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:, 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.


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.


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.



Published by 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.

5 thoughts on “Symplectic Basics III: The Cotangent Bundle

  1. You didn’t go into the proof of Darboux at all, but I think it’s worth mentioning how it’s essentially a restatement of the $d\omega=0$ condition, which otherwise you have not justified (and has no analogue in linear algebra). Using Gram-Schmidt, you can orthonormalize a basis at a point, using a nondegenerate form (symmetric or alternative or neither), but in general cannot assume flat coordinates even in a tiny neighborhood of the point. There is a local geometric obstruction, curvature. Remove that obstruction and of course you have your flat (symplectic) coordinates.

  2. Can you clarify what is the codomain of the Hamiltonian isomorphism? You’ve written it twice as cotangent vectors on the cotangent bundle?

    1. Ok, now the codomain of your Hamiltonian isomorphism is tangent vectors on the cotangent bundle. Can you also confirm the domain? You list it as cotangent vectors on the cotangent bundle. But then you write $H(df)$. $df$ is an element of the cotangent bundle, right? Not the cotangent of the cotangent?

      1. I’m sorry, I misread it. Your function $f$ is a function on the cotangent bundle, not on the manifold. So $df$ is in the domain of $H$.

      2. Sorry, for some reason I didn’t get a notification of your comment. The Hamiltonian isomorphism sends covectors on T^*M (elements of T^*(T^*M) to vectors on T^*M (now in T(T^*M)).

        I thought about doing the proof of Darboux’s theorem, and you present a valid point regarding justifying its existence at all. I suppose I’ve just been rushing to cover (and learn) these symplectic basics, so I can get a little more intuition for the involutivity of the microsupport (Lagrangian in the constructible case), and start to unravel Kashiwara and Schapira’s chapter on characteristic (and the sheaf of Lagrangian) cycles in T^*M.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: