## Classifying the segmentochora

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

### Re: Classifying the segmentochora

Okay, I've tabulated wintersolstice's list, along with my own names and indices for them plus Klitzing's index:

I have not included wintersolstice's "Partially-base snubdis octahedron antiprism", because I have no idea what he meant by that name. (if you can provide me Klitzing's index for that shape, that would help)

I would classify all of these (antiprisms included) as cupolae. However, that still leaves a lot of segmentotopes which are not prisms, pyramids, cupolae (by my definition) or wedges (by Klitzing's definition). I would object to calling these remaining segmentotopes cupolae, unless they are shown to be structurally similar to the ones in this list...

In case you're wondering why it's an image, I didn't have time to re-type it as HTML. I will do that later though.

A cross-check of the Klitzing indices would be appreciated, of course

Keiji

Posts: 1962
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

### Re: Classifying the segmentochora

Keiji wrote:I have not included wintersolstice's "Partially-base snubdis octahedron antiprism", because I have no idea what he meant by that name. (if you can provide me Klitzing's index for that shape, that would help0

it's the "cube||icosahedron"

"snubdis"= alternation of truncate. if you take the octahedron and truncate it, then alternate it you get the icosahedron, so it's based on the cube||octahedron (i.e octahedral/cube antiprism)

I indexed a few more a couple of posts ago but I'll repost them (in Bold

"Truncated tetrahedral cupoliprism" 55
(this is vertex transitive but not uniform because it has Johnson solids for some of it's cells)

gyro 5-gon cupolarotunda ring 146

It's augmentation 139

4-gon bipyramidal bicupolic ring 109

gyrated octahedral prism 13

diminished 3-gon antiprismatic ring/bidimished rectified 5 cell 8

3-gon||tridiminished icosahedron (unnamed) 33

and I'll post the 49 diminishes and gyrations of these above shapes as soon as I can that should classify them all
wintersolstice
Trionian

Posts: 91
Joined: Sun Aug 16, 2009 11:59 am

### Re: Classifying the segmentochora

wintersolstice wrote:
Keiji wrote:I have not included wintersolstice's "Partially-base snubdis octahedron antiprism", because I have no idea what he meant by that name. (if you can provide me Klitzing's index for that shape, that would help0

it's the "cube||icosahedron"

"snubdis"= alternation of truncate. if you take the octahedron and truncate it, then alternate it you get the icosahedron, so it's based on the cube||octahedron (i.e octahedral/cube antiprism)

Ooh!

Can we just call it the "snubdis antiprism" or somesuch then, to keep it short?

I don't really want to name it after any individual Platonic, because it's pretty much cross-family, and unique in that respect.

Keiji

Posts: 1962
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

### Re: Classifying the segmentochora

Keiji wrote:Can we just call it the "snubdis antiprism" or somesuch then, to keep it short?

I don't really want to name it after any individual Platonic, because it's pretty much cross-family, and unique in that respect.

well I don't see any problem with that it is a bit daft anyway call "octahedral" when it doesn't even have an octahdral
cell. (I only used that because I wanted to give it a name!)

anyway I've catorgorised the rest of the segmentoptopes

CRF polychora discovery project segmentochoron,
wintersolstice
Trionian

Posts: 91
Joined: Sun Aug 16, 2009 11:59 am

### Re: Classifying the segmentochora

quickfur wrote:
Klitzing wrote:
quickfur wrote:[...]As for what is/isn't a cupola, I think that may be splitting hairs. As Klitzing says, there's no unique generalization of the 3D notion of cupola. IMO the real solution is to recognize that there is no unique 4D analogue of the 3D cupolae instead of trying to decide which of the equally-valid possibilities should be designated as "true" cupolae.

Well, a cupola in 3d has full symmetrical bases. It has triangles and squares connecting those.
So to extrapolate these figures into 4d, you would have to say what each of those components would become. Top-base polygons (the smaller ones) might become platonic solids. (But I extended that even to quasiregular cases after all.) The triangles most obviously generalize to pyramids. The squares should become somthing with axial symmetry again. So they either could become antiprisms (my choice, as there are more segmentochora which can be classified by that case) or prisms. In the latter case you would have additional things occuring within that new dimension: triangular prisms (kind as line atop square). Those cases do not even apply to ike (it would be hyperbolic!), nor to the quasiregulars. But those OTOH are caps again...
[...]

But again, I think this is just splitting hairs. I could, for example, analyse a 3D cupola as a segmentohedron having a bottom face that is the result of marking an unmarked node in the CD diagram of the top face. For example, a square cupola would have a top face x4o, then the bottom face would be x4x. Then in the 4D case, I can allow any such combination: x4o3o || x4x3o can be considered a cupola, and so can x4x3o || x4x3x, or o4x3o || x4x3o, or any other such combination with the same relationship between the top/bottom cells. This would become a third definition of 4D cupola. (And I'm sure you can come up with more, if you wanted to.)

I could argue that this definition is more encompassing than yours, and therefore "better". But that is just a matter of opinion. My point is that there is more than one way to generalize the 3D concept, which aren't all compatible with each other, so maybe we should just call them all lace prisms and be done with it, instead of trying to argue over which definition of cupola is the "correct" one.

Oh great idea! I like that one. - But your definition of 4cupolae then encompasses a huge class, which even then does NOT include my 4cupolae: those were defined with bases which un-ring the ringed node of the top! (Else no antiprisms were lacing facets.)

So let's revisit 3d first.
We have the following cases: (P some integer)
Code: Select all
`oPo || xPo : pyramidoPo || xPx : (except for P=2 flat or even hyperbolic)xPo || xPo : prismxPo || oPx : antiprismxPo || xPx : cupolaxPx || xPx : prism again`

Now for 4d: (P,Q integers)
Code: Select all
`oPoQo || xPoQo : pyramidoPoQo || oPxQo : (except for P=Q=3 flat or even hyperbolic, but o3x3o = x3o4o!)oPoQo || xPxQo : (except for P=2 hyperbolic)oPoQo || xPxQx : (except for 2=P<=Q hyperbolic)xPoQo || xPoQo : prismxPoQo || oPxQo : what I considered a cupolaxPoQo || oPoQx : what I considered an antiprismxPoQo || xPxQo : (except for P=2 or P=Q=3 flat or even hyperbolic)xPoQo || xPoQx : what I considered a capxPoQo || oPxQx : (except for Q=2 flat or even hyperbolic)xPoQo || xPxQx : (except for Q=2 flat or even hyperbolic)oPxQo || oPxQo : prism againoPxQo || xPxQo : no special nameoPxQo || xPoQx : what I considered a cupola toooPxQo || xPxQx : (except for 2=P<=Q flat or even hyperbolic)xPxQo || xPxQo : prism againxPxQo || xPoQx : no special namexPxQo || oPxQx : no special namexPxQo || xPxQx : no special namexPoQx || xPoQx : prism againxPoQx || xPxQx : no special namexPxQx || xPxQx : prism again`

So either we coin a special name for any class, just as in 3d, or we could cummulate every lace prism except of prisms, antiprisms, or pyramids into a large class of copoloids... (being thus even larger than your proposal).

