\section{ The Pfaff Lattice and a Lie algebra splitting}


Remember the decomposition (0.1) of the Lie algebra into
subalgebras:
\begin{equation}
{\cal D}=gl_{\infty}={\cal D}_-\oplus{\cal D}_0\oplus{\cal D}_+={\cal D}_-\oplus{\cal D}_0^-\oplus{\cal D}_0^+\oplus{\cal D}_+
.\end{equation}
Remember the element $J \in {\cal D}_0^+$, with $J^2=-I$ and the
associated map:
\begin{equation}
{\cal J}:{\cal D}\longrightarrow {\cal D}:  a\longmapsto{\cal J}(a):=Ja^{\top}J.
\end{equation}

\begin{theorem}
The vector fields\footnote{$\nabla$ denotes the gradient in the Lie
algebra ${\cal D} $ with respect to the natural pairing ${\cal D}$ with
${\cal D}^*={\cal D} $, i.e., $\nabla
{\cal H}_i=L^i$.}
\begin{equation}
\frac{\partial L}{\partial t_i}=\left[-\pi_{{\Bbb k}}\nabla {\cal H}_i, L \right]=
\left[\pi_{{\Bbb n}}\nabla {\cal H}_i, L \right]
~~\mbox{with}~~{\cal H}_i=\frac{\mbox{tr} L^{i+1}}{i+1},
\end{equation}
or written out as\footnote{${\cal J} (L^i)_+$ means ${\cal J}
 \left((L^i)_+\right);$ the same for ${\cal J} (L^i)_0$.}
\begin{eqnarray}
\frac{\partial L}{\partial
t_i}&=&\left[-\left((L^i)_--{\cal J} (L^i)_+\right)-\frac{1}{2}
\left((L^i)_0-{\cal J} (L^i)_0 \right),L\right]\nonumber\\
 &=&\left[\left((L^i)_++{\cal J} (L^i)_+\right)+\frac{1}{2}
\left((L^i)_0+ {\cal J} (L^i)_0 \right),L\right],
\end{eqnarray}
all commute.
\end{theorem}

The proof of this theorem hinges on a number of Lemmas and the
Adler-Kostant-Symes theorem \cite{AvM0}:

\begin{lemma} ${\cal J}$ is a Lie algebra isomorphism and an order 2 involution on ${\cal D}$:
\begin{equation}
{\cal J}^2=I,
\end{equation}
with ${\cal N}_+$ and ${\cal N}_-$ being the $\pm 1$ eigenspaces of ${\cal J}$:
\begin{equation}
{\cal N}_{\pm}=\{\mbox{$a$ such that ${\cal J} a=\pm a\}=\{b\pm{\cal J} b$ with
$b\in{\cal D}\}$}.
\end{equation}
Then ${\cal D}$ is decomposed into the Lie algebra ${\cal N}_+$ and the
(symmetric) vector space ${\cal N}_-$
\begin{equation}
{\cal D}={\cal N}_++{\cal N}_-,\mbox{ with }[{\cal N}_{\pm},{\cal N}_{\pm}]
\subset {\cal N}_+\mbox{ and }[{\cal N}_+,{\cal N}_-]\subset {\cal N}_-,
\end{equation}
with
\begin{equation}
{\cal N}_{\pm}={\cal D}^{\pm}_0\oplus({\cal N}_{\pm}\backslash{\cal D}_0),
\end{equation}
and
$$
{\cal N}_{\pm}\backslash{\cal D}_0=\{a\pm {\cal J} a \mbox{ such that }
 a\in{\cal D}_-~\mbox{or}~{\cal D}_+\}
$$
\begin{equation}
{\cal D}_0^{\pm}={\cal D}_0\cap {\cal N}_{\pm}=\{a\pm {\cal J} a \mbox{ such that }a\in{\cal D}_0\}.
\end{equation}
\end{lemma}

\underline{\sl Proof}:  The map ${\cal J}$ is a Lie algebra homomorphism,
 such that ${\cal J}^2=I$; indeed, upon using $J^2=-I$,
$${\cal J}([a,b])=J[a,b]^{\top}J=-J[a^{\top},b^{\top}]J
=[Ja^{\top}J,Jb^{\top}J]=[{\cal J} a,{\cal J} b].
$$
In particular, ${\cal J}(a)=\pm a$ and ${\cal J}(b)=\pm b$ imply
$$
{\cal J}[a,b]=[\pm a,\pm b]=[a,b],
$$
leading to the inclusions (1.7); so ${\cal N}_+$ is a Lie subalgebra of
${\cal D}$, namely an infinite-dimensional version of the symplectic
algebra $\mbox{sp}(\infty)$. The second description (1.6) of
${\cal N}_{\pm}$ follows from
$$
{\cal J}(b\pm{\cal J} b)={\cal J} b\pm{\cal J}^2 b=\pm(b\pm{\cal J} b),
\quad\mbox{for }b\in{\cal D};
$$
and the vector space decomposition ${\cal D}={\cal N}_++{\cal N}_-$:
$$a=\frac{1}{2}(a+{\cal J} a)+\frac{1}{2}(a-{\cal J} a).
$$
Since $J{\cal D}_{\varepsilon},{\cal D}_{\varepsilon}J \subset {\cal D}_{\varepsilon},
{\cal D}_{\varepsilon}^{\top}={\cal D}_{-\varepsilon},$ for $\varepsilon=+,-,0 $, we have
$$
{\cal J}:{\cal D}_{\varepsilon}\longrightarrow{\cal D}_{-\varepsilon},
$$
and so, if $a\in{\cal D}_+$, we have $b:={\cal J} a\in{\cal D}_-$ and
$$
{\cal N}_{\pm}\supset a\pm{\cal J} a={\cal J}^2 a\pm{\cal J} a=\pm(b\pm{\cal J} b)
\quad\mbox{with}\quad b={\cal J} a\in{\cal D}_-.
$$
Therefore
$$
\{b\pm{\cal J} b\mbox{ such that }b\in{\cal D}_-\}=\{a\pm{\cal J} a\mbox{ such that }
b\in{\cal D}_+\}={\cal N}_{\pm}\backslash{\cal D}_0
$$
establishing the first relation (1.9). As to the second, one checks
that
\begin{equation}
\MAT{2}a&b\\c&d\end{array}\right)\pm\MAT{2}0&1\\-1&0
\end{array}\right)\MAT{2}a&b\\c&d\end{array}\right)^{\top}
\MAT{2}0&1\\-1&0\end{array}\right) =\left\{
\begin{array}{c}
\MAT{2}a-d&2b\\
2c&-(a-d)\end{array}\right)\\
 \\
(a+d)I
\end{array}
\right.\,.
\end{equation}
\qed

