There are only a few choices for k, m and n that give elliptic kaleidoscopes. For m=n=2 we can use any value for k that is larger than 1. The center of the circle lies at the intersection of the straight lines and the circle crosses the straight lines at right angles. This gives us rosettes with the same symmetries as discussed earlier in “combinations of mirror symmetries“. But now we use explicit inversion at the circle instead of symmetric mapping function. Thus there is not a hole at the center and less distortion. A typical result looks like that:

More interesting are the remaining kaleidoscopes. They have the same symmetry as the tetrahedron, the octahedron and the icosahedron. An octahedron has a four-fold rotational symmetry at its corners, where four triangles meet. At the center of the triangles we have a three-fold and at the middle of its edges a two-fold rotational symmetry. A corresponding kaleidoscope has the same symmetries. It generates a stereo-graphic projection of a sphere with octahedral decoration as shown in this image:

The yellow triangle is the basic kaleidoscope. The dark brown triangles are its images with an odd number of reflections and inversions and images resulting from an even number are light blue. This is a typical image of this kaleidoscope:

There is a five-fold rotational symmetry at the corners of an icosahedron. The corresponding kaleidoscope has five-, three- and twofold rotational symmetries. The petals of a rose waterlily flower gave this result with icosahedral symmetry:

A tetrahedron has a three-fold rotational symmetry at its corners. A kaleidoscope with tetrahedral symmetry thus uses a triangle with three-fold rotational symmetry at two of its corners and a two-fold rotational symmetry at one corner. Its images look like this one:

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This becomes easy if the intersection is at the origin and using polar coordinates. The mapping does not change the radius of the point. The angle ψ of the mapped point oscillates between 0 and π/k like a triangle wave with a period of 2π/k when the angle φ of the original point goes from 0 to 2π. Thus, in a first step I do the rotational symmetry with ψ=φ mod (2π/k). Then comes the mirror symmetry. For ψ>π/k I put ψ->(2π/k)-ψ.

To get the image of the kaleidoscope on the Poincaré disk I simply alternate this calculation with an inversion from the inside of the circle to its outside. This stops when the point lies inside the triangle of the kaleidoscope. But the resulting mapping has a discontinuous derivative at the mirror lines and at the inversion circle. This makes that sharp angles appear in the output image if the input image has straight lines and other artifacts. We can easily improve the image near the straight lines.

Without inversion at the circle we get a simple rosette. This is a typical result:

You can clearly see the discontinuities, especially the angles. We can improve and remove the discontinuities with an extra mapping that has a vanishing derivatives at the straight lines. In polar coordinates we can use a Fourier expansion of the triangle function of φ, as discussed in the earlier post “Better images from higher harmonics“. With the first three terms we get a smoother image with rounded edges:

This gives kaleidoscopic images like this one:

There is another possibility. We can use the rosette mappings discussed earlier in “How to generate rosettes” and “Rosettes with mirror symmetry“. They are built on powers of exp(i k φ) and cause strong distortions destroying details of the input image. This is a typical example:

At first sight this is a disappointing mush. To my surprise I got nevertheless interesting kaleidoscopic images, such as this one:

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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 disc 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).

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Yes, we simply move the vertical lines! To get an n-fold rotational symmetry the circle has to intersect a straight line at an angle of π/n. The distance between the line and the center of the unit circle is then cos(π/n). We can freely choose the rotational symmetry at the left and at the right of the unit circle as long as the sum of the angles of the kaleidoscopic triangle is less than π.

Here is an example with 2- and 4-fold rotational symmetry:

And here comes a symmetric kaleidoscope with two 3-fold rotational symmetries:

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Farris uses algebra and group theory to discuss the structure of these images. To me this is fascinating and challenging. To see if I really got it I tried to explain these images using geometry.

First, the most important properties of hyperbolic space: The Poincaré plane occupies the upper half plane. Equal distances in the hyperbolic space seem to shrink going to the x-axis because of the projection. A particle with constant speed in the hyperbolic plane would never reach our x-axis. Thus the x-axis is infinity far away in the hyperbolic plane. Straight lines of the hyperbolic space are projected as circles in the Poincaré plane. Their centers lie on the x-axis. If the radius of the circle goes to infinity we get a straight vertical line. A mirror images at a straight line in hyperbolic space becomes an inversion at the corresponding circle in the Poincaré plane. Note that the inversion at a circle maps circles into circles. It has to be like that, because straight lines in the hyperbolic plane have straight lines as mirror images. Similarly, inversion at a circle preserves the angle between intersecting circles.

I have drawn the two vertical lines and the unit circle in another image of the same symmetries:

Note how the vertical lines generate the mirror symmetry and periodicity in x-direction. At the point (0,1) the circle intersects the vertical line at a right angle. Thus we get a two-fold rotational symmetry with mirror symmetries at this point. Going away from this point distortions set in. Repeated copies lie at the line y=1, which is not a straight line in hyperbolic space. Inversion at the unit circle maps this line into a circle of radius 0.5 with its center at the point (0,0.5). Thus images of this center of two-fold rotational symmetry appear on this circle. The first one is at (0.5,0.5). At (0.5,0.866) there is an intersection angle of 60 degrees between the vertical line and the unit circle. This makes a center of three-fold rotational symmetry. Again, copies appear on a horizontal line and on circles.

Let us compare this with kaleidoscopes in Euclidean space, as discussed in “Geometry of kaleidoscopes with periodic images“. In general they have three straight lines that serve as mirror axis and form a triangle. Reflections of its inside cover the plane. Thus the kaleidoscopic image arises. Here we have something similar. The two straight lines and the circle make a triangle in the hyperbolic space if you accept that the parallel lines meet at infinity with a vanishing intersection angle. Again, we can use mirror symmetry and inversion to cover the plane.

To show the structure more clearly I filled the basic triangle region with a dark brown color. Each reflection or inversion switches the color from dark brown to light blue or inversely. In the resulting image we can easily see the different images of the basic region:

Look out for centers of two-fold and three-fold rotational symmetries. How are they mapped by the inversion at the unit circles?

I have created my images using iterated reflections of points and not with symmetric bundles of wave function as Farris. Thus I get discontinuities in the first derivative of the images and my images are not so smooth at the unit circle. But I can generate images much faster and easily. Thus I can rapidly change the elements of symmetry.

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With only the basic sine and cosine waves I get:

The center of perfect 5-fold rotational symmetry lies slightly at the right of the center of the image. Note the concentric shapes inside a pentagon and the large orange bulls-eyes.

Adding the third harmonic I get more details:

At the center we now have star like shapes instead of the pentagon and the circle.

Using the basic wave together with the third and the fifth harmonics gives smaller changes:

Finally, the triangle wave makes:

Now, straight line segments appear. The image has a more crystalline appearance. The rather large orange blobs remain, but they have now a more interesting polygon shape.

I like that I can tune the images changing the wave function, but I am disappointed that the large blobs do not disappear.

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This is a basic result that shows mainly the color-changing structure:

And here is a more subtle image that shows increasingly complex stars of 5-fold symmetry

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Zoom in to see the molten watch faces in this image:

Here I used the portrait of a damselfly, but you wont recognize much:

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