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 $\Sigma$ embedded properly in Y.



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> Why is the dividing set on $\Sigma$ nonempty?
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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](https://mathoverflow.net/questions/98640/why-is-tb-0-for-boundary-of-a-convex-surface?rq=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!



Why is the dividing set nonempty when a convex surface has Legendrian boundary?
When I first read the question, I found it really odd, but now I'm convinced that this is really true.


I'll use the setup of Etnyre's [lecture notes](http://people.math.gatech.edu/~etnyre/preprints/papers/surfaces.pdf) on convex surfaces, around pages 5-6. Just to recap: take a 1-form $\alpha\_1$ defining $\xi$. If $\Sigma$ is convex, we can find coordinates near $\Sigma$ such that $\alpha\_1 = \beta\_1 + u\_1dt$, where $\beta\_1 = \iota^\*\alpha\_1$ is the pull-back of $\alpha\_1$ to $\Sigma$ and $u\_1$ is a function on $\Sigma$ (and $t$ parametrises the direction transverse to $\Sigma$). The function $u\_1$ is such that $u\_1d\beta\_1 + \beta\_1\wedge du\_1 > 0$ (this is half of Lemma 2.10 in the notes).


The dividing set $\Gamma$ is cut by $u\_1$, in the sense that $\Gamma = \{u\_1 = 0\}$. Suppose this is empty: then $u\_1$ is everywhere positive or everywhere negative. I will suppose that $u\_1>0$ everywhere (the other option being symmetric, as seen below).


Then $\alpha = \alpha\_1/u\_1$ is still a contact form for $\xi$, and $\iota^\*\alpha = \beta\_1/u\_1 =: \beta$. It follows that $\alpha = \beta + dt$, and in particular the corresponding function $u \equiv 1$. Since convexity doesn't depend on the contact form, but only on the contact planes, it follows from Lemma 2.10 that $d\beta$ is a volume form on $\Sigma$.


Now you can apply Stokes' theorem to $\beta$ on $\Sigma$: since the boundary of $\Sigma$ is Legendrian, $\beta$ vanishes identically on the boundary, and you'd have
$$ 0 = \int\_{\partial\Sigma} \beta = \int\_\Sigma d\beta > 0.$$
Had I found a negative $u\_1$, I'd have probably found a < instead of a > in the last step, but the arguments runs in this case, too.


