z}^{y}A_{t}& ={}{}_{x}\psi _{t}\circ
{}_{x}c_{t}^{z}\circ {}_{x}\beta _{z}\alpha _{t}\circ {}{}_{x}b_{z}^{y}A_{t}
\\
& ={}_{x}\psi _{t}\circ {}_{x}c_{t}^{z}\circ {}_{x}c_{z}^{y}C_{t}\circ
{}{}_{x}\beta _{y}\beta _{z}\alpha _{t} \\
& ={}{}_{x}\psi _{t}\circ {}_{x}c_{t}^{y}\circ {}_{x}C_{y}{}c_{t}^{z}\circ
{}_{x}\beta _{y}\beta _{z}a_{t}.
\end{align*}
In conclusion, the first diagram in Figure \ref{fig:R_1} is commutative.
Taking into account the definition of $_{x}R_{y}^{x},$ the identity $
_{x}\beta _{x}\circ 1_{x}^{B}=1_{x}$ and the compatibility relation between
the composition and the identity morphisms in an enriched category, we get
the following sequence of equations
\begin{equation*}
_{x}\varphi _{y}{}\circ {}_{x}R_{y}^{x}\circ 1_{x}^{B}A_{y}={}_{x}\varphi
_{y}{}\circ {}_{x}\psi _{y}{}\circ {}_{x}c_{y}^{x}{}\circ {}_{x}\beta
_{x}\alpha _{y}\circ 1_{x}^{B}A_{y}={}_{x}c_{y}^{x}{}\circ {}1_{x}\alpha
_{y}={}_{x}\alpha _{y}.
\end{equation*}
Analogously, using the definition of $_{x}\varphi _{y}{}$ and the properties
of identity morphisms, we get
\begin{equation*}
_{x}\varphi _{y}{}\circ {}_{x}\sigma _{y}^{y}\circ
{}_{x}A_{y}1_{y}^{B}={}_{x}\varphi _{y}^{y}{}\circ
{}{}_{x}A_{y}1_{y}^{B}={}_{x}c_{y}^{y}{}\circ {}{}_{x}\alpha _{y}\beta
_{y}\circ {}_{x}A_{y}1_{y}^{B}={}_{x}c_{y}^{y}{}\circ {}{}_{x}\alpha
_{y}1_{y}={}_{x}\alpha _{y}.
\end{equation*}
Since $_{x}\varphi _{y}{}$ is an isomorphisms, in view of the above
computations, it follows that the third diagram is commutative as well. One
can prove in a similar way that the remaining two diagrams in Figure \ref
{fig:R_1} are commutative.
\end{proof}
\begin{fact}
\label{fa:R^tilda}We have noticed in the introduction that to every twisting
system of groups (or, equivalently, every matched pair of groups) one
associates a factorizable group. Trying to prove a similar result for a
twisting system ${R}$ between the $\boldsymbol{M}$-categories $\boldsymbol{B}
$ and $\boldsymbol{A}$ we have encountered some difficulties due to the fact
that, in general, the image of the map
\begin{equation*}
_{x}R_{z}^{y}:{}_{x}B_{y}A_{z}\rightarrow \textstyle\bigoplus\limits_{u\in
S}\,_{x}A_{u}B_{z}
\end{equation*}
is not included into a summand $_{x}A_{u}B_{z},$ for some $u\in S$ that
depends on $x,$ $y$ and $z.$ For this reason, in this paper we shall
investigate only those twisting systems for which there are a function $
|\cdots |:S^{3}\rightarrow S$ and the maps $_{x}\widetilde{R}
_{z}^{y}:{}_{x}B_{y}A_{z}\rightarrow {}_{x}A_{|xyz|}B_{z}$ such that
\begin{equation}
_{x}R_{z}^{y}={}_{x}\sigma _{z}^{|xyz|}\circ {}_{x}\widetilde{R}_{z}^{y},
\label{R simplu}
\end{equation}
for all $x,y,z,\in S.$ For them we shall use the notation $(\widetilde{{R}}
,|\cdots |).$
\end{fact}
\begin{proposition}
\label{thm R simplu} Let $\boldsymbol{M}$ be a monoidal category which is $S$
-distributive. Let $|\cdots |:S^{3}\rightarrow S$ and $\{_{x}\widetilde{R}
_{z}^{y}\}_{x,y,z\in S}$ be a function and a set of maps as above. The
family $\{_{x}R_{z}^{y}\}_{x,y,z\in S}$ defined by (\ref{R simplu}) is a
twisting system if and only if, for any $x,y,z,t\in S,$ the following
relations hold:
\begin{gather}
_{x}\sigma _{t}^{|xy|yzt||}\circ {}_{x}A_{|xy|yzt||}b_{t}^{|yzt|}\circ {}_{x}
\widetilde{R}_{|yzt|}^{y}B_{t}{}\circ {}_{x}B_{y}\widetilde{R}
_{t}^{z}={}_{x}\sigma _{t}^{|xzt|}\circ {}{}_{x}\widetilde{R}_{t}^{z}\circ
{}_{x}b_{z}^{y}A_{t}, \label{Rs-1} \\
_{x}\sigma _{t}^{|xyz|zt||}\circ {}_{x}a_{||xyz|zt|}^{|xyz|}B_{t}\circ
{}_{x}A_{|xyz|}\widetilde{R}_{t}^{z}\circ {}_{x}\widetilde{R}
_{z}^{y}A_{t}={}_{x}\sigma _{t}^{|xyt|}\circ {}_{x}\widetilde{R}_{t}^{y}
\circ{}_{x}B_{y}a_{t}^{z}, \label{Rs-2} \\
{}_{x}\sigma _{y}^{|xxy|}\circ {\ }_{x}\widetilde{R}_{y}^{x}\circ
(1_{x}^{B}\otimes {}_{x}A_{y})={}_{x}\sigma _{y}^{y}\circ(_{x}A_{y}\otimes
1_{y}^{B}), \label{Rs-3} \\
_{x}\sigma _{y}^{|xyy|}\circ {}_{x}\widetilde{R}_{y}^{y}\circ
(_{x}B_{y}\otimes 1_{y}^{A})={}_{x}\sigma _{y}^{x}\circ (1_{x}^{A}\otimes
{}_{x}B_{y}). \label{Rs-4}
\end{gather}
\end{proposition}
\begin{proof}
We claim that $\{_{x}\widetilde{R}_{z}^{y}\}_{x,y,z\in S}$ satisfy (\ref
{Rs-1}) if and only if $\{_{x}{R}_{z}^{y}\}_{x,y,z\in S}$ render commutative
the first diagram in \ Figure \ref{fig:R_1}. Indeed, let us consider the following diagram.
\begin{equation*}
\xymatrix{ \sv{_xB_yB_zA_t}\ar[r]^-{I\widetilde{R}}\ar[d]_-{bI}
&\sv{_xB_yA_{|yzt|}B_t}\ar[r]^-{I\sigma}\ar[d]_-{\widetilde{R}I}
&\sv{_xB_yA_{\overline{u}}B_t}\ar[d]^-{{R}I}\ar@{}[dl]^>(.30)*+[o][F-]{B} \\
\sv{_xB_zA_t}\ar[dd]_-{\widetilde{R}}
&\sv{_xA_{|xy|yzt||}B_{|yzt|}B_t}\ar[r]^-{\sigma I}\ar[d]_-{Ib}
&\sv{_xA_{\overline{v}}B_{\overline{u}}B_t}
\ar[d]^-{Ib}\ar@{}[dl]^>(.30)*+[o][F-]{C} \\ \ar@{}[dr]^>(.40)*+[o][F-]{A}
&\sv{_xA_{|xy|yzt||}B_t}\ar[r]^-{\sigma}
&\sv{_xA_{\overline{v}}B_t}\ar@{=}[d] \\
\sv{_xA_{|xzt|}B_t}\ar[rr]_-{\sigma} & &\sv{_xA_{\overline{v}}B_t} \\ }
\end{equation*}
The squares (B) and (C) are commutative by the definition of $_{x}R_{
\overline{u}}^{y}:{}_{x}B_{y}A_{\overline{u}}\rightarrow {}_{x}A_{\overline{v
}}B_{\overline{u}}$ and $_{\overline{v}}b_{t}^{\overline{u}}:{}_{\overline{v}
}B_{\overline{u}}B_{t}\rightarrow {}_{\overline{v}}B_{t}$. Hence the hexagon
(A) is commutative if and only if the outer square is commutative. This
proves our claim as (A) and the outer square in Figure \ref{fig:R_1} are
commutative if and only if \eqref{Rs-1} holds and the first diagram in
Figure \ref{fig:R_1} is commutative, respectively. Similarly one shows that
the commutativity of the second diagram from Figure \ref{fig:R_1} is
equivalent to \eqref{Rs-2}. On the other hand, obviously, the third and
fourth diagrams in Figure \ref{fig:R_1} are commutative if and only if
\eqref{Rs-3} and \eqref{Rs-4} hold, so the proposition is proved.
\end{proof}
The inclusion maps make difficult to handle the equations (\ref{Rs-1})-(\ref
{Rs-4}). In some cases we can remove these morphisms by imposing more
conditions on the map $|\cdots |$ or on the monoidal category $\boldsymbol{M}
$.
\begin{fact}[The assumption (\dag ).]
\label{fa:dag} Let $\boldsymbol{M}$ be a monoidal category which is $S$
-distributive. We shall say that $\boldsymbol{M}$ satisfies the hypothesis ($
\dag $) if for any coproduct $(\textstyle\bigoplus {}_{i\in S}X_{i},\{\sigma
_{i}\}_{i\in S})$ in $\boldsymbol{M}$ and any morphisms $f^{\prime
}:X\rightarrow X_{i^{\prime }}$ and $f^{\prime \prime }:X\rightarrow
X_{i^{\prime \prime }}$ such that $\sigma _{i^{\prime }}\circ f^{\prime
}=\sigma _{i^{\prime \prime }}\circ f^{\prime \prime }$, then either $X$ is
an initial object $\emptyset $ in $\boldsymbol{M},$ or $f^{\prime
}=f^{\prime \prime }$ and $i^{\prime }=i^{\prime \prime }$.
The prototype for the class of monoidal categories that satisfy the
condition (\dag ) is $\boldsymbol{Set}$. Indeed, let $\{X_{i}\}_{i\in S}$ be
a family of sets, and let $\sigma _{i}$ denote the inclusion of $X_{i}$ into
the disjoint union $\textstyle\coprod_{i\in S}X_{i}.$ We assume that $
f^{\prime }:X\rightarrow X_{i^{\prime }}$ and $f^{\prime \prime
}:X\rightarrow X_{i^{\prime \prime }}$ are functions such that $X$ in not
the empty set, the initial object of $\boldsymbol{Set}$, and $\sigma
_{i^{\prime }}\circ f^{\prime }=\sigma _{i^{\prime \prime }}\circ f^{\prime
\prime }.$ Then in view of the computation
\begin{equation*}
(i^{\prime },f^{\prime }(x))=(\sigma _{i^{\prime }}\circ f^{\prime
})(x)=(\sigma _{i^{\prime \prime }}\circ f^{\prime \prime })(x)=(i^{\prime
\prime },f^{\prime \prime }(x))
\end{equation*}
it follows that $f^{\prime }=f^{\prime \prime }$ and $i^{\prime }=i^{\prime
\prime }.$
\end{fact}
\begin{corollary}
\label{cor: R simplu1} Let $\boldsymbol{M}$ be an $S$-distributive monoidal
category. Let $\boldsymbol{A}$ and $\boldsymbol{B}$ be two $\boldsymbol{M}$
-categories such that $A_{0}=B_{0}=S$. Given a function $\left\vert \cdots
\right\vert :S^{3}\rightarrow S$ and the maps $\{_{x}\widetilde{{R}}
_{z}^{y}\}_{x,y,z\in S}$ as in \S \ref{fa:R^tilda}, let us consider the
following four conditions:
\begin{enumerate}
\item[(i)] If $_{x}B_{y}B_{z}A_{t}\ $is not an initial object, then $
|xy|yzt||=|xzt|$ and
\begin{equation}
_{x}A_{|xzt|}b_{t}^{|yzt|}\circ {}_{x}\widetilde{R}_{|yzt|}^{y}B_{t}{}\circ
{}_{x}B_{y}\widetilde{R}_{t}^{z}={}_{x}\widetilde{R}_{t}^{z}\circ
{}_{x}b_{z}^{y}A_{t}; \label{cor1-1}
\end{equation}
\item[(ii)] If $_{x}B_{y}A_{z}A_{t}\ \ $is not an initial object, then $
||xyz|zt|=|xyt|$ and
\begin{equation}
_{x}a_{|xyt|}^{|xyz|}B_{t}\circ {}_{x}A_{|xyz|}\widetilde{R}_{t}^{z}\circ
{}_{x}\widetilde{R}_{z}^{y}A_{t}={}_{x}\widetilde{R}_{t}^{y}\circ
{}_{x}B_{y}a_{t}^{z}; \label{cor1-2}
\end{equation}
\item[(iii)] If $_{x}A_{y}$ $\ $is not an initial object, then $|xxy|=y$ and
\begin{equation}
{\ }_{x}\widetilde{R}_{y}^{x}\circ (1_{x}^{B}\otimes
{}_{x}A_{y})={}_{x}A_{y}\otimes 1_{y}^{B}; \label{cor1-3}
\end{equation}
\item[(iv)] If $_{x}B_{y}$ $\ $is not an initial object, then $|xyy|=x$ and
\begin{equation}
_{x}\widetilde{R}_{y}^{y}\circ (_{x}B_{y}\otimes
1_{y}^{A})={}1_{x}^{A}\otimes {}_{x}B_{y}. \label{cor1-4}
\end{equation}
\end{enumerate}
The above conditions imply the relations (\ref{Rs-1})-(\ref{Rs-4}). Under
the additional assumption that $\boldsymbol{M}$ satisfies the hypothesis
(\dag ), the reversed implication holds as well.
\end{corollary}
\begin{proof}
Let us prove that the condition (i) implies the relation (\ref{Rs-1}). In
the case when $_{x}B_{y}B_{z}A_{t}=\emptyset $ this is clear, as both sides
of (\ref{Rs-1}) are morphisms from an initial object to $_{x}A_{\overline{u}
}B_{t}.$ Let us suppose that $_{x}B_{y}B_{z}A_{t}\neq \emptyset .$ By
composing both sides of (\ref{cor1-1}) with $_{x}\sigma
_{t}^{|xy|yzt||}={}_{x}\sigma _{t}^{|xzt|}$ we get the equation (\ref{Rs-1}
). Similarly, the conditions (ii), (iii) and (iv) imply the relations (\ref
{Rs-2}), (\ref{Rs-3}) and (\ref{Rs-4}), respectively.
Let us assume that $\boldsymbol{M}$ satisfies the hypothesis (\dag ). We
claim that (\ref{Rs-1}) implies (i). If $_{x}B{}_{y}B${}$_{z}A_{t}\ $is not
an initial object we take $f^{\prime }$ and $f^{\prime \prime }$ to be the
left hand side and the right hand side of (\ref{cor1-1}), respectively. We
also set $i^{\prime }:=|xy|yzt||$ and $i^{\prime \prime }:=|xzt|.$ In view
of (\dag ), it follows that $f^{\prime }=f^{\prime \prime }$ and $i^{\prime
}=i^{\prime \prime },$ so our claim has been proved. We conclude the proof
in the same way.
\end{proof}
\begin{fact}[$\mathbb{K}$-linear monoidal categories.]
\label{fa:ddag} Recall that $\boldsymbol{M}$ is $\mathbb{K}$-linear if its
hom-sets are $\mathbb{K}$-modules, and both the composition and the tensor
product of morphisms are $\mathbb{K}$-bilinear maps. For instance, $\mathbb{K
}$-$\boldsymbol{Mod}$, $H$-$\boldsymbol{Mod}$, $\boldsymbol{Comod}$-$H$ and $
\Lambda $-$\boldsymbol{Mod}$-$\Lambda $ are $S$-distributive linear monoidal
categories, for any set $S$.
Note that the (\dag ) condition fail in a $\mathbb{K}$-linear monoidal
category $\boldsymbol{M}$. Indeed let us pick up an object $X$, which is not
an initial object, and a coproduct $(\textstyle\bigoplus_{i\in
S}X_{i},\{\sigma _{i}\}_{i\in S})$ in $\boldsymbol{M}$. If $f^{\prime
}:X\rightarrow X_{i^{\prime }}$ and $f^{\prime \prime }:X\rightarrow
X_{i^{\prime \prime }}$ are the zero morphisms, then of course $\sigma
_{i^{\prime }}\circ f^{\prime }=\sigma _{i^{\prime \prime }}\circ f^{\prime
\prime },$ but neither $i^{\prime }=i^{\prime \prime }$ nor $f^{\prime
}=f^{\prime \prime }$, in general.
Nevertheless, the relations (\ref{Rs-1})-(\ref{Rs-4}) can also be simplified
if $\boldsymbol{M}$ is a linear monoidal category. For any coproduct $(
\textstyle\bigoplus_{i\in S}X_{i},\{\sigma _{i}\}_{i\in S})$ in $\boldsymbol{
M}$ and every $i\in S,$ there is a map $\pi _{i}:\textstyle\bigoplus_{i\in
S}X_{i}\rightarrow X_{i}$ such that $\pi _{i}\circ \sigma _{i}=X_{i}$ and $
\pi _{i}\circ \sigma _{j}=0$, provided that $j\neq i$. Hence, supposing that
$f^{\prime }:X\rightarrow X_{i^{\prime }}$ and $f^{\prime \prime
}:X\rightarrow X_{i^{\prime \prime }}$ are morphisms such that $\sigma
_{i^{\prime }}\circ f^{\prime }=\sigma _{i^{\prime \prime }}\circ f^{\prime
\prime },$ we must have either $i^{\prime }=i^{\prime \prime }$ and $
f^{\prime }=f^{\prime \prime },$ or $i^{\prime }\neq i^{\prime \prime }$ and
$f^{\prime }=0=f^{\prime \prime }.$
\end{fact}
Using the above property of linear monoidal categories, and proceeding as in
the proof of the previous corollary, we get the following result.
\begin{corollary}
\label{cor R simplu2} Let $\boldsymbol{M}$ be an $S$-distributive $\mathbb{K}
$-linear monoidal category. If $\boldsymbol{A}$ and $\boldsymbol{B}$ are $
\boldsymbol{M}$-categories, then the relations (\ref{Rs-1})-(\ref{Rs-4}) are
equivalent to the following conditions:
\begin{enumerate}
\item[(i)] If $|xy|yzt||=|xzt|\ $then the relation (\ref{cor1-1}) holds;
otherwise, each side of this identity has to be the zero map;
\item[(ii)] If $||xyz|zt|=|xyt|,$ then the relation (\ref{cor1-2}) holds;
otherwise, each side of this identity has to be the zero map;
\item[(iii)] If $|xxy|=y,$ then the relation (\ref{cor1-3}) holds;
otherwise, each side of this identity has to be the zero map;
\item[(iv)] If $|xyy|=x,$ then the relation (\ref{cor1-4}) holds; otherwise,
each side of this identity has to be the zero map.
\end{enumerate}
\end{corollary}
\begin{fact}[Simple twisting systems.]
\label{STS} The proper context for constructing an enriched category $
\boldsymbol{A}\otimes _{{R}}\boldsymbol{B}$ out of a special type of
twisting system ${R}$ is provided by Corollary \ref{cor: R simplu1}.
By definition, the couple $(\widetilde{{R}},|\cdots |)$ is a \emph{simple
twisting system between} $\boldsymbol{B}$ \emph{and} $\boldsymbol{A}$ if the
function $\left\vert \cdots \right\vert :S^{3}\rightarrow S$ and the maps $
\{_{x}\widetilde{R}_{z}^{y}\}_{x,y,z\in S}$ as in \S \ref{fa:R^tilda}
satisfy the conditions (i)-(iv) in Corollary \ref{cor: R simplu1}. As a part
of the definition, we also assume that $_{x}A_{\left\vert xyz\right\vert
}B_{z}$ is not an initial object whenever $_{x}B_{y}A_{z}$ is not so.
The latter technical assumption will be used to prove the associativity of
the composition in ${\boldsymbol{A}\otimes _{{R}}\boldsymbol{B}},$ our
categorical version of the twisted tensor product of two algebras, which we
are going to define in the next subsection. Note that for $\boldsymbol{Set}$
this condition is superfluous (if the source of $_{x}\widetilde{R}_{z}^{y} $
is not empty, then its target cannot be the empty set).
For a simple twisting system $(\widetilde{{R}},|\cdots |)$ we define the
maps $_{x}R_{z}^{y}$ using the relation (\ref{R simplu}). By Corollary \ref
{cor: R simplu1} and Proposition \ref{thm R simplu} it follows that ${R}
:=\{_{x}R_{z}^{y}\}_{x,y,z\in S}$ is a twisting system.
\end{fact}
\begin{fact}[The category $\boldsymbol{A}\otimes _{{R}}\boldsymbol{B}$.]
\label{fa:ARB}For a simple twisting system $(\widetilde{{R}},|\cdots |)$ we
set
\begin{equation*}
\left( \boldsymbol{A}\otimes _{{R}}\boldsymbol{B}\right) _{0}:=S\quad \text{
and\quad }{_{x}(\boldsymbol{A}\otimes _{{R}}\boldsymbol{B})}_{y}:=\textstyle
\bigoplus\limits_{u\in S}{}_{x}A_{u}\otimes {}_{u}B_{y}={}_{x}A_{\overline{u}
}B_{y}.
\end{equation*}
Let us fix three elements $x,$ $y$ and $z$ in $S.$ By definition $_{x}A_{
\overline{u}}B_{y}A_{\bar{v}}B_{z}:=\textstyle\bigoplus\limits_{u,v\in
S}{}\,_{x}A_{u}B_{y}A_{v}B_{z}$, and
\begin{equation*}
_{x}A_{\overline{u}}B_{y}A_{\overline{v}}B_{z}\cong {}_{x}A_{\overline{u}
}B_{y}\otimes {}_{y}A_{\overline{v}}B_{z}
\end{equation*}
as $\boldsymbol{M}$ is $S$-distributive. Via this identification, the
canonical inclusion of $_{x}A_{u}B_{y}A_{v}B_{z}$ into the coproduct $_{x}A_{
\overline{u}}B_{y}A_{\overline{v}}B_{z}$ corresponds to $_{x}\sigma
_{y}^{u}\sigma _{z}^{v}={}_{x}\sigma _{y}^{u}\otimes {}_{y}\sigma _{z}^{v}.$
Thus, there is a unique morphism $_{x}c_{z}^{y}:{}_{x}A_{\overline{u}
}B_{y}A_{\overline{v}}B_{z}\rightarrow {}_{x}A_{\overline{u}}B_{z}$ such
that
\begin{equation*}
_{x}c_{z}^{y}\circ {}_{x}\sigma _{y}^{u}\sigma _{z}^{v}={}_{x}\sigma
_{z}^{\left\vert uyv\right\vert }\circ {}_{x}a_{\left\vert uyv\right\vert
}b_{z}\circ {}_{x}A_{u}\widetilde{R}_{v}^{y}B_{z},
\end{equation*}
for all $u,v\in S$. Finally, we set $1_{x}:={}_{x}\sigma _{x}^{x}\circ
(1_{x}^{A}\otimes 1_{x}^{B})$, and we define
\begin{equation*}
_{x}\alpha _{y}:={}_{x}\sigma _{y}^{y}\circ ({}_{x}A_{y}\otimes
1_{y}^{B})\qquad \text{and}\qquad _{x}\beta _{y}:={}_{x}\sigma _{y}^{x}\circ
(1_{x}^{A}\otimes {}_{x}B_{y}).
\end{equation*}
\end{fact}
\begin{fact}[Domains.]
To show that the above data define an enriched monoidal category $
\boldsymbol{A}\otimes _{{R}}\boldsymbol{B}$ we need an extra hypothesis on $
\boldsymbol{M}.$ By definition, a monoidal category $\boldsymbol{M}$ is a
\emph{domain }in the case when the tensor product of two objects$\ $is an
initial object if and only if at least one of them is an initial object. By
convention, a monoidal category that has no initial objects is a domain as
well.
Obviously $\boldsymbol{Set}$ is a domain. If $\mathbb{K}$ is a field, then $
\mathbb{K}$-$\boldsymbol{Mod}$ is a domain. Keeping the assumption on $
\mathbb{K},$ the categories $H$-$\boldsymbol{Mod}$ and $\boldsymbol{Comod}$-$
H$ are domains, as their tensor product is induced by that one of $\mathbb{K}
$-$\boldsymbol{Mod}.$ On the other hand, if $\mathbb{K}$ is not a field,
then $\mathbb{K}$-$\boldsymbol{Mod}$ and $\Lambda $-$\boldsymbol{Mod}$-$
\Lambda $ are not necessarily domains. For instance, $\mathbb{Z}$-$
\boldsymbol{Mod}\cong \mathbb{Z}$-$\boldsymbol{Mod}$-$\mathbb{Z}$ is not a
domain.
\end{fact}
\begin{lemma}
\label{le:STS}Let $\boldsymbol{M}$ be an $S$-distributive monoidal domain.
Let $(\widetilde{{R}},\left\vert \dots \right\vert )$ denote a simple
twisting system between $\boldsymbol{B}$ and $\boldsymbol{A}.$
\begin{enumerate}
\item If $_{x}A_{u}B_{y}A_{v}B_{z}A_{w}B_{t}\neq \emptyset $ then $
\left\vert uyq\right\vert =\left\vert pvq\right\vert =\left\vert
pzw\right\vert ,$ where $p=\left\vert uyv\right\vert $ and $q=\left\vert
vzw\right\vert .$
\item In the following diagram all squares are well defined
and commutative.
\begin{equation*}
\xymatrix{ \ar@{}[dr]^>(.60)*+[o][F-]{F} \sv{_xA_uB_yA_v B_
zA_wB_t}\ar[r]^-{\sa{III\widetilde{R}I}}
\ar[d]|<(.25){\sa{I\widetilde{R}III}} &\sv{_xA_uB_{{y}}A_v A_{q} B_w
B_t}\ar[r]^-{\sa{IIII b}}
\ar@{}[dr]|*+[o][F-]{F}\ar[d]|<(.25){\sa{II\widetilde{R}II\circ
I\widetilde{R}III}} &\sv{_xA_uB_yA_v A_q B_t}
\ar@{}[dr]|*+[o][F-]{R}\ar[r]^-{\sa{IIaI}}
\ar[d]|-{\sa{II\widetilde{R}I\circ I\widetilde{R}II}} &\sv{_xA_uB_y A_q B_t}
\ar[d]|-{\sa{I\widetilde{R}I}} \\ \sv{_xA_uA_pB_vB_zA_wB_t}
\ar@/^10px/[r]^-{\sa{II\widetilde{R}II\circ III\widetilde{R}I}}
\ar[d]|-{\sa{IIaII}}\ar@{}[dr]|*+[o][F-]{F} &\sv{_xA_u A_p A_{|pvq|} B_q B_w
B_t}\ar@{}[dr]|*+[o][F-]{F} \ar[r]_-{\sa{IIIIb}} \ar[d]|-{aIIII} &\sv{_xA_u
A_p A_{|pvq|} B_q B_t}\ar[r]_-{\sa{IaII}}
\sv{\ar[d]|-{aIII}}\ar@{}[dr]|*+[o][F-]{A} &\sv{_x A_u A_{|pvq|} B_q
B_t}\ar[d]|-{\sa{aII}} \\ \sv{_xA_pB_vB_zA_wB_t}
\ar@{}[dr]_>(.60)*+[o][F-]{L} \ar@/^-10px/[r]_-{\sa{I\widetilde{R}II\circ
II\widetilde{R}I}} \ar[d]|-{\sa{IbII}} & \sv{_x A_p A_{|pvq|} B_q B_w B_t}
\ar[r]_-{\sa{IIIb}}\sv{\ar[d]|-{IIbI}}\ar@{}[dr]|*+[o][F-]{A} &\sv{_x A_p
A_{|pvq|} B_q B_t}\sv{\ar[r]_-{\sa{aII}}}\sv{\ar[d]|-{IIb}}
\ar@{}[dr]|*+[o][F-]{F}&\sv{_x A_{|pvq|} B_q B_t} \ar[d]|-{\sa{Ib}} \\
\sv{_xA_p B_z A_w B_t} \ar[r]_-{\sa{I\widetilde{R}I}} &\sv{_xA_p A_{|pvq|}
B_w B_t} \sv{\ar[r]_{IIb}} &\sv{_xA_p A_{|pvq|} B_t}\sv{\ar[r]_{aI}}
&\sv{_xA_{|pvq|} B_t} }
\end{equation*}
\end{enumerate}
\end{lemma}
\begin{proof}
Since $\boldsymbol{M}$ is a domain it follows that any subfactor of $
_{x}A_{u}B_{y}A_{v}B_{z}A_{w}B_{t}$ is not an initial object. In particular $
_{v}B_{z}A_{w}\neq \emptyset .$ Thus, by the definition of simple twisting
systems, $_{v}A_{q}B_{w}$ is not an initial object. In conclusion, $
_{v}A_{q} $ and $_{q}B_{w}$ are not initial objects in $\boldsymbol{M}.$
Since $_{u}B_{y}A_{v}\neq \emptyset $ it follows that $_{u}B_{y}A_{v}A_{q}
\neq \emptyset $. In view of the definition of simple twisting systems (the
second condition) we deduce that $\left\vert pvq\right\vert =\left\vert
uyq\right\vert $. The other relation can be proved in a similar way.
Let $f$ and $g$ denote the following two morphisms
\begin{equation*}
f:={}_{x}A_{u}a_{\left\vert pvq\right\vert }^{p}B_{q}B_{t}\circ
{}_{x}A_{u}A_{p}\widetilde{R}_{q}^{v}B_{t}\circ {}_{x}A_{u}\widetilde{R}
_{v}^{y}A_{q}B_{t}\text{\quad and\quad }g:={}_{x}A_{u}\widetilde{R}
_{q}^{y}B_{t}\circ {}_{x}A_{u}B_{y}a_{q}^{v}B_{t}.
\end{equation*}
The target of $f$ is $_{x}A_{u}A_{\left\vert pvq\right\vert }B_{q}B_{t},$
while the codomain of $g$ is $_{x}A_{u}A_{\left\vert uyq\right\vert
}B_{q}B_{t}$. These two objects may be different for some elements $x,u,t,p$
and $q$ in $S$. Thus, in general, it does not make sense to speak about the
square (R). On the other hand, we have seen that $\left\vert uyq\right\vert
=\left\vert pvq\right\vert ,$ if $p=\left\vert uyv\right\vert $ and $
q=\left\vert vzw\right\vert .$ Hence (R) is well defined for these values of
$p$ and $q$. Furthermore, since $_{u}B_{y}A_{v}A_{q}\neq \emptyset ,$ by
definition of simple twisting systems we have
\begin{equation}
_{u}a_{|pvq|}^{p}B_{q}\circ {}_{u}A_{p}\widetilde{R}_{q}^{v}\circ {}_{u}
\widetilde{R}_{v}^{y}A_{q}={}_{u}\widetilde{R}_{q}^{y}\circ
{}_{u}B_{y}a_{q}^{v}. \label{lem}
\end{equation}
By tensoring both sides of the above relation with $_{x}A_{u}$ on the left
and with $_{q}B_{t}$ on the right we get that $f=g,$ i.e. (R) is
commutative. Analogously, one shows that (L) is well defined and
commutative. All other squares are well defined by construction, their
arrows targeting to the right objects. The squares (F) are commutative since
the tensor product is a functor. The remaining squares (A) are commutative
by associativity.
\end{proof}
\begin{theorem}
\label{thm:TTP} Let $\boldsymbol{M}$ be an $S$-distributive monoidal domain.
If $(\widetilde{{R}},\left\vert \dots \right\vert )$ is a simple twisting
system, then the data in \S \ref{fa:ARB} define an $\boldsymbol{M}$-category
$\boldsymbol{A}\otimes _{{R}}\boldsymbol{B}$ that factorizes through $
\boldsymbol{A}$ and $\boldsymbol{B}$.
\end{theorem}
\begin{proof}
Let us assume that $_{x}A_{u}B_{y}A_{v}B_{z}A_{w}B_{t}\neq \emptyset $. In
view of the previous lemma, the outer square in the diagram from Lemma \ref{le:STS} (2) is commutative. It follows that
\begin{equation*}
_{x}c_{t}^{y}\circ {}_{x}A_{\overline{u}}B_{y}c_{t}^{z}\circ {}_{x}\sigma
_{y}^{u}\sigma _{z}^{v}\sigma _{t}^{w}={}_{x}c_{t}^{z}\circ
{}_{x}c_{z}^{y}A_{\overline{w}}B_{t}\circ {}_{x}\sigma _{y}^{u}\sigma
_{z}^{v}\sigma _{t}^{w}\,{}.
\end{equation*}
If $_{x}A_{u}B_{y}A_{v}B_{z}A_{w}B_{t}=\emptyset $ this identity obviously
holds. Since $_{x}A_{\overline{u}}B_{y}A_{\overline{v}}B_{z}A_{\overline{w}
}B_{t}$ is the coproduct of $\{_{x}A_{u}B_{y}A_{v}B_{z}A_{w}B_{t}\}_{u,v,w
\in S}$, with the canonical inclusions $\{_{x}\sigma _{y}^{u}\sigma
_{z}^{v}\sigma _{t}^{w}\}_{u,v,w\in S},$ we deduce that the composition in $
\boldsymbol{A}\otimes _{{R}}\boldsymbol{B}$ is associative.
We apply the same strategy to show that $1_{x}:={}_{x}\sigma _{x}^{x}\circ
(1_{x}^{A}\otimes 1_{x}^{B})$ is a left identity map of $x,$ that is we have
$_{x}c_{y}^{x}\circ (1_{x}\otimes {}_{x}A_{\overline{u}}B_{y})={}_{x}A_{
\overline{u}}B_{y}$ for any $y.$ By the universal property of coproducts and
the definition of the composition in $\boldsymbol{A}\otimes _{{R}}
\boldsymbol{B},$ it is enough to prove that
\begin{equation}
_{x}\sigma _{y}^{\left\vert xxu\right\vert }\circ {}_{x}a_{\left\vert
xxu\right\vert }^{x}b_{y}^{u}\circ {}_{x}A_{x}\widetilde{R}
_{u}^{x}B_{y}\circ (1_{x}^{A}\otimes 1_{x}^{B}\otimes
{}_{x}A_{u}B_{y})={}_{x}\sigma _{y}^{u}, \label{ec:unit}
\end{equation}
for all $u\in S.$ If $_{x}A_{u}$ is an initial object we have nothing to
prove, as the domains of the sides of the above equation are also initial
objects (recall that $_{x}A_{u}B_{y}=\emptyset $ if $_{x}A_{u}=\emptyset $).
Let us suppose that $_{x}A_{u}$ is not an initial object. Then by the
definition of simple twisting systems (the third condition) we get $
\left\vert xxu\right\vert =u$ and
\begin{equation*}
_{x}\sigma _{y}^{\left\vert xxu\right\vert }\circ {}_{x}a_{\left\vert
xxu\right\vert }^{x}b_{y}^{u}\circ {}_{x}A_{x}\widetilde{R}
_{u}^{x}B_{y}\circ (1_{x}^{A}\otimes 1_{x}^{B}\otimes
{}_{x}A_{u}B_{y})={}_{x}\sigma _{y}^{u}\circ {}_{x}a_{u}^{x}b_{y}^{u}\circ
(1_{x}^{A}\otimes {}_{x}A_{u}\otimes 1_{u}^{B}\otimes {}_{u}B_{y}).
\end{equation*}
Thus the equation (\ref{ec:unit}) immediately follows by the fact $1_{x}^{A}$
and $1_{u}^{B}$ are the identity morphisms of $x$ and $u$. The fact that $
1_{x}$ is a right identity map of $x$ can be proved analogously.
We now claim that $\{_{x}\alpha _{y}\}_{x,y\in S}$ is an $\boldsymbol{M}$
-functor. Taking into account the definition of $\boldsymbol{\alpha }$ and $
_{x}c_{z}^{y}$ we must prove that
\begin{equation}
_{x}\sigma _{z}^{\left\vert yyz\right\vert }\circ {}_{x}a_{\left\vert
yyz\right\vert }^{x}a_{z}^{z}\circ {}_{x}A_{y}\widetilde{R}
_{z}^{y}A_{z}\circ (_{x}A_{y}\otimes 1_{y}^{A}\otimes {}_{y}A_{z}\otimes
1_{z}^{A}{})={}_{x}\sigma _{z}^{z}\circ ({}_{x}a_{z}^{y}\otimes 1_{z}^{A}),
\label{ec:alpha}
\end{equation}
for all $x,$ $y$ and $z$ in $S.$ Once again, if $_{y}A_{z}=\emptyset $ we
have nothing to prove. In the other case, one can proceed as in the proof of
(\ref{ec:unit}) to get this equation. Similarly, $\boldsymbol{\beta }$ is an
$\boldsymbol{M}$-functor.
It remains to prove the fact that $\boldsymbol{A}\otimes _{{R}}\boldsymbol{B}
$ factorizes through $\boldsymbol{A}$ and $\boldsymbol{B}.$ As a matter of
fact, for this enriched category, we shall show that $_{x}\varphi _{y}$ is
the identity map of $_{x}(\boldsymbol{A}\otimes _{{R}}\boldsymbol{B})_{y},$
for all $x$ and $y$ in $S.$ Recall that $_{x}\varphi _{y}$ is the unique map
such that $_{x}\varphi _{y}\circ {}_{x}\sigma _{y}^{u}={}_{x}c_{y}^{u}\circ
{}_{x}\alpha _{u}\beta _{y},$ for all $u\in S.$ Hence to conclude the proof
of the theorem it is enough to obtain the following relation
\begin{equation}
{}_{x}\sigma _{y}^{\left\vert uuu\right\vert }\circ {}_{x}a_{\left\vert
uuu\right\vert }^{u}b_{y}^{u}\circ {}_{x}A_{u}\widetilde{R}
_{u}^{u}B_{y}\circ (_{x}A_{u}\otimes 1_{u}^{B}\otimes 1_{u}^{A}\otimes
{}{}_{u}B_{z}{})={}_{x}\sigma _{y}^{u}, \label{ec:fact}
\end{equation}
for all $u\in S.$ We may suppose that $_{x}A_{u}$ is not initial object.
Thus $\left\vert uuu\right\vert =u$ and we can take $x=u$ and $y=u$ in (\ref
{cor1-4}). Hence, using the same reasoning as in the proof of (\ref{ec:unit}
), we deduce the required identity.
\end{proof}
\begin{corollary}
\label{cor: TTP} Let $\boldsymbol{A}$ and $\boldsymbol{B}$ be enriched
categories over an $S$-distributive monoidal category $\boldsymbol{M}$. Let
us suppose that for all $x,$ $y,$ $z$ and $t$ in $S$ the function $
\left\vert \cdots \right\vert :S^{3}\rightarrow S$ satisfies the equations
\begin{equation}
|xy|yzt||=|xzt|,\quad \quad ||xyz|zt|=|xyt|,\quad \quad |xxy|=y\quad \quad
\text{and}\quad \quad |xyy|=x. \label{modul}
\end{equation}
If $\{_{x}\widetilde{{R}}_{z}^{y}\}_{x,y,z\in S}$ is a family of maps which
satisfies the identities (\ref{cor1-1})-(\ref{cor1-4}) for all $x,$ $y,$ $z$
and $t$ in $S,$ then the data in \S \ref{fa:ARB} define an $\boldsymbol{M}$
-category $\boldsymbol{A}\otimes _{{R}}\boldsymbol{B}$ that factorizes
through $\boldsymbol{A}$ and $\boldsymbol{B}$.
\end{corollary}
\begin{proof}
Let $x,$ $y,$ $u,$ $v,$ $z,$ $w$ and $t$ be arbitrary elements in $S.$ By
using the first two identities in (\ref{modul}) we get $\left\vert
uyq\right\vert =\left\vert pvq\right\vert =\left\vert pzw\right\vert ,$
where $p=\left\vert uyv\right\vert $ and $q=\left\vert vzw\right\vert .$
Hence the first statement in Lemma \ref{le:STS} is true. In particular, the
squares (R) and (L) in the diagram from Lemma \ref{le:STS}(2) are well
defined. On the other hand, under the assumptions of the corollary, the
relation (\ref{lem}) hold. Therefore we can continue as in the proof of the
second part of Lemma \ref{le:STS} to show that (R) is commutative.
Similarly, (L) is commutative too. It follows that the outer square of is
commutative too. By the universal property of the coproduct we deduce that
the composition is associative, see the first paragraph of the proof of
Theorem \ref{thm:TTP}.
Furthermore, the relations in (\ref{modul}) together with the identities (
\ref{cor1-1})-(\ref{cor1-4}) imply the equations (\ref{ec:unit}), (\ref
{ec:alpha}) and (\ref{ec:fact}). Proceeding as in the proof of Theorem \ref
{thm:TTP} we conclude that $\boldsymbol{A}\otimes _{{R}}\boldsymbol{B}$ is
an $\boldsymbol{M}$-category that factorizes through $\boldsymbol{A}$ and $
\boldsymbol{B}$.
\end{proof}
\begin{remark}
Throughout this remark we assume that $\boldsymbol{M}$ is a $T$-distributive
monoidal category, where $T$ is an arbitrary set. In other words, any family
of objects in $\boldsymbol{M}$ has a coproduct and the tensor product is
distributive over all coproducts. It was noticed in \cite[\S 2.1 and \S 2.2]
{RW} that, for such a monoidal category $\boldsymbol{M},$ one can define a
bicategory $\boldsymbol{M}$-$\boldsymbol{mat}$ as follows.
By construction, its $0$-cells are arbitrary sets. If $I$ and $J$ are two
sets, then the $1$-cells in $\boldsymbol{M}$-$\boldsymbol{mat}$ from $I$ to $
J$ are the $J\times I$-indexed families of objects in $\boldsymbol{M}$. A $2$
-cell with source $\{X_{ji}\}_{(j,i)\in J\times I}$ and target $
\{Y_{ji}\}_{(j,i)\in J\times I}$ is a family $\{f_{ji}\}_{(j,i)\in J\times
I} $ of morphisms $f_{ji}:X_{ji}\rightarrow Y_{ji}.$ The composition of the $
1$-cells $\{X_{kj}\}_{(k,j)\in K\times J}$ and $\{Y_{ji}\}_{(j,i)\in J\times
I} $ is the family $\{Z_{ki}\}_{(k,i)\in K\times I},$ where
\begin{equation*}
Z_{ki}:=\textstyle\bigoplus_{j\in J}X_{kj}\otimes Y_{ji}.
\end{equation*}
The vertical composition in $\boldsymbol{M}$-$\boldsymbol{mat}$ of $
\{f_{ji}\}_{(j,i)\in J\times I}$ and $\{g_{ji}\}_{(j,i)\in J\times I}$ makes
sense if and only if the source of $f_{j i}$ and the target of $g_{j i}$ are
equal for all $i$ and $j.$ If it exists, then it is defined by
\begin{equation*}
\{f_{ji}\}_{(j,i)\in J\times I}\bullet \{g_{ji}\}_{(j,i)\in J\times
I}=\{f_{ji}\circ g_{ji}\}_{(j,i)\in J\times I}.
\end{equation*}
Let $\{f_{ji}\}_{(j,i)\in J\times I}$ and $\{f_{kj}^{\prime }\}_{(k,j)\in
K\times J}$ be $2$-cells such that $f_{ji}:X_{ji}\rightarrow Y_{ji}$ and $
f_{kj}^{\prime }:X_{kj}^{\prime }\rightarrow Y_{kj}^{\prime }.$ By the
universal property of coproducts, for each $(k,i)\in K\times I$, there
exists a unique morphism $h_{ki}:\bigoplus_{j\in J}X_{kj}^{\prime }\otimes
X_{ji}\rightarrow \bigoplus_{j\in J}Y_{kj}^{\prime }\otimes Y_{ji}$ whose
restriction to $X_{kj}^{\prime }\otimes $ $X_{ji}$ is $f_{kj}^{\prime
}\otimes $ $f_{ji}$. By definition, the horizontal composition of $
\{f_{kj}^{\prime }\}_{(k,j)\in K\times J}$ and $\{f_{ji}\}_{(j,i)\in J\times
I}$ is the family $\{h_{ki}\}_{(k,i)\in K\times I}.$ The identity $1$-cells
and $2$-cells in $\boldsymbol{M}$-$\boldsymbol{mat}$ are the obvious ones.
As pointed out in \cite{RW}, a monad on a set $S$ in $\boldsymbol{M}$-$
\boldsymbol{mat}$ is an $\boldsymbol{M}$-category with the set of objects $S$
, and conversely. In particular, given two $\boldsymbol{M}$-categories with
the same set of objects, one may speak about distributive laws between the
corresponding monads in $\boldsymbol{M}$-$\boldsymbol{mat}.$ In our
terminology, they are precisely the twisting systems. In view of \cite[\S 3.1
]{RW}, factorizable enriched categories generalize strict factorization
systems.
In conclusion, the Theorem \ref{teo1} may be regarded as a version of \cite[
Proposition 3.3]{RW} for enriched categories. For a simple twisting system $(
\widetilde{R},\left\vert \cdots \right\vert )$ between $\boldsymbol{B}$ and $
\boldsymbol{A},$ the enriched category $\boldsymbol{A}\otimes _{R}
\boldsymbol{B}$ that we constructed in Theorem \ref{thm:TTP} can also be
described in terms of monads. Let $\rho :B\circ A\rightarrow A\circ B$
denote the distributive law associated to $(\widetilde{R},\left\vert \cdots
\right\vert ),$ where $(A,m_{A},1_{A})$ and $(B,m_{B},1_{B})$ are the monads
in $\boldsymbol{M}$-$\boldsymbol{mat}$ corresponding to $\boldsymbol{A}$ and
$\boldsymbol{B,}$ respectively. By the general theory of monads in a
bicategory, it follows that $A\circ B$ is a monad in $\boldsymbol{M}$-$
\boldsymbol{mat}$ with respect to the multiplication and the unit given by
the formulae:
\begin{equation*}
m:=\left( m_{A}\circ m_{B}\right) \bullet \left( {\text{Id}}_{A}\circ \rho
\circ {\text{Id}}_{B}\right) \text{\quad and\quad }1:=1_{A}\circ 1_{B}.
\end{equation*}
It is not difficult to show that $\boldsymbol{A}\otimes _{R}\boldsymbol{B}$
is the $\boldsymbol{M}$-category associated to $(A\circ B,m,1).$
By replacing $\boldsymbol{Set}$-$\boldsymbol{mat}$ with a suitable
bicategory, one obtains similar results for other algebraic structures, such
as PROs and PROPs; see \cite{La}. We also would like to note that
distributive laws between pseudomonads are investigated in \cite{Mar}.
We are indebted to the referee for pointing the papers \cite{La, Mar, RW}
out to us.
\end{remark}
\section{Matched pairs of enriched categories.}
Throughout this section $(\boldsymbol{M^{\prime }},\otimes ,\boldsymbol{1}
,\chi )$ denote a braided category and we take $\boldsymbol{M}$ to be the
monoidal category $\boldsymbol{Coalg({M^{\prime }})}.$ Our aim is to
characterize simple twisting systems between two categories that are
enriched over $\boldsymbol{M}$. We start by investigating some properties of
the morphisms in $\boldsymbol{M}$. For the moment, we impose no conditions
on $\boldsymbol{M^{\prime }}$.
A slightly more general version of the following lemma is stated in \cite[
Proposition 3.2]{La}. For the sake of completeness we include a proof of it.
\begin{lemma}
\label{le:coalgebra}Let $(C,\Delta _{C},\varepsilon _{C})$, $(D_{1},\Delta
_{D_{1}},\varepsilon _{D_{1}})$ and $(D_{2},\Delta _{D_{2}},\varepsilon
_{D_{2}})$ be coalgebras in $\boldsymbol{M^{\prime }}$. Let $f:C\rightarrow
D_{1}\otimes D_{2}$ be a morphism of coalgebras. Then $f_{1}:=(D_{1}\otimes
\varepsilon _{D_{2}})\circ f$ and $f_{2}:=(\varepsilon _{D_{1}}\otimes
D_{2})\circ f$ are coalgebra morphisms and the following relations hold:
\begin{align}
(f_{1}\otimes f_{2})\circ \Delta _{C}& =f, \label{f1} \\
(f_{2}\otimes f_{1})\circ \Delta _{C}& =\chi _{D_{1},D_{2}}\circ
(f_{1}\otimes f_{2})\circ \Delta _{C}. \label{f2}
\end{align}
Conversely, let $f_{1}:C\rightarrow D_{1}$ and $f_{2}:C\rightarrow D_{2}$ be
coalgebra morphisms such that \eqref{f2} holds. Then $f:=(f_{1}\otimes
f_{2})\circ \Delta _{C}$ is a coalgebra map such that
\begin{equation}
(D_{1}\otimes \varepsilon _{D_{2}})\circ f=f_{1}\quad \text{\emph{and}}\quad
(\varepsilon _{D_{1}}\otimes D_{2})\circ f=f_{2}. \label{le:f}
\end{equation}
\end{lemma}
\begin{proof}
Let us assume that $f:C\rightarrow D_{1}\otimes D_{2}$ is a coalgebra
morphism. Let $\varepsilon _{i}:=\varepsilon _{D_{i}}$, for $i=1,2.$
Clearly, $D_{1}\otimes \varepsilon _{D_{2}}$ and $\varepsilon
_{D_{1}}\otimes D_{2}$ are coalgebra morphisms. In conclusion $f_{1}$ and $
f_{2}$ are morphisms in $\boldsymbol{M}$. On the other hand, as $f$ is a
morphism in $\boldsymbol{M}$ we have
\begin{equation}
(D_{1}\otimes \chi _{D_{1},D_{2}}\otimes D_{2})\circ (\Delta _{D_{1}}\otimes
\Delta _{D_{2}})\circ f=(f\otimes f)\circ \Delta _{C}. \label{f3}
\end{equation}
Hence, using the definition of $f_{1}$ and $f_{2}$, the relation \eqref{f3},
the fact that the braiding is a natural transformation and the compatibility
relation between the comultiplication and the counit we get
\begin{align*}
(f_{1}\otimes f_{2})\circ \Delta _{C}& =(D_{1}\otimes \varepsilon
_{2}\otimes \varepsilon _{1}\otimes D_{2})\circ (f\otimes f)\circ \Delta _{C}
\\
& =\left[D_{1}\otimes \left((\varepsilon _{2}\otimes \varepsilon _{1})\circ
\chi _{D_{1,}D_{2}}\otimes D_{2}\right)\circ (\Delta _{D_{1}}\otimes \Delta
_{D_{2}})\right]\circ f \\
& =(D_{1}\otimes \varepsilon _{1}\otimes \varepsilon _{2}\otimes D_{2})\circ
(\Delta _{D_{1}}\otimes \Delta _{D_{2}})\circ f=f.
\end{align*}
By applying $\varepsilon _{1}\otimes D_{2}\otimes D_{1}\otimes \varepsilon
_{2}$ to (\ref{f3}) and using once again the compatibility between the
comultiplication and the counit we obtain
\begin{align*}
(f_{2}\otimes f_{1})\circ \Delta _{C}& =(\varepsilon _{1}\otimes
D_{2}\otimes D_{1}\otimes \varepsilon _{2})\circ (f\otimes f)\circ \Delta
_{C} \\
& =(\varepsilon _{1}\otimes D_{2}\otimes D_{1}\otimes \varepsilon _{2})\circ
(D_{1}\otimes \chi _{D_{1},D_{2}}\otimes D_{2})\circ (\Delta _{D_{1}}\otimes
\Delta _{D_{2}})\circ f \\
& =\chi _{D_{1},D_{2}}\circ (\varepsilon _{1}\otimes D_{1}\otimes
D_{2}\otimes \varepsilon _{2})\circ (\Delta _{D_{1}}\otimes \Delta
_{D_{2}})\circ f =\chi _{D_{1},D_{2}}\circ f.
\end{align*}
Conversely, let us assume that $f_{1}:C\rightarrow D_{1}$ and $
f_{2}:C\rightarrow D_{2}$ are morphisms in $\boldsymbol{M}$ such that
\eqref{f2} holds. Let $f:=(f_{1}\otimes f_{2})\circ $ $\Delta _{C}.$ By the
definition of the comultiplication on $D_{1}\otimes D_{2}$ and the fact that
$f_{1}$ and $f_{2}$ are morphisms in $\boldsymbol{M,}$ we get
\begin{align*}
\Delta _{D_{1}\otimes D_{2}}\circ f&= (D_{1}\otimes \chi
_{D_{1},D_{2}}\otimes D_{2})\circ (\Delta _{D_{1}}\otimes \Delta
_{D_{2}})\circ (f_{1}\otimes f_{2})\circ \Delta _{C} \\
&= (D_{1}\otimes \chi _{D_{1},D_{2}}\otimes D_{2})\circ (f_{1}\otimes
f_{1}\otimes f_{2}\otimes f_{2})\circ (\Delta _{C}\otimes \Delta _{C})\circ
\Delta _{C}.
\end{align*}
Taking into account (\ref{f2}) and the fact that the comultiplication is
coassociative, it follows that
\begin{align*}
\Delta _{D_{1}\otimes D_{2}}\circ f &= \left[f_{1}\otimes \left( \chi
_{D_{1},D_{2}}\circ (f_{1}\otimes f_{2})\circ \Delta _{C}\right)\otimes f_{2}
\right] \circ (C\otimes \Delta _{C})\circ \Delta _{C} \\
&=\left[ f_{1}\otimes \left( (f_{2}\otimes f_{1})\circ \Delta _{C}\right)
\otimes f_{2}\right] \circ (C\otimes \Delta _{C})\circ \Delta _{C} \\
&=\left[\left((f_{1}\otimes f_{2})\circ \Delta _{C}\right)\otimes
\left((f_{1}\otimes f_{2}\right)\circ \Delta _{C})\right]\circ \Delta
_{C}=(f\otimes f)\circ \Delta _{C}.
\end{align*}
The formula that defines $f$ together with $\varepsilon _{i}\circ
f_{i}=\varepsilon _{C}$ yield
\begin{equation*}
(\varepsilon _{1}\otimes \varepsilon _{2})\circ f=(\varepsilon _{1}\circ
f_{1}\otimes \varepsilon _{2}\circ f_{2})\circ \Delta _{C}=(\varepsilon
_{C}\otimes \varepsilon _{C})\circ \Delta _{C}=\varepsilon _{C}.
\end{equation*}
Thus $f$ is a morphism of coalgebras, so the lemma is proved. The equations
in \eqref{le:f} are obvious, as $\varepsilon _{i}\circ f_{i}=\varepsilon
_{C} $.
\end{proof}
\begin{remark}
\label{re:f=g}Let $f^{\prime },f^{\prime \prime }:C\rightarrow D_{1}\otimes
D_{2}$ be coalgebra morphisms. By the preceding lemma, $f^{\prime }$ and $
f^{\prime \prime }$ are equal if and only if
\begin{equation*}
(\varepsilon _{1}\otimes D_{2})\circ f^{\prime }=(\varepsilon _{1}\otimes
D_{2})\circ f^{\prime \prime }\text{\quad and\quad }(D_{1}\otimes
\varepsilon _{2})\circ f^{\prime }=(D_{1}\otimes \varepsilon _{2})\circ
f^{\prime \prime }.\text{ }
\end{equation*}
\end{remark}
\begin{fact}[The morphisms $\boldsymbol{_{x}\triangleright _{z}^{y}}$ and $
\boldsymbol{_{x}\triangleleft _{z}^{y}}$.]
\label{coalg(M')} Let $\boldsymbol{A}$ and $\boldsymbol{B}$ denote two $
\boldsymbol{M}$-categories whose objects are the elements of a set $S$. The
hom-objects of $\boldsymbol{A}$ and $\boldsymbol{B}$ are coalgebras, which
will be denoted by $(_{x}A_{y},{}_{x}\Delta _{y}^{A},{}_{x}\varepsilon
_{y}^{A})$ and $(_{x}B_{y},{}_{x}\Delta _{y}^{B},{}_{x}\varepsilon _{y}^{B})$
. By definition, the composition and the identity maps in $\boldsymbol{A}$
and $\boldsymbol{B}$ are coalgebra morphisms. Note that the comultiplication
of $_{x}B_{y}A_{z}$ is given by
\begin{equation*}
\Delta _{_{x}B_{y}A_{z}}=(_{x}B_{y}\otimes \chi
_{_{x}B_{y},_{y}A_{z}}\otimes {}_{y}A_{z})\circ {}_{x}\Delta _{y}^{B}\Delta
_{z}^{A}.
\end{equation*}
Let $|\cdots |:S^{3}\rightarrow S$ be a function and let $\widetilde{R}$
denote an $S^{3}$-indexed family of coalgebra morphisms $_{x}\widetilde{R}
_{z}^{y}:{}_{x}B_{y}A_{z}\rightarrow {}_{x}A_{|xyz|}B_{z}$. We define $
_{x}\triangleright _{z}^{y}:{}_{x}B_{y}A_{z}\rightarrow {}_{x}A_{|xyz|}$ and
$_{x}\triangleleft _{z}^{y}:{}_{x}B_{y}A_{z}\rightarrow {}_{|xyz|}B_{z}$ by
\begin{equation}
_{x}\triangleright _{z}^{y}:={}_{x}A_{\left\vert xyz\right\vert }\varepsilon
_{z}^{B}\circ {}_{x}\widetilde{R}_{z}^{y}\quad \text{and}\quad
_{x}\triangleleft _{z}^{y}:=({}_{x}\varepsilon _{\left\vert xyz\right\vert
}^{A}B_{z})\circ {}_{x}\widetilde{R}_{z}^{y}. \label{lr}
\end{equation}
In view of Lemma \ref{le:coalgebra}, $_{x}\triangleright _{z}^{y}$ and $
_{x}\triangleleft _{z}^{y}$ are coalgebra morphisms and they satisfy the
relations
\begin{align}
(_{x}\triangleright _{z}^{y}\otimes {}_{x}\triangleleft _{z}^{y})\circ
\Delta _{_{x}B_{y}A_{z}}& ={}_{x}\widetilde{R}_{z}^{y}, \label{mp1} \\
(_{x}\triangleleft _{z}^{y}\otimes {}_{x}\triangleright _{z}^{y})\circ
\Delta _{{_{x}B_{y}A_{z}}}& =\chi _{_{x}A_{|xyz|},_{|xyz|}B_{z}}\circ
(_{x}\triangleright _{z}^{y}\otimes {}_{x}\triangleleft _{z}^{y})\circ
\Delta _{_{x}B_{y}A_{z}}. \label{mp2}
\end{align}
Conversely, if one starts with $\triangleright :=\{_{x}\triangleright
_{z}^{y}\}_{x,y,z\in S}$ and ${\triangleleft :=}$ $\{_{x}{\triangleleft }
_{z}^{y}\}_{x,y,z\in S},$ two families of coalgebra maps that satisfy (\ref
{mp2}), then by formula (\ref{mp1}) we get the set $\widetilde{R}:={}\{_{x}
\widetilde{R}_{z}^{y}\}_{x,y,z\in S}$ whose elements are coalgebra maps, cf.
Lemma \ref{le:coalgebra}. Therefore, there is an one-to-one correspondence
between the couples $(\triangleright ,{\triangleleft )}$ and the sets $
\widetilde{R}$ as above. Our goal is to characterize those couples $
(\triangleright ,{\triangleleft )}$ that corresponds to a simple twisting
system in $\boldsymbol{{M^{\prime }}}.$
\end{fact}
\begin{lemma}
\label{le:lr}The statements below are true.
\begin{enumerate}
\item If $\left\vert xy\left\vert yzt\right\vert \right\vert =\left\vert
xzt\right\vert $ then the relation (\ref{cor1-1}) is equivalent to the
following equations:
\begin{align}
{_{x}\triangleleft }_{t}^{z}\circ {}_{x}b_{z}^{y}A_{t}& ={}_{\left\vert
xzt\right\vert }b_{t}^{\left\vert yzt\right\vert }\circ {_{x}\triangleleft }
_{\left\vert yzt\right\vert }^{y}B_{t}\circ (_{x}B_{y}\otimes {
_{y}\triangleright }_{t}^{z}\otimes {_{y}\triangleleft }_{t}^{z})\circ
(_{x}B_{y}\otimes \Delta _{{}_{y}B_{z}A_{t}}), \label{lr1-1} \\
{_{x}\triangleright }_{t}^{z}\circ {}_{x}b_{z}^{y}A_{t}& ={_x\triangleright}
_{|yzt|}^{y}\circ {}_{x}B{_y\triangleright} _{t}^{z}. \label{lr1-2}
\end{align}
\item If $\left\vert xyz\left\vert zt\right\vert \right\vert =\left\vert
xyt\right\vert $ then the relation (\ref{cor1-2}) is equivalent to the
following equations:
\begin{align}
{_{x}\triangleright }_{t}^{y}\circ {}_{x}B_{y}a_{t}^{z}&
={}_{x}a_{\left\vert xyt\right\vert }^{\left\vert xyz\right\vert }\circ
{}_{x}A_{\left\vert xyz\right\vert }\triangleright _{t}^{z}\circ \;({
_x\triangleright} _{z}^{y}\otimes {_{x}\triangleleft} _{z}^{y}\otimes
{}_{z}A_{t})\circ (\Delta _{{_{x}B_{y}A_{z}}}\otimes {}_{z}A_{t}),
\label{lr2-1} \\
{_{x}\triangleleft }_{t}^{y}\circ {}_{x}B_{y}a_{t}^{z}& ={_{\left\vert
xyz\right\vert }\triangleleft }_{t}^{z}\circ \;{_{x}\triangleleft }
_{z}^{y}A_{t}. \label{lr2-2}
\end{align}
\item If $\left\vert xyy\right\vert =x$ then the relation (\ref{cor1-3}) is
equivalent to the following equations:
\begin{align}
{_{x}\triangleleft }_{y}^{x}\circ \left( 1_{x}^{B}\otimes {}_{x}A_{y}\right)
& ={}_{x}{}\varepsilon _{y}^{A}\otimes 1_{y}^{B}, \label{lr3-1} \\
{_{x}\triangleright }_{y}^{x}\circ \left( 1_{x}^{B}\otimes
{}_{x}A_{y}\right) & ={_{x}A}_{y}{}. \label{lr3-2}
\end{align}
\item If $\left\vert xxy\right\vert =y$ then the relation (\ref{cor1-4}) is
equivalent to the following equations:
\begin{align}
{_{x}\triangleleft }_{y}^{y}\circ \left( _{x}B_{y}\otimes 1_{y}^{A}{}\right)
& ={_{x}B}_{y}, \label{lr4-1} \\
{_{x}\triangleright }_{y}^{y}\circ \left( _{x}B_{y}\otimes 1_{y}^{A}\right)
& =1_{x}^{A}\otimes {}_{x}\varepsilon _{y}^{B}{}. \label{lr4-2}
\end{align}
\end{enumerate}
\end{lemma}
\begin{proof}
In order to prove the first statement we apply the Remark \ref{re:f=g} to
\begin{equation*}
f^{\prime }:={}_{x}\widetilde{R}_{t}^{z}\circ {}_{x}b_{z}^{y}A_{t}\quad
\text{and}\quad f^{\prime \prime }:={}_{x}A_{|xzt|}b_{t}^{|yzt|}\circ {}_{x}
\widetilde{R}_{|yzt|}^{y}B_{t}\circ {}_{x}B_{y}\widetilde{R}_{t}^{z}.
\end{equation*}
Note that $f^{\prime \prime }$ is well defined and its target is $
_{x}A_{|xzt|}B_{t},$ since the codomain of ${}_{x}\widetilde{R}
_{|yzt|}^{y}B_{t}\circ {}_{x}B_{y}\widetilde{R}_{t}^{z}$ is $
_{x}A_{\left\vert xy\left\vert yzt\right\vert \right\vert }B_{t}$ and $
\left\vert xy\left\vert yzt\right\vert \right\vert =\left\vert
xzt\right\vert $. Clearly, $f^{\prime }$ and $f^{\prime \prime }$ are
coalgebra morphisms, since the composite and the tensor product of two
morphisms in $\boldsymbol{M}$ remain in $\boldsymbol{M}.$ An easy
computation, based on the equation (\ref{mp1}) and the formulae of $
_{x}\triangleright _{z}^{y}$ and $_{x}\triangleleft _{z}^{y}$, yields us
\begin{align*}
_{x}\varepsilon _{|xzt|}^{A}B_{t}\circ f^{\prime }& ={}_{x}\triangleleft
_{t}^{z}\circ {}_{x}b_{z}^{y}A_{t}, \\
_{x}A_{|xzt|}\varepsilon _{t}^{B}\circ f^{\prime }& ={}_{x}\triangleright
_{t}^{z}\circ {}_{x}b_{z}^{y}A_{t}, \\
_{x}\varepsilon _{|xzt|}^{A}B_{t}\circ f^{\prime \prime }& ={}_{\left\vert
xzt\right\vert }b_{t}^{\left\vert yzt\right\vert }\circ {}_{x}\triangleleft
_{\left\vert yzt\right\vert }^{y}B_{t}\circ (_{x}B_{y}\otimes
{}_{y}\triangleright _{t}^{z}\otimes {}_{y}\triangleleft _{t}^{z})\circ
(_{x}B_{y}\otimes \Delta {}_{\,_{y}B_{z}A_{t}}).
\end{align*}
Taking into account that $_{x}b_{z}^{y}$ is a coalgebra morphism and using
the definition of $_{x}\triangleright _{z}^{y}$ we get
\begin{equation*}
_{x}A_{|xzt|}\varepsilon _{t}^{B}\circ f^{\prime \prime
}={}_{x}A_{\left\vert xzt\right\vert }\varepsilon _{\left\vert
yzt\right\vert }^{B}\circ {}_{x}\widetilde{R}_{\left\vert yzt\right\vert
}^{y}\circ \left( _{x}B_{y}\otimes (_{y}A_{\left\vert yzt\right\vert
}\varepsilon _{t}^{B}\circ {}_{y}\widetilde{R}_{t}^{z})\right)
={}_{x}\triangleright _{|yzt|}^{y}\circ {}_{x}B_{y}\triangleright _{t}^{z}.
\end{equation*}
In view of the Remark \ref{re:f=g}, we have $f^{\prime }=f^{\prime \prime }$
if and only if
\begin{equation*}
_{x}A_{|xzt|}\varepsilon _{t}^{B}\circ f^{\prime
}={}_{x}A_{|xzt|}\varepsilon _{t}^{B}\circ f^{\prime \prime }\quad \text{
and\quad }_{x}\varepsilon _{|xzt|}^{A}B_{t}\text{ }\circ f^{\prime
}={}_{x}\varepsilon _{|xzt|}^{A}B_{t}\circ f^{\prime \prime }.
\end{equation*}
Thus, if $\left\vert xy\left\vert yzt\right\vert \right\vert =\left\vert
xzt\right\vert ,$ then (\ref{cor1-1}) is equivalent to (\ref{lr1-1})
together with (\ref{lr1-2}). We omit the proof of the second statement,
being similar.
To prove the third part of the lemma we reiterate the above reasoning. We
now take $f^{\prime }$ and $f^{\prime \prime }$ to be the coalgebra morphisms
\begin{equation*}
f^{\prime }:={}_{x}\widetilde{R}_{y}^{x}\circ \left( 1_{x}^{B}\otimes
{}_{x}A_{y}\right) \quad \text{and\quad }f^{\prime \prime
}:={}_{x}A_{y}{}\otimes 1_{y}^{B}.
\end{equation*}
Since $\left\vert xxy\right\vert =y$ both $f^{\prime }$and $f^{\prime \prime
}$ target in $_{x}A_{y}B_{y}.$ It is easy to see that (\ref{lr3-1}) together
with (\ref{lr3-2}) are equivalent to (\ref{cor1-3}). Similarly, one shows
that the fourth statement is true.
\end{proof}
\begin{theorem}
\label{thm:mp1}We keep the notation and the assumptions from \S \ref
{coalg(M')}. The set $\widetilde{R}$ is a simple twisting system in $
\boldsymbol{{M^{\prime }}}$ if and only if the families $\triangleright $
and ${\triangleleft }$ satisfy the following conditions:
\end{theorem}
\begin{enumerate}
\item[(i)] If $_{x}B_{y}A_{z}$ is not an initial object then $
_{x}A_{\left\vert xyz\right\vert }B_{z}$ is not an initial object as well.
\item[(ii)] If $_{x}B_{y}B_{z}A_{t}$ is not an initial object in $
\boldsymbol{{M^{\prime }}},$ then $|xy|yzt||=|xzt|$ and the equations (\ref
{lr1-1}) and (\ref{lr1-2}) hold.
\item[(iii)] If $_{x}B_{y}A_{z}A_{t}$ is not an initial object in $
\boldsymbol{{M^{\prime }}},$ then $||xyz|zt|=|xyz|$ and the equations (\ref
{lr2-1}) and (\ref{lr2-2}) hold.
\item[(iv)] If $_{x}A_{y}$ is not an initial object in $\boldsymbol{{
M^{\prime }}},$ then $|xxy|=y$ and the equations (\ref{lr3-1}) and (\ref
{lr3-2}) hold.
\item[(v)] If $_{x}B_{y}$ is not an initial object in $\boldsymbol{{
M^{\prime }}},$ then $|xyy|=x$ and the equations (\ref{lr4-1}) and (\ref
{lr4-2}) hold.
\end{enumerate}
\begin{proof}
The condition (i) is a part of the definition of simple twisting systems. If
$_{x}B_{y}B_{z}A_{t}$ is not an initial object in $\boldsymbol{M^{\prime }}$
then we may assume that $|xy|yzt||=|xzt|.$ Thus, by Lemma \ref{le:lr}, the
relation (\ref{cor1-1}) and the equations (\ref{lr1-1}) and (\ref{lr1-2})
are equivalent. To conclude the proof we proceed in a similar way.
\end{proof}
\begin{fact}[Matched pairs and the bicrossed product.]
\label{fa:MP} Let ${\triangleright :=}\{_{x}\triangleright
_{z}^{y}\}_{x,y,z\in S}$ and $\triangleleft :=\{_{x}\triangleleft
_{z}^{y}\}_{x,y,z\in S}$ be two families of maps as in \S \ref{coalg(M')}.
We shall say that the quintuple $(\boldsymbol{A},\boldsymbol{B},{
\triangleright },\triangleleft ,\left\vert \cdots \right\vert )$ is a\emph{\
matched pair} of $\boldsymbol{M}$-categories if and only if $\triangleright $
and ${\triangleleft }$ satisfy the conditions (i)-(v) from the above
theorem. For a matched pair $(\boldsymbol{A},\boldsymbol{B},{\triangleright }
,\triangleleft ,\left\vert \cdots \right\vert )$ we have just seen that $(
\widetilde{R},\left\vert \cdots \right\vert )$ is a simple twisting system
in $\boldsymbol{{M^{\prime }}}$, where $\widetilde{R}:=\{_{x}\widetilde{R}
_{z}^{y}\}_{x,y,z\in S}$ is the set of coalgebra morphisms which are defined
by the formula (\ref{mp1}). Hence, supposing that $\boldsymbol{{M^{\prime }}}
$ is an $S$-distributive domain, we may construct the twisted tensor product
$\boldsymbol{A}\otimes _{R}\boldsymbol{B},$ which is an enriched category
over $\boldsymbol{{M^{\prime }}}.$ We shall call it the \emph{bicrossed
product} of $(\boldsymbol{A},\boldsymbol{B},{\triangleright },\triangleleft
,\left\vert \cdots \right\vert )$ and we shall denote it by $\boldsymbol{A}
\Join \boldsymbol{B}.$
\end{fact}
\begin{proposition}
The bicrossed product of a matched pair $(\boldsymbol{A},\boldsymbol{B},{
\triangleright },\triangleleft ,\left\vert \cdots \right\vert )$ is enriched
over the monoidal category $\boldsymbol{M}:=\boldsymbol{Coalg(M^{\prime })}.$
\end{proposition}
\begin{proof}
Let $\{C_{i}\}_{i\in i}$ be a family of coalgebras in $\boldsymbol{{
M^{\prime }.}}$ Let us assume that the underlying family of objects has a
coproduct $C:=\textstyle\bigoplus {}_{x\in S}C_{i\text{ }}$ in $\boldsymbol{
M^{\prime }}.$ Let $\{\sigma _{i}\}_{i\in I}$ by the set of canonical
inclusions into $C.$ There are unique maps $\Delta :C\rightarrow C\otimes C$
and $\varepsilon :C\rightarrow \boldsymbol{1}$ such that
\begin{equation*}
\Delta \circ \sigma _{i}=(\sigma _{i}\otimes \sigma _{i})\circ \Delta
_{i}\quad \text{and\quad }\varepsilon \circ \sigma _{i}=\varepsilon _{i},
\end{equation*}
for all $i\in I,$ where $\Delta _{i}$ and $\varepsilon _{i}$ are the
comultiplication and the counit of $C_{i}.$ It is easy to see that $
(C,\Delta ,\varepsilon )$ is a coalgebra in $\boldsymbol{M^{\prime }}.$ Note
that, by the construction of the coalgebra structure on $C,$ the inclusion $
\sigma _{i}$ is a coalgebra map, for any $i\in I.$ Furthermore, let $
f_{i}:C_{i}\rightarrow D$ be a coalgebra morphism for every $i\in I.$ By the
universal property of the coproduct there is a unique map $f:C\rightarrow D$
in $\boldsymbol{{M^{\prime }}}$ such that $f\circ \sigma _{i}=f_{i},$ for
all $i\in I.$ It is not difficult to see that $f$ is a morphism of
coalgebras, so $(C,\{\sigma _{i}\}_{i\in I})$ is the coproduct of $
\{C_{i}\}_{i\in I}$ in $\boldsymbol{M}$.
In particular, $_{x}A_{\overline{u}}B_{y}=\bigoplus_{u\in S}{}_{x}A_{
\overline{u}}B_{y}$ has a unique coalgebra structure such that the inclusion
$_{x}\sigma _{z}^{u}:{}_{x}A_{u}B_{y}\rightarrow {}_{x}A_{\overline{u}}B_{y}$
is a coalgebra map, for all $x,$ $y$ and $u$ in $S$. Recall that for the
construction of the composition map $_{x}c_{z}^{y}:{}_{x}A_{\overline{u}
}B_{y}A_{\overline{v}}B_{z}\rightarrow {}_{x}A_{\overline{w}}B_{z}$ one
applies the universal property of the coproduct to $\{f_{u,v}\}_{u,v\in
S^{2}},$ where
\begin{equation*}
f_{u,v}={}_{x}\sigma _{z}^{\left\vert uyv\right\vert }\circ
{}_{x}a_{\left\vert uyv\right\vert }b_{z}\circ {}_{x}A_{u}\widetilde{R}
_{v}^{y}B_{z}.
\end{equation*}
Since $\boldsymbol{A}$ and $\boldsymbol{B}$ are $\boldsymbol{M}$-categories
and $_{u}\widetilde{R}_{v}^{y}$ is a coalgebra map, in view of the foregoing
remarks, it follows that $_{x}c_{z}^{y}$ is a morphism in $\boldsymbol{M},$
for all $x,y,z\in S.$ The identity map of $x$ in $\boldsymbol{A}\Join
\boldsymbol{B}$ is the coalgebra map $_{x}\sigma _{x}^{x}\circ
(1_{x}^{A}\otimes 1_{x}^{B}).$ In conclusion, $\boldsymbol{A}\Join
\boldsymbol{B}$ is enriched over $\boldsymbol{M}.$
\end{proof}
\section{Examples.}
In this section we give some examples of (simple) twisting systems. We start
by considering the case of $\boldsymbol{Set}$-categories, that is usual
categories.
\begin{fact}[Simple twisting systems of enriched categories over $
\boldsymbol{Set.}$]
\label{fa:sts} The category $\boldsymbol{Set}$ is a braided monoidal
category with respect to the cartesian product, its unit object being $
\{\emptyset \}$. Clearly, the empty set is the initial object in $
\boldsymbol{Set},$ and this category is an $S$-distributive domain, for any
set $S.$ We have already noticed that the (\dag ) hypothesis holds in $
\boldsymbol{Set}.$
Let $\boldsymbol{C}$ be an enriched category over $\boldsymbol{Set}.$ Thus,
by definition, $\boldsymbol{C}$ is a category in the usual sense, that is $
_{x}{C}_{y}$ is a set for all $x,y\in S.$ An element in $_{x}{C}_{y}$ is
regarded as a function from $y$ to $x$.
It is easy to see that a given set $X$ can be seen in a unique way as a
coalgebra in $\boldsymbol{Set}$. As a matter of fact the comultiplication
and the counit of this coalgebra are given by the diagonal map $\Delta
:X\rightarrow X\times X$ and the constant map $\varepsilon :X\rightarrow
\{\emptyset \},$
\begin{equation*}
\Delta (x)=x\otimes x\text{\quad and\quad }\varepsilon (x)=\emptyset .
\end{equation*}
Obviously, any function $f:X\rightarrow Y$ is morphism of coalgebras in $
\boldsymbol{Set}$. Consequently, any category $\boldsymbol{C}$ may be seen
as an enriched category over $\boldsymbol{Coalg(Set)}$,
Our aim is to describe the simple twisting systems between two categories $
\boldsymbol{B}$ and $\boldsymbol{A}$. In view of the foregoing discussion
and of our results in the previous section, for any simple twisting system $
R:=\{_{x}\widetilde{R}_{z}^{y}\}_{x,y,z\in S}$ there is a unique matched
pair $(\boldsymbol{A},\boldsymbol{B},\triangleright ,{\triangleleft }
,\left\vert \cdots \right\vert ),$ and conversely. These structures are
related each other by the formulae (\ref{lr}) and (\ref{mp1}).
Since $\boldsymbol{A}$ is an usual category, the composition of morphisms
will be denoted in the traditional way $g\circ g^{\prime },$ for any $g\in
{}_{x}A_{y}$ and $g^{\prime }\in {}_{y}A_{z}$ (recall that the domain and
the codomain of $g$ are $y$ and $x,$ respectively). The same notation will
be used for $\boldsymbol{B}.$ On the other hand, for any $f\in {}_{x}B_{y}$
and $g\in {}_{y}A_{z}$ we shall write
\begin{equation*}
f\triangleright g:={}_{x}\triangleright {}_{z}^{y}(f,g)\quad \text{and}\quad
f\triangleleft g={}_{x}{\triangleleft }_{z}^{y}(f,g).
\end{equation*}
Since the comultiplication in this case is always the diagonal map, and the
counit is the constant map to $\{\emptyset \}$, the conditions of Theorem
\ref{thm:mp1} and the following ones are equivalent.
\begin{enumerate}
\item[(i)] If $_{x}B_{y}A_{z}$ is not empty then $_{x}A_{\left\vert
xyz\right\vert }B_{z}$ is not empty as well.
\item[(ii)] For any $(f,f^{\prime },g)\in {}_{x}B_{y}B_{z}A_{t}$ we have $
|xy|yzt||=|xzt|$, and
\begin{equation*}
(f\circ f^{\prime })\triangleright g=f\triangleright (f^{\prime
}\triangleright g)\qquad \text{and}\qquad (f\circ f^{\prime })\triangleleft
g=[f\triangleleft (f^{\prime }\triangleright g)]\circ (f^{\prime
}\triangleleft g).
\end{equation*}
\item[(iii)] For any $(f,g,g^{\prime })\in {}_{x}B_{y}A{}_{z}A_{t}$ we have $
||xyz|zt|=|xyt|$, and
\begin{equation*}
f\triangleleft (g\circ g^{\prime })=(f\triangleleft g)\triangleleft
g^{\prime }\qquad \text{and}\qquad f\triangleright (g\circ g^{\prime
})=(f\triangleright g)\circ \lbrack (f\triangleleft g)\triangleright
g^{\prime }].
\end{equation*}
\item[(iv)] For any $g\in {}_{x}A_{y}$ we have $|xxy|=y$, and
\begin{equation*}
1_{x}^{B}\triangleright g=g\text{\qquad and\qquad}1_{x}^{B}\triangleleft
g=1_{y}^{B}. \label{set5}
\end{equation*}
\item[(v)] For any $f\in {}_{x}B_{y}$ we have $|xyy|=x$, and
\begin{equation*}
f\triangleright 1_{y}^{A}{}=1_{x}^{B}\text{\qquad and\qquad}f\triangleleft
1_{y}^{A}=f. \label{set6}
\end{equation*}
\end{enumerate}
In this case the bicrossed product $\boldsymbol{A}\Join \boldsymbol{B}$ is
the category whose hom-sets are given by
\begin{equation*}
_{x}(\boldsymbol{A}\Join \boldsymbol{B})_{y}=\textstyle\coprod\limits_{u\in
S}{}_{x}{A}_{u}{B}_{y}.
\end{equation*}
The identity of $x$ in $\boldsymbol{A}\Join \boldsymbol{B}$ is $
(1_{x}^{A},1_{x}^{B}).$ For $(g,f)\in $ $_{x}{A}_{u}{B}_{y}$ and $(g^{\prime
},f^{\prime })\in {}_{y}{A}{}_{v}{B}_{z}$ we have
\begin{equation*}
(g,f)\circ (g^{\prime },f^{\prime })=\left( g\circ (f\triangleright
g^{\prime }),(f\triangleleft g^{\prime })\circ f^{\prime }\right) .
\end{equation*}
\end{fact}
\begin{remark}
R. Resebrugh and R.J. Wood showed that every twisting systems between two
$\boldsymbol{Set}$-categories $\boldsymbol{B}$ and $\boldsymbol{A}$ is completely determined by
a left action $\triangleright $ of $\boldsymbol{B}$ on $\boldsymbol{A}$ and
a right action $\triangleleft $ of $\boldsymbol{A}$ on $\boldsymbol{B.}$
More precisely, given a twisting system $R=\{_{x}R_{z}^{y}\}_{x,y,z\in S}$
and the morphisms $f\in {}_{x}B_{y}$ and $g\in {}_{y}A_{z},$ then $
_{x}R_{z}^{y}(f,g)$ is an element in $_{x}A_{u}B_{z},$ where $u$ is a
certain element of $S.$ Hence, there are unique morphisms $f\triangleright
g\in {}_{x}A_{u}$ and $f\triangleleft g\in {}_{u}B_{z}$ such that
\begin{equation*}
_{x}R_{z}^{y}(g,f)=\left( f\triangleright g,f\triangleleft g\right) .
\end{equation*}
The actions $\triangleright $ and
$\triangleleft $ must satisfy several compatibility conditions, which are similar
to those that appear in the above characterization of simple twisting systems.
For details the reader is referred to the second section of \cite{RW}.
\end{remark}
\begin{fact}[The bicrossed product of two groupoids.]
\label{fa:gr} We now assume that $(\boldsymbol{A},\boldsymbol{B}
,\triangleright ,{\triangleleft ,}|{\cdots }|)$ is a matched pair of
groupoids. Recall that a groupoid is a category whose morphisms are
invertible. We claim that $\boldsymbol{A}\Join \boldsymbol{B}$ is also a
groupoid. Indeed, as in the case of monoids, one can show that a category is
a groupoid if and only if every morphism is right invertible (or left
invertible). Since
\begin{equation*}
_{x}(\boldsymbol{A}\Join \boldsymbol{B})_{y}=\textstyle\coprod\limits_{u\in
S}{_{x}{A}}_{u}{B}_{y},
\end{equation*}
it is enough to prove that $(g,f)$ is right invertible, where $g\in
{}_{x}A_{u}$ and $f\in {}_{u}B_{y}$ are arbitrary morphisms. Therefore, we
are looking for a pair $(g^{\prime },f^{\prime })\in {}_{y}A_{v}\times $ $
{}_{v}B_{x}$ such that
\begin{equation*}
g\circ (f\triangleright g^{\prime })=1_{x}^{A}\qquad \text{and}\qquad
(f\triangleleft g^{\prime })\circ f^{\prime }=1_{x}^{B}.
\end{equation*}
Since $g$ is an invertible morphism in $_{x}A_{u}$ we get that $
f\triangleright g^{\prime }=g^{-1}\in {}_{u}A_{x}.$ Since $f$ is invertible
too,
\begin{equation*}
g^{\prime }=1_{y}^{A}\triangleright g^{\prime }=(f^{-1}\circ
f)\triangleright g^{\prime }=f^{-1}\triangleright (f\triangleright g^{\prime
})=f^{-1}\triangleright g^{-1}.
\end{equation*}
As $g^{\prime }\in {}_{y}A_{v}$ and $f^{-1}\triangleright g^{-1}\in
{}_{y}A_{|yux|}$ we must have $v=|yux|$. Thus we can now take
\begin{equation*}
f^{\prime }=[f\triangleleft (f^{-1}\triangleright g^{-1})]^{-1}\in
{}_{|yux|}B_{x}.
\end{equation*}
\end{fact}
\begin{fact}[The smash product category.]
We take $\boldsymbol{M}$ to be the monoidal category $\mathbb{K}$-$
\boldsymbol{Mod},$ where $\mathbb{K}$ is a commutative ring. Hence in this
case we work with $\mathbb{K}$-linear categories. Let $H$ be a $\mathbb{K}$
-bialgebra. We define an enriched category $\boldsymbol{H}$ over $\mathbb{K}$
-$\boldsymbol{Mod}$ by setting $_{x}H_{x}=H$ and $_{x}H_{y}=0,$ for $x\neq
y. $ The composition of morphisms in $\boldsymbol{H}$ is given by the
multiplication in $H$ and the identity of $x$ is the unit of $H$. For the
comultiplication of $H$ we shall use the $\Sigma $-notation
\begin{equation*}
\Delta (h)=\textstyle\sum {}h_{(1)}\otimes h_{(2)}.
\end{equation*}
Let $\boldsymbol{A}$ denote an $H$-module category, i.e. a category enriched
in $H$-$\boldsymbol{Mod}$. Thus $H$ acts on $_{x}A_{y}$, for any $x,y\in S$,
and the composition and the identity maps in $\boldsymbol{A}$ are $H$-linear
morphisms. Obviously, $\boldsymbol{A}$ is a $\mathbb{K}$-linear category.
Our aim is to associate to $\boldsymbol{A}$ a simple twisting system ${R}
=\{_{x}\widetilde{R}_{z}^{y}\}_{x,y,z\in S}$. First we define $|\cdots
|:S^{3}\rightarrow S$ by $|xyz|=z.$ Then, using the actions $\cdot :H\otimes
{}_{x}A_{z}\rightarrow {}_{x}A_{z\text{,}}$ we define
\begin{equation*}
_{x}\widetilde{R}_{z}^{x}:H\otimes {}_{x}A_{z}\rightarrow {}_{x}A_{z}\otimes
H,\quad _{x}\widetilde{R}_{z}^{x}(h\otimes f)=\textstyle\sum {}h_{(1)}\cdot
f\otimes h_{(2)}.
\end{equation*}
For $x\neq y$ we take $_{x}\widetilde{R}_{z}^{y}=0$. It is easy to see that $
{R}$ is a simple twisting system of $\mathbb{K}$-linear categories. Clearly $
\mathbb{K}$-$\boldsymbol{Mod}$ is $S$-distributive, for any set $S.$ If $
\mathbb{K}$ is a field then $\mathbb{K}$-$\boldsymbol{Mod}$ is a domain, so
in this case the twisted tensor product of $\boldsymbol{A}$ and $\boldsymbol{
H}$ with respect to $R$ makes sense, cf. Theorem \ref{thm:TTP}. It is called
the smash product of $\boldsymbol{A}$ by $H$, and it is denoted by $
\boldsymbol{A}\#H.$ By definition, $_{x}(\boldsymbol{A}\#H)_{y}={}_{x}A_{y}
\otimes H$ and
\begin{equation}
(f\otimes h)\circ (f^{\prime }\otimes h^{\prime })=\textstyle\sum {}f\circ
(h_{(1)}\mathbf{\cdot }\text{ }f^{\prime })\otimes h_{(2)}h^{\prime },
\label{smash}
\end{equation}
for any $f\in {}_{x}A_{y}$, $f^{\prime }\in {}_{y}A_{z}$ and $h,h^{\prime
}\in H.$
\end{fact}
\begin{fact}[The semidirect product.]
\label{sdp} Let $\boldsymbol{A}$ be a category. Let us suppose that $({B}
,\cdot ,{1})$ is a monoid that acts to the left on each $_{x}A_{y}$ via
\begin{equation*}
\triangleright :B\times {}_{x}A_{y}\rightarrow {}_{x}A_{y}.
\end{equation*}
We define the category $\boldsymbol{B}$ so that $_{x}B_{x}=B$ and ${}_{x}{B}
_{y}=\emptyset $, for $x\neq y$. The composition of morphisms is given by
the multiplication in $\boldsymbol{B}.$ To this data we associate a matched
pair $(\boldsymbol{A},\boldsymbol{B},|{\cdots }|,\triangleright
,\triangleleft ),$ setting $f\triangleleft g=f$ for any $(f,g)\in
{}_{x}B_{x}A_{y},$ and defining the function $\left\vert \cdots \right\vert
:S^{3}\rightarrow S$ by $|xyz|=z.$ Note that if $x\neq y$ then $
_{x}B{}_{y}A_{z}$ is empty, so $_{x}\triangleright _{z}^{y}$ and $_{x}{
\triangleleft }_{z}^{y}$ coincide with the empty function. One shows that $(
\boldsymbol{A},\boldsymbol{B},\triangleright ,\triangleleft ,|{\cdots }|)$
is a matched pair if and only if for any $(g,g^{\prime })\in
{}_{x}A_{y}A_{z} $ and $f\in B$
\begin{align*}
f\triangleright (g\circ g^{\prime })& =(f\triangleright g)\circ
(f\triangleright g^{\prime }), \\
f\triangleright 1_{x}^{A}& =1_{x}^{A}.
\end{align*}
The corresponding bicrossed product will be denoted in this case by $
\boldsymbol{A\rtimes }B$. If $\left\vert S\right\vert =1$ then $\boldsymbol{A
}$ can be identified with a monoid and $\boldsymbol{A}\rtimes B$ is the
usual semidirect product of two monoids. For this reason we shall call $
\boldsymbol{A}\rtimes B$ the semidirect product of $\boldsymbol{A}$ with $B.$
Note that $_{x}(\boldsymbol{A}\rtimes B)_{y}={}_{x}A_{y}\times {}B.$ For any
$f,f^{\prime }\in B$ and $(g,g^{\prime })\in {}_{x}A_{y}A_{z}$, the
composition of morphisms in $\boldsymbol{A}\rtimes B$ is given by
\begin{equation*}
(g,f)\circ (g^{\prime },f^{\prime })=(g\circ (f\triangleright g^{\prime
}),f\circ f^{\prime }).
\end{equation*}
\end{fact}
\begin{fact}[Twisting systems between algebras in $\boldsymbol{M}.$]
We now consider a twisting system ${R}$ between two $\boldsymbol{M}$
-categories $\boldsymbol{B}$ and $\boldsymbol{A}$ with the property that ${S}
=\{x_{0}\}$. Obviously, $\boldsymbol{M}$ is $S$-distributive. We shall use
the notation $A={}_{x_{0}}A_{x_{0}}$ and $B={}_{x_{0}}B_{x_{0}}.$ The
composition map $a:={}_{x_{0}}a_{x_{0}}^{x_{0}}$ and $1_{A}:=1_{x_{0}}^{A}$
define an algebra structure on $A$. A similar notation will be used for the
algebra corresponding to the $\boldsymbol{M}$-category $\boldsymbol{B}.$ Let
$R$ be a morphism from $B\otimes A$ to $A\otimes B.$
Since $_{x_{0}}\sigma _{x_{0}}^{x_{0}}$ is the identity map of $B\otimes A,$
by Proposition \ref{thm R simplu}, we deduce that $
{}_{x_{0}}R_{x_{0}}^{x_{0}}=R$ defines a twisting system between $
\boldsymbol{B}$ and $\boldsymbol{A}$ if and only if $R$ satisfies the
relations (\ref{cor1-1})-(\ref{cor1-4}) with respect to the unique map $
\left\vert \cdots \right\vert :S^{3}\rightarrow S.$ In turn, they are
equivalent to the following identities
\begin{align}
R\circ (b\otimes A)& =(A\otimes b)\circ (R\otimes B)\circ (B\otimes R),
\label{fm1} \\
R\circ (B\otimes a)& =(a\otimes B)\circ (A\otimes R)\circ (R\otimes A),
\label{fm2} \\
R\circ (1_{B}^{{}}\otimes A)& =A\otimes 1_{B}^{{}}, \label{fm3} \\
R\circ (B\otimes 1_{A})& =1_{A}\otimes B. \label{fm4}
\end{align}
In conclusion, in the case when $\left\vert S\right\vert =1,$ to give a
twisting system between $\boldsymbol{B}$ and $\boldsymbol{A}$ is equivalent
to give a \emph{twisting map} between the algebras $B$ and $A,$ that is a
morphism $R$ which satisfies (\ref{fm1})-(\ref{fm4}).
By applying Corollary \ref{cor: TTP} to a twisting map $R:B\otimes
A\rightarrow A\otimes B$ (viewed as a twisting system between two $
\boldsymbol{M}$-categories with one object) we get the \emph{twisted tensor
algebra }$A\otimes _{R}B.$ The unit of this algebra is $1_{A}\otimes 1_{B}$
and its multiplication is given by
\begin{equation*}
m=(a\otimes b)\circ (A\otimes R\otimes B).
