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Andreas Thom
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Does there exist a finitely generated discrete amenable group $G$ that acts on a seperableseparable 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$$\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.

Does there exist a finitely generated discrete amenable group $G$ that acts on a seperable 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 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.

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.

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Vladimir
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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 seperable 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 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.