1
$\begingroup$

Let $H$ be a co-dimension 1 quasiconvex subgroup of a one-ended hyperbolic group $G$. In particular, $\partial G$ is connected but $\partial G - \Lambda H$ is disconnected. The number of components of $\partial G - \Lambda H$ can be identified with the number of co-ends of $H$ in $G$ - i.e. the number of ends of the coset graph $H \backslash G$.

It is easy to see that $H$ acts on $\partial G - \Lambda H$ since $H$ acts on $\Lambda H$. My question is then the following.

Does $H$ stabilise each connected component of $\partial G - \Lambda H$, or does $H$ permute these components?

I think if $H$ has infinitely many co-ends in $G$, then $H$ must necessarily permute these components. However, I cannot think of examples with finitely many co-ends where these pieces are not fixed by $H$.

Any illuminating examples would be appreciated. Thanks!

$\endgroup$
1
  • 1
    $\begingroup$ Not really an interesting example, but a group of the form $(G \ast_H G) \rtimes_\varphi \mathbb{Z}/2\mathbb{Z}$, where $\varphi$ fixes $H$ and switches the two copies of $G$, should work (with $H \oplus \mathbb{Z}/2\mathbb{Z}$ as the codimension one subgroup). $\endgroup$
    – AGenevois
    Commented Aug 4, 2022 at 20:11

1 Answer 1

1
$\begingroup$

Here is a more natural example that the one I gave in the comments.

Given a graph $\Gamma$ and a collection of groups $\mathcal{G}= \{G_u \mid u \in V(\Gamma) \}$ indexed by the vertices of $\Gamma$, define the graph product $$\Gamma \mathcal{G}:= \langle G_u, \ u \in V(\Gamma) \mid [G_u,G_v]=1, \ \{u,v\} \in E(\Gamma) \rangle.$$ In other words, we take the free product of the $G_u$ and we add relations so that every element in $G_u$ commutes with every element in $G_v$ as soon as $u,v$ are adjacent vertices in $\Gamma$. For instance, one can think about a right-angled Coxeter groups.

If all the groups in $\mathcal{G}$ are finite, then $\Gamma \mathcal{G}$ is hyperbolic iff $\Gamma$ is square-free (i.e. no induced cycle of length four). The Cayley graph $\mathrm{QM}(\Gamma ,\mathcal{G}):= \mathrm{Cayl}(\Gamma \mathcal{G}, \bigcup \mathcal{G} )$ has a very nice geometry: it is a quasi-median graph. Like in median graphs (a.k.a. one-skeleta of CAT(0) cube complexes), there are hyperplanes and hyperplane-stabilisers, which are conjugates of subgroups of the form $\langle \mathrm{star}(u) \rangle$ in our case, are codimension-one subgroups (when they have infinite indices).

In the hyperplane-stabiliser $\langle \mathrm{star}(u) \rangle= \langle \mathrm{link}(u) \rangle \times \langle u \rangle$, the factor $\langle u \rangle$ permutes the sectors delimited by the corresponding hyperplane.

I can add more details if needed.

$\endgroup$
4
  • $\begingroup$ Excellent example, thank you. Small follow up question - is there anything that can be said about the number of orbits of these components? The ideal situation for me is that there's always at least two, but I'm not sure that's the case. $\endgroup$
    – jpmacmanus
    Commented Aug 5, 2022 at 8:58
  • $\begingroup$ I would say that there is a single orbit. But it is always possible to take a finite-index subgroup in order to increase the number of orbits. $\endgroup$
    – AGenevois
    Commented Aug 5, 2022 at 18:18
  • $\begingroup$ Thanks, though I'm not quite sure I can see how it's always possible to pass to a finite index subgroup to increase the number of orbits. In particular, if $\partial G - \Lambda H$ has infinitely many components, then a component stabiliser could have infinite index, right? $\endgroup$
    – jpmacmanus
    Commented Aug 7, 2022 at 9:10
  • $\begingroup$ What I wanted to say is that $\langle \mathrm{link}(u) \rangle \times \langle u^k \rangle$ is a codimension-one subgroup and probably the number of orbits here is $k$ (as soon as the order of $u$ is larger than $k$). $\endgroup$
    – AGenevois
    Commented Aug 8, 2022 at 12:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .