Geometric Crystal and Tropical $R$ For $D^{(1)}_n$
A. Kuniba
We construct a geometric crystal for the affine Lie algebra $D^{(1)}_n$ in the sense of Berenstein and Kazhdan. % Based on a matrix realization including a spectral parameter, we prove uniqueness and explicit form of the tropical $R$, the birational map that intertwines products of the geometric crystals. % The tropical $R$ commutes with geometric Kashiwara operators and satisfies the Yang-Baxter equation. % It is subtraction-free and yields a piecewise linear formula of the combinatorial $R$ for crystals upon ultradiscretization.
math0208239.tex
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%      Geometric Crystal and Tropical R for D^{(1)}_n
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%                         by
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%       A.Kuniba, M. Okado, T. Takagi and Y. Yamada
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%last update: 30/Aug/2002 by T.T.

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\begin{document}

\title{Geometric Crystal and Tropical $R$ For 
$D^{(1)}_n$}

% author one information
\author{A. Kuniba}
\address{Institute of Physics, University of Tokyo, Tokyo 153-8902, Japan}
%\curraddr{}
\email{atsuo@gokutan.c.u-tokyo.ac.jp}
%\thanks{}

% author two information
\author{M. Okado}
\address{Department of Informatics and Mathematical Science,
    Graduate School of Engineering Science,
Osaka University, Osaka 560-8531, Japan}
%\curraddr{}
\email{okado@sigmath.es.osaka-u.ac.jp}
%\thanks{}

% author three information
\author{T. Takagi}
\address{Department of Applied Physics, National Defense Academy,
Kanagawa 239-8686, Japan}
%\curraddr{}
\email{takagi@nda.ac.jp}
%\thanks{}

% author four information
\author{Y. Yamada}
\address{Department of Mathematics, Faculty of Science,
Kobe University, Hyogo 657-8501, Japan}
%\curraddr{}
\email{yamaday@math.kobe-u.ac.jp}
%\thanks{}

%%\subjclass{Primary 81R50, 82B23, 37B15; Secondary 05E15}
% \footnote{2000 \textit{Mathematics Subject Classification.} 
% 17B37, 15A23, 14E05}
% 15A23 Factorization of matrices
% 17B37 Quantum groups (quantized enveloping algebras) and related deformations
% 14E05 Rational and birational maps

\begin{abstract}
We construct a geometric crystal 
for the affine Lie algebra $D^{(1)}_n$ 
in the sense of Berenstein and Kazhdan.
%
Based on a matrix realization including a spectral parameter, 
we prove uniqueness and explicit form of the 
tropical $R$, the birational
map that intertwines products of the geometric crystals.
%
The tropical $R$ commutes with geometric Kashiwara operators and 
satisfies the Yang-Baxter equation.
%
It is subtraction-free and yields a piecewise
linear formula of the combinatorial $R$ for crystals 
upon ultradiscretization.
\end{abstract}

\maketitle

% \tableofcontents
% \clearpage
\section{Introduction}
%
In this paper we construct a  
geometric crystal for $D^{(1)}_n$ corresponding to 
the $D^{(1)}_n$-crystal in \cite{KKM}, and 
study the associated tropical $R$.
%
Geometric crystal is a notion introduced by 
Berenstein and Kazhdan in \cite{BK} as an 
algebro-geometric analogue of the crystal theory \cite{K1,K2}.
%
The latter is a theory of quantum groups at $q=0$, which 
involves combinatorial operations described by 
various piecewise linear functions.
%
In the geometric setting in \cite{BK}, 
such a structure corresponds to birational maps 
described by subtraction-free functions.

Let us illustrate these features in the case of $A^{(1)}_{n-1}$
based on the constructions in \cite{BK} and \cite{KNY}.
%
Elements $x$ of the crystal $B_l$ for the $l$-symmetric tensor representation
is specified by the coordinates 
$x=(x_1,\ldots, x_n) \in \Z_{\ge 0}^n$ with $x_1 + \cdots + x_n = l$,
where all the indices are considered to be in $\Z/n\Z$ \cite{KKM}.
%
The action of the Kashiwara operator 
${\tilde e}_i$ up to 2-fold tensor product is given, unless they vanish, by
\begin{align*}
& {\tilde e}^c_i(x) = (\ldots, x_{i-1}, x_i+c, x_{i+1}-c, x_{i+2},\ldots),\\
%
& {\tilde e}_i^c(x \otimes y) 
= {\tilde e}^{c_1}_i(x) \otimes {\tilde e}_i^{c_2}(y),\\
%
&c_1 = \max(x_i+c,y_{i+1}) - \max(x_i,y_{i+1}), \\
&c_2 = \max(x_i,y_{i+1}) - \max(x_i,y_{i+1}-c).
\end{align*}
%
In the geometric crystal, one still has the coordinates 
$x=(x_1,\ldots, x_n) \in \B$ and the corresponding structure looks 
\begin{align*}
&e^c_i(x) = (\ldots, x_{i-1}, cx_i, c^{-1}x_{i+1}, x_{i+2},\ldots),\\
%
&e_i^c(x, y) 
= (e^{c_1}_i(x), e_i^{c_2}(y)),\\
%
&c_1 = \frac{cx_i+y_{i+1}}{x_i+y_{i+1}}, \;\;  
 c_2 = \frac{x_i+y_{i+1}}{x_i+c^{-1}y_{i+1}}.
\end{align*}
We call it the geometric Kashiwara operator.
%
Note that the latter $c_1, c_2$ contain no minus sign, i.e., 
they are subtraction-free.
%
The former $c_1, c_2$  are piecewise linear 
and obtained from the latter by replacing 
$+,\times, /$ with $\max, +, -$, respectively.
%
The procedure, which we call {\it ultradiscretization} in this paper,  
is well defined since the two sides of $p(q+r) = pq+pr$ have the coincident image
$p+\max(q,r) = \max(p+q, p+r)$.  
%
The geometric crystal $\B$ is actually 
associated with the {\it coherent family}  
$\{ B_l \}_{l\ge 1}$ \cite{KKM} rather than the single crystal $B_l$.

To each $x \in \B$ we assign the $n$ by $n$ matrix 
\[M(x,z)=(\sum_{i=1}^{n}\frac{1}{x_i}E_{i,i}-
\sum_{i=1}^{n-1}E_{i+1,i}-z E_{1,n} )^{-1}\]
containing a spectral parameter $z$.
%
Then on the product $M=M(x^{(1)},z)\cdots M(x^{(L)},z)$ 
corresponding to $(x^{(1)},\ldots,x^{(L)}) \in \B^{\times L}$, 
one can realize the geometric Kashiwara operator as the multiplications of 
unipotent matrices:
\[G_i\left(\frac{c-1}{z^{\delta_{i0}}\veps_i(M)} \right)M
G_i\left(\frac{c^{-1}-1}{z^{\delta_{i0}}\vphi_i(M)} \right)=
M(e^{c_1}_i(x^{(1)}),z)\cdots M(e^{c_L}_i(x^{(L)}),z),\]
%
where
$G_i(a)=E+a E_{i,i+1}$, $(1 \leq i \leq n-1)$,
$G_0(a)=E+a E_{n,1}$ and 
\begin{displaymath}
\veps_i(M)=z^{-\delta_{i0}}
\frac{M_{i+1,i}}{M_{i,i}}\Big\vert_{z=0}, \quad
\vphi_i(M)=z^{-\delta_{i0}}
\frac{M_{i+1,i}}{M_{i+1,i+1}}\Big\vert_{z=0}.
\end{displaymath}
%
The product $(e^{c_1}_i(x^{(1)}),\ldots,e^{c_L}_i(x^{(L)})) 
\in \B^{\times L}$ corresponds to the element 
${\tilde e}^c_i(x^{(1)} \otimes \cdots \otimes x^{(L)})$ in 
the tensor product crystal
under the ultradiscretization mentioned above.

One of the axioms of the geometric crystal is the relation 
$e^c_i e^{cd}_j e^d_i = e^d_j e^{cd}_i e^c_j$ in the case of
$\langle\alpha^\vee_i,\alpha_j\rangle
=\langle\alpha^\vee_j,\alpha_i\rangle=-1$.
%
The matrix realization reduces its proof to 
a simple manipulation on the matrices $G_i$'s.


In the crystal theory, the isomorphism of tensor products
$B_l\otimes B_k \simeq B_k \otimes B_l$ is called 
the combinatorial $R$.
%
It is the quantum $R$ matrix at $q=0$, and plays a fundamental role.
%
To incorporate it into the geometric setting we consider 
the matrix equation
$M(x,z)M(y,z)=M(x'z)M(y',z)$.
%
Explicitly, it is a discrete Toda type equation 
$x_iy_i=x_i'y_i'$, $\frac{1}{x_i}+\frac{1}{y_{i+1}}=
\frac{1}{x'_i}+\frac{1}{y'_{i+1}}$.
%
Given $x=(x_1,\ldots, x_n) \in \B$, 
call  $\ell(x) = x_1 \cdots x_n$ the level.
%
It turns out that the equation with the level constraints 
$\ell(x)=\ell(y'),\, \ell(y)=\ell(x')$ characterizes the unique solution:
\begin{eqnarray*}
&&x'_i=y_i \frac{P_i(x,y)}{P_{i-1}(x,y)}, \quad
y'_i=x_i \frac{P_{i-1}(x,y)}{P_i(x,y)},\\
&&P_i(x,y)=\sum_{k=1}^{n} \prod_{j=k}^n x_{i+j} \prod_{j=1}^k y_{i+j}.
\end{eqnarray*}
%
We call the rational map $(x,y) \mapsto (x',y')$ the {\it tropical $R$}, 
and simply denote by $R$.
%
{}From the characterization 
one can derive the geometric analogues of most important properties 
of the combinatorial $R$, e.g., 
commutativity with geometric Kashiwara operators 
$e^c_iR= Re^c_i$, the inversion relation $R(R(x,y))=(x,y)$ 
showing $R$ is birational, the Yang-Baxter relation and 
so forth. 
%
Moreover the ultradiscretization of the tropical $R$ yields an 
explicit piecewise linear formula for the combinatorial $R$.

In this paper we show that all the features explained above along  
type $A$ persist also for $D^{(1)}_n$.
%
What we do is to start from a new geometric crystal 
associated with the coherent family of 
$D^{(1)}_n$-crystals in \cite{KKM}, and develop an 
elementary approach based on explicit calculations. 
%
It is motivated by several recent works; 
geometrization of canonical bases \cite{BFZ,BK,L2},
tropical combinatorics and 
birational representations of affine Weyl groups \cite{Ki,KNY,NoY,Y}, 
integrable cellular automata associated to crystals 
\cite{HKT1,FOY,HKT2,HKOTY},  discrete soliton equations 
arising as their tropical evolution equation \cite{TTMS, HHIKTT,Y}, 
combinatorial $R$ in terms of the bumping algorithm \cite{HKOT1,HKOT2}, etc.
%
We hope to report on applications of the present result
to these issues in near future.

The paper is arranged as follows.
%
In section \ref{sec:B} we begin with the definition and basic facts 
on geometric crystals following \cite{BK}. 
%
The geometric $D^{(1)}_n$-crystal $\B$ is presented, which 
is parametrized with the $2n-1$ coordinates 
$x=(x_1,\ldots,x_n,\xb_{n-1},\ldots,\xb_1)$.
%
See section \ref{subsec:piecewise} for the 
precise correspondence to the coordinates in the crystal.
%
In section \ref{sec:M} we introduce the $2n$ by $2n$ 
matrix $M(x,z)$  that realizes the geometric crystal $\B^{\times L}$.
%
It is quadratic in $z$ in contrast with the 
essentially linear one for type $A$.
%
At $z=0$ it admits a factorization (\ref{eq:factorizationofA}) and 
belongs to the lower triangular Borel subgroup of $D_n$ as in
Proposition \ref{pr:MSMS}, whereas at $z= \ell(x)^{-1}$ 
it shrinks to the rank one matrix as in Proposition \ref{prop:meqdpd}.
%
A key in our approach is to 
make full use of such properties 
controlled by the spectral parameter of affine nature,  
which we have been unable to learn in \cite{BK}.
%
In section \ref{sec:R} we study the tropical $R$.
{}From Theorem \ref{th:uniqueness} 
its uniqueness is immediate, hence our main task 
is to assure the existence.
%
We shall show that the explicit birational map 
$(x,y) \rightarrow (x',y')$ 
in Definition \ref{th:tropicalR} is the answer.
%
By the ultradiscretization we then derive a piecewise linear formula for
the combinatorial $R$ and the associated energy function. 
%
Appendix \ref{sec:appA} provides an alternative proof of 
the relation $e_i^ce_j^{cd}e_i^d=e_j^de_i^{cd}e_j^c$ on $\B^{\times L}$.
%
Appendix \ref{sec:proofofaaeqaa} contains 
a proof of $M(x,z)M(y,z)=M(x'z)M(y',z)$ at $z=0$.
%

We remark that our terminology 
``tropical"  in this paper follows the usage in \cite{Ki,NoY}, 
which is unfortunately the opposite of \cite{BK}.  
%
The phrase ``ultradiscretization" has been used in many
papers inspired by \cite{TTMS}.

\section*{Acknowledgements}
The authors thank M. Noumi, S. Naito, and 
D. Sagaki for useful discussions.
A.K. thanks Shi-shyr Roan 
for the hospitality at Academia Sinica, 
Taipei during ``Workshop on Solvable Models and Quantum Integrable Systems"
in November 2001, where a part of the work was presented.
M.O. and Y.Y. are partially supported by Grant-in-Aid for 
Scientific Research from Ministry of Education, Culture, 
Sports, Science and Technology of Japan.






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Geometric $D^{(1)}_n$-crystal}\label{sec:B}

\subsection{Geometric crystal}\label{subsec:gcrystal}
Let $\geh$ be an affine Lie algebra of simply laced type, i.e., 
$\geh=A^{(1)}_n,D^{(1)}_n$
or $E^{(1)}_{6,7,8}$. Let $I=\{0,1,\ldots,n\}$ be the index set of vertices 
of the Dynkin
diagram of $\geh$. We denote by $\{\alpha_i\mid i\in I\}$ (resp. 
$\alpha^\vee_i\mid i\in I\}$)
the set of simple roots (resp. simple coroots). Note that $\langle 
\alpha^\vee_i,\alpha_j\rangle
=0$ or $-1$.

Following \cite{BK} we introduce a geometric crystal associated to $\geh$. 
Throughout this paper,
{\em variable} means a real variable unless otherwise stated. Let 
$x=(x_1,x_2,\ldots,x_N)$ be
an $N$-tuple of variables. For $i\in I$ let $\veps_i(x),\vphi_i(x)$ be 
rational functions in $x$
and set
\begin{equation} \label{eq:gamma}
\gamma_i(x)=\vphi_i(x)/\veps_i(x).
\end{equation}
For $i\in I$ let $\mathbb{R}\times\mathbb{R}^N\longrightarrow\mathbb{R}^N:
(c,x)\mapsto e_i^c(x)$ be a rational mapping, namely, 
$e_i^c(x)=(x'_1,x'_2,\ldots,x'_N)$
with each $x'_j$ ($1\le j\le N$) being a rational function in $c$ and $x$.
In what follows we consider $c$ to be a parameter and $e_i^c$ to be a 
rational transformation
on $\mathbb{R}^N$. We also abbreviate the symbol $\circ$ of composition. For 
instance
we write $e_i^{c_1}e_j^{c_2}$ instead of $e_i^{c_1}\circ e_j^{c_2}$.

\begin{definition} \label{def:pre-cry}
A {\em geometric pre-crystal} is a family $\B=\{x,\veps_i,\vphi_i,e_i^c\}$ 
satisfying
\begin{itemize}
\item[(i)] $e_i^{c_1}e_i^{c_2}(x)=e_i^{c_1c_2}(x),e_i^1(x)=x$,
\item[(ii)] 
$\veps_i(e_i^c(x))=c^{-1}\veps_i(x),\vphi_i(e_i^c(x))=c\vphi_i(x)$,
\item[(iii)] 
$\gamma_i(e_j^c(x))=c^{\langle\alpha^\vee_i,\alpha_j\rangle}\gamma_i(x)$.
\end{itemize}
\end{definition}

\begin{definition} \label{def:geom-cry}
A geometric pre-crystal $\B$ is called {\em geometric crystal}, if it 
further satisfies
the following relations:
\begin{equation} \label{eq:geom-cry1}
e_i^{c_1}e_j^{c_2}=e_j^{c_2}e_i^{c_1}
\end{equation}
if $\langle\alpha^\vee_i,\alpha_j\rangle=0$,
\begin{equation} \label{eq:geom-cry2}
e_i^{c_1}e_j^{c_1c_2}e_i^{c_2}=e_j^{c_2}e_i^{c_1c_2}e_j^{c_1}
\end{equation}
if 
$\langle\alpha^\vee_i,\alpha_j\rangle=\langle\alpha^\vee_j,\alpha_i\rangle=-  
1$.
\end{definition}

\begin{remark}
The definition of the geometric pre-crystal is slightly different from the 
original one
in \cite{BK}. We adopt it in order to make the formulas parallel with those 
in crystals.
\end{remark}

\begin{remark}
As mentioned in \cite{BK} Remarks after Lemma 2.1, 
\eqref{eq:geom-cry1}--\eqref{eq:geom-cry2}
can be thought of as multiplicative analogues of 
the Verma relations in the 
universal enveloping algebra $U(\geh)$. 
See also \cite{L1} Proposition 39.3.7 for the version
of the quantum enveloping algebra $U_q(\geh)$.
The relations in the cases of $\langle \alpha^\vee_i,\alpha_j \rangle
\langle \alpha^\vee_j,\alpha_i \rangle\ge2$ read as follows:
\[
e_i^{c_1}e_j^{c_1^2c_2}e_i^{c_1c_2}e_j^{c_2}=e_j^{c_2}e_i^{c_1c_2}e_j^{c_1^2  
c_2}e_i^{c_1}
\]
if 
$\langle\alpha^\vee_i,\alpha_j\rangle=-1,\langle\alpha^\vee_j,
\alpha_i\rangle=-2$,
\[
e_i^{c_1}e_j^{c_1^3c_2}e_i^{c_1^2c_2}e_j^{c_1^3c_2^2}e_i^{c_1c_2}e_j^{c_2}
=e_j^{c_2}e_i^{c_1c_2}e_j^{c_1^3c_2^2}e_i^{c_1^2c_2}
e_j^{c_1^3c_2}e_i^{c_1}  
\]
if 
$\langle\alpha^\vee_i,\alpha_j\rangle=-1,\langle\alpha^\vee_j,
\alpha_i\rangle=-3$.
\end{remark}

\begin{prop}[\cite{BK}]
Let $\B$ be a geometric crystal.
Set $s_i(x)=e_i^{\gamma_i(x)^{-1}}(x)$. Then $s_i$ ($i\in I$) generates
a birational action of the Weyl group corresponding to $\geh$.
\end{prop}

\begin{proof}
By definition, $s_i^2(x)=e_i^{c'}e_i^c(x)$ with $c=\gamma_i(x)^{-1},
c'=\gamma_i(e_i^{\gamma_i(x)^{-1}}(x))^{-1}$. From Definition 
\ref{def:pre-cry} (iii),
we have
\[
c'=((\gamma_i(x)^{-1})^2\gamma_i(x))^{-1}=\gamma_i(x).
\]
Thus $s_i^2(x)=x$ by Definition \ref{def:pre-cry} (i).

Similarly, if 
$\langle\alpha^\vee_i,\alpha_j\rangle=\langle\alpha^\vee_j,\alpha_i
\rangle=-1$,
we have
\begin{eqnarray*}
s_is_js_i(x)=e_i^{c''}e_j^{c'}e_i^c(x),\\
s_js_is_j(x)=e_j^{d''}e_i^{d'}e_j^d(x),
\end{eqnarray*}
with 
$c=d''=\gamma_i(x)^{-1},c'=d'=\gamma_i(x)^{-1}\gamma_j(x)^{-1},
c''=d=\gamma_j(x)^{-1}$.
Thus $s_is_js_i(x)=s_js_is_j(x)$ by Definition \ref{def:geom-cry}.

If $\langle\alpha^\vee_i,\alpha_j\rangle=0$, one clearly has 
$s_is_j(x)=s_js_i(x)$.
\end{proof}

\begin{remark}
This Weyl group action is an analogue of the one on crystals 
given in \cite{K2}.
\end{remark}

\begin{example} \label{ex:A}
Let $\geh=A^{(1)}_n,N=n+1,x=(x_1,x_2,\ldots,x_{n+1})$.
For $i\in I=\{0,1,\ldots,n\}$ we define
\begin{eqnarray*}
\veps_i(x)&=&x_{i+1},\quad \vphi_i(x)=x_i,\\
e^c_i(x)&=&(\ldots,cx_i,c^{-1}x_{i+1},\ldots).
\end{eqnarray*}
Here $x_0$ should be understood as $x_{n+1}$. One can check that
$\B=\{x,\veps_i,\vphi_i,e_i^c\}$ is a geometric crystal. $s_i(x)$ is given 
by
\[
s_i(x)=(\ldots,\stackrel{\scriptstyle i}{x_{i+1}},\stackrel{\scriptstyle 
i+1}{x_i},\ldots).
\]
\end{example}

\subsection{Geometric $D^{(1)}_n$-crystal $\B$} \label{subsec:defofB}

We are interested in a particular geometric crystal $\B$ of type $D^{(1)}_n$ 
given below.
Let $N=2n-1$ and consider a set of variables 
$x=(x_1,x_2,\ldots,x_n,\xb_{n-1},\ldots,\xb_1)$.
$\veps_i(x),\vphi_i(x),e_i^c(x)$ ($i\in I=\{0,1,\ldots,n\}$) are given as 
follows.

\begin{eqnarray*}
\veps_0(x)&=&x_1(\frac{x_2}{\xb_2}+1),\quad
\vphi_0(x)=\xb_1(\frac{\xb_2}{x_2}+1),\\
\veps_i(x)&=&\xb_i(\frac{x_{i+1}}{\xb_{i+1}}+1),\quad
\vphi_i(x)=x_i(\frac{\xb_{i+1}}{x_{i+1}}+1)\quad(i=1,\ldots,n-2),\\
\veps_{n-1}(x)&=&x_n\xb_{n-1},\quad \vphi_{n-1}(x)=x_{n-1},\\
\veps_n(x)&=&\xb_{n-1},\quad \vphi_n(x)=x_{n-1}x_n,\\
e^c_0(x)&=&(\xi_2^{-1}x_1,c^{-1}\xi_2x_2,\ldots,\xi_2\xb_2,c\xi_2^{-1}\xb_1)  
,\\
e^c_i(x)&=&(\ldots,c\xi_{i+1}^{-1}x_i,c^{-1}\xi_{i+1}x_{i+1},\ldots,
\xi_{i+1}\xb_{i+1},\xi_{i+1}^{-1}\xb_i,\ldots)\quad(i=1,\ldots,n-2),\\
e^c_{n-1}(x)&=&(\ldots,cx_{n-1},c^{-1}x_n,\ldots),\\
e^c_n(x)&=&(\ldots,cx_n,c^{-1}\xb_{n-1},\ldots),\\
\mbox{where }&&\xi_i=\frac{x_i+c\xb_i}{x_i+\xb_i}\quad(i=1,\ldots,n-1).
\end{eqnarray*}
We set
\[
\ell(x)=x_1x_2\cdots x_n\xb_{n-1}\xb_{n-2}\cdots\xb_1
\]
and call it the {\em level}. Note that $\ell(x)$ is invariant 
under
the transformation $e_i^c$ for any $i$ and $c$.

\begin{remark}
This $\B$ is obtained as a multiplicative analogue of the
coherent family of perfect crystals $\{B_l\}_{l\ge1}$ of type $D^{(1)}_n$ 
given in
\cite{KKM}. Of course, $e_i^c(x)$ corresponds to $\et{i}^cb$.
\end{remark}

The functions $\veps_i(x),\vphi_i(x)$ of our $\B$ satisfy the following good 
properties,
which will be used in the next section.

\begin{lemma} \label{lem:eps-phi-decomp}
\begin{itemize}

\item[(a)] If $\langle\alpha^\vee_i,\alpha_j\rangle=0$, then
\[
\veps_i(e_j^c(x))=\veps_i(x),\quad \vphi_i(e_j^c(x))=\vphi_i(x).
\]

\item[(b)] If 
$\langle\alpha^\vee_i,\alpha_j\rangle=\langle\alpha^\vee_j,
\alpha_i\rangle=-1$, then
there exist rational functions $\veps_{ij}(x),\veps_{ji}(x)$ such that
\begin{eqnarray*}
&&\veps_i(x)\veps_j(x)=\veps_{ij}(x)+\veps_{ji}(x),\\
&&\veps_{ij}(e_i^c(x))=c^{-1}\veps_{ij}(x),\quad 
\veps_{ij}(e_j^c(x))=\veps_{ij}(x),\\
&&\veps_{ji}(e_i^c(x))=\veps_{ij}(x),\quad 
\veps_{ji}(e_j^c(x))=c^{-1}\veps_{ji}(x),
\end{eqnarray*}
and $\vphi_{ij}(x),\vphi_{ji}(x)$ such that
\begin{eqnarray*}
&&\vphi_i(x)\vphi_j(x)=\vphi_{ij}(x)+\vphi_{ji}(x),\\
&&\vphi_{ij}(e_i^c(x))=c\vphi_{ij}(x),\quad 
\vphi_{ij}(e_j^c(x))=\vphi_{ij}(x),\\
&&\vphi_{ji}(e_i^c(x))=\vphi_{ij}(x),\quad 
\vphi_{ji}(e_j^c(x))=c\vphi_{ji}(x).
\end{eqnarray*}

\end{itemize}
\end{lemma}

\begin{remark} \label{rem:eps_ij}
Due to Definition \ref{def:pre-cry} (iii) it suffices to check Lemma
\ref{lem:eps-phi-decomp} only for $\vphi_i(x)$ (or $\veps_i(x)$). The 
functions
$\veps_{ij}(x),\veps_{ji}(x)$ in (b) are obtained as
\[
\veps_{ij}(x)=\vphi_{ji}(x)/(\gamma_i(x)\gamma_j(x)),\quad
\veps_{ji}(x)=\vphi_{ij}(x)/(\gamma_i(x)\gamma_j(x)),
\]
once $\vphi_{ij}(x),\vphi_{ji}(x)$ are obtained.
\end{remark}

\begin{proof}[Proof of Lemma \ref{lem:eps-phi-decomp}.]
(a) is easy. For (b) it suffices to give a list of $\vphi_{ij}(x)$ and 
$\vphi_{ji}(x)$
when 
$\langle\alpha^\vee_i,\alpha_j\rangle=\langle\alpha^\vee_j,
\alpha_i\rangle=-1$.
\begin{eqnarray*}
&&\vphi_{02}(x)=\xb_1\xb_2(\frac{\xb_3}{x_3}+1),\quad
\vphi_{20}(x)=\xb_1 x_2(\frac{\xb_3}{x_3}+1),\\
&&\vphi_{i,i+1}(x)=x_i\xb_{i+1}(\frac{\xb_{i+2}}{x_{i+2}}+1),\quad
\vphi_{i+1,i}(x)=x_i x_{i+1}(\frac{\xb_{i+2}}{x_{i+2}}+1) \quad(1\le i\le 
n-3),\\
&&\vphi_{n-2,n-1}(x)=x_{n-2}\xb_{n-1},\quad 
\vphi_{n-1,n-2}(x)=x_{n-2}x_{n-1},\\
&&\vphi_{n-2,n}(x)=x_{n-2}x_n\xb_{n-1},\quad 
\vphi_{n,n-2}(x)=x_{n-2}x_{n-1}x_n.
\end{eqnarray*}
\end{proof}

