Suppose $(M,\omega)$ is a compact symplectic manifold and $C$ a closed curve in it. Is there a Lagrangian submanifold containing $C$? I have a sequence of $J_i$holomorphic maps from a disk to $M$, and all the maps are identical on the boundary of the disk. So, if I know that the image of the boundary is contained in a Lagrangian submanifold, I can apply Gromov convergence.

1$\begingroup$ In your case, you don't even need to worry about the existence of a totally real submanifold containing $C$  the compactness follows immediately "by hand". The main work in proving Gromov compactness when you have boundary on a Lagrangian/totally real submanifold is to show that boundary singularities can be removed and that boundary bubbles connect. In your case, nothing interesting happens at the boundary, so you can sidestep this issue entirely. $\endgroup$– Sam LisiDec 8, 2011 at 13:06

$\begingroup$ Sam: the derivative of the maps on the boundary are bounded, but I am not able to show that there is no energy escaping to the boundary. i.e I can't show that the derivatives are bounded in neighbourhoods of boundary points. $\endgroup$– AnonDec 8, 2011 at 15:40

$\begingroup$ @Sita: I don't believe this should be a problem, but maybe I don't understand what you mean. Since this is not pertinent to your original question about existence of Lagrangian submanifolds, feel free to contact me (contact info on my user page) if you want to discuss further. $\endgroup$– Sam LisiDec 10, 2011 at 17:04
1 Answer
Yes, such a Lagrangian submanifold does exist. The isotropic neighborhood theorem (see for instance p. 24 of Weinstein's Lectures on Symplectic Manifolds), together with the fact that all symplectic vector bundles over the circle are trivial, implies that any closed curve in a symplectic $2n$manifold has a neighborhood symplectomorphic to the standard product $(S^1\times(\epsilon,\epsilon))\times B^{2n2}(r)$, via a symplectomorphism mapping the curve to $S^1\times \{0\}\times\{0\}$. Here $B^{2n2}(r)$ denotes the standard symplectic ball of dimension $2n2$ and some small radius $r$. Now let $T$ be a Lagrangian torus in $B^{2n2}(r)$ which contains the origin (for instance you could use a product of smallradiuscirles containing the origin in each of the $n1$ factors of $\mathbb{R}^2$ in $\mathbb{R}^{2n2}$). Then the image of $S^1\times 0\times T$ under the Weinstein symplectomorphism will be a Lagrangian submanifold containing your curve.