3
$\begingroup$

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 $G$.

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 7, 2013 at 22:03
  • $\begingroup$ Dense in the topology induced from $C^*(G)$. $\endgroup$
    – Valerie
    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
    Feb 8, 2013 at 0:19

1 Answer 1

6
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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
    Feb 8, 2013 at 0:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.