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WannabeNewton

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*directly*shows that ##\omega \wedge d\omega = 0 \Rightarrow \omega = \alpha d\beta## i.e. ##\omega_{[a}\nabla_{b}\omega_{c]} = 0 \Rightarrow \omega_{a} = \alpha \nabla_{a} \beta##?

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WannabeNewton

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- #2

cianfa72

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Hello, I found this very old post. I believe we have the solution you were looking for (thanks also to @PeterDonis).Most of my texts have the standard proof of Frobenius' theorem (both the vector field and differential forms versions) and through multiple indirect equivalences conclude that ##\omega \wedge d\omega = 0## implies (locally) that ##\omega = \alpha d\beta## where ##\omega## is a 1-form and ##\alpha,\beta## are scalar fields. Does anyone know of a proof wherein onedirectlyshows that ##\omega \wedge d\omega = 0 \Rightarrow \omega = \alpha d\beta## i.e. ##\omega_{[a}\nabla_{b}\omega_{c]} = 0 \Rightarrow \omega_{a} = \alpha \nabla_{a} \beta##?

First of all ##\omega \wedge d\omega=0 \Leftrightarrow d\omega = \alpha \wedge \omega## for some function ##\alpha##. From the latter immediately follows ##d\alpha \wedge \omega=0##. Then as shown here Global simultaneity surfaces - how to adjust proper time - #78 it must be ##d\alpha=0##.

From Poincarè lemma there exist locally a function ##h## such that ##\alpha=dh##, so we get locally ##d\omega= dh \wedge \omega## for some function ##h##.

Whatever the function ##h## is there is a 1-form ##\beta## such that ##e^h\beta=\omega## (just pick ##\beta = \omega / e^h##).

From ##d\omega= d(e^h\beta) = e^hdh \wedge \beta + e^h d\beta = dh \wedge e^h \beta + e^h d\beta## it follows ##d \beta=0## i.e. the one-form ##\beta## is closed hence locally ##\beta = dg## for some function ##g## (thanks to Poincarè lemma again).

But from ##\beta = \omega / e^h = dg## we get ##\omega = e^h dg##. Hence locally (i.e. in a open neighborhood around the point where ##\omega \wedge d\omega=0##) we get the expected result.

- #3

cianfa72

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I was wondering about the following: Frobenius' theorem claims that ##\omega \wedge d\omega = 0## implies *locally* ##\omega = \alpha d\beta## for some ##\alpha, \beta## scalar fields.

Suppose the condition ##\omega \wedge d\omega = 0## holds at each point inside an open set ##A##. Then the following are true:

*locally* actually extend to all points in the open region. So if the condition ##\omega \wedge d\omega = 0## holds everywhere in spacetime (by definition the spacetime as set is open) then the above two properties extend entirely to it.

Suppose the condition ##\omega \wedge d\omega = 0## holds at each point inside an open set ##A##. Then the following are true:

- There exist a scalar function ##\beta## defined inside ##A## such that its level sets are hypersurfaces that match up with the distribution defined by the one-form ##\omega## at each point inside A
- By definition of level set, the above hypersurfaces do not intersect each other and foliate the entire open region A

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- #4

PeterDonis

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Not necessarily. If ##\omega \wedge d \omega = 0## holds for every point in an open set A, then at each point of A we have ##\omega = \alpha d \beta## forSuppose the condition ##\omega \wedge d\omega = 0## holds at each point inside an open set ##A##. Then the following are true:

- #5

cianfa72

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Just to be clear: consider ##\mathbb R^3## as smooth manifold equipped with the standard topology (no metric involved at all). Suppose ##\omega \wedge d\omega=0## holds at point p. Frobenius claims there is an open ball ##B(p,\epsilon)## at p and scalar smooth functions ##\alpha, \beta## defined on all points inside it such that ##\omega = \alpha d \beta## is the local expression of ##\omega## at each point inside the open neighborhood ##B(p,\epsilon)##.But this doesnotguarantee that ##\omega = \alpha d \beta## holds at each point of A for thesame##\alpha## and ##\beta##. The Frobenius theorem does not imply that.

