# Multiplicity-free representations of the principal $A_{1}$-subgroup in a
simple algebraic group

Aluna Rizzoli  EPFL
<EMAIL_ADDRESS>Donna M. Testerman Both authors acknowledge the support
of the Swiss National Science Foundation grant number $200020\textunderscore
207730$. In addition, the authors thank Gunter Malle and Mikko Korhonen for
their helpful remarks on an earlier version of the mansucript. EPFL
<EMAIL_ADDRESS>
###### Abstract

Let $G$ be a simple algebraic group defined over an algebraically closed field
$\mathcal{k}$ of characteristic $p>0$. Here we classify all irreducible
$\mathcal{k}G$-modules for which the principal $A_{1}$ has no repeated
composition factors, extending the work of Liebeck-Seitz-Testerman which
treated the same question when $\mathcal{k}$ is replaced by an algebraically
closed field of characteristic zero.

##  1 Introduction

In this paper, we consider a question in the representation theory and
subgroup structure of simple algebraic groups defined over an algebraically
closed field $\mathcal{k}$ of characteristic $p>0$. The main aim of our work
is to generalise the results of [12], [14], and [13], where the authors
consider so-called “multiplicity-free subgroups” of simple algebraic groups
defined over an algebraically closed field $K$ of characteristic 0. More
precisely, the authors consider triples $(X,Y,V)$ where $X$ and $Y$ are simple
algebraic groups defined over $K$ with $X$ a closed subgroup of $Y$, and $V$
is an irreducible $KY$-module such that $V\downarrow X$ is a sum of non-
isomorphic irreducible $KX$-modules (a so-called “multiplicity-free”
$KX$-module). The above cited articles provide a complete classification of
such triples when either $X$ has rank 1 and does not lie in a proper parabolic
subgroup of $Y$, or $Y$ is a classical group with natural module $W$ and $X$
is of type $A_{\ell}$ acting irreducibly on $W$. Note that the case where $X$
acts irreducibly (and hence multiplicity freely) on $V$ was settled in work of
Dynkin [6], in characteristic zero, and in work of Seitz and Testerman in
positive characteristic [18, 21].

The ultimate far-reaching aim of what we undertake in this paper would be to
investigate the “multiplicity-free” triples $(X,Y,V)$ as in [12], [14], and
[13], described above, replacing the field $K$ by the field of positive
characteristic $\mathcal{k}$, and considering composition factors rather than
summands. The proofs in [14, 13] use induction on the rank of the group $X$;
the case where $X$ is simple of rank 1 is considered in [12]. Here we treat
the rank one case for the groups defined over $\mathcal{k}$, but consider a
slightly more general setting than would strictly speaking be required for use
in an inductive set-up. Namely, we consider all simple algebraic groups $G$
(classical and exceptional), defined over $\mathcal{k}$, and $A$ a closed
$A_{1}$-type subgroup of $G$ containing a regular unipotent element of $G$,
which we will call a “principal $A_{1}$-subgroup” of $G$. (There is at most
one conjugacy class of such subgroups.) We then determine all irreducible
$\mathcal{k}G$-modules $V$ such that $V\downarrow A$ has non-isomorphic
composition factors, and obtain a classification analogous to [12, Thm. 1].
Much of the analysis follows the same line of reasoning as that used in [12];
the main differences and difficulties arise from the lack of precise knowledge
about the dimensions of irreducible $\mathcal{k}G$-modules and the
multiplicities of their weights. In addition, while irreducible
$\mathcal{k}A_{1}$-modules are completely understood, the description of the
set of weights is not as simple as in characteristic zero. In [14, 13],
another essential ingredient of the proof is the work of Stembridge [20],
where he determines when the tensor product of two irreducible modules for a
simple algebraic group defined over the field $K$ has no repeated irreducible
summands. There has been recent progress on the analogous question for the
simple groups defined over fields of positive characteristic in [7]. The
combination of the rank-$1$ theorem proven here and Gruber’s work lays the
foundation for the study of multiplicity-free subgroups of higher rank for the
groups defined over fields of positive characteristic.

In order to state our main result, we introduce some notation; further
notation will be set up in §2. Fix $G$ a simply connected simple algebraic
group of rank $\ell\geq 2$ defined over the algebraically closed field
$\mathcal{k}$. We fix a maximal torus $T$ of $G$, a Borel subgroup $B$ of $G$
with $T\subset B$, the root system $\Phi$ of $G$ with respect to $T$, and a
base $\Pi=\\{\alpha_{1},\dots,\alpha_{\ell}\\}$ of $\Phi$, associated with the
choice of Borel subgroup $B$. Let $X(T)$ denote the associated weight lattice,
with fundamental dominant weights $\\{\omega_{1},\dots,\omega_{\ell}\\}$
defined by the choice of $\Pi$, and fix $\lambda\in X(T)$, a dominant weight.
(We label Dynkin diagrams as in [2].) Throughout we set $V=L(\lambda)$, the
irreducible $\mathcal{k}G$-module with highest weight $\lambda$. Assume that
$p\geq h$, the Coxeter number of $G$, a condition which is necessary and
sufficient for ensuring that a regular unipotent element of $G$ has order $p$
and lies in an $A_{1}$-subgroup of $G$ (see [22]). Let $A\subset G$ be a
principal $A_{1}$-subgroup of $G$. Fix a maximal torus $T_{A}$ of $A$ with
$T_{A}\subset T$ and $T_{A}U_{\alpha}$, a Borel subgroup of $A$, with root
group $U_{\alpha}$, lying in $B$. For $\alpha$ the unique positive root of $A$
(with respect to the given choices), we have
$T_{A}=\alpha^{\vee}(\mathcal{k}^{*})$, the image of the co-root
$\alpha^{\vee}$. Throughout, we identify the weight lattice of $T_{A}$ with
the ring $\mathbb{Z}$. We say that _$V\downarrow A$ is MF_ if there are no
repeated composition factors in the restriction.

We will also require a notation for the corresponding modules and subgroups
for the groups defined over the algebraically closed field $K$ of
characteristic $0$. We write $G_{K}(\Phi)$ for a simply connected simple
algebraic group defined over the field $K$, with root system of type $\Phi$,
and $A_{K}$ for a principal $A_{1}$-subgroup of $G_{K}(\Phi)$. For the weight
$\lambda$ as above, we write $\Delta_{K}(\lambda)$ for the corresponding
irreducible $G_{K}(\Phi)$-module. We will use the same terminology of “MF” for
the action of $A_{K}$ on $\Delta_{K}(\lambda)$. Our main result is

######  Theorem 1.

Suppose that $\lambda$ is $p$-restricted. Then $L(\lambda)\downarrow A$ is
$MF$ if and only if one of the following holds.

  1. (i)

We have that $p>(\lambda\downarrow T_{A})$ and $\Delta_{K}(\lambda)\downarrow
A_{K}$ is MF, or

  2. (ii)

the group $G$ is of type $A_{2}$, $\lambda=\omega_{1}+\omega_{2}$ and $p=3$,
or

  3. (iii)

the group $G$ is of type $B_{2}$, $\lambda=2\omega_{1}$ and $p=5$.

######  Corollary 2.

Let $\lambda=\sum_{i=0}^{t}p^{i}\lambda_{i}$ where each $\lambda_{i}$ is a
$p$-restricted dominant weight. Then $L(\lambda)\downarrow A$ is MF if and
only if one of the following holds.

  1. (i)

We have $p>(\lambda_{i}\downarrow A)$ and $\Delta_{K}(\lambda_{i})\downarrow
A_{K}$ is MF for all $0\leq i\leq t$.

  2. (ii)

The group $G$ is of type $A_{2}$, $p=3$ and there exists $0\leq i\leq t$ such
that $\lambda_{i}=\omega_{1}+\omega_{2}$. Moreover, for all $0\leq j\leq t$ we
have $\lambda_{j}\in\\{0,\omega_{1}+\omega_{2},\omega_{1},\omega_{2}\\}$ and
if $\lambda_{j}=\omega_{1}+\omega_{2}$ for some $0\leq j\leq t-1$, then
$\lambda_{j+1}=0$.

  3. (iii)

The group $G$ is of type $B_{2}$, $p=5$ and there exists $0\leq i\leq t$ such
that $\lambda_{i}=2\omega_{1}$. Moreover, for all $0\leq j\leq t$ we have
$\lambda_{j}\in\\{0,2\omega_{1},\omega_{1},\omega_{2}\\}$ and if
$\lambda_{j}=2\omega_{1}$ for some $0\leq j\leq t-1$, then
$\lambda_{j+1}\in\\{0,\omega_{2}\\}$.

For the reader’s convenience and for completeness, we list in Table 1 below
the $\lambda$ for which $\Delta_{K}(\lambda)\downarrow A_{K}$ is MF, as
obtained in [12].

Throughout the paper, set $r=\lambda\downarrow T_{A}$, that is
$\lambda(\alpha^{\vee}(c))=c^{r}$, for all $c\in\mathcal{k}^{*}$. Note that
the co-cocharacter $\alpha^{\vee}:\mathbb{G}_{m}\to T$ which defines the
maximal torus of $A$ satisfies $\alpha_{i}(\alpha^{\vee}(c))=c^{2}$ for all
$c\in\mathcal{k}^{*}$; that is, $\alpha_{i}\downarrow T_{A}=2$ for all $1\leq
i\leq\ell$. The value for $r$ can then be determined by writing $\lambda$ as a
linear combination of simple roots and then using that each simple root takes
value $2$ on $T_{A}$.

We conclude the introduction with a few remarks about the proof. We first note
that if $p>r$, then one can show that the Weyl module with highest weight
$\lambda$ is an irreducible $\mathcal{k}G$-module (see [11, Cor.2.7.6]), and
then the considerations of [12] for the groups defined over $K$ yield the
result. (See Proposition 2.2) The arguments therefore focus on the cases where
$p\leq r$. Many aspects of the proof follow closely the arguments used in
[12]. In particular, we use the fact that all irreducible
$\mathcal{k}A_{1}$-modules have multiplicity one weight spaces and therefore
considering the set of $T_{A}$-weights and their multiplicities in $V$ can
directly be used to detect multiplicities of composition factors of
$V\downarrow A$. Moreover, there are certain dimension bounds which must be
respected by an MF-module. Thus, many of our preliminary lemmas are inspired
by the results in [12, Section 2]. In addition, we rely on a result from [8]
showing that the principal $A_{1}$-subgroup $A$ of $G$ satisfies the property
that certain tilting modules for $G$ have a filtration by tilting modules for
$A$. Since indecomposable reducible tilting modules for $A_{1}$ necessarily
have repeated composition factors, this result is quite useful for showing
that many $\mathcal{k}G$-modules are not MF as $\mathcal{k}A$-modules.

]l l Multiplicity-free restrictions in characteristic $0$
---
$G_{K}$ |  $\lambda$ , $\lambda\neq 0$
$A_{\ell}$ | $\omega_{1},\,\omega_{2},\,2\omega_{1},\,\omega_{1}+\omega_{\ell}$
| $\omega_{3}\,(5\leq\ell\leq 7)$
| $3\omega_{1}\,(\ell\leq 5),\,4\omega_{1}\,(\ell\leq
3),\,5\omega_{1}\,(\ell\leq 3)$
$A_{3}$ | $110$
$A_{2}$ | $c1,\,c0$
$B_{\ell}$ | $\omega_{1},\,\omega_{2},\,2\omega_{1}$
| $\omega_{\ell}\,(\ell\leq 8)$
$B_{3}$ | $101,\,002,\,300$
$B_{2}$ | $b0,\,0b\,(1\leq b\leq 5),\,11,\,12,\,21$
$C_{\ell}$ | $\omega_{1},\,\omega_{2},\,2\omega_{1}$
| $\omega_{3}\,(3\leq\ell\leq 5)$
| $\omega_{\ell}\,(\ell=4,\,5)$
$C_{3}$ | $300$
$D_{\ell}\,(\ell\geq 4)$ | $\omega_{1},\,\omega_{2}\,(\ell=2m+1),\,2\omega_{1}\,(\ell=2m)$
| $\omega_{\ell}\,(\ell\leq 9)$
$E_{6}$ | $\omega_{1},\omega_{2}$
$E_{7}$ | $\omega_{1},\omega_{7}$
$E_{8}$ | $\omega_{8}$
$F_{4}$ | $\omega_{1},\omega_{4}$
$G_{2}$ | $10,\,01,\,11,\,20,\,02,\,30$

##  2 Preliminary lemmas

Let us fix the notation to be used throughout the paper, in addition to that
which was introduced in the previous section.

We assume throughout that $\ell\geq 2$, respectively $2,3,4$, for $G$ of type
$A_{\ell}$, respectively $B_{\ell},C_{\ell},D_{\ell}$. For $1\leq i\leq\ell$,
let $s_{i}$ denote the simple reflection associated to the root $\alpha_{i}$.
For $\lambda\in X(T)$, a dominant weight, we write $\Delta_{G}(\lambda)$ for
the Weyl module for $G$ of highest weight $\lambda$. We will suppress the $G$
in this notation if there is no ambiguity. For a $\mathcal{k}G$-module $V$ and
$\mu\in X(T)$, we write $V_{\mu}$ for the $\mu$-weight space with respect to
$T$ of the module $V$. For an $A_{1}$-type group, where we identify the set of
dominant weights with the set of nonnegative integers, we will write $(s)$ for
the irreducible $\mathcal{k}A_{1}$-module of highest weight $s$. If we want to
underline that we are talking about the fixed principal $A_{1}$-subgroup $A$,
we may write as well $L_{A}(s)$. Similarly, we write $\Delta(s)$ for the Weyl
module of highest weight $s$ and $T(s)$ for the indecomposable tilting module
of highest weight $s$. We require as well a notation for the Frobenius twist
of a $\mathcal{k}A$-module. Namely for a $\mathcal{k}A$-module $(s)$, we write
$(s)^{(p^{i})}$ for the module whose structure is induced by the composition
of the $p^{i}$-Frobenius map on $A$ and the morphism defining the module
structure on $(s)$.

Table 1: Values of $r=\lambda\downarrow T_{A}$ for $\lambda=\sum_{1}^{\ell}{c_{i}}\omega_{i}$ $G$ | $r$
---|---
$A_{\ell}$ | $\sum_{1}^{\ell}i(\ell+1-i)c_{i}$
$B_{\ell}$ | $\sum_{1}^{\ell-1}i(2\ell+1-i)c_{i}+\frac{\ell(\ell+1)}{2}c_{\ell}$
$C_{\ell}$ | $\sum_{1}^{\ell}i(2\ell-i)c_{i}$
$D_{\ell}$ | $\sum_{1}^{\ell-2}i(2\ell-1-i)c_{i}+\frac{\ell(\ell-1)}{2}c_{\ell-1}+\frac{\ell(\ell-1)}{2}c_{\ell}$
$G_{2}$ | $6c_{1}+10c_{2}$
$F_{4}$ | $22c_{1}+42c_{2}+30c_{3}+16c_{4}$
$E_{6}$ | $16c_{1}+22c_{2}+30c_{3}+42c_{4}+30c_{5}+16c_{6}$
$E_{7}$ | $34c_{1}+49c_{2}+66c_{3}+96c_{4}+75c_{5}+52c_{6}+27c_{7}$
$E_{8}$ | $92c_{1}+136c_{2}+182c_{3}+270c_{4}+220c_{5}+168c_{6}+114c_{7}+58c_{8}$

Here and in Sections 3 and 4, we fix a $p$-restricted dominant weight
$\lambda\in X(T)$ and set $V=L_{G}(\lambda)$. Recall as well that we have
$r=\lambda\downarrow T_{A}$. We list the values of $r$ in Table
LABEL:tab:r_values. In addition, recall that the existence of a principal
$A_{1}$-subgroup implies $p\geq h$, the Coxeter number of $G$. This then
allows us to use [17], which ensures that the set of weights of $L{(\mu)}$ is
the same as the set of weights of $\Delta(\mu)$ for all $p$-restricted
dominant weights $\mu\in X(T)$. We now introduce a shorthand notation for
weights of $V$. For $\lambda-\sum_{i=1}^{\ell}a_{i}\alpha_{i}$, we write
$\lambda-i_{1}^{a_{i_{1}}}\cdots i_{m}^{a_{i_{m}}}$, where $a_{j}=0$ for
$j\not\in\\{i_{1},\dots i_{m}\\}$, and suppress those $a_{j}$ with $a_{j}=1$;
for example, the weight $\lambda-\alpha_{2}-2\alpha_{3}-\alpha_{5}$ will be
written as $\lambda-23^{2}5$. For $G$ of rank 2, we write $\lambda-ab$ for the
weight $\lambda-a\alpha_{1}-b\alpha_{2}$.

The following result is Corollary 2.7.6 from [11]; we include a sketch of the
proof for completeness.

######  Lemma 2.1.

If $p>r$, then $\Delta(\lambda)$ is irreducible.

###### Proof.

By the Jantzen sum formula [10, Part II, 8.19], it suffices to prove that for
all $\alpha\in\Phi^{+}$, we have $r\geq\langle\lambda+\delta,\alpha\rangle-1$,
where $\delta=\sum_{i=1}^{\ell}\omega_{i}$. It is easy to see that
$\langle\lambda,\alpha\rangle$ is maximal when $\alpha$ is the highest root of
the dual root system $\Phi^{\vee}$, i.e. when $\alpha$ is the highest short
root $\beta$ of $\Phi$. By [19, Prop. 5], we have
$\langle\lambda+\delta,\beta\rangle\leq
1+\sum_{\alpha\in\Phi^{+}}\langle\lambda,\alpha\rangle$. It is therefore
sufficient to show that
$r=\sum_{\alpha\in\Phi^{+}}\langle\lambda,\alpha\rangle$, which is a simple
calculation using the fact that for a simple root $\alpha_{i}$ we have
$\sum_{\alpha\in\Phi^{+}\setminus\\{\alpha_{i}\\}}\langle\alpha_{i},\alpha\rangle=\sum_{\alpha\in\Phi^{+}\setminus\\{\alpha_{i}\\}}\langle
s_{i}(\alpha_{i}),s_{i}(\alpha)\rangle=\sum_{\alpha\in\Phi^{+}\setminus\\{\alpha_{i}\\}}\langle-\alpha_{i},\alpha\rangle$
and so
$\sum_{\alpha\in\Phi^{+}\setminus\\{\alpha_{i}\\}}\langle\alpha_{i},\alpha\rangle=0$.
∎

The next proposition establishes Theorem 1 in case $p>r$.

######  Proposition 2.2.

Assume $p>r$. Then $\Delta_{K}(\lambda)\downarrow A_{K}$ is MF if and only if
$V\downarrow A$ is MF.

###### Proof.

By Lemma 2.1, the Weyl module is irreducible and therefore
$V=\Delta(\lambda)$. We have $\Delta_{K}(\lambda)\downarrow
A_{K}=\sum_{0}^{k}\Delta_{A_{K}}(r_{i})$ for some integers $r_{0}\geq
r_{1}\dots\geq r_{k}\geq 0$ with $r_{0}=r$. Since $p>r$, a comparison of
characters gives $\Delta(\lambda)\downarrow
A=\sum_{0}^{k}\Delta_{A}(r_{i})=\sum_{0}^{k}(r_{i})$, which then implies the
result. ∎

As the next result shows, in many cases when $\Delta(\lambda)$ is irreducible
and $r\geq p$, we can directly conclude that $V\downarrow A$ is not MF.

######  Lemma 2.3.

Assume that $\Delta(\lambda)$ is irreducible and if $G$ is of type $B_{\ell}$,
respectively $D_{\ell}$, and $\lambda$ does not lie in the root lattice of
$G$, assume in addition that $p>{{\ell+1}\choose{2}}$, respectively
$p>{\ell\choose 2}$. If $r\geq p$ and $r\not\equiv-1\mod p$, then $V\downarrow
A$ is not MF.

###### Proof.

Here we use [8, Theorems 4.1.2, 4.1.4 and 4.2.1] to see that $A$ is a so-
called “good filtration” subgroup, which then implies that the irreducible
Weyl module $\Delta(\lambda)=V$ has an $A$-filtration by both Weyl modules and
by induced modules. So in particular, $V\downarrow A$ is a tilting module.
Since tilting modules for $A$ are self-dual, no reducible indecomposable
tilting module is MF. The hypotheses on $r$ imply that the indecomposable
tilting module $T(r)$ is reducible (see [9, Theorem 2.1]). As $T(r)$ is a
summand of $V\downarrow A$, this concludes the proof. ∎

We now turn to a sequence of definitions and lemmas which provide tools for
studying the set of composition factors of $V\downarrow A$ based upon knowing
the set of weights of $V$.

######  Definition 2.4.

For $n\in\mathbb{Z}$, let $n_{d}$ be the multiplicity of the $T_{A}$-weight
$r-2d$ in $V\downarrow A$ and let $m_{d}$ be the multiplicity of the
composition factor $(r-2d)$ in $V\downarrow A$. Also, let $\mathbf{S}_{d}$
denote the multiset of composition factors whose highest weight is higher than
$r-2d$ and in which $r-2d$ does not occur as a weight, and let $s_{d}$ denote
the cardinality of $\mathbf{S}_{d}$.

######  Lemma 2.5.

Assume that $V\downarrow A$ is MF. Then $n_{d}\leq d+1$.

###### Proof.

Let $\mathbf{B}$ be the multiset of composition factors of $V\downarrow A$
where $r-2d$ occurs as a weight. Since $V\downarrow A$ is MF, we have
$\mathbf{B}\subseteq\\{(r),(r-2),\dots,(r-2d)\\}$. Therefore
$\lvert\mathbf{B}\rvert\leq d+1$ and we can conclude that $n_{d}\leq d+1$. ∎

######  Lemma 2.6.

For all $0\leq d\leq r$ we have

$m_{d}=n_{d}-n_{d-1}+s_{d}-s_{d-1}.$ (1)

###### Proof.

We prove this by induction on $d$. If $d=0$ the statement holds. Indeed,
$m_{0}=n_{0}=1$ since $r$ is the highest weight and is afforded only by
$\lambda$, and $n_{-1}=s_{-1}=s_{0}=0$. Assume that (1) holds up to an
arbitrary $d$. In general, the multiplicity $m_{d+1}$ can be determined by
taking the difference between $n_{d+1}$ and the number of times the
$T_{A}$-weight $r-2(d+1)$ appears in composition factors with higher highest
weight. Thus,

$m_{d+1}=n_{d+1}-\left(\sum_{0\leq k\leq d}m_{k}-s_{d+1}\right).$

By the inductive hypothesis $m_{k}=n_{k}-n_{k-1}+s_{k}-s_{k-1}$ for all $k\leq
d$. Substituting we get

$m_{d+1}=n_{d+1}+s_{d+1}-\sum_{0\leq k\leq
d}\left(n_{k}-n_{k-1}+s_{k}-s_{k-1}\right)=n_{d+1}-n_{d}+s_{d+1}-s_{d},$

concluding the proof. ∎

######  Lemma 2.7.

For all $1\leq d<p$ we have $\mathbf{S}_{d-1}\subseteq\mathbf{S}_{d}$. In
particular $s_{d}\geq s_{d-1}$.

###### Proof.

We begin by analysing what weights occur in an arbitrary composition factor
$(t)$. We will write a tuple $(a_{0},a_{1},\dots,a_{m})$ to denote the integer
$\sum_{i=0}^{m}a_{i}p^{m}$. Then $t=(a_{0},a_{1},\dots,a_{m})$ where the
$a_{i}$’s are the coefficients in the $p$-adic expansion of $t$. By
Steinberg’s tensor product theorem, we have

$(t)\simeq(a_{0})\otimes(a_{1})^{(p)}\otimes\dots\otimes(a_{m})^{(p^{m})}.$

The weights occurring in $(t)$ are therefore of the form
$(a_{0}-2i_{0},a_{1}-2i_{1},\dots,a_{m}-2i_{m})$ where $0\leq i_{j}\leq
a_{j}$. Let $t-2q\geq 0$, with $q\in\mathbb{N}$, be an integer denoting a
weight not occurring in $(t)$. Then $t-2q$ lies in an open interval of the
form

$\left((a_{0},\dots,a_{j},a_{j+1}-2i_{j+1}-2,\dots,a_{m}-2i_{m}),(-a_{0},\dots,-a_{j},a_{j+1}-2i_{j+1},\dots,a_{m}-2i_{m})\right)$
(2)

where $0\leq i_{j+1}<a_{j+1}$ and $0\leq i_{k}\leq a_{k}$ for $k>j+1$.
Conversely, any integer $t-2q$ lying in such an interval corresponds to a
weight not occurring in $(t)$. We call these intervals the gaps of $(t)$, so
that a composition factor $(t)$ is in $\mathbf{S}_{d}$ if and only if $r-2d$
is in a gap of $(t)$.

Assume for a contradiction that
$(t)\in\mathbf{S}_{d-1}\setminus\mathbf{S}_{d}$ for some $t\leq r$. Then
$t>r-2d+2$ and the composition factor $(t)$ has a gap of the form (2)
containing $r-2d+2$, but not containing $r-2d$. This means that
$r-2d=(a_{0},\dots,a_{j},a_{j+1}-2i_{j+1}-2,\dots,a_{m}-2i_{m})$, implying
that

$2d-(r-t)=t-(r-2d)=t-(a_{0},\dots,a_{j},a_{j+1}-2i_{j+1}-2,\dots,a_{m}-2i_{m})$
$=2p^{j+1}(i_{j+1}+1,i_{j+2},\dots,i_{m})\geq 2p.$

This contradicts the assumption that $d<p$. ∎

######  Lemma 2.8.

Let $1\leq d<{\rm min}\\{\lfloor\frac{r}{2}\rfloor,p\\}$. Then the following
hold:

  1. (i)

If $n_{d}-n_{d-1}=1$ then $r-2d$ is a composition factor of $V\downarrow A$.

  2. (ii)

If $n_{d}-n_{d-1}\geq 2$ then $m_{d}\geq 2$ and $V\downarrow A$ is not MF.

  3. (iii)

If $\lambda=c\omega_{i}$ and $n_{d}\geq d+1$ then $V\downarrow A$ is not MF.

  4. (iv)

If $n_{d}-n_{d-1}=1$ and $\mathbf{S}_{d-1}\neq\mathbf{S}_{d}$, then $m_{d}\geq
2$ and $V\downarrow A$ is not MF.

###### Proof.

Parts (i),(ii) and (iv) follow directly from combining Lemma 2.6 and Lemma
2.7. If $\lambda=c\omega_{i}$ then $n_{1}=1$, and since $n_{d}\geq d+1$, there
exists $2\leq d^{\prime}\leq d$ such that $n_{d^{\prime}}-n_{d^{\prime}-1}\geq
2$, concluding by part (ii). ∎

We can often deduce the value $n_{d}$ from the characteristic zero case.

######  Lemma 2.9.

Assume that $V=\Delta(\lambda)$. Then
$n_{d}=\dim(\Delta_{K}(\lambda)\downarrow A_{K})_{r-2d}$.

###### Proof.

This follows from [10, Part II, 5.8], since $T_{A}$ is uniquely determined by
the property $\alpha_{i}\downarrow\nolinebreak T_{A}=\nolinebreak 2$ for all
$1\leq i\leq\ell$.∎

We now establish two dimension bounds for multiplicity-free
$\mathcal{k}A_{1}$-modules.

######  Definition 2.10.

Given $r\in\mathbb{N}$, define $B(r)$ and $B_{K}(r)$ as

$B(r)=\sum_{r-2k\geq 0}\dim L_{A}(r-2k),$ $B_{K}(r)=\sum_{r-2k\geq
0}\dim\Delta_{A_{K}}(r-2k).$

In particular note that $B_{K}(r)$ is either $(\frac{r}{2}+1)^{2}$ or
$\frac{r+1}{2}\frac{r+3}{2}$ according as $r$ is even or odd, respectively.

######  Lemma 2.11.

Suppose that $V\downarrow A$ is MF. Then $\dim V\leq B(r)\leq B_{K}(r)$.

###### Proof.

Since $V\downarrow A$ is MF, it can have at most one composition factor
$(r-2d)$, i.e. $m_{d}=1$, for every $0\leq d\leq\lfloor\frac{r}{2}\rfloor$.
Therefore $\dim V\leq B(r)$. The fact that $B(r)\leq B_{K}(r)$ is immediate
since $\dim L_{A}(r-2k)\leq\dim\Delta_{A_{K}}(r-2k)$ for all $k$ such that
$r-2k\geq 0$. ∎

######  Lemma 2.12.

Suppose that $\lambda=a\omega_{i}$ and $r\not\equiv 0\mod p$. If $V\downarrow
A$ is MF, then $\dim V\leq B(r)-\dim(r-2)$.

###### Proof.

The $T_{A}$-weight $r-2$ occurs with multiplicity $1$ in $V$, and since
$r\not\equiv 0\mod p$, it occurs as a weight in the composition factor $(r)$.
Therefore $r-2$ does not afford a composition factor of $V\downarrow A$, i.e.
$m_{1}=0$. Since $V\downarrow A$ is MF, we have $m_{d}\leq 1$ for all $d\geq
0$ such that $r-2d\geq 0$. This proves that $\dim V\leq B(r)-\dim(r-2)$. ∎

The following result is our main reduction tool for showing that if
$V\downarrow A$ is MF, then $\lambda$ satisfies some highly restrictive
conditions. The proof follows closely that of [14, Lemma 2.6].

######  Proposition 2.13.

Let $\lambda=\sum_{i=1}^{\ell}c_{i}\omega_{i}$. Assume that there exist $i<j$
with $c_{i}\neq 0\neq c_{j}$ and that $V\downarrow A$ is MF. Then

  1. (i)

$c_{k}=0$ for $k\neq i,j$.

  2. (ii)

If $\alpha_{i}$ and $\alpha_{j}$ are non-adjacent, then $c_{i}=c_{j}=1$.

  3. (iii)

If $\alpha_{i}$ and $\alpha_{j}$ are non-adjacent then they are both end-
nodes.

  4. (iv)

Either $\alpha_{i}$ or $\alpha_{j}$ is an end-node.

  5. (v)

If both $c_{i}>1$ and $c_{j}>1$, then $G$ has rank $2$ and $\lambda-ij$ has
multiplicity $1$.

  6. (vi)

If either $c_{i}>1$ or $c_{j}>1$, then either $G$ has rank $2$, or
$\alpha_{i}$ is adjacent to $\alpha_{j}$ and $\lambda-ij$ has multiplicity
$1$.

###### Proof.

Throughout the proof, we will use the main result of [17] which implies that
for $p\geq h$, we have that all weights occurring in $\Delta_{K}(\lambda)$
occur in $V$.

  1. (i)

If $c_{k}\geq 1$ for $k\neq i,j$, we have $n_{1}\geq 3$ as the $T_{A}$-weight
$r-2$ is afforded by $\lambda-i$, $\lambda-j$ and $\lambda-k$. This
contradicts Lemma 2.5.

  2. (ii)

Suppose $\alpha_{i}$ and $\alpha_{j}$ are not adjacent and that $c_{i}\geq 2$.
Let $k\neq i,j$ such that $\alpha_{k}$ is adjacent to $\alpha_{i}$ and let
$k^{\prime}\neq i,j$ such that $\alpha_{k^{\prime}}$ is adjacent to
$\alpha_{j}$. Then $n_{2}\geq 4$, as the $T_{A}$-weight $r-4$ is afforded by
$\lambda-i^{2}$, $\lambda-ik$, $\lambda-jk^{\prime}$, $\lambda-ij$. This
contradicts Lemma 2.5.

  3. (iii)

Assume that $\alpha_{i}$ and $\alpha_{j}$ are non-adjacent and that
$\alpha_{i}$ is not an end-node. Then there exist distinct simple roots
$\alpha_{k},\alpha_{l}$, both adjacent to $\alpha_{i}$, and a simple root
$\alpha_{m}\neq\alpha_{i}$ adjacent to $\alpha_{j}$. Then $n_{2}\geq 4$, as
the $T_{A}$-weight $r-4$ is afforded by $\lambda-ik$, $\lambda-il$, $\lambda-
jm$ and $\lambda-ij$. This contradicts Lemma 2.5.

  4. (iv)

Assume that neither $\alpha_{i}$ nor $\alpha_{j}$ is an end-node. Then by
$(\Rn{3})$ the roots $\alpha_{i}$ and $\alpha_{j}$ are adjacent. Let $1\leq
k,l\leq\ell$ be distinct indices such that $\\{i,j\\}\cap\\{k,l\\}=\emptyset$
and such that $\alpha_{i}$ is adjacent to $\alpha_{k}$ and $\alpha_{j}$ is
adjacent to $\alpha_{l}$. Then the $T_{A}$-weight $r-8$ is afforded by
$\lambda-kijl,\lambda-kij^{2},\lambda-i^{2}j^{2},\lambda-i^{2}jl,\lambda-
ki^{2}j$ and $\lambda-ij^{2}l$. Therefore $n_{4}\geq 6$, contradicting Lemma
2.5.

  5. (v)

If both $c_{i}>1$ and $c_{j}>1$, then by (ii), the roots $\alpha_{i}$ and
$\alpha_{j}$ are adjacent. If the rank of $G$ is not $2$ we can find $k\neq
i,j$, such that $\alpha_{k}$ is adjacent to either $\alpha_{i}$ or
$\alpha_{j}$. But then the $T_{A}$-weight $r-4$ is afforded by $\lambda-ij$,
$\lambda-i^{2}$, $\lambda-j^{2}$ and either $\lambda-ik$ or $\lambda-jk$.
Therefore $n_{2}\geq 4$, contradicting Lemma 2.5. In addition, $\lambda-ij$
has multiplicity 1, else $n_{2}\geq 4$, again contradicting Lemma 2.5.

  6. (vi)

Assume $c_{i}\geq 2$ and that $G$ has rank at least $3$. Then $\alpha_{i}$ and
$\alpha_{j}$ are adjacent by $(\Rn{2})$, and $r-4$ is afforded by
$\lambda-i^{2}$, $\lambda-ij$ and either $\lambda-ik$ or $\lambda-jk$ for some
$k\neq i,j$. Therefore, Lemma 2.5 implies that $\lambda-ij$ has multiplicity
1, as claimed.

