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\input amssym.def
\input amssym

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%% AUTHORs MACROs
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\def\AA {{\cal A}}
\def\BB {{\cal B}}
\def\CC {{\cal C}}
\def\DD {{\cal D}}
\def\G{{\cal G}}
\def\LLL{\cal L}
\def\LL {{\Lambda}}
\def\SS {\ifmmode {\cal S\!\it h} \else ${\cal S\!\it h}$ \fi}
\def\abs#1{{| #1 |}}
\def\aalpha{{\underline\alpha}}
\def\bbeta{{\underline\beta}}
\def\n{\ifmmode n \else $n$\fi}
\def\dddelta#1#2{\delta_{x_#1,y_#2}}
\def\dx#1#2{\delta_{x_#1,x_#2}}
\def\dy#1#2{\delta_{y_#1,y_#2}}
\def\interval#1#2{z_{{#1},{#2}}}
\def\xi#1{x_{i_{#1}}}
\def\yi#1{y_{i_{#1}}}
\def\d{\partial}
\def\dt{\partial^t}
\def\zz#1{ z_{x_{#1},y_{#1}}}
\def\ww{\wedge}
\def\wdw{\wedge\cdots\wedge}
\def\wdd{\wedge \dots}
\def\ddw{ \dots \wedge}
\def\odo{\otimes \cdots \otimes}
\def\zzeta#1{\zz1\ww\zz2\wdw\zz#1}
\def\PPP{{\Pi}}
\def\CS#1{{\Bbb {C}S_{#1}}}
\def\DONE{{\rule {5pt} {5pt}}}
\def\CrP{{\cal C}_r(P)}
\def\binom#1#2{\mbox{\small $\left(\!\!\begin{array}{c}{#1}\\{#2}\end{array} \!\!\right)$}}
\def\smbinom#1#2{\mbox{\tiny $\left(\!\!\!\!\begin{array}{c}{#1}\\{#2}\end{array} \!\!\!\!\right)$}}
\def\zex#1#2#3#4{z_{1,#1}\ww z_{2,#2}\ww z_{3,#3}\ww z_{4,#4}}
\def\ith{\ifmmode {i^{\mbox{\small\,th}}\/ } \else {$i^{\mbox{\small\,th}}\/ $}
\fi}
\def\nth{\ifmmode {n^{\mbox{\small\,th}}\/ } \else {$n^{\mbox{\small\,th}}\/ $}
\fi}
\def\kth{\ifmmode {k^{\mbox{\small\,th}}\/ } \else {$k^{\mbox{\small\,th}}\/ $}
\fi}
\def\<{\langle}
\def\>{\rangle}
\def\u#1{{\underline #1}}
\def\ch{{\cal C\!\it h}}
\def\taken{\framebox[7pt]{$\cdot$}}
\def\one{\framebox[7pt]{\phantom{$\cdot$}}}
\def\em{\bf}
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\begin{document}
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%% RUNNING HEADS -- ELECTRONIC JOURNAL OF COMBINATORICS
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\baselineskip 15pt 
\pagestyle{myheadings}
\markright{\sc the electronic journal of combinatorics 2 (1995), \#R14\hfill}
\thispagestyle{empty}
%%==============================================================



\newtheorem{theorem}{Theorem}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{cor}[theorem]{Corollary}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{prop}[theorem]{Proposition}
\newtheorem{defi}{Definition}[section]
\newtheorem{fact}[theorem]{Fact}


\begin{center}
{\Large The eigenvalues of the Laplacian for the homology of the 
Lie algebra corresponding to a poset}
\vspace{1.0cm}

{Iztok Hozo}
\vspace{0.5cm}

Department of Mathematics

Indiana University Northwest

Gary, In 46408

{email: ihozo@iunhaw1.iun.indiana.edu}
\bigskip

Submitted: April 6, 1995; Accepted: July 21, 1995.
\end{center}
\bigskip





\begin{abstract}

In this paper  we study the spectral resolution of the Laplacian ${\cal L}$ 
of the Koszul complex of the Lie algebras corresponding to a
certain class of  posets.

Given a poset $P$ on the set $\{1,2,\dots,n\}$, we define the nilpotent
Lie algebra $L_P$ to be the span of all elementary matrices $z_{x,y}$, such
that $x$ is less than $y$ in $P$. In this paper, we make a decisive step 
toward calculating the Lie algebra homology of $L_P$ in the case that 
the Hasse diagram of $P$ is a rooted tree.  

We show that the Laplacian $\LLL$ simplifies significantly  
when the Lie algebra
corresponds to a poset whose Hasse diagram is a tree. The main result of
this paper determines the spectral resolutions of three commuting
linear operators whose sum is  the Laplacian $\LLL$
of the Koszul complex of $L_P$ in the case that the Hasse 
diagram is a rooted tree. 

We show that these eigenvalues are integers, give a
combinatorial indexing of these eigenvalues and describe the corresponding
eigenspaces in  representation-theoretic terms. The homology of $L_P$
is represented by the nullspace of $\LLL$, so in future work, these 
results should allow for the homology to be effectively computed.

\medskip\noindent
AMS Classification Number: 17B56 (primary) 05E25 (secondary)

\end{abstract}

%==================================================
% preliminaries: Definitions and known
% results needed for later work
%-------------------------------------------------------------------------------------------------------------
% 
%
%-------------------------------------------------------------------------------------------------------------

\section{Preliminaries}
\subsection{Definitions}


A {\em partially ordered set}  $P$ (or {\em poset}, for short) is a set (which
by abuse of notation we  also call $P$), together with a binary relation
denoted $\le$ (or $\le_P$ when there is a possibility of confusion),
satisfying the following three axioms:
\begin{enumerate}
\item For all $x \in P$\/, $ x\le x$. {\em (reflexivity)}
\item If $x\le y$ and $y \le x$, then $x = y$. {\em (antisymmetry)}
\item If $x\le y$ and $y\le z$, then $x\le z$. {\em (transitivity)}
\end{enumerate}

A {\em chain} (or {\em totally ordered set} or {\em linearly ordered set}) is
a poset in which any two elements are comparable. A subset $C$ of a poset $P$
is called a {\em chain} if $C$ is a chain when regarded as a subposet of $P$.

\begin{defi}
A poset $P$ is {\em linear} if for any two comparable
elements $x,~y~\in~P$, the interval $[x,y]$ is a chain, i.e., if every interval
has the structure of a chain.
\end{defi}

The {\em length} $l(C)$ of a finite chain $C$ is defined by $l(C)=\abs C  -1$. 

%=======================================================================

\subsection{The homology of a poset}

  The combinatorial approach to a homology theory for posets was
developed by Rota \cite{rota}, Farmer \cite{farmer},
Lakser \cite{lakser}, Mather \cite{mather}, Crapo \cite{crapo}
and others (more references can be found in \cite{walker}).
A systematic development of the relationship between the combinatorial
and topological properties of posets was begun by K.~Baclawski \cite{baclawski}
and A.~Bj\"orner \cite{bjorner} and continued by J.~Walker \cite{walker}.

Define the set $\CrP$ to be
 the set of 0-1 chains of length $r$
 in the poset $P$. By abuse of notation we will use the same name for the
complex vector space $C_r$ or $\CrP$, with basis the set of $r$-chains.
The $C_r$'s are called {\em chain spaces}. The map
$\partial_r : C_r \rightarrow C_{r-1}$,  called the {\em boundary map},
is defined by:
$$
\partial_r (\hat 0 < x_1 < \ldots <x_r <\hat 1) =
\sum_{i=1}^r (-1)^{i-1} (\hat 0 < x_1 < \ldots < \widehat{x_i} < \ldots <x_r <
\hat 1)
$$

It is easy to check that:
\begin{lemma}
$$
\partial_{r-1} \circ \partial_{r} = 0 .
$$
\end{lemma}


This allows us now to define the {\em homology of a poset} to be:
$$
H_r(P) = Ker(\partial_r)/Im(\partial_{r+1})
$$

Later in this work we will talk about an operator, called the  Laplacian
of a complex, for which
we need to identify the transpose of the boundary map. We
are in fact transposing the  matrix of the boundary map with respect to the
basis of $r$-chains.
In this case - the case
of the poset homology, the transpose of the boundary map is not so difficult
to evaluate.


\begin{lemma}
The transpose of the boundary operator (viewed as a linear map), is given
by the following expression:
\begin{eqnarray*}
\lefteqn{\dt (\hat 0 < x_1 < \dots <x_r< \hat 1)} \\
&=&\sum_{i=0}^{r} \sum_{x_i<y<x_{i+1}} (-1)^{i} (\hat 0<x_1<\dots< x_i<y<x_{i+1}<\dots<x_r < \hat
1) ,
\end{eqnarray*}
where $x_0 = \hat 0$ and $x_{r+1} =\hat 1$.
\end{lemma}


\subsection{Lie Algebras}

In this section we will introduce some basic notions from the theory
of Lie algebras, and the homology of Lie algebras.

We will always work over  $\Bbb C$, the field of complex numbers.



%==================================================
%_________________________________________________________
%
%       L I E      A L G E B R A S
%__________________________________________________________
%
%===================================================

Lie algebras arise ``in nature'' as vector spaces of linear transformations
endowed with an operation which is in general neither commutative nor
associative: $$[x,y] = xy - yx.$$ It is possible to describe this kind of system
abstractly in a few axioms.

\begin{defi}
 A vector space $L$ over a field $\Bbb C$, with an operation
$L \times L \rightarrow L$, denoted $(x,y)\rightarrow [x,y]$ and, called the
{\em bracket} or {\em commutator} of $x$ and $y$, is a {\em Lie algebra}
over $\Bbb C$ if the following axioms are satisfied:

\begin{eqnarray*}
\mbox{(L1)} & \mbox{ The bracket operation is bilinear. }\\
\mbox{(L2)} & [x,x] = 0 \mbox{ for all $x\in L$. }\\
\mbox{(L3)} & [x,[y, z]] + [y,[z,x]] + [z,[x,y]] = 0 \mbox{ } (x, y, z \in L).
\end{eqnarray*}
\end{defi}

Axiom (L3) is called {\em Jacobi identity}. The axioms (L1) and (L2) imply (L2'): $ [x,y] = -[y,x]
$. In the field of complex numbers (L2') implies (L2).


%=================================================================
\subsection{Homology of a Lie algebra }

Suppose $L$ is a Lie algebra and $A$ is a module over $L$.
The space $\Gamma_q(L; A)$ of $q$-dimensional chains of the Lie algebra $L$
with coefficients in $A$ is defined as $A \otimes \Lambda^q L$. The
boundary operator $\partial  =
 \partial_q :\Gamma_q(L;A)\rightarrow \Gamma_{q-1}(L;A)$
acts in accordance with
the formula
\begin{eqnarray}
\lefteqn{\partial ( a\otimes (x_1 \wedge \ldots  \wedge x_q)) =} \nonumber \\
&=&\sum_{1\le s<t \le q} (-1)^{s+t-1} a\otimes ([x_s,x_t]\wedge x_1 \wedge
\ldots \hat{x_s}\ldots \hat{x_t} \ldots \wedge x_q) \label{eq:boundary} \\
+& &\sum_{1 \le s \le q} (-1)^{s-1} x_s a \otimes (x_1 \wedge \ldots
\hat{x_s} \ldots \wedge x_q)            \nonumber
\end{eqnarray}

\begin{lemma}
\[\partial_{r-1} \partial_r = 0 \]
\end{lemma}
The proof of this lemma is straightforward. 

Let $\theta$ be the representation of $L$ on
  $A\otimes \Lambda^q L$.
If $y\in L$, we have:

\begin{eqnarray*}
\lefteqn{\theta(y) ( a\otimes x_1\wedge \ldots \wedge x_q)}\\
&=&(y\cdot a \otimes x_1 \wedge \ldots \wedge x_q) + \sum_i (a \otimes
x_1 \wedge \ldots \wedge [y,x_i] \wedge \ldots \wedge x_q)\\
\end{eqnarray*}


It is easy to check:
\begin{lemma}
For $y\in L$:
\[ \partial_q \circ \theta(y) = \theta(y) \circ \partial_q \]
\end{lemma}

The homology of the complex $\{ \Gamma_q (L;A), \partial_q \}$ is referred to as
{\em the homology of the Lie algebra $L$ with coefficients in $A$ } and denoted
by $H_q (L;A)$; if $A$ is the field of complex numbers
viewed as a trivial $L$-module (as in our case),
the second sum in the formula~\ref{eq:boundary} vanishes.
In this case the
notations $\Gamma_q(L;A)$ and $H_q(L;A)$  are abbreviated to $\Gamma_q(L)$ and $H_q(L)$.


\subsection{The Laplacian operator}

Suppose that $\{ \Gamma_r(L), \partial_r\}$ is a  finite dimensional complex.
We will first define an orthogonal inner product
$\<\cdot , \cdot \> $ on the product $\oplus \Gamma_r$,
such that $\< \Gamma_r,\Gamma_s\>~=~0$ whenever $r\not = s$.
We will restrict our attention to the subspaces of  the nilpotent
Lie algebra $T_n({\Bbb C})$ of all  strictly upper triangular matrices
over the complex numbers, with standard basis
 $\{ z_{i,j} ~:~1\le i < j \le n\}$,
so we can define this product naturally:

\begin{defi}
Let $L$ be a Lie algebra, $L\subset T_n({\Bbb C})$. Define an inner product
for  standard basis elements  $v,w \in L$ by:
$$
\< v, w\> = \left \{ \begin{array} {lr}
                        1 & \quad \mbox{ if } v=w \\
                        0 & \mbox{otherwise}\\
                        0 & \mbox{ if $v$ and $w$ have different exterior degrees}\\
                \end{array}
         \right.
$$
\end{defi}

Extend this to the exterior algebra, i.e., to the complexes mentioned
above.
\begin{defi}
Suppose that $ v = v_1 \wdw v_k$ and
$w = w_1 \wdw w_k$. Then define the inner product:
$$
\< v,w\> = det ( \< v_i, w_j\> )_{1\le i,j \le k}
$$
\end{defi}

Note that this can be written also as
$$
\< v,w\> = \sum_{\sigma \in S_n} sgn(\sigma) \prod_i \< v_i, w_{\sigma(i)}\> =
\left \{ \begin{array} {lr}
         sgn(\sigma) & \mbox{ iff } v_i = w_{\sigma(i)} \mbox{ for all $i$}\\
         0      &  \mbox{ otherwise }\\
        \end{array}
        \right.
$$
In other words, the product of two pure wedges of basis elements
is nonzero if and only
if two pure wedges differ only in the order of the elements, and
in that case, the product is just the sign of the permutation that
changes one into another.


                        
 Define $\delta_r$ mapping $\Gamma_r$
into $\Gamma_{r+1}$ by
$$
\<\delta_r v, w\> = \< v, \partial_{r+1} w\>
$$
over all $v \in \Gamma_r$, and all $w\in \Gamma_{r+1}$. It is enough to calculate
$\delta$ on pure wedges (as in our definitions), since the inner product
and $\delta$ are both  linear functions.


\begin{lemma}
The map $\delta$ is given by 
\begin{eqnarray*}
\lefteqn{\delta_r(\zz 1 \wedge \zz 2 \wedge \dots \wedge \zz r)} \\
&=&\sum_{s=1}^r (-1)^{s-1} { \sum_{x_s < l < y_s} \zz 1 \wedge\dots\wedge z_{x_s,l}\wedge
z_{l,y_s}\wedge \dots \wedge \zz r }
\end{eqnarray*}
\end{lemma}

{\em Note:} It is easy to check that $\delta_{r+1} \delta_r = 0$, thus
$\delta_*$ defines a coboundary operator, and so we can define 
the cohomology to be 
$$ H^r(L) = Ker( \delta_r)/Im(\delta_{r-1})$$

Proof: But to prove that, 
it is enough to show that the coefficient of the pure wedge  $z_{x_1,y_1}\wedge z_{x_2,y_2} \wedge \dots \wedge z_{x_r,y_r}$ in $\d (\zz 1 \wdw z_{x_s,l} \ww z_{l,y_s}
\wdw \zz r )$ is $(-1)^{s-1}$ for any $l \in (x_s,y_s)$, i.e., 

\begin{eqnarray*} 
\lefteqn{\d(z_{x_1,y_1}\wedge\dots\wedge z_{x_s,l}\wedge z_{l,y_s}\wedge \dots \wedge z_{x_r,y_r})} \\
&=&\dots  +  (-1)^{s-1} (z_{x_1,y_1}\wedge z_{x_2,y_2} \wedge \dots \wedge z_{x_r,y_r}) + \dots
\end{eqnarray*}

and this is not difficult by the definition of $\d$.