\begin{lemma} The decomposition
$${\cal D}={\Bbb k}\oplus{\Bbb n},
$$
with
$$
{\Bbb k}={\cal D}_-\oplus{\cal D}_0^-,\quad {\Bbb n}={\cal N}_+=
\{ a ~\mbox{such that}~{\cal J} a=a \}=\mbox{sp}(\infty),$$
is a vector space splitting of ${\cal D}$ into Lie subalgebras. Any
element $a\in{\cal D}$ decomposes uniquely into its projections onto
${\Bbb k}$ and ${\Bbb n}$:
\begin{eqnarray}
a&=&\pi_{{\Bbb k}}a+\pi_{{\Bbb n}}a\nonumber\\ &=&\left\{\left(a_--{\cal J}
a_+\right)+\frac{1}{2}\left(a_0-{\cal J} a_0
\right)\right\}+\left\{\left(a_++{\cal J} a_+\right)+\frac{1}{2}\left(a_0+{\cal J}
a_0
\right)\right\}\nonumber.
\end{eqnarray}
\end{lemma}

\underline{\sl Proof}:   Since $[{\cal D}_-,{\cal D}_-]\subset{\cal D}_-$, since
 ${\cal D}_-{\cal D}_0, ~{\cal D}_0{\cal D}_-\subset{\cal D}_-$ and $[{\cal D}^-_0,{\cal D}^-_0]=0$, we have that
 $[{\Bbb k},{\Bbb k}]={\cal D}_-\subset{\Bbb k}$, so that
${\Bbb k} ={\cal D}_-+{\cal D}^-_0$ and
${\Bbb n}={\cal N}_+$ are both Lie subalgebras.

But also ${\Bbb k} \cap {\Bbb n} =0$; indeed, a typical element of ${\cal N}_+$
has the form,
$$
(b_-+{\cal J}(b_-))+(d_0+{\cal J}(d_0))\in({\cal N}_+\backslash{\cal D}_0)+({\cal N}_+\cap{\cal D}_0)=
({\cal N}_+\backslash{\cal D}_0)+{\cal D}_0^+,
$$
with $b_-\in{\cal D}_-$, $d_0\in{\cal D}_0$. If this element
$\in{\Bbb k}={\cal D}_-+{\cal D}^-_0$, then ${\cal J}(b_-)=0$ and thus $b_-=0$, but
then, since $d_0+{\cal J} d_0 \in {\cal D}_0^+ \cap
{\cal D}_0^-=0 $. Finally, any element $a \in {\cal D}$ can be decomposed
into\footnote{$a_{\pm}=\pi_{{\cal D}_{\pm}}a,~a_{0}=\pi_{{\cal D}_{0}}a.$}
\begin{eqnarray}
a&=&a_-+a_0+a_+\nonumber\\ &=&\left\{\left(a_--{\cal J}
a_+\right)+\frac{1}{2}\left(a_0-{\cal J} a_0
\right)\right\}\nonumber\\
& &\quad +\left\{\left(a_++{\cal J} a_+ \right)
+\frac{1}{2}\left(a_0+{\cal J}
a_0
\right)\right\}\nonumber\\
&\in&({\cal D}_-+{\cal D}_0^-)\oplus\left(({\cal N}_+\backslash{\cal D}_0)+{\cal D}^+_0\right)={\Bbb k}\oplus{\Bbb n}.\nonumber
\end{eqnarray}
\qed

We now state the R-matrix version of the AKS-theorem (see
\cite{AvM0} and \cite{RS}). The substance of the R-matrix extension
is that initial conditions $\xi(0)$ can be taken in all of ${\bf
g}$, instead of merely in ${\bf k}$.