∎

######  Lemma 2.14.

Assume that $\lambda=\omega_{i}$ and that there exist
$\\{\beta_{i-3},\dots,\beta_{i+3}\\}\subseteq\Pi$ such that for $i-3\leq
s<t\leq i+3$, $(\beta_{s},\beta_{t})\neq 0$ if and only if $t=s+1$. Then
$V\downarrow A$ is not MF.

###### Proof.

Here $p>5$. It is a simple check to see that $n_{4}\geq 5$, concluding by
Lemma 2.8(3). ∎

######  Lemma 2.15.

Assume that $\lambda=b\omega_{i}$ with $b\geq 2$. If $V\downarrow A$ is MF,
then $\alpha_{i}$ is an end-node.

###### Proof.

If $\alpha_{i}$ is not an end-node, it is easy to see that $n_{2}\geq 3$. As
$p>2$, Lemma 2.8(iii) implies that $V\downarrow A$ is not MF. ∎

The following lemma provides a classification for the second possibility of
Proposition 2.13(vi).

######  Lemma 2.16.

[21, 1.35] Assume that $\lambda=c_{i}\omega_{i}+c_{j}\omega_{j}$ with
$\alpha_{i}$ and $\alpha_{j}$ adjacent and $c_{i}c_{j}\neq 0$. Let $d=\dim
V_{\lambda-ij}$. Then $1\leq d\leq 2$ and the following hold.

  1. (i)

If $(\alpha_{i},\alpha_{i})=(\alpha_{j},\alpha_{j})$, then $d=1$ if and only
if $c_{i}+c_{j}=p-1$;

  2. (ii)

If $(\alpha_{i},\alpha_{i})=2(\alpha_{j},\alpha_{j})$, then $d=1$ if and only
if $2c_{i}+c_{j}+2\equiv 0\mod p$;

  3. (iii)

If $(\alpha_{i},\alpha_{i})=3(\alpha_{j},\alpha_{j})$, then $d=1$ if and only
if $3c_{i}+c_{j}+3\equiv 0\mod p$.

Finally, we conclude this section with two further results on the dimensions
of certain weight spaces in $V$.

######  Lemma 2.17.

[18, 8.6] Let $G=A_{\ell}$. Suppose that
$\lambda=c_{i}\omega_{i}+c_{j}\omega_{j}$ and $1\leq s\leq i<j\leq t\leq k$,
with $c_{i}c_{j}\neq 0$. Let $d=\dim V_{\lambda-s(s+1)\dots(t-1)t}$. Then

  1. (i)

If $a+b+j-i\not\equiv 0\mod p$, then $d=j-i+1$;

  2. (ii)

If $a+b+j-i\equiv 0\mod p$, then $d=j-i$.

######  Lemma 2.18.

[3, Lemma 2.2.8] Let $\lambda=\sum_{i=1}^{\ell}d_{i}\omega_{i}$ and let
$\mu=\lambda-\sum_{\alpha\in S}c_{\alpha}\alpha\in X(T)$ for some subset
$S\subseteq\Pi(G)$. Set $X=\left\langle U_{\pm\alpha}\mid\alpha\in
S\right\rangle$, $\lambda^{\prime}=\lambda\downarrow(T\cap X)$ and
$\mu^{\prime}=\mu\downarrow(T\cap X)$. Then $\dim
V_{\mu}=V^{\prime}_{\mu^{\prime}}$, where
$V^{\prime}=L_{X}(\lambda^{\prime})$.

##  3 The case where $G$ has rank $2$

In this section we establish Theorem 1 in the case where $G$ has rank $2$. Let
us recall our setup. Throughout the section we assume that $\lambda$ is a
$p$-restricted dominant weight, and we let $r=\lambda\downarrow T_{A}$ and
$V=L(\lambda)$. We write $\lambda=ab$ as shorthand notation for
$\lambda=a\omega_{1}+b\omega_{2}$, and $\lambda-ab$ for
$\lambda-a\alpha_{1}-b\alpha_{2}$. Furthermore we assume that $p\leq r$, as
Proposition 2.2 settles the case $r<p$.

###  3.1 The case where $G$ is $A_{2}$

We begin with $G=A_{2}$, where $p\geq h=3$. The following is the main result,
which we will prove after a sequence of lemmas.

######  Proposition 3.1.

Let $G=A_{2}$ and assume $p\leq r$. Then $V\downarrow A$ is MF if and only if
$\lambda=\omega_{1}+\omega_{2}$ and $p=3$.

######  Lemma 3.2.

Let $G=A_{2}$ and $\lambda=ab$. Then $\lambda-ij$, with $i+j\leq a+b$, is a
weight of $\Delta(\lambda)$ if and only if one of the following holds:

  1. (i)

$i\leq j$ and $j-i\leq b$;

  2. (ii)

$i\geq j$ and $i-j\leq a$.

###### Proof.

By [1, VIII, §7, Prop. 10], the weights in $\Delta(\lambda)$ are precisely the
same as those occurring in $\Delta(a0)\otimes\Delta(0b)$. Let $\lambda_{1}=a0$
and $\lambda_{2}=0b$ and recall that $\Delta(c_{i}\omega_{i})$, for $i=1,2$ is
the $c_{i}$-th symmetric power of the natural, respectively, dual module for
$G$. Hence, $\lambda_{1}-i_{1}j_{1}$ is a weight of $\Delta(\lambda_{1})$ if
and only if $i_{1}+j_{1}\leq 2a$ and $0\leq i_{1}-j_{1}\leq a$. Similarly,
$\lambda_{2}-i_{2}j_{2}$ is a weight of $\Delta(\lambda_{2})$ if and only if
$i_{2}+j_{2}\leq 2b$ and $0\leq j_{2}-i_{2}\leq b$. By symmetry it suffices to
show that the statement of the lemma is valid when $i\geq j$. First of all, it
is clear that all weights $\lambda-ij$ of $\Delta(\lambda)$ satisfy $i-j\leq
a$, since a weight $\lambda_{2}-i_{2}j_{2}$ of $L(0b)$ satisfies
$i_{2}-j_{2}\leq 0$, and a weight $\lambda_{1}-i_{1}j_{1}$ of $L(a0)$
satisfies $i_{1}-j_{1}\leq a$.

For the converse, consider a pair $(i,j)$, such that $i\geq j$, $i+j=d\leq
a+b$ and $i-j\leq a$. If $d\leq a$, then $j\leq i\leq a$, so $\lambda_{1}-ij$
is a weight of $\Delta(\lambda_{1})$ and $\lambda_{1}+\lambda_{2}-ij$ is then
a weight of $\Delta(\lambda)$. If $d>a$, write $d=a+k$, where $k\leq b$. To
conclude we show that we can find $(i_{1},j_{1})$ with $i_{1}+j_{1}=a$ and
$(i_{2},j_{2})$ with $i_{2}+j_{2}=k$, such that $\lambda_{1}-i_{1}j_{1}$ is a
weight of $\Delta(\lambda_{1})$, $\lambda_{2}-i_{2}j_{2}$ is a weight of
$\Delta(\lambda_{2})$ and $i_{1}-j_{1}+i_{2}-j_{2}=i-j$. Fix $i_{1}+j_{1}=a$
and $i_{2}+j_{2}=k$. Note that we are allowed to pick $i_{1}-j_{1}$ to be any
integer between $a$ and $1$ if $a$ is odd, and between $a$ and $0$ if $a$ is
even. Similarly, we are allowed to choose $i_{2}-j_{2}$ between $-k$ and $-1$
if $k$ is odd, and between $-k$ and $0$ when $k$ is even, concluding easily. ∎

######  Lemma 3.3.

Let $G=A_{2}$, $\lambda=ab$ with $a\geq b>0$ and $a+b=p-1$. Then

  1. (i)

for $0\leq d\leq b$, we have $n_{d}=d+1$;

  2. (ii)

for $b+1\leq d\leq a$, we have that $n_{d}$ increases alternatingly by
respectively $0$ and $1$ with respect to $n_{d-1}$;

  3. (iii)

for $a<d\leq a+b$, we have that $n_{d}$ alternates between
$\left\lceil\frac{a+b}{2}\right\rceil$ and
$\left\lceil\frac{a+b+1}{2}\right\rceil$.

###### Proof.

Here we use the fact that all $T$-weights in $V$ are of multiplicity 1. (See
[23]. ) Hence, the proof consists of counting the pairs $(i,j)$ with $i+j=d$
and satisfying the conditions of Lemma 3.2.

  1. (i)

Let $0\leq d\leq b$. The statement then follows immediately from noting that
$\lambda-i(d-i)$ is a weight for $0\leq i\leq d$.

  2. (ii)

Let us start from $d=b+1$, where the weights are given by $\lambda-(b-i+1)i$
for $0\leq i\leq b$. This means that $n_{b+1}=b+1=n_{b}$ by part (i). For
$d=b+2$, still assuming that $d\leq a$, we find weights of the form
$\lambda-(b-i+2)i$ for $0\leq i\leq b+1$. The same reasoning continues until
$d=a$, proving the statement.

  3. (iii)

Let $a<d\leq a+b$. We must count the weights of the form $\lambda-i(d-i)$
where $a\geq 2i-d$ and $b\geq d-2i$. The conditions on $i$ are equivalent to
the inequalities $\frac{d-b}{2}\leq i\leq\frac{a+d}{2}$. Considering the
various possibilities for the evenness of the terms in the inequality gives
the result.

∎

######  Lemma 3.4.

Let $G=A_{2}$ and $\lambda=ab$, with $a\geq b>0$ and $a+b=p-1$. Then
$V\downarrow A$ is MF if and only if $a=b=1$.

###### Proof.

Note that $r=2(a+b)<2p$ and that $a-b$ is an even number. For clarity we split
the proof into $3$ cases, depending on whether $a-b\geq 4$, $a-b=2$ or $a=b$.
Suppose first that $a-b\geq 4$, so that $p-5\geq 2b$ and $2a\geq p+4$. By
Lemma 3.3, all weights of the form $r-2d$ with $b+1\leq d\leq a$ follow the
pattern in (ii) of the same lemma, and by our assumptions this includes
weights that restrict to $p+3,p+1,p-1,p-3,p-5$. Therefore by Lemma 2.8(i)
either $(p+3)$ or $(p+1)$ is a composition factor for $V\downarrow A$. In the
first case $p-5$ occurs with multiplicity $1$ more than $p-3$, and does not
occur as a weight in the composition factor $(p+3)$, while $p-3$ does.
Therefore by Lemma 2.8(iv) the module $V\downarrow A$ is not MF. In the second
case $(p-3)$ is similarly a repeated composition factor.

Now suppose that $a=b+2$. Then by Lemma 3.3 we know that $r-2k$ occurs with
multiplicity $k+1$ for $k$ ranging between $0$ and $b$. In particular
$(r-2b)=(p+1)$ is a composition factor by Lemma 2.8(i). Again by Lemma 3.3,
the $T_{A}$-weight $r-2(b+1)=2b+2$ occurs with multiplicity $b+1$ and
$r-2a=2b$ occurs with multiplicity $b+2$. Since $2b=p-5$ does not occur as a
weight in the composition factor $(p+1)$, while $p-3=2b+2$ does, Lemma 2.8(iv)
implies that $V\downarrow A$ is not MF.

Finally assume that $a=b$. Then by Lemma 3.3 and Lemma 2.8(i), the weights
$r=4a,4a-2,\dots,2a$ afford composition factors for $V\downarrow A$, with the
last weight occurring with multiplicity $a+1$. If $a\geq 2$ we find that
$2a-2$ occurs with multiplicity $\left\lceil\frac{a+b}{2}\right\rceil=a$ and
$2a-4$ occurs with multiplicity $\left\lceil\frac{a+b+1}{2}\right\rceil=a+1$.
Since $2a-4=p-5$ does not occur as a weight in the composition factor $(p+3)$,
while $p-3$ does, Lemma 2.8(iv) implies that $V\downarrow A$ is not MF. On the
other hand if $a=b=1$ we find that $V\downarrow A=(4)\oplus(2)$. ∎

###### Proof of Proposition 3.1..

Suppose that $V\downarrow A$ is MF, with $\lambda=ab$ and $a\geq b$. Since the
Weyl module $\Delta(c0)$ is irreducible, the assumption that $r=2a+2b\geq p$,
together with Lemma 2.3, implies that $b\geq 1$. If $\dim V_{\lambda-11}=2$,
then $a+b\neq p-1$ by Lemma 2.16, and $b=1$ by Proposition 2.13(v). In this
case using the Jantzen $p$-sum formula [10, Part II, 8.19] (for example), one
sees that $\Delta(\lambda)$ is irreducible, a contradiction by Lemma 2.3. If
$\dim V_{\lambda-11}=1$, then by Lemma 2.16$a+b=p-1$, and we conclude by Lemma
3.4. ∎

###  3.2 The case where $G$ is $B_{2}$

We proceed with the case $G=B_{2}$, where $p\geq h=4$. The main result is the
following, which we shall prove after a series of lemmas.

######  Proposition 3.5.

Let $G=B_{2}$ and assume that $p\leq r$. Then $V\downarrow A$ is MF if and
only if $\lambda=2\omega_{1}$ and $p=5$.

We begin by recalling some information about the structure of $B_{2}$ Weyl
modules (with $p$-restricted highest weights). Let $\lambda=ab$ be a
$p$-restricted dominant weight; here $\alpha_{1}$ is long.

We consider the following alcoves in which a $p$-restricted weight can lie :

$C_{0}=\\{\lambda\ |\ 2a+b+3<p\\}$

$C_{1}=\\{\lambda\ |\ 2a+b+3>p,a+b+2<p\\}$

$C_{2}=\\{\lambda\ |\ a+b+2>p,b+1<p,2a+b+3<2p\\}$

$C_{3}=\\{\lambda\ |\ 2a+b+3>2p,b+1<p,a+1<p\\}$

######  Lemma 3.6.

The following hold.

  1. (i)

If $\lambda\in C_{i}$ for $i=1,2,3$, then $\Delta(\lambda)$ has exactly two
composition factors, namely $V$ and $L(\mu)$, where
$\mu=(p-a-b-3)\omega_{1}+b\omega_{2}$, respectively
$a\omega_{1}+(2p-2a-b-4)\omega_{2}$, $(2p-a-b-3)\omega_{1}+b\omega_{2}$, when
$i=1,2,3$.

  2. (ii)

For $\lambda=a\omega_{1}+(p-1)\omega_{2}$ with $2a+(p-1)+3>2p$ and $a<p-1$, we
have that $\Delta(\lambda)$ has exactly two composition factors, $V$ and
$L(\mu)$ for $\mu=(p-a-2)\omega_{1}+(p-1)\omega_{2}$.

For $\lambda$ a $p$-restricted dominant weight not lying in
$\cup_{i=1}^{3}C_{i}$ and not of the form described in (ii) above,
$\Delta(\lambda)$ is irreducible.

###### Proof..

This follows from the Jantzen $p$-sum formula [10, Part II, 8.19]. ∎

######  Remark 3.7.

Note that for a $p$-restricted weight $\lambda=ab$, if $\Delta(\lambda)$ is
reducible, then the second composition factor has highest weight of the form
$(a-k)\omega_{1}+b\omega_{2}$ or $a\omega_{1}+(b-k)\omega_{2}$, for some
$k\geq 1$. More precisely, for $\lambda\in C_{i}$, $i=1,2,3$ and $\mu$ as in
the statement of the lemma, we have
$\mu=\lambda-(2a+b+3-p)(\alpha_{1}+\alpha_{2})$, respectively
$\lambda-(a+b+2-p)(\alpha_{1}+2\alpha_{2})$,
$\lambda-(2a+b+3-2p)(\alpha_{1}+\alpha_{2})$. And in case (ii) of the lemma,
$\mu=\lambda-(2a-p+2)(\alpha_{1}+\alpha_{2})$.

We record for convenience the dimension of the Weyl module $\Delta(xy)$,
namely

$\dim\Delta(xy)=\frac{1}{6}(x+1)(y+1)(x+y+2)(2x+y+3).$

######  Lemma 3.8.

Let $G=B_{2}$ and $\lambda=c0$. Then $\lambda-ij$, with $i+j\leq 2c$, is a
weight of $V$ if and only if one of the following holds:

  1. (i)

$i\leq j$ and $j-i\leq i$;

  2. (ii)

$i\geq j$ and $i-j\leq c-\left\lfloor\frac{j+1}{2}\right\rfloor$.

###### Proof..

Since $p\neq 2$, the main result of [17] implies that the set of weights of
$V$ is precisely the same as the set of weights of the corresponding
$KG_{K}(\Phi)$-module $\Delta_{K}(\lambda)$. First we show that all weights
satisfying either (i) or (ii) are weights of $\Delta_{K}(\lambda)$.

Suppose $i\leq j\leq 2i$. Since $i+j\leq 2c$, we have that $i\leq c$. In
particular, $\lambda-i0$ is a weight of $\Delta_{K}(\lambda)$. Now the weight
$s_{\alpha_{2}}(\lambda-i0)=\lambda-i(2i)$ is also a weight of
$\Delta_{K}(\lambda)$ and by saturation, for all $0\leq m\leq 2i$, $\lambda-
im$ is a weight of $\Delta_{K}(\lambda)$. So in particular, $\lambda-ij$ is a
weight of $V$.

