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In reduction theory of arithmetic groups, one has the following finiteness property.

Proposition 1 (Siegel property). Let $G$ be a reductive group over $\mathbb{Q}$ and let $\Gamma\subset G(\mathbb{Q})$ be an arithmetic subgroup. Let $\mathfrak{S}\subset G(\mathbb{R})$ be a Siegel set. Then the set $$\{\gamma\in \Gamma; \mathfrak{S}\cap \gamma. \mathfrak{S}\ne \emptyset\}$$ is finite.

Now I am consider a generalization of the above finiteness property as following.

Question 2 Let $G$ be a reductive group over $\mathbb{Q}$ and let $\Gamma\subset G(\mathbb{Q})$ be an arithmetic subgroup. We fix $\mathfrak{S}\subset G(\mathbb{R})$ be a Siegel set. Take $H\subset G$ be a reductive subgroup over $\mathbb{Q}$. Then is there a Siegel set $\mathfrak{S}_H\subset H(\mathbb{R})$ such that the set $$\{\gamma\in \Gamma; \mathfrak{S}\cap \gamma. \mathfrak{S}_H\ne \emptyset\}$$ is finite?

So is this generalization still true? Could you please provide a proof or a counter example? Thanks very much!

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  • $\begingroup$ If I'm understanding this correctly, then this appears to follow from Theorem 1.2 of this paper: arxiv.org/pdf/1609.01315.pdf $\endgroup$ – Keerthi Madapusi Pera Aug 12 '17 at 19:54
  • $\begingroup$ @KeerthiMadapusiPera Thanks a lot for your comment, however, my question is different from Theorem 1.2 in Orr's paper, because I fix a Siegel set of $G(\mathbb{R})$ in the beginning, while Orr's theorem starts from a Siegel set of $H(\mathbb{R})$. $\endgroup$ – Golden Wave Aug 12 '17 at 20:27
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This is true (at least up to conjugating H). See Lemmas 7.5 and 7.7 in "Arithmetic subgroups of algebraic groups".

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