
%% ProcCrac.tex   Expose Cracovie sept 2000. version Proceedings 
% Corrections 28 sept 2000
%
\documentclass[a4paper,12pt]{article}
\usepackage{amsmath,amsfonts,amsthm}

\begin{document}
\title{\bf Youngs's Representations
 of the Symmetric Group }

\author{ 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 }

\date{September 2000}
\maketitle

\def\s{\scriptstyle }
\def\a{\alpha}
\def\b{\beta}
\def\l{\lambda}
\def\L{\Lambda}
\def\G{\Gamma}
\def\ss{\sigma}
\def\d{\partial}
\def\P{\psi}
\def\bu{$\bullet$\quad}

\def\S{{\mathfrak S}}
\def\Sym{{\mathfrak Sym}}
\def\LR{\Longleftrightarrow}
\def\lr{\leftrightarrow}

\def\N{{\mathbb N}}
\def\Z{{\mathbb Z}}
\def\A{{\mathbb A}}
\def\B{{\mathbb B}}
\def\C{{\mathbb C}}
\def\un{{\overline 1}}

\def\carre{\clubsuit}
\catcode`\@=11
\def\petitematrice#1{\null\vcenter {\normalbaselines \m@th
\ialign {\hfil $##$\hfil &&\thinspace  \hfil $##$\hfil\crcr
\mathstrut \crcr \noalign {\kern -\baselineskip } #1\crcr
\mathstrut \crcr \noalign {\kern -\baselineskip }}}}

\def\moyennematrice#1{\null\vcenter {\normalbaselines \m@th
\ialign {\hfil $##$\hfil &&\   \hfil $##$\hfil\crcr
\mathstrut \crcr \noalign {\kern -\baselineskip } #1\crcr
\mathstrut \crcr \noalign {\kern -\baselineskip }}}}

\newdimen\squaresize \squaresize=14pt
\newdimen\thickness \thickness=0.7pt

\def\square#1{\hbox{\vrule width \thickness
   \vbox to \squaresize{\hrule height \thickness\vss
      \hbox to \squaresize{\hss#1\hss}
   \vss\hrule height\thickness}
\unskip\vrule width \thickness}
\kern-\thickness}

\def\vsquare#1{\vbox{\square{$#1$}}\kern-\thickness}
\def\blank{\omit\hskip\squaresize}

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

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

\begin{abstract}  We describe the different matrices, due to Young,
representing the symmetric group, by reading the same graph 
with various labellings. Orthogonal idempotents are obtained
in the like manner. The only mathematical tools needed for these
constructions are
 comparison of integers and addition of vectors.
\end{abstract}

Young's purpose was to decompose functions -- 
nowadays one would say \lq\lq tensors" --
into components according to their type of symmetry. In other words, he
wanted to generalize the decomposition
$$  f(x,y) = {1\over 2}\bigl( f(x,y) + (f(y,x) \bigr) +
{1\over 2}\bigl( f(x,y) - (f(y,x) \bigr)   $$
for functions of two variables to the case of more variables (cf. [Ro]).

He first rewrote the preceding equation into
$$ 1 = {1\over 2}(1+\sigma)  + {1\over 2}(1-\sigma)  \ ,  $$
where $\sigma$ is the transposition of $x,y$, and thus moved to the study of
equations in the group algebra of the symmetric group. 

He produced different families of idempotents, and, as 
 a by-product, he obtained various
matrices representing the symmetric group (we shall refer to
[JK], [Ru] for the traditional approach). 

A new impetus to the description of representations 
was given by the notions of {\it Gelfand-Tsetlin basis} (cf. [OV]), of 
maximal commutative subalgebras and of Yang-Baxter equation,
which allow to reinterpret the constructions of Young in a simpler
manner.  As a matter of fact, the only mathematical tools which we 
require in this text are

\bu Given two numbers $a$, $b$,  
 decide whether $a\geq b$ or not

\bu compute their difference $a-b$.

\bu add vectors and test whether the result is a permutation.


\medskip
 Now, introducing various initial elements  associated to
a given partition, ones generates labelled graphs from which one reads
the matrices representing simple transpositions,
as well as orthogonal idempotents. 
One also can directly write  other 
matrices of representation from addition of vectors.

We shall illustrate all these constructions on the partition [3,2],
while giving the precise initial data for any partition.
There will be some variations, to simplify labellings, in the way
a partition is represented graphically by  diagrams of boxes
in the different quarters of the plane.

When using the term \lq\lq word" instead of integral vector, 
we refer to computations in the free algebra,
as a proper way of handling
such objects as Young tableaux (cf. [LLT2]). For example,
$[3,5,1,2,4]$ can be seen as the word $x_3 x_5 x_1 x_2 x_4$, as well as
 the Young tableau $\petitematrice{3&5\cr 1&2&4}$\ .


\smallskip 
For a start,
take the word $[1,0,2,1,0]$, vertical reading of 
$\petitematrice{1&2\cr 0&1\cr &0} $,  and given any two adjacent letters,
$a,b$, transpose them iff $a> b$.

\smallskip 
Iterating and recording positions by colors $\ss_1,\ss_2,\ldots$, 
one gets a colored graph.
$$
\begin{matrix}
&& 1\ 0\ 2\ 1\ 0  \cr
&& \Big\downarrow \, {}^{\sigma_2}\cr
&&  1\ 2\ 0\ 1\ 0  \cr
&  {}^{\sigma_1}  \swarrow &&\searrow {}^{\sigma_3} \cr
2\ 1\ 0\ 1\ 0   &&&&  1\ 2\ 1\ 0\ 0   \cr
& {}_{\sigma_3} \searrow  && \swarrow  {}_{\sigma_1} \cr
&& 2\ 1\ 1\ 0\ 0    
\end{matrix}  \eqno(\G1)$$

For a general partition $\b =[\b_1,\b_2,\ldots]$ 
(written increasingly), the starting word 
is $[\cdots (\b_3-1,\ldots,0) (\b_2-1,\ldots,0)(\b_1-1,\ldots,0) ]$
and can be visualized (as we have done for $[2,3]$)
on the north-east diagram with column lengths
the partition, each bottom box being occupied by 0, 
columns being occupied by consecutive numbers.


Now, we keep the same graph, but interpret each word as describing the levels 
occupied by letters $1,2,\ldots$ in the diagram of [3,2]:
$$\begin{matrix}
&& \young{3\cr 2&5\cr1&4\cr}  \cr
&& \Big\downarrow \, {}^{s_3}\cr
&&  \young{4\cr 2&5\cr1&3\cr}  \cr
&  {}^{s_4}\! \swarrow &&\searrow \!{}^{s_2} \cr
\young{5\cr 2&4\cr1&3\cr}   &&&&  \young{4\cr 3&5\cr1&2\cr}   \cr
& {}_{s_2}\! \searrow  && \swarrow \! {}_{s_4} \cr
&& \young{5\cr 3&4\cr1&2\cr}  \cr  
\end{matrix}  \eqno(\G2) $$

The vertices of the graph are now exactly all the {\it standard Young tableaux}
of a given shape, and the edges are the transpositions of consecutive
letters.  The labelling of edges has been changed $s_i \lr \ss_{n-i}$
(we could also have read positions from the right).


\medskip We did not record values which were exchanged.
Label vertices differently once more, and label arrows by the inverse
of the {\bf differences} of the values exchanged
($\un$ stands for -1)~:

$$\begin{matrix}
&& 0\ \un\ 2\ 1\ 0  \cr
&& \Big\downarrow \, {1\over 3}\cr
\noalign{\kern 5pt}
&&  0\ 2\ \un\ 1\ 0    \cr
& {1\over 2} \swarrow &&\searrow {1\over 2} \cr
 2\ 0\ \un\ 1\ 0   &&&&   0\ 2\ 1\ \un\ 0   \cr
& {1\over 2} \searrow  && \swarrow {1\over 2} \cr
&& 2\ 0\ 1\ \un\ 0    
\end{matrix}  \eqno(\G3) $$

\def\long{\hbox to 1.3cm {\rightarrowfill}  }
The initial vector is the \lq\lq vector of contents", that is, the diagonal
positions of the boxes in the diagram 
$$ \raise -10pt \hbox{$\young{0&1&2 \cr \un &0\cr}$}
 \quad {\buildrel{\s read}\over \long} \quad 
[0,1,2,\un,0] \quad 
{\buildrel{\s reverse}\over \long} \qquad  
[0,\un,2,1,0] $$

\medskip 
How to read matrices from the last graph~?

