# Spectra of non-regular elements in irreducible representations of simple
algebraic groups

Donna M. Testerman  Institut de Mathématiques, Station 8, École Polytechnique
Fédérale de Lausanne, CH-1015 Lausanne, Switzerland<EMAIL_ADDRESS>and  Alexander Zalesski  Department of Physics, Mathematics and Informatics,
Academy of Sciences of Belarus, 66 Prospect Nezalejnasti, Minsk 220000,
Belarus<EMAIL_ADDRESS>Dedicated to the memory of Ernest
Vinberg

###### Key words and phrases:

semisimple elements, irreducible representations, eigenvalue multiplicities,
simple linear algebraic groups

###### 2010 Mathematics Subject Classification:

20G05, 20G07, 20E28

Testerman was supported by the Fonds National Suisse de la Recherche
Scientifique grant number 200021-175571. The second author acknowledges the
hospitality of the EPFL Institute of Mathematics during research visits to
Lausanne.

Abstract We study the spectra of non-regular semisimple elements in
irreducible representations of simple algebraic groups. More precisely, we
prove that if $G$ is a simply connected simple linear algebraic group and
$\phi:G\to{\rm GL}(V)$ is a non-trivial irreducible representation for which
there exists a non-regular non-central semisimple element $s\in G$ such that
$\phi(s)$ has almost simple spectrum, then, with few exceptions, $G$ is of
classical type and $\dim V$ is minimal possible. Here the spectrum of a
diagonalizable matrix is called _simple_ if all eigenvalues are of
multiplicity 1, and _almost simple_ if at most one eigenvalue is of
multiplicity greater than 1. This yields a kind of characterization of the
natural representation (up to their Frobenius twists) of classical algebraic
groups in terms of the behavior of semisimple elements.

Keywords: simple algebraic groups, representations, eigenvalue multiplicities,
non-regular elements

##  1\. Introduction

A rather general problem which has received attention in the literature can be
stated as that of classifying irreducible group representations whose image
contains a matrix with a certain specified property. In this paper we
concentrate on a property of the eigenvalue multiplicities of a semisimple
element of simple linear algebraic groups in their irreducible
representations. (Henceforth we will use “algebraic group” to mean “linear
algebraic group”.) Although problems on eigenvalues in group representations
are important for many applications, little can be said in full generality. In
fact, the behavior of individual elements in the image of a representation is
quite unpredictable. For a discussion of this and related questions, we refer
the reader to [26].

Here, we consider matrices with almost simple spectrum, that is, matrices
having at most one eigenvalue of multiplicity greater than 1. More precisely,
we will address the following:

######  Problem 1.1.

Let G be a simple algebraic group defined over an algebraically closed field
$F$. Determine the irreducible representations $\phi$ of G such that $\phi(G)$
contains a non-scalar diagonalizable matrix with almost simple spectrum.

Note that the notion of matrices with almost simple spectrum is a natural
generalization of the similar notion of pseudo-reflections, the latter being
diagonalizable matrices with two eigenvalues, one of which has multiplicity 1.
The classification of irreducible matrix groups generated by pseudo-
reflections was an important project enjoying numerous applications. (See [25,
24, 13, 12].) We note as well that the consideration of Problem 1.1 is an
extension of the analogous question for finite quasi-simple groups of Lie type
and their representations in defining characteristic (see [18, 19]), as well
as the classification (in [11, 17]) of irreducible representations of simple
algebraic groups for which a maximal torus acts with 1-dimensional weight
spaces [11, 17]. A similar problem for irreducible representations of finite
simple groups occurring as subgroups of ${\rm GL}_{n}(\mathbb{C})$ has been
studied in [7].

While Problem 1.1 is a question about semisimple elements, there is a natural
generalization of the notions of simple and almost simple spectra to matrices
that are not diagonalizable. Let $V$ be a finite-dimensional vector space over
a field $F$ and $M\in{\rm GL}(V)$. Then $M$ is called _cyclic_ if, for some
$v\in V$, the space $V$ is spanned by the vectors $v,Mv,M^{2}v,\ldots$, and
_almost cyclic_ if, for some $\lambda\in F$, $M$ is conjugate to a matrix
${\rm diag}(\lambda\cdot{\rm Id},M_{1})$, where $M_{1}$ is a cyclic matrix.
Almost cyclic matrices in the images of irreducible representations of finite
simple groups are studied in [5, 6, 4] (in certain special cases). Now let
$g\in G$ and let $\phi$ be an irreducible representation such that $\phi(g)$
is almost cyclic. If $g$ is not semisimple, then $g=su=us$ with $u\neq 1$
unipotent, and one sees that $\phi(u)$ has a single non-trivial Jordan block.
Such representations have been determined in [16] and [23]. On the other hand,
if $g$ is semisimple, and $\phi(g)$ is almost cyclic, then $\phi(g)$ has
almost simple spectrum; indeed $\phi(g)$ has at most two eigenvalues, one of
which has multiplicity 1.

Let us now return to our considerations of semisimple elements of $G$ whose
spectrum in some irreducible representation of $G$ is almost simple. As every
semisimple element $s\in G$ lies in a maximal torus, the condition for
$\phi(s)$ to have simple spectrum implies that all weight multiplicities of
$\phi$ are equal to 1. The irreducible representations whose set of weights
satifies this property are determined in [11] for tensor-indecomposable
representations and completed in [17]. By analogy, one could expect $\phi$ in
Problem 1 to have all but one weight multiplicity equal to 1. And indeed this
is the case, as the following result, which will be etablished in §3, shows.

######  Theorem 1.2.

Let $G$ be a simple algebraic group defined over an algebraically closed field
$F$ and $\phi$ an irreducible representation of $G$. Then the following
statements are equivalent:

$(1)$ The matrix $\phi(s)$ has almost simple spectrum for some non-central
semisimple element $s\in G$.

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

Theorem 1.2 will be relevant to our consideration of Problem 1, especially as
the irreducible representations of simple algebraic group satisfying (2) have
been determined in [22]. The above theorem is best possible in the sense that
in order to obtain a more precise result one has to specify the nature of the
semisimple element $s$ in question. We recall that an element $g\in G$ is said
to be _regular_ if $\dim(C_{G}(g))$ is equal to the rank of $G$; for $g$
semisimple this is equivalent to $C_{G}(g)^{\circ}$ being abelian [14, Chapter
III, Corollary 1.7]. Our investigations show that, with very few exceptions, a
non-central semisimple element $s$ having an almost simple spectrum in an
irreducible representation $\phi$ must be regular.

######  Theorem 1.3.

Let $G$ be a simply connected simple algebraic group defined over an
algebraically closed field $F$ 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. If the spectrum of $s$ on $V$ is almost simple, then
one of the following holds:

  1. (1)

$G$ 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. (2)

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

  3. (3)

$G=C_{2}$, $p\neq 2$ and $\dim V=5;$

  4. (4)

$G=D_{4}$ and $\dim V=8.$

The irreducible representations of $G$ of the dimensions given in Theorem 1.3
are well known; a description of elements $s$ which have almost simple
spectrum on $V$ is provided in Section 3.

Notation We fix an algebraically closed field $F$ of characteristic $p\geq 0$.

Throughout the paper $G$ is a simple simply connected linear algebraic group
defined over $F$. All $G$-modules considered are rational finite-dimensional
$FG$-modules. For a $G$-module $V$ (or a 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 rational and irreducible.
If $H$ is a subgroup of $G$ then we write $V|_{H}$ for the restriction of a
$G$-module $V$ to $H$.

We fix a maximal torus $T$ in $G$, which in turn defines the roots of $G$ as
well as the weights of $G$-modules and representations. The $T$-weights of a
$G$-module $V$ are the irreducible constituents of the restriction of $V$ to
$T$. As $T$ is fixed, we will omit the reference to $T$ and write “weights” in
place of “$T$-weights”. The set of weights of $V$ is denoted by $\Omega(V)$.
For $\mu\in\Omega(V)$, the dimension of the $\mu$-weight space $\\{v\in
V:tv=\mu(t)v$ for all $t\in T\\}$ is called the multiplicity of $\mu$ in $V$.
The Weyl group of $G$ is denoted by $W(G)$; as $W(G)=N_{G}(T)/T$, the
conjugation action of $N_{G}(T)$ on $T$ yields an action of $W(G)$ on $T$ and
consequently on the set of $T$-weights. The $W(G)$-orbit of $\mu\in\Omega$ is
denoted by $W(G)\mu$. The set $\Omega={\rm Hom}(T,F)$ (the rational
homomorphisms of $T$ to the multiplicative group of $F$) is called the weight
lattice, which is a free $\mathop{\mathbb{Z}}\nolimits$-module of finite rank
called the rank of $G$.

With an algebraic group $H$ is associated the Lie algebra of $H$ denoted here
by ${\rm Lie}(H)$. For the simple group $G$, we denote the set of roots (that
is, the non-zero weights of the $G$-module ${\rm Lie}(G)$) by $\Phi$ or
$\Phi(G)$. For notions of closed subsystems of $\Phi$ and subsystem subgroups
see [9, §13.1]. The $\mathop{\mathbb{Z}}\nolimits$-span of $\Phi$ is called
the root lattice and is denoted here by $R$ or $R(G)$. In $\Phi(G)$, we fix a
base $\Pi=\\{\alpha_{1},\ldots,\alpha_{n}\\}$ and order the simple roots
according to the Dynkin diagrams as in [1]. The weights in $R$ are called
radical. For each root $\alpha\in\Phi(G)$, we choose a non-zero element
$X_{\alpha}$ in the $\alpha$-weight space of $T$ on ${\rm{Lie}}(G)$. Thus,
$FX_{\alpha}$ is the Lie algebra of a $T$-invariant one-dimensional unipotent
subgroup $U_{\alpha}$ of $G$; see [9, Theorem 8.16] for details.

One defines a non-degenerate, $W(G)$-invariant, symmetric bilinear form on
$\Omega\otimes_{\mathop{\mathbb{Z}}\nolimits}{\mathbb{R}}$, which we express
as $(\mu,\nu)$. The elements $\omega_{i}$ satisfying
$2(\omega_{j},\alpha_{i})=(\alpha_{i},\alpha_{i})\delta_{ij}$ for $1\leq
i,j\leq n$ belong to $\Omega$ and are called fundamental dominant weights [1,
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 $\sum
a_{i}\omega_{i}$, for $a_{i}\in\mathbb{Z}$; the set of $\nu$ with
$a_{1},\ldots,a_{n}\geq 0$ is denoted by $\Omega^{+}$, the set of dominant
weights. We set $\Omega^{+}(V)=\Omega^{+}\cap\Omega(V)$, so $\Omega^{+}(V)$ is
the set of dominant weights of $V$. In what follows, we will regularly use so-
called “Bourbaki weights”, when $R(G)$ is of type $A_{r-1},B_{r},C_{r}$ or
$D_{r}$, which are elements of a $\mathop{\mathbb{Z}}\nolimits$-lattice
containing $\Omega$ with basis
$\varepsilon_{1},\varepsilon_{2},\ldots,\varepsilon_{r}$; the explicit
expressions of the fundamental weights and the simple roots of $G$ in terms of
$\varepsilon_{i}$’s are given in [1, Planches I – IV].

There is a standard partial ordering of elements of $\Omega$: for
$\mu,\mu^{\prime}\in\Omega$ we write $\mu\prec\mu^{\prime}$ and
$\mu^{\prime}\succ\mu$ if and only if $\mu\neq\mu^{\prime}$ and
$\mu^{\prime}-\mu\in R^{+}$. (We write $\mu\preceq\mu^{\prime}$ and
$\mu^{\prime}\succeq\mu$ to allow $\mu=\mu^{\prime}$.) If $\mu$ and
$\mu^{\prime}$ are dominant weights such that $\mu^{\prime}\preceq\mu$, we say
$\mu$_is subdominant to_ $\mu$. For the notion of a minuscule weight see [2,
Ch. VIII, §7.3], where they are tabulated. Every irreducible $G$-module has a
unique 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 $\omega$. Suppose that $p>0$; a dominant weight $\sum
a_{i}\omega_{i}$ is called $p$-restricted if $0\leq a_{i}<p$ for all
$i=1,\ldots,n$. For uniformity, we often do not separate the cases with $p=0$
and $p>0$; by convention, when $p=0$, a $p$-restricted weight is simply a
dominant weight. An irreducible $G$-module is called $p$-restricted if its
highest weight is $p$-restricted. For classical groups $G$, that is, those
with root system one of $A_{n}$, $B_{n}$, $C_{n}$ or $D_{n}$, the module with
highest weight $\omega_{1}$ is called the natural module and the associated
representation the _natural representation_. (There is an exceptional case,
when $G=B_{n}$ and $p=2$, where the natural module is the Weyl module of
highest weight $\omega_{1}$.)

