Question. Let $S$ be a closed surface of genus $> 1$. Can $\pi_1(S)$ act faithfully and minimally on a simplicial tree of finite valence? Here "minimal" means that there is no invariant sub-tree.

Things I know related to this: a minimal action on an $\mathbb{R}$-tree $T$ (like a simplicial tree) gives a canonical measured foliation $F$ on $S$, and a measured foliation in turn gives an action on an $\mathbb{R}$-tree $T_F$, with a surjective equivariant map $T_F \to T$. For simplicial trees, the measured foliation will come by taking a finite collection of simple closed curves and expanding them to make a measured foliation. The resulting tree $T_F$ is then an infinite-valence simplial tree, on which $\pi_1(S)$ acts faithfully. So the question is basically whether you can fold that infinite-valence tree to get a finite-valence tree on which the action is still faithful.

Another point of view is to look at the resulting graph of groups (from Bass-Serre theory). But the edge groups cannot be cyclic (Skora proved this), and it's not clear to me how to proceed from this point.

Richard K. Skora, Splittings of surfaces, J. Amer. Math. Soc. 9 (1996), no. 2, 605--616.


Theorem 1.2 of Breuillard-Gelander-Souto-Storm in "Dense embeddings of surface groups" (http://arxiv.org/abs/math/0602635) says the following:

Let $G$ be a locally compact group. Suppose that $G$ contains a nondiscrete free subgroup $F$ of finite rank $r > 1$. Then $G$ has a subgroup $\Gamma$ containing $F$ such that $\Gamma$ is isomorphic to a surface group (of genus $2r$). In particular, if $G$ has a dense free subgroup of finite rank, then it has a dense surface group.

Apply this in the case $G$ being the automorphism group of a locally finite regular tree, which clearly contains a free group. Of course, you can always pass to a minimal sub-tree, or alternatively start with a dense free group and get a dense surface which thus will act minimally on the original tree.

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  • $\begingroup$ Thanks! That's great. I will study the paper. You do have to be careful about losing faithfulness when passing to a sub-tree, so I think you'll need the dense argument. (For instance, your original free group action might fix a vertex of the tree.) $\endgroup$ – Dylan Thurston Apr 13 '16 at 18:53
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    $\begingroup$ Dense surface subgroups in $\mathrm{PSL}_2(\mathbf{Q}_p)$ were previously known. A closed surface subgroup in $\mathrm{PSL}_2(\mathbf{Q}_p)$ is necessarily dense, and many arithmetic surface groups lie in $\mathrm{PSL}_2$ of some real quadratic extension of $\mathbf{Q}$, and this fits in $\mathbf{Q}_p$ for many values of $p$. $\endgroup$ – YCor Apr 13 '16 at 21:03

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