Construction of BT-polytopes via partial Stott-expansion

[student91]

[...]
so Wendy thus contributed to CRFebruary ...

yup, I think we can put this in the discovery index as well.(in fact I did, it is a very short article called D4.14)

[...]

Anyway what I was trying to say by investigating the symmetries, is trying to find out what (reflective) subsymmetries all symmetries have. This image is said to show the hierarchy of symmetries: Yet it doesn't show a line from [5,3,3] to [3^1,1,1]. Therefore I thought nobody really knew the relation between these two (though you two seem to do). Anyway, it might be that there are more not-drawn lines that might give rise to expansions of some uniforms. Therefore it's very important to find all the lines to be able to prove we've found all EPE's of the uniforms.

Besides, I realized the things I've posted before (about the perp-space/para-space of lace-cities etc., introducing the (...)(...)-notation) were exactly the representations of the polytopes in the corresponding (duo)prism-symmetry. Let's take the representation of ex like this:(2f)(o5o3o)+(F)(o5o3x)+(f)(x5o3o)+(x)(o5o3f)+(o)(o5x3o). Actually that is the representation of ex in ike-prism symmetry, it can be written as AFfxo ooxoo5oooox3oxofo&#zx. Similarly (o5o)(x5x)+(f5o)(o5x)+(o5f)(x5o)+(x5o)(o5f)+(o5x)(f5o)+(x5x)(o5o) is the representation of ex in 5×5-duoprism symmetry, as ofoxox5oofoxx xoxofo5xxofoo&#zx. Now apart from (duo)prism-symmetry, most things also have some kind of antiprism symmetry. The .3.3.-representation of the .5.3.3.-uniforms has tetrahedral antiprismatic symmetry, and the .5.-representation of id has pentagonal antiprismatic symmetry. I think a notation for this might be usefull as well, to be able to write down the id=>pent.el.gyrobirotunda partial expansion.

P.S have I already said Wendy's discovery is quite awesome?

[wendy]

In the diagram in the previous post, [5,3,3] directly connects to [3,4,3+] as a subgroup of 25. Lilewise [5,3,3]+ directly connects to [3+,4,3+] of order 288. I don't think Tom Ruen's piccie shows that.

Likewise, [5,3,3] connects to [[3,3,3]] of order 60, and [5,3,3]+ connects to [[3,3,3]]+ of order 60.

There is a connection of [5,3,3] to [[p,2,p]], p=2, 3, 5 the latter group has a order of 8p²

The relationship between [3,3,5] and [3,3,A] (the three-armed one). The arms are indeed equal, but what happens is that the triangle that links the ends of the arms are actually rotational only. The [3,3,A] corresponds to an octant of an octahedral face of {3,4,3}. When you put an ike into this, the three vertices preserve the rotational symmetry of the octahedron-face, but not the reflective symmetry of the triangle. The 'r' mirrors (ie the reversal of x <-> -x etc, is preserved, as is reflection in the octahedron-face from octahedron to octahedron.

[student91]

In the diagram in the previous post, [5,3,3] directly connects to [3,4,3+] as a subgroup of 25. Lilewise [5,3,3]+ directly connects to [3+,4,3+] of order 288. I don't think Tom Ruen's piccie shows that.

Indeed it doesn't. I think that is a pitty. Do you think you can give all lines that aren't in this picture? Did you just do that in your previous post? Anyways, thanks for this post.

Likewise, [5,3,3] connects to [[3,3,3]] of order 60, and [5,3,3]+ connects to [[3,3,3]]+ of order 60.

Now this is interesting! I guess this means the .5.3.3.-polytopes can be representetd as .....3.....3.....3......&#zx. This might also give new expansions. What I don't understand is what the [[..]] mean, Wikipedia says it is extended symmetry. Does this mean you take the symmetry of the compound of the polytope and the dual? (so [[3,3]] would be [4,3], [[3,4,3]] would be the symmetry of xo3oo4oo3ox&#xz and [[3,3,3]] would be the symmetry of xo3oo3oo3ox&#xz) Am I right here?

There is a connection of [5,3,3] to [[p,2,p]], p=2, 3, 5 the latter group has a order of 8p²

I already understood these connections due to the lace-cities of Klitzing. (the ones of ex show all these symmetries)

The relationship between [3,3,5] and [3,3,A] (the three-armed one). The arms are indeed equal, but what happens is that the triangle that links the ends of the arms are actually rotational only. The [3,3,A] corresponds to an octant of an octahedral face of {3,4,3}. When you put an ike into this, the three vertices preserve the rotational symmetry of the octahedron-face, but not the reflective symmetry of the triangle. The 'r' mirrors (ie the reversal of x <-> -x etc, is preserved, as is reflection in the octahedron-face from octahedron to octahedron.

This connection has for ex been investigated completely with your new expansion. Maybe rox might give a central-node expansion though.

Anyway, Thanks for the 333-line, I think it is interesting. Furthermore, do you know a site or something that shows all these connections (so also for 334 etc.)? that would be great, as we will be able to fully investigate the expansions.

[Klitzing]

That is, an expansion may be called partial, or at least not a normal expansion, if either:
1] You are expanding according to a subsymmetry of the whole figure, or
2] You use only the radial vectors of that subsymmetry for expansion directions. This means you use whole complexes (or patches) as limits. (if you use patches, you have to use radial vectors, and you only need to use radial vectors if the direction of expansion is ambiguous. Therefore these two parts have to be together).
sidequestion is if we are going to restrict ourselves to reflective subsymmetry only, or include more symmetries as well

These two give rise to different sets: When you are using 1], you have to discard things with facetings...

that bit in fact in both your choices: no quirks allowed within the transformation (whereas, when applying outside the mere transformation it might come in - but then the faceting might break the symmetry already by itself...)

... or middle-steps of expansion-chains (like esquidpy=>squobcu etc.)

Well, you are partly true in that.

Code: Select all

esquidpy <-> squobcu : neither subsymmetrical transformation
                       nor pushing/pulling single boundary elements


But on the other hand

Code: Select all

oct -> esquidpy  : perfectly valid partial Stott expansion (true subsymmetry)
oct -> squobcu   : perfectly valid partial Stott expansion (true subsymmetry)
sirco -> esquidpy  : perfectly valid partial Stott contraction (true subsymmetry)
sirco -> squobcu   : perfectly valid partial Stott contraction (true subsymmetry)


so we have some overlap here! I.e. it more is like the sequence as such is spanned between 2 extremes, which both have a higher symmetry than the intermediate ones.

... When you are using 2] you include these, but you exclude things like o3x3o->o3x3x etc.

Can't follow you here. What are the radial vectors you are allowing, and which not? - o3x3o->o3x3x clearly is a true Stott expansion (the left triangles of the symbol are pulled, i.e. we consider oct here rather as a tetratetrahedron). It even can be considered a subsymmetrical one, when o3x3o = oct = x3o4o would have been considered as having 8 equivalent triangles.

Therefore 2] does not correspond to the way Stott uses the partial expansions. I think both definitions are valuable,...

(your aimed for distinction is not too clear so far)

... but we have to be able to distinguish between them, to prevent more ambiguities. I suggest we keep calling 1] a partial expansion, (as it is being used like this in Stotts paper), and the call 2] an extended partial expansion, or EPE, or something like that. (I don't like that name, but discussions about names are so horribly useless, much more useless than discussions about definitions. when you have a better proposition I'll stick to that one). Note that it is incredibly difficult to make a comprehensive set of definitions to include both of the sets 1] and 2]. I guess in hindsight that is one of the reasons I was using vertex configuration only, as then more of both sets are included. However, in hindsight this approach indeed seems less perfect.

In fact I also have already considered to distinguish the different found examples into different levels of partiality. I came up with 3 levels so far:

• partiality of 1st degree
  just a single (to be specified) class (transiency under full symmetry!) of (sub-)facets (out of severals) defines the operation. The operation always acts radially from some center. (For euclidean tesselations, this center also would be at infinity, for sure; like considering those as polytopes in one dimension up, having infinite radius.) – This usage already was mentioned by Mrs. Stott herself in her original paper.
• partiality of 2nd degree
  one or more (to be specified) classes (transiency under full symmetry!) of (sub-)facets defines the operation. The operation acts on such sub-groups, and (for D-1 facets) needs no longer be directed along the facet normals. (The latter would be necessary here, when it is just one class, for sure.) – This type of partiality was stressed a lot in the research for CRFs, esp. in 2014. Even those 3 found transitions from ike facetings to bilbiro, to thawro, resp. to pocuro would subsume here.
• partiality of 3rd degree
  either of the former would be applied wrt. some true subsymmetry only. Here usually larger patches do move as such. – This is how the topic of (symmetry based) partial Stott expansions originally came up exactly 100 years after her publication in my distinct researches.

(You are right, we might restrict "subsymmetry" within complete sequences of transformations to be compared to the outermost sequence members only...)

--- rk

[Klitzing]

... I have no idea how things like x5o4o3o etc. work.

Not too hard. On a localized level those work just the same as any other polychoron too. It is only the global effect which forces it to belong to hyperbolic geometry in this specific case.

First of all: x5o4o3o is a regular polytope. (Even such honeycombs can be considered polytopes! Cf. the easier example of tilings: just consider the space below as ist body.) In fact there is just one node ringed and that specific node is at the end of a linear Dynkin diagram.

Next: its only cells are regular hyperbolic polyhedra x5o4o, i.e. a (likewise hyperbolic) relative to the dodecahedron, just that here 4 pentagrams are incident to each vertex.

And then these tiling-like polyhedra would come together at each vertex in the form of a cube: The vertex figure of that polychoron is just . f4o3o (given metrically correct).

Thus, the incidence matrix would read (N, M --> oo)

Code: Select all

5NM |    8 |   12 |  6
----+------+------+---
  2 | 20NM |    3 |  3
----+------+------+---
  5 |    5 | 12NM |  2
----+------+------+---
 5M |  10M |   4M | 6N


--- rk

[wendy]

There's a few groups i have not heard of, and a few other missing lines. I would need to draw up something a bit more convincing though.

This is a list of groups and their known subgroups. I have largely not included the prismic groups, save for the highest one that has a symmetric link to a full group.

Double brackets means that there is a 'wander' involved. In the pentachoron, this is by inversion, which is an even operator in 4d (a c60 clifford will do nicely). This means, that while [[3,3]]+ does not exist, [[3,3,3]]+ does. In the octagonny, it is not an inversion.

The [[,,]] is not a 'specific operator' in the style of mirrors. What it means is that one can make something like o4x3x4o (a tiling of truncated octahedra), the has cells at cubic-corners and cubic-cells. However, even though the cells are identical, the mirrors will not bring one to another. This is why one has to introduce a 'wander' (a centre-less rotation), to bring one to another.

In the case of [[3,3,3]], the wander can be done by a central inversion, so ye get a ten-pointed 'star of david' thing, the convex intersection being o3x3x3o. The x3o3o3x has this symmetry too. In the case of [[3,4,3]], the {3,4,3} already has central inversion, so you just have to do something else to bring one face of the octagonny o3x4x3o onto the other.

+ means removal of mirrors, leaving the even operators. It transmits across all odd numbers. When an even number is involved, one has to plus each set of nodes to be removed separately.

The subgroups are shown as twelfty-size + letter, these are the first and third column. Suggested names are also given, eg 'octagonny'.

Code: Select all

     twe   dec
   10000 14400  a   [3,3,5]      6000a  496a  200a  180a  72a
    6000  7200  a   [3,3,5]+      248a  100b
    1924  2304  a   [[3,4,3]]     972a  972b   72a  "octagonny"
     972  1152  a   [[3,4,3]]+    496c
     972  1152  b   [3,4,3]       496a  496b  324a
     496   576  a   [3,4,3+]      248a  172a        great pyritohedral
     496   576  b   [3,4,3]+      248a
     496   576  c   [[3+,4,3+]]   248a
     324   384  a   [3,3,4]       172a
     248   288  a   [3+,4,3+]      36a
     200   240  a   [[3,3,3]]     100a  100b       inversion pentachoron
     180   200  a   [[5,2,5]]
     172   192  a   [3,3,A]        96a             semi tesseract
     172   192  b   [3,3+,4]       96a             pyritochoral
     172   192  c   [3,3,4]+       96a             rotary tesseract
     100   120  a   [3,3,3]        60a             pentachoral
     100   120  b   [[3,3,3]]+     60a             pentachoral inversion
      96    96  a   [3,3,A]+
      72    72  a   [[3,2,3]]      36a
      60    60  a   [3,3,3]+                       pentachoral rotary
      36    36  a   [3,2,3]                        bi-triangular prism


Note that 3,4,3 does not call down to [[3,2,3]]. This is because in 3,4,3, the actual relation of triangles is 1:q. Octagonny is a class-2 operator, which maps across the 4-branch, does allow the two to co-mingle (it produces an discrete (infinite-dense) tiling in a+bQ, rather as the octagon does).

The symmetries orthogonal to a mirror (necessary for containing (2,p,q) are [3,3,5] -> [3,5], [3,4,3] -> [4,3], [3,3,4] are [2] and [3,4] and for [3,3,3] gives [3].

[Klitzing]

The [[,,]] is not a 'specific operator' in the style of mirrors. What it means is that one can make something like o4x3x4o (a tiling of truncated octahedra), the has cells at cubic-corners and cubic-cells. However, even though the cells are identical, the mirrors will not bring one to another. This is why one has to introduce a 'wander' (a centre-less rotation), to bring one to another.

+ means removal of mirrors, leaving the even operators. It transmits across all odd numbers. When an even number is involved, one has to plus each set of nodes to be removed separately.

Good to be finally introduced to these notations! Always have been speculating about these notations myself.


According to your description it kind of seems to be more appropriate to replace "+" either by a dagger (in the sense that one of the mirrors has died effectively), or alternatively by a minus sign (in the sense of some reduction). Else it would too easily be misinterpreted to mean some increase of symmetry instead (plus sign)...

--- rk

[student91]

[...]
that bit in fact in both your choices: no quirks allowed within the transformation (whereas, when applying outside the mere transformation it might come in - but then the faceting might break the symmetry already by itself...)

Indeed no quirks are allowed within the transformations, you've convinced me about that. Then, the faceting will indeed break the symmetry of the polytope, and thus won't be a partial expansion according to definition 1]. (as an expansion that is subsymmetrical). 2] would allow this to be an EPE, but I'll get back to that later in this post.

[...]

... When you are using 2] you include these, but you exclude things like o3x3o->o3x3x etc.

Can't follow you here. What are the radial vectors you are allowing, and which not? - o3x3o->o3x3x clearly is a true Stott expansion (the left triangles of the symbol are pulled, i.e. we consider oct here rather as a tetratetrahedron). It even can be considered a subsymmetrical one, when o3x3o = oct = x3o4o would have been considered as having 8 equivalent triangles.

Therefore 2] does not correspond to the way Stott uses the partial expansions. I think both definitions are valuable,...

(your aimed for distinction is not too clear so far)

I think I misphrased my defenition 2]. I made my defenitions according to the following:
First of all, I thought the following to be true: "A true expansion is not a partial one". Otherwise the word "partial" is used wrongly.
now what is a true expansion? that's already given by Stott: you take the "limits", which are surtopes of a single kind, and move them apart according to the full symmetry of the polytope.
We could then define partial to differ on two points: 1] says you use a subsymmetry instead of full symmetry. 2] says you move more than one thing as a whole limit. (so you move whole patches). this implies you use radial vectors. (in my previous post I switched these around, I started with the radials. The idea behind definition 2] is that you move whole patches though)
2] is a kind of weird definition for a partial expansion: something non-uniform that gets expanded according to it's full symmetry might still be called "partial". Nevertheless it is a very interesting definition that allows for very interesting expansions (like ike=>bilbiro or esquidpy=>squobcu. Placing these in defenition 1] seems odd to me, but could be done) That's why I suggested to name it differently, EPE for example, but you're free to suggest a good name

[...]
In fact I also have already considered to distinguish the different found examples into different levels of partiality. I came up with 3 levels so far:

• partiality of 1st degree
  just a single (to be specified) class (transiency under full symmetry!) of (sub-)facets (out of severals) defines the operation. The operation always acts radially from some center. (For euclidean tesselations, this center also would be at infinity, for sure; like considering those as polytopes in one dimension up, having infinite radius.) – This usage already was mentioned by Mrs. Stott herself in her original paper.

doesn't seem partial to me. Am I right this is just full expansion?

partiality of 2nd degree
one or more (to be specified) classes (transiency under full symmetry!) of (sub-)facets defines the operation. The operation acts on such sub-groups, and (for D-1 facets) needs no longer be directed along the facet normals. (The latter would be necessary here, when it is just one class, for sure.) – This type of partiality was stressed a lot in the research for CRFs, esp. in 2014. Even those 3 found transitions from ike facetings to bilbiro, to thawro, resp. to pocuro would subsume here.

sounds a bit similar to my definition 2]. I'm not sure what you comletely mean though, but it seems pretty the same as my 2]

partiality of 3rd degree
either of the former would be applied wrt. some true subsymmetry only. Here usually larger patches do move as such. – This is how the topic of (symmetry based) partial Stott expansions originally came up exactly 100 years after her publication in my distinct researches.

(You are right, we might restrict "subsymmetry" within complete sequences of transformations to be compared to the outermost sequence members only...)

--- rk

that sounds the most partial to me, more like 1]. It seems then that we both have similar definitions, though I'm still not sure if you meant what I meant. Anyway it seems we are really close to some consensus here, and can finish this stage of article-writing!!

[Klitzing]

Okay, will have a next run.