 Note that we can change the order of the elements in the 
pure wedges, and obtain a slightly different form for $\delta$:

\begin{eqnarray*}
\lefteqn{\delta_r(z_{x_1,y_1}  \wedge z_{x_2,y_2} \wedge \dots \wedge z_{x_r,y_r})}\\
&=& \sum_{s=1}^r (-1)^{s-1} { \sum_{x_s < l < y_s} z_{x_1,y_1}\wedge\dots\wedge z_{x_s,l}\wedge z_{l,y_s}\wedge \dots \wedge z_{x_r,y_r} }\\
&=&\sum_m \sum_{x_m < l < y_m} (z_{x_m,l} \ww \zz 1 \wdw z_{l,y_m} \wdw \zz k)
\end{eqnarray*}
This is  the form for the $\delta=\dt$ we will use.

\begin{defi}
 Define the {\em Laplacian operator}
 $L_r: \Gamma_r \rightarrow \Gamma_r$ by 
$$
L_r = \delta_{r-1} \partial_r + \partial_{r+1} \delta_r
$$
\end{defi}

\begin{theorem}[Kostant, \cite{kostant} \label{kost1}]
Let $B = \{\beta_1,\ldots,\beta_d\}$ be a basis for $Ker(L_r)$. Then $B$ is
simultaneously a complete set of representatives of 
$H^r(L)$ and $H_r(L)$.
In particular $\dim(H^r(L)) = \dim(H_r(L)) = \dim( Ker(L_r))$.
\end{theorem}


Sometimes, the Laplacian $L_r$ will turn out to be very simple.
In these cases, Theorem~\ref{kost1} is a very efficient method for
 evaluating the homology and 
cohomology of a Lie algebra. One famous result obtained in this way
is given by Kostant~\cite{kostant}.

\subsection{Kostant's Theorem}

We need some preliminary definitions. Suppose $\cal G$ is
a semisimple Lie algebra, with the root system $R$, whose basis is $\Delta$.
Thus ${\cal G} = H \oplus ( \oplus_{\alpha \in R} \< z_\alpha\>)$, where
$H$ is the torus. 
Suppose that $S \subset \Delta$, and let $R_S$ be the set of roots in the
$\Bbb Z$ (integer) module spanned by elements of $S$.
Define $\cal G_S$ to be ${\cal G_S} = H \oplus \< z_\alpha : \alpha \in R_S \>$.
 Define a ${\cal G}_S$ module $N_S$ 
to be $N_S = \< z_\alpha : \alpha \in R^+ \setminus R_S^+ \>$.


We will state a couple of facts without proof:
\begin{itemize}
\item  $N_S$ is a nilpotent subalgebra of $\cal G$.
\item Let $W$ be a $\cal G$-module. Then $W$ is also a $N_S$-module and a
$\cal G_S$-module.
\item Thus we can compute $H(N_S ;W^\mu)$ as $\cal G_S$-module, where $W^\mu$
is an irreducible $\cal G$-module. Kostant used the 
Laplacian operator to prove the following theorem:
\end{itemize}

\begin{theorem} [Kostant, Theorem 5.7,\cite{kostant}]
Let $\lambda$ be a dominant weight for $\cal G$, and let $\mu$ be a minimal
weight for $\cal G_S$. Let $V$ be a $\cal G_S$-invariant subspace of $W^\lambda
\otimes \bigwedge^r N_S$ isomorphic to the $\cal G_S$-irreducible (indexed by
$\mu$) with minimal weight $\mu$.
\begin{itemize}
\item The Laplacian $L=\delta \partial + \partial \delta$ preserves $V$.
\item Then, $L\vert_V$ is a scalar, and the scalar is given by 
$$
\frac{1}{2} (\vert \rho + \lambda \vert^2 - \vert \rho - \mu \vert^2)
$$
where $\rho$ is half of the sum of the positive roots of $\cal G$.
\end{itemize}
\end{theorem}




\subsection{The Lie Algebra corresponding to a Poset}



\begin{defi}
A {\em standard labeling} of the poset $P$ is a total ordering of the elements
of $P$ such that whenever $x <_P y$, $x$ precedes $y$ in that total ordering.
\end{defi}
  
 Since $P$ is a 
partial order, i.e. transitive , there always is  such labeling. Fix a standard labeling of the poset $P$.

We can define a Lie algebra $L_P$  corresponding to the poset $P$
 in the following way.
First, for every relation $x<_P y$ in the poset $P$, i.e., for every
two elements $x,y \in P$ such that $x<_P y$ we can define the matrix
$z_{x,y}$ having all entries equal to zero, except for exactly one entry 
equal to 1, namely the entry at the position $x,y$ in the standard
 labeling of the poset $P$.

All matrices $z_{x,y}$  are  strictly upper triangular because
of our labeling. So $L_P$ is a subalgebra of $T_n$. 
The Lie algebras $L_P$ obtained from  distinct labellings are  isomorphic --
the labeling only specifies embedding of $L_P$ in the $n\times n$ matrices.


%==========================================================
% the main part
%-------------------------------------------------------------------------------------------------------------------
% 
%-------------------------------------------------------------------------------------------------------------

\section{The Formula for Laplacian of a Linear Poset \label{ch4}}

In this section we will present a significant simplification of the
Lie algebra 
Laplacian in the case of linear posets. That will allow us
to prove our main result  on the eigenvalues
of those Laplacians. 


\subsection{Simplification }

Recall the Lie algebra boundary map:
\begin{eqnarray*}
\lefteqn{\d(\zz 1\wedge \dots \wedge \zz k)} \\
&=&\sum_{i<j} (-1)^{i+j-1}  [\zz i,\zz j]\wedge \zz 1 \wedge \dots \wedge \widehat {\zz i} \wedge \dots \wedge \widehat {\zz j} \wedge \dots \wedge \zz k
\end{eqnarray*}

The transpose, $\dt$, is given by the following
formula:
\begin{eqnarray*}
\lefteqn{\dt_r(\zz 1 \wedge \zz 2 \wedge \dots \wedge \zz r)} \\
&=&\sum_{s=1}^r (-1)^{s-1} { \sum_{x_s < l < y_s} \zz 1 
\wedge\dots\wedge z_{x_s,l}\wedge z_{l,y_s}\wedge \dots \wedge \zz r } \\
&=&\sum_m \sum_{x_m < l < y_m} (z_{x_m,l} \ww \zz 1 \wdw z_{l,y_m} \wdw \zz k)
\end{eqnarray*}

To compute the action of $L$ on a basis vector  $\zz 1 \wdw \zz k$ 
of $\Gamma_k (L_P)$ we begin
with the action of $\d \dt$. We have,

\begin{eqnarray*}
\lefteqn{\d \dt (\zz 1 \wdw \zz k)} \\
&=& \sum_m \sum_{x_m < l < y_m}  \d (z_{x_m,l} \ww \zz 1 \wdw z_{l,y_m} \wdw \zz k)\\
&=&\sum_{i<j} \sum_{m\neq i,j} \sum_{x_m < l < y_m} (-1)^{i+1 +j} ([\zz i,\zz j] \ww z_{x_m,l} \ww \zz 1 \ww \dots \\
&& \dots \ww \widehat {\zz i} \wdw z_{l,y_m} \wdw \widehat {\zz j} \wdw \zz k)\\
&+& \sum_m \sum_{j \neq m} \sum_{x_m < l <y_m} (-1)^{1+j+1-1} ([z_{x_m,l},\zz j] \ww \zz 1 \ww \dots \\
&&\dots \ww  \widehat {\zz j} \wdw z_{l,y_m} \wdw \zz k)\\
&+& \sum_{i<m} \sum_{x_m <l <y_m} (-1)^{i+1+m+1-1} ([\zz i,z_{l,y_m}]\ww z_{x_m,l} \ww \zz 1 \ww \dots \\
&&\dots \ww \widehat {\zz i} \wdw \widehat {\zz m} \wdw \zz k)\\
&+& \sum_{m<j} \sum_{x_m <l<y_m} (-1)^{m+1+j+1-1}([z_{l,y_m},\zz j] \ww z_{x_m,l} \ww \zz 1 \ww \dots \\
&&\dots \ww \widehat {\zz m} \wdw \widehat {\zz j} \wdw \zz k)\\
&+& \sum_{m=1}^k \abs{(x_m,y_m)} (\zz 1 \wdw \zz k)
\end{eqnarray*}

which is equal to:
 
\begin{eqnarray*}
&=&\sum_{i<j} \sum_{m\neq i,j} \sum_{x_m < l < y_m} (-1)^{i+j-1} ([\zz i,\zz j] \ww z_{x_m,l} \ww \zz 1 \ww \dots \\
&&\dots \ww \widehat {\zz i} \wdw z_{l,y_m} \wdw \widehat {\zz j} \wdw \zz k)\\
&+& \sum_{i<m} \sum_{x_m < l <y_m} (-1)^{i+1} ([z_{x_m,l},\zz i] \ww \zz 1 \ww
\dots \\
&&\dots \ww  \widehat {\zz i} \wdw z_{l,y_m} \wdw \zz k)\\
&+& \sum_{m<j} \sum_{x_m < l <y_m} (-1)^{j+1} ([z_{x_m,l},\zz j] \ww \zz 1 \ww
\dots \\
&&\dots \ww z_{l,y_m}\wdw \widehat {\zz j} \wdw  \zz k)\\
&+& \sum_{i<m} \sum_{x_m <l <y_m} (-1)^{i+m+1} ([\zz i,z_{l,y_m}]\ww z_{x_m,l} \ww \zz 1\ww \dots \\
&&\dots  \ww \widehat {\zz i} \wdw \widehat {\zz m} \wdw \zz k)\\
&+& \sum_{m<j} \sum_{x_m <l<y_m} (-1)^{m+j+1}([z_{l,y_m},\zz j] \ww z_{x_m,l} \ww \zz 1 \ww \dots \\
&&\dots \ww \widehat {\zz m} \wdw \widehat {\zz j} \wdw \zz k)\\
&+&\sum_{m=1}^k \abs{(x_m,y_m)} (\zz 1 \wdw \zz k)
\end{eqnarray*}

Now use the definition of bracket in this Lie algebra:

\begin{eqnarray*}
[\zz i,\zz j] = \delta_{y_i,x_j} z_{x_i,y_j}-\delta_{x_i,y_j} z_{x_j,y_i}
\end{eqnarray*}

 and we have the following:

\begin{eqnarray*}
\lefteqn{\d \dt (\zz 1 \wdw \zz k)} \\
&=&\sum_{i<j} \sum_{m\neq i,j} \sum_{x_m < l < y_m} (-1)^{i+j-1} ([\zz i,\zz j] \ww z_{x_m,l} \ww \zz 1 \wdw \widehat {\zz i}\ww \dots \\
&&\dots \ww z_{l,y_m} \wdw \widehat {\zz j} \wdw \zz k)\\
&+& \sum_{i<m} \sum_{x_m < l <y_m}\delta_{l,x_i} ( \zz 1 \wdw z_{x_m,y_i} \wdw z_{l,y_m} \wdw \zz k)\\
&-& \delta_{x_m,y_i} (\zz 1 \wdw z_{x_i,l} \wdw z_{l,y_m} \wdw \zz k)\\
&+& \sum_{m<j} \sum_{x_m < l <y_m}\delta_{l,x_j} (\zz 1 \wdw z_{l,y_m}\wdw z_{x_m,y_j} \wdw  \zz k)\\
&-&\delta_{x_m,y_j} (\zz 1 \wdw z_{l,y_m}\wdw z_{x_j,l} \wdw  \zz k)\\ 
&+& \sum_{i<m} \sum_{x_m <l <y_m}\delta_{l,y_i} (\zz 1 \wdw z_{x_i,y_j} \wdw z_{x_m,l} \wdw \zz k)\\
&-&\delta_{x_i,y_m} (\zz 1 \wdw z_{l,y_i} \wdw z_{x_m,l} \wdw \zz k)\\
&+& \sum_{m<j} \sum_{x_m <l<y_m}\delta_{l,y_j} (\zz 1 \wdw z_{x_m,l} \wdw z_{x_j,y_m} \wdw \zz k)\\
&-&\delta_{x_j,y_m} (\zz 1 \wdw z_{x_m,l} \wdw z_{l,y_j} \wdw \zz k)\\
&+& \sum_{m=1}^k  \abs{(x_m,y_m)} (\zz 1 \wdw \zz k)
\end{eqnarray*}

Note that every sum over $x_m<l<y_m$ which has an occurrence 
of $\delta_{l,*}$ 
has only one summand if $*$ really is between 
$x_m$ and $y_m$, and  is  zero otherwise. 
We will use the symbol $\chi$ for denoting 
the truth of some statement, i.e.,

\begin{eqnarray*}
\chi( *) =\left\{ \begin{array}{ll}
                 1, & \mbox{ if $*$ is true}\\
                 0, & \mbox{ if $*$ is false} 
                  \end{array}
           \right.
\end{eqnarray*}

We  label some of the resulting  sums:


\begin{eqnarray}
\lefteqn{\d \dt (\zz 1 \wdw \zz k)} \\
&=&\sum_{i<j} \sum_{ m\neq i,j} \sum_{x_m < l < y_m} (-1)^{i+j-1} ([\zz i,\zz j] \ww z_{x_m,l} \ww \zz 1 \ww \dots \nonumber\\
&&\dots \ww \widehat {\zz i} \wdw z_{l,y_m} \wdw \widehat {\zz j} \wdw \zz k) \label{IV.1}\\
&-&\sum_{i<j} \chi(x_j<x_i<y_j) (\zz 1 \wdw z_{x_i,y_j} \wdw z_{x_j,y_i} \wdw \zz k) \nonumber\\
&-&\sum_{i<j} \sum_{x_j < l <y_j} \delta_{x_j,y_i} ( \zz 1 \wdw z_{x_i,l} \wdw z_{l,y_j} \wdw \zz k) \label{IV.2} \\
&-&\sum_{i<j} \chi(x_i < x_j <y_i) (\zz 1 \wdw z_{x_i,y_j}\wdw z_{x_j,y_i} \wdw  \zz k) \nonumber\\
&-&\sum_{i<j} \sum_{x_i < l <y_i} \delta_{x_i,y_j} ( \zz 1 \wdw z_{l,y_i}\wdw z_{x_j,l} \wdw  \zz k) \label{IV.3} \\
&+&\sum_{i<j} \chi(x_j<y_i<y_j)( \zz 1 \wdw z_{x_i,y_j} \wdw z_{x_j,y_i} \wdw \zz k) \nonumber \\
&-&\sum_{i<j} \sum_{x_j <l <y_j}  \delta_{x_i,y_j} (\zz 1 \wdw z_{l,y_i} \wdw z_{x_j,l} \wdw \zz k) \label{IV.4}\\
&-&\sum_{i<j} \sum_{x_i <l<y_i} \delta_{x_j,y_i} (\zz 1 \wdw z_{x_i,l} \wdw z_{l,y_j} \wdw \zz k)\label{IV.5} \\
&+&\sum_{i<j} \chi(x_i<y_j<y_i) (\zz 1 \wdw z_{x_i,y_j} \wdw z_{x_j,y_i} \wdw \zz k)\nonumber \\
&+& \sum_{m=1}^k \abs{(x_m,y_m)} (\zz 1 \wdw \zz k) \nonumber
\end{eqnarray}

On the other hand:

\begin{eqnarray*}
\lefteqn{\dt \d (\zz 1 \wdw \zz k)} \\
&=&\sum_{i<j} (-1)^{i+j-1} \dt ([\zz i,\zz j] \ww \zz 1 %\ww  \dots \\&&
\dots \ww \widehat {\zz i} \wdw \widehat {\zz j} \wdw \zz k)\\
&=&\sum_{i<j} \sum_{m\neq i,j} \sum_{x_m < l < y_m} (-1)^{i+j-1} (z_{x_m,l}\ww [\zz i,\zz j] \ww \zz 1 \ww \dots \\
&&\dots \ww  \widehat {\zz i} \wdw z_{l,y_m} \wdw \widehat {\zz j} \wdw \zz k)\\
&+& \sum_{i<j}  \sum_{x_m < l < y_m} (-1)^{i+j-1} \delta_{x_j,y_i}(z_{x_i,l}\ww z_{l,y_j} \ww \zz 1 \ww \dots \\
&&\dots \ww \widehat {\zz i} \wdw \widehat {\zz j} \wdw \zz k)\\
&-&\sum_{i<j}  \sum_{x_m < l < y_m} (-1)^{i+j-1} \delta_{x_i,y_j}(z_{x_j,l}\ww z_{l,y_i} \ww \zz 1 \ww \dots \\
&&\dots \ww \widehat {\zz i} \wdw \widehat {\zz j} \wdw \zz k)
\end{eqnarray*}

