Last post I mentioned some types of subsets of the cotangent bundle, associated to the bundle’s natural symplectic structure (i.e., the isotropic, involutive, and Lagrangian subsets). What was I talking about? Back to basics! Today, I want to talk about some “symplectic linear algebra.”
A symplectic vector space is a pair , where is a finite dimensional real vector space (henceforth, all vector spaces for us will be finite dimensional over ), and is a symplectic form on ; that is, is a non-degenerate, alternating, bilinear form on . Let’s play with an example to get acquainted.
Example 1. Let be a vector space, , and let be the canonical pairing of and . Define a bilinear form on by
for . Naturally, I claim that is a symplectic form on . By construction, is alternating and bilinear, so we only need to check non-degeneracy. Let be such that, for all , . That is, for all ,
By non-degeneracy of , setting yields , and setting yields (remember, that equality was assumed to hold for all elements of !). Hence, , implying is non-degenerate. Note that, for , the form looks a lot like the determinant map! ().
Now, for a subspace of a symplectic vector space , we associate its symplectic complement, or symplectic orthogonal.
This is where we get the notions of isotropic, involutive, and Lagrangian subspaces: a subspace of is
- isotropic if ,
- involutive if , and
- Lagrangian if .
Let’s end with some easy examples:
Example 2. A line is always an isotropic subspace.
Let be a non-zero vector, so that every element of is of the form for some . Then, the fact that is bilinear and alternating implies that .
Example 3. A hyperplane is always an involutive subspace.
Let be non-zero. If , then since is a hyperplane, we must have (by , I mean the line spanned by the non-zero vector ), so that every element of is of the form , for some and . But, since , we must have
by bilinearity. Since was arbitrary, the non-degeneracy of yields , a contradiction. Thus, , so is an involutive subset.
Naturally, when we introduce new structures on spaces, we want to identify those morphisms that “preserve” that structure. In this case, it’s the symplectic form. A linear map is called symplectic provided . That is, for all , we have (by definition of the pullback)
A symplectic map that is also invertible is called a symplectomorphism.
Just like every vector space is modeled on for some (upon choosing a basis), all symplectic vector spaces of dimension are symplectomorphic to , where
() for each (cf: example 1). This isn’t TOO hard to show, but it takes a little bit to work through all the necessary details. I don’t feel like writing this one out; you’ll just have to take my word for it (or, you know, work it out yourself).
That being said, there is a similar result that I do want to show you. It’s pretty clear that, for each , the subspace is a Lagrangian subspace of (i.e., ). As it turns out, is the prototype for all Lagrangian subspaces:
Proposition 4: Given any symplectic vector space of dimension , and Lagrangian , there exists a symplectic map sending to .
proof: Assume that we’ve proved the result for all dimensions (for , the Lagrangian subspaces are all just lines through the origin in , and the desired symplectic map is just a rotation about the origin). We want to then show the result for dimension . Okay. Let be a Lagrangian subspace, . Pick some non-zero. Since is non-degenerate, there exists some such that . As is Lagrangian, this gives . Set
with the restriction , is a symplectic space. Of course, from , the only thing to check is that is non-degenerate (if is non-zero, there exists some such that . By the definition of , we must have . It follows that ).
Now, set . We need to show is Lagrangian in , and . Since , and , it follows that is an isotropic subspace. Is it maximally isotropic in (i.e., Lagrangian?). If not, there would exist an isotropic subspace with . But then, would be isotropic in , and . But this is a contradiction, since an isotropic subspace of must have dimension ! (exclamation, not factorial. whoops). Thus, is Lagrangian. For the second part of the claim, we note that , and the above shows , so .
Okay, here’s where we invoke the inductive hypothesis: there exists a symplectic map sending to . Then, the map
is symplectic, and sends to . Oh, by the way: we define the form . By assumption, we know . Since I’m lazy, and this calculation is pretty messy, let’s write , . Then, by algebra:
as , and hence (() and by expanding the term in terms of ). So, is symplectic. , because , and . Done!
Next time, I’ll talk some more about Lagrangian subspaces and some facts about the Lagrangian Grassmanian of a symplectic vector space.
Reference: M. Kashiwara and P. Schapira, Sheaves on Manifolds (Appendix A).