More ideas about searching for (non-EKF) CRF crown jewels

[quickfur]

These few days I've been trying to tackle the problem of non-EKF crown jewels again. It seems that lacking any firm methodology as to how to search for them, and lacking a concrete algorithm for brute-force searching, the next best thing might be to implement a "4D CRF lego" software that lets you build the polytopes piece-by-piece until you find something that closes up.

Another line of thought I have is with the snub antiprism sequence in 3D: digonal snub antiprism = snub disphenoid; trigonal snub antiprism = icosahedron; snub square antiprism; and snub pentagonal antiprism (concave, non-CRF). These all start with some base polygon, with attached triangles ("edge pyramids"), and forming some kind of skew polygon where the other half is joined. In terms of higher dimensions, it seems a better way to see the other half is not merely some rotation of the first half, flipped, but rather as a similar construction of triangles ("edge pyramids") surrounding the dual of the starting base polygon. So in 4D, potentially we could look in the direction of some base polyhedron P, surrounded by some "skirt" of pyramids of appropriate base shapes to fit onto the faces of P, then the dual of P, also with its own set of pyramids, and try to find a configuration where the skew polygon around the unattached boundary of the "pyramid skirts" have the same shape, and therefore can be joined together to form a closed polychoron. (There might also be some other pyramids like tetrahedra to serve as lacing cells that may not directly share a face with P or its dual.) To help with finding more candidates, we could perhaps relax the CRF requirement at first, and study families of such polytopes, to see if we can find what are the conditions for their existence, and afterwards evaluate which ones among them might possibly be CRF-able.

[quickfur]

More concretely, say we start with some base polyhedron P, then ideally around each edge there should be 4 pyramids (of some kind), since if there were only 3, then we will probably only get the P antiprism, which are already well-known. The angle defect around an edge then would be the dihedral angle of P on that edge + the sum of dihedral angles of these 4 pyramids.

If we start with P=cube, for example, we could have 2 tetrahedra straddling the square pyramids attached to P, which gives us a total dihedral angle of 90° + 2*54.7356° + 2*70*5288° = 340.5288°, or an angle defect of 19.47°. Not 100% sure what lacing cells would be needed to fill in the remaining gaps.

On the opposite side, the dual of P would be an octahedron... but unfortunately, the dihedral angle of the octahedron is too big to fit 4 tetrahedra around an edge. So we might need to use square pyramids instead, maybe 2 square pyramids + 2 tetrahedra? Then it would have some kind of correspondence with the cube side of the polytope... though I'm not sure if this means it's possible to fit the two halves together!

Hmm wait, actually 2 square pyramids + 2 tetrahedra + octahedron has zero angle defect.

[quickfur]

Yet another idea: this was inspired by staring at the disphenocingulum (J90). It appears to be a belt of triangles capped by two spheno complexes (pairs of lune complexes, i.e. 2 triangles + 2 squares). If we remove the spheno complexes, we get a distorted concave polyhedron that can be deformed into a hexagonal prism. So this made me think of a possible 4D analogue, where you start with some kind of duoprism, then deform it so that one ring of cells no longer lies on a great circle, causing the cells on the other ring to become prisms of non-uniform polygons. Then cover up these non-uniform prisms with some kind of cell complex such that it becomes CRF again.

If such a CRF exists, it would imply:

1) The existence of some kind of cell complex that ends with a base in the shape of a prism of a non-uniform polygon (specifically, it has to be a prism with unit edge lengths and square lacing faces, but not necessarily regular top/bottom polygons); and:

2) The dichoral angles at the base of this cell complex is small enough that you can at least attach 3 of them in a loop to form regular polygons in the orthogonal ring. So that means the dichoral angle must be ≤ 60°, so that at least it would work with an orthogonal ring of triangular prisms. If it's ≤ 45°, you could form an orthogonal ring of cubes; if it's ≤ 36°, you could form an orthogonal ring of pentagonal prisms. But I'm not holding out much hope for higher-order orthogonal rings; what about at least the ≤ 60° case?

Does anyone know of a 4D cell complex that might potentially have these properties?

