#  Almost cyclic regular elements in irreducible representations of simple
algebraic groups

D.M. Testerman and A.E. Zalesski  Donna Testerman, Ecole Polytechnique Fédéral
de Lausanne,
Institute of Mathematics, Station 8, CH-1015 Lausanne, Switzerland
e-mail<EMAIL_ADDRESS>Alexandre E. Zalesski, Department of Physics,
Mathematics and Informatics,
Academy of Sciences of Belarus, 66 Prospekt Nezalejnasti, Minsk, 220000,
Belarus
e-mail<EMAIL_ADDRESS>
###### Key words and phrases:

simple algebraic groups, representations, eigenvalue multiplicities, regular
elements

###### 1991 Mathematics Subject Classification:

20G05

The first author acknowledges the support of the Swiss National Science
Foundation, research grant number $200020\textunderscore 175571$. In addition,
this work was initiated while the second author visited the EPFL; he
gratefully acknowledges this institute’s support.

Abstract Let $G$ be a simple linear algebraic group defined over an
algebraically closed field of characteristic $p\geq 0$ and let $\phi$ be a
$p$-restricted irreducible representation of $G$. Let $T$ be a maximal torus
of $G$ and $s\in T$. We say that $s$ is _strongly regular_ if
$\alpha(s)\neq\beta(s)$ for all distinct $T$-roots $\alpha$ and $\beta$ of
$G$. Our main result states that if all but one of the eigenvalues of
$\phi(s)$ are of multiplicity 1 then, with a few specified exceptions, $s$ is
strongly regular. This can be viewed as an extension of our earlier result
saying that under the same hypotheses, $s$ must be regular and all non-zero
weights of $\phi$ are of multiplicity 1.

##  1\. Introduction

Fix an algebraically closed field $F$ of characteristic $p\geq 0$, let $G$ be
a simple linear algebraic group defined over $F$ (hereafter referred to as an
_algebraic group_). Let $\phi:G\rightarrow{\rm GL}(V)$ be a rational
irreducible representation (so $V$ is a finite-dimensional $FG$-module).
Recall that $g\in G$ is said to be _regular_ if $\dim C_{G}(g)={\rm rank}(G)$;
so in particular, for a semisimple element $s\in G$, $s$ is regular if and
only if $C_{G}(s)^{\circ}$ is a maximal torus of $G$. A starting point of our
project is the following observation:

$(*)$ Let $s\in G$ be semisimple. If $\phi(s)$ is a regular element in ${\rm
GL}(V)$, then $s$ is regular in $G$.

Note that the assumption on $\phi(s)$ in $(*)$ is equivalent to saying that
all eigenvalue multiplicities of $\phi(s)$ are equal to 1.

Let $T$ be a maximal torus of $G$. As $T$ meets every conjugacy class of
semisimple elements of $G$, for our considerations, we can assume that $s\in
T$. Recall that the roots of $G$ with respect to $T$ are the non-trivial
irreducible constituents of the restriction ${\rm Ad}|_{T}$, where ${\rm Ad}$
is the adjoint representation of $G$. Then $s$ is regular if and only if
$\alpha(s)\neq 1$, for every root $\alpha$ of the root system of $G$, see [10,
Ch. III, §1, Corollary 1.7]. We generalize this in the following

######  Definition 1.1.

Let $s\in T$. We say that $s$ is strongly regular if $\alpha(s)\neq\beta(s)$
whenever $\alpha\neq\beta$ are $T$-roots of $G$.

A strongly regular element is regular as $\alpha(s)=1$ implies
$(-\alpha)(s)=1$; however, there are many elements in $T$ that are regular but
not strongly regular. Our first main result enforces $(*)$ as follows:

######  Theorem 1.2.

Let $G$ be a simply connected simple algebraic group defined over $F$. Let
$s\in G$ be a semisimple element and let $\phi$ be an irreducible
representation of $G$ with $p$-restricted highest weight $\omega\neq 0$.
Suppose that all of the eigenvalues of $\phi(s)$ are of multiplicity $1$. Then
one of the following holds:

  1. (1) the element $s$ is strongly regular;

  2. (2) $G=A_{n}$, $n\geq 1$, with $p\neq 2$ if $n=1$, and $\omega\in\\{\omega_{1},\ldots,\omega_{n}\\}$;

  3. (3) $G=B_{n}$, $n\geq 3$, and $\omega\in\\{\omega_{1},\omega_{n}\\}$;

  4. (4) $G=C_{n}$, $n\geq 2$, $p=2$, and $\omega\in\\{\omega_{1},\omega_{n}\\}$;

  5. (5) $G=C_{n}$, $n\geq 2$, $p\neq 2$, and either

    1. (a) $\omega=\omega_{1}$, or

    2. (b) $n\in\\{2,3\\}$ and $\omega=\omega_{n}$, or

    3. (c) $n\geq 3$, $p=3$, and $\omega\in\\{\omega_{n-1},\omega_{n}\\}$;

  6. (6) $G=D_{n}$, $n\geq 4,$ and $\omega\in\\{\omega_{1},\omega_{n-1},\omega_{n}\\}$;

  7. (7) $G=E_{6}$ and $\omega\in\\{\omega_{1},\omega_{6}\\}$;

  8. (8) $G=E_{7}$ and $\omega=\omega_{7}$;

  9. (9) $G=F_{4}$, $p=3$, and $\omega=\omega_{4}$;

  10. (10) $G=G_{2}$ and either $\omega=\omega_{1}$, or $p=3$ and $\omega=\omega_{2}$.

(We refer the reader to the end of this section for an explanation of the
notation used in the statement. For the reader’s convenience the highest
weights given in items (2)–(10) are collected in Table 4, at this end of this
manuscript.)

The above result, as well as Theorems 1.4, 1.5, 1.7 below, can be viewed as an
identification result, in the following sense: the existence of an element
whose spectrum satisfies a specified criteria on a given representation either
determines the representation or provides structural information about the
element. We expect that strongly regular elements can play a certain role in
the general theory of algebraic groups and finite groups of Lie type. We
mention a paper by Seitz [9, Lemma 2], where he proves that
$\alpha|_{T(q)}\neq\beta|_{T(q)}$ for the maximal tori $T(q)$ of finite
quasisimple groups of Lie type $G(q)$, with $q>5$. This allows him to obtain a
certain upper bound for the homogeneous components of $T(q)$ on irreducible
$G(q)$-modules. It was mentioned in [8, p.179] that an analogous bound holds
for the $s$-eigenspace dimensions, $s\in G$ semisimple, provided $s$ is
strongly regular.

In fact, we are able to obtain nearly the same conclusion as in Theorem 1.2
under a significantly weaker assumption, specifically, by allowing one of the
eigenvalue multiplicities of $\phi(s)$ to be arbitrary (excluding the trivial
case of $s$ being central in $G$). A step in this direction is made in our
paper [12, Theorem 2], where we prove the following:

######  Theorem 1.3.

Let $G$ be a simply connected simple algebraic group defined over $F$ and let
$s\in G$ be a non-regular non-central semisimple element. Let $V$ be a non-
trivial irreducible $FG$-module with associated representation $\phi$. Suppose
that at most one eigenvalue of $\phi(s)$ is of multiplicity greater than $1$.
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) $G$ is of Lie type $B_{n}$, $n\geq 3$, $p=2$ and $\dim V=2n$;

  3. (3) $G$ is of type $A_{3}$ and $\dim V=6;$

  4. (4) $G$ is of type $C_{2}$, $p\neq 2$, $\dim V=5$.

In view of Theorem 1.3, in our consideration of semisimple elements having at
most one eigenvalue mutiplicity greater than $1$ on some irreducible
representation, we may now restrict our attention to regular elements. We have
the following result.

######  Theorem 1.4.

Let $G$ be a simply connected simple algebraic group defined over $F$, $s\in
G$ a regular semisimple element, and let $\phi$ be a non-trivial irreducible
representation with $p$-restricted highest weight $\omega$. Suppose that at
most one eigenvalue multiplicity of $\phi(s)$ is greater than $1$. Then one of
the following holds:

  1. (i) $s$ is strongly regular;

  2. (ii) $G$ and $\omega$ are as in one of the cases $(2)$ \- $(10)$ of the conclusion of Theorem 1.2;

  3. (iii) $G=A_{n}$ and $\omega\in\\{2\omega_{1},2\omega_{2}\\}$, or $(n,p)=(2,3)$ and $\omega=\omega_{1}+\omega_{2}$;

  4. (iv) $G=B_{n}$, $n\geq 3$, $p=2$ and $\omega=\omega_{2}$;

  5. (v) $G=C_{n}$, $n\geq 3$, $(n,p)\neq(3,3)$ and $\omega=\omega_{2}$, or $n=4$, $p\neq 2,3$ and $\omega=\omega_{4};$

  6. (vi) $G=F_{4}$, $p\neq 3$ and $\omega=\omega_{4}$, or $p=2$ and $\omega=\omega_{1}$.

The following result shows that the exceptions in Theorems 1.2 and 1.4 are
genuine.

######  Theorem 1.5.

Let $G$ be as above and let $\phi$ be a non-trivial irreducible representation
of $G$.

$(1)$ Suppose that $\phi$ is as in items $(2)$ \- $(10)$ of Theorem 1.2. Then
there exists a regular, not strongly regular, semisimple element $s\in G$ such
that all eigenvalue multiplicities of $\phi(s)$ are equal to $1$.

$(2)$ Suppose that $\phi$ is as in items ${\rm{(ii)}}$ \- ${\rm{(vi)}}$ of
Theorem 1.4. Then there exists a regular, not strongly regular, semisimple
element $s\in G$ such that at most one eigenvalue multiplicity of $\phi(s)$ is
greater than $1$.

There are many examples of irreducible representations $\phi$ and regular
elements $s\in T$ such that $\phi(s)$ has exactly one eigenvalue of
multiplicity greater than 1. However, these examples are within a rather
narrow class of representations, as the following result shows.

######  Theorem 1.6.

[12, Theorem 2] Let $G$ be a simply connected simple algebraic group defined
over $F$ and $\phi$ a non-trivial irreducible representation of $G$. Then the
following statements are equivalent:

  1. (i) There exists a non-central semisimple element $s\in G$ such that at most one eigenvalue multiplicity of $\phi(s)$ is equal to $1$.

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

The irreducible representations $\phi$ whose non-zero weights are of
multiplicity $1$ have been determined in [11] and those with $p$-restricted
highest weight are reproduced in Tables 2 and 3; see the end of the
manuscript.

One can ask whether the assumption that $\omega$ be $p$-restricted can be
dropped in Theorems 1.2 and 1.4. Here, we have the following result.

######  Theorem 1.7.

Let $G$, $s$ be as in Theorem 1.2, where ${\rm char}(F)=p>0$. Let $\phi$ be an
irreducible representation of $G$ with highest weight $\omega=\sum
p^{i}\lambda_{i}$, where all $\lambda_{i}$ are $p$-restricted weights and at
least two of them are non-zero. Suppose that at most one of the eigenvalue
multiplicities of $\phi(s)$ is greater than $1$. Then either $s$ is strongly
regular or for each $i$ with $\lambda_{i}\neq 0$, we have
$(G,\lambda_{i})=(G,\omega)$ for $(G,\omega)$ as in items $(2)-(10)$ of
Theorem 1.2. Moreover,

  1. (i) if $(G,p)=(C_{n},2)$, then for all $i$, $(\lambda_{i},\lambda_{i+1})\neq(\omega_{n},\omega_{1})$;

  2. (ii) if $(G,p)=(G_{2},2)$, then for all $i$, $(\lambda_{i},\lambda_{i+1})\neq(\omega_{1},\omega_{1})$;

  3. (iii) if $(G,p)=(G_{2},3)$, then for all $i$, $(\lambda_{i},\lambda_{i+1})\neq(\omega_{2},\omega_{1})$.

Notation We fix throughout an algebraically closed field $F$ of characteristic
$p\geq 0$. By abuse of notation, when we write $p>p_{0}$ for some prime
$p_{0}$, we allow the case ${\rm char}(F)=0$.

Throughout the paper $G$ is a simple simply connected linear algebraic group
defined over $F$. (As mentioned before, we will suppress the adjective
“linear”.) All $G$-modules considered are rational finite-dimensional
$FG$-modules.

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. Recall that 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 reference to $T$ and write “weights” in place of
“$T$-weights”. The set of weights of a $G$-module $V$ is denoted by
$\Omega(V)$. Recall that the roots of $G$ are the non-zero weights of the
adjoint representation, equivalently of the $G$-module ${\rm Lie}(G)$. We
denote the set of roots by $\Phi$ or $\Phi(G)$. The $\mathbb{Z}$-span of
$\Phi$ is denoted by $R$ or $R(G)$. The weights in $R$ are called radical. We
fix a base of the root system and the simple roots of $\Phi(G)$ are denoted by
$\alpha_{1},\ldots,\alpha_{n}$ and ordered as in [1]; we denote the associated
set of positive roots by $\Phi^{+}$ and the $\mathbb{Z}_{\geq 0}$-span of the
simple roots by $R^{+}$.

The Weyl group of $G$ is denoted by $W$ (or $W(G)$); 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 $W$-orbit of $\mu\in\Omega$ is denoted by
$W\mu$. Minuscule weights are defined and tabulated in [2, Ch. VIII, §7.3].

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 $G$.
For a $G$-module $V$, $t\in T$ 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 $\nu\in\Omega(V)$. As every semisimple element
of $G$ is conjugate to an element of $T$, in most results which follow, the
assumption that $s$ is semisimple is replaced by the assumption that $s\in T$.

We let $\\{\omega_{1},\dots,\omega_{n}\\}$ be the set of fundamental dominant
weights with respect to the choice of simple roots. These form a
$\mathop{\mathbb{Z}}\nolimits$-basis of $\Omega$, so every $\nu\in\Omega$ can
be expressed in the form $\nu=\sum a_{i}\omega_{i}$
($a_{i}\in\mathop{\mathbb{Z}}\nolimits$); the set of $\nu$ with $a_{i}\geq 0$
is denoted by $\Omega^{+}$. If $p>0$, we set $\Omega^{+}_{p}$ to be the set of
weights $\nu=\sum a_{i}\omega_{i}$ with $0\leq a_{i}<p$. If ${\rm char}(F)=0$
we let $\Omega_{p}^{+}=\Omega^{+}$. The weights of $\Omega^{+}$, respectively
$\Omega^{+}_{p}$ are called dominant, respectively $p$-restricted weights (so
that if ${\rm char}(F)=0$, all dominant weights are by definition
$p$-restricted weights). An irreducible $G$-module is called $p$-restricted if
its highest weight is $p$-restricted.

