Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

up vote 16 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.

share|improve this answer
3  
Very impressing! –  Andreas Thom May 29 '11 at 9:57
1  
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
show 1 more comment

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 $x^* x-x x^*$.

share|improve this answer
1  
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 ams.org/mathscinet-getitem?mr=546503 –  Matthew Daws Jun 16 '11 at 9:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.