My point is that since the above expression of ##\omega## holds for all points inside ##B## then the two functions ##\alpha, \beta##

See also here http://staff.ustc.edu.cn/~wangzuoq/Courses/21F-Manifolds/Notes/Lec16.pdf

- #6

martinbn

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Take for example a closed form i.e. ##d\omega=0##, then the condition ##\omega \wedge d\omega = 0## is satisfied. By Poincare's lemma it is locally (every point has an open neighborhood) exact so ##\omega = d\theta## (##\alpha=1## and ##\beta = \theta##). But it need not be globally exact, so the ##\theta## is not the same in all neighborhoods.Just to be clear: consider ##\mathbb R^3## as smooth manifold equipped with the standard topology (no metric involved at all). Suppose ##\omega \wedge d\omega=0## holds at point p. Frobenius claims there is an open ball ##B(p,\epsilon)## at p and scalar smooth functions ##\alpha, \beta## defined on all points inside it such that ##\omega = \alpha d \beta## is the local expression of ##\omega## at each point inside the open neighborhood ##B(p,\epsilon)##.

My point is that since the above expression of ##\omega## holds for all points inside ##B## then the two functions ##\alpha, \beta##mustbe the same for all points in ##B(p,\epsilon)##.

See also here http://staff.ustc.edu.cn/~wangzuoq/Courses/21F-Manifolds/Notes/Lec16.pdf

- #7

cianfa72

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Yes, it is not the same inBut it need not be globally exact, so the ##\theta## is not the same in all neighborhoods.

- #8

martinbn

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That is not what you said and he replied to what you had said.Yes, it is not the same inallneighborhoods. However if we consider another point inside that particular open neighborhood (where ##\theta## above is defined) then we continue to get ##\omega = d\theta## for that 'specific' ##\theta## scalar function, I believe.

- #9

cianfa72

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Ok sorry (maybe I was unclear ). Yet can you confirm my claim in post #7 ? Thanks.That is not what you said and he replied to what you had said.

Edit: in you example take a closed form ##\omega## (i.e. ##d\omega=0##) in the open neighborhood A of point p. By Poincare's lemma there is a scalar function ##\theta## defined in an open neighborhood B of p (possibly a proper open subset of the open set A) such that ##\omega=d\theta## holds not

That was my point so far....

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- #10

PeterDonis

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Yes.Suppose ##\omega \wedge d\omega=0## holds at point p. Frobenius claims there is an open ball ##B(p,\epsilon)## at p and scalar smooth functions ##\alpha, \beta## defined on all points inside it such that ##\omega = \alpha d \beta## is the local expression of ##\omega## at each point inside the open neighborhood ##B(p,\epsilon)##.

This does not follow from the above.My point is that since the above expression of ##\omega## holds for all points inside ##B## then the two functions ##\alpha, \beta##mustbe the same for all points in ##B(p,\epsilon)##.

What in particular in these notes are you referring to, and what do you think it shows?

- #11

PeterDonis

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Exactly.Yes, it is not the same inallneighborhoods.

Inside that particular open neighborhood, yes. But that particular open neighborhood does not have to be the entire manifold. Inside a different open neighborhood you could have a different ##\theta##.However if we consider another point inside that particular open neighborhood (where ##\theta## above is defined) then we continue to get ##\omega = d\theta## for that 'specific' ##\theta## scalar function, I believe.

- #12

cianfa72

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Maybe I was unclear: I assumed ##\omega \wedge d\omega=0## holds not only at point p but at all points within ##B(p,\epsilon)##.This does not follow from the above.

Yes, of course.Inside that particular open neighborhood, yes. But that particular open neighborhood does not have to be the entire manifold.

Yes, definitely.Inside a different open neighborhood you could have a different ##\theta##.

- #13

PeterDonis

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You don't have to assume that; the Frobenius theorem establishes it forMaybe I was unclear: I assumed ##\omega \wedge d\omega=0## holds not only at point p but at all points within ##B(p,\epsilon)##.

