I was trying to understand a statement in Theorem 1.5 of this where the author seems to imply that if $G$ is a reductive group over $\mathbb{Q}$ such that $G/Z(G)$ is anisotropic, then for any function $f\in C^\infty_c(G(\mathbb{A}_\mathbb{Q}))$ the number of conjugacy classes $[\gamma]\in G(\mathbb{Q})/\sim$ such that $O_\gamma(f)\ne 0$ is finite. This led me to suspect that maybe the following is true/was what was meant:

Please read Edit(2)

Question: Let $G$ be a reductive group over $\mathbb{Q}$ which is anisotropic. Let $C\subseteq G(\mathbb{A}_\mathbb{Q})$ be a compact subset. Is the set of semi-simple $G(\mathbb{Q})$-conjugacy classes which intersect $C$ finite?

If this is not true, can anyone provide insight into what is meant by `that sum is finite'?

Also, if anyone can clarify what happens when only $G/Z(G)$ is assumed to be anisotropic, that would also be super helpful.


EDIT: It was suggested (in a now deleted comment) that there might be a super easy solution that assumes nothing about $G$ (just that it's an algebraic group). The deletion of that comment has me worried. The suggested proof (as far as I understood it) is as follows:

Proof: Note that $G(\mathbb{Q})$ is closed in $G(\mathbb{A}_\mathbb{Q})$. This follows, I believe, from observing that if $G$ embeds into $\mathbf{A}^n_\mathbb{Q}$ (affine space) for some $n$, which gives a closed embedding of $G(\mathbb{A}_\mathbb{Q})$ in to $\mathbb{A}_\mathbb{Q}^n$. But, $\mathbb{Q}^n$ is closed in $\mathbb{A}_\mathbb{Q}^n$ and so $\mathbb{Q}^n\cap G(\mathbb{A}_\mathbb{Q})=G(\mathbb{Q})$ is closed in $G(\mathbb{A}_\mathbb{Q})$. Note then that if $C\subseteq G(\mathbb{A})$ is compact, then $C\cap G(\mathbb{Q})$ is a closed subset of $C$ and thus compact (since $C$ is Hausdorff). But, we also have that $C\cap G(\mathbb{Q})$ is a subspace of $G(\mathbb{Q})$, but $G(\mathbb{Q})$ is discrete in $G(\mathbb{A}_\mathbb{Q})$ so that $C\cap G(\mathbb{Q})$ is discrete. This implies that $G(\mathbb{Q})\cap C$ is finite. Thus the number of rational conjugacy classes that meet $C$ is finite.

Is there an issue with this proof? Any comment for/against it would be greatly appreciated!

EDIT(2): I realized what I actually need is the following:

Let $C\subseteq G(\mathbb{A}_\mathbb{Q})$ is such that $C$ has compact image in $G(\mathbb{A})/A_G(\mathbb{R})^+$ (where $A_G$ is the maximal split component of $Z(G)$ and the $+$ denotes connected component) then $C$ meets only finitely many $G(\mathbb{Q})$-conjugacy classes. You may assume that $G/Z(G)$ is $\mathbb{Q}$-anisostropic, if that somehow helps.

If one only assumes that $C$ has compact image in $G(\mathbb{A}_\mathbb{Q})/A_G(\mathbb{R})^+$ do we still know that $C\cap G(\mathbb{Q})$ is finite? The examples I've done have seem to indicate this. I think I can prove it if $C$ is of the form $C'A_G(\mathbb{R})^+$ where $C'\subseteq G(\mathbb{A}^\infty_\mathbb{Q})$ is compact, and maybe some few more cases. Any comments appreciated (on this or the above proof in the case when $C$ is actually compact).

  • $\begingroup$ You need $G/Z$ to be compact to get that $C$ is compact. E.g., consider a unipotent conjugacy class in GL(2). $\endgroup$ – Kimball Jun 11 at 5:53
  • $\begingroup$ @Kimball What exactly do you mean? I am assuming that $C$ is compact to start with. Like I want for any conjugacy class $X$ in $G(\mathbb{Q})$ (this means $G(\mathbb{Q})$-conjugacy class) and $C$ to be compact that $X\cap C$ is finite. I don't whether the conjugacy class is compactly supported, if that's what you mean. I only want for an element $f$ of the Hecke algebra that $\mathrm{supp}(f)\cap X$ is finite. $\endgroup$ – NumberTheoryQuestions28 Jun 11 at 13:36
  • $\begingroup$ @Kimball I'm not trying to show the conjugacy class is compact, but only that finitely many conjugacy classes can intersect this compact subset. I want to justify why in the case of a function $f$ with compact support modulo $A_G(\mathbb{R})^+$ (see my second edit) only finitely many conjugacy classes have non-trivial orbital integral. $\endgroup$ – NumberTheoryQuestions28 Jun 11 at 15:04
  • $\begingroup$ @LSpice Did I claim that it's dense somewhere? $\endgroup$ – NumberTheoryQuestions28 Jun 11 at 15:06
  • $\begingroup$ Bourbaki shows that a compact subset of $G/H$ can be lifted to the product of $H$ with a compact subset of $G$, under relatively mild hypotheses. I don't have the reference to hand to check. $\endgroup$ – LSpice Jun 11 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.