May 2014 – brainhelper

Symplectic Basics III: The Cotangent Bundle

Now that we’ve gotten a little comfortable with the idea of a symplectic vector space, it shouldn’t take a huge leap of the imagination to say there’s a similar notion for (smooth) manifolds, too: they’re called symplectic manifolds (surprise). A (real, smooth) symplectic manifold of dimension $2n$ consists of a pair $(M,\omega)$ , where $M$ is the manifold, and $\omega$ is a closed, non-degenerate 2-form on $M$ (i.e., for all $p \in M$ , $\omega_p$ is an alternating bilinear form on the tangent space $T_p M$ , $d\omega = 0$ , and $\frac{1}{n!} \omega^n$ is a volume form on $M$ (equivalently, for all $p \in M$ , $\omega_p$ is a non-degenerate bilinear form on $T_pM$ ) ). In analogy with the case with symplectic vector spaces, $\omega$ picks out certain types of submanifolds of $M$ . That is, for a submanifold $N \subseteq (M,\omega)$ , we say $N$ is isotropic (resp., Lagrangian; resp., involutive) if, for all $p \in N$ , $T_p N \subseteq (T_pM, \omega_p)$ is an isotropic (resp., Lagrangian; resp., involutive) subspace. So, at face value, there doesn’t seem to be much of a change from the ordinary case of symplectic vector spaces, at least regarding these special types of submanifolds. Wrong, I was. I hope to talk about some of these things today.

Example 1. So, for baby’s first example of a symplectic manifold, we look at $M = \mathbb{R}^{2n}$ , equipped with the 2-form $\sigma_n \in \Omega^2(\mathbb{R}^{2n})$ (fyi, we write $\Omega^k(M)$ to denote the space of ( $C^\infty$ ) $k$ -forms on $M$ ) which has constant value $(\sigma_n)_p = \sigma_n$ on $T_p \mathbb{R}^{2n} \cong \mathbb{R}^{2n}$ , for all $p \in \mathbb{R}^{2n}$ . If you understood proposition 4 from my post about symplectic linear algebra (here: https://brainhelper.wordpress.com/2014/05/04/symplectic-basics/), you shouldn’t have any trouble with this example. The biggest change here is needing to work with differential forms on $\mathbb{R}^{2n}$ , not just a single bilinear form.

Let $(x;y) = (x_1,\cdots,x_n;y_1,\cdots,y_n) \in \mathbb{R}^{2n}$ be a (global) choice of linear coordinates on $\mathbb{R}^{2n}$ , and let $p = (x_0;y_0) \in \mathbb{R}^{2n}$ . Then,

$(\sigma_n)_p(v_1;v_2) := d_py \wedge d_px (v_1;v_2) = \sum_{i=1}^n (d_p y_i \wedge d_p x_i)(v_1;v_2)$

for $v_1,v_2 \in T_p \mathbb{R}^{2n} \cong \mathbb{R}^{2n}$ .

Example 2. As far as I’m (currently) concerned, these are the most important examples of symplectic manifolds: given any (real, smooth) manifold $M$ of dimension $n$ , the cotangent bundle $T^*M$ has a canonical symplectic structure. This is important, so I’ll go through most of the details on this one.

We’re going to need to conjure up some “canonical” symplectic form $\omega \in \Omega^2(T^*M)$ . There are essentially two ways to do this: first, a coordinate-free definition of $\omega$ ; after that, we’ll see what $\omega$ looks like in a system of coordinates to get a better feel for what we’re doing. The gist is that, on any cotangent bundle, we get a really useful 1-form for free, called the canonical 1-form (or tautological 1-form, or Liouville 1-form, or the Poincar\'{e} 1-form…), $\alpha \in \Omega^1(T^*M)$ .

Let $\pi : T^*M \to M$ be the projection map, $(x;\xi) \in T^*M$ (i.e., $x \in M$ is a point, and $\xi \in T_x^* M$ is a covector at $x$ ). Then, using the pullback $\pi^* : T^*M \to T^*(T^*M)$ , we define the value of $\alpha$ at $(x;\xi)$ to be

$\alpha_{(x;\xi)} := \pi_{(x;\xi)}^* (\xi)$ .

This seems stranger/more abstruse than it actually is. From the definition of $\pi_{(x;\xi)}^*$ , $\pi_{(x;\xi)}^*(\xi) = \xi \circ d_{(x;\xi)} \pi$ (this checks out: $\xi$ is, by definition, a linear map $T_xM \overset{\xi}{\to} \mathbb{R}$ , and $d_{(x;\xi)} \pi : T_{(x;\xi)}(T^*M) \to T_x M$ , so the composition at least makes sense). Finally, we define the “canonical” symplectic form $\omega$ on $T^*M$ to be $\omega := d\alpha$ (NB: some authors take $\omega = -d\alpha$ . It doesn’t really matter, since both choices give you a coordinate-independent construction of a symplectic form).

Now, suppose we have a local system of coordinates $(x) = (x_1,\cdots,x_n) \in M$ near $x_0$ , with associated linear coordinates $(x; \sum_{i=1}^n \xi_i dx_i) \in T^*M$ . What does $\alpha$ look like in these coordinates? Well, we know $\alpha_{(x;\xi)} := \pi_{(x;\xi)}^*( \xi)$ , so we just plug the coordinate expressions in and wind up with:

$\alpha= \pi^* (\sum_{i=1}^n \xi_i d x_i ) = \sum_{i=1}^n \xi_i d ( x_i \circ \pi) = \sum_{i=1}^n \xi_i dx_i$

(since $\pi(x;\xi) = x$ ). Hence, the canonical 1-form looks like $\alpha = \sum_{i=1}^n \xi_i dx_i$ in the local coordinates $(x;\xi)$ ; consequently, the symplectic form $\omega$ has the coordinate expression

$\omega = d\alpha = \sum_{i=1}^n d(\xi_i dx_i) = \sum_{i=1}^n d\xi_i \wedge dx_i$

near $x_0$ . In particular, since $\omega$ is exact, it is automatically a closed 2-form, and in these coordinates, it is easy to show that $\omega = d\alpha$ is non-degenerate, and that (had we started out with using coordinates, the above expression is independent of the coordinates chosen). The rest of the details are yours to check.

What’s the big deal? Why is $\alpha$ important/useful? Well, for one, it satisfies the following universal property:

proposition 1: The canonical 1-form $\alpha$ is uniquely characterized by the property that, for every 1-form $\eta : M \to T^*M$ (i.e., $\eta(x) = (x;\eta_x) \in T^*M$ is the graph of $\eta$ ), one has $\eta^* \alpha = \eta$ .

proof.

First, we note that $\eta^* : T^*(T^*M) \to T^*M$ , and

$\eta^* \alpha = \sum_{i=1}^n \eta^*(\xi_i dx_i) = \sum_{i=1}^n (\xi \circ \eta) d (x_i \circ \eta)$

$= \sum_{i=1}^n \eta_i dx_i = \eta$

is just the expression of $\eta$ in the local coordinates $(x; \sum_{i=1}^n \xi_i dx_i)$ on $T^*M$ . I’ll leave the proof of uniqueness of $\alpha$ to the reader 🙂

end proof.

Secondly, the canonical 1-form gives a really easy way to produce Lagrangian submanifolds of $(T^*M,\omega)$ . Let $\eta$ be a smooth 1-form on $M$ , and denote by $\Lambda_\eta := \{ (x;\eta_x) \in T^*M | x \in M\}$ the graph of $\eta$ . Then,

proposition 2: $\Lambda_\eta$ is a Lagrangian submanifold of $(T^*M,\omega)$ if and only if $\eta$ is closed.

proof:

Clearly, $\Lambda_\eta$ is a smooth submanifold of $T^*M$ of dimension $n$ (it’s diffeomorphic to $M$ itself). Then,

$\omega|_{\Lambda_\eta} = \eta^* \omega = \eta^* d\alpha = d(\eta^* \alpha) = d\eta$ ,

So $\omega$ vanishes on $\Lambda_\eta$ if and only if $d\eta = 0$ , i.e., if $\eta$ is closed.

end proof.