Now use the fact that we are dealing with a linear poset. This implies that for  every interval $(x_m,y_m)$ and every $l$ , $x_m<l<y_m$  we have

\begin{eqnarray*}
(x_m,y_m) = (x_m,l) \cup \{ l\} \cup (l,y_m)
\end{eqnarray*}
Hence 

\begin{eqnarray}
\lefteqn{\dt \d (\zz 1 \wdw \zz k) }\\
&=&\sum_{i<j} \sum_{m\neq i,j} \sum_{x_m < l < y_m} (-1)^{i+j-1} (z_{x_m,l}\ww [\zz i,\zz j] \ww \zz 1 \ww \dots \nonumber \\
&&\dots \ww \widehat {\zz i} \wdw z_{l,y_m} \wdw \widehat {\zz j} \wdw \zz k) \label{IV.I} \\
&+&\sum_{i<j} \sum_{x_i < l < y_i} \delta_{x_j,y_i}(\zz 1 \wdw z_{x_i,l} \wdw z_{l,y_j} \wdw \zz k) \label{IV.II} \\
&+&\sum_{i<j} \sum_{l=x_j=y_i} \delta_{x_j,y_i}(\zz 1 \wdw z_{x_i,l} \wdw z_{l,y_j} \wdw \zz k) \nonumber \\
&+&\sum_{i<j} \sum_{x_j<l<y_j} \delta_{x_j,y_i}(\zz 1 \wdw z_{x_i,l} \wdw z_{l,y_j} \wdw \zz k) \label{IV.III} \\
&+&\sum_{i<j}  \sum_{x_j < l < y_j} \delta_{x_i,y_j}(\zz 1 \wdw z_{l,y_i} \wdw z_{x_j,l} \wdw \zz k) \label{IV.IV}\\
&+&\sum_{i<j}  \sum_{ l = x_i=y_j} \delta_{x_i,y_j}(\zz 1 \wdw z_{l,y_i} \wdw z_{x_j,l} \wdw \zz k)\nonumber \\
&+&\sum_{i<j}  \sum_{x_i < l < y_i} \delta_{x_i,y_j}(\zz 1 \wdw z_{l,y_i} \wdw z_{x_j,l} \wdw \zz k) \label{IV.V}
\end{eqnarray}

Then we have :

\begin{eqnarray*}
(\ref{IV.I}) + (\ref{IV.1}) &=& 0\\
(\ref{IV.II}) + (\ref{IV.5}) &=&0\\
(\ref{IV.III}) + (\ref{IV.2}) &=&0\\
(\ref{IV.IV}) + (\ref{IV.4}) &=&0\\
(\ref{IV.V}) + (\ref{IV.3}) &=&0
\end{eqnarray*}

After these cancellations we obtain the following expression for
the action of the Laplacian $L$:

\begin{eqnarray*}
\lefteqn{L(\zz 1 \wdw \zz k)= (\d \dt + \dt \d) ( \zz 1 \wdw \zz k)}\\
&=& \sum_{m=1}^k \abs{(x_m,y_m)} (\zz 1 \wdw \zz k)\\
&+&\sum_{i<j} (\delta_{x_i,y_j}+\delta_{x_j,y_i})(\zz 1 \wdw \zz k)\\
&+&\sum_{i<j} \chi(x_i<y_j<y_i) (\zz 1 \wdw z_{x_i,y_j} \wdw z_{x_j,y_i} \wdw \zz k)\\
&+&\sum_{i<j} \chi(x_j<y_i<y_j) (\zz 1 \wdw z_{x_i,y_j} \wdw z_{x_j,y_i} \wdw \zz k)\\
&-&\sum_{i<j} \chi(x_j<x_i<y_j)( \zz 1 \wdw z_{x_i,y_j} \wdw z_{x_j,y_i} \wdw \zz k) \\
&-&\sum_{i<j} \chi(x_i < x_j <y_i)( \zz 1 \wdw z_{x_i,y_j}\wdw z_{x_j,y_i} \wdw  \zz k)
\end{eqnarray*}

\subsection{The Formula}

To further simplify our expressions we will introduce some notation.
Define 
\begin{eqnarray*}
 \zeta &=&\zz 1 \wdw \zz k \\ 
\zeta_{i,j} &=&  \zz 1 \wdw z_{x_i,y_j} \wdw z_{x_j,y_i} \wdw \zz k \\
 w(\zeta) &=& \sum_{m=1}^k \abs{(x_m,y_m)} \\ 
 \Delta(\zeta) &=& \sum_{i<j} (\delta_{x_i,y_j}+\delta_{x_j,y_i}) = \sum_{i,j} \delta_{x_i,y_j}
\end{eqnarray*}

Thus, we can reformulate the calculations from the previous
section into:
\begin{theorem}[The Formula]\label{the.formula}
Let $P$ be a linear poset and let $L_P$ be the corresponding Lie algebra.
The action of the Laplacian $L$ on an element $$\zeta = \zzeta k$$ is 
given by the following formula: 
 \begin{eqnarray*}
\lefteqn{ L(\zeta) = (w(\zeta) + \Delta(\zeta))\zeta } \\
&+&\sum_{i<j} ( \chi(x_i<y_j<y_i) +\chi(x_j<y_i<y_j)-\chi(x_j<x_i<y_j)-\chi(x_i < x_j <y_i)) \zeta_{i,j}
\end{eqnarray*}
\end{theorem}

Note that $\zeta_{i,j}$ is obtained from  $\zeta$ by transposing 
a comparable pair of $y$'s or a comparable pair of $x$'s.






\section{Linear poset with a $\hat 0$}




Suppose now that the poset $P$ has a $\hat 0$, the minimum element. That is
the assumption under which we will work in the future. In that 
case, we can further simplify our notation:
\begin{lemma}
\begin{eqnarray*}
\lefteqn{ L(\zeta) = (w(\zeta) + \Delta(\zeta))\zeta } \\
&+&\sum_{i<j} ( \chi(x_i<y_j<y_i) +\chi(x_j<y_i<y_j)-\chi(x_j<x_i<y_j)-\chi(x_i < x_j <y_i)) \zeta_{i,j} \\
&=& (w(\zeta)+\Delta(\zeta))\zeta +\sum_{y_i <_P y_j} \zeta_{i,j}
-\sum_{x_i<_P x_j} \zeta_{i,j}\\
\end{eqnarray*}
\end{lemma}
Proof: We need to prove that we can write 
 $$\chi(y_i<y_j) + \chi(y_j<y_i) - \chi(x_i<x_j) - \chi(x_j<x_i)$$ instead
of  $$\chi(x_i<y_j<y_i) +\chi(x_j<y_i<y_j)-\chi(x_j<x_i<y_j)-\chi(x_i < x_j <y_i)
$$ in the expression for the Laplacian above.

Let $y_i$ and $y_j$ be two comparable distinct $y$'s. Without loss of generality, assume
that $y_i<y_j$. Thus $x_i < y_i <y_j$.     The existence of $\hat 0$ and linearity of the poset implies that the interval
$[\hat 0, y_j]$ must be a chain, and since $x_i, x_j \in [\hat 0, y_j]$,
$x_i$ and $x_j$ must be comparable. 
There are several  possibilities:
\begin{enumerate}
\item $x_j < x_i < y_i < y_j$ 
\item $x_i < x_j < y_i < y_j$ 
\item $x_i < y_i < x_j < y_j$ 
\end{enumerate}
  In all three possibilities, $$ \chi(x_i<y_j<y_i) +\chi(x_j<y_i<y_j)-\chi(x_j<x_i<y_j)-\chi(x_i < x_j <y_i) = 0,$$ and at the
same time 
$$\chi(y_i<y_j) + \chi(y_j<y_i) - \chi(x_i<x_j) - \chi(x_j<x_i) = 0.$$
  
On the other hand, if $y_i$ and $y_j$ are incomparable, then we have one of:
\begin{enumerate}
\item $x_j < x_i < y_i,y_j$  
\item $x_i < x_j < y_i,y_j$ 
\item $x_i < y_i , x_j < y_j$, $x_i$ and $x_j$ are incomparable. 
\end{enumerate}
Now in the first two cases 
$$ \chi(x_i<y_j<y_i) +\chi(x_j<y_i<y_j)-\chi(x_j<x_i<y_j)-
\chi(x_i < x_j <y_i) = -1,$$ with
 $$\chi(y_i<y_j) + \chi(y_j<y_i) - \chi(x_i<x_j) - \chi(x_j<x_i) = -1$$ too.
In the last remaining case both expressions are zero.

Hence, the expression for the Laplacian
above can be rewritten in the following form:
$$
L(\zeta) = (w(\zeta)+\Delta(\zeta))\zeta +\sum_{y_i <_P y_j} \zeta_{i,j}
-\sum_{x_i<_P x_j} \zeta_{i,j}. \DONE
$$
In other words, the meaning of the theorem above is that the Laplacian
only transposes comparable labels of the element $\zzeta k$,
 without introducing
any new indices. This is the key observation for next section. 

\begin{lemma}
Let $\zeta =\zz 1 \wdw \zz n$, and let $\zeta_\sigma = z_{x_1,y_{\sigma(1)}}
\ww z_{x_2, y_{\sigma(2)}} \wdw z_{x_n,y_{\sigma(n)}}$. 
If $\zeta_\sigma \not =  0$, i.e., if $x_i <_P y_{\sigma(i)}$ for all $i$, 
then
\begin{enumerate}
\item $w$ does not depend on $\sigma$, i.e.,
 $$ w(\zeta) = w(\zeta_{\sigma})$$

\item $\Delta$ does not depend on $\sigma$, i.e., 
$$\Delta(\zeta) = \Delta(\zeta_\sigma)$$
\end{enumerate}
\end{lemma}

This lemma actually proves that $w$ and $\Delta$ are dependent only on the
choice of the (multi--)sets 
$X = \{x_1,x_2, \dots, x_k\},Y = \{y_1, y_2, \dots, y_k\}$ 
(and a poset $P$), and not on the specific pure wedge 
constructed from those sets.

Proof: First we will check the claim for $w$.
\begin{eqnarray*}
 w(\zeta) &=& \sum_i \abs{(x_i,y_i)} = \sum_i ht(y_i) - ht(x_i) - 1 ,\\
\end{eqnarray*}
where the $ht(v)$ is the size of the interval $[\hat 0, v]$. 
The sum on the right does not depend on $\sigma$, so 
we can write $w(X,Y)$ instead of
$w(\zeta)$. 

Now we will check the claim for $\Delta$.
\begin{eqnarray*}
 \Delta(\zeta) &=& \sum_{i<j}(\dddelta{i}{j} + \dddelta{j}{i}) \\
&=& \sum_i \mbox{ ( multiplicity of $x_i$ in the set $Y$)} \\
&=& \sum_j \mbox{ ( multiplicity of $y_j$ in the set $X$)}\\
&=& \sum_{i,j} \delta_{x_i,y_j} \\
\end{eqnarray*}
which also does not depend on $\sigma$. Thus we can write $\Delta(X,Y)$ 
instead of
$\Delta(\zeta)$ too. \DONE

We will use both notations, depending whether we want to stress
$\zeta$ or the sets $(X,Y)$. Note that while $\Delta$ is completely 
determined by the sets $(X,Y)$, $w$ also depends on the poset $P$ globally, i.e.,
it counts the sizes of intervals 
$(x_i, y_i)$ not relative to the sets $X$ and $Y$, but
with respect to the whole poset $P$.






The simplicity of this formula is in the way the elements to which we are restricting
the Laplacian, are obtained one from another, by simply transposing the labels. 
In general, this example shows that the 
Laplacian $L$ can be broken down into diagonal blocks, which
are generated by a  pure wedge $\zeta$, and  all pure  wedges obtained 
by  permutations of
the labels of $\zeta$.
Furthermore, since $ a\ww b = - b \ww a$, we can always keep the $x$-labels
in order, i.e, we will always put the element $z_{x_i,*}$ at the $i^{\mbox{th}}$
position of the pure wedge. 




%--------------------------------------------------------------------------------------------
% 
%--------------------------------------------------------------------------------------------


\section{The eigenvalues of the Laplacian}

Let $\zeta = \zzeta n$ be an element of the exterior algebra of the  Lie algebra of $P$.
In the last section  we saw
 that the Laplacian acts on pure wedges of  Lie algebra  elements  
 $\zzeta n$ by summing the action of  switching pairs of comparable $x$'s, 
 and pairs of comparable $y$'s among themselves (plus a scalar).
 
That fact gives us the opportunity to divide our Laplacian into 
diagonal blocks 
 where each block corresponds to all possible permutations of 
the $x$'s and $y$'s for a fixed choice of the element $\zzeta n$,
 i.e., for  the fixed choice of the multisets $X=\{ x_1,x_2,\dots, x_n\}$, and $Y=\{y_1,
y_2,\dots,y_n\}$.  
In other words each block represents the ``action'' of the Laplacian on the 
subspace of the \nth exterior power of
 our Lie algebra spanned by the elements $\{ z_{x_1,y_{\sigma(1)}}\ww
z_{x_2,y_{\sigma(2)}} \wdw z_{x_n, y_{\sigma(n)}} :  \sigma \in S_n \}$.
 Here  the element $ z_{x_1,y_{\sigma(1)}}\ww z_{x_2,y_{\sigma(2)}} \wdw z_{x_n, y_{\sigma(n)}}$ is defined if and only if $x_i <_P y_{\sigma(i)}$  for every $i=1,2,\dots,n$.
Thus each block is of size $n!$, if all the elements are defined, or less, if
some of the elements are not defined which is the case in general. The size of the block
 depends
on the structure of the poset, and in particular, it depends on the relations
in the subposet of $P$ spanned by the sets $X$ and $Y$. More 
formally :

\begin{defi}
The {\em L-block  $V$ spanned by the (multi)-sets $(X,Y)_P$}, subsets of a poset
$P$, is the vector space  with basis 
$$
\{ z_{x_1,y_{\sigma(1)}} \ww z_{x_2,y_{\sigma(2)}} \wdw z_{x_n,y_{\sigma(n)}} :
\sigma \in S_n \}
$$
where $n = \abs{X} = \abs{Y}$, $\sigma $ is a permutation in  $S_n$, and the element
$$z_{x_1,y_{\sigma(1)}} \ww z_{x_2,y_{\sigma(2)}} \wdw z_{x_n,y_{\sigma(n)}}$$ is zero unless  $x_i <_P y_{\sigma(i)}$ for all $i = 1,\dots , n$.

If we want to stress the dependence of the L-block $V$ of the sets $X$
and $Y$ and the poset $P$, we write $V(X,Y)_P$.
\end{defi}
The sets $X$ and $Y$ may be multisets since some of the $x$'s or $y$'s
might appear more than once as a label. In that case the sizes $\abs{X}$ 
and $\abs{Y}$ are 
counting multiplicities as well.

Using this division of the chain space into L-blocks, we can
use the results of the previous section, and state the theorem:

\begin{theorem}
Let $ L_P $ be the Lie algebra corresponding to a linear poset $P$, and
let $C_n(L_P)$ be the \nth chain space. Then
$$
C_n(L_P) = \bigoplus_{(X,Y)} V(X,Y)_P
$$
where the direct sum is over all possible choices of (multi)-sets $X$ and
$Y$ of equal cardinality,
 and each  summand $V(X,Y)_P$ is invariant under the action of the Laplacian.
\end{theorem}


Thus we can now concentrate on the action of the Laplacian  on each of these blocks.



        

\subsection{Embedding of the L-block in $\CS n$}

Write the multisets $X$ and $Y$ as $X= \cup_{i\in A_1} \{ x_i\} \cup 
\dots \cup_{
i\in A_l}\{ x_i\}$,
and $Y = \cup_{j\in B_1}\{y_j\} \cup \dots 
\cup_{j\in B_m}\{ y_j\}$, where the $A_i$'s contain
the sets of indices of equal $x$'s, and $B_i$'s contain the sets of
indices of equal $y$'s.

For example, if $X= \{ x_1, x_2, x_3, x_4, x_5 \}$, where
$x_1 = x_2, x_3 = x_4$, then $A_1 =\{ 1,2 \}$, $A_2=\{ 3,4\}$ and $A_3 =\{5\}$.

Switching two of the $x$'s  will displace $x_i$ from its original position.
To take into account the fact that we have to bring it back (by the choice
of our basis) into the \ith place, we need a minus sign.

Let 
$$
\PPP_x = \sum_{\sigma_1 \in Sym(A_1), \sigma_2 \in Sym(A_2) \dots} (\prod_i sgn(\sigma_i)) \sigma_1 \sigma_2 \cdots \sigma_l
$$

and 
$$
\PPP_y = \sum_{\sigma_1 \in Sym(B_1), \sigma_2 \in Sym(B_2)\dots}  \sigma_1 \sigma_2 \cdots \sigma_m .
$$

Then the L-block $V$ can be identified with a subspace of $ \PPP_x \CS n \PPP_y$.
So  $\PPP_y$ symmetrizes over equal $y$'s and $\PPP_x$ anti-symmetrizes over
equal $x$'s.  In other words, $\PPP_x$ permutes the positions, while
$\PPP_y$ permutes indices.







\subsection{The Laplacian $L_Y$}

Let $X = \{ x_1, \dots , x_n\}$ and $Y=\{ y_1, \dots, y_n\}$ be two 
fixed (multi-)sets
of vertices of the poset $P$, and consider the restriction of the Laplacian $L$ to
L-block $V(X,Y)$. 


To simplify the notation, we will write the Laplacian $L$ as:
$$
L(\zeta) = \left( (w(X,Y) + \Delta(X,Y))Id  + \sum_{x_i<_P x_j} (x_i,x_j) +
 \sum_{y_i <_P y_j} (y_i,y_j)  \right) \cdot \zeta ,
$$
where the ``action'' of $(y_i, y_j)$ or $(x_i, x_j)$   on $\zzeta n$
means switching the corresponding pairs of $y$'s, or $x$'s. 



