Global coordinate chart on a 2-sphere

[cianfa72]

TL;DR Summary

Formal proof that it does not exist a global coordinate chart on a 2-sphere

Hi,

I know there is actually no way to set up a global coordinate chart on a 2-sphere (i.e. we cannot find a family of 2-parameter curves on a 2-sphere such that two nearby points on it have nearby coordinate values on $\mathbb R^2$ and the mapping is one-to-one).

So, from a formal mathematical point of view, how to prove it ? Just because there is not a (global) homeomorphism between the 2-sphere and the Euclidean plane $\mathbb R^2$ ? Thanks.

 

[fresh_42]

You could use the theorem of Borsuk-Ulam. It says that given a continuous function $f:\mathbb{S}^2\rightarrow \mathbb{R}^2$ there is always a point $x\in \mathbb{S}^2$ such $f(x)=f(-x)$ which makes bijectivity impossible.

 

[cianfa72]

You could use the theorem of Borsuk-Ulam. It says that given a continuous function $f:\mathbb{S}^2\rightarrow \mathbb{R}^2$ there is always a point $x\in \mathbb{S}^2$ such $f(x)=f(-x)$ which makes bijectivity impossible.

ok, so the point is: take the 2-sphere $\mathbb{S}^2$ with an atlas. Define a global continuous function from $\mathbb{S}^2$ to $\mathbb{R}^2$. In force of that theorem it cannot be bijective (one to one).

 

[fresh_42]

ok, so the point is: take the 2-sphere $\mathbb{S}^2$ with an atlas. Define a global continuous function from $\mathbb{S}^2$ to $\mathbb{R}^2$. In force of that theorem it cannot be bijective (one to one).

Yes. Simply take your chart $\mathbb{S}^2\rightarrow \mathbb{R}^2$ as $f.$ As $f$ cannot be injective, you will need at least two charts.

 

[cianfa72]

Yes. Simply take your chart $\mathbb{S}^2\rightarrow \mathbb{R}^2$ as $f.$ As $f$ cannot be injective, you will need at least two charts.

ok, honestly it seems to me an egg & chicken problem: to define a global function as continuous on $\mathbb{S}^2$ we actually need -- let me say "in advance" -- an atlas for $\mathbb{S}^2$, don't you ?

 

[fresh_42]

ok, honestly it seems to me an egg & chicken problem: to define a global function as continuous on $\mathbb{S}^2$ we actually need -- let me say "in advance" -- an atlas for $\mathbb{S}^2$, don't you ?

I had an indirect proof in mind. Assume there is a $1$-chart atlas, then Borsuk-Ulam leads to a contradiction. Hence, there must be at least two charts.

Two charts do the job. How would you prove it can't be less? (I mean the structure of a possible proof, not a proof itself.)

 

[martinbn]

The sphere is compact! It cannot be homeomeprhpic to any open set in the plane.

 

[martinbn]

ok, honestly it seems to me an egg & chicken problem: to define a global function as continuous on $\mathbb{S}^2$ we actually need -- let me say "in advance" -- an atlas for $\mathbb{S}^2$, don't you ?

No, you do not need an atlas. You only need a topology.

 

[cianfa72]

Assume there is a $1$-chart atlas, then Borsuk-Ulam leads to a contradiction. Hence, there must be at least two charts.

I believe the contraddiction is as follows: suppose $(\mathbb S^2, \phi)$ is the 1-chart atlas. Take $f= \phi$, then $\phi^{-1} \circ f = I$ hence $f$ globally defined is continuous by definition. It contraddicts that theorem since it is bijective by definition of global chart.

Two charts do the job. How would you prove it can't be less? (I mean the structure of a possible proof, not a proof itself.)

It can'be less since the contradiction above, no ?

 

[cianfa72]

The sphere is compact! It cannot be homeomeprhpic to any open set in the plane.

Surely, that was actually my point !

 

[martinbn]

Surely, that was actually my point !

You had written it as a question why are they not homeomorphic. I just pointed out why: one is compact the other is not. Same reason why any compact manifold cannot be covered by a single chart.

 

[cianfa72]

ok, thank you. Now the next question could be: why two charts do the job ?

 

[martinbn]

ok, thank you. Now the next question could be: why two charts do the job ?

Because you can find two charts that do the job.

 

[cianfa72]

Because you can find two charts that do the job.

ok, so the proof of it is actually a "constructive" proof ?

 

[fresh_42]

ok, so the proof of it is actually a "constructive" proof ?

You still have to prove compactness and its preservation under homeomorphisms. Compactness might be proven with Heine-Borel, which usually uses indirect parts, too.

 

[martinbn]

ok, so the proof of it is actually a "constructive" proof ?

Stereographic projecton covers the whole sphere but a point. So two from different poles will do.

 

[cianfa72]

Sorry for the stupid question: take a 2-sphere embedded in 3D euclidean space with cartesian coordinates $(x,y,z)$.

Slice up the 2-sphere with planes $x=const$ and $z=const$ and assign coordinates $(s,t)$ to each of the resulting family of curves on the 2-sphere. This way each point of the 2-sphere should be mapped to just one $(s,t)$ tuple. Unlike the latitude/longitude coordinate system on the Earth there should be no singularity at all at the poles.

Does it actually make sense ?

 

[fresh_42]

No, you do not need an atlas. You only need a topology.

Íf you consider $\mathbb{S}^2$ as manifold, and not as a subset of $\mathbb{R}^3$, which is the subject here, then you do need an atlas, and the atlas defines the topology, not the other way round. This also makes compactness not evident.

