It is a standard result that closed subgroups of locally compact amenable groups are themselves amenable, so for example $F\_2$, the free group on two generators, cannot be embedded as a closed subgroup of a locally compact amenable group. However, by a result of Pestov, $F\_2$ embeds as a closed subgroup of $\mathrm{Aut}(\Bbb Q,\leq)$ and the latter group is (extremely) amenable.


Are there topological groups that cannot be embedded as a closed subgroup of any amenable group?



Does every topological group embed as a closed subgroup in an amenable group?
At least for Polish groups, which was the case I was most interested in, the answer is positive.


I mentioned this question to Ola Kwiatkowska yesterday and she immediately pointed out that one of the standard universal Polish groups, the group $\mathrm{Iso}(\Bbb U)$ of isometries of the Urysohn space, is in fact not only amenable, but even extremely amenable (and Polish subgroups of Polish groups are closed).


For arbitrary groups I still expect the answer to be positive, but I have no meaningful comments to make about that case, apart from the fact that if a group embeds as a closed subgroup into an amenable group $G$, then it also embeds as a closed subgroup into an extremely amenable group, namely $L^0(G,X,\mu)$.


