Couple of questions about honeycombs

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Couple of questions about honeycombs

Postby Mrrl » Thu Jul 07, 2011 4:23 am

Hi,
1. Consider bitruncated cubic honeycomb in Euclidean space (filling by truncated octahedra). It is uniform, all edges loops have even length, so we can build alternated honeycomb for it. Cells of the original honeycomb are transforming to icosahedra (and we can make them regular), but what will be in space between them? Maze of non-regular tetrahedra? Or something else? What is the vertex figure of this honeycomb? It has four pentagonal faces... or not?
2. Is there any good way to enumerate cells of regular hyperbolic honeycombs (with finite cells and vertices) by exact mathematical objects - like we can use triples of integers for enumeration of Euclidean honeycombs? We can use real numbers and represent cells as coordinates of thier centers in some Poincare model, but I afraid that we'll quickly loose accuracy when go from the center. By "good way" I mean that we can easily check that two cells are the same (given their representation), and calculate action of transitions and rotations of the honeycomb. For {6,3,3} we can use elements of field Q[x]/[x^2+x+1] (or better matrices 2x2 over Z[x]/[x^2+x+1] - they give all movements of the honeycomb), and I hope that for other honeycombs with affine cells or vertices there is something similar. But what can we do with {4,3,5} and three others?
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Re: Couple of questions about honeycombs

Postby wendy » Fri Jul 08, 2011 7:20 am

1. The honeycomb o4s3s4o (which is formed by the alternation of vertices of that of the truncated octahedron), must always contain edges of two or three lengths. The thing has three free edges, and two degrees of freedom, and does not lead to a uniform tiling. On the other hand, there is s3s4o3o3o, which does comprise of uniform tiling with icosahedral walls.

2. Regards the hyperbolic polytopes like {4,3,5} etc. I did enumerate at one stage the first ten rings of {3,3,3,5}, in terms of a sum of polytopes. The thing runs exponentially, which means that at ring x, there are something like 10^(x/k) vertices. The whole thing about hyperbolic geometry is that the circle is asymptotic on the exponent of the radius. This is why, for example, the {6,4} can contain as a subgroup, that group of {3,3,3} or {4,3,4} or {5,3,5} or even {6,3,6} or {7,3,7}.
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Re: Couple of questions about honeycombs

Postby Mrrl » Fri Jul 08, 2011 7:55 am

wendy, thank you for the reply. What I need is not just count cells of the honeycomb, but be able to navigate in its connectivity graph. I found the representation of the symmetry group of {4,3,5} as a subgroup of Moebius functions generated by f1(z)=i*z, f2(z)=(z-1)/(z+1) and f3(z)=a*z where a=phi+sqrt(phi) (these functions act on the "boundary" plane z=0 of half-space Poincare model). Every function of the group defines some cell of the honeycomb with its orientation (so 24 functions show the same cell). Numbers "a" and "i" are elements of the field Q[b,i]/[b^4-b^2-1,i^2+1], so I've got representation of a cell as a point n Q^24 (three numbers from the field define one Moebius function). But it looks a little too complicated.
This represenation gives me a chance to get a periodic patterns on the honeycomb (subgroups of the symmetry group with the finite factor?). So far I see patterns with fundamental areas of 650 and 1015 cells (that is good for the maze but too much for the twisty puzzle). Is there any place where I can find description of smaller patterns?
Digging in Wiki I found that there exists honeycomb made of (8,6,4) bodies (truncated cuboctahedrons). Never met is before. Should be nice thing!
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Re: Couple of questions about honeycombs

Postby wendy » Mon Jul 11, 2011 7:46 am

{4,3,5} is hardly what one might call suitable for this sort of activity. It's rather like deriving the cubic from the body-centred cubic, or something like that. It's not a lattice, but 2/5 of the vertices of the dual {5,3,4}, and double of the {3,5,A}. These are indeed more suitable. Any two sets of {3,5,A} in {5,3,4} make a {4,3,5}.

{5,3,4} of course, contains {5,4}, while {3,5,A} contains sections that equate to {3,10}, {5,6} and {10,5}. Also, it's a section of {3,3,3,5}.
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