Nilpotent Centralizers and Good Filtrations

Abstract

Let G be a connected reductive group over an algebraically closed field $\Bbbk$. Under mild restrictions on the characteristic of $\Bbbk$, we show that any G-module with a good filtration also has a good filtration as a module for the reductive part of the centralizer of a nilpotent element x in its Lie algebra.

1 Introduction

Let G be a connected reductive group over an algebraically closed field $\Bbbk$ of characteristic p > 0, and let H be a connected reductive subgroup. Recall that (G,H) is said to be a Donkin pair or a good filtration pair if every G-module with a good filtration still has a good filtration when regarded as an H-module.

Now let x be a nilpotent element in the Lie algebra of G, and let G^x ⊂ G be its stabilizer. If p is good for G, then the theory of associated cocharacters is available, and this gives rise to a decomposition

$$ G^{x} = {G}_{\text{red}}^{x} \ltimes {G}_{\text{unip}}^{x} $$

where ${G}_{\text {unip}}^{x}$ is a connected unipotent group, and ${G}_{\text {red}}^{x}$ is a (possibly disconnected) group whose identity component $({G}_{\text {red}}^{x})^{\circ }$ is reductive (cf. [11, 5.10]). The main result of this paper is the following.

Theorem 1.1

Let G be a connected reductive group over an algebraically closed field $\Bbbk$ of good characteristic. For any nilpotent element x in its Lie algebra, $(G, ({G}_{\text {red}}^{x})^{\circ })$ is a Donkin pair.

Now suppose that H ⊂ G is a possibly disconnected reductive subgroup, i.e., a group whose identity component H^∘ is reductive. If the characteristic of $\Bbbk$ does not divide the order of the finite group H/H^∘, then the category of finite-dimensional H-modules is a highest-weight category, as shown in [2]. In particular, it makes sense to speak of good filtrations for H-modules, and so the definition of “Donkin pair” makes sense for (G,H).

In order to apply this notion in the case where $H = {G}_{\text {red}}^{x}$, we must impose a slightly stronger condition on p: we require it to be pretty good in the sense of [9, Definition 2.11]. (In general, this condition is intermediate between “good” and “very good.” It coincides with “very good” for semisimple simply connected groups, whereas for GL_n, all primes are pretty good.) This is equivalent to requiring G to be standard in the sense of [14, §4]. It follows from [8, Theorem 1.8] and [3, Lemma 2.1] that when p is pretty good for G, it does not divide the order of $G^{x}/(G^{x})^{\circ } \cong {G}_{\text {red}}^{x}/({G}_{\text {red}}^{x})^{\circ }$ for any nilpotent element x. As an immediate consequence of Theorem 1.1 and Lemma 2.2 below, we have the following result.

Corollary 1.2

Let G be a connected reductive group over an algebraically closed field $\Bbbk$ of pretty good characteristic. For any nilpotent element x in its Lie algebra, $(G, {G}_{\text {red}}^{x})$ is a Donkin pair.

This corollary plays a key role in the proof of the Humphreys conjecture [1].

The paper is organized as follows: Section 2 contains some general lemmas on Donkin pairs, along with a lengthly list of examples (some previously known, and some new). Section 3 gives the proof of Theorem 1.1. The proof consists of a reduction to the quasi-simple case, followed by case-by-case arguments.

Remark 1.3

It would, of course, be desirable to have a uniform proof of Theorem 1.1 that avoids case-by-case arguments, perhaps using the method of Frobenius splittings. Thanks to a fundamental result of Mathieu [13], Theorem 1.1 would come down to showing that the flag variety of G admits a $({G}_{\text {red}}^{x})^{\circ }$-canonical splitting. According to a result of van der Kallen [16], this geometric condition is equivalent to a certain linear-algebraic condition (called the “pairing condition”) on the Steinberg modules for G and $({G}_{\text {red}}^{x})^{\circ }$. Unfortunately, for the moment, the pairing condition for these groups seems to be out of reach.

2 Preliminaries

2.1 General Lemmas on Donkin Pairs

We begin with three easy statements about good filtrations.

Lemma 2.1

Let H be a possibly disconnected reductive group over an algebraically closed field $\Bbbk$. Assume that the characteristic of $\Bbbk$ does not divide |H/H^∘|. An H-module M has a good filtration if and only if it has a good filtration as an H^∘-module.

Proof

According to [2, Eq. (3.3)], any costandard H-module regarded as an H^∘-module is a direct sum of costandard H^∘-modules. Hence, any H-module with a good filtration has a good filtration as an H^∘-module.

For the opposite implication, suppose M is an H-module that has a good filtration as an H^∘-module. To show that it has a good filtration as an H-module, we must show that ${\text {Ext}}_{H}^{1}({-},M)$ vanishes on standard H-modules. As explained in the proof of [2, Lemma 2.18], we have

$$ {\text{Ext}}_{H}^{1}({-},M) \cong ({\text{Ext}}_{H^{\circ}}^{1}({-},M))^{H/H^{\circ}}, $$

and the right-hand side clearly vanishes on standard H-modules (using [2, Eq. (3.3)] again). □

Lemma 2.2

Let G be a connected, reductive group, and let H ⊂ G be a possibly disconnected reductive subgroup. Assume that the characteristic of $\Bbbk$ does not divide |H/H^∘|. Then, (G,H) is a Donkin pair if and only if (G,H^∘) is a Donkin pair.

