I read recently that on a symplectic manifold $M$, the infinitesimal deformations of a Lagrangian manifold $L$ can be identified with closed 1 forms in $T^\*L$ (cotangent bundle of L).


How can this correspondance be made? I suppose that one somehow has to use Weinstein's tubular neighborhood theorem, but I can't write down the required map.


I am sure that this construction is standard in sympletic geometry so if someone knows a good reference please let me know.



Deformation of Lagrangian manifolds
You don't need to use Weinstein's tubular neighborhood theorem to assign closed one forms on L to deformations of L. Here is a construction which makes it clear the assignment is canonical.


A smooth family of Lagrangian submanifolds is given by a pair of smooth maps
$$\mathbb R \xleftarrow{t}X \xrightarrow{f} M$$
so that the map $t$ is a proper submersion and $f$ includes every fiber of $t$ as a Lagrangian submanifold of $M$. 


There is a vertical cotangent bundle of $X$ which is the quotient of $T^\*X$ by the pullback of one forms from $\mathbb R$. This vertical cotangent bundle should be regarded as putting together the cotangent bundles of the fibers of $t$ into a smooth vector bundle over $X$. Each differential form $\theta$ on $X$ has a well defined projection to a section $\pi\theta$ of the wedge of the vertical cotangent bundle, which is the definition of a smooth family of differential forms on the fibers of $t$. The fact that this is a family of Lagrangian submanifolds implies that $\pi(f^\*\omega)=0$.


Choose any smooth vector field $\frac \partial {\partial t} $ on $X$ so that $\frac\partial{\partial t} t=1$. Then $$\pi(\iota\_{\frac \partial{\partial t}} f^\*\omega)$$ is a family of one forms on the fibers of $t$ which does not depend on the choice of $\frac \partial {\partial t}$. It is a family of closed one forms because $\pi$ commutes with $d$ and 
$$\pi L\_{\frac\partial{\partial t}}f^\*\omega=0$$.


This construction reverses the assignment of a deformation of L to a closed one form on L which uses the Weinstein neighborhood theorem.


