# Polarisation in a nighbourhood of a Lagrangian submanifold

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

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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

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
No, there is no reason that the symplectomorphism send your $\alpha$ to the standard 1-form. –  Francois Ziegler Nov 14 '12 at 20:28