The maximal height root of $\Phi(G)$ is denoted by $\omega_{a}$; this is the
highest weight of ${\rm{Lie}}(G)$ and affords a non-trivial composition factor
of the adjoint module ${\rm{Lie}}(G)$. The short root module for $G$ of type
$B_{n},C_{n},F_{4},G_{2}$ is the irreducible $G$-module all of whose non-zero
weights are short roots. This is unique, and the highest weight of the short
root module is maximal among short roots (with respect to $\prec$). An
irreducible $G$-module is called _tensor-decomposable_ if it is a tensor
product of two or more non-trivial irreducible modules, similarly for
representations.

If $h:G\rightarrow G$ is a surjective algebraic group homomorphism and $\phi$
is a representation of $G$ then the $h$-twist $\phi^{h}$ of $\phi$ is defined
as the mapping $g\rightarrow\phi(h(g))$ 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 and
$k$ a nonnegative integer, then the modules $V^{Fr^{k}}$ are called Frobenius
twists of $V$; if $V$ is irreducible with highest weight $\omega$ then the
highest weight of $V^{Fr^{k}}$ (for $k\geq 0$) is $p^{k}\omega$.

If $p=2$, then for every $n$ there is a surjective algebraic group
homomorphism $B_{n}\rightarrow C_{n}$ with trivial kernel (so this is an
abstract group isomorphism); for our purposes, the choice between these two
groups is irrelevant, so we choose to work with $C_{n}$ when $p=2$.

For the natural $2n$-dimensional module $M$ of the group $C_{n}$, $n\geq 2$, a
basis $\\{e_{i},f_{i}\ |\ 1\leq i\leq n\\}$ is called _symplectic_ $M$ if
$\\{e_{i},f_{i}\\}$ is a hyperbolic pair for all $i$ and $M$ is the orthogonal
direct sum of the spaces $\langle e_{i},f_{i}\rangle$, $1\leq i\leq n$.

Finally, we will assume $n\geq 1$ for $A_{n}$, $n>1$ for $C_{n}$, $n>2$ for
$G=B_{n}$, and $n>3$ for $D_{n}$. 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.

##  2\. Preliminaries

######  Lemma 2.1.

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. Suppose that M has
almost simple spectrum. Then

  1. (1) $M_{1}$ and $M_{2}$ have simple spectrum, and

  2. (2) if $M_{i}$ is similar to $M_{i}^{-1}$ for $i=1,2$, then the eigenvalue multiplicities of $M$ do not exceed $2$.

###### Proof.

(1) Suppose that $M_{1}$ has an eigenvalue $e$, say, of multiplicity $r>1$.
Let $b_{1},b_{2}$ be distinct eigenvalues of $M_{2}$. Then $eb_{1},$ $eb_{2}$
are distinct eigenvalues of $M$, each of multiplicity greater than 1. This
implies the claim.

(2) Suppose the contrary, and let $e$ be an eigenvalue of $M$ of multiplicity
at least 3. By (1), $M_{1}$ and $M_{2}$ have simple spectra so $e=a_{i}b_{i}$
for $i=1,2,3$ 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 $M$ has almost simple spectrum and is
similar to $M^{-1}$ by hypothesis, we have $e=e^{-1}$, so
$a_{1}b_{2}=a_{2}^{-1}b_{1}^{-1}$. If
$(a_{2}^{-1},b_{2})\neq(a_{1},b_{1}^{-1})$, then $a_{1}b_{2}$ is an eigenvalue
of $M$ of multiplicity 2 and so is equal to $e$. But this then implies
$a_{1}b_{2}=a_{1}b_{1}$, contradicting that the $b_{i}$ are distinct. Hence
$a_{2}=a_{1}^{-1}$ and $b_{2}=b_{1}^{-1}$. Similarly,
$a_{1}b_{3}=a_{3}^{-1}b_{1}^{-1}$ implies that $a_{3}=a_{1}^{-1}$ and
$b_{3}=b_{1}^{-1}$. But now $a_{2}=a_{3}$ contradicting that the $a_{i}$ are
distinct. ∎

######  Definition 2.2.

Let $V$ be a $G$-module and $\mu,\nu\in\Omega(V)$, $\mu\neq\nu$. We say that
$s\in T$ _separates the weights $\mu$ and $\nu$_ if $\mu(s)\neq\nu(s)$. If
this holds for every pair of distinct weights $\mu,\nu$ of $V$, we say that
$s$ _separates the weights of_ $V$.

If $s$ separates the weights of $V$ then the eigenvalue multiplicities of $s$
acting on $V$ are simply the weight multiplicities of $V$.

######  Lemma 2.3.

Let $V$ be a non-trivial $G$-module. Let $S\subset T$ be the set of all $t\in
T$ that separate the weights of $V$. Then

  1. (1) $S$ is a nonempty Zarisky open subset of $T$.

  2. (2) Suppose that at most one weight of $V$ has multiplicity greater than $1$. Then, for all $s\in S$, the spectrum of $s$ is almost simple.

###### Proof.

$(1)$ Let $\mu,\nu$ be weights of $V$, $\mu\neq\nu$. 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 $T_{\mu,\nu}$, is a
proper closed subset of $T$. Moreover, $S=T\setminus(\cup T_{\mu,\nu})$, and
so (1) follows.

$(2)$ Let $s\in S$, so that $\mu(s)\neq\nu(s)$ whenever $\mu\neq\nu$ are
weights of $V$. Then 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 $\mu$, giving (2). ∎

We will require the following characterization of regular semisimple elements.

######  Proposition 2.4.

[14, Ch. III, §1, Corollary 1.7] Let $G$, $T$ be as usual, and let $s\in T$.
Then the following conditions are equivalent:

$(1)$ $s$ is regular;

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

$(3)$ for all $\alpha\in\Phi(G)$, $\alpha(s)\neq 1$;

$(4)$ $C_{G}(x)^{\circ}$ is a torus.

######  Lemma 2.5.

Let $V,V_{1},V_{2}$ be non-trivial G-modules. Let $s\in T\setminus Z(G)$ have
almost simple spectrum on V.

$(1)$ Suppose that $V=V_{1}\otimes V_{2}$. Then 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 $V_{2}$.

###### Proof.

The first claim of (1) follows from Lemma 2.1. For the second assertion,
suppose that $s$ is not regular. Then by Proposition 2.4, $C_{G}(s)$ 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
Lemma 2.1(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
$(\mu_{1}+\mu)(s)=(\mu_{2}+\mu)(s)$. As $s\notin Z(G)$, the spectrum of $s$ on
$V$ is not almost simple, a contradiction.∎

With regards to applying Lemma 2.5(2), we note that
$\Omega(V)=\Omega(V_{1})+\Omega(V_{2})$ if $V=V_{1}\otimes V_{2}$. For certain
choices of $V,V_{1},V_{2}$, and under certain conditions on $p$, we may deduce
that $\Omega(V)=\Omega(V_{1})+\Omega(V_{2})$, for $V$ different from
$V_{1}\otimes V_{2}$. See Lemma 2.8(2) below.

We recall here some basic facts about the set of weights of irreducible
representations of a simple algebraic group defined over a field of
characteristic $0$ (which are derived from analogous statements about the
weights of irreducible representations of simple Lie algebras defined over
$\mathbb{C}$). Fixing a maximal torus $T_{H}$ of a simple algebraic group $H$
defined over ${\mathbb{C}}$, and adopting the notation fixed earlier, so in
particular, writing $W(H)$ for the Weyl group of $H$ relative to $T_{H}$, let
$\lambda$ be a dominant $T_{H}$-weight. Then the set of weights of the
irreducible ${\mathbb{C}}H$-module with highest weight $\lambda$ is precisely
the set

$\\{w(\mu)\ |\ \mu\in\Omega^{+},\mu\preceq\lambda,w\in W(H)\\},$

that is, the $W(H)$-conjugates of all weights which are subdominant to the
highest weight $\lambda$. From this one directly deduces the following facts:

  1. (1)

Let $\lambda,\mu\in\Omega^{+}$ and $\mu\prec\lambda$. Let $V_{\lambda}$,
respectively $V_{\mu}$ be the associated irreducible ${\mathbb{C}}H$-modules;
then $\Omega(V_{\mu})\subset\Omega(V_{\lambda})$.

  2. (2)

[2, Ch. VIII, §7, Proposition 10] Let $\lambda,\mu\in\Omega^{+}$, with
associated irreducible ${\mathbb{C}}H$-modules $V_{\lambda}$, $V_{\mu}$; then
$\Omega(V_{\lambda+\mu})=\Omega(V_{\lambda}\otimes V_{\mu})$.

  3. (3)

[2, Ch. VIII, §7, Propositions 4 and 6] Let $\lambda\in\Omega^{+}$,
$\lambda\neq 0$. If $\lambda$ is a radical weight, then some root is a weight
of $V_{\lambda}$; otherwise $\Omega(V_{\lambda})$ contains some minuscule
weight.

We now return to the situation where the field $F$ is of arbitrary
characteristic. We will use a fundamental result of Premet, which relies on
the following definition and notation.

######  Definition 2.6.

We set $e(G)=1$ for $G$ of types $A_{n},D_{n},E_{n}$, $e(G)=2$ for $G$ of type
$B_{n},C_{n},F_{4}$, and $e(G)=3$ for $G$ of type $G_{2}$.

######  Theorem 2.7.

[10, Theorem 1] 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 2.7 and the preceding remarks now gives:

######  Lemma 2.8.

Assume $p=0$ or $p>e(G)$. Let $\lambda,\mu\in\Omega^{+}$, where $\lambda$ is
$p$-restricted, and let $V_{\lambda}$, respectively, $V_{\mu}$ be the
associated irreducible $G$-modules. Then the following hold.

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

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

For the following result we introduce an additional notation. Let
$\Psi\subset\Phi$ be a closed subsystem. Then we set $G(\Psi)$ to be the
subgroup generated by the $T$-root subgroups corresponding to roots in $\Psi$.

######  Theorem 2.9.

[20, Theorem 1] Let $G$ be a simple algebraic group with root system $\Phi$.
If $\Phi$ is of type $B_{n}$, assume ${\rm char}(F)\neq 2$. Let
$R_{1},R_{2}\subset\Phi$ be closed subsystems such that the subgroups
$G_{1}:=G(R_{1})$ and $G_{2}:=G(R_{2})$ are simple and $[G_{1},G_{2}]=1$. Let
$\phi$ be an irreducible representation of $G$. Then one of the following
holds:

  1. (1) $\phi|_{G_{1}G_{2}}$ contains a composition factor which is non-trivial for both $G_{1}$ and $G_{2};$

  2. (2) $G$ is classical and $\phi$ is a Frobenius twist of either the natural representation or the dual of the natural representation of $G;$

  3. (3) $G=C_{n}$ with $p=2$, $G=B_{n}$ with $n>2$, or $G=D_{n}$ with $n>4$, and $\phi$ is a Frobenius twist of the irreducible representation of highest weight $\omega_{n}$, or one of $\omega_{n}$ and $\omega_{n-1}$ if $G=D_{n}$.

The following lemma will allow us in some cases to reduce our analysis of
elements with almost simple spectrum to representations all of whose weights
occur with multiplicity one.

######  Lemma 2.10.

Let G be a simple algebraic group of rank greater than $1$ and $s\in
T\setminus Z(G)$. Assume that $p=0$ or $p>e(G)$. Let $\mu\neq 0$ be a
$p$-restricted dominant weight.

  1. (1) Let $\mu_{m}$ be the minimal non-zero weight subdominant to $\mu$. Assume that the spectrum of $s$ on $V_{\mu_{m}}$ is not almost simple. Then the following hold:

    1. (i) if $\mu$ is not radical, then the spectrum of $s$ on $V_{\mu}$ is not almost simple;

    2. (ii) if $\mu$ is radical and the multiplicity of the weight $0$ in $V_{\mu_{m}}$ is at most $1$, then the spectrum of $s$ on $V_{\mu}$ is not almost simple;

    3. (iii) if $\mu$ is radical and the multiplicity of the weight $0$ on both $V_{\mu}$ and $V_{\mu_{m}}$ is greater than $1$, then the spectrum of $s$ on $V_{\mu}$ is not almost simple;

    4. (iv) if $0\prec\mu_{m}\preceq\mu$ and $s$ is non-regular, then the spectrum of $s$ on $V_{\mu}$ is not almost simple.

  2. (2) Suppose that $\omega_{a}\prec\mu$, the multiplicity of the weight $0$ in $V_{\mu}$ is greater than $1$, and the spectrum of $s$ on $V_{\omega_{a}}$ is not almost simple. Then the spectrum of $s$ on $V_{\mu}$ is not almost simple.

  3. (3) Suppose that $\omega_{a}\prec\mu$, $s$ is non-regular, and the spectrum of $s$ on $V_{\omega_{a}}$ is not almost simple. Then the spectrum of $s$ on $V_{\mu}$ is not almost simple.

###### Proof.

By assumption, Theorem 2.7 holds, and we may apply Lemma 2.8. If $\mu_{m}$ is
non-radical, then all weight multiplicities of $V_{\mu_{m}}$ are well known to
be equal to $1$; (i) follows. Together with the hypothesis in (ii) about the
multiplicity of the zero weight, we observe that if the spectrum of $s$ on
$V_{\mu_{m}}$ is not almost simple then there are $4$ distinct weights
$\lambda_{1},\lambda_{2},\mu_{1},\mu_{2}$ of $V_{\mu_{m}}$ such that
$\lambda_{1}(s)=\lambda_{2}(s)\neq\mu_{1}(s)=\mu_{2}(s)$. Then Lemma 2.8(1)
implies that these weights are weights of $V_{\mu}$, and the result follows.