There is a standard partial ordering of elements of $\Omega$; specifically,
for $\mu,\mu^{\prime}\in\Omega$ we write $\mu\prec\mu^{\prime}$ and
$\mu^{\prime}\succ\mu$ if and only if $\mu^{\prime}-\mu\in R^{+}$ and
$\mu\neq\mu^{\prime}$. We write $\mu\preceq\mu^{\prime}$ or
$\mu^{\prime}\succeq\mu$ if and only if $\mu^{\prime}-\mu\in R^{+}$. For
$\mu,\nu\in\Omega^{+}$, with $\mu\preceq\nu$, we say $\mu$ _is subdominant to_
$\nu$. Every irreducible $G$-module $V$ 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 the set of isomorphism classes of irreducible $G$-modules, 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 $s\in S\\}$. The maximal root of $\Phi(G)$ is denoted by
$\omega_{a}$; this is the highest weight of the adjoint module ${\rm Lie}(G)$
and of a composition factor $V_{a}$ of ${\rm Lie}(G)$.

Because of natural isomorphisms between certain indecomposable root systems,
we will assume throughout that the rank of $G$ is as in Table 1 below.

$\begin{array}[]{|l|c|}\hline\cr G&{\rm rank}(G)\\\ \hline\cr A_{n}&n\geq 1\\\
\hline\cr B_{n}&n\geq 3\\\ \hline\cr C_{n}&n\geq 2\\\ \hline\cr D_{n}&n\geq
4\\\ \hline\cr\end{array}$ Table 1. rank(G)

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. For
classical groups $G$ the module with highest weight $\omega_{1}$ is called the
natural module, with one exception, namely the natural module for $G=B_{n}$
when $p=2$ is a reducible $(2n+1)$-dimensional module with two composition
factors $V_{\omega_{1}}$ and $V_{0}$.

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

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
then, for integers $k\geq 0$, the modules $V^{Fr^{k}}$ are called Frobenius
twists of $V$; if $V$ is irreducible with highest weight $\omega$ then the
highest weight of $V^{Fr^{k}}$ is $p^{k}\omega$. 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 $k$, where
$\lambda_{0},\ldots,\lambda_{k}\in\Omega_{p}^{+}$. The Steinberg tensor
product theorem then implies that $V_{\omega}\cong V_{\lambda_{0}}\otimes
V_{\lambda_{1}}^{Fr}\otimes\cdots\otimes V_{\lambda_{k}}^{Fr^{k}}$.

Finally, we recall that a matrix $M\in M_{n\times n}(F)$ is called _cyclic_ if
it is diagonalisable and all eigenvalue multiplicities equal $1$; we will say
$M$ is _almost cyclic_ , if it is diagonalizable and at most one eigenvalue
multiplicity is greater than $1$. This leads us to introduce the following
terminology to be used throughout.

######  Definition 1.8.

Let $s\in T$ and let $V$ be a $G$-module with associated representation
$\phi:G\to{\rm GL}(V)$. We say that $s$ is _cylic on $V$_, respectively
_almost cyclic on $V$_ if $\phi(s)$ is cyclic, respectively almost cyclic.

##  2\. Preliminaries

We start by recalling a characterization of regular semisimple elements, as
mentioned in the Introduction. We will use this frequently without direct
reference.

######  Lemma 2.1.

[10, Ch. III, §1, Corollary 1.7] Let $G$ be a semisimple algebraic group,
$T\leq G$ a maximal torus and $s\in T$. Then the following conditions are
equivalent:

  1. (1) $s$ is regular;

  2. (2) if $gs=sg$ for $g\in G$, then g is semisimple;

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

The following lemma concerns assertion (*) in the Introduction, rephrased in
terms of Definition 1.8.

######  Lemma 2.2.

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 $C_{{\rm GL}(V)}(\rho(s))$ consists of semisimple elements. So the claim
follows from Lemma 2.1(2).

(2) By [12, Lemma 1], $s$ is cyclic on $V$, so the result follows from (1).∎

######  Definition 2.3.

Let $V$ be a $G$-module, $\mu,\nu\in\Omega(V)$, and $s\in T$. We say that $s$
_separates the weights_ $\mu,\nu$ of $V$ if $\mu(s)\neq\nu(s)$. If this holds
for every pair of weights $\mu,\nu\in\Omega(V)$, 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 2.4.

Let $V,V_{1},V_{2}$ be non-trivial G-modules. Let $s\in T\setminus Z(G)$ and
assume that $s$ is almost cyclic on $V$.

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

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

###### Proof.

This follows from [12, Lemma 1] and [12, Lemma 3].∎

The set of weights of a $G$-module $V$ determines the action of a given
semisimple element on the module $V$. We will rely on the following important
result [7, Theorem 1] describing precisely, in many cases, the set
$\Omega(V)$. We require an additional notation, defined in loc.cit.; we write
$e(G)$ for the maximum of the squares of the ratios of the lengths of the
roots in $\Phi(G)$. So $e(G)=1$ 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 2.5.

[7, Theorem 1] Assume that $p=0$ or $p>e(G)$. Let $\lambda$ be a
$p$-restricted weight. Then $\Omega(V_{\lambda})=\\{w(\mu)\ |\
\mu\in\Omega^{+},\mu\preceq\lambda,w\in W\\}.$

As an immediate Corollary of Theorem 2.5, we have:

######  Lemma 2.6.

Assume that $p=0$ 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. (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 radical then some root is a weight of $V_{\lambda}$; otherwise $\Omega(V_{\lambda})$ contains some minuscule weight.

We begin our considerations of strongly regular elements and their action in
certain representations by considering the action on $V_{a}$, the irreducible
$G$-module with highest weight the highest root.

######  Lemma 2.7.

Assume that $p=2$ if $G=A_{1}$, $p\neq 3$ if $G=A_{2}$ or $G_{2}$, and $p\neq
2$ if $G=B_{n}$, $C_{n}$ or $F_{4}$. Let $s\in T\setminus Z(G)$. Then $s$ is
strongly regular if and only if $s$ is almost cyclic on $V_{a}$.

###### Proof.

If $G=A_{1}$ and $p=2$, then all non-identity semisimple elements of $G$ act
cycically on $V_{a}$ (which is just a twist of the natural representation of
$G$) and all non-identity semisimple elements are strongly regular. So here
the statement is clear.

Now turn to the other cases and suppose that $s$ is strongly regular, so that
$\alpha(s)\neq\beta(s)$ for roots $\alpha,\beta\in\Phi(G)$, with
$\alpha\neq\beta$. Then, with one exception, the hypothesis of (1), together
with Theorem 2.5, implies that $\Omega(V_{a})=\Phi(G)\cup\\{0\\}$ and so $s$
is almost cyclic on $V_{a}$. The exception occurs when $G=G_{2}$ and $p=2$.
But here as well, we have $\Omega(V_{a})=\Phi(G)\cup\\{0\\}$; see for example
[5].

Now suppose that $s$ is almost cyclic on $V_{a}$ (and ${\rm rank}(G)>1$).
Then, in all cases considered here, the zero weight has multiplicity at least
2 on $V_{a}$. Thus $\alpha(s)\neq\beta(s)$ for all roots
$\alpha,\beta\in\Phi(G)$, with $\alpha\neq\beta$. That is, $s$ is strongly
regular.∎

######  Remark 2.8.

The exceptions in Lemma 2.7 are genuine. Indeed, if $p\neq 2$ and $G={\rm
SL}_{2}(F)$, then elements of $T$ which are not strongly regular are of the
form $s={\rm diag}(c,c^{-1})$ with $c^{4}=1$, since $\alpha(s)=c^{2}$. We
choose $c$ such that $c^{2}=-1$ and then $s$ is non-central, regular, and
almost cyclic on $V_{a}$. If $G=C_{n},n\geq 2$, with $p=2$, then
$\omega_{a}=2\omega_{1}$. Then $V_{a}$ is the Frobenius twist of
$V_{\omega_{1}}$, so $s$ is almost cyclic on $V_{a}$ if and only if $s$ is
almost cyclic $V_{\omega_{1}}$. However, it is easy to construct non-regular
elements $s\in T$ that are almost cyclic on $V_{\omega_{1}}$. For $G=G_{2}$,
$p=3$, see Proposition 4.9(4), and for $G=F_{4}$, $p=2$, see Remark 4.4(1).
For $G=A_{2}$ with $p=3$ take $s=\mathop{\rm diag}\nolimits(a,-a,-a^{-2})$ for
$a\in F^{\times}$, so $\alpha_{1}(s)=(-\alpha_{1})(s)$. Then the Jordan normal
form of $s$ on $V_{a}$ is $\mathop{\rm
diag}\nolimits(1,-1,-1,a^{3},-a^{3},a^{-3},-a^{-3}).$ This matrix is almost
cyclic if $a^{4}\neq 1$. Finally, for $G=B_{n},n\geq 3,$ and $p=2$, see Lemma
3.15 and Remark 3.16 .

Lemma 2.9 and Remark 2.10 will not be required in what follows, but give some
additional information about the relationship between a semisimple element
being regular and its action on $V_{a}$.

######  Lemma 2.9.

Let $s\in T$. Let $\omega$ be the highest short root and $V=V_{\omega}$.

  1. (1) If $s$ is regular, then $V^{s}=V^{T}$ and $V_{a}^{s}=V_{a}^{T}$.

  2. (2) Suppose that $V_{a}^{s}=V_{a}^{T}$. If $p>e(G)$ or if $G=G_{2}$ and $p=2$, then $s$ is regular.

###### Proof.

By Lemma 2.1, $s$ is regular if and only if $\alpha(s)\neq 1$ for all
$\alpha\in\Phi(G)$. For (1), we use the fact that all non-zero weights of
$V_{a}$ and $V$ are roots. (2) Using Theorem 2.5, one observes that for
$p>e(G)$, $\Omega(V_{a})=\Phi(G)\cup\\{0\\}$. As mentioned in the proof of
Lemma 2.7, this also is true for $G=G_{2}$, when $p=2$. So
$V_{a}^{T}=V_{a}^{s}$ implies $\alpha(s)\neq 1$ for $\alpha\in\Phi(G)$, and
hence $s$ is regular.∎

######  Remark 2.10.

If $p=e(G)$, there exists a non-regular element $s\in T$, such that
$V_{a}^{s}=V_{a}^{T}$. See Proposition 4.9(5), for $G=G_{2}$ with $p=3$ and
Remark 4.4(2) for $G=F_{4}$ with $p=2$. Now let $G=C_{n}$ with $p=2$. Then
$V_{a}=V_{2\omega_{1}}$ and hence $\Omega(V_{a})=\\{\pm
2\varepsilon_{1},\ldots,\pm 2\varepsilon_{n}\\}$. So $V_{a}^{T}=\\{0\\}$, and
$V_{a}^{s}=V_{a}^{T}$ if and only if $V_{a}^{s}=\\{0\\}$, equivalently, if and
only if 1 is not an eigenvalue of $s$ on $V_{\omega_{1}}$. As $n\geq 2$, there
are non-regular elements $s\in T$ such that $V_{\omega_{1}}^{s}=\\{0\\}$. For
$G=B_{n}$, when $p=2$, the non-zero weights of $V_{a}$ are
$\\{\pm\varepsilon_{i}\pm\varepsilon_{j}\ |\ 1\leq i<j\leq n\\}$. Choose $s$
such that $\varepsilon_{1}(s)=1$, so $s$ is not regular, but
$(\varepsilon_{i}\pm\varepsilon_{j})(s)\neq 1$ for all $i\neq j$.

##  3\. Classical groups

In this section we prove Theorems 1.2, 1.4 and 1.5 for classical groups. In
view of Theorem 1.6, we must examine the modules $V_{\omega}$, for
$\omega\in\Omega_{p}^{+}$, all of whose non-zero weights are of multiplicity
1. These are listed in Tables 2 and 3. Recall that we have fixed a maximal
torus $T$ of $G$, and we assume that our semisimple element belongs to $T$.

###  3.1. Groups of type $A_{n}$

Throughout this subsection, $G$ is of type $A_{n}$.

######  Lemma 3.1.

Let $\omega\in\Omega^{+}$ and $n\geq 2$.

  1. (1) If $\omega\succ 2\omega_{1}$ then $\omega\succeq\omega_{1}+\omega_{2}+\omega_{n}$.

  2. (2) If $\omega\succ\omega_{1}+\omega_{n}$, then $\omega\succeq\omega_{2}+\omega_{n-1}$ if $n\geq 3$ and $\omega\succeq\nu$, for some $\nu\in\\{3\omega_{1},3\omega_{2}\\}$ if $n=2$.

###### Proof.

Let $\omega\succ 2\omega_{1}$. By [14, Lemma 3.2], we can assume that
$\omega=2\omega_{1}-\omega_{j-1}+\omega_{j}+\omega_{k}-\omega_{k+1}$ for some
$j,k\in\\{1,\ldots,n\\}$, $j\leq k$. (We set $\omega_{0}=0=\omega_{n+1}$.) It
follows that $k=n$ and $j\leq 2$, and in both cases (1) easily follows.

Now, let $\omega\succ\omega_{1}+\omega_{n}$. As above, we can assume that
$\omega=\omega_{1}+\omega_{n}-\omega_{j-1}+\omega_{j}+\omega_{k}-\omega_{k+1}$.
Then $j\in\\{1,2\\}$ and $k\in\\{n-1,n\\}$. If $n\geq 3$, then
$\omega\in\\{2\omega_{1}+2\omega_{n},\omega_{2}+\omega_{n-1},\omega_{2}+2\omega_{n},2\omega_{1}+\omega_{n-1}\\}$.
As
$2\omega_{1}+2\omega_{n}\succ\omega_{2}+2\omega_{n}\succ\omega_{2}+\omega_{n-1}$
and $2\omega_{1}+\omega_{n-1}\succ\omega_{2}+\omega_{n-1}$, the result
follows. If $n=2$ then
$\omega\in\\{2\omega_{1}+2\omega_{2},3\omega_{2},3\omega_{1}\\}$ and
$2\omega_{1}+2\omega_{2}\succ 3\omega_{i}$, $i=1,2$, and the result of (2)
follows.∎

######  Proposition 3.2.

Let $G=A_{n}$, with $n\geq 1$, $s\in T_{reg}$ and let
$\omega\in\Omega^{+}_{p}$.

