Why is the dividing set nonempty when a convex surface has Legendrian boundary? 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.

Why is the dividing set on $\Sigma$ nonempty?

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 is a related question, showing that the answer to the question above is trivial if ${\rm tb}(L)\neq 0$.
Thanks for your help!
 A: 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 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.
