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.)