Question

Hello! It seems to me that a simpler counter-example than $S_\infty$ is the crossed product group of the integers by When the two-element group , acting by the unique non-trivial automorphism $\Phi$. This is amenableas it is an extension of abelian groups, and, for instance, the conjugacy class of $(0,\Phi)$ is reduced $\{(2b, \Phi):b\in\mathbb{Z}\}$. Under Fourier transformC^\ast$-algebra coincides with the full one, so there is the augmentation homomorphism associated $C^\ast$-algebra consists to the trivial representation. One can restrict it to the closure of two summands the linear subspace spanned by the elements of $C(\mathbb{S}^1)$ with twisted multiplicationan arbitrary conjugacy class, and set it zero on the other conjugacy class in question are the functions in the second summand with even Fourier coefficientsclasses. Thus, the associated densely defined trace does not extend to This only decreases the $C^\ast$-algebra as of course not all continuous functions have summable even Fourier coefficientsnorm, and so the cc can not be detected by result is a continuous trace on the $C^\ast$-algebra. Does that answer which detects the question? given conjugacy class. Best, Mathias