\subsection{Product}\label{subsec:product}
Let $\B_1=\{x=(x_1,\ldots,x_{N_1}),\veps_i,\vphi_i,e_i^c\},
\B_2=\{y=(y_1,\ldots,y_{N_2}),$ $\veps_i,\vphi_i,e_i^c\}$ be geometric 
pre-crystals.
For brevity we use the same symbols $\veps_i,\vphi_i,e_i^c$ for different 
geometric
pre-crystals, since they are distinguished by the sets of variables.
Now consider the set of variables 
$(x,y)=(x_1,\ldots,x_{N_1},y_1,\ldots,y_{N_2})$.
We define a new structure of a geometric pre-crystal $\B_1\times\B_2$ on 
$(x,y)$ by

\begin{eqnarray}
\veps_i(x,y)&=&\veps_i(x)+\veps_i(x)\veps_i(y)/\vphi_i(x),\\
\vphi_i(x,y)&=&\vphi_i(y)+\vphi_i(x)\vphi_i(y)/\veps_i(y),\\
e_i^c(x,y)&=&(e_i^{c_1}(x),e_i^{c_2}(y)), \label{eq:ecixy}\\
\text{where }&&c_1=\frac{c\vphi_i(x)+\veps_i(y)}{\vphi_i(x)+\veps_i(y)},\;
c_2=\frac{\vphi_i(x)+\veps_i(y)}{\vphi_i(x)+c^{-1}\veps_i(y)}.
\label{eq:c1c2}
\end{eqnarray}

\begin{remark} \label{rem:tensor}
These are analogues of formulas for the tensor product of crystals \cite{K1}
given by
\begin{eqnarray}
\veps_i(x\ot y)&=&\max(\veps_i(x),\veps_i(x)+\veps_i(y)-\vphi_i(x)),\\
\vphi_i(x\ot y)&=&\max(\vphi_i(y),\vphi_i(x)+\vphi_i(y)-\veps_i(y)),\\
\et{i}^c(x\ot y)&=&\et{i}^{c_1}(x)\ot\et{i}^{c_2}(y), \label{eq:etcixy}\\
\text{where }&&c_1=\max(c+\vphi_i(x),\veps_i(y))-\max(\vphi_i(x),\veps_i(y)),\\
&&c_2=\max(\vphi_i(x),\veps_i(y))-\max(\vphi_i(x),-c+\veps_i(y)).
\end{eqnarray}
In \eqref{eq:etcixy} $c$ can be a negative integer, in which case we understand
$\et{i}^c=\ft{i}^{-c}$.
\end{remark}

\begin{lemma}
Assume that the functions $\veps_i,\vphi_i$ for both $\B_1$ and $\B_2$ 
satisfy
the properties in Lemma \ref{lem:eps-phi-decomp}.
Then those for $\B_1\times\B_2$ also satisfy the same property.
\end{lemma}

\begin{proof}
If $\langle\alpha^\vee_i,\alpha_j\rangle=0$, the statement is clear.
Suppose 
$\langle\alpha^\vee_i,\alpha_j\rangle=\langle\alpha^\vee_j,\alpha_i\rangle
=-1$. Due to Remark \ref{rem:eps_ij} it suffices to construct desired 
functions
$\vphi_{ij}(x,y),\vphi_{ji}(x,y)$. Define
\[
\vphi_{ij}(x,y)=\vphi_{ij}(y)+\vphi_i(x)\vphi_j(y)\gamma_i(y)
+\vphi_{ij}(x)\gamma_i(y)\gamma_j(y)
\]
and $\vphi_{ji}(x,y)$ by interchanging $i$ and $j$.
To illustrate we calculate $\vphi_{ij}(e_i^c(x,y))$. Let $c_1,c_2$ be
as in \eqref{eq:c1c2}. Using Definition \ref{def:pre-cry} (iii) and
the assumption,
\begin{eqnarray*}
\vphi_{ij}(e_i^c(x,y))&=&
c_2\vphi_{ij}(y)+\frac{c_1\vphi_i(x)(c_2\vphi_{ij}(y)+\vphi_{ji}(y))}{c_2^{-  
1}\veps_i(y)}
+c_1\vphi_{ij}(x)\cdot c_2^2\gamma_i(y)\cdot c_2^{-1}\gamma_j(y)\\
&=&c_2(\vphi_{ij}(y)+\frac{c\vphi_i(x)\vphi_{ij}(y)}{\veps_i(y)})
+c(\frac{\vphi_i(x)\vphi_{ji}(y)}{\veps_i(y)}+\vphi_{ij}(x)\gamma_i(y)\gamma  
_j(y))\\
&=&c(\vphi_{ij}(y)+\frac{\vphi_i(x)\vphi_{ij}(y)}{\veps_i(y)})
+c(\frac{\vphi_i(x)\vphi_{ji}(y)}{\veps_i(y)}+\vphi_{ij}(x)\gamma_i(y)\gamma  
_j(y))\\
&=&c\vphi_{ij}(x,y).
\end{eqnarray*}

The other cases are similar.
\end{proof}

We now consider a multiple product. Let $\B_1,\ldots,\B_L$ be geometric 
pre-crystals
such that $\B_l=\{x^l=(x^l_1,\ldots,x^l_{N_l})\}$ ($l=1,\ldots,L$). Let us 
set
$\boldsymbol{x}=(x^1,\ldots,x^L)$. Then $\veps_i,\vphi_i,e^c_i$ of the 
$L$-fold product
of geometric pre-crystals $\B_1\times\cdots\times\B_L$ is given by

\begin{align}
\label{eq:vepsforxxx}
\veps_i(\boldsymbol{x}) &=
\frac{\sum_{k=1}^L \left( \prod_{j=1}^k \veps_i(x^j) \right)
\left( \prod_{j=k}^{L-1} \vphi_i(x^j) \right)}
{\prod_{j=1}^{L-1} \vphi_i(x^j)}, \\
%
\label{eq:vphiforxxx}
\vphi_i(\boldsymbol{x}) &=
\frac{\sum_{k=1}^L \left( \prod_{j=2}^k \veps_i(x^j) \right)
\left( \prod_{j=k}^{L} \vphi_i(x^j) \right)}
{\prod_{j=2}^{L} \veps_i(x^j)},\\
%
e^c_i(\boldsymbol{x}) &= (e^{c_1}_i(x^1),\ldots, e^{c_L}_i(x^L)),\\
%
\label{eq:def-of-c_l}
\mbox{with}\quad c_l &= \frac{\sum_{k=1}^L c^{\theta(k \leq l)}
\left( \prod_{j=2}^k \veps_i(x^j) \right)
\left( \prod_{j=k}^{L-1} \vphi_i(x^j) \right)}
{\sum_{k=1}^L c^{\theta(k \leq l-1)}
\left( \prod_{j=2}^k \veps_i(x^j) \right)
\left( \prod_{j=k}^{L-1} \vphi_i(x^j) \right)}.
\end{align}
Here $\theta(s)=1$ if $s$ is true and $=0$ otherwise.
%
Note that the product is associative,
$e^c_i(x^1,x^2,x^3) = 
\left(e^{c_1c_2}_i(x^1,x^2),e^{c_3}_i(x^3) \right) =
\left(e^{c_1}_i(x^1),e^{c_2c_3}_i(x^2,x^3) \right)$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Realization by unipotent matrices}\label{sec:M}

\subsection{Matrix $M(x,z)$}
\noindent
Let us introduce $2n$ by $2n$
matrices $G_i(a)$ for $0\le i\le n$ by
%
\begin{displaymath}
G_i(a) = E +
\begin{cases}
a (E_{i,i+1} + E_{2n-i,2n+1-i} ) & \mbox{for} \; 1 \leq i \leq n-1, \\
a (E_{n-1,n+1} + E_{n,n+2} ) & \mbox{for} \; i=n, \\
a (E_{2n-1,1} + E_{2n,2}) & \mbox{for} \; i=0.
\end{cases}
\end{displaymath}
Here $E$ is the identity matrix, $E_{ij}$ is the $(i,j)$ matrix unit,
and $a$ is a parameter.
$G_i(a)$'s have the following properties.
\begin{eqnarray}
&&G_i(a)G_i(b)=G_i(a+b),\\
&&G_i(a)G_j(b)=G_j(b)G_i(a)\quad \mbox{ if 
}\langle\alpha^\vee_i,\alpha_j\rangle=0,\\
&&G_i(a)G_j(b)G_i(c)=G_j(a')G_i(b')G_j(c')\quad \mbox{ if }
\langle\alpha^\vee_i,\alpha_j\rangle=\langle\alpha^\vee_j,\alpha_i\rangle=-1  
,
\label{eq:GiGjGi=GjGiGj}
\end{eqnarray}
where $a'=bc/(a+c),b'=a+c,c'=ab/(a+c)$.

Let $\B=\{x=(x_1,\ldots,\xb_1),\veps_i,\vphi_i,e_i^c\}$
be the geometric $D^{(1)}_n$-crystal given in the previous section.
In this subsection
we introduce a $2n$ by $2n$ matrix
$M(x,z)$ that satisfies
\begin{displaymath}
G_i \left( \frac{c-1}{z^{\delta_{i0}}\veps_i(x)} \right) M(x,z)
G_i \left( \frac{c^{-1}-1}{z^{\delta_{i0}}\vphi_i(x)} \right) =
M(e^c_i (x),z)
\end{displaymath}
for any $i$ ($0 \leq i \leq n$) and an extra parameter $z$.
We call $z$ the {\em spectral parameter}.
%

We give an explicit form of $M(x,z)$.
Matrix elements in the first column
$M(x,z)_{i,1}$ are given as follows.
\begin{equation}
\label{eq:firstcolumnofA}
M(x,z)_{i,1} =
\begin{cases}
x_1/\xb_1 & \mbox{for}  \; i=1, \\
x_1 \cdots x_{i-1} (1+x_i/\xb_i) & \mbox{for} \;  2 \leq i \leq n-1, \\
x_1 \cdots x_{n-1}x_n & \mbox{for} \;  i=n, \\
x_1 \cdots x_{n-1} & \mbox{for} \;  i=n+1, \\
\ell(x) / (\xb_1 \cdots \xb_{2n-i}) & \mbox{for} \;  n+2 \leq i \leq 2n.
\end{cases}
\end{equation}

%
Given the first column as above we define
the other matrix elements of $M(x,z)$
by the relations (\ref{eq:defMno1})-(\ref{eq:defMno6}).
\begin{equation}
\label{eq:defMno1}
M(x,z)_{i,j+1} =
\frac{M(x,z)_{i,j}}{x_j} +
\begin{cases}
(\ell (x) z - 1)/\xb_j & \mbox{for} \; i=j, \\
\ell (x) z - 1 & \mbox{for} \; i=j+1, \\
0 & \mbox{for} \; i \ne j,j+1,
\end{cases}
\end{equation}
for $1 \leq j \leq n-2$.

\begin{equation}
M(x,z)_{i,n} =
\frac{M(x,z)_{i,n-1}}{x_{n-1}} +
\begin{cases}
(\ell (x) z - 1)/\xb_{n-1} & \mbox{for} \; i=n-1, \\
\ell (x) z - 1 & \mbox{for} \; i=n+1, \\
0 & \mbox{for} \; i \ne n-1,n+1.
\end{cases}
\end{equation}

\begin{equation}
M(x,z)_{i,n+1} =
\frac{M(x,z)_{i,n-1}}{x_{n-1}x_n} +
\begin{cases}
(\ell (x) z - 1)/(\xb_{n-1}x_n) & \mbox{for} \; i=n-1, \\
\ell (x) z - 1 & \mbox{for} \; i=n, \\
0 & \mbox{for} \; i \ne n-1,n.
\end{cases}
\end{equation}

\begin{equation}
M(x,z)_{i,2n} =
z \frac{\xb_1 M(x,z)_{i,1}}{x_1} +
\begin{cases}
\frac{\xb_1}{x_1} (1 - \ell (x) z) & \mbox{for} \; i=2n, \\
z(\ell (x) z - 1) & \mbox{for} \; i=1, \\
0 & \mbox{for} \; i \ne 1,2n.
\end{cases}
\end{equation}
Let us denote $1+\xb_i/x_i$ by $\overline{X}_i$.

\begin{equation}
M(x,z)_{i,2n-1} =
z \xb_1 \overline{X}_2 M(x,z)_{i,1} +
\begin{cases}
\xb_1 \overline{X}_2 (1 - \ell (x) z) & \mbox{for} \; i=2n, \\
\frac{\xb_2}{x_2}(1 - \ell (x) z) & \mbox{for} \; i=2n-1, \\
0 & \mbox{for} \; i \ne 2n-1,2n.
\end{cases}
\end{equation}

\begin{equation}
\label{eq:defMno6}
M(x,z)_{i,2n-j} =
\xb_j \frac{\overline{X}_{j+1} M(x,z)_{i,2n+1-j}}{\overline{X}_{j} } +
\begin{cases}
\xb_j \frac{\overline{X}_{j+1}}{\overline{X}_{j} }
(1 - \ell (x) z) & \mbox{for} \; i=2n+1-j, \\
\frac{\xb_{j+1}}{x_{j+1}}(1 - \ell (x) z) & \mbox{for} \; i=2n-j, \\
0 & \mbox{for} \; i \ne 2n+1-j,2n-j,
\end{cases}
\end{equation}
for $2 \leq j \leq n-2$.
%
For later use we also present
matrix elements in the last row.
\begin{equation}
\label{eq:lastrowofA}
M(x,z)_{2n,i} =
\begin{cases}
\ell(x) / (x_1 \cdots x_{i-1}) & \mbox{for} \;  1 \leq i \leq n,\\
\xb_1 \cdots \xb_{n-1} & \mbox{for} \;  i=n+1, \\
\xb_1 \cdots \xb_{2n-i} (1+\xb_{2n+1-i}/x_{2n+1-i})
& \mbox{for} \;  n+2 \leq i \leq 2n-1, \\
\xb_1/x_1 & \mbox{for}  \; i=2n.
\end{cases}
\end{equation}

It can be seen that each matrix element is a polynomial in $z$ of degree at 
most $2$.
We denote the coefficients of $M(x,z)$ of degree $0,1,2$ by 
$A(x),B(x),C(x)$. Hence
we have
\[
M(x,z)=A(x)+zB(x)+z^2C(x).
\]
%
\begin{example}[$n=4$ case]
\label{ex:abc}
\noindent
$x=(x_1,x_2,x_3,x_4,\xb_3,\xb_2,\xb_1)$.
We use the notations
$[1]= x_1, [2]= x_2, \ldots , [\bar{1}] = \xb_1$, and
$[(2)] = 1+x_2/\xb_2, [(\bar{3})]=1+\xb_3/x_3$, etc.
For instance, $[34\bar{3}]=x_3x_4\xb_3$ and 
$[1(\bar{2})]=x_1(1+\xb_2/x_2)$.
Define $8 \times 8$ matrices $A(x), B(x), C(x)$ as follows.
%
\begin{displaymath}
A(x)=\left(
\begin{array}{rrrrrrrr}
\frac{[1]}{[\bar{1}]}  \\
%
 \left[1(2)\right]  & \frac{ [2]}{ [\bar{2}]}  \\
%
 \left[12(3)\right]  &  \left[2(3)\right]  & \frac{ [3]}{ [\bar{3}]}  \\
%
 \left[1234\right]  &  \left[234\right] &  \left[34\right] &  [4]  \\
%
 \left[123\right]  &  \left[23\right] &  [3] & & \frac{ 1}{ [4]}   \\
%
 \left[1234\bar{3}\right] &  \left[234\bar{3}\right]  & 
 \left[34\bar{3}\right] &  \left[4\bar{3}\right] &  [\bar{3}] & \frac{ 
[\bar{3}]}{ [3]}   \\
%
 \left[1234\bar{3}\bar{2}\right] & \left[234\bar{3}\bar{2}\right] & 
 \left[34\bar{3}\bar{2}\right] &  \left[4\bar{3}\bar{2}\right] & 
 \left[\bar{3}\bar{2}\right] &  \left[(\bar{3})\bar{2}\right] & \frac{ 
[\bar{2}]}{ [2]}  \\
%
 \left[1234\bar{3}\bar{2}\bar{1}\right] & 
 \left[234\bar{3}\bar{2}\bar{1}\right] & 
 \left[34\bar{3}\bar{2}\bar{1}\right] &  \left[4\bar{3}\bar{2}\bar{1}\right] 
&  \left[\bar{3}\bar{2}\bar{1}\right] & 
 \left[(\bar{3})\bar{2}\bar{1}\right] &  \left[(\bar{2})\bar{1}\right] & 
\frac{ [\bar{1}]}{ [1]}
%
\end{array}
\right),
\end{displaymath}
%%%%%
%
\begin{eqnarray*}
B(x)&=&\left(
\begin{array}{rrrrrr}
0  &  \left[1234 \bar{3}\bar{2}\right]  & \left[134 \bar{3}\bar{2}\right]  & 
\left[14\bar{3}\bar{2}\right]  & \left[1 \bar{3}\bar{2}\right]  & 
\left[1(\bar{3})\bar{2}\right] \\
%
 & \left[1234 \bar{3}\bar{2}\bar{1}\right] &  \left[1(2)34 
\bar{3}\bar{2}\bar{1}\right] & \left[1(2)4\bar{3}\bar{2}\bar{1}\right]  & 
\left[1(2) \bar{3}\bar{2}\bar{1}\right]  & \left[1(2)(\bar{3})\bar{2}\bar{1} 
\right] \\
%
&  & \left[1234 \bar{3}\bar{2}\bar{1}\right] & 
\left[12(3)4\bar{3}\bar{2}\bar{1}\right]
& \left[12(3) \bar{3}\bar{2}\bar{1}\right] & 
\left[12(3)(\bar{3})\bar{2}\bar{1}\right] \\
%
&  &  & & \left[1234\bar{3}\bar{2}\bar{1}\right] & 
\left[1234(\bar{3})\bar{2}\bar{1}\right] \\
%
&  &  & \left[1234 \bar{3}\bar{2}\bar{1}\right] & & \left[123 
(\bar{3})\bar{2}\bar{1}\right] \\
%
 &  &  & &   &\left[1234 \bar{3}\bar{2}\bar{1}\right] \\
%
 &  & &  &  &  \\
%
 &  &  & &  &
\end{array}
\right. \\
%
&&\hspace{4mm}\left.
\begin{array}{rr}
\left[1(\bar{2})\right] &   \\
\left[1(2)(\bar{2})\bar{1}\right]  &  \left[(2)\bar{1}\right] \\
\left[12(3)(\bar{2})\bar{1}\right] & \left[2(3)\bar{1}\right] \\
\left[1234(\bar{2})\bar{1}\right] & \left[234\bar{1}\right] \\
\left[123 (\bar{2})\bar{1}\right] & \left[23 \bar{1}\right] \\
\left[1234 \bar{3}(\bar{2})\bar{1}\right] & \left[234 \bar{3}\bar{1}\right] 
\\
\left[1234 \bar{3}\bar{2}\bar{1}\right] & \left[234 
\bar{3}\bar{2}\bar{1}\right] \\
 & 0
\end{array}
\right),
\quad
C(x)=\left(
\begin{array}{rrrc}
0  &  \cdots  & 0  & \left[1234 \bar{3}\bar{2}\bar{1}\right] \\
%
0  &  \cdots  & \cdots  & 0 \\
%
\vdots &  &   & \vdots \\
0 & \cdots & \cdots & 0
\end{array}
\right).
\end{eqnarray*}
%%%%%
The matrix $M(x,z)$ for $D_4^{(1)}$ is given by
\begin{math}
M(x,z) = A(x) + z B(x) + z^2 C(x).
\end{math}
%
\end{example}
%

Before going into the main theorem (Theorem \ref{th:GMGequalM})
we present several properties of the matrix $M(x,z) = A(x)+z B(x) + 
z^2C(x)$.
%
The only nonzero element of $C(x)$ is $C(x)_{1,2n}$
that is equal to $\ell(x)$.
The matrix $A(x)$ is lower triangular. 
%
Explicit expressions for the nonzero elements of $A(x)$ are given as follows:
\begin{itemize}
	\item $A(x)_{i,i}=x_i/\xb_i \, (1 \leq i \leq n-1), \, = x_n \, (i=n),
	\, = 1/x_n \, (i=n+1), \, =\xb_{2n+1-i}/x_{2n+1-i} \, (n+2 \leq i \leq 
	2n)$. \\
	\item For $2 \leq i \leq n-1, 1 \leq j \leq i-1$, we have
	\begin{math}
	A(x)_{i,j} = x_j \cdots x_{i-1} (1 + x_i/\xb_i )
	\end{math}
	and 
	\begin{math}
	A(x)_{2n+1-j,2n+1-i} = \xb_j \cdots \xb_{i-1} (1 + \xb_i/x_i ).
	\end{math} \\
	\item For $1 \leq j \leq n-1$ we have
	\begin{math}
	A(x)_{n,j} = x_j \cdots x_n, A(x)_{n+1,j} = x_j \cdots x_{n-1}, 
        \end{math}
        \begin{math}
	A(x)_{2n+1-j,n} = \xb_j \cdots \xb_{n-1} x_n, 
	\end{math}
	and
	\begin{math}
	A(x)_{2n+1-j,n+1} = \xb_j \cdots \xb_{n-1}.
	\end{math} \\
	\item For $1 \leq i,j \leq n-1$ we have
	\begin{math}
	A(x)_{2n+1-i,j} = (x_j \cdots x_n ) \times (\xb_i \cdots \xb_{n-1}).
	\end{math}
\end{itemize}
%
\begin{lemma}
Let $F_i(a)={}^t\!G_i(a)$ and
\begin{displaymath}
d(x) = \mbox{diag} \left( x_1/\xb_1, \ldots, x_{n-1}/\xb_{n-1},x_n,
1/x_n,\xb_{n-1}/x_{n-1},\ldots,\xb_1/x_1 \right).
\end{displaymath}
Then the matrix $A(x)$ has the following factorization.
\begin{eqnarray}
\label{eq:factorizationofA}
A(x) &=& F_1(\xb_1) F_2(\xb_2) \cdots F_{n-2}(\xb_{n-2}) F_n(\xb_{n-1})
d(x) \\
& \times & F_{n-1}(x_{n-1}) F_{n-2}(x_{n-2}) \cdots F_2(x_2)F_1(x_1)
\nonumber
\end{eqnarray}
\end{lemma}
%
\begin{proof}
$F_1(\xb_1)^{-1} A(x)F_1(x_1)^{-1}$ is given by deleting the elements in
the 1st column and the $2n$-th row of $A(x)$ except the diagonal elements.
To illustrate we take the $n=4$ case. See Example \ref{ex:abc}.
%
\begin{displaymath}
F_1(\xb_1)^{-1} A(x)F_1(x_1)^{-1}=\left(
\begin{array}{rrrrrrrr}
\frac{[1]}{[\bar{1}]}  \\
%
  & \frac{ [2]}{ [\bar{2}]}  \\
%
  &  \left[2(3)\right]  & \frac{ [3]}{ [\bar{3}]}  \\
%
  &  \left[234\right] &  \left[34\right] &  [4]  \\
%
  &  \left[23\right] &  [3] & & \frac{ 1}{ [4]}   \\
%
 &  \left[234\bar{3}\right]  &  \left[34\bar{3}\right] & 
 \left[4\bar{3}\right] &  [\bar{3}] & \frac{ [\bar{3}]}{ [3]}   \\
%
 & \left[234\bar{3}\bar{2}\right] &  \left[34\bar{3}\bar{2}\right] & 
 \left[4\bar{3}\bar{2}\right] &  \left[\bar{3}\bar{2}\right] & 
 \left[(\bar{3})\bar{2}\right] & \frac{ [\bar{2}]}{ [2]}  \\
%
 &   &   &   &   &   &   & \frac{ [\bar{1}]}{ [1]}
%
\end{array}
\right).
\end{displaymath}
%
Similarly, $F_2(\xb_2)^{-1} F_1(\xb_1)^{-1} A(x)
F_1(x_1)^{-1} F_2(x_2)^{-1}$ is given by deleting the elements in
the 2nd column and the $(2n-1)$-th row of
$F_1(\xb_1)^{-1} A(x)F_1(x_1)^{-1}$ except the diagonal elements.
This process continues until we obtain $d(x)$.
For example, the final step for the $n=4$ case goes as
\begin{align*}
\left(
\begin{array}{cccc}
1 & & & \\
  & 1 & & \\
-\left[\bar{3}\right] & & 1 & \\
 & -\left[\bar{3}\right] & & 1
\end{array}
\right)
%
&\left(
\begin{array}{rrrr}
\frac{[3]}{ [\bar{3}]}  \\
%
\left[34\right] &  [4]  \\
%
\left[3\right] & & \frac{ 1}{ [4]}   \\
%
\left[34\bar{3}\right] &  \left[4\bar{3}\right] &  [\bar{3}] &
\frac{ [\bar{3}]}{ [3]}
%
\end{array}
\right)
%
\left(
\begin{array}{cccc}
1 & & & \\
-\left[3\right]   & 1 & & \\
& & 1 & \\
& & -\left[3\right] & 1
\end{array}
\right) \\
%
& =
\left(
\begin{array}{rrrr}
\frac{[3]}{ [\bar{3}]}  \\
%
&  [4]  \\
%
& & \frac{ 1}{ [4]}   \\
%
& & & \frac{ [\bar{3}]}{ [3]}
%
\end{array}
\right).
%
\end{align*}
\end{proof}
%
\begin{remark}
We note that the sequence $(1,2,\ldots,n-2,n,n-1,\ldots,1)$ 
extracted from the indices of $F$ in the right hand side of 
\eqref{eq:factorizationofA} is essentially the same as the 
one given in \cite{KMOTU,HKT2}.
\end{remark}
%
\begin{prop} \label{prop:meqdpd}
\begin{equation}
\label{eq:meqdpd}
M(x,\ell(x)^{-1}) = \frac{1}{\ell (x)} {\mathcal D}_2 (x) P
{\mathcal D}_1 (x),
\end{equation}
%
where
%
\begin{eqnarray*}
{\mathcal D}_1 (x) &=& \mbox{diag} \left(
A(x)_{2n,1},A(x)_{2n,2},\ldots,A(x)_{2n,2n} \right),\\
%
{\mathcal D}_2 (x) &=& \mbox{diag} \left(
A(x)_{1,1},A(x)_{2,1},\ldots,A(x)_{2n,1} \right),
\end{eqnarray*}
%
and $P$ is the matrix with all the entries being 1.
\end{prop}

\begin{corollary} \label{cor:B}
\[
B(x)={\mathcal D}_2(x)P{\mathcal D}_1(x)-\ell(x)A(x)-\ell(x)^{-1}C(x).
\]
\end{corollary}

Set $T=(\delta_{i+j,n+1}(-1)^{i-1})_{1 \leq i,j \leq n}$
and define a $2n$ by $2n$ matrix $S$ by
\begin{equation}
\label{eq:signmatrix}
S = \left(
\begin{array}{cc}
   & T \\
{}^t T &
\end{array}
\right).
\end{equation}
It is clear that $S^{-1} = \, ^t \! S = S$.
%
\begin{prop}\label{pr:MSMS}
$M(x,z) S \, ^t \! M(x,z) S = (1-z \ell(x))^2 E$.
\end{prop}
%
\begin{proof}
For brevity we write $M(x,z)=M$, $A(x)=A$, $B(x)=B$,
$C(x)=C$, ${\mathcal D}_1 (x)={\mathcal D}_1$,
${\mathcal D}_2 (x) = {\mathcal D}_2$, and $\ell(x)=\ell$.
For any $2n$ by $2n$ matrix $X$ we denote $S \, ^t \! X S$
by $\check{X}$.
Then we have
\begin{displaymath}
M \check{M} = A \check{A} + z (A \check{B} + B \check{A}) +
z^2 (A \check{C} + B \check{B} + C \check{A}).
\end{displaymath}
First we see that $A \check{A} = E$ since $A$ has the factorization
(\ref{eq:factorizationofA}) and any factor $X$ thereof satisfies
$X \check{X} = E$.
%
By Corollary \ref{cor:B} we have
\begin{math}
B = {\mathcal D}_2 P {\mathcal D}_1 - \ell A - \ell^{-1} C.
\end{math}
Let us denote
$\ell^{-1} C =(\delta_{i,1} \delta_{j,2n})_{1 \leq i,j \leq 2n}$ by $H$.
Then we have
\begin{align*}
A \check{B} &= A \check{{\mathcal D}_1} \check{P} \check{{\mathcal D}_2}
- \ell E - A H, \\
B \check{A} &= {\mathcal D}_2 P {\mathcal D}_1 \check{A}
- \ell E - H \check{A}.
\end{align*}
Here we have used $A \check{A} = E$ and $\check{H}=H$.
It is easy to see that
$A \check{{\mathcal D}_1} \check{P} \check{{\mathcal D}_2}=H \check{A} $,
${\mathcal D}_2 P {\mathcal D}_1 \check{A} = A H$,
and thereby $A \check{B} + B \check{A} = -2 \ell E$.

