Fractal sphere packing from inversion in five spheres

As presented in previous posts, inversion in three touching circles makes a Poincaré disc representation of a hyperbolic plane tiled by a triangle. Adding a fourth circle touching the other three results in a fractal decoration of the hyperbolic disc with discs. Actually, this is an Apollonian gasket. I want to do essentially the same thing, but in three dimensions: Inversion in four circles can generate a Poincaré ball representation of hyperbolic space and an additional fifth sphere could create a fractal packing of spheres in this space. In analogy to the two-dimensional case, I am using four spheres that have their centers on the corners of a tetrahedron and add a fifth sphere at the center of the tetrahedron.

It is important to realize that the limit set of the inversion in these five spheres has to cover completely the surfaces of all the spheres of the sphere packing. We can only get this if the limit set of the inversions in four of them covers the entire surface of the Poincaré ball that they generate. Observing that the space lying outside the inverting spheres does not belong to the limit set, we conclude that the inverting spheres have to cover the entire surface of the Poincaré ball. This can definitely not be achieved with touching spheres, as I naively tried first.

Figure 1: This tetrahedron has the centers of the inverting spheres at its corners. Blue lines: Intersection with the surfaces of the spheres. The light and dark colors show the tiling pattern due to multiple inversion. Red lines: Intersection with the plane of the cross-section shown in Figure 2.

Instead, we need intersecting spheres. To see how large they have to be, we simply look at the surfaces of the tetrahedron that has four spheres at its corners. They are equilateral triangles and should lie inside these spheres. We achieve this if the spheres reach the center of the triangles. This corresponds to an intersection angle of 60 degrees between the spheres, see Figure 1. Multiple inversion in all four spheres creates a Poincaré ball representation of a tiled hyperbolic space. The surface of the ball touches the centers of the sides of the tetrahedron. The four spheres can be seen as curved sides of another tetrahedron. This tetrahedron tiles the hyperbolic space. Its corners lie on the centers of the sides of the tetrahedron that has the spheres at its corners. Thus the tiling tetrahedron is its dual tetrahedron. The spatial angle of its corners vanishes. Note that this is similar to the planar angle at the corners of a triangle made with three touching spheres, which vanishes too. Also, the corners of the tiling tetrahedron lie on the surface of the Poincaré ball. Thus this tetrahedron is an ideal tetrahedron.

Figure 2: Cross section at a plane that contains the centers of two spheres and the center of the tetrahedron, as indicated by the red lines in Figure 1. The limit set of the inversions is approximated by white color. The intersection with the tiling tetrahedron is shown in black. A small part of its inverted image is shown in grey.
Figure 3: Planar cross section similar to Figure 3. The plane goes through the center of the tetrahedron and lies parallel to one of its surfaces.

The space outside the ball is an inverted image of its inside. On the triangle surfaces we get an inverted Euclidean tiling of equilateral triangles, see Figure 1. This is simply a particular cross section of this inverted hyperbolic space. The limit set of the inversions in the four spheres covers the entire surface of the Poincaré ball, see Figures 2 and 3.

Figure 4: As Figure 2, but with an additional inverting sphere at the center. The limit set is made of the surfaces of a sphere packing. We see cross-sections of the spheres.
Figure 5: As Figure 3, but with an additional inverting sphere at the center.

To get a fractal packing of Poincaré balls we add a fifth inverting sphere at the center of the tetrahedron. It too intersects the other spheres at an angle of 60 degrees. This creates new groups of four intersecting spheres, always three of the first four spheres together with the new one. Each of these groups creates by multiple inversion a similar hyperbolic space as presented in the first part of this post. Thus we get touching Poincaré balls covered by the limit set of the inversions. Inversion in all five spheres multiplies them and fills in the gaps between the spheres. Thus the entire hyperbolic space gets covered by Poincaré balls. We see this in Figures 4 and 5. They are planar cross-sections of this sphere packing. Instead of spheres we see circles as their cross-section. The sphere packing appears thus as a fractal packing of circles in these figures.

