Johnsonian Polytopes

Discussion of known convex regular-faced polytopes, including the Johnson solids in 3D, and higher dimensions; and the discovery of new ones.

Re: Johnsonian Polytopes

Postby Klitzing » Tue Mar 26, 2013 3:34 pm

Not so clear on what you after, quickfur - except of potential applications onto duoprsmatic symmetries, which I indeed so far not have looked at.

If you'd were after thickening that toroidal layer of squares, i.e. pulling apart those 2 cycles of prisms, I'd think that this cannot work for some reason, might be asking the cubes to be non-CRF-ly distorted. Else you should be able to do that esp. for any n,n-duoprism as well (both prism types identical). But then you should re-enter the realm of uniforms: all used elements should be uniform themselves, and all vertices would be equivalent by symmtery. But such a n,n-dip with an intermediate layer of n times n cubes is not a known one... - or am I wrong?

In euclidean space that one clearly exists: a 2D grid of cubes, running into either infinity, on the one side glued by infinite stripes (representing infinity-prisms) in N-S direction, while on the other glued by similar stripes in E-W direction.

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Re: Johnsonian Polytopes

Postby Klitzing » Tue Mar 26, 2013 3:39 pm

Yesterday, I came up with 2 new CRFs.
Here are their incidence matrices:

A = poxic (partially octa-expanded ico) = phic srico (partially hexadeca-contracted srico)

Code: Select all
8  * |  8  0  0 | 12  0  0 |  6  0  0  cube vf.
* 64 |  1  3  3 |  3  6  3 |  3  3  1
-----+----------+----------+---------
1  1 | 64  *  * |  3  0  0 |  3  0  0
0  2 |  * 96  * |  1  2  0 |  2  1  0  ortho
0  2 |  *  * 96 |  0  2  2 |  1  2  1  para
-----+----------+----------+---------
1  2 |  2  1  0 | 96  *  * |  2  0  0
0  4 |  0  2  2 |  * 96  * |  1  1  0
0  3 |  0  0  3 |  *  * 64 |  0  1  1
-----+----------+----------+---------
2  8 |  8  8  4 |  8  4  0 | 24  *  *  esquidpy
0  6 |  0  3  6 |  0  3  2 |  * 32  *  trip
0  4 |  0  0  6 |  0  0  4 |  *  * 16  tet


and
B = phixic (partially hexadeca-expanded ico) = pocsric (partially octa-contracted srico)

Code: Select all
64  * |  3  1   3   0 |  3   6  3  0  0 |  1  3  3 0  tet vertices
 * 96 |  0  0   2   4 |  0   4  1  2  2 |  0  2  2 1  co vertices
------+---------------+-----------------+-----------
 2  0 | 96  *   *   * |  2   2  0  0  0 |  1  1  2 0  edges of tet
 2  0 |  * 32   *   * |  0   0  3  0  0 |  0  3  0 0  tet connectings
 1  1 |  *  * 192   * |  0   2  1  0  0 |  0  2  1 0
 0  2 |  *  *   * 192 |  0   1  0  1  1 |  0  1  1 1
------+---------------+-----------------+-----------
 3  0 |  3  0   0   0 | 64   *  *  *  * |  1  0  1 0
 2  2 |  1  0   2   1 |  * 192  *  *  * |  0  1  1 0
 2  1 |  0  1   2   0 |  *   * 96  *  * |  0  2  0 0
 0  4 |  0  0   0   4 |  *   *  * 48  * |  0  1  0 1
 0  3 |  0  0   0   3 |  *   *  *  * 64 |  0  0  1 1
------+---------------+-----------------+-----------
 4  0 |  6  0   0   0 |  4   0  0  0  0 | 16  *  * *  tet
 8  8 |  4  4  16   8 |  0   8  8  2  0 |  * 24  * *  squobcu
 3  3 |  3  0   3   3 |  1   3  0  0  1 |  *  * 64 *  trip
 0 12 |  0  0   0  24 |  0   0  0  6  8 |  *  *  * 8  co


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Re: Johnsonian Polytopes

Postby Klitzing » Thu Mar 28, 2013 2:38 pm

poxic, btw., can be considered a sidpith with the 8 cubes, which have full symmetrical surroundings, being augmented by cubepys: It is that the squippys thereof then become corealmic to the remaining cubes (which have prismatic surroundings only), i.e. those thus blend into those esquidpys, which were recently mentioned in the incidence matrix.

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Re: Johnsonian Polytopes

Postby Marek14 » Sat Mar 30, 2013 11:19 am

In past months, several interesting polytopes were found, but we might be no closer to our ultimate goal of enumeration. So, just to remind, here is my current best guess for algorithm:

1) Enumerate possible CRF vertices, i.e. polyhedra whose whose all vertices lie on a unit hypersphere and whose faces are verfs of Johnson solids/uniforms/archimedeans. Since the polygons can be skew, the polyhedra might be skew as well and have to be searched in full 4 dimensions.

2) Connect the vertices into a network. Each kind of vertex has only limited amount of other vertices that can be attached to it in every orientation.
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Re: Johnsonian Polytopes

Postby Klitzing » Sat Mar 30, 2013 2:24 pm

Ad 1)
Considering CRF polychora the verfs in fact might be taken orbiform. But then its faces nedn't be planar any longer.
Truely speaking, verfs generally are the intersection of the vertex surroundings with a small sphere. Therefore vertex figure polytopes are per se spherical tesselations.

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Re: Johnsonian Polytopes

Postby Marek14 » Sat Mar 30, 2013 4:00 pm

Klitzing wrote:Ad 1)
Considering CRF polychora the verfs in fact might be taken orbiform. But then its faces nedn't be planar any longer.
Truely speaking, verfs generally are the intersection of the vertex surroundings with a small sphere. Therefore vertex figure polytopes are per se spherical tesselations.

--- rk


In this case, though, I'd prefer a bit different definition: intersection with unit hypersphere. Since the polytope is CRF, the unit hypersphere centred at a vertex will, by definition, contain all neighbouring vertices.
Then, instead of checking the edge lengths and comparing them to chords, it might be arguably easier to check angles - a triangle means that two edges from central vertex have angle of 60 degrees, square means 90 degrees, and so on. And if, say, four triangles meet at one vertex and form a quadrangle on the vertex figure, the angles corresponding to diagonals can be checked as well -- two angles of 90 degrees signify a square verf corresponding to octahedron, square pyramid, or similar, while trigonal bipyramid, for example, would have different diagonal angles.
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Re: Johnsonian Polytopes

Postby Klitzing » Sat Mar 30, 2013 4:19 pm

Marek14 wrote:
Klitzing wrote:Ad 1)
Considering CRF polychora the verfs in fact might be taken orbiform. But then its faces nedn't be planar any longer.
Truely speaking, verfs generally are the intersection of the vertex surroundings with a small sphere. Therefore vertex figure polytopes are per se spherical tesselations.

--- rk


In this case, though, I'd prefer a bit different definition: intersection with unit hypersphere. Since the polytope is CRF, the unit hypersphere centred at a vertex will, by definition, contain all neighbouring vertices.
Then, instead of checking the edge lengths and comparing them to chords, it might be arguably easier to check angles - a triangle means that two edges from central vertex have angle of 60 degrees, square means 90 degrees, and so on. And if, say, four triangles meet at one vertex and form a quadrangle on the vertex figure, the angles corresponding to diagonals can be checked as well -- two angles of 90 degrees signify a square verf corresponding to octahedron, square pyramid, or similar, while trigonal bipyramid, for example, would have different diagonal angles.


different?
you just use the implicite "hyper" prefix explicitely. And the size (radius) does not matter, as long it is lower or equal to unity (edge length).
And sure "tesselation" likewise was meant to be the intersection of those vertex incident elements with the surface of that "sphere".

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Re: Johnsonian Polytopes

Postby Marek14 » Sat Mar 30, 2013 6:29 pm

Klitzing wrote:
Marek14 wrote:
Klitzing wrote:Ad 1)
Considering CRF polychora the verfs in fact might be taken orbiform. But then its faces nedn't be planar any longer.
Truely speaking, verfs generally are the intersection of the vertex surroundings with a small sphere. Therefore vertex figure polytopes are per se spherical tesselations.

