MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $G$.

share|cite|improve this question
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. – Gabor Szabo Feb 7 '13 at 22:03
Dense in the topology induced from $C^*(G)$. – Valerie Feb 7 '13 at 23:58
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. – Yemon Choi Feb 8 '13 at 0:19

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 $I\cap\ell^1(G)=I\cap C_c(G)=\{0\}$.

share|cite|improve this answer
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. – Valerie Feb 8 '13 at 0:03

Your Answer


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.