--- rk
Klitzing
Pentonian

Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Classifying the segmentochora

wintersolstice wrote:
Klitzing wrote:You should be careful when applying names to kind of property-extrapolations into 4d, which do not conform with the very meaning of the word itself. This is a great deal esp. of Wendys polygloss, to try to cut all that historically wrong applied even. - The very word gyro just means rotated. Sure a rotated polygon looks like its dual, thus for segmentohedra this would be the same. But in 4d a rotated cube does not become an octahedron!

because a rotated square becomes it's dual this could mean that gyration could be redifined to mean both in 4D (like there being more than one meaning of "antiprism" in 4D) I'm not saying it should only that it could

I created a thread for a proposal to modify the definition of "Gyrated" for both analogies

it is only a suggestion and maybe there is a better one

No! gyro is greek and means rotating. Consider a gyroscope, which is spinning. Nothing within this stem which would somehow relate to dualizing. (The latter is a mere topological concept.)
--- rk
Klitzing
Pentonian

Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Classifying the segmentochora

Keiji wrote:I have not included wintersolstice's "Partially-base snubdis octahedron antiprism", because I have no idea what he meant by that name. (if you can provide me Klitzing's index for that shape, that would help)
[...]

Somewhere came apon that one already too. This is again a misnomer! There is nothing like a snubdis-operator!!!
The operation is snubbing, so the operator is snub-.
There has been some discussion in the past about the snub names:
The snub cube (s3s4s) could equally well be called a snub octahedron. And a snub dodecahedron (s3s5s) could equally well called a snub icosahedron.
This is why Norman Johnson was proposing snub cuboctahedron, resp. snub icosidodecahedron instead.
And the icosahedron likewise could be called a snub tetratetrahedron
In fact this operation is well-defined as well. It is kind of alternating a truncated form: the truncation of a rectified form especially.
The truncated icosidodecahedron is just the omnitruncated form, etc.
Likewise the snub icositetrachoron was called snub dis-icositetrachoron (as that symmetry group does look equal from both sides). "dis" is greak and just means 2, double. Like "bi" in latin.
--- rk
Klitzing
Pentonian

Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Classifying the segmentochora

Klitzing wrote:
wintersolstice wrote:[...]
because a rotated square becomes it's dual this could mean that gyration could be redifined to mean both in 4D (like there being more than one meaning of "antiprism" in 4D) I'm not saying it should only that it could
[...]
it is only a suggestion and maybe there is a better one

No! gyro is greek and means rotating. Consider a gyroscope, which is spinning. Nothing within this stem which would somehow relate to dualizing. (The latter is a mere topological concept.)
--- rk

I agree that gyro is misleading in this context. Maybe we should coin a new term for it. Since dualizing is kinda like substituting a polytope with its counterpart, I looked up "counterpart" in various languages on Google Translate:

Galician: cambio (counterpart) - cambiotope, cambiated? Sounds nice. But the meaning is non-obvious.
Italian: controparte - controtope, contrated? Maybe.
Slovenian: nasprotna - nasprotope, nasprated? Doesn't sound too good.
Turkish: karşılık (I don't know Turkish, but I assume the anglicized form is something like carsilic or carshilic). Carsilated, maybe?

I omitted the languages that give hard-to-anglicize words or something too similar to "counterpart". The possibilities don't look too good, so I tried "conjugate" instead (ala conjugate of a complex number):

Belarusian: spaluchany - spaluchated? Doesn't sound good and looks weird.
Bulgarian: spergnat - spergiated? Doesn't sound good.
Czech: časovat - chasovated? Looks weird.
Dutch: vervoegen - vervogated? verviated? Maybe.
Estonian: paaris - pariated? parisated? Looks OK but sounds obscure.
Greek: συζευγνύω - suzeugnio - looks weird, let's anglicize it as suzated? suzegiated? None of them looks any good.
Latvian: kopoties - copotiated? copated? Hmm. What about syncopated? A nice existing English word that sorta kinda implies the meaning of shifting (a polytope into its dual), an off-beat (using the dual instead of the polytope).
Russian: сопряженный (sopryazhennyi) - sopriated? soprigiated? May be workable but doesn't look promising.
Slovak: časovať (chasovat) - chasovated? chasiated? Nah...
Swahili: nyambua - nambiated? nambulated? Sounds weird.
Vietnamese: liên hợp - linopated, maybe? Seems workable, but sounds to close to "linear" which would give the wrong idea.
Welsh: cyfun - cyphonated? Too close to "siphon". Cyphated? Sounds too obscure, but may be usable.

Hmm. Looks like the best candidates are syncopated (existing English word!) and cambiated. I vote for syncopated.
quickfur
Pentonian

Posts: 2482
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: Classifying the segmentochora

Klitzing wrote:[...]
So either we coin a special name for any class, just as in 3d, or we could cummulate every lace prism except of prisms, antiprisms, or pyramids into a large class of copoloids... (being thus even larger than your proposal). [...]

I prefer not to coin special names for different subclasses of things, unless there's a good reason for it. For example, if a specific subclass has special interest due to some special properties, or is most-studied, etc.. Coining a new name just for the sake of coining a new name seems to be a waste of word-stems that could otherwise be reused for other, more useful notions.

So IMO, let's call them all cupoloids, and leave it at that.
quickfur
Pentonian

Posts: 2482
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: Classifying the segmentochora

This is getting a bit silly now, so I'd like to just step in and say this:

The focus of this thread was to classify the segmentochora, not to name them.

Names are useful tools, but I really don't care whether you call it K4.21, the cube||icosahedron, the snubdis antiprism, the syncopated whatsit or anything else.

The important thing is to find out which segmentochora are similar to others, so that we can group things together (like we did a while ago for the bicupolic rings), understand what things are more easily, and overall make it easier to see important errors such as counting things twice or thinking something can be CRF when it can't.

I don't know, maybe everyone else on this forum is able to just picture one polyhedron alongside another, imagine connecting it up and go "Ah, this is one of those!", but I sure can't. And I would hope that this exercise was not just for my own benefit.

Klitzing wrote:There is nothing like a snubdis-operator!!!

I hate to speak out against a mathematician I respect, however, there is no need for this. wintersolstice defines, and I quote, "snubdis = alternation of truncate". I am not sure of the etymology of the latter half of the word myself, but as above, names are tools, tools for communicating a concept, and what the name is does not matter so long as it is useful and not misleading. Similarly, my "snubdis antiprism" - which is still tentative, by the way - was only conjured up because it rolls off the tongue nicely and it certainly doesn't mean anything else. If anyone has a better name for the shape, I'm all ears.

On the gyro vs dual vs syncopated issue though, this is absolutely unnecessary. gyro means rotated and has nothing to do with dual except for the (perhaps misleading to some) overlap in 2D. We already have a word for describing duals, this word is dual and there's nothing wrong with it. Why write "the syncopated X" when I can just write "the dual of X"?

