Let $G$ be a non-abelian locally compact group, $M(G)$ be the measure algebra and $B(G)$ be the Fourier Stieltjes algebra of $G$..
Question. When are $M(G)$ and $B(G)$ semi-simple?
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3$\begingroup$ Both are always semisimple. I do not have time right now to hunt for the relevant references. In the case of B(G), one can deduce this from the fact that it is by definition an algebra of functions on G; a function in the radical of B(G) would have to be in the kernel of every character, and point evaluations at points of G are all characters, so the function would have to be zero. $\endgroup$– Yemon ChoiCommented Jan 25, 2017 at 22:39
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3$\begingroup$ Here is a reference for $B(G)$: sciencedirect.com/science/article/pii/0022123672900778 $\endgroup$– T. AmdeberhanCommented Jan 26, 2017 at 0:44
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