In case (iii), $\mu_{m}$ is the maximal height short root and the multiplicity
of any non-zero weight in $V_{\mu_{m}}$ is equal to $1$. Saying that the
spectrum of $s$ on $V_{\mu_{m}}$ is not almost simple means that there exist
weights $\lambda_{1},\lambda_{2}$ of $V_{\mu_{m}}$ such that
$\lambda_{1}(s)=\lambda_{2}(s)\neq 1$. As these weights are weights of
$V_{\mu}$ (again by Lemma 2.8(1)) and, by hypothesis, the weight $0$ occurs in
$V_{\mu}$ with multiplicity greater than $1$, the result follows.

(iv) As $s$ is non-regular, there exists $\alpha\in\Phi(G)$ such that
$\pm\alpha(s)=1$ (Proposition 2.4). Since the spectrum of $s$ on $V_{\mu_{m}}$
is not almost simple, there are distinct short roots $\beta,\gamma$ such that
$\beta(s)=\gamma(s)\neq 1$. Then Lemma 2.8 implies that
$\pm\alpha,\beta,\gamma$ are weights of $V_{\mu}$, and the result follows.

For (2), first note that the multiplicity of the weight $0$ in
$V_{\omega_{a}}$ is greater than $1$ unless $(G,p)=(A_{2},3)$ (here we again
rely on the prime restrictions in the hypotheses). This case is considered in
(ii). In all other cases, saying that the spectrum of $s$ on $V_{\omega_{a}}$
is not almost simple means that there are two roots $\alpha,\beta$ such that
$\alpha(s)=\beta(s)\neq 1$. As the weights of $V_{\omega_{a}}$ occur as
weights of $V_{\mu}$ and the weight $0$ occurs in $V_{\mu}$ with multiplicity
greater than $1$, the result follows.

Finally, the case (3) follows as (iv) above, where one has to replace
$V_{\mu_{m}}$ by $V_{\omega_{a}}$ and “short roots” by “roots”. ∎

We complete this section with a straightforward observation about the natural
modules for classical groups.

######  Lemma 2.11.

Let $G$ be a classical type group and assume $p\neq 2$ when $G$ is of type
$B_{n}$. Let $V=V_{\omega_{1}}$ and $s\in G$ be a non-central semisimple
element.

$(1)$ For $G=A_{n}$ or $C_{n}$, if $s$ is regular, then $s$ has simple
spectrum on $V$

$(2)$ Let $G=B_{n}$. 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 $1$.

$(3)$ Let $G=D_{n}$. 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 the spectrum of $s$ on $V$ is
not almost simple then that of $s$ on $V_{\omega_{2}}$ is not almost simple.

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

###### Proof.

(1) This is straightforward and well known.

For the remainder of the proof, we take $T$ to be the maximal torus consisting
of the diagonal matrices in the image of the natural representation of $G$. We
now turn to (2) and the first statement of (3). Observe that $\Omega(V)$
consists of the weights $\pm\varepsilon_{i}$, $1\leq i\leq n$, together with
the weight $0$ in case $G=B_{n}$. In addition, $s$ is regular if and only if
$\alpha(s)\neq 1$ for every root $\alpha$. Set $a_{i}=\varepsilon_{i}(s)$ and
recall that $\Phi(D_{n})=\\{\pm\varepsilon_{i}\pm\varepsilon_{j}\ |\ 1\leq
i<j\leq n\\}$ and
$\Phi(B_{n})=\\{\pm\varepsilon_{i}\pm\varepsilon_{j},\pm\varepsilon_{r}\ |\
1\leq i<j\leq n,1\leq r\leq n\\}$. So $s$ is regular if and only if $a_{i}\neq
a_{j}$ and $a_{i}\neq a_{j}^{-1}$ for every $i\neq j$, and if in addition, for
$G=B_{n}$, $a_{i}\neq 1$ for all $1\leq i\leq n$. So if $G=B_{n}$, we see that
$s$ is regular if and only if either all of the eigenvalues $a_{1}^{\pm
1},a_{2}^{\pm 1},\dots,a_{n}^{\pm 1}$ are distinct and distinct from $1$, or
there exists a unique $i$ with $a_{i}=a_{i}^{-1}$. If $a_{i}=a_{i}^{-1}=-1$,
then $s$ is regular if and only if all eigenvalues of $s$ on $V$ different
from $-1$ occur with multiplicity $1$, and $-1$ occurs with multiplicity at
most $2$. Now if $G=D_{n}$, then $s$ is regular if and only if
$a_{1}^{\pm},\dots,a_{n}^{\pm}$ are distinct or there exists $1\leq i\leq n$
such that $a_{i}=a_{i}^{-1}$. In the latter case, $s$ is regular if and only
if all eigenvalues different from $a_{i}$ occur with multiplicity $1$ and the
eigenvalue $a_{i}$ can occur with multiplicity at most $2$, as claimed.

For the final statement of (3), let $G=D_{n}$ and suppose that the spectrum of
$s$ on $V$ is not almost simple. Then, without loss of generality, we may
assume $a_{i}=a_{j}$ for some $1\leq i\neq j\leq n$. Then
$(\varepsilon_{i}-\varepsilon_{k})(s)=(\varepsilon_{j}-\varepsilon_{k})(s)$
and
$(-\varepsilon_{i}-\varepsilon_{k})(s)=(-\varepsilon_{j}-\varepsilon_{k})(s)$
for every $k\neq i,j$. Recall that the non-zero weights of $V_{\omega_{2}}$
are the roots in $\Phi(G)$, and the zero weight occurs with multiplicity at
least $2$. Assume for a contradiction that the spectrum of $s$ on
$V_{\omega_{2}}$ is almost simple. Then
$(\varepsilon_{i}-\varepsilon_{k})(s)=(\varepsilon_{j}-\varepsilon_{k})(s)=(-\varepsilon_{i}-\varepsilon_{k})(s)=(-\varepsilon_{j}-\varepsilon_{k})(s)=1$,
whence $-\varepsilon_{i}(s)=\varepsilon_{i}(s)=\varepsilon_{k}(s)$ for all
$1\leq k\leq n$. As $s\notin Z(G)$, we get a contradiction.

(4) This follows from $(1),(2)$ and $(3)$. ∎

##  3\. Reduction theorem, and proof of Theorem 1.2

Let $S\subset{\rm GL}(V)$ be an abelian subgroup and let $\mathop{\rm
Irr}\nolimits(S)$ denote the set of irreducible $F$-linear representations of
$S$ and write $1_{S}$ for the trivial representation. For $\eta\in\mathop{\rm
Irr}\nolimits S$, set $V_{S}(\eta)=\\{v\in V:sv=\eta(s)v$ for all $s\in S\\}$.
If $V_{S}(\eta)\neq\\{0\\}$, we say $\eta$ is an $S$-weight of $V$ and we call
$V_{S}(\eta)$ the _$\eta$ -weight space for $S$_. As throughout $G$ is a
simple algebraic group defined over $F$ and $T\subset G$ is a maximal torus of
$G$. If $V$ is a rational $G$-module then $V$ is a direct sum of $T$-weight
spaces and for any subgroup $S\subseteq T$, these weight spaces are
$S$-invariant. Thus for $\eta\in\mathop{\rm Irr}\nolimits(S)$, $V_{S}(\eta)$
is a sum of $T$-weight spaces of $V$. We establish here a result about such
subgroups $S$ of $T$, and later will apply this to the case where $S$ is the
subgroup generated by an element $s\in T$.

Recall (see for instance [9, §7]) that for any rational representation
$\rho:G\to{\rm GL}(V)$, we have a corresponding representation of
${\rm{Lie}}(G)$, namely $d\rho:{\rm{Lie}}(G)\to{\rm{Lie}}({\rm GL}(V))$. For
$g\in G$, let $t_{g}:G\to G$ denote the automorphism induced by conjugation by
$g$. Then using the basic definitions and properties of the differential, we
have that $t_{\rho(g)}\circ\rho=\rho\circ t_{g}$ and so

${\rm Ad}(\rho(g))\circ d\rho=d\rho\circ{\rm Ad}(g).$

######  Theorem 3.1.

(Reduction theorem) Let $G$ be a simple algebraic group, $T$ a maximal torus
of $G$, and $S\subseteq T$ a subgroup such that $C_{G}(S)\neq G$. Let $V$ be
an irreducible $G$-module with $p$-restricted highest weight. Let
$V_{S}(\eta)$ be an $S$-weight space of $V$, for some $\eta\in\mathop{\rm
Irr}\nolimits S$. Suppose that $\dim V_{S}(\eta)=k>1$ and that all other
$S$-weight spaces on $V$ are of dimension $1$. Then all non-zero $T$-weights
of $V$ are of multiplicity $1$.

###### Proof.

Set $E=V_{S}(\eta)$. For $\mu\in\Omega(V)$, write $M_{\mu}$ for the $T$-weight
space of $V$ associated to $\mu$. Suppose that $\dim M_{\mu}\geq 2$, for some
$\mu\in\Omega(V)$. Then $M_{\mu}\subset E$. As $\dim M_{\mu}=\dim M_{w(\mu)}$
for any $w\in W$, we necessarily have $M_{w(\mu)}\subset E$. Now let
$\rho:G\to{\rm GL}(V)$ be the corresponding rational representation of $G$.
For a root $\alpha\in\Phi$, $\alpha$ induces a $1$-dimensional representation
$\lambda_{\alpha}$ of the group $S$.

Consider first the case where $\lambda_{\alpha}\neq 1_{S}$, for all
$\alpha\in\Phi$. Recall the notation $X_{\alpha}\in{\rm{Lie}}(G)$, a root
vector associated to the root $\alpha$, a fixed element which spans the Lie
algebra of the associated root group. Then $d\rho(X_{\alpha})E\subset
V_{S}(\eta\lambda_{\alpha})$. Since $\eta\lambda_{\alpha}\neq\eta$, this
latter $S$-weight space is of dimension at most $1$. Hence
$K_{\alpha}:=\ker((d\rho(X_{\alpha}))\mid_{E})$ is of dimension at least
$k-1$. Setting $K_{1}=\cap_{\alpha\in\pm\Pi}K_{\alpha}$, we see that
$K_{1}\subset V$ is a proper ${\rm{Lie}}(G)$-submodule on which
${\rm{Lie}}(G)$ acts trivially. But by [3], $V$ is an irreducible
${\rm{Lie}}(G)$-module, and so $K_{1}=\\{0\\}$. Therefore, $k=\dim E\leq 2n$,
where $n$ is the rank of $G$. We can now show that $\mu=0$; for otherwise the
$W$-orbit of $\mu$ is of length at least $n+1$ (the exact values are in the
[26, Table 1]). Therefore, $\dim\sum M_{w(\mu)}\geq 2(n+1)$, which is a
contradiction.

Consider now the case where there exists $\alpha\in\Phi$ such that
$\lambda_{\alpha}=1_{S}$. Set $M^{\prime}:=\sum_{w\in W}M_{w(\mu)}$, so that
$M^{\prime}\subseteq E$. Let
$R_{0}=\\{\alpha\in\Phi:\lambda_{\alpha}=1_{S}\\}$, $R_{2}=\Phi\setminus
R_{0}$. Since $S$ is non-central, $R_{0}\neq\Phi$ and $R_{2}\neq\emptyset$.
Let $R_{1}$ be the set of roots $\alpha$ such that
$\dim(d\rho(X_{\alpha})M^{\prime})\leq 1.$ By the considerations of the first
case above, $R_{2}\subseteq R_{1}$. Moreover, we claim that $R_{1}$ is
$W$-stable. Indeed for $w\in W$, choose $\dot{w}\in N_{G}(T)$ such that
$w=\dot{w}T$. Then

$\rho(\dot{w})d\rho(X_{\alpha})M^{\prime}=\rho(\dot{w})d\rho(X_{\alpha})\rho(\dot{w})^{-1}\rho(\dot{w})M^{\prime}={\rm
Ad}(\rho(\dot{w}))(d\rho(X_{\alpha}))M^{\prime}.$

By the remarks preceding the statement of the result, this latter is equal to

$d\rho({\rm Ad}(\dot{w})X_{\alpha})M^{\prime}=d\rho(X_{w(\alpha)})M^{\prime}$

and since
$\dim(\rho(\dot{w})(d\rho(X_{\alpha})M^{\prime}))=\dim(d\rho(X_{\alpha})M^{\prime})$,
we have the claim. Now, if all roots of $\Phi$ are of the same length then
$R_{1}=\Phi$, and we conclude as in the first case.

