Johnsonian Polytopes

[student91]

Conjecture (almost certainly a theorem, but I don't have a proof): in any dimension n, (n-1)-simplex || (n-1)-simplex = n-cross.

Proof is rather easy. First we note that any (n-1)-simplex (with edge-length sqrt(2)) can be given n-coordinats as follows: (1,0,0,...) (0,1,0,...) (0,0,1,...) etc.
Furthermore any n-cross-polytope can be given coordinates (1,0,0,...) (-1,0,0,...) (0,1,0,...) (0,-1,0,...) (0,0,1,...) (0,0,-1,...) etc.
These can be divided into two sets of coordinates, namely the positives ( (1,0,0,...) (0,1,0,...) (0,0,1,...) etc.) and the negatives ( (-1,0,0,...) (0,-1,0,...) (0,0,-1,...) etc. )
These are the coordinates of two simpices in dual orientation.
If you want me to prove that they are in dual orientation, here i go:
The middles of the simplices have coordinates (1/n,1/n,1/n,...) and respectively (-1/n,-1/n,-1/n,...). If we want to make them origin-centered, we have to subtract these vectors, and we get (n-1/n,-1/n,-1/n,...) (-1/n,n-1/n,-1/n,...) (-1/n,-1/n,n-1/n,...) and respectively (-(n-1/n),1/n,1/n,...) (1/n,-(n-1/n),1/n,...) (1/n,1/n,-(n-1/n),...). First we note that a vertex of a simplex is always opposite to the middle of a facet, thus the coordinates of a vertex of a origin-centered simplex multiplied with a specific negative constant n, gives the middle of a facet. If we scale one simplex down with some positive constant p, and it's vertices lie in the middle of the facets of the other simplex, we've proven that they're in dual orientation. And it's easily seen that if we choose p to be -n^-1, n*p*(n-1/n,-1/n,-1/n,...)= (-(n-1/n),1/n,1/n,...), and for all the other vertices a similar procedure can be followed, which means it is proven.

EDIT: middle coordinates of simplex corrected

[quickfur]

Nice proof, thanks!

[wendy]

The AB type lattices, use a pair of simplexes with different size lacing. This is the general simplex-antiprism. One can see that the diagonal of the antiprism is simply the rss(side of simplex, lacing).

[quickfur]

While studying what kind of CRFs may be obtained from various combinations of square pyramids and tetrahedra, I stumbled across this one:



This is basically the cube antiprism with 1 vertex deleted from the octahedral cell, resulting in a CRF with 1 square pyramid, 1 square antiprism, 4+4+4 = 12 tetrahedra, 1 cube, and 1+1+4=6 square pyramids. In the image above, the red square pyramid is the remaining half of the original octahedral cell, and the yellow cell is the new square antiprism cell produced by the deletion. It can also be obtained by inserting a new vertex into the cube||gyro square (equivalent to augmenting it with a square antiprism pyramid).

Now, since this is a segmentochoron, I'm pretty sure it should be in Klitzing's list of segmentochora; but I can't seem to find it...? I tried looking for cube||4pyr or 4ap||4pyr, but didn't find it in the index. Is it in the list, and if it is, what's the entry number?

Anyway, along with my policy of always including full Cartesian coordinates for CRF renders that I post, here they are:

Code: Select all

<±sqrt(2), 0, 0, 0>
<0, ±sqrt(2), 0, 0>
<0, 0,  sqrt(2), 0>
<±1, ±1, ±1, sqrt(2*sqrt(2)-1)>


[student91]

It's K4.16, cube is seen as a 4prism, and then it is 4p || gyro 4pyr

[quickfur]

It's K4.16, cube is seen as a 4prism, and then it is 4p || gyro 4pyr

Ahhh, it's gyro 4pyr, that's why I missed it in the index. I was looking for 4p || 4pyr and didn't think to look at the gyro entries as well. Thanks!

[Klitzing]

Found a further non-obvious 5D CRF. Not only rico || sadi is possible, but so is ico || sadi too!

Here is its incidence matrix, again being obtained as a (full dimensional) snubbing with respect to alternate bottom base vertices only:

Code: Select all

os3os4oo3xo&#x
(ico || sadi)

