A Hamiltonian formulation of classical mechanics as symplectic manifold

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In the Hamiltonian formulation of classical mechanics, the phase space is identified as a symplectic manifold characterized by a closed non-degenerate 2-form, denoted as ##\omega##. The discussion clarifies that the non-degeneracy of ##\omega## implies it can be expressed locally as a linear combination of wedge products of covector bases, leading to local canonical coordinates. The non-degenerate condition is equivalent to the determinant of the associated matrix ##a_{ij}## being non-zero. Additionally, it is confirmed that this matrix is skew-symmetric due to the properties of the symplectic structure. Understanding these properties is crucial for grasping the implications of Darboux's theorem in the context of symplectic geometry.
cianfa72
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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.
 
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.
 
martinbn said:
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.
 
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Yes.
 
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