text:
\section{Introduction}

The purpose of this Letter is to give new descriptions  of the Yang-Mills theory \cite{YM54} on a lattice, which are expected to give an efficient framework to explain  quark confinement based on the dual superconductivity picture.
The dual superconductivity \cite{dualsuper} is conjectured to occur due to the condensation of magnetic monopoles, just as the ordinary superconductivity is caused by the condensation of the Cooper pairs.
In the dual superconductor, the dual Meissner effect forces the color electric flux between a quark and an antiquark to be squeezed into a tube like region forming the hadronic string.

In view of these, we wish to construct new descriptions which enable us to extract the dominant degrees of freedom that are relevant to quark confinement according to the Wilson criterion in such a way that they reproduce almost all the string tension of the linear inter-quark potential.
In a previous paper \cite{KSM08}, we have considered the decomposition in which the original $SU(N)$ Yang-Mills field $\mathscr{A}_\mu(x)$ is decomposed into two parts, $\mathscr{V}_\mu(x)$ and $\mathscr{X}_\mu(x)$, i.e.,
\begin{equation}
 \mathscr{A}_\mu(x)=\mathscr{V}_\mu(x)+\mathscr{X}_\mu(x)
  ,
  \label{decomp1}
\end{equation}
such that
$\mathscr{V}_\mu$ transforms under the gauge transformation just like the original gauge field $\mathscr{A}_\mu$, while $\mathscr{X}_\mu$ transforms like an adjoint matter field:
\begin{subequations}
\begin{align}
  \mathscr{V}_\mu(x) & \rightarrow \mathscr{V}_\mu^\prime(x) = \Omega(x) (\mathscr{V}_\mu(x) + ig^{-1} \partial_\mu) \Omega^{-1}(x)
 ,
 \label{V-ctransf}
  \\
  \mathscr{X}_\mu(x) & \rightarrow \mathscr{X}_\mu^\prime(x) = \Omega(x)  \mathscr{X}_\mu(x) \Omega^{-1}(x)
 \label{X-ctransf}
 .
\end{align}
In the decomposition (\ref{decomp1}), a crucial role is played by a unit vector field $\bm{n}(x)$ called the \textit{color field}.
The color field is defined by the following property.
It must be a functional or composite operator of the original Yang-Mills field  $\mathscr{A}_\mu(x)$ such that it transforms under the gauge transformation according to the adjoint representation:
\begin{align}
  \bm{n}(x) & \rightarrow \bm{n}^\prime(x) = \Omega(x)  \bm{n}(x) \Omega^{-1}(x)
   .
 \label{n-ctransf}
\end{align}
\end{subequations}
The prescription for obtaining such a color field is to be specified separately.
The usefulness of the color field is as follows.
It enables one to give a gauge-invariant definitions of magnetic monopole with the integer-valued magnetic charges subject to a quantization condition analogous to the Dirac type.
For $SU(2)$, indeed, once such a \textit{single} color field $\bm{n}(x)$ of unit length is introduced, we can define
the \textit{gauge-invariant} two-form by
\begin{align}
 f_{\alpha\beta}(x)
  :=  2{\rm tr}\{ \partial_\alpha [\bm{n}(x) \mathscr{A}_\beta(x)]  -
  \partial_\beta  [\bm{n}(x)  \mathscr{A}_\alpha(x) ]
+  ig^{-1}  \bm{n}(x) [  \partial_\alpha \bm{n}(x) ,  \partial_\beta \bm{n}(x)]  \}
   ,
\end{align}
and the gauge-invariant ``magnetic-monopole current''   $k$  by
\begin{align}
 k_\mu = \frac12 \epsilon_{\mu\nu\alpha\beta} \partial_\nu f_{\alpha\beta}
 ,
\end{align}
which is conserved in the sense that
$\partial_\mu k_\mu=0$.
Consequently, the magnetic charge $q_m$ is gauge-invariant and obeys the quantization condition:
\begin{align}
 q_m :=\int d^3 \tilde{\sigma}_\mu k_\mu
 = 4\pi g^{-1} n, \quad n \in \mathbb{Z} := \{ 0, \pm 1, \pm 2, \cdots  \}
  ,
\end{align}
where $\bar{x}^\mu$ denotes a parameterization of the 3-dimensional volume $V$ and
 $d^3 \tilde{\sigma}_{\mu} $ is the dual of the 3-dimensional volume element $d^3 \sigma^{\gamma_1\gamma_2\gamma_3}$.

 For $SU(2)$,  $\mathscr{V}_\mu$ and $\mathscr{X}_\mu$ are specified by two defining equations by way of the color field:
\begin{subequations}
\begin{align}
 0 =&  \mathscr{D}_\mu[\mathscr{V}] \bm{n}(x) :=  \partial_\mu \bm{n}(x)-ig [ \mathscr{V}_\mu(x), \bm{n}(x) ]   ,
\\
  0 =& {\rm tr}[\mathscr{X}_\mu(x)  \bm{n}(x) ]
 .
\end{align}
\end{subequations}
In fact, by solving these defining equations, $\mathscr{V}_\mu(x)$ and $\mathscr{X}_\mu(x)$ in $SU(2)$ are determined uniquely as functions of $\bm{n}(x)$ and $\mathscr{A}_\mu(x)$:
\begin{subequations}
\begin{align}
  \mathscr{V}_\mu(x) =&   2{\rm tr}[\mathscr{A}_\mu(x) {\bm n}(x)] {\bm n}(x)   +i g^{-1}  [ {\bm n}(x) , \partial_\mu {\bm n}(x) ]
 ,
\\
  \mathscr{X}_\mu(x) =&   -ig^{-1}  [\bm{n}(x) , \mathscr{D}_\mu[\mathscr{A}]\bm{n}(x) ]
 .
\end{align}
 \label{V-X}
\end{subequations}

 From these point of view, we have already given a new framework for $SU(2)$ Yang-Mills theory \textit{on a lattice} and subsequently presented numerical evidences supporting the validity of the new lattice framework by performing numerical simulations \cite{KKMSSI05,IKKMSS06,SKKMSI07}.
The relevant lattice description could be regarded as a lattice version of the continuum formulation \cite{KMS06} which was known as the Cho-Faddeev-Niemi-Shabanov (CFNS) decomposition \cite{Cho80,FN98,Shabanov99}.
In previous papers \cite{KKMSSI05,IKKMSS06,SKKMSI07}, we have given a lattice version of the decomposition (\ref{decomp1}) and (\ref{V-X}) for $SU(2)$ Yang-Mills theory.
We have shown that the gauge-invariant magnetic monopole can be constructed in terms of $\mathscr{V}_\mu$ alone in such a way that the magnetic charge is integer-valued and subject to the Dirac quantization condition and that the magnetic contribution to the string tension defined from the variable $\mathscr{V}_\mu$ reproduces almost all the string tension (90\% of the full string tension) through a non-Abelian Stokes theorem for the  Wilson loop operator \cite{DP89,KondoIV}, yielding the magnetic monopole dominance \cite{IKKMSS06}.  Moreover, we have shown \cite{SKKMSI07} that   $\mathscr{V}_\mu(x)$ part is dominant in the infrared region, while  the remaining degrees of freedom represented by $\mathscr{X}_\mu(x)$ are suppressed by acquiring the large mass $M_X \cong 1.2 \sim 1.3$GeV, yielding the infrared ``Abelian" dominance \cite{tHooft81,EI82}.
These are the gauge-invariant confirmations for remarkable results \cite{SY90,SNW94,AS99} which were first observed in the maximal Abelian gauge (MAG) \cite{KLSW87}, although the MAG breaks the original gauge group $SU(2)$ into U(1) explicitly.

Now we turn our attention to the $SU(N)$ case.
The issue of generating magnetic monopoles in Yang-Mills theory has been investigated so far   under the MAG which breaks the original gauge group $SU(N)$ into the maximal torus subgroup $H=U(1)^{N-1}$. Therefore, MAG yields $(N-1)$ types of magnetic monopole, in agreement with the observation due to the Homotopy group:
$\pi_2(SU(N)/U(1)^{N-1})=\pi_1(U(1)^{N-1})=\mathbb{Z}+\cdots+\mathbb{Z}$.
In the continuum formulation,  a possible extension of the CFNS decomposition from $SU(2)$ to $SU(N)$ group was worked out already in \cite{Cho80c,FN99a} where $N-1$ color fields $\bm{n}^j(x)$ ($j=1, \cdots, N-1)$ were introduced where $N-1$ is the rank of $SU(N)$, i.e., the dimension of the maximal torus subgroup $H=U(1)^{N-1}$.
For $SU(N)$ Yang-Mills theory, therefore, it tends to assume that magnetic monopoles of $(N-1)$ types are necessary to cause the dual Meissner effect for realizing quark confinement.
To the best of our knowledge, however, it is not yet confirmed  whether or not  $(N-1)$ type of magnetic monopole are necessary to achieve confinement in $SU(N)$ Yang-Mills theory.
Rather, we have a conjecture that \textit{a single type of magnetic monopole is sufficient to achieve quark confinement even in $SU(N)$ Yang-Mills theory}, once it is defined in a gauge-invariant way.  In fact, this scenario was originally proposed in \cite{Kondo99Lattice99,KT99} based on a non-Abelian Stokes theorem for Wilson loop operator \cite{KT99,DP01}.   However, at that time, this idea was not substantiated because there did not exist the relevant lattice formulation which enables one to prove or disprove this conjecture.

From this perspective, we consider new lattice descriptions of $SU(N)$ Yang-Mills theory in this Letter.  We must emphasize that possible ways of extending the $SU(2)$ machinery to $SU(N)$ case is not unique and there are several options.
For the SU(3) gauge group, indeed, we have two options, maximal or minimal, although  there occur more involved  intermediate cases other than the maximal and minimal cases  for $N \ge 4$ in $SU(N)$ Yang-Mills theory \cite{KSM08}.
In the maximal case, we can introduce $(N-1)$ unit vector fields $\bm{n}^j(x)$ which enable us to define $(N-1)$ types of magnetic monopoles.  This option corresponds to a straightforward extension of the $SU(2)$ lattice formulation given in previous papers \cite{KKMSSI05,IKKMSS06,SKKMSI07}.  Then the naive continuum limit of the maximal case on a lattice reduces to the $SU(N)$ CFNS decomposition in the continuum formulation \cite{Cho80c,FN99a}.
Our gauge-invariant formulation in this option can reproduce the conventional results based on MAG by choosing the specific gauge, in which all the color fields are fixed to be the constant Cartan subalgebra on the whole lattice points, $\bm{n}^j(x) \rightarrow H^j$.
In the minimal case, on the other hand, we introduce a single color field  yielding a single magnetic monopole even for $SU(N)$ Yang-Mills theory, as suggested from
$\pi_2(SU(N)/U(N-1))=\pi_1(U(N-1))=\mathbb{Z}$.
Such a possibility seems to be overlooked in the previous works, although the continuum version of the minimal case was proposed and examined recently by three of us in a separate paper \cite{KSM08}.
The lattice version given in this Letter is constructed so  as to agree with the continuum formulation \cite{KSM08} in the naive continuum limit.
For this machinery to work, we give a prescription for constructing such a color field from the original Yang-Mills theory along the line shown in the continuum version.

\section{Maximal case}

For the Yang-Mills gauge theory with a gauge group $G=SU(N)$, it is convenient
\footnote{
But it is not essential to introduce $r$ fields $ {\bf n}^j(x)$, since it is enough to introduce a single color field $\bm{n}(x)$, see \cite{KSM08}.
}
 to introduce a set of  $(N^2-1)$-dimensional unit vector fields $ {\bf n}^j(x)$
($j=1, \cdots, r$) with the components $n_j^A(x)$, i.e.,
$
 {\bf n}^j(x) \cdot  {\bf n}^j(x) := n_j^A(x) n_j^A(x) = 1
$
($ A =1, 2, \dots, {\rm dim}G=N^2-1 $)
where $r:={\rm rank}G=N-1$ is the rank  of the gauge group $G=SU(N)$.
We omit the summation symbol for $A$ in what follows.
The   $\bm n^j(x)$ fields having the value in the Lie algebra $\mathscr{G}$ are constructed according to
\begin{equation}
 \bm n^j(x) = n_j^A(x) T^A  = U^\dagger(x) H_j U(x)
 , \quad U(x) \in G
 ,
\end{equation}
where $H_j$ are generators in the Cartan subalgebra in the generators $T^A$ of the Lie algebra $\mathscr{G}=su(N)$ of  $G=SU(N)$ and $U(x)$ is a group element of $G=SU(N)$.
We adopt  the normalization ${\rm tr}(T_A T_B)= \frac12 \delta_{AB}$.

\subsection{Continuum: maximal case}

In the maximal case,  it has been shown \cite{KSM08} that the $su(N)$ Lie algebra valued Yang-Mills field  $\mathscr{A}_\mu(x)$ is decomposed into two parts:
\begin{align}
  \mathscr{A}_\mu(x) = \mathscr{V}_\mu(x) + \mathscr{X}_\mu(x) ,
\end{align}
 such  that all fields $\bm{n}^j(x)$ are covariantly constant in the background  field $\mathscr{V}_\mu(x)$:
\begin{subequations}
\begin{align}
 0 = \mathscr{D}_\mu[\mathscr{V}] \bm{n}^j(x)
:= \partial_\mu \bm{n}^j(x) -i g [\mathscr{V}_\mu(x) , \bm{n}^j(x) ]
\quad (j=1,2, \cdots, r)
 ,
 \label{defVL}
\end{align}
and that the remaining field $\mathscr{X}_\mu(x)$ is orthogonal to all fields ${\bm n}^j(x)$:
\begin{align}
  0 = (\mathscr{X}_\mu(x), \bm{n}^j(x))   := 2{\rm tr}(\mathscr{X}_\mu(x) \bm{n}^j(x))  :=  \mathscr{X}_\mu^A(x) n_j^A(x)
\quad (j=1,2, \cdots, r)
\label{defXL}
 .
\end{align}
\end{subequations}
Both  $\mathscr{A}_\mu(x)$ and ${\bm n}^j(x)$ are Hermitian fields. This is also the case for $ \mathscr{V}_\mu(x)$ and $\mathscr{X}_\mu(x)$.

