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The following assertion appears in a paper I am reading, and I can't seem to verify it.

Let $\text{Gr}_{n,m}$ denote the set of pairs $(V,W)$ where $V$ and $W$ are as follows.

  1. $V$ is an $n$-dimensional subspace of $\mathbb{C}^{\infty}$.
  2. $W$ is an $m$-dimensional subspace of $\mathbb{C}^{\infty}$.
  3. $V$ and $W$ are orthogonal.

The space $\text{Gr}_{n,m}$ has an obvious topology. If $\text{Gr}_n$ and $\text{Gr}_m$ are the usual Grassmannians of $n$ and $m$ planes in $\mathbb{C}^{\infty}$, then there is an obvious map $\psi : \text{Gr}_{n,m} \rightarrow \text{Gr}_n \times \text{Gr}_m$.

The map $\psi$ is almost a homeomorphism, but not quite because of condition 3 above. The paper claims that $\psi$ is a homotopy equivalence.

Thanks for any help!

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Is $\mathbb{C}^\infty$ $\bigoplus_{i=1}^{\infty} \mathbb{C}$ or $\bigtimes_{i=1}^{\infty} \mathbb{C}$? – David Roberts Nov 2 '11 at 4:47
Oh, and what paper? – David Roberts Nov 2 '11 at 4:50
It is the direct sum of infinity many copies of $\mathbb{C}$. The paper is actually the 10th lecture (on Bott periodicity) from the following set of lecture notes : – Ralph Nov 2 '11 at 4:57
@Ralph Thanks for the link, btw, — very interesting. – Grigory M Nov 12 '11 at 18:51
up vote 10 down vote accepted

The forgetful map $Gr_{n,m} \to Gr_n$ that drops $W$ is a fiber bundle (exercise), and the map $Gr_{n,m} \to Gr_n \times Gr_m$ is a map of fiber bundles. It's an equivalence on the (connected) base space, so it suffices to check that the map of fibers is an equivalence.

The fibers over $V$ are, respectively: $m$-dimensional subspaces in $V^\perp \subset \mathbb{C}^\infty$, and $m$-dimensional subspaces in $\mathbb{C}^\infty$.

The inclusion of one infinite-dimensional complex vector space in another induces a homotopy equivalence of Grassmannians; you could construct an explicit homotopy equivalence by choosing an appropriate basis, or you could argue that the associated map of Stiefel manifolds is a homotopy equivalence (both are contractible, so this is easy) and so it passes to an equivalence after taking the quotient by the general linear group.

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Beautiful, thanks! – Ralph Nov 2 '11 at 5:06
Yes, very nice. But where I come from "hyperplane" means a linear subspace of codimension one. – Tom Goodwillie Nov 2 '11 at 5:15
@Tom: That's a bad habit that I picked up for no apparent reason, and it's proved difficult to drop. Edited. – Tyler Lawson Nov 2 '11 at 5:33

Look at the canonical principal $U(n)\times U(m)$ bundle over $Gr_{n,m}$ given by pairs of orthonormal frames $(v_1,\ldots, v_n), (w_1,\ldots w_m)$. Its total space is the set of all orthonormal $n+m$-frames in $\mathbb C^\infty$. It's contractible (that's well-known) and hence $Gr_{n,m}$ is a homotopy $B_{U(n)\times U(m)}$ which is clearly homotopy equivalent to $B_{U(n)}\times B_{U(m)}=Gr_n\times Gr_m$. To see that the natural map $Gr_{n,m}\to Gr_n\times Gr_m$ is the one inducing an equivalence notice that it's obviously covered by a map of principal bundles and hence the result follows by 5-lemma since total spaces are contractible and the fibers are the same.

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