Symplectic Basics I: Symplectic Linear Algebra

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 (V,\sigma), where V is a finite dimensional real vector space (henceforth, all vector spaces for us will be finite dimensional over \mathbb{R}), and \sigma is a symplectic form on V; that is, \sigma is a non-degenerate, alternating, bilinear form on V.  Let’s play with an example to get acquainted.

Example 1. Let V be a vector space, E := V \oplus V^*, and let \langle \cdot, \cdot \rangle : E \to \mathbb{R} be the canonical pairing of V and V^*. Define a bilinear form \sigma on E by

\sigma((x_1,\xi_1);(x_2,\xi_2)) := \langle x_2, \xi_1 \rangle - \langle x_1 , \xi_2 \rangle

for (x_i,\xi_i) \in E.  Naturally, I claim that \sigma is a symplectic form on E. By construction, \sigma is alternating and bilinear, so we only need to check non-degeneracy.  Let (x_1,\xi_1) \in E be such that, for all (x_2,\xi_2) \in E, \sigma((x_1,\xi_1);(x_2,\xi_2)) = 0.  That is, for all (x_2,\xi_2) \in E,

\langle x_2, \xi_1 \rangle = \langle x_1, \xi_2 \rangle.

By non-degeneracy of \langle \cdot, \cdot \rangle, setting x_2 = 0 yields \xi_1 = 0, and setting \xi_2 = 0 yields x_1 = 0 (remember, that equality was assumed to hold for all elements of E!).  Hence, (x_1,\xi_1) = (0,0), implying \sigma is non-degenerate.  Note that, for V = \mathbb{R}, the form \sigma looks a lot like the determinant map! (\sigma((x_1,y_1);(x_2,y_2)) = x_2 y_1 - x_1 y_2).

Now, for a subspace W of a symplectic vector space (V,\sigma), we associate its symplectic complement, or symplectic orthogonal.

W^\perp := \{ x \in V | \sigma(x,y) = 0 \text{ for all }y \in W\}.

This is where we get the notions of isotropic, involutive, and Lagrangian subspaces: a subspace W of (V,\sigma) is

  • isotropic if W \subseteq W^\perp,
  • involutive if W^\perp \subseteq W, and
  • Lagrangian if W = W^\perp.

Let’s end with some easy examples:

Example 2.  A line \ell is always an isotropic subspace.

Let x \in \ell be a non-zero vector, so that every element of \ell is of the form tx for some t \in \mathbb{R}.  Then, the fact that \sigma is bilinear and alternating implies that \ell \subseteq \ell^\perp.

Example 3. A hyperplane H is always an involutive subspace.

Let x \in H^\perp be non-zero.  If x \notin H, then since H is a hyperplane, we must have \langle x \rangle + H = V (by \langle x \rangle, I mean the line spanned by the non-zero vector x), so that every element y of V is of the form y = tx + z, for some t \in \mathbb{R} and z \in H.  But, since x \in H^\perp, we must have

\sigma(x,y) = \sigma(x,tx + z) = t\sigma(x,x) + \sigma(x,z) = 0

by bilinearity.  Since y \in V was arbitrary, the non-degeneracy of \sigma yields x = 0, a contradiction.  Thus, x \in H, so H 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 \varphi : (V_1,\sigma_1) \to (V_2,\sigma_2) is called symplectic provided \sigma_1 = \varphi^* \sigma_2.  That is, for all x,y \in V_1, we have (by definition of the pullback)

\sigma_1(x,y) = \sigma_2(\varphi(x),\varphi(y)).

A symplectic map that is also invertible is called a symplectomorphism.

Just like every vector space is modeled on \mathbb{R}^n for some n (upon choosing a basis), all symplectic vector spaces of dimension 2n are symplectomorphic to (\mathbb{R}^{2n},\sigma_n), where

\sigma_n((x,y);(x',y')) := \sum_{j=1}^n (x_j'y_j - x_j y_j')

(x = (x_1,\cdots,x_n), y= (y_1,\cdots,y_n)) for each n \geq 1 (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 n \geq 1, the subspace Z_n := \mathbb{R}^n \oplus \{0 \} is a Lagrangian subspace of (\mathbb{R}^{2n},\sigma_n) (i.e., Z_n = \{(x_1,\cdots,x_n,y_1,\cdots,y_n) | y_i = 0, 1 \leq i \leq n \}).  As it turns out, Z_n is the prototype for all Lagrangian subspaces:

Proposition 4: Given any symplectic vector space (V,\omega) of dimension 2n, and Lagrangian \lambda \subseteq (V,\omega), there exists a symplectic map \psi : (\mathbb{R}^{2n},\sigma_n) \to (V,\omega) sending Z_n to \lambda.

proof: Assume that we’ve proved the result for all dimensions \leq n-1 (for n=1, the Lagrangian subspaces are all just lines through the origin in \mathbb{R}^2, and the desired symplectic map is just a rotation about the origin).  We want to then show the result for dimension n.  Okay.  Let \lambda \subseteq (V,\omega) be a Lagrangian subspace, \dim V = 2n.  Pick some e_1 \in \lambda non-zero.  Since \omega is non-degenerate, there exists some f_1 \in V such that \omega(e_1,f_1)=1.  As \lambda is Lagrangian, this gives f_1 \notin \lambda.  Set

