Darboux theorem for symplectic manifold

In summary: It seems to me that definition of rank actually applies just to 2-forms on fields with characteristic 0 (e.g. the field of real numbers ##\mathbb R##).
  • #1
cianfa72
1,692
192
TL;DR Summary
Application of Darboux theorem to symplectic manifold
Hi,

I am missing the point about the application of Darboux theorem to symplectic manifold case as explained here Darboux Theorem.

We start from a symplectic manifold of even dimension ##n=2m## with a symplectic differential 2-form ##w## defined on it. Since by definition the symplectic 2-form is closed (##d\omega=0##) then from Poincaré lemma there exist locally a 1-form ##\theta## such that ##\omega=d\theta##.

The point I'm missing is why ##\theta \wedge (d\theta)^m## is identically the null form.

Thank you.

p.s. ##(d\theta)^m## should be ##d\theta \wedge d\theta \wedge d\theta \wedge ...## ##m## times
 
Last edited:
Physics news on Phys.org
  • #2
It will be a ##2m+1## form on a ##2m## dimensional manifold.
 
  • #3
martinbn said:
It will be a ##2m+1## form on a ##2m## dimensional manifold.
If I understand correctly, you mean ##(d\theta)^m## is actually a ##2m## form just because it is the wedge product of ##m## 2-forms. Then ##\theta \wedge (d\theta)^m## is a ##2m +1## form on an ##2m## dimensional manifold hence it vanishes.

By the way...when we say it vanishes we actually mean it is the null form, right ?
 
Last edited:
  • Like
Likes martinbn
  • #4
Another point related to the above.

Take the ##\Lambda^{m} V^*## vector space where the dimension of the underlying dual space ##V^*## is ##n##. That means the rank ##r## of a generic ##m##-form is ##r \leq dim \Lambda^{m} V^* = \begin{pmatrix} n \\ m \end {pmatrix}##.

In Darboux theorem's hypothesis the ##2##-form ##d\theta## is assumed to be with constant rank ##m=n/2##.
Does it actually mean ##(d\theta)^m \neq 0 ## and ##(d\theta)^{m+1} = 0## ?
 
Last edited:
  • #5
cianfa72 said:
Another point related to the above.

Take the ##\Lambda^{m} V^*## vector space where the dimension of the underlying dual space ##V^*## is ##n##. That means the rank ##r## of a generic ##m##-form is ##r \leq dim \Lambda^{m} V^* = \begin{pmatrix} n \\ m \end {pmatrix}##.

In Darboux theorem's hypothesis the ##2##-form ##d\theta## is assumed to be with constant rank ##m=n/2##.
Does it actually mean ##(d\theta)^m \neq 0 ## and of course ##(d\theta)^{m+1} = 0## ?
Yes, that's way I understood it, that is how the rank is defined. It is the largerst ##l## such that ##\omega^l\not = 0##.
 

1. What is the Darboux theorem for symplectic manifolds?

The Darboux theorem states that any symplectic manifold can be locally described by a set of canonical coordinates, such that the symplectic form is given by a standard expression. In other words, it provides a way to locally simplify the description of a symplectic manifold.

2. What is a symplectic manifold?

A symplectic manifold is a smooth manifold equipped with a symplectic form, which is a non-degenerate, closed, and skew-symmetric 2-form. It is a fundamental concept in symplectic geometry, which studies the geometric properties of systems with a conserved quantity called the symplectic form.

3. How does the Darboux theorem relate to Hamiltonian mechanics?

The Darboux theorem is closely related to Hamiltonian mechanics, which is a mathematical framework for describing the dynamics of physical systems. In Hamiltonian mechanics, the symplectic form plays a crucial role in determining the equations of motion, and the Darboux theorem allows us to locally simplify the description of the symplectic manifold on which the system evolves.

4. Is the Darboux theorem for symplectic manifolds unique?

No, the Darboux theorem is not unique. There are different versions of the theorem, depending on the specific properties of the symplectic manifold, such as its dimension and whether it is compact or non-compact. However, all versions of the theorem share the same fundamental idea of locally simplifying the description of a symplectic manifold.

5. How is the Darboux theorem used in practical applications?

The Darboux theorem has many practical applications in physics and engineering. It is used in the study of classical mechanics, quantum mechanics, and statistical mechanics, as well as in the analysis of dynamical systems and control theory. It also has applications in symplectic topology, a branch of mathematics that studies the topological properties of symplectic manifolds.

Similar threads

I Contact manifold and Darboux's theorem
    • Differential Geometry
Replies
14
Views
749
A Hamiltonian formulation of classical mechanics as symplectic manifold
    • Classical Physics
Replies
5
Views
281
I Frobenius theorem for differential one forms
    • Differential Geometry
Replies
7
Views
479
A Parallel transport on a symplectic space
    • Differential Geometry
Replies
1
Views
1K
I Differential operator vs one-form (covector field)
    • Differential Geometry
Replies
10
Views
642
I Differential k-form vs (0,k) tensor field
    • Differential Geometry
Replies
8
Views
2K
Symplectic form on a 2-sphere
    • Differential Geometry
Replies
5
Views
9K
I Integration of differential forms
    • Differential Geometry
Replies
5
Views
2K
I The Road to Reality - exercise on scalar product
    • Differential Geometry
Replies
2
Views
514
A Lorentz Invariant Phase Space: Symplectic Geometry
    • Special and General Relativity
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top