# Can an amenable group have a weak mixing unitary representation without almost invariant vectors?

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.

• Yes, with $G=\mathbf{Z}$. Just split the regular representation $\ell^2(\mathbf{Z})$ as $H_+\oplus H_-$ where $H_-$ corresponds to the part of the spectrum with negative real part. Then $H_-$ is your representation. Equivalently, consider $U_-$, the subset of the unit complex circle with negative real part and Lebesgue meausre, and consider the unitary operator of $L^2(U_-)$ given by $f\mapsto (z\mapsto zf(z))$. – YCor Apr 18 '17 at 9:01
• You are right, I deleted this as it was nonsense. – Andreas Thom Apr 18 '17 at 10:42

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.
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.