\documentstyle[a4,12pt]{article}
\title{The composition factors of the Weyl modules with
fundamental weights for the symplectic group}
\author{A.E. Brouwer}
\date{November 25, 1992}
\begin{document}
\maketitle

\newcommand{\Sym}{{\rm Sym}}
\newcommand{\End}{{\rm End}}
%\newcommand{\Ff}{{\Bbb F}}
\newcommand{\Ff}{{\bf F}}
\newcommand{\Cc}{{\bf C}}
\newcommand{\Zz}{{\bf Z}}
\newtheorem{Theorem} {Theorem} [section]
\newtheorem{Proposition} [Theorem] {Proposition}
\newtheorem{Lemma} [Theorem] {Lemma}
\newtheorem{Corollary} [Theorem] {Corollary}
\newcommand{\Proof}{ \noindent{\bf Proof:}\quad }
\newcommand{\Remark}{ \noindent{\bf Remark:}\quad }
\newcommand{\qed  }{\hfill$\Box$\medskip}
\newcommand{\eps}{\epsilon}
\newcommand{\isom}{\,\simeq\,}
\newcommand{\ti}{t.i.\ }
\newcommand{\ts}{t.s.\ }

\begin{abstract}
We investigate the structure of the Weyl modules $W(\lambda)$
for the symplectic group and
for fundamental weights $\lambda$, in arbitrary characteristic.
\end{abstract}

\section{The modules spanned by isotropic vectors}
Let $V$ be a vector space of dimension $n$ over the field $\Ff$, and let
$E := EV = \oplus_{i=0}^n \bigwedge^i V$ be its exterior algebra.
We have $\dim \bigwedge^i V = { n \choose i }$ and (hence) $\dim EV = 2^n$.

Let $f$ be a nondegenerate symplectic form on $V$ (so that $n$ is even,
say $n = 2d$), and let $G = Sp(V,f)$ be the group consisting of
the linear transformations of $V$ preserving this form.

Let $F_m$ be the span in $E$ (that is, in $\bigwedge^m V$)
of all vectors representing totally isotropic (t.i.) $m$-spaces.

\begin{Theorem}
\label{WEYL}
We have $\dim F_m = {n \choose m} - {n \choose m-2}$.
The $\Ff G$-module $F_m$ is the Weyl module $W(\lambda_m)$
of $G$ for the fundamental weight $\lambda_m$.
\end{Theorem}

This result can be found in \cite{B}, Chap. 8, \S 13 for
fields of characteristic 0, and in \cite{PS} for the case
of odd characteristic. Here we shall give a proof
(similar to the proof in \cite{PS})
that is valid for arbitrary characteristic.

Choose a basis $\{e_1 , ... , e_d, e_{d'}, ..., e_{1'} \}$ in $V$ such that
$f = \sum_j e_j^* e_{j'}^*$. We shall write $e_{I,J}$ for the product in $E$
(in natural order) of the $e_i$ for $i \in I$ and the $e_{j'}$ for $j \in J$.
The set of $e_{I,J}$ with $|I|+|J| = m$ forms a basis of $\bigwedge^m V$.

