pxoPoxp&#xt and friends.

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

pxoPoxp&#xt and friends.

Postby wendy » Wed Jun 26, 2013 6:53 am

This is a new class of non-wythoff uniforms i discovered. It's quite startling, really.

We start x4o4o4x. This has a vertex-figure qo4oq&qt, a square antiprism of edge q. This means that we can further derive figures, such as the truncate and rectate of this, (ie xxoo!~ and oxoo!~), and more startlingly, the semiate sooo!~. The truncate and rectate are pretty ordinary, but non-wythoff just the same. The notation xxoo! is a prefix-operator that implements the general form of Conway-Hart, which converts on a flag-by-flag basis, the regular xooo into the equate of whatever is before the !. The ~ simply means as in dictionaries, the head figure, here x4o4o4x.

We turn to the sooo! form. This is formed by alternating the verticies of x4o4o4x, which has the further effect that alternating vertices are replaced by the vertex-figure. The overall vertex figure then is the lace-tower qxo4oxq&xt. This means that there are eight square-antiprisms, eight tetra (s4o3o) and two s4o4o at each vertex. The vertex figure is essentially, a rectified square-antiprism.

Of interest is the middle section of this: x4x, This means that it contains a {3,8}. Now, this x4x is equatorial on the vf, and no further octagons cross the vertex figure, so the half-vertex is then to belong to a laminatope with {3,8} etchings (ie bounded by flat {3,8}. This means that we can do several things here.

1. We can rotate the two halves, to get eg a vertex-figure qxq4oxo&xt, which gives the equate of the hexagonal close-pack.

2. We can replace one half of the vertex-figure by a x4x3o8o, which likewise has flat x3o8o. This is similar to elongating the oct-tet by a layer of triangular prisms. However, the thing is not a prism, but a truncated cube. No problems. This is the first implementation of a figure that seems to need the modified 'atop' lacing, ie "q4o || x4x |k| o4o".

3. We can do both, to get the equate of an elongated hexagonal close-pack.

It turns out, that the general class here is x4oPo4x, which gives a vertex figure of pxo4oxp&xt. When x=4 or x=3, one gets the three instances above. One can evaluate x=2, which gives rxo2oxr&xt, with opposite vertices of an octahedron replaced by linekins. But while it gives equates of the three above, they all fold in together to give a simple octahedron.

With x4o5o4x, one can see the necessary vertex figure, by placing a unit ID o3x5o on its pentagon face, and drawing its tower. This gives, eg ofxox5xoxfo&xt. Dropping the top and bottom pentagon gives fxo5oxf&xt.

Early calculations on x4o6o4x, gives hxo6oxh&xt, which can be summounted by a pyramid of slope-lacing h. However, the day has been too noisy to derive this further.
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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Tue Jul 09, 2013 4:54 pm

Wendy recently (2013/6/26) came up with some interest in x4oPo4x. So let's have a look, whether we can follow her explanatory notes.

In itself x4oPo4x (P=2,3,4,5,...) is some 4D uniform polytope with localized structure within according 3D space.

P=2 is just x4o o4x, the bi-square duoprism with finite spherical space structure. The only cells here are x4o . x (cube) and x . o4x (cube again). So in fact it would be equivalent to the tesseract, just using a different coloring: 2 instead of just 1.

P=3 is then x4o3o4x, the cubical honeycomb, i.e. of euclidean space structure. Cells here are x4o3o . (cube), x4o . x (cube), x . o4x (cube), and . o3o4x (cube). So it would be equivalent to x4o3o4o, just using a different coloring: 4 instead of 1.

Her starting point was P=4, i.e. x4o4o4x, a hyperbolic structure, which uses euclidean tilings in the sense of horotopes, i.e. infinite
"bodies", which become tangential to the sphere of infinity. In fact, the cells here are x4o4o . (euclidean square tiling, here seen as horotope), x4o . x (cube), x . o4x (cube), and . o4o4x (euclidean square tiling again).

The same structure then would remain for P>4 too, just that we would use hyperbolic tilings x4oPo (at either end of the Dynkin diagram), i.e. hyperbolic order P square tilings, which are used here in a sense of bolotopes, i.e. as infinite "bodies",
which would cut the sphere of infinity.

We would have a quick look onto the vertex figure here. For non-quasiregular diagrams (those would have just 1 node ringed) we have to eleminate all but 1 ringed node, take the vertex figure therefrom, and stack figures in the sense of a lace simplex by lacing edges, which in turn are the vertex figures of the corresponding rectangle, which occurs as subdiagram of the corresponding 2 ringed nodes. Thus here, starting with x4oPo4x we have to consider x4oPo . and . oPo4x. Their vertex figures then are . qPo . resp. . oPq ..
Those are laced by the vertex figure of x . . x, i.e. by an edge of size q = sqrt(2). Thus concluding, we generally get as vertex figure of x4oPo4x the figure qoPoq&#q, i.e. a q-scaled P-gonal antiprism. Esp. for P=2 this is a (scaled) tetrahedron, resp. for P=3 a (scaled) octahedron.

Then she considers operators. xooo!~ is just the identity applied on those structures. Thus the incidence matrix of xooo!x4oPo4x too is the same as that of x4oPo4x (provided for P>2 only as else we'd have to handle degenerate elements):

Code: Select all
x4oPo4x (M=2 for P=3; resp. M->oo for P>3)

. . . .    | 12NM |    2P |   2P   2P |  2   2P  vf: sPs2s
-----------+------+-------+-----------+--------
x . . .  & |    2 | 12NMP |    2    2 |  1    3
-----------+------+-------+-----------+--------
x4o . .  & |    4 |     4 | 6NMP    * |  1    1
x . . x    |    4 |     4 |    * 6NMP |  0    2
-----------+------+-------+-----------+--------
x4oPo .  & |   4M |   2MP |   MP    0 | 6N    *
x4o . x  & |    8 |    12 |    2    4 |  * 3NMP


For P=3 (and thus M=2) we not only could insert those special numbers, but further get cubes as either one type of cells. This allows for additional symmetry here, finally resulting in x4o3o4x = x4o3o4o (up to identified color symmetries).

For the missing case P=2 we further could derive the matrix directly. But again we note that we have x4o2o4x = x4o3o3o (up to identified color symmetries), so this might be neglected here.

xxoo!~ is the truncation operator. This operator uses the central parts of the original edges, but introduces small copies of the above mentioned original vertex figure at the former vertex positions, choping off the former cells accordingly. Thus, the cells of xxoo!x4oPo4x generally would be x4xPo (at either end of the diagram), x4x3o (truncated cube) from either tetragonal prism, and sPs2s (P-gonal antiprism). - The according incidence matrix would read (again P>2):

Code: Select all
xxoo!x4oPo4x (M=2 for P=3; resp. M->oo for P>3)

24NMP |     1     2     2 |    2    2    1     3 |  1    3    1
------+-------------------+----------------------+-------------
    2 | 12NMP     *     * |    2    2    0     0 |  1    3    0
    2 |     * 24NMP     * |    1    0    1     1 |  1    1    1
    2 |     *     * 24NMP |    0    1    0     2 |  0    2    1
------+-------------------+----------------------+-------------
    8 |     4     4     0 | 6NMP    *    *     * |  1    1    0
    8 |     4     0     4 |    * 6NMP    *     * |  0    2    0
    P |     0     P     0 |    *    * 24NM     * |  1    0    1
    3 |     0     1     2 |    *    *    * 24NMP |  0    1    1
------+-------------------+----------------------+-------------
  4MP |   2MP   4MP     0 |   MP    0   4M     0 | 6N    *    *  x4xPo
   24 |    12     8    16 |    2    4    0     8 |  * 3NMP    *  x4x3o
   2P |     0    2P    2P |    0    0    2    2P |  *    * 12NM  sPs2s


In the euclidean case, i.e. P=3 (and thus M=2) we could insert those numbers and additionally add further symmetries (P <-> 3), but the above equality would derive the outcome directly: xxoo!x4o3o4x = xxoo!x4o3o4o = x4x3o4o.

For the finite case we similarily have xxoo!x4o2o4x = xxoo!x4o3o3o = x4x3o3o.

oxoo!~ is the rectification operator. This operator reduces the original edges down to zero (Wendy's r-edge). I.e. the former cells would become correspondingly rectified. The former vertex positions would use slightly larger copies of the above mentioned vertex figures of the original. Therefore the cells of oxoo!x4oPo4x generally would be o4xPo (at either end of the diagram), o4x3o (cuboctahedron) from either tetragonal prism, and sPs2s (P-gonal antiprism) again. - The according incidence matrix would read (again P>2):

Code: Select all
oxoo!x4oPo4x (M=2 for P=3; resp. M->oo for P>3)

12NMP |     4     4 |    2    2    2     6 |  1    3    2
------+-------------+----------------------+-------------
    2 | 24NMP     * |    1    0    1     1 |  1    1    1
    2 |     * 24NMP |    0    1    0     2 |  0    2    1
------+-------------+----------------------+-------------
    4 |     4     0 | 6NMP    *    *     * |  1    1    0
    4 |     0     4 |    * 6NMP    *     * |  0    2    0
    P |     P     0 |    *    * 24NM     * |  1    0    1
    3 |     1     2 |    *    *    * 24NMP |  0    1    1
------+-------------+----------------------+-------------
  2MP |   4MP     0 |   MP    0   4M     0 | 6N    *    *  o4xPo
   12 |     8    16 |    2    4    0     8 |  * 3NMP    *  o4x3o
   2P |    2P    2P |    0    0    2    2P |  *    * 12NM  sPs2s


For P=3 we clearly would get again oxoo!x4o3o4x = oxoo!x4o3o4o = o4x3o4o.
For P=2 similarily oxoo!x4o2o4x = oxoo!x4o3o3o = o4x3o3o.

sooo!~ finally, as she states, is meant to be the operator of alternated vertex faceting. (The term faceting already implies that the individual truncation would be at the level of the neighbouring vertices each.) Thus the tetragonal prisms (cubes) would become q-scaled tetrahedra. The former x4oPo poly-/horo-/bolohedra clearly become s4oPo, which in fact are nothing but qPoPo, i.e. mere q-edges for P=2, (scaled) tetrahedra for P=3 (as alternated faceting of the corresponding unit cube), and s4o4o becomes a rotated (and scaled) x4o4o again. As additional cells here again the full q-scaled original vertex figure, i.e. the P-gonal antiprism qoPoq&#q, comes in.

Obviously all edges here are of the same size (q), thus those snubbed figures are uniform again! Coming back to unit size we'd have for cells thus: x3o3o, xPoPo, and sPs2s. - The according incidence matrix then would read (again P>2):

Code: Select all
sooo!x4oPo4x (M=2 for P=3; resp. M->oo for P>3)

6NM |   2P   2P |   2P    6P |  2   2P  2P
----+-----------+------------+------------
  2 | 6NMP    * |    2     2 |  1    1   2
  2 |    * 6NMP |    0     4 |  0    2   2
----+-----------+------------+------------
  P |    P    0 | 12NM     * |  1    0   1
  3 |    1    2 |    * 12NMP |  0    1   1
----+-----------+------------+------------
 2M |   PM    0 |   2M     0 | 6N    *   *  xPoPo
  4 |    2    4 |    0     4 |  * 3NMP   *  x3o3o
 2P |   2P   2P |    2    2P |  *    * 6NM  sPs2s


For P=3 we get more directly sooo!x4o3o4x = sooo!x4o3o4o = s4o3o4o = x3o3o *b4o.
For P=2 we get additionally sooo!x4o2o4x = sooo!x4o3o3o = s4o3o3o = x3o3o4o.

Next she consideres the vertex figure of that last one, i.e. of sooo!x4oPo4x. - Well, the element numbers are already provided within the superdiagonal elements of the first line of the above matrix: it obviously has 2P+2P vertices, 2P+6P edges, and 2+2P+2P faces. The lengths of those edges resp. the shapes of those faces would be derived as the vertex figures of the provided (uniform) hyperbolic polytope:
vf( {P} ) = p-sized edge, vf( {3} ) = x-sized edge; vf( xPoPo ) = pPo, vf( x3o3o ) = x3o, vf( sPs2s ) = xp&#x (i.e. the trapezium with edge sequence x-x-x-p).

We might derive the coresponding incidences of elements directly. But we could get a clue for that from the known euclidean case (P=3). Here the sooo!x4o3o4x just is the alternated cubical honeycomb, a.k.a. tetrahedral-octahedral honeycomb. Its vertex figure is just the cuboctahedron. - Providing that along 3-fold axial symmetry results in xxo3oxx&#xt = x3o || x3x || o3x (with unit lacings). - The generalized form (P>3) of that vertex figure thus is indeed (as Wendy proclaimed) pxoPoxp&#xt = pPo || xPx || oPp (with unit lacings). This figure could be considered topologically as a P-gonal antiprism, bulging out its equatorial section a bit (thereby folding the lacing triangles into a pair of a trapezium and a triangl each).

Would this hold true for P=2 too? - Here we first would have to consider what p then would be: In fact, p becomes the zero length edge (r), or just a vertex (o). Thus we could write here rxo2oxr&#xt = o2o || x2x || o2o, which is nothing but an octahedron, the asked for vertex figure of the hexdacachoron. Q.e.d.

Further she points out, that the equatorial section here is an xPx, being itself known as the vertex figure of the hyperbolic tiling x3o2Po = x3oPo3*a. So, yes, this tiling might fit here in the sense that 2P triangles would be incident to each vertex, which all lie within the same plane. And this hyperbolic plane would have the same curvature as the whole structure itself (because it is defined by the equator of the vertex figure). - But so far I cannot wrap my mind around whether this plane extends beyond this finite range to infinity, or rather would cut somewhere into an xPoPo.

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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Tue Jul 09, 2013 9:07 pm

Btw., just for record:
the height h of half of that verf, i.e. of pPo || xPx, generally equates to:
Code: Select all
h^2 = (c + 1/2) / (c + 1),
c = cos(pi/P)

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Re: pxoPoxp&#xt and friends.

Postby wendy » Wed Jul 10, 2013 7:40 am

More hyperbolica, although a revisit of the Coxeter snubs in s4s4o3o.

One notes by the laws of symmetry that the equality: saPsePsaAPse = saPsePsa4o = sePsa4o3o exists.

By the rule of Coxeter, se should note that there is a uniform snub of the form sPs4o[3o], for any number of 3's, this corresponding to a radial arms of the form sPs. The equality above is 'three-arms separate = two arms equal = three arms equal'.

When P=3, a=e=1, this gives the snub 24choron, and the snub 3433 tiling.

