higher-dimensional CRFs

[quickfur]

You're right, of course. I'm getting rusty at this! haven't wrangled with 4D for a while now...

What of (3), though? Any ideas how/whether it might generalize to 5D and beyond?

[username5243]

One other thing - in 5D and higher, you might get ursatopes with a uniform base, if it has triangular faces only. For instance, oox3xfo3oox3ooo&#xt could be called the rap ursateron (rap = rectified pentachoron) with 1 rap, 5 tet-ursachora, 5 oct-ursachora, 10 triangular prism pyramids, and 1 srip (small rhombated pentachoron, x3o3x3o). The only other polychora with only triangular faces are sadi (snub icositetrachoron) and rox (rectified hexacosichoron) - but I get the feeling neither of these will work. The sadi case would be a diminishing of the ex (hexacosichoron) ursateron, so if that one doesn't work, the sadi case won't either. Rox probably won't work either, especially if the ex case is degenerate...

With regards to the expansions, in 5D for the regular based ones, there are several types. The first kind is of the type xfo3oox3ooo3xxx&#xt and has teddi-prisms connecting the ursachoral facets. The other kind is of type xfo3oox3xxx3ooo&#xt and replaces the ursachoral cells with expanded-ursachoral cells. There is also a combined expansion xfo3oox3xxx3xxx&#xt with expanded-ursachoron cells with teddi-prisms connecting them.

[quickfur]

One other thing - in 5D and higher, you might get ursatopes with a uniform base, if it has triangular faces only. For instance, oox3xfo3oox3ooo&#xt could be called the rap ursateron (rap = rectified pentachoron) with 1 rap, 5 tet-ursachora, 5 oct-ursachora, 10 triangular prism pyramids, and 1 srip (small rhombated pentachoron, x3o3x3o). The only other polychora with only triangular faces are sadi (snub icositetrachoron) and rox (rectified hexacosichoron) - but I get the feeling neither of these will work. The sadi case would be a diminishing of the ex (hexacosichoron) ursateron, so if that one doesn't work, the sadi case won't either. Rox probably won't work either, especially if the ex case is degenerate...

Rap ursateron sounds like it might actually exist! That's very exciting. I've never thought of it that way, that the only requirement is that the base facet have only triangular 2-faces. (Well, besides the decagon radius restriction as Wendy explained. (And also the expanded form must also be CRF, since that's what ends up being the base facet.))

I'm pretty sure the sadi case won't work, since as you said it's a diminishing of the ex ursateron, which I think somebody has already proven is either flat or hyperbolic.

The rox case is unlikely to work, but it's hard to tell for sure: the "bad" thing is that the dichoral angles in rox are even wider than in ex, so it leaves even less room for the lacing teddy cells to fit in spherical curvature. However, the 600 octahedral teddies lacing the octahedral cells do have a narrower dichoral angle, so perhaps they might compensate just enough for it to be CRF? It's highly unlikely, but might be worth a more careful investigation before we rule it out.

With regards to the expansions, in 5D for the regular based ones, there are several types. The first kind is of the type xfo3oox3ooo3xxx&#xt and has teddi-prisms connecting the ursachoral facets. The other kind is of type xfo3oox3xxx3ooo&#xt and replaces the ursachoral cells with expanded-ursachoral cells. There is also a combined expansion xfo3oox3xxx3xxx&#xt with expanded-ursachoron cells with teddi-prisms connecting them.

Cool. I think these should exist, as per Klitzing's note on Stott expansions. But I'd like to see proof before I commit to it.

[username5243]

One other thing - in 5D and higher, you might get ursatopes with a uniform base, if it has triangular faces only. For instance, oox3xfo3oox3ooo&#xt could be called the rap ursateron (rap = rectified pentachoron) with 1 rap, 5 tet-ursachora, 5 oct-ursachora, 10 triangular prism pyramids, and 1 srip (small rhombated pentachoron, x3o3x3o). The only other polychora with only triangular faces are sadi (snub icositetrachoron) and rox (rectified hexacosichoron) - but I get the feeling neither of these will work. The sadi case would be a diminishing of the ex (hexacosichoron) ursateron, so if that one doesn't work, the sadi case won't either. Rox probably won't work either, especially if the ex case is degenerate...

Rap ursateron sounds like it might actually exist! That's very exciting. I've never thought of it that way, that the only requirement is that the base facet have only triangular 2-faces. (Well, besides the decagon radius restriction as Wendy explained. (And also the expanded form must also be CRF, since that's what ends up being the base facet.))

