Darboux theorem for symplectic manifold

  • #1
873
92
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
 
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Answers and Replies

  • #2
martinbn
Science Advisor
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It will be a ##2m+1## form on a ##2m## dimensional manifold.
 
  • #3
873
92
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 ?
 
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  • #4
873
92
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## ?
 
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  • #5
martinbn
Science Advisor
2,573
954
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##.
 
  • #6
873
92
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##.
I found this wiki entry Rank of a k-vector. It refers to ##k##-vectors elements of ##\Lambda^{k} V## however it should be the same as for ##k##-forms elements of ##\Lambda^{k} V^*##.

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##).
 
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