# Monogamous subvarieties of the nilpotent cone

Simon M. Goodwin  School of Mathematics, University of Birmingham, Birmingham,
B15 2TT, UK<EMAIL_ADDRESS>(Goodwin) ,  Rachel Pengelly  Department
of Mathematics, The University of Manchester, Oxford Road, Manchester, M13
9PL, UK<EMAIL_ADDRESS>(Pengelly) ,  David I. Stewart
Department of Mathematics, The University of Manchester, Oxford Road,
Manchester, M13 9PL, UK<EMAIL_ADDRESS>(Stewart) and  Adam
R. Thomas  Department of Mathematics, University of Warwick, Coventry, CV4
7AL, UK<EMAIL_ADDRESS>(Thomas) In memory of Gary, who
influenced us greatly

###### Abstract.

Let $G$ be a reductive algebraic group over an algebraically closed field
$\Bbbk$ of prime characteristic not $2$, whose Lie algebra is denoted
$\mathfrak{g}$. We call a subvariety $\mathfrak{X}$ of the nilpotent cone
$\mathcal{N}\subset\mathfrak{g}$ _monogamous_ if for every $e\in\mathfrak{X}$,
the $\mathfrak{sl}_{2}$-triples $(e,h,f)$ with $f\in\mathfrak{X}$ are
conjugate under the centraliser $C_{G}(e)$. Building on work by the first two
authors, we show there is a unique maximal closed $G$-stable monogamous
subvariety $\mathcal{V}\subset\mathcal{N}$ and that it is an orbit closure,
hence irreducible. We show that $\mathcal{V}$ can also be characterised in
terms of Serre’s $G$-complete reducibility.

###### 2010 Mathematics Subject Classification:

17B45

##  1\. Introduction

Let $\Bbbk$ be an algebraically closed field of characteristic $p\neq 2$, and
$G$ a simple algebraic $\Bbbk$-group with Lie algebra
$\mathfrak{g}=\operatorname{Lie}(G)$. Theorems of Jacobson–Morozov and Kostant
say that if $\Bbbk$ is of characteristic $0$, then for any nilpotent
$e\in\mathfrak{g}$ there exists an $\mathfrak{sl}_{2}$-triple $(e,h,f)$ in
$\mathfrak{g}$ which is unique up to conjugacy by the centralizer of $e$ in
$G$, see [Mor42, Jac51, Kos59]. We continue the investigation into
generalising Kostant’s uniqueness theorem to fields of small characteristic.
Let $\mathfrak{X}$ be a subset of the nilpotent cone
$\mathcal{N}\subset\mathfrak{g}$. We say that $\mathfrak{X}$ is _monogamous_
if the following property holds:

Let $(e,h,f)$ and $(e,h^{\prime},f^{\prime})$ be $\mathfrak{sl}_{2}$-triples
with $e,f,f^{\prime}\in\mathfrak{X}$. Then $(e,h,f)$ is $C_{G}(e)$-conjugate
to $(e,h^{\prime},f^{\prime})$.

The main theorem of [ST18] proves that $\mathcal{N}$ is monogamous if and only
if $p>h(G)$, where $h(G)$ is the Coxeter number for $G$. When $G$ is of
classical type, the first two authors [GP22] found a unique maximal $G$-stable
closed subvariety of $\mathcal{N}$ that is monogamous. This paper completes
the story by treating the exceptional types. Define the following subset of
$\mathcal{N}$:

$\mathcal{V}:=\left\\{x\in\mathcal{N}\ \left|\ \begin{array}[]{l}x^{[p]}=0,\\\
x\text{ is not regular in a Levi subalgebra with a factor of type
}A_{p-1},\text{ and}\\\ x\text{ is not subregular if }G\text{ is of type
}G_{2}\text{ and }p=3.\end{array}\right\\}\right.$

######  Theorem 1.1.

Let $G$ be a simple algebraic group over an algebraically closed field $\Bbbk$
of characteristic $p>2$. Then $\mathcal{V}$ is the unique maximal $G$-stable
closed monogamous subvariety of $\mathcal{N}$. Furthermore, $\mathcal{V}$ is
irreducible, being the closure of a single orbit as specified in Tables 1 and
2 below.

In [Ste10b], a close relationship was found between uniqueness of
$\mathfrak{sl}_{2}$-subalgebras and the existence of so-called non-$G$-cr
$\mathfrak{sl}_{2}$-subalgebras. The notion of $G$-complete reducibility for
subgroups of $G$ is due to Serre [Ser05], and the natural generalisation to
subalgebras of $\mathfrak{g}$ was introduced by McNinch [McN07]. Given a
subalgebra $\mathfrak{h}\subseteq\mathfrak{g}$, we say that $\mathfrak{h}$ is
_$G$ -completely reducible_ ($G$-cr for short) if for every parabolic
subalgebra $\mathfrak{p}$ such that $\mathfrak{h}\subseteq\mathfrak{p}$ we
have that there exists some Levi subalgebra $\mathfrak{l}$ of $\mathfrak{p}$
with $\mathfrak{h}\subseteq\mathfrak{l}$.

We say $\mathfrak{X}\subseteq\mathcal{N}$ is _$A_{1}$ -$G$-cr_ if every
subalgebra generated by an $\mathfrak{sl}_{2}$-triple $(e,h,f)$ with
$e,f\in\mathfrak{X}$ is $G$-cr.

######  Theorem 1.2.

Let $G$ be a simple algebraic group over an algebraically closed field $\Bbbk$
of characteristic $p>2$. Then $\mathcal{V}$ is the unique maximal $G$-stable
closed $A_{1}$-$G$-cr subvariety of $\mathcal{N}$.

The proof follows very quickly from Theorem 1.1; see Section 4.

######  Remark 1.3.

It would be interesting to know more about the geometry of the nilpotent
variety $\mathcal{V}$. In type $A$, Donkin [Don90] showed that the closure of
all orbits are normal. Orbit closures in the remaining classical types are
considered by Xiao and Shu [XS15]. For exceptional types
$G_{2},F_{4},\ldots,E_{8}$, results of Thomsen [Tho00] show that our varieties
$\mathcal{V}$ are in fact Gorenstein normal varieties with rational
singularities as long as $p\geq 5,11,7,11,13$, respectively.

### Acknowledgments

Part of this work contributed to the second author’s PhD thesis at the
University of Birmingham, they were supported by the EPSRC during this period.
The second author also gratefully acknowledges the financial support of both
the LMS and the Heilbronn institute. The third author is supported by a
Leverhulme Trust Research Project Grant RPG-2021-080 and the fourth author is
supported by an EPSRC grant EP/W000466/1. For the purpose of open access, the
authors have applied a Creative Commons Attribution (CC BY) licence to any
Author Accepted Manuscript version arising from this submission.

##  2\. Preliminaries

Throughout, $\Bbbk$ is an algebraically closed field of characteristic $p>2$
and $G$ is a simple $\Bbbk$-group with $\mathfrak{g}=\operatorname{Lie}(G)$.
There is an inherited $[p]$-map on $\mathfrak{g}$ and we use $x^{[p]}$ to
denote the image of $x\in\mathfrak{g}$ under this map. The variety of all
nilpotent elements in $\mathfrak{g}$, often called the nilpotent cone, is
denoted by $\mathcal{N}$. The restricted nullcone is the subvariety of
$\mathcal{N}$ consisting of elements $x$ such that $x^{[p]}=0$ and we denote
it by $\mathcal{N}_{p}$. The distribution of nilpotent elements among
$\mathfrak{sl}_{2}$-subalgebras of $\mathfrak{g}$ is insensitive to central
isogeny, and so we assume that whenever $G$ is classical, it is one of
$\mathrm{SL}(V)$, $\mathrm{Sp}(V)$ or $\mathrm{SO}(V)$ and write
$G=\operatorname{Cl}(V)$ for brevity; if $G$ is exceptional, we take it to be
simply connected.

Recall that a prime $p$ is bad for $G$ if $p=2$ and $G$ is of type $B$, $C$ or
$D$; if $p\leq 3$ and $G$ is exceptional; or if $p\leq 5$ and $G$ is of type
$E_{8}$; otherwise it is good. In some examples we require a choice of base
for the root system associated to $\mathfrak{g}$; we use Bourbaki notation
[Bou05]. Finally, we fix a maximal torus $T$ of $G$.

###  2.1. Nilpotent orbits and Hasse diagrams

The orbits for the action of $G$ on $\mathcal{N}$ are called nilpotent orbits.
There are finitely many such and they are classified. In case $G$ is of
exceptional type, we describe an orbit $\mathrm{O}=G\cdot x$ by a label
indicating a Levi subalgebra in which $e$ is distinguished; for these labels
we refer to [LS12].

