# 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?

• I've re-tagged the question. – Marco Golla Jun 2 '12 at 9:58

## 1 Answer

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: ${\rm tb}(L) = -|\Gamma \cap L|/2$, where $|\cdot|$ is the number of points (counted without sign).

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

(Added Dec 07 '13): I've corrected the sign in the formula above. I also wanted to expand a bit on the basic idea of the proof.

A surface is convex if there is a contact vector field $v$ transverse to it: in particular, the twisting number of the contact structure $\xi$ along $L$ with respect to $v$ is the same as the twisting number with respect to $\Sigma$. The latter is ${\rm tb}(L)$, by definition; the former, on the other hand, counts how many times $v$ belongs to $\xi$ with sign: since $\Gamma$ is the set $\{x\mid v_x\in \xi\}$, this twisting is counting $\Gamma \cap L$ with some sign. The only thing we need to argue for is that all intersections count as negative: they all come with the same sign because the contact structure "always twists in the same direction", and now pinning down which one is a local computation (see the notes for details).