alasdairforsythe/text-english-code-fiction-nonfiction

text stringlengths 0 634k
X^{n_2} & P(g) (1-q^{1-{n_1}}g) \\
X^{n_2-1} & P(g) (1-q^{1-{n_1}}g) (1-q^{1-n_2})\\
\vdots & \\
X^{n_3} & P(g) (1-q^{1-{n_1}}g) (1-q^{1-{n_2}}g) \\
X^{n_3-1} & P(g) (1-q^{1-{n_1}}g) (1-q^{1-{n_2}}g) (1-q^{1-n_3})\\
\vdots & \\
& P(g) (1-q^{1-{n_1}}g) (1-q^{1-n_2}g) (1-q^{1-n_3}g) \ldots (1-q^{1-n_m}g)
\end{array}
\]
We see that the integers $n_1,\ldots,n_m$ uniquely determine the shape
of this picture. The polynomial $P(g)$ on the other hand can be
shifted (by $g$ and $g^{-1}$) or renormalised. To determine $M$
uniquely we shift and normalise $P$ in such a way that it contains no
negative powers
and has unit constant coefficient. $P$ can then be viewed as a
polynomial $\in\k(q)[g]$.

We see that the codimension of $M$ is the sum of the lengths of the
polynomials in $g$ over all degrees in $X$ in the above
picture. Finite codimension corresponds to $P=1$. In this
case the codimension is the sum
$n_1+\ldots +n_m$.

(b) We observe that polynomials of the form $1-q^{j}g$
have no common divisors for distinct $j$. Therefore,
finite codimensional crossed
submodules are maximal if and only if
there is just one integer ($m=1$). Thus, the maximal left
crossed submodule of
codimension $k$ is generated by $X^k$ and $1-q^{1-k}g$.
For an infinite codimensional crossed submodule we certainly need
$m=0$. Then, the maximality corresponds to irreducibility of
$P$.

(c) This is again due to the distinctness of factors $1-q^j g$.
\end{proof}


\begin{cor}
\label{cor:cqbp_eclass}
(a) Left crossed \ensuremath{C_q(B_+)}-submodules $M\subseteq\ker\cou\subset\ensuremath{C_q(B_+)}$
are in one-to-one correspondence to pairs
$(P,I)$ as in lemma \ref{lem:cqbp_class}
with the additional constraint $(1-g)$ divides $P(g)$ or $1\in I$.
$\codim M<\infty$ iff $P=1$. In particular $\codim M=(\sum_{n\in I}n)-1$
if $P=1$.

(b) The finite codimensional maximal $M$
correspond to the pairs
$(1,\{1,n\})$ with $n\ge 2$ the
codimension. The infinite codimensional maximal $M$ correspond to pairs
$(P,\{1\})$ with $P$ irreducible and $P(g)\neq 1-q^{-k}g$ for any
$k\in\mathbb{N}_0$.

(c) Crossed submodules $M$ of finite
codimension are intersections of maximal ones.
In particular $M=\bigcap_{n\in I} M^n$, with $M^n$ corresponding to
$(1,\{1,n\})$.
\end{cor}
\begin{proof}
First observe that $\sum_n X^n P_n(g)\in \ker\cou$ if and only if
$(1-g)$ divides $P_0(g)$. This is to say that that $\ker\cou$
is the crossed submodule corresponding to the pair $(1,\{1\})$ in
lemma \ref{lem:cqbp_class}. We obtain the classification
from the one of lemmas \ref{lem:cqbp_class} by intersecting
everything with this crossed submodule. In particular, this reduces
the codimension by one in the finite codimensional case.
\end{proof}



\begin{lem}
\label{lem:uqbp_class}
(a) Left crossed \ensuremath{U_q(\lalg{b_+})}-submodules $L\subseteq\ensuremath{U_q(\lalg{b_+})}$ via the left adjoint
action and left
regular coaction are in one-to-one correspondence to the set
$3^{\mathbb{N}_0}\times2^{\mathbb{N}}$.
Finite dimensional $L$ are in one-to-one correspondence to
finite sets $I\subset\mathbb{N}$ and $\dim L=\sum_{n\in I}n$.

(b) Finite dimensional irreducible $L$ correspond to $\{n\}$
with $n$ the dimension.

(c) Finite dimensional $L$ are direct sums of irreducible ones. In
particular $L=\oplus_{n\in I} L^n$ with $L^n$ corresponding to $\{n\}$.
\end{lem}
\begin{proof}
(a) The action takes the explicit form
\[g\triangleright X^n g^k = q^{-n} X^n g^k\qquad
X\triangleright X^n g^k = X^{n+1}g^k(1-q^{-(n+k)})\]
while the coproduct is
\[\cop(X^n g^k)=\sum_{r=0}^{n} \binomq{n}{r}
q^{-r(n-r)} X^{n-r} g^{k+r}\otimes X^r g^k\]
which we view as a left coaction here.
Let now $L\subseteq\ensuremath{U_q(\lalg{b_+})}$ be a crossed \ensuremath{U_q(\lalg{b_+})}-submodule via this action
and coaction. For $\sum_n X^n P_n(g)\in L$ invariance under
the action by
$g$ clearly means that \mbox{$X^n P_n(g)\in L\ \forall n$}. Then from
invariance under the coaction we can conclude that
if $X^n \sum_j a_j g^j\in L$ we must have

