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I would like to prove the following (without the Frobenius Theorem):

On $\mathbb{R}^n$, if $\omega$ is nowhere vanishing $1$-form such that $d\omega\wedge\omega=0$ then there exists (at least locally) $f$ a positive function such that $d(f\omega)=0$.

In fact, I am looking for an easier proof of the Frobenius theorem for distribution of hyperplanes. Indeed the proof I know are inductive, we start with distribution of lines, then inductively we arrive to distribution of hyperplanes.

My statement implies the Frobenius theorem for hyperplanes, so I wonder if we can prove it with as less as technology as possible?

Thanks for your help

Added: It is easy to prove that there exists a 1-form $\theta$ such that $d\omega =\theta \wedge \omega$, then $d(f\omega)=(df+f\theta)\wedge \omega$, so the problem can be rephrased as, does there exisst $f$ and $g$ such that $df+f\theta =g \omega$? I stuck here.

Paul
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  • The problem here is that what you're trying to prove is not just implied by the Frobenius Theorem. It is equivalent to the Frobenius Theorem. So, if there were a simpler non-inductive proof known, we would all be teaching it, instead of the usual one.

    More specifically, what you're trying to prove is the last step in the proof of the Frobenius Theorem. The condition $d\omega\wedge\omega = 0$ is equivalent to the condition that you have $n-1$ linearly independent vector fields whose Lie brackets lie in the span of the vector fields. This isn't easy to deal with directly.

     – Deane May 30 '21 at 16:22
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    This question, may be of interest, as it asks for a proof in a similar vein. – Kajelad May 30 '21 at 18:57
  • This is a deep theorem. You're not going to bypass it with some algebraic manipulation. Much as I am a devotee of differential forms, Frobenius needs a substantial proof. – Ted Shifrin Jun 01 '21 at 01:05

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