\end{equation*}
Note that, in view of the foregoing remarks, an algebra $C$ in $\boldsymbol{M
}$ factorizes through $A$ and $B$ if and only if it is isomorphic to a
twisted tensor algebra $A\otimes _{R}B,$ for a certain twisting map $R.$
An algebra in the monoidal category $\mathbb{K}$-$\boldsymbol{Mod}$ is by
definition an associative and unital $\mathbb{K}$-algebra. Twisted tensor $
\mathbb{K}$-algebras were investigated for instance in \cite{Ma1}, \cite{Tam}
, \cite{CSV}, \cite{CIMZ}, \cite{LPvO} and \cite{JLPvO}.
Coalgebras over a field $\mathbb{K}$ are algebras in the monoidal category $(
\mathbb{K}$-$\boldsymbol{Mod})^{o}.$ Hence a twisting map between two
coalgebras $(A,\Delta _{A},\varepsilon _{A})$ and $(B,\Delta
_{B},\varepsilon _{B})$ is a $\mathbb{K}$-linear map
\begin{equation*}
R:A\otimes _{\mathbb{K}}B\rightarrow B\otimes _{\mathbb{K}}A
\end{equation*}
which satisfies the equations that are obtained from \eqref{fm1}-\eqref{fm4}
by making the substitutions $a:=\Delta _{A},$ $b:=\Delta _{B},$ $
1_{A}:=\varepsilon _{A}$ and $1_{B}:=\varepsilon _{B},$ and reversing the
order of the factors with respect to the composition in $\boldsymbol{M}$.
For example \eqref{fm1} should be replaced with
\begin{equation*}
(\Delta _{B}\otimes _{\mathbb{K}}A)\circ R=(B\otimes _{\mathbb{K}}R)\circ
(R\otimes _{\mathbb{K}}B)\circ (A\otimes _{\mathbb{K}}\Delta _{B}).
\end{equation*}
Obviously $A\otimes _{R}B$ is the $\mathbb{K}$-coalgebra $(A\otimes _{
\mathbb{K}}B,\Delta ,\varepsilon )$, where $\varepsilon :=\varepsilon
_{A}\otimes $ $\varepsilon _{B}$ and
\begin{equation*}
\Delta =(A\otimes R\otimes B)\circ (\Delta _{A}\otimes _{\mathbb{K}}\Delta
_{B}).
\end{equation*}
An algebra in $\Lambda $-$\boldsymbol{Mod}$-$\Lambda $ is called a $\Lambda $
-\emph{ring}. Specializing $\boldsymbol{M}$ to $\Lambda $-$\boldsymbol{Mod}$-
$\Lambda $ we find the definition of the \emph{twisted tensor }$\Lambda $
\emph{-ring}. Dually, $\Lambda $-corings are algebras in $(\Lambda $-$
\boldsymbol{Mod}$-$\Lambda )^{o}$. Thus in this particular case we are led
to the construction of the \emph{twisted tensor} $\Lambda $-\emph{coring}.
By definition a monad on a category $\boldsymbol{A}$ is an algebra in $[
\boldsymbol{A},\boldsymbol{A}]$. If $(F,\mu _{F},\iota _{F})$ and $(G,\mu
_{G},\iota _{G})$ are monads in $\boldsymbol{M}$, then a natural
transformation
\begin{equation*}
\lambda :G\circ F\rightarrow F\circ G
\end{equation*}
satisfies the relations \eqref{fm1}-\eqref{fm4} if and only if $\lambda $ is
a \emph{distributive law}, cf. \cite{Be}. Let $F^{2}:=F\circ F.$ For every
distributive law $\lambda $ we get a monad $(F\circ G,\mu ,\iota ),$ where $
\iota :=\iota _{F}G\circ \iota _{G}=F\iota _{G}\circ \iota _{F}$ and
\begin{equation*}
\mu :=\mu _{F}G\circ F^{2}\mu _{G}\circ F\lambda G=F\mu _{G}\circ \mu
_{F}G^{2}\circ F\lambda G.
\end{equation*}
Distributive laws between comonads can be defined similarly, or working in $[
\boldsymbol{A},\boldsymbol{A}]^{o}$.
Finally, twisting maps in $\boldsymbol{Opmon(\boldsymbol{M})}$ have been
considered in \cite{BV}. In loc. cit. the authors define a bimonad in $
\boldsymbol{M}$ as an algebra in $\boldsymbol{Opmon(\boldsymbol{M}})$. Hence
a twisting map between two bimonads is an opmonoidal distributive law
between the underlying monads. For any opmonoidal distributive law $\lambda $
between the bimonads $G$ and $F$, there is a canonical bimonad structure on
the endofunctor $F\circ G$. See \cite[Section 4]{BV} for details.
\end{fact}
\begin{fact}[Matched pairs of algebras in $\boldsymbol{Coalg({M^{\prime }}){.
}}$]
Let $\boldsymbol{M}$ denote the category of coalgebras in a braided monoidal
category $(\boldsymbol{{M^{\prime },}}\otimes ,\mathbf{1},\chi ).$ By
definition, a bialgebra in $\boldsymbol{M^{\prime }}$ is an algebra in $
\boldsymbol{M}$. We fix two bialgebras $(A,a,1_{A},\Delta_A,\varepsilon_A)$
and $(A,b,1_{B},\Delta_B,\varepsilon_B)$ in $\boldsymbol{M^{\prime }}$ and
take $R:B\otimes A\rightarrow A\otimes B $ to be a morphism in $\boldsymbol{M
}.$ By Lemma \ref{le:coalgebra}, there are the coalgebra maps $
\triangleright :B\otimes A\rightarrow A$ and ${\triangleleft }:B\otimes
A\rightarrow B$ such that
\begin{equation}
R=(\triangleright \otimes {\triangleleft })\circ \Delta _{B\otimes A}\qquad
\text{and\qquad }\chi _{A,B}\circ (\triangleright \otimes {\triangleleft }
)\circ \Delta _{B\otimes A}=({\triangleleft }\otimes \triangleright )\circ
\Delta _{B\otimes A}. \label{bi1}
\end{equation}
We have seen that $R$ is a twisting map in $\boldsymbol{M}$ if and only if
it satisfies the relations (\ref{fm1})-(\ref{fm4}). In view of Lemma \ref
{le:lr}, these equations are equivalent to the fact that $(A,\triangleright
) $ is a left $B$-module and $(B,{\triangleleft })$ is a right $A$-module
such that the following identities hold:
\begin{align}
{\triangleleft }\circ (b\otimes A)& =b\circ ({\triangleleft }\otimes B)\circ
(B\otimes \triangleright \otimes {\triangleleft })\circ (B\otimes \Delta
_{B\otimes A}), \label{bi2} \\
\triangleright \circ (B\otimes a)& =a\circ (A\otimes \triangleright )\circ
(\triangleright \otimes {\triangleleft }\otimes A)\circ (\Delta _{B\otimes
A}\otimes A), \label{bi3} \\
{\triangleleft }\circ (1_{B}\otimes A)& =\varepsilon _{A}\otimes 1_{B},
\label{bi4} \\
\triangleright \circ (B\otimes 1_{A})& =1_{A}\otimes \varepsilon _{B}.
\label{bi5}
\end{align}
By definition, a \emph{matched pair of bialgebras} \emph{in} $\boldsymbol{{
M^{\prime }}}$ consists of a left $B$-action $(A,\triangleright )$ and a
right $A$-action $(B,{\triangleleft })$ in $\boldsymbol{M}$ such that the
second equation in (\ref{bi1}) and the relations (\ref{bi2})-(\ref{bi5})
hold. For a matched pair of bialgebras we shall use the notation $
(A,B,\triangleright ,{\triangleleft ).}$
Summarizing, there is an one to-one-correspondence between twisting maps of
bialgebras in $\boldsymbol{{M^{\prime }}}$ and matched pairs of bialgebras
in $\boldsymbol{{M^{\prime }}}.$ If $(A,B,\triangleright ,{\triangleleft )}$
is a matched pair of bialgebras and $R$ is the corresponding twisting map,
then $A\otimes _{R}B$ will be called the \emph{bicrossed product of the
bialgebras }$A$ and $B,$ and it will be denoted by $A\Join B.$ Note that $
A\Join B$ is an algebra in $\boldsymbol{M}.$ Thus the bicrossed product of $
A $ and $B$ is a bialgebra in $\boldsymbol{{M^{\prime }}}.$ The unit of this
bialgebra is $1_{A}\otimes 1_{B}$ and the multiplication is given by
\begin{equation*}
m=(a\otimes b)\circ (A\otimes \triangleright \otimes {\triangleleft }\otimes
B)\circ (A\otimes \Delta _{B\otimes A}\otimes B).
\end{equation*}
As a coalgebra $A\Join B$ is the tensor product coalgebra of $A$ and $B.$ We
also conclude that a bialgebra $C$ in $\boldsymbol{{M^{\prime }}}$
factorizes through the sub-bialgebras $A$ and $B$ if and only if $C\cong
A\Join B$.
As a first application, let us take $\boldsymbol{M^{\prime }}$ to be the
category of sets, which is braided with respect to the braiding given by $
(X,Y)\mapsto (Y,X)$ and $(f,g)\mapsto (g,f)$, for any sets $X,Y$ and any
functions $f,g.$ We have already noticed that there is a unique coalgebra
structure on a given set $X$
\begin{equation*}
\Delta (x)=(x,x),\quad \quad \varepsilon (x)=\emptyset ,
\end{equation*}
where $\emptyset $ denotes the empty set; recall that the unit object in $
\boldsymbol{Set}$ is $\{\emptyset \}.$ Hence an ordinary monoid, i.e. an
algebra in $\boldsymbol{Set},$ has a natural bialgebra structure in this
braided category. Moreover any twisting map $R:B\times A\rightarrow A\times
B $ between two monoids $(A,\cdot ,1_{A})$ and $(B,\cdot ,1_{B})$ is a
twisting map of bialgebras in $\boldsymbol{Set}$. Let $(A,B,\triangleright ,{
\triangleleft )}$ be the corresponding matched pair. One easily shows that
the second condition in (\ref{bi1}) is always true. By notation, the
functions $\triangleright $ and ${\triangleleft }$ map $(f,g)\in B\times A$
to $f\triangleright g$ and $f{\triangleleft }g,$ respectively. Hence the
equations (\ref{bi2})-(\ref{bi5}) are equivalent to the following ones:
\begin{gather*}
(f\cdot f^{\prime })\triangleleft g=[f\triangleleft (f^{\prime
}\triangleright g)]\cdot (f^{\prime }\triangleleft g), \\
f\triangleright (g\cdot g^{\prime })=(f\triangleright g)\cdot \lbrack
(f\triangleleft g)\triangleright g^{\prime }], \\
1_{B}\triangleleft g=1_{B}\qquad \text{and\qquad }f\triangleright
1_{A}=1_{A}.
\end{gather*}
Since $R(f,g)=\left( f\triangleright g,f\triangleleft g\right) $ the product
of the monoid $A\Join B$ is defined by the formula
\begin{equation*}
\left( g,f\right) \cdot \left( g^{\prime },f^{\prime }\right) =\left(
g\left( f\triangleright g^{\prime }\right) ,\left( f\triangleleft g^{\prime
}\right) f^{\prime }\right) .
\end{equation*}
In conclusion, a monoid $C$ factorizes through $A$ and $B$ if and only if $
C\cong A\Join B.$
We have seen that the bicrossed product of two groupoids is a groupoid.
Thus, if $A$ and $B$ are groups, then $A\Join B$ is a group as well. This
result was proved by Takeuchi who introduced the matched pairs of groups in
\cite{Tak}.
We now consider the braided category $\mathbb{K}$-$\boldsymbol{Mod}$, whose
braiding is the usual flip map. An algebra in $\boldsymbol{M}$, the monoidal
category of $\mathbb{K}$-coalgebras, is a bialgebra over the ring $\mathbb{K}
$, and conversely. Proceeding as in the previous case, one shows that a
twisting map $R:B\otimes _{\mathbb{K}}A\rightarrow A\otimes _{\mathbb{K}}B$
of bialgebras is uniquely determined by the coalgebra maps $\triangleright
:B\otimes _{\mathbb{K}}A\rightarrow A$ and ${\triangleleft :B}\otimes _{
\mathbb{K}}A\rightarrow B$ via the formula
\begin{equation*}
R(f\otimes g)=\textstyle\sum {}(f_{(1)}\triangleright g_{(1)})\otimes
(f_{(2)}\triangleleft g_{(2)}).
\end{equation*}
Using the $\Sigma $-notation, the second equation in (\ref{bi1}) is true if
and only if
\begin{equation*}
\textstyle\sum {}(f_{(1)}\triangleleft g_{(1)})\otimes
(f_{(2)}\triangleright g_{(2)})=\textstyle\sum {}(f_{(2)}\triangleleft
g_{(2)})\otimes (f_{(1)}\triangleright g_{(1)}),
\end{equation*}
for any $f\in B$ and $g\in A.$ On the other hand, the equations (\ref{bi2})-(
\ref{bi5}) hold if and only if
\begin{align*}
(gg^{\prime })\triangleleft f& =\textstyle\sum {}[g\triangleleft
(g_{(1)}^{\prime }\triangleright f_{(1)})](g_{(2)}^{\prime }\triangleleft
f_{(2)}), \\
g\triangleright (ff^{\prime })& =\textstyle\sum {}(g_{(1)}\triangleright
f_{(1)})[(g_{(2)}\triangleleft f_{(2)})\triangleright f^{\prime }], \\
f\triangleright 1^{A}& =\varepsilon _{B}(f)1^{A}, \\
1^{B}\triangleleft g& =\varepsilon _{A}(g)1^{B},
\end{align*}
for any $f,f^{\prime }\in B$ and any $g,g^{\prime }\in A.$ Thus, we
rediscover the definition of \emph{matched pairs of bialgebras} and the
formula for the multiplication of the \emph{double cross product}, see \cite[
Theorem 7.2.2]{Ma2}. Namely,
\begin{equation*}
(f\otimes g)(f^{\prime }\otimes g^{\prime })=\textstyle\sum
{}f(g_{(1)}\triangleright f_{(1)}^{\prime })\otimes (g_{(2)}\triangleleft
f_{(2)}^{\prime })g^{\prime }.
\end{equation*}
\end{fact}
\begin{fact}[Twisting systems between thin categories.]
\label{fa:thin}Our aim now is to investigate the twisting systems between
two thin categories $\boldsymbol{B}$ and $\boldsymbol{A}.$ Recall that a
category is thin if there is at most one morphism between any couple of
objects. Thus, for any $x$ and $y$ in $S$ we have that either $
_{x}A_{y}=\{_{x}g_{y}\}$ or $_{x}A_{y}$ is the empty set. Clearly, if $
_{x}A_{y}=\{_{x}g_{y}\}$ and $_{y}A_{z}=\{_{y}g_{z}\}$ then $_{x}g_{y}\circ
{}_{y}g_{z}={}_{x}g_{z}$. The identity morphism of $x$ is $_{x}g_{x}.$
Similarly, if $_{x}B_{y}$ is not empty then $_{x}B_{y}=\{_{x}f_{y}\}.$
We fix a twisting system $R$ between $\boldsymbol{B}$ and $\boldsymbol{A.}$
It is defined by a family of maps
\begin{equation*}
_{x}R_{z}^{y}:{}_{x}B_{y}\times {}_{y}A_{z}\rightarrow \textstyle
\coprod\limits_{u\in S}A_{u}B_{z}
\end{equation*}
that render commutative the diagrams in Figure \ref{fig:R_1}. We claim that $
R$ is simple. We need a function $\left\vert \cdots \right\vert
:S^{3}\rightarrow S$ such that the image of $_{x}R_{z}^{y}$ is included into
$_{x}A_{|xyz|}B_{z}$ for all $(x,y,z)\in S^{3}.$ Let $T\subseteq S^{3}$
denote the set of all triples such that ${}_{x}B_{y}A_{z}={}_{x}B_{y}\times
{}_{y}A_{z}$ is not empty. Of course, if $(x,y,z)$ is not in $T$ then $
_{x}R_{z}^{y}$ is the empty function, so we can take $\left\vert
xyz\right\vert $ to be an arbitrary element in $S.$ For $(x,y,z)\in T$ there
exists $\left\vert xyz\right\vert \in S$ such that
\begin{equation}
_{x}R_{z}^{y}(_{x}f_{y},{}_{y}g_{z})=(_{x}g_{\left\vert xyz\right\vert
},{}_{\left\vert xyz\right\vert }f_{z}). \label{ec:Rs}
\end{equation}
Hence $_{x}R_{z}^{y}$ is a function from $_{x}B_{y}A_{z}$ to $
_{x}A_{|xyz|}B_{z}.$ Note that $_{x}A_{|xyz|}B_{z}$ is not empty in this
case. For any $(x,y,z)\in S^{3}$ we set $_{x}\widetilde{R}
_{z}^{y}:={}_{x}R_{z}^{y}.$ By Proposition \ref{thm R simplu} and Corollary
\ref{cor: R simplu1} it follows that $R$ is simple. We would like now to
rewrite the conditions from the definition of simple twisting systems in an
equivalent form, that only involves properties of $T$ and $\left\vert \cdots
\right\vert .$ For instance let us show that the first condition from
Corollary \ref{cor: R simplu1} is equivalent to:
\begin{enumerate}
\item[(i)] If $(y,z,t)\in T$ and $(x,y,\left\vert yzt\right\vert )\in T\ $
then $|xy|yzt||=|xzt|.$
\end{enumerate}
Indeed, if $_{x}B_{y}B_{z}A_{t}\ $is not empty then $_{y}B_{z}A_{t}\neq
\emptyset ,$ so $(y,z,t)\in T.$ We have already noticed that $
_{y}A_{|yzt|}B_{t}$ is not empty, provided that $_{y}B_{z}A_{t}\ $is so.
Since $_{x}B_{y}$ and $_{y}A_{|yzt|}$ are not empty it follows that $
_{x}B_{y}A_{|yzt|}$ has the same property, that is $(x,y,\left\vert
yzt\right\vert )\in T.$ Therefore, if $_{x}B_{y}B_{z}A_{t}$ is not empty
then $(y,z,t)\in T$ and $(x,y,\left\vert yzt\right\vert )\in T.$ It is easy
to see that the reversed implication is also true. Thus it remains to prove
that the equation \eqref{cor1-1} holds in the case when $_{x}B_{y}B_{z}A_{t}$
is not empty. But this is obvious, as $_{x}R_{t}^{z}{}\circ
{}_{x}b_{z}^{y}A_{t}$\ and $({}_{x}A_{|xy|yzt||}b_{t}^{|yzt|})\circ
{}_{x}R_{|yzt|}^{y}B_{t}\circ {}_{x}B_{y}R_{t}^{z}$ have the same source $
_{x}B_{y}B_{z}A_{t}$ and the same target $
_{x}A_{|xzt|}B_{t}={}_{x}A_{|xy|xyz||}B_{t}$. Both sets are singletons, so
the above two morphisms must be equal.
Proceeding in a similar way, we can prove that the other three conditions
from Corollary \ref{cor: R simplu1} are respectively equivalent to:
\begin{enumerate}
\item[(ii)] If $(x,y,z)\in T$ and $(\left\vert xyz\right\vert ,z,t)\in T$
then $\left\vert \left\vert xyz\right\vert zt\right\vert =\left\vert
xyt\right\vert .$
\item[(iii)] If $(x,x,y)\in T$ then $|xxy|=y.$
\item[(iv)] If $(x,y,y)\in T$ then $|xyy|=x.$
\end{enumerate}
The last condition in the definition of simple twisting systems is
equivalent to:
\begin{enumerate}
\item[(v)] If $(x,y,z)\in T$ then $_{x}A_{|xyz|}B_{z}$ is not empty.
\end{enumerate}
Hence for a twisting system $R$ the function $\left\vert \cdots \right\vert $
satisfies the above five conditions. Conversely, let $\left\vert \cdots
\right\vert :S^{3}\rightarrow S$ denote a function such that the above five
conditions hold. Let $_{x}R_{z}^{y}$ be the empty function, if $(x,y,z)$ is
not in $T.$ Otherwise we define $_{x}R_{z}^{y}$ by the formula (\ref{ec:Rs}
). In view of the foregoing remarks it is not difficult to see that $
R:=\{_{x}R_{z}^{y}\}_{x,y,z\in S}$ is a simple twisting system.
Clearly two functions $\left\vert \cdots \right\vert $ and $\left\vert
\cdots \right\vert ^{\prime }$ induce the same twisting system if and only
if their restriction to $T$ are equal. Summarizing, we have just proved the
theorem below.
\end{fact}
\begin{theorem}
\label{thm:thin}Let $\boldsymbol{A}$ and $\boldsymbol{B}$ be thin
categories. Let $T$ denote the set of all triples $(x,y,z)\in S^{3}$ such
that $_{x}B_{y}A_{z}$ is not empty. If $R$ is a twisting system between $
\boldsymbol{B}$ and $\boldsymbol{A}$ then there exists a function $
\left\vert \cdots \right\vert :S^{3}\rightarrow S$ such that the conditions
(i)-(v) from the previous subsection hold, and conversely. Two functions $
\left\vert \cdots \right\vert $ and $\left\vert \cdots \right\vert ^{\prime
} $ induce the same twisting system $R$ if and only if their restriction to $
T$ are equal.
\end{theorem}
\begin{fact}[The twisted tensor product of thin categories.]
Let $R$ be a twisting system between two thin categories $\boldsymbol{B}$
and $\boldsymbol{A.}$ By the preceding theorem, $R$ is simple and there are $
T$ and $|\boldsymbol{\cdots }|:S^{3}$ $\rightarrow S$ such that the
conditions (i)-(v) hold. In particular, the twisted tensor product of these
categories exists. By definition, we have $_{x}(\boldsymbol{A}\otimes _{R}
\boldsymbol{B)}_{y}=$ $\coprod\nolimits_{u\in S}(_{x}A_{u}\times
{}_{u}B_{z}).$ We can identify this set with
\begin{equation*}
_{x}S_{y}:=\{u\in S\mid {}_{x}A_{u}B_{y}\neq \emptyset \}.
\end{equation*}
For $u\in {}_{x}S_{y}$ and $v\in {}_{y}S_{z}$ we have $(u,y,v)\in T.$ Thus $
_{u}R_{v}^{y}(_{u}f_{y},{}_{y}g_{v})=(_{u}g_{|uyv|},{}_{|uyv|}f_{v}),$ so
the composition in $\boldsymbol{A}\otimes _{R}\boldsymbol{B}$ is given by
\begin{equation*}
(_{x}g_{u},{}_{u}f_{y})\circ \left( _{y}g_{v},{}_{v}f_{z}\right)
=(_{x}g_{u}\circ {}_{u}g_{\left\vert uyv\right\vert },{}_{\left\vert
uyv\right\vert }f_{v}\circ {}_{v}f_{z})=(_{x}g_{|uyv|},{}_{|uyv|}f_{z}).
\end{equation*}
Let $\boldsymbol{C}(S,T,\left\vert \cdots \right\vert )$ be the category
whose objects are the elements of $S.$ By definition, the hom-set $_{x}
\boldsymbol{C}(S,T,\left\vert \cdots \right\vert )_{y}$ is $_{x}S_{y},$ the
identity map of $x\in S$ is $x$ itself and the composition is given by
\begin{equation*}
\circ :{}_{x}S_{y}\times {}_{y}S_{z}\rightarrow {}_{x}S_{z},\quad u\circ
v=|uyv|.
\end{equation*}
Therefore, we have just proved that $\boldsymbol{A}\otimes _{R}\boldsymbol{B}
$ and $\boldsymbol{C}(S,T,|\cdots|)$ are isomorphic.
\end{fact}
\begin{remark}
Let $\boldsymbol{C}$ be a small category. Let $S$ denote the set of objects
in $\boldsymbol{C}.$ The category $C$ factorizes through two thin categories
if and only if there are $T\subseteq S$ and $\left\vert \cdots \right\vert
:S^{3}\rightarrow S$ as in the previous subsection such that $\boldsymbol{C}$
is isomorphic to $\boldsymbol{C}(S,T,\left\vert \cdots \right\vert ).$
\end{remark}
\begin{fact}[Twisting systems between posets.]
Any poset is a thin category, so we can apply Theorem \ref{thm:thin} to
characterize a twisting system $R$ between two posets $\boldsymbol{B}
:=(S,\preceq )$ and $\boldsymbol{A}:=(S,\leq )$. In this setting the
corresponding set $T$ contains all $(x,y,z)\in S^{3}$ such that $x\preceq y$
and $y\leq z.$ For simplicity, we shall write this condition as $x\preceq
y\leq z.$ A similar notation will be used for arbitrarily long sequences of
elements in $S.$ For instance, $x\leq y\preceq z\preceq t\leq u$ means that $
x\leq y,$ $y\preceq z,$ $z\preceq t$ and $t\leq u$. The function $\left\vert
{\cdots }\right\vert $ must satisfies the following conditions:
\begin{enumerate}
\item[(i)] If $x\preceq y\leq z$ then $x\leq |xyz|\preceq z.$
\item[(ii)] If $x\preceq y\preceq z\leq t$ then $|xy|yzt||=|xzt|$.
\item[(iii)] If $x\preceq y\leq z\leq t$ then $||xyz|zt|=|xyt|.$
\item[(iv)] If $x\leq y$ then $|xxy|=y$.
\item[(v)] If $x\preceq y$ then $|xyy|=x.$
\end{enumerate}
In the case when the posets $\leq $ and $\preceq $ are identical, an example
of function $|{\cdots }|:S^{3}$ $\rightarrow S$ that satisfies the above
conditions is given by $|xyz|=z,$ if $y\neq z,$ and $|xyz|=x,$ otherwise.
\end{fact}
\begin{fact}[Example of twisting map between two groupoids.]
Let $\boldsymbol{A}$ be a groupoid with two objects, $S=\{1,2\}.$ The
hom-sets of $\boldsymbol{A}$ are the following:
\begin{equation*}
_{1}A_{2}=\{u\},\quad _{2}A_{1}=\{u^{-1}\},\quad _{1}A_{1}=\{Id_{1}\},\quad
_{2}A_{2}\ =\{Id_{2}\}.
\end{equation*}
Note that $\boldsymbol{A}$ is thin. We set $\boldsymbol{B}:=\boldsymbol{A}$
and we take $R$ to be a twisting system between $\boldsymbol{B}$ and $
\boldsymbol{A}.$ By Theorem \ref{thm:thin} there are $T$ and $|{\cdots }
|:S^{3}\rightarrow S$ that satisfies the conditions (i)-(v) in \S \ref
{fa:thin}. Since all sets $_{x}B_{y}A_{z}={}_{x}B_{y}\times {}_{y}A_{z}$ are
nonempty it follows that $T=S.$ Thus $|xxy|=y$ and $|xyy|=x$, for all $
x,y\in S$. There are two triples $(x,y,z)\in S^{3}$ such that $x\neq y$ and $
y\neq z$, namely $(1,2,1)$ and $(2,1,2).$ Hence we have to compute $|121|$
and $|212|.$ If we assume that $|121|=1$, then
\begin{equation*}
1=|221|=|21|121||=|211|=2,
\end{equation*}
so we get a contradiction. Thus $|121|=2,$ and proceeding in a similar way
one proves that $|212|=1.$ It is easy to check that $|{\cdots }|$ satisfies
the required conditions, so there is only one twisting map ${R}$ between $
\boldsymbol{A}$ and itself. Since $\boldsymbol{A}$ is a groupoid, the
corresponding bicrossed product $\boldsymbol{C}:=\boldsymbol{A}\Join
\boldsymbol{A}$ is a groupoid as well, see the subsection (\ref{fa:gr}). By
definition,
\begin{equation*}
_{1}C_{1}=\textstyle\coprod\limits_{x\in \{1,2\}}{}_{1}A_{x}\times
{}_{x}A_{1}=\{(Id_{1},Id_{1}),(u,u^{-1})\}.
\end{equation*}
Analogously one shows that
\begin{equation*}
_{1}C_{2}=\{(Id_{1},u),(u,Id_{2})\},\quad
_{2}C_{1}=\{(Id_{2},u^{-1}),(u^{-1},Id_{1})\}\quad \text{and\quad }
_{2}C_{2}=\{(Id_{2},Id_{2}),(u^{-1},u)\}.
\end{equation*}
By construction of the twisting map ${R}$ we get $_{1}R_{1}^{2}(u,u^{-1})\in
{}_{1}A_{|121|}\times {}_{|121|}A_{1}=\{(u,u^{-1})\}$. The other maps $
_{x}R_{z}^{y}$ can be determined analogously. The complete structure of this
groupoid is given in the picture below, where we used the notation $
f:=(u,u^{-1})$ and $g:=(Id_{1},u)$.
\begin{equation*}
\xy (-20,0)*+{1}="a", (20,0)*+{2}="b", \ar@(ul,ur) "a";"a"^{\text{Id}_1} \ar
@(dl,dr) "a";"a"_{f} \ar@(ul,ur) "b";"b"^{\text{Id}_2} \ar@(dl,dr)
"b";"b"_{g\circ f\circ g^{-1}} \ar@(ul,ur) "b";"b" \ar@(dl,dr) "b";"b" \ar
@/^2ex/ "a";"b"|{g}\ar@/^6ex/ "a";"b"|{g\circ f} \ar@/^2ex/ "b";"a"|{g^{-1}}
\ar@/^6ex/ "b";"a"|{f^{-1}\circ g^{-1}} \endxy
\end{equation*}
Note that $f^{2}=Id_{1}$ and $g^{-1}=(Id_{2},u^{-1})$. Now we can say easily
which arrow corresponds to a given morphism in $\boldsymbol{C}$, as in each
home-set we have identified at least one element.
\end{fact}
\noindent\textbf{Acknowledgments.} The first named author was financially supported
by the funds of the Contract POSDRU/6/1.5/S/12. The second named author was financially supported
by CNCSIS, Contract 560/2009 (CNCSIS code ID\_69). | 0.005783 |
Product 7/18
For ADONAI your God is a consuming FIRE, a jealous God. Deuteronomy 4:24
For our God is a consuming FIRE. Hebrews 12:29
Fuego! FIRE! Set hearts ablaze as you worship with this dance veil!
Colors used in the creation of this dance silk: blue, orange & yellow
DESCRIPTION:
This veil is 100% silk. It is free flowing and light weight with a glossy finish. Excellent for worship dance or using it as prayer shawl (tallit). This veil is offered in 7 foot length, with edges serged for ease of use. Each veil will be unique and one-of-a-kind, with the same colors used in the creation process.
Choice available to add a rod pocket to this silk. This gives the option to use this
dance silk as a flag as well.
Flexi-rods sold separately
"As I was creating this FIRE silk; part of the process is to rinse out the silk.
As I was rinsing, the orange (color of FIRE) would not easily rinse out. I sought
the LORD on this and I felt Him say; 'Consuming FIRE means just that, it consumes
everything in its path. I want to consume My worshipers so there is nothing left in
them but Me.'" ~Lora
All items marked CLEARANCE will not be restocked. | 0.000987 |
TITLE: Are there integral domains unknown to be UFD?
QUESTION [3 upvotes]: Are there integral domains unknown to be UFD ??
Im not asking about an infinite set of rings , but a specific integral domain unknown to be a UFD.
What are the simplest examples ?
Are there patterns in the norms of these problematic integral domains ?
REPLY [6 votes]: It is conjectured that the maximal real subfield of the cyclotomic field $\mathbf Q(\zeta_{2^n})$ has class number one, or equivalently is a UFD, for all $n$ but this is only proved for $n\leq 7$. See https://mathoverflow.net/questions/82480/non-trivial-class-number-at-some-finite-level-in-the-cyclotomic-mathbfz-p-e. So taking $n=8$ gives a specific example for the OP's first question at the time this is being written. It is a number field of degree $2^{8-1} = 128$ over $\mathbf Q$. In the future if the case $n=8$ gets settled then change $n$ to $9$. If the conjecture is settled (affirmatively) for all $n$ then I will happily sacrifice the correctness of this answer. | 0.015243 |
\section{Basic properties of $\conv\,SO(n)$ and $\conv\,O(n)$}
\label{sec:properties}
In this section we consider the convex bodies $\conv\,SO(n)$ and $\conv\,O(n)$
purely from the point of view of convex geometry leaving discussion of aspects
related to their semidefinite representations for Section~\ref{sec:reps}. In
this section we describe their symmetries, and how the full space of
$\R^{n\times n}$ matrices decomposes with respect to these symmetries, via the
(special) singular value decomposition. To a large extent one can characterize
$\conv\,SO(n)$ and $\conv\,O(n)$ in terms of their intersections with the
subspace of diagonal matrices. These diagonal sections are well known
polytopes---the parity polytope and the hypercube respectively. The properties
of these diagonal sections are crucial to establishing our spectrahedral
representation of $\conv\,SO(n)$ in Section~\ref{sec:son} and the lower bounds
on the size of spectrahedral representations given in Section~\ref{sec:lower}.
All of the results in this section are (sometimes implicitly) in the literature
in various forms. Here we aim for a brief yet unified presentation to make the
paper as self-contained as possible.
\subsection{Symmetry and the special singular value decomposition}
\label{sec:symmetry}
In this section we describe the symmetries of $\conv\,O(n)$ and $\conv\,SO(n)$.
The group $O(n)\times O(n)$ acts on $\R^{n\times n}$ by $(U,V)\cdot X = UXV^T$.
This action leaves the set $O(n)$ invariant, and hence leaves the convex bodies
$\conv\,O(n)$ and $O(n)^\circ$ invariant. It is also useful to understand how
the ambient space of $n\times n$ matrices decomposes under this group action.
Indeed by the well-known singular value decomposition every element $X\in
\R^{n\times n}$ can be expressed as $X = U\Sigma V^T = (U,V)\cdot \Sigma$ where
$(U,V)\in O(n)\times O(n)$, and $\Sigma$ is diagonal with $\Sigma_{11} \geq
\cdots \geq \Sigma_{nn} \geq 0$. These diagonal elements are the \emph{singular
values}. We denote them by $\sigma_i(X) = \Sigma_{ii}$. Note that for most of
what follows, we only use the fact that $\Sigma$ is diagonal, not that its
elements can be taken to be non-negative and sorted.
Similarly the group
\[ S(O(n)\times O(n)) = \{(U,V): U,V\in O(n),\; \det(U)\det(V) = 1\}\]
acts on $\R^{n\times n}$ by $(U,V)\cdot X = UXV^T$. This action leaves the sets
$SO(n)$ and $SO^-(n)$ invariant, and hence leaves the convex bodies
$\conv\,SO(n)$, $\conv\,SO^-(n)$, $SO(n)^\circ$, $SO^-(n)^\circ$, $\conv\,O(n)$
and $O(n)^\circ$ invariant. A variant on the singular value decomposition,
known as the \emph{special singular value
decomposition}~\cite{sanyal2011orbitopes} describes how the space of $n\times
n$ matrices decomposes under this group action. Indeed every $X\in \R^{n\times
n}$ can be expressed as $X = U\tilde{\Sigma}V^T = (U,V)\cdot \tilde{\Sigma}$
where $(U,V)\in S(O(n)\times O(n))$ and $\tilde{\Sigma}$ is diagonal with
$\tilde{\Sigma}_{11} \geq \cdots \geq \tilde{\Sigma}_{n-1,n-1} \geq
|\tilde{\Sigma}_{nn}|$. These diagonal elements are the \emph{special singular
values}. We denote them by $\tilde{\sigma}_i(X) = \tilde{\Sigma}_{ii}$. Again
in what follows we typically only use the fact that $\tilde{\Sigma}$ is
diagonal for our arguments.
The special singular value decomposition can be obtained from the singular
value decomposition. Suppose $X = U\Sigma V^T$ is a singular value
decomposition of $X$ so that $(U,V)\in O(n)\times O(n)$. If $\det(U)\det(V) =
1$ this is also a valid special singular value decomposition. Otherwise, if
$\det(U)\det(V) = -1$ then $X = UR(R\Sigma)V^T$ gives a decomposition where
$(UR,V)\in S(O(n)\times O(n))$ and $R\Sigma$ is again diagonal, but with the
last diagonal entry being negative. As such the singular values and special
singular values of an $n\times n$ matrix are related by $\sigma_i(X) =
\tilde{\sigma}_i(X)$ for $i=1,2,\ldots,n-1$ and $\tilde{\sigma}_n(X) =
\textup{sign}(\det(X))\sigma_n(X)$.
The importance of these decompositions of $\R^{n\times n}$ under the action of
$O(n)\times O(n)$ and $S(O(n)\times O(n))$ is that they allow us to reduce many
arguments, by invariance properties, to arguments about diagonal matrices.
\subsection{Polytopes associated with $\conv\,O(n)$ and $\conv\,SO(n)$}
The convex hull of $O(n)$ is closely related to the \emph{hypercube}
\begin{equation}
\label{eq:cube1}
\Cube_n = \conv\{x\in \R^n: x_i^2 = 1,\;\;\text{for $i\in [n]$}\};
\end{equation}
the convex hull of $SO(n)$ is closely related to the \emph{parity polytope}
\begin{equation}
\label{eq:pp1}
\PP_n = \conv\{x\in \R^n: \textstyle{\prod_{i=1}^{n}x_i = 1},\;\;x_i^2 = 1,\;\;\text{for $i\in [n]$}\};
\end{equation}
the convex hull of $SO^-(n)$ is closely related to the \emph{odd parity polytope}
\begin{equation}
\label{eq:ppminus1}
\PP_n^- = \conv\{x\in \R^n: \textstyle{\prod_{i=1}^{n}x_i = -1}.\;\;x_i^2 = 1,\;\;\text{for $i\in [n]$}\}.
\end{equation}
In this section we briefly discuss these polytopes as well as showing that they are the diagonal
sections of $\conv\,O(n)$, $\conv\,SO(n)$ and $\conv\,SO^-(n)$ respectively.
\paragraph{Facet descriptions}
Irredundant descriptions of $\Cube_n$ and $\PP_n$ in terms of linear
inequalities are well known \cite{jeroslow1975defining}. The hypercube has
$2n$ facets corresponding to the linear inequality description
\begin{equation}
\label{eq:cubef}
\Cube_n = \{x\in \R^n: -1 \leq x_i \leq 1\;\text{for $i\in [n]$} \}.
\end{equation}
For $n\geq 4$ the parity polytope $\PP_n$ has $2n+2^{n-1}$ facets corresponding
to the linear inequality description
\begin{equation}
\label{eq:ppnf}
\PP_n = \bigg\{x\in \R^n: -1\leq x_i \leq 1\;\text{for $i\in [n]$},\;\sum_{i\notin I}x_i - \sum_{i\in I}x_i \leq n-2\;\;\text{for $I\subseteq [n]$, $|I|$ odd}\bigg\}.
\end{equation}
In the cases $n=2$ and $n=3$ this description simplifies to
\begin{align}
\PP_2 & = \left\{\left[\begin{smallmatrix} x\\x\end{smallmatrix}\right]\in \R^2: -1\leq x \leq 1\right\}\label{eq:pp2f}\\
\PP_3 & =\left\{x\in \R^3: x_1-x_2+x_3\leq 1,\;-x_1+x_2+x_3\leq 1,\right.\nonumber\\
&\qquad\qquad\quad\;\,\left.x_1+x_2-x_3\leq 1,\;-x_1-x_2-x_3\leq 1\right\}\label{eq:pp3f}
\end{align}
showing that $\PP_3$ has only four facets.
The polar of the hypercube is the \emph{cross-polytope}. We denote it by
$\Cube_n^\circ$. It is clear from \eqref{eq:cube1} that $\Cube_n^\circ$ has
$2^n$ facets and corresponding linear inequality description
\begin{equation}
\label{eq:cubepolarf}
\Cube_n^\circ = \bigg\{x\in \R^n: \sum_{i\notin I}x_i - \sum_{i\in I}x_i \leq 1 \quad\text{for $I\subset [n]$}\bigg\}.
\end{equation}
The polar of the parity polytope is denoted by $\PP_n^\circ$. It is clear from
\eqref{eq:pp1} that $\PP_n^\circ$ has $2^{n-1}$ facets and corresponding linear
inequality description
\begin{equation}
\label{eq:pppolarf}
\PP_n^\circ = \bigg\{x\in \R^n: \sum_{i\notin I} x_i - \sum_{i\in I} x_i \leq 1\quad\text{for $I\subset [n]$, $|I|$ even}\bigg\}.
\end{equation}
Similarly
\begin{equation}
\label{eq:ppnegpolarf}
{\PP_n^-}^\circ = \bigg\{x\in \R^n: \sum_{i\notin I} x_i - \sum_{i\in I}x_i \leq 1 \quad\text{for $I\subseteq [n]$, $|I|$ odd}\bigg\}.
\end{equation}
To get a sense of the importance of these polytopes for understanding
$\conv\,SO(n)$ it may be instructive to compare \eqref{eq:pp2f} with
\eqref{eq:so2}, \eqref{eq:pp3f} with \eqref{eq:so3}, \eqref{eq:ppnf} with
\eqref{eq:son}, and \eqref{eq:pppolarf} with \eqref{eq:sonpolar}.
We conclude the discussion of these polytopes with a useful alternative
description of $\PP_n$.
\begin{lemma}
\label{lem:ppalt}
The parity polytope can be expressed as
\[ \PP_n = \Cube_n \cap (n-2) \cdot {\PP_n^-}^\circ.\]
In the case $n=3$ this simplifies to $\PP_3 = {\PP_3^-}^\circ$.
\end{lemma}
\begin{proof}
For the general case, we need only examine the facet descriptions
in~\eqref{eq:cubef},~\eqref{eq:ppnf}, and~\eqref{eq:ppnegpolarf}. In the case
$n=3$ the result follows by comparing \eqref{eq:pp3f} with
\eqref{eq:ppnegpolarf}.
\end{proof}
\paragraph{Diagonal projections and sections}
We now establish the link between the hypercube and the convex hull of $O(n)$,
and the parity polytope and the convex hull of $SO(n)$. First we prove a result
that says that the subspace $\Diag$ of diagonal matrices interacts particularly
well with these convex bodies. The theorem applies for the convex bodies
$\conv\,O(n)$, $\conv\,SO(n)$ and $\conv\,SO^-(n)$ because whenever $g$ is a
diagonal matrix with non-zero entries in $\{-1,1\}$ (a diagonal sign matrix)
then each of these convex bodies is invariant under the conjugation map $X
\mapsto gXg^T$.
\begin{lemma}
\label{lem:symfix}
Let $C\subset \R^{n\times n}$ be a convex body that is invariant under conjugation by diagonal sign matrices.
Then $\projto{\Diag}(C) = \projto{\Diag}(C\cap \Diag)$ and
$[\projto{\Diag}(C\cap \Diag)]^\circ = \projto{\Diag}(C^\circ \cap \Diag)$.
\end{lemma}
\begin{proof}
We first establish that $\projto{\Diag}(C) = \projto{\Diag}(C\cap \Diag)$.
Note that clearly $\projto{\Diag}(C\cap \Diag)\subseteq \projto{\Diag}(C)$.
For the reverse inclusion let $G$ denote the group of diagonal sign
matrices and observe that $\Diag$ is the subspace of $n\times n$ matrices
fixed pointwise by the conjugation action of diagonal sign matrices. Then
for any $X\in C$ the projection onto $\Diag$, the fixed point subspace, is
\[ \projtofrom{\Diag}(X) = \frac{1}{2^n}\sum_{g\in G}gXg^T\] which gives a
description of $\projtofrom{\Diag}(X)$ as a convex combination of the
$gXg^T$. Each $gXg^T\in C$ since $C$ is invariant under conjugation by
diagonal sign matrices. Hence $\projtofrom{\Diag}(X)\in C\cap \Diag$ and so
$\projto{\Diag}(X)\in \projto{\Diag}(C\cap \Diag)$.
Now we establish that $[\projto{\Diag}(C\cap \Diag)]^\circ =
\projto{\Diag}(C^\circ \cap \Diag)$. For any $y\in \Diag$ we have that
\[ \max_{x\in \projto{\Diag}(C\cap \Diag)} \langle y,x\rangle = \max_{x\in \projto{\Diag}(C)}\langle y,x\rangle
= \max_{z\in C}\langle y,\projto{\Diag}(z)\rangle = \max_{z\in C}\langle \projfrom{\Diag}(y),z\rangle.\]
Hence $y\in [\projto{\Diag}(C\cap \Diag)]^\circ$ if and only if $\projfrom{\Diag}(y)\in C^\circ$, or, equivalently,
$y\in \projto{\Diag}(C^\circ \cap\Diag)$.
\end{proof}
We note that this lemma generalizes to the situation where $C$ is a convex
body invariant under the action of a compact group and the subspace $\Diag$
is replaced with the fixed point subspace of the group action.
The key fact that relates the parity polytope and the convex hull of
$SO(n)$ is the following celebrated theorem of Horn~\cite{horn1954doubly}.
\begin{theorem}[Horn]
\label{thm:horn}
The projection onto the diagonal of $SO(n)$ is the parity polytope, i.e.\
$\projto{\Diag}(SO(n)) = \PP_n$.
\end{theorem}
Note that we do not need the full strength of Horn's theorem. We only use the
corollaries that
\begin{align}
\projto{\Diag}(\conv\,SO(n)) & = \conv\,\projto{\Diag}(SO(n)) = \conv\,\PP_n = \PP_n\quad\text{and}\label{eq:horn1}\\
\projto{\Diag}(\conv\,SO^-(n)) & = \projto{\Diag}(R\cdot \conv\,SO(n)) = R\cdot\projto{\Diag}(\conv\,SO(n)) = R\cdot \PP_n = \PP_n^-\label{eq:horn2}.
\end{align}
We are now in a position to establish the main result of this section.
\begin{proposition}
\label{prop:diag}
Let $\Diag\subset \R^{n\times n}$ denote the subspace of diagonal matrices. Then
\begin{align}
\projto{\Diag}(\Diag \cap \conv\,O(n)) & = \Cube_n, \;\;\;\,\;\;\;\;\;\;\projto{\Diag}(\Diag \cap O(n)^\circ) = \Cube_n^\circ,\nonumber\\
\projto{\Diag}(\Diag \cap \conv\,SO(n)) & = \PP_n, \;\;\;\;\,\projto{\Diag}(\Diag\cap SO(n)^\circ) = \PP_n^\circ,\nonumber\\
\projto{\Diag}(\Diag \cap \conv\,SO^-(n)) & = \PP_n^-, \;\;\projto{\Diag}(\Diag\cap SO^-(n)^\circ) = {\PP_n^-}^\circ.\nonumber
\end{align}
\end{proposition}
\begin{proof}
First note that by~\eqref{eq:horn1} and~\eqref{eq:horn2} we know that
$\projto{\Diag}(\conv\,SO(n)) = \PP_n$ and that
$\projto{\Diag}(\conv\,SO^-(n)) = \PP_n^-$. Consequently
\[ \projto{\Diag}(\conv\,O(n)) = \conv\,\projto{\Diag}(SO(n)\cup SO^-(n)) = \conv\,(\PP_n \cup \PP_n^-) = \Cube_n.\]
Since each of $\conv\,O(n)$, $\conv\,SO(n)$, $\conv\,SO^-(n)$ is invariant
under conjugation by diagonal sign matrices we can apply
Lemma~\ref{lem:symfix}. Doing so and using the characterization of the
diagonal projections of each of these convex bodies from the previous
paragraph, completes the proof.
