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 second-year math postdoc at the University of Wisconsin-Madison, and I think math is pretty neat. 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.

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