      o.3o.4o.3o.      | 24  * |  8  12   0   0 | 12  24   6  24  0   0  0 |  6 12   8  12  24 12  0  0  0 | 1  1  8  6 0 12
demi( .o3.o4.o3.o    ) |  * 96 |  0   3   3   6 |  0   3   3  12  3   9  3 |  0  1   6   9   6  3  3  1  4 | 0  3  3  1 1  4
-----------------------+-------+----------------+--------------------------+-------------------------------+----------------
      .. .. .. x.      |  2  0 | 96   *   *   * |  3   3   0   0  0   0  0 |  3  3   0   0   3  3  0  0  0 | 1  0  1  3 0  3
demi( oo3oo4oo3oo&#x ) |  1  1 |  * 288   *   * |  0   2   1   4  0   0  0 |  0  1   2   3   4  2  0  0  0 | 0  1  2  1 0  3
      .. .s4.o ..      |  0  2 |  *   * 144   * |  0   0   1   0  0   2  2 |  0  0   0   2   0  2  1  1  2 | 0  1  0  1 1  2
sefa( .s3.s .. ..    ) |  0  2 |  *   *   * 288 |  0   0   0   2  1   2  0 |  0  0   2   2   1  0  2  0  1 | 0  2  1  0 1  1
-----------------------+-------+----------------+--------------------------+-------------------------------+----------------
      .. .. o.3x.      |  3  0 |  3   0   0   0 | 96   *   *   *  *   *  * |  2  1   0   0   0  1  0  0  0 | 1  0  0  2 0  1
demi( .. .. .. xo&#x ) |  2  1 |  1   2   0   0 |  * 288   *   *  *   *  * |  0  1   0   0   2  1  0  0  0 | 0  0  1  1 0  2
      .. os4oo ..&#x   |  1  2 |  0   2   1   0 |  *   * 144   *  *   *  * |  0  0   0   2   0  2  0  0  0 | 0  1  0  1 0  2
sefa( os3os .. ..&#x ) |  1  2 |  0   2   0   1 |  *   *   * 576  *   *  * |  0  0   1   1   1  0  0  0  0 | 0  1  1  0 0  1
      .s3.s .. ..      |  0  3 |  0   0   0   3 |  *   *   *   * 96   *  * |  0  0   2   0   0  0  2  0  0 | 0  2  1  0 1  0
sefa( .s3.s4.o ..    ) |  0  3 |  0   0   1   2 |  *   *   *   *  * 288  * |  0  0   0   1   0  0  1  0  1 | 0  1  0  0 1  1
sefa( .. .s4.o3.o    ) |  0  3 |  0   0   3   0 |  *   *   *   *  *   * 96 |  0  0   0   0   0  1  0  1  1 | 0  0  0  1 1  1
-----------------------+-------+----------------+--------------------------+-------------------------------+----------------
      .. o.4o.3x.      |  6  0 | 12   0   0   0 |  8   0   0   0  0   0  0 | 24  *   *   *   *  *  *  *  * | 1  0  0  1 0  0  oct
demi( .. .. oo3xo&#x ) |  3  1 |  3   3   0   0 |  1   3   0   0  0   0  0 |  * 96   *   *   *  *  *  *  * | 0  0  0  1 0  1  tet
      os3os .. ..&#x   |  1  3 |  0   3   0   3 |  0   0   0   3  1   0  0 |  *  * 192   *   *  *  *  *  * | 0  1  1  0 0  0  tet
sefa( os3os4oo ..&#x ) |  1  3 |  0   3   1   2 |  0   0   1   2  0   1  0 |  *  *   * 288   *  *  *  *  * | 0  1  0  0 0  1  tet
sefa( os3os .2 xo&#x ) |  2  2 |  1   4   0   1 |  0   2   0   2  0   0  0 |  *  *   *   * 288  *  *  *  * | 0  0  1  0 0  1  tet
sefa( .. os4oo3xo&#x ) |  3  3 |  3   6   3   0 |  1   3   3   0  0   0  1 |  *  *   *   *   * 96  *  *  * | 0  0  0  1 0  1  oct
      .s3.s4.o ..      |  0 12 |  0   0   6  24 |  0   0   0   0  8  12  0 |  *  *   *   *   *  * 24  *  * | 0  1  0  0 1  0  ike
      .. .s4.o3.o      |  0  4 |  0   0   6   0 |  0   0   0   0  0   0  4 |  *  *   *   *   *  *  * 24  * | 0  0  0  1 1  0  tet
sefa( .s3.s4.o3.o    ) |  0  4 |  0   0   3   3 |  0   0   0   0  0   3  1 |  *  *   *   *   *  *  *  * 96 | 0  0  0  0 1  1  tet
-----------------------+-------+----------------+--------------------------+-------------------------------+----------------
      o.3o.4o.3x.      | 24  0 | 96   0   0   0 | 96   0   0   0  0   0  0 | 24  0   0   0   0  0  0  0  0 | 1  *  *  * *  *  ico
      os3os4oo ..&#x   |  1 12 |  0  12   6  24 |  0   0   6  24  8  12  0 |  0  0   8  12   0  0  1  0  0 | * 24  *  * *  *  ikepy
      os3os .2 xo&#x   |  2  3 |  1   6   0   3 |  0   3   0   6  1   0  0 |  0  0   2   0   3  0  0  0  0 | *  * 96  * *  *  pen (= line || perp {3})
      .. os4oo3xo&#x   |  6  4 | 12  12   6   0 |  8  12   6   0  0   0  4 |  1  4   0   0   0  4  0  1  0 | *  *  * 24 *  *  rap (= oct || tet)
      .s3.s4.o3.o      |  0 96 |  0   0 144 288 |  0   0   0   0 96 288 96 |  0  0   0   0   0  0 24 24 96 | *  *  *  * 1  *  sadi
sefa( os3os4oo3xo&#x ) |  3  4 |  3   9   3   3 |  1   6   3   6  0   3  1 |  0  1   0   3   3  1  0  0  1 | *  *  *  * * 96  octpy (= {3} || gyro tet)


The height here can be calculated as sqrt[(-3+2 sqrt(2)-sqrt(5)+sqrt(10))/2] = 0.614262; and the circumradius as sqrt[(9+5 sqrt(2)+3 sqrt(5)+2 sqrt(10))/8] = 1.907349.
Edit: cf. http://teamikaria.com/hddb/forum/viewtopic.php?f=25&t=1468&p=19547#p19547

--- rk

[quickfur]

Found a further non-obvious 5D CRF. Not only rico || sadi is possible, but so is ico || sadi too! [...]

That's 24-cell||snub 24-cell? Wow!! This one is unexpected! What are the facets? Oh nevermind, it's in the incmat.

[Klitzing]

Found a further non-obvious 5D CRF. Not only rico || sadi is possible, but so is ico || sadi too! [...]

That's 24-cell||snub 24-cell? Wow!! This one is unexpected! What are the facets? Oh nevermind, it's in the incmat.

Oops

Please rewind! Got some algebraic calculations wrong...

The structure of rico || sadi still is correct, but its height was wrong. It should be quite simply: Height^2 = 1/tau, i.e Height = 0.786151. Then the circumradius too evaluates to a much simpler value: Circumradius^2 = (13+5 sqrt(5))/8, i.e. Circumradius = 1.738546.

For ico || sadi I sadly made the same error. The structure again is correct, but its height here now comes out to be zero! So, in fact, this is no longer a true segmentoteron, it rather would be a 4D decomposition of sadi into a central ico plus several further bits (as described)!

Sorry for that.

--- rk

[student91]

I saw the tetahedral expanded ursachoron said "to be verified" so i calculated cartesian coordinates for it.
(0,sqrt(1/2),sqrt(1/2),-sqrt((11-3tau)/8))
(0,-sqrt(1/2),sqrt(1/2),-sqrt((2+3tau)/8))
(0,sqrt(1/2),-sqrt(1/2),-sqrt((2+3tau)/8))
(0,-sqrt(1/2),-sqrt(1/2),-sqrt((2+3tau)/8))
(sqrt(1/2),0,sqrt(1/2),-sqrt((2+3tau)/8))
(sqrt(1/2),0,-sqrt(1/2),-sqrt((2+3tau)/8))
(sqrt(1/2),sqrt(1/2),0,-sqrt((2+3tau)/8))
(sqrt(1/2),-sqrt(1/2),0,-sqrt((2+3tau)/8))
(-sqrt(1/2),0,sqrt(1/2),-sqrt((2+3tau)/8))
(-sqrt(1/2),0,-sqrt(1/2),-sqrt((2+3tau)/8))
(-sqrt(1/2),sqrt(1/2),0,-sqrt((2+3tau)/8))
(-sqrt(1/2),-sqrt(1/2),0,-sqrt((2+3tau)/8))