  1. (1) Assume that $\omega\notin\\{0,\omega_{1},\ldots,\omega_{n},2\omega_{1},2\omega_{n}\\}$ and that $s$ is almost cyclic on $V_{\omega}$. Then $s$ is strongly regular, unless $(n,p)=(2,3)$ and $\omega=\omega_{1}+\omega_{2}$.

  2. (2) If $\omega\in\\{2\omega_{1},2\omega_{n}\\}$ and $s$ is cyclic on $V_{\omega}$, then $s$ is strongly regular.

  3. (3) If $(n,p)=(2,3)$ and $\omega=\omega_{1}+\omega_{2}$, with $s$ cyclic on $V_{\omega}$, then $s$ is strongly regular.

###### Proof.

Set $V=V_{\omega}$. Recall that the $T$-roots of $G$ are
$\varepsilon_{i}-\varepsilon_{j}$ for $i,j\in\\{1,\ldots,n+1\\}$, $i\neq j$.
Then $s$ regular implies that $\alpha(s)\neq 1$ for all $\alpha\in\Phi(G)$.

For (1), suppose the contrary, that is, $s$ is almost cyclic on $V_{\omega}$
and $s$ is not strongly regular, so that $\alpha(s)=\beta(s)$ for some roots
$\alpha\neq\beta$. As $W$ acts transitively on the roots, we can assume that
$\alpha=\alpha_{1}=\varepsilon_{1}-\varepsilon_{2}$.

If $n=1$ then $\beta=-\alpha_{1}$ and $\alpha(s)=\beta(s)$ implies
$(2\alpha_{1})(s)=1$, whence $\alpha_{1}(s)=-1$. Let $\omega=a\omega_{1}$,
where by hypothesis $3\leq a\leq p-1$. Then the weights of $V$ are
$\omega-i\alpha_{1}$ for $0\leq i\leq a$, and
$(\omega-i\alpha_{1})(s)=(-1)^{i}\omega(s)$, contradicting that $s$ is almost
cyclic on $V$.

Now let $n\geq 2$. Recall that $W$ acts on
$\varepsilon_{1},\ldots,\varepsilon_{n+1}$ as the symmetric group $S_{n+1}$
does. Since $\beta=\varepsilon_{i}-\varepsilon_{j}$ for some $i\neq j$, we can
assume that
$\beta\in\\{-(\varepsilon_{1}-\varepsilon_{2}),\pm(\varepsilon_{1}-\varepsilon_{3}),\pm(\varepsilon_{2}-\varepsilon_{3}),\varepsilon_{3}-\varepsilon_{4}\\}$
(where $\varepsilon_{3}-\varepsilon_{4}$ occurs only if $n\geq 3$). As $s$ is
regular, $\alpha-\beta$ is not a root, so
$\beta\notin\\{\varepsilon_{1}-\varepsilon_{3},-\varepsilon_{2}+\varepsilon_{3}\\}$,
and hence
$\beta\in\\{-(\varepsilon_{1}-\varepsilon_{2}),-(\varepsilon_{1}-\varepsilon_{3}),\varepsilon_{2}-\varepsilon_{3},\varepsilon_{3}-\varepsilon_{4}\\}=\\{-\alpha_{1},-\alpha_{1}-\alpha_{2},\alpha_{2},\alpha_{3}\\}$.
In addition, as $s$ is regular, if $\beta=-\alpha$ then $p\neq 2$ and
$\alpha_{1}(s)=-1$.

By [12, Lemma 11] (see Definition 3, loc.cit.) and Theorem 2.5, there exists a
weight
$\mu\in\Omega(V)\cap\\{\omega_{1}+\omega_{i},\omega_{i}+\omega_{n}:i=1,\ldots,n\\}$
and all weights subdominant to $\mu$ also lie in $\Omega(V)$. We consider in
turn each possibility for the weight $\mu$.

Suppose first that $\mu:=\omega_{1}+\omega_{i}$ for $1<i<n$. Then
$\mu-\alpha_{1}-\alpha_{2}-\cdots-\alpha_{i}=\omega_{i+1}\in\Omega(V)$ and
$\omega_{i+1}-\alpha$, $\omega_{i+1}-\beta\in W\mu$. In addition,
$\omega_{i+1}-\alpha_{i+1}-\alpha$ and
$\omega_{i+1}-\alpha_{i+1}-\beta\in\Omega(V)$, as long as
$\beta\in\\{-\alpha_{1},\alpha_{2}\\}$ when $i=2$. Since $s$ is almost cyclic
on $V$, we deduce that either $i=2$ and
$\beta\in\\{-\alpha_{1}-\alpha_{2},\alpha_{3}\\}$ or
$\omega_{i+1}-\alpha=\omega_{i+1}-\alpha-\alpha_{i+1}$, so
$\alpha_{i+1}(s)=1$, contradicting the fact that $s$ is regular. So it remains
to consider the case where $i=2$, $\mu=\omega_{1}+\omega_{2}$, $n\geq 3$, and
$\beta\in\\{-\alpha_{1}-\alpha_{2},\alpha_{3}\\}$. We give the details for the
case $\beta=-\alpha_{1}-\alpha_{2}$. Here
$\eta=\mu-\alpha_{1}-2\alpha_{2}-\alpha_{3}\in\Omega(V)$ and
$\eta-\alpha,\eta-\beta\in\Omega(V)$. Hence we deduce as above that
$\omega_{i+1}=\eta$ and so $(\alpha_{2}+\alpha_{3})(s)=1$, contradicting that
$s$ is regular. The case $i=2$ and $\beta=\alpha_{3}$ is entirely similar.

Suppose now that $\mu=\omega_{1}+\omega_{n}$. Then all roots are weights of
$V_{\omega}$, and since $\alpha(s)=\beta(s)$, we have
$(-\alpha)(s)=(-\beta)(s)$, whence $\alpha(s)=\beta(s)=-1$ and
$(\alpha+\beta)(s)=1$. Then $\alpha+\beta$ is not a root as $s\in T_{reg}$.
Therefore, $\beta\in\\{-\alpha_{1},\alpha_{3}\\}$. If $\beta=\alpha_{3}$, so
$n\geq 3$, then $\alpha_{1}+\alpha_{2}+\alpha_{3}$ is a root and
$(\alpha_{1}+\alpha_{2}+\alpha_{3})(s)=\alpha_{2}(s)$. As before, we deduce
that $\alpha_{2}(s)=\alpha_{1}(s)=-1$, and then
$(\alpha_{1}+\alpha_{2})(s)=1$, whence a contradiction. So we are left with
the case where $\beta=-\alpha$ and $\alpha(s)=(-\alpha)(s)=-1$. By Lemma 2.7
(and the hypothesis of (1)), we may assume that
$\omega\neq\omega_{1}+\omega_{n}$. By Lemma 3.1, $\omega\succeq\nu$, where
$\nu\in\\{3\omega_{1},3\omega_{2}\\}$ if $n=2$, and
$\nu=\omega_{2}+\omega_{n-1}$ if $n\geq 3$. As usual, Theorem 2.5 implies that
all weights subdominant to $\nu$ lie in $\Omega(V)$.

Now if $n\geq 3$, so $\nu=\omega_{2}+\omega_{n-1}$, then set
$\eta=\nu-\alpha_{1}-2\alpha_{2}-\alpha_{3}-\cdots-\alpha_{n-1}$. Note that
$\eta\pm\alpha_{1}$ and $\eta-\alpha_{n}\pm\alpha_{1}$ lie in $\Omega(V)$. But
then we deduce that $\alpha_{n}(s)=1$, giving the usual contradiction. If
$n=2$ and $\nu=3\omega_{1}$, we argue similarly, setting
$\eta=\nu-\alpha_{1}$, and deduce that $\alpha_{2}(s)=1$. The case
$\nu=3\omega_{2}$ is analogous.

Note that the cases considered above cover also the weights
$\mu=\omega_{i}+\omega_{n}$, for $2\leq i\leq n$, by considering the dual
$V^{*}$; so (again invoking duality), it remains to consider the case
$\mu=2\omega_{1}$. Then Lemma 3.1 shows that
$\omega\succeq\omega_{1}+\omega_{2}+\omega_{n}$, which then implies that
$\omega\succ\omega_{3}+\omega_{n}$, if $n\geq 3$, and
$\omega\succeq\omega_{1}+2\omega_{2}$, if $n=2$. In the former case the result
follows from previously considered cases if $n\geq 4$. For the case $n=3$ and
$\omega\succeq\omega_{1}+\omega_{2}+\omega_{3}$, we consider the different
possibilities for $\beta$, as before; recall that $\alpha=\alpha_{1}$ and
$\beta\in\\{-\alpha_{1},-\alpha_{1}-\alpha_{2},\alpha_{2},\alpha_{3}\\}$.
Setting $\nu=\omega_{1}+\omega_{2}+\omega_{3}$, and
$\eta=\nu-\alpha_{1}-\alpha_{2}-\alpha_{3}$ and
$\eta^{\prime}=\nu-\alpha_{1}-2\alpha_{2}-\alpha_{3}$, we have that
$\eta-\alpha,\eta-\beta,\eta^{\prime}-\alpha,\eta^{\prime}-\beta\in\Omega(V)$
and deduce that $\eta(s)=\eta^{\prime}(s)$ and so $\alpha_{2}(s)=1$ giving the
usual contradiction. Finally, let $n=2$, and set $\nu=\omega_{1}+2\omega_{2}$
and assume that $\omega\succeq\nu$. We argue as in the previous cases. As
before, since $s$ is regular and $\alpha(s)=\beta(s)$, $\alpha-\beta$ is not a
root. So taking $\alpha=\alpha_{1}$, we have that
$\beta\in\\{-\alpha_{1},\alpha_{2},-\alpha_{1}-\alpha_{2}\\}$. Set
$\eta=\nu-\alpha_{1}-\alpha_{2}$ and
$\eta^{\prime}=\nu-2\alpha_{1}-2\alpha_{2}$. Then one checks that
$\eta-\alpha,\eta-\beta,\eta^{\prime}-\alpha,\eta^{\prime}-\beta\in\Omega(V)$.
So we deduce that $\eta(s)=\eta^{\prime}(s)$ and so
$(\alpha_{1}+\alpha_{2})(s)=1$, a contradiction.

Turning to the proof of (2), we assume now that $s$ is regular and cyclic on
$V=V_{\omega}$, with $\omega\in\\{2\omega_{1},2\omega_{n}\\}$. By duality, we
may assume that $\omega=2\omega_{1}$; note that $p\neq 2$. Suppose the
contrary, that is, $s$ is not strongly regular, and let $\alpha,\beta$ be as
in the proof of (1). Here, we only need to find distinct weights
$\eta,\eta^{\prime}\in\Omega(V)$ such that $\eta(s)=\eta^{\prime}(s)$. It is
straightforward to find such a pair for each choice of $\beta$ (recall that
$\alpha=\alpha_{1}$). For example, if $\beta=-\alpha_{1}$, then
$2\alpha_{1}(s)=0$ and so we may take $\eta=\omega$ and
$\eta^{\prime}=\omega-2\alpha_{1}$. Or if $\beta=\alpha_{2}$ (so $n\geq 2$),
we take $\eta=\omega-2\alpha_{1}$ and
$\eta^{\prime}=\omega-\alpha_{1}-\alpha_{2}$. The remaining cases are left to
the reader.

To conclude, we turn to (3). Here, as $\Omega(V)=\\{0\\}\cup\Phi(G)$, and $s$
is cyclic on $V$, we see that $\alpha(s)\neq\beta(s)$ for all
$\alpha,\beta\in\Phi(G)$, $\alpha\neq\beta$; thus $s$ is strongly regular as
claimed.∎

######  Corollary 3.3.

Theorem 1.2 and Theorem 1.4 hold for $G=A_{n}$.

###### Proof.

For Theorem 1.2, we assume that $s\in T$ is cyclic on $V_{\omega}$, for some
$\omega\in\Omega_{p}^{+}$, $\omega\neq 0$. In addition, we assume that either
$p=2$ when $n=1$ or $\omega\not\in\\{\omega_{1},\dots,\omega_{n}\\}$. For
$n=1$, we have $\omega=\omega_{1}$ and $s\in T$ is cyclic on $V_{\omega}$ if
and only if $s\neq 1$, if and only if $s$ is strongly regular, as claimed. So
we now assume $n\geq 2$; Theorem 1.3 implies that $s$ is regular and then we
use Proposition 3.2 to see that $s$ is strongly regular.

For Theorem 1.4, we assume that $s\in T_{reg}$ is almost cyclic on
$V_{\omega}$, for some $\omega\in\Omega_{p}^{+}$, $\omega\neq 0$. We suppose
further that either $p=2$ if $n=1$ or
$\omega\not\in\\{2\omega_{1},2\omega_{n},\omega_{1},\dots,\omega_{n}\\}$. In
addition, we assume that if $n=2$ and $\omega=\omega_{1}+\omega_{2}$, then
$p\neq 3$. Now if $n=1$, with $p=2$ and so $\omega=\omega_{1}$, then $s$ is
almost cyclic on $V_{\omega}$ if and only if $s\neq 1$, if and only if $s$ is
cyclic on $V_{\omega}$, if and only if $s$ is strongly regular. For $n\geq 2$,
Proposition 3.2 gives the result.∎

We now investigate the weights in Theorem 1.2(2) and Theorem 1.4(iii).

######  Lemma 3.4.

Let $G=A_{n}$ and $\omega\in\Omega_{p}^{+}$, $\omega\neq 0$.

  1. (1) Assume that $p\neq 2$ if $n=1$. If $\omega\in\\{\omega_{1},\ldots,\omega_{n}\\}$, then there exists $s\in T$ with $s$ not strongly regular and $s$ cyclic on $V_{\omega}$.

  2. (2) If $\omega\in\\{2\omega_{1},2\omega_{n}\\}$, then there exists $s\in T_{reg}$, with $s$ not strongly regular and $s$ almost cyclic on $V_{\omega}$.

