by **wendy** » Wed May 13, 2009 8:57 am

You should understand that the dynkin symbol is not so much just a handy way of writing some polytopes, but an actual coordinate system, no less than the usual x,y,z system. A figure like x3o3x corresponds to 1,0,1 (all changes of sign). The big difference here is that instead of 8 changes of sign, ye can have 24 or 120 or whatever changes of sign.

It then becomes possible to think of a shape that is 20% cube and 80% cuboctahedron or whatever. This is what makes "progressions" tick. You can even have something that's 30% C + 50% CO + 20% O. It's a kind of truncated-cuboctahedron, but the three kinds of edges are not equal.

The standard coordinates for a cube is (1,1,1) ACHS. This gives a edge of 2 (eg 1,1,1 to 1,1,-1). For a dynkin symbol, the standard is to make the edges 2, which makes the plane v=0 to v=1 differ by unit-thickness. Since 0,0,0 to 1,0,0 is a sloping line, this distance gives no direct relation to the thickness.

The relevant axies for the octahedral group are: [q=sqrt(2)] O = (q,0,0), CO = (q,q,0) and C = (1,1,1). The effect of symmetry is all sign, all permutations, applied after the sum. One can see that the tC is o3x3x = 0,1,1 = CO + C = (q+1,1,1). Of course, you can go entirely gradually, so that one goes from C to CO smoothly. The trick here is that it's a progression of x* C + y*z*CO, where x+y = 1, and z is some free scale-factor. Going from the CO to the O, requires some fiddling, but the same thing occurs.

You can of course progress any figure to any other, since at x% of the change you have x% (TOP) + y% (BOTTOM), where x+y=100.

The class of figures formed by this form of progression are 'lace prisms', and their duals are 'lace tegums'. In 3D, you can have T, B as any of (point, figure, dual, truncate). The relevant lace-prisms are point-figure = pyramid, figure-figure = prism, figure-dual = antiprism, figure-truncate = cupola. Of course, if T and B can be different sizes, you can have figure-big figure = rostrum.

For example, the progression of a dodecahedron, pentagon first is:

x5o [f] f5o [1] o5f [f] o5x. f = 1.61803398875&c.

This is a pentagonal (rostrum + antiprism + rostrum), which can be seen by drawing in the three layers. An icosahedron goes pentagonal (pyramid + antiprism + pyramid). Once you get the different heights right (here shown [f], [1]), it's quite easy to draw something like a pentagon of size 1.309017 for a distance of 0.309017 of the total height of 4.236067.

[we note that 0.309017 is half of f, so the figure is 50% (x5o) + 50% (f5o), or (0.5+0.5f, 0) on the pentagonal coordinates. This is a pentagon of that size (ie 2.618 / 2 = 1.309017).]

One can calculate the kinds of faces of any lace prism, because all the sloping faces are lace-prisms. You apply the same rule as for ordinary dynkin-symbol, but match out the corresponding nodes top and bottom, eg

x3o5x || x3x5o

x3o5x || ----- top = rhombo-ID

------- "|| x3x5o bottom = trunc icosa

* o5x || * x5o anti-prisms (12)

x * x || x * o triangular prism (30), literally rectangle on-top line

x3o * || x3x * triangular cupola (triangle-hexagon of cuboctahedron)

This is the usual "cover a node to reveal a face".

The dream you dream alone is only a dream

the dream we dream together is reality.