((tau-1)sqrt(2)/4,(tau+1)sqrt(2)/4,(tau+1)sqrt(2)/4,0)
((tau-1)sqrt(2)/4,-(tau+1)sqrt(2)/4,-(tau+1)sqrt(2)/4,0)
(-(tau-1)sqrt(2)/4,-(tau+1)sqrt(2)/4,(tau+1)sqrt(2)/4,0)
(-(tau-1)sqrt(2)/4,(tau+1)sqrt(2)/4,-(tau+1)sqrt(2)/4,0)
((tau+1)sqrt(2)/4,(tau-1)sqrt(2)/4,(tau+1)sqrt(2)/4,0)
((tau+1)sqrt(2)/4,-(tau-1)sqrt(2)/4,-(tau+1)sqrt(2)/4,0)
(-(tau+1)sqrt(2)/4,-(tau-1)sqrt(2)/4,(tau+1)sqrt(2)/4,0)
(-(tau+1)sqrt(2)/4,(tau-1)sqrt(2)/4,-(tau+1)sqrt(2)/4,0)
((tau+1)sqrt(2)/4,(tau+1)sqrt(2)/4,(tau-1)sqrt(2)/4,0)
((tau+1)sqrt(2)/4,-(tau+1)sqrt(2)/4,-(tau-1)sqrt(2)/4,0)
(-(tau+1)sqrt(2)/4,-(tau+1)sqrt(2)/4,(tau-1)sqrt(2)/4,0)
(-(tau+1)sqrt(2)/4,(tau+1)sqrt(2)/4,-(tau-1)sqrt(2)/4,0)

(sqrt(2)/4,3*sqrt(2)/4,-3*sqrt(2)/4,tau*sqrt(2)/4)
(sqrt(2)/4,-3*sqrt(2)/4,3*sqrt(2)/4,tau*sqrt(2)/4)
(-sqrt(2)/4,3*sqrt(2)/4,3*sqrt(2)/4,tau*sqrt(2)/4)
(-sqrt(2)/4,-3*sqrt(2)/4,-3*sqrt(2)/4,tau*sqrt(2)/4)
(3*sqrt(2)/4,sqrt(2)/4,-3*sqrt(2)/4,tau*sqrt(2)/4)
(3*sqrt(2)/4,-sqrt(2)/4,3*sqrt(2)/4,tau*sqrt(2)/4)
(-3*sqrt(2)/4,sqrt(2)/4,3*sqrt(2)/4,tau*sqrt(2)/4)
(-3*sqrt(2)/4,-sqrt(2)/4,-3*sqrt(2)/4,tau*sqrt(2)/4)
(3*sqrt(2)/4,3*sqrt(2)/4,-sqrt(2)/4,tau*sqrt(2)/4)
(3*sqrt(2)/4,-3*sqrt(2)/4,sqrt(2)/4,tau*sqrt(2)/4)
(-3*sqrt(2)/4,3*sqrt(2)/4,sqrt(2)/4,tau*sqrt(2)/4)
(-3*sqrt(2)/4,-3*sqrt(2)/4,-sqrt(2)/4,tau*sqrt(2)/4)

tau= (1+sqrt(5))/2

I'm not sure if those are correct, but i think they are. (calculations are based on the demicube-orientation of the tetahedron, first three coordinates (x, y and z) give x3o3x, f3o3x and o3x3x respectively, 4th coordinate (w) gives the height)
in shorter form:
(x3o3x) all perm. and changes of sign of (0,sqrt(2)/2,sqrt(2)/2) with 4th coordinate -sqrt((2+3tau)/8)
(f3o3x) all perm. and even number of minus-signs of ((tau-1)sqrt(2)/4, (tau+1)sqrt(2)/4, (tau+1)sqrt(2)/4) with 4th coordinate 0
(o3x3x) all perm. and even number of plus-signs of (sqrt(2)/4,3*sqrt(2)/4,3*sqrt(2)/4) with 4th coordinate tau*sqrt(2)/4

[Klitzing]

tau*sqrt(2)/4 = sqrt[3+sqrt(5)]/4 = 0.572061   thus this height seems to be correct.
but   sqrt((11-3tau)/8) = 0.876491 < sqrt[7+3 sqrt(5)]/4 = 0.925615,
cf. http://bendwavy.org/klitzing/incmats/tut=fx-co=co.htm

--- rk

[quickfur]

[...]
For ico || sadi I sadly made the same error. The structure again is correct, but its height here now comes out to be zero! So, in fact, this is no longer a true segmentoteron, it rather would be a 4D decomposition of sadi into a central ico plus several further bits (as described)!

Sorry for that.

--- rk

That's still interesting, though! Because it implies that there's a decomposition of the 600-cell into a 24-cell plus some CRF pieces!

[Klitzing]

[...]
For ico || sadi I sadly made the same error. The structure again is correct, but its height here now comes out to be zero! So, in fact, this is no longer a true segmentoteron, it rather would be a 4D decomposition of sadi into a central ico plus several further bits (as described)!

Sorry for that.

--- rk

That's still interesting, though! Because it implies that there's a decomposition of the 600-cell into a 24-cell plus some CRF pieces!

Quite easy, that one: my degenerate partially snubbed lace prism os3os4oo3xo&#x, i.e. that decomposition of sadi into a central ico and some few orbiting bits, does use 24 ikepy to connect the tips of the central ico to the flat icosahedral facets of sadi.

On the other hand sadi is derived from ex by chopping off 24 of its vertices right down to exactly those icosahedra. That is, one clearly would be lead to think, that the replacement of these 24 ikepy by corresponding ikedpy (dipyramids), and replacing sadi by ex for sure, would solve the bet.

One easy can check this too: the circumradius of ico is the edge unity. The height of the ikepy is (sqrt(5)-1)/4 = 0.309017. Accordingly the height of ikedpy is twice that value. If we further add that value to unity, we clearly end by the circumradius value of ex.


Thus, we just have proven:
There is a decomposition of ex into 1x ico + 24x ikedpy + 96x pen + 96x octpy + 24x rap.

