7
$\begingroup$

I am looking for a proof that a finitely generated hyperbolic groups is non-amenable [unless it is virtually cyclic] which is as "metric/combinatoric" as possible.

Here are the two proofs I am aware of: 1) use the arguments similar to "the centraliser of an element of infinite order is virtually cyclic" to show there is actually a free group [unless the group is virtually cyclic], 2) construct a Floyd boundary (or other ideal boundary) and ping-pong a free group out of the action on the boundary.

To make things slightly more precise, for a [finite] generating set $S$ and for a finite set $F$, define $\partial_S F$ to be the set of edges between $F$ and its complement in the Cayley graph with respect to $S$.

$\mathbf{Question:~}$ What is a reference for a "as direct as possible" proof that a finitely generated hyperbolic groups satisfy: given $S$ as above, there is a $K>0$ such that for all $F$ finite, $|\partial_S F| \geq K |F|$?

Both proofs above use quite heavily the algebraic side of the problem. And there is a minimal algebraic content to any proof of the above implication: it is easy to construct a non-Cayley graph counter-example to the above question (take any hyperbolic Cayley graph and add a half-line to it).

[Edit: just ignore this PS] PS: If it helps, one could assume that the group is finitely presented (i.e. it is the $\pi_1$ of some compact manifold). In fact, I would be delighted to see that this extra assumption can really be put to use.

[Edit: as suggested by Yves in the comments...]

$\mathbf{Possible~motivation~and~alternate~question:~}$ Given a quasi-transitive hyperbolic graph $\Gamma$, does it satisfy a "strong" isoperimetry, i.e. $|\partial F| \geq K |F|$ for some $K>0$?

Definition of quasi-transitive: there are constants $C$ and $L$ so that the set $SQI_{C,L}$ of all $(C,L)$-quasi-isometries from $\Gamma$ to itself is transitive, i.e. for any vertices $x$ and $y$ there is a $\phi \in SQI_{C,L}$ so that $\phi(x) =y$.

Improve this question
$\endgroup$
14
  • 1
    $\begingroup$ I am slightly puzzled, because if you are talking about Gromov/word hyperbolic groups, then they are all finitely presented, so why are you writing "finitely generated hyperbolic group" and why might one not assume that they are finitely presented? $\endgroup$
    – Derek Holt
    Mar 18, 2016 at 11:03
  • 1
    $\begingroup$ oh yeah, my bad! I got confused while writing this down. I think I meant that one could take this for granted (even if the proof of this fact might feel algebraic and not combinatoric to some). As for the "finitely generated", there are some people looking at hyperbolic groups which are not finitely generated, see "Amenable hyperbolic groups" by Caprace, Cornulier, Monod & Tessera. But yes, I was pedantic in writing this. $\endgroup$
    – ARG
    Mar 18, 2016 at 12:12
  • $\begingroup$ The classical books about word hyperbolic groups are written for finitely generated groups, and finite presentability is one of the first theorems proved about them. So writing "finitely generated hyperbolic groups" is perfectly natural, and writing "finitely presented hyperbolic group" would either sound as an unnecessary restriction, or as redundant. (By the way, Gromov used hyperbolic groups to mean a group endowed with a left-invariant distance that is a hyperbolic metric space, and specified to finitely generated groups with word metric as word hyperbolic groups). $\endgroup$
    – YCor
    Mar 18, 2016 at 12:39
  • 1
    $\begingroup$ Antoine, you could ask about a quasi-transitive, finite valency graph (in the sense that there exist $C,C'$ such that $(C,C')$-self-quasi-isometries are transitive on the graph. I'm not sure it's know that under these conditions, (hyperbolic with at least 3 boundary points) implies non-amenable. It's true for transitive graphs. $\endgroup$
    – YCor
    Mar 18, 2016 at 13:04
  • 1
    $\begingroup$ @Mikael: take a generic enough connected Lie group and consider a metric lattice (=Delone subset) inside. By generic enough, I mean: not quasi-isometric to any finitely generated group (this holds for semidirect products $\mathbf{R}\ltimes_{(e^t,e^{bt})}\mathbf{R}^2$ for $b\notin\{1,0,-1\}$, which are hyperbolic for $b>0$ and not (metrically) amenable, or for some one-parameter family $(G_t)$ 7-dimensional 3-step-nilpotent simply connected nilpotent Lie groups when $t$ is transcendent, using Pansu's theorem). $\endgroup$
    – YCor
    Mar 18, 2016 at 14:16

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.