 

[martinbn]

Íf you consider $\mathbb{S}^2$ as manifold, and not as a subset of $\mathbb{R}^3$, which is the subject here, then you do need an atlas, and the atlas defines the topology, not the other way round. This also makes compactness not evident.

What is your definition of a manifold? What is your definition of the two-sphere? In any case for a continuous map you only need topology, no additional structure.

 

[fresh_42]

What is your definition of a manifold? What is your definition of the two-sphere? In any case for a continuous map you only need topology, no additional structure.

https://ncatlab.org/nlab/show/manifold

Definitions 2.1. to 2.3.

 

[martinbn]

https://ncatlab.org/nlab/show/manifold

Definitions 2.1. to 2.3.

In these definitions $M$ is a topological space. The topology comes before the atlas.

 

[cianfa72]

What do you think about my post#17, is it really a one-to-one map ?
Maybe the problem is that even if it is one-to-one it is not (bi) continuous however (i.e. it is not a global homeomorphism)

Thank you.

 

[martinbn]

Sorry for the stupid question: take a 2-sphere embedded in 3D euclidean space with cartesian coordinates $(x,y,z)$.

Slice up the 2-sphere with planes $x=const$ and $z=const$ and assign coordinates $(s,t)$ to each of the resulting family of curves on the 2-sphere. This way each point of the 2-sphere should be mapped to just one $(s,t)$ tuple. Unlike the latitude/longitude coordinate system on the Earth there should be no singularity at all at the poles.

Does it actually make sense ?

If you slice with a plane $x=const$ you get a circle, same with $z=const$. Those two circles can intersect at two points. In your construction those two points will have the same coordinates. For exanple $x=0$ and $z=0$ will intersect at the points $(0,1,0)$ and $(0,-1,0)$ so they have to have coordinates $(s,t)=(0,0)$.

 

[cianfa72]

If you slice with a plane $x=const$ you get a circle, same with $z=const$. Those two circles can intersect at two points. In your construction those two points will have the same coordinates.

You're right, I was too sloppy

For each circle on $x=const$ planes assign different $s$ coordinate values to each half-circle of it. To get a bijective map, points on the circle you get slicing the 2-sphere with the $y=const$ plane along the poles, belong to one alone of the two half-circles above.

Basically the $t=const$ coordinate lines are the latitude coordinate lines while the $s=const$ coordinate lines are akin to the longitude coordinate lines, however they do not result in a singularity at poles.

 

[cianfa72]

Any feedback about my last post ? Thanks

 

[martinbn]

Any feedback about my last post ? Thanks

To me it is not clear what the map is. You just say pick different values, how? In any case what you end up with is not a chart, because the image is not an open set.

 

[cianfa72]

To me it is not clear what the map is. You just say pick different values, how? In any case what you end up with is not a chart, because the image is not an open set.

Let's try to visualize it in 3D space. Suppose the 2-sphere is placed in 3D such that the plane $x=0$ is tangent to it on the "left side". Then, starting from the left, for each half-circle on $x=c , c>0$ planes assign coordinate $s=c$ to one half-circle and $s=-c$ to the other (proceed this way up to the 2-sphere "right side").

As above I believe it should result in a one-to-one map, however as you pointed out the image on $\mathbb R^2$ should be not an open set in $\mathbb R^2$ standard topology.

The goal was try to build a one-to-one map for the 2-sphere. As we know, however, it can never be a (global) chart for it.

Make sense ? Thank you in advance.

 

[WWGD]

What do you think about my post#17, is it really a one-to-one map ?
Maybe the problem is that even if it is one-to-one it is not (bi) continuous however (i.e. it is not a global homeomorphism)

Thank you.

But if it's not 1-1, its inverse is not even defined.

 

[cianfa72]

But if it's not 1-1, its inverse is not even defined.

Why not ? In post #27 I specified better the map: for each point in $\mathbb R^2$ there is 0 or 1 point on the 2-sphere.

 

[fresh_42]

Why not ? In post #27 I specified better the map: for each point in $\mathbb R^2$ there is 0 or 1 point on the 2-sphere.

Have a look at the stereographic projection. That is how it is usually done.

 

[cianfa72]

Have a look at the stereographic projection. That is how it is usually done.

It seems to me it is not bijective because for example the North Pole is not mapped on any point of the plane. Mine was just a "proposal" of a bijective map on its image nevertheless it is not a (global) chart.

 

[martinbn]

Mine was just a "proposal" of a bijective map on its image nevertheless it is not a (global) chart.

Bijection is not hard if it is not continuous.

 

[cianfa72]

Bijection is not hard if it is not continuous.

So do you agree it is actually a bijection on its image ?

 

[fresh_42]

It seems to me it is not bijective because for example the North Pole is not mapped on any point of the plane. Mine was just a "proposal" of a bijective map on its image nevertheless it is not a (global) chart.

Yes, that is the reason why we need two charts. One, if we allow infinity to be the image of the North Pole.

 

[jbergman]

Summary:: Formal proof that it does not exist a global coordinate chart on a 2-sphere

Hi,

I know there is actually no way to set up a global coordinate chart on a 2-sphere (i.e. we cannot find a family of 2-parameter curves on a 2-sphere such that two nearby points on it have nearby coordinate values on $\mathbb R^2$ and the mapping is one-to-one).

So, from a formal mathematical point of view, how to prove it ? Just because there is not a (global) homeomorphism between the 2-sphere and the Euclidean plane $\mathbb R^2$ ? Thanks.

Another route to prove this would be the following.

If there is a single chart for the 2-sphere then we could use that chart to define a non-vanishing vector field for every point on the sphere, i.e., just a vector in the direction of one of the coordinate axes.

We know this is impossible by the hairy ball theorem.