Timeline for When is non amenablity witnessed by a single non measurable set?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 5, 2011 at 15:36 | vote | accept | Justin Moore | ||
| Apr 1, 2011 at 20:41 | answer | added | Justin Moore | timeline score: 11 | |
| Apr 1, 2011 at 16:34 | comment | added | Juris Steprans | The non amenability of Olshanskii's group is established by seeing it as a quotient of a free group and then looking at the limit of the number of words of length n in the kernel. (Actually, you look at the $n^{th}$ root of the number of words of length 2n.) This is unlikely to establish, one way or the other, the answer to Justin's question in that group. | |
| Apr 1, 2011 at 5:35 | comment | added | Stefan Geschke | @David: I heard the same in the same talk by Steprans, and he said that there is an example by Olshanskii of a non-amenable group that doesn't have a copy of $\mathbb F_2$. But what Justin is asking is whether it is the case that if the non-amenability is witnessed by a single set $E$ (the same set witnesses for all finitely additive probability measures that they are not translation invariant), then the group contains the free group on two generators? $\mathbb F_2$ is non-amenable for the reason Justin describes, but I have no idea what the situation with Olshanskii's groups is. | |
| Apr 1, 2011 at 3:13 | comment | added | David Fernandez-Breton | As far as I know, the question of whether every non-amenable group must contain $\mathbb F_2$ remains open (I heard it just last week in a talk by Steprans). | |
| Apr 1, 2011 at 1:16 | history | asked | Justin Moore | CC BY-SA 2.5 |