--- rk


In this case, though, I'd prefer a bit different definition: intersection with unit hypersphere. Since the polytope is CRF, the unit hypersphere centred at a vertex will, by definition, contain all neighbouring vertices.
Then, instead of checking the edge lengths and comparing them to chords, it might be arguably easier to check angles - a triangle means that two edges from central vertex have angle of 60 degrees, square means 90 degrees, and so on. And if, say, four triangles meet at one vertex and form a quadrangle on the vertex figure, the angles corresponding to diagonals can be checked as well -- two angles of 90 degrees signify a square verf corresponding to octahedron, square pyramid, or similar, while trigonal bipyramid, for example, would have different diagonal angles.


different?
you just use the implicite "hyper" prefix explicitely. And the size (radius) does not matter, as long it is lower or equal to unity (edge length).
And sure "tesselation" likewise was meant to be the intersection of those vertex incident elements with the surface of that "sphere".

--- rk


What I meant was that for uniform polytopes, vertex figure can be inscribed in a lower-dimensional sphere, which is not true in general case.
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Apr 02, 2013 3:00 pm

Just for record, we could apply those partial Stott expansions (resp. partial Stott contractions) clearly to non-spherical geometries as well. Just as the Sott expansion (respectively Stott contraction) itself did apply there too.

We could consider 3D euclidean space kind as a huge spherical polychoron with infinte circumradius, such that the local curvature becomes flat. Thus I'll describe new CRF honeycombs in that sense in this thread too, if you'd allow.

Consider the symmetry o4o3o4o, i.e. the full symmetry of the cubical lattice.
Sure, we could well consider the brick subsymmetry as well, i.e o-infin-o o-infin-o o-infin-o.
Obviously that latter one allows for a (classical) Stott expansion in any of the orthogonal direction separately.

Ie. we have
Code: Select all
o-infin-o  o-infin-o  o-infin-o  = point
o-infin-o  o-infin-o  o-infin-x  = aze (integer line)
o-infin-o  o-infin-x  o-infin-x  = squat (integer plane, square grid)
o-infin-x  o-infin-x  o-infin-x  = chon (cubical honeycomb)

But in view of the full symmetry of that cubical honeycomb we just have
Code: Select all
o4o3o4o  = point
? : aze
? : squat
o4o3o4x  = chon

That is, with respect to this higher symmetry, those intermediate steps would be partial Stott expansions.

Similarily we could apply that sequence onto chon itself (all described Dynkin symbols would equate to chon again, just being seen as different colorings). Either as classical one again:
Code: Select all
x-infin-o  x-infin-o  x-infin-o
x-infin-o  x-infin-o  x-infin-x
x-infin-o  x-infin-x  x-infin-x
x-infin-x  x-infin-x  x-infin-x

Or as partial one:
Code: Select all
x4o3o4o
?
?
x4o3o4x


Thus so far we have not described anything new. We just saw that this alternating elongation in the differnet directions independently can be described in closed form in symmetry o-infin-o o-infin-o o-infin-o, but not in closed form within o4o3o4o. In the latter all 3 alternating elongations have to be applied at once, if one wants to have a closed description as classical Stott expansion.

We further know that the cubical honeycomb has several Wythoffian relatives (according to its full symmetry). Whereas that lower symmetry here already would be exhausted. This then is were new things will occur!

Consider to start with
Code: Select all
o4x3o4o  = rich
? : pexrich (partially expanded rich)
? : pacsrich (partially contracted srich)
o4x3o4x  = srich

Here we get as cells frequencies (N being assumed to run to infinity at equal speed)
Code: Select all
  rich  pexrich   pacsrich  srich
N oct   esquidpy  squobcu   sirco
N co    co        co        co
N {4}   {4}       {4}       cube
N {4}   {4}       cube      cube
N {4}   cube      cube      cube

(Count of subdimensional elements here and thereafter are only provided if to be continued into full dimensional ones. There would be other ones additionally for sure.)

Or consider to start with
Code: Select all
x4x3o4o  = tich
? : pextich (partially expanded tich)
? : pacprich (partially contracted prich)
x4x3o4x  = prich

Here we get as cells frequencies (N being assumed to run to infinity at equal speed)
Code: Select all
  tich  pextich   pacprich  prich
N oct   esquidpy  squobcu   sirco
N tic   tic       tic       tic
N {8}   {8}       {8}       op
N {8}   {8}       op        op
N {8}   op        op        op
N edge  edge      {4}       cube
N edge  {4}       {4}       cube
N edge  {4}       cube      cube


We even could consider alternated facetings here!
That is, let's start with
Code: Select all
s4o3o4o  = octet ( = o3o3x *4o )
? : pextoh (partially extended tetrahedral-octahedral honeycomb)
? : pacratoh (partially contracted ratoh)
s4o3o4x  = ratoh ( = o3o3x *b4x )

Here we get as cells frequencies (N being assumed to run to infinity at equal speed)
Code: Select all
  octet pextoh    pacratoh  ratoh
N tet   tet       tet       tet
N tet   tet       tet       tet
N oct   esquidpy  squobcu   sirco
N point edge      {4}       cube

The interesting in these 2 last new CRF honeycombs (i.e. pextoh and pacratoh) is, that those indeed use Johnson solids for cells, but still have a single vertex class only. That is, those are scaliform figures!

Finally you could ask about starting with s4x3o4o. But this is not anything new at all. This is due to the below displayed equivalences (which in turn are due to this special underlaying symmetry only):
Code: Select all
s4x3o4o = o4x3o4o
?
?
s4x3o4x = o4x3o4x

that is, those intermediate figures were already described above too.

Concluding: we thus have found 6 new CRF honeycombs, 2 of which are even scaliform (while the other 4 would need more than 1 vertex class).

--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Mon Apr 08, 2013 1:50 pm

Have now elaborated the so far omitted intermediate cases of those 4D CRF partial Stott expansion series which were designed with respect to mutually orthogonal axial subsymmetries:

series A)
o3o3o4o = point
edge
square
cube
o3o3o4x = tes

series B)
x3o3o4o = hex
pex hex (NEW: partially (mono-)expanded hex)
quawros
pacsid pith (NEW: partially (mono-)contracted sidpith)
x3o3o4x = sidpith

series C)
o3x3o4o = ico
pexic (NEW: partially (mono-)expanded ico)
bicyte ausodip
pacsrit (NEW: partially (mono-)contracted srit)
o3x3o4x = srit

series D)
x3x3o4o = thex
pex thex (NEW: partially (mono-)expanded thex)
pabex thex
pacprit (NEW: partially (mono-)contracted prit)
x3x3o4x = prit

Series A thereby is clearly a partial one with respect to the full o3o3o4o symmetry. But it can be given in closed form as a classical one with respect to the subsymmetry o2o2o2o for sure.

Now to Dynkin representations (lace towers) and the cell counts of the new fellows:

pex hex = oxxo3oooo4oooo&#xt with 16 tet + 8 trip
pacsid pith = oxo3ooo4xxx&#xt with 2+12 cube + 16 tet + 24 trip
pexic = xoox3oxxo4oooo&#xt with 6 esquidpy + 2+16 oct + 8 trip
pacsrit = xox3oxo4xxx&#xt with 16 oct + 2 sirco + 6 squobcu + 24 trip
pex thex = xuxxux3ooxxoo4oooooo&#xt with 6 esquidpy + 8 hip + 2 oct + 16 tut
pacprit = xuxux3ooxoo4xxxxx&#xt with 12 cube + 24 hip + 2 sirco + 6 squobcu + 16 tut

Interestingly the first 2 of those do not contain any Johnson solid. Moreover they fall into clean stacks of segmentochora. - The next 2 might still be considered as "glued" stacks of segmentochora, i.e. some of the cells of those would become co-realmic and thus are considered here as the corresponding exterior blends. - For sure the existance of those tuts (respectively the occurance of the letter "u" in the Dynkin symbol) in the last 2 cases clearly makes sure to surpass that approach.

