I Manifold hypersurface foliation and Frobenius theorem

[lavinia]

Can you help me ?

Yes. But give it a shot to start the conversation.
I have never done this myself. We can work on it together.

 

[lavinia]

Locally it will be, I believe (it is basically the content of Frobenius theorem since a foliation is basically the same as the existence of a completely integrable distribution over the manifold). The point is that, in general, one is not guaranteed that there will be a global function with that property at each point on the manifold.

What about the 1 form in the plane ydx? Its exterior derivative is dy^dx which is not zero but its kernel at each point is the tangent space to the line that is parallel to the y-axis.

 

[lavinia]

Locally it will be, I believe (it is basically the content of Frobenius theorem since a foliation is basically the same as the existence of a completely integrable distribution over the manifold). The point is that, in general, one is not guaranteed that there will be a global function such that the 1-form is the differential of at any point on the manifold.

Interestingly, an example of what you are saying is the 1 form dx. It does not project to the differential of a function on the Möbius band. (So it is no surprise that its integral over one of the foliating circles is not zero). But locally it is the differential of the local function x.

 

[cianfa72]

What about the 1 form in the plane ydx? Its exterior derivative is dy^dx which is not zero but its kernel at each point is the tangent space to the line that is parallel to the y-axis.

Sorry, I was sloppy. The above means $\omega = ydx$ isn't closed in the plane hence it can't be exact even locally. In this case $\omega \wedge d\omega = 0$, therefore Frobenius claims $\omega = fdg$ for some smooth functions $f$ and $g$ in a neighborhood of each point of the foliation. Hence, even though $\omega$ is not "exactly" the differential of a function, nevertheless the kernel of $dg$ gives the distribution tangent at any point to the foliation lines (lines parallel to y-axis).

 

[cianfa72]

Yes. But give it a shot to start the conversation.
I have never done this myself. We can work on it together.

From Wikipedia there is a complete parametrization of the open Mobius band that is continuous from $\mathbb R^2$ to $\mathbb R^3$. Then if one restricts it to the domain $0 \leq u < 2\pi, -1 < v < 1$ with the subspace topology, it is 1-1 on the image and continuous as well.

The inverse map is 1-1 on the image and continuous with the subspace topology on it, hence we have established an homeomorphism with the respective subspace topologies.

 

[WWGD]

Just to comment that a Contact Structure is dual/converse in the sense of Frobenius. It is a nowhere integrable distribution, and it's defined as the kernel of a 1-form.

 

[lavinia]

On the Möbius band a 1 form that kills the tangent spaces to the foliation by circles
pulls back to the Euclidean plane to a form f(x,y)dy that is invariant under the action of the group of covering transformations.

This means that f(x+1/2,-y)=-f(x,y)

If this form is closed then the partial derivative of f with respect to x must be zero so f(x,-y) =-f(x,y).

f(x,y)dy must be identically zero along the equatorial circle and this shows that there is no closed 1 form whose kernel is the tangent spaces to the foliation by circles at all points.

Also since f depends only on y, the differential of its antiderivative is equal to fdy. And since f is an odd function its antiderivative is even and thus descends to a function on the Möbius band.

Comment:This exact analysis extends to the Klein bottle which is made by adding integer translations in the y direction to the group of covering transformations of the Möbius band. The vertical lines of the Möbius band are now made into circles and the Klein bottle is the total space of a circle bundle over the circle.

For those who know some homology theory, the above analysis shows that the vertical circles do not represent non-zero real homology classes since there is no real cohomology class that is dual to them. On the other hand, these circles are not homologous to zero over the integers. They are examples of torsion classes. That is: finite multiples of them (in this case their doubles) are homologous to zero. Real homology has no torsion.

From Wikipedia there is a complete parametrization of the open Mobius band that is continuous from $\mathbb R^2$ to $\mathbb R^3$. Then if one restricts it to the domain $0 \leq u < 2\pi, -1 < v < 1$ with the subspace topology, it is 1-1 on the image and continuous as well.