With regards to the post ending

Klitzing wrote:So either we coin a special name for any class, just as in 3d, or we could cummulate every lace prism except of prisms, antiprisms, or pyramids into a large class of copoloids... (being thus even larger than your proposal).

I shall point you to the table I have added at CRF polychora discovery project#Cupolae_of_regular_polyhedra. I believe this covers all the listed cases. They are all named as either cupolae or antiprisms, but with prefixes, mostly reused from my uniform polytope naming conventions. This does not waste roots, and upon seeing the name you have an idea of the kind of shape it is, because you already know what cupolae and antiprisms are.

Keiji

Posts: 1962
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

### Re: Classifying the segmentochora

Keiji wrote:I shall point you to the table I have added at CRF polychora discovery project#Cupolae_of_regular_polyhedra. I believe this covers all the listed cases. They are all named as either cupolae or antiprisms, but with prefixes, mostly reused from my uniform polytope naming conventions. This does not waste roots, and upon seeing the name you have an idea of the kind of shape it is, because you already know what cupolae and antiprisms are.

is is OK if I put my names for then aswell (like under yours to refer to variations of the names)

and have you seen my update on the segmentochora? (I fully classified then on both lists )
wintersolstice
Trionian

Posts: 91
Joined: Sun Aug 16, 2009 11:59 am

### Re: Classifying the segmentochora

Keiji wrote:This is getting a bit silly now, so I'd like to just step in and say this:

The focus of this thread was to classify the segmentochora, not to name them.

Names are useful tools, but I really don't care whether you call it K4.21, the cube||icosahedron, the snubdis antiprism, the syncopated whatsit or anything else.

The important thing is to find out which segmentochora are similar to others, so that we can group things together (like we did a while ago for the bicupolic rings), understand what things are more easily, and overall make it easier to see important errors such as counting things twice or thinking something can be CRF when it can't.

And that is exactly why I've been saying that we should group the cupoloids together, because they are basically the convex hull of two parallel polyhedra from the same symmetry group. I singled out the pyramids and wedges because they are special cases where one of the polyhedra is subdimensional or degenerate (i.e. a point or a polygon, respectively). You can further single out antiprisms and other such shapes if you like to treat them specially, but I mean, ultimately, what's the difference between cube||octahedron and truncated_cube||truncated_octahedron? You basically get some kind of lacing cells that connect corresponding elements of the CD diagram to each other between the top/bottom cells. If you have a point on the top and a triangle on the bottom, that makes a tetrahedron. If you have an edge on top and an anti-aligned edge on the bottom, you get a line antiprism (i.e., sideways-oriented tetrahedron). If you have a square on top and an octagon on the bottom you get a 3D cupola. If you have pentagon on top and an antialigned pentagon on the bottom, you get a pentagonal antiprism. And vice versa, and so forth. Each correponding element on the CD diagram will generate one of these connecting cells between the top and the bottom.

I daresay grouping the cupoloids together actually helps understanding, because you can very clearly see the pattern of how the side cells are generated based on which CD diagram elements are ringed/unringed between the top/bottom cells, regardless of whatever analogy one may wish to draw with the 3D cases. Making arbitrary subgroups among them only serves to obfuscate the unifying notion, that is, the pairing up of the corresponding nodes of the CD diagram of the top and bottom cells, or in other words, a lace prism.

I don't know, maybe everyone else on this forum is able to just picture one polyhedron alongside another, imagine connecting it up and go "Ah, this is one of those!", but I sure can't. And I would hope that this exercise was not just for my own benefit.

The easiest way to think about this is to use the CD diagram. Take cubic symmetry, as an example. If you start with a cuboctahedron, you have 8 triangles that correspond with the cube's vertices and 6 diamonds (i.e. dual squares) corresponding with the cube's faces. Put it inside a rhombicuboctahedron, which has 8 triangles corresponding with the cube's vertices, 6 squares corresponding with the cube's 6 faces, and another 12 squares corresponding with the cube's edges. So what are the side cells?

1) The 8 triangles of the cuboctahedron correspond with the 8 triangles of the rhombicuboctahedron (since they both correspond with the cube's vertices). They are anti-aligned (compare an octahedron vs. a cuboctahedron, both lined up in cubic symmetry orientation: if you look straight down the octahedron's face, you'll find that the cuboctahedron's triangular face is dual to it). So you have 8 cells that are segmentohedra of two anti-aligned triangles. What could that be? An octahedron, of course (i.e., a triangular antiprism).

2) The 12 off-axial faces of the cuboctahedron correspond with the 12 vertices of the cuboctahedron (because both correspond with the cube's edges). So you have vertex || square, which is a square pyramid. Therefore, there are 12 square pyramids.

3) The 6 square faces of the cuboctahedron correspond with the rhombicuboctahedron's 6 square faces (both correspond with the cube's faces). These are again, anti-aligned, so you have square||dual_square, i.e., a square antiprism. So there are 6 square antiprisms.

Therefore, cuboctahedron||rhombicuboctahedron has cells: 1 cuboctahedron, 1 rhombicuboctahedron, 8 octahedra, 12 square pyramids, and 6 square antiprisms. Since all elements of the CD diagram are accounted for, these are the only cells there are.

What about octahedron||truncated_cube? Well, a first hint is that the octahedron's vertices correspond with the truncated cube's octagons, so to connect them would require 6 octagonal pyramids, which, as we know, cannot be CRF. So the result cannot be CRF either. Then the truncated cube's 8 triangles correspond with the octahedron's faces, anti-aligned, so you get 8 more octahedra (but most likely non-CRF, because they flank the octagonal pyramids). Then you have the octahedron's edges corresponding with 12 of the truncated cube's edges, again anti-aligned. So there must be 12 tetrahedra. So octahedron||truncated_cube is non-CRF, and consists of 1 (regular) octahedron, 1 truncated cube, 6 octagonal pyramids, 8 (non-regular) octahedra, and 12 (non-regular) tetrahedra.

You can repeat this exercise with any two polyhedra from the same symmetry group, and by matching up the corresponding elements according to the CD diagram (i.e., according to the symmetry group) you can easily derive the shape and number of side cells.

[...]
On the gyro vs dual vs syncopated issue though, this is absolutely unnecessary. gyro means rotated and has nothing to do with dual except for the (perhaps misleading to some) overlap in 2D. We already have a word for describing duals, this word is dual and there's nothing wrong with it. Why write "the syncopated X" when I can just write "the dual of X"?

Because wintersolstice was using "gyro/gyrated" as a prefix/adjective. A gyrated rhombicuboctahedron is not a dual rhombicuboctahedron; the word "gyrated" implies that a subpart of the polyhedron was modified by rotation. In 4D, you can cut off some bicupolic rings off some uniform polychora X, and stick them back rotated the "wrong" way -- so those would sensibly be called "gyrated X". But wintersolstice's idea was that not only you can stick the part you cut off back rotated; you can also replace the part with something generated from a dual polytope. So it's not the case that X becomes its dual, nor that X has a part P cut off and have the dual of P glued back on, but that P itself is generated from an underlying shape Q, and now we use the dual of Q to generate a (adjective) version of P to glue back onto X. The issue at hand is, what adjective should we apply to X to indicate such a modification? Simply calling it "dual X" is obviously wrong: we're not taking the dual of X. Calling it "gyrated" is misleading, because nothing is being rotated. Hence the need for a new adjective to describe cutting off a subpart generated from Q and gluing back a subpart generated from the dual of Q.

