%  VolSchur.tex     Corrections 4 juillet01   

\documentclass[a4paper,12pt]{article}
\usepackage{amsmath,amsfonts, amssymb}
\input{xypic}

\title{ DOUBLE CRYSTAL GRAPHS }

\author{\small Alain Lascoux\\
\small CNRS, Institut Gaspard Monge, Universit\'e de Marne-la-Vall\'ee\\
\small 77454 Marne-la-Vall\'ee Cedex, France\\
\small  Alain.Lascoux@univ-mlv.fr }

\long\def\bee#1\ee{\begin{equation}#1\end{equation}}

\def\bea{\begin{eqnarray}}
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\def\s{\scriptstyle }
\def\a{\alpha}
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\def\l{\lambda}
\def\L{\Lambda}
\def\e{\epsilon}
\def\g{\gamma}
\def\ss{\sigma}
\def\d{\partial}

\def\S{{\mathfrak S}}


\def\N{{\mathbb N}}
\def\Z{{\mathbb Z}}
\def\Q{{\mathbb Q}}
\def\C{{\mathbb C}}

\def\cB{{\mathcal B}}
\def\cC{{\mathcal C}}
\def\cBh{\widehat{\mathcal B}}
\def\cBB{{\mathcal B\mathcal B}}
\def\cCB{{\mathcal C\mathcal B}}
\def\Pl{{\mathcal P\ell}}
\def\Cl{{\mathcal C\ell}}

\def\K{{\mathbb K}}
\def\Kh{\widehat K}
\def\KKh{\widehat{\mathbb K}}
\def\bp{\widehat{\pi}}

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\def\qed{\hfill QED}
\def\QED{\hfill QED}

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\unskip\vrule width \thickness}
\kern-\thickness}

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\def\blank{\omit\hskip\squaresize}

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\halign{&\vsquare{##}\cr #1}}}}
%%%%%%%%%%%%%%%%%%%%%
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\kern-\thickness}

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\def\blank{\omit\hskip\squaresize}

\def\bigyoung#1{\vcenter{
\vbox{\smallskip\offinterlineskip
\halign{&\vsquare{##}\cr #1}}}}

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



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\begin{document}
\date{}
\maketitle

\begin{abstract}
We show how to expand a non-symmetric Cauchy kernel\\ 
$\prod_{i+j\leq n} (1-x_iy_j)^{-1}$ in the basis of 
Demazure characters for type $A_n$. The construction involves
using the left and right structure of crystal graphs on words,
and mostly reduces to properties of the jeu de taquin.

We give the expansion of a similar kernel in the nilplactic algebra,
and define characters in two sets of variables. 
\end{abstract}

%%%%%%%%%%%%%%%%%%%%%%%%
% \input{Kernel}

\section{\large Kernels}

The best characterization of Schur functions, as symmetric functions,
relies on the diagonalization of the Cauchy kernel
\bee  \label{eq:Ker1}
\prod_i \prod_j  (1-x_iy_j)^{-1}   \ .
\ee
Indeed,
\bee  \label{eq:Ker2}
\prod_i \prod_j  (1-x_iy_j)^{-1}= 
\sum_\l S_\l(X)\, S_\l(Y)   
\ee
defines the Schur functions $S_\l$ uniquely (choosing them to be positive).

One could as well have started from the resultant 
$ \prod_i \prod_j  (x_i-y_j)$ of the two polynomials in $z$, 
$\prod_i (z-x_i)$ and  $\prod_j (z-y_j)$.

Relaxing symmetry, one still wants to ``separate variables'' in\\ 
$\prod_{i+j\leq n} (x_i-y_j)$ and $\prod_{i+j\leq n} (1-x_iy_j)^{-1}$, or more
generally, in products corresponding to the diagram of a partition. 


Schubert polynomials solve the first problem \cite{La82}. For the second one,
given a partition $\l$, let
$$ F_\l:= \prod_{(i,j)\in\l} (1-x_iy_j)^{-1} \ ,  $$
product over all the boxes of the diagram of $\l$, 
noting $(i,j)$ the coordinates of a box. 
We shall see that the cases 
$\l=\rho:=[n,\ldots,1]$, $n\in \N$,  allows to recover all the $F_\l$,
and therefore we shall mostly comment the
case 
$$F_\rho = \prod_{i+j\leq n+1} (1-x_iy_j)^{-1} \ .$$

One way to understand the Cauchy kernel is to have recourse to Schur functors
instead of Schur functions, and to use representations of linear groups
as Schur did in his thesis \cite{Schur}.

One can equivalently consider spaces of sections of ample line bundles over
the flag manifold ${\mathcal F}(\C^n)$. The non-symmetric case involves
using sections over Schubert varieties, and this leads
 to {\it Demazure characters} \cite{Dem}.
However, the geometrical study of line bundles over Schubert varieties is more
complicated 
than the study of characters only. Instead of representations, we shall rather 
use spaces of words (which can be thought as ``canonical bases''). 

 In that framework, Cauchy formula (\ref{eq:Ker2})
is interpreted as a bijection, due mostly to Schensted, between the terms in
the expansion of the left side of (\ref{eq:Ker2}) (words in  
commutative biletters $(x_i,y_j)$), and pairs of tableaux of the same shape.

Since $\prod_{i+j\leq n+1} (1-x_iy_j)^{-1}$ expands into a subfamily of
monomials,
we already know that $F_\rho$ can be interpreted as a family of pairs of
tableaux.

To characterize it,  we need some combinatorial tools associated to a two-sided
structure of crystal
graph on words. In fact, we do  not need to prove new properties of crystal
graphs, but have mostly to gather results stated in different languages.
We shall use words and tableaux as the simplest concepts in the type $A_n$ case.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% input{Tools}
\section{\large  Combinatorial Tools}

Let $X=\{ x_1,\, x_2,\ldots \}$ be a totally ordered alphabet 
(of non commuting letters). 

A {\it row} is a weakly increasing word,
a {\it column} is a strictly decreasing one.
A pair of  rows $a_1\cdots a_\ell$, $b_1\cdots b_k$ , $a_i,\, b_j\in X$,
is a 
\begin{itemize}  
  \item {\it 2-rows tableau }\quad  iff $\ell \leq k$ \quad \& 
           \quad $a_i>b_i,\, i=1,\ldots,\, \ell$
  \item {\it 2-rows contretableau}\quad   iff $\ell\geq k$ \quad \& \quad 
                 $a_{i+\ell-k}> b_i,\, i=1,\ldots,\, k$
\end{itemize}

A sequence  of rows is a {\it tableau} (resp. a {\it contretableau})
if every pair of consecutive rows is so.  Both can be planarly 
represented, stacking rows on top of each other, and left justifying
in the case of tableaux,  right justifying them in the case of 
contetableaux.  We shall identify  planar objects with 
their row reading, and with the sequence of rows composing them.  

