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following a suggestion by Zoran, I have created a stub (nothing more) for Kuiper’s theorem
Looks good – being precise :)
while we are at it: what is the center of $U(\infty) = \Omega {\lim_\to}_n B U(n)$?
I have no idea. Mike, Todd ?
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.
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.
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.
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.
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