4
$\begingroup$

Let $\omega$ be a smooth 1-form on a smooth manifold $M$. A smooth positive function $\mu$ on some open subset $U\subset M$ is called an integrating factor for $\omega$ if $\mu\omega$ is exact on U.

In Lee's ISM, there is a necessary and sufficient condition for a nonvanishing 1-form to admit integrating factor.

If $\omega$ is nowhere-vanishing, then $\omega$ admits an integrating factor in a neighborhood of each point if and only if $d\omega \wedge \omega \equiv 0$.

I have no idea how to prove the "if" part. Assume $d\omega \wedge \omega \equiv 0$ and that $\omega$ is nowhere-vanishing. What does these conditions tell us about $\omega$ and how can we construct a smooth positive function $\mu$ on some open neighborhood $U$ of any point $p$?

Partial solution:

Ted indicates that I need to use the Frobenius theorem for the proof. I assume the version of Frobenius theorem is the following one in Warner's book:

Let $\mathcal I \subset E^*(M)$ be a differential ideal locally generated by $d-p$ independent 1-forms. Let $m\in M$. Then there exists a unique maximal, connected, integral manifold of $\mathcal I$ through $m$, and this integral manifold has dimension $p$.

For any $m\in M$, we want to prove that there exists a smooth positive function $\mu$ on some open subset $m\in U\subset M$ such that $\mu\omega$ is exact on U. To use the Frobenius theorem, we need a involutive distribution $\mathcal D$ and let $\mathcal I:=\mathcal {I(D)}$(annihilator of $\mathcal D$). But what is our $\mathbb D$? What good can the existence of a unique maximal, connected, integral manifold of $\mathcal I$ through $m$ do to us in order to find $\mu$?

Thanks in advance.

$\endgroup$
3
  • 5
    $\begingroup$ This is the simplest case of the Frobenius Theorem. It's far from an immediate result. $\endgroup$ Commented May 2, 2016 at 21:27
  • $\begingroup$ @TedShifrin Thanks, I added your idea in my question. But I still don't know how to use the Frobenius Theorem to find the $\mu$ we want. $\endgroup$
    – No One
    Commented May 3, 2016 at 0:18
  • 1
    $\begingroup$ You need a stronger version of the Frobenius theorem than the one you quoted -- for example, Theorem 19.12 in my ISM. $\endgroup$
    – Jack Lee
    Commented May 4, 2016 at 2:31

1 Answer 1

4
$\begingroup$

Once you know the Frobenius Theorem, you know that the integral manifolds of $\omega = 0$ are (locally) given by level surfaces of some function $f$. This means that $df = \mu\omega$ for some nonzero function $\mu$.

$\endgroup$
7
  • $\begingroup$ The distribution is given by the vectors annihilated by $\omega$, i.e., the hyperplane $\ker\omega$. It is involutive because $d\omega$ is in the ideal generated by $\omega$ (Warner's differential ideal). $\endgroup$ Commented May 3, 2016 at 6:46
  • $\begingroup$ Thanks! Could you explain why "...by level surfaces of some function f. This means that df=uw" ? $\endgroup$
    – No One
    Commented May 3, 2016 at 7:01
  • $\begingroup$ Because $df$ and $\omega$ annihilate the same subspace at each point. (If you prefer, you can argue that $df\wedge\omega=0$ by evaluating on any pair of vectors.) $\endgroup$ Commented May 4, 2016 at 6:05
  • $\begingroup$ @Ted, if I understand correctly, you are claiming that the ideal $I := \langle\omega\rangle$ is a differential ideal, but that's not always tree. Taken from here, we can define the 1-form on $\mathbb{R}^2$ $\omega = x\ dy$, which is a counterxample. $\endgroup$ Commented Dec 2, 2018 at 0:12
  • 1
    $\begingroup$ @Guillermo: Cartan's Lemma is one of my favorites ;) $\endgroup$ Commented Dec 2, 2018 at 0:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .