# Convex boundary of a symplectic manifold

Given a symplectic manifold $(M,\omega)$, suppose that $\partial M$ is of contact type. A Liouville field on a symplectic manifold is a vector field $X$ such that $\mathcal L_X \omega = \omega$. We say that $\partial M$ is convex if there is an outward-pointing Liouville field near $\partial M$ (it will usually not be everywhere defined). We say that $\partial M$ is concave if there is an inward-poiting Liouville field. (We say that a hypersurface is of contact type if there is a Liouville vector field defined on a neighborhood of the hypersurface that is transverse to the surface.)

An exercise (3.60) in McDuff-Salamon says that any two Liouville vector fields near a hypersurface of contact type must point in the same direction. I can't see why this is true, and what's more, I seem to have found a counterexample. So it's not clear to me that the boundary of a symplectic manifold cannot be both convex and concave.

Consider $S^1$ embedded in any symplectic surface. Because this circle is automatically Lagrangian, we know that a neighborhood of it is symplectomorphic to a neighborhood of the zero section in $T^* S^1 \cong S^1 \times (-\varepsilon,\varepsilon)$, which has symplectic form $-dt\wedge d\theta$. Given a vector field $X = a\frac{\partial}{\partial t} + b\frac{\partial}{\partial \theta}$, the Liouville condition says that $a$ is never zero on the zero section, and that $\left(\frac{\partial a}{\partial t} + \frac{\partial b}{\partial \theta}\right) = 1$.

Now pick $X = (\pm 1 + t)\frac{\partial}{\partial t}$. These are two different Liouville vector fields that point in opposite directions.

Have I misunderstood the definition of convex boundary? Can the boundary of a higher-dimensional symplectic manifold, whose boundary has contact type, be both concave and convex, or is this a fluke of 2-dimensional symplectic manifolds? If not, what's up with the McDuff-Salamon exercise that says Liouville fields must all point in the same direction?

• You mean $X=(\cdots)\frac{\partial}{\partial t}$, right?
– Kyle
Commented May 21, 2015 at 4:05
• @squirrel Yes. Thank you for the correction.
– user98602
Commented May 21, 2015 at 4:06

Let $S$ be a hypersurface of contact type bounding a symplectic manifold $M$. We say $S$ is convex if there is a Liouville vector field near $S$ that points outwards, and concave if there is one that points inwards.
The case of a surface with boundary $S^1$ is given as - and, as above, is - an example of a boundary that is both convex and concave.
Now let $\alpha$ be a compatible contact structure on $S$ - that is, that $d\alpha = \omega$ restricted to $S$. Call the Reeb field for this contact form $\delta$. Gromov says that if $G$ is a null-homologous closed Reeb orbit for $\alpha$, then $\int_G \alpha' > 0$ for all compatible contact forms $\alpha'$. This implies that $S$ cannot be simultaneously convex and concave. (From Liouville fields pointing in opposite directions, we can produce contact forms, compatible with $\omega$, that are homologically negative of each other.)
In particular, consider $S^3$ as the boundary of your favorite symplectic manifold with boundary $S^3$. Then it cannot be both convex and concave. In addition, there is a symplectically fillable contact structure on $T^3$ with no null-homologous Reeb orbits; so this does not provide an obstruction to being both convex and concave. In fact, it's both with respect to a certain symplectic filling. (Note that every closed 3-manifold with a given contact form has a closed Reeb orbit; this was proved by Taubes in 2007, and is a special case of the Weinstein conjecture.)