When P=4, a=e=1, this gives a corresponding tiling, whose vertex figure is best described as "x3o |q| h3o |x| o3x". It's kind of like the teddi, but it has a verf of x4x4o for the pentagons.

When P=4, one can nave also a=1,e=0, and a=0,e=1. The second of these is a wythoff polytope "x4o4oAo". The first gives rise to some interesting things, since we have o4s4o3o. This is also non-wythoffian.

Regarding Richard's comments over the xo4ox&#xt etc, one might note that the equators of the vertex-figures never cross at any instance, and thus must continue throughout. In the cubic case, this amounts to the equators giving rise to layers, being only in planes like z=2n. What's more statling is the appearence of a Z4 system in the form of a x3o8o, as the etching on these planes.
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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Wed Jul 10, 2013 12:56 pm

wendy wrote:... Regarding Richard's comments over the xo4ox&#xt ...

really that one, or rather pxoPoxp&#xt ?
... etc, one might note that the equators of the vertex-figures never cross at any instance, and thus must continue throughout. ...

Might be, so far I just do not see the clue/proof for that reasoning.
... In the cubic case, this amounts to the equators giving rise to layers, being only in planes like z=2n. What's more statling is the appearence of a Z4 system in the form of a x3o8o, as the etching on these planes.

Well, as I wrote, more generally (general P) you'd get (at least locally) a x3o(2P)o : above (and also gyratedly below) the equator you'd have P copies of P-gonal antiprisms and further P copies of tetrahedra within an alternating circuit, all showing their trigonal faces to match that equatorial structure. (The opposite incidence then would be that to the xPoPo.)

To be honest, stutling I already found the general identity s4oPo = xPoPo. Even so, it can be readily visualized directly starting 2 instances of M.Green/D.Hatch's Tyler applet, comparing the according designs of x4oPo resp. xPoPo.

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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Wed Jul 10, 2013 8:55 pm

wendy wrote:... One notes by the laws of symmetry that the equality: saPsePsaAPse = saPsePsa4o = sePsa4o3o exists. ...


Okay, your glyphes "saPsePsaAPse = saPsePsa4o = sePsa4o3o" would just compress the 3 following equation lines into one:

Code: Select all
sPsPs *bPs = sPsPs4o = sPs4o3o
sPoPs *bPo = sPoPs4o = oPs4o3o
oPsPo *bPs = oPsPo4o = sPo4o3o

(i.e. "sa" resp. "se" could independently vary between an s- and an o-node.)

The first line here is evidently true for P=2: replacing first each s- by an x-node would provide thrice the tesseract. Thus, re-substitution them then would generate thrice the hexadecachoron. (For the other lines P=2 is not defined or at least would result in degenerate figures.)

The same trick for P=3 would give thrice the truncated icositetrachoron / the rectified ico. / resp. ico itself, and so their snubs would accordingly be the same too (in fact the uniform snub ico. / a non-uniform holosnub which is the alternating vertex faceting of rico / a Grünbaumian structure which looks like a compound of ico with the edge-inscribed 3-tesseract compound).

I'd suppose that the same trick of base equivalences works for all the other P's too, though we then have joined the realm of non-compact hyperbolics already. And so far I neither have looked into all those related x-polytopes, nor in the resulting s-ones. So I can't say.

But after all, it seems that the symmetry o4o4o4o is your current favorit pet, ain't it? :)

So let's just have a quick look into what is needed here:

Code: Select all
o4x4o4o (N,M,K->oo)

. . . . | NMK |    8 |   4    8 |   4   2  vf:q-cube
--------+-----+------+----------+--------
. x . . |   2 | 4NMK |   1    2 |   2   1
--------+-----+------+----------+--------
o4x . . |   4 |    4 | NMK    * |   2   0
. x4o . |   4 |    4 |   * 2NMK |   1   1
--------+-----+------+----------+--------
o4x4o . |  2M |   4M |   M    M | 2NK   *  squat
. x4o4o |   K |   2K |   0    K |   * 2NM  squat


Code: Select all
x4o4x4o (N,M,K->oo)

. . . . | 4NMK |    2    4 |   1    4    2    2 |   2   2   1  vf:q-trip
--------+------+-----------+--------------------+------------
x . . . |    2 | 4NMK    * |   1    2    0    0 |   2   1   0
. . x . |    2 |    * 8NMK |   0    1    1    1 |   1   1   1
--------+------+-----------+--------------------+------------
x4o . . |    4 |    4    0 | NMK    *    *    * |   2   0   0
x . x . |    4 |    2    2 |   * 4NMK    *    * |   1   1   0
. o4x . |    4 |    0    4 |   *    * 2NMK    * |   1   0   1
. . x4o |    4 |    0    4 |   *    *    * 2NMK |   0   1   1
--------+------+-----------+--------------------+------------
x4o4x . |   4M |   4M   4M |   M   2M    M    0 | 2NK   *   *  squat
x . x4o |    8 |    4    8 |   0    4    0    2 |   * NMK   *  cube
. o4x4o |   2K |    0   4K |   0    0    K    K |   *   * 2NM  squat


Code: Select all
x4x4x4o (N,M,K->oo)

. . . . | 8NMK |    1    1    2 |   1    2    2    1 |   2   1   1  vf:oq3oo&#k
--------+------+----------------+--------------------+------------
x . . . |    2 | 4NMK    *    * |   1    2    0    0 |   2   1   0
. x . . |    2 |    * 4NMK    * |   1    0    2    0 |   2   0   1
. . x . |    2 |    *    * 8NMK |   0    1    1    1 |   1   1   1
--------+------+----------------+--------------------+------------
x4x . . |    8 |    4    4    0 | NMK    *    *    * |   2   0   0
x . x . |    4 |    2    0    2 |   * 4NMK    *    * |   1   1   0
. x4x . |    8 |    0    4    4 |   *    * 2NMK    * |   1   0   1
. . x4o |    4 |    0    0    4 |   *    *    * 2NMK |   0   1   1
--------+------+----------------+--------------------+------------
x4x4x . |   8M |   4M   4M   4M |   M   2M    M    0 | 2NK   *   *  tosquat
x . x4o |    8 |    4    0    8 |   0    4    0    2 |   * NMK   *  cube
. x4x4o |   4K |    0   2K   4K |   0    0    K    K |   *   * 2NM  tosquat

(where q = sqrt(2) = shortchord of square, w = sqrt(2+sqrt(2)) = shortchord of octagon)

Now let's consider the respective snubs (or better: alternated vertex-wise facetings, i.e. no rescaling intended here at the moment).

Code: Select all
o4s4o4o (N,M,K->oo)

demi( . . . . ) | NMK |    4    8 |   16    8 |   4   2   8  vf:q-co
----------------+-----+-----------+-----------+------------
      o4s . .   |   2 | 2NMK    * |    4    0 |   2   0   2
      . s4o .   |   2 |    * 4NMK |    2    2 |   1   1   2
----------------+-----+-----------+-----------+------------
sefa( o4s4o . ) |   4 |    2    2 | 4NMK    * |   1   0   1
sefa( . s4o4o ) |   4 |    0    4 |    * 2NMK |   0   1   1
----------------+-----+-----------+-----------+------------
      o4s4o .   |   M |    M    M |    M    0 | 4NK   *   *  squat
      . s4o4o   |   K |    0   2K |    0    K |   * 2NM   *  squat
sefa( o4s4o4o ) |   8 |    4    8 |    4    2 |   *   * NMK  cube

So, yes, identification of the square tiling (= squat) classes, and thus too of square clases and of edge classes, would result in the same
structure as x4o4o *b3o would describe.


