Following this question, I'm looking for a coordinate transformation which leaves distances unchanged. Does such a transformation exist? The isometries for the poincaré disk looked promising, but only conserve angles, not distances.

Edit: I'm using the hyperbolic law of cosines to get the distances between two points in polar coordinates. By constructing a triangle with the origin and the two query points, the length of the third side, which is the desired distance, can be calculated.

I tried a circle inversion with a circle centered outside the disk. A circle inversion on the poincaré disc

Assuming I have point $a$ near the left border of $P$ and a point $b$ immediately to the left of the arc segment $c$ marked in red. After mirroring both points at $C$, they are closer than they were before. Do I have to adjust the metric or am I doing something wrong?

Ultimately I'm looking for a lower bound for the distances of a point to all points in a (convex) shape.

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    $\begingroup$ Your question is confusing. Which distance are you talking about? The reason that isometries for the Poincare disc are called isometries is that they preserve the distance formula for the hyperbolic metric on the Poincare disc. That's what "isometry" means, if you are specific about you distance formula. $\endgroup$ – Lee Mosher Jul 9 '14 at 14:38
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    $\begingroup$ Even after your edit, your question is still confusing. Circle inversions preserve distance in the hyperbolic metric, not in the ordinary Euclidean metric which our eyes are trained to perceive. $\endgroup$ – Lee Mosher Jul 9 '14 at 15:27

Hyperbolic isometries in the Poincaré disk preserve angles (both hyperbolic and Euclidean, since they are the same) and hyperbolic distances (i.e. distances measured using the hyperbolic metric), but in general not Euclidean distances. The only hyperbolic isometries which also preserve Euclidean distances are reflections in the diameters of the Poincaré disk.

Note that for a circle inversion to be an isometry, you need the circle to be orthogonal to the boundary of the Poincaré disk. Otherwise it's not a hyperbolic line (or a geodesic, if you prefer that term) but instead some equidistance curve which is not a line in hyperbolic geometry.

Any Möbius or anti-Möbius transformation which preserves the unit circle is an isometry of the Poincaré disk model.

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  • $\begingroup$ " you need the circle to be orthogonal to the boundary of the Poincaré disk" - this was the critical detail I forgot, it works now. Thanks! $\endgroup$ – user155598 Jul 16 '14 at 16:11

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