\begin{Lemma}
\label{SPAN}
Let $d_{kl} = e_{k,k} - e_{l,l}$, and $d_{k_1,l_1,...,k_t,l_t} =
d_{k_1,l_1} ... d_{k_t,l_t}$. Then the
vector space $F_m$ is spanned by the vectors $e_{I,J} d_K$, where
$I$, $J$ and $K$ are pairwise disjoint sequences and $|I|+|J|+|K|=m$.
(By definition $|K|$ is even.)
\end{Lemma}
\Proof
Let $A$ be this set of vectors, and let $S = \langle A \rangle$
be their $\Ff$-span.
Since $G$ is transitive on the set of \ti $m$-spaces, it suffices
to show that $A \subseteq F_m$,
that at least one element of $A$ represents a \ti $m$-space,
and that $S$ is $G$-invariant.
Now
$$d_{kl} = (e_k + e_l)(e_{k'} - e_{l'}) + e_k e_{l'} - e_l e_{k'}$$
is in $F_2$, and the factors of $e_{I,J} d_K$ lie in mutually
ortogonal subspaces of $V$, so $e_{I,J} d_K \in F_m$.
Moreover, if $K = \emptyset$, then this vector represents a \ti
$m$-space. Remains to show that $S$ is $G$-invariant.
Now clearly $S$ is invariant under the Weyl group $W \isom 2^d:\Sym (d)$,
and under the diagonal group $H$. Remains to show that $S$ is $U$-invariant.
But $U$ is generated by elements $g_j (\alpha) = 1 + \alpha x_j$
($1 \le j \le d$, $\alpha \in \Ff$)
where $x_j \in \End (V)$ kills all $e_i$ and $e_{i'}$
except that $x_j e_{j+1} = e_j$ and $x_j e_{j'} = - e_{(j+1)'}$
for $1 \le j \le d-1$, and $x_d e_{d'} = e_d$.
Consider $z = x_j e_{I,J} d_K$.
If $j = d$, then $z = 0$ unless $d \in J$, and then $z \in A$.
If $j+1 \not\in I \cup K$ and $j \not\in J \cup K$, then $z = 0$.
If precisely one of $j,j+1$ occurs in $I \cup J \cup K$, then
either it occurs in $I \cup J$, and $z \in A$, or it occurs
in $K$, and $z \in \pm A$ since
$x_j d_{j+1,k} = e_{j,j+1}$ and $x_j d_{j,k} = - e_{j,j+1}$.
If both $j+1 \in I \cup K$ and $j \in J \cup K$, then we find $z \in S$
since either both occur in $I \cup J$, and then 
$(g_j (\alpha) - 1) e_{j+1,j} = \alpha d_{j,j+1} - \alpha^2 e_{j,j+1}$,
or both occur in $K$, and then $x_j d_{j,j+1} = -2 e_{j,j+1}$ and
$x_j d_{j,k} d_{j+1,l} = e_{j,j+1} d_{l,k}$,
or $j+1 \in I$ and $j \in K$, and then
$x_j e_{j+1} d_{j,k} = e_j d_{j+1,k}$,
or $j \in J$ and $j+1 \in K$, and then
$x_j e_{j'} d_{j+1,k} = e_{(j+1)'} d_{k,j}$.
\qed

\Remark
This proof shows that if $\Ff$ has characteristic 2, then
$N_d := \langle e_{I,J} d_K \in F_d | K \neq \emptyset \rangle$
is the unique maximal submodule of $F_d$, and the quotient
$M_d := F_d / N_d$ is irreducible of dimension $2^d$, the
{\em spin module}.

\medskip
We can be more precise, and give a basis for $F_m$,
and hence find its dimension
$\dim F_m = {n \choose m} - {n \choose m-2}$.

Let us put $c_k = e_{k,k}$, so that the $c_k$ commute,
and let $c_K = \prod_{k \in K} c_k$.
Now $d_{k,l} = c_k - c_l$.
Let $C_{R,t} = \langle c_K | K \subseteq R,~ |K|=t \rangle$, and
$D_{R,t} = \langle d_K | K \subseteq R,~ |K|=2t \rangle$.
Write $C_{r,t}$ and $D_{r,t}$ in case $R = \{1,...,r\}$.

Clearly, elements $e_{I,J} d_K$ with different $I,J$ are
independent, so $F_m$ is a direct sum $\oplus e_{I,J} D_{R,t}$,
where $R = \{1,...,d\} \setminus (I \cup J)$
and $2t = m - |I| - |J|$.
Thus, it suffices to give a basis for $D_{R,t}$
and to show that $\dim D_{R,t} = {r \choose t} - {r \choose t-1}$
where $r = |R|$.
(Indeed, we want to show that $\bigwedge^m V / F_m$ has
dimension ${n \choose m-2}$, so that this quotient has
a basis in one-to-one correspondence with the $(m-2)$-subsets
of $\{1,...,d,d',...,1'\}$. This will follow when
$C_{R,t} / D_{R,t}$ has a basis in one-to-one correspondence
with the $(t-1)$-subsets of $R$.)

\begin{Lemma}
$\dim D_{r,t} = {r \choose t} - {r \choose t-1}$.
\end{Lemma}
\Remark
This is a very special case of Theorem 20.1 (and given
explicitly as Example 14.4) in \cite{J}.

