Constructible sheaves on nilpotent cones in rather good characteristic

Abstract

We study some aspects of modular generalized Springer theory for a complex reductive group G with coefficients in a field $\Bbbk$ under the assumption that the characteristic $\ell$ of $\Bbbk$ is rather good for G, i.e. $\ell$ is good and does not divide the order of the component group of the centre of G. We prove a comparison theorem relating the characteristic-$\ell$ generalized Springer correspondence to the characteristic-0 version. We also consider Mautner’s characteristic-$\ell$ ‘cleanness conjecture’; we prove it in some cases; and we deduce several consequences, including a classification of supercuspidal sheaves and an orthogonal decomposition of the equivariant derived category of the nilpotent cone.

Appendix: Central characters

In this appendix, we collect some standard facts concerning central characters of objects in an equivariant derived category. Some of these facts are well known and were used in [2], but since we could not find a convenient reference for the versions needed in the present paper, we include proofs. See also [21, Appendix A] for related results.

Let $\mathbb {F}$ be an arbitrary field. Let H be a connected algebraic group, and let X be an H-variety. Let $\mathrm {For}: D^{\mathrm {b}}_H(X, \mathbb {F}) \rightarrow D^{\mathrm {b}}(X,\mathbb {F})$ be the forgetful functor. Then if $a,p : H \times X \rightarrow X$ are the action and the projection, for any $\mathcal {F}$ in $D^{\mathrm {b}}_H(X,\mathbb {F})$ there exists a canonical isomorphism

$$\begin{aligned} a^* \mathrm {For}(\mathcal {F}) \xrightarrow {\sim }p^*\mathrm {For}(\mathcal {F}). \end{aligned}$$

(7.6)

Indeed, the inverse image $p^*$ induces an equivalence of categories $\varphi _p : D^{\mathrm {b}}(X,\mathbb {F}) \xrightarrow {\sim }D^{\mathrm {b}}_H(H \times X, \mathbb {F})$. Hence the object $a^* \mathcal {F}$ is isomorphic to $\varphi _p(\mathcal {G})$ for a unique object $\mathcal {G}$ of $D^{\mathrm {b}}(X,\mathbb {F})$. Taking the restriction to $\{1\} \times X$, we see that $\mathcal {G}=\mathrm {For}(\mathcal {F})$, and taking the image under the forgetful functor $D^{\mathrm {b}}_H(H \times X, \mathbb {F}) \rightarrow D^{\mathrm {b}}(H \times X,\mathbb {F})$, we deduce (7.6).

Now, let $Z \subset H$ be a closed subgroup, and assume that Z acts trivially on X. Again let $\mathcal {F}$ be an object of $D^{\mathrm {b}}_H(X,\mathbb {F})$. Taking the restriction of (7.6) to $\{z\} \times X \cong X$ for all $z \in Z$, we obtain a (functorial) action of Z on the object $\mathrm {For}(\mathcal {F})$. By standard arguments, this action factors through $Z/Z^\circ$. If $\chi : Z/Z^\circ \rightarrow \mathbb {F}^\times$ is a character, we say that $\mathcal {F}$ has Z-character $\chi$ if Z acts on $\mathrm {For}(\mathcal {F})$ via $\chi$. When Z is the centre of H, we will rather say that $\mathcal {F}$ has central character $\chi$.

Lemma A.1

Let $\chi , \chi '$ be distinct characters of $Z/Z^\circ$, and let $\mathcal {F}$, $\mathcal {G}$ be objects of $D^{\mathrm {b}}_H(X,\mathbb {F})$. If $\mathcal {F}$ has Z-character $\chi$ and $\mathcal {G}$ has Z-character $\chi '$, then ${{\mathrm{Hom}}}^\bullet _{D^{\mathrm {b}}_H(X,\mathbb {F})}(\mathcal {F}, \mathcal {G})=0$.

