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!



Homotopy equivalence between the Grassmannian Gr_{n,m} and Gr_n \times Gr_m.
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.


