Regular compounds in 4D

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

Regular compounds in 4D

Postby wendy » Tue Apr 30, 2019 1:40 pm

The current list of regular compounds in 4d is 186 items, that is 140 more than that of Coxeter's regular polytopes.

The list is like this.

1: 4h

This is in the simplex group, where each number from 1 to x represents a polytope, and the compounds are identified with the vertices, edges, tringles, etc of a simplex.

Coxeter's notation, which we severly modify is v{vertex-shell}[p {figure}]f{face-shell},

A compound of 'p' figures may have all their vertices in a vertex-shell, each vertex of the shell taken v times, and likewise, may use the faces of an inner core, the face of the inner core yielding a compound of f elements or faces of the outer core. In 3d, the compound of ten tetrahedra (40 vertices), is 2{5,3}[10{3,3}]2{3,5}, uses the vertices of the outer dodecahedron twice each, and the faces lie by pairs in the face-planes of the core icosahedron, though this is not a regular hexagon (it is x3f ),

[2] Order-2 of this is the {4,p}[2 {p,p}] {p,4} eg stella octangula and {2p/2} eg hexagram.
[3] Order-3 is represented by subgroups of {3,4,3} and {6,3}. We have 1={3,3,4}, 2 = {4,3,3} and 3 = {3,4,3} or 1 = h3o6o, 2 = x6o3o, and 3 = x3o6o. {2}[2{1,2}], [2{2,1}]2 and the two regular stars 2{3}[3{2,1}]{3} and {3}[3[1,2]2{3},
[4] Order-4 occurs in the {3,4,3,3}, contains the various combinations of vertex, edge, triangle, tetrahedron. Note that you can have just two edges or four edges of the tetrahedron as a valid compound. The pair of opposite edges corresponds to the self-dual ((4,3,3,4)), ie a cubic and its body-centre. The four edges form a compound representing the four quarter-cubics.

The h/f or LR group

The totem for the LR groups 3r/f and 4h/f is a array of 5×5 squares. The whole array is one figure, each row and column is a copy of a second figure, and the squares form a third figure. In 3d, the central inversion transforms rows into columns, and so there is a definite diagonal not present in 2D or 4D.

In 3D, the cells are vertices, the rows or columns are tetrahedra, the row and column that passes through the same diagonal cell form a cube. The total is the dodecahedron vertices or icosahedron cells. The five diagonals represent the 5 copies of 3r = [2,2] that make 3f = [3,5].

In 4D, each square represents the point-group of 24 (ie vertices or faces of the {3,4,3}, the actual {3,4,3} is (24,24), meaning that the vertex-rays and face-rays point in different directions.) The vertex-ray and face-rays of the stella-prismata and stella-tegmata point in the same direction, so these are represented by (24). A row or column is 120, representing the 120 verticies of the {3,3,5} and most of the stars, or the faces of {5,3,3} and most of the stars. In other words, what we write as (120) represents 13 different polyhedra.

So as before, we have (five cells in a row), or (five rows in a grid) or (ten rows in a grid), or (25 cells in a grid).

This gives (120)[5(24,24)](120) = (120,120), (600)[5(120,600)], 2(600)[10(120,600)], and (600)[25(24,24)]5(120) = 600[5(120,120)]5(120). of 3, 12, 12, 6 compounds respectively.

The (120,600) represent the vertex-rays of the {3,3,5} and {5,3,3}, as well as the face-rays of {3,3,5/2} and {5/2,3,3}. The remaining stars {5/2,5,3}, {3,5,5/2}, {5,5/2,5}, {5/2,3,5}. {5,3,5/2}, {5/2,5,5/2}, {5,5/2,3}, and {3,5/2,5} have vertex and face-rays pointing to the same (120).

The point-group of (24) represents the vertex-ray of a x3o4o3o, the face-ray of o3o4o3x, and the vertex- and face-rays of x3+3o4o and o3+3o4x (+ = klitzing-sytle branch *a4/2*c). So if the (24,24) is in a compound with numbers at both ends, then 3 are given if the two ends are the same, and 6 otherwise. The face-rays of one direction become the vertex-rays of the other.

The compound of five 24ch, represents the five squares in a row. But note that there is a compound of (600)[5(120,120)]5(120), This means, that in the first half, there is a unique 25-colouring of the (600), each representing one of the inscribed (600)[25(24,..., and at the same time, each of the vertices is represented five-times over in the 5(120)[25(24.. . In other words, for a given diametric axis, it appears once in each row and column. The common element between two cells, not in the same row or column is a hexagon. One can see by fixing one cell, that there is a 4×4 array representing the 96 edges of the {3,4,3} as two sets of separate hexagons.

Since each row of the array represents both a subset of 600 and a group of 120, we can take n rows to get

n(120)[5n(24,24)]. four each for n=2,3,4. 1 and 5 have already bean dealt with.

The figures with 24n vertices on the right-hand side, are respectively, the figure bounded by 72 weimholt-hexahedra, of 48 teddys, and the snub-24 cell, none of which are regular. This is 12 new compounds. None of these are in coxeter.

Given further that each row and column represents (120), we can 'flip' the array so it stands above the plane with five vertical towers (intersecting at a row), and downwards on the columns. This creates nine rows, the combination of 5400 verticies makes nothing unless only the middle row is taken. But the compounds are all set in the 600 rays of the central layer.

The symmetry is that each set of above and below are chirally alike, (ie 100 = 001), and there is no ordered series here. In other words, the combinations that represent different compounds here are [1-4][0-1][0-a], where 'a' is the value in the first set. So 100 represents a layer above the grid, while 001 is a layer below the grid. These two are identical, and are counted once. 010 is the layer of the grid, and is entirely different.

So we get 100, 110, 101, 200, 111, 210, 201, 300, 211, 310, 202, 301, 400, 212, 311, 410, 302, 401, 302, 411, 312, 303, 402, 313, 412, 403, 413, 404, 414, or 28 different sets.

Coxeter list only 400 and 404. The transcription of his list gives 410 and 414, the first by the process of (600)[(5(120,... and 5(120)[25(24,.. gives 5(600)[125(24,...).

None of these vertices of the duals point to any regular, so we get 4 compounds for each, of the form

n(600)[25n(24,24)] = n{5,3,3}[25n{3,4,3}], n{5,3,3}[75{3,3,4}]2n{3,3,5}, 2n{5,3,3}[75{4,3,3}]n{3,3,5} and [25{3,4,3}]n{3,3,5}.

where n is the sum of digits in the string above, eg n=5 for 212, 311, 410, amd 302. Coxeter's notation represents these five distinct compounds.

The group 4s/f "mete-star" (3,3,3).

The root compound here is n(600)[120n(5) = n{5,3,3}[120n{3,3,3}]{3,3,5}, for n = 1 to 7 (some repeats). The totem here for the non-special group is a six-by-six array.

Each row and column of the array represents 100 vertices, being a bi-decagonal prism. It is the same as the centres of the 100 tetrahedra of the grand antiprism, that do not touch the pentagonal antiprisms. The six array is represented by the six diagonals of the icosahedron, we can take these left- and right- parallel.

Since each row of the totem is a partition of the same (600), we can take 1, 2, 3, 4, 5, 6 of these, and add in the special to each, to give 1s, 2s, 3s, 4s, 5s, 6s.

We now investigate the totem, and see that all 720 possible simplexes are indicated in right-turning of the 36 inscribed bidecagon-prisms. That means that for a row, the 120 corresponding left-turning figures are scattered through the remaining five rows. We can, for example, take it that if '1' represents the partition of (600) into 6 sets of (100), and that set is further partitioned into 5(20), the common element of intersection of a row and column is (0), and the cells off the intersection is then (20) = five pentachora, apparing left in one direction and right in the other. Then we can take the left- and right- of a set of 6 clifford-parallel prisms (ie a row and column), and these form a compound of n=2, different to 2 rows.

We now have 1x, and its complement 5x, different to the rows of 2 and 4 above. Each of these can also couple with the special group s, giving 1xs, 5xs,

All together, we add here 16 new compounds, over and above coxeter's listing of 's'.

All together, there are 4 + 33 + 12 + 116 + 1 + 16, giving 186 compounds, and a lesson in colour-groups or ray-sets, to boot.
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Re: Regular compounds in 4D

Postby ndl » Wed May 08, 2019 2:31 am

Would there be a good way to reduce these compounds into categories, lead by the compound that represents all the iterations of a sub-symmetric polychoron in the larger symmetry?
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Re: Regular compounds in 4D

Postby ndl » Sun May 19, 2019 2:32 am

Would all the 4D compounds be able to be created by faceting the regular and rectified polychora and stellating the duals like can be done in 3D? Klitzing already has the facetings of the regulars on his website and that does provide many of the compounds already.
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Re: Regular compounds in 4D

Postby wendy » Sun May 19, 2019 12:08 pm

Some are faceting, some are stellations (ie extending the faces), and some are both.

The greater part come from that a 24-point group contains the vertices and faces of 3 tesseracts, and the vertices and faces of 3 16chora, or the group of 5 contains 5 vertices and 5 faces of a pentachoron. But the greatest of these has 5400 vertices, which really correspond to nothing in particular, nor do anything 600n, where n>1.

The real supprise here is that the bulk of the s/f groups (pentachora), pass through a bidecagon duoprism, the vertices of {5,3,3} split neatly into six of these in twelve different ways.

Only some of the simpler ones would appear as simple facetings. For example, if you take a {5,3,3}, and replace its faces with 5 cubes or 10 tetrahedra, you will end up with the compounds of 75 {3,3,4} and {4,3,3}. The five octahedra will lead to 25 {3,4,3}. But those in the non-special groups don't lead to much. Some of the vertex-lists are given: the compounds of 5n {3,4,3} are given as known CRF's. But the situation is worse than in 3d, as many of the figures in the extraordinary groups have many more vertices.

You can derive these by faceting, which is what Coxeter did. But unless you wrangle the colour-groups, you won't find anything like 1/2 of the total.
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Re: Regular compounds in 4D

Postby ndl » Tue Jun 04, 2019 12:29 am

I found two very interesting uniform compounds while playing around with Stella4D: A 25 and 50 hex compounds. Both have the convex hull of gap and the 50hex is also a compound of 25 haddets. The vertex figures are compounds of 2 and 4 octs respectively. They really are subsets of the 25 stico compound removing 1 or 2 hexes from each stico. It's just cool they have a uniform hull. Anyone have any good names for them?
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Re: Regular compounds in 4D

Postby Klitzing » Fri Aug 30, 2019 12:46 pm

Yesterday I was considering the lace city of rox = o3x3o5o, cf. its upper half:
Code: Select all
                                            o5o                                           
                                    x5o             x5o                                   
                                            o5f                                           
                        o5x                                     o5x                       
                o5o             x5x                     x5x             o5o               
                                                                                          
                    f5o                     x5f                     f5o                   
            o5x                     F5o             F5o                     o5x           
                                o5F                     o5F                               
                        f5x                                     f5x                       
                                                                                          
            x5x                     f5f             f5f                     x5x           
    x5o             o5F                     V5o                     o5F             x5o   
                                                                                          
                F5o             o5V                     o5V             F5o               
o5o                     f5f                                     f5f                     o5o
    o5f                                     x5F                                     o5f   
            x5f                     F5x             F5x                     x5f           
                                                                                          
x5o                     V5o                                     V5o                     x5o
                                x5F                     x5F                               
            F5o                                                             F5o           
    x5x             f5f                    oA5Ao                    f5f             x5x   

where x=1, f=(1+sqrt(5))/2, F=ff=x+f, V=2f, A=F+x. The central oA5Ao clearly denotes the compound of o5A and A5o.

Further I tried to spot a vertex inscribed d-scaled ex = d3o3o5o, d=x(10,2)=sqrt[(5+sqrt(5))/2] and found
Code: Select all
                                            o5o                                           
                                                                                          
                                                                                          
                                                                                          
                o5o             d5o                     o5d             o5o               
                                                                                          
                                                                                          
                                                                                          
                                                                                          
                                                                                          
                                                                                          
            o5d                     o5D             D5o                     d5o           
                                                                                          
                                                                                          
                                                                                          
o5o                     D5o                                     o5D                     o5o
                                                                                          
                                                                                          
                                                                                          
                                                                                          
                                                                                          
                                                                                          
    d5o             o5D                     d5d                     D5o             o5d   

where D=df and dd=(5+sqrt(5))/2=2+(1+sqrt(5))/2=u+f=A.

That is you obviously could have a {3;3,5}[6{3,3,5}] (which is thus similar to {5,3}[5{3,3}], the 5 tet compound within doe), thereby providing a 6 ex compound within rox. This is as d5o and o5d both are the alternate x(10,2)-chord pentagons inscribed into x5x, and D5o and o5D both are the f(10,2)-chord pentagons inscribed into f5f, just as A5o and o5A happen to be the d(10,2)-chord pentagons inscribed into d5d.

