So, now that we have a notion of symplectic manifolds, we can talk about the Poisson bracket. This gives a way of making the smooth functions into a Lie algebra. The first step in the story is to use the symplectic form to associate to a function a vector field (obtained by duality from ). These Hamiltonian vector fields have many nice properties: for instance, their Lie bracket is of the same type. Moreover, they (and, locally, only they) are the vector fields whose flows preserve the symplectic form. In mechanics, the flows of the Hamiltonian field associated to the energy function trace out the paths of a particle acted on by a conservative force.

Let be a symplectic manifold. Given a smooth function , we have a 1-form on . The self-duality of induced by can be used to “lower indices” (kind of like how one gets a gradient on a Riemannian manifold) so that we get a vector field. Call its opposite , the **Hamiltonian vector field** associated to .

In other words,

By nondegeneracy, this uniquely determines .

Using this, we can talk about the **Poisson bracket** of two functions :

which makes it clear that the Poisson bracket is antisymmetric.

There are a bunch of useful properties satisfied by these brackets, of which I want to talk about a few. Also, I’ll see if I can try to say something about the applications to physics.

Anyway, we should probably do an example here. Take the canonical symplectic form on (to which we know every symplectic manifold is locally symplectomorphic). Then

**The Lie bracket of Hamiltonian vector fields **

I now claim:

Proposition 1

We could of course prove this by a computation in local coordinates. I think it is more fun to give an invariant proof.

First, let us recall a basic fact about the exterior derivative of a 2-form. If is a 2-form, then

One way to see this is to check that both sides are linear over smooth functions and then check when commute. (There is actually a more general such formula.)

Now, back to the proposition. Since is closed, we take and get that the following two functions are equal:

and

The first (*) can be written as

and the second as

or

by the antisymmetry of in . Comparing these two shows that satisfies the characteristic property of .

**The Jacobi identity for the Poisson bracket **

Now I claim:

Proposition 2 (Jacobi identity)

So, the vector space of smooth functions on a symplectic manifold is a Lie algebra under the Poisson bracket.

(**Edit: **Apologies to those readers who saw this initially; I had forgotten to actually put in the proof. It’s now fixed.)

This is now a simple corollary of what has already been proved. By the definitions, antisymmetry, and the Lie bracket identity, the displayed term equals:

**Invariance of the symplectic and volume forms **

Proposition 3

i.e. the flow from a Hamiltonian vector field leaves invariant the symplectic form.

This follows from the magic formula

by assumption. The vanishing of the Lie derivative means that for the flow of has identically zero derivative, hence is constant.

In fact, if the flows of an arbitrary vector field preserve the symplectic form, then by the same calculation

which means is closed, hence locally exact—and globally if is simply connected. Thus locally is Hamiltonian, and globally if is simply connected.

On a -dimensional symplectic manifold with 2-form , we can consider the closed -form

which can be checked to be nonzero. So a symplectic manifold can be given an orientation and a **symplectic volume**.

Indeed, in local coordinates, if , then is just up to a sign the usual volume form.

Corollary 4 (Liouville)The flows of leave invariant the symplectic volume.

As lewallen observes, the analogy in Riemannian geometry is a Killing field (whose flow preserves the Riemannian metric), which I have not yet discussed.

**Example in mechanics **

The idea of a Hamiltonian vector field arises in classical mechanics as follows. Consider a particle of mass moving in (say). This is parametrized by a curve . At each time , the particle has both a position in and a vector momentum in , so we take the curve to take this account.

The position is specified by the first -coordinates , and the momentum by . By abuse of notation, we also use as coordinates in .

Now suppose we have a conservative force in . So we have a potential energy function such that the force at is .

The particle must satisfy:

and

I claim that the above two conditions state that the curve is necessarily an integral curve of the Hamiltonian vector field associated to the **energy function**

Incidentally, since a function must be constant along the integral curves of its Hamiltonian vector field—because —this implies conservation of energy. Now

It is now clear that the two conditions correspond to being an integral curve of .

See also this. Next time we’ll use these ideas to prove local existence to certain PDEs.

December 27, 2009 at 11:28 am

Oooh, excellent. When I start rambling about integrable systems (mostly ACIH systems, in fact) I’ll definitely be linking back to you.

December 28, 2009 at 1:57 am

Re: Proposition 2, what Lie group is it a Lie algebra of?

December 28, 2009 at 7:41 am

The Hamiltonian symplectomorphisms since the Poisson Lie algebra is isomorphic to the Lie algebra of Hamiltonian fields. Though this is from Wikipedia. I don’t actually know whether the symplectomorphisms (or Hamiltonian symplectomorphisms) form a Lie group, and if not I don’t know what that means rigorously.

I guess the analogy is that Killing fields are the Lie algebra of the isometry group of a Riemannian manifold (which in this case is a Lie group).

August 25, 2011 at 11:32 am

Im guessing its the Symplectic group Sp(n).