The underlying vector space has a basis coded by the vertices of the graph. 
To represent  any simple transposition $s_i$, one first
erase all edges which are not labelled by $s_i$. One is left 
with isolated vertices (corresponding to 1-dimensional representations
of $\S_2$), and pairs of vertices connected by an edge, 
corresponding to 2-dimensional representations.  
In this last case, if $\b$ is the parameter written on the edge, then 
the matrix representing $s_i$ on the subspace generated by the two vertices is 
$$\left[\, \petitematrice{ -\b & \sqrt{1-\b^2}\cr
        \sqrt{1-\b^2} & \b\cr}\, \right] \ .$$

In the one-dimensional case, if $i,i+1$ is a subword of the vertex, then
the restriction of the representation is trivial (i.e. the submatrix is 1),
otherwise it is the alternating representation (submatrix $= -1$).

The full matrix $M_i$ representing $s_i$ is composed of these elementary matrices.
The reader can check that the elements $\neq \pm 1$ of the following matrices
constitute elementary $2\times 2$ matrices with entries $\pm \b$ or 
$\sqrt{1-\b^2}$ given by the graph $(\Gamma3)$.

$$
M_1= \left[
\begin{matrix}
-1& 0& 0& 0& 0\cr\cr 0& -1& 0& 0& 0\cr\cr 0& 0& 1&
0& 0\cr\cr 0& 0& 0& 1& 0\cr\cr 0& 0& 0& 0& 1
\end{matrix}  \right] 
\qquad M_2= \left[ 
\begin{matrix}
 {1\over 2}& 0& 0& 0& {\sqrt{3}\over 2}
\cr\cr 0& {1\over 2}& {\sqrt{3}\over 2}
& 0& 0\cr\cr 0& {\sqrt{3}\over 2} & {-1\over 2}& 0& 0\cr\cr 0& 0& 0& 1& 
0\cr\cr {\sqrt{3}\over 2} & 0& 0& 0&  {-1\over 2} 
\end{matrix}\right]$$

\medskip
$$M_3 = \left[
\begin{matrix}
-1& 0& 0& 0& 0\cr\cr 0& 1& 0& 0& 0\cr\cr 0& 0& {1\over 3}& 
 {\sqrt{8}\over 3} & 0\cr\cr 0& 0& {\sqrt{8}\over 3} & -{1\over 3}& 
0\cr\cr 0& 0 & 0& 0& 1
\end{matrix} \right]  
\qquad 
M_4= \left[ 
\begin{matrix}
{1\over 2}& {\sqrt{3}\over 2}
& 0& 0& 0\cr\cr {\sqrt{3}\over 2}& -{1\over 2}
& 0& 0& 0\cr\cr 0& 0& -{1\over 2}& 0& {\sqrt{3}\over 2}\cr\cr 0& 0& 0& 1
& 0\cr\cr 0& 0& {\sqrt{3}\over 2} & 0& {1\over 2}
\end{matrix}  \right]
$$

Young's theorem is that these matrices represent the symmetric group,
i.e. satisfy the braid relations. 

Only the relations $M_i M_{i+1}M_i = M_{i+1}M_i M_{i+1}$ 
are not immediate, but their proof
 amounts to check Yang-Baxter equation (see below), and finally reduces
to the fact that the \lq\lq contents" choosen by Young are a
\lq\lq distance" satisfying the additivity: 
$$ (a-b) +(b-c)= (a-c)$$ 
as required by Yang-Baxter's rule of choosing spectral parameters
(cf. [Ya], [LLT1]).