But then note that one could reflect each symbol of the lower partial lace city individually, That is, both reflections could be inscribed simultaneously into rox. Again this should be applicable to the various orientations to, giving rise to a different 6 ex compound within rox, kind its opposite Enantiomere. Therefore, if I'm not missled, there ought be a 12 ex compound within rox as well, then becoming 2{3;3,5}[12{3,3,5}] (then being kind the 4D equivalent to 2{5,3}[10{3,3}]).

Was the reasoning of my latter find without any flaws?
What are the full names of those compounds?
What are/would be the according OBSAs?

--- rk
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Re: Regular compounds in 4D

Postby wendy » Sat Aug 31, 2019 7:34 am

No flaw in the reasoning. Compounds of x3o3o4o, {3,4,3}, m3o4o3m, and x3o3o5o should be all that common, since these are the base poincare groups in quarterions.

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Re: Regular compounds in 4D

Postby Polyhedron Dude » Thu Sep 05, 2019 1:09 pm

Richard, you really found a winner with this one!

After investigating this 12-ex compound, it lead me to find all sorts of possibilities. Of course there would also be 12-sishi compounds and 12-gishi compounds along with their regiment members and 6-compound counterparts. I also found that there's a 6-sisp and a 12-sisp compound (as well as gisp ones). For a while, I thought there would be two 6-sisps and three 12-sisps, but it turned out to be a false alarm - there's only one of each. The 12-sisp is actually achiral where its combocells are 10/2 APs (10/2 being the compound of 2 pentagons). The 12-ex regiment has 20 uniform members (if we include the 6-ex ones and the mixed 'combo-uniform' compounds such as '6-ex 6-fix') plus 7 scaliform compounds based on spysp. The verf of 12-ex is a compound of two ikes meeting at opposite vertices - aka point || 10/2 || 10/2 || point. I decided to call it Pedisna for pental disnubbachoron. The 6-ex is Kepisna for chiropental snubbachoron. The 12-gishi regiment will have the same number of members. The 12-sishi on the other hand will be much larger and should have 350 members plus 55 scaliform compounds (assuming my quick calculation was correct).

And this was the beginning of new discoveries. We all know that there's a 10-rox blend which forms the sidtaps - therefore there will be a 10 pedisna compound - aka a 120-ex compound with the same vertices as the sidtaps as well as its regiment, there will also be a 120-gishi regiment also. Then there's the horrificly huge 120-sishi regiment that would likely number near a half a million or more - I'll need to analyse this one in more detail. Mixed compounds could take on four different polytopes - 30A 30B 30C 30D. The 120-ex compound is also a compound of 12 10-ex as well as 10 12-ex.

Yet more finds. The 6-ex can lose a few exes and be uniform. The 5-ex has the same vertices as spidrox. The 4-ex was biform instead of uniform. The 3-ex was uniform as well as the 2-ex. So there will also be 2,3, and 5 - sishis and gishis as well as their regiments among the uniform compounds. After rendering the 6 ex in Stella, I took its dual which will be the 6-hi - it was uniform! It has the same vertices as a golden ratio srix (fox^o) - this is also the vertices of the sidtaps. The 12-hi will also have the same vertices with a verf which looks like an X || 'X rotated 90 degrees'. There's also 6 and 12 gogishi compounds. It gets worse, there are also 6 and 12 sidtaxhi, dattady, and gadtaxady compounds along with their regiments. There was also an unexpected surprize - the 12-sidtaxhi and 12-gadtaxady verfs fused and formed actual scaliform polychora that were not compounds - this will lead to two new categories of scaliform polychora - S10 and S11. There are 66 scaliforms and one fissary scaliform in each regiment. I did get some of them to render in Stella, they are as wild as the idcossids. The fissary scaliforms are the blend of 12 sitphis and the blend of 12 gitphis.

Below are some more names for the 6 and 12 compounds:

6-hi Kadhep (the dh pronounced like a soft th) chirododecahedral hyperprismatochoron
12-hi Dadhep disdodecahedral hyperprismatochoron
6-fix Kihp chiroicosahedral hyperprismatochoron
12-fix Dihp disicosahedral hyperprismatochoron
6-gohi Gikadhep great kadhep
12-gohi Gedadhep great dadhep
6-gahi Gakidhep grand kadhep
12-gahi Gadidhep grand dadhep
6-sishi Skadhep stellated kadhep
12-sishi Sidadhep stellated dadhep
6-gaghi Gigakidhep great grand kadhep
12-gaghi Gigadidhep great grand dadhep
6-gishi Giskadhep great stellated kadhep
12-gishi Gisdadhep great stellated dadhep
6-gashi Gaskadhep grand stellated kadhep
12-gashi Gasdadhep grand stellated dadhep
6-gofix Gikihp great kihp
12-gofix Gidihp great dihp
6-gax Gakpisna grand kepisna
12-gax Gapdisna grand pedisna
6-gogishi Gogaskadhep great grand stellated kadhep
12-gogishi Gogasdadhep great grand stellated dadhep
6-sisp Kepap chiropentagonal antiprismatochoron
12-sisp Dipap dipentagonal antiprismatochoron
6-gisp Kiparp chiropentagrammatic retroprismatochoron
12-gisp Diparp dipentagrammatic retroprismatochoron
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Re: Regular compounds in 4D

