# Is there an countable amenable dense subgroup of $U(\ell^2 \mathbb N)$?

Question: Does the unitary group $U(\ell^2 \mathbb N)$, equipped with the strong operator topology, contain a countable dense subgroup which is amenable as a discrete group?

I would be also interested in answers to this for similar groups, such as for example for the Fredholm unitary group $U_C(\ell^2 \mathbb N)$ (with the uniform topology) or ${\rm Aut}(\mathbb Q)$ (with the topology of pointwise convergence). For ${\rm Aut}(\mathbb Q)$, this may be related (or even equivalent) to the question if Thompson's group $F$ is amenable; hence, this is probably a hard problem. For $U(\ell^2 \mathbb N)$, I do not even see a candidate for an amenable subgroup.

It is known that the set of pairs in $U(\ell^2 \mathbb N)$ which generate a finite subgroup is SOT-dense, so that there is no "local" obstruction to the generation of an amenable group; unlike for $U(\ell^2 \mathbb N)$ in the norm topology. Indeed, a pair of elements which is sufficiently norm-close to the generators of the left-regular representation of a non-abelian free group cannot generate an amenable group.

For $U(n)$, every amenable subgroup must be virtually solvable, so that it cannot be dense.

• Out of curiosity, do you know if it has a locally finite dense subgroup? – YCor Aug 27 '16 at 16:24
• @YCor: I do not know. I tried to show this starting with the density of $n$-tuples that generate a finite group, but I failed. – Andreas Thom Aug 27 '16 at 18:11
• Just to answer Yves' question: it is a known open problem to decide whether the unitary group has a dense locally finite subgroup. I think it is mentioned in Kechris' book on ergodic theory, though I'm unsure. In the case of permutation groups, density of n-tuples generating a finite group gives you a dense locally finite subgroup, but in general I see no reason why this should be true. – user97781 Aug 27 '16 at 18:27