It remains to check the desired relation for one special value
of $z$ other than $z=0$.
Let us put $z=\ell^{-1}$.
By Proposition \ref{prop:meqdpd} we have
\begin{math}
M \check{M} = \ell^{-2} {\mathcal D}_2 P {\mathcal D}_1
\check{{\mathcal D}_1} \check{P} \check{{\mathcal D}_2}.
\end{math}
By using $A \check{A}=E$ we see
\begin{math}
P {\mathcal D}_1 \check{{\mathcal D}_1} \check{P} = O,
\end{math}
where $O$ is the zero matrix.
The proof is completed.
\end{proof}
\begin{corollary}\label{coro:detM}
$\det M(x,z) = (1-z \ell(x))^{2n}$.
\end{corollary}

On the set of variables of the geometric crystal $\B$ we introduce involutive 
automorphisms $\sigma_1$, $\sigma_n$ and $\tau$ by
%
\begin{align}
\label{eq:sigma1o}
\sigma_1 &: x_1 \longleftrightarrow \xb_1, \\
%
\label{eq:sigmano}
\sigma_n &: x_{n-1} \rightarrow x_{n-1}x_n, \quad
             \xb_{n-1} \rightarrow \xb_{n-1}x_n, \quad
             x_n \rightarrow 1/x_n,\\
%             
\label{eq:tauo}
\tau &: x_i \longleftrightarrow \xb_i, \quad
        \xb_i \longleftrightarrow x_i, \quad (1 \leq i \leq n-2) \\
     &\;\;\; x_{n-1} \rightarrow \xb_{n-1}x_n, \quad
             \xb_{n-1} \rightarrow x_{n-1}x_n, \quad
             x_n \rightarrow 1/x_n. \nonumber
\end{align}
%
Note that $\sigma_1$, $\sigma_n$ and $\tau$ are mutually commutative. 
For simplicity we write $x^{\sigma_1}$ to mean $\sigma_1(x)$, etc. 
We can easily check

\begin{lemma} \label{lem:sigma}
\begin{align}
\label{eq:sigma_1}
&\veps_1(x^{\sigma_1})=\veps_0(x),\quad \vphi_1(x^{\sigma_1})=
\vphi_0(x),\quad e_1^c(x^{\sigma_1})=(e^c_0(x))^{\sigma_1},\\
\label{eq:sigma_n}
&\veps_{n-1}(x^{\sigma_n})=\veps_n(x),\quad 
\vphi_{n-1}(x^{\sigma_n})=\vphi_n(x),\quad
e_{n-1}^c(x^{\sigma_n})=(e^c_n(x))^{\sigma_n}.
\end{align}
The other $\veps_i$ and $\vphi_i$ are invariant under $\sigma_1$ and 
$\sigma_n$.
\end{lemma}

Let us introduce the $2n$ by $2n$ matrices:
\begin{align}
\label{eq:JMatrixA}
J_1(z) &= z E_{1,2n} + z^{-1} E_{2n,1} + {\textstyle \sum_{i=2}^{2n-1}}
 E_{i,i}, \\
 \label{eq:JMatrixB}
J_n &= E_{n,n+1} + E_{n+1,n} + {\textstyle \sum_{i=1}^{n-1}}
(E_{i,i}+E_{2n+1-i,2n+1-i}), \\
%
\label{eq:JMatrixC}
J &= {\textstyle \sum_{i=1}^{2n}} E_{i,2n+1-i}.
\end{align}
The following lemma is immediate.

\begin{lemma} \label{lem:J}
\begin{align}
\label{eq:J_1}
J_1(z)G_1(a)&=G_0(\frac{a}{z})J_1(z),\\
\label{eq:J_n}
J_nG_{n-1}(a)&=G_n(a)J_n.
\end{align}
\end{lemma}
%
\begin{lemma}
\label{lem:JMJequalsM}
\begin{align}
\label{eq:JMJequalsMno1}
J_1(z) M(x,z) J_1(z) &= M(x^{\sigma_1},z), \\
\label{eq:JMJequalsMno2}
J_n M(x,z) J_n &= M(x^{\sigma_n},z), \\
\label{eq:JMJequalsMno3}
J \,^t \!M(x,z) J &= M(\tau(x),z).
\end{align}
\end{lemma}
%
\begin{proof}
First we prove (\ref{eq:JMJequalsMno1}).
In this case, only
the first and $2n$-th rows and columns
are changed.
Let us consider $J_1(z) A(x) J_1(z)$.
All the elements in the first column and the $2n$-th row of
this matrix are zero except the $(1,1)$ and
$(2n,2n)$ elements.
And we have
\begin{align*}
\left( J_1(z) A(x) J_1(z) \right)_{1,i} &=
\begin{cases}
\xb_1/x_1 & \mbox{for}\,i=1,\\
\ell(x)z^2 & \mbox{for}\,i=2n,\\
A(x)_{2n,i}z & \mbox{for}\,i \ne 1,2n,
\end{cases} \\
%
\left( J_1(z) A(x) J_1(z) \right)_{i,2n} &=
\begin{cases}
\ell(x)z^2 & \mbox{for}\,i=1,\\
x_1/\xb_1 & \mbox{for}\,i=2n,\\
A(x)_{i,1}z & \mbox{for}\,i \ne 1,2n.
\end{cases} \\
\end{align*}
%
Note that $\xb_1/x_1 = A(x^{\sigma_1})_{1,1}$,
$A(x)_{2n,i} = B(x^{\sigma_1})_{1,i} \, (i \ne 1,2n)$,
$x_1/\xb_1 = A(x^{\sigma_1})_{2n,2n} $, and
$A(x)_{i,1} = B(x^{\sigma_1})_{i,2n} \, (i \ne 1,2n)$.
Here we have used
\begin{align*}
B(x)_{1,i} &=
\begin{cases}
0 & \mbox{for}\,i=1,2n,\\
A(x^{\sigma_1})_{2n,i} & \mbox{for}\,i \ne 1,2n,
\end{cases} \\
%
B(x)_{i,2n} &=
\begin{cases}
0 & \mbox{for}\,i=1,2n,\\
A(x^{\sigma_1})_{i,1} & \mbox{for}\,i \ne 1,2n,
\end{cases} \\
\end{align*}
which is obtained from Corollary \ref{cor:B}.
Let us consider $J_1(z) B(x) J_1(z)$.
{}From Corollary \ref{cor:B} we find
$B(x)_{2n,i} = B(x)_{i,1} =0$ for $1 \leq i \leq 2n$.
Therefore all elements in the first row and the $2n$-th column of
this matrix are zero.
%
And we have
\begin{align*}
\left( J_1(z) B(x) J_1(z) \right)_{2n,i} &=
\begin{cases}
0 & \mbox{for}\,i=1,2n,\\
B(x)_{1,i}z^{-1} & \mbox{for}\,i \ne 1,2n,
\end{cases} \\
%
\left( J_1(z) B(x) J_1(z) \right)_{i,1} &=
\begin{cases}
0 & \mbox{for}\,i=1,2n,\\
B(x)_{i,2n} z^{-1} & \mbox{for}\,i \ne 1,2n.
\end{cases} \\
\end{align*}
Note that $B(x)_{1,i}=A(x^{\sigma_1})_{2n,i}$ and
$B(x)_{i,2n} = A(x^{\sigma_1})_{i,1}$ for $i \ne 1,2n$.
%
Let us consider $J_1(z) C(x) J_1(z)$.
The only nonzero element of this matrix is the $(2n,1)$ element
that takes the value $\ell(x) z^{-2}$.
Putting all the above relations together we obtain 
(\ref{eq:JMJequalsMno1}).
%%%%%%%%%%%%%%%%

Next let us consider (\ref{eq:JMJequalsMno2}).
{}From (\ref{eq:factorizationofA}) we have
\begin{eqnarray*}
J_n A(x)J_n &=& F_1(\xb_1) F_2(\xb_2) \cdots F_{n-2}(\xb_{n-2})
F_{n-1}(\xb_{n-1})
d(x^{\sigma_n}) \\
& \times & F_{n}(x_{n-1}) F_{n-2}(x_{n-2}) \cdots F_2(x_2)F_1(x_1).
\end{eqnarray*}
Since $F_{n-1}(\xb_{n-1}) d(x^{\sigma_n})=
d(x^{\sigma_n}) F_{n-1}(x_{n-1}x_n)$
%, $F_{n-1}(x_{n-1}x_n) F_{n}(x_{n-1}) =
%F_{n}(x_{n-1}) F_{n-1}(x_{n-1}x_n)$, and
and $d(x^{\sigma_n}) F_{n}(x_{n-1}) = F_{n}($ 
\par\noindent
$\xb_{n-1}x_n) d(x^{\sigma_n})$ 
we obtain $J_n A(x)J_n = A(x^{\sigma_n})$.
Note that $F_{n-1}$ and $F_n$ are commutative.
Clearly we have
$J_n C(x)J_n = C(x) = C(x^{\sigma_n})$.
Then by Corollary \ref{cor:B} we obtain
$J_n B(x)J_n = B(x^{\sigma_n})$.
The relation (\ref{eq:JMJequalsMno2}) is proved.

Finally we show (\ref{eq:JMJequalsMno3}).
{}From (\ref{eq:factorizationofA}) we have
\begin{eqnarray*}
J  \,^t \!A(x)J &=& F_1(x_1) F_2(x_2) \cdots F_{n-2}(x_{n-2})
F_{n-1}(x_{n-1})
d(\tau(x)) \\
& \times & F_{n}(\xb_{n-1}) F_{n-2}(\xb_{n-2}) \cdots 
F_2(\xb_2)F_1(\xb_1).
\end{eqnarray*}
Since $F_{n-1}(x_{n-1}) d(\tau(x))=
d(\tau(x)) F_{n-1}(\xb_{n-1}x_n)$
and $d(\tau(x)) F_{n}(\xb_{n-1}) = F_{n}(x_{n-1}$
\par\noindent
$x_n) d(\tau(x))$
we obtain
$J \,^t \!A(x)J = A(\tau(x))$.
Clearly we have
$J \,^t \!C(x)J = C(x) = C(\tau(x))$.
Then by Corollary \ref{cor:B} we can obtain
$J \,^t \!B(x)J = B(\tau(x))$.
The relation (\ref{eq:JMJequalsMno3}) is proved.
%%%%%%%%%%%%%%%%
\end{proof}

The following theorem states that the transformation $e^c_i$ of the 
geometric
$D^{(1)}_n$-crystal $\B$ is realized as the multiplications of unipotent 
matrices.
\begin{theorem}
\label{th:GMGequalM}
For $0 \leq i \leq n$ we have
\begin{equation}
G_i \left( \frac{c-1}{z^{\delta_{i0}}\veps_i(x)} \right) M(x,z)
G_i \left( \frac{c^{-1}-1}{z^{\delta_{i0}}\vphi_i(x)} \right) =
M(e^c_i (x),z).
\end{equation}
\end{theorem}

\begin{proof}
First note that the $i=n$ case can be derived from the $i=n-1$ case as 
follows.
\begin{align*}
M(e^c_n (x),z) &= J_n M(e^c_{n-1} (x^{\sigma_n}),z) J_n \\
%
&= J_n G_{n-1} \left( \frac{c-1}{\veps_{n-1}(x^{\sigma_n})}
\right) M(x^{\sigma_n},z)\;
G_{n-1} \left( \frac{c^{-1}-1}{\vphi_{n-1}(x^{\sigma_n})} \right) J_n \\
%
%&= J_n G_{n-1} \left( \frac{c-1}{\veps_{n}(x)}
%\right) M(x^{\sigma_n},z)\;
%G_{n-1} \left( \frac{c^{-1}-1}{\vphi_{n}(x)} \right) J_n \\
%
&= G_{n} \left( \frac{c-1}{\veps_{n}(x)}
\right) J_n M(x^{\sigma_n},z) J_n\;
G_{n} \left( \frac{c^{-1}-1}{\vphi_{n}(x)} \right) \\
%
&= G_{n} \left( \frac{c-1}{\veps_{n}(x)}
\right) M(x,z)\;
G_{n} \left( \frac{c^{-1}-1}{\vphi_{n}(x)} \right).
\end{align*}
Here we have used 
\eqref{eq:sigma_n},\eqref{eq:J_n},
\eqref{eq:JMJequalsMno2}.
In a similar way the $i=0$ case can be derived from the $i=1$ case 
by using \eqref{eq:sigma_1},\eqref{eq:J_1},\eqref{eq:JMJequalsMno1}.
%
Thus we can assume $1 \leq i \leq n-1$.
The identity to be proved is of degree 2 with respect to $z$.
However,
it is easy to see that the coefficients of $z^2$ in the both hand sides 
coincide.
Thus it remains to check the identity with two distinct values of $z$,
which we take to be $\ell(x)^{-1}$ and $0$.

First let us put $z = \ell(x)^{-1}$. From Proposition \ref{prop:meqdpd}
the identity to be proved reads as
\begin{equation}
G_i \left( \frac{c-1}{\veps_i(x)} \right)
{\mathcal D}_2(x) P {\mathcal D}_1(x)
G_i \left( \frac{c^{-1}-1}{\vphi_i(x)} \right) =
{\mathcal D}_2(e^c_i (x)) P {\mathcal D}_1(e^c_i (x)).
\end{equation}
This equality indeed holds since we have
\begin{eqnarray}
G_i \left( \frac{c-1}{\veps_i(x)} \right)
{\mathcal D}_2(x) P &=&
{\mathcal D}_2(e^c_i (x)) P, \\
%
P {\mathcal D}_1(x)
G_i \left( \frac{c^{-1}-1}{\vphi_i(x)} \right) &=&
P {\mathcal D}_1(e^c_i (x)),
\end{eqnarray}
%
or equivalently
\begin{eqnarray*}
&&A(e^c_i (x))_{k,1} = \begin{cases}
A(x)_{k,1} + A(x)_{k+1,1} \frac{c-1}{\veps_i(x)}
& \mbox{for}\; k=i, 2n-i,\\
A(x)_{k,1} & \mbox{otherwise},
\end{cases} \\
%
&&A(e^c_i (x))_{2n,k} = \begin{cases}
A(x)_{2n,k-1} \frac{c^{-1}-1}{\vphi_i(x)}+ A(x)_{2n,k}
& \mbox{for}\; k=i+1, 2n+1-i,\\
A(x)_{2n,k} & \mbox{otherwise}.
\end{cases}
\end{eqnarray*}
One can check them from (\ref{eq:firstcolumnofA}) and 
(\ref{eq:lastrowofA}).

Next let us put $z=0$.
We consider the $i=1$ case. The other cases are similar.
The identity to be proved reads as
\begin{equation}
\label{eq:GAGequalA}
G_1 \left( \frac{c-1}{\veps_1(x)} \right) A(x)
G_1 \left( \frac{c^{-1}-1}{\vphi_1(x)} \right) =
A(e^c_1 (x)).
\end{equation}
Substitute the factorized form of $A(x)$
(\ref{eq:factorizationofA}) into the left hand side.
Exchange the leftmost two matrices as
\begin{displaymath}
G_1 \left( \frac{c-1}{\veps_1(x)} \right) F_1(\xb_1) =
F_1(\xi_2^{-1} \xb_1) \tilde{G}_1 \left( \frac{c-1}{\veps_1(x)} \right).
\end{displaymath}
Here $\tilde{G}_1(a)$ is the matrix whose 1st and $(2n-1)$-th (resp.~2nd 
and
$2n$-th) diagonal elements are $\xi_2$ (resp.~$\xi_2^{-1}$) and the other
matrix elements are the same as those of $G_1(a)$.
Recall $\xi_2$ is defined in section \ref{subsec:defofB}.
Then the next step is to exchange the second and third matrices as
\begin{displaymath}
\tilde{G}_1 \left( \frac{c-1}{\veps_1(x)} \right) F_2(\xb_2) =
F_2(\xi_2 \xb_2) \tilde{G}_1 \left( \frac{c-1}{\veps_1(x)} \right).
\end{displaymath}
Similarly, for the rightmost three matrices we have
\begin{displaymath}
F_2(x_2)F_1(x_1) G_1 \left( \frac{c^{-1}-1}{\vphi_1(x)} \right) =
\hat{G}_1 \left( \frac{c^{-1}-1}{\vphi_1(x)} \right) F_2(c^{-1}\xi_2 x_2)
F_1(c \xi_2^{-1} x_1).
\end{displaymath}
Here $\hat{G}_1(a)$ is the matrix whose 1st and $(2n-1)$-th (resp.~2nd and
$2n$-th) diagonal elements are $c \xi_2^{-1}$ (resp.~$c^{-1} \xi_2$) and the 
other
matrix elements are the same as those of $G_1(a)$.
The other $F_i$'s commute with $\tilde{G}_1$ and $\hat{G}_1$.
Thus the left hand side of (\ref{eq:GAGequalA}) leads to
\begin{displaymath}
F_1(\xi_2^{-1} \xb_1) F_2(\xi_2 \xb_2) \cdots
\tilde{G}_1 \left( \frac{c-1}{\veps_1(x)} \right) d(x)
\hat{G}_1 \left( \frac{c^{-1}-1}{\vphi_1(x)} \right) \cdots
F_2(c^{-1} \xi_2 x_2) F_1(c \xi_2^{-1} x_1).
\end{displaymath}
Thanks to
\begin{displaymath}
d(x) \hat{G}_1 \left( \frac{c^{-1}-1}{\vphi_1(x)} \right) =
\tilde{G}_1 \left( \frac{c-1}{\veps_1(x)} \right)^{-1} d(e^c_1(x)),
\end{displaymath}
we obtain the right hand side of \eqref{eq:GAGequalA}.

The proof is completed.
\end{proof}

\subsection{Product of $M(x,z)$}

Throughout this subsection we consider the $L$-fold product $\B^{\times L}$ 
of
the geometric $D^{(1)}_n$-crystal $\B$ given in section 
\ref{subsec:defofB}.
Our main theorem here is Theorem \ref{th:ijieqjij} which states that
$\B^{\times L}$ also becomes a geometric crystal.

\begin{theorem}
\label{th:MMMequalGMMMG}
Let $\boldsymbol{x} = (x^1,\ldots,x^L)$ the set of variables of $\B^{\times 
L}$ and
set $e^c_i(\boldsymbol{x})=(y^1,\cdots,y^L)$.
Then we have
\begin{displaymath}
M(y^1,z) \cdots M(y^L,z) =
G_i\left(\frac{c-1}{z^{\delta_{i0}}\veps_i(\boldsymbol{x})} \right)
M(x^1,z) \cdots M(x^L,z) \;
G_i\left(\frac{c^{-1}-1}{z^{\delta_{i0}}\vphi_i(\boldsymbol{x})} \right),
\end{displaymath}
for $0 \leq i \leq n$.
\end{theorem}
\begin{proof}
{}From Theorem \ref{th:GMGequalM} we have
\begin{displaymath}
M(y^l,z) = M(e^{c_l}_i(x^l),z) =
G_i\left(\frac{c_l-1}{z^{\delta_{i0}}\veps_i(x^l)} \right)
M(x^l,z) \;
G_i\left(\frac{c_l^{-1}-1}{z^{\delta_{i0}}\vphi_i(x^l)} \right),
\end{displaymath}
for $1 \leq l \leq L$.
Using \eqref{eq:vepsforxxx}--\eqref{eq:def-of-c_l} we can check
\begin{eqnarray*}
&& \frac{c_{l-1}^{-1}-1}{\vphi_i(x^{l-1})} +
\frac{c_l-1}{\veps_i(x^l)} = 0 \quad \mbox{for $2 \leq l \leq L$}, \\
%
&&\frac{c_1-1}{\veps_i(x^1)} =
\frac{c-1}{\veps_i(\boldsymbol{x})}, \quad
\frac{c_L^{-1}-1}{\vphi_i(x^L)}=
\frac{c^{-1}-1}{\vphi_i(\boldsymbol{x})},
\end{eqnarray*}
which finish the proof.
\end{proof}
%

The following theorem plays a crucial role in the subsequent part of this 
paper.
It asserts that we can retrieve components $x^l$ from the product of 
matrices
$M(x^1,z)\cdots M(x^L,z)$.

\begin{theorem}
\label{th:uniqueness}
Let $\boldsymbol{x} = (x^1,\ldots,x^L)$ and
$\boldsymbol{y} = (y^1,\ldots,y^L)$ be sets of variables of $\B^{\times 
L}$.
Suppose
\begin{eqnarray*}
\ell(x^i)&=&\ell(y^i)\quad(1\le i\le L),\\
\ell(x^i)&\neq&\ell(x^j)\quad (i\neq j),
\end{eqnarray*}
and
\begin{equation}
\label{eq:mdotsmeqmdosm}
M(x^{1},z) \cdots M(x^{L},z) =M(y^{1},z) \cdots M(y^{L},z).
\end{equation}
Then we have $x^{i} = y^{i}\;(1 \leq i \leq L)$.
\end{theorem}
%
\begin{proof}
Denote $\ell(x^{1}) = \ell(y^{1})$ by $a^{-1}$.
Let us put $z=a$ in (\ref{eq:mdotsmeqmdosm}). Using Proposition 
\ref{prop:meqdpd}
we obtain
\begin{align*}
&{\mathcal D}_2 (y^{1})^{-1} {\mathcal D}_2 (x^{1}) P \\
&=P {\mathcal D}_1 (y^{1}) M(y^{2},a) \cdots M(y^{L},a)
M(x^{L},a)^{-1} \cdots M(x^{2},a)^{-1} {\mathcal D}_1 (x^{1})^{-1}.
\end{align*}
Here we used the property that $\det M(x,z) \ne 0$ unless 
$z=\ell(x)^{-1}$.
See Corollary \ref{coro:detM}.
We see that the both sides of this equation should be a scalar multiple of
$P$.
{}From the $2n$-th row of the left hand side
we see that the scalar should be $1$. Hence we have
${\mathcal D}_2 (x^{1}) = {\mathcal D}_2 (y^{1})$,
that implies $x^{1} = y^{1}$.
In the same way we can deduce $x^{i} = y^{i}$ for all $i$.
\end{proof}

We prepare lemmas. Let $\boldsymbol{x} = (x^1,\ldots,x^L)$ be the set of 
variables
of $\B^{\times L}$. Define $\boldsymbol{x}^{\sigma_a}
=((x^1)^{\sigma_a},\ldots,(x^L)^{\sigma_a})$ for $a=1,n$.

