1
$\begingroup$

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:

enter image description here

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?

$\endgroup$
  • $\begingroup$ I think your calculation of the ideal points does not work the mu can be further simplified and r becomes plus min 1 and almost meaningless $\endgroup$ – Willemien Mar 30 '17 at 19:06
  • $\begingroup$ @Willemien Why does it become plus minus 1? $\endgroup$ – user76284 Mar 30 '17 at 20:36
  • $\begingroup$ Just calculate $\mu$ , $r_1$ and $r_2$ but maybe I made a mistake will work it out in the weekend $\endgroup$ – Willemien Mar 30 '17 at 22:33
1
$\begingroup$

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.

enter image description here

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

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

$\endgroup$
  • $\begingroup$ @Carlos: Then I don't understand the question. Do we not want that $d(p,q)$ be $\|v\|$? $d(0,q)=d(0,\lambda v)=d(0,v)$ So $\lambda=1$. What's wrong with this argumentation? $\endgroup$ – zoli Apr 5 '17 at 20:01
  • $\begingroup$ $\lvert \mathbf{v} \rvert = d(\mathbf{p}, \mathbf{q}) \neq \lvert \mathbf{p} - \mathbf{q} \rvert$. $\lvert \mathbf{p} - \mathbf{q} \rvert$ is the distance in the ambient Euclidean space we are using to model the hyperbolic plane, whereas $d(\mathbf{p}, \mathbf{q})$ is the distance inside the hyperbolic plane itself. $d(0, q) = d(0, \lambda v)$, but $d(0, \lambda v) \neq d(0, v)$. $\endgroup$ – user76284 Apr 6 '17 at 0:00
  • $\begingroup$ Think of it this way: The ambient distance (from the origin) to any point on the circle is unity, but the hyperbolic distance to that same point is infinity. We are representing the entire hyperbolic plane on the interior of the circle, whereas the boundary is infinitely far away. $\endgroup$ – user76284 Apr 6 '17 at 0:06
  • $\begingroup$ @Carlos: I edited my answer. $\endgroup$ – zoli Apr 6 '17 at 7:20
  • $\begingroup$ That works. $\lambda = \frac{\tanh \lvert v \rvert}{\lvert v \rvert}$ so $\exp_p(v) = \hat{v} \tanh \lvert v \rvert$. What if $p \neq 0$? $\endgroup$ – user76284 Apr 6 '17 at 16:59
0
$\begingroup$

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 :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.