Is the property of not containing the free group on two generators invariant under quasiisometry? Amenability is, so if there is a counterexample it is also a solution to the von NeumannDay problem (which of course already has a solution).
1 Answer
It is a famous open problem. Akhmedov in MR2424177 claimed he could prove that the answer is "no". No proof exists, so I guess he discovered a gap in his argument.

$\begingroup$ Mark, is the supposed proof contained in that Thompson F preprint, or is it something separate? $\endgroup$ Feb 1, 2012 at 2:35

5$\begingroup$ @Yemon: That is separate. The paper MR2424177 (see MathSci) actually contains the claim, but proves a much weaker (still nice, though!) result where "free subgroups" are replaced by "free subsemigroups" or "no nontrivial law". He says that the "big example" will be in the sequel of that paper but the sequel never happened. $\endgroup$– user6976Feb 1, 2012 at 2:55


$\begingroup$ @YemonChoi do you know if that "or" can be taken for two separate statements, or for one; i.e. are "no free subsemigroups" QIinvariant and "no nontrivial law" QIinvariant, or is "no free subsemigroups and no nontrivial law" QIinvariant? (sorry for asking, I don't have access to the paper) $\endgroup$– ARGApr 21, 2021 at 11:04

$\begingroup$ @ARG I'm afraid I never looked at the paper which Mark mentions, and I don't have immediate access to it although I can probably get hold of it through my university's VPN or similar if you need $\endgroup$ Apr 21, 2021 at 13:00