7
$\begingroup$

I'd like to implement an interactive simulation of an actor controlled by the user moving around in the Poincaré disc model of the hyperbolic plane.

I need to know how to perform translation and rotation on the Poincaré disc. If i represent points as regular Euclidean points in the unit circle, I believe that hyperbolic translation and rotation would then be Möbius transformations if you think of the points as complex numbers. Is this correct? If not what is a better way of performing translation and rotation.

Assuming Möbius transformations are the right approach, however, I am unable to find defined anywhere how specific translations and rotations correspond to specific Möbius transformations. Can someone clear this up for me? I'm looking for functions of complex variables that correspond to "translate z in direction d on the hyperbolic plane represented as the unit circle" and "rotate z around c by $\theta$ on the hyperbolic plane represented as the unit circle"

$\endgroup$
6
$\begingroup$

Yes, orientation-preserving rigid body motions in the Poincaré disc are Möbius transformations which fix the unit disc. A hyperbolic translation along the real axis by a distance of $x\in\mathbb R$ can be expressed as

$$\begin{pmatrix}e^x+1 & e^x-1 \\ e^x-1 & e^x+1\end{pmatrix}$$

A rotation by an angle of $\varphi\in\mathbb R$ around the origin can be expressed as

$$\begin{pmatrix}e^{i\varphi} & 0 \\ 0 & 1\end{pmatrix}$$

All other rotations and translations can be composed from these.

The formulation of a Möbius transformation as a matrix is based on viewing the Poincaré disc as part of the real projective line, $\mathbb C\mathrm P^1$. If you prefer fractions, you can translate as follows:

\begin{align*} \begin{pmatrix}a&b\\c&d\end{pmatrix} \cdot\begin{pmatrix}z\\1\end{pmatrix} &\mapsto\begin{pmatrix}p\\q\end{pmatrix} \sim\begin{pmatrix}\tfrac pq\\1\end{pmatrix} & \text{corresponds to}\qquad z\mapsto\frac{az+b}{cz+d} \end{align*}

$\endgroup$
  • $\begingroup$ So the x in the translation matrix is a hyperbolic distance, right? $\endgroup$ – jwezorek Feb 20 '13 at 18:41
  • $\begingroup$ @jwezorek, that's right. $\endgroup$ – MvG Feb 20 '13 at 20:10

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.