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!



Existence of Liouville vector fields on symplectic manifolds
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 $dA\_{S^2} + d\vartheta \wedge dt$. 


