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Jun 1, 2020 at 15:53 comment added Pengfei @MikeMiller Thank you!
Jun 1, 2020 at 14:20 history edited Pengfei CC BY-SA 4.0
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Jun 1, 2020 at 14:12 vote accept Pengfei
Jun 1, 2020 at 14:11 comment added mme 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.
Jun 1, 2020 at 14:06 history edited Pengfei CC BY-SA 4.0
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Jun 1, 2020 at 13:59 history edited Pengfei CC BY-SA 4.0
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Jun 1, 2020 at 13:32 answer added Ben McKay timeline score: 2
Jun 1, 2020 at 13:26 history asked Pengfei CC BY-SA 4.0