Johnsonian Polytopes

[Marek14]

I don't feel too well toght and part is a CRF idea that won't let me sleep, so I share it in a misguided attempt to fall asleep

I was thinking about various tetrahedra as vertex figures and one of them was this: a tetrahedron with two opposite edges of length sqrt(2) and other four 1.

This is a shape where four square pyramids fit to one vertex. I can imagine two more points of this figure (opposite points of the square faces), but I can't go further while trying to sleep. Is there a way to add more edges and end up with a CRF figure?

[quickfur]

I don't feel too well toght and part is a CRF idea that won't let me sleep, so I share it in a misguided attempt to fall asleep

I was thinking about various tetrahedra as vertex figures and one of them was this: a tetrahedron with two opposite edges of length sqrt(2) and other four 1.

This is a shape where four square pyramids fit to one vertex. I can imagine two more points of this figure (opposite points of the square faces), but I can't go further while trying to sleep. Is there a way to add more edges and end up with a CRF figure?

Are you sure this tetrahedron exists? Wouldn't it just be a square with edge length 1? I.e., the distance between two opposite edges of the tetrahedron is zero; they are just diagonals of the square (length = sqrt(2)).

[Marek14]

Right... I must be in worse shape than I thought... I didn't realize this limits potential tetrahedra.

[Klitzing]

I rather would try to lift the prove of completeness of Normans original research directly, by means of a brute force computer research sticking allowed polyhedra face to face, with lots of freedoms of deformation until those complexes would close finally. - Just could run into problems with the sheer number of to be found polychora, so that such a program wont end within a human lifetime...

I've been thinking about that, but the major difficulty is the sheer number of 3D CRFs, which means a giant combinatorial explosion, coupled with the complexity of representing partial CRF complexes in an efficient way that would allow algorithmic detection of when something can be closed up to make a CRF polychoron.

Here is a way how to set up that brute force algorithm:
1. enlist all to be used 3d solids: Archimedeans, Prismatics, Jonsonians.
2. enlist all vertex figures of those as pairs: verf + solid it comes from (possibly even counting its orientation!) Note, all those verfs are rigid!
3. try to build up all solids out of those verfs (and multiply those by all those pairings), resulting in a list of all possible vertex surrounding complexes. Note, again all those constructed potential verfs of 4d solids, as being built from rigid faces, and additionally being closed, will be rigid as well!
4. now gradually build them up: take any edge emanating from an already completely surrounded vertex to an uncompletely surrouded vertex. Try to find all counterparts out of the list of step 4, which would complete that vertex. Thus the complexes iteratively would grow. You might run into a blind end, but else you should get it finally closed: thus you would have constructed a biesty you were looking for.

EDIT1: Suppose a good try for this algorithmus wood be to calculate all 4d CUCs. Here list1 wood be smaller already. But the main benefit wood be at list2: no different orientations have to be taken in account! So this procedure wood be a good test for the program, while its application to search for the 4d CRFs = 4d CUHs then would just become a question of memory space and calculation time.
--- rk

[Marek14]

I posted list of 3D verfs yesterday, so now next step should be building 4D verfs, right?

[Klitzing]

I posted list of 3D verfs yesterday, so now next step should be building 4D verfs, right?

Dont forget list2, the pairings, including all possible orientations:
- the [3,4,5,4] of srid is clear, no orientation to be encountered.
- but with respect to the johnsonian diminished srid versions or its gyrated versions, the question arises, which [3,4,5,4]-vertex was taken, that is, how the johnson solid has to be oriented, when applying that specific verf.
This is why not only verfs as such are relevant, but those pairings.
--- rk

[Marek14]

I posted list of 3D verfs yesterday, so now next step should be building 4D verfs, right?

Dont forget list2, the pairings, including all possible orientations:
- the [3,4,5,4] of srid is clear, no orientation to be encountered.
- but with respect to the johnsonian diminished srid versions or its gyrated versions, the question arises, which [3,4,5,4]-vertex was taken, that is, how the johnson solid has to be oriented, when applying that specific verf.
This is why not only verfs as such are relevant, but those pairings.
--- rk

For now, the list includes how many different ways there is to fit.

It's kinda hard to describe, though. In Stella, each model has its vertices numbered, but not everyone here has it, so I can't use those numbers to explain the orientations.

