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

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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.

Answer 1

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 $\lambda$.