\begin{proposition} {\bf (Adler-Kostant-Symes, Russian flavor)}  Let ${\bf g} = {\bf k} + {\bf n}$ be a
(vector space) direct sum of a Lie algebra  ${\bf g}$ in terms of
Lie subalgebras  ${\bf k}$ and  ${\bf n}$, with ${\bf g}$ paired
with itself via a non-degenerate {\rm ad}-invariant inner
product\footnote{$\langle{\rm Ad}_gX;Y\rangle =\langle X,{\rm Ad}_{g-1}Y\rangle$,
$g\in G$, and thus $\langle [z,x],y\rangle =\langle x,-[z,y]\rangle.$} $\langle\,
,\,\rangle$; this in turn induces a decomposition ${\bf g} = {\bf
k}^{\bot} + {\bf n}^{\bot}$ and isomorphisms  ${\bf g}\simeq {\bf
g}^*$, ${\bf k}^{\bot}\simeq {\bf n}^*$, ${\bf n}^{\bot}\simeq {\bf
k}^*$.  It also leads to a second Lie algebra  ${\bf g}_R
\simeq {\bf g}_R^*$ derived from ${\bf g}$, namely:
$$ {\bf g}_R : [x,y]_R = \frac{1}{2}[Rx,y] +
\frac{1}{2}[x,Ry] = [\pi_{{\bf k}}x,\pi_{{\bf k}}y] -
[\pi_{{\bf n}}x,\pi_{{\bf n}}y],  $$ with  $R =
\pi_{{\bf k}}-\pi_{{\bf n}}$.
 {\rm Ad}$^* \simeq$ {\rm Ad}-invariant functions $\varphi $
 on ${\bf g}^* \simeq {\bf g}$ Poisson commute for
  the
Kostant-Kirillov  Poisson structure
$$
\{f,h\}_R (\xi) :=\langle \xi,[\nabla h(\xi),
\nabla f(\xi)]_R\rangle
~~\mbox{ on} ~~{\bf g}^*_R\simeq{\bf g}_R.
$$
 The associated
Hamiltonian flows are expressed in terms of the Lax
pair\footnote{$\nabla\varphi$ is defined as the element in
${\bf g}^*$
such that $d\varphi(\xi)=\langle\nabla\varphi,d\xi\rangle$,
$\xi\in{\bf g}$.}
$$
\dot\xi =[-\pi_{{\bf
k}}\nabla\varphi(\xi),\xi]=[\pi_{{\bf
n}}\nabla\varphi(\xi),\xi]~~\mbox{\,for\,}~~\xi\in{\bf g}_R.
$$
The systems has a solution expressible in two different
ways\footnote{naively written Ad$_{K(t)}\xi_0=K(t)\xi_0K(t)^{-1}$,
Ad$_{N^{-1}}\xi_0=N^{-1}(t)\xi_0N(t)$.}:
\begin{equation}
\xi(t)={\rm Ad}_{K(t)}\xi_0={\rm Ad}_{N^{-1}(t)}\xi_0
\end{equation}
with\footnote{with regard to the group factorization
$A=\pi_{{\cal G}_{{\Bbb k}}}A~
\pi_{{\cal G}_{{\Bbb n}}}A$.}
$$
K(t) = \pi_{{\cal G}_{{\Bbb k}}} e^{t \nabla\varphi(\xi_0)},\quad
 \mbox{and}\quad
N(t)
=
 \pi_{{\cal G}_{{\Bbb n}}} e^{t \nabla\varphi(\xi_0)}.
$$
\end{proposition}

\noindent \underline{{\em Proof of Theorem 1.1}}: Identifying
${\Bbb k}$ and ${\Bbb n}$ of Lemma 1.3, with those of proposition 1.4 and
unraveling the projection $\pi_{{\Bbb k}}$, according to Lemma 1.3,
imply that the vector fields (1.3) and so (1.4) all commute.\qed



\section{The vector fields $\partial m / \partial t_k=\Lambda^k m
+m \Lambda^{\top k} $}

The main claim of this section can be summarized in the following
statement:

\begin{theorem}
Consider the skew-symmetric solution
 $$m_{\infty}(t)=e^{\sum t_k \Lambda^k}m_{\infty}(0)e^{\sum t_k \Lambda^{\top k}}$$
 to the commuting equations
\begin{equation}
\frac{\partial m_{\infty} }{ \partial t_k}=\Lambda^k m_{\infty}
+m_{\infty} \Lambda^{\top k}, \end{equation}
 with skew-symmetric initial condition $m(0)$ and
 its ``skew-Borel decomposition"
\footnote{Remember ${\cal G}_{{\Bbb k}}$ denotes the group of invertible elements
 in ${\Bbb k}$.}
\begin{equation}
m_{\infty}=Q^{-1} J Q^{-1 \top}, ~~\mbox{with}~~Q \in {\cal G}_{{\Bbb k}}.
\end{equation}
Then the matrix $Q$ evolves according to the equations
\begin{equation}
\frac{\partial Q}{\partial t_i}Q^{-1} =-\pi_{{\bf k}}
\left( Q \Lambda^i Q^{-1} \right)
\end{equation}
and the matrix $L:=Q \Lambda Q^{-1}$ provides a solution to the Lax
pair
\begin{equation}
\frac{\partial L}{\partial t_i}=\left[-\pi_{{\bf k}} L^i,L    \right]
=\left[\pi_{{\bf n}} L^i,L    \right].
\end{equation}
Conversely, if $Q \in {\cal G}_{{\Bbb k}}$ satisfies (2.3),
 then $m_{\infty}$, defined by (2.2),
satisfies (2.1).
\end{theorem}

The proof of this theorem hinges on the following proposition:

\begin{proposition} For the matrices
$$
L:=Q\Lambda Q^{-1}\quad\mbox{and}\quad m:=Q^{-1}J Q^{-1\top}, \mbox{
with }Q\in {\cal G}_{{\Bbb k}},
$$
the following three statements are equivalent\medbreak\indent (i) $\displaystyle{\frac{\partial Q}{\partial
t_i}Q^{-1}=-\pi_{{\Bbb k}}L^i}$
\medbreak\indent (ii) $L^i+\displaystyle{\frac{\partial Q}{\partial
t_i}}Q^{-1}\in{\Bbb n}$\medbreak\indent (iii) $\displaystyle{\frac{\partial m}{\partial
t_i}}=\Lambda^im+m\Lambda^{\top^i}.$
\medbreak\noindent Whenever the vector fields on $Q$ or $m$
satisfy (i), (ii) or (iii), then the matrix $L=Q\Lambda Q^{-1}$ is a
solution of the Lax pair
$$
\frac{\partial L}{\partial t_i}=[-\pi_{{\Bbb k}}L^i,L]=
 \left[\pi_{{\bf n}} L^i,L    \right].
$$
\end{proposition}

\underline{\sl Proof}:  Written out, proposition 2.2 amounts to showing the
equivalence of the three formulas:
\medbreak
\noindent(I) $\displaystyle{\frac{\partial Q}{\partial t_i}Q^{-1}+\left((L^i)_--J(L^i_+)^{\top}J\right)+\frac{1}{2}
\left((L^i)_0-J((L^i)_0)^{\top}J\right)=0}$

\medbreak

\noindent(II) $\displaystyle{\left(L^i+\frac{\partial Q}{\partial t_i}Q^{-1}\right)-J
\left(L^i+\frac{\partial Q}{\partial t_i}Q^{-1}\right)^{\top}J=0}$

\medbreak

\noindent(III) $\Lambda^im+m\Lambda^{\top i}-\displaystyle{\frac{\partial m}{\partial
t_i}}=0$.

