# Upper bounds for eigenvalue multiplicities
of almost cyclic elements in irreducible representations
of simple algebraic groups

A.E. Zalesski

###### Abstract

We study the rational irreducible representations of simple algebraic groups
in which some non-central semisimple element has at most one eigenvalue of
multiplicity greater than $1$. We bound the multiplicity of this eigenvalue in
terms of the rank of the group. 1112020 Mathematics subject classification:
20G05, 15A18 222 Keywords: simple algebraic groups, representations, regular
elements, eigenvalue multiplicities.

Dedicated to the memory of Irina Suprunenko

##  1 Introduction

In this paper we study a special case of a general problem on eigenvalue
multiplicities of semisimple elements in irreducible representations of
algebraic groups. All representations of algebraic groups and their modules
are assumed to be rational. This problem is connected with a similar problem
on weight multiplicities. In [23] we determine the irreducible representations
of simple algebraic groups $G$ whose all non-zero weights are of multiplicity
1, and compute the zero weight multiplicity in each case. Note that weights
are one-dimensional representations of a maximal torus $T$, say, of $G$, and
the multiplicity of a weight $\omega$, say, in a representation $\rho$ of $G$
is the multiplicity of $\omega$ as a constituent of $\rho(T)$. Turning to
individual non-central semisimple elements $g\in G$, an analog of this result
could be (i) obtaining a list of the irreducible representations $\rho$ of $G$
such that at most one eigenvalue multiplicity of $\rho(g)$ is greater than 1,
and then (ii) determining the multiplicity of the exceptional eigenvalue. An
essential step of understanding problem (i) is made in [25], where it is
proved that all non-zero weights of $\rho$ must be of multiplicity 1. This
does not mean however that the eigenvalue multiplicities of $\rho(g)$ are the
same as the weight multiplicities. This can be seen on examples with $G$ to be
of classical type and $\rho$ the natural representation of $G$. Moreover, in
these examples $m_{g}(\rho)$, the maximum eigenvalue multiplicity of
$\rho(g)$, is almost arbitrary (subject to the obvious condition
$m_{g}(\rho)\leq\dim\rho$), whereas all weight multiplicities of $\rho$ are
equal to 1. However, it is shown below that $m_{g}(\rho)$ can be bounded in
terms of the rank of $G$. An initial step in this direction is made in [25],
where, for a non-central non-regular semisimple element $g\in G$, it is shown
that the natural representation of a classical group $G$ is essentially the
only one in which at most one eigenvalue multiplicity of $\rho(g)$ is greater
than 1; that result is quoted below as Theorem 3.17. To be more precise,
$\rho$ is a twist of the natural representation. (We call a representation
$\rho^{\prime}$ a twist of $\rho$ if $\rho^{\prime}(g)=\rho(\sigma(g))$ for
$g\in G$ and a surjective homomorphism $\sigma:G\rightarrow G$.) Therefore,
the current paper mainly deals with regular semisimple elements.

The main result of this paper is the following theorem:

######  Theorem 1.1.

Let $G$ be a simple algebraic group of rank n and $V$ an irreducible
$G$-module. Let $s\in G$ be a regular semisimple element and let $m_{s}(V)$ be
the maximum of the eigenvalue multiplicities of $s$ on V. Suppose that at most
one eigenvalue multiplicity of $s$ on V exceeds $1$. Then $m_{s}(V)\leq n+1.$

In fact, we can give more details as follows:

######  Theorem 1.2.

Let $G$ be a simple algebraic group of rank n and $V$ an irreducible
$G$-module. Let $s\in G$ be a regular semisimple element and let $m_{s}(V)$ be
the maximum of the eigenvalue multiplicities of $s$ on V. Suppose that at most
one eigenvalue multiplicity of $s$ on V exceeds $1$ and $m_{s}(V)>1$. Then
either $m_{s}(V)=2$ or one of following holds.

$(1)$ $G=A_{n}$ and $m_{s}(V)\leq n+1;$

$(2)$ V is a twist of a G-module from Table $1$ and $m_{s}(V)$ equals the
weight zero multiplicity of V recorded there; in particular $n-2\leq
m_{s}(V)\leq n+1;$

$(3)$ $G=E_{6}$ and $m_{s}(V)\leq 3.$

The bound in item (1) of Theorem 1.2 is sharp for $n>1$, at least for $p\neq
0$, see Example 4.36. In this example for $G=A_{n}$ we construct a tensor-
decomposable $G$-module $V$ with $m_{s}(V)=n+1$. For $E_{6}$ see Example 4.16.

######  Corollary 1.3.

Let G be a simple algebraic group in defining characteristic $p\geq 0$, $s\in
G$ a regular semisimple element and let V be an irreducible G-module with
highest weight $\omega\neq 0$. Suppose that $\dim V>2$ and at most one
eigenvalue multiplicity of $s$ on V exceeds $1$. Then $m_{s}(V)<\dim V/2$,
unless $G=A_{1}$, $\dim V=3,p\neq 2$ or $\dim V=4$ and either
$\omega=3p^{i}\omega_{1},p\neq 2,3$ or $\omega=(p^{i}+p^{j})\omega_{1}$ with
$1\leq i<j$. (If $p=0$ and $n=4$ then $\omega=3\omega_{1}$.)

The fact that the multiplicity $m_{s}(V)$ in question is of certain
significance can be observed from [8], in which the authors study irreducible
subgroups $X$ of $GL_{n}(q)=GL(V)$, containing a semisimple element $g$ of
prime order acting irreducibly on $(\mathop{\rm Id}\nolimits-g)V$. They assume
that $\dim(\mathop{\rm Id}\nolimits-g)V>\dim V/2$, equivalently,
$m_{s}(V)=\dim V-\dim(\mathop{\rm Id}\nolimits-g)V<\dim V/2$, and we
complement this result by dropping the latter assumption for $X$ a quasi-
simple group of Lie type in defining characteristic $p|q$. More precisely, we
use Corollary 1.3 to prove the following result. In the statement below
$\overline{V}=V\otimes F$, where $F$ is an algebraically closed field
containing $\mathbb{F}_{q}$.

######  Theorem 1.4.

Let G be a quasi-simple group of Lie type in defining characteristic $p>0$,
and let V be a non-trivial irreducible $\mathbb{F}_{q}G$-module where $p|q$.
Let $s\in G$ be a regular semisimple element. Suppose that $\dim V>2$ and at
most one eigenvalue multiplicity of $s$ on $\overline{V}$ exceeds $1$. Then
$m_{s}(\overline{V})<\dim V/2$, unless $G=SL_{2}(q^{\prime})$ or
$PSL_{2}(q^{\prime})$ for some p-power $q^{\prime}$, and either
$m_{s}(\overline{V})=\dim V/2$ is even or $\dim V=3,p\neq 2$ and $s^{2}\in
Z(G)$.

Conversely, if V is a vector space over $\mathbb{F}_{q}$ and $\dim V\equiv
0\pmod{4}$ or $p\neq 2$ and $\dim V=3$ then there exists a group $G\cong
SL_{2}(q^{\prime})$, a semisimple element $s\in G$ and irreducible
representation $\rho:SL_{2}(q^{\prime})\rightarrow GL(V)$ such that at most
one eigenvalue multiplicity of $\rho(s)$ on $\overline{V}$ exceeds $1$ and
$m_{s}(\overline{V})=\max\\{\dim V/2,2\\}$.

In Theorem 1.4 saying that an element $g\in G$ is regular means that $g$ is
regular as an element of a simple algebraic group $\mathbf{G}$ from which $G$
is obtained as the fixed point set of a Steinberg (Frobenius) endomorphism of
$\mathbf{G}$, see [16, §21.1]. For a semisimple element $g\in G$ this is
equivalent to saying that $C_{G}(g)$ has no non-identity unipotent element. If
$g\in G$ is semisimple and not regular then we have a result similar to that
of [8]. To state it, it is convenient to introduce a notion of a nearly
natural module of a classical group as follows.

Let $G$ be a quasi-simple group of classical Lie type in defining
characteristic $p>0$. Let $V$ be an irreducible $\mathbb{F}_{q}G$-module for
some $p$-power $q$ and $F$ an algebraically closed field containing
$\mathbb{F}_{q}$, and $\overline{V}=V\otimes F$. Let
$G=\mathbf{G}^{\sigma}=\\{g\in\mathbf{G}:\sigma(g)=g\\}$, where $\sigma$ is a
Steinberg endomorphism of $\mathbf{G}$. We say that $V$ is nearly natural if
the composition factor dimensions of $G$ on $\overline{V}$ are the same as the
dimension of the natural $\mathbf{G}$-module. (See Notation for the formal
definition of the natural module.) This notion unfortunately depends on the
Lie type of $\mathbf{G}$. For instance, the dimensions of the natural module
for $A_{3}\cong D_{3}$ are 4 and 6, respectively, and these for $B_{2}\cong
C_{2}$ with $p\neq 2$ are 5,4, respectively.

The following statement is a special case of the main theorem of [8].

######  Theorem 1.5.

Let $G$ be a quasi-simple group of Lie type in defining characteristic $p>0$
and $q$ a $p$-power. Let $s\in G$ be a non-central semisimple element and let
$\rho:G\rightarrow GL_{n}(q)$, $n>9$, be an irreducible representation such
that $\rho(g)$ is similar to $\mathop{\rm diag}\nolimits(\mathop{\rm
Id}\nolimits_{m},M)$, where $M\in GL_{n-m}(q)$ is irreducible. Suppose that
$m<n/2$. Then G is classical and $\rho$ is nearly natural.

Note that the cases with $n\leq 9$ are listed in [8, Table 6].

Theorem 1.4 deals with regular elements $g\in G$. The case where $g$ is not
regular follows from Theorem 3.17 obtained in [25]:

######  Theorem 1.6.

Let G be a quasi-simple group of Lie type in defining characteristic $p$ and
$g\in G$ a non-central semisimple element. Let $\rho:G\rightarrow GL_{n}(q)$
be an irreducible representation. Suppose that g is not regular and at most
one eigenvalue multiplicity of $\rho(g)$ exceeds $1$. Then $\rho$ is nearly
natural for $G$ or for a group $G_{1}\cong G$. (Here $G_{1}=Spin^{+}_{6}(q)$
if $G=SL_{4}(q)$; $G_{1}=Spin^{-}_{6}(q)$ if $G=SU_{4}(q)$; $G_{1}=C_{2}(q)$
if $G=B_{2}(q)$; $G_{1}=B_{2}(q)$ if $G=C_{2}(q)$.)

##  2  Notation

The cardinality of a finite set $A$ is denoted by $|A|$. The ring of integers
is denoted by $\mathop{\mathbb{Z}}\nolimits$, and by
$\mathop{\mathbb{Z}}\nolimits^{+}$ the subset of non-negative integers. If
$0\neq a,b$ are natural numbers then we write $a|b$ to say that $b$ is a
multiple of $a$ and we denote by $(a,b)$ the greatest common divisor of $a,b$.
We fix $F$ for an algebraically closed field of characteristic $p\geq 0$; the
multiplicative group of $F$ is denoted by $F^{\times}$. The field of complex
numbers is denoted by $\mathop{\mathbb{C}}\nolimits$. We write $Z(X)$ for the
center of a group $X$. For $x\in X$ the group generated by $x$ is denoted by
$\langle x\rangle$, by $C_{X}(x)$ the centralizer of $x$ in $X$ and by $|x|$
the order of $x$. A diagonal $(n\times n)$-matrix with sequent diagonal
entries $d_{1},\ldots,d_{n}$ is denoted by $\mathop{\rm
diag}\nolimits(d_{1},\ldots,d_{n})$.

For our convenience we call an element $g\in GL_{n}(F)$ cyclic if $g$ is
diagonalizable and all eigenvalues of $g$ are of multiplicity 1. This is
equivalent to saying that the characteristic polynomial of $g$ has no multiple
roots. We call $g$ almost cyclic if $g$ is diagonalizable and at most one of
the eigenvalues of $g$ is of multiplicity greater than 1. If $g\in
GL_{n}(F_{1})$, where $F_{1}$ is a subfield of $F$ then saying that $g$ is
almost cyclic means that $g$ is almost cyclic in $GL_{n}(F)$. (See [6] and [7]
for more comments on these notions.)

In Section 3 $G$ is a non-trivial simply connected simple linear algebraic
group over $F$ (unless otherwise explicitly stated); sometimes we write $G(F)$
to make it precise. (In Section 5 we use $\mathbf{G}$ for a simple algebraic
group and $G$ for a finite group of Lie type.) All $G$-modules considered are
rational finite-dimensional $FG$-modules. For a $G$-module $V$ (or a rational
representation $\rho$ of $G$) we write $V\in\mathop{\rm Irr}\nolimits(G)$ (or
$\rho\in\mathop{\rm Irr}\nolimits(G)$) to mean that $V$ (or $\rho$) is
irreducible. If $H$ is a subgroup of $G$ then $V|_{H}$ means the restriction
of a $G$-module $V$ to $H$.

We fix a maximal torus $T$ in $G$ and a Borel subgroup containing $T$, which
in turn define the roots and simple roots of $G$, as well as the weights of
$G$-modules and representations. The set $\Omega={\rm Hom}(T,F^{\times})$ (the
rational homomorphisms of $T$ to $F^{\times}$) is called the weight lattice,
which is a free $\mathop{\mathbb{Z}}\nolimits$-module of finite rank called
the rank of $G$. The elements of $\Omega$ are called $T$-weights of $G$. As
$T$ is fixed we write ”weights” in place of $T$-weights. For $t\in T$ and
$\mu\in\Omega(V)$, $\mu(t)$ is an eigenvalue of $t$ on $V$; then
$(-\mu)(t)=t^{-1}$ and $(\mu+\nu)(t)=\mu(t)\nu(t)$ for $\nu\in\Omega(V)$.

The $T$-weights of a $G$-module $V$ are the irreducible constituents of the
restriction of $V$ to $T$. When $T$ is fixed (which we assume throughout the
paper), we omit the indication to $T$ and write ”weights” in place of
”$T$-weights”. The set of weights of $V$ is denoted by $\Omega(V)$. If $\mu$
is a weight of $V$, the $\mu$-weight space is $\\{v\in V:tv=\mu(t)v$ for all
$t\in T\\}$; the dimension of the $\mu$-weight space is called the
multiplicity of $\mu$ on $V$. The Weyl group of $G$ is denoted by $W(G)$; in
fact $W(G)=N_{G}(T)/T$ so the conjugation action of $N_{G}(T)$ on $T$ yields
an action of $W(G)$ on $T$ and on the $T$-weights. The $W(G)$-orbit of
$\mu\in\Omega$ is denoted by $W\mu$.

As every semisimple element of $G$ is conjugate to an element of $T$, in the
most of our results the assumption $g\in G$ semisimple is equivalent to saying
$g\in T$, the latter is more rigorous when we deal with $\mu(g)$ for a weight
$\mu$.

We use $G_{reg}$ to denote the set of semisimple regular elements of $G$, in
particular, $T_{reg}$ is the set of regular elements of $T$. The semisimple
regular elements $g$ of $G$ are exactly those centralizing no non-identity
unipotent element (Lemma 3.8), and see [26, Introduction] for more comments on
such elements and their properties.

With an algebraic group is associated the Lie algebra of $G$ denoted here by
$L(G)$. The group $G$ acts on $L(G)$ in the natural way making it a
$G$-module. If $G$ is simple (or reductive) then the non-zero weights of $T$
on $L(G)$ are called roots. We denote the set of roots by $\Phi$ or $\Phi(G)$.
The $\mathop{\mathbb{Z}}\nolimits$-span of $\Phi$ is called the root lattice
and denoted by $R$ or $R(G)$. In $\Phi(G)$ one can choose a
$\mathop{\mathbb{Z}}\nolimits$-basis $\Pi$ of $R$ whose elements are denoted
by $\alpha_{1},\ldots,\alpha_{n}$ and called simple roots (here $n$ is the
rank of $G$). The ordering of simple roots and the Dynkin diagrams of $G$ are
as in [3]. The weights in $R$ are called radical.

One defines a non-degenerate symmetric bilinear form on $\Omega$, which we
express as $(\mu,\nu)$. The elements $\omega_{i}$ satisfying
$(\omega_{j},\alpha_{i})=2(\alpha_{i},\alpha_{i})\delta_{ij}$ (the Kronecker
delta) belong to $\Omega$ and called fundamental weights [3, Ch. VI,§1,
no.10]. These form a $\mathop{\mathbb{Z}}\nolimits$-basis of $\Omega$ so every
$\nu\in\Omega$ can be expressed in the form $\sum a_{i}\omega_{i}$ with
$a_{1},\ldots,a_{n}\in\mathop{\mathbb{Z}}\nolimits;$ the set of $\nu$ with
$a_{1},\ldots,a_{n}\geq 0$ is denoted by $\Omega^{+}$. The weights of
$\Omega^{+}$ are called dominant. Suppose that $p>0$; a (dominant) weight
$(a_{1},\ldots,a_{n})$ is called $p$-restricted if $0\leq a_{i}<p$ for all
$i=1,\ldots,n$; the set of them is denoted by $\Omega^{+}_{p}$. For
uniformity, we often do not separate the cases with $p=0$ and $p>0$; by
convention, a $p$-restricted weight is simply a dominant weight when $p=0$. An
irreducible $G$-module is called $p$-restricted if so is the highest weight of
it.

The weight with $a_{1}=\cdots=a_{n}=0$ is denoted by $\omega_{0}$ and called
the zero weight (also the weight 0). We often simplify $\mu\neq\omega_{0}$ by
writing $\mu\neq 0$. We can (and sometimes do) write the elements of $\Omega$
as $(a_{1},\ldots,a_{n})$ and in some cases, especially when we compare
representations of $G(\mathop{\mathbb{C}}\nolimits)$ with $G(F)$, saying that
weights of $G(\mathop{\mathbb{C}}\nolimits)$ and $G(F)$ are the same means the
coincidence of the set of strings $(a_{1},\ldots,a_{n})$ for these groups. We
set $\Omega^{+}(V)=\Omega^{+}\cap\Omega(V)$, so $\Omega^{+}(V)$ is the set of
dominant weights of $V$.

There is a standard partial ordering of elements of $\Omega$; specifically,
for $\mu,\mu^{\prime}\in\Omega$ we write $\mu\preceq\mu^{\prime}$ and
$\mu^{\prime}\succeq\mu$ if and only if $\mu^{\prime}-\mu\in R^{+}$.
(Sometimes we write $\mu\prec\mu^{\prime}$ and $\mu^{\prime}\succ\mu$ to
indicate that $\mu\neq\mu^{\prime}$.) Every non-trivial irreducible $G$-module
has a weight $\omega$ such that $\mu\prec\omega$ for every $\mu\in\Omega(V)$
with $\mu\neq\omega$. This is called the highest weight of $V$. There is a
bijection between $\Omega^{+}$ and $\mathop{\rm Irr}\nolimits(G)$, so for
$\omega\in\Omega^{+}$ we denote by $V_{\omega}$ the irreducible $G$-module
with highest weight $\omega$. The zero weight space of $V$ is sometimes
denoted by $V^{T}$. In general, for $S\subseteq G$ we set $V^{S}=\\{v\in
V:sv=v$ for all $s\in S\\}$. $L(G)$ is a $G$-module called adjoint; it is
irreducible if $p=0$; if $p>0$ it may be reducible but has a highest weight
denoted here by $\omega_{a}$. We denote by $V_{a}$ the composition factor of
$L(G)$ with highest weight $\omega_{a}$.

Our notation for simple algebraic groups is standard, in particular those of
types $A_{n},B_{n},C_{n},D_{n}$ are called classical,
$E_{6},E_{7},E_{8},F_{4},G_{2}$ are called exceptional, the subscript is the
rank of the group. Here $n\geq 1$ for $A_{n}$, and we usually assume $n>1$ for
$B_{n},C_{n}$, and $n>3$ for $D_{n}$, due to isomorphisms $D_{3}\cong A_{3},$
$D_{2}\cong A_{1}\times A_{1}$ and $A_{1}\cong B_{1}\cong C_{1}$. For brevity
we write $G=A_{n}$ to say that $G$ is a simple simply connected algebraic
group of type $A_{n}$, and similarly for the other types. There is also an
isomorphism $B_{2}\cong C_{2}$, and we usually choose $C_{2}$ for the standard
representative of this algebraic group. (In Section 4 we allow to view $A_{3}$
as $D_{3}$ for uniformity of some statements.) If $p=2$ then for every $n$
there are surjective algebraic group homomorphisms $B_{n}\rightarrow C_{n}$
and $C_{n}\rightarrow B_{n}$ with trivial kernel (so these are abstract group
isomorphisms); for our purpose the choice between these two groups is not
essential, so we choose to work with $C_{n}$.

For an algebraic group $G$ of classical type the module (and the
representation) with highest weight $\omega_{1}$ is called natural. (We keep
this definition even for types $B_{n}$ with $p=2$.)

In our reasonings we regularly use so called ”Bourbaki weights”, which are
elements of a $\mathop{\mathbb{Z}}\nolimits$-lattice containing $\Omega(G)$
with basis $\varepsilon_{1},\varepsilon_{2},...$; the explicit expressions of
the fundamental weights and the simple roots of $G$ in terms of
$\varepsilon_{i}$’s is given in [3, Planches 1 – XIII].

If $\sigma:G\rightarrow G$ is a surjective algebraic group homomorphism and
$\phi$ is a representation of $G$ then the $\sigma$-twist $\phi^{\sigma}$ of
$\phi$ is defined as the mapping $g\rightarrow\phi(\sigma(g))$ for $g\in G$.
Of fundamental importance is the Frobenius mapping $Fr:G\rightarrow G$ arising
from the mapping $x\rightarrow x^{p}$ $(x\in F)$ when $p>0$. If $V$ is a
$G$-module then the modules $V^{Fr^{k}}$ ($k\geq 1$ an integer) are called
Frobenius twists of $V$; if $V$ is irreducible with highest weight $\omega$
then the highest weight of $V^{Fr^{k}}$ is $p^{k}\omega$ for some integer
$k\geq 0$.

Every $\omega=\sum a_{i}\omega_{i}\in\Omega^{+}$ has a unique $p$-adic
expansion $\omega=\lambda_{0}+p\lambda_{1}+\cdots+p^{k}\lambda_{k}$ for some
integer $k\geq 0$, where $\lambda_{0},\ldots,\lambda_{k}\in\Omega_{p}^{+}$ are
$p$-restricted dominant weights. By Steinberg’s tensor product theorem [21,
Theorem 41], if $G$ is simply connected then
$V_{\omega}=V_{\lambda_{0}}\otimes V_{\lambda_{1}}^{Fr}\otimes\cdots\otimes
V_{\lambda_{k}}^{Fr^{k}}$, which we refer as the Steinberg expansion of $V$.

A surjective algebraic group homomorphism $\sigma:G\rightarrow G$ is called
Steinberg if the subgroup $G^{\sigma}=\\{g\in G:\sigma(g)=g\\}$ is finite.
Equivalently, $\sigma$ is Steinberg if and only if $\sigma^{m}=Fr^{k}$ for
some integers $m,k>0$, see [16, p. 183]. If $G$ is semisimple and $\sigma$ is
Steinberg then $G^{\sigma}$ is called a finite group of Lie type.

Let $G$ be an algebraic group, $V$ a $G$-module and let $T$ be a maximal torus
of $G$. We say that $s\in T$ separates a weight $\omega$ of $V$ if the
multiplicity of the eigenvalue $\omega(s)$ equals the multiplicity of
$\omega$. In addition, we say that $s$ separates the weights of $V$ if every
eigenspace of $s$ on $V$ is a weight space. It is well known that $s$ is
regular if and only if it separates weight zero of the adjoint module. If $s$
separates all weights of the adjoint module, we say that $s$ is strictly
regular.

The set of eigenvalues of an element $g\in G$ on a $G$-module $V$
(disregarding the multiplicities) is denoted by $E_{V}(g)$. The notation
$m_{s}(V)$ is introduced in Theorem 1.2.

##  3 Preliminaries

In this section $G$ is a simple simply connected linear algebraic group over
an algebraically closed field of characteristic $p\geq 0$. For our purposes we
can assume that $G=G(F)$, the group of $F$-points of $G$, where $F$ is an
algebraically closed field of characteristic $p$.

Recall that a diagonalizable matrix is called _cyclic_ if all of its
eigenspaces are one-dimensional, and _almost cyclic_ if at most one of its
eigenspaces has dimension greater than $1$.

In Lemma 3.1 we denote by ${}^{t}M$ the transpose of a matrix $M$.

######  Lemma 3.1.

$(1)$ Let $M\in M_{r\times r}(F)$ be an almost cyclic diagonal matrix such
that $M$ is conjugate to ${}^{t}M^{-1}$ and let $e\in F$ be an eigenvalue of
$M$. If $e\notin\\{\pm 1\\}$, then the multiplicity of $e$ equals $1$.

$(2)$ Let $G$ be a semisimple algebraic group, $s\in G$ a semisimple element,
and V a G-module. Suppose that $V$ is self-dual and $s$ is almost cyclic on
$V$. Then all eigenvalues of $s$ on $V$, not equal to $1$ or $-1$, occur with
multiplicity $1$. In particular, this holds if $V$ is irreducible and $G$ is
simple of type $A_{1},B_{n},C_{n},D_{n}$ ($n$ even), $E_{n}$, $F_{4}$ or
$G_{2}$.

###### Proof.

The first claim is clear, whereas the second one follows from a well known
fact that every irreducible representation of $G$ listed is self-dual.∎

######  Lemma 3.2.

Let $M=M_{1}\otimes M_{2}$ be a Kronecker product of diagonal non-scalar
matrices $M_{1},M_{2}$ of sizes $m\leq n$, respectively.

$(1)$ The eigenvalue multiplicities of $M$ do not exceed $mt$, where t is the
maximal eigenvalue multiplicity of $M_{2}$.

$(2)$ If $M_{1},M_{2}$ are cyclic then the eigenvalue multiplicities of $M$ do
not exceed $m$.

$(3)$ Suppose that M is almost cyclic. Then $M_{1}$ and $M_{2}$ are cyclic.

$(4)$ Suppose that M is almost cyclic and $M_{i}$ is similar to $M_{i}^{-1}$
for $i=1,2$. Then the eigenvalue multiplicities of $M$ do not exceed $2$. In
addition, if $e$ is an eigenvalue of M of multiplicity $2$ then $e\in\\{\pm
1\\}$.

###### Proof.

(1) Let $e$ be an eigenvalue of $M$. Then $e=a_{i}b_{j}$, where $a_{i},b_{j}$
are some eigenvalue of $M_{1},M_{2}$, respectively. Then the multiplicity of
$e$ equals $\sum m(a_{i})m(b_{j})$, where $m(a_{i}),m(b_{j})$ is the
multiplicity of $a_{i},b_{j}$, respectively, and the sum ranges over the pairs
$a_{i},b_{j}$ with $e=a_{i}b_{j}$. We have $\sum m(a_{i})m(b_{j})\leq(\sum
m(a_{i}))\cdot\max_{j}m(b_{j})\leq mt$, as claimed.

(2) Suppose the contrary, and let $r>m$ be the multiplicity of some eigenvalue
$e$ of $M$. Let $a_{1},\ldots,a_{m}$, $b_{1},\ldots,b_{n}$ be the eigenvalues
of $M_{1}$, $M_{2}$, respectively. Then there exist $a_{i},b_{j},b_{k}$
$(1\leq i\leq m,1\leq j<k\leq n)$ such that $a_{i}b_{j}=a_{i}b_{k}$ and hence
$b_{j}=b_{k}$. This is a contradiction.

(3) Suppose that $M_{1}$ has an eigenvalue $a$, say, of multiplicity $r>1$.
Let $b_{1},b_{2}$ be distinct eigenvalues of $M_{2}$. Then $ab_{1},$ $ab_{2}$
are distinct eigenvalues of $M$, each of multiplicity greater than 1. So $M$
is not almost cyclic, thus establishing the claim.

(4) Suppose the contrary, and let $e$ be an eigenvalue of $M$ of multiplicity
at least 2. By (3), $M_{1}$ and $M_{2}$ are cyclic so $e=a_{i}b_{i}$ for
$i=1,2$ and some (distinct) eigenvalues $a_{i}$ of $M_{1}$ and $b_{i}$ of
$M_{2}$. Then $e^{-1}=a_{i}^{-1}b_{i}^{-1}$ is an eigenvalue of $M$ of the
same multiplicity as that of $e$. As $M$ is almost cyclic, $e=e^{-1}$ and
$e\in\\{\pm 1\\}$. Then $a_{1}b_{1}=a_{2}^{-1}b_{2}^{-1}$, whence
$a_{1}b_{2}=a_{2}^{-1}b_{1}^{-1}$. If $a_{2}^{-1}\neq a_{1}$ then $a_{1}b_{2}$
is an eigenvalue of $M$ of multiplicity at least 2; this is not equal to
$e=a_{1}b_{1}$, hence $s$ is not almost cyclic on $M$, a contradiction. So
$a_{2}=a_{1}^{-1}$. If the multiplicity of $e$ is at least 3, say,
$e=a_{3}b_{3}$ then, as above, this implies $a_{3}=a_{1}^{-1}$. Then
$a_{3}=a_{2}$, contrary to $(3)$. ∎

Note that the assertion (1) of Lemma 3.2 is known, see for instance [14, Lemma
3.7].

######  Lemma 3.3.

Let $V,M$ be non-trivial G-modules and $s\in T$.

$(1)$ If $s$ is cyclic on $V$ then $s$ is regular.

$(2)$ If $s$ is almost cyclic on $V\otimes M$ then $s$ is regular.

###### Proof.

(1) Let $\rho$ be the representation afforded by $V$. If $s$ is cyclic on $V$
then $C_{GL(V)}(\rho(s))$ is a torus. Then so is $C_{G}(s)^{\circ}$, giving
the claim.

(2) By Lemma 3.2(3), $s$ is cyclic on $V$, so the result follows from (1).∎

In Lemma 3.4 below and elsewhere, for $A,B\subset\Omega$ we define
$A+B=\\{a+b:a\in A,b\in B\\}$.

######  Lemma 3.4.

Let $V,V_{1},V_{2}$ be non-trivial G-modules. Let $s\in G$ be a non-central
semisimple element and assume that $s$ is almost cyclic on V.

$(1)$ Suppose that $V=V_{1}\otimes V_{2}$. Then the matrices of $s$ on $V_{1}$
and $V_{2}$ are cyclic, all weights of $V_{1}$ and $V_{2}$ are of multiplicity
$1$, and $s$ is regular.

$(2)$ Suppose that $\Omega(V_{1})+\Omega(V_{2})=\Omega(V)$. Then $s$ separates
the weights of $V_{1}$ and $V_{2}$.

$(3)$ Suppose that $\Omega(V_{1})+\Omega(V_{2})=\Omega(V)$ and the weights of
$V$ are of multiplicity $1$. Then the multiplicity of every eigenvalue of $s$
on $V$ does not exceed $|\Omega(V_{1})|$, the number of weights of $V_{1}$.

###### Proof.

Assertion (1) follows from Lemma 3.2, except the last claim, so suppose that
$s$ is not regular. Then $C_{G}(s)$ contains a unipotent element $u\neq 1$. As
$u$ stabilizes every eigenspace of $s$ on $V_{1}$, at least one of them is of
dimension greater than $1$, contradicting that $s$ is cyclic on $V_{1}$.

(2) Suppose the contrary, that the weights of $V_{1}$, say, are not separated
by $s$, so there exist distinct weights $\mu_{1},\mu_{2}\in\Omega(V_{1})$ such
that $\mu_{1}(s)=\mu_{2}(s)$. Then for every $\lambda,\mu\in\Omega(V_{2})$,
$\mu_{i}+\lambda,\mu_{i}+\mu\in\Omega(V)$ for $i=1,2$ and
$(\mu_{1}+\lambda)(s)=(\mu_{2}+\lambda)(s)$ and
$(\mu_{1}+\mu)(s)=(\mu_{2}+\mu)(s)$. This implies $\lambda(s)=\mu(s)$ as $s$
is almost cyclic on $V$. Since $\mu$ is an arbitrary weight of $V_{2}$, we
conclude that $s$ is scalar on $V_{2}$, and hence $s\in Z(G)$, a
contradiction.

(3) Suppose the contrary. Let $e$ be an eigenvalue of $s$ on $V$ whose
multiplicity exceeds $m=|\Omega(V_{1})|$. As all weight spaces of $V$ have
dimension at most one, there are at least $m+1$ distinct weights
$\sigma_{1},\ldots,\sigma_{m+1}$ of $V$ such that
$e=\sigma_{1}(s)=\cdots=\sigma_{m+1}(s)$. We have
$\sigma_{i}=\mu_{i}+\lambda_{i}$ for $\mu_{i}\in\Omega(V_{1})$,
$\lambda_{i}\in\Omega(V_{2})$ for $i=1,\ldots,m+1.$ By hypothesis,
$\mu_{i}=\mu_{j}$ for some $1\leq i\neq j\leq m+1$, which yields both
$\lambda_{i}\neq\lambda_{j}$ and $\lambda_{i}(s)=\lambda_{j}(s)$,
contradicting (2).∎

Obviously, $\Omega(V_{1})+\Omega(V_{2})=\Omega(V)$ if $V=V_{1}\otimes V_{2}$;
however in many instances the former equality holds for $V\neq V_{1}\otimes
V_{2}$, see Lemma 3.11(2) below. The next example illustrates the use of Lemma
3.4(3).

######  Example 3.5.

(1) Let $G=A_{n}$, $p=0$, $V_{1}=V_{\omega_{1}}$, $V_{2}=V_{m\omega_{1}}$,
$V=V_{(m+1)\omega_{1}}$. Then all weight multiplicities of $V$ are known to be
equal to 1 and $\Omega(V_{1})+\Omega(V_{2})=\Omega(V)$ by [4, Ch. VIII, §7,
Proposition 10]. So if $s$ is almost cyclic on $V$ then the multiplicity of
any eigenvalue of $s$ on $V$ is at most $n+1$.

(2) Let $G=A_{n}$, $V_{1}=V_{\omega_{1}}$, $V_{2}=V_{\omega_{n}}$ and
$V=V_{\omega_{1}+\omega_{n}}$. Then $\Omega(V)=\Omega(V_{1})+\Omega(V_{2})$.
As $V$ satisfies the hypotheses of Lemma 3.12(2), some $s$ is almost cyclic on
$V$. Here the zero weight of $V$ is of multiplicity $n$ if $n+1$ is not a
multiple of $p$, and $n-1$ otherwise. So the bound in Lemma 3.4(3) is close to
the best possible one. In Example 4.36 below we show that the bound is in fact
sharp.

For reader’s convenience we record here the following well known observations
on the natural modules for classical groups. Note that we add the case $n=3$
for $D_{n}$ for our convenience here.

######  Lemma 3.6.

[25, Lemma 6] Let $G\in\\{A_{n}$, $B_{n},p\neq 2$, $C_{n},n>1$, $D_{n},n\geq
3\\}$, $V=V_{\omega_{1}}$ and let $s\in G$ be a non-central semisimple
element.

$(1)$ If $s$ is cyclic on $V$, then $s$ is regular. For $G=A_{n}$ or $C_{n}$,
the converse is true.

$(2)$ If $G=B_{n}$, then $s$ is regular if and only if the multiplicity of the
eigenvalue $-1$ on $V$ is at most $2$ and the other eigenvalue multiplicities
are equal to $1$.

$(3)$ If $G=D_{n}$, then $s$ is regular if and only if the multiplicities of
the eigenvalues $1$ and $-1$ on $V$ are at most $2$ and the other eigenvalue
multiplicities are equal to $1$. In addition, if $s$ is not almost cyclic on
$V$, then $s$ is not almost cyclic on $V_{\omega_{2}}$.

$(4)$ If $s$ is regular, then $s$ is almost cyclic on $V$ unless
$G=D_{n},p\neq 2$ and $1,-1$ are eigenvalues of $s$ on V, each of multiplicity
$2$.

######  Lemma 3.7.

Let $G=B_{2}$, $p\neq 2$, and let $s\in T_{reg}$. Then s is cyclic on
$V_{\omega_{2}}$ and almost cyclic on $V_{\omega_{1}}$.

###### Proof.

This follows from [26, Lemma 3.10(1)]. (Note that there is an algebraic group
homomorphism $h:Spin_{5}(F)\rightarrow Sp_{4}(F)$; under this homomorphism the
natural module for $B_{2}$ has highest weight $\omega_{2}$ as a
$C_{2}$-module. The natural module for $C_{2}$ viewed as a module for $B_{2}$
has highest weight $\omega_{2}$.) ∎

The following lemma is frequently used with no explicit reference, especially
when $S$ consists of a single element.

######  Lemma 3.8.

[20, Ch. II, §4, Theorem 4.1] Let G be a connected semisimple algebraic group,
T a maximal torus and $S\subset T$ a subset. Then the following conditions are
equivalent:

$(1)$ $C_{G}(S)^{\circ}=T;$

$(2)$ $C_{G}(S)$ consists of semisimple elements;

$(3)$ $\alpha(S)\neq 1$ for every root $\alpha$ of $G$.

The following theorem proved in [25] is of fundamental significance for this
paper, this is used many times, often without an explicit reference.

######  Theorem 3.9.

[25, Theorem 1] Let $G$ be a simply connected simple algebraic group and
$\phi$ a non-trivial irreducible representation of $G$. Then the following
statements are equivalent:

$(1)$ there exists a non-central semisimple element $s\in G$ such that
$\phi(s)$ is almost cyclic;

$(2)$ All non-zero weights of $\phi$ are of multiplicity $1$.

Recall the notation $e(G)$, which denotes the maximum of the squares of the
ratios of the lengths of the roots in $\Phi(G)$. We will rely on the following
important result [18, Theorem 1]:

######  Theorem 3.10.

Assume $p=0$ or $p>e(G)$. Let $\lambda$ be a $p$-restricted dominant weight.
Then
$\Omega(V_{\lambda})=\\{w(\mu)\,|\,\mu\in\Omega^{+},\mu\preceq\lambda,w\in
W(G)\\}.$

An application of Theorem 3.10 then gives:

######  Lemma 3.11.

[26, Lemma 2.6] Assume $p=0$ or $p>e(G)$. Let $\lambda,\mu\in\Omega^{+}$ with
$\lambda$ $p$-restricted, and $V_{\lambda}$, respectively, $V_{\mu}$ the
associated irreducible $G$-modules. Then the following hold.

(1) If $\mu\prec\lambda$ then $\Omega(V_{\mu})\subseteq\Omega(V_{\lambda})$.

(2) If $\lambda+\mu$ is $p$-restricted then
$\Omega(V_{\lambda+\mu})=\Omega(V_{\lambda}\otimes
V_{\mu})=\Omega(V_{\lambda})+\Omega(V_{\mu})$.

(3) If $\lambda\neq 0$ is radical then some root is a weight of $V_{\lambda}$;
otherwise $\Omega(V_{\lambda})$ contains some minuscule weight.

A dominant weight $\nu\neq 0$ is called minuscule if $W\nu=\Omega(V_{\nu})$,
that is, the weights of $V_{\nu}$ form a single orbit under the Weyl group.
The minuscule weights are well known and tabulated in [4, Ch. VIII,§3].

######  Lemma 3.12.

Let V be a non-trivial $G$-module. Let $T^{\prime}\subset T$ be the subset of
all $t\in T$ that separate the weights of V.

(1) $T^{\prime}$ is nonempty Zarisky open in T.

(2) Suppose that at most one weight of $V$ has multiplicity greater than $1$.
Then all elements of $T^{\prime}$ are almost cyclic on V.

###### Proof.

$(1)$ Let $\mu,\nu$ be weights of $V$. Then $T_{\mu,\nu}:=\\{x\in
T\,|\,\mu(x)=\nu(x)\\}$ is a Zarisky closed subset $T_{\mu,\nu}$ of $T$. The
set of elements of $T$ that do not separate some pair of weights of $V$, being
the finite union of all $T_{\mu,\nu}$ with $\mu\neq\nu$, is a proper closed
subset of $T$. So $T^{\prime}=T\setminus(\cup T_{\mu,\nu})$ is nonempty open.

$(2)$ If $s\in T^{\prime}$ then $\mu(s)\neq\nu(s)$ whenever $\mu\neq\nu$ are
weights of $V$. Therefore, the eigenvalues of $s$ on $V$ are exactly $\mu(s)$,
where $\mu$ runs over the weights of $V$, and the multiplicity of $\mu(s)$
equals that of $\mu$, whence the claim. ∎

We often use the following lemma with no explicit reference.

######  Lemma 3.13.

$(1)$ Let V,$V_{1}$ be FG-modules (for some group G) that are Galois
conjugates of each other. Let $s\in G$ be a $p^{\prime}$-element. Then
$m_{s}(V)=m_{s}(V_{1})$.

$(2)$ Let G be an algebraic group and $s\in G$ a semisimple element. Let V be
a G-module and $V_{1}$ a Frobenius twist of V. Then $\dim V^{s}=\dim
V_{1}^{s}$ and $m_{s}(V)=m_{s}(V_{1})$.

###### Proof.

(1) is obvious. (2) A Frobenius twist, as it is defined in Section 2, arises
from a so called standard Frobenius mapping $Fr:G\rightarrow G$; the latter
can be defined in terms of the mapping
$h_{k}:(a_{ij})\rightarrow(a^{p^{k}}_{ij}))$ for every matrix $(a_{ij})\in
GL_{n}(F)$ as follows, see [5, §1.17]. A mapping $Fr:G\rightarrow G$ is called
standard Frobenius if there exists a faithful rational representation
$\rho:G\rightarrow GL_{n}(F)$ such that $\rho(Fr(g))=h_{k}(\rho(g))$ for every
$g\in G$. If $(a_{ij})$ is a diagonal matrix then
$h_{k}(a_{ij})=(a_{ij})^{p^{k}}$. As $s$ is semisimple, so is $\rho(s)$, and
hence $x\rho(s)x^{-1}$ is diagonal for some $x\in GL_{n}(F)$. So
$h_{k}(x\rho(s)x^{-1})=h_{k}(x)\rho(Fr(g))h_{k}(x)^{-1})=(x\rho(s)x^{-1})^{p^{k}}=x\rho(s^{p^{k}})x^{-1}$,
whence $\rho(Fr(s))$ is a conjugate of $\rho(s^{p^{k}})=\rho(s)^{p^{k}}$ for
every representation $\rho$. In particular, this holds for the representation
afforded by $V$. Since $\rho(s)$ is semisimple, the eigenvalue 1 multiplicity
of $\rho(s)^{p^{k}}$ coincides with that of $\rho(s)$. If $e_{1},e_{2}$ are
distinct eigenvalues of $\rho(s)$ then $e^{p^{k}}_{1},e_{2}^{p^{k}}$ are
distinct eigenvalues of $\rho(s)^{p^{k}}$. So the result follows. (It is
rather customary to use the term ”standard Frobenius” only in the case with
$k=1$; in the above reasoning it is convenient not to assume $k=1$.) ∎

Note that we cannot drop ”Frobenius” from the assumption in (2); indeed, let
$G$ be a simple algebraic group of type $D_{4}$, and the highest weights of
$V,V^{\prime}$ are $\omega_{1},\omega_{4}$, respectively. Then $V^{\prime}$ is
a twist of $V$ (by a graph automorphism of order 3) but, if $p\neq 3,5$ then
$m_{s}(V)=2$, $m_{s}(V^{\prime})=0$ for a suitable element of order 15.

######  Lemma 3.14.

Let $G$ be a simple algebraic group in characteristic p and $V$ a non-trivial
p-restricted irreducible $G$-module. Let $p=0$ or $p>e(G)$. If $0\in\Omega(V)$
then one of the following holds:

$(1)$ $\Phi(G)\subseteq\Omega(V);$

$(2)$ $V$ is the short root module;

$(3)$ $G$ is of type $C_{n}$ and $\omega=\omega_{i}$ for $i>2$ even.

###### Proof.

If all root lengths are the same then the result is known, see for instance
[22, Proposition 2.3]. So we are left with groups $G$ of type
$B_{n},C_{n},F_{4}$ or $G_{2}$, and we have to show that the maximal root
$\omega_{a}$ lies in $\Omega(V)$ unless (2) or (3) holds. We proceed case-by-
case. Let $\omega=\sum a_{i}\omega_{i}$ be the highest weight of $V$.

