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Consider the metric space $(\mathbb{H}²,d_{\mathbb{H}^2})$, where $d_{\mathbb{H}^2}$ is the hyperbolic Cayley Klein metric, i.e., $ d_{\mathbb{H}^2}(A,B) = |log ((AA_{\infty}. BB_{\infty}) / (BA_{\infty}. AB_{\infty}))| $

Here $A_{\infty}$ and $ B_{\infty}$ are the points of $r_{\infty}$ that make a diameter for the (upper) half circle trough them, where the points (except $A_{\infty}$ and $B_{\infty}$) of this half circle are the points of an hyperbolic line. If $B_{\infty} = \infty $, then the hyperbolic line is represented by the points of the semi-line that are between $A_{\infty}$ and $\infty$.

Proof that a circle in $d_{\mathbb{H}^2}$ metric is also a circle in the canonical euclidean metric.

Well, I have no clue how to proceed in this problem. I tried considering an hyperbolic circle centered in $p$, which is the set $ \left \{x \in \mathbb{H}^2 ; d_{\mathbb{H}^2}(x,p)=k, \right\} $ for some $k \in \mathbb{R}$ and then tried to use Euclidean Analytic Geometry to describe those points and then find how they are points in some euclidean circle, but the calculations are getting bigger and uglier... can someone give me a hint?

Thanks!

  • Thanks! I edited the name. But are you sure it is only for the Disk Model? Because my book constructs the Half Plane Model, and uses it. –  Feb 29 '16 at 19:10
  • Could you please give us the name of your book? – Korf Mar 01 '16 at 08:51
  • the formula is correct for both Poincare models (although the formula you give is always negative before you take the absolute value). Still puzzeling on how to prove it, I think the problem is that the centres are not on the same point see https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model : "A circle (curves equidistant from a central point) with center ( x , y ) and radius r is modeled by: a circle with center $ ( x , y \cosh ⁡ ( r ) ) $ and radius $y \sinh ⁡( r )$ maybe you need someting like that the radians cut the circle orthogonal, but not sure yet, what have you tried ? – Willemien Mar 01 '16 at 11:56
  • I posted my attempt as an answer. –  Mar 11 '16 at 14:52
  • My book is Geometria Hiperbólica, by Luiz Fernando Rocha –  Mar 11 '16 at 14:52

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As Poincaré Disk Model is symetric by rotations in the origin (i.e., Rotations through origin are isometries), then hyperbolic circles with center in the origin of the Disk are represented by euclidean circles with same center and vice-versa. This imply that, as conformal circles and hyperbolic circles are invariant through Möbius transformations and Lobachevski transformations (Respectively), then hyperbolic circles are represented by conformal circles of the Hyperbolic Plane and vice-versa. As hyperbolic circles are entirely over the line $r_{\infty}$, then hyperbolic circles are represented by conformal circles that are euclidean circles.

The following shows that the euclidean radius and the hyperbolic radius are different.

Let $C$ be an euclidean circle in the Half-Plane, with center $Oe$. Let $r$ be the perpendicular to $r_{\infty}$ through $Oe$ and call the intersections with $C$ by $A$ and $B$, where $a<b$ (euclidean coordinates in $r$). Since $C$ is a circle that is entirely over $r_{\infty}$, it is an hyperbolic circle. So $AB$ is a diameter.

The hyperbolic middle point of $A$ and $B$ is a point located at the $\sqrt{ab}$ coordinate of $r$. Then the euclidean coordinate in $r$ of the hyperbolic center $Oh$ is $a + \sqrt{ab}$.

Since $r$ is a hyperbolic line that is also an euclidean semiline, the hyperbolic radius of $C$ is $\log{\frac{b}{\sqrt{ab}}}$.