(@hedrondude: hope I stroke your ring with those new OBSAs again...)

--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Sun Apr 14, 2013 5:07 pm

Here it comes now, both
- the comprehensive walk through of so far found Partial Stott Expansions,
- and its application to tetracombs:


In spherical 3D we have:

Transition sequence according to o3o3o subsymmetry ("a" being any node symbol, i.e. "o" or "x"):
Code: Select all
   | o4o3a    patex-o4o3a  o4x3a  :: tetrahedral
---+----------------------------
its 1st type of cells (in the right case for any "a", in the left only for "a" = "x"):
 4 | . o3a    . o3a     -> . x3a  :: triangular
 4 | . o3a -> . x3a        . x3a
---+----------------------------
its 2nd type of cells (only for "a" = "x"):
12 | o . a    o . a        o . a  :: none
---+----------------------------
its 3rd type of cells (in the right case ever, in the left one never):
 6 | o4o . -> pex-o4o   -> o4x .  :: axial



In spherical 4d we have:

Transition sequence according to o3o3o *b3o subsymmetry ("a" and "b" being any node symbol, i.e. "o" or "x"):
Code: Select all
   | a3o4o3b    pox-a3o4o3b    poc-a3o4x3b    a3o4x3b  :: hexadecachoral
---+-------------------------------------------------
its 1st type of cells (in the right case for any "b", in the left only for "b" = "x"):
 8 | . o4o3b -> . o4o3b        patex-o4o3b    . o4x3b  :: tetrahedral
 8 | . o4o3b    patex-o4o3b -> patex-o4o3b    . o4x3b
 8 | . o4o3b    patex-o4o3b    . o4x3b     -> . o4x3b
---+-------------------------------------------------
its 2nd type of cells (in the right case for any "b" and "a" = "x", in the left only for "a" = "b" = "x"):
32 | a . o3b    a . o3b        a . o3b     -> a . x3b  :: triangular
32 | a . o3b    a . o3b     -> a . x3b        a . x3b
32 | a . o3b -> a . x3b        a . x3b        a . x3b
---+-------------------------------------------------
its 3rd type of cells (only for "a" = "b" = "x"):
96 | a3o . b    a3o . b        a3o . b        a3o . b  :: none
---+-------------------------------------------------
its 4th type of cells (in the right case ever, in the left only for "a" = "x"):
24 | a3o4o . -> pex-a3o4o   -> pac-a3o4x   -> a3o4x .  :: axial



In euclidean 4D we have:

Transition sequence according to o3o3o *b3o *b3o subsymmetry ("a", "b", and "c" being any node symbol, i.e. "o" or "x"):
(Here "N" is meant to run to infinity, the pre-factors show their relative frequencies.)
Code: Select all
    | a3b3o4o3c    phextex-a3b3o4o3c  pabhextex-a3b3o4o3c  phextco-a3b3o4x3c  a3b3o4x3c  :: hexadecachoric-tetracombal
----+----------------------------------------------------------------------------------
its 1st type of cells (in the right case for any "b" and "c", in the left only for "b" = "x" or "c" = "x"):
  N | . b3o4o3c -> . b3o4o3c          pox-b3o4o3c          poc-b3o4x3c        . b3o4x3c  :: hexadecachoral
  N | . b3o4o3c    pox-b3o4o3c     -> pox-b3o4o3c          poc-b3o4x3c        . b3o4x3c
  N | . b3o4o3c    pox-b3o4o3c        poc-b3o4x3c       -> poc-b3o4x3c        . b3o4x3c
  N | . b3o4o3c    pox-b3o4o3c        poc-b3o4x3c          . b3o4x3c       -> . b3o4x3c
----+----------------------------------------------------------------------------------
its 2nd type of cells (in the right case only for "a" = "x", in the left only for "a" = "c" = "x"):
 8N | a . o4o3c    a . o4o3c          a . o4o3c         -> patex-o4o3a-p   -> a . o4x3c  :: tetrahedral
 8N | a . o4o3c    a . o4o3c       -> patex-o4o3a-p        patex-o4o3a-p      a . o4x3c
 8N | a . o4o3c    a . o4o3c          patex-o4o3a-p        patex-o4o3a-p      a . o4x3c
 8N | a . o4o3c -> patex-o4o3a-p      patex-o4o3a-p     -> a . o4x3c          a . o4x3c
 8N | a . o4o3c    patex-o4o3a-p      patex-o4o3a-p        a . o4x3c          a . o4x3c
 8N | a . o4o3c    patex-o4o3a-p   -> a . o4x3c            a . o4x3c          a . o4x3c
----+----------------------------------------------------------------------------------
its 3rd type of cells (in the right case only for "a" = "x" or "b" = "x", in the left only for ("a" = "x" or "b" = "x") and "c" = "x"):
32N | a3b . o3c    a3b . o3c          a3b . o3c            a3b . o3c       -> a3b . x3c  :: triangular
32N | a3b . o3c    a3b . o3c          a3b . o3c         -> a3b . x3c          a3b . x3c
32N | a3b . o3c    a3b . o3c       -> a3b . x3c            a3b . x3c          a3b . x3c
32N | a3b . o3c -> a3b . x3c          a3b . x3c            a3b . x3c          a3b . x3c
----+----------------------------------------------------------------------------------
its 4th type of cells (only for ("a" = "x" or "b" = "x") and "c" = "x"):
96N | a3b3o . c    a3b3o . c          a3b3o . c            a3b3o . c          a3b3o . c  :: none
----+----------------------------------------------------------------------------------
its 5th type of cells (in the right case ever, in the left only for "a" = "x" or "b" = "x"):
12N | a3b3o4o . -> pex-a3b3o4o     -> pabex-a3b3o4o     -> pac-a3b3o4x     -> a3b3o4x .  :: axial


In there the following prefix terms were used:
axials:
pex = partially (mono-)expanded
pac = partially (mono-)contracted
pabex = partially bi-(mono-)expanded

tetrahedrals (o3o3o):
patex = partially tetra-expanded

hexadecachorals (o3o3o *b3o):
pox = partially octa-expanded
poc = partially octa-contracted

hexadecachoric-tetracombic (o3o3o *b3o *b3o):
phextex = partially hexadecachoric-tetracomb-expanded
pabhextex = partially bi-hexadecachoric-tetracomb-expanded
phextco = partially hexadecachoric-tetracomb-contracted

Individual cases (i.e. choosing "a", "b", "c") are meant to use those namings too, at least for alternate names (i.e. if those would have been named otherwise already before - by being constructed in a different, independent way).

Obviously the cases with "a" arbitrary and all other conditional nodes ("b" and "c") being "o" generally would result in possible rewritings in the mentioned subsymmetry too. Therefore those cases can be seen both: as partial Stott sequence (as shown) and as classical one (with respect to that subsymmetry). Any other cases ("b" and/or "c" being "x") clearly is new in this respect, i.e. can only be given as a partial one.

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Re: Johnsonian Polytopes

Postby Marek14 » Sun Apr 14, 2013 7:54 pm

Would that mean that 4-3-3-5 hyperbolic tetracomb could be partially reduced according to the branched subsymmetry as well?
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Re: Johnsonian Polytopes

Postby Klitzing » Mon Apr 15, 2013 9:46 am

Marek14 wrote:Would that mean that 4-3-3-5 hyperbolic tetracomb could be partially reduced according to the branched subsymmetry as well?


Well, sorry, I have not yet looked into that symmetry.

Clearly there has to be at least some o3o4o Dynkin subsymbol to apply. ("necessary")
But so far I don't know whether this then would be possible always. ("sufficient")

Up to now I've found that the following pairings of symmetry to subsymmetry would work. And those most probably for any number of nodes, i.e. through various dimensions and curvatures:

Code: Select all
o3o...o3o4o    with  o o ... o o o
o4o3o...o3o4o  with  o-infin-o o-infin-o ... o-infin-o o-infin-o
o3o...o3o4o3o  with  o3o3o *b3o ... *b3o


(I'm in the run to set up a new extra page on that topic for my IncMats website. So far it is not much deeper than what is spread so far here as well. And it is not uploaded so far. But I'll attach it in zipped form here - hope that works -, so that you can get a glimpse of what will be contained then. - Sure, none of the page-external links would work for that stand-alone html-file, nor the appropriate css is contained.)