Proof

By standard arguments, there exists a spectral sequence converging to ${{\mathrm{Hom}}}_{D^{\mathrm {b}}_H(X,\mathbb {F})}^{\bullet }(\mathcal {F}, \mathcal {G})$, and with $E_2$-term

$$\begin{aligned} \mathsf {H}^\bullet _H(\mathrm {pt}) \otimes _\mathbb {F}{{\mathrm{Hom}}}^\bullet _{D^{\mathrm {b}}(X,\mathbb {F})}(\mathrm {For}\mathcal {F}, \mathrm {For}\mathcal {G}). \end{aligned}$$

Therefore, it is enough to prove that ${{\mathrm{Hom}}}_{D^{\mathrm {b}}(X,\mathbb {F})}^\bullet (\mathrm {For}\mathcal {F}, \mathrm {For}\mathcal {G})=0$. However, the Z-actions on $\mathrm {For}\mathcal {F}$ and $\mathrm {For}\mathcal {G}$ induce an action on ${{\mathrm{Hom}}}_{D^{\mathrm {b}}(X,\mathbb {F})}^\bullet (\mathrm {For}\mathcal {F}, \mathrm {For}\mathcal {G})$, for which Z acts via the character $\chi '/\chi$. On the other hand, this action can be extended to H, as follows. For any $h \in H$, restricting isomorphism (7.6) to $\{h\} \times X$ we obtain an isomorphism $\phi ^{\mathcal {F}}_h : \iota _h^* (\mathrm {For}\mathcal {F}) \xrightarrow {\sim }\mathrm {For}\mathcal {F}$, where $\iota _h : X \xrightarrow {\sim }X$ is the action of h. We also have similar isomorphisms for $\mathcal {G}$. Then we can define the action of H on ${{\mathrm{Hom}}}_{D^{\mathrm {b}}(X,\mathbb {F})}^\bullet (\mathrm {For}\mathcal {F}, \mathrm {For}\mathcal {G})$ by declaring that h acts via the composition

$$\begin{aligned}&{{\mathrm{Hom}}}^\bullet _{D^{\mathrm {b}}(X,\mathbb {F})}(\mathrm {For}\mathcal {F}, \mathrm {For}\mathcal {G}) \xrightarrow {\iota _{h^{-1}}^*} {{\mathrm{Hom}}}^\bullet _{D^{\mathrm {b}}(X,\mathbb {F})}(\iota _{h^{-1}}^* \mathrm {For}\mathcal {F}, \iota _{h^{-1}}^*\mathrm {For}\mathcal {G}) \\&\qquad \qquad \qquad \qquad \qquad \quad \xrightarrow {\phi _{h^{-1}}^{\mathcal {G}} \circ (-) \circ (\phi _{h^{-1}}^{\mathcal {F}})^{-1}} {{\mathrm{Hom}}}^\bullet _{D^{\mathrm {b}}(X,\mathbb {F})}(\mathrm {For}(\mathcal {F}), \mathrm {For}(\mathcal {G})). \end{aligned}$$

Since H is connected this action is trivial, and we deduce that the Z-action considered above is also trivial. It follows that necessarily ${{\mathrm{Hom}}}_{D^{\mathrm {b}}(X,\mathbb {F})}^\bullet (\mathrm {For}\mathcal {F}, \mathrm {For}\mathcal {G})=0$, which finishes the proof. $\square$

Lemma A.2

Let $\chi$ be a character of $Z/Z^\circ$.

• (1) If $\mathcal {F}$ is an object of $\mathsf {Perv}_H(X,\mathbb {F})$ with Z-character $\chi$, then any subquotient of $\mathcal {F}$ has Z-character $\chi$.

• (2) Assume that the characteristic of $\mathbb {F}$ does not divide $|Z/Z^\circ |$. Let $\mathcal {F}$, $\mathcal {G}$ be objects of $\mathsf {Perv}_H(X,\mathbb {F})$ with Z-character $\chi$, and consider an exact sequence

  $$\begin{aligned} 0 \rightarrow \mathcal {F}\rightarrow \mathcal {H}\rightarrow \mathcal {G}\rightarrow 0 \end{aligned}$$

  in $\mathsf {Perv}_H(X,\mathbb {F})$. Then $\mathcal {H}$ has Z-character $\chi$.