Proof

This is an immediate consequence of Lemma 2.1. □

Lemma 2.3

Let G be a connected, reductive group, and let $G^{\prime }$ be its derived subgroup. Let H ⊂ G be a connected, reductive subgroup. Then, (G,H) is a Donkin pair if and only if $(G^{\prime },(G^{\prime } \cap H)^{\circ })$ is a Donkin pair.

Proof

Let T ⊂ B ⊂ G denote a maximal torus and Borel subgroup, respectively, and suppose that $G^{\prime \prime }$ is any closed connected subgroup satisfying $G^{\prime } \subseteq G^{\prime \prime } \subseteq G$. Now let $T^{\prime \prime } = G^{\prime \prime }\cap T$, $B^{\prime \prime } = G^{\prime \prime } \cap B$, and observe that by [10, I.6.14(1)], we have

$$ {\text{Res}}_{G^{\prime\prime}}^{G} {\text{Ind}}_{B}^{G} M \cong {\text{Ind}}_{B^{\prime\prime}}^{G^{\prime\prime}} {\text{Res}}_{B^{\prime\prime}}^{B} M $$

(2.1)

for any B-module M. Thus, for any dominant weight λ ∈X(T)⁺ where we set $\lambda ^{\prime \prime } = {\text {Res}}_{T^{\prime \prime }}^{T} (\lambda ) \in \mathbf {X}(T^{\prime \prime })^{+}$, it follows that ${\text {Res}}_{G^{\prime \prime }}^{G} {\text {Ind}}_{B}^{G}(\lambda ) \cong {\text {Ind}}_{B^{\prime \prime }}^{G^{\prime \prime }} (\lambda ^{\prime \prime })$. Thus, $(G,G^{\prime \prime })$ is always a Donkin pair. Furthermore, if we let $H^{\prime } \subseteq H$ be the derived subgroup and $H^{\prime \prime } = (G^{\prime }\cap H)^{\circ }$, then $H^{\prime } \subseteq H^{\prime \prime } \subseteq H$. We can therefore apply [10, I.6.14(1)] again to show that $(H,H^{\prime \prime })$ is a Donkin pair.

Now suppose (G,H) is a Donkin pair. In this case, it immediately follows from above that $(G, H^{\prime \prime })$ is a Donkin pair. Moreover, if we let $T^{\prime } = G^{\prime }\cap T$ and $B^{\prime } = G^{\prime }\cap B$, then Eq. 2.1 actually implies that for any $\lambda ^{\prime } \in \mathbf {X}(T^{\prime })^{+}$, there exists λ ∈X(T)⁺ with $\lambda ^{\prime } = \text {Res}^{T}_{T^{\prime }}(\lambda )$ such that

$$ {\text{Res}}_{G^{\prime}}^{G} {\text{Ind}}_{B}^{G} (\lambda) \cong {\text{Ind}}_{B^{\prime}}^{G^{\prime}} (\lambda^{\prime}). $$

In particular,

$$ {\text{Res}}_{H^{\prime\prime}}^{G^{\prime}} {\text{Ind}}_{B^{\prime}}^{G^{\prime}} (\lambda^{\prime}) \cong {\text{Res}}_{H^{\prime\prime}}^{G} {\text{Ind}}_{B}^{G}(\lambda) $$

has a good filtration as an $H^{\prime \prime }$-module, and hence, $(G^{\prime }, H^{\prime \prime })$ is also a Donkin pair.

Conversely, suppose that $(G^{\prime }, H^{\prime \prime })$ is a Donkin pair. We can first deduce that $(G, H^{\prime \prime })$ is a Donkin pair from the fact that $(G,G^{\prime })$ is a Donkin pair. Also, by similar arguments as above we can see that for any $\mu ^{\prime \prime } \in \mathbf {X}(H^{\prime \prime }\cap T)^{+}$, there exists some μ ∈X(H ∩ T)⁺ with $\mu ^{\prime \prime } = {\text {Res}}_{H^{\prime \prime }\cap T}^{H\cap T}(\mu )$, such that

$$ {\text{Res}}_{H^{\prime\prime}}^{H} {\text{Ind}}_{H\cap B}^{H} (\mu) \cong {\text{Ind}}_{H^{\prime\prime}\cap B}^{H^{\prime\prime}}(\mu^{\prime\prime}). $$

This implies that an H-module M has a good filtration if and only if the $H^{\prime \prime }$-module ${\text {Res}}_{H^{\prime \prime }}^{H} M$ has a good filtration. Therefore, (G,H) is also a Donkin pair. □

2.2 Examples of Donkin Pairs

The following proposition collects a number of known examples of Donkin pairs. The last five parts of the proposition deal with various examples where G is quasi-simple and simply connected. For pairs of the form (Spin_n,H), it is usually more convenient to describe the image $H^{\prime }$ of H under the map π : Spin_n →SO_n. Of course, H can be recovered from $H^{\prime }$, as the identity component of π^− 1(H). We use the notation that

$$ (\text{SO}_{n},H^{\prime})^{\sim} = (\text{Spin}_{n}, H). $$

It should be noted that the following proposition does not exhaust the known examples in the literature: for instance, according to [5], there is a Donkin pair of type (B₃,G₂), but this example is not needed in the present paper.

