Uniqueness of hyperbolic geodesics

The Poincare’ model of hyperbolic geometry consists of the unit disk \mathbb{D}=\{z\in\mathbb{C}: |z|<1\} where the geodesics are arcs of circles perpendicular to the unit circle \partial\mathbb{D}=\{|z|=1\}. It turns out that given two points z,w\in\mathbb{D} there is a unique circle orthogonal to \partial\mathbb{D} passing through z and w.

A nice way to prove the uniqueness of geodesics is to use the stereographic projection. Let \mathcal{S} be the unit sphere in \mathbb{R}^3 and N its north pole. The stereographic projection puts in one-to-one correspondence each point z\in\mathbb{C} with the unique point z^*\in \mathcal{S}\setminus\{N\} that lies on the line through N and z. For example, the unit disk \mathbb{D} corresponds to the southern hemisphere on the sphere \mathcal{S}.

One of the main properties of the stereographic projection is that it is a conformal map, i.e., it preserves angles between curves. Proving this is a nice exercise in geometry.

Another property is that the stereographic projection maps circles to “circles”, in the sense that circles on \mathcal{S} through the north pole are mapped to lines in \mathbb{C} and those that don’t go through N are mapped to actual circles in \mathbb{C}.

Armed with these two facts we can now easily establish the uniqueness of hyperbolic geodesics. Given two points z,w\in\mathbb{D} consider their images z^*, w^* on the southern hemisphere of \mathcal{S}. It is clear that z^* and w^* do not lie on a vertical line. Therefore, there is a unique plane through z^* and w^* perpendicular to the (x,y)-plane in \mathbb{R}^3, i.e. \mathbb{C}. The intersection of this plane with \mathcal{S} is a circle that is perpendicular to the equator of \mathcal{S}. Under the stereographic projection it will be mapped to a “circle” through the points z,w and, by conformality, its image will be orthogonal to the image of the equator, i.e., \partial\mathbb{D} (the equator is fixed by the projection).

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