\medbreak

\noindent The point is to show that
$$
(\mbox{I})_+=0,~~~~(\mbox{I})_-=(\mbox{II})_-
=-J~(\mbox{II}_+)^{\top} ~J,~~~~
(\mbox{I})_0=\frac{1}{2}\left(\mbox{II}\right)_0,
$$
\begin{equation}
Q^{-1}(\mbox{II})JQ^{-1^{\top}}
=(\mbox{III}).
\end{equation}
\noindent The fact that $Q\in{\cal G}_{{\Bbb k}}\subset
{\Bbb k}={\cal D}_-\oplus{\cal D}_0^-={\cal D}_-\oplus({\cal D}_0 \cap {\cal N}^-)
 $ amounts to
$$
Q_+=0\quad\mbox{and}\quad(Q+JQ^{\top}J)_0=0.
$$
The latter statement implies that $Q$ is lower-triangular and,
along the diagonal, is composed of $2\times 2$ blocks proportional
to the identity. Hence
$$
Q_0,\dot Q_0\in{\cal D}_0^-,\quad\mbox{and so }~~(\dot Q
Q^{-1})_0\in{\cal D}_0^-
$$
and thus $J$ commutes with $(\dot Q Q^{-1})_0$, yielding:
\begin{equation}
J(\dot Q Q^{-1})_0J=J^2(\dot Q Q^{-1})_0=-(\dot Q Q^{-1})_0.
\end{equation}
Also notice that for any matrix $A$
$$
(JAJ)_{\pm}=JA_{\pm}J\quad\mbox{and}\quad  (JAJ)_0=JA_0J.
$$
At first, one observes that
\begin{equation}
(\mbox{I})_+=0.
\end{equation}
Also
\begin{eqnarray}
(\mbox{II})_-&=&\left(L^i+\dot Q Q^{-1}-J(L^i+\dot Q
Q^{-1})^{\top}J\right)_-\nonumber\\ &=&\left(L^i+\dot Q
Q^{-1}-J((L^i)_+)^{\top}J\right)_-,\quad\mbox{using }(\dot Q
Q^{-1})_+=0\nonumber\\ &=&(\mbox{I})_-
\end{eqnarray}

\begin{eqnarray}
(\mbox{II})_0&=&\left(L^i+\dot Q Q^{-1}-J\left(L^i+\dot Q
Q^{-1}\right)^{\top}J\right)_0\nonumber\\
&=&\left(L^i\right)_0-J\left(L^{i\top}\right)_0 J+\left(\dot Q
Q^{-1}\right)_0-J\left((\dot Q Q^{-1})^{\top}\right)_0 J\nonumber\\
&=&(L^i)_0-J(L^{i
\top})_0 J+2(\dot Q Q^{-1})_0,\quad\mbox{using (2.6)}\nonumber\\
&=&2(\mbox{I})_0
\end{eqnarray}
and
\begin{eqnarray}
(\mbox{II})_+&=&\left(L^i+\dot Q Q^{-1}-J(L^i+\dot Q
Q^{-1})^{\top}J\right)_+\nonumber\\
&=&(L^i)_+-J((L^i)_-)^{\top}J-J((\dot Q
Q^{-1})_-)^{\top}J\nonumber\\
&=&\left(((L^i)_+)^{\top}-J(L^i)_-J-J(\dot Q
Q^{-1})_-J\right)^{\top}\nonumber\\
&=&J\left(J((L^i)_+)^{\top}J-J^2(L^i)_-J^2-J^2(\dot Q
Q^{-1})_-J^2\right)^{\top}J\nonumber\\ &=&-J\left((L^i+\dot Q
Q^{-1}-J(L^i_+)^{\top}J)_-\right)^{\top}J\nonumber\\
&=&-J((\mbox{I})_-)^{\top}J.
\end{eqnarray}

\noindent Using the definitions $L=Q\Lambda Q^{-1}$, $m=Q^{-1}JQ^{-1\top}$
and $J^2=-I$, one finds
\begin{eqnarray}
(\mbox{III})&=&\Lambda^im+m\Lambda^{\top^{i}}-\frac{\partial m}{\partial
t_i}\nonumber\\ &=&\Lambda^im+m\Lambda^{\top^{i}}+(Q^{-1}\dot
Q)m+m(Q^{-1}\dot Q)^{\top},\mbox{ setting
}m=Q^{-1}JQ^{-1^{\top}}\nonumber\\ &=&Q^{-1}(Q\Lambda^i
Q^{-1})JQ^{-1^{\top}}-Q^{-1}(JQ^{-1^{\top}}
\Lambda^{{\top}^i}Q^{\top}J)JQ^{-1^{\top}}\nonumber\\ & &\quad +\,Q^{-1}(\dot
QQ^{-1})JQ^{-1^{\top}}-Q^{-1}\left(JQ^{-1^{\top}}\dot
Q^{\top}J\right)JQ^{-1^{\top}}\nonumber\\ &=&Q^{-1}\left(L^i+\dot
QQ^{-1}-J(L^i+\dot QQ^{-1})^{\top}J\right)JQ^{-1^{\top}}\nonumber\\
&=&Q^{-1}(\mbox{II})JQ^{-1^{\top}}.
\end{eqnarray}
The five facts (2.7) up to (2.11) imply (2.5) and so the desired
equivalences.\qed











\section{Skew-orthogonal polynomials}

As before, consider the skew-symmetric solution
$$m_{\infty}(t)=e^{\sum t_k \Lambda^k}m_{\infty}(0)
e^{\sum t_k \Lambda^{\top k}}$$ to the commuting equations (2.1). We
now perform the ``{\em Borel decomposition}" of $m_{\infty}$:
\begin{equation}
m_{\infty}=Q^{-1} J Q^{\top -1},
\end{equation}
where $Q\in {\cal G}_{{\Bbb k}}$ (lower-triangular invertible matrices in
${\Bbb k}$). To carry out this factorization, we found the following
useful algorithm, based on the observation that the skew-Borel
decomposition is tantamount to finding skew-orthogonal polynomials,
in the same way that Borel decomposing a H\"ankel matrix is
tantamount to finding orthogonal polynomials.


