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Let $G$ be a unimodular locally compact group (my main examples are algebraic groups over local fields. Thefore we can assume $G$ is Type I, if necessary). Then there are at least three group algebras associated to $G$ : the algebra $C_c(G)$ (with convolution) of continuous functions of compact support, the algebra $L^1(G)$ (with convolution), and the $C^\star$-algebra of $G$, denoted by $C^\star(G)$.

Let $I$ be a primitive ideal in $C^*(G)$.

(a) Is it true that $I\cap L^1(G)$ is dense in $I$?

(b) Is it true that $I\cap C_c(G)$ is dense in $I$? If not, then is it true that $I\cap C(G)$ is dense in $I$? (Here $C(G)$ denotes the space of continuous functions on $G$.

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    $\begingroup$ You should be careful with taking these intersections so naively. (a) does not really make that much sense to me, since $L^1(G)$ and $C^*(G)$ have different structure coming from different norm completions. So it is not clear in what sense "dense" has to be understood. Taking the intersection with $C(G)$ is even more problematic, since that does not have a natural topology on it. $\endgroup$
    – Gabor Szabo
    Commented Feb 7, 2013 at 22:03
  • $\begingroup$ Dense in the topology induced from $C^*(G)$. $\endgroup$
    – Valerie
    Commented Feb 7, 2013 at 23:58
  • $\begingroup$ I have to say I have no problem taking these intersections (since they always contain $0$) and I thought it was clear from context that $I$ is closed and equipped with the subspace norm from the full group $C^\ast$ algebra. $\endgroup$
    – Yemon Choi
    Commented Feb 8, 2013 at 0:19

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This is certainly false in the non-type I case. Let $G$ be the free group on 2 generators, and let $\pi$ be an irreducible representation of the reduced $C^*$-algebra $C^*_r(G)$. By simplicity of $C^*_r(G)$, the rep $\pi$ is faithful. Now let $\lambda_G:C^*(G)\rightarrow C^*_r(G)$be the surjective homomorphism corresponding to the left regular representation. View $\pi\circ\lambda_G$ as an irreducible representation of $C^*(G)$, let $I$ be its kernel (it is non-zero, as $G$ is non-amenable). Since $\lambda_G|_{\ell^1(G)}$ is faithful, we have $I\cap\ell^1(G)=I\cap C_c(G)=\{0\}$.

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  • $\begingroup$ Thanks Alain Valette. On the positive side, I just found a paper of Boidol, Leptin, Schurman, and Vahle (Math. Ann. 1978) which states that the answer to my Question (a) is affirmative in the case of groups with polynomial growth. I am wondering if the answer to my Question (b) is also affirmative in this case. Any hints/remarks would be helpful. $\endgroup$
    – Valerie
    Commented Feb 8, 2013 at 0:03

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