Mathematic enumeration of stellated dodecaplexes

[pentagonalpolytope747]

So I have been wondering lately what would happen if H_n coxeter group, {5,3,3,3,3,...} was in every dimension.This is my favourite polytope group of all, and I made a list of stellations enumerated using power setting and operations. Although geometrically not possible in dimensions higher than 5, the notation itself is fun to play with.

I call the stellated forms "gothic polytopes". I use symbol s to denote stellation with edges, faces, etc.

First, in 2D, there is the pentagon, and stellated pentagon - pentagram. Our list is:
{5} - s0{5}
{5/2}- s1{5}

This is the power set of dimension one below, that is, dimension of the polytope facet.
In 3D, there are Kepler-Poinsot solids, power set enumeration gives 4, including the unstellated convex shape.

{5,3} - s0{5,3} - dodecahedron
{5/2,5} - s1{5,3} - [small] stellated dodecahedron
{5,5/2} - s2{5,3} - great dodecahedron
{5/2,3} - s1,2{5,3} - great stellated dodecahedron
-------
{3,5/2} - x{3,5} - great icosahedron. Checking the dual polytopes gives us 5 total forms, or 4 stellated forms. Stellation with edges is just stellation, stellation with faces is called greatening.
In 4D, there are 2^(4-1) - 1 = 7 stellations of 120-cell, and checking the dual forms gives us total of 10. A new operation is introduced, aggrandizement.

{5,3,3} - s0{5,3,3} - 120-cell
{p,5,3} - s1{5,3,3} - [small] stellated 120-cell
{5,p,5} - s2{5,3,3} - great 120-cell
{5,3,p} - s3{5,3,3} - grand 120-cell
{p,3,5} - s1,2{5,3,3} - stellated great 120-cell
{p,5,p} - s1,3{5,3,3} - stellated grand 120-cell
{5,p,3} - s2,3{5,3,3} - great grand 120-cell
{p,3,3} - s1,2,3{5,3,3} - stellated great grand 120-cell
------- Dual-checking gives 3 new forms:
{3,5,5/2} - x0{5,3,3} - icosahedral 120-cell
{3,5/2,5} - x1{5,3,3} - great icosahedral 120-cell
{3,3,5/2} - x{3,3,5} - grand 600-cell

A fun thought experiment would be enumerating these failed poytopes-honeycombs for every dimension N.

So I tried writing 5D list of these, but I am not sure if I am correct. I will use "p" for pentagram. I will call the {5,3,3,...} polytope dodecaplex, and {...3,3,5} polytope icosaplex. Although these are honeycombs in geometry. The 4th operation I will call awesome-ing, 5th will call pentellizing, then hexellizing, heptellizing, etc.
By power set, there should be 15 stellated dodecaplexes in 5D. However, as I was enumerating these, I noticed a few problems, such as {5,3,p,5} being both grand and grand awesome polytopes. It can't be {5,3,p,3} because {3,p,3} is technically from {3,5,3} family, icosahedral honeycomb family, thus doesn't belong here.

{5,3,3,3} - s0{5,3,3,3} - 5-dodecaplex
{p,5,3,3} - s1{5,3,3,3} - [small] stellated 5-dodecaplex
{5,p,5,3} - s2{5,3,3,3} - great 5-dodecaplex
{5,3,p,5} - s3{5,3,3,3} - grand 5-dodecaplex
{5,3,3,p} - s4{5,3,3,3} - awesome 5-dodecaplex
{p,3,5,p} - s1,2{5,3,3,3} - stellated great 5-dodecaplex
{p,5,p,5} - s1,3{5,3,3,3} - stellated grand 5-dodecaplex
{p,5,3,p} - s1,4{5,3,3,3} - stellated awesome 5-dodecaplex
{5,p,3,5} - s2,3{5,3,3,3} - great grand 5-dodecaplex
{5,p,5,p} - s2,4{5,3,3,3} - great awesome 5-dodecaplex
{5,3,x,x} - s3,4{5,3,3,3} - grand awesome 5-dodecaplex - i can't find correct symbol for this one because its 4-facet should be grand 120-cell but eh
{p,3,3,5} - s1,2,3{5,3,3,3} - stellated great grand 5-dodecaplex
{p,3,5,p} - s1,2,4{5,3,3,3} - stellated great awesome 5-dodecaplex
{p,5,p,3} - s1,3,4{5,3,3,3} - stellated grand awesome 5-dodecaplex
{5,p,3,3} - s2,3,4{5,3,3,3} - great grand awesome 5-dodecaplex
{p,3,3,3} - s1,2,3,4{5,3,3,3} - stellated great grand awesome 5-dodecaplex
---------
Using the icosahedra, we get 5 new polyshapes:
{3,3,3,p} - awesome 5-icosaplex
{3,3,5,p} - 600-cellic 5-dodecaplex
{3,3,p,5} - grand 600-cellic 5-dodecaplex
{3,5,p,5} - icosahedral 120-cellic 5-dodecaplex
{3,5,p,3} - grand icosahedral 120-cellic 5-dodcaplex
{3,p,5,3} - great icosahedral 120-cellic 5-dodecaplex

