Geometric approach to parabolic induction

Abstract

In this paper we construct a “restriction” map from the cocenter of a reductive group G over a local non-archimedean field F to the cocenter of a Levi subgroup. We show that the dual map corresponds to parabolic induction and deduce that parabolic induction preserves stability. We also give a new (purely geometric) proof that the character of normalized parabolic induction does not depend on the parabolic subgroup. In the appendix, we use a similar argument to extend a theorem of Lusztig–Spaltenstein on induced unipotent classes to all infinite fields. We also prove a group version of a theorem of Harish-Chandra about the density of the span of regular semisimple orbital integrals.

Appendices

Appendix A: A generalization of a theorem of Lusztig–Spaltenstein

A.1. Notation. Let F be an infinite field. All algebraic varieties and all morphisms of algebraic varieties are over F.

(a) Let $\mathbf {G}$ be a connected reductive group, $\mathbf {P}\subset \mathbf {G}$ a parabolic subgroup, ${\mathbf {U}}\subset \mathbf {P}$ the unipotent radical, and $\mathbf {M}\subset \mathbf {P}$ a Levi subgroup.

(b) For an algebraic variety $\mathbf {X}$, we denote the set $\mathbf {X}(F)$ by X. In particular, we have $G=\mathbf {G}(F)$, $\widetilde{G}_P=\widetilde{\mathbf {G}}_{\mathbf {P}}(F)$, etc. (compare 3.8).

(c) For an ${\text {Ad}}\, P$-invariant subset $D\subset P$, we set ${\text {Ind}}_P^G(D):=G\overset{P,{\text {Ad}}}{\times }D\subset \widetilde{G}_P$.

(d) For an ${\text {Ad}}\, M$-invariant subset $C\subset M$, we set ${\text {Ind}}_M^G(C):=G\overset{M,{\text {Ad}}}{\times }C\subset \widetilde{G}_M$, $C_P:=U\cdot C\subset P$, ${\text {Ind}}_P^G(C_P)\subset \widetilde{G}_P$ and $C_{P;G}:=a_{P,G}({\text {Ind}}_P^G(C_P))\subset G$, where $a_{P,G}:\widetilde{G}_P\rightarrow G$ was defined in 1.4.

From now on, we assume that $C\subset M$ is a unipotent M-conjugacy class.

A.2. Question. Does the set $C_{P;G}$ depend on the choice of $\mathbf {P}\supset \mathbf {M}$?

A.3. Remarks. (a) $C_{P;G}$ is a union of unipotent conjugacy classes in G.

(b)Let F be algebraically closed. By a theorem of Chevalley, $C_{P;G}\subset G$ is a constructible set, whose Zariski closure $\overline{C}_{P;G}$ is irreducible. This case was considered by Lusztig and Spaltenstein in [6], and they showed that $\overline{C}_{P;G}$ does not depend on $\mathbf {P}$, using representation theory. A simpler proof of this fact was given later by Lusztig [7, Lem 10.3(a)].

The goal of this appendix is to generalize the result of [6] to other fields.

A.4. Saturation. Let $\mathbf {X}$ be an algebraic variety over F, and let $A\subset X$ be a subset.

(a) We denote by ${\text {sat}}'(A)={\text {sat}}'_\mathbf {X}(A)\subset X$ the union $\cup _{(\mathbf {V},x,f)}f(x)$, taken over triples $(\mathbf {V},x,f)$, where $\mathbf {V}\subset \mathbb A^1$ is an open subvariety, $x\in V$, $\mathbf {V}':=\mathbf {V}\smallsetminus \{x\}$, and $f:\mathbf {V}\rightarrow \mathbf {X}$ is a morphism such that $f(V')\subset A$.

(b) We say that a subset $A\subset X$ is saturated, if ${\text {sat}}'(A)=A$.

(c) Let ${\text {sat}}(A)\subset X$ be the smallest saturated subset, containing A.

Theorem A.5

The saturation ${\text {sat}}(C_{P;G})$ does not depend on $\mathbf {P}$.

A.6. Remarks. (a) The notion of saturation is only reasonable, if the variety $\mathbf {X}$ is rationally connected.