The inverse map is 1-1 on the image and continuous with the subspace topology on it, hence we have established an homeomorphism with the respective subspace topologies.

OK.But the problem was to figure it out on on one's own.

 

[cianfa72]

On the Möbius band a 1 form that kills the tangent spaces to the foliation by circles
pulls back to the Euclidean plane to a form f(x,y)dy that is invariant under the action of the group of covering transformations.

This means that f(x+1/2,-y)=-f(x,y)

If this form is closed then the partial derivative of f with respect to x must be zero so f(x,-y) =-f(x,y).

Say $\omega = f(x,y)dy$, then $$d\omega = df \wedge dy =\frac {\partial f} {\partial x} dx \wedge dy$$ Therefore if $\omega$ is closed $d\omega = 0=\frac {\partial f} {\partial x}$.
Why it follows that $f(x,-y) = - f(x,y)$ ?

f(x,y)dy must be identically zero along the equatorial circle and this shows that there is no closed 1 form whose kernel is the tangent spaces to the foliation by circles at all points.

Yes, since the equatorial circle has equation $y=0$, hence $f(x,0)=-f(x,0)$.

Also since f depends only on y, the differential of its antiderivative is equal to fdy. And since f is an odd function its antiderivative is even and thus descends to a function on the Möbius band.

Sorry, $f()$ depends only on $y$ since $\frac {\partial f} {\partial x}=0$ ?

 

[lavinia]

Say $\omega = f(x,y)dy$, then $$d\omega = df \wedge dy =\frac {\partial f} {\partial x} dx \wedge dy$$ Therefore if $\omega$ is closed $d\omega = 0=\frac {\partial f} {\partial x}$.
Why it follows that $f(x,-y) = - f(x,y)$ ?


Yes, since the equatorial circle has equation $y=0$, hence $f(x,0)=-f(x,0)$.


Sorry, $f()$ depends only on $y$ since $\frac {\partial f} {\partial x}=0$ ?

Right.

A general moral is that a 1 form might not be cohomologous to zero. In this case it can not be the differential of a function.

For instance the torus is foliated by circles but none of these are homologous to zero so the 1 forms that are dual to them can not be cohomologous to zero. This depends on De Rham's Theorem and the theorem that real cohomology is the dual vector space to real homology.

 

[WWGD]

Right.

A general moral is that a 1 form might not be cohomologous to zero. In this case it can not be the differential of a function.

For instance the torus is foliated by circles but none of these are homologous to zero so the 1 forms that are dual to them can not be cohologous to zero. This depends on de rams Theorem and the theorem that real cohomology is the dual vector space to real homology.

Then the form lives in a space with non-trivial (co)homology. I can't remember the groups of the Mobius Strip. Not even that of 1-forms. Edit: Never mind. It's homotopically equivalent to $S^1$, so its first group over $\mathbb Z=\mathbb Z$.

 

[WWGD]

Sorry, I didn't get yet why the above holds true. Thanks.

You're quoting your own post.

 

[cianfa72]

Yes, what about my last question

Why it follows that $f(x,-y) = - f(x,y)$ ?

Is it just because $f(\text{ },)$ actually doesn't depend on its first "slot" (Since the partial derivative w.r.t. it is null) ?

 

[jbergman]

Yes, what about my last question

Is it just because $f(\text{ },)$ actually doesn't depend on its first "slot" (Since the partial derivative w.r.t. it is null) ?

@lavinia is probably more qualified to answer, but this follows from the fact that he claimed that $f(x+1/2, -y) = -f(x, y)$ and the fact that $\partial_x f = 0$.

 

[cianfa72]

Just to help to visualize the mapping of the open Mobius band given by the atlas in post #31, I drew the following picture. $(U_1,Id)$ and $(U_2,f)$ are the two atlas's chart mapping. Vertical red lines's points are identified with a sign flip.