I'am wondering whether there exists a non-discrete hyperbolic totally disconnected locally compact group such that the boundary is a finite-dimensional sphere. If the answer is positive, could you please provide few examples?



Examples of (non-discrete) hyperbolic totally disconnected locally compact groups whose boundaries are spheres
Non-discrete is not a reasonable assumption: for any hyperbolic group take the direct product with a compact group. Every hyperbolic group locally compact group $G$ has a unique maximal compact normal subgroup $W(G)$, which for $G$ non-elementary is the kernel of the $G$-action on the boundary $\partial G$. A reasonable assumption is to assume that $W(G)=1$, and in this case "non-discrete" is a reasonable assumption. I'll assume so, to make the question non-trivial. I will also assume that by "is a sphere" you mean "is homeomorphic to a sphere" (this is not an anecdotical point: for instance I don't know if there's a single (discrete) hyperbolic group whose boundary is homeomorphic, but not Hölder homeomorphic to a round sphere).


The Hilbert-Smith conjecture asserts that a non-discrete totally disconnected locally compact group cannot act continuously and faithfully on a connected topological manifold.


If one believes this conjecture, then the answer to your question is no. Indeed assume that $W(G)=1$, and that $\partial G$ is homeomorphic to a sphere (which is equivalent to $\partial G$ being a topological manifold):


* if $G$ is non-elementary, then it acts faithfully and continuously on its boundary $\partial G$, by the Hilbert-Smith conjecture we deduce that $G$ is discrete;
* if $G$ is any elementary hyperbolic group with $W(G)=1$, then it is known that $G$ has an open subgroup of index $\le 2$ that is isomorphic to one of the following three groups: $\{1\}$, $\mathbf{Z}$ and $\mathbf{R}$. Hence for $G$ totally disconnected, this again implies discrete.


For small-dimensional spheres the Hilbert-Smith conjecture is known to hold:


* for the 1-sphere it's an exercise;
* for the 2-sphere it's done in Montgomery-Zippin (although I think the proof has a gap, but anyway using facts from a [2006 survey of Kolev](https://arxiv.org/abs/math/0304106) [(MR link)](https://mathscinet.ams.org/mathscinet-getitem?mr=2300606) in l'Ens. Math., it's fine)
* for the 3-sphere it follows from the more solution in dimension 3 by J. Pardon.



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> So the answer (of the corrected question) is no for spheres of dimension $\le 3$, and no in general assuming the Hilbert-Smith conjecture.
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It is maybe possible to prove the Hilbert-Smith conjecture in the very specific setting of such actions: they are bilipschitz actions for some (non-Riemannian) metric, they are minimal, etc. Probably the natural setting would be convergence groups, so the question becomes whether if $G$ is a totally disconnected locally compact group acts continuously and faithfully on a sphere so that the action on the set of distinct triples is proper and cocompact, is $G$ discrete?