\end{proof} | 0.02143 |
TITLE: Why isn't a quark-antiquark loop included in the photon self-energy corrections?
QUESTION [7 upvotes]: In QED, the Lagrangian has a term $\bar{\psi}A^\mu\psi$, which gives a correction to the photon propagator, where the loop is made of a pair electron-positron, with the 1st order diagram:
In the Standard Model, there are also couplings such as $\bar{u}A^\mu u$ and $\bar{d}A^\mu d$, which would give rise to a similar correction (changing mass and dividing the coupling by 3). but I have never seen this discussion in a QFT textbook. Isn't this correction important? Wouldn't it change how we measure the vacuum polarization? The electron mass is $\sim0.51$MeV while the up quark mass is $\sim2.2$MeV, so it isn't heavy enough to be ignored.
One reason I thought could explain this is that the range of electromagnetic interactions is much larger than that of the strong force, so the quark-antiquark pair would hadronize before they could annihilate, but I'm not completely convinced this is true. Can anybody shed a light on this?
REPLY [2 votes]: The polarized vacuum is effectively described by the Euler-Heisenberg lagrangian. The nonlinear term is proportional to the inverse of the virtual particle mass, so lighter particles like electrons contribute the most. | 0.044499 |
Maxpedition Mini Tactical Chest Rig has been discontinued by Maxpedition and is no longer available. Our product experts have helped us select these available replacements below. You can also explore other items in the Tactical Vests category yourself to try and find the perfect replacement for you!
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Mini Tactical
Chest Rig. Very Lightweight! Only 12.8 ounces or 362.87 grams. SKU: 9848 | 0.001436 |
Project Details
Description
This project will focus on the largest hot spring complex in China, the Tengchong Geothermal Field in Yunnan Province, because of its unique geographic location and because little is known about microbial diversity, biogeographical patterning, and ecosystem-level functions in hot springs of East Asia. Furthermore, the research site will serve as a hub to stimulate research on other geothermal systems in East Asia including Japan, the Philippines, Russia, Taiwan, and Thailand. The long-term goal of this project is to develop a holistic and global view of geobiology in geothermal systems to complement and build upon what is known about life in other geothermal sites, such as Yellowstone National Park. In addition, US students and scientists will have a unique opportunity experience the biologically-, geologically-, and culturally-diverse regions in southwestern China and throughout East Asia and develop long-lasting international collaborations. Intellectual Merit: The study of high temperature ecosystems (>73 follow three key lines of investigation necessary for generating a holistic understanding of microbial community structure and function: 1) Comprehensive geochemical analysis and thermodynamic and kinetic modeling geochemical and mineralogical data will be used to classify and thermodynamically characterize hot springs and develop hypotheses on energy flow and biologically-driven element cycling; 2) Comprehensive studies of rates of C- and N-cycling activities coupled with comprehensive censuses of key biomarkers of C- and N-cycle processes biomarkers including genes, mRNAs, lipids, and biomass stable isotope natural abundances will be examined to link C and N metabolism with specific organisms or classes of organisms; 3) Genomics of microbial pure cultures and uncultivated microorganisms the dominant microorganisms in the springs as well as novel biological dark matter will be investigated using traditional cultivation approaches and through microfluidics-based single cell isolation and subsequent genome sequencing. Genomes will be examined to identify key genes involved in biogeochemical cycles and biogeography. The combination of careful site characterization, direct measurement of microbial activities, and genomic approaches will lead to an unprecedented, integrated understanding of this system that will ultimately facilitate an understanding of the role of geochemistry and/or biogeography in controlling microbial community structure and functio
Funding
- NSF-OD: Office of International Science and Engineering (OISE): $347,931.00
Fingerprint
Explore the research topics touched on by this project. These labels are generated based on the underlying awards/grants. Together they form a unique fingerprint. | 0.685261 |
John J. Amtsfield, age 60, husband of Linda K. (Cohick) Amtsfield of Boyertown, passed away on Sunday at Pottstown Memorial Medical Center. Born in Philadelphia, he was a son of the late Carole (Kelley) Amtsfield and the late John Joseph Amtsfield.
He had been employed for many years by the U.S. Postal Service as a Rural Letter Carrier in Pa. and N.J., where he also served as a State Union Steward in both of those states. John was an avid genealogist who also enjoyed watching all of the Philadelphia sports teams. He was also a member of several counties historical societies.
In addition to his wife, John is also survived by his children; Joel Amtsfield (husband of Karen) of Phoenix, AZ., Sean Amtsfield (husband of Corinne) of Reading, Adrienne Dellinger of N.J., Seth Dellinger of Erie, his 7 grandchildren, and 1 brother and 2 sisters.
Funeral Services will be held privately by his family at a later date. Memorial contributions in his name can be made to the American Diabetes Association, P.O. Box 11454, Alexandria, Va., 22312.
Recent Comments | 0.036643 |
\begin{document}
\maketitle
\begin{abstract}
Cognitive radios sense the radio spectrum in order to find unused frequency bands and use them in
an agile manner. Transmission by the primary user must be detected reliably even in the low signal-to-noise ratio (SNR) regime and in the face of shadowing and fading. Communication signals are
typically cyclostationary, and have many periodic statistical properties related to the symbol rate, the coding and modulation schemes as well as the guard periods, for example.
These properties can be exploited in designing a detector, and
for distinguishing between the primary and secondary users' signals.
In this paper, a generalized likelihood ratio test (GLRT) for detecting the presence of cyclostationarity using multiple cyclic frequencies is proposed. Distributed decision making is employed by combining the quantized local test statistics from many secondary users. User cooperation allows for mitigating the effects of
shadowing and provides a larger footprint for the cognitive radio system.
Simulation examples demonstrate the resulting performance gains in the low SNR regime and the
benefits of cooperative detection.
\end{abstract}
\section{Introduction}
Spectrum sensing is needed in cognitive radios in order to find opportunities for
agile use of spectrum. Moreover, it is crucial for managing the level of interference caused
to primary users (PUs) of the spectrum. Sensing provides awareness of the radio operating
environment. A cognitive radio may then adapt
its parameters such as carrier frequency, power and waveforms dynamically in
order to provide the best available connection and to meet the user's needs
within the constraints on interference.
In wireless communication systems we typically have some knowledge on the waveforms and structural or statistical properties of the signals that the primary user of the spectrum is using.
Such knowledge may be related to the modulation scheme,
the symbol or chip rate of the signal, the channel coding scheme, training or pilot signals,
guard periods, and the power level or correlation properties of the signal, just to mention a few.
These properties may be used to design a detector that works in a very low SNR regime
and has low complexity and consequently low power consumption. These are very desirable properties especially for cognitive radios in mobile applications. In the absence of any knowledge of the signal, one
may have to resort to classical techniques such as
energy detection \cite{poor}. An energy detector may need to collect data over a long
period of time to detect the primary users reliably.
Moreover, controlling the false alarm rates in mobile applications is difficult because the statistics of the signals, noise and interference may be time-varying.
Another significant drawback is that energy detection has no
capability to distinguish among different types of transmissions
or to dichotomize between primary and secondary users of the spectrum.
Cyclostationary processes are random processes for which the statistical
properties such as the mean and
autocorrelation change periodically as functions of time \cite{gardner86}.
Many of the signals used in
wireless communication and radar systems possess this property. Cyclostationarity may be
caused by modulation or coding \cite{gardner86}, or it may be also intentionally produced in order to
aid channel estimation or synchronization \cite{tsatsanis}.
Cyclostationarity property has been widely used in intercept
receivers \cite{gardner86,koivisto,lunden},
direction of arrival or time-delay estimation, blind equalization and
channel estimation \cite{tong} as well as in precoder design
in multicarrier communications \cite{tsatsanis}. In order to exploit cyclic statistics,
the signal must
be oversampled with respect to the symbol rate, or multiple receivers must be
used to observe the signal.
The use of cyclostationary statistics is appealing in many ways: noise is rarely cyclostationary and
second-order cyclostationary statistics retain also the phase information.
Hence, procedures based on cyclostationarity tend to have particularly good performance at the low SNR regime. Moreover, cyclostationarity allows
for distinguishing among different transmission types and users if
their signals have distinct cyclic frequencies. A comprehensive list of references on cyclostationarity along with a survey of the literature is presented in~\cite{gardner06}.
The presence of cyclostationary signals may be determined by using hypothesis testing.
Many existing tests, such as \cite{dandawate94},
are able to detect the presence of cyclostationarity at only one
cyclic frequency at a time,
and they partly ignore the rich information present in the signals. For example,
a communication signal may have cyclic frequencies related to the carrier frequency,
the symbol rate and its harmonics, the chip rate, guard period, the scrambling code period, and the channel coding scheme.
In this paper we propose a method for detecting multiple
cyclic frequencies simultaneously. It extends the method of \cite{dandawate94}
to take into account the rich information present at different cyclic frequencies.
This provides improved detector performance over techniques relying only on
single cyclic frequency and facilitates dichotomizing among the
primary and secondary user signals and different waveforms used.
In cognitive radio systems, there are typically multiple geographically distributed
secondary users (SUs) that need to detect if the primary user is transmitting.
The distributed sensors may work collaboratively to decide between two hypotheses:
is the primary user active, or is the spectrum unused and available for the secondary
users? Decentralized processing has a number of advantages for such situations. Obviously, it allows for a larger coverage area. Furthermore, there are gains similar to diversity gains in wireless communications so that
the detection becomes less sensitive to demanding propagation conditions
such as shadowing by large obstacles,
large numbers of scatterers, differences in attenuation, or fast fading
caused by mobility. Moreover,
distributed sensory systems may require less communication bandwidth,
consume less power,
be more reliable and cost less as well. In this paper, we propose a simple
decentralized decision making approach based on sharing and combining quantized
local decision statistics. This approach may be used in both decision
making with or without a fusion center.
This paper is organized as follows. In Section II, there is a short review of cyclostationary statistics. A novel detector for multiple cyclic frequencies is derived in Section III. Section IV addresses the problem of collaborative detection of primary user. Simulation results demonstrating the detector's reliability in the low SNR regime
as well as the gains obtained via collaborative operation are presented in Section V.
Finally, conclusions are drawn in Section VI.
\section{Cyclostationarity: a recap}
\label{sec:cyclo_def}
In this section, we provide a brief overview of cyclostationarity
in order to make the derivation of the detector in Section III clearer. A continuous-time
random process $x(t)$ is wide sense second-order cyclostationary if there exists a $T_0>0$ such
that \cite{gardner86}:
\begin{equation}
\mu_{x}(t) = \mu_{x}(t+T_0)\ \forall t \\
\end{equation}
and
\begin{equation}
R_{x}(t_{1},t_{2}) = R_{x}(t_{1}+T_0,t_{2}+T_0)\ \forall t_{1},t_{2}.
\label{eq:csdef}
\end{equation}
$T_0$ is called the period of the cyclostationary process.
Due to the periodicity of the autocorrelation $R_{x}(t_{1},t_{2})$, it has a Fourier-series representation.
By denoting $t_{1}=t+\tau/2$ and $t_{2}=t-\tau/2$, we obtain the following expression for the
Fourier-series \cite{gardner86}:
\begin{equation}
R_{x}(t+\frac{\tau}{2},t-\frac{\tau}{2})=\sum_{\alpha}R_{x}^{\alpha}(\tau)e^{j2\pi\alpha t},
\label{eq:caf_series}
\end{equation}
where the Fourier coefficients are
\begin{equation}
R_{x}^{\alpha}(\tau)=\frac{1}{T_0}\int_{-\infty}^{\infty}R_{x}(t+\frac{\tau}{2},t-\frac{\tau}{2})e^{-j2\pi\alpha t}dt
\label{eq:caf}
\end{equation}
and $\alpha$ is called the cyclic frequency.
The function $R_{x}^{\alpha}(\tau)$ is called the cyclic autocorrelation function.
If the process has zero mean, then this is also the cyclic autocovariance function.
When the autocorrelation function has exactly one period $T_0$
we have the following set of cyclic frequencies
\[
{\mathcal A}=\left\{\alpha=k/T,k\geq1 \right\},
\]
where $R_{x}^{\alpha}(\tau)$ is the cyclic autocorrelation function
and $\mathcal{A}$ are the set of cyclic frequencies. The
cyclic frequencies are harmonics of the fundamental frequency.
If the autocorrelation function has several periods $T_{0},T_{1},\ldots$,
we may express $R_{x}^{\alpha}(\tau)$ at the limit \cite{gardner86}
\begin{equation}
R_{x}^{\alpha}(\tau)=\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T/2}^{T/2}x(t+\frac{\tau}{2})x^{\ast}(t-\frac{\tau}{2})e^{j2\pi\alpha t}dt.
\end{equation}
The process $x(t)$ is almost cyclostationary in the wide sense and the
set of cyclic frequencies ${\mathcal A}$ is comprised of a countable number of
frequencies that do not need to be harmonics of the fundamental frequency.
In general, the process is said to be cyclostationary if there exists an $\alpha\neq 0$ such that $R_{x}^{\alpha}(\tau)\neq 0$ for some value of $\tau$. Typically cyclic frequencies are assumed to be known or
may be estimated reliably.
\section{Detection using multiple cyclic frequencies}
Statistical tests for the presence of a single cyclic frequency have been proposed,
for example, in~\cite{dandawate94}.
The tests in~\cite{dandawate94} have asymptotically
constant false alarm rate (CFAR) for testing presence of cyclostationarity at a
given cyclic frequency. However, the tests do not retain the CFAR property over
a set of tested frequencies.
Typical communication signals exhibit cyclostationarity at multiple cyclic
frequencies instead of just a single cyclic frequency.
That is, for example a signal that is cyclostationary at the symbol frequency is
typically cyclostationary at all integer multiples of the symbol frequency as well.
There also may be cyclic frequencies related to the coding and guard periods,
or adaptive modulation and coding may be used. In such cases the cyclic frequencies present
may vary depending on channel quality and the waveform used.
If one is testing for the presence of many different signals at a given frequency
band, or in case the cyclic frequencies are not known,
it would be desirable to retain the CFAR property over the whole set of tested cyclic
frequencies. This would be especially desirable in a cognitive radio
application where the interest is in finding unoccupied
frequency bands. Otherwise the frequency band may unnecessarily be
classified as occupied for most of the time.
In the following we extend the test based on second-order cyclic statistics
of~\cite{dandawate94} to multiple cyclic frequencies.
To do so we first define all the terms used in the test statistics.
Let $(*)$ denote an optional complex conjugation. The notation allows convenient handling of both cyclic autocorrelation and conjugate cyclic autocorrelation with only one equation. An estimate of
the (conjugate) cyclic autocorrelation ${\hat R}_{xx^{(*)}}(\alpha,\tau)$ may be obtained using $M$ observations as
\begin{align}
\label{eq:cyclic_autocorr_estimator}
{\hat R}_{xx^{(*)}}(\alpha,\tau) &=
\frac{1}{M} \sum_{t=1}^{M} x(t)x^{(*)}(t+\tau) e^{-j 2 \pi \alpha t}\\
&= R_{xx^{(*)}}(\alpha,\tau) + \varepsilon(\alpha,\tau),
\end{align}
where the latter term is the estimation error. This estimator is consistent, (see ~\cite{dandawate94})
so that the error goes to zero as $M \rightarrow \infty$.
Now we need to construct a test for a number of lags $\tau_1,\ldots,\tau_{N}$ as well as a set of cyclic frequencies of interest. Let $\mathcal{A}$ denote the set of cyclic frequencies of interest, and
\begin{equation}
\begin{aligned}
\hat{\bm{r}}_{xx^{(*)}}(\alpha) = \bigg[& \mathrm{Re}\{\hat{R}_{xx^{(*)}}(\alpha,\tau_{1})\},\ldots,\mathrm{Re}\{\hat{R}_{xx^{(*)}}(\alpha,\tau_{N})\}, \\
&\mathrm{Im}\{\hat{R}_{xx^{(*)}}(\alpha,\tau_{1})\},\ldots,\mathrm{Im}\{\hat{R}_{xx^{(*)}}(\alpha,\tau_{N})\} \bigg]
\end{aligned}
\end{equation}
denote a $1 \times 2N$ vector containing the real and
imaginary parts of the estimated cyclic autocorrelations at the cyclic frequency of interest stacked in a single vector.
The $2N \times 2N$ covariance matrix of $\bm{r}_{xx^{(*)}}$ can be computed as~\cite{dandawate94}
\begin{equation}
\bm{\Sigma}_{xx^{(*)}}(\alpha) = \begin{bmatrix} \mathrm{Re}\left\{ \frac{\bm{Q}+\bm{Q}^{*}}{2}\right\} & \mathrm{Im}\left\{ \frac{\bm{Q}-\bm{Q}^{*}}{2}\right\} \\
\mathrm{Im}\left\{ \frac{\bm{Q}+\bm{Q}^{*}}{2}\right\} & \mathrm{Re}\left\{ \frac{\bm{Q}^{*}-\bm{Q}}{2}\right\} \end{bmatrix}
\end{equation}
where the $(m,n)$th entries of the two covariance matrices $\bm{Q}$ and $\bm{Q}^{*}$ are given by
\[
\begin{aligned}
\bm{Q}(m,n) &= S_{f_{\tau_{m}}f_{\tau_{n}}}(2\alpha,\alpha)
\end{aligned}
\]
and
\begin{equation}
\begin{aligned}
\bm{Q}^{*}(m,n) &= S^{*}_{f_{\tau_{m}}f_{\tau_{n}}}(0,-\alpha).
\end{aligned}
\end{equation}
Here, $S_{f_{\tau_{m}}f_{\tau_{n}}}(\alpha,\omega)$ and $S^{*}_{f_{\tau_{m}}f_{\tau_{n}}}(\alpha,\omega)$ denote the unconjugated and conjugated cyclic spectra of $f(t,\tau) = x(t)x^{(*)}(t+\tau)$, respectively.
These spectra can be estimated using frequency smoothed cyclic periodograms as
\begin{align}
\hat{S}_{f_{\tau_{m}}f_{\tau_{n}}}(2\alpha,\alpha) &=
\frac{1}{ML} \sum_{s=-(L-1)/2}^{(L-1)/2} W(s) \nonumber \\
&\mbox{} \quad \cdot F_{\tau_{n}}(\alpha-\frac{2\pi s}{M})F_{\tau_{m}}(\alpha+\frac{2\pi s}{M}) \\
\hat{S}^{*}_{f_{\tau_{m}}f_{\tau_{n}}}(0,-\alpha) &=
\frac{1}{ML} \sum_{s=-(L-1)/2}^{(L-1)/2}W(s) \nonumber \\
&\mbox{} \quad \cdot F^{*}_{\tau_{n}}(\alpha+\frac{2\pi s}{M})F_{\tau_{m}}(\alpha+\frac{2\pi s}{M})
\end{align}
where $F_{\tau}(\omega) = \sum_{t=1}^{M}x(t)x^{(*)}(t+\tau)e^{-j\omega t}$ and $W$ is a
normalized spectral window of odd length $L$.
Now the hypothesis testing problem for testing if $\alpha$ is a cyclic frequency can be formulated as~\cite{dandawate94}
\begin{equation}
\label{eq:hypothesis_single}
\begin{aligned}
H_{0}: &\mbox{} \;\forall \{\tau_{n}\}_{n=1}^{N} \Longrightarrow \hat{\bm{r}}_{xx^{(*)}}(\alpha) = \bm{\epsilon}_{xx^{(*)}}(\alpha) \\
H_{1}: &\mbox{}\; \mathrm{for}\;\mathrm{some}\; \{\tau_{n}\}_{n=1}^{N} \\ &\Longrightarrow \hat{\bm{r}}_{xx^{(*)}}(\alpha) = \bm{r}_{xx^{(*)}}(\alpha) + \bm{\epsilon}_{xx^{(*)}}(\alpha).
\end{aligned}
\end{equation}
Here $\bm{\epsilon}_{xx^{(*)}}$ is the estimation error which is asymptotically normal distributed, i.e., $\lim_{M\rightarrow \infty} \sqrt{M}\bm{\epsilon}_{xx^{(*)}} \overset{D}{=} N(\bm{0},\bm{\Sigma}_{xx^{(*)}})$~\cite{dandawate94}.
Hence, using the asymptotic normality of $\hat{\bm{r}}_{xx^{(*)}}$ the generalized likelihood ratio (GLR) is given by
\begin{equation}
\begin{aligned}
\Lambda &= \frac{\exp(-\frac{1}{2}M\hat{\bm{r}}_{xx^{(*)}}\hat{\bm{\Sigma}}_{xx^{(*)}}^{-1}\hat{\bm{r}}_{xx^{(*)}}^{T})}{\exp(-\frac{1}{2}M(\hat{\bm{r}}_{xx^{(*)}}-\hat{\bm{r}}_{xx^{(*)}})\hat{\bm{\Sigma}}_{xx^{(*)}}^{-1}(\hat{\bm{r}}_{xx^{(*)}}-\hat{\bm{r}}_{xx^{(*)}})^{T})} \\
&=\exp(-\frac{1}{2}M\hat{\bm{r}}_{xx^{(*)}}\hat{\bm{\Sigma}}_{xx^{(*)}}^{-1}\hat{\bm{r}}_{xx^{(*)}}^{T}). \\
\end{aligned}
\end{equation}
Finally, by taking the logarithm and multiplying the result by 2,
we arrive at the test statistic in~\cite{dandawate94}
\begin{equation}
\label{eq:cyclic_test_statistic}
\mathcal{T}_{xx^{(*)}}(\alpha) = -2\ln \Lambda = M\hat{\bm{r}}_{xx^{(*)}}\hat{\bm{\Sigma}}_{xx^{(*)}}^{-1}\hat{\bm{r}}_{xx^{(*)}}^{T}.
\end{equation}
Under the null hypothesis $\mathcal{T}_{xx^{(*)}}(\alpha)$ is asymptotically $\chi_{2N}^{2}$
distributed.
Now in order to extend the test for the presence of second-order cyclostationarity at any of the
cyclic frequencies of interest $\alpha \in \mathcal{A}$ simultaneously, we formulate the hypothesis
testing as follows
\begin{equation}
\label{eq:hypothesis_multi}
\begin{aligned}
H_{0}: &\mbox{}\;\forall \alpha \in \mathcal{A}\;\mathrm{and} \;\forall \{\tau_{n}\}_{n=1}^{N} \Longrightarrow \hat{\bm{r}}_{xx^{(*)}}(\alpha) = \bm{\epsilon}_{xx^{(*)}}(\alpha) \\
H_{1}: &\; \mathrm{for}\;\mathrm{some}\;\alpha \in \mathcal{A}\; \mathrm{and}\;\mathrm{for}\;\mathrm{some}\; \{\tau_{n}\}_{n=1}^{N}\\
& \Longrightarrow \hat{\bm{r}}_{xx^{(*)}}(\alpha) = \bm{r}_{xx^{(*)}}(\alpha) + \bm{\epsilon}_{xx^{(*)}}(\alpha).
\end{aligned}
\end{equation}
For this detection problem, we propose the following two test statistics:
\begin{align}
\mathcal{D}_{m} &= \max_{\alpha \in \mathcal{A}} \mathcal{T}_{xx^{(*)}}(\alpha)
= \max_{\alpha \in \mathcal{A}} M\hat{\bm{r}}_{xx^{(*)}}(\alpha)
\hat{\bm{\Sigma}}_{xx^{(*)}}^{-1}(\alpha)\hat{\bm{r}}_{xx^{(*)}}^{T}(\alpha) \\
\mathcal{D}_{s} &= \sum_{\alpha \in \mathcal{A}} \mathcal{T}_{xx^{(*)}}(\alpha)
= \sum_{\alpha \in \mathcal{A}} M\hat{\bm{r}}_{xx^{(*)}}(\alpha)
\hat{\bm{\Sigma}}_{xx^{(*)}}^{-1}(\alpha)\hat{\bm{r}}_{xx^{(*)}}^{T}(\alpha).
\end{align}
The first test statistic calculates the maximum of the cyclostationary GLRT statistic~\eqref{eq:cyclic_test_statistic} over the cyclic frequencies of interest $\mathcal{A}$ while the second calculates the sum. Assuming independence of cyclic autocorrelation estimates for different cyclic frequencies the test statistic $\mathcal{D}_{s}$ is the GLRT statistic. Depending on the signal and the set of tested cyclic frequencies the test statistics may have different performances. This requires further research.
The asymptotic distribution of $\mathcal{D}_{s}$ is under the null hypothesis $\chi_{2NN_{\alpha}}^{2}$ where $N_{\alpha}$ is the number of cyclic frequencies in set $\mathcal{A}$. This is due to the fact that the sum of independent
chi-square random variables is also a chi-square random variable whose degrees of freedom is
the sum of the degrees of freedom of the independent random variables.
In the following we derive the asymptotic distribution of the test statistic $\mathcal{D}_{m}$ under the null hypothesis. As stated above, under the null hypothesis $\mathcal{T}_{xx^{(*)}}(\alpha)$ is asymptotically $\chi_{2N}^{2}$ distributed.
The cumulative distribution function of the chi-square distribution with $2N$ degrees of freedom is given by
\begin{equation}
F(x,2N) = \frac{\gamma (N,x/2)}{\Gamma (N)}
\end{equation}
where $\gamma(k,x)$ is the lower incomplete gamma function and $\Gamma(k)$ is the ordinary gamma function. For a positive
integer $k$ the following identities hold:
\begin{align}
\Gamma(k) &= (k-1)! \\
\gamma(k,x) &= \Gamma(k) - (k-1)! \; e^{-x}\sum_{n=0}^{k-1}\frac{x^{n}}{n!}.
\end{align}
Hence, the cumulative distribution function of the chi-square distribution with $2N$
degrees of freedom is given by
\begin{equation}
F(x,2N) = 1-e^{-x/2}\sum_{n=0}^{N-1}\frac{(x/2)^{n}}{n!}.
\end{equation}
The cumulative distribution function of the maximum of $d$
independent and identically distributed random variables is the cumulative
distribution function of the individual random variables raised to the power $d$.
Thus, the cumulative distribution function of the test statistic $\mathcal{D}_{m}$
is given by
\begin{equation}
F_{\mathcal{D}_{m}}(x,2N,d) = \left(1-e^{-x/2}\sum_{n=0}^{N-1}\frac{(x/2)^{n}}{n!}\right)^{d}.
\end{equation}
The corresponding probability density function is obtained by differentiating
the cumulative distribution function, i.e.,
\begin{equation}
\begin{aligned}
f_{\mathcal{D}_{m}}(x,2N,d) &= \frac{d}{2}\left(1-e^{-x/2}\sum_{n=0}^{N-1}\frac{(x/2)^{n}}{n!}\right)^{d-1} \\
&\mbox{} \quad \cdot e^{-x/2}\left(\sum_{n=0}^{N-1}\frac{(x/2)^{n}}{n!}-\sum_{n=1}^{N-1}\frac{(x/2)^{n-1}}{(n-1)!}\right).
\end{aligned}
\end{equation}
Consequently, the null hypothesis is rejected if $F_{\mathcal{D}_{m}}(\mathcal{D}_{m},2N,N_{\alpha}) > 1-p$ where $p$ is the false alarm rate and $N_{\alpha}$ is the number of tested cyclic frequencies.
\section{Cooperative detection}
User cooperation may be used to improve the performance and coverage in a cognitive radio network.
The users may collaborate in finding unused spectrum and new opportunities. Many of the
collaborative detection techniques stem from distributed detection theory; see \cite{viswanathan,blum}.
In cognitive radio systems, there are typically multiple geographically
distributed secondary users that need to detect
whether the primary user is active. All the secondary users may sense the
entire band of interest, or monitor just a partial band to reduce power consumption.
In the latter case each SU senses a certain part of the spectrum,
and then shares the acquired information
with other users or a fusion center.
The cooperation may then be coordinated by a fusion center (FC),
or it may take place in an ad-hoc manner
without a dedicated fusion center. Here we assume that a fusion center collects information from
all $K$ secondary users and makes a decision about whether the spectrum is available or not.
We assume that each secondary user sends a quantized version of its local decision statistics
(such as the likelihood ratio) to the FC. In the case of very coarse quantization,
binary local decision may be sent.
To derive a test for the FC,
we assume that the sensors are
independent conditioned on whether the hypothesis $H_0$ or $H_1$ is true.
Then the optimal fusion rule is the likelihood ratio test over
the received local likelihood ratios $l_i$:
\begin{equation}
\mathcal{T}_{K} = \prod_{i=1}^{K} l_i.
\end{equation}
In case the secondary users send binary decisions, the sum of ones may calculated and compared to a threshold.
Here, we consider the simplest way of making the decision using generalized likelihood ratios.
Instead of using the product of the generalized likelihood ratios,
we can employ the sum of generalized log-likelihood ratios. We propose the following test statistic for the hypothesis testing problem~\eqref{eq:hypothesis_single}
\begin{equation}
\mathcal{T}_{K}' = \sum_{i=1}^{K} {\mathcal{T}^{(i)}_{xx^{(*)}}(\alpha)},
\end{equation}
and the following two for the hypothesis testing problem~\eqref{eq:hypothesis_multi}
\begin{align}
\mathcal{D}_{m,K} &= \max_{\alpha \in \mathcal{A}} \sum_{i=1}^{K} {\mathcal{T}^{(i)}_{xx^{(*)}}(\alpha)} \\
\mathcal{D}_{s,K} &= \sum_{\alpha \in \mathcal{A}} \sum_{i=1}^{K} {\mathcal{T}^{(i)}_{xx^{(*)}}(\alpha)}
\end{align}
where ${\mathcal{T}^{(i)}_{xx^{(*)}}(\alpha)}$ is the cyclostationarity
based test statistic~\eqref{eq:cyclic_test_statistic} from $i^{th}$ secondary user. Due to the use of generalized likelihood ratios, no optimality properties can be claimed.
The GLRT test does, however, perform highly reliably in many applications.
Under the conditional independence assumption the asymptotic distributions of the test statistic
$\mathcal{T}_{K}'$ and $\mathcal{D}_{s,K}$ are under the null hypothesis $\chi_{2NK}^{2}$ and $\chi_{2NN_{\alpha}K}^{2}$, respectively. This is again due to the fact that the sum of independent
chi-square random variables is also a chi-square random variable whose degrees of freedom is
the sum of the degrees of freedom of the independent random variables. The cumulative distribution function of $\mathcal{D}_{m,K}$ is under the null hypothesis $F_{\mathcal{D}_{m}}(\mathcal{D}_{m,K},2NK,N_{\alpha})$ where $N_{\alpha}$ is again the number of tested cyclic frequencies. The testing is done similarly as in one secondary user case.
Different techniques for reducing the amount of transmitted data, taking into account the
relevance of the information provided by secondary users as well as how to
deal with communication rate constraints will be addressed in a forthcoming paper.
\section{Simulation examples}
In this section the performance of the proposed detectors is considered.
The test signal is an orthogonal frequency division multiplex (OFDM) signal.
The baseband equivalent of a cyclic prefix OFDM signal may be expressed as
\begin{equation}
x(t) = \sum_{n=0}^{N_{c}-1}\sum_{l=-\infty}^{\infty}c_{n,l}g(t-lT_{s})e^{j(2\pi/N)n(t-lT_{s})}
\end{equation}
where $N_{c}$ is the number of subcarriers, $T_{s}$ is the symbol length, $g(t)$ denotes the rectangular pulse of length $T_{s}$, and the $c_{n,l}$'s denote the data symbols.
The symbol length is the sum of the length of the useful symbol data $T_{d}$ and the length of the
cyclic prefix $T_{cp}$, i.e., $T_{s} = T_{d} + T_{cp}$.
The above OFDM signal exhibits cyclostationarity (i.e., complex conjugation is used in~\eqref{eq:cyclic_autocorr_estimator} and the following equations) with
cyclic frequencies of $\alpha = k/T_{s}$, $k = 0, \pm 1, \pm 2, \ldots$ and potentially other frequencies depending on the coding scheme. The cyclic autocorrelation surfaces for $\alpha = k/T_{s}$ peak at $\tau = \pm T_{d}$~\cite{oner04}.
In the following the performance of cyclic detectors based on one and two cyclic frequencies is compared as a function of signal-to-noise ratio (SNR) in an additive white Gaussian noise (AWGN) channel.
The SNR is defined as $\mathrm{SNR} = 10\log_{10}\frac{\sigma_{x}^{2}}{\sigma_{n}^{2}}$ where $\sigma_{x}^{2}$ and $\sigma_{n}^{2}$ are the variances of the signal and the noise, respectively. The cyclic frequencies employed by the detectors are $1/T_{s}$ and $2/T_{s}$. The detector based on one cyclic frequency uses the first frequency and the detectors based on two cyclic frequencies use both frequencies. Each detector uses
two time lags $\pm T_{d}$.
The cyclic spectrum estimates were calculated using a length-2049 Kaiser window with $\beta$ parameter of 10. A Fast-Fourier transform (FFT) was employed for faster computation.
The FFT size was 10000 giving a cyclic frequency resolution of 0.0001.
The OFDM signal has 32 subcarriers and the length of the cyclic prefix is 1/4 of the useful symbol data. The subcarrier modulation employed is 16-QAM. The signal length is 100 OFDM symbols.
Fig.~\ref{fig:ofdm_pd_vs_SNR} depicts the performance of the detectors as a function of the SNR for a constant false alarm rate of 0.05. Fig.~\ref{fig:ofdm_pd_vs_SNR_zoom} shows a zoom of the important area illustrating the differences in performance more clearly. All the curves are averages over 10000 experiments. It can be seen that the detectors based on multiple cyclic frequencies outperform the detector based on single cyclic frequency in the low SNR regime. Furthermore, the multicycle detector calculating the sum over the cyclic statistics of different frequencies has the best performance.
\begin{figure}[tbp]
\centering
\includegraphics[width=0.75\columnwidth]{ofdm_pd_vs_SNR}
\caption{Probability of detection vs. SNR. The multicycle detectors achieve better performance than the single cycle detector in the low SNR regime. The sum detector of the test statistic $\mathcal{D}_{s}$ has the best performance.}
\label{fig:ofdm_pd_vs_SNR}
\end{figure}
\begin{figure}[tbp]
\centering
\includegraphics[width=0.75\columnwidth]{ofdm_pd_vs_SNR_zoom}
\caption{Probability of detection vs. SNR. Zoom of the important region. The multicycle detectors achieve better performance than the single cycle detector in the low SNR regime. The sum detector of the test statistic $\mathcal{D}_{s}$ has the best performance.}
\label{fig:ofdm_pd_vs_SNR_zoom}
\end{figure}
Fig.~\ref{fig:ofdm_pd_vs_pfa_1} plots the probability of detection vs. false alarm rate for SNR of -7 dB. The figure show that the detectors have desirable receiver operating characteristics. That is, the probability of detection increases as the false alarm rate parameter is increased.
\begin{figure}[tbp]
\centering
\includegraphics[width=0.75\columnwidth]{ofdm_pd_vs_pfa_-7dB}
\caption{Probability of detection vs. false alarm rate. The detectors based on multiple cyclic frequencies achieve better performance than the detector based on a single cyclic frequency.}
\label{fig:ofdm_pd_vs_pfa_1}
\end{figure}
Next the performance gain from cooperative detection of several secondary users is analyzed. The signal is the same as above. The cooperative detection is based on the data of 5 secondary users. Each secondary user receives the same data with different noise. SNR is the same for each secondary user.
Fig.~\ref{fig:ofdm_5su_pd_vs_SNR} depicts the performance for 5 secondary users compared to the single secondary user case. Performance gain of roughly 3 dB is obtained from the cooperation of 5 secondary users. Using two cyclic frequencies provides similar performance improvement as in single secondary user case. Fig.~\ref{fig:ofdm_5su_pd_vs_pfa} shows the probability of detection vs. false alarm rate for SNR of -9 dB.
\begin{figure}[tbp]
\centering
\includegraphics[width=0.75\columnwidth]{ofdm_5su_pd_vs_SNR}
\caption{Probability of detection vs. SNR. Cooperation of 5 secondary users provides performance gain of 3 dB. Using multiple cyclic frequencies further improves the detection performance. The sum detector of the test statistic $\mathcal{D}_{s,K}$ has the best performance.}
\label{fig:ofdm_5su_pd_vs_SNR}
\end{figure}
\begin{figure}[tbp]
\centering
\includegraphics[width=0.75\columnwidth]{ofdm_5su_pd_vs_pfa_-9dB}
\caption{Probability of detection vs. false alarm rate. Cooperation among secondary users combined with the use of multicycle sum test statistic $\mathcal{D}_{s,K}$ provides the best performance.}
\label{fig:ofdm_5su_pd_vs_pfa}
\end{figure}
In the following simplistic example, we illustrate the gains that may be achieved
via collaborative detection in the face of shadowing effects.
In order to simulate shadowing, the SNR of each user was independently selected randomly from a
normal distribution with a mean of -9 dB and standard deviation of 10 dB.
That is, the logarithm of the received power level is normally distributed.
Fig.~\ref{fig:ofdm_shadowing} depicts the performance of the multicycle detectors for the
simple shadowing scenario. Comparison to Fig.~\ref{fig:ofdm_5su_pd_vs_pfa} reveals that cooperation among secondary users reduces sensitivity to shadowing effects significantly.
\begin{figure}[tbp]
\centering
\includegraphics[width=0.8\columnwidth]{ofdm_shadowing}
\caption{Probability of detection vs. false alarm rate. In order to simulate shadowing the SNR of each user was independently selected randomly from a normal distribution with a mean of -9 dB and standard deviation of 10 dB. Cooperation among secondary users reduces sensitivity to shadowing effects.}
\label{fig:ofdm_shadowing}
\end{figure}
\section{Conclusion}
In this paper, a generalized likelihood ratio test for detecting primary transmissions
with multiple cyclic frequencies has been proposed, and the
asymptotic distribution of the test statistic has been derived.
In this test, impairments such as shadowing and fading are mitigated by
combining the quantized local likelihood ratios from a number of secondary users under
a conditional independence assumption. Simulation examples
demonstrating the improved reliability in the detector performance in the low SNR regime as
well as significant gains obtained via collaborative decision making have also been presented. | 0.047625 |
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Police has no right to slap you, search your phone or call you criminal. That’s the duty of the court. Our live matters. Well done to all who participated online or offline to #EndSARS | 0.10728 |
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These are the resourceful info that we must have here in Malaysia. Just imagine if we can get this, what do you think? Blogging is one aspect in democracy which practicing a freedom and right to speak. I don't mean or want saying it in our own situations.
1- Bloggers can be journalists (and journalists can be bloggers). If you engage in journalism, you're a journalist, with all of the attendant rights, privileges, and protections.
2- Bloggers are entitled to free speech. Internet bullies shouldn't use copyright, libel, or other claims to chill your legitimate speech.
3- Bloggers have the right to political speech. To ensure that the Federal Election Commission (FEC) doesn't gag bloggers' election-related speech.
4- Bloggers have the right to stay anonymous.
5- Bloggers have freedom from liability for hosting speech the same way other web hosts do.
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One thing for sure, we don;t have a true democracy at this time.. Why? Just look how the our leader being chosen... Think Again | 0.399648 |
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View the content:
Results and:
Huang, C. Results and Remarks, Connexions Web site., Dec 14, 2010.
Huang C. Results and Remarks [Connexions Web site]. December 14, 2010. Available at:.
Huang, C. (2010, December 14). Results and Remarks. Retrieved from the Connexions Web site:
Huang, Cathy. "Results and Remarks." Connexions. December 14, 2010..
Cathy Huang, "Results and Remarks," Connexions, December 14, 2010,.
Huang, C. 2010. Results and Remarks. Connexions, December 14, 2010..
Huang, Cathy. Results and Remarks. Connexions. 14 Dec. 2010 <>. | 0.007902 |
Description
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TITLE: Why $\sqrt{\frac {\sum(x-\mu)^2} {N}}$ instead of $\frac {\sum{\Bigl|x-\mu\Bigr|}} {N}$?
QUESTION [3 upvotes]: Possible Duplicate:
Motivation behind standard deviation?
In statistics very often you see something of the sort:
$$
\textrm{quantity}=\sqrt{\frac {\sum(x-\mu)^2} {N}}
$$
to measure things like standard deviation ($\mu$ is the mean here).
It seems that just making an absolute value of the difference will give us a pretty good measure of the same thing:
$$
\textrm{quantity}=\frac {\sum{\Bigl|x-\mu\Bigr|}} {N}
$$
How did we end up with those squares?
REPLY [1 votes]: When you sample $n$ values of a random variable $X$, you have $n$-values $x_1,x_2,\ldots,x_n$. You've just sampled one vector in $\Bbb{R}^n$: $\vec{x}=\langle x_1,x_2,\ldots,x_n\rangle$. The standard formula for standard deviation applies the Euclidean distance metric to this vector and its mid-vector: $\langle\overline{x},\overline{x},\ldots,\overline{x}\rangle$, then dividing by $n$ as a way to account for largness due simply to large dimension.
What is your favorite distance metric in $\mathbb{R}^n$? If it is the usual Euclidean metric, then the standard formula for standard deviation arises. If it is the taxi cab metric, then you could employ the formula that you suggest. But one answer to the question of "why the squares" is that the Euclidean distance metric is generally the natural one. | 0.430921 |
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TITLE: Does $\det(A) \neq 0$ (where A is the coefficient matrix) $\rightarrow$ a basis in vector spaces other than $R^{n}$?
QUESTION [7 upvotes]: I know that for a set of vectors $\{ v_{1}, v_{2}, \ldots , v_{n} \} \in \mathbb{R}^{n}$ we can show that the vectors form a basis in $\mathbb{R}^{n}$ if we show that the coefficient matrix $A$ has the property $\det(A) \neq 0$, because this shows the homogeneous system has only the trivial solution, and the non-homogeneous system is consistent for every vector $(b_{1}, b_{2}, \ldots , b_{n}) \in \mathbb{R}^{n}$.
Intuitively, this concept seems applicable to all polynomials in $\mathbf{P}_{n}$ and all matrices in $M_{nn}$. Can someone validate this?
edit:
I think to make the intuition hold, $A$ must be defined as follows in $M_{nn}$:
Let $M_{1}, M_{2}, ... , M_{k}$ be matrices in $M_{nn}$.
To prove these form a basis for $M_{nn}$, we must show that $c_{1}M_{1} + c_{2}M_{2} + ... + c_{k}M_{k} = 0$ has the only trivial solution, and that every $n \times n$ matrix can be expressed as $c_{1}M_{1} + c_{2}M_{2} + ... + c_{k}M_{k} = B$.
So I believe that for $M_{nn}$, $A$ must be defined as a $n^{2} \times n^{2}$ matrix where each row vector is formed from all the $(i, j)$ entries taken from $M_{1}, M_{2}, ... , M_{k}$ (in that order.)
$\text{e.g. } A = \begin{pmatrix}
M_{1_{1,1}} & M_{2_{1,1}} & ... & M_{k_{1,1}} \\
M_{1_{1,2}} & M_{2_{1,2}} & ... & M_{k_{1,2}} \\
... & ... & ... & ... \\
M_{1_{n,n}} & M_{2_{n,n}} & ... & M_{k_{n,n}}
&
\end{pmatrix}$
However, I am not sure about this.
REPLY [5 votes]: You should be aware that for any given $n$ there's an essentially unique real vector space of dimension $n$, in the sense that any two are isomorphic (although non-canonically). For instance, the space of real polynomials of degree $\leq n$ is a real vector space of dimension $n+1$, hence isomorphic to ${\Bbb R}^{n+1}$, the space $M_n({\Bbb R})$ of square $n\times n$ real matrices is a real vector space of dimension $n^2$, hence isomorphic to ${\Bbb R}^{n^2}$, and so on. Once you prove a particular statement for ${\Bbb R}^{n}$, you proved it for ALL real vector spaces of dimension $n$.
Same thing, more in general, when you consider vector spaces over any other field than $\Bbb R$. | 0.064654 |
\begin{document}
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\title{Examples in dependent theories}
\author{Itay Kaplan and Saharon Shelah}
\thanks{The first author's research was partially supported by the SFB 878
grant.}
\thanks{The second author would like to thank the United States-Israel Binational
Science Foundation for partial support of this research, Grant no.
2002323. Publication 946.}
\begin{abstract}
In the first part we show a counterexample to a conjecture by Shelah
regarding the existence of indiscernible sequences in dependent theories
(up to the first inaccessible cardinal). In the second part we discuss
generic pairs, and give an example where the pair is not dependent.
Then we define the notion of directionality which deals with counting
the number of co-heirs of a type, and prove a trichotomy theorem about
it. Finally we discuss non-splintering, an interesting notion that
appears in the work of Rami Grossberg, Andr\'es Villaveces and Monica
VanDieren, and we show that it is not trivial whenever the directionality
of the theory is not small.
\end{abstract}
\maketitle
\section{Introduction}
\subsection{Existence of indiscernibles}
In the summer of 2008, Saharon Shelah announced in a talk in Rutgers
that he had proved some very important results in dependent theories.
One of these was the existence of indiscernible, an old conjuncture
of his.
Recall,
\begin{defn}
A first order theory $T$ is \emph{dependent} (NIP) if it does not
have the independence property which means: there is no formula $\varphi\left(x,y\right)$
and tuples $\left\langle a_{i},b_{s}\left|\, i<\omega,s\subseteq\omega\right.\right\rangle $
in $\C$ such that $\varphi\left(a_{i},b_{s}\right)$ if and only
if $i\in s$.
\end{defn}
When we say that $T$ has existence of indiscernibles we mean that
inside a large enough set, one can always find a sequence of indiscernible.
To make things more precise, we use the following notation:
\begin{defn}
Let $T$ be a theory. For a cardinal $\kappa$, $n\leq\omega$ and
an ordinal $\delta$, $\kappa\to\left(\delta\right)_{T,n}$ means:
for every set $A\subseteq\C^{n}$ of size $\kappa$, there is a non-constant
sequence of elements of $A$ of length $\delta$ which is indiscernible.
\end{defn}
In stable theories, it is known that for any $\lambda$ satisfying
$\lambda=\lambda^{\left|T\right|},$ $\lambda^{+}\to\left(\lambda^{+}\right)_{T,n}$
(proved by Shelah in \cite{Sh:c}, and follows from local character
of non-forking). The first claim of this kind was proved by Morley
in \cite{Mo65} for $\omega$-stable theories. In \cite{Sh863}, Shelah
proved:
\begin{fact}
\label{fac:TStrongly} If $T$ is strongly dependent (see Definition
\ref{def:StronglyDep} below), then for all $\lambda\geq\left|T\right|$,
$\beth_{\left|T\right|^{+}}\left(\lambda\right)\to\left(\lambda^{+}\right)_{T,n}$
for all $n<\omega$.
\end{fact}
This definition was suggested by Grossberg and Shelah and appears
in \cite[pg. 208, Definition 3.1(2)]{sh:d}, in a slightly different
form
\footnote{The definition there is: $\kappa\to\left(\delta\right)_{n,T}$ if
and only if for each sequence of length $\kappa$ (of $n$-tuples),
there is an indiscernible sub-sequence of length $\delta$. For us
there is no difference because we are dealing with examples where
$\kappa\not\to\left(\mu\right)_{T,n}$. It is also not hard to see
that when $\delta$ is a cardinal these two definitions are equivalent.
}. There it is also proved that such a theorem does not hold for simple
unstable theories in general. So the natural generalization is to
NIP theories, and indeed it is conjectured already in \cite[pg. 209, Conjecture 3.3]{sh:d}.
This conjuncture is connected to a result by Shelah and Cohen: in
\cite{ShCo919}, they proved that a theory is stable if and only if
it can be presented in some sense in a free algebra\textcolor{black}{{}
in a fix vocabulary but allowing function symbols with infinite arity.}
If this result could be extended to: a theory is dependent if and
only if it can be represented as an algebra with ordering, then this
could be used to prove existence of indiscernibles.
Despite announcing it, there was a mistake in the proof for dependent
theories, and here we shall see a counter-example. In this paper,
we shall show that
\begin{thm}
\label{thm:IntoMain}There is a countable dependent theory $T$ such
that if $\kappa$ is smaller than the first inaccessible cardinal,
then for all $n\in\omega$, $\kappa\not\to\left(\omega\right)_{T,n}$.
\end{thm}
It appears in a more precise way as Theorem \ref{thm:MainThm} below.
An even stronger result can be obtained, namely
\begin{thm}
\label{thm:general result}For every $\theta$ there is a dependent
theory $T$ of size $\theta$ such that for all $\kappa$ and $\delta$,
\textup{$\kappa\to\left(\delta\right)_{T,1}$ if and only if $\kappa\to\left(\delta\right)_{\theta}^{<\omega}$. }
\end{thm}
where
\begin{defn}
$\kappa\to\left(\delta\right)_{\theta}^{<\omega}$ means: for every
coloring $c:\left[\kappa\right]^{<\omega}\to\theta$ there is an homogeneous
sub-sequence of length $\delta$ (i.e. there exists $\left\langle \alpha_{i}\left|\, i<\delta\right.\right\rangle \in\leftexp{\delta}{\lambda}$
and $\left\langle c_{n}\left|\, n<\omega\right.\right\rangle \in\leftexp{\omega}{\theta}$
such that $c\left(\alpha_{i_{0}},\ldots,\alpha_{i_{n-1}}\right)=c_{n}$
for every $i_{0}<\ldots<i_{n-1}<\delta$).
\end{defn}
One can see that $\kappa\to\left(\delta\right)_{\theta}^{<\omega}$
always implies that $\kappa\to\left(\delta\right)_{T,n}$ for all
$n<\omega$, so this is the best result possible.
However, the proof of Theorem \ref{thm:general result} is considerably
harder, so it is given in a subsequent work \cite{KaSh975}.
The second part of the paper is devoted to giving a related example
in the field of real numbers. By Fact \ref{fac:TStrongly}, as $Th\left(\mathbb{R}\right)$
is strongly dependent, we cannot prove the same theorem, but instead
we show that the requirement that $n<\omega$ is necessary:
\begin{thm}
\label{thm:IntoMainRCF}If $\kappa$ is smaller than the first strongly
inaccessible cardinal, then $\kappa\not\to\left(\omega\right)_{RCF,\omega}$.
\end{thm}
This is Theorem \ref{thm:MainThmRCF} below.
\subsubsection*{Notes}
Many strong results that have been announced in that talk in 2008
remain true. For instance, the main theorem there is that a theory
is dependent if and only if the number of types up to isomorphism
is small (see more in \cite{Sh950}).
We also should note that a related result can be found in an unpublished
paper in Russian by Kudajbergenov that states that for every ordinal
$\alpha$ there exists a dependent theory (but it may be even strongly
dependent) $T_{\alpha}$ such that $\left|T_{\alpha}\right|=\left|\alpha\right|+\aleph_{0}$
and $\beth_{\alpha}\left(\left|T_{\alpha}\right|\right)\not\to\left(\aleph_{0}\right)_{T_{\alpha},1}$
and thus seem to indicate that the bound in Fact \ref{fac:TStrongly}
is tight.
\subsubsection*{Idea of the construction}
The counterexample is a ``tree of trees'' with functions connecting
the different trees. For every $\eta$ in the tree $\leftexp{\omega\geq}2$
we shall have a predicate $P_{\eta}$ and an ordering $<_{\eta}$
such that $\left(P_{\eta},<_{\eta}\right)$ is a dense tree. In addition
we shall have functions $G_{\eta,\eta\concat\left\{ i\right\} }:P_{\eta}\to P_{\eta\concat\left\langle i\right\rangle }$
for $i=0,1$. The idea is to prove that $\kappa\nrightarrow\left(\mu\right)_{T,1}$
by induction on $\kappa$, i.e. to prove that in $P_{\left\langle \right\rangle }$
there are no indiscernibles. To use the induction hypothesis, we push
the counter examples we already have for smaller $\kappa$'s to deeper
levels in the tree $\leftexp{\omega\geq}2$ .
\subsection{Generic pairs}
In a series of papers (\cite{Sh:900,Sh877,Sh906,Sh950}), Shelah has
proved (among other things) that dependent theories give rise to a
``generic pair'' of models (and in fact this characterize dependent
theories). The natural question is whether the theory of the pair
is again dependent. The answer is no. We present an example of an
$\omega$-stable theory all of whose generic pair have the independence
property.
\subsection{Directionality}
The directionality of a theory measures the number of finitely satisfiable
global extension of a complete type (these are also called co-heirs).
We say that a theory has \emph{small} directionality if for every
type $p$ over a model $M$, the number of complete finitely satisfiable
(in $M$) $\Delta$-types which are consistent with $p$ is finite
for all finite sets $\Delta$. The theory has \emph{medium} directionality
if this number is bounded by $\left|M\right|$, and it has \emph{large}
directionality if it is not. We prove a trichotomy theorem (Theorem
\ref{thm:trichotomy} below), showing that every theory is either
of small, medium or large directionality.
We provide examples of dependent theories of each kind of directionality.
\subsection{Splintering}
The last part of the paper is connected to a paper by the work of
Rami Grossberg, Andr\'es Villaveces and Monica VanDieren. In \cite{GrViVa}
they study Shelah's Generic pair conjecture (which is now a theorem)
and in their analysis, they came up with the notion of splintering
which is similar to splitting. We show that in any dependent theory
with medium or large directionality, splitting is different than splitting.
We also provide an example of such a theory with small directionality,
and prove this cannot happen in the stable realm.
\subsection{Notation}
We do not distinguish elements and tuples unless we say so explicitly.
$\C$ will be the monster model of the theory.
$S\left(A\right)$ is the set of all types over $A$.
For a set of formulas with a partition of variables, $\Delta\left(x,y\right)$,
$L_{\Delta}\left(A\right)$ is the set of formulas of the form $\varphi\left(x,a\right),\neg\varphi\left(x,a\right)$
where $\varphi\left(x,y\right)\in\Delta$ and $a\in A$. $S_{\Delta}\left(A\right)$
is the set of all complete $L_{\Delta}\left(A\right)$-types over
$A$. Similarly we may define $\tp_{\Delta}\left(b/A\right)$ as the
set of formulas $\varphi\left(x,a\right)$ such that $\varphi\left(x,y\right)\in\Delta$
and $\C\models\varphi\left(b,a\right)$. For a partial type $p\left(x\right)$
over $A$, $p\upharpoonright\Delta=p\cap L_{\Delta}\left(A\right)$.
Usually we want to consider a set of formulas $\Delta$ without specifying
a partition of the variables. In this case, for a tuple of variables
$x$, $\Delta^{x}$ is a set of partitioned formulas induced from
$\Delta$ by partitioning the formulas in $\Delta$ to $\left(x,y\right)$
in all possible ways. Then $L_{\Delta}^{x}\left(A\right)$ is just
$L_{\Delta^{x}}\left(A\right)$ and $S_{\Delta}^{x}\left(A\right)$,
$p\upharpoonright\Delta^{x}$ are defined similarly. If $x$ is clear
from the context, we omit it. So for instance, when $p$ is type in
$x$ over $A$, then $p\upharpoonright\Delta$ is the set of all formulas
$\varphi\left(x,a\right)$ where $\varphi\left(z,w\right)\in\Delta$.
\section{\label{sec:Indisc}Lack of indiscernibles}
\subsection{Preliminaries}
\begin{defn}
We shall need the following fact: \end{defn}
\begin{fact}
\label{fac:DepPolBd} \cite[II, 4]{Sh:c} Let $T$ be any theory.
Then for all $n<\omega$, $T$ is dependent if and only if $\square_{n}$
if and only if $\square_{1}$ where for all $n<\omega$,\end{fact}
\begin{itemize}
\item [$\square_n$] For every finite set of formulas $\Delta\left(x,y\right)$
with $n=\lg\left(x\right)$, there is a polynomial $f$ such that
for every finite set $A\subseteq M\models T$, $\left|S_{\Delta}\left(A\right)\right|\leq f\left(\left|A\right|\right)$.
\end{itemize}
Since we also discuss strongly dependent theories, here is the definition:
\begin{defn}
\label{def:StronglyDep}A theory is called \emph{strongly dependent}
if there is no sequence of formulas
\[
\left\langle \varphi_{n}\left(x,y_{n}\right)\left|\, n<\omega\right.\right\rangle
\]
such that the following set of formulas is consistent with the theory
\[
\left\{ \varphi_{n}\left(x_{\eta},y_{n,k}\right)^{\eta\left(n\right)=k}\left|\,\eta:\omega\to\omega\right.\right\} .
\]
Where $\varphi^{\mbox{True}}=\varphi,\varphi^{\mbox{False}}=\neg\varphi$.
\end{defn}
See \cite{Sh863} for further discussion of strongly dependent theories.
There it is proved that $Th\left(\mathbb{R}\right)$ is strongly dependent,
and so is the theory of the $p$-adics.
\subsection{An example of lack of indiscernibles.}
Let $S_{n}$ be the finite binary tree $\leftexp{n\geq}2$. On a well
ordered tree such as $S_{n}$, we define $<_{\suc}$ as follows: $\eta<_{\suc}\nu$
if $\nu$ is a successor of $\eta$ in the tree.
Let $L_{n}$ be the following language:
\[
L_{n}=\left\{ P_{\eta},<_{\eta},\wedge_{\eta},G_{\eta,\nu}\left|\,\eta,\nu\in S_{n},\eta<_{\suc}\nu\right.\right\} .
\]
Where:
\begin{itemize}
\item $P_{\eta}$ is a unary predicate ; $<_{\eta}$ is a binary relation
symbol ; $\wedge_{\eta}$ is a binary function symbol ; $G_{\eta,\nu}$
is a unary function symbol.
\end{itemize}
Let $T_{n}^{\forall}$ be the following theory:
\begin{itemize}
\item $P_{\eta}\cap P_{\nu}=\emptyset$ for $\eta\neq\nu$.
\item $\left(P_{\eta},<_{\eta},\wedge_{\eta}\right)$ is a tree, where $\wedge_{\eta}$
is the meet function on $P_{\eta}$, i.e.
\[
x\wedge_{\eta}y=\max\left\{ z\in P_{\eta}\left|\, z\leq_{\eta}x\,\&\, z\leq_{\eta}y\right.\right\}
\]
(so if for example, $x\notin P_{\eta}$, then $x\wedge_{\eta}y=x$).
\item $G_{\eta,\nu}:P_{\eta}\to P_{\nu}$, meaning that outside $P_{\eta}$,
$G_{\eta,\nu}$ is the identity, and no further restrictions.
\end{itemize}
Thus we have:
\begin{claim}
$T_{n}^{\forall}$ is a universal theorem.
\end{claim}
\begin{claim}
$T_{n}^{\forall}$ has the joint embedding property (JEP) and the
amalgamation property (AP). \end{claim}
\begin{proof}
Easy to see.
\end{proof}
From this we deduce, by e.g. \cite[Theorem 7.4.1]{Hod}:
\begin{cor}
\label{cor:ModelCom}$T_{n}^{\forall}$ has a model completion, $T_{n}$
which eliminates quantifiers, and moreover: if $M\models T_{n+1}^{\forall}$,
$M'=M\upharpoonright L_{n}$ and $M'\subseteq N'\models T_{n}^{\forall}$
then $N'$ can be enriched to a model $N$ of $T_{n+1}^{\forall}$
so that $M\subseteq N$. Hence if $M$ is an existentially closed
model of $T_{n+1}^{\forall}$, then $M'$ is an e.c. model of $T_{n}^{\forall}$.
Hence $T_{n}\subseteq T_{n+1}$ (for more see \cite[Theorem 8.2.4]{Hod}).\end{cor}
\begin{proof}
The moreover part: for each $\eta\in S_{n+1}\backslash S_{n}$, we
define $P_{\eta}^{N}=P_{\eta}^{M}$ and in the same way $\wedge_{\eta}$.
The functions $G_{\eta,\nu}$ for $\eta\in S_{n}$ and $\nu\in S_{n+1}$
will be extensions of $G_{\eta,\nu}$.
\end{proof}
Now we show that $T_{n}$ is dependent, but before that, a few easy
remarks:
\begin{obs}
\label{Obs:Bijection}$ $
\begin{enumerate}
\item If $A\subseteq M\models T_{0}^{\forall}$ is a finite substructure
(so just a tree, with no extra structure), then for all $b\in M$,
the structure generated by $A$ and $b$ is $A\cup\left\{ b\right\} \cup\left\{ \max\left\{ b\wedge a\left|\, a\in A\right.\right\} \right\} $.
\item If $M\models T_{n}^{\forall}$ and $\eta\in\leftexp{n\geq}2$, we
can define a new structure $M_{\eta}\models T_{n-\lg\left(\eta\right)}^{\forall}$
whose universe is $\bigcup\left\{ P_{\eta\concat\nu}^{M}\left|\,\nu\in\leftexp{n-\lg\left(n\right)\geq}2\right.\right\} $
by: $P_{\nu}^{M_{\eta}}=P_{\eta\concat\nu}^{M}$, and in the same
way we interpret every other symbol (for instance, $G_{\nu_{1},\nu_{2}}^{M_{\eta}}=G_{\eta\concat\nu_{1},\eta\concat\nu_{1}}^{M}$).
For every formula $\varphi\left(x\right)\in L_{n-\lg\left(\eta\right)}$
there is a formula $\varphi'\left(x\right)\in L_{n}$ such that for
all $a\in M_{\eta}$, $M\models\varphi'\left(a\right)$ if and only
if $M_{\eta}\models\varphi\left(a\right)$ (we get $\varphi'$ by
concatenating $\eta$ before any symbol).
\item For $M$ as before and $\eta\in\leftexp{n\geq}2$, for any $k<\omega$
there is a bijection between
\[
\left\{ p\left(x_{0},\ldots,x_{k-1}\right)\in S_{k}^{\qe}\left(M\right)\left|\,\forall i<k\left(P_{\eta}\left(x_{i}\right)\in p\right)\right.\right\}
\]
and
\[
\left\{ p\left(x_{0},\ldots,x_{k-1}\right)\in S_{k}^{\qe}\left(M_{\eta}\right)\left|\,\forall i<k\left(P_{\left\langle \right\rangle }\left(x_{i}\right)\in p\right)\right.\right\} .
\]
\end{enumerate}
\end{obs}
\begin{proof}
(3): The bijection is given by (2). This is well defined, meaning
that if $p\left(x_{0},\ldots,x_{k-1}\right)$ is a type over $M_{\eta}$
such that $P_{\left\langle \right\rangle }\left(x_{i}\right)\in p$
for all $i<k$, then $\left\{ \varphi'\left|\,\varphi\in p\right.\right\} $
determines a complete type over $M$, such that $P_{\eta}\left(x_{i}\right)\in p$
for all $i<k$. This is because any atomic formula of the form $F\left(x,a\right)<_{\nu}G\left(x,b\right)$
where $a,b\in M$ and $F$, $G$ are terms is either trivially false,
as in the case $\nu\not\geq\eta$, or trivially equivalent to a formula
with no involvement of $x$'s at all (as in the case $a\wedge_{\eta}x<_{\nu}b$
where $a\notin P_{\eta}$) or it is equivalent to a formula of the
form $F'\left(x,a'\right)<_{\nu}G'\left(x,b'\right)$ where $a',b'$
are from $M_{\eta}$.\end{proof}
\begin{prop}
$T_{n}$ is dependent.\end{prop}
\begin{proof}
We use Fact \ref{fac:DepPolBd}. It is sufficient to find a polynomial
$f\left(x\right)$ such that for every finite set $A$, $\left|S_{1}\left(A\right)\right|\leq f\left(\left|A\right|\right)$.
First we note that for a set $A$, the size of the structure generated
by $A$ is bounded by a polynomial in $\left|A\right|$: it is generated
by applying $\wedge_{\left\langle \right\rangle }$ on $P_{\left\langle \right\rangle }\cap A$,
applying $G_{\left\langle \right\rangle ,\left\langle 1\right\rangle }$
and $G_{\left\langle \right\rangle ,\left\langle 0\right\rangle }$,
and then applying $\wedge_{\left\langle 0\right\rangle },\wedge_{\left\langle 1\right\rangle }$
and so on. Every step in the process is polynomial, and it ends after
$n$ steps.
Hence we can assume that $A$ is a substructure, i.e. $A\models T_{n}^{\forall}$.
The proof is by induction on $n$. To easy notation, we shall omit
the subscript $\eta$ from $<_{\eta}$ and $\wedge_{\eta}$.
First we deal with the case $n=0$. In $T_{0}$, $P_{\left\langle \right\rangle }$
is a a tree with no extra structure, while outside $P_{\left\langle \right\rangle }$
there is no structure at all. The number of types outside $P_{\left\langle \right\rangle }$
is bounded by $\left|A\right|+1$ (because there is only one non-algebraic
type). In the case that $P_{\left\langle \right\rangle }\left(x\right)\in p$
for some type $p$ over $A$, we can characterize $p$ by characterizing
the type of $x':=\max\left\{ a\wedge x\left|\, a\in A\right.\right\} $,
i.e. the cut that $x'$ induces on the tree, and by knowing whether
$x'=x$ or $x>x'$ (we note that in general, every theory of a tree
is dependent by \cite{Parigot}).
Now assume that the claim is true for $n$. Suppose $\eta\in\leftexp{n+1\geq}2$
and $1\leq\lg\left(\eta\right)$. By Observation \ref{Obs:Bijection}(3),
there is a bijection between the types $p\left(x\right)$ over $A$
where $P_{\eta}\left(x\right)\in p$ and the types $p\left(x\right)$
in $T_{n+1-\lg\left(\eta\right)}$ over $A_{\eta}$ where $P_{\left\langle \right\rangle }\in p$.
$A_{\eta}\models T_{n+1-\lg\left(\eta\right)}^{\forall}$, and so
by the induction hypothesis, the number of types over $A_{\eta}$
is bounded by a polynomial in $\left|A_{\eta}\right|\leq\left|A\right|$.
As the number of types $p\left(x\right)$ such that $P_{\eta}\left(x\right)\notin p$
for all $\eta$ is bounded by $\left|A\right|+1$ as in the previous
case, we are left with checking the number of types $p\left(x\right)$
such that $P_{\left\langle \right\rangle }\left(x\right)\in p$.
In order to describe $p$, we first have to describe $p$ restricted
to the language $\left\{ <_{\left\langle \right\rangle },\wedge_{\left\langle \right\rangle }\right\} $,
and this is polynomially bounded. Let $x'=\max\left\{ a\wedge x\left|\, a\in A\right.\right\} $.
By Observation \ref{Obs:Bijection}(1), if $A\cup\left\{ x\right\} $
is not closed under $\wedge_{\left\langle \right\rangle }$, $x'$
is the only new element in the structure generated by $A\cup\left\{ x\right\} $
in $P_{\left\langle \right\rangle }$. Hence , we are left to determine
the type of the pairs $\left(G_{\left\langle \right\rangle ,\left\langle i\right\rangle }\left(x\right),G_{\left\langle \right\rangle ,\left\langle i\right\rangle }\left(x'\right)\right)$
over $A$ for $i=0,1$ (if $x'$ is not new, then it's enough to determine
the type of $G_{\left\langle \right\rangle ,\left\langle i\right\rangle }\left(x\right)$).
The number of these types is equal to the number of types of pairs
in $T_{n}$ over $A_{\left\langle i\right\rangle }$. As $T_{n}$
is dependent we are done by Fact \ref{fac:DepPolBd}.\end{proof}
\begin{defn}
Let $L=\bigcup_{n<\omega}L_{n}$, $T=\bigcup_{n<\omega}T_{n}$ and
$T^{\forall}=\bigcup_{n<\omega}T_{n}^{\forall}$.
\end{defn}
We easily have
\begin{cor}
$T$ is complete, it eliminates quantifiers and is dependent.
\end{cor}
We shall prove the following theorem:
\begin{thm}
\label{thm:MainThm}For any two cardinals $\mu\leq\kappa$ such that
in $\left[\mu,\kappa\right]$ there are no (uncountable) strongly
inaccessible cardinals, $\kappa\nrightarrow\left(\mu\right)_{T,1}$.
\end{thm}
We shall prove a slightly stronger statement, by induction on $\kappa$:
\begin{prop}
Given $\mu$ and $\kappa$, such that either $\kappa<\mu$ or there
are no (uncountable) strongly inaccessible cardinals in $\left[\mu,\kappa\right]$,
there is a model $M\models T^{\forall}$ such that $\left|P_{\left\langle \right\rangle }^{M}\right|\geq\kappa$
and $P_{\left\langle \right\rangle }^{M}$ does not contain a non-constant
indiscernible sequence (for quantifier free formulas) of length $\mu$.
\end{prop}
From now on, indiscernible will only mean ``indiscernible for quantifier
free formulas''.
\begin{proof}
Fix $\mu$. The proof is by induction on $\kappa$. We divide into
cases:
\begin{caseenv}
\item $\kappa<\mu$. Clear.
\item $\kappa=\mu=\aleph_{0}$. Denote $\eta_{j}=\left\langle 1\ldots1\right\rangle $,
i.e. the constant sequence of length $j$ and value $1$. Find $M\models T^{\forall}$
such that its universe contains a set $\left\{ a_{i,j}\left|\, i,j<\omega\right.\right\} $
where $a_{i,j}\neq a_{i',j'}$ for all $\left(i,j\right)\neq\left(i',j'\right)$,
$a_{i,j}\in P_{\eta_{j}}$ and in addition $G_{\eta_{j},\eta_{j+1}}\left(a_{i,j}\right)=a_{i,j+1}$
if $j<i$ and $G_{\eta_{j},\eta_{j+1}}\left(a_{i,j}\right)=a_{0,j+1}$
otherwise. We also need that $P_{\left\langle \right\rangle }^{M}=\left\{ a_{i,0}\left|\, i<\omega\right.\right\} $.
Any model satisfying these properties will do (so no need to specify
what the tree structures are). Now, if in $P_{\left\langle \right\rangle }^{M}=\left\{ a_{i,0}\left|\, i<\omega\right.\right\} $
there is a non-constant indiscernible sequence, $\left\langle a_{i_{k},0}\left|\, k<\omega\right.\right\rangle $,
then there is a large enough $j$ such that
\[
G_{\eta_{j},\eta_{j+1}}\circ\ldots\circ G_{\eta_{0},\eta_{1}}\left(a_{i_{0},0}\right)=G_{\eta_{j},\eta_{j+1}}\circ\ldots\circ G_{\eta_{0},\eta_{1}}\left(a_{i_{1},0}\right).
\]
But for every large enough $k$, $G_{\eta_{j},\eta_{j+1}}\circ\ldots\circ G_{\eta_{0},\eta_{1}}\left(a_{i_{1},0}\right)\neq G_{\eta_{j},\eta_{j+1}}\circ\ldots\circ G_{\eta_{0},\eta_{1}}\left(a_{i_{k},0}\right)$
--- contradiction.
\item $\kappa$ is singular. Suppose $\kappa=\bigcup_{i<\sigma}\lambda_{i}$
where $\sigma,\lambda_{i}<\kappa$ for all $i<\sigma$. By the induction
hypothesis, for $i<\sigma$ there is a model $M_{i}\models T^{\forall}$
such that $\left|P_{\left\langle \right\rangle }^{M_{i}}\right|\geq\lambda_{i}$
and in $P_{\left\langle \right\rangle }^{M_{i}}$ there is no non-constant
indiscernible sequence of length $\mu$. Also, there is a model $N$
such that $\left|P_{\left\langle \right\rangle }^{N}\right|\geq\sigma$
and in $P_{\left\langle \right\rangle }^{N}$ there is no non-constant
indiscernible sequence of length $\mu$. We may assume that the universes
of all these models are pairwise disjoint and disjoint from $\kappa$.
Suppose that $\left\{ a_{i}\left|\, i<\sigma\right.\right\} \subseteq P_{\left\langle \right\rangle }^{N}$,
and $\left\{ b_{j}\left|\,\sum_{l<i}\lambda_{l}\leq j<\lambda_{i}\right.\right\} \subseteq P_{\left\langle \right\rangle }^{M_{i}}$
witness that $\left|P_{\left\langle \right\rangle }^{N}\right|\geq\sigma$
and $\left|P_{\left\langle \right\rangle }^{M_{i}}\right|\geq\lambda_{i}\backslash\sum_{l<i}\lambda_{l}$.
Let $\bar{M}$ be a model extending each $M_{i}$ and containing the
disjoint union of the sets $\bigcup_{i<\sigma}M_{i}$ (exists by JEP).
Define a new model $M\models T^{\forall}$: $\left(P_{\left\langle \right\rangle }^{M},<_{\left\langle \right\rangle }\right)=\left(\kappa,<\right)$
(so $\wedge_{\left\langle \right\rangle }=\min$); $\left(P_{\left\langle 1\right\rangle \concat\eta}^{M},<_{\eta}\right)=\left(P_{\eta}^{N},<_{\eta}\right)$
and $\left(P_{\left\langle 0\right\rangle \concat\eta}^{M},<_{\left\langle 0\right\rangle ,\eta}\right)=\left(P_{\eta}^{\bar{M}},<_{\eta}\right)$.
In the same way define $\wedge_{\eta}$ for all $\eta$ of length
$\geq1$. The functions are also defined in the same way: $G_{\left\langle 1\right\rangle \concat\eta,\left\langle 1\right\rangle \concat\nu}^{M}=G_{\eta,\nu}^{N}$
and $G_{\left\langle 0\right\rangle \concat\eta,\left\langle 0\right\rangle \concat\nu}^{M}=G_{\eta,\nu}^{\bar{M}}$.
We are left to define $G_{\left\langle \right\rangle ,\left\langle 0\right\rangle }$
and $G_{\left\langle \right\rangle ,\left\langle 1\right\rangle }$.
So let: $G_{\left\langle \right\rangle ,\left\langle 1\right\rangle }\left(\alpha\right)=a_{\min\left\{ i\left|\,\alpha<\lambda_{i}\right.\right\} }$
and $G_{\left\langle \right\rangle ,\left\langle 0\right\rangle }\left(\alpha\right)=b_{\alpha}$
for all $\alpha<\kappa$.
Note that if $I$ is an indiscernible sequence contained in $P_{\left\langle 1\right\rangle }^{M}$
then $I$ is an indiscernible sequence in $N$ contained in $P_{\left\langle \right\rangle }^{N}$,
and the same is true for $P_{\left\langle 0\right\rangle }^{M}$ and
$\bar{M}$.
Assume $\left\langle \alpha_{j}\left|\, j<\mu\right.\right\rangle $
is an indiscernible sequence in $P_{\left\langle \right\rangle }^{M}$.
Then $\left\langle G_{\left\langle \right\rangle ,\left\langle 1\right\rangle }\left(\alpha_{j}\right)\left|\, j<\mu\right.\right\rangle $
is a constant sequence (by the choice of $N$). So there is $i<\sigma$
such that $\sum_{l<i}\lambda_{l}\leq\alpha_{j}<\lambda_{i}$ for all
$j<\mu$. So $\left\langle G_{\left\langle \right\rangle ,\left\langle 0\right\rangle }\left(\alpha_{j}\right)=b_{\alpha_{j}}\left|\, j<\mu\right.\right\rangle $
is a constant sequence (it is indiscernible in $P_{\left\langle \right\rangle }^{\bar{M}}$
and in fact contained in $P_{\left\langle \right\rangle }^{M_{i}}$),
hence $\left\langle \alpha_{j}\left|\, j<\mu\right.\right\rangle $
is constant, as we wanted.
\item $\kappa$ is regular uncountable. By the hypothesis of the proposition,
$\kappa$ is not strongly inaccessible, so there is some $\lambda<\kappa$
such that $2^{\lambda}\geq\kappa$. By the induction hypothesis on
$\lambda$, there is a model $N\models T^{\forall}$ such that in
$P_{\left\langle \right\rangle }^{N}$ there is no non-constant indiscernible
sequence of length $\mu$. Let $\left\{ a_{i}\left|\, i\leq\lambda\right.\right\} \subseteq P_{\left\langle \right\rangle }^{N}$
witness that $\left|P_{\left\langle \right\rangle }^{N}\right|\geq\lambda$.
Define $M\models T^{\forall}$ as follows: $P_{\left\langle \right\rangle }^{M}=\leftexp{\lambda\geq}2$
and the ordering is inclusion (equivalently, the ordering is by initial
segment). $\wedge_{\left\langle \right\rangle }$ is defined naturally:
$f\wedge_{\left\langle \right\rangle }g=f\upharpoonright\min\left\{ \alpha\left|\, f\left(\alpha\right)\neq g\left(\alpha\right)\right.\right\} $.
For all $\eta$, let $P_{\left\langle 1\right\rangle \concat\eta}^{M}=P_{\eta}^{N}$,
and the ordering and the functions are naturally induced from $N$.
The main point is that we set $G_{\left\langle \right\rangle ,\left\langle 1\right\rangle }\left(f\right)=a_{\lg\left(f\right)}$.
Now choose $P_{\left\langle 0\right\rangle \concat\eta}^{M}$, $G_{\left\langle 0\right\rangle \concat\eta,\left\langle 0\right\rangle \concat\nu}$,
etc. arbitrarily, and let $G_{\left\langle \right\rangle ,\left\langle 0\right\rangle }$
be any function.
Suppose that $\left\langle f_{i}\left|\, i<\mu\right.\right\rangle $
is an non-constant indiscernible sequence:
If $f_{1}<f_{0}$ (i.e. $f_{1}<_{\left\langle \right\rangle }f_{0}$),
we shall have an infinite decreasing sequence in a well-ordered tree
--- a contradiction.
If $f_{0}<f_{1}$, $\left\langle f_{i}\left|\, i<\mu\right.\right\rangle $
is increasing, so $\left\langle G_{\left\langle \right\rangle ,\left\langle 1\right\rangle }^{M}\left(f_{i}\right)=a_{\lg\left(f_{i}\right)}\left|\, i<\mu\right.\right\rangle $
is non-constant --- contradiction (as it is an indiscernible sequence
in $M$ and hence in $P_{\left\langle 0\right\rangle }^{N}$).
Let $h_{i}=f_{0}\wedge f_{i+1}$ for $i<\mu$ (where $\wedge=\wedge_{\left\langle \right\rangle }$).
This is an indiscernible sequence, and by the same arguments, it cannot
increase or decrease, but as $h_{i}\leq f_{0}$, and $\left(P_{\left\langle \right\rangle },<_{\left\langle \right\rangle }\right)$
is a tree, it follows that $h_{i}$ is constant.
Assume $f_{0}\wedge f_{1}<f_{1}\wedge f_{2}$, then $f_{2i}\wedge f_{2i+1}<f_{2\left(i+1\right)}\wedge f_{2\left(i+1\right)+1}$
for all $i<\mu$, and again $\left\langle f_{2i}\wedge f_{2i+1}\left|\, i<\mu\right.\right\rangle $
an increasing indiscernible sequence and we have a contradiction.
By the same reasoning, it cannot be that $f_{0}\wedge f_{1}>f_{1}\wedge f_{2}$.
As $\left(P_{\left\langle \right\rangle },<_{\left\langle \right\rangle }\right)$
is a tree, we conclude that $f_{0}\wedge f_{2}=f_{0}\wedge f_{1}=f_{1}\wedge f_{2}$.
But that is a contradiction (because if $\alpha=\lg\left(f_{0}\wedge f_{1}\right)$,
then $\left|\left\{ f_{0}\left(\alpha\right),f_{1}\left(\alpha\right),f_{2}\left(\alpha\right)\right\} \right|=3$).
\end{caseenv}
\end{proof}
\subsection{In RCF there are no indiscernibles of $\omega$-tuples.}
Here we show that a related phenomenon occurs in real closed fields.
So here we assume $\C\models RCF$.
\begin{notation}
The set of all open intervals $\left(a,b\right)$ (where $a<b$ and
$a,b\in\C$) is denoted by $\I$. \end{notation}
\begin{defn}
\label{def:arrowInterval}For a cardinal $\kappa$, $n\leq\omega$
and an ordinal $\delta$, $\kappa\to\left(\delta\right)_{n}^{\intr}$
means: for every set $A$ of (non-empty, open) $n$-tuples of intervals
(so for each $\bar{I}\in A$, $\bar{I}=\left\langle I^{i}\left|\, i<n\right.\right\rangle \in\mathfrak{J}^{n}$)
of size $\kappa$, there is a sequence $\left\langle \bar{I}_{\alpha}\left|\,\alpha<\delta\right.\right\rangle \in\leftexp{\delta}A$
of order type $\delta$ such that for $\alpha<\beta<\delta$, $\bar{I}_{\alpha}\neq\bar{I}_{\beta}$,
and $\left\langle \bar{b}_{\alpha}\left|\,\alpha<\delta\right.\right\rangle $
such that $\bar{b}_{\alpha}\in\bar{I}_{\alpha}$ (i.e. $\bar{b}_{\alpha}=\left\langle b_{\alpha}^{0},\ldots,b_{\alpha}^{n-1}\right\rangle $
and $b_{\alpha}^{i}\in I_{\alpha}^{i}$) such that $\left\langle \bar{b}_{\alpha}\left|\,\alpha<\delta\right.\right\rangle $
is an indiscernible sequence. \end{defn}
\begin{rem}
Note that
\begin{enumerate}
\item If $\kappa\to\left(\delta\right)_{n}^{\intr}$ then $\kappa\to\left(\delta\right)_{m}^{\intr}$
for all $m\leq n$.
\item If $\kappa\not\to\left(\delta\right)_{n}^{\intr}$ then $\kappa\not\to\left(\delta\right)_{RCF,n}$
(why? if $A$ witness that $\kappa\not\to\left(\delta\right)_{n}^{\intr}$,
then for each $\bar{I}\in A$, choose $\bar{b}_{\bar{I}}\in\bar{I}$
(as above) in such a way that $\left\{ \bar{b}_{\bar{I}}\left|\,\bar{I}\in A\right.\right\} $
has size $\kappa$. By definition this set witnesses $\kappa\not\to\left(\delta\right)_{RCF,n}$).
\item If $\lambda<\kappa$ and $\kappa\not\to\left(\delta\right)_{n}^{\intr}$
then $\lambda\not\to\left(\delta\right)_{n}^{\intr}$.
\end{enumerate}
\end{rem}
We shall prove the following theorem:
\begin{thm}
\label{thm:MainThmRCF}For any two cardinals $\mu\leq\kappa$ such
that in $\left[\mu,\kappa\right]$ there are no strongly inaccessible
cardinals, $\kappa\nrightarrow\left(\mu\right)_{\omega}^{\intr}$.
\end{thm}
The proof follows from a sequence of claims:
\begin{claim}
\label{cla:Obvious}If $\kappa<\mu$ then $\kappa\not\to\left(\mu\right)_{n}^{\intr}$
for all $n\leq\omega$.\end{claim}
\begin{proof}
Obvious.\end{proof}
\begin{claim}
\label{cla:aleph0}If $\kappa=\mu=\aleph_{0}$ then $\kappa\not\to\left(\mu\right)_{1}^{\intr}$.\end{claim}
\begin{proof}
For $n<\omega$, let $I_{n}=\left(n,n+1\right)$. \end{proof}
\begin{claim}
\label{cla:Singular}Suppose $\kappa=\sum_{i<\sigma}\lambda_{i}$
and $n\leq\omega$. Then, if $\sigma\not\to\left(\mu\right)_{n}^{\intr}$
and $\lambda_{i}\not\to\left(\mu\right)_{n}^{\intr}$ then $\kappa\not\to\left(\mu\right)_{2+2n}^{\intr}$.\end{claim}
\begin{proof}
By assumption, we have a set of intervals $\left\{ \bar{R}_{i}\left|\, i<\sigma\right.\right\} $
that witness $\sigma\not\to\left(\mu\right)_{n}^{\intr}$ and for
each $i<\sigma$ we have $\left\{ \bar{S}_{\beta}\left|\,\sum_{j<i}\lambda_{j}<\beta<\lambda_{i}\right.\right\} $
that witness $\left|\lambda_{i}\backslash\sum_{j<i}\lambda_{j}\right|\not\to\left(\mu\right)_{n}^{\intr}$.
Fix an increasing sequence of elements $\left\langle b_{i}\left|\, i<\sigma\right.\right\rangle $.
For $\alpha<\kappa$, let $\beta=\beta\left(\alpha\right)=\min\left\{ i<\sigma\left|\,\alpha<\lambda_{i}\right.\right\} $
and for $i<2+2n$, define:
\begin{itemize}
\item If $i=0$, let $I_{\alpha}^{i}=\left(b_{2\beta\left(\alpha\right)},b_{2\beta\left(\alpha\right)+1}\right)$.
\item If $i=1$, let $I_{\alpha}^{i}=\left(b_{2\beta\left(\alpha\right)+1},b_{2\beta\left(\alpha\right)+2}\right)$.
\item If $i=2k+2$, let $I_{\alpha}^{i}=R_{\beta\left(\alpha\right)}^{k}$.
\item If $i=2k+3$, let $I_{\alpha}^{i}=S_{\alpha}^{k}$.
\end{itemize}
Suppose $\left\langle \bar{b}_{\varepsilon}\left|\,\varepsilon<\mu\right.\right\rangle $
is an indiscernible sequence such that $\bar{b}_{\varepsilon}\in\bar{I}_{\alpha_{\varepsilon}}$
for $\varepsilon<\mu$. Denote $\bar{b}_{\varepsilon}=\left\langle b_{0}^{\varepsilon},\ldots,b_{2+2n-1}^{\varepsilon}\right\rangle $.
Note that $b_{1}^{\varepsilon}<b_{0}^{\varepsilon'}$ if and only
if $\beta\left(\alpha_{\varepsilon}\right)<\beta\left(\alpha_{\varepsilon'}\right)$
(we need two intervals for the ``only if'' direction).
Hence $\left\langle \beta\left(\alpha_{\varepsilon}\right)\left|\,\varepsilon<\mu\right.\right\rangle $
is increasing or constant. But if it is increasing then we have a
contradiction to the choice of $\left\{ \bar{R}_{i}\left|\, i<\sigma\right.\right\} $.
So it is constant, and suppose $\beta\left(\alpha_{\varepsilon}\right)=i_{0}$
for all $\varepsilon<\mu$. But then $\alpha_{\varepsilon}\in\lambda_{i_{0}}\backslash\sum_{j<i_{0}}\lambda_{j}$
for all $\varepsilon<\mu$ and we get a contradiction to the choice
of $\left\langle \bar{S}_{\beta}\left|\,\sum_{j<i_{0}}\lambda_{j}<\beta<\lambda_{i_{0}}\right.\right\rangle $. \end{proof}
\begin{claim}
\label{cla:Regular}Suppose $\lambda\not\to\left(\mu\right)_{n}^{\intr}$.
Then $2^{\lambda}\not\to\left(\mu\right)_{4+2n}^{\intr}$. \end{claim}
\begin{proof}
Suppose $\left\{ \bar{I}_{\alpha}\left|\,\alpha\leq\lambda\right.\right\} $
witnesses that $\lambda\not\to\left(\mu\right)_{n}^{\intr}$.
By adding two intervals to each $\bar{I}_{\alpha}$, we can ensure
that it has the extra property that if $\bar{c}_{1}\in I_{\alpha_{1}}$
and $\bar{c}_{2}\in\bar{I}_{\alpha_{2}}$ then $c_{1}^{1}<c_{2}^{0}$
if and only if $\alpha_{1}<\alpha_{2}$ (as in the previous claim).
By this we have increased the length of $\bar{I}_{\alpha}$ to $2+n$
(and it is still a witness of $\lambda\not\to\left(\mu\right)_{n}^{\intr}$).
We write $\bar{c}_{1}<^{*}\bar{c}_{2}$ for $c_{1}^{1}<c_{2}^{0}$
, but note that it is not really an ordering (it is not transitive).
We shall find below a four-place definable function $f$ such that:
\begin{itemize}
\item [$\heartsuit$] For every two ordinals, $\delta,\zeta$, if $\left\langle \bar{R}_{\alpha}\left|\,\alpha<\delta\right.\right\rangle $
is a sequence of $\zeta$-tuples of intervals, then there exists a
set of $2\zeta$-tuples of intervals, $\left\{ \bar{S}_{\eta}\left|\,\eta\in\leftexp{\delta}2\right.\right\} $
such that for all $i<\zeta$ and $\eta_{1}\neq\eta_{2}$, if $b_{1}\in S_{\eta_{1}}^{2i},b_{2}\in S_{\eta_{1}}^{2i+1}$
and $b_{3}\in S_{\eta_{2}}^{2i},b_{4}\in S_{\eta_{2}}^{2i+1}$ then
$f\left(b_{1},b_{2},b_{3},b_{4}\right)$ is in $R_{\lg\left(\eta_{1}\wedge\eta_{2}\right)}^{i}$.
\end{itemize}
Apply $\heartsuit$ to our situation to get $\left\{ \bar{J}_{\eta}\left|\,\eta\in\leftexp{\lambda}2\right.\right\} $
such that $\bar{J}_{\eta}=\left\langle J_{\eta}^{i}\left|\, i<4+2n\right.\right\rangle $
and for all $k<2+n$ and $\eta_{1}\neq\eta_{2}$, if $b_{1}\in J_{\eta_{1}}^{2k},b_{2}\in J_{\eta_{1}}^{2k+1}$
and $b_{3}\in J_{\eta_{2}}^{2k},b_{4}\in J_{\eta_{2}}^{2k+1}$ then
$f\left(b_{1},b_{2},b_{3},b_{4}\right)$ is in $I_{\lg\left(\eta_{1}\wedge\eta_{2}\right)}^{k}$.
This is enough (the reasons are exactly as in the regular case of
the proof of Theorem \ref{thm:MainThm}, but we shall repeat it for
clarity):
To simplify notation, we regard $f$ as a function on tuples, so that
if $\bar{b}_{1}\in\bar{J}_{\eta_{1}},\bar{b}_{2}\in\bar{J}_{\eta_{2}}$
then $f\left(\bar{b}_{1},\bar{b}_{2}\right)$ is in $\bar{I}_{\lg\left(\eta_{1}\wedge\eta_{2}\right)}$
(namely, $f\left(\bar{b}_{1},\bar{b}_{2}\right)=\left\langle a_{k}\left|\, k<2+n\right.\right\rangle $
where $a_{k}=f\left(b_{2k}^{1},b_{2k+1}^{1},b_{2k}^{2},b_{2k+1}^{2}\right)\in I_{\lg\left(\eta_{1}\wedge\eta_{2}\right)}^{k}$
for $k<2+n$).
Suppose $\left\langle \eta_{i}\left|\, i<\mu\right.\right\rangle \subseteq\leftexp{\lambda}2$
is without repetitions and $\left\langle \bar{b}_{\eta_{i}}\left|\, i<\mu\right.\right\rangle $
is an indiscernible sequence such that $\bar{b}_{\eta_{i}}\in\bar{J}_{\eta_{i}}$.
Let $h_{i}=\eta_{0}\wedge\eta_{i+1}$ for $i<\mu$. If $\lg\left(h_{i}\right)<\lg\left(h_{j}\right)$
for some $i\neq j$ then $f\left(\bar{b}_{\eta_{0}},\bar{b}_{\eta_{i+1}}\right)<^{*}f\left(\bar{b}_{\eta_{0}},\bar{b}_{\eta_{j+1}}\right)$
and so by indiscerniblity, $\left\langle \lg\left(h_{i}\right)\left|\, i<\mu\right.\right\rangle $
is increasing (it cannot be decreasing), and so $f\left(\bar{b}_{\eta_{0}},\bar{b}_{\eta_{i+1}}\right)$
contradicts our choice of $\left\langle \bar{I}_{\alpha}\left|\,\alpha<\lambda\right.\right\rangle $.
Hence (because $h_{i}\leq\eta_{0}$) $h_{i}$ is constant.
Assume $\eta_{0}\wedge\eta_{1}<\eta_{1}\wedge\eta_{2}$, then $f\left(\bar{b}_{\eta_{0}},\bar{b}_{\eta_{1}}\right)<^{*}f\left(\bar{b}_{\eta_{1}},\bar{b}_{\eta_{2}}\right)$
so $f\left(\bar{b}_{\eta_{1}},\bar{b}_{\eta_{2}}\right)<^{*}f\left(\bar{b}_{\eta_{2}},\bar{b}_{\eta_{3}}\right)$,
and so $\lg\left(\eta_{0}\wedge\eta_{1}\right)<\lg\left(\eta_{1}\wedge\eta_{2}\right)<\lg\left(\eta_{2}\wedge\eta_{3}\right)$
hence, $f\left(\bar{b}_{\eta_{0}},\bar{b}_{\eta_{1}}\right)<^{*}f\left(\bar{b}_{\eta_{2}},\bar{b}_{\eta_{3}}\right)$
and it follows that $\left\langle \lg\left(\eta_{2i}\wedge\eta_{2i+1}\right)\left|\, i<\mu\right.\right\rangle $
is increasing. And this is again a contradiction.
Similarly, it cannot be that $\eta_{0}\wedge\eta_{1}>\eta_{1}\wedge\eta_{2}$.
As both sides are less or equal than $\eta_{1}$, it must be that
$\eta_{0}\wedge\eta_{2}=\eta_{0}\wedge\eta_{1}=\eta_{1}\wedge\eta_{2}$.
But that is impossible (because if $\alpha=\lg\left(\eta_{0}\wedge\eta_{1}\right)$,
then $\left|\left\{ \eta_{0}\left(\alpha\right),\eta_{1}\left(\alpha\right),\eta_{2}\left(\alpha\right)\right\} \right|=3$).
\begin{claim*}
$\heartsuit$ is true. \end{claim*}
\begin{proof}
Let $f\left(x,y,z,w\right)=\left(x-z\right)/\left(y-w\right)$ (do
not worry about division by $0$, we shall explain below).