\begin{lemma}
\label{lem:vepsvphisigma}
\begin{align*}
&\veps_{1}(\boldsymbol{x}^{\sigma_1})
=\veps_0(\boldsymbol{x}), \quad
\vphi_{1}(\boldsymbol{x}^{\sigma_1})
=\vphi_0(\boldsymbol{x}),\\
&\veps_{n-1}(\boldsymbol{x}^{\sigma_n})
=\veps_n(\boldsymbol{x}), \quad
\vphi_{n-1}(\boldsymbol{x}^{\sigma_n})
=\vphi_n(\boldsymbol{x}).
\end{align*}
%Other $\veps_i$ and $\vphi_i$ are invariant under $\sigma_n$ and 
%$\sigma_1$.
\end{lemma}

\begin{proof}
The $L=1$ case is given in Lemma \ref{lem:sigma}.
By (\ref{eq:vepsforxxx}) and (\ref{eq:vphiforxxx}) we see that
the claim is also true for $L \geq 2$.
\end{proof}

\begin{lemma}
\label{lem:Dynkinrightmost}
\begin{align*}
e^c_{1} (\boldsymbol{x}^{\sigma_1}) &=
(e^c_0 (\boldsymbol{x}))^{\sigma_1}, \\
e^c_{n-1} (\boldsymbol{x}^{\sigma_n}) &=
(e^c_n (\boldsymbol{x}))^{\sigma_n}.
\end{align*}
\end{lemma}

\begin{proof}
Denote the left (resp.~right) hand side of the 
second relation by
$\boldsymbol{y}=(y^1,\cdots,y^L)$
(resp.~$\boldsymbol{w}=(w^1,\cdots,w^L)$).
We have
\begin{align*}
&M(y^1,z) \cdots M(y^L,z) \\
&=
G_{n-1}\left(\frac{c-1}{\veps_{n-1}(\boldsymbol{x}^{\sigma_n})} \right)
M((x^1)^{\sigma_n},z) \cdots M((x^L)^{\sigma_n},z) \;
G_{n-1}\left(\frac{c^{-1}-1}{\vphi_{n-1}(\boldsymbol{x}^{\sigma_n})}
\right) \\
%
&=
G_{n-1}\left(\frac{c-1}{\veps_{n}(\boldsymbol{x})} \right)
J_n M(x^1,z) \cdots M(x^L,z) J_n\;
G_{n-1}\left(\frac{c^{-1}-1}{\vphi_{n}(\boldsymbol{x})}
\right) \\
%
&=
J_n \; G_{n}\left(\frac{c-1}{\veps_{n}(\boldsymbol{x})} \right)
M(x^1,z) \cdots M(x^L,z) \;
G_{n}\left(\frac{c^{-1}-1}{\vphi_{n}(\boldsymbol{x})}
\right) J_n\\
%
&= M(w^1,z) \cdots M(w^L,z) .
\end{align*}
Here we have used Lemma \ref{lem:J}, \ref{lem:JMJequalsM}, 
\ref{lem:vepsvphisigma}
and Theorem \ref{th:MMMequalGMMMG}.
Now Theorem \ref{th:uniqueness} tells 
$\boldsymbol{y}=\boldsymbol{w}$.
The first relation is similar.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Our main theorem in this section is

\begin{theorem}
\label{th:ijieqjij}
$\B^{\times L}$ is a geometric crystal. Namely,
\begin{equation}
\label{eq:ijeqji}
e^d_j e^c_i (\boldsymbol{x}) =
e^c_i e^d_j (\boldsymbol{x})
\end{equation}
for $\langle\alpha^\vee_i,\alpha_j\rangle=0$, and
\begin{equation}
\label{eq:ijieqjij}
e^d_i e^{cd}_j e^c_i (\boldsymbol{x}) =
e^c_j e^{cd}_i e^d_j (\boldsymbol{x})
\end{equation}
for 
$\langle\alpha^\vee_i,\alpha_j\rangle=\langle\alpha^\vee_j,\alpha_i\rangle=-  
1$.
\end{theorem}
%
\begin{proof}
We prove (\ref{eq:ijieqjij}) first.
Denote the left (resp.~right) hand side of (\ref{eq:ijieqjij}) by
$\boldsymbol{y}=(y^1,\cdots,y^L)$
(resp.~$\boldsymbol{w}=(w^1,\cdots,w^L)$).
It suffices to prove
$M(y^1,z) \cdots M(y^L,z) = M(w^1,z) \cdots M(w^L,z)$
by Theorem \ref{th:uniqueness}.
Let us define
\begin{align}
\label{eq:PPQQ1}
P &= G_i\left(\frac{d-1}{\veps_{i}(e^{cd}_j e^c_i(
\boldsymbol{x}))} \right)
G_j\left(\frac{cd-1}{\veps_{j}(e^c_i(
\boldsymbol{x}))} \right)
G_i\left(\frac{c-1}{\veps_{i}(\boldsymbol{x})} \right), \\
%
P' &= G_j\left(\frac{c-1}{\veps_{j}(e^{cd}_i e^d_j(
\boldsymbol{x}))} \right)
G_i\left(\frac{cd-1}{\veps_{i}(e^d_j(
\boldsymbol{x}))} \right)
G_j\left(\frac{d-1}{\veps_{j}(\boldsymbol{x})} \right), \\
%
Q &= G_i\left(-\frac{d^{-1}-1}{\vphi_{i}(e^{cd}_j e^c_i(
\boldsymbol{x}))} \right)
G_j\left(-\frac{(cd)^{-1}-1}{\vphi_{j}(e^c_i(
\boldsymbol{x}))} \right)
G_i\left(-\frac{c^{-1}-1}{\vphi_{i}(\boldsymbol{x})} \right), \\
%
\label{eq:PPQQ4}
Q' &= G_j\left(-\frac{c^{-1}-1}{\vphi_{j}(e^{cd}_i e^d_j(
\boldsymbol{x}))} \right)
G_i\left(-\frac{(cd)^{-1}-1}{\vphi_{i}(e^d_j(
\boldsymbol{x}))} \right)
G_j\left(-\frac{d^{-1}-1}{\vphi_{j}(\boldsymbol{x})} \right).
\end{align}
% Then we have $P=P'$ due to Proposition \ref{prop:eee} and
% Lemma \ref{lem:epsandphiforaaa}.
For notational simplicity we have supposed $i,j \ne 0$.
Otherwise the associated arguments should be divided by $z$.
{}From Theorem \ref{th:MMMequalGMMMG} we have
\begin{align*}
M(y^1,z) \cdots M(y^L,z) &= P M(x^1,z) \cdots M(x^L,z) Q^{-1},\\
M(w^1,z) \cdots M(w^L,z) &= P' M(x^1,z) \cdots M(x^L,z) {Q'}^{-1}.
\end{align*}
We show $Q=Q'$ only, for $P=P'$ follows from this by using
\eqref{eq:gamma} and Definition \ref{def:pre-cry} (iii).
{}From \eqref{eq:GiGjGi=GjGiGj} it suffices to check
\begin{align*}
&\vphi_{i}(\boldsymbol{x}) \vphi_{j}(e^c_i(
\boldsymbol{x})) = \vphi_{i}(e^d_j(
\boldsymbol{x})) \vphi_{j}(e^{cd}_i e^d_j(
\boldsymbol{x})), \\
%
&\vphi_{j}(e^c_i(
\boldsymbol{x})) \vphi_{i}(e^{cd}_j e^c_i(
\boldsymbol{x})) =
\vphi_{j}(\boldsymbol{x}) \vphi_{i}(e^d_j(
\boldsymbol{x})), \\
%
&\frac{c^{-1}-1}{\vphi_{i}(\boldsymbol{x})} +
\frac{d^{-1}-1}{\vphi_{i}(e^{cd}_j e^c_i(
\boldsymbol{x}))} =
\frac{(cd)^{-1}-1}{\vphi_{i}(e^d_j(
\boldsymbol{x}))}.
\end{align*}
%
Let us prove the first identity.
Using Lemma \ref{lem:eps-phi-decomp} (b) for the multiple product case we 
have
\begin{align*}
\vphi_{j}(e^{cd}_i e^d_j(
\boldsymbol{x})) &=
\frac{
\vphi_{ij}(e^{cd}_i e^d_j(\boldsymbol{x})) +
\vphi_{ji}(e^{cd}_i e^d_j(\boldsymbol{x}))}
{\vphi_i(e^{cd}_ie^d_j(\boldsymbol{x}))}\\
&=
\frac{
cd \vphi_{ij}(\boldsymbol{x}) +
d \vphi_{ji}(\boldsymbol{x})}
{cd \vphi_{i}( e^d_j(\boldsymbol{x}))} .
\end{align*}
Therefore
\begin{align*}
\vphi_{i}( e^d_j(\boldsymbol{x}))
\vphi_{j}(e^{cd}_i e^d_j(
\boldsymbol{x})) &=
c^{-1} (c \vphi_{ij}(\boldsymbol{x}) + \vphi_{ji}(\boldsymbol{x})) \\
%
& =
c^{-1} \vphi_{i}(e^{c}_i (\boldsymbol{x}))
\vphi_{j}(e^{c}_i (\boldsymbol{x})) \\
& =
\vphi_{i}(\boldsymbol{x})
\vphi_{j}(e^{c}_i (\boldsymbol{x})).
\end{align*}

The proof of the second identity is similar.
Let us consider the third one.
By similar calculation we have
\[
\vphi_{i}(e^{d}_j (
\boldsymbol{x})) =\frac{
\vphi_{ij}(\boldsymbol{x}) +
d \vphi_{ji}(\boldsymbol{x})}
{d \vphi_{j}(\boldsymbol{x})},\quad
%
\vphi_{i}(e^{cd}_j e^c_i(
\boldsymbol{x})) =
\frac{
\vphi_{ij}(\boldsymbol{x}) +
d \vphi_{ji}(\boldsymbol{x})}{d}
\times
\frac{
c\vphi_i(\boldsymbol{x})}
{c\vphi_{ij}(\boldsymbol{x}) +
\vphi_{ji}(\boldsymbol{x})}.
\]
%
Therefore the identity to be proved is written as
\begin{displaymath}
\frac{c^{-1}-1}{\vphi_{i}(\boldsymbol{x})} +
\frac{(d^{-1}-1)d}{
\vphi_{ij}(\boldsymbol{x}) +
d \vphi_{ji}(\boldsymbol{x})} \cdot
%
\frac{
c \vphi_{ij}(\boldsymbol{x}) +
\vphi_{ji}(\boldsymbol{x})}
{c \vphi_{i}(\boldsymbol{x})} =
\frac{((cd)^{-1}-1)d \vphi_{j}(\boldsymbol{x})}{
\vphi_{ij}(\boldsymbol{x}) +
d \vphi_{ji}(\boldsymbol{x})},
\end{displaymath}
or equivalently
\begin{displaymath}
(c^{-1}-1) (\vphi_{ij}(\boldsymbol{x}) +
d \vphi_{ji}(\boldsymbol{x})) +
(1-d) (\vphi_{ij}(\boldsymbol{x}) +
c^{-1} \vphi_{ji}(\boldsymbol{x})) =
(c^{-1}-d ) \vphi_i(\boldsymbol{x}) \vphi_j(\boldsymbol{x}).
\end{displaymath}
This equality holds since $\vphi_i(\boldsymbol{x}) \vphi_j(\boldsymbol{x}) 
=
\vphi_{ij}(\boldsymbol{x}) + \vphi_{ji}(\boldsymbol{x})$.

The proof of \eqref{eq:ijeqji} is much simpler due to
Lemma \ref{lem:eps-phi-decomp} (a) for the multiple product case.
The proof is completed.
\end{proof}
We present an alternative proof of this theorem
in appendix \ref{sec:appA}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Tropical $R$}\label{sec:R}

\subsection{Definition and basic properties}
\label{subsec:properties}
Let $M(x,z)=A(x)+zB(x)+z^2C(x)$ be the $2n$ by $2n$ matrix defined in
the previous section.
Consider the relation
\begin{equation}
M(x,z)M(y,z)=M(x',z)M(y',z).
\label{eq:mmeqmmx}
\end{equation}
Given $x, y$, we regard it as a system of simultaneous equations for 
the $4n-2$ unknowns $x', y'$ 
containing a generic parameter $z$.
%
\begin{theorem}
\label{pr:mainmain}
There is a unique solution of (\ref{eq:mmeqmmx})
under the constraints $\ell(x)=\ell(y') \ne \ell(y)=\ell(x')$.
The map $(x,y) \rightarrow (x',y')$ is a birational transformation.
\end{theorem}
%

\begin{definition}\label{def:RRRR}
The unique solution $(x',y')$ of 
(\ref{eq:mmeqmmx}) specified in Theorem 
\ref{pr:mainmain} is denoted by $R(x,y)$, and the 
birational transformation $(x,y) \rightarrow R(x,y)$ is called 
the {\em tropical $R$} for the geometric $D^{(1)}_n$-crystal $\B$.
\end{definition}

In Theorem \ref{pr:mainmain} uniqueness is immediate 
{}from Theorem \ref{th:uniqueness}. 
%
To show the existence, we introduce an explicit 
birational transformation ${\tilde R}$ 
in section \ref{subsec:tilde_R}.
%
We then prove in section \ref{subsec:tilde_R_equals_R} 
that ${\tilde R}$ actually solves (\ref{eq:mmeqmmx}), hence 
$R = {\tilde R}$.

Here we exhibit basic properties of the
tropical $R$ by assuming its existence.
%
Let $x=(x_1,\ldots,\xb_1)$ and
$y=(y_1,\ldots,\yb_1)$ be sets of variables for the geometric $D^{(1)}_n$-crystal
$\B$.
%
On the pair $(x,y)$ we introduce mutually commuting 
involutive automorphisms:
%
\begin{equation}\label{eq:pairauto}
(x,y)^{\sigma_1} = (x^{\sigma_1},y^{\sigma_1}), \quad
(x,y)^{\sigma_n} = (x^{\sigma_n},y^{\sigma_n}), \quad
(x,y)^{*} = (x^{*},y^{*})
\end{equation}
in terms of $\sigma_1, \sigma_n$ specified in 
(\ref{eq:sigma1o}) and (\ref{eq:sigmano}).
%
The automorphism $\ast$ is the only one that 
mixes $x$ and $y$ and defined as
\begin{align}
\label{eq:ast}
\ast &: x_i \longleftrightarrow \yb_i, \quad
\xb_i \longleftrightarrow y_i \quad (1 \leq i \leq n-1),
\quad x_n \longleftrightarrow y_n.
\end{align}
%
\begin{lemma}
\label{lem:JMJequalsMxxx}
Set 
\begin{equation*}
J_* = E_{n,n} + E_{n+1,n+1} + \sum_{i=1}^{n-1} 
(E_{i,2n+1-i}+E_{2n+1-i,i}).
\end{equation*}
Then we have
\begin{align*}
&J_* \,^t\!M(y,z) J_* = M(x^*,z),\;J_* \,^t\!M(x,z) J_* = M(y^*,z),\\
%
&J_*\, {\mathcal D}_1(y) J_* = {\mathcal D}_2(x^\ast), \; 
J_*\, {\mathcal D}_1(x) J_* = {\mathcal D}_2(y^\ast).
\end{align*}
\end{lemma}
\begin{proof}
By noting that  $x^* = \tau(y^{\sigma_n})$,
$y^* = \tau(x^{\sigma_n})$, and $J_* = J J_n$, 
the first two relations follow from 
(\ref{eq:JMJequalsMno2}) and (\ref{eq:JMJequalsMno3}).
The latter relations are derived by considering 
the 1st column of the former.
\end{proof}
%

Let us present basic properties of $R$.
\begin{prop}
\label{lem:sigmaonexx}
\begin{align}
\label{eq:sigma_1_commute_with_R}
R((x,y)^{\sigma_1}) &=(R(x,y))^{\sigma_1}, \\
\label{eq:sigma_n_commute_with_R}
R((x,y)^{\sigma_n}) &=(R(x,y))^{\sigma_n}, \\
\label{eq:ast_commute_with_R}
R((x,y)^*) & =(R(x,y))^*.
\end{align}
\end{prop}
%
\begin{proof}
Write the left (resp.~right) side 
of (\ref{eq:sigma_1_commute_with_R}) as
$((x^{\sigma_1})',(y^{\sigma_1})')$
(resp.~$((x')^{\sigma_1},(y')^{\sigma_1})$).
Then we have
\begin{align*}
M((x^{\sigma_1})',z) M((y^{\sigma_1})',z) &=
M(x^{\sigma_1},z) M(y^{\sigma_1},z) \\
&= J_1(z) M(x,z) M(y,z) J_1(z) \\
&= J_1(z) M(x',z) M(y',z) J_1(z) \\
&= M((x')^{\sigma_1},z) M((y')^{\sigma_1},z).
\end{align*}
Here we have used Lemma \ref{lem:JMJequalsM}.
Due to Theorem \ref{th:uniqueness} we have 
(\ref{eq:sigma_1_commute_with_R}).
(\ref{eq:sigma_n_commute_with_R}) can be shown similarly.
%
Write the left (resp.~right) side of
(\ref{eq:ast_commute_with_R}) as
$((x^{*})',(y^{*})')$
(resp.~$((x')^{*},(y')^{*})$).
Then we have
\begin{align*}
M((x^{*})',z) M((y^{*})',z) &=
M(x^{*},z) M(y^{*},z) \\
&= J_* \,^t\!M(y,z) \,^t\!M(x,z) J_* \\
& = J_* \,^t\!(M(x,z) M(y,z)) J_* \\
& = J_* \,^t\!(M(x',z) M(y',z)) J_* \\
&= J_* \,^t\!M(y',z) \,^t\!M(x',z) J_* \\
&= M((x')^{*},z) M((y')^{*},z).
\end{align*}
Here we have used Lemma \ref{lem:JMJequalsMxxx}.
Again Theorem \ref{th:uniqueness} implies 
(\ref{eq:ast_commute_with_R}).
\end{proof}
%
\begin{prop}
\label{th:conservesignatures}
$\veps_i(R(x,y)) = \veps_i(x,y), \quad \vphi_i(R(x,y)) = \vphi_i(x,y)$.
\end{prop}
%
\begin{proof}
For $1 \leq i \leq n-1$, Lemma \ref{lem:epsandphiforaaa} states that
\begin{align*}
\veps_i(x,y) &= \frac{(M(x,0)M(y,0))_{i+1,i}}{(M(x,0)M(y,0))_{i,i}}, \\
\vphi_i(x,y) &= \frac{(M(x,0)M(y,0))_{i+1,i}}{(M(x,0)M(y,0))_{i+1,i+1}},
\end{align*}
hence the claim follows.
%
The case $i=0$ is reduced to $i=1$ as
\begin{align*}
\veps_0(R(x,y)) &= \veps_0 ((R(x^{\sigma_1},y^{\sigma_1}))^{\sigma_1}) \\
&= \veps_1(R(x^{\sigma_1},y^{\sigma_1})) \\
&= \veps_1(x^{\sigma_1},y^{\sigma_1}) \\
&= \veps_0(x,y).
\end{align*}
Here we have used Lemma \ref{lem:vepsvphisigma} and
Proposition \ref{lem:sigmaonexx}.
%
Similarly $i=n$ case follows from  $i=n-1$ case.
\end{proof}
%
\begin{prop} \label{th:eRequalsRe}
\begin{displaymath}
e^c_i R = R e^c_i.
\end{displaymath}
\end{prop}
\begin{proof}
For $e_i^c(x,y)=(e_i^{c_1}(x),e_i^{c_2}(y))$ we write 
\begin{displaymath}
R(e_i^c(x,y))=((e_i^{c_1}(x))',(e_i^{c_2}(y))'),
\end{displaymath}
and for $R(x,y) = (x',y')$,
\begin{displaymath}
e_i^c(R(x,y))=(e_i^{c_1'}(x'),e_i^{c_2'}(y')),
\end{displaymath}
where
\begin{math}
c_1'=\frac{c\vphi_i(x')+\veps_i(y')}{\vphi_i(x')+\veps_i(y')},\;
c_2'=\frac{\vphi_i(x')+\veps_i(y')}{\vphi_i(x')+c^{-1}\veps_i(y')}.
\end{math}
By using Theorem \ref{th:MMMequalGMMMG} and Proposition 
\ref{th:conservesignatures}
we have
\begin{align*}
&M((e_i^{c_1}(x))',z)M((e_i^{c_2}(y))',z) =
M(e_i^{c_1}(x),z)M(e_i^{c_2}(y),z) \\
%
&=G_i\left( \frac{c-1}{z^{\delta_{i0}}\veps_i(x,y)} \right)
M(x,z)M(y,z) \;
G_i\left( \frac{c^{-1}-1}{z^{\delta_{i0}}\vphi_i(x,y)} \right)\\
%
&=G_i\left( \frac{c-1}{z^{\delta_{i0}}\veps_i(x',y')} \right)
M(x',z)M(y',z) \;
G_i\left( \frac{c^{-1}-1}{z^{\delta_{i0}}\vphi_i(x',y')} \right)\\
%
&=M(e_i^{c_1'}(x'),z)M(e_i^{c_2'}(y'),z).
\end{align*}
%
Then Theorem \ref{th:uniqueness} tells
$((e_i^{c_1}(x))',(e_i^{c_2}(y))') = (e_i^{c_1'}(x'),e_i^{c_2'}(y'))$.
\end{proof}
%

\begin{prop} \label{th:RRequalsId}
$R(R(x,y))=(x,y)$.
\end{prop}
\begin{proof}
Writing $R(x,y)=(x',y')$ and $R(x',y')=(x'',y'')$, we get
$M(x,z)M(y,z) = M(x',z)M(y',z)=M(x'',z)M(y'',z)$ by definition.
Then $(x,y)=(x'',y'')$ owing to Theorem \ref{th:uniqueness}.
\end{proof}
%

Finally we prove the Yang-Baxter equation.
Let $\B\times\B\times\B$ be the 3-fold product of the 
geometric $D^{(1)}_n$-crystal $\B$.
Denote by $R_{12}$ (resp.~$R_{23}$)
the birational map on $\B \times \B \times \B$
that acts on the first (resp. last) two components as $R$ and the other
single component trivially.
\begin{prop}
\label{th:yang-baxter}
We have $R_{12} R_{23} R_{12} = R_{23} R_{12} R_{23}$
as a birational map on $\B \times \B \times \B$.
\end{prop}
\begin{proof}
Let $(x^1,x^2,x^3) \in \B \times \B \times \B$.
Writing $(u^{1},u^{2},u^{3}) = R_{12} R_{23} R_{12} (x^1,x^2,x^3)$ 
and $(v^{1},v^{2},v^{3}) = R_{23} R_{12} R_{23}(x^1,x^2,x^3)$, 
one has
\begin{align*}
M(u^{1},z) M(u^{2},z) M(u^{3},z) &=
M(x^1,z) M(x^2,z) M(x^3,z) \\
&=M(v^{1},z) M(v^{2},z) M(v^{3},z).
\end{align*}
Thus $(u^{1},u^{2},u^{3}) = (v^{1},v^{2},v^{3})$
follows {}from Theorem \ref{th:uniqueness}.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Birational map $\tilde{R}$}
\label{subsec:tilde_R}
\noindent
Let $V_0=(A(x)A(y))_{2n,1}$. Explicitly we have
\begin{align}
\label{eq:16}
V_0 &= \ell (x) \frac{y_1}{\yb_1} + \ell (x) \sum_{m=2}^{n-1}
\left( \prod_{i=1}^{m-1} \frac{y_i}{x_i} \right)
\left( 1+\frac{y_{m}}{\yb_{m}} \right) +
%% \ell (x) \left( \prod_{i=1}^{n-1} \frac{y_i}{x_i} \right) y_n \\
\left( \prod_{i=1}^{n-1} \xb_i y_i \right) x_n y_n \\
%
\nonumber
&+ \ell (y) \frac{\xb_1}{x_1} + \ell (y) \sum_{m=2}^{n-1}
\left( \prod_{i=1}^{m-1} \frac{\xb_i}{\yb_i} \right)
\left( 1+\frac{\xb_{m}}{x_{m}} \right) +
%%\ell (y) \left( \prod_{i=1}^{n-1} \frac{\xb_i}{\yb_i} \right) 
\left( \prod_{i=1}^{n-1} \xb_i y_i \right).
\end{align}
%
Define $V_i \, (1 \leq i \leq n-1)$ by
\begin{align}
&V_i = \frac{\yb_i}{\xb_i}V_{i-1} + (\ell (x) - \ell (y))\left( 
1+\frac{\yb_i}{x_i} \right),
\quad (1 \leq i \leq n-2)
\label{eq:vind}\\
%
&V_{n-1} = \frac{\yb_{n-1}}{\xb_{n-1}}V_{n-2} +
(\ell (x) - \ell (y)) \left( \frac{1}{y_n } +
\frac{\yb_{n-1}}{ x_{n-1} } \right).
\label{eq:unv}
\end{align}
%
%
Define $W_i \, (1 \leq i \leq n-1)$ by
\begin{eqnarray}
&& W_i = V_i V_i^* + (\ell (y) - \ell (x))V_i^* + (\ell (x) - \ell 
(y))V_i,
\quad (1 \leq i \leq n-2)
\label{eq:27c}\\
%
&& W_{n-1} = V_{n-1} V_{n-1}^{*}.
\label{eq:27b}
\end{eqnarray}
%
Here $V_i^\ast = V_i((x,y)^\ast)$, and similar notations 
will be used for $\sigma_1, \sigma_n$ from now on.

\begin{definition}
\label{th:tropicalR}
A rational transformation specified below is 
denoted by $\tilde{R}(x,y) = (x',y')$:
\begin{eqnarray*}
&& x_1' = y_1 \frac{V_0^{\sigma_1}}{V_1}, \quad
\xb_1' = \yb_1 \frac{V_0}{V_1}, \\
%
&& x_i' = y_i \frac{V_{i-1}W_i}{V_iW_{i-1}}, \quad
\xb_i' = \yb_i \frac{V_{i-1}}{V_i}, \quad (2 \leq i \leq n-1)\\
%
&& x_n' = y_n \frac{V_{n-1}}{V_{n-1}^{*}},  \\
%
%
&& y_1' = x_1 \frac{V_0}{V_1^*}, \quad
\yb_1' = \xb_1 \frac{V_0^{\sigma_1}}{V_1^*}, \\
%
&& y_i' = x_i \frac{V_{i-1}^*}{V_i^*}, \quad
\yb_i' = \xb_i \frac{V_{i-1}^* W_i}{V_i^* W_{i-1}}, \quad (2 \leq i \leq 
n-1)\\
%
&& y_n' = x_n \frac{V_{n-1}^*}{V_{n-1}}.
\end{eqnarray*}
\end{definition}
%

To provide an explicit formula for $V_i$'s, we introduce 
\begin{align*}
\theta_{i,j}(x,y) &=
\begin{cases}
\ell(x) {\displaystyle
\prod_{k=j+1}^{i} \frac{\yb_k}{\xb_k}} & \mbox{for} \quad
1 \leq j \leq i, \\
\ell(y) {\displaystyle
\prod_{k=i+1}^{j} \frac{\xb_k}{\yb_k}} & \mbox{for} \quad
i+1 \leq j \leq n-2,
\end{cases}
\\
\theta'_{i,j}(x,y) &= \ell(x) \left( \prod_{k=1}^{i} \frac{\yb_k}{\xb_k}
\right) \left(
\prod_{k=1}^{j} \frac{y_k}{x_k} \right) \quad
\mbox{for} \quad j=1,\ldots,n-2,\\
%
\end{align*}
\begin{align*}
\eta_{i,j}(x,y) &=
\begin{cases}
\ell(x) {\displaystyle 
\left( \prod_{k=j+1}^{i} \frac{\yb_k}{\xb_k} \right) 
\left( \frac{\yb_{j}}{x_{j}} \right)} &\mbox{for} \quad 1 \leq j \leq i,\\
\ell(y) {\displaystyle 
\left( \prod_{k=i+1}^{j} \frac{\xb_k}{\yb_k} \right) 
\left( \frac{\yb_{j}}{x_{j}} \right)} &\mbox{for} \quad i+1 \leq j \leq 
n-1,\\
\ell(y) {\displaystyle \left( \prod_{k=i+1}^{n-1} \frac{\xb_k}{\yb_k} 
\right)
x_n }&\mbox{for} \quad j=n,
\end{cases}
\\
\eta'_{i,j}(x,y) &=
\begin{cases}
\ell(x) {\displaystyle \left( \prod_{k=1}^{i} \frac{\yb_k}{\xb_k}
\right) \left(
\prod_{k=1}^{j} \frac{y_k}{x_k} \right) \left( \frac{x_{j}}{\yb_{j}} \right) 
}
\quad \mbox{for} \quad 1 \leq j \leq n-1,\\
%
\ell(x) {\displaystyle \left( \frac{\ell(x)}{\ell(y)} \right)
^{\delta_{i,n-1}}
\left( \prod_{k=1}^{i} \frac{\yb_k}{\xb_k}
\right) \left(
\prod_{k=1}^{n-1} \frac{y_k}{x_k} \right) \left( \frac{1}{x_{n}} \right) }
\quad \mbox{for} \quad j=n.
\end{cases}
\end{align*}
%
\begin{prop}
\label{th:subtractionfree}
For $0 \leq i \leq n-1$ one has
\begin{displaymath}
V_i = \sum_{j=1}^{n-2} \left( \theta_{i,j}(x,y) + \theta'_{i,j}(x,y) 
\right)
+ \sum_{j=1}^{n} \left( \eta_{i,j}(x,y) + \eta'_{i,j}(x,y) \right).
\end{displaymath}
\end{prop}
\begin{proof}
For $i=0$ the above expression agrees with (\ref{eq:16}).
%
Note that $\eta_{0,n}(x,y)=
\left( \prod_{i=1}^{n-1} \xb_i y_i \right) x_n y_n$ and
$\eta'_{0,n}(x,y)=
\left( \prod_{i=1}^{n-1} \xb_i y_i \right)$.
For $1 \leq i \leq n-2$ the recursion relation 
(\ref{eq:vind}) is valid since the nonzero 
contribution to the difference $V_i - V_{i-1}\yb_i/\xb_i$ only comes 
{}from $\theta_{i,i}(x,y)-\frac{\yb_i}{\xb_i}\theta_{i-1,i}(x,y) =
\ell(x)-\ell(y)$ and
$\eta_{i,i}(x,y)-\frac{\yb_i}{\xb_i}\eta_{i-1,i}(x,y) =
(\ell(x)-\ell(y))\frac{\yb_i}{x_i}$.  
%
Similarly (\ref{eq:unv}) is checked by 
$\eta_{n-1,n-1}(x,y)-\frac{\yb_{n-1}}{\xb_{n-1}}\eta_{n-2,n-1}(x,y) =
(\ell(x)-\ell(y))\frac{\yb_{n-1}}{x_{n-1}}$ and
$\eta'_{n-1,n}(x,y)-\frac{\yb_{n-1}}{\xb_{n-1}}\eta'_{n-2,n}(x,y) =
(\ell(x)-\ell(y))\frac{1}{y_{n}}$.
\end{proof}
%
Proposition \ref{th:subtractionfree} with $i=n-1$ reads
\begin{align}\label{eq:Vnminusoneexplicit}
V_{n-1}
&= x_1 y_1 x_n \left( \prod_{m=2}^{n-1} x_m \yb_m \right)
+ \ell (y) \sum_{m=2}^{n-1}
\left( \prod_{i=m}^{n} \frac{x_i}{y_i} \right)
\left( 1+\frac{y_{m}}{\yb_{m}} \right) +
\ell (y) x_n \\
%
&+ \xb_1 \yb_1 x_n \left( \prod_{m=2}^{n-1} x_m \yb_m \right)
+ \ell (x) \sum_{m=2}^{n-1}
\left( \prod_{i=m}^{n-1} \frac{\yb_i}{\xb_i} \right)
\left( 1+\frac{\xb_{m}}{x_{m}} \right) +
\ell (x) \frac{1}{y_n} ,
\nonumber
\end{align}
{}from which 
$V_{n-1} = V_{n-1}^{\sigma_1} = (V_{n-1}^*)^{\sigma_n}$ is easily seen.
%
Similarly from (\ref{eq:16}) one finds 
$V_0 = V_0^{\sigma_n} = V^\ast_0$.
%
Starting from these properties one can use 
(\ref{eq:vind})--(\ref{eq:27b}) to figure out 
the transformation property of 
$V_i, W_i$  under the commuting automorphisms 
$\sigma_1, \sigma_n$ and $\ast$. 
The result is summarized in