I'm pretty sure the sadi case won't work, since as you said it's a diminishing of the ex ursateron, which I think somebody has already proven is either flat or hyperbolic.

The rox case is unlikely to work, but it's hard to tell for sure: the "bad" thing is that the dichoral angles in rox are even wider than in ex, so it leaves even less room for the lacing teddy cells to fit in spherical curvature. However, the 600 octahedral teddies lacing the octahedral cells do have a narrower dichoral angle, so perhaps they might compensate just enough for it to be CRF? It's highly unlikely, but might be worth a more careful investigation before we rule it out.

With regards to the expansions, in 5D for the regular based ones, there are several types. The first kind is of the type xfo3oox3ooo3xxx&#xt and has teddi-prisms connecting the ursachoral facets. The other kind is of type xfo3oox3xxx3ooo&#xt and replaces the ursachoral cells with expanded-ursachoral cells. There is also a combined expansion xfo3oox3xxx3xxx&#xt with expanded-ursachoron cells with teddi-prisms connecting them.

Cool. I think these should exist, as per Klitzing's note on Stott expansions. But I'd like to see proof before I commit to it.

If you mean the radius has to be smaller than that of the regular decagon to work, then rox ursateron shouldn't work - if thefigure on Klitzing's site is correct, rox's circumradius is greater than 3, whereas the radius of a decagon is the golden ratio...

[Mercurial, the Spectre]

So I decided to investigate on other ursatopes based on uniform polytope bases.

The concept of a CRF ursatopic representation, as discussed by username5243 earlier, must be based on an alternation of a polytope whose verf has topological triangular faces and nothing else so that its alternation has triangular faces only, ensuring the CRFness of that derived ursatope. It may not be always possible because some, such as s3s3s3s, won't work.

Again, for convenience I will design a short representation for these pyramids that can be used as vertex figures: ^q|(base polytope). This represents the pyramid of the base polytope (considered to have lengths of sqrt(2)) and lateral sides sqrt(3). Then the polytope/honeycomb that has this as its vertex figure can be adjusted and uniformly alternated to get something whose verf is the ursatopic representation of the verf.

Adding on username5243's comment, the restriction is that the base polytope must have a circumradius of less than 1 (if the base polytope has length 1).

In short, as proven above, the existence of a CRF ursatope entirely depends on the base polytope's faces and its circumradius measured in proportion to its edge length.

2D: one valid representation
^q|line is toe's verf and its ursatopic representation is the regular pentagon.

3D: one valid representation
^q|x3o is tico's verf and its ursatopic representation is teddi.
Others, such as ^q|x4o and ^q|x5o will not have CRF ursatopic representations.

4D: three valid representations
^q|x3o3o is t[5]'s or ticot's ver, and its ursatopic representation is the tetrahedral ursachoron.
^q|x3o4o is t4[5]'s or x3x4o3o4o's verf and its ursatopic representation is the octahedral ursachoron.
^q|x3o5o is x3x4o3o5o's verf (x3x4o3o5o has these facets: x3x4o3o (tico) and x4o3o5o (order-5 cubic honeycomb)) and its ursatopic representation is the icosahedral ursachoron.

5D: three valid representations
^q|x3o3o3o is t[6]'s or x3x4o3o3o3o's verf (facets: x3x4o3o3o and x4o3o3o3o), and its ursatopic representation is the pentachoric ursateron.
^q|x3o3o4o is t4[6]'s or x3x4o3o3o4o's verf (facets: x3x4o3o3o and x4o3o3o4o), and its ursatopic representation is the 16-cell ursateron.
^q|o3x3o3o actually has a CRF ursatopic representation. This is because rap's circumradius is less than 1, so from the honeycomb with ^q|o3x3o3o as its vertex figure it can be rescaled to give a CRF rappy as its verf, and then alternated to form a honeycomb whose verf is the rap ursateron.
Others, such as ico ursateron, ex ursateron, sadi ursateron, and rox ursateron don't work because their bases either have a circumradius equal (for ico) or greater than 1.