The {\it plactic relations} (due to Knuth\cite{Knuth}) are the congruences
\begin{eqnarray}
cab \equiv acb \qquad (a\leq b<c) \label{eq:Plac1a}\, , \\
bac \equiv bca \qquad  (a <b\leq c) \label{eq:Plac1b}\, .
\end{eqnarray}
that can be planarly written
\begin{eqnarray}
\young{  c\cr a & b \cr} \quad \equiv \quad \young{a & c\cr \blank & b \cr } 
\qquad \qquad (a\leq b<c) \\
\young{  b\cr a & c \cr} \quad \equiv\quad  \young{b & c\cr \blank & a \cr}
\qquad \qquad (a <b\leq c)
\end{eqnarray}

% \parbox{2cm}{$ \begin{smallmatrix}  c\\ a & b \end{smallmatrix} \\ cab $}
% \equiv \parbox{2cm}{$ \begin{smallmatrix}  c\\ a & b \end{smallmatrix} \\ cab $}

The set of words congruent modulo the plactic relations to a given word $w$
is called the {\it plactic class} of $w$.

The following theorem is an algebraic reinterpretation \cite{Schutz77}
\cite{Napoli} of
a construction due to Schensted \cite{Schensted}.

\begin{theorem}  \label{th:Plac2}
In the plactic class $\Pl(w)$
 of a word $w$, there exists a tableau and only one tableau,
denoted $P(w)$. 

\end{theorem} 

A {\it right string} is the ordered string of two-rows words in the plactic class
of a 2-rows tableau (with the tableau as its origin, and the contretableau
as its end; edges are denoted $f$)~:
\bee \label{eq:String1}
\begin{smallmatrix}  3&6\cr 1&2 & 4&5&7     \end{smallmatrix}
\stackrel{f}{\longrightarrow}
 \begin{smallmatrix} 1&3&6\cr  &2 & 4&5&7  \end{smallmatrix} 
\stackrel{f}{\longrightarrow}   
\begin{smallmatrix} 1&3&4&6\cr & &2 &5&7 \end{smallmatrix} 
\stackrel{f}{\longrightarrow}
\begin{smallmatrix} 1&3&4&6&7\cr && &2 &5 \end{smallmatrix} 
\ee

Left strings are described in terms of words on an alphabet of two letters,
and it is convenient to take the two parentheses ( , ) as letters.
Given a word $w$ in such an alphabet, pair successive parentheses.
The initial word $w$ is the shuffle of two words, its {\it paired subword},
and a subword of the type $)\cdots )\, (\cdots ($.

A {\it left string} is a maximal sequence of words of the same degree which
coincide on their paired subwords. Here is a left string,
indicating below each word  the
positions of letters~:
\bee  \label{eq:String2}
{) \atop \s 7}{( \atop\s 6 }{) \atop\s 5 }{) \atop\s 4 }
            {( \atop\s 3 }{) \atop\s 2 }{)\atop\s 1}
\ \stackrel{f}{\longrightarrow}\ 
{)\atop\s 7}{(\atop\s 6}{)\atop\s 5}{)\atop\s 4}
                    {(\atop\s 3}{)\atop\s 2}{(\atop\s 1}
\ \stackrel{f}{\longrightarrow}\
{)\atop\s 7}{(\atop\s 6}{)\atop\s 5}{(\atop\s 4}
                  {(\atop\s 3}{)\atop\s 2}{(\atop\s 1}
\ \stackrel{f}{\longrightarrow}\ 
{(\atop\s 7}{(\atop\s 6}{)\atop\s 5}{(\atop\s 4}
                   {(\atop\s 3}{)\atop\s 2}{(\atop\s 1}  \ .
\ee

One exchanges the two notions of strings by reading the positions
of the ('s as the entries of the top rows of the elements of 
(\ref{eq:String1}), and the 
)'s as the entries of the bottom rows.  We shall later give a better
interpretation 
in terms of Schensted bijection. 

 
\medskip
Given two totally ordered alphabets  
$X=\{ x_1,\, x_2,\ldots \}$, $Y= \{ y_1,\, y_2,\ldots \}$, a
monomial  $w$ in the commutative variables  ${\bl y|x|}$, $x\in X$, $y\in Y$,
is called a {\it biword}. 

Ordering lexicographically $w$, with priority to $Y$, one writes
\bee
  w= {\bl y_1|a_1|} \cdots {\bl y_1|b_1|}
     {\bl y_2|a_2|} \cdots {\bl y_2|b_2|}
      {\bl y_3|a_3|} \cdots {\bl y_3|b_3|}  \cdots 
\ee
where $a_1\cdots b_1$, $a_2\cdots b_2,\, \ldots$ are rows in $X^*$,
and $a_i, b_j$ belong to $X$.

One can erase the $y_j$'s and interpret $w$ as a sequence of rows~:
\bee
  w \mapsto w^x:= [ a_1\cdots b_1,\, a_2\cdots b_2,\, a_3\cdots b_3,\ldots]
\ee 
We shall write $a_1\cdots b_1= row_1(w)$, $a_2\cdots b_2=row_2(w),\ldots $

Ordering lexicographically $w$ with priority to $X$, one 
obtains similarly a sequence $w^y$ of rows in $Y^*$.

Schensted's bijection, as reinterpreted by Knuth \cite{Knuth}, is described by
the following theorem.

\begin{theorem} \label{th:Plac3}
The map  $w\mapsto \bigl(P(w^x)\, ,\, P(w^y)   \bigr)$ is a bijection between
biwords and pairs of tableaux of the same shape.
\end{theorem}

The algebraic interpretation of theorem (\ref{th:Plac3}) is the following
non commutative version of Cauchy formula \cite{Bender}~:
\bee \label{eq:Plac4}
\prod_{x\in X}\prod_{y\in Y} \Bigl(1- {\bl y|x|}  \Bigr)^{-1} 
 = \sum \ (t,t')  \ ,
\ee
sum over all pairs of tableaux $t\in X^*$, $t'\in Y^*$ of the same shape
(we have written an equality instead of a bijection).

Strings can now be defined at the level of biwords.

Given an integer $i$, a {\it right $i$-string} is a sequence of biwords
$w_1,\ldots,\, w_\ell$ such that 
$$ [row_i(w_1),\, row_{i+1}(w_1)], [row_i(w_2),\, row_{i+1}(w_2)],
\ldots,\, [row_i(w_\ell),\, row_{i+1}(w_\ell)]   $$
is a right string, and such that 
$row_j(w_1)= \cdots = row_j(w_\ell)$, $\forall j\neq i,i\!+\!1$.

Similarly, a {\it left $i$-string} is a sequence of biwords
$w_1,\ldots,\, w_\ell$
such that $w_1,\ldots,\, w_\ell$ coincide under the specialization
$x_i=x_{i+1}$, and such that 
$w_1^x,\ldots,\, w_\ell^x$ becomes a string under the specialization 
$x_j=1,\, \forall j\neq i,i\!+\!1$, $x_{i+1}=(\, $, $x_i=)\, $.

