Further hyperbolic kaleidoscopes

In the last post I have used reflections at two parallel lines and a circle to get a Poincaré plane that shows a periodic decoration of hyperbolic space. What happens if the straight lines are not parallel and intersect? Then their mirror symmetries do not generate translations, instead we get rotations around their point of intersection. Thus together with the inversion at the circle we now have a Poincaré disk.

The two straight lines and the circle make a hyperbolic triangle, which we use as a kaleidoscope. We get a k-fold rotational symmetry if π/k is the angle of the intersection between the lines. As in the last post n- and m-fold rotational symmetries arise at the corners between the circle and the lines for angles of π/n and π/m. The sum of these three angles is π(1/k+1/n+1/m), which has to be smaller than π for hyperbolic space. Thus the condition 1/k+1/n+1/m<1 limits the choices for the rotational symmetries.

Now let us compare this with the usual Euclidean space. The sum of the angles is then equal to π and 1/k+1/n+1/m=1. Up to irrelevant permutations there are only three solutions with integer numbers of this equation. They are (k,l,m)=(3,3,3), (4,2,4) and (6,2,3). You can see the resulting kaleidoscopes in my earlier post “Geometry of kaleidoscopes with periodic images“.

To get a kaleidoscope for hyperbolic space we simply increase some numbers. An interesting choice is (k,n,m)=(5,2,4). The resulting Poincaré disk looks like this:

Here the yellow triangle is the inside of the kaleidoscope. Dark brown triangles show images with an odd number of reflections and light blue is for even number of reflections. The borders between triangles are circles with a right angle to the border of the disc. These are straight lines in hyperbolic space. Note that this is a periodic tiling of hyperbolic space with five-fold rotational symmetry.

And now for some images. From (5,2,4) I got this:

And the (4,3,3) kaleidoscope gave that:

The circle seems to act like a convex mirror and reduces the apparent size of the reflected images in the Poincaré disc. But actually, inversion at the circle is not the same as true optics and in the hyperbolic space all images have the same size. If you zoom in at the border you will see essentially the same image. Only the radius of the disc appears to increase in comparison to the distance between images. In the end we approach a Poincaré plane. The kaleidoscopes of the last post are thus actually limits of the kind (∞,n,m).

Note added on the 16 December 2017: These kaleidoscopes use Möbius triangles, which are Schwarz triangles with integer rotation number. Reflections at the sides of a Möbius triangle generate the triangle group and make a tiling of an elliptic, Euclidean or hyperbolic plane.





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