# Is the commensurator of a tree lattice a simple group?

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?

• $\mathrm{Aut}(T)$ 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.
– YCor
Oct 24 '19 at 19:15
– HJRW
Oct 24 '19 at 20:21
• @YCor Ah yes, sorry I should have stated that. Oct 24 '19 at 20:38
• @YCor -- yes, I'm aware that they're different groups.
– HJRW
Oct 25 '19 at 11:35
• @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. Nov 14 '19 at 15:28

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 $$[W_m : B \cap W_m] < \infty$$.