Out of these, only one is a stellation of the icosaplex, in every dimension, others are stellations of dodecaplex, just with faces extended peculiarly.

So in N dimensions, we have 2^(N-1)-1 dodecaplex stellations, + (N-2)(N-1)/2 icosaplexes, of which only 1 is a pure icosaplex stellation {3,3,3,3,...,3,3,5/2}. This is just my estimate for the icosaplex stellations.

I am not trying to pass these off as real polytopes, this is just a thought experiment that I had lately.

Thoughts?

[quickfur]

I'm pretty sure these could be made to exist by immersing them in a suitable hyperbolic space.

[pentagonalpolytope747]

I'm pretty sure these could be made to exist by immersing them in a suitable hyperbolic space.

4 of these can already be made in a normal hyperbolic space. {p,5,3,3}, {5,p,5,3}, {3,5,p,5}, {3,3,5,p}.

Other than that, some of these have dihedral angles that actually match properly, like {3,3,3,p}, and therefore would be an infinite coverage of a sphere instead. Interestingly enough, the awesome 5-dodecaplex {5,3,3,p} would actually be Euclidean since 120-cell's dichoral angle is 144 degrees, but since a piece of 120-cell can't tesselate 4D space, it would be an infinite looping of a tiling.

[wendy]

The numbering system for stellated dodecaplexes goes like this.

You replace 5/2, 3, 5 with 1, <. 0. The < is replaced by the 1 or 0 it first points to. The number is read in reverse, as a binary number. So 5,3,3,5/2 becomes 0<<1. = 0001, read in reverse, gives 8.

For four dimensions (three symbols), this gives a range from 0 to 7. For four symbols, 0-2 are hyperbolic tilings, 3 is {5/2.3.5.3}, has diverging vertices, but shows us that {3,5,3} is a cross-section of {5,3,3,3}. Stars 4 to 11 are pentagonal tilings related to the tilings in {5,3,3,5/2} [#8 in the list]. The remaining four are dense polytopes. This works quite neatly, since the stellations 1, 2, 3 derive as simplex-stellations of {3,4,3,3}, where the tetrahedra are the walls between the cells of {3,3,4,3}. Of course, this does not take to account such joys as o3o3o5/2z3o3x, a tiling of regular pentachora and grand 600ch (ie {3,3,5/2}), 600 pentachora and 120 g600ch at each vertex, giving a nine-fold coverage.

For the trigonal subgroup, we find placing '3' in front of earlier stars might do the trick, so < at the lead points to a letter, counted in order, so 3,3,3,5 gives <<< 3 = C3.

Given that {5,3,3,...} is in the simplex group (ie the tail after the '5' is a simplex), there is an order of simplex-faceting, which preserves a surtope of the face of {3,3,...,5} to 'n' vertices. The order here is to suppose {....,3.5.5/2,5,3,....} is the count staff, and you start with 0 as the second 5, and count backwards, so 5/2,5,3. is one, 5,5/2,5,3 is two, 3,5,5/2,5,3... is 3. This works for any number of P,P/2,P. The nth simplex-star preserves the simplex with n vertices on the {3,3,3,...,P}. You can see this with the 3d case, {5/2,5}, {5,5/2}, {3,5} preserve 1, 2, 3 vertices of the icosahedron.