By solving the defining equation (\ref{defVL}) and (\ref{defXL}), $\mathscr{V}_\mu(x)$ and $\mathscr{X}_\mu(x)$ have been expressed in terms of $\mathscr{A}_\mu(x)$ and $\bm{n}^j(x)$ in the form:
\footnote{
In what follows, we use the script e.g., $\mathscr{A}_\mu$, $\bm{n}^j$ to express the $su(N)$ Lie algebra valued field and the boldface to express the vector field e.g., ${\bf n}^j$.
}
\begin{subequations}
\begin{align}
  \mathscr{V}_\mu(x)
  =& \sum_{j=1}^{r}  (\mathscr{A}_\mu(x),{\bm n}^j(x)) {\bm n}^j(x)   +i g^{-1} \sum_{j=1}^{r} [ {\bm n}^j(x) , \partial_\mu {\bm n}^j(x) ] ,
  \label{Vdef}
\\
 \mathscr{X}_\mu(x) =& -ig^{-1} \sum_{j=1}^{r}  [\bm{n}^j(x), \mathscr{D}_\mu[\mathscr{A}]\bm{n}^j(x) ]
 .
 \label{Xdef}
\end{align}
\end{subequations}
In the maximal case, all $\bm{n}^j(x)$ fields are constructed from a single color field $\bm{n}(x)$.
 Therefore, we have a  change of variables from $\mathscr{A}_\mu(x)$ to  ($\bm{n}(x)$, $c_\mu^j(x)$,  $\mathscr{X}_\mu(x)$) with $c_\mu^j(x):=(\mathscr{A}_\mu(x), \bm{n}^j(x))$, once    $\bm{n}(x)$ is given as a functional of $\mathscr{A}_\mu(x)$, as shown in \cite{KSM08}.

\subsection{Lattice: maximal case}

We try to decompose the $SU(N)$ link variable $U_{x,\mu}$ on a lattice into the product of two $SU(N)$ variables $X_{x,\mu}$ and $V_{x,\mu}$ on the same lattice:
\begin{equation}
 U_{x,\mu} = X_{x,\mu} V_{x,\mu} \in SU(N) ,
 \quad X_{x,\mu}, V_{x,\mu} \in SU(N)
  .
  \label{decomp-1}
\end{equation}
The link variable $U_{x,\mu}$ obeys the well-known lattice gauge transformation:
\begin{equation}
  U_{x,\mu}  \rightarrow \Omega_{x} U_{x,\mu} \Omega_{x+\mu}^\dagger = U_{x,\mu}^\prime
  , \quad \Omega_{x} \in SU(N)
  \label{U-transf}
 .
\end{equation}
We require that $V_{x,\mu}$ transforms like a usual gauge variable as
\begin{subequations}
\begin{equation}
  V_{x,\mu}  \rightarrow \Omega_{x} V_{x,\mu} \Omega_{x+\mu}^\dagger = V_{x,\mu}^\prime
  , \quad \Omega_{x} \in SU(N)
   .
   \label{V-transf}
\end{equation}
Consequently, $X_{x,\mu}$ must transform like an adjoint matter field:
\footnote{
It is possible to consider another decomposition of the form:
$
 U_{x,\mu} = V_{x,\mu} X_{x,\mu}
  .
  \label{decomp-2}
$
Then  $X_{x,\mu}$ must have the gauge transformation:
$
  X_{x,\mu} (=  V_{x,\mu}^\dagger U_{x,\mu} )  \rightarrow \Omega_{x+\mu} X_{x,\mu} \Omega_{x+\mu}^\dagger
  .
$
}
\begin{equation}
  X_{x,\mu} (=  U_{x,\mu}V_{x,\mu}^\dagger)
 \rightarrow \Omega_{x} X_{x,\mu} \Omega_{x}^\dagger = X_{x,\mu}^\prime
  , \quad \Omega_{x} \in SU(N)
  .
  \label{X-transf}
\end{equation}
At the same time, we introduce Hermitian variables ${\bm n}_{x}^j$  $(j=1, \cdots, r=N-1)$ on   sites of the lattice, $({\bm n}^j_{x})^\dagger={\bm n}^j_{x}$.
Suppose that each site variable ${\bm n}^j_{x}$ transforms under the gauge transformation according to the adjoint representation as (\ref{n-ctransf}) suggests:
\begin{align}
  {\bm n}_{x}^j \rightarrow \Omega_{x} {\bm n}_{x}^j \Omega_{x}^\dagger = {\bm n}_{x}^j{}' , \quad \Omega_{x} \in SU(N)
\label{n-transf}
 .
\end{align}
\end{subequations}
Once such $SU(N)$ lattice variables $V_{x,\mu}$ and $X_{x,\mu}$ are constructed together with ${\bm n}^j_{x}$, they are related to the Lie algebra valued fields $\mathscr{V}_\mu$ and $\mathscr{X}_\mu$ in the form:
\begin{align}
  V_{x,\mu} = \exp [-i\epsilon g \mathscr{V}_\mu(x+\epsilon\mu/2)]
  , \quad
  X_{x,\mu} = \exp [-i\epsilon g \mathscr{X}_\mu(x)]
 ,
\end{align}
just as the original link variable $U_{x,\mu}$ is related to the gauge potential $\mathscr{A}_\mu(x)$:
\begin{align}
U_{x,\mu} = \exp [ -i \epsilon g \mathscr{A}_\mu(x+\epsilon\mu/2)]
 ,
\label{def-U}
\end{align}
where $\epsilon$ is the lattice spacing.
The transformation property (\ref{V-transf}) for the group element $V_{x,\mu}$ in $SU(N)$ is required from   (\ref{V-ctransf}) of the $su(N)$ valued Hermitian  variable $\mathscr{V}_\mu(x)$,
which is the background field $\mathscr{V}_\mu(x)$ to be identified with the continuum  variable (\ref{Vdef}) in the continuum limit $\epsilon \rightarrow 0$.
Similarly, the transformation (\ref{X-transf}) for the other variable  $X_{x,\mu}$ is  consistent with  (\ref{X-ctransf})  of the $su(N)$ valued Hermitian  variable $\mathscr{X}_\mu(x)$ to be identified with the continuum  variable (\ref{Xdef}) in the continuum limit.
For details of the continuum formulations, see \cite{KSM08} for $SU(N)$ and especially \cite{KMS06} for $SU(2)$ case.
We first try to impose the following conditions as the lattice versions  of the defining equations (\ref{defVL}) and (\ref{defXL}) for $SU(N)$, which is a straightforward extension of the $SU(2)$ case \cite{IKKMSS06,SKKMSI07}.
\begin{subequations}
\begin{align}
 & \text{(a)} \quad
 \bm{n}_{x}^j V_{x,\mu}  = V_{x,\mu} \bm{n}_{x+\mu}^j
  ,
 \label{Lcc}
\\
 & \text{(b)} \quad
 {\rm tr}(\bm{n}_{x}^j U_{x,\mu} V_{x,\mu}^\dagger)
    =  0 = {\rm tr}(\bm{n}_{x}^j V_{x,\mu} U_{x,\mu}^\dagger)
 .
  \label{cond2m}
\end{align}
\end{subequations}
The last one is equivalent to
$
 {\rm tr}(\bm{n}_{x}^j X_{x,\mu} )
  = 0 = {\rm tr}(\bm{n}_{x}^j X_{x,\mu}^\dagger)
 .
$
Both defining equations must be solved to determine new variables $V_{x,\mu}$ and $X_{x,\mu}$ for a given set of $\bm{n}_{x}$ and $U_{x,\mu}$.
The first defining equation (\ref{Lcc}) for the link variable $V_{x,\mu}$  is form-invariant under the gauge transformation for new variables, i.e.,
$ \bm{n}_{x}^{j \prime} V_{x,\mu}^\prime  = V_{x,\mu}^\prime \bm{n}_{x+\mu}^{j \prime}
$.
This is also the case for the second defining equation (\ref{cond2m}).
The  (\ref{defVL}) comes from the lattice covariant derivative in the adjoint representation under the background ${\bf V}_\mu(x)$ \cite{SKKMSI07}:
\begin{align}
 D_\mu^{(\epsilon)}[V] \bm{n}_{x}^j := \epsilon^{-1}[V_{x,\mu} \bm{n}_{x+\mu}^j - \bm{n}_{x}^j V_{x,\mu}]
  .
 \label{cderivative}
\end{align}
The  (\ref{cond2m}) is suggested from a lattice version of the orthogonality equation (\ref{defXL}):
${\rm tr}({\bm n}_{x}^j \mathscr{X}_\mu(x))=0$ or
\begin{equation}
 {\rm tr}({\bm n}^j_{x} \exp \{-i\epsilon g \mathscr{X}_\mu(x)\})
  =  {\rm tr}({\bm n}^j_{x}  \{ {\bf 1}-i\epsilon g \mathscr{X}_\mu(x) \} ) + O(\epsilon^2) = 0 + O(\epsilon^2) .
  \label{cond2}
\end{equation}
This implies that the trace vanishes  up to   $O(\epsilon)$.
Remembering  the relation $\mathscr{X}_\mu(x)=\mathscr{A}_\mu(x)-\mathscr{V}_\mu(x)$, we can rewrite   (\ref{cond2}) into (\ref{cond2m}) in terms of ${\bm n}_{x}$ and $U_{x,\mu}$.

\subsection{How to solve the lattice defining equations?}

First, we consider how the defining equation (\ref{Lcc}) is solved to express the link variable $V_{x,\mu}$ in terms of the site variable ${\bm n}^j_{x}$ and the original link variable $U_{x,\mu}$,
just as the continuum variable $\mathscr{V}_\mu(x)$ is expressed in terms of ${\bm n}^j(x)$ and $\mathscr{A}_\mu(x)$ in (\ref{Vdef}).
In order to solve the matrix equation (\ref{Lcc}), we adopt an ansatz:
\begin{align}
 \tilde V_{x,\mu}
 = c U_{x,\mu} + \sum_{j=1}^{r} \alpha_j {\bm n}^j_{x} U_{x,\mu} + \sum_{j=1}^{r} \beta_j U_{x,\mu} {\bm n}^j_{x+\mu}
 + \sum_{j=1}^{r} \sum_{k=1}^{r} \gamma_{jk} {\bm n}^j_{x} U_{x,\mu} {\bm n}^k_{x+\mu}
  ,
 \label{ansatz}
\end{align}
where $c, \alpha_j, \beta_k, \gamma_{jk}$ ($j,k=1,\cdots,r$) are complex numbers to be determined.
In fact, $\tilde V_{x,\mu}$ has the transformation property (\ref{V-transf}) required for the link variable $V_{x,\mu}$, provided that the original link variable $U_{x,\mu}$ and the color field ${\bm n}^j_x$ transform  according to (\ref{U-transf}) and (\ref{n-transf}), respectively.
At first glance, this ansatz is rather specific in that it is linear in the link variable $U_{x,\mu}$ and contains the color field ${\bm n}^j_x$ up to quadratic in all ${\bm n}^j$.
However, this is the most general form within the general linear matrix $GL(N, \mathbb{C})$, since  the ansatz involves $1+r+r+r^2=N^2$ complex parameters which is equal to the dimension of a matrix in $GL(N, \mathbb{C})$.
Moreover, the first equation (\ref{Lcc}) is linear in  $V_{x,\mu}$ and the normalization of $\tilde{V}_{x,\mu}$ cannot be determined by this equation alone.  In what follows, therefore, we set $c=1$ hereafter without loss of generality.

We show that the solution of the defining equation based on the ansatz (\ref{ansatz}) for $\tilde V_{x,\mu}$ is sufficient to obtain the special unitary link variable $V_{x,\mu}$ satisfying the defining equations.
First, we show that the unitary link variable $\underbar{V}_{x,\mu}$ can be constructed from $\tilde V_{x,\mu}$ based on the {\it polar decomposition theorem},
\footnote{
This is a complex version of the polar decomposition theorem:  Any real regular (invertible) matrix $M$ is uniquely expressed by an orthogonal matrix $U$ and a positive symmetric matrix $P$ as
$
 M =PU = UQ, \quad P = \sqrt{MM^T}, \quad Q = \sqrt{M^T M}
    .
$
}
although the unitarity of the link variable $\tilde V_{x,\mu}$ of the ansatz (\ref{ansatz}) is not guaranteed.