(b) For every closed subvariety $\mathbf {Y}\subset \mathbf {X}$, the subset $\mathbf {Y}(F)\subset X$ is saturated. Also, if F is a local field, then every closed subset of X is saturated.

(c) If $\mathbf {X}=\mathbb A^1$, then a subset $A\subset X$ is saturated if and only if either $A=X$ or $X\smallsetminus A$ is infinite.

(d) By (c), saturated subsets of X are not closed under finite unions. Therefore the set X does not have a topology, whose closed subsets are saturated subsets. On the other hand, our proof of Theorem A.5 indicates that in some respects saturated sets behave like closed subsets in some topology.

Lemma A.7

Let $\mathbf {X}$ and $\mathbf {Y}$ be algebraic varieties.

(a) For a morphism $f:\mathbf {X}\rightarrow \mathbf {Y}$ and a subset $A\subset X$, we have an inclusion $f({\text {sat}}'(A))\subset {\text {sat}}(f'(A))$.

(b) For every $A\subset X$ and $B\subset Y$, we have ${\text {sat}}'(A\times B)={\text {sat}}'(A)\times {\text {sat}}'(B)$.

(c) Let $\mathbf {H}$ be an algebraic group, and let $f:\mathbf {X}\rightarrow \mathbf {Y}$ be a principal $\mathbf {H}$-bundle, locally trivial in the Zariski topology. Then for every subset $A\subset Y$ we have the equality ${\text {sat}}'(f^{-1}(A))=f^{-1}({\text {sat}}'(A))$.

(d) For an ${\text {Ad}}\, P$-invariant subset $A\subset P$, the corresponding subset ${\text {Ind}}_P^G(A)\subset {\text {Ind}}_P^G(X)$ satisfies ${\text {sat}}'({\text {Ind}}_P^G(A))={\text {Ind}}_P^G({\text {sat}}'(A))$.

(e) Let $\mathbf {Y}\subset \mathbf {X}=\mathbb A^n$ be an open dense subvariety. Then ${\text {sat}}'_X (Y)=X$.

Proof

(a) is clear.

(b) The inclusion $\subset$ follows from (a). Conversely, assume that $a\in {\text {sat}}'(A)$ and $b\in {\text {sat}}'(B)$ are defined using triples $(\mathbf {V}_a,x_a,f_a)$ and $(\mathbf {V}_b,x_b,f_b)$, respectively, where $\mathbf {V}_a$ and $\mathbf {V}_b$ are open subsets of $\mathbb A^1$. Then we can assume that $\mathbf {V}_a=\mathbf {V}_b\subset \mathbb A^1$ and $x_a=x_b$, which implies that $(a,b)\in {\text {sat}}'(A\times B)$.

(c) Since the saturation ${\text {sat}}'$ is local in the Zariski topology, we can assume that $\mathbf {X}=\mathbf {Y}\times \mathbf {H}$. In this case the assertion follows from (b).

(d) Arguing as in 3.7(b), the assertion follows from (c).

(e) It suffices to show that for every $x\in \mathbb A^n(F)$, there exists a line $\mathbf {L}\subset \mathbb A^n$, defined over F, such that $x\in \mathbf {L}$ and $\mathbf {L}\cap \mathbf {Y}\ne \emptyset$. Consider the variety $\mathbb P_x$ of lines $\mathbf {L}\subset \mathbb A^n$ such that $x\in \mathbf {L}$. Since $\mathbf {Y}\subset \mathbb A^n$ is Zariski dense, the set of $\mathbf {L}\in \mathbb P_x$ such that $\mathbf {L}\cap \mathbf {Y}\ne \emptyset$, is Zariski dense. Since $\mathbb P_x\cong \mathbb P^{n-1}$, while F is infinite, the subset $\mathbb P_x(F)\subset \mathbb P_x$ is Zariski dense, and the assertion follows. $\square$

A.8. Remark. All the properties of ${\text {sat}}'$, formulated in Lemma A.7, have natural analogs for ${\text {sat}}$.

A.9. Relative saturation. Let $h:\mathbf {X}\rightarrow \mathbf {Y}$ be a morphism and $A\subset X$.

(a) Denote by ${\text {sat}}'(h;A)\subset {\text {sat}}'(h(A))$ the union $\cup _{(\mathbf {V},x,f)}f(x)$, taken over all triples $(\mathbf {V},x,f)$ in the definition of ${\text {sat}}'(h(A))$ (see A.4(a)) such that $f|_{\mathbf {V}'}:\mathbf {V}'\rightarrow \mathbf {Y}$ has a lift $\widetilde{f}':\mathbf {V}'\rightarrow \mathbf {X}$ with $\widetilde{f}'(V')\subset A$.

(b) If h is proper, then ${\text {sat}}'(h;A)= h({\text {sat}}'(A))$. Indeed, the valuative criterion implies that every pair $(f,\widetilde{f}')$ as in (a) defines a unique morphism $\widetilde{f}:\mathbf {V}\rightarrow \mathbf {X}$ such that $h\circ \widetilde{f}=f$ and $\widetilde{f}|_{\mathbf {V}'}=\widetilde{f}'$.

(c) Let $\mathbf {X}'\subset \mathbf {X}$ be an open subvariety such that $A\subset X'$, and let $h':=h|_{\mathbf {X}'}:\mathbf {X}'\rightarrow \mathbf {Y}$ be the restriction. By definition, ${\text {sat}}'(h';A)={\text {sat}}'(h;A)$.

A.10. Notation. (a) Let $a_{\mathbf {M},\mathbf {G}}:\widetilde{\mathbf {G}}_\mathbf {M}^{{\text {reg}}}\rightarrow \mathbf {G}$ be the map defined in 1.4 and 1.5, let $\nu _{\mathbf {G}}:\mathbf {G}\rightarrow \mathbf {c}_\mathbf {G}$ be the Chevalley map (see 1.1(a)), and set $e_\mathbf {G}:=\nu _\mathbf {G}(1)\in \mathbf {c}_\mathbf {G}$.

(b) Let $\mathbf {G}^{{\text {der}}}\subset \mathbf {G}$ be the derived group of $\mathbf {G}$, $\mathbf {Z}(\mathbf {M})$ the center of $\mathbf {M}$, and set $\mathbf {Z}_\mathbf {M}:=(\mathbf {Z}(\mathbf {M})\cap \mathbf {G}^{{\text {der}}})^0$. Then $\mathbf {Z}_\mathbf {M}$ is a split torus over F. Set $\mathbf {Z}^{{\text {reg}}}_{\mathbf {M}}:=\mathbf {Z}_\mathbf {M}\cap \mathbf {M}^{{\text {reg}}/\mathbf {G}}$. Notice that since $\mathbf {Z}_\mathbf {M}$ acts on ${\text {Lie}}\,\mathbf {G}/{\text {Lie}}\,\mathbf {M}$ by a direct sum of non-trivial characters, the subset $\mathbf {Z}^{{\text {reg}}}_{\mathbf {M}}\subset \mathbf {Z}_\mathbf {M}$ is open and dense.

(c) Set $C^{{\text {reg}}}_Z:=C\cdot Z^{{\text {reg}}}_M\subset M$. Since $C\subset M$ consists of unipotent elements, and $Z_M\subset Z(M)$, we have $C^{{\text {reg}}}_Z\subset M^{{\text {reg}}/G}$. Also $C^{{\text {reg}}}_Z$ is ${\text {Ad}}\, M$-invariant, so we can form a subset ${\text {Ind}}_M^G(C^{{\text {reg}}}_Z)\subset \widetilde{G}_M^{{\text {reg}}}$.

(d) Set $C^{{\text {reg}}}_{P,Z}:=p^{-1}(C^{{\text {reg}}}_Z)=C_P\cdot Z^{{\text {reg}}}_M\subset P^{{\text {reg}}/G}$ (see Lemma 1.9(d)).