Suppose now that $j\leq i\leq c+j-\lfloor\frac{j+1}{2}\rfloor$. As $i+j\leq
2c$, we have that $j\leq c$ and so $\lfloor\frac{j+1}{2}\rfloor\leq c$ and
$\mu=\lambda-(\lfloor\frac{j+1}{2}\rfloor)\alpha_{1}$ is a weight of $V$.
Hence, $s_{\alpha_{2}}(\mu)=\mu-2\lfloor\frac{j+1}{2}\rfloor\alpha_{2}$ is a
weight of $\Delta_{K}(\lambda)$, which by saturation implies that
$\nu=\lambda-(\lfloor\frac{j+1}{2}\rfloor)j$ is a weight of
$\Delta_{K}(\lambda)$. Further, we have that
$\langle\nu,\alpha_{1}\rangle=c+j-2\lfloor\frac{j+1}{2}\rfloor$ and by
saturation, for all $0\leq m\leq c+j-2\lfloor\frac{j+1}{2}\rfloor$,
$\nu-m\alpha_{1}=\lambda-(m+\lfloor\frac{j+1}{2}\rfloor)j$ is a weight of
$\Delta_{K}(\lambda)$, giving that $\lambda-ij$ is a weight of $V$.

We now show that any weight $\lambda-ij$ with $i+j\leq 2c$ satisfies either
(i) or (ii). To this end, we use the fact that the set of weights of
$\Delta_{K}(\lambda)$ is the same as the set of weights of the $c$-fold tensor
product $\otimes^{c}V_{1}$, where $V_{1}$ is the $G$-module with highest
weight $\omega_{1}$. (See [1, VIII, §7, Prop. 10].) Let $\mu$ be a weight of
$\otimes^{c}V_{1}$, so that
$\mu=c\omega_{1}-a_{1}\alpha_{1}-a_{2}(\alpha_{1}+\alpha_{2})-a_{3}(\alpha_{1}+2\alpha_{2})-a_{4}(2\alpha_{1}+2\alpha_{2})=\lambda-(a_{1}+a_{2}+a_{3}+2a_{4})\alpha_{1}-(a_{2}+2a_{3}+2a_{4})\alpha_{2}$,
with $a_{i}\in\mathbb{N}$ such that $a_{1}+a_{2}+a_{3}+a_{4}\leq c$.

There are two cases to consider; suppose first that $a_{1}\leq a_{3}$, so that
$a_{1}+a_{2}+a_{3}+2a_{4}\leq a_{2}+2a_{3}+2a_{4}$. Then
$a_{2}+2a_{3}+2a_{4}\leq
2a_{1}+2a_{2}+2a_{3}+4a_{4}=2(a_{1}+a_{2}+a_{3}+2a_{4})$ and the weight $\mu$
satisfies the conditions of (i).

Now suppose $a_{1}\geq a_{3}$, so that $a_{1}+a_{2}+a_{3}+2a_{4}\geq
a_{2}+2a_{3}+2a_{4}$ and as usual

$a_{1}+a_{2}+a_{3}+2a_{4}+a_{2}+2a_{3}+2a_{4}=a_{1}+2a_{2}+3a_{3}+4a_{4}\leq
2c,$ (3)

and

$a_{1}+a_{2}+a_{3}+a_{4}\leq c.$ (4)

Note that $j-\lfloor\frac{j+1}{2}\rfloor=\lfloor\frac{j}{2}\rfloor$. If
$a_{1}+a_{2}+a_{3}+2a_{4}>c+\lfloor\frac{a_{2}+2a_{3}+2a_{4}}{2}\rfloor=c+a_{3}+a_{4}+\lfloor\frac{a_{2}}{2}\rfloor$,
then $a_{1}+a_{2}-\lfloor\frac{a_{2}}{2}\rfloor+a_{4}>c$ and
$2a_{1}+a_{2}+1+2a_{4}>2c$. If $2a_{1}+2a_{2}+2a_{3}+2a_{4}\geq
2a_{1}+a_{2}+1+2a_{4}$ we obtain a contradiction to inequality (4). Hence we
may now assume $2a_{2}+2a_{3}<a_{2}+1$, that is $a_{2}=0=a_{3}$. Now
inequality (4) becomes $a_{1}+a_{4}\leq c$ and so $a_{1}+2a_{4}\leq
c+\lfloor\frac{2a_{4}}{2}\rfloor$ and the weight satisfies condition (ii). ∎

######  Lemma 3.9.

Let $G=B_{2}$ and $\lambda=c1$, $c<p$. Then $\lambda-ij$, with $i+j\leq 2c$,
is a weight of $V$ if and only if one of the following holds:

  1. (i)

$i\leq j$ and $j-i\leq i+1$;

  2. (ii)

$i\geq j$ and $i-j\leq c-\left\lfloor\frac{j}{2}\right\rfloor$.

###### Proof..

By [1, VIII, §7, Prop. 10], the weights occurring in $V$ are the same as the
weights occurring in $c0\otimes 01$. The statement then follows from Lemma
3.8. ∎

######  Lemma 3.10.

Let $G=B_{2}$ and $\lambda=ab$ with $p>a\geq 1$, $p>b\geq 2$ and $2a+b+2\equiv
0\mod p$. Then $V\downarrow A$ is not MF.

###### Proof..

Since $2a+b+2\equiv 0\mod p$, by Lemma 3.6 we have

$\dim V=\dim L(ab)=\dim\Delta(ab)-\dim
L((a-1)b)\geq\dim\Delta(ab)-\dim\Delta((a-1)b).$

Using the Weyl character formula, we have that

$\displaystyle\dim V$
$\displaystyle\geq\frac{1}{6}(1+b)(6+6a^{2}+5b+b^{2}+12a+6ab).$ (5)

Since $p>a$ and $p>b$, there are exactly $2$ possibilities for $p$, either
$p=2a+b+2$ and $b$ is odd, or $p=a+1+\frac{b}{2}$ and $b$ is even. Let us
start with the first case, namely $p=2a+b+2$. Assume that $b=3$ and $a\geq 2$.
Then $r=4a+3b=2p-1$ and

$\displaystyle
B(r)=2\sum_{k=1}^{\frac{p+1}{2}}(2k-1)+\sum_{k=1}^{\frac{p-1}{2}}2k=\frac{3p^{2}+4p+1}{4}.$
(6)

Plugging in $p=2a+5$ and combining (6) with (5) gives $\dim V>B(r)$ and Lemma
2.11 implies that $V\downarrow A$ is not MF. The case $b=3$, $a=1$ and $p=7$
can be handled directly; we observe that $n_{1}=n_{2}=2$, while $n_{3}=4$, as
the weight space $\lambda-12$ is $2$-dimensional (see [16]). Then Lemma
2.8(ii) implies that $V\downarrow A$ is not MF.

Next assume that $b\geq 5$, in which case $r=2p+(b-4)<3p$. Then

$\displaystyle
B(r)=3\sum_{k=1}^{\frac{b-3}{2}}2k+2\sum_{k=1}^{\frac{p+1}{2}}(2k-1)+\sum_{k=1}^{\frac{p-1}{2}}2k=\frac{3p^{2}+4p+10-12b+3b^{2}}{4}.$
(7)

Plugging in $p=2a+b+2$ and combining (7) with (5) gives

$\dim V-B(r)\geq-39-36a-12a^{2}+5b+6a^{2}b-3b^{2}+6ab^{2}+b^{3}.$

As $b\geq 5$ and $a\geq 1$, this means that $\dim V-B(r)>0$, and Lemma 2.11
implies that $V\downarrow A$ is not MF.

We now consider the second case, where $p=a+1+\frac{b}{2}$. Here we have
$r=4p+b-4$. Suppose that $b=2$, so that $a=p-2\geq 3$ and $r=3p+a<4p$. If $a$
is even, we have

$\displaystyle
B(r)=4\sum_{k=1}^{\frac{a+2}{2}}(2k-1)+3\sum_{k=1}^{\frac{p-1}{2}}2k+2\sum_{k=1}^{\frac{p+1}{2}}(2k-1)+\sum_{k=1}^{\frac{p-1}{2}}2k=\frac{3p^{2}+2p+7+8a+2a^{2}}{2}.$
(8)

Plugging in $p=a+2$ and combining (8) with (5) gives

$\dim V-B(r)\geq\frac{1}{2}(a^{2}+2a-3).$

Therefore $\dim V-B(r)>0$, and Lemma 2.11 implies that $V\downarrow A$ is not
MF. If $a$ is odd, we have

$\displaystyle
B(r)=4\sum_{k=1}^{\frac{a+1}{2}}2k+3\sum_{k=1}^{\frac{p+1}{2}}(2k-1)+2\sum_{k=1}^{\frac{p-1}{2}}2k+\sum_{k=1}^{\frac{p+1}{2}}(2k-1)=\frac{3p^{2}+4p+7+8a+2a^{2}}{2}.$
(9)

Plugging in $p=a+2$ and combining (9) with (5) gives

$\dim V-B(r)\geq\frac{1}{2}(a^{2}-7).$

As $a\geq 3$, Lemma 2.11 implies that $V\downarrow A$ is not MF.

Now suppose that $b\geq 4$, in which case $r=4p+b-4<5p.$ Then

$\displaystyle
B(r)=5\sum_{k=1}^{\frac{b-2}{2}}(2k-1)+4\sum_{k=1}^{\frac{p-1}{2}}2k+3\sum_{k=1}^{\frac{p+1}{2}}(2k-1)+2\sum_{k=1}^{\frac{p-1}{2}}2k+\sum_{k=1}^{\frac{p+1}{2}}(2k-1)=\frac{10p^{2}+8p+5b^{2}-20b+18}{4}.$
(10)

Plugging in $p=a+1+\frac{b}{2}$ and combining (10) with (5) gives

$\dim
V-B(r)\geq\frac{1}{24}(-192-120a-36a^{2}+80b+12ab+24a^{2}b-21b^{2}+24ab^{2}+4b^{3}).$

We can write this is

$\dim
V-B(r)\geq\frac{1}{24}(-192+80b-21b^{2}+4b^{3}+12a^{2}(-3+2b)+12a(-10+b+2b^{2})).$

Treating the right-hand-side as a quadratic polynomial in $a$, it is easy to
see that since $b\geq 4$, we must again have $\dim V-B(r)>0$, concluding by
Lemma 2.11. ∎

######  Lemma 3.11.

Let $G=B_{2}$ and $\lambda=1b$ with $b\geq 2$. Then $V\downarrow A$ is not MF.

###### Proof..

By Lemma 3.6, we have that one of the following holds.

  1. (i)

$p>b+5$ and $V=\Delta(1b)$;

  2. (ii)

$b=p-4$; or

  3. (iii)

$b=p-2$ and $\dim V\geq\dim\Delta(1b)-\dim\Delta(1(b-2))$.

In the first case, for $b\geq 3$, $\dim V$ exceeds $B_{K}(r)$ and Lemma 2.11
then implies that $V\downarrow A$ is not MF. For $b=2$, we have $p\geq 11>r$,
contradicting our assumption that $p\leq r$.

The second case is covered by Lemma 3.10.

Finally, we consider the third case. Here $\dim V\geq
2(3+4b+b^{2})=2(p-1)(p+1)$, $r=3p-2>p$ and $b\geq 3$. In addition, we have
$B(r)=\frac{1}{2}(p+1)(3p-1)$ so that $\dim V>B(r)$ and $V\downarrow A$ is not
MF by Lemma 2.11. ∎

######  Lemma 3.12.

Let $G=B_{2}$ and $\lambda=c1$ with $c\geq 1$ and $p=2c+3$. Then $V\downarrow
A$ is not MF.

###### Proof..

Note that all $T$-weight spaces of $V$ are $1$-dimensional, see [23]. By Lemma
3.9, for $1\leq d\leq c$ and $0\leq k\leq d$, we have that $\lambda-(d-k)k$ is
a weight if and only if $0\leq k\leq 2d-2k+1$. We then find that
$n_{d}=\lfloor\frac{2d+1}{3}\rfloor+1$, for $1\leq d\leq c$. Since
$\lambda-(c+1-k)k$ is a weight if and only if $1\leq k\leq 2c+2-2k+1$, we find
that $n_{c+1}=\lfloor\frac{2c}{3}\rfloor+1$. Similarly
$n_{c+2}=\lfloor\frac{2c+2}{3}\rfloor+1$. There are now two cases to consider:
either $c\equiv 1\mod 3$ or $c\equiv-1\mod 3$. In the first case
$n_{c-1}>n_{c-2}$ (note that $n_{c-2}=0$ if $c=1$), and therefore by Lemma
2.8(i) we know that $(p+2)$ is a composition factor of $V\downarrow A$. We
have $n_{c+2}=n_{c+1}+1$, and the $T_{A}$-weight $r-2(c+2)=p-4$ does not occur
in the composition factor $(p+2)$, but the $T_{A}$-weight $p-2$ does.
Therefore Lemma 2.8(iv) implies that $V\downarrow A$ is not MF. The second
case, when $c\equiv-1\mod 3$, follows similarly. ∎

######  Lemma 3.13.

Let $G=B_{2}$ and $\lambda=c1$ with $2c+3\neq p$ and $c\geq 1$. Then
$V\downarrow A$ is not MF.

###### Proof..

Note that $r=4c+3$. First assume $c=1$, so that $r=7$; hence $p=7$ and the
result follows from Lemma 2.3. When $c=2$ we have $p\neq 7$ and by Lemma 3.6
$\Delta(\lambda)$ is irreducible. Since $p=5$ or $p=11$, the hypotheses of
Lemma 2.3 are satisfied, and $V\downarrow A$ is not MF.

We henceforth assume that $c\geq 3$ and will show that $V\downarrow A$ is not
MF. Suppose first that $p>2c+4$, so that by Lemma 3.6 $\Delta(\lambda)$ is
irreducible. Since $p\geq 5$ and $r=4c+3$, the hypotheses of Lemma 2.3 are
satisfied and $V\downarrow A$ is not MF.

Assume now that $p\leq 2c+4$, so that in fact $p\leq 2c+1$. Then by Lemma 3.6
and Remark 3.7, either $p-3\leq c\leq p-1$ and $\Delta(\lambda)$ is
irreducible and so $\dim V=\dim\Delta(\lambda)$; or $c\leq p-4$ and $\dim
V\geq\dim\Delta(c1)-\dim\Delta((c-k)1)$, for $k=2c+4-p\geq 3$. In particular,
$\dim V\geq\dim\Delta(c1)-\dim\Delta((c-3)1)=4+6c+6c^{2}$. In both cases, one
checks that $\dim V>B_{K}(r)$ so that $V\downarrow A$ is not MF by Lemma 2.11.
∎

######  Lemma 3.14.

Let $G=B_{2}$ and $\lambda=c0$ with $c>1$ and $p=2c+1$. Then $V\downarrow A$
is MF if and only if $c=2$.

###### Proof..

First note that $L(20)\downarrow A=(8)+(4)$. Now assume that $c\geq 3$, so
that $p\geq 7$. By [23], all $T$-weight spaces of $V$ are $1$-dimensional. Let
$1\leq l\leq c$. Then $n_{l}=\lfloor\frac{2l}{3}\rfloor+1$, since by Lemma 3.9
we have that $\lambda-(l-k)k$ is a weight if and only if $0\leq k\leq 2l-2k.$
Similarly $n_{c+1}=\lfloor\frac{2c+2}{3}\rfloor$,
$n_{c+2}=\lfloor\frac{2c+1}{3}\rfloor$, $n_{c+3}=n_{c}$. Therefore by Lemma
2.8(i) either both $(p+1)$ and $(p+5)$ or both $(p+3)$ and $(p+5)$ are
composition factors of $V\downarrow A$, and by Lemma 2.8(iii), $(p-7)$ must
occur as a repeated composition factor. ∎

######  Lemma 3.15.

Let $G=B_{2}$ and $\lambda=c0$ with $c>1$ and $p\neq 2c+1$. Then $V\downarrow
A$ is not MF.

###### Proof..

First assume that $p>2c+3$, so that Lemma 3.6 implies that $\Delta(c0)$ is
irreducible. Since $r=4c$ and by hypothesis $r\geq p$ and $p\geq 2c+4$, we
have that $p$ does not divide $r+1$ and Lemma 2.3 gives the result.

So we now assume that $p\leq 2c+3$. If $\Delta(\lambda)$ is irreducible, then
one checks that $\dim V=\dim\Delta(\lambda)$ exceeds $B_{K}(r)$ for all $c\geq
9$. For $c\leq 8$ we combine the information from the tables in [15] with the
criteria of Lemma 2.3 to reduce to the case $c=5$ and $p=7$. But then we have
$\dim V=91$ and $B(r)=88$, so we conclude by applying Lemma 2.11.

Now assume that $p\leq 2c+3$ and $\Delta(c0)$ is reducible. Then by Lemma 3.6,
$c+2<p$ and $2c+3>p$, and so in particular, $4c=r>p$. By Remark 3.7, $\dim
V=\dim\Delta(c0)-\dim\Delta((c-k)0)$ for $k=2c+3-p$, so $k$ is even. Assume
that $c\geq k\geq 6$. We find that

$\dim V-B_{K}(r)=c^{2}(-4+k)-c(4-3k+k^{2})+1/6(-6+13k-9k^{2}+2k^{3}).$

Treating this as a quadratic polynomial in $c$, we find that $\dim V-B_{K}(r)$
is certainly strictly positive if $-44+47k-16k^{2}+k^{3}>0$. Therefore if
$k>12$, by Lemma 2.11 we have that $V\downarrow A$ is not MF. If $k=6,8,10$ or
$12$, we have $\dim V-B_{K}(r)>0$ when $c\geq 10$. Therefore the only
possibilities for $(c,k)$ are $(8,6)$ with $p=13$, $(7,6)$ with $p=11$,
$(8,8)$ with $p=11$ or $(9,6)$ with $p=13$. In each of these cases, we find
that $\dim V-B(r)>0$, concluding by Lemma 2.11.

Note that $k\neq 2$ as $p\neq 2c+1$. So finally we consider the case $k=4$ and
$p=2c-1$. Now $r=4c=2p+2$ and a direct computation shows that $\dim
V=4c^{2}-4c+6$ while $B(r)=3c^{2}-2c+12$. Now we have $c\geq 4$ (since $k=4$)
and hence $\dim V$ exceeds $B(r)$, showing as before that $V\downarrow A$ is
not MF. ∎

######  Lemma 3.16.

Let $G=B_{2}$ and $\lambda=0c$ with $c>1$ and $p\leq r$. Then $V\downarrow A$
is not MF.

###### Proof..

We have $V=\Delta(0c)$ (see [18, Table 1]). One checks that $\dim
V=\frac{1}{6}(1+c)(2+c)(3+c)>B_{K}(r)$ for $c\geq 9$; by Lemma 2.11, the
module $V\downarrow A$ is not MF in these cases. For $2\leq c\leq 8$, we may
apply Lemma 2.3 to conclude that $V\downarrow A$ is not MF except for the
pairs $(c,p)=(3,5)$ and $(c,p)=(7,11)$. Here we apply Lemma 2.12 to again
conclude that $V\downarrow A$ is not MF. ∎

###### Proof of Proposition 3.5..

Suppose that $V\downarrow A$ is MF, with $\lambda=ab$ and $r\geq p$. By
Proposition 2.13 and Lemma 2.16, if $a,b\geq 2$ we must have $2a+b+2\equiv
0\mod p$. Therefore Lemma 3.10 implies that either $a\leq 1$ or $b\leq 1$ and
Lemma 3.12 and Lemma 3.13 show that $\lambda\neq 11$. By Lemma 3.11 we
conclude that if $a=1$, then $b=0$ contrary to our assumption that $r\geq p$
and another application of Lemma 3.12 and Lemma 3.13 shows that if $b=1$ then
$a=0$, again contrary to our assumption on $p$ and $r$. We therefore reduce to
the case $a=0$ or $b=0$, the first being ruled out by Lemma 3.16. If $b=0$, by
Lemma 3.14 and Lemma 3.15, and the above remarks, we conclude that $a=2$ with
$p=5$. Finally observe that when $\lambda=20$ and $p=5$, we have $V\downarrow
A=(8)+(4)$. ∎

###  3.3 The case where $G$ is $G_{2}$

We now move on to the final case where $G$ has rank $2$, i.e. $G=G_{2}$. Our
main result, to be proven in a sequence of lemmas, is the following

######  Proposition 3.17.

Let $G=G_{2}$ and $\lambda=ab$ with $p\leq r$. Then $V\downarrow A$ is not MF.

Set $\lambda=ab$, with $0\leq a,b<p$, where we take $\alpha_{1}$ to be short,
$(\alpha_{2},\alpha_{2})=1$. Here we have $r=6a+10b$, and $p\geq 7$ since
$p\geq h$. In addition we set $\mu=\lambda-11$ throughout the entire section
and note that $\mu=(a+1)\omega_{1}+(b-1)\omega_{2}$. We require an additional
notation: for $\alpha\in\Phi$, we let $e_{\alpha},f_{\alpha}$ denote the
$T$-weight vectors in the Lie algebra of $G$ associated with the root
$\alpha$, respectively $-\alpha$.

We will use a result from [18], which we state here only for the group
$G_{2}$:

######  Lemma 3.18.

[18, (6.2)] Assume $p>3$. Let $\nu$ be a dominant weight such that $L(\nu)$
affords a composition factor of $\Delta(\lambda)$. Then

$2(\lambda+\rho,\lambda-\nu)-(\lambda-\nu,\lambda-\nu)\in\frac{p}{6}\mathbb{Z}.$

In view of applying Lemma 3.18, we record the results of some computations for
particular subdominant weights in $\Delta(\lambda)$ in Table 3.3.

]—c—c—c—c— Weight multiplicities for $G_{2}$-modules
---
$(a,b)$ | $\nu$ | $\dim\Delta(\lambda)_{\nu}$ | $2(\lambda+\rho,\lambda-\nu)-(\lambda-\nu,\lambda-\nu)$
$a\geq 1,b\geq 1$ | $\lambda-21$ | $3-\delta_{a,1}$ | $\frac{2a+3b+4}{3}$
$a\geq 1,b\geq 2$ | $\lambda-12$ | $2$ | $\frac{a+6b}{3}$
$a\geq 1,b\geq 1$ | $\lambda-22$ | $4-\delta_{a,1}-\delta_{b,1}$ | $\frac{2a+6b+4}{3}$
$a=0,b\geq 2$ | $\lambda-22$ | $2$ | $\frac{6b+4}{3}$
$a\geq 1,b\geq 2$ | $\lambda-13$ | $2-\delta_{b,2}$ | $\frac{a+9b-9}{3}$
$a\geq 1,b\geq 2$ | $\lambda-32$ | $7-2\delta_{a,1}-\delta_{a,2}$ | $a+2b+2$
$a\geq 1,b\geq 3$ | $\lambda-23$ | $4-\delta_{a,1}$ | $\frac{2a+9b-2}{3}$
$a=1,b\geq 3$ | $\lambda-14$ | $2-\delta_{b,3}$ | $\frac{a+12b-24}{3}$

######  Lemma 3.19.

Assume $a\geq 2$, $b\geq 1$ and set $\nu=\lambda-21$. The following hold.

  1. (i)

If $a+3b+3\not\equiv 0\mod p$ and $2a+3b+4\not\equiv 0\mod p$, then $\dim
V_{\nu}=3$.

  2. (ii)

If $a+3b+3\equiv 0\mod p$, then $\dim V_{\nu}=2$.

  3. (iii)

If $2a+3b+4\equiv 0\mod p$, then $\dim V_{\nu}=2$.

###### Proof..

Using [16] and [5, Proposition A], one checks that
$\dim\Delta(\lambda)_{\nu}=3$. Note as well that if $\eta$ is a dominant
weight satisfying $\nu\prec\eta\prec\lambda$, then
$\eta\in\\{\lambda-10,\lambda-01\ ({\text{if }}b\geq 2),\lambda-20\ ({\text{if
}}a\geq 4),\mu\\}$. The weight $\eta$ does not afford a composition factor of
$\Delta(\lambda)$.

First consider the case where $a+3b+3\not\equiv 0\mod p$. In this case, $\mu$
does not afford a composition factor of $\Delta(\lambda)$ (see Lemma 2.16) and
the vectors $f_{\alpha_{1}+\alpha_{2}}v^{+}$ and
$f_{\alpha_{2}}f_{\alpha_{1}}v^{+}$ are linearly independent, where $v^{+}$ is
a highest weight vector. The weight space $V_{\nu}$ is spanned by the vectors
$v_{1}=f_{2\alpha_{1}+\alpha_{2}}v^{+}$,
$v_{2}=f_{\alpha_{1}+\alpha_{2}}f_{\alpha_{1}}v^{+}$ and
$v_{3}=f_{\alpha_{2}}f_{\alpha_{1}}^{2}v^{+}$. Suppose
$\sum_{i=1}^{3}a_{i}v_{i}=0$, for $a_{i}\in\mathcal{k}$. Then applying
$e_{\alpha_{1}}$ and $e_{\alpha_{2}}$ respectively, and using the fact that
$f_{\alpha_{1}}^{2}v^{+}\neq 0$, we obtain the following system of equations :

$2a_{1}+aa_{2}=0,\quad 3a_{2}+a_{3}(2a-2)=0\mbox{ and }a_{3}(b+2)-a_{2}=0.$

(These computations depend on a choice of structure constants; we have used
those given in [4, §12.5].) We then have that $v_{1},v_{2},v_{3}$ are linearly
dependent if and only if $a_{3}\neq 0$. If $a_{3}\neq 0$, then we deduce that
$2a+3b+4\equiv 0\mod p$. Moreover, if $2a+3b+4\equiv 0\mod p$ the three
vectors are linearly dependent and it is easy to check that $v_{1}$ and
$v_{2}$ are linearly independent. This gives (i).

Now consider the case where $a+3b+3\equiv 0\mod p$, so that $\mu$ affords a
composition factor of $\Delta(\lambda)$ and one checks that
$bf_{\alpha_{1}+\alpha_{2}}v^{+}+f_{\alpha_{1}}f_{\alpha_{2}}v^{+}=0$. Now if
$a=p-1$ (so that $2a+3b+4\equiv 0\mod p$), then $\nu$ does not occur in the
composition factor afforded by $\mu$. In addition, arguing as above, we see
that $v_{1}\in\langle v_{2},v_{3}\rangle$ and $v_{2}$ and $v_{3}$ are linearly
independent, so that $\dim V_{\nu}=2$. While if $a\neq p-1$, then $\nu$ occurs
in the composition factor afforded by $\mu$, with multiplicity 1. Moreover,
$2a+3b+4\not\equiv 0\mod p$, and Lemma 3.18 implies that the weight $\nu$ does
not afford a composition factor of $\Delta(\lambda)$ and so $\dim V_{\nu}=2$.
These arguments give the conclusions of (ii) and (iii).∎

######  Lemma 3.20.

Let $a=1$, $b\geq 1$, and set $\nu=\lambda-21$. Then $\dim V_{\nu}=1$ if
$3b+4\equiv 0\mod p$ and $\dim V_{\nu}=2$ otherwise.

###### Proof..

Using [16] and [5, Proposition A], we have $\dim\Delta(\lambda)_{\nu}=2$ and
$V_{\nu}$ is spanned by $v_{1}=f_{2\alpha_{1}+\alpha_{2}}v^{+}$ and
$v_{2}=f_{\alpha_{1}+\alpha_{2}}f_{\alpha_{1}}v^{+}$. If $3b+4\equiv 0\mod p$,
then by Lemma 2.16, $\mu$ affords a composition factor of $\Delta(\lambda)$
and $\nu$ occurs with multiplicity 1 there. So by [17], we have $\dim
V_{\nu}=1$.

If $3b+4\not\equiv 0\mod p$, then $\mu$ does not afford a composition factor
and $f_{\alpha_{1}+\alpha_{2}}v^{+}$ and $f_{\alpha_{2}}f_{\alpha_{1}}v^{+}$
are linearly independent. If $a_{1}v_{1}+a_{2}v_{2}=0$ for
$a_{i}\in\mathcal{k}$, then applying $e_{\alpha_{1}}$ and $e_{\alpha_{2}}$, we
deduce that $2a_{1}+a_{2}=0=3a_{2}$. Hence the two vectors are linearly
independent and $\dim V_{\nu}=2$.∎

######  Lemma 3.21.

Assume $b\geq 2$, $a\geq 1$, and set $\nu=\lambda-12$. Then $\dim V_{\nu}=1$
if $a+3b+3\equiv 0\mod p$ and $\dim V_{\nu}=2$ otherwise.

###### Proof..

As in the preceding lemmas, we find that $\dim\Delta(\lambda)_{\nu}=2$. If
$a+3b+3\equiv 0\mod p$, then $\mu$ affords a composition factor of
$\Delta(\lambda)$ and using [17] we deduce that $\dim V_{\nu}=1$.

So assume that $a+3b+3\not\equiv 0\mod p$ and then
$f_{\alpha_{1}+\alpha_{2}}v^{+}$ and $f_{\alpha_{2}}f_{\alpha_{1}}v^{+}$ are
linearly independent. The $\nu$ weight space is spanned by
$v_{1}=f_{\alpha_{1}+\alpha_{2}}f_{\alpha_{2}}v^{+}$ and
$v_{2}=f_{\alpha_{2}}^{2}f_{\alpha_{1}}v^{+}$. Suppose
$a_{1}v_{1}+a_{2}v_{2}=0$ for $a_{i}\in\mathcal{k}$. Applying $e_{\alpha_{1}}$
and $e_{\alpha_{2}}$ and using that $f_{\alpha_{2}}^{2}v^{+}\neq 0$, we deduce
that $3a_{1}+aa_{2}=0$ and $a_{1}b=0$. Hence the two vectors are linearly
independent, giving the result. ∎

We are now ready to prove the main proposition.

###### Proof of Proposition 3.17..

We treat various cases separately below. In Cases 1 to 4, we use Proposition
2.13(v) and Lemma 2.16 to reduce to the case where $a+3b+3\equiv 0\mod p$ (as
otherwise $V\downarrow A$ is not MF and neither is
$\Delta_{K}(\lambda)\downarrow A_{K}$). Throughout we rely on the Tables in
[16].

Case 1: $a\geq 3$ and $b\geq 3$. The $T_{A}$-weight $r-6$ is afforded by
$\lambda-30$, $\lambda-03$, $\lambda-21$ and $\lambda-12$. An application of
Lemma 3.19 then shows that $n_{3}\geq 5$ and Lemma 2.5 then shows that
$V\downarrow A$ is not MF.

Case 2: $a=2$ and $b\geq 4$. Here $3b+5\equiv 0\mod p$, so Lemma 3.18 implies
that neither of the weights $\lambda-21$ and $\lambda-12$ affords a
composition factor of $\Delta(\lambda)$. Now, the $T_{A}$-weight $r-8$ is
afforded by $\lambda-31$, $\lambda-22$, $\lambda-13$ and $\lambda-04$. By
Lemma 3.18, none of these weights affords a composition factor of
$\Delta(\lambda)$. Therefore $\dim V_{\lambda-31}=\dim V_{\lambda-22}=2$,
Indeed, for the weight $\lambda-31$ we note that the only dominant weights
$\mu^{\prime}$ with $\lambda-31\prec\mu^{\prime}$ are $\mu$, $\lambda-01$,
$\lambda-10$, $\lambda-21$. We have assumed that $\mu$ affords a composition
factor of $\Delta(\lambda)$, but the second and third weights occur with
multiplicity 1 in $\Delta(\lambda)$ and so do not afford a composition factor
of $\Delta(\lambda)$ (and as mentioned above, neither does the fourth weight).
This then allows us to determine $\dim V_{\lambda-31}$ and similarly for
$V_{\lambda-22}$. We conclude that $n_{4}\geq 6$ and apply Lemma 2.5 to see
that $V\downarrow A$ is not MF.

Case 3: $a\geq 3$ and $b=2$. Here we have $a+9\equiv 0\mod p$. Lemma 3.18
implies that neither of the weights $\lambda-21$, $\lambda-12$ affords a
composition factor of $\Delta(\lambda)$. Now consider the $T_{A}$-weight
$r-8$, afforded by $\lambda-31$, $\lambda-22$ and $\lambda-13$, none of which
affords a composition factor of $\Delta(\lambda)$. Counting the occurrences of
these weights in the irreducible $L(\mu)$, we see that $n_{4}\geq 6$ and then
use Lemma 2.5 to see that $V\downarrow A$ is not MF.

Case 4: $a\geq 2$ and $b\geq 2$. Here it remains to consider the pairs
$(a,b)\in\\{(2,2),(2,3)\\}$. In addition, we have $a+3b+3\equiv 0\mod p$.
Consider first the weight $\lambda=2\omega_{1}+2\omega_{2}$ with $p=11$; here
$\dim V=295$ and $r=32$. One then checks that $B(r)=204$. For the weight
$\lambda=2\omega_{1}+3\omega_{2}$, with $p=7$, we have $r=42$, $\dim V=532$
and $B(r)=295$. In both cases, Lemma 2.11 then implies that $V\downarrow A$ is
not MF.

We now turn to the cases where one or both of $a$ and $b$ is less than $2$, in
which case we no longer deduce that $\dim V_{\mu}=1$.

Case 5: $a\geq 3$ and $b=1$. If $a=p-6$, then $n_{2}=2$, while the
$T_{A}$-weight $r-6$ is afforded by $\lambda-21$, $\lambda-12$ and
$\lambda-30$; using Lemma 3.18 we have that $n_{3}=4$ and so $V\downarrow A$
is not MF by Lemma 2.8.