\noindent
{\bf Theorem} (Polar decomposition theorem)
{\it Any complex regular matrix $M$ is uniquely decomposed into a unitary matrix $U$ and a positive-definite Hermitian matrix $P$ as
 \footnote{
Any complex regular matrix $M$ is also decomposed into another  positive-definite Hermitian matrix $Q$ as
$
 M = UQ
$,
$Q = \sqrt{M^\dagger M}
$.
}
\begin{equation}
 M = PU  , \quad P = \sqrt{MM^\dagger }
  .
\end{equation}
}
Thus the unitary link variable $\underbar{V}_{x,\mu} \in U(N)$ is obtained in terms of $\tilde{V}_{x,\mu}$ as
\footnote{
Here
$\underbar{V}_{x,\mu} = \tilde V_{x,\mu} Q_{x,\mu}^{-1}$ must be used for another case
$
 U_{x,\mu} = V_{x,\mu} X_{x,\mu}
$
where
$
 Q_{x,\mu} = \sqrt{\tilde V_{x,\mu}^\dagger  \tilde V_{x,\mu}}
    .
$
}
\begin{equation}
\underbar{V}_{x,\mu}
= P_{x,\mu}^{-1} \tilde V_{x,\mu}
 ,
\quad
 P_{x,\mu} = \sqrt{\tilde V_{x,\mu}  \tilde V_{x,\mu}^\dagger}
  .
\label{polar}
\end{equation}
Moreover,  the special unitary link variable $V_{x,\mu} \in SU(N)$ is obtained by the normalization which guarantees $\det V_{x,\mu}=1$:
\footnote{
For $SU(2)$, in particular, $ \tilde V_{x,\mu}  \tilde V_{x,\mu}^\dagger$ (or $ \tilde V_{x,\mu}^\dagger  \tilde V_{x,\mu}$) is proportional to the unit matrix and  the operation $P^{-1}$ (or $Q^{-1}$) is  equal to multiplying a normalization factor
$1/\sqrt{\frac12 {\rm tr}( \tilde V_{x,\mu}^\dagger  \tilde V_{x,\mu})}=1/\sqrt{\frac12 {\rm tr}( \tilde V_{x,\mu} \tilde V_{x,\mu}^\dagger)}$  \cite{IKKMSS06}.
This is not the case for $SU(N)$, $N \ge 3$.
}
\begin{equation}
 V_{x,\mu} = \underbar{V}_{x,\mu}/(\det \underbar{V}_{x,\mu})^{1/N}
  .
  \label{speciality}
\end{equation}

Next, we show that the first defining equation (\ref{Lcc}) is automatically satisfied for $V_{x,\mu}$ which are made unitary by (\ref{polar}), if   $\tilde V_{x,\mu}$ satisfy the same equation (a'):
\begin{align}
\text{(a')}  \quad\quad
  \bm{n}_{x}^j \tilde V_{x,\mu}
= \tilde V_{x,\mu} \bm{n}_{x+\mu}^j
 .
 \label{Lcc2}
\end{align}
  If $\tilde V_{x,\mu}$ satisfies the defining equation (a'), $\tilde V_{x,\mu} \tilde V_{x,\mu}^\dagger$ commutes with $\bm{n}_{x}^j$, since
\begin{align}
 {\bf n}_{x}^j \tilde V_{x,\mu} \tilde V_{x,\mu}^\dagger
= \tilde V_{x,\mu} {\bf n}_{x+\mu}^j  \tilde V_{x,\mu}^\dagger
=  \tilde V_{x,\mu}  \tilde V_{x,\mu}^\dagger {\bf n}_{x}^j
 \Longleftrightarrow  [ \tilde V_{x,\mu} \tilde V_{x,\mu}^\dagger , {\bf n}_{x}^j ] = 0
\label{VVd}
  ,
\end{align}
while $\tilde V_{x,\mu}^\dagger \tilde V_{x,\mu}$ commutes with ${\bf n}_{x+\mu}^j$:

   \tilde V_{x,\mu}^\dagger  \tilde V_{x,\mu} {\bf n}_{x+\mu}^j
= \tilde V_{x,\mu}^\dagger {\bf n}_{x}^j \tilde V_{x,\mu}
=   {\bf n}_{x+\mu}^j \tilde V_{x,\mu}^\dagger \tilde V_{x,\mu}
 \Longleftrightarrow  [ \tilde V_{x,\mu}^\dagger  \tilde V_{x,\mu} , {\bf n}_{x+\mu}^j ] = 0
  .
$
It is also possible to prove (see Appendix~\ref{appendix:spectral-resolution}) that  ${\bf n}_{x}^j$   commutes  with the square root $P_{x,\mu}$ and its inverse   square root $P_{x,\mu}^{-1}$,
\footnote{
Whereas

  [ \tilde V_{x,\mu}^\dagger \tilde V_{x,\mu}  , {\bf n}_{x+\mu}^j ] = 0 \Longleftrightarrow
 [ \sqrt{\tilde V_{x,\mu}^\dagger \tilde V_{x,\mu} } , {\bf n}_{x+\mu}^j ] = 0
 \Longleftrightarrow
  [ Q_{x,\mu}^{-1} , {\bf n}_{x}^j ] = 0
 .
$
}
\begin{equation}
  [ \tilde V_{x,\mu} \tilde V_{x,\mu}^\dagger , {\bf n}_{x}^j ] = 0
\Longleftrightarrow
 [ \sqrt{\tilde V_{x,\mu} \tilde V_{x,\mu}^\dagger} , {\bf n}_{x}^j ] = 0
 \Longleftrightarrow
  [ P_{x,\mu}^{-1} , {\bf n}_{x}^j ] = 0
 .
\label{VdV2}
\end{equation}
To obtain  $V_{x,\mu}$ satisfying (a), therefore, it is sufficient to find a solution $\tilde V_{x,\mu}$ of   (a'), since
(\ref{VdV2}) leads to
\footnote{
Whereas

 {\bf n}_{x}^j  \underbar{V}_{x,\mu}
=  {\bf n}_{x}^j \tilde V_{x,\mu} Q_{x,\mu}^{-1}
= \tilde V_{x,\mu} {\bf n}_{x+\mu}^j  Q_{x,\mu}^{-1}
= \tilde V_{x,\mu}  Q_{x,\mu}^{-1} {\bf n}_{x+\mu}^j
=  \underbar{V}_{x,\mu} {\bf n}_{x+\mu}^j
 .

}
\begin{align}
 {\bf n}_{x}^j  \underbar{V}_{x,\mu}
=&  {\bf n}_{x}^j  P_{x,\mu}^{-1} \tilde V_{x,\mu}
=   P_{x,\mu}^{-1} {\bf n}_{x}^j  \tilde V_{x,\mu}
=  P_{x,\mu}^{-1} \tilde V_{x,\mu}   {\bf n}_{x+\mu}^j
=  \underbar{V}_{x,\mu} {\bf n}_{x+\mu}^j
  .
\end{align}

For the second defining equation, we adopt the following  (b') instead of (b).
\begin{equation}
\text{(b')}  \quad \quad
 {\rm tr}(\bm{n}_{x} \tilde X_{x,\mu})=0, \quad
   \tilde  X_{x,\mu} = U_{x,\mu} \tilde V_{x,\mu}^{-1}
( \ne U_{x,\mu} \tilde V_{x,\mu}^\dagger )
  ,
  \label{def-2b}
\end{equation}
or more calculable form:
$\text{(b'')}$
$   {\rm tr}(\bm{n}_{x} \tilde X_{x,\mu}^{-1})=0,
$
$\tilde X_{x,\mu}^{-1} = \tilde  V_{x,\mu} U_{x,\mu}^{-1} = \tilde  V_{x,\mu} U_{x,\mu}^\dagger
  .
$
Although (b') is written for $\tilde X_{x,\mu}$ and it does not guarantee the validity of   (b) for $X_{x,\mu}$,  two conditions (b) and (b') have no difference in the level of the Lie algebra which is relevant in the continuum limit and that (b') yield (\ref{defXL}) in the continuum limit.

For $SU(N)$ ($N \ge 3$),  thus, the decomposition of the original link variable is achieved, once we know $\tilde V_{x,\mu}$ satisfying the defining equation (a'), (b'):
\footnote{
The identification
$X_{x,\mu}
=  Q_{x,\mu} \tilde X_{x,\mu}$ must be used for another case
$
 U_{x,\mu} = V_{x,\mu} X_{x,\mu}
$.
}
\begin{subequations}
\begin{align}
SU(N)  \ni U_{x,\mu}
=& X_{x,\mu} V_{x,\mu}
=  \tilde X_{x,\mu} \tilde V_{x,\mu}    ,
  \quad \tilde X_{x,\mu}, \tilde V_{x,\mu} \notin SU(N)
  ,
  \label{decomp-f}
\\
  V_{x,\mu}  =&  P_{x,\mu}^{-1} \tilde V_{x,\mu}/(\det (P_{x,\mu}^{-1} \tilde V_{x,\mu}))^{1/N}  \in SU(N)
    , \quad
    P_{x,\mu} := \sqrt{\tilde V_{x,\mu}  \tilde V_{x,\mu}^\dagger}
\label{polar11}
\\
   X_{x,\mu}
=&  U_{x,\mu} \tilde V_{x,\mu}^{-1} P_{x,\mu} (\det (P_{x,\mu}^{-1} \tilde V_{x,\mu}))^{1/N} \in SU(N)
  .
\label{polar-X}
\end{align}
\end{subequations}
Thus, solving the defining equations (a'):(\ref{Lcc2}) and (b'):(\ref{def-2b}) under the ansatz (\ref{ansatz}) is sufficient to achieve the desired decomposition through the identification (\ref{decomp-f}).

\subsection{Solving the lattice defining equations}

In order to see when the first defining equation (a') meets, we calculate
\begin{align}
 {\bm n}^\ell_{x} {\tilde V}_{x,\mu}
 =& {\bm n}^\ell_{x} U_{x,\mu} + \sum_{j=1}^{r} \alpha_j {\bm n}^\ell_{x} {\bm n}^j_{x} U_{x,\mu} + \sum_{j=1}^{r} \beta_j {\bm n}^\ell_{x} U_{x,\mu} {\bm n}^j_{x+\mu}
 + \sum_{j=1}^{r} \sum_{k=1}^{r} \gamma_{jk} {\bm n}^\ell_{x} {\bm n}^j_{x} U_{x,\mu} {\bm n}^k_{x+\mu}
  \nonumber\\
 =&
  \alpha_\ell
  \frac{1}{2N}  U_{x,\mu}
+ \left[ {\bm n}^\ell_{x} U_{x,\mu}
+ \sum_{j=1}^{r} \alpha_j  \frac{1}{2} (\bm n^\ell * \bm n^j)_{x}  U_{x,\mu} \right]
 +  \sum_{k=1}^{r} \gamma_{\ell k}
 \frac{1}{2N} U_{x,\mu} {\bm n}^k_{x+\mu}
 \nonumber\\
 &+ \left[ \sum_{j=1}^{r} \beta_j {\bm n}^\ell_{x} U_{x,\mu} {\bm n}^j_{x+\mu}
  + \sum_{j=1}^{r} \sum_{k=1}^{r} \gamma_{jk}  \frac{1}{2} (\bm n^\ell * \bm n^j)_{x}
U_{x,\mu} {\bm n}^k_{x+\mu} \right]
 .
 \label{ansatz1}
\end{align}
Here we have used
$ T^A T^B
=  \frac12 [ T^A , T^B  ] +  \frac12 \{ T^A , T^B  \}
=  \frac{i}{2} f^{ABC} T^C + \frac{1}{2N} \delta^{AB} {\bf 1} + \frac{1}{2} d^{ABC} T^C
$
and
$
 \bm n^j \bm n^k
 =  \frac12 [ \bm n^j ,\bm n^k ] +  \frac12 \{ \bm n^j ,\bm n^k \}
= \frac{1}{2N} \delta_{jk} {\bf 1} + \frac{1}{2} (\bm n^j * \bm n^k)
  ,
$
since $[\bm n^j ,\bm n^k ]=0$,
where we have defined
$
\bm n^j * \bm n^k
:= ({\bf n}^j * {\bf n}^k)^C T^C
:= d^{ABC} n_j^A n_k^B T^C
 .
$
On the other hand, we have
\begin{align}
 {\tilde V}_{x,\mu} {\bm n}^\ell_{x+\mu}
 =& U_{x,\mu} {\bm n}^\ell_{x+\mu} + \sum_{j=1}^{r} \alpha_j {\bm n}^j_{x} U_{x,\mu} {\bm n}^\ell_{x+\mu} + \sum_{j=1}^{r} \beta_j U_{x,\mu} {\bm n}^j_{x+\mu} {\bm n}^\ell_{x+\mu}
 + \sum_{j=1}^{r} \sum_{k=1}^{r} \gamma_{jk} {\bm n}^j_{x} U_{x,\mu} {\bm n}^k_{x+\mu} {\bm n}^\ell_{x+\mu}
 \nonumber\\
 =&
 \frac{1}{2N}  \beta_\ell U_{x,\mu}
+ \sum_{j=1}^{r}
 \frac{1}{2N} \gamma_{j\ell} {\bm n}^j_{x} U_{x,\mu}
+ \left[ U_{x,\mu} {\bm n}^\ell_{x+\mu}
+  \sum_{j=1}^{r} \beta_j \frac{1}{2} U_{x,\mu} (\bm n^j * \bm n^\ell)_{x+\mu}   \right]
 \nonumber\\
 & + \left[ \sum_{j=1}^{r} \alpha_j {\bm n}^j_{x} U_{x,\mu} {\bm n}^\ell_{x+\mu}
+ \sum_{j=1}^{r} \sum_{k=1}^{r} \gamma_{jk} \frac{1}{2}  {\bm n}^j_{x} U_{x,\mu}  (\bm n_k * \bm n^\ell)_{x+\mu}  \right]
  .
 \label{ansatz2}
\end{align}
The four terms in the right-hand sides of (\ref{ansatz1}) and (\ref{ansatz2}) are independent.  In order to fulfill (a'), therefore, the following conditions must be satisfied simultaneously:
\begin{subequations}
\begin{align}
  \text{(i)} \quad &
\alpha_\ell = \beta_\ell
  ,
\\
  \text{(ii)} \quad &
 {\bm n}^\ell_{x} U_{x,\mu}
+ \sum_{j=1}^{r} \alpha_j  \frac{1}{2} (\bm n^\ell * \bm n^j)_{x}  U_{x,\mu}
=\sum_{j=1}^{r}
 \frac{1}{2N} \gamma_{j\ell} {\bm n}^j_{x} U_{x,\mu}
 ,
\\
  \text{(iii)} \quad &
 \sum_{k=1}^{r} \gamma_{\ell k}
 \frac{1}{2N} U_{x,\mu} {\bm n}^k_{x+\mu}
= U_{x,\mu} {\bm n}^\ell_{x+\mu}
+ \sum_{j=1}^{r} \beta_j \frac{1}{2}  U_{x,\mu} (\bm n^j * \bm n^\ell)_{x+\mu}
 ,
\\
  \text{(iv)} \quad &
  \sum_{j=1}^{r} \beta_j {\bm n}^\ell_{x} U_{x,\mu} {\bm n}^j_{x+\mu}
  + \sum_{j=1}^{r} \sum_{k=1}^{r} \gamma_{jk}  \frac{1}{2} (\bm n^\ell * \bm n^j)_{x}
U_{x,\mu} {\bm n}^k_{x+\mu}
\nonumber\\
&= \sum_{j=1}^{r} \alpha_j {\bm n}^j_{x} U_{x,\mu} {\bm n}^\ell_{x+\mu}
+ \sum_{j=1}^{r} \sum_{k=1}^{r} \gamma_{jk}\frac{1}{2}  {\bm n}^j_{x} U_{x,\mu}  (\bm n^k * \bm n^\ell)_{x+\mu}
 ,
\end{align}
\end{subequations}
where the symmetric product of two $\bm n^j$ fields is written as a linear combination of some of $\bm n^j$ fields.
For $SU(N)$, the relevant information is given in \cite{KSM08}.

For $SU(2)$, the symmetric product is identically zero: $d^{ABC}=0$:
\begin{align}
 \bm n  \bm n   = (1/4) {\bf 1}  , \quad
 \bm n * \bm n  = 0
  .
\end{align}
Therefore, we immediately find the solution:
$\alpha_1=\beta_1$, $\gamma_{11}=4$,
and
\begin{align}
 {\tilde V}_{x,\mu}
 = U_{x,\mu} +  \alpha_1  ({\bm n}_{x} U_{x,\mu} +   U_{x,\mu} {\bm n}_{x+\mu} )
 +  4{\bm n}_{x} U_{x,\mu} {\bm n}_{x+\mu}
  .
 \label{SU2}
\end{align}
Therefore, one parameter $\alpha_1$ is undetermined.
This result reproduces the previous one \cite{IKKMSS06}. (This result for the value $\gamma_{11}=4$ is apparently different from the previous one due to the fact that we adopt the different normalization for the color field.)

For $SU(3)$, we have three relations for two  fields $\bm{n}^1$ and $\bm{n}^2$,
\begin{align}
{\bf n}^1 * {\bf n}^1  =& (1/\sqrt{3}) {\bf n}^2
 ,
\nonumber\\
{\bf n}^1 * {\bf n}^2 =& {\bf n}^2 * {\bf n}^1 = (1/\sqrt{3}){\bf n}^1
 ,
\nonumber\\
{\bf n}^2 * {\bf n}^2 =& -(1/\sqrt{3}) {\bf n}^2
\label{nSU3}
 .
\end{align}
For $SU(3)$, the ansatz reads
\begin{align}
{\tilde V}_{x,\mu} =& {\tilde V}_{x,\mu}^{(+)} + {\tilde V}_{x,\mu}^{(-)} ,
 \nonumber\\
{\tilde V}_{x,\mu}^{(+)}
 &=U_{x,\mu}
   +\alpha_2 {\bm n^2_x}U_{x,\mu}
   +\beta_2 U_{x,\mu}{\bm n^2_{x+\mu}}
   +\gamma_{11}{\bm n^1_x}U_{x,\mu}{\bm n^1_{x+\mu}}
   +\gamma_{22} {\bm n^2_x}U_{x,\mu}{\bm n^2_{x+\mu}}
 ,
 \nonumber\\
{\tilde V}_{x,\mu}^{(-)}
 &=\alpha_1 {\bm n^1_x}U_{x,\mu}
   +\beta_1 U_{x,\mu}{\bm n^1_{x+\mu}}
   +\gamma_{12} {\bm n^1_x}U_{x,\mu}{\bm n^2_{x+\mu}}
   +\gamma_{21} {\bm n^2_x}U_{x,\mu}{\bm n^1_{x+\mu}}
    ,
\end{align}
where we have separated ${\tilde V}_{x,\mu}$ into two parts according to the parity under the global transformation $\bm n^1_x \rightarrow-\bm n^1_x$, under which $\bm n^2_x$ does not change.

The requirement (ii) reads
for $\ell=1$,
\begin{align}
0 =&  {\bm n}^1_{x} U_{x,\mu}
+  \alpha_1  \frac{1}{2} (\bm n_1 * \bm n_1)_{x}  U_{x,\mu}
+  \alpha_2  \frac{1}{2} (\bm n_1 * \bm n_2)_{x}  U_{x,\mu}
- \frac{1}{6} \gamma_{11} {\bm n}^1_{x} U_{x,\mu}
 - \frac{1}{6} \gamma_{21} {\bm n}^2_{x} U_{x,\mu}
 \nonumber\\
=&  {\bm n}^1_{x} U_{x,\mu}
+  \alpha_1  \frac{1}{2}  \frac{1}{\sqrt{3}} \bm n^2_{x}  U_{x,\mu}
+  \alpha_2  \frac{1}{2} \frac{1}{\sqrt{3}} \bm n^1_{x} U_{x,\mu}
- \frac{1}{6} \gamma_{11} {\bm n}^1_{x} U_{x,\mu}
 - \frac{1}{6} \gamma_{21} {\bm n}^2_{x} U_{x,\mu}
 ,
\end{align}
while for $\ell=2$,
\begin{align}
0 =&  {\bm n}^2_{x} U_{x,\mu}
+  \alpha_1  \frac{1}{2} (\bm n_2 * \bm n_1)_{x}  U_{x,\mu}
+  \alpha_2  \frac{1}{2} (\bm n_2 * \bm n_2)_{x}  U_{x,\mu}
- \frac{1}{6} \gamma_{12} {\bm n}^1_{x} U_{x,\mu}
 - \frac{1}{6} \gamma_{22} {\bm n}^2_{x} U_{x,\mu}
 \nonumber\\
=&  {\bm n}^2_{x} U_{x,\mu}
+  \alpha_1  \frac{1}{2}  \frac{1}{\sqrt{3}} \bm n^1_{x}  U_{x,\mu}
-  \alpha_2  \frac{1}{2} \frac{1}{\sqrt{3}} \bm n^2_{x} U_{x,\mu}
- \frac{1}{6} \gamma_{12} {\bm n}^1_{x} U_{x,\mu}
 - \frac{1}{6} \gamma_{22} {\bm n}^2_{x} U_{x,\mu}
 .
\end{align}
Then we easily find
\begin{align}
& \alpha_1=\beta_1, \quad \alpha_2=\beta_2,
 \nonumber\\
 & \gamma_{11} = 6 +\sqrt{3} \alpha_2 , \quad \gamma_{21} = \sqrt{3} \alpha_1
 , \quad
 \gamma_{22} = 6 -\sqrt{3} \alpha_2 , \quad \gamma_{12} = \sqrt{3} \alpha_1
 .
 \label{SU3-parameter}
\end{align}
Thus the requirement (a') yields the result:
\begin{align}
{\tilde V}_{x,\mu}^{(+)}
 &=U_{x,\mu}  +\alpha_2({\bm n^2_x}U_{x,\mu}+U_{x,\mu}{\bm n^2_{x+\mu}})
   \nonumber\\
 &   +(6+\sqrt3\alpha_2){\bm n^1_x}U_{x,\mu}{\bm n^1_{x+\mu}}
   +(6-\sqrt3\alpha_2){\bm n^2_x}U_{x,\mu}{\bm n^2_{x+\mu}}
 ,
 \nonumber\\
{\tilde V}_{x,\mu}^{(-)}
 &=\alpha_1 ({\bm n^1_x}U_{x,\mu}
   + U_{x,\mu}{\bm n^1_{x+\mu}}
   +\sqrt3 {\bm n^1_x}U_{x,\mu}{\bm n^2_{x+\mu}}
   +\sqrt3 {\bm n^2_x}U_{x,\mu}{\bm n^1_{x+\mu}})
 .
\label{SU3-1}
\end{align}
Therefore, two parameters $\alpha_1$ and $\alpha_2$ are still undetermined and should be fixed by imposing the requirement (b') or (b''), in the similar way to the $SU(2)$ case.

Next, we examine the second defining equation (b'). We rewrite the product
\begin{align}
 {\bm n}^\ell_{x} {\tilde V}_{x,\mu}U_{x,\mu}^\dagger
 =&{\bm n}^\ell_{x} + \sum_{j=1}^{r} \alpha_j {\bm n}^\ell_{x}{\bm n}^j_{x}  + \sum_{j=1}^{r} \beta_j {\bm n}^\ell_{x} U_{x,\mu} {\bm n}^j_{x+\mu} U_{x,\mu}^\dagger
 + \sum_{j=1}^{r} \sum_{k=1}^{r} \gamma_{jk} {\bm n}^\ell_{x} {\bm n}^j_{x} U_{x,\mu} {\bm n}^k_{x+\mu}U_{x,\mu}^\dagger
\nonumber\\
 =&   \alpha_\ell   \frac{1}{2N} {\bf 1} + {\bm n}^\ell_{x} +  \sum_{j=1}^{r}  \alpha_j  \frac12 ({\bm n}^\ell*{\bm n}^j)_{x}
+   \sum_{k=1}^{r} \gamma_{\ell k}   \frac{1}{2N}    U_{x,\mu} {\bm n}^k_{x+\mu}U_{x,\mu}^\dagger
\nonumber\\
 & + \sum_{j=1}^{r} \beta_j {\bm n}^\ell_{x} U_{x,\mu} {\bm n}^j_{x+\mu} U_{x,\mu}^\dagger
+ \sum_{j=1}^{r} \sum_{k=1}^{r} \gamma_{jk} \frac12 ({\bm n}^\ell*{\bm n}^j)_{x} U_{x,\mu} {\bm n}^k_{x+\mu}U_{x,\mu}^\dagger
  .
\end{align}
The trace of ${\bm n}^\ell_{x} {\tilde V}_{x,\mu}U_{x,\mu}^\dagger$ reads
\begin{align}
 {\rm tr}({\bm n}^\ell_{x} {\tilde V}_{x,\mu}U_{x,\mu}^\dagger)

 =&
 \alpha_\ell   \frac{1}{2}
  + \sum_{j=1}^{r} \beta_j {\rm tr}({\bm n}^\ell_{x} U_{x,\mu} {\bm n}^j_{x+\mu} U_{x,\mu}^\dagger)
\nonumber\\&
 + \sum_{j=1}^{r} \sum_{k=1}^{r} \frac{1}{2} \gamma_{jk} {\rm tr}(({\bm n}^\ell_{x}*{\bm n}^j_{x}) U_{x,\mu} {\bm n}^k_{x+\mu}U_{x,\mu}^\dagger)
  ,
\end{align}
where we have used
${\rm tr}({\bm n}^\ell_{x})=0$ and ${\rm tr}({\bm n}^\ell_{x}*{\bm n}^j_{x})=0$.
The term ${\bm n}^k_{x} U_{x,\mu} {\bm n}^j_{x+\mu} U_{x,\mu}^\dagger$ has the expansion in the Lie algebra:
\begin{align}
& {\bm n}^k_{x} U_{x,\mu} {\bm n}^j_{x+\mu} U_{x,\mu}^\dagger
\nonumber\\
&= {\bm n}^k_{x}[{\bf 1}-i\epsilon g\mathscr{A}_\mu(x)+ O((\epsilon^2)][{\bm n}^j_{x}+\epsilon \partial_\mu {\bm n}^j_{x}+ O((\epsilon^2)][1+i\epsilon g\mathscr{A}_\mu(x)+ O((\epsilon^2)]
\nonumber\\
&= {\bm n}^k_{x} {\bm n}^j_{x} + i {\bm n}^k_{x} {\bm n}^j_{x} \epsilon g\mathscr{A}_\mu(x) + \epsilon {\bm n}^k_{x} \partial_\mu {\bm n}^j_{x} -i {\bm n}^k_{x} \epsilon g\mathscr{A}_\mu {\bm n}^j_{x} +  O((\epsilon^2)
  .
\end{align}
Then its trace is given by
\begin{align}
& {\rm tr}({\bm n}^k_{x} U_{x,\mu} {\bm n}^j_{x+\mu} U_{x,\mu}^\dagger)
\nonumber\\
&= {\rm tr}({\bm n}^k_{x} {\bm n}^j_{x}) + i {\rm tr}([{\bm n}^k_{x}, {\bm n}^j_{x}] \epsilon g\mathscr{A}_\mu(x)) + {\rm tr}(\epsilon {\bm n}^k_{x} \partial_\mu {\bm n}^j_{x})  +  O((\epsilon^2)
\nonumber\\
&= \frac{1}{2} \delta_{kj}    +  O((\epsilon^2)
\label{tr(nUnU)}
  ,
\end{align}
where we have used
$[{\bm n}^k_{x}, {\bm n}^j_{x}]=0$ and ${\rm tr}({\bm n}^\ell_{x})=0$.
Taking into account this result, thus, we obtain another set of conditions:
\begin{align}
0   =  {\rm tr}({\bm n}^\ell_{x} {\tilde V}_{x,\mu}U_{x,\mu}^\dagger) =   \alpha_\ell
+ \sum_{j=1}^{r} \sum_{k=1}^{r} \frac{1}{4} \gamma_{jk}
 {\rm tr}(({\bm n}^\ell_{x}*{\bm n}^j_{x}) U_{x,\mu} {\bm n}^k_{x+\mu}U_{x,\mu}^\dagger)
  ,
  \label{nVU}
\end{align}
where the last trace is calculated from (\ref{tr(nUnU)}) after applying (\ref{nSU3})  to  $({\bm n}^\ell_{x}*{\bm n}^j_{x})$.

For $SU(2)$, we immediately have
$
 \alpha_1 = 0
$
and reproduce the solution \cite{IKKMSS06}, see also \cite{CGI98}:
\begin{align}
 {\tilde V}_{x,\mu}
 = U_{x,\mu}
 +  4{\bm n}_{x} U_{x,\mu} {\bm n}_{x+\mu}
  .
 \label{SU2-sol}
\end{align}

For $SU(3)$, for $\ell=1$, (\ref{nVU}) is
\begin{align}
0  =     \alpha_1
+ \sum_{j=1}^{2} \sum_{k=1}^{2} \frac{1}{4}\gamma_{jk}
 {\rm tr}(({\bm n}^1_{x}*{\bm n}^j_{x}) U_{x,\mu} {\bm n}^k_{x+\mu}U_{x,\mu}^\dagger)
 =  \alpha_1
+  \frac{1}{4}\gamma_{12}
 \frac{1}{\sqrt{3}}\frac{1}{2}
+  \frac{1}{4}\gamma_{21}
 \frac{1}{\sqrt{3}}\frac{1}{2}
  =   \frac{3}{2} \alpha_1
   ,
\end{align}
where we have used
(\ref{nSU3}), (\ref{SU3-parameter}) and (\ref{tr(nUnU)}).
Similarly, for $\ell=2$, (\ref{nVU}) is
\begin{align}
0  =&    \alpha_2
+ \sum_{j=1}^{2} \sum_{k=1}^{2} \frac{1}{4}\gamma_{jk}
 {\rm tr}(({\bm n}^2_{x}*{\bm n}^j_{x}) U_{x,\mu} {\bm n}^k_{x+\mu}U_{x,\mu}^\dagger)
=   \frac{3}{2} \alpha_2
 .
\end{align}
For $SU(3)$, thus, we have determined the parameters as
\begin{align}
 \alpha_1=\beta_1 = \alpha_2=\beta_2 = 0  , \quad
 \gamma_{11} =  \gamma_{22} = 6 , \quad \gamma_{21} = \gamma_{12} = 0
 ,
\end{align}
and have obtained the solution of the defining equations:
\begin{align}
{\tilde V}_{x,\mu}  = U_{x,\mu} +  6{\bm n}^1_{x}U_{x,\mu}{\bm n}^1_{x}  +  6{\bm n}^2_{x}U_{x,\mu}{\bm n}^2_{x}
   .
\end{align}

Thus we have shown that the solution for $\tilde{V}_{x,\mu}$ is given for $N=2,3$ in the form:
\begin{align}
 \tilde{V}_{x,\mu}
= U_{x,\mu} + \sum_{j=1}^{N-1} 2N {\bm n}^j_{x} U_{x,\mu} {\bm n}^j_{x+\mu}
  .
  \label{Vtilde}
\end{align}
It should be remarked that (\ref{Vtilde}) is also a solution for $SU(N)$ with $N >3$, since it is possible to check that (\ref{Vtilde}) fulfils the defining equations (a') and (b') for any $N$ of $SU(N)$.
Furthermore, it is possible to prove that it is a unique solution of the defining equations which is slightly modified for (b).  This result will be published elsewhere, because of limitations of space.

The unitary link variable $V_{x,\mu}$ is obtained  from ${\tilde V}_{x,\mu}$ (\ref{Vtilde})  following  (\ref{polar}) and (\ref{speciality}).
Moreover, it is not difficult to show that the lattice condition (b')
$
 {\rm tr}({\bm n}_{x}^\ell \tilde{X}_{x,\mu})
=    {\rm tr}({\bm n}_{x}^\ell (\tilde{V}_{x,\mu}U_{x,\mu}^{-1})^{-1} ) = 0
$
reproduces the continuum version
$
2 {\rm tr}({\bm n}_{x}^\ell \mathscr{X}_\mu(x) )
={\bf n}^\ell(x) \cdot \mathbb{X}_\mu(x) =0
$ in the naive continuum limit.

\section{Minimal case}

Now we discuss the minimal case  in which   $\mathscr A_\mu$ is decomposed into
$\mathscr{V}_\mu(x)$ and $\mathscr{X}_\mu(x)$, i.e.,
$
 \mathscr{A}_\mu(x) = \mathscr{V}_\mu(x) + \mathscr{X}_\mu(x)
$,
using a single color  field
\begin{align}
 \bm h(x):=\bm n_r(x) =U^\dagger(x) H_r U(x)
  ,
\end{align}
where the final Cartan matrix is given by
$
 H_r
 = \frac{1}{\sqrt{2N(N-1)}} {\rm diag}(1,\cdots,1,-N+1)
 .
$

\subsection{Continuum: minimal case}

The respective components $\mathscr V_\mu(x)$ and $\mathscr X_\mu(x)$ are specified by two defining equations:

\noindent
(I)  $\bm{h}(x)$ is covariantly constant in the background $\mathscr{V}_\mu(x)$:
\begin{subequations}
\begin{align}
  0 = \mathscr{D}_\mu[\mathscr{V}] \bm{h}(x)
:=\partial_\mu \bm{h}(x) -  ig [\mathscr{V}_\mu(x), \bm{h}(x)]
 ,
\label{defVL2}
\end{align}
(II)  $\mathscr{X}^\mu(x)$  has the vanishing $\tilde{H}$-commutative part, i.e., $\mathscr{X}^\mu(x)_{\tilde{H}}=0$:
\begin{align}
  \mathscr{X}^\mu(x)_{\tilde{H}} := \left( {\bf 1} -   2\frac{N-1}{N}  [\bm{h} , [\bm{h} ,  \cdot ]]
\right) \mathscr{X}_\mu(x)   = 0
\label{defXL2}
 .
\end{align}
\end{subequations}
By solving the defining equations, it has  been shown \cite{KSM08} that the new variables are written in terms of $\bm h$ and $\mathscr A_\mu$:
\begin{subequations}
\begin{align}
\mathscr A_\mu(x)
 =& \mathscr V_\mu(x)
  +\mathscr X_\mu(x)
 =\mathscr C_\mu(x)
  +\mathscr B_\mu(x)
  +\mathscr X_\mu(x)
 ,
\\
  \mathscr{C}_\mu(x)
=& \mathscr{A}_\mu(x) - \frac{2(N-1)}{N}   [\bm{h}(x), [ \bm{h}(x), \mathscr{A}_\mu(x) ] ]
,
\\
 \mathscr{B}_\mu(x)
=& ig^{-1} \frac{2(N-1)}{N}[\bm{h}(x) , \partial_\mu  \bm{h}(x) ]
,
\\
 \mathscr{X}_\mu(x) =& -ig^{-1}  \frac{2(N-1)}{N}  [\bm{h}(x), \mathscr{D}_\mu[\mathscr{A}]\bm{h}(x) ]
 ,
\end{align}
 \label{NLCV-minimal}
\end{subequations}
once a single color field $\bm{h}(x)$ is given as a functional of $\mathscr A_\mu$.

\subsection{Lattice: minimal case}

In order to obtain the defining equations on a lattice and to solve them, we  repeat the same argument as that in the maximal case.
Consequently, the lattice versions  of the defining equations in the minimal choice for $SU(N)$ are  given by
\footnote{
Note that (\ref{cond2m-min}) is sufficient to determine the ansatz ${\tilde V}_{x,\mu}$ of the form (\ref{ansatzmini}), although we could impose a stronger condition on the lattice which is equivalent to (\ref{defXL2}) in the continuum form.
}
\begin{subequations}
\begin{align}
 & \text{(a')} \quad
 \bm{h}_{x} V_{x,\mu}  = V_{x,\mu} \bm{h}_{x+\mu}
  ,
 \label{Lcc-min}
\\
 & \text{(b')} \quad
 {\rm tr}(\bm{h}_{x} U_{x,\mu} V_{x,\mu}^\dagger)
    =  0 = {\rm tr}(\bm{h}_{x} V_{x,\mu} U_{x,\mu}^\dagger)
 .
  \label{cond2m-min}
\end{align}
\end{subequations}

In the minimal case, we adopt the ansatz for $V_{x,\mu}$:
\begin{align}
 {\tilde V}_{x,\mu}
 = U_{x,\mu} +  \alpha  {\bm h}_{x} U_{x,\mu} +  \beta U_{x,\mu} {\bm h}_{x+\mu}
 +  \gamma {\bm h}_{x} U_{x,\mu} {\bm h}_{x+\mu}
  .
 \label{ansatzmini}
\end{align}
Then we calculate
\begin{align}
 {\bm h}_{x} {\tilde V}_{x,\mu}
 =&   \frac{\alpha }{2N}  U_{x,\mu}
+ \left( 1 +  \frac{\kappa}{2}  \alpha  \right) {\bm h}_{x} U_{x,\mu}
 +
 \frac{\gamma}{2N} U_{x,\mu} {\bm h}_{x+\mu}
+ \left( \beta +  \frac{\kappa}{2}  \gamma  \right)  {\bm h}_{x} U_{x,\mu} {\bm h}_{x+\mu}
 ,
\nonumber\\
 {\tilde V}_{x,\mu} {\bm h}_{x+\mu}
=& \frac{\beta}{2N}   U_{x,\mu}
+
 \frac{\gamma}{2N}  {\bm h}_{x} U_{x,\mu}
+  \left( 1 +  \frac{\kappa}{2}  \beta  \right)  U_{x,\mu} {\bm h}_{x+\mu}
+  \left( \alpha +  \frac{\kappa}{2}  \gamma  \right)  {\bm h}_{x} U_{x,\mu} {\bm h}_{x+\mu}
  ,
 \label{ansatz2a}
\end{align}
where we have used
$
  {\bm h}_{x} {\bm h}_{x}
=  \frac{1}{2N}  {\bf 1} + (\kappa/2) {\bm h}_{x}
$
with $\kappa :=\frac{2(2-N)}{\sqrt{2N(N-1)}}$.
By comparing the right-hand sides of two equations,
it turns out that (a') is satisfied when
\begin{equation}
\alpha=\beta,
\quad
1+(\kappa/2) \alpha=\gamma/(2N).
\label{eq:parameter-eq1}
\end{equation}

Moreover, we calculate
\begin{align}
{\bm h_x}{\tilde V}_{x,\mu}U_{x,\mu}^\dagger
 &=\frac\alpha{2N}\bm1
   +\left(1+\frac\kappa2\alpha\right){\bm h}_x
   +\frac\gamma{2N}U_{x,\mu}{\bm h}_{x+\mu}U_{x,\mu}^\dagger
   \nonumber\\
 &\qquad
   +\left(\beta+\frac\kappa2\gamma\right){\bm h}_x U_{x,\mu}{\bm h}_{x+\mu}U_{x,\mu}^\dagger
    .
\end{align}
Therefore,  (b')
  yields
\begin{equation}
 {\rm Tr}({\bm h}_x {\tilde V}_{x,\mu}U_{x,\mu}^\dagger)
= \alpha+ (\kappa/4) \gamma=0
 ,
\label{eq:parameter-eq2}
\end{equation}
where we have used $\alpha=\beta$,
${\rm Tr}({\bf 1})=N$,
${\rm Tr}({\bm h}_x)=0$,
and ${\rm Tr}({\bm h_x}U_{x,\mu}{\bm h}_{x+\mu}U_{x,\mu}^\dagger)=1/2$.
Solving (\ref{eq:parameter-eq1})
 and (\ref{eq:parameter-eq2})
 simultaneously, we obtain
\begin{equation}
\alpha=\beta=\frac{N-2}{N^2-2N+2}\sqrt{2N(N-1)},
\quad
\gamma=\frac{4N(N-1)}{N^2-2N+2}
 .
\label{eq:parameters}
\end{equation}
In fact, the ansatz (\ref{ansatzmini}) for ${\tilde V}_{x,\mu}$ correctly reproduces the continuum counterpart $\mathscr{V}_\mu(x)$ in the naive continuum limit.
\begin{align}
 {\tilde V}_{x,\mu}
 =& \left(1+\frac\gamma{2N}\right)\bm1
   -i\epsilon g[{\mathscr A_{x,\mu}}+\alpha\{{\bm h_x},{\mathscr A_{x,\mu}}\}+\gamma{\bm h_x}{\mathscr A_{x,\mu}}{\bm h_x}]
   +\epsilon[\alpha\partial_\mu{\bm h_x}+\gamma{\bm h_x}\partial_\mu{\bm h_x}]
+O(\epsilon^2)
\nonumber\\
 =&  \left(1+\frac\gamma{2N}\right) [\bm1-i \epsilon g \mathscr{V}_\mu(x) +O(\epsilon^2) ]
  .
\end{align}

For $SU(2)$, $\alpha=\beta=0$ and the linear terms in ${\bm h}$ vanish and (\ref{ansatzmini}) reduces to the same form (\ref{SU2-sol}) as the maximal case $\gamma=4$.  This is reasonable, since there is no distinction between maximal and minimal for $SU(2)$.

For $SU(3)$, the parameters are decided as
\begin{equation}
\alpha=(2/5) \sqrt3,
\quad
\gamma=24/5
 ,
\end{equation}
which leads to the link variable (up to the normalization):
\begin{equation}
{\tilde V}_{x,\mu}=U_{x,\mu}+\frac25\sqrt3({\bm h_x}U_{x,\mu}+U_{x,\mu}\bm h_{x+\mu})+\frac{24}5{\bm h_x}U_{x,\mu}\bm h_{x+\mu}
 .
\end{equation}
In particular, the large $N$ limit $N \rightarrow \infty$ is taken in the minimal case:
\begin{equation}
{\tilde V}_{x,\mu}=U_{x,\mu}+ \sqrt2({\bm h_x}U_{x,\mu}+U_{x,\mu}\bm h_{x+\mu})+4{\bm h_x}U_{x,\mu}\bm h_{x+\mu}
 .
 \label{tildeV2}
\end{equation}
The unitary link variable $V_{x,\mu}$ is obtained from ${\tilde V}_{x,\mu}$ (\ref{tildeV2}) following  (\ref{polar}) and (\ref{speciality}).

\section{Construction of the color field}

Now we discuss how to obtain the desired color field $\bm{n}_x$ from the original lattice Yang-Mills theory with a gauge group $G=SU(N)$ written in terms of link variables $U_{x,\mu}$.
Such a prescription has already been given in the continuum reformulation of Yang-Mills theory \cite{KMS06,KSM08}.
Here we present the lattice version.