Thus, five intersecting spheres really can make a fractal packing of spheres in three dimensions similarly as four circles generate an Apollonian gasket. Further, we proceed as in the last post and think of the three-dimensional space as being the surface of a Poincaré half-“hyper”-space representation of a four-dimensional hyperbolic space. Using inversion in a four-dimensional sphere, we can get five intersecting spheres of equal radius. Their centers lie on the corners of a 4-simplex, which is a “hyper”-tetrahedron in four-dimensional space. They generate a Poincaré “hyper”-ball representation of hyperbolic space. The fractal sphere packing then lies in its surface, a spherical three-dimensional space. It has the symmetries of the 4-simplex. Thus the packing of spheres is also a fractal decoration of three-dimensional spherical space. Both views are related by a stereographic projection from the surface of the four dimensional sphere to three-dimensional Euclidean space. Unfortunately, it is not possible to extend this scheme one dimension higher.

Posted in Fractals, Tilings | Tagged , , , | Leave a comment

Apollonian Gasket from Inversion in Four Spheres

In the earlier post “Apollonian gasket as a fractal in tiled hyperbolic space” I have already discussed the Apollonian gasket. There, I began with three touching circles that define an ideal triangle with vanishing corner angles. Multiple inversion at the circles creates a Poincaré disc model of a hyperbolic space. It is tiled by images of the triangle. The limit set of the inversions is exactly the boundary of the disc. Adding a fourth circle ,which touches the other three, gives the Apollonian gasket as the limit set of the inversions in all four circles. Seen as a two-dimensional object, it is a fractal decoration of the tiled hyperbolic disc that resulted from the first three circles. On the other hand, in “Apollonian gasket as a spherical fractal with tetrahedral symmetry” I have shown that the Apollonian gasket is also a fractal decoration of a spherical surface with tetrahedral symmetry. Both are related by a stereographic projection.

This becomes more interesting if we think in three dimensions. We can see the four touching circles as equators of spheres. The plane going through their centers is the boundary of two Poincaré half-space models of three-dimensional hyperbolic space. These models lie on the two sides of the plane. The four spheres are flat planes in the corresponding hyperbolic spaces. Multiple inversion in the spheres creates a periodic tiling of these spaces and a non-periodic tiling of their mutual boundary. The limit set is obviously the Apollonian gasket.

The red and blue circles represent spheres. Inversion (orange lines) at the red sphere is the same as a stereographic projection of the blue sphere.

As mentioned before, we can project the Apollonian gasket on the surface of a sphere using the stereographic projection. But this projection only maps two-dimensional surfaces. It would be useful if we could somehow use the stereographic projection for the entire three-dimensional space. Considering only the surface of a sphere, its stereographic projection to a plane gives the same result as an inversion in a mapping sphere, which has the same center as the stereographic projection. The radius of the mapping sphere is then directly related to the position of the plane of projection. Typically, the mapping sphere intersects the projected spherical surface in this plane. The first figure shows an example. The center of projection is the north pole of the sphere and the plane of projection goes through its center. The intersection with the mapping sphere is then the equator of the spherical surface and the radius of the mapping sphere is larger by a factor of the square root of two. Using the inversion in the sphere, we can extend the stereographic projection to the entire three-dimensional space. That is exactly what we need.

The four inverting spheres (blue areas) have their centers on the corners of the tetrahedron. The red circles correspond to the surface of the resulting Poincaré ball model.

We now use an additional inverting sphere to map the four spheres and the hyperbolic model they generate. The center of this sphere lies above the center of the Apollonian gasket. It is closer to the smaller sphere than the three other spheres. Thus, inversion in the additional mapping sphere increases the size of the smaller sphere in comparison to the three larger spheres and it is possible to make that all four spheres get the same size after inversion. As they are touching each other, the distances between their centers are then all the same. Then, the centers of the three spheres lie on the corners of a tetrahedron, see the second figure. The spheres touch in the middle of its edges. A small part of the space inside the tetrahedron lies between the four spheres. It is useful to think that this is a tetrahedron with curved sides, which are parts of the surfaces of the spheres. This tetrahedron has no corners. The black regions show its intersection with the large tetrahedron. Yet, these tetrahedrons are dual to each other. Multiple inversion in the four spheres creates a Poincaré ball model of hyperbolic space. It is tiled by the curved tetrahedron between the four spheres. The surface of the ball model is drawn in the second figure in red color. Note that the tiling tetrahedron has no corners inside the ball or on its surface. Thus it is a “hyperideal” tetrahedron.

Surface of a Poincare ball model of hyperbolic space tiled by images of a hyper-ideal tetrahedron (black color). The limit set is an Apollonian gasket (approximated by white color).

The last figure shows the surface of the ball model. The intersection with the tiling tetrahedron is again shown in black and the intersection with the four inverting spheres in dark blue. Note that all points get mapped into the black region. Points requiring more iterations get a lighter shade of blue. Thus the white color approximates the limit set, which is indeed an Apollonian gasket.

We see that using four spheres is the most symmetric way for getting the Apollonian gasket. Actually, this is the starting point for further research. What can we get from inverting spheres at the corners of other (regular) polyhedrons? This gives interesting variations on the Apollonian gasket. Does an additional inverting sphere at the center result in a fractal space filling structure of Poincaré ball models? We need a tiling tetrahedron with real corners to do that. This might also be seen as five four-dimensional inverting spheres at the corners of a four-dimensional simplex, which is sometimes called a “hyper-thetrahedron”. It gives a much more complicated structure than a three-dimensional “Apollonian gasket”. More results in future posts.

Posted in Fractals, Self-similarity, Tilings, Uncategorized | Tagged , , , , | Leave a comment

Yet another tiling with 12-fold rotational symmetry

While writing together with Theo Schaad the paper “A Quasiperiodic Tiling With 12-Fold Rotational Symmetry and Inflation Factor 1 + Sqrt(3)”, see, about the post “Another tiling of dodecagonal symmetry” I found an interesting variation. The Ammann-Beenker tiling of 8-fold rotational symmetry has a similar inflation factor of 1+sqrt(2), see . However, the substitution rule for the square is asymmetric and has the same tiles on the border as for the rhomb. Thus, I thought that it might be possible to do something similar for 12-fold rotational symmetry and a factor of 1+sqrt(3). But again, this has been difficult, although rhombs and squares have now the same substitutions at their borders. Note that the substitutions at the border have two different orientations. Thus it is simply not enough to find dissections of the tiles into smaller rhombs, squares and triangles. Instead I had also to take into account the substitutions of the next generation to make them fit across their boundaries as for the earlier tiling.

Trying to have as much rhombs as possible I got these substitution rules:

Substitution rules for the new tiling.

Here I had to use rhombs of 60 degrees acute angle at the borders and it was not possible to respect the form of the inflated tile. Further, the substitution rule for this rhomb is not mirror symmetric because of triangle tiles that have themselves different substitution rules. The deviation from mirror symmetry is rather small and thus the tiling has only a small chirality. It looks like this:

The tiling resulting from the substitution rules shown above. The center of 12-fold rotational symmetry lies at the lower left.

The tiling has a surprising structure at large scale. You can see it better in a more symmetric image with less colors:

Posted in Quasiperiodic design, Tilings | Tagged | Leave a comment

Regular polygons as kaleidoscopes

We are using reflection at the sides of a regular polygon to get a space filling periodic image. Its symmetries depend on the symmetry of the image, which lies inside the polygon. As an example, let us look at the equilateral triangle. Patching an image without any particular symmetry on the triangle we get this:

The black lines show the sides of the triangle. We see that the reflections generate three-fold rotational symmetries at each corner of the triangle. Each corner has a different appearance. Thus the symmetry of the image has the orbifold symbol *333. A square gives a similar result, which you can work out yourselves.

In general, we want that the reflections at the sides of the polygon generate dihedral groups with n-fold rotational symmetry. The intersection angles of the sides then has to be an integer fraction of 180 degrees. This means that the corner angles cannot be larger than 90 degrees. This makes it impossible to use polygons with more than 4 corners, if they have straight lines as sides and we can only have a three-fold rotational symmetry at the corner of triangles. Can we get more ?

We can use circle arcs as sides of polygons and inversion in these circles as reflections. This allows us to use polygons with any number m of corners getting any n-fold rotational symmetry at the corners. An example is a regular triangle with corner angles of 45 degrees. Again patching an image without symmetry on the triangle gives us :

The inversions reduce the size of the basic triangle and distort its image. Thus the resulting image is limited to the inside of a disc, which is actually a representation of a periodic decoration of bent hyperbolic space. The reflections now make four-fold rotational symmetries at the three corners. Each is different and thus this image has the symmetry *444.

For a more symmetric image we can put a rosette with three-fold rotational symmetry on the basic triangle. This is such a result :

We see that the image two different centers of symmetry : A three-fold rotational symmetry at the center of the triangle and a three-fold dihedral symmetry at its corners, which have all the same design. The orbifold symbol is thus 3*4.