When $G=\operatorname{Cl}(V)$, the classification or orbits in terms of
actions on $V$ is well-known and can be found in [Jan04, Section 1], but we
recap it here for ease of reference. Set $m=\dim V$. If $G=\mathrm{SL}(V)$,
orbits are parameterised by partitions of $m$ according to the Jordan
decomposition of their elements’ actions on $V$; we write
$x\sim(\lambda_{1},\dots,\lambda_{r})$ where
$\lambda_{1}\geq\dots\geq\lambda_{r}$ is the partition of $m$ corresponding to
$x$. In types $B$ and $C$ orbits are parameterised by partitions of $m$ with
an even number of even parts and an even number of odd parts, respectively. In
type $D$ it is slightly more complicated. A partition is called very even if
it only has even parts and they all occur with even multiplicity. There is one
orbit for each partition of $m$ with an even number of even parts that is not
very even; and two orbits for every very even partition of $m$.

To check that $\mathcal{V}$ is a closed subvariety of $\mathcal{N}$ we require
information about the Hasse diagrams for the closure relation on nilpotent
orbits. For classical types, apart from type $D$, the closure order on orbits
is precisely the dominance order on partitions. In type $D$ we start with the
Hasse diagram for the dominance order on partitions with an even number of
even parts. Then we replace each very even partition $\lambda$ with two nodes
$\lambda_{1},\lambda_{2}$ and replace each edge from $\lambda$ to $\mu$ with
two edges from $\lambda_{i}$ to $\mu$. For exceptional types the picture is
actually incomplete in general. But if $p$ is good for $G$, the existence of
Springer morphisms implies that the Hasse diagrams remain the same as those in
characteristic $0$; [Spa82, Thèoréme III 5.2]. In bad characteristic, there
are not even the same number of nilpotent and unipotent orbits; for certain
bad primes there are more nilpotent orbits than in characteristic $0$. To deal
with this, Hesselink gives a partition of $\mathcal{N}$ into strata, one for
each nilpotent orbit of $G$ in characteristic $0$. These strata are locally
closed and $G$-stable.

The following forthcoming theorem of Premet [Pre] is sufficient for our needs;
it reduces the determination of the Hasse diagram of nilpotent orbit closures
to establishing the extra edges needed to accommodate when the Hesselink
strata split into multiple orbits.

######  Theorem 2.1.

Let $S_{i}\subset\mathcal{N}$ be a Hesselink stratum and let $\mathrm{O}_{i}$
be the orbit of maximal dimension contained in $S_{i}$. Then the Hasse diagram
of the closure relation on the nilpotent orbits $\\{\mathrm{O}_{i}\\}$ is the
same as the Hasse diagram for the closure relation on all nilpotent orbits in
characteristic $0$.

Since $p>2$, there are precisely two cases where the Hesselink strata contains
more than one orbit, namely when $p=3$ and $G$ is of type $G_{2}$ or $E_{8}$.
In these cases precisely one stratum of each splits into two orbits. Away from
those cases then, the Hasse diagrams are the same as in characteristic $0$;
these can be found in [Car93, Section 13.4]. However, two edges are missing in
the $E_{8}$ diagram: there should be edges between the pairs of labels
($E_{6}+A_{1}$, $E_{8}(b_{6})$) and $(A_{3}+A_{1}$, $A_{3}$) (see [Spa82, p.
249]). The Hasse diagram for $G_{2}$ when $p=3$ can be deduced from [Stu71]
and is reproduced in the left of Figure 1.

$G_{2}$$G_{2}(a_{1})$$(\tilde{A_{1}})_{(3)}$$\tilde{A_{1}}$$A_{1}$$0$

$A_{1}^{2}A_{2}$$A_{1}A_{2}$$A_{1}^{4}$$A_{2}$$A_{1}^{3}$$A_{1}^{2}$$A_{1}$$0$

Figure 1. Full Hasse diagram for $G_{2}$ when $p=3$ and partial Hasse diagram
for $E_{8}$ when $p=3$.

We can now prove part of Theorem 1.1.

######  Lemma 2.2.

The subset $\mathcal{V}\subseteq\mathcal{N}$ is a closed $G$-stable
subvariety; moreover, it is the closure of a single orbit in every case, as
specified in Tables 1 and 2.

$G$ | $m$ | $\lambda$ |
---|---|---|---
$A_{m-1}$ | $a(p-1)+r$ | $((p-1)^{a},r)$ |
$B_{\frac{m-1}{2}}$ |  $p+a(p-1)+r$ $(r>0)$ | $(p,(p-1)^{a},r-1,1)$ |  $a$ even
|  | $(p,(p-1)^{a-1},p-2,r+1)$ |  $a$ odd
| $p+a(p-1)$ | $(p,(p-1)^{a})$ |
| $\leq p$ | $(m)$ |
$C_{\frac{m}{2}}$ | $a(p-1)+r$ | $((p-1)^{a},r)$ |
$D_{\frac{m}{2}}$ | $p+a(p-1)+r$ | $(p,(p-1)^{a},r)$ |  $a$ even
|  | $(p,(p-1)^{a-1},p-2,r,1)$ |  $a$ odd
| $\leq p$ | $(m-1,1)$ |
Table 1. Partition $\lambda$ corresponding to the orbit $\mathrm{O}_{\lambda}$ such that $\mathcal{V}=\overline{\mathrm{O}}_{\lambda}$ in the classical types, where $a\geq 0$ and $0\leq r<p-1$.  $G$ | $p$ | $\mathrm{O}$ | $G$ | $p$ | $\mathrm{O}$ | $G$ | $p$ | $\mathrm{O}$ | $G$ | $p$ | $\mathrm{O}$
---|---|---|---|---|---|---|---|---|---|---|---
$G_{2}$ | $3$ | $\tilde{A}_{1}^{(3)}$ | $E_{6}$ | $3$ | $A_{1}^{3}$ | $E_{7}$ | $3$ | $A_{1}^{4}$ | $E_{8}$ | $3$ | $A_{1}^{4}$
| $5$ | $G_{2}(a_{1})$ |  | $5$ | $D_{4}(a_{1})$ |  | $5$ | $A_{3}A_{2}A_{1}$ |  | $5$ | $A_{3}^{2}$
| $\geq 7$ | $G_{2}$ |  | $7$ | $E_{6}(a_{3})$ |  | $7$ | $E_{7}(a_{5})$ |  | $7$ | $A_{4}^{2}$
$F_{4}$ | $3$ | $A_{1}\tilde{A_{1}}$ |  | $11$ | $E_{6}(a_{1})$ |  | $11$ | $E_{7}(a_{3})$ |  | $11$ | $E_{8}(a_{6})$
| $5$ | $F_{4}(a_{3})$ |  | $\geq 13$ | $E_{6}$ |  | $13$ | $E_{7}(a_{2})$ |  | $13$ | $E_{8}(a_{5})$
| $7$ | $F_{4}(a_{2})$ |  |  |  |  | $17$ | $E_{7}(a_{1})$ |  | $17$ | $E_{8}(a_{4})$
| $11$ | $F_{4}(a_{1})$ |  |  |  |  | $\geq 23$ | $E_{7}$ |  | $19$ | $E_{8}(a_{3})$
| $\geq 13$ | $F_{4}$ |  |  |  |  |  |  |  | $23$ | $E_{8}(a_{2})$
|  |  |  |  |  |  |  |  |  | $29$ | $E_{8}(a_{1})$
|  |  |  |  |  |  |  |  |  | $\geq 31$ | $E_{8}$
Table 2. Orbit $\mathrm{O}$ such that $\mathcal{V}=\overline{\mathrm{O}}$ in
the exceptional types.

###### Proof.

Suppose $G=\operatorname{Cl}(V)$ with $\dim V=m$. An orbit corresponding to a
partition $\lambda$ of $m$ is contained in the restricted nullcone if and only
if the largest part of $\lambda$ is at most $p$. Let $G=\mathrm{SL}(V)$ or
$\mathrm{Sp}(V)$ (resp. $\mathrm{SO}(V)$), and let $x\in\mathcal{N}$ with
partition represented by $\lambda$. Then $x$ is not regular in a Levi
subalgebra with a factor of type $A_{p-1}$ if $\lambda$ contains no parts of
size $p$ (resp. at most one part of size $p$). Now every orbit represented in
Table 1 represents a single orbit in $\mathcal{V}$: for $G$ of type $D$, each
$\lambda$ given in Table 1 is not very even. Observe that any other orbit in
$\mathcal{V}$ must be represented by a partition lower than $\lambda$ in the
dominance ordering, and hence is contained in
$\overline{\mathrm{O}}_{\lambda}$; and vice-versa, by definition of
$\mathcal{V}$.

