Global coordinate chart on a 2-sphere

[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.

 

[wrobel]

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

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 ? Thanks.

the sphere is a compact set and a chart is not

 

[cianfa72]

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.

Just to add some detail to your claim. Suppose there is a single chart for the 2-sphere: we can use such chart to define a (one-chart) differentiable atlas for the 2-sphere (by definition a chart is compatible with itself).

Now, by definition, a vector field over the 2-sphere equipped with that 'one-chart atlas differentiable structure' is assigned through differentiable functions as its components. Take as non-vanishing vector field the vector field having the constant function 1 as the component in one of the coordinate axes and zero otherwise.

As you pointed out that is actually impossible for the 2-sphere by the hairy ball theorem.

 

[WWGD]

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.

I haven't read the details but by Borsuk Up an, it is not possible.