Manifold hypersurface foliation and Frobenius theorem

[jbergman]

Just to avoid confusion, your function $f$ is given in $(U_2,\phi)$ chart as $f(u,v)=v$. The level set of $f(u,v)= \text{c}$ is the set of points $A \cup B$ where $$A = \begin {cases} (x, c) & \text{if } 0< x< \frac12 \\ (x, -c) & \text{if } \frac12 < x < 1 \end {cases}$$ and $B=[0,c]=[1,-c]$.

However such a level set is not a single leaf of the foliation given by "circles" going around twice around the strip for $c \neq 0$.

Good point.

 

[jbergman]

Yes, your function was given by $f(u,v)=v^2$ in $(U_1,\theta)$ and $(U_2,\phi)$ charts. Its level sets give the foliation's curves other than the 0-leaf. The kernel of gradient/1-form $df$ gives at each point of both coordinate patches (other than the 0-leaf) the 1D distribution associated to that foliation.

I don't think so. As I mentioned before, I don't believe it smooth as a global function on $M$. It's derivative in the covering space is $\frac{\partial f(x,y)}{\partial y} = 2y$.

So we can see that $\frac{\partial f([x,y])}{\partial y} = 2y$ is not smooth as we approach the point $p= [1, 1]$ from the right and the left. From the right the limit is 2 and from the left it is -2.

 

[cianfa72]

So we can see that $\frac{\partial f([x,y])}{\partial y} = 2y$ is not smooth as we approach the point $p= [1, 1]$ from the right and the left. From the right the limit is 2 and from the left it is -2.

Sorry, the point $p=[1,1] = [0,-1]$ on the Mobius strip in $(U_2,\phi)$ chart has coordinates $(u=1/2,v=-1)$. Therefore whether we approach it from the right or from the left the value of the derivative (evaluated in that chart) is the same.

Edit: perhaps the point is that in $(U_1, \theta)$ chart as we approach to $p=[1,1]=[0,1]$ from the right and from the left the limits of the derivative are different (note, however, that $(U_1,\theta)$ chart doesn't cover the identified edges of the strip).

 

[WWGD]

I think it's too hard to explain in a forum, but basically there is a well known foliation of the Mobius strip consisting of loops except that you loop around twice before returning to the starting point except for the base loop.

My point was that I believe your initial conclusion was too strong. I am not 100% sure, but I don't think you can always describe a single leaf of a foliation as a level set of a function defined in a neighborhood of the entire leaf. I think that is only true locally.

In the context of GR and spacetime the situation may be simpler since the topology is $R^4$.

Still, the topology on spacetime , IIRC $\mathbb R^3 \times t$, where $t$ is time, and this isn't topologically equivalent to Euclidean topology on $\mathbb R^4$, if that's what you meant.

 

[WWGD]

I take it like this: slice up the Moebius strip with planes parallel to the base. One gets a foliation via circles that go around twice all but at the center. Such a foliation is given as the level set of a global function $t$ defined over the entire strip/manifold (the height from center circle).

There is not, however, a global chart for the stript (at least two). Therefore your point is that there is also no globally defined non-zero differential 1-form $\omega$ describing the distribution (as its kernel).

If there was a single chart, the Mobius Strip would be a 1-manifold.

 

[jbergman]

Still, the topology on spacetime , IIRC $\mathbb R^3 \times t$, where $t$ is time, and this isn't topologically equivalent to Euclidean topology on $\mathbb R^4$, if that's what you meant.

Minkowski space os topologically equivalent to $\mathbb R^4$ and even diffeomorphic to it. It just has a different metric. In GR, I always assumed that this was also the case...

 

[jbergman]

Sorry, the point $p=[1,1] = [0,-1]$ on the Mobius strip in $(U_2,\phi)$ chart has coordinates $(u=1/2,v=-1)$. Therefore whether we approach it from the right or from the left the value of the derivative (evaluated in that chart) is the same.

Edit: perhaps the point is that in $(U_1, \theta)$ chart as we approach to $p=[1,1]=[0,1]$ from the right and from the left the limits of the derivative are different (note, however, that $(U_1,\theta)$ chart doesn't cover the identified edges of the strip).

My intuition could be wrong as you say I didn't actually calculate this, but I defined $f([x,y])=y^2$ on the manifold $M$. It's definition in a chart would be $f \circ \phi^{-1}$. Did you actually work out this computation.

 

[WWGD]

Minkowski space os topologically equivalent to $\mathbb R^4$ and even diffeomorphic to it. It just has a different metric. In GR, I always assumed that this was also the case...

Space-time is a Lorentzian Manifold, Euclidean 4-space is not ( I don't have a good proof of this at the moment, other than Timespace is not technically metrizable, as we may have d(x,y)=0 without having x=y). The metrics also differ, the one in Euclidan Space being positive-definite, while Minkowski, is not, though that's more about Geometry than about Topology. Maybe @PeterDonis can chime in on them not being different.

