% SchurSeries.tex
\documentclass[a4paper,12pt]{article}
\usepackage{amsmath,amsfonts,amssymb}


\def\s{\scriptstyle }
\def\a{\alpha}
\def\b{\beta}
\def\l{\lambda}
\def\e{\epsilon}
\def\L{\Lambda}
\def\N{{\mathbb N}}
\def\Z{{\mathbb Z}}
\def\A{{\mathbb A}}
\def\B{{\mathbb B}}
\def\C{{\mathbb C}}
\def\X{{\mathbb X}}
\def\Y{{\mathbb Y}}
\def\AB{ {\mathbb A}\moins {\mathbb B}}
\def\ss{\sigma}
\def\Sym{{\mathfrak Sym}}
\def\mfS{{\mathfrak S}}

\def\bu{\scriptstyle \bullet}
\def\moins{\raise 1pt\hbox{$\s -$}}
\def\plus{\raise 1pt \hbox{ $\s +$}}
\def\Rt{\widetilde{\mathcal R}}
\def\cR{\mathcal{R}}
\def\cO{{\mathcal O}}  


\newtheorem{theorem}{Theorem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\def\proof{\noindent{\it Proof.}\ }

\begin{document}
\title{Schur Functions Series}
\maketitle

\section{Symmetrizing Operators}

To follow usual conventions, operators act on their right.

Def. $\pi_\omega$

Factorization corresponding to a Young subgroup 
$\mfS_k\times \mfS_{n-k}$.

Define, for two sets of positive integers,
$$ \cR(I,J)= \prod_{i\in I,\, j\in J} (x_i-x_j) \ ,  $$
$$ \Rt(I,J)  = \prod_{i\in I,\, j\in J} (1- x_j/x_i)  \ . $$

Then 
$$ \pi_\omega = \Theta_{k|n-k}\,  \pi_{\omega'\times \omega''} \ ,  $$
where $\omega'\times \omega''$ is the maximal element of
$\mfS_k\times \mfS_{n-k}$, and  
\begin{equation}  \label{Theta}
  \Theta_{k|n-k}(f) = \sum \sigma\left(
  \frac{f}{\Rt([1,\ldots,k], [k\plus 1,\ldots n])} 
  \right) \ , 
\end{equation}
where the sum is over all the permutations which are
shuffles of $[1,\ldots,k]$ and $[k\plus 1,\ldots n]$.

Equivalently, $\Theta_{k|n-k}$ can be understood as a summation
over complementary subsets $X_I:\{ x_i,\, i\in I\}$, $X_J$,
of $\{x_1,\ldots, x_\}$~:
$$ \Theta_{k|n-k}(f) = \sum
  \frac{f(X_I, X_J)}{\Rt(I,J)}
  \ , $$
$f$ being a function in $\Sym(k|n\moins k)$,
i.e. a function of $n$ variables symmetrical in the first 
$k$ ones, and in the last $n\moins k$ ones.

Notice that $\Theta_{1|n\moins 1}$,
restricted to  functions of one variable,
 is the Lagrange-type operator
$$ f\to   \sum f(a)\, \prod_{b\in \A-a} (1-b/a)^{-1}  \ .$$

\section{ $S$-series by Symmetrization}

Given any series $\sum  c_\lambda x^\lambda$, its image under
$\pi_\omega$  is $\sum  c_\lambda S_\lambda(X)$.
This will be interesting if all the Schur functions thus obtained
are different. In particular, if all the $\lambda$ are dominant,
then the functions obtained are indexed by partitions
and there are no cancellations.

For example, one obtains
\begin{multline*} 
 \prod_i (1+x_i) \prod_{i<j}(1 -x_ix_j)  = 
  \pi_\omega\bigl( (1 +x_1)(1+x_2^3)(1+x_3^5)\cdots\bigr) \\
= \pi_\omega\bigl( 1+ x^{10..}+x^{03..}+ x^{13..} \bigr)
= 1+ S_1 -S_{21} -S_{22}+\cdots  \ ,
\end{multline*}
\begin{multline*}
\prod_{i\leq j}(1-x_ix_j)= 
 \pi_\omega\bigl( (1-x_1^2)(1-x_2^4)(1-x_3^6)\cdots\bigr) \\
 = \pi_\omega\bigl( 1 -x^{20..} -x^{04..}+ x^{24..} \bigr)=
 1 -S_2 +S_{31} -S_{33} +\cdots 
\end{multline*} 
in this way. The simplest proof 
of these expansions is using Vandermonde determinants,
as stated in the next section.

\begin{verbatim}
Litt:=proc(n)  local i,fr; fr:=w[];
  for i from 1 to n do
    fr:=FreeConcat(fr,w[0]+w[2*i-1])
  od;
  subs(w=s,fr)
end:
Litt2:=proc(n)  local i,fr; fr:=w[];
  for i from 1 to n do
    fr:=FreeConcat(fr,w[0]-w[2*i])
  od;
  subs(w=s,fr)
end:
ACE> CLG_n(3):   # fixes the number of variables  