6D: five valid representations
^q|x3o3o3o3o's ursatopic representation is the 5-simplex (or hix) ursapeton.
^q|x3o3o3o4o's ursatopic representation is the 5-orthoplex (or tac) ursapeton.
^q|s4o3o3o3o's ursatopic representation exists. This is because hin's circumradius is less than 1 (0.791 to be exact). The hin ursapeton will have these facets: hin, nit (rectified hin), 16-cell ursateron, 5-cell ursateron and rappy (rectified 5-cell pyramid).
^q|o3x3o3o3o works. The resulting rix ursapeton will have rix, srix, pen ursateron, rap ursateron, and tepepy (tetrahedral-prism pyramid) for facets.
^q|o3o3x3o3o, ditto. Dot ursapeton has dot, sibrid, rap ursateron, and triddippy (3-3 duoprism pyramid) for facets.

7D: six valid representations
^q|x3o3o3o3o3o and ^q|x3o3o3o3o4o are trivial.
^q|s4o3o3o3o3o's ursatope is the hax ursaexon, with hax, brox, hin ursapeton, hix ursapeton, and rixpy (rectified 5-simplex pyramid) for facets.
^q|o3x3o3o3o3o's ursatope is the ril ursaexon, with ril, sril, rix ursapeton, hix ursapeton, and penppy (pentachoron-prism pyramid) for facets.
^q|o3o3x3o3o3o's ursatope is the bril ursaexon, with bril, sabril, rix ursapeton, dot ursapeton, and tratetpy (triangle-tetrahedron duorism pyramid) for facets.
^q|x3o3o3o3o *c3o's ursatope is the jak ursaexon, with jak, rojak, tac ursapeton, hix ursapeton, and hinpy (demipenteract pyramid) for facets.
The ^q|o3o3o3o3o *c3x, which is tied to the mo ursaexon, is flat since mo has a circumradius equal to its edge length.

8D: six valid representations
Since it's getting too long, here are the possible bases: 7-simplex, 7-orthoplex, 7-demicube, rectified 7-simplex, birectified 7-simplex, and o3o3o3o *c3o3o3x (naq).

9D: three valid representations.
Possible bases: 8-simplex, 8-orthoplex, rectified 8-simplex
The 8-demicube and the birectified 8-simplex has a circumradius equal to its edge length.

10D and above: three valid representations.
Possible bases: n-simplex, n-orthoplex, rectified n-simplex.
The rectified n-simplex is always possible since it is the vertex figure of the (n+1)-demicube, which exists in any dimension, and the fact that an (n+1)-demicube has a finite circumradius.

[Klitzing]

It appears that the series of ursatopes is infinite and orbiform, although only ones based on the simplex and the orthoplex.
To consider, look at them as vertex figures.
...

Conclusion:
s[x]'s verf happens to be the (x-1)-dimensional ursachoron. Its infinitude (and orbiformity) is due to it being represented as a verf of s[x], which can be derived from t[x], from which t[x]'s verf can always be deformed to equal lengths! Though its circumradius does not hold, it keeps on increasing so that at 4D, you'd have an ursatope with a circumradius equal to its edge length, and in 5D, it is more than 1, so it can only be represented as a verf of a hyperbolic honeycomb. Same applies to the hyperoctahedral ursatopes.

I think all of you missed a simple trick, haha!
But that's ok , I'm contributing to your efforts.

Cheers,
Mercurial

Thus you are saying that these ursachora do exist just because they are vertex figures of according (possibly hyperbolic) hyperhoneycombs.
Moreover you relate xfo3oox(3ooo)*&#xt (N+2 nodes) to s3s4o(3o)* (N+3 nodes) and furthermore xfo4oox(3ooo)*&#xt ((N+1)+2 nodes) to s3s4o(3o)*4o (N+4 nodes).

But by mere formal analogy then xfo5oox(3ooo)*&#xt ((N+1)+2 nodes) ought occur as vertex figure of s3s4o(3o)*5o (N+4 nodes).
So the stated non-existance here of the formers (for any N>0) thus then should imply the non-existance of the latter - is that correct?

How could one understand that?

--- rk

[wendy]

Of course, the rap ursulatope exists.

For the uniform polytopes, the base can consist of a singly marked dynkyn-graph, with the x-node connected only by '3' branches.

You then get by the Conway-Hart rule, that the rectate consists of unmarking the marked node, and marking the adjacent nodes. This will turn a figure to a rectate, or a n-rectate into an n-cantelate (eg o3x3o3o -> x3o3x3o. ). From this, one makes the ursulatope between these two sections.

There are an awful lot. All of the rectate simplexes up to something like 19 dimensions are in the list, and a good few of the rectates of the cross-polytope. The half-cube is in there, as well as many of the Elte-Gosset polyhedra.

