Wythoff Polytopes

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

Wythoff Polytopes

This is really an annex to the CRF project, but we're not really going to find any crf's here. Instead, it's intended to lay down some formulae.

Wythoff Polytopes

Basically, this is any polytope written in a dynkin symbol, using node-marks that represent directly, the size of the edge. so F3x is a WP. This distinction means that you can use the coordinates as a vector, and use the fancy vector arithmetic to calculate size and distances. In essence, a polytope like x5o3x is just a fancy coordinate (1,0,1), "AICO" (all icosahedral coordinates).

We can directly make a matrix that allows you to do a dot product between two vectors specified as AICO, or even find the length of a given vector.

Wythoff Lace Prisms

These are wythoff polytopes "laced together". This seems to be the mainstay of the polytope project, and part of the hassle is finding coordinates from lacing lengths. The first bit is relatively straight-forward: L² = H² + D². L is given in the "&#xt" bit. H is sought, and we shall find D from our vector project.

The relevant vector we are going to look at, is then (T-B), where these are the polytope vector at the top, less the similar one at the bottom. So we have, eg for xo3ox5of, t = xoo = (1,0,0) b = o,x,f = (0,1,f), d = (1,-1,-f). We then calculate d·d, and subtract this from L², to get the height. ie

H² = L² - (t-b)·(t-b).

Because the coordinates are oblique, we can not do the usual dot product, but must either set the thing into an orthogonal schema, or do a "matrix-dot" product, ie D² = Sij di dj, where Sij is the relevant stott matrix.

Stott Matrices

The two ways to get the stott matrix are to use the relation Sij Dij = I, where Dij is the dynkin matrix, or the time-honoured way of writing the stott matrix out from the Schläfli vector and the resident animal. The latter is very efficient if you are using hand calculations.

The Dynkin matrix D<sub>ij</sub> is simply a matrix, where the main diagonal Dii is filled in with '1', and the rest of the elements are filled in with Dij = -cos( pi/[ij]), where [ij] is the branch-mark between nodes i, j (counting unconnected nodes as '2'. It's a dynkin matrix, because you can load it directly from the dynkin symbol. When i do this by hand, i normally double all the entries, and put a leading '1/2' out the front. This saves having to remember both 1.61803398875 and 0.809017&c.

You can do it by hand: just write out the required matrix by using the Schläfli vector and the resident animal. When you do this in front of someone, like 'just use this technique', it looks very cool etc.

The vector is written in the first column, starting at the bottom. It always takes the same pattern for ,3, etc, so once you learnt the general pattern, it's pretty easy to write these things down. The order is icosahedral, which means we put the non-threes bit down the bottom end (ie we do 3,3,5, not 5,3,3).

The schläfli vector is derived pretty much from a kind of 'continued fraction' derived from the schläfli symbol. For the first few, you can see that the animal is a single cell, occupied by the dimension-number. Tilings are the surface of an n+1 polytope, so {3,6} is a polyhedron, not a 2d thing, so n=3 here.

• 3...3 1, 2, 3, 4, 5, 6, ... Animal = [n]
• 3...4 q, 2, 2, 2, 2, 2, ... Animal = [n]
• 3...5 f, 2, 3-f, 4-2f, 5-3f, 6-4f, Animal = [n]
• 3...6 h, 2, 1, 0 Animal = [n]
• 3...P a, 2, 3-b, 4-2b, 5-3b, ... Animal = [n] a=shortchord, b=third-chord = a²-1.
• 3...3 q, 2q, 3, 2, 1, 0
• 3...A 2, 2, 4, 4, 4, 4, 4, 4, ...
• 3...B 3, 2, 4, 6, 5, 4, 3, 2, 1, 0

The animals for the larger symmetries like {3,4,3} and {3,3,3,3,B} = 2_21 are given here. There's only three of them.

Code: Select all
`            Animals for the non-regular symmetries    3..4,3           3...A                  3.....B                      k_11                     k_21   [ 2n-2  n-1 ]   [  n   n-2 ]     [ 2n-2  n-1  2n-6 ]   [ n-1    2  ]   [ n-2    n ]     [  n-1   4    n-3 ]                                    [ 2n-6  n-3    n  ]`

The rest of the Sij is filled out by Sij = j Si1, as long as j <= i. The matrix is symmetric, so Sij = Sji.

Here are some worked examples, of writing these things directly. The order is to fill in the animal first, then the schlafli vector (which must be found to get the correct denominator).

