Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
2  
Why is $G_{\otimes}$ closed in the $\sigma$-weak topology? –  Matthew Daws Apr 12 '12 at 15:58
1  
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
3  
Shouldn't the title be "Where is the error?" –  Martin Brandenburg Apr 12 '12 at 18:06
1  
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
7  
$G_\otimes$ 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
show 8 more comments

1 Answer

up vote 5 down vote accepted

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.

share|improve this answer
    
This is a good reference, thank you! But the real answer is what Jesse Peterson said. –  Sergio A. Yuhjtman Apr 13 '12 at 15:33
    
@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
    
OK, I'll accept the answer to be polite and grateful (actually I want to thank everyone who got involved!) but I still think as before. –  Sergio A. Yuhjtman Apr 14 '12 at 1:28
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.