I Manifold hypersurface foliation and Frobenius theorem

[cianfa72]

TL;DR Summary

About the foliation of smooth manifolds and Frobenius's theorem condition, including the existence of a globally defined function $t$ such that its level curves are the hypersurfaces of the foliation

Hi,
starting from this thread, I'd like to clarify some mathematical aspects related to the notion of hypersurface orthogonality condition for a congruence.

Let's start from a congruence filling the entire manifold (e.g. spacetime). The condition to be hypersurface orthogonal basically means that the 3D distribution orthogonal to the direction of the congruence's tangent vector field "evaluated" at each point (such a 3D distribution defines a 3D linear subspace within the tangent space "attached" at each point) is integrable.

If we call $\omega$ the differential 1-form such that its kernel defines the 3D distribution, the above condition about integrability boils down to the Frobenius's theorem -- namely the 3D distribution defined by $\omega$ is integrable iff $\omega \wedge d\omega=0$ everywhere.

Suppose the above condition holds true for a given vector field $V$ on the manifold. Then I believe the entire manifold is actually foliated by hypersurfaces (the integral submanifolds) that do not intersect each other in any point.

If the above is correct, why one cannot always find a smooth function $t$ globally defined on the manifold such that its level curves are the hypersurfaces of the foliation ?

 

[cianfa72]

I found an interesting insight on John Lee's book "Introduction to Smooth Manifold". In chapter 19 section Foliations he states the Global Frobenius Theorem, namely:

For any involutive distribution D on a smooth manifold M, the collection of all maximal connected integral manifolds of D forms a foliation of M.

A foliation of $M$ is actually a partition of $M$ into disjoint, connected, maximal immersed submanifolds.

I think the point I raised in the OP has the following answer: it is true that any single hypersurface of the foliation can be given as the level set of a smooth function for a given constant value of the function itself, however the point is that, in general, might not exist a global function such that all its level sets w.r.t. all its constant values are the desired hypersurfaces of the foliation.

Only in special cases there exists such a globally defined smooth function $t$ (it also means there exists a smooth function $h$ defined on $M$ such that globally $\omega = hdt$ ).

Does it make sense ? Thanks.

 

[jbergman]

I found an interesting insight on John Lee's book "Introduction to Smooth Manifold". In chapter 19 section Foliations he states the Global Frobenius Theorem, namely:


A foliation of $M$ is actually a partition of $M$ into disjoint, connected, maximal immersed submanifolds.

I think the point I raised in the OP has the following answer: it is true that any single hypersurface of the foliation can be given as the level set of a smooth function for a given constant value of the function itself, however the point is that, in general, might not exist a global function such that all its level sets w.r.t. all its constant values are the desired hypersurfaces of the foliation.

Only in special cases there exists such a globally defined smooth function $t$ (it also means there exists a smooth function $h$ defined on $M$ such that globally $\omega = hdt$ ).

Does it make sense ? Thanks.

I don't think so. Looking at Lee's text most of the theorems are local in nature. In other words, only in a local neighborhood of each point of p are the defining forms non-zero. See proposition 19.8.

 

[cianfa72]

I don't think so. Looking at Lee's text most of the theorems are local in nature. In other words, only in a local neighborhood of each point of p are the defining forms non-zero. See proposition 19.8.

Yes, Lee's proposition 19.8 is an involutivity criterion for locally defined differential forms defining a distribution.

For the $n$ dimensional case, assuming an integrable $n-1$ dimensional distribution, then only in special cases will exist a global smooth function $t$ such that the integral (immersed) submanifolds of the foliation are level sets of. In that case one can define a non-zero everywhere differential 1-form $\omega$ having as kernel that distribution.

 

[jbergman]

Yes, Lee's proposition 19.8 is an involutivity criterion for locally defined differential forms defining a distribution.

For the $n$ dimensional case, assuming an integrable $n-1$ dimensional distribution, then only in special cases will exist a global smooth function $t$ such that the integral (immersed) submanifolds of the foliation are level sets of. In that case one can define a non-zero everywhere differential 1-form $\omega$ having as kernel that distribution.

Just to give a specific example, an interesting case to think about is the Mobius strip which can be foliated by circles, apparently, that go around twice except for the center circle. In this case, even for a single circle/leaf, I don't there is a non-singular one form for the entire leaf.

 

[cianfa72]

Just to give a specific example, an interesting case to think about is the Mobius strip which can be foliated by circles, apparently, that go around twice except for the center circle. In this case, even for a single circle/leaf, I don't there is a non-singular one form for the entire leaf.

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).

 

[jbergman]

I take it as follows: 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 on the entire strip (the height from center circle).

No, I am disputing this point. I do not believe it is the level set of a $t$ defined on the entire strip.

 

[cianfa72]

No, I am disputing this point. I do not believe it is the level set of a $t$ defined on the entire strip.

Why not ? The "signed height" w.r.t. the center circle is a function defined on the entire manifold (even though its representation is given in at least two charts covering the strip).

 

[jbergman]

Why not ? The "signed height" w.r.t. the center circle is a function defined on the entire manifold (even though its representation is given in at least two charts covering the strip).

I am not sure what you are even saying here. The mobius is not a trivial bundle, i.e. it isn't equal to $S^1\times R$. There doesn't exist a smooth section of the vector bundle that is non-zero for every point in the base manifold of $S^1$.

 

[cianfa72]

The mobius is not a trivial bundle, i.e. it isn't equal to $S^1\times R$. There doesn't exist a smooth section of the vector bundle that is non-zero for every point in the base manifold of $S^1$.

Sorry, maybe I didn't get your point. How do you plan to foliate the entire Moebius strip using circles going around?

 

[jbergman]

Sorry, maybe I didn't get your point. How do you plan to foliate the entire Moebius strip using circles going around?

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 $R^4$.

 

[jbergman]

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 $R^4$.

Ok, the simplest way for me to visualize this as follows. Let our mobius strip be $[0,1]\times[0,1]$ with the following equivalence relation, $(0,x) \sim (1, 1-x)$.

Then a leaf is basically a straight line across the square that returns to where it starts.

If we start at $(0, 1/2)$ we go straight across and when we get to $(1, 1/2)$ we are back where started.

However if we start at $(0, 1)$ and move across to $(1, 1)$, we end up at $(0, 0)$ by our equivalence relation and we now follow the bottom of the square to $(1, 0)$ which is equivalent to where we started. So if we follow "straight lines" across the square we will be on one of these "circles".

 

[cianfa72]

So if we follow "straight lines" across the square we will be on one of these "circles".

Ok yes. Using such "straight lines" or "circles" we can foliate the entire Mobius strip.

The next point is: does exist a global smooth function $t$ defined on the Mobius manifold such that the above foliation's leaves are level sets of ? By global smooth function I mean a function defined on all manifold's points such that it is smooth w.r.t. the differential structure given by an Atlas.

Btw a "circle" going around twice doesn't count as a smooth section, right ?

 

[jbergman]

Ok yes. Using such "straight lines" or "circles" we can foliate the entire Mobius strip.

The next point is: does exist a global smooth function $t$ defined on the Mobius manifold such that the above foliation's leaves are level sets of ? By global smooth function I mean a function defined on all manifold's points such that it is smooth w.r.t. the differential structure given by an Atlas.

Btw a "circle" going around twice doesn't count as a smooth section, right ?

Remember the square quotienting opposite sides with a flip I described before.

Imagine you draw a line across the square starting at (0, 1). This line has to end at (1,0) to be continuous as those two points are identified. How can you draw a line from (0,1) to (1,0) without crossing the line y=1/2 (the zero line in this description)?

 

[cianfa72]

How can you draw a line from (0,1) to (1,0) without crossing the line y=1/2 (the zero line in this description)?

As said in the other thread, that line must cross the y=1/2 line necessarily.

Coming back to main topic, we can foliate the Mobius strip using the "circle" y=1/2 and other "circles" going around twice around the strip. Such "circles" are continuous map into $\mathbb E$ and can be given as the level set of the global function $f=|y - 1/2|$ (e.g. $f=0$ gives the circle y=1/2). However this map is not smooth.

 

[jbergman]

As said in the other thread, that line must cross the y=1/2 line necessarily.

Coming back to main topic, we can foliate the Mobius strip using the "circle" y=1/2 and other "circles" going around twice around the strip. Such "circles" are continuous map into $\mathbb E$ and can be given as the level set of the global function $f=|y - 1/2|$ (e.g. $f=0$ gives the circle y=1/2). However this map is not smooth.

Yes, that works, but as you say it isn't smooth.

 

[cianfa72]

Yes, that works, but as you say it isn't smooth.

Ok, so there is no way to define a global smooth function $f$ on the Mobius strip such that the foliation is given through a differential 1-form $\omega = df$.

From a local point of view, what is a smooth function that gives locally as level set a piece of the base circle?

 

[cianfa72]

From a local point of view, what is a smooth function that gives locally as level set a piece of the base circle?

How can one write down (locally) the differential 1-form that give that foliation over the Mobius strip ? Any foliation, locally, can always be given as the level set of a smooth function.

 

[jbergman]

How can one write down (locally) the differential 1-form that give that foliation over the Mobius strip ? Any foliation, locally, can always be given as the level set of a smooth function.

I think this is a good hood exercise to try and work out.

 

[cianfa72]

I think this is a good hood exercise to try and work out.

The topology of Mobius strip is obtained as follows: take the square $[0,1]\times[0,1]$ with the subspace topology from $\mathbb R^2$. Then using the equivalence relation $(0,x) \sim (1, 1-x)$ one gets the Mobius strip with the quotient topology.

In any local chart around the center that wraps at the edges of the identification, I have no idea of which could be a locally defined smooth function such that its level sets give locally the foliation.

 

[jbergman]

The topology of Mobius strip is obtained as follows: take the square $[0,1]\times[0,1]$ with the subspace topology from $\mathbb R^2$. Then using the equivalence relation $(0,x) \sim (1, 1-x)$ one gets the Mobius strip with the quotient topology.

In any local chart around the center that wraps at the edges of the identification, I have no idea of which could be a locally defined smooth function such that its level sets give locally the foliation.

I don't want to do all the work for you. The first and probably most important step is to define your charts. Then it should be obvious how to define your one form in an open set for a single chart.

 

[cianfa72]

The first and probably most important step is to define your charts.

One can define two charts as here. In the first $U_1$ the identified edges are excluded. This chart maps the open set $U_1$ to the open square $0<x<1,0<y<1$. Then $\omega = dy$ is the one-form that gives locally the foliation, hence $y=$const gives the foliation's curves in that chart. The same for $U_2$ chart domain.

 

[cianfa72]

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.

Ok, so coming back to post #11, we found the neighborhood $U_1$ that includes only a part of the $y=1/2$ leaf given as the level set of such coordinate function $y$ in that local chart.

However in the neighborhood/chart $U_2$ the base/central "circle" is given by $y=1/2$ as well. Hence the complete base circle leaf is given globally as $y=1/2$. Am I wrong ?

 

[jbergman]

Ok, so coming back to post #11, we found the neighborhood $U_1$ that includes only a part of the $y=1/2$ leaf given as the level set of such coordinate function $y$ in that local chart.

However in the neighborhood/chart $U_2$ the base/central "circle" is given by $y=1/2$ as well. Hence the complete base circle leaf is given globally as $y=1/2$. Am I wrong ?

That looks correct. More complicated cases are the leaves that wrap around twice. Honestly, this is something I'd have to think about more carefully to answer more surely.

 

[jbergman]

That looks correct. More complicated cases are the leaves that wrap around twice. Honestly, this is something I'd have to think about more carefully to answer more surely.

I also thought about trying to define a leaf as the level sets of $f([x,y]) = y^2$ with the mobius strip centered at $y=0$. However, this isn't smooth on the Mobius strip as $df = 2ydy$ which has two possible values at $[0 ,y], y\ne 0$.

 

[cianfa72]

I also thought about trying to define a leaf as the level sets of $f([x,y]) = y^2$ with the mobius strip centered at $y=0$. However, this isn't smooth on the Mobius strip as $df = 2ydy$ which has to possible values at $[0 ,y], y\ne 0$.

Sorry, here with $[x,y]$ do you mean the equivalence class of the point $(x,y)$ with $x,y \in [0,1]$ ?

In other words is $f([x,y]) = y^2$ to be understood as a global function defined on the Manifold itself and not as its representative in a given local chart ?

 

[jbergman]

Sorry, here with $[x,y]$ do you mean the equivalence class of the point $(x,y)$ with $x,y \in [0,1]$ ?

I was working with a shifted mobius strip so $x \in [0,1]$ and $y \in \mathbb{R}$.

In other words is $f([x,y]) = y^2$ to be understood as global function defined on the Manifold itself and not as its representative in a given local chart ?

Yes.

 

[jbergman]

TLDR, I don't think we can make more than local statements in the general case, even for a single leaf.

 

[cianfa72]

I was working with a shifted mobius strip so $x \in [0,1]$ and $y \in \mathbb{R}$.

Ah ok, so in this case using $(x,y)$ as labels for points in $[0,1] \times \mathbb R$ the Mobius strip is defined by the equivalence relation $(0,y) \sim (1, -y)$ and $y=0$ is the central "circle" leaf.

Now, does your $df=2ydy$ expression make sense only in a local chart or is it actually a coordinate independent one-form object ?

 

[jbergman]

Ah ok, so in this case using $(x,y)$ as labels for points in $[0,1] \times \mathbb R$ the Mobius strip is defined by the equivalence relation $(0,y) \sim (1, -y)$ and $y=0$ is the central "circle" leaf.

Now, does your $df=2ydy$ expression make sense only in a local chart or is it actually a coordinate independent one-form object ?

Only in a specific chart because as mentioned above, $f$ isn't smooth as a global function.

 

[cianfa72]

Only in a specific chart because as mentioned above, $f$ isn't smooth as a global function.

Sorry to be pedantic. The differentiability of a function $f$ on a manifold is always defined w.r.t. a given differential structure on it.

The differential structure on the Mobius strip can be given by the two charts $U_1$ and $U_2$ as in post #22 where now $y \in \mathbb R$. In particular for $(U_2,\phi)$

$$(u,v) = \phi (x, y) = \begin {cases} (x + \frac12, y) & \text{if } x< \frac12 \\ (x - \frac12, -y) & \text{if } x > \frac12 \end {cases}$$
$U_2$ covers the identified edges, hence your point is that the representative of $f$ in this chart is not differentiable at $[0,y]=[1,y], y \neq 0$ -- i.e. at $u=1/2, v \neq 0$.

 

[cianfa72]

I also thought about trying to define a leaf as the level sets of $f([x,y]) = y^2$ with the mobius strip centered at $y=0$. However, this isn't smooth on the Mobius strip as $df = 2ydy$ which has two possible values at $[0 ,y], y\ne 0$.

Re-thinking about it, in $U_2$ chart your $f$ is indeed everywhere differentiable. I believe the point is that, however, $df=0$ at points with chart's coordinate $v=0$ (note the use of labels $u,v$ for the coordinates of $(U_2,\phi)$ just to avoid confusion). Hence at points that map to $v=0$ $df$ doesn’t define a 1D distribution.

 

[jbergman]

Re-thinking about it, in $U_2$ chart your $f$ is indeed everywhere differentiable. I believe the point is that, however, $df=0$ at points with chart's coordinate $v=0$ (note the use of labels $u,v$ for the coordinates of $(U_2,\phi)$ just to avoid confusion). Hence at points that map to $v=0$ $df$ doesn’t define a 1D distribution.

Right, but it is easy to define an f within a single chart. Just let $f(x,y) = y$. I was trying to define a function s.t. a single leaf (other than the 0-leaf) was a level set of it.

 

[cianfa72]

Right, but it is easy to define an f within a single chart. Just let $f(x,y) = y$.

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 $c \neq 0$.

 

[cianfa72]

I was trying to define a function s.t. a single leaf (other than the 0-leaf) was a level set of it.

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.

 

[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?