\overset{\thicksim}{V} := \{x \in V | \omega(x,e_1)=\omega(x,f_1) = 0\};

with the restriction \overset{\thicksim}{\omega} := \omega|_{\overset{\thicksim}{V}}, (\overset{\thicksim}{V},\overset{\thicksim}{\omega}) is a symplectic space.  Of course, from \omega, the only thing to check is that \overset{\thicksim}{\omega} is non-degenerate (if x \in \overset{\thicksim}{V} \cap (\overset{\thicksim}{V})^\perp is non-zero, there exists some y \in V such that \omega(x,y) \neq 0.  By the definition of \overset{\thicksim}{V}, we must have y \notin \overset{\thicksim}{V}.  It follows that x = 0).

Now, set \overset{\thicksim}{\lambda} := \lambda \cap \overset{\thicksim}{V}.  We need to show \overset{\thicksim}{\lambda} is Lagrangian in \overset{\thicksim}{V}, and \lambda = \overset{\thicksim}{\lambda} + \langle e_1 \rangle.    Since \overset{\thicksim}{\omega}|_{\overset{\thicksim}{\lambda}} = \omega|_{\overset{\thicksim}{\lambda}}, and \lambda = \lambda^\perp, it follows that \overset{\thicksim}{\lambda} is an isotropic subspace.  Is it maximally isotropic in \overset{\thicksim}{V} (i.e., Lagrangian?).  If not, there would exist an isotropic subspace \mu with \overset{\thicksim}{\lambda} \subset \mu \subseteq \overset{\thicksim}{V}.  But then, \mu + \langle e_1 \rangle would be isotropic in (V,\omega), and n = \dim \lambda < \dim (\mu + \langle e_1 \rangle.  But this is a contradiction, since an isotropic subspace of V must have dimension \leq n! (exclamation, not factorial. whoops).  Thus, \overset{\thicksim}{\lambda} is Lagrangian.  For the second part of the claim, we note that \lambda \subseteq \overset{\thicksim}{\lambda} + \langle e_1 \rangle, and the above shows \dim \lambda = \dim (\overset{\thicksim}{\lambda} + \langle e_1 \rangle ) = n, so \lambda = \overset{\thicksim}{\lambda} + \langle e_1 \rangle.

Okay, here’s where we invoke the inductive hypothesis: there exists a symplectic map \varphi_{n-1} : (\mathbb{R}^{2n-2},\sigma_{n-1}) \to (\overset{\thicksim}{V},\overset{\thicksim}{\omega}) sending Z_{n-1} to \overset{\thicksim}{\lambda}.  Then, the map

\varphi_n : (\mathbb{R}^2,\sigma_1) \oplus (\mathbb{R}^{2n-2},\sigma_{n-1}) \to (V,\omega) via

(x,y;z) \mapsto x e_1 + y f_1 + \varphi_{n-1}(z)

is symplectic, and sends Z_n to \lambda.  Oh, by the way: we define the form \sigma_1 \oplus \sigma_{n-1}((x_1,y_1;z_1);(x_2,y_2;z_2)) := \sigma_1((x_1,y_1);(x_2,y_2)) + \sigma_{n-1}(z_1,z_2).  By assumption, we know \sigma_{n-1} = \varphi_{n-1}^* \overset{\thicksim}{\omega}. Since I’m lazy, and this calculation is pretty messy, let’s write X_1 = x_1 e_1 + y_1 f_1, X_2 = x_2 e_1 + y_2 f_1.  Then, by algebra:

\omega(X_1+ \varphi_{n-1}(z_1),X_2 + \varphi_{n-1}(z_2)) = \omega(X_1,X_2) + \omega(X_1,\varphi_{n-1}(z_2) + \omega(\varphi_{n-1}(z_1),X_2) + \omega(\varphi_{n-1}(z_1),\varphi_{n-1}(z_2))

= \omega(X_1,X_2) + \omega(\varphi_{n-1}(z_1),\varphi_{n-1}(z_2) = \sigma_1((x_1,y_1);(x_2,y_2)) + \sigma_{n-1}(z_1,z_2)

as \varphi_{n-1}(z_i) \in \overset{\thicksim}{V}, and hence \omega(\varphi_{n-1}(z_i),e_1) \omega(\varphi_{n-1}(z_i),f_1) = 0 ((i= 1,2) and by expanding the term \omega(X_1,X_2) in terms of e_1,f_1).  So, \varphi_n is symplectic.  \varphi_n(Z_n) = \lambda, because \varphi_n(x_1,0;z) = x_1 e_1 + \varphi_{n-1}(z) \in \lambda = \overset{\thicksim}{\lambda} + \langle e_1 \rangle, and \varphi_{n-1}(Z_{n-1}) = \overset{\thicksim}{\lambda}.  Done!

end proof.

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).

Author: brianhepler

I'm a third-year math postdoc at the University of Wisconsin-Madison, where I work as a member of the geometry and topology research group. Generally speaking, I think math is pretty neat; and, if you give me the chance, I'll talk your ear off. Especially the more abstract stuff. It's really hard to communicate that love with the general population, but I'm going to do my best to show you a world of pure imagination.

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