Darboux theorem for symplectic manifold

[cianfa72]

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

 

[martinbn]

It will be a $2m+1$ form on a $2m$ dimensional manifold.

 

[cianfa72]

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 ?

 

[cianfa72]

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 $(d\theta)^{m+1} = 0$ ?

 

[martinbn]

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 $\omega^l\not = 0$.

 

[cianfa72]

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 $\mathbb R$).