Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
I've re-tagged the question. –  Marco Golla Jun 2 '12 at 9:58
add comment

1 Answer

up vote 3 down vote accepted

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).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.