Before we proceed with Lagrangian stuff, I should really talk about a fundamental property of (finite dimensional, real!) symplectic vector spaces:
proposition 1: They’re always even dimensional!
Let be a symplectic vector space of dimension m. I claim then that m is even. Choose some basis of , so that we get a matrix representative of , i.e., for all . Since is alternating, is skew-symmetric, giving the relation . Hence,
So, if is odd, we must have . But, since is non-degenerate, , which forces to be even in the above equation. Done!
A quick application of the “rank-nullity” theorem for a subspace applied to implies ; hence, all Lagrangian subspaces of are of dimension . Similarly, if and , then is Lagrangian.
Of course, from plain old linear algebra, we know that, given any Lagrangian subspace of , there exists an dim subspace such that . The question now is:
proposition 2: Can the complementary subspace be chosen such that is also Lagrangian?
Given Lagrangian, choose an isotropic space such that ( is a proper subspace, so we can always at least choose a line that satisfies this property, and lines are always isotropic). If , is not a subset of . Indeed, if it were, then we’d have (by “symplectic duality” given by regarding as the annihilator of w.r.t. ) . Since (so ), we contradict the assumption that is isotropic: i.e., we assumed . Thus, does not contain . Choose then some . Then, is an isotropic subspace (as is bilinear and alternating) satisfying . Arguing by induction on , we get our Lagrangian space . Perfect.
Why is this relevant?
Recall the space of -dim subspaces of , called the Grassmannian and denoted . It is a smooth manifold of dimension . My main goal is to introduce the Lagrangian Grassmannian of Lagrangian planes in , a closed (smooth) submanifold of of dimension . Also, it’s compact! (implicitly, this is proposition 3)
Let . I claim that there is a bijection between the spaces (which is open in the subspace topology of in ) and the space of (real) quadratic forms on (a real vector space of dimension ).
Set (where is the algebraic dual of ), equipped with the standard symplectic form . With respect to , the subspace is Lagrangian (follows immediately from the definition of and the assumption ). By proposition 4 from the last post, we know there exists a symplectic map with . So, we might as well work inside . Let . Then, (it’s an easy exercise to show) that is the “graph” of a (unique!) linear map , that is, we can write (this is actually very similar to how you construct the smooth atlas on the ordinary Grassmannian manifold; check it out!).
Since , . Using the fact that is the graph of the matrix (suppose we’ve chosen a basis), this tells us
i.e., is a symmetric matrix. With a basis fixed for (giving a basis of by taking the dual basis for ), we know that the collection of real symmetric matrices is linearly isomorphic to the space of real quadratic forms on (cf. the wiki page on quadratic forms, or whatever linear algebra reference you hold dear), which has the desired dimension. Since this map is linear, it’s (a fortiori) smooth. Compactness follows trivially from the fact that is compact and is closed in . The only thing left to check is that the transition maps between different charts are smooth. Screw that; left to the reader.
In the up and coming posts, we generalize our “symplectic vector spaces” to get symplectic manifold: these are smooth (real, at least for us) manifolds equipped with a closed, non-degenerate 2-form that gives each tangent space the structure of a symplectic vector space. Neat, right? Will there be some sort of “standard” symplectic manifold, like we saw last post (cf. example 1)?
(The answer is yes).
- M. Kashiwara and P. Schapira, Sheaves on Manifolds (Appendix A).