With regards to the post ending

Klitzing wrote:So either we coin a special name for any class, just as in 3d, or we could cummulate every lace prism except of prisms, antiprisms, or pyramids into a large class of copoloids... (being thus even larger than your proposal).

I shall point you to the table I have added at CRF polychora discovery project#Cupolae_of_regular_polyhedra. I believe this covers all the listed cases. They are all named as either cupolae or antiprisms, but with prefixes, mostly reused from my uniform polytope naming conventions. This does not waste roots, and upon seeing the name you have an idea of the kind of shape it is, because you already know what cupolae and antiprisms are.

I don't find those names very helpful at all -- maybe you do, and that's OK, but I find it much easier to understand these shapes as lace prisms as I described above (matching corresponding elements of the underlying CD diagrams to generate the lacing cells between the top and bottom). For this, Klitzing's X||Y notation seems to be almost ideal (and definitely more pronunciable than Wendy's CD diagram based notation -- which OTOH does have the virtue of being precise). I have a hard time convincing myself that there's a compelling argument for inventing more naming conventions for what already has a perfectly suitable naming convention.
quickfur
Pentonian

Posts: 2482
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: Classifying the segmentochora

quickfur wrote:
Because wintersolstice was using "gyro/gyrated" as a prefix/adjective. A gyrated rhombicuboctahedron is not a dual rhombicuboctahedron; the word "gyrated" implies that a subpart of the polyhedron was modified by rotation. In 4D, you can cut off some bicupolic rings off some uniform polychora X, and stick them back rotated the "wrong" way -- so those would sensibly be called "gyrated X". But wintersolstice's idea was that not only you can stick the part you cut off back rotated; you can also replace the part with something generated from a dual polytope. So it's not the case that X becomes its dual, nor that X has a part P cut off and have the dual of P glued back on, but that P itself is generated from an underlying shape Q, and now we use the dual of Q to generate a (adjective) version of P to glue back onto X. The issue at hand is, what adjective should we apply to X to indicate such a modification? Simply calling it "dual X" is obviously wrong: we're not taking the dual of X. Calling it "gyrated" is misleading, because nothing is being rotated. Hence the need for a new adjective to describe cutting off a subpart generated from Q and gluing back a subpart generated from the dual of Q.

this is probably my fault for suggesting it , but that was based on analogy only. but I do agree that a new word is needed to remove confusion

maybe something based on the word dual?
wintersolstice
Trionian

Posts: 91
Joined: Sun Aug 16, 2009 11:59 am

### Re: Classifying the segmentochora

I recall wrangling this list into someting very simple. I'll give it another shot.
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1811
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Classifying the segmentochora

I still think of ike || cube as a 'crown jewel'. It's the only one that has two pyritohedral figures in its figure.

By the new notation: 3xo*xx2o%&t

Ok, here's the pyritohedral group. It has an orbifold of 3*2. This is a kind of symmetry group akin to the dynkin graph, but is the handy work of conway and thurston (no, not the cowboy's footballer).

There is an extra kind of edge, a 'swallowed edge'. This is an edgekin that is reabsorbed as an internal chord of a polygon, say like the diagonal of a square.

3x*o2o Octahedron
3x*%2o Cuboctahedron (the % is down the diagonal of the square)
3o * x2% Cube. (the squares are actually x % rectangles, the edges are both x and % coinciding
3x*x2o Icosahedron (there's triangles 3x, and iscolese triangles x*x
3x*x2% tCO. The octagon is divided into three faces, by a pair of parallel %, which are parallel edges
3x*x2x rCO The cube faces are the x2x (A,B). There are the rD faces, which generalised to trapeziums ACBC. The triangles are CCC always regular

None of the others stack up like this...
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1811
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Classifying the segmentochora

wendy wrote:I still think of ike || cube as a 'crown jewel'. It's the only one that has two pyritohedral figures in its figure.

By the new notation: 3xo*xx2o%&t

Ok, here's the pyritohedral group. It has an orbifold of 3*2. This is a kind of symmetry group akin to the dynkin graph, but is the handy work of conway and thurston (no, not the cowboy's footballer).

Dear Wendy,
would you mind to outline those Conway-Thurston symbols for symmetry groups once more,
I still have not digested those in full...
(Meanwhile I do know that 3*2 denotes the pyritohedral group, but still I cannot dechiffre those symbols and their actions.)
There is an extra kind of edge, a 'swallowed edge'. This is an edgekin that is reabsorbed as an internal chord of a polygon, say like the diagonal of a square.

3x*o2o Octahedron
3x*%2o Cuboctahedron (the % is down the diagonal of the square)
3o * x2% Cube. (the squares are actually x % rectangles, the edges are both x and % coinciding
3x*x2o Icosahedron (there's triangles 3x, and iscolese triangles x*x
3x*x2% tCO. The octagon is divided into three faces, by a pair of parallel %, which are parallel edges
3x*x2x rCO The cube faces are the x2x (A,B). There are the rD faces, which generalised to trapeziums ACBC. The triangles are CCC always regular

Even worse about your invention of edge decorations applied to those group symbols. Probably kind of similar to Dynkin symbols, but obviously just "kind of"...
None of the others stack up like this...

Well, there surely are such things like oct || co, oct || cube, cube || co, cube || sirco, co || sirco, ... (but I assume you were hunting after ike || *).

Infact, even with a mere Stott-Coxeter-Dynkin symbol you would be able to display that jewel: o3o4x || s3s4o. Thus, kind of os3os4xo&#x could serve for its lace prism notation.

--- rk
Klitzing
Pentonian

Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Classifying the segmentochora

Here's the classification of segmentochora as promised.

The bulk of segmentotopes fall into these catogries.

• Wythoff mirror-edge figures
• Wythoff lace prisms, of 2 and 3 bases, belonging to the general polygons
• Wythoff lace prisms of the type 3,P
• Pseudo-symmetrics of the first kind
• Pseudo-symmetrics of the second kind
• ungrouped

It's best for the analysis, to start at 4.34, and go around to 4.33, because 4.20 to 4.33 is a rather hard analysis by this method (both the ungrouped ones are here).

The plan of attack is as follows. First, the 173 figures are divided according to circumradius (this is Klitzing's order). Anything that has just one member is counted as a 'sundry', since it is not simply related to others. Outside the sundries, there are some 20 groups of between 2 and 15 or so, members.

The sundries are entirely lace-prisms of both kinds. The following table gives the WLPs by polygon, and the list of sundry WLPs and small pairs. The ones in brackets occur elsewhere, but are shown for completeness.

