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I am reading "Information Geometry and its Application" by Shun-ichi Amari. The example of a sphere as a 2-dimensional manifold says that, and I quote:

A sphere is the surface of a three-dimensional ball. The surface of the earth is regarded as a sphere, where each point has a two-dimensional neighborhood so that we can draw a local geographic map on a flat sheet. The pair of latitude and longitude gives a local coordinate system. However, a sphere is topologically different from a Euclidean space and it cannot be covered by one coordinate system. At least two coordinate systems are required to cover it. If we delete one point, say the north pole of the earth, it is topologically equivalent to a Euclidean space. Hence, at least two overlapping coordinate neighborhoods, one including the north pole and the other including the south pole, for example, are necessary and they are sufficient to cover the entire sphere.

My question is: It seems to me that by specifying one coordinate system using a pair of longitude and latitude, we can uniquely describe all points (including the north pole and the south pole) in the sphere. Why does the text says that we need two coordinate neighborhoods to cover all points in the sphere? Do I misunderstand something?

Sorry for my ignorance and thanks in advance for your explanation!

user21820
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TNg
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    If one chart was enough, the sphere would be homeomorphic to an open subset of $\Bbb R^2$ (namely the domain of the chart), which is not the case, for instance because the sphere is compact.. – Berci May 11 '19 at 13:56
  • @Berci: You are probably right. A sphere is a closed set so we need at least 2 open sets to cover a closed set. These two open sets would then be homeomorphic to two Euclidean spaces. If you like, you can post this comment as an official answer and I will accept this answer. – TNg May 12 '19 at 12:50

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While the map (lat,lon) $\mapsto$ surface point is continuous, the inverse is not. The discontinuities lead to special parties on cruise ships passing the date line and to funny puzzles involving people walking south, then west, then north and arriving at their starting point.

If by coordinate system (aka. chart) we mean an open subset $U$ of $\Bbb R^2$ plus an open subset $V$ of the sphere plus a bijective map $\phi\colon U\to V$ such that both $\phi$ and $\phi^{-1}$ are continuous, then the latitude/longitude system does not fall into this category.

(On the other hand, if you are willing to sacrifice continuity in one direction, why not sacrifice continuity in the other direction as well? In that case, one can use a space-filling curve or just any bijection $\Bbb R\to \Bbb R^2$ to use only a single coordinate system with a single real coordinate.)

Hagen von Eitzen
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  • Thanks for your answer but I am not sure I get it. Why is the inverse map discontinuous? Given a point on the sphere with (lat, lon)=(phi, theta) coordinate, its corresponding 3-dimensional coordinate is (x,y,z) = (Rsin(theta)cos(90° - phi), Rsin(theta)sin(90° - phi), Rcos(theta) ) where R is the radius of the sphere. Similarly, given a point in the sphere with 3-dimensional coordinates (x,y,z), we can always compute (theta, phi), and this computation seems continuous because when (x,y,z) approaches (x',y',z'), (theta,phi) also approaches (theta', phi'). – TNg May 12 '19 at 12:43
  • Even if the notion of the number of coordinate systems to fully cover a manifold has something to do with the continuity of the map and its inverse between two spaces, why the sphere without the north pole can be covered by one Euclidean space (as stated in the text)? – TNg May 12 '19 at 12:43
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For a local chart on $S^2$ each point $p\in S^2$ covered by the chart has to be a "normal point" of the chart, i.e., a full two-dimensional neighborhood of $p$ has to lie in the interior of the corresponding chart domain on ${\mathbb R}^2$. Geographical longitude and latitude on $S^2$ forms a chart on $S^2\setminus\{{\rm north\ pole, south\ pole}\}$, whereby concerning longitude we still have the uncertitude with $\pm2k\pi$. Here we have restricted the latitude to $-{\pi\over2}<\theta<{\pi\over2}$, whereby $\theta=0$ is the equator. Adding $\theta=\pm{\pi\over2}$ to the chart domain in ${\mathbb R}^2$ does not create north pole and south pole as legal chart points.

You can invent charts covering the "larger north hemisphere" from the north pole inclusive until some negative latitude south of the equator, and similarly the "larger southern hemisphere". Both these "larger hemispheres" are homeomorphic to two-dimensional discs. They intersect in an annular band around the equator.

Christian Blatter
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  • This is a good intuitive explanation. But after I read the comment of @Berci, I believe the core thing here for the explanation is about how many open sets we can cover the sphere, thus the number of coordinate systems needed. – TNg May 12 '19 at 13:00
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If $S^2$ could be covered by a single chart, it would be homeomorphic to an open subset of the plane. On the other hand, $S^2$ is compact, so by the Heine-Borel theorem, you get a contradiction.

Zbigniew Fiedorowicz
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