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Is it known that: Any countable Property (T) group (or more generally, a non-amenable group) has a faithful, weakly-mixing representation which is NOT weakly included in its left regular representation?

If this is true, then we would have a Gaussian action which is free and ergodic and not weakly contained (as an action) in the Bernoulli action. So, the second question:

Are there countable Property (T) groups (or more generally, non-amenable groups), that have a free and ergodic Gaussian action not weakly contained in the Bernoulli action?

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    $\begingroup$ Faithful sounds weird here and it's indeed unnecessary: if $\pi$ is a a weakly mixing rep not weakly contained in $\ell^2(G)$, then $\pi\oplus\ell^2(G)$ is faithful, weakly mixing (if $G$ is infinite), and not weakly contained in $\ell^2(G)$. $\endgroup$
    – YCor
    Oct 30, 2014 at 21:47
  • $\begingroup$ Indeed, thank you. I said faithful because I wanted the corresponding Gaussian action to be free. $\endgroup$ Oct 30, 2014 at 21:58
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    $\begingroup$ I think it is an open problem whether there exists such a non-amenable group that every unitary representation of it is weakly contained in the regular representation plus trivial. $\endgroup$ Oct 30, 2014 at 23:55
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    $\begingroup$ Any non-amenable group without property (T) has such a representation. By Theorem 1 in [Bekka, Valette: Kazhdan's property (T) and amenable representations. Math. Z. 212, 293-299 (1993)] any group without property (T) has a weak mixing amenable representation, which cannot be weakly contained in the left-regular representation if the group is non-amenable. $\endgroup$ Oct 31, 2014 at 17:26

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