# Can the full and reduced group $C^*$-algebras be “noncanonically” isomorphic?

Is there a locally compact group $G$ such that the canonical map from $C^{*}(G)$ to $C^{*}_{red} G$ is not isomorphism, hence $G$ is not amenable but these two $C^{*}$ algebras are isomorphic $C^{*}$ algebras via another morphism?

• It is known that if $C^*(G)$ and $C^*_r(G)$ are isomorphic (through any -isomorphism) then $G$ is amenable (and vice-versa of course). This should be no surprise I think, as C-algebras are quite rigid because of the C*-identity and hence the existence of a unique C*-norm on a C*-algebra. – Phoenix87 Jan 7 '15 at 16:53
• @Phoenix87 Can you provide a reference? (Alain Valette has already given a justification.) – Yemon Choi Jan 7 '15 at 18:29
• @YemonChoi please reread my comment carefully. I've never said they are intuitive, but that it shouldn't be surprising. As for a reference there must be something in Brown-Ozawa (although just for the discrete case), but I can't check as I don't have a copy of it with me right now. There is a mention to this fact at en.wikipedia.org/wiki/…, although there is no reference cited there – Phoenix87 Jan 7 '15 at 18:51
• @Phoenix87 I appreciate your point, and I have read your comments, but I think "This should be no surprise I think, as $C^*$-algebras are quite rigid" is not very solid reasoning. I agree that the statement sounds plausible, but many things sound plausible in mathematics without being true. – Yemon Choi Jan 7 '15 at 18:58
• For the interest of those reading: in general $C_r^*(G)$ does not even "remember" if $G$ is unimodular or not: see, for instance, the remarks on page 190 of J. Rosenberg's paper The $C^*$-algebras of some real and $p$-adic solvable gropups, Pac. J. M. 65 (1976) – Yemon Choi Jan 7 '15 at 20:30

If $C^*(G)$ is isomorphic to $C^*_r(G)$, then $C^*_r(G)$ has a 1-dimensional representation, i.e. $G$ has a 1-dimensional representation $\chi$ weakly contained in the regular representation $\lambda_G$. Then $1_G=\chi\otimes\overline{\chi}$ is weakly contained in $\lambda_G\otimes\overline{\lambda_G}\simeq \infty.\lambda_G$, hence $G$ is amenable.