The discussion focuses on proving Frobenius' theorem directly, specifically the implication that if ##\omega \wedge d\omega = 0##, then locally ##\omega = \alpha d\beta## for scalar fields ##\alpha## and ##\beta##. A participant outlines a proof using the relationship ##d\omega = \alpha \wedge \omega##, leading to the conclusion that ##d\alpha = 0## and thus locally defining ##\alpha = dh##. The proof illustrates that while local expressions hold, they do not guarantee global consistency across the manifold.
PREREQUISITESThis discussion is beneficial for mathematicians, particularly those specializing in differential geometry, as well as students and researchers interested in the applications of Frobenius' theorem and differential forms in manifold theory.
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 said:we are basically saying that the Leminscate curve cannot be a leaf of a foliation associated with a possible 1-dimensional smooth distribution
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 said:It is not a submanifold, so it cannot be an integral manifold of a vector field.
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 said:My concern is that Frobenius theorem (as stated here) involves immersed submanifolds and the Lemniscate curve is an immersed submanifold in ##\mathbb R^2##.
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.martinbn said: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 a smooth vector field ##X## that restricts to the tangent field along the curve.martinbn said: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.
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 said:I'm not saying it cannot be, and you certainly have not proved it.
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.cianfa72 said:My concern is that Frobenius theorem (as stated here) involves immersed submanifolds and the Lemniscate curve is an immersed submanifold in ##\mathbb R^2##.
Can you prove this or are you just waving your hands?cianfa72 said:there is not any smooth vector field that restricted to the Lemniscate gives the tangent vector at each point along it.
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.)martinbn said:there isn't a smooth vector field that restricts to the tangent field along the curve.
Consider an open neighborhood of the origin ##(0,0)## in ##\mathbb R^2## and apply what @martinbn pointed out in his post #29.PeterDonis said: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.)
Ah, got it.cianfa72 said:Consider an open neighborhood of the origin ##(0,0)## in ##\mathbb R^2## and apply what @martinbn pointed out in his post #29.
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.PeterDonis said: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.
What "must" exist in that way? Why is this even an issue?cianfa72 said: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.
Yes, fine, what does that have to do with the lemniscate?cianfa72 said:Otherwise the use of the word "immersed submanifold" in the theorem's statement does not make any sense.
Indeed. So why are we still discussing it?cianfa72 said: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.
Nothing, mine was just an example to help me in clarify the real content of the theorem.PeterDonis said:Yes, fine, what does that have to do with the lemniscate?
A curve that turns and touches itself so that there is just one tangent line will be an example.cianfa72 said: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.
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 said:A curve that turns and touches itself so that there is just one tangent line will be an example.
Injective only on tangent vectors, not the map itself.cianfa72 said: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).
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 said:Injective only on tangent vectors, not the map itself.
No, read it more carefully. Injective diferential, not the map. If the map is also injective then it is an embedding.cianfa72 said: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).
See Definition 4.7 a) in section 4.2.2:martinbn said:Injective diferential, not the map. If the map is also injective then it is an embedding.
If ##f## is an injective immersion, then ##N## is called immersed smooth submanifold of M through ##f##.
Not necessarily, the irrational curve on the torus is typically not considered an embedding.martinbn said:No, read it more carefully. Injective diferential, not the map. If the map is also injective then it is an embedding.