# Abelian subgroups and semisimple elements in $2$-modular representations of
the symplectic group $Sp_{2n}(2)$

Alexandre Zalesski

###### Key words and phrases:

Symplectic group, Characteristic $2$, Modular representations, Semisimple
elements, Eigenvalue $1$

###### 2000 Mathematics Subject Classification:

20G05, 20G40

e-mail: E-mail<EMAIL_ADDRESS>(Alexandre Zalesski)

Abstract We determine the irreducible 2-modular representations of the
symplectic group $G=Sp_{2n}(2)$ whose restrictions to every abelian subgroup
has a trivial constituent. A similar result is obtained for maximal tori of
$G$. There is significant information on the existence of eigenvalue 1 of
elements of $G$ in a given irreducible representations of $G$.

##  1\. Introduction

The primary motivation of this work lies in the study of eigenvalues of
semisimple elements of groups of Lie type in their representations over fields
of the natural characteristic. This problem is not treatable in full
generality, however, a number of special cases of it were discussed in
literature. In [23] there appeared a number of useful observations, and
certain comments are available in surveys [24] and [25]. Occurrance of
eigenvalue 1 deserves a particular attention, as this carries some geometrical
meaning, and is significant in applications. See [7], where one can learn the
level of difficulty of the eigenvalue 1 problem. In general, relatively little
is known.

Other motivations come from the study of finite groups with disconected prime
graphs [12, 13], from a characterization of groups by the set of their element
orders [29, Lemma 10], from the study of decomposition numbers [27], from
other applications described in [7] and [25], in each case with a specific
aspect of the general problem. The complexity of the problem in general
justifies considerations of special cases, one of these is dealt with in the
current paper, where we specify the problem to the symplectic group
$Sp_{2n}(2)$, $n>1$. Note that the group $GL_{n}(2)$ is dealt with in [28].

To state our results, we need a parametrization of the irreducible
representations of $G$. This is available in terms of algebraic groups. So we
first recall some fact of algebraic group representation theory, specified
here to the group $\mathbf{G}=Sp_{2n}(F)$, where $F$ is algebraically closed
field of characteristic 2.

Note that every 2-modular irreducible representation of $G=Sp_{2n}(2)$ lifts
to a representation of the algebraic group $\mathbf{G}=Sp_{2n}(F)$. The
irreducible representations $\phi$ of $\mathbf{G}$ are parameterized by so
called highest weights, equivalently, by the strings of non-negative integers
$(a_{1},\ldots,a_{n})$. An irreducible representation of $\mathbf{G}$ (and the
highest weight of it) is called 2-restricted if $0\leq a_{1},\ldots,a_{n}\leq
1$. If $\phi$ is 2-restricted then the restriction $\phi|_{G}$ is irreducible
and all irreducible representations of $G$ are obtainable in this way.
Moreover, the irreducible representations $\rho$ of $G$ are parameterized by
the strings $\omega=(a_{1},\ldots,a_{n})$ with $0\leq a_{1},\ldots,a_{n}\leq
1$. So we write $\rho_{\omega}$ to specify the parameter. Set
$\omega_{i}=(0,\ldots,0,1,0,\ldots,0)$ with 1 at the $i$-th position.

For a group $X$ let $1_{X}$ be the trivial one-dimensional representation of
$X$.

######  Theorem 1.1.

Let $G=Sp_{2n}(2)$, and let $\rho$ be an irreducible F-representation of G
with highest weight $\omega=(a_{1},\ldots,a_{n})$. Then $1_{A}$ occurs as a
constituent of the restriction $\rho|_{A}$ for every abelian subgroup A of G
if and only if one of the following holds:

$(1)$ $a_{n}=0$ and either $\sum a_{i}$ is even or $\sum a_{i}>2;$

$(2)$ $a_{n}=1$ and $\sum a_{i}i\geq 2n$.

Theorem 1.1 yields a sufficient condition for an arbitrary element $g\in G$ to
have eigenvalue 1 in a given irreducible representation $\rho$. Obtaining a
sharp necessary condition for an arbitrary $g$ does not seem to be a realistic
task; in Theorem 1.3 we provide a criterion for $\rho(g)$ to have eigenvalue 1
for every $g\in G$.

######  Definition 1.2.

Let $n$ be a natural number. Then the Singer height ${\rm Si}(n)$ of n is the
maximum number l such that there are natural numbers $n_{1},\ldots,n_{l}$ such
that $n_{1}+\cdots+n_{l}\leq n$ and the numbers
$2^{n_{1}}+1,\ldots,2^{n_{l}}+1$ are coprime to each other.

Obviously, $1\leq{\rm Si}(m)\leq{\rm Si}(n)$ for $m<n$, but in general the
behavior of ${\rm Si}(n)$ is quite irregular.

######  Theorem 1.3.

Let $\rho_{\omega}$ be an irreducible representation of $G=Sp_{2n}(2)$ with
highest weight $\omega=(a_{1},\ldots,a_{n})$. Then $1$ is an eigenvalue of
$\rho(g)$ for every $g\in G$ if and only if one of the following holds:

$(1)$ $a_{n}=0$ and $\omega\neq\omega_{i}$ for $i$ odd;

$(2)$ $a_{n}=1$ and $\sum a_{i}i\geq n+{\rm Si}(n)$.

Using terminology of [7, Definition 1.2], a representation $\rho$ of a group
$G$ is called unisingular if $\rho(g)$ has eigenvalue 1 for every $g\in G$. So
Theorem 1.3 classifies irreducible unisingular representations of
$G=Sp_{2n}(2)$ in characteristic 2. Theorem 1.4 refines this result for
elements of prime-power order.

######  Theorem 1.4.

Let $G=Sp_{2n}(2)$ and let $\rho$ be an irreducible $F$-representation of $G$
with highest weight $\omega$. Suppose that $\omega\neq\omega_{i}$ for $i$ odd
or $i=n$. If $g\in G$ is of prime power order then $1$ is an eigenvalue of
$\rho(g)$.

We have rather precise information on eigenvalues of $\rho_{\omega_{n}}(g)$
for an arbitrary semisimple element $g\in G$, see Lemma 3.11, and also of
$\rho_{\omega}(g)$ for $a_{n}=1$ (Theorem 4.5). However, the occurrence of
eigenvalue 1 of $\rho_{\omega_{i}}(g)$ with $i<n$ odd for an arbitrary $g\in
G$ does not reveal good regularities, and we have no uniform result even for
$p$-elements $g$. If $a_{n}=0$ then $\rho(g)$ has eigenvalue 1 for every $g\in
G$ whenever so has $\rho(t)$ for $t\in G$ of order $2^{n}+1$. Similar
”testing” elements exist also when $a_{n}=1$, they are constructed in terms of
${\rm Si}(n)$, see Proposition 4.6.

Another test is Proposition 4.7: for $g\in G$ given, 1 is an eigenvalue of
$\rho(g)$ if 1 is an eigenvalue of $\rho_{\omega_{1}}(g)$ and
$\rho_{\omega_{1}}(g)$.

As a by-product we obtain the following result on the occurrence of weight 0
in an irreducible representation $\rho_{\omega}$ of the algebraic group
$Sp_{2n}(\overline{\mathop{\mathbb{F}}\nolimits}_{2})$.

######  Theorem 1.5.

Let $\omega=(a_{1},\ldots,a_{n})$ be a $2$-restricted dominant weight of
${\mathbf{G}}=Sp_{2n}(\overline{\mathop{\mathbb{F}}\nolimits}_{2})$, and let
$\omega^{\prime}=(a_{1},\ldots,a_{n-1},0)$. Suppose that $a_{n}=1$. Then the
following statements are equivalent:

$(1)$ V has weight $0;$

$(2)$ $\sum a_{i}i$ is even and greater than $2n-1;$

$(3)$ $\omega_{n}\prec\omega^{\prime}.$

Note that, in view of [25, Theorem 15], the case considered in Theorem 1.5,
was the only one where the occurrence of weight 0 in an infinitesimally
irreducible representation of a simple algebraic group was not known.

Another result of independent interest is

######  Theorem 1.6.

Let F be an algebraically closed field of any characteristic,
$\mathbf{G}=Sp_{2n}(F)$ and $n=n_{1}+\cdots+n_{k}$, where $n_{1},\ldots,n_{k}$
are positive integer. Let
$\mathbf{H}=\mathbf{H}_{1}\times\cdots\times\mathbf{H}_{k}$, where
$\mathbf{H}_{i}\cong Sp_{2n_{i}}(F)$ for $i=1,\ldots,k$. Let $\rho$ be an
irreducible representation of G with p-restricted highest weight
$\omega=(a_{1},\ldots,a_{n})$. Then the following are equivalent:

$(1)$ $\tau(\mathbf{H}_{i})=1$ for some $i\in\\{1,\ldots,k\\}$ whenever $\tau$
is a composition factor of $\rho|_{\mathbf{H}};$

$(2)$ $\sum a_{i}i<k$.

Note that the case with $k=2$ in Theorem 1.6 follows from the main result of
[19].

Recall that every element of $G$ of odd order lies in a maximal torus of $G$.
So one can ask when the trivial representation $1_{T}$ of a maximal torus $T$
of $G$ occurs as a constituent of the restriction of $\rho_{\omega}$ to $T$.
The answer is similar to the result of Theorem 1.1, see Theorem 3.7.

In [28] we classify unisingular irreducible representations of
$H=GL_{n}(2)=SL_{n}(2)$. For some applications one needs a more precise
information for specific elements $h\in H$. We apply the above results to deal
with the real elements of $H$. Recall that an element $x$ of a group $X$ is
called real if $x$ is conjugate to $x^{-1}$.

######  Theorem 1.7.

Let $H=SL_{2m}(2)$ or $SU_{2m}(2)$, and $\rho$ an irreducible $2$-modular
representation of $H.$ Let $h\in H$ be real. Then $\rho(h)$ has eigenvalue $1$
unless, possibly, $\rho$ is an odd exterior power of the natural (that is,
defining) representation of $H$.

Note that $h\in SL_{2m}(q)$ is real whenever $|h|$ divides $q^{i}+1$ for some
$i$ (Lemma 5.9), and $h\in SU_{2m}(q)$ is real whenever $|h|$ divides
$q^{i}-1$ for some odd $i$ (Lemma 5.10).

Observe that if $H=SL_{n}(F)$ with $n$ odd and $G=Sp_{n-1}(F)$ for any
algebraically closed field $F$ then $\rho|_{G}$ has weight 0 for an arbitrary
irreducible representation $\rho$ of $H$ (Lemma 5.11(3)); therefore $\rho(g)$
has eigenvalue 1 whenever $g\in H$ is real (Lemma 5.11).

Kondrat’ev [12] raises the problem of determining the irreducible modular
representations of simple groups $G$ over a prime field such that 1 is not an
eigenvalue of $\rho(g)$ for $g$ to be a prime order. (In [14] the problem was
specified to irreducible elements $g\in G=SL_{n}(q)$ such that $|g|$ is a
prime dividing $(q^{n}-1)/(q-1)$ and $q$ a $p$-power.) So Theorems 1.4 and 1.7
contribute to this problem.

Other results of the paper refine the above results for some special cases. We
mention the following:

######  Theorem 1.8.

††margin:  th2

Let $\phi$ be an irreducible $F$-representation of $\mathbf{G}$ with highest
weight $\omega=(a_{1},\ldots,a_{n})$. Let s be a Singer cycle of $G$. Then $1$
is not an eigenvalue of $\phi(s)$ if and only if $\omega=2^{k}\omega_{n}$ or
$\omega=2^{k}\omega_{i}$ for some odd $i\in\\{1,\ldots,n-1\\}$ and an integer
$k\geq 0$.

An element of order $2^{n}+1$ in $G=Sp_{2n}(2)$ is called a Singer cycle [9].
This generates a maximal torus of $G$ and plays a role in some applications.

Notation $\mathop{\mathbb{C}}\nolimits$ is the field of complex numbers,
$\mathop{\mathbb{Z}}\nolimits$ is the ring of integers and
$\mathop{\mathbb{N}}\nolimits$ the set of natural numbers. For integers
$a,b>0$ we write $(a,b)$ for the greatest common divisor of $a,b$, and $a|b$
means that $b$ is a multiple of $a$. For sets $A,B$ in a
$\mathop{\mathbb{Z}}\nolimits$-lattice we write $A+B$ for the set $\\{a+b:a\in
A,b\in B\\}$. A diagonal $(n\times n)$-matrix with subsequent entries
$x_{1},\ldots,x_{n}$ is denoted by $\mathop{\rm
diag}\nolimits(x_{1},\ldots,x_{n})$. By $\mathop{\mathbb{F}}\nolimits_{q}$ we
denote the finite field of $q$ elements and by
$\overline{\mathop{\mathbb{F}}\nolimits}_{q}$ the algebraic closure of
$\mathop{\mathbb{F}}\nolimits_{q}$. All representations below are over an
algebraically closed field $F$ of characteristic 2 (unless otherwise is stated
explicitly). $F^{\times}$ is the multiplicative group of $F$. If $\phi$ is a
representation of a group $X$ and $Y\subset X$ is a subgroup then $\phi|_{Y}$
means the restriction of $\phi$ to $Y$. The tensor product of representations
$\phi,\psi$ is denoted by $\phi\otimes\psi$.

Notation and terminology for weight systems are introduced in Section 2.1, for
those for algebraic groups see Sections 2.2 and 2.3.

##  2\. Preliminaries

###  2.1. The weight system of type $C_{n}$

For a precise definition of weight system of type $C_{n}$ see [1]. We recall
here some notions of the theory.

A weight lattice $\Omega$ is formed by strings of integers
$(a_{1},\ldots,a_{n})$, which are called weights of $\Omega$, and those with
non-negative entries $a_{1},\ldots,a_{n}$ are called dominant weights. The
subset of dominant weights is denoted by $\Omega^{+}$. The weights
$\omega_{i}=(0,\ldots,0,1,0,\ldots,0)$ $(1\leq i\leq n)$, where $1$ occupies
the $i$-th position, are called fundamental, and the weight $(0,\ldots,0)$ is
called the zero weight. We usually simplify this by denoting this by 0. So
arbitrary weights can be written as $\sum_{i=1}^{n}a_{i}\omega_{i}$ with
$a_{1},\ldots,a_{n}\in\mathop{\mathbb{Z}}\nolimits$. For an integer $m>1$ a
dominant weight $(a_{1},\ldots,a_{n})$ or $\sum_{i=1}^{n}a_{i}\omega_{i}$ is
called $m$-restricted if $a_{i}<m$ for $i=1,\ldots,n$. (We use this mainly for
$m=2.$) We denote by $\Omega^{\prime}$ the subset of weights
$(a_{1},\ldots,a_{n-1},0)$.

The theory of weight system of type $C_{n}$ singles out a sublattice
$\mathcal{R}$ of index $2$ in $\Omega$ called the root lattice. In addition,
one singles out some elements $\alpha_{1},\ldots,\alpha_{n}\in\mathcal{R}$
called simple roots, which form a $\mathop{\mathbb{Z}}\nolimits$-basis of
$\mathcal{R}$. Denote by $\mathcal{R}^{+}$ the set of non-negative linear
combinations of simple roots. For weights $\omega,\mu$ one writes
$\mu\preceq\omega$ if $\omega-\mu\in\mathcal{R}^{+}$; if additionally
$\omega\neq\mu$ we write $\mu\prec\omega$. If $\mu\prec\omega$ are dominant
then we say that $\mu$ is a subdominant weight for $\omega$. The weights
$\omega\in\mathcal{R}$ are called radical. We write $0\prec\mu$ to state that
the weight 0 is a subdomininant weight of $\mu$.

The theory of weights can be viewed as a part of linear algebra, and it is
somehow independent from the theory of algebraic groups, see for instance [1],
where detailed data on relations between fundamental weights and simple roots
are tabulated in [1, Planche III].

In our analysis an essential role is plaid by another basis
$\varepsilon_{1},\ldots,\varepsilon_{n}$ of $\Omega$; specifically
$\varepsilon_{1}=\omega_{1}$ and $\varepsilon_{i}=\omega_{i}-\omega_{i-1}$ for
$i=2,\ldots,n$.

We have $\omega_{i}=\varepsilon_{1}+\cdots+\varepsilon_{i}$ for
$i=1,\ldots,n$, and $\omega_{i}$ is radical if and only if $i$ is even. The
simple roots are $\alpha_{i}=\varepsilon_{i}-\varepsilon_{i+1}$ for $i<n$, and
$\alpha_{n}=2\varepsilon_{n}$. Note that
$\alpha_{i}=2\omega_{i}-\omega_{i-1}-\omega_{i+1}$ for $i<n$, and
$\alpha_{1}=2\omega_{1}-\omega_{2}$, $\alpha_{n}=2\omega_{n}-2\omega_{n-1}$.

The Weyl group $W$ of $\mathbf{G}$ (or the weight system $\Omega$) acts on
$\Omega$ as a group of linear transformations. Specifically, $W$ transitively
permutes $\pm\varepsilon_{1},\ldots,\pm\varepsilon_{n}$. In fact, $W$ has an
elementary abelian normal subgroup $W_{1}$ which acts on
$\pm\varepsilon_{1},\ldots,\pm\varepsilon_{n}$ by changing the signs, and $W$
is a semidirect product $W_{1}\cdot S_{n}$, where $S_{n}$ is the symmetric
group; the latter acts on $\varepsilon_{1},\ldots,\varepsilon_{n}$ by
permutations. It follows that the $W$-orbit of
$\omega_{i}=\varepsilon_{1}+\cdots+\varepsilon_{i}$ consists of weights
$\pm\varepsilon_{j_{1}}\pm\cdots\pm\varepsilon_{j_{i}}$, where
$\varepsilon_{j_{1}},\ldots,\varepsilon_{j_{i}}\in\\{\varepsilon_{1},\ldots,\varepsilon_{n}\\}$
and $j_{1}<\cdots<j_{i}$.

For a weight $\mu=\sum a_{i}\omega_{i}$ set $\delta(\mu)=\sum a_{i}i$ and
$\gamma(\mu)=\sum a_{i}$.

######  Lemma 2.1.

$(1)$ $\delta(\alpha_{i})=0$ for $i=1,\ldots,n-1$, and $\delta(\alpha_{n})=2$.

$(2)$ $\gamma(\alpha_{i})=0$ for $i=2,\ldots,n-1$, and
$\gamma(\alpha_{1})=1=\gamma(\alpha_{n})$.