Define the skew-inner product $\langle f,g\rangle$ between formal Taylor
series,
 induced by
$$\langle y^i,z^j\rangle :=\mu_{ij}(t)=(m_{\infty}(t))_{ij}.$$
 Monic polynomials
$p_k$ are said to be {\em skew-orthogonal}, when \begin{equation}(\langle
 p_i,
p_j\rangle)_{0\leq i,j<\infty}=J\tilde h, ~~\mbox{with}~~
\tilde h\in
{\cal D}_0^-.\end{equation}
They, in turn, lead to {\em
 skew-orthonormal} polynomials $q_k(z)$,
 \begin{equation}
q_{2n}(t,z)=\frac{1}{\sqrt{\tilde h_{2n}}}p_{2n}(t,z)~~\mbox{and}~~
q_{2n+1}(t,z)=\frac{1}{\sqrt{\tilde h_{2n}}}p_{2n+1}(t,z),\end{equation}
 thus satisfying
\begin{equation}(\langle q_i,q_j\rangle)_{0\leq i,j<\infty}=J.\end{equation}

Indeed, finding a set of skew-orthonormal polynomials is tantamount
to the ``{\em skew-Borel decomposition}" (3.1).
 Using the lower-triangular matrix
$Q$ of coefficients of the polynomials $q(z)=Q
\chi(z)$, we have

\begin{equation} J=\left(\langle q_i,q_j   \rangle  \right)_{i,j\geq 0}
=\left(\langle (Q\chi)_i,(Q\chi)_j   \rangle  \right)_{i,j\geq 0}
=Q\left(\langle y^i,z^j   \rangle  \right)_{i,j\geq 0}Q^{\top}=Qm_{\infty}
 Q^{\top}
,\end{equation} from which (3.1) follows, and vice-versa.


Given a skew-symmetric matrix
$$
A=(a_{ij})_{0\leq i,j\leq n-1},\quad\mbox{even $n$},
$$
we define the Pfaffian by means of the formula\footnote{In the formula below $
 (i_0,i_1, ...
,i_{n-2},i_{n-1})=\sigma (0,1,...,n-1)$, where $\sigma$ is a
permutation and $\varepsilon(\sigma) $ its parity.}

\medbreak


\noindent $\displaystyle pf(A)dx_0\wedge ...\wedge dx_{n-1}$
 \hfill
 \begin{eqnarray}
&=&\frac{1}{(n/2)!}\left(
\sum_{0\leq i<j\leq n-1}
a_{ij}dx_i\wedge dx_j\right)^{n/2}\nonumber\\
&=&\frac{1}{2^{n/2}(n/2)!}\left(\sum_{\sigma}\varepsilon(\sigma)a_{i_0,i_1}a_{i_2,i_3}...
a_{i_{n-2},i_{n-1}}
\right)dx_0\wedge  ...\wedge dx_{n-1},\nonumber\\
\end{eqnarray}
so that $pf(A)^2=\det A$, but with a sign specification for
$pf(A)$. Henceforth, we shall be using the notation
$$
pf(i_0,i_1,...,i_{\ell-1}),~0\leq i_0< i_1<...<i_{\ell-1},~~\ell
~\mbox{even},
$$
to denote the pfaffian of the skew-symmetric matrix, formed by the
$i_0, i_1,...,i_{\ell-1}$th rows and columns of $m_{\infty}$.


\begin{theorem}
Consider a semi-infinite skew-symmetric matrix $m_{\infty}$,
satisfying (2.1); then the monic polynomials $p_k$
\begin{eqnarray}
p_{2n}(t,z)&=&\frac{1}{\tilde\tau_{2n}(t)}
 ~pf
\pmatrix{0&
\mu_{01}& ~~...~~&\mu_{0,2n}&1\cr
-\mu_{01}& 0 & ~~...~~& \mu_{1,2n}&z\cr
\vdots& \vdots &   & \vdots&\vdots\cr
-\mu_{0,2n}&-\mu_{1,2n} & ~~...~~&0 &z^{2n} \cr
-1& -z & ~~...~~&-z^{2n}&0\cr}\nonumber\\
&&\nonumber\\
 &&\nonumber\\ &&\nonumber\\
 p_{2n+1}(t,z)&=&\frac{1}{\tilde\tau_{2n}} ~pf
\pmatrix{0&
\mu_{01}& ~~...~~&1&\mu_{0,2n+1}\cr
-\mu_{01}& 0 & ~~...~~&z& \mu_{1,2n+1}\cr
\vdots& \vdots &   & \vdots&\vdots\cr
-1& -z & ~~...~~&0&-z^{2n+1}\cr
-\mu_{0,2n+1}&-\mu_{1,2n+1} & ~~...~~&z^{2n+1} &0 \cr}\nonumber\\
 &=&\frac{1}{\tilde\tau_{2n}(t)}
 \left( z+\frac{\partial}{\partial t_1}\right)\tilde\tau_{2n}(t)
 p_{2n}(t,z)
. \nonumber\\
\end{eqnarray}
form a skew-orthogonal sequence, $\langle  p_i,  p_j\rangle_{0\leq
i,j<\infty}=J\tilde h$, where
$$  \tilde h=(\tilde h_0,\tilde h_0,\tilde h_2,\tilde
h_2,...)\in
{\cal D}_0^-,~~\mbox{with}~~\tilde h_{2n}=
\frac{\tilde \tau_{2n+2}}{\tilde \tau_{2n}},
~~\tilde \tau_{2n}= pf (0,...,2n-1).$$
 whereas the $q_k$, defined in (3.3) from the
$p$'s, form an orthonormal sequence.
The matrix $Q$
defined by $q(z)=Q\chi(z)$ is the unique solution
(modulo signs)
 to the skew-Borel decomposition of $m_{\infty}$:
\begin{equation}
m_{\infty}(t)=Q^{-1}JQ^{\top -1}, ~~\mbox{with}~~Q\in {\Bbb k}. \end{equation} The
matrix $L=Q\Lambda Q^{-1}$, also defined by $$z q(t,z)=L q(t,z),$$
 and the diagonal matrix $\tilde h$ satisfy the equations
\begin{equation}
\frac{\partial L}{\partial t_i}=\left[-\pi_{{\Bbb k}}L^i, L \right].
~~\mbox{and}~~
\tilde h^{-1}\frac{\partial\tilde h}{\partial t_i}
=2\pi_{{\Bbb k}} (L^i)_0  .
\end{equation}
\end{theorem}