We could have used the symmetry in $X$ and $Y$ to exchange left and right 
strings. 

The oriented edges of strings are called {\it lowering operators}. 

A {\it double crystal graph} for $\S_n$ is a colored graph,
with biwords as vertices, such that if it contains a biword $w$,
then it contains all left and right $i$-strings, 
$i=1,\ldots, n\!-\!1$, passing through $w$.

The symmetric group $\S_n$ acts by permuting the vertices
of a double  crystal graph \cite{Napoli}. The left (resp. right)
 action of a simple transposition $s_i$
restricts on a left (resp. right)
 $i$-string to the symmetry with respect to the middle of the string 
\cite{Littelmann94}
 \cite{LLT2}.  We shall note $w\mapsto s_i(w)$ the left action.

Instead of taking symmetry with respect to middle of strings, one can define 
operators $\pi_i$ and $\bp_i$, $i=1,\ldots,\, n\!-\!1$, 
 on the free algebra in $X$ as follows \cite{Keys} \cite{Kashiwara}: 
\begin{itemize}  
\item if $w$ is the first part of the left $i$-string containing it, 
     then $\pi_i(w)$ is the sum of all words between $w$ and $s_i(w)$,
     and $\bp_i(w)$ is the same sum minus $w$.
\item the action is extended to all words by  requiring that 
     $\pi_i(w +s_i(w)) = w +s_i(w)$ and $\bp_i(w +s_i(w)) =0$.
\end{itemize}

To relate biwords (or tableaux) to permutations, 
I introduced with M.P. Sch\"utzenberger the following
two notions which are in fact equivalent.

The first one consists in singling out some element $Vice(w)$
in each plactic class
of a word $w\in X^*$, that we called a {vice-tableau}
(we used it to give a non-commutative version of Schubert polynomials,
but finally reformulated this construction  in terms of ``keys'' in 
\cite{NCSchub}). 

$Vice(w)$ can be recursively defined as follows. It is the unique word 
$a_1\cdots a_\ell$ in a plactic class $\Pl(w)$ such that
\begin{itemize}
\item $x_\ell$ is the suffix of the contretableau in $\Pl(w)$
\item The shape of $P(a_1\cdots a_{\ell-1})$ is maximum 
among the words in $\Pl(w)$ having suffix $a_\ell$.
\item $a_1\cdots a_{\ell-1}$  is a vice-tableau. 
\end{itemize}

This definition implies an algorithm by reversing the Schensted algorithm
on contretableaux~: at each step, the box which is erased is the highest
which factors out on the right the suffix of the contretableau. 

For example 
\bee \label{eq:Ex1}
 \begin{smallmatrix}
  1&3&3&4&5\\ && 1&2&4\\ &&& 1&2 
\end{smallmatrix}
\equiv \begin{smallmatrix}
  1&3&3&4&5\\ && 1&2&4\\ &&& \blacksquare&1
\end{smallmatrix}  \cdot 2 
\equiv
\begin{smallmatrix}
  1&3&3&4&5\\ && \blacksquare &2&4\\ &&& 1&1
\end{smallmatrix}\cdot 2 
\equiv \begin{smallmatrix}
  \blacksquare &1&3&3&5\\&& 1&2&4\\ &&& 1&2 
\end{smallmatrix} \cdot 4  
\ee
shows that 
$$ Vice( 13345\, 124\, 12) = Vice( 13345\, 24\, 11)\cdot 2 $$
(iterating, one finds $Vice= [35\, 12344\, 112]$).

Factoring $Vice(w)$ as a product of maximal rows, on can consider it as
a biword $v$, with $row_1(v)\, row_2(v)\cdots = Vice(w)$. 
One can show that the sequence of lengths $\ell(row_1(v)),\, 
\ell(row_2(v)), \ldots$ is a permutation of the shape $\l$ of 
the tableau $P(w)$. Thus $Vice(w)$ determines a permutation $\ss(w)$
(which is unique when using only minimum representatives of $\S_n$
modulo the stabilizer of $\l$).

One can also show that  $Vice(w)$ can be obtained
by choosing among all the 
words obtained by permuting the rows of the contretableau 
$t$ in $\Pl(w)$ the farther away from $t$ (with respect to the Bruhat order), 
such that the sequence of suffixes of $row_1(v)$, $row_2(v), \ldots$
is the same as for $t$. 

For example,
\bee \label{eq:Ex2}
 \begin{smallmatrix}
  1&3&3&4&5\\ && 1&2&4\\ &&& 1&2
\end{smallmatrix}
\equiv   \begin{smallmatrix}
  1&3&3&4&5\\ &&&2&4\\ &&& 1&1&2
\end{smallmatrix}
\equiv   \begin{smallmatrix}
      3& 5\\ 1& 2& 3& 4& 4\cr && 1&1&2
\end{smallmatrix} 
\equiv   \begin{smallmatrix}
            3& 5\\  2& 3& 4\\ 1& 1& 1& 2& 4
\end{smallmatrix}
\equiv   \begin{smallmatrix}
        3& 3& 5\\ & 2& 4\\ &1& 1& 1& 2& 4
\end{smallmatrix}
\equiv   \begin{smallmatrix}
   3& 3& 5\\ 1& 1& 2& 4& 4\\ &&&1&2
\end{smallmatrix}
\ee
and the vice-tableau is the third element  $35\, 12344\, 122$, being the
farthest element having suffixes $5,4,2$
(and $\ss(w)= [42513]$). 

Instead of permuting rows, one can use the action of the symmetric group
on columns of a tableau $t$ which generates $k!$ words 
$c_1, c_2, \ldots, c_{k!}$  in the plactic class of $t$, if $t$ has $k$
columns \cite{Keys}. 

The {\it key} $\Cl(t)$ is the tableau of the same shape as $t$ such that its
columns appears as a right column in one of the words $c_i$'s. 
By construction, the set of columns of  $\Cl(t)$ (columns interpreted as sets)
is a flag of sets (with respect to inclusion). Writing the letters in the order
they appear in the successive columns 
(ordering them increasingly when they appear
several at a time, and adding a complete column $[n,\ldots,1]$), one obtains 
a permutation which is equal to $\ss(t)$, i.e. to the permutation determined
by the vice-tableau. Continuing with the same example, 
reading the objects columnwise,
\bee \label{eq:Ex3}
 \begin{smallmatrix}
  1&3&3&4&5\\ && 1&2&4\\ &&& 1&2
\end{smallmatrix}
\equiv  \begin{smallmatrix}
    1&3&3&5\\ && 2&4&4\\ && 1&1&2
\end{smallmatrix}
\equiv  \begin{smallmatrix}
    1&3&5\\ & 2&4&4\\ & 1&1&2&4
\end{smallmatrix}
\ \Rightarrow \ \Cl(t)= 
\begin{smallmatrix}   5&5\\ 4&4&4\\ 2&2&2&4&4
\end{smallmatrix}
\ee 
and $\ss(t)= [4\, 2\, 5\, 13]$. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% \input{Crystal.tex}
\section{ \large Crystal bases}

Given a dominant weight $\l=[\l_1\geq \l_2\geq \cdots \geq \l_n\geq 0]$, and
a permutation $\ss\in \S_n$, there exists a {\it crystal base} 
$\cB(\l,\ss)$, which is a collection of tableaux which can be described in
many ways. It is in fact more convenient to index  bases by weights, i.e. to
write $\cB(\l^\ss)$ rather than $\cB(\l,\ss)$, to take into account
the stabilizer of $\l$ in $\S_n$.

