Question

Why is $tb(K)$ (Thurston-Bennequin invariant) of a Legendrian knot $K$ which is the boundary of a convex surface $\Sigma$ is negative in a contact 3 manifold?

Answer 1

If the boundary of a convex surface is Legendrian, then we can see the Thurston-Bennequin number directly from the dividing curves $\Gamma$, in the sense that: $tb(L) = \#(\Gamma \cap L)/2$.

This is proved, for example, in Etnyre's notes, Theorem 2.30.