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{table}[h]
	\caption{Transformation by automorphisms and $\tilde{R}$.}
	\begin{tabular}[h]{|c|c|c|c|c|}
		\hline
		& $V_0$ & $V_i \, (1 \leq i \leq n-2)$ & $V_{n-1}$ & 
		$W_i \, (1 \leq i \leq n-1)$ \\
		\hline
	$\sigma_1$& $V_0^{\sigma_1}$ & $V_i$ & $V_{n-1}$ & $W_i$ \\
	$\sigma_n$& $V_0$ & $V_i$ & $V_{n-1}^*$ & $W_i$ \\
	$\ast$ & $V_0$ & $V_i^*$ & $V_{n-1}^*$ & 
	$W_i$ \\
	$\tilde{R}$ & $V_0$ & $W_i/V_i^*$& $V_{n-1}$& $W_i$ \\
		\hline
	\end{tabular}
	\label{tab:one}
\end{table}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent
The transformation properties under $\tilde{R}$ will be shown in
Lemma \ref{lem:invariances}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{prop}
\label{lem:rstar}
\begin{align}
\label{eq:commute_with_sigma_one}
\tilde{R}((x,y)^{\sigma_1})&=(\tilde{R}(x,y))^{\sigma_1},\\
%
\label{eq:commute_with_sigma_n}
\tilde{R}((x,y)^{\sigma_n}) &= (\tilde{R}(x,y))^{\sigma_n},\\
%
\label{eq:commute_with_ast}
\tilde{R}((x,y)^*)&=(\tilde{R}(x,y))^*.
\end{align}
\end{prop}
%
\begin{proof}
Apply Table 1 to Definition \ref{th:tropicalR}.
\end{proof}
%

%
\begin{lemma}
\label{lem:factorW1}
$W_1 = V_0 V_0^{\sigma_1}$.
\end{lemma}
\begin{proof}
Consider the identities
\begin{math}
V_0 = \frac{\xb_1}{\yb_1} V_1 + 
(\ell (y) - \ell (x))  \frac{\xb_1}{\yb_1} 
\left( 1 +  \frac{\yb_1}{x_1} \right),
\end{math}
%
\begin{math}
V_0^{\sigma_1}  =\frac{\yb_1}{\xb_1} V_1^* + (\ell (x) - \ell (y)) 
 \frac{\yb_1}{\xb_1}
\left( 1 +  \frac{\xb_1}{y_1} \right),
\end{math}
which are equivalent to (\ref{eq:vind}) and
$* \circ \sigma_1$ of (\ref{eq:vind}) with $i=1$.
%
Thus we obtain
\begin{math}
V_0 V_0^{\sigma_1} - V_1 V_1^* -
(\ell (y) - \ell (x)) V_1^* - (\ell (x) - 
\ell (y)) V_1
% \nonumber\\
% && \quad 
= (\ell (y) - \ell (x)) \frac{\yb_1}{x_1}V_1^*
         + (\ell (x) - \ell (y)) \frac{\xb_1}{y_1}V_1-
         (\ell (x) - \ell (y))^2 \left( 1 +  \frac{\yb_1}{x_1} \right)
         \left( 1 +  \frac{\xb_1}{y_1} \right).
\end{math}
The right hand side vanishes due to (\ref{eq:vind}) and
$*$ of (\ref{eq:vind}) for $i=1$.
\end{proof}
%
%
\begin{lemma}
%\label{lem:5}
\begin{equation}
W_i = \frac{x_i \yb_i}{\xb_i y_i}
V_{i-1} V_{i-1}^* + (\ell (x) - \ell (y)) \frac{\yb_i}{y_i}
V_{i-1}^* + (\ell (y) - \ell (x)) \frac{x_i}{\xb_i} V_{i-1},
\label{eq:40b}
\end{equation}
for $1 \leq i \leq n-1$.
\end{lemma}
\begin{proof}
For $i=1$ this relation reduces to
\begin{math}
V_0^{\sigma_1} = \frac{x_1 \yb_1}{\xb_1 y_1} +
(\ell (x) - \ell (y)) \frac{\yb_1}{y_1} +
(\ell (y) - \ell (x)) \frac{x_1}{\xb_1},
\end{math}
which follows from (\ref{eq:vind}) and $\sigma_1$ of
(\ref{eq:vind}) with $i=1$.
For $2 \leq i \leq n-2$ the relation is shown by
substituting (\ref{eq:vind}) and $*$ of (\ref{eq:vind}) into (\ref{eq:27c}). 
Next consider $i=n-1$ case.
{}From (\ref{eq:unv}) and $*$ of (\ref{eq:unv}) one has
\begin{math}
\mbox{LHS} - \mbox{RHS}
= (\ell (x) - \ell (y)) \frac{x_{n-1}  }{y_{n-1} y_n} V_{n-2}^* +
  (\ell (y) - \ell (x)) \frac{\yb_{n-1}  }{\xb_{n-1} x_n} V_{n-2}
 - (\ell (x) - \ell (y))^2
  \left( 1 + \frac{x_{n-1}  }{\yb_{n-1} y_n} \right)
  \left( 1 + \frac{\yb_{n-1}  }{x_{n-1} x_n} \right).
\end{math}
The right hand side vanishes because of
(\ref{eq:unv}) and $* \circ \sigma_n$ of (\ref{eq:unv}).
\end{proof}
%
%
\begin{lemma}
\label{lem:W_subtraction_free}
\begin{equation}
\left( \frac{1}{x_i} + \frac{1}{\yb_i} \right) W_i =
\frac{1}{y_i} V_i V_{i-1}^* + \frac{1}{\xb_i} V_{i-1} V_i^*
\quad (1 \leq i \leq n-2).
\label{eq:44b}
\end{equation}
\end{lemma}
\begin{proof}
For $i=1$ this relation reduces to
\begin{math}
\left( \frac{1}{x_1} + \frac{1}{\yb_1} \right)V_0^{\sigma_1}=
\frac{V_1^*}{\xb_1} +\frac{V_1}{y_1},
\end{math}
which follows from $\sigma_1$ of (\ref{eq:vind})
and $\sigma_1 \circ *$ of (\ref{eq:vind}) with $i=1$.
To see the other cases,
substitute (\ref{eq:vind}) and $*$ of (\ref{eq:vind}) into the right
hand side of (\ref{eq:44b}). 
%
The result is the right hand side of 
(\ref{eq:40b}) multiplied by 
$\left( \frac{1}{x_i} + \frac{1}{\yb_i} \right)$.
\end{proof}
%
\begin{remark}\label{rem:subtractionfree}
We call a rational function {\it subtraction-free},
if its denominator and numerator are polynomials with 
nonnegative coefficients. 
%
{}From Proposition \ref{th:subtractionfree}, 
Lemma \ref{lem:factorW1}, \ref{lem:W_subtraction_free}
and (\ref{eq:27b}),  all the functions $V_i, W_i$ 
appearing in Definition \ref{th:tropicalR} 
are subtraction-free Laurent polynomials in $x,y$. 
Thus ${\tilde R}$ is also subtraction-free.
This property will be used in section \ref{subsec:piecewise}.
\end{remark}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
\begin{prop}
\label{th:main}
$\tilde{R}(x,y) =(x',y')$ solves
the equation $A(x)A(y)=A(x')A(y')$.
\end{prop}
\noindent
A proof by a direct but lengthy calculation is available 
in appendix \ref{sec:proofofaaeqaa}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}
\label{lem:invariances}
$V_0, V_0^{\sigma_1}, V_{n-1}, V_{n-1}^{*},
W_i \quad (1 \leq i \leq n-1)$ are invariant under $\tilde{R}$.
$\tilde{R}$ acts as $V_i \rightarrow W_i/V_i^*$ and
$V_i^* \rightarrow W_i/V_i \quad (1 \leq i \leq n-1)$.
\end{lemma}
\begin{proof}
Proposition \ref{th:main} tells that 
$V_0 = (A(x)A(y))_{2n,1}$ is invariant.
Then Proposition \ref{lem:rstar} tells $V_0^{\sigma_1}$
and thereby $W_1 = V_0 V_0^{\sigma_1}$ are also invariant.
Consider $V_1$ and $V_1^*$.
By applying $\tilde{R}(x,y)=(x',y')$
to (\ref{eq:vind}) with $i=1$ we have
\begin{math}
\tilde{R}(V_1) =\frac{\yb_1'}{\xb_1'}V_0 + (\ell (x') - \ell (y'))
\left( 1+\frac{\yb_1'}{x_1'} \right)
%
=\frac{V_0^{\sigma_1}}{V_1^*} \left[
\frac{\xb_1}{\yb_1 } V_1+ (\ell (y) -\ell (x))
\frac{\xb_1}{V_0^{\sigma_1}}
\left( \frac{V_1^*}{\xb_1} +\frac{V_1}{y_1} \right) \right].
\end{math}
By substituting (\ref{eq:44b}) and (\ref{eq:vind})
with $i=1$ into its right hand side
we find $\tilde{R}(V_1)=V_0^{\sigma_1} V_0/V_1^* = W_1/V_1^*$.
Hence by Proposition \ref{lem:rstar} 
we also have $\tilde{R}(V_1^*)= W_1/V_1$.

The claim for $V_i$ and $V_i^*$ with $2 \leq i \leq n-2$
is proved by induction on $i$.
%
Suppose that the claim is true for $V_{i-1}$.
By applying $\tilde{R}(x,y)=(x',y')$
to (\ref{eq:vind}) we have
\begin{math}
\tilde{R}(V_i) =
\frac{\yb_i'}{\xb_i'}\frac{W_{i-1}}{V_{i-1}^*} +
(\ell (x') - \ell (y'))
\left( 1+\frac{\yb_i'}{x_i'} \right) 
=\frac{W_i}{V_{i-1} V_i^*} \left[
\frac{\xb_i}{\yb_i } V_i+ (\ell (y) - \ell (x))
\frac{\xb_i}{W_i}
\left( \frac{V_{i-1} V_i^*}{\xb_i} 
\right.\right.
\end{math}
\par\noindent
\begin{math}
\left. \left.
+\frac{V_{i-1}^* V_i}{y_i} \right) 
\right] .
\end{math}
By substituting (\ref{eq:44b}) and (\ref{eq:vind})
into the right hand side
we obtain $\tilde{R}(V_i) = W_i/V_i^*$.
Hence by Proposition \ref{lem:rstar} we also
have $\tilde{R}(V_i^*)= W_i/V_i$ for $2 \leq i \leq n-2$.

Let us consider $W_i$ for $2 \leq i \leq n-1$.
By applying $\tilde{R}(x,y)=(x',y')$ to (\ref{eq:40b}) we have
\begin{math}
\tilde{R}(W_i) = \frac{y_i \xb_i}{\yb_i x_i}
\frac{(W_i)^2}{V_{i-1} V_{i-1}^*} + (\ell (y) - \ell (x)) 
\frac{\xb_i}{x_i}
\frac{W_i}{V_{i-1}} + (\ell (x) - \ell (y)) \frac{y_i}{\yb_i}
\frac{W_i}{V_{i-1}^*}.
\end{math}
Dividing both hand sides by $W_i$ and substituting
(\ref{eq:40b}) into the right hand side we find
$\tilde{R}(W_i)/W_i = 1$ for $2 \leq i \leq n-1$.

{}From (\ref{eq:unv}) and its $\sigma_n \circ *$, 
the identity
\begin{math}
V_{n-1} \left( \frac{1}{x_{n-1}x_n} + \frac{1}{\yb_{n-1}} \right)
=\frac{V_{n-2}}{\xb_{n-1}} +\frac{V_{n-2}^*}{y_{n-1}y_n}
\end{math} can be derived.
%
Applying $\tilde{R}$ to this and using the identity
once again we get $\tilde{R}(V_{n-1})=V_{n-1}$.
The $\tilde{R}$ also leaves
$V_{n-1}^*$ invariant since
$\tilde{R}$ and $*$ are commutative.
\end{proof}
\begin{remark}
{}From the explicit formulas (\ref{eq:16}) and
(\ref{eq:Vnminusoneexplicit}) we find that
$V_0^{\sigma_1} = (A(x)C(y))_{1,2n} + (B(x)B(y))_{1,2n}
+(C(x)A(y))_{1,2n}$,
$V_{n-1} = (A(x)B(y))_{n,n+1}+(B(x)A(y))_{n,n+1}$, and
$V_{n-1}^* = (A(x)B(y))_{n+1,n}+(B(x)A(y))_{n+1,n}$.
Therefore they are invariant under the tropical $R$, and thus 
should be so also under ${\tilde R}$.
\end{remark}
%
\begin{prop}
\label{th:inversion}
$\tilde{R}(\tilde{R}(x,y)) = (x,y).$
\end{prop}
\begin{proof}
Let us illustrate the $x_2$ case.  The other cases are similar.
Under $\tilde{R}(x,y)=(x',y')$ 
$x_2$ becomes $x_2' = y_2 V_1 W_2/(V_2 W_1)$.
By applying it once again it changes into
\begin{math}
y_2' \frac{R(V_1)R(W_2)}{R(V_2)R(W_1)} =
x_2 \frac{V_1^*}{V_2^*} \frac{(W_1/V_1^*)W_2}{(W_2/V_2^*)W_1} =
x_2.
\end{math}
\end{proof}
Proposition \ref{th:inversion} is the inversion relation of  ${\tilde R}$, 
and will play a role in the proof of Lemma \ref{th:afterward}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Set
\begin{equation}
\tilde{V}_{j} =
\frac{1}{V_{j}^*}
\left( W_{j} + \frac{\xb_{j+1}}{x_{j+1}}\, W_{j+1} \right), \quad
(1 \leq j \leq n-2).
\label{eq:67apr}
\end{equation}
%
%
%\noindent
%{}From (\ref{eq:vind})-(\ref{eq:27b}) we have:
\begin{lemma}
\label{lem:26apr}
\begin{eqnarray*}
\tilde{V}_1 &=& V_0 \frac{\yb_1}{\xb_1} \overline{Y}_2
+ (\ell (x) - \ell (y)) \left[
\frac{\yb_2}{y_2} \overline{X}_2 +
\frac{\yb_1}{x_1} \overline{Y}_2 \right],\\
%
\tilde{V}_j &=& \tilde{V}_{j-1} \frac{\yb_j}{\xb_j}
\frac{\overline{Y}_{j+1}}{\overline{Y}_j} +
(\ell (x) - \ell (y)) \left[
\frac{\yb_{j+1}}{y_{j+1}} \overline{X}_{j+1} +
\frac{\yb_j}{\xb_j}
\frac{\overline{X}_j}{\overline{Y}_j} 
\overline{Y}_{j+1} \right],
\\
&& \qquad \qquad \qquad \qquad (2 \leq j \leq n-2)
\nonumber
\end{eqnarray*}
where $\overline{X}_i = 1+\xb_i/x_i$, 
$\overline{Y}_i = 1+\yb_i/y_i$ for $1 \leq i \leq n-1$.
\end{lemma}
%
\begin{proof}
{}From (\ref{eq:27c}) and (\ref{eq:40b}) we get
\begin{displaymath}
\tilde{V}_j =
\overline{Y}_{j+1} V_{j} + (\ell (x) - \ell (y))
\left( \frac{\xb_{j+1} \yb_{j+1}}{x_{j+1} y_{j+1}} -1 \right).
\end{displaymath}
This allows us to write $V_j$ in terms of $\tilde{V}_j$.
Substitution of them into (\ref{eq:vind}) leads to the desired 
relations.
\end{proof}

\begin{remark}
\label{rem:tildeV}
The equation 
(\ref{eq:67apr}) gives an alternative definition of $W_j$'s in terms of
$\tilde{V}_j$'s.
%
Given $W_1$ and $\tilde{V}_{j}$'s one can determine all the other 
$W_j$'s from (\ref{eq:67apr}).
%
This fact will be used in 
the proof of Lemma \ref{th:yprimesolve}.
\end{remark}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Proof of Theorem \ref{pr:mainmain}}
\label{subsec:tilde_R_equals_R}
\noindent
As mentioned after Definition \ref{def:RRRR}, 
Theorem \ref{pr:mainmain} is established 
at the same time with 
the explicit form $R = {\tilde R}$ once it is shown that 
$(x',y') = {\tilde R}(x,y)$ 
solves the defining equation (\ref{eq:mmeqmmx}) 
of the tropical $R$.
%
In view of $B(x)C(y)+C(x)B(y)=C(x)C(y)=O$, we have the expansion
\begin{align*}
M(x,z)M(y,z) &=A(x)A(y) + z (A(x)B(y)+B(x)A(y)) \\
&+ z^2 (A(x)C(y)+B(x)B(y)+C(x)A(y)).
\end{align*}
Since the equation (\ref{eq:mmeqmmx}) is quadratic with respect to $z$,
it suffices to check it for three distinct values of $z$.
%
We have already done it at  $z=0$ in Proposition \ref{th:main}.
%
In what follows we treat the cases 
$z=\ell(x)^{-1}$ and $z=\ell(y)^{-1}$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let us put $z= \ell(x)^{-1}=\ell(y')^{-1}$ in (\ref{eq:mmeqmmx}).
 Proposition \ref{prop:meqdpd} leads to
%
\begin{equation}
P {\mathcal D}_1 (x) M(y, \ell(x)^{-1}) {\mathcal D}_1 (y')^{-1}=
{\mathcal D}_2 (x)^{-1} M(x', \ell(x)^{-1}) {\mathcal D}_2 (y') P.
\label{eq:pdmdeqdmdp}
\end{equation}
Thus the both sides is a scalar multiple of $P$.
Denoting the scalar by $\alpha$ we see that 
(\ref{eq:pdmdeqdmdp}) is equivalent to the simultaneous equations
\begin{eqnarray}
&&P {\mathcal D}_1 (x) M(y, \ell(x)^{-1}) =
\alpha P {\mathcal D}_1 (y'),
\label{eq:mateq1} \\
%
&&M(x', \ell(x)^{-1}) {\mathcal D}_2 (y') P =
\alpha {\mathcal D}_2 (x) P.
\label{eq:mateq2}
\end{eqnarray}
Equation (\ref{eq:mateq2})
will be considered later (Lemma \ref{th:afterward}).
%
\begin{lemma}
\label{pr:August4_2}
The following $y_i'$'s and $\yb_i'$'s solve the equation (\ref{eq:mateq1}).
\begin{eqnarray}
&& y_i' = \frac{Q_i}{Q_{i+1}} \quad (1 \leq i \leq n), \quad
\yb_1' = \ell (x) \frac{Q_{2n}}{Q_{2}}, \\
&& \label{eq:QQoverQQ}
\frac{\yb_i'}{y_i'} = - \frac
{\sum_{k=1}^{i} (-1)^k Q_{k} Q_{2n+1-k}}
{\sum_{k=1}^{i-1} (-1)^k Q_{k} Q_{2n+1-k}}, \quad (2 \leq i \leq n-1)
\end{eqnarray}
where $Q_i = (A(x) M(y,\ell (x)^{-1}))_{2n,i}$.
\end{lemma}
\begin{proof}
All elements in the $i$th column of the LHS (resp.~RHS) 
of (\ref{eq:mateq1}) are equal to
$Q_i$ (resp.~$\alpha A(y')_{2n,i}$).
Considering the first column fixes the scalar as
$\alpha = Q_1/A(y')_{2n,1}=Q_1 / \ell (x)$.
Taking the ratios of the $i$th and the $i+1$th column
for $1 \leq i \leq n$ leads to $y_i' = \frac{Q_i}{Q_{i+1}}$.
Similarly the second column and the last column imply
$\yb_1' = \ell (x) \frac{Q_{2n}}{Q_{2}}$, and 
the $(2n-1)$th column does 
$Q_{2n-1} = \alpha \left( 1+ \frac{\yb_2'}{y_2'} \right)\yb_1'
= \left( 1+ \frac{\yb_2'}{y_2'} \right) \frac{Q_1 Q_{2n}}{Q_2}$.
Therefore we have $1+ \frac{\yb_2'}{y_2'}=
\frac{Q_2 Q_{2n-1}}{Q_1 Q_{2n}}$ or equivalently
(\ref{eq:QQoverQQ}) for $i=2$.
Now let us show
\begin{equation}
\label{eq:QQoverQQx}
1+\frac{\yb_i'}{y_i'} = (-1)^{i+1} \frac
{Q_{i} Q_{2n+1-i}}
{\sum_{k=1}^{i-1} (-1)^k Q_{k} Q_{2n+1-k}},
\end{equation}
or equivalently (\ref{eq:QQoverQQ}) for $3 \leq i \leq n-1$.
This claim is checked by induction on $i$.
%
We have already shown $i=2$ case.
Assume the identity (\ref{eq:QQoverQQ})
with $i$ replaced by $i-1$.
Then we have
\begin{displaymath}
\frac{Q_{2n+1-i}}{Q_{2n+2-i}} =
\frac{\left( 1+\frac{\yb_i'}{y_i'} \right)\frac{\yb_{i-1}'}{y_{i-1}'}}
{\left( 1+\frac{\yb_{i-1}'}{y_{i-1}'} \right)} y_{i-1}'
%
= \frac{\left( 1+\frac{\yb_i'}{y_i'} \right)
\sum_{k=1}^{i-1} (-1)^k Q_{k} Q_{2n+1-k}}
{(-1)^{i+1} Q_{i-1} Q_{2n+2-i}} \frac{Q_{i-1}}{Q_{i}},
\end{displaymath}
which is equivalent to (\ref{eq:QQoverQQx}).
\end{proof}
%
\begin{lemma}
\label{th:yprimesolve}
The $y_i'$'s and $\yb_i'$'s specified in Lemma \ref{pr:August4_2}
coincide with those
given by $\tilde{R}(x,y)=(x',y')$
in Definition \ref{th:tropicalR}.
\end{lemma}
%
%
\begin{proof}
Set
\begin{eqnarray*}
&& ^{\triangle}\!V_0 = Q_1, \quad
^{\triangle}\!V_i^* = x_1 \ldots x_i Q_{i+1} \quad (1 \leq i \leq n-1),
\\
&&
^{\triangle}\!V_{n-1} = x_1 \ldots x_{n-1} x_n Q_{n+1}, \quad
^{\triangle}\!V_0^{\sigma_1} = \ell (x) \frac{x_1}{\xb_1} Q_{2n}, \\
&&
^{\triangle}\!W_i = \ell (x) \frac{x_1 \ldots x_i}{\xb_1 \ldots \xb_i}
(-1)^i {\sum_{k=1}^{i} (-1)^k Q_{k} Q_{2n+1-k}}
\quad (1 \leq i \leq n-1).
\end{eqnarray*}
%
We show that these quantities coincide with the
same symbols without $^\triangle$.

First note 
$^{\triangle}\!V_0 =(A(x)A(y))_{2n,1}=V_0$ from (\ref{eq:16}).
It is not difficult to see that $^{\triangle}\!V_0$, 
$^{\triangle}\!V_0^{\sigma_1}$,
$^{\triangle}\!V_i^*$, and $^{\triangle}\!V_{n-1}$
satisfy the same recursion relations
as those without $^\triangle$.
For instance consider
\begin{displaymath}
^{\triangle}\!V_i^* - \frac{x_i}{y_i} \,^{\triangle}\!V_{i-1}^*
= x_1 \ldots x_i \sum_{k=1}^{2n} A(x)_{2n,k}
\left[ M(y,\ell (x)^{-1})_{k,i+1} -
\frac{M(y,\ell (x)^{-1})_{k,i}}{y_i} \right].
\end{displaymath}
By using (\ref{eq:defMno1}) and  $A(x)_{2n,i} = \ell(x)/(x_1\cdots x_{i-1})$, 
this becomes 
\begin{displaymath}
^{\triangle}\!V_i^* = \frac{x_i}{y_i}\,^{\triangle}\!V_{i-1}^* + (\ell (y) - 
\ell (x))\left( 1+\frac{x_i}{\yb_i} \right),
\end{displaymath}
which is $*$ of (\ref{eq:vind}).
The other cases are similar.
%
Note that $V_0^{\sigma_1}$ is related to $V_1^*$
by $* \circ \sigma_1$ of (\ref{eq:vind}) and $V_{n-1}$ to $V_{n-2}^*$
by $* \circ \sigma_n$ of (\ref{eq:unv}).
By repeated use of the recursion relations,
$^{\triangle}\!V_0^{\sigma_1}$, 
$^{\triangle}\!V_i^*$'s and $^{\triangle}\!V_{n-1}$ are related to
$^{\triangle}\!V_0$ in the same way as those without $^{\triangle}$
are to $V_0$, proving the coincidence.