Define the {\it top component} 
\bee  \label{eq:Cry1}
 \cBh( \l^\ss) = \cB(\l^\ss) \setminus \cup_{\nu<\ss:\l^\nu\neq \l^\ss}
   \cB(\l^\nu)  \ .
\ee 

Moreover, instead of using tableaux only, one can choose a tableau $Q$ of shape
$\l$, and call {\it crystal base} $\cB(\l^\ss,\, Q)$ the collection of 
biwords
$$w: Q(w)=Q,\, P(w)\in \cB(\l^\ss) \ .$$ 

Figure 1 gives $\cB([2431], Q)$ for 
$Q= \begin{smallmatrix}
 4 \\ 3 &4\\ 2&3&3\\ 1&1&1&3 
\end{smallmatrix}$,
the arrows being left crystal operations (increasing some $i$ into  $i+1$,
$i=1,2,3$). 

The bottom element is $4^3\cdot 3^1\cdot 2^4\cdot 1^2$. However, there exists
another minimum element, $244\cdot 3\cdot 1222\cdot 11$, i.e. another element
which has no descendant with respect to 
the lowering operators $f_i$  inside $\cB([2431], Q)$. This observation
implies that $\cB([2431], Q)$ cannot be obtained from its lowest 
element by raising operators.  

Lakshmibai and Seshadri \cite{LakSesh} described crystal bases in terms 
of increasing paths (with respect to the Ehresmann-Bruhat order).
Crystal operators to generate crystal bases were used by Littelmann
\cite{Littelmann94} \cite{Littelmann96}, Kashiwara \cite{Kashiwara},
\cite{Kashiwara2}, 
Kashiwara and Nakashima \cite{KashiNaka}.
With M.P. Sch\"utzenberger, I have prefered using divided differences, keys
and flag conditions on words \cite{Keys}
(th. 4.3, prop. 5.6).  
The following theorem summarizes
these different points of view for what concerns type $A_n$. 

\begin{theorem}  \label{th:Crys1}
Let $n$ be an integer, $\l\in \N^n$ be a dominant weight,
$\ss\in \S_n$ be minimum in its class modulo the stabilizer of $\l$, 
and let $\mu=\l^\ss$.

Then the crystal base $\cB(\mu)$ (resp. $\cBh(\mu)$) can be characterized
by any of the following properties~:

1) It is the set of all tableaux $t$ of shape $\l$, with 
$\Cl(t)\leq \ss$ (resp. $\Cl(t)=\ss$).

2) Given any reduced decomposition $s_i\cdots s_j=\ss$, then
\bea
 \cB(\mu) &=& \pi_i\cdots \pi_j \bigl( \cdots x_2^{\l_2} x_1^{\l_1} \bigr)\ , \\
\cBh(\mu) &=& \bp_i\cdots \bp_j \bigl( \cdots x_2^{\l_2} x_1^{\l_1} \bigr)\ . 
\eea

3) Let $W(\mu)$ (resp. $\widehat{W}(\mu)$) be the set of all biwords 
(resp. biwords such that $row_1(w)\, row_2(w)\cdots$ is a vice-tableau)
such that $\ell(row_1(w))=\mu_1$, $\ell(row_2(w))=\mu_2,\, \ldots$
and such that $\forall i,\, \forall x\in row_i(w)$, then $x \leq x_{n+i-1}$.

Let moreover $Q=P( \cdots y_2^{\mu_2}\, y_1^{\mu_1})$.

Then 
\bee
  W(\mu) = \cB(\mu,Q) \quad \text{and} \quad 
   \widehat{W}(\mu) = \cBh(\mu,Q)
\ee

\end{theorem}

The elements of a crystal base are connected by left crystal operations. 
If a biword $w$ belongs to a crystal base $\cB(\mu,Q)$,
and $i$ is such that $w$ is not the origin of the left $i$-string 
passing through it, then the full $i$-string is
contained in $\cB(\mu,Q)$. Thus, for any $i$, $\cB(\mu,Q)$
is composed of full $i$-strings and of origins of $i$-strings \cite{Kashiwara}.

\smallskip
{\it Demazure characters} $K(\mu,X)$ and $\Kh(\mu,X)$ are
 the images of respectively $\cB(\mu,Q)$ and $\cBh(\mu,Q)$ (considered as sums)
in the ring of polynomials in commutative variables $x_1,\ldots,\, x_n$.
They depend only on $\mu$ and not on $Q$. 

\smallskip
Crystal bases admit a dual description, using right crystal operations. 
We shall need only one characterization. 

Given $\l$, $\ss$, $\mu=\l^\ss$, and any tableau $P$ of shape $\l$, 
call {\it cobase} $\cCB(\mu, P)$
the set of biwords $w$: $P(w)=P,\, Q(w)\in \cB(\mu)$. 

\begin{theorem}  \label{th:Crys2} 
Let $n$ be an integer, $\l\in \N^n$ be a dominant weight,
$\ss\in \S_n$, $\mu=\l^\ss$.  Let $P$ be a tableau in 
$\cBh(\mu)$.  Then $\cCB(\mu, P)$ is the set of biwords  $w$
\bee  
  P(w)=P \quad \& \quad \forall i,\, \forall x\in row_i(w)
\ \text{then}\  x\leq x_{n+1-i} \ .
\ee
\end{theorem}

Figure 1 and Figure 2 have the same underlying colored graph. In Figure 2, 
crystal operations are the jeu de taquin on two consecutive rows, and vertices
are all the 4-rows words in the plactic class of 
$\begin{smallmatrix}
 4 \\ 3 &4\\ 2&3&3\\ 1&1&1&3
\end{smallmatrix}$
which satisfy the flag conditions 
$$ row_1(w) \in \{1,2,3,4\}^*,\ row_2(w) \in \{1,2,3\}^*,\ 
 row_3(w) \in \{1,2\}^*,\ row_4(w) \in \{1\}^* \ . $$

One cannot add to Figure 3 
any word by the jeu de taquin, without breaking the flag
conditions. For example, the second lowering operator has the effect,
on the two minimum elements
$$ \petitematrice{3&4&4\cr 1&3&3&3\cr &&&2\cr&&&1&1 }
\stackrel{f_2}{\longrightarrow} 
\petitematrice{3&4&4\cr 1&3&3\cr &&2&3\cr &&&1&1 } \quad , \quad 
\petitematrice{4&4\cr3&3&3&3\cr &&&2\cr &&&1&1&1 }
\stackrel{f_2}{\longrightarrow}
  \petitematrice{4&4\cr 3&3&3\cr &&2&3\cr &&&1&1&1 }\ , $$
to push a 3 on the third row. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% \input{NonSymm.tex}
\section{\large Non symmetric Cauchy kernels}

The function $F_\rho = \prod_{i+j\leq n+1} (1-x_iy_j)^{-1}$, rewritten\\
 $F_\rho = \prod_{i+j\leq n+1} \Bigl(1-{\bl y_j|x_i|}  \Bigr)^{-1}$,
is the sum of all biwords in the commutative variables 
${\bl y_j|x_i|}$, such that $i+j\leq n+1$.