Hence we may assume that $\Phi$ has two root lengths and that the roots of
$R_{1}$ are of a single length. Note that $R_{0}=-R_{0}$ and $\beta,\gamma\in
R_{0}$ implies $\beta+\gamma\in R_{0}$ provided $\beta+\gamma$ is a root. This
implies (see for example [9, B.14]) that $R_{0}$ is a root system, that is,
$R_{0}$ is a closed subsystem of $\Phi$. Moreover, $R_{0}$ is of maximal rank
(equal to the rank of $\Phi$) as otherwise, by [9, B.18], $R_{0}$ lies in some
subsystem corresponding to a proper subset of $\Pi$, in which case $R_{2}$,
and so $R_{1}$ has roots of both lengths. So $R_{0}$ is a subsystem of maximal
rank, and by the classification of such, [9, B.18], one checks that in every
case $\Phi\setminus R_{0}=R_{2}$ again contains roots of both lengths and we
conclude as above.∎

######  Remark 3.2.

If $\omega=p^{k}\omega^{\prime}$, with $\omega^{\prime}$ $p$-restricted, then
the weights of $V_{\omega}$ are $p^{k}\mu$ for $\mu$ a weight of
$V_{\omega^{\prime}}$. Then $p^{k}\mu(s)=\mu(s^{p^{k}})$. As the mapping
$x\rightarrow x^{p}$ for $x\in F$ is bijective on $F$, the spectrum of $s$ on
$V_{\omega}$ is almost simple if and only if the spectrum of $s$ on
$V_{\omega^{\prime}}$ is almost simple.

We now take $S$ to be generated by a single element $s\in T$ and consider the
case of tensor-decomposable irreducible representations.

######  Lemma 3.3.

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 weight. Suppose that the spectrum of $s$ on $V_{\omega}$ is
almost simple. Then all weights of $V_{\omega}$ are of multiplicity $1$.

###### Proof.

By Steinberg’s tensor product theorem, $V_{\omega}=V_{p^{k_{1}}\mu_{1}}\otimes
V_{p^{k_{2}}\mu_{2}}\otimes\cdots\otimes V_{p^{k_{t}}\mu_{t}}$, where $t>1$
and $\mu_{1},\ldots,\mu_{k}$ are non-zero $p$-restricted weights and
$(k_{1},\dots,k_{t})$ are distinct non-negative integers. Then Lemma 2.5
implies that the spectrum of $s$ on each tensor factor is simple so the
weights of each tensor factor have multiplicity $1$. Furthermore, [17,
Proposition 2] implies that the weights of $V_{\omega}$ are of multiplicity 1
unless there exists $1\leq j<t$ such that $k_{j+1}=k_{j}+1$ and one of the
following holds:

  1. (i) $G=C_{n}$, $p=2$, $\mu_{j}=\omega_{n}$, $\mu_{j+1}=\omega_{1};$

  2. (ii) $G=G_{2}$, $p=2$, $\mu_{j}=\omega_{1}$, $\mu_{j+1}=\omega_{1};$

  3. (iii) $G=G_{2}$, $p=3$, $\mu_{j}=\omega_{2}$, $\mu_{j+1}=\omega_{1}$.

Moreover, in each of the cases (i), (ii) and (iii), the module
$V_{\mu_{j}}\otimes V_{p\mu_{j+1}}$ has a weight of multiplicity greater than
$1$. Hence if one of the three cases occurs, we deduce that $t=2$ and so we
can also assume that $j=1$ and $k_{1}=0$, that is,
$V_{\omega}=V_{\mu_{1}}\otimes V_{p\mu_{2}}$. We consider the above cases in
detail.

Case (i): Take $T$ to be the set of diagonal matrices in the image of the
natural representation of $G$. Here
$\Omega(V_{\omega_{n}})=\\{\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n}\\}$
and $\Omega(V_{2\omega_{1}})=\\{\pm 2\varepsilon_{1},\ldots,\pm
2\varepsilon_{n}\\}$. (As usual, we have adopted the notation of [1, Planche
III].) Let $\nu$ be a weight of $V_{\omega_{n}}$ with positive signs of both
$\varepsilon_{i}$ and $\varepsilon_{j}$, for some $1\leq i,j\leq n$, $i\neq
j$. As $\nu-2\varepsilon_{i}$ and $\nu-2\varepsilon_{j}$ are weights of
$V_{\omega_{n}}$, it follows that $\nu$ is also a weight of $V_{\omega}$ with
multiplicity at least 2. This remains true for weights where both
$\varepsilon_{i}$ and $\varepsilon_{j}$ have coefficient $-1$ or have opposite
coefficients. It follows that the restriction of $V_{\omega_{n}+2\omega_{1}}$
to $T$ contains a direct sum of at least two copies of $V_{\omega_{n}}|_{T}$.
Therefore, every eigenvalue of $s$ on $V_{\omega_{n}}$ is also an eigenvalue
of $s$ on $V_{\omega_{n}+2\omega_{1}}$, and occurs with multiplicity at least
$2$. So this case is ruled out as the spectrum of $s$ on $V_{\omega_{n}}$ is
simple.

Case (ii): Here the weights of $V_{\omega_{1}}$ are the short roots of $\Phi$,
and the following weights occur with multiplicity $2$ in $V_{\omega}$:
$3\alpha_{1}+\alpha_{2}$, $3\alpha_{1}+2\alpha_{2}$. Since the spectrum of $s$
on $V_{\omega}$ is almost simple, these roots must all take equal value on
$s$. In particular, $\alpha_{2}(s)=1$. But now the eigenvalue $5\alpha_{1}(s)$
occurs with multiplicity $2$ as well as $3\alpha_{1}(s)$, implying that
$\alpha_{1}(s)=1$ as well, contradicting the fact that $s$ is non-central.

Case (iii): This case is similar. Here the weights of $V_{\omega_{2}}$ are the
long roots of $\Phi$ and the zero weight, and the weights of $V_{\omega_{1}}$
are the short roots and the zero weight. We find that each of the weights
$3\alpha_{1}+\alpha_{2}$ and $\alpha_{2}$ occur with multiplicity $2$, and
deduce that $\alpha_{1}(s)=1$. But now the eigenvalue $\alpha_{2}(s)$ occurs
with multiplicity greater than $1$, as well as the eigenvalue $1$, and so
$\alpha_{2}(s)=1$ as well, again contradicting $s$ non-central. ∎

###### Proof of Theorem 1.2.

Using Lemma 2.3, we see that assertion (1) follows from assertion (2). We
apply Theorem 3.1, Remark 3.2 and Lemma 3.3 to obtain the reverse implication.
∎

##  4\. Commuting subgroups and a partial proof of Theorem 1.3

An essential element of our proof of Theorem 1.3 is an application of Theorem
2.9, which allows us to treat many of the groups and representations in a
uniform way. (See Proposition 4.5 below.) Let $s\in G$ be a non-regular
semisimple element. In order to apply Theorem 2.9, we need to find a pair of
subsystem subgroups $K,Y$ such that $[K,Y]=1$, $[K,s]=1$ and $[s,Y]\neq 1$.
For technical reasons, it will suffice to do this for groups other than
$B_{n},D_{n}$, and $G_{2}$.

######  Lemma 4.1.

Let $G={\rm SL}_{n}(F)$, $n>3$, and let $s\in T\setminus Z(G)$ be a non-
regular element. Then there are simple subsystem subgroups $K,Y$, normalized
by $T$, such that $[K,Y]=1$, $[K,s]=1$ and $[s,Y]\neq 1$, unless $n=4$ and, up
to conjugacy in $G$, $s=\mathop{\rm diag}\nolimits(a,a,a^{-1},a^{-1})$ or
$s=\mathop{\rm diag}\nolimits(a,a,-a^{-1},-a^{-1})$.

###### Proof.

We take $T$ to be the torus of diagonal matrices in $G$. As $s$ is non-regular
and non-central, we may assume that $s=\mathop{\rm
diag}\nolimits(b,b,a_{3},\ldots,a_{n})$, where $a_{3}\neq b$. Suppose first
that $a_{3}\neq a_{i}$ for some $i>3$. Set $K=\mathop{\rm diag}\nolimits({\rm
SL}_{2}(F),\mathop{\rm Id}\nolimits_{n-2})$, $Y=\mathop{\rm
diag}\nolimits({\rm Id}_{2},{\rm SL}_{n-2}(F))$. Next, suppose
$a_{3}=\cdots=a_{n}$. If $n>4$ then we can take $Y=\mathop{\rm
diag}\nolimits(1,{\rm SL}_{2}(F),\mathop{\rm Id}\nolimits_{n-3})$ and
$K=\mathop{\rm diag}\nolimits(\mathop{\rm Id}\nolimits_{n-2},{\rm
SL}_{2}(F))$. If $n=4$, then $s=\mathop{\rm diag}\nolimits(b,b,a,a)$ and
$b^{2}a^{2}=1$, whence $b=\pm a^{-1}$.∎

######  Remark 4.2.

If $G={\rm SL}_{4}(F)$, and $s=\mathop{\rm
diag}\nolimits(\lambda,\lambda,\lambda^{-1},\lambda^{-1})$ or $s=\mathop{\rm
diag}\nolimits(\lambda,\lambda,-\lambda^{-1},-\lambda^{-1})$, for $\lambda\in
F$, $\lambda^{4}\neq 1$, then $s$ is non-regular, non-central, and it is
impossible to find a pair of subsystem subgroups $K$, $Y$ such that $[s,K]=1$
and $[s,Y]\neq 1$. Moreover, the Jordan form of $s$ on the exterior square of
the natural $4$-dimensional module is $\mathop{\rm
diag}\nolimits(\lambda^{2},\lambda^{-2},1,1,1,1)$, which is non-central with
almost simple spectrum.

######  Lemma 4.3.

Let $G=C_{n}$, $n>1$, and let $s\in T\setminus Z(G)$ be a non-regular element.
Then there are simple subsystem subgroups $K$, $Y$ of $G$, normalized by $T$,
such that $[K,Y]=1$, $[K,s]=1$ and $[s,Y]\neq 1$, unless $n=2$ and with
respect to an ordered symplectic basis $(e_{1},f_{1},e_{2},f_{2})$ of
$V_{\omega_{1}}$, the Jordan form of s on the natural G-module is either
$\mathop{\rm diag}\nolimits(a,a^{-1},a,a^{-1})$, for $\pm 1\neq a\in F$, or
$s=\pm\mathop{\rm diag}\nolimits(1,1,-1,-1)$, for $p\neq 2$.

###### Proof.

The group $G=C_{n}={\rm Sp}_{2n}(F)$ contains a maximal rank subsystem
subgroup $H$ isomorphic to ${\rm Sp}_{2}(F)\times\cdots\times{\rm Sp}_{2}(F)$,
so every semisimple element is conjugate to an element of $H$. Therefore, we
can write the matrix of $s$ with respect to a suitable basis of the natural
$G$-module $V_{\omega_{1}}$ as $\mathop{\rm
diag}\nolimits(a_{1},a_{1}^{-1},\ldots,a_{n},a_{n}^{-1})$ for some
$a_{1},\ldots,a_{n}\in F$. By Lemma 2.11, the diagonal entries of $s$ are not
distinct. Hence either $a_{i}=\pm 1$ for some $i\in\\{1,\ldots,n\\}$, or,
replacing some $a_{i}$ by $a_{i}^{-1}$, we can assume that $a_{i}=a_{j}$ for
some $1\leq i<j\leq n$.

Suppose first that $a_{i}=\pm 1$ for some $i\in\\{1,\ldots,n\\}$ and assume
without loss of generality that $i=1$. If there exists $j$ such that
$a_{j}\neq\pm 1$, we can assume $j=n$ and then take $K=\mathop{\rm
diag}\nolimits({\rm Sp}_{2}(F),\mathop{\rm Id}\nolimits_{2n-2})$,
$Y=\mathop{\rm diag}\nolimits(\mathop{\rm Id}\nolimits_{2n-2},{\rm
Sp}_{2}(F))$. Otherwise, $s^{2}=1$ and $p\neq 2$. We can reorder
$a_{1},\ldots,a_{n}$ so that $a_{1}\neq a_{2}$, and if $n>2$ we take
$Y=\mathop{\rm diag}\nolimits({\rm Sp}_{4}(F),\mathop{\rm
Id}\nolimits_{2n-4})$, $K=\mathop{\rm diag}\nolimits(\mathop{\rm
Id}\nolimits_{2n-2},{\rm Sp}_{2}(F))$. If $n=2$, $s^{2}=1$ and $p\neq 2$, such
a choice is not possible and we have $s$ as in the final statement.

Now suppose that $a_{i}\neq\pm 1$ for all $i\in\\{1,\ldots,n\\}$, so there
exists $1\leq i<j\leq n$ such that $a_{i}=a_{j}$. In this case, there exists a
$2$-dimensional totally isotropic subspace of the underlying $2n$-dimensional
symplectic space on which $s$ acts as scalar multiplication. If $n>2$, then
$s$ is contained in a Levi subgroup $L=L_{1}\times L_{2}$ of $G$, where
$L_{1}\cong{\rm GL}_{2}(F)$ and $L_{2}\cong{\rm Sp}_{2n-4}(F)$. Moreover
$[s,L_{1}]=1$, so we can take $K=L_{1}$, $Y=L_{2}$. If $n=2$ then
$s=\mathop{\rm diag}\nolimits(a,a^{-1},a,a^{-1})$ as in the statement of the
result. ∎

######  Lemma 4.4.

Let $G\in\\{E_{6},E_{7},E_{8},F_{4}\\}$. Let $s\in T\setminus Z(G)$ be a non-
regular element. Then there exist simple subsystem subgroups $K$, $Y$,
normalized by $T$, such that $K$ is of type $A_{1}$, $[K,Y]=1$, $[K,s]=1$,
$[s,Y]\neq 1$.

