Fairly simple question: does there exist an area-preserving map from the hyperbolic plane to the Euclidean plane?

If not, does there exist an area-preserving map from an arbitrarily large subset of the hyperbolic plane, to an arbitrarily large subset of the Euclidean plane?

If so, what does the map look like? It would basically be similar to the "Mollweide projection."

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    $\begingroup$ But of course the Mercator projection is not area-preserving. $\endgroup$ – Lubin Aug 7 at 0:47
  • $\begingroup$ Oops, meant Mollweide! $\endgroup$ – Mike Battaglia Aug 7 at 1:15
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    $\begingroup$ I think you should find roguetemple.com/z/hyper/models.php interesting -- it includes azimuthal and cylindrical equi-area and hyperbolic cosinusoidal (as explained in Magma's answer) and many other hyperbolic projections. $\endgroup$ – Zeno Rogue Aug 7 at 19:03
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    $\begingroup$ I played that game for hours last night! I assume you made it? Really great, and I did see the equal-area projection. I have so many questions about this game... $\endgroup$ – Mike Battaglia Aug 7 at 21:40
  • $\begingroup$ Thanks! Probably the best place to ask questions specifically about HyperRogue is our Discord server: discord.gg/8G44XkR $\endgroup$ – Zeno Rogue Aug 7 at 23:03

There does in fact exist an area-preserving map, as demonstrated in this video at 11:20: the Lambert azimuthal equal-area projection.

The idea is that you take polar coordinates of the hyperbolic plane and map them to polar coordinates of the euclidean plane via a map $(r, \theta) \mapsto (f(r), \theta)$ where $f$ is chosen such that area is preserved.

Let's derive $f$:

Let $h(r)$ be the area of a hyperbolic ball of radius $r$, and $e(r)$ be the area of a euclidean ball of radius $r$.

Assuming that the hyperbolic plane has curvature $-1$, it is true that $h(r) = 2\pi(\cosh(r)-1)$ and $e(r) = \pi r^2$.

We want to ensure $h(r) = e(f(r))$. Substitution and rearrangement yields $f(r) = \sqrt{2\cosh(r)-2}$.

Another area-preserving map, which is analogous to the sinusoidal projection:

Pick any geodesic $g$ in the hyperbolic plane, and mark a special point $O$ on it.

For any point $P$ on the hyperbolic plane, project $P$ perpendicularly onto $g$ to obtain a point $Q$. Let $x$ denote the signed distance $OQ$, let $y$ denote the signed distance $QP$.

Then the map that takes $P$ to the point $(x \cosh y, y)$ is equal-area.

The similarity is obtained as such: If you replace the hyperbolic plane with a sphere, and let $g$ be the equator in particular, and replace $\cosh$ by $\cos$, you get the sinusoidal projection.

Finally, let's do an analogue of the Lambert cylindrical equal-area projection of the sphere, by taking the previous setup and mapping $P$ to the point $(x, \sinh y)$ instead. Again, this map is equal-area.

The analogue: In the spherical case, the map maps $P$ to $(x, \sin y)$.

  • $\begingroup$ Thank you, this is great! $\endgroup$ – Mike Battaglia Aug 7 at 2:17
  • $\begingroup$ The second projection seems more analogous to the sinusoidal projection than to Mollweide. In particular, the projection you get in the last paragraph is sinusoidal, not Mollweide. $\endgroup$ – Zeno Rogue Aug 7 at 19:04
  • $\begingroup$ Ah, that's true. $\endgroup$ – Magma Aug 7 at 19:13

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