Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer
2  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.