A.11. Proof of Theorem A.5. Consider the subset $D_P:=a_{P,G}({\text {sat}}'({\text {Ind}}_P^G(C_P)))$ of G. Since $C_{P;G}$ is equal to $a_{P,G}({\text {Ind}}_P^G(C_P))$, we have inclusions $C_{P;G}\subset D_P\subset {\text {sat}}(C_{P;G})$ (see Lemma A.7(a)), thus ${\text {sat}}(D_P)={\text {sat}}(C_{P;G})$. It suffices to show that $D_P$ does not depend on $\mathbf {P}$. But this follows from the following description of $D_P$. $\square$

Claim A.12

We have the equality $D_P={\text {sat}}'(a_{M,G};{\text {Ind}}_M^G(C^{{\text {reg}}}_Z))\cap \nu _G^{-1}(e_\mathbf {G})$.

Proof

Recall (see 1.5(d)) that morphism $a_{M,G}$ factors as $\widetilde{G}_M^{{\text {reg}}}\overset{a_{M,P}}{\longrightarrow }\widetilde{G}_P\overset{a_{P,G}}{\longrightarrow } G$. Notice that $a_{M,P}:\widetilde{G}_M^{{\text {reg}}}\rightarrow \widetilde{G}_P$ is an open embedding (use Lemma 1.9(e)) and it satisfies $a_{M,P}({\text {Ind}}_M^G(C^{{\text {reg}}}_Z))={\text {Ind}}_P^G(C^{{\text {reg}}}_{P,Z})$. Therefore, by A.9(c), we have the equality

$$\begin{aligned} {\text {sat}}'(a_{M,G};{\text {Ind}}_M^G(C^{{\text {reg}}}_Z))={\text {sat}}'(a_{P,G};{\text {Ind}}_P^G(C^{{\text {reg}}}_{P,Z})). \end{aligned}$$

Next, since $a_{P,G}$ is proper, we conclude from A.9(b) that

$$\begin{aligned} {\text {sat}}'(a_{P,G};{\text {Ind}}_P^G(C^{{\text {reg}}}_{P,Z}))=a_{P,G}({\text {sat}}'({\text {Ind}}_P^G(C^{{\text {reg}}}_{P,Z}))). \end{aligned}$$

Thus it suffices to show the equality

$$\begin{aligned} a_{P,G}({\text {sat}}'({\text {Ind}}_P^G(C_P)))= a_{P,G}({\text {sat}}'({\text {Ind}}_P^G(C^{{\text {reg}}}_{P,Z})))\cap \nu _G^{-1}(e_\mathbf {G}). \end{aligned}$$

Using the commutative diagram from 1.5(b) for $\mathbf {H}=\mathbf {P}$ and equality $\pi _{P,G}^{-1}(e_\mathbf {G})=e_\mathbf {P}$ (see Lemma 1.9(f)), we conclude that $a_{P,G}(A)\cap \nu _G^{-1}(e_\mathbf {G})=a_{P,G}(A\cap \nu _{\widetilde{G}_P}^{-1}(e_\mathbf {P}))$ for every subset $A\subset \widetilde{G}_P$. Thus it suffices to show the equality

$$\begin{aligned} {\text {sat}}'({\text {Ind}}_P^G(C^{{\text {reg}}}_{P,Z}))\cap \nu _{\widetilde{G}_P}^{-1}(e_\mathbf {P})= {\text {sat}}'({\text {Ind}}_P^G(C_P))\subset \widetilde{G}_P. \end{aligned}$$

Using Lemma A.7(d), it suffices to show the equality

$$\begin{aligned} {\text {sat}}'(C^{{\text {reg}}}_{P,Z})\cap \nu _P^{-1}(e_\mathbf {P})= {\text {sat}}'(C_P)\subset P. \end{aligned}$$

(A.1)