To see that Young's matrices 
can be interpreted as equations in the group algebra of $\S_n$,
 one needs now another reading of the graph. 
The vertices, whatever equivalent labelling is choosen
(tableaux, contents, etc.), will be interpreted
as element of the group algebra of $\S_n$, and an oriented edge of color
$i$ and parameter $\b$ will be

\smallskip
\centerline{\bf \lq\lq multiplication by $(s_i+\b) \frac{1}{\sqrt{ 1-\b^2}}$" }

\medskip Non-physicists usually require $\b \neq {1\over 0},\, 1 $.

The braid relation $M_i M_{i+1}M_i = M_{i+1}M_i M_{i+1}$ is now seen
graphically as

\setbox2=\hbox{$\bullet$}
\setbox31=\hbox{$ {1\over v} $}
\setbox32=\hbox{$\, \nearrow$}
\setbox3=\hbox{\kern 8mm\box31\box32} 
\setbox41=\hbox{$ \nwarrow$}
\setbox42=\hbox{$\, {1\over u}$}
\setbox4=\hbox{{\box41 \box42}\kern 8mm}
\setbox5=\hbox{$\bullet$}
\setbox6=\hbox{$\bullet $}
\setbox7=\hbox{${1\over u+v} \, \bigg\uparrow$\kern 8.8mm}

\setbox8=\hbox{\kern 8.8mm$\bigg\uparrow \, {1\over u+v}$}
\setbox9=\hbox{$\bullet$}
\setbox10=\hbox{$\bullet$}
\setbox11=\hbox{\kern 8mm$ {1\over u} \ \nwarrow$}
\setbox12=\hbox{$ \nearrow \ {1\over v} $\kern 8mm}
%\setbox13=\hbox{$\bullet$}
{\def\quad{\kern -2mm}
$$ 
\begin{matrix}    &\copy2    \cr \box3 & &\box4\cr
\box5  & &\box6 \cr\noalign{\vskip 1mm}\box7  &
&\box8 \cr\noalign{\vskip 1mm}\box9  & &\box10 \cr
 {\raise -3pt\box11} & &{\raise -3pt \box12} \cr & \box2 \cr  
\end{matrix}$$
}

that is ({\it Yang-Baxter equation})

$$ (s+ {1\over u})\, (s'+{1\over u+v})\, (s+ {1\over v})
= (s'+ {1\over v})\, (s+{1\over u+v})\, (s'+ {1\over u}) \eqno(\star)$$ 

which is checked directly in the group algebra of $\S_3$.

\smallskip Instead of using the basis $1,s,s',ss', s's, ss's$ of
the group algebra of $\S_3$, one prefers the 
Yang-Baxter basis  \quad $1,\ (s+ {1\over u}),\ (s'+ {1\over v}),\
(s+ {1\over u})\, (s'+{1\over u+v}),\hfill\break
 (s'+ {1\over v})\, (s+{1\over u+v}),\
(s+ {1\over u})\, (s'+{1\over u+v})\, (s+ {1\over v})$,
the change of basis being triangular.

\medskip 
We shall  keep the same edges, but instead of starting from 
a bottom element equal to 1, 
we shall use a more sophisticated element to get irreducible representations.

\smallskip Let us first notice that 
there are 2 distinguished elements in $\C[\S_n]$, which correspond
to the two 1-dimensional representations:

$$ \left\{\  
\begin{matrix}
 \carre &= &\sum_{\ss\in \S} \ss \cr
                     \nabla &= &\sum_{\ss \in \S} (-1)^{\ell(\ss)}\, \ss 
\end{matrix} \right. $$

One has more general elements by taking sums over a Young subgroup
(that is, a direct product of symmetric groups diagonally embedded).

