For $G$ a discrete group, there is a canonical inclusion $g\mapsto u_g$ of $G$ into the unitary group of the reduced $C^*$-algebra $C^*_r(G)$. Denote by $[u_g]$ the class of $u_g$ in the (topological) $K$-theory group $K_1(C^*_r(G))$. It is well-known that, if $g$ is a commutator in $G$, then $[u_g]=0$ in $K_1(C^*_r(G))$: indeed the $2\times 2$ matrix $u_g\oplus 1$ is connected to the identity in the unitary group of $M_2(C^*_r(G))$.

Now, let $G$ be the free group on two generators $a,b$, and let $g$ be the commutator of $a$ and $b$. I recently bumped into this nice paper by Haagerup, Dykema and Rordam:

http://arxiv.org/pdf/funct-an/9608001.pdf

At the top of page 3, they mention as a consequence of their main result, that $u_g$ is connected to $1$ already in the unitary group of $C^*_r(G)$. Partly out of curiosity, I was wondering whether an explicit path of unitaries between $u_g$ and $1$ had been written down somewhere.

`$C(S^1)\simeq C^*_r(\mathbb{Z})\to C^*_r(G)$`

. $\endgroup$ – Alain Valette Jan 5 '13 at 19:39