Code: Select all
`    {3,3,5}        5-3f      [ 4-2f                   ]           [ 4-2f  3-f     2    f  ]  2   [ 3-f                    ]      2    [ 3-f  6-2f     4   2f  ] --   [  2                     ]  -> ---   [  2    4       6   3f  ]      [  f       .     .    4  ]     5-3f  [  f    2f     3f    4  ]       Schlafli            Animal     2_21 = .3.3.3.3.B.   n=6         3  (overall demom)      [  4     5    6    4   2   3   ]   2  [  5    10   12    8   4   6   ]   -  [  6    12   18   12   6   9   ]   3  [  4     8   12 ( 10   5   6 ) ]      [  2     4    6 (  5   4   3 ) ]      [  3     6    9 (  6   3   6 ) ]        S.v           (    animal  )`

Canonical Polytopes etc

We can write a prism of size l*b*h as (l)2(b)2(h). The brackets mean that the letters are just algebraic variables, without any special meaning. In any case, while this is a wythoff-polytope representing a prism of l long, b broad, and h high, the standard coordinate gives l,b,h ACS (all change of sign). In other words, putting l directly into the coordinate, gives a prism whose size is 2l, 2b, 2h.

Since we often just feed the 'size-as-written' from the dynkin symbol into a program, there are the odd 2 and 4 to be met as we eliminate this doubling.

So one has to be wary of this when one finds the value of d from lacing and height.

For example, the subtraction of eg (1,1,0) from (0,1,1) gives a vector (1,0,1), but this applies to a polytope whose side is *2*. (it gets confusing, and i need to think this one through, its always one of my trouble-spots).

APACS and EPACS etc

The task of finding coordinates from a wythoff polytope, involves passing a single coordinate say (1,1,0) AICO, through a number of filters to get a larger number out the bottom. Much can be saved by way of expanding filters. But basically, AICO represents a corresponding vector in every element of the icosahedral symmetry. You get one of those black-and-white symmetry groups, and think, the white ones are even octants, the black ones are odd quadrants.

Most polytopes, and all Wythoff polytopes, have vertices in every quadrant.

The routine AICO -> EPACS replaces a single point with up to five points.

The routine EPACS -> ACS replaces a single epac point (1,0,0) with three points (1,0,0), (0,1,0), (0,0,1)

The routine ACS -> ALL replaces a point of acs with up to eight standard points.

There are other coordinate systems.

EICO is just the white sectors of the icosahedral. It converts five-fold into EPECS

AICO is the icosahedral group, * 2 3 5 it five-folds down EPACS.
EICO is the rotary icosahedral group 2 3 5 (snub dodeca), it is five-fold to EPECS
APACS is the ordinary cubic symmetry. *2 3 4
EPACS is the pyritohedral symmetry 3*2
APECS is the tetrahedral symmetry *2 3 4
EP&CS is the octahedral-rotary group 2 3 4 (ie snub cube)
EPECS is the tetrahedral-rotary group 2 3 3
ACS is the rectangular group * 2 2 2

In a lace tower, the bit after the &# is another coordinate axis. So xo3oo5ox&#x actually is 4d, three over the base by AICO, and the fourth by height.
The dream you dream alone is only a dream
the dream we dream together is reality.

\(Latex\) at https://greasyfork.org/en/users/188714-wendy-krieger wendy
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Re: Wythoff Polytopes

This got me thinking, what about non-Wythoffian polytopes? One way to generalize this that occurred to me, is the idea of "AICO": basically, it takes a reference vector V, and generates a set of vectors based on a set of transformations applied to V. In the most general case, these transformations do not have to map to any symmetry group. One could, conceivably, map V to the vertices of an irregular polytope. As long as subsequent layers of the lace tower / lace city can be expressed in terms of these transformations, we can represent all vertices with a single reference vector V.

Granted, it will be more difficult to calculate lacing heights, but in a sense, it is no different from what we're already -- at least for me, when I build the models of the CRFs, it boils down to finding two reference vectors V1 and V2 that have a lacing edge between them, and computing the required lacing height. Since V1 and V2 automatically generate the other vertices, this is sufficient to build the entire stratum. So as one builds the lace tower, which is just a matter of computing the heights between each layer, one essentially only needs to look at a single vertex per layer and everything else sorts itself out via the transforms. (Provided, of course, that the lace tower construction is sound.)