Code: Select all
`                           3    4      5    6    8     10    P   xxoPoox   P   // Pap   (6)  (14)   22   46    58    93   174   xxxPooo   P   // Pp    10   (18)   34   47    59    94   175   xoPox2xx  Pap // Pap   11    19   (39)  53    65    96   176   xxPoo2xx  Pp  // Pp   (18)  (20)   42  (54)  (63)   97   177   3,X       35  57  60  74  75  76  89  95  98  110  125  126 127 130 173             54 55  56 ;  128 129 ; 150 151`

The pseudo-symmetries arise from 3,P in that in any x3oPo, it is possible to remove a peak (diminish), and this extends to the x3oPx (where the peak forms a P || 2P cupola. Also, it is possible to restore the cupola in reverse. O means untuched, D is removed, and G is turned (gyrated).

The first kind means that there is only one side involved in the slicing, usually I or rID, or O or rCO or T or rTT(=CO). A second kind edits both of the bases, not necessarily symmetric. This is the main hassle in the lower order groups (4.1 to 4.34 with exceptions).

In 3,3, it's possible just to remove just one peak (ie D). In 3,4 one can have two opposites only, ie OD, DD. In 3,5 the are the opposite set OD vs DD, and the three adjacent positions (meta = ODD, tri = DDD).

This is the icosahedral set. The first column shows what the icosahedron comes as. The 'X' is a kind of extended diminished, which takes the antiprism as well as the pyramid. The columns 111 and 159 corresponds to a group on rCO, which allows a gyrate in the same position as a removal.

Instead of treating the meta- and tri- as distinct groups, we consider the three sites of the tri- as possible action places. The various kinds of action are then pretty much independent, which allows us to handle a great number of variances with a triplet of letters.

Code: Select all
`                   p  ic do    O  icosa   36   84  78     111     131  137  (*)   152   159   OD  py+ap   37   85  79     116     132  138  140   153   164   OX  py      38   86         117    .133       139   154  .165   DD    ap    39   80  81     121     134  142  144   155   169   DX   (5)                                      146  ODD  meta-   40   87  82     122     135  143  158   156   170  DDD  tri-    41   88  83     124     136  147  158   157   172                                *                             **              OG   GG  OGG  GGG   DG  ODG  DGG  DDG         *   112  113  114  115  118  119  120  123        **   160  161  162  163  166  167  168  171`

This is the corresponding octahedral group. The symmetry can be guaged from the rCO, presented in the o4o axis. The difference here is that instead of gyrating the truncated cube in 101 etc, we treat the rCO into a double-gyration or gyration. This is what OG and GG mean.

Code: Select all
`       <-- o4x -->    O  ox xx xx ox     100   107       66  71      61   OG  ox xx xx xo     101             67   GG  xo xx xx xo     102   OD  ox xx xx        103   108       68  72      62   GD  xo xx xx        104   OX  xo xx           105   109       69  73      63   DD     xx xx        106             70   XG     xx                                       64                         + 99`

We then can sort out the remaining examples from 4.20 to 4.33, but i leave this and a few other ends, to a later post.
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1811
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Classifying the segmentochora

Richard asks over the orbifold notation.

This is a description of a symmetry group, similar to the dynkin symbol, but it has a more geometric implementation. It apparently covers every symmetric group in hyberbolic hedrix. Conway has been wrangling archifolds (edges of different end-numbers and parities, so an edge [1,7] goes out on arm one and returns without reflection on arm 7. An arm (1,7) does the same with reflection. Edge (1) is the same as edge (1,1). A third kind of edge <1,2> lies in the plane of the mirror, going out on arm 1, and returning on edge 2. An edge like this is the edge 3x * o2o in the octahedron.

• n : a number before any symbol represents represents a pure rotation or cone. The idea is that the cone is that for (eg) 3, that things are repeated every 120 degrees.
• * : This is a mirror. Numbers n after this are angles of the form pi/n. For example, the standard dynkin groups like oPoQo is * 2 P Q. Mirrors in separate chains are represented by separate *'s, so ** represents two infinite mirrors which do not connect.
• × : This is a miricale. It's a mirror-reflecting action without a mirror involved (like central inversion in 3d).
• O : This is a wander. It's a movement of symmetry without the aid of rotation, for example.

On top of these, one can decorate this kaleidoscope in the same way that the dynkin graph is decorated. o, x, etc have the same meaning in both. There's a new kind of 'edge' (%) possible in uniforms, which produces an edge that disappears (or is swallowed up). For example, there is a triangle whose sides are x,x,q. If you put two of these against the q, and make the margin anlge 180 degrees, the q edge disappears, so you get triangles x,x,%.

example: 3 * 2

3 There is a 3-cone, or an isolated rotation group of 3.

* This starts a wall of mirrors.
2 There's a right angle in there.

So what we get is a right-angle of mirrors, carried around a triangle by an isolated centric 3-cone. This is the pyritohedral group.

When we have 3x * x2% or 3/ * /2%

The triangle governs an edge, which is not mirror-edge. There are two kinds of edge on the mirror. The ends of these are on the triangle edge. One of these edges is exposed, the other is chordal. There's a central rectangle / × %, which is bounded by two trapeziums /2 /1 /% /1. The /1 and /2 are the first

The general orbifolds for polyhedra are 2 3 p (rotational), and * 2 3 p (reflectional groups). One sees that the nodes for * 2 3 p fall between the numbers, but since 'p' here is an angle between two mirrors, then * 2a3ePo would correspond to what is written as a3ePo.

Much work has to be done on drawings for other groups, such as 2 3 P and * 2 3 P, before comment is made thereon.
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1811
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Classifying the segmentochora

I know s3s4o // o3o4x works, eg so3so4ox would give the cube // ike, but really, it defeats the whole point of both the dynkin symbol and the c/i.
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1811
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Classifying the segmentochora

It is well possible to use the wythoff-lace-prism symbol to determine the elements, exactly as one can use the dynkin graph. Also, cupolae etc do not really work in higher dimensions, because there are more ways to add things. Here is the surtope-analysis for a wythoff-lace-prism, using the \$ graph.

This is the analysis for the lace-prism tO // C (which is not equal-edged, but this does not matter, since it has hexagonal pyramids among its surchora.

v, e, h, c stand for vertices, edges, hedra, chora.

The 's' gives the nodes that count towards the face. These equal, in number the vertices of a simplex of the same dimension, eg three = triangle. The order of the mirror-group value is the value S (not shown).

The 'a' mirrors are those mirrors (numbers), not directly connected to the 's' mirrors. The figure falls exactly on these mirrors. One has to count the mirror-cells that these include, that is, one treats 'a' as a second-mirror-group, of order 'A'. A value '-' means that no mirrors belong here, and that the value of A is '1' (the identity itself).