Now suppose $a\neq p-6$ so that $n_{2}=3$ and $\mu$ does not afford a
composition factor of $\Delta(\lambda)$. Let $\nu=\lambda-21$. Suppose first
that $2a+7\equiv 0\mod p$; then by Lemma 3.19 we have $\dim V_{\nu}=2$, which
implies that $[\Delta(\lambda):L(\nu)]=1$ and $n_{3}=4$. Now count the
occurrences of the $T_{A}$-weight $r-8$ which is afforded by $\lambda-31$,
$\lambda-22$ and $\lambda-40$, the latter only if $a\geq 4$. If $a\geq 4$,
Lemma 3.18 implies that $n_{4}\geq 6$, giving the usual contradiction. The
case where $2a+7\not\equiv 0\mod p$ is easier; here $\nu$ does not afford a
composition factor of $\Delta(\lambda)$ and $n_{3}=5=n_{2}+3$ (even if $a=3$).

So we are left with the case $a=3$, $b=1$ and $p=13$, where $\dim V=259$ and
$r=28$. But as above, one checks that $\dim V>B(r)$, and Lemma 2.11 implies
that $V\downarrow A$ is not MF.

Case 6: $a=1$ and $b\geq 3$.

Consider first the case where $3b+4\equiv 0\mod p$, when $\mu$ affords a
composition factor of $\Delta(\lambda)$. Moreover, we note that $b\neq 4$. We
claim that $n_{4}=4-\delta_{b,3}$ and $n_{5}\geq 6-\delta_{b,3}$, which then
shows that $V\downarrow A$ is not MF.

The $T_{A}$-weight $r-8$ is afforded by $\lambda-31$, $\lambda-22$,
$\lambda-13$ and $\lambda-04$ (the latter only if $b\geq 4$). The first of
these is conjugate to $\mu$ and so has multiplicity 1 in $V$ and the last of
these has multiplicity $1-\delta_{b,3}$. For the remaining two weights, we use
repeatedly Lemma 3.18 and note that

  * •

$\lambda-21$ and $\lambda-12$ do not afford composition factors of
$\Delta(\lambda)$;

  * •

$\lambda-22$ does not afford a composition factor of $\Delta(\mu)$ and so
occurs with multiplicity 2 in $L(\mu)$;

  * •

$\lambda-13$ and $\lambda-22$ do not afford composition factors of
$\Delta(\lambda)$.

We then deduce that the weights $\lambda-22$ and $\lambda-13$ each occur with
multiplicity 1 in $V$. Hence $n_{4}=4-\delta_{b,3}$ as claimed.

Now we turn to $n_{5}$; the $T_{A}$-weight $r-10$ is afforded by $\lambda-41$,
$\lambda-32$, $\lambda-23$, $\lambda-14$ and $\lambda-05$ (the latter only if
$b\geq 5$). The first of these is conjugate to $\lambda$. We now argue that
$\nu=\lambda-32$ has multiplicity $2$ in $V$, which establishes the claim on
$n_{5}$. Note that $\nu$ does not afford a composition factor of
$\Delta(\lambda)$ nor of $\Delta(\mu)$. Applying Lemma 3.19, we deduce that
$\nu=\mu-21$ has multiplicity 3 in $L(\mu)$ and so has multiplicity $2$ in
$V$, as claimed.

Now consider the case where $3b+4\not\equiv 0\mod p$ and so $n_{2}=3$. By
Lemma 3.20 and Lemma 3.21 we have $\dim V_{\lambda-21}=\dim V_{\lambda-12}=2$,
which means that $n_{3}=5$, so that $V\downarrow A$ is not MF.

Case 7: $a=1$ or $b=1$ and $ab\neq 0$.

By the above arguments, it remains to consider the cases $(a,b)=(1,1),(1,2)$
and $(2,1)$, where $r=16$, respectively $26$, $22$. If $p\neq 7$, respectively
$p\neq 7$, $p\neq 11$, the Weyl modules are irreducible and we may apply Lemma
2.3. For the primes $p=7,7,11$, respectively, an application of Lemma 2.12
shows that $V\downarrow A$ is not MF in the second and third cases. Now for
the case $\lambda=\omega_{1}+\omega_{2}$ and $p=7$, we must argue more
carefully. Here, one checks that the weights $r,r-2,r-4,r-6$ occur with
multiplicities $1,2,1,2$ respectively. Since $r-6$ does not occur as a weight
in $(r)$, while $r-4$ does, by Lemma 2.8 we conclude that $V\downarrow A$ is
not MF.

Case 8: $b=0$. Here we view $G$ as a subgroup of $B_{3}$ via the
$7$-dimensional irreducible representation afforded by $L(\omega_{1})$. Then
we have that $A\subseteq G$ is the principal $A_{1}$-subgroup of $B_{3}$ and
moreover the $B_{3}$-module $L_{B_{3}}(a\omega_{1})$ remains irreducible upon
restriction to $G$, and affords the module $V$. (See [18], Table 1.) Hence, we
can use the $B_{3}$ analysis, which is given in Proposition 4.16, to conclude.

Case 9: $a=0$, $b\geq 4$. Here the $T_{A}$-weight $r-6$ is afforded by
$\lambda-21$, $\lambda-12$ and $\lambda-03$, each of which has multiplicity
$1$ in $\Delta(\lambda)$ and so $n_{3}=3$. In particular, none of the listed
weights affords a composition factor of $\Delta(\lambda)$, nor does
$\lambda-11$. Now we separate into two cases. First suppose that $\lambda-22$
does not afford a composition factor of $\Delta(\lambda)$; then $n_{4}\geq 5$
and $V\downarrow A$ is not MF.

Now suppose that $\nu=\lambda-22$ affords a composition factor of
$\Delta(\lambda)$ and so by Lemma 3.18 we have $3b+2\equiv 0\mod p$. We first
treat the case where $b\geq 6$. We claim that $n_{5}=4$. The $T_{A}$-weight
$r-10$ is afforded by $\lambda-23$, $\lambda-32$, $\lambda-14$, and
$\lambda-05$. The first two occur in the composition factor afforded by $\nu$,
each with multiplicity 1, and using the multiplicities in the Weyl module and
[17], we see that each of the four listed weights occurs with multiplicity 1
in $V$, establishing the claim. The $T_{A}$-weight $r-12$ is afforded by
$\lambda-42$, $\lambda-33$, $\lambda-24$, $\lambda-15$ and $\lambda-06$. The
second of these weights has multiplicity $4$ in $\Delta(\lambda)$ and occurs
with multiplicity 2 in $L(\nu)$. Moreover, this weight does not afford a
composition factor of $\Delta(\lambda)$ (nor does any dominant weight
$\eta\neq\nu$ with $\lambda-33\prec\eta\prec\lambda$) and so occurs with
multiplicity 2 in $V$. Hence, $n_{6}\geq 6$ and $V\downarrow A$ is not MF.

It remains to consider the cases $b=4$ and $b=5$ with $p=7$, respectively
$p=17$ and $\dim V=267$, respectively 546. In both cases, an application of
Lemma 2.12 shows that $V\downarrow A$ is not MF.

Case 10: $a=0$ and $1\leq b\leq 3$. When $b=1$, the Weyl module is irreducible
and the result follows from Lemma 2.3. If $b=2$ and $p\neq 7$, we may apply
Lemma 2.3 to conclude. When $(b,p)=(2,7)$, we use Lemma 2.9 and the proof of
[12, Lemma 4.5] to deduce that
$n_{0}=1,n_{1}=1,n_{2}=2,n_{3}=2,n_{4}=3,n_{5}=4$ , so that $V\downarrow A$
has composition factors $(20),(16)$ and $(12)$. Since the $T_{A}$-weight $12$
lies in the composition factor $(16)$ but the $T_{A}$-weight $10$ does not,
Lemma 2.8 implies that $V\downarrow A$ is not MF.

Finally, we consider the case $b=3$, where $r=30$. Here the Weyl module is
irreducible unless $p=11$. If $p\neq 11$, the result follows from Lemma 2.3.
If $p=11$, we use [16] to see that
$n_{0}=1,n_{2}=1,n_{2}=2,n_{3}=3,n_{4}=3,n_{5}=3$. We then deduce that
$V\downarrow A$ has no composition factor $(22)$, nor $(20)$. But then $\dim
V=148>B(30)-\dim(20)-\dim(22)$, so that $V\downarrow A$ is not MF.∎

##  4 The case where $G$ has rank at least $3$

In this section we handle the case where $G$ has rank at least $3$,
establishing the following proposition.

######  Proposition 4.1.

Suppose that $G$ has rank at least $3$ and $p\leq r$. Then $V\downarrow A$ is
not MF.

We assume throughout §4 that $p\leq r$. Furthermore by Lemma 2.13(i) we only
need to consider the case $\lambda=c_{i}\omega_{i}+c_{j}\omega_{j}$ (with
$c_{i}$ or $c_{j}$ possibly $0$).

###  4.1 The case where $\lambda=c_{i}\omega_{i}+c_{j}\omega_{j}$

We begin by treating the case where $\lambda=c_{i}\omega_{i}+c_{j}\omega_{j}$
with $c_{i}c_{j}\neq 0$.

######  Lemma 4.2.

Suppose that $G$ has rank at least $4$ and
$\lambda=c_{i}\omega_{i}+c_{j}\omega_{j}$ with $\alpha_{i}$ and $\alpha_{j}$
adjacent and $c_{i},c_{j}\geq 1$. Then $V\downarrow A$ is not MF.

###### Proof..

Since $p\geq h$ we have $p\geq 5,11,11,7$ respectively for
$G=A_{\ell},B_{\ell},C_{\ell},D_{\ell}$ and $p\geq 13,13,19,31$ respectively
for $G=F_{4},E_{6},E_{7},E_{8}$. By Proposition 2.13(iv) and (v), we can
assume that $c_{i}=1$ or $c_{j}=1$, and $\alpha_{i}$ or $\alpha_{j}$ is an
end-node. If $c_{i}\geq 3$ or $c_{j}\geq 3$, it is easy to see that
$V\downarrow A$ is not MF. For example if $\lambda=c_{1}\omega_{1}+\omega_{2}$
and $c_{1}\geq 3$, the $T_{A}$-weight $r-6$ is afforded by $\lambda-123$,
$\lambda-234$, $\lambda-12^{2}$, $\lambda-1^{2}2$ and $\lambda-1^{3}$.
Therefore $n_{3}\geq 5$, and Lemma 2.5 implies that $V\downarrow A$ is not MF.
If $c_{i}=2$ or $c_{j}=2$, by Lemma 2.16 we have $\dim
V_{\lambda-\alpha_{i}-\alpha_{j}}>1$, and by Proposition 2.13(vi) the module
$V\downarrow A$ is not MF. Thus, we reduce to $c_{i}=c_{j}=1$.

Consider $\lambda=\omega_{1}+\omega_{2}$ for $G$ classical. The weights
$\lambda-123=(\lambda-12)^{s_{3}},\lambda-234,\lambda-1^{2}2=(\lambda-2)^{s_{1}},\lambda-12^{2}=(\lambda-1)^{s_{2}}$
occur with multiplicities $2$, $1$, $1$, $1$ by Lemma 2.16. Therefore
$n_{3}\geq 5$ and Lemma 2.5 implies that $V\downarrow A$ is not MF. The same
argument, with the appropriate relabelling of indices, handles all remaining
cases where $(\alpha_{i},\alpha_{i})=(\alpha_{j},\alpha_{j})$.

Therefore we reduce to $G=B_{\ell}$ or $G=C_{\ell}$ with
$\lambda=\omega_{\ell-1}+\omega_{\ell}$. Suppose $G=B_{\ell}$. Since $p\geq
h$, by Lemma 2.16 we have $\dim V_{\lambda-(\ell-1)\ell}=2$. The
$T_{A}$-weight $r-6$ is afforded by
$\lambda-(\ell-1)\ell^{2}=(\lambda-(\ell-1)\ell)^{s_{\ell}}$,
$\lambda-(\ell-2)(\ell-1)\ell$ and $\lambda-(\ell-1)^{2}\ell$. If $\ell=4$,
the first two weight spaces have dimension $2$ by [16]. By Lemma 2.18, for any
$\ell\geq 4$, we have $n_{3}\geq 5$, and Lemma 2.5 implies that $V\downarrow
A$ is not MF. The $C_{\ell}$ case is handled similarly. ∎

######  Lemma 4.3.

Let $G=A_{3}$ and $\lambda=c\omega_{1}+\omega_{2}$ or
$\omega_{1}+c\omega_{2}$, with $c>1$. Then $V\downarrow A$ is not MF.

###### Proof..

By Proposition 2.13(vi) we can assume that the weight space $\lambda-12$ is
$1$-dimensional. In particular we must have $c=p-2$ by Lemma 2.16. Let us
start with $\lambda=\omega_{1}+(p-2)\omega_{2}$. Since $p\geq 5$, the
$T_{A}$-weight $r-6$ is afforded by $\lambda-123$, $\lambda-12^{2}$,
$\lambda-2^{2}3$, $\lambda-2^{3}$ and $\lambda-1^{2}2$. Therefore Lemma 2.5
implies that $V\downarrow A$ is not MF.

Now assume $\lambda=(p-2)\omega_{1}+\omega_{2}$. This will be a dimension
argument. We refer to the discussion in [10, Part II, 8.20], where the weight
$\lambda$ satisfies the conditions of the weight $\lambda_{2}$, with $s=1=t$
and $r=p-3$. Then one has

$\dim
V=\dim\Delta(\lambda)-\dim\Delta(\lambda-\alpha_{1}-\alpha_{2})+\dim\Delta(\lambda-2\alpha_{1}-2\alpha_{2}-\alpha_{3}).$

Then using the Weyl degree formula we find that $\dim
V=\frac{(p-1)}{6}(p^{2}+7p+18)$. We have $r=3p-2$, and a simple calculation
shows that $B(r)=\frac{(p+1)(3p-1)}{2}$. Since $B(r)<\dim V$ for all $p>3$, by
Lemma 2.11 we conclude that $V\downarrow A$ is not MF. ∎

######  Lemma 4.4.

Let $G=B_{3}$ or $C_{3}$ and let $\lambda\in\\{c\omega_{1}+\omega_{2}$,
$\omega_{1}+c\omega_{2}$, $c\omega_{2}+\omega_{3}$,
$\omega_{2}+c\omega_{3}\\}$, with $c>1$. Then $V\downarrow A$ is not MF.

###### Proof..

By Proposition 2.13(vii) we can assume that the weight space $\lambda-ij$ is
$1$-dimensional, where $\lambda=c_{i}\omega_{i}+c_{j}\omega_{j}$. Note that
$p\geq 7$ as $p\geq h$.

Case 1 : $\lambda=c\omega_{1}+\omega_{2}$. As the weight space $\lambda-ij$ is
$1$-dimensional, we have $c=p-2$ by Lemma 2.16. In particular $c\geq 5$. The
$T_{A}$-weight $r-6$ is afforded by $\lambda-1^{3}$, $\lambda-1^{2}2$,
$\lambda-12^{2}$, and $\lambda-123$; in addition, for $G=B_{3}$, $r-6$ is
afforded by $\lambda-23^{2}$ and if $G=C_{3}$, by $\lambda-2^{2}3$. Hence
$n_{3}\geq 5$ and $V\downarrow A$ is not MF by Lemma 2.5.

Case 2 : $\lambda=\omega_{1}+c\omega_{2}$. As in the previous case, we reduce
to $c=p-2$, so $c\geq 5$. The $T_{A}$-weight $r-6$ is afforded by
$\lambda-12^{2}$, $\lambda-1^{2}2$, $\lambda-2^{2}3$, $\lambda-123$, and
$\lambda-2^{3}$. Therefore $n_{3}\geq 5$ and $V\downarrow A$ is not MF by
Lemma 2.5.

Case 3 : $G=B_{3}$ and $\lambda=c\omega_{2}+\omega_{3}$. The $T_{A}$-weight
$r-8$ is afforded by $\lambda-12^{2}3$, $\lambda-1^{2}2^{2}$,
$\lambda-2^{3}3$, $\lambda-23^{3}$, $\lambda-123^{2}$, $\lambda-2^{2}3^{2}$
which implies $n_{4}\geq 6$ and $V\downarrow A$ is not MF by Lemma 2.5.

Case 4 : $G=C_{3}$ and $\lambda=c\omega_{2}+\omega_{3}$. By Lemma 2.16 we may
assume that $c+4\equiv 0\mod p$, implying $c\geq 3$. The $T_{A}$-weight $r-6$
is afforded by $\lambda-2^{3}$, $\lambda-12^{2}$, $\lambda-2^{2}3$,
$\lambda-123$, $\lambda-23^{2}$, which implies $n_{3}\geq 5$ and $V\downarrow
A$ is not MF by Lemma 2.5.

Case 5 : $G=B_{3}$ and $\lambda=\omega_{2}+c\omega_{3}$. As above we reduce to
$c+4\equiv 0\mod p$ and so $c\geq 3$. The $T_{A}$-weight $r-6$ is afforded by
$\lambda-123$, $\lambda-2^{2}3$, $\lambda-3^{3}$ and $\lambda-23^{2}$. By
Lemma 3.6, the Weyl module $\Delta_{B_{2}}(1c)$ has exactly two composition
factors $L_{B_{2}}(1c)$ and $L_{B_{2}}(0c)$, the latter afforded by
$\lambda-11$. Therefore the multiplicity of the weight $\lambda-23^{2}$ in $V$
is $2$ and so $n_{3}\geq 5$, which by Lemma 2.5 implies that $V\downarrow A$
is not MF.