Now suppose $G$ is of exceptional type. We use the tables in [Ste16] to
determine the orbits in the restricted nullcone. Note that the $A_{7}$ orbit
is not restricted in $E_{8}$ when $p=3$ so we may appeal to Theorem 2.1 and
inspect the Hasse diagrams. ∎

Let $X$ be a node in the Hasse diagram for the closure order on nilpotent
orbits of $\mathfrak{g}$. We call $X$ a _neighbour of $\mathcal{V}$_ if
$\mathcal{V}$ does not contain the orbit corresponding to $X$ but there is an
edge from $X$ to some orbit contained in $\mathcal{V}$. We say $Y$ is a
_minimal neighbour of $\mathcal{V}$_ if $Y$ is a neighbour of $\mathcal{V}$
and the closure of $Y$ contains no other neighbours of $\mathcal{V}$.

######  Example 2.3.

Let $G$ be of type $E_{8}$, with $p=3$. The bottom of the Hasse diagram is as
shown on the right of Figure 1. By Lemma 2.2, $\mathcal{V}$ is the closure of
the $4A_{1}$ orbit. Therefore $A_{2}$ and $A_{2}+A_{1}$ are the only
neighbours of $\mathcal{V}$. As $A_{2}$ is in the closure of the $A_{2}+A_{1}$
orbit, $A_{2}$ is the only minimal neighbour of $\mathcal{V}$.

######  Lemma 2.4.

Let $G$ be of exceptional type. Table 3 lists the minimal neighbours of
$\mathcal{V}$.

$G$ | $p$ | min. orbs. | $G$ | $p$ | min. orbs. | $G$ | $p$ | min. orbs. | $G$ | $p$ | min. orbs.
---|---|---|---|---|---|---|---|---|---|---|---
$G_{2}$ | $3$ | $G_{2}(a_{1})$ | $E_{6}$ | $3$ | $A_{2}$ | $E_{7}$ | $3$ | $A_{2}$ | $E_{8}$ | $3$ | $A_{2}$
| $5$ | $G_{2}$ |  | $5$ |  $A_{4}$, $D_{4}$ |  | $5$ |  $A_{4}$, $D_{4}$ |  | $5$ |  $A_{4}$, $D_{4}$
$F_{4}$ | $3$ |  $A_{2}$, $\tilde{A}_{2}$ |  | $7$ | $D_{5}$ |  | $7$ |  $D_{5}$, $A_{6}$ |  | $7$ |  $D_{5}$, $A_{6}$
| $5$ |  $C_{3}$, $B_{3}$ |  | $11$ | $E_{6}$ |  | $11$ | $E_{6}$ |  | $11$ |  $E_{6}$, $D_{7}$
| $7$ | $F_{4}(a_{1})$ |  |  |  |  | $13$ | $E_{7}(a_{1})$ |  | $13$ | $E_{7}(a_{1})$
| $11$ | $F_{4}$ |  |  |  |  | $17$ | $E_{7}$ |  | $17$ | $E_{7}$
|  |  |  |  |  |  |  |  |  | $19$ | $E_{8}(a_{2})$
|  |  |  |  |  |  |  |  |  | $23$ | $E_{8}(a_{1})$
|  |  |  |  |  |  |  |  |  | $29$ | $E_{8}$
Table 3. Minimal neighbours of $\mathcal{V}$.

###### Proof.

When the Hasse diagram is known, the result follows by inspection. For $G$ of
type $E_{8}$ with $p=3$ we need to rule out $A_{7}$ as a minimal neighbour of
$\mathcal{V}$. However, since an element in this orbit is regular in a Levi
subalgebra of type $A_{7}$, it must be connected to $A_{6}$, hence cannot be a
minimal neighbour. ∎

###  2.2. $G$-cr subalgebras

######  Proposition 2.5.

Suppose $e\in\mathcal{N}_{p}$. If $e$ is contained in an
$\mathfrak{sl}_{2}$-triple then there exists a $G$-cr subgroup $X\leq G$ of
type $A_{1}$ such that $\operatorname{Lie}(X)$ contains $e$.

###### Proof.

If $G=\mathrm{SL}(V)$ then $e^{[p]}=0$ implies $e$ has Jordan blocks of size
at most $p$, which means $e$ is regular in a Levi subalgebra of type
$A_{r_{1}}\times\dots\times A_{r_{i}}$ with each $r_{i}\leq p-1$. The image of
$X=\mathrm{SL}_{2}$ under the completely reducible representation given by
$L(r_{1})\oplus\dots\oplus L(r_{i})$ satisfies the demands of the theorem,
where $r_{j}$ now represents a (restricted) high weight. So assume $G$ is not
of type $A$. Then if $p$ is good for $G$, it is very good, and the result
follows from [McN05, Propositions 33, 52].

So we may assume $p$ is bad, and therefore that $G$ is exceptional. As before,
the orbits of $\mathcal{N}_{p}$ can be worked out from the tables in [Ste16]
and there are not very many. By inspection, it follows that the label of every
restricted nilpotent class is denoted by sums of $A_{r}$ for $r<p$ and
$D_{4}(a_{1})$ if $G=E_{8}$, $p=5$ or is $G_{2}(a_{1})$ when $G=G_{2}$, $p=3$;
note that the class $(\tilde{A}_{1})_{(3)}$ is excluded since it is not
contained in an $\mathfrak{sl}_{2}$-triple.

We first deal with the final case. The subsystem subgroup $A_{2}<G_{2}$
contains an $A_{2}$-irreducible subgroup $X$ of type $A_{1}$. By [Ste10a,
Theorem 1], all simple subgroups of $G_{2}$ are $G_{2}$-cr when $p=3$. The
restriction of the nontrivial $7$-dimensional $G_{2}$-module to $X$ is
$L(2)^{2}+L(0)$. It follows that the nilpotent elements contained in
$\operatorname{Lie}(X)$ have Jordan blocks of size $3^{2},1$ and thus are in
the $G_{2}(a_{1})$ class by [Ste16, Table 4].

In the remaining cases, every class is a distinguished element in
$\mathfrak{l}=\operatorname{Lie}(L)$ for some Levi subgroup $L$ with simple
factors only of type $A_{r}$ with $r<p$ or $D_{4}$. By [Ser05, Proposition
3.2], a subgroup $X$ of $L$ is $G$-cr if and only if it is $L$-cr. Furthermore
a subgroup $X$ of a central product $L=L_{1}L_{2}$ is $L$-cr if and only if
the projection of $X$ to both $L_{1}$ and $L_{2}$ is $L$-cr. Therefore, it
suffices to deal with the cases where $L$ is simple and simply connected of
type $A_{r}$ $(r<p)$ or $D_{4}$–but these cases have already been tackled. ∎

If $X$ is $G$-cr then so is $\operatorname{Lie}(X)$ by [McN07, Theorem 1]; so
we get the following.

######  Corollary 2.6.

Suppose $e\in\mathcal{N}_{p}$. Then there exists a $G$-cr subalgebra
$\mathfrak{s}\cong\mathfrak{sl}_{2}$ of $\mathfrak{g}$ containing $e$.

The following is used a couple of times, and is [McN07, Lemma 4].

######  Lemma 2.7.

Let $L$ be a Levi factor of a parabolic subgroup of $G$. Suppose that we have
a Lie subalgebra $\mathfrak{s}\subset\mathfrak{l}=\operatorname{Lie}(L)$. Then
$\mathfrak{s}$ is $G$-cr if and only if $\mathfrak{s}$ is $L$-cr.

######  Proposition 2.8.

Suppose $e\in\mathcal{N}$ is distinguished in a Levi subalgebra
$\mathfrak{l}=\operatorname{Lie}(L)$ with a factor of type $A_{p-1}$. Then
there is an $\mathfrak{sl}_{2}$-triple $(e,h,f)$ such that
$\mathfrak{s}:=\langle e,h,f\rangle$ is non-$G$-cr and $f\in\overline{L\cdot
e}$.

###### Proof.

By Lemma 2.7 it suffices to treat the case that $L=\mathrm{SL}(V)$ with $\dim
V=p$. In that case, let $\mathfrak{s}=\langle e,h,f\rangle$ be the image of
$\mathfrak{sl}_{2}$ under the representation given by the $p$-dimensional baby
Verma module $Z_{0}(0)$; cf. [Jan98, Section 5.4]. As $V\downarrow X=Z_{0}(0)$
is a non-trivial extension of the irreducible module $L(p-2)$ by the trivial
module we have that $\mathfrak{s}$ is not $L$-cr. It is easy to see that one
of $e$ or $f$ has a full Jordan block on $V$ and is therefore regular. But the
whole of $\mathcal{N}(L)$ is the closure of a regular nilpotent element so we
are done. ∎

######  Lemma 2.9.

Let $(e,h,f)$ be an $\mathfrak{sl}_{2}$-triple with $e,f\in\mathcal{N}$.
Suppose that $e$ and $f$ are distinguished in Levi subalgebras of
$\mathfrak{g}$ with no factors of type $A_{p-1}$. If $\mathfrak{s}:=\langle
e,f\rangle$ is $G$-cr then $\mathfrak{s}$ is a $p$-subalgebra.

###### Proof.

Suppose $\mathfrak{s}$ is not a $p$-subalgebra. Then by [ST18, Lemma 4.3],
$\mathfrak{s}$ must be $L$-irreducible in a Levi subalgebra
$\mathfrak{l}=\operatorname{Lie}(L)$ with $L=L_{1}L_{2}\ldots L_{r}$ and
$L_{1}$ of type $A_{p-1}$, say. Therefore, the projection
$\overline{\mathfrak{s}}$ of $\mathfrak{s}$ to
$\mathfrak{l}_{1}=\operatorname{Lie}(L_{1})$ is also $L_{1}$-irreducible, so
that $\overline{\mathfrak{s}}$ acts irreducibly on the $p$-dimensional natural
$L_{1}$-module. However, the classification of $p$-dimensional irreducible
$\mathfrak{sl}_{2}$-modules in [Jan98, Section 5.4] shows that the image of
$e$ or $f$ in $\overline{\mathfrak{s}}$ is regular in $L_{1}$, a
contradiction. ∎

##  3\. Monogamy of $\mathcal{V}$

We start with an observation that $\mathcal{V}$ can be characterised using the
following partial order on $\mathcal{N}$.

######  Definition 3.1.

Let $x,y\in\mathcal{N}$. We say $x\preceq y$ (resp. $x\prec y$) if
$\operatorname{rank}(\operatorname{ad}(x)^{p-1})\leq\operatorname{rank}(\operatorname{ad}(y)^{p-1})$
(resp.
$\operatorname{rank}(\operatorname{ad}(x)^{p-1})<\operatorname{rank}(\operatorname{ad}(y)^{p-1})$).

Note that $\operatorname{rank}(\operatorname{ad}(x)^{p-1})$ can be calculated
from the adjoint Jordan blocks of $x$ of size at least $p$, and if $G$ is
exceptional, this can be done by reference to [Ste16, Section 3.1]. Now if $G$
is classical and $e,f\in\mathcal{V}$ one always has $e\in G\cdot f$.
Therefore, the next lemma follows from a simple case-by-case check, using
Tables 1 & 2, the Hasse diagrams for nilpotent orbit closures and [Ste16,
Section 3.1].

######  Lemma 3.2.

Let $x,y\in\mathcal{N}$ such that $x\in\mathcal{V}$, and $y\notin\mathcal{V}$.
Then $x\prec y$.

######  Remark 3.3.

Comparing ranks of $(p-1)$-th powers is necessary for the partial order to
differentiate nilpotent orbits contained in $\mathcal{V}$. For example, let be
$G$ of type $E_{6}$, $p=5$, and take $x,y\in\mathcal{N}$ to be representatives
of the $D_{4}(a_{1})$ and $A_{4}$ classes respectively. Then we have
$x\in\mathcal{V}$ and $y\notin\mathcal{V}$. Using [Ste16, Table 16] we see
that
$\operatorname{rank}(\operatorname{ad}(x))=\operatorname{rank}(\operatorname{ad}(y))=78$,
however
$\operatorname{rank}(\operatorname{ad}(x)^{p-1})=11<15=\operatorname{rank}(\operatorname{ad}(y)^{p-1})$.

Let $\mathfrak{X}\subseteq\mathcal{N}$. We say that $\mathfrak{X}$ is
_partially monogamous_ if the following holds.

Whenever $(e,h,f)$ and $(e,h^{\prime},f^{\prime})$ are two
$\mathfrak{sl}_{2}$-triples with $e,f,f^{\prime}\in\mathfrak{X}$ and
$f,f^{\prime}\preceq e$, then $f$ and $f^{\prime}$ are conjugate under the
action of $C_{G}(e)$.

######  Lemma 3.4.

Let $\mathfrak{X}$ be a subvariety of $\mathcal{N}_{p}$. Then $\mathfrak{X}$
is monogamous if and only if it is partially monogamous.

###### Proof.

One direction is trivial. Suppose $\mathfrak{X}$ is partially monogamous but
not monogamous. Then there exist $\mathfrak{sl}_{2}$-triples $(e,h,f)$ and
$(e,h^{\prime},f^{\prime})$ with $e,f,f^{\prime}\in\mathfrak{X}$ such that
$(e,h,f)$ is not $C_{G}(e)$-conjugate to $(e,h^{\prime},f^{\prime})$. Since
$\mathfrak{X}$ is partially monogamous it follows that either $f\not\preceq e$
or $f^{\prime}\not\preceq e$; without loss of generality we assume the former.
Thus
$\operatorname{rank}(\operatorname{ad}(e)^{p-1})<\operatorname{rank}(\operatorname{ad}(f)^{p-1})$,
and in particular, $e$ and $f$ are not conjugate.

Let $(f,\tilde{h},\tilde{e})$ be an $\mathfrak{sl}_{2}$-triple with $f$
conjugate to $\tilde{e}$, which exists by Proposition 2.5. Then the two
$\mathfrak{sl}_{2}$-triples $(f,-h,e)$ and $(f,\tilde{h},\tilde{e})$ satisfy
$f,e,\tilde{e}\in\mathfrak{X}$ and $e,\tilde{e}\preceq f$. But as
$\mathfrak{X}$ is partially monogamous, we must have that $f$ conjugate to
$\tilde{e}$, which is in turn conjugate to $e$, a contradiction. ∎

Theorem 1.1 for classical types follows from Lemma 2.2 and the main theorem of
[GP22]. For the remainder of this section we suppose $G$ is of exceptional
type.

###  3.1. Bad characteristic

We first treat the case when $p$ is bad for $G$. Fix $0\neq e\in\mathcal{V}$
for the remainder of this section. We use the representatives as in [LS04],
presented in [Ste16]. If $G$ is of type $G_{2}$ and $p=3$, then the element
$e$ with label $(\tilde{A}_{1})_{(3)}$ cannot be extended to an
$\mathfrak{sl}_{2}$-triple by [ST18, Theorem 1.7]. So we exclude that case
from now on.

######  Lemma 3.5.

The normaliser $N_{G}(\langle e\rangle)$ (and centraliser $C_{G}(e)$) is
smooth if and only if the class of $e$ does not occur in the following table.

$G$ | $p$ |  class of $e$
---|---|---
$G_{2}$ | $3$ | $G_{2}(a_{1})$
$F_{4}$ | $3$ |  $F_{4}$, $\tilde{A}_{2}A_{1}$
$E_{6}$ | $3$ |  $E_{6}$, $E_{6}(a_{1})$, $E_{6}(a_{3})$, $A_{5}$, $A_{2}^{2}A_{1}$, $A_{2}^{2}$
$E_{8}$ | $3$ |  $E_{8}$, $E_{8}(a_{1})$, $E_{8}(a_{3})$, $E_{7}$, $E_{6}A_{1}$, $E_{8}(b_{6})$, $A_{7}$, $E_{6}$, $E_{6}(a_{3})A_{1}$, $A_{5}A_{1}$, $A_{2}^{2}A_{1}^{2}$, $A_{2}^{2}A_{1}$
| $5$ |  $E_{8}$, $A_{4}A_{3}$

###### Proof.

Every element $e$ has a cocharacter $\tau$ for which $\operatorname{im}(\tau)$
is contained in $N_{G}(\langle e\rangle)$ but not $C_{G}(e)$. Therefore, the
dimension of $N_{G}(\langle e\rangle)$ is precisely $\dim C_{G}(e)+1$.
Similarly, $\dim\mathfrak{n}_{\mathfrak{g}}(\langle
e\rangle)=\dim\mathfrak{c}_{g}(\langle e\rangle)+1$ thanks to the existence of
$\mathfrak{sl}_{2}$-triples. Therefore $N_{G}(\langle e\rangle)$ is smooth
precisely when $C_{G}(e)$ is smooth.

It is straightforward to use Magma to calculate the dimension of
$\mathfrak{c}_{\mathfrak{g}}(e)$. Comparing these dimensions with the
dimension of $C_{G}(e)$ presented in [LS04, Tables 22.1.1–22.1.5] completes
the proof. ∎

We may now deduce an important reduction.

######  Proposition 3.6.

There exists an $\mathfrak{sl}_{2}$-triple $(e,\overline{h},\overline{f})$
with $\overline{f}\in\mathcal{V}$ and
$\overline{h}\in\mathfrak{t}=\operatorname{Lie}(T)$ . Moreover, if $(e,h,f)$
is also an $\mathfrak{sl}_{2}$-triple then $h$ is $C_{G}(e)$-conjugate to
$\overline{h}$.

