# Where is the error in this argument?

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.

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Why is $G_{\otimes}$ closed in the $\sigma$-weak topology? – Matthew Daws Apr 12 '12 at 15:58
Because the functional $<(-)(\pi) \alpha, \beta>$ is sigma-weakly continuous. This can be found in the first two links in the question. – Sergio A. Yuhjtman Apr 12 '12 at 17:48
Shouldn't the title be "Where is the error?" – Martin Brandenburg Apr 12 '12 at 18:06
The product condition may be a closed condition, but being non-zero doesn't look like a closed condition. So what we can say is that $G_\otimes\cup \{ 0\}$ is closed. That would mean, we get the one-point compactification, that's all. Anyway, have you tried looking at the example $G$ equals the integers with the discrete topology? – Anton Apr 12 '12 at 19:18
$G_\otimes$ is not a group. You make a mistake in your proof when you assume that for an infinite dimensional Hilbert space the map $\xi \otimes \overline{\eta} \mapsto \langle \xi, \eta \rangle$ extends to a bounded linear operator. – Jesse Peterson Apr 13 '12 at 2:43

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.
@Sergio, that seems a slightly unfair comment. Matthew has stated exactly the same reason that Jesse gave, namely that $G_\otimes$ is NOT a group, and he provides a link to a paper which shows this and gives details. – Yemon Choi Apr 13 '12 at 21:41