Let $G$ be a discrete group. Let $C^*(G)$ denote the full group C*-algebra of $G.$ Let $\pi:C^*(G)\rightarrow \mathbb{C}$ be the *-homomorphism associated with the trivial representation of $G.$

Does there exist a group $G$ so that the kernel of $\pi$ is the unique non-trivial maximal ideal of $C^*(G)$?

Notice that if there is a $G$ with this property, then the trivial representation of $G$ must be weakly contained in the left regular representation of $G$, so $G$ must be amenable. It feels that no such group should exist, but I've had no luck constructing a second maximal ideal in a general $C^*(G).$

  • $\begingroup$ What exactly do you mean by a "non-trivial" maximal ideal? Is there a "trivial" maximal ideal that you have in mind? $\endgroup$ Apr 14, 2014 at 19:36
  • $\begingroup$ Manny, The trivial ideal I have in mind is the whole algebra $C^*(G).$ $\endgroup$ Apr 14, 2014 at 22:12
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    $\begingroup$ What about the universal locally finite group? This is a countable (discrete) locally finite group with the property that its ordinary group algebra $C G$ has exactly one ideal, the augmentation ideal. Since $G$ is lf, its C*-group algebra is AF with $CG$ dense and locally finite dimensional, and the closed ideals are in bijection with the ideals of $CG$. In particular, the group C*-algebra has a unique closed ideal. Very likely, there are no other ideals.{\par}Other groups to try are algebraically closed (ordinary group ring has unique proper ideal). $\endgroup$ Apr 15, 2014 at 2:44
  • $\begingroup$ David, Thanks a lot! It sounds like this group answers my question, right? If $CG$ has a unique closed ideal, then it must have a unique maximal ideal ($C^*(G)$ is unital so every maximal ideal is closed). Am I missing something? $\endgroup$ Apr 15, 2014 at 3:15
  • $\begingroup$ Yes, although I was hoping it could be proved to have exactly one proper nonzero ideal. Algebraically closed groups are also worth looking at. $\endgroup$ Apr 15, 2014 at 3:52


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