The basic idea is the following. For the sake of simplicity, I henceforth assume that every function does not depend explicitly on time (with a little effort, everything could be generalized dealing with a suitable fiber bundle over the axis of time whose fibers are spaces of phases at time $t$).
On a symplectic $2n$ dimensional manifold (a space of phases), $(M, \omega)$, where $\omega$ is the symplectic 2-form, you first define the Hamiltonian vector field $X_f$ associated to a smooth function $f: M \to \mathbb R$ as the unique vector field such that:
$$\omega(X_f, \cdot) = df\:.\tag{1}$$
The definition is well-posed because $\omega$ is non degenerate by hypotheses. $\omega$ is also antisymmetric and closed by hypotheses, therefore due to a theorem due to Darboux, on a suitable atlas which always exists $\omega = \sum_{i=1}^n dq^i \wedge dp_i$. In this picture you recover the standard $q-p$ formulation of Hamiltonian mechanics.

Next, you have the **Poisson bracket** of two smooth functions defined as
$$\{f, g\}:= \omega(X_f,X_g)\:.\tag{2}$$

The (generally local) one-parameter group of diffeomorphisms generated by $X_f$
turns out to be made of **canonical transformations** in the standard sense of Hamiltonian mechanics. $f$ is said to be the **Hamiltonian generator** of that transformation. Using Darboux' atlas, i.e. coordinates, $q^1,\ldots, q^n, p_1,\ldots, p_n$, both Hamilton equations and Poisson brackets assume the standard form more familiar to physicists.

If $H$ is a preferred function $f$ called the **Hamiltonian function**, the integral lines of $X_H$ are nothing but the solution of Hamilton equations.

With these definition it turns out that, if $[\:.,\:.]$ is the standard commutator of vector fields,
$$[X_f,X_g] = X_{\{f,g\}}\:.\tag{3}$$

As an immediate consequence of (3), you see that if $\{f,H\}=0$, then the integral lines of $X_H$ remains integral lines of $X_H$ also under the action of the group generated by $X_f$. In this case you have a dynamical symmetry.
Moreover, from (1) and (2), $$X_H(f) = \{f,H\}$$
so that $\{f,H\}=0$ also implies that $f$ is invariant under the Hamiltonian flow, i.e., it is a *constant of motion*.

*The fact that $f$ is a constant of motion and that it generates (canonical) transformations which preserve the evolution of the system are equivalent facts.*

This fantastic equivalence does not hold within the Lagrangian formulation of mechanics.

In this scenario, suppose that the $N$ dimensional Lie group $G$ freely acts on $M$ in terms of diffeomorphisms bijective. The one-parameters of the group define corresponding one parameter groups of diffeomorphisms whose generators have the same Lie algebra as that of $G$. So if $e_1,\ldots, e_n$ is a basis of $\mathbb g$ (the Lie algebra of $G$), with
$$[e_i,e_j] = \sum_{k=1}^N c^k_{ij}e_k$$
you correspondingly find, for the associated vector fields defining th corresponding one-parameter groups of diffeomorphisms
$$[X_i,X_j] = \sum_k c^k_{ij}X_k\:.$$

Suppose eventually that each $X_i$ can be written as $X_{f_i}$ for a corresponding smooth function $f_i : M \to \mathbb R$. In this case, the one-parameter group of diffeomorphisms generated by $X_f$ is a one-parameter group of *canonical transformations*. (This automatically happens when the action of $G$ preserves the symplectic form.)
Consequently,
$$X_{\{f_i,f_j\}} = [X_{f_i},X_{f_j}] = \sum_k c^k_{ij}X_{f_k}\:.$$
Using the fact that the Poisson brakes are bi-linear:
$$X_{\{f_i,f_j\} - \sum_k c^k_{ij}f_k} =0$$
and thus, since $f \mapsto X_f$ is injective up an additive constant to $f$,
$$\{f_i,f_j\} = q_{ij} +\sum_k c^k_{ij}f_k$$
the constants $q_{ij}$ generally appear, sometime (as it happens if $G=SO(3)$) you can re-absorb them in the definition of the $f_i$ which, in turn, are defined up to additive constants. (It is a co-homological problem depending on $G$).

This post imported from StackExchange Physics at 2015-12-18 22:27 (UTC), posted by SE-user Valter Moretti