Proving that $S\mapsto\text{Hom}(\bigwedge^2S,\mathbb R)$ defines a vector bundle over a Grassmanian. My question has to do with the most voted answer to this post. There, it is suggested that the map
$$S\in G_n(\mathbb R^{2n})\mapsto\text{Hom}(\wedge^2S,\mathbb R)$$
determines a vector bundle over the Grassmanian $G_n(\mathbb R^{2n})$. However, I don't know how to prove this rigorously. Specifically, I am having trouble with the maps one should take as trivializations.
Lets recall that $G_n(\mathbb R^{2n})$ has a natural smooth manifold structure, in which the coordinate domains are of the form
$$U_T=\{V\in G_n(\mathbb R^{2n})\,|\,V\cap T=\{0\}\}$$
I would like to give trivializations based on these open sets. More specifically, I would like to find maps
$$\Phi_T:\pi^{-1}(U_T)\rightarrow U_T\times\mathbb R^{\bullet}$$
that are bijective and isomorphisms in its restriction to each fiber. However, for this we should have some way of parametrizing bases of vector spaces $V$ such that $V\cap T=\{0\}$ in terms of a canonical, fixed basis of $T$. My idea is that we could complete such a basis of $T$ to a basis of $\mathbb R^{2n}$, then apply rotations in suitable directions to parametrize all such vector spaces. This would give a decomposition of $\pi^{-1}(U_T)$ into a product of the form $U_T\times\mathbb R^\bullet$, where some part of the $\mathbb R^\bullet$ part has to encode the information about the rotations performed.
I don't know if this is the way to think about this problem, but thanks in advance for your answers.
 A: Here's a totally abstract way to do this without using bases at all. Let's do it in the general case. Given an $m$-dimensional vector space $V$ and $1 \le k \le m$, let $G_k(V)$ be the space of $k$-dimensional linear subspaces of $V$. I will leave the proofs as a long exercise consisting mainly of unwinding all the abstraction.
$\newcommand\Hom{\operatorname{Hom}}$ Given an $(m-k)$-dimensional subspace $T$, let
$$ U_T = \Hom(V/T, T). $$
This is a $k(m-k)$-dimensional vector space.
Each $L \in \Hom(V/T,T)$ defines an injective linear map
\begin{align*}
\phi_L: V/T &\rightarrow V\\
[v] &\mapsto v + L([v])
\end{align*}
This in turn defines an injective map
\begin{align*}
\Phi_T: \Hom(V/T,T) &\rightarrow G_k(V)\\
L &\mapsto \phi_L(V/T).
\end{align*}
You can now check that
$$ \{ \Phi_T: U_T \rightarrow G_k(V)\ :\ T \in G_{m-k}(V)\} $$
is a smooth atlas of $G_k(V)$.
The trivialization of the vector bundle $E$, where $E_S = \Hom(\bigwedge S,\mathbb{R})$, over each coordinate chart $U_T$ can now be defined as follows: First, observe that using the isomorphism $\phi_L: V/T \rightarrow \Phi_T(L)$, we get an isomorphism
$$
(\phi_L^{-1})^*: \Hom(\wedge^2(V/T),\mathbb{R}) \rightarrow \Hom(\wedge^2S,\mathbb{R}),
$$
where $S = \Phi_T(L)$. The trivialization is now given by
\begin{align*}
U_T\times \Hom(\wedge^2(V/T),\mathbb{R}) &\rightarrow \left.E\right|_{\Phi_LU_T}\\
(L, \ell) &\mapsto (\Phi_T(L), (\phi_L^{-1})^*\ell).
\end{align*}
To do this using bases, it suffices, for each $T \in G_{m-k}(V)$, to fix a transversal subspace $S \in G_k(V)$ and choose a basis $(e_1, \dots, e_m)$ such that $(e_1, \dots, e_k)$ is a basis of $S$ and $(e_{k+1}, \dots, e_m)$ is a basis of $T$. Then there is an isomorphism between $\Hom(V/T,T)$ and the vector space of $(m-k)$-by-$k$ matrices. I recommend doing it this way instead of trying to decipher what I wrote above.
If you want to do this only for $V = \mathbb{R}^m$, then it suffices to restrict $S$ to subspaces spanned by $k$ standard coordinate basis vectors and $T$ to subspaces spanned by the other $m-k$ standard coordinate basis vectors.