As a last possibility we can use an image with a three-fold dihedral symmetry the center of the triangle :

Now we have points of eight-fold, three-fold rotational symmetry and two-fold rotational symmetry. They all have mirror symmetry too and thus the orbifold symbol is now *832.

Note that regular polygons always result in periodic images. Most belong to hyperbolic space, but there are two in flat Euclidean and one in elliptic space. Using an irregular polygon we can get more interesting fractal images. As in this post, we can add symmetries compatible with the polygon. This results in images of different appearance, although they all use the same tiling.

Posted in Kaleidoscopes, Tilings | Tagged , , , , | Leave a comment

A rosette in a roman mosaic is an exponential transform of a periodic tiling

In the depot of the museum of Avenches (Switzerland) lies this interesting fragment of a large roman mosaic :You see immediately that this is a rosette with rotational symmetry, except for the fruit at the center. Looking closer we see an additional symmetry. The black and white shapes are all similar. They grow exponentially in size, going away from the center.

An inverse transformation of this image with the complex exponential function exp(x+i*y)=exp(x)*(cos(y)+i*sin(y)) results in :The rosette becomes now a vertical strip with a periodic image. The periodicities in x- and y-direction correspond to radial and azimutal motions in the rosette. The rosette is thus an exponential transform of a periodic tiling. Note the high precision of the design of the mosaic.



Posted in Anamorphosis, Tilings | Tagged , , | Leave a comment

Fractal tiling of a sphere with octahedral two-colour symmetry

The octahedron can have a nice two-colour symmetry. We get it from putting two tetrahedrons together, making a stellated octahedron. It is an eight-pointed star and has already been discussed by Pacioli in his book “de divina proportione” in the 16th century. Leonardo da Vinci did the drawings for this book. Here is his image for the stellated octahedron with some additional colouring:

Drawing of the stellated octahedron by Leonardo da Vinci with additional colouring.

One of the tetrahedrons is coloured green and the other blue to show you how they are put together. You can see that each point of the star is surrounded by points of the other colour. The points are obviously smaller tetrahedrons. Cutting them away you get an octahedron. Give its sides the same colours as the points you have cut away from them. Then, the octahedron has a two-colour symmetry as a green triangle side now shares its edges only with blue triangles, and vice versa. Can we use inversion in circles and reflection at straight lines to obtain a spherical tiling with a similar two-colour symmetry?

We start with a spherical triangle that generates a tiling of the sphere with tetrahedral symmetry using reflection at its sides. It has an angle of 90 degrees and two angles of 60 degrees, which makes it isosceles. We add an inverting circle that crosses the two legs of the triangle at angles of 60 degrees and touches the hypothenuse. Then, multiple reflections at these elements gives us this:

Fractal tiling with two-colour symmetry obtained from multiple reflection at the yellow circles and lines. The dotted line shows the projection of the equator of a matching sphere.

Here, the triangle and circle are shown in yellow. You can see that the right angle of the basic triangle lies at the center of the image. The additional circle covers most of it, except for two small triangles. Multiple reflection at these four elements maps every point of the plane into one of these small triangles, which is discussed more in detail in “Fractal Images from Multiple Inversion in Circles”. In this image I have used green colours for points that get mapped to one triangle and blue colours for points that go to the other one.

The entire plane is covered by blue and green discs. They are Poincaré representations of hyperbolic space tiled by images of these triangles. Every disc touches only discs of the other colour and none of the same colour. Two green and two blue discs can form a closed chain with a gap inside. The gap contains two green and two blue discs that touch its sides. Together with the surrounding four discs we get five chains of discs and five gaps,  which again will contain four discs. Thus arises a fractal self-similar structure with a two colour symmetry, as a rotation by 90 degrees exchanges green and blue colours.

The octahedral symmetry becomes obvious if we make an inverse stereographic projection to a sphere. Its equator is shown in the image above as a dotted yellow line. Matching two input images to the two tiling triangles we get a fractal kaleidoscopic image, such as this one:

Sphere with a kaleidoscopic decoration of octahedral two-colour symmetry.

On this hemisphere you see four large discs. Their borders are incircles of the triangular sides of an octahedron with two-colour symmetry. The gaps in-between are filled by images of these circles, making a fractal decoration of the sphere with octahedral two-colour symmetry. The opposite hemisphere looks alike except for an exchange of the blue and green discs.