\smallskip  Thus for any composition $J$ of $n$ (denoting the order
of the successive symmetric group composing a Young subgroup), one has
two elements
$$\carre_J \ ,\  \nabla_J \ . $$

Studying the product (that he denoted $P(t)N(t)$) of the sum of all
permutations preserving the rows of a given tableau, multiplied by the
alternating sum of the permutations fixing the columns of the same tableau,
Young ([Y], QS1)
found 1-dimensional  spaces from which it is easy 
to write idempotents. This construction
can be rewritten  as follows.

\smallskip\noindent 
Lemma(Young).\  {\sl For two conjugate compositions $I$, $J$, then
 the spaces \hfill\break  $\carre_J \, \C[\S_n]\, \nabla_I$ 
and $ \nabla_I \, \C[\S_n]\, \carre_J$  are 1-dimensional.}

\medskip Take $J$ to be a partition $\l$, and let $\mu$ 
be the conjugate partition.
Then 
$$ \carre_\l \ \C[\S_n]\ \nabla_\mu\ \C[\S_n]\ \carre_\l $$
is still 1-dimensional.  It contains a unique idempotent $\xi_\l$. 
Similarly 
$$  \nabla_\mu \carre_\l \ \C[\S_n]\ \carre_\l\ \C[\S_n]\ \nabla_\mu $$
contains another unique idempotent $\xi_\l^{top}$. 

\smallskip To have a precise expression of these two idempotents (i.e. to
give their decomposition in the basis of permutations), one needs to
handle double cosets. Non reduced expressions are easier to write :
Young took, up to scalars,
 $P(t) N(t) P(t)$ and $N(t') P(t') N(t')$, with $t$ and $t'$
the bottom and top tableaux respectively (in the second graph). 
One can also use a coding of elements of the group algebra as
polynomials to obtain compact expressions (cf. [La?]).


\smallskip Now we are ready to describe the orthogonal idempotents.
Take the graph with vertices labelled by the contents 
of Young tableaux 
of shape $\l$, put $\xi_\l^0:= \xi_\l$ 
in the bottom place and play the Yang-Baxter game ~:
a path in the graph means a product of 
factors $(s_i+\b) /\sqrt{1-\b^2}$, as indicated by the labelling of edges. 

$$
\begin{matrix}
&& \xi_\l^4\quad  \cr\cr
&& \Big\uparrow\! s_3\!+\! {1\over 3}\cr\cr
&&  \xi_\l^3\quad    \cr\cr
& s_4+{1\over 2} \nearrow &&\nwarrow s_2\!+\!{1\over 2} \cr\cr
\xi_\l^1   &&&&   \xi_\l^2   \cr\cr
& s_2\!+\!{1\over 2} \nwarrow  && \nearrow s_4\!+\! {1\over 2} \cr\cr
&& \xi_\l^0\quad  \cr  
\end{matrix}  \eqno(\G4) $$

\medskip   For any pair of vertices $t,t'$, let $\phi(t,t')$ be the evaluation
of a path from $t$ to $t'$ in the Yang-Baxter graph (with proper
normalization, i.e. adding the factors $1/\sqrt{1-\b^2}$ that we have not
written). 

Define $$e_{t,t'}:= \phi(t,t^0)\, \xi_\l\, \phi(t^0,t') $$

Then the $e_{t,t'}$ are {\bf Young's orthogonal units}, the $e_{t,t}$ are 
{\bf Young's orthogonal idempotents} 
([Y],QS4); they are a complete system of mutually
orthogonal idempotents. Thus, the orthogonal idempotents
(written $\xi^i_\l$ instead of $e_{t,t}$) are obtained from
the bottom one by following any path (even non reduced), and multiplying
left and right, at each step, by the factor $(s_i+\b) /\sqrt{1-\b^2}$ or
$(s_i-\b) /\sqrt{1-\b^2}$ according to the orientation of the edge. 

One could also have started from the top of the graph, using the idempotent
$\xi_\l^{top}$ instead. As we have said,
both these idempotents were explicited by Young.
He obtained the others by a not too easy orthonormalization process
that we shall bypass ([Y], QS3).
Notice that expanding the idempotents
 in the basis of permutations gives the entries of
Young's matrices.