For example, in metagyrate diminished rhombicosidodecahedron, I counted 35 different orientations how it can fit to 3,4,5,4 vertex, but we need some standard to describe these orientations.

[Marek14]

Actually, thinking about it now, we don't actually NEED the list of various ways a polyhedron can fit a verf. We only need those verfs themselves.

Why? Well, verf can be imagined in various ways, but let's go with the most "solid" one -- let's interpret the verf as a chunk cut off of the polychoron, with full edges and parts of faces (full faces for triangles).

Fitting two verfs next to each other means you overlap an edge and blend the faces around it. It's like interlocking of lego blocks. And as for cells, every verf around a cell will contribute something to its outline and since all faces are regular and co-hyperplanarity is easy to check, we'll automatically get CRF cells as a result; we don't have to enter them. So instead of trying all polyhedra in all possible orientations, we might just add other verfs around and some polyhedron or other will appear.

As an aside, I wonder how versatile the various gyrate/diminished rhombicosidodecahedra are. They should figure in a family of srahi (small rhombated 120-cell) derivatives since you can cut a 5g||10p cupola and either leave it or glue it back gyrated. In addition to more traditional diminishings by removing rhombicosidodecahedron||truncated dodecahedron caps.

[Klitzing]

Actually, thinking about it now, we don't actually NEED the list of various ways a polyhedron can fit a verf. We only need those verfs themselves.

Why? Well, verf can be imagined in various ways, but let's go with the most "solid" one -- let's interpret the verf as a chunk cut off of the polychoron, with full edges and parts of faces (full faces for triangles).

Fitting two verfs next to each other means you overlap an edge and blend the faces around it. It's like interlocking of lego blocks. And as for cells, every verf around a cell will contribute something to its outline and since all faces are regular and co-hyperplanarity is easy to check, we'll automatically get CRF cells as a result; we don't have to enter them. So instead of trying all polyhedra in all possible orientations, we might just add other verfs around and some polyhedron or other will appear.

As an aside, I wonder how versatile the various gyrate/diminished rhombicosidodecahedra are. They should figure in a family of srahi (small rhombated 120-cell) derivatives since you can cut a 5g||10p cupola and either leave it or glue it back gyrated. In addition to more traditional diminishings by removing rhombicosidodecahedron||truncated dodecahedron caps.

You are right, we dont need the emanating cells, verfs do suffice. But interlocking 2 verfs of both ends of an edge is horribly more investigation. You then would have to redo what else would be granted: check whether those could fit by running through all polyhedra which could fit either (face-)verf, and whether there is some which belongs to both such sets (of either side).

If you would deal with complexes of cells instead those just have to overlap in some cells. Nothing else to be checked. - That is, my buildingblocks are just a bit larger than the mere verf...

Sure, you would come around all this in the sub-research for CUCs. There any verf belongs to exactly one solid. And any such solid additionally has full symmetry (uniform, that is: vertex transitive). Thus orientation does not matter here. So our 2 setups become fully equivalent.

--- rk

[Marek14]

So it's just a trade-off between our "alphabet" and "grammar"? (i.e. building blocks and rules for their connections). In that case, for computer search it would be probably best to have a big alphabet and simple grammar, right?

[quickfur]

[...]
Here is a way how to set up that brute force algorithm:
1. enlist all to be used 3d solids: Archimedeans, Prismatics, Jonsonians.
2. enlist all vertex figures of those as pairs: verf + solid it comes from (possibly even counting its orientation!) Note, all those verfs are rigid!
3. try to build up all solids out of those verfs (and multiply those by all those pairings), resulting in a list of all possible vertex surrounding complexes. Note, again all those constructed potential verfs of 4d solids, as being built from rigid faces, and additionally being closed, will be rigid as well!
4. now gradually build them up: take any edge emanating from an already completely surrounded vertex to an uncompletely surrouded vertex. Try to find all counterparts out of the list of step 4, which would complete that vertex. Thus the complexes iteratively would grow. You might run into a blind end, but else you should get it finally closed: thus you would have constructed a biesty you were looking for.

Excellent!

