Frobenius theorem for differential one forms

[cianfa72]

TL;DR

Frobenius theorem for differential one forms - equivalence of conditions

Hi, starting from this old PF thread I've some doubts about the Frobenius condition for a differential 1-form $\omega$, namely that $d\omega = \omega \wedge \alpha$ is actually equivalent to the existence of smooth maps $f$ and $g$ such that $\omega = fdg$.

I found this About Frobenius's theorem for differential forms where the OP asks for an "algebraic" proof of the equivalence $d\omega = \omega \wedge \alpha \Leftrightarrow\omega = f dg$. The implication $\Leftarrow$ is algebraically clear, just take $\alpha = - df/f$.

The other implication $\Rightarrow$ seems to be, instead, not algebraically straightforward (the fourth comment there shows a counterexample regarding the fact that the OP's proposal algebraic proof does not work).

Is basically this latter implication the actual content of Frobenius's theorem ? Thanks.

 

[cianfa72]

Said in other terms: in the simple case of just 1-form $\omega$ defined on a smooth manifold $M$, the Frobenius condition $\omega \wedge d\omega= 0$ (that is equivalent to the existence of a 1-form $\alpha$ such that $d\omega = \omega \wedge \alpha$) tell us that the $n-1$ dimensional smooth distribution is integrable (i.e. there exist an integral immersed submanifold of $M$ for each point $p \in M$).

The above does mean there exist a complete foliation of $M$ via such maximal connected immersed submanifolds (see also here Lec11).

My point is that such leaves of the foliation can be always given as level sets of a smooth map $t$ defined on the manifold $M$. Then $\omega$ can be always "recovered" from such a map $t$ using another smooth map $f$ as $\omega=fdt$.

Does it make sense ?

 

[martinbn]

Yes, it is Frobenius theorem.

 

[cianfa72]

I checked also Lee's book Theorem 19.12 (Global Frobenius theorem): the foliation via maximal connected integral submanifolds does exist. Therefore, I believe, in case of an $n-1$ smooth involutive distribution assigned as the kernel of a 1-form $\omega$, such submanifolds can be given as the level sets of a smooth function/map $t$ defined on the entire manifold $M$. Hence it seems to me $\omega = fdt$ for smooth functions/maps $f$ and $t$ globally defined on $M$.

 

[graphking]

can't see how to use Frobenius theorem, I think this is just a hard exercise. you can search in John lee, gtm218 for the true meaning of Frobenius theorem

 

[cianfa72]

you can search in John lee, gtm218 for the true meaning of Frobenius theorem

Yes, I've seen it on Jon Lee's Introduction on smooth manifolds.

 

[graphking]

Yes, I've seen it on Jon Lee's Introduction on smooth manifolds.

so how to use? can't see they are relevant