Partiallity of 1st degree is what Mrs Stott already uses herself, when extending her own definition, which is given for regular polytopes only, towards iterative application, i.e. thus applying it to the Archimedeans. Then partiallity is needed in the sense that global symmetry distinguishes several classes of non-equivalent elements of some dimensionality. And there expansion resp. contraction applies then to one of these classes only. NB, this might be triangles and squares, as for the co, but even squares A and squares B in the case of the sirco. - If I remember correctly, she even uses that very term (partial) already in that context.

Partiallity of 2nd degree indeed is meant to mean what you are trying to tell with radial vectors. In fact, it generalizes from single facets, which had to be normal to the displacement vector by Stott's definition (which is trivially achieved for regulars) towards facets, which can have any slope. It further generalizes from a single class of elements (under full symmetry of the starting figure) towards non-overlapping patches with elements from an arbitrary amount of classes (>=1) of a given dimension (again >=1, as specified by Mrs Stott!).

E.g. in the ike-faceting towards bilbiro expansion it is still a single class of elements which gets displaced, the faceting pentagons. Even so it uses patches of 2 such connected ones each, which move as a single unit. Moreover the displacement vector is not normal to any of these pentagons.

Partiallity of 3rd degree then finally introduces the consideration of arbitrary subsymmetries as well. That then generally asks for a decomposition in larger patches in general. So one could consider this case also kind a partiallity of 2nd degree, when being applied to a starting figure which aforehand had been colored. It is that coloring which then breaks the global symmetry.

It is not that we want to set up a new definition and thereby want to exclude all what does not litterally bow to it. Instead we just want to extend the formerly too narrow definition into several directions. And moreover, most series of examples would more or less stumble right between all of these provided degrees to and fro. - It is just that a 100 years history cannot be rolled back. So if we try to extend her definition, we cannot do that right under the same name. There are already (then) narrower usages out there. This is why I took over her in those days informally only being used adjective to settle into a now formal one. And I would like it to cover all these extensions. Kind of allowing of according displacements in a much wider range of setups.

Yes, we both are catching the same fly indeed.


We well might consider your x -> (-x) things in the same article, when considering those then as a subject to some partial Stott expansions. We just will have to make sure, that this will be some independent a priori transformation. And we then should make clear additionally, what new facet structure would be the true outcome of that mere edge inversion! E.g. we would not be allowed to call your fo(-x)2xfo2oxf&#zx any longer an ike. Instead it is that specific ike faceting, I already had shown somewhere. Only to that one then the partial Stott expansion will apply.

--- rk

[Klitzing]

The thing Wendy suggests then is foxo3xxxF3xfoo *b3oxfo&#zx. That is, a demicubic expansion of ex using the middle node. Using one of the other nodes gives the ico-augmented D4.11 (D4.11.1 or so?). I think it is weird we have never tried expanding according to this node.

We start from ex = foxo3ooof3xfoo *b3oxfo&#zx here. Wendy's ico-augmented prissi (how about "icau prissi" then, Jonathan?) is a true Stott expansion wrt. THIS symmetry display, i.e. a true partial one wrt. the symmetry of ex (3rd degree).


That D4.11.1 was "icau pretasto" (ico-augmented pentagonorhombic-trisnub-trisoctachoron) = Fxox3ooof3xfoo*b3oxfo&#zx, which thus can be seen as a true partial Stott expansion of fo(-x)o3ooof3xfoo *b3oxfo&#zx (2nd degree: full symmetry, use of non-orthogonal displacement of faceting faces).


It ought to be noted here that whereas the edge-flip x -> (-x) in ike = fxo2xof2ofx -> fxo2xof2of(-x) does not change the circumradius of that pseudo "layer" (a f2x rectangle is equivalent to a f2(-x) rectangle), this is not true in general. The former pseudo "layer" of ex was x3o3o *b3f, a non-uniform variant of rit, having circumradius f. But (-x)3o3o *b3f just has a circumradius of 1!

This likewise can be seen from my recently provided alternate description of inverse edges, which transported the inversity of edges wrt. to normal symmetry towards normal prograde edges wrt. a different symmetry with larger fundamental domain (multiple dihedral angles): [-x]-n/d- = [x]-n/(n-d)-. For the case of n/d=2 we clearly get n/(n-d)=2 again. But for n/d=3 we will get n/(n-d)=3/2. Sure there is no difference within x3/2o and x3o when considered as mere geometrical object without orientation. But there is a clear difference between x3/2x = (-x)3x and x3x, or between x3/2f = (-x)3f and x3f, or between x3/2o3f = (-x)3o3f and x3o3f - not only wrt. the dihedral angles, but also wrt. the circumradius of that object.

In fact x3o3f results by Stott expansion from an f-tet, pulling the large triangles out, providing new prograde small triangles and rectangles. Whereas (-x)3o3f results the same way from that f-tet by pushing these large triangles a bit in, providing retrograde small triangles and rectangles. - Well, both would be representations of the same abstract polyhedron. They both have the combinatorics of co.

x3o3o *b3f has cells being x3o3o   ., x3o   *b3f, and . o3o *b3f, i.e. small tets, convex cuboctahedra variants, and large tets. Thus this is some rectified tesseract variant with only demitesseractic symmetry. Now reducing x first to o would reduce the small tets to nothing, those co variants become further large tets, and the old large tets remain as is. Now going on, and reducing those edges to negative values then would transform the formerly prograde small tetrahedra into now retrograde small tetrahedra, the former convex co-variants now become these above mentioned "pushed in large tets", i.e. non-convex co-variants, and the other large tets still remain unchanged in form and orientation (but not in position for sure).

Sure, x3o3o *b3f is just a pseudo element of ex. And likewise (-x)3o3o *b3f will be only a psedo element of fo(-x)o3ooof3xfoo*b3oxfo&#zx. But at least that changed circumradius shows that we here no longer have an orbiform figure anymore. And therefore it neither can be uniform.


Conclusion:
Your statement "Using one of the other nodes gives the ico-augmented D4.11 (D4.11.1 or so?)." definitely is wrong. There you do NOT start from ex (in the demitessic symmetry display). You rather start with a quite different non-convex figure. (The complete structure thereof still have to be evaluated. The above remarks just provide some hints here!) That figure not even is a mere faceting of ex. (As facetings would respect vertex positions at least.)

A true partial Stott expansion of ex = foxo3ooof3xfoo *b3oxfo&#zx at one of the other nodes clearly gives Fxux3ooof3xfoo *b3oxfo&#zx, which, because of the edge u (which most probably would survive at the outside) then is no longer CRF.

--- rk

[student91]

[...]It ought to be noted here that whereas the edge-flip x -> (-x) in ike = fxo2xof2ofx -> fxo2xof2of(-x) does not change the circumradius of that pseudo "layer" (a f2x rectangle is equivalent to a f2(-x) rectangle), this is not true in general. The former pseudo "layer" of ex was x3o3o *b3f, a non-uniform variant of rit, having circumradius f. But (-x)3o3o *b3f just has a circumradius of 1!

Whenever you change a node, you have to follow the rules of node-switching in order to keep the vertex-arrangements the same. Tose rules are that if you change a node with value A from A to -A, all nodes will have an addition of value of A*str(n), where str(n) gives the shortchord of an n-gon where n is the value of the node. otherwise the vertex-arrangements won't be the same , and thus your node-changing will result in something completely different, rather than a different polytope for the same vertex set. Thus f2x"="f2(-x), because str(2)=0, but x3o3o*b3f "=" (-x)3x3o*b3f, and not (-x)3o3o*b3f, because str(3)=1. I cant go in full detail about this, cause I'm going to leave for the weekend, and can only talk after that.

[Klitzing]

 
Edit: this post was writen in parallel to the above reply of student91 and so contains the same error stated there. Cf. to the next post of mine, which then accordingly corrects the new ideas, contained here!


Just guessing:

you transformed ex = foxo3ooof3xfoo *b3oxfo&#zx into some fo(-x)o3ooof3xfoo *b3oxfo&#zx for a then to be applied partial Stott expansion, resulting in D4.11.1 (icau pretasto). - Resp. sadi = fox3ooo3xfo *b3oxf&#zx into some fo(-x)3ooo3xfo *b3oxf&#zx for a then to be applied partial Stott expansion, resulting in D4.11 = pretasto.

What about a bi-dental such transformations? I.e. starting with fo(-x)o3ooof3(-x)foo *b3oxfo&#zx (or fo(-x)3ooo3(-x)fo *b3oxf&#zx) and applying then a partial Stott expansion of the form (+x)3.3(+x) *b3.?

Or even about a tri-dental one? I.e. starting with fo(-x)o3ooof3(-x)foo *b3o(-x)fo&#zx (or fo(-x)3ooo3(-x)fo *b3o(-x)f&#zx) and applying then a partial Stott expansion of the form (+x)3.3(+x) *b3(+x)?

(So, I have not checked so far, whether such multi-edge-inversions might get into some troubles somewhere.)

