Does $\pi \otimes \overline{\pi}$ contain weakly the identity representation $1_G$? Let $G$ be a locally compact group and $\pi \colon G \to B(H)$ be a strongly continuous unitary representation on a Hilbert space $H$. If $H$ is finite-dimensional and non-zero then it is well-known that $\pi \otimes \overline{\pi}$ contains the identity representation $1_G$.
Now suppose that $H$ is infinite dimensional. Does $\pi \otimes \overline{\pi}$ contain weakly the identity representation $1_G$?
 A: Sometimes and not always.
If $\pi$ contains a nonzero finite-dimensional representation, the answer is yes. If $\pi$ weakly contains the 1-dimensional trivial representation, the answer is also yes, and still more generally if $\pi$ weakly contains a nonzero finite-dimensional representation.
If $G$ has Kazhdan's Property T and $\pi$ does not contain any nonzero finite-dimensional representation then $\pi\otimes\bar{\pi}$ does not contain the trivial 1-dimensional representation, and hence does not weakly contain it, by Property T.
More generally if $V$ is a subgroup of $G$ and $(G,V)$ has relative Property T, and $\pi|_V$ does not contain any nonzero finite-dimensional representation then $(\pi\otimes\bar{\pi})|_V$ does not contain the identity 1-dimensional representation, and hence $\pi\otimes\bar{\pi}$ does not weakly contain it, by relative Property T.
I'm not sure what the general picture is. I do not know any case where $\pi$ does not weakly contain any nonzero finite-dimensional representation, while $\pi\otimes\bar{\pi}$ weakly contains the 1-dimensional trivial representation.
