I am an undergrad and curious about the following question. Let (Y,ξ) be a contact manifold, and L⊂(Y,ξ) be a Legendrian knot which is the boundary of a convex surface Σ embedded properly in Y.
Why is the dividing set on Σ nonempty? I know you can use Stokes' theorem to prove the statement for the closed case, but I don't see how it helps for the Legendrian boundary case.
Thanks for your help!
The same argument works in the relative case. Consider the integral of $d\alpha$ over the convex surface. Since the boundary is Legendrian, $\alpha$ vanishes identically on it. Thus $d\alpha$ has to change sign.