Set $\mathbf {P}_{{\text {un}}}:=\nu _\mathbf {P}^{-1}(e_\mathbf {P})\subset \mathbf {P}$, and $\mathbf {P}_{\mathbf {Z}_\mathbf {M}}:=\nu _\mathbf {P}^{-1}(\nu _\mathbf {P}(\mathbf {Z}_\mathbf {M}))\subset \mathbf {P}$. Since the map $\nu _\mathbf {P}|_{\mathbf {Z}_\mathbf {M}}:\mathbf {Z}_\mathbf {M}\rightarrow \mathbf {c}_\mathbf {P}\cong \mathbf {c}_\mathbf {M}$ is a closed embedding, the multiplication map induces an isomorphism $\mathbf {P}_{{\text {un}}}\times \mathbf {Z}_\mathbf {M}\overset{\thicksim }{\rightarrow }\mathbf {P}_{\mathbf {Z}_\mathbf {M}}$. Moreover, it induces a bijection $C_P\times Z^{{\text {reg}}}_M\overset{\thicksim }{\rightarrow }C^{{\text {reg}}}_{P,Z}$. Thus, formula (A.1) follows from the equality

$$\begin{aligned} {\text {sat}}'_{\mathbf {P}_{{\text {un}}}\times \mathbf {Z}_\mathbf {M}}(C_P\times Z^{{\text {reg}}}_M)={\text {sat}}'_{\mathbf {P}_{{\text {un}}}}(C_P)\times {\text {sat}}'_{\mathbf {Z}_\mathbf {M}}(Z^{{\text {reg}}}_M)={\text {sat}}'_{\mathbf {P}_{{\text {un}}}}(C_P)\times Z_M, \end{aligned}$$

which follows from Lemma A.7(b),(e). $\square$

Corollary A.13

(a) If F is algebraically closed, then the closure ${\text {cl}}(C_{P;G})\subset G$ of $C_{P;G}$ in the Zariski topology does not depend on $\mathbf {P}$.

(b) If F is a local field, then the closure ${\text {cl}}(C_{P;G})\subset G$ of $C_{P;G}$ in the analytic topology does not depend on $\mathbf {P}$.

Proof

In both cases, every closed subset in G is saturated. Therefore we have inclusions $C_{P;G}\subset {\text {sat}}(C_{P;G})\subset {\text {cl}}(C_{P;G})$, which imply that ${\text {cl}}(C_{P;G})={\text {cl}}({\text {sat}}(C_{P;G}))$. Thus the assertion follows from Theorem A.5. $\square$

A.14. Notation. For an ${\text {Ad}}\, G$-invariant subset $D\subset G$, we denote by $D^{\heartsuit }\subset D$ the union of G-conjugacy classes that are Zariski dense in (the Zariski closure of) D.

A.15. Question. Is it true that $C^{\heartsuit }_{P;G}$ is independent of $\mathbf {P}$?

A.16. Remarks. (a) Let F be algebraically closed. Since the number of unipotent conjugacy classes in G is finite, we conclude that $C^{\heartsuit }_{P;G}$ is a single conjugacy class.

(b) Let F be general. Then, by (a), $C^{\heartsuit }_{P;G}$ is a union of unipotent conjugacy classes, belonging to a single conjugacy class over $\overline{F}$.

Lemma A.17

Let F be either algebraically closed or local. Then for every ${\text {Ad}}\, G$-invariant subset $D\subset G$, we have $D^{\heartsuit }={\text {sat}}(D)^{\heartsuit }$.

Proof

Let ${\text {cl}}(D)\subset G$ be the closure of D in the Zariski topology if F is algebraically closed, and in the analytic topology if F is local. Then, as in the proof of Corollary A.13, we have $D\subset {\text {sat}}(D)\subset {\text {cl}}(D)$. Thus, it suffices to show that $D^{\heartsuit }={\text {cl}}(D)^{\heartsuit }$.

Let $O\subset {\text {cl}}(D)$ be a Zariski dense G-conjugacy class, and let $\mathbf {D}\subset \mathbf {G}$ be the Zariski closure of D. Choose $x\in O$. Then the morphism $\mathbf {G}\rightarrow \mathbf {D}:g\mapsto gxg^{-1}$ is dominant. Therefore, in both cases, the corresponding map $G\rightarrow {\text {cl}}(D)$ is open. Thus $O\subset {\text {cl}}(D)$ is open, hence $O\subset D$. $\square$

Corollary A.18

Let F be either algebraically closed or local. Then the subset $C^{\heartsuit }_{P;G}\subset G$ does not depend on $\mathbf {P}$.

