Given a Legendrian $\Lambda$ in a contact manifold $(Y,\alpha)$, suppose one has a rigid J-holomorphic curve into the symplectization $(M,\omega)=(Y\times\mathbb{R},\mathrm{d}(e^t\alpha))$. As considered in Legendrian contact homology, the domain of the curve should be the boundary punctured unit disc, at boundary punctures it should converge to connecting Reeb chords and the boundary should be mapped to the Lagrangian $\Lambda\times\mathbb{R}$. Does this curve persist to exist under small deformations of the Legendrian or the Reeb flow?

## 1 Answer

Yes, if the linearized operator is also surjective. If the linearized operator is not surjective (i.e. you don't have transversality), what I say is not applicable.

To prove this, you want to use an implicit function theorem argument. You can, for instance, set up the problem so that you are looking for sections of the normal bundle to your curve that solve a certain equation. In the case without boundary, this is done in Hofer-Wysocki-Zehnder *Properties of pseudoholomorphic curves in symplectizations 3*. I believe the case with boundary is done in Ekholm, Etnyre & Sullivan's two papers on contact homology -- they avoid some of the difficulties by considering a special class of contact manifolds, but the Fredholm theory is completely general. The deformations of Legendrian and of Reeb flow (i.e. deformations of the contact form) can be realized as small deformations of the equation. The surjectivity of the linearized operator is now precisely what you need to apply the implicit function theorem.

Note that this proof says that for sufficiently small deformations, there exists a solution. It's much more delicate to estimate how large a small deformation you are allowed.