gt.geometric topology - Formulas for vector fields on Grassmannians?

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 $omega:Votimes Vtomathbb R$ by extending it from $Lambda^2V$ to $Votimes V$ as zero on the symmetric part. Fix also an inner product $langlemathord-,mathord-rangle$ in $V$.

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

Let $Wsubseteq V$ be an $k$-dimensional subspace, and let $W^perp$ and $W^{perpomega}$ be the subspaces orthogonal to $W$ with respect to $langlemathord-,mathord-rangle$ and to $omega$, respectively. We know that $dim W^perp=dim W^{perpomega}=dim V-dim W$.

The restriction $omega|_{Wotimes W^perp}:Wotimes W^perptomathbb R$, 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 $W^{perpomega}$, so in fact these two orthogonal subspaces would be equal, and in consequence we would have that $Wcap W^{perpomega}=Wcap W^perp=0$. 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 $hom(Wotimes W^perp,mathbb R)$, which identifies canonically with $hom(W,hom(W^perp,mathbb R))$. The inner product $langlemathord-,mathord-rangle$ 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 $Wmapsto omega_W$ gives a non-zero tangent vector field.