Since we're allowing arbitrary transforms, the lace "tower" no longer needs to be an actual tower in the sense of having layers of vertices on parallel hyperplanes; conceivably, one could from a single point V generate a duoprism of some order. This then allows us to describe polytopes with duocylindrical symmetry as a pseudo-"lace tower" in which the symbols employed are not simple CD diagrams, but something that describes concentric duoprisms. Perhaps a "nested" CD diagram will be able to describe such a construction, where you have the notation AnB, where A and B are nested CD symbols, and n is the order of the duoprism ring. So a tesseract could be described as (x4o)4o, and a 6,6-duoprism could be described as (x6o)6o. We can then describe a duoprismic generalized Stott expansion (x6o)6x, meaning that the x6o's are expanded outwards to form a "runcinated" 6,6-duoprism, in which the two rings of hexagonal prisms are interfaced by an additional layer of hexahedra. I haven't sat down to rigorously work out this new notation, so there may be some parts that need some ironing out, but these give a rough idea of what is possible.

Conceivably, we could also have a notation for swirlprism symmetry, in which a single vector V generates all the vertices of some underlying discrete Hopf fibration. Perhaps such a notation could be a powerful tool for discovering CRFs with swirlprism symmetry.
quickfur
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Re: Wythoff Polytopes

If you are ultimately going to evaluate cartesian coordinates for the points, there's really no point in constructing a stott matrix. That's really an avoiding route.

Anyway, i banged together a routine AICO to EPACS. The first column is a series of constants. These are representative vectors corresponding to (1,0,0) etc for the ICO system. The second column is the expansion of the five vectors in terms of the references given in the dynkin-graph eg (x3o5o) and the six vectors in column 1. This produces five vectors P0 to P4. If any of A, B, C is zero, you only have to evaluate P0, P2, and P4. Column 3 is a reference to the numbers 1, f, F in col 1.

Feeding in from x3o5o, eg the vector (1,0,0), will give the canonical vertices of the icosahedron: that is, the one of edge of 2. This is important, because you must multiply any non-reflected edges (like lacings), by two also. But if we suppose the unit is 2, the heights are correct as given, (since it's really 2h for 2e), but the lacings have to be doubled before being used.

Code: Select all
`    Given (A)3(B)5(C)    e1  =  2f,  0, 0        P0 = A v1 + B  e1 + C h1        f = 1.61803398875    h1  =  F,   1, 0        P1 = A v1 + B  e2 + C h1        F = 2.61803398875    v1, =  f,   0, 1        P2 = A v1 + B  e2 + C h2    e2  =  F,   f, 1        P3 = A v2 + B  e2 + C h2    h2  =  f,   f, f        P4 = A v2 + B  e2 + C h1    v2  =   1,  f, 0`

When any of A, B, C are zero, only P0, P2 and P4 need be found. This gives 3 points, which expand to 60 by way of EPACS.

When any two of A, B, C are zero, the set corresponds to v1, or e1, e2 or h1, h2. This will give 12, 30 and 20 points under EPACS.

To calculate heights, only P0 needs to be evaluated.

The height of the layers are found by evaluating P0 for the difference between the top and bottom vectors, and subtracting this from 4L².

Suppose we want to find the lacing-height of xo3oo5ox. The top vector is (1,0,0). The bottom vector is (0,0,1). The difference is (1,0,-1), all ICO.

The conversion by P0 gives v1 - h1 = f-F, 0-1, 1-0 = (-1, -1, 1). This is in right-angle format, so the size of this vector is x²+y²+z² gives 3. The lacing is 1, so 4L² is 4. The indicated height is then sqrt(4-3)=1. This is actually the height for an edge of 2, but we can keep this number as it is, because we're keeping the edge at two.
The dream you dream alone is only a dream
the dream we dream together is reality.

\(Latex\) at https://greasyfork.org/en/users/188714-wendy-krieger wendy
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Re: Wythoff Polytopes

The reason that i hived off the wythoffs like i did, is because the maths is easy.

You can pretty much do 'nested' dynkin symbols by using a '2' branch, eg x6o2x6o is your hexagonal duoprism. The usual dynkin/stott stuff works here too. But actually 'nesting' them, i suppose defetes the whole point of having a vector coordinate system read directly from the dynkin graph.

Cosines and Sines and matricies are about the limit of my maths at the moment. I rarely use tangents, unless the calculator tells me that atan(x) is a neater value of x than asin(x) or whatever. The books on non-euclidean geometry go on with a lot of un-neeeded rubbish about tanh and cosh etc, most of which i know you don't have to do this. The dynkin/stott matrices work just as well in hyperbolic space for example. But i rely heavily on insight and a good mindhoarde.

