Does there exist an area-preserving map from the hyperbolic plane to the Euclidean plane? [duplicate]

Solution 1:

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 $f(r) = \sqrt{2\cosh(r)-2}$.

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.

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