TITLE: Is the commensurator of a tree lattice a simple group?
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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,

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