And I note that in step 3, building all possible solids from the verfs is equivalent to brute-force search for the 3D case, where the input polygons are the verfs. Which, in turn, can be solved by applying the same algorithm except in 3D. This requires verfs of the polygons; for triangular verfs, it will just be the opposite edge; for other polygons, it's the chord between two vertices adjacent to the vertex in question. So this gives us a bunch of line segments of various lengths, and the task is to find all polygons we can build from them. We don't have to worry about being able to make arbitrarily large polygons, since any valid figure must have circumradius strictly less than the length of the input line segments, since otherwise it cannot possibly be a verf.

However, I see a potential problem here: the circumradius requirement does not give us enough constraints to fix quadrilaterals or higher. So we still have the problem of non-rigid vertices. Or am I missing something here?

EDIT1: Suppose a good try for this algorithmus wood be to calculate all 4d CUCs. Here list1 wood be smaller already. But the main benefit wood be at list2: no different orientations have to be taken in account! So this procedure wood be a good test for the program, while its application to search for the 4d CRFs = 4d CUHs then would just become a question of memory space and calculation time.
--- rk

We can even just try it with the Blind polychora first (input is just the platonic solids), which can then be checked against known results.

[Marek14]

We shouldn't have a problem with nonrigid vertices because the vertex figures are built from a finite, discontinuous set of polygons (including skew polygons). That should mean they can't vary that much.
Note that this is markedly different from 3D case where searching for possible skew vertex figures is much harder because you don't have any structure to build on.

[quickfur]

We shouldn't have a problem with nonrigid vertices because the vertex figures are built from a finite, discontinuous set of polygons (including skew polygons). That should mean they can't vary that much.

But what if one of the polygons is, say, an isosceles triangle with a very narrow base? Then we can fit many copies of it around a vertex (e.g. to make a shape like a 10-gonal pyramid), and once you get past 3 polygons per vertex, it becomes non-rigid.

[Marek14]

We shouldn't have a problem with nonrigid vertices because the vertex figures are built from a finite, discontinuous set of polygons (including skew polygons). That should mean they can't vary that much.

But what if one of the polygons is, say, an isosceles triangle with a very narrow base? Then we can fit many copies of it around a vertex (e.g. to make a shape like a 10-gonal pyramid), and once you get past 3 polygons per vertex, it becomes non-rigid.

You mean past 3 polygons per edge -- there must be always at least 4 per vertex.

But no, that won't make the vertex become nonrigid. You're concentrating on that one edge and forgetting that the edge can't exist alone -- you also need the rest of the vertex figure and that will put additional constraints. In case of 10-polygon edge, there will have to be at least 5-6 other polygons at the same vertex...

[quickfur]

We shouldn't have a problem with nonrigid vertices because the vertex figures are built from a finite, discontinuous set of polygons (including skew polygons). That should mean they can't vary that much.

But what if one of the polygons is, say, an isosceles triangle with a very narrow base? Then we can fit many copies of it around a vertex (e.g. to make a shape like a 10-gonal pyramid), and once you get past 3 polygons per vertex, it becomes non-rigid.

You mean past 3 polygons per edge -- there must be always at least 4 per vertex.

But no, that won't make the vertex become nonrigid. You're concentrating on that one edge and forgetting that the edge can't exist alone -- you also need the rest of the vertex figure and that will put additional constraints. In case of 10-polygon edge, there will have to be at least 5-6 other polygons at the same vertex...

I think you misread me. I wasn't talking about the 4D case; I was talking about the preliminary step of generating the 3D verfs. So you have a bunch of polygons (which are verfs of CRF polyhedra) and you want to assemble them together to form verfs of potential 4D CRFs. How do you find all such verfs? You have to assemble these polygons into polyhedra, of course. But how do you find all polyhedra that can be made from these polygons? One way is the reuse the algorithm to brute force search all such polyhedra -- except you're running it for the 3D case now. So you make verfs of the polygons (verfs of the verfs), i.e., take the chords of the polygons, and then try to assemble possible verfs from those line segments. The radius of the polygon is restricted, of course, since it must be smaller than the length of edges emanating from the vertex corresponding with the verf.

But this is still not enough to fully define the polygon. A short chord (say from a narrow isosceles triangle) allows you to make a big polygon as verf. So that's equivalent to making a large n-gonal pyramid from the isosceles triangles. When n>3, the apex of the pyramid is non-rigid; you can vary the dihedral angles of the isosceles triangles. So how do we know which angles will lead to a closed 4D verf? We have to solve the non-rigid peak (vertex) problem for the 3D case.

