I Manifold hypersurface foliation and Frobenius theorem

[jbergman]

What kind of topology are you usIng , to speak of continuity, or of differentiable structure , to speak of differentiability?

For smooth manifolds, the manifold is locally homeomorphic to $\mathbb R^n$. A chart is such a homeomorphism. In the example given above is the image of the manifold in an open set of $\mathbb R^n$. Since it isn't continuous in that chart it's not continuous on the manifold.

 

[cianfa72]

For smooth manifolds, the manifold is locally homeomorphic to $\mathbb R^n$. A chart is such a homeomorphism. In the example given above is the image of the manifold in an open set of $\mathbb R^n$. Since it isn't continuous in that chart it's not continuous on the manifold.

Yes, the topology on the Mobius strip is given from its atlas (a set is open in the Mobius strip topology iff its image is open in any atlas's chart). The function $f$ at $u=1/2$ is continuous only at $v=0$ in $(U_2, \phi)$ chart, hence it is continuous at $[0,0]$ however is not differentiable there.

 

[WWGD]

For smooth manifolds, the manifold is locally homeomorphic to $\mathbb R^n$. A chart is such a homeomorphism. In the example given above is the image of the manifold in an open set of $\mathbb R^n$. Since it isn't continuous in that chart it's not continuous on the manifold.

Yes, thanks for refreshing my memory. I've been away from the topic for too long.

 

[lavinia]

Suppose that there was a 1-form on the Möbius band that is perpendicular to the foliation by circles. As @jbergman pointed out this would mean that the Möbius band, viewed as a line bundle over the equatorial circle, would have a non-zero section. But It does not so this is impossible. Any function that defines the circles as its level sets must have a zero differential somewhere.

On the other hand the Möbius band is foliated by the fibers of this non-trivial line bundle and the 1 form that points parallel to the foliation by circles is well defined. This means that the tangent spaces to the circles form a trivial sub-bundle of the tangent bundle of the Möbius band.

In bundle language, the tangent bundle to the Möbius band is a Whitney sum of these two line bundles. One of these line bundles is trivial while the other is not.

If one identifies the boundary circle of the Möbius band to it self by a two fold covering then the quotient space is a Klein bottle and it is foliated by circles in both directions. The tangent spaces to the circles again form a Whitney sum of two line bundles only one of which is trivial.

In general, there is no non-orientable manifold whose tangent bundle is a Whitney sum of trivial line sub-bundles.

 

[cianfa72]

Suppose that there was a 1-form on the Möbius band that is perpendicular to the foliation by circles.

A 1-form is not a vector field. Therefore I guess you mean the specific 1-form $\omega$ one gets "lowering" the index of the vector field $v$ that gives the foliation by circle (i.e. the relevant flow), using the metric tensor $g$ defined on the strip/band.

As @jbergman pointed out this would mean that the Möbius band, viewed as a line bundle over the equatorial circle, would have a non-zero section. But It does not so this is impossible. Any function that defines the circles as its level sets must have a zero differential somewhere.

Sorry, I lost why that would mean that Möbius band would have a non-zero (smooth) section.

On the other hand the Möbius band is foliated by the fibers of this non-trivial line bundle and the 1 form that points parallel to the foliation by circles is well defined. This means that the tangent spaces to the circles form a trivial sub-bundle of the tangent bundle of the Möbius band.

Here the relevant 1-form $\delta$ is such that its kernel is orthogonal to the foliation by circles ?

 

[lavinia]

@cianfa72

- Right I meant the 1 form whose dual vector field is perpendicular the foliation by circles under the Euclidean metric.

The Euclidean metric in the plane descends naturally to the Möbius band. This can be seen intuitively because one can make a Möbius band out of paper by bending it then pasting the two ends with a half twist. Bending does not change the metric.

Mathematically, the Mobius band is the quotient space of the Euclidean plane by the action of the group of isometries generated by integer length translations in the x direction together with the transformation
(x+1/2,-y)

- The Möbius band can be viewed as the total space of a one dimensional real vector bundle(called a line bundle) over the circle. The set of zero vectors of this bundle is the equatorial circle and a section is a continuous choice of a vector in each line. The equatorial circle is the section made up entirely of zero vectors. By the Intermediate Value Theorem every section must cross the equatorial circle and so must contain a zero vector.

 

[cianfa72]

The Euclidean metric in the plane descends naturally to the Möbius band. This can be seen intuitively because one can make a Möbius band out of paper by bending it then pasting the two ends with a half twist. Bending does not change the metric.

Ok, since the metric is a local concept and it isn't affected from global topology.

Mathematically, the Mobius band is the quotient space of the Euclidean plane by the action of the group of isometries generated by integer length translations in the x direction together with the transformation $(x+1/2,-y)$

So, is the Mobius band topology the same as the subspace topology from its "embedding" into $\mathbb R^3$ with standard topology ?

Btw, as discussed here, a path that goes around two times along the Mobius band should not count as a section (even though it is a continuous map from a real line segment into the Mobius vector bundle).

 

[lavinia]

Ok, since the metric is a local concept and it isn't affected from global topology.


So, is the Mobius band topology the same as the subspace topology from its "embedding" into $\mathbb R^3$ with standard topology ?

Btw, as discussed here, a path that goes around two times along the Mobius band should not count as a section (even though it is a continuous map from a real line segment into the Mobius vector bundle).

Right.If one thinks of a piece of paper as rigid in the sense that it can not be stretched then bending it does not change distances or angles locally.

Yes the topology is the same. Proof?

Right again. The circle that goes around two times intersects each line twice. A section is a continuous choice of one vector on each line.

 

[lavinia]

It is not true that a 1 form whose kernel is the tangent space to a foliation is always the differential of a function.

The 1 form on the Möbius band whose kernel is the tangent spaces to the vertical lines is an example but the same is also true of the I form whose kernel is the vertical lines on a cylinder. No need for the half twist, just bend the paper around.

If one integrates these 1 forms over one of the circles one does not get zero for an answer. But by Stokes Theorem if the form was a differential its integral would be zero.

Both of these examples apply verbatim to the Klein bottle and the torus. Each can be made by extending the covering group to include the integer translations in the y direction.

BTW: The torus and Klein bottle made in this way can not be made from a piece of paper. Their metrics are still Euclidean since translations in the y direction are also isometries of the Euclidean plane but they can not be embedded Euclidean 3 space. A theorem of Hilbert says that every closed surface in 3 space must have a point of positive Gauss curvature. Proof?

If one could bend paper in Euclidean 4 space then one could make a flat torus. Not sure about the flat Klein bottle.

Caveat: It is assumed in this aside that the embeddings are smooth. Basically this means that a unit normal varies smoothly as a mapping from the surface into the unit sphere. If the normal is assumed to be only continuous but is not differentiable then there is an incredible embedding of the flat torus in Euclidean 3 space.

 

[cianfa72]

It is not true that a 1 form whose kernel is the tangent space to a foliation is always the differential of a function.

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.

 

[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.