## Uniform lattices in semisimple Lie groups

Let $\Gamma$ be a uniform lattice in a semisimple Lie group $G$.

1. Must $\Gamma$ be virtually torsion-free?
2. If (1) is true, then does this work more generally if $G$ is reductive?

I am motivated by a prima facie knowledge of Theorem B of Armand Borel's paper, "Compact Clifford--Klein forms of symmetric spaces" (1963).

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A tag lie-groups would be more appropriate here, since gr is much too broad a classification. – Jim Humphreys Aug 26 2011 at 21:33

If $G$ is linear (which would be the case, for instance, if it is centerless) then it is special case of the more general fact that any finitely generated subgroup of $GL_n(F)$ for a field $F$ of characteristic zero is virtually torsion-free.

So all you need to know is that the lattices are finitely generated. This, for cocompact lattices, is easy, and for instance follows from what is usually called Milnor-Schwarz lemma, which is a very general lemma about cocompact isometric actions of groups on spaces. You can find a version of it in Pierre de la Harpe's book. For general lattices, in higher rank, this follows from property T and in rank 1 by some more geometric methods.

If not, this may fail. See this for instance: http://people.uleth.ca/~dave.morris/talks/deligne-torsion.pdf

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 Okay, I see, the first paragraph is Selberg-type lemma. The case of $F=\mathbb{C}$ is found in Proposition 2.1 and Section 5.1 of Borel's paper. Morris' note about covers of $Sp(2n,\mathbb{Z})$ in $Sp(2n,\mathbb{R})$ is very interesting, but it is not a counter-example since I required above that $G/\Gamma$ is compact. So my question now reduces to the case where $G$ is non-linear. Does an expert knowledge of Margulis' book shed any light on this? – Qayum Khan Sep 5 2011 at 1:34 See Raghunathan's paper "Torsion in cocompact lattices in coverings of Spin(2,n)" Mathematische Annalen, 266 (4). pp. 403-419, for a large supply of examples of uniform lattices (in finite covers over $Spin(2,n)$) which are not residually finite. No, Margulis' book will not help you with this. – Misha Mar 26 2012 at 21:44