\smallskip Modern proof is very easy, because it was realized by Jucys [Ju] 
that Young had found in fact a 
  {\it Gelfand-Tsetlin basis} of the group algebra.  The $e_{t,t'}$
are simultaneous two-sided eigenvectors of the {\it Jucys-Murphy} elements
which constitute a maximal commutative subalgebra of $\C[\S_n]$
( Jucys-Murphy elements are just sums of transpositions: 
  $J_k:= \sum_{i<k} (i,k)$).

In fact, Thrall  gave an induction rule for the $e_{t,t}$ which renders
their characterization immediate. Denote by $u$ the tableau obtained by erasing
the last letter $n$ of $t$. Then one has (up to normalization; [Th])
$$ e_{t,t} = e_{u,u}\, P(t) N(t)\, e_{u,u} \eqno(\star\star) $$
This is checked by having the Jucis-Murphy elements act on both 
sides of $(\star\star)$. 
For $J_1,\ldots, J_{n-1}$, it is known by induction on $n$ that
$e_{u,u}$, and thus $e_{t,t}$, is a two-sided eigenvector.
To add the action of last element $J_n$, 
one can as well take $J_1+\cdots +J_n$,
but it  is a central element, and thus easy to handle. 

\medskip I shall finish by a
last description, more in spirit of Young who, as I have already said,
 was motivated by the
classification of polynomials according to their type of symmetry. 

Young ([Y], QS3)
described  what was later called \lq\lq Specht modules"
[Sp], [Ga],  by having 
the symmetric group act on products of Vandermonde determinants associated
to tableaux.  One can simplify his construction by having recourse to the
 quotient ring $\C[x_1,\ldots,x_n]/\Sym_+$ 
($\Sym_+$ = ideal generated by symmetric functions without constant term).
The {\it Specht polynomials} are in this interpretation replaced by
mere monomials. 

The quotient ring, as a representation of $\S_n$,
 is isomorphic to the regular representation, i.e., to
$\C[\S_n]$.  Thus all the constructions given above can be
interpreted in terms of polynomials. Not only it is 
 easier to manipulate  polynomials in commutative indeterminates,
but moreover, one has many tools on polynomials which can be put into use.
I shall refrain from introducing {\it divided differences}, though they
are perfectly fit to describe representations and shall only use
a canonical scalar product (with $\Delta:=\prod_{i<j}(x_i-x_j)$)~: 
$$(f,g) := \sum (-1)^{\ell(\ss)}\, (fg)^\ss \, {1\over \Delta} \,
|_{x_1=0=x_2=\cdots}  $$  
(more generally, one does not need to specialize variables to 0, one should
work in the ring of polynomials as a free module over the ring of symmetric
polynomials, with quadratic form 
$ \sum (-1)^{\ell(\ss)}\, (fg)^\ss /\, \Delta$).

Explicitely, the scalar product boils down to
$$ (x^u,x^v)=(x^{u+v},1)=\left\{ 
\begin{matrix}
(-1)^{\ell(\ss)} &if\ u+v &=[\ldots 210]^\ss\cr
       0 & {\rm otherwise} \cr
\end{matrix} \right. $$
that is, one has to test whether the vector $u+v$ is a permutation
of \\ 
$[n-1,\ldots, 1,0]$ or not, and if so, keep the sign of the permutation.

The reader will not be surprised that we have to change the labelling
of vertices of our graph : the 
 top vertex is now $[00033]$ (in general, it would be 
$[0^{\l_1}\, \l_1^{\l_2}\, (\l_1+\l_2)^{\l_3}\ (\l_1+\l_2+\l_3)^{\l_4} \cdots]$
for a decreasing partition $\l$)~:

$$\begin{matrix}
&& [00033]  \cr
&& \Big\downarrow  {}^{s_3}\cr
&& [00303]  \cr
&  {}^{s_4}\!\!  \swarrow &&\searrow\! {}^{s_2} \cr
[03003]   &&&& [00330]   \cr
& {}_{s_2}\!\! \searrow  && \swarrow\!  {}_{s_4} \cr
&& [03030]  \cr  
\end{matrix}  \eqno(\G5)$$

Interpret the vertices as {\bf exponents} of monomials. Then one has 
the following theorem, for which we refer to [CLL]
( a priori one should expect a \lq\lq permutation representation",
but evaluating modulo $\Sym_+$ entails irreducibility -- in characteristics 0,
of course). 