\underline{\sl Remark}:  It is of interest to mention that, except for the
second expression for $p_{2n+1}$ in (3.7), the first part of
theorem 3.1 is actually {\em $t$-independent}. Namely, (3.7) gives
a geometric map, but also an effective algorithm, to perform the
skew-Borel decomposition of a skew-symmetric matrix $m_{\infty}$, by
applying formulae (3.7).

\bigbreak

\begin{corollary}
The polynomials $(p_0(t,z),p_1(t,z),...)$ have the following
explicit form in terms of Pfaffians of the moment matrix $m$ and
the Pfaffian $\tilde\tau$-functions:
\begin{eqnarray*}
 p_{2n}(z)&=&
\sum_{0\leq k\leq 2n}(-z)^k \frac{pf(0,...,\hat k,...,2n)}{pf(0,...,2n-1)}
\nonumber\\
&&\\
  p_{2n+1}(z)&=&
  z^{2n+1} +
 \sum_{0\leq k\leq 2n-1}(-z)^k  \frac{pf(0,...,\hat
k,...,\widehat{2n},2n+1) }{pf(0,...,2n-1)}.\\
\end{eqnarray*}
\begin{equation}
\end{equation}
\end{corollary}


\underline{\sl Remark}:  Taking into account the identity $pf($odd set$)=0$ and
the convention $$pf(0,...,\widehat
k,...,\widehat{2n},2n+1)=-pf(0,...,2n-1) ~~\mbox{ for}~~ k=2n+1, $$
the polynomials can be written as follows:
\begin{eqnarray}
 p_{2n}(t,z)&=& \frac{1}{\tilde\tau_{2n}(t)}\sum_{0\leq k\leq
2n}(-z)^k pf(0,...,\hat k,...,2n,\widehat{2n+1})\nonumber\\
&&\nonumber\\
  p_{2n+1}(t,z)&=&
 \frac{1}{\tilde\tau_{2n}(t)}\sum_{0\leq k\leq 2n+1}(-z)^k pf(0,...,\hat
k,...,\widehat{2n},2n+1). \nonumber\\
\end{eqnarray}

 The proof of this theorem hinges on a number of lemmas,
involving properties of Pfaffians:



\begin{lemma} Consider an arbitrary skew-symmetric
 matrix $a=(a_{ij})_{i,j\geq 0}$. For
 odd $\ell$ and fixed $i\geq \ell$, the following holds:
$$
\sum_{0\leq k\leq\ell-1}(-1)^k a_{ ki}
pf(0,...,\hat k,...,\ell-1)=pf(0,...,\ell-1,i).
$$
\end{lemma}

\underline{\sl Proof}:  On the one hand
\begin{equation}
\det(a_{ij})_{0\leq i,j\leq\ell}=pf^2(0,1,...,\ell)
\end{equation}
and, on the other hand, using
 $$
\det \pmatrix{0& ... & a_{0r}&\Bigr| &a_{0k}\cr
          \vdots  &      &\vdots& \Bigr|& \vdots\cr
          -a_{0r} & ... & 0&\Bigr| &a_{rk}\cr
          --&--&--&--&--\cr
           a_{\ell 0} & ... & a_{\ell r}&\Bigr| &*
           }=pf(0,...,r,k) pf(0,...,r,{\ell}),
 $$
 and expanding according to the last row
and column

\bigbreak
\noindent$\displaystyle{\det(a_{ij})_{0\leq
i,j\leq\ell}}$
\begin{eqnarray}
&=&-\sum_{0\leq i,j\leq\ell-1}(-1)^{i+j}a_{\ell
i}a_{j\ell}pf(0,...,\hat i,...,\ell-1)pf(0,...,\hat
j,...,\ell-1)\nonumber\\
 &=&\left(\sum_{0\leq i\leq\ell-1}(-1)^i
a_{\ell i}pf(0,...,\hat i,...,\ell-1)\right)\nonumber\\ &&
\hspace{5cm} \left(\sum_{0\leq j\leq\ell-1}(-1)^j a_{\ell j}pf(0,...,\hat
j,...,\ell-1)\right)\nonumber\\
 &=&\left(\sum_{0\leq
i\leq\ell-1}(-1)^i a_{i\ell }pf(0,...,\hat
i,...,\ell-1)\right)^2\quad.
\end{eqnarray}
Comparing the two expressions (3.12) and (3.13) and taking into
account the sign, lead to the claim of lemma 3.3.\qed








\underline{\sl Remark}:  Lemma 3.3 will be applied as follows (for odd $\ell$):
$$
\sum_{0\leq k\leq\ell-1}(-1)^k a_{ i_k,i_{\ell}}
pf(i_0,...,\hat
i_k,...,i_{\ell-1})=pf(i_0,...,i_{\ell-1},i_{\ell}).
$$

\vspace{0.2cm}

\begin{lemma} The polynomials $p_k(z)$ defined by the determinantal expression (3.7) equal the
 expressions (3.10) of Corollary 3.2 and
satisfy
\begin{equation} \left\langle p_i(z),p_j(z)\right\rangle_{i,j\geq 0}=J \tilde h
\end{equation}
\end{lemma}