We can introduce  a single color field  $\bm{n}_x$  according to $\bm{n}_x=U_x^\dagger T U_x$ ($U_x \in G$) for a certain diagonal matrix $T$ \cite{KSM08} in addition to link variables $U_{x,\mu}$. Then we have an extended gauge theory,
called the master Yang-Mills theory in which fields  transform under the enlarged gauge transformation:
the color field transforms as
\begin{align}
  {\bm n}_{x} \rightarrow \Theta_{x} {\bm n}_{x} \Theta_{x}^\dagger = {\bm n}_{x}' ,  \quad
 \Theta_{x} \in G/\tilde{H}
\label{n-transf2}
 ,
\end{align}
when the link variable $U_{x,\mu}$ obeys the well-known lattice gauge transformation:
\begin{equation}
  U_{x,\mu}  \rightarrow \Omega_{x} U_{x,\mu} \Omega_{x+\mu}^\dagger = U_{x,\mu}^\prime
  , \quad \Omega_{x} \in G
  \label{U-transf2}
 .
\end{equation}
Therefore, the master Yang-Mills theory has the enlarged gauge symmetry $\tilde{G}:=G_{\Omega} \times (G/\tilde{H})_{\Theta}$.
Here $G/\tilde{H}$ denotes the target space of the color field: $\bm{n}_x \in G/\tilde{H}$ and the subgroup $\tilde{H}$ depends on the chosen matrix $T$.
The maximal case  corresponds to $\tilde{H}=U(1)^{N-1}$, while the minimal case  to $\tilde{H}=U(N-1)$.
For $SU(2)$, in particular, a single color field is unique, and $\tilde H$ agrees with the maximal torus subgroup: $\tilde H=H=U(1)$ and $\bm n_x \in SU(2)/U(1)$.

However, the master Yang-Mills theory has more degrees of freedom than the original Yang-Mills theory by construction.
We impose a constraint called the reduction condition (mimicking the continuum version \cite{KMS06,KSM08}) to reduce
$\tilde{G}:=G_{\Omega} \times (G/\tilde{H})_{\Theta}$ in the master Yang-Mills theory to the subgroup: $G^\prime=SU(N)_{\Omega^\prime}$:
\begin{equation}
 G=SU(N)^{\rm local}_{\Omega}
 \nearrow
\tilde{G}:=SU(N)^{\rm local}_{\Omega} \times [SU(N)/\tilde{H}]^{\rm local}_{\Theta}
 \searrow
G^\prime :=SU(N)^{\rm local}_{\Omega^\prime}
 ,
\end{equation}
so that the resulting theory has the same local gauge symmetry $G'$ (equipollent gauge symmetry) as the original Yang-Mills theory.
Thus we have a new description of Yang-Mills theory for a choice of $T$.
See Fig.~\ref{fig:MYM}.

This procedure should be compared with the conventional MAG. The conventional MAG  is the partial gauge fixing $G^{\rm local} \rightarrow H^{\rm local}$ which breaks the local gauge symmetry $G^{\rm local}$ to  a maximal torus group $H^{\rm local}=U(1)^{N-1}$  and leaves both local $H^{\rm local}=U(1)^{N-1}$ and  global $H^{\rm global}=U(1)^{N-1}$ unbroken, but breaks the global $SU(N)^{\rm global}$, i.e., color symmetry.
The reduction condition leaves both local $G'^{\rm local}=SU(N)'$ and global $SU(N)'^{\rm global}$ unbroken (color rotation invariant).
In order to fix the local gauge symmetry $G'^{\rm local}$, we can impose an overall gauge fixing condition (e.g., Landau gauge) with the color symmetry being preserved.
This is an advantage of the new procedure.

\begin{figure}[ptb]
\begin{center}
\includegraphics[height=2.9in, angle=-90]{maximal.eps}
\includegraphics[height=2.9in, angle=-90]{minimal.eps}
\end{center}
\caption{The original Yang-Mills (YM) theory with a gauge group $G_{\Omega}$ is first extended to the master Yang-Mills theory (M-YM) with the enlarged gauge symmetry $\tilde{G}:=G_{\Omega} \times (G/\tilde{H})_{\Theta}$ and subsequently reduced by imposing reduction conditions to the equipollent Yang-Mills (YM') theory with $G^\prime_{\Omega^\prime}$. The overall gauge fixing condition e.g., Lattice Landau gauge (LLG) can be imposed without breaking color symmetry.
(Left panel) SU(3) maximal case. (Right panel) SU(3) minimal case.
}%
\label{fig:MYM}%
\end{figure}

In numerical simulations,
 the configurations of $SU(N)$ link variables
$\{ U_{x,\mu} \}$ are generated using the standard procedure for the  Wilson action \cite{CM82}.
For $SU(3)$
the distinction between the maximal and minimal cases is attributed to the method of constructing the color field by imposing appropriate reduction conditions given below.

In the maximal case, the variables $\bm{n}^j_{x}$ are constructed as follows.
Consider  two functionals $F_{\rm rc:max}$ and $F_{LLG}$ written in terms of   $U_{x,\mu}$ and $\bm{n}^j_{x}$:
\begin{align}
 F_{\rm rc:max}[U,\bm{n};\Omega,\Theta]
 :=& \epsilon^D \sum_{x,\mu,j} \frac12 {\rm tr}\left[ (D_\mu^{(\epsilon)}[{}^{\Omega}U]{}^{\Theta}\bm{n}^j_x)^\dagger (D_\mu^{(\epsilon)}[{}^{\Omega}U]{}^{\Theta}\bm{n}^j_x)\right]
\nonumber\\
=& - \sum_{x,\mu,j} \epsilon^{D-2}  {\rm tr}
  \left[  {}^{\Theta}\bm{n}_{x}^{j} {}^{\Omega}U_{x,\mu} {}^{\Theta}\bm{n}_{x+\mu}^{j}  {}^{\Omega}U_{x,\mu}^\dagger -\frac{N-1}{2}    \right]
   ,
\label{Lattice-nMAGf}
\\
F_{LLG}[U;\Omega]
:=& - \epsilon^{D}  \sum_{x,\mu}\mathrm{tr}( {}^{\Omega} U_{x,\mu})
 .
\end{align}
Then we  minimize simultaneously the two functionals $F_{\rm rc:max}$ and $F_{LLG}$ with respect to two gauge transformations:
${}^{\Omega}{}U_{x,\mu}:=\Omega_{x}U_{x,\mu}\Omega_{x+\mu}^{\dagger}$ for the
link variable $U_{x,\mu}$ and
${}^{\Theta}\bm{n}^j_{x}:=\Theta_{x} \bm{n}_{x}^{j(0)}\Theta_{x}^{\dagger}$ for a given initial site variable
$\bm{n}^j_{x}{}^{(0)}$ (we can choose the initial value $\bm{n}^j_{x}{}^{(0)}=H_{j}$)
where  gauge group elements $\Omega_{x}$ and $\Theta_{x}$
are {\it independent} $SU(N)$ matrices on a site $x$.
Then we can determine  the configurations  ${}^{\Theta^{*}}\bm{n}^j_{x}$ and ${}^{\Omega^{*}}U_{x,\mu}$ realizing simultaneously the minimum of two functionals, just as in the $SU(2)$ case \cite{IKKMSS06}.

It is easy to show that the continuum limit of the lattice functional (\ref{Lattice-nMAGf}) reduces to the continuum one in \cite{KSM08}:
Expanding the link variable in power series of $\epsilon$, we obtain
\begin{align}
 F_{\rm rc:max}[U, \bm{n}]
  =&  \epsilon^D  \sum_{x,\mu,j} \frac12 {\rm tr} \{ (D_\mu[\mathscr{A}] \bm{n}_{x}^{j} )^2 \}
  + O(\epsilon^2)
 ,
\end{align}
where  we have used
$
  \bm{n}_{x+\mu}^{j} = \bm{n}_{x}^{j} + \epsilon \partial_\mu \bm{n}_{x}^j + O(\epsilon^2) ,
$

and
$
 {\bm n}^j_{x} {\bm n}^j_{x}
 = \frac{N-1}{2N} {\bf 1}
$.
Indeed, this result reproduces the continuum functional in the maximal case\cite{KSM08}:
$
R= \int d^Dx \frac12 (D_\mu[\mathscr{A}]\bm{n}^j(x))^2
  .
$
There is the other choice for the $F_{\rm rc:max}[U, \bm{n}]$, see  \cite{KSM08} .

The minimal case is achieved by minimizing two functionals simultaneously:
\begin{align}
 F_{\rm rc:min}[U,\bm{h};\Omega,\Theta]
 :=& \epsilon^D \sum_{x,\mu,j} \frac12 {\rm tr}\left[ (D_\mu^{(\epsilon)}[{}^{\Omega}U]{}^{\Theta}\bm{h}_x)^\dagger (D_\mu^{(\epsilon)}[{}^{\Omega}U]{}^{\Theta}\bm{h}_x)\right]
 \nonumber\\
  =& - \sum_{x,\mu} \epsilon^{D-2} {\rm tr}
 \left[  {}^{\Theta}\bm{h}_{x}  {}^{\Omega}U_{x,\mu} {}^{\Theta}\bm{h}_{x+\mu}  {}^{\Omega}U_{x,\mu}^\dagger  -\frac{1}{2}  \right]
 ,
\\
F_{LLG}[U;\Omega]
:=& - \epsilon^{D-4} \sum_{x,\mu}\mathrm{tr}({}^{\Omega} U_{x,\mu}) .
\end{align}
The naive continuum limit of this functional is calculated as in the maximal case:
\begin{align}
   F_{\rm rc:min}[U, \bm{h}]
  =&  \epsilon^{D}  \sum_{x,\mu}   \left[  \frac12 \epsilon^2 {\rm tr} \{ (D_\mu[\mathscr{A}] \bm{h}_{x} )^2 \}
\right]
  + O(\epsilon^2)
   ,
\end{align}
where we have used
$
 H_r H_r
 = \frac{1}{2N(N-1)} {\rm diag}(1,\cdots,1,(N-1)^2)
$
and
$
  \bm{h}_{x+\mu}  = \bm{h}_{x}  + \epsilon \partial_\mu \bm{h}_{x}  + O(\epsilon^2)
$.
Indeed, this result reproduces the continuum functional in the minimal case \cite{KSM08}:
$
 F = \int d^Dx \frac12 (D_\mu[\mathscr{A}]\bm{h}(x))^2
$
 apart from a constant term.
More practical procedures necessary in numerical simulations will be given in subsequent papers.

\section{Conclusion and discussion}

In this Letter we have given new descriptions for lattice Yang-Mills theory in terms of new variables.
We expect that they are quite useful to give a better understanding of dual superconductivity and to clarify the true mechanism for quark confinement in $SU(N)$ Yang-Mills theory.
In fact, some applications of this framework were already presented in the continuum formulation \cite{Kondo08,KS08,Kondo08b}.
The results of relevant numerical simulations on a lattice will be reported soon, although  preliminary results were given in  \cite{SKKMSI07b} for $SU(3)$ maximal choice.

We have presented a set of defining equations for specifying the decomposition of lattice variables to obtain new lattice variables. However, it is not the unique way for the decomposition.
The first defining equation for $V_{x,\mu}$ has the intrinsic meaning even on a lattice without referring to the continuum limit, since it implies that the color field $\bm n_x$ at a site $x$ can be moved to the next site $x+\mu$ without being affected by the background field expressed as the link variable $V_{x,\mu}$.
On the other hand, the second defining equation (the extra condition) for $X_{x,\mu}$  will not be the best form, although it is enough to reproduce the continuum form expressed in the Lie algebra,
${\rm tr}(\bm{n}(x) \mathscr{X}_\mu(x))=0$.
This issue will be discussed in a separate paper.

redpajama_set_name: RedPajamaArXiv


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text:
My Autobiography (eBOOK)
Rating:4.11 (Votes: 27)
Author:Rory Best
Publisher:Hodder & Stoughton
Categories:Biography, Sports Books, True Stories
Overview:My Autobiography - 'Extraordinary... great fun' - Barry Egan, Irish Sunday Independent''Throughout it all though there is a feeling of warmth for the sport and for others. Above all there is a sense of achievement... Best was never one of the glamour boys, but he deserves star billing.' - Daily Telegraph Rory Best is widely-regarded as one of Ireland's greatest ever captains. Entrusted by Joe Schmidt to lead the side that looked on the wane following the 2015 World Cup, Best's inspirational leadership skills and abrasive qualities proved to be the foundation stones for the most successful period in Ireland's history.His first year in charge saw Ireland complete a hat-trick of victories against the southern hemisphere 'Big Three', including leading his side to a first ever victory over world champions New Zealand in Chicago, a feat that etched Best's place in Irish sporting folklore and ended the All Blacks' record-winning streak of 18 Test victories.Ireland's annus mirabilis under Best's captaincy would come in 2018 however, when he led the side to only their third Grand Slam title, culminating with a famous victory over England at Twickenham, and a record-breaking run of 12 successive Test victories.When he steps down as Ireland captain at the age of 37 following the World Cup in Japan, his fourth tournament, history will no doubt also judge Best to be one of their greatest forwards.A hugely-popular figure across the game, Best will finish his career as Ireland's most capped forward, behind only Brian O'Driscoll and Ronan O'Gara in the all-time records, and also made over 200 appearances for his province Ulster.
This book is on page /3574b176dc725218de3835184742780c/book/1603725640-9781529362404. It was written by the following authors: Rory Best. Book My Autobiography, which can be read online, published by the company: Hodder & Stoughton. Other books on similar topics can be found in sections: Biography, Sports Books, True Stories. The book was published on 2020-03-05 00:00:00. It has 368 pages and is published in Hardback format and weight 620 g. File for download My Autobiography has PDF format and is called my-autobiography.pdf. Other books you can download below. Our koopbitcoin.eu site is not responsible for the content of PDF files.
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Autobiography and Authorship in Renaissance Verse: Chronicles of the Self - Early Modern Literature in History
Autobiography, Letters and Literary Remains of Mrs. Piozzi (Thrale) (2nd ed.) (2 vols.) Edited with notes and Introductory Account of her life and writings
Autobiography of a Werewolf Hunter Trilogy: Autobiography of a Werewolf Hunter, Heart of Scars, The Lineage - Autobiography of a Werewolf Hunter
Autobiography in Verse
Autobiography of James L. Smith: Including Also, Reminiscences of Slave Life, Recollections of the War, Education of Freedmen, Causes of the Exodus, Etc - War College Series
Autobiography of a Nobody
Autobiography of Santa Claus T
Autobiography of a Yogi: Reprint of the Philosophical Library 1946 First Edition
Autobiography of a Yogi (Latvian)
Autobiography of a Restless Mind: Reflections on the Human Condition Volume 1
Autobiography Of Seventy Years (Volume II): In Two Volumes, Vol. II.
Autobiography Of Mikl+S Bethlen
Autobiography of Mark Twain, Volume 3: The Complete and Authoritative Edition - Mark Twain Papers 12
Autobiography of an Also-ran
Autobiography of a Yogi (Croatian)
Autobiography of a Country Parson: The Donegal Man (1814-96), Author of 'The Dill Worthies', Who Was the First Presbyterian Minister of Dromore, County Tyrone - Ascona S.
Autobiography of a Vampire
Autobiography Writing for Seniors
Autobiography of John Malvin
Autobiography of Death
Autobiography of Samuel D. Gross M.D. V1: With Reminiscences of His Times and Contemporaries
Autobiography Of Adin Ballou, 1803-1890 - Containing An Elaborate Record And Narrative Of His Life From Infancy To Old Age
Autobiography of Swami Sivananda
Autobiography - A Friend in the Library - Volume XII - A Practical Guide to the Writings of Ralph Waldo Emerson, Nathaniel Hawthorne, Henry Wadsworth Longfellow, James Russell Lowell, John Greenleaf Whittier, Oliver Wendell Holmes - In Twelve Volumes
Autobiography of a Female Slave
Autobiography Keller Paul Madere in Words and Pictures
Autobiography of the Lower East Side: A Novel in Short Stories
Autobiography: Truth and Fiction Relating to My Life (Dodo Press)
Autobiography and the Construction of Identity and Community in the Middle East
Autobiography of Rev. James B. Finley: Or, Pioneer Life in the West - American Philosophy and Religion
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redpajama_set_name: RedPajamaCommonCrawl