[Mercurial, the Spectre]

It appears that the series of ursatopes is infinite and orbiform, although only ones based on the simplex and the orthoplex.
To consider, look at them as vertex figures.
...

Conclusion:
s[x]'s verf happens to be the (x-1)-dimensional ursachoron. Its infinitude (and orbiformity) is due to it being represented as a verf of s[x], which can be derived from t[x], from which t[x]'s verf can always be deformed to equal lengths! Though its circumradius does not hold, it keeps on increasing so that at 4D, you'd have an ursatope with a circumradius equal to its edge length, and in 5D, it is more than 1, so it can only be represented as a verf of a hyperbolic honeycomb. Same applies to the hyperoctahedral ursatopes.

I think all of you missed a simple trick, haha!
But that's ok , I'm contributing to your efforts.

Cheers,
Mercurial

Thus you are saying that these ursachora do exist just because they are vertex figures of according (possibly hyperbolic) hyperhoneycombs.
Moreover you relate xfo3oox(3ooo)*&#xt (N+2 nodes) to s3s4o(3o)* (N+3 nodes) and furthermore xfo4oox(3ooo)*&#xt ((N+1)+2 nodes) to s3s4o(3o)*4o (N+4 nodes).

But by mere formal analogy then xfo5oox(3ooo)*&#xt ((N+1)+2 nodes) ought occur as vertex figure of s3s4o(3o)*5o (N+4 nodes).
So the stated non-existance here of the formers (for any N>0) thus then should imply the non-existance of the latter - is that correct?

How could one understand that?

--- rk

In practice, yes. I've been researching on this for several days now.
I'll give t4[6] - x3x4o3o3o4o. Facets are then (as stated before) x3x4o3o3o (truncated 24-cell honeycomb) and x4o3o3o4o (tesseractic honeycomb). The verf is a regular 16-cell pyramid (facets: 16 regular-tetrahedral pyramids and one 16-cell). The 16-cell base corresponds to x4o3o3o4o and is of length sqrt(2), and the tetrahedral pyramids correspond to x3x4o3o3o. All other lateral sides are of length sqrt(3). Then you alternate t4[6] and get s4[6] (s3s4o3o3o4o). Now the x3x4o3o3o gets alternated to s3s4o3o3o (snub 24-cell honeycomb), x4o3o3o4o into s4o3o3o4o (16-cell honeycomb), and the verf of t4[6] is introduced as new facets of that polytope. Note that it can be made scaliform because one can resize the hexpy's edges to unity by adjusting geometries of t4[6]. Then s4[6]'s verf is formed by replacing the 16-cell base with a 24-cell base (verf of s4o3o3o4o or x3o3o4o3o), then the 16 tetrahedral pyramids are replaced by 16 tetrahedral ursachora (s3s4o3o3o's verf). Then there will be one 16-cell and eight octahedral pyramids (corresponding to hexpy's verfs).

Indeed, s4[x]'s verf is similar to t4[x]'s verf, formed by putting s[x-1]'s verf into the lateral cells of t4[x]'s verf, then taking its convex hull (which happens to be an ursachoron itself). In fact ursatopes can be discovered this way.

Let's assume your latter part as t5[x]. Now if we try to start with t5[5] (x3x4o3o3o5o) one has these facets: x3x4o3o3o and x4o3o3o5o. Then the verf is a 600-cell pyramid. But the 600-cell pyramid cannot be made CRF, in fact the ratio of circumradius to edge length is in the golden ratio. This is also the reason why {3,3,3,5} is not representable as a polyteron. Simply put, s5[5] wouldn't have a scaliform realization if we stick to euclidean space only. The former part works because the verfs of both the n-cube and its honeycomb variant are (n-1)-simplices and (n-1)-orthoplexes, which always has a circumradius of less than 1 assuming unit edges, but there is no infinite polytope family that corresponds to {3^n,5}, since it stops at {3^2,5} or {3,3,5}.

If one tried to construct a 600-cell ursateron, you'd be better going to hyperbolic space if you want unit edges. It doesn't imply non-existence of both, their existence depends on the base polytope (which I mentioned earlier).

[Klitzing]

Of course, the rap ursulatope exists.

For the uniform polytopes, the base can consist of a singly marked dynkyn-graph, with the x-node connected only by '3' branches.

You then get by the Conway-Hart rule, that the rectate consists of unmarking the marked node, and marking the adjacent nodes. This will turn a figure to a rectate, or a n-rectate into an n-cantelate (eg o3x3o3o -> x3o3x3o. ). From this, one makes the ursulatope between these two sections.

