Hamiltonian formulation of classical mechanics as symplectic manifold

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• cianfa72
cianfa72
TL;DR Summary
About the definition of symplectic manifold structure employed in the hamiltonian formulation of classical mechanics
Hi, in the Hamiltonian formulation of classical mechanics, the phase space is a symplectic manifold. Namely there is a closed non-degenerate 2-form ##\omega## that assign a symplectic structure to the ##2m## even dimensional manifold (the phase space).

As explained here Darboux's theorem since ##\omega## is by definition closed from Poincare lemma there exist locally a 1-form ##\theta## such that locally ##\omega = d\theta##.

However I've not a clear understanding why such ##d\theta## fulfills the Darboux's theorem hypothesis hence there are local canonical coordinates such that ##\omega## can be written as
$$\omega = dq_i \wedge dp_i$$
If ##\omega## was a rank ##m## form then by definition ##(d\theta)^m \neq 0## and of course ##\theta \wedge (d\theta)^m = 0## since it would be a ##2m+1## form defined on a 2m-dimensional manifold.

So the question is: why ##\omega## is assumed to be a 2-form with constant rank ##m## ? Thanks.

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cianfa72 said:
So the question is: why ##\omega## is assumed to be a 2-form with constant rank ##m## ? Thanks.
It is not assumed, it follows from the fact that it is non-degenerate. Write it locally as ##\omega=\sum_{i,j} a_{ij}\theta^i\wedge\theta^j##, then ##\wedge^m \omega = det(a_{ij})\theta^1\wedge\cdots\wedge\theta^{2m}## is non-zero exactly when ##det(a_{ij})## is non-zero exactly when ##\omega## is non-degenerate.

cianfa72
martinbn said:
Write it locally as ##\omega=\sum_{i,j} a_{ij}\theta^i\wedge\theta^j##
Ah ok, so every 2-form can be always written as linear combination of wedge products of covector (1-form) basis of the dual space at each point on the manifold.

Then the definition of non-degenerate is equivalent (iff condition) to ##det(a_{ij}) \neq 0##.

cianfa72 said:
Ah ok, so every 2-form can be always written as linear combination of wedge products of covector (1-form) basis of the dual space at each point on the manifold.

Then the definition of non-degenerate is equivalent (iff condition) to ##det(a_{ij}) \neq 0##.
Yes, if the ##\theta^i## form a basis of 1-forms, then the ##\theta^i\wedge\theta^j## form a basis of 2-forms.

cianfa72
If the system's Hamiltonian does not explicitly depend on time, then Hamilton's equations actually claim that the system evolves in phase space on hypersurfaces of constant total energy (i.e. the Hamiltonian function evaluated along the solution's integral path does not change in time).

What if the system is not energy conservative ? I believe it always evolves along an integral curve of the vector field defined by the Hamiltonian on the phase space, however this time it is not just on a hypersurface of constant Hamiltonian value. Also there could be other "constants of motion" though, right ?

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Btw, I believe that in the wiki section Frobenius theorem there is a typo: the rank ##p## of ##d\theta## should be ##p=1## and then we get:
$$\theta = x_1dy_1$$ in some local coordinate system ##(x_1,...,x_{n-1},y_1)##.

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