Interesting polytopes with different edge sizes

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

Interesting polytopes with different edge sizes

Postby Klitzing » Mon Dec 19, 2016 5:56 pm

Thought to make up this thread, 'cos there are lots of interesting polytopes out there, which do not qualify as Wythoffian, uniform, scaliform, CRF, etc.
Special examples probably are found in the Catalans, cf. this thread, or the perfect Gévay polytopes mentioned here.

Today I managed to wrap my mind around an interesting convex polychoron of this type. It just uses 2 different edge sizes. These are x (unity) and v (=1/f = 0.618...).
The cells of this polychoron are regular unit dodecahedra x5o3o and two further convex polyhedra, which use both edge types.

The first additional cell type could be described as x3v3o, i.e. a truncated tetrahedron with only semiregular hexagons. Here its semiregularity does not affect the angles between the sides, those still are all 120 degrees. Rather it affects the edge sizes, which would alternate. The triangle then are again regular, but using the smaller edge size v only.

The other additional polyhedron then is best understood, when refering to pero (pentagonal rotunda). That one clearly has regular pentagonal lacing faces, a further regular pentagon at the top and a regular decagon at the bottom. The remainder then are regular triangles. In the figure to be used here we only apply trigonal axial symmetry. Then the figure could be described accordingly as a triangular variant of pero, as vov3ofx&#xt = v3o || o3f || v3x (where f=1.618...).

The idea to the to be described here polychoron in fact was applying the Gévay ideas a bit beyond his usages. I tried to place other polyhedra into the positions of cells of well-known polychora in some symmetrical way and then apply the convex hull to this. So I started here with the f-scaled rectified icositetrachoron (rico) o3f4o3o. And then I placed vertex incident unit dodecahedra at the places of the former f-cubes. Here the dodecahedra at the one hand would break the full symmetry of the cubes down into a pyritohedral one only, and on the other hand will reach beyond. Therefore the other cell type of rico, the cuboctahedra, get augmented by some kind of cupola. Thus we will obtain these truncated tetrahedra atop the former cuboctahedra, will have these dodecahedra at the places of the former cubes, and additionally would need some further figure to fill in the remainder, esp. in order to match these regular unit pentagons, these smaller regular triangles and these semiregular hexagons. This is how this trigonal variant of the pentagonal rotunda comes into play. Here the medial vertex layer, the f-scaled triangle, then will be just the remainder of the "old" triangular faces of o3f4o3o. The extension thereatop resp. therebelow then are the lacing parts of these asked for cupolae, which happen here to be co-realmic, one side then attaching to the small v3o triangle of these truncated tetrahedra, while the other one would attach to the semiregular hexagons of those. The latteral regular pentagons then clearly attach to the regular unit dodecahedra, while the acute golden triangles ov&#x = o || v just happen to attach to further such cells.

The full incidence matrix then just is:
Code: Select all
96   * |   6   0   0 |   6   3  0  0 |  2  3  0
 * 288 |   2   1   2 |   3   2  2  1 |  1  3  1
 1   1 | 576   *   * |   2   1  0  0 |  1  2  0  x
 0   2 |   * 144   * |   2   0  2  0 |  1  2  1  x
 0   2 |   *   * 288 |   0   1  1  1 |  0  2  1  v
 2   3 |   4   1   0 | 288   *  *  * |  1  1  0  x5o
 1   2 |   2   0   1 |   * 288  *  * |  0  2  0  ov&#x
 0   6 |   0   3   3 |   *   * 96  * |  0  1  1  x3v
 0   3 |   0   0   3 |   *   *  * 96 |  0  1  1  v3o
 8  12 |  24   6   0 |  12   0  0  0 | 24  *  *  x5o3o
 3   9 |  12   3   6 |   3   6  1  1 |  * 96  *  vov3ofx&#x
 0  12 |   0   6  12 |   0   0  4  4 |  *  * 24  x3v3o

Would anyone have an idea on how to call that found polychoron?

--- rk
Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Interesting polytopes with different edge sizes

Postby Klitzing » Mon Feb 12, 2018 4:44 pm

Today I want to describe a recent find of mine, which is a 6D polytope (aka polypeton) with 2 different edge sizes. So it might be best described in this thread.

The most exiting figure of 6D surely is the Gosset polytope 221 (aka jak). Jak has 27 vertices and 2 types of facets. 27 of those are 5D crosspolytopes (aka tac) and 72 are 5D simplices (aka hix). This figure surely is uniform and therefore still uses a single edge size, say unity.