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Re: Johnsonian Polytopes

Postby Klitzing » Wed Apr 17, 2013 10:17 pm

Marek14 wrote:Would that mean that 4-3-3-5 hyperbolic tetracomb could be partially reduced according to the branched subsymmetry as well?


More generally you might ask:
Would that mean that 4-3-... could be partially reduced according to the branched subsymmetry as well?


Answer:
for sure! This is because of the general identity
Code: Select all
o4y3z.. = y3z3y *b...

Thus the corresponding transition series would run:
Code: Select all
o4o3z.. = o3z3o *b.. <-> x3z3o *b.. = pex-o4o3z.. = pac-o4x3z.. <-> o4x3z.. = x3z3x *b..

(y,z being any node symbols like x,o)

So: yes, it is possible.
Just that this subsymmetry is kind of too close, so that we don't seem to come beyond those already known uniform ones...

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Re: Johnsonian Polytopes

Postby Klitzing » Fri Apr 19, 2013 1:49 pm

 
Ah, Marek14, now I got your idea.
I kind of hooked to a parallel thread in my last mail, sorry. :oops:


All the so far mentioned / found CRF ways for partial Stott expansion (resp. contraction) could be condensed into the following description:
Code: Select all
(1)...o4o...(2) <-> (several partial expansion/contraction steps) <-> (1)...o4x...(2)

were (1) is any (possibly empty) pre-fixed linear sequence from o3 or x3 (possibly being ended at the very left by o4 or x4) and (2) is either the empty post-fixed sequence or just 3o resp. 3x (in the non-empty cases the former o4 or x4 ending was not considered so far).

It further turned out that in all these cases the number of transitions in any specific sequence depends only on (1) and equals the number of the there contained -3-'s + 2.

Esp. I neither have considered (decorations of) o4o3o4o3o nor o3o3o4o3o3o so far. (And most probably won't in the next few days, 8) .)

My recent answer now considered arbitrary (2) post-fixes (which start at least by 3o or 3x) and further were subject to the (1) sequence being empty. There I showed that the above given transition number formula then holds again. (Even so, as I showed there too, we would not get anything new in those.)


Now I realised that you might have been asking about ending the (linear) (1) sequence instead in o5 or x5.
As there already has been stated too, I have not considered those so far.

Now i just had a quick investigation on possible transition steps between x5o4o to x5o4x.

It results as follows:

Whenever we pre-fix that o4o by any number of (linarily aligned) -3- links (and use no post-fix), we have an axial extension at each application. This will be applied at the equatorial vertex layer only. We can over and over apply that extension in orthogonal directions again. - Note that the transition steps clearly would no longer be spherical. Only the final one reaches that state again.

When ending this (1) with an -4- link, we enter euclidean geometry, and then could well consider the axial extension at a single "equatorial layer" (i.e. euclidean pseudo face), but usually would do so at all parallel ones simultanuously.

Now considering x5o4o. Here we again could introduce a single elongation. This too would ask for a discontinuity of curvature, as for the spherical case. We even could use a single such aze (vertex layer) as a ladder, and apply "parallel" elongations simultanuously to all those azes, which are perpendicular to that chosen one. We even could apply the same then again to a perpendicular aze (ladder). But the 5th side of those 4 central pentagons (opposite to the crossing of the ladders) now never could get reached by such a set of parallel elongations any more. Thus we could not finally reach were we aimed to: x5o4x!

(And the same type of argument surely would then apply likewise to 534 or esp. 5334 - the one of your quest.)

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Re: Johnsonian Polytopes

Postby Klitzing » Wed Apr 24, 2013 8:39 pm

Hy all Partial-Stott-Expansion enthusiasts!

Today I've got a question. A question not in the sense of a riddle, rather I'd ask for your opinion / advice.

First some remark. Consider the Partial expansion series
Code: Select all
octahedron <-> elongated square dipyramid <-> square orthobicupola <-> small rhombicuboctahedron

This clearly is just one of those partial expansion / contraction series. This time according to mutually orthogonal axial directions.
I'd like to point out, that both the beginning and the end do have a unique circumradius, while the intermediate steps do not. I'd say that this would mean, that both the starting figure and the final one have an own specific curvature throughout, if considered as a mere 2D manifold each, while the intermediates do not: those would locally change that value.

Now to my question. I'd like to apply that partial Stott expansion within hyperbolic space too. So let's start with H2. Would that be valid? could we have similar local changes of curvature? - Consider the following hyperbolic tiling for toy model:
x3o8o-4colored.png
(37.56 KiB) Not downloaded yet
i.e. x3o8o = x3o4o3*a, with one set of (alternate) triangles being consistently 4-colored (red / green / brown / black).

My aim would be to have intermediate (partial) steps when hunting for the (single-step) classical Stott expansion towards x3o4x3*a (with vertex figure [3,6,4,6]). That latter one would be done by simultanuously expanding all non-yellow triangles into hexagons, while the former vertices would blow up to new squares.

So far up to classical Stott expansion in hyperbolic space. - I'd rather would do that step by step, i.e. expanding the red triangles only into hexagons, thereby the the corresponding incident vertices (of x3o8o) would deform into new edges. I.e. the tiling would deform into a laminate of connected infinite regions with the former unchanged curvature (those containing the unchanged vertices, i.e. those with no incident red triangle) and red laminas of always bifurcating hexagons, pushing the former ones apart (and thereby locally breaking the former curvature).

My question now is: :arrow: Would that be an allowed process for hyperbolic space too?

If yes, you then could apply the same to the green triangles in a second step. Again blowing up those former vertices (of x3o8o) into new edges, which are incident to green triangles, but not to red ones. Those which are incident to red ones, but not to green ones, have been blown up similarily in the former step. Those, which are incident to red and green ones, now get blown up again, this time into new squares. - And then you could do that again for the brown ones. And finally also for the black ones. That is, we would have implemented a 4 step Partially Stott Expansion!

So, all now depends on that quite independent question: Is it allowed for hyperbolic laminates to locally break the respective curvature (at the borders of the laminas)?

Any input / insight to that?

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Re: Johnsonian Polytopes

Postby wendy » Thu Apr 25, 2013 8:00 am

If you are working with tilings, then everything falls flat, and it's pretty hard to do. In essence, you have to prove that the replacement polygons have vertices in the same circle.

If you suppose that it's a polytope, eg {3,8} as a cell of {3,8,?}, then you can pretty much remove the restriction that the vertices have to fall in the same bollosphere, and thus you can pretty much do anything. You should remember that the edges are infinite when the vertex figure is a euclidean tiling {8,8/3}, so the margin-angle of your {p,8} can not exceed 135 degrees = 360 * 3/8.

So, for example, if you're adding octahedra to the faces of {3,8}, if the margin-angle of that is much less than 45 degrees, then it can be glued onto faces without making it non-convex.
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Apr 25, 2013 9:53 am

Basically, I see it like this:

A polyhedron can be seen as a spherical tiling or as a sum of flat polygons in space -- and, of course, the curvature of that space can determine what is a valid polyhedron, two cubes stuck together is not a Johnson solid, yet it might qualify as one if it happened in hyperbolic space where the dihedral angle between equatorial square faces would be less than 180 degrees.

But: there are polyhedra that can't be easily inscribed into spheres, and partial Stott expansions belong there. You could try to inscribe them into ellipsoids, but I think that it would be actually better to inscribe them in compound shapes formed out of pieces of spheres and cylinders. Ellipsoid has constantly-varying curvature, a compound shape has constant curvature with sharp discontinuities, and I think that this describes these shapes better.

So, how does this work in hyperbolic space? Key is that once again, a hyperbolic tiling can be also seen as an infinite polyhedron. If you take {3,8} tiling and try to make a {3,7} tiling with the same triangles, you'll construct a bollohedron, as the vertices become convex. This particular bollohedron can be, of course, inscribed into a pseudosphere, but, and this is the key, noone is saying that this has to be true for EVERY bollohedron.