###### Proof.

As $s$ is not regular, $C_{G}(s)$ contains root subgroups $U_{\pm\alpha}$ for
some root $\alpha\in\Phi$. Clearly, we can assume $\alpha$ to be a simple
root. Moreover, we can assume that $\alpha=\alpha_{1}$ if $G\neq F_{4}$,
otherwise, that $\alpha=\alpha_{1}$ or $\alpha_{4}$.

Denote by $R_{\alpha}$ the set of roots orthogonal to $\alpha$, and observe
that $R_{\alpha}$ is not empty. Set $Y=\langle U_{\pm\beta}:\beta\in
R_{\alpha}\rangle$ and $K=\langle U_{\pm\alpha}\rangle$. Then $[Y,K]=1$ and
$[K,s]=1$. If $[Y,s]\neq 1$, replacing $Y$ by a suitable simple subgroup of
$Y$, we are done.

We now assume $[s,U_{\beta}]=1$ for all $\beta\in R_{\alpha}$. In this
situation, as $s$ is non-central, $[s,U_{\gamma}]\neq 1$ for some simple root
adjacent to $\alpha$ in the Dynkin diagram. Moreover, the Dynkin diagram of
the above groups contains a node $\beta$, not adjacent to each of
$\alpha,\gamma$. In particular, $\beta\in R_{\alpha}$ and so
$[s,U_{\beta}]=1$, while $[s,U_{\gamma}]\neq 1$. So now we can take $K=\langle
U_{\pm\beta}\rangle$ and $Y=\langle U_{\pm\gamma}\rangle$.

This completes the proof.∎

We now apply the previous three lemmas and Theorem 2.9 to establish Theorem
1.3 for certain groups.

######  Proposition 4.5.

Let $G$ be of type $A_{n}$ for $n>3$, $C_{n}$ for $n>2$, or of type
$F_{4},E_{6},E_{7}$, or $E_{8}$. Let $V$ be a non-trivial irreducible
$G$-module and $s\in T\setminus Z(G)$. Suppose that the spectrum of s has on V
is almost simple. Then one of the following holds:

  1. (1) $s$ is regular,

  2. (2) $G=C_{n}$ with $p=2$ and the highest weight of V is $2^{m}\omega_{n}$, or

  3. (3) $G$ is classical and $V$ is a Frobenius twist of the natural or the dual of the natural module for $G$.

###### Proof.

Suppose that $s$ is not regular. By Lemma 4.1 for $A_{n}$, Lemma 4.3 for
$C_{n}$, and Lemma 4.4 for the other groups in the statement, there are simple
subsystem subgroups $K$, $Y$, normalized by $T$, such that $[K,Y]=1$,
$[K,s]=1$ and $[Y,s]\neq 1$. Then we apply Theorem 2.9 to $K$, $Y$ in place of
$G(R_{1}),G(R_{2})$ to conclude that either (2) or (3) holds or there is a
$KY$-composition factor $M$ of $V$ afforded by an irreducible representation
$\tau$ of $KY$, such that $\tau$ is non-trivial on both $K$ and $Y$. So we
assume neither (2) nor (3) holds, so we are in the latter situation, and aim
for a contradiction.

We first note that $TY=Y\cdot Z(TY)$, as $Y$ is simple. Therefore, as $s\in
T$, $s=s_{1}s_{Y}$ for some $s_{1}\in Z(TY)\subset T$ and $s_{Y}\in(T\cap Y)$.
As $[s,K]=1$ and $[Y,K]=1$, we have $[s_{1},K]=1$ and $[s_{1},YK]=1$. Also, as
$[s,Y]\neq 1$, we have $[s_{Y},Y]\neq 1$.

Now $M$ is a direct sum of eigenspaces for $s_{1}$. It follows that $\tau$ is
realized in one of the $s_{1}$-eigenspaces $M_{1}$, say, and hence the
spectrum of $s$ on $M_{1}$ is almost simple if and only if that of $s_{Y}$ on
$M_{1}$ is almost simple. Therefore, it suffices to show that the spectrum of
$\tau(s_{Y})$ is not almost simple.

Now $\tau=\tau_{K}\otimes\tau_{Y}$, where $\tau_{K}$, $\tau_{Y}$ are non-
trivial irreducible representations of $K$, $Y$, respectively. As
$[s_{Y},Y]\neq 1$, there are at least two distinct $s_{Y}$-eigenspaces on the
representation space corresponding to $\tau_{Y}$, each of them is of dimension
at least $2$ as $\tau_{K}(K)$ acts on each eigenspace and all $\tau_{K}(K)$
composition factors of $M$ are of dimension strictly greater than $1$. Hence,
the spectrum of $s_{Y}$ on $M$ is not almost simple, giving the desired
contradiction. ∎

######  Remark 4.6.

$(1)$ Let $G=C_{2}$, $p$ odd. If s is not as described in the exceptional
cases of Lemma 4.3 then Proposition 4.5 remains valid.

$(2)$ Note that the irreducible representation of $G=C_{2}$ with highest
weight $\omega_{2}$ induces an isomorphism between ${\rm PSp}_{4}(F)$ and
${\rm SO}_{5}(F)$, and the element $s=\pm\mathop{\rm
diag}\nolimits(1,1,-1,-1)$ in Lemma 4.3 acts as $\mathop{\rm
diag}\nolimits(1,-1,-1,-1,-1)$, hence has almost simple spectrum. Similarly,
the element $s=\mathop{\rm diag}\nolimits(a,a^{-1},a,a^{-1})$ acts as
$\mathop{\rm diag}\nolimits(a^{2},1,1,1,a^{-2})$, which has almost simple
spectrum provided $a^{2}\neq\pm 1$.

(3) In view of Lemma 2.5 and Proposition 4.5, to complete the proof of Theorem
1.3, it remains to consider $p$-restricted representations (of highest weight
$\lambda$) of the groups $B_{n}$ for $n>2$, $D_{n}$ for $n>3$, $C_{n}$ for
$p=2$ and $\lambda=\omega_{n}$, and the small rank groups $A_{2}$, $A_{3}$,
$C_{2}$, and $G_{2}$. We will handle the small rank groups in Section 6.1 and
complete the proof in Section 6.2 by dealing with the remaining groups.

##  5\. Weight levels

Set $\Lambda=\sum_{i=1}^{n}{\mathbb{Z}}\omega_{i}$, the weight lattice
associated with $\Phi$, and $\Lambda^{+}$ the set of dominant weights in
$\Lambda$. In this section we establish some results on $\Lambda$ in view of
applying the results in Section 2. Recall that a weight is radical if it is an
integral linear combination of roots. The irreducible $G$-module whose highest
weight is the maximal height short root is called the short root module. If
all weights are of the same length then any root is regarded as short, and the
short root module is $V_{\omega_{a}}$.

######  Definition 5.1.

Let

$\Lambda_{1}=\\{\mu\in\Lambda^{+}\ |\ \mbox{ if }\nu\preceq\mu\mbox{ for some
}\nu\in\Lambda^{+}\mbox{ then }\mu=\nu\ \\}.$

For $i>1$, let

$\Lambda_{i}=\\{\mu\in\Lambda^{+}\
,\mu\notin\Lambda_{1}\cup\cdots\cup\Lambda_{i-1}\\}\,|\ \mbox{ if
}\nu\prec\mu\mbox{ for some }\nu\in\Lambda^{+}\mbox{ then
}\nu\in\Lambda_{1}\cup\cdots\cup\Lambda_{i-1}\\}.$

The elements of $\Lambda_{i}$ are called _weights of level $i$_.

######  Lemma 5.2.

The sets $\Lambda_{1}$ and $\Lambda_{2}$ for the root systems of types
$A_{n}$, $B_{n}$, $C_{n}$ and $D_{n}$ are given in the table below. In
addition, we have

  1. (1) for $\Phi=B_{n}$, $n>2$, $\omega_{2}$ is the only radical weight in $\Lambda_{3}$;

  2. (2) for $\Phi=C_{n}$, $n>3$, $2\omega_{1},\omega_{4}$ are the only radical weights in $\Lambda_{3};$

  3. (3) for $\Phi=C_{2}$ or $C_{3}$, $2\omega_{1}$ is the only radical weight in $\Lambda_{3}$.

$\begin{array}[]{|l|c|c|}\hline\cr\Phi&\Lambda_{1}&\Lambda_{2}\\\ \hline\cr
A_{n},n\geq
1&0,\omega_{1},\dots,\omega_{n}&2\omega_{1},2\omega_{n},\omega_{1}+\omega_{n},\omega_{1}+\omega_{i},\omega_{i}+\omega_{n},i=2,\dots,n-1\\\
\hline\cr B_{n},n\geq 3&0,\omega_{n}&\omega_{1},\omega_{1}+\omega_{n}\\\
\hline\cr C_{n},n>2&0,\omega_{1}&\omega_{2},\omega_{3}\\\ \hline\cr
C_{2}&0,\omega_{1}&\omega_{2},\omega_{1}+\omega_{2}\\\ \hline\cr
D_{n},n>4&0,\omega_{1},\omega_{n-1},\omega_{n}&\omega_{2},\omega_{3},\omega_{1}+\omega_{n-1},\omega_{1}+\omega_{n}\\\
\hline\cr
D_{4}&0,\omega_{1},\omega_{3},\omega_{4}&\omega_{2},\omega_{1}+\omega_{3},\omega_{1}+\omega_{4},\omega_{3}+\omega_{4}\\\
\hline\cr\end{array}$

###### Proof.

By Lemma 2.8(3), $\Lambda_{1}$ consists of minuscule weights and the weight
$0$, justifying the entries in the column headed $\Lambda_{1}$ of the above
table. Furthermore, $\Lambda_{2}$ contains a unique radical weight, which is
the maximal short root (see for instance [21, Proposition 10]).

Let now $\omega=\sum a_{i}\omega_{i}\in\Lambda_{2}$ be a non-radical weight.
Suppose that $a_{i}\geq 2$ for some $i$. Then
$\omega^{\prime}=\omega-\alpha_{i}\in\Lambda^{+}$, so
$\omega^{\prime}\in\Lambda_{1}$. Inspecting $\Lambda_{1}$ and the expressions
of simple roots in terms of fundamental dominant weights, we observe that
$\omega^{\prime}+\alpha_{i}$ (for $\omega^{\prime}\in\Lambda_{1}$) is dominant
only if $\Phi$ is of type $A_{n}$ and
$\omega\in\\{2\omega_{1},2\omega_{n}\\}$; furthermore, it is straightforward
to see that in this latter case, we have
$2\omega_{1},2\omega_{n}\in\Lambda_{2}$. So we can assume that $a_{i}\leq 1$
for all $i$. Next we proceed case-by-case, still assuming
$\omega\in\Lambda_{2}$ a non-radical weight.

Consider first the case where $\Phi=A_{n}$. If $n=1,2$ then the result is
clear, so assume now $n>2$. Note that
$\omega_{i}+\omega_{j}\succ\omega_{i-1}+\omega_{j+1}$ for $1\leq i<j\leq n$ as
$\omega_{i}+\omega_{j}-\omega_{i-1}-\omega_{j+1}=\alpha_{i}+\cdots+\alpha_{j}$.
(Here $\omega_{0}$ and $\omega_{n+1}$ are understood to be zero.) So if
$a_{i},a_{j}\neq 0$ for some $i\neq j$, then
$\omega=\omega^{\prime}+\omega_{i-1}+\omega_{j+1}$ with
$\omega^{\prime}\in\Lambda_{1}$. Using the same reasoning for different pairs
of non-zero coefficients, we see that either $i=1$ and
$\omega^{\prime}=\omega_{1}$ or $j=n$ and $\omega^{\prime}=\omega_{n}$.
Finally, one observes that no weight obtained is subdominant to another one.
So $\Lambda_{2}$ is as in the table. This completes the consideration of
$\Phi=A_{n}$.

For $\Phi\neq A_{n}$, the argument differs, as some fundamental dominant
weights are radical. Recall that $\omega=\sum a_{i}\omega_{i}\in\Lambda_{2}$
is a non-radical weight and we have seen that $a_{i}\leq 1$ for all $i$. If
$\omega_{i}$ is a radical weight and $a_{i}>0$, then $\omega-\omega_{i}$ is
subdominant to the weight $\omega$, and hence
$0\neq\omega-\omega_{i}\in\Lambda_{1}$. So $\omega=\nu+\omega_{i}$, for some
$\nu\in\Lambda_{1}$, $\nu\neq 0$. Moreover, $\omega_{i}=\mu$, where $\mu$ is
the maximal height short root, as otherwise $\nu+\mu$ is subdominant to
$\omega$ and $\omega\notin\Lambda_{2}$. So either $\omega=\nu+\mu$, for some
$\nu\in\Lambda_{1}$, or $a_{i}=0$ for all $i$ such that $\omega_{i}$ is
radical. For each root system, we determine when $\nu+\mu$ lies in
$\Lambda_{2}$.

