Proof of Frobenius' Theorem: Directly Showing ##\omega \wedge d\omega = 0##

[WannabeNewton]

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

 

[cianfa72]

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

Hello, I found this very old post. I believe we have the solution you were looking for (thanks also to @PeterDonis).

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.

 

[cianfa72]

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:

1. 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
2. By definition of level set, the above hypersurfaces do not intersect each other and foliate the entire open region A

My point is that the qualifier 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.

 

[PeterDonis]

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

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 $\omega = \alpha d \beta$ for some $\alpha$ and $\beta$, where $\alpha$ and $\beta$ are scalar functions. But this does not guarantee that $\omega = \alpha d \beta$ holds at each point of A for the same $\alpha$ and $\beta$. The Frobenius theorem does not imply that.

 

[cianfa72]

But this does not guarantee that $\omega = \alpha d \beta$ holds at each point of A for the same $\alpha$ and $\beta$. The Frobenius theorem does not imply that.

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 $\alpha, \beta$ must be the same for all points in $B(p,\epsilon)$.

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

 

[martinbn]

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 $\alpha, \beta$ must be the same for all points in $B(p,\epsilon)$.

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

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.

 

[cianfa72]

But it need not be globally exact, so the $\theta$ is not the same in all neighborhoods.

Yes, it is not the same in all neighborhoods. 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.

 

[martinbn]

Yes, it is not the same in all neighborhoods. 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.

That is not what you said and he replied to what you had said.

 

[cianfa72]

That is not what you said and he replied to what you had said.

Ok sorry (maybe I was unclear ). Yet can you confirm my claim in post #7 ? Thanks.

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 only at point p but at all points within B.

That was my point so far...

 

[PeterDonis]

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 $B(p,\epsilon)$.

Yes.

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

This does not follow from the above.

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

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

 

[PeterDonis]

Yes, it is not the same in all neighborhoods.

Exactly.

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.

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 $\theta$.

 

[cianfa72]

This does not follow from the above.

Maybe I was unclear: I assumed $\omega \wedge d\omega=0$ holds not only at point p but at all points within $B(p,\epsilon)$.

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

Yes, of course.

Inside a different open neighborhood you could have a different $\theta$.

Yes, definitely.

 

[PeterDonis]

Maybe I was unclear: I assumed $\omega \wedge d\omega=0$ holds not only at point p but at all points within $B(p,\epsilon)$.

You don't have to assume that; the Frobenius theorem establishes it for some open ball $B(p, \epsilon)$. The theorem just does not establish that $B$ is the entire manifold. You seem to agree with that so I don't have any objection now.

 

[cianfa72]

You don't have to assume that; the Frobenius theorem establishes it for some open ball $B(p, \epsilon)$. The theorem just does not establish that $B$ is the entire manifold.

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 a given point p in the manifold.

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

 

[PeterDonis]

start with the one-form field $\omega$ defined at each point in the manifold M such that $\omega \wedge d\omega=0$ holds at a given point p in the manifold.

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

Yes, for some $B(p, \epsilon)$.

 

[cianfa72]

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

At the end of pag 4 the global Frobenius theorem claims that for an involutive distribution (i.e. the kernel of $\omega$ such that $\omega \wedge d\omega=0$ on the entire manifold M) the collection of (unique) maximal connected integral (sub)manifolds form a foliation on the entire manifold (leaves of the foliation).

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 entire manifold. Hence it does mean there is a global scalar function defined on all the entire manifold M such that all the leaves are the level sets of.

 

[PeterDonis]

an involutive distribution (i.e. the kernel of $\omega$ such that $\omega \wedge d\omega=0$ on the entire manifold M)

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.

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

Yes.

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 entire manifold.

Not necessarily, no. It might take more than one scalar function to fully parameterize the leaves.

 

[cianfa72]

Not necessarily, no. It might take more than one scalar function to fully parameterize the leaves.

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 ?

 

[PeterDonis]

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 ?

Yes.

 

[cianfa72]

Since from global Frobenius theorem the maximal connected integral submanifolds (leaves) of the foliation are basically immersed submanifolds of the 'ambient' manifold M, I was thinking about the following scenario in $\mathbb R^2$

The curve above is a Lemniscate and it is an injective immersion of the real line $\mathbb R$ into $\mathbb R^2$ -- note that the curve does not intersect itself at the origin $(0,0)$.

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 must lie within each of them (i.e. each one cannot extend to points outside of the relevant region).

Does it make sense ? Thank you.

 

[martinbn]

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.

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.)

 

[cianfa72]

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.)

The curve above is an injective immersion hence only the point $x=0$ of the Real line is mapped to the origin $(0,0)$ of $\mathbb R^2$. So, I believe, the curve at the origin has only one tangent vector hence it does exist a vector field in the plane that assigns that value at the origin -- i.e. it basically assigns only one vector in the tangent vector space 'attached' at the origin.

 

[cianfa72]

@martinbn maybe your concern is about the 'smoothness' of a possibly vector field $X$ in a neighborhood of the origin. In other words, even though the tangent vector is unique at origin, however, there could not exist a smooth vector field $X$ defined on $\mathbb R^2$ such that restricted to the Lemniscate curve gives its tangent vector at each point.

 

[martinbn]

No, here is the standard parametrization (see the wiki page)

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

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.

 

[martinbn]

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

 

[cianfa72]

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

Ah yes, you are right.

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 injective immersion and the tangent vector at the origin $(0,0)$ is unique.

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

 

[PeterDonis]

does exist a smooth vector field $X$ defined on $\mathbb R^2$ such that its restriction to the above Lemniscate curve gives its tangent vector at each point ?

The Frobenius theorem does not say that any such thing must exist. The Frobenius theorem does not say anything about extending a vector field from some submanifold to the entire manifold--i.e., it does not say that any vector field on a submanifold must be a leaf in a foliation on the entire manifold. It only says that if you already have a smooth vector field on the entire manifold that meets the Frobenius condition, then there will be a corresponding foliation.

 

[cianfa72]

It only says that if you already have a smooth vector field on the entire manifold that meets the Frobenius condition, then there will be a corresponding foliation.

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 $\mathbb R^2$).

 

[martinbn]

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 an injective immersion and the tangent vector at the origin $(0,0)$ is unique.

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

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.

 

[martinbn]

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 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.

 

[PeterDonis]

we are basically saying that the Leminscate curve cannot be a leaf of a foliation associated with a possible 1-dimensional smooth distribution

I'm not saying it cannot be, and you certainly have not proved it. I am just saying that the Frobenius theorem has nothing to say about the question.

 

[cianfa72]

It is not a submanifold, so it cannot be an integral manifold of a vector field.

My concern is that Frobenius theorem (as stated here) involves immersed submanifolds and the Lemniscate curve is an immersed submanifold in $\mathbb R^2$.

 

[martinbn]

My concern is that Frobenius theorem (as stated here) involves immersed submanifolds and the Lemniscate curve is an immersed submanifold in $\mathbb R^2$.

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.

 

[cianfa72]

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 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.

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.

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 a smooth vector field $X$ that restricts to the tangent field along the curve.

Anyway, I believe there should be some example of immersed submanifold such that there is a smooth vector field that restricts to the tangent vector field along it.

 

[cianfa72]

I'm not saying it cannot be, and you certainly have not proved it.

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.