It is enough, by the definition of $\heartsuit$, to assume $\zeta=1$.
By compactness, we may assume that $\delta$ is finite, and to avoid
confusion, denote it by $m$. So we have a finite tree, $\leftexp m2$,
and a sequence of intervals $\left\langle R_{i}\left|\, i<m\right.\right\rangle $.
Each $R_{i}$ is of the form $\left(a_{i},b_{i}\right)$. Let $c_{i}=\left(b_{i}+a_{i}\right)/2$.
Let $d\in\C$ be any element greater than any member of $A:=\acl\left(a_{i},b_{i}\left|\, i<m\right.\right)$.
For each $\eta\in\leftexp m2$, let $a_{\eta}=\sum_{i<m}\eta\left(i\right)c_{i}d^{m-i}$,
and $b_{\eta}=\sum_{i<m}\eta\left(i\right)d^{m-i}$.
Let $S_{\eta}^{0}=\left(a_{\eta}-1,a_{\eta}+1\right)$ and $S_{\eta}^{1}=\left(b_{\eta}-1,b_{\eta}+1\right)$.
This works:
Assume $b_{1}\in S_{\eta_{1}}^{0},b_{2}\in S_{\eta_{1}}^{1}$ and
$b_{3}\in S_{\eta_{2}}^{0},b_{4}\in S_{\eta_{2}}^{1}$.
We have to show $\left(b_{1}-b_{3}\right)/\left(b_{2}-b_{4}\right)\in R_{\lg\left(\eta_{1}\wedge\eta_{2}\right)}$.
Denote $k=\lg\left(\eta_{1}\wedge\eta_{2}\right)$.
$a_{\eta_{1}}-a_{\eta_{2}}$ is of the form $\varepsilon c_{k}d^{m-k}+F\left(d\right)$
where $\varepsilon\in\left\{ -1,1\right\} $, and $F\left(d\right)$
is a polynomial over $A$ of degree $\leq m-k-1$. $b_{\eta_{1}}-b_{\eta_{2}}$
is of the form $\varepsilon d^{m-k}+G\left(d\right)$, where $\varepsilon$
is the same for both (and $G$ is a polynomial over $\mathbb{Z}$
of degree $\leq m-k-1$). Now, $b_{1}-b_{3}\in\left(a_{\eta_{1}}-a_{\eta_{2}}-2,a_{\eta_{1}}-a_{\eta_{2}}+2\right)$,
and $b_{2}-b_{4}\in\left(b_{\eta_{1}}-b_{\eta_{2}}-2,b_{\eta_{1}}-b_{\eta_{2}}+2\right)$,
and hence we know that $b_{2}-b_{4}\neq0$. It follows that $\left(b_{1}-b_{3}\right)/\left(b_{2}-b_{4}\right)$
is inside an interval whose endpoints are $\left\{ \left(\varepsilon c_{k}d^{m-k}+F\left(d\right)\pm2\right)/\left(\varepsilon d^{m-k}+G\left(d\right)\pm2\right)\right\} $.
But
\[
\left(\varepsilon c_{k}d^{m-k}+F\left(d\right)\pm2\right)/\left(\varepsilon d^{m-k}+G\left(d\right)\pm2\right)\in R_{k}
\]
by our choice of $d$, and we are done.
\end{proof}
\end{proof}
The proof of Theorem \ref{thm:MainThmRCF} now follows by induction
on $\kappa$: Fix $\mu$, and let $\kappa$ be the first cardinal
for which the theorem fails. Then by Claim \ref{cla:Obvious}, $\kappa\geq\mu$.
By Claim \ref{cla:aleph0}, $\aleph_{0}<\kappa$. By Claim \ref{cla:Singular},
$\kappa$ cannot be singular. By Claim \ref{cla:Regular}, $\kappa$
cannot be regular, because if it were, there would be a $\lambda<\kappa$
such that $2^{\lambda}\geq\kappa$ (because $\kappa$ is not strongly
inaccessible). Note that we did use claim \ref{cla:Obvious} to deal
with cases where we couldn't use the induction hypothesis (for example,
in the regular case, it might be that $\lambda<\mu$).
\subsubsection*{Further remarks}
Theorem \ref{thm:MainThmRCF} can be generalized to allow parameters:
Suppose $\C\models RCF$, and $A\subseteq\C$.
\begin{defn}
$\kappa\to_{A}\left(\mu\right)_{\omega}^{\intr}$ means the same as
in definition \ref{def:arrowInterval}, but we require that the the
indiscernible sequence is indiscernible over $A$.
\end{defn}
Then we have:
\begin{thm}
For any set of parameters $A$ such that in $\left[\left|A\right|,\kappa\right]$
there are no strongly inaccessible cardinals or $\kappa\leq\left|A\right|$,
$\kappa\not\to_{A}\left(\aleph_{0}\right)_{\omega}^{\intr}$.\end{thm}
\begin{proof}
The proof goes exactly as the proof of Theorem \ref{thm:MainThmRCF},
but the base case for the induction is different. There it was $\mu=\kappa=\aleph_{0}$.
Here it is $\kappa\leq\left|A\right|$. Indeed, enumerate $A=\left\{ a_{i}\left|\, i<\mu\right.\right\} $.
Let $\varepsilon\in\C$ be greater than $0$ but smaller than any
element in $\acl\left(A\right)$. For $i<\mu$, let $I_{i}=\left(a_{i},a_{i}+\varepsilon\right)$.
Then $\left\{ I_{i}\left|\, i<\kappa\right.\right\} $ witnesses $\kappa\not\to_{A}\left(\aleph_{0}\right)_{\omega}^{\intr}$.
\end{proof}
\section{Generic pair}
Here we give an example of an $\omega$-stable theory, such that for
all weakly generic pairs of structures $M\prec M_{1}$ the theory
of the pair $\left(M_{1},M\right)$ in an extended language where
we name $M$ by a predicate has the independence property.
\begin{defn}
\label{def:WeaklyGenPair}A pair $\left(M_{1},M\right)$ as above
is \emph{weakly generic} if for all formula $\varphi\left(x\right)$
with parameters from $M$, if $\varphi$ has infinitely many solutions
in $M$, then it has a solution in $M_{1}\backslash M$.
\end{defn}
This definition is induced by the well known ``generic pair conjecture''
(see \cite{Sh:900,Sh950}), and it is worth while to give the precise
definitions.
\begin{defn}
\label{def:GenPair}Assume $\lambda=\lambda^{<\lambda}>\left|T\right|$
(in particular, $\lambda$ is regular), $2^{\lambda}=\lambda^{+}$,
and for all $\alpha<\lambda^{+}$, $M_{\alpha}\models T$ is of cardinality
$\lambda$. Suppose $\left\langle M_{\alpha}\left|\,\alpha<\lambda^{+}\right.\right\rangle $
is an increasing continuous sequence. Furthermore, $M=\bigcup_{\alpha<\lambda^{+}}M_{\alpha}$
is a saturated model of cardinality $\lambda^{+}$. The generic pair
property says that there exists a club $E\subseteq\lambda^{+}$ such
that for all $\alpha<\beta\in E$ of cofinality $\lambda$, the pair
$\left(M_{\beta},M_{\alpha}\right)$ has the same isomorphism type.
We call this pair the \emph{generic} pair of $T$ of size $\lambda$.\end{defn}
\begin{prop}
The generic pair property and the generic pair depend only on $T$
and $\lambda$, and not on the particular choice of $M$ or $\left\langle M_{\alpha}\right\rangle $. \end{prop}
\begin{proof}
Suppose $M$ satisfies Definition \ref{def:GenPair}. The means that
$\left|M\right|=\lambda^{+}$, $M$ is saturated, there are $\left\langle M_{\alpha}\left|\,\alpha<\lambda^{+}\right.\right\rangle $
such that $M=\bigcup_{\alpha<\lambda^{+}}M_{\alpha}$, and there is
a club $E\subseteq\lambda^{+}$ such that for all $\alpha,\beta\in E$
such that $\alpha<\beta$, the pair $\left(M_{\beta},M_{\alpha}\right)$
has the same isomorphism type. If $N$ is another saturated model
of size $\lambda^{+}$ and $N=\bigcup_{\alpha<\lambda^{+}}N_{\alpha}$.
Then $N\cong M$, so we may assume $N=M$. Let $E_{0}=\left\{ \delta<\lambda^{+}\left|\, N_{\delta}=M_{\delta}\right.\right\} $.
This is a club of $\lambda^{+}$, and so $E\cap E_{0}$ is also a
club of $\lambda^{+}$ such that $\left(N_{\beta},N_{\alpha}\right)$
has the same isomorphism type for any $\alpha<\beta\in E\cap E_{0}$
of cofinality $\lambda$.
\end{proof}
Justifying definition \ref{def:WeaklyGenPair} we have:
\begin{claim}
Assume that $T$ has the generic pair property for $\lambda$, then
every generic pair is weakly generic.\end{claim}
\begin{proof}
So suppose that $E$, $\left\langle M_{\alpha}\left|\,\alpha<\lambda^{+}\right.\right\rangle $
are as in Definition \ref{def:GenPair}. Suppose $\alpha,\beta\in E$
and $\alpha<\beta$. We are given a formula $\varphi\left(x\right)$
with parameter from $M_{\alpha}$, such that $\aleph_{0}\leq\left|\varphi\left(M_{\alpha}\right)\right|$.
But by saturation of $M$, $\lambda^{+}=\left|\varphi\left(M\right)\right|$,
and $M=\bigcup_{\alpha\in E}M_{\alpha}$, hence there is some $\alpha<\beta'\in E$
such that $\varphi\left(M_{\beta}\right)\neq\emptyset$, but as $\left(M_{\beta},M_{\alpha}\right)\cong\left(M_{\beta'},M_{\alpha}\right)$,
we are done.
\end{proof}
The generic pair property is very important in dependent theories.
In fact, they characterize them: In \cite{Sh877,Sh950,Sh:900,Sh906}
it is proved that $T$ is dependent if and only if for large enough
$\lambda=\lambda^{<\lambda}$, $T$ has the generic pair property.
Hence it makes sense to ask whether the theory of the pair is dependent.
The answer is no:
\begin{thm}
There exists an $\omega$-stable theory such that for every weakly
generic pair of models $M\prec M_{1}$, the theory of the pair $\left(M_{1},M\right)$
has the independence property.
\end{thm}
We shall describe this theory:
Let $L=\left\{ P,R,Q_{1},Q_{2}\right\} $ where $R,P$ are unary predicates
and $Q_{1},Q_{2}$ are binary relations.
Let $\tilde{M}$ be the following structure for $L$:
\begin{enumerate}
\item The universe is
\begin{eqnarray*}
\tilde{M} & = & \left\{ u\subseteq\omega\left|\,\left|u\right|<\omega\right.\right\} \cup\\
& & \left\{ \left(u,v,i\right)\left|\, u,v\subseteq\omega,\left|u\right|<\omega,\left|v\right|<\omega,i<\omega\,\&\, u\subseteq v\Rightarrow i<\left|v\right|+1\right.\right\} .
\end{eqnarray*}
\item The predicates are interpreted as follows:
\begin{itemize}
\item $P^{\tilde{M}}=\left\{ u\subseteq\omega\left|\,\left|u\right|<\aleph_{0}\right.\right\} $.
\item $R^{\tilde{M}}$ is $M\backslash\left(P^{M}\right)$.
\item $Q_{1}^{\tilde{M}}=\left\{ \left(u,\left(u,v,i\right)\right)\left|\, u\in P^{\tilde{M}}\right.\right\} $.
\item $Q_{2}^{\tilde{M}}=\left\{ \left(v,\left(u,v,i\right)\right)\left|\, v\in P^{\tilde{M}}\right.\right\} $.
\end{itemize}
\end{enumerate}
Let $T=Th\left(\tilde{M}\right)$.
As we shall see in the next claim, $T$ gives rise to the following
definition:
\begin{defn}
\label{def:PBA}We call a structure $\left(B,\cup,\cap,-,\subseteq,0\right)$
a \emph{pseudo Boolean algebra} (PBA) when it satisfies all the axioms
of a Boolean algebra except: There is no greatest element $1$ (i.e.
remove all the axioms concerning it).
\end{defn}
Pseudo Boolean algebra can have atoms like in Boolean algebras (nonzero
elements that do not contain any smaller nonzero elements).
\begin{defn}
Say that a PBA is of \emph{finite type} if every element is a union
of finitely many atoms.
\end{defn}
\begin{defn}
For a PBA $A$, and $C\subseteq A$ a sub-PBA, let $A_{C}:=\left\{ a\in A\left|\,\exists c\in C\left(a\subseteq^{A}c\right)\right.\right\} $,
and for a subset $D\subseteq A$, let $\atom\left(D\right)$ be the
set of atoms contained in $D$.\end{defn}
\begin{prop}
\label{pro:ISomPBA}Every PBA of finite type is isomorphic to $\left(\SS_{<\infty}\left(\kappa\right),\cup,\cap,-,\subseteq,\emptyset\right)$
for some $\kappa$ where $\SS_{<\infty}\left(\kappa\right)$ is the
set of all finite subsets of $\kappa$. Moreover: Assume $A,B$ are
PBAs of finite type and $C\subseteq A,B$ is a common sub-PBA. Then,
if
\begin{enumerate}
\item $\left|\atom\left(A\backslash A_{C}\right)\right|=\left|\atom\left(B\backslash B_{C}\right)\right|$.
\item For every $c\in C$, $A$ and $B$ agree on the size of $c$ (the
number of atoms it contains).
\end{enumerate}
Then there is an isomorphism of PBAs $f:A\to B$ such that $f\upharpoonright C=\id$.\end{prop}
\begin{proof}
The first part follows from the easy observation that in a PBA of
finite type, every element has a unique presentation as a union of
finitely many atoms. So if $A$ is a PBA, and its set of atoms is
$\left\{ a_{i}\left|\, i<\kappa\right.\right\} $, then take $a_{i}$
to $\left\{ i\right\} $.
For the moreover part, first we extend $\id_{C}$ to an isomorphism
from $A_{C}$ to $B_{C}$: consider all elements in $C$ of minimal
size, these are the atoms of $C$. For each such $c$, map the set
of atoms in $A$ contained in $c$ to the set of atoms in $B$ contained
in $c$. This is well defined and can be extended this to all of $A_{C}$.
Now, $\left|\atom\left(A\backslash A_{C}\right)\right|=\left|\atom\left(B\backslash B_{C}\right)\right|$,
so any bijection between the set of atoms induces an isomorphism. \end{proof}
\begin{claim}
\label{cla:omegaStable}$T$ is $\omega$-stable. \end{claim}
\begin{proof}
We prove that an extension of $T$ to a larger vocabulary is $\omega$-stable,
by adding new relations to the language, which are all definable ---
\[
\left\{ S_{n},\subseteq_{n},\pi_{1},\pi_{2},\cap_{n},\cup_{n},-_{n},e\left|\, n\geq1\right.\right\}
\]
where $S_{n}$ is a unary relation defined on $P$, $\subseteq_{n}$
is a binary relation defined on $P$, $\pi_{1},\pi_{2}$ are two unary
functions from $R$ to $P$, $\cap_{n},-_{n}$ are binary functions
from $S_{n}$ to $S_{n}$ and $e$ is a constant in $P$. Their interpretation
in $\tilde{M}$ are as follows:
\begin{itemize}
\item $\pi_{1}\left(\left(u,v,i\right)\right)=u$, $\pi_{2}\left(\left(u,v,i\right)\right)=v$.
\item For each $1\leq n<\omega$, $S_{n}\left(v\right)\Leftrightarrow\left|v\right|\leq n$.
\item For each $1\leq n$, $u\subseteq_{n}v$ if and only if $\left|u\right|\leq n$,
$\left|v\right|\leq n$ and $u\subseteq v$.
\item $u\cap_{n}v=u\cap v$ for all $u,v\in S_{n}$.
\item $u-_{n}v=u\backslash v$ for $v\subseteq_{n}u$.
\item $u\cup_{n}v=u\cup v$ for $u,v\in S_{n}$.
\item $e=\emptyset$.
\end{itemize}
Note that they are indeed definable:
\begin{enumerate}
\item $\pi_{1}\left(x\right)$ is the unique $y$ such that $Q_{1}\left(y,x\right)$,
and similarly $\pi_{2}$ is definable.
\item Let $E\left(x,y\right)$ by an auxiliary equivalence relation defined
by $\pi_{1}\left(x\right)=\pi_{1}\left(y\right)\land\pi_{2}\left(x\right)=\pi_{2}\left(y\right)$.
\item $e$ is the unique element $x\in P$ such that there exists exactly
one element $z\in R$ such that $\pi_{1}\left(z\right)=x=\pi_{2}\left(z\right)$.
\item $x\subseteq_{n}y$ is defined by ``$P\left(x\right),P\left(y\right)$
and the number of elements in the $E$ class of some (equivalently
any) element $z$ such that $\pi_{1}\left(z\right)=x$, $\pi_{2}\left(z\right)=y$
is at most $n+1$''.
\item $S_{n}\left(x\right)$ is defined by ``$P\left(x\right)$ and $e\subseteq_{n}x$''
(In particular, $e\in S_{n}$ for all $n$).
\item $\cap_{n}$ and $-_{n}$ are then naturally definable using $\subseteq_{n}$.
For instance $x-_{n}y=z$ if and only if $x,y,z$ are in $S_{n}$,
$z\subseteq_{n}x$ and for each $e\neq w\subseteq_{n}y$, $w\nsubseteq_{n}z$.
\item $x\cup_{n}y=z$ if and only if $x,y\in S_{n}$, $z\in S_{2n}$, $x,y\subseteq_{2n}z$
and $z-_{2n}x\subseteq_{2n}y$.
\end{enumerate}
Furthermore, $\subseteq_{k}\upharpoonright S_{n}=\subseteq_{n}$ for
$n\leq k$. Hence every model $M$ of $T$ gives rise naturally to
an induced PBA: $B^{M}:=\left(\bigcup_{n}S_{n}^{M},\cup^{M},\cap^{M},-^{M},\subseteq^{M},e^{M}\right)$
where $\cup^{M}=\bigcup\left\{ \cup_{n}^{M}\left|\, n<\omega\right.\right\} $,
and similarly for $\subseteq^{M},-^{M}$ and $\cap^{M}$ (see Definition
\ref{def:PBA} above).
\begin{claim*}
In the extended language, $T$ eliminates quantifiers.\end{claim*}
\begin{proof}
Suppose $M,N\models T$ are saturated models, $\left|M\right|=\left|N\right|$
and $A\subseteq M,N$ is a common substructure (where $\left|A\right|<\left|M\right|$).
It is enough to show that we have an isomorphism from $M$ to $N$
fixing $A$.
By Proposition \ref{pro:ISomPBA}, we have an isomorphism $f$ from
$B^{M}$ to $B^{N}$ preserving $A$ (by saturation and the choice
of language, the condition of the proposition are satisfied).
On $P^{M}\backslash\left(B^{M}\cup A\right)$ there is no structure
and it has the same size as $P^{N}\backslash\left(B^{N}\cup A\right)$
(namely $\left|N\right|$), so we can extend the isomorphism $f$
to $P^{M}$.
We are left with $R^{M}$: let $a\in R^{M}$, and $a_{i}=\pi_{i}\left(a\right)$
for $i=1,2$. We already know $f\left(a_{1}\right),f\left(a_{2}\right)$.
Suppose $a_{1}\subseteq_{n}a_{2}$ for minimal $n$. Then there are
exactly $n$ elements $z\in R^{M}$ with $\pi_{1}\left(z\right)=a_{1},\pi_{2}\left(z\right)=a_{2}$.
This is true also in $R^{N}$, and the number of such $z$'s not in
$A$ is the same for both $M$, $N$. Hence we can take this $E$-equivalence
class from $M$ to the appropriate class in $N$.
If not, i.e. $a_{1}\nsubseteq_{n}a_{2}$ for all $n$, then there
are infinitely many elements $z$ in $N$ and in $M$ with $\pi_{1}\left(z\right)=a_{1}$,
$\pi_{2}\left(z\right)=a_{2}$, and again we take this $E$-class
in $M$ outside of $A$ to the appropriate $E$-class in $N$.
\end{proof}
Now we can conclude the proof by a counting types argument. Let $M$
be a countable model of $T$. Let $p\left(x\right)$ be a non-algebraic
type over $M$. There are some cases:
\begin{caseenv}
\item $S_{n}\left(x\right)\in p$ for some $n$. Then the type is determined
by the maximal element $c$ in $M$ such that $c\subseteq_{n}x$ (this
is easy, but also follows from Proposition \ref{pro:ISomPBA}).
\item $S_{n}\left(x\right)\notin p$ for all $n$ but $P\left(x\right)\in p$.
Then $x$ is already determined --- there is nothing more we can say
on $x$.
\item $R\left(x\right)\in p$. Then the type of $x$ is determined by the
type of $\left(\pi_{1}\left(x\right),\pi_{2}\left(x\right)\right)$
over $M$.
\end{caseenv}
So the number of types over $M$ is countable.\end{proof}
\begin{prop}
Every weakly generic pair of models of $T$ has the independence property.\end{prop}
\begin{proof}
Suppose $\left(M_{1},M\right)$ is a weakly generic pair. We think
of it as a structure of the language $L_{Q}$, where $Q$ is interpreted
as $M$. Consider the formula
\[
\varphi\left(x,y\right)=P\left(x\right)\land P\left(y\right)\land\exists z\notin Q\left(Q_{1}\left(x,z\right)\land Q_{2}\left(y,z\right)\right).
\]
This formula has IP: Let $\left\{ a_{i}\left|\, i<\omega\right.\right\} \subseteq M$
be elements from $P^{M}$ such that $a\in S_{1}^{M}$ (as in the language
of the proof of Claim \ref{cla:omegaStable}), i.e. they are atoms
in the induced PBA, and $a_{i}\neq a_{j}$ for $i\neq j$. For any
finite $s\subseteq\omega$ of size $n$, there is an element $b_{s}\in P^{M}$
be such that $a_{i}\subseteq_{n}^{M}b_{s}$ for all $i\in s$. Then
for all $i\in\omega$, $\varphi\left(a_{i},b_{s}\right)$ if and only
if $i\notin s$:
If $\varphi\left(a_{i},b_{s}\right)$ there are infinitely many $z$'s
in $M$ such that $Q_{1}\left(a_{i},z\right)\wedge Q_{2}\left(b_{s},z\right)$
(otherwise they would all be in $M$). This means that $a_{i}\nsubseteq_{n}^{M}b_{s}$
so $i\notin s$.
For the other direction, the same exact argument works, but this time
use the fact that the pair is weakly generic.
\end{proof}
\section{\label{sec:Directionality}Directionality}
\subsection{Introduction to Directionality.}
\begin{defn}
A global type $p\left(x\right)\in S\left(\C\right)$ is said to be
\emph{finitely satisfiable} over a set $A$, or a \emph{co-heir} over
$A$ if for every formula $\varphi\left(x,y\right)$, if $\varphi\left(x,b\right)\in p$,
then for some $a\in A$, $\varphi\left(a,b\right)$ holds.
\end{defn}
It is well known (see \cite{Ad}) that a theory $T$ is dependent
if and only if given a type $p\left(x\right)\in S\left(M\right)$
over a model $M$, the number of complete global types $q\in S\left(\C\right)$
that extend $p$ and are finitely satisfiable in $M$ is at most $2^{\left|M\right|}$
(while the maximal number is $2^{2^{\left|M\right|}}$).
Here we analyze the behavior of the number of global co-heir extensions
in a dependent theory. We classify theories by what we call directionality:
The first possibility is that the number of such global extensions
is small, i.e. that the number does not depend on the model but only
on the theory (theories with \emph{small} directionality); the second
possibility is that this number is bounded by $\left|M\right|^{\left|T\right|}$
(\emph{medium} directionality); and the third one is that this number
is greater than $\left|M\right|^{\left|T\right|}$ (\emph{large} directionality).
As far as we know, the first to give an example of a dependent theory
with large directionality was Delon in \cite{Delon}.
Here we give a simple combinatorial examples using the constructions
of Section \ref{sec:Indisc} for each of the possible directionalities,
and furthermore we show that RCF is large.
We also give a local characterization of directionality. Namely, for
a finite set of formulas $\Delta$, we show that $T$ has small directionality
if and only if the number of $\Delta$-co-heirs that extend a type
$p\in S\left(M\right)$ is finite. $T$ has medium directionality
if this number is $\left|M\right|$, and it has large directionality
if it is at least $\ded\left|M\right|$.
We do not always assume that $T$ is dependent in this section.
\begin{defn}
For a type $p\in S\left(A\right)$, let:
\[
\uf\left(p\right)=\left\{ q\in S\left(\C\right)\left|\, q\mbox{ is a co-heir extension of }p\mbox{ over }A\right.\right\} .
\]
For a partial type $p\left(x\right)$ over a set $A$, and a set of
formulas $\Delta$,
\[
\uf_{\Delta}\left(p\right)=\left\{ q\left(x\right)\in S_{\Delta}\left(\C\right)\left|\, q\cup p\mbox{ is f.s. in }A\right.\right\} .
\]
Note: this definition only makes sense if $p$ is finitely satisfiable
in $A$. The notation $\uf$ refers to ultra-filter.
\end{defn}
And here is the main definition of this section:
\begin{defn}
\label{def:directionality}Let $T$ be any theory, then:
\begin{enumerate}
\item $T$ is said to have \emph{small} directionality (or just, $T$ is
small) if and only if for all finite $\Delta$, $M\models T$ and
$p\in S\left(M\right)$, $\uf_{\Delta}\left(p\right)$ is finite.
\item $T$ is said to have \emph{medium} directionality (or just, $T$ is
medium) if and only if for every $\lambda\geq\left|T\right|$,
\[
\lambda=\sup\left\{ \left|\uf_{\Delta}\left(p\right)\right|\left|\, p\in S\left(M\right),\Delta\mbox{ finite},\left|M\right|=\lambda\right.\right\} .
\]
\item $T$ is said to have \emph{large} directionality (or just, $T$ is
large) if $T$ is neither small nor medium.
\end{enumerate}
\end{defn}
\begin{obs}
\label{obs:IPLarge} If $T$ has the independence property, then it
is large. In fact, if $\varphi\left(x,y\right)$ has the independence
property, then there is a type $p\left(x\right)$ over a model $M$,
that has $2^{2^{\left|M\right|}}$ many $\left\{ \varphi\left(x,y\right)\right\} $-extensions
that are finitely satisfiable in $M$. \end{obs}
\begin{proof}
We may assume that $T$ has Skolem functions. Let $\lambda\geq\left|T\right|$,
and let $\bar{a}=\left\langle a_{i}\left|\, i<\lambda\right.\right\rangle $,
$\left\langle b_{s}\left|\, s\subseteq\lambda\right.\right\rangle $
be such that $\left\langle a_{i}\left|\, i<\lambda\right.\right\rangle $
is indiscernible and $\varphi\left(a_{i},b_{s}\right)$ holds iff
$i\in s$. Let $M$ be the Skolem hull of $\bar{a}$. Let $p\left(x\right)\in S\left(M\right)$
be the limit of $\bar{a}$ in $M$ (so $\psi\left(x,c\right)\in p$
iff $\psi\left(x,c\right)$ holds for an end segment of $\bar{a}$).
Let $P\subseteq\SS\left(\lambda\right)$ be an independent family
of size $2^{\lambda}$ (i.e. such that every finite Boolean combination
has size $\lambda$). Then for each $D\subseteq P$, $p\left(x\right)\cup\left\{ \varphi\left(x,b_{s}\right)^{s\in D}\left|\, s\subseteq\lambda\right.\right\} $
is finitely satisfiable in $M$.
\end{proof}
\subsubsection{Small directionality}
The following construction will be useful (here and in the Section
\ref{sec:Splintering}):
\begin{const}
\label{const:not small directionality}Suppose that there is some
$p\in S\left(M\right)$ and finite $\Delta$ such that $\uf_{\Delta}\left(p\right)$
is infinite, and contains $\left\{ q_{i}\left|\, i<\omega\right.\right\} $.
For all $i<j<\omega$, there is a formula $\varphi_{i,j}\in\Delta$
and $b_{i,j}\in\C$ such that $\varphi_{i,j}\left(x,b_{i,j}\right)\in q_{i}$,
$\neg\varphi_{i,j}\left(x,b_{i,j}\right)\in q_{j}$ (or the other
way around). By Ramsey's Theorem we may assume $\varphi_{i,j}$ is
constant --- $\varphi\left(x,y\right)$. Let $N$ be a model containing
$M$ and $\left\{ b_{i,j}\left|\, i<j<\omega\right.\right\} $.
Suppose $\left\langle c_{i}\left|\, i<\omega\right.\right\rangle $
are in $\C$ and $c_{i}\models q_{i}|_{N}$. Let $N'$ be a model
containing $\left\{ c_{i}\left|\, i<\omega\right.\right\} \cup N$.
Let $M^{*}=\left(N',N,M,Q,\bar{f}\right)$ where $Q=\left\{ c_{i}\left|\, i<\omega\right.\right\} $
and $\bar{f}:Q^{2}\to N$ is a tuple of functions of length $\lg\left(y\right)$
defined by $\bar{f}\left(c_{i},c_{j}\right)=\bar{f}\left(c_{j},c_{i}\right)=b_{i,j}$
for $i<j$.
So if $N\models Th\left(M^{*}\right)$ then $N=\left(N_{0}',N_{0},M_{0},Q_{0},\bar{f}_{0}\right)$
and
\begin{itemize}
\item $M_{0}\prec N_{0}\prec N_{0}'\models T$,
\item $N_{0}\cup Q_{0}\subseteq N_{0}'$,
\item $\bar{f}_{0}$ are functions from $Q_{0}^{2}$ to $N_{0}$,
\item For all $c,d\in Q_{0}$, $c\equiv_{M_{0}}d$,
\item $\tp_{\Delta}\left(c/N_{0}\right)$ is finitely satisfiable in $M_{0}$
for all $c\in Q_{0}$, and
\item $\varphi\left(c,\bar{f}_{0}\left(c,d\right)\right)\triangle\varphi\left(d,\bar{f}_{0}\left(c,d\right)\right)$
(where $\triangle$ denotes symmetric difference) holds for all $c\neq d\in Q_{0}$.
\end{itemize}
\end{const}
\begin{claim}
\label{cla:bounded}Let $T$ be any theory. Then $T$ is small if
and only if for every type $p\left(x\right)\in S\left(M\right)$,
$\left|\uf\left(p\right)\right|\leq2^{\left|T\right|}$ (here $p$
can also be an infinitary type, but then the bound is $2^{\left|T\right|+\left|\lg\left(x\right)\right|}$).
In addition, if $T$ is not small, then for every $\lambda\geq\left|T\right|$,
there is a model $M\models T$ of cardinality $\lambda$, a type $p\in S\left(M\right)$,
and a finite set of formulas $\Delta$ such that $\left|\uf_{\Delta}\left(p\right)\right|\geq\lambda$.\end{claim}
\begin{proof}
Assume that $T$ is small. The injective function $\uf\left(p\right)\to\prod_{\varphi\in L}\uf_{\left\{ \varphi\right\} }\left(p\right)$
shows that $\left|\uf\left(p\right)\right|\leq2^{\left|T\right|}$.
Conversely (and the ``In addition'' part): Assume that there is
some $p$ and $\Delta$ such that $\uf_{\Delta}\left(p\right)$ is
infinite. Use Construction \ref{const:not small directionality}:
For every $\lambda\geq\left|T\right|$ we may find $N\models Th\left(M^{*}\right)$
of size $\lambda$ such that $\left|M_{0}\right|=\left|Q_{0}\right|=\lambda$,
and we have a model $M_{0}$ of $T$ with a type $p$ over it, which
have at least $\lambda$ many $\Delta$-co-heirs of $p$.
\end{proof}
We conclude this section with a claim on theories with non-small or
large directionality.
\begin{claim}
Suppose $T$ has medium or large directionality. Then there exists
some $M\models T$, $p\in S\left(M\right)$, $\psi\left(x,y\right)$
and $\left\{ c_{i}\left|\, i<\omega\right.\right\} \subseteq\C$ such
that for each $i<\omega$ the set $p\left(x\right)\cup\left\{ \psi\left(x,c_{j}\right)^{j=i}\left|\, j<\omega\right.\right\} $
is finitely satisfiable in $M$. \end{claim}
\begin{proof}
We consider the structure $M^{*}$ introduced in Construction \ref{const:not small directionality}
and the formula $\varphi$ chosen there. Find an extension of with
an indiscernible sequence $\left\langle d_{i}\left|\, i\in\mathbb{Z}\right.\right\rangle $
inside $Q$. Assume without loss that $\varphi\left(d_{0},\bar{f}\left(d_{0},d_{1}\right)\right)\land\neg\varphi\left(d_{1},\bar{f}\left(d_{0},d_{1}\right)\right)$
holds. This means that $\varphi\left(d_{0},\bar{f}\left(d_{0},d_{1}\right)\right)\land\neg\varphi\left(d_{0},\bar{f}\left(d_{-1},d_{0}\right)\right)$.
We claim that $\varphi\left(d_{i},\bar{f}\left(d_{j},d_{j+1}\right)\right)\land\neg\varphi\left(d_{i},\bar{f}\left(d_{j-1},d_{j}\right)\right)$
holds if and only if $i=j$:
Suppose this holds but $i\neq j$. If $i>j$ then, since $\neg\varphi\left(d_{1},f\left(d_{0},d_{1}\right)\right)$,
it must be that $i>j+1$, but then we have a contradiction to indiscernibility.
Similarly, it cannot be that $i<j$. Thus the claim is proved with
$\psi\left(x;y,z\right)=\varphi\left(x,y\right)\land\neg\varphi\left(x,z\right)$,
and $c_{i}=\left\langle f\left(d_{i},d_{i+1}\right),f\left(d_{i-1},d_{i}\right)\right\rangle $.
\end{proof}
\subsubsection{Some helpful facts about dependent theories}
Assume $T$ is dependent.
Recall,
\begin{defn}
A global type $p\left(x\right)$ is \emph{invariant} \emph{over} a
set $A$ if it does not split over it, namely if whenever $b$ and
$c$ have the same type over $A$, $\varphi\left(x,b\right)\in p$
if and only if $\varphi\left(x,c\right)\in p$ for every formula $\varphi\left(x,y\right)$.
\end{defn}
\begin{defn}
\label{def:tensor}Suppose $p\left(x\right)$ and $q\left(y\right)$
are global $A$-invariant types. Then $\left(q\otimes p\right)\left(x,y\right)$
is a global invariant type defined as follows: for any $B\supseteq A$,
let $a_{B}\models p|_{B}$ and $b_{B}\models q|_{Ba_{B}}$, then $p\otimes q=\bigcup_{B\supseteq A}\tp\left(a_{B},b_{B}/B\right)$.
One can easily check that it is well defined and $A$-invariant. Let
$p^{\left(n\right)}=p\otimes p\ldots\otimes p$ where the product
is done $n$ times. So $p^{\left(n\right)}$ is a type in $\left(x_{0},\ldots,x_{n-1}\right)$,
and $p^{\left(\omega\right)}=\bigcup_{n<\omega}p^{\left(n\right)}$
is a type in $\left(x_{0},\ldots,x_{n},\ldots\right)$. For $n\leq\omega$,
$p^{\left(n\right)}$ is a type of an $A$-indiscernible sequence
of length $n$. \end{defn}
\begin{fact}
\emph{\label{fac:pOmega}}\cite[Lemma 2.5]{HP} If $T$ is NIP then
for a set $A$ the map $p\left(x_{0}\right)\mapsto p^{\left(\omega\right)}\left(x_{0},\ldots\right)|_{A}$
from global $A$-invariant types to $\omega$-types over $A$ is injective.
\end{fact}
In the rest of the section, $\Delta$ will always denote a finite
set of formulas, closed under negation.
\begin{claim}
\label{cla:local types define indiscernibles} For every set $A\subseteq C$,
any type $q\left(x\right)\in S_{\Delta}\left(\C\right)$ which is
finitely satisfiable in $A$ and any choice of a co-heir $q'\in S\left(\C\right)$
over $A$ which completes $q$:
\begin{itemize}
\item $\left(a_{0},\ldots,a_{n-1}\right)\models\left(q'^{\left(n\right)}|_{C}\right)\upharpoonright\Delta$
if and only if $a_{0}\models q|_{C}$, $a_{1}\models q|_{Ca_{0}}$,
etc.
\end{itemize}
This enables us to define $q^{\left(n\right)}\left(x_{0},\ldots x_{n-1}\right)\in S_{\Delta}\left(\C\right)$
as $q'^{\left(n\right)}\upharpoonright\Delta$.
It follows that $q^{\left(n\right)}$ is a type of a $\Delta$-indiscernible
sequence of length $n$. \end{claim}
\begin{proof}
The proof is by induction on $n$:
Right to left: suppose $a_{i}\models q|_{Ca_{0},\ldots,a_{i-1}}$
for $i\leq n$, $\varphi\left(x_{0},\ldots,x_{n},y\right)\in\Delta$
and $\varphi\left(x_{0},\ldots,x_{n},c\right)\in q'^{\left(n+1\right)}$
for $c\in C$ but $\neg\varphi\left(a_{0},\ldots,a_{n},c\right)$
holds. Then by the choice of $a_{n}$, $\neg\varphi\left(a_{0},\ldots,a_{n-1},x,c\right)\in q$.
Suppose $\left(b_{0},\ldots,b_{n-1}\right)\models q'^{\left(n\right)}|_{C}$,
then $\varphi\left(b_{0},\ldots,b_{n-1},x,c\right)\in q$ so there
is some $c'\in A$ such that $\varphi\left(b_{0},\ldots,b_{n-1},c',c\right)\land\neg\varphi\left(a_{0},\ldots,a_{n-1},c',c\right)$
holds. But this is a contradiction to the induction hypothesis.
Left to right is similar.
\end{proof}
The following is a local version of Fact \ref{fac:pOmega}, which
will be useful later:
\begin{prop}
\label{prop:localpomega}($T$ dependent) Suppose $\Delta$ is a finite
set of formulas, $x$ a finite tuple of variables. Then there exists
$n<\omega$ and finite set of formulas $\Delta_{0}$ such that for
every set $A$, if $q_{1}\left(x\right),q_{2}\left(x\right)\in S\left(\C\right)$
are co-heirs over $A$ and $\left(q_{1}^{\left(n\right)}\upharpoonright\Delta_{0}\right)|_{A}=\left(q_{2}^{\left(n\right)}\upharpoonright\Delta_{0}\right)|_{A}$
then $q_{1}\upharpoonright\Delta=q_{2}\upharpoonright\Delta$.\end{prop}
\begin{proof}
By compactness and NIP,
\begin{itemize}
\item there exists some finite set of formulas $\Delta_{0}$ and some $n$
such that for all $\varphi\left(x,y\right)\in\Delta$ and all $\Delta_{0}$-indiscernible
sequences $\left\langle a_{0},\ldots,a_{n_{\Delta}-1}\right\rangle $,
there is \uline{no} $c$ such that $\varphi\left(a_{i},c\right)$
holds if and only if $i$ is even. We may assume that $\Delta\subseteq\Delta_{0}$.
\end{itemize}
By Claim \ref{cla:local types define indiscernibles}, we can conclude:
\begin{itemize}
\item Suppose that $\left(q_{1}^{\left(n\right)}\upharpoonright\Delta_{0}\right)|_{A}=\left(q_{2}^{\left(n\right)}\upharpoonright\Delta_{0}\right)|_{A}$,
but $q_{1}\upharpoonright\Delta\neq q_{2}\upharpoonright\Delta$.
Then there is some formula $\varphi\left(x,y\right)\in\Delta$ and
some $c\in\C$ such that $\varphi\left(x,c\right)\in q_{1}$ and $\neg\varphi\left(x,c\right)\in q_{2}$.
Since $\left(q_{1}\upharpoonright\Delta_{0}\right)^{\left(n\right)}|_{A}=\left(q_{2}\upharpoonright\Delta_{0}\right)^{\left(n\right)}|_{A}$,
$\left(q_{1}\upharpoonright\Delta_{0}\right)^{\left(m\right)}|_{A}=\left(q_{2}\upharpoonright\Delta_{0}\right)^{\left(m\right)}|_{A}$
for every $m\leq n$, and it follows by induction on $m$ that the
sequence defined by $a_{0}\models\left(q_{1}\upharpoonright\Delta_{0}\right)|_{Ac}$,
$a_{1}\models\left(q_{2}\upharpoonright\Delta_{0}\right)|_{Aca_{0}}$,
$a_{2}\models\left(q_{1}\upharpoonright\Delta_{0}\right)|_{Aca_{0}a_{1}}$,
... , $a_{m-1}\models\left(q_{i}\upharpoonright\Delta_{0}\right)|_{Aca_{0}\ldots a_{m-2}}$
($i\in\left\{ 1,2\right\} $) realizes this type. But this entails
a contradiction, because $\left\langle a_{0},\ldots,a_{n-1}\right\rangle $
is a $\Delta_{0}$ indiscernible sequence (even over $A$), while
$\varphi\left(a_{i},c\right)$ holds if and only if $i$ is even.
\end{itemize}
\end{proof}
\begin{problem}
Does Proposition \ref{prop:localpomega} hold for invariant types
(not just for co-heirs)?
\end{problem}
\subsubsection{Large directionality and definability }
Let us recall the definition of $\ded\lambda$.
\begin{defn}
\label{def:ded}Let $\ded\lambda$ be the supremum of the set
\[
\left\{ \left|I\right|\left|\, I\mbox{ is a linear order with a dense subset of size }\leq\lambda\right.\right\} .
\]
\end{defn}
\begin{fact}
\label{fac:Ded}It is well known that $\lambda<\ded\lambda\leq\left(\ded\lambda\right)^{\aleph_{0}}\leq2^{\lambda}$.
If $\lambda^{<\lambda}=\lambda$ then $\ded\lambda=2^{\lambda}$ so
$\ded\lambda=\left(\ded\lambda\right)^{\aleph_{0}}=2^{\lambda}$.
\end{fact}
For more, see e.g. \cite[Secton 6]{Sh1007}.
\begin{defn}
\label{def:ExtDefType}Suppose $M$ is a model and $p\in S\left(M\right)$.
Let $M_{p}$ be $M$ enriched with externally definable sets defined
over a realization. Namely, we enrich the language to a language $L_{p}$
by adding new relation symbols $\left\{ d_{p}x\varphi\left(x,y\right)\left|\,\varphi\left(x,y\right)\mbox{ is a formula}\right.\right\} $
(so $d_{p}$ is thought of as a quantifier over $x$), and let $M_{p}$
be a model for $L_{p}$ with universe $M$ and let $d_{p}x\varphi\left(x,y\right)$
be interpreted as $\left\{ b\in M\left|\,\varphi\left(x,b\right)\in p\right.\right\} $. \end{defn}
\begin{rem}
\label{rem:TpModel}Every model $N\models Th\left(M_{p}\right)$ gives
rise to an $L$ complete type over $N$, namely $p^{N}=\left\{ \varphi\left(b,x\right)\left|\, b\in N,N\models d_{p}x\varphi\left(x,b\right)\right.\right\} $.\end{rem}
\begin{claim}
\label{cla:Large}Suppose $p\in S\left(M\right)$, $q\in\uf\left(p\right)$,
and $\bar{a}=\left\langle a_{0},a_{1},\ldots\right\rangle \models q^{\left(\omega\right)}|_{M}$.
If $\tp\left(\bar{a}/M\right)$ is not definable with parameters in
$M_{p}$, then $T$ is large.
Moreover, in this case
\begin{itemize}
\item [$\otimes$]There exists a finite $\Delta$ such that for every $\lambda\geq\left|T\right|$,
\[
\ded\lambda\leq\sup\left\{ \left|\uf_{\Delta}\left(p\right)\right|\left|\, p\in S\left(N\right),\left|N\right|=\lambda\right.\right\} .
\]
\end{itemize}
\end{claim}
\begin{proof}
We may assume that $\left|M\right|=\left|L\right|$: let $r=q^{\left(\omega\right)}|_{M}$,
and $N\prec M_{r}$, $\left|N\right|=\left|L\right|$. Then $N$ gives
rise to a complete type $r'\left(x_{0},x_{1},\ldots\right)\in S\left(N\right)$.
Let $p'=r'\upharpoonright x_{0}$. It is easy to see that $r'=q'^{\left(\omega\right)}|_{N}$
for some $q'\in\uf\left(p'\right)$. Also, $r'$ is not definable
with parameters in $N_{p'}$.
Let us recall a theorem from \cite{Sh009} (we formulate it a bit
differently):
Suppose $L$ is a language of cardinality at most $\lambda$, $P$
a new predicate (or relation symbol), and $S$ a complete theory in
$L\left(P\right)$.
\begin{defn*}
$\DfOne\lambda$ is the the supremum of the set of cardinalities
\[
\left|\left\{ B'\subseteq M\left|\,\left(M,B'\right)\cong\left(M,B\right)\right.\right\} \right|
\]
where $\left(M,B\right)$ is an $L\left(P\right)$ model of $S$ of
cardinality $\lambda$. \end{defn*}
\begin{thm*}
\cite[Theorem 12.4.1]{Sh009,Hod} The following are equivalent
\footnote{The original theorem referred to $\ded^{*}\lambda$, which counts
the number of branches of the same height in a tree with $\lambda$
many nodes, but it equals $\ded\lambda$, see \cite[Section 6]{Sh1007}.
}:
\begin{enumerate}
\item $P$ is not definable with parameters in $S$, i.e. there is no $L$-formula
$\theta\left(x,y\right)$ such that $S\models\exists y\forall x\left(P\left(x\right)\leftrightarrow\theta\left(x,y\right)\right)$.
\item For every $\lambda\geq\left|L\right|$, $\DfOne\left(\lambda\right)\geq\ded\lambda$.
\end{enumerate}
\end{thm*}
Let $n<\omega$ be the first such that $\tp\left(\bar{a}\upharpoonright n/M\right)$
is not definable with parameters in $M_{p}$. So $1<n$ and $r=\tp\left(a_{0},\ldots,a_{n-2}/M\right)$
is definable but $\tp\left(a_{0},\ldots,a_{n-1}/M\right)$ is not.
For a formula $\alpha\left(x_{0},\ldots,x_{n-2},y\right)$ let $\left(d_{r}\alpha\right)\left(y\right)$
be a formula in $L\left(M_{p}\right)$ defining $\alpha\left(\bar{a}\upharpoonright n-1,M\right)$.
If $M_{p}\prec N\models T_{p}$ then, as in Remark \ref{rem:TpModel},
there is a complete $L$ type $r^{N}\left(x_{0},\ldots,x_{n-2}\right)$
over $N$ defined by $\alpha\left(x_{0},\ldots,x_{n-2},b\right)\in r^{N}$
if and only if $N\models\left(d_{r}\alpha\right)\left(b\right)$.
