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Is the property of not containing the free group on two generators invariant under quasi-isometry? Amenability is, so if there is a counterexample it is also a solution to the von Neumann-Day problem (which of course already has a solution).

edited Feb 1 2012 at 2:24

Yemon Choi
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asked Feb 1 2012 at 1:22

Justin Moore
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Mark Sapir · Accepted Answer · 2012-02-01 02:29:25Z UTC

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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.

answered Feb 1 2012 at 2:29

Mark Sapir
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Mark, is the supposed proof contained in that Thompson F preprint, or is it something separate? – Yemon Choi Feb 1 2012 at 2:35

Thanks Mark.-- Justin – Justin Moore Feb 1 2012 at 2:43

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@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 non-trivial law". He says that the "big example" will be in the sequel of that paper but the sequel never happened. – Mark Sapir Feb 1 2012 at 2:55

@Mark: thank you for the information. – Yemon Choi Feb 1 2012 at 3:00

Is the property of not containing $\mathbb{F}_2$ invariant under quasi-isometry?

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