Proposition 2.4

Let G be a connected, reductive group, and let H ⊂ G be a closed, connected, reductive subgroup. If the pair (G,H) satisfies one of the following conditions, then it is a Donkin pair.

• (1) G = H ×⋯ × H, and H↪G is the diagonal embedding.

• (2) H is a Levi subgroup of G.

For the remaining parts, assume that G is quasi-simple and simply connected.

• (3) G is of simply laced type, and H is the fixed-point set of a diagram automorphism of G:

  $$ \begin{array}{lc} (\mathrm{A}_{2n-1}, \mathrm{C}_{n}) = (\text{SL}_{2n},\text{Sp}_{2n}) & \qquad (\mathrm{D}_{4},\mathrm{G}_{2}) = (\text{Spin}_{8}, \mathrm{G}_{2}) \\ ~~(\mathrm{D}_{n}, \mathrm{B}_{n-1}) = (\text{SO}_{2n}, \text{SO}_{2n-1})^{\sim} & \qquad (\mathrm{E}_{6}, \mathrm{F}_{4}) \end{array} $$

• (4) Certain embeddings of classical groups:

  $$ \begin{array}{@{}rcl@{}} \left.\begin{array}{c} (\mathrm{A}_{2n}, \mathrm{B}_{n}) \\ (\mathrm{A}_{2n-1}, \mathrm{D}_{n}) \end{array}\right\} &=& (\text{SL}_{r}, \text{SO}_{r}) \qquad (p > 2) \\ (\mathrm{A}_{2n-1}, \mathrm{C}_{n}) &=& (\text{SL}_{2n}, \text{Sp}_{2n}) \\ \left.\begin{array}{c} (\mathrm{B}_{n+m}, \mathrm{B}_{n}\mathrm{D}_{m}) \\ (\mathrm{D}_{n+m}, \mathrm{D}_{n}\mathrm{D}_{m}) \\ (\mathrm{D}_{n+m+1}, \mathrm{B}_{n}\mathrm{B}_{m}) \end{array}\right\} &=& (\text{SO}_{r+s}, \text{SO}_{r} \times \text{SO}_{s})^{\sim} \quad(p > 2) \\ (\mathrm{C}_{n+m}, \mathrm{C}_{n}\mathrm{C}_{m}) &=& (\text{Sp}_{2n+2m}, \text{Sp}_{2n} \times \text{Sp}_{2m}) \end{array} $$

• (5) Certain maximal-rank subgroups of exceptional groups:

  $$ \begin{array}{ll} & \qquad\qquad (\mathrm{E}_{8} , \mathrm{A}_{2}\mathrm{E}_{6})\quad (p > 5) \\ (\mathrm{E}_{8} , \mathrm{D}_{8}) \quad (p > 2) & \qquad\qquad (\mathrm{E}_{8} , \mathrm{A}_{1}\mathrm{A}_{2}\mathrm{A}_{5})\quad (p > 5) \\ (\mathrm{E}_{8} , \mathrm{A}_{1}\mathrm{E}_{7}) \quad (p > 2 ) & \qquad\qquad (\mathrm{E}_{8} , \mathrm{A}_{3}\mathrm{D}_{5})\quad (p > 5) \\ (\mathrm{E}_{7} , \mathrm{A}_{1}\mathrm{D}_{6}) \quad (p > 2) & \qquad\qquad (\mathrm{E}_{8} , \mathrm{A}_{4}\mathrm{A}_{4})\quad (p > 5) \\ (\mathrm{F}_{4} , \mathrm{B}_{4}) \quad (p > 2) & \qquad\qquad (\mathrm{F}_{4} , \mathrm{A}_{3}\mathrm{A}_{1})\quad (p > 3) \\ & \qquad\qquad (\mathrm{G}_{2} , \mathrm{A}_{1}\mathrm{A}_{1}) \end{array} $$

• (6) Certain restricted irreducible representations:

  $$ \begin{array}{@{}rcl@{}} (\mathrm{A}_{n} , \mathrm{A}_{1}) \quad (p > n) \\ (\mathrm{A}_{7} , \mathrm{A}_{2}) \quad (p > 3) \\ (\mathrm{A}_{6} , \mathrm{G}_{2}) \quad (p > 3) \end{array} $$

• (7) Tensor product embeddings of classical groups (p > 2):

  $$ \begin{array}{@{}rcl@{}} \left.\begin{array}{c} (\mathrm{C}_{(2n+1)m}, \mathrm{B}_{n})\\ (\mathrm{C}_{2nm} , \mathrm{D}_{n}) \end{array}\right\} &=& (\text{Sp}_{2rm}, \text{SO}_{r}) \qquad \left.\begin{array}{c} (\mathrm{B}_{n+m+2nm}, \mathrm{B}_{n}) \\ (\mathrm{D}_{(2n+1)m}, \mathrm{B}_{n})\\ (\mathrm{D}_{nm} , \mathrm{D}_{n}) \end{array}\right\} = (\text{SO}_{rs}, \text{SO}_{r})^{\sim} \\ (\mathrm{D}_{2nm}, \mathrm{C}_{n}) &=& (\text{SO}_{4nm}, \text{Sp}_{2n})^{\sim} \qquad\qquad\quad (\mathrm{C}_{nm}, \mathrm{C}_{n}) = (\text{Sp}_{2nm}, \text{Sp}_{2n}) \end{array} $$

The details of the embeddings in parts (6) and (7) will be described below.