[quickfur]

[...]
Looks crazy to me if xfo3ooxPooo&#x would have true teddis for P=3, 5; but not for P=4.

You're right, I made another mistake. The icosidodecahedron is o3x5o, not x3o5x.

Unless I made a mistake in my calculations for the octahedron/cuboctahedron case, it's pretty crazy.

Just for the record, yes I did make a mistake; the bistratic exists for P=4 too. Here are the renders from that post:




The coordinates are posted in the linked post as well.

[quickfur]

I've calculated the coordinates of the "expanded octahedral 4-teddy", the bistratic xxx4oox3xfo&#xf. It consists of 1 rhombicuboctahedron, 8 tridiminished icosahedra, 12 pentagonal prisms, 6 cubes, 6 square cupolae, and 1 truncated cube. Here's a render of it:



Since the tridiminished icosahedra are well-separated, this makes it easier to color them all together in the projection:



I rotated the 3D viewpoint a bit so that we can get a clearer view at the 8 tridiminished icosahedra. I also changed the texture of the rhombicuboctahedron in the center back to the light transparent color so that all 8 cells are clearly visible.

The coordinates are:

Code: Select all

# x4o3x
< 1,  1,  (1+sqrt(2)),  sqrt(2*phi)>
< 1,  1, -(1+sqrt(2)),  sqrt(2*phi)>
< 1, -1,  (1+sqrt(2)),  sqrt(2*phi)>
< 1, -1, -(1+sqrt(2)),  sqrt(2*phi)>
<-1,  1,  (1+sqrt(2)),  sqrt(2*phi)>
<-1,  1, -(1+sqrt(2)),  sqrt(2*phi)>
<-1, -1,  (1+sqrt(2)),  sqrt(2*phi)>
<-1, -1, -(1+sqrt(2)),  sqrt(2*phi)>
< 1,  (1+sqrt(2)),  1,  sqrt(2*phi)>
< 1,  (1+sqrt(2)), -1,  sqrt(2*phi)>
< 1, -(1+sqrt(2)),  1,  sqrt(2*phi)>
< 1, -(1+sqrt(2)), -1,  sqrt(2*phi)>
<-1,  (1+sqrt(2)),  1,  sqrt(2*phi)>
<-1,  (1+sqrt(2)), -1,  sqrt(2*phi)>
<-1, -(1+sqrt(2)),  1,  sqrt(2*phi)>
<-1, -(1+sqrt(2)), -1,  sqrt(2*phi)>
< (1+sqrt(2)),  1,  1,  sqrt(2*phi)>
< (1+sqrt(2)),  1, -1,  sqrt(2*phi)>
< (1+sqrt(2)), -1,  1,  sqrt(2*phi)>
< (1+sqrt(2)), -1, -1,  sqrt(2*phi)>
<-(1+sqrt(2)),  1,  1,  sqrt(2*phi)>
<-(1+sqrt(2)),  1, -1,  sqrt(2*phi)>
<-(1+sqrt(2)), -1,  1,  sqrt(2*phi)>
<-(1+sqrt(2)), -1, -1,  sqrt(2*phi)>

# x4o3f
< 1,  1,  (1+phi*sqrt(2)),  0>
< 1,  1, -(1+phi*sqrt(2)),  0>
< 1, -1,  (1+phi*sqrt(2)),  0>
< 1, -1, -(1+phi*sqrt(2)),  0>
<-1,  1,  (1+phi*sqrt(2)),  0>
<-1,  1, -(1+phi*sqrt(2)),  0>
<-1, -1,  (1+phi*sqrt(2)),  0>
<-1, -1, -(1+phi*sqrt(2)),  0>
< 1,  (1+phi*sqrt(2)),  1,  0>
< 1,  (1+phi*sqrt(2)), -1,  0>
< 1, -(1+phi*sqrt(2)),  1,  0>                                                         
< 1, -(1+phi*sqrt(2)), -1,  0>
<-1,  (1+phi*sqrt(2)),  1,  0>
<-1,  (1+phi*sqrt(2)), -1,  0>
<-1, -(1+phi*sqrt(2)),  1,  0>
<-1, -(1+phi*sqrt(2)), -1,  0>
< (1+phi*sqrt(2)),  1,  1,  0>
< (1+phi*sqrt(2)),  1, -1,  0>
< (1+phi*sqrt(2)), -1,  1,  0>
< (1+phi*sqrt(2)), -1, -1,  0>
<-(1+phi*sqrt(2)),  1,  1,  0>
<-(1+phi*sqrt(2)),  1, -1,  0>
<-(1+phi*sqrt(2)), -1,  1,  0>
<-(1+phi*sqrt(2)), -1, -1,  0>