This is an especially symmetric result of multiple reflections. But it is quite similar to other less symmetric configurations of the reflecting elements. You can create similar images with my public browser app at and GIMP.

Posted in Fractals, Kaleidoscopes, Self-similarity, Tilings | Tagged , , , , | Leave a comment

A fractal tiling of both octahedral and icosahedral symmetry

I want to show you a fractal tiling which can be seen as a decoration of a sphere with octahedral symmetry and at the same time as another decoration of a sphere with icosahedral symmetry. It arises as the limit set of a dihedral group and inversion in two circles. You can reproduce these results with my public browser app at The basic setup is shown in figure 1.

Figure 1, upper and lower part:
The black lines are the reflecting elements that generate the tiling. The yellow regions are the target of the iterated reflections.

In both parts of figure 1 two straight lines meet at an angle of 60 degrees. Reflections at these pairs of lines generate a dihedral group with three fold rotational symmetry. In the upper part, a circle arc of large radius intersects the lines at angles of 90 and 45 degrees. This makes an elliptic triangle with a sum of angles of 195 degrees. Inversions at the circle arc together with the dihedral group create a tiling with octahedral symmetry. The additional smaller circle touches the circle arc and intersects the lines at angles of 90 and 36 degrees. This creates a very small elliptic triangle at the intersection of the straight lines, which is a reduced copy of the large triangle in the lower part of figure 1. On itself, it would create a tiling with icosahedral symmetry. The two circles and the horizontal straight line define a hyperbolic triangle. It is coloured yellow and has angles of 45 and 36 degrees and a corner with a vanishing angle. Reflections at its sides make a Poincaré disc representation of hyperbolic space tiled by images of this triangle. Note that the lower part of figure 1 is simply an inverted image of the upper part. The center of the inversion lies at the intersection of the two straight lines. This exchanges the angles of 36 and 45 degrees as well as the octahedral and icosahedral symmetries.

We use repeated inversion in the two circles and the elements of the dihedral group of the two straight lines to map any point into the hyperbolic triangle. Thus we get a fractal tiling of the plane by images of the Poincaré disc mentioned above. We can best see the structure of the tiling if we draw only the borders of the discs. To do this exactly would be very difficult, but it is easy to find an approximation. We simply highlight all pixels that need a large number of inversions to go into the hyperbolic triangle. These pixels lie near the border of Poincaré discs as their hyperbolic distance to the center of the disc diverges at the border and the number of inversions is related to this distance. Figures 2 and 3 show the result at different scales. We see an interesting recursive structure made of two different steps. In each step there are triangular gaps with vanishing corner angles. They are filled by several discs, which leaves smaller gaps to be filled in the next step. This is similar to the Apollonian gasket.

In one step, three discs are put into a gap. Together with the surrounding three discs they make a stereographic projection of circles inscribed on the sides of a cube. They touch each other:

Figure 2, recursive structure of the limit set (approximated by the white regions): Three discs fill the triangular region. This leaves smaller gaps, which are filled in as shown in figure 3. The yellow circle is the stereographic projection of a matching sphere.

This step is followed by putting nine discs in each gap. Together with the surrounding three discs they are a stereographic projection of circles drawn on the sides of a dodecahedron. Again, they are touching:

Figure 3: As figure 2, but with nine discs filling the triangular gap. The gaps between these discs are filled as shown in figure 2.

We can now make inverse stereographic projections to spheres. From Figure 2 we get a decoration with octahedral symmetry:

Figure 4: Fractal decoration of a sphere with octahedral symmetry resulting from an inverse stereographic projection of figure 2.. The upper hemisphere is shown in yellow and the lower hemisphere in dark blue.

Here I have used different colours to distinguish the upper and lower hemispheres. Figure 3 gives a decoration with icosahedral symmetry:

Figure 5: As figure 4, but with icosahedral symmetry resulting from an inverse stereographic projection of figure 3.

Although the two spheres have different decorations they show the same fractal covering of the plane.

