Talk:Classifying space for U(n)

For the case N=1,it is incorrect to say EU(1) = CP^\infty. This is not contractible,for one thing, and contradicts what is said above about the general case, where EU(n) is the set of orthonormal N-frames in Hilbert space. So for N=1, this reduces to the case of the unit sphere in Hilbert space - which is contractible - see e.g. http://www.math.ucr.edu/home/baez/week151.html Cgwaldman 06:23, 29 September 2007 (UTC)[reply]

I removed the stuff that ES^1 is a unit sphere in Hilbert space (kinda projective limit of finite-dim spaces), instead inserting CP^\infty (inductive limit). 18.87.0.72 04:02, 31 October 2005 (UTC)[reply]

Seems like the same thing to me, isn't it? linas 07:04, 1 November 2005 (UTC)[reply]

Not at all. The only complex projective space that is ismorphic to a shpere is CP(1). If I remember well, the unit sphere in a Hilbert space is contractible (please correct me if I'm wrong or write a reference if you know one) whereas

CP^{\infty }

has its cohomology isomorphic to

\mathbb {R} [X^{2}]

. S.racaniere 23:47, 30 December 2006 (UTC)[reply]

"Case of n = 1 (first construction)" is not very clear

In particular, what is the action of S^1 on S^{\infty}? It's not really evident for someone who is new to this matter. Kromsson (talk) 08:22, 30 April 2009 (UTC)[reply]

Notation?

I'm confused about the notation for EU(n). It seems to me to be saying that this space is just n orthogonal vectors. What is meant? Perhaps the set of the spans of all such things? Sorry if I'm just not familiar with the notation and I'm being dumb!

Jjw19 (talk) 15:02, 11 March 2010 (UTC)[reply]

Bogus Proof

The computation of the cohomology of BU(n) is almost complete nonsense. What one can deduce from the argument is that there is a homomorphism from H^*(BU(n)) to the ring of W-invariant polynomials. This homomorphism is given by applying the functor B to the inclusion of T in U(n), and observing that it commutes with the action of the Weyl group by conjugation. However, the whole point is to show that it is a bijection, which the "proof" certainly does not accomplish. —Preceding unsigned comment added by 76.119.236.237 (talk) 14:35, 12 March 2010 (UTC)[reply]