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?



An example of a non-amenable exact group without  free subgroups.
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. 