(Aside: for a symplectic vector space $(V,\sigma)$ (of dimension 2n), a basis $\{e_1,\cdots,e_n;f_1,\cdots, f_n\} \subset V$ is called a symplectic basis, provided $\sigma = \sum_{i=1}^n f_i^* \wedge e_i^*$ . In such case, we obtain the relations

$\sigma(e_i,e_j) = \sigma(f_i,f_j) = 0$ ,
$\sigma(e_i,f_j) = -\sigma(f_j,e_i) = -\delta_{ij}$ .

( $1 \leq i,j \leq n$ ). Additionally, in a symplectic basis, the Hamiltonian isomorphism $H : V^* \to V$ is given by

$H(e_j^* ) = -f_j$ .
$H(f_j^*) = e_j$ .

for $1\leq i \leq n$ (in the general case, $H : V^* \to V$ is defined by the formula $\langle \theta , v \rangle = \sigma(v,H(\theta))$ , where $\langle \bullet, \bullet \rangle : V^* \times V \to \mathbb{R}$ is the canonical pairing, and $\theta \in V^*, v \in V$ ). I forgot to talk about this in previous posts, and the Hamiltionian isomorphism has a really neat analogue for symplectic manifolds.

end aside.)

So, I’m giving you fair warning now: for the majority of examples/ situations I’ll talk about, the symplectic manifolds will always be of the form $(T^*M,\omega)$ for some smooth manifold $M$ . Okay. I was talking (secretively) about symplectic bases of symplectic vector spaces, and what the Hamiltonian isomorphism looks like when expressed in such a basis. For vector spaces, all we needed was for the (duals of) the basis elements to fit together in a regular way to form the original symplectic form (i.e., $\sigma = \sum_{i=1}^n f_i^* \wedge e_i^*$ ). But, for symplectic manifolds, the symplectic linear algebra takes place on the tangent space to every point; we don’t have the same freedom to choose “global” symplectic bases. The next best thing would be, of course, if we can at least always locally construct some analogue of symplectic bases, spanned by some frame of vector fields which would be a symplectic basis on each tangent space in a neighborhood of a point. Praise be upon us, for this is in fact the case: Darboux’s theorem guarantees that, for a smooth, 2n-dimensional symplectic manifold $(M,\omega)$ , for all $p \in M$ , there exists an open neighborhood $U$ of $p$ with smooth coordinate chart $(x_1,\cdots,x_n;y_1,\cdots,y_n) : U \overset{\thicksim}{\to} \mathbb{R}^{2n}$ (actually, this is a symplectomorphism as well!) such that $\omega_q = \sum_{j=1}^n d_q y_j \wedge d_q x_j$ for all $q \in U$ (one sometimes calls $(x;y)$ a system of Darboux coordinates, instead of/in addition to their symplectic properties). Luckily for us and our cotangent bundles, on $(T^*M,\omega)$ , any local system of cotangent coordinates $(x; \xi)$ on $T^*M$ serves as a system of Darboux coordinates, by our construction of $\omega$ . The most important thing to take away from Darboux’s theorem, however, is that all symplectic manifolds of a given dimension are (locally) the same! No distinguishing local invariants to be found here.

On $(T^*M, \omega)$ , the Hamiltonian isomorphism is the (fiber-wise) linear isomorphism $H : T^*(T^*M) \overset{\thicksim}{\to} T(T^*M)$ , and we’ll use it to construct some of the most important objects in symplectic geometry: Hamiltonian vector fields (check out Noether’s theorem, Hamiltonian dynamics, etc.). Precisely, given a smooth (real-valued) function $f$ on an open subset $U \subseteq T^*M$ , we define the Hamiltonian vector field of f, denoted by $H_f$ , to be the image of the differential $df$ under the Hamiltonian isomorphism $H : T^*(T^*M) \overset{\thicksim}{\to} T(T^*M)$ .

