# The number of rational semisimple conjugacy class/the Arthur-Selberg trace formula

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:

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.

Thanks!

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).

• You need $G/Z$ to be compact to get that $C$ is compact. E.g., consider a unipotent conjugacy class in GL(2). – Kimball Jun 11 at 5:53
• @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. – NumberTheoryQuestions28 Jun 11 at 13:36
• @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. – NumberTheoryQuestions28 Jun 11 at 15:04
• @LSpice Did I claim that it's dense somewhere? – NumberTheoryQuestions28 Jun 11 at 15:06
• 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. – LSpice Jun 11 at 15:06