Timeline for A "direct" proof that hyperbolic groups are not amenable
Current License: CC BY-SA 3.0
18 events
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| Dec 7, 2016 at 17:45 | comment | added | Cusp | Don't know direct or not but one can construct nontrivial quasimorphisms explicitly on hyperbolic groups (see the work of Hamenstatd or Calegari and Fujiwara). This means that the second bounded cohomology is nontrivial. Hence the group is not amenable. | |
| Mar 18, 2016 at 22:04 | comment | added | YCor | @MikaeldelaSalle: btw, for an arbitrary hyperbolic connected Lie group I'm not sure that my claim that "QI to any transitive graph" and "QI to any f.g. group" are indeed equivalent. This would follow from the Hilbert-Smith conjecture (using the action on the boundary). But actually this is also conjecturally equivalent to the very restrictive "acts geometrically on some rank 1 symmetric space of noncompact type". | |
| Mar 18, 2016 at 22:00 | comment | added | YCor | @MikaeldelaSalle (...) In the hyperbolic case, the arguments are in 6C of my survey arxiv.org/abs/1212.2229v2. In short: for $b>0$, $b\neq 1$, Pansu's results imply that any CGLC group QI to the 3-dimensional Lie group I mentioned, say $H_b$ is also focal; but a totally disconnected focal group has a totally disconnected boundary while the boundary of $H_b$ is a 2-sphere. So $H_b$ is not QI to any CGTDLC group. | |
| Mar 18, 2016 at 21:57 | comment | added | YCor | @MikaeldelaSalle First, "QI to a transitive (finite valency) graph" is the same as "QI to a CGTDLC group" (Compactly Generated Totally Disconnected Locally Compact). Second, Trofimov proved that any transitive graph of polynomial growth fibers with finite fibers to a graph with discrete automorphism group (that is, a f.g. group modulo a finite subgroup) and the LC counterpart (Lozert) is that any CGTDLC group of polynomial growth is compact-by-discrete. (...) | |
| Mar 18, 2016 at 20:46 | comment | added | Mikael de la Salle | Thanks Yves. What is a reference why it is the same? | |
| Mar 18, 2016 at 20:38 | comment | added | YCor | @MikaeldelaSalle PS: in my previous comment I mean "not QI to any transitive graph". This is stronger than "not QI to any f.g. group", but for hyperbolic Lie groups and groups with polynomial growth it's the same, so in particular the examples I gave are fine (at least for $b>0, b\neq 1$ in the first one). | |
| Mar 18, 2016 at 14:16 | comment | added | YCor | @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). | |
| Mar 18, 2016 at 13:57 | comment | added | ARG | @Mikael See Elek & Tardos "On roughly transitive amenable graphs and harmonic dirichlet functions" section 3. But this particular example is "amenable". | |
| Mar 18, 2016 at 13:53 | comment | added | Mikael de la Salle | @Yves: what are examples of quasi-transitive finite valency graphs which are not quasi-isometric to transitive graphs? | |
| Mar 18, 2016 at 13:52 | history | edited | ARG | CC BY-SA 3.0 |
Added the question suggested by YCor
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| Mar 18, 2016 at 13:43 | comment | added | ARG | @Yves, oh, yes that would make a much more interesting question! I'll edit it into the post. | |
| Mar 18, 2016 at 13:04 | comment | added | YCor | 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. | |
| Mar 18, 2016 at 12:39 | comment | added | YCor | 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). | |
| Mar 18, 2016 at 12:14 | history | edited | ARG | CC BY-SA 3.0 |
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| Mar 18, 2016 at 12:12 | comment | added | ARG | 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. | |
| Mar 18, 2016 at 11:53 | history | edited | ARG | CC BY-SA 3.0 |
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| Mar 18, 2016 at 11:03 | comment | added | Derek Holt | 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? | |
| Mar 18, 2016 at 10:20 | history | asked | ARG | CC BY-SA 3.0 |