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Is it known whether any element of trace 0 in the reduced $C^*$-algebra of a non-abelian free group, is a limit of sums of (additive) commutators?

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up vote 18 down vote accepted

This is true. Perhaps it is known whether this property for a group $G$ is equivalent to $C^*_\lambda G$ having a unique trace? In any case, the same proof of Powers which shows unique trace for free groups, can be adapted to this question.

First note that if $F = F(x_1, x_2, \ldots, x_n)$ is the free group on $n$ generators, then by a trick of Bob Powers, in $\mathcal B(\ell^2 F)$ we have $\| \Sigma_{i = 1}^n \lambda(x_i) \| \leq 2 \sqrt{n}$. Indeed, if we denote by $P_i$ the projection in $\mathcal B(\ell^2 F)$ onto the subspace generated by Dirac functions on all words which in reduced form begin with the letter $x_i$, then we have $(1 - P_i) \lambda(x_i) (1 - P_i) = 0$ and hence $$ \Sigma_{i = 1}^n \lambda(x_i) = \Sigma_{i = 1}^n P_i \lambda(x_i) + ( \Sigma_{i = 1}^n P_i \lambda(x_i) (1 - P_i))^*. $$ Since $P_i$ have orthogonal ranges we have that $\| \Sigma_{i = 1}^n P_i \lambda(x_i) \| \leq \sqrt{n}$ and $\| \Sigma_{i = 1}^n P_i \lambda(x_i) (1 - P_i) \| \leq \sqrt{n}$.

Next note that the limits of sums of commutators forms a subspace and hence it is enough in the reduced group $C^*$-algebra to show that the non-trivial group elements can be written as a limit of a sums of commutators. Moreover, in a free non-abelian group, every element has an element which is free from it, and hence by restricting to a subgroup it is enough to show that in the free group on 2 generators $F(a, b)$ we can write $\lambda(a)$ as a limit of sums of commutators.

For this we consider an arbitrary $n \in \mathbb N$ and easily verify the formula $$ \frac{1}{n} \Sigma_{i = 0}^{n -1} \lambda( b^{-i} a b^i ) = \lambda(a) - \frac{1}{n}\Sigma_{i = 1}^{n - 1} [ \lambda(b^i), \lambda(b^{-i} a) ]. $$

Since $x_i \mapsto b^{-(i - 1)} a b^{(i - 1)}$ extends to an isomorphism between $F(x_1, \ldots, x_n)$ to the subgroup $\langle a, b^{-1} a b, \ldots, b^{-(n - 1)}a b^{(n - 1)} \rangle$ it follows from Powers trick that $$ \| \lambda(a) - \frac{1}{n}\Sigma_{i = 1}^{n - 1} [ \lambda(b^i), \lambda(b^{-i} a) ] \| \leq 2/ \sqrt{n}. $$ Since $n$ was arbitrary, this finishes the proof.

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Very impressing! – Andreas Thom May 29 '11 at 9:57
You are right with your suggestion. Let $C$ be the norm closure of the span of additive commutators, then any linear functional on $(C^{\ast}_{\lambda}G)/C$ defines a trace on $C^{\ast}_{\lambda}G$. Hence, if there is a unique trace, then $C \subset C^{\ast}_{\lambda}G$ has codimension one. Clearly, $C$ has to coincide with the space of those elements with vanishing trace. – Andreas Thom May 29 '11 at 10:02
This will give a linear functional with the tracial property, but it's not quite clear to me why it would give a state. I may be missing something. – Jesse Peterson May 29 '11 at 10:10
No, it will not be unital or positive in general. I thought "uniqueness of trace" means uniqueness of continuous linear functionals with the tracial property. – Andreas Thom May 29 '11 at 10:36
I always interpreted that as unique tracial state, but perhaps these two uniqueness conditions are equivalent on reduced group $C^*$-algebras. – Jesse Peterson May 29 '11 at 10:50

More is true: an element of a C*-algebra is a norm limit of sums of commutators if and only if it is 0 on any bounded trace. For selfadjoints, this was proven by Cuntz and Pedersen (in their only paper together I think). One reduces from arbitrary elements to selfadjoints by writing $c=a+bi$, with $a$ and $b$ selfadjoint. So if $C^*_r(G)$ has a unique tracial state then any element vanishing on it is a limit of sums of commutators.

A couple more facts proven by Cuntz and Pedersen: (1) the commutators can be arranged to form a convergent series, (2) for selfadjoints, the commutators can be chosen of the form $x^* x-x x^*$.

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The paper is: Cuntz, Joachim; Pedersen, Gert Kjaergȧrd "Equivalence and traces on $C^{\ast} $C∗-algebras." J. Funct. Anal. 33 (1979), no. 2, 135–164. (They wrote 2 papers together!) See – Matthew Daws Jun 16 '11 at 9:13

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