Deformation of Lagrangian manifolds - MathOverflow

Deformation of Lagrangian manifolds

2012-06-01T15:38:03Z

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.

Answer by Francois Ziegler for Deformation of Lagrangian manifolds

2012-06-01T16:18:20Z

You already know that the pair $(M,L)$ of a symplectic manifold and a Lagrangian submanifold is locally isomorphic to $(T^*L, L)$. This is the beginning of Corollary 6.2 of Weinstein, who continues: "and the lagrangian submanifolds of $M$ "near" $L$ are in 1-1 correspondence with "small" closed forms on $L$."

The correspondence in question (explained on the previous page of Weintein's paper) is that "a submanifold of $T^*L$ transversal to the fibres is locally the graph of a 1-form $\sigma:L\to T^*L$. The graph of $\sigma$ is isotropic if and only if... $\sigma$ is a closed 1-form."

In short, the map you want attaches to a closed 1-form (on $L$!) its graph in $M\simeq T^*L$.

Update: This construction identifies a neighborhood of $f_0:L\hookrightarrow M$ in the space of embeddings (Whitney C$^1$ topologized), with a neighborhood of zero in the space of closed 1-forms on $L$. See Thm II.3.8 in Michèle Audin's notes (available here). She concludes that $Z^1(L)$ "can be considered as a neighbourhood of $f_0$ in the “manifold” of deformations of $f_0$, or as its tangent space at $f_0$."

Answer by Richard Montgomery for Deformation of Lagrangian manifolds

2012-06-01T19:24:08Z

In general, deformations of a submanifold L of an ambient space M are identified with sections of L's normal bundle: $TM|_{L}/TL$. For your case, the normal bundle is canonically isomorphic to $T^*L$ by way of the symplectic form. To be more concrete: look at just the exact' deformations, deformations whose one-form is exact and so given by function on $L$. Take such a function $f$. Extend it arbitrarily to a function $F$ on M. Take the Hamiltonian vector field $X_F$ of $F$, restricted to $L$. That $X_F$ defines a vector field which tells you which way to push $L$ into $M$. Note that if $F, G$ are two different extensions of $f$ then they differ by a function which vanishes on $L$, so that their Hamiltonian vector fields $X_F, X_G$ differ by a vector field tangent to $L$: the vector field is well defined as a section of the normal bundle. In other words, we can think of$X_f$' as a section of $L$'s normal bundle.

You seem to want to go `the other way' and directly concoct a vector field out of ``$dL_t/ dt$''. How are you going to do that in the general case?

Answer by Brett Parker for Deformation of Lagrangian manifolds

2012-06-06T04:03:56Z

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.

Answer by Jay for Deformation of Lagrangian manifolds

2012-11-15T17:17:50Z

generally calculation is like this:

you write down the tubular neighborhood and the exp map there;
you do re-parametrization, such that your symplectic form comes in the "darboux type"

then the section of the normal bundle will be a nearby lagrangian.

there are some simple examples you can do the calculation explicitly, for example: you consider the unit circle in R^2 with the standard symplectic form, then you choose the polar coordinate to write down the exp map in the tubular neighborhood, you will find you need a simple substitution to make the symplectic form in the "darboux type"