Let $G$ be of type $B_{n}$. Then $2\omega_{1}\succ\omega_{a}=\omega_{2}$ and
$2\omega_{n}\succ\omega_{n-1}\succ\omega_{n-2}\succ\cdots\succ\omega_{2}\succ\omega_{1}$,
see [3, Planche II]. As $\omega\succ 0$, the coefficient $a_{n}$ must be even.
Therefore, $\omega\succeq(a_{1}+\cdots+a_{n-1}+\frac{a_{n}}{2})\omega_{1}$. If
$a_{1}+\cdots+a_{n-1}+\frac{a_{n}}{2}>1$ then $\omega\succeq
2\omega_{1}\succ\omega_{2}$, as desired. If
$a_{1}+\cdots+a_{n-1}+\frac{a_{n}}{2}=1$ and $\omega\neq\omega_{1}$ then
$\omega\in\\{\omega_{2},\ldots,\omega_{n-1},2\omega_{n}\\}$. So the result
again follows.

Let $G$ be of type $C_{n}$. Then $\omega_{a}=2\omega_{1}$,
$\omega_{i}\succeq\omega_{2}$ for $i$ even, and $\omega_{i}\succeq\omega_{1}$
for $i$ odd. Therefore,
$\omega\succeq(\sum_{i\,\,odd}a_{i})\omega_{1}+(\sum_{i\,\,even}a_{i})\omega_{2}$.
Here $\sum_{i\,\,odd}a_{i}$ is even as $\omega\succ 0$; if this is non-zero
then the result follows. Otherwise, since $2\omega_{2}\succ 2\omega_{1}$, the
result again follows unless $\omega=\omega_{i}$ for $i$ even. Moreover,
$2\omega_{1}=\omega_{a}$ is not a subdominant weight of $\omega_{i}$ as one
can conclude by comparing the $\alpha_{1}$-coefficients of the expansions of
$\omega_{i}$’s in terms of simple roots in [3, Planche III].

Let $G$ be of type $F_{4}$. Then
$\omega_{2}\succ\omega_{3}\succ\omega_{1}=\omega_{a}\succ\omega_{4}$ and
$2\omega_{4}\succ\omega_{1}$, see [3, Planche XIII]. So
$\omega\succeq(a_{1}+a_{2}+a_{3}+a_{4})\omega_{4}\succeq
2\omega_{4}\succ\omega_{1}$, unless $a_{1}+a_{2}+a_{3}+a_{4}=1$. If
$a_{1}+a_{2}+a_{3}>0$ then the result follows, otherwise
$a_{1}=a_{2}=a_{3}=0$, $a_{4}=1$, that is, $\omega=\omega_{4}$ is a short
root.

Let $G$ be of type $G_{2}$. Then
$2\omega_{1}\succ\omega_{a}=\omega_{2}\succ\omega_{1}$ so
$\omega\succeq(a_{1}+a_{2})\omega_{1}$. If $a_{2}>0$ or $a_{2}=0,a_{1}\geq 2$
then the result follows, otherwise $\omega=\omega_{1}$, which is a short
root.∎

######  Lemma 3.15.

Let G be a simple algebraic group and V a G-module. Let $s\in T_{reg}$.
Suppose that $\mu(s)=\nu(s)$ for distinct weights $\mu,\nu\in\Omega(V)$ and
$\mu-\alpha,\nu-\alpha\in\Omega(V)$ for some $\alpha\in\Phi(G)$. Then s is not
almost cyclic on V.

###### Proof.

Suppose the contrary. Observe that $\mu(s)=\nu(s)$ implies
$(\mu-\alpha)(s)=(\nu-\alpha)(s)$. If $s$ is almost cyclic on $V$ then
$(\mu-\alpha)(s)=\mu(s)$, whence $\alpha(s)=1$. By Lemma 3.8(3), this is a
contradiction as $s$ is regular. ∎

######  Lemma 3.16.

Let $s\in G=SL_{n}(F)$ be a strictly regular element. Then $s$ is cyclic on
$V_{\omega_{2}}$.

###### Proof.

The weights of $V_{\omega_{2}}$ are $\varepsilon_{i}+\varepsilon_{l}$ for
$i<l$, so if $s$ is not cyclic on $V_{\omega_{2}}$ then
$(\varepsilon_{i}+\varepsilon_{l})(s)=(\varepsilon_{j}+\varepsilon_{k})(s)$
for some $i<l$, $j<k$ and $(i,l)\neq(j,k)$. It follows that
$(\varepsilon_{i}-\varepsilon_{j})(s)=(\varepsilon_{k}-\varepsilon_{l})(s)$.
As $\varepsilon_{i}-\varepsilon_{j}$ with $i\neq j$ is a root, and $s$ is
regular, we have $i\neq j$. Similarly, we have $k\neq l$. Then
$\varepsilon_{i}-\varepsilon_{j}$, $\varepsilon_{k}-\varepsilon_{l}$ are
distinct roots, which is a contradiction as $s$ is strictly regular. ∎

For reader’s convenience we quote here [25, Theorem 2]:

######  Theorem 3.17.

Let $G$ be a simply connected simple algebraic group defined over a field of
characteristic $p\geq 0$ and let $s\in G$ be a non-regular non-central
semisimple element. Let $V$ be a non-trivial irreducible $G$-module. Then
either $s$ is not almost cyclic on $V$ or one of the following holds:

(1) $G$ is classical, that is, of Lie type $A_{n},B_{n}\,\,(p\neq 2)$, $C_{n}$
or $D_{n}$ and $\dim V=n+1,2n+1,2n,2n$, respectively;

(2) $G=B_{n},p=2$ and $\dim V=2n;$

(3) $G=A_{3}$ and $\dim V=6;$

(4) $G=C_{2}\cong B_{2}$, $p\neq 2$ and $\dim V=5.$

##  4 Eigenvalue multiplicities: a proof of Theorem 1.2

In this section we bound the eigenvalue multiplicities of a regular semisimple
element $s\in G$ on a $G$-module $V$ provided $s$ is almost cyclic on $V$. Of
course, in this case there is at most one eigenvalue of multiplicity greater
than 1, so we bound the multiplicity of this single eigenvalue.

###  4.1 A uniform bound

######  Lemma 4.1.

Let G be a simple algebraic group in characteristic $p=0$ or $p>e(G)$, and let
$0\neq\omega\in\Omega^{+}$. If $p>0$, suppose that $\omega$ is not of the form
$p\omega^{\prime}$ for some $\omega^{\prime}\in\Omega^{+}$. Let $V=V_{\omega}$
and $\lambda\in\Omega(V)$. Then there is $\alpha\in\pm\Pi$ such that
$\lambda-\alpha\in\Omega(V)$.

###### Proof.

If $p=0$ then the result is true by [4, Ch. VIII, §7, Prop. 5]. If $p>e(G)$
and $\omega$ is $p$-restricted then the result follows from that for $p=0$ and
Lemma 3.11. Suppose that $\omega$ is not $p$-restricted. Then $V=M\otimes
V^{\prime}$, where $M,V^{\prime}$ are irreducible $G$-modules, and $M$ is
$p$-restricted. Every weight $\lambda$ of $V$ is of shape $\mu+\nu$ with
$\mu\in\Omega(M)$ and $\nu\in\Omega(V^{\prime})$. As $M$ is $p$-restricted, we
have $\mu-\alpha\in\Omega(V)$ for some $\alpha\in\pm\Pi$ by the above. Then
$\lambda-\alpha=(\mu-\alpha)+\nu\in\Omega(V)$, as required. ∎

######  Theorem 4.2.

Let G be a simple algebraic group of rank n in characteristic $p$. Assume
$p=0$ or $p>e(G)$. Let $S\subset T$ be a subgroup such that $C_{G}(S)=T$, and
$V$ an irreducible $G$-module with $p$-restricted highest weight $\omega\neq
0$. Assume in addition that some homogeneous component U of $V|_{S}$ is of
dimension greater than $1$ and all other homogeneous component of $V|_{S}$ are
$1$-dimensional. Then

$(1)$ if $V^{T}\neq 0$, then $V^{S}=V^{T}$ and $\dim V^{S}\leq n+1;$

$(2)$ $\dim U\leq 2n$.

###### Proof.

By [25, Theorem 5], all non-zero weights of $V$ are of multiplicity 1.
Therefore, $\omega$ appears in Tables 1,2, and in particular $\dim V^{T}\leq
n+1$. By Lemma 3.8 we have $\alpha(S)\neq 1$ for every $\alpha\in\Phi(G)$.

(1) Here we have $V^{T}\subseteq V^{S}$. Suppose the contrary, that $V^{T}\neq
V^{S}$, so $\dim V^{S}>1$. Note that $V^{S}$ is a sum of weight spaces of $V$,
so $\mu(S)=1$ for some non-zero weight of $V$.

Suppose first that $\omega$ is a root. Then all non-zero weights of $V$ are
also roots. (This is well known.) Since $\alpha(S)\neq 1$ for every
$\alpha\in\Phi(G)$, we conclude that $V^{S}=V^{T}$, a contradiction.

Next suppose that $\omega$ is not a root. By the above, $\mu$ is not a root.
By replacing $S$ by a conjugate if necessary, we may assume that $\mu$ is
dominant. By Lemma 4.1, $\mu+\beta$ is a weight of $V$ for some
$\beta\in\pm\Pi$. Then $\mu+\beta\neq\beta$, while $\mu+\beta$ and $\beta$
restricts to the same (non-zero) character of $S$. If $\beta$ is not a weight
of $V$ then $C_{G}(S)=T$ implies $\beta(S)=1$, a contradiction. So $\beta$ is
not a weight of $V$. But $0\prec\omega$ so Lemma 3.11(3) implies that some
root is a weight of $V$. We deduce that $\Phi(G)$ has two root lengths and
either the long roots or the short roots are not weights of $V$. As $p>e(G)$
if $p>0$, by Lemma 3.14, either $\omega$ is a short root, a contradiction, or
$G=C_{n}$ and $\omega=\omega_{i}$ with $i$ even. In the latter case
$\omega_{j}\in\Omega(V)$ for $i-j>0$ even, and these are the only dominant
weights of $\Omega(V)$. So $\mu=\omega_{r}$ with $i-r>0$ even. Here $r>2$ as
$\omega_{2}$ is a root. Then $r-2\neq 0$ and
$\omega_{r}-\omega_{r-2}=\varepsilon_{r}+\varepsilon_{r-1}\in\Phi(G)$. Set
$\beta=-\varepsilon_{r}-\varepsilon_{r-1}$. As $\beta$ is a short root, it
lies in the $W$-orbit of $\omega_{2}=\varepsilon_{1}+\varepsilon_{2}$, and
hence $\beta\in\Omega(V)$. However, by the above, we have a contradiction.

(2) In view of (1) we can assume that $\dim V^{T}\leq 1$, so all weights of
$V$ are of multiplicity 1. If $U=V^{S}$ then the claim follows from (1). So we
now assume that $U\cap V^{S}=0$ hence $U$ is the sum of $T$-weight spaces
which are of dimension $1$ by Theorem 3.9. Assume for a contradiction that
there exist distinct $T$-weights $\mu_{1},\dots,\mu_{2n+1}$ such that
$\mu_{j}|_{S}=\eta$ for some character $\eta$ of $S$ and all $1\leq j\leq
2n+1$. For each $j$ there exists $\gamma_{j}\in\pm\Pi$ such that
$\mu_{j}+\gamma_{j}$ is a weight of $V$. Then there exists $i\neq j$ such that
$\gamma_{i}=\gamma_{j}$. But then $\mu_{i}+\gamma_{i}$ and
$\mu_{j}+\gamma_{j}$ are distinct weights of $V$ and have equal restrictions
to $S$. By the hypothesis, this then means that
$\mu_{i}|_{S}=\eta|_{S}=(\mu_{i}+\gamma_{i})|_{S}$, whence $\gamma_{i}(S)=1$,
a contradiction. ∎

###  4.2 $G\neq A_{n}$: Special cases

A more detailed case-by-case analysis allows us to improve the bound $\dim
V_{S}(\eta)\leq 2n$ in Theorem 4.2 to $\dim V_{S}(\eta)\leq n+1$. We prefer to
assume $S=\langle s\rangle$ to be cyclic, the general case is similar. We
first perform this for groups $G\neq A_{n}$, and then consider the case of
$G=A_{n},n>1,$ which turns out to be more complex. We denote by $m_{s}(V)$ the
maximum of eigenvalue multiplicities of $s\in G$ on $V$. In this section we
obtain a sharp bound $m_{s}(V)\leq 2$ in a number of special cases.

Recall (Theorem 3.9) that the non-zero weights of an irreducible $G$-module
$V$ are of multiplicity 1 whenever a semisimple element $s\in G$ is almost
cyclic on $V$. Such modules are classified in [23] and the highest weights of
$V$, for $V$ $p$-restricted, are reproduced in Tables 1,2 at the end of this
paper.

######  Lemma 4.3.

Let $G=B_{n}$, $p\neq 2$, or $C_{n}$, $p=2$, and let $s\in T_{reg}$. Suppose
that s is almost cyclic on $V_{\omega_{n}}$. Then $m_{s}(V_{\omega_{n}})\leq
2$.

###### Proof.

The weights of $V$ are
$\frac{1}{2}(a_{1}\varepsilon_{1}+\cdots+a_{n}\varepsilon_{n})$ for $G=B_{n}$
and $a_{1}\varepsilon_{1}+\cdots+a_{n}\varepsilon_{n}$ for $G=C_{n}$, where
$a_{1},\ldots,a_{n}\in\\{\pm 1\\}.$ Let $\mu,\nu\in\Omega(V)$ be distinct
weights such that $\mu(s)=\nu(s)$. Note that $\pm\varepsilon_{i}$,
respectively, $\pm 2\varepsilon_{i}$ for $i=1,\ldots,n$ are roots of $G$ of
type $B_{n}$, respectively, $C_{n}$. If $\mu\neq-\nu$ then, obviously, there
is $i$ such that $\mu-\alpha,\nu-\alpha\in\Omega(V)$ for
$\alpha\in\\{\pm\varepsilon_{i}:1\leq i\leq n\\}$, respectively,
$\alpha\in\\{\pm 2\varepsilon_{i}:1\leq i\leq n\\}$. This contradicts Lemma
3.15.

Now suppose the contrary, and let $m_{s}(V_{\omega_{n}})>2$. Then there are
distinct weights $\lambda,\mu,\nu$ of $V$ such that
$\lambda(s)=\mu(s)=\nu(s)$. By the above, $\lambda=-\mu$, $\lambda=-\nu$ and
$\nu=-\mu$. This is a contradiction.∎

######  Lemma 4.4.

Let $G=B_{n}$, $n>2$, $p\neq 2$, or $C_{n}$, $p=2$, and let $V$ be a non-
trivial irreducible p-restricted G-module of highest weight $\omega$. Let
$s\in T_{reg}$. Suppose that s is almost cyclic on $V$. Then one of the
following holds:

$(1)$ $\omega\in\\{\omega_{1},\omega_{n}\\}$ and $m_{s}(V)\leq 2;$

$(2)$ $\omega=\omega_{2}$, $n>2$, and $n-1\leq m_{s}(V)\leq n;$

$(3)$ $G=B_{n},\omega=2\omega_{1}$ and $n\leq m_{s}(V)\leq n+1$.

###### Proof.

By Theorem 3.9, all non-zero weights of $V$ are of multiplicity 1, and hence,
by Theorem 3.9 and Tables 1,2,
$\omega\in\\{\omega_{1},\omega_{n},\omega_{2},2\omega_{1}\\}$, the latter for
$p\neq 2$. For $\omega=\omega_{1}$ see Lemma 3.6, for $\omega=\omega_{n}$ see
Lemma 4.3. If $\omega=\omega_{2}$, $n>2$, or $2\omega_{1}$ with $p\neq 2$ then
$\dim V^{T}\geq 2$ by Table 1. Then $m_{s}(V)=\dim V^{s}$, and $V^{T}=V^{s}$
by Theorem 4.2. So the result follows from Table 1. ∎

The case $G=B_{2}\cong C_{2}$, $p\neq 2$, will be considered later in Lemma
4.11.

######  Lemma 4.5.

Let $G$ be an algebraic group of type $D_{n},n>3$, and let $V=V_{\omega}$,
where $\omega\in\\{\omega_{1},\omega_{n-1},\omega_{n}\\}$. Let
$\mu,\nu\in\Omega(V)$. Then $\mu-\alpha,\nu-\alpha\in\Omega(V)$ for some root
$\alpha$ of G, unless one of the following holds:

$(1)$ $\mu=-\nu$ and either n is even or $\omega=\omega_{1};$

$(2)$ n is odd, $\omega\neq\omega_{1}$ and $\mu=-\nu\pm\varepsilon_{i}$ for
some $i\in\\{1,\ldots,n\\}$.

###### Proof.

Suppose that $\omega=\omega_{1}$. Then the weights of $V$ are
$\pm\varepsilon_{1},\ldots,\pm\varepsilon_{n}$. We can assume that
$\mu=\varepsilon_{i}$ for $i\in\\{1,\ldots,n\\}$. If $\nu=\varepsilon_{j}$
$(1\leq j\leq n,j\neq i)$ then set $\alpha=\varepsilon_{i}+\varepsilon_{j}$.
If $\nu=-\varepsilon_{j}$, $j\neq i$, then set
$\alpha=\varepsilon_{i}-\varepsilon_{j}$. Then
$\mu-\alpha,\nu-\alpha\in\Omega(V)$ in each case.

Suppose that $\omega=\omega_{n}$ or $\omega_{n-1}$. Then the weights of $V$
are $\frac{1}{2}(\sum a_{i}\varepsilon_{i})$ with $a_{i}\in\\{1,-1\\}$, and
the number of indices $i$ with $a_{i}=-1$ is even if $\omega=\omega_{n}$,
otherwise it is odd. Let $\mu=\frac{1}{2}(\sum a_{i}\varepsilon_{i})$,
$\nu=\frac{1}{2}(\sum b_{i}\varepsilon_{i})$ be weights of $V$, and let
$L=\\{i:a_{i}=b_{i}\\}$. If $|L|>1$ then
$\alpha=a_{i}\varepsilon_{i}+a_{j}\varepsilon_{j}\in\Phi(G)$ for $i,j\in L$,
$i\neq j$ and $\mu-\alpha,\nu-\alpha\in\Phi(G)$, as required.

Suppose that $|L|=1$. Let $i\in L,$ $j\notin L$, $1\leq j\leq n$. Then
$\mu+\nu=\pm\varepsilon_{i}$ and $n$ is odd (otherwise $-\nu\in\Omega(V)$ and
$\mu-(-\nu)=\pm\varepsilon_{i}$ is not a radical weight).

Suppose that $|L|=0$. Then $\mu=-\nu$ and $n$ is even. ∎

######  Lemma 4.6.

Let $G=D_{n},n>3,$ let $s\in T_{reg}$. Suppose that s is almost cyclic on
$V=V_{\omega_{i}}$ for $i\in\\{1,n-1,n\\}$. Then $m_{s}(V)\leq 2$.

###### Proof.

Suppose the contrary, and let $\mu_{1},\mu_{2},\mu_{3}$ be distinct weights of
$V$ such that $\mu_{1}(s)=\mu_{2}(s)=\mu_{3}(s)$. If
$\mu_{i}-\alpha,\mu_{j}-\alpha\in\Omega(V)$ for some root $\alpha$ and some
distinct $i,j\in\\{1,2,3\\}$ then the result follows from Lemma 3.15.

Suppose that for every distinct $i,j\in\\{1,2,3\\}$ and any root $\alpha$
either $\mu_{i}-\alpha\notin\Omega(V)$ or $\mu_{j}-\alpha\notin\Omega(V)$. By
Lemma 4.5, this implies that $n$ is odd and
$\mu_{i}+\mu_{j}=\pm\varepsilon_{k(i,j)}$ with $k(i,j)\in\\{1,\ldots,n\\}$ for
every pair $i,j\in\\{1,2,3\\}$ with $i\neq j$. If $k(1,2)\neq k(1,3)$ then
$\mu_{3}-\mu_{2}=(\mu_{1}-\mu_{2})-(\mu_{1}-\mu_{3})=\pm\varepsilon_{k(1,2)}\mp\varepsilon_{k(1,3)}$
is a root, a contradiction. So $k(1,2)=k(1,3)$, and similarly,
$k(1,2)=k(2,3)$. This means that $\mu_{i}+\mu_{j}=a(i,j)\varepsilon_{k}$ for
some fixed $k\in\\{1,\ldots,n\\}$ and $a(i,j)\in\\{1,-1\\}$ for every pair
$i,j\in\\{1,2,3\\}$ with $i\neq j$. By reordering of $\mu_{1},\mu_{2},\mu_{3}$
we can assume that $\mu_{1}+\mu_{2}=\mu_{1}+\mu_{3}$, whence
$\mu_{2}=\mu_{3}$, a contradiction. ∎

######  Lemma 4.7.

Let $G=D_{n},n>3,$ let M be an irreducible G-module with highest weight
$\omega\in\\{\omega_{1},\omega_{n},\omega_{n-1}\\}$ and let $V$ an arbitrary
irreducible G-module. Let $s\in T_{reg}$. Suppose that s is almost cyclic on
$M\otimes V$. Then $m_{s}(M\otimes V)\leq 2$.

###### Proof.

Suppose the contrary, and let $\lambda_{i}$, $i=1,2,3$ be distinct weights of
$M\otimes V$ such that $\lambda_{i}(s)=e\in E(M\otimes V)$. Then
$\lambda_{i}=\mu_{i}+\nu_{i}$, where $\mu_{i},\nu_{i}$ are weights of $M,V$,
respectively. By Lemma 3.2(3), $s$ is cyclic on $M$ and $V$. If (*) there
exists a root $\alpha$ of $G$ such that
$\mu_{i}-\alpha,\mu_{j}-\alpha\in\Omega(M)$ for some $i,j\in\\{1,2,3\\}$ then
$\mu_{i}-\alpha+\nu_{i},\mu_{j}-\alpha+\nu_{j}\in\Omega(V)$, and
$(\mu_{i}-\alpha+\nu_{i})(s)=(\mu_{j}-\alpha+\nu_{j})(s)=e\cdot\alpha(s)^{-1}$.
So the result follows from Lemma 3.15.

Suppose that (*) does not hold. In this case, by Lemma 4.5, $n$ is odd and
$\mu_{i}+\mu_{j}=\pm\varepsilon_{k(i,j)}$ with $k(i,j)\in\\{1,\ldots,n\\}$ for
every pair $i,j\in\\{1,2,3\\}$ with $i\neq j$. Then we repeat the argument in
Lemma 4.6 to get a contardiction.

∎

######  Lemma 4.8.

Let $G=C_{3}$, $p\neq 2$. Let $s\in T_{reg}$. Suppose that s is almost cyclic
on $V=V_{\omega_{3}}$. Then $m_{s}(V)\leq 2$.

###### Proof.

Note that every weight of $V$ has multiplicity 1 (see Table 2) and
$\Omega(V)=X\cup Y$, where
$X=\\{\pm\varepsilon_{1},\pm\varepsilon_{2},\pm\varepsilon_{3}\\}$ and
$Y=\\{\pm\varepsilon_{1}\pm\varepsilon_{2}\pm\varepsilon_{3}\\}$. Suppose the
contrary, and let $\mu_{i}$, $i=1,2,3$, be distinct weights of $V$ such that
$\mu_{i}(s)=e$. By Lemma 3.1, $e\in\\{\pm 1\\}$. If $\mu_{i}\in X$ for some
$i$ then $(-\mu_{i})(s)=e$, which is false as $X=\Omega(V_{\omega_{1}})$ and
then $s$ is not regular by Lemma 3.6(1). So $\mu_{i}\in Y$ for every
$i=1,2,3$. Then
$\mu_{i}=a_{i1}\varepsilon_{1}+a_{i2}\varepsilon_{2}+a_{i3}\varepsilon_{3}$
for $a_{ik}\in\\{\pm 1\\}$ for $1\leq i,k\leq 3$. By reordering $\mu_{i}$ we
can assume $a_{11}=a_{21}$, and then
$\mu_{1}-\alpha,\mu_{2}-\alpha\in\Omega(V)$ for $\alpha\in\\{\pm
2\varepsilon_{1}\\}$. This contradicts Lemma 3.15. ∎

######  Lemma 4.9.

Let $G=C_{n}$, $n>2$, $p=3$, and $\omega\in\\{\omega_{n-1},\omega_{n}\\}$. Let
$s\in T_{reg}$ be almost cyclic on $V_{\omega}$. Then $m_{s}(V)\leq 2$.

###### Proof.

The non-zero weights of $V$ are
$\pm\varepsilon_{i_{1}}\pm\cdots\pm\varepsilon_{i_{r}}$ for
$r\in\\{1,\ldots,n\\}$, where $r-n$ is odd if $\omega=\omega_{n-1}$, and even
if $\omega=\omega_{n}$.

Suppose the contrary, let $m_{s}(V)>2$ and let $\lambda,\mu\in\Omega(V)$,
$\lambda(s)=\mu(s)=e$, $\lambda\neq\mu$, where we can assume that
$\lambda\neq-\mu$. By Lemma 3.1, $e\in\\{\pm 1\\}$. Set $\lambda=\sum
a_{i}\varepsilon_{i}$, $\mu=\sum b_{i}\varepsilon_{i}$, where
$a_{i},b_{i}\in\\{0,1,-1\\}$.

Observe that $\lambda,\mu$ are non-zero. Indeed, otherwise,
$\lambda(s)=\mu(s)=1$ so $\dim V^{s}\geq 2$. Then $m_{s}(V)=\dim V^{s}$ and
$\dim V^{T}=\dim V^{s}$ by Theorem 4.2(1). However, $\dim V^{T}=1$, which is a
contradiction.

If $a_{i}=b_{i}\neq 0$ for some $i$ then $\lambda-2a_{i}\varepsilon_{i}$,
$\mu-2a_{i}\varepsilon_{i}\in\Omega(V)$ and
$(\lambda-2a_{i}\varepsilon_{i})(s)=(\mu-2a_{i}\varepsilon_{i})(s)$. As
$2a_{i}\varepsilon_{i}$ is a root, we have a contradiction by Lemma 3.15.

If $a_{i}=-b_{i}\neq 0$ for some $i$ then we replace $\lambda$ by $-\lambda$;
as $e\in\\{\pm 1\\}$, we have $(-\lambda)(s)=\lambda(s)$, so we arrive at the
case considered above.

So $a_{i}b_{i}=0$ for every $i=1,\ldots,n$. Using the Weyl group, we can
assume that
$\lambda=\varepsilon_{1}+\cdots+\varepsilon_{r-1}-\varepsilon_{r}$,
$\mu=\varepsilon_{r+1}+\cdots+\varepsilon_{r^{\prime}}$ for some
$0<r<r^{\prime}\leq n$. Then
$\lambda+(\varepsilon_{r}-\varepsilon_{r+1}),\mu+(\varepsilon_{r}-\varepsilon_{r+1})\in\Omega(V)$,
and
$(\lambda+(\varepsilon_{r}-\varepsilon_{r+1}))(s)=(\mu+(\varepsilon_{r}-\varepsilon_{r+1}))(s)$.
As above, this contradicts Lemma 3.15. ∎

######  Lemma 4.10.

Let $G=C_{n}$, $n\geq 2,p>3$, and
$\omega\in\\{\omega_{n-1}+\frac{p-3}{2}\omega_{n},\frac{p-1}{2}\omega_{n}\\}$.
Let $s\in T_{reg}$ be almost cyclic on $V_{\omega}$. Then
$m_{s}(V_{\omega})\leq 2$.

###### Proof.

The weights of $V_{\omega}$ are
$l_{1}\varepsilon_{1}+\cdots+l_{n}\varepsilon_{n}$, where $-(p-1)/2\leq
l_{1},\ldots,l_{n}\leq(p-1)/2$, see [27, proof of Lemma 1.6]; moreover, if
$\mu$ is of this form then $\mu$ is a weight of $V_{\omega}$ with
$\omega\in\\{\omega_{n-1}+\frac{p-3}{2}\omega_{n},\frac{p-1}{2}\omega_{n}\\}$.
Let $\lambda,\mu,\nu$ be three distinct weights of $V_{\omega}$ such that
$\lambda(s)=\mu(s)=\nu(s)$. As $(-\lambda)(s)=(-\mu)(s)$, we have
$\lambda(s)=\mu(s)=\nu(s)\in\\{\pm 1\\}$. Let

$\lambda=l_{1}\varepsilon_{1}+\cdots+l_{n}\varepsilon_{n},\,\,\,\,\mu=m_{1}\varepsilon_{1}+\cdots+m_{n}\varepsilon_{n},\,\,\,\,\nu=k_{1}\varepsilon_{1}+\cdots+k_{n}\varepsilon_{n}.$

By Lemma 3.15, either $\lambda-2\varepsilon_{i}$ or $\mu-2\varepsilon_{i}$ is
not a weight of $V_{\omega}$ for every $i=1,\ldots,n.$ Observe that
$\lambda-2\varepsilon_{i}$, respectively, $\mu-2\varepsilon_{i}$ is a weight
of $\Omega(V_{\omega})$ unless $l_{i}\leq 1-\frac{p-1}{2}$, respectively, or
$m_{i}\leq 1-\frac{p-1}{2}$, and similarly for the pairs $\lambda,\nu$ and
$\mu,\nu$. So there are at least two pairs with $i$-th coefficients at most
$1-\frac{p-1}{2}$. By reordering $\lambda,\mu,\nu$ we can assume that
$l_{i},m_{i}\leq 1-\frac{p-1}{2}$. Then $-\lambda,-\mu$ are weights
$V_{\omega}$, $(-\lambda)(s)=(-\mu)(s)$ and
$-l_{i},-m_{i}\geq\frac{p-1}{2}-1$. Then the $i$-th coefficient of
$-\lambda-2\varepsilon_{i}$ equals
$-l_{i}-2\geq\frac{p-1}{2}-3\geq-\frac{p-1}{2}$ for $p>3$. So
$-\lambda-2\varepsilon_{i}\in\Omega(V_{\omega})$; similarly,
$-\mu-2\varepsilon_{i}\in\Omega(V_{\omega})$. Then Lemma 3.15 yields a
contradiction. ∎

######  Lemma 4.11.

Let $G=B_{2}\cong C_{2}$, $s\in T_{reg}$ and let $V$ be a non-trivial
irreducible p-restricted G-module of highest weight $\omega$. If s is almost
cyclic on $V$ then $m_{s}(V)\leq 2$.

###### Proof.

Let $G=C_{2}$. As $s$ is almost cyclic on $V$, $\omega$ occurs in Tables 1,2.
For those in Table 1 we have $\dim V^{T}=2$, and hence $m_{s}(V)=\dim
V^{s}=\dim V^{T}$ by Theorem 4.2; then $\dim V^{T}=2$ by Table 1. Those in
Table 2 are $\omega_{1},\omega_{2}$ and, for $p>3$, also
$\omega_{1}+\frac{p-3}{2}\omega_{2},\frac{p-1}{2}\omega_{2}$. In the two last
cases the result is contained in Lemma 4.10. The module $V_{\omega_{1}}$ is
the natural module for $G$, and $m_{s}(V)=1$ by Lemma 3.6(1). The module
$V_{\omega_{2}}$ is the natural module for $B_{2}\cong C_{2}$, so the result
follows from Lemma 3.6(2). ∎

######  Remark 4.12.

If $\dim V=4$ then $m_{s}(V)=1$. Indeed, if $G=C_{2}$ then either $V$ is a
Frobenius twist of the natural module or $p=2$ and $V$ is a Frobenius twist of
$V_{\omega_{2}}$. The former case is ruled out as $s$ is regular (see Lemma
3.6), the latter case can be checked straightforwardly. (Say, let
$s=\mathop{\rm diag}\nolimits(a,b,b^{-1},a^{-1})\in Sp_{4}(q)$, where the
diagonal entres are distinct; then the eigenvalues of $s$ on $V_{\omega_{2}}$
are $a^{\pm 1}b^{\pm 1}$; if $m_{s}(V)>0$ then $a^{i}b^{j}=1$ for some
$i,j\in\\{\pm 1\\}$, which is a contradiction.) If $G=B_{2}$ then
$\omega=p^{k}\omega_{2}$; as $V_{\omega_{2}}$ is the natural module for
$C_{2}$, it follows that $m_{s}(V)=1$ for a regular element $s\in G$.

ptB$\omega_{1}$pt1Cpt3DFpt2EHpt2GIJpt4Kpt4Npt3MIpt2Ppt3Lpt3RTQpt1Opt1Spt1Upt1Vpt1Wpt3Xpt3Ypt4Zpt5Apt6Ept6456655442662525

Figure 1. Weight diagram for $V_{\omega_{1}}$ for $E_{6}$

######  Lemma 4.13.

Let $G=E_{6}$ and let $\mu_{1},\mu_{2},\mu_{3},\mu_{4}$ be distinct weights of
$V=V_{\omega_{1}}$. Then $\mu_{i}-\alpha,\mu_{j}-\alpha\in\Omega(V)$ for some
root $\alpha$ and $i,j\in\\{1,2,3,4\\}$, $i\neq j$.

###### Proof.

We use the weight diagram of $V$ given on Figure 1 above. (This is a more
precise version of a diagram provided in [17].) The weights of $V$ are shown
as the nodes denoted by capital letters $A,\ldots,Z$ and $\omega_{1}$, and two
nodes are linked with an arrow with label $i$ if the second one is obtained
from the first one by substracting simple weight $\alpha_{i}$. For instance,
$F-\alpha_{5}=H$, $Z-\alpha_{6}=A$.

All weights of $V_{\omega_{1}}$ lie in the Weyl group orbit of $\omega_{1}$,
so we can assume that $\mu_{1}=H$. Suppose the contrary.

Let $\Phi=\Phi(G)$. For a weight $\lambda$ of $V$ set
$[\lambda,\Phi]=\\{\nu\in\Omega(V):\lambda-\alpha,\nu-\alpha\in\Omega(V)$ for
some $\alpha\in\Phi\cup 0\\}$. In addition, for $\alpha\in\Phi$ we set
$[\lambda,\alpha]=\lambda\cup\\{\nu\in\Omega(V):\lambda-\alpha,\nu-\alpha\in\Omega(V)\\}$.
So $[\lambda,\Phi]=\cup_{\alpha\in\Phi}[\lambda,\alpha]$. Note that
$\lambda\in[\lambda,\Phi]$.

One can easily observe, using the weight diagram, that $[H,\Phi]$ consists of
17 weights, and the remaining weights are
$\Delta_{1}=\\{D,G,K,N,Q,R,U,W,Y,A\\}$. (It is less obvious that
$B,X\in[H,\Phi]$; however, $B\in[H,\beta]$, where
$\beta=\alpha_{3}+\alpha_{4}+\alpha_{5}+\alpha_{6}\in\Phi$. Similarly,
$Z\in[H,-\gamma]$, where
$\gamma=-\alpha_{3}-\alpha_{4}-\alpha_{5}-\alpha_{2}\in\Phi$.)

It follows that $\mu_{2},\mu_{3},\mu_{4}\in\Delta_{1}$. We show that
$\mu_{i}-\alpha,\mu_{j}-\alpha\in\Omega(V)$ for some root $\alpha$ and
$i,j\in\\{2,3,4\\}$, $i\neq j$.

Suppose first that $R,Q\in\\{\mu_{2},\mu_{3},\mu_{4}\\}$. Then we can assume
that $R=\mu_{2},Q=\mu_{3}$. Then $\Delta_{1}\subset[R,\Phi]\cup[Q,\Phi]$, so
the result follows.

So we assume that either $R$ or $Q$ are not in
$\\{\mu_{2},\mu_{3},\mu_{4}\\}$. Observe that $D\succ G\succ K\succ N\succ
R\succ U\succ W\succ Y\succ A$ and $D\succ G\succ K\succ N\succ Q\succ U\succ
W\succ Y\succ A$. Therefore, we can assume that
$\mu_{2}\succ\mu_{3}\succ\mu_{4}$.

Suppose that $\mu_{4}=A$. As $G,K,N,R,U,W,Y\in[A,\Phi]$, we have
$\mu_{2}=D,\mu_{3}=Q$, a contradiction as $D\in[Q,\alpha_{5}]$.

Suppose that $\mu_{4}=Y$. As $D,K,N,Q,W\in[Y,\Phi]$, we observe that
$\mu_{2},\mu_{3}\in\\{G,R,U,W\\}$. However,
$G-\alpha_{2},R-\alpha_{2},U-\alpha_{2},W-\alpha_{2}\in\Omega(V)$, which
yields a contradiction.

Suppose that $\mu_{4}=W$. As $D,G,N,Q,R,U\in[W,\Phi]$, we conclude that
$\mu_{2},\mu_{3}\in\\{G,K,R,U\\}$. Since
$G-\alpha_{2},R-\alpha_{2},U-\alpha_{2}\in\Omega(V)$, we have a contradiction.

Suppose that $\mu_{4}=U$. As $D,K,R,G\in[U,\Phi]$, we get a contradiction.

Suppose that $\mu_{4}=R$. We have $D,G,K,N\in[R,\Phi]$ so
$\mu_{2}=D,\mu_{3}=Q$, a contradiction.

Suppose that $\mu_{4}=Q$. We have $D,G,K,N\in[Q,\Phi]$ so
$\mu_{2}=D,\mu_{3}=R$, a contradiction.

Finally, suppose that $\mu_{4}\in\\{K,N\\}$. Then $[K,\Phi]$ and $[N,\Phi]$
contain $D,G,K,N$; this yields a contradiction. ∎

######  Lemma 4.14.

Let $G=E_{6}$, $V=V_{\omega_{1}}$ and let $M$ be an irreducible $G$-module.
Let $s\in T_{reg}$. Suppose that s is almost cyclic on $M\otimes V$. Then
$m_{s}(M\otimes V)\leq 3$.

###### Proof.

Suppose the contrary, and let $e$ be an eigenvalue of $s$ on $V$ of
multiplicity at least 4.

(i) Suppose first that $M$ is trivial. Let $\mu_{i}$, $i=1,2,3,4$, be distinct
weights of $V$ such that $\mu_{1}(s)=\mu_{i}(s)$ for every $i$. By Lemma 4.13,
there are distinct $i,j\in\\{1,2,3,4\\}$ and $\alpha\in\Phi(G)$ such that
$\mu_{i}-\alpha,\mu_{j}-\alpha\in\Omega(V_{\omega_{1}})$. This contradicts
Lemma 3.15.

(ii) Suppose that $M$ is not trivial. By Lemma 3.2(3), $s$ is cyclic on $M$
and $V$. Then $e=a_{i}b_{i}$ for $i=1,2,3,4$, where $a_{1},a_{2},a_{3},a_{4}$
are distinct eigenvalues of $s$ on $V$ and $b_{1},b_{2},b_{3},b_{4}$ are
distinct eigenvalues of $s$ on $M$. We have $a_{i}=\mu_{i}(s)$ for some
distinct weights $\mu_{1},\mu_{2},\mu_{3},\mu_{4}$ of $V$. As above, there is
a root $\alpha\in\Phi$ and $i,j\in\\{1,2,3,4\\}$, $i\neq j$, such that
$\mu_{i}-\alpha,\mu_{j}-\alpha\in\Omega(V_{\omega_{1}})$. Then
$c_{i}=(\mu_{i}-\alpha)(s)$ and $c_{j}:=(\mu_{j}-\alpha)(s)$ are eigenvalues
of $s$ on $V=V_{\omega_{1}}$, which are distinct as $s$ is cyclic on $V$. In
addition, $c_{i}\neq a_{i}$, $c_{j}\neq a_{j}$ as otherwise $\alpha(s)=1$,
which is false as $s$ is regular. So $c_{i}b_{i}=c_{j}b_{j}\neq
a_{i}b_{i}=a_{j}b_{j}$. This is a contradiction as $s$ is almost cyclic on
$M\otimes V$.∎

The following example shows that the bound in Lemma 4.14 is attained.

######  Example 4.15.

Let $G=E_{6}$. Then $G$ contains a subgroup $H$, say, of type $F_{4}$. Let
$\eta_{1},\eta_{2},\eta_{3},\eta_{4}$ be the fundamental weights of $F_{4}$.
Let $V$ be an irreducible $G$-module of highest weight $\omega_{1}$. It is
well known that $V|_{H}$ has a composition factor $U$ of highest weight
$\eta_{4}$ whose dimension is $\dim V-1$ or $\dim V-2$, the latter for $p=3$.
Then the other composition factors of $V_{\omega_{1}}|_{H}$ are trivial. Let
$s\in F_{4}$ be a regular semisimple element in general position, that is, the
cyclic group $\langle s\rangle$ is dense in a maximal torus $S$, say, of
$F_{4}$. Then $s$ is regular and separates the weights of $U$, and hence $s$
is almost cyclic on $V$. In addition, $\dim U^{s}=\dim V^{s}=2$ or 1, the
latter for $p=3$. It follows that $\dim V^{s}=3$.

######  Lemma 4.16.

Let $G=E_{6}$, let $V$ be an irreducible $G$-module and $s\in T_{reg}$.
Suppose that s is almost cyclic on $V$. Then $m_{s}(V)\leq 3$ unless the
highest weight of $V$ is $p^{t}\omega_{2}$.

###### Proof.

Let $\omega$ be the highest weight of $V$. Recall that the non-zero weights of
$V$ must be of multiplicity 1 (Theorem 3.9) and $s$ is regular by Theorem
3.17. Therefore, either $\omega=p^{t}\omega^{\prime}$, where
$\omega^{\prime}\in\\{\omega_{1},\omega_{6},\omega_{2}\\}$ by [23, Theorem 2]
or $V=V^{\prime}\otimes M$, where $M$ is a non-trivial irreducible $G$-module
and the highest weight of $V^{\prime}$ is $p^{t}\omega_{1}$ or
$p^{t}\omega_{6}$ (here $\omega^{\prime}\neq\omega_{2}$ by Lemma 3.2(3)). Note
that the irreducible representations with highest weights $p^{t}\omega_{1}$
and $p^{t}\omega_{6}$ are dual to each other. So the result follows from Lemma
4.14. (If $\omega^{\prime}=\omega_{2}$ then $m_{s}(V)=6$ by Theorem 4.2(1) and
the fact that $V_{\omega_{2}}$ is the adjoint $G$-module.) ∎

######  Lemma 4.17.

Let $G=E_{7}$, $V=V_{\omega_{7}}$ and $s\in T_{reg}$. Suppose that $s$ is
almost cyclic on $V$. Then $m_{s}(V)\leq 2$.

###### Proof.

Let $H$ be a subgroup of $G$ isomorphic to a simple algebraic group of type
$A_{7}$, so the maximal tori of $H$ are maximal in $G$. Hence we can assume
that $T\subset H$. Let $\eta_{1},\ldots,\eta_{7}$ be the fundamental weights
of $H$. Then the restriction $V|_{H}$ is $V_{\eta_{2}}+V_{\eta_{6}}$. The
result follows from the following lemma: ∎

######  Lemma 4.18.

Let H be a simple algebraic group of type $A_{n-1}$, $n\geq 4$. Let
$U:=V_{\eta_{2}}\oplus V_{\eta_{n-2}}$. Let $s\in T_{reg}$. Suppose that $s$
is almost cyclic on U. Then $m_{s}(U)\leq 2$.

###### Proof.

We can assume that $g=\mathop{\rm diag}\nolimits(a_{1},\ldots,,a_{n})\in
SL_{n}(F)$, where $a_{1},\ldots,a_{n}\in F$ and $a_{1}\cdots a_{n}=1$. Then
the eigenvalues of $s$ on $U$ are $a_{i}a_{j},a_{k}^{-1}a_{l}^{-1}$ for $1\leq
i,j,k,l\leq n$ and $i<j,k<l$. Suppose that $s$ is not cyclic on $U$. Then
either $a_{i}a_{j}=a_{k}a_{l}$ for $(i,j)\neq(k,l)$ or
$a_{i}a_{j}=a_{k}^{-1}a_{l}^{-1}$ for some $i,j,k,l$. Then $(a_{i}a_{j})^{-1}$
is an eigenvalue of $s$ of multiplicity at least 2, so
$a_{i}a_{j}=(a_{i}a_{j})^{-1}$ as $s$ is almost cyclic. Then
$a_{i}^{2}=a_{j}^{-2}$ and $a_{j}=\theta a_{i}^{-1}$ with $\theta=1$ or $-1$.
It follows that $\theta$ is an eigenvalue of $s$ on $U$ of multiplicity at
least 2. We show that the multiplicity of $\theta$ is exactly 2. Suppose that
contrary. Then $\theta=a_{k}a_{l}$ for some $k<l$ with $(k,l)\neq(i,j)$. Then
$a_{j}a_{l}=a_{i}^{-1}a_{k}^{-1}$ and these are eigenvalues of $s$ on $U$. As
$s$ is almost cyclic, we must have
$a_{i}^{-1}a_{k}^{-1}=a_{j}a_{l}=\theta=a_{k}a_{l}=a_{k}^{-1}a_{l}^{-1}$,
whence $a_{j}=a_{k},a_{i}=a_{l}$. As $s$ is regular, we have $j=k,i=l$. This
is a contradiction. ∎

######  Lemma 4.19.

Let $G=E_{7}$, let $V$ be an irreducible $G$-module and $s\in T_{reg}$.
Suppose that s is almost cyclic on $V$. Then $m_{s}(V)\leq 2$ unless the
highest weight of $V$ is $p^{t}\omega_{1}$.

###### Proof.

Let $\omega$ be the highest weight of $V$. The non-zero weights of $V$ are of
multiplicity 1 (Theorem 3.9) and $s$ is regular by Theorem 3.17. Therefore,
either $\omega\in\\{p^{t}\omega_{7},p^{t}\omega_{1}\\}$ by [23, Theorem 2] or
$V=V^{\prime}\otimes M$, where $M$ is a non-trivial irreducible $G$-module and
the highest weight of $V^{\prime}$ is $p^{t}\omega_{7}$ (this cannot be
$p^{t}\omega_{1}$ by Lemma 3.2(3)). So the result follows from Lemma 4.17. (If
$\omega=p^{t}\omega_{1}$ then $m_{s}(V)=7$ by Theorem 4.2(1) and the fact that
$V_{\omega_{1}}$ is the adjoint $G$-module.) ∎

######  Lemma 4.20.

Let $G=F_{4}$, $p=3$, let $V$ be a p-restricted irreducible $G$-module with
highest weight $\omega_{4}$ and $s\in T_{reg}$. Suppose that s is almost
cyclic on $V$. Then $m_{s}(V)\leq 2$.

###### Proof.

In this case $\dim V^{T}=1$ and hence $V^{s}=V^{T}$ by Theorem 4.2. Therefore,
if $m_{s}(V)>1$ and $e$ is the eigenvalue of $s$ on $V$ of multiplicity
$m_{s}(V)$ then $e=-1$ by Lemma 3.1(2).

The non-zero weights of $V=V_{\omega_{4}}$ are short roots, which form a
single orbit of the Weyl group of $G$. So we can assume that
$\omega_{4}(s)=-1=(-\omega_{4})(s)$. Let $\mu\succ 0$ for $\mu\in\Omega(V)$.
One can check that $\omega_{4}-\mu$ is a root. If $m_{s}(V)>2$ then
$\mu(s)=-1=(-\mu)(s)$ for some $0\neq\mu\in\Omega(V)$, $\mu\neq\pm\omega_{4}$.
So we can assume that $\mu\succ 0$. Then $\omega_{4}-\mu$ is a root and
$(\omega_{4}-\mu)(s)=1$, This is a contradiction as $s$ is regular.∎

The case of $G=G_{2}$ is essentially treated in [26]:

######  Lemma 4.21.

[26, Proposition 4.9] Let $G=G_{2}$, $V_{i}=V_{\omega_{i}}$ for $i=1,2$ and
let $s\in T_{reg}$.

$(1)$ Suppose that s is not cyclic on $V_{1}$. Then $p\neq 2$ and the
eigenvalues of s on $V_{1}$ are $\\{1,-1,-1,-b,b^{-1},b,-b^{-1}\\}$, where
$b\in F,b^{2}\neq 1$. In particular, s is almost cyclic on $V_{1}$ if and only
if $b^{2}\neq-1$. In addition, $\\{\alpha(s):\alpha$ a long
root$\\}=\\{b,b^{-1},-b,-b^{-1},b^{2},b^{-2}\\}$.

$(2)$ Let $p=3$. Then s is almost cyclic on $V_{2}$ and cyclic on $V_{1}$.
More precisely, if s is not cyclic on $V_{2}$ then the eigenvalues of s on
$V_{2}$ are $\\{1,-1,-1,a^{3},-a^{3},a^{-3},-a^{-3}\\}$ for some $a\in
F,a^{4}\neq 1$, and those on $V_{1}$ are
$\\{1,a,a^{-1},-a,-a^{-1},-a^{2},-a^{-2}\\}$.

######  Corollary 4.22.

Theorem 1.2 is true for G of type $G_{2}$.

###### Proof.

Suppose first that $V$ is $p$-restricted. Then, by Theorem 3.9, we need to
deal with $FG$-modules $V$ of highest weight
$\omega\in\\{\omega_{1},\omega_{2}\\}$. The case $\omega=\omega_{1}$ is
treated in Lemma 4.21(1). Let $\omega=\omega_{2}$. The case with $p=3$ is
examined in Lemma 4.21(2). Let $p\neq 3$. Then $\dim V^{T}=2$ and hence $\dim
V^{s}\geq 2$. If $\dim V^{s}>2$ then $s$ is not regular as the non-zero
weights of $V$ are roots. This contradicts Theorem 3.17.

Consider then the case where $V$ is not $p$-restricted. Then the result
follows from Lemma 3.2(4).∎

###  4.3 Groups of type $A_{n}$

Let $G$ be a simple algebraic group of type $A_{n}$ and
$\omega_{1},\ldots,\omega_{n}$ the fundamental weights of $G$. As above, for a
dominant weight $\omega$ we denote by $V_{\omega}$ an irreducible $G$-module
with highest weight $\omega$.

###  4.4 Tensor-indecomposable modules

######  Lemma 4.23.

Let $V=V_{\omega_{i}}$, $i\leq(n+1)/2$, and $n+1<(k+1)i$ for some integer
$k>0$. Suppose that $s\in T_{reg}$ and s is almost cyclic on V. Then
$m_{s}(V)\leq k$. In particular, $m_{s}(V)\leq(n+1)/2$ for $i>1$.

###### Proof.

The weights of $V$ are $\varepsilon_{j_{1}}+\cdots+\varepsilon_{j_{i}}$ for
$1\leq j_{1}<...<j_{i}\leq n+1$. If
$\lambda=\varepsilon_{j_{1}}+\cdots+\varepsilon_{j_{i}}$ then we set ${\rm
supp}(\lambda)=\\{j_{1},\ldots,j_{i}\\}$. Suppose the contrary, and let
$\mu_{1},\ldots,\mu_{k+1}$ be distinct weights of $V$ such that
$\mu_{1}(s)=\cdots=\mu_{k+1}(s)$. As $n+1<(k+1)i$, it follows that there is
$d\in\\{1,\ldots,n+1\\}$ and a pair $\mu_{r},\mu_{t}$ with $1\leq r<t\leq k+1$
such that $d\in{\rm supp}(\mu_{r})\cap{\rm supp}(\mu_{t})$. In addition,
$|{\rm supp}(\mu_{r})\cup{\rm supp}(\mu_{t})|\leq 2i-1$ and $2i-1\leq n$ by
assumption. So $l\notin{\rm supp}(\mu_{r})\cup{\rm supp}(\mu_{t})$ for some
$l\in\\{1,\ldots,n+1\\}$. Set $\alpha=\varepsilon_{d}-\varepsilon_{l}$. It
follows that $\mu_{r}-\alpha,\mu_{t}-\alpha$ are weights of $V$. This
contradicts Lemma 3.15.

For the additional claim, take $k=(n+1)/2$. Then $(k+1)i>n+1$ as $i>1$. So the
claim follows from the above. ∎

######  Lemma 4.24.

Let $G=A_{n},n>2$, $M=V_{\omega_{i}}$ with $2\leq i\leq(n+1)/2$,
$M_{1}=V_{\omega_{1}}$, $M_{2}=V_{\omega_{i-1}}$. Let
$\lambda,\lambda^{\prime}\in\Omega(M_{2})$, $\lambda\neq\lambda^{\prime}$.
Then there are distinct weights $\mu,\mu^{\prime}\in\Omega(M_{1})$ such that
$\lambda+\mu,\lambda+\mu^{\prime},\lambda^{\prime}+\mu,\lambda^{\prime}+\mu^{\prime}\in\Omega(M)$.
In addition, $\Omega(M)\subset\Omega(M_{1})+\Omega(M_{2})$.

###### Proof.

The weights of $M_{1}$ are $\varepsilon_{1},\ldots,\varepsilon_{n+1}$ and
those of $M_{2}$ are $\varepsilon_{j_{1}}+\cdots+\varepsilon_{j_{i-1}}$ for
$1\leq j_{1}<\cdots<j_{i-1}\leq n+1$. We can assume (using the Weyl group)
that $\lambda=\varepsilon_{1}+\cdots+\varepsilon_{i-1}$ and
$\lambda^{\prime}=\varepsilon_{j_{1}}+\cdots+\varepsilon_{j_{i-1}}$ with
$j_{1}<\cdots<j_{i-1}\leq 2i-2\leq n-1$. Then we take
$\mu=\varepsilon_{n},\mu^{\prime}=\varepsilon_{n+1}$.

For the additional claim, we have
$\Omega(V_{\omega_{1}})+\Omega(V_{\omega_{i-1}})=\Omega(V_{\omega_{1}+\omega_{i-1}})\supset\Omega(V_{\omega_{i}})$
as $\omega_{i}\prec\omega_{1}+\omega_{i-1}$, see Lemma 3.11. Whence the
result. ∎

######  Lemma 4.25.

Let $G=A_{n},n>2,$ and let $s\in T_{reg}$.

$(1)$ If s is almost cyclic on $V_{\omega_{i}}$ for $2\leq i\leq(n+1)/2$, then
s is cyclic on $V_{\omega_{j}}$ for every $j<i$.

$(2)$ If s is almost cyclic on $V_{a\omega_{1}}$ for $1<a\leq p-1$, then s is
cyclic on $V_{b\omega_{1}}$ for $1\leq b<a$.

###### Proof.

Set $M_{1}=V_{\omega_{1}}$, and in case (1) set $M=V_{\omega_{i}}$,
$M_{2}=V_{\omega_{i-1}}$, in case (2) set $M=V_{a\omega_{1}}$,
$M_{2}=V_{(a-1)\omega_{1}}$.

$(i)$ $s$ separates the weights of $M_{2}$.

In case (2) this follows from Lemma 3.11(2) and Lemma 3.4(1).

In case (1) suppose the contrary. Then there are distinct weights
$\lambda_{1},\lambda_{2}$ of $M_{2}$ such that
$\lambda_{1}(s)=\lambda_{2}(s)$. By Lemma 4.24, there are distinct weights
$\mu_{1},\mu_{2}$ of $M_{1}$ such that $\lambda_{k}+\mu_{l}$ are weights of
$M$ for $k,l\in\\{1,2\\}$. Then
$(\lambda_{1}+\mu_{1})(s)=(\lambda_{2}+\mu_{1})(s)$ and
$(\lambda_{1}+\mu_{2})(s)=(\lambda_{2}+\mu_{2})(s)$. As $s$ is almost cyclic
on $M$, it follows that $(\lambda_{1}+\mu_{1})(s)=(\lambda_{1}+\mu_{2})(s)$,
and then $\mu_{1}(s)=\mu_{2}(s)$. This is a contradiction, as every regular
semisimple element is cyclic on $V_{\omega_{1}}=M_{1}$.

$(ii)$ $s$ is cyclic on $M_{2}$.

Indeed, the weights of $V_{\omega_{j}}$ and of $V_{b\omega_{1}}$ are well
known to be of multiplicity 1, so the claim follows from $(i)$.

$(iii)$ The result follows by induction on $i$ in case (1) and on $a$ in case
(2).∎

We use Lemma 4.24 to prove

######  Lemma 4.26.

Let $G=A_{n}$, $n>1$, $M_{1}=V_{\omega_{1}}$,
$M=V_{a\omega_{i}+(p-1-a)\omega_{i+1}}$ with $1\leq i\leq(n+1)/2$,
$M_{2}=V_{\omega_{i-1}+(a-1)\omega_{i}+(p-1-a)\omega_{i+1}}$ with $1\leq a\leq
p-1$. (If $i=1$ then $\omega_{i-1}$ is interpreted as the zero weight.)

$(1)$ Let $\lambda,\lambda^{\prime}\in\Omega(M_{2})$,
$\lambda\neq\lambda^{\prime}$. Then there are distinct weights
$\mu,\mu^{\prime}\in\Omega(M_{1})$ such that
$\lambda+\mu,\lambda+\mu^{\prime},\lambda^{\prime}+\mu,\lambda^{\prime}+\mu^{\prime}\in\Omega(M)$.
In addition, $\Omega(M)\subseteq\Omega(M_{1})+\Omega(M_{2})$.

$(2)$ If s is almost cyclic on M then s separates the weights of $M_{2}$.

###### Proof.

(1) We have
$\Omega(M)=\Omega(V_{\omega_{i}})+\Omega(V_{(a-1)\omega_{i}+(p-1-a)\omega_{i+1}})$
by Lemma 3.11, and
$\Omega(V_{\omega_{i}})\subset\Omega(V_{\omega_{1}})+\Omega(V_{\omega_{i-1}})$
by Lemma 4.24. So

$\Omega(M)\subset\Omega(M_{1})+\Omega(V_{\omega_{i-1}})+\Omega(V_{(a-1)\omega_{i}+(p-1-a)\omega_{i+1}})=\Omega(M_{1})+\Omega(M_{2})$

as
$\Omega(M_{2})=\Omega(V_{\omega_{i-1}})+\Omega(V_{(a-1)\omega_{i}+(p-1-a)\omega_{i+1}})$
again by Lemma 3.11.

Let $i>1$. Then we can write $\lambda=\beta+\gamma$,
$\lambda^{\prime}=\beta^{\prime}+\gamma^{\prime}$ with
$\beta,\beta^{\prime}\in\Omega(V_{\omega_{i-1}})$,
$\gamma,\gamma^{\prime}\in\Omega(V_{(a-1)\omega_{i}+(p-1-a)\omega_{i+1}})$. By
Lemma 4.24, there are distinct weights
$\mu,\mu^{\prime}\in\Omega(V_{\omega_{1}})$ such that
$\mu+\beta,\mu^{\prime}+\beta,\mu+\beta^{\prime},\mu^{\prime}+\beta^{\prime}\in\Omega(V_{\omega_{i}})$.
As
$\Omega(V_{\omega_{i}})+\Omega(V_{(a-1)\omega_{i}+(p-1-a)\omega_{i+1}})=\Omega(V_{a\omega_{i}+(p-1-a)\omega_{i+1}})$,
we have
$\mu+\beta+\gamma,\mu^{\prime}+\beta+\gamma,\mu+\beta^{\prime}+\gamma^{\prime},\mu^{\prime}+\beta^{\prime}+\gamma^{\prime}\in\Omega(M)$,
and the result follows.

Let $i=1$ so $V_{\omega_{i-1}}$ is the trivial module. Then
$\Omega(M)=\Omega(M_{1})+\Omega(M_{2})$, and the claim is obvious.

(2) We mimic the proof of Lemma 4.25. Suppose the contrary, and let
$\lambda_{1},\lambda_{2}$ be distinct weights of $M_{2}$ such that
$\lambda_{1}(s)=\lambda_{2}(s)$. By the above, there are distinct weights
$\mu_{1},\mu_{2}$ of $M_{1}$ such that $\lambda_{j}+\mu_{k}$ are weights of
$M$ for $i,k\in\\{1,2\\}$. Then
$(\lambda_{1}+\mu_{j})(s)=(\lambda_{2}+\mu_{j})(s)$ for $j=1,2$. As $s$ is
almost cyclic on $M$, it follows that
$(\lambda_{1}+\mu_{1})(s)=(\lambda_{1}+\mu_{2})(s)$, and then
$\mu_{1}(s)=\mu_{2}(s)$. This is a contradiction, as $s$ is regular. ∎

######  Proposition 4.27.

For $G=A_{n}$ let $M\in\mathop{\rm Irr}\nolimits(G)$ be p-restricted and $s\in
T_{reg}$. Suppose that s is almost cyclic on M. Then $m_{s}(M)\leq n+1$.

###### Proof.

Suppose first that the weight zero multiplicity of $M$ is greater than $1$.
Suppose the contrary; as $s$ is almost cyclic, only the eigenvalue 1
multiplicity is greater than 1. Then the result follows from Theorem 4.2(1)
and Table 1.

Suppose that the weight zero multiplicity of $M$ is at most $1$. By Theorem
3.9, all weight multiplicities of $M$ equal $1$. By Table 2, the highest
weight $\omega$ of $M$ is in the following list: $0,\omega_{i}$ $(1\leq i\leq
n)$, $a\omega_{1},$ $a\omega_{n}$, $a\omega_{i}+(p-1-a)\omega_{i+1}$, where
$1\leq i\leq n-1$ and $1\leq a<p$.

Note that we assume $\omega\neq 0,\omega_{1},\omega_{n}$ as these cases are
trivial. Since the list of the eigenvalue multiplicities of $M$ is the same as
for the dual of $M$, it suffices to deal with
$\omega_{i},a\omega_{1},a\omega_{i}+(p-1-a)\omega_{i+1}$ for $i\leq(n+1)/2$.

Suppose the contrary. Let $M_{1}=V_{\omega_{1}}$ and

$M_{2}=\begin{cases}V_{\omega_{i-1}}&if\,\,\omega=\omega_{i},i>1;\cr
V_{(a-1)\omega_{1}}&if\,\,\omega=a\omega_{1},a>1;\cr
V_{\omega_{i-1}+(a-1)\omega_{i}+(p-1-a)\omega_{i+1}}&if\,\,\omega=a\omega_{i}+(p-1-a)\omega_{i+1}.\end{cases}$

(If $i=1$ then $\omega_{i-1}$ is understood to be the zero weight.) By Lemmas
4.25 and 4.26(2), $s$ separates the weights of $M_{2}$.

Note that $M$ is a constituent of $M_{1}\otimes M_{2}$. Indeed,
$V_{\omega_{i}}$ is a constituent of $V_{\omega_{1}}\otimes V_{\omega_{i-1}}$
and $M_{2}$ is a constituent of $V_{\omega_{i-1}}\otimes
V_{(a-1)\omega_{i}+(p-1-a)\omega_{i+1}}$, whence the claim.

Furthermore, $\Omega(M)\subseteq\Omega(M_{1})+\Omega(M_{2})$ (see Lemma 4.26).
As all eigenvalue multiplicities of $M$ equal 1, there are $n+2$ distinct
weights $\nu_{1},\ldots,\nu_{n+2}$ of $M$ such that $\nu_{i}(s)=e$. Then we
can write $\nu_{i}=\lambda_{i}+\mu_{i}$, where $\lambda_{i},\mu_{i}$ are
weights of $M_{1},M_{2}$, respectively, for $i=1,\ldots,n+2$. Then
$e=\lambda_{i}(s)\mu_{i}(s)$. As $M_{1}=V_{\omega_{1}}$ has exactly $n+1$
distinct weights, we have $\lambda_{i}=\lambda_{j}$ for some
$i,j\in\\{1,\ldots,n+1\\}$ with $i\neq j$. Then $\mu_{i}\neq\mu_{j}$ as
otherwise
$\nu_{i}=\lambda_{i}+\mu_{i}=\lambda_{j}+\mu_{j}=\lambda_{j}+\mu_{j}=\nu_{j}$,
which is false. However,
$\lambda_{j}(s)\mu_{j}(s)=e=\lambda_{i}(s)\mu_{i}(s)=\lambda_{j}(s)\mu_{i}(s)$,
whence $\mu_{i}(s)=\mu_{j}(s)$. This is a contradiction as $s$ separates the
weights of $M_{2}$. ∎

The bound in Proposition 4.27 is not probably sharp in general, at least, for
$G=A_{n}$ we have no example of a $p$-restricted $G$-module $M$ with
$m_{s}(M)>n$ for $s$ regular and almost cyclic on $M$. Moreover, the equality
$m_{s}(M)=n$ probably holds only for $M$ adjoint, and we have no sharp bound
for other $p$-restricted modules. In Lemmas 4.23 and 4.28 we consider some
special cases. Example 4.29 shows that, for $G=A_{n}$, $n>3$ odd, there exists
a regular semisimple element $s\in G$ and a $p$-restricted irreducible
$G$-module $V$ which is not a twist of the natural or adjoint $G$-module such
that $s$ is almost cyclic on $V$ and $m_{s}(V)=(n+1)/2$. (If $V$ is not a
twist of a $p$-restricted $G$-module then there are examples with
$m_{s}(V)=n+1$, see Example 4.36.)

######  Lemma 4.28.

Let $G=A_{n}=SL_{n+1}(F)$ and $p\neq 2$. If $s\in T_{reg}$ then
$m_{s}(V_{2\omega_{1}})\leq(n+3)/2$.

###### Proof.

By [26, Theorem 1.2], if $s$ is almost cycic on $V_{2\omega_{1}}$ then $s$ is
strictly regular. Suppose that
$\mu_{1},\ldots,\mu_{l}\in\Omega(V_{2\omega_{1}})$ be distinct weights such
that $\mu_{1}(s)=\cdots=\mu_{l}(s)=e$. The weights of $V$ are
$2\varepsilon_{i}$ ($i=1,\ldots,n+1$) and $\varepsilon_{i}+\varepsilon_{j}$
$(i,j\in\\{1,\ldots,n+1\\},i<j)$; this can be recorded as
$\varepsilon_{i}+\varepsilon_{j}$ $(i,j\in\\{1,\ldots,n+1\\},i\leq j)$.
Suppose first that
$\mu_{i},\mu_{j},\mu_{k}\in\\{2\varepsilon_{1},\ldots,2\varepsilon_{n+1}\\}$
for some distinct $i,j,k\in\\{1,\ldots,l\\}$. By reordering of
$\mu_{1},\ldots,\mu_{l}$ we can assume that
$\mu_{1},\mu_{2},\mu_{3}\in\\{2\varepsilon_{1},\ldots,2\varepsilon_{n+1}\\}$,
and let $\mu_{i}=2\varepsilon_{r(i)}$ for $i=1,2,3$. Then for
$i,j\in\\{1,2,3\\}$ we have
$1=(\mu_{i}-\mu_{j})(s)=(2\varepsilon_{r(i)}-2\varepsilon_{r(j)})(s)=\varepsilon_{r(i)}(s)^{2}\varepsilon_{r(j)}(s)^{-2}=(\varepsilon_{r(i)}(s)\varepsilon_{r(j)})(s)^{-1})^{2},$
whence $\varepsilon_{r(i)}(s)\varepsilon_{r(j)}(s)^{-1}=\pm 1$. In fact,
$\varepsilon_{r(i)}(s)\varepsilon_{r(j)}(s)^{-1}=-1$, as
$\varepsilon_{r(i)}(s)=\varepsilon_{r(j)}(s)$ implies $s\notin T_{reg}$. So
$\varepsilon_{r(i)}(s)=-\varepsilon_{r(j)}(s)$. However, this cannot happen
for three choices $i,j\in\\{1,2,3\\},i\neq j$. Therefore, at most two weights
$\mu_{i}$ $(1\leq i\leq l)$ are in the set
$2\varepsilon_{1},\ldots,2\varepsilon_{n+1}$.

Note that
$\Omega(V_{\omega_{2}})=\Omega(V_{2\omega_{1}})\setminus\\{2\varepsilon_{1},\ldots,2\varepsilon_{n}\\}$.
Let $l^{\prime}$ be the number of weights $\mu_{1},\ldots,\mu_{l}$ that lie in
$\Omega(V_{\omega_{2}})$. By the above $l^{\prime}\geq l-2$. By Lemma 4.23, we
have $l^{\prime}\leq m_{s}(V_{\omega_{2}})\leq(n+1)/2$. If $l^{\prime}=l$ or
$l-1$ then the result follows. Suppose that $l^{\prime}=l-2$, and we can
assume that $\mu_{1},\ldots,\mu_{l-2}\in\Omega(V_{\omega_{2}})$ and that
$\mu_{l-1}=2\varepsilon_{n}$, $\mu_{l}=2\varepsilon_{n+1}$. Then
$\varepsilon_{n},\varepsilon_{n+1}$ do not lie in the support of any $\mu_{t}$
with $t=1,\ldots,l-2$. Indeed, otherwise
$(\varepsilon_{i}+\varepsilon_{k})(s)=2\varepsilon_{k}(s)$ for
$k\in\\{n,n+1\\}$ and some $i$, whence
$(\varepsilon_{i}-\varepsilon_{k})(s)=1$, which is false as $s$ is regular.
Therefore, we can apply Lemma 4.23 to $G=A_{n-2}$ in place of $A_{n}$ to
conclude that $l-2\leq(n-1)/2$, and hence $l\leq(n+3)/2$, as required.
(Alternatively, we can obtain this conclusion from the proof of Lemma 4.23 as
there are at most $(n-1)/2$ weights $\mu_{1},\ldots,\mu_{l}$ whose supports
are pairwise disjoint.) ∎

######  Example 4.29.

$(1)$ Let $G=SL_{2m}(F)$ and $s=\mathop{\rm
diag}\nolimits(a_{1},\ldots,a_{m},a_{m}^{-1},\ldots,a_{1}^{-1})\in G$, where
the elements $1,a_{1},\ldots,a_{m}\in F$ are algebraically independent and
non-zero. Let $\omega=\omega_{2}$. Then $V_{\omega}^{T}=0$, $s$ is almost
cyclic on $V_{\omega}$ and $\dim V_{\omega}^{s}=m=(n+1)/2$, where $n=2m-1$ is
the rank of $G$.

If $p\neq 2$ and $s=\mathop{\rm
diag}\nolimits(-a_{1},\ldots,-a_{m},a_{m}^{-1},\ldots,a_{1}^{-1})$ then $-1$
has multiplicity $m=(n+1)/2$ and the other eigenvalues of $s$ on
$V_{\omega_{2}}$ are of multiplicity $1$.

$(2)$ Let $G=SL_{2m+1}(F)$, $p\neq 2$ and $s=\mathop{\rm
diag}\nolimits(1,a_{1},\ldots,a_{m},a_{m}^{-1},\ldots,a_{1}^{-1})\in G$, where
the elements $a_{1},\ldots,a_{m}\in F$ are as above. Let $\omega=2\omega_{1}$.
Then $V_{\omega}^{T}=0$ and $\dim V_{\omega}^{s}=m+1=(n+2)/2$, where $n=2m$ is
the rank of $G$. In this case $s\in H\cong B_{m}$.

$(3)$ Let $G=SL_{2m}(F)$ and $s=\mathop{\rm
diag}\nolimits(b,-b,b^{2}a_{1},a_{1}^{-1},\ldots,b^{2}a_{m-1},a_{m-1}^{-1})\in
G$, where the elements $a_{1},\ldots,a_{m-1}\in F$ are as above and
$b^{n+1}=(-1)^{(n+1)/2}$, $b\in F$. Then the eigenvalue $b^{2}$ has
multiplicity $m+1=(n+3)/2$ on $V_{2\omega_{1}}$ and the other eigenvalues have
multiplicity $1$. (Here $n=2m-1$ is the rank of $G$.)

###  4.5 Tensor-decomposable modules: special cases

Proposition 4.27 states that Theorem 1.2 is true for $p$-restricted
irreducible $A_{n}$-modules. We next prove it for tensor-decomposable modules.

######  Lemma 4.30.

[25, Lemma 7] Let $s\in T$ be a non-central element. Let $\omega$ be a
dominant weight which is not $p$-restricted and not of the form $p^{k}\mu$ for
$\mu$ a $p$-restricted dominant weight. If $s$ is almost cyclic on
$V_{\omega}$ then all weights of $V_{\omega}$ are of multiplicity $1$.

For a $G$-module $V$ and $s\in G$ denote by $\deg_{V}(s)$ the degree of the
minimum polynomial of $\rho(s)$, where $\rho$ is the representation afforded
by $V$. If $s$ is semisimple then $\deg_{V}(s)=|E_{V}(s)|$ is the number of
distinct eigenvalues of $s$ on $V$. Note that $s$ is cyclic on $V$ if and only
if $\deg_{V}(s)=\dim V$.

For $A,B\subset F^{\times}$ we set $AB=\\{ab:a\in A,b\in B\\}$.

######  Lemma 4.31.

Let $M,M_{1},M_{2},V$ be non-trivial G-modules and $s\in T$. Suppose that s is
not scalar on both $M_{1}$, $M_{2}$, $E_{M}(s)=E_{M_{1}}(s)E_{M_{2}}(s)$ and
$s$ is almost cyclic on $M\otimes V$.

$(1)$ $\deg_{M_{i}\otimes V}(s)=\deg_{M_{i}}(s)\cdot\dim V$ for $i=1,2$.

$(2)$ the eigenvalue multiplicities of s on $M\otimes V$ do not exceed $d=\dim
M_{1}$.

###### Proof.

(1) We can assume $i=1$. By Lemma 3.2, $s$ is cyclic on $M$ and $V$. Then
$\dim V=\deg_{V}(s)$. Suppose the contrary, that $\deg_{M_{1}\otimes
V}(s)<\deg_{M_{1}}(s)\cdot\dim V$. Then there are distinct eigenvalues
$\mu,\mu^{\prime}\in E_{M_{1}}(s)$ and $\nu,\nu^{\prime}\in E_{V}(s)$ such
that $\mu\nu=\mu^{\prime}\nu^{\prime}$. Then
$\mu\lambda\nu=\mu^{\prime}\lambda\nu^{\prime}\in E_{M\otimes V}(s)$ for every
$\lambda\in E_{M_{2}}(s)$. So $s$ is not almost cyclic on $M\otimes V$ unless
$s$ is scalar on $M_{2}$, which is not the case by assumption.

(2) This follows from Lemma 3.2(1) as the eigenvalue multiplicities of $s$ on
$V$ equal 1. ∎

To illustrate the use of Lemma 4.31, we state the following:

######  Corollary 4.32.

Let $G=A_{n}$, $\omega=a\omega_{1}$ or $a\omega_{1}+(p-1-a)\omega_{2}$, where
$1<a<p$. Let $M=V_{\omega}$ and let $V$ be a non-trivial irreducible G-module.
Let $s\in(T\setminus Z(G))$. Suppose that s is almost cyclic on $M\otimes V$.
Then the eigenvalue multiplicities of s on $M\otimes V$ do not exceed $n+1$.

###### Proof.

Set $M_{1}=V_{\omega_{1}}$ and $M_{2}=V_{(a-1)\omega_{1}}$ or
$V_{(a-1)\omega_{1}+(p-1-a)\omega_{2}}$, respectively. Then
$\Omega(M)=\Omega(M_{1})+\Omega(M_{2})$ by Lemma 3.11. So the result follows
from Lemma 4.31.∎

Note that Corollary 4.32 is not implied by Proposition 4.27. In addition, here
and in Lemmas 4.33, 4.34 below we do not assume $s$ to be regular as this
automatically holds due to Lemma 3.3(2) when $s\notin Z(G)$.

######  Lemma 4.33.

Let $M,M_{1},M_{2},V$ be non-trivial G-modules. Let $s\in T\setminus Z(G)$.
Suppose that

$(i)$ for every $f,f^{\prime}\in E_{M_{2}}(s)$ there exist distinct
eigenvalues $e,e^{\prime}\in E_{M_{1}}(s)$ such that
$ef,ef^{\prime},e^{\prime}f,e^{\prime}f^{\prime}\in E_{M}(s);$

$(ii)$ $E_{M}(s)\subseteq E_{M_{1}}(s)\cdot E_{M_{2}}(s);$

$(iii)$ $s$ is almost cyclic on $M\otimes V$.

Then $\deg_{M_{2}\otimes V}(s)=\deg_{M_{2}}(s)\cdot\dim V$, and
$m_{s}(M\otimes V)\leq\dim M_{1}$.

###### Proof.

By Lemma 3.2, $s$ is cyclic on $V$, so $\dim V=\deg_{V}(s)$. To prove the
first statement of the lemma, suppose (the contrary), that $\deg_{M_{2}\otimes
V}(s)<\deg_{M_{2}}(s)\cdot\dim V$. Then there are distinct eigenvalues
$f,f^{\prime}\in E_{M_{2}}(s)$ and eigenvalues $v,v^{\prime}\in E_{V}(s)$ such
that $fv=f^{\prime}v^{\prime}$.

By $(i)$, there are distinct eigenvalues $e,e^{\prime}\in E_{M_{1}}(s)$ such
that $ef,ef^{\prime},e^{\prime}f,e^{\prime}f^{\prime}\in E_{M}(s),$ and hence
$efv=ef^{\prime}v^{\prime},e^{\prime}fv=e^{\prime}f^{\prime}v^{\prime}\in
E_{M\otimes V}(s)$. So $a:=efv$ and $b:=e^{\prime}fv$ are distinct eigenvalues
of $s$ on $M\otimes V$, each of multiplicity at least 2. As $s$ is almost
cyclic on $M\otimes V$, we must have $a=b$, whence $e=e^{\prime}$, a
contradiction.

Turn to the second statement. Set $d=\dim M_{1}.$ Suppose the contrary, that
the multiplicity of some eigenvalue $u$ of $s$ on $M\otimes V$ is greater than
$d$. By Lemma 3.2(3), $s$ is cyclic on $M$ and $V$, so there are distinct
eigenvalues $w_{1},\ldots,w_{d+1}$ of $s$ on $M$ and distinct eigenvalues
$v_{1},\ldots,v_{d+1}$ of $s$ on $V$ such that $u=w_{i}v_{i}$ for
$i=1,\ldots,d+1$. Furthermore, by $(ii)$, $w_{i}=e_{i}f_{i}$ for
$i=1,\ldots,d+1$ for some eigenvalues $e_{i}\in E_{M_{1}}(s)$ and some
eigenvalues $f_{i}\in E_{M_{2}}(s)$. Therefore, $e_{i}=e_{j}$ for some
$i,j\in\\{1,\ldots,d+1\\}$. Then $w_{j}=e_{i}f_{j}$, where $f_{i}\neq f_{j}$
as $w_{i}\neq w_{j}$, and $u=e_{i}f_{i}v_{i}=e_{i}f_{j}v_{j}$, whence
$f_{i}v_{i}=f_{j}v_{j}$. Then $\deg_{M_{2}\otimes
V}(s)<\deg_{M_{2}}(s)\deg_{V}(s)$ contradicting the first statement of the
lemma. ∎

######  Lemma 4.34.

Let $G=A_{n}$ and $s\in(T\setminus Z(G))$. Let $M=V_{\omega_{i}}$ for $1<i<n$
or $V_{a\omega_{i}+(p-1-a)\omega_{i+1}}$ for $1<i<n-1$, and let V be a non-
trivial G-module. Suppose that $M\otimes V$ is irreducible and $s$ is almost
cyclic on $M\otimes V$. Then $m_{s}(M\otimes V)\leq n+1$.

###### Proof.

We can assume (by replacing $M$ by its dual) that $i\leq(n+1)/2.$ Set
$M_{1}=V_{\omega_{1}}$.

Suppose first that $M=V_{\omega_{i}}$. Set $M_{2}=V_{\omega_{i-1}}$. The
result follows from Lemma 4.33 as soon as we show that the assumptions
$(i),(ii)$ of that lemma hold in this case (as $(iii)$ is in the assumption).
By Lemma 3.11(2), $\Omega(M)\subseteq\Omega(M_{1})+\Omega(M_{2})$, whence
$(ii)$. To prove $(i)$, let $f,f^{\prime}\in E_{M_{2}}(s)$. Then there are
weights $\lambda,\lambda^{\prime}\in\Omega(M_{2})$ such that
$\lambda(s)=f,\lambda^{\prime}(s)=f^{\prime}$. By Lemma 4.24, there are
weights $\mu,\mu^{\prime}\in\Omega(M_{1})$ such that
$\lambda+\mu,\lambda+\mu^{\prime},\lambda^{\prime}+\mu,\lambda^{\prime}+\mu^{\prime}\in\Omega(M)$.
Set $e=\mu(s),e^{\prime}=\mu^{\prime}(s)$. To get $(i)$, we need $e\neq
e^{\prime}$. Suppose $e=e^{\prime}$; then
$(\lambda+\mu)(s)=(\lambda+\mu^{\prime})(s)$ and
$\lambda+\mu\neq\lambda+\mu^{\prime}$. This is a contradiction as $s$ is
cyclic on $M$ by Lemma 3.2.

Suppose that $M=V_{a\omega_{i}+(p-1-a)\omega_{i+1}}$. Then we set
$M_{2}=V_{\omega_{i-1}+(a-1)\omega_{i}+(p-1-a)\omega_{i+1}}$. The result
follows from Lemma 4.33 if the assumptions $(i),(ii)$ of that lemma hold in
this case (since $(iii)$ is in the assumption). Lemma 4.26 yields $(ii)$.

To prove $(i)$, let $f,f^{\prime}\in E_{M_{2}}(s)$. Then there are weights
$\lambda,\lambda^{\prime}\in\Omega(M_{2})$ such that
$\lambda(s)=f,\lambda^{\prime}(s)=f^{\prime}$. By Lemma 4.26, there are
weights $\mu,\mu^{\prime}\in\Omega(M_{1})$ such that
$\lambda+\mu,\lambda+\mu^{\prime},\lambda^{\prime}+\mu,\lambda^{\prime}+\mu^{\prime}\in\Omega(M)$.
Set $e=\mu(s),e^{\prime}=\mu^{\prime}(s)$. To get $(i)$, we need $e\neq
e^{\prime}$. Suppose $e=e^{\prime}$; then
$(\lambda+\mu)(s)=(\lambda+\mu^{\prime})(s)$ and
$\lambda+\mu\neq\lambda+\mu^{\prime}$. This is a contradiction as $s$ is
cyclic on $M$ by Lemma 3.2.∎

######  Theorem 4.35.

Let G be of type $A_{n}$ and N an irreducible G-module. Let $s\in T_{reg}$.
Suppose that s is almost cyclic on N. Then $m_{s}(N)\leq n+1$.

###### Proof.

Let $N=N_{1}\otimes\cdots\otimes N_{k}$, where $N_{1},\ldots,N_{k}$ are
tensor-indecomposable non-trivial irreducible $G$-modules. Then every $N_{i}$
for $i=1,\ldots,k$ is a Frobenius twist of an infinitesimally irreducible
$G$-module. The problem is well known to be equivalent to the case where one
of $N_{i}$ is infinitesimally irreducible and we can assume that this is
$N_{1}$ by reordering the terms.

If $k=1$ then the result is contained in Proposition 4.27. Let $k>1$. Then, by
Lemma 3.2, $s$ is cyclic on every $N_{i}$. Therefore, all weights of $N_{i}$
are of multiplicity 1. By [19, 6.1], $N_{i}$ is a Frobenius twist of an
irreducible $G$-module, whose highest weight belongs to the list
$\\{\omega_{j},1\leq j\leq n$, $a\omega_{1},a\omega_{n}\,\,(1<a<p)$,
$a\omega_{i}+(p-1-a)\omega_{i+1},\,\,i=1,\ldots,n-1,0\leq a<p\\}$. Set
$M=N_{1}$, $V=N_{2}\otimes\cdots\otimes N_{k}$, so $V_{2}$ is irreducible and
non-trivial, whereas the highest weight $\omega$ of $N_{1}$ lies in the above
list. Then the result follows from Corollary 4.32 and Lemma 4.34.∎

One can expect that the upper bound for $m_{s}(V)$ for $V$ tensor-decomposable
must be significantly lower than for $V$ tensor-indecomposable. The following
example makes evident that this is not so. In fact, we see that the bound in
Theorem 4.35 is sharp.

######  Example 4.36.

Let $F$ be an algebraically closed field of characteristic $p>0$, $G=A_{n}$,
$n\geq 2$, and $V$ an irreducible $G$-module with highest weight
$\omega_{1}+p\omega_{n}$. Then there exists $s\in G$ such that $s$ is almost
cyclic on $V$ and $m_{s}(V)=n+1$ for some $s\in G$.

Indeed, set $s=\mathop{\rm diag}\nolimits(a,a^{p},\ldots,a^{p^{n}})\in
SL_{n+1}(F)$, where $a\in F$ is a primitive $((p^{n+1}-1)/(p-1))$-root of
unity. Then the matrix of $s$ on $V$ is $s\otimes s^{-p}$ and the eigenvalues
of this matrix are $a^{p^{i}-p^{j+1}}$ with $0\leq i,j\leq n$. If $i=j+1$ then
this equals 1, as well as if $i=0,j=n$ as then $a^{1-p^{n+1}}=1$. So 1 is an
eigenvalue of $s\otimes s^{-p}$ of multiplicity at least $n+1$. In fact, this
is exactly $n+1$ as $a^{p^{i}-p^{j+1}}=1$ implies $p^{i}-p^{j+1}\equiv
0\pmod{p^{n}+p^{n-1}+\cdots+p+1}$ which is false unless $i=j+1$ or $i=0,j=n$.
We show that all other eigenvalues are of multiplicity 1. Suppose the
contrary. Then $a^{p^{i}-p^{j+1}}=a^{p^{k}-p^{l+1}}$ for some choice of $0\leq
i,j,k,l\leq n$ and $i\neq j+1$, $k\neq l+1$ and $(i,j)\neq(0,n)\neq(k,l)$. As
above, we observe that $i\neq k,j\neq l$. We have
$p^{i}-p^{j+1}-p^{k}+p^{l+1}\equiv 0\pmod{p^{n}+p^{n-1}+\cdots+p+1}$. Let
$m=\min(i,j+1,k,l+1)$. If $m>0$ then
$p^{i-m}-p^{j+1-m}-p^{k-m}+p^{l+1-m}\equiv 0\pmod{p^{n}+p^{n-1}+\cdots+p+1}$
and now $d:=\max(i-m,j+1-m,k-m,l+1-m)$ does not exceed $n$. If $d<n$ then the
absolute value of the left hand side is at most $2p^{n-1}$ and the congruence
does not hold. (If $p=2$ then $2^{n}+2^{n-1}+\cdots+1=2^{n+1}-1$ and the
congruence fails too.) Suppose that $d=n$. Then $m=1$ and $\max(j,l)=n$. We
can assume $l=n$ and then $j<n$. Then the absolute value of
$p^{i-m}-p^{j+1-m}-p^{k-m}+p^{n}$ does not exceed
$p^{n}+p^{n-1}<p^{n}+p^{n-1}+\cdots+p+1$, hence the congruence above cannot
hold.

So $m=0$. As above, the congruence fails unless possibly $d\in\\{n,n+1\\}$.

Suppose that $d=n$. Then the absolute value of $p^{i}-p^{j+1}-p^{k}+p^{l+1}$
does not exceed $bp^{n}-p^{c}-1$, where $b\in\\{0,2\\},0\leq c\leq n$. Then
the congruence cannot hold.

Suppose that $d=n+1$. Then we can replace $p^{d}$ by 1, and then the absolute
value of $p^{i}-p^{j+1}-p^{k}+p^{l+1}$ does not exceed $2+p^{n}+p^{n-1}$. If
$n>2$ then the congruence above cannot hold. Let $n=2$. We can assume that
$l=2$ and then $s\otimes s^{-p}=\mathop{\rm
diag}\nolimits(1,1,1,a^{1-p},a^{2+p},a^{1+2p},a^{p-1},a^{-1-2p},a^{-2-p})$.
One checks straightforwardly that this matrix is almost cyclic.

###### Proof of Theorem 1.2.

If $G$ is of type $A_{n}$ then the result is contained in Theorem 4.35. For
$G$ of type $E_{6}$ and $E_{7}$ see Lemmas 4.16 and 4.19, respectively. The
case with $G=G_{2}$ is covered by Corollary 4.22.

Suppose first that $V=V_{1}\otimes V_{2}$ is tensor-decomposable, that is,
both $V_{1},V_{2}$ are non-trivial. Then $s$ is cyclic on $V_{1}$ and on
$V_{2}$ (Lemma 3.2(4)). If $G$ is not of type $A_{n},n>1$, $D_{n},n$ odd, or
$E_{6}$ then every semisimple element of $G$ is real and the result follows
from Lemma 3.2(4).

Let $G=D_{n},n>3$ odd. Let $\omega=\sum p^{i}\mu_{i}$ be the highest weight of
$V$, where $\mu_{i}$’s are $p$-restricted. Then
$V=\otimes_{i}V_{p^{i}\mu_{i}}$. By Lemma 4.30, the weights of
$V_{p^{i}\mu_{i}}$, and hence of $V_{\mu_{i}}$ are of multiplicity 1, so, by
Table 2, $\mu_{i}\in\\{\omega_{1},\omega_{n-1},\omega_{n}\\}$. Then
$m_{s}(V)\leq 2$ by Lemma 4.7.

Next suppose that $V$ is tensor-indecomposable. By Steinberg’s theorem, $V$ is
a Frobenius twist of a $p$-restricted $G$-module. So we can assume that $V$
itself is $p$-restricted.

Suppose first that $\omega$ occurs in Table 1. Then $\dim V^{T}>1$ and $\dim
V^{T}=\dim V^{s}=m_{s}(V)$ by Theorem 4.2(1), so (2) follows from the data in
Table 1.

Suppose that $\omega$ does not occur in Table 1. By Theorem 3.9, $\omega$
occurs in Table 2. If $(G,p,\omega)=(F_{4},3,\omega_{4})$ then the result
follows from Lemma 4.20. This completes considerations of groups of the
exceptional Lie types.

So we are left with classical groups (other than of type $A_{n}$). The case
with $G=D_{n},n\geq 4$ is settled in Lemma 4.6 for $\omega=\omega_{n-1}$ or
$\omega_{n}$, for $\omega=\omega_{1}$ see Lemma 3.6(3). This implies the
result for an arbitrary $p$-restricted $\omega$ in view of Theorem 3.9 and
Table 2. A similar argument works for $G=B_{n}$, $n>2,p\neq 2$, with use of
Lemmas 4.10 and 4.4. The case $B_{2}\cong C_{2}$ is considered in Lemma 4.11.

The case $G=C_{n},p=2$ has been examined in Lemma 4.4. Suppose that $G$ is of
type $C_{n}$ with $p\neq 2$. By Theorem 3.9 and Tables 1,2 we have to inspect
the cases with
$\omega\in\\{\omega_{1},\omega_{n-1}+\frac{p-3}{2}\omega_{n},\frac{p-1}{2}\omega_{n}\\}$
as well as $n=3,p\neq 2,\omega=\omega_{3}$. The case with $\omega=\omega_{1}$
is trivial, the other cases are dealt with in Lemmas 4.10, 4.9 and the
remaining case (where $G=C_{3},p\neq 2,\omega=\omega_{3})$ is settled in Lemma
4.8. ∎

One easily observes that the exceptions in Theorem 1.2 are genuine. It
suffices to show this for $\omega\in\\{3\omega_{1},p\neq
2,(1+p^{k})\omega_{1}\\}$. Let $\mathop{\rm diag}\nolimits(a,a^{-1})\in
SL_{2}(F)$ be the matrix of $s$ on $V_{\omega_{1}}$. Then the matrices of $s$
on $V_{2\omega_{1}}$, $V_{3\omega_{1}}$ and $V_{(1+p^{k})\omega_{1}}$ are
$\mathop{\rm diag}\nolimits(a^{2},1,a^{-2})$, $\mathop{\rm
diag}\nolimits(a^{3},a,a^{-1},a^{-3})$ and $\mathop{\rm
diag}\nolimits(a^{p+1},a^{p-1},a^{-p-1},a^{-p+1})$, respectively. If
$a^{2}=-\mathop{\rm Id}\nolimits$, $a^{3}=1\neq a$ then $s$ is almost cyclic
on $V_{2\omega_{1}}$, $V_{3\omega_{1}}$, respectively, and
$m_{s}(V_{2\omega_{1}})=m_{s}(V_{3\omega_{1}})=2$. If $a$ is a primitive
$(p+1)$-root of unity then $s$ is almost cyclic on $V_{(p+1)\omega_{1}}$ and
$m_{s}(V_{(p+1)\omega_{1}})=2$.

###### Proof of Theorem 1.1.

The result follows from Theorem 1.2. ∎

###### Proof of Corollary 1.3.

Suppose the contrary, that $m_{s}(V)\geq\dim V/2>1$. By Theorem 1.2, either
$m_{s}(V)\leq 2$, hence $\dim V\leq 4$, or $V$ is as in item $(1),(2)$ or
$(3)$ of Theorem 1.2.

Suppose first that $\dim V\leq 4$. If $\dim V=3$ then $G=A_{1}$, $p\neq 2$,
$\omega=2p^{i}\omega_{1}$ or $G=A_{2}$,
$\omega\in\\{p^{i}\omega_{1},p^{i}\omega_{2}\\}$. In the former case
$m_{s}(V)=2$ for $s\in SL_{2}(F)$ implies $s^{2}=-\mathop{\rm Id}\nolimits$ as
recorded in the corollary. The latter case is ruled out by Lemma 3.6 or by a
straightforward calculation. Let $\dim V=4$. Then
$G\in\\{A_{1},A_{3},B_{2},C_{2}\\}$. In fact, if $G=A_{3}$ then $V$ is a twist
of the natural $G$-module, on which $m_{s}(V)\leq 1$ for a regular semisimple
element $s\in G$ (Lemma 3.6). The cases with $G=B_{2}$ and $C_{2}$ are ruled
out by Remark 4.12. Let $G=A_{1}$. Then $\omega$ is well known to be either
$3p^{i}\omega_{1}\,\,(p\neq 2,3),$ or $(p^{i}+p^{j})\omega_{1}$ as recorded in
the statement.

The result is obvious if item (3) of Theorem 1.2 holds as $\dim V\geq 27$ for
$G=E_{6}$.

Suppose that (2) of Theorem 1.2 holds. Then $m_{s}(V)=\dim V^{T}$ and
inspection of $\dim V/\dim V^{T}$ in Table 1 yields a contradiction.

Suppose that item (1) of Theorem 1.2 holds so $G$ is of type $A_{n}$. We can
assume $n>1$ as otherwise $m_{s}(V)\leq 2$ by Theorem 1.1 and $\dim V\leq 4$;
this case has been examined above. In addition, we can assume that $V$ is not
a twist of the adjoint $G$-module as this is included in case (2), unless
$n=2,p=3$. In the latter case $\dim V=7$ and $m_{s}(V)\leq 2$ by Theorem 1.1.

By the above, we can assume that $m_{s}(V)>2$ and $\dim V>5$.

So we assume that $n>1$. If $V$ is a twist of the irreducible $G$-module with
highest weight $\omega_{1}$ then we get a contradiction with Lemma 3.6.

If $V$ is not a twist of the irreducible $G$-module with highest weight
$\omega_{1}$ then $\dim V\geq n(n+1)/2$ so $\dim V/m_{s}(V)\geq n/2$ and
$n/2>2$ for $n>4$. Suppose that $n\leq 4$ and $\dim V\leq 2(n+1)$. If $n=2$,
this implies $p\neq 2$ and $\omega\in\\{2\omega_{1},2\omega_{2}\\}$. However,
in these cases $\dim V=6$ and $m_{s}(V)\leq 2$ by Lemma 4.28, so this case is
ruled out. By [15, Appendices A6,A7,A8], the highest weight $\omega$ of $V$ is
$p^{t}\omega_{2}$ if $n=3$ and $\\{p^{t}\omega_{2}$ or $p^{t}\omega_{3}\\}$ if
$n=4$ for some integer $t\geq 0$. So either $G=SL_{4}(F)$, $V=V_{\omega_{2}}$
or $G=SL_{5}(F)$, $V=V_{\omega_{2}}$ or $V_{\omega_{3}}$. As $V_{\omega_{2}}$,
$V_{\omega_{3}}$ are dual for $n=4$, it suffices to examine
$\omega=\omega_{2}$ in both the cases. If $n=3$ then $V$ is a twist of the
natural module for $D_{3}\cong A_{3}$ on which $m_{V}(s)\leq 2$ as $s$ is
regular (see Lemma 3.6). If $n=4$ then $m_{s}(V_{\omega_{2}})\leq 2$ by Lemma
4.23 and $\dim V_{\omega_{2}}=10$, so this case is ruled out. This completes
the proof. ∎

##  5 An application to a recognition problem for linear groups over finite
fields

There is a problem in the theory of linear groups over finite fields aimed to
determining the subgroups $H$ of $GL_{n}(q)$ containing a matrix $\mathop{\rm
diag}\nolimits(\mathop{\rm Id}\nolimits_{m},M)$, where $M\in GL_{n-m}(q)$ is
irreducible. The latest and most significant contribution to this project is
made in [8], where the authors solve this problem for $m<n/2$. The result of
[8] can be also interpreted as a classification of subgroups of $GL_{n}(q)$
containing a cyclic Sylow $\ell$-subgroup of $GL_{n}(q)$ for some prime
divisor $\ell$ of $GL_{n}(q)$, see [1]. Observe that the case where $H$ is
irreducible in $GL_{n}(q)$ is the bulk of the problem.

In this section we contribute to the problem in the case where $H$ is a quasi-
simple group of Lie type in defining characteristic $p>0$ for $p|q$. Our
purpose is to consider the complementary case with $m\geq n/2$. Observe that
the case with $p\not|q$ is considered in [7].

Note that specializing the main theorem of [8] to the case where $H$ is a
quasi-simple group of Lie type leads one to Examples 2.1, 2.4 and 2.9 in [8].
Then it is easy to observe that Examples 2.1, 2.4 yield nearly natural
representations of $H$ (whereas Examples 2.9 are irregular cases arising for
$n\leq 9$). The converse is also true, that is, if $H$ is a quasi-simple group
of Lie type in defining characteristic $p|q$ and $\rho:H\rightarrow GL_{n}(q)$
is a nearly natural irreducible representation then $\rho(H)$ is as one of
these examples. However, we do not provide here the proof of this fact.

###  5.1 Representation of $SL_{2}(q)$ containing almost cyclic elements

Our primary goal here is to give an example with $m=n/2$ (see Lemma 5.6),
which is essential for what follows.

######  Lemma 5.1.

Let $\mathbf{G}=SL_{2}(F)$, let $\rho$ be an irreducible representation of
$\mathbf{G}$ with highest weight $(1+p^{i})\omega_{1}$ for some integer $i\geq
1$. Let $s\in\mathbf{G}$ be a non-central semisimple element such that
$\rho(s)$ is almost cyclic and $m_{s}(\rho)>1$. Then $\rho(s)\in
PSL_{2}(p^{i})\subset GL_{4}(p^{i})$ and s is conjugate in $\mathbf{G}$ to an
element of $SL_{2}(p^{i})$, unless $p\neq 2$ and $s^{2}$ is conjugate to an
element of $SL_{2}(p^{i})$. In addition, the group $\rho(SL_{2}(p^{2i}))$ is
irreducible whereas $\rho(SL_{2}(p^{i}))$ is reducible.

###### Proof.

Let $s=\mathop{\rm diag}\nolimits(a,a^{-1})\in SL_{2}(F)$. Then
$\rho(s)=\mathop{\rm
diag}\nolimits(a^{p^{i}+1},a^{p^{i}-1},a^{-p^{i}+1},a^{-p^{i}-1})$, and some
of these eigenvalues coincide. Here $a\in F^{\times},$ $a^{2}\neq 1$ as
$s\notin Z(\mathbf{G})$. This implies $a^{p}\neq a^{-p}$, $a^{p^{i}+1}\neq
a^{p^{i}-1}$ and $a^{-p^{i}+1}\neq a^{-p^{i}-1}$. So either $a^{2(p^{i}+1)}=1$
or $a^{2(p^{i}-1)}=1$, and hence either $s^{p^{i}+1}\in Z(\mathbf{G})$ or
$s^{p^{i}-1}\in Z(\mathbf{G})$. If $p=2$ then $Z(\mathbf{G})=1$, and the group
$SL_{2}(2^{i})$ contains elements of order $|s|$; these are obviously
conjugate to elements of $\langle s\rangle$. Let $p>2$. If $a^{p^{i}+1}=1$ or
$a^{p^{i}-1}=1$ then, similarly, $s$ is conjugate to an element of
$SL_{2}(p^{i})$. Let $a^{p^{i}+1}=-1$ or $a^{p^{i}-1}=-1$; then
$\rho(s)=(-1,-1,-a^{-2},-a^{2})\in GL_{4}(p^{i})$ as $s^{2}\in SL_{2}(p^{i})$.
So $\rho(s)\in GL_{4}(p^{i})$, as claimed.

Turn to the additional claim of the lemma. If $q=p^{2i}$ then the weight
$(1+p^{i})\omega_{1}$ is $q$-restricted, so $\rho$ remains irreducible under
restriction to $SL_{2}(q)$ (see [9, §2.11]). If $q=p^{i}$ then
$\rho|_{SL_{2}(q)}\cong(\tau\otimes\tau)|_{SL_{2}(q)}$, where $\tau$ is the
irreducible representation of $\mathbf{G}$ with highest weight $\omega_{1}$.
It is well known that $\tau\otimes\tau$ is reducible. ∎

######  Lemma 5.2.

Let $\mathbf{G}=SL_{2}(F)$, $G=SL_{2}(q)$, $q=p^{d}>p$, and let $\rho$ be an
irreducible representation of $\mathbf{G}$ with highest weight
$(1+p^{i})\omega_{1}$ for some integer $i\geq 1$. Suppose that $\rho(G)$ is
conjugate in $GL_{4}(F)$ to an irreducible subgroup of $GL_{4}(p^{m})$, where
$m$ is minimal possible. Then $d\not|i$ and either d is even, $i\pmod{d}=d/2$
and $m=d/2$ or $i\pmod{d}\neq d/2$ and $m=d$.

###### Proof.

Obviously, $\rho(G)\subset GL_{4}(q)$. Suppose that the group $\rho(G)$ is
conjugate to a subgroup of $GL_{4}(F_{0})$, where $F_{0}$ is a subfield of
$\mathbb{F}_{q}$ and $|F_{0}|=p^{m}$. Then $m|d$. (As $\mathbb{F}_{q}$ is a
vector space over $F_{0}$, we have $|\mathbb{F}_{q}|=|F_{0}|^{j}$ for some
integer $j>0$, so $q=p^{d}=(p^{m})^{j}$, whence $mj=d$.) As $\rho(G)$ is
irreducible, $i$ is not a multiple of $d$. (Otherwise $\rho|_{H}\cong
V_{\omega_{1}}|_{H}\otimes V_{\omega_{1}}|_{H}$ is reducible.)

Suppose that $F_{0}\neq\mathbb{F}_{q}$, so $m<d$.

By Zsigmondy’s theorem (see [11, 5.2.14]), if $q^{2}=p^{2d}>p^{2}$ (this holds
by assumption) and $q^{2}\neq 64$ then there is a prime
$\ell|(q^{2}-1)=p^{2d}-1$ such that $(\ell,p^{j}-1)=1$ for every $j<2d$.
Therefore, if $F_{0}\neq\mathbb{F}_{q}$ (hence $m<d$) and $q\neq 8$ then
$(\ell,p^{2m}-1)=1$. Moreover, $(\ell,p^{4m}-1)=1$. Indeed, otherwise $\ell$
divides $(p^{4m}-1,p^{2d}-1)=p^{(4m,2d)}-1$ by [10, Hilfsatz 2(a)]. As
$(4m,2d)\leq 2d$, the claim is true unless $(4m,2d)=2d$, whence $d|2m$, and
hence $2m=d$. Therefore, $\mathbb{F}_{q}$ is the quadratic extension of
$F_{0}$. Let $q^{2}=64$ so $G=SL_{2}(8)$, $d=3$ and $|s|=9$. Then
$\rho(G)\subset GL_{4}(8)$. As $2^{m}|8$, it follows that either $2^{m}=8$, as
required, or $m=1$; in the latter case $\phi(G)\subset GL_{4}(2)$. However,
$GL_{4}(2)$ has no element of order 9, so we have a contradiction.

Finally, we show that $i$ is a multiple of $d/2$. Let $\gamma:F\rightarrow F$
be the mapping defined by $x\rightarrow x^{p^{m}}$ for $x\in F$. This extends
to the mapping $\tau:GL_{4}(F)\rightarrow GL_{4}(F)$. Then $\tau(y)=y$ for
$y\in GL_{4}(F_{0})$ and $\tau$ acts as an automorphism of order 2 on
$GL_{4}(q)$. Then the highest weight of the representation
$\rho^{\tau}:h\rightarrow\tau(\rho(h))$ for $h\in SL_{2}(F)$ is
$p^{m}(1+p^{i})\omega_{1}$. As $\tau$ acts trivially on $GL_{4}(F_{0})$ and
hence on $\rho(G)$, it follows that $i$ is a multiple of $d/2$, as claimed.

Let $F_{0}=\mathbb{F}_{q}$. By the above, $d$ does not divide $i$, and we are
left to show that $i$ is not a multile of $d/2$. Suppose the contrary. Then
the mapping $x\rightarrow x^{p^{i}}$ for $x\in F$ yields the Galois
automorphism of $\mathbb{F}_{q}$ of order 2. It follows that the trace of
every element $\rho(h),h\in H$ lies in the subfield $F_{1}$, say, of index 2
in $\mathbb{F}_{q}$. Then $\rho(H)$ can be realized over $F_{1}$, a
contradiction. ∎

######  Corollary 5.3.

Let $K=GL_{4}(p^{m})=GL(V)$, where $m>1$. Then for every $j|2m$, $j>1$, there
is an absolutely irreducible subgroup $G_{j}\cong PSL_{2}(p^{j})$ of K. In
addition we have:

$(1)$ if $p\neq 3$ and $t\in G_{j}$, $j>1$, is of order $(p+1)/(2,p+1)$ then t
is almost cyclic and $m_{t}(V)=2;$

$(2)$ if $p=3$, $j$ is even and $t\in G_{j}$ of order $4$ then t is almost
cyclic and $m_{t}(V)=2$.

###### Proof.

Let $\mathbf{G}$ be as in Lemma 5.1, and let $\rho$ be an irreducible
representation of $\mathbf{G}$ with highest weight $(1+p)\omega_{1}$. If $j|m$
then $G_{j}:=\rho(SL_{2}(p^{j}))\cong PSL_{2}(p^{j})$ is irreducible for $j>1$
by Lemma 5.1. Obviously, $G_{j}\subset GL_{4}(p^{j})\subset GL_{4}(p^{m}).$ If
$j\not|\,m$ then $j$ is even. Take $i=j/2$ in Lemma 5.1. Then
$G_{j}=\rho(SL_{2}(p^{j}))$ is irreducible and $G_{j}\subset
GL_{4}(p^{j/2})\subset GL_{4}(p^{m})$ as $j/2$ divides $m$.

If $p\neq 3$ then let $s\in SL_{2}(p)\subset SL_{2}(p^{j})$ be of order $p+1$.
By Lemma 5.1, $s$ is almost cyclic on $V$ and $m_{t}(V)=2.$

If $p=3$ then $SL_{2}(9)\subset SL_{2}(p^{j})$ has an element of order 8. As
above, take $i=1$ in Lemma 5.1. Then $t=\rho(s)\in G_{j}$ is is almost cyclic
and $m_{t}(V)=2$. ∎

######  Remark 5.4.

It is not true that the irreducible subgroups of $GL_{4}(F)$ isomorphic to
$G_{j}\cong PSL_{2}(p^{j})$ are conjugate even if they contain elements $s$
with $m_{s}(V)=2$. Indeed, let $\rho_{i}$ be an irreducible representation of
$\mathbf{G}\cong SL_{2}(F)$ with highest weight $(1+p^{i})\omega_{1}$, $i|j$,
$i<j$. Set $G_{ij}=\rho_{i}(SL_{2}(p^{j}))$. Then $G_{ij}$ is irreducible and
$G_{ij}\cong G_{j}$. Let $s_{i}\in SL_{2}(p^{j})$ be of order $p^{i}+1$ and
$t_{i}=\rho_{i}(s_{i})\in G_{ij}$. Then $t_{i}$ has eigenvalue 1. Let
$i^{\prime}|j,$ and $i\neq i^{\prime}<j$. Then $\rho_{i^{\prime}}(s_{i})\in
G_{ij}$ does not have eigenvalue $1$. Therefore, $\rho_{i}$ and
$\rho_{i^{\prime}}$ are not equivalent and, moreover, $G_{ij}$ and
$G_{i^{\prime}j}$ are not conjugate in $GL_{4}(F)$.

######  Lemma 5.5.

Let q be a p-power, and $r\geq 1$ an integer. If $r\neq 1$ then there exists
an odd prime $\ell$ such that $\ell$ divides $q^{r}+1$ and does not divide
$q^{j}-1$ for $j\leq r$. This remains true for $r=1$ unless $q=8$ or $q+1$ is
a $2$-power. In the exceptional cases $q+1$ is an $\ell$-power.

###### Proof.

By Zsigmondy’s theorem (see [11, 5.2.14]), if $q^{2}=p^{2d}>p^{2}$ and
$q^{2}\neq 64$ then there is a prime $\ell|(q^{2}-1)=p^{2d}-1$ such that
$\ell\not|(p^{j}-1)$ for every $j<2d$. Clearly, $\ell$ is odd here. If
$q^{2}=64$ then take $\ell^{2}=9$. If $q=p$ then let $\ell$ be the maximal
prime dividing $q+1$. If $\ell\neq 2$ then $\ell\not|(q-1)$ as desired.
Otherwise, $\ell=2$ and $q+1$ is a 2-power, and the claim follows.∎

######  Lemma 5.6.

Let $H=GL_{4r}(q)=GL(V)$, $q^{r}\neq 3$. Then there exists an irreducible
subgroup $Y\cong PSL_{2}(q^{2r})$ of H and $h\in PSL_{2}(q^{r})\subset Y$ such
that $|h|$ is a prime power, h is almost cyclic on V and the multiplicity of
eigenvalue $1$ of h equals $2r$. Moreover, $V=V_{1}+V_{2}$, where $V_{1}$ is
the $1$-eigenspace of $h$, $hV_{2}=V_{2}$ and $h$ acts irreducibly on $V_{2}$.
If $r\neq 1$ then h can be chosen of prime order.

