Generating the {3,3,5}/{5,3,3}

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

Generating the {3,3,5}/{5,3,3}

Postby Keiji » Sat May 29, 2010 11:25 pm

In 3D, one can construct an icosahedron via the snub tetrahedron.

Is there any analog to this construction in 4D? It'd be nice to have a way to construct all the (Wythoffian) 4D uniforms using only the pentachoron, tesseract and truncations.
User avatar
Keiji
Administrator
 
Posts: 1984
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

Re: Generating the {3,3,5}/{5,3,3}

Postby quickfur » Sun May 30, 2010 5:21 am

Have you read Coxeter's book, Regular Polytopes? He gives a really nice derivation there.

The basic idea is this: start with a 16-cell, truncate its edges at midpoints: that gives you the 24-cell. Now, just as you can have a "coherent indexing" of an octahedron (i.e., assign a direction to every edge such that every face is a consistent direction), you can also extend this "coherent indexing" to the 24-cell by choosing directions so that each face is a clockwise face of one cell and an anticlockwise face of the other cell. (Exactly how this is chosen is a bit involved---see Coxeter's book.) Once you have this coherent indexing, you partition every edge in the golden ratio and truncate, and each octahedral cell becomes an icosahedron, and the gaps produced can be filled in with regular tetrahedra. The result is the snub 24-cell.

Once you have the snub 24-cell, make 24 icosahedral pyramids (each consisting of 20 tetrahedra) and stick them on the icosahedral cells, and you get the 600-cell. Inverting that gives you the 120-cell.

QED.
quickfur
Pentonian
 
Posts: 2935
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Re: Generating the {3,3,5}/{5,3,3}

Postby wendy » Sun May 30, 2010 7:05 am

Coherent indexing of the edges can be done as follows.

1. The verticies of the octahedron, and of the {3,4,3} divide into three groups, each connected only to verticies of the other two:

X (x,0,0) or (2,0,0,0)+ ; E = (0,x,0) or (1,1,1,1)+ ; O = (0,0,x) or (1,1,1,-1)+
where + means all permutations, even change of sign.

The distance between each vertex of X or E or O is 2.sqrt(2), but between XE or EO or OX just 2. Every triangle has then exactly one vertex of each type.

The edges can be given arrows X > E > O > X, which means that every triangle has a simple arrow-ring in it.

We can now divide these edges from 0 to 1, in a scale, so that eg 0.6 gives rise to a single unambiguous point on every edge.

In the case of the octahedron, eight of the faces are always regular. Twelve new faces form by pairs around new edges, which connect the vertices near the octahedron vertex.

When the coherent value is set to 0.38196601125, the new edges are equal to those in the octahedral faces, and the octahedron becomes regular.

In the four-dimensional case, we see the following faces: [from the general 3,4,3 of vertices (2,2,0,0) EPAC]

1. 24 generally irregular icosahedron, in the faces of the {3,4,3} triangular edges are AAA, and AAB
2. 24 always regular tetrahedron, with triangles BBB.
3. 96 triangular tetrahedra, of base BBB, and sides AAB.

When A=B, there are 120 tetrahedra, and 24 icosahedra = snub {3,4,3} of gosset et al. The vertices are (0,1,f,v) EPAC [even permutations, all change of sign].

Since the icosahedron radius is less than the edge, one can place a pyramid of triangles on top of it. This replaces each icosahedron with 20 tetrahedra, the new verticies fall at the centres of the old {3,4,3} faces. These points then fall at (2,0,0,0), (1,1,1,1), EPAC. These are the vertices used by Coxeter et al.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: Generating the {3,3,5}/{5,3,3}

Postby Keiji » Sun May 30, 2010 11:50 am

quickfur wrote:Have you read Coxeter's book, Regular Polytopes?


I haven't, I don't tend to read books too much.

The basic idea is this: [...]


So basically, in Conway operator style, it would be kshA, where A is the aerochoron (loving these new names!), h is what I call hemication which is apparently what everyone else calls rectification, s is snub, and k is kis.

Hmm, for a while I was wondering why the xylochora don't disappear, but then I thought about what kis does in 3D. :D

I shall add this construction to the wiki! Thanks!
User avatar
Keiji
Administrator
 
Posts: 1984
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

Re: Generating the {3,3,5}/{5,3,3}

Postby Keiji » Sat Dec 18, 2010 3:16 pm

I bump this topic to ask, is there any way you can construct the aerochoron from the pyrochoron, or vice versa?

Also, can the runcinate and omnitruncate operators be described from simpler operators (dual, truncate, hemicate, meso)?