\Proof
A spanning set for $D_{r,t}$ is given by the $d_K$ for sequences
$K$ such that for odd $i$ the element $k_i$ is the smallest
positive integer not in $\{k_j | j < i\}$, and for even $i > 2$
we have $k_i > k_{i-2}$.
(Indeed, we may assume that $K$ is chosen such that the $k_i$
for odd $i$ are increasing. Since
$$(c_1-c_4)(c_2-c_3) = (c_1-c_3)(c_2-c_4) - (c_1-c_2)(c_3-c_4)$$
we may also assume that the $k_i$ for even $i$ are increasing.
Finally, since $(c_2-c_3) = (c_1-c_2) - (c_1-c_3)$, we may
assume that each $k_i$ for odd $i$ is the smallest number
distinct from earlier $k_j$.)
This spanning set is a basis. Indeed, if $d_K$ is in our
spanning set, then we have $d_{2j-1} \le 2j-1$ and
$d_{2j} \ge 2j$ for all $j$, so that the product
$c_{k_2} ... c_{k_{2t}}$ is a term in $d_L$ for precisely
one $d_L$ in our spanning set (namely $d_L = d_K$).

Remains to count spanning vectors. We shall do this
by computing a basis for $C_{r,t} / D_{r,t}$ and showing
that it has size ${r \choose t-1}$ by explicitly indicating
a correspondence with $(t-1)$-subsets of $\{1,...,r\}$.
So, let $S$ be an increasing sequence of length $t-1$ in
$\{1,...,r\}$. Let $j$ be minimal such that $s_j \ge 2j$,
or $j = t$ if there is no such $s_j$.
Let $K = \{1,...,2j-1\} \setminus \{s_1,...,s_{j-1}\} \cup
\{ s_j, ... , s_{t-1} \}$.
Then $K$ is an increasing sequence with the property that
there is a $j$ such that $k_j = 2j-1$. Conversely, if $K$
is such a sequence, and we take $j$ minimal, then we can
retrieve $S$ as
$S = \{1,...,2j-1\} \setminus \{k_1,...,k_j\} \cup
\{ k_{j+1}, ... , k_t \}$.
Now the $c_K$ for such sequences $K$ form a basis of
$C_{r,t} / D_{r,t}$: given any increasing sequence $L$
such that $l_j \ge 2j$ for all $j$, there is a unique
element in the basis of $D_{r,t}$ in which it occurs as
a factor, while all other factors are lexicographically smaller.
\qed

Thus, we now know $\dim F_m$.
We want to show that $F_m$ is the Weyl module $W(\lambda_m)$.
Now all of the above was for arbitrary $\Ff$, and we can consider
$\Ff = \Cc$. Let $U$ be the universal enveloping algebra,
and let $U_{\Zz}$ be a $\Zz$-form.
If we let $F_{m,\Zz}$ be the $\Zz$-span of the $e_{I,J} d_K$,
then the above proof of Lemma \ref{SPAN} shows that $F_{m,\Zz}$
is a standard cyclic $U_{\Zz}$-module, spanned by
$e_{\{1,...,m\}}$. Now let $\Ff$ be arbitrary again.
By definition $W(\lambda_m) = F_{m,\Zz} \otimes_{\Zz} \Ff$,
and clearly the map sending $e_{I,J} d_K$ to
$e_{I,J} d_K \otimes 1$
extends to an isomorphism from $F_m$ onto
$F_{m,\Zz} \otimes_{\Zz} \Ff$.

This completes the proof of Theorem \ref{WEYL}.

\section{The composition factors of $F_m$}
The Weyl module $F_m$ has a unique maximal submodule $N_m$
and the quotient $M_m := F_m / N_m$ is the irreducible $G$-module
with highest weight $\lambda_m$.
This follows from general theory; however, it turns out to be
useful to determine $N_m$ and $M_m$ more explicitly.

