Proving that the Lagrangian Grassmanian is compact I am trying to prove that the Lagrangian Grassmanian $\Lambda_n$, that is, the set of all Lagrangian subspaces of $\mathbb R^{2n}$ equiped with the standard symplectic tensor $\omega$, is a compact manifold. I have observed that since a subspace $S$ is Lagrangian if and only if $\omega|_S\equiv0$ and $\dim S=n$, one can characterize $\Lambda_n$ as the following subset of the Grassmanian $G_n(\mathbb R^{2n})$:
$$\Lambda_n=\{S\in G_n(\mathbb R^{2n})\,|\,\omega|_S\equiv0\}\text{.}$$
Since the Grassmanians are compact manifolds, we only need to prove that $\Lambda_n$ is closed in $G_n(\mathbb R^{2n})$; but the above characterization suggests that we should be able to express such a spaces as the zero locus of some continuous function, hence guaranteeing that $\Lambda_n$ is indeed a closed subspace. However, I am struggling to find such a function, and I am worrying that this approach might not work.
Thanks in advance for your answers.
 A: There is a vector bundle on the grassmanian that takes each point $V$ to $\hom(\Lambda^2V,R)$, and a 2-form $\omega$ on the ambient space determines a section of that bundle. You are looking for its zero set
A: When one cannot see a closed subset as the zero set of a continuous function, one might try to see it as the preimage of a more general closed subset.
Here, one cannot consider "$V-V^{\perp}$": such an operation is not well defined in the context we want, and it seems pointless to search for a zero set.
Here is the trick.
First, assume that the map
$$
V\in G_n(\Bbb R^{2n}) \longmapsto V^{\perp} \in G_n(\Bbb R^{2n})
$$
is continuous, where $V^{\perp}$ is the symplectic complement of $V^{\perp}$, defined as $V^{\perp} = \{w \in \Bbb R^{2n} \mid \forall v \in V, \omega(v,w) = 0\}$.
Then, consider the map
$$
\begin{array}{r|ccc}
f\colon & G_n(\Bbb R^{2n}) & \longrightarrow & G_n(\Bbb R^{2n})\times G_n(\Bbb R^{2n}) \\
& V & \longmapsto & (V,V^{\perp})
\end{array}.
$$
Then $f$ is continuous from the previous point.
Let $\Delta$ be the diagonal of $G_n(\Bbb R^{2n})\times G_n(\Bbb R^{2n})$.
Since $G_n(\Bbb R^{2n})$ is Hausdorff, $\Delta$ is closed in $G_n(\Bbb R^{2n})\times G_n(\Bbb R^{2n})$.
It follows that $\Lambda_n = f^{-1}(\Delta)$ is closed in $G_n(\Bbb R^{2n})$.
Since $G_n(\Bbb R^{2n})$ is compact, so is $\Lambda_n$.
So it all boils down to show that $V \in G_n(\Bbb R^{2n}) \longmapsto V^{\perp}\in G_n(\Bbb R^{2n})$ is continuous, which I leave you as an exercise.