So far we have considered $^{\triangle}\!V_0$, 
$^{\triangle}\!V_0^{\sigma_1}$,
$^{\triangle}\!V_i^*$, and $^{\triangle}\!V_{n-1}$.
Next we treat $^{\triangle}\!W_i$.
Note $^{\triangle}\!W_1 = \, ^{\triangle}\!V_0
\,^{\triangle}\!V_0^{\sigma_1} = V_0 V_0^{\sigma_1} = W_1$.
To verify $^{\triangle}\!W_i = W_i$ for $i \geq 2$
we define
$^{\triangle}\!\tilde{V}_j=\frac{\ell(x)}{\xb_1 \ldots \xb_j}Q_{2n-j}$
for $1 \leq j \leq n-2$.
Then 
\begin{displaymath}
^{\triangle}\!\tilde{V}_{j} =
\frac{1}{^{\triangle}\!V_{j}^*}
\left( ^{\triangle}\!W_{j} + \frac{\xb_{j+1}}{x_{j+1}}\,
^{\triangle}\!W_{j+1} \right)
\end{displaymath}
holds for $1 \leq j \leq n-2$.
This is (\ref{eq:67apr}).
Since we have already shown $^{\triangle}\!W_1 = W_1$
and $^{\triangle}\!V_j^* = V_j^*$, it suffices to show
$^{\triangle}\!\tilde{V}_j=\tilde{V}_j$ to verify
$^{\triangle}\!W_i = W_i$ for $i \geq 2$.
(See Remark \ref{rem:tildeV}.)
Thus it remains to check that 
$^{\triangle}\!V_0$ and
$^{\triangle}\!\tilde{V}_j$'s satisfy the recursion
relations in Lemma \ref{lem:26apr}.
For instance consider
\begin{displaymath}
^{\triangle}\!\tilde{V}_j -
\,^{\triangle}\!\tilde{V}_{j-1} \frac{\yb_j}{\xb_j}
\frac{\overline{Y}_{i+1}}{\overline{Y}_{i}}
= \frac{\ell(x)}{\xb_1 \ldots \xb_j}
\sum_{k=1}^{2n} A(x)_{2n,k} {\mathcal M}(y,\ell (x)^{-1})_{k,j},
\end{displaymath}
where
\begin{math}
{\mathcal M}(y,z)_{k,j} =
M(y,z)_{k,2n-j} -
\yb_j 
\frac{\overline{Y}_{i+1}}{\overline{Y}_{i}}
M(y,z)_{k,2n+1-j}.
\end{math}
%
By using (\ref{eq:defMno6}) and 
$A(x)_{2n,2n+1-j} = \xb_1 \ldots \xb_{j-1} 
\left(1+\xb_j/x_j\right)$ we obtain the desired relation.
\end{proof}
%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%

Let us proceed to $z= \ell(y)^{-1}$ case.
Setting $z= \ell(y)^{-1}=\ell(x')^{-1}$ in (\ref{eq:mmeqmmx})
leads to
%
\begin{equation}
{\mathcal D}_2 (x')^{-1} M(x, \ell(y)^{-1}) {\mathcal D}_2 (y) P=
P {\mathcal D}_1 (x') M(y', \ell(y)^{-1}) {\mathcal D}_1 (y)^{-1}.
\label{eq:pdmdeqdmdp2}
\end{equation}
As (\ref{eq:pdmdeqdmdp}), (\ref{eq:pdmdeqdmdp2}) is 
equivalent to the simultaneous equations
\begin{eqnarray}
&&M(x, \ell(y)^{-1}) {\mathcal D}_2 (y) P =
\beta {\mathcal D}_2 (x') P,
\label{eq:mateq3} \\
%
&&P {\mathcal D}_1 (x') M(y', \ell(y)^{-1}) =
\beta P {\mathcal D}_1 (y)
\label{eq:mateq4}
\end{eqnarray}
for some scalar $\beta$.
Equation (\ref{eq:mateq4})
will be considered later (Lemma \ref{th:afterward}).
%
\begin{lemma}
\label{th:xprimesolve}
The relation $\tilde{R}(x,y)=(x',y')$ 
solves the equation (\ref{eq:mateq3}) with $\beta = Q^*_1 / \ell (y)$.
\end{lemma}
%
\begin{proof}
Take the transposition of
(\ref{eq:mateq1}) and
multiply $J_*$'s from the both
sides.  Due to Lemma \ref{lem:JMJequalsMxxx}
the result becomes 
\begin{displaymath}
M(x^*,\ell(x)^{-1}) \mathcal{D}_2 (y^*) P =
\alpha \mathcal{D}_2 ((x')^*) P.
\end{displaymath}
{}From Proposition \ref{lem:rstar} we know $(x')^* = (x^*)'$.
Thus replacing $(x,y)$ with 
$(x,y)^\ast$ in this relation yields (\ref{eq:mateq3}).
\end{proof}
%
\begin{lemma}
\label{th:afterward}
The relation $\tilde{R}(x,y)=(x',y')$
also solves (\ref{eq:mateq2}) with
$\alpha = V_0/\ell(x)$ and (\ref{eq:mateq4}) with
$\beta = V_0/\ell(y)$.
\end{lemma}
\begin{proof}
Recall $Q_1 =Q^*_1= V_0$ and its invariance under $\tilde{R}$
(Lemma \ref{lem:invariances}).
Due to Proposition \ref{th:inversion} we can 
exchange $(x,y)$ with $(x',y')$ in (\ref{eq:mateq1}) and
(\ref{eq:mateq3}), which become (\ref{eq:mateq4}) and
(\ref{eq:mateq2}), respectively.
Then the claim follows from Lemmas 
\ref{th:yprimesolve} and \ref{th:xprimesolve}.
\end{proof}
%
\begin{proof}[Proof of Theorem \ref{pr:mainmain}]
Lemmas \ref{pr:August4_2}, \ref{th:yprimesolve}, \ref{th:xprimesolve}, 
and \ref{th:afterward} establish that $(x',y')=\tilde{R}(x,y)$
solves the equation (\ref{eq:mmeqmmx})
at $z = \ell(x)^{-1}$ and $\ell(y)^{-1}$.
%
With the $z=0$ case verified in Proposition \ref{th:main}, 
we conclude that it solves (\ref{eq:mmeqmmx}) for any $z$ 
as argued in the beginning of the present subsection.
%
The level constraints $\ell(x)=\ell(y') \ne \ell(y)=\ell(x')$
are obviously satisfied. 
%
Due to Theorem \ref{th:uniqueness} there is no other
solution.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Properties of $V_i, W_i$ under $e^c_i$}
Here we list the transformation property of 
the functions $V_i, W_i$ under $e^c_i$.
%
Set
\begin{align*}
&\omega_i = \frac{c \vphi_i(x) + \veps_i(y)}{\vphi_i(x) + \veps_i(y)},
\quad
\psi_i = \frac{c \vphi_i(x) + c \veps_i(y)}{c \vphi_i(x) + \veps_i(y)}
\quad (0 \leq i \leq n),\\
%
&\Omega_i = 
\begin{cases}
\frac{x_{i+1} + \omega_i \xb_{i+1}}{x_{i+1} + \xb_{i+1}} &
(1 \leq i \leq n-1), \\
\frac{x_2 + \omega_0 \xb_2}{x_2 +\xb_2} &( i=0 ),
\end{cases}
\quad
\Psi_i = 
\begin{cases}
\frac{y_{i+1} + \psi_i \yb_{i+1}}{y_{i+1} + \yb_{i+1}} &
(1 \leq i \leq n-1), \\
\frac{y_2 + \psi_0 \yb_2}{y_2 +\yb_2} &( i=0 ).
\end{cases}
\end{align*}
For a function $F=F((x,y))$
in the variables $(x,y)$
we write 
$R(F) = F(R(x,y))$ and $e^c_i (F) = F(e^c_i(x,y))$.

\begin{prop}\label{pr:V0energylike}
Let $(x',y') = R(x,y)$.
Then we have
\begin{equation*}
V_0(e^c_i(x,y)) = V_0(x,y) 
\left(
\frac{c \vphi_0(x')+ \veps_0(y')}{\vphi_0(x') + \veps_0(y')}
\frac{c \vphi_0(x) + \veps_0(y)}{c \vphi_0(x) + c \veps_0(y)}
\right)^{\delta_{i 0}} .
\end{equation*}
\end{prop}
%
\begin{proof}
First suppose $1 \le i \le n$.
{}From Theorem \ref{th:MMMequalGMMMG} the relation 
$({\tilde x}, {\tilde y}) = e^c_i(x,y)$  implies 
\begin{math}
M(\tilde{x},z) M(\tilde{y},z) = 
G_i(a)
M(x,z) M(y,z)
G_i(b)
\end{math} for some $a,b$.
%
But \begin{math}V_0 = (M(x,z) M(y,z))_{2n,1}\end{math} is unchanged 
by the multiplication of the $G_i$'s, proving 
$e^c_iV_0 = V_0$.
%
To show the $i=0$ case, note 
{}from Lemma \ref{lem:Dynkinrightmost} that 
$e^c_0((x,y)^{\sigma_1}) = (e^c_1(x,y))^{\sigma_1}$.
%
Thus we have
$e^c_0(V_0^{\sigma_1}) = V_0(e^c_0((x,y)^{\sigma_1})) 
= V_0((e^c_1(x,y))^{\sigma_1}) = (e^c_1V_0)^{\sigma_1}
= V_0^{\sigma_1}$.
%
Therefore, 
\begin{math}
e^c_0 R \left( \frac{x_1}{\xb_1} \right) =
e^c_0 \left(  \frac{y_1}{\yb_1} \frac{V_0^{\sigma_1}}{V_0} \right) =
\frac{y_1}{\yb_1 \psi_0} \frac{V_0^{\sigma_1}}{e^c_0(V_0)}
\end{math}
and
\begin{math}
R e^c_0 \left( \frac{x_1}{\xb_1} \right) =
R \left( \frac{x_1}{\xb_1 \omega_0} \right) =
\frac{y_1}{\yb_1 R(\omega_0)} \frac{V_0^{\sigma_1}}{V_0}.
\end{math}
Now the assertion follows from 
$e^c_0 R = R e^c_0$ (Proposition \ref{th:eRequalsRe}).
\end{proof}
%
Proposition \ref{pr:V0energylike} indicates that 
$V_0$ is a tropical analogue of the energy function in 
crystal theory, which will be argued in Remark \ref{rem:energy}.

Similar properties can be derived for the other $V_i, W_i$
with the help of Proposition \ref{th:eRequalsRe} and 
Definition \ref{th:tropicalR}.
%
Besides {\sc Table} \ref{tab:two}, 
the result is summarized as 
\begin{displaymath}
e^c_i(V_j) = 
\begin{cases}
V_j & (j \ne i), \\
V_i \frac{R(\Omega_i)}{\Psi_i} & ( j = i), 
\end{cases}
\quad
%
e^c_i(W_j) = 
\begin{cases}
W_j & (j \ne i), \\
W_i \frac{R(\omega_i)}{\psi_i} & ( j = i), 
\end{cases}
\end{displaymath}
for $2 \leq j \leq n-2$ and $0 \leq i \leq n$.
For instance the $j=i$ case of the former (resp.~the latter) relation
can be obtained by applying $e^c_i R = R e^c_i$ on
$\xb_1 \cdots \xb_i$ (resp.~$(\xb_1 \cdots \xb_i)/(x_1 \cdots x_i)$).




\begin{table}[h]
	\caption{Transformation under $e^c_i$.}
	\begin{tabular}[h]{|c|c|c|c|c|c|}
		\hline
		& $V_0$ & $V_0^{\sigma_1}$ & $V_1$ & $V_{n-1}^*$& $V_{n-1}$\\
		\hline
$e^c_0$		& $V_0 \frac{R(\omega_0)}{\psi_0}$ & $V_0^{\sigma_1}$ &
 $V_1\frac{R(\Omega_0)}{\Psi_0}$  & $V_{n-1}^*$ & $V_{n-1}$\\
$e^c_1$		&  $V_0$ &  $V_0^{\sigma_1}\frac{R(\omega_1)}{\psi_1}$& 
 $V_1\frac{R(\Omega_1)}{\Psi_1}$  & $V_{n-1}^*$  & $V_{n-1}$\\
$e^c_i \, (2 \leq i \leq n-2)$		& $V_0$ & $V_0^{\sigma_1}$ & 
$V_1$ & $V_{n-1}^*$& $V_{n-1}$\\
$e^c_{n-1}$		&  $V_0$ &  $V_0^{\sigma_1}$ &  $V_1$ &  
$V_{n-1}^* \frac{R(\omega_{n-1})}{\psi_{n-1}}$& $V_{n-1}$\\
$e^c_n$		& $V_0$ & $V_0^{\sigma_1}$ & $V_1$ & $V_{n-1}^*$& 
$V_{n-1} \frac{R(\omega_n)}{\psi_n}$\\
		\hline
	\end{tabular}
	\label{tab:two}
\end{table}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Piecewise linear formula for
the combinatorial $R$}
\label{subsec:piecewise}
Suppose a rational function $F=F(x,y;c)$ is subtraction-free 
in the sense of Remark \ref{rem:subtractionfree} 
with respect to the variables $x=(x_1,x_2,\ldots,\xb_1)$,
$y=(y_1,y_2,\cdots,\yb_1)$ and $c$.
%
By {\it ultradiscretization} of $F$ we mean the expression obtained by 
replacing $+, \times, /$ with $\max, +, -$, respectively in $F$.
%
(It is called ``tropicalization" in \cite{BK} on the contrary.)
The procedure is well-defined particularly because 
the both sides of $p(q+r) = pq + pr$ have the equal image
$p+\max(q,r) = \max(p+q,p+r)$.
%
A way to substantiate it is to consider the limit
$\lim_{\epsilon \rightarrow +0} \epsilon
\log F(e^{x/\epsilon},e^{y/\epsilon};e^{c/\epsilon})$
after replacing $x,y,c$ by the real positive arrays 
$e^{x/\epsilon} :=(e^{x_1/\epsilon},e^{x_2/\epsilon},
\ldots,e^{\xb_1/\epsilon})$,
$e^{y/\epsilon} :=(e^{y_1/\epsilon},e^{y_2/\epsilon},
\cdots,e^{\yb_1/\epsilon})$ and $e^{c/\epsilon}$ 
\cite{TTMS}.
%
Note that ultradiscretized functions are piecewise linear.
%
In this subsection we use the same symbol to stand for 
the original (tropical) and 
the ultradiscretized (piecewise linear) objects, 
supposing no confusion might arise.
%
\begin{example}
\label{ex:August6_2}
\begin{displaymath}
F=\frac{c \vphi_0(x)+\veps_0(y)}{\vphi_0(x)+\veps_0(y)} =
\frac{c \xb_1 (1+\xb_2/x_2) + y_1 (1+y_2/\yb_2)}
{\xb_1 (1+\xb_2/x_2) + y_1 (1+y_2/\yb_2)}
\end{displaymath}
is subtraction-free. Its ultradiscretization reads
\[
F = \max \left( c + \xb_1 + (\xb_2 - x_2)_+,
y_1 + (y_2 - \yb_2)_+ \right)
- \max \left(\xb_1 + (\xb_2 - x_2)_+,
y_1 + (y_2 - \yb_2)_+ \right),
\]
where $(x)_+ := \max(x,0)$.
Moreover if $c=1$, 
\[
F = \theta \left( \xb_1 + (\xb_2 - x_2)_+
\geq y_1 + (y_2 - \yb_2)_+ \right).
\]

\end{example}
%

Let $(x',y') = R(x,y)$ be the ultradiscretization of 
our tropical $R$. Namely, 
\begin{equation}\label{eq:combiR}
\begin{split}
& x_1' = y_1 +V_0^{\sigma_1} -V_1, \quad
\xb_1' = \yb_1 +V_0 -V_1, \\
%
& x_i' = y_i +V_{i-1}+W_i -V_i -W_{i-1}, \quad
\xb_i' = \yb_i +V_{i-1} -V_i, \quad (2 \leq i \leq n-1)\\
%
& x_n' = y_n +V_{n-1} -V_{n-1}^{*},  \\
%
%
& y_1' = x_1 +V_0 -V_1^*, \quad
\yb_1' = \xb_1 +V_0^{\sigma_1} -V_1^*, \\
%
& y_i' = x_i +V_{i-1}^* -V_i^*, \quad
\yb_i' = \xb_i +V_{i-1}^* +W_i -V_i^* -W_{i-1}, \quad (2 \leq i \leq 
n-1)\\
%
& y_n' = x_n +V_{n-1}^* -V_{n-1}.
\end{split}
\end{equation}
Here $V_i, W_i$ are given by 
\begin{align}
\label{eq:piecewise}
&V_i =  \max
\left( \{ \theta_{i,j}(x,y),
\theta'_{i,j}(x,y) | 1 \leq j \leq n-2 \} \right.
\\
\nonumber
& \left. \qquad \qquad \cup
\{ \eta_{i,j}(x,y), \eta'_{i,j}(x,y) | 1 \leq j \leq n \} \right),
\nonumber\\
%
&W_1 = V_0 + V_0^{\sigma_1},\;\; W_{n-1}=V_{n-1}+V_{n-1}^*, 
\nonumber\\
%
&W_i=\max \left( V_i+V_{i-1}^*-y_i, V_{i-1}+V_i^*-\xb_i \right)
+ \min (x_i, \yb_i ),\; (2 \leq i \leq n-2),  \nonumber
\end{align}
%
where 
\begin{align*}
\theta_{i,j}(x,y) &=
\begin{cases}
\ell(x) +{\displaystyle
\sum_{k=j+1}^{i} (\yb_k - \xb_k)} & \mbox{for} \quad
1 \leq j \leq i, \\
\ell(y) +{\displaystyle
\sum_{k=i+1}^{j} (\xb_k - \yb_k)} & \mbox{for} \quad
i+1 \leq j \leq n-2,
\end{cases}
\\
\theta'_{i,j}(x,y) &= \ell(x) + \sum_{k=1}^{i} (\yb_k - \xb_k)+
\sum_{k=1}^{j} (y_k - x_k)  \quad
\mbox{for} \quad j=1,\ldots,n-2,
\end{align*}
\begin{align*}
\eta_{i,j}(x,y) &=
\begin{cases}
\ell(x) +{\displaystyle  \sum_{k=j+1}^{i} (\yb_k -\xb_k)  +
\left( \yb_{j}- x_{j} \right)} &\mbox{for} \quad 1 \leq j \leq i,\\
\ell(y) +{\displaystyle  \sum_{k=i+1}^{j} (\xb_k - \yb_k)  +
\left( \yb_{j}- x_{j} \right)} &\mbox{for} \quad i+1 \leq j \leq n-1,\\
\ell(y) +{\displaystyle  \sum_{k=i+1}^{n-1} (\xb_k - \yb_k) +
x_n }&\mbox{for} \quad j=n,
\end{cases}
\\
\eta'_{i,j}(x,y) &=
\begin{cases}
\ell(x)+ {\displaystyle  \sum_{k=1}^{i} (\yb_k -\xb_k)+
\sum_{k=1}^{j} (y_k -x_k)  + \left( x_{j} -\yb_{j} \right) }
\quad \mbox{for} \quad 1 \leq j \leq n-1,\\
%
\ell(x)+ {\displaystyle
\delta_{i,n-1} \left( \ell(x) - \ell(y) \right)+
 \sum_{k=1}^{i} (\yb_k -\xb_k)+
\sum_{k=1}^{n-1} (y_k -x_k)  -x_{n} }
\quad \mbox{for} \quad j=n,
\end{cases}
\end{align*}
\begin{equation*}
\ell(x) = \sum_{k=1}^{n} x_k + \sum_{k=1}^{n-1} \xb_k ,\quad
\ell(y) = \sum_{k=1}^{n} y_k + \sum_{k=1}^{n-1} \yb_k .
\end{equation*}
The operations $\sigma_1, \sigma_n$ and $\ast$ in (\ref{eq:combiR}) 
are still defined by (\ref{eq:pairauto}) and (\ref{eq:ast}).

Let us recall the $D^{(1)}_n$-crystal $B_l \; (l \in \Z_{>0})$
{}from \cite{KKM}. It is a finite set 
\[
B_l = \{ (\zeta_1,\ldots,\zeta_n,\overline{\zeta}_n,\ldots,
\overline{\zeta}_1) \in \Z_{\ge 0}^{2n} \mid 
\sum_{i=1}^n(\zeta_i + \overline{\zeta}_i) = l, \; \;
\zeta_n\overline{\zeta}_n = 0 \}
\]
endowed with certain maps ${\tilde e}_i, {\tilde f}_i\, (0 \le i \le n)$
as described in section 5.4 of the mentioned paper.
%
The set $B_l$ is in one to one correspondence with 
another set
\begin{align*}
B'_l = \{ (x_1,\ldots,x_n,\overline{x}_{n-1},\ldots,
\overline{x}_1)& \in \Z^{2n-1} \mid 
x_i,\xb_i\ge0\text{ for }1\le i\le n-1,\\
&x_n\ge-\min(x_{n-1},\xb_{n-1}),
\sum_{i=1}^{n-1}(x_i + \overline{x}_i) + x_n = l\}
\end{align*}
via $x_i = \zeta_i, \;\;  \overline{x}_i 
= \overline{\zeta}_i \;  (1 \le i \le n-2),\; 
x_{n-1}=\zeta_{n-1}+\overline{\zeta}_{n}, \;
x_{n}=\zeta_{n}-\overline{\zeta}_{n}, \; 
\xb_{n-1}=\overline{\zeta}_{n-1}+\overline{\zeta}_{n},\; 
\zeta_n = \max(0,x_n),\; \overline{\zeta}_n=\max(0,-x_n),\; 
\zeta_{n-1} = x_{n-1} + \min(0,x_n),\; 
\overline{\zeta}_{n-1} = \xb_{n-1} + \min(0,x_n)$.
%
Note that $x_n$ can be negative. Below we give the explicit crystal structure
of $B'_l$, which is also obtained by ultradiscretizing the geometric crystal
structure of $\B$ in section \ref{subsec:defofB}.
%
\begin{eqnarray*}
\veps_0(x)&=&x_1+(x_2-\xb_2)_+,\quad
\vphi_0(x)=\xb_1+(\xb_2-x_2)_+,\\
\veps_i(x)&=&\xb_i+(x_{i+1}-\xb_{i+1})_+,\quad
\vphi_i(x)=x_i+(\xb_{i+1}-x_{i+1})_+\quad(i=1,\ldots,n-2),\\
\veps_{n-1}(x)&=&x_n+\xb_{n-1},\quad \vphi_{n-1}(x)=x_{n-1},\\
\veps_n(x)&=&\xb_{n-1},\quad \vphi_n(x)=x_{n-1}+x_n,\\
\et{0}^c(x)&=&(x_1-\xi_2,x_2+\xi_2-c,\ldots,\xb_2+\xi_2,\xb_1-\xi_2+c)  
,\\
\et{i}^c(x)&=&(\ldots,x_i-\xi_{i+1}+c,x_{i+1}+\xi_{i+1}-c,\ldots,
\xb_{i+1}+\xi_{i+1},\xb_i-\xi_{i+1},\ldots)\\
&&\hspace{7.5cm}\quad(i=1,\ldots,n-2),\\
\et{n-1}^c(x)&=&(\ldots,x_{n-1}+c,x_n-c,\ldots),\\
\et{n}^c(x)&=&(\ldots,x_n+c,\xb_{n-1}-c,\ldots),\\
\mbox{where }&&\xi_i=\max(x_i,\xb_i+c)-\max(x_i,\xb_i)\quad(i=1,\ldots,n-1).
\end{eqnarray*}
Here $c$ is an integer. If $c$ is negative, we understand $\et{i}^c$ as 
$\ft{i}^{-c}$. Also, if $\et{i}^c(x)$ does not belong to $B'_l$, one should
assume it to be $0$.

We are interested in the bijection $R:B'_l \otimes B'_k 
\rightarrow B'_k \otimes B'_l$ which commutes with $\et{i}^c$. It is called 
the {\it combinatorial $R$}. See Remark \ref{rem:tensor} for the action of
$\et{i}^c$ on the tensor product. Below we write $(x,y)$ for an element
$x\ot y$ of $B'_l\ot B'_k$. 
We remark that the combinatorial $R: B_l\ot B_k\rightarrow B_k\ot B_l$ has
been given in terms of a certain insertion procedure \cite{HKOT2}.

\begin{theorem}
\label{th:combR}
Restriction of (\ref{eq:combiR}) 
to $x \in B'_{\ell_1}, y \in B'_{\ell_2}$ yields an explicit formula 
$(x,y) \mapsto (x',y')$ 
for the combinatorial 
$R: B'_{\ell_1} \otimes B'_{\ell_2} \rightarrow 
B'_{\ell_2} \otimes B'_{\ell_1}$.
\end{theorem}

\begin{proof}
Set
\[
\tilde{B}'_l=\{(x_1,\ldots,x_n,\overline{x}_{n-1},\ldots,
\overline{x}_1)
\in\Z^{2n-1}\mid
\sum_{i=1}^{n-1}(x_i+\xb_i)+x_n=l\}.
\]
Let $E_i^c$ be the map on $\tilde{B}'_{\ell_1}\ot \tilde{B}'_{\ell_2}$ obtained 
by ultradiscretizing $e_i^c$ on $\B\times\B$, let $R_0$ be the map 
$\tilde{B}'_{\ell_1}\ot\tilde{B}'_{\ell_2}\rightarrow
\tilde{B}'_{\ell_2}\ot\tilde{B}'_{\ell_1}$ given by ultradiscretizing the tropical
$R$, and let $R:B'_{\ell_1}\ot B'_{\ell_2}\rightarrow 
B'_{\ell_2}\ot B'_{\ell_1}$, be the combinatorial $R$. We are to show $R(x,y)=
R_0(x,y)$ if $(x,y)\in B'_{\ell_1}\ot B'_{\ell_2}$. 

First recall that $R$ is uniquely characterized by
\begin{itemize}
\item[(i)] $R((\ell_1,0,\ldots,0),(\ell_2,0,\ldots,0))=
           ((\ell_2,0,\ldots,0),(\ell_1,0,\ldots,0))$,
\item[(ii)] If $(x,y),\et{i}^c(x,y)\in B'_{\ell_1}\ot B'_{\ell_2}$, then
            $R\et{i}^c(x,y)=\et{i}^cR(x,y)$.
\end{itemize}
Let us check (i) for $R_0$. Assume $x=x_0=(\ell_1,0,\ldots,0)$, 
$y=y_0=(\ell_2,0,\ldots,0)$. 
Then we have
$\theta_{i,j} = \ell_1 \, (1 \leq j \leq i), \, = \ell_2 \, (i+1 \leq
j \leq n-2)$;
$\theta'_{i,j}=\ell_2 \, (1 \leq j \leq n-2)$;
$\eta_{i,j} = 0 \, (j=1, i \ne 0), \, = \ell_2 - \ell_1 \,
(j=1, i=0), \, = \ell_1 \, (1 < j \leq i), \, = \ell_2 \,
(i+1 \leq j \leq n)$, and
$\eta'_{i,j} = \ell_1 + \ell_2 \,(j=1), \, = \ell_2 \,
(1 < j \leq n-1 \, \mbox{or} \, j=n, i \ne n-1),
\, =\ell_1 \, (j=n, i=n-1)$.
Therefore we have
$V_0 = V_0^{\sigma_1} = V_i = V_i^* = \ell_1 + \ell_2$ and
$W_i = 2(\ell_1 + \ell_2)$ for $1 \leq i \leq n-1$.
Thus we obtain
$R(x_0,y_0)=((\ell_2 ,0,\ldots,0),(\ell_1 ,0,\ldots,0))$.