Consider the case $\Phi=B_{n}$, $n\geq 3$. Following the notation of the
previous paragraph, we have $\nu=\omega_{n}$, $\mu=\omega_{1}$. Moreover,
$\omega_{i}$ is radical for every $i<n$. So $\omega\in\Lambda_{2}$ non-radical
implies that $\omega=\omega_{1}+\omega_{n}$. It is straightforward to verify
that $\omega_{1}+\omega_{n}\in\Lambda_{2}$. We deduce that
$\Lambda_{2}=\\{\omega_{1},\omega_{1}+\omega_{n}\\}$. For the final claim of
(1), let $\omega\in\Lambda_{3}$ be a radical weight. If $a_{i}\geq 2$ for some
$i$, then $\omega-\alpha_{i}$ is a radical dominant weight which must lie in
$\Lambda_{2}$. We deduce that $\omega-\alpha_{i}=\omega_{1}$ and we find that
$n=2$ and $\omega=2\omega_{2}$. It is then straightforward to verify that
$2\omega_{2}\in\Lambda_{3}$. So we may now assume $a_{i}\leq 1$ for all $i$.
In particular, as $\omega$ is radical, $a_{n}=0$. In addition,
$\omega_{i}=\omega_{i-1}+\alpha_{i}+\cdots+\alpha_{n}$, see [1, Planche II],
i.e. $\omega_{i-1}\prec\omega_{i}$. So $\omega\in\Lambda_{3}$ then implies
that $\omega=\omega_{2}$.

Consider now the case $\Phi=C_{n}$, for $n\geq 2$. If $a_{i}\neq 0$ or some
$i$ such that $\omega_{i}$ is radical (as above), we find that
$\nu=\omega_{1}$, $\mu=\omega_{2}$. In this case
$\mu+\nu=\omega_{1}+\omega_{2}$. But
$\omega_{1}+\omega_{2}-\alpha_{1}-\alpha_{2}$ is subdominant to $\omega$ and
lies in $\Lambda_{1}$ only if $n=2$. We may now assume $a_{i}=0$ if
$\omega_{i}$ is radical, so $a_{i}=0$ for $i$ even. Also by the preliminary
remarks, $a_{i}\leq 1$ for all $i$. It is easy to observe that
$\omega_{i}\succ\omega_{i-2}$ for $i>1$, which implies the result on
$\Lambda_{2}$. We now turn to the final claims of (2) and (3), so let
$\omega\in\Lambda_{3}$ be a radical weight. If $a_{i}\geq 2$ for some $i$,
then $\omega-\alpha_{i}\in\Lambda_{2}$ if only if $\omega=2\omega_{1}$. So we
now assume $a_{i}\leq 1$ for all $i$. Let $1\leq i\leq n$ be maximal such that
$a_{i}=1$. Since the dominant weight
$\omega-\omega_{i}+\omega_{i-2}\prec\omega$ must lie in
$\Lambda_{1}\cup\Lambda_{2}$ and is a radical weight, we find that $n\geq 4$
and $\omega=\omega_{4}$. Finally, one checks that $\omega_{4}$ lies in
$\Lambda_{3}$.

Finally consider the case $\Phi=D_{n}$, $n\geq 4$. Here, in the case where
$a_{i}\neq 0$ for some $i$ with $\omega_{i}$ radical, we have (in the
previously defined notation) $\nu\in\\{\omega_{1},\omega_{n-1},\omega_{n}\\}$
and $\mu=\omega_{2}$, so
$\mu+\nu\in\\{\omega_{1}+\omega_{2},\omega_{2}+\omega_{n-1},\omega_{2}+\omega_{n}\\}$.
Now $\omega_{2}+\omega_{n}\succ\omega_{1}+\omega_{n-1}\notin\Lambda_{1}$ and
$\omega_{2}+\omega_{n-1}\succ\omega_{1}+\omega_{n}\notin\Lambda_{1}$ so
$\omega_{2}+\omega_{n},\omega_{2}+\omega_{n-1}\not\in\Lambda_{2}$.
Furthermore, as
$\omega_{1}+\omega_{2}-\alpha_{1}-\alpha_{2}=\omega_{3}+\delta_{n,4}\omega_{4}$,
it follows that $\omega_{1}+\omega_{2}\not\in\Lambda_{2}$. So we now assume
that $a_{i}=0$ for all $i$ such that $\omega_{i}$ is radical, that is,
$a_{i}=0$ if $i<n-1$ is even and as established earlier $a_{j}\leq 1$ for all
$j$. Moreover, there are at most two $a_{j}$ which are non-zero, as otherwise
there exists $\beta\in\Phi$ with $\omega-\beta$ dominant and not lying in
$\Lambda_{1}$. Suppose $a_{i}=1$ for some (odd) $i<n-1$. Then
$\omega-(\omega_{i}-\omega_{i-2})\prec\omega$ must lie in $\Lambda_{1}$ and so
$i=3$. So finally, recalling that $\omega$ is non-radical we have
$\omega\in\\{\omega_{3}(n>4),\omega_{3}+\omega_{n-1}(n>4),\omega_{3}+\omega_{n}(n>4),\omega_{1}+\omega_{n},\omega_{1}+\omega_{n-1},\omega_{n-1}+\omega_{n}\\}$.
It is straightforward to see that $\omega_{3}(n>4),\omega_{1}+\omega_{n-1}$
and $\omega_{1}+\omega_{n}$ all lie in $\Lambda_{2}$. In addition,
$\omega_{n-1}+\omega_{n}\succ\omega_{n-3}$, and the latter lies in
$\Lambda_{1}$ if and only if $n=4$. So it remains to show that
$\omega_{3}+\omega_{n},\omega_{3}+\omega_{n-1}\notin\Lambda_{2}$ for $n>4$.
This is clear since $\omega_{2}+\omega_{n-1}$, respectively
$\omega_{2}+\omega_{n}$, is subdominant to the given weight and does not lie
in $\Lambda_{1}$. ∎

Now combining Lemma 5.2 with Lemma 2.8, we obtain some information about the
weight lattice of certain irreducible $G$-modules.

######  Lemma 5.3.

Assume $p=0$ or $p>e(G)$. Let $\omega\neq 0$ be a $p$-restricted dominant
weight for $G$. If $\omega\not\in\Lambda_{1}\cup\cdots\cup\Lambda_{i}$ for
some $i>0$, then there are weights $\nu_{1},\ldots,\nu_{i}$ of $V_{\omega}$
such that $\nu_{j}\in\Lambda_{j}$ for $j=1,\ldots,i$. In addition, the weights
of $V_{\nu_{j}}$ occur as weights of $V_{\omega}$, for $1\leq j\leq i$.

###### Proof.

This follows from the definition of $\Lambda_{j}$ and Lemma 2.8. ∎

##  6\. Proof of Theorem 1.3

In this section, we prove Theorem 1.3, so in particular we are concerned with
the action of non-central non-regular semisimple elements on certain specific
representations (as shown by Theorem 1.2). As noted earlier, in remark 4.6(3),
we must handle some small rank groups as well as the groups $B_{n}$, $D_{n}$,
and $C_{n}$ when $p=2$ and for certain highest weights; we do this in two
separate subsections.

###  6.1. Groups of small rank

######  Lemma 6.1.

Let $G=A_{2}$ and let $s\in T\setminus Z(G)$ be a non-regular element. Let
$V=V_{\omega}$ be the irreducible $G$-module of $p$-restricted highest weight
$\omega\neq 0$. Then the spectrum of $s$ on $V_{\omega}$ is almost simple if
and only if $\omega=\omega_{1}$ or $\omega_{2}$.

###### Proof.

We take $T$ to be the torus of diagonal matrices in ${\rm SL}_{3}(F)$. Since
$s$ is non-regular non-central, with respect to an appropriate choice of base
of $V_{\omega_{1}}$, we may assume $s=\mathop{\rm diag}\nolimits(a,a,a^{-2})$,
for some $a\in F$ with $a^{3}\neq 1$. Clearly the spectrum of $s$ on
$V_{\omega_{1}}$ and $V_{\omega_{2}}$ is indeed almost simple. So we now
assume $\omega\notin\\{0,\omega_{1},\omega_{2}\\}$. In particular, Lemma 5.2
implies $\omega\notin\Lambda_{1}$ and by Lemmas 5.3 and 5.2,
$\Omega(V_{\omega})$ has some weight from
$\Lambda_{2}=\\{\omega_{1}+\omega_{2},2\omega_{1},2\omega_{2}\\}$.

Suppose first that $\omega=2\omega_{1}$, and so $p\neq 2$. The weights of
$V_{\omega_{1}}$ are $\\{\varepsilon_{1},\varepsilon_{2},\varepsilon_{3}\\}$,
so the weights of $V_{\omega_{1}}\otimes V_{\omega_{1}}$ are
$2\varepsilon_{1},2\varepsilon_{2},2\varepsilon_{3},\varepsilon_{1}+\varepsilon_{2},\varepsilon_{1}+\varepsilon_{3},\varepsilon_{2}+\varepsilon_{3}$,
which by Lemma 2.8(2) coincide with the weights of $V_{\omega}$. Now,
$2\varepsilon_{1}(s)=2\varepsilon_{2}(s)=a^{2}$, and
$(\varepsilon_{1}+\varepsilon_{3})(s)=(\varepsilon_{2}+\varepsilon_{3})(s)=a^{-1}$.
As $a^{3}\neq 1$, the eigenvalues $a^{2},a^{-1}$ are distinct, so the spectrum
of $s$ on $V_{2\omega_{1}}$ is not almost simple, as claimed. Since
$V_{2\omega_{2}}$ is dual to $V_{2\omega_{1}}$, the spectrum of $s$ on
$V_{2\omega_{2}}$ is not almost simple as well.

Suppose now that $\omega=\omega_{1}+\omega_{2}$. Then the weights of
$V_{\omega}$ are the roots and the zero weight. Then
$(\alpha_{1}+\alpha_{2})(s)=\alpha_{2}(s)=(\varepsilon_{2}-\varepsilon_{3})(s)=a^{3}\neq
1$ and $-(\alpha_{1}+\alpha_{2})(s)=-\alpha_{2}(g)=a^{-3}$. If $p\neq 3$, the
eigenvalue 1 is also of multiplicity $2$, and we are done. If $p=3$ and
$a^{3}\neq a^{-3}$ then we are done as well. So suppose $p=3$ and $a^{6}=1$
and hence $a^{3}=-1$, that is $a=-1$. Note that $\pm\alpha_{2}(s)=-1$ and
$\pm\alpha_{1}(s)=1$, so the result also follows in this case.

We now appeal to Lemma 5.3 to conclude. ∎

######  Lemma 6.2.

Let $G=A_{3}$ and let $s\in T\setminus Z(G)$ be a non-regular element. Let
$V=V_{\omega}$ be the irreducible $G$-module of $p$-restricted highest weight
$\omega\neq 0$. If the spectrum of $s$ on $V$ is almost simple, then either
$\omega=\omega_{1}$ or $\omega_{3}$, or $\omega=\omega_{2}$ and there exists
$a\in F$, $a^{4}\neq 1$ such that with respect to a suitably chosen basis,
$s=\mathop{\rm diag}\nolimits(a,a,\pm a^{-1},\pm a^{-1})$, $a\in F$.

###### Proof.

Without loss of generality, we take $T$ to be the set of diagonal matrices in
${\rm SL}_{4}(F)$. We may assume $s=\mathop{\rm diag}\nolimits(a,a,b,c)$ for
some $a,b,c\in F$ such that $a^{2}bc=1$. Fix the base of $\Phi$ such that
$\alpha_{i}(\mathop{\rm
diag}\nolimits(a_{1},a_{2},a_{3},a_{4}))=a_{i}a_{i+1}^{-1}$ for $1\leq i\leq
3$; in particular $\alpha_{1}(s)=1$. It is clear that if $a^{2}b^{2}\neq 1$
then $s$ has almost simple spectrum on $V_{\omega_{1}}$ and on
$V_{\omega_{3}}$. If $\omega=\omega_{2}$, then the matrix of $s$ on $V$ is
conjugate to $s_{1}=\mathop{\rm
diag}\nolimits(a^{2},a^{-2},ab,ab,(ab)^{-1},(ab)^{-1})$, so the spectrum of
$s$ is almost simple only if $b=\pm a^{-1}$ and $a^{4}\neq 1$, and the result
easily follows.