There are an awful lot. All of the rectate simplexes up to something like 19 dimensions are in the list, and a good few of the rectates of the cross-polytope. The half-cube is in there, as well as many of the Elte-Gosset polyhedra.

In fact, you needn't add the cross-polytopal symmetry extra. Because of  o4o3o(3o)*  =  o3o(3o)* *b3o.
And the 4-adjacent nodes weren't allowed in the former anyway.

Thus to conclude:
The lower base polytope must be any Dynkin symbol with 3-linkages only, having a single node being ringed.
The next layer then will be a f-scaled copy of the former.
And the final layer of the tower then is the rectification (according to the Conway-Hart rule) of the base layer.

You then simply will have to check in addition, whether the stacking heights evaluate to be positive.

E.g. wrt. o3o3o3o *c3o3o3o3o symmetry the nodes at positions a, d, g, and h would work!
I.e. resulting in the following 4 ursayotta with axial E8 symmetry:

Code: Select all

ofx3xoo3ooo3ooo *c3ooo3ooo3ooo3ooo
ooo3ooo3xoo3ofx *c3ooo3ooo3ooo3ooo
ooo3ooo3ooo3ooo *c3ooo3xoo3ofx3xoo
ooo3ooo3ooo3ooo *c3ooo3ooo3xoo3ofx


And all of which would be heavily Stott expandable as well.

--- rk

[Klitzing]

The last conclusion btw. also states that

Code: Select all

xoo3ofx3xoo *b3xoo&#xt  =  ofx3xoo4ooo3ooo&#xt


should exist as well.

At least the heights of the 2 segments there calculate to:
height(1,2) = sqrt[sqrt(5)-2] = 0.485868272,
height(2,3) = sqrt[(srt(5)-1)/2] = 0.786151378

So we just lack the proof that the teddies won't get broken at the medial layer...

(Was that perhaps already somewhere immanent in your restriction to only 3-linkages, Wendy?
Or should we have to check this for any potential candidate separately?
I.e. esp. for that F4 one as well as for the former E8 ones...)

--- rk

[wendy]

The ursulate polytopes derive from a cone where the edges of the base become sides of a pyramid with sides ffx. That is, ox#tf.

If you construct this pyramid of sides 2x, and lacing 2f, it contains the pyramid of sides 1x and lacing 1f, and a rostrum or cupola from 1f to 2f. You can see these triangles are equal, since the f:f:x triangle on one side, will in one direction be part of an inscribed pentagram, and on the outside, the stellation pentagram.

If you now take a trapezoid xu&#tf, you can "clip the toes" by going down the lacing edge unit distance (x), and then draw a line from that point to the middle of the base. This converts it into xfo&#tx, which is of course, a pentagon.

The base-clipping in effect rectifies every polygon uPo into oPa, where 'a' is the shortchord of the polygon. So the clippings have triangles xxa (two pentagon-edges, and an edge of the runcinate. This only gives an x3o if all the values of a are x, ie all surhedra (2d surtopes), are triangles.

We see then that the f:f:x condition limits the overall size of the top to less than a decagon, since this triangle has a 36° apex. The creation of the xxa triangles in the toe-clipping, limits the feet to a=x, ie all polygons are triangles.

Using two marked nodes in the dynkin symbol produces squares, in the form of the surtope made from both marked nodes, and this will produce 'q' edges in the runcinate. eg x3o3x, CO, produces squares with the two nodes "x . x". That's why you can only use one node.

There is no chance of buckling unless the rectate vertices do not fall in the middle of the edges of the top, when scaled to edge-2. This is because in the pentagon, if you draw a line of 2, bisected by a vertex, and parallel to the opposite side of the pentagon, gives points that are on the remaining two sides.

So all you have to do, is ferret out any triangle-only wythoff polytope, whose circum-radius is less than a decagon.

This means:
1: only 1 marked node (produces no squares via x..x)
2: only '3' branches join the marked node.
3: the circumradius is less than f.

The rectate consists of marking those nodes directly connected to the marked node, but not the marked node itself. This is the base.

The remaining features at the bottom, outside of ursulates running from top to bottom, are pyramids corresponding to the vertex-figure of the base.

So the rectate of o3o3x3o3o is o3x3o3x3o, and the vertex-figure is o3x . x3o

The conditions are necessary and sufficient to form an ursulate.