Proof Proofs for parts (2)–(5)

Parts (1) and (2) are due to Mathieu [13] (following earlier work of Donkin [6] that covered most cases). Parts (3) and (4), with the exception of the pair (E₆,F₄), are due to Brundan [5]. The pair (G₂,A₁A₁) in part (5) is also due to Brundan [5]. The pair (E₆,F₄) and the pairs in the first column of part (5) are due to van der Kallen [16]. The pairs in the second column of part (5) are due to Hague–McNinch [7]. □

Proof Proof of part (6)

Each pair (A_n,H) = (SL_n+ 1,H) in this statement arises from some (n + 1)-dimensional representation of H. Call that representation V. The representations V are as follows:

• (A_n,A₁): the dual Weyl module for SL₂ of highest weight n

• (A₇,A₂): the adjoint representation of PGL₃

• (A₆,G₂): the 7-dimensional dual Weyl module whose highest weight is the short dominant root

According to [5, Lemma 3.2(iv)] or [7, §3.2.6], to prove the claim, we must show that each exterior algebra $\bigwedge ^{\bullet } V$ has a good filtration as an H-module. For (A_n,A₁), this is shown in [7, §3.4.3]. For (A₇,A₂) and (A₆,G₂), explicit calculations using the LiE software package [17] show that the character of $\bigwedge ^{\bullet } V$ is the sum of characters of dual Weyl modules whose highest weights are restricted weights when p > 3. □

Proof Proof of part (7)

To define the group embeddings in this statement, we will assume that G is either Sp_2n or SO_n. However, in the latter case, the proof that (G,H) is a Donkin pair will also imply the corresponding statement for G = Spin_n.

Let V₁ be a vector space equipped with a nondegenerate bilinear form B₁ satisfying B₁(v,w) = ε₁B₁(w,v), where ε₁ = ± 1, and let Aut(V₁,B₁)^∘ be the connected group of linear automorphisms of V₁ that preserve B₁. This group is either $\text {SO}_{\dim V}$ or $\text {Sp}_{\dim V}$, depending on ε₁. Let V₂, B₂, and ε₂ be another collection of similar data. Then, B₁ ⊗ B₂ is a nondegenerate pairing on V₁ ⊗ V₂, with sign ε₁ε₂. We obtain an embedding

$$ \text{Aut}(V_{1},B_{1})^{\circ} \times \text{Aut}(V_{2},B_{2})^{\circ} \hookrightarrow \text{Aut}(V_{1} \otimes V_{2}, B_{1} \otimes B_{2})^{\circ}. $$

Now restrict to just one factor:

$$ \text{Aut}(V_{1},B_{1})^{\circ} \hookrightarrow \text{Aut}(V_{1} \otimes V_{2}, B_{1} \otimes B_{2})^{\circ}. $$

(2.2)

The four kinds of pairs listed in the statement are all instances of this embedding, depending on the signs ε₁ and ε₂. We will now prove that

$$ (G,H) = (\text{Aut}(V_{1} \otimes V_{2}, B_{1} \otimes B_{2})^{\circ}, \text{Aut}(V_{1},B_{1})^{\circ}) $$

is a Donkin pair. Let $r = \dim V_{1}$ and $s = \dim V_{2}$.

Suppose first that ε₂ = 1. Then, the embedding (2.2) corresponds to either (SO_rs,SO_r) or (Sp_rs,Sp_r). In this case, V₂ admits an orthonormal basis x₁,…,x_s, where

$$ B_{2}(x_{i},x_{j}) = \delta_{ij}. $$

Then, the group Aut(V₁,B₁)^∘ preserves each V₁ ⊗ x_i ⊂ V₁ ⊗ V₂. In this case, the embedding (2.2) can be factored as

$$ \text{Aut}(V_{1},B_{1})^{\circ} \overset{1}{\hookrightarrow} \underbrace{\text{Aut}(V_{1},B_{1})^{\circ} \times {\cdots} \times \text{Aut}(V_{1},B_{1})^{\circ}}_{s\text{ copies}} \overset{4}{\hookrightarrow} \text{Aut}(V_{1} \otimes V_{2}, B_{1} \otimes B_{2})^{\circ}. $$

The first map is a diagonal embedding; it results in a Donkin pair by part (1) of the proposition. The second embedding gives a Donkin pair by part (4).

Next, suppose that ε₂ = − 1, and assume for now that $\dim V_{2} = 2$. Choose a basis {x,y} for V₂ such that B₂(x,y) = 1. Then, V₁ ⊗ x and V₁ ⊗ y are both maximal isotropic subspaces of V₁ ⊗ V₂. Define an action of GL(V₁) on V₁ ⊗ V₂ as follows:

$$ \begin{array}{l} g \cdot (v \otimes x) = (gv) \otimes x, \\ g \cdot (v \otimes y) = ((g^{\mathrm{t}})^{-1}v) \otimes y \end{array} \qquad \text{for } g \in \text{GL}(V_{1}), $$

where g^t denotes the adjoint operator to g with respect to the nondegenerate form on V₁. This action defines an embedding of GL(V₁) in Aut(V₁ ⊗ V₂,B₁ ⊗ B₂)^∘. In fact, it identifies GL(V₁) with a Levi subgroup of Aut(V₁ ⊗ V₂,B₁ ⊗ B₂)^∘. (This is the usual embedding of GL_r as a Levi subgroup in either SO_2r or Sp_2r.) The embedding (2.2) then factors as

$$ \text{Aut}(V_{1},B_{1})^{\circ} \overset{4}{\hookrightarrow} \text{GL}(V_{1}) \overset{2}{\hookrightarrow} \text{Aut}(V_{1} \otimes V_{2}, B_{1} \otimes B_{2})^{\circ}. $$

The first embedding gives a Donkin pair by part (4) of the proposition, and the second by part (2).

Finally, suppose ε₂ = − 1 and $s = \dim V_{2} > 2$. This dimension must still be even, say s = 2m. Choose a basis x₁,…,x_m,y₁,…,y_m for V₂ such that

$$ B_{2}(x_{i},x_{j}) = B_{2}(y_{i},y_{j}) = 0, \qquad B_{2}(x_{i},y_{j}) = \delta_{ij}. $$

Let ${V}_{2}^{(i)}$ be the 2-dimensional subspace spanned by x_i and y_i. Then, B₂ restricts to a nondegenerate symplectic form $B_{2}^{(i)}$ on ${V}_{2}^{(i)}$. We factor the map (2.2) as follows:

$$ \begin{array}{@{}rcl@{}} \text{Aut}(V_{1},B_{1})^{\circ} &\overset{1}{\hookrightarrow}& \underbrace{\text{Aut}(V_{1},B_{1})^{\circ} \times {\cdots} \times \text{Aut}(V_{1},B_{1})^{\circ}}_{m\text{ copies}} \\ &\hookrightarrow& \text{Aut}(V_{1} \otimes {V}_{2}^{(1)}, B_{1} \otimes {B}_{2}^{(1)})^{\circ} \!\times\! {\cdots} \!\times\! \text{Aut}(V_{1} \otimes {V}_{2}^{(m)}, B_{1} \otimes {B}_{2}^{(m)})^{\circ} \\ &&\!\!\!\!\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \overset{4}{\hookrightarrow} \text{Aut}(V_{1} \otimes V_{2}, B_{1} \otimes B_{2})^{\circ}. \end{array} $$

Here, the first arrow is a diagonal embedding (part (1) of the proposition); the second arrow is several instances of the embedding from the previous paragraph (since $\dim {V}_{2}^{(i)} = 2$); and the last arrow comes from part (4) of the proposition. We thus again obtain a Donkin pair. □

3 Proof of Theorem 1.1

3.1 Reduction to the Quasi-Simple Case

Let G be an arbitrary connected reductive group in good characteristic. For any nilpotent element x ∈Lie(G), there exists a cocharacter $\tau : \mathbb {G}_{m} \rightarrow G$ and a Levi subgroup L_τ ⊂ G such that L_τ is the centralizer of the subgroup $\tau (\mathbb {G}_{m})$, where ${G}_{\text {red}}^{x} = L_{\tau } \cap G^{x}$ (cf. [11, 5.10]).

If we let $G^{\prime }$ be the derived subgroup of G, then any nilpotent element x for G also satisfies $x \in \text {Lie}(G^{\prime })$, and by [11, 5.9], $\tau (\mathbb {G}_{m}) \subset G^{\prime } \subseteq G$. In particular, $(G^{\prime })^{x} = G^{\prime }\cap G^{x}$ and ${L}_{\tau }^{\prime } = G^{\prime }\cap L_{\tau }$ is the centralizer of $\tau (\mathbb {G}_{m})$ in $G^{\prime }$. Thus,

$$ (G^{\prime})^{x}_{\text{red}} = G^{\prime} \cap {G}_{\text{red}}^{x}. $$

It now follows from Lemma 2.3 that $(G,({G}_{\text {red}}^{x})^{\circ })$ is a Donkin pair if and only if $(G^{\prime }, ((G^{\prime })^{x}_{\text {red}})^{\circ })$ is a Donkin pair. So we can reduce to the case where G is semisimple.

Suppose now that $\pi : G \twoheadrightarrow \bar {G}$ is an isogeny (i.e., surjective with finite central kernel), where G is an arbitrary connected reductive group in good characteristic. Then, by [11, Proposition 2.7(a)], π induces a bijection between the nilpotent elements in Lie(G) and those in $\text {Lie}(\bar {G})$, and for any nilpotent element x ∈Lie(G), we have $\pi (G^{x}) = \bar {G}^{\pi (x)}$. Moreover, by similar arguments as above, we can also deduce that $\pi ({G}_{\text {red}}^{x}) = \bar {G}^{\pi (x)}_{\text {red}}$ (cf. [11, 5.9]). In particular,

$$ \pi(({G}_{\text{red}}^{x})^{\circ}) = \pi({G}_{\text{red}}^{x})^{\circ} = ({\bar{G}}_{\text{red}}^{\pi(x)})^{\circ}, $$

since any surjective morphism of algebraic groups takes the identity component to the identity component.

