# 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).