In 3D, we have expand(X) = meso(meso(X)) and bevel(X) = truncate(meso(X)), but nothing can be made from 4D mesotruncates...
User avatar
Keiji
Administrator
 
Posts: 1984
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

Re: Generating the {3,3,5}/{5,3,3}

Postby quickfur » Thu Jan 27, 2011 9:52 pm

Keiji wrote:[...]is there any way you can construct the aerochoron from the pyrochoron, or vice versa?

As far as I know, the pyrochoron has no direct connections with the other polychora. The connection in 3D is due to the coincidence of the pyrohedron with the alternated geohedron. In 4D, the alternated geochoron is the aerochoron, so you're out of luck there. You'll have to use the geoteron or aeroteron to get the pyrochoron as a vertex figure/cell, and project it back to 4-space.

(But speaking of alternations... the fact that the aerochoron is not only the dual but also the alternation of the geochoron has interesting consequences. One can, for example, decompose a geochoron into two aerochora of the same circumradius, and add a third aerochoron such that the convex hull of the three form the xylochoron. The convex hull of any pair of aerochora in the group is then a geochoron, and thus you have both the compound of 3 aerochora and the compound of 3 geochora in 4D, both spanning the vertices of the xylochoron. Furthermore, the dual xylochoron also gives rise to 3 aerochora and consequently 3 geochora, in an orientation complementary to the first set of 3's, so you can form also the compound of 6 aerochora and the compound of 6 geochora in 4D. The aerochora between the two sets, of course, are not in the right orientation to form geochora, so you can't add more geochora to the compound this way.)

Also, can the runcinate and omnitruncate operators be described from simpler operators (dual, truncate, hemicate, meso)?

In 3D, we have expand(X) = meso(meso(X)) and bevel(X) = truncate(meso(X)), but nothing can be made from 4D mesotruncates...

Keep in mind that these are topological operations, because they don't actually yield uniform polyhedra.

If you want a system that spans all uniform polytopes, nothing can beat Dynkin. The expand(X) operator is but one member from the full set of Dynkin operators. I have found that it's profitable to think in terms of contraction rather than expansion, mostly because expansion is hard to define for surtopes of dimension less than N-1. Expanding the facets of a polytope is equivalent to contracting said facets within their hyperplane (i.e., instead of pushing them outwards radially, confine them within their hyperplanes and shrink them). Such a contraction will automatically introduce gaps between vertices, since the vertices will split into k vertices where k is the in-degree of the original vertex, which gaps can be filled with new facets.

Instead of merely shrinking facets, we can allow shrinkage of elements of any dimension. For example, shrinking the pentagonal faces of the 120-cell turns each dodecahedral cell into a rhombicosidodecahedron, while keeping adjacent cells still joined, thus producing the cantellated 120-cell. Shrinking the edges to 2/3 length, instead, causes the pentagonal faces to turn into decagons, while still keeping all the cells joined together as before. This thus produces the truncated 120-cell. If we then shrink the decagonal faces of the truncated 120-cell, the cells turn into great rhombicosidodecahedra, while still keeping the cells joined as before, thus producing the cantitruncated 120-cell. Shrinking these great rhombicosidodecahedra produces the omnitruncated 120-cell. If the 120-cell's edges are shrunk to a point, then the dodecahedra become icosidodecahedra, and you get the rectified 120-cell. Shrinking the pentagons to a point turns the dodecahedra into icosahedra, while keeping the cells joined at these shrunk points, so that's the rectified 600-cell. If you shrink the rectified 120-cell's pentagons instead, the icosidodecahedra turn into truncated icosahedra which are joined to each other by their pentagonal faces: this is the bitruncated 120-cell. So here you have it: the mesotruncate. (You start with a 120-cell, shrink its edges to a point to get the rectified 120-cell, then shrink its pentagonal faces so that its cells turn into truncated icosahedra.)

Note that unlike the expand/truncate operators, these shrinkage operations always produce uniform polytopes (provided the degrees of shrinkages is correct). You never need to deform the resulting polytope in any way to make it uniform, unlike the topological derivation: truncate(icosidodecahedron) = non-uniform great rhombicosidodecahedron.
quickfur
Pentonian
 
Posts: 2935
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Re: Generating the {3,3,5}/{5,3,3}

Postby wendy » Fri Sep 14, 2012 10:39 am

In any dimension, the rectified ...o3x3o.... is ....o3x3o3x3o..., and the truncate of the same is ..o3x3x3x3o.. This is the only case where one can construct a higher order from a lower order.

In other cases, it is best to use Stott's methods. You can take anything centred on a symmetry, and expand it outwards. When the same size is kept for the bit, one can add new faces etc, that bridge between the old elements. In fact, one can make all of the dynkin polytopes by adding coordinates of the single-cross polytopes.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
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 15 guests

cron