\smallskip\noindent 
Theorem. \sl For a given partition $\l$, then the set of monomials $x^u$
read from the
graph associated to $\l$ constitute
 a basis of a copy of the irreducible representation
of $\S_n$ of index $\l$, the symmetric group
acting by permutations of the indeterminates $x_i$'s.   \rm

\medskip Matrices of representations, or decompositions of elements in
the standard basis ({\it straightening}, [DKR], [Ga]) reduce to computing
scalar products.  


One can in fact describe the adjoint basis,
which generates the \lq\lq Specht representation". 
As a first approximation, it is sufficient to take the 
\lq\lq complementary monomials", applying the same transpositions
to the top vector $[ (0,\ldots,\l_1-1), (0,\ldots, \l_2-1), \ldots]$
(or, equivalently,
 replacing repeated letters by $0,1,2,\ldots$, from left to right)


$$\begin{matrix}
&& [00033]  \cr
&& \Big\downarrow \, {}^{s_3}\cr
&& [00303]  \cr
&  {}^{s_4}\!\!  \swarrow &&\searrow\!\! {}^{s_2} \cr
\noalign{\kern 5pt}
&[03003]   && [00330]   \cr
& {}_{s_2}\!\! \searrow  && \swarrow\!\!  {}_{s_4} \cr
&& [03030]  \cr  
\end{matrix} \quad ; \quad 
\begin{matrix}
&& [01201]  \cr
&& \Big\downarrow \, {}^{s_3}\cr
&& [01021]  \cr
&  {}^{s_4}\!\!  \swarrow &&\searrow\!\! {}^{s_2} \cr\cr
\noalign{\kern 5pt}
&[00121]   && [01012]   \cr
& {}_{s_2}\!\! \searrow  && \swarrow\!\!  {}_{s_4} \cr
&& [00112]  \cr  
\end{matrix}  \eqno(\G6) $$

\smallskip  It is not difficult to check that 
the matrix of scalar products between these two bases is 
triangular, with $\pm 1$ in the diagonal.
One has choosen the second top vector to be such that it sums with the
first one is $[0123\ldots]$, and acting by the same permutations
on both vectors generate the diagonal of the matrix.  The proof of the above
theorem mostly relies on this elementary fact [CLL].
In fact, this matrix was given by Rutherford [Ru], [DKR], 
with another interpretation.

\smallskip For shape [3,2], here are the scalar products between the two
monomial bases:

$$Q:= 
\left[
\petitematrice{
1& \cdot & \cdot& \cdot& \cdot\cr \cdot& -1& \cdot& \cdot& \cdot\cr 
\cdot& \cdot& 1& \cdot
& \cdot\cr \cdot& \cdot& \cdot& 1& \cdot\cr -1& \cdot& \cdot& \cdot& -1
} \right]
 $$

\smallskip Because there are some entries outside the diagonal,
one needs to invert the preceding sparse matrix to
get the adjoint basis. 
The matrix representing a permutation $\ss$
 will simply be obtained by multiplying the matrix of scalar products
  $((x^u)^\ss\, ,\, x^v)$ by the inverse of $Q$.

For example, the matrix of scalar products with the image of the basis
under $\ss=[54321]$, and the matrix representing this permutation
are 
$$ \left[ \moyennematrice{
1& -1& \cdot & 1& \cdot \cr  \cdot & \cdot & 1& \cdot & -1\cr \cdot &
\cdot & \cdot & 
-1& 1\cr \cdot & 1& -1& -1& 1\cr -1& \cdot & \cdot & \cdot & -1
} \right]\ , \ 
 \left[
\moyennematrice{
 1& -1& \cdot & 1& \cdot \cr \cdot & \cdot & -1& \cdot 
& 1\cr \cdot & \cdot & \cdot &
-1& 1\cr \cdot & 1& -1& -1& 1\cr \cdot & 1& \cdot& -1& 1
} \right]
 $$

\smallskip
One has needed only addition of vectors to compute the entries of the 
matrices of scalar products ! 


