I'm reading a book on differential geometry and some part of the introduction I do not understand but I'm curious to understand it. Maybe someone can try to explain those parts to me.

"Each point on the earth has a latitude and a longitude which determines its position. The coordinates are not global."

Q: What does it mean that the coordinates are not global? Intuitively, I would call a coordinate system 'global' if every point on earth can be described by them.

Then it continues: "There are four domains on the earth, bounded by the Equator and the Greenwich meridian, which are naturally coordinated. The points on the Equator and the Greenwich meridian admit two different coordinatizations."

I do not understand what the author has in mind when he talks about "naturally coordinated" domain. Further, the points on the Equator should have also unique coordinates, one latitude and a longitude coordinate. Am I wrong?

Any help will be very appreciated!

Best wishes

  • $\begingroup$ It's hard to say. Could you give the source of the book? In principle the only two points on the sphere with coordinates not entirely defined are the poles, for which there is no well defined longitude. $\endgroup$ – MyUserIsThis Jun 6 '15 at 18:14
  • 4
    $\begingroup$ Longitude/Latitude are a nice example. Notice that to hit the north/south pole in those coordinates there are infinitely many ways to do so. e.g. the North Pole has zero latitude but every longitude. The mapping from latitude/longitude to the sphere is not a bijection. $\endgroup$ – muaddib Jun 6 '15 at 18:14

There is a smooth "spherical coordinates parametrization" $\Phi$ from the closed rectangle $[-\pi, \pi] \times [0, \pi]$ onto the unit sphere $S^{2}$ defined by $$ \Phi(\theta, \phi) = (\cos\theta \sin\phi, \sin\theta \sin\phi, \cos\phi). $$ The restriction of $\Phi$ to the open rectangle $(-\pi, \pi) \times (0, \pi)$ is a diffeomorphism to its image. The boundary of the closed rectangle maps to the "International Date Line" $C$, a closed great semi-circle between the north and south poles.

In the complement $S^{2} \setminus C$ of the date line, $\theta$ and $\phi$ are "(smooth local) coordinates". They're not "global", however, because longitude is not continuous and uniquely-defined everywhere on the sphere. (Longitude has a double-valued ambiguity $\pm\pi$ along the date line, and is completely ambiguous at each pole. A bit more correctly, $\theta$ is smooth and single-valued on the complement of $C$, but has no continuous extension to any point of $C$.)

(Incidentally, there's a technical definition of a (local) coordinate system on the sphere, and it turns out there exists no single coordinate system defined on all of $S^{2}$.)

The second assertion is more puzzling (to me). The complement of the equator comprises the northern and southern hemispheres; the complement of the great circle containing the Greenwich meridian comprises the eastern and western hemispheres. Those are presumably the "four domains" the author mentions.

If that's correct, "two different coordinatizations" might mean: A point on the equator is both $0$ degrees north and $0$ degrees south; a point on the Greenwich meridian is both $0$ degrees east and west; a point on the date line is both $180$ degrees east and west.

The main problem is, longitude is in no sense a coordinate at the poles, so I see no geographical sense in which the northern and southern hemispheres are "naturally coordinated". (Certainly there exist perfectly good coordinate systems at the poles, but longitude is not one of the coordinates.)

It's possible the author had in mind that each hemisphere can be expressed mathematically as a graph of a function whose domain is a disk, but even that isn't a trouble-free interpretation:

  • If the disks are open, the four hemispheres do not cover the entire sphere. (The points on the equator at longitude $0$ and $\pi$ are omitted.)

  • If the disks are closed, this description fails to satisfy the technical conditions for a "coordinate system" at the boundary of each disk.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.