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 \alphais 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:

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 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:

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).