To simplify our examination, we will split it into these three
parts:
$$
L=L_D + L_X +L_Y  ,
$$


 where

\begin{itemize}
\item $L_D$ is the scalar matrix, $ L_D= w(X,Y) + \Delta(X,Y)$
\item $L_X$ is the ``action'' of the Laplacian on the set of the $x$'s,
i.e. 
$$
L_X = \sum_{x_i<_P x_j} (x_i,x_j)
$$

\item $L_Y$ is the ``action'' of the Laplacian on the set of the $y$'s
$$
L_Y = \sum_{y_i <_P y_j} (y_i, y_j)
$$

\end{itemize}
on our L-block $V(X,Y)$.

In the embedding of this L-block into $\CS n$, the notation for the Laplacian
$L_Y$ would be $ L_Y =  \sum_{i<j: y_i <_P y_j} (i,j)$, where the actual multiplication is from the right. The proper notation for
$L_X$ in the $\PPP_x \CS n \PPP_y$ is $L_X = \sum_{i<j: x_i <_P x_j} (i,j)$, but
the multiplication in this case is from the left. 

\begin{lemma}\label{PPPandLY}
$L_Y$ and $\PPP_y$ commute, i.e.,
 $$ L_Y \cdot \PPP_y = \PPP_y \cdot L_Y .$$
\end{lemma}
 

Proof:  It is sufficient to prove that the Laplacian $L_Y$ commutes with every 
transposition of the form $(i, k)$, where $y_i = y_k$, because every
permutation in $\PPP_y$ can be written as a product of those permutations.
So, let $y_i = y_k$. That means that $\PPP_y$ has transposition
$(i,k)$ as one of its summands.
 Let $y_j \in Y$ be comparable to $y_i$ (thus it is
comparable to $y_k$). In that case, the Laplacian $L_Y$ contains both 
transpositions, $(i,j)$, and $(k,j)$, i.e., $L_Y = \cdots + (i,j) + (k,j) +
\cdots$. 

 But, $(i,k) \cdot (i,j) = (k,j) \cdot (i,k)$, which shows that
$\PPP_y \cdot L_Y =L_Y \cdot \PPP_y$.\DONE


Using exactly same argument we see that $L_X$ and $\PPP_x$ also commute.















We know from section \ref{ch4}  that $L_D$ is a scalar matrix on each block, and thus it  commutes with $L_X$ and $L_Y$.

As for the $L_X$ and $L_Y$, we have the following Lemma:
\begin{lemma}\label{lemma5.1}
 $$L_X \cdot L_Y \cdot (\zzeta n) = L_Y \cdot L_X \cdot (\zzeta n)
$$
\end{lemma}

Proof:

The absence of certain relations in the poset
may cause terms in the Laplacian to be missing. That is why this lemma
is not obvious, and needs to be proved.

Let $\zeta = \zzeta n$. 
Without loss of generality we can assume that all of the $x$'s and all of the $y$'s 
are distinct, because if they were not,  we would just apply the same reasoning to 
each appearance of an observed element.
Let $(x_i,x_j)$ be a transposition of the operator $L_X$, and let $(y_k,y_l)$ be a 
transposition of the operator $L_Y$. 
If all of the numbers $i,j,k,l$ are
distinct, we have nothing to prove since it would not make any difference 
which transposition  was applied first. On the other hand, if $i=k$ and $j=l$, 
again there is nothing to prove, since their combined action would amount to 
multiplying with -1 no matter in which order they are applied.

Therefore assume that $i \not = k$ but $j=l$, i.e.,  we have
two transpositions, $(x_i,x_j)$ and $(y_j,y_k)$ in $L_X$ and $L_Y$ respectively, 
which overlap at one position.
Without loss of generality assume that $n=3$.
There are only three elements of the pure
wedge, call them $\zz 1 \ww \zz 2 \ww \zz 3$, i.e., $i=1, j=2, n=k=3$.

Let
\begin{eqnarray*}
\AA&=&(x_1,x_2)\cdot(y_2,y_3)\cdot (\zz 1 \ww \zz 2 \ww \zz 3) \\
&=&(x_1,x_2)\cdot(\zz 1 \ww z_{x_2,y_3}\ww z_{x_3,y_2})\\
&=& - ( z_{x_1,y_3} \ww z_{x_2,y_1} \ww z_{x_3,y_2}) 
\end{eqnarray*}

and 
 
\begin{eqnarray*}
\BB &=&(y_2,y_3) \cdot (x_1,x_2) \cdot (\zz 1 \ww \zz 2 \ww \zz 3) \\
&=& - (y_2,y_3) \cdot (z_{x_1,y_2} \ww z_{x_2,y_1} \ww \zz 3 )\\
&=& - ( z_{x_1,y_3} \ww z_{x_2,y_1} \ww z_{x_3,y_2}).
\end{eqnarray*}


 
Thus $$L_X \cdot L_Y\cdot (\zz 1 \ww \zz 2 \ww \zz 3) = L_Y \cdot L_X \cdot (\zz 1 \ww \zz 2 \ww \zz 3 ), $$ 
whenever all of the relations used above are present, i.e., whenever
every $x_i$ is beneath each $y_j$.
 That can be explained by the fact that $L_X$ is
acting on the $x$-indices and $L_Y$ is
acting on the $y$-indices.  


The question remains whether the answer would be the same if some of the
relations needed above were missing, and only one of the expressions above gets
annulled.  
The final expressions in both $\AA$ and $\BB$ are 0 unless:
$$
x_1 < y_3, \quad \quad x_2 < y_1, \quad \quad x_3 < y_2 .
$$
Suppose (without loss of generality)
 that $\BB$ above survives the  procedure, i.e.,  
we have the relation $ x_1 < y_2$.
On the other hand if $\AA$ is annulled in the middle step, the only
possible conflict left is $x_2 \not < y_3$. 
We have that $y_2$ and $y_3$ are comparable, otherwise the transposition 
$(y_2, y_3)$ wouldn't be a summand of  
$L_Y$. If $y_2 < y_3$, then $x_2 < y_2 <y_3$, which is 
contrary to the just stated
 assumption. Thus, we must have $y_3 < y_2$. Also $x_1 < y_3$ 
by our assumption above, and $(x_1, x_2)$ is a  transposition in $L_X$, so they 
 must also  be  comparable. By the same argument as above, $x_2$ must be larger than $x_1$. Hence in the interval $(x_1,y_2)$ there are two
elements $x_2 $ and $ y_3$. Since the poset is linear -- those two elements
must be comparable, and since we assumed that $x_2 \not < y_3$, it must
be that $x_2 > y_3$.

All together, the relations are:
\begin{eqnarray*}
\begin{array}{c}
y_2 \\
y_1 \\
\end{array} > x_2 > y_3 > \begin{array}{c} x_3 \\ x_1 \\ \end{array}
\end{eqnarray*}
So in this case we have $y_1 > y_3$, and $x_2 > x_3$.
Let
\begin{eqnarray*}
\CC &=&(x_2,x_3)\cdot(y_1,y_3)\cdot (\zz 1 \ww \zz 2 \ww \zz 3) \\
&=&(x_2,x_3)\cdot(z_{x_1,y_3} \ww z_{x_2,y_2}\ww z_{x_3,y_1})\\
&=& - ( z_{x_1,y_3} \ww z_{x_2,y_1} \ww z_{x_3,y_2}) 
\end{eqnarray*}


and 
 
\begin{eqnarray*}
\DD&=&(y_1,y_3) \cdot (x_2,x_3) \cdot (\zz 1 \ww \zz 2 \ww \zz 3) \\
&=& - (y_1,y_3) \cdot (z_{x_1,y_1} \ww z_{x_2,y_3} \ww z_{x_3,y_2} )\\
&=& 0
\end{eqnarray*}
since $x_2 > y_3$.

The expressions $\AA$ and $\CC$ are two summands of the product
$L_X L_Y$, while $\BB$ and $\DD$ are two summands of the product
$L_Y L_X$. As we can see, $\AA + \CC = \BB + \DD$.
Thus $L_X \cdot L_Y = L_Y \cdot L_X$.
\DONE

\bigskip

In view of Lemma~\ref{lemma5.1}, $L_X$, $L_Y$ and $L_D$ are commuting linear
transformations. So, to analyze the spectrum of their sum, we can compute the eigenvalues and eigenspaces of each separately. We will begin with
$L_Y$.


\subsection{ A  poset  tableau of type $(X,Y)_P$}




\begin{defi}
The {\em diagram of the L-block}, $P[X,Y]$,
 spanned by the sets $(X,Y)_P$, is the 
Hasse diagram of the subposet $X \cup Y$ with order inherited
from the poset $P$. Furthermore every vertex of $P$, which is in the
intersection $X \cap Y$ is split into two nodes, with the $x$-node above
the $y$-node.
\end{defi}





\begin{defi}
Given a node $v$ in $P[X,Y]$ define the {\em repetition number of $v$ }, $k(v)$,
to be the number of times that $v$ appears in the multiset $X$ if $v$ is an
$x$-node of $P[X,Y]$, or the multiset $Y$ if $v$ is a $y$-node of $P[X,Y]$.
\end{defi}


Let $C(v)$ be the set of covers of node $v$ in $P[X,Y]$. If $v$ is a maximal
node, than $C(v) = \emptyset$. 









\begin{defi}\label{def:poset}
A {\em  poset  tableau of type $(X,Y)_P$ } (or just of type $(X,Y)$) is 
any  labeling  $\LL$ of the diagram, $P[X,Y]$, of the L-block $V$ spanned by $(X,Y)$, where the labels are partitions $\LL(v)$, such that $\LL(v)$ is 
a partition of the number  $\sum_{w\ge v} \epsilon(w) k(w)$, where
$$\epsilon(w) = \left\{ \begin{array}{ll}
                        +1 & \mbox{ if $w$ is a $y$-node} \\
                        -1 & \mbox{ if $w$ is an $x$-node.}
                        \end{array}
                \right.
$$
\end{defi}








Given a poset tableau $\LL$ we will define the {\em multiplicity of $\LL$, $m(\LL)$},
and the {\em eigenvalues of $\LL$, $e(\LL)$.}  
\begin{defi}

\begin{itemize}
\item
Let $v$ be a $y$-node of the diagram $P[X,Y]$, labeled with the
partition $\LL(v)$ and with repetition number $k(v)$. Let $C(v)=
\{ v_1, v_2, \dots , v_l\}$
be the set of covers of $v$.
Let $\lambda_i$ denote $\LL(v_i)$, and let $k_i$ denote the 
repetition numbers, $k(v_i)$. The 
  {\em multiplicity of $\LL$ at $v$}  is defined to be 
$$
 m_v(\LL) = c_{\lambda_1, \dots, \lambda_l, k(v)}^{\LL(v)} 
$$


\item
Let $v$ be an $x$-node of the diagram $P[X,Y]$, labeled with the
partition $\LL(v)$ and with repetition number $k(v)$. Let $C(v)=
\{ v_1, v_2, \dots , v_l\}$
be the set of covers of $v$.
Let $\lambda_i$ denote $\LL(v_i)$, and let $k_i$ denote the 
repetition numbers, $k(v_i)$. The 
  {\em multiplicity of $\LL$ at $v$}  is defined to be 
$$
 m_v(\LL) = \sum_{\mu} c_{\lambda_1, \dots, \lambda_l}^{\mu} c^{\mu}_{\LL(v), 1^{k(v)}}.
$$
\end{itemize}


\end{defi}

If the multiplicity $m_v(\LL) = 0$ then we know that that particular 
labeling is not valid.












Now, we will define the $y$-eigenvalues for each $y$-node $v$ of the diagram $P[X,Y]$.
We want to have as many $y$-eigenvalues as the value
of  multiplicity.
From the representation theory of the symmetric group, we know that 
\begin{equation}\label{eig1}
c_{\lambda_1, \dots, \lambda_l, k(v)}^{\LL(v)} = \sum_{
\mbox{\tiny
$\LL(v)/\mu=k(v)${--horizontal strip}}
}
c_{\lambda_1, \dots, \lambda_l}^{\mu}.
\end{equation}
The {\em node--eigenvalue, $e_v(\LL)$, } for each node $v$,  
 is the set of the sums of the content over all
squares in $\LL(v)/\mu$ for all possible $\mu$ for which 
$\Lambda(v)/\mu$ is a $k(v)$--horizontal strip
 minus the binomial coefficient  $\binom{k(v)}{2}$.

Recall that the content of a square is given by $c(i,k) = k-i$ if 
the square is at position $(i,k)$ in a partition (\ith row and \kth column).


This gives $m_v(\LL)$ eigenvalues at each $y$-node $v$. We now define
{\em $y$-eigenvalue of $\LL$}, $e_y(\LL)$, to be the set of numbers obtained
by taking a sum of one element of $e_v(\LL)$ for each $y$-node $v$.
So $\abs{e_y(\LL)} = \prod_{\mbox{$y$-nodes $v$}} m_v(\LL)$. 


\subsection{Example \label{exa44}}

Let the poset $P = \{ 1,2,3,4,5 \}$ with the relations $1<_P 2$, 
$2<_P 3$, $3<_P 4$ and $4<_P 5$. The Hasse diagram of this poset is
given in figure~\ref{V--9}.
%===================================================================
%\begin{figure} 
%\centerline{\picture{1}{1.208}{2.083}{pic51.1.ps}}
%\caption{Example: The poset $P$ from Example~\ref{exa44}}\label{V--9}
%\end{figure}
\begin{figure}[t]
\begin{center}
\begin{picture}(60,140)(30,0)

%----------------------------------------

\put(0,0){\line(0,1){120}}
\put(3,-1){1}
\put(3,29){2}
\put(3,59){3}
\put(3,89){4}
\put(3,119){5}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(0,90){\circle*{3}}
\put(0,120){\circle*{3}}

%----------------------------------------------

\end{picture}
\caption{ Example: poset $P$}\label{V--9}
\end{center}
\end{figure}



 Let $X$ and $Y$ be  the sets $X=\{1,2,3\}$ and $Y=\{ 4,4,5\}$. So the 
 node $4$, is a node with non-trivial repetition number $k(4) = 2$.
The L-block $V$ is spanned by the following pure wedges:
\begin{eqnarray*}
       \zeta &=& z_{1,4}\ww z_{2,4} \ww z_{3,5} \\
        \tau &=& z_{1,4}\ww z_{2,5} \ww z_{3,4} \\
        \eta &=& z_{1,5}\ww z_{2,4} \ww z_{3,4} .
\end{eqnarray*}
Thus the L-block $V$ is 3-dimensional.
We calculate the Laplacian $L_Y$ on these three elements. Note that the Laplacian
$L_Y$ is in fact $L_Y = (4,5)$, since those are the only two comparable $y$'s.

\begin{eqnarray*}
L(\zeta)  &=& \tau + \eta \\ 
L(\tau)  &=& \zeta + \eta \\
L(\eta) &=& \zeta + \tau
\end{eqnarray*}
The matrix representation of $L_Y$ with respect to the basis $<\zeta, \tau, \eta
>$ is thus 
\begin{eqnarray*}
L_Y = \left( \begin{array}{ccc}
                0 & 1 & 1 \\
                1 & 0 & 1 \\
                1 & 1 & 0 \\
                \end{array} \right)
\end{eqnarray*}





So the eigenvalues of the Laplacian  are $-1, -1, +2$.

Now we will evaluate the $y$--eigenvalue for each of the  poset 
tableaux for this L-block. 
The only nontrivial node is node $4$.
Thus, the $y$--eigenvalue, $e_y(\Lambda)$, is the node-eigenvalue, 
$e_4(\Lambda) = c(1,2) + c(1,3) - \binom{2}{2}$.  
The result is given in figure~\ref{51.9}.
%==========================================================
%\begin{figure} 
%\centerline{\picture{1}{3.250}{2.444}{pic51.9.ps}}
%\caption{Example: the $y$-eigenvalues}\label{51.9}
%\end{figure}
\begin{figure}[t]
\begin{center}
\begin{picture}(300,200)(0,0)

%----------------------------------------
\put(50,50){\begin{picture}(60,140)(30,0)
\put(0,0){\line(0,1){120}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
\put(8,59){\makebox(0,0)[bl]{\one\one}}
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\put(8,119){\makebox(0,0)[bl]{\one}}
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\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(0,90){\circle*{3}}
\put(0,120){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e_y(\Lambda) = 2$}}
\end{picture}}
%----------------------------------------------
%----------------------------------------
\put(150,50){\begin{picture}(60,140)(30,0)
\put(0,0){\line(0,1){120}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
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\put(8,89){\makebox(0,0)[bl]{\one\one}}
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\put(8,119){\makebox(0,0)[bl]{\one}}
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\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(0,90){\circle*{3}}
\put(0,120){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e_y(\Lambda) = -1$}}
\end{picture}}
%----------------------------------------------
%----------------------------------------
\put(250,50){\begin{picture}(60,140)(30,0)
\put(0,0){\line(0,1){120}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
\put(8,59){\makebox(0,0)[bl]{\one}}
\put(8,48){\makebox(0,0)[bl]{\one}}
\put(8,89){\makebox(0,0)[bl]{\one\one}}
\put(8,78){\makebox(0,0)[bl]{\one}}
\put(8,119){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(0,90){\circle*{3}}
\put(0,120){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e_y(\Lambda) = -1$}}
\end{picture}}
%----------------------------------------------

\end{picture}
\caption{Example: the $y$-eigenvalues}\label{51.9}
\end{center}
\end{figure}


Note that the $y$--eigenvalues of this labeling
 give exactly the same numbers as 
 the eigenvalues of the Laplacian $L_Y$.
In the next section we will show that this is not coincidental.