--- rk

[student91]

tau*sqrt(2)/4 = sqrt[3+sqrt(5)]/4 = 0.572061   thus this height seems to be correct.
but   sqrt((11-3tau)/8) = 0.876491 < sqrt[7+3 sqrt(5)]/4 = 0.925615,

true, made some calculaton errors , the right height should be sqrt((2+3tau)/8). I'll edit that in my previous post.
btw,if so much is known about this polytope,then why does it say "to be verified"?

[Klitzing]

[...]
For ico || sadi I sadly made the same error. The structure again is correct, but its height here now comes out to be zero! So, in fact, this is no longer a true segmentoteron, it rather would be a 4D decomposition of sadi into a central ico plus several further bits (as described)!

Sorry for that.

--- rk

That's still interesting, though! Because it implies that there's a decomposition of the 600-cell into a 24-cell plus some CRF pieces!

Well, then you might be interested in that further, likewise degenerate, partially snubbed lace prism os3os3os4xo&#x too, i.e. in tes || sadi. It likewise will have the height = 0. And so comes out rather as an other decomposition of sadi into several bits around a central tesseract.

The incidence (and the to be used bits) can be read from its matrix:

Code: Select all

os3os3os4xo&#x
(tes || sadi)

      o.3o.3o.4o.      | 16  * |  4  12  0  0  0   0 |  6 12  6  12  12 12  0  0   0  0  0 | 4  4  4 12  12 12 12 12  0  0 0  0 | 1  1  4  6 4 0 12
demi( .o3.o3.o4.o    ) |  * 96 |  0   2  2  1  2   4 |  0  1  2   4   4  2  1  2   6  3  3 | 0  2  2  1   6  2  3  3  2  1 1  4 | 0  2  1  1 1 1  4
-----------------------+-------+---------------------+-------------------------------------+------------------------------------+------------------
      .. .. .. x.      |  2  0 | 32   *  *  *  *   * |  3  3  0   0   0  3  0  0   0  0  0 | 3  0  0  6   0  3  3  3  0  0 0  0 | 1  0  1  3 3 0  3
demi( oo3oo3oo4oo&#x ) |  1  1 |  * 192  *  *  *   * |  0  1  1   2   2  1  0  0   0  0  0 | 0  1  1  1   3  2  2  2  0  0 0  0 | 0  1  1  1 1 0  3
      .s .2 .s ..      |  0  2 |  *   * 96  *  *   * |  0  0  1   0   0  0  0  0   2  2  0 | 0  0  0  0   2  0  2  0  1  1 0  2 | 0  1  0  1 0 1  2
      .. .. .s4.o      |  0  2 |  *   *  * 48  *   * |  0  0  0   0   0  2  0  0   0  2  2 | 0  0  0  1   0  0  2  2  0  1 1  2 | 0  0  0  1 1 1  2
sefa( .s3.s .. ..    ) |  0  2 |  *   *  *  * 96   * |  0  0  0   2   0  0  1  0   2  0  0 | 0  2  0  0   2  1  0  0  2  0 0  1 | 0  2  1  0 0 1  1
sefa( .. .s3.s ..    ) |  0  2 |  *   *  *  *  * 192 |  0  0  0   0   1  0  0  1   1  0  1 | 0  0  1  0   1  0  0  1  1  0 1  1 | 0  1  0  0 1 1  1
-----------------------+-------+---------------------+-------------------------------------+------------------------------------+------------------
      .. .. o.4x.      |  4  0 |  4   0  0  0  0   0 | 24  *  *   *   *  *  *  *   *  *  * | 2  0  0  2   0  0  0  0  0  0 0  0 | 1  0  0  1 2 0  0
demi( .. .. .. xo&#x ) |  2  1 |  1   2  0  0  0   0 |  * 96  *   *   *  *  *  *   *  *  * | 0  0  0  1   0  2  1  1  0  0 0  0 | 0  0  1  1 1 0  2
      os .2 os ..&#x   |  1  2 |  0   2  1  0  0   0 |  *  * 96   *   *  *  *  *   *  *  * | 0  0  0  0   2  0  2  0  0  0 0  0 | 0  1  0  1 0 0  2
sefa( os3os .. ..&#x ) |  1  2 |  0   2  0  0  1   0 |  *  *  * 192   *  *  *  *   *  *  * | 0  1  0  0   1  1  0  0  0  0 0  0 | 0  1  1  0 0 0  1
sefa( .. os3os ..&#x ) |  1  2 |  0   2  0  0  0   1 |  *  *  *   * 192  *  *  *   *  *  * | 0  0  1  0   1  0  0  1  0  0 0  0 | 0  1  0  0 1 0  1
sefa( .. .. os4xo&#x ) |  2  2 |  1   2  0  1  0   0 |  *  *  *   *   * 96  *  *   *  *  * | 0  0  0  1   0  0  1  1  0  0 0  0 | 0  0  0  1 1 0  1
      .s3.s .. ..      |  0  3 |  0   0  0  0  3   0 |  *  *  *   *   *  * 32  *   *  *  * | 0  2  0  0   0  0  0  0  2  0 0  0 | 0  2  1  0 0 1  0
      .. .s3.s ..      |  0  3 |  0   0  0  0  0   3 |  *  *  *   *   *  *  * 64   *  *  * | 0  0  1  0   0  0  0  0  1  0 1  0 | 0  1  0  0 1 1  0
sefa( .s3.s3.s ..    ) |  0  3 |  0   0  1  0  1   1 |  *  *  *   *   *  *  *  * 192  *  * | 0  0  0  0   1  0  0  0  1  0 0  1 | 0  1  0  0 0 1  1
sefa( .s .2 .s4.o    ) |  0  3 |  0   0  2  1  0   0 |  *  *  *   *   *  *  *  *   * 96  * | 0  0  0  0   0  0  1  0  0  1 0  1 | 0  0  0  1 0 1  1
sefa( .. .s3.s4.o    ) |  0  3 |  0   0  0  1  0   2 |  *  *  *   *   *  *  *  *   *  * 96 | 0  0  0  0   0  0  0  1  0  0 1  1 | 0  0  0  0 1 1  1
-----------------------+-------+---------------------+-------------------------------------+------------------------------------+------------------
      .. o.3o.4x.      |  8  0 | 12   0  0  0  0   0 |  6  0  0   0   0  0  0  0   0  0  0 | 8  *  *  *   *  *  *  *  *  * *  * | 1  0  0  0 1 0  0  cube
      os3os .. ..&#x   |  1  3 |  0   3  0  0  3   0 |  0  0  0   3   0  0  1  0   0  0  0 | * 64  *  *   *  *  *  *  *  * *  * | 0  1  1  0 0 0  0  tet
      .. os3os ..&#x   |  1  3 |  0   3  0  0  0   3 |  0  0  0   0   3  0  0  1   0  0  0 | *  * 64  *   *  *  *  *  *  * *  * | 0  1  0  0 1 0  0  tet
      .. .. os4xo&#x   |  4  2 |  4   4  0  1  0   0 |  1  2  0   0   0  2  0  0   0  0  0 | *  *  * 48   *  *  *  *  *  * *  * | 0  0  0  1 1 0  0  trip
sefa( os3os3os ..&#x ) |  1  3 |  0   3  1  0  1   1 |  0  0  1   1   1  0  0  0   1  0  0 | *  *  *  * 192  *  *  *  *  * *  * | 0  1  0  0 0 0  1  tet
sefa( os3os .2 xo&#x ) |  2  2 |  1   4  0  0  1   0 |  0  2  0   2   0  0  0  0   0  0  0 | *  *  *  *   * 96  *  *  *  * *  * | 0  0  1  0 0 0  1  tet
sefa( os .2 os4xo&#x ) |  2  3 |  1   4  2  1  0   0 |  0  1  2   0   0  1  0  0   0  1  0 | *  *  *  *   *  * 96  *  *  * *  * | 0  0  0  1 0 0  1  squippy
sefa( .. os3os4xo&#x ) |  2  3 |  1   4  0  1  0   2 |  0  1  0   0   2  1  0  0   0  0  1 | *  *  *  *   *  *  * 96  *  * *  * | 0  0  0  0 1 0  1  squippy
      .s3.s3.s ..      |  0 12 |  0   0  6  0 12  12 |  0  0  0   0   0  0  4  4  12  0  0 | *  *  *  *   *  *  *  * 16  * *  * | 0  1  0  0 0 1  0  ike
      .s .2 .s4.o      |  0  4 |  0   0  4  2  0   0 |  0  0  0   0   0  0  0  0   0  4  0 | *  *  *  *   *  *  *  *  * 24 *  * | 0  0  0  1 0 1  0  tet
      .. .s3.s4.o      |  0 12 |  0   0  0  6  0  24 |  0  0  0   0   0  0  0  8   0  0 12 | *  *  *  *   *  *  *  *  *  * 8  * | 0  0  0  0 1 1  0  ike
sefa( .s3.s3.s4.o    ) |  0  4 |  0   0  2  1  1   2 |  0  0  0   0   0  0  0  0   2  1  1 | *  *  *  *   *  *  *  *  *  * * 96 | 0  0  0  0 0 1  1  tet
-----------------------+-------+---------------------+-------------------------------------+------------------------------------+------------------
      o.3o.3o.4x.      | 16  0 | 32   0  0  0  0   0 | 24  0  0   0   0  0  0  0   0  0  0 | 8  0  0  0   0  0  0  0  0  0 0  0 | 1  *  *  * * *  *  tes
      os3os3os ..&#x   |  1 12 |  0  12  6  0 12  12 |  0  0  6  12  12  0  4  4  12  0  0 | 0  4  4  0  12  0  0  0  1  0 0  0 | * 16  *  * * *  *  ikepy
      os3os .2 xo&#x   |  2  3 |  1   6  0  0  3   0 |  0  3  0   6   0  0  1  0   0  0  0 | 0  2  0  0   0  3  0  0  0  0 0  0 | *  * 32  * * *  *  pen
      os .2 os4xo&#x   |  4  4 |  4   8  4  2  0   0 |  1  4  4   0   0  4  0  0   0  4  0 | 0  0  0  2   0  0  4  0  0  1 0  0 | *  *  * 24 * *  *  {4} || tet
      .. os3os4xo&#x   |  8 12 | 12  24  0  6  0  24 |  6 12  0   0  24 12  0  8   0  0 12 | 1  0  8  6   0  0  0 12  0  0 1  0 | *  *  *  * 8 *  *  cube || ike
      .s3.s3.s4.o      |  0 96 |  0   0 96 48 96 192 |  0  0  0   0   0  0 32 64 192 96 96 | 0  0  0  0   0  0  0  0 16 24 8 96 | *  *  *  * * 1  *  sadi
sefa( os3os3os4xo&#x ) |  2  4 |  1   6  2  1  1   2 |  0  2  2   2   2  1  0  0   2  1  1 | 0  0  0  0   2  1  1  1  0  0 0  1 | *  *  *  * * * 96  squippypy