###### Proof.

We know from Proposition 2.5 that there is an $\mathfrak{sl}_{2}$-triple
$(e,\overline{h},\overline{f})$ with $\overline{f}$ in the same nilpotent
class as $e$. By Lemma 3.5, the group $N_{G}(\langle e\rangle)$ is smooth.
Therefore, all maximal tori in $\mathfrak{n}_{\mathfrak{g}}(\langle e\rangle)$
are $N_{G}(\langle e\rangle)$-conjugate. A computation in Magma shows that
$\mathfrak{n}_{\mathfrak{g}}(\langle e\rangle)\cap\mathfrak{t}$ is a maximal
torus of $\mathfrak{n}_{\mathfrak{g}}(\langle e\rangle)$. So we may assume
that $\overline{h}$ is contained in $\mathfrak{t}$.

For the final part, first note that since $[h,e]=2e$ we have
$[h^{[p]},e]=\operatorname{ad}(h)^{p}e=2e$ thanks to Fermat’s Little Theorem.
Therefore $\mathfrak{h}=\langle h^{{[p]}^{r}}\mid r=0,1,\ldots\rangle$ is an
abelian $p$-closed subalgebra of $\mathfrak{n}_{\mathfrak{g}}(\langle
e\rangle)$. It follows from [SF88, Chapter 2, Corollary 4.2] that
$\mathfrak{h}=\mathfrak{t}^{\prime}\oplus\mathfrak{n}^{\prime}$ where
$\mathfrak{t}^{\prime}$ is the set of semisimple elements of $\mathfrak{h}$.
Since $\mathfrak{t}^{\prime}$ is a torus, the above argument shows that up to
$N_{G}(\langle e\rangle)$-conjugacy we may assume that $\mathfrak{t}^{\prime}$
is contained in $\mathfrak{t}$. In particular,
$\overline{h}\in\mathfrak{t}^{\prime}$.

Because $\mathfrak{c}_{\mathfrak{g}}(\langle e\rangle)$ has codimension $1$ in
$\mathfrak{n}_{\mathfrak{g}}(\langle e\rangle)$ and
$\overline{h}\not\in\mathfrak{c}_{\mathfrak{g}}(\langle e\rangle)$ we see that
the torus $\mathfrak{t}^{\prime}$ decomposes as
$\mathfrak{t}^{\prime}=\mathfrak{c}_{\mathfrak{t}^{\prime}}(e)\oplus\langle\overline{h}\rangle$.
Furthermore, $\mathfrak{n}^{\prime}\subset\mathfrak{c}_{\mathfrak{g}}(\langle
e\rangle)$. It follows that $h=\overline{h}+h^{\prime}$ for some
$h^{\prime}\in\mathfrak{c}_{\mathfrak{g}}(e)\cap\mathfrak{c}_{\mathfrak{g}}(\overline{h})$.

Since $h=[e,f]$ and $\overline{h}=[e,\overline{f}]$ we also have
$h^{\prime}\in\operatorname{im}(\operatorname{ad}(e))$. Thus

$h^{\prime}\in W=\mathfrak{c}_{\mathfrak{g}}(\langle
e,h\rangle)\cap\operatorname{im}(\operatorname{ad}(e)).$

Another Magma check shows that every element in $W$ is $p$-nilpotent.

In particular, all eigenvalues of $h^{\prime}$ are $0$. Since
$h=\overline{h}+h^{\prime}$ and $[h,f]=-2f$ we must have
$[\overline{h},f]=-2f$. Therefore, $f\in
F=\ker(\operatorname{ad}(\overline{h})+2I_{\dim\mathfrak{g}})$ and so
$h=[e,f]\in\operatorname{im}(\operatorname{ad}(e))(F)$. Note that
$\overline{f}\in F$ also, so
$\overline{h}\in\operatorname{im}(\operatorname{ad}(e))(F)$ and hence
$h^{\prime}\in\operatorname{im}(\operatorname{ad}(e))(F)$.

Thus $h^{\prime}\in W\cap\operatorname{im}(\operatorname{ad}(e))(F)$. A final
easy check in Magma shows that
$W\cap\operatorname{im}(\operatorname{ad}(e))(F)=0$, as required. ∎

We now describe an ad-hoc method to prove that if $(e,h,f^{\prime})$ is an
$\mathfrak{sl}_{2}$-triple with $f^{\prime}\in\mathcal{V}$ and
$f^{\prime}\preceq e$ then $f^{\prime}$ is uniquely determined up to
$C:=(C_{G}(e)\cap C_{G}(h))$-conjugacy. In principle, this can be implemented
by hand, but for speed and accuracy we have used the Magma algebra system.
Applying Proposition 3.6 and Lemma 3.4 then completes the proof that
$\mathcal{V}$ is monogamous.

Setup:

By Proposition 3.6, there exists an $\mathfrak{sl}_{2}$-triple $(e,h,f)$ with
$h\in\mathfrak{t}=\operatorname{Lie}(T)$ and $f\in\mathcal{V}$ in the same
nilpotent class as $e$. Let $(e,h,f^{\prime})$ be an
$\mathfrak{sl}_{2}$-triple with $f^{\prime}\in\mathcal{V}$ and
$f^{\prime}\preceq e$. Since

(1) $[h,f^{\prime}]=-2f^{\prime}$

we have $f^{\prime}\in F:=\ker(\operatorname{ad}(h)+2I_{\dim(\mathfrak{g})})$.
We setup a generic element of the subspace $F$, namely $\tilde{f}=\sum
x_{i}v_{i}\in\mathfrak{g}$ where the $x_{i}$ are variables and
$v_{1},\ldots,v_{\dim(F)}$ is a basis for $F$. One can view $\tilde{f}$ as
describing a $\dim(F)$-dimensional subvariety $\mathcal{F}$ of $\mathfrak{g}$.

Step 1: The equation

(2) $[e,\tilde{f}]=h,$

yields a set of linear equations among the $x_{i}$. We use these to constrain
$\tilde{f}$ and thus reduce the dimension of $\mathcal{F}$. Now every element
of $\mathcal{F}$ forms an $\mathfrak{sl}_{2}$-triple with $e$.

######  Example 3.7.

We give an example where Step 1 is sufficient. Let $G$ be of type $E_{7}$,
$p=3$ and $e=e_{\alpha_{2}}+e_{\alpha_{5}}+e_{\alpha_{7}}$. Then $e$ is a
representative of the $(A_{1}^{3})^{(1)}$ orbit and $e\in\mathcal{V}$ by Lemma
2.2. On this occasion it is obvious that $(e,h,f)$ is an
$\mathfrak{sl}_{2}$-triple with $h=h_{2}+h_{5}+h_{7}\in\mathfrak{t}$ and
$f=e_{-\alpha_{2}}+e_{-\alpha_{5}}+e_{-\alpha_{7}}$.

Let $F:=\ker(\operatorname{ad}(h)+2I_{\dim(\mathfrak{g})})$. A straightforward
calculation shows that the space $F$ is $27$-dimensional with a basis of root
vectors $v_{1}=e_{r_{1}},\ldots,v_{27}=e_{r_{27}}$ for some set of roots
$r_{1},\ldots,r_{27}$; in particular $r_{12}=-\alpha_{2}$,
$r_{13}=-\alpha_{5}$ and $r_{14}=-\alpha_{7}$.

We let $\tilde{f}=\sum_{i}x_{i}v_{i}$ as above. We then compute
$[e,\tilde{f}]=h$. For $i\neq 12,13,14$ we find that the left hand side has a
coordinate of the form $\lambda x_{i}$ for $\lambda=1$ or $2$. Thus $x_{i}=0$
for $i\neq 12,13,14$. On the other hand the coordinate of $h_{2}$ is seen to
be equal to $x_{14}+2$. Thus $x_{14}$ must be $1$. Similarly, the coordinates
of $h_{5}$ and $h_{7}$ are $x_{13}+2$ and $x_{12}+2$, respectively. We have
therefore determined all the variables in $\tilde{f}$ and found it must be
$f$, as required.

Step 2: Consider the action of $C$ on $\mathcal{F}$ by applying elements $g\in
C$ to $\tilde{f}$. Find a set of variables $\\{x_{i}\mid i\in Z\\}$ such that
every $C$-orbit in $\mathcal{F}$ contains a representative with $x_{i}=0$ for
$i\in Z$. Thus we may assume that these variables are zero in $\tilde{f}$,
further reducing $\mathcal{F}$.

######  Example 3.8.

We give an example where Steps 1 and 2 are sufficient. Let $G$ be of type
$G_{2}$ and $p=3$. Consider $e=e_{10}$ which is a representative of the
$\tilde{A_{1}}$ orbit, thus contained in $\mathcal{V}$ by Lemma 2.2.