Among the stellations, one finds the regular compounds, being face-regular. A tetrahedron of {3,3,5} can produce 9 octahedra or 18 cubes, in the extreme case, the result having the shared symmetry of 3*2 "pytitohedral".

[wendy]

For this table,
'dense' means that the figure is infinitely dense, but piecewise discrete (ie you can completely construct a surtope and everything attached to it).
'open cell' means that there are cells that contain complete planes as in the hyperbolic sense.

The number in the second column is the reversed binary value of the first column, these are converted into the Schläfli symbol by the rule that if the digit is 0, a 5 results, if a digit is >0, a 5/2 results, and if the digit to be written is the same as the last non-3, a '3' is written.

The basis of stellation is that the pentagon gives rise to a pentagram, this in general creates a new vertex-figure. Note however the edges are preserved, so

5,A,B,C -> 5/2 D,B,C where 5,3 -> 5/2,5 and 5,5/2 gives 5/2,3

Replacing the pentagram with the enclosing pentagon causes the vertex-figure to be stellated if there is a pentagon is at D.

The second step of stellation is to replace the 5/2,3 with 5,3. As before we see 5/2,3,A,B becomes 5,3,D,B, where this preserves the hedron-figure. Here, the edge of the former is f² of the latter, and the vertex-figure is also shortened. In reality, if A is 5, then the AB is stelated into DB (ie 5/2,3,3,5) becomes (5,3,3,5/2). Even Coxeter found this extremely tricky, and used the ultimate patch of using the dual from (5/2,3,5) to (5,3,5/2). The dodecahedra arising from (5/2,3,3) do not close.




So the pentagonal stars in 4d go:

0 000 5,3,3
1 100 5/2,5,3
2 020 5,5/2,5
3 120 5/2,3,5
4 004 5,3,5/2
5 104 5/2,5,5/2
6 024 5,5/2,3
7 124 5/2,3,3
SI 3,3,5 dual of 0
S 3,5,5/2 dual of 1
G 3,5,5/2 dual of 6
GI 3,3,5/2 dual of 7

The 5d groups go

0 0000 5.3.3.3 hyperbolic
1 1000 5/2,3,3 do.
2 0200 5,5/2,5,3 do.
3 1200 5/2,3,5,3 ideal vertex

3,5,3,3 is an open-cell hyperbolic tiling,

4 0040 5,3,5/2,5 euclidean tiling (dense)
5 1040 5/2,5,5/2,5 do.
6 0240 5,5/2,3,5 do.
7 1240 5/2,3,3,5 do.
8 0008 5,3,3,5/2 do.
9 1008 5/2,3,5,5/2 do.
10 0208 5,5/2,5,5/2 do.
11 1208 5/2,3,5,5/2 do.
S 3,5,5/2,3
G 3,5/2,5,3

Dense polychora
12 0048 5,3,5/2,3 [vertex does not close]
13 1048 5/2,5,5/2,3
14 0248 5,5/2,3,3
15 1248 5/2,3,3,3.

[pentagonalpolytope747]

For this table,
'dense' means that the figure is infinitely dense, but piecewise discrete (ie you can completely construct a surtope and everything attached to it).
'open cell' means that there are cells that contain complete planes as in the hyperbolic sense.

...

"Even Coxeter found this extremely tricky,"

WOAH. It will take me a while to grasp this concept, but I get it. I never thought that these would be THIS complex, when all I needed is the filling for grand awesome dodecaplex and a sequence.

You know, this is the high quality content I visit the internet for. I have even considered writing my own book on polyshapes in general.

I also considered using the Euler's formula for toroidal polyhedra V - E + F = 2 - 2*number of holes to create hyperbolic polyhedra, but this is a topic for another time. On the other hand, such holed surfaces can be used to create the stellations described here as closed polytopes.

[pentagonalpolytope747]

3 1200 5/2,3,5,3 ideal vertex

And another woah. Is {5/2,3,5,3} an actual stellation of the {5,3,3,3}? I read the statement that {3,5,3} is a cross-section of {5,3,3,3}, so therefore it must be true?

I thought that vertex/edge/face figures can only come from the previous star polychora, while {3,5,3} is a different piece of symmetry.

So now {3,5,3,5/2} is also a thing?

