4
$\begingroup$

Let $G$ be a group and let $K$ be a subgroup. Say $K$ is commensurated in $G$ if $gKg^{-1} \cap K$ has finite index in $K$ for all $g \in G$. Commensurated subgroups are an inherent feature of totally disconnected, locally compact groups: namely, if $G$ is t.d.l.c. and neither discrete nor compact, then $G$ has an infinite open profinite subgroup $K$ of infinite index, which is then commensurated in $G$. Moreover, given any group $G$ with a commensurated subgroup $K$, there is a natural localised profinite completion $\hat{G}_K$ of $G$ at $K$, which is a t.d.l.c. group such that the closure of the image of $K$ is an open compact subgroup of $\hat{G}_K$.

What I am looking for is examples of classes of countable groups where interesting commensurated subgroups arise naturally and/or have some significance, so that t.d.l.c. methods might be useful (similar to the way countable residually finite groups are sometimes studied via profinite groups). Here 'interesting' basically means 'not commensurate to a normal subgroup'. I know for instance they show up in linear groups over number fields, e.g. $\mathrm{SL}_n(\mathbb{Z})$ inside $\mathrm{SL}_n(\mathbb{Q})$. (In this vein, there is an article by Yehuda Shalom and George Willis on the arXiv about commensurated subgroups of arithmetic groups.) Any more examples coming from different contexts?

$\endgroup$
2
  • 2
    $\begingroup$ In the context of noncommutative geometry and operator algebras, commensurated subgroups are called almost normal. Regarding your question about the possibility of a T.D.L.C. approach to this class of subgroups, you may find some interesting notions and discussion in Tzanev's paper, titled "Hecke C*-algebras and amenability", J. Operator Theory, 50, (2003), 169–178. $\endgroup$
    – user23860
    Commented Aug 20, 2012 at 18:41
  • $\begingroup$ Interesting. I was not aware of the term 'Hecke pair'. I should also cite Schlichting's 1979 article in my document on the arXiv. $\endgroup$
    – Colin Reid
    Commented Aug 21, 2012 at 1:55

0

You must log in to answer this question.

Browse other questions tagged .