Frobenius theorem for differential one forms

  • #1
cianfa72
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TL;DR Summary
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.
 
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  • #2
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 ?
 
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  • #3
Yes, it is Frobenius theorem.
 
  • #4
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##.
 
  • #5
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
 
  • #6
graphking said:
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.
 
  • #7
cianfa72 said:
Yes, I've seen it on Jon Lee's Introduction on smooth manifolds.
so how to use? can't see they are relevant
 

What is the Frobenius Theorem for differential one forms?

The Frobenius Theorem for differential one forms is a fundamental result in differential geometry that characterizes when a set of one-forms can be simultaneously integrated to define smooth submanifolds. Specifically, the theorem states that a collection of one-forms on a manifold defines an integrable distribution if and only if the exterior derivatives of these forms can be expressed as a linear combination of the forms themselves.

How is the Frobenius Theorem used in practice?

In practice, the Frobenius Theorem is used to determine whether certain geometric and physical systems have solutions that can be described by smooth functions. For example, in fluid dynamics, the theorem helps in finding potential flow regions. It is also essential in the study of foliations on manifolds and the analysis of differential equations on manifolds.

What does it mean for a distribution to be integrable?

A distribution, which is a collection of tangent vectors at each point in a manifold, is said to be integrable if there exists a foliation of the manifold into submanifolds such that the tangent space to each submanifold at any point is exactly the distribution at that point. In simpler terms, a distribution is integrable if you can "stitch" together the vectors into smooth submanifolds.

What are the conditions for integrability according to Frobenius Theorem?

According to the Frobenius Theorem, a necessary and sufficient condition for a distribution to be integrable is that the Lie bracket (or commutator) of any two vector fields within the distribution also lies within the distribution. For differential one forms, this translates to the condition that the exterior derivative of any form in the set can be expressed as a linear combination of the forms themselves.

Can you give an example of applying the Frobenius Theorem?

An example of applying the Frobenius Theorem is in examining the flow of a fluid. If you have a vector field that represents the velocity of the fluid, the corresponding one-form is the dual vector field. The Frobenius Theorem can be applied to determine if there exist surfaces in the fluid such that the fluid's velocity is tangent to these surfaces at every point, indicating no flow across these surfaces and thus defining potential flow regions.

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