Does there exist a finitely generated discrete amenable group $G$ that acts on a separable Hilbert space $\mathcal{H}$ by unitary transformations, and where (1) $\mathcal{H}$ has no finite dimensional $G$-invariant closed subspaces and (2) there does not exist a sequence $(v\_n)\_n$ of unit vectors in $\mathcal{H}$ such that $\lim\_n \|g.v\_n - v\_n\|=0$ for all $g \in G$?


This is impossible if $\mathcal{H}$ is $L^2$ of an invariant probability measure.



Can an amenable group have a weak mixing unitary representation without almost invariant vectors?
If $G$ is amenable, then every weakly mixing representation $\pi$ has almost invariant finite-dimensional subspaces. This means that $\pi \otimes \bar \pi$ has almost invariant vectors.


Results like this can be found in


M.E.B. Bekka. *Amenable unitary representations of locally compact groups*. Invent. Math. **100** (1990), 383–401.


A concrete example of a weakly mixing representation without almost invariant vectors would be given by the Heisenberg group $H(\mathbb Z)$ acting on $\ell^2(\mathbb Z^2)$ with generator of the center acting by multiplication with some irrational $\theta \in S^1$. It is easy to see that there are no finite-dimensional subrepresentations and since the generator of the center acts by $\theta$ (and not by $1$), there cannot be any almost invariant vectors.