ACE> Litt(3);
s[0,0,5]+s[1,0,5]+s[0,3,5]+s[1,3,5]+s[0,0,0]+s[1,0,0]
                                   +s[0,3,0]+s[1,3,0]
ACE> factor(Tox_n(Tos(%))); 
        # One reorders the indices, then evaluate in $x$
    -(1+x3)(1+x2)(-1+x2x3)(x1+1)(-1+x1x3)(-1+x1x2)
ACE> factor(Tox_n(Tos(Litt2(3))));
  (x3-1)(1+x3)(x2-1)(1+x2)(-1+x2x3)(x1-1)(x1+1)
                             (-1+x1x3)(-1+x1x2)

\end{verbatim}



\section{$S$-series from Vandermonde determinants}%%%%%%%%%%%%%%%%%%%%%%

The expansion of functions of the type
$$ \prod_{i<j}(1-x_ix_j) \ , \ \prod_{i\leq j}(1-x_ix_j)
\ ,\ \prod_i (1-x_i)\prod_{i<j}(1-x_ix_j)\, ,\ \ldots $$
can be obtained by taking Vandermonde determinants
in the variables $x_i+1/x_i$.  In that case, the action
of $\pi_\omega$ can be written as a determinant, and one does not need
any sophisticated theory to expand these functions in 
terms of Schur functions  (cf. notes with Fumei).

For example, we have already used in the preceding section that 
\begin{equation}
 \pi_\omega\Bigl((1+x_1)(1+x_2^3)(1+x_3^5)\cdots \Bigr)
= \sum_\epsilon S_{1\epsilon_1, 3\epsilon_2, 5\epsilon_3, ..}   \ ,
\end{equation}
sum over all $\epsilon \in \{0,1\}$. In the right hand side,
the Schur functions are not indexed by (decreasing) partitions
and their indices must be reordered.

On the other hand, one can write
$\pi_\omega\Bigl((1+x_1)(1+x_2^3)(1+x_3^5)\cdots \Bigr)$ as a determinant
(this is the case for all $\pi_\omega( f_1(x_1) f_2(x_2) f_3(x_3)\cdots)$).
One then recognizes that it can be transformed into a Vandermonde
for the variables $x_i+1/x_i$, and one obtains that the function
computed above is
\begin{equation}
  \prod_i( (1+x_i) \, \prod_{i<j}  (1-x_ix_j)   \ .
\end{equation}

\section{Functoriality}

The argument of a symmetric function does not need to be restricted
to a single alphabet, one can for example use differences of
alphabets. If one shows an identity on symmetric functions
$f(\A) = g(\A)$ valid for any $\A$, then it will remain
valid for $A-B$.

This has the consequence that knowing the expansion,
in terms of Schur functions, of such products as
\begin{equation*}
\prod ( 1-a_i)^{\pm 1} \prod_{i<j} (1- a_ia_j)^{\pm 1} \quad \& \quad
\prod ( 1-a_i)^{\pm 1} \prod_{i\leq j} (1- a_ia_j)^{\pm 1} \ ,
\end{equation*}
implies that these expansions remain valid for products like
\begin{equation*}
\prod ( 1-a_i)^{\pm 1} \prod ( 1-b_i)^{\mp 1}
\prod_{i\leq j} (1- a_ia_j)^{\pm 1} \prod_{i< j} (1- b_ib_j)^{\pm 1}
\prod (1-a_i b_j)^{\mp 1}
\end{equation*}
involving a second alphabet $\B=\{ b_1,b_2,\ldots\}$.

Moreover, distinguishing between letters, i.e. taking $A+X$ instead
of $A$, where $X$ is a finite alphabet, and expanding in terms
of the letters in $X$, allows to put more parameters
in the usual series for a single alphabet $A$.

I explain this point of view in the article
``Littlewood's formulas for characters of orthogonal and
  symplectic groups''.
\section{Using $\pi_\omega$ for Induction}  %%%%%%%%%%%%%%%%%%%%%%%%%

Let us give an example how to compute one of the Littlewood formulas
by induction on the number of variables, using $\pi_\omega$.

Let us take $\sigma_1(X+\L^2X)$.

Given $n$, let 
$$ F(n) = \sigma_1(X_n+\L^2X_n)= \prod_{1\leq i\leq n}( 1-x_i)^{-1}
 \, \prod_{1\leq i<j\leq n}( 1-x_ix_j)^{-1}  $$

The image of $F(n-1)$ under $\pi_\omega$, with $\omega=[n,\ldots,1]$
is given by writing 
$$ F(n-1)=F(n)\, (1-x_n) \prod_{1\leq i< n}( 1-x_ix_n)  =
\lambda_{-1}( x_n(1+X_{n-1})) \ ,$$
and computing the image of  $ \lambda_{-1}( x_n(1+X_{n-1}))$
under $\pi_\omega$.

One finds that this image is $1-x_1\cdots x_n$,
and therefore
$$ \pi_\omega\bigl( F(n-1)  \bigr) = (1-x_1\cdots x_n)\, F(n)  \ .$$

Supposing known that $ F(n-1)= \sum_{\l: \ell(\l)\leq n-1}
   S_\lambda(X_{n-1})$, then one has, by taking the image of
this equality  under $\pi_\omega$~:
\begin{multline}
 F(n) = \frac{1}{1-x_1\cdots x_n}\sum_{\l: \ell(\l)\leq n-1}
   S_\lambda(X_n)  \\
  =\sum_{\l: \ell(\l)\leq n} S_\lambda(X_n)  \ .
\end{multline}

\section{Filtering by the biggest part}

In the case where a function $F(n)$ of the preceding type 
has a complicated expression, a recursion on $n$ will not be feasible.

It will be better to start from a sum $\sum c_\lambda x^\lambda$.

Let us take the problem of finding an expression for
$$ F(u,[x_1,\ldots,x_n]):=  \sum u^{\lambda_1}\, S_\lambda(X_n)   \ , $$
i.e. of filtering the sum of Schur functions by their
greatest part, the sum being over all partitions with $\ell(\lambda)\leq n$.

Evidently,
$$ F(u,[x_1,\ldots,x_n])  =\pi_\omega\left(\frac{1}{
  (1-ux_1)(1-ux_1x_2)\cdots (1-ux_1\cdots x_n)}   \right) \ ,
$$
the power in $x_1$ being the first part of the partition.

However, $F(u,[x_1,\ldots,x_n])$ does not factorize. 
To write a simple expression 
of it, one must decompose it into simple fractions (relative to $u$).

This is the method followed by Desarm\'enien (S\'eminaire Lotharingien)   
and Macdonald (see Ex.16, p.83; Example 5, p.232).

Let us add a factor $(1-u)$ and start with 
$$ G(u,[x_1,\ldots,x_n]) := \Bigl((1-u)(1-ux_1)(1-ux_1x_2)\cdots 
       (1-ux_1\cdots x_n)\bigr)^{-1}
$$

For $i=1,\ldots,n$, let $\xi_i$ be the operator,
acting only on $x_i$ and $u$, such that 
$$  \xi_i :  x_i\to 1/x_i \ , \ u\to ux_i \ . $$

\begin{theorem}
\begin{multline}
 (1+\xi_1)\cdots (1+\xi_n)\Bigl(
   (1-u)^{-1}\, F(1,[x_1,\ldots,x_n])\Bigr)   \\
= \pi_\omega\left(\frac{1}{
  (1-u)(1-ux_1)(1-ux_1x_2)\cdots (1-ux_1\cdots x_n)}   \right) \\
 = (1-u)^{-1}\, F(u,[x_1,\ldots,x_n]) 
  = \sum_k\sum_{\lambda: \lambda_1\leq k} u^k\, S_\lambda(X_n)  
\end{multline}
\end{theorem}

\begin{proof}   
We shall show the required identity by 
determining  the residues, relative to $u$, of the first two functions.
To avoid accumulation of indices,
we shall compute only the residue for $u=1/(x_1x_2x_3)$, 
which is sufficiently representative of the general case.

The coefficient of
$(1-ux_1x_2x_3)^{-1}$ in the first function is
\begin{multline*}
  \xi_1\xi_2\xi_3\, \Bigl(F(1,[x_1,\ldots,x_n])\Bigr)  \\
= F\left(1,[\frac{1}{x_1},\frac{1}{x_2},\frac{1}{x_3}]\right)\, 
    F(1,[x_4,\ldots, x_n]) \Rt(123,4\ldots n)^{-1}   \\
% \prod_{i=1}^3\prod_{j=4}^n (1-x_j/x_i)    \ .
 = F(1,[x_1,x_2,x_3])\, F(1,[x_4,\ldots, x_n])
      \frac{(x_1 x_2 x_3)^n}{ R(123, 4\ldots n)}   \ . 
\end{multline*}

The second function is the image under $\pi_\omega$ 
of a $ G(u,[x_1,\ldots,x_n])$. 
Lagrange tells us how to decompose a function of the type $\prod 1/(1-au)$.
Therefore, we can decompose $ G(u,[x_1,\ldots,x_n])$ 
as a sum of simple fractions $\sum \frac{f(a)}{R(a,A-a)} \frac{1}{1-ua}$,
with the alphabet $A= \{1, x_1, x_1x_2,\ldots, x_1\cdots x_n\}$.

Explicitely, the coefficient of the term $1/(1-a u)$,
with $a=x_1x_2x_3$, in $G(u,[x_1,\ldots,x_n])$   is 
$$   a^n / R( a ,\, A-a)  \ .
$$
The residue in $u=1/a$ is transformed by $\pi_{321\, n\ldots 4}$
as follows.

The  symmetrisation in $x_1,x_2,x_3$ is the one of the preceding section
for the variables $1/x_3,1/x_2,1/x_1$. One gets  
$$ \pi_{321}\Bigl(\bigl( (a-1)(a-x_1)(a-x_1x_2)\bigr)^{-1}\Bigr)  
    =   F(1,[x_1,x_2,x_3])  \ .
$$ 

The symmetrisation in $x_4,\ldots, x_n$ gives
\begin{multline*}
\pi_{123\,  n\ldots 4}\Bigl(a^{n-3}/
                 R([a],[ax_4,\ldots, ax_4\cdots x_n])\Bigr)\\
= \pi_{123 n\ldots 4}\Bigl(1/R([1],[x_4,\ldots, x_4\cdots x_n])\Bigr)\\
 = F(1,[x_4,\ldots, x_n])  \ .
\end{multline*}  

In total, the coefficient of $1/(1-ux_1x_2x_3)$   is now
$$ (x_1x_2x_3)^3\, F(1,[x_1,x_2,x_3])\, F(1,[x_4,\ldots, x_n])  \ .   $$

% facteur (x1x2x3) par rapport au residu:  (u -1/(x1x2x3))
\begin{verbatim}

# Case n=5; residue of u=1/(x1x2x3)
ACE> factor(residue(1/(1-u)/(1-u*x1)/(1-u*x1*x2)
                  /(1-u*x1*x2*x3)/(1-u*x1*x2*x3*x4)
                  /(1-u*x1*x2*x3*x4*x5), u=1/x1/x2/x3));
                                 2
                            x2 x3
  - ------------------------------------------------------
     (x3 - 1)(x2 x3 - 1)(x1 x2 x3 - 1)(-1 + x4)(x4 x5 - 1)

# Symmetrisation by \pi_{321 54} : reduced decomposition [121 4] 
ACE> factor(PiRdOnPol(%,[1,2,1,4]));
  2   2   2
x1  x2  x3 /((x1 x2-1)(x2 -1)(x1 x3 -1)(x4 x5 -1)(-1 + x4)(x3 -1)

    (x2 x3 - 1) (-1 + x1) (-1 + x5))

\end{verbatim}

I remains to symmetrize between the variables in the two blocks 
$[x_1,x_2,x_3]$ and $[x_4,\ldots, x_n]$.
This can be done with the operator $\Theta_{3|n-3}$ of the first
section. This operator will transform a rational function
$$ \frac{f(\{x_1,x_2,x_3\}\, ,\, \{x_4,\ldots, x_n\})}{1-ux_1x_2x_3}  $$
into a rational function having simple poles 
$u= 1/(x_1 x_2 x_3)^{\sigma}$, for any 
$\sigma \in \mfS_n/\mfS_k\times \mfS_{n-k}$.
According to the expression (\ref{Theta}) of 
$\Theta_{3|n-3}$, the operator has for effect to divide the residue
in $u= 1/(x_1 x_2 x_3)$ by
$\Rt([1,\ldots,3], [4,\ldots n])  $.

In final,
the coefficient of $1/(1-u x_1x_2x_3)$ in
\begin{multline*} 
\pi_\omega\Bigl((x_1x_2x_3)^2\, F(1,[x_1,x_2,x_3])\, F(1,[x_4,\ldots, x_n])
\Bigr)   \\
= \Theta_{3|n-3}
\Bigl((x_1x_2x_3)^2\, F(1,[x_1,x_2,x_3])\, F(1,[x_4,\ldots, x_n])
\Bigr)   
\end{multline*}
is 
\begin{multline*}
  (x_1x_2x_3)^3\,  F(1,[x_1,x_2,x_3])\, F(1,[x_4,\ldots, x_n])/
                    \Rt([1,\ldots,3], [4,\ldots n])  \\
 = (x_1x_2x_3)^n\,  F(1,[x_1,x_2,x_3])\, F(1,[x_4,\ldots, x_n])/
   \cR([1,\ldots,3], [4,\ldots n]) \, .
\end{multline*}

One has found the same residue, for the two
functions, in $u=1/(x_1x_2x_3)$, and, by symmetry,
in $u=1/(x_ix_jx_k)$. This finishes the proof of the theorem. 
\end{proof}

\medskip
Let us detail the case $n=2$.

We start with 
$$  1/ (1-u)(1-ux_1) (1-ux_1x_2)  \ .$$
By Lagrange, it is written
\begin{multline*}
 \frac{1^2}{(1-u)(1-x_1)(1-x_1x_2)}  
  + \frac{x_1^2}{(1-ux_1)(x_1-1)(x_1-x_1x_2)}  \\
 +  \frac{x_1^2 x_2^2}{(1-u x_1x_2)(x_1x_2-1)(x_1x_2-x_1)}  \ .
\end{multline*}

The image of the first term under 
 $\pi_\omega=\pi_1$ is
$$  \frac{1^2}{(1-u)(1-x_1)(1-x_2)(1-x_1x_2)}  $$
(one factor added), and similarly, the last term becomes
$$   \frac{x_1^2 x_2^2}{(1-ux_1x_2)(1-x_1x_2 )
 (1-x_1)(1-x_2)}
$$

The image of the middle term does not factorize, and we must write
it as a sum over $\mfS_2$~:
$$  \frac{1}{(1-ux_1)(1-\frac{1}{x_1}) (1-x_2) (1-\frac{x_2}{x_1})}
+ \frac{1}{(1-ux_2)(1-\frac{1}{x_2}) (1-x_1) (1-\frac{x_1}{x_2})}\, .
$$

The final step is just to recognize that these four terms are 
indeed the four terms in the expansion of 
$$ (1+\xi_1)(1+\xi_2)\, \left(
\frac{1}{ (1-u)(1-x_1)(1-x_2) (1-x_1x_2)}\right) \ .  $$

\newpage
\section{Generalizing Gordon's Determinant}

(See notes with Hou and Mu.)