Code: Select all
s4o4s4o (N,M,K->oo)

demi( . . . . ) | 2NMK |   1    4    2    2 |    8    6    4 |   2   2   1    6
----------------+------+--------------------+----------------+-----------------
      s4o . .   |    2 | NMK    *    *    * |    4    0    0 |   2   0   0    2
      s . s .   |    2 |   * 4NMK    *    * |    2    2    0 |   1   1   0    1
      . o4s .   |    2 |   *    * 2NMK    * |    2    0    2 |   1   0   1    2
      . . s4o   |    2 |   *    *    * 2NMK |    0    2    2 |   0   1   1    2
----------------+------+--------------------+----------------+-----------------
sefa( s4o4s . ) |    4 |   1    2    1    0 | 4NMK    *    * |   1   0   0    1
sefa( s . s4o ) |    3 |   0    2    0    1 |    * 4NMK    * |   0   1   0    1
sefa( . o4s4o ) |    4 |   0    0    2    2 |    *    * 2NMK |   0   0   1    1
----------------+------+--------------------+----------------+-----------------
      s4o4s .   |   2M |   M   2M    M    0 |   2M    0    0 | 2NK   *   *    *  squat
      s . s4o   |    4 |   0    4    0    2 |    0    4    0 |   * NMK   *    *  tet
      . o4s4o   |    K |   0    0    K    K |    0    0    K |   *   * 2NM    *  squat
sefa( s4o4s4o ) |    6 |   1    4    2    2 |    2    2    1 |   *   *   * 2NMK  trip

Here too all edges are of the same length (diagonals of original squares) and there is a single vertex class. Faces are all regular (triangles and squares) and cells are all uniform again. Thus this structure would be uniform itself. (Whether this even would be a Wythoffian structure eludes me in the moment.)


Code: Select all
s4s4s4o (N,M,K->oo)

demi( . . . . ) | 4NMK |    2    1    2    4 |   1    2    6    3    3 |   2   1   1    4
----------------+------+---------------------+-------------------------+-----------------
      s . s .   |    2 | 4NMK    *    *    * |   0    0    2    2    0 |   1   1   0    2
      . . s4o   |    2 |    * 2NMK    *    * |   0    0    0    2    2 |   0   1   1    2
sefa( s4s . . ) |    2 |    *    * 4NMK    * |   1    0    2    0    0 |   2   0   0    1
sefa( . s4s . ) |    2 |    *    *    * 8NMK |   0    1    1    0    1 |   1   0   1    1
----------------+------+---------------------+-------------------------+-----------------
      s4s . .   |    4 |    0    0    4    0 | NMK    *    *    *    * |   2   0   0    0
      . s4s .   |    4 |    0    0    0    4 |   * 2NMK    *    *    * |   1   0   1    0
sefa( s4s4s . ) |    3 |    1    0    1    1 |   *    * 8NMK    *    * |   1   0   0    1
sefa( s . s4o ) |    3 |    2    1    0    0 |   *    *    * 4NMK    * |   0   1   0    1
sefa( . s4s4o ) |    3 |    0    1    0    2 |   *    *    *    * 4NMK |   0   0   1    1
----------------+------+---------------------+-------------------------+-----------------
      s4s4s .   |   4M |   2M    0   4M   4M |   M    M   4M    0    0 | 2NK   *   *    *  non-uniform snasquat
      s . s4o   |    4 |    4    2    0    0 |   0    0    0    4    0 |   * NMK   *    *  regular tet
      . s4s4o   |   2K |    0    K    0   4K |   0    K    0    0   2K |   *   * 2NM    *  non-uniform snasquat
sefa( s4s4s4o ) |    4 |    2    1    1    2 |   0    0    2    1    1 |   *   *   * 4NMK  non-uniform tet

That one clearly is non-uniform per se. (I have no idea for now, whether it could be made such.)

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Re: pxoPoxp&#xt and friends.

Postby wendy » Thu Jul 11, 2013 7:00 am

I was asked a question from the Wiki editor 'tomruen', on the uniformity of the various snubs in o4o4o4o and kindred groups. A lot more of these appear to be uniform than first suspected. One uses the equities to take out nine of the figures, and then s4o4o4s was a bit of a supprise. I expected octagonal numbers, but a fully fledged x3o8o?

sa and se are as you correctly guessed, not a stack, but variations on 's', here a=e=1 is the same as using s in all positions. They are to be read as a single snub, not a double-snub (where the alternation on sa and se are independent).
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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Thu Jul 11, 2013 10:18 am

As I already had pointed out in my snub paper, contrary to their historical intend the snub notation by effect is more related with the actual alternating faceting operation (applied to the corresponding Wythoffian starting figure), than to the therefrom derived uniform variant! This is because the former does exist in any circumstance, while the existance of the latter one depends on the number of inequivalent edge classes, which (by generation of mere alternated faceting) still differ in size, and the degrees of freedom, which can be used for re-scaling.

Even so, for your new pet, the group o4o4o4o, this behaviour is rather exceptional: Here quite often only diagonals of former squares will be in use. So, independent of the number of such edge classes, if all have the same size by construction, i.e. a priori, then just a global scaling (q -> x) would be needed to derive the final uniform snub variant, which surely is applicable generally.

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Re: pxoPoxp&#xt and friends.

Postby wendy » Thu Jul 11, 2013 10:50 am

This new crowd form 'coxeter snubs'. Coxeter noted that in xPo2Qo....o, you could put arrows on the edges coherently, so that you can divide them into any ratio. In essence, the x2Qo.... has verticies of alternating colours, and in xPo2Qo... you come from a black vertex and go to a white one. You can then divide such edges into any real ratio, consistently.

The vertex figure of sPs2Qo... is a pyramid, whose apex is a point, and whose base is a x... (where ... is whatever comes after the Q). Because the division is variable, one sees that the lacing can vary. So you could support a lacing of 1 or q or whatever. The figure will generate uniforms only when the base of the vertex figure is a simplex, so sPs2Qo, sPs2Qo3o, sPs2Qo3o3o all give uniforms but sPs2Qo3o4o does not (it's a octahedral pyramid).

When the base is a simplex, wythoff's construction applies, and one must have a solution where all sides are equal.

So, finding s4o4o3o, o4s4o3o, and s4s4o3o, is hardly surprising in these conditions. Using the laws of symmetry to dismantle these to the other forms is likewise unsurprising. s4s4o3o actually has a teddi-form vertex figure. It's sort of like the teddi, but the pentagon is actually formed from a dodecagon, keeping vertices at 1, 4, 6, 8, 11 oclock (where 12-oclock is up). The top is the single triangle, and the base is a cluster of 4 triangles.

The novelty is x4o4o4x, which like x4o3o4x, and other figures in this list, give a figure whose edges are all 'q', If you replace x4o3o4x to s4o3o4s, you get an oct-tet truss. However, one can suppose that the 'xy' plane is primary, and insert prisms and reflect layers, to get the three laminate tilings in 3d, (ie LB2, LPA2, LPB2). These have quite definite bands (in much the same way that bands of triangles and squares form LC1, and that you can have in 3d, LC2 and LPC2 as well). This means, that there are LA2, LB2, LPA2 and LPB2 for the q(square antiprism) as well.

The figure o4x4x4o gives a vertex figure of a simplex, with two degrees of freedom, and three kinds of edge, which typically does not have a solution. You get here as a vertex-figure qo2oq&#kt. You would have to dink it somewhat to make it work, but the raw vf does not lend itself to what is called of it. If the 'q' segment that comes from s4s4o (vf is a polygon sqsqs = pentagon), falls in the middle of the line, it should be OK, but i doubt it.

Likewise, i have resevations over s3s4o4o, which among other things, imply a square pyramid as one of the faces.
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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Fri Jul 12, 2013 3:58 pm

Klitzing wrote:
Code: Select all
s4o4s4o (N,M,K->oo)

demi( . . . . ) | 2NMK |   1    4    2    2 |    8    6    4 |   2   2   1    6
----------------+------+--------------------+----------------+-----------------
      s4o . .   |    2 | NMK    *    *    * |    4    0    0 |   2   0   0    2
      s . s .   |    2 |   * 4NMK    *    * |    2    2    0 |   1   1   0    1
      . o4s .   |    2 |   *    * 2NMK    * |    2    0    2 |   1   0   1    2
      . . s4o   |    2 |   *    *    * 2NMK |    0    2    2 |   0   1   1    2
----------------+------+--------------------+----------------+-----------------
sefa( s4o4s . ) |    4 |   1    2    1    0 | 4NMK    *    * |   1   0   0    1
sefa( s . s4o ) |    3 |   0    2    0    1 |    * 4NMK    * |   0   1   0    1
sefa( . o4s4o ) |    4 |   0    0    2    2 |    *    * 2NMK |   0   0   1    1
----------------+------+--------------------+----------------+-----------------
      s4o4s .   |   2M |   M   2M    M    0 |   2M    0    0 | 2NK   *   *    *  squat
      s . s4o   |    4 |   0    4    0    2 |    0    4    0 |   * NMK   *    *  tet
      . o4s4o   |    K |   0    0    K    K |    0    0    K |   *   * 2NM    *  squat
sefa( s4o4s4o ) |    6 |   1    4    2    2 |    2    2    1 |   *   *   * 2NMK  trip

Here too all edges are of the same length (diagonals of original squares) and there is a single vertex class. Faces are all regular (triangles and squares) and cells are all uniform again. Thus this structure would be uniform itself. (Whether this even would be a Wythoffian structure eludes me in the moment.)


There might be some further class unifications by applying an higher symmetry, resulting in just
Code: Select all
2NM |   3   6 |  12   6 |  3  2   6
----+---------+---------+----------
  2 | 3NM   * |   4   0 |  2  0   2
  2 |   * 6NM |   2   2 |  1  1   2
----+---------+---------+----------
  4 |   2   2 | 6NM   * |  1  0   1
  3 |   0   3 |   * 4NM |  0  1   1
----+---------+---------+----------
  M |   M   M |   M   0 | 6N  *   *  squat
  4 |   0   6 |   0   4 |  * NM   *  tet
  6 |   3   6 |   3   2 |  *  * 2NM  trip


It further comes out that the vertex figure here is a clear relative to the cuboctahedron:
The latter one is xox4oqo&#xt = x4o || o4q || x4o, while the vertex figure here is xox3ouo&#qt = x3o |q| o3u |q| x3o, that is, just a (streched) three-fold version of the former!

I further realized that this snub is truely a non-Wythoffian uniform. Because the symbols x3o3o (tet), x3o . x (trip), and x4o4o (squat - or any other according symbol) cannot be combined into a single 4-node diagram in such a way, that any hiding of a single node would just re-produce one of those given symbols (possibly in addition one with un-ringed nodes only).

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Re: pxoPoxp&#xt and friends.

Postby wendy » Sat Jul 13, 2013 7:09 am

Richard describes o4s4o3o in the previous post. It is not a wythoff-snub, because the symmetry is a right triangular tegum: it has six triangular faces, at right angles to each other. The face consist is q.q3o (triangular prims at alternating vertices), and o4s4o (quartic) and s4o3o (tetrahedra). This follows from the laws of symmetry.
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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Sat Jul 20, 2013 10:58 pm

Klitzing wrote:
Klitzing wrote:
Code: Select all
s4o4s4o (N,M,K->oo)

demi( . . . . ) | 2NMK |   1    4    2    2 |    8    6    4 |   2   2   1    6
----------------+------+--------------------+----------------+-----------------
      s4o . .   |    2 | NMK    *    *    * |    4    0    0 |   2   0   0    2
      s . s .   |    2 |   * 4NMK    *    * |    2    2    0 |   1   1   0    1
      . o4s .   |    2 |   *    * 2NMK    * |    2    0    2 |   1   0   1    2
      . . s4o   |    2 |   *    *    * 2NMK |    0    2    2 |   0   1   1    2
----------------+------+--------------------+----------------+-----------------
sefa( s4o4s . ) |    4 |   1    2    1    0 | 4NMK    *    * |   1   0   0    1
sefa( s . s4o ) |    3 |   0    2    0    1 |    * 4NMK    * |   0   1   0    1
sefa( . o4s4o ) |    4 |   0    0    2    2 |    *    * 2NMK |   0   0   1    1
----------------+------+--------------------+----------------+-----------------
      s4o4s .   |   2M |   M   2M    M    0 |   2M    0    0 | 2NK   *   *    *  squat
      s . s4o   |    4 |   0    4    0    2 |    0    4    0 |   * NMK   *    *  tet
      . o4s4o   |    K |   0    0    K    K |    0    0    K |   *   * 2NM    *  squat
sefa( s4o4s4o ) |    6 |   1    4    2    2 |    2    2    1 |   *   *   * 2NMK  trip

Here too all edges are of the same length (diagonals of original squares) and there is a single vertex class. Faces are all regular (triangles and squares) and cells are all uniform again. Thus this structure would be uniform itself. (Whether this even would be a Wythoffian structure eludes me in the moment.)


There might be some further class unifications by applying an higher symmetry, resulting in just
[...]

Just realised :idea: that this one, then being provided as mere incidence matrix of some greater symmetry, is nothing but (i.e. using this specific symmetry):
Code: Select all
o3o4s4o (N,M->oo)

demi( . . . . ) | 2NM |   6   3 |   6  12 |  2  3   6  vf:xox3ouo&#qt
----------------+-----+---------+---------+----------
      . o4s .   |   2 | 6NM   * |   2   2 |  1  1   2
      . . s4o   |   2 |   * 3NM |   0   4 |  0  2   2
----------------+-----+---------+---------+----------
sefa( o3o4s . ) |   3 |   3   0 | 4NM   * |  1  0   1
sefa( . o4s4o ) |   4 |   2   2 |   * 6NM |  0  1   1
----------------+-----+---------+---------+----------
      o3o4s .   |   4 |   6   0 |   4   0 | NM  *   *  tet
      . o4s4o   |   M |   M   M |   0   M |  * 6N   *  squat
sefa( o3o4s4o ) |   6 |   6   3 |   2   3 |  *  * 2NM  trip

[...]
It further comes out that the vertex figure here is a clear relative to the cuboctahedron:
The latter one is xox4oqo&#xt = x4o || o4q || x4o, while the vertex figure here is xox3ouo&#qt = x3o |q| o3u |q| x3o, that is, just a (streched) three-fold version of the former!