Wendy's consideration then would allow also (+x)3(+x)3(+x) *b3. resp. (+x)3(+x)3(+x) *b3(+x) to the respective ones. And surely also (+x)3(+x)3. *b3. to the so far known cases!

If these all come out possible (who would be the first to check? ), we should have a total of 10 new members of CRFebruary then ... (At least provided these don't come out to result in already known ones.)

--- rk

[wendy]

The '+' thing in [3,4,3+] u.s.w. is a coxeter thing, it supposes to represent an ion. In any case, i use s nodes of several kinds.

[Klitzing]

Whenever you change a node, you have to follow the rules of node-switching in order to keep the vertex-arrangements the same. Tose rules are that if you change a node with value A from A to -A, all nodes will have an addition of value of A*str(n), where str(n) gives the shortchord of an n-gon where n is the value of the node. otherwise the vertex-arrangements won't be the same , and thus your node-changing will result in something completely different, rather than a different polytope for the same vertex set. Thus f2x"="f2(-x), because str(2)=0, but x3o3o*b3f "=" (-x)3x3o*b3f, and not (-x)3o3o*b3f, because str(3)=1. I cant go in full detail about this, cause I'm going to leave for the weekend, and can only talk after that.

Okay, understood!
So you where changing ex = foxo3ooof3xfoo *b3oxfo&#zx NOT into some fo(-x)o3ooof3xfoo *b3oxfo&#zx, but rather into some fo(-x)o3ooxf3xfoo *b3oxfo&#zx! (Indeed, I should have spotted that, my incmats file of icau pretasto (the partial Stott expansion of that symbol) did already contain that additional x at the middle node... )

This then show as well, that my intended part

Wendy's consideration then would allow also (+x)3(+x)3(+x) *b3. resp. (+x)3(+x)3(+x) *b3(+x) to the respective ones. And surely also (+x)3(+x)3. *b3. to the so far known cases!

can not work directly, as these would produce u-edges at the middle node.

But still the other 4 multidentals should be in the run as the transformations in these arms each would apply to a different layer! Thus the accordingly corrected starting versions now should read:

• fo(-x)o3xoxf3(-x)foo *b3oxfo&#zx + (+x)3(+o)3(+x) *b3(+o) --> Fxox3xoxf3oFxx *b3oxfo&#zx
• resp. fo(-x)3xox3(-x)fo *b3oxf&#zx + (+x)3(+o)3(+x) *b3(+o) --> Fxo3xox3oFx *b3oxf&#zx
• fo(-x)o3xxxf3(-x)foo *b3o(-x)fo&#zx + (+x)3(+o)3(+x) *b3(+x) --> Fxox3xxxf3oFxx *b3xoFx&#zx
• resp. fo(-x)3xxx3(-x)fo *b3o(-x)f&#zx + (+x)3(+o)3(+x) *b3(+x) --> Fxo3xxx3oFx *b3xoF&#zx

But then, we might apply your x --> (-x) rule to the middle node again. This then results in:

• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3ooxf3xfoo *b3oxfo&#zx --> foxo3oo(-x)f3xfxo *b3oxFo&#zx, which then allows for a further pSe at the central node, then resulting in: foxo3xxoF3xfxo *b3oxFo&#zx
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3oox3xfo *b3oxf&#zx --> fox3oo(-x)3xfx *b3oxF&#zx, which then allows for a further pSe at the central node, then resulting in: fox3xxo3xfx *b3oxF&#zx
• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3xoxf3(-x)foo *b3oxfo&#zx --> Foxo3(-x)o(-x)f3ofxo *b3xxFo&#zx, which then allows for a further pSe at the central node, then resulting in: Foxo3oxoF3ofxo *b3xxFo&#zx
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3xox3(-x)fo *b3oxf&#zx --> Fox3(-x)o(-x)3ofx *b3xxF&#zx, which then allows for a further pSe at the central node, then resulting in: Fox3oxo3ofx *b3xxF&#zx
• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3xxxf3(-x)foo *b3o(-x)fo&#zx --> Fxoo3ooof3oFxo *b3xoFo. That one as such ought be interesting itself! And it then allows for a further pSe at the central node, then resulting in: Fxoo3xxxF3oFxo *b3xoFo
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3xxx3(-x)fo *b3o(-x)f&#zx --> Fxo3ooo3oFx *b3xoF. That one as such ought be interesting itself! And it then allows for a further pSe at the central node, then resulting in: Fxo3xxx3oFx *b3xoF

Hehe, this finally increases the count of potential new CRFs up to 12 !

--- rk

[Klitzing]

• fo(-x)o3xoxf3(-x)foo *b3oxfo&#zx + (+x)3(+o)3(+x) *b3(+o) --> Fxox3xoxf3oFxx *b3oxfo&#zx
• resp. fo(-x)3xox3(-x)fo *b3oxf&#zx + (+x)3(+o)3(+x) *b3(+o) --> Fxo3xox3oFx *b3oxf&#zx
• fo(-x)o3xxxf3(-x)foo *b3o(-x)fo&#zx + (+x)3(+o)3(+x) *b3(+x) --> Fxox3xxxf3oFxx *b3xoFx&#zx
• resp. fo(-x)3xxx3(-x)fo *b3o(-x)f&#zx + (+x)3(+o)3(+x) *b3(+x) --> Fxo3xxx3oFx *b3xoF&#zx

But then, we might apply your x --> (-x) rule to the middle node again. This then results in:

• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3ooxf3xfoo *b3oxfo&#zx --> foxo3oo(-x)f3xfxo *b3oxFo&#zx, which then allows for a further pSe at the central node, then resulting in: foxo3xxoF3xfxo *b3oxFo&#zx
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3oox3xfo *b3oxf&#zx --> fox3oo(-x)3xfx *b3oxF&#zx, which then allows for a further pSe at the central node, then resulting in: fox3xxo3xfx *b3oxF&#zx
• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3xoxf3(-x)foo *b3oxfo&#zx --> Foxo3(-x)o(-x)f3ofxo *b3xxFo&#zx, which then allows for a further pSe at the central node, then resulting in: Foxo3oxoF3ofxo *b3xxFo&#zx
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3xox3(-x)fo *b3oxf&#zx --> Fox3(-x)o(-x)3ofx *b3xxF&#zx, which then allows for a further pSe at the central node, then resulting in: Fox3oxo3ofx *b3xxF&#zx
• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3xxxf3(-x)foo *b3o(-x)fo&#zx --> Fxoo3ooof3oFxo *b3xoFo. That one as such ought be interesting itself! And it then allows for a further pSe at the central node, then resulting in: Fxoo3xxxF3oFxo *b3xoFo
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3xxx3(-x)fo *b3o(-x)f&#zx --> Fxo3ooo3oFx *b3xoF. That one as such ought be interesting itself! And it then allows for a further pSe at the central node, then resulting in: Fxo3xxx3oFx *b3xoF

Hehe, this finally increases the count of potential new CRFs up to 12 !

Couldn't resist to start looking into those myself.
Managed so far the first 2:

Ad Fxo3xox3oFx *b3oxf&#zx :
That one sadly does not work. This is because of the needed starting with faceting of ike:
fox 2 xfo 2 oxf &#zx --> fo(-x) 2 xfo 2 oxf &#zx --> fo(-x) 2 (-x)fo 2 oxf &#zx
In the first step you'd introduce regular pentagonal facets underneath the vertices of the inverted edges. These pentagons in fact here are described by ... 2 xfo 2 ... &#zx. The second edge inversion then would swap the base edges (each) of these pentagons (where those are mutually connected) and thus would produce new edges running through that pentagonal vertex set, i.e. introducing f-edges. (The pentagons then become fish shaped figures.) These f-edges then just would get displaced in the partial Stott expansion to be applied, not changed in size. I.e. we would not get rid of them.
Thus it clearly is a valid transformation, so. Just not resulting in a CRF.

Ad Fxox3xoxf3oFxx *b3oxfo&#zx :
Surprisingly that one does not suffer from that above mentioned problem. The former ikes here where dissected into (or rather: augmented by) smaller parts and so the needed edge inversions then become possible too. The finally resulting CRF figure will have a total of 336 vertices (4 classes). For cells it will have 8 coes, 72 octs (3 classes), 288 squippies (3 classes), 288 tets (2 classes), 32 tricues, 96 trips (2 classes), and 8 tuts (as well as a quite huge incidence matrix).
(Have not checked yet whether that one happens to be a known one so. )

--- rk

Edit: cf. this post for the corrected cell list!

[Klitzing]

• fo(-x)o3xoxf3(-x)foo *b3oxfo&#zx + (+x)3(+o)3(+x) *b3(+o) --> Fxox3xoxf3oFxx *b3oxfo&#zx
• resp. fo(-x)3xox3(-x)fo *b3oxf&#zx + (+x)3(+o)3(+x) *b3(+o) --> Fxo3xox3oFx *b3oxf&#zx
• fo(-x)o3xxxf3(-x)foo *b3o(-x)fo&#zx + (+x)3(+o)3(+x) *b3(+x) --> Fxox3xxxf3oFxx *b3xoFx&#zx
• resp. fo(-x)3xxx3(-x)fo *b3o(-x)f&#zx + (+x)3(+o)3(+x) *b3(+x) --> Fxo3xxx3oFx *b3xoF&#zx

Couldn't resist to start looking into those myself.
Managed so far the first 2:

And now the next 2 follow:

Ad Fxo3xxx3oFx *b3xoF&#zx :
As the intended bi-faceting of the in sadi being used ikes did not become regular faced, this does not become better for a tri-faceting.
Thus this again clearly is a valid transformation, but again it is not resulting in a CRF, so.

Ad Fxox3xxxf3oFxx *b3xoFx&#zx :
Again that one does not suffer from that above mentioned problem. The finally resulting CRF figure will have a total of 480 vertices. For cells it will have 24 cubes, 288 squippies, 192 tets, 96 tricues, 240 trips, and 24 tut.
(Again not yet checked whether that one happens to be a known one. )

--- rk

Edit: cf. that post which then states that the latter one after all fails to be CRF: cells are only abstractly valid, but not all geometrically also.

[Klitzing]

Ad Fxox3xxxf3oFxx *b3xoFx&#zx :
The finally resulting CRF figure will have a total of 480 vertices. For cells it will have 24 cubes, 288 squippies, 192 tets, 96 tricues, 240 trips, and 24 tut.

Here it becomes possible to use the additional symmetry of the arms of that Dynkin symbol description.
So we can boil down the huge incidence matrix in that specific case into just:

Code: Select all

Fxox3xxxf3oFxx *b3xoFx&#zx

o...3o...3o... *b3o...    & | 288   * |   2   1   4   2   0 |  1  2   4   3   1   2   2   4   0 |  1  2   1  2   3   2  0
...o3...o3...o *b3...o      |   * 192 |   0   0   0   3   3 |  0  0   0   0   3   3   0   3   3 |  0  0   3  0   3   1  1
----------------------------+---------+---------------------+-----------------------------------+------------------------
.... x... ....    ....    & |   2   0 | 288   *   *   *   * |  1  1   2   0   0   0   0   0   0 |  1  2   0  1   0   0  0
.... .... ....    x...    & |   2   0 |   * 144   *   *   * |  0  2   0   2   0   2   0   0   0 |  1  2   1  0   2   0  0
oo..3oo..3oo.. *b3oo..&#x & |   2   0 |   *   * 576   *   * |  0  0   1   1   0   0   1   1   0 |  0  1   0  1   1   1  0
o..o3o..o3o..o *b3o..o&#x & |   1   1 |   *   *   * 576   * |  0  0   0   0   1   1   0   2   0 |  0  0   1  0   2   1  0
...x .... ....    ....    & |   0   2 |   *   *   *   * 288 |  0  0   0   0   1   1   0   0   2 |  0  0   2  0   1   0  1
----------------------------+---------+---------------------+-----------------------------------+------------------------
.... x...3o...    ....    & |   3   0 |   3   0   0   0   0 | 96  *   *   *   *   *   *   *   * |  1  1   0  0   0   0  0
.... x... .... *b3x...    & |   6   0 |   3   3   0   0   0 |  * 96   *   *   *   *   *   *   * |  1  1   0  0   0   0  0
.... xx.. ....    ....&#x & |   4   0 |   2   0   2   0   0 |  *  * 288   *   *   *   *   *   * |  0  1   0  1   0   0  0
.... .... ....    xo..&#x & |   3   0 |   0   1   2   0   0 |  *  *   * 288   *   *   *   *   * |  0  1   0  0   1   0  0
.... .... o..x    ....&#x & |   1   2 |   0   0   0   2   1 |  *  *   *   * 288   *   *   *   * |  0  0   1  0   1   0  0
.... .... ....    x..x&#x & |   2   2 |   0   1   0   2   1 |  *  *   *   *   * 288   *   *   * |  0  0   1  0   1   0  0
ooo.3ooo.3ooo. *b3ooo.&#x   |   3   0 |   0   0   3   0   0 |  *  *   *   *   *   * 192   *   * |  0  0   0  1   0   1  0
oo.o3oo.o3oo.o *b3oo.o&#x & |   2   1 |   0   0   1   2   0 |  *  *   *   *   *   *   * 576   * |  0  0   0  0   1   1  0
...x .... ...x    ....    & |   0   4 |   0   0   0   0   4 |  *  *   *   *   *   *   *   * 144 |  0  0   1  0   0   0  1
----------------------------+---------+---------------------+-----------------------------------+------------------------
.... x...3o... *b3x...    & |  12   0 |  12   6   0   0   0 |  4  4   0   0   0   0   0   0   0 | 24  *   *  *   *   *  * tut
.... xx.. .... *b3xo..&#x & |   9   0 |   6   3   6   0   0 |  1  1   3   3   0   0   0   0   0 |  * 96   *  *   *   *  * tricu
.... .... o..x    x..x&#x & |   2   4 |   0   1   0   4   4 |  0  0   0   0   2   2   0   0   1 |  *  * 144  *   *   *  * trip
.... xxx. ....    ....&#x   |   6   0 |   3   0   6   0   0 |  0  0   3   0   0   0   2   0   0 |  *  *   * 96   *   *  * trip
.... .... ....    xo.x&#x & |   3   2 |   0   1   2   4   1 |  0  0   0   1   1   1   0   2   0 |  *  *   *  * 288   *  * squippy
oooo3oooo3oooo *b3oooo&#x   |   3   1 |   0   0   3   3   0 |  0  0   0   0   0   0   1   3   0 |  *  *   *  *   * 192  * tet
...x .... ...x    ...x      |   0   8 |   0   0   0   0  12 |  0  0   0   0   0   0   0   0   6 |  *  *   *  *   *   * 24 cube


--- rk

Edit: cf. edit of previous post!

[Klitzing]

Btw., just for those doubting, this recent research on further CRFs truely belongs to this thread - even so it does NOT contain any bilbiro nor thawro.

Why?

They all are derived from ex (or sadi) by the same means, which derive bilbiro from the ike! They just happen to have no bilbiroes so far (but plenty of other stuff, eg. squippies and tricues).

--- rk

[Klitzing]

But then, we might apply your x --> (-x) rule to the middle node again. This then results in:

• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3ooxf3xfoo *b3oxfo&#zx --> foxo3oo(-x)f3xfxo *b3oxFo&#zx, which then allows for a further pSe at the central node, then resulting in: foxo3xxoF3xfxo *b3oxFo&#zx
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3oox3xfo *b3oxf&#zx --> fox3oo(-x)3xfx *b3oxF&#zx, which then allows for a further pSe at the central node, then resulting in: fox3xxo3xfx *b3oxF&#zx
• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3xoxf3(-x)foo *b3oxfo&#zx --> Foxo3(-x)o(-x)f3ofxo *b3xxFo&#zx, which then allows for a further pSe at the central node, then resulting in: Foxo3oxoF3ofxo *b3xxFo&#zx
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3xox3(-x)fo *b3oxf&#zx --> Fox3(-x)o(-x)3ofx *b3xxF&#zx, which then allows for a further pSe at the central node, then resulting in: Fox3oxo3ofx *b3xxF&#zx
• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3xxxf3(-x)foo *b3o(-x)fo&#zx --> Fxoo3ooof3oFxo *b3xoFo. That one as such ought be interesting itself! And it then allows for a further pSe at the central node, then resulting in: Fxoo3xxxF3oFxo *b3xoFo
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3xxx3(-x)fo *b3o(-x)f&#zx --> Fxo3ooo3oFx *b3xoF. That one as such ought be interesting itself! And it then allows for a further pSe at the central node, then resulting in: Fxo3xxx3oFx *b3xoF

Continuing with the first 2 of those:

Ad fox3xxo3xfx *b3oxF&#zx :
This one again looks to be impossible with unit edges only.

Ad foxo3xxoF3xfxo *b3oxFo&#zx :
This one again looks like being CRF. It has 312 vertices (4 classes). Cells would be 8 coes, 32 octs, 240 squippies (3 classes), 384 tets (3 classes), 64 tricues (2 classes), and 16 tuts (2 classes).

--- rk

[student91]

Whenever you change a node, you have to follow the rules of node-switching in order to keep the vertex-arrangements the same. Tose rules are that if you change a node with value A from A to -A, all nodes will have an addition of value of A*str(n), where str(n) gives the shortchord of an n-gon where n is the value of the node. otherwise the vertex-arrangements won't be the same , and thus your node-changing will result in something completely different, rather than a different polytope for the same vertex set. Thus f2x"="f2(-x), because str(2)=0, but x3o3o*b3f "=" (-x)3x3o*b3f, and not (-x)3o3o*b3f, because str(3)=1. I cant go in full detail about this, cause I'm going to leave for the weekend, and can only talk after that.