It somehow is related to the formerly discussed ico || sadi, as tes can be vertex inscribed into a subset of the ico vertices. Thus for these remaining 16 vertices the ikepies, connecting these to the flat facets of sadi, do remain in place here as well. But for the other 8 icosahedral facets a much more interesting segmentochoron comes in as replacement: those are now of the form cube || ike!

--- rk

[quickfur]

[...]
Well, then you might be interested in that further, likewise degenerate, partially snubbed lace prism os3os3os4xo&#x too, i.e. in tes || sadi. It likewise will have the height = 0. And so comes out rather as an other decomposition of sadi into several bits around a central tesseract.
[...]
It somehow is related to the formerly discussed ico || sadi, as tes can be vertex inscribed into a subset of the ico vertices. Thus for these remaining 16 vertices the ikepies, connecting these to the flat facets of sadi, do remain in place here as well. But for the other 8 icosahedral facets a much more interesting segmentochoron comes in as replacement: those are now of the form cube || ike!
[...]

Wow. The appearance of cube||ike here is pretty awesome!! So we can also cut the 600-cell into CRF pieces that include a central tesseract (just add 24 icosahedral pyramids). Cool stuff!

I didn't look at the incmat carefully yet, but since we can make a 24-cell by augmenting the tesseract with 8 cube pyramids, does that mean the sadi||ico decomposition is reducible to this one (by cutting the ico into 1 tesseract + 8 cube pyramids)? Or would it give rise to a different decomposition that also has a central tesseract?

[Klitzing]

Sure those are related as those share the same space. But several of its "bits" are dissected differently. E.g. we use raps (tet || oct) in one case and only tet || (diametral) {4} in the other. Also, whereas octpies are used in one case, their halves, i.e. squippypies are used in the other. And for sure: those cube || ike are subdivided in the other one accordingly.

