Exponential map of Beltrami-Klein model of hyperbolic geometry

Question

In the Betrami-Klein model of hyperbolic geometry, geodesics are represented as straight lines. Hence the exponential map of a tangent vector $\mathbf{v}$ at a point $\mathbf{p}$ is $\mathbf{p} + \lambda \mathbf{v}$, where $\lambda$ is a scalar that depends on $\mathbf{p}$ and $\mathbf{v}$. For example, suppose $\mathbf{p} = 0$. Then the exponential map is

$$ \exp_\mathbf{p}(\mathbf{v}) = \hat{\mathbf{v}} \tanh \lvert \mathbf{v} \rvert $$

Notice that $\tanh \lvert \mathbf{v} \rvert < 1$ always, since we cannot reach the circle at infinity with any finite $\mathbf{v}$.

Now consider the diagram below:

where $\mathbf{q} = \mathbf{p} + \lambda \mathbf{v}$. The hyperbolic distance between $\mathbf{p}$ and $\mathbf{q}$ must be equal to the length of the tangent vector, that is, $d(\mathbf{p}, \mathbf{q}) = \lvert \mathbf{v} \rvert$. This hyperbolic distance is

$$ d(\mathbf{p}, \mathbf{q}) = \frac{1}{2} \log \frac{\lvert\mathbf{p} - \mathbf{r_1}\rvert \lvert\mathbf{q} - \mathbf{r_2}\rvert}{\lvert\mathbf{p} - \mathbf{r_2}\rvert \lvert\mathbf{q} - \mathbf{r_1}\rvert}$$

where $\mathbf{r_1}$ and $\mathbf{r_2}$ are the two ideal points intersected by the straight line connecting $\mathbf{p}$ and $\mathbf{q}$. We can find these ideal points as follows:

\begin{align} 1 &= \mathbf{r} \cdot \mathbf{r} \\ &= (\mathbf{p} + \mu \mathbf{v}) \cdot (\mathbf{p} + \mu \mathbf{v}) \\ &= \mathbf{p} \cdot \mathbf{p} + 2 \mu \mathbf{p} \cdot \mathbf{v} + \mu^2 \mathbf{v} \cdot \mathbf{v} \end{align}

Hence

\begin{align} \mu &= \frac{-2 \mathbf{p} \cdot \mathbf{v} \pm \sqrt{(2 \mathbf{p} \cdot \mathbf{v})^2 - 4 (\mathbf{v} \cdot \mathbf{v})(\mathbf{p} \cdot \mathbf{p} - 1)}}{2 \mathbf{v} \cdot \mathbf{v}} \\ &= \frac{-\mathbf{p} \cdot \mathbf{v} \pm \sqrt{(\mathbf{p} \cdot \mathbf{v})^2 - (\mathbf{v} \cdot \mathbf{v}) (\mathbf{p} \cdot \mathbf{p} - 1)}}{\mathbf{v} \cdot \mathbf{v}} \end{align}

so that $\mathbf{r} = \mathbf{p} + \mu \mathbf{v}$ for each root. Hence we have

\begin{align} \lvert \mathbf{v} \rvert &= \frac{1}{2} \log \frac{\lvert\mathbf{p} - (\mathbf{p} + \mu_1 \mathbf{v})\rvert \lvert(\mathbf{p} + \lambda \mathbf{v}) - (\mathbf{p} + \mu_2 \mathbf{v})\rvert}{\lvert\mathbf{p} - (\mathbf{p} + \mu_2 \mathbf{v})\rvert \lvert(\mathbf{p} + \lambda \mathbf{v}) - (\mathbf{p} + \mu_1 \mathbf{v})\rvert} \\ &= \frac{1}{2} \log \frac{\lvert\mu_1 \mathbf{v}\rvert \lvert(\lambda - \mu_2) \mathbf{v}\rvert}{\lvert\mu_2 \mathbf{v}\rvert \lvert(\lambda - \mu_1) \mathbf{v}\rvert} \\ &= \frac{1}{2} \log \frac{\lvert\mu_1\rvert \lvert\lambda - \mu_2\rvert}{\lvert\mu_2\rvert \lvert\lambda - \mu_1\rvert} \end{align}

or equivalently

\begin{align} \left\lvert \frac{\lambda - \mu_2}{\lambda - \mu_1} \right\rvert = \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert} \\ \frac{\lambda - \mu_2}{\lambda - \mu_1} = \pm \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert} \\ \lambda - \mu_2 = \pm \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert} (\lambda - \mu_1) \\ \lambda \left( 1 \mp \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert} \right) = \mu_2 \mp \mu_1 \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert} \\ \lambda = \frac{\mu_2 \mp \mu_1 \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert}}{1 \mp \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert}} \end{align}

We pick the root $\lambda > 0$. Is this result correct? Can it be simplified? Can anyone find a source for it?

Answer 1

The hyperbolic plane doesn't have a center. However, the Klein model does. Because of the isotropy of the hyperbolic plane we can do our calculations from the center of the model. All the results will be the same. Let $p=(0,0)$ and let $\|v\|$ be arbitrary, and let $v$ be originated at $p$ (see the blue vector). $r=p+\lambda v$ is the red vector. We can replace the vectors by their lengths in this setup.

I would be surprised if the answer wouldn't be $\lambda=\pm1$.

EDITED

I guess, I understand the question now. Let $v$ be the Euclidean distance of a point from the origin in the Klein model. Also, let $\lambda$ be a real variable. We are looking for a $\lambda$ for which the hyperbolic distance of the point at $\lambda v$ equals $v$.

The hyperbolic distance (from the origin) of the point at $\lambda v$ is $\operatorname {atanh} (\lambda v)$. And we are looking for a $\lambda$ for which

$$\operatorname {atanh} (\lambda v)=v.$$

Take the $\operatorname {tanh}$ of both sides. We have now

$$\lambda =\pm\frac{\operatorname {tanh}(v)}v.$$

Answer 2

I did some calculation and so and if

$P = (p_x,p_y)$ and $Q=(p_x+v_x,p_y+v_y)$ are two points on a chord the ideal points (points on the unit circle are:

$$I_1 = \left ( \frac{v_y \ c + v_x \sqrt{v_x^2 +v_y^2 -c^2}} {v_x^2 +v_y^2} \ , \ \frac{-v_x \ c + v_y \sqrt{v_x^2 +v_y^2 -c^2} }{v_x^2 +v_y^2} \right) $$

$$I_2 = \left ( \frac{v_y \ c - v_x \sqrt{v_x^2 +v_y^2 -c^2}} {v_x^2 +v_y^2} \ , \ \frac{-v_x \ c - v_y \sqrt{v_x^2 +v_y^2 -c^2} }{v_x^2 +v_y^2} \right) $$

with $ c =p_x v_y - v_x p_y $

I don't know if this will help you and also I don't know how to write this as a formula of complex numbers (but will make a seperate question for that :)