Clearly, if $h=h_{10},f=e_{-10}$, then $(e,h,f)$ is an
$\mathfrak{sl}_{2}$-triple with $f\in\mathcal{V}$. Define
$F:=\ker(\operatorname{ad}(h)+2I_{\dim(\mathfrak{g})})$. This is 3-dimensional
and we build $\tilde{f}$ as above:

$\tilde{f}=x_{1}e_{-11}+x_{2}e_{-10}+x_{3}e_{21}.$

After Step 1 we find

$\tilde{f}=x_{1}e_{-11}+e_{-10}+x_{3}e_{21}.$

Now we apply elements of $C=C_{G}(e)\cap C_{G}(h)$ to $\tilde{f}$. First
consider $x_{-01}(t)\in C$. We calculate that

$x_{-01}(t)\cdot\tilde{f}=(t+x_{1})e_{-11}+e_{-10}+x_{3}e_{21}.$

Therefore, by setting $t=-x_{1}$, we see that every $C$-orbit in $\mathcal{F}$
contains a representative with $x_{1}=0$. We’re down to

$\tilde{f}=e_{-10}+x_{3}e_{21}.$

Finally, conjugation by $x_{31}(t)$ sends $\tilde{f}$ to
$e_{-10}+(t+x_{3})e_{21}$. Thus we may conclude that $\tilde{f}=f$, as
required.

Step 3: Finally, we impose the condition that $\tilde{f}$ should represent an
element $f^{\prime}\in\mathcal{V}$ with $f^{\prime}\preceq e$. In particular,
every element in $\mathcal{V}$ is $p$-nilpotent. Therefore, the equation

(3) $\operatorname{ad}(\tilde{f})^{p}=0.$

yields further polynomial equations we want the $x_{i}$ to satisfy.

Forcing $\mathcal{F}$ to only contain elements $f^{\prime}$ with
$f^{\prime}\preceq e$ is slightly more subtle since we cannot simply calculate
the ‘rank’ of $M=\operatorname{ad}(\tilde{f})^{p-1}$. Let
$R=\text{rank}(\operatorname{ad}(e)^{p-1})$ and $\epsilon$ be a map evaluating
the remaining variables to choices in $\Bbbk$ (so each
$f^{\prime}\in\mathcal{F}$ is simply some $\epsilon(\tilde{f})$). We find a
subset $r_{1},\ldots,r_{R}$ of rows and subset $c_{1},\ldots,c_{R}$ of columns
such that, up to the reordering of rows and columns, the corresponding
submatrix $S$ of $M$ is upper triangular and all diagonal entries are elements
of $\mathbb{F}_{p}^{*}$. Then any element $f^{\prime}\in\mathcal{F}$ will
satisfy $\operatorname{rank}(\operatorname{ad}(f^{\prime})^{p-1})\geq R$. We
only want those elements $f^{\prime}\preceq e$ which means
$\operatorname{rank}(\operatorname{ad}(f^{\prime})^{p-1})\leq R$. Thus, given
any row $r$ of $M$ we must have $\epsilon(r)$ is in the span of
$\epsilon(r_{1}),\ldots,\epsilon(r_{R})$. In particular, a row $r^{\prime}$ of
$M$ with zeroes at all columns $c_{1},\ldots,c_{R}$ must evaluate to zero.
This final set of conditions is enough to force all remaining variables to be
$0$.

######  Example 3.9.

We give an example where we require Step 3. Let $G$ be of type $G_{2}$ and
$p=3$. Consider $e=e_{01}$ which is a representative of the $A_{1}$ orbit,
thus contained in $\mathcal{V}$ by Lemma 2.2.

Take $h=h_{01},f=e_{-01}$, then $(e,h,f)$ is an $\mathfrak{sl}_{2}$-triple in
$\mathfrak{g}$ with $f\in\mathcal{V}$. Define
$F:=\ker(\operatorname{ad}(h)+2I_{\dim(\mathfrak{g})})$. This is 5-dimensional
and we build $\tilde{f}$ as above:

$\tilde{f}=x_{1}e_{-32}+x_{2}e_{-01}+x_{3}e_{-10}+x_{4}e_{11}+x_{5}e_{32}.$

After Step 1 we find

$\tilde{f}=x_{1}e_{-32}+e_{-01}+x_{3}e_{-10}+x_{4}e_{11}+x_{5}e_{32}.$

There are no elements of $C=C_{G}(e)\cap C_{G}(h)$ which we can use to reduce
$\tilde{f}$, so we move onto Step 3.

The equation $\operatorname{ad}(\tilde{f})^{p}=0$ gives many relations amongst
the remaining variables but none that allow us to conveniently reduce
$\tilde{f}$. Consider the matrix $M=\operatorname{ad}(\tilde{f})^{p-1}$. The
first, eighth, tenth and thirteenth column of $M$ consist only of zeroes, so
we remove them, leaving the matrix $M^{\prime}$ as follows.

$\small\begin{pmatrix}x_{1}x_{5}&0&0&x_{5}&2x_{4}^{2}&0&0&x_{4}x_{5}&0&x_{5}^{2}\\\
0&2x_{4}&0&0&0&x_{5}&0&0&0&0\\\ 0&0&2x_{4}&0&0&0&x_{5}&0&0&0\\\
0&0&0&0&2x_{1}x_{5}+x_{3}x_{4}&0&0&x_{3}x_{5}+x_{4}^{2}&0&0\\\
0&2x_{1}x_{4}+2x_{3}^{2}&0&0&0&x_{1}x_{5}+2x_{3}x_{4}&0&0&x_{3}x_{5}+x_{4}^{2}&0\\\
0&x_{3}&0&0&0&2x_{4}&0&0&x_{5}&0\\\ 0&0&x_{3}&0&0&0&x_{4}&0&0&0\\\
0&0&0&0&0&0&0&0&0&0\\\
0&0&0&0&x_{1}x_{4}+x_{3}^{2}&0&0&2x_{1}x_{5}+x_{3}x_{4}&0&0\\\
x_{1}&0&0&1&x_{3}&0&0&x_{4}&0&x_{5}\\\ 0&2x_{1}&0&0&0&2x_{3}&0&0&2x_{4}&0\\\
0&0&2x_{1}&0&0&0&x_{3}&0&0&0\\\
x_{1}^{2}&0&0&x_{1}&x_{1}x_{3}&0&0&2x_{3}^{2}&0&x_{1}x_{5}\\\
0&0&0&0&0&x_{1}&0&0&x_{3}&0\end{pmatrix}$

We calculate that $R=\text{rank}(\operatorname{ad}(e)^{p-1})=1$. Therefore, if
$\epsilon(\tilde{f})=f^{\prime}\preceq e$ for some evaluation map $\epsilon$,
we must have that the rank of $\epsilon(M^{\prime})$ is at most one. Observe
that $M^{\prime}_{10,4}=1$ and so the rank of $\epsilon(M^{\prime})$ is at
least one. It follows that every row of $\epsilon(M^{\prime})$ must be a
multiple of the tenth row of $\epsilon(M^{\prime})$.

Consider the sixth row of $M^{\prime}$. This only has nonzero entries in
columns 2, 6 and 9, namely $x_{3},2x_{4}$ and $x_{5}$. Since the tenth row is
zero in columns 2, 6 and 9 it must be the case that the sixth row of
$\epsilon(M^{\prime})$ is zero. Hence $x_{3}=x_{4}=x_{5}=0$.

Similarly, row 11 of $\epsilon(M^{\prime})$ must be zero. Thus $x_{1}=0$, and
we may conclude that $\tilde{f}=f$.

###  3.2. Good characteristic

Suppose $p$ is a good prime for $G$. As in the bad characteristic case, we
describe an algorithm to deduce that $\mathcal{V}$ is monogamous. In good
characteristic there is a considerable amount of theory at our disposal. In
particular, every $e\in\mathcal{N}$ has an associated cocharacter: that is a
homomorphism $\tau:{\mathbb{G}}_{m}\to G$ such that under the adjoint action,
we have $\tau(t)\cdot e=t^{2}e$ and $\tau$ evaluates in the derived subgroup
of the Levi subgroup in which $e$ is distinguished.

######  Lemma 3.10.

Suppose $p$ is good for $G$. Let $(e,h_{1},f_{1})$ and $(e,h_{2},f_{2})$ be
$\mathfrak{sl}_{2}$-triples with $e,f_{1},f_{2}\in\mathcal{V}$. Then we may
assume that $h_{1}=h_{2}=h$, up to $C_{G}(e)$-conjugacy. More precisely, there
exists a cocharacter $\tau$ associated to $e$ such that
$\operatorname{Lie}(\tau({\mathbb{G}}_{m}))=\langle h\rangle$. Furthermore if
$\mathfrak{g}=\bigoplus_{i}\mathfrak{g}(i)$ is the grading of $\mathfrak{g}$
with respect to $\tau$ we have

$f-f^{\prime}\in\bigoplus_{r>0}\mathfrak{g}_{e}(-2+rp),$

where
$\mathfrak{g}_{e}(i):=\mathfrak{c}_{\mathfrak{g}}(e)\cap\mathfrak{g}(i)$.

###### Proof.

We start by proving that $h_{i}$ is toral. By Lemma 2.9, the subalgebra
$\mathfrak{s}_{i}=\langle e,h_{i},f_{i}\rangle$ is either a $p$-subalgebra or
non-$G$-cr. In the former case, we are done. In the latter case, the argument
in the proof of [ST18, Lemma 6.1] applies, showing $h_{i}$ is toral.

Now we apply [ST18, Proposition 2.8]. This yields cocharacters $\tau_{i}$
associated to $e$ such that
$\operatorname{Lie}(\tau_{i}({\mathbb{G}}_{m}))=\langle h_{i}\rangle$. By
[Jan03, Lemma 5.3], any two cocharacters associated to $e$ are
$C_{G}(e)$-conjugate. Therefore, $h_{1}$ and $h_{2}$ are $C_{G}(e)$-conjugate
and so up to $C_{G}(e)$-conjugacy we may assume they are equal. Set
$h=h_{1}=h_{2}$.

Since $[e,f_{1}-f_{2}]=h-h=0$ we know
$f_{1}-f_{2}\in\mathfrak{c}_{\mathfrak{g}}(e)$. Furthermore,
$[h,f_{1}-f_{2}]=-2(f_{1}-f_{2})$ and hence
$f_{1}-f_{2}\in\bigoplus_{r}\mathfrak{g}(-2+rp)$. The conclusion follows by
noting that $\mathfrak{c}_{\mathfrak{g}}(e)$ is contained in the nonnegative
graded part of $\mathfrak{g}$. ∎

Fix $0\neq e\in\mathcal{V}$ for the remainder of this section. Choose a
cocharacter $\tau$ associated to $e$ such that
$h\in\operatorname{Lie}(\tau({\mathbb{G}}_{m}))\subset\mathfrak{t}$ with
$[h,e]=2e$. In practice, we use the representatives and associated
cocharacters given in [LT11]. We know from Pommerening [Pom77, Pom80] and
Lemma 3.10 that there exists a unique $\overline{f}\in\mathfrak{g}(-2)$ such
that $(e,h,\overline{f})$ is an $\mathfrak{sl}_{2}$-triple. Furthermore, if
$(e,h,f)$ is another $\mathfrak{sl}_{2}$-triple then
$f=\overline{f}+f^{\prime}$ with
$f^{\prime}\in\bigoplus_{r>0}\mathfrak{g}_{e}(-2+rp)$. Therefore, we need to
prove that if $f\in\mathcal{V}$ then up to $C=C_{G}(e)\cap C_{G}(h)$-conjugacy
we have $f=\overline{f}$, i.e. that $f^{\prime}=0$.

To do this we use the ad-hoc method from Section 3.1. Indeed, by Lemma 3.4 it
suffices to prove that $f=\overline{f}$ when $f\preceq e$. We now apply Steps
1–3 starting with the space $F=f+\bigoplus_{r>0}\mathfrak{g}_{e}(-2+rp)$.

######  Example 3.11.

We give a final example, this time in good characteristic. Let $G$ be of type
$E_{7}$ and $p=7$. Consider $e=e_{\begin{subarray}{c}1\\\ \
\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
0\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \
\end{subarray}}+e_{\begin{subarray}{c}0\\\ \
\end{subarray}\begin{subarray}{c}1\\\ \ \end{subarray}\begin{subarray}{c}0\\\
0\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \
\end{subarray}}+e_{\begin{subarray}{c}0\\\ \
\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}1\\\
0\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \
\end{subarray}}+e_{\begin{subarray}{c}0\\\ \
\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
0\end{subarray}\begin{subarray}{c}1\\\ \ \end{subarray}\begin{subarray}{c}0\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \
\end{subarray}}+e_{\begin{subarray}{c}0\\\ \
\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
0\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}1\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}}$ which is a
representative of the $(A_{5})^{(2)}$ orbit; thus $e\in\mathcal{V}$ by Lemma
2.2. Furthermore, by [LT11, p. 108], $e$ has the associated cocharacter
$\tau={\footnotesize\begin{array}[]{c c r c c c c}2&2&2&2&2&-5\\\
&&-9\end{array}}.$ It follows that
$h=2h_{1}+6h_{3}+5h_{4}+6h_{5}+2h_{6}\in\operatorname{Lie}(\tau(\mathbb{G}_{m}))$.

The unique $\overline{f}\in\mathfrak{g}(-2)$ such that $(e,h,\overline{f})$ is
an $\mathfrak{sl}_{2}$-triple is then given by
$\overline{f}=2e_{\begin{subarray}{c}-1\\\ \
\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
0\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \
\end{subarray}}+6e_{\begin{subarray}{c}-0\\\ \
\end{subarray}\begin{subarray}{c}1\\\ \ \end{subarray}\begin{subarray}{c}0\\\
0\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \
\end{subarray}}+5e_{\begin{subarray}{c}-0\\\ \
\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}1\\\
0\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \
\end{subarray}}+6e_{\begin{subarray}{c}-0\\\ \
\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
0\end{subarray}\begin{subarray}{c}1\\\ \ \end{subarray}\begin{subarray}{c}0\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \
\end{subarray}}+2e_{\begin{subarray}{c}-0\\\ \
\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
0\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}1\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}}$.

Let $F=f+\bigoplus_{r>0}\mathfrak{g}_{e}(-2+rp)$, which is $6$-dimensional. We
build a generic element $\tilde{f}$ of $F$ as in Section 3.1 with six
variables. Following Step 1 by enforcing the linear equations from
$[e,\tilde{f}]=h$ yields

$\tilde{f}=\overline{f}+x_{1}e_{\begin{subarray}{c}-1\\\ \
\end{subarray}\begin{subarray}{c}2\\\ \ \end{subarray}\begin{subarray}{c}3\\\
2\end{subarray}\begin{subarray}{c}2\\\ \ \end{subarray}\begin{subarray}{c}1\\\
\ \end{subarray}\begin{subarray}{c}1\\\ \
\end{subarray}}+x_{2}e_{\begin{subarray}{c}-0\\\ \
\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}1\\\
1\end{subarray}\begin{subarray}{c}1\\\ \ \end{subarray}\begin{subarray}{c}0\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \
\end{subarray}}+x_{2}e_{\begin{subarray}{c}-0\\\ \
\end{subarray}\begin{subarray}{c}1\\\ \ \end{subarray}\begin{subarray}{c}1\\\
1\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \
\end{subarray}}+x_{3}e_{\begin{subarray}{c}-0\\\ \
\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
0\end{subarray}\begin{subarray}{c}0\\\ \ \end{subarray}\begin{subarray}{c}0\\\
\ \end{subarray}\begin{subarray}{c}1\\\ \
\end{subarray}}+x_{4}e_{\begin{subarray}{c}1\\\ \
\end{subarray}\begin{subarray}{c}1\\\ \ \end{subarray}\begin{subarray}{c}1\\\
0\end{subarray}\begin{subarray}{c}1\\\ \ \end{subarray}\begin{subarray}{c}1\\\
\ \end{subarray}\begin{subarray}{c}1\\\ \
\end{subarray}}-x_{5}e_{\begin{subarray}{c}1\\\ \
\end{subarray}\begin{subarray}{c}2\\\ \ \end{subarray}\begin{subarray}{c}2\\\
1\end{subarray}\begin{subarray}{c}1\\\ \ \end{subarray}\begin{subarray}{c}1\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \
\end{subarray}}+x_{5}e_{\begin{subarray}{c}1\\\ \
\end{subarray}\begin{subarray}{c}1\\\ \ \end{subarray}\begin{subarray}{c}2\\\
1\end{subarray}\begin{subarray}{c}2\\\ \ \end{subarray}\begin{subarray}{c}1\\\
\ \end{subarray}\begin{subarray}{c}0\\\ \
\end{subarray}}+x_{6}e_{\begin{subarray}{c}2\\\ \
\end{subarray}\begin{subarray}{c}3\\\ \ \end{subarray}\begin{subarray}{c}4\\\
2\end{subarray}\begin{subarray}{c}3\\\ \ \end{subarray}\begin{subarray}{c}2\\\
\ \end{subarray}\begin{subarray}{c}1\\\ \ \end{subarray}}.$

On this occasion $C:=C_{G}(e)\cap C_{G}(h)$ is finite and we move onto Step 3.

Let $M=\operatorname{ad}(\tilde{f})^{p-1}$. We calculate that
$R=\text{rank}(\operatorname{ad}(e)^{p-1})=13$. So if
$\epsilon(\tilde{f})=f^{\prime}\preceq e$ for some evaluation map $\epsilon$,
we must have that the rank of $\epsilon(M)$ is at most 13.