It happens that the 27 tacs are situated right opposite to the vertices in jak. That is, when considering the compound of 2 jaks, then this figure is possibly made fully similar to the stella octangula in 3D, i.e. the compound of 2 dual tetrahedra. In fact this compound of 2 alternate jaks, to which has been given the acronym madek (m=54, d=dis, k=27), would have 2*27=54 vertices, all of which happen to lie exactly above those tac facets. Further it happens in this arangement, that the hixes too form subdimensional compounds, as those would align by 2 in mutually dual orientation in a single affine subspace. That 5D compound of 2 hixes was given the acronym stade (st=stellated, d=dot=dodecateron).

Now I tried to consider the convex hull of madek. As it happens, all edges of both jaks would survive in this hull as its long edges. But there come in others too, which are shorter by a factor of sqrt(2/3), as connections of the vertices of both jaks.

Obviously we too have to consider for that purpose the subdimensional convex hull of stade as well. Wrt. the edge sizes nothing has to be said here any more, those are dictated by the 6D scenario already. Right from the definition of stade, it was derived from 2 dual hixes. Thus its kernel (intersection) is nothing but a dodecateron (dot). Accordingly, as the vertices of stade happen to be atop the facets, the 5D hull of stade is nothing but the dual of dot.

Thus the 6D hull of madek surely would have 72 duals of dot for some of its facets. And, as it turns out, those are already all facets being used. The fattening from stade to its hull, the dual of dot, already eats up all remaining space. Thus we are left for this 6D hull of madek with a polyteron with 54 vertices and 72 facets, having edges of 2 different sizes.

But what was even more surprising, this convex hull of madek happens to be nothing but the dual of the Gosset polytope 122 (aka mo).

Accordingly we just have provided a tegum sum description each for the dual of dot and for the dual of mo.

- Has anyone come up with these before?

--- rk
Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Interesting polytopes with different edge sizes

Postby username5243 » Mon Feb 12, 2018 9:53 pm

Wow, interesting. Didn't see that one coming.

What's interesting, to me, is that not only is this one noble (ie, identical vertices and identical facets), its facet is too. The dual of dot should have 20 facets in the shape of bi-triangle tegums, if I'm not mistaken. And even those 4-D elements are noble too, having equal vertices and 9 disphenoid cells.
Posts: 56
Joined: Sat Mar 18, 2017 1:42 pm

Re: Interesting polytopes with different edge sizes

Postby wendy » Tue Feb 13, 2018 9:55 am

The 4B\ as Richard describes, is made from two 2_21 polytopes, inverted, ie /4B + 4/B = xo3oo3oo3oo3oxBoo. The seventytwo faces come from the 72 simplexes or hexaterons, which are likewise inverted. Doing this makes the triangles of these intersect perpendicularly, which leaves margins of bi-triangular tegums.

4B\, of course, tiles, as 4B1\ or the vornelli cells of the 2_22. The tri-triangular tegum also tiles (as does the prism), there are 720 tegums, or 216 prisms at a corner. Likewise, the 2_21 itself tiles. The relation between 4B\ and 2_21 is a doubling, preserving the simplex-faces, so it is much like the 3d cube -> rhombic dodecahedron process. It is known in many different instances.

The polytope can be made of edges sqrt(2) and sqrt(3), which means it can serve as a vertex-figure of a hyperbolic tiling. The tiling then consists of cells of one type of cell (The margin-tegums turn into bi-hex-lat combs (ie euclidean-tile-prisms),), there are 20 of these at a vertex.

The general case for the first-order catalans (ie the duals of figures with one marked node), have all been studied. Mostly they are bounded by tegums at some level, or polytopes that descend down to tegums.

The dual of the gosset polytopes E3 to E9, start with a 2:2:3 triangle. Six of these form a triangular tegum (or bipyramid in 3d), the result is equiangular. Putting 10 of these tegums along the ten edges of a pentachoron, leads to the dual of E4. And so it goes all the way up.

The 3_21 or /5B is a sum of two polytopes 1/4 + 4/1, ie oo3xo3oo3oo3ox3oo. The second is a central inversion, so it means that the 3_21 vertices split consistantly into a set of 28 axies, which we can fix a direction to each.
The dream you dream alone is only a dream
the dream we dream together is reality.
User avatar
Posts: 1814
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Return to Other Polytopes

Who is online

Users browsing this forum: No registered users and 1 guest