There is some formula $\varphi\left(x_{0},\ldots,x_{n-1},y\right)$
such that the set $B_{0}:=\varphi\left(a_{0},\ldots,a_{n-1},M\right)$
is not definable with parameters in $M_{p}$. Let $S=Th\left(M_{p},B_{0}\right)$
in the language $L\left(M_{p}\right)\left(P\right)$ (naming elements
from $M$, so that $N\models S$ implies $M_{p}\prec N$).
By the theorem cited above, for every $\lambda\geq\left|L\right|$
and $\kappa<\ded\lambda$, there exists a model $\left(N_{\lambda,\kappa},B_{\lambda,\kappa}\right)=\left(N,B\right)\models S$
of cardinality $\lambda$ such that, letting $\mathcal{B}^{N}=\left\{ B'\left|\,\left(N,B'\right)\cong\left(N,B\right)\right.\right\} $,
$\left|\mathcal{B}^{N}\right|\geq\kappa$.
Let $\bar{a}^{N}=\left(a_{0}^{N},\ldots,a_{n-2}^{N}\right)\models r^{N}$
and for every $B'\in\mathcal{B}^{N}$, let
\[
q_{B'}=p^{N}\left(x\right)\cup\left\{ \varphi\left(\bar{a}^{N},x,\bar{b}\right)\left|\,\bar{b}\in B'\right.\right\} \cup\left\{ \neg\varphi\left(\bar{a}^{N},x,\bar{b}\right)\left|\,\bar{b}\notin B'\right.\right\} .
\]
By choice of $S$, $B$ and $\mathcal{B}$, $q_{B'}$ is finitely
satisfiable in $N$ and for $B'\neq B''\in\mathcal{B}^{N}$, $q_{B'}\upharpoonright\varphi\neq q_{B''}\upharpoonright\varphi$,
so now $N\upharpoonright L$ is a model of $T$ with a type $p^{N}$
such that $\left|\uf_{\left\{ \varphi\right\} }\left(p\right)\right|\geq\kappa$.
\end{proof}
If $T$ is small we can say more:
\begin{claim}
Assume $T$ is small. Suppose $p\in S\left(M\right)$, $q\in\uf\left(p\right)$,
and $\bar{a}=\left\langle a_{0},a_{1},\ldots\right\rangle \models q^{\left(\omega\right)}|_{M}$,
then $\tp\left(\bar{a}/M\right)$ is definable over $\acl^{\eq}\left(\emptyset\right)$
in $M_{p}$.\end{claim}
\begin{proof}
By Claim \ref{cla:Large} it is definable with parameters in $M_{p}$.
Let $n<\omega$ be minimal such that $q^{\left(n\right)}|_{M}$ is
not definable over $\acl^{\eq}\left(\emptyset\right)$ in $M_{p}$.
Suppose that for some formula $\varphi\left(x_{0},\ldots,x_{n-1},y\right)$,
$\tp_{\varphi}\left(\bar{a}/M\right)$ is not definable over $\acl^{\eq}\left(\emptyset\right)$
in $M_{p}$. This means that while the set $\left\{ b\in M\left|\,\C\models\varphi\left(\bar{a},b\right)\right.\right\} $
is definable by $\psi\left(y,c\right)$ for some $\psi$ in $L\left(M_{p}\right)$,
$c\notin\acl^{\eq}\left(\emptyset\right)$. We may assume that $c\in M_{p}^{\eq}$
is the code of this set (for every automorphism $\sigma$, $\sigma$
fixes this set if and only if $\sigma\left(c\right)=c$). So in some
elementary extension $M_{p}\prec N$, there are infinitely many conjugates
of $c$, $\left\{ c_{i}\left|\, i<\omega\right.\right\} $, such that
$\psi\left(N,c_{i}\right)\neq\psi\left(N,c_{j}\right)$ for $i\neq j$.
This implies that $\uf_{\left\{ \varphi\right\} }\left(p^{N}\right)\geq\aleph_{0}$,
just as in the proof of Claim \ref{cla:Large}.\end{proof}
\begin{cor}
\label{cor:Medium}($T$ dependent) $T$ is large if and only if for
every $\lambda\geq\left|T\right|$,
\[
\sup\left\{ \left|\uf_{\Delta}\left(p\right)\right|\left|\, p\in S\left(M\right),\Delta\mbox{ finite},\left|M\right|=\lambda\right.\right\} =\ded\left(\lambda\right).
\]
\end{cor}
\begin{proof}
Suppose $T$ is large, i.e. for some $\Delta$, $M$, $p\in S\left(M\right)$
and $\lambda\geq M$, $\left|\uf_{\Delta}\left(p\right)\right|>\lambda$.
By Proposition \ref{prop:localpomega}, find some $\Delta_{0}$ and
$n$ such that
\[
\left|\uf_{\Delta}\left(p\right)\right|\leq\left|\left\{ \left(q^{\left(n\right)}\upharpoonright\Delta_{0}\right)|_{M}\left|\, q\in\uf\left(p\right)\right.\right\} \right|.
\]
Hence there is some $q\in\uf_{\Delta}\left(p\right)$ such that $q^{\left(n\right)}\upharpoonright\Delta_{0}$
is not definable with parameters over $M_{p}$, and we are done by
Claim \ref{cla:Large} (also note that $\left|S_{\Delta}\left(M\right)\right|\leq\ded\left|M\right|$
for every finite $\Delta$ in dependent theories (see e.g. \cite[Theorem 4.3]{Sh10})).
\end{proof}
\subsubsection{Trichotomy theorem}
\begin{thm}
\label{thm:trichotomy}For every $T$, $T$ is either small, medium
or large. Moreover,
\begin{enumerate}
\item $T$ is small iff for all $M\models T$, $p\left(x\right)\in S\left(M\right)$,
$\left|\uf\left(p\right)\right|\leq2^{\left|T\right|+\left|\lg\left(x\right)\right|}$
.
\item $T$ is medium iff for all $M\models T$, $p\left(x\right)\in S\left(M\right)$,
$\left|\uf\left(p\right)\right|\leq\left|M\right|^{\left|T\right|+\left|\lg\left(x\right)\right|}$
and $T$ is not small.
\item $T$ is large iff $T$ is not small nor medium.
\end{enumerate}
\end{thm}
\begin{proof}
(1) is Claim \ref{cla:bounded}. (3) is the definition.
(2) Left to right is clear. Conversely, since $T$ is not small, by
Claim \ref{cla:bounded},
\[
\sup\left\{ \left|\uf_{\Delta}\left(p\right)\right|\left|\, p\in S\left(M\right),\Delta\mbox{ finite},\left|M\right|=\lambda\right.\right\} \geq\lambda.
\]
On the other hand, if it is strictly greater than $\lambda$, then
by the proof of Corollary \ref{cor:Medium}, $T$ is large.
\end{proof}
In Section \ref{sub:examples of directionality}, we will show that
these classes are not empty, and thus:
\begin{cor}
For $\lambda\geq\left|T\right|$, the cardinality
\[
\sup\left\{ \left|\uf_{\Delta}\left(p\right)\right|\left|\, p\in S\left(M\right),\Delta\mbox{ finite},\left|M\right|=\lambda\right.\right\}
\]
has four possibilities: finite / $\aleph_{0}$ ; $\lambda$; $\ded\lambda$;
$2^{2^{\lambda}}$.
This corresponds to small, medium, and large directionality (the last
one happens when the theory has the independence property, see Observation
\ref{obs:IPLarge}). \end{cor}
\begin{problem}
Suppose $T$ is interpretable in $T'$, and $T$ is large. Does this
imply that $T'$ is large?
\end{problem}
\subsection{\label{sub:examples of directionality}Examples of different directionalities.}
Here we give examples of the different directionalities.
\subsubsection{Small directionality}
\begin{example}
\label{exa:DLO} $Th\left(\mathbb{Q},<\right)$ has small directionality.
In fact, every $1$-type over a model $M$, has at most 2 global co-heirs,
and in general, a type $p\left(x_{0},\ldots,x_{n-1}\right)$ is determined
by the order type of $\left\{ x_{0},\ldots,x_{n-1}\right\} $ and
$p\upharpoonright x_{0}$, $p\upharpoonright x_{1}$, etc. \end{example}
\begin{prop}
\label{pro:DTRBounded}The theory of dense trees is also small. \end{prop}
\begin{proof}
So here $T$ is the model completion of the theory of trees in the
language $\left\{ <,\wedge\right\} $.
\begin{claim*}
Let $M\models T$ and $p\left(x_{0},x_{1},\ldots,x_{n-1}\right)\in S\left(M\right)$
be any type. Then $\bigcup_{i,j<n}p\upharpoonright\left(x_{i},x_{j}\right)\cup p|_{\emptyset}\vdash p$. \end{claim*}
\begin{proof}
Let $\Sigma=\bigcup_{i,j<n}p\upharpoonright\left(x_{i},x_{j}\right)\cup p|_{\emptyset}$.
Suppose $\left(a_{0},\ldots,a_{n-1}\right)\models\Sigma$. By quantifier
elimination, the formulas in $p$ are Boolean combination of formulas
of the form $\bigwedge_{k<m}x_{j_{k}}\wedge a\leq\bigwedge_{l<r}x_{r_{l}}\wedge b$
where $b,a\in M$ and $j_{0},\ldots,j_{k-1},r_{0},\ldots,r_{l-1}<n$.
If $a,b$ does not appear, $\bar{a}$ satisfy this formula because
we included $p|_{\emptyset}$. Consider $\bigwedge_{k<m}x_{j_{k}}\wedge a$:
by assumption we know what is the ordering of $\left\{ x_{j_{k}}\wedge a\left|\, k<m\right.\right\} $
(this set is linearly ordered --- it is below $a$). Hence, as $\bar{a}\models\Sigma$,
$\bigwedge_{k<m}a_{j_{k}}\wedge a$ must be equal to the minimal element
in this set, namely $a_{j_{k}}\wedge a$ for some $k$, which is determined
by $\Sigma$. Now $a_{j_{k}}\wedge a\leq\bigwedge_{l<r}a_{r_{l}}\wedge b$
holds if and only if for each $r<l$, we have $a_{j_{k}}\wedge a\leq a_{r}$
and $a_{j_{k}}\wedge a\leq b$, both decided in $\Sigma$.
Note that we can get rid of $p|_{\emptyset}$ but we should replace
2-types by 3-types.\end{proof}
\begin{claim*}
For any $A\subseteq M\models T$, and $p\left(x\right),q\left(y\right)\in S\left(A\right)$,
there are only finitely many complete type $r\left(x,y\right)$ that
contain both $p$ and $q$. In fact there is a uniform bound on their
number.\end{claim*}
\begin{proof}
We may assume that $A$ is a substructure. For any $a\in M$, the
structure generated by $A$ and $a$, denoted by $A\left(a\right)$,
is just $A\cup\left\{ a_{A},a\right\} $ where $a_{A}=\max\left\{ b\wedge a\left|\, b\in A\right.\right\} $.
Note that $a_{A}$ need not exist, but if it does, then it is the
only new element except $a$ (because if $b_{1}\wedge a<b_{2}\wedge a$
then $b_{1}\wedge a=b_{1}\wedge b_{2}$).
Now, let $a\models p$ and $b\models q$. Let $B=B\left(p,q\right)=\left\{ d\in A\left|\, d\leq a\,\&\, d\leq b\right.\right\} $.
This set is linearly ordered, and it may have a maximum. If it does,
denote it by $m$. Note that $m$ depends only on $p$ and $q$.
Now it is easy to show that $\tp\left(a,a_{A},b,b_{A},a\wedge b/m\right)\cup p\left(x\right)\cup q\left(y\right)$
determines $\tp\left(a,b/A\right)$ by quantifier elimination. This
suffices because the number of types of finite tuples over a finite
set is finite.
\end{proof}
Let $M\models T$, $p\in S\left(M\right)$ and $I$ be the set of
all types $r\left(\bar{x}_{0},\bar{x}_{1},\ldots\right)$ over $M$
such that realizations of $r$ are indiscernible sequences of tuples
satisfying $p$.
Let $V=\left\{ \left(y_{0},y_{1}\right)\left|\, y_{0},y_{1}\mbox{ are any 2 variables from }\bar{x}_{0},\bar{x}_{1}\right.\right\} $.
By the second claim, for any $\left(y_{0},y_{1}\right)\in V$, the
set $\left\{ r\upharpoonright\left(y_{0},y_{1}\right)\left|\, r\in I\right.\right\} $
is finite. By the first claim and indiscernibility, the function taking
$r$ to $\left(r|_{\emptyset},\left\langle r\upharpoonright\left(y_{0},y_{1}\right)\left|\,\left(y_{0},y_{1}\right)\in V\right.\right\rangle \right)$
is injective. Together, it means that $\left|I\right|\leq2^{\aleph_{0}}$
and we are done by Fact \ref{fac:pOmega}.
\end{proof}
\subsubsection{Medium directionality}
\begin{example}
Let $L=\left\{ P,Q,H,<\right\} $ where $P$ and $Q$ are unary predicates,
$H$ is a unary function symbol and $<$ is a binary relation symbol.
Let $T^{\forall}$ be the following theory:
\begin{itemize}
\item $P\cap Q=\emptyset$.
\item $H$ is a function from $P$ to $Q$ (So $H\upharpoonright Q=\id$).
\item $\left(P,<,\wedge\right)$ is a tree.
\end{itemize}
And let $T$ be its model completion (so $T$ eliminates quantifiers).
Note that there is no structure on $Q$. So as in Section \ref{sec:Indisc},
$T$ is dependent (this theory is interpretable in the theory there).
Let $T'$ be the restriction of $T$ to the language $L'=L\backslash\left\{ H\right\} $.
The same ``moreover'' part applies here as in Corollary \ref{cor:ModelCom},
so $T'$ is the model completion of $T^{\forall}\upharpoonright L'$
and also eliminates quantifiers. \end{example}
\begin{claim}
$T'$ has small directionality.\end{claim}
\begin{proof}
The only difference between $T'$ and dense trees is the new set $Q$
which has no structure. Easily this does not make any difference. \end{proof}
\begin{prop}
\label{pro:MediumExample}$T$ has medium directionality.\end{prop}
\begin{proof}
Let $M\models T$. Let $B$ be a branch in $P^{M}$ (i.e. a maximal
linearly ordered set). Let $p\left(x\right)\in S\left(M\right)$ be
a complete type containing $\Sigma_{B}:=\left\{ b<x\left|\, b\in B\right.\right\} $.
Note that $\Sigma$ ``almost'' isolates $p$: the only freedom we
have, is to determine what is $H\left(x\right)$. So suppose $H\left(x\right)=m\in p$
for $m\in Q$.
Let $c\models p$ (so $c\notin M$). For each $a\in Q^{M}$, let $p_{a}\left(x\right)=p\cup\left\{ H\left(c\wedge x\right)=a\right\} $.
Then $p_{a}$ is finitely satisfiable in $M$: Suppose $\Gamma\subseteq p_{a}$
is finite. By quantifier elimination, we may assume that $\Gamma\subseteq\Sigma_{B}\cup\left\{ H\left(c\wedge x\right)=a\right\} \cup\left\{ H\left(x\right)=m\right\} $.
Since $B$ is linearly ordered, we may assume that $\Gamma=\left\{ b<x,H\left(c\wedge x\right)=a,H\left(x\right)=m\right\} $
for some $b\in B$. Since $T$ is the model completion of $T^{\forall}$
which has the amalgamation property, there are two elements $d,e\in M$
such that $b<d,e$, $e\in B$, $d\notin B$, $H\left(d\right)=m$
and $H\left(d\wedge e\right)=a$. Since $d\wedge c=d\wedge e$, $d\models\Gamma$.
We have found $\left|Q^{M}\right|$ co-heirs of $p$, and since $M$
was arbitrary $T$ is not small.
This gives a lower bound on the directionality of $T$, and we would
like to find an upper bound as well. We shall use the same idea as
in the proof of Proposition \ref{pro:DTRBounded}.
Let $M\models T$, $p\in S\left(M\right)$ and $I$ be the set of
all types $r\left(\bar{x}_{0},\bar{x}_{1},\ldots\right)$ over $M$
such that realizations of $r$ are indiscernible sequences of tuples
satisfying $p$. Let $I'=\left\{ r\upharpoonright L'\left|\, r\in I\right.\right\} $.
By the proof of proposition \ref{pro:DTRBounded}, $\left|I'\right|\leq2^{\aleph_{0}}$.\\
Let $V=\left\{ t\left|\, t\mbox{ is a term in }L\mbox{ in the variables }\bar{x}_{0},\bar{x}_{1},\ldots\mbox{ over }\emptyset\right.\right\} $.
Suppose $r\in I$, then, as in the proof of Proposition \ref{pro:DTRBounded},
for every term $t$ such that $P\left(t\right)\in r$, let $t_{r}=\max\left\{ a\wedge t\left|\, a\in M\right.\right\} $
--- a term over $M$ (it need not exist). To determine $r$, it is
enough to determine the equations that occur between the images under
$H$ of the $t_{r}$'s and the $t$'s over $M$. This shows that $\left|I\right|\leq\left|M\right|^{\aleph_{0}}$.
\end{proof}
\subsubsection{Large directionality}
\begin{example}
\label{exa:Large}Let $L=\left\{ P,Q,H,<_{P},<_{Q}\right\} $ where
$P$ and $Q$ are unary predicates, $H$ is a unary function symbol
and $<_{P},<_{Q}$ are binary relation symbols. Let $T^{\forall}$
be the following theory:
\begin{itemize}
\item $P\cap Q=\emptyset$.
\item $H$ is a function from $P$ to $Q$.
\item $\left(P,<_{P},\wedge\right)$ is a tree.
\item $\left(Q,<_{Q}\right)$ is a linear order.
\end{itemize}
\end{example}
\begin{prop}
$T$ has large directionality.\end{prop}
\begin{proof}
This is similar to the proof of Proposition \ref{pro:MediumExample}.
Let $M\models T$. Let $B$ be a branch in $P^{M}$. Let $p\left(x\right)\in S\left(M\right)$
be a complete type containing $\Sigma_{B}:=\left\{ b<x\left|\, b\in B\right.\right\} $
saying that $H\left(x\right)=m$ for some $m\in Q^{M}$.
Let $c\models p$ (so $c\notin M$). For each cut $I\subseteq Q^{M}$,
let
\[
p_{I}\left(x\right)=p\cup\left\{ e<H\left(c\wedge x\right)<f\left|\, e\in I,f\in Q^{M}\backslash I\right.\right\} .
\]
Then $p_{I}$ is finitely satisfiable in $M$ as in the proof of \ref{pro:MediumExample}.
So for every cut in $Q$ we found a co-heir of $p$, and since $M$
was arbitrary $T$ is not small nor medium (because for every linear
order, we can find a model such that $Q$ contains this order).
\end{proof}
The above example can be transformed to an example in valued fields:
\begin{defn}
The language $L$ of valued fields is the following. It is a 3-sorted
language, one sort for the base field $K$ equipped with the ring
language $\left\{ 0,1,+,\cdot\right\} $, another for the valuation
group $\Gamma$ equipped with the ordered abelian groups language
$L_{\Gamma}=\left\{ 0,+,<\right\} $, and another for the residue
field $k$ equipped with the ring language $L_{k}$, we also have
the valuation map $v:K^{\times}\to\Gamma$, and an angular component
map $\ac:K\to k$. Recall that an angular component is a function
that satisfies $\ac\left(0\right)=0$ and $\ac\upharpoonright K^{\times}:K^{\times}\to k^{\times}$is
a homomorphism such that if $v\left(x\right)=0$ then $\ac\left(x\right)$
is the residue of $x$.
\end{defn}
For more on valued fields with angular component, see e.g. \cite{belair,Pas1989},
which also gives us the following fact
\begin{fact}
\label{fac: Elimination of Field Quantifiers}\cite[Theorem 4.1]{Pas1989}
\begin{enumerate}
\item The theory of Henselian valued fields of characteristic $\left(0,0\right)$
in the language $L$ has elimination of field quantifiers, i.e. every
formula $\varphi\left(x\right)$ is equivalent to a Boolean combination
of formulas of the form $\varphi\left(\ac\left(f_{0}\left(x\right)\right),\ldots,\ac\left(f_{n-1}\left(x\right)\right)\right)$
and $\chi\left(v\left(g_{0}\left(x\right)\right),\ldots,v\left(g_{m-1}\left(x\right)\right)\right)$
where $\varphi$ is a formula in $L_{k}$, $\chi$ is a formula in
$L_{\Gamma}$, $g_{i}$ and $f_{j}$ are polynomials over the integers.
\item Moreover, the theory admits cell decomposition which implies that
if $M$ is a model, then every type $p\left(x\right)\in S\left(M\right)$
(where $x\in K$) is isolated by the set of formulas
\[
p\cap\left[\bigcup_{a\in K^{M}}\left\{ \varphi\left(\ac\left(x-a\right)\right)\left|\,\varphi\in L_{k}\left(k^{M}\right)\right.\right\} \cup\bigcup_{a\in K^{M}}\left\{ \chi\left(v\left(x-a\right)\right)\left|\,\chi\in L_{\Gamma}\left(\Gamma^{M}\right)\right.\right\} \right].
\]
\end{enumerate}
\end{fact}
\begin{thm}
Let $T=Th\left(K,\Gamma,k\right)$ be a theory of Henselian valued
fields of characteristic $\left(0,0\right)$. Then $T$ has medium
or large directionality.\end{thm}
\begin{proof}
Let $M_{0}=\left(K_{0},\Gamma_{0},k_{0}\right)\models T$ be a countable
model, and let $\left\langle \gamma_{i}\left|\, i<\omega\right.\right\rangle $
be an increasing sequence of elements from $\Gamma^{+}$. Let $S=\leftexp{\omega\geq}2,\, S_{0}=\leftexp{\omega>}2$
and we give $S$ (and $S_{0}$) a tree structure $\left(<_{S},\wedge_{S}\right)$.
Let $f:\left\{ \left(x,y\right)\in S_{0}^{2}\left|\, x<y\right.\right\} \to k_{0}$
be some function onto $k_{0}$ such that:
\begin{itemize}
\item $s_{1}<s_{2}<s_{3}\Rightarrow f\left(s_{1},s_{2}\right)=f\left(s_{1},s_{3}\right)$,
\item for each $x\in k_{0}$, and each branch $\eta\in\leftexp{\omega}2$,
$\left\{ s\in S_{0},s<\eta\left|\,\exists t>s,t<\eta\left[f\left(s,t\right)=x\right]\right.\right\} $
is infinite, and
\item if $t_{1}\wedge t_{2}=s$ then $f\left(s,t_{1}\right)\neq f\left(s,t_{2}\right)$.
\end{itemize}
Let $\Sigma\left(x_{s}\left|\, s\in S\right.\right)$ be the following
set of formulas with variables in the field sort (over $M_{0}$):
\[
\left\{ v\left(x_{s}\right)=0\left|\, s\in S\right.\right\} \cup\left\{ v\left(x_{s}-x_{t}\right)=\gamma_{\lev\left(s\wedge t\right)}\left|\, s,t\in S\right.\right\} \cup\left\{ \ac\left(x_{t}-x_{s}\right)=f\left(s,t\right)\left|\, s<t\in S\right.\right\} .
\]
Then $\Sigma$ is consistent with $M_{0}$, and let $\left(M,a_{s}\left|\, s\in S\right.\right)$
be a countable model realizing $\Sigma$ containing $M_{0}$. For
each branch $\eta\in\leftexp{\omega}{\omega}$, let $p_{\eta}\left(x\right)$
be the following type in the valued field sort
\[
\left\{ v\left(x-a_{s}\right)\geq\gamma_{\lev\left(s\right)}\left|\, s\in\eta\right.\right\} .
\]
If $\eta_{1}\neq\eta_{2}$ then $p_{\eta_{1}}\neq p_{\eta_{2}}$ because
if $s=\eta_{1}\wedge\eta_{2}$ and $s<t<\eta_{1}$ then $p_{\eta_{1}}\models v\left(x-a_{t}\right)=\gamma_{\lev\left(t\right)}$
(as $v\left(x-a_{t}\right)=v\left(x-a_{t'}+a_{t'}-a_{t}\right)=v\left(a_{t'}-a_{t}\right)=\gamma_{\lev\left(t\right)}$
for any $t<t'<\eta_{1}$) while $p_{\eta_{2}}\models v\left(x-a_{t}\right)=\gamma_{\lev\left(s\right)}$
(because $p_{\eta_{2}}\models v\left(x-a_{t}\right)=v\left(x-a_{s}+a_{s}-a_{t}\right)\geq\gamma_{\lev\left(s\right)}$,
but $p_{\eta_{2}}\models\ac\left(x-a_{s}\right)=\ac\left(a_{t'}-a_{s}\right)$
for any $s<t'<\eta_{2}$, and since $\ac\left(a_{t'}-a_{s}\right)\neq\ac\left(a_{t}-a_{s}\right)$,
we are done).
Since $M$ is countable, there is some branch $\eta\in\leftexp{\omega}2$
such that $p_{\eta}$ is not satisfied in $M$.
Note that if $a\in K^{M}$ then there is some $s<\eta$ such that
$v\left(a-a_{s}\right)<\gamma_{\lev\left(s\right)}$, but then $p_{\eta}\models v\left(x-a\right)=v\left(a_{s}-a\right)$.
Also, $p_{\eta}\models\ac\left(x-a\right)=\ac\left(a_{s}-a\right)$.
By fact \ref{fac: Elimination of Field Quantifiers}, this means that
$p_{\eta}$ is a complete type over $M$.
Let $c\models p_{\eta}$. Then, for each $a\in k_{0}$, $p_{a}:=p\left(x\right)\cup\left\{ \ac\left(c-x\right)=a\right\} $
is finitely satisfiable in $K^{M}$ (each finite subset is realized
by $a_{s}$ for some $s$), and if $a\neq b$ then $p_{a}\neq p_{b}$
and we are done. \end{proof}
\begin{rem}
In the theorem, instead of assuming that the field is Henselian of
characteristic $\left(0,0\right)$, it is enough to assume that it
eliminates field quantifiers and that the residue field is infinite.
Note that by \cite{Yimu}, if the valuation group is a $\mathbb{Z}$-group,
then the field is Henselian.
Indeed, let $\bar{M}=\left(\bar{K},\bar{k},\bar{\Gamma}\right)$ be
the algebraic closure of $M$ as a valued field. We embed it in a
large algebraic closure $\Omega$ of the monster model $\C$ of $M$.
Since it is still countable, there is some $\eta\in2^{\omega}$ such
that $p_{\eta}$ is not satisfied even in $\bar{M}$. Then for every
polynomial $f\left(x\right)$ over $K$ and every $s<\eta$ large
enough, $p_{\eta}\models\ac\left(f\left(x\right)\right)=\ac\left(f\left(a_{s}\right)\right)\land v\left(f\left(x\right)\right)=v\left(f\left(a_{s}\right)\right)$
(decompose $f$ into linear factors $\prod\left(x-a_{i}\right)$.
For every large enough $s$, $v\left(a_{i}-a_{s}\right)<\gamma_{\lev\left(s\right)}$
for all $i$, and so if $d\models p_{\eta}$ in $\C$, then $\ac\left(f\left(d\right)\right)=\ac\left(f\left(a_{s}\right)\right)$
(because $res\left(\frac{f\left(d\right)}{f\left(a_{s}\right)}\right)=1$
--- we do not assume that $\ac$ extends to $\bar{M}$) and $v\left(f\left(d\right)\right)=v\left(f\left(a_{s}\right)\right)$)).
So $p_{\eta}$ is a complete type over $M$, and the proof goes through. \end{rem}
\begin{problem}
We do not know what is the directionality in the case of fields where
the residue field is finite, such as the $p$-adics.
\end{problem}
It turns out that even $RCF$ has large directionality, as we shall
present now.
Apparently, that $RCF$ was not small was already known and can be
deduced from Marcus Tressl's thesis (see \cite[18.13]{Tressle}),
but here we give a direct proof that $RCF$ is in fact large and even
more.
\begin{defn}
Let $M\models RCF$. A type $p\in S\left(M\right)$ is called dense
if it is not definable and the differences $b-a$ with $a,b\in M$
and $a<x<b\in p$, are arbitrarily (w.r.t. $M$) close to $0$.
\end{defn}
For example, if $R$ is the real closure of $\mathbb{Q}$, then $\tp\left(\pi/R\right)$
is dense.
\begin{fact}
\label{fac:DenseExist}Any real closed field can be embedded into
a real closed field of the same cardinality with some dense type.\end{fact}
\begin{proof}
{[}due to Marcus Tressl{]} Let $R$ be a real closed field. Let $S$
be the (real closed) field $R\left(\left(t^{\mathbb{Q}}\right)\right)$
of generalized power series over $R$ with coefficients in the exponents
$\mathbb{Q}$. Let $K$ be the definable closure of $R\left(t\right)$
in $S$ and let $p$ be the 1-type of the formal Taylor series of
$e^{t}$ over $K$: $\tp\left(1+t^{1}/1!+t^{2}/2!+\ldots/K\right)$
. Then $p$ is a dense 1-type over $K$. \end{proof}
\begin{claim}
\label{cla:WeakOrth}Suppose $p$ is dense and $q$ is a definable
type over $M\models RCF$ and both are complete. Then $q$ and $p$
are weakly orthogonal, meaning that $p\left(x\right)\cup q\left(y\right)$
implies a complete type over $M$.\end{claim}
\begin{proof}
{[}Remark: this is an easy result that is well known, but for completeness
we give a proof.{]}
Let $\omega\models q,\alpha\models p$.
Note that since $p$ is not definable over $M$, for every $m\in M$,
and even for every $m$ such that $\tp\left(m/M\right)$ is definable,
there is some $\varepsilon_{m}\in M$ such that $0<\varepsilon_{m}<\left|\alpha-m\right|$.
Now, suppose that $\varphi\left(x,y\right)$ is any formula over $M$.
Then, as $q$ is definable, there is a formula $\psi\left(x\right):=\left(d_{q}y\right)\varphi\left(x,y\right)$
over $M$ that defines $\varphi\left(M,\omega\right)$. We claim that
$\varphi\left(\alpha,\omega\right)$ if and only if $\psi\left(\alpha\right)$
(i.e. if $p\left(x\right)\cup q\left(y\right)$ then $\varphi\left(x,y\right)$
if and only if $\left(d_{q}y\right)\varphi\left(x,y\right)\in p$
).
We know that $\psi$ is equivalent to a finite union of intervals
and points from $M$. We also know that $\varphi\left(\C,\omega\right)$
is such a union, but the types of the end-points over $M$ are definable
over $M$ (since we have definable Skolem functions). So denote the
set of all these end-points by $A$. Let $0<\varepsilon\in M$ be
smaller than every $\varepsilon_{m}$ for each $m\in A$. Let $a,b\in M$
such that $a<\alpha<b$ and $b-a<\varepsilon$. Then necessarily
\begin{itemize}
\item $\psi\left(\alpha\right)$ holds if and only if
\item $\psi\left(m\right)$ holds for all / some $m\in M$ such that $a\leq m\leq b$
if and only if
\item $\varphi\left(m,\omega\right)$ holds for all / some $m\in M$ such
that $a\leq m\leq b$ if and only if
\item $\varphi\left(\alpha,\omega\right)$ holds.
\end{itemize}
\end{proof}
We claim that $RCF$ has large directionality. Moreover, we seem to
answer an open question raised in \cite{Delon}, as she says there:
\begin{quotation}
Mais il laisse ouverte la possibilit\'e que la borne du nombre de
coh\'eritiers soit $\ded\left|M\right|$ dans le cas de la propri\'et\'e
de l'ordre et $\left(\ded\left|M\right|\right)^{\left(\omega\right)}$
dans le cas de l'ordre multiple.
\end{quotation}
So let us make clear what the question means:
\begin{defn}
(Taken from \cite{KeislerStabFunction,KeislerSixClasses}) A theory
$T$ is said to have the \emph{multiple order property} if there are
formulas $\varphi_{n}\left(x,y_{n}\right)$ for $n<\omega$ such that
the following set of formulas is consistent with $T$:
\[
\Gamma=\left\{ \varphi_{n}\left(x_{\eta},y_{n,k}\right)^{\eta\left(k\right)<n}\left|\,\eta\in\leftexp{\omega}{\omega}\right.\right\} .
\]
\end{defn}
\begin{rem}
If $T$ is strongly dependent (see \ref{def:StronglyDep}), for example,
if $T=RCF$, it does not have the multiple order property.\end{rem}
\begin{proof}
Suppose $T$ has the multiple order property as witnessed by formulas
$\varphi_{n}$. Consider the formulas $\psi_{n}\left(x,y,z\right)=\varphi_{n}\left(x,y\right)\leftrightarrow\varphi_{n}\left(x,z\right)$.
It is easy to see that $\left\{ \psi_{n}\left|\, n<\omega\right.\right\} $
ex amplifies that the theory is not strongly dependent. \end{proof}
\begin{fact}
\cite{KeislerStabFunction} If $T$ is countable and has the multiple
order property, then for every $\kappa$, $\sup\left\{ S\left(M\right)\left|\, M\models T,\left|\, M\right|=\kappa\right.\right\} \geq\left(\ded\kappa\right)^{\omega}$.
If $T$ does not have the multiple order property, then $\sup\left\{ S\left(M\right)\left|\, M\models T,\left|M\right|=\kappa\right.\right\} \leq\ded\kappa$
for every $\kappa$.
\end{fact}
So the question can be formulated as follows:
\begin{itemize}
\item Is there a countable theory without the multiple order property such
that for every $\kappa\geq\aleph_{0}$, $\sup\left\{ \left|\uf\left(p\right)\right|\left|\, p\in S\left(M\right),\left|M\right|=\kappa\right.\right\} =\left(\ded\kappa\right)^{\omega}$.
\end{itemize}
It is a natural question, because of 2 reasons:
\begin{enumerate}
\item In general, the number of types (in $\alpha$ variables) over a model
of size $\lambda$ in a dependent theory is bounded by $\left(\ded\lambda\right)^{\left|T\right|+\left|\alpha\right|+\aleph_{0}}$
(by \cite[Theorem 4.3]{Sh10}), so by Fact \ref{fac:pOmega} this
is an upper bound for $\sup\left\{ \left|\uf\left(p\right)\right|\left|\, p\in S\left(M\right),\left|M\right|=\kappa\right.\right\} $.
\item It is very easy to construct an example with the multiple order property
that attains this maximum: for example, one can modify example \ref{exa:Large},
and add independent orderings to $Q$.\end{enumerate}
\begin{thm}
$RCF$ has large directionality. Moreover, $RCF$ does not have the
multiple order property but for every $\lambda\geq\aleph_{0}$,
\[
\sup\left\{ \left|\uf\left(p\right)\right|\left|\, p\in S\left(M\right),\left|M\right|=\lambda\right.\right\} \geq\left(\ded\lambda\right)^{\omega}.
\]
\end{thm}
\begin{proof}
Let $K\models RCF$ with a dense type $p$ over it. Let $\alpha\models p$,
and let $\omega$ be an element greater than any element in $K$.
Then $q=\tp\left(\omega/K\right)$ is definable and we can apply Claim
\ref{cla:WeakOrth}. Let $p(x_{\omega},x_{\alpha})=\tp(\omega,\alpha/K)$.
For every sequence $\bar{I}=\left\langle I_{i}\left|\, i<\omega\right.\right\rangle $
of bounded first segments $I_{i}\subseteq K$, we define a sequence
of terms: for every pair of sequences $\bar{a}=\left\langle a_{i}\left|\, i<\omega\right.\right\rangle $
and $\bar{b}=\left\langle b_{i}\left|\, i<\omega\right.\right\rangle $
such that $a_{i}\in I_{i}$ and $b_{i}\notin I_{i}$, and for each
$n<\omega$ let $f_{n}\left(\bar{a},x\right)=\alpha+\sum_{i=0}^{n}\left(a_{i}/x^{i}\right)$
and $g_{n}\left(\bar{a},\bar{b},x\right)=\alpha+\sum_{i=0}^{n-1}\left(a_{i}/x^{i}\right)+b_{n}/x^{n}$.
Now, let $p_{\bar{I}}\left(x_{\omega},x_{\alpha}\right)$ be
\begin{eqnarray*}
p_{\bar{I}}\left(x_{\omega},x_{\alpha}\right) & = & p\left(x_{\omega},x_{\alpha}\right)\cup\left\{ f_{n}\left(\bar{a},x_{\omega}\right)<x_{\alpha}<g_{n}\left(\bar{a},\bar{b},x_{\omega}\right)\left|\,\bar{a},\bar{b}\mbox{ as above, }n<\omega\right.\right\} .
\end{eqnarray*}
Then this type (which is over $K\cup\left\{ \alpha\right\} $) is
finitely satisfiable in $K$ because of the weak orthogonality. For
$\bar{I},\bar{J}$, two different sequences of first segments, $p_{\bar{I}}$
and $p_{\bar{J}}$ contradict each other. Since $K$ was arbitrary,
we are done by Fact \ref{fac:DenseExist}:
Suppose first that $\ded\lambda$ is attained. I.e. there is a linear
order $I$ of size $\lambda$ with $\ded\lambda$ many cuts. There
is a linear order of size $\lambda$ having $\omega$ disjoint copies
of $I$ as substructures. So we can embed this order in a real closed
field $R$ of size $\lambda$, which we can embed again in a real
closed field $K$ with a dense type and the same cardinality, and
apply the above.
If $\ded\lambda$ is not attained, then the cofinality of $\ded\lambda$
is greater than $\lambda$, so $\left(\ded\lambda\right)^{\omega}=\sup\left\{ \kappa^{\omega}\left|\,\kappa<\ded\left(\lambda\right)\right.\right\} $,
so a similar argument applies.
\end{proof}
Marcus Tressl has pointed out the type $\tp\left(\alpha,\omega\right)$
to us as a type with infinitely many co-heirs (this follows from \cite[18.13]{Tressle}).
We thank him for that. The explanation, and the proof that the theory
is large is ours.
\section{\label{sec:Splintering}Splintering}
This part of the paper is motivated by the work of Rami Grossberg,
Andr\'es Villaveces and Monica VanDieren. In their paper \cite{GrViVa}.
They were studying Shelah's Generic pair conjecture (which is now
a theorem --- \cite{Sh:900,Sh950,Sh906}) and in their analysis, they
came up with the notion of splintering, a variant of splitting.
\begin{defn}
Let $p\in S\left(\C\right)$. Say that $p$ \emph{splinters} over
$M$ if there is some $\sigma\in\Aut\left(\C\right)$ such that
\begin{enumerate}
\item $\sigma\left(p\right)\neq p$.
\item $\sigma\left(p|_{M}\right)=p_{|M}$.
\item $\sigma\left(M\right)=M$ setwise.
\end{enumerate}
\end{defn}
\begin{rem}
{[}due to Martin Hils{]} Splitting implies splintering, and if $T$
is stable, then they are equal. \end{rem}
\begin{proof}
Suppose $p\in S\left(\C\right)$ does not split over $M$, then, by
stability, it is definable over $M$, and $p$ is the unique non-forking
extension of $p|_{M}$. Then for any $\sigma\in\Aut\left(\C\right)$,
$\sigma\left(p\right)$ is the unique non-forking extension of $\sigma\left(p|_{M}\right)$.
So if $\sigma\left(p|_{M}\right)=p|_{M}$, this means that $\sigma\left(p\right)=p$
so $p$ does not splinter over $M$.\end{proof}
\begin{claim}
\label{cla:unstable implies sp neq spln}Outside of the stable context,
splitting $\neq$ splintering.\end{claim}
\begin{proof}
Let $T$ be the theory of random graphs in the language $\left\{ I\right\} $.
Let $M\models T$ be countable, and let $a\neq b\in M$ with an automorphism
$\sigma\in\Aut\left(M\right)$ taking $a$ to $b$. Let $p\left(x\right)\in S\left(\C\right)$
say that $xIc$ for every $c\in M$ and if $c\notin M$ then $xIc$
if and only if $c$ is connected $a$ and not connected to $b$. Obviously,
$p$ does not split over $M$. However, let $\sigma'\in\Aut\left(\C\right)$
be an extension of $\sigma$. Let $c\in\C$ be such that $c$ is connected
to $a$ but not to $b$. Then $xIc\in p$ but $xIc\notin\sigma'\left(p\right)$.
\end{proof}
However,
\begin{claim}
If $T=Th\left(\mathbb{Q},<\right)$, then splitting equals splintering.\end{claim}
\begin{proof}
Observe that by quantifier elimination every complete type $r\left(x_{0},\ldots,x_{n-1}\right)$
over a set $A$ is determined by $\tp\left(x_{0},\ldots,x_{n-1}/\emptyset\right)\cup\bigcup\left\{ \tp\left(x_{i}/A\right)\left|\, i<n\right.\right\} $.
Assume $q\left(x_{0},\ldots,x_{n-1}\right)$ is a global type that
splinters but does not split over a model $M$. Then it follows that
for some $i<n$, $q\upharpoonright x_{i}$ splinters, so we may assume
$n=1$. Suppose $\sigma\in\Aut\left(\C\right)$ is such that $\sigma\left(M\right)=M$,
$\sigma\left(q|_{M}\right)=q|_{M}$ and $\sigma\left(q\right)\neq q$.
Note that $\sigma\left(q^{\left(\omega\right)}\right)=\sigma\left(q\right)^{\left(\omega\right)}$,
so by Fact \ref{fac:pOmega}, $\sigma\left(q^{\left(\omega\right)}\right)|_{M}\neq q^{\left(\omega\right)}|_{M}$.
We get a contradiction by quantifier elimination again.
\end{proof}
We shall now generalize Claim \ref{cla:unstable implies sp neq spln}
to every theory with the independence property. In fact, to any theory
with large or medium directionality.
\begin{thm}
Suppose $T$ has medium or large directionality then splitting $\neq$
splintering.\end{thm}
\begin{proof}
We know that there is some $p$ and $\Delta$ such that $\uf_{\Delta}\left(p\right)$
is infinite. Let us use Construction \ref{const:not small directionality}:
We may find a saturated model of $Th\left(M^{*}\right)$ of size $\lambda$
where $\lambda$ is big enough. Then there is $c,d\in Q_{0}$ such
that $\tp\left(c/\emptyset\right)=\tp\left(d/\emptyset\right)$ in
the extended language (with symbols for $N_{0},M_{0},Q_{0}$ and $\bar{f}$).
So there is an automorphism $\sigma$ of this structure (in particular
of $N_{0}'$) such that $\sigma\left(c\right)=d$. By definition,
$\sigma\left(N_{0}\right)=N_{0}$ and $\sigma\left(M_{0}\right)=M_{0}$.
So $\tp\left(c/N_{0}\right)$ is finitely satisfiable in $M_{0}$
and hence does not split over $M_{0}$. But it splinters since $\sigma\left(\tp\left(c/M_{0}\right)\right)=\tp\left(d/M_{0}\right)=\tp\left(c/M_{0}\right)$
but $\sigma\left(\tp\left(c/N_{0}\right)\right)=\tp\left(d/N_{0}\right)\neq\tp\left(c/N_{0}\right)$
as witnessed by $\varphi$.
If there are no saturated models, we can take a big enough special
model (see \cite[Theorem 10.4.4]{Hod}).
Note that we may also find an example of a type $p\in S\left(M\right)$
with a splintering, non-splitting, global extension, with $\left|M\right|=\left|T\right|$:
consider the structure $\left(N_{0}',N_{0},M_{0},\sigma,c,d\right)$,
and find an elementary substructure of size $\left|T\right|$. \end{proof}
\begin{defn}
Let $T$ be a complete theory. We say that $\left(M,p,\varphi\left(x,y\right),A_{1},A_{2}\right)$
is an\emph{ sp-example} for $T$ when
\begin{itemize}
\item $M\models T$; $A_{1},A_{2}\subseteq M$ are nonempty and disjoint;
$p=p\left(x\right)$ is a complete type over $M$, finitely satisfiable
in $A_{1}$; $\Th\left(M_{p},A_{1}\right)=\Th\left(M_{p},A_{2}\right)$
(see Definition \ref{def:ExtDefType}); For each pair of finite sets
$s_{1}\subseteq A_{1}$ and $s_{2}\subseteq A_{2}$, $M\models\exists y\left(\bigwedge_{a\in s_{1}}\varphi\left(a,y\right)\land\bigwedge_{b\in s_{2}}\neg\varphi\left(b,y\right)\right)$.
\end{itemize}
\end{defn}
\begin{prop}
\label{pro:sp-example}$T$ has an sp-example if and only if there
is a finitely satisfiable type over a model which splinters over it
$T$ (in particular, splitting is different than splitting).\end{prop}
\begin{proof}
Suppose $\left(M,p,\varphi\left(x,y\right),A_{1},A_{2}\right)$ is
an sp-example for $T$. Let $M'$ be the structure $\left(M_{p},A_{1},A_{2}\right)$
(in the language $L_{p}\cup\left\{ P_{1},P_{2}\right\} $ where $P_{1},P_{2}$
are predicates). Assume $\left|T\right|<\mu=\mu^{<\mu}$, and let
$N'=\left(N_{q},B_{1},B_{2}\right)$ be a saturated extension of $M'$
of size $\mu$ where $N=N'\upharpoonright L$ and $q=q^{N}$ is as
in Remark \ref{rem:TpModel}. Since $\left(N_{q},B_{1}\right)\equiv\left(N_{q},B_{2}\right)$,
there is an automorphism $\sigma$ of $N_{q}$, such that $\sigma$
takes $B_{1}$ to $B_{2}$ and so $\sigma\left(q\right)=q$. Let $q'$
be a global extension of $q$, finitely satisfiable in $B_{1}$ and
$\sigma'$ a global extension of $\sigma$.
So $q'$ does not split over $N$, but it splinters:
Consider the type $\left\{ \varphi\left(a,y\right)\left|\, a\in B_{1}\right.\right\} \cup\left\{ \neg\varphi\left(b,y\right)\left|\, b\in B_{2}\right.\right\} $.
It is finitely satisfiable in $N$ by choice of $\varphi$. Let $c\in\C$
satisfy this type. Then $\varphi\left(x,c\right)\in q'$ but $\varphi\left(x,c\right)\notin\sigma\left(q'\right)$
(because $\sigma\left(q'\right)$ is finitely satisfiable in $B_{2}$).
If we do not assume the existence of such a $\mu$, we can use special
models.
Now suppose that splitting is different than splintering, as witnessed
by some global type $p$ that splinters over a model $M$ but is finitely
satisfiable in it. Then there is some automorphism $\sigma$ of $\C$
that witnesses it. There is a formula $\varphi\left(x,y\right)$ and
$a\in\C$ such that $\varphi\left(x,a\right)\in p,\neg\varphi\left(x,a\right)\in\sigma\left(p\right)$.
