---
tag: 1WN9B18H86
kind: example
name: Beltrami–Klein model
msc: [51M10, 51N15]
instance_of: [0VKD1TP7F3]
uses: [GN4N94GQ0Z, J7ZE3EGRYX, 85MM2EA0TK]
sources:
  - url: https://en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model
same_as:
  wikidata: Q1189065
  wikipedia: Beltrami–Klein model
---

## Example

The *Beltrami–Klein model* of the [[hyperbolic plane|GN4N94GQ0Z]] is the open unit disc $D = \set{x \in \R^2 : \abs{x} < 1}$ with the following distance.
For distinct points $p, q \in D$, the line through $p$ and $q$ meets the unit circle in two points $a$ and $b$.
Label them so that the order along the line is $a, p, q, b$.
Then
$$d(p, q) = \frac{1}{2} \ln \frac{\abs{aq} \, \abs{pb}}{\abs{ap} \, \abs{qb}},$$
where $\abs{xy}$ is the Euclidean distance, and $d(p, p) = 0$.
This is the [[Cayley–Klein metric|J7ZE3EGRYX]] whose absolute is the unit circle, and it makes $D$ a [[metric space|0VKD1TP7F3]].
The lines of the model are the open chords of the disc.

Check on a diameter: take $p = (0, 0)$ and $q = (r, 0)$ with $0 < r < 1$.
Then $a = (-1, 0)$ and $b = (1, 0)$, so $\abs{aq} = 1 + r$, $\abs{pb} = 1$, $\abs{ap} = 1$ and $\abs{qb} = 1 - r$.
The formula gives $d(p, q) = \frac{1}{2} \ln \frac{1 + r}{1 - r} = \operatorname{artanh} r$, which tends to $\infty$ as $r \to 1$.
So the boundary circle lies at infinite distance, and the disc is a whole infinite plane.

## Remarks

Through a point $p$ not on a chord $\ell$ there pass infinitely many chords that do not meet $\ell$ inside the disc, so the parallel postulate fails.
The isometries of the model are the projective transformations of the plane that map the disc onto itself.
The model is not conformal: Euclidean angles between chords are in general not the hyperbolic angles.
The map $u \mapsto 2u / (1 + \abs{u}^2)$ is an isometry from the [[Poincaré disk model|85MM2EA0TK]] onto the Beltrami–Klein model.
Indeed, a point at Poincaré distance $t$ from the centre has Euclidean radius $\tanh(t/2)$ in the Poincaré disc and $\tanh t$ in the Klein disc, and $\tanh t = 2 \tanh(t/2) / (1 + \tanh^2(t/2))$.

## Explanations

### L1 human

The Beltrami–Klein model draws the hyperbolic plane inside a round disc, with straight chords as its lines.
Because lines look straight, problems about lines and their crossings become problems of ordinary straight-line drawing.
The price is that distances and angles are distorted: lengths blow up near the rim, and right angles do not look right.
For example, the segment from the centre to the point at Euclidean radius $0.9$ has hyperbolic length $\operatorname{artanh} 0.9 \approx 1.47$.