# x4x3o
< 1,  (1+sqrt(2)),  (1+sqrt(2)), -sqrt(2/phi)>
< 1,  (1+sqrt(2)), -(1+sqrt(2)), -sqrt(2/phi)>
< 1, -(1+sqrt(2)),  (1+sqrt(2)), -sqrt(2/phi)>
< 1, -(1+sqrt(2)), -(1+sqrt(2)), -sqrt(2/phi)>
<-1,  (1+sqrt(2)),  (1+sqrt(2)), -sqrt(2/phi)>
<-1,  (1+sqrt(2)), -(1+sqrt(2)), -sqrt(2/phi)>
<-1, -(1+sqrt(2)),  (1+sqrt(2)), -sqrt(2/phi)>
<-1, -(1+sqrt(2)), -(1+sqrt(2)), -sqrt(2/phi)>
< (1+sqrt(2)),  1,  (1+sqrt(2)), -sqrt(2/phi)>
< (1+sqrt(2)),  1, -(1+sqrt(2)), -sqrt(2/phi)>
< (1+sqrt(2)), -1,  (1+sqrt(2)), -sqrt(2/phi)>
< (1+sqrt(2)), -1, -(1+sqrt(2)), -sqrt(2/phi)>
<-(1+sqrt(2)),  1,  (1+sqrt(2)), -sqrt(2/phi)>
<-(1+sqrt(2)),  1, -(1+sqrt(2)), -sqrt(2/phi)>
<-(1+sqrt(2)), -1,  (1+sqrt(2)), -sqrt(2/phi)>
<-(1+sqrt(2)), -1, -(1+sqrt(2)), -sqrt(2/phi)>
< (1+sqrt(2)),  (1+sqrt(2)),  1, -sqrt(2/phi)>
< (1+sqrt(2)),  (1+sqrt(2)), -1, -sqrt(2/phi)>
< (1+sqrt(2)), -(1+sqrt(2)),  1, -sqrt(2/phi)>
< (1+sqrt(2)), -(1+sqrt(2)), -1, -sqrt(2/phi)>
<-(1+sqrt(2)),  (1+sqrt(2)),  1, -sqrt(2/phi)>
<-(1+sqrt(2)),  (1+sqrt(2)), -1, -sqrt(2/phi)>
<-(1+sqrt(2)), -(1+sqrt(2)),  1, -sqrt(2/phi)>
<-(1+sqrt(2)), -(1+sqrt(2)), -1, -sqrt(2/phi)>


[Keiji]

Excellent!

Do you know how many CRF ursachora there are?

We know of 6 already (the tetrahedral, octahedral and icosahedral teddies and their expanded versions), but are there any other 3D bases that produce CRF ursachora?

[Marek14]

We shouldn't have a problem with nonrigid vertices because the vertex figures are built from a finite, discontinuous set of polygons (including skew polygons). That should mean they can't vary that much.

But what if one of the polygons is, say, an isosceles triangle with a very narrow base? Then we can fit many copies of it around a vertex (e.g. to make a shape like a 10-gonal pyramid), and once you get past 3 polygons per vertex, it becomes non-rigid.

You mean past 3 polygons per edge -- there must be always at least 4 per vertex.

But no, that won't make the vertex become nonrigid. You're concentrating on that one edge and forgetting that the edge can't exist alone -- you also need the rest of the vertex figure and that will put additional constraints. In case of 10-polygon edge, there will have to be at least 5-6 other polygons at the same vertex...

