I am an undergrad and curious about the following question. Let $(Y,\xi)$ be a contact manifold, and $L\subset (Y,\xi)$ be a Legendrian knot which is the boundary of a convex surface $\Sigma$.

> Why is the dividing set on $\Sigma$ 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.

[Here][1] is a related question, showing that the answer to the question above is trivial if ${\rm tb}(L)\neq 0$.

Thanks for your help!

[1]: http://mathoverflow.net/questions/98640/why-is-tb-0-for-boundary-of-a-convex-surface?rq=1