###### Proof.

(2) Here we have $p\neq 2$ and the result follows from Remark 2.8 when $n=1$.
So we now take $n\geq 2$. By duality, it suffices to consider
$\omega=2\omega_{1}$. Choose $T\leq{\rm SL}_{n+1}(F)$ to be the group of
diagonal matrices $t=\mathop{\rm diag}\nolimits(t_{1},\ldots,t_{n+1})$ of
determinant $1$, and set $\varepsilon_{i}(t)=t_{i}$, for $i=1,\ldots,n+1$.

Suppose first that $n\geq 3$. Let $s=\mathop{\rm
diag}\nolimits(a^{3},a^{2},a,b_{1},\ldots,b_{n-2})\in{\rm SL}_{n+1}(F)$ for
some $a,b_{1},\ldots,b_{n-2}\in F^{\times}$. Then
$\alpha_{1}(s)=\alpha_{2}(s)=a$ so $s$ is not strongly regular. The weights of
$V_{\omega}$ are $2\varepsilon_{i}$ and $\varepsilon_{i}+\varepsilon_{j}$ for
$i,j\in\\{1,\ldots,n\\}$, $i\neq j$. So the eigenvalues of $s$ on $V$ are
$a^{6},a^{4},a^{2},b_{i}^{2}$ for $i=1,\ldots,n-2$ and
$a^{5},a^{4},a^{3},a^{k}b_{i},b_{i}b_{j}$ for $1\leq i<j\leq n-2$, $k=1,2,3$.
It is straightforward to see that there exist $a,b_{1},\ldots,b_{n-2}\in
F^{\times}$ such that $a^{4}$ is the only eigenvalue of $s$ on $V$ of
multiplicity greater than 1, whence the result in this case.

If $n=2$, then let $s=\mathop{\rm diag}\nolimits(a,-a,-a^{-2})$. Then
$\alpha_{1}(s)=(-\alpha_{1})(s)=-1$ and the eigenvalues of $s$ on
$V_{2\omega_{1}}$ are $a^{2},a^{2},a^{-4},-a^{2},-a^{-1},a^{-1}$, so $s$ is
not strongly regular and almost cyclic on $V_{\omega}$ for a suitable choice
of $a\in F^{\times}$.

(1) A similar construction can be used when
$\omega\in\\{\omega_{1},\ldots,\omega_{n}\\}$. The assumption on the
characteristic in case $n=1$ is necessary as all non-identity semisimple
elements in ${\rm SL}_{2}(F)$ are strongly regular when ${\rm char}(F)=2$. ∎

We note that Remark 2.8 and Lemma 3.4 establish the statements of Theorem 1.5
for the groups of type $A_{n}$.

###  3.2. Groups of type $B_{n}$, $C_{n}$ and $D_{n}$

In this section, we prove the main results for classical groups not of type
$A_{n}$. Recall that we assume that $n\geq 3$ when $G=B_{n}$, $n\geq 2$ when
$G=C_{n}$, and $n\geq 4$ when $G=D_{n}$.

The weights of $V_{\omega_{1}}$ are
$\pm\varepsilon_{1},\ldots,\pm\varepsilon_{n}$ and, additionally, 0 when $G$
is of type $B_{n}$ and $p\neq 2$. So every element $s\in T$ can be written as
$s=\mathop{\rm
diag}\nolimits(a_{1},\ldots,a_{n},1,a_{n}^{-1},\ldots,a_{1}^{-1})$, with
respect to some basis of $V_{\omega_{1}}$, where 1 must be dropped if
$G=B_{n}$ with $p=2$.

######  Lemma 3.5.

Let $G=B_{n}$, $n\geq 3$, or $D_{4}$, $n\geq 4$, or $G=C_{n}$, $n\geq 3$ and
$p=2$. Let $\omega\in\Omega_{p}^{+}$, $\omega\neq 0$, and $s\in T_{reg}$.
Suppose that $s$ is almost cyclic on $V_{\omega}$. Then one of the following
holds:

  1. (i) $s$ is strongly regular,

  2. (ii) $G=B_{n}$ and $\omega\in\\{\omega_{1},\omega_{n}\\}$,

  3. (iii) $G=B_{n}$, $p=2$ and $\omega=\omega_{2}$,

  4. (iv) $G=C_{n}$ and $\omega\in\\{\omega_{1},\omega_{2},\omega_{n}\\}$, or

  5. (v) $G=D_{n}$ and $\omega\in\\{\omega_{1},\omega_{n-1},\omega_{n}\\}$.

###### Proof.

By Theorem 1.6, all non-zero weights of $V_{\omega}$ are of multiplicity 1.
The non-zero $p$-restricted irreducible $G$-modules with this property are
listed in Tables 2 and 3. We find that
$\omega\in\\{0,\omega_{1},2\omega_{1},\omega_{2},\omega_{n}\\}$ for $G=B_{n}$,
$\omega\in\\{0,\omega_{1},2\omega_{1},\omega_{2},\omega_{n-1},\omega_{n}\\}$
for $G=D_{n}$, and $\omega\in\\{0,\omega_{1},\omega_{n},\omega_{2}\\}$ for
$G=C_{n}$. Lemma 2.7 then gives the result unless $G=B_{n}$ or $D_{n}$ and
$\omega=2\omega_{1}$.

So suppose that $p\neq 2$, $\omega=2\omega_{1}$ and $s$ is almost cyclic on
$V_{\omega}$. Note that $\omega_{2}=2\omega_{1}-\alpha_{1}$, so by Lemma
2.6(1), $\Omega(V_{\omega_{2}})\subset\Omega(V_{2\omega_{1}})$. Moreover, the
$0$ weight of $V_{2\omega_{1}}$ has multiplicity strictly greater than 1. (See
Table 3.) Finally, as $V_{a}=V_{\omega_{2}}$ and $p\neq 2$, we have
$\Phi(G)\subset\Omega(V_{2\omega_{1}})$. Now suppose that $s$ is not strongly
regular; then Lemma 2.7 implies that $s$ is not almost cyclic on
$V_{\omega_{2}}$, and since here as well the 0 weight occurs with multiplicity
at least $2$, we have that there exist $\alpha,\beta\in\Phi(G)$,
$\alpha\neq\beta$ with $\alpha(s)=\beta(s)\neq 1$. But this then implies that
$s$ is not almost cyclic on $V_{2\omega_{1}}$, giving the result. ∎

######  Corollary 3.6.

Theorem 1.2 and Theorem 1.4 hold for $G=B_{n}$, $n\geq 3$, $G=D_{n}$, $n\geq
4$, and $G=C_{n}$ with $n\geq 3$ and $p=2$.

###### Proof.

For Theorem 1.2, we take $s\in T$ and $\omega\in\Omega_{p}^{+}$, $\omega\neq
0$, such that $s$ is cyclic on $V_{\omega}$. Then all weights of $V_{\omega}$
occur with multiplicity 1 and so $\omega$ is as in Table 2. As all of these
weights appear in the conclusion of Theorem 1.2, the result holds.

For Theorem 1.4, we take $\omega$ as before and suppose that $s\in T_{reg}$,
with $s$ almost cyclic on $V_{\omega}$. Then Lemma 3.5 gives the result.∎

We now turn to the consideration of the group $G=C_{n}$, when $p\neq 2$. Our
first result gives some information about the associated weight lattice
$\Omega$.

######  Lemma 3.7.

Let $\Phi$ be of type $C_{n}$ and let $\omega\in\Omega^{+}$. Suppose that
$\omega\not\in\\{0,2\omega_{1}\\}$ and $\omega$ is not a fundamental dominant
weight. Then one of the following holds:

  1. (i) $\omega\succeq\omega_{1}+\omega_{2}$,

  2. (ii) $n\geq 3$ and $\omega\succeq\omega_{1}+\omega_{3}$, or

  3. (iii) $n=2$ and $\omega\succeq 2\omega_{2}$.

###### Proof.

If $\omega$ is not radical (so $\omega\succ\omega_{1}$ by [12, Lemma 12]) then
the result is available in [13, Lemma 2.2]. So suppose now that $\omega$ is
radical and set $\omega=\sum c_{i}\omega_{i}$, with $\sum c_{i}\geq 2$.
Consider first the case $n=2$. If $c_{2}\geq 2$ then $\omega-2\omega_{2}$ is a
radical dominant weight, so $\omega\succeq 2\omega_{2}$. If $c_{2}=1$ then
$c_{1}$ is even and $c_{1}\geq 2$ by hypothesis, so again
$\omega\succ\omega-\alpha_{1}=(c_{1}-2)\omega_{1}+2\omega_{2}\succeq
2\omega_{2}$. So finally, suppose that $c_{2}=0$, so that $c_{1}\geq 4$ is
even. Then
$\omega\succeq\omega-2\alpha_{1}=(c_{1}-4)\omega_{1}+2\omega_{2}\succeq
2\omega_{2}$.

Consider now the case $n\geq 3$. If $c_{1}c_{3}\neq 0$ then
$\omega-\omega_{1}-\omega_{3}$ is a dominant radical weight, so
$\omega-\omega_{1}-\omega_{3}\succeq 0$, whence the result. Next suppose that
$c_{i}c_{j}\neq 0$ for some odd $i<j$, $(i,j)\neq(1,3)$. As
$\omega_{i}\succ\omega_{i-2}$ for $i\geq 2$, it follows that
$\omega\succ\omega^{\prime}=\omega+\omega_{1}+\omega_{3}-\omega_{i}-\omega_{j}$,
and $\omega^{\prime}$ is a dominant radical weight having non-zero coefficient
of $\omega_{1}$ and $\omega_{3}$. By the previous argument,
$\omega^{\prime}\succeq\omega_{1}+\omega_{3}$, whence the result in this case.
The same argument works if $c_{i}\geq 2$ for some odd $i\geq 3$, as
$\omega_{i-2}=\omega_{i}-\beta$ for some $\beta\in\Phi^{+}$, and so
$c_{i}\omega_{i}=(c_{i}-1)\omega_{i}+\omega_{i}=(c_{i}-1)\omega_{i}+\omega_{i-2}+\beta$,
and we are in the previous case. We are left with the case where $c_{i}=0$ for
all odd $i$ and either $c_{i}\geq 2$ for some even $i$ or $c_{i}c_{j}\neq 0$
for some even $i<j$. In each case, it is a strightforward check to see that
$\omega\succeq 2\omega_{2}\succ\omega_{1}+\omega_{3}$ as required.∎

######  Proposition 3.8.

Let $G=C_{n}$, $n\geq 3,$ $p\neq 2$, and $s\in T$. Let
$\omega\in\Omega_{p}^{+}$, $\omega\neq 0$, and suppose that $s$ is almost
cyclic on $V_{\omega}$. Then $s$ is strongly regular unless one of the
following holds:

  1. (1) $\omega\in\\{\omega_{1},\omega_{2}\\}$;

  2. (2) $n=3,4$ and $\omega=\omega_{n};$ or

  3. (3) $n\geq 4$, $p=3$, and $\omega\in\\{\omega_{n-1},\omega_{n}\\}$.

###### Proof.

We assume that $\omega\not\in\\{0,\omega_{1}\\}$. Then by Theorem 1.3, $s$ is
regular, so $\alpha(s)\neq 1$ for every root $\alpha$. By Theorem 1.6,
$\omega$ occurs in Table 2 or 3. The case $\omega=2\omega_{1}=\omega_{a}$ is
settled in Lemma 2.7, so we assume as well that $\omega\neq 2\omega_{1}$. The
other entries of Tables 2 and 3 either appear in cases (1) - (3) of the
statement or $p>3$ and
$\omega\in\\{\omega_{n-1}+\frac{p-3}{2}\omega_{n},\leavevmode\nobreak\
\frac{p-1}{2}\omega_{n}\\}$. So consider these remaining configurations. By
Theorem 3.7, either $\omega\succeq\omega_{1}+\omega_{2}$ or
$\omega\succeq\omega_{1}+\omega_{3}$. By Lemma 2.5, we have
$\Omega(V_{\omega_{1}+\omega_{2}})\subseteq\Omega(V)$, respectively
$\Omega(V_{\omega_{1}+\omega_{3}})\subseteq\Omega(V)$

Suppose that $s$ is not strongly regular, so that we have $\alpha(s)=\beta(s)$
for some $\alpha,\beta\in\Phi(G)$, $\alpha\neq\beta$. The roots of $G$ are
$\pm 2\varepsilon_{i},\pm\varepsilon_{j}\pm\varepsilon_{k}$ for
$i,j,k\in\\{1,\ldots,n\\}$. As $s$ is regular, we have
$\varepsilon_{i}(s)\neq\pm 1$ and $(\varepsilon_{j}\pm\varepsilon_{k})(s)\neq
1$, for all $i$ and $j\neq k$. Using the Weyl group, we can assume that $s$
satisfies one of:

  1. (i)

$(2\varepsilon_{1})(s)=(2\varepsilon_{2})(s)$,

  2. (ii)

$(2\varepsilon_{1})(s)=(\varepsilon_{2}-\varepsilon_{1})(s)$,

  3. (iii)

$(2\varepsilon_{1})(s)=(\varepsilon_{2}-\varepsilon_{3})(s)$,

  4. (iv)

$n\geq 4$ and
$(\varepsilon_{1}-\varepsilon_{2})(s)=(\varepsilon_{3}-\varepsilon_{4})(s)$,
or

  5. (v)

$(2\varepsilon_{1})(s)=(-2\varepsilon_{1})(s)$.

We now consider the two cases $\omega\succeq\omega_{1}+\omega_{2}$ or
$\omega\succeq\omega_{1}+\omega_{3}$ separately.