Now consider the general case, where
$\omega\not\in\\{\omega_{1},\omega_{2},\omega_{3}\\}$. Assume $s$ has almost
simple spectrum on $V_{\omega}$. Factor $s$ as

$s=\mathop{\rm
diag}\nolimits(a\gamma,a\gamma,a^{-2}\gamma^{-2},1)\cdot\mathop{\rm
diag}\nolimits(\gamma^{-1},\gamma^{-1},\gamma^{-1},c),$

where $\gamma,c\in F$ with $\gamma^{3}=c$. Then viewing $s$ as lying in the
maximal parabolic $P=LQ$, $Q=R_{u}(P)$, corresponding to the root
$\alpha_{3}$, we see that the second factor acts as a scalar on the fixed
point space $V_{\omega}^{Q}$. Hence the eigenvalue multiplicities of $s$ on
this fixed point space are determined by those of the first factor. We now
apply Lemma 6.1 to the element $h=\mathop{\rm diag}\nolimits(\gamma a,\gamma
a,(\gamma a)^{-2})$ and the weight $\omega\downarrow L^{\prime}$, which is the
highest weight of the irreducible $L^{\prime}$-module $V_{\omega}^{Q}$. In
addition, we apply Lemma 6.1 to $(V_{\omega}^{*})^{Q}$. By Lemma 6.1, the only
$p$-restricted irreducible representations of ${\rm SL}_{3}(F)$ on which $h$
has an almost simple spectrum are the natural representation and its dual.
Writing $\omega=m_{1}\omega_{1}+m_{2}\omega_{2}+m_{3}\omega_{3}$, we deduce
that $(m_{1},m_{2}),(m_{2},m_{3})\in\\{(0,0),(1,0),(0,1)\\}$. We are therefore
reduced to considering the case $\omega=\omega_{1}+\omega_{3}$, (a quotient
of) the adjoint representation. The multiplicity of the weight $0$ is at least
$2$ and $\alpha_{1}(s)=1$. Therefore,
$(\alpha_{1}+\alpha_{2})(s)=\alpha_{2}(s)$, so $\alpha_{2}(s)=1$ as well. But
then $(\alpha_{2}+\alpha_{3})(s)=\alpha_{3}(s)\neq 1$, as $s$ is non-central;
hence $s$ is not almost cyclic on $V_{\omega_{1}+\omega_{3}}$.∎

######  Lemma 6.3.

Let $G=C_{2}$, $p=2$, and let $\omega$ be a non-zero $2$-restricted dominant
weight. Let $s\in T$ be a non-regular element. Suppose that the spectrum of
$s$ on $V_{\omega}$ is almost simple. Then
$\omega\in\\{\omega_{1},\omega_{2}\\}$ and the spectrum of $s$ is almost
simple on precisely one of the modules $V_{\omega_{1}}$ and $V_{\omega_{2}}$.
Assume moreover that $T$ is the torus of diagonal matrices in the group ${\rm
Sp}_{4}(F)$, written with respect to a fixed symplectic basis
$(e_{1},e_{2},f_{2},f_{1})$ of the natural module $V_{\omega_{1}}$. Let $g\in
T$ be non-regular. If the spectrum of $g$ on $V_{\omega_{1}}$ is almost simple
then, up to conjugacy, $\varepsilon_{1}(g)=a$, $\varepsilon_{2}(g)=1$ for
$1\neq a\in F$; if the spectrum of $g$ on $V_{\omega_{2}}$ is almost simple
then, up to conjugacy $\varepsilon_{1}(g)=\varepsilon_{2}(g)=a$ for $1\neq
a\in F$.

###### Proof.

As $\omega$ is $2$-restricted, if $\omega\not\in\\{\omega_{1},\omega_{2}\\}$
then $\omega=\omega_{1}+\omega_{2}$, and [15, §12, Corollary of Theorem 41]
implies that $V_{\omega}=V_{\omega_{1}}\otimes V_{\omega_{2}}$. By Lemma 2.5,
the spectrum of $s$ is simple on $V_{\omega_{1}}$, and hence $s$ is regular,
contradicting our hypothesis. One easily verifies the validity of the
additional assertions. ∎

######  Lemma 6.4.

Let $G=C_{2}$, $p\neq 2$, and fix an ordered symplectic basis
$(e_{1},e_{2},f_{2},f_{1})$ of the natural module of $G$ and let $T$ be the
torus of diagonal matrices of $G$ in the natural representation. Let $s\in
T\setminus Z(G)$ be a non-regular element and let $V_{\omega}\in\mathop{\rm
Irr}\nolimits(G)$ be non-trivial $p$-restricted $G$-module. Then $s$ has
almost simple spectrum on $V_{\omega}$ if and only if one of the following
holds:

  1. (i) $\omega=\omega_{1}$, and up to conjugacy, $\varepsilon_{1}(s)=1$, $\varepsilon_{2}(s)=a$ or $\varepsilon_{1}(s)=-1$, $\varepsilon_{2}(s)=a$, where $a\in F$, $a^{2}\neq 1;$

  2. (ii) $\omega=\omega_{2}$, and up to conjugacy, $\varepsilon_{1}(s)=1$, $\varepsilon_{2}(s)=-1$ or $\varepsilon_{1}(s)=\varepsilon_{2}(s)=a$, where $a\in F$, $a^{2}\neq\pm 1.$

###### Proof.

Let $\varepsilon_{1}(s)=b$, $\varepsilon_{2}(s)=a$, that is $s={\rm
diag}(b,a,a^{-1},b^{-1})$.

We first consider $\omega=\omega_{1}$, so
$\Omega(V_{\omega})=\\{\pm\varepsilon_{1},\pm\varepsilon_{2}\\}$. Since $s$ is
non-regular, we may assume that either $a=b$ or $b^{2}=1$. In the first case,
$s$ does not have almost simple spectrum on $V_{\omega}$, while in the second
case $s$ has almost simple spectrum on $V_{\omega}$ if and only if $a^{2}\neq
1$.

We now turn to the cases $\omega\neq\omega_{1}$. By Remark 4.6(1), we are left
with the exceptional cases described in Lemma 4.3, $s_{1}=\mathop{\rm
diag}\nolimits(a,a,a^{-1},a^{-1})$ with $a^{2}\neq 1$, or
$s_{2}=\pm\mathop{\rm diag}\nolimits(1,-1,-1,1)$. Note that
$\alpha_{1}(s_{1})=1$ and $\alpha_{2}(s_{2})=1$. By Lemma 5.2,
$\Lambda_{1}=\\{0,\omega_{1}\\}$, and
$\Lambda_{2}=\\{\omega_{1}+\omega_{2},\omega_{2}\\}$ and $2\omega_{1}$ is the
only radical weight in $\Lambda_{3}$. We consider these weights in turn,
before turning to the general case.

The weights of $V_{\omega_{2}}$ are $0,\pm\varepsilon_{1}\pm\varepsilon_{2}$.
The remarks of the preceding paragraph imply that the cases in the statement
are the only possible ones, and they yield the matrices of $s_{1},s_{2}$ on
$V_{\omega_{2}}$ (with respect to a suitable basis) $\mathop{\rm
diag}\nolimits(a^{2},1,1,1,a^{-2})$ and $\mathop{\rm
diag}\nolimits(-1,-1,1,-1,-1)$, respectively.

Suppose $\omega=\omega_{1}+\omega_{2}$. Then
$\Omega(V_{\omega})=\Omega(V_{\omega_{1}}\otimes V_{\omega_{2}})$, by Lemma
2.8. In terms of Bourbaki weights, the weights in $\Omega(V_{\omega})$ are
$\pm\varepsilon_{1}+(\pm\varepsilon_{1}\pm\varepsilon_{2})$,
$\pm\varepsilon_{2}+(\pm\varepsilon_{1}\pm\varepsilon_{2})$,
$\pm\varepsilon_{1}$, and $\pm\varepsilon_{2}$. Then
$(\pm\varepsilon_{1}+(\pm\varepsilon_{1}\pm\varepsilon_{2}))(s_{2})=-1$,
$(\pm\varepsilon_{2}+(\pm\varepsilon_{1}\pm\varepsilon_{2})(s_{2})=1$, so the
spectrum of $s_{2}$ on $V_{\omega}$ is not almost simple. Furthermore,
$(\varepsilon_{1}+(-\varepsilon_{1}+\varepsilon_{2}))(s_{1})=a=(\varepsilon_{2}+(\varepsilon_{1}-\varepsilon_{2}))(s_{1})$
and
$(-\varepsilon_{1}+(\varepsilon_{1}-\varepsilon_{2}))(s_{1})=a^{-1}=(-\varepsilon_{2}+(-\varepsilon_{1}+\varepsilon_{2}))(s_{1})$.
So the spectrum of $s_{1}$ on $V_{\omega}$ is not almost simple.

Finally, suppose $\omega=2\omega_{1}$. Then by Lemma 2.8, the weights of
$V_{\omega}$ are the same as those of $V_{\omega_{1}}\otimes V_{\omega_{1}}$.
These are $\pm\varepsilon_{i}\pm\varepsilon_{j}$, for $i,j\in\\{1,2\\}$. But
now it is easy to see that neither $s_{1}$ nor $s_{2}$ has almost simple
spectrum on $V_{\omega}$.

We now turn to the general case and suppose that $\omega$ differs from the
weights examined above. Then $\omega\notin\Lambda_{1}\cup\Lambda_{2}$ and
$\omega\neq 2\omega_{1}$. Recall that if $\mu\in\Lambda_{i}$ for some $i$ then
$V_{\mu}$ has a weight from $\Lambda_{j}$ for every $j=1,\ldots,i-1$ (Lemma
5.3). Then Lemma 5.2 implies that either $2\omega_{1}$ or
$\omega_{1}+\omega_{2}$ is a weight of $V_{\omega}$ and by Lemma 2.8, the
weights of $V_{2\omega_{1}}$ or $V_{\omega_{1}+\omega_{2}}$ are weights of
$V_{\omega}$. The above considerations of $V_{\omega_{1}+\omega_{2}}$ and
$V_{2\omega_{1}}$ show then that, given $s=s_{1}$ or $s_{2}$, there are $4$
distinct weights $\lambda_{1},\lambda_{2},\nu_{1},\nu_{2}$ in
$\Omega(V_{\omega})$ such that
$\lambda_{1}(s)=\lambda_{2}(s)\neq\nu_{1}(s)=\nu_{2}(s)$. So $s$ is not almost
cyclic on $V_{\omega}$, which completes the proof of the result. ∎

######  Lemma 6.5.

Theorem 1.3 is true for G of type $G_{2}$.

###### Proof.

Let $V$ be a non-trivial $G$-module and $1\neq s\in T$ a non-regular element.
We have to show that the spectrum of $s$ on $V$ is not almost simple. Let
$\omega$ be the highest weight of $V$. Suppose first that $\omega=\omega_{1}$
or $p=3,\omega=\omega_{2}$, so $\dim V=7$, or 6 for $p=2$. The group $G$
contains a maximal rank closed subgroup $H$ isomorphic to $A_{2}$ such that
the restriction of $V_{\omega_{1}}$ to $H$ is completely reducible; the
irreducible constituents are the natural module for ${\rm SL}_{3}(F)$, and its
dual and, if $p\neq 2$, an additional trivial summand. So the matrix of $s$ on
$V_{\omega_{1}}$ can be written as $\mathop{\rm
diag}\nolimits(a,b,c,1,a^{-1},b^{-1},c^{-1})$ if $p\neq 2$, otherwise
$\mathop{\rm diag}\nolimits(a,b,c,a^{-1},b^{-1},c^{-1})$, where $abc=1$ in
both cases. This is also true if $p=3$ and $V=V_{\omega_{2}}$. If all the
entries are distinct, this matrix is a regular element in ${\rm SL}(V)$, and
hence in $G$, contrary to the assumption.

Suppose that the entries are not distinct. As any permutation of $a,b,c$ can
be realized by an inner automorphism of $G$, we may assume that $a$ equals
some other diagonal entry and by the same reasoning, we may ignore the
possibilities $a=c$ and $a=c^{-1}$. So we examine the cases $a=b$, $a=a^{-1}$,
and $a=b^{-1}$.

Let $a=b$. Then $s$ has almost simple spectrum on $V_{\omega}$ only if
$a=a^{-1}$. But then $c=1$ and $s$ is not almost cyclic on $V_{\omega}$.

Let $a=a^{-1}\neq b$, so $a=\pm 1$, $c=\pm b^{-1}$. If $a=1$, then $b\neq 1$,
$s$ acts on $V_{\omega}$ as $\hat{s}=\mathop{\rm
diag}\nolimits(1,b,b^{-1},1,1,b^{-1},b)$ (where we drop the 1 in the middle if
$p=2$) which does not have almost simple spectrum. If $a=-1$ then $p\neq 2$
and $s$ acts on $V_{\omega}$ as $\hat{s}=\mathop{\rm
diag}\nolimits(-1,b,-b^{-1},1,-1,b^{-1},-b)$. If $b=\pm 1$ then the spectrum
of $\hat{s}$ is not almost simple. Let $b\neq\pm 1$. As $V$ is an orthogonal
space, $s$ is a regular element of $SO(V)$ (Lemma 2.11), and hence in $G$,
contrary to the assumption.

Let $a=b^{-1}$. Then $c=1$. By reordering $a,c$, we arrive at the case $a=1$,
considered above. This completes the analysis of the cases
$\omega=\omega_{1}$, and $(\omega,p)=(\omega_{2},3)$.