I further realized that this snub is truely a non-Wythoffian uniform. Because the symbols x3o3o (tet), x3o . x (trip), and x4o4o (squat - or any other according symbol) cannot be combined into a single 4-node diagram in such a way, that any hiding of a single node would just re-produce one of those given symbols (possibly in addition one with un-ringed nodes only).

--- rk

which as such is nothing new here :\ as such, Wendy already gmentioned that, cf. the last equation sign in the central row of:
Code: Select all
sPsPs *bPs = sPsPs4o = sPs4o3o
sPoPs *bPo = sPoPs4o = oPs4o3o
oPsPo *bPs = oPsPo4o = sPo4o3o


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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Thu Aug 08, 2013 8:52 pm

"... and friends"

Well, here is a friend.
(It was so quiet lately, so I dreamed that one up.)

Start with a cube as a tetragonal prism. When being rectified (vertices being chopped off down to the center of the edges) then you'd end with a cuboctahedron: xox4oqo&#xt (or, scaled up by q: qoq4ouo&#qt). - Now increase the axial symmetry to any larger even number. Let x(2Q) be the shortchord length of a 2Q-gon, then you'd get a figure according to x(2Q),o,x(2Q)-2Q-o,u,o&#qt or, provided in my atop-notation, x(2Q)-2Q-o |q| o-2Q-u |q| x(2Q)-2Q-o. (It uses again 2Q lateral squares, but those reside now on their tips instead!)

Okay, here it is how you'd get involved with such a figure.

Start with the following Dynkin diagram:
Code: Select all
*a   o----4---(o)  *b
     |  \      |
     4    Q    4
     |      \  |
*d  (o)---4----o   *c

or in linearized form: o4x4o4x4*aQ*c. (Here Q being any integer greater 1.) This hypercompact (resp. paracompact for Q=2) hyperbolic polychoron has for cells the order 4 square tiling (used as horohedra with euclidean curvature) and order 2Q square tilings (used as bolohedra with hyperbolic curvature themselves). Its vertex figure is just a uniform 2Q-prism. - So far we are clearly in the region of Wythoffian polytopes.

Next consider the snub thereof, i.e. its vertex alternated faceting. That then would have for Dynkin diagram:
Code: Select all
*a   o----4---( )  *b
     |  \      |
     4    Q    4
     |      \  |
*d  ( )---4----o   *c

or in linearized form: o4s4o4s4*aQ*c. Its cells then would be derived of those of the former as their snubs, and additionally the former vertex figure would come in as further cell type. - Thus we have 2Q-prisms, order 4 square tilings (those are the snubs of themselves), and order 2Q {2Q} tilings (x2Qo2Qo being the snub of x4o2Qo).

So far this is topologically valid. But then consider the new edges. All are derived as diagonals of former squares. Therefore this mere alternated faceting already has just a single edge size! (And that one just needs an always possible global scaling back to unity.) Therefore this snub truely is a further uniform one (just as your original one).

Finally: what is the vertex figure of this new uniform snub? Well, that is just the figure described above, i.e. that 2Q-gonal variant of the cuboctahedron.

PS: I could provide the incidence matrices of both hyperbolic polychora, if required.

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Re: pxoPoxp&#xt and friends.

Postby wendy » Fri Aug 09, 2013 9:40 am

Let's see. The laws of symmetry reduce *a and *c to (here, *C), and *b, *d, to *B, it gives o4s4o2Qo. This has the snub-cells that are oq2Qoo&#qt, (the verf of o4x4o2Qo which are pyramids. It certianly can be made equilateral, but it does not come uniform. It's one of those figures that come to 'made of polygons'. the generalised johnson thingies.
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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Fri Aug 09, 2013 11:31 am

You are certainly right, that o4x4o4x4*aQ*c was not the simplest symmetry description possible. It just was the one I started with observation. You further are right in that this structure can be more easily given as o4x4o2Qo.

But you are clearly wrong in your derived vertex figure!

Your own vertex figure derivation rules would result, when being applied to my dynkin symbol:
verf of o4x4o . *aQ*c = q . q . *aQ*c = qQq
verf of o . o4x4*aQ*c = q . q . *aQ*c = qQq (in same orientation)
verf of . x . x = q (lacings)
i.e. the 2Q-prism (scaled by q), as I proposed - and definitely not a pyramid

When being applied to your Dynkin symbol it is even more direct:
verf of o4x4o2Qo = q . q2Qo

And here are the incidence matrices (provided wrt your much easier Dynkin description):
Code: Select all
o4x4o2Qo (Q parametrisable, N,M,K->oo)

. . .  . | 2NMK |    4Q |   2Q    4Q |   2Q   2
---------+------+-------+------------+---------
. x .  . |    2 | 4QNMK |    1     2 |    2   1
---------+------+-------+------------+---------
o4x .  . |    4 |     4 | QNMK     * |    2   0
. x4o  . |    4 |     4 |    * 2QNMK |    1   1
---------+------+-------+------------+---------
o4x4o  . |   2M |    4M |    M     M | 2QNK   *
. x4o2Qo |   2K |   2QK |    0    QK |    * 2NM


respectively the therefrom derived uniform, non-Wythoffian snub
Code: Select all
o4s4o2Qo (Q parametrisable, N,M,K->oo)

demi( . . .  . ) | NMK |   2Q    4Q |    8Q   4Q |   2Q   2  4Q
-----------------+-----+------------+------------+-------------
      o4s .  .   |   2 | QNMK     * |     4    0 |    2   0   2
      . s4o  .   |   2 |    * 2QNMK |     2    2 |    1   1   2
-----------------+-----+------------+------------+-------------
sefa( o4s4o  . ) |   4 |    2     2 | 2QNMK    * |    1   0   1
sefa( . s4o2Qo ) |  2Q |    0    2Q |     * 2NMK |    0   1   1
-----------------+-----+------------+------------+-------------
      o4s4o  .   |   M |    M     M |     M    0 | 2QNK   *   *
      . s4o2Qo   |   K |    0    QK |     0    K |    * 2NM   *
sefa( o4s4o2Qo ) |  4Q |   2Q    4Q |    2Q    2 |    *   * NMK


PS: just for clearity (to an arbitrary reader): "q" is being used here for edges of length sqrt(2), as being introduced by Wendy. "Q" is just an arbitrary parameter for the respective link mark. There is no correlation between those letters intended (in contrast to Wendy's usage of "p" and "P" in the title of this thread, where "p" described the shortchord length of the regular unit-edged polygon {P} - if I recall correctly).

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Re: pxoPoxp&#xt and friends.

Postby wendy » Sat Aug 10, 2013 7:06 am

Richard K is indeed right about the vertex figure. What i gave belongs to s4s4oPo, not o4s4oPo.

In fact, the is no reqiorement for P to be even: it works with all values of P, such as o4s4oPo

o4s4oPo verf = popPouo&#qt p = shortchord of P edge2 = 4p/(4-p).

Cells are o4s4o = x4o4o, s4oPo = xPoPo, and q2qPo = pgonal prism.

When p=4, the outcome is x4o4oAo. When p=U, the outcome is a slice of x4o4o4o.

For P=3 and 4, the tiling is type 2 (ie have finite content). This means they have horospheric cells.

For P>4, the tiling is type 4, ie might occur in a frieze that has horospheric cells.

