Now I want to present color symmetry upon rotation for periodic and quasi-periodic kaleidoscopes. We have n different versions how to show the pixel colors of the input image in the new output image. For a color symmetry we have to make good choices of these versions. As you can get from the book “Creating Symmetry” of Farris and as discussed in my earlier post “n-fold color symmetry” we make a mapping W(x,y) from the position (x,y) of a point in the output image to a complex plane. We divide the plane into n equal sectors. The sector containing the mapped point W(x,y) determines the color version we have to use.

We combine the n-fold color symmetry with an r-fold rotational symmetry. A rotation of the image plane by an angle of 2π/r maps to the same position of the input image X[R(2π/r)(x,y)]=X(x,y) and Y[R(2π/r)(x,y)]=Y(x,y). But in the complex plane of color choices we get a rotation by 2π/n resulting in another color choice. Thus

where multiplying with the exponential function makes a rotation. Clearly, r must be an integer multiple of n to get unique choices.

As for the mapping X(x,y) and Y(x,y) to the input image we make symmetric packages of waves. As discussed in the last post, except for additional phase factors:

Finally, the mapping is a sum of such packages

with constants C of complex value.

If the order r of the rotational symmetry is odd, then it equals the dimension p=r of the embedding space. The package of waves are then trivially given by

Even orders r of the rotational symmetry are double the dimension p=r/2 of the embedding space and we can simplify the calculation using

which results in

Here we have two different cases. First, if 2p/n is an even number, then exp(i2πp/n)=1 and thus

Second, if 2p/n is an odd number, then exp(i2πp/n)=-1 and thus

These equations are particularly easy to evaluate for k-vectors that have only one nonzero component or two neighboring nonzero components.

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