The symplectic form $f$ on $V$ determines a nondegenerate
bilinear form on $\bigwedge^m V$ defined by
$(x_1 ... x_m , y_1 ... y_m) = \det ( \, f(x_i,y_j) \, )$.
We shall use $\perp$ for orthogonality with respect to this form.
Note that this form is $G$-invariant.

\begin{Lemma}
Let $U$ be a $G$-submodule of $\bigwedge^m V$.
Then $F_m \subseteq U$ or $U \subseteq F_m^{\perp}$.
\end{Lemma}
\Proof
Suppose that $U$ is not contained in $F_m^{\perp}$.
Then there is an isotropic vector $v \in F_m$
and a $u \in U$ such that $(u,v) \neq 0$.
Since $G$ is transitive on isotropic vectors, we may
assume that $v = e_{1'} ... e_{m'}$.
Now $(u,v) \neq 0$ means that $u$, written on the standard basis,
has a nonzero term $c e_1 ... e_m$. Let $g_j$ and $g_{j'}$
be the elements $x \mapsto x + f(e_j,x) e_j$ and
$x \mapsto x + f(e_{j'},x) e_{j'}$. Then
$\prod_{j=1}^m (g_j - 1)(g_{j'} - 1) ~ u = \pm c e_1 ... e_m \in U$,
and we see that $F_m \subseteq U$.
\qed

Using the same argument (applying the $g_j$ and $g_{j'}$), one may
prove that every submodule $U$ of $\bigwedge^m V$ is a direct sum
of weight spaces.

\begin{Lemma}
$N_m = F_m \cap F_m^{\perp}$ is the unique maximal submodule of $F_m$.
\end{Lemma}
\Proof
We only have to check that $F_m$ is not contained in $F_m^{\perp}$,
and that is obvious.
\qed

In characteristic 0 the module $F_m$ is irreducible.
Let us now determine its decomposition in characteristic $p$.
For nonnegative integers $a,b$, let us say that $b$
{\em contains} $a$ {\em to base} $p$ when $b > 2a$ and in the
base $p$ expansions $a = \sum a_i p^i$, $b = \sum b_i p^i$
(with $0 \le a_i,b_i \le p-1$) we have $a_i = 0$ or $a_i = b_i$
for all $i$.

\begin{Theorem}
$F_m$ has a composition factor $M_j$ if and only if
$0 \le j \le m$, $m-j$ is even, and $d-j+1$ contains ${1 \over 2} (m-j)$
to base $p$. When this is the case, $M_j$ occurs with multiplicity one.
\end{Theorem}
\Remark
The case $p > 2$ of this result is the main result of \cite{PS}.

\Proof
We reduce the problem to that of Specht modules for $\Sym (r)$
defined by partitions $\{r-t,t\}$ into two parts.

For each ordered partition $\lambda = (I,J,R)$ of $\{1,...,d\}$
into three possibly empty parts where $m-|I|-|J|$ is even, let
$F_m^{\lambda} = e_{I,J} D_{R,t}$ (where $|I|+|J|+2t = m$),
so that $F_m = \oplus_{\lambda} F_m^{\lambda}$.
Then $F_m^{\lambda}$ is a $\Sym (R)$-module, the Specht module
for the partition $\{r-t,t\}$, where $r = |R|$.

The bilinear form on $\bigwedge^m V$ determines a symmetric
bilinear form on $F_m^{\lambda}$, given by
$(e_{I,J} u, e_{I,J} v)_{\lambda} := (-1)^{|I|.|J|} (e_{I,J} u, e_{J,I} v)$.
The resulting form on the Specht module is the canonical one, and
it follows from \cite{J} (Theorem 4.9 and Corollary 12.2) that
$F_m^{\lambda} \cap F_m^{\perp}$ is the unique maximal
$\Sym (R)$-submodule of $F_m^{\lambda}$ unless $p = 2$ and $m = d$
and $R \neq \emptyset$, in which case $F_m^{\lambda} \subseteq F_m^{\perp}$.

