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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeApr 20th 2011

following a suggestion by Zoran, I have created a stub (nothing more) for Kuiper’s theorem

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeApr 20th 2011

Looks good – being precise :)

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeApr 20th 2011

while we are at it: what is the center of $U(\infty) = \Omega {\lim_\to}_n B U(n)$?

• CommentRowNumber4.
• CommentAuthorzskoda
• CommentTimeApr 20th 2011

I have no idea. Mike, Todd ?

• CommentRowNumber5.
• CommentAuthorDavidRoberts
• CommentTimeApr 20th 2011

You need to specify the topology to say it is contractible. The strength of the theorem is that it is contractible in several topologies, none of which I can recall at the moment.

And I would define U(oo) as lim U(n), but the inclusion maps don’t preserve the centre, so I’m not sure about Urs’ question.

• CommentRowNumber6.
• CommentAuthorAndrew Stacey
• CommentTimeApr 20th 2011

The original paper proved that it was contractible in the operator topology. Atiyah and Segal note in their paper on twisted K-theory that there is an easy proof of contractibility in the weak topology. One major difference in the topologies is that with the operator topology then it is a CW complex but with the weak topology then it isn’t even an ANR (something that annoys me intensely).

I would also define $U(\infty)$ as $\lim U(n)$ and I believe that its centre is trivial: any $A \in U(\infty)$ must be in some $U(n)$, but then there is an element in $U(2n)$ which rotates the first $n$ directions to the last $n$ directions and which therefore maps $(A,I_n)$ to $(I_n,A)$, hence if $A$ is in the centre it must be the identity.

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeApr 21st 2011

Thanks, Andrew. Somebody should add that to the entry.

About the center: that’s what I was thinking, too, but I thought I must be missing something. Maybe not.

• CommentRowNumber8.
• CommentAuthorjim_stasheff
• CommentTimeApr 21st 2011
there is an element in
U(2n) which rotates the first
n directions to the last
n directions and which therefore maps
(A,I n) to
(I n,A), hence
U(oo) is homotopy commutative, the first step on the way to being an infinite loop space
• CommentRowNumber9.
• CommentAuthorDavidRoberts
• CommentTimeAug 16th 2017

I’ve updated Kuiper’s theorem slightly to point out that in fact most of the various topologies on $U(H)$ weaker than the norm topology agree. This is due to Espinoza-Uribe, which an earlier partial contribution by Schottenloher. So it seems that Andrew’s complaint in #6 was probably directed at $B(H)$ or $GL(H)$, since the weak operator and strong operator topologies agree on $U(H)$, but not on $GL(H)$.

I’ve also edited unitary group to add this reference, and I found that the page claimed that $U(H)$ was the maximal compact subgroup of $GL(H)$ even in the infinite-dimensional setting (!). So I definitely fixed that.

A note to myself to add later: $U(H)$ is a Banach Lie group in the norm topology, but not a Lie group in the strong topology; conversely, the left regular representation of a compact topological group with Haar measure is not continuous in the norm topology, but is continuous in the strong topology.