However, from theorems (\ref{th:Crys1}, \ref{th:Crys2}), one knows that
for such a biword $w$,
if $P(w)=P$, $P$ of shape $\l$,  $\ss(P)=\ss$, $\mu=\l^\ss$, then 
$Q(w)\in \cCB(\mu,P) $.  Therefore, the conditions $i+j\leq n+1$
are equivalent to the following expansion of $F_\rho$, writing
$\mu^\omega=[\mu_n,\ldots,\mu_1]$ when $\mu=[\mu_1,\ldots,\mu_n]$.

\begin{theorem} \label{th:Ker3}
Let $n$ be a positive integer. Then, under Schensted's bijection, 
$F_\rho$ can be written
$$ \prod_{i+j\leq n+1} \Bigl(1-{\bl y_j|x_i|}  \Bigr)^{-1} 
= \sum\,  (t,t')  \ ,$$
sum over all  weights $\mu\in \N^n$,  
all $t\in \cBh(\mu,X)$, $t'\in\cB(\mu^\omega,Y)$.
The commutative image of the preceding identity is
$$ F_\rho= \sum_{\mu\in \N^n}\,  \Kh(\mu,X)\, K(\mu^\omega,Y) \ .$$
\end{theorem}

Let us see how expansions for staircases imply expansions
for all partitions.

Given a partition $\l$, 
let $\rho(\l)=[k,k-1,\ldots,1]$ be the biggest staircase, the
diagram of which is contained in the one of $\l$.

Take any box in the staircase $[k+1,\ldots,1]$ which does not belong to $\l$.
The diagonal passing through this box cuts the diagram of $\l/\rho(\l)$
into two pieces that we call respectively the {\it North-West part}
and the {\it South-East part} of $\l/\rho$. 
Fill now each bow of row $r$, $r=2,\ldots$ of the North-West 
part with the number $r-1$. Similarly, fill each box of column $c$ of the
South-East part with the number $c-1$.

Reading the North-West part by columns, from right to left,
and interpreting $r$ as
the simple transposition $s_r$, gives a reduced decomposition of
a certain permutation $\ss(\l,NW)$; similarly, reading rows, from right to left,
and from top to bottom, of the South-East part, 
 gives a permutation $\ss(\l,SE)$ (we erased the dependency upon the
choice of the cut of $\l/\rho$).


\begin{theorem}  \label{th:Ker4}
 Let $\l$ be a partition, $\rho(\l)=[k,\ldots,1]$ the maximal
staircase contained in the diagram of $\l$, $\ss(\l,NW)$, $\ss(\l,SE)$
be the two permutations obtained by cutting the diagram of
$\l/\rho$ as explained above.  
Then
$$ F_\l(X,Y) = \sum\nolimits_{\mu\in \N^k}   
 \pi^x_{\ss(\l,NW)}\Bigl(\Kh(\mu,X) \Bigr)\  
\pi^y_{\ss(\l,SE)}\Bigl( K(\mu^\omega ,Y) \Bigr) \ ,  $$
sum over all weights $\mu\in \N^k$.

\end{theorem}

\Proof
Starting with $F_\rho$ (as given by theorem (\ref{th:Ker3}))
and applying to it $\pi^x_{\ss(\l,NW)}$,
then $\pi^y_{\ss(\l,SE)} $, one obtains $F_\l$, because, the two reduced
decompositions of $\ss(\l,NW)$ and $\ss(\l,SE)$ 
that we have chosen are such that  each step is of the type
$$ F\, \frac{1}{ 1-x_ky}  \stackrel{\pi_k^x}{\longrightarrow}          
  F\, \frac{1}{(1-x_ky)(1-x_{k+1}y)}   \, $$
with $F$ symmetrical in $x_k$ and $x_{k+1}$, or it is of the similar type  
with $X$ and $Y$ exchanged.  Therefore, the action of simple isobaric divided
differences gives a sequence of functions $F_\nu$ with one box added 
at a time to the diagram of each successive $\nu$.   \QED

Note that the image of $K(\mu,Y)$ under a $\pi_i^y$ is some $K(\nu,Y)$,
and that the image of $\Kh(\mu,X)$ under a $\pi_i^x = \bp_i^x+1$ 
is either 0, or the sum of two $\Kh(\nu,X)$.  We have not 
made totally explicit the possible pairs of weights appearing in the
expansion of $F_\l(X,Y)$.  

\def\bs{\blacksquare}

In the case $\l=[7,5,2,2,2,2]$, one has $\rho=[4,3,2,1]$,
$$ \begin{smallmatrix}  5&5\\ 4&4\\ \bs &3\\ \bs&\bs\\  \bs&\bs&\bs &3&4\\
          \bs&\bs&\bs &\bs &4&5&6
  \end{smallmatrix}  
\Longrightarrow   
\ \ss(\l,NW)= s_5s_4s_3\, s_5s_4  \ \& \ 
  \ss(\l, SE)= s_4s_3 \, s_6s_5s_4 
 $$