[...]
But then, we might apply your x --> (-x) rule to the middle node again. This then results in:

• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3ooxf3xfoo *b3oxfo&#zx --> foxo3oo(-x)f3xfxo *b3oxFo&#zx, which then allows for a further pSe at the central node, then resulting in: foxo3xxoF3xfxo *b3oxFo&#zx
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3oox3xfo *b3oxf&#zx --> fox3oo(-x)3xfx *b3oxF&#zx, which then allows for a further pSe at the central node, then resulting in: fox3xxo3xfx *b3oxF&#zx
• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3xoxf3(-x)foo *b3oxfo&#zx --> Foxo3(-x)o(-x)f3ofxo *b3xxFo&#zx, which then allows for a further pSe at the central node, then resulting in: Foxo3oxoF3ofxo *b3xxFo&#zx
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3xox3(-x)fo *b3oxf&#zx --> Fox3(-x)o(-x)3ofx *b3xxF&#zx, which then allows for a further pSe at the central node, then resulting in: Fox3oxo3ofx *b3xxF&#zx
• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3xxxf3(-x)foo *b3o(-x)fo&#zx --> Fxoo3ooof3oFxo *b3xoFo. That one as such ought be interesting itself! And it then allows for a further pSe at the central node, then resulting in: Fxoo3xxxF3oFxo *b3xoFo
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3xxx3(-x)fo *b3o(-x)f&#zx --> Fxo3ooo3oFx *b3xoF. That one as such ought be interesting itself! And it then allows for a further pSe at the central node, then resulting in: Fxo3xxx3oFx *b3xoF

[...]

You have to follow the rules all the time. this means that when you did foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3ooxf3xfoo *b3oxfo&#zx, and you want to do ....3..x.3....*b3.... --> ....3..(-x).3....*b3...., this becomes ..[+x].3..(-x).3..[+x].*b3..[+3]., and thus the (-x) in fo(-x)o.... becomes o and not x. This means the pS's above should rather be foo3xxo3xfx *b3oxF&#zx, Fooo3oxoF3ofxo *b3xxFo&#zx (here you make this mistake only once), Foo3oxo3ofx *b3xxF&#zx.
In the last two you make a transition I'm unable to follow, fo(-x)3xxx3(-x)fo *b3o(-x)f&#zx --> Fxo3ooo3oFx *b3xoF. This is not a normal change-nodes transition. I think you ment fo(-x)3xxx3(-x)fo *b3o(-x)f&#zx --> Fxo3(-x)(-x)(-x)3oFx *b3xoF.

Furthermore I've found quite an interesting property of the node-changing, though I'll split that to another topic.

Ad Fxox3xoxf3oFxx *b3oxfo&#zx :
Surprisingly that one does not suffer from that above mentioned problem. The former ikes here where dissected into (or rather: augmented by) smaller parts and so the needed edge inversions then become possible too. The finally resulting CRF figure will have a total of 336 vertices (4 classes). For cells it will have 8 coes, 72 octs (3 classes), 288 squippies (3 classes), 288 tets (2 classes), 32 tricues, 96 trips (2 classes), and 8 tuts (as well as a quite huge incidence matrix).
(Have not checked yet whether that one happens to be a known one so. )

Ad Fxox3xxxf3oFxx *b3xoFx&#zx :
Again that one does not suffer from that above mentioned problem. The finally resulting CRF figure will have a total of 480 vertices. For cells it will have 24 cubes, 288 squippies, 192 tets, 96 tricues, 240 trips, and 24 tut.
(Again not yet checked whether that one happens to be a known one. )

Those are interesting!! I always thought the nonexistence of a multi-node expansion of ike meant the impossibility of a corresponding 4D expansion, but you say it is possible!!. this opens new possibilities, what about multi-node expansions of ex in axial symmetry, or other interesting things! these new discoveries keep being awesome

[Klitzing]

But then, we might apply your x --> (-x) rule to the middle node again. This then results in:

• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3ooxf3xfoo *b3oxfo&#zx --> foxo3oo(-x)f3xfxo *b3oxFo&#zx, which then allows for a further pSe at the central node, then resulting in: foxo3xxoF3xfxo *b3oxFo&#zx
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3oox3xfo *b3oxf&#zx --> fox3oo(-x)3xfx *b3oxF&#zx, which then allows for a further pSe at the central node, then resulting in: fox3xxo3xfx *b3oxF&#zx
• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3xoxf3(-x)foo *b3oxfo&#zx --> Foxo3(-x)o(-x)f3ofxo *b3xxFo&#zx, which then allows for a further pSe at the central node, then resulting in: Foxo3oxoF3ofxo *b3xxFo&#zx
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3xox3(-x)fo *b3oxf&#zx --> Fox3(-x)o(-x)3ofx *b3xxF&#zx, which then allows for a further pSe at the central node, then resulting in: Fox3oxo3ofx *b3xxF&#zx
• foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3xxxf3(-x)foo *b3o(-x)fo&#zx --> Fxoo3ooof3oFxo *b3xoFo. That one as such ought be interesting itself! And it then allows for a further pSe at the central node, then resulting in: Fxoo3xxxF3oFxo *b3xoFo
• fox3ooo3xfo *b3oxf&#zx --> fo(-x)3xxx3(-x)fo *b3o(-x)f&#zx --> Fxo3ooo3oFx *b3xoF. That one as such ought be interesting itself! And it then allows for a further pSe at the central node, then resulting in: Fxo3xxx3oFx *b3xoF

[...]

You have to follow the rules all the time. this means that when you did foxo3ooof3xfoo *b3oxfo&#zx --> fo(-x)o3ooxf3xfoo *b3oxfo&#zx, and you want to do ....3..x.3....*b3.... --> ....3..(-x).3....*b3...., this becomes ..[+x].3..(-x).3..[+x].*b3..[+3]., and thus the (-x) in fo(-x)o.... becomes o and not x. This means the pS's above should rather be foo3xxo3xfx *b3oxF&#zx, Fooo3oxoF3ofxo *b3xxFo&#zx (here you make this mistake only once), Foo3oxo3ofx *b3xxF&#zx.
In the last two you make a transition I'm unable to follow, fo(-x)3xxx3(-x)fo *b3o(-x)f&#zx --> Fxo3ooo3oFx *b3xoF. This is not a normal change-nodes transition. I think you ment fo(-x)3xxx3(-x)fo *b3o(-x)f&#zx --> Fxo3(-x)(-x)(-x)3oFx *b3xoF.

Damn! Again typos.
(Even so I understood your prior advice, and meant to apply it correctly now... )

Here is what I meant to do - a bit more slowly, and hopefully less error prone:

Code: Select all

| : quirks application = (x -> (-x))
^ : quirks implication = [+P]
. : unchanged
E : expansion

In all these (then error prone) cases I tried to apply 2 or more quirks modes next to each other. (The 2 cases already correctly worked out in fact where the cases with 2 or more quirks modes being mutually non-adjacent.)

Again we start with ex = a and derive d:

Code: Select all

a = foxo3ooof3xfoo *b3oxfo&#zx
      |    ^    .       .         quirks appl. in 3rd layer
becomes
b = fo(-x)o3ooxf3xfoo *b3oxfo&#zx
        ^     |    ^       ^      quirks appl. in 3rd layer
becomes
c = fooo3oo(-x)f3xfxo *b3oxFo&#zx
         EE  E E                  Stott appl. to central node
becomes
d = fooo3xxoF3xfxo *b3oxFo&#zx


We then apply the same to the next arm, so we start with c and derive g:

Code: Select all

c = fooo3oo(-x)f3xfxo *b3oxFo&#zx
    .    ^       |       .        quirks appl. in 1st layer
becomes
e = fooo3xo(-x)f3(-x)fxo *b3oxFo&#zx
    ^    |         ^        ^     quirks appl. in 1st layer
becomes
f = Fooo3(-x)o(-x)f3ofxo *b3xxFo&#zx
           E E  E E               Stott appl. to central node
becomes
g = Fooo3oxoF3ofxo *b3xxFo&#zx


Finally we apply the same to the 3rd arm, i.e. starting with f and derive j:

Code: Select all

f = Fooo3(-x)o(-x)f3ofxo *b3xxFo&#zx
     .       ^       .       |    quirks appl. in 2nd layer
becomes
h = Fooo3(-x)x(-x)f3ofxo *b3x(-x)Fo&#zx
     ^       |       ^         ^  quirks appl. in 2nd layer
becomes
i = Fxoo3(-x)(-x)(-x)f3oFxo *b3xoFo&#zx
           E   E   E E            Stott appl. to central node
becomes
j = Fxoo3oooF3oFxo *b3xoFo&#zx


Cases d, g, and j then have none of these reverse edges any longer. Thus they are candidates for CRFs.

The other originally mentioned cases always were just the diminishing of the 4th layer each. But as I then outlined and as student91 again stated, there a multi quirks mode seems not to work. It always is the cell introduced at these diminishings, which will get the problems.