X^{n_2} & P(g) (1-q^{1-{n_1}}g) \\

X^{n_2-1} & P(g) (1-q^{1-{n_1}}g) (1-q^{1-n_2})\\

\vdots & \\

X^{n_3} & P(g) (1-q^{1-{n_1}}g) (1-q^{1-{n_2}}g) \\

X^{n_3-1} & P(g) (1-q^{1-{n_1}}g) (1-q^{1-{n_2}}g) (1-q^{1-n_3})\\

\vdots & \\

& P(g) (1-q^{1-{n_1}}g) (1-q^{1-n_2}g) (1-q^{1-n_3}g) \ldots (1-q^{1-n_m}g)

\end{array}

\]

We see that the integers $n_1,\ldots,n_m$ uniquely determine the shape

of this picture. The polynomial $P(g)$ on the other hand can be

shifted (by $g$ and $g^{-1}$) or renormalised. To determine $M$

uniquely we shift and normalise $P$ in such a way that it contains no

negative powers

and has unit constant coefficient. $P$ can then be viewed as a

polynomial $\in\k(q)[g]$.

We see that the codimension of $M$ is the sum of the lengths of the

polynomials in $g$ over all degrees in $X$ in the above

picture. Finite codimension corresponds to $P=1$. In this

case the codimension is the sum

$n_1+\ldots +n_m$.

(b) We observe that polynomials of the form $1-q^{j}g$

have no common divisors for distinct $j$. Therefore,

finite codimensional crossed

submodules are maximal if and only if

there is just one integer ($m=1$). Thus, the maximal left

crossed submodule of

codimension $k$ is generated by $X^k$ and $1-q^{1-k}g$.

For an infinite codimensional crossed submodule we certainly need

$m=0$. Then, the maximality corresponds to irreducibility of

$P$.

(c) This is again due to the distinctness of factors $1-q^j g$.

\end{proof}

\begin{cor}

\label{cor:cqbp_eclass}

(a) Left crossed \ensuremath{C_q(B_+)}-submodules $M\subseteq\ker\cou\subset\ensuremath{C_q(B_+)}$

are in one-to-one correspondence to pairs

$(P,I)$ as in lemma \ref{lem:cqbp_class}

with the additional constraint $(1-g)$ divides $P(g)$ or $1\in I$.

$\codim M<\infty$ iff $P=1$. In particular $\codim M=(\sum_{n\in I}n)-1$

if $P=1$.

(b) The finite codimensional maximal $M$

correspond to the pairs

$(1,\{1,n\})$ with $n\ge 2$ the

codimension. The infinite codimensional maximal $M$ correspond to pairs

$(P,\{1\})$ with $P$ irreducible and $P(g)\neq 1-q^{-k}g$ for any

$k\in\mathbb{N}_0$.

(c) Crossed submodules $M$ of finite

codimension are intersections of maximal ones.

In particular $M=\bigcap_{n\in I} M^n$, with $M^n$ corresponding to

$(1,\{1,n\})$.

\end{cor}

\begin{proof}

First observe that $\sum_n X^n P_n(g)\in \ker\cou$ if and only if

$(1-g)$ divides $P_0(g)$. This is to say that that $\ker\cou$

is the crossed submodule corresponding to the pair $(1,\{1\})$ in

lemma \ref{lem:cqbp_class}. We obtain the classification

from the one of lemmas \ref{lem:cqbp_class} by intersecting

everything with this crossed submodule. In particular, this reduces

the codimension by one in the finite codimensional case.

\end{proof}

\begin{lem}

\label{lem:uqbp_class}

(a) Left crossed \ensuremath{U_q(\lalg{b_+})}-submodules $L\subseteq\ensuremath{U_q(\lalg{b_+})}$ via the left adjoint

action and left

regular coaction are in one-to-one correspondence to the set

$3^{\mathbb{N}_0}\times2^{\mathbb{N}}$.

Finite dimensional $L$ are in one-to-one correspondence to

finite sets $I\subset\mathbb{N}$ and $\dim L=\sum_{n\in I}n$.

(b) Finite dimensional irreducible $L$ correspond to $\{n\}$

with $n$ the dimension.

(c) Finite dimensional $L$ are direct sums of irreducible ones. In

particular $L=\oplus_{n\in I} L^n$ with $L^n$ corresponding to $\{n\}$.

\end{lem}

\begin{proof}

(a) The action takes the explicit form

\[g\triangleright X^n g^k = q^{-n} X^n g^k\qquad

X\triangleright X^n g^k = X^{n+1}g^k(1-q^{-(n+k)})\]

while the coproduct is

\[\cop(X^n g^k)=\sum_{r=0}^{n} \binomq{n}{r}

q^{-r(n-r)} X^{n-r} g^{k+r}\otimes X^r g^k\]

which we view as a left coaction here.

Let now $L\subseteq\ensuremath{U_q(\lalg{b_+})}$ be a crossed \ensuremath{U_q(\lalg{b_+})}-submodule via this action

and coaction. For $\sum_n X^n P_n(g)\in L$ invariance under

the action by

$g$ clearly means that \mbox{$X^n P_n(g)\in L\ \forall n$}. Then from

invariance under the coaction we can conclude that

if $X^n \sum_j a_j g^j\in L$ we must have