\medskip
\centerline{\bf References}

\medskip
\parindent 0mm

[BR] {\small  H. BARCELLO, A. RAM},
Combinatorial representation theory,
New perspectives in algebraic combinatorics,
Math. Sci. Res. Inst. Publ., {\bf 38},
Cambridge Univ. Press, Cambridge (1999) 23--90.

[CLL] {\small  C. CARRE,  A. LASCOUX, B. LECLERC},
Turbo straightening, 
{\it Int J. of Algebra and Comp.} (1992) 275-290.

[DKR]\ {\small J. DESARMENIEN, J. KUNG,  G.C . ROTA},
Invariant Theory, Young bitableaux and combinatorics, Adv. in M. {\bf 27}
(1978) 63--92.

[JK]\ {\small JAMES,  A. KERBER}, The representation theory of
the symmetric group, Encyclopedia of Maths, vol {\bf 16} Addison-Wesley (1981).

[Ju]\ {\small A.A. JUCYS}, 
 On the Young operators on the symmetric groups,
Lietuvos fizikos rinkinys VI:2 (1966) 163-180;
Factorization of Young projection operators for the
symmetric group,
ibidem, X1:1 (1971) 5--10.

 
[Ga]\ {\small H. GARNIR},\rm Th\'eorie de la repr\'esentation des
groupes sym\'etriques, M\'emoires de la Soc. Royale des Sc. de Li\`ege
{\bf 10} (1950).

[La]\ { A. LASCOUX},
{\it Young's natural idempotents as polynomials},
 Annals of Combinatorics {\bf 1} (1997) 91--98.


[LLT1]\ {\small A.  LASCOUX,  B.  LECLERC, J-Y THIBON}, 
{\it Flag Varieties and the Yang-Baxter Equation },
Letters in Math. Phys. {\bf 40} (1997) 75--90.

[LLT2]\ {\small A.  LASCOUX,  B.  LECLERC, J-Y THIBON},
   The Plactic Monoid, in
  "Combinatorics On Words", M.Lothaire,
Acad. Press (to appear). 

[0V]\ {\small A. OKOUNKOV,  A.  VERSHIK},  A new approach to representation
theory of symmetric groups, Selecta Math. {\bf 2} (1996)  581--605.

[Ro]\ {\small  G. C. ROTA},  Combinatorial Theory and Invariant Theory,
 Notes by L.Guibas, Bowdoin College (1971).

[Ru]\ {\small D. RUTHERFORD}, Substitutional Analysis, Edinburgh University
Press (1948).

[Sp]\ {\small W. SPECHT}, Die irreduziblen darstellungen der symmetrischen
gruppe, Math. Z. {\bf 39} (1935) 606--711.


[Th]\ {\small R. M.  THRALL},  Young's semi  normal representation of
the symmetric group, Duke M. {\bf 8} (1941) 611--624.

[Ya]\ {\small C.N. YANG}, Some exact results for the many-body
problem in one dimension with repulsive delta-function interaction\rm,
Phys. Rev. Lett. {\bf 19} (1967), 1312-1315.


[Y]\ {\small A. YOUNG},  The Collected Papers of Alfred Young, 
  University of Toronto Press (1977). Representations are
described in the eight articles 
{\it On Quantitative Substitutional Analysis}, published in
the Proc. London Math. Soc. We refer more specially to 
QS1, {\bf 33} (1900) 97--146; 
QS3, {\bf 28} (1928) 255--292;
QS4, {\bf 31} (1930) 253--272.  

\end{document}

\bibitem{Y} \sc C.N. Yang, \it Some exact results for the many-body
problem in one dimension with repulsive delta-function interaction\rm,
Phys. Rev. Lett. {\bf 19} (1967), 1312-1315.














