Let $G$ be a locally compact group. Is the Fourier-Stieltjes algebra $B(G)$ semi-simple? 



Is the Fourier-Stieltjes algebra of a locally compact group semi-simple?
(The following was originally a comment, but I've moved it to an answer. Please don't TeXify this.)




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Recall that for commutative unital algebras, the Jacobson radical is exactly the set of quasi-nilpotent elements. By definition (see e.g. [Eymard's original 1964 article](http://www.numdam.org/item?id=BSMF_1964__92__181_0)), B(G) is an algebra of functions – you do not need to define it as a completion – and for each x in G, "evaluation at x" is a character on the unital, commutative Banach algebra B(G). So if f is a quasi-nilpotent element of B(G) then it lies in the kernel of every character, hence f(x)=0 for all x in G, hence f = 0.