For $(x,y)\in B'_{\ell_1}\ot B'_{\ell_2}$ let $m$ be the minimum such that
$\et{i_p}^{c_p}\cdots\et{i_1}^{c_1}(x_0,y_0)\in B'_{\ell_1}\ot B'_{\ell_2}$
($1\le p\le m$), $(x,y)=\et{i_m}^{c_m}\cdots\et{i_1}^{c_1}(x_0,y_0)$
for some integers $c_1,\ldots,c_m$. We prove $R=R_0$ on $B'_{\ell_1}\ot 
B'_{\ell_2}$ by induction on $m$. The $m=0$ case is proven above. Assume 
the $m-1$ case. Note that 
\begin{eqnarray*}
&&E_i^c(x,y)=\et{i}^c(x,y)\text{ if }(x,y),\et{i}^c(x,y)\in 
B'_{\ell_1}\ot B'_{\ell_2},\\
&&E_i^cR_0(x,y)=R_0E_i^c(x,y)\text{ for }(x,y)\in 
\tilde{B}'_{\ell_1}\ot\tilde{B}'_{\ell_2}.
\end{eqnarray*}
The second equality is obtained by ultradiscretizing Proposition 
\ref{th:eRequalsRe}. Thus we get 
\begin{eqnarray*}
R_0(x,y)&=&R_0E_{i_m}^{c_m}\et{i_{m-1}}^{c_{m-1}}\cdots\et{i_1}^{c_1}(x_0,y_0)\\
&=&E_{i_m}^{c_m}R_0\et{i_{m-1}}^{c_{m-1}}\cdots\et{i_1}^{c_1}(x_0,y_0)\\
&=&\et{i_m}^{c_m}R\et{i_{m-1}}^{c_{m-1}}\cdots\et{i_1}^{c_1}(x_0,y_0)\\
&=&R(x,y).
\end{eqnarray*}
The proof is completed.
\end{proof}
%

\begin{remark}\label{rem:energy}
By the ultradiscretization of 
Proposition \ref{pr:V0energylike}, we have 
\begin{align*}
V_0(\tilde{e}_i(x, y)) &= V_0(x, y) +
\left( \theta (\vphi_0(x') \geq \veps_0(y')) -
\theta(\vphi_0(x) < \veps_0(y)) \right)\delta_{i 0} \\
%
&= 
\begin{cases}
V_0(x, y) +1&\mbox{ if }i=0,\ \vphi_0(x)\geq\veps_0(y),\ 
\vphi_0(x')\geq\veps_0(y'),\\
V_0(x, y) -1&\mbox{ if }i=0,\ \vphi_0(x)<\veps_0(y),\ 
\vphi_0(x')<\veps_0(y'),\\
V_0(x, y)&\mbox{ otherwise},
\end{cases}
\end{align*}
where $\vphi_0(x) = \xb_1 + (\xb_2 - x_2)_+, \; 
\veps_0(y) = y_1 + (y_2 - \yb_2)_+$.
%
See Example \ref{ex:August6_2}.
%
This recursion relation is known to characterize 
an important function on $B'_{\ell(x)} \otimes B'_{\ell(y)}$ called 
the {\em energy function} up to an additive constant.
%
See [NaY, Definition 3.2].
%
Thus we conclude that (\ref{eq:16}) is the tropical analogue 
of the energy function. 
%
In our formula (\ref{eq:piecewise}), it is so normalized that
$\min\{V_0(x,y) \mid (x,y) \in B'_{\ell(x)} \otimes B'_{\ell(y)} \}
= | \ell(x) - \ell(y)|$.
%
If we put $\ell(x)=\ell(y)=0$, the expression 
(\ref{eq:piecewise}) with $i=0$ reduces to 
the one in \cite{KKM}
associated to the crystal $B_\infty \otimes B_\infty$.
\end{remark}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\appendix
\section{Alternative proof of Theorem \ref{th:ijieqjij}}\label{sec:appA}
\noindent
Let $X=\sum_{1 \leq j \leq i \leq 2n} x_{i,j} E_{i,j}$ be a lower 
triangular
matrix satisfying
\begin{equation}
\label{eq:Liegr}
X S \, ^t \! X S =S \, ^t \! X S X = E,
\end{equation}
where $S$ is the matrix given by (\ref{eq:signmatrix}).
% This condition is also written as
% \begin{equation}
% \label{eq:Liegr}
% \sum_{j=k}^i x_{i,j} x_{2n+1-k,2n+1-j} (-1)^j =
% \sum_{j=k}^i x_{j,k} x_{2n+1-j,2n+1-i} (-1)^j =
% (-1)^k \delta_{i,k}.
% \end{equation}
%%%%%%
\begin{lemma}
$G_k(u)XG_k(v)^{-1}$ for $1 \leq k \leq n-1$
is lower-triangular if and only if
\begin{align}
\label{eq:uvrel1}
u x_{k+1,k+1} &= v x_{k,k}+u v x_{k+1,k},\\
\label{eq:uvrel2}
u x_{2n+1-k,2n+1-k} &= v x_{2n-k,2n-k}+u v x_{2n+1-k,2n-k}.
\end{align}
\end{lemma}
%
\begin{proof}
Put $X'=G_k(u)XG_k(v)^{-1}$. The elements of the matrix $X'=(x'_{i,j})$
read
\begin{align*}
x'_{i,j} &= x_{i,j}+u (\delta_{i,k}x_{k+1,j} +
\delta_{i,2n-k}x_{2n+1-k,j})
-v (\delta_{k,j-1}x_{i,k}+\delta_{2n+1-k,j}x_{i,2n-k}) \\
%
&-uv (
\delta_{i,k}\delta_{j,k+1}x_{k+1,k} +
\delta_{i,2n-k}\delta_{j,k+1}x_{2n+1-k,k} +
\delta_{i,2n-k}\delta_{j,2n+1-k}x_{2n+1-k,2n-k}).
\end{align*}
The only non-zero upper-triangular elements are
\begin{align*}
x'_{k,k+1} &= u x_{k+1,k+1}-v x_{k,k}-u v x_{k+1,k},\\
x'_{2n-k,2n+1-k} &= u x_{2n+1-k,2n+1-k}-v x_{2n-k,2n-k}-u v
x_{2n+1-k,2n-k}.
\end{align*}
Hence the relations (\ref{eq:uvrel1}) and (\ref{eq:uvrel2}) follow.
\end{proof}
%
Note that the following parameterization solves the
relation (\ref{eq:uvrel1}):
\begin{equation}
\label{eq:uvc}
u=\frac{x_{k,k}}{x_{k+1,k}}(c-1), \quad
v=\frac{x_{k+1,k+1}}{x_{k+1,k}}\frac{(c-1)}{c}.
\end{equation}
It also solves the relation (\ref{eq:uvrel2})
due to (\ref{eq:Liegr}).
In view of this, we define the action $e_i^c \; (1 \leq i \leq n-1)$
on $X$ by
\begin{equation}\label{eq:harry}
e_i^c(X)=G_i\left(\frac{x_{i,i}}{x_{i+1,i}}(c-1)\right)X
G_i\left(\frac{x_{i+1,i+1}}{x_{i+1,i}}\frac{(c-1)}{c}\right)^{-1}.
\end{equation}
Note that we have $e_i^c(X) S \, ^t \! e_i^c(X) S =
S \, ^t \! e_i^c(X) S e_i^c(X) = E$.


\begin{prop}
\label{prop:eee}
\begin{equation*}
e_i^d e_{i+1}^{cd} e_i^c(X)=e_{i+1}^c e_i^{cd} e_{i+1}^d(X)
\quad \hbox{for }\;  1 \leq i \leq n-2.
\end{equation*}
\end{prop}
%
\begin{proof}
Setting \begin{math}
Y=e_i^c(X), \, Z=e_{i+1}^{cd}(Y), \, Y'=e_{i+1}^d(X), \,
Z'=e_{i}^{cd}(Y'), 
\end{math}
we introduce 
\begin{equation}\label{eq:uv}
\begin{split}
&u_1=\frac{X_{i,i}}{X_{i+1,i}}(c-1), \quad
u_2=\frac{Y_{i+1,i+1}}{Y_{i+2,i+1}}(cd-1), \quad
u_3=\frac{Z_{i,i}}{Z_{i+1,i}}(d-1),\\
%
&u'_1=\frac{X_{i+1,i+1}}{X_{i+2,i+1}}(d-1), \quad
u'_2=\frac{Y'_{i,i}}{Y'_{i+1,i}}(cd-1), \quad
u'_3=\frac{Z'_{i+1,i+1}}{Z'_{i+2,i+1}}(c-1),\\
%
&v_1=\frac{X_{i+1,i+1}}{X_{i+1,i}}\frac{(c-1)}{c}, \quad
v_2=\frac{Y_{i+2,i+2}}{Y_{i+2,i+1}}\frac{(cd-1)}{cd}, \quad
v_3=\frac{Z_{i+1,i+1}}{Z_{i+1,i}}\frac{(d-1)}{d},\\
%
&v'_1=\frac{X_{i+2,i+2}}{X_{i+2,i+1}}\frac{(d-1)}{d}, \quad
v'_2=\frac{Y'_{i+1,i+1}}{Y'_{i+1,i}}\frac{(cd-1)}{cd}, \quad
v'_3=\frac{Z'_{i+2,i+2}}{Z'_{i+2,i+1}}\frac{(c-1)}{c}.
\end{split}
\end{equation}
{}From (\ref{eq:harry}) we know
\begin{align*}
&e_i^d e_{i+1}^{cd} e_i^c(X)=G_i(u_3)G_{i+1}(u_2)G_i(u_1)X
\left(G_i(v_3)G_{i+1}(v_2)G_i(v_1)\right)^{-1},\\
%
&e_{i+1}^c e_{i}^{cd} e_{i+1}^d(X)=G_{i+1}(u'_3)G_{i}(u'_2)G_{i+1}(u'_1)X
\left(G_{i+1}(v'_3)G_{i}(v'_2)G_{i+1}(v'_1)\right)^{-1}.
\end{align*}
A direct calculation leads to
\begin{equation*}
\begin{array}l
\displaystyle u_1=\frac{x_{i,i}}{x_{i+1,i}}(c-1),\quad
\displaystyle u_2=\frac{x_{i+1,i}x_{i+1,i+1}}{w_1}(cd-1),\quad
\displaystyle u_3=\frac{x_{i,i}}{x_{i+1,i}}\frac{w_1}{w_2}(d-1),\\[4mm]
\displaystyle u'_1=\frac{x_{i+1,i+1}}{x_{i+2,i+1}}(d-1),\quad
\displaystyle u'_2=\frac{x_{i,i}x_{i+2,i+1}}{w_2}(cd-1),\quad
\displaystyle u'_3=\frac{x_{i+1,i+1}}{x_{i+2,i+1}}\frac{w_2}{w_1}(c-1),\\[4mm]
%
w_1=(1-c) x_{i+2,i}x_{i+1,i+1}+cx_{i+1,i}x_{i+2,i+1}, \\[3mm]
w_2=(d-1)x_{i+2,i}x_{i+1,i+1}+x_{i+1,i}x_{i+2,i+1}.
\end{array}
\end{equation*}
Then the relations
\begin{equation*}
u'_1=\frac{u_2 u_3}{u_1+u_3}, \quad
u'_2=u_1+u_3, \quad
u'_3=\frac{u_1 u_2}{u_1+u_3},
\end{equation*}
can be checked, and from (\ref{eq:GiGjGi=GjGiGj}), 
\begin{equation}\label{eq:GGG=GGG1}
G_i(u_3)G_{i+1}(u_2)G_i(u_1)=G_{i+1}(u'_3)G_{i}(u'_2)G_{i+1}(u'_1).
\end{equation}
Similarly, one can check
\begin{equation}\label{eq:GGG=GGG2}
G_i(v_3)G_{i+1}(v_2)G_i(v_1)=G_{i+1}(v'_3)G_{i}(v'_2)G_{i+1}(v'_1).
\end{equation}
\end{proof}
%%%%%%

\begin{lemma}
\label{lem:epsandphiforaaa}
Let $X=\left( X_{i,j} \right)_{1 \leq i,j \leq 2n} = A(x^1)\cdots A(x^L)$.
Then we have
\begin{displaymath}
\frac{X_{i+1,i}}{X_{i,i}} = \veps_i (\boldsymbol{x}), \quad
\frac{X_{i+1,i}}{X_{i+1,i+1}} = \vphi_i (\boldsymbol{x})
\quad \hbox{for } \; 1 \leq i \leq n-1.
\end{displaymath}
\end{lemma}
\begin{proof}
We prove by induction on $L$.
If $L=1$ the claim can be checked directly.
Suppose $L > 1$.
Let $\tilde{X}=\left( \tilde{X}_{i,j}
\right)_{1 \leq i,j \leq 2n} = A(x^1)\cdots A(x^{L-1})$.
Since $\tilde{X}$ and $A(x^L)$ are lower-triangular we have
\begin{align*}
X_{i,i} &= \tilde{X}_{i,i} A(x^L)_{i,i}, \\
X_{i+1,i} &= \tilde{X}_{i+1,i} A(x^L)_{i,i}+
\tilde{X}_{i+1,i+1} A(x^L)_{i+1,i}, \\
X_{i+1,i+1} &= \tilde{X}_{i+1,i+1} A(x^L)_{i+1,i+1}.
\end{align*}
Denote $(x^1,\ldots,x^{L-1})$ by $\tilde{\boldsymbol{x}}$.
Suppose
\begin{displaymath}
\frac{\tilde{X}_{i+1,i}}{\tilde{X}_{i,i}} =
\veps_i (\tilde{\boldsymbol{x}}), \quad
\frac{\tilde{X}_{i+1,i}}{\tilde{X}_{i+1,i+1}} =
\vphi_i (\tilde{\boldsymbol{x}}).
\end{displaymath}
Then we have
\begin{align*}
\frac{X_{i+1,i}}{X_{i,i}} &= \frac{\tilde{X}_{i+1,i}}{\tilde{X}_{i,i}}+
\frac{\tilde{X}_{i+1,i+1}A(x^L)_{i+1,i}}{\tilde{X}_{i,i}A(x^L)_{i,i}}
= \veps_i (\tilde{\boldsymbol{x}})+
\frac{\veps_i (\tilde{\boldsymbol{x}}) \veps_i(x^L)}
{\vphi_i (\tilde{\boldsymbol{x}})}
=\veps_i (\boldsymbol{x}),\\
\frac{X_{i+1,i}}{X_{i+1,i+1}} &=
\frac{A(x^L)_{i+1,i}}{A(x^L)_{i+1,i+1}}+
\frac{\tilde{X}_{i+1,i}A(x^L)_{i,i}}{\tilde{X}_{i+1,i+1}A(x^L)_{i+1,i+1}}
= \vphi_i (x^L)+
\frac{\vphi_i (\tilde{\boldsymbol{x}}) \vphi_i(x^L)}
{\veps_i (x^L)}
=\vphi_i (\boldsymbol{x}).
\end{align*}
\end{proof}


\begin{proof}[Proof of Theorem \ref{th:ijieqjij}]
We concentrate on the nontrivial case (\ref{eq:ijieqjij}) with $i<j$.
Note that $(i,j)=(0,2)$ case reduces to
$(i,j)=(1,2)$ as
\begin{align*}
e^d_0 e^{cd}_2 e^c_0 (\boldsymbol{x}) &=
(e^d_1 e^{cd}_2 e^c_1 (\boldsymbol{x}^{\sigma_1}))^{\sigma_1} \\
&=
(e^c_2 e^{cd}_1 e^d_2 (\boldsymbol{x}^{\sigma_1}))^{\sigma_1} \\
&=e^c_2 e^{cd}_0 e^d_2 (\boldsymbol{x}).
\end{align*}
Similarly  $(i,j)=(n-2,n)$ case reduces to 
$(i,j)=(n-2,n-1)$.
%
Thus it suffices to prove the relations
$P=P'$ and $Q=Q'$ $(1 \leq i \leq n-2)$ for 
$P,P',Q$ and $Q'$ specified in (\ref{eq:PPQQ1})-(\ref{eq:PPQQ4}).

Let $\boldsymbol{x}=(x^1,\ldots,x^L)$ and $X= A(x^1)\cdots A(x^L)$.
Since $X$ is lower triangular and satisfies (\ref{eq:Liegr}), 
we can apply Lemma \ref{prop:eee} keeping all the notations in its proof.
%
Then (\ref{eq:GGG=GGG1}) and 
(\ref{eq:GGG=GGG2}) are nothing but $P=P'$ and $Q=Q'$, 
respectively.
%
Let us show the former for example.
%
By Lemma \ref{lem:epsandphiforaaa} we have
$u_1 = \frac{c-1}{\veps_{i}(\boldsymbol{x})}$ and 
$u'_1=\frac{d-1}{\veps_{i+1}(\boldsymbol{x})}$.
%
Next we are to show
$u_2=\frac{cd-1}{\veps_{i+1}(e^c_i(
\boldsymbol{x}))}$ and 
$u'_2=\frac{cd-1}{\veps_{i}(e^d_{i+1}(
\boldsymbol{x}))}$.
%
We illustrate the way along $u_2$.
%
Setting  $e^c_i(\boldsymbol{x}) = (y^1,\ldots,y^L)$, we get
\begin{align*}
Y = e^c_i(X) &=
G_i\left( u_1 \right)
X \;
G_i\left( v_1 \right)^{-1} \\
&=
G_i\left( u_1 \right)
A(x^1) \cdots A(x^L) \;
G_i\left( v_1 \right)^{-1} \\
&= A(y^1) \ldots A(y^L),
\end{align*}
where the last equality is due to Theorem \ref{th:MMMequalGMMMG}.
Now Lemma \ref{lem:epsandphiforaaa} tells that 
\begin{math}
\frac{Y_{i+2,i+1}}{Y_{i+1,i+1}} = \veps_{i+1} (e^c_i(\boldsymbol{x})).
\end{math}
%
The remaining relations are similarly verified.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of Proposition \ref{th:main}}\label{sec:proofofaaeqaa}
\noindent
%
Set $I=A(x)A(y)$, which should not be confused with the identity matrix $E$.
\begin{lemma}
\begin{align}
0 &= \sum_{m=i}^n I_{m,i} I_{2n+1-m,i} (-1)^m,
\quad (1 \leq i \leq n)
%
\label{eq:1c} \\
%
I_{j,i} &= y_i \left( 1+\frac{x_i}{\yb_i} \right)
\left( 1 + \frac{x_j}{\xb_j} \right) \left( \prod_{m=i+1}^{j-1} x_m 
\right)
+ y_i \left( I_{j,i+1} - \delta_{j,i+1} \right),
\label{eq:reccurent1} \\
& \qquad (1 \leq i \leq j-1 \leq n-2)
\nonumber\\
%
I_{n,n-1} &= \frac{y_{n-1}}{\yb_{n-1}} x_{n-1} x_n +
y_{n-1} I_{n,n},
%
\label{eq:4c} \\
%
I_{n,i} &= y_i \left( 1+\frac{x_i}{\yb_i} \right)
\left( \prod_{m=i+1}^{n} x_m \right)+ y_i I_{n,i+1}, \quad
(1 \leq i \leq n-2)
%
\label{eq:reccurent2}
%
\\
%
I_{j,i} &= y_i \left( 1+\frac{x_i}{\yb_i} \right)
\left( \prod_{m=i+1}^{n} x_m \right)
\left( \prod_{m=2n+1-j}^{n-1} \xb_m \right) + y_i I_{j,i+1},
\label{eq:reccurent3} \\
&\qquad (1 \leq i \leq 2n-j \leq n-2)
\nonumber \\
%
I_{2n+1-i,i} &= y_i \xb_i 
\left( \frac{1}{x_i} + \frac{1}{\yb_i} \right)
\left( x_i x_n \prod_{m=i+1}^{n-1} x_m \xb_m + \yb_i y_n
\prod_{m=i+1}^{n-1} y_m \yb_m
\right) 
\label{eq:reccurent4} \\
& + y_i \xb_i I_{2n-i,i+1}. \qquad (1 \leq i \leq n-2) \nonumber
\end{align}
\end{lemma}

\begin{proof}
%
{}From Proposition \ref{pr:MSMS}, one has $I S \, ^t \! I S = E$.
%
In particular 
$\sum_{m=i}^{2n+1-i} I_{m,i} I_{2n+1-m,i}$ $S_{m,2n+1-m} = 0$,
which is equivalent to (\ref{eq:1c}).

To show  (\ref{eq:reccurent1}), (\ref{eq:reccurent2}) and
(\ref{eq:reccurent3}), we consider
\begin{align*}
I_{j,i} - y_i I_{j,i+1} &= \sum_{k=1}^{2n} A(x)_{j,k}
\left( A(y)_{k,i} - y_i A(y)_{k,i+1} \right) \\
&= A(x)_{j,i} \frac{y_i}{\yb_i} + A(x)_{j,i+1} y_i \\
&= y_i \left( 1+\frac{x_i}{\yb_i} \right) \frac{A(x)_{j,i}}{x_i}
-y_i \delta_{j,i+1},
\end{align*}
where (\ref{eq:defMno1}) is used twice.
Repeated use of (\ref{eq:defMno1}) leads to 
$A(x)_{j,i} = A(x)_{j,1}/(x_{i-1}\cdots x_1)$.
Substituting the explicit form (\ref{eq:firstcolumnofA}) of
$A(x)_{j,1}$ we obtain the desired relations.

Note that $I_{n,n-1}=A(x)_{n,n}A(y)_{n,n-1} + A(x)_{n,n-1}A(y)_{n-1,n-1}
=x_n y_{n-1}y_n + x_{n-1}x_n \frac{y_{n-1}}{\yb_{n-1}}$.
(\ref{eq:4c}) follows from this and $I_{n,n}=x_n y_n$.

Let us consider (\ref{eq:reccurent4}).
{}From (\ref{eq:defMno1}) and (\ref{eq:JMJequalsMno3})
it follows that
\begin{align*}
A(x)_{2n+1-i,k} &= A(x)_{2n-i,k} \xb_i +
\begin{cases}
\xb_i/x_i & \mbox{for} \, k=2n+1-i,\\
\xb_i     & \mbox{for} \, k=2n-i,\\
0         & \mbox{for} \, k \ne 2n+1-i,2n-i,
\end{cases}\\
%
A(y)_{k,i} &= A(y)_{k,i+1} y_i +
\begin{cases}
y_i/\yb_i & \mbox{for} \, k=i,\\
y_i       & \mbox{for} \, k=i+1,\\
0         & \mbox{for} \, k \ne i,i+1.
\end{cases}
\end{align*}
%
Substitute these relations into
\begin{math}
I_{2n+1-i,i} = \sum_{k=1}^{2n} A(x)_{2n+1-i,k} A(y)_{k,i}.
\end{math}
The result reads 
\begin{align*}
I_{2n+1-i,i} - y_i \xb_i I_{2n-i,i+1} &=
\xb_i y_i \left( A(x)_{2n-i,i} \frac{1}{\yb_i} + A(x)_{2n-i,i+1} \right) 
\\
&+ \xb_i y_i
\left( A(y)_{2n+1-i,i+1} \frac{1}{x_i} + A(y)_{2n-i,i+1} \right) \\
%
& = \xb_i y_i \left( \frac{1}{x_i} + \frac{1}{\yb_i} \right)
\left( A(x)_{2n-i,i}  + A(y)_{2n+1-i,i+1}  \right),
\end{align*}
where 
$A(x)_{2n-i,i+1}  = A(x)_{2n-i,i}/x_i$
and $A(y)_{2n-i,i+1} = A(y)_{2n+1-i,i+1}/\yb_i$ are used 
in the last equality.
It remains to show
$A(x)_{2n-i,i} = x_i x_n \prod_{m=i+1}^{n-1} x_m \xb_m$ and
$A(y)_{2n+1-i,i+1} = \yb_i y_n \prod_{m=i+1}^{n-1} y_m \yb_m $.
Repeated use of (\ref{eq:defMno1}) leads to 
$A(x)_{2n-i,i} = A(x)_{2n-i,1}/(x_{i-1}\cdots x_1)$.
The former relation is derived 
by substituting  (\ref{eq:firstcolumnofA}) into $A(x)_{2n-i,1}$ here.
%
The latter follows from the former by (\ref{eq:JMJequalsMno3}).
\end{proof}
%
%
\begin{lemma}
\label{lem:August4_1}
\begin{align}
\label{eq:August4_1_1}
I_{j,i} &= I_{2n+1-i,2n+1-j}^{* \circ \sigma_n}
& (1 \leq i,j \leq n), \\
\label{eq:August4_1_2}
I_{n+1,i} &= I_{n,i}^{\sigma_n} & (1 \leq i \leq n-1).
\end{align}
\end{lemma}
\begin{proof}
{}From the proof of Lemma \ref{lem:JMJequalsMxxx}, we know 
$(x,y) = \tau((x^{\sigma_n},y^{\sigma_n})^\ast)$.
Therefore by using (\ref{eq:JMJequalsMno3}) we have
\begin{math}
A(x)A(y) = 
J\,^t \!A(y^{\ast \circ \sigma_n})
\,^t \!A(x^{\ast \circ \sigma_n}) J =
J\,^t \!(A(x^{\ast \circ \sigma_n} )
A(y^{\ast \circ \sigma_n})) J.
\end{math}
This yields the former relation.
%
The latter relation follows from
$A(x)A(y) = J_n A(x^{\sigma_n})A(y^{\sigma_n}) J_n$ due to 
(\ref{eq:JMJequalsMno2}).
\end{proof}
%
The following expressions can be checked directly:
\begin{align}
I_{n+1,n} &= 0,\\
I_{i,i} &= \frac{x_iy_i}{\xb_i\yb_i} \quad ( 1 \leq i \leq n-1),
\quad I_{n,n} = x_n y_n,
\label{eq:diag}
\\
%
I_{n,n-1} &= x_{n-1}x_n y_{n-1}y_n \left( \frac{1}{x_{n-1} } + 
\frac{1}{\yb_{n-1} y_n } \right),
\label{eq:12a}\\
%
\label{eq:a11}
I_{n+2,n-1} &= y_{n-1}y_n \xb_{n-1}
\left( \yb_{n-1} + x_{n-1}x_n \right)
\left( \frac{1}{x_{n-1} } + \frac{1}{\yb_{n-1} y_n } \right).
%
\end{align}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}
\label{lem:1}
Let $(x',y') = \tilde{R}(x,y)$.
Then we have
\begin{equation*}
1+\frac{x_j'}{\xb_j'}=
\frac{V_{j-1}}{W_{j-1}} \left[
\left( 1+\frac{x_j}{\xb_j} \right) V_{j-1}^* + (\ell (y) - \ell (x))
\left( \frac{x_j y_j}{\xb_j \yb_j} -1 \right) \right],
\end{equation*}
for $2 \leq j \leq n-1$.
\end{lemma}
%
\begin{proof}
By definition of $\tilde{R}$ we obtain
\begin{equation}
1+\frac{x_j'}{\xb_j'} = 
\frac{1}{W_{j-1}} \left( W_{j-1} + \frac{y_j}{\yb_j} W_j \right).
\end{equation}
In the parentheses of the right side, apply (\ref{eq:27c}) to
(\ref{eq:40b}) to rewrite $W_{j-1}$ and $W_j$.
\end{proof}
%
\begin{lemma}
\label{lem:2}
Let $(x',y') = \tilde{R}(x,y)$.
Then we have
\begin{equation}
y_i' \left( 1 + \frac{x_i'}{\yb_i'} \right) =
y_i \left( 1 + \frac{x_i}{\yb_i} \right) \frac{W_i}{V_i V_i^*},
\end{equation}
for $1 \leq i \leq n-2$.
\end{lemma}
\begin{proof}
First we suppose $2 \leq i \leq n-2$.
By definition of $\tilde{R}$ we have
\begin{eqnarray}
1 + \frac{x_i'}{\yb_i'} &=&
\frac{y_i}{V_i V_{i-1}^*}
\left( \frac{V_i V_{i-1}^*}{y_i} +
\frac{V_{i-1} V_i^*}{\xb_i} \right) \nonumber\\
%
&=& \frac{y_i}{V_i V_{i-1}^*}
\left( \frac{1}{x_i} + \frac{1}{\yb_i} \right) W_i,
\end{eqnarray}
%
where the last step is due to (\ref{eq:44b}).
Multiplying the both sides by 
$y_i' = x_i \frac{V_{i-1}^*}{V_i^*}$ we get the
desired relation.
%
For  $i=1$, we have
\begin{eqnarray}
1 + \frac{x_1'}{\yb_1'} &=&
\frac{y_1}{V_1}
\left( \frac{V_1}{y_1} +
\frac{V_1^*}{\xb_1} \right) \nonumber\\
%
&=& \frac{y_1}{V_1}
\left( \frac{1}{x_1} + \frac{1}{\yb_1} \right) V_0^{\sigma_1},
\end{eqnarray}
%
where the last step is due to 
$\sigma_1$ and $* \circ \sigma_1$ of (\ref{eq:vind}).
%
Now multiply the both sides by 
$y_1' = x_1 \frac{V_0}{V_1^*}$ and 
simplify the result by Lemma \ref{lem:factorW1}.
\end{proof}
%
%
\begin{lemma}
\label{lem:3}
Let $(x',y') = \tilde{R}(x,y)$.
Then we have
\begin{equation}
y_i' = y_i \left[ 1 + \frac{\ell (x) - \ell (y)}{V_i^*} \left( 1 + 
\frac{x_i}{\yb_i} \right)
\right],
\end{equation}
for $1 \leq i \leq n-2$.
\end{lemma}
\begin{proof}
This identity is a consequence of
$y_i' = x_i \frac{V_{i-1}^*}{V_i^*}$ and
$*$ of (\ref{eq:vind}).
\end{proof}
%
It is straightforward to check 
\begin{lemma}
\label{lem:4}
Let $(x',y') = \tilde{R}(x,y)$.
Then we have
%
\begin{align*}
\left( \prod_{m=i+1}^{j-1} x'_m \right) &=
\left( \prod_{m=i+1}^{j-1} y_m \right) \frac{V_i W_{j-1}}{V_{j-1}W_i},
\quad (1 \leq i \leq j-1 \leq n-2) \\
%
%\left( \prod_{m=i+1}^{n} x'_m \right) &=
%\left( \prod_{m=i+1}^{n} y_m \right) \frac{V_i V_{n-1}}{W_i}, \\
%
\left( \prod_{m=i+1}^{n} x'_m \right)
\left( \prod_{m=2n+1-j}^{n-1} \xb'_m \right)
&=
\left( \prod_{m=i+1}^{n} y_m \right)
\left( \prod_{m=2n+1-j}^{n-1} \yb_m \right)
\frac{V_i V_{2n-j}}{W_i}.\\
\quad (n+1 \leq j \leq 2n-1)
\end{align*}
\end{lemma}
%