If we’ve chosen some local coordinates $(x;\xi)$ on $T^*M$ , our above discussion implies the vector field $H_f$ is given by

$H_f = \sum_{j=1}^n \frac{\partial f}{\partial \xi_j} \frac{\partial}{\partial x_j} - \frac{\partial f}{\partial x_j} \frac{\partial }{\partial \xi_j}$

Indeed, $\omega = \sum_{j=1}^n d \xi_j \wedge dx_j$ , so $H(dx_j ) = -\frac{\partial }{\partial \xi_j}$ , and $H(d\xi_j) = \frac{\partial}{\partial x_j}$ (for $1 \leq j \leq n$ ). Therefore,

$H_f := H(df) = \sum_{j=1}^n (\frac{\partial f}{\partial x_j}H(dx_j) + \frac{\partial f}{\partial \xi_j} H(d \xi_j) )$

from which result is immediate.

I really want to tie all this material together to talk about isotropic,Lagrangian, and involutive submanifolds of $(T^*M,\omega)$ , but I’ve already rambled on for quite a while in this post. If you’ve been following along these past few posts, you’ll remember that my original motivation for talking about these symplectic objects in $T^*M$ was for their relationship with the microsupport $SS(F)$ of sheaves in $D^b(M)$ . More precisely, for all $F \in D^b(M)$ , $SS(F) \subseteq T^*M$ is an involutive subset; if $F$ is also $\mathbb{R}$ -constructible, then $SS(F)$ is a Lagrangian subset of $T^*M$ . A little alarm should be going off in your head right now; there’s no reason at all for $SS(F)$ to be a smooth submanifold of $T^*M$ (take, for example, $F$ to be the constant sheaf supported on a singular subset of $M$ ), so how do these definitions apply? What details of the definition would we need to modify, and what changes? Let’s find out, next time.

References:

M. Kashiwara and P. Schapira, Sheaves on Manifolds (Appendix A).
A. Cannas de Silva, Lectures on Symplectic Geometry (http://www.math.ist.utl.pt/~acannas/Books/symplectic.pdf)
Alex Zamorzaev, notes from a course on introductory symplectic topology (http://math.stanford.edu/~alzaor/257a.html).

Symplectic Basics II : The Lagrangian Grassmanian

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!

proof:

Let $(E,\sigma)$ be a symplectic vector space of dimension m. I claim then that m is even. Choose some basis of $E$ , so that we get a matrix representative $A$ of $\sigma$ , i.e., $\sigma(x,y) = x^T A y$ for all $x, y \in E$ . Since $\sigma$ is alternating, $A$ is skew-symmetric, giving the relation $A^T = -A$ . Hence,

$det(A) = det(A^T) = det(-A) = (-1)^m det(A)$ .

So, if $m$ is odd, we must have $det(A) = 0$ . But, since $\sigma$ is non-degenerate, $det(A) \neq 0$ , which forces $m$ to be even in the above equation. Done!

end proof.

A quick application of the “rank-nullity” theorem for a subspace $W \subseteq E$ applied to $\sigma$ implies $\dim W + \dim W^\perp = m = 2n$ ; hence, all Lagrangian subspaces of $E$ are of dimension $n = \frac{1}{2} \dim E$ . Similarly, if $\dim W = n$ and $W \subseteq W^\perp$ , then $W$ is Lagrangian.

Of course, from plain old linear algebra, we know that, given any Lagrangian subspace $\lambda_0$ of $E$ , there exists an $n-$ dim subspace $\lambda_1 \subseteq E$ such that $E = \lambda_0 \oplus \lambda_1$ . The question now is:

proposition 2: Can the complementary subspace $\lambda_1$ be chosen such that $\lambda_1$ is also Lagrangian?

proof:

Given $\lambda_0$ Lagrangian, choose an isotropic space $\rho \subseteq E$ such that $\lambda_0 \cap \rho = \{0\}$ ( $\lambda_0$ is a proper subspace, so we can always at least choose a line that satisfies this property, and lines are always isotropic). If $\rho^\perp \neq \rho$ , $\rho^\perp$ is not a subset of $\lambda_0 + \rho$ . Indeed, if it were, then we’d have (by “symplectic duality” given by regarding $\rho^\perp$ as the annihilator of $\rho$ w.r.t. $\sigma$ ) $\lambda_0 \cap \rho^\perp \subseteq \rho$ . Since $\rho^\perp \cap \lambda_0 \subseteq \rho \cap \lambda_0 = \{0\}$ (so $\rho^\perp \cap \lambda_0 = \{0\}$ ), we contradict the assumption that $\rho$ is isotropic: i.e., we assumed $\dim \rho^\perp > 0$ . Thus, $\rho + \lambda_0$ does not contain $\rho^\perp$ . Choose then some $x \in \rho^\perp \backslash (\lambda_0 + \rho)$ . Then, $\rho + \langle x \rangle$ is an isotropic subspace (as $\sigma$ is bilinear and alternating) satisfying $(\rho + \langle x \rangle) \cap \lambda_0 = \{0\}$ . Arguing by induction on $\dim \rho$ , we get our Lagrangian space $\lambda_1$ . Perfect.

end proof.

Why is this relevant?

Recall the space of $n$ -dim subspaces of $E$ , called the Grassmannian and denoted $G(E,n)$ . It is a smooth manifold of dimension $n^2$ . My main goal is to introduce the Lagrangian Grassmannian $\Lambda (E)$ of Lagrangian planes in $E$ , a closed (smooth) submanifold of $G(E,n)$ of dimension $\frac{n(n+1)}{2}$ . Also, it’s compact! (implicitly, this is proposition 3)

proof:

Let $\lambda \in \Lambda(E)$ . I claim that there is a bijection between the spaces $\Lambda_\lambda(E) := \{ \mu \in \Lambda(E) | \lambda \cap \mu = \{0\} \}$ (which is open in the subspace topology of $\Lambda(E)$ in $G(E,n)$ ) and the space of (real) quadratic forms on $\mathbb{R}^n$ (a real vector space of dimension $\frac{n(n+1)}{2}$ ).

Set $W = \lambda \oplus \lambda^*$ (where $\lambda^* := \text{Hom}_\mathbb{R}(\lambda,\mathbb{R})$ is the algebraic dual of $\lambda$ ), equipped with the standard symplectic form $\omega((x,\xi);(x',\xi')) := \langle x', \xi \rangle - \langle x,\xi' \rangle$ . With respect to $\omega$ , the subspace $\lambda \cong \lambda \oplus \{0 \}$ is Lagrangian (follows immediately from the definition of $\omega$ and the assumption $\lambda \in \Lambda(E)$ ). By proposition 4 from the last post, we know there exists a symplectic map $\psi : (E,\sigma) \to (W,\omega)$ with $\psi(\lambda) = \lambda \oplus \{0\}$ . So, we might as well work inside $W$ . Let $\mu \in \Lambda_\lambda(W)$ . Then, (it’s an easy exercise to show) that $\mu$ is the “graph” of a (unique!) linear map $A : \lambda^* \to \lambda$ , that is, we can write $\mu = \{ (Ay^*,y^*) | y^* \in \lambda^*\}$ (this is actually very similar to how you construct the smooth atlas on the ordinary Grassmannian manifold; check it out!).

Since $\mu \in \Lambda_\lambda(W) \subseteq \Lambda(W)$ , $\omega|_{\mu} = 0$ . Using the fact that $\mu$ is the graph of the matrix $A$ (suppose we’ve chosen a basis), this tells us

$\langle Ay_2^*,y_1^* \rangle = \langle y_2^*, Ay_1^* \rangle$

i.e., $A$ is a symmetric matrix. With a basis fixed for $E$ (giving a basis of $W$ by taking the dual basis for $\lambda^*$ ), we know that the collection of $n \times n$ real symmetric matrices is linearly isomorphic to the space of real quadratic forms on $\mathbb{R}^n$ (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 $G(E,n)$ is compact and $\Lambda(E)$ is closed in $G(E,n)$ . The only thing left to check is that the transition maps between different charts are smooth. Screw that; left to the reader.

end proof.

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).

References:

M. Kashiwara and P. Schapira, Sheaves on Manifolds (Appendix A).
Alex Zamorzaev, notes from a course on introductory symplectic topology (http://math.stanford.edu/~alzaor/257a.html).

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!