List of uniform honeycombs

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Re: List of uniform honeycombs

Postby polychoronlover » Thu Jun 01, 2017 6:04 am

wendy wrote:A number of stary hyperbolic tilings of finite density are known. But there are not a lot of star-subgroups up there, and we mostly rely on Coxeter's theorm that starry symmetries are nade from regular ones.


There's a theorem that symmetries of nonconvex Wythoffian figures can always be described by Dynkin diagrams for convex figures? I'm not surprised; this seems true in practice but it's nice that there's a proof.

I wonder how well this extends to compound polytopes. It's probably still true if the compound is formed by replacing each instance of an element or element-figure (whose symmetry is given by the Dynkin diagram with one node removed) and replacing it by a compound that preserves the symmetry of each component. Such compounds are the only ones I consider to be "true extensions" of Wythoffian polytopes. For example, the rhombihedron (compound of 5 cubes) doesn't count because the triangular symmetry around each vertex of a cube degrades into chiral triangular symmetry in the compound.

There is a similar subclass of uniform polytopes that I call "locally uniform", where each element is not only uniform but also has the property that symmetries of the original shape can map a vertex from one element onto every other vertex of the same element, while still mapping the element onto itself. This filters out almost all polytopes whose vertex configurations can't be made by kaleidoscopical construction; it excludes the antiprisms and snubs in 3 dimensions and almost everything from Bowers' "miscellaneous" categories in 4 and 5 dimensions.
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Re: List of uniform honeycombs

Postby wendy » Thu Jun 01, 2017 8:27 am

Wythoffian simply means that it has a construct from a Dynkin symbol. Wythoff discovered the mirror-edge reflections that the marked nodes of the Dynkin symbol means.

There are a number of non-wythoffian hyperbolic and euclidean tilings, and a number of star-tilings as well, although the star-tilings are all wythoffian.

For example, there is a tiling of truncated cubes, 16 at a corner. It derives from o8o3x4x, but the cell o8o3x is perfectly flat (ie concentric with the tiling), and can be used as a mirror. This leads to space filled with truncated cubes o3x4x, the vertex figure is a octagonal tegum oxo8ooo&#tq. There is a tiling of rhombo-cuboctahedra x4o3x, and triangle prisms x2x3o, sixteen of each at a vertex: the resulting tiling has a vertex figure of oxqxo8ooooo&#tq (the #tq means a tower with lacing of sqrt(2), rather than 1).

The usual form for non-wythoff star-tilings in the euclidean plane is the pleat-fold. You take some sort of tiling like LPA2, which is an alternating layers of the oct-tet tiling and a triangular prism. But instead of making the oct-tet on top of the triangle prisms, like ---o----, you make the common plane into a pleat, like o====. The oct-tet tiling then faces into the paper, while the prisms face outwards.

There's an ASN x4x4x, whose vertex figure is a pentagram (5/2), with an array of 4,3,4,3,3.

The group o5/2o5o5/2o3*a (d4) and o5/2o5o5/2o3o3*a (d8), both produce stars, as well as the regular x5/2o5o3o3o, and x5o5/2o5o3o. I have not really looked into o4o3o3o3o3o3*a to see if it makes stars yet, but the edges on it are quite long.
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Re: List of uniform honeycombs

Postby wendy » Thu Jun 01, 2017 8:51 am

The proof of Coxeter's theorm, is that when you divide a symmetry group into two parts with a mirror, these are also symmetry groups, since the mirrors reflect what's on one side. The final destination here is usually a simplex-group that has its angles as fractions of a half-circle 'pi/n'.

The 'Wythoffian' group uses simplex mirror groups, like triangles, tetrahedra, etc. You can show that a vertex is equidistant from each of two walls, by bisecting the angle between the walls. Likewise, the intersection of two bisectors will give a set of points where A=B, C=D, or A=B=C. And so forth. The Mirror-edge construction relies on the transport of a vertex over a mirror, eg v----|----i v = vertex, | is mirror, i is image of v, -------- is the line.

While you can construct mirror-edge polytopes in nearly any symmetry, you can only guarrantee that the thing is uniform in the simplex groups. Moreover, the various matricies I wrote for Richard rely on the simplex groups. These are 'wythoff symmetries', and we usually use a 'dynkin' graph for it. A 1:q rectangle supports mirror edge figures, but they are not uniform.

WME are 'wythoff mirror-edge figures'. These consist of all of the available dottings of a Dynkin symbol.