\underline{\sl Proof}:  At first notice that the leading coefficients of the
pfaffians in the polynomials (3.7) equal $pf(0,1,...,2n-1)=\tilde
\tau_{2n}(t)$. Then apply Lemma 3.3 (remark following the lemma)
 to the first case, setting
the $a$'s equal to $\mu$'s from $0$ to $2n$ and
$a_{k,2n+1}=(1-\delta_{k,2n+1})z^k$. Also apply Lemma 3.3 to the
second case, by setting the $a$'s equal to $\mu$, from $0$ to
$2n+1$, skipping $2n$, and $a_{k,2n}=(1-\delta_{k,2n})z^k$. Then, we
find:
\begin{eqnarray}
\tilde\tau_{2n}(t)  p_{2n}(t,z)&=& \sum_{0\leq k\leq
2n}(-z)^k pf(0,...,\hat k,...,2n-1,2n,\widehat{2n+1})\nonumber\\
 &=&
\sum_{0\leq k\leq 2n}(-z)^k pf(0,...,\hat
k,...,2n)\nonumber\\
 &&\nonumber\\
 \tilde\tau_{2n}(t) p_{2n+1}(t,z)&=&
 \sum_{0\leq k\leq 2n+1}(-z)^k pf(0,...,\hat
k,...,2n-1,\widehat{2n},2n+1) \nonumber\\
 &=&
 \sum_{0\leq k\leq 2n-1}(-z)^k pf(0,...,\hat
k,...,\widehat{2n},2n+1) \nonumber\\
 && ~~~~~+ z^{2n+1} pf(0,...,2n-1) .\nonumber\\
\end{eqnarray}
Using the expressions (3.15) and Lemma 3.3, one computes the
inner-product
\begin{eqnarray*}
\langle p_{2n}(z),z^i\rangle&=&\sum_{0\leq k\leq
2n}(-1)^k\mu_{ki}\frac{pf(0,...,\hat k,...,2n)}{pf(0,...,2n-1)}\\
&=&\frac{pf(0,...,2n,i)}{pf(0,...,2n-1)},
\end{eqnarray*}
which vanishes for $0\leq i\leq 2n$, and so
$$
\langle p_{2n}(z),z^i\rangle=0 ~~\mbox{for}~~0\leq i \leq 2n;
$$
therefore
$$
\langle p_{2n}(z),p_i(z)\rangle=0
 ~~\mbox{for}~~0\leq i \leq 2n.
$$
Moreover, using the expression for $p_{2n+1}$ and Lemma 3.3,
\begin{eqnarray*}
\langle p_{2n+1}(z),z^i\rangle&=&\mu_{2n+1,i} +\sum_{0\leq k\leq
2n-1}(-1)^k\mu_{ki}\frac{pf(0,...,\hat
k,...,\widehat{2n},...,2n+1)}{pf(0,...,2n-1)}\\ & &\quad\quad
\\ &=&\frac{pf(0,...,\widehat{2n},2n+1,i)}{pf(0,...,2n-1)}\quad
\\ &=&\left\{\begin{array}{l}
0\mbox{\,\,for\,\,}0\leq i\leq 2n+1,i\neq 2n\\
 \\
\frac{pf(0,...,\widehat{2n},2n+1,2n)}{pf(0,...,2n-1)}
=-\frac{p(0,...,2n,2n+1)}{pf(0,...,2n-1)}
=-\frac{\tilde\tau_{2n+2}(t)}{\tilde\tau_{2n}(t)}=-\tilde h_{2n}\\
\hspace{7.5cm}\mbox{\,\,for\,\,} i=2n.
\end{array}\right.
\end{eqnarray*}
and so
$$
\langle p_{2n+1}(z),p_i(z)\rangle=-\delta_{i,2n}\tilde h_{2n}
 ~~\mbox{for}~~0\leq i \leq 2n+1;
$$
the skew-symmetry of $\langle,\rangle$ does the rest. \qed

\begin{lemma} We have
$$
\frac{\partial}{\partial t_k}pf(i_0,...,i_{\ell})=\sum^{\ell}_{r=0}
pf(i_0,...,i_r+k,...,i_{\ell}).
$$
\end{lemma}


\underline{\sl Proof}:  The differential equations (2.1) for $m_{\infty}(t)$ read
in terms of the elements $\mu_{ij}$:
\begin{equation}
\frac{\partial \mu_{ij}}{\partial t_k}=\mu_{i+k,j}+\mu_{i,j+k};
\end{equation}
hence, setting $\sigma (i)=i'$, we compute

\noindent $\displaystyle{\frac{\partial}{\partial t_k}pf(i_0,...,i_{n-1})}$
\begin{eqnarray*}
&=&\frac{1}{(n/2)!
2^{n/2}}\sum_{\sigma}\varepsilon(\sigma)\sum_{{r\in\{0,...,n-2\} }\atop {r
~\mbox{\tiny{even}}}}\mu_{i'_0i'_1}...
(\mu_{i'_r+k,i'_{r+1}}+\mu_{i'_r,i'_{r+1}+k})...\mu_{i'_{n-2},i'_{n-1}}\\
\\
&=&\sum_{r=0}^{n-1}pf(i_0,...,i_{r}+k,...,i_{n-1}),
\end{eqnarray*}
upon reorganizing the first sum on the right hand side
appropriately.\qed


\begin{lemma}
The polynomials $p_k$ of Theorem 3.1 satisfy:
$$
\frac{1}{\tilde\tau_{2n}(t)}\left(\frac{\partial}{\partial t_1}+z\right)
\tilde\tau_{2n}(t)p_{2n}(z)=p_{2n+1}(z).
$$
\end{lemma}

\underline{\sl Proof}: 
Using the explicit form (3.10) of $p_{2n}$ and Lemma 3.5, one
computes