Case 6 : $G=C_{3}$ and $\lambda=\omega_{2}+c\omega_{3}$. This is entirely
similar. Here we may assume $2c+3\equiv 0\mod p$. If $c\geq 4$, the
$T_{A}$-weight $r-8$ is afforded by $\lambda-12^{2}3$, $\lambda-2^{3}3$,
$\lambda-23^{3}$, $\lambda-2^{2}3^{2}$, $\lambda-123^{2}$, and
$\lambda-3^{4}$. Therefore $n_{4}\geq 6$, and $V\downarrow A$ is not MF by
Lemma 2.5. If $c\leq 3$, we must have $c=2$ and $p=7$. By [16], all weight
spaces of $V$ are $1$-dimensional. The $T_{A}$-weight $r-8$ is again afforded
by the first $5$ weights listed above. On the other hand, the $T_{A}$-weight
$r-6$ is afforded precisely by $\lambda-123,\lambda-23^{2}$ and
$\lambda-2^{2}3$, implying that $n_{3}=3$. Therefore $n_{4}-n_{3}\geq 2$, and
by Lemma 2.8 we conclude that $V\downarrow A$ is not MF. ∎

######  Lemma 4.5.

Let $G=A_{3}$ or $B_{3}$ or $C_{3}$ and $\lambda=\omega_{1}+\omega_{2}$ or
$\omega_{2}+\omega_{3}$. Then $V\downarrow A$ is not MF.

###### Proof..

Consider $G=A_{3}$. Then $r=7$ and $p=5$ or $p=7$ as $r\geq p\geq h$. In both
cases the Weyl module is irreducible and the conditions for Lemma 2.3 are
satisfied, implying that $V\downarrow A$ is not MF.

Now consider $G=B_{3}$. Since $p\geq 7$, the Weyl module is irreducible. The
conditions for Lemma 2.3 are satisfied, implying that $V\downarrow A$ is not
MF.

Finally consider $G=C_{3}$. If $(\lambda,p)\neq(\omega_{1}+\omega_{2},7)$ we
can conclude as for $B_{3}$. Therefore assume that
$\lambda=\omega_{1}+\omega_{2}$ with $p=7$. By Lemma 2.16 we have $\dim
V_{\lambda-123}=2$, and it is straightforward to see that $n_{3}\geq 5$, which
by Lemma 2.5 implies that $V\downarrow A$ is not MF. ∎

######  Lemma 4.6.

Assume that $G$ has rank at least $3$ and $\lambda=\omega_{i}+\omega_{j}$
where $\alpha_{i}$ and $\alpha_{j}$ are end-nodes. Then $V\downarrow A$ is not
MF.

###### Proof..

Consider first the case where $G=A_{\ell}$, where $p\geq h=\ell+1$. By [15]
the Weyl module $\Delta(\lambda)$ is irreducible if and only if $p\neq\ell+1$.
If $p\neq\ell+1$, the conditions of Lemma 2.3 are satisfied and therefore
$V\downarrow A$ is not MF. We therefore reduce to the case $p=\ell+1$, where
$V\simeq\Delta(\lambda)/L(0)$. For $d<\ell$, it is straightforward to see that
$n_{d}=d+1$. Therefore by Lemma 2.8(i) we know that $(p+1)$ is a composition
factor of $V\downarrow A$. Now the $T_{A}$-weight $p-3$ occurs with
multiplicity one more than the $T_{A}$-weight $p-1$, and it does not occur as
a weight in $(p+1)$, while $p-1$ does. Therefore Lemma 2.8(iv) implies that
$V\downarrow A$ is not MF.

If $G=B_{\ell}$ or $C_{\ell}$ and $\ell\geq 4$, the first paragraph of the
proof of [12, Lemma 3.5] shows that $n_{3}\geq 5$, so $V\downarrow A$ is not
MF by Lemma 2.5. If $G=C_{3}$, we can apply Lemma 2.3 to conclude that
$V\downarrow A$ is not MF.

Now assume $G=B_{3}$. We have $p=7$ or $p=11$, as $r=12$. If $p=11$, the Weyl
module is irreducible by [16], and the conditions of Lemma 2.3 are satisfied,
implying that $V\downarrow A$ is not MF. When $p=7$, again using [16], we find
that $n_{2}=3$ and $(r-4)$ is therefore a composition factor by Lemma 2.8(i).
Furthermore, we have $n_{3}=3$, $n_{4}=4$ and the $T_{A}$-weight $r-8$ does
not occur as a weight in $(r-4)$, while the $T_{A}$-weight $r-6$ does.
Therefore Lemma 2.8(iv) implies that $V\downarrow A$ is not MF.

Now consider $G=D_{\ell}$, with $\ell\geq 4$. If
$\lambda=\omega_{1}+\omega_{\ell-1}$, the $T_{A}$-weight $r-2(\ell-1)$ is
afforded by $\lambda-1\dots(\ell-1)$, $\lambda-2\dots\ell$, and
$\lambda-1\dots(\ell-2)\ell$. Since $p\geq h$ we have $p>\ell$, and therefore
by Lemma 2.17 and Lemma 2.18 we have $\dim V_{\lambda-1\dots(\ell-1)}=\ell-1$.
Therefore $n_{\ell-1}\geq\ell+1$ and Lemma 2.5 implies that $V\downarrow A$ is
not MF. It is also easy to see that if
$\lambda=\omega_{\ell-1}+\omega_{\ell}$, we have $n_{3}\geq 5$. Again Lemma
2.5 implies that $V\downarrow A$ is not MF.

Finally, if $G$ is exceptional, the arguments used in the proof of [12, Lemma
3.6] in characteristic zero allow us to conclude, as [16] shows that the
relevant weight spaces in $V$ have the same dimension as the corresponding
weight spaces in $\Delta_{K}(\lambda)$. ∎

######  Proposition 4.7.

Suppose that $G$ has rank at least $3$ and
$\lambda=c_{i}\omega_{i}+c_{j}\omega_{j}$ with $c_{i},c_{j}\geq 1$. Then
$V\downarrow A$ is not MF.

###### Proof..

By Lemma 4.2, if $\alpha_{i}$ and $\alpha_{j}$ are adjacent and $G$ has rank
at least $4$, the module $V\downarrow A$ is not MF. If $\alpha_{i}$ and
$\alpha_{j}$ are adjacent and $G$ has rank $3$, Lemma 4.3, Lemma 4.4 and Lemma
4.5 combine to imply that $V\downarrow A$ is not MF.

We have therefore reduced to $\alpha_{i}$ and $\alpha_{j}$ not adjacent, in
which case by Proposition 2.13 we can assume that $c_{i}=c_{j}=1$ and
$\alpha_{i}$ and $\alpha_{j}$ are both end-nodes. In this case, by Lemma 4.6
we conclude that $V\downarrow A$ is not MF. ∎

###  4.2 The case where $\lambda=b\omega_{i}$

We now consider the case $\lambda=b\omega_{i}$. Note that if $G$ is classical,
then $\lambda\neq\omega_{1}$, as we are assuming that $p\leq r$, and
necessarily $p\geq h$.

######  Lemma 4.8.

Assume that $G=A_{\ell},B_{\ell},C_{\ell}$ with $\ell\geq 3$ or $G=D_{\ell}$
with $\ell\geq 4$. Let $\lambda=b\omega_{1}$, with $b\geq 2$. Then
$V\downarrow A$ is not MF.

###### Proof..

We first consider the case $b=2$ and start by assuming that
$(G,p)\neq(B_{\ell},2\ell+1)$. By [15] and since $p\geq h$, the Weyl module is
irreducible. A simple check shows that the conditions of Lemma 2.3 are
satisfied, implying that $V\downarrow A$ is not MF. Consider now the case
$G=B_{\ell}$ and $p=2\ell+1$, where $V\simeq\Delta(\lambda)/L(0)$. By [12,
Lemma 4.2], we have $\Delta_{K}(\lambda)\downarrow
A_{K}=(4\ell)+(4\ell-4)+\dots$, which by a slight modification of Lemma 2.9
implies $n_{d}=d+1$ for $d$ even with $d<2\ell$, and $n_{d+1}=n_{d}$ for $d$
even with $d+1<2\ell$. By Lemma 2.8(i), for all $0\leq d<2\ell$ we have that
$(r-2d)$ is a composition factor of $V\downarrow A$. In particular, either
$(p+1)$ or $(p+3)$ is a composition factor of $V\downarrow A$. In the first
case, the $T_{A}$-weight $p-3$ occurs with multiplicity one more than the
$T_{A}$-weight $p-1$, but does not occur as a weight in $(p+1)$. Therefore
Lemma 2.8(iv) implies that $V\downarrow A$ is not MF. Similarly, if $(p+3)$ is
a composition factor of $V\downarrow A$, then the $T_{A}$-weight $p-5$ (note
that $p>5$ since $\ell\geq 3$) occurs with multiplicity one more than the
$T_{A}$-weight $p-3$, but does not occur as a weight in $(p+3)$, concluding in
the same way.

Now consider the case $b\geq 3$. Start with $G=A_{\ell}$. Here
$V=\Delta(\lambda)$ by [18, 1.14].

If $b\geq 3$ and $\ell\geq 6$, the first paragraph of the proof of [12, Lemma
4.4] shows that $n_{6}\geq 7$, which by Lemma 2.8(iii) implies that
$V\downarrow A$ is not MF. If $b=4$ and $\ell=4$ or $\ell=5$, we similarly
have $n_{4}\geq 5$. If $b\geq 5$ and $(b,\ell)\neq(5,3)$, we have
$n_{6}-n_{5}\geq 2$. Therefore by Lemma 2.8(iii) and (ii), we reduce down to
the cases $(b,\ell)=(4,3)$, $(5,3)$, $(3,3)$, $(3,4)$, $(3,5)$. For these
cases we can conclude using Lemma 2.3, unless $\ell=b=3$ and $p=5$. In this
case $r=9$ and the weights $9,7,5,3$ occur respectively with multiplicities
$1,1,2,3$. By Lemma 2.8(i), we have that $(5)$ is a composition factor, and in
addition $r-6$ does not occur as weight in this composition factor. Therefore
by Lemma 2.8(iv) the module $V\downarrow A$ is not MF.

The $C_{\ell}$ case follows from the $A_{2\ell-1}$ case since
$A<C_{\ell}<A_{2\ell-1}$ is a principal $A_{1}$ of $A_{2\ell-1}$ and
$V=S^{b}(L_{C_{\ell}}(\omega_{1}))=L_{A_{2\ell-1}}(b\omega_{1})\downarrow
C_{\ell}$. Now consider $G=B_{\ell}$. If $b\geq 4$, it is straightforward to
check that $n_{4}\geq 5$, which by Lemma 2.8(iii) implies that $V\downarrow A$
is not MF. If $b=3$ and $\ell\geq 4$, by the proof of [12, Lemma 4.4], we have
$n_{6}\geq 7$, concluding in the same way. If $\ell=b=3$ we have $p\leq r=18$
and by Lemma 2.3 the module $V\downarrow A$ is not MF. If $G=D_{\ell}$, we
have $A\leq B_{\ell-1}<G$. Since $\Delta_{B_{\ell-1}}(b\omega_{1})$ is a
composition factor of $\Delta_{D_{\ell}}(b\omega_{1})$, if
$\Delta_{D_{\ell}}(b\omega_{1})\downarrow A$ is MF, so is
$\Delta_{B_{\ell-1}}(b\omega_{1})$. Therefore by the $B_{\ell}$ result, we
conclude that $V\downarrow A$ is not MF. ∎

######  Lemma 4.9.

Assume that $G=B_{\ell},C_{\ell}$ with $\ell\geq 3$ or $G=D_{\ell}$ with
$\ell\geq 4$. Let $\lambda=b\omega_{i}$, with $i>1$ and $b>1$. Then
$V\downarrow A$ is not MF.

###### Proof..

By Lemma 2.15 we can assume that $\lambda=b\omega_{\ell}$. Assume for now that
$b\geq 3$. If $G=C_{\ell}$, the $T_{A}$-weight $r-6$ is afforded by
$\lambda-\ell^{3}$, $\lambda-(\ell-2)(\ell-1)\ell$,
$\lambda-(\ell-1)^{2}\ell$, $\lambda-(\ell-1)\ell^{2}$. If $G=B_{\ell}$, the
$T_{A}$-weight $r-6$ is afforded by $\lambda-(\ell-1)\ell^{2}$,
$\lambda-(\ell-2)(\ell-1)\ell$ and $\lambda-\ell^{3}$, and using the fact that
the $B_{2}$-module $\Delta(b\omega_{2})$ is irreducible, by Lemma 2.18, we
have that the first of these weights has multiplicity $2$. Hence for both of
the groups $C_{\ell}$ and $B_{\ell}$, we have $n_{3}\geq 4$. By Lemma
2.8(iii), the module $V\downarrow A$ is not MF. We will treat the case
$G=D_{\ell}$ at the end of the proof.

We now consider the case $b=2$ when $G=C_{\ell}$ and first assume that
$\ell\geq 5$. The $T_{A}$-weight $r-10$ is afforded by
$\lambda-(\ell-1)^{3}\ell^{2}$, $\lambda-(\ell-2)(\ell-1)^{2}\ell^{2}$,
$\lambda-(\ell-1)^{2}\ell^{3}$, $\lambda-(\ell-2)^{2}(\ell-1)^{2}\ell$,
$\lambda-(\ell-3)(\ell-2)(\ell-1)\ell^{2}$ and
$\lambda-(\ell-4)(\ell-3)(\ell-2)(\ell-1)\ell$. Again, by Lemma 2.8(iii) the
module $V\downarrow A$ is not MF.

Now consider the cases $C_{\ell}$, for $\lambda=2\omega_{\ell}$ and $\ell=3,4$
where $r=18$, respectively $32$, and $p\geq 7$, respectively $11$. In both
cases we have that $\Delta(2\omega_{\ell})$ is irreducible by [16]. For
$\ell=3$, the conditions of Lemma 2.3 are satisfied, implying that
$V\downarrow A$ is not MF. If $\ell=4$, by the first paragraph of [12, Lemma
4.3] we have $\dim(\Delta_{K}(\lambda)\downarrow
A_{K})_{r-12}\geq\dim(\Delta_{K}(\lambda)\downarrow A_{K})_{r-10}+2$.
Therefore by Lemma 2.9 we find that $n_{6}-n_{5}\geq 2$, concluding by Lemma
2.8(ii).

Turn now to the case $G=B_{\ell}$ and $b=2$. Here the $T_{A}$-weight $r-8$ is
afforded by $\lambda-(\ell-2)(\ell-1)\ell^{2}$,
$\lambda-(\ell-1)^{2}\ell^{2}$, $\lambda-(\ell-1)\ell^{3}$ and if $\ell\geq
4$, $\lambda-(\ell-3)(\ell-2)(\ell-1)\ell$. The first of these weights is
conjugate to $\lambda-(\ell-1)\ell^{2}$ and so has multiplicity $2$ by the
first paragraph of this proof. Therefore if $\ell\geq 4$, $V\downarrow A$ is
not MF by Lemma 2.8(iii). So finally, we reduce to $b=2$ and $\ell=3$, where
$p\geq 7$ and $r=12$. The Weyl module is irreducible by [15]. The conditions
of Lemma 2.3 are satisfied, implying that $V\downarrow A$ is not MF.

Finally suppose that $G=D_{\ell}$ and $\lambda=b\omega_{\ell}$. Since $A\leq
B_{\ell-1}\leq D_{\ell}$ and $V\downarrow B_{\ell-1}\cong
L_{B_{\ell-1}}(b\omega_{\ell-1})$, we may use the $B_{\ell-1}$ result to
conclude. ∎

######  Lemma 4.10.

If $G=E_{\ell}$ and $\lambda=b\omega_{i}$ with $b>1$, then $V\downarrow A$ is
not MF.

###### Proof..

This follows verbatim from the proof of [12, Lemma 4.6], unless $i=\ell$ and
$G=E_{7}$ or $E_{8}$ with $b=2$ or $b=3$, in which case it is not difficult to
check that we have $n_{6}\geq n_{5}+2$ (as stated in the proof of [12, Lemma
4.6]), as this count relies on $1$-dimensional weight spaces. By Proposition
2.13 the module $V\downarrow A$ is not MF. ∎

######  Lemma 4.11.

If $G=F_{4}$ and $\lambda=b\omega_{i}$ with $b>1$, then $V\downarrow A$ is not
MF.

###### Proof..

By Lemma 2.15 the simple root $\alpha_{i}$ corresponds to an end-node. If
$i=1$ we can conclude as in the first paragraph of the proof of [12, Lemma
4.7].

Assume $i=4$. If $b\geq 3$, like in [12, Lemma 4.7] we have $n_{4}\geq 5$,
concluding by Lemma 2.8(iii). If $b=2$ we have $V=\Delta(\lambda)$ by [15],
and since $r=32$ and $r\geq p>11$, the conditions of Lemma 2.3 are satisfied.
Thus, $V\downarrow A$ is not MF. ∎

It remains to consider the case $\lambda=\omega_{i}$. Recall that for $G$
classical, we have $\lambda\neq\omega_{1}$.

