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The Mercator projection is the unique conformal projection which maps parallels to horizontal lines and meridians to vertical lines. There is an interesting hemispheric analogue of this, which maps parallels to parallels and meridians to meridians, which is called the Gilbert's two-world projection.

Question 1: does there also exist an equal-area map from the sphere to the hemisphere which maps parallels to parallels and meridians to meridians?

Question 2: if so, would either the equal area or conformal map preserving meridians and parallels be unique, possibly given some extra details (such as a central point on the sphere that maps to itself)?

My thinking is that one possibility would be to just naively halve the longitudes and preserve the latitudes, choosing some arbitrary central point to preserve along the equator. It is pretty easy to see this will be equal area, and I thought that this would probably be the unique solution, but then I thought you could also try to halve the latitudes and preserve longitudes, such that one of the two poles is preserved, basically mapping the entire sphere onto the northern or the southern hemisphere. This wouldn't preserve areas as a small circle around the south pole would become a huge ring around the equator, but perhaps there is some way to adjust latitudes such that the result is equal area and meridians and parallels are preserved. That would be three solutions: one equatorial and two polar. Are these the only three?

Mike Battaglia
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tl;dr: "Yes" and "no" respectively.

A mapping from a unit sphere to a unit hemisphere naturally cannot preserve area (because the hemisphere has only half the area), but for posterity let's agree to use "equal-area" to mean area-halving.

If we fix an arbitrary hemisphere $H$ on the unit sphere, there exist infinitely many equal-area mappings to $H$. These include:

  • The longitude-halving mappings mentioned in the question. There are infinitely many of these because we can choose an arbitrary axis $A$ whose endpoints lie on the boundary of $H$, choose an arbitrary "date line" determined by the "poles" (endpoints of $A$), halve longitude with respect to this date line, and finally rotate the image hemisphere about $A$ onto $H$.
  • There is another equal-area mapping to $H$: Here let $A$ denote the axis for which $H$ is the northern hemisphere, and let $C$ be the cylinder with axis $A$ and circumscribed about the sphere. Projection away from $A$ is an area-preserving mapping from the sphere to $C$. (Mathematicians sometimes call this fact Archimedes' (hat box) theorem.) There is an obvious equal-area mapping from $C$ to the half-cylinder corresponding to the northern hemisphere; the composition is an equal-area mapping from the sphere to $H$. This is not exactly "latitude-halving", but is qualitatively similar. Technically we can then rotate about $A$, again obtaining infinitely many mappings.

These mappings send latitudes to latitudes and longitudes to longitudes if and only if $A$ is the geographical axis, but if I understand the geographic constraints that still leaves a one-parameter family of ambiguity (the choice of date line) even if $H$ is a fixed hemisphere bounded by two "antipodal" longitudes.

(If it matters, there is no conformal mapping from a once- or twice-punctured sphere to a hemisphere, e.g., because of Liouville's theorem in complex analysis or by the Koebe-Poincaré uniformization theorem.)

Andrew D. Hwang
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    Thanks. Before I get into the equal area thing, are you saying Gilbert's map (which should be from a once punctured sphere to a hemisphere) cannot be conformal? Because that is what it is claimed to be. One can build it from the usual Mercator projection of the globe onto a cylinder: simply place two Mercator projections side by side, then roll back into a cylinder and project back conformally onto the sphere. Here is an article talking about the conformal projection used: https://www.researchgate.net/publication/279930504_Gilbert_Two-World_Projection – Mike Battaglia Aug 06 '22 at 20:47
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    I was using "conformal" like a mathematician, which is to say, being Extremely Fussy about domains. :) By your and Davies' descriptions, Gilbert's map is conformal from a slit sphere (entire date line removed) to a hemisphere. <> Something is irksome about placing two Mercator projections side-by-side then mapping back to the sphere, however: Loosely, because the width is halved, the height must also be halved to ensure conformality, i.e., regions are pushed toward the equator. This looks incompatible with Davies' image (but is mathematically consistent/correct). – Andrew D. Hwang Aug 06 '22 at 21:06
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    I take back the "irksome" claim: On further inspection, Davies' continents do look to be pushed toward the equator. – Andrew D. Hwang Aug 06 '22 at 21:09
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    Thanks, I think it makes sense. Does something like this exist for half of an ellipsoid? Do we have conformal or equal area mappings? Kind of interested in the situation where the ellipsoid is twice as long as it is tall, for instance. – Mike Battaglia Aug 06 '22 at 22:12
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    Conformal and equal-area mappings exist for half a non-spherical ellipsoid, but to my knowledge no explicit example is known, and latitudes and longitudes are not preserved. Relevant search terms include "uniformization theorem," "isothermal parameters," and "symplectomorpism," but the results I know of are all pretty technical (e.g., cannot be shown using just multivariable calculus). – Andrew D. Hwang Aug 07 '22 at 00:54
  • Do you mean equal area and conformal at the same time? Just equal area and parallel/meridian preserving seems pretty easy to build - just take the equal area map from before to the hemisphere and squish the entire thing along the geographic axis by a factor of 2. This isn't equal area, but I am pretty sure the area distortion is just a function of the latitude, so some adjustment of latitudes ought to be able to uniformize it. Then you should have an equal area and meridian/parallel preserving map. Not quite sure how to build any kind of conformal map though... – Mike Battaglia Aug 07 '22 at 08:19
  • "Equal area" and "conformal" are being considered separately. :) <> By an "ellipsoid" I was thinking of three distinct axis lengths. For a spheroid (ellipsoid of rotation, i.e. with two equal axes), your scheme works, though the necessary squishing involves inverting definite integrals that are probably not in closed form. The mathematical scheme is outlined here for a torus; I'd be happy to translate details as needed. – Andrew D. Hwang Aug 07 '22 at 13:02
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    Oh yes, meant spheroid. I will take a look... I have also moved this to another question, one which is hopefully general enough to answer these kinds of things, if you have any thoughts on that: https://math.stackexchange.com/q/4507872/52694 – Mike Battaglia Aug 07 '22 at 17:33