$(3)$ Let $\omega,\nu\in\Omega$. If $\omega>\mu$ then
$\delta(\omega)\geq\delta(\mu)$ and $\delta(\omega)-\delta(\mu)$ is even. In
particular, $\omega\in{\mathcal{R}}$ if and only if $\delta(\omega)$ is even.

Proof. (1) and (2) follow straightforwardly from the above expressions of the
simple roots in terms of the dominant weights, and (3) follows from (1) and
the definition of the partial order $\omega>\mu$.

######  Lemma 2.2.

Let $\omega$ be a dominant weight of $C_{n}$.

$(1)$ If $0\prec\omega$ then $\omega_{2}\preceq\omega.$

$(2)$ If $\omega$ is not radical then $\omega_{1}\preceq\omega.$

$(3)$ Let $\omega=\sum a_{i}\omega_{i}$ and $c=\sum a_{i}$. Then
$c\omega_{1}\preceq\omega$ or
$(c-1)\omega_{1}\prec(c-1)\omega_{1}+\omega_{2}\prec\omega.$ In addition, if
$\omega_{1}\prec\omega$ and $c>2$ then $3\omega_{1}\preceq\omega$.

$(4)$ $\omega_{i-2}\prec\omega_{i}$ for $i=3,\ldots,n$. Moreover,
$\omega\prec\omega_{i}$ if and only if $i<j$ and $j-i$ is odd.

$(5)$ Suppose that $\omega\neq\omega_{i}$ for $1\leq i\leq n$. If
$\omega_{1}\prec\omega$ then $\omega_{1}+\omega_{2}\preceq\omega$; if
$0\prec\omega$ then $2\omega_{1}\preceq\omega.$

$(6)$ Let $l=\delta(\omega)$, and $l=j+kn$ with $0\leq j<n$. If $j=0$ then
$k\omega_{n}\preceq\omega$, otherwise $\omega_{i}+k\omega_{n}\preceq\omega$.
In addition, if $k>0$ then $\omega_{n}\preceq\omega$ if $l-n$ is even, and
$\omega_{1}+\omega_{n}\preceq\omega$ if $l-n$ is odd.

Proof. The statements (1),(2) are well known.

(3) Set $c_{1}=\sum_{i~{}odd}a_{i}$, $c_{2}=\sum_{i~{}even}a_{i}$. By (1),(2)
we have $\omega_{1}\preceq\omega_{i}$ for $i$ odd and
$\omega_{2}\preceq\omega_{i}$ for $i$ even. Therefore,
$c_{1}\omega_{1}+c_{2}\omega_{2}\preceq\omega$. As $2\omega_{1}\prec
2\omega_{2}$, the result follows if $c_{2}$ is even, otherwise we have
$(c-1)\omega_{1}\prec(c-1)\omega_{1}+\omega_{2}\prec$. If
$\omega_{1}\prec\omega$ then $c$ is odd in the former case, whence the claim,
and $c$ is even in the latter case. Then $c\geq 4$, and the result follows.

(4) The first claim is obvious as
$\omega_{i}-\omega_{i-2}=\varepsilon_{i}+\varepsilon_{i-1}$ is a positive root
[1, Planche III]. For the second one follows from Lemma 2.1(3).

(5) By (3), $c\omega_{1}\preceq\omega$ or
$(c-1)\omega_{1}+\omega_{2}\preceq\omega$. If $\omega_{1}\prec\omega$ then $c$
is odd in the former case and even in the latter one. As $(c-2)\omega_{1}\prec
c\omega_{1}$ for $c>2$, we have $\omega_{1}+\omega_{2}\prec
3\omega_{1}\preceq\omega$ in the former case, and
$\omega_{1}+\omega_{2}\preceq\omega$ in the latter case. If $0\prec\omega$
then $c$ is even in both the cases, whence $2\omega_{1}\preceq\omega$ or
$2\omega_{1}\prec 2\omega_{1}+\omega_{2}\preceq\omega$, as required.

(6) It suffices to prove this for $\omega$ minimal, in the sense that there is
no dominant weight $\mu\prec\omega$ such that $\delta(\mu)=\delta(\omega)$. As
$\delta(\alpha_{i})=0$ for $i<n$, it follows that $\omega-\alpha_{i}$ is not a
dominant weight for every $i=1,\ldots,n-1$. This implies $a_{i}\leq 1$ for
$i=1,\ldots,n-1$.

Suppose that there are $i<j<n$ such that $a_{i}=a_{j}=1$. As
$\omega_{i}+\omega_{j}=\omega_{i-1}+\omega_{j+1}+\alpha_{i}+\cdots+\alpha_{j}$,
we have $\omega_{i-1}+\omega_{j+1}\prec\omega_{i}+\omega_{j}$ and
$\delta(\omega_{i}+\omega_{j})=\delta(\omega_{i-1}+\omega_{j+1})$ (because
$j<n$). This is a contradiction. So there is at most one $i<n$ such that
$a_{i}>0$, so $\omega=k\omega_{n}$ or $\omega_{i}+k\omega_{n}$, as required.

This also implies the additional statement if $l\leq n+1$. Suppose $l>n+1$.
Now choose $\omega$ to be minimal in the sense that there is no dominant
weight $\mu\prec\omega$ such that $\delta(\mu)>n+1$. If $a_{n}>1$ then
$\delta(\omega-\alpha_{n})=l-2$, contradicting the minimality of $\omega$
unless $l-2\leq n+1$, that is, $l\leq n+3$. In this case $n+3\geq
l=\delta(\omega)\geq\delta(2\omega_{n})=2n$, whence $n\leq 3$. If $n=3$ then
$\omega=2\omega_{3}$ and $\omega-\alpha_{3}=2\omega_{2}>\omega_{1}+\omega_{3}$
as required. If $n=2$ then $\omega=\omega_{1}+2\omega_{2}$ or $2\omega_{2}$,
and the lemma is true in these cases as $\omega_{2}\prec 2\omega_{1}\prec
2\omega_{2}$.

Let $a_{n}=1$. Then $\omega=\omega_{i}+\omega_{n}$ and $0\prec\omega_{i}$ if
$i$ is even, otherwise $\omega_{1}\prec\omega$, whence the result. $\Box$

###  2.2. Abelian subgroups and maximal tori: generalities

Let $V$ be the natural $\mathop{\mathbb{F}}\nolimits_{2}G$-module and $A$ an
abelian subgroup of $G$ of odd order. By Maschke’s theorem,
$V=U_{1}\oplus\cdots\oplus U_{m}$, where $U_{1},\ldots,U_{m}$ are irreducible
$F_{2}A$-modules. Each $U_{i}$, $i=1,\ldots,m$, is either non-degenerate or
totally isotropic; in the latter case there is a unique $j$ such that
$U_{i}+U_{j}$ is non-degenerate. Moreover, $U_{j}$ is dual to $U_{i}$ as
$F_{2}A$-modules. We can rearrange the above decomposition and write
$V=V_{1}\oplus\cdots\oplus V_{k}$, where each $V_{i}$ is non-degenerate and
either irreducible or a sum of two irreducible submodules dual to each other.
Let $2n_{i}=\dim V_{i}$ and let $A_{i}$ be the restriction of $A$ to $V_{i}$.
Then $A_{i}\subset Sp(V_{i})$ is a cyclic group and $|A_{i}|$ divides
$2^{n_{i}}+1$ if $V_{i}$ is irreducible, otherwise $|A_{i}|$ divides
$2^{n_{i}}-1$. In addition, $n_{1}+\cdots+n_{k}=n$. If $A$ is a maximal
abelian subgroup of odd order then $A\cong A_{1}\times\cdots\times A_{k}$ and
$|A_{i}|=2^{n_{i}}+1$ or $2^{n_{i}}-1$, in the latter case $n_{i}>2$. If $k=1$
and $|A_{1}|=2^{n}+1$ then $A$ is called a Singer subgroup of $G$, and the
generators of $A$ are called Singer cycles.

In the theory of finite reductive groups an important role belongs to maximal
tori, which are abelian groups of order coprime to the characteristic of the
ground field. Every maximal abelian subgroup of odd order is a maximal torus,
but not conversely. The definition and general properties of maximal tori is
available in [4] and [3], for specific properties of maximal tori in classical
groups used in this paper one can consult with [27], [26] or [8]. The maximal
tori in $G=Sp_{2n}(2)$ are in bijective correspondence with the decompositions
$V=V_{1}\oplus\cdots\oplus V_{k}$ defined above, and the conjugacy classes of
maximal tori in $G$ are in bijective correspondence with the $G$-orbits of the
decompositions. These can be parameterized by the strings $(\pm
n_{1},\ldots,\pm n_{k})$ up to the ordering, the sign minus is chosen whenever
the torus acts on $V_{i}$ irreducibly ($i=1,\ldots,k$). The string
$(1,\ldots,1)$ corresponds to the torus of one element $\mathop{\rm
Id}\nolimits$, the string $(-1,\ldots,-1)$ labels the torus denoted below by
$T^{\\#}$ (so $|T^{\\#}|=3^{n}$). A maximal torus of order $2^{n}+1$ is called
a Singer torus; it corresponds to $(-n)$. A maximal torus corresponding to
$(n)$ is a cyclic group of order $2^{n}-1$. Every maximal torus $T$ of $G$ can
be written as $T=T_{1}\times\cdots\times T_{k}$, where $|T_{i}|=2^{n_{i}}+1$
if the sign of $n_{i}$ is minus, otherwise $|T_{i}|=2^{n_{i}}-1$. (The above
condition $n_{i}>2$ is irrelevant now.) The decomposition is unique up to the
ordering of $T_{1},\ldots,T_{k}$, and we call it the cyclic decomposition
(reflecting the fact that $T_{1},\ldots,T_{k}$ are cyclic groups). Observe
that the multiples $T_{i}$ with $|T_{i}|=1$ ($1\leq i\leq k$) are not dropped
from the expression $T=T_{1}\times\cdots\times T_{k}$.

Sometimes it is convenient to use a pair of strings $(n_{1},\ldots,n_{k})$,
$(\eta_{1},\ldots,\eta_{k})$, where $\eta_{1},\ldots,\eta_{k}\in\\{1,-1\\}$.
Then $|T_{i}|=2^{n_{i}}-\eta_{i}$ for $1\leq i\leq k$. It follows that every
maximal torus is contained in a direct product
$G_{T}=Sp_{2n_{1}}(2)\times\cdots\times Sp_{2n_{k}}(2),$ where $T_{i}$ is a
maximal torus of $Sp_{2n_{i}}(2)$ for $i=1,\ldots,n$. In addition, $T_{i}$ is
a Singer torus in $Sp_{2n_{i}}(2)$ if and only if $\eta_{i}=-1$, otherwise
$T_{i}$ is of order $2^{n_{i}}-1$ and reducible in $Sp_{2n_{i}}(2)$. The
number of indices $i$ such that $\eta_{i}=-1$ is here called the Singer index
of $T$.

Let $T=\langle t\rangle$ be a maximal torus of order $2^{n}+1$ and $2^{n}-1$,
respectively. (These are known to be unique up to conjugation, see for
instance [26, Lemma 2.1] and [5, Lemma 7.1].) Then $t$ is conjugate in
$\mathbf{G}$ to an element $d\in{\mathbf{T}}$ such that
$\varepsilon_{i}(d)=\zeta^{2^{i-1}}$ for $i=1,\ldots,n$, where $\zeta$ is a
primitive $|t|$-root of unity. (This is well known and explained in detail in
[27].) Under a certain basis of $V,$ the matrix of $J$ on $V$ is diagonal of
shape

(1) $d:=\mathop{\rm
diag}\nolimits(J,J^{-1})\in\mathbf{G}=Sp_{2n}(\overline{\mathop{\mathbb{F}}\nolimits}_{2}),~{}{\rm
where}~{}J=\mathop{\rm
diag}\nolimits(\zeta,\zeta^{2},\ldots,\zeta^{2^{n-1}}).$

Therefore, $\varepsilon_{i}(d)=\zeta^{2^{i-1}}$ for $i=1,\ldots,n$. This
implies

(2) $2\varepsilon_{i}(d)=\varepsilon_{i+1}(d)~{}{\rm
for}~{}i=1,\ldots,n-1~{}{\rm
and}~{}2\varepsilon_{n}(d)=\zeta^{2^{n}}=\begin{cases}-\varepsilon_{1}(d)&if~{}|t|=2^{n}+1\\\
~{}\varepsilon_{1}(d)&if~{}|t|=2^{n}-1.\end{cases}$

This can be extended to arbitrary maximal tori. Let $V=\oplus V_{i}$ and let
$T_{i}$ be a maximal torus in $Sp(V_{i})\cong Sp_{2n_{i}}(2)$, so $T$ is
conjugate to the group $D:=\\{\mathop{\rm
diag}\nolimits(J^{m_{i}}_{1},\ldots,J^{m_{k}}_{k},J^{-m_{k}}_{k},\ldots,J^{-m_{i}}_{1}\\}$,
where $J_{i}$ $(i=1,\ldots,k)$ is constructed as above, $1\leq
m_{i}\leq|T_{i}|$. We call $D$ a canonical form of $T$. Note that there is a
conceptual way to construct $D$ for arbitrary finite group of Lie type, see
the proof of [3, Proposition 3.3.6].

######  Lemma 2.3.

[17, p. 209, Corollary 5.10(a)] Let $G=Sp_{2n}(2)$ and $A\subset G$ an abelian
subgroup of odd order. Then A is contained in a maximal torus of G. In
particular, if A is a maximal abelian subgroup of odd order then A is a
maximal torus of G.

The following lemma is trivial, but we state it explicitly for reader’s
convenience.

######  Lemma 2.4.

††margin:  aa1

Let A be an abelian subgroup of G, and $\rho$ a representation of G over a
field of characteristic $2$. Let B be a maximal subgroup of A of odd order.
Then $1_{A}$ is a constituent of $\rho|_{A}$ if and only if $1_{B}$ is a
constituent of $\rho|_{B}$.

###  2.3. Representations and their weights

The above parametrization of the irreducible representations of $G=Sp_{2n}(2)$
is based on Steinberg’s theorem [10, §2.7] saying that the irreducible
representations of $Sp_{2n}(2)$ are in bijection with the $2$-restricted
dominant weights of $\mathbf{G}$. In fact, the representation theory of
$\mathbf{G}$ plays a significant role in proving the above result. The
representation theory of the algebraic group $Sp_{2n}(F)$ is described on the
language of the theory of the weight system of type $C_{n}$.

Let $F$ be an algebraically closed field of characteristic $p>0$. The group
$\mathbf{G}=Sp_{2n}(F)$ is defined as the group of $(2n\times 2n)$-matrices
over $F$ preserving a non-degenerate alternating form $(*,*)$ on $V=F^{2n}$. A
basis $e_{1},\ldots,e_{2n}$ of $V$ is called a Witt basis if
$(e_{i},e_{2n+1-i})=1$ and $(e_{i},e_{j})=0$ if $j\neq 2n+1-i$ for
$i=1,\ldots,n$ and $1\leq j\leq 2n$. The subgroup $\mathbf{T}$ of diagonal
matrices in $\mathbf{G}$ under a Witt basis is of shape $\mathop{\rm
diag}\nolimits(t_{1},\ldots,t_{n},t_{n}^{-1},\ldots,t_{1}^{-1})$
($t_{1},\ldots,t_{n}\in F^{\times}$). We refer to this group as a reference
torus, and define a maximal tori in $\mathbf{G}$ as those conjugate to
$\mathbf{T}$ in $\mathbf{G}$. The weights are defined as rational
homomorphisms $\mathbf{T}\rightarrow F^{\times}$, and $\varepsilon_{i}$ are
defined as those sending $\mathop{\rm
diag}\nolimits(t_{1},\ldots,t_{n},t_{n}^{-1},\ldots,t_{1}^{-1})$ to $t_{i}$.

The Weyl group $W$ of $\mathbf{G}$ is defined as
$W:=N_{\mathbf{G}}(\mathbf{T})/\mathbf{T}$. The conjugation action of
$N_{\mathbf{G}}(\mathbf{T})$ on $\mathbf{T}$ yields the action of $W$ on
$\Omega={\rm Hom}(\mathbf{T},F^{\times})$. In this action $W$ preserves the
set $\\{\pm\varepsilon_{1},\ldots,\pm\varepsilon_{n}\\}$ and acts on this set
transitively. Note that the $W$-orbit of any weight $\omega\in\Omega$ contains
a unique dominant weight.

For a $\mathbf{G}$-module $V$ or a representation $\rho$ of $\mathbf{G}$ we
denote by $\Omega(V)$, $\Omega(\rho)$ the set of weights of $V,\rho$,
respectively. Then $\Omega(V)$ is $W$-invariant, in particular,
$\Omega(V)=-\Omega(V)$.

Let $\mu$ be a dominant weight and $W$ the Weyl group of $\mathbf{G}$. Denote
by $\chi_{\mu}$ the sum of distinct weights $w(\mu)$ with $w\in W$. For a
semisimple element $g\in G$ let $g^{\prime}$ be a conjugate of $g$ in
$\mathbf{T}$ (it is well known that every semisimple element of $\mathbf{G}$
is conjugate to one in $\mathbf{T}$). Then $g^{\prime}$ is not unique but the
value $\chi_{\mu}(g^{\prime})$ does not depend from the choice of
$g^{\prime}\in\mathbf{T};$ so $\chi_{\mu}$ is a well defined function on the
semisimple elements of $G$ called an orbit character. (This is well known to
be a generalized Brauer character of $G$, see [10, §5.7].)

If $F=\overline{\mathop{\mathbb{F}}\nolimits}_{p}$ and $\beta_{\omega}$ is the
Brauer character of $V_{\omega}$ then $\beta_{\omega}=\sum m_{\mu}\chi_{\mu}$,
where $\mu$ runs over the dominant weights of $V_{\omega}$ and $m_{\mu}$ is
the multiplicity of $\mu$ in $V_{\omega}$, see [10, §2.2 and §5.6] for
details. Therefore, $\beta_{\omega}|_{G}=\sum m_{\mu}\chi_{\mu}|_{G}$; this is
used below for computations of eigenvalue 1 of elements $g\in G$ on
$V_{\omega}$.

######  Lemma 2.5.

Let F be an algebraically closed field of characteristic $p>0$ and
$\mathbf{G}=SL_{n+1}(F)$ or $Sp_{2n}(F)$. Let V be an irreducible
$\mathbf{G}$-module with $p$-restricted highest weight
$\omega=(a_{1},\ldots,a_{n})$.

$(1)$ The weights of V are the same as those of the irreducible L-module $M$
of the Lie algebra over the complex numbers of the same type as $\mathbf{G}$
with with highest weight $(a_{1},\ldots,a_{n})$ (but the multiplicities of the
same weight in V and M may differ), unless possibly $p=2$,
$\mathbf{G}=Sp_{2n}(F)$ and $a_{n}=1$.

$(2)$ Let $\mu\prec\omega$ be dominant weights. Then $\mu\in\Omega(V)$.
Moreover, all weights of $\Omega(V_{\mu})\subset\Omega(V_{\omega})$ provided
$a_{n}=0$ for $p=2$ and $\mathbf{G}=Sp_{2n}(F)$.

$(3)$ Let $\lambda$ be a dominant weights for $\mathbf{G}$ such that
$\lambda+\omega$ is p-restricted. Then
$\Omega(V_{\lambda+\omega})=\Omega(V_{\lambda})+\Omega(V_{\omega})$.

Proof. For (1) see [25, Theorem 15]. (2) follows from (1) and [2, Ch. VIII,
Prop 5]. (3) Clearly, either $\lambda\in\Omega^{\prime}$ or
$\omega\in\Omega^{\prime}$ (or both). By swapping $\lambda,\omega$ we can
assume that $\lambda\in\Omega$. (i) Suppose first that we are not in the
exceptional situation of (1). Then the result follows from (1) and [2, Ch.
VIII, Prop 10]. (ii) Let $p=2$, $\mathbf{G}=Sp_{2n}(F)$ and $a_{n}=1$. Set
$\mu^{\prime}=\omega-\omega_{n}$. Then
$V_{\lambda+\omega}=V_{\lambda+\mu^{\prime}+\omega_{n}}=V_{\lambda+\mu^{\prime}}\otimes
V_{\omega_{n}}$, and then
$\Omega(V_{\lambda+\omega})=\Omega(V_{\lambda+\mu^{\prime}}\otimes
V_{\omega_{n}})=\Omega(V_{\lambda+\mu^{\prime}})+\Omega(V_{\omega_{n}})$. By
(i),
$\Omega(V_{\lambda+\mu^{\prime}})=\Omega(V_{\lambda})+\Omega(V_{\mu^{\prime}})$,
so
$\Omega(V_{\lambda+\omega})=\Omega(V_{\lambda})+\Omega(V_{\mu^{\prime}})+\Omega(V_{\omega_{n}})$.
As $V_{\mu^{\prime}}\times V_{\omega_{n}}=V_{\mu^{\prime}+\omega_{n}}$, we
have
$\Omega(V_{\mu^{\prime}})+\Omega(V_{\omega_{n}})=\Omega(V_{\mu^{\prime}+\omega_{n}})$,
whence the result. $\Box$

If $\omega\notin\Omega^{\prime}$ then the statement (2) of Lemma 2.5 is not
valid anymore; Lemma 2.6 and Theorem 2.7 below describe some special cases.

######  Lemma 2.6.

Let $\omega=(a_{1},\ldots,a_{n})$ be a $2$-restricted dominant weight of
$\mathbf{G}=Sp_{2n}(F)$. Suppose that $a_{n}=1$ and $\delta(\omega)=\sum
a_{i}i\geq 2n$.

$(1)$ If $\omega$ is radical then $0$ is a weight of $V_{\omega};$

$(2)$ if $\omega$ is not radical then
$\omega_{1},\omega_{1}+\omega_{2},3\omega_{1}$ are weights of $V_{\omega}$ as
well as $\omega_{3}$ for $n>2$. In addition, all weights of
$V_{\omega_{1}+\omega_{2}}$ as weights of $V_{\omega}$.

Proof. Let $\omega^{\prime}=\omega-\omega_{n}$. Then
$V_{\omega}=V_{\omega^{\prime}}\otimes V_{\omega_{n}}$. Then
$\omega^{\prime}\in\Omega^{\prime}$, and hence $\mu\prec\omega^{\prime}$ for
$\mu$ dominant implies $\mu$ to be a weight of $V_{\omega^{\prime}}$ (Lemma
2.5(2)).

(1) As $\delta(\omega^{\prime})\geq n$ and $\omega$ is radical, we have
$\omega^{\prime}-\omega_{n}\in\mathcal{R}$, and then
$\omega_{n}\preceq\omega^{\prime}$ by Lemma 2.2(6). So $\omega_{n}$ is a
weight of $V_{\omega^{\prime}}$. As $\omega_{n}$ is a weight of
$V_{\omega_{n}}$, the claim follows.

(2) As $\omega$ is not radical, we have
$\omega^{\prime}-\omega_{n}\notin\mathcal{R}$, so $\omega_{n}$ is not a weight
of $V_{\omega^{\prime}}$. By Lemma 2.2(6),
$\omega_{1}+\omega_{n}\preceq\omega^{\prime}$ (as $\delta(\omega^{\prime})\geq
n+1$). So $\omega_{1}+\omega_{n}\in\Omega(V_{\omega^{\prime}})$ by Lemma 2.5,
and hence $\omega_{1}\in\Omega(V_{\omega})$.

We have
$\omega_{1}+\omega_{n}=2\varepsilon_{1}+\varepsilon_{2}+\cdots+\varepsilon_{n}$
and $\omega_{n}=\varepsilon_{1}+\varepsilon_{2}+\cdots+\varepsilon_{n}$. As
$\mu=\varepsilon_{1}-\varepsilon_{2}-\cdots-\varepsilon_{n}$ is a weight of
$V_{\omega_{n}}$, we conclude that $3\omega_{1}=\omega_{1}+\omega_{n}+\mu$ is
a weight of $V_{\omega}$.

Similarly,
$\nu=\varepsilon_{1}+2\varepsilon_{2}+\varepsilon_{3}+\cdots+\varepsilon_{n}\in\Omega(V_{\omega^{\prime}})$,
so
$\nu+\mu=2\varepsilon_{1}+\varepsilon_{2}=\omega_{1}+\omega_{2}\in\Omega(V_{\omega})$.

Finally, if $n>2$ then $\omega_{1}+\omega_{n-2}\prec\omega_{1}+\omega_{n}$,
and hence $\omega_{1}+\omega_{n-2}\in\Omega(V_{\omega^{\prime}})$ by Lemma
2.5. As
$\omega_{1}+\omega_{n-2}=2\varepsilon_{1}+\varepsilon_{2}+\cdots+\varepsilon_{n-2}$,
it follows that
$\varepsilon_{1}-\varepsilon_{n-1}-\varepsilon_{n}\in\Omega(V_{\omega})$. This
weight is in the $W$-orbit of
$\omega_{3}=\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3}$, so
$\omega_{3}\in\Omega(V_{\omega})$.

The additional statement follows as the weights $\omega_{1}$, and $\omega_{3}$
for $n>2$, are the only subdominant weights of $V_{\omega_{1}+\omega_{2}}$.
$\Box$

######  Theorem 2.7.

Let $\omega=(a_{1},\ldots,a_{n})$ be a $2$-restricted weight and $a_{n}=1$.
Then $V=V_{\omega^{\prime}}\otimes V_{\omega_{n}}$, where
$\omega^{\prime}=\omega-\omega_{n}$, and the following statements are
equivalent:

$(1)$ V has weight $0;$

$(2)$ $\omega_{n}\prec\omega^{\prime};$

$(3)$ $\delta(\omega)$ is even and greater than $2n-1$.

Proof. Note that
$\Omega(V)=\Omega(V_{\omega^{\prime}})+\Omega(V_{\omega_{n}})$.

$(2)\rightarrow(1)$. By Lemma 2.5(1), $\omega_{n}$ is a weight of
$V_{\omega^{\prime}}$, and hence so is $-\omega_{n}$. Whence the claim.

$(1)\rightarrow(2)$. The weights of $V_{\omega_{n}}$ are $W$-conjugate. It
follows that if a weight $\mu\in\Omega(V_{\omega_{n}})$ then $\pm\omega_{n}$
is a weight of $V_{\omega^{\prime}}$. Then $\omega_{n}$ is a weight of
$V_{\omega^{\prime}}$ too, whence $\omega_{n}\prec\omega^{\prime}$.

$(2)\rightarrow(3)$. We have $\omega^{\prime}-\omega_{n}\in\mathcal{R}$ as
$\omega_{n}\prec\omega^{\prime}$, and hence
$\omega^{\prime}+\omega_{n}\in\mathcal{R}$ as $2\omega_{n}\in\mathcal{R}$. So
the result follows from Lemma 2.1(3).

$(3)\rightarrow(2)$ As $\delta(\omega)$ is even,
$\delta(\omega^{\prime})-n=\delta(\omega)-2n$ is even too. Since
$\omega^{\prime}$ is dominant and $\omega^{\prime}\neq\omega_{n}$ by
assumption, the result follows from Lemma 2.2(6) applied to $\omega^{\prime}$.
$\Box$

Remark. Under the assumption of Theorem 2.7, if $V$ has weight 0 then
$\omega_{i}\in\Omega(V)$ for every $i\leq n$ even. Indeed, by Lemma 2.5.
$\omega_{n-i}\in\Omega(V_{\omega^{\prime}})$ for $i<n$ even. So
$\mu:=-\varepsilon_{i+1}-\cdots-\varepsilon_{n}\in\Omega(V_{\omega^{\prime}})$
so
$\mu+\omega_{n}=\varepsilon_{1}+\cdots+\varepsilon_{i}\in\Omega(V_{\omega})$.

##  3\. Symplectic groups over a field of two elements

In this section $G=Sp_{2n}(2)$ and ${\mathbf{G}}=Sp_{2n}(F)$, where $n>1$ and
$F$ is an algebraically closed field of characteristic 2. We fix a reference
torus ${\mathbf{T}}$ of ${\mathbf{G}}$ which determine the weight system of
${\mathbf{G}}$. Recall that the conjugacy classes of maximal tori of $G$ are
in a canonical bijection with the conjugacy classes of the Weyl group $W$ of
${\mathbf{G}}$. So one can write $T_{w}$ for a maximal torus of $G$ labeled by
the conjugacy class of $w\in W$. When we consider properties of $T$ in a
representation of ${\mathbf{G}}$ we choose a canonical conjugate
$D\subset{\mathbf{G}}$ as defined in Section 2.2. Let $T$ be a maximal torus
of $G$. Then $T=T_{1}\times\cdots\times T_{k}$ for some $k\leq n$, and let
$(\pm n_{1},\ldots,\pm n_{k})$ be the label of $T$, see Section 2.2. Abusing
notation, we often assume that $D=T$.

The following lemma is a special case of [27, Lemma 3.9].

######  Lemma 3.1.

Let $\chi_{\omega_{i}}$ be the orbit character of $G$ and $T=T_{w}$ a maximal
torus labeled by an element $w\in W$. Then $1_{T}$ is a constituent of
$\chi_{\omega_{i}}|_{T}$ if and only if w is conjugate in W to an element of
$W_{i}$, where $W_{i}$ is the stabilizer of
$\varepsilon_{1}+\cdots+\varepsilon_{i}$ in W.

Proof. In [27, Lemma 3.9] the result is stated in a more precise form,
specifically, it states that the multiplicity of $1_{T}$ in
$\chi_{\omega_{i}}|_{T}$ equals the value of the induced character
$1_{W_{i}}^{W}$ at $w$, where $W_{i}$ is the stabilizer of
$\omega_{i}=\varepsilon_{1}+\cdots+\varepsilon_{j}$ in $W$. It is well known
that this is non-zero if and only if $w$ is conjugate to an element of
$W_{i}$. $\Box$

######  Lemma 3.2.

Let $s$ be a Singer cycle of G. Then s has no fixed point on $V_{\omega_{i}}$
if and only if $i=n$ or $i<n$ odd.

Proof. Set $S=\langle s\rangle$. If $i<n$ is even then $V_{\omega_{i}}$ has
weight 0, whence the claim for this case. Let $i<n$ be odd. By Lemma 2.2(4),
$\omega_{j}$ with $j<i<n$ odd are the only subdominant weights of
$\omega_{i}$. In addition, the weights of $V_{\omega_{n}}$ are in the
$W$-orbit of $\omega_{n}.$ Therefore, it suffices to show that $1_{S}$ is not
a constituent of the orbit character $\chi_{\omega_{j}}$ for $j$ odd and
$j=n$. This follows from Lemma 3.1. Indeed, $S=T_{w}$, where $w\in W$ is such
that $w(\varepsilon_{i})=\varepsilon_{i+1}$ for $i=1,\ldots,n-1$, and
$w(\varepsilon_{n})=-\varepsilon_{1}$. Note that $W$ acts on the set
$\\{\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n}\\}$ in the natural way as
explained in Section 2.3, and the $W$-orbit of
$\omega_{i}=\varepsilon_{1}+\cdots+\varepsilon_{i}$ consists of the elements
$\pm\varepsilon_{j_{1}}\pm\cdots\pm\varepsilon_{j_{i}}$ ($1\leq
j_{i}\leq\cdots\leq j_{i}\leq n$). Then it is clear that $w$ fixes none of
these elements. It follows that $w$ is not conjugate to any element of $W_{j}$
for $j=1,\ldots,n$. $\Box$

Denote by $T^{\\#}$ a maximal torus of $G$ whose cyclic decomposition is
$T=T_{1}\times\cdots\times T_{n}$ and $|T_{1}|=\cdots=|T_{n}|=3$. The torus
$T^{\\#}$ is unique up to conjugacy as the Singer index of it must be $n$.
Then a canonical torus $D\subset\mathbf{T}$ consists of elements $\mathop{\rm
diag}\nolimits(t_{1},\ldots,t_{n},t_{n}^{-1},\ldots,t_{1}^{-1})$ with
$t_{i}\in F,t_{i}^{3}=1$ for $i=1,\ldots,n$. It follows that $D$ is invariant
under the Weyl group of $\mathbf{T}$. In addition, $\varepsilon_{i}(t)=t_{i}$
for $t\in T^{\\#}$ and $i=1,\ldots,n$.

######  Lemma 3.3.

Let $T=T_{1}\times\cdots\times T_{k}$ be a maximal torus of G of type $(\pm
n_{1},\ldots,\pm n_{k})$ with $n_{1}\geq n_{2}\geq\cdots\geq n_{k}$. Let
$\lambda=\omega_{1}+\omega_{2}$ and let $\chi_{\lambda}$ be for the orbit
character of $\lambda$.

$(1)$ $(\chi_{\lambda}|_{T},1_{T})=0$ if and only if $k=n$, $n_{i}=1$ for
$i=1,\ldots,n$ and $|T_{1}|=\cdots=|T_{n-1}|=3$.

$(2)$ $1_{T}$ is not a constituent of $V_{\lambda}|_{T}$ if and only if $k=n$
and $T=T^{\\#}$.

$(3)$ Let $g\in G$ be a semisimple element. Then g has eigenvalue $1$ on
$V_{\lambda}$.

Proof. Let $D$ be a canonical form of $T$, see Section 2.2. To simplify
notation, we keep $T$ for $D$ and $t_{i}$ for $J_{i}$ for $i=1,\ldots,k$.

(1) Suppose first that $n_{1}>1$ and let $t\in T$. Then
$\varepsilon_{1}(t)=\zeta$, $\varepsilon_{2}(t)=\zeta^{2}$, where $\zeta\in F$
with $|\zeta|=|T_{1}|$. Then the $W$-orbit of
$\lambda=2\varepsilon_{1}+\varepsilon_{2}$ contains weight
$\mu=2\varepsilon_{1}-\varepsilon_{2}$, and $\mu(t)=1$.

Next, suppose that $n_{1}=1$. Then $n_{i}=1$ and $|T_{i}|\in\\{3,1\\}$ for
$i=1,\ldots,n$. If $|T_{n-1}|=1$ then $|T_{n}|=1$ and
$\varepsilon_{n-1}(t)=\varepsilon_{n}(t)=1$. As the $W$-orbit of $\lambda$
contains weight $\mu=2\varepsilon_{n-1}+\varepsilon_{n}$, we again get
$\mu(t)=1$.

Let $|T_{n-1}|\neq 1$. The $W$-orbit of $\lambda$ consists of weights $\pm
2\varepsilon_{i}\pm\varepsilon_{j}$ for any choice of $i,j$ with $i\neq j$.
Then $(\pm 2\varepsilon_{i}\pm\varepsilon_{j})(T)\neq 1$. Indeed, otherwise
$1=(\pm 2\varepsilon_{i}\pm\varepsilon_{j})(T)=\\{t_{i}^{\pm 2}t_{j}^{\pm
1}:t_{i}\in T_{i},t_{j}\in T_{j}\\}$ for some choice of the signs, which is
false.

(2) By (1), we are left to inspect the cases where $k=n$, $n_{i}=1$ for
$i=1,\ldots,n$ and $|T_{1}|=\cdots=|T_{n-1}|=3\geq|T_{n}|$.

The only subdominant weights of $\lambda$ are $\omega_{1},\omega_{3}$ for
$n>2$ and $\omega_{1}$ for $n=2$. If $n>2$ then $\lambda\in\Omega^{\prime}$,
and hence $\omega_{1},\omega_{3}\in\Omega(V_{\lambda})$ by Lemma 2.5. If $n=2$
then $V_{\lambda}=V_{\omega_{1}}\otimes V_{\omega_{2}}$, and hence
$\pm\varepsilon_{2}+(\pm\varepsilon_{1}\pm\varepsilon_{2})\in\Omega(V_{\lambda})$.
So $\omega_{1}=\varepsilon_{1}\in\Omega(V_{\lambda})$. The $W$-orbit of
$\omega_{1}$ (respectively, of $\omega_{3}$ if $n>2$) consists of weights
$\pm\varepsilon_{1},...,\pm\varepsilon_{n}$ (respectively,
$\varepsilon_{i_{1}}+\varepsilon_{i_{2}}+\varepsilon_{i_{3}}$ with $1\leq
i_{1}<i_{2}<i_{3}\leq n$). If $|T_{n}|=1$ then $\varepsilon_{n}(T)=1$. Let
$|T_{n}|=3$. It is clear that $\varepsilon_{i}(T)\neq 1_{T}$ for every
$i=1,\ldots,n$, as well as
$(\varepsilon_{i_{1}}+\varepsilon_{i_{2}}+\varepsilon_{i_{3}})(T)\neq 1_{T}$
for $n>2$. So the result follows.