Postby Polyhedron Dude » Sat Sep 07, 2019 4:04 am

There are even more finds. I started looking at the compound of two icoes in dual position and noticed that there are several hex compounds: a 6-hex, a swirly 6-hex, a swirly 12-hex, a different 12-hex, an 18-hex, a swirl 18-hex, and a 24-hex. The 24-hex's verf looks like an oct with three more octs, each attached to the original one at opposite vertices and spun half way - I call this 4-oct compound a vertex inensified oct. The dual are also uniform leading to tes compounds. There's also a uniform 6-ico that is chiral - I suspect there's a full 12-ico, but the rectified 12-hex lead to a non-uniform. The 6-ico's verf is in the spic army. I then trucated the swirl-6, swirl-12, and 18-tes compounds and all of them are uniform as well as the quasitruncates. Therefore, there is a swirl compound of 6 quitits that is uniform. I suspect the truncated forms of the other tes compounds are uniform also, but Stella wouldn't render those. There are also rit compounds, but only the 6 and 18 ones are uniform, I suspect the 24 one is also, but was unable to render it with Stella. The greatest surprise was with the sidpith and gittith based compounds, for they fused together to form new scaliforms. The verf of the 24-sidpith looks like a trigon antipodium with three more blended into the bottom walls (representing square prisms) at their bases. I got several of them to render on Stella, but I couldn't get the swirly mixed blends to render, but they should work. One of these blends should be a compound of ondips. These should add one or two more scaliform categories S12 and S13 which I may call the 'tetrasidpiths' and 'tetragittiths' due to the shape of their verfs.
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Re: Regular compounds in 4D

Postby ndl » Sun Sep 08, 2019 4:38 am

You mentioned that you rendered the 6-ex compound in stella. How did you do that? I've not been able to figure out an easy way to render compounds in stella. Thanks.
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Re: Regular compounds in 4D

Postby Polyhedron Dude » Sun Sep 08, 2019 1:16 pm

I had to render the verf of a 6-ex 6-sisp compound to get it to work. It looks like a compound of two ikes where the ikes meet at opposite vertices and are turned exactly halfway, but with one of the ikes faceted to resemble a sisp vertex figure. Align the sisp verf where the star shaped vertices are on the top and bottom where they meet the other ike. This figure will render in Stella, then you can remove the sisps to get the 6-ex compound, or remove the exes to get a 6-sisp compound. Strangely, the 12-sisp verf will render a polychoron, make sure that you have a right handed and left handed sisp verf that form 5 X like structures on the equator of the verf. The 6 and 12 gisp compounds can render in a similar manner and they look awesome. One way to get the 2-ike compound is to look at the sections of rox at the vertices starting from a single vertex, one of the sections will have the 2-ike vertex arrangement, turn it to a 3-D polyhedron and facet out the two ikes.
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Re: Regular compounds in 4D

Postby ndl » Sun Sep 08, 2019 6:38 pm

I got it to work, thanks! Did you have names for the 50 and 25 hex compounds in gap I mentioned above?
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Re: Regular compounds in 4D

Postby Polyhedron Dude » Mon Sep 09, 2019 5:33 am

I took a look at them and got some names. The 50-hex has 20 10-tet (e) and 100 6-tet (crazy looking compound) and the 25-hex has 20 5-tet (ki) and 100 3-tet (another crazy looking combo) for its combo-cells, so I gave them the following names:

50-hex: hexal snub decatetrahedral icosachoron - Hesadti
25-hex: chirotrital snub pentatetrahedral icosachoron - Ketspatti
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Re: Regular compounds in 4D

Postby ndl » Tue Sep 10, 2019 3:13 am

Great! Thanks again. If you have any stella files of compounds you created I would love to have a look at them. I attached a zip of some of the ones I made:

Stella compounds.zip
(367.22 KiB) Downloaded 6 times
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