I think you misread me. I wasn't talking about the 4D case; I was talking about the preliminary step of generating the 3D verfs. So you have a bunch of polygons (which are verfs of CRF polyhedra) and you want to assemble them together to form verfs of potential 4D CRFs. How do you find all such verfs? You have to assemble these polygons into polyhedra, of course. But how do you find all polyhedra that can be made from these polygons? One way is the reuse the algorithm to brute force search all such polyhedra -- except you're running it for the 3D case now. So you make verfs of the polygons (verfs of the verfs), i.e., take the chords of the polygons, and then try to assemble possible verfs from those line segments. The radius of the polygon is restricted, of course, since it must be smaller than the length of edges emanating from the vertex corresponding with the verf.

But this is still not enough to fully define the polygon. A short chord (say from a narrow isosceles triangle) allows you to make a big polygon as verf. So that's equivalent to making a large n-gonal pyramid from the isosceles triangles. When n>3, the apex of the pyramid is non-rigid; you can vary the dihedral angles of the isosceles triangles. So how do we know which angles will lead to a closed 4D verf? We have to solve the non-rigid peak (vertex) problem for the 3D case.

How do I find all polyhedra that can be assembled from these polygons? Luckily, I don't have to! That was done almost 50 years ago, all we have to do now is take the list and find the verfs in there

[Keiji]

Another point to ponder: knowing the existence of the BXD (bixylodiminished hydrochoron) as well as the higher-dimensional ursatopes, does this mean there are equivalents to the BXD in 5D and up, one for each ursatope, which are also chiral, CRF, vertex- and cell-transitive? Or perhaps they would only exist in even dimensions 6D, 8D etc, due to the spiralling structure required?

[quickfur]

Excellent!

Do you know how many CRF ursachora there are?

We know of 6 already (the tetrahedral, octahedral and icosahedral teddies and their expanded versions), but are there any other 3D bases that produce CRF ursachora?

You could try it with any uniform polychoron with triangular faces, and attach 3-teddies to them, perhaps with suitable prisms or other filler cells for the non-triangular faces, and see if the thing closes up. I tried it with the icosidodecahedron but it made gaps with adjacent pentagons, which I think requires diminished dodecahedra to fill in (which is non-CRF). The restriction to tetrahedron/octahedron/icosahedron may be an artifact of the 3-teddy being the only CRF ursahedron.

I think we may be jumping the gun a bit in naming the ursachora, though... this morning it occurred to me that similar constructions exist by attaching truncated tetrahedra instead of teddies, in which case you get the truncated 5-cell from the tetrahedron, the bisected truncated 16-cell (i.e. octahedral rotunda) from the 16-cell, and a kind of icosahedral rotunda with pentagonal pyramids lacing the bottom from the icosahedron (I don't remember if this was already counted with the existing rotunda, if not, it's another one to add to the list -- it's basically something you can slice off a truncated 600-cell).

More interestingly, the icosidodecahedral rotunda may be considered a kind of variation in which pentagonal rotundae replace the teddies, and the top cell is kinda expanded but not with prisms, but in a way that lets you stick tetrahedra in it to fill up the gaps between adjacent pentagonal rotunda. I tried to make a cubical rotunda by attaching cuboctahedra and filling the gaps with tetrahedra, but it turns out that the shape is non-CRF unless you keep extending it and adding more cells until you end up with the rectified 24-cell. From a non-CRF perspective, though, this construction gives you something with a cube, 8 diminished cubes, 6 cuboctahedra, rhombicuboctahedra, and square pyramids (I'm not sure which of them can be made uniform). OTOH if you bisect the cuboctahedra and then "squeeze" them a little, they turn into square antiprisms, and you get the segmentochoron cube||cuboctahedron.

IOW these constructions are just multistratic versions of the segmentochora, with diverse variations.

[quickfur]

[...]
How do I find all polyhedra that can be assembled from these polygons? Luckily, I don't have to! That was done almost 50 years ago, all we have to do now is take the list and find the verfs in there

Oh really? lol... and here I thought we had to start from scratch! My ignorance is showing.

[quickfur]

Another point to ponder: knowing the existence of the BXD (bixylodiminished hydrochoron) as well as the higher-dimensional ursatopes, does this mean there are equivalents to the BXD in 5D and up, one for each ursatope, which are also chiral, CRF, vertex- and cell-transitive? Or perhaps they would only exist in even dimensions 6D, 8D etc, due to the spiralling structure required?