Suppose first that $\omega\succeq\omega_{1}+\omega_{2}\succ\omega_{1}$. Then
$V_{\omega}$ has weights $\pm\varepsilon_{i}$ ($i=1,\ldots,n$) and $\pm
2\varepsilon_{i}\pm\varepsilon_{j}$, for $1\leq i\neq j\leq n$. If (i) holds,
$(2\varepsilon_{1})(s)=(2\varepsilon_{2})(s)$, so
$(2\varepsilon_{1}\pm\varepsilon_{3})(s)=(2\varepsilon_{2}\pm\varepsilon_{3})(s)$.
Since $s$ is almost cyclic on $V_{\omega}$, we have
$(2\varepsilon_{1}+\varepsilon_{3})(s)=(2\varepsilon_{1}-\varepsilon_{3})(s)$,
whence $\varepsilon_{3}(s)=\pm 1$, contradicting $s$ regular. If (ii) holds,
$(2\varepsilon_{1})(s)=(\varepsilon_{2}-\varepsilon_{1})(s)$ , so that
$(2\varepsilon_{1}-\varepsilon_{2})(s)=(-\varepsilon_{1})(s)$. As
$\pm(2\varepsilon_{1}-\varepsilon_{2})$, $\pm\varepsilon_{1}$ are weights of
$V_{\omega}$, it follows that $(-\varepsilon_{1})(s)=-1=\varepsilon_{1}(s)$,
again a contradiction. If (iii) holds,
$(2\varepsilon_{1})(s)=(\varepsilon_{2}-\varepsilon_{3})(s)$, so that
$(2\varepsilon_{1}-\varepsilon_{2})(s)=(-\varepsilon_{3})(s)$ and we get a
contradiction as in (ii). Note that $\omega_{1}+\omega_{2}\succ\omega_{3}$, so
$V_{\omega}$ contains the weights of $V_{\omega_{3}}$. These include the
weights $\pm(\varepsilon_{1}-\varepsilon_{3}+\varepsilon_{4})$. If (iv) holds,
then
$(-\varepsilon_{1}+\varepsilon_{3}-\varepsilon_{4})(s)=(-\varepsilon_{2})(s)$,
and
$(+\varepsilon_{1}-\varepsilon_{3}+\varepsilon_{4})(s)=\varepsilon_{2}(s)$; as
usual, $s$ almost cyclic on $V_{\omega}$ implies that
$\varepsilon_{2}(s)=(-\varepsilon_{2})(s)$, again a contradiction. Finally, if
(v) holds, so that $(2\varepsilon_{1})(s)=(-2\varepsilon_{1})(s)$, then
$(2\varepsilon_{1}+\varepsilon_{2})(s)=(-2\varepsilon_{1}+\varepsilon_{2})(s)$
and
$(-2\varepsilon_{1}+\varepsilon_{3})(s)=(2\varepsilon_{1}+\varepsilon_{3})(s)$,
which then implies that $\varepsilon_{2}(s)=\varepsilon_{3}(s)$ so
$(\varepsilon_{2}-\varepsilon_{3})(s)=1$, contradicting that $s$ is regular.

Suppose now that $\omega\succeq\omega_{1}+\omega_{3}\succ
2\omega_{1}\succ\omega_{2}$. Then
$\Omega(V_{\lambda})\subseteq\Omega(V_{\omega})$, for
$\lambda\in\\{\omega_{1}+\omega_{3},2\omega_{1},\omega_{2}\\}$; in particular,
$\pm 2\varepsilon_{i}\pm\varepsilon_{j}\pm\varepsilon_{k}$ are weights of
$V_{\omega}$ for $i,j,k$ distinct, as are $\pm 2\varepsilon_{i}$ for $1\leq
i\leq n$. If (i) holds then $(2\varepsilon_{1})(s)=(2\varepsilon_{2})(s)$ and
$(-2\varepsilon_{1})(s)=(-2\varepsilon_{2})(s)$. As usual, $s$ almost cyclic
on $V_{\omega}$ implies that $(2\varepsilon_{1})(s)=(-2\varepsilon_{2})(s)$,
so that we have
$(2\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3})(s)=(\varepsilon_{3}-\varepsilon_{2})(s)$.
Similarly, we find that
$(-2\varepsilon_{1}+\varepsilon_{2}-\varepsilon_{3})(s)=(-\varepsilon_{2}-\varepsilon_{3})(s)$
and we deduce that $\varepsilon_{3}(s)=(-\varepsilon_{3})(s)$ contradicting
that $s$ is regular. If (ii) holds,
$(2\varepsilon_{1})(s)=(\varepsilon_{2}-\varepsilon_{1})(s)$, then
$(2\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3})(s)=(-\varepsilon_{1}+2\varepsilon_{2}+\varepsilon_{3})(s)$
and
$(2\varepsilon_{1}+\varepsilon_{2}-\varepsilon_{3})(s)=(-\varepsilon_{1}+2\varepsilon_{2}-\varepsilon_{3})(s)$.
As $2\varepsilon_{1}+\varepsilon_{2}\pm\varepsilon_{3}$ and
$-\varepsilon_{1}+2\varepsilon_{2}\pm\varepsilon_{3}$ are weights of
$V_{\omega_{1}+\omega_{3}}$, it follows that
$(2\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3})(s)=(2\varepsilon_{1}+\varepsilon_{2}-\varepsilon_{3})(s)$,
whence $(2\varepsilon_{3})(s)=1$, a contradiction. If (iii) holds,
$(2\varepsilon_{1})(s)=(\varepsilon_{2}-\varepsilon_{3})(s)$, then
$(2\varepsilon_{1}-\varepsilon_{2}+\varepsilon_{3})(s)=1$ and
$(2\varepsilon_{1}-\varepsilon_{2}-\varepsilon_{3})(s)=(-2\varepsilon_{3})(s)$.
The former equality means that the multiplicity of the eigenvalue 1 is greater
than 1 and the second equality implies that $(-2\varepsilon_{3})(s)=1$, a
contradiction. If (iv) holds,
$(\varepsilon_{1}-\varepsilon_{2})(s)=(\varepsilon_{3}-\varepsilon_{4})(s)$,
$n\geq 4$, then
$(2\varepsilon_{1}-\varepsilon_{2}-\varepsilon_{3})(s)=(\varepsilon_{1}-\varepsilon_{4})(s)$
and
$(-\varepsilon_{1}+\varepsilon_{3}-2\varepsilon_{4})(s)=(-\varepsilon_{2}-\varepsilon_{4})(s)$.
As $s$ is almost cyclic on $V_{\omega}$, we have
$(\varepsilon_{1}-\varepsilon_{4})(s)=(-\varepsilon_{2}-\varepsilon_{4})(s)$,
whence $\varepsilon_{1}(s)=(-\varepsilon_{2})(s)$, giving the usual
contradiction. Finally, if (v) holds, then
$(2\varepsilon_{1})(s)=(-2\varepsilon_{1})(s)$, both of which occur in
$V_{2\omega_{1}}$. We have as well
$(2\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3})(s)=(-2\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3})(s)$.
So we deduce that
$(2\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3})(s)=(2\varepsilon_{1})(s)$
giving $(\varepsilon_{2}+\varepsilon_{3})(s)=1$ and the final contradiction.∎

######  Corollary 3.9.

Theorem 1.2 and Theorem 1.4 hold for $G=C_{n}$, $n\geq 3$ and $p\neq 2$.

###### Proof.

We first let $s\in T$ and $\omega\in\Omega_{p}^{+}$, $\omega\neq 0$ such that
$s$ is cyclic on $V_{\omega}$. Then all weights of $V_{\omega}$ occur with
multiplicity 1 and so $\omega$ appears in Table 2. In addition, Proposition
3.8 applies. So either $s$ is strongly regular or one of the following holds:

  1. (i)

$\omega=\omega_{1}$,

  2. (ii)

$\omega=\omega_{2}$ and $(n,p)=(3,3)$,

  3. (iii)

$n=3$ and $\omega=\omega_{3}$, or

  4. (iv)

$n\geq 4$, $p=3$ and $\omega\in\\{\omega_{n-1},\omega_{n}\\}$.

These are precisely the weights listed in the statement of Theorem 1.2.

Now for Theorem 1.4, we let $\omega$ be as above, but this time we take $s\in
T_{reg}$ and suppose that $s$ is almost cyclic on $V_{\omega}$. Then
Proposition 3.8 applies and we find that either $s$ is strongly regular or
$\omega$ is as listed in (i) - (iv) above, or $\omega=\omega_{2}$, for
$(n,p)\neq(3,3)$, or $n=4$ and $\omega=\omega_{4}$ with $p\neq 3$. Hence we
have the result of Theorem 1.4.∎

We now consider the group $C_{2}$.

######  Lemma 3.10.

Let $G=C_{2}$, and let $s\in T_{reg}$. Then

  1. (1) $s$ is cyclic on $V_{\omega_{1}}$ and almost cyclic on $V_{\omega_{2}}$. If $s$ is not cyclic on $V_{\omega_{2}}$ then $p\neq 2$ and $-1$ is an eigenvalue of $s$ on $V_{\omega_{2}}$ of multiplicity $2$.

  2. (2) There exists $s^{\prime}\in T_{reg}$, with $s^{\prime}$ is not strongly regular, and $s^{\prime}$ cyclic on $V_{\omega_{1}}$ and on $V_{\omega_{2}}.$

  3. (3) Let $\omega\in\Omega_{p}^{+}$, $p\neq 2$, $\omega\not\in\\{0,\omega_{1},\omega_{2}\\}$. If $s$ is almost cyclic on $V_{\omega}$, then $s$ is strongly regular.

###### Proof.

The proof of (1) is straightforward and left to the reader. For (2), define
$s^{\prime}$ by $\varepsilon_{1}(s^{\prime})=a$,
$\varepsilon_{2}(s^{\prime})=a^{3}$, where $a\in F^{\times}$ and $a^{24}\neq
1$. Then $s^{\prime}$ is cyclic on $V_{\omega_{1}}$ (and hence regular)
$s^{\prime}$ and is not strongly regular as
$2\varepsilon_{1}(s^{\prime})=(\varepsilon_{2}-\varepsilon_{1})(s^{\prime})$.
The eigenvalues of $s^{\prime}$ on $V_{\omega_{2}}$ are $1,a^{\pm 2},a^{\pm
4}$ (where 1 is to be dropped if $p=2$), so $s^{\prime}$ is cyclic on
$V_{\omega_{2}}$.

(3) As usual, Theorem 1.6 implies that $\omega$ appears in Table 2 or Table 3;
we have to examine the cases with $p\neq 2$ and
$\omega\in\\{2\omega_{2},\omega_{1}+\frac{p-3}{2}\omega_{2},\frac{p-1}{2}\omega_{2}\\}$.

Suppose that $s$ is not strongly regular. As in the proof of Proposition 3.8,
we can assume that one of the following holds:

  1. (i)

$(2\varepsilon_{1})(s)=(2\varepsilon_{2})(s)$,

  2. (ii)

$(2\varepsilon_{1})(s)=(\varepsilon_{2}-\varepsilon_{1})(s)$, or

  3. (iii)

$(2\varepsilon_{1})(s)=(-2\varepsilon_{1})(s)$.

By Lemma 3.7, either $\omega\succeq\omega_{1}+\omega_{2}\succ\omega_{1}$ or
$\omega\succeq 2\omega_{2}$.

Suppose first that $\omega\succeq\omega_{1}+\omega_{2}$. Note that $\pm
2\varepsilon_{1}\pm\varepsilon_{2},\pm\varepsilon_{1}\pm
2\varepsilon_{2},\pm\varepsilon_{1},\pm\varepsilon_{2}\in\Omega(V_{\omega_{1}+\omega_{2}})$
If (i) holds, then $(2\varepsilon_{1}-\varepsilon_{2})(s)=\varepsilon_{2}(s)$
and $(-2\varepsilon_{1}+\varepsilon_{2})(s)=-\varepsilon_{2}(s)$. As $s$ is
almost cyclic on $V_{\omega}$, it follows that
$\varepsilon_{2}(s)=-\varepsilon_{2}(s)$, which is a contradiction as
$2\varepsilon_{2}$ is a root. If (ii) holds then
$(2\varepsilon_{1})(s)=(\varepsilon_{2}-\varepsilon_{1})(s)$ implies
$(2\varepsilon_{1}-\varepsilon_{2})(s)=(-\varepsilon_{1})(s)$ and
$(-2\varepsilon_{1}+\varepsilon_{2})(s)=\varepsilon_{1}(s)$. As $s$ is almost
cyclic on $V_{\omega}$ and $\pm(2\varepsilon_{1}-\varepsilon_{2})$,
$\pm\varepsilon_{1}\in\Omega(V_{\omega_{1}+\omega_{2}})$, it follows that
$\varepsilon_{1}(s)=(-\varepsilon_{1})(s)$, whence $(2\varepsilon_{1})(s)=1$,
which contradicts $s$ being regular. Finally, if (iii) holds, so that
$(2\varepsilon_{1})(s)=(-2\varepsilon_{1})(s)$, then
$(2\varepsilon_{1}\pm\varepsilon_{2})(s)=(-2\varepsilon_{1}\pm\varepsilon_{2})(s)$
which implies as before that $\varepsilon_{2}(s)=(-\varepsilon_{2})(s)$,
giving the usual contradiction.

Suppose that $\omega\succeq 2\omega_{2}$. As $2\omega_{2}\succ
2\omega_{1}\succ\omega_{2}$, we observe that $\pm 2\varepsilon_{1}\pm
2\varepsilon_{2}$, $\pm 2\varepsilon_{1}$, $\pm 2\varepsilon_{2}$,
$\pm\varepsilon_{1}\pm\varepsilon_{2}$ are weights of $V_{\omega}$. If (i)
holds then $\pm(2\varepsilon_{1}-2\varepsilon_{2})(s)=1$ and hence the
multiplicity of the eigenvalue 1 of $s$ on $V_{\omega}$ is at least 2. In
addition, $(2\varepsilon_{1})(s)=(2\varepsilon_{2})(s)$, so
$(2\varepsilon_{2})(s)=1$ as $s$ is almost cyclic on $V_{\omega}$. This is a
contradiction. If (ii) holds, that is,
$(2\varepsilon_{1})(s)=(\varepsilon_{2}-\varepsilon_{1})(s)$ then
$(-2\varepsilon_{1})(s)=(-\varepsilon_{2}+\varepsilon_{1})(s)$. So
$(\varepsilon_{2}-\varepsilon_{1})(s)=-1$ and the multiplicity of $-1$ as an
eigenvalue of $s$ on $V_{\omega}$ is at least 2. As
$\pm(2\varepsilon_{1}-2\varepsilon_{2})\in\Omega(V_{2\omega_{2}})$ and
$(2\varepsilon_{1}-2\varepsilon_{2})(s)=1$, we conclude that 1 is an
eigenvalue of $s$ on $V_{\omega}$ is of multiplicity at least 2. This is a
contradiction, as $s$ is almost cyclic on $V_{\omega}$. Finally, if (iii)
holds, then we have $(2\varepsilon_{1})(s)=(-2\varepsilon_{1})(s)$ , and so
$(2\varepsilon_{1}+2\varepsilon_{2})(s)=(-2\varepsilon_{1}+2\varepsilon_{2})(s)$
we deduce that $(2\varepsilon_{2})(s)=1$, a contradiction.∎

######  Corollary 3.11.

Theorem 1.2 and Theorem 1.4 hold for $G=C_{2}$.