I suppose it would be interesting to actually do a swirl prism or something. You could do something in what johnson calls 'unitary space', and i call 'CE2' (complex euclidean hedrix), where at least a common slope gives a common swirl. I get strange ideas about this, but i really have to look through CE2 to see if i can get a consistant angle, say by way of feeding in a polytope. We can look at the complex-polytopes, and see if one can make sense of what coxeter wrote, but it's more a case of calculations.
The dream you dream alone is only a dream
the dream we dream together is reality.

\(Latex\) at https://greasyfork.org/en/users/188714-wendy-krieger wendy
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Re: Wythoff Polytopes

Ths particular thread was rather silent the last years.

Today I obtained an interesting fact on the 421 Gosset fugure, called fy by Jonathan Bowers. And I searched for a place to post it.

There are already several subsymmetric lace tower representations known, cf. to the ones provided in the above given link.
Code: Select all
`o3o3o3o *c3o3o3o = pto3o3o3o *c3o3o3x = naqx3o3o3o *c3o3o3o = laqo3o3o3o *c3o3o3x = naqo3o3o3o *c3o3o3o = pt`
or
Code: Select all
`o3o3o *b3o3o3o3x = zeex3o3o *b3o3o3o3o = hesao3o3o *b3o3o3x3o = rezo3o3x *b3o3o3o3o = alt. hesao3o3o *b3o3o3o3x = zee`

Today I now found the following further representation
Code: Select all
`o3o3o *b3o3o x3o = {3}x3o3o *b3o3o o3o = hino3o3o *b3o3x o3x = trataco3o3x *b3o3o x3o = trahino3o3o *b3x3o o3o = rat  +  o3o3o *b3o3o x3x = {6}x3o3o *b3o3o o3x = trahino3o3o *b3o3x x3o = trataco3o3x *b3o3o o3o = hino3o3o *b3o3o o3x = {3}`

--- rk
Klitzing
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Re: Wythoff Polytopes

The representation of fy = 4_21 of my former post was wrt. axial stacking, i.e. to be understood as lace towers.

Today I like to provide an other found representation of a different 8D polytope, in fact you can
dissect the vertex set of bay = 2_41 = x3o3o3o *c3o3o3o3o into 3 subsets, which then
provide an according description of bay as the tegum sum oxo3ooo3ooo *b3ooo3xoo3ooo3ooo3oxu&#zxt.
That is bay is the convex hull of the compound of o3o3o *b3o3x3o3o3o + x3o3o *b3o3o3o3o3x + o3o3o *b3o3o3o3o3u.
These components respectively are tark, spuho, and a u-scaled version of ek (u=2x). - Or, taken the other way round:
these 3 polyzetta all are vertex inscribable into bay, and, taken in the given respective alignments, then
those define a dissection of its vertex set with neither gaps nor overlaps: every vertex thereby is being hit exactly once.

--- rk
Klitzing
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Re: Wythoff Polytopes

Today I was asked privately, whether it would be possible to
• reduce 221 somehow symmetrically by 6 vertices
• reduce 321 somehow symmetrically by 14 vertices
• reduce 421 somehow symmetrically by 72 vertices
This quest originated from some DNA/RNA research, where some findings could be mapped to the vertices of those polytopes, and where, by means of some special viewpoint considerations, just those subsets would remain...

The main issue here is, even so those mentioned polytopes belong to the exceptional En symmetries, those figures well can be represented as axial stacks of vertex layers wrt. some common across symmetry. E.g. consider a more accessible figure from C3 symmetry, the cube. When considered vertex first, we have an axial stack of topmost vertex, a regular triangle underneath, then a triangle in dual orientation, and finally the diametral opposite vertex. Thus we have represented this figure wrt. the axial A2 subsymmetry.

Within the quest it is that we will have to represent the En figures each wrt. their axial An-1 subsymmetry. This then provides individual layers (or combinations of such) with vertex counts as required. (Those will be highlighted below.)

• fy = o3o3o3o *c3o3o3o3x = 421 (wrt. E8) might be represented in axial A7 subsymmetry as xooxooo3oooooxo3ooxoooo3ooooooo3ooooxoo3oxooooo3oooxoox&#xt = hull of stack of x3o3o3o3o3o3o || o3o3o3o3o3o3x3o || o3o3x3o3o3o3o || x3o3o3o3o3o3x || o3o3o3o3x3o3o || o3x3o3o3o3o3o || o3o3o3o3o3o3x
• naq = o3o3o3o *c3o3o3x = 321 (wrt. E7) might be represented in axial A6 subsymmetry as xooo3ooxo3oooo3oooo3oxoo3ooox&#xt = hull of stack of x3o3o3o3o3o || o3o3o3o3x3o || o3x3o3o3o3o || o3o3o3o3o3x
• jak = o3o3o3o *c3o3x = 221 (wrt. E6) might be represented in axial A5 subsymmetry as xox3ooo3ooo3oxo3ooo&#xt = hull of stack of x3o3o3o3o || o3o3o3x3o || x3o3o3o3o

Other axial orientations exist for sure as well, but would not match the above cited quest.