The count is always for each element, G/SA, where G is the total order of the symmetry. So, for the edge A1. The s mirror is the nodes in A1 (ie 1), the symmetry value for 1 is S. The around-mirrors 'a' are those values 1,2,3 which are not part of the group (ie not 1), nor are connected by a branch to it (ie 2). It's only '3' that is not connected, so the edge falls in mirror-3. The value of the 'a' group is then A=2. The total count is G/SA, is 48/2/2 = 12. There are 12 edges of type A1.

Code: Select all
`      xo3xo4ox    tO // C       A                    s      a      count    always   A // B      / \             v     A      3       24   =  48/1/2  top vertices     1---2====3             B     1,2       8   =  48/1/6  bottom vert              |       e    AB      -       48      lacing              B            A1      3       12      octa hex-tri edge                           A2      -       24      octa hex-hex edge                           B1      3       12      cube edge                     h    AB1      -       24      triangle = line // point                          AB2      -       24      triangle = line // pt                          AB3      -       24      triangle = pt // line                          A12      -        8      hexagon                          A23      -        6      square                          B32      -        6      square                      c   A123     -        1       tO                          A12B     -        8      hex pyr = hexa // point                          A13B     -       12      tetra  = line x2o // line o2x                          A23B     -        6      square =  x4o // o4x                          B321     -        1      cube`
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1811
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Classifying the segmentochora

While i am not over fussed on creating terms, some terms can be introduced to carry meaning.

Lace prisms can exist whenever a progression between two figures exist. All one needs to do is to be able to orientate them in the desired way, and describe how each surtope of A becomes onto B: in other words, how A would transform into B. Neither A or B need to have dynkin-graphs. The dual of a lace-prism is a lace-tegum, which also has a regular construction.

When the bases of the lace-prism are written on a dynkin graphs in o,x, etc, then it's a Wythoff lace-prism. The name 'wythoff' is used to suggest that the deep coordinate-theory that comes with vector analysis, etc, applies. The wythoff scheme works on mirrors in the shape of a simplex, such as might be represented by a dynkin-graph. So things like xx3oo5oo&t is a wythoff lace-prism.

The quasisymetrics (diminished rCO, diminished Ike &c), are not generated on the dynkin symbol, and while a clear progression exists between top and bottom, the shapes are not generated by a wythoff-construction, and therefore calling them wythoff-lace-prisms is wrong. Instead, they are quasi-symmetric LPs.

A 'lace-tower' is simply a stack of lace-prisms, although in order to keep convexity, additional edges might be called for, which might run from non-consecutive layers. In the table of qs lace-prisms, one finds the first three rings of 3,3,5, which gives pairs of Klitzing's segmentochora. The wythoff example is

oxo3ooo5oox&t, or pt // ike // dodeca This is pairs 4.84 + 4.78.

In my table, i give examples like 85+79, 80+81, 87+82, 88+83 are corresponding qs variations of this group. In essence, the progression is through a diminished icosa to a dodeca, etc. In any case, these are not WLt, but QS Lace-towers.

The classification of the segmentochora is here in the same order i use the uniforms, viz

1, wythoffian, based on a simple group
2, wythoffian, based on product-groups
3, non-wythoffian, with large classes divvied out.
3a, the quasi-symmetrics,
3b, ungrouped.
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1811
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Classifying the segmentochora

Seems awfully quiet around here recently. Part of it may be my fault. But in any case, my silence is due to having discovered something very interesting about lace prisms, but not having the full results yet. But maybe a teaser is due.

I have discovered that CRF lace prisms with n-cube symmetry, that is, segmentotopes of the form .4.3.3....||.4.3.3..., have a rich structure only up to 19 dimensions. From 20D and upwards, there are only 3 kinds: those derived(*) from the bisected n-cross, those whose top and bottom facets differ in exactly two nodes in their CD diagrams, and the uniform prisms (i.e., top/bottom facets are the same). This means that from 20D and upwards, all lace prisms with n-cube symmetry have either height 2 or sqrt(2), without exception.

(*) By "derived", I mean adding a ring(s) to the same node(s) in both top and bottom facets.

For 19D and below, the "rich structure" involves those lace prisms for which the top/bottom facets differ in ringedness of the first node in the CD diagram (the one connected to the rest of the diagram by the edge labelled 4). These have decreasing height with increasing dimension, and 19D is the last dimension where they are CRF. The lace prisms with maximal height in this category are the ones derived from x4o3o...||o4x3o....

I have the proof of these results in a little document that I'm working on; I'm refraining from posting it at the moment as I'm researching the lace prisms with n-simplex symmetry right now, and they look to also collapse into a relatively simple pattern at a high-enough dimension; I'm hoping to also have general results for lace prisms with n-simplex symmetry before I post the document.
quickfur
Pentonian

Posts: 2482
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: Classifying the segmentochora

Great work quickfur I look forward to seeing the finished work.

Keiji

Posts: 1962
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

### Re: Classifying the segmentochora

Do you calculate all those lace prism heights manually? This would become teddious. It would be easier to have some calculation routine by hand, inserting the top resp. bottom Dynkin diagram, and getting out the appropriate height.

In fact this calculation "engine" once was done by Wendy. This "engine" moreover can calculate the circumradius of any polytope, provided its Dynkin symbol.

I'll attach the excel spreadsheet (zipped, as "xls" is a not allowed format to be uploaded) I once got from her (with some minor changes).

In row 3 you could enter the nodes. In fact you would enter the edge lengths of the respective nodes: "1" for a unit edge, "0" for no edge, "1.41428" for a sqrt(2) edge, etc.
In the triangle matrix below, marked "symmetry" you would have to enter the link numbers of the required Dynkin graph. Those are to read in the way: cell C5 relates to the link mark between the nodes of B5 and C6 (that diagonal underneath is to be identified with the elements entered in row 3 at the respective columns)
Having just entered those numbers you would get the circumradius of that polytope at "radius", ie. in cell B30
Alternatively you could use the same engine to get the heights of lace prisms! Just enter the difference of the top and bottom layer Dynkin symbols in row 3: Use positive numbers for the edge lengths of the top graph, and negative ones for the respective lengths of the bottom graph. If both are to be used at the same position you clearly would have to calculate the result and enter that (or let excel do that stuff, entering "=+1-1.41428" etc.)
Having entered those numbers (i.e. the Dynkin symbol of the actual lace prism), you would get out the corresponding height at "root(1-radius^2)", ie. cell E30.

That's all. Have fun!

--- rk
Attachments
calculation engin for circumradii and lace prism heights
Klitzing
Pentonian

Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Classifying the segmentochora

Klitzing wrote:Do you calculate all those lace prism heights manually? This would become teddious. It would be easier to have some calculation routine by hand, inserting the top resp. bottom Dynkin diagram, and getting out the appropriate height.

In fact this calculation "engine" once was done by Wendy. This "engine" moreover can calculate the circumradius of any polytope, provided its Dynkin symbol.

I'll attach the excel spreadsheet (zipped, as "xls" is a not allowed format to be uploaded) I once got from her (with some minor changes).