It is interesting to note that the shape of the vertex figure is what i was originally thinking of when i was thinking of 'exotic prisms' all those years ago. These became 'lace prisms' in the end.
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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Sat Aug 10, 2013 2:04 pm

wendy wrote:Richard K is indeed right about the vertex figure. What i gave belongs to s4s4oPo, not o4s4oPo.
:)

In fact, the is no reqiorement for P to be even: it works with all values of P, such as o4s4oPo

Good observation, Wendy! So this is valid only for your o4o4oPo symmetry, not for my o4o4o4o4*aQ*c symmetry, because the latter just transforms into o4o4o2Qo. - And you could not use e.g. Q=3/2 in there, as this would produce a Grünbaumian 6/2-prism rather than a 3-prism!

o4s4oPo   verf = popPouo&#qt   p = shortchord of P   edge2 = 4p/(4-p).

Cells are o4s4o = x4o4o, s4oPo = xPoPo, and q2qPo = pgonal prism.

Well, either those "x" (in this last line) should be "q" as well, or those "q" (there) would be "x" too. All cells do have the same edge size.

When p=4, the outcome is x4o4oAo.

Yes, x4o4o *b3o = o3o4o4s = o4s4o4o = o4s4o *b4o = s4o4s *b4o = s4o4s4o4*a = ...

When p=U, the outcome is a slice of x4o4o4o.

Suppose that p=u would require, because of |u|=2, some infinity symbol link mark?

For P=3 and 4, the tiling is type 2 (ie have finite content). This means they have horospheric cells.

Cells then are tet, trip, and squat.

For P>4, the tiling is type 4, ie might occur in a frieze that has horospheric cells.

Cells then are cube and squat.

It is interesting to note that the shape of the vertex figure is what i was originally thinking of when i was thinking of 'exotic prisms' all those years ago. These became 'lace prisms' in the end.
:)
Nearly, the vertex figure of o4s4oPo is popPouo&#qt, as you already wrote correctly above. That one clearly is a (bistratic) lace tower, rather than a (monostratic) lace prism.

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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Sun Aug 11, 2013 7:57 pm

Rethought our new uniform snubs a bit.

Thought about xNoPoNx. But then it occured to me, that its vertex figure is noPon&#q. And that one would be uniform itself only if n (shortchord of xNo) equals q. That revolves us back to N=4. Therefore your snub is kind of static (at least with respect to the legs connected to the ringed node).

Then I tried oNxNoPo. Here the vertex figure is n . nPo, i.e. a uniform prism for any N!

So we could consider oNsNoPo for any even N (and any P). Each one would result in a uniform snub!

N now is even, say N = 2M, then let m (in abuse of your notation) here denote the shortchord of xMo.

Having introduced this notation we can describe the elements of oNsNoPo. The cells are oNsNo = xMo4o4*a, further sNoPo = xMoPoP*a, and finally x . xPo. The vertex figure here is no longer a rectification of the P-prism, instead it is in general a truncation of it. In fact its 2 polar bases are mPp, its 2P hemispherical (truncational) triangles are clearly op&#q, and the P lateral octagons are m4q. (For N=4 we clearly get M=2 and thus m=r, thus those octagons get back to the so far latteral squares.)

It then striked me to consider a rather special choice of parameters, i.e. using esp. P=M=4, thus N=8. Then we would get to consider o8s8o4o. - What is special about that one? - Let's see: Its cells then are o8s8o = x4o4o4*a, s8o4o = x4o4o4*a again, and x . x4o = cube. - Okay, 2 of its cell types get equivalent, we have a higher symmetry than usual. - But what comes out even more impressive, because of p=m=q, is that its vertex figure becomes itself a uniform polyhedron: it just would be tic (the truncated cube)!

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Re: pxoPoxp&#xt and friends.

Postby wendy » Mon Aug 12, 2013 7:11 am

Of course, while xPoQoPx to have a uniform vertex, requires P=4, the general form is xPoQoPxP/2z, (z=*a), which has a uniform Q-antiprism as a vertex figure. Most notable is x6o3o6x3z, which is the same as x6o3o4o. In fact, this equity is used to devolve the symmetries of the latter into something more.

One could look at the groups like x8o4o8xPz, because these are likely to occur in the frieze-groups based on o8o3o4o.

But there are folk coming in from the outback soon, which will generally curb my activities for a week or so.
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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Mon Aug 12, 2013 1:01 pm

wendy wrote:Of course, while xPoQoPx to have a uniform vertex, requires P=4, the general form is xPoQoPxP/2z, (z=*a), which has a uniform Q-antiprism as a vertex figure.

Right you are, my fault! By use of that additional link of half the magnitude the antiprism becomes uniform again indeed.

Most notable is x6o3o6x3z, which is the same as x6o3o4o. In fact, this equity is used to devolve the symmetries of the latter into something more.

Interesting. Will come back to that later, I suppose...

One could look at the groups like x8o4o8xPz, because these are likely to occur in the frieze-groups based on o8o3o4o.


Well, have now fiddled out that snub in general, i.e. s-2R-o-Q-o-2R-s-R-*a.
Its cells then are s-2R-o-Q-o = xRoQoQ*a, further s-R-s-2R-o = xRo3o3*a, finally sQs2s = Q-antiprism.
Its vertex figure is kind a truncation of a Q-antiprism. In fact, using x(.) as function for the shortchord length of that regular polygon with the argument number of edges, then its base polygons are x(Q)-Q-x(R), the truncational faces are x,x(Q)&#x, and the latterals are x-3-x(R).
This shows the importance of the one you mentioned above already: the vertex figure of s6o3o6s3*a is a toe (truncated octahedron)!

But there are folk coming in from the outback soon, which will generally curb my activities for a week or so.

Well, that's life!

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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Mon Aug 12, 2013 4:53 pm

Klitzing wrote:
wendy wrote:Most notable is x6o3o6x3z, which is the same as x6o3o4o. In fact, this equity is used to devolve the symmetries of the latter into something more.

Interesting. Will come back to that later, I suppose...


Okay, here it comes.

There has been a long way from x4oPo4x, generalizing to x2RoQo2RxR*a, specializing to x6o3o6x3*a, and finding its equivalent x6o3o4o.

I meanwhile observed o4x4o4x4*aP*c; Wendy then gave its equivalent o4x4o2Po, then being generalized to o2Rx2RoQo, and finally specialized to o8x8o4o.

Both final ones then proved to show up rather special, and still uniform snubs: in the first case I showed this morning that it would have a truncated octahedron for vertex figure, and in the second one I already showed yesterday, that it would have a truncated cube for vertex figure.

Well, 'interesting' is that cited equivalence indeed. It provided to me the hint, that, given such a close relation between our 2 final structures, that there ought to be a further Dynkin diagrammal description, which shows up this relationship more directly. - I searched a few minutes and quickly digged it out!

Here it is:
Consider the general hyperbolic structure x2PoPoQo.

It is obvious, that its vertex figure will be readily some regular polyhedron. Therefore the snub therefrom can be derived as usual, and it would turn out, that the vertex figure of s2PoPoQo always happens to become xPxQo! The cells of the snub also can be given directly as xPoQo (= verf( x2PoPoQo ) ) and xPo2Po (= s2PoPo ).

Further we can see thus directly: as long as P=3 and Q<=6, then the hyperbolic honeycomb will be paracompact, else hypercompact.


