Poisson bracket

For a symplectic manifold

Definition
Given a symplectic manifold $(M, ω)$ , the Poisson bracket on $C^{\infty} (M)$ , the space of smooth functions on $M$ , is a binary operation ${\cdot, \cdot} : C^{\infty} (M) \times C^{\infty} (M) \to C^{\infty} (M)$ defined as follows: for any two functions $f, g \in C^{\infty} (M)$ , the Poisson bracket is given by

{f, g} (p) = ω (X_{f}, X_{g}) (p),

where $X_{f}$ and $X_{g}$ are the Hamiltonian vector fields associated to $f$ and $g$ respectively, and $p$ is a point in $M$ .
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In the canonical coordinates of a classical Hamiltonian system is written as

{f, g} = \sum_{i = 1}^{N} (\frac{\partial f}{\partial q_{i}} \frac{\partial g}{\partial p_{i}} - \frac{\partial f}{\partial p_{i}} \frac{\partial g}{\partial q_{i}})

The Poisson brackets of the canonical coordinates are

\begin{array}{l} {q_{i}, q_{j}} = 0 \\ {p_{i}, p_{j}} = 0 \\ {q_{i}, p_{j}} = δ_{i j} \end{array}

which are reminiscent of canonical commutation relations.

Generalization

The Poisson bracket induced by a symplectic form satisfies the following properties:

Bilinearity
Antisymmetry: ${f, g} = - {g, f}$ .
Leibniz rule: ${f, g h} = {f, g} h + g {f, h}$ .
Jacobi identity: ${f, {g, h}} + {g, {h, f}} + {h, {f, g}} = 0$ .
These properties make the Poisson bracket a Lie bracket, that is, it makes $(C^{\infty} (M), {\cdot, \cdot})$ into a Lie algebra. Also is a Poisson algebra. If we abstract these properties we can define a Poisson manifold.

For every $f$ we have that ${-, f}$ is a differential operator, so it is a vector field. Indeed, if the Poisson bracket is coming from a symplectic form, it is ${-, f} \equiv X_{f}$ since for any smooth function $g$

{g, f} = ω (X_{g}, X_{f}) = d g (X_{f}) = X_{f} (g) .

Symmetrically, $X_{g} (f) = {f, g}$ , so together

{f, g} = ω (X_{f}, X_{g}) = X_{g} (f) = - X_{f} (g) .

In particular, a function $f$ is a first integral of the Hamiltonian vector field $X_{H}$ iff ${f, H} = 0$ . Indeed, $X_{H} (f) = {f, H}$ , so $X_{H} (f) = 0 ⟺ {f, H} = 0$ .

Relation with Lie bracket

On the other hand, observe that we can rewrite Jacobi identity in this way:

{f, {g, h}} = {{h, f}, g} - {h, {f, g}}

Leaving an slot instead of $f$ we get

{-, {g, h}} = {- X_{h} (-), g} - {h, X_{g} (-)} =

= - X_{g} (X_{h} (-)) + X_{h} (X_{g} (-))

expression that can be rewritten as

X_{{g, h}} = [X_{h}, X_{g}] = X_{ω (X_{g}, X_{h})}

In local coordinates

It can be shown (see @olver86 page 393 section "The structure functions") that in local coordinates $(x^{1}, \dots, x^{m})$ the Poisson bracket can be expressed:

{F, H} = \sum_{i = 1}^{m} \sum_{j = 1}^{m} {x^{i}, x^{j}} \frac{\partial F}{\partial x^{i}} \frac{\partial H}{\partial x^{j}}

The functions ${x^{i}, x^{j}}$ are called the structure functions of the Poisson bracket in this coordinates, and they can be arranged into a skew-symmetric matrix denoted $J (x)$ . If we denote by $\nabla F$ the usual gradient we have that

{F, H} = \nabla F^{T} J \nabla H

Example. In the natural example given in Poisson manifold the matrix is

J = (\begin{array}{rrr} 0 & - I & 0 \\ I & 0 & 0 \\ 0 & 0 & 0 \end{array})

in the $(p, q, z)$ -coordinates.
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I think this matrix is a kind of degenerate symplectic form.

In this context, since

X_{H} = {-, H} = \sum_{i = 1}^{m} \sum_{j = 1}^{m} J^{i j} (x) \frac{\partial H}{\partial x^{j}} \frac{\partial}{\partial x^{i}}

Hamiltonian equations for a function $H$ are given by

\frac{d x}{d t} = J (x) \nabla H (x) .