(3) If $g$ is contained in a maximal torus distinct from $T^{\\#}$ then $g$
has eigenvalue 1 on $V_{\lambda}$ by (2). This is obviously the case if $g$
has eigenvalue 1 on $V_{\omega_{1}}$, the natural module for $G$. Otherwise,
$g$ has no fixed point on $V_{\omega_{1}}$. Then $V_{\omega_{1}}$ is a direct
sum of $n$ two-dimensional $g$-stable subspaces, isomorphic to each other as
$\mathop{\mathbb{F}}\nolimits_{2}\langle g\rangle$-modules. In particular,
$V_{\omega_{1}}$ is a homogeneous $\mathop{\mathbb{F}}\nolimits_{2}\langle
g\rangle$-module. Then $C_{G}(g)\cong U_{n}(3)$, see for instance [5, Lemma
6.6]. Since $g$ is in the center of $U_{n}(3)$, this lies in every maximal
torus of $U_{n}(3)$. As every maximal torus of $U_{n}(3)$ is a maximal torus
of $G$ and $n>1$, there is a maximal torus $T\neq T^{\\#}$ of $G$ with $g\in
T$, so (3) follows from (2). $\Box$

######  Lemma 3.4.

Let $\omega=\sum a_{i}\omega_{i}\in\Omega^{+}$ be a $2$-restricted weight.
Suppose that $a_{n}=1$ and $\delta(\omega)\geq 2n$. Then $1_{T}$ is a
constituent of $V_{\omega}|_{T}$ for every maximal torus T of G.

Proof. If $\omega$ is radical then $V_{\omega}$ has weight 0 (Lemma 2.6(1)),
whence the result. If $\omega$ is not radical then, by Lemma 2.6(2),
$3\omega_{1}\in\Omega(V_{\omega})$. If $T=T^{\\#}$ then $t^{3}=1$ for every
$t\in T^{\\#}$, and hence $(3\omega_{1})(t)=\omega_{1}(t^{3})=1$, whence the
result.

Let $T\neq T^{\\#}$. By Lemma 2.6(2),
$\Omega(V_{\omega_{1}+\omega_{2}})\subseteq\Omega(V_{\omega})$. So the result
follows from Lemma 3.3. $\Box$

######  Lemma 3.5.

Let $\omega=\sum a_{i}\omega_{i}$ be a non-radical $2$-restricted dominant
weight and $a_{n}=0$. Suppose that $\omega\neq\omega_{i}$ for $i=1,\ldots,n-1$
odd.

$(1)$ Let $T\neq T^{\\#}$ be a maximal torus of $G$. Then $1_{T}$ is a
constituent of $V_{\omega}|_{T}$.

$(2)$ Let $g\in G$ be a semisimple element. Then g has eigenvalue $1$ on
$V_{\omega}$.

Proof. By Lemma 2.2, $\lambda=\omega_{1}+\omega_{2}\preceq\omega$. In
addition, all weights of $\Omega(V_{\lambda})\subseteq\Omega(V_{\omega})$
(Lemma 2.5). So the result follows from Lemma 3.3. $\Box$

######  Lemma 3.6.

Let $\omega=\sum a_{i}\omega_{i}$ be a $2$-restricted dominant weight and
$T=T^{\\#}$.

$(1)$ Suppose that $a_{n}=0$. Then $1_{T}$ is a constituent of
$V_{\omega}|_{T}$ if and only if either $0\preceq\omega$ or $\sum a_{i}>2$.

$(2)$ Suppose that $a_{n}=1$. Then $1_{T}$ is a constituent of
$V_{\omega}|_{T}$ if and only if $\delta(\omega)\geq 2n$.

Proof. Taking $T^{\\#}$ to the canonical form (see Section 2.2), we can assume
that $t=\mathop{\rm diag}\nolimits(t_{1},\ldots,t_{n},\\\
t_{n}^{-1},\ldots,t_{1}^{-1})$, where $t_{1}^{3}=\cdots=t_{n}^{3}=1$; then
$\varepsilon_{i}(t)=t_{i}$ for $i=1,\ldots,n$.

(1) The ‘only if’ part. Suppose the contrary. So $\omega$ is not radical and
$\sum a_{i}\leq 2$. Let $\mu=\sum a_{i}^{\prime}\omega_{i}$ be a dominant way
such that $\mu\prec\omega.$ By Lemma 2.1(3), we have $\sum a_{i}^{\prime}\leq
2$. Let $\mu^{\prime}$ be in the $W$-orbit of $\mu$. We claim that
$\mu^{\prime}(T^{\\#})\neq 1$ (getting a contradiction). We can assume that
$\mu=\omega_{i}+\omega_{j}$ $(i<j)$ as the cases with $\mu=\omega_{i}$ and
$\mu=2\omega_{i}$ ($1\leq i\leq n-1$) are trivial. Then
$\mu=2\varepsilon_{1}+\cdots+2\varepsilon_{i}+\varepsilon_{i+1}+\cdots+\varepsilon_{j}$
and $\mu^{\prime}=2(\sum_{k\in K}\pm\varepsilon_{k})+(\sum_{l\in
L}\pm\varepsilon_{l})$, where $K,L$ are disjoint subsets of $\\{1,\ldots,n\\}$
and $|K|=i,|L|=j-i$. Now the claim is obvious.

The ‘if’ part. The case where $\omega$ is radical is trivial. Suppose that
$\omega$ is not radical. By Lemma 2.2(3), $3\omega_{1}$ is a subdominant
weight of $\omega$. Therefore, by Lemma 2.5(2), $3\omega_{1}$ is a weight of
$V_{\omega}$. Obviously, $3\omega_{1}(T^{\\#})=1$, whence the result.

(2) Set $\omega^{\prime}=\omega-\omega_{n}$. Then we have
$V_{\omega}=V_{\omega^{\prime}}\otimes V_{\omega_{n}}$. So the weights of
$V_{\omega}$ are of shape
$\nu+(\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n})$, where $\nu$ is a weight
of $V_{\omega^{\prime}}$. Note that the $W$-orbit of $\nu$ contains a dominant
weight and $w(T^{\\#})=T^{\\#}$. So we can assume that $\nu$ is dominant.

The ‘only if’ part. Suppose that $\delta(\omega)<2n$. Then
$\delta(\omega^{\prime})<n$. By Lemma 2.2(6), $\delta(\nu)<n$ whenever
$\nu\prec\omega^{\prime}$ so $\omega_{n}$ is not a subdominant weight of
$\omega^{\prime}$. Let $t\in T^{\\#}$. Then
$((\nu+(\varepsilon_{1}\pm\cdots\pm\varepsilon_{n-1}))\pm\varepsilon_{n})(t)=x\cdot(\varepsilon_{n}(t_{n})^{\pm
1}$, where $x=(\nu+(\varepsilon_{1}\pm\cdots\pm\varepsilon_{n-1}))(t)$. We can
fix $t_{1},\ldots,t_{n-1}$ and vary $t_{n}$ to be any $3$-root of 1. So we
cannot have $x\cdot(\varepsilon_{n}(t_{n}))^{\pm 1}=1$ for arbitrary $t$.

The ‘if’ part follows from Lemma 3.4. $\Box$

######  Theorem 3.7.

Let $\omega=\sum a_{i}\omega_{i}$ be a $2$-restricted dominant weight. Then
$1_{T}$ is a constituent of $V_{\omega}|_{T}$ for every maximal torus of G if
and only if one of the following holds:

$(1)$ $a_{n}=0$ and either $\sum a_{i}>2$ or $\delta(\omega)=\sum a_{i}i$ is
even;

$(2)$ $a_{n}=1$ and $\delta(\omega)\geq 2n.$

Proof. The ‘if’ part. In (1) if $\delta(\omega)$ is even then $\omega$ is
radical, and $V_{\omega}$ has weight 0 by Lemma 2.5 (as $a_{n}=0$). This also
holds in (2) by Lemma 2.6. So the result follows in these cases. If $\sum
a_{i}>2$ then the result follows from Lemma 3.5(1) if $T\neq T^{\\#}$ and from
Lemma 3.6(1) if $T=T^{\\#}$.

The ‘only if’ part. Let $a_{n}=0$, and suppose that $\sum a_{i}i$ is odd; then
$\omega$ is not radical. So $\sum a_{i}>2$ by Lemma 3.6(1). If $a_{n}=1$ then
the result follows from Lemma 3.6(2). $\Box$

Proof of Theorem 1.1. By Lemma 2.4, we can assume that $|A|$ is odd.
Therefore, it suffices to prove the result for maximal abelian groups of odd
order. By Lemma 2.3, these are maximal tori of $G$. So the result follows from
Theorem 3.7. $\Box$

Next we turn to proving Theorem 1.3. For this we study the problem of the
occurrence of the eigenvalue 1 for individual elements $g\in G$ in irreducible
representations $\rho$ that do not satisfy the condition (1), (2) of Theorem
1.1 for $A=\langle g\rangle$. We start with the case where the highest weight
of $\rho$ is $\omega_{n}$.

Let $g\in G$ be a semisimple element and $V$ the natural
$\mathop{\mathbb{F}}\nolimits_{2}G$-module. Set $A=\langle g\rangle$, and let
$V=V_{1}\oplus\cdots\oplus V_{k}$ be a decomposition of $V$ as a orthogonal
sum of minimal non-degenerate $A$-stable subspaces, see Section 2.2. Let
$g_{i}$ be the restriction of $g$ to $V_{i}$ for $i=1,\ldots,k$, so we can
write $g=\mathop{\rm diag}\nolimits(g_{1},\ldots,g_{k})$. Let $d_{i}=\dim
V_{i}/2$. The decomposition $V=V_{1}\oplus\cdots\oplus V_{k}$ is unique up to
reordering of $V_{1},\ldots,V_{k}$ (if $|g|=3$ and $V_{1}$ is the sum of two
totally isotropic $g$-stable subspaces then $V_{1}$ is also the sum of two
non-degenerate subspaces, so $V_{1}$ is not minimal.) Note that $g\in T$ for
some maximal torus $T=T_{1}\times\cdots\times T_{k}$ of $G$ such that
$TV_{i}=V_{i}$, $g_{i}\in T_{i}$ for $i=1,\ldots,k$, and $|T_{i}|=1$ if
$g_{i}=1$. Then such a torus is below called compatible with $g$.

Define a graph $\Gamma(g)$ with vertices $1,\ldots,k$, where the vertices
$i,j\in\\{1,\ldots,k\\}$ are linked if and only if $(|g_{i}|,|g_{j}|)\neq 1$.
A vertex $i$ is called singular if it is isolated and $|g_{i}|=2^{d_{i}}+1$;
then $V_{i}$ is irreducible $\mathop{\mathbb{F}}\nolimits_{2}A$-module. The
set of singular vertices is denoted by $\Gamma_{0}$ and the number ${\rm
Si}\,(g)$ of them is called the Singer index of $g$. So ${\rm
Si}\,(g)=|\Gamma_{0}|$. If $i$ is singular then $\langle g_{i}\rangle$ is a
Hall subgroup of $A$ (that is, $(|g_{i}|,|A|/|g_{i}|)=1$). If $T$ is a maximal
torus compatible with $g$ then $|g_{i}|=|T_{i}|$ whenever $i$ is singular.

######  Lemma 3.8.

[8, Proposition 4.12] Let $\rho=\rho_{\omega_{n}}$ and let
$T=T_{1}\times\cdots\times T_{k}$ be a maximal torus of G. Set
$m_{i}=\eta_{i}+1$. Then
$\rho|_{T}=\otimes(\rho^{reg}_{T_{i}}+(-1)^{m_{i}}1_{T_{i}})$, where
$\rho^{reg}_{T_{i}}$ denotes the regular representation of $T_{i}$ and
$1_{T_{i}}$ the trivial one. In particular, an irreducible representation
$\tau$ of $T$ occurs in $\rho|_{T}$ if and only if $\tau(T_{i})\neq 1_{T_{i}}$
whenever $\eta_{i}=-1$.

######  Corollary 3.9.

Under notation of Lemma 3.8 set
$\rho_{i}^{\prime}=\rho^{reg}_{T_{i}}+(-1)^{m_{i}}1_{T_{i}}$. Let $g\in T$ and
$g=g_{1}\cdots g_{k}$ with $g_{i}\in T_{i}$ for $i\in\\{1,\ldots,k\\}$.

$(1)$ $\rho_{i}^{\prime}(g_{i})$ has less than $|g_{i}|$ distinct eigenvalues
if and only if $\eta_{i}=-1$ and $|g_{i}|=|T_{i}|$.

$(2)$ If $1$ is an eigenvalue of $\rho_{i}^{\prime}(g_{i})$ then
$\rho_{i}^{\prime}(g_{i})$ has $|g_{i}|$ distinct eigenvalues.

$(3)$ the eigenvalues of $\rho_{\omega_{n}}(g)$ are the products $e_{1}\cdots
e_{k}$, where $e_{i}$ runs over the eigenvalues of $\rho_{i}^{\prime}(g_{i})$
(or $\rho(g_{i}))$. Consequently, every primitive $|g|$-root of unity is an
eigenvalue of $\rho_{\omega_{n}}(g)$.

Proof. This follows straightforwardly from Lemma 3.8. $\Box$

######  Lemma 3.10.

Let $g\in G$ be a semisimple element and let $\Gamma(g)$ be a graph defined
above. Then the following are equivalent:

$(1)$ $\rho_{\omega_{n}}(g)$ has $|g|$ distinct eigenvalues;

$(2)$ $\rho_{\omega_{n}}(g)$ has eigenvalue $1;$

$(3)$ $\Gamma(g)$ has no singular vertex.

Proof. Set $\rho=\rho_{\omega_{n}}$. For a connected component $J$ of
$\Gamma(g)$ set $g_{J}=\Pi_{i\in J}g_{i}$ and $\rho^{\prime}_{J}=\otimes_{i\in
J}\rho^{\prime}_{i}$. We first prove

$(*)$ $\rho^{\prime}_{J}(g_{J})$ has $|g_{J}|$ distinct eigenvalues unless
$J=\\{j\\}$ is a singular vertex.

Indeed, if $|J|=1$ then $(*)$ follows from Corollary 3.9(1). Let $|J|>1$.
Suppose first that $|J|=2$, and we can assume $J=\\{1,2\\}$. By Lemma 3.8, all
non-trivial $|g_{1}|$-roots of unity $\lambda_{1},\ldots,\lambda_{m}$ are
eigenvalues of $\rho^{\prime}_{1}(g_{1})$ (so $m=|g_{1}|-1$), and, by a
similar reason, $\rho^{\prime}_{2}(g_{2})$ has at least 2 distinct eigenvalues
$\eta_{1},\eta_{2}$, say, that are $|g_{1}|$-roots of unity. Then
$\lambda_{i}\eta_{j}$ $(1\leq i\leq m,j=1,2)$ yield all $|g_{1}|$-roots of
unity, and they are eigenvalues of
$\rho^{\prime}_{1}(g_{1})\rho^{\prime}_{2}(g_{2})$. Similarly, all
$|g_{2}|$-roots of unity are eigenvalues of
$\rho^{\prime}_{1}(g_{1})\rho^{\prime}_{2}(g_{2})$. As
$\rho^{\prime}_{1}(g_{1})\rho^{\prime}_{2}(g_{2})=(\rho^{reg}_{T_{1}}(g_{1})+(-1)^{m_{i}}1_{T_{1}}(g_{1}))(\rho^{reg}_{T_{2}}(g_{2})+(-1)^{m_{i}}1_{T_{2}}(g_{2}))$,
it follows that all other $|g_{1}g_{2}|$-roots of unity are eigenvalues of
$\rho^{\prime}_{1}(g_{1})\rho^{\prime}_{2}(g_{2})$, whence the claim. Let
$|J|>2$. Then we argue similarly using induction on $|J|$.

$(3)\rightarrow(1)$ follows from $(*)$ and Corollary 3.9(3), and
$(1)\rightarrow(2)$ is trivial.

$(2)\rightarrow(3)$ Suppose the contrary, and let $i$ be a singular vertex.
Then $|g_{i}|=|T_{i}|$ and $\eta_{i}=-1$. It follows that 1 is not an
eigenvalue of $\rho_{i}^{\prime}(g_{i})$, as well as of $\rho(g)$ by Corollary
3.9(3) as $|g_{i}|$ is coprime to $gg_{i}^{-1}$. $\Box$

The following lemma describes the set of eigenvalues of
$\rho_{\omega_{n}}(g)$:

######  Lemma 3.11.

Let $g\in G$ be a semisimple element and $\Gamma_{0}$ the set of singular
vertices on $\Gamma(g)$. For a $|g|$-root of unity
$\zeta\in\overline{\mathop{\mathbb{F}}\nolimits}_{q}$ let $|\zeta|$ denote the
order of $\zeta$ in $\overline{\mathop{\mathbb{F}}\nolimits}_{q}^{\times}$.
Then $\zeta$ is an eigenvalue of $\rho_{\omega_{n}}(g)$ if and only if
$(|\zeta|,|g_{i}|)>1$ for every $i\in\Gamma_{0}$.

Proof. Let $T=T_{1}\times\cdots\times T_{k}$ be a maximal torus compatible
with $g$. Suppose $\zeta$ is an eigenvalue of $\rho(g)$. By Corollary 3.9,
$\zeta=e_{1}\cdots e_{k}$, where $e_{i}$ is an eigenvalue of
$\rho^{\prime}_{i}(g_{i})$ for $1\leq i\leq k$. So $e_{i}=\lambda_{i}(g_{i})$
for some irreducible constituent $\lambda_{i}$ of
$\rho^{\prime}_{i}|_{T_{i}}$. Let $i\in\Gamma_{0}$. Then $\langle
g_{i}\rangle=T_{i}$ and $\eta_{i}=-1$; therefore 1 is not an eigenvalue of
$\rho_{i}^{\prime}(g_{i})$ (Corollary 3.9). So $e_{i}\neq 1$. Furthermore,
$(|g_{i}|,|g_{j}|)=1$ for $j\neq i$, and hence $(|e_{i}|,|e_{j}|)=1$.
Therefore, $|e_{i}|$ divides $|\zeta|$, as required.