So, start not with {3,8} tiling, but with {3,8} bollohedron -- as you'll expand it, the new sections will have different curvature than the old ones, but why would that pose any problems if the thing was never flat to begin with? You can glue various parts of bollohedra together, even if they don't have the same curvature. But you won't be able to flatten it any more than you can make a correct spherical representation of biagumented square pyramid.
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Re: Johnsonian Polytopes

Postby Klitzing » Thu Apr 25, 2013 10:39 am

Just to describe again, what I've in mind.

I start with x3o8o = x3o4o3*a. That latter Dynkin symbol shows, that the triangles would fall into 2 alternating sets. One being displayed in yellow, 4 per vertex, and inbetween further 4 which are given in non-yellow colors: It happens that if you would color opposite non-yellow triangles in one color, while those which are orthogonal to the formers in an other one, that you just need 4 non-yellow colors to do this job consistently. - This is what I've provided in that attached picture (using red, green, brown and black for that purpose). - The vertex angle of each triangle here is clearly 360°/8 = 45°.

In the frist step expansion I'd choose those red triangles to become hexagons each, by pulling the sides (which do connect to yellow triangles) slightly apart, while those vertices (which formerly would contain 4 yellow, 2 red, and 2 ones of a different color) get elongated into new edges. Each end of those new edges thus would be a vertex containing then 2 (unchanged) yellow triangle corners, 1 (unchanged) triangle corner of that different color, and 2 red hexagon corners. Accordingly the vertex angle of those red hexagons would be 360°/8*(8-3)/2=112.5°. Further all vertices of those former red triangles were alike. Thus now all vertices of these new red hexagons are alike. Clearly also, the pulling apart could be done to any extend, providing a free variable, the size of those new edges. That free variable here would be chosen to use the same length as all the former (unchanged) edges. Therefore those new red hexagons would be regular faces as well. - It only is that the curvature, taken as a hyperbolic 2D manifold, would not be uniform any longer. Just at this new tiling itself is not a uniform one either.

The next step of expansion would apply the same then to the green (so far still) triangles: extending those into hexagons as well. As far as formerly vertices are concerned which neither contain a red or a green corner, those remain still as such. As far as original vertices are concerned which had red triangles, but no green ones, then those were expanded into new edges in the previos step, but now remain as they have become then. As far as original vertices are concerned which contained green triangles but no red ones, those behave as in the privious step those with red incidences did. But there were also vertices of the original triangle tiling, which used red as well as green triangles. Those have been deformed in the previous step into new edges, having 2 red hexagons adjacent, and 1 incident green triangle at either end. Those edges now will become squares, the sides of which now being adjacent alternatingly to (old) red and (new) green hexagons.

So far I just did a euclidean space expansion (i.e. a locally 0-curvature one): The calculation of the amount of hexagon angle clearly is valid, as long as the first step transformation is done only as abstract step, maintaining the size of the new introdruced edges at zero size (but already doubling those vertices). - That is just as an elongated square dipyramid could be considered as a 2D manifold obtained from the octahedron, as a projection onto the sphere, which then is cut in 2 hemispheres, and a zero-curvatute intermediate space (zylinder) would be introduced for elongation, this could be understood here similarily. - Then, in this second expansion step, the vertex angle of those introduced squares clearly is: 360° - 1*45° (yellow triangle) - 2*112.5° (one red and one green hexagon) = 90°.

But one clearly aproximate that elongated square dipyramid also by a longuish ellipsoid, thereby changing slightly those measurings in a form, which adopts that continously changing curvature. The same could most probably applied in that hyperbolic case too. Then those new squares would become non-euclidean too. We will come back to this point later, which at the moment seems to be mere accademical only.

In a third step we then do the same expansion with respect to the brown (so far still) triangles additionally. This adds further new edges between those new brown hexagons (when the other non-yellow triangles were black ones), respectively would expand in privious steps introduced edges into further squares (in those cases where in the beginning red and brown triangles respectively green and brown triangles had been incident).

And finally, in the 4th step of partial expansion, we finally would extend the black triangles into hexagons as well.

This now is where the former point comes back into play: we now are left with all the privious yellow triangles. All non-yellow ones have become identical (but differently colored) regular hexagons. And any former vertex (of x3o8o) now has become a square. We well can this achieved tiling consider in that discontinuous curvature, i.e using regular triangles with 45° as corner angle, regular hexagons with 112.5°for corner angle, and euclidean squares (with 90° for corner angle). But we could alternatively iron down the curvature to become continuous. In fact here it even becomes possible to have a single specific value of curvature again. In fact, our final outcome would be nothing but the tiling x3o4x3*a. (That one clearly uses different angle sizes each - but I haven't considered their specific amounts so far.)


The main problem in those curvature considerations most probably would be the tiling aspect itself. I.e. the assumption that all tiles would have to live in some 2D manifold itself. But in case of octahedron, elongated square dipyramid, etc. we are used to do different: we consider those shapes having flat (euclidean) regular polygons, assembling within some more or less spherical arrangement within a 3D euclidean embedding space.

The same should best be done for hyperbolic things here too, I think. I.e. considering those hyperbolic space tilings as kind of scrambled quilt structure within embedding 3D space, and all used tiles being normal euclidean regular polygons again. Then there will be no effect at all for all that stuff of dicontinuous curvature, making that continuous, according deformation of measurements, etc. Esp. that at least at the end needed deformation to become again x3o4x3*a (in the usual sense) then would completely be superfluous.

For the other geometries this was no theme so far: for spherical geometries this is usually is being assumed to apply generally. And in euclidean spaces that expansion effects would not add any different curvature at all.

So this theme of partial Stott expansion finally comes out to be a propagation of a different model of hyperbolic spaces as well. In fact, this does not belongs to that new idea alone, it more generally would apply to any consideration of non-uniform hyperbolic structures.

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Re: Johnsonian Polytopes

Postby wendy » Fri Apr 26, 2013 9:07 am

I think i see your game.

What you are trying to do, is to deconstruct a stott-expansion over the whole symmetry into several bits. For example, in going from x3o4o to x3o4x, one adds the X axis, and then the Y axis, and then the Z axis.

If you stop working with bollo-apeirotopes (tilings), and work with real polytopes (eg {3,8} in 3D) then yes, it will work. The resulting polytopes won't be co-radial (that is a bollosphere through the vertices), but the will pass by the other rules. It should work over any value of 2P, not just 4, but some of the intermediate steps might not give regular polygons, now thinking of x3o6o to x3o6x.

I'm trying to think of the {5,3,4}. If you look at the dodecahedron, by way of example, there are six edges which are parallel/perpendicular. These are the six edges that do not connect to an inscribed cube. In the {5,3,4}, one can colour these in three different colours (eg red, blue, green), which gives one some room to work with. (see 'Not Knot' on youtube, part 2, stimly).

If you are looking at a polytope, say a cell of {5,3,4,3}, then it should be possible to cut the planes that contain red lines, and insert a prismatic layer there. It won't be concentric on the vertices as such, but it is possible, because you can insert struts between the fragments to keep the general curvature. This would eventually introduce pentagonal prisms, and square prisms along the red edges. Yes, that would work. The second progression would introduce pentagonal prisms along the the second. Yes, it would be possible to convert x5o3o4o into x5o3o4x in three steps, akin to x3o3o4x and x4o3o4x.
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Re: Johnsonian Polytopes

Postby Klitzing » Fri Apr 26, 2013 7:52 pm

wendy wrote:I think i see your game.

What you are trying to do, is to deconstruct a stott-expansion over the whole symmetry into several bits.

Right! That's why I'm using "Partial Stott Expansion" as its working title.

For example, in going from x3o4o to x3o4x, one adds the X axis, and then the Y axis, and then the Z axis.

If you stop working with bollo-apeirotopes (tilings), and work with real polytopes (eg {3,8} in 3D) then yes, it will work.