[Klitzing]

Great news, Wendy, thanx for explanation!

So I deduce that the yesterday mentioned "a" and "h" positional cases of E8 ought work, as these base on 3-linkage only diagrams and those positions will provide a polytopal circumradius (of that layer) < f.
(While those there also mentioned positions "d" and "g" would have radius > f. Thus your mentioned pyramid there would become hyperbolic.)

Further I note that you just refer to the actual 2D faces of the non-rectified quasiregular starting figure. That is, you just restrict to diagrams with the marked node having incident links marked 3 only. Esp. nothing is thus said to all the other links.

Therefore both representations of ico (o3x3o *b3o = x3o4o3o) would work: adjacent 3-linkages only, radius = x < f.
Or that already known ofx3xoo5ooo&#xt could be given here too. (Whereas a potential ofx3xoo3ooo5ooo&#xt would fail because of r(x3o3o5o) = f; i.e. your mentioned pyramid would become flat there.)

And of course, we have the freedom for across-dimensional Stott expansions, which e.g. at these E8 figures provides lots of applications, hehe.

--- rk

[wendy]

The nature of the other links does not matter. In x3o3o5o, the 'toe' pyramids are ox3oo5oo&#xt, but these are CRF in any case. Where this example falls is that the pentagons would lie flat in the hedrix containing the girthing decagon, and thus all of the teddies developed of the tetrahedral faces of x3o3o5o, would be in the same space as the top. The thing is flat.

When you consider your example of o3x3o *b3o, the rules of symmetry still apply with ursulates, so it is not just an ursulate of the named figure, but also x3o4o3o, and o3x3o4o, since the base is simply the o3x4o3o, x3o3x4o, and x3o3x *b3ox.

Yes, the assorted stott-expansions work here too. In something like G8 4_21, you could make 64 possible expansions here.

[Klitzing]

One more interesting thing about the urs(ul)atopes.

Sure the pentagon and teddi itself are orbiform.
For the tet-urs and oct-urs Mercurial showed that these occur as vertex figures of snub hyperbolic tetracombs, thus those are orbiform too.
The ike-urs is a diminishing of ex, so it clearly is orbiform.

Both for the rap-urs and the ico-urs I just calculated the circumradii of the bottom and the top segments each. They too do come out to be the same values.

So I suspect that the non-expanded ursatopes might be orbiform generally.
Could that conjecture be proved?

--- rk

[wendy]

Your assertion about being an orbiform (read as all vertices are cotangential to a sphere), is indeed correct.

The top and base are both orbiforms, and thus you can place these inside a sphere of any size (think segmentotope). You can then intersect this sphere with a hedrix, through an edge of the top and the matching vertex of the bottom, this contains the pentagon. Thus the middle layer is thus on the same sphere.

The circumradius of this beast can be derived from the top and middle layer, which forms an isoceles trapezoid, with a = d, b = fd, and c = 1.

https://en.wikipedia.org/wiki/Isosceles_trapezoid \( R^2 = (ab+c^2)/(4c^2 - (a-b)^2)\).

R^2 = f^2(f D^2 + 1)/(4f^2-D^2) : where D is the diameter of the base.

When D^2 = 3.618, then the centre of radius is in the middle of the base, and the ursulate and its reflect form a common orbifold. You need to cover the toe-points with five tetrahedra each.

When D^2 = 5.236 = 3+2f, then the toe-faces are vertical. This means that if you join a second figure to the base, the result is still a CRF. Larger values will make the angle between the toe-faces and the base into acute angles. You can join the same or different ursulates together. An example would be the expanded usulates

o3x3o3o3e *c3a and e3o3o3x3o *3a, which share a common base x3o3x3o3x *3a, (where "e" is a required expand, and "a" an optional expand), the result is then

ooxxx3xfooo3oox3ooofx3xxxoo *c3aaaaa a=o | x. This is not orbiform.

[Klitzing]

Orbiform = all unit edges + all vertices on a circumsphere

 
Is your argument truely correct, Wendy? Cause when applying across expansions I've the feeling that the former circumsphere would become prolate or oblate, i.e. orbiformity ought get lost. But by your argument these should all be orbiform as well. -

--- rk

[wendy]

It's the basis of circle-drawing, that if you intersect a sphere with an isocurve (such as a euclidean plane), the result is a sphere. Because we know that the points of a pentagon are colpanar, the five vertices fall on the same sphere that the top and bottom fall on. Therefore the figure is itself an orbiform.