[ubersketch]

It's very hard to find crown jewels without stott-expansions.

[student91]

Hmm, I think with the discovery of the EKFs, we could start working on some kind of brute-force searching algorithm, along the following lines:
Marek showed somewhere that vertex figures in 4D are rigid, and he started listing all possible such vertex figures.
His listing became quite big, so I though now that we know the three main groups
-(Diminishings of) Uniform polychora
-Segmentochora
-EKFs
and then we could search for vertex figures that do not belong to one of these groups.
Maybe this is not very concrete, but at least it would limit the ways in which a crown jewel could occur.
Note that all 3D johnson solids either fall into one of these groups or are algebraically "more complex"
(although the sphenocorona seems to fall between "simple" and "really complex")

[Marek14]

Yes, if I still remember the result (it's here somewhere), the polygons that serve as faces of 4D verfs (and which may be skew polygons, for Johnson solids as cells) are always rigid because there is only a finite number of polygons with given edge lengths that correspond to a valid face. I think I managed to enumerate all possible verfs that are tetrahedra or quadragonal pyramids, but then it grew a bit too much. Next step, trigonal bipyramids, is still unfinished.

[quickfur]

Hey Marek14, good to see you again!

I've been thinking about this whole rigid vertex thing, and I think there's still something about it that I don't quite understand yet. Not that I doubt your proof, but I'm still having a hard time fully understanding how it works. If the verfs all lie on a 3D hyperplane, then it's obvious that the polyhedron must be rigid. But since we're talking about potentially skew polyhedra, I'm still having a hard time understanding why it must be rigid too.

Clearly, continuous deformation is not possible, but have we truly ruled out the possibility of multiple (discrete) skew polyhedra solutions to a given topology of connected verfs? For example, in the 3D case, if we have a skew hexagon, say, with fixed angles between every pair of edges, there is still the possibility of different conformations: a particular angle can be skewed upwards or downwards, so there are two configurations that don't break the constraints, even though continuous deformation is not allowed (because of the angle restriction). An example of this is in chemistry, e.g., hexane, which consists of a ring of 6 carbon atoms at tetrahedral angles w.r.t each other. It can twist between two "chair" conformations that have equal energy, and also two higher-energy "boat" configurations. All 4 configurations have tetrahedral angles. Is it possible that in 4D, a skew polyhedron might have multiple possible configurations? Since a skew polyhedron's faces can bend "upwards" or "downwards" off the 3D hyperplane, is it possible for the same set of faces to have different "conformations"? If so, this would imply the possibility that the same set of cells around a given 4D vertex may have multiple solutions, even though continuous deformation at the vertex is not possible.

Not trying to challenge your proof, but just trying to figure out exactly what's going on with the rigid vertex thing. I've been thinking about a brute force algorithm as well, and issues like this are pretty important in order to have confidence that the algorithm will actually work correctly.

Also, for generating the set of possible verfs, I wonder if it's possible to automate it, since given the large number of 3D CRFs, there's going to be a lot of possible verfs, and doing it by hand will be too tedious and rather error-prone.

[quickfur]

[...]
-(Diminishings of) Uniform polychora
-Segmentochora
-EKFs
and then we could search for vertex figures that do not belong to one of these groups.
Maybe this is not very concrete, but at least it would limit the ways in which a crown jewel could occur.
Note that all 3D johnson solids either fall into one of these groups or are algebraically "more complex"
(although the sphenocorona seems to fall between "simple" and "really complex")

Don't forget the augmentations of uniform polychora, and cut-n-pastes of the segmentochora, polychora, EKFs, etc..

This week I transcribed the list of 92 Johnson solids for my website, and I noticed that the two largest groups of Johnsons are the cut-n-pastes of the 3D segmentochora and prismatoids (pyramids, prisms, cupolae), and the augmentations/diminishings of the 3D uniforms. This seems to correspond with our currently-known 4D CRF counts: with Klitzing's >100 segmentochora plus non-orbiform monostratics, many of which can be pasted with each other, there's going to be a huge number of combinations. And my last count of duoprism augmentations (still not independently verified ) ended up with almost 12 million distinct CRFs up to symmetry. Then on the uniform side, we have the 300+ million basic 600-cell diminishings, plus an unknown number of non-trivial diminishings (including the hemi-600-cell, the lunae, and various other such things), plus unexplored analogues of the 600-cell diminishings among the other 120-cell family uniforms.

Then we have the EKFs, which in 3D are only a few (J32, J91, J92, maybe one or two others that coincide with something else), but in 4D quite a sizable number, and AFAIK still not fully enumerated. That's not counting the probably huge set of their augmentations, diminishings, cut-n-pastes, etc..

With this huge array of CRFs in currently-known categories, there's going to be a huge number of possible verfs, particularly once you take into account the non-elementary CRFs produced by cut-n-paste operations on the basic ones. How do we go about searching for something outside this set? I mean, if the brute force program is faced with some verf V, how do we determine whether or not V belongs to the set? (I.e., whether to discard V as something already known, or continue with something that may be a crown jewel.) Since even we ourselves don't know all the members of this set yet.

[Marek14]

I think that skew verfs are counted in the proof. Basically (I could be wrong, it's been a long time), whenever you get a polygon with more than 3 sides, you assign one of the several possible conformations to it (by setting the diagonal distances between its vertices).

Let me think about it. The proof, IIRC, is based on number of constraints for a set of geometric relations (since all vertices must lie on the same hypersphere and we get a bunch of constraints for distances between pairs of vertices). Yes, I believe that in order to work, the vertex figure must be composed solely of triangles -- and forcing a conformation on a higher polygon basically triangulates it.

[quickfur]

Hmm. Distance constraints are quadratics... meaning there exists a possibility of up to 2 distinct solutions, unless excluded by other constraints / considerations. Seems to correspond with the idea of different conformations (but without intermediate continuously-deformed states). Just a thought.

[quickfur]

This is a (very) long shot, and maybe it's just grasping at straws, or possibly a completely insane idea, but recently, while computing coordinates for various Johnson solids to fill in the Johnson solids section of my website, I noticed that the augmented triangular prism (J51) / augmented hexagonal prism (J57) have coordinates that involve √6, and the sphenocorona (J86) also happens to have coordinates involving √6 (albeit nested under an outer radical). Furthermore, the 5-cell also sports coordinates that involve √6 (though in the denonimator, so √(1/6)).

Given that the coordinates of a pentagon in 2D involves nested radicals containing √5, which seems to have √5 in common with the golden ratio that features so prominently in the pentagonal polytopes, I couldn't help wondering if somehow there is some 4D (or higher?) configuration of the sphenocorona might somehow be such that its coordinates will line up with an augmented triangular prism or augmented hexagonal prism? Might there exist a CRF crown jewel somewhere out there that contains J86 and J51/J57 cells? Maybe they're somehow related to the 5-cell? (5-cell symmetry? Or some kind of construct involving cut-and-paste 5-cell family polychora with some modifications?)

Of course, I could be completely off my rocker with this wild idea. But I don't think we've previously considered combining J86 with J51/J57 in the past in our search for CRF crown jewels. Perhaps there is something to be found in this direction?

[student5]

Hello everyone

I've had my fair share of staring at the disphenocingulum as well now, and an idea popped up, which might give some (AFAIK new) CRF polytope

Yet another idea: this was inspired by staring at the disphenocingulum (J90). It appears to be a belt of triangles capped by two spheno complexes (pairs of lune complexes, i.e. 2 triangles + 2 squares). If we remove the spheno complexes, we get a distorted concave polyhedron that can be deformed into a hexagonal prism. So this made me think of a possible 4D analogue, where you start with some kind of duoprism, then deform it so that one ring of cells no longer lies on a great circle, causing the cells on the other ring to become prisms of non-uniform polygons. Then cover up these non-uniform prisms with some kind of cell complex such that it becomes CRF again.

I thought of that ring as an antiprism, If you'd have a prism there, It'd become the square orthobicupola. For this construction, you start with a hexagonal prism, and remove the hexagonal faces. You are now left with a ring of 6 squares, on which you want to place the spheno complex.
Since we are looking at the hexagonal prism, which has orthogonality all over the place, all 6 top points will stay in the same plane. Therefore, we deform the spheno complex until all outer edges lie on the same plane, and paste it to the irregular hexagonal prism on both sides, to obtain the square orthobicupola.

If we'd start with the hexagonal antiprism, a deformation would result in a shift from the plane on which all points were. If we now stick the spheno complex to one side, the antiprism is deformed, such that it nicely fits a 90 degree rotated spheno complex on the other side.

If such a CRF exists, it would imply:

1) The existence of some kind of cell complex that ends with a base in the shape of a prism of a non-uniform polygon (specifically, it has to be a prism with unit edge lengths and square lacing faces, but not necessarily regular top/bottom polygons); and:

2) The dichoral angles at the base of this cell complex is small enough that you can at least attach 3 of them in a loop to form regular polygons in the orthogonal ring. So that means the dichoral angle must be ≤ 60°, so that at least it would work with an orthogonal ring of triangular prisms. If it's ≤ 45°, you could form an orthogonal ring of cubes; if it's ≤ 36°, you could form an orthogonal ring of pentagonal prisms. But I'm not holding out much hope for higher-order orthogonal rings; what about at least the ≤ 60° case?

Does anyone know of a 4D cell complex that might potentially have these properties?

Since I only saw an antiprism in the bispenocingulum, I did not get this part. I also had some difficulty finding a satisfactory 4D analogue of the antiprism. This is also where the likely new CRF will pop up . The first way to generalize the concept of an antiprism, is to consider the top polygon as the dual of the bottom polygon. This would result in 4D in segmentochora such as tet||inverted tet (16-cell), cube||oct and ike||doe. However, Having square or pentagonal faces greatly diminishes our freedom of squishing, which we do not want if we want to create some disphenocingulum-like CRF.
we also do not want a regular prism, because they are too orthogonal, and will not do stuff that is interesting enough. (the vertices remain in the same hyperplane upon squishing)

What I then considered, is placing a 90 degrees rotated ike on top of another ike (resp oct, tet). In lace-city-ness:

Code: Select all

prism: (ike, oct, tet)
x20 x20   x20 x20   x20 x20
02f 02f   02q 02q   02x 02x
f2x f2x   x20 x20
02f 02f
x20 x20

'antiprism':
x20 02x   x20 02x   x20 02x
02f f20   02q q20   02x x20
f2x x2f   x20 02x
02f f20
x20 02x


the nice thing here is that, because of the symetry of ike, the 90-degree rotation will result in quite a symetric constellation . It is easiest to visualize with the 3-interlocking-golden-rectangles way of viewing ike. now, if we rotate ike, and look at the two wide rectangles, we see that the horizontal and vertical rectangles will find each other, meaning that we could also have rotated around their long axes

For the tet, since it's dual is its 90-degree-rotated self, we have that this antiprism-thing is indeed CRF.
With the oct-constellation, an additional 4 octs and 6 tets will fill the gaps, for a total of 6 octs and 6 tets. I do not know how to verify whether this is CRF, or know of any polytope of which this is a snippet?
The ike gaps are then filled with 6 tets (on the x-edges of the rectangles) and AFAI could see, 20 octs :O edit: 30 tets and 8 octs (you could also see this constellation as snub oct || dual snub oct, on the basis of which I start to doubt the CRF-ness )

if, and that's a big if, these things (especially ike-like) happen to be CRF, we could start with the process which leads to the bisphenocingulum in 3D (finally). We only need to find some complex which has a distorted version of ike as a convex hull. For this, I looked at the crown jewels in 3D a lot, and recalled my post in the snub disphenoid based CRF thread, which showed:
oct -> pentagonal bipyramid (adding an edge)
ike -> sphenocorona (deleting an edge and squishing until we have two squares)

Code: Select all


        o  o  |    o
         f    |     B  o
      x    (x)| x
         f    |     B  o
        o  o  |    o

      (o) (o) |   
         f    |     x
      x     x |  x     x
         f    |     D
        o  o  |   o   o


ike -> sphenomegacorona (stretching a point to become a line)
ike -> hebesphenomegacorona (stretching two points to become lines)

Now, especially the sphenocorona is a diminishing of ike, and therefore could be equipped with two squippies and five tets to make the ike-hull. The constellation would then be:

Code: Select all

ike -> ike-hull || other ike->hull
02x   |   02x   ||   x20
f20   |   x20   ||   02D
x2f   |   x2B   ||   B2x
f20   |   D20   ||   02x
02x   |   02x   ||   x20


Here I chose for opposing-oriented sphenocoronas, hoping that they will exactly cancel each other's curvature in the ring-of-20-octs-and-6-tets I know this is also a (very) long shot, and might be proven rong from like step 1, but I considered the symetry-lining-up thingy in ike too beatiful to not share

[Klitzing]

tet||inverted tet (24-cell?)

tet || dual tet = hex (16-cell)

--- rk

[student5]

tet||inverted tet (24-cell?)

tet || dual tet = hex (16-cell)

--- rk

thanks for pointing that out, I edited it. Also, I found this nice image of the compound of two icosahedra. In this you can see that the coplanar triangles (that would be connected by octahedra) are actually not each other's duals, but have a weird angle between them...

Does this mean that there is no possible CRF-ness?

[Klitzing]

Does this mean that there is no possible CRF-ness?

sorry, but yes. The mere stacking of those ikes would therefore ask for lacing edges with different lengths.

--- rk

[Klitzing]

You even could dream up stacking 2 n-antiprisms in mutually gyrated form atop of each other. The top bases then would truely connect by a further n-antiprism, and so do the bottom bases. The latteral triangles of the starting 2 antiprisms each would connect to a vertex of the other one. And there would be further simplices defined by any of the lacing edges of either base antiprism, in fact those would match pairwise in the bottom/top antiprism in an mutually inclined fashion. - The only problem here is, that generally those mutually inclined edges won't be at a right angle in general. Therefore the set of lacing edges between those 2 antiprisms necessarily would be of different length, i.e. none of those 6n simplices could become a regular tetrahedron. - The single exception here would be n=2, resulting then in the hexadecachoron.

--- rk

[Klitzing]

In fact, any such a stacking of an enantiomorph snub pair would just qualify as a near-miss segmentochoron for the same reason.
(And even that nearness attribute might be disputable in general.)

--- rk

[student5]

Thanks for your clarifying answer Klitzing!

Also, I have found Marek's proof, it's here: http://hi.gher.space/forum/viewtopic.php?f=32&t=1978&p=22213&hilit=+rigid#p22213 (under derivation of 3d crown jewels)

Heuristically it says: since in 3d we have vertex figures up to a pentagon (which you can deform), we can dp squishy deformations, that result in cool Johnson solids. However, in 4d there are vertex figures up to an icosahedron, which you cannot deform. Therefore, any squishy deformation will result in a broken CRF.

In fact, this basically proves the impossibility of any squishy deformations resulting in a CRF :/ I think this then rules out the possibility of any analogue of squishy polyhedron :'( (snubs and crown jewels).

Too bad, the search here (as well as under snub disphenoid) ends (unless someone has an idea? )

Putting it this way; In 3d we were playing with papercraft, now we are building lego

[quickfur]

I think the ultimate solution is to write software that will let you assemble polytopes cell-by-cell, with automatic calculation of joining angles, convexity, etc.. It's the only way I can think of to discover more higher-dimensional crown jewels.

[Klitzing]

Yes, kind of some 4D Lego software with non-orthogonal bricks.
Added by lots of measuring possibilities, possibilities of folding up of partial nets, etc.

Idealy with various visualisation possibilities:
not only with some globally projective view, even if stereographic (like quickfur's pics),
but additionally also with some local projection onto the tangent hyperplane.
That is, some slight deformation of the correspondingly inclined cells,
such that the angular defect gets removed, and the projected cells just fill that 3-space without gaps.

--- rk