## Why is tb < 0 for boundary of a convex surface?

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?

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 I've re-tagged the question. – Marco Golla Jun 2 at 9:58

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: $tb(L) = \#(\Gamma \cap L)/2$.