Proof

This follows immediately from Theorem A.5 and Lemma A.17. $\square$

A.19. Remark. We do not expect that the conclusion Lemma A.17 holds for an arbitrary field F. We wonder whether the equality ${\text {sat}}(C_{P;G})^{\heartsuit }=C^{\heartsuit }_{P;G}$ always holds.

Appendix B: On a theorem of Harish-Chandra

The goal of this section is to explain the proof of the following result, usually attributed to Harish-Chandra.

Theorem B.1

Let F be a local non-archimedean field of characteristic zero, and let $h\in \mathcal {H}(G)_G$ be such that $O_{\gamma }(h)=0$ for every $\gamma \in G^{{\text {rss}}}$. Then $h=0$.

B.2. The Lie algebra analog. Let $\mathfrak {g}$ be the Lie algebra of $\mathbf {G}$, equipped with the adjoint action of G, and let $h\in \mathcal {H}(\mathfrak {g})_G$ be such that $O_{x}(h)=0$ for every $x\in \mathfrak {g}^{{\text {rss}}}$. Then the original theorem of Harish-Chandra ([3, Thm 3.1]) asserts that $h=0$.

The goal of this section is to deduce Theorem B.1 from its Lie algebra analog.

B.3. G -domains. (a) Let X be a smooth analytic variety over F equipped with an action of G, let $\mathcal {H}(X)$ be the space of locally constant measures with compact support (see 3.4) and let $\mathcal {H}(X)_G$ be the space of G-coinvariants.

(b) By a G -domain in X, we mean an open and closed G-invariant subset $U\subset X$. Then $\mathcal {H}(U)\subset \mathcal {H}(X)$ is a G-invariant subspace, and the map $h\mapsto 1_U\cdot h$ is a G-equivariant projection $\mathcal {H}(X)\rightarrow \mathcal {H}(U)$. Taking G-coinvariants, we get an inclusion $\mathcal {H}(U)_G\hookrightarrow \mathcal {H}(X)_G$ and a projection $\mathcal {H}(X)_G\rightarrow \mathcal {H}(U)_G\subset \mathcal {H}(X)_G:h\mapsto h|_U$.

Lemma B.4

Let $f:X\rightarrow Y$ be a proper, surjective G-equivariant local isomorphism between smooth analytic varieties. Then the pullback map $f^*:\mathcal {H}(Y)_G\rightarrow \mathcal {H}(X)_G$ (see 3.5(b)) is injective.

Proof

For every $m\in \mathbb N$, we denote by $Y_m\subset Y$ the set of all $y\in Y$ such that the cardinality of $f^{-1}(y)$ is m. The assumptions on f imply that every $Y_m\subset Y$ is a G-domain, and that Y is the disjoint union of the $Y_m$’s. Then every $X_m:=f^{-1}(Y_m)\subset X$ is a G-domain as well, and it suffices to show that the induced map $f^*:\mathcal {H}(Y_m)_G\rightarrow \mathcal {H}(X_m)_G$ is injective. Since for every $h\in \mathcal {H}(Y_m)$ we have $f_!f^*(h)=mh$, we are done. $\square$

Proof of Theorem B.1

We carry out the proof in five steps.

Step 1. There exists a G-domain $U\ni 1$ in G such that $h|_U=0$.

Proof

Observe first that there exist G-domains $\mathfrak {u}\ni 0$ in $\mathfrak {g}$ and $U\ni 1$ in G such that the exponential map induces an ${\text {Ad}}\, G$-equivariant analytic isomorphism $\epsilon :\mathfrak {u}\overset{\thicksim }{\rightarrow }U$. Namely, if $\mathbf {G}=\mathbf {GL}_n$, the assertion is straightforward, and the general case follows from it.

We claim that this U satisfies the required property. Indeed, consider the pullback $h':=\epsilon ^*(h|_U)\in \mathcal {H}(\mathfrak {u})_G\subset \mathcal {H}(\mathfrak {g})_G$. It suffices to show that $h'=0$. For $x\in \mathfrak {g}^{{\text {rss}}}$ we have $O_{x}(h')=0$ if $x\notin \mathfrak {u}$, since $h'\in \mathcal {H}(\mathfrak {u})_G$, and $O_{x}(h')=O_{\epsilon (x)}(h)=0$ if $x\in \mathfrak {u}$ by our assumption on h. Then $h'=0$ by [3, Thm 3.1] (see B.2). $\square$

Step 2. For every $s\in Z(G)$ there exists a G-domain $U\ni s$ in G such that $h|_U=0$.

Proof

Since the map $g\mapsto gs:G\rightarrow G$ is ${\text {Ad}}\, G$-equivariant, the assertion follows from the $s=1$ case shown in Step 1. More precisely, if $U\ni 1$ is the G-domain constructed in Step 1, then $sU\ni s$ is the G-domain such that $h|_U=0$. $\square$

Step 3. For every semisimple $s\in G\smallsetminus Z(G)$ there exists a G-domain $U\ni s$ in G such that $h|_U=0$.

Proof

Let $\mathbf {H}:=\mathbf {G}_s^0$ be the connected centralizer of s. Then $\mathbf {H}\subsetneq \mathbf {G}$, and by induction, we may assume that Theorem B.1 is valid for $\mathbf {H}$.

Let $\mathbf {H}^{{\text {reg}}/\mathbf {G}}\subset \mathbf {H}$ and $\mathbf {c}_{\mathbf {H}}^{{\text {reg}}/\mathbf {G}}\subset \mathbf {c}_\mathbf {H}$ be the open subschemes defined in 1.2(b). Note that $\mathbf {H}\subset \mathbf {G}$ is an equal rank subgroup (see 1.6(a)), and $s\in \mathbf {H}^{{\text {reg}}/\mathbf {G}}$. Indeed, let $\mathbf {T}\ni s$ be a maximal torus of $\mathbf {G}$. Then $s\in \mathbf {T}\subset \mathbf {H}$, and $\mathbf {Z}_{\mathbf {G}}(s)^0=\mathbf {H}=\mathbf {Z}_{\mathbf {H}}(s)^0$. Hence $s\in \mathbf {H}^{{\text {reg}}/\mathbf {G}}$ by 1.6(b).

Let $\nu _{\mathbf {H}}:\mathbf {H}^{{\text {reg}}/\mathbf {G}}\rightarrow \mathbf {c}_{\mathbf {H}}^{{\text {reg}}/\mathbf {G}}$ be the Chevalley map (see 1.1(a) and 1.2(b)), and we denote by $\nu _H:H^{{\text {reg}}/G}\rightarrow c_{H}^{{\text {reg}}/G}$ the induced map on F-points (compare 3.8).

Choose an open and compact neigbourhood $V\subset c_{H}^{{\text {reg}}/G}$ of $\nu _H(s)$, and consider its preimage $U':=\nu _{H}^{-1}(V)\subset H^{{\text {reg}}/G}\subset H$. Then $U'\subset H$ is an H-domain, so we can form the induced space ${\text {Ind}}_H^G(U')$ (see 3.7) and the G-equivariant morphism $f:=a_{H,G}|_{{\text {Ind}}_H^G(U')}:{\text {Ind}}_H^G(U')\rightarrow G:[g,x]\mapsto gxg^{-1}$ (compare 1.4(c)).

Recall that the subset ${\text {Ind}}_H^G(U')\subset {\text {Ind}}_{\mathbf {H}}^{\mathbf {G}}(\mathbf {H}^{{\text {reg}}/\mathbf {G}})(F)$ is open and closed (by 5.2(a)). Since $a_{\mathbf {H},\mathbf {G}}:{\text {Ind}}_{\mathbf {H}}^{\mathbf {G}}(\mathbf {H}^{{\text {reg}}/\mathbf {G}})\rightarrow \mathbf {G}$ is étale (see Corollary 2.6), we conclude that f is a local isomorphism. On the other hand, since both morphisms $\iota _{\mathbf {H},\mathbf {G}}:{\text {Ind}}_{\mathbf {H}}^{\mathbf {G}}(\mathbf {H}^{{\text {reg}}/\mathbf {G}})\rightarrow \mathbf {G}\times _{\mathbf {c}_\mathbf {G}} \mathbf {c}_\mathbf {H}^{{\text {reg}}/\mathbf {G}}$ (see Corollary 1.11) and $\pi _{\mathbf {H},\mathbf {G}}:\mathbf {c}_{\mathbf {H}}\rightarrow \mathbf {c}_\mathbf {G}$ (see 1.1(b)) are finite, and $V\subset c_H$ is a compact subset, the composition