###### Proof.

Recall that for $s\in T$ and $\omega\in\Omega_{p}^{+}$, $\omega\neq 0$, if $s$
is almost cyclic on $V_{\omega}$, then $\omega$ appears in Table 2 or Table 3.
In particular, for $p=2$, the only weights which need to be considered are
$\omega_{1}$ and $\omega_{2}$ and both of these are in the statements of the
theorems. Henceforth we assume that $p\neq 2$. If $s\in T$ is cyclic on
$V_{\omega}$ for some $\omega\in\Omega_{p}^{+}$, $\omega\neq 0$, then again
consulting Table 2, we find that $\omega=\omega_{1}$ or $\omega=\omega_{2}$,
as in the conclusion of Theorem 1.2. So finally, we are left to consider the
case where $s\in T_{reg}$, $\omega\in\Omega_{p}^{+}$, and $s$ is almost cyclic
on $V_{\omega}$. Then here Lemma 3.10(3) shows that either $s$ is strongly
regular or $\omega\in\\{0,\omega_{1},\omega_{2}\\}$ as claimed in Theorem
1.4.∎

To complete our consideration of the classical type groups, we must establish
the existence of elements as claimed in Theorem 1.5.

######  Lemma 3.12.

Let $G=B_{n}$ with $n\geq 3$, or $G=C_{n}$, with $p=2$ and $n\geq 3$. Then
there exists $s\in T_{reg}$ such that $s$ is not strongly regular, but cyclic
on each of the modules $V_{\omega_{1}}$ and $V_{\omega_{n}}$. Moreover, if
$G=B_{n}$, then we may choose $s$ so that $s$ is cyclic on
$V_{\omega_{1}}\oplus V_{\omega_{n}}\oplus\delta_{p,2}V_{0}$.

###### Proof.

For $0\leq i\leq n-3$, let $b_{i}\in F^{\times}$ be a primitive $p_{i}$-th
root of unity where $p_{0},\ldots,p_{n-3}$ are distinct primes greater than
$11$. Set $X=\mathop{\rm
diag}\nolimits(b_{0},(b_{0})^{3},(b_{0})^{5},b_{1},\ldots,b_{n-3})$. Now set
$t=\mathop{\rm diag}\nolimits(X,1,X^{-1})$, where the 1 is to be dropped if
$G=C_{n}$. For a suitable choice of basis for the natural module for $G$, we
have $t\in{\rm SO}_{2n+1}(F)$, respectively $t\in{\rm Sp}_{2n}(F)$, for
$G=B_{n}$, respectively $G=C_{n}$.

We first note that $t$ is cyclic on $V_{\omega_{1}}$. In addition,
$(\varepsilon_{2}-\varepsilon_{1})(t)=b_{0}^{2}=(\varepsilon_{3}-\varepsilon_{2})(t)$,
so $t$ is not strongly regular.

Consider first the case $G=B_{n}$ and let $s\in G$ be a semisimple element
acting on $V_{\omega_{1}}$ as $t$ does. In particular, $s$ is regular (since
cyclic on $V_{\omega_{1}}$), but $s$ is not strongly regular. Now the weights
of $V_{\omega_{1}}\oplus V_{\omega_{n}}\oplus\delta_{p,2}V_{0}$ are
$0,\pm\varepsilon_{i}\ (1\leq i\leq
n),\frac{1}{2}(\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n})$. Hence, the
remaining claim will follow if we show that the elements
$\pm\varepsilon_{i}(s),1,(\frac{1}{2}(\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n})(s)$
are distinct. As $|t|$ is odd, we may assume that $|s|$ is odd as well. Hence,
it suffices to show that the eigenvalues
$2\varepsilon_{i}(s),1,(\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n})(s)$ are
distinct. The elements $(\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n})(s)$
have the form $(b_{0})^{\pm m}x$, for $m\in\\{1,3,7,9\\}$ and for some
$x\in\\{b_{1}^{\pm 1}\cdots b_{n-3}^{\pm 1}\\}$. As $p_{i}>11$, all these
elements are distinct, and differ from $1,(b_{0})^{\pm 2},(b_{0})^{\pm 6}$,
$b_{0}^{10}$, and $b_{1}^{\pm 2},\ldots,b_{n-3}^{\pm 2}$, whence the result.

The argument for $G=C_{n}$ is entirely similar as the weights of
$V_{\omega_{n}}$ are $\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n}$.∎

######  Lemma 3.13.

Let $G=D_{n}$, $n\geq 4$. Then there exists $s\in T_{reg}$ such that $s$ is
not almost cyclic on $V_{\omega_{2}}$ (equivalently, $s$ is not strongly
regular), but cyclic on $V_{\omega_{1}}\oplus V_{\omega_{n-1}}\oplus
V_{\omega_{n}}\oplus V_{0}$.

###### Proof.

As $\omega_{a}=\omega_{2}$, by Lemma 2.7, $s^{\prime}\in T$ is almost cyclic
on $V_{\omega_{2}}$ if and only if $s^{\prime}$ is strongly regular. We now
exhibit the existence of the $s$ as claimed in the statement.

The group $G=D_{n}$ is a maximal rank subgroup of $B_{n}$ generated by root
subgroups corresponding to long roots. The element $s$ introduced in Lemma
3.12 is therefore contained in a subgroup $G=D_{n}$, and $s$ is regular as an
element of $D_{n}$, since it is regular in $B_{n}$. Moreover, we have seen in
the proof of Lemma 3.12 that
$(\varepsilon_{2}-\varepsilon_{1})(s)=(\varepsilon_{3}-\varepsilon_{2})(s)$;
as $\varepsilon_{2}-\varepsilon_{1}$ and $\varepsilon_{3}-\varepsilon_{2}$ are
long roots of $B_{n}$, they are roots of the subgroup $D_{n}$. So $s$ is not
strongly regular in $D_{n}$ and by Lemma 2.7, it follows that $s$ is not
almost cyclic on $V_{\omega_{2}}$. Furthermore, for the purposes of this
proof, let $\eta_{1},\dots,\eta_{n}$ denote the fundamental dominant weights
of $B_{n}$. Then, the irreducible module for $B_{n}$ with highest weight
$\eta_{n}$ restricts to $D_{n}$ as $V_{\omega_{n-1}}\oplus V_{\omega_{n}}$. In
addition, $V_{\eta_{1}}|_{H}=V_{\omega_{1}}\oplus(1-\delta_{p,2})V_{0}$. Now
Lemma 3.12 shows that $s$ is cyclic on
$V_{\eta_{1}}+V_{\eta_{n}}+\delta_{p,2}V_{0}$, and so $s$ is cyclic on
$V_{\omega_{1}}+V_{\omega_{n-1}}+V_{\omega_{n}}+V_{0}$, as claimed. ∎

######  Remark 3.14.

Note that the above two results provide a proof of Theorem 1.5(1) for
$G=B_{n}$, $n\geq 3$, $G=D_{n}$ with $n\geq 4$, and $G=C_{n}$ for $n\geq 3$
and $p=2$.

######  Lemma 3.15.

Let $G=C_{n}$, $n\geq 2$. Then there exists $s\in T_{reg}$, not strongly
regular, with $s$ almost cyclic on $V_{\omega_{2}}$.

###### Proof.

Define $s$ by $\varepsilon_{1}(s)=a$, $\varepsilon_{2}(s)=a^{3}$,
$\varepsilon_{i}(s)=b_{i}$, for $a,b_{1},\ldots,b_{n-2}\in F^{\times}$. Then
$s$ is not strongly regular as
$(2\varepsilon_{1})(s)=a^{2}=(\varepsilon_{2}-\varepsilon_{1})(s)$, and
$(2\varepsilon_{1})(s)$, $\varepsilon_{2}-\varepsilon_{1}$ are roots of $G$.
Moreover, one may choose the $a,b_{i}$ such that $s$ is regular. The
eigenvalues of $s$ on $V_{\omega_{2}}$ are $1,a^{\pm 2},a^{\pm 4},a^{\pm
1}b_{i}^{\pm 1},a^{\pm 3}b_{i}^{\pm 1},b_{i}^{\pm 1}b_{j}^{\pm 1}$, $i\neq j$.
We may choose $a,b_{1},\ldots,b_{n-2}\in F^{\times}$ such that the eigenvalues
different from $1$ are distinct. Indeed, the non-zero weights of
$V_{\omega_{2}}$ are roots and by choosing $s$ regular, we know that the
multiplicity of the eigenvalue $1$ is precisely the multiplicity of the zero
weight in $V_{\omega_{2}}$, while all other weights have multiplicity 1. Hence
$s$ is almost cyclic on $V_{\omega_{2}}$. ∎

######  Remark 3.16.

A similar argument works for $G=B_{n}$, with $p=2$; for this case, set
$\varepsilon_{1}(s)=a$ and $\varepsilon_{2}(s)=a^{2}$ and the remaining values
as for $C_{n}$.

######  Lemma 3.17.

Let $G=C_{n}$, $n\geq 2$. Then there exists $s\in T_{reg}$, $s$ not strongly
regular, which separates the weights of $V_{\omega_{k}}$, for $k=1,\ldots,n$.

###### Proof.

Let $a,b_{i}\in F^{\times}$, for $1\leq i\leq n$, and define $s\in T$ by
$\varepsilon_{i}(s)=b_{i}$ for $1\leq i\leq n$ and in addition $b_{1}=a$ and
$b_{2}=a^{3}$. We first note that $s$ is not strongly regular as
$2\varepsilon_{1}(s)=(\varepsilon_{2}-\varepsilon_{1})(s)$. We now show that
by choosing appropriate values for the $b_{i}$, the element $s$ separates the
weights of $V_{\omega_{k}}$, as claimed. This will also show that $s$ is
regular as $s$ is then cyclic on $V_{\omega_{1}}$.

The non-zero weights of $V_{\omega_{k}}$ are
$\pm\varepsilon_{i_{1}}\pm...\pm\varepsilon_{i_{j}}$ for $j\leq k$ with $k-j$
even and $1\leq i_{1}<...<i_{j}\leq n$. This follows from the fact that
$V_{\omega_{j}}$ occurs as a composition factor of
$\wedge^{j}(V_{\omega_{1}})$, for all $1\leq j\leq n$, and the subdominant
weights of $\omega_{k}$, for $k\geq 2$, are of the form $\omega_{k-2i}$ for
$0\leq i\leq\lfloor\frac{k}{2}\rfloor$. A weight
$\pm\varepsilon_{i_{1}}\pm...\pm\varepsilon_{i_{j}}$ takes the value
$b_{i_{1}}^{\pm 1}\cdots b_{i_{j}}^{\pm 1}=xy$, on $s$, where $x=a^{\pm m}$
for some $0\leq m\leq 4$ and $y$ is a product of certain $b_{i}^{\pm 1}$, with
$i>2$. (In particular, if $n=2$ then the weights of $V_{\omega_{k}}$ take
values among $a^{\pm 1},a^{\pm 3},a^{\pm 2},a^{\pm 4},1$ on $s$.) For a
suitable choice of $a$ and $b_{i}$, these values are distinct. For example, we
can choose $b_{i}$ to be a primitive $p_{i}$-th root of unity for $3\leq i\leq
n$ for primes $p_{3}>...>p_{n}$, and $a$ to be a primitive $p_{0}$-th root of
unity for a prime $p_{0}>{\rm max}\\{p_{3},7\\}$. Therefore, $s$ separates the
weights of $V_{\omega_{k}}$ for every fixed $k$.∎

######  Proposition 3.18.

Let $G=C_{n},n\geq 2$, $p\neq 2$.

  1. (1) Let $\omega\in\Omega_{p}^{+}$ be as follows:

    1. (i) $\omega=\omega_{1}$,

    2. (ii) $\omega\in\\{\omega_{n-1},\omega_{n}\\}$, $n\geq 3$, $p=3$, or

    3. (iii) $\omega=\omega_{3}$, $n=3$, $p\neq 3$.

Then there exists $s\in T_{reg}$, with $s$ not strongly regular but cyclic on
$V_{\omega}$.

  2. (2) Let $\omega\in\Omega_{p}^{+}$ be one of:

    1. (i) $\omega=\omega_{4}$, $n=4$ and $p\neq 3$, or

    2. (ii) $\omega=\omega_{2}$ with $(n,p)\neq(3,3)$.

Then there exists $s\in T_{reg}$, with $s$ not strongly regular but almost
cyclic on $V_{\omega}$.

###### Proof.

This follows from Lemma 3.17 and Tables 2 and 3. Indeed, if $s$ separates the
weights of $V$ then $s$ is cyclic on $V$ whenever all weights of $V$ are of
multiplicity 1, and almost cyclic if exactly one weight is of multiplicity
greater than 1.∎

######  Lemma 3.19.

Theorem 1.5 is true if G is of classical Lie type.

###### Proof.

For $G$ of type $A_{n}$ this was established in Section 3.1. For Theorem
1.5(1), we refer to Remark 3.14 and note that it remains to consider $G=C_{2}$
and $G=C_{n}$ with $n\geq 3$ and $p\neq 2$. For $G=C_{2}$, we must establish
the existence of $s\in T$, cyclic on $V_{\omega_{1}}$ and on $V_{\omega_{2}}$,
with $s$ not strongly regular. This follows from Lemma 3.10(2).

For $G=C_{n}$, $n\geq 3$ and $p\neq 2$, we must establish the existence of
$s\in T$, $s$ not strongly regular, with $s$ cyclic on

  1. (a)

$V_{\omega_{1}}$,

  2. (b)

$V_{\omega_{3}}$ when $n=3$, and

  3. (c)

$V_{\omega_{n-1}}$ or $V_{\omega_{n}}$, when $p=3$.

(If necessary, we may have different $s$ for each of the modules.) This is
precisely the statement of Proposition 3.18(1).

Now we turn to the list of modules to be treated for Theorem 1.5(2). For
$n\geq 3$, $\omega=\omega_{2}$, and $G=C_{n}$, or $G=B_{n}$ with $p=2$, the
result follows from Lemma 3.15 and Remark 3.16. For $G=C_{4}$ with
$\omega=\omega_{4}$, $p\neq 2,3$, the result follows from Proposition
3.18(2).∎

##  4\.  Groups of exceptional types and Theorem 1.7

In this section we will establish all of the main results in case $G$ is of
exceptional type and deduce Theorem 1.7.