###### Proof.

In assumptions of Lemma 5.1 set $G=SL_{2}(q^{2r})$ and $i=r$. By Lemma 5.5,
there exists $s\in SL_{2}(q^{r})\subset G$ be such that $|s|$ is a prime
power, $|s|$ divides $q^{r}+1$ and does not divide $q^{2t}-1$ for any positive
integer $t<r$. If $r\neq 1$ then we can choose $s$ to be of a prime order. In
addition, we can choose $s$ to be of odd order unless $r=1$ and $q+1$ is a
2-power.

Set $h=\rho(s)$. By Lemma 5.1, $\rho(G)$ is irreducible and contained in
$GL_{4}(q^{r})$ (see Lemma 5.2). Let $V^{\prime}$ be the underlying space of
$GL_{4}(q^{r})$. Then $h=\rho(s)$ is conjugate in $GL_{4}(q^{r})$ to a matrix
$\mathop{\rm diag}\nolimits(1,A,1)$, where $A\in GL_{2}(q^{r})$. As $\det
A=1$, it follows that $A$ does not have eigenvalue 1. Therefore,
$V^{\prime}=V^{\prime}_{1}+V^{\prime}_{2}$, where $V^{\prime}_{1}$ is the
1-eigenspace of $h$ on $V^{\prime}$ and $sV^{\prime}_{2}=V^{\prime}_{2}$.

Next embed $GL_{4}(q^{r})$ into $GL_{4r}(q)$ by viewing $\mathbb{F}_{q^{r}}$
as a vector space over $\mathbb{F}_{q}$. Then $V^{\prime}_{1},V_{2}^{\prime}$
and $V^{\prime}$ can be viewed as vector spaces $V_{1},V_{2},V$, say, over
$\mathbb{F}_{q}$, and we have $\dim V_{1}=\dim V_{2}=2r$. We claim that $h$ is
irreducible on $V_{2}$. Indeed, if $h$ is reducible on $V_{2}$ then $|h|$
divides $q^{j}-1$ for $j<2r$ and also $q^{2r}-1$, hence $q^{(2r,j)}-1$ by [10,
Hilfsatz 2(a)]. So we can assume $j|2r$, hence $j\leq r$. As the order of $h$
is equal to $|s|$ or $|s|/2$, we conclude that $h$ is irreducible on $V_{2}$
unless $|h|=|s|/2$, in particular, $|s|$ is a 2-power. This happens only when
$q+1$ is a 2-power and then we can choose $s$ with $|s|=q+1$. Then $(q-1)/2$
is odd, so $|h|=(q+1)/2$ divides $q-1$, which implies $q+1=4$. This
contradicts the assumption. ∎

###  5.2 Regular semisimple elements in representations of quasi-simple finite
groups of Lie type

Recall that the notions of cyclic and almost cyclic elements are meaningful
for elements $g\in GL_{n}(F_{1})$ for any subfield $F_{1}$ of $F$, that is, we
simply apply the definition to $g$ as an element of $GL_{n}(F)$.

######  Lemma 5.7.

Let $G$ be a finite quasi-simple group of Lie type in defining characteristic
$p>0$, let $V$ be a non-trivial irreducible $\mathbb{F}_{q}G$-module for some
p-power $q$, $\dim V>2$ and $\overline{V}=V\otimes F$. Let $s\in G$ be a
regular semisimple element and let $m_{s}(V)$ be the maximal eigenvalue
multiplicities of $s$ on $\overline{V}$. Suppose that $s$ is almost cyclic on
$\overline{V}$ and $m_{s}(\overline{V})\geq\dim V/2.$ Then $G$ is isomorphic
to $SL_{2}(q^{\prime})$, or $PSL_{2}(q^{\prime})$ for some p-power
$q^{\prime}$. Let $\rho$ be the representation afforded by V. Then we can
assume that $G\cong SL_{2}(q^{\prime})$ and one of the following holds:

$(1)$ $\dim V=3$, $\rho(G)\cong PSL_{2}(q^{\prime})$, $q$ is a power of
$q^{\prime}$, $p\neq 2$ and $s^{2}\in Z(G)$;

$(2)$ $\dim V=4$, $p\neq 2,3$, $\rho(G)\cong SL_{2}(q^{\prime})$, $q$ is a
power of $q^{\prime}$ and $s^{3}\in Z(G);$

$(3)$ $\dim V=4$, $\rho(G)\cong PSL_{2}(q^{\prime})$, $q^{\prime}>p$, $q^{2}$
is a $q^{\prime}$-power, and if $q^{\prime}=3^{l}$ then $l$ is not a prime;

$(4)$ $\dim V=4r$, $r>1$, $\rho(G)\cong PSL_{2}(q^{\prime})$, $q^{2r}$ is a
power of $q^{\prime}$, and $m_{s}(V)=\dim V/2.$

###### Proof.

Let $\mathbf{G}$ be a simply connected simple algebraic group such that
$G=\mathbf{G}^{\sigma}$ and $\sigma$ is some Steinberg endomorphism of
$\mathbf{G}$. Saying that $s\in G$ is a regular semisimple element means that
$s$ is regular in $\mathbf{G}$.

(i) Suppose first that $V$ is absolutely irreducible. Then $V$ extends to a
$\mathbf{G}$-module $\overline{V}$. By Corollary 1.3, $\mathbf{G}$ is of type
$A_{1}$, so $\mathbf{G}\cong SL_{2}(F)$ and $3\leq\dim\overline{V}=\dim V\leq
4$. We will also use $\rho$ to denote the corresponding representation of
$\mathbf{G}$.

Then $G\cong SL_{2}(q^{\prime})$ for some $p$-power $q^{\prime}$. Let $s\in G$
and let $\mathop{\rm diag}\nolimits(a,a^{-1})$ with $a^{2}\neq 1$ be a
conjugate of $s$ in $SL_{2}(F)$.

If $\dim\overline{V}=3$ then $p\neq 2$ and $\overline{V}$ is a Frobenius twist
of the irreducible $\mathbf{G}$-module with highest weight $2\omega_{1}$; the
matrix of $\rho(s)$ is $\mathop{\rm diag}\nolimits(a^{2},1,a^{-2})$, which is
cyclic unless $a^{2}=a^{-2}$, and hence $s^{2}=-\mathop{\rm Id}\nolimits\in
Z(G)$. One observes that such $s$ is conjugate to an element of $SL_{2}(p)$.
So $s\in SL_{2}(q^{\prime})$ whenever $SL_{2}(q^{\prime})\subseteq SL_{2}(q)$,
equivalently, $q$ is a $q^{\prime}$-power.

Let $\dim\overline{V}=4$. Then $\overline{V}$ is a Frobenius twist of the
irreducible $\mathbf{G}$-module with highest weight $3\omega_{1}$ (for $p\neq
3$) or $(1+p^{i}\omega_{1})$ for some $i>0$. By Lemma 3.13, we can assume that
the highest weight of $\overline{V}$ is $3\omega_{1}$ or
$(1+p^{i}\omega_{1})$. In the former case $\rho(G)\cong SL_{2}(q^{\prime})$
and the matrix of $s$ on $\overline{V}$ is $\mathop{\rm
diag}\nolimits(a^{3},a,a^{-1},a^{-3})$. If this matrix is not cyclic then
either $a^{6}=1,p\neq 3$, and hence $s^{3}\in Z(G)$, as required, or $a^{4}=1$
in which case $s$ is not almost cyclic. In the latter case $q^{\prime}>p$ ,
$\rho(G)\cong PSL_{2}(q^{\prime})$ and the matrix of $s$ on $\overline{V}$ is
$\mathop{\rm
diag}\nolimits(a^{p^{i}+1},a^{p^{i}-1},a^{-p^{i}+1},a^{-p^{i}-1})$. If this
matrix is not cyclic then either $a^{2p^{i}+2}=1$ or $a^{2p^{i}-2}=1$ or
$a^{2p^{i}}=1$; the latter implies $a^{2}=1$, which is false as $s$ is
regular. If $a^{2p^{i}+2}=a^{2p^{i}-2}$ then $a^{4}=1$, whence $p\neq 2$ and
$a^{2}=-1$. One checks that $s$ is not almost cyclic on $\overline{V}$ in this
case. In two other cases $s$ is almost cyclic on $\overline{V}$. By Lemma 5.2,
$\rho(G)$ is contained in $GL_{4}(q^{\prime})$ or $GL_{4}(\sqrt{q^{\prime}})$,
whence $q^{2}$ is a power of $q^{\prime}$. If $p\neq 3$ then
$SL_{2}(q^{\prime})$ has an element $s$ in question for every $q^{\prime}$
(see Corollary 5.3), so (3) holds in this case (provided $q^{\prime}>p$).

Let $p=3$ and let $q=3^{m}$ for some $m>1$. By Corollary 5.3, if $q^{\prime}$
is a square then $SL_{2}(q^{\prime})$ contains an element $s$ with properties
required. Suppose that $q^{\prime}$ is not a square. Then $q^{\prime}=3^{l}$
with $l$ odd, and $l>1$ (as $q^{\prime}>p=3$). By Lemma 5.2, $q^{\prime}$ is a
minimal 3-power $x$ such that $\rho(SL_{2}(q^{\prime}))\subseteq GL_{4}(x)$.
So $GL_{4}(q^{\prime})\subseteq GL_{4}(q)$, and hence $q$ is a power of
$q^{\prime}$.

Suppose that $l$ is a prime. By Lemma 5.1, $s$ is contained in a subgroup
isomorphic to $SL_{2}(p^{i})$, so $|s|$ divides $3^{2i}-1$, and also
$3^{2l}-1$ as $s\in SL_{2}(3^{l})$. By [10, Hilfsatz 2(a)], $|s|$ divides
$3^{(2i,2l)}-1$. As $l$ is a prime, $(2i,2l)\in\\{2,l,2l\\}$, so either
$(i,l)=1$ or $l|i$. In the latter case $\rho(SL_{2}(q^{\prime}))$ is reducible
by Lemma 5.2. So $(i,l)=1$ so $|s|$ divides $3^{2}-1=8$. As $l$ is odd, we
have $(3^{l}\pm 1,8)\neq 4$,whence $|s|=4$, but then $\rho(s)$ is not almost
cyclic.

Suppose that $l$ is not a prime, and let $l^{\prime}$ be a prime with
$l^{\prime}|l$. Let $i=l^{\prime}$ and let $s\in SL_{2}(q^{\prime})$ be of
order $3^{l^{\prime}}+1$. Then the group $\rho(SL_{2}(q^{\prime}))$ is
irreducible by Steinberg’s theorem (see [9, §2.11]), $\rho(s)$ is almost
cyclic and $m_{s}(\overline{V})=2$ by Lemma 5.1. This completes the proof of
the case with $\dim\overline{V}=4$.

(ii) Suppose that $V$ is not absolutely irreducible. Then, by Schur’s lemma,
the centralizer of $\rho(G)$ in ${\rm End}(V)$ is a field $F_{0}$, say, and
$|F_{0}|=q^{r}$ for some $r>1$. It follows that $\rho(G)\subset
GL_{n/r}(q^{r})\subset GL(V)$ ([11, Lemma 2.10.1]).

Therefore, $\overline{V}=V\otimes F$ as an $FGL_{n/r}(q^{r})$-module is a
direct sum $\overline{V}_{1}\oplus\cdots\oplus\overline{V}_{r}$, where
$\overline{V}_{1},\ldots,\overline{V}_{r}$ are irreducible
$FGL_{n/r}(q^{r})$-modules that are Galois conjugate to each other (by ${\rm
Gal}(\mathbb{F}_{q^{r}}/\mathbb{F}_{q})$), and each
$\overline{V}_{1},\ldots,\overline{V}_{r}$ is absolutely irreducible as
$FG$-module, see [11, Lemma 2.10.2]. Then $\dim V=r\dim\overline{V}_{1}$. Note
that $\dim\overline{V}_{1}>2$. (Indeed, if $\dim\overline{V}_{i}=2$ then the
eigenvalues of $s$ on $\overline{V}_{i}$ are $a_{i},a_{i}^{-1}$, where $\pm
1\neq a_{i}\in F$. As $m_{s}(\overline{V})>1$, we have
$a_{i}\in\\{a_{j},a_{j}^{-1}\\}$, hence $s$ is not almost cyclic on $V$, a
contradiction.) So $r<\dim V/2$.

Clearly, $s$ is almost cyclic on each
$\overline{V}_{1},\ldots,\overline{V}_{r}$. As these are Galois conjugate to
each other, it follows that $m_{s}(V)\leq
m_{s}(\overline{V}_{1})+\cdots+m_{s}(\overline{V}_{r})=r\cdot
m_{s}(\overline{V}_{1})$.

As above, each $\overline{V}_{1},\ldots,\overline{V}_{r}$ extends to a
$\mathbf{G}$-module. By Corollary 1.3,
$m_{s}(\overline{V}_{i})<\dim\overline{V}_{i}/2$ for every $i=1,\ldots,r$
(which is a contradiction so the result follows), unless possibly when
$\mathbf{G}$ is of type $A_{1}$ and $\dim\overline{V}_{1}=3,4$. So we can
assume that $\mathbf{G}\cong SL_{2}(F)$ and then $G=SL_{2}(q^{\prime})$ for
some $p$-power $q^{\prime}$.

Since $s$ is almost cyclic on $\overline{V}$ and $s,s^{-1}$ are conjugate in
$G$, the eigenvalue of $s$ of multiplicity greater than one is 1 or $-1$
(Lemma 3.2(4)).

If $\dim\overline{V}_{1}=3$ and $m_{s}(\overline{V}_{1})>1$ then $s^{2}\in
Z(\mathbf{G})$, and $s$ is almost cyclic on $V$ implies $\dim V=3$, which is a
contradiction as $r>1$.

Let $\dim\overline{V}_{1}=4$. If the highest weight of $\overline{V}_{1}$ is
$3p^{i}\omega_{1}$ with $p\neq 2,3$ then $s^{3}\in Z(\mathbf{G})$. Then $s$ is
conjugate to an element of a subgroup $SL_{2}(p)$. The restrictions of
$\overline{V}_{j}$ to $SL_{2}(p)$ for $j=1,\ldots,r$ yield isomorphic modules.
As $s$ is almost cyclic and $r>1$, this is a contradiction.

Suppose that the highest weight of $\overline{V}_{1}$ is
$(1+p^{i})\omega_{1}$. Then $\rho_{j}(\mathbf{G})\cong PSL_{2}(F)$ for every
$j=1,\ldots,r$, so $Z(G)$ acts trivially on $V$.

By the above, $\overline{V}_{1}$ is an irreducible $FGL_{n/r}(q^{r})$-modules
and absolutely irreducible as $FG$-module, that is,
$\rho_{1}(G)=\rho_{1}(SL_{2}(q^{\prime}))$ is absolutely irreducible subgroup
of $GL_{4}(q^{r})$. By (3), $q^{2r}$ is a power of $q^{\prime}$, as claimed in
(4). In addition, $m_{s}(\overline{V}_{1})\in\\{1,2\\}$, so
$m_{s}(\overline{V})\in\\{1,2r\\}$ as required. ∎

Observe that the case (4) is genuine. Indeed, by Lemma 5.6, if
$q^{2r}=q^{\prime}$ then $GL_{4r}(q)$ contains an irreducible subgroup
isomorphic to $PSL_{2}(q^{\prime})$ in which some element $s$ of prime order
is almost cyclic on $V$, $m_{s}(V)=\dim V/2$ and $s$ is irreducible on
$(1-s)V$.

######  Remark 5.8.

One can wish to further refine item (4) of Lemma 5.7 by determining
$q^{\prime}$ and the irreducible groups $X\cong PSL_{2}(q^{\prime})\subset
GL_{4r}(q)$ up to conjugacy. However, we shall not work in this direction.
Observe that, in notation of Lemma 5.7(4), $\rho(G)$ remains irreducible in
some larger groups $GL_{4r}(q^{t})$. For instance, let $t$ be an arbitrary
prime such that $(4r,t)=1$. If $\rho$ is reducible in $GL_{4r}(q^{t})$ then
$\rho(PSL_{2}(q^{\prime}))=\rho_{1}(PSL_{2}(q^{\prime}))\oplus\cdots\oplus\rho_{t}(PSL_{2}(q^{\prime}))$,
where $t|4r$. This is a contradiction.

###### Proof of Theorem 1.4.

The main assertion of the theorem follows from Lemma 5.7.

We will then prove the converse claim. If $4|\dim V$ then this follows from
Lemma 5.6. If $\dim V=3$, $p\neq 2$ then consider an irreducible
representation $\rho$ of $SL_{2}(F)$ with highest weight $2\omega_{1}$, and
set $\rho^{\prime}=\rho|_{SL_{2}(q^{\prime})}$. Then for $s\in
SL_{2}(q^{\prime})$ with $s^{2}=-\mathop{\rm Id}\nolimits$ we have
$\rho^{\prime}(s)=\mathop{\rm diag}\nolimits(-1,1,-1)$ under a suitable basis
of $V$.∎

######  Corollary 5.9.

In assumptions of Lemma 5.7 suppose that $q=2$. Then item $(3)$ or $(4)$ Lemma
5.7 holds and $q^{\prime}=2^{2r}$. In addition, $\rho(G)\subset
SO^{-}_{4r}(q)$.

###### Proof.

It is obvious that (1) and (2) of Lemma 5.7 do not hold. Let $q^{\prime}$ be
as in Lemma 5.7. If (3) holds then $q^{\prime}|4$ and $q>2$, so $q^{\prime}=4$
as required. Let (4) hold. Suppose the contrary, that $q^{\prime}<2^{2r}$. We
have seen in the proof of Lemma 5.7 that $V\otimes
F=\overline{V}_{1}\oplus\cdots\oplus\overline{V}_{r}$, where
$\overline{V}_{1},\ldots,\overline{V}_{r}$ are irreducible $SL_{2}(F)$-modules
of dimension 4, $s$ is almost cyclic on each $\overline{V}_{l}$ and
$m_{s}(\overline{V}_{l})=2$ for $l=1,\ldots,r$. Let
$\omega=(2^{j}+2^{i})\omega_{1}$ be the highest weight of $V_{1}$, where
$1\leq j<i$. By Lemma 3.13, we can assume here that $j=1$. By Lemma 5.1, $s$
is conjugate to an element of $SL_{2}(2^{i})$ and $\rho_{l}(s)$ is conjugate
to an element of $GL_{4}(2^{i})$, where $\rho_{l}$ is the representation
afforded by $\overline{V}_{l}$. As $s$ is almost cyclic on $V$, the modules
$\overline{V}_{l}|_{SL_{2}(2^{i})}$ for $l=1,\ldots,r$ are pairwise non-
isomorphic. In addition, these are twisted (or conjugate) to each other by
elements of the Galois group ${\rm Gal}(\mathbb{F}_{2^{r}}/\mathbb{F}_{2})$.
It follows that $i=r$ and $\omega=(1+2^{r})\omega_{1}$. So $q^{\prime}$ is a
power of $2^{r}$ and $2^{2r}$ is a power of $q^{\prime}$ by Lemma 5.7(4). Here
$q^{\prime}\neq 2^{r}$ as $\rho_{l}(G)$ is irreducible (see Lemma 5.1). So
$q^{\prime}=2^{2r}$ and $G=SL_{2}(2^{2r})$ as claimed. In addition,
$\rho_{1}(G)\subset O^{-}_{4}(2^{r})$, and hence $\rho(G)\subset
O^{-}_{4r}(2)$ by [10, Hilfsatz 1(d) and Bemerkung at page 148]. ∎

###  5.3 Non-regular semisimple elements in representations of quasi-simple
finite groups of Lie type

In order to deal uniformly with $p=2$ and $p>2$ we denote by $SO_{8}(F)$ the
connected component of the orthogonal group $O_{8}(F)$.

######  Lemma 5.10.

Let $s\in{\mathbf{G}}=Spin_{8}(F)$ be a non-central semisimple element, and V
the natural $G$-module. Suppose that s is almost cyclic on V.

$(1)$ If $\lambda$ is an eigenvalue of s on V of multiplicity $m_{s}(V)>1$
then $\lambda\in\\{1,-1\\}$ and either $1$ or $-1$ is not an eigenvalue of
$s$.

$(2)$ Let $G\cong{}^{3}D_{4}(q)$ and $s\in G$. Then s is regular.

###### Proof.

Set $\mathbf{H}=SO_{8}(F)$ and let $h:=\mathbf{G}\rightarrow\mathbf{H}$ be the
representation of $\mathbf{G}$ with highest weight $\omega_{1}$.

(1) By Lemma 3.1, $\lambda^{-1}$ is an eigenvalue of $s$ on $V$ of
multiplicity equal to that of $\lambda.$ As $s$ is almost cyclic on $V$, we
have $\lambda\in\\{1,-1\\}$. The $1$-eigenspace and the $-1$-eigenspace are
well known to be non-degenerate and of even dimension. As $s$ is almost cyclic
on $V$, both $1,-1$ cannot be eigenvalues of $s$. It follows that
$C_{\mathbf{H}}(s)$ is connected, see [28, Lemma 2.2].

(2) Let $V_{1}\subset V$ be the eigenspace of dimension $m_{s}(V)$. We have
$V=V_{1}\oplus V_{2}$ for some $s$-stable subspace $V_{2}$ of $V$, and $s$ is
cyclic on $V_{2}$. Then $C_{\mathbf{H}}(h(s))$ contains the subgroup
$\mathbf{Y}_{1}:=\mathop{\rm diag}\nolimits(SO(V_{1}),\mathop{\rm
Id}\nolimits)$.

Suppose the contrary, that $s$ is not regular. Then $C_{G}(s)$ and hence
$C_{\mathbf{H}}(h(s))$ contains a unipotent element $u\neq 1$. If $\dim
V_{1}=2$ then $u$ is a transvection, however, $\mathbf{H}$ contains no
transvection. So $\dim V_{1}\in\\{6,4\\}$.

It is known that $C_{\mathbf{G}}(s)$ is a reductive group and contains a
maximal torus $T$ of $\mathbf{G}$ [20, Theorem 5.8].

Note that $\mathbf{G}=Spin_{8}(F)$ is the universal group of type $D_{4}$. Let
$\sigma$ be the Steinberg endomorphism of $\mathbf{G}$ such that