Conversely, we have to show that $\zeta=e_{1}\cdots e_{k}$, where $e_{i}$ is
an eigenvalue of $\rho^{\prime}_{i}(g_{i})$ for $1\leq i\leq k$. Set
$A=\langle g\rangle$. As $|\zeta|$ divides $|g|$ and $\langle g_{i}\rangle$ is
a Hall subgroup of $A$ for $i\in\Gamma_{0}$, it follows that $\zeta$ is
uniquely expressed as $\zeta_{1}\zeta_{2}$, where $|\zeta_{2}|$ divides
$|g_{i}|$, and $(|\zeta_{1}|,|g_{i}|)=1$. Then $\zeta_{2}\neq 1$ and we may
set $e_{i}=\zeta_{2}$ (as all non-trivial $|g_{i}|$-roots of unity are
eigenvalues of $\rho_{i}^{\prime}(g_{i}))$. Furthermore, by reordering of
$T_{1},\ldots,T_{k}$ we can assume that $\Gamma_{0}=\\{1,\ldots,r\\}$, where
$r$ is the Singer index of $g$. Set $D=\langle g_{1},\ldots,g_{r}\rangle$;
then $D$ is a Hall subgroup of $A$ and $D=T_{1}\times\cdots\times T_{r}$. So
$A=D\times D_{1}$, where $(|D|,|D_{1}|)=1$. Then $g=g_{0}h$, where
$g_{0}=g_{1}\cdots g_{r}\in D$ and $h=g_{r+1}\cdots g_{k}\in D_{1}$. The
eigenvalues of $\rho(g)$ are product of those of
$\Pi_{i=1}^{r}\rho_{i}^{\prime}(g_{i})$ and of
$\Pi_{i=r+1}^{k}\rho_{i}^{\prime}(g_{i})$. It follows from $(*)$ that
$\Pi_{i=r+1}^{k}\rho_{i}^{\prime}(g_{i})$ has $|h|$ distinct eigenvalues. In
turn, the eigenvalues of $\Pi_{i=1}^{r}\rho_{i}^{\prime}(g_{i})$ are exactly
the products of non-trivial $|g_{i}|$-roots of unity for $i=1,\ldots,r$.
Whence the result. $\Box$

Next we consider representations of shape $\tau\otimes\rho_{\omega_{n}}$,
where $\tau\in\mathop{\rm Irr}\nolimits{\mathbf{G}}$. We first observe the
following:

######  Lemma 3.12.

Let $\tau$ be a representation of
${\mathbf{G}}=Sp_{2n}(\overline{\mathop{\mathbb{F}}\nolimits}_{2})$ and let
$T$ be a maximal torus of G. Then $\rho|_{T}$ contains $1_{T}$ if and only if
$\tau$ has weight $\lambda$ such that $\lambda|_{T}=\omega_{n}|_{T}$.

Proof. Clearly, $1_{T}$ occurs in $(\tau\otimes\rho_{\omega_{n}})|_{T}$ if and
only if there is an irreducible constituent $\nu$, say, of
$\rho_{\omega_{n}}|_{T}$ such that $\nu^{-1}$ occurs as an irreducible
constituent of $\tau|_{T}$. Let $\mu$ be a weight of $\tau$ such that
$\nu=\mu(T)$. The Weyl group of $Sp_{2n}(F)$ contains $-\mathop{\rm
Id}\nolimits$, so $-\mu$ is a weight of $\tau$, and then
$(-\mu)|_{T}\cong\nu^{-1}$. Therefore, $\nu$ is a constituent of $\tau|_{T}$
too. As the weights of $\rho_{\omega_{n}}$ are exactly those occurring in the
$W$-orbit of $\omega$, the lemma follows. $\Box$

Moreover, if $\tau$ has weight $\omega_{n}$ then
$(\tau\otimes\rho_{\omega_{n}})|_{T}>\rho^{reg}_{T}$ for any maximal torus $T$
of $G$. (We write $\sigma>\tau$ for representations of $T$ if
$\sigma=\tau+\nu$ for a proper representation $\nu$ of $T$.) This essentially
follows from Lemma 3.8 as
$(\tau\otimes\rho_{\omega_{n}})|_{T}>\otimes_{i}(\rho^{\prime}_{i}\otimes\rho_{i}^{\prime})$
and $(\rho^{\prime}_{i}\otimes\rho_{i}^{\prime})_{T_{i}}>\rho_{T_{i}}^{reg}$.

Similarly we have:

######  Lemma 3.13.

Let ${\mathbf{G}},\tau,\rho$ be as in Lemma 3.12 and $g\in G=Sp_{2n}(2)$ a
semisimple element. Then $\rho(g)$ has eigenvalue $1$ if and only if $\tau(g)$
and $\rho_{\omega_{n}}(g)$ have a common eigenvalue. The latter holds if and
only if $\tau(g)$ has eigenvalue $\zeta$ such that $(|\zeta|,|g_{i}|)>1$ for
every $i\in\Gamma_{0}$.

Proof. The eigenvalues of $(\tau\otimes\rho_{\omega_{n}})(g)$ are products of
those of $\tau(g)$ and $\rho_{\omega_{n}}(g)$. It is well known that all
elements of $G$ are real, so if $\zeta$ is an eigenvalue of $\tau(g)$ then so
is $\zeta^{-1}$. Whence the first statement of the lemma. The second one
follows from Lemma 3.11. $\Box$

So we are faced to decide when $\rho_{\omega_{n}}(g)$ and $\tau(g)$ have a
common eigenvalue. This will be done in Section 4. We complete this section
with a few useful observations.

######  Corollary 3.14.

Let ${\mathbf{G}},\tau,\rho,g$ be as in Lemma 3.13. Let h be the product of
all $g_{i}$ with $i\in\Gamma_{0}$. Then $\rho(g)$ has eigenvalue $1$ if and
only if $\rho(h)$ has eigenvalue $1$.

Proof. As $(|g_{i}|,|gg_{i}^{-1}|)=1$ for $i\in\Gamma_{0}$, it follows that
$h\in\langle g\rangle$. Moreover, $\Gamma_{0}=\Gamma(h)$. So the conditions in
Lemma 3.11 for $\rho(g)$ and $\rho(h)$ to have eigenvalue 1 coincide, and the
result follows. $\Box$

######  Corollary 3.15.

Let ${\mathbf{G}},\tau,\rho,g$ be as in Lemma 3.13. If $\tau$ has weight
$\omega_{n}$ then $\rho(g)$ has exactly $|g|$ distinct eigenvalues.

Proof. Let $Q,P$ be the sets of all, all primitive $|g|$-roots of unity,
respectively. By Corollary 3.9(3), every $e\in P$ is an eigenvalue of
$\rho_{\omega_{n}}$. As $e_{1}e_{2}\in P$ whenever $e_{1}\in P$, $e_{2}\in
Q\setminus P$, it follows that all non-primitive $|g|$-roots of unity are
eigenvalues of $\sigma(g)$.

As $|g|$ is odd, it follows that every $e\in P$ is a product $e_{1}e_{2}$ with
$e_{1},e_{2}\in P$. Indeed, let $e_{3}\in Q\setminus P$. Then
$e_{3}e,e_{3}^{-1}e\in P$ and $(e_{3}e)(e_{3}^{-1}e)=e^{2}\in P$. There is a
2-power $k$, say, such that $e^{2k}=e$. Then
$(e_{3}e)^{2k},(e_{3}^{-1}e)^{2k}\in P$ and
$(e_{3}e)^{2k}(e_{3}^{-1}e)^{2k}=e$, as stated. $\Box$

Remark. In fact we have shown that if $r$ is odd then every $r$-root of unity
is a product of two primitive $r$-roots of unity.

Proof of Theorem 1.4. If $a_{n}=0$ then the result follows from Lemma 3.5(2).
If $a_{n}=1$ then this follows from Lemma 3.13, as $|\Gamma_{0}|\leq 1$
whenever $|g|$ is a prime power. (More precisely, if $|\Gamma_{0}|=1$ then
$\tau(g_{i})\neq\mathop{\rm Id}\nolimits$ (as $\tau$ is non-trivial by
assumption), and hence $\tau(g_{i})$ has an eigenvalue $\zeta$ with
$(|\zeta|,|g_{i}|)>1$ as required in Lemma 3.13.) $\Box$

##  4\. Tensor products

In this section $F$ is an algebraically closed field of arbitrary
characteristic $p\geq 0$. Note that the results of Subsection 2.1 are
independent from characteristic, so we are free to use them here.

Let ${\mathbf{G}}=Sp_{2n}(F)$ and let ${\mathbf{T}}$ be the reference torus,
that is, a maximal torus ${\mathbf{T}}=(t_{1},\ldots,t_{n})$
$(t_{1},\ldots,t_{n}\in F^{\times})$ such that
$\varepsilon_{i}({\mathbf{T}})=t_{i}$. Set
${\mathbf{T}}_{i}=\\{(1,\ldots,1,t_{i},1,\ldots,1):t_{i}\in F^{\times}\\}$.

######  Lemma 4.1.

Let $\omega\in\Omega^{+}$, $l=\delta(\omega)$ and $k=\min(l,n)$. Let
$V_{\omega}$ be an irreducible ${\mathbf{G}}$-module with p-restricted highest
weight $\omega$. Then there is a p-restricted dominant weight $\mu$ of
$V_{\omega}$ such that $\mu({\mathbf{T}}_{i})\neq 1$ for $i=1,\ldots,k$.

Proof. For uniformity set $\omega_{n+1}=\omega_{1}+\omega_{n}$. By Lemma
2.2(6), $\omega_{l}\preceq\omega$ if $\l\leq n$, otherwise
$\omega_{n}\preceq\omega$ or $\omega_{n+1}\preceq\omega$ if $l-n$ is even or
odd, respectively.

Recall that $\omega_{i}=\varepsilon_{1}+\cdots+\varepsilon_{i}$ for
$i=1,\ldots,n$ and $\omega_{n+1}=2\varepsilon_{1}+\cdots+\varepsilon_{n}$. If
$p\neq 2$ or $p=2$ and the $\omega_{n}$-coefficient of $\omega$ is 0, then, by
Lemma 2.5, $\omega_{l}$, $\omega_{n}$ or $\omega_{n+1}$ is a weight of
$\phi_{\omega}$, provided, respectively, $l\leq n$, $l>n$ with $l-n$ even or
$l>n$ with $l-n$ odd. We choose this weight for $\mu$. Then
$\mu({\mathbf{T}}_{i})\neq 1$ for $i=1,\ldots,k=\min(l,n)$, as stated.

Suppose that $p=2$ and the $\omega_{n}$-coefficient of $\omega$ is 1. Then
$V_{\omega}=V_{\omega^{\prime}}\otimes V_{\omega_{n}}$, where
$\omega^{\prime}=\omega-\omega_{n}\in\Omega^{\prime}$. Then either
$0\preceq\omega^{\prime}$ or $\omega_{1}\preceq\omega^{\prime}$, so
$V_{\omega^{\prime}}$ has weight 0 or $\omega_{1}$ by Lemma 2.5. Then
$V_{\omega}$ has weight $\mu=\omega_{n}=\varepsilon_{1}+...+\varepsilon_{n}$
or $\mu=\omega_{1}+\omega_{n}=2\varepsilon_{1}+...+\varepsilon_{n}$. So again
$\mu({\mathbf{T}}_{i})\neq 1$ for $i=1,\ldots,n$. $\Box$

######  Theorem 4.2.

Let $\mathbf{G}=Sp_{2n}(F)$ and let $n_{1},\ldots,n_{k}$ be positive integers
with $n_{1}+\cdots+n_{k}\leq n$. Let
$\mathbf{H}={\mathbf{H}}_{1}\times\cdots\times{\mathbf{H}}_{k}$, where
${\mathbf{H}}_{i}\cong Sp_{2n_{i}}(F)$ for $i=1,\ldots,k$. Let
$\phi=\phi_{\omega}$ be an irreducible representation of $\mathbf{G}$ with
p-restricted highest weight $\omega=\sum a_{i}\omega_{i}$. Then the following
are equivalent:

$(1)$ $\rho({\mathbf{H}}_{i})=1$ for some $i\in\\{1,\ldots,k\\}$ whenever
$\rho$ is a composition factor of $\phi|_{\mathbf{H}};$

$(2)$ $\delta(\omega)=\sum a_{i}i<k$.

In addition, the result is valid for $\tilde{\phi}_{\omega}$, the
representation afforded by the Weyl module with highest weight $\omega$.

Proof. It suffices to prove the theorem for the case where
$n_{1}=\cdots=n_{k}=1$. Indeed, we can choose a subgroup $X_{i}=Sp_{2}(F)$ in
every $\mathbf{H}_{i}$, and set $X=X_{1}\times\cdots\times X_{k}$. Then
$\rho({\mathbf{H}}_{i})=1$ if and only if $\rho(X_{i})=1$, whence the claim.

$(2)\rightarrow(1)$ Let ${\mathbf{T}}^{\prime}$ be a maximal torus of
$\mathbf{H}={\mathbf{H}}_{1}\times\cdots\times{\mathbf{H}}_{k}$ (where now
${\mathbf{H}}_{1}\cong\cdots\cong{\mathbf{H}}_{k}\cong Sp_{2}(F)$). Let
${\mathbf{T}}_{i}^{\prime}$ be a maximal torus of ${\mathbf{H}}_{i}$, so
${\mathbf{T}}^{\prime}={\mathbf{T}}_{1}^{\prime}\times\cdots\times{\mathbf{T}}_{k}^{\prime}$.
We can assume that ${\mathbf{T}}^{\prime}\subset{\mathbf{T}}$ and
${\mathbf{T}}^{\prime}=(t^{\prime}_{1},\ldots,t^{\prime}_{k})$ in the sense
that $\varepsilon_{i}((t^{\prime}_{1},\ldots,t^{\prime}_{k}))=t^{\prime}_{i}$
for $i\leq k$ and $\varepsilon_{i}((t^{\prime}_{1},\ldots,t^{\prime}_{k}))=1$
for $i>k$. Let $\sigma$ be a weight of $\phi$, and write
$\sigma=\sum_{i=1}^{n}c_{i}\varepsilon_{i}$ for some integers $c_{i}$. Then
$\sigma|_{{\mathbf{T}}^{\prime}}=(c_{1}\varepsilon_{1},\ldots,c_{k}\varepsilon_{k})$,
that is, $\varepsilon_{i}$ as a function on ${\mathbf{T}}_{i}^{\prime}$ is in
fact the fundamental weight of ${\mathbf{H}}_{i}$ for $i=1,\ldots,k$. If
$\sigma$ is such that $\sigma|_{{\mathbf{T}}^{\prime}}$ is a highest weight of
an irreducible constituent $\rho$ of $\phi|_{\mathbf{H}}$ then $c_{i}\geq 0$
for $i=1,\ldots,k$. Let $\nu=\sum d_{i}\varepsilon_{i}$ be a dominant weight
in the $W$-orbit $W\sigma$. As $\nu$ is a weight of $\phi$, we have
$\nu\leq\omega$, and then $\delta(\nu)\leq\delta(\omega)<k$ by Lemma 2.1(3).
Note that $\delta(\nu)=d_{1}+\cdots+d_{n}$, so $d_{1}+\cdots+d_{n}<k$. The
action of $W$ on the weights preserves the set
$\pm\varepsilon_{1},\ldots,\pm\varepsilon_{n}$; it follows that the set
$(|c_{1}|,\ldots,|c_{n}|)$ coincides with $(d_{1},\ldots,d_{n})$ up to
reordering, where $|c_{i}|$ means the absolute value of $c_{i}$. So
$\sum_{i}|c_{i}|\cdot i<k$. As $c_{1},\ldots,c_{k}$ are non-negative, we have
$\sum_{i=1}^{k}c_{i}i<k$, and hence at least one of $c_{1},\ldots,c_{k}$
equals 0. This means that $\sigma({\mathbf{H}}_{i})=1$ for this $i$, and hence
${\mathbf{H}}_{i}$ is in the kernel of $\rho$, as required.

$(1)\rightarrow(2)$ follows from Lemma 4.1 as ${\mathbf{T}}_{i}$ is a maximal
torus of ${\mathbf{H}}_{i}$.

For additional statement, $(1)\rightarrow(2)$ follows as $\phi_{\omega}$ is a
composition factor of $\tilde{\phi}_{\omega}$. As $(2)\rightarrow(1)$ is true
for $F=\mathop{\mathbb{C}}\nolimits$, and the weights of $\tilde{\phi}$ are
the same as those of the irreducible representation of
$Sp_{2n}(\mathop{\mathbb{C}}\nolimits)$ with highest weight $\omega$, the
above reasoning remains valid for $\tilde{\phi}_{\omega}$. $\Box$

######  Corollary 4.3.

In notation of Theorem 4.2 let $p>0$ and
$\tau=\phi_{1}\otimes\cdots\otimes\phi_{m}$, where $\phi_{j}$ is an
irreducible representation of ${\mathbf{G}}$ with $p$-restricted highest
weight $\mu_{j}$, $j=1,\ldots,m$. Let
$l=\delta(\mu_{1})+\cdots+\delta(\mu_{m})$. Suppose that $l<n$. Let $\rho$ be
an irreducible constituent of $\tau|_{{\mathbf{H}}}$ and
$J=\\{i\in\\{1,\ldots,k\\}:\rho({\mathbf{H}}_{i})\neq 1\\}$. Then $|J|\leq l$.

Proof. Note that $\tau|_{{\mathbf{H}}}=\otimes\phi_{j}|_{{\mathbf{H}}}$, so
$\rho=\otimes\sigma_{j}$, where $\sigma_{j}$ is an irreducible constituent of
$\phi_{j}|_{{\mathbf{H}}}$. In turn,
$\sigma_{j}=\nu_{1}\otimes\cdots\otimes\nu_{k}$, where $\nu_{i}\in\mathop{\rm
Irr}\nolimits{\mathbf{H}}_{i}$ for $i=1,\ldots,k$. By Theorem 4.2, this
product has at most $\delta(\mu_{j})$ non-trivial multiples. (Indeed, let
$l_{j}=\delta(\mu_{j})$; if the claim is false then $\sigma_{j}$ has at least
$l_{j}+1$ non-trivial multiples, so $k>l_{j}$. We can assume that these are
$\nu_{1},\ldots,\nu_{l_{j}+1}$. Set
${\mathbf{H}}^{\prime}={\mathbf{H}}_{1}\times\cdots\times{\mathbf{H}}_{l_{j}+1}$
and $\rho^{\prime}=\nu_{1}\cdots\nu_{l_{j}+1}$. By Theorem 4.2 applied to
${\mathbf{H}}^{\prime}$ in place of ${\mathbf{H}}$ and $\rho^{\prime}$ in
place of $\rho$, there is $i\in\\{1,\ldots,l_{j+1}\\}$ such that
$\sigma^{\prime}({\mathbf{H}}_{i})=1$, which is a contradiction.) Therefore,
$\rho$ has at most $l=\sum_{j}\delta(\mu_{j})$ non-trivial multiples. $\Box$

Recall that the Singer index of a maximal torus $T=T_{1}\times\cdots\times
T_{k}$ of $G$ is the number of $i\in\\{1,\ldots,k\\}$ with $\eta_{i}=-1$.