######  Lemma 4.1.

Theorem 1.4 holds for $G$ of exceptional type.

###### Proof.

Let $\omega\in\Omega_{p}^{+}$ and $s\in T$ be almost cyclic on $V_{\omega}$.
Then Theorem 1.6 and Tables 2 and 3 imply that either $\omega=\omega_{a}$, or
$\omega$ is the highest short root (in case $G=G_{2}$ or $F_{4}$), or $\omega$
is a miniscule weight. The miniscule weights appear in the conclusion of
Theorem 1.4, as well as the highest short roots. Moreover, Lemma 2.7 shows
that if $s$ is almost cyclic on $V_{a}$, then either $s$ is strongly regular
or $(G,p)=(G_{2},3)$ or $(F_{4},2)$. The latter configurations are also in the
list of exceptions in the statement of Theorem 1.4, hence the result.∎

######  Lemma 4.2.

Theorem 1.2 holds for $G$ of exceptional type.

###### Proof.

Let $\omega\in\Omega_{p}^{+}$ and $s\in T$ be cyclic on $V_{\omega}$. Then all
weights of $V_{\omega}$ have multiplicity $1$ and now comparing the statement
of Theorem 1.2 with Table 2 gives the result. ∎

We are left with proving Theorem 1.5; we treat each of the groups in turn.

######  Proposition 4.3.

Let $G$ be of type $F_{4}$. There exists $s\in T_{reg}$ such that $s$ is
almost cyclic on $V_{\omega_{4}}$ but not strongly regular. If $p=3$ then $s$
can be chosen to be cyclic on $V_{\omega_{4}}$.

###### Proof.

Let $V=V_{\omega_{4}}$. Recall that $G$ contains a subsystem subgroup $H$ of
type $B_{4}$, generated by the long root subgroups of $G$, so $T\leq H$. For
the purposes of this proof, let $\eta_{i}$, $1\leq i\leq 4$ denote the
fundamental dominant weights of $H=B_{4}$ (with respect to an appropriate
choice of Borel subgroup of $H$). Then the non-zero weights of $V|_{H}$ are
exactly the non-zero weights of the $FH$-module $V_{\eta_{1}}+V_{\eta_{4}}$.
By comparison of the dimensions we conclude that the non-trivial composition
factors of $V|_{H}$ are $V_{\eta_{1}}$ and $V_{\eta_{4}}$, and there are
$(1-\delta_{p,3})(1+\delta_{p,2})$ trivial composition factors.

Now let $s\in H$ be as in Lemma 3.12; then, $\alpha(s)=\beta(s)$, for $\alpha$
and $\beta$ distinct roots of $H$, so $s$ is not strongly regular in $F_{4}$.
Moreover, $s$ is cyclic on $V_{\eta_{1}}+V_{\eta_{4}}$ and so almost cyclic on
$V$. Then Theorem 1.3 implies that $s$ is regular in $G$. If $p=3$, then
$V|_{H}=V_{\eta_{1}}+V_{\eta_{4}}$ and Lemma 3.12 shows that $s$ is cyclic on
$V$. ∎

######  Remark 4.4.

Let $G=F_{4}$, $p=2$.

  1. (1)

There exists an element $s\in T_{reg}$, $s$ not strongly regular, which is
almost cyclic on $V_{a}=V_{\omega_{1}}$ (and not almost cyclic on
$V_{\omega_{4}}$). Recall that, identifying $T$ with $(F^{\times})^{4}$ and
$\varepsilon_{i}$ the $i$-th coordinate function for $1\leq i\leq 4$, the
roots of $F_{4}$ are
$\pm\varepsilon_{i},\frac{1}{2}(\pm\varepsilon_{1}\pm\varepsilon_{2}\pm\varepsilon_{3}\pm\varepsilon_{4})$
(short) and $\pm\varepsilon_{i}\pm\varepsilon_{j}$, $1\leq i<j\leq 4$ (long).
Let $a_{i}\in F^{\times}$ be a primitive $p_{i}$-root of unity, where
$p_{1}=3,p_{2}=5,p_{3}=7$. Define $s\in T$ by $\varepsilon_{1}(s)=a_{1},$
$\varepsilon_{2}(s)=a_{2},$ $\varepsilon_{3}(s)=a_{1}a_{2},$
$\varepsilon_{4}(s)=a_{3}$. The non-zero weights of $V_{a}$ are
$\pm\varepsilon_{i}\pm\varepsilon_{j}$ $(1\leq i<j\leq 4)$ so $s$ is almost
cyclic on $V_{a}$. Then
$(\varepsilon_{1}+\varepsilon_{2}-\varepsilon_{3}+\varepsilon_{4})(s)=(-\varepsilon_{1}-\varepsilon_{2}+\varepsilon_{3}+\varepsilon_{4})(s)=\varepsilon_{4}(s)=a_{3}$
so $s$ is not almost cyclic on $V_{\omega_{4}}$. Note that $s$ is regular as
$\alpha(s)\neq 1$ for every root $\alpha$, but it is not strongly regular.
This justifies one of the claims of Remark 2.8, and together with Proposition
4.3, completes the proof of Theorem 1.5 for the group $G=F_{4}$.

  2. (2)

There exists a non-regular element $s\in T$ such that $V^{s}_{a}=V_{a}^{T}$.
Indeed, let $\varepsilon_{i}(s)=t_{i}\in F^{\times}$, where $t_{1}=1$. Then
$s$ is not regular. The weights of $V_{a}=V_{\omega_{1}}$ are
$0,\pm\varepsilon_{i}\pm\varepsilon_{j}$, so the eigenvalues of $s$ on the
non-zero weights of $V_{a}$ are $1,(\pm\varepsilon_{i}\pm\varepsilon_{j})(s)$,
that is, $1,t_{i}^{\pm 1},t_{i}^{\pm 1}t_{j}^{\pm 1}$, where $i,j=2,3,4,i\neq
j$. For an appropriate choice of $t_{i}\in F^{\times}$, these eigenvalues
differ from 1, and the claim follows. This proves a claim in Remark 2.8 for
the group $G=F_{4}$.

######  Proposition 4.5.

Let $G=E_{6}$. Then there exists $s\in T$, not strongly regular, which is
cyclic on $V_{\omega_{1}}$ and on $V_{\omega_{6}}$.

###### Proof.

It suffices to consider $V=V_{\omega_{1}}$. Let $H<G$ be a subsystem subgroup
of type $D_{5}$. For the purposes of this proof, let $\eta_{i}$, for $1\leq
i\leq 5$, denote the fundamental dominant weights of $H$ (with respect to an
appropriate choice of maximal torus $T_{H}$ and Borel subgroup of $H$). Then,
we have that $V|_{H}=V_{\eta_{1}}\oplus V_{\eta_{4}}\oplus V_{0}$. (Here we
have made a choice of the labeling of the roots of $H$.)

Now let $s\in H$ be as in Lemma 3.13. Then $s$ is cyclic on
$V_{\eta_{1}}+V_{\eta_{4}}+V_{\eta_{5}}+V_{0}$, so in particular on $V$. As
well, $s$ is not strongly regular in $H$ and since roots of $H$ are roots of
$G$, $s$ is not strongly regular in $G$. ∎

######  Proposition 4.6.

Let $G$ be of type $E_{6}$. Then there exists $s\in T_{reg}$ such that $s$ is
not strongly regular but almost cyclic on $V_{\omega_{1}}$ and on
$V_{\omega_{6}}$.

###### Proof.

It suffices to establish the result for $V_{\omega_{1}}$. Let $H\leq G$ be of
type $F_{4}$. Let $s$ be as constructed in the proof of Proposition 4.3. For
the purposes of this proof, let $\eta_{i}$, $1\leq i\leq 4$ denote the
fundamental dominant weights of $H$ (with respect to a fixed choice of Borel
subgroup and maximal torus). Then
$V_{\omega_{1}}|_{H}=V_{\eta_{4}}\oplus(1+\delta_{p,3})V_{0}$. Since $s$ is
almost cyclic on $V_{\eta_{4}}$, and cyclic if $p=3$, and the zero weight has
multiplicity 2 in $V_{\eta_{4}}$ if $p\neq 3$, we see that $s$ is almost
cyclic on $V_{\omega_{1}}$. We claim that $s$ is regular but not strongly
regular. Recall that $V_{a}=V_{\omega_{2}}$. We note that

$V_{a}|_{H}=\begin{cases}V_{\eta_{1}}\oplus V_{\eta_{4}},&\mbox{if }p\neq
2,3;\cr V_{\eta_{1}}\oplus V_{\eta_{4}}\oplus V_{0},&\mbox{if }p=3;\cr
V_{\eta_{1}}\oplus V_{\eta_{4}}\oplus V_{\eta_{4}},&\mbox{if }p=2.\end{cases}$

To see that $s$ is regular, we need to show that the multiplicity of the
eigenvalue $1$ for $s$ on $V_{a}$ is equal to $\dim T=6$. Note that $s$ is
regular in $H$ and so the eigenvalue $1$ on $V_{\eta_{1}}$ has multiplicity
$4-2\delta_{p,2}$. Also, the non-zero weights of $V_{\eta_{4}}$ are roots of
$H$ and the zero weight has multiplicity $2-\delta_{p,3}$ in $V_{\eta_{4}}$;
so again the eigenvalue 1 has multiplicity $2-\delta_{p,3}$ in $V_{\eta_{4}}$.
In each of the above decompositions, we find that the eigenvalue 1 for $s$ on
$V_{a}$ has multiplicity 6.

Finally, to see that $s$ is not strongly regular, we appeal to Lemma 2.7. It
suffices to see that $s$ is not almost cyclic on $V_{a}$. Recall that
$\eta_{4}\prec\eta_{1}$. Using Theorem 2.5 and the above decompositions of
$V_{a}|_{H}$, we find that all non-zero weights of $V_{\eta_{4}}$ occur with
multiplicity 2 in $V_{a}$, whence the result.∎

We will use the following result when treating the group $G=E_{7}$.

######  Proposition 4.7.

Let $G=A_{n}$, $n\geq 4$. Then there exists $s\in T_{reg}$, $s$ not strongly
regular, such that $s$ is not almost cyclic on $V_{\omega_{1}+\omega_{n}}$ and
cyclic on $V=V_{\omega_{2}}\oplus V_{\omega_{n-1}}$.

###### Proof.

We start by considering a certain subvariety of $F^{m}$. Let $m\geq 4$. Set
$X=\\{(x_{1},\dots,x_{m})\in F^{m}\ |\ x_{1}\cdots x_{m}=1\\}$. Then $X$ is
isomorphic to the principal open subset $D(f)\subset F^{m-1}$ defined by the
function $f\in F[T_{1},\dots,T_{m-1}]$, $f=T_{1}\cdots T_{m-1}$. Indeed, the
associated $F$-algebra of $X$ is isomorphic to
$F[T_{1},\dots,T_{m-1},(T_{1}\cdots T_{m-1})^{-1}]$ which is (isomorphic to)
the localization of $F[T_{1},\dots,T_{m-1}]$ with respect to the
multiplicative set $S=\\{(T_{1}\cdots T_{m-1})^{\ell}\ |\ \ell\geq 0\\}$. In
particular, $X$ is irreducible of dimension $m-1$. Now consider the closed
subset $Y$ of $X$, $Y=\\{(x_{1},\dots,x_{m})\in X\ |\
x_{1}^{2}=x_{2}x_{3}\\}$. Then the associated affine $F$-algebra is
$F[X]/(T_{1}^{2}-T_{2}T_{3})$. Since $F[X]$ is a localization of
$F[T_{1},\dots,T_{m-1}]$ and $T_{1}^{2}-T_{2}T_{3}$ is an irreducible
polynomial function in $F[T_{1},\dots T_{m-1}]$, $Y$ is also an irreducible
subvariety of $F^{m}$, of dimension $m-2$.

We now apply the above reasoning to a maximal torus of $G={\rm SL}_{n+1}(F)$,
namely the usual torus of diagonal matrices of determinant 1. Set $s={\rm
diag}(d_{1},\dots,d_{n+1})\in T$, where $d_{i}\in F^{\times}$ and
$\prod_{i=1}^{n+1}d_{i}=1$. If we assume further that $d_{1}^{2}=d_{2}d_{3}$
then
$(\varepsilon_{1}-\varepsilon_{2})(s)=(\varepsilon_{3}-\varepsilon_{1})(s)$
and so $s$ is not almost cyclic on $V_{\omega_{1}+\omega_{n}}$. Let $Y=\\{s\in
T\ |\ d_{1}^{2}=d_{2}d_{3}\\}$ so that by the discussion of the first
paragraph, $Y$ is an irreducible closed subvariety of $T$ of dimension $n-1$.
The module $V$ has $1$-dimensional $T$-weight spaces. For weights
$\lambda,\mu\in\Omega(V)$, with $\lambda\neq\mu$, let
$K_{\lambda\mu}=\ker(\lambda-\mu)$. The set
$K:=\cup_{\lambda,\mu}K_{\lambda\mu}$ is a proper closed subset of $T$. Since
$Y$ is irreducible, if $Y\subseteq K$, then there exist
$\lambda,\mu\in\Omega(V)$, with $\lambda\neq\mu$, such that
$Y\subseteq\ker(\lambda-\mu)^{\circ}$ and dimension considerations show that
these two sets must be equal. But the weights of $V$ are of the form
$\pm(\varepsilon_{i}+\varepsilon_{j})$, $1\leq i<j\leq n+1$. So the possible
weight differences $\lambda-\mu$ (up to a sign) are
$\varepsilon_{i}+\varepsilon_{j}\pm(\varepsilon_{k}+\varepsilon_{\ell})$,
$\varepsilon_{j}-\varepsilon_{\ell}$,
$2\varepsilon_{i}+\varepsilon_{j}+\varepsilon_{\ell}$ and
$2(\varepsilon_{i}+\varepsilon_{j})$, for $i,j,k,\ell$ distinct. So there are
no distinct weights $\lambda,\mu$ with $Y=\ker(\lambda-\mu)^{\circ}$. Hence
$Y$ does not lie in $K$ and the elements in $Y\setminus K$ are almost cyclic,
indeed cyclic, on $V$ and not almost cyclic on $V_{\omega_{1}+\omega_{n}}$. So
these elements are regular by Lemma 2.2 and not strongly regular by Lemma
2.7.∎

We now use the preceding result to establish the following

######  Proposition 4.8.

Let $G=E_{7}$. There exists an element $s\in T_{reg}$, with $s$ not strongly
regular, and such that $s$ is cyclic on $V_{\omega_{7}}$.

