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Let $(M^{2n},\omega)$ be a symplectic manifold. Then $\omega^n$ is a volume form. Suppose that $H \in C^\infty(U)$, where $U \subseteq M$ is open and $dH \neq 0$ on $U$. Why exactly can I decompose $$\omega^n = dH \wedge \alpha$$ for a $\alpha \in \Omega^{2n - 1}(U)$? I mean locally I can express both sides in a coordinate induced basis and choose for example a nonzero coefficient of $dH$ in a small neighbourhood.

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    $\begingroup$ Are you familiar with the Hodge star operator? $\endgroup$ – Or Eisenberg May 28 '19 at 19:32
  • $\begingroup$ @OrEisenberg No, not really. But probably I should get familiar with it. How would it help? $\endgroup$ – TheGeekGreek May 28 '19 at 19:37
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    $\begingroup$ The Hodge star is essentially defined to provide solutions to the sort of problem you've posed here, so yes. =) $\endgroup$ – Or Eisenberg May 28 '19 at 19:50
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    $\begingroup$ Why do you think $\alpha$ could be unique? You can obviously add a product of $dH$ with any $(2n-2)$-form. $\endgroup$ – Ted Shifrin May 28 '19 at 21:51
  • $\begingroup$ @TedShifrin I am sorry. The uniqueness was a typo. $\endgroup$ – TheGeekGreek May 29 '19 at 7:19
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You can indeed decompose the volume form but the decomposition is definitely not unique. Consider pretty much the simplest example with $n = 1, M = \mathbb{R}^2,\omega = dx \wedge dy$, $U = M$ and $H(x,y) = x$. Then $dH = dx$ and if we take $\alpha = f dx + dy$ where $f$ is any smooth function, we'll get

$$ dH \wedge \alpha = dx \wedge (f dx + dy) = dx \wedge dy = \omega. $$

In general, to show that such a decomposition is possible, write everything down in local coordinates, show that it exists locally and use a partition of unity argument to patch the local solutions to a global one.

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