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.)
Best Answer
This is false without compactness. Every $2$-form on $\Bbb R^2-\{0\}$ is exact, but the infamous $1$-form $d\theta$ is not.