######  Theorem 4.4.

Let $G=Sp_{2n}(2)$ and let $T=T_{1}\times\cdots\times T_{k}$ be a maximal
torus of G of Singer index $m$. Let $\omega=(a_{1},\ldots,a_{n-1},a_{n})$ with
$a_{n}=1$ be a $2$-restricted weight and $\phi_{\omega}$ be an irreducible
representation of G with highest weight $\omega$. Then $1_{T}$ is a
constituent of $\phi_{\omega}|_{T}$ if and only if $\sum a_{i}i\geq n+m$.

Proof. Set $\omega^{\prime}=\omega-\omega_{n}$. By Lemma 3.12, $1_{T}$ is a
constituent of $\phi|_{T}$ if and only if $\phi_{\omega_{n}}|_{T}$ and
$\phi_{\omega^{\prime}}|_{T}$ have a common irreducible constituent. By Lemma
3.8, an irreducible representation $\tau$ of $T$ occurs in
$\phi_{\omega_{n}}|_{T}$ if and only if $\tau(T_{i})\neq 1$ whenever
$\eta_{i}(T_{i})=-1$ $(1\leq i\leq k)$. We show that $\tau$ with this property
is a constituent of $\phi_{\omega^{\prime}}|_{T}$ if and only if
$\delta(\omega)\geq\sum_{i}a_{i}i\geq m+n$, equivalently,
$\delta(\omega^{\prime})\geq m$.

By reordering of $T_{1},\ldots,T_{k}$ we can assume that $\eta_{i}=-1$ for
$i=1,\ldots,m$. One easily observes that there is a subgroup
${\mathbf{H}}={\mathbf{H}}_{1}\times\cdots\times{\mathbf{H}}_{m}$ of
$Sp_{2n}(\overline{\mathop{\mathbb{F}}\nolimits}_{2})$ such that
${\mathbf{H}}_{i}=Sp_{2n_{i}}(\overline{\mathop{\mathbb{F}}\nolimits}_{2})$
for $i=1,\ldots,m$, and $T_{i}$ is a maximal torus of $Sp_{2n_{i}}(2)$. Let
$\overline{\phi}_{\omega^{\prime}}$ be a representation of
${\mathbf{G}}=Sp_{2n}(\overline{\mathop{\mathbb{F}}\nolimits}_{2})$ with
highest weight $\omega^{\prime}$. Then
$\overline{\phi}_{\omega^{\prime}}|_{G}=\phi_{\omega^{\prime}}$.

Suppose that $\delta(\omega^{\prime})\geq m$. By Theorem 4.2, there is a
composition factor of $\overline{\phi}_{\omega^{\prime}}|_{\mathbf{H}}$ which
is non-trivial on each ${\mathbf{H}}_{i}$ for $i=1,\ldots,k$. Then
$\overline{\phi}_{\omega^{\prime}}|_{T_{i}}\neq 1$ for $i=1,\ldots,m$ as such
$T_{i}\neq 1$. (Note that $\eta_{i}=-1$ implies $|T_{i}|\neq 1$.) So $1_{T}$
is a constituent of $\phi_{\omega}|_{T}$ by the above.

Conversely, if $\phi_{\omega^{\prime}}|_{T}$ has a constituent $\tau$, say,
non-trivial on each $T_{i}$ with $1\leq i\leq m$ then
$\overline{\phi}_{\omega^{\prime}}|_{\mathbf{H}}$ has an irreducible
constituent that is non-trivial on ${\mathbf{H}}_{i}$ for $i=1,\ldots,m$. Then
$\sum_{i=1}^{n-1}a_{i}i>m$ by Theorem 4.2, and the result follows. $\Box$

######  Theorem 4.5.

Let $g\in G$ be a semisimple element and let
$\rho=\rho_{\omega^{\prime}}\otimes\rho_{\omega_{n}}$ be an irreducible
representation of G. Then $1$ is an eigenvalue of $\rho(g)$ if and only if
$\delta(\omega)\geq n+{\rm Si}(g)$.

Proof. Let $h$ be the product of $g_{i}$ with $i\in\Gamma_{0}$. Then ${\rm
Si}(h)={\rm Si}(g)$. By Corollary 3.14, $\rho(g)$ has eigenvalue 1 if and only
if $\rho(h)$ has eigenvalue 1. So we can assume that $g=h$. Then $T=\langle
g\rangle$ as in this case $(|g_{i}|,|g_{j}|)=1$ for all $i\in\Gamma(g)$ and
$g_{i}=1$ if $i\notin\Gamma_{0}$. Therefore, the Singer index of $g$ equals
that of $T$. Then the result follows from Theorem 4.4. $\Box$

Proof of Theorem 1.3. If $a_{n}=0$ then the result follows from Lemma 3.5 (the
”if” part) and Lemma 3.2 (the ”only if” part).

Let $a_{n}=1$. Then ${\rm Si}(g)\leq{\rm Si}(n)$, and the equality holds for
some $g\in G$. So the result follows from Theorem 4.5. $\Box$

Observe that there are no elements whose Singer index is greater than ${\rm
Si}(n)$, and there are semisimple elements $g\in G$ whose Singer index equals
${\rm Si}(n)$. We refer to such elements as those of maximal Singer index.

######  Proposition 4.6.

Let $G=Sp_{2n}(2)$ and let $t\in G$ be an element of maximal Singer index. Let
$\rho$ be an irreducible representation of G with highest weight
$\omega=(a_{1},\ldots,a_{n})$, $a_{n}=1$. If $1$ is an eigenvalue of $\rho(t)$
then $1$ is an eigenvalue of $\rho(g)$ for every $g\in G$.

Proof. This follows from Theorem 4.5. $\Box$

######  Proposition 4.7.

Let $g\in G=Sp_{2n}(2)$ and
${\mathbf{G}}=Sp_{2n}(\overline{\mathop{\mathbb{F}}\nolimits}_{2})$. Suppose
that $\rho_{\omega_{i}}(g)$ has eigenvalue $1$ for $i=1,n$. Then $\rho(g)$ has
eigenvalue $1$ for every $\rho\in\mathop{\rm Irr}\nolimits{\mathbf{G}}$.

Proof. Let $\omega=(a_{1},\ldots,a_{n})$ be the highest weight of $\rho$.
Suppose first that $\omega$ is 2-restricted. If $a_{n}=0$ and $\omega$ is
radical then $\rho$ has weight 0 by Lemma 2.6(1), and the claim is trivial. If
$\omega$ is not radical then $\rho$ has weight $\omega_{1}$ and
$\Omega(\rho_{\omega_{1}})\subset\Omega(\rho)$, whence the result.

Let $a_{n}=1$. By assumption, $\rho_{\omega_{n}}(g)$ has eigenvalue 1, so
$\rho_{\omega_{n}}(g)$ has $|g|$ distinct eigenvalues by Lemma 3.10. Let
$\omega^{\prime}=\omega-\omega_{n}$. Then
$\rho_{\omega}=\rho_{\omega^{\prime}}\otimes\rho_{\omega_{n}}$, so
$\rho_{\omega}(g)$ has $|g|$ distinct eigenvalues, whence the result in this
case.

Finally suppose that $\rho$ is not 2-restricted. Then $\rho|_{G}$ is a tensor
product of irreducible representations of $G$, and we conclude similarly.
$\Box$

######  Lemma 4.8.

††margin:  tp1

Let $V=V_{\omega_{i}}\otimes V_{\omega_{j}}$ for $1\leq i<j\leq n$. Let $g\in
G$ be a semisimple element. If $j<n$ or $j=n$ and $i\geq{\rm Si}(g)$ then g
has eigenvalue $1$ on V.

Proof. It is well known that $V$ has a composition factor
$V_{\omega_{i}+\omega_{j}}$. If $j<n$ then the result follows by Theorem
1.3(1). If $j=n$ then $V_{\omega_{i}}\otimes V_{\omega_{n}}\cong
V_{\omega_{i}+\omega_{n}}$, and the result follows by Theorem 1.3(2). $\Box$

Note that ${\rm Si}(g)=|\Gamma_{0}(g)|\leq{\rm Si}(n)$ and ${\rm Si}(n)<n$. In
addition, ${\rm Si}(3)=2={\rm Si}(4)={\rm Si}(5)={\rm Si}(6)$, ${\rm
Si}(7)=3={\rm Si}(8)={\rm Si}(9)={\rm Si}(10)={\rm Si}(11)$, ${\rm Si}(12)=4$.

######  Proposition 4.9.

Let $G=Sp_{2n}(2)$,
${\mathbf{G}}=Sp_{2n}(\overline{\mathop{\mathbb{F}}\nolimits}_{2})$ and let
$\rho$ be an irreducible representation of ${\mathbf{G}}$ with highest weight
$\omega=\sum a_{i}\omega_{i}$. Let $g\in G$ be a semisimple elt. Then one of
the following holds:

$(1)$ $1$ is an eigenvalue of $\rho_{\omega}(g);$

$(2)$ $\omega=2^{i}\omega_{j}$ for $j$ odd or $j=n$;

$(3)$ $\omega=\omega^{\prime}+2^{k}\omega_{n}$ with
$0\neq\omega^{\prime}\in\Omega^{\prime}$ and $k\geq 0$. In addition, if
$\omega^{\prime}=\sum 2^{i}\nu_{i}$, where $\nu_{i}$’s are $2$-restricted,
then $\delta(\sum\nu_{i})<{\rm Si}(g)$.

Proof. Suppose that (1),(2) do not hold. Let $\rho=\otimes_{i\geq
0}\rho_{2^{i}\mu_{i}}$, where $\mu_{i}$ are 2-restricted dominant weights.
Note that $\rho_{2^{i}\mu_{i}}|_{G}\cong\rho_{\mu_{i}}|_{G}$. So
$\rho(g)\cong(\otimes\rho_{\mu_{i}})(g)$. Set $\tau=\otimes\rho_{\mu_{i}}$. If
$\mu_{i}\notin\Omega^{\prime}$ then $\mu=\nu_{i}+\omega_{n}$, where
$\nu_{i}\in\Omega^{\prime}$, and then
$\rho_{\mu_{i}}=\rho_{\nu_{i}}\otimes\rho_{\omega_{n}}$. So
$\tau=\tau^{\prime}\otimes\sigma$, where $\tau^{\prime}=\otimes\rho_{\nu_{i}}$
with $\nu_{i}\in\Omega^{\prime}$ and $\sigma$ is either trivial or the tensor
product of $k>0$ copies of $\rho_{\omega_{n}}$; the former is expressed as
$k=0$. As (1) does not holds, $\tau$ does not have weight 0.

If $k>1$ then $(\rho_{\omega_{n}}\otimes\rho_{\omega_{n}})(g)$ has $|g|$
distinct eigenvalues (Corollary 3.15), and hence so is $\rho(g)$. This remains
true if $k=1$ and $\omega_{n}$ is a weight of the product of any multiples of
$\otimes\rho_{\nu_{i}}$ (as $\Omega(\rho_{\omega_{n}})=W\omega_{n}$). So, to
prove (3), we can assume that $k\leq 1$.

Let $k=1$. So $\tau=\tau^{\prime}\otimes\rho_{\omega_{n}}$, where
$\tau^{\prime}=\otimes\rho_{\nu_{i}}$, and we can assume that $\omega_{n}$ is
not a weight of any product of $\rho_{\nu_{i}}$; in particular,
$\delta(\nu_{i})<n$ by Lemmas 2.2(6) and 2.5 (as $\nu_{i}\in\Omega^{\prime}$).
Set $l_{i}=\delta(\nu_{i})$, so $l_{i}<n$. Again by Lemma 2.2(6),
$\rho_{\nu_{i}}$ has weight $\omega_{l_{i}}$, and hence, by Lemma 2.5, the
weights of $\rho_{\omega_{l_{i}}}$ are also weights of $\rho_{\nu_{i}}$. Note
that $\delta(\nu_{i})=\delta(\omega_{l_{i}})$.

Observe that $\omega_{j},\omega_{j-2},\omega_{j-4}\ldots$ are weights of
$\rho_{\omega_{j}}$ for $j<n$. It follows that
$\rho_{\omega_{i}}\otimes\rho_{\omega_{j}}$ for $i,j<n$ has all weights
$\omega_{r}$ for $r\leq\min\\{n,i+j\\}$ and $i+j-r$ even. (This is clear if
$i\neq j$ and for $i=j$ with $2i>n$. Let $2i\leq n$. Then
$\rho_{\omega_{i}}\otimes\rho_{\omega_{i}}$ has weights $2\omega_{i}$ and
$\omega_{i-1}+\omega_{i+1}$ as well as $\omega_{2i}$.)

Suppose that $\sum l_{i}\geq n$, $k=1$. Then either $\omega_{n}$ or
$\omega_{n-1}$ is a weight of $\tau^{\prime}$. In the former case 1 is
observed above to be an eigenvalue of $\tau(g)$. In the latter case
$\omega_{n-3},\omega_{n-5},\ldots$ are weights of $\tau^{\prime}$; this
implies $\Omega(\rho_{\omega_{n-1}})\subseteq\Omega(\tau^{\prime})$. So it
suffices to deal with the case where $\tau^{\prime}=\rho_{\omega_{n-1}}$,
$\tau=\rho_{\omega_{n-1}}\otimes\rho_{\omega_{n}}$ (ignoring that
$\delta(\omega_{n-1})<n$). So $\delta(\tau)=2n-1$. By Theorem 4.5, if
$\delta(\tau)=2n-1\geq n+{\rm Si}(g)$ then 1 is an eigenvalue of $\tau$.
However, ${\rm Si}(g)\leq{\rm Si}(n)\leq n-1$ for $n>1$. So we are done.

Next assume that $k=1$, $\sum l_{i}<n$, and let $l=\sum l_{i}$. The above
observation on weights of $\rho_{\omega_{i}}\otimes\rho_{\omega_{j}}$ for
$i,j<n$ implies $\omega_{l}$ to be a weight of $\tau^{\prime}$ as well as all
$\omega_{l-2},\omega_{l-4}\ldots$. This means
$\Omega(V_{\omega_{l}})\subseteq\Omega(\tau^{\prime})$. By Lemma 4.8, if
$l\geq{\rm Si}(g)$ then 1 is an eigenvalue of $(V_{\omega_{l}}\otimes
V_{\omega_{n}})(g)$, and hence $\tau(g)$, whereas we have assumed that (1)
does not hold. Therefore, $l<{\rm Si}(g)$, whence (3) in this case.

Let $k=0$ so $\tau=\tau^{\prime}$ and $\mu_{i}=\nu_{i}$. By reordering of the
multiples of $\tau$ we can assume that $\pm\omega_{1}$ are weights of
$\rho_{\mu_{1}}$ (as $\tau$ does not have weight 0). If $\tau=\rho_{\mu_{1}}$
then the result follows from Theorem 1.4, so we assume that
$\sigma:=\Pi_{i>1}\rho_{\mu_{i}}$ is not trivial. By Lemmas 2.2 and 2.5,
either $\pm\omega_{1}$ or $0,\omega_{2}$ are weights of every multiple. The
tensor product of two multiples with the former property has weight 0. It
follows that that $0,\omega_{2}\in\Omega(\sigma)$, and hence
$\omega_{1}+\omega_{2},\omega_{1}\in\Omega(\tau)$. If $n>2$ then
$\varepsilon_{2}+\varepsilon_{3}\in\Omega(\tau)$, and hence
$\omega_{3}\in\Omega(\tau)$. So
$\Omega(\rho_{\omega_{1}+\omega_{2}})\subseteq\Omega(\tau)$. Then 1 is an
eigenvalue of $\tau$ by Lemma 3.3(3). $\Box$

Proof of Theorem 1.8. The ”if” part follows from Lemma 3.2. The ”only if” part
follows from Proposition 4.9. Indeed, we only have to show that (3) of
Proposition 4.9 does not hold. For this, in notation of 4.9 we have ${\rm
Si}(g)=1$ (as defined prior to Lemma 3.8), and $\omega^{\prime}\neq 0$. Then
$\sum\nu_{i}\neq 0$, and hence $\delta(\sum\nu_{i})\geq 1$. It follows that
(3) fails, and we are done. $\Box$

##  5\. An application

###  5.1. Remarks on Weyl modules

For a definition of Weyl module we refer to [10, §3.1]. Here we mainly use the
fact that, for a simple algebraic group $\mathbf{X}$, a Weyl module is an
$\mathbf{X}$-module (in general, reducible) whose weight set coincides with
the weight set of some irreducible representation of the Lie algebra of
$\mathbf{X}$, equivalently, of the simple algebraic group over the complex
numbers of the same type as that of $\mathbf{X}$ (regarding the multiplicities
of weights). (The word ”coincides” here means the coincidence of the
respective sets of strings $(a_{1},\ldots,a_{n})$.) By [2, Ch. VIII,
Proposition 5], this implies

######  Lemma 5.1.

Let $\mathbf{X}$ be a simple algebraic group, $\tilde{V}_{\omega}$ a Weyl
module of $\mathbf{X}$ with highest weight $\omega$ and $\mu\prec\omega$ a
dominant weight of $\mathbf{X}$. Then $\mu$ is a weight of
$\tilde{V}_{\omega}$.

Let $\mathbf{G}=Sp_{2n}(F)$ and the characteristic of $F$ equal 2. Then the
weight set of $\tilde{V}_{\omega}$, $\omega=\sum a_{i}\omega_{i}$, in contrast
with irreducible $\mathbf{G}$-module $V_{\omega}$, has no singularity when
$a_{n}\neq 0$. In particular, we have the following version of Lemma 3.5(2)
for Weyl modules:

######  Lemma 5.2.

Let $\omega$ be a dominant weight of $\mathbf{G}$ and
$\tilde{V}=\tilde{V}_{\omega}$ a Weyl module of highest weight $\omega$. Let
$g\in G$ be a semisimple element. Then g has eigenvalue $1$ on $\tilde{V}$
unless possibly $\omega=\omega_{i}$ for i odd.