Representing a function $F_\l$ by the diagram of $\l$ filled with $\bs$,
theorem (\ref{th:Ker4}) provides the following sequence to
pass from $F_{4321}$ to $F_{752222}$~:
\begin{multline*}
$$  \begin{smallmatrix} \\ {} \\ {} \bs \\ \bs&\bs\\  \bs&\bs&\bs \\
          \bs&\bs&\bs &\bs
\end{smallmatrix}
\stackrel{\pi_4^x}{\longrightarrow}
\begin{smallmatrix} {}\\ \bs \\ \bs \\ \bs&\bs\\  \bs&\bs&\bs \\
          \bs&\bs&\bs &\bs
\end{smallmatrix}
\stackrel{\pi_5^x}{\longrightarrow}
\begin{smallmatrix} \bs\\ \bs \\ \bs \\ \bs&\bs\\  \bs&\bs&\bs \\
          \bs&\bs&\bs &\bs
\end{smallmatrix}
\stackrel{\pi_3^x}{\longrightarrow}
\begin{smallmatrix} \bs\\ \bs \\ \bs&\bs \\ \bs&\bs\\  \bs&\bs&\bs \\
          \bs&\bs&\bs &\bs
\end{smallmatrix}
\stackrel{\pi_4^x}{\longrightarrow}
\begin{smallmatrix} \bs\\ \bs&\bs \\ \bs&\bs \\ \bs&\bs\\  \bs&\bs&\bs \\
          \bs&\bs&\bs &\bs
\end{smallmatrix}
\stackrel{\pi_5^x}{\longrightarrow}
\begin{smallmatrix} \bs&\bs\\ \bs&\bs \\ \bs&\bs \\ \bs&\bs\\  \bs&\bs&\bs \\
          \bs&\bs&\bs &\bs
\end{smallmatrix}   \\
\stackrel{\pi_4^y}{\longrightarrow}
\begin{smallmatrix} \bs&\bs\\ \bs&\bs \\ \bs&\bs \\ \bs&\bs\\  \bs&\bs&\bs \\
          \bs&\bs&\bs &\bs&\bs 
\end{smallmatrix}
\stackrel{\pi_5^y}{\longrightarrow}
\begin{smallmatrix} \bs&\bs\\ \bs&\bs \\ \bs&\bs \\ \bs&\bs\\  \bs&\bs&\bs \\
          \bs&\bs&\bs &\bs&\bs&\bs
\end{smallmatrix}
\stackrel{\pi_6^y}{\longrightarrow}
\begin{smallmatrix} \bs&\bs\\ \bs&\bs \\ \bs&\bs \\ \bs&\bs\\  \bs&\bs&\bs \\
          \bs&\bs&\bs &\bs&\bs&\bs&\bs
\end{smallmatrix}
\stackrel{\pi_3^y}{\longrightarrow}
\begin{smallmatrix} \bs&\bs\\ \bs&\bs \\ \bs&\bs \\ \bs&\bs\\  \bs&\bs&\bs&\bs \\
          \bs&\bs&\bs &\bs&\bs&\bs&\bs
\end{smallmatrix}
\stackrel{\pi_4^y}{\longrightarrow}
\begin{smallmatrix}\bs&\bs\\ \bs&\bs \\ \bs&\bs \\ \bs&\bs\\ \bs&\bs&\bs&\bs&\bs\\
          \bs&\bs&\bs &\bs&\bs&\bs&\bs
\end{smallmatrix}
\end{multline*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% \input{Nilplaxique.tex}
\section{ \large Nilplactic kernel}

The two sets of indeterminates $X$ and $Y$ played a symmetrical role in the
preceding sections. We present now an example where it is not the case.

The $x_i$'s are commutative variables, and commute with the $y_j$'s. 
The $y_j$'s obey the nilplactic relations \cite{LS82b}, \cite{EG}
\bea \label{eq:Nil1a}
y_i^2 =0 & , & y_iy_{i+1}y_i = y_{i+1}y_iy_{i+1} \\  
\label{eq:Nil1b}
y_j y_i y_k =  y_j y_k y_i & , & y_i y_k y_j = y_k y_i y_j
  \quad (i<j<k) 
\eea 

These relations lift Coxeter relations. In particular, one has a notion of {\it
reduced word} (the image under $y_i\mapsto s_i$, the $s_i$'s  being simple
transpositions, is a reduced word).  For reduced words, 
Schensted's construction can be adapted to the nilplactic case.
The basic objects are now sequences of columns $w= [u_1,\ldots,\, u_\ell]$.

\begin{theorem} \label{th:Nil2}
There exists a bijection between $\ell$-uples of columns 
$w= [u_1,\ldots,\, u_\ell]$, such that $u_1\cdots u_\ell$ is reduced, 
and pairs of tableaux $t=P^{nil}(w)$,   
$t'=Q^{nil}(w)\in \{ x_1,\ldots,\, x_\ell \}^*$ 
of conjugate shape, $t$ being reduced.
\end{theorem}

The tableau $P^{nil}(w)$ is the unique tableau in the nilplactic class
of $u_1\cdots u_\ell$. Multiplying $P^{nil}(u_1\cdots u_{\ell-1})$ 
by the column $u_\ell$ increase  the shape  by a vertical string, which is
recorded as the set of boxes occupied by the letter $x_\ell$  in the tableau
$Q^{nil}(w)$.

For example, the product 
$\begin{smallmatrix}  5\\1   \end{smallmatrix}\cdot 
\begin{smallmatrix}   6\\5\\2\\1   \end{smallmatrix}\cdot 
\begin{smallmatrix}   4\\3\\2  \end{smallmatrix}$
produces the following sequence of tableaux and transposed shapes
$$ {
\begin{smallmatrix}        5\\1                \end{smallmatrix}  \atop
\begin{smallmatrix}  \blacksquare\\  \blacksquare&\blacksquare&\blacksquare\\
          1&1&\blacksquare&\blacksquare&\blacksquare   \end{smallmatrix}}  
\quad \longrightarrow\quad  {
\begin{smallmatrix}     6\\  5\\ 2&6\\ 1 &2  \end{smallmatrix}  \atop
\begin{smallmatrix}  \blacksquare\\  2&2&\blacksquare\\
          1&1&2&2&\blacksquare   \end{smallmatrix}}
\quad \longrightarrow\quad {
\begin{smallmatrix}     6\\  5\\ 4&6\\2&3\\ 1 &2&3  \end{smallmatrix}  \atop
\begin{smallmatrix}  3\\  2&2&3\\ 1&1&2&2&3 \end{smallmatrix}}$$

As before, instead of taking all possible reduced words, one wants to
characterize the possible pairs of tableaux which correspond to the expansion of
some kernel.

Instead of taking a general partition, we restrict to $\rho=[n,\ldots,\, 1]$
for simplicity, and define the {\it NilCauchy kernel} to be
\begin{multline} \label{eq:Nil3}
NilC(n) = 
\Bigl( (1+x_1 y_{n})\cdots(1+x_1y_1)\Bigr) \\
            \Bigl( (1+x_2 y_{n})\cdots (1+x_2y_2)\Bigr)\cdots
                     \cdots \Bigl( (1+x_n y_{n}) \Bigr)
\end{multline}

One notices that each block expands into a sum of columns~:
\bee \label{eq:Nil4}
   (1+x y_n)\cdots (1+x y_k)=  \sum_u x^{\ell(u)} u \ ,
\ee
sum over all possible columns $u$ in $y_n,\ldots,\, y_k$.

