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Let $G$ be a group and $l^2(G)$ the Hilbert space on $G$. The complex group algebra $CG$ can be imbedded in $B(l^2(G))$, the set of all bounded linear operators, by left translation. The reduced group $C^*$ algebra $C_r^*(G)$ is the operator norm completion of $CG$. It's clear that $l^1(G)$ lies in $C_r^*(G)$.

Q1. Are there some typical elements in $C_r^*(G)\setminus l^1(G)$?

Q2. For a normed algebra $A$, let $C$ be the completion of the subalgebra $\{ab-ba|a,b\in A\}$ and $T(A)=A/C$. It's not hard to see $T(l^1(G))$ is the $l^1$ completion of $C[conj(G)]$, the set of all finitely supported functions on the conjugacy classes $conj(G)$. How to describle $T(C_r^*(G))$? In particular, if $g$ is an element of infinite order, can $g$ have the same image in the quotient algebra as $0$ or a finite order element in $G$? Thank you very much.

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But Jonas, what will people do now you have taken a c-star algebra question/topic and taken the NCG tag off? how will we all cope? :) – Yemon Choi Mar 28 '10 at 5:44
Funny, but I honestly did hesitate on that one, because I know that some have a different view of what NCG should mean. But because I was already removing the apparently irrelevant von Neumann algebras tag, it seemed best. @yeshengkui: I hope that you don't mind the tag edit. If there are connections to von Neumann algebras or NCG that you have in mind for these specific questions, please let us know. – Jonas Meyer Mar 28 '10 at 6:06
OK. Actually, there is some relation with the group von Neumann algebra. All the $l^1(G)$ and $C_r^*(G)$ are subalgbra of the group von Neumann algebra $NG$, which is the weak operator norm completion of $CG$. Furthermore, $T(NG)$ is just its center $Z(NG).$ – yeshengkui Mar 28 '10 at 12:46
But the set of all commutators is not closed with respect to products, is it? So it is itself not an algebra. – Ulrich Pennig Mar 28 '10 at 13:17
If $I$ denotes the commutator ideal, then you have a map $A/C \to A/I$ of Banach spaces (since $C \subset I$). No $g \in G$ belongs to the class of the zero element in $A/I$ since this is an algebra and $g$ is invertible there. Therefore $g$ is not represented by the zero class in $A/C$ either. – Ulrich Pennig Mar 28 '10 at 14:27
up vote 4 down vote accepted

Q1 seems to be related to symmetry of $\ell^1(G)$ (a Banach $^*$-algebra $A$ is symmetric if, for every $a\in A$, the spectrum of $a^*a$ is contained in $\mathbb{R}^+$). If $\ell^1(G)$ is not symmetric, then for an element $a^*a$ with non-positive $\ell^1$-spectrum, since the $C^*$-spectrum is clearly positive, the resolvent of $a^*a$ will define elements in $C^*_r(G)\backslash\ell^1(G)$. For examples of groups with non-symmetric algebras, see papers by J. Boidol, H. Leptin and D. Poguntke from the 1980's.

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A couple of quick comments.

Firstly, I'd advise you to think about what you know about Q1 in the case $G= {\mathbb Z}^d$ (or even just the case $G={\mathbb Z}$) -- because in such settings you can think of the group C*-algebra as the algebra of continuous functions on the Pontryagin dual. In that case, your question becomes one about examples of continuous functions which don't have absolutely summable Fourier coefficients.

Secondly, I think your definition of C is not quite right -- don't you need to take the ideal generated by all the commutators, not just the set of all commutators? (It feels like what you want to be asking about is the centre of $C_r^*(G)$ -- I think Losert has a paper on this -- but perhaps I have misunderstood.)

Added 28-05-2010: The paper of Losert which I was dimly recalling is

[ Zbl 1088.22003] On the center of group C * -algebras. (English)

V. Losert, J. Reine Angew. Math. 554, 105-138 (2003).

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hi Yemon, thanks for your comments. As for the definition, I think it's just a subalgebra, not necessary an ideal. The quotient may be not an algebra anymore. Would you like to give me more details of Losert's paper, title for example? Thanks a lot. – yeshengkui Mar 28 '10 at 12:53
The reference is good. Thank you very much for that. I think it's helpful for me. – yeshengkui Mar 29 '10 at 1:55

In Q2 a couple of things are mixed up. If you consider $\ell^1 G$ with the point-wise multiplication, then $\ell^1 G$ is a commutative non-unital Banach algebra and the closure of the space of commutators (with respect to the multiplication which is induced by the group multiplication) is indeed a two-sided ideal. The quotient is given by $\ell^1$-functions on the space of conjugacy classes.

However, if you consider $\ell^1 G$ only with the multiplication coming from the group, then it is a unital (typically) non-commutative Banach algebra and the ideal generated by the commutators is larger that just the closure of the space of commutators. The quotient is $\ell^1 H$, where $H$ is the abelianization of $G$.

For the reduced $C_{red}^*$-algebra, the ideal structure can be quite different compared with $\ell^1 G$. For example, if $G$ is a non-abelian free group, then $C_{red}^\star G$ is simple and there is only the trivial quotient. In particular, the ideal generated by the commutators is everything. However, if $G$ is amenable, the quotient by the commutator ideal can be identified with the reduced $C_{red}^\star$-algebra of the abelianization of $G$.

Another way to read your question is the following: Can an element of finite order in $G$ become equal to an element of infinite order modulo commutators in $C_{red}^\star G$? That is now a question about traces on group $C_{red}^*$-algebras. Equivalently, you could ask: Is there a trace on $C_{red}^\star G$ which distinguishes element of finite order from those of infinite order. For many group (for example a free product of finite groups; not both $C_2$) there exists a unique trace on $C_{red}^\star G$. In particular, traces cannot distinguish between non-trivial conjugacy classes.

If $G$ is amenable, the situation is different. I do not know whether the reduced $C^\star$-algebra of an amenable group supports sufficiently many traces in order to distinguish conjugacy classes.

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I asked the question arising from the last two lines of my answer as… – Andreas Thom Aug 8 '10 at 9:32

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