Proof

(1) is obvious. Let us consider (2). For $z \in Z$, let us denote by $\phi _z : \mathcal {H}\xrightarrow {\sim }\mathcal {H}$ the action of z on $\mathcal {H}$. Since z acts on $\mathcal {F}$ and $\mathcal {G}$ via $\chi (z)$, the morphism $\chi (z)^{-1} \phi _z - \mathrm {id}_\mathcal {H}: \mathcal {H}\rightarrow \mathcal {H}$ factors through a morphism $\psi _z : \mathcal {G}\rightarrow \mathcal {F}$. For any $n \ge 1$, using the factorization $X^n-1=(1+X + \cdots + X^{n-1})(X-1)$ we obtain that $\psi _{z^n}=n \psi _z$. Since $\psi _{z'}=0$ for all $z'\in Z^\circ$ and the order of $zZ^\circ$ in $Z/Z^\circ$ is invertible in $\mathbb {F}$, it follows that $\psi _z=0$, which proves that $\mathcal {H}$ has Z-character $\chi$. $\square$

Lemma A.3

Assume that Z is finite and central, that the characteristic of $\mathbb {F}$ does not divide |Z|, and that $\mathbb {F}$ is a splitting field for Z. Then the forgetful functor $\mathsf {Perv}_{H/Z}(X,\mathbb {F}) \rightarrow \mathsf {Perv}_H(X,\mathbb {F})$ identifies $\mathsf {Perv}_{H/Z}(X,\mathbb {F})$ with the full subcategory of $\mathsf {Perv}_H(X,\mathbb {F})$ consisting of objects with trivial Z-character.

Proof

Since the forgetful functors $\mathsf {Perv}_{H/Z}(X,\mathbb {F}) \rightarrow \mathsf {Perv}(X,\mathbb {F})$ and $\mathsf {Perv}_{H}(X,\mathbb {F}) \rightarrow \mathsf {Perv}(X,\mathbb {F})$ are fully faithful, our functor $\mathsf {Perv}_{H/Z}(X,\mathbb {F}) \rightarrow \mathsf {Perv}_H(X,\mathbb {F})$ is also fully faithful. Clearly, all the objects in its essential image have trivial Z-character. Conversely, let $\mathcal {F}$ be an object of $\mathsf {Perv}_H(X,\mathbb {F})$ with trivial Z-character. For simplicity, we denote similarly its image in $\mathsf {Perv}(X,\mathbb {F})$. Let $a,p : H \times X \rightarrow X$ be the action and the projection, respectively, and let $a',p' : H/Z \times X \rightarrow X$ be the similar morphisms for H / Z. Let also $\xi : H \times X \rightarrow H/Z \times X$ be the projection, so that $a=a'\circ \xi$ and $p=p' \circ \xi$. To show that $\mathcal {F}$ is H / Z-equivariant, we have to show that the objects $(a')^* \mathcal {F}$ and $(p')^* \mathcal {F}$ of $D^{\mathrm {b}}(H/Z \times X, \mathbb {F})$ are isomorphic. Consider the space

$$\begin{aligned} {{\mathrm{Hom}}}_{D^{\mathrm {b}}(H \times X, \mathbb {F})}(a^* \mathcal {F}, p^* \mathcal {F})\cong & {} {{\mathrm{Hom}}}_{D^{\mathrm {b}}(H \times X, \mathbb {F})}(\xi ^* (a')^* \mathcal {F}, \xi ^* (p')^* \mathcal {F}) \\\cong & {} {{\mathrm{Hom}}}_{D^{\mathrm {b}}(H/Z \times X, \mathbb {F})}((a')^* \mathcal {F}, \xi _* \xi ^* (p')^* \mathcal {F})\\\cong & {} {{\mathrm{Hom}}}_{D^{\mathrm {b}}(H/Z \times X, \mathbb {F})}((a')^* \mathcal {F}, (p')^* \mathcal {F}\otimes \xi _* \underline{\mathbb {F}}_{H \times X}). \end{aligned}$$