Yes, this was my feeling too. This is, why a spoke of "scrubled quilts".

The resulting polytopes won't be co-radial (that is a bollosphere through the vertices), but the will pass by the other rules. It should work over any value of 2P, not just 4, but some of the intermediate steps might not give regular polygons,

Yep, I saw this too, but when I met it first it was by singular examples of quickfurs findings right in this thread, i.e. all was restricted to CRF. That one I kept so far. So I'm kind of struggling about the restriction on the form of Dynkin symbols this would give, when we'd ask for all steps should remain within CRF at least.

now thinking of x3o6o to x3o6x.

Well there a 60° rhomb would be needed... - In other cases, when considering the octagon, a longish hexagon would occure. But all these odd things I brushed away so far.

I'm trying to think of the {5,3,4}. If you look at the dodecahedron, by way of example, there are six edges which are parallel/perpendicular. These are the six edges that do not connect to an inscribed cube. In the {5,3,4}, one can colour these in three different colours (eg red, blue, green), which gives one some room to work with. (see 'Not Knot' on youtube, part 2, stimly).

Ey, you are right indeed! So far I did not succeed, with anything having a 5 in its symbol, but you are right indeed. That Not Knot video clearly shows, that this x5o3o4o structure has vertex layers, which are also face "planes" and do form in fact a x5o4o tiling. So we could cut the former structure at these "red edge layers", and introduce a layer of pentagonal prisms inbetween. And as those edges all have 4fold symmetry, this would be possible in 2 directions each. Thereby the red edges themselves would be blown up into cubes. The same holds true for the green and the blue edges. And yes, in fact, any pentagon thus gets enlarged into a pentagonal prism, and, the other way round, no pentagon would be used twice, as far as I can see. - At least if we'd do both orthogonal expansions of any color at the same time. Then we would result in a 3 step expansion from x5o3o4o up to x5o3o4x. - It's not clear to me, whether there would be a six step transition too, using either layer (of any of these 3 colors) separately. -> Help?

If you are looking at a polytope, say a cell of {5,3,4,3}, then it should be possible to cut the planes that contain red lines, and insert a prismatic layer there. It won't be concentric on the vertices as such, but it is possible, because you can insert struts between the fragments to keep the general curvature. This would eventually introduce pentagonal prisms, and square prisms along the red edges. Yes, that would work. The second progression would introduce pentagonal prisms along the the second. Yes, it would be possible to convert x5o3o4o into x5o3o4x in three steps, akin to x3o3o4x and x4o3o4x.


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Re: Johnsonian Polytopes

Postby Klitzing » Fri Apr 26, 2013 8:26 pm

Klitzing wrote:Ey, you are right indeed! So far I did not succeed, with anything having a 5 in its symbol, but you are right indeed.


My meanwhile "success" with allowed "Partial Stott Expansions" (i.e. keeping CRF) so far was as follows:
  • Let S1 be any linear prefix Dynkin subsymbol from "o3" or "x3" constituents.
  • Let S2 be any linear postfix Dynkin subsymbol from "3o" or "3x" constituents.
  • Then S1-o4o-S2 would allow for a partial Stott expansion towards S1-o4x-S2.
  • Also o4-S1-o4o would allow for a partial Stott expansion towards o4-S1-o4x.
  • Also x4-S1-o4o would allow for a partial Stott expansion towards x4-S1-o4x.
  • Also o4o-S2-4o would allow for a partial Stott expansion towards o4x-S2-4o.
  • Also o4o-S2-4x would allow for a partial Stott expansion towards o4x-S2-4x, here provided that S2 would not be empty.
  • Let further |S1| be the length of S1, i.e. the number of -3-'s of that subsymbol (possibly 0).
  • Then any of the above partial Stott series would have exactly |S1|+2 transitions.
  • Note that this result is independent of S2!

Sure, that list is an "at least" proposition. I was already aware of some few examples for partial Stott expansion, not fitting in there. E.g. my recently asked for x3o8o = x3o4o3*a (4 transition steps up to x3o4x3*a). And now your Not Knot 2/2 one, i.e. x5o3o4o (3 steps for sure up to x5o3o4x, perhaps each one further splittable into 2? - or would that "split" perhaps just might not be allowed for the simple purpose that there would not be a subsymmetry selecting a consistent choice throughout?).

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Re: Johnsonian Polytopes

Postby wendy » Sat Apr 27, 2013 7:44 am

There's quite a few of these partial stott expansions. The sufficient condition is the segment ..o4o... to ..o4x..

One can find the number of independent types of mirrors in a symmetry by removing all of the even branches. So, in the group o5o3o4o, there are two kinds of mirror, those internal to the dodecahedron, and the walls between the dodecahedron. Each of these groups stands freely. What it does not tell you, is that the resulting group r (those mirrors o5o3o4r), also degenerates into free subsets, each subset or combination of subset stands freely.

In the case of o3oo3o4r, the group r is the rectangularoid group, each combination of W, X, Y, Z, ... stands freely. This is why there are four steps in this group.

In the case of o4o3o4r, there are six mirrors, which can be represented by eX, oX, eY, oY, eZ, oZ, ..., and this can be expanded in six steps.

In the case of o5o3o4r, the group r divides into no fewer than 12 separate mirrors. It's not apparent as yet, but consider marking the faces of a dodecahedral cell with numbers from 1 to 12. Now, just reflecting these in the walls of the dodecahedron, one sees that there are planes of pentagons marked '1', and different planes marked '2', usw. None of thes are images of each other, so we have deconstructed this 'rectododecahedral' group into the intersection of 12 separate laminatopic (polytopes bounded by unbounded faces) groups, each one presenting just a single pentagon to a given cell.

Likewise o8o4r, gives an rectioctagonal group, divided into eight laminaohedra. And o7o4r gives a rectoheptagonal group, which divides into seven laminatotopic groups. You get the idea.

Now, we see that R Klitzing's various partial stott expansions, then consist of raising stilts, or mirror edges to free combinations of these laminatopic groups, so, zB

1. octahedron has no stilts.
2. elongated square bipyramid has stilts in X axis
3. bi-square cupola has stilts in X, Y axis
4. rhombo-cuboctahedron has stilts in X, Y, Z axis.

The x5o3o4o then does the same sort of thing, but in place of three steps, twelve. You erect stilts on what's inside the dodecahedron, one face at a time, in any order. There is no need for faces to be adjacent, or side by side, or whatever. So, there are three different arrangements of 'two faces', and several more for 'three faces'.

The progression from x8o4o to x8o4x, takes eight steps, and from xPo4o to xPo4x takes no fewer than P steps. The x5o3o3o4o has 120 steps and many more intermediates, while the x3o4o3o4o has 24 steps to x3o4o3o4r.
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Re: Johnsonian Polytopes

Postby wendy » Sat Apr 27, 2013 8:00 am

It goes further.

If one considers the group x5o3o4r, for example, then the 'r' divides to 12 separate nodes, i..xii, each of which can be marked separately, in much the same way that x3o3o4r corresponds to dividing r to four separate nodes w,x,y,z. But each of i-xii and w-z correspond to faces of the dodecahedron and tetrahedron separately, the number of unique figures is much less than 2^12 = 4096, or 2^4 = 16. In the latter, there are just 5 combinations, the former gives at least
1 for 1 and 11, 3 for 2 and 10, and various different numbers for 3,9, and 4,8 and 5,7 and 6.

In short, the partial stott expansion has a wythoff-mirror-edge explaination, to the defragment of the recto-(face of xPo4xRo) group.
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Re: Johnsonian Polytopes

Postby Klitzing » Sat Apr 27, 2013 2:49 pm

wendy wrote:There's quite a few of these partial stott expansions. The sufficient condition is the segment ..o4o... to ..o4x..

Surely not!
I don't know about quite general prefixings. But there will be problems at general postfixings. Just consider the easy example o4o4x, i.e. the euclidean square tiling. There clearly is a partial Stott expansion transition series towards x4o4x. But there is none towards o4x4x. - Well, in fact there might be one too, but that one would clearly run beyond CRF, which I aimed to reject. (In order to expand an square towards an octagon in a multiple step expansion, you clearly would have to run through some crude (90°,135°,135°,90°,135°,135°)-hexagon, which obviously is no regular polygon...

