further wallpapers for hyperbolic space

An equilateral triangle gives us a kaleidoscope of three-fold rotational symmetry. With a square we get two-fold rotational symmetry. Would reflection at the sides of other regular polygons too give periodic images with rotational symmetry ?

To get an h-fold rotational symmetry at a corner we need an angle of 180°/h between the sides meeting at the corner. This is not possible for polygons with more than four corners in Euclidean space because they would have angles of more than 90°. However, in hyperbolic space the corner angles can have any value smaller than (1-2/k)180°, where k is the number of corners of the polygon and the limit results from Euclidean geometry.

We use the Poincaré disc model for hyperbolic geometry. You can find a very interesting detailed discussion at http://www.malinc.se/math/noneuclidean/mainen.php. The most important facts are: An open disc represents the infinite hyperbolic space in our Euclidean space, straight lines in hyperbolic space become circles that intersect the boundary of the disc at a right angle and the inversion at these circles is equivalent to the mirror image at the straight lines in the hyperbolic space.

A regular polygon that has its center at the center of the Poincaré disc shows its dihedral symmetry in our Euclidean drawing plane. Its sides are arcs of equal width and they belong to circles of equal radius. These circles are straight lines in hyperbolic space. The angle between the two arcs at a corner is simply the angle between their tangents. Using basic trigonometry we get their radius from the desired value 180°/h for the angles, the number k of corners of the polygon and the radius of the Poincaré disc. This defines the reflection at the sides of the polygon as inversions at the circles bearing the arcs. Thus we get an easy generalization of the usual regular polygon with straight edges.

As in the earlier posts, we have to map points of the Poincaré disc into the polygon using reflections at its sides. The symmetries of the regular polygons make this easier than for triangles. As long as the point lies outside the polygon we use inversion at the circle with the arc that lies closest to the point. The required number of inversions increases going to the border of the Poincaré disc and finally diverges at its border. Thus we have to impose an upper limit and disregard points that take too many inversions.

We rapidly find the circle to use for inversion using a coordinate system with its origin at the center of the Poincaré disc and polar coordinates. A polygon with k corners has its corners at polar angles of i*(360°/k) where i goes from 1 to k. The sides of the polygon are arcs connecting the two corners at angles i*(360°/k) and (i+1)*(360°/k). Now, if the point that we are mapping has a polar angle φ we have to find the arc that covers this angle. Thus we have to find an integer number i with i*(360°/k) < φ < (i+1)*(360°/k), all obviously modulo 360°. This defines the arc that is closest to the point. It is part of a circle which has its center at the polar angle (i+1/2)*(360°/k). If the point lies outside this circle then it has to be inside the polygon and we can get the colour for the pixel at the starting point from an input image at this mapped position. If the point lies inside the circle we invert it at the circle and repeat this procedure.

For a pentagon with angles of 90° we get a result with this structure:The original pentagon is shown in yellow. All points that need an odd number of inversions are coloured light blue and those with an even number are shown dark brown. Away from the center we see strongly distorted pentagons, shrinking towards the border of the disc. But in hyperbolic space, they are all regular pentagons of the same size, tiling the hyperbolic plane. Actually, we get tilings for all k>2 and h>1. They are made of regular polygons with k corners and each corner belongs to 2*h polygons.

Using an asymmetric input image for the pentagon at the center we get kaleidoscopic images that look like this one:It has five different motifs of two-fold rotational symmetry, one for each corner of the pentagons. Often, such images seem to lack any order. Here we see strips of repeating green swirls with mirror symmetry. In hyperbolic space they are straight infinite friezes with horizontal and vertical reflection lines resulting from the mirror symmetries at two corners of the pentagon. These friezes are separated by narrow branching lines of lighter colour.

With an input image of rotational symmetry we get wallpapers with the same rotational symmetry if the symmetry is compatible with the polygon. The same pentagon as before and a rosette with five-fold symmetry gave this:Note that an input image with dihedral symmetry would give the same image as a kaleidoscope using the (5,2,4) triangle instead of this pentagon. In general, a regular polygon with k corners and angles of 180°/h using an input image of k-fold dihedral symmetry is equivalent to using a Möbius triangle (k 2 2*h) with corner angles of 180°/k, 90° and 90°/h.

For more images of higher resolution have a look at my Flickr album “hyperbolic wallpapers”.

This entry was posted in Anamorphosis, Kaleidoscopes, Tilings and tagged , , , , . Bookmark the permalink.

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