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?



Commutators in the reduced C*-algebra of the free group
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. 