Suppose now that $\omega$ is an arbitrary $p$-restricted weight. If $p\neq
2,3$ then the weights of $V_{\omega_{1}}$ occur as weights of $V$ (Lemma 2.8),
so the result follows from that for $V_{\omega_{1}}$. Let $p=2$; now
$0,\omega_{1},\omega_{2},\omega_{1}+\omega_{2}$ are the only $2$-restricted
dominant weights of $G$. By [26, Theorem 15], the weights of $V_{\omega}$ are
the same as in characteristic $0$, in particular all weights of
$V_{\omega_{1}}$ are weights of $V_{\omega}$, and we conclude as above.

Now turn to the case $p=3$ and $\omega$ still $p$-restricted. By [26, Theorem
15], if $\omega\neq 2\omega_{2}$ then the weights of $V_{\omega}$ are the same
as in characteristic $0$, and in particular all weights of $V_{\omega_{1}}$
are weights of $V_{\omega}$. So the result follows as above. For $p=3$ and
$\omega=2\omega_{2}$, we use the tables of [8] to see that the weights of
$V_{\omega_{2}}$ are weights of $V_{2\omega_{2}}$, and then conclude as
before.

Finally, suppose that $\omega$ is not $p$-restricted. By Remark 3.2, we may
assume that $V$ is tensor-decomposable, say, $V=V_{1}\otimes V_{2}$, where the
highest weight of $V_{1}$ is of the form $p^{k}\omega^{\prime}$ for some $k$.
Then the result follows by Lemma 2.5.∎

###  6.2. Groups $B_{n}$ with $n>2$, $D_{n}$ with $n>3$, and $C_{n}$ with
$p=2$ and $n>2$

In this section, we consider the groups as indicated in the heading of the
section. Recall that when $G=B_{n}$, we may assume $p\neq 2$. Note that for
groups $G$ of type $B_{n}$ and of type $D_{n}$, the multiplicity of the $0$
weight in the adjoint representation $V_{\omega_{2}}$ is greater than $1$.
Therefore, if $\omega$ is a dominant weight such that $\omega_{2}\prec\omega$
then, by Lemma 2.10(2), it suffices to observe that a non-central non-regular
semisimple element $s\in G$ is not of almost simple spectrum on
$V_{\omega_{2}}$. This is done in Lemma 6.6 below. The condition
$\omega_{2}\prec\omega$ holds provided $\omega$ is a radical weight and
$\omega\neq 0,\omega_{1},\omega_{2}$ for $G$ of type $B_{n}$, and $\omega\neq
0,\omega_{2}$ for $G$ of type $D_{n}$.

######  Lemma 6.6.

Let $G=B_{n}$, $n>2$, $p\neq 2$, $\omega\in\\{\omega_{2},\omega_{n}\\}$ or
$G=D_{n},n>3$, $\omega\in\\{\omega_{2},\omega_{n-1},\omega_{n}\\}$. Let $s\in
T\setminus Z(G)$ be a non-regular element. Then the spectrum of $s$ on
$V_{\omega}$ is not almost simple, unless $G=D_{4}$,
$\omega\in\\{\omega_{3},\omega_{4}\\}$.

###### Proof.

Here we take $T$ to be the preimage in $G$ of the set of diagonal matrices in
the image of $G$ under the natural representation. We take $s\in T$ and assume
the spectrum of $s$ on $V_{\omega}$ is almost simple. Since $s$ is not
regular, there exists a root $\alpha$ with respect to $T$ such that
$\alpha(s)=1$. We will assume without loss of generality that either
$\alpha=\alpha_{1}$, or $G=B_{n}$ and $\alpha=\alpha_{n}$.

Suppose first that $\omega=\omega_{2}$. Set $R_{0}=\\{\alpha\in\Phi\ |\
\alpha(s)=1\\}$. Since $s$ is non-central, there exists $\beta\in\Phi\setminus
R_{0}$. Moreover, since $\Phi$ is an irreducible root system, there exists
$\beta\in\Phi\setminus R_{0}$ which is not orthogonal to $R_{0}$. So for some
$\alpha\in R_{0}$, $w_{\beta}(\alpha)\neq\beta$. Let $w_{\alpha}\in W(G)$ be
the reflection corresponding to $\alpha.$ Then
$\beta(s)=w_{\alpha}(\beta)(s)\neq 1$, while $\alpha(s)=-\alpha(s)=1$. So $s$
is not almost cyclic on $\omega_{2}$.

Let $\omega\in\\{\omega_{n-1},\omega_{n}\\}$, for $G=D_{n}$ and $n>4$, or
$\omega=\omega_{n}$ for $G=B_{n}$. Then $\mu=\frac{1}{2}(\alpha_{1}+\nu)$ is a
weight of $V_{\omega}$, for
$\nu\in\\{\pm\varepsilon_{3}\pm\cdots\pm\varepsilon_{n}\\}$, with certain
condition on the parity of the number of minus signs in the $D_{n}$-case.
Suppose that $\alpha=\alpha_{1}$. Then $\mu-\alpha_{1}$ is a weight of
$V_{\omega}$ for any admissible choice of the signs. As the spectrum of $s$ on
$V_{\omega}$ is almost simple, we deduce that $\mu(s)$ does not depend on the
choice of $\nu$ and so $\varepsilon_{3}(s)=\cdots=\varepsilon_{n}(s)=1$.
Similarly, this then implies that
$(\frac{1}{2}(\varepsilon_{1}+\varepsilon_{2}+\nu))(s)$ does not depend on the
choice of $\nu$, so again this value must be equal to
$(\frac{1}{2}(\pm(\varepsilon_{1}-\varepsilon_{2})+\nu))(s)$, whence
$\varepsilon_{1}(s)=1=\varepsilon_{2}(s)$ as well. This implies $s\in Z(G)$, a
contradiction.

Finally, suppose that $G=B_{n}$, $\omega=\omega_{n}$ and $\alpha=\alpha_{n}$.
Then for all $1\leq i\leq n-1$, we have the two distinct weights of
$V_{\omega}$, $\omega-\alpha_{i}-\alpha_{i+1}-\cdots-\alpha_{n-1}-\alpha_{n}$
and $\omega-\alpha_{i}-\alpha_{i+1}-\cdots-\alpha_{n-1}-2\alpha_{n}$, taking
the same value on $s$, and therefore deduce that $\alpha_{i}(s)=1$ for all
$i$, again contradicting the fact that $s$ is non-central.∎

######  Remark 6.7.

If $G=D_{4}$ then there exist non-central non-regular semisimple elements $s$
with almost simple spectrum on $V_{\omega_{3}}$ or $V_{\omega_{4}}$. Indeed,
one easily observes that there are non-regular elements $s\in T\setminus Z(G)$
whose spectrum is almost simple on $V_{\omega_{1}}$. Let $\sigma$ be the
triality automorphism of $G$. Then $\sigma(s)$ has almost simple spectrum on
$V_{\omega_{1}}^{\sigma}$, whence the claim.

######  Lemma 6.8.

Let $G=B_{n}$, $n>2$, $p\neq 2$, or $G=D_{n}$, $n>4$ , and let
$V_{\omega}\in\mathop{\rm Irr}\nolimits(G)$, where $\omega\neq 0$ is
p-restricted. Let $s\in T\setminus Z(G)$ be a non-regular element with almost
simple spectrum on $V_{\omega}$. Then $\omega=\omega_{1}$. If $G=D_{4}$ then
$\omega\in\\{\omega_{1},\omega_{3},\omega_{4}\\}$.

###### Proof.

If $\omega$ is radical, this follows from Lemmas 6.6, 2.10(2) and 2.8, both
for $B_{n}$ and $D_{n}$.

Suppose that $\omega$ is not radical. If $G=B_{n}$ then
$\omega_{n}\preceq\omega$ by Lemma 2.8(2), so again the result follows from
Lemmas 6.6 and 2.10(1)(i). Let $G=D_{n}$. By Theorem 3.1, all non-zero weights
of $V_{\omega}$ are of multiplicity 1. Then, by [22, Tables 1,2],
$\omega\in\\{\omega_{1},2\omega_{1},\omega_{2},\omega_{n-1},\omega_{n}\\}$,
where the radical weights $2\omega_{1},\omega_{2}$ are to be dropped. Whence
the result for $n=4$. If $n>4$ then the spectrum of $s$ on $V_{\omega}$ is not
almost simple by Lemma 6.6. ∎

We now handle the case $G=C_{n}$, for $n>2$ and $p=2$, which is excluded in
Proposition 4.5. Moreover, we only need to consider $V_{\omega_{n}}$ (see
Proposition 4.5).

######  Lemma 6.9.

Let $G=C_{n}$, $n>2$, $p=2$. Let $1\neq s\in T$ be a non-regular element. Then
the spectrum of $s$ on $V_{\omega_{n}}$ is not almost simple.

###### Proof.

We argue as in the proof of Lemma 6.6. We can assume that $\alpha(s)=1$ for
$\alpha=\alpha_{1}$ or $\alpha=2\varepsilon_{1}$. The weights of
$V_{\omega_{n}}$ are $\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n}$. Then
$\mu:=\varepsilon_{1}\pm\varepsilon_{2}+\nu$ are weights of $V_{\omega_{n}}$
for any $\nu=\pm\varepsilon_{3}\pm\cdots\pm\varepsilon_{n}$. If
$\alpha=2\varepsilon_{1}$ then $\mu-\alpha$ is a weight of $V_{\omega_{n}}$,
and we conclude (as in the proof of Lemma 6.6) that
$(2\varepsilon_{1})(s)=\cdots=(2\varepsilon_{n})(s)$. As $p=2$, we have
$\varepsilon_{1}(s)=\cdots=\varepsilon_{1}(s)$, whence $s\in Z(G)=1$.

If $\alpha=\alpha_{1}$ then for $\mu=\varepsilon_{1}-\varepsilon_{2}+\nu$ we
have $\mu-\alpha_{1}\in\Omega(V_{\omega_{n}})$, whence
$(2\varepsilon_{3})(s)=\cdots=(2\varepsilon_{n})(s)=1$. This implies that
$(\varepsilon_{1}+\varepsilon_{2}+\nu)(s)$ does not depend on $\nu$, nor does
$(\epsilon_{1}-\epsilon_{2}+\nu)(s)$, whence $(2\varepsilon_{1})(s)=1$, and
again we conclude that $s=1$.∎

###  6.3. Proof of Theorem 1.3

###### Proof.

Let $G,s$ be as in the statement of Theorem 1.3. Note that ${\rm rank}(G)\geq
2$.

Suppose first that $\lambda$ is $p$-restricted. The groups of rank $2$ have
been examined in Lemmas 6.1, 6.3, 6.4 and 6.5, and the group of type $A_{3}$
in Lemma 6.2. In Proposition 4.5, we handled the groups $A_{n},n>3$, $F_{4}$,
$E_{6},E_{7},E_{8}$, and all $p$-restricted weights for the group $C_{n}$,
$n>2$, except the weight $\omega=\omega_{n}$ when $p=2$. The latter is handled
in Lemma 6.9.

Groups of type $B_{n},n>2,p\neq 2$ and $D_{n}$ are dealt with in Lemma 6.8.

By remark 3.2, we may now assume that $V$ is tensor-decomposable. Let
$V=V_{1}\otimes V_{2}$ be a non-trivial tensor decomposition of $V$. By Lemma
2.5, the spectra of $s$ on $V_{1}$ and $V_{2}$ are simple. This contradicts
Lemma 2.5. ∎

Finally, we conclude with a straightforward corollary of Theorem 1.3.

######  Corollary 6.10.

Let $s\in T\setminus Z(G)$ be a non-regular element and V an irreducible
$G$-module. Suppose that the spectrum of s on V is almost simple. Then the
eigenvalue multiplicities of $s$ on V do not exceed $m=m_{V}(s)$, where either
$m\leq{\rm rank}(G)$ or one of the following holds:

$(1)$ $G=A_{3}$, $\dim V=6$, $m=4;$

$(2)$ $G=B_{n}$, $n>2$, $p\neq 2$, $\dim V=2n+1$, $m=2n$;

$(3)$ $G=C_{n}$, and either $\dim V=2n$ and $m=2n-2$ or $n=2$, $p\neq 2$,
$\dim V=5$ and $m=4$;

$(4)$ $G=D_{n}$, $n>3$, $\dim V=2n$, $m=2n-2$.

###### Proof.

This will follow from Theorem 1.3; we discuss each of the cases of the
theorem. To get (1) above, we additionally use Lemma 6.2. For $G=C_{2}$,
$p\neq 2$, we use Lemma 6.4. The modules of dimensions indicated in Theorem
1.3(1) are obtained by Frobenius twisting of $V_{\omega_{1}}$ (where the
statement is clear); $m_{V}(s)$ remains unchanged under such a twist. This
leaves us with $G=D_{4}$ and $\dim V=8$. The modules $V_{\omega_{1}},$
$V_{\omega_{3}},$ $V_{\omega_{4}},$ are obtained from each other by a graph
automorphism of $G$, and the other modules of dimension $8$ as in Theorem
1.3(4) are Frobenius twists of these. The result follows.∎

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