Explicit path in the unitary group of a $C^*$-algebra - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T21:19:39Z http://mathoverflow.net/feeds/question/118115 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118115/explicit-path-in-the-unitary-group-of-a-c-algebra Explicit path in the unitary group of a $C^*$-algebra Alain Valette 2013-01-05T12:08:23Z 2013-01-05T12:08:23Z <p>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 <code>$C^*_r(G)$</code>. Denote by $[u_g]$ the class of $u_g$ in the (topological) $K$-theory group <code>$K_1(C^*_r(G))$</code>. It is well-known that, if $g$ is a commutator in $G$, then $[u_g]=0$ in <code>$K_1(C^*_r(G))$</code>: indeed the $2\times 2$ matrix $u_g\oplus 1$ is connected to the identity in the unitary group of <code>$M_2(C^*_r(G))$</code>. </p> <p>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:</p> <p><a href="http://arxiv.org/pdf/funct-an/9608001.pdf" rel="nofollow">http://arxiv.org/pdf/funct-an/9608001.pdf</a></p> <p>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 <code>$C^*_r(G)$</code>. Partly out of curiosity, I was wondering whether an explicit path of unitaries between $u_g$ and $1$ had been written down somewhere.</p>