Posted in Fractals, Tilings | Tagged , , , , , | Leave a comment

A variant of the Apollonian gasket with icosahedral symmetry

We modify the Apollonian gasket presented in the earlier post Apollonian gasket as a spherical fractal with tetrahedral symmetry. In an icosahedron, five triangles meet at their corners, which gives us a fivefold rotational symmetry. At the centers of the triangles we have a threefold rotational symmetry and at the middle of their edges a twofold one. The corresponding kaleidoscopic triangle thus has corner angles of 36, 60 and 9’0 degrees. Their sum is 186 degrees, just enough to make the triangle elliptic. Multiple reflection at the sides of this triangle makes a stereographic projection of a sphere with an icosahedral tiling :

Spherical tiling with icosahedral symmetry generated by reflections at the blue lines and circle shown in stereographic projection. The dashed line marks the projection of the equator.

Adding an inverting circle we get a fractal. It is a decoration of the icosahedral tiling :

Icosahedral fractal, shown in black, resulting from reflection at two lines and two circles. It decorates the icosahedral tiling, which is shown in yellow. This is a stereographic projection.

The icosahedral symmetry becomes evident using an inverse stereographic projection to a sphere. The normal projections of the upper and lower hemispheres are the same, except for a rotation by 36 degrees. This is a view of the lower hemisphere :

Normal view of a hemisphere with an icosahedral tiling (yellow) decorated by a variant of the Apollonian gasket (black).

Drawing both hemispheres of the gasket together we get :

Superposition of both hemispheres of a variant of the Apollonian gasket with icosahedral symmetry.

You can see how they nicely fit together.

The stars of five-fold rotational symmetry contain small copies of themselves. This results from a recursive packing of discs, quite similar to the Apollonian gasket. But now it is based on ideal pentagons instead of triangles:

The five discs, shown in green, leave a gap in the shape of a pentagon. Putting fifteen discs, which are coloured blue, in this gap we get eleven gaps. They are again pentagons and are filled in the same way.

Posted in Kaleidoscopes, Self-similarity, Tilings | Tagged , , , , | Leave a comment

Apollonian gasket as a fractal in tiled hyperbolic space

Reading the fascinating book « Indra`s Pearls », written by David Mumford, Caroline Series and David Wright, you discover that the Apollonian gasket can be created by multiple inversions at four touching circles. Three of the circles are of equal size. The points lying outside these circles belong to two different regions. One lies in between the circles. It is a hyperbolic triangle with vanishing angles. The second region surrounds the circles and is an inverted image of the first region. It includes the point at infinity. We get a hyperbolic tiling of the plane by multiple inversion in these three circles. For each pixel we take its position and invert it at any circle if it lies inside. We repeat this until the position lies outside of all three circles. Then we colour the pixel depending on the number of reflections and its final position. We get this :

Blue colours : Poincaré disc representation of a tiled hyperbolic space. Yellow : Surrounding inverted disc. The generating circles are shown in black.

Here pixels that are mapped into the inner triangle get a light blue colour for an even number of reflections and a dark blue colour for an odd number. Pixels going to the outer inverted triangle are similarly coloured in light and dark yellow. In blue colours we see a Poincaré disc representation of tiled hyperbolic space inside an inverted disc of the same geometry in yellow colours.

We get a generator for the Apollonian gasket by adding a fourth circle inside the inner triangle. The fourth circle touches the three outer circles. Note that this cuts the inner triangle into three smaller triangles. Each of them has this additional circle as one of its sides and two of the three larger circles as other sides. Repeated inversion on the sides of one of the small triangles creates a Poincaré disc representation of tiled hyperbolic space similarly as for the figure above. Inversion in the three larger circles makes an inverted Poincaré disc, which is shown in blue in the above figure. By the way, this inverted disc is also neeeded for the tetrahedral symmetry of the Apollonian gasket. Thus we get four basic discs as building blocks. Inversion in all four circles generates infinitely many images of these discs. They cover the yellow disc of the above figure. It is important that the number of inversions becomes infinite at the border of each of these discs. Thus, we can approximate the Apollonian gasket by colouring those pixels that require many inversions. Superimposing on the image above we get :

Apollonian gasket shown in black. It is generated by inversion in the circles shown in green. Blue and yellow : Poincaré disc representations of tiled hyperbolic space generated by the three larger circles.

We see that the Apollonian gasket nicely fits the tiled space as each hyperbolic triangle is decorated by the same fractal triangle. By the way, these triangles have a close similarity to the Sierpinsky triangle. Note that the projection of hyperbolic space into the Euclidean drawing plane distorts the tiles and their decorations in the same way.

