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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!

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    $\begingroup$ 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. $\endgroup$ – Mike Miller Eismeier Jun 1 '20 at 14:11
  • $\begingroup$ @MikeMiller Thank you! $\endgroup$ – Pengfei Jun 1 '20 at 15:53
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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 $dA_{S^2} + d\vartheta \wedge dt$.

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  • $\begingroup$ This is in fact an equivalence, since every 2-dimensional homology class is represented by a closed surface. $\endgroup$ – Mike Miller Eismeier Jun 1 '20 at 15:39
  • $\begingroup$ Thank you for the clarification! $\endgroup$ – Pengfei Jun 1 '20 at 15:56

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