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

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

2011-11-02T04:25:36Z

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.

$V$ is an $n$-dimensional subspace of $\mathbb{C}^{\infty}$.
$W$ is an $m$-dimensional subspace of $\mathbb{C}^{\infty}$.
$V$ 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!

Answer by Tyler Lawson for Homotopy equivalence between the Grassmannian Gr_{n,m} and Gr_n \times Gr_m.

2011-11-02T05:05:01Z

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.

Answer by Vitali Kapovitch for Homotopy equivalence between the Grassmannian Gr_{n,m} and Gr_n \times Gr_m.

2011-11-02T05:09:03Z

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.