\medbreak
\noindent$\displaystyle{
\left(z+\frac{\partial}{\partial t_1}
\right)\tilde\tau_{2n}(t)p_{2n}(z)}$
\begin{eqnarray*}
&=&-\sum_{k=0}^{2n}(-z)^{k+1}pf(0,1,...,\hat k,...,2n)\\ &
&\quad\quad
+\sum_{k=0}^{2n}(-z)^k(pf(0,...,\widehat{k-1},...,2n)+pf(0,...,\hat
k,...,\widehat{2n},2n+1))\\ &=&z^{2n+1}pf(0,1,...,2n-1)+z^{2n}
\left(-pf(0,...,\widehat{2n-1},2n)+
pf(0,...,\widehat{2n-1}, 2n)\right)\\ & &\quad +\sum_1^{2n-1}(-z)^k
pf(0,...,\hat k,...,\widehat{2n},2n+1)+pf(1,...,\widehat{2n},
2n+1)\\ &=&
\tilde\tau_{2n}(t)p_{2n+1}(z),
\end{eqnarray*}
ending the proof.\qed



\begin{lemma}
The decomposition
\begin{equation}
m_{\infty}=Q^{-1}J Q^{\top -1},~~\mbox{with}~~Q\in {\cal G}_{{\Bbb k}},
\end{equation}
is unique (modulo signs) and is completely given by the
skew-orthonormal polynomials $q_k(z)$.
\end{lemma}

\underline{\sl Proof}:  In view of relations (3.5), the skew-orthonormal polynomials
(3.3) provide a solution to the Borel decomposition
$m_{\infty}=Q^{-1}J Q^{\top -1}$, with $Q\in {\cal G}_{{\Bbb k}}$. We now show
this $Q\in
{\cal G}_{{\Bbb k}}$ is unique! To do so, we proceed by induction on
$q_k=(Q\chi(z))_k$. Assuming the existence of another matrix
$\tilde Q \in {\cal G}_{{\Bbb k}}$, such that $m_{\infty}=\tilde Q^{-1}J
\tilde Q^{\top
-1}$, we must show $q=\tilde q$. Indeed $m_{\infty}=Q^{-1}J Q^{\top -1}$
leads unambiguously (except for the sign) to
$$
q_0=\mu_{01}^{-1/2},~~\mbox{and}~~q_1=\mu_{01}^{-1/2}z.
$$
But, for some $\alpha_i$ and $\beta_i$, we would have, taking into
account the absence of the $z^{2n}$-term in $\tilde q_{2n+1}$,
\begin{equation}
\tilde q_{2n}=\sum_0^{2n} \alpha_i q_i~~\mbox{and}~~
\tilde q_{2n+1}=\alpha_{2n}q_{2n+1}+\sum_0^{2n-1} \beta_i q_i.
\end{equation}
Then using the inductive step, namely $\tilde q_{k}=q_{k}$ for $0
\leq k \leq 2n-1$, and the skew-orthonormality, we compute for $\ell < n$,
\begin{eqnarray}
0&=&\langle \tilde q_{2n} ,\tilde q_{2\ell} \rangle=\langle \tilde q_{2n} ,
q_{2\ell} \rangle=-\alpha_{2\ell+1}\nonumber\\
 0&=&\langle \tilde q_{2n},\tilde q_{2\ell +1} \rangle=
 \langle \tilde q_{2n}, q_{2\ell +1} \rangle =\alpha_{2\ell}\nonumber\\
0&=&\langle \tilde q_{2n+1},\tilde q_{2\ell} \rangle =\langle \tilde q_{2n+1},
q_{2\ell} \rangle =-\beta_{2\ell+1}\nonumber\\ 0&=&\langle
\tilde q_{2n+1},\tilde q_{2\ell+1}\rangle=\langle
\tilde q_{2n+1}, q_{2\ell+1} \rangle=\beta_{2\ell},
\nonumber
\end{eqnarray}
and so, since also $\beta_{2n}=0$,
$$
\tilde q_{2n}=\alpha_{2n}  q_{2n}~~\mbox{and}~~\tilde
 q_{2n+1}=\alpha_{2n}  q_{2n+1}.
 $$
 Finally
 $$
 1=\langle \tilde q_{2n}, \tilde q_{2n+1} \rangle
 =\alpha_{2n}^2 \langle  q_{2n}, q_{2n+1} \rangle = \alpha_{2n}^2,
 $$
leading to  $\alpha_{2n}^2=1$. \qed




\underline{\sl Proof of Theorem 3.1 and Corollay 3.2}: The proof
follows at once from lemmas 3.3 up to 3.7. The fact that $Q$ and
$L$ satisfy the differential equations (2.3) and (2.4) follows from
the fact that the matrix $m_{\infty}$ satisfies the differential
equations
$$\partial m / \partial t_k=\Lambda^k m
+m \Lambda^{\top k}, $$ according to Theorem 2.1.

The second relation in (3.9) is established as follows: defining
the semi-infinite lower-triangular matrix $P$ (with $1$'s along the
diagonal) by means of the monic skew-orthogonal polynomials $p(z)=P
\chi(z)$ and differentiating $P=\tilde h^{1/2} Q$ with regard to
$t_k$, one
 obtains
$$
\tilde h^{1/2} \dot Q Q^{-1}\tilde h^{-1/2}=
-\frac{1}{2}(\log \tilde h)^. +  \dot P P^{-1}.
$$
Then, using proposition 2.2, (formula (II)), one finds:
\begin{eqnarray*}
0=\tilde h^{1/2}(\mbox{II})_0 \tilde h^{-1/2}&=&
 \tilde h^{1/2}\left((L^k)_0-J(L^{k
\top})_0 J+2(\dot Q Q^{-1})_0\right)\tilde h^{-1/2}\\
&=&\tilde h^{1/2}2 \pi_{{\Bbb k}}(L^k)_0\tilde h^{-1/2}-
  (\log \tilde h)^. +2(\dot P~P^{-1})_0\\
&=& 2 \pi_{{\Bbb k}}(L^k)_0 -
  (\log \tilde h)^.~,
\end{eqnarray*}
using the commutation of $\pi_{{\Bbb k}}(L^k)_0$ with $\tilde h^{1/2}\in
{\cal D}_0^-$, together with $(\dot P~P^{-1})_0=0$.
\qed