$$\begin{aligned} f:{\text {Ind}}_H^G(U')\overset{\iota _{H,G}}{\longrightarrow } G\times _{c_G} V \overset{\pi _{H,G}}{\longrightarrow }G \end{aligned}$$

is proper. Therefore $U:={\text {Im}}\,f$ is a G-domain containing s, and we claim that $h|_U=0$.

By Lemma B.4, the induced map $f^*:\mathcal {H}(U)_G\rightarrow \mathcal {H}({\text {Ind}}_{H}^{G}(U'))_G$ is injective. Let $\phi :\mathcal {H}(U)_G\rightarrow \mathcal {H}(U')_H$ be the composition of $f^*$ and the isomorphism $\varphi _H^G:\mathcal {H}({\text {Ind}}_{H}^{G}(U'))_G\overset{\thicksim }{\rightarrow }\mathcal {H}(U')_H$ from 3.7(c). Then $\phi$ is injective, thus it remains to show that $h':=\phi (h|_U)\in \mathcal {H}(U')_H\subset \mathcal {H}(H)_H$ is zero.

Since Theorem B.1 is valid for H, it suffices to show that $0_{\gamma }(h')=0$ for all $\gamma \in H^{{\text {rss}}}$. This is clear for $\gamma \notin U'$, since $h'\in \mathcal {H}(U')_H$. By construction, for every $\gamma \in U'\cap G^{{\text {rss}}}$ we have $O_{\gamma }(h')=O_{\gamma }(h)$. Hence $O_{\gamma }(h')=0$ by the assumption on h. This shows that $0_{\gamma }(h')=0$ for all $\gamma \in H^{G-{\text {rss}}}:=H\cap G^{{\text {rss}}}$.

Finally, since $S\cap H^{G-{\text {rss}}}\subset S\cap H^{{\text {rss}}}$ is dense for every maximal torus $\mathbf {S}\subset \mathbf {H}$, the equality $0_{\gamma }(h')=0$ for every $\gamma \in H^{{\text {rss}}}$ now follows from 6.10. $\square$

Step 4. Let $U\subset G$ be a G-domain, and let $g=su$ be the Jordan decomposition of $g\in G$. Then $g\in U$ is and only if $s\in U$.

Proof

Set $H:=G_s^0$. It suffices to show the closure of the ${\text {Ad}}\, H$-orbit of u contains 1, hence the closure of the ${\text {Ad}}\, G$-orbit of g contains s.

Since u is a unipotent element of H, the Zariski closure of the ${\text {Ad}}\, \mathbf {H}$-orbit of u contains 1. Hence the assertion follows from a theorem of Kempf [5, Cor. 4.3]. $\square$

Step 5: Completion of the proof. By Steps 3 and 4, for every $g\in G$ there exists a G-domain $U\ni g$ in G such that $h|_U=0$. From this the assertion follows. Indeed, choose a lift $\widetilde{h}\in \mathcal {H}(G)$ of h, and let $K\subset G$ be the support of $\widetilde{h}$. Since K is compact, there is a finite collection of G-domains $U_i,i=1,\ldots ,n$ such that $K\subset \cup _{i}U_i$ and each $h_i:=h|_{U_i}$ is zero. Moreover, replacing $U_j$ by $U_j{\smallsetminus }(\cup _{i=1}^{j-1}U_i)$ we can assume that the $U_i$’s are disjoint. Then $h=\sum _{i=1}^n h_i=0$. $\square$