Thus: using both subdivisions simultanuously would provide a further dissection of sadi with central tes, but using much more and much smaller "bits".

--- rk

[wendy]

A good deal of the {3,3,5} segmentotopes end up as slices of {5,3,3,5/2}. You knew that. There's a flat lace tower that runs through þirteen of the fifteen uniforms of {3,3,5}.

[student91]

I was recently thinking about cartesian products, and came up with an interesting idea.
a cartesian product can be seen as putting polytopes on the vertices of another polytope, and then connecting them.
when thinking about it this way, you might wonder what would happen if you'd put different polytopes on the vertices of the "base" polytope. like both squares and octagons on a pentagonal cupola as base. you can choose to put either octagons or squares on the decagon, and you can choose to put either octagons or squares on the pentagon. This results in 4 "new" polytera: the cartesian product of a pentagonal cupola and a square (both the pentagon and the decangon have squares), the cartesian product of a pentagonal cupola and a octagon (both the pentagon and the decagon have octagons), a (4,10)-duoprism||(8,5)-duoprism (decagon has squares, pentagon has octagons), and a (8,10)-duoprism||(4,5)-duoprism (decagon has octagons, pentagon has squares).
a certain set of sechmentochora can also be seen in this way, namely the set of bicupolic rings,the prisms of the cupolas and the infinite families of sechmentochora.
the bicupolic rings and 4.174 and 4.175 are placed on a triangle: xPo xPx xPo (orthobicupolic ring), xPo xPx oPx (gyrobicupolic ring), xPx xPo xPx (magnabicupolic ring), xPo xPo oPx (K4.174) and xPo xPo xPo (K4.175).
The prisms of cupolas and K4.176 and K4.177 are placed on a square: xPx xPo xPo xPx (cupolic prism) xPo oPx oPx xPo (K4.176) and xPo xPo xPo xPo (K4.177).
I wondered why the squares are always divided into two pairs of the same kind, and not altered (always like xyyx instead of xyxy). I noted that the distance between x and x was 1, and so was the distance between y and y, and the distance between y and x was equal to the height of the corresponding sechmentotope. this ment that xyyx looked somewhat like a rectangle, and xyxy would look somewhat like a rhomb. furthermore the diagonals of the romb has to be equal to 1 in order to fill the rest of the polytope with CRF cells. this means that xy should be sqrt(1/2). this is true for the digonal antiprism (tetahedron), and hence oxox2xoxo&#xr gives a CRF-polytope (in fact it's the demicube, or the 16-cell). Other sechmentochora with height sqrt(1/2) are the square pyramid and the square cupola, hence, I think oxox4oooo&#xr and oxox4xxxx&#xr might give other CRF's as well.

student91

[wendy]

When you make something like xoxo4xxxx&xr, the xoxo produces edges of length q. You have to consider the diagonals.

[Klitzing]

A good deal of the {3,3,5} segmentotopes end up as slices of {5,3,3,5/2}. You knew that. There's a flat lace tower that runs through þirteen of the fifteen uniforms of {3,3,5}.

I knew that? - No, at least I cannot remember.
Which geometry does that one belong to? Would it be finite at all?
(At the moment I cannot access my copy of "Regular Polytopes"; and "Regular Complex Polytopes" I've never got a fetch...)
So please elaborate!
--- rk

[Klitzing]

I was recently thinking about cartesian products, and came up with an interesting idea.
a cartesian product can be seen as putting polytopes on the vertices of another polytope, and then connecting them.
when thinking about it this way, you might wonder what would happen if you'd put different polytopes on the vertices of the "base" polytope. like both squares and octagons on a pentagonal cupola as base. you can choose to put either octagons or squares on the decagon, and you can choose to put either octagons or squares on the pentagon. This results in 4 "new" polytera: the cartesian product of a pentagonal cupola and a square (both the pentagon and the decangon have squares), the cartesian product of a pentagonal cupola and a octagon (both the pentagon and the decagon have octagons), a (4,10)-duoprism||(8,5)-duoprism (decagon has squares, pentagon has octagons), and a (8,10)-duoprism||(4,5)-duoprism (decagon has octagons, pentagon has squares).
a certain set of sechmentochora can also be seen in this way, namely the set of bicupolic rings,the prisms of the cupolas and the infinite families of sechmentochora.
the bicupolic rings and 4.174 and 4.175 are placed on a triangle: xPo xPx xPo (orthobicupolic ring), xPo xPx oPx (gyrobicupolic ring), xPx xPo xPx (magnabicupolic ring), xPo xPo oPx (K4.174) and xPo xPo xPo (K4.175).
The prisms of cupolas and K4.176 and K4.177 are placed on a square: xPx xPo xPo xPx (cupolic prism) xPo oPx oPx xPo (K4.176) and xPo xPo xPo xPo (K4.177).
I wondered why the squares are always divided into two pairs of the same kind, and not altered (always like xyyx instead of xyxy). I noted that the distance between x and x was 1, and so was the distance between y and y, and the distance between y and x was equal to the height of the corresponding sechmentotope. this ment that xyyx looked somewhat like a rectangle, and xyxy would look somewhat like a rhomb. furthermore the diagonals of the romb has to be equal to 1 in order to fill the rest of the polytope with CRF cells. this means that xy should be sqrt(1/2). this is true for the digonal antiprism (tetahedron), and hence oxox2xoxo&#xr gives a CRF-polytope (in fact it's the demicube, or the 16-cell). Other sechmentochora with height sqrt(1/2) are the square pyramid and the square cupola, hence, I think oxox4oooo&#xr and oxox4xxxx&#xr might give other CRF's as well.

student91

The main point here, as far as I can see, is that you just reverse the order of operations. In fact you'd like to build duoprisms and then stack those as bases of lace prisms (or segmentotopes) atop of each other. By reversion of operations, this equates to using one face type each, build the lace prism of those first, and thereafter attach the other face type each to the top, resp. bottom base polygon of that lace prism.

Even so this surely can be viewed this way in several of the known cases, it would be a rather cumbersome task in general. In fact, you would have to set up kind a theorem, restricting the possibilities in advance to those which finally come out to work, i.e. connect correctly. That is, if you could settle those restrictions, and would manage to formulate those not only for the known cases, but for arbitrary dimension, then your idea surely would be apreciated! Providing then an additional way to get an access to those resulting figures.

Sure, meanwhile it at most could serve in the way of a conjecture: One might think about individual outcomes, being derived in your new way, but allways will have to check them additionally in the so far known one.