 

[cianfa72]

My intuition could be wrong as you say I didn't actually calculate this, but I defined $f([x,y])=y^2$ on the manifold $M$. It's definition in a chart would be $f \circ \phi^{-1}$. Did you actually work out this computation.

Since the point $[0,-1]=[1,1]$ is covered from $(U_2,\phi)$ alone, it makes no sense calculate the derivate of the $f$ representative at that point in $(U_1,\theta)$ chart.

As said before, the representative of your $f([x,y])= y^2$ in $(U_2,\phi)$ chart has derivative $2v$ along $v$, therefore it is $C^{\infty}$ everywhere in the $U_2$ chart's domain.

Btw, in $(U_1,\theta)$ chart if one calulates the directional derivative of $f$ representative near $[0,-1]$ from the right in positive $v$ direction, he gets the same value as the directional derivative near $[1,1]$ from the left in negative $v$ direction.I think it makes sense due to the edge identification that flips the sign along $y$.

 

[jbergman]

Since the point $[0,-1]=[1,1]$ is covered from $(U_2,\phi)$ alone, it makes no sense calculate the derivate of the $f$ representative at that point in $(U_1,\theta)$ chart.

As said before, the representative of your $f([x,y])= y^2$ in $(U_2,\phi)$ chart has derivative $2v$ along $v$, therefore it is $C^{\infty}$ everywhere in the $U_2$ chart's domain.

Btw, in $(U_1,\theta)$ chart if one calulates the directional derivative of $f$ representative near $[0,-1]$ from the right in positive $v$ direction, he gets the same value as the directional derivative near $[1,1]$ from the left in negative $v$ direction.I think it makes sense due to the edge identification that flips the sign along $y$.

Yes. This looks correct. My intuition was wrong and shows the dangers of trying to jump to conclusions without calculations.

And you are also correct in that the 0-leaf that only takes 1 loop to traverse appears to be the one that is difficult to find a function such that it is a level set of and the corresponding 1-form is non-zero.

Interestingly, this may be related to the fact that when you cut a mobius strip along the middle 0-leaf you end up with a normal orientable cylinder that is twice as long!

 

[cianfa72]

And you are also correct in that the 0-leaf that only takes 1 loop to traverse appears to be the one that is difficult to find a function such that it is a level set of and the corresponding 1-form is non-zero.

Ok, so the main result is that there is a global smooth function $f([x,y]) = y^2$ defined on the Mobius strip s.t. its level sets define the foliation given by "circles" going around twice including the 0-leaf that only takes 1 loop. However there is not a global smooth function $f$ that gives the above foliation as its level sets and s.t. the corresponding 1-form $df$ is non-zero everywhere.

 

[jbergman]

Ok, so the main result is that there is a global smooth function $f([x,y]) = y^2$ defined on the Mobius strip s.t. its level sets define the foliation given by "circles" going around twice including the 0-leaf that only takes 1 loop. However there is not a global smooth function $f$ that gives the above foliation as its level sets and s.t. the corresponding 1-form $df$ is non-zero everywhere.

I would add that we haven't found a function with a non-zero defining form for the 0-leaf.

 

[cianfa72]

I would add that we haven't found a function with a non-zero defining form for the 0-leaf.

Consider the writing $f([x,y]) = y$. As function it isn't well-defined at points $[0,y]=[1,-y]$ (indeed, which is its value at that points ?)

Define the following $f$ as $$f([x,y]) = \begin {cases} y & \text{if } x \neq 1 \\ - y & \text{if } x=1 \end {cases}$$
The representative of $f$ in $(U_1,\theta)$ chart is $f (u,v) = v$. In $(U_2,\phi)$ it is defined from $$f(u,v) = \begin {cases} v & \text{if } \frac 12 \le u < 1 \\ -v & \text{if } 0 < u < \frac12 \end {cases}$$ The level set $f=0$ defines the 0-leaf, however $f$ isn't differentiable at $x = 0$ (i.e. $u=1/2$).

 

[jbergman]

Consider the writing $f([x,y]) = y$. As function it isn't well-defined at points $[0,y]=[1,-y]$ (indeed, which is its value at that points ?)

Define the following $f$ as $$f([x,y]) = \begin {cases} y & \text{if } x \neq 1 \\ - y & \text{if } x=1 \end {cases}$$
The representative of $f$ in $(U_1,\theta)$ chart is $f (u,v) = v$. In $(U_2,\phi)$ it is defined from $$f(u,v) = \begin {cases} v & \text{if } \frac 12 \le u < 1 \\ -v & \text{if } 0 < u < \frac12 \end {cases}$$ The level set $f=0$ defines the 0-leaf, however $f$ isn't differentiable at $x = 0$ (i.e. $u=1/2$).

It's not even continuous at $u = 1/2$.

 

[WWGD]

It's not even continuous at $u = 1/2$.

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

 

[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.