Manifold: Why a sphere requires two $2$-dimensional coordinate systems to fully describe it, not just one?

Question

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!

Answer 1

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.)

Answer 2

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.

Answer 3

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.