Proof. If $\omega$ is radical then the zero weight is a weight of $\tilde{V}$,
and the claim follows. Suppose $\omega$ is not radical. Then
$\omega_{1}\preceq\omega$. As $\omega\neq\omega_{i}$, we have
$\omega_{1}+\omega_{2}\preceq\omega$ by Lemma 2.2(5). By Lemma 2.5(2),
$\omega_{1}+\omega_{2}\in\Omega(\tilde{V})$ and
$\Omega(V_{\omega_{1}+\omega_{2}})\subseteq\Omega(\tilde{V})$. So Lemma 3.3(3)
yields the result.

###  5.2. $SL_{2n}(F)$-modules viewed as $Sp_{2n}(F)$-modules

Let $F$ be an algebraically closed field and $\mathbf{H}=SL_{2n}(F)$,
$\mathbf{G}=Sp_{2n}(F)$. Let $\mathbf{T}$, $\mathbf{T}^{\prime}$ be maximal
tori of $\mathbf{H}$, $\mathbf{G}$, respectively. Under a Witt basis
$e_{1},\ldots,e_{2n}$ of the natural $F\mathbf{G}$-module, we assume that
$\mathbf{T}$ consists of matrices $t=\mathop{\rm
diag}\nolimits(t_{1},\ldots,t_{2n})$ with $t_{1},\ldots,t_{2n}\in F^{\times}$
and $t_{1}\cdots t_{2n}=1$, and $\mathbf{T}^{\prime}$ consists of matrices
$t^{\prime}=\mathop{\rm
diag}\nolimits(t_{1},\ldots,t_{n},t_{n}^{-1},\ldots,t_{1}^{-1}$)
($t_{1},\ldots,t_{n}\in F^{\times})$. So
$\mathbf{T}^{\prime}\subset\mathbf{T}$. The Bourbaki weights
$\varepsilon_{1},\ldots,\varepsilon_{2n}$ for $\mathbf{H}$ are defined by
$\varepsilon_{i}(t)=t_{i}$ ($i=1,\ldots,2n$). We denote by
$\varepsilon^{\prime}_{1},\ldots,\varepsilon^{\prime}_{n}$ the Bourbaki
weights for $\mathbf{G}$, which are defined by
$\varepsilon^{\prime}_{i}(t^{\prime})=t_{i}$ for $i=1,\ldots,n$.

Then $\varepsilon_{i}|_{\mathbf{T}^{\prime}}=\varepsilon^{\prime}_{i}$ if
$i\leq n$ and
$\varepsilon_{i}|_{\mathbf{T}^{\prime}}=-\varepsilon^{\prime}_{2n+1-i}$ if
$i>n$ (or
$\varepsilon_{2n-i}|_{\mathbf{T}^{\prime}}=-\varepsilon^{\prime}_{i+1}$ for
$1,\ldots,n-1$). We denote $\omega_{1},\ldots,\omega_{2n-1}$ the fundamental
weights for $\mathbf{H}$ and $\omega^{\prime}_{1},\ldots,\omega^{\prime}_{n}$
the fundamental weights for $\mathbf{G}$.

In Lemmas 5.3, 5.4 and 5.5 $F$ is a field of arbitrary characteristic $p>0$.

######  Lemma 5.3.

Let p be the characteristic of F. Let V be an irreducible $\mathbf{H}$-module
with $p$-restricted highest weight, and $\mu$ a dominant weight of
$V|_{\mathbf{G}}$ (not necessarily $p$-restricted). Then all weights of
$\tilde{V}_{\mu}$ occur as weights of $V|_{\mathbf{H}}$.

Proof. By Lemma 5.1 and [20], the set of weights of $V$ coincides with that of
an irreducible $\tilde{\mathbf{H}}$-module $\tilde{V}$ whose highest weight is
the same as that of $V$. So $\mu$ is a weight of $\tilde{V}|_{\mathbf{G}}$.
Let $L_{\mathbf{H}}$ and $L_{\mathbf{G}}$ be the Lie algebra over the complex
numbers of the same type as $\mathbf{H}$ and $\mathbf{G}$, respectively. (So
$L_{\mathbf{H}}$ is of type $A_{2n-1}$ and $L_{\mathbf{G}}$ is of type
$C_{n}$.) Then $\mu$ as a weight of $L_{\mathbf{G}}$ is a dominant weight of
some irreducible representation $\Lambda$, say, of $L_{\mathbf{G}}$, and all
weights of $\Lambda$ are weights of $\tilde{V}|_{\mathbf{G}}$. By [2, Ch.
VIII, Prop. 5], all weights of irreducible representation of $L_{\mathbf{G}}$
with highest weight $\mu$ are weights of $\Lambda$. Therefore, all weights of
$\tilde{V}_{\mu}$ are weights of $\tilde{V}|_{\mathbf{G}}$.

######  Lemma 5.4.

Let $\omega=\sum a_{i}\omega_{i}$ be a dominant weight of $\mathbf{H}$. Then
$\omega|_{\mathbf{T}^{\prime}}=\sum b_{i}\omega_{i}^{\prime}$, where
$b_{i}=a_{i}+a_{2n-i}$ for $i=1,\ldots,n-1$ and $b_{n}=a_{n}$.

Proof. It suffices to prove the lemma for $\omega=\omega_{i}$. If $i\leq n$
then
$\omega_{i}|_{T^{\prime}}=(\varepsilon_{1}+\cdots+\varepsilon_{i})|_{T^{\prime}}=\varepsilon^{\prime}_{1}+\cdots+\varepsilon^{\prime}_{i}=\omega_{i}^{\prime}$.
Let $i>n$. Then
$\omega_{i}|_{T^{\prime}}=(\varepsilon_{1}+\cdots+\varepsilon_{i})|_{T^{\prime}}=\varepsilon^{\prime}_{1}+\cdots+\varepsilon^{\prime}_{n}-\varepsilon^{\prime}_{n-1}-\cdots-\varepsilon_{2n-i+1}=\varepsilon_{1}^{\prime}+\cdots+\varepsilon_{2n-i}^{\prime}=\omega_{2n-i}^{\prime}$.
$\Box$

######  Lemma 5.5.

Let V be an irreducible $\mathbf{H}$-module with highest weight $\omega_{k}$
for $1\leq k\leq 2n-1$, and let $V^{\prime}$ be a composition factor of
$V|_{\mathbf{G}}$. Then the highest weight of $V^{\prime}$ is
$\omega_{j}^{\prime}$ for some $j$ with $1\leq j\leq\min(k,2n-k)$ and $k-j$
even.

Proof. We can assume $k\leq n$ by replacing $V$ by the dual of it. The weights
of $V$ form the $W$-orbit of
$\omega_{k}=\varepsilon_{1}+\cdots+\varepsilon_{k}$, where $W$ is the Weyl
group of $\mathbf{H}$. So if $\mu$ is a weight of $V$ then
$\mu=\varepsilon_{i_{1}}+\cdots+\varepsilon_{i_{k}}$, where $1\leq
i_{1}\leq...\leq i_{k}\leq 2n$. Let $l$ be an integer such that $i_{l-1}\leq
n<i_{l}$ (it exists unless $i_{k}\leq n$ or $i_{1}>n$). Then
$\mu|_{\mathbf{T}^{\prime}}=\varepsilon^{\prime}_{i_{1}}+\cdots+\varepsilon^{\prime}_{i_{l-1}}-\varepsilon^{\prime}_{2n+1-i_{l}}-\cdots-\varepsilon_{2n+1-i_{k}}$
(with an obvious refining for the exceptional cases). After canceling the
terms occurring here with opposite signs, one obtains a similar expression
$\mu^{\prime}=\sum b_{r}\varepsilon^{\prime}_{r}$ with $-1\leq b_{r}\leq 1$
for $1\leq r\leq k$. Moreover, $k-s$ is even, where $s$ is the number of non-
zero coefficients $b_{r}$. This is a weight of some irreducible constituent of
$V|_{\mathbf{G}}$. Recall that $W^{\prime}$, the Weyl group of $\mathbf{G}$,
acts transitively on every set
$\pm\varepsilon^{\prime}_{m_{1}}\pm\cdots\pm\varepsilon^{\prime}_{m_{s}}$ for
any fixed $s$ and integers $1\leq m_{1}\leq...\leq m_{s}\leq n$. Then the
dominant weight in the orbit $W^{\prime}\mu^{\prime}$ is
$\varepsilon^{\prime}_{1}+\cdots+\varepsilon^{\prime}_{s}$. We conclude that
the highest weight of $V^{\prime}$ must be
$\varepsilon^{\prime}_{1}+\cdots+\varepsilon^{\prime}_{j}=\omega_{j}$ for some
$j$, as stated.

######  Corollary 5.6.

Let $\omega=\sum a_{i}\omega_{i}$ be a $2$-restricted dominant weight and
$\omega^{\prime}=\omega|_{\mathbf{T}^{\prime}}=\sum
a^{\prime}_{i}\omega_{i}^{\prime}$.

$(1)$ $a_{i}^{\prime}\leq 2$ for $i=1,\ldots,n-1$ and $a_{n}^{\prime}\leq 1;$

$(2)$ if $\omega$ is not radical then $\omega^{\prime}\neq
2\omega_{j}^{\prime}$ for any $j=1,\ldots,n;$

$(3)$ if $\omega^{\prime}=\omega^{\prime}_{j}$ for some j then
$\omega=\omega_{j}$ or $\omega_{2n-j}$.

Proof. (1) follows from Lemma 5.4. (2) If $j=n$ then the claim is a part of
(1). Let $j<n$. Then $\omega^{\prime}=2\omega_{j}^{\prime}$ yields
$a_{j}+a_{2n-j}=2$ and $a_{i}+a_{2n-i}=0$ for $i\neq j$ by Lemma 5.4. This
means that $\omega=\omega_{j}+\omega_{2n-j}$, which is a radical weight. This
is a contradiction.

(3) If $j<n$ then $a_{j}+a_{2n-j}=1$ and $a_{i}+a_{2n-i}=0$ for $i\neq j$,
$a_{n}=0$. If $j=n$ then $a_{n}=1$ and $a_{i}+a_{2n-i}=0$ for $i<n$. So the
claim follows. $\Box$

######  Lemma 5.7.

††margin:  do1

Let $V$ be an irreducible $\mathbf{H}$-module with $2$-restricted highest
weight $\omega$. If $\omega\neq\omega_{i}$ for any $i\in\\{1,\ldots,2n-1\\}$
then $V|_{\mathbf{G}}$ has either weight $0$ or all weights of
$V_{\omega^{\prime}_{1}+\omega_{2}^{\prime}}.$

Proof. Set $\omega^{\prime}=\omega|_{\mathbf{T}^{\prime}}$. It follows from
[18, Theorem 39] that $V|_{\mathbf{G}}$ has a composition factor $U$, say, of
highest weight $\omega^{\prime}$. (Indeed, let $v^{+}$ be a vector of weight
$\omega$. Then $v^{+}$ is stabilized by the unipotent radical of certain Borel
subgroup $B$ of $\mathbf{H}$, and hence by the unipotent radical of a Borel
subgroup $B^{\prime}$ of $\mathbf{G}$ contained in $B$. Set
$V^{\prime}=\mathbf{G}v^{+}$; by [18, Theorem 39], $v^{+}$ is a highest weight
vector of $V^{\prime}$. Clearly, $v^{+}$ does not lie in any maximal
$\mathbf{G}$-submodule $M$ of $V^{\prime}$, so $V^{\prime}/M$ is an
irreducible $\mathbf{G}$-module of highest weight $\omega^{\prime}$.) By
Corollary 5.6(3), $\omega^{\prime}\neq\omega^{\prime}_{i}$ for any
$i=1,\ldots,n$. By Lemma 2.2(2), either $0\prec\omega^{\prime}$, or
$\omega^{\prime}_{1}\preceq\omega^{\prime}$ and hence
$\omega^{\prime}_{1}\preceq\mu\preceq\omega^{\prime}$ for every dominant
weight $\mu$ of $U$; moreover, as $\omega^{\prime}\neq\omega^{\prime}_{i}$ for
$i=1,\ldots,n$, we have
$\omega^{\prime}_{1}\prec\omega^{\prime}_{1}+\omega^{\prime}_{2}\preceq\omega^{\prime}$
(Lemma 2.2(5)). By Lemma 5.3, $V|_{\mathbf{G}}$ has all weights of the Weyl
module $\tilde{V}_{\omega^{\prime}}$. In turn, all weights of
$V_{\omega_{1}^{\prime}+\omega_{2}^{\prime}}$ are weights of
$\tilde{V}_{\omega^{\prime}}$, as stated. $\Box$

###  5.3. Real elements in $SL_{n}(q)$ and $SU_{n}(q)$

######  Lemma 5.8.

Let $H=SL_{n}(q)$ or $SU_{n}(q)$, where $q$ is an arbitrary prime power, and
let $g\in H$ be a semisimple element. Suppose that g is real. Then g is
conjugate to a subgroup isomorphic to $Sp_{n}(q)$ or $Sp_{n-1}(q)$ (depending
on parity of n).

Proof. The multiplicity of every eigenvalue $e$ of $g$ as an element of
$GL_{n}(\overline{\mathop{\mathbb{F}}\nolimits}_{q})$ equals that of $e^{-1}$.
By determinant reason, the multiplicity of $-1$ is even. If $n$ is odd then 1
is an eigenvalue of $g$ on the natural module $V$ for $H$, hence $g$ is
contained in a subgroup isomorphic to $SL_{n-1}(q)$ or $SU_{n-1}(q)$. (In the
unitary case observe that the 1-eigenspace $V^{g}$ of $g$ on $V$ is non-
degenerate. So $V^{g}$ contains an anisotropic vector, and the claim follows
again.) So $n$ can be assumed to be even. Then the result is due to Wall [21,
p.36]. (This is straightforwardly for $H=SL_{n}(q);$ for $SU_{n}(q)$ this is
explained in [6, p. 594]. For $n,q$ even the result is explicitly stated in
[6, Proposition 2.4].) $\Box$

######  Lemma 5.9.

Let $h\in H=GL_{n}(q)$. If $|h|$ divides $q^{i}+1$ for some integer $i>0$ then
$h$ is real.

Proof. It suffices to prove this for the case where $h$ is irreducible and
$|h|>2$. Let $E$ be the enveloping algebra of $\langle h\rangle$ over
$\mathop{\mathbb{F}}\nolimits_{q}$. By Schur’s lemma, $E$ is a field, and
$E\cong\mathop{\mathbb{F}}\nolimits_{q^{n}}$ as $h$ is irreducible. Then $|h|$
divides $q^{n}-1$, and hence $(q^{n}-1,q^{i}+1)$, and also
$(q^{n}-1,q^{2i}-1)$. By [9, Hilfsatz 2], $(q^{n}-1,q^{2i}-1)=q^{(n,2i)}-1$.
If $n$ is odd then $q^{(n,2i)}-1=q^{(n,i)}-1=(q^{n}-1,q^{i}-1)$, so $|h|$
divides $q^{i}-1$. As $(q^{i}-1,q^{i}+1)\leq 2$, we have $|h|\leq 2$, a
contradiction.

Let $n=2m$ be even and $k=(m,i)$. Then $(q^{n}-1,q^{i}+1)=(q^{2m}-1,q^{i}+1)$
which divides $(q^{2m}-1,q^{2i}-1)=q^{2k}-1$. So $|h|$ divides $q^{2k}-1$. As
$2k|n$, it follows that $\mathop{\mathbb{F}}\nolimits_{q^{2k}}$ is a subfield
of $E\cong\mathop{\mathbb{F}}\nolimits_{q^{n}}$, and hence $h$ lies in the
subfield $E_{1}$ of $E$ of order $q^{2k}$. This implies $E_{1}=E$ and $2k=n$,
$k=m$. Then $m|i$ (as $k=(m,i)$), and $i/m$ is odd. Indeed, if $i=2mj=nj$ then
$q^{i}+1=q^{nj}+1$; as $|h|$ divides $q^{n}-1$ and hence $q^{nj}-1$, it
follows that $|h|\leq 2$, a contradiction. So $i/m$ is odd, and hence
$q^{m}+1$ divides $q^{i}+1$. In addition, $(q^{i}+1)/(q^{m}+1)$ is odd, so
$|h|$ divides $q^{m}+1$. (Indeed, $(|h|,q^{i}-1)\leq 2$, and hence
$(|h|,q^{m}-1)\leq 2$ as $m|i$ and $q^{m}-1$ divides $q^{i}-1$. So $|h|$
divides $2(q^{m}+1)$. As $|h|$ divides $q^{i}+1$ and $q^{i}+1=a(q^{m}+1)$ with
$a$ odd, we conclude that $|h|$ divides $q^{m}+1$, as claimed.)

Let $N=N_{H}(E)$. Then the Galois group
$\mathop{\mathbb{F}}\nolimits_{q^{n}}/\mathop{\mathbb{F}}\nolimits_{q}$ is
isomorphic to a subgroup of $N/C_{H}(E)$ by the Noëther-Skolem theorem [15,
§12.6]. The Galois group
$\mathop{\mathbb{F}}\nolimits_{q^{n}}/\mathop{\mathbb{F}}\nolimits_{q^{m}}$ is
of order 2, and the non-trivial Galois automorphism of
$\mathop{\mathbb{F}}\nolimits_{q^{n}}/\mathop{\mathbb{F}}\nolimits_{q^{m}}$
sends $x\in\mathop{\mathbb{F}}\nolimits_{q^{n}}$ to $x^{q^{m}}$. As $|h|$
divides $q^{m}+1$, we have $h^{q^{m}}=h^{q^{m}+1}h^{-1}=h^{-1}$. So $h$ is
real in $GL_{n}(q)$. $\Box$

######  Lemma 5.10.

Let $g\in G=SU_{n}(q)$. Suppose that $|g|$ divides $q^{i}-1$ for some $i$ odd.
Then $g$ is real.

Proof. It suffices to prove the lemma for the case where $g$ stabilizes no
non-degenerate proper subspace of the underlying space $V$ of $G$. Then either
$g$ is irreducible or $n$ is even and $V=V_{1}+V_{2}$, where $V_{1},V_{2}$ are
$g$-stable totally isotropic subspaces of dimension $n/2$. In the former case
$n$ is odd and $|g|$ divides $q^{n}+1$, in the latter case $g$ is irreducible
on $V_{1}$ and $|g|$ divides $q^{n}-1$. Note that
$(q^{i}-1,q^{j}-1)=q^{(i,j)}-1$ by [9, Hilfsatz 2].