To express the sum of all Schur functions indexed by partitions of
length $\leq 2n$,
Gordon wrote a determinant of order $n$, whose entries are linear
combination of the  
   $c_k:=\sum_i  S_i(\A) S_{i+k}(\A)$.

One can consider the same determinants,
but using two alphabets $\A,\B$ and 
$$ c_k:=\sum_i  S_i(\B) S_{i+k}(\A)  \quad ,\quad k\in\Z \ .  $$
The new determinant is easier to evaluate than the preceding one,
by induction on the cardinality of $\A$.

For example, for $n=2$, one finds
$$  \sum_\lambda  S_\lambda B\, \bigl(1+S_1B -S_{21}B-S_{22}B\bigr)\, 
  S_\lambda \A  =  
  \sum_\lambda  S_\lambda \B\, \bigl( S_\lambda \A +
   S_{\lambda/1} \A -S_{\lambda/21} \A  -S_{\lambda/22} \A \bigr)\, 
$$
where the sum is over partitions with two parts at most.

Taking $\B=\A$, it gives an expression of the
sum of all Schur functions indexed by partitions
with $4$ parts at most.

Let us see the construction in more details.

The determinants that one considers are determinants of order 
$n$, which can be expanded in sums of
Jacobi-Trudi determinants with entries 
$$  c_k = \sum_i  S_i(\B) S_{i+k}(\A) $$ 
instead of $h_k=S_k$.
If expanded in terms of $\A$, such a determinant is a linear 
combination of Schur functions $S_\l(\A)$, with $\ell(\A)\leq n$.
Therefore, \emph{one can restrict to an alphabet $\A$ of cardinality $n$}.
However, one can replace Jacobi-Trudi determinants by
determinants of powers  of the variables $a\in \A$ (divided by the
Vandermonde $\Delta(\A)$).
Thus, a determinant with a typical column 
$$ c_k, c_{k-1},\ldots, c_{k-n+1} \ ,\ k\geq 0  $$ 
is equal, up to division by the Vandermonde, to a determinant 
with column 
$$\left[\, \sum_i S_i(\B) a_1^{i+k},\ldots, \sum_i S_i(\B) a_n^{i+k}
\, \right] $$
 i.e.
$$\left[\,  a_1^k \sigma_1( a_1\B),\ldots, a_n^k \sigma_1( a_n\B)\,\right]$$

Such a determinant will reduce to a determinant of powers of the variables
$a_i$ times the function 
$$ \sigma_1(a_1\B)\cdots \sigma_1(a_n\B)  = \sigma_1(\A\B)   \ .$$

Let us now take the two types of determinants used by Gordon.

The first type is 
$$ \Big| c_{n-i+j-1} + c_{n+i+j}    \Big|_{i,j=0\ldots n-1} \ . $$
From what we said, when $|\A|=n$, then Gordon's determinant
becomes
\begin{equation}
 \sigma_1(\A \B)\,   \frac{1}{\Delta(\A)}\, 
 \Big|  a^{n-1}+a^n,\, a^{n-2}+a^{n+1}, \ldots ,\, 
     a^{0}+a^{2n-1}  \Big|_{a\in\A}  \ .
\end{equation}  

We recognize the determinant of powers, from our considerations 
about Littlewood formulas.

\def\Sp{\mathfrak{S\hspace{-0.06 em}p}}

In final, Gordon's determinant is
$$ \sigma_1(\A\B)\,  \sigma_1\bigl(\A -\L^2(\A)\bigr)  \ .$$


% According to Littlewood (see section 10), this is the generating function of \emph{symplectic Schur functions} $$ \sigma_1\bigl(\A\B -\L^2(\A)\bigr) = \sum_{\l:\ell(\l)\leq n} S_\l(\B)\, \Sp_\l(\A)  \ .$$

% Therefore, for any cardinality, Gordon's determinant is equal to $$  \sum_{\l:\ell(\l)\leq n} S_\l(\B)\, \Sp_\l(\A)  \ .$$  

For example, for $n=2$, one obtains
\begin{multline}
 \begin{vmatrix}   c_1+c_2 & c_0+c_3 \\ c_0+c_1 & c_{-1}+c_2 
  \end{vmatrix} \\    
% = \sum_{\l:\ell(\l)\leq 2} S_\l(\B)\, \Sp_\l(\A) \\
= \sum_{\l:\ell(\l)\leq 2} S_\l(\B)\, \bigl(S_\l(\A)
  + S_{\l/1}(\A) - S_{\l/21}(\A) - S_{\l/22}(\A) \bigr)  \ .
\end{multline}

Specializing $\B=\A$, thanks to the interpretation of the determinant
due to Gordon, Bender-Knuth, one gets 
\begin{equation}
\sum_{\mu:\, \ell(\mu)\leq 4}\, S_\mu(\A)   =
\sum_{\l:\ell(\l)\leq 2} S_\l(\A)\, \bigl(S_\l(\A)
  + S_{\l/1}(\A) - S_{\l/21}(\A) - S_{\l/22}(\A) \bigr)  \ .
\end{equation}