If we take the root system of type $C_d$ to consist of the $2d^2$
vectors $\pm \eps_i \pm \eps_j$ ($1 \le i < j \le d$) and
$\pm 2 \eps_i$ ($1 \le i \le d$), with fundamental system of roots
$\alpha_i = \eps_i - \eps_{i+1}$ ($1 \le i \le d-1$), $\alpha_d = 2 \eps_d$,
then the fundamental weights are $\lambda_i = \eps_1 + ... + \eps_i$
($1 \le i \le d$).
Let $M_{\mu}$ be the irreducible $G$-module with highest weight $\mu$.
If $M_{\mu}$ occurs in a composition series of $F_m$, then
$\mu \le \lambda_m$, i.e., $\lambda_m - \mu$ is a sum of positive roots,
and it follows that $\mu = \lambda_j$ (i.e., $M_{\mu} = M_j$)
for some $j$ with $0 \le j \le m$, $m - j$ even.

Let $F_m = V_0 \supset V_1 \supset ... \supset V_l \supset 0$
be a composition series for the $G$-module $F_m$,
with $V_i / V_{i+1} \isom M_{j_i}$.
Each $V_i$ is a direct sum of weight spaces, and for any given
weight $\kappa$ we find
$F_m^{\kappa} = V_0^{\kappa} \supseteq V_1^{\kappa} \supseteq ... \supseteq
V_l^{\kappa} \supseteq 0$.
Take $\kappa = 0$ if $m$ is even, and $\kappa = \eps_1$ if $m$ is odd,
and put $|\kappa|=0$ resp. 1 in these cases.
(Note, that if $p = 2$ and $m = d$ then
$\dim F_d^{\lambda_d} = 1$, so that $M_d$ occurs precisely once,
and $j_0 = d$, $j_i < d$ for $i > 0$.)
Then $V_i^{\kappa} / V_{i+1}^{\kappa} \isom M_{j_i}^{\kappa} =
F_{j_i}^{\kappa} / (F_{j_i}^{\kappa} \cap F_{j_i}^{\perp})$
is an irreducible $\Sym (d-|\kappa|)$-module (or zero in case
$p = 2$, $m = d > 0$, $i = 0$),
and we see that $F_m^{\kappa} = V_0^{\kappa} \supseteq V_1^{\kappa} \supset ...
\supset V_l^{\kappa} \supset 0$ is a composition series of the
$\Sym (d-|\kappa|)$-module $F_m^{\kappa}$.
Thus, $M_j$ occurs in $F_m$ with the same multiplicity with which
$M_j^{\kappa}$ occurs in $F_m^{\kappa}$, and by \cite{J}, Theorem 24.15,
we find that this multiplicity is at most one, and equals one
if and only if $(d-|\kappa|)-(j-|\kappa|)+1$
contains ${1 \over 2} (m-j)$ to base $p$.
\qed

Note that since $\dim F_m$ is known, and $M_m$ occurs in $F_m$
but not in $F_j$ for $j < m$, we can determine $\dim M_m$ for
$0 \le m \le d$.

\section{A characterization of the isotropic vectors}
For a sequence $M = (m_1, ... , m_l)$ consisting of distinct integers,
let $x_M = x_{m_1} ... x_{m_l}$.
For a sequence $R$ consisting of distinct integers all occurring in $M$,
let $x_{M \setminus R}$ be the element $sgn (\sigma) \, x_N$,
where $\sigma$ is any permutation of $M$ such that the sequence
$\sigma M$ has $R$ as initial subsequence, and $N$ is the tail
following $R$ in $\sigma M$.
Define for $t \ge 0$ a linear transformation $\phi_t \colon E \to E$ by
$$ \phi_t (x_M) = \sum_R ~ (\,\prod_{i=1}^t f(x_{r_i},x_{s_i}) \,) ~
x_{M \setminus R} $$
where the sum is over all sequences $R = (r_1,s_1,r_2,s_2,...,r_t,s_t)$
of distinct integers contained in $M$ such that
$r_1 < r_2 < ... < r_t$ and $r_i < s_i$ for all $i$.

In order to verify that $\phi_t$ is well-defined, we have to check that
the right-hand side vanishes when $x_a = x_b$ for two indices $a,b \in M$.
Consider the permutation $\tau = (a,b)$ of the elements of $M$.
It induces an involution on the terms on the right. The orbits of length
one are those with either $a,b$ both in $M \setminus R$ (and then
$x_{M \setminus R} = 0$) or $\{a,b\} = \{r_i,s_i\}$ (and then
$f(r_i,s_i) = 0$).
The orbits of length two consist of two terms with different signs that cancel.
So indeed the right hand side vanishes for $x_a = x_b$.