###### Proof.

Set $V=V_{\omega_{7}}$. Let $H\leq G$ be a maximal rank subgroup of type
$A_{7}$ and for the purposes of this proof, let $\eta_{i},1\leq i\leq 7$,
denote the fundamental dominant weights of $H$ (with respect to an appropriate
choice of maximal torus $T_{H}$ and Borel subgroup). Then
$V|_{H}=V_{\eta_{2}}+V_{\eta_{6}}$. Let $s\in T_{H}$, be as in Proposition
4.7, so that $s$ is cyclic on $V$ and not strongly regular in $H$. Then $s$ is
not strongly regular in $G$ as the roots of $H$ are roots of $G$. Finally,
Lemma 2.2 shows that $s$ is regular.∎

For the proof of Theorem 1.5, it remains to treat the group $G=G_{2}$. We
investigate in detail the two modules corresponding to the fundamental
dominant weights.

######  Proposition 4.9.

Let $G=G_{2}$, $V_{i}=V_{\omega_{i}}$ for $i=1,2$.

  1. (1) Let $s\in T_{reg}$ and 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 $b\in F^{\times}$, $b^{2}\neq 1$. In particular, $s$ is almost cyclic on $V_{1}$ if and only if $b^{2}\neq-1$.

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

  3. (3) There exists $s\in T_{reg}$, not strongly regular, with $s$ cyclic on $V_{1}$.

  4. (4) Let $p=3$. Then there exists $s\in T_{reg}$, not strongly regular, with $s$ cyclic on $V_{2}$.

  5. (5) Let $p=3$. Then there exists a non-regular element $s\in T$, with $V_{a}^{s}=V_{a}^{T}$.

###### Proof.

Set $\alpha_{1}(s)=a$, $\alpha_{2}(s)=b$, for some $a,b\in F^{\times}$.

(1) Here we suppose $s$ to be regular and not cyclic on $V_{1}$. The weights
of $V_{1}$ are the short roots, and the weight 0 if $p\neq 2$, all with
multiplicity 1. Then $(\alpha_{1}+\alpha_{2})(s)=ab$,
$(2\alpha_{1}+\alpha_{2})(s)=a^{2}b$ so the eigenvalues of $s$ on $V_{1}$ are
$1$ (if $p\neq 2$), $a^{\pm 1},(ab)^{\pm 1},(a^{2}b)^{\pm 1}$. Since $s$ is
not cyclic on $V_{1}$, at least two of the eigenvalues are equal. As $s$ is
regular, $\alpha(s)\neq 1$ for every root $\alpha$, in particular,
$1\notin\\{a,ab,a^{2}b\\}$. As $W(G)$ is transitive on the short roots, we can
assume that $a\in\\{a^{-1},(ab)^{\pm 1},(a^{2}b)^{\pm 1}\\}$. Now if
$a\in\\{(ab)^{\pm 1},(a^{2}b)^{\pm 1}\\}$, then
$1\in\\{b,a^{2}b,ab,a^{3}b\\}$. But $\alpha_{2},2\alpha_{1}+\alpha_{2}$,
$\alpha_{1}+\alpha_{2},3\alpha_{1}+\alpha_{2}$ are roots, so this contradicts
$s$ being regular. So we have $a=a^{-1}$. Then $a^{2}=1$ and
$a=\alpha_{1}(s)\neq 1$, so $p\neq 2$ and $a=-1$. Then the eigenvalues of $s$
on $V_{1}$ are $1,-1,-1,-b,-b^{-1},b,b^{-1}$, where $b\neq\pm 1$ as $s$ is
regular, establishing the first claim. Finally, $s$ is almost cyclic on
$V_{1}$ if and only if $b^{2}\neq-1$.

(2) Here we suppose $s$ to be regular and $p=3$. The weights of $V_{2}$ are
the long roots and the weight 0, so the eigenvalues of $s$ on $V_{2}$ are
$1,b^{\pm 1},(a^{3}b)^{\pm 1},(a^{3}b^{2})^{\pm 1}$. Note that $s$ regular
implies that $b\neq 1$. As above, if $s$ is not cyclic on $V_{2}$, we can
assume that
$\alpha_{2}(s)\in\\{-\alpha_{2}(s),\pm(3\alpha_{1}+\alpha_{2})(s),\pm(3\alpha_{1}+2\alpha_{2})(s)\\}$.
As $s$ is regular, $\alpha_{2}(s)\neq(-3\alpha_{1}-\alpha_{2})(s)$ and
$\alpha_{2}(s)\neq(3\alpha_{1}+2\alpha_{2})(s)$. If
$\alpha_{2}(s)=(3\alpha_{1}+\alpha_{2})(s)$ then $(3\alpha_{1})(s)=1$, so
$a^{3}=1$, and if $\alpha_{2}(s)=(-3\alpha_{1}-2\alpha_{2})(s)$, then
$(ab)^{3}=1$. But since $p=3$, this implies that $a=1$, respectively $ab=1$,
contradicting that $s$ is regular. So these two cases are ruled out, and we
are left with the case where $\alpha_{2}(s)=-\alpha_{2}(s)$. So $b^{2}=1$, and
hence $b=-1$, and $a^{2}\neq 1$ since $s$ is regular.

We now have that the eigenvalues of $s$ on $V_{2}$ are
$\\{1,-1,-1,a^{3},-a^{3},a^{-3},-a^{-3}\\}$. Suppose for a contradiction that
$s$ is not almost cyclic on $V_{2}$; then either $a^{3}=-1$ or $a^{6}=-1$. In
the first case, $a=-1$, and hence $ab=1$ contradicting $s$ regular. Thus
$a^{6}=-1$, equivalently, $a^{2}=-1$. But then,
$(2\alpha_{1}+\alpha_{2})(s)=a^{2}b=1$, again contradicting that $s$ is
regular. Hence $s$ is almost cyclic on $V_{2}$ as claimed. Note that
$a^{2}\neq-1$, else $(2\alpha_{1}+\alpha_{2})(s)=1$. Now the eigenvalues of
$s$ on $V_{1}$ are $\\{1,a,a^{-1},-a,-a^{-1},-a^{2},-a^{-2}\\}$, and hence $s$
is cyclic on $V_{1}$.

(3) Let $a=b$, so that the eigenvalues of $s$ on $V_{1}$ are $\\{a^{\pm
3},a^{\pm 2},a^{\pm 1}$, $1\\}$, where 1 is to be dropped if $p=2$; they are
distinct if $a^{k}\neq 1$ for $k\leq 6$; for instance choose $a$ to be a
primitive $7$-th root of unity. Then $s$ is cyclic on $V_{1}$ (hence regular)
and is not strongly regular as $\alpha_{1}(s)=\alpha_{2}(s)=a$.

(4) Take again $a=b$, so that the eigenvalues of $s$ on $V_{2}$ are $a^{\pm
1},a^{\pm 4},a^{\pm 5},1$. Then choosing $a$ appropriately, we see that $s$ is
cyclic on $V_{2}$, regular and not strongly regular, as in (3).

The final statement (5) is straightforward, as we may take $a=1$ so that $s$
is not regular, while the non-zero weights of $V_{a}$ are
$\pm(3\alpha_{1}+2\alpha_{2})$, $\pm(3\alpha_{1}+\alpha_{2})$ and
$\pm\alpha_{2}$; so that choosing $b$ such that $b^{2}\neq 1$, we have
$V_{a}^{s}=V_{a}^{T}$.∎

We have now completed the proof of Theorem 1.5 for the exceptional groups.
Finally, we conclude the article by proving Theorem 1.7:

###### Proof of Theorem 1.7.

By Lemma 2.4, $\phi_{i}(s)$ is cyclic for every $i$. Therefore, by Theorem
1.2, $s$ is strongly regular unless, possibly, for every $i$,
$(G,\lambda_{i})$ is as in cases $(2)-(10)$ there. Also, Theorem 1.6 implies
that all non-zero weights of $V_{\omega}$ have multiplicity one, and we then
apply [11, Theorem 2(2)] and [15, Proposition 2] to justify the conditions
given in (i) - (iii) of the statement.∎

##  5\. Tables

Tables 2 and 3 are taken from [11, Tables 1,2]. We recall here our convention
for reading the tables in case ${\rm char}(F)=0$: for a natural number $a$ the
expressions $p>a$, $p\geq a$ or $p\neq a$ are to be interpreted as the absence
of any restriction, that is, $a$ is allowed to be any natural number. Note
further that when a weight $\lambda$ has coefficients expressed in terms of
$p$, we are assuming that ${\rm char}(F)=p>0$.

$\begin{array}[]{|l|c|}\hline\cr\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ G&\\\ \hline\cr\hline\cr
A_{1}&a\omega_{1},1\leq a<p\\\ \hline\cr A_{n},n>1&a\omega_{1},b\omega_{n},\
1\leq a,b<p\\\ &\omega_{i},\ 1<i<n\\\ &c\omega_{i}+(p-1-c)\omega_{i+1},\ 1\leq
i<n,\ 0\leq c<p\\\ \hline\cr B_{n},n>2&\omega_{1},\ \omega_{n}\\\ \hline\cr
C_{n},n>1,p=2&\omega_{1},\ \omega_{n}\\\ \hline\cr C_{2},p>2&\omega_{1},\
\omega_{2},\ \omega_{1}+\frac{p-3}{2}\omega_{2},\ \frac{p-1}{2}\omega_{2}\\\
\hline\cr C_{3}&\omega_{3}\\\ \hline\cr C_{n},n>2,p>2&\omega_{1},\
\omega_{n-1}+\frac{p-3}{2}\omega_{n},\ \frac{p-1}{2}\omega_{n}\\\ \hline\cr
D_{n},n>3&\omega_{1},\ \omega_{n-1},\ \omega_{n}\\\ \hline\cr
E_{6}&\omega_{1},\ \omega_{6}\\\ \hline\cr E_{7}&\omega_{7}\\\ \hline\cr
F_{4},p=3&\omega_{4}\\\ \hline\cr G_{2},p\neq 3&\omega_{1}\\\ \hline\cr
G_{2},p=3&\omega_{1},\ \omega_{2}\\\ \hline\cr\end{array}$ Table 2. Non-
trivial irreducible $p$-restricted $G$-modules with all weights of
multiplicity 1

$\begin{array}[]{|l|l|c|c|}\hline\cr G&\mbox{conditions}&&\mbox{weight
}0\mbox{ {multiplicity }}\\\ \hline\cr\hline\cr
A_{n},&n>1,(n,p)\neq(2,3)&\omega_{1}+\omega_{n}&\begin{cases}n-1&\mbox{ if
}p|(n+1)\\\ n&\mbox{ if }p\\!\not|(n+1)\end{cases}\\\ &&&\\\
A_{3}&p>3&2\omega_{2}&2\\\ \hline\cr B_{n}&n>2,p\neq 2&\omega_{2}&n\\\
&n>2,p=2&\omega_{2}&n-{\rm gcd}(2,n)\\\ &&&\\\
&-&2\omega_{1}&\begin{cases}n&\mbox{ if }p|(2n+1)\\\ n+1&\mbox{ if
}p\\!\not|(2n+1)\end{cases}\\\ \hline\cr C_{n}&n>1&2\omega_{1}&n\\\
&n>2,(n,p)\neq(3,3)&\omega_{2}&\begin{cases}n-2&\mbox{ if }p|n\\\ n-1&\mbox{
if }p\\!\not|n\end{cases}\\\ C_{2}&p\neq 5&2\omega_{2}&2\\\ C_{4}&p\neq
2,3&\omega_{4}&2\\\ \hline\cr D_{n}&n>3&2\omega_{1}&\begin{cases}n-2&\mbox{ if
}p|n\\\ n-1&\mbox{ if }p\\!\not|\,n\end{cases}\\\ &n>3,p\neq 2&\omega_{2}&n\\\
D_{n}&n>3,p=2&\omega_{2}&n-\gcd(2,n)\\\ \hline\cr
E_{6}&&\omega_{2}&\begin{cases}5&\mbox{ if }p=3\\\ 6&\mbox{ if }p\neq
3\end{cases}\\\ \hline\cr E_{7}&&\omega_{1}&\begin{cases}6&\mbox{ if }p=2\\\
7&\mbox{ if }p\neq 2\end{cases}\\\ \hline\cr E_{8}&&\omega_{8}&8\\\ \hline\cr
F_{4}&&\omega_{1}&\begin{cases}2&\mbox{ if }p=2\\\ 4&\mbox{ if }p\neq
2\end{cases}\\\ &p\neq 3&\omega_{4}&2\\\ \hline\cr G_{2}&p\neq
3&\omega_{2}&2\\\ \hline\cr\end{array}$ Table 3. Irreducible $p$-restricted
$G$-modules with non-zero weights of multiplicity 1 and whose zero weight has
multiplicity greater than $1$.
$\begin{array}[]{|l|l|l|l|l|l|l|}\hline\cr&&&&&&\\\
A_{1}&A_{n}&B_{n}&C_{2}&C_{n}&C_{3}&C_{n}\\\ p\neq 2&n\geq 2&n\geq 3&&n\geq
3,p=2&p\neq 2,3&n\geq 3,p=3\\\ &&&&&&\\\
\hline\cr\omega_{1}&\omega_{1},\ldots,\omega_{n}&\omega_{1},\omega_{n}&\omega_{1},\omega_{n}&\omega_{1},\omega_{n}&\omega_{1},\omega_{3}&\omega_{1},\omega_{n-1},\omega_{n}\\\
\hline\cr\hline\cr&&&&&&\\\ D_{n},n\geq 4&E_{6}&E_{7}&F_{4},p=3&G_{2},p\neq
3&G_{2},p=3&\\\ &&&&&&\\\
\hline\cr\omega_{1},\omega_{n-1},\omega_{n}&\omega_{1},\omega_{6}&\omega_{7}&\omega_{4}&\omega_{1}&\omega_{1},\omega_{2}&\\\
\hline\cr\end{array}$ Table 4. Highest weights of the non-trivial
$p$-restricted modules in cases $(2)-(10)$ of Theorem 1.2

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