To see how the Apollonian gasket is composed of discs representing hyperbolic spaces, we can colour the pixels depending on the triangle its position gets mapped to and on the number of inversions used. Using red, green and blue for the inner triangles and yellow for the surrounding inverted one we get :

The Apollonian gasket as a covering of the plane with discs representing tiled hyperbolic space. Discs of the same colour are images of each other. White lines show the circles generating the gasket. Black pixels indicate the borders of the discs.

Note that discs touching each other never get the same colour. Discs of the same colour are inverted images of each other. Their tilings are all the same and they only appear to be different. That’s because of the inversion mapping.

The Cayley transform changes the Poincaré disc into a Poincaré plane representation of hyperbolic space. Actually, an inversion in a circle that has its center on the border of the gasket gives an equivalent result. Applying it to the figure that superimposes the gasket and the embedding hyperbolic space, we get :

Poincaré plane representation of hyperbolic space decorated with an Apollonian gasket.

The triangles of the tiling again match the Apollonian gasket. In comparison, their sizes vary much stronger than in the figure above. Thus, at first sight the decorations of the tiles seem to be different. But this is only an effect of their different sizes. Because of the fractal nature of the gasket we see more or less details resulting in a different appearance. Again, we can show the discs of the gasket in more detail :

Poincaré plane representation of an Apollonian gasket made of discs and two planes representing tiled hyperbolic space.

Note that in this image we see the gasket as a periodic frieze resulting from mirror images in two vertical lines and inversions in a string of touching circles.

I conclude that the Apollonian gasket is a fractal covering of a tiled hyperbolic space by Poincaré disc representations of hyperbolic space. But this is only one of its many faces.

Posted in Fractals, Kaleidoscopes, Tilings | Tagged , , , , | Leave a comment

Apollonian gasket as a spherical fractal with tetrahedral symmetry

Before discussing the relation between the Apollonian gasket and tilings of the sphere, I want to present briefly the spherical kaleidoscope with tetrahedral symmetry.

A tetrahedron has three different kinds of points with rotational symmetry. Four equilateral triangles make up the tetrahedron. It has three-fold rotational symmetry with respect to the points lying at the center of triangles. Another set of points with the same rotational symmetry is made of the points at the corners of the triangles, where three triangles meet. Points of two-fold rotational symmetry lie in the middle of the sides of the triangles.

Note that reflections in two mirrors that intersect at an angle of 180/n degrees result in an n-fold rotational symmetry. Thus, a kaleidoscope with three planar mirrors makes an image with tetrahedral symmetry if its angles between the mirrors are 90 degrees and twice 60 degrees. These mirrors are not parallel, instead they make up a pyramid. To create a two-dimensional image with these symmetries, we replace the mirrors by mirror lines and the kaleidoscope becomes a triangle. However, the sum of its angles is 210 degrees. This is larger than 180 degrees and thus it is a spherical triangle. At least one of its sides has then to be a circle arc instead of a straight line.

To get the kaleidoscopic image we repeatedly mirror the position of a pixel at straight lines and invert it at circle arcs until it lies inside the kaleidoscopic triangle. The colour of the pixel can depend on the number of reflections and on its final position. A typical result for the tiling generated by the kaleidoscope looks like this :

Stereoscopic projection of the spherical tiling with tetrahedral symmetry.

The solid black lines mark the kaleidoscopic triangle. To show the structure of the tiling I coloured pixels already lying inside the triangle in light yellow. Pixels needing an even number of reflections to get into the triangle are shown in primary yellow and dark yellow shows those with an odd number. Here you can see how images of the kaleidoscopic triangle are put together to cover the entire plane. Isn’t it surprising that the triangles have such different shapes and sizes ? Shouldn’t they all be equal ?

What you actually see is a stereographic projection of the tiled sphere. This projection causes strong distortions. The tiled sphere can better be recognized in a normal projection. To show this, we first make an inverse stereographic projection from the plane to the sphere. The dashed line in the figure above shows the equator of the sphere. Its inside gets projected to the lower hemisphere with the south pole at its center. The south pole corresponds to the intersection point of the straight lines. The outside of this circle goes to the upper hemisphere having the north pole at its center. Stereographic projection maps the north pole to infinity in the plane. We now see that all triangles of the tiling have really the same shape :

Normal projection of the tiled sphere. Lower hemisphere at left, upper at the right. Both as seen from above.

