x4oW(1/2)o and x4oW(3/2)o

Higher-dimensional geometry (previously "Polyshapes").

x4oW(1/2)o and x4oW(3/2)o

Postby wendy » Sun May 24, 2015 7:27 am

I've been mainly looking after people, and my polytope thoughts have been with these pair of interesting polyhedra. They both have squares as faces, the dihedral or margin-angle being 60 degrees and 120 degrees respectively. [These are regular figures of infinite densities].

What makes these pair interesting, is that they correspond to an infinite cover of squares from the triangular and hexagonal prism, respectively. This means that if you put six x4oW(1/2)o around an edge, you get the same ring of squares as an x4oW(3/2)o. If we suppose that Proffessor Coxeter is right, and that a polytope whose faces are centre-inversion, also is centre-inversion, it means that the 1/2 form forms the squares of a hexagram prism, and that three of these stacked on opposite squares, will form the opposite squares of 3/2 form.

What is even more interesting is that x4oW(1/2)x = x4oW(3/2)o. One can see from the bridging constants for these groups, that the nodes of the first are at the ratios of x, q, x, and of the second, x, q, h. This means that there are no further subgroups of these figures. I thought one could expand the edges of the first out to the sphere of the second, but this would not fit well either, because it would imply that x12o is a subgroup, when it clearly isn't.

Both of these are of course, infinitely dense, as neither of xW(1/2)o or xW(3/2)o close. The vertex figure, for an edge of root-2, of these polygons are q:q:1. and q:q:h respectively. I calculated the van Oss polyhedra for these as W(20/7) and W(16/5) respectively, these are fairly near the hexagon, the first slightly smaller, the second slightly larger.

Later i was able to calculate the Schläfli Index for these two groups (3 and 1 respectively), which leads to the correct vertex-diameters for both of these figures D2 = 7/3, and D2 = 5 respectively).

But there are some interesting things i am still unsure of, as i have not as yet walked the surface of x4oW(1/2)o far enough to verify a direct connection to imply central inversion, or that at least three edges cross in right angles, that one can state catergorically that central inversion holds. It is certainly true that the number system known for this system does not support three equal edges crossing at right angles (although they do so in pairs), because the third axis is at ratio of h to the first two.
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wendy
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