Let $H = ({G}_{\text {red}}^{x})^{\circ }$ and $\bar {H} = (\bar {G}^{\pi (x)}_{\text {red}})^{\circ }$, and note that for any $\bar {G}$-module M, there is a natural isomorphism

$$ {\text{Res}}_{H}^{G} {\text{Res}}_{G}^{\bar{G}} M \cong {\text{Res}}_{H}^{\bar{H}} {\text{Res}}_{\bar{H}}^{\bar{G}} M. $$

From this, we can see that if (G,H) is a Donkin pair, then $(\bar {G},\bar {H})$ must also be a Donkin pair, since it is straightforward to check that a $\bar {G}$-module M (resp. an $\bar {H}$-module N) has a good filtration if and only if ${\text {Res}}_{G}^{\bar {G}} M$ (resp. ${\text {Res}}_{H}^{\bar {H}} N$) has a good filtration. This allows us to reduce to the case where G is semisimple and simply connected.

Finally, suppose that that G = G₁ × G₂ where G₁, G₂ are connected reductive groups in good characteristic. Let x = (x₁,x₂) ∈Lie(G₁) ⊕Lie(G₂) be an arbitrary nilpotent element. We can immediately see that

$$ ({G}_{\text{red}}^{x})^{\circ} = ((G_{1})^{x_{1}}_{\text{red}})^{\circ} \times ((G_{2})^{x_{2}}_{\text{red}})^{\circ}. $$

It now follows from the general properties of induction for direct products (see [10, I.3.8]) that $(G,({G}_{\text {red}}^{x})^{\circ })$ is a Donkin pair if and only if $(G_{1},((G_{1})^{x_{1}}_{\text {red}})^{\circ })$ and $(G_{2},((G_{2})^{x_{2}}_{\text {red}})^{\circ })$ are Donkin pairs. Therefore, by the well-known fact that any simply connected semisimple group is a direct product of quasi-simple simply connected groups, we can reduce the proof of Theorem 1.1 to the case where G is quasi-simple.

3.2 Proof for Classical Groups

We now prove the theorem for the groups GL_n, Sp_n, and Spin_n. For the last case, we will actually describe the group $({G}_{\text {red}}^{x})^{\circ }$ and its embedding in G for SO_n instead, but the proof of the Donkin pair property will also apply to Spin_n.

Let x be a nilpotent element in the Lie algebra of one of GL_n, Sp_n, or SO_n. Let $\mathbf {s} = [{s}_{1}^{r_{1}}, {s}_{2}^{r_{2}}, \ldots , {s}_{k}^{r_{k}}]$ be the partition of n that records the sizes of the Jordan blocks of x. (This means that x has r₁ Jordan blocks of size s₁, and r₂ Jordan blocks of size s₂, etc.) The vector space $V = \Bbbk ^{n}$ can be decomposed as

$$ V = V^{(1)} \oplus V^{(2)} \oplus {\cdots} \oplus V^{(k)} $$

where each V⁽ⁱ⁾ is preserved by x, and x acts on V⁽ⁱ⁾ by Jordan blocks of size s_i. (Thus, $\dim V^{(i)} = r_{i}s_{i}$.) When G is Sp_n or SO_n, the nondegenerate bilinear form on V restricts to a nondegenerate form of the same type on each V⁽ⁱ⁾.

The description of $({G}_{\text {red}}^{x})^{\circ }$ in [12, Chapter 3] shows that it factors through the appropriate embedding below:

$$ \begin{array}{@{}rcl@{}} \text{GL}(V^{(1)}) \times {\cdots} \times \text{GL}(V^{(k)}) &\hookrightarrow& \text{GL}(V) \\ \text{Sp}(V^{(1)}) \times {\cdots} \times \text{Sp}(V^{(k)}) &\hookrightarrow& \text{Sp}(V) \\ \text{SO}(V^{(1)}) \times {\cdots} \times \text{SO}(V^{(k)}) &\hookrightarrow& \text{SO}(V) \end{array} $$

All three of these embeddings give Donkin pairs: in the case of GL_n, it is an inclusion of a Levi subgroup (Proposition 2.4(2)); and in the case of Sp_n or SO_n, it falls under Proposition 2.4(4).

We can therefore reduce to the case where x has Jordan blocks of a single size. Suppose from now on that s = [s^r]. Then, there exists a vector space isomorphism

$$ V \cong V_{1} \otimes V_{2} $$

where $\dim V_{1} = r$ and $\dim V_{2} = s$, and such that x corresponds to $\text {id}_{V_{1}} \otimes N$, where N : V₂ → V₂ is a nilpotent operator with a single Jordan block (of size s).

Suppose now that G = GL(V ). Then, according to [12, Proposition 3.8], we have $({G}_{\text {red}}^{x})^{\circ } \cong \text {GL}(V_{1})$. Choose a basis {v₁,…,v_s} for V₂. The embedding of $({G}_{\text {red}}^{x})^{\circ }$ in G factors as

$$ \text{GL}(V_{1}) \hookrightarrow \text{GL}(V_{1} \otimes v_{1}) \times {\cdots} \times \text{GL}(V_{1} \otimes v_{s}) \hookrightarrow \text{GL}(V). $$

The first map above is a diagonal embedding (Proposition 2.4(1)), and the second is the inclusion of a Levi subgroup (Proposition 2.4(2)), so $(G, ({G}_{\text {red}}^{x})^{\circ })$ is a Donkin pair in this case.

