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The reduced $C^*$-algebra of a non-abelian free group $G$ has a unique trace. Hence, there is no chance to separate conjugacy classes of group elements using traces on $C^\star_{red} G$. On the other side, for the group ${\mathbb Z}$, separation is clearly possible.

Question: Let $G$ be an amenable group. Does the reduced group $C^\star$-algebra of $G$ support sufficiently many traces to distinguish between conjugacy classes of group elements?

EDIT: The question seems already interesting for $S_{\infty} = \cup_n S_n$. Let's get explicit and pick $g \in S_n$ (for some $n$) and consider the canonical trace $\tau_{g,n}$ which sends every conjugate of $g$ to $1$ and all other elements to zero. (This can be done for any $n' \geq n$ since $S_n \subset S_{n'}$.) The function $\tau_{g,n} \colon {\mathbb C}S_n \to {\mathbb C}$ is a conjugation invariant function and hence, it must be a linear combination of the normalized traces of irreducible representations of $S_n$.

Question: What is the sum of the absolute values of the coefficients that come up in this linear combination of traces?

This is (as one can check) the norm of $\tau_{g,n}$, call it $c(g,n)$. So, we see that the compatible family of maps $\tau_{g,n}$ extends from ${\mathbb C}S_{\infty}$ to $C^* S_{\infty}$ if and only if $c(g,n)$ remains bounded.

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As you mention in the question, the answer fails for $\mathbb{Z}$. And for groups with centre I would also expect to be able to define different traces. So maybe you want to ask for $G$ to be ICC? –  Martin Argerami Aug 10 '10 at 3:33
The question is correct as it is. But you are right, the infinite conjugacy classes are more problematic. The first case that I do not really understand (but did not think too much about it) is $S_{\infty}$. –  Andreas Thom Aug 10 '10 at 5:15

3 Answers 3

A group $G$ is conjugacy separable if every two elements which are not conjugate in $G$, remain non conjugate in some finite quotient of $G$. Since characters of a finite group separate conjugacy classes, we see that if $G$ is amenable and conjugacy separable, then $C^*(G)$ has enough traces to separate conjugacy classes in $G$.

It is known that polycyclic-by-finite groups are conjugacy separable.

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up vote 5 down vote accepted

There is an easy example, namely $SL_3(F)$ where $F$ is the algebraic closure of some finite field. This group does not admit non-trivial characters (a result of Kirillov) and is locally finite, hence amenable. This gives a negative answer to the first question.

Sorry, I deleted my previous answer since it contained a mistake.

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Hello! When the group is amenable, the reduced $C^\ast$-algebra coincides with the full one, so there is the augmentation homomorphism associated to the trivial representation. One can restrict it to the closure of the linear subspace spanned by the elements of an arbitrary conjugacy class, and set it zero on the other conjugacy classes. This only decreases the norm, so the result is a continuous trace which detects the given conjugacy class. Best, Mathias

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It is not so clear to me, why setting it zero on the other conjugacy classes only decreases the norm. –  Andreas Thom Sep 28 '10 at 11:41
As Andreas has (implicitly observed), when your group is the integers you seem to be claiming that truncation of the Fourier series of a continuous function is norm-decreasing. This does not look right to me –  Yemon Choi Oct 7 '10 at 22:06

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