# If $\omega \wedge d\omega = 0$ then $\omega = \lambda\,df$ for some real functions $\lambda, f$

differential-formsdifferential-geometrysmooth-manifolds

I found the following problem on an old qualifying exam and wasn't able to solve it, and I was wondering if anyone could help:

Let $$\omega$$ be a smooth 1-form on a smooth manifold $$M$$.

Suppose $$\omega \wedge d\omega = 0$$.

The problem asks to show that for every $$p \in M$$, there are an open neighborhood $$U \ni p$$
and smooth real functions $$\lambda, f$$ on $$U$$
such that $$\omega = \lambda df$$ on $$U$$.

Here I am stuck. I know that (if $$\omega$$ is non-vanishing then) $$\omega \wedge d\omega = 0$$ implies that $$\mathrm{ker}\omega$$ is involutive (closed under Lie brackets) and the corresponding distribution is integrable, but other than that I'm not sure how to proceed.

You will find this question answered in various forms all over this site. If $$\omega=0$$ is integrable, then locally (on some neighborhood $$U$$ of $$p$$) there are coordinates $$(x^1,\dots,x^n)$$ so that the leaves of $$\omega=0$$ are the level sets $$x^n=c$$. This means that $$\omega = \lambda\,dx^n$$ for some smooth function $$\lambda$$.