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Kuiper's theorem is well-known to give the triviality of the homotopy groups of ${\rm GL}(\mathcal{H})$ for $\mathcal{H}$ a (separable) infinite-dimensional complex Hilbert space. Work of Palais later gives sufficient control over the infinite-dimensional manifold structure so that we know ${\rm GL}(\mathcal{H})$ is in fact contractible. Mathai Varghese asked me about ten years ago (as a student) if I would be interested in proving that ${\rm GL}(\mathcal{H})$ was holomorphically contractible. This was not a challenge I took up at the time, though it was suggested to me that a first step would be to think about whether $\mathcal{H}\setminus\{0\}$ was holomorphically contractible.

Is this something reasonable to expect? The first issue is to figure out in what topology we even have a complex infinite-dimensional manifold, if any.

The reason that it would be useful to know is that ${\rm GL}(\mathcal{H})/\mathbb{C}^\times$ would be a holomorphic $K(\mathbb{Z},2)$. I've always wanted to know if this is possible...

David Roberts
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    What do "holomorphically contractible" and "holomorphic $K(\mathbb{Z}, 2)$" mean? – Qiaochu Yuan May 02 '16 at 04:16
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    I would interpret holomorphic contractibility of $X=\mathcal{H}\setminus {0}$ as meaning that any holomorphic map from a complex manifold $M$ to $X$ is homotopic to a point, via a holomorphic homotopy $M\times\mathbb{C}\to X$. The holomorphic mappings $M\to X$ would be given by a colimit procedure from the finite-dimensional approximations of $\mathcal{H}$. With this interpretation, it should be possible to see that the classical homotopy which shows that $S^\infty$ is contractible gives rise to a holomorphic contraction. – Matthias Wendt May 02 '16 at 07:21
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    This is part of the question. Perhaps one wants contractible with respect to the cylinder object using the complex numbers with basepoints 0,1, perhaps. By holomorphic K(Z,2) I mean a natural complex manifold that is homotopically a K(Z,2). Feel free to interpret these definitions as you see fit. – David Roberts May 02 '16 at 08:01
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    @MatthiasWendt well, it's got to be more complicated than that, otherwise Kuiper's theorem would be a corollary of the contractibility of S^oo – David Roberts May 02 '16 at 08:03
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    @DavidRoberts I did not say I knew how to go from the contractibility of $\mathcal{H}\setminus{0}$ to the contractibility of $\operatorname{GL}(\mathcal{H})$, and I think that the complications reside more in this latter step. For example, the $(2n-2)$-connectedness of $\mathbb{C}^n\setminus{0}$ does not imply $(2n-2)$-connectedness of $\operatorname{GL}_n(\mathbb{C})$. – Matthias Wendt May 02 '16 at 08:16
  • So for instance, the Eilenberg swindle (one step of the proof) is surely holomorphic, but the rotation into the decomposition needs to be checked etc etc, assuming GL(H) is even a complex manifold. – David Roberts May 02 '16 at 08:23
  • I was prompted into asking this by the nice references in this comment: http://mathoverflow.net/questions/8800/proofs-of-bott-periodicity/54163?noredirect=1#comment588034_54163 – David Roberts May 02 '16 at 08:26

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