[wendy]

Things get rather complex in hyperbolic geometry. You can have an {8,3}, octagons, three at a corner. As per the usual rule with {p,3}, it stellates into {p/2,p}. So there is a stellation of {8,3} that is {8/2,8}, and {8,8/2}, rather like the small stellated dodecahedron and great dodecahedron. These are compounds of {4,8} and {8,4}, respectively, of three-fold cover.

The next stellation is the same as {p,4} -> {p,p}, the octahedron giving a double-tetrahedron, or stella octangula. So {8,4} stellates into two {8,8}, and {8,3} gives rise to a stellation of six {8,8}.

Nearly all of the stars have been enumerated in hyperbolic geometry. We use some fancy set theory to chuck out 99.99999999999% of the data, and look at the the two or three that survive this simple test. I've asked on quora about 'what is this called', and they don't have a word for it, or grasp the utility of the idea. So we chugging through some new stuff here.

You can take the octagons of {8,4}, and make an {8,3}, in much the same way you can cut squares from {4,4} amd make {4.3}. In both examples, you will get a polyhedron with a right-angle margin (dihedral angle = pi/2). You can make a number of these and 'fill space' with them, to give {8,3,4} and {4,3,4} resp. {8,3,4} "stellates" into {8/2,8.4}, ie a compund of {4,8,4}'s that fill space. I have not really walked this one as yet, but i imagine it's something like {4/2,4,4}, the filling of space with square columns in three different directions. The vertex-figure is far away, and has {4,4} symmetry.

There are several pentagonal tilings in hyperbolic geometry, apart from the usual regulars {5,3,5}, {3,5,3}, {5,3,4}. {5,3,A}, {3,3,3,5:}, and {3,5,3,5:}. Apart from {5,3,4} and {5,3,A}, these groups are unrelated. {5,3,4} gives rise to o5o5/2o5o3z (a dynkin symbol in the shape of a square, the : and z mean 'return to node 0 here). So this produces, for example, a tiling x5o5/2o5o3z, gives great dodecahedra, icosahedra and icoasdodecahedra. The density is 4 (ie a four-fold covering).

Coxeter's forays are fairly limited in his published works. I spent some time correcting this, and tidying up the notations. It's the mainstay of the forum here, though not because of my effort.

The existance of stars like {5/2,3,5,3} tells us something about cross-sections of {5,3,3,3}. But the same cross-section can be derived from o5x3o5o3o and o5o3o5x3o, both of which occur as cross-sections of x3o3o3o5o, they have the same edge, and the vertex-figure is nested in that of {3,3,3,5}.

[wendy]

Euler's formula breaks down in 4D, because holes come in both positive and negative forms, the simplest hole has a net Euler-characteristic of 0.

What is done to evaluate that the genus of {5,5/2} is 4, is this. You start with an icosahedron n+12v+30e+20h+c, where n,v,e,h,c,t,... are alternately -1 and +1.

You remove the 20 triangles, and replace these with 12 pentagons, giving n+12v+30e+12h+c = -2g (where g is the number of holes in 3d). This equates to 12h + 2g = 20h, or g=4.

When you do this in 4D, you can remove what Coxeter calls 'j-circults that do not bound', of both 2d and 3d. These act in opposite directions, so a pair of 2d circuits = +1 genus, and a pair of 3d circuits is -1. In the icosahedron example, a '2-circuit' is a closed polygon. It on a convex bounds an area. But in the great icosahedron, the 2-circuit forms a triangle with no interior. This interior is to be found in the three pentagons that dip into that area. So a 'fence' built around these triangles will not stop one wandering around the surface, whereas a fence built around a pentagon will. So the twelve pentagonal circuits bind, but the 20 triangle-circuits don't.

In four dimensions, you can remove both 2-circuits (like a pentagon), or a 3-circuit, like a tetrahedron. You can remove any combination of these, as long as the figure is enclosed in joining 3-circuits, and the 3-circuits are bounded by 2-circuits. etc. Restructuring the polytope like this is 'faceting', the opposite to 'stellation'. Each 2-circuit removed in net, adds ½ to the genus, and each 3-circuit takes ½ from the circuit. To make something with a hole through it, you have to remove two polygons somewhere.

[pentagonalpolytope747]

Wow.

Also it is surprising that someone was so active in this topic. I thought that it would be just left to rot on this site.

