Let $G$ be an amenable group. I wonder whether it is true that the reduced group $C^*$algebra $C_r^*(G)$ is quasidiagonal.
1 Answer
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This is true for countable discrete groups by the celebrated Quasidiagonality Theorem of TikuisisWhiteWinter (Quasidiagonality of nuclear C∗algebras. Ann. of Math. (2) 185 (2017), no. 1, 229–284.)

$\begingroup$ If G is a uncoutable discrete group, does the above conclusion also hold? $\endgroup$ Commented May 13, 2020 at 14:28

1$\begingroup$ Quasidiagonality is a local property for $C^\ast$algebras (this follows for instance by Voiculescu's characterisation of quasidiagonality combined with Arveson's extension theorem). By writing your uncountable group as a direct limit of countable subgroups, you also obtain quasidiagonality for uncountable discrete groups. $\endgroup$ Commented May 14, 2020 at 4:43

$\begingroup$ Thanks, Pro Gabe. Please forgive my stupidity. If the amenable group is not discrete, does there exist a counterexample to show that the associated reduced group $C^*$algebra is quasidiagonal? $\endgroup$ Commented May 14, 2020 at 8:05

1$\begingroup$ There exist nondiscrete amenable groups with nonquasidiagonal C*algebras, see [Beltiţă, Ingrid; Beltiţă, Daniel, Quasidiagonality of C*algebras of solvable Lie groups. Integral Equations Operator Theory 90 (2018)] for an example. $\endgroup$ Commented May 14, 2020 at 12:51

$\begingroup$ Pro Gabe, may I ask you another question? My teacher told me that separable simple nuclear C∗algebras have been completely classified . I wonder which problems do most operator algebraists concern about nowadays? $\endgroup$ Commented May 17, 2020 at 16:59