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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:


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.

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Do we know that the spectrum of $u_g$ is the entire unit circle? (Since otherwise the answer is easy.) –  Nik Weaver Jan 5 '13 at 18:46
@Nik: if $g$ is an element of infinite order in a group $G$, then $u_g$ has full spectrum. Reason: the inclusion $<g>\to G$ induces an inclusion $C(S^1)\simeq C^*_r(\mathbb{Z})\to C^*_r(G)$. –  Alain Valette Jan 5 '13 at 19:39
Oh, I see. So we need to move out of that $C(S^1)$. –  Nik Weaver Jan 5 '13 at 20:14
@AlainValette - did you get anywhere with this question? –  David Roberts Aug 16 '13 at 1:27
@DavidRoberts: No progress whatsoever... –  Alain Valette Aug 16 '13 at 21:27
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This question has an open bounty worth +200 reputation from Ali Taghavi ending tomorrow.

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