I believe you need even dimensions to get the higher-dimensional equivalent of the Hopf fibration (which is what the BXD is based on). But this is just a wild guess, so I might be completely wrong.

And teddies aren't required for these things; all the regular polychora (except perhaps the 5-cell) exhibit this spiralling structure. The bitruncated 24-cell does, too (and probably the bitruncated 5-cell as well). The tesseractic family mainly exhibit the 2-ring structure of the duoprisms (which is a subset of this spiralling symmetry -- think the first ring and the equatorial ring). The 24-cell family exhibits the 8-ring structure you see in the BXD. The 120-cell family exhibits more spiralling rings -- 12 of them, IIRC. But we just don't notice it because there are too many other interesting symmetries that obscure it.

There are some known non-convex uniforms that show the 12-ring structure of the 120-cell -- I think the first one was discovered by Olshevsky or maybe Bowers, and was dubbed the "swirlprism". Later a few more were found. Bowers then extended the idea to smooth shapes (i.e., take polygonal rods, bend them around these rings and then twist them so that their ridges fit into each other), and the result was the polytwisters, a number of which are regular (all of the bent and twisted rods, called "twisters", are transitive).

[Marek14]

[...]
How do I find all polyhedra that can be assembled from these polygons? Luckily, I don't have to! That was done almost 50 years ago, all we have to do now is take the list and find the verfs in there

Oh really? lol... and here I thought we had to start from scratch! My ignorance is showing.

That's what Johnson solids are... I even posted the list of their vertex figures few days ago. Lots of them are skew, though.

[quickfur]

[...]
How do I find all polyhedra that can be assembled from these polygons? Luckily, I don't have to! That was done almost 50 years ago, all we have to do now is take the list and find the verfs in there

Oh really? lol... and here I thought we had to start from scratch! My ignorance is showing.

That's what Johnson solids are... I even posted the list of their vertex figures few days ago. Lots of them are skew, though.

Wait, what? I thought we were talking about taking Johnson solids' vertex figures and putting them (the verfs) together to make verfs for potential CRFs. Or did you mean dual Johnson solids? I'm confused.

[Keiji]

Another point to ponder: knowing the existence of the BXD (bixylodiminished hydrochoron) as well as the higher-dimensional ursatopes, does this mean there are equivalents to the BXD in 5D and up, one for each ursatope, which are also chiral, CRF, vertex- and cell-transitive? Or perhaps they would only exist in even dimensions 6D, 8D etc, due to the spiralling structure required?

I believe you need even dimensions to get the higher-dimensional equivalent of the Hopf fibration (which is what the BXD is based on). But this is just a wild guess, so I might be completely wrong.

And teddies aren't required for these things; all the regular polychora (except perhaps the 5-cell) exhibit this spiralling structure. The bitruncated 24-cell does, too (and probably the bitruncated 5-cell as well). The tesseractic family mainly exhibit the 2-ring structure of the duoprisms (which is a subset of this spiralling symmetry -- think the first ring and the equatorial ring). The 24-cell family exhibits the 8-ring structure you see in the BXD. The 120-cell family exhibits more spiralling rings -- 12 of them, IIRC. But we just don't notice it because there are too many other interesting symmetries that obscure it.

That all makes sense, but see my bolded conditions above - the regular polytopes aren't chiral.

Also, it's interesting that the duoprisms are a special case with 2 rings - what are the possibilities with respect to number of rings? Does it have to be even, is there a maximum, are there polychora with 6 or 10 rings?

[Marek14]

[...]
How do I find all polyhedra that can be assembled from these polygons? Luckily, I don't have to! That was done almost 50 years ago, all we have to do now is take the list and find the verfs in there

Oh really? lol... and here I thought we had to start from scratch! My ignorance is showing.

That's what Johnson solids are... I even posted the list of their vertex figures few days ago. Lots of them are skew, though.

Wait, what? I thought we were talking about taking Johnson solids' vertex figures and putting them (the verfs) together to make verfs for potential CRFs. Or did you mean dual Johnson solids? I'm confused.

Yes, taking vertex figures of Johnson solids and building CRF verfs out of them.