These images are somewhat underwhelming and we do not even see any difference between the points of three-fold rotational symmetry! We get more interesting kaleidoscopic images from patching a part of a photography on the triangles. For each output pixel we simply read the colour of the input photography at the mapped position of the pixel position inside the kaleidoscopic triangle. This is an example :

Stereographic projection of a sphere covered by a kaleidoscopic image with tetrahedral symmetry. Solid white lines show the kaleidoscopic triangle. The dashed line is the projected equator.

We now see a difference between the two kinds of points with three-fold rotational symmetry. Some points lie inside a grey shape and the other points lie inside a dark brown shape. The distortion of the stereographic projection appears to be even more pronounced in this image. Again, the normal projections of the decorated sphere show the symmetries and equivalencies much better :

Normal projection of the sphere with a kaleidoscopic image. Lower hemisphere at left, upper at the right. Both as seen from above.

The Apollonian gasket too is a decoration of the tiled sphere. It results from multiple reflection similar to the tetrahedral tiling. Adding a circle to the kaleidoscopic triangle we get :

Stereographic projection of the tiled sphere together with an Apollonian gasket. Solid green lines show the generating elements. The dotted line is a projection of the equator.

Now, reflections at the two lines and two circles are repeated until a point gets mapped into the small triangle made of two circles and one straight line. This triangle has two angles of 60 degrees and an angle of 0 degrees. It is a hyperbolic triangle, because the sum of its angles is less than 180 degrees. On itself, it would only create a Poincaré disc representation of a tiled hyperbolic space. All four reflecting elements together make a fractal covering of the entire plane by non-overlapping images of this disc. The image above shows essentially the borders of those discs superimposed on the tetrahedral tiling. I am using that the number of reflections required to map a point into the kaleidoscopic triangle becomes infinite at the border of the discs. Thus pixels are shown in black if they require more than a certain number of reflections. This indicates the borders.

In normal projection we see the symmetry of the gasket much better:

Normal projection of the Apollonian gasket on a sphere. Lower hemisphere at left, upper at the right.

This view shows that each triangle really bears the same decoration.

We get more from these images if we relate them to the standard way of drawing the Apollonian gasket. It starts with a single circle. Then, three circles of equal radius are drawn inside this circle, touching each other and also the first circle. You can easily identify these four circles in the two figures above. Note that they all have the same size in the normal projection. Their centers lie on points of three-fold rotational symmetry. All these points lie in the grey shapes of the other kaleidoscopic image. One of these points is the north pole of the sphere and does not appear in the stereoscopic image. Note that there are four triangular gaps between the four circles. The all have the same size and shape in the normal projection. In a second step, a circle is drawn inside each of these gaps such that it touches its borders. The centers of these circles lie at other points of three-fold rotational symmetry. In the other kaleidoscopic image, they are in the center of the dark brown shapes. One of these points is the south pole of the lower hemisphere. These additional circles leave more gaps, which are filled again in the same manner. Repeating this procedure gives the same fractal decoration for all triangles of the tetrahedral tiling, as you can see in the images above.

We get a nice and instructive image if we draw the sphere as a black shadow with the Apollonian gasket in light colour. It shows how the two hemispheres fit together :

Normal projection of the Apollonian gasket. Lower hemisphere in pale blue and upper hemisphere in pale yellow.

To get another view of the Apollonian gasket we rotate the sphere, such that a point of two-fold rotational symmetry lies at the north pole and another one at the south pole. This results in an Apollonian gasket, that looks like a fractal decoration of an Euclidean frieze :

Stereographic projection of the rotated tiled sphere together with an Apollonian gasket.

This periodic repetition is an effect of the distorting stereographic projection. Note that the parallel lines and the periodicity do not match the tiling of the sphere. In the normal projection we see that the sphere is really only rotated :

Normal projection of the Apollonian gasket on a sphere. Lower hemisphere at left and upper at the right.

The combined view shows that now the upper and lower hemispheres have the same decoration up to a rotation by 90 degrees :

Normal projection of the Apollonian gasket. Lower hemisphere in pale blue and upper hemisphere in pale yellow.

I have made all these images using my public browser app at and with a little help of GIMP. Try out this browser app. It allows you to zoom into the gasket without limits, except for computer time. In this blog and on you find more information on kaleidoscopes.

Posted in Anamorphosis, Fractals, Kaleidoscopes, Tilings | Tagged , , , , , | Leave a comment