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I am interested in a reference and proof for some version of the following (folklore?) statement:

``Let $G$ be a (semi)simple Lie group (with no compact factors and trivial centre) and let $\Gamma$ be an (irreducible) arithmetic lattice. Then $\mathrm{Comm}_G(\Gamma)$ is a simple group."

I have yet been unable to locate a precise statement or proof of this result. However, it is briefly alluded to on page 2 of this open problem list.

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  • $\begingroup$ I doubt this. Let $G$ be an $\mathbf{R}$-split $\mathbf{Q}$-form of $\mathrm{SL}_n$ that is anisotropic over $\mathbf{Q}_p$ for some prime $p$. So $G(\mathbf{Z})$ is an arithmetic cocompact lattice in $\mathrm{SL}_n(\mathbf{R})$. Then I'd expect that ($\ast$) the commensurator in $G(\mathbf{R})$ of $G(\mathbf{Z})$ to be contained in $G(\mathbf{Q})$ . But the latter is residually finite, since it is contained in $G(\mathbf{Q}_p)$ which is profinite. Still ($\ast$) has to be double checked. $\endgroup$ – YCor Oct 29 '19 at 17:47
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Hmm, that's not exactly true. For example, $Comm_{SL_2(\mathbb{R})}(SL_{2}(\mathbb{Z}))$ contains the normal subgroup $\pm I$.

For a special case describing the commensurator (when the complexified algebraic group has trivial center), see Ex. 4, §5.2. https://arxiv.org/abs/math/0106063 For the general case, map to the product of the projective groups of the simple factors, which will have trivial center when complexified, and hence satisfy the conditions of the exercise. Then pull back to get the full description. I think then that sometimes the commensurator will simple modulo its center.

Under certain hypotheses this is known. See Tits' Simplicity Theorem. But as indicated in Yves de Cornulier's comment about, the commensurator will not always be simple.

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  • $\begingroup$ Yes of course we need to take $\mathrm{Comm}_G(\Gamma)/Z(\mathrm{Comm}_G(\Gamma))$. ls it obvious why there should be no other quotients of $\textrm{Comm}_G(\Gamma)$? $\endgroup$ – Sam Hughes Oct 29 '19 at 16:40

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