In characteristic 0 we have $\phi_t = {1 \over t!} (\phi_1)^t$.
In all characteristics we have $\phi_i \phi_j = {i+j \choose i} \phi_{i+j}$.

\begin{Theorem}
Let $F = \oplus_m F_m$ be the span in $E$ of all isotropic vectors.
Then $$F = \bigcap_{t \ge 1} \ker \phi_t .$$
\end{Theorem}
\Proof
Clearly $\phi_t$ vanishes (for $t \ge 1$)
on each isotropic vector $x = x_1 ... x_m$ since all inner products
$f(x_i,x_j)$ are zero. It follows that $\phi_t$ vanishes on all of $F$
and $F \subseteq \bigcap_{t \ge 1} \ker \phi_t$.
We have $\phi_t (e_{I,J} c_K) = e_{I,J} \phi_t(c_K)$ and
$\phi_t (c_K) = \sum_{T \subseteq K} c_{K \setminus T}$
where the sum is over $t$-sets $T$.
(This time no signs are involved.)
It remains to show that if $u \in C_{r,m} / D_{r,m}$ and all
$\phi_t$ vanish on $u$, then $u = 0$.
Write $u$ in the canonical basis (of $c_K$ with increasing
$K = (k_1,...,k_m)$ and $k_j = 2j-1$ for some $j$):
$u = \sum \alpha_K c_K$.
That $\phi_t (u) = 0$ for all $t \ge 1$ means that
$\sum_{L \subset K} \alpha_K = 0$ for all $L$ with $|L| < m$.
In particular, $\sum_{L \subset K} \alpha_K = 0$ for all such $L$
containing $r$, so that
$\phi_t ( \sum_{K \ni r} \alpha_K c_{K \setminus r} ) = 0$.
By induction on $m$ it follows that
$\sum_{K \ni r} \alpha_K c_{K \setminus r} \in D_{r-1,m-1}$.
But if $r > 2m-1$, then the $c_{K \setminus r}$ form the standard basis for
$C_{r-1,m-1} / D_{r-1,m-1}$, so either there are no terms $\alpha_K$
with $r \in K$ (and we are done by induction on $r$) or $r \le 2m-1$.
Thus, we may assume that $r = 2m-1$ and $D_{r,m} = 0$.
Now our statement follows from the following lemma.

\begin{Lemma}
The 0-1 matrix $M$ with rows indexed by subsets $A$ of a $(2m-1)$-set $X$
where $|A| \le m-1$ and columns indexed by subsets $B$ of $X$, where
$|B| = m$ and with entries $M_{AB} = 1$ if and only if $A \subset B$,
has full column rank over any field.
\end{Lemma}
\Proof
Fix an $m$-subset $B$ of $X$.
Let $r_A$ be row $A$ of $M$ and consider the row vector
$r = \sum_{A \subset B} \, (-1)^{|A|} r_A$.
It has $C$-entry $\sum_{A \subseteq B \cap C} \, (-1)^{|A|}$
and this sum (over $A$ with $|A|<m$)
vanishes unless $B = C$, in which case it is 1.
\qed
\qed

As an application, we can determine in what cases the vector
$f_m := \sum_{|M|=m} c_M$ belongs to $F_{2m}$.

\begin{Proposition}
If $\Ff$ has characteristic zero, then only $f_0$ is isotropic.
If $\Ff$ has positive characteristic $p$, then for
$0 \le m \le d$ we have $f_m \in F_{2m}$ if and only if
$d+1 \equiv m$ (mod $p^a$), where $p^a$ is the smallest power
of $p$ larger than $m$.
\end{Proposition}
\Proof
We have $\phi_t f_m = {d-m+t \choose t} f_{m-t}$,
so $f_m \in F_{2m}$ if and only if ${d-m+t \choose t}= 0$
(in $\Ff$) for $1 \le t \le m$. Now the result follows from
Lucas' theorem.
\qed

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\end{document}