- #14

cianfa72

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Ah ok, I take it as follows: start with the one-form field ##\omega## defined at each point in the manifold M such that ##\omega \wedge d\omega=0## holds at aYou don't have to assume that; the Frobenius theorem establishes it forsomeopen ball ##B(p, \epsilon)##. The theorem just does not establish that ##B## is the entire manifold.

Since local Frobenius theorem establishes that there exist smooth functions ##\alpha,\beta## defined in

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- #15

PeterDonis

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Yes, forstart with the one-form field ##\omega## defined at each point in the manifold M such that ##\omega \wedge d\omega=0## holds at agivenpoint p in the manifold.

Since local Frobenius theorem establishes that there exist smooth functions ##\alpha,\beta## defined insomeopen ball ##B(p, \epsilon)## such that ##\omega = \alpha d \beta## atallpoints within it, it follows that ##\omega \wedge d\omega = \alpha d \beta \wedge d(\alpha d \beta) = 0## at all points within ##B(p, \epsilon)##.

- #16

cianfa72

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At the end of pag 4 theWhat in particular in these notes are you referring to, and what do you think it shows?

Now, as far as I can understand, each of these leaves (i.e. connected immersed submanifolds) can be given as the level set of a scalar function defined on the

- #17

PeterDonis

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Note that the reference you give isn't even using the language of differential forms. It's talking about distributions, which are generalizations of vector fields.an involutive distribution (i.e. the kernel of ##\omega## such that ##\omega \wedge d\omega=0## on the entire manifold M)

Yes.the collection of (unique) maximal connected integral (sub)manifolds form a foliation on the entire manifold (leaves of the foliation).

Not necessarily, no. It might take more than one scalar function to fully parameterize the leaves.Now, as far as I can understand, each of these leaves (i.e. connected immersed submanifolds) can be given as the level set of a scalar function defined on theentiremanifold.

- #18

cianfa72

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Do you mean more than one scalar function, for instance 2 scalar functions to fully parameterize an ##(n-2)## dimensional submanifold in an n-dimensional manifold ?Not necessarily, no. It might take more than one scalar function to fully parameterize the leaves.

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- #19

PeterDonis

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Yes.Do you mean more than one scalar function, for instance 2 scalar functions to fully parameterize an ##(n-2)## dimensional submanifold in an n-dimensional manifold ?

- #20

cianfa72

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The curve above is a Lemniscate and it is an

My point is: we can define of a smooth vector field ##X## on the entire ##\mathbb R^2## such that the (unique) maximal connected integral submanifold passing for ##(0,0)## is the Lemniscate above. Since the global Frobenius theorem guarantees that a foliation of the entire ##\mathbb R^2## does exist, the Leminscate above basically defines 3 regions: the region within the right lobe, the region within the left lobe and the region outside of it.

Hence the maximal connected integral sumbanifolds (i.e. the integral curves) for the points in each of those 3 regions

Does it make sense ? Thank you.

- #21

martinbn

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Not sure if this makes sense, but may be I missunderstood you. The curve has two different tangent vectors at the origin, and there is no vector field in the plane that has two values at a point! (The tangent vector field along the curve is not the restriction of a vector field in the plane.)My point is: we can define of a smooth vector field ##X## on the entire ##\mathbb R^2## such that the (unique) maximal connected integral submanifold passing for ##(0,0)## is the Lemniscate above.

- #22

cianfa72

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The curve above is anNot sure if this makes sense, but may be I missunderstood you. The curve has two different tangent vectors at the origin, and there is no vector field in the plane that has two values at a point! (The tangent vector field along the curve is not the restriction of a vector field in the plane.)

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- #23

cianfa72

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martinbn

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##x(t)={a\sqrt{2}\cos t\over 1+\sin^2 t},\quad y(t)={a\sqrt{2}\cos t\sin t\over 1+\sin^2 t},\quad t\in [0, 2\pi]##

For ##t=\frac\pi2## and ##t=\frac{3\pi}2## you get ##(x,y)=(0,0)##.