Even in all these undiminished ones (plus the 2 already worked out ones), the existance of a multi quirks mode not truely was proven. Rather to the converse: I just tried out what should come out, if being applied. Always under the premise that it should be checked individually. But the so far derived outcomes seem to suggest that it ought to get allowable then.

--- rk

[Klitzing]

I always thought the nonexistence of a multi-node expansion of ike meant the impossibility of a corresponding 4D expansion, but you say it is possible!!. this opens new possibilities, what about multi-node expansions of ex in axial symmetry, or other interesting things! these new discoveries keep being awesome

Wrt. Stott expansions (partial or not) applications at several nodes (in sequence or at once) is always possible. We just will have to check that all resulting true edges will have (zero or) unit size then, if we were searching for uniforms or CRFs.

And the same then would have to apply to your quirks stuff.

--- rk

[Klitzing]

Here comes d = fooo3xxoF3xfxo *b3oxFo&#zx now:

it would have 248 vertices (4 classes) and for cells there are
192 squippies (2 classes), 328 tets (4 classes), 64 tricues (2 classes), and 16 tuts (2 classes).

--- rk

Edit: cf. this post for its corrected cell list.

[student91]

I haven't tried to confirm the things you have suggested, but I have investigated the double axial expansion of ex. note that axial symmetry can be represented with a lace city, a lace tower, a lace-universe or a &#zx-thing, because (A2B2C2D)=(A B C D)(.)[&#zx]=(A B C)(D)[tower]=(A B)(C D)[city]=(A)(B C D)[universe].
The first axial expansion of ex has already been found, and is given by x3o5o||o3o5x||o3x5o||f3o5o||o3x5o||o3o5x||x3o5o. it's lace-city looks like this:

Code: Select all

                    o2o                   
                                           
          o2x f2o   x2f   f2o o2x         
                                           
                                           
    x2o   f2f o2F   F2x   o2F f2f   x2o   
                                           
                                           
                                           
                                           
o2o f2x   x2F F2f  Vo2oV  F2f x2F   f2x o2o
                                           
                                           
    o2f   F2o       f2F       F2o   o2f   
                                           
                                           
o2o f2x   x2F F2f  Vo2oV  F2f x2F   f2x o2o
                                           
                                           
                                           
                                           
    x2o   f2f o2F   F2x   o2F f2f   x2o   
                                           
                                           
          o2x f2o   x2f   f2o o2x         
                                           
                    o2o                   

But can also be given as

Code: Select all

                    x2o                   
                                           
          x2x F2o   o2f   F2o x2x         
                                           
                                           
    o2o   F2f x2F   A2x   x2F F2f   o2o   
                                           
    x2f   A2o       F2F       A2o   x2f   
                                           
                                           
x2o F2x   o2F A2f  Bx2oV  A2f o2F   F2x x2o
                                           
                                           
    x2f   A2o       F2F       A2o   x2f   
                                           
    o2o   F2f x2F   A2x   x2F F2f   o2o   
                                           
                                           
          x2x F2o   o2f   F2o x2x         
                                           
                    x2o                   
A=F+x=f+2x
B=V+x=2f+x

This is a polytope with quite some tetrahedra, 24 icosahedra, 20 triangular prisms and 60 squippies. When we want to apply the second expansion, we first want to see the limit/moving patch, so we can see if it is convex. The limit looks like this:

Code: Select all

                    o2o     
                             
          o2x       x2f   f2o
                             
                             
    x2o   f2f       F2x   o2F
                             
                             
                             
                             
o2o f2x   x2F      Vo2oV  F2f
                             
                             
    o2f   F2o       f2F     
                             
                             
o2o f2x   x2F      Vo2oV  F2f
                             
                             
                             
                             
    x2o   f2f       F2x   o2F
                             
                             
          o2x       x2f   f2o
                             
                    o2o     

This limit has true bilbiro cells. This means the double expansion will have true bilbiro cells as well. These bilbiros occur because some vertices that did belong to the full polytope are not part of this patch (you can see an expanded f-ike is taken away), and these vertices formed the apices of some (4 to be exact) bilbiro pseudopyramids. Because these bilbiro's are quite hard to spot, I have drawn the bilbiro in the lace-city:

kiwii.png (8.94 KiB) Viewed 114562 times

(the top o2o has been chopped off). The appearance of these bilbiros make that some icosahedra become (metabi)diminished icosahedra. Furthermore some other things get rearranged a bit. I'm quite confident it is CRF, but not totally sure. It would be great to have some renders of this thing, as it is the first true pSe (3rd kind) of ex that has bilbiro's. Pitifully quickfur hasn't been very active for a while now, (is he having a long holiday or something?) and thus I'm afraid it might take a while to get these renders. (Furthermore, how will we get nice images in the paper without him, where is he?) I have heard of things like stella4d etc. Is it possible to get (ugly) renders from these?
Coordinates for the doubly expanded ex can be read off directly from these two lace-cities:

Code: Select all

                    o2o         o2o                   
                                           
          o2x       x2f   f2o   x2f      o2x         
                                           
                                           
    x2o   f2f       F2x   o2F   F2x      f2f   x2o   
                                           
   
                                           
                                           
o2o f2x   x2F      Vo2oV  F2f  Vo2oV     x2F   f2x o2o


    o2f   F2o       f2F         f2F      F2o   o2f   
                                           
                                           
o2o f2x   x2F      Vo2oV  F2f  Vo2oV     x2F   f2x o2o
                                           
                                           
   
                                           
    x2o   f2f       F2x   o2F   F2x      f2f   x2o   
                                           
                                           
          o2x       x2f   f2o   x2f      o2x         
                                           
                    o2o         o2o           

Code: Select all

                    x2x                   
                                           
          x2o F2x   o2F   F2x x2o         
                                           
                                           
    o2x   F2F x2A   A2o   x2A F2F   o2x   
                                           
    x2F   A2x       F2A       A2x   x2F   
                                           
                                           
x2x F2o   o2A A2F  Bx2xB  A2F o2A   F2o x2x
                                           
                                           
    x2F   A2x       F2A       A2x   x2F   
                                           
    o2x   F2F x2A   A2o   x2A F2F   o2x   
                                           
                                           
          x2o F2x   o2F   F2x x2o         
                                           
                    x2x                   

Klitzing, how do you check these polytopes for CRF-ity? you seem to have found a quick method, and my analyzing is giving me headaches. Do you check it differently, or don't you get headaches that quickly?

[Klitzing]

Coordinates for the doubly expanded ex can be read off directly from these two lace-cities:

Code: Select all

                    o2o         o2o                   
                                           
          o2x       x2f   f2o   x2f      o2x         
                                           
                                           
    x2o   f2f       F2x   o2F   F2x      f2f   x2o   
                                           
   
                                           
                                           
o2o f2x   x2F      Vo2oV  F2f  Vo2oV     x2F   f2x o2o


    o2f   F2o       f2F         f2F      F2o   o2f   
                                           
                                           
o2o f2x   x2F      Vo2oV  F2f  Vo2oV     x2F   f2x o2o
                                           
                                           
   
                                           
    x2o   f2f       F2x   o2F   F2x      f2f   x2o   
                                           
                                           
          o2x       x2f   f2o   x2f      o2x         
                                           
                    o2o         o2o           

Code: Select all

                    x2x                   
                                           
          x2o F2x   o2F   F2x x2o         
                                           
                                           
    o2x   F2F x2A   A2o   x2A F2F   o2x   
                                           
    x2F   A2x       F2A       A2x   x2F   
                                           
                                           
x2x F2o   o2A A2F  Bx2xB  A2F o2A   F2o x2x
                                           
                                           
    x2F   A2x       F2A       A2x   x2F   
                                           
    o2x   F2F x2A   A2o   x2A F2F   o2x   
                                           
                                           
          x2o F2x   o2F   F2x x2o         
                                           
                    x2x                   

Klitzing, how do you check these polytopes for CRF-ity? you seem to have found a quick method, and my analyzing is giving me headaches. Do you check it differently, or don't you get headaches that quickly?

Looks highly interesting, your new find. Could you bring it into some Dynkin style notation too? (I agree with you that these things always make headache!)

Yes I'm doing it differently. Yours is a graphical approach (based on lace cities). Thus you would be quite near to coordinates. This would then make it easy to calculate some edge distances to check them whether or not those are unity. - My approach on the other hand is completely based on the Dynkin diagram Display. I then elaborate the often quite huge incidence matrix therefrom. And a incidence matrix in fact is nothing different to the abstract polytopal description! Thus I easily can read off all elements. But, what I cannot check this way, is whether lacing edges would be truely of the right size, that its realization would become convex, etc...

--- rk

[student91]

Coordinates for the doubly expanded ex can be read off directly from these two lace-cities:

Code: Select all

                    o2o         o2o                   
                                           
          o2x       x2f   f2o   x2f      o2x         
                                           
                                           
    x2o   f2f       F2x   o2F   F2x      f2f   x2o   
                                           
   
                                           
                                           
o2o f2x   x2F      Vo2oV  F2f  Vo2oV     x2F   f2x o2o


    o2f   F2o       f2F         f2F      F2o   o2f   
                                           
                                           
o2o f2x   x2F      Vo2oV  F2f  Vo2oV     x2F   f2x o2o
                                           
                                           
   
                                           
    x2o   f2f       F2x   o2F   F2x      f2f   x2o   
                                           
                                           
          o2x       x2f   f2o   x2f      o2x         
                                           
                    o2o         o2o           

Code: Select all

                    x2x                   
                                           
          x2o F2x   o2F   F2x x2o         
                                           
                                           
    o2x   F2F x2A   A2o   x2A F2F   o2x   
                                           
    x2F   A2x       F2A       A2x   x2F   
                                           
                                           
x2x F2o   o2A A2F  Bx2xB  A2F o2A   F2o x2x
                                           
                                           
    x2F   A2x       F2A       A2x   x2F   
                                           
    o2x   F2F x2A   A2o   x2A F2F   o2x   
                                           
                                           
          x2o F2x   o2F   F2x x2o         
                                           
                    x2x                   

Klitzing, how do you check these polytopes for CRF-ity? you seem to have found a quick method, and my analyzing is giving me headaches. Do you check it differently, or don't you get headaches that quickly?

Looks highly interesting, your new find. Could you bring it into some Dynkin style notation too? (I agree with you that these things always make headache!)

This notation can easilly be read out of the two provided lace-cities. those lace-cities are connected in the way that the para-space of one lacecity is perp-space of the other and vice-versa. this means that when you see x2x in one lace-city, you look at the other lace-city what element makes a x2x-shape (or you look what shape is made by the x2x's), and then you conclude that (x2x)(Vo2oV) is part of the polytope, or x2x2V2o and x2x2o2V when written differently. To help you prevent headaches, I will do this. you then have:

Code: Select all

(x2o)(A2F)+(o2x)(F2A) + (F2f)(o2x)+(f2F)(x2o)+
(V2o)(x2x)+(o2V)(x2x) + (o2o)(B2x)+(o2o)(x2B)+
(F2o)(F2o)+(o2F)(o2F) + (x2f)(x2A)+(f2x)(A2x)+
(o2f)(o2A)+(f2o)(A2o) + (F2x)(x2F)+(x2F)(F2x)+
(f2f)(F2F)

These should be all. If you want it even more Dynkin-like, you can replace the parentheses with 2's. If you really want to, you could even put them in a single line, to destroy any readability:

Code: Select all

x2o2A2F
o2x2F2A
F2f2o2x
f2F2x2o

V2o2x2x
o2V2x2x
o2o2B2x
o2o2x2B

F2o2F2o
o2F2o2F
x2f2x2A
f2x2A2x

o2f2o2A
f2o2A2o
F2x2x2F
x2F2F2x

f2f2F2F

=> xoFfVoooFoxfofFxf2oxfFoVoooFfxfoxFf2XFoxxxBxFoxAoAxFF2FAxoxxxBoFAxAoFxF&#zx (Completely unreadable )

Yes I'm doing it differently. Yours is a graphical approach (based on lace cities). Thus you would be quite near to coordinates. This would then make it easy to calculate some edge distances to check them whether or not those are unity.

You don't even need the full coordinates to calculate the edge-length, IF I want to, your spreadsheet always suffices. most of the time though I don't use this (with the pSe's, the earlier discoveries were all done with the spreadsheet), but I use my knoledge of the polytope I'm expanding to bypass the actual calculation, and cause a little bit more headache with return of some time.

- My approach on the other hand is completely based on the Dynkin diagram Display. I then elaborate the often quite huge incidence matrix therefrom. And a incidence matrix in fact is nothing different to the abstract polytopal description! Thus I easily can read off all elements. But, what I cannot check this way, is whether lacing edges would be truly of the right size, that its realization would become convex, etc...

--- rk

Sounds like more work, but more info. My method still works using the same two restrictions:
1. the limits must be CRF
2. the things of import/export must be CRF.
I haven't fully checked 1 in this case, and just assumed 2 to be true, so it's very likely that this polytope is CRF.

[Klitzing]

=> xoFfVoooFoxfofFxf2oxfFoVoooFfxfoxFf2XFoxxxBxFoxAoAxFF2FAxoxxxBoFAxAoFxF&#zx (Completely unreadable )

I agree
Even so this is what my method would need for.

I haven't fully checked 1 in this case, and just assumed 2 to be true, so it's very likely that this polytope is CRF.

Similar to my recent posts. The matrices come out correctly, so at least some such structure should exist. But I cannot say much about the geometric realization.

E.g. meanwhile run through g = Fooo3oxoF3ofxo *b3xxFo&#zx as well. That one then has 32+96+32+24 = 184 vertices. Cells ought to be 8+32+96+96+192+32+8 = 464 tets, 32 tricues, 96 squippies, and 8 tuts. - But then I spotted an edge class in the matrix, which shall have 6 incident tets. - So either I made some further error in calculation, or it simply is not possible with unit edges only (e.g. by having longer lacing ones, and so dihedral angles would change accordingly). ...

Even so the other ones provided so far did not show up such a prominent fault, we still ought to crosscheck these independently. Best by coordinates or other means for edge size calculations - at least for all lacing types. Or even better by Quickfur's picture program.

--- rk

Edit: in that post the above guess on non-unit lacings gets assured. Thus the above mentioned cell list remains abstractly possible, but not metrically correct. The mere count of longer lacing types then shows that no CRF would result in that case.

[student91]

=> xoFfVoooFoxfofFxf2oxfFoVoooFfxfoxFf2XFoxxxBxFoxAoAxFF2FAxoxxxBoFAxAoFxF&#zx (Completely unreadable )

I agree
Even so this is what my method would need for.

I just realized this awful-to-read thing can be made a bit more readable by grouping the things differently:
Line||bilbiro||doe-thing||id-thing||f-ike-thing||id-thing etc.
(x)(Fxo)(AxoF)(BAFxxo)(AFx)2(o)(xfo)(oxFf)(oxFVof)(foF)2(o)(oxf)(xFof)(ofxoVF)(oFf)2(B)(A)(F)(x)(o)
Note the last node has less classes. You should just group things between parentheses together.

Similar to my recent posts. The matrices come out correctly, so at least some such structure should exist. But I cannot say much about the geometric realization.

E.g. meanwhile run through g = Fooo3oxoF3ofxo *b3xxFo&#zx as well. That one then has 32+96+32+24 = 184 vertices. Cells ought to be 8+32+96+96+192+32+8 = 464 tets, 32 tricues, 96 squippies, and 8 tuts. - But then I spotted an edge class in the matrix, which shall have 6 incident tets. - So either I made some further error in calculation, or it simply is not possible with unit edges only (e.g. by having longer lacing ones, and so dihedral angles would change accordingly). ...

you could of course find out what the length of that edge is with the spreadsheet. You should just find out what the lacing-length between the two parts is with height 0, and you have the edge-length. (assuming the edge is formed by "vertex of part 1"+"vertex of part 2"&#zx. If not, the edge-length should be derivable trivially)

Even so the other ones provided so far did not show up such a prominent fault, we still ought to crosscheck these independently. Best by coordinates or other means for edge size calculations - at least for all lacing types. Or even better by Quickfur's picture program.

--- rk

now we need quickfur or some kind of program that can process raw coordinates to polytopes in eg Stella format. Do you know someone who has time to build a convex-hull program, or do you know a way to make quickfur active again? I can't wait to see the full structure of these new things.

[Klitzing]

Similar to my recent posts. The matrices come out correctly, so at least some such structure should exist. But I cannot say much about the geometric realization.

you could of course find out what the length of that edge is with the spreadsheet. You should just find out what the lacing-length between the two parts is with height 0, and you have the edge-length. (assuming the edge is formed by "vertex of part 1"+"vertex of part 2"&#zx. If not, the edge-length should be derivable trivially)

Seems that there ought to be some mystical ether for consciousness:
Inventions or other mental break-throughs occur always within a short time interval - otherwise uncorrelated - at several places all over of the world.
That hint of yours also dawned on me this morning independently ...

--- rk

[wendy]

I have been thinking about the problem of very long strings that has come up in the list. There are several ways out of this. It's after all, a table of coordinates, and if the string is getting too long, ye might have to reformat the table.

I use commas, eg ox,q,xo8oo,o,oo&#qt. But ye could use something along the line of Klitzing's atop stuff, with a lead symmetry, eg

8: | oo | xo | qo | xo | oo &#qt

It's more useful when ye have longer strings at hand.

I prefer to use () to denote a single symbol, made from an expression, eg (F=f+1) is perfectly acceptable and self explaining (to introduce other F symbols where it is not expected.

Ideally, the whole spirit of the thing is that one should be able to cut and paste the string into some command, to do something.