# How can I think about Poisson commutation geometrically?

In the Hamiltonian formulation of classical mechanics, we have the result that, in a system with Hamiltonian $H(q_i,p_i),\ i=1,\dots,N$, a quantity $f(q_i,p_i)$ time-evolves according to $$\frac{\mathrm{d} f}{\mathrm{d} t} = \{f,H\} = \frac{\partial f}{\partial q_i}\frac{\partial H}{\partial p_i} - \frac{\partial f}{\partial p_i}\frac{\partial H}{\partial q_i}$$ and so $f$ is conserved if and only if it Poisson commutes with $H$, i.e. $\{f,H\} = 0$.

In the case $N=1$ so that the phase space is 2-dimensional, and contour surfaces of functions are curves. Contours of $H$ are trajectories, and $f$ is a conserved quantity if and only if the contours of $f$ align with those of $H$. But for $N\geq2$, the contour surfaces of $H$ and $f$ are $(2N-1)$-dimensional manifolds in the $2N$-dimensional phase space. What can you say about these surfaces when $\{f,H\} = 0$?

In quantum mechanics, when the functions $f$ and $H$ are replaced by operators $\hat{f}$ and $\hat{H}$ and the Poisson bracket is replaced by the commutator, then $[\hat{f},\hat{H}] = 0$ means that the two operators have coinciding eigenspaces. Is there such a nicely geometric way of visualising the classical equivalent?

So your velocity in phase-space $$(\dot q_1, \dot q_2, ..., \dot q_N,\dot p_1,...,\dot p_N )$$ is an 2N -dim tangent vector lying on the kinematically constant 2N -1 -dim hypersurfaces specified by the k independent conserved quantities, the ones Poisson-bracket commuting with the Hamiltonian (which is one of them) and thus invariant in time.