%--------------------------------------------------------------------------------------------
%--------------------------------------------------------------------------------------------
%  
%--------------------------------------------------------------------------------------------

\section{Centerpiece Theorem for $L_Y$}


\begin{theorem}[$L_Y$-Centerpiece]
Let $P$ be a linear poset with a minimum element, $\hat 0$. Let
$X$ and $Y$ be two (multi-)sets, subsets of $P$.
For every labeling $\LL$ of positive multiplicity,
 each
element in $e_y(\LL)$ is an eigenvalue of $L_Y$ with multiplicity
$\prod_{\mbox{$x$-nodes $v$}} m_v(\LL)$.

\end{theorem}
  



Proof:

The proof of this theorem will be by induction on the sizes of
the (multi)-sets $X$ and $Y$. So let $n=\abs{X} = \abs{Y}$ (counting
multiplicities).

If $n=1$ --- there is nothing to prove  as the Laplacian $L_Y$ has 
no pairs to switch, and  the only $y$-node is the maximal element for the diagram of
the L-block. The Laplacian $L_Y$ is the one-by-one zero matrix
and the eigenvalue of this unique pair is zero.


Suppose $n=2$. There are several different possible combinations of 
relations between sets $X=\{x_1,x_2\}$ and $Y=\{y_1, y_2\}$. 
\begin{itemize}
\item The most obvious one is $x_1 < x_2 < y_1 < y_2$. In that case
        the Laplacian $L_Y = (y_1, y_2)$, and the two possible elements
        are $\zeta_1 = \zz 1 \ww \zz 2$, 
        $\zeta_2 = z_{x_1,y_2} \ww z_{x_2,y_1}$. The Laplacian $L_Y$
        has the following matrix representation with
         respect to the basis $\< \zeta_1 ,
        \zeta_2 \>$:
        \begin{eqnarray*}
        L_Y = \left( \begin{array}{cc}
                        0 & 1\\
                        1 & 0
                        \end{array}
                        \right)
        \end{eqnarray*}
        The eigenvalues of $L_Y$ are $+1$ and $-1$.
        The eigenvalue of the   poset  tableau of type $(X,Y)_P$
        is given in figure~\ref{sl51.11}. It also gives values
        $+1$ and $-1$, so the claim of the theorem holds.
\item The second case is when 
\begin{eqnarray*}
        x_1 < x_2 < \begin{array}{c}{y_1}\\{y_2}\end{array} 
\end{eqnarray*} 


%\begin{figure}
%\centerline{\picture{1}{1.930}{1.944}{pic51.11.ps}}
%\caption{poset  tableaux}\label{sl51.11}
%\end{figure}
%==========================================================
\begin{figure}[t]
\begin{center}
\begin{picture}(200,120)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,110)(30,0)
\put(0,0){\line(0,1){90}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
\put(8,59){\makebox(0,0)[bl]{\one\one}}
\put(8,89){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(0,90){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e_y(\Lambda) = +1$}}
\end{picture}}
%----------------------------------------------
%----------------------------------------
\put(150,30){\begin{picture}(60,110)(30,0)
\put(0,0){\line(0,1){90}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
\put(8,59){\makebox(0,0)[bl]{\one}}
\put(8,48){\makebox(0,0)[bl]{\one}}
\put(8,89){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(0,90){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e_y(\Lambda) = -1$}}
\end{picture}}
%----------------------------------------------


\end{picture}
\caption{ poset  tableaux}\label{sl51.11}
\end{center}
\end{figure}


In that case the Laplacian $L_Y$ does nothing (since $y_1$ and $y_2$
are not comparable), thus the eigenvalues of $L_Y$ are $0$. The 
dimension of the L-block spanned by $(X,Y)$ is two.
The eigenvalues of the   poset  tableaux give the same values 
(figure~\ref{sl51.12}), where the ''$\cdot$'' in a box 
denotes which square was deleted in that step.

%\begin{figure}
%\centerline{\picture{1}{2.875}{1.750}{pic51.12.ps}}
%\caption{poset  tableaux}\label{sl51.12}
%\end{figure}
%==========================================================
\begin{figure}[t]
\begin{center}
\begin{picture}(200,120)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,110)(30,0)
\put(0,0){\line(0,1){30}}
\put(0,30){\line(1,1){30}}
\put(0,30){\line(-1,1){30}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one\taken}}
\put(38,59){\makebox(0,0)[bl]{\one}}
\put(-38,59){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(30,60){\circle*{3}}
\put(-30,60){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e_y(\Lambda) = 0$}}
\end{picture}}
%----------------------------------------------
%----------------------------------------
\put(150,30){\begin{picture}(60,110)(30,0)
\put(0,0){\line(0,1){30}}
\put(0,30){\line(1,1){30}}
\put(0,30){\line(-1,1){30}}
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\put(8,29){\makebox(0,0)[bl]{\one}}
\put(8,18){\makebox(0,0)[bl]{\taken}}
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\put(-38,59){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(30,60){\circle*{3}}
\put(-30,60){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e_y(\Lambda) = 0$}}
\end{picture}}
%----------------------------------------------


\end{picture}
\caption{ poset  tableaux}\label{sl51.12}
\end{center}
\end{figure}

Thus in this case the theorem checks too.

\item $x_1 < y_1 < x_2 < y_2$ or equivalently 
(for our purpose) $x_1 < y_1 = x_2 < y_2$.\\
 There is only one poset  tableau spanned by these
sets $X$ and $Y$, namely the one shown on the figure~\ref{sl504a}.

%\begin{figure}
%\centerline{\picture{0.7}{0.680}{1.944}{pic504.ps}}
%\caption{\label{sl504a} poset  tableau}
%\end{figure}
%==========================================================
\begin{figure}
\begin{center}
\begin{picture}(100,120)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,110)(30,0)
\put(0,0){\line(0,1){90}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
\put(8,59){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,89){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(0,90){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e_y(\Lambda) = 0$}}
\end{picture}}
%----------------------------------------------

\end{picture}
\caption{\label{sl504a} poset  tableau}
\end{center}
\end{figure}




The $y$-eigenvalue for the poset tableau is zero in both cases.

The Laplacian $L_Y$ can not switch the $y$'s, since that would
produce the element $z_{x_2,y_1}$ which doesn't exist.
So the Laplacian $L_Y$ also acts as zero.


\item $x_1 < x_2 < y_1 = y_2$.\\
There is only one poset  tableau spanned by these
sets $X$ and $Y$, namely the one shown on the figure~\ref{sl505a}.

%\begin{figure}
%\centerline{\picture{0.7}{0.680}{1.444}{pic505.ps}}
%\caption{\label{sl505a} poset  tableau}
%\end{figure}
%==========================================================
\begin{figure}
\begin{center}
\begin{picture}(100,90)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,80)(30,0)
\put(0,0){\line(0,1){60}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
\put(8,59){\makebox(0,0)[bl]{\one\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e_y(\Lambda) = 0$}}
\end{picture}}
%----------------------------------------------

\end{picture}
\caption{\label{sl505a} poset  tableau}
\end{center}
\end{figure}


The $y$-eigenvalue is again zero (contents of the partition $(2)$ minus the binomial coefficient  $\binom{2}{2}$).
The Laplacian $L_Y$ doesn't have two distinct $y$'s to switch, thus, it
is zero.

\item $x$'s are the same.
$$x_1 = x_2 < \begin{array}{c} y_1 \\ y_2 \end{array}$$
There is only one poset  tableau spanned by these
sets $X$ and $Y$, namely the one shown on the figure~\ref{sl506a}.

%\begin{figure}
%\centerline{\picture{0.7}{1.125}{1.250}{pic506.ps}}
%\caption{\label{sl506a} poset  tableau}
%\end{figure}
%==========================================================
\begin{figure}
\begin{center}
\begin{picture}(100,50)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,60)(30,0)
\put(0,0){\line(1,1){30}}
\put(0,0){\line(-1,1){30}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(36,29){\makebox(0,0)[bl]{\one}}
\put(-40,29){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(30,30){\circle*{3}}
\put(-30,30){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e_y(\Lambda) = 0$}}
\end{picture}}
%----------------------------------------------

\end{picture}
\caption{\label{sl506a} poset  tableau}
\end{center}
\end{figure}


The $y$-eigenvalue is zero.
The Laplacian $L_Y$ has no comparable $y$'s to switch - thus $L_Y=0$.

\item $x_1 = x_2 < y_1 < y_2$.
There is only one poset  tableau spanned by these
sets $X$ and $Y$, namely the one shown on the figure~\ref{sl507a}.

%\begin{figure}
%\centerline{\picture{0.7}{0.680}{1.444}{pic507.ps}}
%\caption{\label{sl507a} poset  tableau}
%\end{figure}
%==========================================================
\begin{figure}
\begin{center}
\begin{picture}(100,90)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,80)(30,0)
\put(0,0){\line(0,1){60}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one\one}}
\put(8,59){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e_y(\Lambda) = 1$}}
\end{picture}}
%----------------------------------------------

\end{picture}
\caption{\label{sl507a} poset  tableau}
\end{center}
\end{figure}

The $y$-eigenvalue is equal to 1.
The Laplacian $L_Y$ can switch $y_1$ and $y_2$ but the result
would be the same element, since the $x$'s are indistinguishable.
Thus the Laplacian $L_Y = Id$, with the eigenvalue 1.


\item In the trivial case when the $x$'s are not comparable the $y$'s 
are not comparable because of linearity and the 
existence of a minimal element. So we have
$$ 
\begin{array}{ccc} x_1 & < & y_1 \\x_2 & < & y_2 \end{array}.
$$
The poset tableau again gives zero as the $y$-eigenvalue, and since
the $y$'s are not comparable, the Laplacian $L_Y$ is also zero.



So the theorem holds for  the case $n=2$.
\end{itemize}

Now, we will treat the general case $n>2$. 



\begin{itemize}

\item  Label the $y$-nodes of the diagram of the L-block
                using the depth-first algorithm:
        \begin{enumerate}
        \item Start with a leftmost minimal $y$-element $v$. 
        \item{\label{step0}} If $v$ is not the maximal unlabeled $y$-node 
                go to the leftmost unlabeled cover of $v$, and repeat this step.
                Otherwise label $v$ with next available number
                from the set $\{1,2,\dots, \abs{Y}\}$.
        \end{enumerate}
        From this labeling we see  that,
        $y_i > y_j   \Rightarrow i<j$. 


\item Let $P[X_1,Y_1], P[X_2,Y_2], \dots , P[X_c,Y_c]$ be the connected
components of $P[X,Y]$. In that case the L-block
$V$ is the tensor product of the L-blocks of the $P[X_i,Y_i]$. 
The Laplacian $L_Y$ switches only comparable $y$'s, and two $y$'s from
different connected components are not comparable. Thus 
$$
L_Y(v_1 \odo  v_c) = \sum_{i=1}^c v_1 \odo (L_{Y_i}v_i) \odo v_c .
$$
So if $v_1, \dots , v_c$ are eigenvectors of $L_Y$ with
the eigenvalues $e_1, \dots , e_c$, then $v_1 \odo v_c$ is an 
eigenvector with eigenvalue $e_1 + \cdots + e_c$. Thus, by induction, 
we can label each of the components of $P[X,Y]$ to get the eigenvalues
of $L_Y$ on the total L-block $V$.




\item
Suppose that $P[X,Y]$ is connected. In that case, there must 
be a minimal element in $P[X,Y]$, which must be an $x$-node.

Call the minimal element $x_n$. Then define $x_{n-1}, x_{n-2}, \dots ,
x_{a}$ by:
\begin{enumerate}
        \item   $x_a > x_{a+1} > \cdots > x_n $, where all $>$ are covering
                relations.
        \item Either
                \begin{itemize}
                \item[Case 1: ] There is more than one element covering $x_a$.
                \item[Case 2: ] $x_a$ has unique cover in $P[X,Y]$ but it
                                is a $y$-element.
                \end{itemize}
        
\end{enumerate}

Let $B = \{ x_a, \dots , x_n\}$, and let $G = $Sym($B$).

\begin{lemma}\label{first.one.lemma}
Let $\sigma \in G$, and let $\zeta = z_{x_{i_1},y_1} \ww z_{x_{i_2},y_2} \wdw 
z_{x_{i_n},y_n}$ be non-zero. Then
$$
\zeta^{\sigma} = z_{\sigma(x_{i_1}),y_1} \ww z_{\sigma(x_{i_2}),y_2} \wdw 
z_{\sigma(x_{i_n}),y_n}
$$
is also non-zero.
\end{lemma}

Proof: It is sufficient to prove the lemma for the transposition
$(x_{i_k},x_{i_l}) \in G$.
$$
\zeta^{(x_{i_k},x_{i_l})} = z_{x_{i_1},y_1} \wdw z_{x_{i_l},y_l} 
\wdw z_{x_{i_k},y_k} \wdw 
z_{x_{i_n},y_n}.
$$
Now, since $(x_{i_k},x_{i_l}) \in B$, i.e., $x_{i_k},x_{i_l}$ are both less or equal to
$x_a$, which is below all of the $y$'s -- the lemma is clear.\DONE

\begin{lemma}\label{second.one.lemma}
This action of $G$ commutes with $L_Y$.
\end{lemma}

Proof: Let $\zeta = z_{x_{i_1},y_1} \ww z_{x_{i_2},y_2} \wdw 
z_{x_{i_n},y_n}$ be in our L-block, $V$, let $y_k <_P y_l$
and let $\sigma \in G$. Then
$$
\sigma \cdot (y_k,y_l) \cdot \zeta = z_{\sigma(x_{i_1}),y_1} \wdw
z_{\sigma(x_{i_k}),y_l} \wdw z_{\sigma(x_{i_l}),y_k} \wdw
z_{\sigma(x_{i_n}),y_n}
$$
and
$$
(y_k,y_l)\cdot \sigma \cdot \zeta = z_{\sigma(x_{i_1}),y_1} \wdw
z_{\sigma(x_{i_k}),y_l} \wdw
z_{\sigma(x_{i_l}),y_k} \wdw z_{\sigma(x_{i_n}),y_n} .
$$
So they are equal unless one of the expressions above is zero, and the other
is not. The only way for that to happen is in the middle step, i.e., either
$\sigma \cdot \zeta =0$ or $(y_k,y_l)\cdot \zeta = 0$. But we assumed the
$\zeta \not = 0$, and by our lemma above $\sigma \cdot \zeta \not = 0$.
So the only possible conflict is $(y_k,y_l) \cdot \zeta =0$, and
$(y_k,y_l) \cdot \sigma \cdot \zeta \not = 0$. But since $\sigma \in G$, it
only moves elements of $B$ which are allowed to be paired with 
any $y_k$.\DONE



Let $C(x_a)=\{v_1, v_2, \dots, v_l\}$ be the set of covers of $x_a$.
We will prove the following generalization of the Centerpiece Theorem:

\begin{theorem}

Let $\LL$ be a poset tableau of positive multiplicity, and let
$\lambda_i$ be the label of $v_i$ in $\LL$. Let $\lambda$ be a partition
such that
$$
        c_{\lambda_1, \dots, \lambda_l}^\lambda c_{\LL(x_a), 1^{k(x_a)}}^{\lambda} \not = 0.
$$
Then the occurrences of $G$-irreducibles $S^\lambda$ in $V$ can be indexed 
by such poset tableaux $\LL$, and the Laplacian $L_Y$ acts on $S^\lambda$
as one of the scalars in $e_Y(\LL)$. 

\end{theorem}

%
%
%======================   C A S E 1   ===================
%
%

\subsubsection{Case 1}


%\begin{figure}[t]
%\centerline{\picture{0.66}{2.750}{3.972}{pic811.ps}}
%\caption{Case 1}\label{slucaj1}
%\end{figure}
\begin{figure}
\begin{center}
\begin{picture}(100,160)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,150)(30,0)
\put(0,0){\line(0,1){20}}
\put(0,60){\line(0,1){60}}
\put(0,100){\line(2,1){40}}
\put(0,100){\line(-1,1){20}}
\put(0,0){\circle*{3}}
\put(0,20){\circle*{3}}\put(0,30){\circle*{3}}\put(0,40){\circle*{3}}
\put(0,50){\circle*{3}}\put(0,60){\circle*{3}}\put(0,80){\circle*{3}}
\put(0,100){\circle*{3}}\put(0,120){\circle*{3}}
\put(-20,120){\circle*{3}}
\put(40,120){\circle*{3}}
\put(10,120){\circle*{3}}\put(20,120){\circle*{3}}\put(30,120){\circle*{3}}
\put(8,-3){\makebox(0,0)[bl]{$x_n$}}
\put(8,77){\makebox(0,0)[bl]{$x_{a+1}$}}
\put(8,97){\makebox(0,0)[bl]{$x_{a}$}}
\put(-22,112){\makebox(0,0)[bl]{$\mu_1$}}
\put(4,112){\makebox(0,0)[bl]{$\mu_2$}}
\put(44,112){\makebox(0,0)[bl]{$\mu_r$}}
\put(-20,120){\line(0,1){10}}\put(-20,130){\makebox(0,0)[bl]{$T_1$}}
\put(0,120){\line(0,1){10}}\put(0,130){\makebox(0,0)[bl]{$T_2$}}
\put(40,120){\line(0,1){10}}\put(40,130){\makebox(0,0)[bl]{$T_r$}}
\end{picture}}
%----------------------------------------------

\end{picture}
\caption{Case 1}\label{slucaj1}
\end{center}
\end{figure}


Suppose there are two or more subtrees above the node $x_a$, in our poset
$P$ (as in figure~\ref{slucaj1}). 
Label the subtrees above the $x_a$ by $T_1, T_2, \dots , T_r$.
 Let $Y = Y_1 \cup Y_2 \cup \cdots \cup Y_r$  ($Y_i \subset T_i$), where
$Y_i \cap Y_j = \emptyset$.   Let  $k_i = \abs{Y_i}$, and let
$$
Y_i = \{ y_{k_1+k_2+\cdots +k_{i-1}+1} , \dots , y_{k_1+\cdots +k_{i-1}+k_i} \}.
$$ 
Because of our labeling, we know  that all relations
between $y$'s are contained within the sets $Y_i$, i.e., $y_i <_P y_j$ implies
that both  $y_i, y_j$ are in the same $Y_k$.



Let $b_i$ be the number of $y_j$'s in $T_i$ minus the number
of $x_j$'s in $T_i$ (note that in general $T_i$ will have more
$y_j$'s than $x_j$'s). In other words, $b_i = \abs{Y_i} - \abs{X \cap T_i}$.

Consider element $\zeta = z_{x_{i_1},y_1} \ww z_{x_{i_2},y_2} \wdw 
z_{x_{i_n},y_n}$. If we want $\zeta$ to be non-zero, there will be
exactly $b_i$ $x_j$'s from $B$ in $\zeta$ paired up with the $y_j$'s
of $T_i$. 


Split the L-block 
\begin{equation}
V = \oplus_{ ( S_1,S_2,\dots,S_r) } V[S_1,S_2,\dots,S_r] , \label{spliting:eq}
\end{equation}
where $S_1 \cup S_2 \cup \cdots \cup S_r = B$, $\abs{S_i}=b_i$, and
 $V[S_1,\dots,S_r]$ is the span of all 
$\zeta$ with exactly the elements of $S_i$ paired with the $y$'s from $T_i$.

\begin{lemma}\label{LLLEEEMMMAAA}
\begin{enumerate}
\item[1.]
        As a vector space 
$$
V[S_1,\dots, S_r] \cong V(X_1\cup S_1, Y_1) \otimes
V(X_2\cup S_2, Y_2) \odo 
V(X_r\cup S_r, Y_r),
$$
where $X_i$ and $Y_i$ are the multisets of the $x$ and $y$ elements of the subtree $T_i$.

\item[2.]
With respect to the decomposition in 1., the Laplacian $L_Y$ acts as:
$$
L_Y(v_1 \odo v_r) = \sum_i v_1 \odo (L_{Y_i}v_i) \odo v_r .
$$

\end{enumerate}
\end{lemma}

Proof: Statement 1. is clear by the definition of $V[S_1,\dots, S_r]$.
 To prove statement 2., we only need to recall that the Laplacian $L_Y$
switches comparable $y$'s, and that $y$'s in different subtrees 
can not be comparable because of linearity of the poset. $L_Y$ can
switch only $y$'s in the same subtree $T_i$.\DONE







Let $G_i = \mbox{Sym($S_i$)}$. Note that 
$G_1 \times G_2 \times \dots
\times G_r$ acts on $V[S_1,S_2, \dots ,S_r]$.
Let $s_i$ be the minimal node of subtree $T_i$. 

Now apply the induction hypothesis to L-blocks, $V_i = V(X_i\cup S_i, Y_i)$.
According to our theorem this gives the decomposition of the L-block
$V_i$ as $G_i$-module, and the eigenvalues of $L_{Y_i}$ are indexed
by poset tableau of shape $\mu_i \vdash \abs{S_i}$.  Moreover
each poset tableau of shape $\mu_i$ with eigenvalue $e_i$ represents a
copy of the irreducible $S^{\mu_i}$ in the $e_i$-eigenspace.  

Now, as a $G_1 \times \dots \times G_r$ module we know the eigenspaces
of $L_Y$ are given by our labeling up to the points $s_i$ where
the partitions $\mu_i$  come together at $x_a$.






\begin{lemma}
As a $G = $Sym($B$)-module, the space $V$ is
$$
V  \cong \mbox{\rm ind}_{(\mbox{\tiny Sym($S_1^0$) $\times \cdots \times $
Sym($S_r^0$) )}}^{\mbox{\tiny Sym($B$)}} (V[S_1^0, \dots, S_r^0]) ,
$$
where $(S_1^0, \dots , S_r^0)$ is any fixed  ordered partition of
$B$.
\end{lemma}

Proof:
Choose $S_i^0 = \{ x_{a+b_1+\cdots+b_{i-1}}, x_{a+b_1+\cdots+b_{i-1}+1}, \dots
, x_{a+b_1+\cdots+b_{i}-1} \}$. Let \SS denote the set of permutations
$\sigma \in \mbox{Sym}(B)$ such that  $\sigma(u) < \sigma(v)$ whenever
$u,v$ are in the same set $S_i^0$ for some $i$. There is 1--1 correspondence
between  the $\sigma \in \SS$ and  the sequences indexing the summands 
in the~\ref{spliting:eq}, namely 
$$
\sigma \leftrightarrow (S_1^\sigma, \dots , S_r^\sigma)
$$
where $S_i^\sigma = \{ \sigma(u) : u \in S_i^0 \} $. Also \SS is a collection of coset representatives for $\mbox{Sym}(S_1^0) \times \cdots \times \mbox{Sym}(S_r^0)$ in Sym($B$).
Thus we have a natural vector space  isomorphism between $V$ and
$$
V[S_1^0, \dots, S_r^0] \otimes_{(\mbox{\tiny Sym($S_1^0$) $\times 
\cdots \times $
Sym($S_r^0$)})} \mbox{Sym}(B).
$$

It is straightforward to check that this isomorphism commutes with the 
action of Sym($B$).\DONE




 Let $\LL_i$ be a  poset  tableau of type $(X_i\cup S_i, Y_i)$ , where $\mu_i \vdash b_i$ is the label of the vertex $s_i$. By our inductive hypothesis the Laplacian $L_{Y}$ acts as a scalar
on the irreducible $S^{\mu_i}$, i.e., $L_{Y}\vert_{S^{\mu_i}}
 = e_Y(\LL_i)$.
Applying Lemma~\ref{LLLEEEMMMAAA} part 2., we have 


\begin{eqnarray*}
L_Y(v_1 \odo v_r) &=& \sum_i v_1 \odo (L_{Y_i}v_i) \odo v_r \\
&=& \sum_i v_1 \odo e_Y(\LL_i)v_i \odo v_r \\
&=& (\sum_i e_Y(\LL_i)) (v_1 \odo v_r).
\end{eqnarray*}

Now, we will use the fact (\cite{james,james-kerber,macdonald}) that 
$$
(S^{\mu_1}\odo S^{\mu_r})\uparrow_{G_1\times \cdots \times G_r}^G =\oplus_{\lambda \vdash 
\abs{B}} c_{\mu_1, \mu_2, \dots , \mu_r}^{\lambda} S^\lambda.
$$

Thus we have $c_{\mu_1, \mu_2, \dots , \mu_r}^{\lambda}$ copies of
the $G$-module $S^\lambda$ which explains why this is the 
multiplicity of the label $\lambda$ on node $x_a$ in our 
labeling.

\item 
Now we have to decide what is the dimension of each eigenspace.
But that is something we will have to do in the second case too - 
so we will do it for both cases at the end.


%
%
%===============      C A S E 2 =================================
%
%

\subsubsection{Case 2}

%\begin{figure}[h]
%\centerline{\picture{0.7}{1.708}{5.138}{pic55.1.eps}}
%\caption{Case 2}\label{prikaz.55.1}
%\end{figure}
%============================================================

\begin{figure}
\begin{center}
\begin{picture}(100,160)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,150)(30,0)
\put(0,0){\line(0,1){120}}
\put(0,120){\line(1,1){10}}
\put(0,120){\line(-1,1){10}}
\put(0,0){\circle*{3}}
\put(0,10){\circle*{3}}\put(0,20){\circle*{3}}\put(0,30){\circle*{3}}
\put(0,40){\circle*{3}}
\put(0,50){\circle*{3}}\put(0,60){\circle*{3}}
\put(0,70){\circle*{3}}\put(0,85){\circle*{3}}
\put(0,100){\circle*{3}}\put(0,115){\circle*{3}}
\put(0,120){\circle*{3}}
\put(8,-3){\makebox(0,0)[bl]{$x_n$}}
\put(8,27){\makebox(0,0)[bl]{$x_{a}$}}
\put(8,37){\makebox(0,0)[bl]{$y_n$}}
\put(8,67){\makebox(0,0)[bl]{$y_{n-k+1}$}}
\put(8,82){\makebox(0,0)[bl]{$t$}}
\put(20,83){\makebox(0,0)[bl]{$ \rangle NB$}}
\put(8,97){\makebox(0,0)[bl]{$s$}}
\put(20,98){\makebox(0,0)[bl]{$\rangle \hat{A}$}}
\put(8,112){\makebox(0,0)[bl]{$v_0$}}
\put(20,30){\line(2,-3){10}}
\put(20,0){\line(2,3){10}}\put(31,12){\makebox(0,0)[bl]{$B$}}
\put(20,70){\line(2,-3){10}}
\put(20,40){\line(2,3){10}}\put(31,52){\makebox(0,0)[bl]{$A$}}

\put(-2,123){\makebox(0,0)[bl]{$T$}}

\end{picture}}
%----------------------------------------------

\end{picture}
\caption{Case 2}\label{prikaz.55.1}
\end{center}
\end{figure}


Let $A = \{y_{n-k+1}, \dots , y_n \}$ be the largest possible set so 
that $$x_a \le_P y_n \le_P y_{n-1} \le_P \cdots \le_P y_{n-k+1}$$
and there are no $x_i$'s with $y_n \le_P x_i <y_{n-k+1}$. Call $A$ 
the "terminal $Y$-set of $V$" (figure~\ref{prikaz.55.1}).
Note that $\abs{B} \ge \abs{A}$. 
Split the L-block 
$$V = \oplus_{(a_1, \dots , a_k)} V(a_1,\dots, a_k)$$
for $(a_1,\dots a_k)$, a sequence of distinct elements of length $k$ 
from $B$, and the vector space $V(a_1,a_2,\dots , a_k)$ represents the span of all $\zeta$
which are of the form
$$
\zeta = \zz 1 \wdw \zz{{n-k}} \ww z_{a_1,y_{n-k+1}} \wdw z_{a_k,y_n}
$$

For the moment assume that all of the elements $x_a, \dots, x_n$ are distinct
and all of $y_{n-k+1}, \dots , y_n$ are distinct. Then we will look 
back at how we must modify the argument when some of the $x_i$'s and $y_j$'s 
are equal.

Fix the sequence $(a_1,\dots, a_k)$, let $B' = B \backslash \{a_1,\dots ,a_k\}$
and let $G' $ be the subgroup $G'= \mbox{Sym($B'$)} \leq G$. Note that $G'$ acts on $V(a_1,\dots
, a_k)$.

\begin{lemma}
$V(a_1,\dots, a_k)$ is isomorphic to the L-block $V_0$ given by the
following sets:\par
$X'= X \backslash \{a_1, a_2, \dots, a_k\}$ and $ Y' = Y \backslash \{y_{n-k+1}, \dots, y_n\}$.
\end{lemma}

Proof: The isomorphism $\psi: V(a_1, \dots, a_k) \rightarrow V_0$ is an
obvious one
$$
\psi(\zz 1 \wdw \zz{n-k} \ww z_{a_1,y_{n-k+1}} \wdw z_{a_k,y_n}) =
\zz 1 \wdw \zz{n-k}.
$$
It is clearly a bijective linear map.\DONE



\item
Now let's examine the L-block given by $X'$ and $Y'$. Let $\hat A$ be the 
terminal $Y$-set for $X',Y'$. Let $\hat B = B' \cup \{\mbox{new $x_i$'s} \}$
be the $x_i$'s below $\hat A$. Let $NB$ denote the set of those new $x_i$'s.
 Let $\hat G = \mbox{Sym($\hat B$)}$. Note that
$G' \leq \hat G$.

If $\hat A = \emptyset$ then there are no $y$'s in the interval
$(y_{n-k+1}, v_0]$. In that case the group $\hat G$ is the group
described in the case 1., i.e., $\hat G = G_1 \times G_2 \times \cdots
\times G_r$, where $G_i$ is acting on the subtree $T_i$ above $x_a$.

Now apply the induction hypothesis to $X', Y'$. This gives the decomposition
of the L-block as a $\hat G$-module. The theorem says that the irreducible
summands are indexed by the $(X',Y')$ poset tableaux $\hat 
\LL$ of shape $\lambda'$
(where $\lambda' \vdash \abs{\hat B}$), and "shape" means that the minimal
element $s$ of $\hat A$ is labeled with $\lambda'$. Also the theorem tells
 us that the $Y$-Laplacian $L_Y$ for $(X',Y')$ acts
like the scalar $e_Y(\hat \LL)$ on this copy of the irreducible $S^{\lambda'}$.
From this we can deduce by restriction from $\hat G$ to $G'$ the following

\begin{lemma}
As a $G'$-module, the space $V(a_1,\dots, a_k)$ decomposes as a sum over
all $(X',Y')$ tableaux $\hat \LL$ of shape $\lambda'$ of the module
$$
S^{\lambda'}\downarrow_{G'}^{{\hat{G}}}
$$
Moreover the Laplacian $L_Y$ for $(X',Y')$ acts like the scalar $e_Y(\hat \LL)$
on this entire restriction.
\end{lemma}

Let $\ch(\lambda', \mu')$ be    
the set  of  chains $\lambda' \ge \lambda_1 \ge \cdots \ge \mu'$
where the steps in the chain of the partitions  are all of size 1.

Now using the fact \cite{james} that 
$$
(S^{\lambda'})\downarrow^{{\hat{G}}}_{G'} = \oplus_{\mu' \vdash \abs{B'}} 
S^{\mu'} \abs{\ch(\lambda', \mu')},
$$
we can rewrite this lemma to say that the sum is over all extensions of $\hat \LL$ to a labeling of the new $x_i$'s of $S^{\mu'}$ where the extension
gives   the label $\mu'$ to $t$, the minimal element of the set $NB$.

\item
Now we need one last lemma.
\begin{lemma}
As a $G$-module $ V \cong {\mbox{\rm ind}}_{G'}^{G} (V(x_{n-k+1},\dots, x_n))$.
\end{lemma}

Proof: For each sequence $\aalpha = (a_1,\dots, a_k)$ let $\pi_\aalpha$
be the permutation in $G$ which maps $a_i$ to $x_{n-k+i}$ and which
leaves the elements of $B\backslash \{a_1,\dots, a_k\}$
in increasing order. Then $\pi_\aalpha$ is a set of coset representatives
for $G \backslash G'$.
  Since there is one for every $\aalpha$ this shows that as vector spaces

$$
V\cong V(x_{n-k+1},\dots,x_n) \otimes_{G'} {\Bbb C}(G).
$$
Let $g\in G$, let $(a_1,\dots, a_k) = \aalpha$ be a sequence, and let $b_i
=g(a_i), \bbeta = (b_1,\dots,b_k)$. Then 
\begin{eqnarray*}
\lefteqn{ g(\zz 1 \wdw \zz {n-k} \ww z_{a_1,y_{n-k+1}} \wdw
z_{a_k,y_n}) = }\\
& = &z_{g(x_1),y_1}\wdw z_{g_(x_{n-k}),y_{n-k}} \ww z_{b_1,y_{n-k+1}}\wdw z_{b_k,
y_n}\\
& = & (\pi_\bbeta^{-1} g \pi_\aalpha) z_{x_1',y_1}\wdw z_{x_{n-k}',y_{n-k}}\ww
z_{x_{n-k+1},y_{n-k+1}} \wdw z_{x_n,y_n},
\end{eqnarray*}
where $x_i'$ is obtained by replacing the elements of $\{x_a,\dots,x_n\} 
\backslash \{a_1,\dots,a_k\}$ by the elements of the set 
$\{x_a,\dots, x_{n-k}\}$ in order.
This computation shows that the vector space isomorphism above
is a $G$-module isomorphism.\DONE



Now putting all the claims together with the fact(\cite{james}) that
\begin{equation}
\mbox{ind}_{\mbox{\tiny Sym$(B')$}}^{\mbox{\tiny Sym$(B)$}}
(S^{\mu'}) = \oplus_{\ch(\lambda, \mu')} S^{\lambda} , \label{two-stars}
\end{equation}
 shows
that as a module for $G$, $V$ decomposes as a sum, over all $(X,Y)$ poset
tableaux $\LL$ of shape $\lambda$, of a copy of $S^\lambda$. 
In the expression~\ref{two-stars} 
we know by induction that the Laplacian $L_{Y'}$ involving all switches 
which do not involve the terminal set, i.e., $L_{Y'} = L_Y - L_0$ 
(where $L_0$ is the sum of the switches involving the terminal $Y$-set), 
acts like the scalar
$e_Y(\LL')$, where $\LL'$ is the labeling as far as the point $t$ (the
minimal $x$ above $A$). 
The switches involving the terminal $Y$-set $A$
must be studied. 
But those $y_i$ in $A$ can either be switched with each other or with 
$y_j$ that are above an $x_i \in B$. By the choice of the terminal set
$A$, $y_i$ is comparable to $y_j$ for all $y_i \in A$.
It follows that $L_0$ acts on  $V = \oplus V(a_1,\dots, a_k)$
by the sum of all transpositions $(x_i,x_j)$ for $x_i \in B, x_j \in
\{ a_1,\dots, a_k\}$.
In terms of our induced module,
$L_0$ acts on $$\mbox{ind}_{G'}^G(S^{\mu'}) = S^{\mu'} \otimes_{{\Bbb C}G'} 
{\Bbb C}G
$$
like left multiplication on ${\Bbb C}G$ by $\sum_{x_i,x_j \in B} (x_i,x_j) -
\sum_{x_i,x_j \in B'} (x_i,x_j)$.