Next, suppose G = Sp(V ) or SO(V ). According to [12, Proposition 3.10], both V₁ and V₂ can be equipped with nondegenerate bilinear forms B₁ and B₂ such that B₁ ⊗ B₂ agrees with the given bilinear form on V. Moreover, $({G}_{\text {red}}^{x})^{\circ } = \text {Aut}(V_{1},B_{1})^{\circ }$. We are thus in the setting of Proposition 2.4(7).

3.3 Proof for E₈

When x is distinguished, $({G}_{\text {red}}^{x})^{\circ }$ is the trivial group; and when x = 0, $({G}_{\text {red}}^{x})^{\circ } = G$. For all remaining nilpotent orbits, we rely on the very detailed case-by-case descriptions of $({G}_{\text {red}}^{x})^{\circ }$ given in [12, Chapter 15]. In each case, that description shows that the embedding $({G}_{\text {red}}^{x})^{\circ } \hookrightarrow G$ factors as a composition of various cases from Proposition 2.4.

These factorizations are shown in Tables 1 and 2. Here is a brief explanation of the notation used in these tables. Nearly all groups mentioned are semisimple, and they are recorded in the tables by their root systems. However, the notation “T₁” indicates a 1-dimensional torus; this is used to indicate a reductive group with a 1-dimensional center. In a few cases, nonstandard names for root systems—such as B₁ or C₁, in place of A₁—are used when it is convenient to emphasize the role of a certain classical group. The notation D₁ (meant to evoke SO₂) is occasionally used as a synonym for T₁.

Table 1 Nilpotent centralizers in E₈

Table 2 Nilpotent centralizers in E₈, continued

Finally, we remark that there are two orbits—labeled by A₆ and by A₆A₁—where the information given in [12] is insufficient to finish the argument. In each of these cases, $({G}_{\text {red}}^{x})^{\circ }$ contains a copy of A₁ that is the centralizer of a certain copy of G₂ inside E₇, and [12] does not give further details on this embedding A₁↪E₇. However, according to [15, §3.12], this A₁ is in fact (the derived subgroup of) a Levi subgroup of E₇.

3.4 Proof for E₇ and E₆

Recall that if $H \subseteq G$ is a closed subgroup, then there is a natural embedding $\mathcal {N}_{H} \hookrightarrow \mathcal {N}_{G}$ of nilpotent cones. We also recall that a subgroup $L \subseteq G$ is a Levi subgroup if and only if it is the centralizer of a torus S ⊂ G.

Lemma 3.1

Let G be a reductive group, $S \subseteq G$ a torus with L = C_G(S) a Levi subgroup, and suppose $x \in \mathcal {N}_{L} \subseteq \mathcal {N}_{G}$ is such that $S \subseteq ({G}_{\text {red}}^{x})^{\circ }$. Then, $({L}_{\text {red}}^{x})^{\circ }$ is a Levi subgroup of $({G}_{\text {red}}^{x})^{\circ }$.

Proof

We clearly have $C_{G^{x}}(S) = G^{x} \cap L = L^{x}$. Moreover, the identity component (L^x)^∘ must be contained in $(G^{x})^{\circ } \cap L = C_{(G^{x})^{\circ }}(S)$. But since the centralizer of a torus in a connected group is connected, we actually have

$$ (L^{x})^{\circ} = C_{(G^{x})^{\circ}}(S). $$

Next, consider the semidirect product decomposition $(G^{x})^{\circ } = ({G}_{\text {red}}^{x})^{\circ } \ltimes {G}_{\text {unip}}^{x}$. Let g ∈ G^x, and write it as g = (g_r,g_u), with $g_{r} \in ({G}_{\text {red}}^{x})^{\circ }$, $g_{u} \in G^{x}_{\text {unip}}$. If g centralizes S, then g_r and g_u must individually centralize S as well. In other words,

$$ C_{(G^{x})^{\circ}}(S) = C_{({G}_{\text{red}}^{x})^{\circ}}(S) \ltimes C_{{G}_{\text{unip}}^{x}}(S). $$

(3.1)

Here, $C_{({G}_{\text {red}}^{x})^{\circ }}(S)$ is a connected reductive group, and $C_{{G}_{\text {unip}}^{x}}(S)$ is a normal unipotent group (which must be connected, because $C_{(G^{x})^{\circ }}(S)$ is connected). We conclude that Eq. 3.1 is a Levi decomposition of (L^x)^∘. In particular, we see that $({L}_{\text {red}}^{x})^{\circ } = C_{({G}_{\text {red}}^{x})^{\circ }}(S)$. □

We now let G denote the simple, simply connected group of type E₈. If G₀ is the simple, simply connected group of type E₇ or E₆, then as explained in [12, Lemma 11.14], there is a simple subgroup H of type A₁ or A₂, respectively, such that G₀ = C_G(H). Moreover, by [12, 16.1.2], there exists a torus S ⊂ H ⊂ G such that G₀ is the derived subgroup of the Levi subgroup L = C_G(S). Explicitly, let α₁,…,α₈ be the simple roots for G, labelled as in [4], and let α₀ be the highest root. The groups H and G₀ can be described as follows.

