## Tegum polytopes

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

### Re: Tegum polytopes

To imagine the groups here.

Take an octahedron from x3o4o3o. This is divided by three mirrors in the planes perpendicular to the diagonals. So you divide the octahedron into 8 triangle-pyramids. These are the cells of the symmetry o3o3o o3*b. There is a 60-60-60 triangle, and three 90-45-45 ones. We take the 60-60-60 triangle, and can apply different symmetries. The order is 8 * 24 = 192.

Dividing the triangle, so A = B gives the symmetry [3,3,4] or the tesseract symmetry. This can be done in three different ways, giving the compound of three tesseract in x3o4o3o. The order is 2 * 192 = 384.

The second symmetry of the triangle is to rotate A to B to C to A. This converts the octahedra into pyritohedral symmetry 3*2. The relevant symmetry is 3/*/2, that is, a rotated triangle with three legs, like the manx thing. This is the symmetry of the s3s4o3o. The o3o bit are the preserved mirrors, the s3s bit are the rotational-only bit. This is [3,4,3+].

The third symmetry is full permutations of A,B,C, which leads to a subgroup of 6, or a group of 1152. This is the full ico symmetry [3,4,3] by Coxeter.
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wendy
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### Re: Tegum polytopes

Okay then, the hex symmetry is [31,1,1].
The 3-hex compound has hull = ico and thus has full symmetry [3,4,3]. That is, it ought cycle AND reflect.
The ex has symmetry [3,3,5].
But the common subsymmetry of [3,3,5] and [3,4,3] is NOT [3,4,3] itself, even so a f-scaled ico is vertex inscribable into ex, it rather is [3,4,3]+ only.
Therefore it IS clear that sadi = ico-dim. ex has a symmetry which just cyclically changes the legs of [31,1,1].

This is like in 3D too: [3,5] and [3,4] have for common subsymmetry [3,4]+, the pyritohedral one, even so a cube is vertex-inscribable into doe.

But now onto the other part of my mail:
The according hi representation within [31,1,1] symmetry still seems to be open.

--- rk
Klitzing
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### Re: Tegum polytopes

Klitzing wrote:[...]But now onto the other part of my mail:
The according hi representation within [31,1,1] symmetry still seems to be open.

--- rk

That was the main point of this post (click on the [31,1,1]-link, then look at number 14, in binary 14 is 1110, which translates to o3o3o5x)
student91
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### Re: Tegum polytopes

Ah yes, forgot about that one.
Thus you tell:
hi = Coo|foB|xoF|f-3-ooo|ooo|fff|x-3-oCo|oBf|oFx|f  *b3-ooC|Bfo|Fxo|f-&#zx
where C = 2f+2x, B = 2f+x, F = f+x, and f : x=1.618 : 1.

Great, thanx!
--- rk
Klitzing
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### Re: Tegum polytopes

Fiddled it out, finally!

This is, how quidex could be designed directly bottom-up as tegum sum, rather than the top-down description so far known only (eg. as dual of sadi, as quatri-ico-diminished ex, as ico-stellated hi):
Code: Select all
`ooV|xoF|o-3-ooo|ooo|F-3-Voo|Fxo|o *b3-oVo|oFx|o-&#z(v,f,F)   → existing heights=0                                                               V = 2f = 3.236067977                                                               F = ff = 2.618033989                                                               v = 1/f = 0.618033989                                                               f = 2 cos(36°) = 1.618033989o.. ... .-3-o.. ... .-3-o.. ... . *b3-o.. ... .            | 8 * *  *  *  *  * |  4  0  0  0  0  0  0 |  6  0  0  0  0  0 |  4  0  0.o. ... .-3-.o. ... .-3-.o. ... . *b3-.o. ... .            | * 8 *  *  *  *  * |  0  4  0  0  0  0  0 |  0  6  0  0  0  0 |  0  4  0..o ... .-3-..o ... .-3-..o ... . *b3-..o ... .            | * * 8  *  *  *  * |  0  0  4  0  0  0  0 |  0  0  6  0  0  0 |  0  0  4... o.. .-3-... o.. .-3-... o.. . *b3-... o.. .            | * * * 32  *  *  * |  1  0  0  3  0  0  0 |  3  0  0  3  0  0 |  3  0  1... .o. .-3-... .o. .-3-... .o. . *b3-... .o. .            | * * *  * 32  *  * |  0  1  0  0  3  0  0 |  0  3  0  0  3  0 |  1  3  0... ..o .-3-... ..o .-3-... ..o . *b3-... ..o .            | * * *  *  * 32  * |  0  0  1  0  0  3  0 |  0  0  3  0  0  3 |  0  1  3... ... o-3-... ... o-3-... ... o *b3-... ... o            | * * *  *  *  * 24 |  0  0  0  4  4  4  8 |  2  2  2  8  8  8 |  4  4  4-----------------------------------------------------------+-------------------+----------------------+-------------------+---------o.. o.. .-3-o.. o.. .-3-o.. o.. . *b3-o.. o.. .-&#v        | 1 0 0  1  0  0  0 | 32  *  *  *  *  *  * |  3  0  0  0  0  0 |  3  0  0  v-edges.o. .o. .-3-.o. .o. .-3-.o. .o. . *b3-.o. .o. .-&#v        | 0 1 0  0  1  0  0 |  * 32  *  *  *  *  * |  0  3  0  0  0  0 |  0  3  0  v-edges..o ..o .-3-..o ..o .-3-..o ..o . *b3-..o ..o .-&#v        | 0 0 1  0  0  1  0 |  *  * 32  *  *  *  * |  0  0  3  0  0  0 |  0  0  3  v-edges... o.. o-3-... o.. o-3-... o.. o *b3-... o.. o-&#f        | 0 0 0  1  0  0  1 |  *  *  * 96  *  *  * |  1  0  0  2  0  0 |  2  0  1  f-edges... .o. o-3-... .o. o-3-... .o. o *b3-... .o. o-&#f        | 0 0 0  0  1  0  1 |  *  *  *  * 96  *  * |  0  1  0  0  2  0 |  1  2  0  f-edges... ..o o-3-... ..o o-3-... ..o o *b3-... ..o o-&#f        | 0 0 0  0  0  1  1 |  *  *  *  *  * 96  * |  0  0  1  0  0  2 |  0  1  2  f-edges... ... .   ... ... F   ... ... .     ... ... .            | 0 0 0  0  0  0  2 |  *  *  *  *  *  * 96 |  0  0  0  1  1  1 |  1  1  1  F-edges-----------------------------------------------------------+-------------------+----------------------+-------------------+---------o.. x.. o   ... ... .   ... ... .     ... ... .-&#(v,f)t   | 1 0 0  2  0  0  1 |  2  0  0  2  0  0  0 | 48  *  *  *  *  * |  2  0  0  kite... ... .   ... ... .   .o. .x. o     ... ... .-&#(v,f)t   | 0 1 0  0  2  0  1 |  0  2  0  0  2  0  0 |  * 48  *  *  *  * |  0  2  0  kite... ... .   ... ... .   ... ... .     ..o ..x o-&#(v,f)t   | 0 0 1  0  0  2  1 |  0  0  2  0  0  2  0 |  *  * 48  *  *  * |  0  0  2  kite... ... .   ... o.. F   ... ... .     ... ... .-&#f        | 0 0 0  1  0  0  2 |  0  0  0  2  0  0  1 |  *  *  * 96  *  * |  1  0  1  golden triangle... ... .   ... .o. F   ... ... .     ... ... .-&#f        | 0 0 0  0  1  0  2 |  0  0  0  0  2  0  1 |  *  *  *  * 96  * |  1  1  0  golden triangle... ... .   ... ..o F   ... ... .     ... ... .-&#f        | 0 0 0  0  0  1  2 |  0  0  0  0  0  2  1 |  *  *  *  *  * 96 |  0  1  1  golden triangle-----------------------------------------------------------+-------------------+----------------------+-------------------+---------o.. xo. o-3-o.. oo. F   ... ... .     ... ... .-&#(v,f,f)t | 1 0 0  3  1  0  3 |  3  0  0  6  3  0  3 |  3  0  0  3  3  0 | 32  *  *  tower a-d-g-e... ... .   .o. .oo F-3-.o. .xo o     ... ... .-&#(v,f,f)t | 0 1 0  0  3  1  3 |  0  3  0  0  6  3  3 |  0  3  0  0  3  3 |  * 32  *  tower b-e-g-f... ... .   ..o o.o F   ... ... . *b3-..o o.x o-&#(v,f,f)t | 0 0 1  1  0  3  3 |  0  0  3  3  0  6  3 |  0  0  3  3  0  3 |  *  * 32  tower c-f-g-d`

Obviously this incidence matrix shows up the same cyclical symmetry of the legs of this symmetry representation as in the according representation sadi does. That is, both are [3,4,3]+ symmetrical figures, which is the common subgroup of [3,3,5] and [3,4,3].

The first 6 vertex types display tetrahedral vertex figures, the 7th then is a dodecehardal one. The first 3 face types are a kite, which happens to be nothing but a mono-stellated regular pentagon. The other 3 then are a large obtuse golden triangle, which happens to be a bistellated regular pentagon. The cells then are all tri-trigonally diminished icosahedra, or, conversely described, a mono-trigonally stellated dodecahedron. A picture of that cell looks like this:

Quidex is the one but last member of that multi-ico-diminishing of ex:
0-idex = ex,
1-idex = sadi,
2-idex = bidex (noble, cells being teddies),
3-idex = tridex,
4-idex = quidex,
5-idex = hi.
Moreover we have: k-idex is dual to (5-k)-idex ! (More details on that can be found here (scroll a bit down).

--- rk
Klitzing
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