It is possible to not only stellate things as {p,p/2}, but as {p/3}, {p/4}, etc. The number of these would be N * ceil(p/2)-2, or number of stars * number of stellations of a polygon. So for {7/2,7,3,3} there would be {7/3,7,3,3}, and so on and so forth.

Interestingly enough, in odd dimensions there are no self-dual gothic polytopes, while in even dimensions there are.

[wendy]

There are no stellations of {p,3} that run to p/3 or p/4. You can see the case for p/3, that the edges would run down the centre of the cells, and out the far side. Even when p=7, they are diverging rapidly.

p/3 and p/4 do occur in some non-regular stars. For example, there are symmetry groups o3oPoP/3z, and oPoPoP/4z, of densities 4 and 6, derived from {p,3}. The dynkin symbols for these are triangles. We already met the second of these as {8,8} of density 6. The first as o3o9o3z, occurs as a subgroup of order 2 of {3,18} and of order four of {3,9}.

In odd dimensions, such as 3, ye have {p,p} would imply 4, but the only value of {p,q,r} in 3d that is a euclidean tiling, is where the shortchord-squares of p and r become 4, (ie 4, and 4), and thus q is 3. There is no other regular polyhedron with a rational angle than the cube.

[pentagonalpolytope747]

There are no stellations of {p,3} that run to p/3 or p/4. You can see the case for p/3, that the edges would run down the centre of the cells, and out the far side. Even when p=7, they are diverging rapidly.

Interesting, I see.

In my opinion, we are just inventing this universe ourselves. And all the rules of which polytope is "real" or which is "not" are just reliant of different levels of things that we allow to exist. This is math and artificially created universe, not physics or real-world science. While edges are "diverging rapidly", same happens with our {5,3,3,3} polyshape. This is the reason why only 4 star-honeycombs exist in H4-space, not 22. You can't have things beyond {5/2,5,3,3} and {5,5/2,5,3} and their duals, because, once again, edges would go through center and through far side. We can gatekeep hyperbolic geometry as "not being real geometry", but only "real" geometry then is Euclidean geometry.

So the reason why {7/3,7,2,2} exists, is because we say it is. It is just a product of set theory. Just invent a space where it is possible and done. A small fragment of geometry describes real world interactions, while a panfinitely large section of geometry describes everything else.

[wendy]

All of the different methods for inventing hyperbolig geometry lead to the same place.

I built a hyperbolic geometry as a slightly distorted euclidean one, measured up with a crooked (ie euclidean) ruler. It's pretty much as if you measured distances on a sphere by the embedded chord. The results are exactly the same as the 'standard' geometry, as far as i can tell. The main difference here is that you really can have a 1:1:3 triangle, as long as the edges are long enough.

The pictures are the same, the meainigs are different.

For example, a plane in hyperbolic space behaves as a kind of 'complex' point. Where you can have ordinary vertex-figures around a point, the same function can be carried out by a plane to which the 'converging' lines are orthogonal. Unlike euclidean geometry, a distant plane vanishes into the horizon in exactly the same way as a point does. So {3,5,3,5/2} can indeed exist. It appears as a faceting of something like {3,3,3,5}, but the icoahedra that come never really close up. In a real sense, you can't tell something like a {3,5,3} at a distance, from an ordinary polytope. This is because if you are not actually on the space concerned, the thing closes up in a fairly small distance.

Have a look at download/file.php?id=74 . Each of the circles in the central circle, is a complete plane, of which you can see every point, be mostly crowded towards the horizon. They are circles, because the boundary of a complete hyperbolic space is not 'straight': a sphere of infinite radius has a definite curvature relative to straight. (a horosphere is an infinite-radius circle). Those claw-like things are planes vertical to the large circle, and you can follow the circle right through the picture, as it intersects other circles.

The viewtopic.php?f=3&t=1842 contains another graphic, which is vertical to the first. The purple shapes represent the figure inbetween each of the planes. It's a laminatope, bounded by unbounded flat planes (rather like an infinite slab with two sides). Each circle, even the tiny dots, in the central disk is a complete plane, and virtually every direction leads you to a circle. The only infinite lines is to suppose you stood in the centre of a truncated cube, (it's made by sticking the truncated cubes of x4x3o8o together), there are six directions through the cube-face centres, which run to infinity without exiting.