One can find the number of independent types of mirrors in a symmetry by removing all of the even branches. So, in the group o5o3o4o, there are two kinds of mirror, those internal to the dodecahedron, and the walls between the dodecahedron. Each of these groups stands freely. What it does not tell you, is that the resulting group r (those mirrors o5o3o4r), also degenerates into free subsets, each subset or combination of subset stands freely.

Can't follow that...
So, the ...o4o bit of x5o3o4o does play an essential role here, as this means that the vertex figure is . x3o4o and then the edge figure is . . x4o, i.e. there are 4 dodecahedra around each edge. That is, the pentagonal faces of that hyperbolic honeycomb would come in edge-opposite pairs, thereby providing layers to cut the honeycomb into 2, resp. introduce an additional prism layer (elongation).

Btw., I am used in that context, when there would be then co-realmic cells, that those would be joined in the sense of an external blend. E.g. the elongation of the octahedron does not constitute of 2 square pyramids plus 1 cube (square prism), but rather would become a single unit, the Johnson solid called elongated square dipyramid.

In the case of o3oo3o4r, the group r is the rectangularoid group, each combination of W, X, Y, Z, ... stands freely. This is why there are four steps in this group.

From "4 steps" I deduce that you meant "o3o3o4r" here.

In the case of o4o3o4r, there are six mirrors, which can be represented by eX, oX, eY, oY, eZ, oZ, ..., and this can be expanded in six steps.

Don't know how you'd get 6 here. I just come up to 3: o4o3o4o = point = o-infin-o o-infin-o o-infin-o <-> aze = o-infin-o o-infin-o x-infin-o <-> squat = o-infin-o x-infin-o x-infin-o <-> o4o3o4x = chon = x-infin-o x-infin-o x-infin-o.

Oh, while writing I get a glance of what you could have in mind: not to combine opposite pairs of sides / mirrors into 1, but use a further medial step: o-infin-o (ignoring those mirrors) <-> x-infin-y (using them both, but not uniting those classes) <-> x-infin-x (again using both, but now uniting those classes), i.e thus getting = x-infin-o. - Is that what you meant here?

In the case of o5o3o4r, the group r divides into no fewer than 12 separate mirrors. It's not apparent as yet, but consider marking the faces of a dodecahedral cell with numbers from 1 to 12. Now, just reflecting these in the walls of the dodecahedron, one sees that there are planes of pentagons marked '1', and different planes marked '2', usw. None of thes are images of each other, so we have deconstructed this 'rectododecahedral' group into the intersection of 12 separate laminatopic (polytopes bounded by unbounded faces) groups, each one presenting just a single pentagon to a given cell.

Oh, yeah, I see what you are after here.

In fact, with respect to axial expansions, I performed those so far as follows: In squerical geometry just cut at an equatorial vertex layer, and elongate here. Within euclidean space geometries, i.e. honeycombs etc. I'd rather use all parallel layers. - But you now would rather propagate to use alternate ones instead, only. Thus doubling the transition steps.

But this does not longer allow for commutativity. For 3 mutually orthogonal directions, you'd get essentially those combinations of elongations: 0, x, x+x', x+y, x+x'+y, x+y+z, x+x'+y+y', x+x'+y+z, x+x'+y+y'+z, x+x'+y+y'+z+z'. This provides a much higher degree of asymmetry of the outcomes. (I' so far just considered here: 0, x+x', x+x'+y+y', x+x'+y+y'+z+z', i.e. just 3 transitions.)

Likewise o8o4r, gives an rectioctagonal group, divided into eight laminaohedra. And o7o4r gives a rectoheptagonal group, which divides into seven laminatotopic groups. You get the idea.

Now, we see that R Klitzing's various partial stott expansions, then consist of raising stilts, or mirror edges to free combinations of these laminatopic groups, so, zB

1. octahedron has no stilts.
2. elongated square bipyramid has stilts in X axis
3. bi-square cupola has stilts in X, Y axis
4. rhombo-cuboctahedron has stilts in X, Y, Z axis.

The x5o3o4o then does the same sort of thing, but in place of three steps, twelve. You erect stilts on what's inside the dodecahedron, one face at a time, in any order. There is no need for faces to be adjacent, or side by side, or whatever. So, there are three different arrangements of 'two faces', and several more for 'three faces'.

Even when unifying the classes of opposite faces, as I was used to so far, you still get 6 independent directions. This would lead to combinations: 0 (none), 1, 1+2 (necessarily neighbours), 1+2+3 (linear sequence with 2 pairs of neighbours only), 1+2+4 (all incident to a single dodecahedron vertex), 1+2+3+4 (necessarily all incident to a dodecahedron edge = all but 2 directions), 1+2+3+4+5 (all but 1 direction), 1+2+3+4+5+6 (all). I.e. there still are 8 combinations. Esp. the expansion series no longer is a linear one, it rather becomes a net with alternate paths! - Obviously even worse, if no such unification would apply...

The progression from x8o4o to x8o4x, takes eight steps, and from xPo4o to xPo4x takes no fewer than P steps. The x5o3o3o4o has 120 steps and many more intermediates, while the x3o4o3o4o has 24 steps to x3o4o3o4r.


Note that I usually go like this:
Take any "easy" pair of Dynkin symbols with a classical 1-step Stott expansion. Consider a subsymmetry display of the same pair of polytopes, which allows for a multi-step classical Stott expansion. Then apply that subsymmetry multi-step expansion directions to any further (allowed, i.e keeping that o4o to o4x transition) decoration of the former Dynkin symbol:
E.g.
o3o3o4o = point = o o o o <-> edge = o o o x <-> square = o o x x <-> cube = o x x x <-> o3o3o4x = tes = x x x x.
Then you could apply that multi-directional expansion step-wise likewise to
x3o3o4o = hex <-> pex hex <-> quawros <-> pacsid pith <-> x3o3o4x = sidpith
or to
o3x3o4o = ico <-> pexic <-> bicyte ausodip <-> pacsrit <-> o3x3o4x = srit
or to
x3x3o4o = thex <-> pex thex <-> pabex thex <-> pacprit <-> x3x3o4x = prit
(Those inbetween fellows already had been individual topics within this very thread. So those series clearly imply their shapes directly as well.)

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Re: Johnsonian Polytopes

Postby Klitzing » Sat Apr 27, 2013 4:48 pm

wendy wrote:It goes further.

If one considers the group x5o3o4r, for example, then the 'r' divides to 12 separate nodes, ...

I remember, "r" is your node symbol for a zero sized edge, i.e. you just split the related vertices into 2 with all related topological changes into the figure which would there contain an "x", but the look would still remain that of the corresponding "o" figure.

Bringing in this "r" to the context of partial Stott expansions, this relates kind of going right onto the topology of the completely expanded member of the series, but keeping the look of the first member. Then you will have to choose some (somehow?) implied subsymmetry, which separates this set of r-edges into subsets, and then finally you would apply scalings of those r-edges into x-edges independently. - Well, yes, that is exactly what I have in mind.

So yes, this should be a rather general concept indeed, independant on any "4"-link. Just subject to the question what implied subsets could exist.

My intend even so, where to connect this to the topic of CRF additionally. I.e. to deduce sufficient and necessary arguments on the Dynkin symbol, such that
- such true subsets would exist
- all relevant polygons would be regular

(The convexity of the CRF subject would be easy here: when the completely contracted figure is convex, then too is the completely expanded one, and even all ones inbetween. And the other way round too: if the completely expanded one is, then too the others. And as those extreme ones are uniform, that is easy to determine from the given Dynkin symbol. - This is why I'd ignore that for the remainder of the research. But would stick with examples to convex ones for convenience.)

... i..xii, each of which can be marked separately, in much the same way that x3o3o4r corresponds to dividing r to four separate nodes w,x,y,z. But each of i-xii and w-z correspond to faces of the dodecahedron and tetrahedron separately, ...