Our assumptions on $\mathbb {F}$ imply that we have a decomposition $\xi _* \underline{\mathbb {F}}_{H \times X} \cong \bigoplus _\chi \mathcal {L}_\chi$, where $\chi$ runs over the characters $Z \rightarrow \mathbb {F}^\times$, and each $\mathcal {L}_\chi$ is a rank-1 H-equivariant local system with Z-character $\chi$. We deduce an isomorphism

$$\begin{aligned} {{\mathrm{Hom}}}_{D^{\mathrm {b}}(H \times X, \mathbb {F})}(a^* \mathcal {F}, p^* \mathcal {F}) \cong \bigoplus _\chi {{\mathrm{Hom}}}_{D^{\mathrm {b}}(H/Z \times X, \mathbb {F})}((a')^* \mathcal {F}, (p')^* \mathcal {F}\otimes \mathcal {L}_\chi ). \end{aligned}$$

Now, since $a'$ is an H-equivariant morphism (where H acts on $H/Z \times X$ via its action on the first factor), $(a')^* \mathcal {F}$, considered as an object of $D^{\mathrm {b}}_H(H/Z \times X, \mathbb {F})$, has trivial Z-character, and it is clear that $(p')^* \mathcal {F}$ also has trivial Z-character. We deduce that

$$\begin{aligned} {{\mathrm{Hom}}}_{D^{\mathrm {b}}(H/Z \times X, \mathbb {F})}((a')^* \mathcal {F}, (p')^* \mathcal {F}\otimes \mathcal {L}_\chi ) =0 \qquad \text {if }\chi \ne 1 \end{aligned}$$

(see the proof of Lemma A.1) and then that

$$\begin{aligned} {{\mathrm{Hom}}}_{D^{\mathrm {b}}(H \times X, \mathbb {F})}(a^* \mathcal {F}, p^* \mathcal {F})\cong & {} {{\mathrm{Hom}}}_{D^{\mathrm {b}}(H/Z \times X, \mathbb {F})}((a')^* \mathcal {F}, (p')^* \mathcal {F}\otimes \mathcal {L}_1) \\= & {} {{\mathrm{Hom}}}_{D^{\mathrm {b}}(H/Z \times X, \mathbb {F})}((a')^* \mathcal {F}, (p')^* \mathcal {F}). \end{aligned}$$

Since $\mathcal {F}$ is H-equivariant, there exists an isomorphism $a^* \mathcal {F}\xrightarrow {\sim }p^* \mathcal {F}$. The image of this morphism under the isomorphism constructed above provides a morphism $(a')^* \mathcal {F}\rightarrow (p')^* \mathcal {F}$, which is easily shown to be an isomorphism. Thus, $\mathcal {F}$ is H / Z-equivariant. $\square$

