# Global coordinate chart on a 2-sphere

• I
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
Mentor
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.

• jbergman, dextercioby and 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
Mentor
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.

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
Mentor
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.

• jbergman and WWGD
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.

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 ?

• fresh_42
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.

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.

Because you can find two charts that do the job.
ok, so the proof of it is actually a "constructive" proof ?

fresh_42
Mentor
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.

• cianfa72
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 ?

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fresh_42
Mentor
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.

• weirdoguy
fresh_42
Mentor
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
• weirdoguy
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.

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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)##.

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.

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Any feedback about my last post ? Thanks