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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:

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.

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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
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@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
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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
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@AlainValette Is there some obvious obstruction to such a path in the full Cstar algebra of the free group? –  Yemon Choi Apr 20 at 18:08

1 Answer 1

Here are some thoughts about a construction of an explicit path. Unfortunately, I could not carry it over, so take it as a long comment. I'd be glad if somebody is able to complete it, or maybe to get some useful hint!

Any skew-symmetric unitary element $J$ of a $C^*$ algebra $A$ can be connected to the identity in the unitary group of $A$ by the path $\exp(tJ)=\cos(t)I+\sin(t)J$, for $0\le t\le \pi/4$. This suggests to consider a factorization of $U:=u_g$ as a product of two skew-symmetric unitary operators $J$ and $K$ on $\ell_2(G) $ (this can be done in many ways, indeed): $$J\in B(\ell_2(G)),\quad K\in B(\ell_2(G))\, ,$$ $$J^*=J^{-1}=-J,\quad K^*=K^{-1}=-K\, ,$$ $$U=JK\,\quad (\,\mathit{so\, that}\quad K=-JU)\, .$$ Then $U(t):=\exp(tJ)\exp(tS)=\cos(t)^2I+\sin(t)\cos(t)J(I-U)+\sin(t)^2U$ for $0\le t\le \pi/4$ is a path in the unitary group of $B(\ell_2(G))$ connecting $I$ to $U$. So, if we can choose the above factorization in such a way that we also have $J(I-U)\in C^*_{\mathrm {r}}(G)$, the whole path $U(t)$ would be in the reduced algebra for all $t$. Of course, $J \in C^*_{\mathrm {r}}(G)$ would do it, but this doesn't quite seem to be needed, as $I-U$ is not invertible.

I've thus tried the simplest Ansatz for $J$, namely $Jf(x):=\alpha(x)f(\sigma(x))$, for all $f\in \ell_2(G)$ and $x\in G$, with a convenient $\alpha\in\ell_\infty(G)$ and a $\sigma:G\to G$ bijective. However, it seems the last requirement, $J(I-U)\in C^*_{\mathrm {r}}(G)$ can't be fulfilled in this case. A slightly more general form, $Jf(x):=\alpha(x)f(\sigma(x))+\beta(x)f(\tau(x))$ seems more promising (yet there could be some good reason why the whole approach is hopeless, though).

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