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

[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.

 

[PeterDonis]

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

So what? The theorem does not say that any immersed submanifold must be a leaf of a foliation. Nor does it give any conditions that tell you whether an immersed submanifold is or is not a leaf of a foliation. So the theorem has nothing to say about your lemniscate curve.

The fact that the words "immersed submanifold" appear in your reference's statement of the theorem does not mean that the theorem must automaticallly decide all questions you have about immersed submanifolds.

 

[PeterDonis]

there is not any smooth vector field that restricted to the Lemniscate gives the tangent vector at each point along it.

Can you prove this or are you just waving your hands?

 

[PeterDonis]

there isn't a smooth vector field that restricts to the tangent field along the curve.

Why do you think this? (Note, I'm not saying you're wrong, I'm just asking for a more explicit argument if you have one.)

 

[cianfa72]

Why do you think this? (Note, I'm not saying you're wrong, I'm just asking for a more explicit argument if you have one.)

Consider an open neighborhood of the origin $(0,0)$ in $\mathbb R^2$ and apply what @martinbn pointed out in his post #29.

 

[PeterDonis]

Consider an open neighborhood of the origin $(0,0)$ in $\mathbb R^2$ and apply what @martinbn pointed out in his post #29.

Ah, got it.

 

[cianfa72]

The fact that the words "immersed submanifold" appear in your reference's statement of the theorem does not mean that the theorem must automaticallly decide all questions you have about immersed submanifolds.

Maybe I was unclear: my point was that it must exist at least an "immersed submanifold" (that however is not a regular embedded submanifold) which is an integral submanifold of some smooth vector field $X$ defined on the 'ambient' manifold. Otherwise the use of the word "immersed submanifold" in the theorem's statement does not make any sense.

The lemniscate curve was just an example of immersed submanifold that is not a regular embedded submanifold and we found that it is not actually a good example though.

 

[PeterDonis]

my point was that it must exist at least an "immersed submanifold" (that however is not a regular embedded submanifold) which is an integral submanifold of some smooth vector field $X$ defined on the 'ambient' manifold.

What "must" exist in that way? Why is this even an issue?

Otherwise the use of the word "immersed submanifold" in the theorem's statement does not make any sense.

Yes, fine, what does that have to do with the lemniscate?

The lemniscate curve was just an example of immersed submanifold that is not a regular embedded submanifold and we found that it is not actually a good example though.

Indeed. So why are we still discussing it?

 

[cianfa72]

Yes, fine, what does that have to do with the lemniscate?

Nothing, mine was just an example to help me in clarify the real content of the theorem.

Edit: Btw do you know an example of immersed submanifold (that is not however a regular embedded submanifold) that is a leaf of the foliation associated to some smooth vector field ?

 

[cianfa72]

Searching for it I found this interesting example of immersed submanifold (a 1D curve) that is not an embedding such that there is a smooth vector field $X$ that restricts to the tangent field along the curve.

The example involves a torus $\mathbb R^2/ \mathbb Z^2$ and a constant smooth vector field $X$ defined on the torus such that its integral curves are straight lines with slope an irrational number.

 

[martinbn]

Searching for it I found this interesting example of immersed submanifold (a 1D curve) that is not an embedding such that there is a smooth vector field $X$ that restricts to the tangent field along the curve.

The example involves a torus $\mathbb R^2/ \mathbb Z^2$ and a constant smooth vector field $X$ defined on the torus such that its integral curves are straight lines with slope an irrational number.

A curve that turns and touches itself so that there is just one tangent line will be an example.

 

[cianfa72]

A curve that turns and touches itself so that there is just one tangent line will be an example.

But such a curve that turns and touches itself cannot be an immersed submanifold of the real line $\mathbb R$ (AFAIK the definition of immersed submanifold requires an injective map).

 

[martinbn]

But such a curve that turns and touches itself cannot be an immersed submanifold of the real line $\mathbb R$ (AFAIK the definition of immersed submanifold requires an injective map).

Injective only on tangent vectors, not the map itself.

 

[cianfa72]

Injective only on tangent vectors, not the map itself.

It depends on the definition used: for example some source requires also the injectivity of the map -- see for instance here at section 4.2.2 (pag 48).

 

[martinbn]

It depends on the definition used: for example some source requires also the injectivity of the map -- see for instance here at section 4.2.2 (pag 48).

No, read it more carefully. Injective diferential, not the map. If the map is also injective then it is an embedding.

 

[cianfa72]

Injective diferential, not the map. If the map is also injective then it is an embedding.

See Definition 4.7 a) in section 4.2.2:

If $f$ is an injective immersion, then $N$ is called immersed smooth submanifold of M through $f$.

 

[jbergman]

No, read it more carefully. Injective diferential, not the map. If the map is also injective then it is an embedding.

Not necessarily, the irrational curve on the torus is typically not considered an embedding.