Thus, $NilC(n)$ can be studied using theorem(\ref{th:Nil2}). All the tools 
presented in the first sections can be adapted to the nilplactic case. For
example, there are two symmetric groups acting on a tableau,
permuting rows, or permuting columns. One can choose to record 
the columns appearing on the left, the columns appearing on the right,
the rows appearing upwards or the rows appearing downwards. One has 
indeed four possible
keys in the nilplactic case.  To describe $NilC(n)$, one needs the {\it bottom
key}, which is illustrated on the following example, which will be used 
in Figure 3.

\begin{equation*}
\petitematrice{5\cr4&6\cr2&3&5\cr1&2&3&4} \mapsto 
\petitematrice{5\cr4&6\cr 1&2&3&5\cr &1&2&4} \mapsto
\petitematrice{5\cr 1&2&4&6\cr && 3&5\cr &&1&2&4} \mapsto
 \petitematrice{5\cr 1&2&4&6\cr & 1&3&5\cr  &&2&4} \mapsto 
 \petitematrice{ 1&2&5&6\cr &&&4\cr &&& 1&3&5\cr  &&&&2&4} \mapsto
\petitematrice{ 1&2&5&6\cr & 1&4&5\cr &&& 3\cr &&& 2&4} \mapsto
 \petitematrice{ 1&2&5&6\cr & 1&4&5\cr && 3&4\cr &&& 2}
\end{equation*}

In that case, the bottom key is
\bee \label{eq:Nil5}
 \{\, [2],\, [2,4],\, [1,2,4],\, [1,2,3,4]\,  \} \ ,
\ee
with weight $\mu=[2,4,1,3]$ (number of occurrences of $1,2,3,4$ respectively).

We shall just state, but not prove, the following theorem. The proof requires 
giving specific properties of keys, vice-tableaux and strings in the nilplactic
case, which would carry us too far in a combinatorics of tableaux 
only slightly different than the one used in the first sections.

\begin{theorem} \label{th:Nil6} 
The NilCauchy kernel expands, in the nilplactic algebra, as
\bee \label{eq:Nil6}
  NilC(n) \cap {\mathcal Reduced\ Words} \equiv 
 \sum_t  K(\mu(t),\, X)\ t   \ ,
\ee
sum over all tableaux (read by columns, and which are reduced words)
that appear as subwords of
$$ (y_n\cdots y_1)\,  (y_n\cdots y_2) \cdots (y_n) \ ,$$
$\mu(t)$ being the weight of the bottom key of $t$.
\end{theorem}

Figure 3 shows all the terms in $NilC(6)$ belonging to  the nilplactic 
class of
$t=
\begin{smallmatrix} 5 \\ 4&6 \\  2&3&5\\ 1&2&3&4
\end{smallmatrix}$. 
They constitute a crystal base,
the crystal operations being the jeu de taquin on two consecutive columns. 

The factor of $t$ in theorem (\ref{th:Nil6})
is obtained by replacing each vertex by $x_1^{m_1} x_2^{m_2}\cdots$,
$m_i$, $i=1,\ldots,\, 4$, being the number of boxes in column $i$,
and taking the sum of these monomials.

Thus the study of $NilC(n)$ essentially reduces to recognize that the
intersection of the nilplactic class of any tableau with the set of reduced words
appearing in the expansion of $NilC(n)$ is isomorphic to some
crystal base.

Theorem (\ref{th:Nil6}) implies the expansion of Schubert polynomials
in terms of Demazure characters given in \cite{NCSchub}. 

Projecting expression  (\ref{eq:Nil6}) on the NilCoxeter algebra defined by
\bee  \label{eq:Nil7}
y_i^2 \eqCox 0 \quad , \quad y_iy_{i+1}y_i \eqCox  y_{i+1}y_iy_{i+1} \quad ,
\quad y_i y_k \eqCox y_k y_i \, ,\, |k-i|\neq 1 
\ee
one gets  Fomin-Kirillov expansion \cite{FK} of the nilCoxeter kernel
involving Schubert polynomials (see also \cite{LLT}, \cite{FG}) 
\bee  \label{eq:Nil8}
 NilC(n) \eqCox \sum_{\sigma \in \S(n+1)} Schub_\sigma(X)\, \sigma \ .
\ee
For example
$$NilC(2) \eqCox x_1^2x_2 \, s_2s_1s_2 + x_1^2\, s_2s_1 + x_1x_2\, s_1s_2 
+ x_1s_1 + (x_1+x_2)\, s_2 +1 \ .$$




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% \input{Double.tex}
\section{ \large Double characters}

Having a combinatorial interpretation of characters in terms of tableaux
and crystal graphs, it is easy to modify their definition to handle
functions in two sets of indeterminates.  Let $w$ be a word in 
non commuting letters $\binom{j}{i}$, and let $w_x$ be the word obtained
by sending $\binom{j}{i}$ to $x_i$, $\forall j$.  
We have seen that  a left 
crystal operator $f_i$ sends a word $w_x$ to 0, or change in it 
a single precise occurrence of the letter $x_i$ into a $x_{i+1}$.

Define accordingly the action of $f_i$ on $w$ to be
\bee
\left\{\  
\begin{matrix}  f_i(w_x) =0 &\Rightarrow &f_i(w) =0 \\
   f_i(w'_x\, x_i\, {w''}_x) = w'_x\, x_{i+1}\, {w''}_x &\Rightarrow &
   f_i\bigl(w'\, \binom{j}{i}\, w'' \bigr) = w'\, \binom{j+1}{i+1}\, w''  
\end{matrix}
\right.
\ee

One extends similarly the action of $\pi_i$ from
words in the $x_j$'s to words in the  $\binom{j}{i}$'s. 

Let $\l\in \N$ be a dominant weight, $\ss\in \S_n$, and 
$\mu= \l^\ss$.  Let now
\begin{equation*}
 w^\l := \begin{matrix} {\bl 1|n|} &\ldots & {\bl \l_n|n|}\\[3pt]
  {\bl 1|n-1|} &\ldots & \ldots &{\bl \l_{n-1}|n-1|}\\
     \vdots & & &\vdots                             \\
  {\bl 1|1|} &\ldots & \ldots&\ldots  &{\bl \l_1|1|}
\end{matrix}  \quad \text{ (read by rows)}
\end{equation*}


Let $s_i\cdots s_j= \ss$ be a reduced decomposition of $\ss$ 
($\ss$ can be supposed to be minimum in its class modulo the stabilizer
of $\l$).   

Then one defines the {\it double crystal base} $\cBB(\mu)$ to be 
\bee
  \cBB(\mu) = \pi_i\cdots \pi_j \left( w^\l \right)
\ee

This sum 
of words  does not depend on the choice of the reduced decomposition of
$\ss$, as shows the characterization given in the next theorem.

Given a tableau $t\in X^*$, let $\varphi(t)$ be the tableau in the ${\bl j|i|}$'s
obtained by changing an entry $x_i$ in row $r$, column $c$, into
${\bl i+c-r|i|}$, and extends $\varphi$ by linearity to sums of tableaux.
Considering $\cB(\mu,X)$ to be a sum of tableaux rather than a set, one has~:

\begin{theorem}
Let $\l\in \N$ be a dominant weight,  $\mu$ be  a permutation of $\l$.
then
\bee
  \varphi\Bigl(\cB(\mu,X)  \Bigr) = \cBB(\mu)  
\ee

\end{theorem}

The proof consists just in noticing that the action of
any $f_i$ preserves the differences between subscripts
and superscripts:  $ {\bl j|i|} \mapsto {\bl j+1|i+1|}$.
Thus, recording the initial superscripts in $w^\l$, one can 
as well erase them and operate on tableaux in the $x_i$'s,
reintroducing the superscripts at the last step.