Oh, I see where you get into that subject of using opposite faces separately (for cubes, dodecahedra, etc.), as this was done for the tetrahedron too: all faces are used independently; just that one does not have parallel ones, either.

... the number of unique figures is much less than 2^12 = 4096, or 2^4 = 16. In the latter, there are just 5 combinations, the former gives at least
1 for 1 and 11, 3 for 2 and 10, and various different numbers for 3,9, and 4,8 and 5,7 and 6.

In short, the partial stott expansion has a wythoff-mirror-edge explaination, to the defragment of the recto-(face of xPo4xRo) group.

Would be good to elaborate on that latter point more deeply. So that we get a clearer sight onto that recent topic of interest.

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Re: Johnsonian Polytopes

Postby Klitzing » Sat Apr 27, 2013 7:15 pm

wendy wrote:The progression from x8o4o to x8o4x, takes eight steps, and from xPo4o to xPo4x takes no fewer than P steps. The x5o3o3o4o has 120 steps and many more intermediates, while the x3o4o3o4o has 24 steps to x3o4o3o4r.


I've made the corresponding 8-coloring of x4o8x, which Wendy has in mind here.
x4o8x-8colored.png
(35.09 KiB) Not downloaded yet

All strips of squares of any chosen color have simultanuously to be removed. This obviously can be done independently for any subset out of those 8 colors.

"No fewer" would be correct, if P is prime. Else there are fewer ones too. E.g. in this x4o8x you need not color all the squares attached to any octagon differently, you could color the opposite ones by the same one. Or even you could just alternate those colors: 4 divides 8, and 2 does so too. - Btw. the divisor 1 would then represent the classical Stott contraction.

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Re: Johnsonian Polytopes

Postby Klitzing » Sat Apr 27, 2013 8:19 pm

Klitzing wrote:
wendy wrote:now thinking of x3o6o to x3o6x.

Well there a 60° rhomb would be needed... - In other cases, when considering the octagon, a longish hexagon would occure. But all these odd things I brushed away so far.


Re-thought that one: Sure, when one is dealing with bilateral reduction, then the hexagon would be reduced to a rhomb. And sure too, 3 such bilateral ones (mutually in 120°) would reduce the hexagon to a point. - But then, what about tripesic reductions at the hexagonal centers, i.e applying kind a hexagonal lattice on top of those hexagon centers of x3o6x? Then 2/3 of these hexagons would be reduced to triangles (and the corresponding squares between would be reduced to mere edges), but 1/3 of the hexagons would remain in place.

How to go on? - Well you cannot connect those remaining hexagons for any reduction. But there is an other way out: The triangle tiling of all those former hexagonal centers in fact consist out of an overlay of 3 mutually shifted hexagonal tilings. Thereby any one uses 2/3 of the vertices, thus any 2 of those hexagonal combs reuses half of either others vertices. This implies a 3coloring of those hexagons: the mentioned first reduction step would maintain the red hexagons, but reduces the yellow ones to triangles with their tip pointing up, while the green ones are reduced to triangles with their tip pointing down.

Then the next tripesic reduction (again along a hexagonal lattice) would reuse the yellow triangles, reducing those to points, and using now the red hexagons instead, which become then triangles (pointing up). Here the formerly produced green triangles (pointing down) will be maintained.

Finally we could apply a third tripesic reduction, maintaining the yellow points unchanged, but reducing the red triangles and the green ones to points as well. Thus finally all squares are reduced as well. So we are left with the final tiling x3o6o. - I.e. we just have described a partial Stott contraction from x3o6x down to x3o6o, using 3 independent steps!

Hurray! - Similarily we surely could apply a tetrapesic reduction to octagon centers, pentapesic reductions to decagon centers, etc. I.e. there is no longer the necessity to have some ...o4o... to ...o4x... total change for partial Stott expansions / contractions, we just showed that this is possible for any ...o-2N-o... to ...o-2N-x...

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Re: Johnsonian Polytopes

Postby Klitzing » Sun Apr 28, 2013 7:24 am

Klitzing wrote:
Klitzing wrote:
wendy wrote:now thinking of x3o6o to x3o6x.

[...]


Re-thought that one: [...] - But then, what about tripesic reductions at the hexagonal centers, i.e applying kind a hexagonal lattice on top of those hexagon centers of x3o6x? Then 2/3 of these hexagons would be reduced to triangles (and the corresponding squares between would be reduced to mere edges), but 1/3 of the hexagons would remain in place.

[...] - I.e. we just have described a partial Stott contraction from x3o6x down to x3o6o, using 3 independent steps! [...]
--- rk


And here are the corresponding steps being depicted:
trat-pextrat-pacrothat-rothat.png
(125.02 KiB) Not downloaded yet


- 'Well', you might ask, 'and how about your recent comment about "easier" cases which then would allow for lesser symmetry description, and thereby would allow for a corresponding classical Stott transition?' - Yes, this would work here too! Consider:

x3o6o = trat <-> pextrat <-> pacrothat <-> x3o6x = rothat
is the series, I yesterday night had described, and now was being depicted.

The corresponding "easier" family member then would be
o3o6o = point = o3o3o3*a <-> trat = x3o3o3*a <-> that = x3x3o3*a <-> o3o6x = hexat = x3x3x3*a

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Re: Johnsonian Polytopes

Postby wendy » Sun Apr 28, 2013 7:52 am

Let's take a little trip through hyperbolic space. Here is the Not_Knot poster, suitable for downloading, and the topic of this duscussion.

http://www.geom.uiuc.edu/graphics/pix/Video_Productions/Not_Knot/NKposter.1500.html

Let's look at first a {5,4} that forms one of the planes of faces. There's a big circle which is actually a complete plane. One can see that it is possible to select any number of edges of a pentagon, in any order. For a single line, there are the red lines. Then there are the pairs of edges that point to a blue line, representing a pair of adjoining edges. The other white sets that join to the red edges make a set of two non-adjacent edges. Combinations of three and four edges are then the complements of these sets.

So we can select, eg 1, 2, 3, 4 or 5 edges of a pentagon of {5,4} in any combination, and consistantly progress it across the whole plane.

We can then expand these by adding squares for each line.

When i said about 'using polytopes' rather than tilings. Consider first the line of red edges. This is a particular kind of hyperbolic polygon "w5.236". The W notation gives the square of the short-chord (the base of a triangle of two edges), so the square would be eg w2. The line is straight, but we can likewise have girthing decagons and dodecagons on a sphere.

We find the ratio between the radius and diameter of a circle, of radius=edge. This is done by way of using the circumferences of circles, which are the intersection of a euclidean and non-eculidean space. The first circle is that around a pentagon, which tells us that the line across the two edges at right angles is 1.618033 edge. The second is around the central vertex, where we have the inscribed edge of a square. The diameter is then 2.288, and its square is 5.236. We then call this polygon a w5.236. The polygons that criss-cross the {6,4} for example gives a w6, which is larger.

If we now look at the polygon, whose vertices are where the blue lines cross, we see that its edges are two of the w5.236 in arc, the lines are straight (through the non-blue points), but turn at right angles at the blue points. It is a right-angle "w5.236". This means that for example, we can push edges of the red line out here (without changing size), and leave enough room for extra bits to be added in.

Let's now look at "planes" equidestant from the x5o4o. This is like a 'vertex first projection', except we're stating at a plane. The first section is f5o4o, formed by raising the pentagons up the edges vertical to the plane. The second section has again the slice o5f of the dodecahedron, and squares of this same edge of the form we met calculating the short-chord two paras ago. This section is an o5f4o. One can see that these have margin-angles less than 180deg, and that even though the pentagons are the same size, there is room enough in the third section to add squares.

If you remove all of the coloured edges, and their edges, the original dodecahedra becomes cubes, and new cubes form where the coloured line runs down the centres of them, and the four lines radiating from the coloured are a cross on opposite faces of a cube. That is {5,3,4} contains the vertices of {4,3,5}.

The {5,3,3,4} has an edge of the same length as this, the diameter of a cell {5,3,3} is the same as four edge-lengths end to end.
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