\begin{lemma}
\label{th:1}
Let $i,j$ be integers such that $1 \leq i \leq j-1 \leq n-2$.
Suppose $I_{j,i+1}$ is invariant under $\tilde{R}$.
Then $I_{j,i}$ is also invariant under $\tilde{R}$.
\end{lemma}
\begin{proof}
Let $(x',y') = \tilde{R}(x,y)$.
By applying Lemmas \ref{lem:1}, \ref{lem:2}, \ref{lem:3} and \ref{lem:4}, 
the relation (\ref{eq:reccurent1}) is written as 
\begin{align}
I_{j,i}(x',y') &= y_i \left( 1+\frac{x_i}{\yb_i} \right)
\frac{1}{V_i^*}
\left[
\left( 1+\frac{x_j}{\xb_j} \right) V_{j-1}^* + (\ell (y) - \ell (x))
\left( I_{j,j} -1 \right) \right]
\left( \prod_{m=i+1}^{j-1} y_m \right) \nonumber\\
%
& +
y_i \left[ 1 + \frac{\ell (x) - \ell (y)}{V_i^*} \left( 1 + 
\frac{x_i}{\yb_i} \right)
\right]
\left( I_{j,i+1} - \delta_{j,i+1} \right).
\end{align}
Thus it reduces to the following Lemma.
\end{proof}
%
\begin{lemma}
\begin{align}
& \quad
\left[
\left( 1+\frac{x_j}{\xb_j} \right) V_{j-1}^* + (\ell (y) - \ell (x))
\left( I_{j,j} -1 \right) \right]
\left( \prod_{m=i+1}^{j-1} y_m \right)\nonumber\\
%
& = \left( \prod_{m=i+1}^{j-1} x_m \right)
\left( 1+\frac{x_j}{\xb_j} \right) V_i^* +
(\ell (y) - \ell (x)) \left( I_{j,i+1} - \delta_{j,i+1} \right),
\label{eq:56b}
\end{align}
for $1 \leq i \leq j-1 \leq n-2$.
\end{lemma}
\begin{proof}
By applying
(\ref{eq:reccurent1}) and $*$ of (\ref{eq:vind})
this is transformed into the same relation with $i$ replaced by $i-1$.
Thus it suffices to check (\ref{eq:56b}) for 
$i=j-1$, which is straightforward.
\end{proof}
%
%
\begin{lemma}
\label{th:2}
Let $i$ be an integer such that $1 \leq i \leq n-2$.
Suppose $I_{n,i+1}$ is invariant under $\tilde{R}$.
Then $I_{n,i}$ is also invariant under $\tilde{R}$.
\end{lemma}
\begin{proof}
Let $(x',y') = \tilde{R}(x,y)$.
By applying Lemmas \ref{lem:2}, \ref{lem:3} and \ref{lem:4}, 
the relation (\ref{eq:reccurent2}) is written as
\begin{equation}
I_{n,i}(x',y') = y_i \left( 1+\frac{x_i}{\yb_i} \right)
\frac{V_{n-1}}{V_i^*}
\left( \prod_{m=i+1}^{n} y_m \right)
+
y_i \left[ 1 + \frac{\ell (x) - \ell (y)}{V_i^*} \left( 1 + 
\frac{x_i}{\yb_i} \right)
\right] I_{n,i+1}.
\end{equation}
Thus it reduces to the following Lemma.
\end{proof}
%
\begin{lemma}
\label{lem:01}
For $1 \leq i \leq n-2$
\begin{equation}
\left( \prod_{m=i+1}^n y_m \right) V_{n-1} =
\left( \prod_{m=i+1}^n x_m \right) V_i^* + 
(\ell (y) - \ell (x)) I_{n,i+1}. 
\label{eq:21a}
\end{equation}
\end{lemma}
%
\begin{proof}
By using
(\ref{eq:reccurent2}) and $*$ of (\ref{eq:vind})
this is transformed into the same relation with $i$ replaced by $i-1$.
Thus it suffices to check (\ref{eq:21a}) for $i=n-2$, 
which reads
\begin{displaymath}
y_{n-1}y_n V_{n-1} = x_{n-1}x_n V_{n-2}^* + (\ell (y) - \ell (x)) 
I_{n,n-1}.
\end{displaymath}
This follows from (\ref{eq:12a}) and $* \circ \sigma_n$ of (\ref{eq:unv}).
\end{proof}
%%%%




\begin{lemma}
\label{th:3}
Let $i,j$ be integers such that $1 \leq i \leq 2n-j \leq n-2$.
Suppose $I_{j,i+1}$ is invariant under $\tilde{R}$.
Then $I_{j,i}$ is also invariant under $\tilde{R}$.
\end{lemma}
%
\begin{proof}
Let $(x',y') = \tilde{R}(x,y)$.
By applying Lemmas \ref{lem:2}, \ref{lem:3} and  \ref{lem:4}, 
the relation (\ref{eq:reccurent3}) is written as
\begin{align}
I_{j,i}(x',y') &= y_i \left( 1+\frac{x_i}{\yb_i} \right)
\frac{V_{2n-j}}{V_i^*}
\left( \prod_{m=i+1}^{n} y_m \right)
\left( \prod_{m=2n+1-j}^{n-1} \yb_m \right)
\nonumber \\
& +
y_i \left[ 1 + \frac{\ell (x) - \ell (y)}{V_i^*} \left( 1 + 
\frac{x_i}{\yb_i} \right)
\right] I_{j,i+1}.
\end{align}
Thus it reduces to the following Lemma.
\end{proof}
%
\begin{lemma}
\begin{align}
V_{2n-j} \left( \prod_{m=i+1}^{n} y_m \right)
\left( \prod_{m=2n+1-j}^{n-1} \yb_m \right) &=
V_{i}^* \left( \prod_{m=i+1}^{n} x_m \right)
\left( \prod_{m=2n+1-j}^{n-1} \xb_m \right)
\nonumber \\
%
& + (\ell (y) - \ell (x)) I_{j,i+1},
%
\label{eq:59}
%
\end{align}
for $1 \leq i \leq 2n-j \leq n-2$.
\end{lemma}
\begin{proof}
By applying
$*$ of (\ref{eq:vind}) and (\ref{eq:reccurent3})
this is transformed into the same relation with $i$ replaced by $i-1$.
Thus it suffices to check (\ref{eq:59}) for $i=2n-j$, which 
is done in the next Lemma.
\end{proof}
%
\begin{lemma}
\begin{equation}
V_{i} \left( y_n \prod_{m=i+1}^{n-1} y_m \yb_m \right) =
V_{i}^* \left( x_n \prod_{m=i+1}^{n-1} x_m \xb_m \right)
+ (\ell (y) - \ell (x)) I_{2n-i,i+1},
%
\label{eq:60}
\end{equation}
for $1 \leq i \leq n-2$.
\end{lemma}
\begin{proof}
By applying (\ref{eq:vind}), $*$ of (\ref{eq:vind}) and 
(\ref{eq:reccurent4})
this is transformed into the same relation with $i$ replaced by $i-1$.
Thus it suffices to check (\ref{eq:60}) for $i=1$.
Consider the trivial identity:
\begin{equation*}
V_0 \left( y_n \prod_{m=1}^{n-1} y_m \yb_m \right) =
V_0 \left( x_n \prod_{m=1}^{n-1} x_m \xb_m \right)
+ (\ell (y) - \ell (x)) I_{2n,1}.
%
\end{equation*}
The $i=1$ case of (\ref{eq:60}) is shown by 
applying (\ref{eq:vind}), $*$ of (\ref{eq:vind}) and 
(\ref{eq:reccurent4}) to this identity.
\end{proof}
%
\begin{lemma}
\label{th:4}
Let $i$ be an integer such that $1 \leq i \leq n$.
Then $I_{i,i}$ is invariant under $\tilde{R}$.
\end{lemma}
\begin{proof}
This is clear from (\ref{eq:diag}).
\end{proof}
%
\begin{lemma}
\label{th:5}
$I_{n,n-1}$ is invariant under $\tilde{R}$.
\end{lemma}
\begin{proof}
Let $(x',y') = \tilde{R}(x,y)$.
{}From (\ref{eq:4c}) and Theorem \ref{th:4} we have
\begin{equation}
I_{n,n-1}(x',y') = \frac{x_{n-1}}{\xb_{n-1}} y_{n-1}y_n
\frac{V_{n-2}}{V_{n-1}^{*}} +
x_{n-1}  \frac{V_{n-2}^*}{V_{n-1}^{*}} I_{n,n}.
\end{equation}
By using $\sigma_n$ and $*$ of (\ref{eq:unv})
we recover the right hand side of (\ref{eq:4c}).
\end{proof}
%
\begin{proof}[Proof of Proposition \ref{th:main}]
It should be shown that all $I_{j,i}$'s are invariant under $\tilde{R}$.
In view of $I_{n+1,n}=0$, $I_{j,i}=0\, (j < i)$, 
(\ref{eq:August4_1_1}) and Proposition \ref{lem:rstar}, 
it suffices to check the case $i+j \leq 2n+1$.
%
Suppose $1 \leq j \leq n-1$.
Then the claim follows from Lemmas \ref{th:1} and \ref{th:4}.
%
Suppose $j=n$.
Then the claim follows from Lemmas \ref{th:2}, \ref{th:4} and 
\ref{th:5}.
%
Suppose $j=n+1$ and $1 \leq i \leq n-1$.
Then the claim follows from $j=n$ 
due to (\ref{eq:August4_1_2}) 
and Proposition \ref{lem:rstar}.
%
We have already proved the invariance of
$I_{n-1,n-1}, I_{n,n-1}$ and $I_{n+1,n-1}$.
%
Because of the identity (\ref{eq:1c}),
$I_{n+2,n-1}$ in (\ref{eq:a11}) is also invariant.
Then the claim for $j=n+2$ follows from Lemma \ref{th:3}.
%
Similarly the invariance of
$I_{n-a,n-a}, I_{n-a+1,n-a},\cdots,I_{n+a,n-a}
\quad (1 \leq a \leq n-1)$
implies the invariance of $I_{n+a+1,n-a}$ by (\ref{eq:1c}), and
the claim for $j=n+a+1, 1 \leq i \leq n-a$ follows from 
Lemma \ref{th:3}. By induction on $a$ the proof is completed.
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{thebibliography}{99}


\bibitem[BFZ]{BFZ}
A.~Berenstein, S.~Fomin and A.~Zelevinsky,
\textit{Parametrizations of canonical bases and totally positive matrices},
Adv. in Math. {\bf 122} (1996) 49--149.


\bibitem[BK]{BK}
A.~Berenstein and D.~Kazhdan,
\textit{Geometric and unipotent crystals},
GAFA 2000 (Tel Aviv, 1999), 
Geom. Funct. Anal. 2000, Special Volume, Part I, 188--236.
%math.QA/9912105.


\bibitem[FOY]{FOY}
K.~Fukuda, M.~Okado, Y.~Yamada, 
\textit{Energy functions in box ball systems},
Int.\ J.\ Mod.\ Phys.\ A {\bf 15} (2000) 1379--1392.


\bibitem[HHIKTT]{HHIKTT}
G. Hatayama, K. Hikami, R. Inoue, A. Kuniba, T. Takagi and T. Tokihiro,
\textit{The $A^{(1)}_M$ Automata related to crystals of symmetric tensors},
J.\ Math.\ Phys.\ {\bf 42} (2001) 274-308.

\bibitem[HKOT1]{HKOT1}
G. Hatayama, A. Kuniba, M. Okado and T. Takagi,
\textit{Combinatorial R matrices for a family of  crystals: 
$C^{(1)}_n$ and $A^{(2)}_{2n-1}$ cases}, in
``Physical Combinatorics" (Eds 
M. Kashiwara and T. Miwa)
Prog. in Math.\ {\bf 191}, pp 105--139, Birkh\"auser, Boston, 2000.

\bibitem[HKOT2]{HKOT2}
G.~Hatayama, A.~Kuniba, M.~Okado and T.~Takagi,
\textit{Combinatorial $R$ Matrices for a Family of Crystals:
$B^{(1)}_n, D^{(1)}_n, A^{(2)}_{2n}$, and $D^{(2)}_{n+1}$ cases},
J. of Algebra\ {\bf 247} (2002) 577--615.


\bibitem[HKOTY]{HKOTY}
G. Hatayama,  A. Kuniba, M. Okado,  T. Takagi and Y. Yamada,
\textit{Scattering rules in soliton cellular automata associated with 
crystal bases}, in
``Recent Developments in Infinite-Dimensional Lie Algebras and
Conformal Field Theory" (Eds
S. Berman, P. Fendley, Y-Z. Huang, 
K.C. Misra, and B. Parshall)
Contemporary Math.\ {\bf 297}, (to appear), 
American Mathematical Society, 2002;
preprint, math.QA/0007175.


\bibitem[HKT1]{HKT1}
G. Hatayama, A. Kuniba and T.  Takagi,
\textit{Soliton cellular automata associated with crystal bases},
Nucl. Phys. B{\bf 577}[PM] (2000) 619--645.

\bibitem[HKT2]{HKT2}
G.~Hatayama, A.~Kuniba, and T.~Takagi,
\textit{Factorization of combinatorial $R$ matrices and associated 
cellular automata},
J. Stat. Phys. {\bf 102} (2001) 843--863.


\bibitem[Kac]{Kac}
V.~G.~Kac,
Infinite dimensional Lie algebras, 3rd edition,
Cambridge Univ. Press. Cambridge, 1990.

\bibitem[K1]{K1}
M.~Kashiwara,
{\em Crystallizing the $q$-analogue of universal enveloping
algebras}, Comm. Math. Phys. {\bf 133} (1990) 249--260.


\bibitem[K2]{K2}
M.~Kashiwara,
{\em Crystal bases of modified quantized enveloping algebra},
Duke Math. J. {\bf 73} (1994) 383--413.


\bibitem[Ki]{Ki}
A. N. Kirillov, 
\textit{Introduction to tropical combinatorics},
in ``Physics and Combinatorics 2000" (Eds. A. N. Kirillov and N. Liskova)
Proceedings of the Nagoya 2000 International Workshop, pp.~82--150,
World Scientific, 2001.

\bibitem[KKM]{KKM}
S-J.~Kang, M.~Kashiwara and K.~C.~Misra,
\textit{Crystal bases of Verma modules for quantum affine Lie algebras},
Compositio Math.\ {\bf 92} (1994) 299--325;
RIMS preprint {\bf 887} (1992).

\bibitem[KMOTU]{KMOTU}
A.~Kuniba, K.C.~Misra, M.~Okado, T.~Takagi and J.~Uchiyama,
{\em Crystals for Demazure modules of classical affine Lie algebras},
J. of Algebra {\bf 208} (1998) 185--215.

\bibitem[KNY]{KNY}
K.~Kajiwara, M.~Noumi and Y.~Yamada,
\textit{Discrete dynamical systems with $W(A^{(1)}_{m-1}\times 
A^{(1)}_{n-1})$ symmetry},
nlin.SI/0106029, to appear in Lett. Math. Phys. (2002).



\bibitem[L1]{L1}
G.~Lusztig,
Introduction to Quantum Groups,
Prog. in Math.\ {\bf 110},
Birkh\"auser, Boston, 1993.


\bibitem[L2]{L2}
G.~Lusztig,
\textit{Total positivity in reductive Groups},
in ``Lie theory and geometry", Prog. in Math. {\bf 123}, 
pp. 531--568, Birkh\"auser, Boston, 1994.

%
\bibitem[NaY]{NaY}
A.~Nakayashiki and Y.~Yamada,
\textit{Kostka polynomials and energy functions in 
solvable lattice models},
Selecta Mathematica, New Ser. {\bf 3} (1997) 547-599.


\bibitem[NoY]{NoY}
M.~Noumi and Y.~Yamada,
\textit{Tropical Robinson-Schensted-Knuth correspondence and
birational Weyl group actions},
math-ph/0203030.

\bibitem[TTMS]{TTMS} 
T.~Tokihiro, D.~Takahashi, J.~Matsukidaira and J.~Satsuma,
\textit{From soliton equations to integrable cellular automata through
a limiting procedure},
Phys. Rev. Lett. {\bf 76} (1996) 3247--3250.


\bibitem[Y]{Y}
Y. Yamada,
\textit{A birational representation of Weyl group, combinatorial $R$-matrix
and discrete Toda equation},
in ``Physics and Combinatorics 2000" (Eds. A. N. Kirillov and N. Liskova)
Proceedings of the Nagoya 2000 International Workshop, pp.~305--319,
World Scientific, 2001.
%
\end{thebibliography}
\end{document}

math0208239.extracted_bib
\begin{thebibliography}{99}


\bibitem[BFZ]{BFZ}
A.~Berenstein, S.~Fomin and A.~Zelevinsky,
\textit{Parametrizations of canonical bases and totally positive matrices},
Adv. in Math. {\bf 122} (1996) 49--149.


\bibitem[BK]{BK}
A.~Berenstein and D.~Kazhdan,
\textit{Geometric and unipotent crystals},
GAFA 2000 (Tel Aviv, 1999), 
Geom. Funct. Anal. 2000, Special Volume, Part I, 188--236.
%math.QA/9912105.


\bibitem[FOY]{FOY}
K.~Fukuda, M.~Okado, Y.~Yamada, 
\textit{Energy functions in box ball systems},
Int.\ J.\ Mod.\ Phys.\ A {\bf 15} (2000) 1379--1392.


\bibitem[HHIKTT]{HHIKTT}
G. Hatayama, K. Hikami, R. Inoue, A. Kuniba, T. Takagi and T. Tokihiro,
\textit{The $A^{(1)}_M$ Automata related to crystals of symmetric tensors},
J.\ Math.\ Phys.\ {\bf 42} (2001) 274-308.

\bibitem[HKOT1]{HKOT1}
G. Hatayama, A. Kuniba, M. Okado and T. Takagi,
\textit{Combinatorial R matrices for a family of  crystals: 
$C^{(1)}_n$ and $A^{(2)}_{2n-1}$ cases}, in
``Physical Combinatorics" (Eds 
M. Kashiwara and T. Miwa)
Prog. in Math.\ {\bf 191}, pp 105--139, Birkh\"auser, Boston, 2000.

\bibitem[HKOT2]{HKOT2}
G.~Hatayama, A.~Kuniba, M.~Okado and T.~Takagi,
\textit{Combinatorial $R$ Matrices for a Family of Crystals:
$B^{(1)}_n, D^{(1)}_n, A^{(2)}_{2n}$, and $D^{(2)}_{n+1}$ cases},
J. of Algebra\ {\bf 247} (2002) 577--615.


\bibitem[HKOTY]{HKOTY}
G. Hatayama,  A. Kuniba, M. Okado,  T. Takagi and Y. Yamada,
\textit{Scattering rules in soliton cellular automata associated with 
crystal bases}, in
``Recent Developments in Infinite-Dimensional Lie Algebras and
Conformal Field Theory" (Eds
S. Berman, P. Fendley, Y-Z. Huang, 
K.C. Misra, and B. Parshall)
Contemporary Math.\ {\bf 297}, (to appear), 
American Mathematical Society, 2002;
preprint, math.QA/0007175.


\bibitem[HKT1]{HKT1}
G. Hatayama, A. Kuniba and T.  Takagi,
\textit{Soliton cellular automata associated with crystal bases},
Nucl. Phys. B{\bf 577}[PM] (2000) 619--645.

\bibitem[HKT2]{HKT2}
G.~Hatayama, A.~Kuniba, and T.~Takagi,
\textit{Factorization of combinatorial $R$ matrices and associated 
cellular automata},
J. Stat. Phys. {\bf 102} (2001) 843--863.


\bibitem[Kac]{Kac}
V.~G.~Kac,
Infinite dimensional Lie algebras, 3rd edition,
Cambridge Univ. Press. Cambridge, 1990.

\bibitem[K1]{K1}
M.~Kashiwara,
{\em Crystallizing the $q$-analogue of universal enveloping
algebras}, Comm. Math. Phys. {\bf 133} (1990) 249--260.


\bibitem[K2]{K2}
M.~Kashiwara,
{\em Crystal bases of modified quantized enveloping algebra},
Duke Math. J. {\bf 73} (1994) 383--413.


\bibitem[Ki]{Ki}
A. N. Kirillov, 
\textit{Introduction to tropical combinatorics},
in ``Physics and Combinatorics 2000" (Eds. A. N. Kirillov and N. Liskova)
Proceedings of the Nagoya 2000 International Workshop, pp.~82--150,
World Scientific, 2001.

\bibitem[KKM]{KKM}
S-J.~Kang, M.~Kashiwara and K.~C.~Misra,
\textit{Crystal bases of Verma modules for quantum affine Lie algebras},
Compositio Math.\ {\bf 92} (1994) 299--325;
RIMS preprint {\bf 887} (1992).

\bibitem[KMOTU]{KMOTU}
A.~Kuniba, K.C.~Misra, M.~Okado, T.~Takagi and J.~Uchiyama,
{\em Crystals for Demazure modules of classical affine Lie algebras},
J. of Algebra {\bf 208} (1998) 185--215.

\bibitem[KNY]{KNY}
K.~Kajiwara, M.~Noumi and Y.~Yamada,
\textit{Discrete dynamical systems with $W(A^{(1)}_{m-1}\times 
A^{(1)}_{n-1})$ symmetry},
nlin.SI/0106029, to appear in Lett. Math. Phys. (2002).



\bibitem[L1]{L1}
G.~Lusztig,
Introduction to Quantum Groups,
Prog. in Math.\ {\bf 110},
Birkh\"auser, Boston, 1993.


\bibitem[L2]{L2}
G.~Lusztig,
\textit{Total positivity in reductive Groups},
in ``Lie theory and geometry", Prog. in Math. {\bf 123}, 
pp. 531--568, Birkh\"auser, Boston, 1994.

%
\bibitem[NaY]{NaY}
A.~Nakayashiki and Y.~Yamada,
\textit{Kostka polynomials and energy functions in 
solvable lattice models},
Selecta Mathematica, New Ser. {\bf 3} (1997) 547-599.


\bibitem[NoY]{NoY}
M.~Noumi and Y.~Yamada,
\textit{Tropical Robinson-Schensted-Knuth correspondence and
birational Weyl group actions},
math-ph/0203030.

\bibitem[TTMS]{TTMS} 
T.~Tokihiro, D.~Takahashi, J.~Matsukidaira and J.~Satsuma,
\textit{From soliton equations to integrable cellular automata through
a limiting procedure},
Phys. Rev. Lett. {\bf 76} (1996) 3247--3250.


\bibitem[Y]{Y}
Y. Yamada,
\textit{A birational representation of Weyl group, combinatorial $R$-matrix
and discrete Toda equation},
in ``Physics and Combinatorics 2000" (Eds. A. N. Kirillov and N. Liskova)
Proceedings of the Nagoya 2000 International Workshop, pp.~305--319,
World Scientific, 2001.
%
\end{thebibliography}