# Existence of Liouville vector fields on symplectic manifolds

Let $$(M, \omega)$$ be a symplectic manifold. A vector field $$V: M \to TM$$ is Liouville if $$L_{X} \omega=\omega$$. The existence of a Liouville vector field implies that $$(M, \omega)$$ is exact: the one-form $$\lambda = i_V \omega$$ satisfies $$d\lambda=d\circ i_V\omega = L_V\omega=\omega$$. In particular, there is no Liouville vector field on any closed (compact and boundaryless) symplectic manifold.

My question is about the existence of Liouville vector fields. Is it a sufficient condition that $$\partial M\neq \varnothing$$?

Thanks!

• Given any 1-form $\alpha$, there is a vector field $V$ so that $i_V \omega = \alpha$. This is non-degeneracy. So if $\omega = d\lambda$ then the unique $V$ so that $i_V \omega = \lambda$ is Liouville. The existence of a Liouville field is equivalent to exactness. For surfaces this is equivalent to nonempty boundary. – Mike Miller Eismeier Jun 1 '20 at 14:11
• @MikeMiller Thank you! – Pengfei Jun 1 '20 at 15:53

If the symplectic form integrates to a nonzero quantity on a compact surface in your manifold, it is not exact. For example, on $$M=S^2\times S^1\times [0,1]$$ with symplectic form $$dA_{S^2} + d\vartheta \wedge dt$$.