# Analogy of spherical and hyperbolic geometry projection

I have heard of the following analogy:

• The Poincare disc model of the hyperbolic plane is analogous to the stereographic projection of the sphere, and
• The Beltrami-Klein model of the hyperbolic plane is analogous to the gnomonic projection of the sphere.

Is it possible to make this connection more rigorous? Perhaps by exploiting some kind of trick, like considering the hyperbolic plane a "sphere of imaginary radius"? If so, is there an analogous spherical-geometry projection for the Poincare half-plane model of the hyperbolic plane?

• The Poincare disk model and the stereographic projection preserve angles (i.e. they are conformal). The Beltrami-Klein model and the gnomonic projection preserve geodesics (i.e. they map geodesics to straight lines). – Rahul Dec 30 '16 at 4:20

## 2 Answers

$\newcommand{\Reals}{\mathbf{R}}$If $P$ is a plane in $\Reals^{3}$ and $c$ is a point not on $P$, define projection to $P$ with center $c$ to be the map that sends each point $x \not\in P$ to $\overline{cx} \cap P$, the point where the line from $c$ to $x$ crosses $P$.

The table below gives a quantitative analogy, in which $(x, y, z)$ denote Cartesian coordinates and the columns are related by the Wick rotation $(x, y) \mapsto i(x, y)$: $$\begin{array}{l|ll} & \text{Elliptic} & \text{Hyperbolic} \\ \hline \text{Plane} & x^{2} + y^{2} + z^{2} = 1 & -(x^{2} + y^{2}) + z^{2} = 1,\ z > 0 \\ \text{Gnomonic} & c = (0, 0, 0),\ P = \{z = 1\} & c = (0, 0, 0),\ P = \{z = 1\} \\ \text{Conformal} & c = (0, 0, -1),\ P = \{z = 0\} & c = (0, 0, -1),\ P = \{z = 0\} \\ \end{array}$$

To expand the analogy:

• The center of gnomonic projection of the unit sphere is the center (i.e., the origin), and maps each open hemisphere (for example, the "northern hemisphere" $\{z > 0\}$) onto the tangent plane at the "center" (here, the point $(0, 0, 1)$).

The center of gnomonic projection for the hyperbolic plane is the origin (the "center" of sorts for the hyperboloid of two sheets), and the image of projection is the open unit disk.

As Rahul notes, the geodesics of each plane model are the intersections of the corresponding unit sphere with a plane through the origin; projection from the origin consequently sends geodesics to lines (or line segments).

• The center of conformal projection of the unit sphere is a point of tangency (e.g., the "south pole" $(0, 0, -1)$), and the image plane is the "orthogonal plane" through the center (here, $\{z = 0\}$).

The center of conformal projection for the hyperbolic plane can also be taken to be $(0, 0, -1)$, not a point of the hyperbolic plane, but the "antipode of the center $(0, 0, 1)$", lying on the bottom sheet, and the image of projection is again the open unit disk.

The two gnomonic projections, and stereographic projection of the sphere, are readily-visualized. Probably the least well-known projection is the conformal projection from the hyperbolic plane to the unit disk in the plane $\{z = 0\}$, using $c = (0, 0, -1)$: This is not a full answer, but it is the only way I can include graphical information which comes from p. 426 of a French book.

I think this figure is self-explanatory.

If you read French, this (excellent) book is "Initiation à la géométrie" by Daniel LEHMANN and Rudolphe BKOUCHE, Presses Universitaires de France,1988. 