Weinstein's local classification of Lagrangian foliations

Question

In the paper "Symplectic manifolds and their Lagrangian submanifolds", Weinstein showed that locally all the Lagrangian foliations are symplectomorhic to the fiber foliation of a cotangent bundle.

I have a tiny problem in one step in his argument. Let $N$ be a smooth manifold and $U$ a small neighborhood of the zero section of $T^*N$ with the canonical symplectic structure on $T^*N$. Let $F$ be a Lagrangian foliation on $U$ such that it is transversal to $N$ at each point of $N$. Weinstein asserts that there exist a diffeomorphism $h$ from a neighborhood of $N$ to some other neighborhood of $N$ such that $h$ is the identity map on $N$ and $h$ maps leaves of $F$ onto cotangent fibers. I am wondering how this $h$ could be constructed. Probably it has something to do with the Whitney extension theorem. Maybe this is something really standard, I would like to know the details anyway.

Thank you for your attention.

Answer 1

From the assumption that the intersections are transverse, together with the fact that the foliation is smooth, it is possible to find a smooth map $\phi \colon T^*N \to TU$ which maps each fibre $T^*_pN$ by a linear isomorphism onto the plane inside $T_pU$ that is tangent to the foliation.

Fix a Riemannian metric $g$ on $U$. In a neighbourhood of the zero-section of $T^*N$ we can compose $\phi$ with the exponential map defined on each leaf that is induced by the Riemannian metric given by the restriction of $g$ to the leaf. This is the required diffeomorphism.

The crucial thing is of course to check that the above constructions indeed are smooth. To see this, it is advisable to use smooth local coordinates $(x_1,...,x_{2n})$ in which the foliation is of the form $\{x_1=...=x_n=0\}$ (such coordinates exist by the definition of a smooth foliation).