Classical angular momentum components are numbers. Can they be generators of some symmetry group?

Question

In Quantum Mechanics (QM), angular momentum turn out to be the generator of rotational symmetry. This is trivial to see because in QM, angular momenta are defined by the commutation relations $$[J_j,J_k]=i\hbar\epsilon_{jkl}J_l.$$ One immediately recognises these as the generators of the rotation group. But in classical mechanics, angular momenta $L_i$ are numbers, or at best, functions of $x_j$ and $p_k$ as $L_i=\epsilon_{ijk}x_j p_k$. Can they be called the generators of some symmetry group because generators are usually differential operators or matrices?

Answer 1

As you mention, they're not "just numbers" - they are functions of the coordinates and momenta. And, as such, they can indeed be used as generators of some symmetry group via the usual tool for the job in hamiltonian mechanics: the Poisson bracket.

Here, you won't be much surprised to learn that the mutual relationships between the angular momentum components are $$ \{L_i,L_j\}=\epsilon_{ijk}L_k, $$ where the Poisson bracket is defined as $$ \{f,g\} = \sum_i \left[ \frac{\partial f}{\partial x_i} \frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial g}{\partial x_i} \right], $$ and the group action generated by the angular momentum on a function $f$ works as $$ f \mapsto f+\{L_i,f\} $$ in its infinitesimal version, and as $$ f \mapsto\exp\left(\theta\{L_i,\cdot\}\right) f $$ for a finite angle $\theta$.

And, of course, the group that they generate is simply $\rm SO(3)$, acting on the space of real functions on phase space.