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I am looking for an example of a group $G$ that acts (cocompactly and) acylindrically on a hyperbolic graph $\Gamma$, such that
a) the graph $\Gamma$ is fine,
b) $\Gamma$ is not a tree,
c) not all edges have finite stabilizer,
d) the action of $G$ on $\Gamma$ is minimal, i.e. there is no proper, connected $G$-invariant subgraph of $\Gamma$.

Here I use the definition that $G$ acts acylindrical on $\Gamma$ iff there is a $k\in\mathbb{N}$ such that all geodesic segments of length $k$ in $\Gamma$ have finite (pointwise) stabilizer.

Examples are known if one condition is dropped: If I drop c), then any relative hyperbolic group (in the sense of Bowditch) is an example. If I drop b) it is easy enough to construct examples by taking say a malnormal subgroup $C\leq A$ and letting $A\ast_C B$ act on its Bass-Serre tree. If I drop a), then any acylindrically hyperbolic group will do by definition (at least if I drop cocompactly as well).

Does anyone know where to find an example in the literature satisfying a)-d) and/or how to construct one?

Edit (Oct 23rd 2015): Condition d) was added.
Edit (Oct 24th 2015): added "connected" in Condition d)

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  • $\begingroup$ Could you define what a fine graph is? $\endgroup$
    – YCor
    Commented Oct 22, 2015 at 14:36
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    $\begingroup$ One could easily concoct artificial examples from known examples. For instance, take a relatively hyperbolic example, so c) fails, let $V$ be a vertex with infinite stabilizer, and for each vertex in the orbit $G \cdot V$, attach an edge with both endpoints at that vertex. $\endgroup$
    – Lee Mosher
    Commented Oct 22, 2015 at 14:41
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    $\begingroup$ "Fine" is defined by Bowditch, meaning that for each edge $E$ and integer $L \ge 1$ the graph has only finitely many circuits of length $L$ passing through $E$. homepages.warwick.ac.uk/~masgak/papers/bhb-relhyp.pdf $\endgroup$
    – Lee Mosher
    Commented Oct 22, 2015 at 14:42
  • $\begingroup$ @LeeMosher thanks for the example! When asking the question I had a minimal action of $G$ on $\Gamma$ in mind though. I edited the question accordingly. $\endgroup$
    – Svenja Knopf
    Commented Oct 23, 2015 at 8:24
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    $\begingroup$ @Misha Replacing one edge in a tree with infinitely many edges between its two endpoints results in a graph that is not fine. So this is excluded by a). $\endgroup$
    – Svenja Knopf
    Commented Oct 24, 2015 at 6:14

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