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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()=Ωlim nBU(n)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()U(\infty) as limU(n)\lim U(n) and I believe that its centre is trivial: any AU()A \in U(\infty) must be in some U(n)U(n), but then there is an element in U(2n)U(2n) which rotates the first nn directions to the last nn directions and which therefore maps (A,I n)(A,I_n) to (I n,A)(I_n,A), hence if AA 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)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)B(H) or GL(H)GL(H), since the weak operator and strong operator topologies agree on U(H)U(H), but not on GL(H)GL(H).

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

    A note to myself to add later: U(H)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.

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