Kathleen Hoye, who grew up in Louisville, KY, began her songwriting career in Northern Ireland , where she was strongly influenced by celtic music and Irish urban culture. She returned to the U.S. and shared the stage with relevant artists such Leon Russel, Gillian Welch, Patti Loveless, Patti Griffin, and Iris DeMent.
Her music, which has been alternately described as both vibrant and somber, has taken on many faces, influences, textures, and moods over the years.
Her first album, Delilah, is "a point on a broad continuum between ambient mood music, white noise, and all-out alternative rock abandon." The lyrics are impressionistic rather than narrative, relying on archetypal imagery and the logic of dreams to explore themes of artistic struggle, addiction, release, spirit, and the solace of time.
Her second EP project, Unbridled, is a groovy, relaxed ride through songs that "will kill you pretty" and that "linger like a sweet kiss goodbye" according to one music critic. Here the songs are about love, horses, and the value of spending time alone.
Although we have known Kathleen for many years, this will be her first visit to Little Wing Hollow. We cant wait to hear her play her beautiful music on the hill.


ROCK HILL, S.C. (TheSAC.com) -- Catawba's Darius Turner was named the South Atlantic Conference AstroTurf Men's Outdoor Track Athlete of the Week, while Coker's Alex Berry was named AstroTurf Men's Outdoor Field Athlete of the Week for meets contested March 18 through March 24. It was announced by the conference office on Tuesday.
Turner, a freshman from Burlington, N.C., posted the fourth best time in NCAA Division II in the 100-meter dash at 10.48 seconds. His time was an NCAA provisional mark and it earned him second place honors at the High Point VertKlasse meet last week. In addition to his second place finish in the 100m, Turner anchored the 4x100 relay team that finished second with a time of 41.20 seconds.
Berry, a sophomore from Latta, S.C., won both the high jump and the triple jump at last week's CIU Invitational. His high jump mark of 1.87 meters was a personal best and ranks fifth all-time at Coker, while his leap of 12.82 meters in the triple jump was also a new person best.
AstroTurf is an official corporate partner of the South Atlantic Conference and title sponsor of the Conference's Player of the Week awards program.


Is your child going to bed at the 'right' time?
What time should you put your children to bed each night?
It's not a question with an easy answer - every child is different and, depending what he or she has been doing during the day, the 'right' time for a child to sleep can be something of a moving target.
However, the Canadian parenting blog Mommy Matters recently produced this list of suggested bedtimes for children aged from two to 18.
The list has been a viral hit, appearing on numerous Facebook groups and sparking conversation among parents over what they consider to be the appropriate time for their child to hit the sack.
We'd like to hear what you think. Does this list of bedtimes look about right to you, or wildly off target?
In the comments below, tell us what time is bedtime for your children.


Cutest Lord of the Sith ever!
Yesterday, Mache went to a Super Hero party. Yes, we dressed our little beauty as the Lord of the Sith and took her to attend a Super Hero party 🙂 Our daughter, it seems, took her role seriousy though and the method actress she is, she thought she'd work at sound like Darth Vadar as well.
Over the past week Mache's nose became quite the spout, but Friday night she caught a cough. We played it off, as we would with ourselves – nasal drip, trying to clear up what's in her system – but as overnight it got worse. Her temperature started to rise and the nasal drip got really bad. She wouldn't lay down for more than 15 minutes, unable to clear her throat. She wanted to be cuddeled. She wanted to sleep but couldn't get comfortable.
We finally took her to bed with us. I let her sleep on me and we got a solid hour – but then – then it happened. She coughed her usual cough of the day; it hurt and she cried. At least, she tried to cry. A half gasp of air followed by a rasp cough, a second of silence, and a wheeze as she tried to breath again. The pattern continued.
During the day we talked about doctors but we both thought – maybe hoped – that she'd last until Monday and even be better, but certainly stay well enough to avoid the emergency room. Yet, here we found ourselves. Not quite 5am, and a 14 month old who could barely breath.
To E.R. (Oh look! It's a verb now!) is human; at least, that's what I told myself as I started to tear up. Most kids find their way to the emergency room once or twice – broken legs, alergic reactions, great falls, and over reactive parents – that last one a line I have no doubt we come close to. Here I was; 5am and I'm telling my wife it's time to take our baby girl to the emergency room. The place where sickly and injured go. Our baby was sickly. Maybe we're over reacting but it's past the time of caution. We need to act. For what it's worth, I'm a giant baby about these sorts of things!
The story is short, thankfully. After an hour in the hospital everything looked good: Mache's sinuses drained on their own and she was smiles and giggles again; her fever wasn't too bad, hovering around 100 degrees; her lungs are clear, which I couldn't be more thankful for. The beginnings of a middle ear infection and a viral cold. Antibiotics for the infection, motrin for the fever, fluids and rest. She still cries a little when she coughts but it only lasts a few seconds before she gets distracted by something good.
My wife is great! She kept calm, asked questions, and knew everything we needed to know about her medical history for the doctor and nurses. My little method actress no longer sounds like Darth Vader and is napping, no doubt dreaming about something a long time ago in a galaxy far, far away.
Your description of how she took her job seriously as a Sith Lord is uncanny – love it!
Thanks Dan! She really was the cutest damn Vader ever!
The first urgent care or ER visit is always the hardest and kids ALWAYS get super sick the minute doctors offices are closed. I'm glad she's getting better. Good thing it wasn't croup – nasty nasty cough.


A vaporizer pipe is not really a real dry herb vaporizer pipe. There is no pipe shaped vaporizer that you can just pack your dry herbs, press a button and start vaping them. However there are other types of vaporizers that can come off or passed as a vaporizer pipe. The Vapor Genie Pipe Vaporizer and Click n Vape Torch Butane Vaporizer are the two that automatically come to mind. One of definitely shaped like a pipe but need a lighter to be used and the other is a straight pipe filled with butane and your favorite herbs. Ironically, they both will probably need butane to work and no batteries are needed.
The Vapor Genie Pipe Vaporizer is a pipe looking piece that lets you pack a bowl and use a lighter to carefully vaporize the herbs. It has a two piece design where the upper bowl is a flame filter which helps it vaporize as it is the main reason why it vaporizes. When you unscrew it you can pack your bowl of dry herbs. The best part we like this over a vaporizer is only because running out of batteries is impossible. Just go to a liquor store and purchase another lighter. The main disadvantage to the vapor genie vaporizer pipe is that you can't see your herbal bowl so you have to keep unscrewing the chamber to check it. Another downside is that you have no temperature control like you do with other vaporizers that are a couple bucks more like the G Pro Vaporizer. Want to know the major advantage of it? It has a lifetime warranty on it even with a small price point.
So how the Vapor Genie Pipe Vaporizer works is very simple but can take some time to get used too. Make sure your herbs are ground up and pack them into your herbal chamber. Allow the flame to go into the top bowl while you SLOWLY inhale from the mouthpiece. Make sure the flame from the lighter never touches the silicon carbide disc which sits in the top bowl. The disc dissipates heat which if it hits the disc, it will reduce the effectiveness. If you keep having the flame touch the disc over and over, it will clog the filter. It is also recommended not to use a butane lighter and use an electric torch.
The Click n Vape Torch Butane Vaporizer is a portable dry herb vaporizer that looks like a pipe. Matter of fact when you unscrew the middle part, you realize it's something Mario and Luigi might use for plumbing. The Click n Vape starts the vaporization process with a click of a button like many of our other portable vapes. The difference with this one is it requires butane and no batteries. Many people question the butane part and wonder if it's safe. With vapes like this, yes it is because the inner flame kills all butane before it gets through the mouthpiece. You will not even taste or smell it. It even has 4 holes for the butane to escape through as a last resort.
The Click n Vape is another easy vaporizer to use. First fill up the bottom with your butane. Make sure to put the nozzle of the butane directly into the copper middle piece of the Click n Vape. Let it soak in for about a minute then go ahead and click the button to light it. Make sure the safety isn't in the off position so the button easily clicks down. You will see a pipe that screws open and becomes two. Leave the bottom one screwed into the portable dry herb vaporizer and open the top one. The mouthpiece should be attached to the bottom chamber on top which is now exposed. This is your herbal chamber which will have a mesh screen on the bottom. This mesh screen is where your ground up herbs will go. Make sure it's not too fine or else it will seep through. Keeping your click n vape right side up at all times, close the top chamber and light away.
Vaporizer pipes are not very common in vaporizers but they do make for some great vapes. A vaporizer isn't necessary an electronic gadget that vaporizes your dry herbs, concentrates or vape juices but rather one that vaporizes it efficiently. Vaporizers are here to stay and they are only getting better.


text:
Q: SVC Classifier support vector classes in python Sklearn How can I find out which support vectors belong to which class in sklearn SVC?
model = clf.fit(X,y)

vectors = model.support_vectors_

Which vector belongs to which decision boundary?

A: You can use the SVC.support_ attribute. The support_ attribute provides the index of the training data for each of the support vectors in SVC.support_vectors_. You can retrieve the classes for each support vector as follows (given your example):
X[model.support_]

A more complete example:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.colors import ListedColormap
from sklearn.model_selection import train_test_split
from sklearn.preprocessing import StandardScaler
from sklearn.datasets import make_classification
from sklearn.svm import SVC

svc = SVC(kernel='linear', C=0.025)
X, y = make_classification(n_samples=500, n_features=2, n_redundant=0, n_informative=2, random_state=1, n_clusters_per_class=1)
rng = np.random.RandomState(2)
X += 2 * rng.uniform(size=X.shape)
X = StandardScaler().fit_transform(X)
X_tr, X_te, y_tr, y_te = train_test_split(X, y, test_size=.4, random_state=42)
cm_bright = ListedColormap(['#FF0000', '#0000FF'])
fig, ax = plt.subplots(figsize=(18,12))
ax.scatter(X_tr[:, 0], X_tr[:, 1], c=y_tr, cmap=cm_bright)
svc.fit(X_tr, y_tr)
y_tr[svc.support_]
array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1])
fig2, ax2 = plt.subplots(figsize=(18,12))
ax2.scatter(X_tr[:, 0], X_tr[:, 1], c=y_tr, cmap=cm_bright)
ax2.scatter(svc.support_vectors_[:, 0], svc.support_vectors_[:, 1])
fig3, ax3 = plt.subplots(figsize=(18,12))
ax3.scatter(X_tr[:, 0], X_tr[:, 1], c=y_tr, cmap=cm_bright)
ax3.scatter(svc.support_vectors_[:, 0], svc.support_vectors_[:, 1], c=y_tr[svc.support_], cmap=cm_bright)

