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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!

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!

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 $\Sigma$.

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!

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!

I am an undergrad and curious why dividing set of a convex surface $\Sigma$ is nonempty when it has a Legendrian boundary?

I know you can use Stokes theorem to contrapositive the statement for closed case but I don't see why for Legendrian boundary case.

Thanks for your help!

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 $\Sigma$.

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!