WMM are 'wythoff mirror-margins', that is, figures whose faces (N-1) reflect into each other through the margins (N-2). The margin angles and cell inspheres are all identical.

WSN are the 'wythoff snubs'. All of these can be derived from deriving the omnitruncate of some WME. This group includes the antiprisms, the snubs in 3 and 4d, and snub tilings. Unlike the WME and WMM, these are generally irregular, because you are trying to position N vertices of a simplex so that the N(N-1)/2 edges are all equal. When N-1=2, this is always possible, but in 4D, you have N=4, and edges = 6.

Then you have the laminate tilings. These are made by taking uniform layers, like a band of triangles or a band of squares, and alternating them. The non-wythoff LPC1 is a tiling of strips of squares (P = prismatic layers), and a 1-dimensional coordinate swap C1. (this gives the triangles).

Many, but not all, of the hyperbolic tilings are of these kinds. There's an infinite class of tiling in H3 not listed here, the borrochoral group.

Then you list
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Re: List of uniform honeycombs

Postby Klitzing » Fri Jun 02, 2017 2:12 pm

wendy wrote:For example, there is a tiling of truncated cubes, 16 at a corner. It derives from o8o3x4x, but the cell o8o3x is perfectly flat (ie concentric with the tiling), and can be used as a mirror. This leads to space filled with truncated cubes o3x4x, the vertex figure is a octagonal tegum oxo8ooo&#tq. There is a tiling of rhombo-cuboctahedra x4o3x, and triangle prisms x2x3o, sixteen of each at a vertex: the resulting tiling has a vertex figure of oxqxo8ooooo&#tq (the #tq means a tower with lacing of sqrt(2), rather than 1).

The first verf there is wrong. It rather is oxo8ooo&#kt. Here k=x(8,2) is the chord of an octagon.

The second verf, oxqxo8ooooo&#qt, was known to me as polyhedron (octagonal ball). But never thought about it as being used as a verf. Thought about it. Found x4o3x *b8o. That then describes half of the rhombicuboctahedra at least. The corresponding square tiling of order 8 should work in a similar way as a mirror. Thus obtaining the other rhombicuboctahedra too. But when thinking about the triangular tilings of order 8, then those have a different curvature. So you could cap these, say by a full triangular prisms layer - as your verf would imply - but thereafter? I would not assume, that this so far constructed hyperbolic honeycomb could be made uniform, i.e. having a single vertex type only.

Or am I missing something?

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Re: List of uniform honeycombs

Postby wendy » Sat Jun 03, 2017 8:12 am

The first style is in the original lacing form, where a lacing-edge of q is a vertex-figure of x4x. But we use real lengths now, so it should be a(8) here. So you are right.

The second vertex-figure (octagon-ball = oxqxo8ooooo&#tq), is indeed correct. The polar caps are a prismatic layer of x3o8o, of which you see only the eight prisms at the end of a prism. The middle two layers, can be variously read as x4o8o *b3x, where the equatorial plane is both a tiling of x4o8o, and a mirror.

But the octagon ball consists of two cuboctahedra, rotated by 45 degrees around an axis through the square-centres. So there are two intersecting o4o8x *b3x, whose octagons appear as the octagonal layers of the rCO x4o3x, and there is a common dividing symmetry that turns one of the four-fold axies into an eight-fold one.

The lines of longitude are also x4o8o, these are variously the square faces of x4o3x (particularly the set of 12, the set of 8 are the equatorial planes), and the square faces of the triangular prisms.

The first one involves a lot of similar crossing symmetries of x8o3o4o. The vertex-figure is comprised of a pair of octahedra, being the vertex figure of x8o3o4o, some octagons appear in full as the faces of tC, the others are eight-triangles of x3o8o. You can feed a second o8o3o4x whose vertices are the centres of cells. This one never crosses the layer of triangles, and is rather like the horizontal planes in x4o3o4o, reflected in the vertical.
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Re: List of uniform honeycombs

Postby Klitzing » Sat Jun 03, 2017 11:56 am

wendy wrote:The second vertex-figure (octagon-ball = oxqxo8ooooo&#tq), is indeed correct. The polar caps are a prismatic layer of x3o8o, of which you see only the eight prisms at the end of a prism. The middle two layers, can be variously read as x4o8o *b3x, where the equatorial plane is both a tiling of x4o8o, and a mirror.

But the octagon ball consists of two cuboctahedra, rotated by 45 degrees around an axis through the square-centres. So there are two intersecting o4o8x *b3x, whose octagons appear as the octagonal layers of the rCO x4o3x, and there is a common dividing symmetry that turns one of the four-fold axies into an eight-fold one.