(3.2)

Now it is explained in [12, 16.1.1], that if $x \in \mathcal {N}_{G_{0}} = \mathcal {N}_{L} \subset \mathcal {N}_{G}$, then the subgroup $({G}_{\text {red}}^{x})^{\circ }$ must contain a conjugate of H. Without loss of generality we can assume that x is chosen so that $H \subseteq ({G}_{\text {red}}^{x})^{\circ }$. Hence, we can also assume that $S \subseteq ({G}_{\text {red}}^{x})^{\circ }$. Thus, by Lemma 3.1, $({L}_{\text {red}}^{x})^{\circ }$ is a Levi subgroup of $({G}_{\text {red}}^{x})^{\circ }$ and we also have

$$ ({L}_{\text{red}}^{x})^{\circ \prime} \subseteq (({G}_{0}^{x})_{\text{red}})^{\circ} \subseteq ({L}_{\text{red}}^{x})^{\circ}. $$

By Lemma 2.3, Proposition 2.4(2) and Section 3.3 we deduce that $(G,({G}_{0}^{x})_{\text {red}}))^{\circ })$ is a Donkin pair.

Finally, to show that $(G_{0}, (({G}_{0}^{x})_{\text {red}})^{\circ })$ is a Donkin pair, it will be sufficient to show that every fundamental tilting module for G₀ is a summand of the restriction of a tilting module for G. In more detail, let $\pi : \mathbf {X}_{G} \twoheadrightarrow \mathbf {X}_{G_{0}}$ be the map on weight lattices. It is well known that if λ is a dominant weight for G, then the G₀-tilting module $T_{G_{0}}(\pi (\lambda ))$ occurs as a direct summand of $\text {Res}^{G}_{G_{0}}(T_{G}(\lambda ))$. So it is enough to show that every fundamental weight for G₀ occurs as π(λ) for some dominant G-weight λ. Let ϖ₁,…,ϖ₈ be the fundamental weights for G. A short calculation with the table (3.2) shows that π(ϖ₁),…,π(ϖ₇) are precisely the fundamental weights for E₇, and that π(ϖ₁),…,π(ϖ₆) are the fundamental weights for E₆.

Remark 3.2

One can also prove the theorem for E₇ and E₆ directly by writing down the embedding of each centralizer, as we did for E₈. Here is a brief summary of how to carry out this approach. Let G be of type E₈, and let G₀ and H be as in the discussion above. As explained in [12, §16.1], we have

$$ (({G}_{0}^{x})_{\text{red}})^{\circ} = C_{({G}_{\text{red}}^{x})^{\circ}}(H). $$

The computation of $C_{({G}_{\text {red}}^{x})^{\circ }}(H)$ is explained in [12, §16.1.4], and the results are recorded in Tables 3, 4, and 5, following the same notational conventions as in the E₈ case.

Table 3 Nilpotent centralizers in E₇

Table 4 Nilpotent centralizers in E₇, continued

Table 5 Nilpotent centralizers in E₆

3.5 Proof for F₄

Let G be the simple, simply connected group of type E₈. Then, [12, Lemma 11.7] implies that G contains a simple subgroup H of type G₂, and that its centralizer G₀ = C_G(H) is a simple group of type F₄. The embeddings of centralizers of nilpotent elements for G₀ = F₄ can then be computed using the method explained in Remark 3.2. One caveat is that the name (i.e., the Bala–Carter label) of a nilpotent orbit usually changes when passing from F₄ to E₈. The correspondence between these names is given in [12, Proposition 16.10].

We remark that in some cases, the book [12] does not quite give enough details about embeddings of subgroups to establish our result, but in these cases, the relevant details can be found in [15, §3.16]. Here is an example illustrating this. The F₄-orbit labelled $\tilde {\mathrm {A}}_{1}$ corresponds (by [12, Proposition 16.10]) to the E₈-orbit labelled ${\mathrm {A}}_{1}^{2}$. Let x be an element of this orbit. We have see that in E₈, $({G}_{\text {red}}^{x})^{\circ } = \mathrm {B}_{6}$, which embeds in the Levi subgroup D₇ ⊂E₈. The group $({G}_{\text {red}}^{x})^{\circ }$ has a subgroup of type D₃B₃, which embeds in D₃D₄ ⊂D₇ ⊂D₈. The explicit construction of H = G₂ in [15, §3.16] shows that it is contained in the second factor in each of D₃B₃ ⊂D₃D₄. It follows that D₃ = A₃ is contained in $(({G}_{0}^{x})_{\text {red}})^{\circ } = C_{({G}_{\text {red}}^{x})^{\circ }}(H)$, and then a dimension calculation shows that in fact $(({G}_{0}^{x})_{\text {red}})^{\circ } = \mathrm {A}_{3}$.

The results of these calculations are recorded in Table 6.

Table 6 Nilpotent centralizers in F₄

3.6 Proof for G₂

In this case, there are only two nilpotent orbits that are neither distinguished nor trivial. From the classification, both of these orbits meet the maximal reductive subgroup $\mathrm {A}_{1}\tilde {\mathrm {A}}_{1} \subset \mathrm {G}_{2}$, and an argument explained in [12, §16.1.4] shows that if x belongs to either of these orbits, then the reductive part of its centralizer in $\mathrm {A}_{1}\tilde {\mathrm {A}}_{1}$ is equal to the reductive part of its centralizer in G₂. See Table 7.

Table 7 Nilpotent centralizers in G₂