Though, I'm not calculating the height of every lace prism manually. What I'm doing is to use the coordinate generation scheme (that I posted in another thread) to produce general coordinates for any lace prism of a given symmetry, then use the CRF criterion to derive an inequality constraining its height. For the n-cube lace prisms, I use this inequality to derive constraints on the CD diagrams of the top/bottom faces, and prove that after 19D, they cannot differ in the leftmost node, and that for any dimension, lace prisms that do not differ in the first node can only be a prism, a bisected n-cross, or a lace prism whose top/bottom faces differ in precisely two adjacent nodes.

I'm still working on the n-simplex lace prisms, they are a bit more complicated because more possibilities are allowed unlike the n-cube family. But I think I may have discovered a pattern that is probably general across dimensions. I'll post everything once I have the proof.
quickfur
Pentonian

Posts: 2482
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

### Re: Classifying the segmentochora

If you are doing the simplex groups, there is a quick way of "calculating" the stott matrix. You write the numbers upwards from 1 to N, and write N+1 as the denominator. You then fill out to the diagonal with multiples of the first number, and then make the matrix symmetric (a_ij = a_ji).

The next step is to create a vector from the dynkin matrix (o = 0, x=1, etc), as they fall on the graph. You multiply this matrix by the vector ie S_ij . v_i = w_j, and then take the dot product of v_j and S_ij v_i. This gives you S_ij v_i v_j which is the circumradius of the figure, or the length of the vector, when one subtracts two vectors. Polytopes with dynkin graphs in the same group are position-polytopes, and they have radial vectors which gives the position vectors. Because the coordinate system is oblique, the dot of base vectors (such as a_i . a_j ), is not just 1 or 0, but the cosine of the angle between them. This matrix takes care of all of this.

Code: Select all
`    (6         (6                                 (6   5   4   3    2   1 )    (5         (5  10                             (5  10   8   6    4   2 ) 2  (4      2  (4   8  12                      2  (4   8  12   9    6   3 ) -  (3      -  (3   6   9  12                  -  (3   6   9  12    8   4 ) 7  (2      7  (2   4   6   8   10             7  (2   4   6   8   10   5 )    (1         (1   2   3   4    5   6            (1   2   3   4    5   6 )`

The spreadsheat that richard klitzing put up does this up to about six dimensions, but the same idea can be pushed out to 19 or 20 dimensions without difficulty. The recriprocal of the stott matrix is the dynkin matrix. The main diagonal of this consists of '2', and the value a_ij is -2 cos(pi / p). What you do is enter these values into the input field, it fills in a_ji for you, and finds the value of the stott matrix, and then the matrix-dot of the entered vector.
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1811
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Classifying the segmentochora

wendy wrote:[...]The spreadsheat that richard klitzing put up does this up to about six dimensions, but the same idea can be pushed out to 19 or 20 dimensions without difficulty. The recriprocal of the stott matrix is the dynkin matrix. The main diagonal of this consists of '2', and the value a_ij is -2 cos(pi / p). What you do is enter these values into the input field, it fills in a_ji for you, and finds the value of the stott matrix, and then the matrix-dot of the entered vector.

In fact, I already have pushed that up to 10D.

Moreover I normalized the outcomes to unit edges (instead of your previous size 2 edges), i.e. by entering "1" at a node position, the radius would be for a corresponding unit size edge, etc.

Further I've added the height calculations for lace prisms. Even so, they are restricted to unit size lacings ("root(1-radius^2)") so far.

--- rk
Klitzing
Pentonian

Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Classifying the segmentochora

You could point the lacing-length to a cell, and then make it a free variable.
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
Pentonian

Posts: 1811
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Classifying the segmentochora

wendy wrote:You could point the lacing-length to a cell, and then make it a free variable.

This was exactly what I had in mind, when pointing it out.
Just that I hadn't any necessity for implementation so far, as I either was looking for heighs of segmentotopes or alternatively for distances of vertex layers in uniform polytopes. Both do require unit lacings only.
But yes, you are right for sure...
--- rk
Klitzing
Pentonian

Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Classifying the segmentochora

Alright, I'm having some trouble proving some things about lace prisms with n-simplex symmetry, which is why I haven't posted anything yet. But since I haven't been able to make any headway in that direction, I thought I might as well post my results for lace prisms with n-cube symmetry first.

Fundamental lace prisms

To make the problem of classifying lace prisms simpler, I make use of Klitzing's observation that if a lace prism A||B exists and is CRF, then if both A and B have an unringed node in their CD diagram in position i, then adding a ring to the respective nodes will produce a CRF with the same height. IOW, this is just the Stott expansion of the lacing cells of A||B. Therefore, given any lace prism A||B, we may remove rings that occur in the same position in both A and B, until A's and B's CD diagram have no more ringed nodes in common. A lace prism of this form is called a fundamental lace prism.

Given a fundamental lace prism, we can easily derive the corresponding non-fundamentals by adding rings to the same nodes in both A and B.

A fundamental lace prism also has the property that if the i'th node of A is ringed, then the i'th node of B must be unringed, and vice versa.

Note that it is legal for A or B to be a point (all nodes in CD diagram are unringed); this produces a B-pyramid (resp. A-pyramid). Also note that ooo...||ooo..., which is the line segment, is the fundamental lace prism for all uniform polytope prisms.

Fundamental lace prisms with n-cube symmetry

Let A||B be a fundamental lace prism where A and B are both uniform polytopes with n-cube symmetry. Then we may derive coordinates for A||B by using my method of reading off the coordinates from the CD diagram:

- Start with the node at the end of the diagram connected to the edge labelled 4. If this node is ringed, then construct an n-vector whose first coordinate is 1; otherwise, the first coordinate of the vector is 0.
- For each subsequent node: if the node is ringed, then the corresponding coordinate in the vector is the previous coordinate plus sqrt(2), otherwise it is a copy of the previous coordinate.
- Take all permutations of coordinate and changes of sign of the vector: this generates the vertices of the corresponding polytope.

Example: x4o3x --> apacs(1,1,1+sqrt(2)). This is a rhombicuboctahedron of edge length 2.

Using this method of deriving coordinates for A and B, we may thus write the coordinates of A||B as:

(x1,x2,x3,...,0)
(y1,y2,y3,...,H)

where (x1,x2,...) and (y1,y2,...) are the coordinates generated by the above method (together with all permutations of coordinate and changes of sign thereof), and H is the height of the lace prism.

Since A and B are vertex-transitive, we may take the vectors generated for A and B (without the permutations/sign changes) as representative points on A and B, respectively. Then there must be an edge between (x1,x2,...) and (y1,y2,...). Which means that if A||B is CRF, then:

||(x1,x2,...,0) - (y1,y2,...,H)|| = 2

since the edge lengths of A and B are 2. Squaring both sides of this equation and solving for H, we get:

H^2 = 4 - ((x1-y1)^2 + (x2-y2)^2 + ...)

Since H>0 (otherwise A||B is either subdimensional or not CRF), this implies that:

((x1-y1)^2 + (x2-y2)^2 + ...) < 4

This inequality gives us a simple way of determining if a given lace prism with n-cube symmetry is CRF or not. I'll call this Inequality_1, as I'll be referring to it repeatedly below.

