# Decomposition of volume form

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.

• Are you familiar with the Hodge star operator? – Or Eisenberg May 28 '19 at 19:32
• @OrEisenberg No, not really. But probably I should get familiar with it. How would it help? – TheGeekGreek May 28 '19 at 19:37
• The Hodge star is essentially defined to provide solutions to the sort of problem you've posed here, so yes. =) – Or Eisenberg May 28 '19 at 19:50
• Why do you think $\alpha$ could be unique? You can obviously add a product of $dH$ with any $(2n-2)$-form. – Ted Shifrin May 28 '19 at 21:51
• @TedShifrin I am sorry. The uniqueness was a typo. – TheGeekGreek May 29 '19 at 7:19

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