An extra condition on Frobenius theorem for $1$-forms

Question

Consinder a smooth manifold $M$ and $\omega$ a smooth $1$-form on $M$. Assume that there is an open set $U\subset M$ such that $\omega$ never vanishes on $U$. One can define a smooth distribution $\cal F$ on $TU$ via the kernel of $\omega$. Frobenius theorem tells us that such distribution promotes a foliation on $M$ provided if $d\omega\wedge\omega =0$, that happens to be equivalent to the existence of a smooth $1$-form $\alpha$ such that $d\omega = \alpha\wedge \omega.$

My questions are, can one assume that $\alpha$ is closed in some case? If yes, what can one conclude something else about the smooth distribution? How restrictive is this condition?

I am particularly interested on this question because this can provide simple proofs for Calabi-Honda theorems of intrinsically harmonic 1-forms.

Answer 1

Locally, yes. By the Frobenius theorem, if $\omega\wedge d\omega=0$ then there is a local coordinate system, near each point, in which $\omega=f \, dx$ for some function $f$, and then we can take $\alpha=df/f$.

Answer 2

The existence of a global closed $\alpha$ imposes strong restrictions on the foliation. The foliation would transversely affine, in the sense that there would exist an open covering of the ambient manifold and a first integral over each open set of the covering such that, over non-empty intersections, the first integrals differ by an element of the affine group.

The first integrals are of the form $$ \int \exp \left( \pm \int \alpha \right) \cdot \omega . $$