In an after-post I was asked whether this is "new" or already has been published. - I'd say that the figures themselves are old. Esp. the kij notations usually are attributed to Gosset, as far as I can say. That polytopes can be displayed in axial subsymmetries is obvious. But when exactly the above given representations popped out into light for the first time, I don't know. In fact, some of them I did know of already for long, others I just derived (rediscovered) from skretch.

--- rk
Klitzing
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Re: Wythoff Polytopes

Klitzing wrote:Today I was asked privately, whether it would be possible to
• reduce 221 somehow symmetrically by 6 vertices
• reduce 321 somehow symmetrically by 14 vertices
• reduce 421 somehow symmetrically by 72 vertices
[...]
• fy = o3o3o3o *c3o3o3o3x = 421 (wrt. E8) might be represented in axial A7 subsymmetry as xooxooo3oooooxo3ooxoooo3ooooooo3ooooxoo3oxooooo3oooxoox&#xt = hull of stack of x3o3o3o3o3o3o || o3o3o3o3o3o3x3o || o3o3x3o3o3o3o || x3o3o3o3o3o3x || o3o3o3o3x3o3o || o3x3o3o3o3o3o || o3o3o3o3o3o3x
• naq = o3o3o3o *c3o3o3x = 321 (wrt. E7) might be represented in axial A6 subsymmetry as xooo3ooxo3oooo3oooo3oxoo3ooox&#xt = hull of stack of x3o3o3o3o3o || o3o3o3o3x3o || o3x3o3o3o3o || o3o3o3o3o3x
• jak = o3o3o3o *c3o3x = 221 (wrt. E6) might be represented in axial A5 subsymmetry as xox3ooo3ooo3oxo3ooo&#xt = hull of stack of x3o3o3o3o || o3o3o3x3o || x3o3o3o3o
[...]
--- rk

• that 6-reduced 221 for sure is nothing but the known segmentopeton hixalrix = xo3oo3oo3ox3oo&#x
• that 14-reduced 321 for sure is nothing but the known scaliform segmentoexon rilalril = oo3xo3oo3oo3ox3oo&#x
• more interesting then is just the 72-reduced 421. Meanwhile it got the Bowers acronym kadify. It can be provided as lace tower oooo3xooo3ooxo3oooo3oxoo3ooox3oooo&#xt. Note that this tower is NOT a mere stack of segemtozettons, as there are cross-layer lacings too. Or stated differently: the mere such stack would not be convex. - It should be noted here additionally that the vertex set of fy = 421 can be separated into those of 2 mutually inverse brene plus one soxeb. The 72-diminishing then just omits the soxeb. Therefore kadify also can be rewritten as tegum sum oo3oo3xo3oo3oo3ox3oo3oo&#zx - and as such it qualifies obviously as a scaliform polyzetton too.
Klitzing
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Re: Wythoff Polytopes

Since the dawn of 2018 I was in conversations with David Halitsky. Then he finally asked me about subsymmetrical diminishings of the Gosset polytopes of E6 (212 = jak), E7 (312 = laq), E8 (412 = fy). As it turned out, all of those diminishings, which he proposed - in fact then being based on empirical data from molecular biological DNA/RNA researches - could be found. This was achieved by orienting those En figures as an axial stack of An-1 cross-sectional vertex layers and by therefrom omitting just every third vertex layer.

The 6-dimensional and 7-dimensional diminished polytopes had been given very easily, as those had been already pre-calculated within different contexts. But although the 8-dimensional polytope surely had to exist, its true detailed structure still eluded me so far. But now, 6 month and lots of huge incidence matrices later, I finally managed to get hold of this 8-dimensional diminishing of 412 = fy.

There are 72 vertices, situated in 3 different vertex layers of fy, which are to be omitted. This number of 72 Jonathan Bowers associates in different polytopal acronyms with the letter “K”. This is why I then proclaimed, what meanwhile got supported by him, the acronym “kadify” (short for: “72 vertices diminished fy”). Quite early I managed already to show that this polyzetton itself should be a scaliform polytope. And it was obvious that there ought several 7d simplicial (oca) facets, as well as rectified simplices (roc). But not only their exact counts was not known then, neither there were the number and shapes of the remaining ones.