--- rk

[wendy]

{5,3,3,5/2} is a plain old euclidean tiling in the same way that {5,10/3} gives penroses and rescalinf-symmetries.

It's just 4d, and the pentagonal version of E8. {5/2,5,A} is the 3d version.

[student91]

EDIT: this post is full of bullshit. my next post will be bullshit-free (I hope)

The main point here, as far as I can see, is that you just reverse the order of operations. In fact you'd like to build duoprisms and then stack those as bases of lace prisms (or segmentotopes) atop of each other. By reversion of operations, this equates to using one face type each, build the lace prism of those first, and thereafter attach the other face type each to the top, resp. bottom base polygon of that lace prism.

this is indeed true for my example (and for most of the cases, to be honest), but this way of looking at it ables you to place different polytopes on irregular johnson solids as well (e.g. the snub disphenoid)

Even so this surely can be viewed this way in several of the known cases, it would be a rather cumbersome task in general. In fact, you would have to set up kind a theorem, restricting the possibilities in advance to those which finally come out to work, i.e. connect correctly. That is, if you could settle those restrictions, and would manage to formulate those not only for the known cases, but for arbitrary dimension, then your idea surely would be apreciated! Providing then an additional way to get an access to those resulting figures.

Sure, meanwhile it at most could serve in the way of a conjecture: One might think about individual outcomes, being derived in your new way, but allways will have to check them additionally in the so far known one.

--- rk

I already thought of some of the restrictions, but unfortunately there are some exceptions to those restrictions.
the way I looked at placing multiple topes at the same base, is first saying what tope would come on what vertex, next you adjust the length of the edges corresponding to the length of the sechmentotope (so if i'd distribute triangles and hexagons over a snub disphenoid, the edges connecting a hexagon with a triangle would get lenght sqrt(2/3), and the edges connecting two of a kind with each other (wich gives a prism) would have length 1)
next I'd build this "new" polytope (so a modified snub disphenoid), check wether it's convex, and if that's true, I could do a cartesian product-like construction to get the final polytope.
now those restrictions:
a triangle can have any polytopes distributed over it's vertices.
a square can only have distributed polytopes over it's vertices in a xyyx-fashion
a pentagon and up can have only the same polytopes on it's vertices.
the time I tiped my other post, I realized those restrictions do not always suffice, as the demicube was possible to write as oxox2xoxo&#xr, meaning it distributed digons over a square in a xyxy-fashion. the connections of those digons go diagonally across the square, so I had to draw a square with diagonals. now the edges have x2o||o2x and the diagonals have x2o||x2o and o2x||o2x. the height of o2x||x2o is equal to sqrt(1/2), and the height of x2o||o2x is 1(=height prism). a square drawn with edge-length sqrt(1/2) give diagonals of 1, which means oxox2xoxo&#xr is possible. so now I could write a more general restriction for squares (I should be calling them quadiliterals instead of squares now): a quadiliteral should be flat(2-dimensional) with the given connections and the given heights of those connections.
It is easily seen that quadiliterals that do not look like xyyx are quite hard to find, and so we could say that in most cases something other than xxxx or xyyx is not possible.
EDIT:This assumption is where I gotwrong(I guess)->for three-dimensional base-polytopes it's quite easy: it should consist of possible faces(see above), and in order to be CRF, the "adjusted base" (i.e. the base with adjusted edge-lenghts) should be convex.
for four(and up)-dimensional base-polytopes it can be possible that the adjusted base-polytope is impossible. In fact, I think it's only possible if you scale the whole thing down (which is wat you do is you distribute orthogonal digons over a hypercube, in order to get a demicube),or change the height of a sechmentochoron (wich would yield in other sechmentotopes, and as you said, this isn't be the best way of looking for those)
so three-dimensional irregular johnson-solids (i.e.the last few) are the base-polytopes I think I should be looking for. let's for example take the snub disphenoid. the snub disphenoid can be written as xo(s)o2o(s)ox&#xt where (s) is a specific distance. As this is completely build up of triangles (i.e. it's a deltahedron), we can choose to alter it anyhow. let's choose to distribute p-gons and dual p-gons over the snub disphenoid. we can choose to have every .2. have both a p-gon and a dual p-gon, so we get
(xoPox&#x)||orthogonal (oxPxo&#(t) ) ||orth. (xoPox&#(t) ) ||orth. (oxPxo&#x ). (t) is some value other than (s). This polyteron can't be found through sechmentochora-stacking. (EDIT: those aren't really orthogonal, they're off with a slight angle)
In the same way we might search for polychora, but then we'd use 1-dimensional things instead of 2-dimensional,so we'duse points and lines. we will get:
(xo&#x )||orth. (xo&#(v) )|| orth. (ox&#(v) )|| orth. (ox&#x ). I think this is yet another new CRF-polychoron.

When you make something like xoxo4xxxx&xr, the xoxo produces edges of length q. You have to consider the diagonals.

I think, because the base-polytope is a square with edge-lenght sqrt(1/2), these edges would be of length q*srqt(1/2)=1, I chose xoxo4xxxx&#xt because xo4xx&#x has height sqrt(1/2), giving the "base-square" a diagonal of length 1.





Besides all this,last night I came up with another idea, while thinking of ursachora (I think ursachora are awesome)
ursachora are based on the oox3xfo&#xt-construction of a trid. icosahedron.
I thought in the same way we could make polytopes based on the oox3xux&#xt-construction of a truncated tetahedron (u being 2), so we'd get ooo3oox3xux&#xt for a tetahedral thingy (comes out to be the same as x3x3o3o), xxx3oox3xux&#xt for a expanded tetahedral thingy, and so on.

[Klitzing]

{5,3,3,5/2} is a plain old euclidean tiling in the same way that {5,10/3} gives penroses and rescalinf-symmetries.

It's just 4d, and the pentagonal version of E8. {5/2,5,A} is the 3d version.

Okay, I see. So I take for granted, that those all have the right intrinsic curvature to be euclidean. (At least {5,10/3} is most easily recognized to bow to that condition.) But OTOH, those probably all would provide ultimately dense modules. This is why those usually were discarded.

--- rk

[Klitzing]

{5,3,3,5/2} is a plain old euclidean tiling in the same way that {5,10/3} gives penroses and rescalinf-symmetries.

It's just 4d, and the pentagonal version of E8. {5/2,5,A} is the 3d version.

