Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hallo,

Let $(X, \omega)$ be a symplectic manifold of dimension $2n$ and $\omega$ is an exact symplectic form i.e. $\omega = -d\alpha$. Let furthermore $M \subset X$ be a Lagrangian submanifold such that $\alpha = 0$ of $TX|_{M}$. I am interested in the following questions:

  1. Is there a unique polarisation defined on $X$ near $M$ which is transversal to $M$ and whose one form is $\alpha$ ? By polarisation I mean the following: A polarisation of a symplectic manifold $X$, with symplectic form $\omega$, is a smooth assignment of a Lagrangian subspace of $T_{x}X$ to each $x \in X$ in such a way that this assignment is integrable.

  2. If 1. is true, is there a symplectic diffeomorpism $\Phi$ of a neigbourhood of $M$ is $X$ with a neigbourhood of $M$ in its cotangent bundle which carries the leaves of the polarisation into the standard cotangent fibration of $T^{*}M$ ?

Actually I know that these results are true. I would like to see the proof of them. Are there any references where I can look them up? If so, can you please tell me where these references can be found? Thanks a lot!

hapchiu

share|improve this question
    
Assume also that $M$ is compact. –  hapchiu Nov 13 '12 at 13:38
    
Are you assuming $X$ has a boundary? If not, then $X$ can't be compact (since $\omega^n = -d(\alpha\wedge\omega^{n-1})$ is an exact volume form on $X$). Like Francois, I'm also puzzled by the statement "whose one form is $\alpha$". Do you mean the polarization is contained in the null space of $\alpha$ (which is $2n-1$-dimensional at points where $\alpha$ is non-degenerate, and $2n$-dimensional at points where $\alpha$ is degenerate, such as along $M$)? –  user17945 Nov 14 '12 at 4:26
    
My bad on the first part - I read your comment as "$X$ is compact". –  user17945 Nov 14 '12 at 4:27

1 Answer 1

up vote 1 down vote accepted

Arnol'd (p. 3314) puts it that way:

Weinstein's Theorem. Some neighborhood of any Lagrangian submanifold in any symplectic manifold is symplectomorphic to some neighborhood of this Lagrangian submanifold in any other symplectic manifold, for instance in its own cotangent bundle space.

(The resulting neighborhood then has the obvious transverse polarization by fibers of the cotangent bundle.) Unless I am mistaken, Weinstein proves this in Symplectic manifolds and their lagrangian submanifolds, Theorem 6.1 and Corollary 6.2 (which he points out goes back to Souriau).

share|improve this answer
    
does this diffeomorphism preserve $\alpha$ ? –  hapchiu Nov 13 '12 at 16:56
    
I don't know. By "preserve" do you mean "map $\alpha$ to the standard cotangent bundle 1-form"? Why should it? I will admit that I couldn't tell what you mean by a polarization "whose one form is $\alpha$". That may also be why I don't see why you'd expect uniqueness of the transverse polarization. –  Francois Ziegler Nov 13 '12 at 19:08
    
yes I mean the standard cotangebt bundle 1-form. Is it preserved? –  hapchiu Nov 14 '12 at 5:55
    
No, there is no reason that the symplectomorphism send your $\alpha$ to the standard 1-form. –  Francois Ziegler Nov 14 '12 at 20:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.