This is somewhere between a "reference request" and "ask an expert", but I hope it is not too trivial or off-topic.

Anyway. There has been a lot of attention given to showing that for certain discrete groups $G$, the reduced group C*-algebra $C_r^*(G)$ is an exact C*-algebra in the sense of Kirchberg/Wassermann. For something I am working on, I want to consider certain classes of nondiscrete groups. (I am actually thinking of the case where the group von Neumann algebra is finite, but it's not clear to me how much that helps.)

This leads me to my question: does anyone know of results for when either of the following group C*-algebras are exact?

1) $C_r^*(G)$ where G is not discrete . If $G$ is amenable or connected then this algebra will be nuclear, hence in particular exact.

2) $C_d^*(G)$, which is defined to be the norm-closed subalgebra of ${\mathcal B}(L^2(G))$ generated by all left translation operators $\lambda_t$, with $\lambda_t(\xi)(s) = \int_G \xi(t^{-1}s)\\,ds$.

The second case would be of most interest to me right now (the problem is that for the groups I am interested in, $C^*(G_d)$ is too big to be exact, as $G_d$ will contain a nonabelian free subgroup).

I suspect that I am simply not looking in the right places or at the right parts of the standard literature, so a pointer to the right places would be enough.

Update October 2012: In a partial vindication for the old-school way of doing things, involving not this new-fangled InterWeb2.0 stuff but "giving a conference talk and raising a question": I have been informed by Narutaka Ozawa that for many examples of compact non-abelian connected Lie groups $G$, the algebra denoted above by $C^*_d(G)$ fails to be exact. In cases such as $SO(n)$ for $n$ large enough this follows from the existence of dense subgroups that have Property (T) as discrete groups; in the case $G=SU(2)$ one needs some spectral gap results that have been recently been obtained in hard work of Bourgain--Gamburd.

(I am going from memory of our conversation here, so if anything in the above is faulty, then it is almost surely a mistake on my part and not on Ozawa's.)

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    $\begingroup$ This paper by Claire Anantharaman-Delaroche seems relevant for 1): univ-orleans.fr/mapmo/membres/anantharaman/publications/…, see in particular Theorems 7.2 and 7.3. I have nothing to say about 2). $\endgroup$ Apr 2, 2011 at 4:11
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    $\begingroup$ Often the best way to show that some $C^\ast$-algebra is exact is to embed it into a nuclear one. By Kirchberg's work, this is actually a characterization of exactness. This is also the key to analyzing when reduced $C^\ast$-algebras of discrete groups are exact (in that case the nuclear $C^\ast$-algebra often comes from some action on a group boundary). Perhaps this may be helpful to your situation. About your comment after question 2), is $C^\ast(G_d)$ meant to be $C^\ast_d(G)$? Note that the reduced group $C^\ast$-algebra of a non-abelian free group is exact. $\endgroup$ Apr 2, 2011 at 5:56
  • $\begingroup$ Thank you, Theo. A quick look at the Anatharaman-Delaroche paper suggests that the results you refer to might not be directly helpful for what I'm trying to do; but I will think on this. $\endgroup$
    – Yemon Choi
    Apr 2, 2011 at 6:32
  • $\begingroup$ Thanks also, Dima. I hadn't thought about trying to embed into something nuclear via a crossed-product construction. To answer your second point: yes, in my comment I did mean $C^*(G_d)$. Obviously this quotients onto $C^*_d(G)$, so if the full $C*$-algebra of $G_d$ were exact, then $C^*_d(G)$ would also be exact. However, $G_d$ might contain a nonabelian free subgroup in cases I care about; I think this will stop $C^*(G)d)$ being exact (by combining various hereditary properties and known counter-examples). $\endgroup$
    – Yemon Choi
    Apr 2, 2011 at 6:37
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    $\begingroup$ @m07kl It is not easy (for me at least) so there is no "stupidity" on your part. I will try to update my post when I have time. But you may prefer to see a stronger result, that $C^*_d(G)$ is not even locally reflexive, which was obtained recently by Ruan and Wiersma: see Section 4 of arxiv.org/abs/1505.00805 $\endgroup$
    – Yemon Choi
    Aug 10, 2016 at 11:44

1 Answer 1


This might not be directly relevant, but it is known that $C^*_d(G)$ is nuclear if and only if $G_d$ is amenable: see Theorem 3 in Erik Bedos, ON THE C*-ALGEBRA GENERATED BY THE LEFT REGULAR REPRESENTATION OF A LOCALLY COMPACT GROUP, Proceedings of the American Mathematical Society, Vol. 120, No. 2 (Feb., 1994), pp. 603-608. The references in that paper might be interesting too.

  • $\begingroup$ Thanks. I was aware of Bedos's paper (which was partial motivation for my question) but will have another look at the references, following your suggestion. $\endgroup$
    – Yemon Choi
    Apr 26, 2011 at 21:00
  • $\begingroup$ See above for breaking news. Light shed in Luminy... $\endgroup$
    – Yemon Choi
    Oct 24, 2012 at 14:31

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