7
$\begingroup$

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.

$\endgroup$
0

4 Answers 4

3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Is this how the deformation of Lagrangian manifolds are defined? It would sure make life easier if you just wrote "X = \Bbb R \times L" or something like that. Also, this is sort of what I asked. Where can I read more about this stuff? $\endgroup$
    – Hammerhead
    Jun 6, 2012 at 12:35
  • $\begingroup$ If I wrote $X=\Bbb R\times L$, that would mean deformations of Lagrangian submanifolds parametrized by $L$. This is a different, and much larger space, because it has more structure. The tangent space would then also have to include vectorfields on $L$. $\endgroup$ Jun 7, 2012 at 1:14
  • $\begingroup$ I wan't to learn more about this stuff. You seem to knowledgeable. where can I read about this? $\endgroup$
    – Hammerhead
    Jun 7, 2012 at 9:22
  • $\begingroup$ I'm afraid I don't know any reference where this kind of stuff is explained explicitly. $\endgroup$ Jun 7, 2012 at 12:02
5
$\begingroup$

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 $f_0$."

$\endgroup$
2
  • $\begingroup$ I am aware of the things you wrote. I was asking for the correspondence between infinitesimal deformations of a Lagrangian manifold L and closed 1 forms in $T^∗L$. So if you have a family $L_t$ of symplectic manifolds in $T*L$ close to $L$(closed 1-forms basically), then how does one associate "canonically" to "$\frac{d}{dt}L_t$" a closed 1-from? One can just differentiate with respect to $t$ for each $x \in L$ in $L_t_x$ but I am thinking that this might not be the canonical way, since along the smooth deformation $L_t$ one might afford variations that do not respect fibers! $\endgroup$
    – Hammerhead
    Jun 1, 2012 at 17:42
  • $\begingroup$ Being transverse to the fibers is an open condition. I have added some details and a reference in my answer above; if this still doesn't answer your question, then please edit to make it more precise. $\endgroup$ Jun 6, 2012 at 1:05
2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ "You seem to want to go `the other way' and directly concoct a vector field out of ``dLt/dt''. How are you going to do that in the general case?" That still seems to be the question ... do you know a good reference for deformation of Lagrangian submanifolds? $\endgroup$
    – Hammerhead
    Jun 1, 2012 at 21:12
0
$\begingroup$

generally calculation is like this:

  1. you write down the tubular neighborhood and the exp map there;

  2. 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"

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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