In the former case $|g|$ divides $(q^{i}-1,q^{n}+1)$, hence also
$(q^{i}-1,q^{2n}-1)=q^{(i,2n)}-1=q^{(i,n)}-1$. So $|g|$ divides $q^{n}-1$. As
$(q^{n}-1,q^{n}+1)\leq 2$, it follows $|g|\leq 2$, and then $g$ is real.

In the latter case $g\in Y$, where $Y=\\{s\in G:sV_{1}=V_{1}\\}$. It is well
known that $Y\cong GL(V_{1})\cong GL_{n/2}(q^{2})$. Furthermore, there is a
basis $B$, say, in $V$ such that $B\cap V_{i}$ is a basis of $V_{i}$ for
$i=1,2$, and the matrix of every $y\in Y$ is $\mathop{\rm
diag}\nolimits(x,({}^{t}x^{-1})^{J})$ with $x\in GL(V_{1})$, where ${}^{t}x$
means the transpose of $x$ and $J$ is the Galois automorphism of
$\mathop{\mathbb{F}}\nolimits_{q^{2}}/\mathop{\mathbb{F}}\nolimits_{q}$ of
order 2 extended to $GL_{n/2}(q^{2})$.

Let $g=\mathop{\rm diag}\nolimits(h,({}^{t}h^{-1})^{J}))$, so $h$ is
irreducible in $GL(V_{1})$. If $h$ is real in $GL(V_{1})$ then $g$ is real in
$G$, so we assume that $h$ is not real in $GL(V_{1})$. In addition, the
characteristic polynomial of $h$ is irreducible as well as that of
$({}^{t}h^{-1})^{J};$ and they are distinct (otherwise $V$ is a homogeneous
reducible $F_{q^{2}}\langle g\rangle$-module, and then $V$ contains a non-
degenerate $g$-stable subspace (see [5, Lemma 6.3]). Therefore,
$C_{G}(g)V_{1}=V_{1}$, so $C_{G}(g)\subset Y$; as the group $C_{GL(V_{1})}(h)$
is cyclic, so is $C_{G}(g)$.

As $i$ is odd, $n$ is even, we have $(i,n)=(i,n/2)$. Therefore, $|h|$ divides
$(q^{i}-1,q^{n}-1)=q^{(i,n)}-1=q^{(i,n/2)}-1$ by the above. As $h$ is
irreducible on $V_{1}$, the $\mathop{\mathbb{F}}\nolimits_{q^{2}}$-enveloping
algebra $E$, say, of $\langle h\rangle$ is a field of order $q^{n}$, so the
multiplicative group $E^{\times}$ of $E$ is of order $q^{n}-1$. The elements
of $E^{\times}$ whose order divides $q^{n/2}-1$ together with 0 form a
subfield $E_{1}$, say, of order $q^{n/2}$. So $h\in E_{1}$. In addition, $n/2$
is odd (otherwise $E_{1}$ contains $\mathop{\mathbb{F}}\nolimits_{q^{2}}$,
which is false as $h$ is irreducible). Let $d_{1}$ be a generator of
$E_{1}^{\times}$ and $d=\mathop{\rm
diag}\nolimits(d_{1},({}^{t}d_{1}^{-1})^{J}))$ and $D=\langle d\rangle$. Then
$g\in D$.

Note that $Sp_{n}(q)$ contains a cyclic subgroup of order $q^{n/2}-1$. So the
result follows if we show that every cyclic subgroup of $G$ of order
$q^{n/2}-1$ is conjugate to $D$. For this it suffices to show that $D$
contains a Sylow $r$-subgroup $R$ of $G$ for some prime $r$ such that
$C_{G}(R)=E^{\times}$. We can assume $n>2$, as $SU_{2}(q)\cong Sp_{2}(q)$, and
$g\in Sp_{2}(q)$ is real whenever $(|g|,q)=1$.

By [11, Theorem 5.2.14], $q^{n}-1$ is divisible by a prime $r$ such that $r$
does not divide $q^{j}-1$ for any $j<n$, unless $q=2,n=6$ or $n=2$. By the
above, $n>2$; if $(q,n)=(2,6)$ then we take $r=7$. Then $|R|=7$ and one can
easily check that $C_{G}(R)=E^{\times}$.

In the non-exceptional case let $R$ be the Sylow $r$-subgroups of $D$. Then
$R$ is a Sylow $r$-subgroups of $GL_{n/2}(q^{2})$. Comparing the orders of
$GL_{n/2}(q^{2})$ and $G$ we conclude that $R$ remains a Sylow $r$-subgroups
of $G$. (Note that $(r,q^{j}+1)=1$ for $0<j<n$ as otherwise $r$ divides
$q^{2j}-1$ and $(q^{2j}-1,q^{i}-1)=q^{(i,j)}-1$ as $i$ is odd; this
contradicts the choice of $r$ as $(i,j)<n$.) Finally, one easily observe that
$C_{G}(R)=E^{\times}$. $\Box$

######  Lemma 5.11.

Let $\mathbf{L}=SL_{n+1}(F)$, $n$ even, and let $\mathbf{H}=SL_{n+1}(F)$. Let
$\phi$ be an irreducible representation of $\mathbf{L}$ with $p$-restricted
highest weight $\lambda$. Let $\lambda_{1},\ldots,\lambda_{n}$ be the
fundamental weights of $\mathbf{L}$ and $\omega_{1},\ldots,\omega_{n-1}$ be
those of $\mathbf{H}$.

$(1)$ Suppose that $n>2$ if $p=2$ and if $\nu$ is the highest weight of a
composition factor of $\phi|_{\mathbf{H}}$ and $\nu$ is $p$-restricted then
$\nu\in\\{0,\omega_{1},\ldots,\omega_{n-1}\\}$. Then
$\lambda\in\\{\lambda_{1},\ldots,\lambda_{n}\\}$.

$(2)$ Suppose that $\lambda=\lambda_{i}$ with $i\leq n$. Then
$\phi_{\lambda}|_{\mathbf{H}}$ contains an irreducible constituent of highest
weight $\omega_{i-1}$. (If $i=1$ then $\omega_{i-1}$ is interpreted as the
zero weight.)

$(3)$ Let $\psi$ be any representation of $\mathbf{L}$ and
${\mathbf{G}}=Sp_{n}(F)$. Then $\psi|_{{\mathbf{G}}}$ has weight $0$.

Proof. We can view $\mathbf{H}$ as a subgroup of $\mathbf{L}$ of shape
$\mathop{\rm diag}\nolimits(\mathbf{H},1)$ or $\mathop{\rm
diag}\nolimits(1,\mathbf{H})$. Then the reference torus
$\mathbf{T}_{\mathbf{H}}$ of $\mathbf{H}$ (chosen to be the group of diagonal
matrices) is contained in that of $\mathbf{L}$ and a Borel subgroup $B$ of
$\mathbf{H}$ containing $\mathbf{T}_{\mathbf{H}}$ is contained in a Borel
subgroup of $\mathbf{L}$ that contains the reference torus $\mathbf{T}$ of
$\mathbf{L}$. Then $\varepsilon_{n}(\mathbf{T}_{\mathbf{H}})=1$ in the former
case and $\varepsilon_{1}(\mathbf{T}_{\mathbf{H}})=1$ in the latter case.

(1) Let $\lambda=\sum a_{i}\lambda_{i}$. Then the restriction
$\phi|_{\mathbf{H}}$ has composition factors with highest weights
$a_{1}\omega_{1}+\cdots+a_{n-1}\omega_{n-1}$ and
$a_{2}\omega_{1}+\cdots+a_{n}\omega_{n-1}$. These weights are $p$-restricted
so must lie in $\\{0,\omega_{1},\ldots,\omega_{n-1}\\}$. This implies the
claim unless possibly $\lambda=\lambda_{1}+\lambda_{n}$. Then
$\phi_{\lambda_{1}+\lambda_{n}}$ is well known to be the unique non-trivial
irreducible constituent of the adjoint module (when $\mathbf{L}$ acts by
conjugation of the $((n+1)\times(n+1))$-matrices with trace 0). For $n>3$ one
can easily observe that $\phi_{\omega_{1}+\omega_{n-1}}$ is a constituent of
the adjoint module and hence of $V_{\lambda_{1}+\lambda_{n}}$, whence the
result. If $n=2$ then the restriction of $\phi_{\lambda_{1}+\lambda_{2}}$ to
$\mathbf{H}$ has an irreducible constituent of highest weight $2\omega_{1}$,
which is $p$-restricted for $p\neq 2$. So this option is rules out.

(2) $V_{\lambda}$ has weight
$\mu:=\varepsilon_{1}+\cdots+\varepsilon_{i-1}+\varepsilon_{n}$. Let $0\neq
v\in V_{\lambda}$ be a vector in the $\mu$-weight space. Then $v$ is a
primitive vector for $\mathbf{H}$, that is, $Bv=\langle v\rangle$. Therefore,
$V^{\prime}:=\langle\mathbf{H}v\rangle$ is a $\mathbf{H}$-module (possibly
reducible) with highest weight $\varepsilon_{1}+\cdots+\varepsilon_{i-1}$ [18,
Theorem 39]. Then one can factorize $V^{\prime}$ by a maximal submodule to
obtain a composition factor with highest weight $\omega_{i-1}$, as claimed.

(3) It suffices to prove this for tensor-indecomposable irreducible
representations, and then we can assume that $\psi$ has a $p$-restricted
highest weight $\mu$, say. Then either $0\prec\mu$ or $\lambda_{i}\preceq\mu$
for some $i\in\\{1,\ldots,n\\}$. If $i$ is even then
$\psi_{\lambda_{i}}|_{\mathbf{G}}$ has weight 0 by Lemma 5.5. If $i$ is odd
then $\psi_{\lambda_{i}}|_{\mathbf{H}}$ has a constituent $\rho$, say, of
highest weight $\omega_{i-1}$; as $i-1$ is even, $\rho|_{\mathbf{G}}$ has
weight 0, by Lemma 5.5. $\Box$

######  Corollary 5.12.

Let $H=SL_{n+1}(q)$ or $SU_{n+1}(q)$, where n is even, p a prime dividing q,
and let $\rho$ be a p-modular representation of H. Then $\rho(g)$ has
eigenvalue $1$ for every real element $g\in H$.

Proof. It suffices to prove the theorem for $\rho$ irreducible, and with
$p$-restricted highest weight. By Lemma 5.8, we can assume that $g\in G\cong
Sp_{n}(q)$. So the result follows from Lemma 5.11(3). $\Box$

######  Theorem 5.13.

Let $\mathbf{H}=SL_{n+1}(\overline{\mathop{\mathbb{F}}\nolimits}_{2})$, let
$H=SL_{n+1}(2)$ or $SU_{n+1}(2)$, and let $g\in H$ be a semisimple element.
Let $\phi$ be an irreducible representation of $\mathbf{H}$ with highest
weight $\lambda$.

$(1)$ Suppose that $g\in H$ is real. Then $\phi(g)$ has eigenvalue $1$ unless
(possibly) $n+1$ is even and $\lambda=2^{i}\lambda^{\prime}$, where
$\lambda^{\prime}$ is some fundamental weight of $\mathbf{H}$.

$(2)$ Suppose that $H=SL_{n+1}(2)$ and $|g|$ divides $2^{i}+1$ for some i, or
$H=SU_{n+1}(2)$ and $|g|$ divides $2^{i}-1$ for some i odd. Then the
conclusion of $(1)$ holds.

Proof. It suffices to prove (1) as $g$ in (2) is real by Lemmas 5.9 and 5.10.
Furthermore, if $n+1$ is odd then the argument in the proof of Corollary 5.12
works and yields the result. So we assume that $n+1$ is even. By Lemma 5.8, we
can assume that $g\in G=Sp_{n+1}(2)$.

Let $\lambda_{1},\ldots,\lambda_{n}$ be the fundamental weights of
$\mathbf{H}$ and $\omega_{1},\ldots,\omega_{(n+1)/2}$ those of
$\mathbf{G}=Sp_{n+1}({\overline{\mathop{\mathbb{F}}\nolimits}_{2}})$. Suppose
first that $\lambda$ is 2-restricted. By Lemma 5.7, $\rho|_{\mathbf{G}}$ has
either weight 0 (and we are done) or all weights of an irreducible
representation of $\mathbf{G}$ with highest weight $\omega_{1}+\omega_{2}$. In
the latter case the result follows by Lemma 3.3(3).

Suppose that $\lambda$ is not 2-restricted. Then $\lambda=\sum 2^{j}\mu_{j}$,
where $\mu_{j}$ are 2-restricted dominant weights. Then
$\phi=\otimes_{j}\rho_{2^{j}\mu_{j}}$ and
$\phi|_{H}=\otimes_{j}\rho_{\mu_{j}}|_{H}$. Clearly, we only need to examine
the case where this product has at least two terms. If $\mu_{j}$ is not
radical, denote by $\mu_{j}^{\prime}$ the fundamental weight such that
$\mu_{j}^{\prime}\preceq\mu_{j}$, otherwise set $\mu_{j}^{\prime}=\mu_{j}$.
Then
$\Omega(\otimes_{j}\rho_{\mu^{\prime}_{j}})\subseteq\Omega(\otimes_{j}\rho_{\mu_{j}})$,
so it suffices to prove the theorem for the case with
$\mu_{j}=\mu^{\prime}_{j}$. Set
$\nu_{j}=\mu^{\prime}_{j}|_{\mathbf{T}^{\prime}}$, where $\mathbf{T}^{\prime}$
is a maximal torus of $\mathbf{G}$ as in Lemma 5.4, which tells us that
$\nu_{j}\in\\{\omega_{1},\ldots,\omega_{n}\\}$ whenever $\mu_{j}^{\prime}$ is
not radical. For such $j$ the weights of the Weyl module $\tilde{V}_{\nu_{j}}$
are weights of $\rho_{\mu_{j}}$ (Lemma 5.3). In addition, either $\nu_{j}$ is
radical or $j$ is odd and $\tilde{V}_{\nu_{j}}$ has weights $\pm\omega_{1}$
(Lemmas 2.2 and 2.5 ). So the tensor product of two such modules has weight 0.
It follows that that we can write
$\phi|_{\mathbf{G}}=\phi_{1}\otimes\phi_{2}$, where $\phi_{1}|_{\mathbf{G}}$
has weight 0 and $\phi_{2}=\rho_{\mu_{j}}$ for some $j$, where
$\rho_{\mu_{j}}$ contains all weights of $\tilde{V}_{\nu_{j}}$ for some
$\nu_{j}\in\\{\omega_{1},\ldots,\omega_{n}\\}$ with $j$ odd. If $\phi_{1}$ is
trivial then the result follows by Corollary 5.6(3). Otherwise,
$\phi_{1}|_{\mathbf{G}}$ has weight $\omega_{2}$. Indeed, the weights of
$\sigma\in\Omega(\phi_{1})$ have the same residue modulo the radical weights
(as so are the weights of every multiple $\rho_{\mu_{j}}$), and
$\sigma|_{\mathbf{T}^{\prime}}=0$ for some $\sigma$. It follows that
$\sigma|_{\mathbf{T}^{\prime}}$ is radical for every
$\sigma\in\Omega(\phi_{1})$. In particular, this is the case for
$\sigma|_{\mathbf{T}^{\prime}}$ to be the highest weight of a non-trivial
composition factor of $\phi_{1}|_{\mathbf{G}}$. The latter has weight
$\omega_{2}$ by Lemma 2.5, and hence the weight
$\varepsilon_{2}+\varepsilon_{3}$ if $n>2$. So all weights of
$V_{\omega_{1}+\omega_{2}}$ are weights of $\phi|_{\mathbf{G}}$. Then the
result follows by Lemma 3.3. $\Box$

Proof of Theorem 1.7. The result is a special case of Theorem 5.13.

The following Lemmas 5.14 and 5.15 show that the result of Theorem 5.13 is in
a sense best possible.

######  Lemma 5.14.

Let $h\in H=SL_{2n}(2)$. Suppose that $|h|=p$, where $p=2^{n}+1$ is a prime.
Let $\phi$ be an $i$-th exterior power of the natural
$\mathop{\mathbb{F}}\nolimits_{2}H$-module, where i is odd. Then $\phi(h)$
does not have eigenvalue $1$.

Proof. Note that $h$ is conjugate to a Singer cycle in $G=Sp_{n}(2)$. By Lemma
5.5, the composition factors of $\phi|_{\mathbf{G}}$ are of highest weights
$\omega_{j}^{\prime}$ for $j$ odd. So Lemma 3.2 yields the result. $\Box$

######  Lemma 5.15.

Let $h\in H=SU_{n}(2)$. Suppose that $|h|=p$, where $p=2^{n/2}-1$ is a prime.
Let $\phi$ be an $i$-th exterior power of the natural
$\mathop{\mathbb{F}}\nolimits_{4}H$-module, where $i<n-1$ is odd. Then
$\phi(h)$ does not have eigenvalue $1$.

Proof. This follows from Lemma 3.1. Indeed, $n/2$ is odd, so $h$ is conjugate
to an element of $G=Sp_{n}(2)$ (Lemmas 5.10 and 5.8). Moreover, $h\in G$ is a
Singer cycle, so $h$ is a generator of a (cyclic) maximal torus $T_{w}$ of $G$
labeled by an element $w$ of the Weyl group which transitively permutes
$\varepsilon_{1},\ldots,\varepsilon_{n}$. Let $W_{i}$ be as in Lemma 3.1. Then
$w$ is conjugate to an element of $W_{i}$ if and only if $i=n$. As in Lemma
5.14 one observes that the composition factors of $\phi|_{G}$ are of highest
weights $\omega_{j}^{\prime}$ for $j\in\\{1,\ldots,n\\}$ odd. Then the
reasoning in the proof of Lemma 3.2 shows that $h$ does not have eigenvalue 1
on $V_{\omega^{\prime}_{i}}$ whenever $i<n-1$ odd. $\Box$

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Alexandre E. Zalesski

Department of Physics, Mathematics and Informatics,

National Academy of Sciences of Belarus,

66 Prospekt Nezalejnasti,

220072 Minsk,

Belarus

e-mail<EMAIL_ADDRESS>