# example of a non-amenable l.c. group such that $C_r^*(G)$ satisfies the UCT

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.

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 $$$$C^\ast(G) \oplus_{\mathbb C} C^\ast(H) = \{ (x,y) \in C^\ast(G) \oplus C^\ast(H) : t_G(x) = t_H(y)\}$$$$ via the trivial representations $$t_G$$ and $$t_H$$, so it suffices to show that this $$C^\ast$$-algebra satisfies the UCT. It fits into a short exact sequence $$$$0 \to I(G) \to C^\ast(G) \oplus_{\mathbb C} C^\ast(H) \to C^\ast(H) \to 0$$$$ where $$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)$$ fits into the split short exact sequence $$$$0 \to I(G) \to C^\ast(G) \to \mathbb C \to 0,$$$$ and as $$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.