Question

Let $G$ be a locally compact Hausdorff group. It is known that $G$ can be topologically embedded in $W^{\ast}(G)$ , its universal $W^{\ast}$-algebra (with the $\sigma$-weak topology). An element $T \in W^{\ast}(G)$ is a function assigning to each representation $\pi$ a bounded operator $T(\pi) \in B(H_{\pi})$. This $T$ must be compatible with interwiners and $T(\pi)$ must be uniformly bounded.

This was done (in a slightly different language) by J. Ernest here: http://www.jstor.org/stable/2373020

Now define $G_{\otimes}= \{ T \in W^*(G)_{\neq 0} / T(\pi_1 \otimes \pi_2) = T(\pi_1) \otimes T(\pi_2) \}$

It's not hard to see that elements in $G_{\otimes}$ are unitaries. This is briefly proven here: http://cms.dm.uba.ar/Members/sergioyuhjtman/WG2.pdf/download (proposition 4.2).

Now Tatsuuma's duality theorem applies (Tatsuuma, proposition 2) so $G=G_\otimes$. But $G_\otimes$ is closed and inside the unit ball, so it is compact (always $\sigma$-weak topolgy). Therefore $G$ is compact.

Answer 1

A link to the literature: I think of $C^*(G)^*$ as being $B(G)$, the Fourier-Stieltjes algebra, realised as a (non-closed) algebra of continuous functions on $G$. Any member of $C^*(G)^*$ can be realised as the composition of a representation $\pi$ on $H$ with a vector functional $\omega_{\xi,\eta}$ on $H$. The resulting function in $B(G)$ is $g\mapsto (\pi(g)\xi|\eta)$.

Then $W^*(G)$ is $B(G)^*$. As the tensor product of representations corresponds to the product in $B(G)$, it follows that $G_{\otimes}$ is actually just the collection of characters on $B(G)$, namely algebra homomorphisms $B(G)\rightarrow\mathbb C$. Such things were explored by Walter in his paper "On the structure of the Fourier-Stieltjes algebra"

It's shown that $G_{\otimes}$ is not a group, and that it contains proper partial isometries and projections; it is a semigroup though.