Hamiltonian formulation of classical mechanics as symplectic manifold

[cianfa72]

TL;DR

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.

 

[martinbn]

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]

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 $det(a_{ij}) \neq 0$.

 

[martinbn]

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]

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.

Sorry to resume this old thread, in the definition of the 2-form $\omega$, is the matrix $a_{ij}$ (with even dimension) assumed to be skew-symmetric (with even dimension)? Thanks.

 

[martinbn]

Yes.