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 $df$).  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 ${M, \omega}$ be a symplectic manifold. Given a smooth function ${f: M \rightarrow \mathbb{R}}$, we have a 1-form ${df}$ on ${M}$. The self-duality of ${TM}$ induced by ${\omega}$ 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 ${H_f}$, the Hamiltonian vector field associated to ${f}$.

In other words,

$\displaystyle \sigma( H_f, V) = -df(V) = -Vf.$

By nondegeneracy, this uniquely determines ${H_f}$.

Using this, we can talk about the Poisson bracket of two functions ${f,g}$:

$\displaystyle \{f,g\} := H_f g = \sigma(H_f, H_g) = - \sigma(H_g, H_f )$

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 ${\sum dx^i \wedge d\xi^i}$ on ${\mathbb{R}^{2n}}$ (to which we know every symplectic manifold is locally symplectomorphic). Then

$\displaystyle H_f = \sum_i \frac{\partial f }{\partial \xi_i} \frac{\partial}{\partial x_i} - \frac{\partial f }{\partial x_i} \frac{\partial}{\partial \xi_i}.$

The Lie bracket of Hamiltonian vector fields

I now claim:

Proposition 1

$\displaystyle H_{\{f,g\}} = [H_f, H_g].$

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 ${\sigma}$ is a 2-form, then

$\displaystyle d\sigma(A,B,C) = A\sigma(B,C)-B\sigma(A,C)+C\sigma(A,B) -\sigma([A,B], C)$

$\displaystyle + \sigma([A,C],B) - \sigma([B,C],A).$

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

Now, back to the proposition. Since ${\omega}$ is closed, we take ${A=H_f,B=H_g,C=V}$ and get that the following two functions are equal:

$\displaystyle H_f \omega(H_g,V) - H_g\omega(H_f,V) + V\omega(H_f, H_g) \quad (*)$

and

$\displaystyle \omega([H_f,H_g],V) - \omega([H_f,V], H_g) + \omega([H_g,V],H_f) \quad (**).$

The first (*) can be written as

$\displaystyle -H_f Vg + H_g Vf + V \{f,g\}$

and the second as

$\displaystyle -V(\{f,g\}) - (H_f V-VH_f)g + (H_gV - VH_g)f$

or

$\displaystyle -V(\{f,g\}) - H_f Vg + H_g Vf$

by the antisymmetry of ${H_fg}$ in ${f,g}$. Comparing these two shows that ${[H_f,H_g]}$ satisfies the characteristic property of ${H_{\{f,g\}}}$.

The Jacobi identity for the Poisson bracket

Now I claim:

Proposition 2 (Jacobi identity)

$\displaystyle \{ f, \{g, h\} \} + \{ h, \{ f, g \} \} + \{ g, \{ h, f \} \} = 0.$

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:

$\displaystyle H_f H_g h - (H_f H_g h - H_g H_f h) + H_g H_h f = H_g H_f h + H_g H_h f = 0.$

Invariance of the symplectic and volume forms

Proposition 3

$\displaystyle L_{H_f} \omega = 0,$

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

This follows from the magic formula

$\displaystyle L_{H_f} \omega = d i_{H_f} \omega + i_{H_f} d \omega = d i_{H_f} \omega = d(-df) = 0$

by assumption. The vanishing of the Lie derivative means that ${F_t^*\omega}$ for ${F_t}$ the flow of ${H_f}$ has identically zero derivative, hence is constant.

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

$\displaystyle d i_X \omega = 0$

which means ${i_X\omega}$ is closed, hence locally exact—and globally if ${M}$ is simply connected. Thus locally ${\omega}$ is Hamiltonian, and globally if ${M}$ is simply connected.

On a ${2n}$-dimensional symplectic manifold with 2-form ${\omega}$, we can consider the closed ${2n}$-form

$\displaystyle V = \omega \wedge \dots \wedge \omega = \omega^n$

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 ${\omega = \sum dx^i \wedge d \xi^i}$, then ${V}$ is just up to a sign the usual volume form.

Corollary 4 (Liouville) The flows of ${H_f}$ 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 ${m}$ moving in ${\mathbb{R}^n}$ (say). This is parametrized by a curve ${c}$. At each time ${t}$, the particle has both a position in ${\mathbb{R}^n}$ and a vector momentum in ${\mathbb{R}^n}$, so we take the curve ${c: I \rightarrow \mathbb{R}^{2n}}$ to take this account.

The position is specified by the first ${x}$-coordinates ${x^1(t),\dots, x^n(t)}$, and the momentum by ${\xi^1(t), \dots, \xi^n(t)}$. By abuse of notation, we also use ${x^1, \dots, x^n, \xi^1, \dots, \xi^n}$ as coordinates in ${\mathbb{R}^{2n}}$.

Now suppose we have a conservative force in ${\mathbb{R}^n}$. So we have a potential energy function ${V: \mathbb{R}^n \rightarrow \mathbb{R}}$ such that the force at ${\mathbf{x} \in \mathbb{R}^n}$ is ${F(\mathbf{x}) = -\mathrm{grad} \ V(\b{x})}$.

The particle must satisfy:

$\displaystyle \frac{ d \mathbf{x} }{dt} = \frac{1}{m} \mathbf{\xi}$

and

$\displaystyle \frac{d \mathbf{\xi}}{dt} = F( \mathbf{x}, \mathbf{\xi}).$

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

$\displaystyle E(x, \xi) = \frac{1}{2m} |\mathbf{\xi}|^2 + V(\b{x}).$

Incidentally, since a function must be constant along the integral curves of its Hamiltonian vector field—because ${H_f f = \{f,f\} = 0}$—this implies conservation of energy. Now

$\displaystyle H_E = \sum_i \frac{\xi}{m} \frac{\partial}{\partial x_i} + F_i(x) \frac{\partial}{\partial \xi_i}.$

It is now clear that the two conditions correspond to ${c}$ being an integral curve of ${H_E}$.

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