Now the entries of the initial tableau $w^\l$ are of the type
$ {\bl i+c-r|i|}$, for a letter in column $c$, row $r$. \QED 

For example, 
$$
\cBB([1,0,2]) =
 \bigyoung{ {\bl 1|2|}\cr   {\bl 1|1|} &  {\bl2|1|} \cr }
+  \bigyoung{ {\bl 1|2|}\cr  {\bl 1|1|} &  {\bl 3|2|}\cr}
+ \bigyoung{ {\bl 2|3|}\cr   {\bl 1|1|} &  {\bl2|1|} \cr } 
+ \bigyoung{ {\bl 1|2|}\cr  {\bl 1|1|} &  {\bl 4|3|} \cr}
+ \bigyoung{ {\bl 2|3|}\cr  {\bl 1|1|} &  {\bl 4|3|} \cr}
$$


The characters $K(\mu,X)$ are obtained by taking the sum 
of the commutative images of the tableaux in $\cB(\mu,X)$.  
Similarly, one defines a {\it double character}
$\K(\mu,X,Y)$ in two sets of commutative indeterminates by
taking the image of a word  under the morphism
$$ {\bl j|i|} \mapsto (x_i-y_j) \ . $$
By construction, one recovers $K(\mu,X)$ by specializing the $y_j$'s to 0~:
$$ K(\mu, X) = \K(\mu,X,0) \ .$$
But now, the polynomials $\K(\mu,X,Y)$ have some
specific properties (e.g. vanishing properties) that make them easier to
characterize than the polynomials $K(\mu,X)$. 

For example, if $\mu$ is the {\it code} of a {\it vexillary} permutation $\ss$
(cf. \cite{Mac_Schub}), then one can show (but we shall not) that 
$\K(\mu,X,Y)$ coincides with the double Schubert
polynomial $Schub_\ss$.  As such, it is the only polynomial $P(X,Y)$ 
satisfying the following conditions~:

\begin{itemize}
\item It belongs to the linear span of monomials $x^\nu$,
$\nu\leq[n-1,\ldots,0]$, of degree $\leq \ell(\ss)$.
\item $P(Y^\zeta, Y)=0 \quad \forall \zeta\neq \ss$, $\ell(\zeta)\leq \ell(\ss)$.
\item $P(X,Y) = x^\mu + \cdots$ ,
  \end{itemize}
where $X=Y^\zeta$ stands for the specialization $x_1=y_{\zeta_1}$, 
$x_2=y_{\zeta_2},\, \ldots $, and $x^\mu + \cdots$ means that the coefficient of 
$x^\mu$ in $P(X,Y)$ is 1.  

In that case, $\K(\mu,X,Y)$ inherits properties of Schubert polynomials
which would be difficult to check directly on the list of tableaux composing
$\cBB(\mu,X,Y)$: 
 the two alphabets $X$ and $Y$ play a symmetrical role in Schubert polynomials;
these polynomials have partial symmetries in consecutive $x_i$'s. 
Schubert polynomials, moreover, satisfy vanishing properties induced by
the Ehresmann-Bruhat order \cite{Mac_Schub} which are more general 
than the vanishing of the individual tableaux composing $\cBB(\mu)$. 


A special case, where $\mu$ is of the type
$$ \mu= [\mu_1 \leq \mu_2\leq \cdots \leq \mu_r , 0,0,\ldots] $$
is known since long. Indeed, specializing 
$y_1=0$, $y_2=1$, $y_3=2, \, \ldots$,  one has that 
$\K(\mu,X,\{0,1,2,\ldots\})$ is a {\it factorial Schur function}
of Biedenharn and Louck \cite{BiedLouck} in the variables
$x_1,\ldots,\, x_r$. The more general case were the $y_j$'s 
are not specialized has been obtained by several authors
( \cite{Mac_Loth}, \cite{Molev}, \cite{OO}), but was
already used, following Giambelli, 
 in \cite{La74b} to describe Schubert cycles for Grassmann
varieties.   











%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\medskip
\small
\noindent {\bf Some historical comments.}\ 
I shall end this text evoking why Schur's work was personally important 
for me, even before I was aware of it, restricting to my work in 1974. 

In \cite{La74a} I considered symmetric functions as operators on isobaric
determinants, Schur functions were not named and were noted $t_I$.
Shortly after \cite{La74b}, multi-Schur functions appeared as Schubert cycles,
and were denoted $D_I$. In \cite{La74c} Schur functions were
``Schur-Littlewood'' functions, sums of all the tableaux of
a given shape \cite{Littlewood}. 
Finally, in \cite{La74d}, Schur functions became elements
of the plactic algebra, this interpretation being needed to explain
their products. It took me some time to pass from $t_I$ to $S_i$,
and from Schur functions to Schur functors (in my thesis \cite{La77}). 
I could add
that all my further work with M.P. Sch\"utzenberger
can be interpreted as reflecting about   
what remains of Schur functions when allowing non-symmetry 
\cite{LS82a} and non-commutativity \cite{NCSchub}.  

\normalsize
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A.~Okounkov and G.~Olshanski.
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Strasbourg 1975},
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\end{thebibliography}

\bigskip 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% dimension de la boite contenant les sommets du graphe
%%%
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\vskip 20pt
\centerline{\fbox{ \bf Figure 1}}

\setlength{\fboxsep}{3mm}
\medskip
\centerline{\fbox{ {\bf Crystal base} $\cB([2413], Q)$ \quad with  
$ Q=\begin{smallmatrix}
  4\\ 3&4\\ 2&3&3\\ 1&1&1&3
\end{smallmatrix}$  }} 

%\centerline{\fbox{ Crystal base $\cB([2413], Q)$}\quad with
%$ Q=\begin{smallmatrix}
%  4\\ 3&4\\ 2&3&3\\ 1&1&1&3
%\end{smallmatrix}$  }
 

\vfill\eject


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\medskip 
\centerline{\large Figure 2:  Crystal Cobase $\cCB([2413])$} 
\vfill\eject

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

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    \centerline{\bf Figure 3}
 
   \medskip 
    A copy of the crystal base ${\mathcal{B}}([2,4,1,3])$,
    with reduced words as vertices.

    Read each vertex as a product
    of four columns, interpreting  it
    as a reduced subword  of
     $ (6\cdots 1) (6\cdots 2) (6\cdots 3) (6\cdots 4)$.

    The arrows are given by the jeu de taquin on columns 1,2,
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  }}
% \end{center}
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\end{document}

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Read each tableau as a product
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 $ (6\cdots 1) (6\cdots 2) (6\cdots 3) (6\cdots 4)$.

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columns 2,3 or columns 3,4.
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\medskip
\centerline{ Figure 2: \large Crystal base $\cCB([2413])$}
\vfill\eject

