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A countable discrete group $\Gamma$ is said to be exact if it admits an amenable action on some compact space.

So clearly amenable groups are exact, but large familes of non-amenable groups are as well.

For many of the families that I know of (ex. linear groups, hyperbolic groups) that are exact, they also satisfy the von Neumann conjecture (i.e. that if they are non-amenable then they have subgroup isomorphic to a free group.)

So my questions is:

Are there examples of exact groups that are non-amenable and do not contain free subgroups?

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Hi Owen, this is funny...the example I had in mind when asking about a nonamenable i.c.c. simple group that is not inner amenable is the Olshanskii monster... an infinite simple group such that every element has order p (with p a large prime).The counterexamples to von Neumann's conjecture are all similarly horrible objects to this Olshanskii monster.The free Burnside groups is a limit (in the space of marked groups) of hyperbolic groups. If you can show that exactness survives such limit processes, then you'd have an example. This may be tough, though. –  Jon Bannon Jun 13 '10 at 13:37
    
In the case of the free Burnside groups, the limit is known to be aspherical...so some information is still there after the limit process, but this may be delicate. –  Jon Bannon Jun 13 '10 at 18:15
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Thompson's Group $F$ be may a good bet. To begin with, it does not contain any free subgroups (see the proof that it is not elementary amenable). However, the questions of ameanbility and exactness are (to the best of my knowledge) both still open. As a note on exactness, it does have Hilbert Space Compression equal to 1/2 (if it was >1/2 then it would be exact) by a paper of Arzhantseva, Guba and Sapir, so it is epsilon-close... –  user6503 Jun 15 '10 at 14:54
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@Agol: Exactness of $Gamma$ is equivalent to the topological amenability of some action on sa compact space. The definition is the following : the action of $Gamma$ on $X$ is exact if there is a map $\mu_n$ from $X$ to the set of probability measures on $\Gamma$, such that $\Vert \mu_n(sx)-s\mu_n(x)\Vert$ converges to 0 uniformly in $x$, as $n$ goes to infinity. See icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_74.pdf for references. –  Jean Lecureux Sep 15 '10 at 14:56
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@Justin: There is no non-triviality requirement. In fact, it is not true that for every group the trivial action is amenable. Because (following Jean's definition above) if $X$ is a point then the $\mu_n$ will be just a choice of probability measure on $\Gamma$ and so the condition is that $\Gamma$ admits an almost invariant sequence of probability measures (i.e. that $\Gamma$ is amenable). Actually there is a more general fact that if $\Gamma$ has a topological amenable action on some compact space, and there is an invariant measure on the space, then $\Gamma$ is amenable. –  Owen Sizemore Dec 13 '10 at 0:59
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3 Answers

up vote 11 down vote accepted

I did not check the details, but most probably Gromov's random groups can be made torsion as well (and thus will not contain $F_2$). Just impose the relations $u^{n_u}$ on the steps with even numbers and do Gromov's construction of embedding the next graph of an expander on steps with odd numbers. Here $u$ runs over all words of the free group (more precisely, $u_k$ is the smallest length word that has infinite order in the group number $k-1$, and $n_{u_{k}}>>1$, odd, is chosen after the word $u_k$ is determined). Thus there are non-exact groups without free subgroups. If one does only the even steps of this construction, then the resulting group will be torsion and non-amenable (it will have the previous group as its factor) and most probably exact, although I am not as sure about it as about the non-exact example above. So the answer to the original question is most probably "yes". The way to prove exactness can be through finiteness of asymptotic dimension.

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Owen, I'm a bit late to the party, but I think the answer to your question is ``no", to the best of my knowledge. To phrase it properly, I believe it is not known whether any of the known counterexamples to von Neumann's conjecture is exact.

Jon, one has to be careful with limits hyperbolic groups, for example Gromov's random groups which are not exact are such limits (they are lacunary hyperbolic, in the sense of Olshanskii, Osin and Sapir).

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That's good to know, Piotr, thanks! –  Jon Bannon Aug 4 '10 at 17:37
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It has been some years since this question was posted, but maybe (if you haven't seen it yet) you'll enjoy reading the new geometric solution for the von Neumann problem.

http://www.math.cornell.edu/~justin/Ftp/vN_fp.pdf

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I don't think the group above is known to be exact. –  Dan Sălăjan Dec 11 '13 at 21:57
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