Okay, I see. So I take for granted, that those all have the right intrinsic curvature to be euclidean. (At least {5,10/3} is most easily recognized to bow to that condition.) But OTOH, those probably all would provide ultimately dense modules. This is why those usually were discarded.

--- rk

Hmm, looked just into that 4D tesselation, you proclaimed: {5,3,3,5/2}.

It looks to me rather that you got it the wrong way: it should be rather {3,3,5,5/2}, as hi = {5,3,3} has a dihedral angle of 144 degrees at its pentagons between the dodecahedra. That is, you have to read it like this: 3 3 5 <- | -> 5/2. Cf. also the opposite version: 3 <- | -> 3 5 5/2, i.e. fix = {3,5,5/2}, which has a dihedral angle of 120 degrees at its triangles between the icosahedra.
Sorry for that. Wendy you were right. It was my ad hoc reading of the symbols which i got wrong! It would have be rather -> 5 3 3 | -> 5/2, i.e that dihedral angle around the pentagon is (still) 144 degrees, which is 5/2 of the full circuit (360 degrees). The correctnes of this reading (i.e. the changed placement of pointers) can be proved by the corresponding application to the well-known tesseractic honeycomb {4,3,3,4}: -> 4 3 3 | -> 4, i.e. the dihedral angle of the tesseract between the cubes around their squares surely is 90 degrees. Thus in euclidean space there can be 4 (not 3) of those be placed to complete a full circuit. - In contrast: the hexadecachoral honeycomb {3,3,4,3} would then evaluate as -> 3 3 4 | -> 3, and indeed: the dihedral angle in the hex at the triangle is known to be 120 degrees, i.e. there can be 3 (not 4) in a full circuit.

On the other hand, the dihedral angle in ex = {3,3,5} at the triangles between the tetrahedra evaluates to arccos(-(1+3rt5)/8) = 164.478 degrees, and thus is not rational.
(The deleted part here would not be wrong but simply is irrelevant now.) As 164.478 > 144 it follows that for {3,3,5,5/2} we would have -> 3 3 5 | -> 5/2 and therefore will have to multiply that dihedral angle by 5/2, which accordingly evaluates as more than a single circuit. And indeed, {3,3,5,5/2} is already known to belong to hyperbolic curvature. - This once more disproves my above (now deleted) snap decision.

Btw. the other, now deleted example shows, done correctly, that {3,5,5/2,3} likewise would be euclidean!

--- rk

[wendy]

In 5v, {5,3,3,5/2}, {5/2,5,3,5/2}, {5,5/2,5,5/2}, {5,3,5/2,5} and {3,5,5/2,3} are regular, and {3,3,5/2:3,3}, {3,3/2,5:3,3} are semiregular.

The margin-angle of the {3,3,5} is indeed irrational, as it consists of unbalanced oblique factors, and also the wrong power for the underlying prime (19). Either is sufficient.

The other group that provides some interest is 5bb, consists of a non-wythoff group corresponding to two equal size 5q groups (3,3,A,B), at equal sizes. In {3,4,3,3} the ratios are 1:q.

[Klitzing]

Okay, so we get some first rather easy degenerate lace prisms (zero height) from those:

• (3D) dodecadodecahedral pyramid oo5ox5/2oo&#x with cells being: 12x peppy + 12x stippy + 1x did
• (3D) great-dodecahedral cupola xx5ox5/2oo&#x with cells being: 1x gad + 12x pecu + 12x stippy + 1x tigid
• (4D) great-grand-hecatonicosachoral pyramid ox5oo5/2oo3oo&#x: 120x gadpy + 1x gaghi
• (4D) small-stellated-hecatonicosachoral pyramid ox5/2oo5oo3oo&#x: 120x sissidpy + 1x sishi
• (4D) small-ditrigonary-hexacosihecatonicosachoral pyramid ox3oo3oo3oo5/2*b&#x: 600x pen + 120x gikepy + 1x sidtixhi
• (4D) great-ditrigonary-hexacosihecatonicosachoral pyramid ox3oo3oo3/2oo5*b&#x: 600x pen + 120x ikepy + 1x gidtixhi

--- rk

[student91]

Cleared all the bullshit out of my previous post, and now I can introduce you to:

the sechmentotopical product (naming suggestions?)

It works as follows:
1. take two sechmentotopes.(wedges and pyramids etc. are also allowed)
2. "multiply" (e.g. take the cartesian product) the top of sechmentotope 1 with the top of sechmentotpe 2, and the bottom of sech.1 with the bottom of sech.2
3. put those products on top of each other. the height of the new polytope will be sqrt( [height of sech.1]² + [height of sech.2]² -1). If this height is greater than 0, the new polytope exists.

now what has this to do with my previous post?. In my previous post I was talking about squares with an xyxy distribution of polytopes. none of those came out to be 4-dimensional CRF's. this was, because the things I proposed were degenerate polytera! To be more specific: they were sechmentotopical products of a digonal antiprism and a square pyramid respectively a square cupola. (their height evaluates to sqrt(1/2+1/2-1)=0, so they're degenerate)
The first two restrictions in my previous post still are true: a triangle can be altered anyhow, a square only in an xyyx-fashion (then it's the sech.prod. of a digonal prism and a x||y-polytope) the triangle in the first restriction can be replaced with any simplex. (this way you can see sechmentotopes are alternations of 1-simplices (except for the highlighted cases in my first post about this) ). those simplices keep bothering me, because I can't find a way to "tame" them (i.e. put them in a system of operations)

but aren't we forgetting something??
yes we are!!, we haven't enumerated the 4-CRF-polytopes yet!!
luckily I gave another motivation in my previous post: new CRF-polychora!!
we have: ooo5oox3xux&#xt xxx5oox3xux&#xt ooo4oox3xux&#xt xxx4oox4xux&#xt xxx3oox3xux&#xt

student91

[Klitzing]

but aren't we forgetting something??
yes we are!!, we haven't enumerated the 4-CRF-polytopes yet!!
luckily I gave another motivation in my previous post: new CRF-polychora!!
we have: ooo5oox3xux&#xt xxx5oox3xux&#xt ooo4oox3xux&#xt xxx4oox4xux&#xt xxx3oox3xux&#xt

Sorry for being a party pooper, but those have been known already.
Cf. e.g. http://bendwavy.org/klitzing/explain/johnson.htm#axials.
--- rk