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Let $T$ be an $n$-regular tree ($n\geq3$). Let $\operatorname{Aut}^+(T)$ be the subgroup of index 2 of $\operatorname{Aut}(T)$ preserving the bicoloring of the tree for which adjacent vertices have distinct colours.

Let $\Gamma$ be a lattice in $G:=\operatorname{Aut}^+(T)$.

There are a number of open problems concerning whether the commensurator $\operatorname{Comm}_G(\Gamma)$ is discrete, or dense.

In the case where $\Gamma$ is uniform, the commensurator is known to be dense in $\operatorname{Aut}^+(T)$. This is also true for non-uniform lattices of Nagao type (Abramenko–Rémy).

In any of these cases or in any specific examples, is the commensurator $\operatorname{Comm}_G(\Gamma)$ known to be simple?

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    $\begingroup$ $\mathrm{Aut}(T)$ surjects on the cyclic group $C_2$, and hence so do its dense subgroups, so no dense subgroup can be simple. Maybe you want to ask about the the commensurator in $\mathrm{Aut}(T)^+$, the subgroup of index 2 (preserving the bicoloring of the tree for which adjacent vertices have distinct colors)— by the way, biregular trees is probably the right setting for such a question. $\endgroup$
    – YCor
    Oct 24 '19 at 19:15
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    $\begingroup$ I'd be tempted to e-mail Pierre-Emmanuel Caprace about this. This abstract seems relevant: warwick.ac.uk/fac/sci/maths/research/events/2017-18/symposium/… . $\endgroup$
    – HJRW
    Oct 24 '19 at 20:21
  • $\begingroup$ @YCor Ah yes, sorry I should have stated that. $\endgroup$
    – Sam Hughes
    Oct 24 '19 at 20:38
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    $\begingroup$ @YCor -- yes, I'm aware that they're different groups. $\endgroup$
    – HJRW
    Oct 25 '19 at 11:35
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    $\begingroup$ @HJRW I've added a partial answer based on an appendix of a paper of Caprace, which I believe is related to that talk abstract. $\endgroup$
    – Sam Hughes
    Nov 14 '19 at 15:28
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This is only a partial answer which follows from Henry Wilton's (@HJRW) comment. In the appendix of the paper here (arXiv 1712.01091) Caprace proves:

For $m\geq 3$, consider the commensurator ${\rm Comm}_{{\rm Aut}(T)}(W_m)$ of the free Coxeter group $W_m$ of rank $m$, where $T$ is the Cayley tree. Then,

  1. ${\rm Comm}_{{\rm Aut}(T)}(W_m)$ is almost simple
  2. ${\rm Comm}_{{\rm Aut}(T)}(W_m)$ contains a simple subgroup $B$ such that $[W_m : B \cap W_m] < \infty$.
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