Formulas for vector fields on Grassmannians? - MathOverflow

Formulas for vector fields on Grassmannians?

2009-12-31T19:49:56Z

The Wikipedia article on (real) Grassmannians 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 $M$.

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.

Answer by jvp for Formulas for vector fields on Grassmannians?

2009-12-31T20:46:32Z

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 $G_{n,r}$.

Answer by Mariano Suárez-Alvarez for Formulas for vector fields on Grassmannians?

2009-12-31T20:48:31Z

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 $V$.

Let $k$ be odd and such that $1\leq k\leq n$.

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 $\dim W^\perp=\dim W^{\perp\omega}=\dim V-\dim W$.

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 were 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.

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 $\hom(W,W^\perp)$.

Now, as explained in an answer 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 $W$.

We thus see that the rule $W\mapsto \omega_W$ gives a non-zero tangent vector field.