Finally providing the various examples explicitely by means of their incidence matrices:

Code: Select all
s6o3o3o

demi( . . . . ) | NM |  12 |   6  12 |  4  4  tut
----------------+----+-----+---------+------
sefa( s6o . . ) |  2 | 6NM |   1   2 |  2  1
----------------+----+-----+---------+------
      s6o . .   |  3 |   3 | 2NM   * |  2  0
sefa( s6o3o . ) |  3 |   3 |   * 4NM |  1  1
----------------+----+-----+---------+------
      s6o3o .   |  M |  3M |   M   M | 4N  *  x3o6o
sefa( s6o3o3o ) |  4 |   6 |   0   4 |  * NM  tet


Code: Select all
s6o3o4o

demi( . . . . ) | NM |   24 |  12  24 |  8  6  toe
----------------+----+------+---------+------
sefa( s6o . . ) |  2 | 12NM |   1   2 |  2  1
----------------+----+------+---------+------
      s6o . .   |  3 |    3 | 4NM   * |  2  0
sefa( s6o3o . ) |  3 |    3 |   * 8NM |  1  1
----------------+----+------+---------+------
      s6o3o .   |  M |   3M |   M   M | 8N  *  x3o6o
sefa( s6o3o4o ) |  6 |   12 |   0   8 |  * NM  oct


Code: Select all
s6o3o5o

demi( . . . . ) | NM |   60 |   30   60 |  20 12  ti
----------------+----+------+-----------+-------
sefa( s6o . . ) |  2 | 30NM |    1    2 |   2  1
----------------+----+------+-----------+-------
      s6o . .   |  3 |    3 | 10NM    * |   2  0
sefa( s6o3o . ) |  3 |    3 |    * 20NM |   1  1
----------------+----+------+-----------+-------
      s6o3o .   |  M |   3M |    M    M | 20N  *  x3o6o
sefa( s6o3o5o ) | 12 |   30 |    0   20 |   * NM  ike


Code: Select all
s8o4o3o

demi( . . . . ) | NM |   24 |  12  24 |  6  8  tic
----------------+----+------+---------+------
sefa( s8o . . ) |  2 | 12NM |   1   2 |  2  1
----------------+----+------+---------+------
      s8o . .   |  4 |    4 | 3NM   * |  2  0
sefa( s8o4o . ) |  4 |    4 |   * 6NM |  1  1
----------------+----+------+---------+------
      s8o4o .   |  M |   4M |   M   M | 6N  *  x4o8o
sefa( s8o4o3o ) |  8 |   12 |   0   6 |  * NM  cube


Code: Select all
s10o5o3o

demi( .  . . . ) | NM |   60 |  30   60 |  12 20  tid
-----------------+----+------+----------+-------
sefa( s10o . . ) |  2 | 30NM |   1    2 |   2  1
-----------------+----+------+----------+-------
      s10o . .   |  5 |    5 | 6NM    * |   2  0
sefa( s10o5o . ) |  5 |    5 |   * 12NM |   1  1
-----------------+----+------+----------+-------
      s10o5o .   |  M |   5M |   M    M | 12N  *  x5o10o
sefa( s10o5o3o ) | 20 |   30 |   0   12 |   * NM  doe


As those are as outstanding, I can't belief that anyone wouldn't have considered those before. - Any reference?

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Re: pxoPoxp&#xt and friends.

Postby wendy » Tue Aug 13, 2013 7:40 am

The last five incidence matrices describe the polytopes s2PoPoQo = oPxPoPzQo, which comes from the laws of symmetry.

I'm not really sure if there are references. It's all pretty much new ground here, because we made available to ourselves, some pretty fancy tools. Coxeter did much the same when he loaded Wythoff's construction onto the Lie-group symbols we call Coxeter-Dynkin symbols, and Galileo did when he poked his spyglass at the skys. Lots to see, but you got to go outback to see it, so to speak.

One of the main reasons that you won't find many references to what we're doing is because the bulk of what's here are "type four" tesselations. This means that they have both bollotopic and horocyclic cells. There's plenty of them out there, many can not be rendered uniform.

Coxeter, Johnson and others do not stray past type two, because the way they approach hyperbolic mathematics conspires against random forays that lead to dead ends in the main (which is what type 3 and higher do). We have a pretty fancy approacy to this which gets rid of all of that ugly mathematics, and so it's not too hard to find these things. So the sort of thing that Johnson had to resort to for days to find my calculations were correct, were in response to pretty ordinary mental arithmetic. I don't use things like sinh or cosh or wotnots. Most of it comes pretty much by euclidean geometry.

The group o6o3o6o3z is size 40, consists of two identical mirror-sets of size "240", which corresponds to the cells of x4o3o6o. Apart from the regular division into 48 that x4o3o confers, there is a second division of order 5, because the cube is made of five regular tetrahedra, (rather like diminishing the cube to get a tetrahedron, by alternating corners, but the bits that come off are also regular. So this mirror group divides into five sets of cells of x3o3o6o, which in turn divides into 24 of the symmetry [3,3,6] (of size '2'). I however misread 240 as an octahedron, and missed bits of it until i revisited.
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Re: pxoPoxp&#xt and friends.

Postby Klitzing » Tue Aug 13, 2013 12:28 pm

wendy wrote:The last five incidence matrices describe the polytopes s2PoPoQo = oPxPoPzQo, which comes from the laws of symmetry.


Oh damn, you are right!
s2PoPo is not only the same as xPo2Po, but also as xPoPoP*a. And this one makes free for your equality: s2PoPoQo = oQoPoPxP*b.
That is, those found snubs turn out to be Wythoffian again after all. :(

Hope this is true only for those more special ones, i.e. where the vertex figures have higher degrees of symmetry, and not for the general family members of those 2 investigated cases with truncated or rectified antiprisms resp. with truncated or rectified prisms for vertex figures. (Of which co/toe resp. co/tic where just special cases.)

I'm not really sure if there are references. It's all pretty much new ground here, because we made available to ourselves, some pretty fancy tools. Coxeter did much the same when he loaded Wythoff's construction onto the Lie-group symbols we call Coxeter-Dynkin symbols, and Galileo did when he poked his spyglass at the skys. Lots to see, but you got to go outback to see it, so to speak.

One of the main reasons that you won't find many references to what we're doing is because the bulk of what's here are "type four" tesselations. This means that they have both bollotopic and horocyclic cells. There's plenty of them out there, many can not be rendered uniform.

Coxeter, Johnson and others do not stray past type two, because the way they approach hyperbolic mathematics conspires against random forays that lead to dead ends in the main (which is what type 3 and higher do). We have a pretty fancy approacy to this which gets rid of all of that ugly mathematics, and so it's not too hard to find these things. So the sort of thing that Johnson had to resort to for days to find my calculations were correct, were in response to pretty ordinary mental arithmetic. I don't use things like sinh or cosh or wotnots. Most of it comes pretty much by euclidean geometry.

That was my feeling too, but as I have no unversity library at reach, I better asked for.

The group o6o3o6o3z is size 40, consists of two identical mirror-sets of size "240", which corresponds to the cells of x4o3o6o. Apart from the regular division into 48 that x4o3o confers, there is a second division of order 5, because the cube is made of five regular tetrahedra, (rather like diminishing the cube to get a tetrahedron, by alternating corners, but the bits that come off are also regular. So this mirror group divides into five sets of cells of x3o3o6o, which in turn divides into 24 of the symmetry [3,3,6] (of size '2'). I however misread 240 as an octahedron, and missed bits of it until i revisited.

Completely incomprehensible, what you are telling here, sorry.

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