redpajama_set_name: RedPajamaStackExchange


text:
THE LIFESPAN OF A FACT TO CHANGE WEEKLY PERFORMANCE SCHEDULE BEGINNING THURSDAY, OCTOBER 25
A New Play by Jeremy Kareken & David Murrell
And Gordon Farrell
Based on the Essay/Book by John D'Agata & Jim Fingal
Directed by Tony Award nominee Leigh Silverman
Strictly Limited 16-Week Engagement
Currently Running on Broadway
"A smart, rib-bustingly funny play!"
– Wall Street Journal
"Terrifically engaging & fascinating! It's what you call a good time."
"Daniel Radcliffe is perfection, Cherry Jones is terrific, and Bobby Cannavale is compellingly dynamic."
– NY1
(New York, NY) – The world premiere play, THE LIFESPAN OF A FACT, has announced a new performance schedule on Broadway. Beginning Thursday, October 25, 2018, THE LIFESPAN OF A FACT will play Monday, Friday, and Saturday at 8pm, Tuesday, Thursday, and Sunday at 7pm, and Saturday and Sunday at 2pm at Studio 54 (254 West 54th Street). The production opened on Broadway on October 18 to critical acclaim.
The world premiere production stars Daniel Radcliffe (Privacy, "Harry Potter"), Cherry Jones (The Glass Menagerie, Doubt, "24"), and Bobby Cannavale (The Mother with the Hat, "I, Tonya"). Written by Jeremy Kareken & David Murrell and Gordon Farrell, the play is directed by Tony nominee Leigh Silverman.
Based on the book written by John D'Agata and Jim Fingal, THE LIFESPAN OF A FACT is playing a limited engagement through Sunday, January 13, 2019.
The determined young fact checker (Daniel Radcliffe) is about to stir up trouble. The demanding editor (Cherry Jones) has given him a big assignment: apply his skill to a groundbreaking piece by the unorthodox author (Bobby Cannavale). Together, they take on the high-stakes world of publishing in this new comedy of conflict. The ultimate showdown between fact and fiction is about to begin – with undeniably delicious consequences.
The creative team includes Mimi Lien (Scenic Design), Linda Cho (Costume Design), Jen Schriever (Lighting Design), Palmer Hefferan (Original Music & Sound Design) and Lucy Mackinnon (Projection Design).
THE LIFESPAN OF A FACT is produced by Jeffrey Richards, Norman and Deanna Twain, Will Trice, Barbara H. Freitag, Suzanne Grant, Gold/Ross Productions, Jamie deRoy, Jennifer Manocherian, Barbara Manocherian, ManGol Productions, Carl Moellenberg/Wendy Federman, Ken Greiner, Van Kaplan, Dominick LaRuffa Jr., Marc David Levine, WitzEnd Productions, Eric Falkenstein/Moreland Mott, Caiola Productions, Remmel T. Dickinson, Jayne Baron Sherman, and Jacob Soroken Porter.
TICKET INFORMATION: Tickets range from $59 - $169. For more information, please visit www.LifespanOfaFact.com or www.Telecharge.com (212.239.6200).
Beginning Monday, December 17, 2018 through Wednesday, January 2, 2019, THE LIFESPAN OF A FACT will play Wednesday at 2pm and 8pm, Thursday at 7pm, Friday at 8pm, Saturday at 2pm and 8pm, Sunday at 2pm and 7pm.
Please note the following exceptions to the performance calendar:
There will be a 2pm and 8pm performance on Wednesday, November 21.
There will be no performance on Thursday, November 22 in observance of Thanksgiving.
There will be a 2pm performance on Thursday, December 27.
There will be no matinee performance on Wednesday, January 2, 2019.
Though it is being performed at Studio 54, this production of THE LIFESPAN OF A FACT is not a Roundabout Theatre Company production.
www.LifespanOfaFact.com
Facebook: The Lifespan of a Fact
Twitter: @LifespanOfaFact
Instagram: @LifespanOfaFact


WASHINGTON (May 30, 2011) – Congressman Raul Grijalva, D-AZ, submitted a Resolution in the U. S. House of Representatives recognizing the birthday of the Pryor Mountain wild stallion, Cloud—for his role in enhancing the appreciation of all wild horses and burros in the American West.
Documented from the day of his birth, May 29, 1995, by filmmaker Ginger Kathrens, Cloud puts a face on wild horses living in the West. Kathrens' films for PBS's Nature <http://www.pbs.org/wnet/nature/episodes/cloud-challenge-of-the-stallions/introduction/936/> series focus on the challenges of growing up wild and the complexity and occasional brutality of wild horse society. Through Cloud we learn about wild horse behavior, family dynamics, and the courage it takes for even a bold young stallion to win mares.
Even though Cloud is beloved around the world, the Bureau of Land Management Billings Field Office was quoted in an October 2010 Billings Gazette article, stating they could "no longer protect Cloud and his family." This appears to be a thinly veiled threat to remove the wild stallion and his progeny from their wilderness home in the Pryor Mountains.
"Thousands of wild horses and burros are slated to lose their freedom this summer," states Kathrens who also serves as Executive Director of the Cloud Foundation, a non-profit organization dedicating to wild horse and burro preservation. "Most will be warehoused for life at taxpayer expense. Meanwhile, millions of head of livestock continue to monopolize the same public lands also at taxpayer expense." Livestock permits on public lands loses an average of $123 million <http://www.sagebrushsea.org/pdf/factsheet_Grazing_Fiscal_Costs.pdf> per year for administrative costs alone.
Other pressures are also on the rise, as well. Secretary of Interior, Ken Salazar, is fast-tracking permits for oil and gas drilling as well as other extractive uses of public lands. Many of these projects will destroy pristine wilderness lands which serves as vital habitat for wild horses and burros and other endangered wildlife species.


"In spite of difference of soil and climate, of language and manners, of laws and customs, in spite of things silently gone out of mind and things violently destroyed, the Poet binds together by passion and knowledge the vast empire of human society, as it is spread over the whole earth, and over all time. The objects of the Poet's thoughts are every where; though the eyes and senses of man are, it is true, his favorite guides, yet he will follow wheresoever he can find an atmosphere of sensation in which to move his wings. Poetry is the first and last of all knowledge--it is as immortal as the heart of man."
--William Wordsworth, "Preface to Lyrical Ballads "
ALL RIGHTS RESERVED. This material is protected under International and Federal Copyright Laws and Treaties. Any unauthorized reprint or use of this material is prohibited. No part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system without express written permission from the author.
Google + is closing soon..so I come back to my roots..


text:
March 20, 2016 by muaythaibloguk	in Home, My Muay Thai Journey and tagged jordan watson, liam harrison, muay thai, thai boxing, world champion, yokkao
So last night there were two new world champions crowned! Last night Facebook was a buzz with fighters and spectators alike all sharing their predictions and thoughts. Yokkao is renowned for its consistency is putting out fantastic shows. Yokkao 17 & 18 was no different.
I wasn't there so I can't pass judgement on what happened with any of the fights but the news of the two main events of the show have been circulating my news feed all day! First of all, fighting for the 65kg YOKKAO world title, was Liam Harrison from the world famous Bad Company gym in Leeds and his very worth opponent was Tetsuya Yamato from Japan. Tetsuya recently fought Pakorn who actually beat Liam last time they met and won! So everything was on the line for this upcoming encounter.
The title was only put on the line last week as Pakorn's last three fights were lost which means he was made, by YOKKAO rules, to vacate his world title. This left the belt up for grabs by these two warriors! Through reading Facebook, I have heard it was a close bout, some say that either fighter could have took it but Liam deserved the win. He mentioned in his post-fight video that he adapted his game plan, rather than hurting them with leg kicks, he would kick and move, kick and move. Which in turn got him the win over the Jap!
I can imagine it being a very proud moment for Liam, this was his sixth world title win and one he most definitely deserved!
Who do you think he will have to defend it against next?
The main event for YOKKAO 18 was not only a revenge bout but one that held a lot of substance for Jordan Watson. With the last encounter that Jordan had with the Swede even he can admit, Sanny made an example of him really. He gassed early and in turn, was over turned but Dahlbeck.
Since the loss, the rematch was wanted by all the fans (and Jordan and Sanny) and this time round the fighters definitely didn't disappoint. As I said before, I wasn't there to watch it but I can imagine it was all kinds of a war! I can't wait to see the videos that YOKKAO will no doubt be posting very soon for all of our viewing pleasure!
On the run up to the fight Jordan recently went out to Thailand to train for the fight at the new YOKKAO Training centre in Bangkok to train with Saenchai and Singdam being just two names that were there to help him prepare. He said in an interview before the fight that this was the hardest he has train for a fight and he was totally ready to give everything in the ring to get the belt back.
And that he did! He has now being crowned the 70kg YOKKAO world champion! Both the fighters looked over the moon with their wins. It was definitely a good day at the office for Bad Co. I will post the videos of the fights on here when they eventually go live on YouTube for you to watch.
I would like to say a massive well done and congratulations to these two awesome fighters! They deserve everything and are a prime example of hard work paying off! Catch them here saying a few words after their fights last night!
Head over to Facebook and give us a thumbs up and also to twitter and give us a follow! All the support is greatly appreciated. Hopefully will be getting a bit more content published again!
Take it easy and keeping working hard!
← 5 rounds of sparring… God I've missed training!


text:
/**
 * Thermal compensation for barometric pressure sensors.
 *
 * @group Thermal Compensation
 * @reboot_required true
 * @boolean
 */
PARAM_DEFINE_INT32(TC_B_ENABLE, 0);

redpajama_set_name: RedPajamaGithub


Are You Ready to Recruit Generation Z?
Is there a crisis looming over the workforce?
Future Forward is the outlook of February. Are you also in Miami for SIA?
Is Your Recruitment Niche or Specialty Safe?
The post Workplace Culture Predictions for 2030 appeared first on Recruiting Daily Advisor. At the start of a new year, it's common to make predictions about trends for the coming months. Predictions are based on emerging trends, sociopolitical and market factors, etc.
In a previous post , we looked at a few workplace culture predictions you should be prepared for by the time 2030 rolls around. In our last post on this subject, we'll look at two more predictions on how work culture will change by 2030.
In two previous posts, we've discussed a few workplace culture trends expected to impact employers by 2030. The post Workplace Culture: Is Your Company Ready for 2030? tumsasedgars / iStock / Getty Images Plus.
Below are my thoughts on talent sourcing in 2030: Remote & virtual based teams: Employees can work virtually anywhere as long they have a wifi and cell phone connection. The post Predictions on Talent Sourcing in 2030 appeared first on WizardSourcer.
Some 85% of the jobs that today's students will be doing in 2030 haven't been invented yet. The big question for talent acquisition teams everywhere: What will those jobs be?
Millennial hiring strategy is now important more than ever, with studies stating that 75 percent of the workforce will be taken over by millennials in 2030. Who are millennials and why are they everywhere? A millennial is anyone who is born anywhere between the years 1980 and 2000.
By 2030, it's predicted that over 75% of the workforce around the world will be made up of the Millennial generation. Businesses beware. When you compare how different the current workforce is to this new generation, recruiters have certainly got their work cut out for them.
As some jobs are destroyed, new jobs are created, and still more are changed dramatically, McKinsey Global Institute projects as many as 375 million workers may need to switch occupational categories by 2030. Technologies like machine learning, automation, and artificial intelligence are here, and they are causing massive digital disruption in the business world.
It is the year 2030 and R2C2, the first recruiting robot, is being introduced to the world. R2C2 is using artificial intelligence to make any human recruiter obsolete. In just nanoseconds R2C2 can find the perfect candidate for a job wherever he or she is … Read more.
A new Korn Ferry study found that there will be a global talent shortage of more than 85 million people by 2030. In 2006, Joel Spolsky wrote that the great software developers are never on the open market.
Considering that 85 percent of the jobs of 2030 don't even exist yet , it might be unwise for new entrants to the workforce to stubbornly commit themselves to one career.
All indicators agree: until 2030, the number of skilled jobs needed in the global economy will exceed the number of available candidates. The war for talent is raging and companies are not sure they will win it. What should firms be doing to attract the best young talent?
But, if it's true that 85 percent of the jobs of 2030 don't yet exist, this could be a major problem. Behavioral economics tells us that humans prefer the comfort of the status quo over the effort of adaptation.
It's exciting to think where we could be by 2030. When AI springs to mind, you can't help but think of the endless possibilities it could bring. With machine learning, robotics, and more, breaking new grounds year after year.
By 2030, according to the report, financial and business services will see a 10.7 No degree? No problem! Research carried out by job search site indeed has revealed that a candidate's lack of degree need not be a cause for worry for recruiters.
As birth rates continue to steadily decline, Strack foresees a workforce crisis breaking in 2030, as Baby Boomers retire and there simply aren't enough people to replace them.
By the year 2030, 75% of the workforce will be made up of Millennials. Millennials, millennials, millennials. The buzzword of the moment, isn't it? To be fair, it's for good reason. So if your job includes recruiting Millennials, it's probably worthwhile to get a better understanding of what this generation is looking for in terms of work.
These concerns are valid— a report from McKinsey global institute predicts that automation could take over 800 million jobs globally by 2030. The public sector is facing a crisis.
For example, Musk believes that in his lifetime there will be a city of one million people living on Mars, or that by 2030 the first child to live to 1000 will be born. Last week I was at a conference of independent recruiters from around the world.
Because of this, few job seekers are interested in pursuing home health care roles and as a result, the industry is projected to see a shortage of 151,000 paid care workers by 2030. Many job seekers have the misconception that working in home health care leads to limited career opportunities.
The Workforce Crisis of 2030—and How to Start Solving It Now Rainer Strack You think talent acquisition is competitive now? By 2030, the majority of today's active workforce will retire—meaning there will be more open positions than workers available to fill them, Rainer Strack argues.
The home health care industry is projected to see a shortage of 150,000 paid care workers by 2030, meaning home health care businesses need to be more strategic than ever when it comes to attracting and hiring top talent.
This comes as no surprise—in fact, 85 percent of jobs that will exist in 2030 haven't been created yet.
It is predicted that by 2030 the largest company online is going to be education based. Maybe you're asking yourself. What is Gen Z?
In this digital era, where automation could replace up to 800 million jobs by 2030 , leadership needs to be more human than ever.
Manufacturing will face a deficit of nearly 8 million workers by 2030. With the rise of artificial intelligence and automation, many industries are facing talent shortages right now and will continue to do so over the next decade or so.
jobs by 2030. Which sounds more like your perfect job: managing a construction team, or analyzing high-volume data sets? Everyone will have a unique answer. However you define it, two factors are universally important: there need to be open positions, and they need to have appealing salaries.
12 Jobs You'll Be Recruiting for in 2030 -- by Bruce Anderson. In case you need a stimulating read, here's a great list for you. We took a look through some of the best articles published this past week for talent professionals and posted them below.
Staffing 2030: What Will the Landscape Look Like? It's the 27th year of Staffing Industry Analysts' Annual Executive Forum Conference and the agenda and theme is announced.
I'm not saying that come 2030 that engineers and software developers won't be in strong demand, but we will be surprised at the kinds of creative and teaching jobs that come along as well.
We estimate that between 400 million and 800 million individuals could be displaced by automation and need to find new jobs by 2030.
Predicting what the Global Workplace Will Look Like in 2030 by China Gorman ( @ChinaGorman ). Here is our weekly feature in which we share the top articles we enjoyed from the past week about recruitment marketing and other recruitment trends. This week we'll be covering the candidate experience, what it means to be a recruiter, social sourcing and the future of the global workforce.