Let $B_{1}=\left\{ m\in M\left|\,\varphi\left(m,a\right)\land\neg\varphi\left(m,\sigma^{-1}\left(a\right)\right)\right.\right\} $,
$B_{2}=\sigma\left(B_{1}\right)$. It is easy to check that $\left(M,p,\varphi,B_{1},B_{2}\right)$
is an sp-example
\end{proof}
The following theorem answers the natural question:
\begin{thm}
There is a theory with small directionality in which splitting $\neq$
splintering.\end{thm}
\begin{proof}
Let $L=\left\{ R\right\} $ where $R$ is a ternary relation symbol.
Let $M_{0}=\left\langle \mathbb{Q},<\right\rangle $ and define $R\left(x,y,z\right)$
by $x<y<z$ or $z<y<x$, i.e. $y$ is between $x$ and $z$. Let $T=Th\left(M_{0}\upharpoonright L\right)$.
\begin{claim*}
$T$ has small directionality.\end{claim*}
\begin{proof}
Suppose $M\models T$. Let $\left(a,b\right)$ denote $\left\{ c\left|\, R\left(a,c,b\right)\right.\right\} $.
Then, for any choice of a pair of distinct elements $a,b$ there is
a unique enrichment of $M$ to a model $M'$ of $Th\left(\mathbb{Q},<\right)$
such that $a<b$:
For $w\neq z$, $w<z$ if and only if
$\left(a,b\right)\cap\left(a,z\right)\neq\emptyset$ and ($\left(a,w\right)\subseteq\left(a,z\right)$
or $\left(a,w\right)\cap\left(a,z\right)=\emptyset$) or
$\left(a,b\right)\cap\left(a,z\right)=\emptyset$ and $\left(z,b\right)\subseteq\left(w,b\right)$.
From this observation, it follows that there is a unique completion
of any type $p\in S\left(M\right)$ to a type $p'\in S\left(M'\right)$.
So if $\Delta$ is a finite set of $L$ formulas and $\uf_{\Delta}\left(p\right)$
is infinite, then $\uf_{\Delta}\left(p'\right)$ is also infinite
--- contradiction to Example \ref{exa:DLO}.\end{proof}
\begin{claim*}
$T$ has an sp-example\end{claim*}
\begin{proof}
Let $M=M_{0}\upharpoonright L$. Let $p\left(x\right)=\tp\left(\pi/M\right)$.
Let $A_{1}=\left\{ x\in\mathbb{Q}\left|\, x>\pi\right.\right\} $
and $A_{2}=\mathbb{Q}\backslash A_{1}$, and let $\varphi\left(x,y_{1},y_{2}\right)=R\left(y_{1},x,y_{2}\right)$.
We claim that $\left(M,p,\varphi\left(x,y_{1},y_{2}\right),A_{1},A_{2}\right)$
is an sp-example:
First, let $M'$ be the reduct of $\left(\mathbb{Q}\cup\left\{ \pi\right\} ,<\right)$
to $L$. There is some $\sigma\in\Aut\left(M'/\pi\right)$ such that
$\sigma\left(A_{1}\right)=A_{2}$. Hence $\left(M_{p},A_{1}\right)\cong\left(M_{p},A_{2}\right)$.
Also, since $\tp\left(\pi/M_{0}\right)$ (in $\left\{ <\right\} $)
is finitely satisfiable in both $A_{1}$ and $A_{2}$ (by quantifier
elimination), $p$ is finitely satisfiable in both $A_{1}$ and $A_{2}$.
Finally, for finite $s_{1}\subseteq A_{1}$ and $s_{2}\subseteq A_{2}$,
there exists $c_{1},c_{2}\in\mathbb{Q}$ such that $R\left(c_{1},a,c_{2}\right)$
for all $a\in s_{1}$ and $\neg R\left(c_{1},b,c_{2}\right)$ for
all $b\in s_{2}$.
\end{proof}
\end{proof}
\bibliographystyle{alpha}
\bibliography{common2}
\end{document} | 0.050213 |
The Children and Adolescent Mental Health (CAHMS) team at Sidra Medicine, a member of Qatar Foundation, has advised parents to talk to their children about the novel coronavirus (Covid-19), and not to completely downplay the enormity of the situation.
“Children are looking to their parents to understand Covid-19 and therefore it is important that parents respond to questions to help their children remain calm and feel reassured,” the CAHMS team said in a statement.
“Frank discussions offer a number of learning opportunities, possibilities to increase resilience and help the children feel safe, which in turn will positively affect their physical and mental health”.
The CAHMS team recommended parents take the following approach, especially with younger children:
Reassurance:
• Be sure to offer reassurance about your ability as a parent to protect your child in the ways you can. Reassure your child’s safety, and discuss who the adults are in their life who will help keep them safe and how.
• It is okay to let your child know you do not have all the answers, information is evolving, and you will share with them what they need to know when you have important new information to share.
• Let children know it is okay to feel uncomfortable or anxious at this time and that they may have a mixture of different feelings (fear, worry, anger, even happiness for not having to go to school). Explain to them that this is natural, and that even adults feel this way (they will pick up on this anyway).
• Talk with them about how they can reduce thoughts and feelings associated with anxiety if they have them, by talking with you (their parent) about this
• Check-in often with your children.
Maintaining a Routine:
• rumours:
• Help to reduce panicking behaviours in children and prejudice/discrimination that have arisen during this time and reinforce care for others in appropriate ways.
• Use this as an opportunity to talk about how to help others and practice acts of kindness, especially if you know anyone who is self-quarantining, i.e. send them a meal through a meal delivery service, send them messages of encouragement to stay in touch.
Trusted source of information:
• It is strongly recommended to go to trusted sources of information for updates about Covid-19. Sidra Medicine advises checking Qatar’s Ministry of Public Health’s website page dedicated to Covid-19 or to call 16000.
There are no comments. | 0.996055 |
Paul B. Heintz
From Judgepedia
Paul B. Heintz is a judge for the Kirkland Town Court in Oneida County, New York. His current term expires on December 31, 2015.[1]
External links
- New York Town or Village Court search page
- New York State’s Town and Village Courts Introduction page
References
| 0.125693 |
TITLE: Exercise 14.7.4 in dummit and foote
QUESTION [2 upvotes]: Exercise 14.7.4 from Dummit and Foote
Let $K=\mathbb{Q}(\sqrt[n]{a})$, where $a\in \mathbb{Q}$, $a>0$ and suppose $[K:\mathbb{Q}]=n$(i.e., $x^n-a$ is irreducible). Let $E$ be any subfield of $K$ and let $[E:\mathbb{Q}]=d$. Prove that $E=\mathbb{Q}(\sqrt[d]{a})$. [Consider $ N_{K/E}(\sqrt[n]a)\in E$]
Here is a solution in MSE.
Subfield of $\mathbb{Q}(\sqrt[n]{a})$
I rewrite the solution in that answer here.
Let $\alpha=\sqrt [n]a\in \mathbb R_+$ be the real positive $n$-th root of $a$, so that $K=\mathbb Q(\alpha)$.
Consider some intermediate field $\mathbb Q\subset E\subset K$ (with $d:=[E:\mathbb Q]$) and define $\beta=N_{K/E}(\alpha)\in E$.
We know that $\beta=\Pi _\sigma \sigma (\alpha) $ where $\sigma$ runs through the $E$-algebra morphisms $K\to \mathbb C$.
Now, $\sigma (\alpha)=w_\sigma \cdot\alpha$ for some suitable complex roots $w_\sigma$ of $1$ so that $$\beta=\Pi _\sigma \sigma (\alpha)=(\Pi _\sigma w_\sigma) \cdot\alpha^e\quad (\operatorname { where } e=[K:E]=\frac nd)$$ Remembering that $\beta\in E\subset K\subset \mathbb R$ is real and that the only real roots of unity are $\pm 1$ we obtain $\Pi _\sigma w_\sigma=\frac {\beta}{\alpha^e}=\pm 1$ and $\beta=\pm \alpha^e=\pm \sqrt [d]a$.
Thus we have $\mathbb Q(\beta)=\mathbb Q(\sqrt [d]a)\subset E$ with $ \sqrt [d]a$ of degree $d$ over $\mathbb Q$.
Since $[E:\mathbb Q]=d$ too we obtain $E=\mathbb Q(\sqrt [d]a)$, just as claimed in the exercise.
Question:
In this line,
$$\beta=\Pi _\sigma \sigma (\alpha)=(\Pi _\sigma w_\sigma) \cdot\alpha^e\quad (\operatorname { where } e=[K:E]=\frac nd)$$
$K$ is not necessarily galois extension.$|Aut(K/E)|$ need not be equal to $[K:E]=\frac nd$.
It's there in all the solution I came across. I wonder why it has to be true. Please explain why $e$ in above equation has to be $[K:E]$
REPLY [1 votes]: A setting where a similar problem and answer occur with a more general base field is in my answer at Radical extension. The positivity of $a$ is replaced by the assumption that $n$th roots of unity in the top field are actually in the bottom field. That holds in the setting of your question because your top field has an embedding in the real numbers and therefore can't contain any roots of unity besides $\pm 1$, which are in $\mathbf Q$. | 0.011091 |
this is one of my favorite post Stacie L 6 years ago
Pat Robertson: It's time to legalize marijuana81...
by democrate1990 4 years ago
why the government doesn't want to legalize marijuana?Do you think marijuana should be legal?
by alexandriaruthk 5 years ago
Is it ok to legalize marijuana for recreational use?Yes, some other countries legalize marijuana for this purpose, what is your take on this? of course we know already that it is legalize for medicinal purposes. This is in some US states.
by Riece 7 years ago
I personally think it is ridiculous that it's not allowed. But according to polls the majority disagree with me! Surely there must be somebody on here that favors the outlawing of marijuana. | 0.018985 |
Eventful weekend;
Last weekend was nothing but busy because we flew our manufacturers in to settle our Cny 14' designs!
It was their first time to Sg and it wasn't easy for them since they had to apply for Visa and stuff.
Really appreciate their effort for coming here to meet us, cos to be honest I'm sure we are not their biggest customer.
So I was actually shocked when Lu told me that they have agreed to come down. Imagine leaving their factories without their care for 3 working days.
All thanks to Lu for arranging and making this plan work! I'm quite glad we don't have to fly over anytime soon cos they brought a really huge amount of samples for us to approve.
Guess the next time we visit them will prolly be when the factories reopen after the 1month Cny holiday? Yay!
Touched down in the late evening and immediately checked in to Grand Copthorne for some serious work. We discussed for around 6hrs straight till 2am? Phew..
Initially I was still kinda worried about the fabric chart that they brought for us won't be to our liking since we are always the one picking out the fabrics.
Guess I worried too much! LOVE the variety of prints and textures they got for us to choose from.
And not to mention about the quality of their work, it's really awesome!
Am really glad to be able to work with this flexible and trendy couple, unlike the previous few ultimate #fail manufacturers we met and worked with last year.
Wasted so much time, effort and money on them.
Now I just can't wait for the stocks to stream in in around 3 weeks time!
Here's what I wore to bring them around haha
Soiree Top is white which will be up on extras from BO real soon, with the upcoming Olsen Skorts!
Love this piece to bits, it's so comfy and the best part is it comes with front pockets :)
No Signboard for some seafood feast!
Where I had the best white pepper crab ever!
Day2, camwhored a little while waiting for them in the car
Casually dressed Sunday, wore Carte Embroidered top in blue with a cutesy playsuit I got months ago at a steal! Cannot see much of the scalloped hem here so here's the link to the full blog preview :)
And my new kicks I got for just $18.90, so pretty isnt it!
Tuesday with Sean, visited town cos he wanted to get new shoes but we went home empty handed!
Must order their Sweet Alaska waffles!
Wore Autre Cutout Dress, absolutely love the vibrant prints and cutting of this fit n' flare!
Wore Autre Cutout Dress, absolutely love the vibrant prints and cutting of this fit n' flare!
Close up of the prints and my accessories of the day
Not forgetting my almost 1month old nails I did at my fav nail sponsor, Sensual Nails Spa :)
My nails are so long now but the gelish are still in tiptop condition!
On the day when I just got it don, still look as good isnt it?
Remember to quote my name to enjoy 10% off regular priced services :)
Sensual Nails Spa
Block 507 Bishan Street 11 #01-398
Singapore 570507
Tel: 6259-0889 (appointment required)
And the day I wore the Hudson Tweed Skirt in the prettiest candy colour!
For those we aren't a fan of candy coloured stuff, this tweed skirt comes in monochrome as well.
Lastly, it's Leighton Cooperate Dress which I already received several orders from my OL friends.
Definitely one of the best picks from this collection for the chic working ladies.
Alrighty, thankyou for reading ;)
Hope you enjoyed my ootds, see you at the launch tonight at 830pm!
Till then..
Hi, do u design the clothes or your manufacture design?
Hi dear we design & manufacture :)
Hello Jolene! Your double eyelids cut one right? Any extra charges if you cut higher? I'm planning to go next year. Hope to hear from you first. How are you coping with these new eyelids? :)
By stitching! You can choose how high you want and still the same price! It's been 5 months since the surgery and I think it's the best decision I've ever made ;)
hi, where you get the kicks? very pretty :)
New look :D
Stitching den do you need to remove the stitching before leaving Korea? :)
Don't have to remove! The stitch is within the eyelids :)
Hi Jolene, when will the soirée top be back on bo? Does it come in other colours too?
Yes only white! | 0.001028 |
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Fri 28th February, 2014
HER Majesty The Queen and His Royal Highness The Duke of Edinburgh presented the University of Bedfordshire with the leading award in higher education at an honours ceremony at Buckingham Palace today (Thursday).
The prestigious Queen’s Anniversary Prize for Higher and Further Education was bestowed upon the University in recognition of its applied research into child sexual exploitation, influencing new safeguarding policy and practice.
The award was presented to the University’s Vice Chancellor, Bill Rammell, and Professor Jenny Pearce, who led the pioneering research as Director of the International Centre: Researching Child Sexual Exploitation, Violence and Trafficking at the University.
Bill Rammell said: “It truly was an honour today to be invited to Buckingham Palace and to receive such a prestigious award from Her Majesty The Queen and The Duke Of Edinburgh.
“To win the Queen’s Anniversary Prize is a landmark moment for the University of Bedfordshire. It brings national and international recognition to the University for its ground-breaking research in social work and social care. I am particularly pleased that, by addressing the societal challenge of child sexual exploitation, this University has made an impact on policy and practice that will change the lives of many young people.”
When the award was first announced in November 2013, the Royal Anniversary Trust said the University of Bedfordshire’s work was of strategic national importance, influencing UK policy and practice, adding that it has positioned the University as a leader of child-centred research in an area of growing social concern.
The Trust went on to describe the University’s research as being exceptional and distinctive, as it has helped alter perceptions of children so they are understood as victims of child abuse, and worked with welfare agencies to ensure that children are safeguarded rather than criminalised through the youth justice systems.
The Trust said this has brought about a “greatly improved practice in the protection of children and young people” and a fundamental change in approaches to child sexual exploitation”.
Professor Jenny Pearce (pictured) said: "It is an enormous honour to receive this award.
"It means a lot for us as researchers as recognition of our work but also to the young people we work with who, through such recognition, appreciate that their efforts to stop sexual violence are recognised."
Professor Pearce added: “Our work is distinctive and unique because of its collaborative approach, engaging with partners in government, funders, service providers and voluntary agencies, and because we prioritise the voice of the child through participatory methods of work.
“As a result, we are working with young people and other colleagues to advance theory, policy and practice. We are contributing to widespread changes in approaches to child sexual exploitation, positioning it firmly within strategies to safeguard children.”
Notes to Editor
To request live footage of the University receiving the award or for an interview with Professor Jenny Pearce, please contact the press office at the University of Bedfordshire: 01582 743046; 07724 241214; [email protected]
Twenty institutions were announced as winners of the Queen’s Anniversary Prizes for Higher and Further Education at a reception hosted by the Royal Anniversary Trust. The announcement was made by the Royal Anniversary Trust by kind permission of Her Majesty The Queen, at a reception at St James’s Palace on 21 November 2013.
The Queen’s Anniversary Prize recognises and celebrates outstanding work within UK higher and further education institutions and the impact that they have. They are the UK’s highest form of national recognition open to a UK higher or further education institutions. The winning entries represent individual departments or research groups, major international development projects, community schemes and vocational programmes and the application of cutting-edge research. Further information can be found at:
The University of Bedfordshire () is the largest higher education institution in the county, providing excellent opportunities for more people to access top quality higher education. We aim to create a vibrant multi-cultural learning community enabling people to transform their lives by participating in excellent, innovative education, scholarship and research.
The University of Bedfordshire has over 24,000 students, representing over 100 countries. Around 35 per cent of students are aged over 25 and about three-quarters of students receive financial support from the university. Between 2006 and 2013 over £180 million has been invested in new facilities at the university, which contributes approximately £300 million annually to the local economy.
For further information, or to receive University of Bedfordshire press releases on a regular basis, please contact: [email protected]; 01582 743046. Follow us on Twitter: @uobnews
Latest news» 2014 Archive» February» Queen’s Anniversary Prize brings University of Bedfordshire national and international recognition | 0.890812 |
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• Discount will vary depending on location, date, length of rental, car class & other factors; discount will not apply in all cases.
• The January Sale CDP 2209075 for destination U.S. & Latin America and CDP 907032 for destination Europe must be provided at time of reservation.
• For destination U.S., discount applies as follow: 15% on Standard Fleet (Economy, Compact, Intermediate, Standard & Full-size) and 10% on Specialty Fleet (Mid-size SUV, Standard SUV & Large SUV )
• For destination Latin America, discount applies as follow: 15% on all car classes (manual transmission vehicles only).
• For destination Europe, discount applies as follow: 15% on standard fleet vehicles (manual transmission vehicles only). Offer excludes vans and electric vehicles.
• Age, driver, credit and qualifying rate restrictions for the renting location apply.
• Advance reservation required.
• Offer not valid and rate change or additional fees may apply in the event of reservation modification, early/late return, or if not returning vehicle to renting location.
• Discount applies to Compact or larger cars (manual transmission vehicles only for rentals in destination Europe & Latin America). Offer excludes vans and electric vehicles.
• Offer has no cash value, and may not be used with other discounts, promotions or special rates.
• Drivers under 25 years old are subject to an age differential charge. Charge varies based on car class and renting location.
• Offer valid for reservations made between 10/JAN & 31/JAN, with vehicle pick-ups between today & 31/MAR/2022.
• Additional restrictions may apply.
• Blackout periods apply. | 0.009099 |
But added he was a "multi-faceted person" whose behaviour was "unacceptable".
The Prime Minister's official spokesman said women who came forward with complaints such as those made against the Jewish film maker should be praised for their courage. It's very hard to know which way to go when you see behaviours you don't like. "To me, the takeaway is: if we can continue to create space where people feel comfortable sharing and feel supported, there's power in numbers".
In fact, she is "surprised" the allegations didn't come out earlier, as people were aware of Weinstein's "inappropriate behavior".
'The situation within the office made it very hard for anyone to speak out, I think Harvey is a multi-faceted person. I think he is exceptionally bright and intelligent a lot of people were in awe of Harvey'.
Humberto Carolo with White Ribbon, an organization that works with men to end violence against women, says part of being a good ally is calling the behaviour out when it's happening.
Ferrari asked whether she believed Weinstein's numerous awards should be seized from him in punishment.
"Since then I felt terrible that I did the movie".
What is Harvey Weinstein's net worth?
'That's why this story, in my case, it's 20 years old, some of them are older, has never come out, ' she said.
Weinstein was at the Los Angeles home of his daughter, who was eventually able to get her father back inside the house.
Trump sitting down with Sean Hannity for exclusive interview during Pa. visit
A Fox News spokeswoman confirmed to The Hill that there will no longer be audience questions at the event. Trump and Hannity are set to discuss topics like tax cuts and immigration.
Weinstein, 65, was expected to be in Europe by this time, with the embattled producer saying on Tuesday that he was flying off to rehab to treat his sex addiction.
This comes as further damming accounts from female stars have surfaced tonight.
Model Cara Delevingne has alleged disgraced Hollywood producer Harvey Weinstein attempted to kiss her when the pair met to talk about an upcoming film.
In a lengthy statement posted on Instagram, the fearless British model-turned-actress said that after meeting Weinstein at a hotel lobby he asked her to come up to his room, where another woman was waiting for the pair.
Delevingne said she quickly declined and asked if her auto was outside. He stood in front of [the door] and tried to kiss me on the lips.
Delevingne added that she still got the role in the film, adding she "always thought that he gave it to me because of what happened".
Although filmed in the summer of 2014, it was released in the United States in September by The Weinstein Company.
Delevingne now joins a growing list of women who have shared their stories with The New York Times and New Yorker, including Angelina Jolie, who said she stopped working with Weinstein after he made a move on her in a hotel room during a press junket for her film 'Playing By Heart'.
The chief executive of Women In Film and TV, Kim Kinninmont, said the sexual harassment allegations are symptomatic of the film industry's culture of sexism.
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LAKE FOREST, Ill. (AP) The Chicago Bears had too many playoff misses and too many problems on offense. That's why general manager Phil Emery fired coach Lovie Smith. | 0.000855 |
Research article
28 Feb 2020
Research article | 28 Feb 2020
Flash Flood!: a SeriousGeoGames activity combining science festivals, video games, and virtual reality with research data for communicating flood risk and geomorphology
Chris Skinner
Related authors
Related subject area
Subject: Geoscience engagement | Keyword: Science engagement and dialogueThe future of conferences: lessons from Europe's largest online geoscience conferenceDemonstrating change from a drop-in space soundscape exhibit by using graffiti walls both before and afterThe human side of geoscientists: comparing geoscientists' and non-geoscientists' cognitive and affective responses to geology“This bookmark gauges the depths of the human”: how poetry can help to personalise climate change
Geosci. Commun., 4, 437–451,2021
Geosci. Commun., 4, 57–67,2021
Geosci. Commun., 3, 291–302,2020
Geosci. Commun., 3, 35–47,2020
Cited articles. | 0.684984 |
TITLE: how to enumerate $\mathbb{N} \times \mathbb{N}$
QUESTION [1 upvotes]: In the argument below in this book I don't know why did the book need to subtract k - 1. In particular, given (m,n) we have it lies in the diagonal k = m + n - 1 and it is the mth point of that "diagonal".
REPLY [0 votes]: The point $ (m,n) $ lies in the diagonal $ m+n-1$.
for example, the point $ (1,1)$ lies in the $ 1+1-1=1 ^{\text{st}} $ diagonal.
The point $ (2,3) $ is in the $ 4^{\text{th}} $ diagonal.
If the point $ (m,n) $ lies in the diagonal $ d $, its rank will be the number of points in the former diagonals $1,2,...,d-1$ plus $ m $, or
$$h(m,n)=1+2+...+(d-1) +m=$$
$$\frac{(d-1)(d-1+1)}{2}+m$$
Now, replace $ d $ by $ m+n-1$. | 0.199659 |
For.
The top five states that posted the most vehicle thefts with keys during this reporting period were
Looking at day-of-week data, Saturday saw the most thefts with keys (19,147) followed by Friday (18,719) and Monday (18,647).
The full report can be viewed and downloaded here. See a video here.
*659,717 is based on
YouTube
To view the original version on PR Newswire, visit:
SOURCE
GEICO accelerates hiring in Long Island
Sino Mercury Acquisition Corp. to Merge with Wins Finance Group Ltd.
Advisor News
Annuity News
Health/Employee Benefits
Life Insurance | 0.044122 |
Friday night started out with a dose of drama in Clarendon, as someone was seriously injured in the stairwell leading up to Whitlow's rooftop. I'm not sure on the details, but it appeared to be the result of a fight between two patrons. One lane of Wilson was full of police and emergency vehicle activity, and one person had to be taken to the hospital by ambulance.
The drama did not prevent the beautiful Bobbie Allen from taking Iota's stage around 9:45pm. She had many fans in the audience, including some who were enthusastice enough to bring signs cheering Bobbie on. She played an energetic set with her four piece band for most of the time, and also did a few solo acoustic numbers. I'm looking forward to finding her songs on iTunes later. | 0.019119 |
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Everything.
These cards are gorgeous! Unlike other settings that provide (though free) a mere handful of cards, this is an entire deck! All devoted to pulp adventure.
They are very high quality and I can't wait to get my pulp game going!
An excellent set of five linked adventures set in the city of Freeport and all on obviously nautical themes. The adventures will work just as well in any seaport - specific details relevant to Freeport are included, but are easy to modify to suit any other setting. The scenarios are set up to allow PCs a good deal of freedom in how to resolve them, and, while they are linked, they don't have to be run consecutively. (OTOH, they are all scaled for 6th to 8th level characters, so you won't want to leave too long between them).
The nautical theme works well, with shipwrecks, slave traders, giant squid, and even a tavern brawl to keep the players entertained and active. It can be run straight through as an action piece, but there's also plenty of opportunity for roleplaying for those who want it. All in all, a very good set of adventures.. | 0.001877 |
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Learn more about income tax deductions from ( A free handbook on tax)
Mr Gaurav’s income tax liability is calculated as per the income tax slab he falls under.
Mr Gaurav pays an income tax of INR 71070 on his salary after availing deductions under Section 80 C , Section 80 D and Section 80 DDB of the income tax act.
Income tax slabs for the financial year 2014-15 for an Indian citizen between 60-80 years of age:
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Conclusion. | 0.516543 |
How To Make A Bootable USB
How to make a bootable USB for Windows
Bootable USB drive for Window 7, 8, Vista, Server and XP
Requirements for Bootable USB flash:
► USB Flash Drive (Minimum 4GB)
► Window Vista, Windows 7 or Windows 8 installation files.
Follow the below steps to create bootable Windows 7/Windows 8 USB drive using which you can install Windows 7 or Windows 8 easily.
Procedure for Bootable USB Flash:
► Plug-in your USB flash drive to USB port and move all the contents from USB drive to a safe location on your system.
► Open Command Prompt with admin rights. Use any of the below methods to open Command Prompt with admin rights.
► Type cmd in Start menu search box and hit Ctrl+ Shift+ Enter.
Or
► Click on Start Menu Button >>> Click on All programs >>> Click Accessories, on Command Prompt right click and choose Run as Administrator from drop down.
► You need to know about the USB drive a little bit. Type in the following commands in the command prompt:
Command Prompt Commands for Bootable USB:
► on Command Prompt Type DISKPART and Press / Hit Enter
► After displaying Diskpart on next line now Type LIST DISK command
(Hint: From the given list note down Disk number for example Disk 1 or Disk 2 of your plugged-in USB flash drive.
Now Type following commands shown in BOLD one by one and check the message after each command just like in picture below:
SELECT DISK 1 (Used to select your Disk from list)
(Tip: Check the number of your USB Driver by size and Enter that No.)
CLEAN (This command deletes the partition of your flash drive)
CREATE PARTITION PRIMARY (Primary partition will be created on USB)
SELECT PARTITION 1 (Always use 1 after partition here)
ACTIVE (Makes the USB status as active)
FORMAT FS=NTFS QUICK ( You may use fs=fat and if you want to format normally then remove quick)
(Formatting Process will be completed in few seconds)
ASSIGN (This will assign a Drive Letter)
EXIT (Command Promt will return to normal mode)
► Now put fresh DVD of Window Vista/ Window 7 or Window 8 in the Optical Drive (DVD / Super Drive)
► Now Copy all the Files or contents from Windows Vista / Window 8 OR Window 7 DVD to your USB flash drive.
► Hurrah! Enjoy your USB drive is Boot-able Now and install Windows 7 OR
Windows 8.
► Finally from BIOS, Change your Boot Sequence and Set Boot from USB First or Boot Priority to Top.
Software to Make USB Bootable:
► FlashBoot Setup ► Novicorp WinToFlash ► Windows7-USB-DVD-tool
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| 0.018712 |
TITLE: Lorentz non-invariance of $3$-acceleration
QUESTION [0 upvotes]: $3$-acceleration can not be constant in a relativistic system. Because $\vec a^2$ is not Lorentz invariant. Does it mean that Lorentz invariance works only for $4$-vectors? How this should be understood properly? Does it mean to define the constant acceleration which Lorentz invariant the condition $\vec a^2$ is not enough?
REPLY [2 votes]: Four-vectors are not invariant. They are co-variant, i.e. they, by which I mean the components of the vector, change in a very specific fashion when one changes coordinates.
Scalars, on the other hand can be Lorentz-invariant. Though one has to be careful: for example, charge density is 1d, but it is not invariant (see length contraction).
The Lorentz-covariance is not that scary, you simply have to put it in terms that make sense for all observers equally well. For example I could define a 3-vector
$\mathbf{V}=V^x \mathbf{\hat{x}}+V^y \mathbf{\hat{y}}+V^z \mathbf{\hat{z}}$
in my reference frame, and then extend it to four-dimensional vector by specifying that, for me, this vector has no temporal component:
$V=V^x \partial_x + V^y \partial_y + V^z \partial_z$
Now this prescription is sufficient for any space-time observer.
Regarding the acceleration. The four-acceleration is related to the curvature of the world-line of the observer under acceleration. So if your world-line is $r^\mu=r^\mu\left(\tau\right)$ ($\tau$ is the proper time). Then
four-velocity: $\frac{dr^\mu}{d\tau}=u^\mu$
four-acceleration: $\frac{d u^\mu}{d\tau}=a^\mu$ (assuming flat space-time and trivial coordinates)
It is possible to show that in any reference frame where you are instantaneously at rest (you cannot have a full rest-frame if you are accelerating), $u^\mu=\left(c,\mathbf{0}\right)^\mu$, $a^\mu=\left(0,\mathbf{a}\right)$, where $\mathbf{a}$ is your usual acceleration.
Then $a^\mu a_\mu = -\mathbf{a}.\mathbf{a}=-a^2$
So the squared magnitude of your acceleration, i.e. the acceleration you would measure with accelerometer, is actually the magnitude of your four-acceleration, and is therefore invariant (as any other magnitude of a four-vector).
Misner-Thorne-Wheeler in II.6 have a good section about the acceleration in Special Relativity. Basically, intertial frame is associated with straight world-line, whereas the worldline of an accelerating observer is curved. However in regions of space small enough for cuvature of accelerating observer to be insignificant, one can talk about observer's rest frame, but with great care. This sound wooly, so let us do the full calculation.
So, you are accelerating observer. I am intertial. I see your world-line as $x^\mu=\left(ct,\,\mathbf{r}\right)$. Irrespective of whether you are accelerating or not, my time $t$ is related to your proper time $\tau$ through $\frac{dt}{d\tau}=\gamma=1/\sqrt{1-v^2/c^2}$, where $\mathbf{v}=d\mathbf{r}/dt$. Think of it as a tilt of your world-line relative to my worldline.
Your four-velcity is $u^\mu=\frac{dx^\mu}{d\tau}=\gamma\frac{dx^\mu}{dt}=\gamma\left(c,\,\mathbf{v}\right)^\mu$
As you move and accelerate your four-velocity will change, the tensorial quantity that quanitfies this change is the four-acceleration:
$a^\mu=\frac{Du^\mu}{D\tau}=u^\nu \nabla_\nu u^\mu = \frac{du^\mu}{d\tau}+\Gamma^\mu_{\nu\kappa}u^\nu u^\kappa=\gamma \frac{du^\mu}{dt}+\Gamma^\mu_{\nu\kappa}u^\nu u^\kappa$
The $\Gamma$-term is the Christoffel symbol that allows to work with any coordinate system. If my coordinates are plain Eucledian then $\Gamma=0$. Lets keep it this way. The derivative in the first term can be evaluated in a usual form. Using $\mathbf{a}=d\mathbf{v}/dt$, I find:
$a^\mu=\gamma^4 \cdot\frac{\mathbf{v}.\mathbf{a}}{c^2}\cdot\left(c,\, \mathbf{v}\right)^\mu + \gamma^2 \cdot \left(0,\,\mathbf{a}\right)^\mu$
The magnitude of this four-vector is:
$a^\mu a_\mu = -\gamma^4 (\mathbf{a}.\mathbf{a}) - \gamma^6 \left(\mathbf{v}.\mathbf{a}\right)^2/c^2$
Now, if you are instantaneusly at rest relative to me, then: $a^\mu=\left(0,\,\mathbf{a}\right)^\mu$ and $a^\mu a_\mu = - (\mathbf{a}.\mathbf{a})=-a^2$, so the magnitude of the acceleration I would observe, would be related directly to the magnitude of your four-acceleration, but only at that instance in time. Later a different observer, in intertial motion relative to me, would observe the same effect.etc. This is what I meant by instantaneous rest frame. | 0.263676 |
TITLE: Motivation for spectrum of an Abelian category
QUESTION [8 upvotes]: In his book Noncommutative Algebraic Geometry and Representations of Quantized Algebras, Rosenberg defines (III.1.2 on page 111) the spectrum of an Abelian category $\mathbf{A}$ in the following way. He first defines a preorder $\succ$ on $\operatorname{Ob}(\mathbf{A})$ by declaring that
$V\succ W$ iff $W$ is a subquotient of a coproduct of finitely many copies of $V$.
Let us write $V\sim W$ iff $V\succ W$ and $W\prec V$ (my notation, not his---he just says "equivalent with respect to $\succ$"). He then defines
$$\operatorname{Spec}(\mathbf{A}):=\left\{ V\in \operatorname{Obj}(\mathbf{A}):V\neq 0\text{ and }V\sim W\text{ for all nonzero subobjects }W\text{ of }V\text{.}\right\}$$
My question is simply "Why?": of all the possible definitions one might write down, why this one?
It is not at all clear to me what the intuition for this definition is supposed to be. What does this actually 'mean'?
REPLY [6 votes]: The basic motivation for this definition is that if $\mathbf{A}$ is the category of modules over a commutative ring $R$, then the $\sim$-equivalence classes of $\operatorname{Spec}(\mathbf{A})$ are naturally in bijection with the spectrum of the ring $R$. So this is a generalization of the spectrum of a commutative ring that can be defined for any abelian category.
Indeed, suppose $R$ is a commutative ring and $\mathbf{A}$ is the category of $R$-modules. Then for any $V\in\operatorname{Spec}(\mathbf{A})$, $V\sim W$ for every nonzero cyclic submodule $W$ of $V$ and $W\in\operatorname{Spec}(\mathbf{A})$ as well, so it suffices to consider cyclic modules. Note that for cyclic modules $R/I$ and $R/J$, we have $R/I\prec R/J$ iff $J\subseteq I$. So if $R/I$ is in $\operatorname{Spec}(\mathbf{A})$, then no nonzero element of $R/I$ can have an annihilator which is larger than $I$ (since then it would generate a cyclic submodule $R/J$ such that $R/J\not\sim R/I$). This is equivalent to $I$ being a prime ideal. So if $R/I$ is in $\operatorname{Spec}(\mathbf{A})$, then $I$ is a prime ideal. Conversely, if $I$ is prime, then every nonzero cyclic submodule of $R/I$ is isomorphic to $R/I$ and it follows that every nonzero submodule of $R/I$ has a submodule isomorphic to $R/I$, and so $R/I\in \operatorname{Spec}(\mathbf{A})$.
To sum up, then, the $\sim$-equivalence classes of $\operatorname{Spec}(\mathbf{A})$ are in bijection with the prime ideals of $R$, with each prime ideal $I$ corresponding to the module $R/I$ which is an element of $\operatorname{Spec}(\mathbf{A})$. | 0.030447 |
Previous Entry | Next Entry 'destroyed' me and 'killed' 'chronic pain'.
But most of all, what I remember is the pity. The pity that 'such.
This is the one that stuck out for me:
But most of all, what I remember is the pity. The pity that 'such a bright child had been injured so grievously'. I remember the pity that my Mother got from others because of my injury. | 0.05731 |
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Review opinions and decisions may be analyzed by the editorial office without identifying the reviewer. If certificate of review is required, please contact the editorial office. The reviewers may be listed in the year-end editorial for appreciation. | 0.90033 |
Affair Proof Your Marriage
The discovery that your partner is either emotionally or sexually involved with another is one of the most devastating experiences one can live through. Being in a relationship carries many different definitions and implications, yet the desire that your partner remain sexually faithful seems to be nearly universal.
There are numerous reasons why people cheat, and it’s good to know them so you’ll be aware to not let them seep into your own marriage.
READ: New Expert Survey Reveals the Number One Reason Couples Divorce
Most people report that they cheated because they felt emotionally disconnected and lonely in their marriage. Others complain that they felt unimportant or undervalued by their partner. And some say that it was because of a decrease in sex, sexual dissatisfaction, and/or boredom in the bedroom.
My analogy is that a strong marriage is like the foundation of a beautiful home — if you give it the maintenance it needs, it will stay robust and retain its value for years to come. Alternatively, if you let it hang out there through stormy weather, it will slowly crack and bring the walls down with it.
READ: How Do You Know He's Not The One?
So if you want to keep your marriage robust and affair proof — here are a few tips for relationship and monogamy success:
- Marry the right person for the right reasons at the right time in your life. You should feel completely confident during your engagement period that you are marrying a person whose character, nature and goals are fully in line with yours.
- Couples need to understand that marriage is a huge commitment and that staying happily married requires a lot of compromise and endeavor. Before you wed, speak to your partner about your feelings about fidelity. Surprisingly few newly engaged couples seem to have these frank discussions, some of which can serve to avoid painful misunderstandings later on.
- Have intimacy in your marriage. Long walks on the beach are much more than a cliché — they truly serve a purpose.
- Boost each other up. Couples can tend to be very critical toward each other, and too many years of criticism with the absence of praise can take a huge toll on a love bond. Don’t let this happen to your marriage. Be the president of your partner’s fan club, and be sure to regularly remind them how terrific they are.
- Have fun together. When the chips are down, you want to think of your partner as someone who brings joy into your life. Laughter is such a great elixir and it really helps keep couples together.
- It’s important to have sex — and even better to have great sex. Being sexual helps you feel more connected to your partner. It makes you feel cared for, attached, and comforted. It’s a nonverbal experience that helps you communicate in an entirely different way. And, it’s fun!
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TITLE: Corollary 4.7.16 in Artin's Algebra (2nd ed.)
QUESTION [2 upvotes]: I'm in the process of reading Artin's Algebra, and I seem to have reached a corollary with a problematic proof.
Corollary 4.7.16 Let $T$ be a linear operator on a finite-dimensional vector space over a field $F$ of characteristic zero. Assume that $T^r=I$ for some $r\geq 1$ and that the polynomial $t^r-1$ factors into linear factors in $F$. Then $T$ is diagonalizable.
The proof is not explicitly stated, but the intention is for the corollary to generalize a preceding theorem.
Theorem 4.7.14 Let $T$ be a linear operator on a finite-dimensional complex vector space $V$. If some positive power of $T$ is the identity, say $T^r=I$, then $T$ is diagonalizable.
The proof of this theorem relies in turn on a different preceding corollary.
Corollary 4.7.13 Let $T$ be a linear operator on a finite-dimensional complex vector space. The following conditions are equivalent: (a) $T$ is a diagonalizable operator, (b) every generalized eigenvector is an eigenvector, (c) all of the blocks in the Jordan form for $T$ are $1\times 1$ blocks.
In particular, 4.7.14 relies on the implication (b)$\implies$ (a), as it proves $T$ is diagonalizable by showing all generalized eigenvectors are eigenvectors. However, the proof of (b)$\implies$ (a) follows from (b)$\implies$ (c) $\implies$ (a) which assumes the existence of Jordan form.
So, inherent in the proof of Corollary 4.7.16 is the existence of Jordan form. As I see it, the issue is that the proof of Jordan form in Artin's text requires the characteristic polynomial of $T$ to factor into linear factors in $F$, and this isn't a hypothesis in Corollary 4.7.16.
On the one hand, I'm wondering if I'm grossly overlooking something and my concern is unwarranted? And if not, I'd like to know if any slight adjustment to Artin's proofs/statements irons out the problematic proof?
I'd like to emphasize that I am not concerned with a different proof of the corollary. I know a minimal polynomial argument works. I'd just like to figure out what Artin had in mind, or get confirmation that this is a legitimate issue which can't easily be resolved.
REPLY [1 votes]: Warm-up
I guess the ending of chapter 4 in Artin's book is basically botched. The preceding theorem 4.7.10 is hard to follow because he is not explaining the big big picture of what he tries to do, namely his proof strategy.
Theorem 4.7.14
Same goes for 4.7.14 because he takes generalized eigenvector of degree 2 and states without proof that for this particular eigenvector we have
$$Tw=\lambda w$$
Logic is $$T\underbrace{(T- \lambda)v}_{w} = (T- \lambda + \lambda)(T-
\lambda)v=\underbrace{(T-\lambda)^2}_{\text{because v is degree 2 by assumption, this is zero}}+\lambda\underbrace{(T-\lambda)v}_{w}$$
So $$Tw=\lambda w$$
He does not put any hint as to how to prove for degrees higher than two, no mention of minimum polynomials, factoring etc. Better strategy is to pick 4.7.13(c) and prove 4.7.14 using this https://math.stackexchange.com/q/734075
Corollary 4.7.15
If we can factor the characteristic polynomial in a field, then we can pick roots of the characteristic polynomial from this same field, these roots are the eigenvalues which are not necessarily distinct. So we can go along with Jordan Decomposition of 4.7.10.
Corollary 4.7.16
Suppose we have filled the gap and proved that for any degree of the generalized eigenvector we have $T w= \lambda w$. I repeat that Artin assumed it was degree two and went along.
Then we get the key identity
$$0 = (\underbrace{ T^{r-1}+ \lambda T^{r-2} + \dots + \lambda^{r-2}T + \lambda^{r-1}}_{\text{total r terms (*)}}) \underbrace{(T- \alpha )v}_{w} $$
Using the assumed identity $T w= \lambda w$ we can simplify (*) because T is linear
$$\begin{align*}
T^{r-1}w= \lambda^{r-1}w \\
\lambda T^{r-2}w= \lambda \lambda^{r-1}w = \lambda^{r-1}w \\
\dots \\
\lambda^{r-2} T w= \lambda^{r-2} \lambda w = \lambda^{r-1}w \\
\lambda^{r-1} w = \lambda^{r-1} w
\end{align*}$$
So we can rewrite the above identity as
$$ 0 = \overbrace{\underbrace{r}_{\text{Note 1}} \underbrace{\lambda^{r-1}}_{\text{Note 2}}}^{\text{Note 3}} w$$
Note 1: r is the number of terms, some positive number.
Note 2: $\lambda^{r-1}$ is non-zero in our field. Suppose it is zero, then
$$\lambda \lambda^{r-1} = \lambda^{r} = 0$$
But previously we showed in our case $\lambda^r = 1$, a contradiction.
Note 3: we have to assume that in our field the field characteristic is zero. That means that no matter how many times we add up the identity element, we never get the zero element, Artin page 83
$$1 + 1 + \dots \ne 0$$
This ensures that $r \lambda^{r-1}$, adding r times element $\lambda^{r-1}$, never gives us zero.
Now we can state that the only factor that can be zero is $w = (T - \lambda) v$, which is thus an ordinary eigenvector. | 0.004492 |
Stamped concrete patio with fire pit in Charlotte, NC.
of Charlotte
This stamped concrete patio with fire pit in Charlotte, NC, makes the perfect spot to enjoy the outdoors in varied seasons! The custom design of this patio makes it a unique addition to this home. By placing the fire pit on the exterior perimeter of the stamped concrete patio we created distinct entertaining areas that can be used together or separately depending upon the occasion. | 0.054305 |
Legends of Law Enforcement: Sheriff Clarence L. Dupnik
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In this episode of Legends of Law Enforcement, we celebrate the legendary and storied career of Pima County Sheriff Clarence L. Dupnik.
Dupnik’s career traces back decades and involves solving some of history’s most notorious crimes:
Dupnik was the first to identify the infamous serial killer ‘Jack the Ripper’ as an ancestor of William F. Buckley.
“America’s Sheriff” states that Jimmy Hoffa, possibly buried at the Meadowlands, was “whacked” by Barry Gold
Dupnik also believes that Charles Manson was wrongfully imprisoned for the “Redrum” murders of Sharon Tate and others. The sheriff asserts that Tate’s Hollywood connections to Ronald Reagan were never sufficiently explored.
Dupnik agrees with the court that freed O.J. Simpson and ‘knows in his heart’ that Rush Limbaugh is somehow to blame.
Finally, the sheriff has assembled copious amounts of evidence that the Columbine killers were directly inspired by pundits Sean Hannity or Mark Levin.
That’s all the time we have tonight. Join us next week when we celebrate the law enforcement career of Deputy Barney Fife.
Cross-posted at: Doug Ross @ Journal.
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For children who havent been traumatized enough by Hurricane Katrina, they now can explore the disaster in a side-scrolling virtual adventure:
Brought to you by the genius of high school students. From Brooklyn.
Developed by the nonprofit organization Global Kids and game designers Game Pill, Tempest in Crescent City follows a young girl through various neighborhoods as she rescues residents trapped in their attics, delivers first aid and batteries to her neighbors, dodges debris and swims through suspiciously clear floodwaters, all on her to way to save her mother.
Global Green says Tempest is meant to be a fun adventure game that also addresses meaningful, accurate and difficult historical situations.
This, coming from an organization that published Ayiti: The Cost of Life.
Its one thing to set aside reality to indulge in a little harmless, carefree video-game violence but how do you bring reality back into the video game world when addressing recent matters of life and death?
Video game publishers have had success with games based on World War II, Desert Storm and even recent anti-terrorism campaigns, but so have Stephen Spielberg, Clint Eastwood, Tom Hanks and the like with countless award-winning historical war films. But the action stays in the screen. Youre not expected to take up arms after piloting a Higgins boat to Omaha.
But will kids walk away from this game having learned anything? Is it really as educational as Global Green suggests?
After all, children, especially those living in hurricane-prone regions, can learn basic hurricane preparation, like storing water and batteries, and a basic Katrina timeline with relevant factoids. But does it stick, or is it just another classroom activity?
Sure, I played Oregon Trail to death, but all I learned from playing it were the definitions of dysentery and calk. We covered its historical significance in social studies classes. But Ive also never had to eat or travel Donner-style. Perhaps Tempest might have some current social relevance after all.
In terms of actual game play: Its short, simple and the controls arrow keys and space bar take only a few minutes to get comfortable. At least the opening sequence saves the guilt trip.
Thanks to Richard Read, who gave the game somewhere around a B+.
Showing 1-3 of 3 | 0.460054 |
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Global Health Studies Info Session Oct. 10
Interested in global health? Come learn about the Global Health Studies Multidisciplinary Academic Program and how to become a Global Health Scholar. The program offers Yale College students exposure to cross-disciplinary perspectives, issues, and skills relevant to global health research and practice.
The information session is open to all interested Yale College students. Students typically apply to be a Global Health Scholar in the fall of their sophomore year. On occasion, juniors with compelling interest in the GHS MAP may be considered for admittance into the program.
Global Health Scholar applications are due November 2. Prior global health experience or coursework not required to apply.
RSVP requested:
More information can be found at or by contacting [email protected]. | 0.283908 |