Is the unitary group of $l^2(A)$ with the strict topology contractible? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T20:38:21Z http://mathoverflow.net/feeds/question/58747 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58747/is-the-unitary-group-of-l2a-with-the-strict-topology-contractible Is the unitary group of $l^2(A)$ with the strict topology contractible? Ulrich Pennig 2011-03-17T13:27:46Z 2011-03-17T17:59:08Z <p>Let $A$ be a $C^*$-algebra with countable approximate unit. Let $\mathbb{K}$ denote the compact operators on a separable Hilbert space. Mingo and later Cuntz and Higson have shown that the unitary group in the multiplier algebra $M(A \otimes \mathbb{K})$ is contractible in the norm topology. It was then shown by Troitsky in a paper with the title </p> <p>Geometry and Topology of operators on Hilbert $C^*$-modules</p> <p>that $U(M(A \otimes \mathbb{K}))$ is also contractible, if it is equipped with the left strict topology, i.e. the topology generated by the semi-norms $\lVert xa \rVert$ for $x \in M(A \otimes \mathbb{K})$ and $a \in A \otimes \mathbb{K}$. Is the theorem still true, if we change from left strict to strict (which is the topology that includes the semi-norms $\lVert ax \rVert$)? </p> http://mathoverflow.net/questions/58747/is-the-unitary-group-of-l2a-with-the-strict-topology-contractible/58772#58772 Answer by Leonel Robert for Is the unitary group of $l^2(A)$ with the strict topology contractible? Leonel Robert 2011-03-17T17:59:08Z 2011-03-17T17:59:08Z <p>I would post this as a comment but as it just happens I can't do that. I do think that the exercise that you mention proves strict contractibility. The same formula for the homotopy, $$(u,t)\mapsto w_tuw_t^*+(1-w_tw_t^*),$$ is given in Proposition 12.2.2 of Blackadar's book on K-theory, although the statement only says that the unitary group of $M(A\otimes K)$ is path connected. This formula goes back to Dixmier and Douady's paper on fields of Hilbert spaces, applied to $B(H)$.</p>