[quickfur]

Another point to ponder: knowing the existence of the BXD (bixylodiminished hydrochoron) as well as the higher-dimensional ursatopes, does this mean there are equivalents to the BXD in 5D and up, one for each ursatope, which are also chiral, CRF, vertex- and cell-transitive? Or perhaps they would only exist in even dimensions 6D, 8D etc, due to the spiralling structure required?

I believe you need even dimensions to get the higher-dimensional equivalent of the Hopf fibration (which is what the BXD is based on). But this is just a wild guess, so I might be completely wrong.

And teddies aren't required for these things; all the regular polychora (except perhaps the 5-cell) exhibit this spiralling structure. The bitruncated 24-cell does, too (and probably the bitruncated 5-cell as well). The tesseractic family mainly exhibit the 2-ring structure of the duoprisms (which is a subset of this spiralling symmetry -- think the first ring and the equatorial ring). The 24-cell family exhibits the 8-ring structure you see in the BXD. The 120-cell family exhibits more spiralling rings -- 12 of them, IIRC. But we just don't notice it because there are too many other interesting symmetries that obscure it.

That all makes sense, but see my bolded conditions above - the regular polytopes aren't chiral.

Well, then you've found a new direction to search for CRF polychora in.

Also, it's interesting that the duoprisms are a special case with 2 rings - what are the possibilities with respect to number of rings? Does it have to be even, is there a maximum, are there polychora with 6 or 10 rings?

I'm not 100% certain on this, but according to what I understand, the idea behind the Hopf fibration is to map every point on a sphere to a distinct great circle on the 3-sphere. None of these great circles intersect each other, but they fill up the surface of the 3-sphere in a swirling manner. Of course, there are a continuum of points on the sphere, so this doesn't give us a discrete structure; Bowers' insight was to notice that the swirlprism corresponded with a discrete version of the Hopf fibration, in the sense that the base shape divided the surface of the sphere into equal regions, so that one may now group the great circles according to which region of the sphere would fall on them. This produces a finite number of groups of circles (which you can think of as a finite number of rings, if you consider all circles in the same group as being represented by the ring associated with that group). The cells of the swirlprism thus lies along these rings.

Since the discrete rings arise from the division of the surface of the sphere into regions, you would get a regular structure out of it if you used a regular division of the sphere's surface, that is, if you started with the regular polyhedra. You can also start with the digon, which (I believe) gives you duoprism symmetry, that is, two orthogonal rings. The tetrahedron gives a 4-ring structure; the cube gives a 6-ring structure, the octahedron gives an 8-ring structure, the dodecahedron, 12 rings, and the icosahedron, 20 rings.

These are just the regular divisions, of course. If you start with uniform polyhedra, you'll get other ring combinations; the truncated cube, for example, will give you a 14-ring structure, grouped into two distinct set of rings each with 8 and 6 rings, respectively. You can start with other polyhedra as well, including non-symmetric ones, in which case you'll just get non-equivalent rings. You may or may not be able to get CRFs out of these non-regular rings. The upper limit is really the continuum, i.e., the Hopf fibration itself, in which you start from the sphere (as a "polyhedron" with infinite number of faces). But as far as regular rings are concerned, you only have the 5 Platonic solids and the digon, so the maximum there is 20 rings (which, I suspect, will correspond with something derived from the 120-cell family).

[quickfur]

[...]
How do I find all polyhedra that can be assembled from these polygons? Luckily, I don't have to! That was done almost 50 years ago, all we have to do now is take the list and find the verfs in there

Oh really? lol... and here I thought we had to start from scratch! My ignorance is showing.

That's what Johnson solids are... I even posted the list of their vertex figures few days ago. Lots of them are skew, though.

Wait, what? I thought we were talking about taking Johnson solids' vertex figures and putting them (the verfs) together to make verfs for potential CRFs. Or did you mean dual Johnson solids? I'm confused.

Yes, taking vertex figures of Johnson solids and building CRF verfs out of them.

So you're saying we already have the list of these verfs (which are polyhedra with Johnson solid verfs as faces)?

[Keiji]

Well, then you've found a new direction to search for CRF polychora in.

Not quite - it would be polytopes in 5D and higher.

Hopf fibration

Right, thank you for the excellent explanation! After reading that I read through the Wikipedia page and was able to understand that too - I'd heard of the Hopf fibration before but never understood it until now.