# Liouville measure conservation and smoothness of the symplectic structure

Suppose we have hamiltonian $H$ of some system. Suppose we fix the coordinate $q$ and momentum $p$, which are coordinates in the phase volume of the system. In general, they are not canonical. This statement is convenient to represent on the language of Poisson brackets. Introducing the coodinate $\epsilon = (q,p)$, we can write the Poisson brackets as $$\{\epsilon_a,\epsilon_b\} = \omega^{ij}\partial_{i}\epsilon_{a}\partial_{j}\epsilon b,$$ where $\omega^{ij}$ is the inversed symplectic form. If $\epsilon$ are canonical, then $$\omega_{ij} = \text{antidiag}(-1,1)$$ In other cases, there are deviations.

The phase volume element is $$d\Gamma = \sqrt{\text{det}\omega}\prod \frac{dqdp}{(2\pi)^{d}}$$ Recently I've read that the Liouville measure $\sqrt{\text{det}\omega}$ is conserved, if the symplectic form is smooth: $$\frac{d\sqrt{\text{det}\omega}}{dt}= 0$$ However, if it is singular, than the conservation law fails. Possible illustration of this statement is the semiclassical fermion in presence of adiabatically changing EM field.

How to prove this?

If the closed 2-form $\omega$ is degenerate, it is a presymplectic (rather than a symplectic) structure, and there is no longer a well-defined corresponding Poisson bracket. Hence there may be no notion of Hamiltonian vector fields and no Liouville's theorem.