I'm still trying to look at expansions. I'm told that xfo3oox5xxx gives the tri-diminished x3o5x in there. I need to fiddle around with some models in my head to see if the expansions work as indicated.

It should work as marked for non-adjacent nodes, but i need to make some obvious thing there too.

[Klitzing]

Thought of replacing teddies by tuts. We already had that somewhere in the past. Those ursatope like friends already were named by someone. I don't recall how, by whom, nor when.
There you will have to stack a quasiregular, a u-scaled copy, and the truncate of that quasiregular (i.e. ringing the directly neighbouring nodes in addition).

There you get by analogy to Wendy's rules for ursulates

1: only 1 marked node (produces no squares via x..x)
2: only '3' branches join the marked node.
3: the circumradius is less than f.

an according rule here too:

1: only 1 marked node (produces no squares via x..x)
2: only '3' branches join the marked node.
3: the circumradius is less than x.

Moreover it then seems to happen that these then just are the bistratic quasiregular first caps of that truncate polytope, which adds to that single marked node of that quasiregular a further leg marked 3, and that end node will be ringed as well.

E.g.

Code: Select all

o3x3o                         x 
o3u3o  is bistratic cap of    3 
x3x3x                       o3x3o


and therefore all are orbiform trivially.

Again we could apply across dimensional expansions to these tower symbols, e.g. xux3oox3ooo&#xt --> xux3oox3xxx&#xt.
But still I have no feeling whether such an expansion respects or breakes the former orbiformity. (Wendy, did you make some progress here?)

--- rk

[username5243]

I'm pretty sure these expanded cases are orbiform, as they are also caps of uniform polytopes. For example:

Code: Select all

xux3oox3ooo&#xt = x3x3o3o
xux3oox3xxx&#xt = bistratic x3o3x-cap of x3x3o3x

xux3oox3ooo3ooo&#xt = x3x3o3o3o
xux3oox3xxx3ooo&#xt = bistratic x3o3x3o-cap of x3x3o3x3o
xux3oox3ooo3xxx&#xt = bistratic x3o3o3x-cap of x3x3o3o3x
xux3oox3xxx3xxx&#xt = bistratic x3o3x3x-cap of x3x3o3x3x


Can you see a pattern?

The ones based on cross-polytopes (xux3oox3ooo...4ooo&#xt) and their expansions behave similarly - just change the last 3 in each case above to a 4. And upon checking the expansion of the rap case (oox3xux3oox3xxx&#xt) it seems to be a x3o3x3o-cap of x3x3o *b3o3x...

So these are probably orbiform too...

[wendy]

I'm pretty sure too, since all the vertices move in the same angle according to rays from the centre. That's the effect of stott expansion.

The whole ursulation is happening in a very tiny figure, and as long as you have a chain of polygons >3 from top to bottom, you pretty much have proof that they are a progression on the same sphere.

[Mercurial, the Spectre]

Turns out I was wrong with the circumradius thing. Wendy pointed out that if the pentagons are flat, the base must have a circumradius of phi. We can disregard the pyramid limitations given by the bases. Turns out the ico ursateron exists after all...

Again, for the tutsatopes (ursatopes with tut-derived lacings instead of teddi), one considers all bases with a circumradius of less than one. The erroneous list (which I mentioned earlier) could be the bases of such polytopes.

In fact, I've been exploring tetrahedron|gyrobifastigium segmentochora, with both placed in D2d symmetry. Cells are then 1 gyrobifastigium, 1 base tetrahedron, 4 lateral tetrahedra, and 4 lateral augmented triangular prisms (J49). Is it possible to make this object CRF disregarding orbiformity? One interesting fact is that they are used as vertex figures for the 3-{3,3} duoantiprism (alternation of the 6-{4,3} duoprism).

[Klitzing]

I'm pretty sure too, since all the vertices move in the same angle according to rays from the centre. That's the effect of stott expansion.

This is exactly what I was doubting! For the full dimensional expansion acts radially outwards, while the subdimensional one acts radially within each layer, i.e. tangentially wrt. the full dimension!
But then the tiny list from username5243 wrt. the "tutsatopes" at least shows that in those cases it nevertheless works.
So I'm a bit torn...
--- rk

[quickfur]

Thought of replacing teddies by tuts. We already had that somewhere in the past. Those ursatope like friends already were named by someone. I don't recall how, by whom, nor when.
There you will have to stack a quasiregular, a u-scaled copy, and the truncate of that quasiregular (i.e. ringing the directly neighbouring nodes in addition).

