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Let $\omega$ be a closed $k$-form. Then:

If $\omega$ is exact, for every closed form $\beta$, the form $\omega\wedge\beta$ is exact.

Proof: Let $\omega=d\alpha$. Now $d(\alpha\wedge\beta) = d\alpha\wedge\beta +(-1)^k\alpha\wedge d\beta = \omega\wedge\beta + 0$, since $d\beta=0$ by hypothesis.

I want to prove the converse. Let $\omega$ be a closed $k$-form, again. Then:

If for every closed form $\beta$ the form $\omega\wedge\beta$ is exact, then $\omega$ is exact.

How can I prove the latter statement?

(If one uses de Rham currents, this question is linked to this question.)

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    $\begingroup$ If the underlying manifold is compact then this follows from (or rather is) the perfectness of the intersection pairing in Poincaré duality. In the general case I'm not sure what happens. $\endgroup$ – Gunnar Þór Magnússon Apr 26 '14 at 17:22
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    $\begingroup$ Does it hold for $\beta$ a constant $0$-form? $\endgroup$ – Ted Shifrin Apr 26 '14 at 17:45
  • $\begingroup$ @TedShifrin Ha, that is neat. But I would rather need $\beta$ to be of degree $n-k$ for the linked question... $\endgroup$ – geodude Apr 26 '14 at 17:50
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This is false without compactness. Every $2$-form on $\Bbb R^2-\{0\}$ is exact, but the infamous $1$-form $d\theta$ is not.

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  • $\begingroup$ Right. I need a slightly different hypothesis. When I've found it, I'll post a new question. $\endgroup$ – geodude Apr 26 '14 at 18:10

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