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I've been reading a proof concerning S-arithmetic subgroups of algebraic groups and I'm having trouble determining why the following step should be true. First, the setup:

Let $G$ be a connected absolutely simple adjoint algebraic group defined over a nonarchimedean local field $F$. Let $K \subset F$ be a global field and let $v_0$ be the pullback of the normalized valuation on $F$ using the embedding $K \hookrightarrow F$. Let $S$ be a set of discrete valuations on $K$ such that $S$ contains all archimedean places and does not contain any place where $G$ is anisotropic. Let $\mathcal{O}_K(S)$ be the ring of $S$-integers in $K$. Let $\Gamma = G(\mathcal{O}_K(S))$.

Let $\displaystyle G_S = \prod_{v \in S} G(K_v)$.

We are given that $\Gamma$ is discrete in $G(K_{v_0})$. We know from Margulis' "Discrete Subgroups of Semisimple Lie Groups," Theorem 3.2.5, that $\Gamma$ is a lattice in $G_S$.

The paper I'm reading concludes that $G(K_v)$ must be compact for all $v \in S \setminus \{ v_0 \}$. The thing that I'm having difficulty understanding is how this follow from the discreteness of $\Gamma$ and the fact that $\Gamma$ is a lattice in $G_S$. Can someone point me in the right direction?

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That $\Gamma$ is a lattice is due to Borel and Harish-Chandra in the early 60's. – YCor Mar 5 '13 at 21:38
I think that you need to elaborate a bit more (maybe point to the article? or at-least say something about $char(F)$?). Anyway, it sounds very similar to one of the steps in the proof of the Arithmeticity theorem, where Margulis' shows that the $p$-adic embeddings must be pre-compact. – Asaf Mar 5 '13 at 22:29
up vote 4 down vote accepted

This is strong approximation. $\Gamma$ is an $irreducible$ lattice in the product group $G_S$. If there is more than one place $v$ in $S$ other than $v_0$ where $G(K_v)$ is not compact, then the projection of $\Gamma$ to $G(K_{v_0})$ is dense, and therefore cannot be discrete. Margulis' book itself may contain this result that $\Gamma$ is an irreducible lattice. Otherwise, you may look at the book by Platonov and Rapinchuk "Algebraic Groups and Number Theory" where these facts are explained and proved.

I have assumed, in using strong approximation, that $G$ is $simply \quad connected$. You are assuming that $G$ is of adjoint type. However, once you conclude that $G(K_v)$ is compact for all other places for th simply connceted group, the same follows for the adjoint group.

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strong approximation again...seems like something worth knowing. – J. Martel Mar 6 '13 at 0:47
Thank you very much, this is exactly what I was looking for. – JSchw Mar 6 '13 at 1:36
@joshschw: No problem. Glad to be of help. @ Martel: yes, strong approximation is used $p$-adically, tp prove versions of the "Chinese remainder theorem" (strong approximation for unipotent groups), and in this case, could be used to prove denseness of images at archimedean places. – Venkataramana Mar 6 '13 at 3:09

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