Ordering the basis of $\mathfrak{g}$ as in Magma, we use the $13\times 13$
submatrix $S$ of $M$ corresponding to the rows $r$ and columns $c$ where

$\displaystyle r$
$\displaystyle=\\{75,125,62,94,87,129,120,97,42,82,23,34,108\\},$
$\displaystyle c$
$\displaystyle=\\{37,100,24,52,50,109,92,60,14,40,5,9,72\\}.$

The submatrix $S$ is upper triangular and all diagonal entries are elements of
$\mathbb{F}_{p}^{*}$. The only other nonzero entries in $S$ can be found in
row one, which is

$(1\ \ 0\ \ 4x_{2}\ \ 0\ \ 0\ \ 0\ \ 5x_{5}\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0).$

We find that 42 rows of $M$ have zero entries in every column in $c$, so each
of these rows must be zero. An example of such a row is the eighth row of $M$.
In row 8 we find $x_{4},3x_{5}$ and $-x_{6}$ in columns $11,15$ and $70$
respectively. It follows that $x_{4}=x_{5}=x_{6}=0$. Similarly the 133rd row
of $M$ then allows us to deduce that $x_{1}=x_{2}=x_{3}=0$. Thus $\tilde{f}=f$
as required.

##  4\. Proof of Theorems 1.1 and 1.2

Proposition 2.5 shows that for each $e\in\mathcal{V}$ there exists an
$\mathfrak{sl}_{2}$-triple $(e,h,f)$ with $\mathfrak{s}=\langle
e,h,f\rangle=\operatorname{Lie}(X)$ for a $G$-cr subgroup $X<G$ of type
$A_{1}$. Thus $f$ must be $G$-conjugate to $e$ and hence $f\in\mathcal{V}$. We
have demonstrated in Section 3 that any other $\mathfrak{sl}_{2}$-triple
$(e,h^{\prime},f^{\prime})$ with $f^{\prime}\in\mathcal{V}$ is
$C_{G}(e)$-conjugate to $(e,h,f)$. Therefore $\mathfrak{s}^{\prime}=\langle
e,h^{\prime},f^{\prime}\rangle$ is $G$-conjugate to $\mathfrak{s}$ and hence
$G$-cr.

Now all that remains is to prove that $\mathcal{V}$ is the maximal closed
$G$-stable subvariety of $\mathcal{N}$ satisfying both the monogamy and
$A_{1}$-$G$-cr conditions.

For $G$ of classical type, it follows from [GP22, Theorem 1] that
$\mathcal{V}$ is maximal with respect to being monogamous. For the
$A_{1}$-$G$-cr property, the ingredients are there but let us spell out the
details, as these essentially make up the strategy for the groups of
exceptional type used below.

######  Proposition 4.1.

Let $G$ be a simple algebraic group of classical type. Then $\mathcal{V}$ is
the maximal closed $G$-stable $A_{1}$-$G$-cr subvariety of $\mathcal{N}$.

###### Proof.

Suppose $\mathfrak{X}$ is a closed $G$-stable $A_{1}$-$G$-cr variety. Let
$e\in\mathfrak{X}\setminus\mathcal{V}$. If $e$ is distinguished in a Levi
subalgebra $\mathfrak{l}=\operatorname{Lie}(L)$ with $L$ having a factor of
type $A_{p-1}$ then Proposition 2.8 shows that $e$ is contained in an
$\mathfrak{sl}_{2}$-triple generating a non-$G$-cr subalgebra (these
non-$G$-cr subalgebras are also exhibited in [GP22, Section 2.4]).

Therefore we may assume that $e^{[p]}\neq 0$. The discussion before
Proposition 2.2 in _ibid._ exhibits an $\mathfrak{sl}_{2}$-triple $(e,h,f)$
with $f^{[p]}=0$ and $f$ in $\overline{G\cdot e}$, thus $f\in\mathfrak{X}$. By
Lemma 2.9, the $\mathfrak{sl}_{2}$-subalgebra generated by $(e,h,f)$ must be
$G$-cr. ∎

For the remainder of the section we assume $G$ is of exceptional type.

######  Proposition 4.2.

The variety $\mathcal{V}$ is the maximal closed $G$-stable subvariety of
$\mathcal{N}$ satisfying both the monogamy and $A_{1}$-$G$-cr conditions.

###### Proof.

Let $\mathfrak{X}$ be a $G$-stable closed subvariety of $\mathcal{N}$
satisfying either the monogamy or $A_{1}$-$G$-cr condition and
$\mathfrak{X}\not\subset\mathcal{V}$. By Lemma 2.4, we may assume that
$\mathfrak{X}$ contains an orbit from Table 3.

First suppose that there exists $e\in\mathfrak{X}$ which is distinguished in a
Levi subalgebra $\mathfrak{l}=\operatorname{Lie}(L)$ with a factor of type
$A_{p-1}$. Then Propositions 2.5 and 2.8 furnish us with two
$\mathfrak{sl}_{2}$-triples $(e,h,f)$ and $(e,h^{\prime},f^{\prime})$ such
that the first generates a $G$-cr subalgebra and the second generates a
non-$G$-cr subalgebra. Moreover, $f$ is in the same $G$-class as $e$ and
$f^{\prime}$ is in the closure of the $G$-class of $e$. Hence $\mathfrak{X}$
does not satisfy either condition, a contradiction.

Now suppose there is some minimal neighbour $e\in\mathfrak{X}$ such that
$e^{[p]}\neq 0$. By consideration of Table 3 we see in each such case, $e$ is
distinguished in some Levi subalgebra $\mathfrak{l}=\operatorname{Lie}(L)$ of
$\mathfrak{g}$ for which $p$ is good for $L$ and $L$ has no factors of type
$A_{p-1}$.

From [PS19, Section 2.4] we find an $\mathfrak{sl}_{2}$-triple $(e,h,f)$ of
$\mathfrak{l}$ with $f^{[p]}=0$. Since $L$ has no factor of type $A_{p-1}$ it
follows that $f\in\mathcal{V}\subseteq\mathfrak{X}$. Furthermore,
$\mathfrak{s}=\langle e,f\rangle\cong\mathfrak{sl}_{2}$ is a non-$L$-cr
subalgebra by Lemma 2.9. Hence by Lemma 2.7, $\mathfrak{X}$ does not satisfy
the $A_{1}$-$G$-cr condition. Proposition 2.5 yields an
$\mathfrak{sl}_{2}$-triple $(f,h^{\prime},e^{\prime})$ which generates a
$G$-cr $\mathfrak{sl}_{2}$-subalgebra. Therefore, $f$ is contained in two non-
conjugate $\mathfrak{sl}_{2}$-triples. Thus $\mathfrak{X}$ does not satisfy
the monogamy condition either.

Finally, let $G$ be of type $G_{2}$ and $p=3$. The only minimal neighbour of
$\mathcal{V}$ is the subregular orbit $G_{2}(a_{1})$. A representative for
this orbit is $e=e_{\alpha_{2}}+e_{-3\alpha_{1}-\alpha_{2}}$. This is regular
in $\mathfrak{m}=\operatorname{Lie}(M)$ where $M$ is the standard subsystem
subgroup of type $A_{2}$ corresponding to the subsystem
$\pm\alpha_{2},\pm(3\alpha_{1}+2\alpha_{2}),\pm(3\alpha_{1}+\alpha_{2})$.

As in the proof of Proposition 2.8, there exists an $\mathfrak{sl}_{2}$-triple
$(e,h,f)$ in $\mathfrak{m}$ such that $\mathfrak{s}=\langle e,f\rangle$ is
non-$M$-cr. Furthermore, $f$ is in the orbit labelled $A_{1}$ (both as an
$A_{2}$-orbit and $G_{2}$-orbit). We claim that $\mathfrak{s}$ is non-$G$-cr.
By Proposition 2.5, the element $f$ is contained in an
$\mathfrak{sl}_{2}$-triple generating a $G$-cr subalgebra and by the claim,
the $\mathfrak{sl}_{2}$-triple $(f,-h,e)$ generates a non-$G$-cr subalgebra.
Hence $\mathfrak{X}$ does not satisfy either condition.

For the claim, note that $\mathfrak{s}$ is certainly $G$-reducible since it is
non-$M$-cr. All $G$-cr $\mathfrak{sl}_{2}$-subalgebras which are $G$-reducible
are contained in a Levi subalgebra. In this low-rank case, it immediately
follows that all such $\mathfrak{sl}_{2}$-subalgebras are $G$-conjugate to
either $\mathfrak{l}_{1}=\langle e_{\pm\alpha_{1}}\rangle$ or
$\mathfrak{l}_{2}=\langle e_{\pm\alpha_{2}}\rangle$. Therefore a $G$-cr
$\mathfrak{sl}_{2}$-subalgebra only contains nilpotent elements in the $A_{1}$
or $\tilde{A}_{1}$ classes. The claim follows since $\mathfrak{s}$ contains
$e$ which is in the subregular class.

∎

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