The second type is
$$ \Big| c_{n-i+j-1} - c_{n+i+j+1}    \Big|_{i,j=0\ldots n-1} \ . $$

For $\A$ of cardinality $n$, we factorize it into
\begin{equation}
 \sigma_1(\A \B)\,   \frac{1}{\Delta(\A)}\,
 \Big|  a^{n-1} -a^{n+1},\, a^{n-2}+a^{n+2}, \ldots ,\,
     a^{0}+a^{2n}  \Big|_{a\in\A}  \ .
\end{equation}

This is also one of the determinants of Weyl and Littlewood,
and Gordon's determinant  is equal to
\begin{equation}
 \sigma_1(\A \B)\,  \sigma_1\bigl( -S^2\A \bigr)  \ .
\end{equation}

This can be rewritten in  terms of orthogonal Schur functions
(Littlewood, p.240)~:
\begin{equation}
 \sigma_1(\A \B)\,  \sigma_1\bigl( -S^2\A \bigr)=
\sum_\l S_\l(\B)\, \cO_\l(\A) \ .
\end{equation}

The expression
\begin{equation}
\sum_{\l:\, \ell(\l)\leq n}  S_\l(\B)\, \cO_\l(\A)
\end{equation}
remains valid for any $\A$, $\B$, without restricting cardinalities.

For example,
$$ c_0-c_2= \sum_i S_i(\B) \cO_i(\A) = \sum S_i(\B)\, 
      \bigl( S_i(\A)-S_{i-2}(\A)  \bigr)  \ ,
$$
\begin{multline}
  \begin{vmatrix}  c_1-c_3 & c_0-c_4 \\ c_0-c_2 & c_{-1}-c_3
\end{vmatrix} =   \sum_{\l:\, \ell(\l)\leq 2} 
  S_\l(\B)\, \cO(\A)   \\
 = \sum   S_\l(\B)\,\bigl(S_\l(\A)- S_{\l/2}(\A) +
    S_{\l/31}(\A) -S_{\l/33}(\A)   \bigr)  \ .
\end{multline}

\newpage          %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
We have used 
\begin{equation*}
  \sigma_1(-S^2)  \qquad \&\qquad  \sigma_1( S^1 -\L^2 )  \ .
\end{equation*}
Let us see the series and the determinants corresponding to
$$  \sigma_1(-S^1 -S^2)   \ .$$

Since $S^2(\A+y) -y^2 = S^2(\A) +yS^1(\A)$,
one can use the series 
$$ \sigma_1(-S^2(\A+y) +y^2)  \ . $$

Dividing the determinant associated to $\sigma_1(-S^2)$ by
$(1-y^2)\prod_a (1-a^2)$, one obtains, after some
transformation, the determinant of order $n+1$
$$ \begin{vmatrix}
 a^0+a^{2n}  & a^1+a^{2n-1} & \cdots & a^n+a^n \\
   \vdots    &     \vdots   &        &\vdots \\  
    1        &    1         &        &  1 
\end{vmatrix}
$$
Subtracting the last column to all of them, one gets the 
determinant of order $n$
$$ \Bigl|\, a^0(1-a^n)^2,\,  a^1(1-a^{n-1})^2,\,
  a^2(1-a^{n-2})^2,\, \ldots,\,  a^{n-1}(1-a)^2  \, \Bigr|$$
which should be equal to
$$ \Delta(\A)\, \prod (1-a)^2 \prod_{i<j} (1-a_ia_j)   $$
from our computation for $\sigma_1(-S^2)$.

Indeed, subtracting each column to the one on its left,
one gets in the case $n=3$,
$$\Bigl|1-2a^3+a^6,\, a-2a^3+a^5,\, a^2-2a^3+a^4   \Bigr|
  \to  (1-a)\, \Bigl|1\moins a^5,\, a(1\moins a^3),\, a^2(1\moins a) \Bigr|
$$
that we already computed.


Another determinant will be more convenient for 
expanding $\sigma_1(-S^1 -S^2)$ in terms of Schur functions.
It is 
$$ \Bigl|\,  
  a^{n-1}(1\moins a)(1\moins a^2),\, 
  a^{n-2}(1\moins a^2)(1\moins a^3),\,
  a^{n-3}(1\moins a^3)(1\moins a^4),\,
  \ldots,
  a^0(1\moins a^n)(1\moins a^{n+1} \, \Bigr|
$$

One evaluates this determinant by subtracting each column to the next one,
and one finds that it is equal to 
$\Delta(\A)\, \sigma_1(-S^1(\A) -S^2(\A))$.

We can now use this determinant to generalize Gordon, the powers 
appearing in the first row of the determinant being the indices
in Gordon's determinant.
For example, for $n=2$, one has
$$ \begin{vmatrix} 
   c_1-c_2-c_3+c_4  &  c_0 -c_2-c_3+c_5  \\
   c_0 -c_1-c_2+c_3 & c_{-1}-c_1-c_2+c_4
\end{vmatrix}
$$

The expansion of such determinant,  will be
$$   \sum_{\ell\leq n}  S_\lambda(\B)  
              D_{\sigma_1(-S^1-S^2)} S_\lambda(\A)   $$
where 
$$  D_{\sum c_\mu S_\mu} \bigl(S_\lambda\bigr)  =
  \sum c_\mu   S_{\lambda/\mu}  \ .  $$

We are specially interested in the case $\B=\A$, 
which will give sums of Schur functions of length $\leq 2n$.

Here are the results for $n=2$. Only multiplicities $2$
for some precise partitions. Find the law !
\begin{verbatim}
Litt4:=proc(n)  local i,fr;
  fr:=w[];
  for i from 1 to n do
    fr:=FreeConcat(fr,w[0]-w[i]-w[i+1]+w[2*i+1]);
  od;
 Tos(subs(w=s,fr))
end: #   \sigma_1 (-S^1 -S^2)
#   
FiltreSchur:=proc(sf)  local tt;#   Filter according to weight
 tt:=Sf2Table(sf,s);
 convert([seq( tt[pa]*z^convert(pa,`+`)*s[op(pa)],
             pa=map(op,[indices(tt)]))], `+`)
end:
CauchyLitt4:=proc(N)   local i,sf,pa;
 sf:=subs(z=-1/z,FiltreSchur(Litt4(2)));  # Sign Changed  !!!
 taylor(
  1+convert([seq(seq(
      s[op(pa)]*SfDiff(z^(2*convert(pa,`+`))*sf,s[op(pa)]),
     pa=ListPart(i,maxlg=2)),i=1..N)],`+`),z,N+1)
end:
#
CauchyLitt4(9):
s[]+s[1]*z
+(2*s[1,1])*z^2
+(s[2,1]+s[1,1,1])*z^3
+(2*s[2,2]+s[2,1,1]+s[1,1,1,1])*z^4
+(s[3,2]+s[2,2,1]+s[2,1,1,1])*z^5
+(2*s[3,3]+s[3,2,1]+2*s[2,2,1,1])*z^6
+(s[3,3,1]+s[3,2,1,1]+s[2,2,2,1]+s[4,3])*z^7
+(s[4,3,1]+2*s[3,3,1,1]+s[3,2,2,1]+s[2,2,2,2]+2*s[4,4])*z^8
+(s[4,4,1]+s[5,4]+s[4,3,1,1]+s[3,2,2,2]+s[3,3,2,1])*z^9

s[4,3,2,1]+2*s[3,3,2,2]+s[5,4,1]+2*s[4,4,1,1]+2*s[5,5]

s[5,5,1]+s[5,4,1,1]+s[4,4,2,1]+s[3,3,3,2]+s[6,5]+s[4,3,2,2]

s[6,5,1]+s[5,4,2,1]+2*s[6,6]+2*s[5,5,1,1]+2*s[4,4,2,2]
  +s[4,3,3,2]+s[3,3,3,3]

s[6,5,1,1]+s[5,5,2,1]+s[4,3,3,3]+s[5,4,2,2]+s[7,6]
  +s[6,6,1]+s[4,4,3,2]

2*s[5,5,2,2]+s[5,4,3,2]+2*s[6,6,1,1]+s[6,5,2,1]
+2*s[4,4,3,3]+2*s[7,7]+s[7,6,1]

s[7,7,1]+s[7,6,1,1]+s[8,7]+s[4,4,4,3]+s[6,6,2,1]
  +s[6,5,2,2]+s[5,5,3,2]+s[5,4,3,3]              z^15

s[6,5,3,2]+2*s[5,5,3,3]+s[5,4,4,3]+s[4,4,4,4]+2*s[8,8]
  +s[8,7,1]+2*s[7,7,1,1]+s[7,6,2,1]+2*s[6,6,2,2]      z^16

s[8,8,1]+s[8,7,1,1]+s[7,7,2,1]+s[7,6,2,2]+s[6,6,3,2]
 +s[6,5,3,3]+s[5,5,4,3]+s[9,8]+s[5,4,4,4]              z^17
-------
2*s[9,9]+s[9,8,1]+s[8,7,2,1]+s[6,5,4,3]+2*s[5,5,4,4]
 +2*s[6,6,3,3]+2*s[7,7,2,2]+s[7,6,3,2]+2*s[8,8,1,1]
-------
s[10,9]+s[7,6,3,3]+s[8,7,2,2]+s[7,7,3,2]+s[8,8,2,1]
  +s[9,8,1,1]+s[9,9,1]+s[6,6,4,3]+s[6,5,4,4]+s[5,5,5,4]
-------
[9,8,2,1]+s[10,9,1]+2*s[8,8,2,2]+2*s[9,9,1,1]+2*s[10,10]
+s[8,7,3,2]+s[6,5,5,4]+s[5,5,5,5]+2*s[7,7,3,3]+s[7,6,4,3]+2*s[6,6,4,4]
-------
s[11,10]+s[8,7,3,3]+s[9,8,2,2]+s[8,8,3,2]+s[10,9,1,1]+s[9,9,2,1]
+s[10,10,1]+s[7,7,4,3]+s[7,6,4,4]+s[6,5,5,5]+s[6,6,5,4]
-------
2*s[6,6,5,5]+2*s[7,7,4,4]+2*s[10,10,1,1]+2*s[11,11]
+2*s[9,9,2,2]+2*s[8,8,3,3]+s[8,7,4,3]+s[9,8,3,2]
+s[11,10,1]+s[10,9,2,1]+s[7,6,5,4]
-------
s[6,6,6,5]+s[12,11]+s[11,11,1]+s[8,8,4,3]+s[8,7,4,4]
+s[7,7,5,4]+s[7,6,5,5]+s[11,10,1,1]+s[10,10,2,1]
+s[10,9,2,2]+s[9,9,3,2]+s[9,8,3,3]