The lines of longitude are also x4o8o, these are variously the square faces of x4o3x (particularly the set of 12, the set of 8 are the equatorial planes), and the square faces of the triangular prisms.

Yes, so far I'm fully with you. But kind of I have to insist:

Your x4o8o *b3x clearly is the same as I'd put it: x4o3x *b8o. That one surely is a uniform hyperbolic honeycomb (or "polytope" if you'd like to put it). The x4o8o tilings (or "cells") then play the role of a mirror. Also already accepted. Thus the mirror image of the so far obtained laminate (i.e. having only x3o4x and x3o8o cells left) still would be uniform. And its vertex figure so far clearly is xqx8ooo&#q.

But then you want to add some caps onto these clearly curved surfaces x3o8o. In fact there you want to add a full layer of triangular prisms each. Sure, this reflects within the vertex figure as getting the full octagonal ball oxqxo8ooooo&#q. But what about the thereby added vertices "inside" or "atop" the x3o8o cells?

My guess here would be, that these wouldn't any longer be part of a further uniform honeycomb. Rather you probably would obtain a biform one. As there you would have for further vertex figure some ox8oo&#q so far. - Sure, you then could say, that these again are parts of the other ones (the octagon balls), so the laminate there might be somehow continued there the other way round again.

Might be. But I don't see how that could take place. For the x3o8o has a different curvature than x4o8o. The latter shall work as a mirror to x4o3x *8o. Thus I assume that the curvature of the latter is the same as that of x4o8o. (And indeed: the "circumradius" of x4o3x *b8o evaluates as 0.321797126 i and that of x4o8o too. But the one for x3o8o evaluates as 0.594603558 i. Each wrt. unit edges, for sure. Simply put it into your spreadsheet, hehe.)

So: how could the prism layer be used likewise as to mime a 2nd mirror?

Well, that 2nd mirror type (if truely existing) then won't be actually the x3o8o of x4o3x *b8o, rather the midway height section of these prisms, so there is still some space for arguments here ... - But at least I cannot see how that one could be supported with any logics ... - Could you help me out here?

Hope, I made my point better understandable now.

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Re: List of uniform honeycombs

Postby wendy » Sat Jun 03, 2017 12:42 pm

The mirror runs through the half-height of the prism.

The edge that runs from the centre of the vertex-figure to the pole, runs between eight triangular prisms, (tips, if you will), this edge has a mirror bisecting it so the image of the centre is the pole-vertex.

You are looking at a flat x3o8o.

If you take the equator of x3o3o5o, which is o3x5o, you would find that the distances are identical in 3d and 4d. This is because the polytopes are concentric with the centre. If you move the o3x5o so that it's a face of o3o3x5o, then the edges are shorter, but the ratio of chords remain the same. The same applies in hyperbolic space, but the edges get bigger.

The trick is to look at the guage theorm, which is about working the edges of x4oPo and o4oPx. In xPo4o, the van Oss polygon (ie the girthing polygon by going 180 degrees across the vertex), is straight. But the edge of x4oPo is longer, but if you move it a half-edge at the centre, the centres of the vertical edges would exactly match the vertices of xPo4o, and the vertices of x4oPo would then correspond to the projection of the first line onto two curving bolospheres.

Thus you can use the spreadsheet, to verify, that x3o8o has the same edge length as x8o4o. This is in the mirror at the waist of the prisms. Now calculate the x4o8o, which replaces the edges of x8o4o with squares, bisected from edge to opposite edge. This turns the triangles into uniform triangle prisms x2x3o, You will see that this edge is the same as that of E x4o8o = E x8o6o. So indeed, the edges of the 'tip' are the same as the rest.

So you really end up with a slab of prisms, with a mirror running entirely in the interior of the layer. And the prisms are uniform.

What happens is that in hyperbolic space, the x3o8o here is curved like a line of lattitude. On the sphere, the line of lattitude is concave to the equator, the great circles are on the other side of the line. In hyperbolic space, an equidistant to a straight plane, is convex to the plane, the straight line descends closer to the plane. The x3o8o in this figure is clearly convex towards the pole, that is, the angles formed by the two tips is less than the angles of the two rCOs. This means that the figure, without the poles would be concaved, and adding the prism layers makes it convex. But without repeating the middle lattitudes again at the pole, it has different kinds of vertices, and hence not uniform.
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