Are there known any examples of non-amenable locally compact (or more restrictive, non-amenable discrete) groups $G$ for which the reduced group $C^*$-algebra $C_r^*(G)$ satisfies the universal coefficient theorem (UCT)? In this case, $C_r^*(G)$ is non-nuclear. For example,$G=\mathbb{F}_2$ the free non-abelian group in 2 generators is known to be non-amenable. However, $C^*(\mathbb{F}_2)$ is exact. However, I don't know whenever or not this $C^*$-algebra satisfies the UCT.

I am happy about any references, comments etc... Thanks


Both $C^\ast(\mathbb F_2)$ and $C^\ast_r(\mathbb F_2)$ satisfy the UCT. This is the special case of the following:

$\mathbf{Theorem}$. If $G$ and $H$ are countable, discrete, amenable groups, then $C^\ast(G\ast H)$ and $C^\ast_r(G \ast H)$ are $KK$-equivalent and satisfy the UCT.

$\mathbf{Proof}$. By Theorem 2.4 (c) in Cuntz' paper "$K$-theoretic amenability for discrete groups" it follows that $G\ast H$ is $K$-amenable and thus $C^\ast(G\ast H)$ and $C^\ast_r(G\ast H)$ are $KK$-equivalent. In the beginning of Section 3 of the same paper, Cuntz shows/remarks that $C^\ast(G\ast H)$ is $KK$-equivalent to the pull-back \begin{equation} C^\ast(G) \oplus_{\mathbb C} C^\ast(H) = \{ (x,y) \in C^\ast(G) \oplus C^\ast(H) : t_G(x) = t_H(y)\} \end{equation} via the trivial representations $t_G$ and $t_H$, so it suffices to show that this $C^\ast$-algebra satisfies the UCT. It fits into a short exact sequence \begin{equation} 0 \to I(G) \to C^\ast(G) \oplus_{\mathbb C} C^\ast(H) \to C^\ast(H) \to 0 \end{equation} where $I(G) = \mathrm{ker} \, t_G$ is the augmentation ideal. As $C^\ast(H)$ is nuclear the sequence is semi-split, so $C^\ast(G) \oplus_{\mathbb C} C^\ast(H)$ satisfies the UCT provided that $C^\ast(H)$ and $I(G)$ satisfy the UCT, by the 2-out-of-3-property for satisfying the UCT. $C^\ast(H)$ and $C^\ast(G)$ satisfy the UCT by Tu's theorem. As $I(G)$ fits into the split short exact sequence \begin{equation} 0 \to I(G) \to C^\ast(G) \to \mathbb C \to 0, \end{equation} and as $C^\ast(G)$ and $\mathbb C$ satisfy the UCT, so does $I(G)$ which completes the proof. QED

Note that for the free group $\mathbb F_2 = \mathbb Z \ast \mathbb Z$ one does not need to use Tu's deep theorem to obtain UCT.

To my knowledge, the only known group $C^\ast$-algebras which do not satisfy the UCT, are $C^\ast_r(G)$ when $G$ is a countable, discrete group with property (T) and the Akemann-Ostrand property, see Skandalis' "Une Notion de Nuclearite en K-Theorie". By also assuming that the group is residually finite, one can show that $C^\ast(G)$ does not satisfy UCT.


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