# Kuiper's theorem via approximation

Kuiper's theorem says that the unitary group $U(H)$ of a separable infinite dimensional Hilbert space $H$ is contractible, if it is equipped with the norm topology.

Let's suppose, I do not know this theorem, but I do know that $U^{st*}(H)$ is contractible, where the latter group is $U(H)$ equipped with the strong$^*$ topology generated by the semi-norms $p_v(a) = \lVert av \rVert$ and $q_v(a) = \lVert a^*v\rVert$ for all $v \in H$. I have a continuous map $U(H) \to U^{st*}(H)$.

Is there any way to see that this is a weak equivalence by an approximation argument, thereby proving Kuiper's theorem?

side remark:

The motivation for this question comes from a setup which looks completely different, but is from a certain point of view surprisingly similar: Let $\mathcal{O}_{\infty}$ be the Cuntz algebra on countably infinite generators. $Aut(\mathcal{O}_{\infty})$ carries two topologies: One from the norm via the inclusion into bounded maps on Banach spaces $\mathcal{O}_{\infty} \to \mathcal{O}_{\infty}$, the other is the so-called point-norm topology generated by the semi-norms $p_a(\alpha) = \lVert \alpha(a) \rVert$. It is known that $Aut(\mathcal{O}_{\infty})$ is weakly contractible in the latter topology, but I would like to know it for the first.

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Small observation: outside the $C^ast$-setting this kind of approximation can be problematic. IIRC, the nest algebra of N (bounded operators on $\ell^2$ that are upper-triangular wrt the standard basis) is known to have connected invertible group in the point-norm topology, but it is a mildly notorious open problem whether said group is connected in the norm topology. Of course one might reasonably hope that there are better approximation techniques available in the self-adjoint setting –  Yemon Choi May 15 '12 at 18:11
What is a "weak equivalence"? –  Nik Weaver May 15 '12 at 18:30
@Nik: "weak equivalence" means that the map $U(H) \to U^{st*}(H)$ induces an isomorphism on all homotopy groups. All groups above are considered to have the identity as their basepoint. –  Ulrich Pennig May 15 '12 at 18:37
And do you have a simple proof of the fact that $U^{st*}(H)$ is contractible? Here I mean simpler than Kuiper's one for $U(H)$. –  Alberto Abbondandolo May 16 '12 at 16:27
@Alberto: The proof of this can be found for example in Proposition A2.1 in the appendix of the paper "Twisted K-Theory" by Atiyah and Segal and takes (almost) less than a page. –  Ulrich Pennig May 16 '12 at 16:53

This is not an answer but too long for a comment.

It was shown in

Popa, S. and Takesaki,M., The Topological Structure of the Unitary and Automorphism Groups of a Factor, Commun. Math. Phys. 155, 93-101 (1993)

that the unitary group $U(R)$ of the hyperfinite $II_1$-factor $R$ is contractible in the strong topology (which is equal to the strong-$*$-topology in this case).

Years before, it was shown in

Araki, H., Smith, M.-S.B. and Smith, L., On the homotopical significance of the type of von Neumann algebra factors, Commun. Math. Phys. 22, 71-88 (1971)

that the first homotopy group of $U(R)$ in the norm toplogy is isomorphic to $(\mathbb R,+)$. Under this isomorphism, the element $\lambda \in [0,1]$ is represented by the loop $$[0,1] \ni t \mapsto \exp(2\pi i\cdot t p) \in U(R),$$ where $p\in R$ is some projection of trace $\lambda$.

This shows that the homotopy type depends heavily on the chosen topology and there is no reason (at least in general) that one should find an easy approximation argument.

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I knew those two papers, but you're right. The more I think about it, it seems to be a (lucky) coincidence that $U(H)$ is contractible in both topologies. –  Ulrich Pennig Jul 10 '12 at 7:52