Now I can report this part to be solved finally. In fact there are for facets of kadify the following 7d boundary polytopes:
• 630 oca (7d simplices, itself regular)
• 1260 octepe (octahedron-pyramid-pyramid-pyramid-pyramids = pyramid product of octahedron and tetrahedron, obviously biform)
• 72 rilalril (7d convex segmentotope of 2 mutually inverted rectified 6d simplices, happening to be scaliform polytopes)
• 18 roc (7d rectified simplices, itself uniform)

The single type of vertex figure - at any of the 168 vertices of kadify - provides:
• 30 incidences of oca
• 75=30+45 incidences of octepe (splitted into both vertex types of those)
• 18 incidences of rilalril
• 3 incidences of roc

--- rk
Klitzing
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Re: Wythoff Polytopes

In the sequel of the above discussions Wendy came up with an other interesting representation of fy = 421. (Just today I learned that this one was already known to Jonathan (aka PolyhedronDude) for some years, so I don't know to whom that one should be attributed so.) This one uses the subsymmetry of D4 x D4 of E8. In fact we can write fy = 421 as the hull of the compound of
1. ico x {}
2. {} x ico
3. hex x hex
4. +gyro hex x +gyro hex
5. -gyro hex x -gyro hex
From this representation we further learn not only the subdimensional inscription of ico into fy (cf. a. or b.), but we also have the full-dimensional vertex-inscription of the uniform duoprism of hex, i.e. hexdip (cf. c. or d. or e.).

But what is even more interesting wrt. the scaliform fy-diminishing kadify of the above posts, we get 2 more such scaliform polyzetta out of this representation! Those are (acronyms already being provided by Jonathan in a private mail):
1. gecdify = hull of compound of d. + e., i.e. the (64+48)-diminished fy with 128 vertices
2. codify = hull of compound of c. + d. + e., i.e. the 48-diminished fy with 192 vertices
Within those all vertices obviously are equivalent, therefore the vertex transitivity requirement is met. The unit edge only requirement is deduced from their fy derivation. And as those are members of spherical geometry, the final requirement for finiteness of elements is trivially met. Thus those truely are 2 new scaliform polyzetta.

The further details on those, esp. their incidence matrices, still pend to be worked out so. - Just wanted to inform you in the meantime. --- rk
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Re: Wythoff Polytopes

fy = 4_21 has rather a lot of interesting compounds. The one that Richard mentions is more fun, in that any two of c,d,e make a half-8cube, and the remander + a,b make the rectifed 8-cross.

Another compound worth looking at is xo3oo3oo3xo . ox3ox3ox3ox combines with four pentachora - rectified pentacora to make fy. That's from A4×A4 qua 5A4A4.
The dream you dream alone is only a dream
the dream we dream together is reality.

\(Latex\) at https://greasyfork.org/en/users/188714-wendy-krieger wendy
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Re: Wythoff Polytopes

Interesting stuff. Sure hocto (hemi-octeract) and gecdify (64+48-diminished fy, where fy = 4_21) both have 128 vertices, and it was more the quest to investigate that all the elements too could be recognized under that symmetry representation again. - Now finally I succeeded. And in the run found quite a lot more of such tegum sum representations of all these involved hemihypercubes!

• Code: Select all
`hex = x3o3o *b3o`
is the tegum sum of 2 concentrical, orthogonal squares. Thus it well is representable also as
Code: Select all
`hex = xo ox xo ox&#zx`
• Code: Select all
`hin = x3o3o *b3o3o`
likewise is representable by the tegum sum of 2 concentrical, mutually dual and laceing-edge orthogonal tepes as
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`hin = xo3oo3ox xo ox&#zx`
• Code: Select all
`hax = x3o3o *b3o3o3o`
likewise is representable either as the tegum sum of 2 concentrical, mutually gyrated and laceing-edge orthogonal hexips as
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`hax = xo3oo3ox *b3oo xo ox&#zx`
or alternatively as the tegum sum of 2 concentrical, mutually bidual tetdips as
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`hax = xo3oo3ox xo3oo3ox&#zx`
• Code: Select all
`hesa = x3o3o *b3o3o3o3o`
likewise is at least representable as the tegum sum of 2 concentrical, mutually gyro-dual tethexes as
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`hesa = xo3oo3ox *b3oo xo3oo3ox&#zx`
and finally, as the starting point of this investigation,
• Code: Select all
`hocto = x3o3o *b3o3o3o3o3o`
is the tegum sum of 2 concentrical, mutually bigyrated hexdips as
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`hocto = xo3oo3ox *b3oo xo3oo3ox *f3oo&#zx = gecdify`

You see the general pattern here? It's cool, ain't it? Esp. any single "layer" each, i.e. any single component of the underlaying compound of these convex hulls, kind of is some further hemiation of those hemihypercubes and therefore - in that very sense - could be called a quarterhypercube, hehe. But be aware, from the 2 given alternate according representations of hax, and the probably existing further more alike representations beyond that dimensionality, this process indeed is some quateration, but it clearly does not end with a single unique result! (E.g. hesa then too ought be representable as the tegum sum of 2 concentrical, mutually gyrated and lacing-edge orthogonal hinnips, I'd suppose. Etc.)

--- rk
Last edited by Klitzing on Fri Nov 09, 2018 4:22 pm, edited 1 time in total.
Klitzing
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Re: Wythoff Polytopes

wendy wrote:... and the remainder [= c] + a,b make the rectifed 8-cross. ...

Indeed, I just finished the according incmats of rek (= o3x3o3o3o3o3o4o) = oxo3xoo3ooo4ooo oxo3oox3ooo4ooo&#zx. (Will be contained within my next upload.)
--- rk
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Re: Wythoff Polytopes

Klitzing wrote:...
1. ico x {}
2. {} x ico
3. hex x hex
4. +gyro hex x +gyro hex
5. -gyro hex x -gyro hex

1. gecdify = hull of compound of d. + e., i.e. the (64+48)-diminished fy with 128 vertices
2. codify = hull of compound of c. + d. + e., i.e. the 48-diminished fy with 192 vertices
...
--- rk

So far we've seen that gecdify is not new at all, it simply is nothing but the uniform hocto = x3o3o *b3o3o3o3o3o.

Today I tried to get a closer look on the remaining new scaliform codify.
And I found that it can be described as a lace tower too:

For simplicity let's use Wendy's notation for a moment: N = hex, E = +gyro hex, O = -gyro hex.

Then it is obvious, as we are looking for NxN + ExE + OxO, that we could depict the hull of that compound within 8D=4D+4D, taking the first 4D subspace as representation space for a 4D lace hyper city and the second 4D subspace for the to be placed objects. But as the hull of the compound of N+E+O is nothing but a 1/q-ico, we could in turn represent that one as point || 1/q-cube || oct || 1/q-cube || point, or, more precise in this setup, as N-point || compound of E-tet + dual O-tet || N-oct || compound of O-tet + dual E-tet || N-point.

Thus putting this all together we get
Code: Select all
`codify (= xoo3ooo3oxo *b3oox xoo3ooo3oxo *f3oox&#zx)= o(xo)o(xo)o3o(oo)x(oo)o3o(ox)o(ox)o o(xo)o(ox)o3o(oo)o(oo)o3o(ox)o(xo)o *e3x(oo)x(oo)x&#xt`

The first, defining representation as a tegum sum clearly shows its scaliformity. And the according incidence matrix therefore could be much more compact.

The second just figured out representation as a lace tower on the other hand is, even so looking more clumsy and yielding a much larger incidence matrix, a much easier to evaluate representation, simply because we just have to consider the various segmentotopes one by one and then just have to investigate whether 2 consecutive elements would be coplanar or not. Thus the set of to be used subelements here is much more obvious than in the other representation.

For the derivation of the according incidence matrix in the first representation this set of to be used subelements in fact is to be known in advance. else you would possibly consider pseudo elements as well. Thus hopefully this new representation will provide the required totals of subelements within the days to come.

--- rk
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Re: Wythoff Polytopes

I'm still in the run of trying to understand that 48-diminished version of the 8D Gosset polytope 421, aka codify (where c=48, d=diminished, fy=421).

From what the construction was defined, and esp. in my last post was described in perhaps a bit too sparse words, we can provide a 4D lace hypercity of that 8D beasty which furthermore decomposes into an axial stack of the following 3D lace hypercity series:
Code: Select all
`                                                                                                        N                                                                                                        `
atop
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`                                          _E---------_O    O----------E  |    |  |       |  |    |  |       |  |    |  |       |  |    | _O-------|-_E    E----------O                                           `
atop
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`         N                /  \\            /  |  \ \        /    _ - N_ \    / _ - |    |  -_\ N _           _ - N \  -_|  _ +     /    \  N        /        \ \  |  /            \\  /                N         `
atop
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`                                          _O---------_E    E----------O  |    |  |       |  |    |  |       |  |    |  |       |  |    | _E-------|-_O    O----------E                                           `
atop
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`                                                                                                        N                                                                                                        `
where:
Code: Select all
`N = o3o3o *b3x (hex)E = x3o3o *b3o (gyro hex)O = o3o3x *b3o (alt. gyro hex)`

