TITLE: Why are Lagrangian subspaces in a symplectic vector space interesting?
QUESTION [11 upvotes]: A subspace in a symplectic vector space could be one of two extremes: either symplectic (meaning the form is nondegenerate there) or Lagrangian. Or it could be something between the two, meaning a degenerate nonzero form.
What makes Lagrangian subspaces particularly interesting? For example, we consider the Lagrangian Grassmannian. But why not the symplectic Grassmannian?

REPLY [3 votes]: If $L$ is a Langrangian in a symplectic vector space $V$, then $V$ is isomorphic to $L\oplus L^*$ with the standard symplectic form. This is a linear analogue of Weinstein's neighborhood theorem.
Recall that, given a vector space $L$ with dual $L^*$, the vector space $L\oplus L^*$ can be endowed with a symplectic form via the formula
$$
\omega_L((v,\xi),(v',\xi'))=\xi(v')-\xi'(v). 
$$
Note that $L$ and $L^*$ are Lagrangian subspaces of $L\oplus L^*$. Now let $L$ be a Lagrangian inside a symplectic vector space $(V,\omega)$. Then $L$ admits a vector space complement $L'$ that is also Lagrangian. This defines a bijection $\varphi:L'\rightarrow L^*$ via the formula $\varphi(v)(w)=\omega(v,w)$. Then the map $\Phi:V=L\oplus L'\rightarrow L\oplus L^*$ with $\Phi(v,v')=v+\varphi(v')$ is a symplectomorphism between $(V,\omega)$ and $(L\oplus L^*,\omega_L)$.