From the picture it is even obvious, that the curve selfintersect, and is not empbedded in the plane. It is just immersed, but not injectively.

- #25

martinbn

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One of the suggested links below leads me to think that you pobably meant this curve?

- #26

cianfa72

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Ah yes, you are right.One of the suggested links below leads me to think that you probably meant this curve?

The above is actually the flipped version of the curve in #20. Here the immersion is ##(-\pi,\pi) \mapsto \mathbb R^2##.

As you can check it is an

However, as said in #23, does exist a

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- #27

PeterDonis

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The Frobenius theorem does not say that any such thing must exist. The Frobenius theorem does not say anything aboutdoes exist asmoothvector field ##X## defined on ##\mathbb R^2## such that its restriction to the above Lemniscate curve gives its tangent vector at each point ?

- #28

cianfa72

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Yes, that was actually my point. So we are basically saying that the Leminscate curve cannot be a leaf of a foliation associated with a possible 1-dimensional smooth distribution (i.e. of a smooth vector field ##X## defined on ##\mathbb R^2##).It only says thatifyou already have a smooth vector field on the entire manifold that meets the Frobenius condition,thenthere will be a corresponding foliation.

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- #29

martinbn

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I don't think so. The value at the origin needs to be the tangent vector to the right sloping part of the curve (the one that contains the point at the origin). On the other hand since it is smooth (or just continuous) it will be the limit of the tangent vectors to the left part of the curve (the one with the arrows). And those are not the same.Ah yes, you are right.

View attachment 298394

The above is actually the flipped version of the curve in #20. Here the immersion is ##(-\pi,\pi) \mapsto \mathbb R^2##.

As you can check it is aninjectiveimmersion and the tangent vector at the origin ##(0,0)## is unique.

However, as said in #23, does exist asmoothvector field ##X## defined on ##\mathbb R^2## such that its restriction to the above Lemniscate curve gives its tangent vector at each point ?

- #30

martinbn

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It is not a submanifold, so it cannot be an integral manifold of a vector field. On the other hand Frobenius is trivial for 1-dimensional distributions. By trivial I mean that the condition is always true in the 1D case.Yes, that was actually my point. So we are basically saying that the Leminscate curve cannot be a leaf of a foliation associated with a possible 1-dimensional smooth distribution (i.e. of a smooth vector field ##X## defined on ##\mathbb R^2##).

- #31

PeterDonis

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I'm not saying itwe are basically saying that the Leminscate curve cannot be a leaf of a foliation associated with a possible 1-dimensional smooth distribution

- #32

cianfa72

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My concern is that Frobenius theorem (as stated here) involvesIt is not a submanifold, so it cannot be an integral manifold of a vector field.

- #33

martinbn

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I see, I still think it is not possible, because there isn't a smooth vector field that restricts to the tangent field along the curve.My concern is that Frobenius theorem (as stated here) involvesimmersed submanifoldsand the Lemniscate curve is an immersed submanifold in ##\mathbb R^2##.

- #34

cianfa72

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Yes of course, for example in an 'ambient' 2D manifold whatever 1-dimensional smooth distribution given as the kernel of the one-form ##\omega## is such that ##\omega \wedge d\omega## vanishes identically.On the other hand Frobenius is trivial for 1-dimensional distributions. By trivial I mean that the condition is always true in the 1D case.

Yes, that's my point too. The tangent vector at each point along the Lemniscate exists and is unique however, as you pointed out, there is not aI see, I still think it is not possible, because there isn't a smooth vector field that restricts to the tangent field along the curve.

Anyway, I believe there should be some example of

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- #35

cianfa72

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Suppose there was some smooth vector field ##X## defined on ##\mathbb R^2## such that the Lemniscate curve was an integral manifold. Then if we restricted such vector field ##X## on that curve we would get the tangent vectors at each point along it. However it cannot be the case since there is not any smooth vector field that restricted to the Lemniscate gives the tangent vector at each point along it.I'm not saying itcannotbe, and you certainly have not proved it.

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