I then already mentioned that codify might be better understood when considering the topmost segmentotope, i.e. its hex-first cap. (This is because codify already is known to be scaliform, esp. all vertices would be equivalent.) So, within the 4D position subspace that one will be nothing but a 1/sqrt(2) down scaled version of cubpy. But we have the 4D perp space too, where all those mutually gyrated hexes would live. Thus this cap happens to be the 8D polytope hexalhesa = hex||(alt.) hesa (i.e. a segmentotope of a 4D hemihypercube atop a 7D hemihypercube). The "alternation" here refers to the fact that the top hex is of type N, while the hexes within the cubical base (of position subspace) are E or O. - That segmentotope meanwhile is being understood.

Sadly it occurs that hexalhesa contains haxpies for facets, i.e. the pyramids with hax as their bases, and hax in turn is the 6D hemihypercube. But those haxpies cannot be facets of codify! This is best understood in the following lace city display of hexalhesa:
Code: Select all
`o---O---o     = hex (layer of N-point) \     /   h---H       = hesa (layer of E/O-cube)   \ /       .         (where the layer of N-oct would be)`
where:
Code: Select all
`o = point/vertexO = octahedronh = haxH = alternated hax. = center of codify`

That is, the equatorial layer of codify, the N-oct, well would reach out in that displayed direction again, so that those haxes (h, H), and therefore those lacing haxpies (o-h, o-H) too, would get burried somewhere underneath.

None the less I've made a further step here already. I've managed to find - as well as to evaluate - that facet at least, which covers that stuff underneath. That one will be odinaq. - What the hell is that again, you might ask? Okay, here it comes.

You might be acquainted with the 7D Gosset polytope 321=naq. That one allows for a lace hypercity display as
Code: Select all
`     _p-------------p   /      N     _/  |p-----+-------p     ||            O|     ||  E  |       |  E  ||      O      |     ||     p - - - + - - p|  /      N   | _/   p-------------p      `
where:
Code: Select all
`p = o3o3o *b3o (point)N = x3o3o *b3o (hex)E = o3o3x *b3o (gyro hex)O = o3o3o *b3x (alt. gyro hex)`

Thus you might ask, what the diminishing thereof would "look like", when those 8 p-vertices would be chopped off. That is, we are interested in the following lace hypercity instead:
Code: Select all
`        N             //|\         / / |  \     / _O--+----_E   - x3o3o *b3o3o  (hin) /_-  |  | _- /E-----+--O  /      - o3o3x *b3o3o  (alt. hin)   \  | / /         \|//             N       `
This then is "odinaq" (where o=8(=octo), d=diminished, naq=321). As is readily be "seen" from this lace hypercity, all vertices of odinaq would be equivalent. And it is all unit edged. Thus it is a new 7D scaliform polytope itself! Some of its 6D faces are already directly viewable. In fact we have 8 copies of the lace city
Code: Select all
` N E O`
throughout as triangular covers of the octahedron of position space. Those in turn are already known to be tedjak, the tridiminished versions of the 6D Gosset polytope 212=jak, cf. the following lace city display of jak:
Code: Select all
`p           where:    E       N = x3o3o *b3o ("normal" hex)N       p   E = o3o3x *b3o ("even" hex)    O       O = o3o3o *b3x ("odd" hex)p           p = o3o3o *b3o (point)`

Tedjak is a scaliform polytope as well, and already is known to have 3 hins (5D hemihypercubes) + 24 hexpies (pyramids on 4D hemihypercubes = pyramids on 4D crosspolytopes) + 24 hixes (5D simplices) for facets. The facet total of odinaq now has been evaluated to 8 tedjaks (as already mentioned) + 96 hexascs (quarters of 6d crosspolytopes = pyramid of pyramid of 4D crosspolytope = scalene of 4D hemihypercube) + 24 gees (6D crosspolytopes) + 192 hops (6D simplices).

So, stay tuned for the contiuation of this story … --- rk
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Re: Wythoff Polytopes

Ha! Finally solved it. Yes, indeed, codify truely uses the above described odinaqs for facets.
In fact that scaliform 48-diminished 421 has for its facet total:
• 384 ocas (7D simplices)
• 48 odinaqs (8-diminished 321s)
• 1536 hexetes (16-cell tettenes = hex-pyr-pyr-pyrs)
(In my next website update you'll see its full incmats too.)

--- rk
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