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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