But $\sum_{x_i,x_j \in B'} (x_i,x_j)$ passes through $\otimes_{{\Bbb C}G'}$
to act on $S^{\mu'}$ like the scalar $\sum_{x \in\mu'} c_x$ \cite{macdonald}.
And $\sum_{x_i,x_j \in B} (x_i,x_j)$ acts on each $G$-irreducible $S^\lambda$
in $S^{\mu'}\otimes_{{\Bbb C}G'} {\Bbb C}G$ like $\sum_{y \in \lambda} c_y$.
The result is that $L_0$ acts on each copy of $S^\lambda$ in $\mbox{ind}_{G'}^G
(S^{\mu'})$ like $\sum_{y\in \lambda /\mu'}c_y$. This explains the scalars
$e_Y(\LL)$ and their multiplicity. 

In order to be able to add the eigenvalues of $L_{Y'}$ and $L_0$, we
need the following lemma.
\begin{lemma}
$$
L_0 \cdot L_{Y'} = L_{Y'} \cdot L_0
$$
\end{lemma}

Proof:
Let $(x_i,x_j)$ be a transposition in Sym($B$), where $x_i \in B'$, and
let $(x_i,x_k)$ be a transposition of $L_{Y'}$. By the choice of
$B$ we know that $(x_j,x_k)$ is also transposition in $L_{Y'}$,
and since
$$
(x_i,x_j)(x_i,x_k) = (x_j,x_k)(x_i,x_j)
$$
the lemma is clear.\DONE





\item
  Now 
we want to consider the case where some of the elements
of the sets $A$, $B$, and $NB = \{\mbox{new $x$'s}\}$ are equal.
So let's write
\begin{eqnarray*}
A &=& \alpha_1 \cup \alpha_2 \cup \dots \cup \alpha_l, \\
B &=& \beta_1 \cup  \beta_2  \cup \dots \cup \beta_m ,\\
NB&=& \gamma_1 \cup \gamma_2 \cup \dots \cup \gamma_n, \\
\end{eqnarray*}
where $\abs{\alpha_i} = a_i$, $\abs{\beta_i} = b_i$ and $\abs{\gamma_i}
=c_i$, and  the $y$'s in each of the $\alpha_i$ are equal, the $x$'s in
each of the $\beta_i$ are equal, and the $x$'s in each $\gamma_i$ are equal.

Let 
\begin{eqnarray*}
\PPP_A &=& \sum_{\sigma_i \in S_{\alpha_i}} \sigma_1 \times \sigma_2 \times
\dots \times \sigma_l \\
\PPP_B &=& \sum_{\sigma_i \in S_{\beta_i}} (\prod \mbox{sgn}(\sigma_i))
\sigma_1 \times \dots \times \sigma_m \\
\PPP_{NB} &=& \sum_{\sigma_i \in S_{\gamma_i}} (\prod \mbox{sgn}(\sigma_i))
\sigma_1 \times \dots \times \sigma_n .\\
\end{eqnarray*}

These projections commute with the Laplacian $L_Y$ according to 
lemma~\ref{PPPandLY}. 

Let $\PPP$ be the projection $\PPP = \PPP_A \times \PPP_B \times \PPP_{NB}$, 
where the $\PPP_A$ acts on the $y$'s while the other two projections
act on the set of the $x$'s.
Then the projection $\PPP $ projects our original space (with 
all of the $x$'s and $y$'s distinct) examined in the previous 
paragraph to the space where $y_i$'s in each $\alpha_i$, $x$'s in each
$\beta_i$ and $\gamma_i$ are equal.
Each of the components of $\PPP$ commutes with $L_Y$ so we know that 
$\PPP$ and $L_Y$ commute.


\begin{lemma}
The projection $\PPP$ maps the L-block $V$ with all $x$'s and $y$'s
distinct to the L-block $\PPP V$, where $\alpha_i, \beta_i$
and $\gamma_i$ are sets of equal elements.
\end{lemma}

Proof:
Let $\zeta = \zz{1} \wdw z_{a_1,y_{i_1}} \wdw z_{a_k, y_{i_k}} \wdw \zz{n}
$. Note that if $y_i =y_j$, the element $\zeta^{(i,j)}$ and $\zeta$ are the
same. In general, if $\sigma_i \in S_{\alpha_i}$, then $\zeta^{\sigma_i} =
\zeta$. So we need to identify all equal elements. This is accomplished
by projecting with a  symmetrizer, i.e., we identify the class of elements
$\cup_{\sigma \in S_{\alpha_i}} {\zeta^\sigma}$ with the sum 
$\sum_{\sigma \in S_{\alpha_i}} {\zeta^\sigma}$. But $\PPP_A$ does
exactly this identification, $\PPP_A = \sum_{\sigma \in S_{\alpha_i}} {\zeta^\sigma}$. The same is true for $\PPP_B$ and $\PPP_{NB}$.\DONE


{\bf NOTE:}
When we deal with the projection $\PPP_A$ there is one important point
we have to make. The Laplacian $L_Y$ switches all comparable pairs
of $y$'s. If two of the $y$'s are the same -- they
would not get switched. 

Therefore, when we observed the Laplacian $L_0$, we have to subtract 
all switches involving  two $y$'s from the same $\alpha_k$. Each
of these transpositions doesn't move any of the $y$'s (or $x$'s), 
and there are exactly $\binom{a_k}{2}$ of them. Thus, we have
to subtract $\binom{a_k}{2}$ from the eigenvalue 
$\sum_{y\in \lambda /{\mu'}} c_y$ of $L_0$.


 Thus we can write our space $V$ as a direct sum of the eigenspaces
$$
V = \oplus_{w} V_w,
$$
where the sum is over all eigenvalues $w$ of $L_Y$.

We want to know the eigenvalues of $L_Y$ on the image $\PPP V$. We will use
that fact that the multiplicity of $w$ as an eigenvalue on $\PPP V$ is
the dimension of $\PPP V_w$.

So we need to compute the dimension of $\PPP V_w$. At present we have $V_w$
written in terms of labellings of  poset tableaux. So pick such a 
labeling
which at the end has a $\lambda'$ at vertex $s$ coming down to a $\mu'$ at
vertex $t$ then back up 
to a $\lambda$  at vertex $y_n$ (using the notation of this proof). This represents
a piece of the eigenspace of the corresponding eigenvalue $w$ where
$\hat G = \mbox{Sym}(\hat B)$ acts like $S^{\lambda'}$ and 
$G'=\mbox{Sym}(B')$ acts like $S^{\mu'}$. We need the following lemma.
\begin{lemma}
Let $B'\subset \hat B$, $NB = \hat B \backslash B'$. Then the multiplicity
of $S^{\mu'}$ in $\PPP_{NB} S^{\lambda'}$ is equal to the number
of ways to get $\mu'$ from $\lambda'$ by successfully removing vertical 
strips of lengths $c_1, c_2, \dots $.
\end{lemma}

Note: As a consequence of this lemma, when we came to an $x$-node $v$ of 
the repetition number $k(v)$ we have to remove a vertical strip of length
$k(v)$.

Proof of lemma: Recall that $NB= \gamma_1 \cup \gamma_2 \cup \dots \cup 
\gamma_n$. The projection $\PPP_{NB}$ projects onto the
$\mbox{sgn}_{\gamma_1}\otimes \mbox{sgn}_{\gamma_2} \odo \mbox{sgn}_{\gamma_n}$
isotypic component of $S^{\lambda'}$ considered as a $S_{\gamma_1} \times
\dots S_{\gamma_n}$ module. Thus, for $\mu'$ a partition of $\abs{B'}$,

\begin{eqnarray*}
\< \PPP_{NB} S^{\lambda'}, S^{\mu'}\>_{S_{B'}} &=& 
\< S^{\lambda'}, S^{\mu'}\otimes \mbox{sgn}_{\gamma_1} \odo 
\mbox{sgn}_{\gamma_n} \>_{S_{B'}\times S_{\gamma_1} \times \dots \times S_{
\gamma_n}} \\
&=& \mbox{
\begin{minipage}[t]{3.82in} 
(\# of ways to get $\mu'$ from $\lambda'$ by successively removing 
 vertical strips of lengths $c_1, c_2, \dots$)
\end{minipage}
} \\
\end{eqnarray*}


Now consider the next step of going up from $\mu'$ to $\lambda$. At this point
we have a piece of the $w$-eigenspace on which $B'$ acts like $S^{\mu'}$.

The multiplicity coming from this $S^{\mu'}$ is 
$\mbox{dim}(\PPP_A \mbox{ind}_{G'}^G (S^{\mu'}) \PPP_B)$.
We need to check how $\PPP_A$ acts on the induction $\mbox{ind}_{G'}^G(S^{\mu'})
= S^{\mu'} \otimes_{{\Bbb C}G'} {\Bbb C}G$.
The $\sigma \in \PPP_A$ permutes the $y$'s. We identified the induction by
identifying the sequence of the $x_i$'s that are paired with  the set $A$.
So switching $y_i$'s has the effect in Sym($B$) of switching the positions
corresponding to $B\backslash B'$. In other words, $\PPP_A - \PPP_B$
has the effect of projecting  onto the (trivial~$\otimes$~sgn) characters of
$(S_{\alpha_1} \times \dots \times S_{\alpha_l}) \times (S_{\beta_1}\times
\dots \times S_{\beta_m}) \subset S_{B\backslash B'} \times S_B$ where
$S_{B\backslash B'} \times S_B$ is acting on $S^{\mu'} 
\otimes_{{\Bbb C}G'} {\Bbb C}G$ via left multiplication on ${\Bbb C}G$ by 
 $ S_{B\backslash B'} $        and right multiplication on ${\Bbb C}G$
by $S_B$. So to determine $\mbox{dim}(\PPP_A \mbox{ind}_{G'}^G (S^{\mu'}) \PPP_B)$ it will be helpful to know the decomposition of the induction
$\mbox{ind}_{G'}^G(S^{\mu'})$ as a $S_{B\backslash B'} \times S_B$-module.


\begin{lemma}[\cite{hanlon}]
Let $\mu' \vdash m$, $G=S_r$, $G'=S_m$ and $H=S_{r-m}$ (acting on
$\{m+1, \dots, r\}$). Then as a $H\times G$ module, the induced representation
$S^{\mu'}\otimes_{{\Bbb C}G'} {\Bbb C}G$ decomposes as
$$
\mbox{ind}_{G'}^G(S^{\mu'})=
\oplus_{\lambda \vdash r, \mu\subset\lambda} S^{\lambda /\mu} \otimes S^\lambda.
$$
\end{lemma}

Now armed with that lemma, let us return to the dimension count.

\end{itemize}

\begin{eqnarray*}
\lefteqn{
\mbox{dim}(\PPP_A \mbox{ind}_{G'}^G (S^{\mu'}) \PPP_B) =}\\
&=& \< (\mbox{ind}_{G'}^G(S^{\mu'}))\downarrow , (\epsilon_{\alpha_1} \odo
\epsilon_{\alpha_l})\otimes (\mbox{sgn}_{\beta_1} \odo \mbox{sgn}_{\beta_m})\> \\
&=& \sum_{\lambda \vdash n, \mu' \subset \lambda} \<
(S^{\lambda /{\mu'}})\downarrow , (\epsilon_{\alpha_1} \odo
\epsilon_{\alpha_l})\>
\< S^\lambda\downarrow ,\mbox{sgn}_{\beta_1}\odo \mbox{sgn}_{\beta_m}\> \\
&=&  \sum_{\mu'\subset\lambda \vdash n} \<
(S^{\lambda /{\mu'}} \otimes S^\lambda)\downarrow ,
(\epsilon_{\alpha_1}\odo\epsilon_{\alpha_l})\otimes
(\mbox{sgn}_{\beta_1}\odo\mbox{sgn}_{\beta_m})\>\\
&=&
\sum_{\lambda\vdash n} 
\mbox{ \begin{minipage}[t]{4.75in}
(\# of ways to get $\lambda$ from $\mu'$ by adding horizontal strips 
 of lengths $\alpha_1,\alpha_2, \dots$ ) $\cdot$ (\# of ways to get 
 $\lambda$ from $\emptyset$  by removing a vertical strips of lengths 
$\beta_1, \beta_2, \dots $ )
\bigskip
\end{minipage}}\\
&=& \mbox{\begin{minipage}[t]{5.045in} 
 (\# of poset tableaux labellings from $\mu'$ up to $\lambda$ 
then down to $\emptyset$, which add a horizontal strip of length 
$k$ for every $y$-vertex of repetition number $k$ and subtract a 
vertical strip of length $k$ for every $x$-vertex of repetition 
number $k$.)
\end{minipage}} \\
\end{eqnarray*}


This completes the proof of the theorem.\DONE




%===========KRAJ TEOREMA======================================



%--------------------------------------------------------------------------------------------
%--------------------------------------------------------------------------------------------

%--------------------------------------------------------------------------------------------
%--------------------------------------------------------------------------------------------

%============================================================================
%  A  D  I  N  G         T  H  E           Lx  
%=====================================================================   

\subsection{Adding the $L_X$}

Consider the Laplacian $L_X$. 
Since we have identified the L-block $V$ with a subspace of the symmetric group
algebra ${\Bbb C}S_n$, by fixing the order on the $x$'s, every time
the Laplacian $L_X$ switches a pair of $x$'s, it is actually putting a 
minus sign in front of the corresponding basis element, with the $x$'s ordered.
Since $L_X$ acts as a sum of transpositions, every eigenvalue we obtain from
the $L_X$, will have a minus sign. 


Recall the multiplicity of the $x$-node.
Let $v$ be an $x$-node of the diagram $P[X,Y]$, labeled with the
partition $\LL(v)$ and with repetition number $k(v)$. Let $C(v)=
\{ v_1, v_2, \dots , v_l\}$
be the set of covers of $v$.
Let $\lambda_i$ denote $\LL(v_i)$, and let $k_i$ denote the 
repetition numbers, $k(v_i)$. The 
  {\em multiplicity of $\LL$ at $v$}  is defined to be 
$$
 m_v(\LL) = \sum_{\mu} c_{\lambda_1, \dots, \lambda_l}^{\mu} c^{\mu}_{\LL(v), 1^{k(v)}}.
$$

The {\em node--eigenvalue, $e_v(\LL)$, } for each node $v$,  
 is the set of sums of the content over all
squares in $\mu /\LL(v)$ above for a given $\mu$ minus the binomial
coefficient  $\binom{k(v)}{2}$.

This gives $m_v(\LL)$ eigenvalues at each $x$-node $v$. We now define the
{\em $x$-eigenvalue of $\LL$}, $e_x(\LL)$, to be the set of numbers obtained
by taking a sum of one element of $e_v(\LL)$ for each $x$-node $v$.
So $\abs{e_x(\LL)} = \prod_{\mbox{$x$-nodes $v$}} m_v(\LL)$. 



\begin{theorem}[$L_X$-Centerpiece]
Let $P$ be a linear poset with a minimum element, $\hat 0$. Let
$X$ and $Y$ be two (multi-)sets, subsets of $P$.
For every labeling $\LL$ of positive multiplicity,
 each
element in $e_x(\LL)$ is an eigenvalue of $L_X$.

\end{theorem}


Proof:

The proof of this theorem is similar to the proof of the $L_Y$-Centerpiece 
Theorem. We will omit the details here.\DONE





Since $L_X$ and $L_Y$ commute (as established in Lemma~\ref{lemma5.1} ), the 
eigenvalues of $L_X + L_Y$ will be the sum of the eigenvalues on the 
corresponding irreducibles of the eigenspaces. 


Recall that the complete
Laplacian $L$ is the sum of three things (from the beginning of this 
section):
$$
L=L_D + L_X + L_Y
$$

The $L_D$ component is the diagonal matrix, which on the L-block $V$
spanned by the sets $(X,Y)$, has value:
\begin{eqnarray*}
e_D (X,Y) &=& w(X,Y) + \Delta(X,Y) \\
w(X,Y) &=& \sum_{m=1}^k \abs{(x_m,y_m)} \\ 
 \Delta(X,Y) &=& \sum_{i,j} \delta_{x_i,y_j}
\end{eqnarray*}

The computer evidence strongly supports the following 
conjecture:


\section{Complete Centerpiece Conjecture}

\begin{conjecture}
Let $P$ be a linear poset with a minimum element, $\hat 0$. Let $X$ and $Y$
be two multisets of vertices of $P$. Let $e_D(X,Y)$ be defined as above.
For every poset tableau $\LL$ of positive multiplicity, let $\tau_Y(\LL) \in e_Y(\LL)$ and let $\tau_X(\LL) \in e_X(\LL)$. Then
the scalars $e(\LL) = \tau_Y({\LL}) - \tau_X({\LL}) +
e_D(X,Y)$ are the complete set of eigenvalues of the Laplacian $L$.
\end{conjecture}

This conjecture claims that the same poset tableau will work simultaneously
for both Laplacians ($L_X$ and $L_Y$),i.e., that the eigenvalues 
of the Laplacian $L$ are the sum of the eigenvalues of $L_Y$ and 
the eigenvalues of $L_X$ evaluated simultaneously with the same
poset tableau.

%=========================================================================
%               H  O  M  O  L  O  G  Y
%=========================================================================

\section{Homology}

The object of the paper is to get a step closer to
evaluating the homology of any Lie algebra corresponding to a linear
poset, using only combinatorial properties of the poset. In these two
small cases (n=1 and n=2) we had no difficulty. 
For larger n, we need some extra results.


\subsection{$H_1$}
 
For example, if we want to evaluate the homology $H_1(L_P)$ of a Lie
algebra $L_P$ corresponding to a linear poset $P$, with $\hat 0$,
 our construction
gives an immediate answer. 

An L-block $V$ of size 1, is determined by the sets $(X,Y)$, $X=\{x\}$, and
$Y=\{y\}$. Obviously, if we want $V$ to be non-zero, $x<_P y$. So
the corresponding diagram of this L-block is given in figure~\ref{sl501}.

%\begin{figure}[h]
%\centerline{\picture{1}{0.333}{0.625}{pic501.ps}}
%\caption{\label{sl501} A diagram of the L-block}
%\end{figure}
\begin{figure}
\begin{center}
\begin{picture}(100,50)(0,0)
%----------------------------------------
\put(50,30){\begin{picture}(60,60)(30,0)
\put(0,0){\line(0,1){30}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\end{picture}}
%----------------------------------------------

\end{picture}
\caption{\label{sl501} A diagram of the L-block}
\end{center}
\end{figure}

Both indicators $e_Y$ and $e_X$ are zero, so the eigenvalues are given by
$e_D$. But $\Delta(X,Y) = 0$ too, since $X\cap Y = \emptyset$.
Thus $L(z_{x,y}) = w(X,Y) z_{x,y} = \abs{(x,y)} z_{x,y}$. In other words,
the eigenvectors of the Laplacian  $L_1$ are the basis vectors $z_{x,y}$,
and  the corresponding eigenvalues are 
 $\abs{(x,y)}$, i.e.,  the number of the vertices
in the poset $P$, between $x$ and $y$.

The dimension of the homology is the number of zero eigenvalues, i.e., the
number of the intervals $z_{x,y}$, such that $y$ covers $x$.

Thus 
$$
\dim (H_1(L_P)) = \mbox{ (\# of covering relations in $P$)}.
$$



\subsection{$H_2$}

In this case the L-block $V$ 
in question is spanned by the (multi--)sets $(X,Y)$, each of
 size 2, i.e., $X=\{ x_1, x_2 \}$ and $Y=\{ y_1, y_2\}$. There are several
possibilities for the L-block.

\begin{enumerate}


\item All four elements are comparable, and $x$'s are below the $y$'s.
        $$x_1 < x_2 < y_1 < y_2 .$$
        All possible poset  tableaux are shown in figure~\ref{sl502}.

%\begin{figure}
%\centerline{\picture{0.7}{1.930}{1.944}{pic502.ps}}
%\caption{\label{sl502} poset  tableaux}
%\end{figure}
\begin{figure}[t]
\begin{center}
\begin{picture}(200,120)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,110)(30,0)
\put(0,0){\line(0,1){90}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
\put(8,59){\makebox(0,0)[bl]{\one\one}}
\put(8,89){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(0,90){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e(\Lambda) = 0$}}
\end{picture}}
%----------------------------------------------
%----------------------------------------
\put(150,30){\begin{picture}(60,110)(30,0)
\put(0,0){\line(0,1){90}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
\put(8,59){\makebox(0,0)[bl]{\one}}
\put(8,48){\makebox(0,0)[bl]{\one}}
\put(8,89){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(0,90){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e(\Lambda) = 0$}}
\end{picture}}
%----------------------------------------------


\end{picture}
\caption{\label{sl502} poset  tableaux}
\end{center}
\end{figure}


   As we can see, both  $e_Y$ and $e_X$ eigenvalues  are  zero. So we
don't have to worry how to add them up - we will always get zero.
 $\Delta$ is also 
zero, since the sets $X$ and $Y$ are disjoint. Thus again, the Laplacian
is $L(\zz 1 \ww \zz 2) = w(X,Y) (\zz 1 \ww \zz 2)$. But in this case, both
intervals contain at least one element, so $w(X,Y)>0$. Thus in this
case we never get a zero eigenvalue, which might contribute to the 
homology $H_2$.  



\item $y$'s are not comparable.
$$
x_1 < x_2 < \begin{array}{c} y_1 \\ y_2 \end{array}.
$$
All possible poset  tableaux are shown in figure~\ref{sl503}.

%

%\begin{figure}
%\centerline{\picture{0.7}{2.875}{1.750}{pic503.ps}}
%\caption{\label{sl503} poset  tableaux}
%\end{figure}
\begin{figure}[t]
\begin{center}
\begin{picture}(200,120)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,110)(30,0)
\put(0,0){\line(0,1){90}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
\put(8,59){\makebox(0,0)[bl]{\one\one}}
\put(8,89){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(0,90){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e(\Lambda) = -1$}}
\end{picture}}
%----------------------------------------------
%----------------------------------------
\put(150,30){\begin{picture}(60,110)(30,0)
\put(0,0){\line(0,1){90}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
\put(8,59){\makebox(0,0)[bl]{\one}}
\put(8,48){\makebox(0,0)[bl]{\one}}
\put(8,89){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(0,90){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e(\Lambda) = +1$}}
\end{picture}}
%----------------------------------------------


\end{picture}
\caption{\label{sl503} poset  tableaux}
\end{center}
\end{figure}



The value of  $e_Y$ is zero, while the values of $e_X$ are +1 and -1.
 Since $e_D$ is always non-negative,
the value +1 cannot contribute to homology $H_2$. The other value can, but
only if $e_D$ is +1. That means that all of the 
 relations  indicated: $x_1 < x_2$, $x_2 < y_1$ and $x_2< y_2$ are covering
relations in the poset $P$. 
Whenever we have a four-element subset of the poset $P$, with covering 
relations $x_1 < x_2$, $x_2 < y_1$ and $x_2< y_2$, we will call
that a ''Y``-configuration.
Thus in this case every  occurrence of ''Y``-configuration (described 
as above) in the Hasse 
diagram of the poset contributes to the homology $H_2$.

  
\item $x_1 < y_1 < x_2 < y_2$ or equivalently (for our purpose)
 $x_1 < y_1 = x_2 < y_2$.\\
 There is only one poset  tableau spanned by these
sets $X$ and $Y$, namely the one shown in figure~\ref{sl504}.
%\begin{figure}
%\centerline{\picture{0.7}{0.680}{1.944}{pic504.ps}}
%\caption{\label{sl504} poset  tableau}
%\end{figure}
\begin{figure}
\begin{center}
\begin{picture}(100,120)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,110)(30,0)
\put(0,0){\line(0,1){90}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
\put(8,59){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,89){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(0,90){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e(\Lambda) = 0$}}
\end{picture}}
%----------------------------------------------

\end{picture}
\caption{\label{sl504} poset  tableau}
\end{center}
\end{figure}


Since the space is one dimensional, we will add the eigenvalues
in the only possible way. 
In the first case, 
the eigenvalue will be zero, if and only if both relations,
$x_1<y_1$ and $x_2 < y_2$, are covering
relations, i.e., every occurrence of a distinct (all vertices are distinct)
 pair of covering
relations contributes to the dimension of the homology $H_2$.
If the second case occurs, i.e., if $y_1=x_2$ then $\Delta$ will contribute
to the eigenvalue, and it won't be zero anymore.




\item $x_1 < x_2 < y_1 = y_2$.\\
There is only one poset  tableau spanned by these
sets $X$ and $Y$, namely the one shown on the figure~\ref{sl505}.
%\begin{figure}[p]
%\centerline{\picture{0.7}{0.680}{1.444}{pic505.ps}}
%\caption{\label{sl505} poset  tableau}
%\end{figure}
\begin{figure}
\begin{center}
\begin{picture}(100,90)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,80)(30,0)
\put(0,0){\line(0,1){60}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one}}
\put(8,59){\makebox(0,0)[bl]{\one\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e(\Lambda) = -1$}}
\end{picture}}
%----------------------------------------------

\end{picture}
\caption{\label{sl505} poset  tableau}
\end{center}
\end{figure}




Since the space is one dimensional, we will add the eigenvalues
in the only possible way.
The eigenvalue will be zero, if and only if $x_2$ covers $x_1$, and 
both $y_1$ and $y_2$ cover $x_2$. Note that this is a degenerate 
letter ''Y``,  with a joint top vertex. Every such occurrence
contributes to the homology $H_2$. 

\item $x$'s are the same.
$$x_1 = x_2 < \begin{array}{c} y_1 \\ y_2 \end{array}.$$
There is only one poset  tableau spanned by these
sets $X$ and $Y$, namely the one shown in figure~\ref{sl506}.

%\begin{figure}
%\centerline{\picture{0.7}{1.125}{1.250}{pic506.ps}}
%\caption{\label{sl506} poset  tableau}
%\end{figure}
\begin{figure}
\begin{center}
\begin{picture}(100,50)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,60)(30,0)
\put(0,0){\line(1,1){30}}
\put(0,0){\line(-1,1){30}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(38,29){\makebox(0,0)[bl]{\one}}
\put(-38,29){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(30,30){\circle*{3}}
\put(-30,30){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e(\Lambda) = 0$}}
\end{picture}}
%----------------------------------------------

\end{picture}
\caption{\label{sl506} poset  tableau}
\end{center}
\end{figure}




Since the space is one dimensional, we will add the eigenvalues
in the only possible way.
The eigenvalue will be zero, if and only if both $y_1$ and $y_2$ cover 
$x_1 = x_2$. Again, it is a pair of distinct covering relations
with joint vertex, this time the $x$-vertex.  Every such occurrence
contributes to the homology $H_2$. 




\item $x_1 = x_2 < y_1 < y_2$.
There is only one poset  tableau spanned by these
sets $X$ and $Y$, namely the one shown on the figure~\ref{sl507}.
%\begin{figure}
%\centerline{\picture{0.7}{0.680}{1.444}{pic507.ps}}
%\caption{\label{sl507} poset  tableau}
%\end{figure}
\begin{figure}
\begin{center}
\begin{picture}(100,90)(0,0)

%----------------------------------------
\put(50,30){\begin{picture}(60,80)(30,0)
\put(0,0){\line(0,1){60}}
\put(8,-1){\makebox(0,0)[bl]{$\emptyset$}}
\put(8,29){\makebox(0,0)[bl]{\one\one}}
\put(8,59){\makebox(0,0)[bl]{\one}}
\put(0,0){\circle*{3}}
\put(0,30){\circle*{3}}
\put(0,60){\circle*{3}}
\put(-15,-15){\line(1,0){30}}
\put(-20,-30){\makebox(0,0)[bl]{$e(\Lambda) = 1$}}
\end{picture}}
%----------------------------------------------

\end{picture}
\caption{\label{sl507} poset  tableau}
\end{center}
\end{figure}



Since the space is one dimensional, we will add the eigenvalues
in the only possible way.
Since $e_D$ is always non-negative,
the value +1 cannot contribute to homology $H_2$.



\item The trivial case when $x$'s are not comparable (so because of 
linearity and the minimum element neither are $y$'s).
$$ 
\begin{array}{ccc} x_1 & < & y_1 \\x_2 & < & y_2 \end{array}.
$$
This is equivalent to case 3. We have two distinct covering relations, so
it contributes to the homology $H_2$.
\end{enumerate}


All together, the dimension of the  homology $H_2$ is in fact the number
of distinct pairs of covering relations + the number of occurrences
of letter ``Y'' in the poset $P$. In other words, if $\cal H$ is a Hasse
diagram of a poset $P$, with $e$ edges, and $\gamma$ letter {``Y''}'s (degenerate or not), we have $H_1(L_P) = {\Bbb C}^e$ and $H_2 = {\Bbb C}^{\smbinom{e}{2} + \gamma}$.





\subsection{Examples}
\begin{itemize}
 
\item
For example, suppose that we are dealing with the chain poset
on $n$ vertices ($1<2<\cdots <n$), $T_n$. The number
of non-degenerate Y's is zero. The number of degenerate letters
 Y is (n-2).
The number of edges in the Hasse diagram is (n-1). Hence
$$H_1(T_n) = {\Bbb C}^{n-1},$$ and 
$$H_2(T_n) = {\Bbb C}^{n-2 + \smbinom {n-1}{2} } = 
{\Bbb C}^{\frac {(n+1)(n-2)}{2}}
$$





\item
Let the poset $P$ be given in figure~\ref{sl602}, where the length of
the chains are $m$ and $n$. This is in fact equivalent to having two
disjoint chains, of length $m$ and $n$. Thus the corresponding
homologies will be
$H_1 = {\Bbb C}^{n-1 + m-1}  $, and $H_2(T_n) ={\Bbb C}^{\frac{(n+1)(n-2)}{2} + \frac{(m+1)(m-2)}{2}}$



%\begin{figure}
%\centerline{\picture{0.7}{5.277}{3.083}{pic602.ps}}
%\caption{\label{sl602} poset $P$}
%\end{figure}
\begin{figure}
\begin{center}
\begin{picture}(200,90)(0,0)

%----------------------------------------
\put(100,30){\begin{picture}(120,80)(60,0)
\put(0,0){\line(1,1){20}}\put(0,0){\line(-1,1){20}}
\put(40,40){\line(1,1){20}}\put(-40,40){\line(-1,1){40}}
\put(0,0){\circle*{3}}
\put(20,20){\circle*{3}}\put(25,25){\circle*{3}}
\put(35,35){\circle*{3}}\put(40,40){\circle*{3}}
\put(-20,20){\circle*{3}}\put(60,60){\circle*{3}}
\put(-25,25){\circle*{3}}\put(-30,30){\circle*{3}}
\put(-35,35){\circle*{3}}
\put(-40,40){\circle*{3}}\put(-60,60){\circle*{3}}
\put(-80,80){\circle*{3}}

\put(8,-1){\makebox(0,0)[bl]{$1$}}
\put(68,59){\makebox(0,0)[bl]{$n$}}
\put(-88,79){\makebox(0,0)[bl]{$m$}}

\end{picture}}
%----------------------------------------------

\end{picture}
\caption{\label{sl602} poset $P$}
\end{center}
\end{figure}

\end{itemize}

\section {Conclusion}

These results have several interesting corollaries that are of a  
combinatorial nature. We will state one. Let $P$ be a rooted tree
on $n$ nodes and let $\Sigma$ be the sum in the group algebra of
$S_n$ of all transpositions $(i,j)$ such that $i$ is on the unique
path from $j$ to the root in $P$. Then $\Sigma$ acting on ${\Bbb C}S_n$
by left multiplication has non-negative integer eigenvalues and the 
corresponding eigenspaces can be identified in representation-theoretic
terms.

There is still a lot to do in this area. Although my work provides 
partial answers  and a conjecture
for all rooted trees, the question is still open for other posets.
What happens in those cases is very difficult to control, since the 
expression for the Laplacian becomes more complicated. In the tree case
I wish to examine a twisting of the Laplacian by a parameter $\alpha$ which,
my advisor has shown, is related to Jack polynomials and the Krawtchouk
polynomials in certain special cases. Lastly I would like to see if the 
more algebraic consequences of  Kostant's theorem have sensible analogues in
my case. 

%% <<=====================================================>>
%%
%%  ACKNOWLEDGMENTS
%%
%% <<====================================================>>
\vspace{1.0cm}


\noindent{\bf Acknowledgement.} I would like to thank my advisor, 
Professor Phil Hanlon, 
for the guidance, help, and for his invaluable suggestions.

\vspace{1.0cm}


%=======================================================
% BIBLIOGRAPHY FILE
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%=======================================================
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\end{document}