Homotopy equivalence between the Grassmannian Gr_{n,m} and Gr_n \times Gr_m. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:37:43Z http://mathoverflow.net/feeds/question/79799 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79799/homotopy-equivalence-between-the-grassmannian-gr-n-m-and-gr-n-times-gr-m Homotopy equivalence between the Grassmannian Gr_{n,m} and Gr_n \times Gr_m. Ralph 2011-11-02T04:25:36Z 2011-11-02T05:33:21Z <p>The following assertion appears in a paper I am reading, and I can't seem to verify it.</p> <p>Let $\text{Gr}_{n,m}$ denote the set of pairs $(V,W)$ where $V$ and $W$ are as follows.</p> <ol> <li>$V$ is an $n$-dimensional subspace of $\mathbb{C}^{\infty}$.</li> <li>$W$ is an $m$-dimensional subspace of $\mathbb{C}^{\infty}$.</li> <li>$V$ and $W$ are orthogonal.</li> </ol> <p>The space <code>$\text{Gr}_{n,m}$</code> has an obvious topology. If <code>$\text{Gr}_n$</code> and <code>$\text{Gr}_m$</code> are the usual Grassmannians of $n$ and $m$ planes in $\mathbb{C}^{\infty}$, then there is an obvious map $\psi : \text{Gr}_{n,m} \rightarrow \text{Gr}_n \times \text{Gr}_m$.</p> <p>The map $\psi$ is almost a homeomorphism, but not quite because of condition 3 above. The paper claims that $\psi$ is a homotopy equivalence.</p> <p>Thanks for any help!</p> http://mathoverflow.net/questions/79799/homotopy-equivalence-between-the-grassmannian-gr-n-m-and-gr-n-times-gr-m/79801#79801 Answer by Tyler Lawson for Homotopy equivalence between the Grassmannian Gr_{n,m} and Gr_n \times Gr_m. Tyler Lawson 2011-11-02T05:05:01Z 2011-11-02T05:33:21Z <p>The forgetful map <code>$Gr_{n,m} \to Gr_n$</code> that drops $W$ is a fiber bundle (exercise), and the map <code>$Gr_{n,m} \to Gr_n \times Gr_m$</code> 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.</p> <p>The fibers over $V$ are, respectively: $m$-dimensional subspaces in $V^\perp \subset \mathbb{C}^\infty$, and $m$-dimensional subspaces in $\mathbb{C}^\infty$.</p> <p>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.</p> http://mathoverflow.net/questions/79799/homotopy-equivalence-between-the-grassmannian-gr-n-m-and-gr-n-times-gr-m/79802#79802 Answer by Vitali Kapovitch for Homotopy equivalence between the Grassmannian Gr_{n,m} and Gr_n \times Gr_m. Vitali Kapovitch 2011-11-02T05:09:03Z 2011-11-02T05:28:47Z <p>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.</p>