\end{verbatim}

\newpage
\subsection{Case of Bessel functions}
One takes Gordon's determinants, 
with $c_m=I_m:=  \sum_j x^{m+2j}/(j! (m+j)!)   $.
( for Maple: BesselI(m,2x)).

The determinant of order $1$ is a generating function
of Catalan numbers.
$$ I_0-I_2 = \sum x^{2n}/(2n)!\,  C_n   $$

The determinant of order 2 is the generating function 
of $C_{n+1}^2 -C_n C_{n+2}$.

Easy to study with the polarization:
$$ \sigma_1(B A) (1-S_2(A) + S_{13}(A) -S_{33}(A) $$
with $|A| =2$. 

But $\sigma_1(BA) =exp(x (a_1+a_2))$,
$S_{ij}(B)$ specializes in $x^{i+j} dim_{i,j}/(i+j)!$

In total, when one specializes $A$, one has to evaluate
$$  \sum_\mu   x^{2n}  d_\mu/n!( d_\mu/n! -d_{\mu/2}/x^2/(n-2)! 
  +d_{\mu/13}/x^4/(n-4)!  -d_{\mu/33}/x^6/(n-6)! )
$$


\newpage         %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Pfaffians and Representations of the Symmetric Group}

(? note with Fumei)

One wants to compute sums of the type 
  $\sum_\lambda S_\lambda$,
sum over partition  $\lambda$ with even parts, such that 
their conjugate has also even parts.
This problem has been solved by Sundquist using Pfaffians,
and generalized by Ishikawa.

We find it more fruitful, and simpler, to 
deduce such generating series from  the representations
of the symmetric group indexed by partitions $[n,n]$.

The functions to compute appear now as ``diagonal elements''
in these representations.


\section{Weyl groups of classical type}

All the identities of Littlewood have an interpretation 
in terms of root systems of type $,B,C$ or $D$, and 
are related to Weyl character formula.

This is how Macdonald prove them in the exercises of his book.

At our level (symmetrizing operators), it means that we have to use
some extra divided differences, which for the type $B_n$ are~:
$$   f \to \frac{ f(x_1,x_2,\ldots) - f(1/x_1,x_2,\ldots) }
   {x_1 -1/x_1}  $$
or 
$$   f \to \frac{ x_1f(x_1,x_2,\ldots) - f(1/x_1,x_2,\ldots)/x_1 }
   {x_1 -1/x_1} \ .  $$

These new operators act on symmetric functions in a non trivial
way, and thus allow to find identities on symmetric functions,
when using them, and symmetrizing again.

Alternating summations over the full group play a fundamental
role, as in the case of the symmetric group.
In fact these summations factor, one factor
being our usual summation over the symmetric group,
that is to say  our operator $\pi_\omega$.

\section{Variations about Cauchy formula}

Given two alphabets $\A,\B$, Cauchy formula asserts that
$$ \sigma_1(\A\B)  = \sum_\lambda  S_\lambda(\A) S_\lambda(\B)  \ .$$

Using other functions that Schur functions, one has more general
formulas.

\def\cO{\mathcal O}
For example,
 the \emph{orthogonal Schur function} $\cO_\l(\A)$ are defined
by the generating series\footnote{cf. Littlewood, p.240.}~:
\begin{equation}
\prod_{i\leq j} (1-a_ia_j) \prod_i\prod_j (1-a_i b_j)^{-1} =
 \sum_\l \cO_\l(\A)\, S_\l(\B)  \ ,
\end{equation}
that is,
\begin{equation}
  \ss_1\bigl( \A\B -S^2(\A) \bigr)
        = \sum_\l \cO_\l(\A)\, S_\l(\B)  \ .
\end{equation}

These functions can be expressed by a determinant of complete functions,
due to Weyl (th.7.9.A). For example, for a partition with three parts,
this determinant will be
$$\cO_{\l_1 \l_2\l_3} =
\begin{vmatrix}
 S_{\l_1} -S_{\l_1-2} & S_{\l_1+1} -S_{\l_1+1-4} & S_{\l_1+2} -S_{\l_1+2-6}\,
\\
\, S_{\l_2-1} -S_{\l_2-1-2} & S_{\l_2} -S_{\l_2-4} & S_{\l_2+1} -S_{\l_2+1-6}\\
 S_{\l_3-2} -S_{\l_3-2} & S_{\l_3-1} -S_{\l_3-1-4} & S_{\l_3} -S_{\l_3-6}
\end{vmatrix}  $$












\end{document}