######  Lemma 4.12.

Assume that $G$ has rank at least $3$, $\lambda=\omega_{i}$ and that one of
the following holds:

  1. (i)

$G=A_{\ell},B_{\ell},C_{\ell}$ or $G=D_{\ell}$ with $\ell\geq 4$, and $4\leq
i\leq\ell-3$;

  2. (ii)

$G=A_{\ell}$, $i=3$, and $\ell\geq 5$;

  3. (iii)

$G=A_{\ell},B_{\ell},C_{\ell}$ or $G=D_{\ell}$ with $\ell\geq 4$ and $i=2$.

Then $V\downarrow A$ is not MF.

###### Proof..

  1. (i)

By Lemma 2.14 the module $V\downarrow A$ is not MF.

  2. (ii)

Assume that $l\geq 8$. Here $V=\bigwedge^{3}(L(\omega_{1}))$, and therefore
the $T_{A}$-weight $r-12$ is afforded by the wedge of weight vectors for each
of the following triples of $T_{A}$-weights: $\ell(\ell-2)(\ell-16)$,
$\ell(\ell-4)(\ell-14)$, $\ell(\ell-6)(\ell-12)$, $\ell(\ell-8)(\ell-10)$,
$(\ell-2)(\ell-4)(\ell-12)$, $(\ell-2)(\ell-6)(\ell-10)$,
$(\ell-4)(\ell-6)(\ell-8)$. Therefore $n_{6}\geq 7$, and Lemma 2.8(iii)
implies that $V\downarrow A$ is not MF.

Now assume that $5\leq l\leq 7$. In all cases $\Delta(\lambda)=V$ and a quick
check shows that the conditions of Lemma 2.3 are satisfied, implying that
$V\downarrow A$ is not MF.

  3. (iii)

Assume $\lambda=\omega_{2}$. As $p>\ell$, we have $V=\Delta(\lambda)$ [15,
Table 2]. We have $r=2\ell-2,4\ell-2,4\ell-4$ or $r=4\ell-6$ according to
whether $G=A_{\ell},B_{\ell},C_{\ell}$ or $G=D_{\ell}$. Furthermore we have
$p$ greater than $\ell,2\ell-1,2\ell,2\ell-2$ respectively. It is then an easy
check to see that the conditions of Lemma 2.3 are satisfied, implying that
$V\downarrow A$ is not MF.

∎

######  Lemma 4.13.

Assume that $G=B_{\ell},C_{\ell}$ with $\ell\geq 3$ or $G=D_{\ell}$ with
$\ell\geq 4$, and in addition that $\lambda=\omega_{i}$ for $i\geq 3$ and $V$
is not a spin module for $B_{\ell}$ or $D_{\ell}$. Then $V\downarrow A$ is not
MF.

###### Proof..

If $G=B_{\ell}$ or $D_{\ell}$, then $V=\bigwedge^{i}(\omega_{1})$ by [18] and
the result follows from Lemma 4.12(i)(ii) for $G=A_{2\ell}$ or $A_{2\ell-1}$.
Indeed, if $G=B_{\ell}$, then $A$ is regular in $A_{2\ell}$ and
$V=L_{A_{2\ell}}(\omega_{i})\downarrow G$, while if $G=D_{\ell}$, then
$A<B_{\ell-1}<D_{\ell}$ and by the $B_{\ell}$ case there is a
$B_{\ell-1}$-composition factor of $V$ (namely $L_{B_{\ell-1}}(\omega_{i})$)
that is not multiplicity-free in its restriction to $A$, implying that
$V\downarrow A$ is not MF.

We now consider the case $G=C_{\ell}$. By part (i) of Lemma 4.12 we can
furthermore assume that $i=3$ or $i>\ell-3$. If $i=\ell-2>3$, the
$T_{A}$-weight $r-8$ has multiplicity at least $5$ as it is afforded by $5$
different weights as in the proof of [12, Lemma 5.3]. Therefore Lemma 2.8(iii)
implies that $V\downarrow A$ is not MF.

Assume $i=\ell-1>3$. As $\ell\geq 5$, the $T_{A}$-weight $r-12$ is afforded by
$\lambda-(\ell-3)(\ell-2)(\ell-1)^{3}\ell$,
$\lambda-(\ell-2)(\ell-1)^{3}\ell^{2}$,
$\lambda-(\ell-2)^{2}(\ell-1)^{3}\ell$,
$\lambda-(\ell-4)(\ell-3)(\ell-2)(\ell-1)^{2}\ell$,
$\lambda-(\ell-3)(\ell-2)^{2}(\ell-1)^{2}\ell$. When $\ell=5$, the last two
weights have multiplicity $2$ in $V$ by [16], and therefore the same holds for
$\ell\geq 5$ by Lemma 2.18. Thus, $n_{6}\geq 7$, and by Lemma 2.8(iii) the
module $V\downarrow A$ is not MF.

Now assume $i=\ell>3$. Start with $\ell=4$ or $5$. In both cases the Weyl
module is irreducible. We have $r=16$ if $\ell=4$, and $r=25$ if $\ell=5$. The
conditions of Lemma 2.3 are satisfied, unless $\ell=5$ and $p=13$. In this
case, we find that $B(r)-\dim(r-2)=118<\dim V=132$. Therefore by Lemma 2.12,
the module $V\downarrow A$ is not MF.

Now consider the case $\ell\geq 6$ with $\lambda=\omega_{\ell}$. Here the
$T_{A}$-weight $r-10$ has multiplicity $4$ as it is afforded precisely by
$\lambda-(\ell-4)(\ell-3)(\ell-2)(\ell-1)\ell$,
$\lambda-(\ell-2)^{2}(\ell-1)^{2}\ell$,
$\lambda-(\ell-3)(\ell-2)(\ell-1)^{2}\ell$,
$\lambda-(\ell-2)(\ell-1)^{2}\ell^{2}$. The $T_{A}$-weight $r-12$ has
multiplicity at least $6$ as it is afforded by
$\lambda-(\ell-5)(\ell-4)(\ell-3)(\ell-2)(\ell-1)\ell$,
$\lambda-(\ell-2)^{2}(\ell-1)^{2}\ell^{2}$,
$\lambda-(\ell-4)(\ell-3)(\ell-2)(\ell-1)^{2}\ell$,
$\lambda-(\ell-3)(\ell-2)^{2}(\ell-1)^{2}\ell$,
$\lambda-(\ell-3)(\ell-2)(\ell-1)^{2}\ell^{2}$,
$\lambda-(\ell-2)(\ell-1)^{3}\ell^{2}$. By Lemma 2.8(ii) $V\downarrow A$ is
not MF.

To conclude assume that $i=3$. If $\ell\geq 6$, we have $n_{6}\geq 7$, since
by the last paragraph of the proof of [12, Lemma 5.3] there are $7$ distinct
weights of $V$ affording the $T_{A}$-weight $r-12$. Therefore $V\downarrow A$
is not MF by Lemma 2.8(iii).

In the remaining cases the Weyl module is irreducible. For
$\ell\in\\{3,4,5\\}$ we can apply Lemma 2.3, unless $\ell=5$ and $p=11$, in
which case $r=21\equiv-1\mod p$. In this case, we find that
$B(r)-\dim(r-2)=84<\dim V=110$. Therefore by Lemma 2.12, the module
$V\downarrow A$ is not MF. ∎

######  Lemma 4.14.

Assume that $V$ is a spin module for $B_{\ell}$ with $\ell\geq 3$ or
$D_{\ell}$ with $\ell\geq 4$. Then $V\downarrow A$ is not MF.

###### Proof..

We have $V=\Delta(\lambda)$. If $G=D_{\ell}$, then $A\leq B_{\ell-1}<G$ and
$V\downarrow B_{\ell-1}$ is the spin module for $B_{\ell-1}$; therefore it
suffices to prove the lemma for $G=B_{\ell}$. We have $r=\ell(\ell+1)/2$. If
$V\downarrow A$ is MF the dimension of $V$ is at most $B_{K}(r)$, by Lemma
2.11. This implies that if $\ell\geq 10$, the module $V\downarrow A$ is not
MF.

Now assume $\ell\leq 9$. Since $p>h=2\ell$ we know that $p\nmid r$. Therefore
if $V\downarrow A$ is MF the dimension of $V$ is at most $B(r)-\dim(r-2)$, by
Lemma 2.12. This then reduces our considerations to the pairs $(n,p)$ among
the following: $(5,11)$, $(5,13)$, $(6,13)$, $(6,17)$, $(6,19)$, $(7,17)$,
$(7,19)$, $(7,23)$, $(8,31)$. For every $3\leq\ell\leq 8$, by Lemma 2.9 we can
read the dimension of the $T_{A}$-weight space $r-2k$ off the table in the
proof of [12, Lemma 5.4]. In each case we apply part (iii) of Lemma 2.8 to
find that $V\downarrow A$ is not MF. The first repeated composition factors
are of highest weight respectively $5,9,9,11,15,14,14,16,24$. ∎

######  Lemma 4.15.

Assume $G=E_{\ell}$ or $F_{4}$ and $\lambda=\omega_{i}$. Then $V\downarrow A$
is not MF.

###### Proof..

If $G=E_{\ell}$ and $\lambda=\omega_{4}$, the $T_{A}$-weight $r-4$ is afforded
by $\lambda-34$, $\lambda-24$, $\lambda-45$. Therefore $n_{2}\geq 3$, and by
Lemma 2.8(iii), the module $V\downarrow A$ is not MF.

If $G=E_{8}$ and $\lambda=\omega_{3}$ or $\omega_{6}$, then $r=182$
respectively $r=168$, giving $B_{K}(r)=8464$ and $7225$ respectively. By [15],
$\dim V>B_{K}(r)$ and therefore by Lemma 2.11 the module $V\downarrow A$ is
not MF. If $G=E_{8}$ and $\lambda=\omega_{5}$, by Lemma 2.14 the module
$V\downarrow A$ is not MF.

In all remaining cases, by [15] we have that $V=\Delta(\lambda)$. Lemma 2.11
then allows us to reduce to the case where $V$ is the minimal module for $G$,
or the adjoint module for $E_{6},E_{7}$ or $F_{4}$. The conditions of Lemma
2.3 are satisfied, implying that $V\downarrow A$ is not MF. ∎

######  Proposition 4.16.

Suppose that $G$ has rank at least $3$ and $\lambda=b\omega_{i}$ with $b\geq
1$. Then $V\downarrow A$ is not MF.

###### Proof..

If $G$ is classical and $b\geq 2$, this is a direct consequence of Lemma 4.8
and Lemma 4.9. If $G$ is exceptional and $b\geq 2$, we similarly conclude by
Lemma 4.10 and Lemma 4.11.

If $b=1$, we reach the same conclusion by Lemmas 4.12, 4.13, 4.14, 4.15. ∎

###### Proof of Proposition 4.1..

If $V$ is MF, then by Proposition 2.13, the weight $\lambda$ is of the form
$c_{i}\omega_{i}+c_{j}\omega_{j}$. If $c_{i}c_{j}\neq 0$, then the conclusion
follows from Proposition 4.7. If $c_{i}c_{j}=0$, then the conclusion follows
from Proposition 4.16. ∎

##  5 Proof of Corollary 2

In this Section we prove Corollary 2, thereby extending Theorem 1 to the case
where $\lambda$ is not $p$-restricted. The following lemma serves as an
inductive tool.

######  Lemma 5.1.

Let $\lambda=\sum_{i=0}^{t}p^{i}\lambda_{i}$ where $\lambda_{i}$ is a
$p$-restricted dominant weight for all $0\leq i\leq t$. Assume that for some
$s$, $1\leq s<t$, we have
$\left(\sum_{i=0}^{s}p^{i}\lambda_{i}\right)\downarrow T_{A}<p^{s+1}$. Then
$V\downarrow A$ is MF if and only if the following two conditions hold:

  1. (i)

$L(\sum_{i=0}^{s}p^{i}\lambda_{i})\downarrow A$ is MF, and

  2. (ii)

$L(\sum_{i=s+1}^{t}p^{i}\lambda_{i})\downarrow A$ is MF.

###### Proof..

Let $V_{1}=L(\sum_{i=0}^{s}p^{i}\lambda_{i})$ and
$V_{2}=L(\sum_{i=s+1}^{t}p^{i}\lambda_{i})$, so that $V=V_{1}\otimes V_{2}$.
If $V_{2}=L(0)$, the statement is trivial. Thus, assume $V_{2}\neq L(0)$.

One direction is clear. If either $V_{1}\downarrow A$ or $V_{2}\downarrow A$
is not MF, then $V\downarrow A$ is not MF. Assume now that both
$V_{1}\downarrow A$ and $V_{2}\downarrow A$ are MF, and let $V_{1}\downarrow
A=(r_{0})+(r_{1})+\dots+(r_{m})$, so that by the assumption on $s$,
$p^{s+1}>r_{0}>r_{1}>\dots>r_{m}$. Similarly let $V_{2}\downarrow
A=(v_{0})+(v_{1})+\dots+(v_{n})$ where $v_{0}>v_{1}>\dots>v_{n}\geq p^{s+1}$.
Then for all $0\leq i\leq m$ and $0\leq j\leq n$ we have
$(r_{i})\otimes(v_{j})\simeq(r_{i}+v_{j})$, since $r_{i}<p^{s+1}$ and
$s_{j}\geq p^{s+1}$. This implies that the composition factors of
$V_{1}\otimes V_{2}$ are precisely of the form $(r_{i}+v_{j})$, which are all
clearly distinct. Therefore $V\downarrow A$ is MF. ∎

Let us restate, and prove, Corollary 2.

######  Corollary 5.2.

Let $\lambda=\sum_{i=0}^{t}p^{i}\lambda_{i}$ where each $\lambda_{i}$ is a
$p$-restricted dominant weight and set $r_{i}=\lambda_{i}\downarrow T_{A}$,
for $0\leq i\leq t$. Then $V\downarrow A$ is MF if and only if one of the
following holds:

  1. (i)

We have $p>r_{i}$ and $\Delta_{K}(\lambda_{i})\downarrow A_{K}$ is MF for all
$0\leq i\leq t$;

  2. (ii)

$G$ is of type $A_{2}$, $p=3$ and there exists $0\leq i\leq t$ such that
$\lambda_{i}=\omega_{1}+\omega_{2}$. Moreover, for all $0\leq j\leq t$ we have
$\lambda_{j}\in\\{0,\omega_{1}+\omega_{2},\omega_{1},\omega_{2}\\}$ and if
$\lambda_{j}=\omega_{1}+\omega_{2}$ for some $0\leq j\leq t-1$, then
$\lambda_{j+1}=0$;

  3. (iii)

$G$ is of type $B_{2}$, $p=5$ and there exists $0\leq i\leq t$ such that
$\lambda_{i}=2\omega_{1}$. Moreover, for all $0\leq j\leq t$ we have
$\lambda_{j}\in\\{0,2\omega_{1},\omega_{1},\omega_{2}\\}$ and if
$\lambda_{j}=2\omega_{1}$ for some $0\leq j\leq t-1$, then
$\lambda_{j+1}\in\\{0,\omega_{2}\\}$.

###### Proof..

Without loss of generality, we may assume that $\lambda_{0}\neq 0$. We use
induction on $t$. If $t=0$, then $\lambda$ is $p$-restricted and the statement
follows from Theorem 1. Suppose that the statement is valid for all $0\leq
t\leq N$ for some $N\in\mathbb{N}$. Let $t=N+1$ and $V_{1}=L(\lambda_{0})$,
$V_{2}=L(\sum_{i=1}^{t}p^{i}\lambda_{i})$, so that $V=V_{1}\otimes V_{2}$.

Suppose first that $V\downarrow A$ is MF. Then certainly $V_{1}\downarrow A$
and $V_{2}\downarrow A$ are both MF. If $r_{0}<p$, by Lemma 5.1 and the
inductive assumption, we conclude that $(G,\lambda,p)$ is as in one of the
three conclusions of the statement. Therefore assume that
$\lambda_{0}\downarrow A\geq p$. By Theorem 1 we have $G=A_{2}$, $p=3$ and
$\lambda_{0}=11$, or $G=B_{2}$, $p=5$ and $\lambda_{0}=20$.

Consider first the case $G=A_{2}$. By Theorem 1 and Table 1, we must have
$\lambda_{i}\in\\{0,11,10,01\\}$ for all $0\leq i\leq t$. If $\lambda_{1}=0$,
we conclude by the inductive assumption combined with Lemma 5.1 for $s=1$. If
$\lambda_{1}\in\\{\omega_{1}+\omega_{2},\omega_{1},\omega_{2}\\}$ then
$V_{1}\downarrow A\otimes L(p\lambda_{1})\downarrow A$ has $(4)$ as a repeated
composition factor. The $B_{2}$ case follows similarly. Note that here
$\lambda_{1}$ is allowed to be $0$ or $01$.

Suppose now that (i) holds, then $V\downarrow A$ is MF by the inductive
assumption combined with Lemma 5.1.

If (ii) or (iii) holds, it is easy to verify that the conditions of Lemma 5.1
with $s=1$ are satisfied, concluding again by the inductive assumption.

∎

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