Finally, we consider the setting of the body of the paper, namely the case of the G-action on the nilpotent cone $\mathscr {N}_G$, with $Z=Z(G)$, and with coefficient field $\Bbbk$ satisfying (2.1). In particular, since $Z(G) / Z(G)^\circ$ is isomorphic to $A_G(x)$ if x is a regular nilpotent element, this implies that all the irreducible $\Bbbk$-representations of $Z(G) / Z(G)^\circ$ are characters. For any $\chi \in {{\mathrm{Irr}}}(\Bbbk [Z(G)/Z(G)^\circ ])$, we denote by $\mathsf {Perv}_G(\mathscr {N}_G,\Bbbk )_\chi$ the full subcategory of $\mathsf {Perv}_G(\mathscr {N}_G,\Bbbk )$ whose objects have central character $\chi$. We also denote by $D^{\mathrm {b}}_G(\mathscr {N}_G,\Bbbk )_\chi$ the full subcategory of $D^{\mathrm {b}}_G(\mathscr {N}_G,\Bbbk )$ whose objects are the complexes $\mathcal {F}$ such that ${}^p \mathcal {H}^n(\mathcal {F})$ belongs to $\mathsf {Perv}_G(\mathscr {N}_G,\Bbbk )_\chi$ for any $n \in \mathbb {Z}$. It follows from Lemma A.2 that $\mathsf {Perv}_G(\mathscr {N}_G,\Bbbk )_\chi$ is a Serre subcategory of $\mathsf {Perv}_G(\mathscr {N}_G,\Bbbk )$ and that $D^{\mathrm {b}}_G(\mathscr {N}_G,\Bbbk )_\chi$ is a triangulated subcategory of $D^{\mathrm {b}}_G(\mathscr {N}_G,\Bbbk )$.

If $\mathcal {F}$ is a simple object in $\mathsf {Perv}_G(\mathscr {N}_G,\Bbbk )$, then ${{\mathrm{End}}}(\mathcal {F})=\Bbbk$ under our assumption (2.1), so that $\mathcal {F}$ has a central character. This central character can be described more explicitly as follows (see [2, §5.1]). Let $(\mathscr {O},\mathcal {E}) \in \mathfrak {N}_{G, \Bbbk }$, and let $x \in \mathscr {O}$. Then $\mathcal {E}$ corresponds to an absolutely irreducible representation V of $A_G(x)$. Consider the composition $Z(G) \rightarrow G_x \rightarrow A_G(x)$. This morphism is trivial on $Z(G)^\circ$, and its image is central in $A_G(x)$. Hence, by Schur’s lemma, Z(G) acts on V via a character $\chi$ of $Z(G)/Z(G)^\circ$. Then $\mathcal {IC}(\mathscr {O},\mathcal {E})$ has central character $\chi$.

Lemma A.4

Suppose that $\ell \not \mid |Z(G)/Z(G)^\circ |$. Then we have

$$\begin{aligned} \mathsf {Perv}_G(\mathscr {N}_G,\Bbbk )= & {} \bigoplus _{\chi \in {{\mathrm{Irr}}}(\Bbbk [Z(G)/Z(G)^\circ ])} \mathsf {Perv}_G(\mathscr {N}_G,\Bbbk )_\chi \quad \text {and}\quad \\ D^{\mathrm {b}}_G(\mathscr {N}_G,\Bbbk )= & {} \bigoplus _{\chi \in {{\mathrm{Irr}}}(\Bbbk [Z(G)/Z(G)^\circ ])} D^{\mathrm {b}}_G(\mathscr {N}_G,\Bbbk )_\chi . \end{aligned}$$

Proof

Let us first consider the category $\mathsf {Perv}_G(\mathscr {N}_G,\Bbbk )$. Lemma A.1 implies that any morphism and any extension between objects with distinct central characters are trivial. Since any simple object has a central character, we deduce the decomposition as stated.

To prove the decomposition for the category $D^{\mathrm {b}}_G(\mathscr {N}_G,\Bbbk )$, since this category is generated by $\mathsf {Perv}_G(\mathscr {N}_G,\Bbbk )$ (as a triangulated subcategory), it suffices to prove that if $\chi \ne \chi '$, for $\mathcal {F}$ in $D^{\mathrm {b}}_G(\mathscr {N}_G,\Bbbk )_\chi$ and $\mathcal {G}$ in $D^{\mathrm {b}}_G(\mathscr {N}_G,\Bbbk )_{\chi '}$ we have ${{\mathrm{Hom}}}(\mathcal {F},\mathcal {G})=0$. This follows again from Lemma A.1, using an induction on the number of nonzero perverse cohomology objects of $\mathcal {F}$ and $\mathcal {G}$. $\square$