Now observe that if the first node of either A or B (without loss of generality, say it's A) is ringed, then the (xi-yi) terms will all be of the form 1+k*sqrt(2) for some integer k. (Proof: the first nodes of A and B cannot both be ringed simultaneously, since otherwise A||B is not fundamental; so the first node of B is unringed. Then by the above method of coordinate derivation, all B's coordinates are multiples of sqrt(2), but A's coordinates are of the form 1+k*sqrt(2). So their difference will also be of the form 1+k*sqrt(2).) Let's call this the First Category of lace prisms with n-cube symmetry.

OTOH, if both A and B have their first nodes unringed, then the (xi-yi) terms will all be multiples of sqrt(2). Let's call this the Second Category of lace prisms with n-cube symmetry.

First Category: x4...||o4...

Let's consider the first category first. Since the (xi-yi) terms are of the form 1+k*sqrt(2), then Inequality_1 can be written:

1 + (1+k1*sqrt(2))^2 + (1+k2*sqrt(2))^2 + ... < 4

for some integers k1, k2, .... (The first term in the sum is 1 because the first node of A is ringed, and the second node of B is unringed, so the difference in coordinate is precisely 1.)

Now, since sqrt(2) is irrational, 1+k*sqrt(2) is never zero, and so (1+k*sqrt(2))^2 is always positive. Furthermore, the smallest number of this form is (1-sqrt(2))^2. Therefore, by the Archimedean property of the real numbers, the number of terms in the sum cannot exceed some constant number K. Since the number of terms corresponds with the dimension of A||B, this means that there is an upper limit K such that if the dimension n of A||B is greater than K, then A||B cannot be CRF.

What's the value of K? It's basically the maximum number of terms in the sum. Since the sum must be less than 4, maximizing the number of terms is equivalent to minimizing each term. As mentioned above, the smallest number of the form (1+k*sqrt(2))^2 is (1-sqrt(2))^2. So in the ideal case, we'd like to choose A and B such that every term (except the first) is (1-sqrt(2))^2. This ideal case is in fact attained by the lace prism x4o3o3o...||o4x3o3o.... As you can check for yourself, the coordinates for A and B are respectively (1,1,1,...) and (0,sqrt(2),sqrt(2),...), so the squared magnitude of their difference is exactly 1+(1-sqrt(2))^2+(1-sqrt(2))^2+... .

How many terms can we have in this sum before it exceeds 4? Let A||B be n-dimensional. Then A and B are (n-1)-dimensional, and so will have (n-1) coordinates. Therefore:

1+(n-2)*(1-sqrt(2))^2 < 4
(n-2)*(1-2*sqrt(2)+2) < 3
(n-2)*(3-2*sqrt(2)) < 3

Since (3-2*sqrt(2)) > 0, we can divide both sides without changing the sense of the inequality:

n-2 < 3 / (3-2*sqrt(2))
n-2 < (3 * (3+2*sqrt(2))) / ((3-2*sqrt(2)) * (3+2*sqrt(2))
n-2 < (9 + 6*sqrt(2)) / (9-8)
n-2 < 9 + 6*sqrt(2)
n < 11 + 6*sqrt(2)

11+6*sqrt(2) is approximately 19.485. Since n must be an integer, this means n≤19. Therefore K=19. IOW, lace prisms with n-cube symmetry in the First Category are only CRF up to 19D; from 20D onwards, they are no longer CRF.

Second Category: o4...||o4...

Now we consider the 2nd category of lace prisms with n-cube symmetry. Since the first nodes of both A and B are unringed, the coordinates of A and B are all multiples of sqrt(2). Therefore, we may write Inequality_1 as:

(k1*sqrt(2))^2 + (k2*sqrt(2))^2 + ... < 4

for some integers k1, k2, .... But this is simply equivalent to:

2*k1^2 + 2*k2^2 + ... < 4
k1^2 + k2^2 + ... < 2

But since k1, k2, ... are all integers, this means that there cannot be more than 1 non-zero term in the left-hand side sum, since otherwise it will be ≥2. This, in turn, means that A and B can differ in not more than 1 coordinate.

Case 1: both A and B have no ringed nodes, so A||B is the line-segment. This is the fundamental lace prism that gives rise to all the uniform polytope prisms.

Case 2: A has a single ringed node, and B has no ringed nodes. In this case, A's ringed node can only occur in the last node; i.e., A's coordinates are (0,0,...,0,sqrt(2)). If the ringed node occurs anywhere else, the difference in coordinates between A and B will be 2*sqrt(2), and the squared sum of coordinates will be 8, which violates Inequality_1. So A||B in this case is o4o...o3x||o4o...o3o, that is, it is just the (n-1)-cross pyramid (or bisected (n-1)-cross).

Case 3: Both A and B have a single ringed node. In this case, if A's ringed node occurs at position i, then without loss of generality B's ringed node must occur at position (i+1). In this case, A's coordinates will be (0,0,...,0,sqrt(2),sqrt(2),...,sqrt(2)), and B's coordinates will be (0,0,...,0,0,sqrt(2),...,sqrt(2)), so the difference is exactly (0,0,...,0,sqrt(2),0,...0). Any other relative position of ringed nodes will cause there to be more than 1 difference in coordinate, which will violate Inequality_1.

There are no other CRF cases. Why? Because if, for example, A has two or more ringed nodes, then since A||B is fundamental, B cannot also have ringed nodes in the same positions, and so this introduces at least two differences in coordinate, again violating Inequality_1.

For case 1, the height of A||B is 2 (since A=B); for cases 2 and 3, the height of A||B is sqrt(2), since H^2 = 4 - (sqrt(2))^2 = 2, so H=sqrt(2). Furthermore, all 3 cases are independent of dimension, so they exist in all dimensions.

Summary

In summary, CRF fundamental lace prisms A||B with n-cube symmetry are of one of the following forms:

1) x4...||o4...: these fundamental lace prisms only go up to 19D.

2) o4o*x||o4o+ (using Keiji's notation based on regular expressions[**]): these fundamental lace prisms exist in all dimensions, and are just the bisected (n-1)-cross. They have height sqrt(2).

3) o4o*||o4o*: this is just the line segment (of length 2); it exists in all dimensions, and give rise to the uniform polytope prisms (with n-cube symmetry).

4) o4oixo+||o4oioxo* (for some integer i): these fundamental lace prisms exist in all dimensions, and have height sqrt(2).

[**] Superscript * means "zero or more instances of the preceding node", and superscript + means "one or more instances of the preceding node". Superscript i means "exactly i instances of the preceding node".

All other lace prisms with n-cube symmetry are derived from these fundamentals by adding rings to nodes in the same positions in A and B (equivalently, by Stott expansions of these fundamentals).

Further Work

I didn't look into the details of the fundamentals of the form x4...||o4.... There may be other interesting trends within that category, even if they are only CRF up to 19D.
quickfur
Pentonian

Posts: 2482
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

PreviousNext