Formulas for vector fields on Grassmannians? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:09:31Z http://mathoverflow.net/feeds/question/10315 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10315/formulas-for-vector-fields-on-grassmannians Formulas for vector fields on Grassmannians? Ryan Budney 2009-12-31T19:49:56Z 2009-12-31T20:51:40Z <p>The Wikipedia article on (real) <a href="http://en.wikipedia.org/wiki/Grassmannian#Schubert%5Fcells" rel="nofollow">Grassmannians</a> gives a simple argument that the Euler characteristic satisfies a recurrence relation $$\chi G_{n,r} = \chi G_{n-1,r-1} + (-1)^r \chi G_{n-1,r}$$. This implies that the euler characteristic is zero if and only if $n$ even and $r$ odd. So $\chi G_{6,3} = 0$, for example. Basic obstruction theory on manifolds tells us that if $\chi M=0$ then there is a non-vanishing vector field on $M$. </p> <p><br></p> <p>Does anyone have any simple, explicit examples of such vector fields? Let's say $G_{n,1}$ for $n$ even does not count. Not to say those aren't interesting examples -- I'd love solutions that simple. But it's not immediately clear how they can be generalized. </p> http://mathoverflow.net/questions/10315/formulas-for-vector-fields-on-grassmannians/10321#10321 Answer by jvp for Formulas for vector fields on Grassmannians? jvp 2009-12-31T20:46:32Z 2009-12-31T20:51:40Z <p>Identify $\mathbb R^2$ with $\mathbb C$ and consider the $S^1$ action on $\mathbb R^{2n} \simeq \mathbb C^n$ induced by cordinatewise complex multiplication. These of course lead to the trivial examples on $G_{2n,1}$. For $n$ even and $r$ odd the very same examples do the trick. One has just to observe that these $S^1$ actions have no invariant odd dimensional subspaces, and therefore induce $S^1$-actions without fixed points on $G_{n,r}$.</p> http://mathoverflow.net/questions/10315/formulas-for-vector-fields-on-grassmannians/10322#10322 Answer by Mariano Suárez-Alvarez for Formulas for vector fields on Grassmannians? Mariano Suárez-Alvarez 2009-12-31T20:48:31Z 2009-12-31T20:48:31Z <p>Pick an even dimensional real vector space $V$ of dimension $n$ and fix a symplectic form $\omega$ on $V$. Look at it as a map $\omega:V\otimes V\to\mathbb R$ by extending it from $\Lambda^2V$ to $V\otimes V$ as zero on the symmetric part. Fix also an inner product $\langle\mathord-,\mathord-\rangle$ in $V$.</p> <p>Let $k$ be odd and such that $1\leq k\leq n$.</p> <p>Let $W\subseteq V$ be an $k$-dimensional subspace, and let $W^\perp$ and $W^{\perp\omega}$ be the subspaces orthogonal to $W$ with respect to $\langle\mathord-,\mathord-\rangle$ and to $\omega$, respectively. We know that $\dim W^\perp=\dim W^{\perp\omega}=\dim V-\dim W$. </p> <p>The restriction $\omega|_{W\otimes W^\perp}:W\otimes W^\perp\to\mathbb R$, which I will write for simplicity just $\omega_W$, is not zero. Indeed, if it <em>were</em> zero, we would have that $W^\perp$ is contained in $W^{\perp\omega}$, so in fact these two orthogonal subspaces would be equal, and in consequence we would have that $W\cap W^{\perp\omega}=W\cap W^\perp=0$. This would tell us that $W$ is in fact a symplectic subspace of $V$, which is absurd because it is odd dimensional.</p> <p>Now $\omega_W$ is an element of $\hom(W\otimes W^\perp,\mathbb R)$, which identifies canonically with $\hom(W,\hom(W^\perp,\mathbb R))$. The inner product $\langle\mathord-,\mathord-\rangle$ restricts to an inner product on $W^\perp$ which allows us to identify canonically (because the inner product is fixed!) $\hom(W^\perp,\mathbb R)$ with $W^\perp$. After all these identifications, we have a non zero vector $\omega_W$ in $\hom(W,W^\perp)$.</p> <p>Now, as explained in an <a href="http://mathoverflow.net/questions/10301/how-many-parameters-are-needed-to-specify-a-k-dimensional-subspace-of-rd/10305#10305" rel="nofollow">answer</a> to a MO question, $\hom(W,W^\perp)$ parametrizes a neighborhood of $W$ in $G(n,k)$, so it also can be identified with the tangent space to $G(n,k)$ at $W$. </p> <p>We thus see that the rule $W\mapsto \omega_W$ gives a non-zero tangent vector field.</p>