> Wasn't it Anton Sherwood who wrote:
 >> (What I have in mind is to illustrate the various point symmetry
 >> groups with seamless pigments that show the relevant symmetries.)

Mike Williams wrote:
 > Now you've lost me.

May I always be so fortunate, that the most obscure parts of my writing 
are in the footnotes.  ;)

Consider the Mercator projection: the surface of a sphere is mapped to 
that of an endless cylinder.  I've heard that someone once made a globe 
on which the reverse Mercator projection was applied twice, i.e. two 
copies of a six-inch cylinder (say) were used to form a twelve-inch 
cylinder, and the projection applied in reverse to a globe, with the 
result that an orthographic eye in the plane of the equator would see a 
map of the entire world.  The oddity was said to be hard to spot - 
partly because the mapping is conformal (shapes are locally preserved) 
and partly because the doubling is seamless (except for a singularity at 
the poles).

Well, given a `globe' pigment, this multiplication effect is 
accomplished with the following functions (if my algebra is good):

	Xnew = Re[(x+i*y)^n] / ((1+z)^n + (1-z)^n)
	Ynew = Im[(x+i*y)^n] / ((1+z)^n + (1-z)^n)
	Znew = ((1+z)^n - (1-z)^n) / ((1+z)^n + (1-z)^n)

(n=2 for the above example.)  The pigment of the resulting globe has the 
symmetry group C[n].  Other symmetry groups can arise from other sorts 
of functions.  Simple examples, not conformal:

	Xnew = Re[(x+i*y)^n]
	Ynew = Im[(x+i*y)^n]
	Znew = z^2

This gives C[n]h, i.e. the same nfold symmetry as above plus a 
reflection plane perpendicular to the axis.

	Xnew = Re[(x+i*y)^n]
	Y1  =  Im[(x+i*y)^n]
	Ynew = Y1^2 - z^2
	Znew = z*Y1

This gives D[n], the rotation group of an n-gonal prism.

...Thanks for the code!  I'll play with it when I'm more awake.

-- 
Anton Sherwood, http://www.ogre.nu/

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