At least for the case where the base cell is an n-simplex, replacing the teddies (n-teddies) with tuts produces nothing other than the truncation of the n-simplex x3x3o3...3o.

In 4D, if you start with an octahedral base cell, then you get the bisected truncated 16-cell (i.e., half of o4o3x3x, with the equatorial octs cut into square pyramids). I'm not 100% certain, but I think starting with an n-cross base cell and replacing the (n-1)-teddies with (n-1)-simplex truncates would also give you a bisected n-cross.

The icosahedral case in 4D gives you a bistratic cap of the truncated 600-cell o5o3x3x, which I think we named a "rotunda" in the past (IIRC there was some debate over whether "rotunda" should be reserved for hemi-n-spherical polytopes and polystratic caps should just be called caps).

I think I've considered these cases before... and came to the conclusion that you would always get some kind of section of a uniform polytope, i.e., nothing new unlike the ursatope family. But I don't have a formal proof of this, so it's possible I missed some cases that might be more interesting.

[Klitzing]

Thought of replacing teddies by tuts. We already had that somewhere in the past. Those ursatope like friends already were named by someone. I don't recall how, by whom, nor when.
There you will have to stack a quasiregular, a u-scaled copy, and the truncate of that quasiregular (i.e. ringing the directly neighbouring nodes in addition).

At least for the case where the base cell is an n-simplex, replacing the teddies (n-teddies) with tuts produces nothing other than the truncation of the n-simplex x3x3o3...3o.

In 4D, if you start with an octahedral base cell, then you get the bisected truncated 16-cell (i.e., half of o4o3x3x, with the equatorial octs cut into square pyramids). I'm not 100% certain, but I think starting with an n-cross base cell and replacing the (n-1)-teddies with (n-1)-simplex truncates would also give you a bisected n-cross.

The icosahedral case in 4D gives you a bistratic cap of the truncated 600-cell o5o3x3x, which I think we named a "rotunda" in the past (IIRC there was some debate over whether "rotunda" should be reserved for hemi-n-spherical polytopes and polystratic caps should just be called caps).

I think I've considered these cases before... and came to the conclusion that you would always get some kind of section of a uniform polytope, i.e., nothing new unlike the ursatope family. But I don't have a formal proof of this, so it's possible I missed some cases that might be more interesting.

Yes, "rotunda" I use solely for hemisphere, else cap.

And yes again, you are right, the "tutsatopes" are all bistratic caps of uniforms.
Cf. the description and code I already gave

Moreover it then seems to happen that these then just are the bistratic quasiregular first caps of that truncate polytope, which adds to that single marked node of that quasiregular a further leg marked 3, and that end node will be ringed as well.

E.g.

Code: Select all

o3x3o                         x 
o3u3o  is bistratic cap of    3 
x3x3x                       o3x3o


Thus the xPoQo...-based tutsatope is just the bistratic cap of x3xPoQo..., the oPxQo...-based tutsatope is just the bistratic cap of oPx3x *bQo..., etc.

So you are right that these won't produce something completely new.
But the other way round it becomes interesting:

• All those truncates thus allow for a bistratic cap cut off, which happens to be CRF in general. (I.e. no slantedly dissected cells can occur.)
• And also: All these tutsatopes are trivially orbiform, even without application of the consideration of Wendy, just because of being a cap of a uniform.
• And finally: There is a complete sufficient and necessary characterization of the tutsatopes directly from their base polytope, i.e. without taking refuge to that relation to the truncate, cf.
  

  1: only 1 marked node (produces no squares via x..x)
  2: only '3' branches join the marked node.
  3: the circumradius is less than x.
  

 
What I'm not fully clear here is whether the Stott expansions of across-symmetry would translate into those of the truncate.
E.g. would it be correct that we obtain from the true sentence "oox3xux3oox3ooo&#xt is a bistratic cap of o3x3x *b3o3o" a further true sentence - via correspondingly applied Stott expansion on either side - i.e. something along the lines "oox3xux3oox3xxx&#xt is(?) a bistratic cap of o3x3x *b3o3x"?

--- rk

[Klitzing]

What I'm not fully clear here is whether the Stott expansions of across-symmetry would translate into those of the truncate.
E.g. would it be correct that we obtain from the true sentence "oox3xux3oox3ooo&#xt is a bistratic cap of o3x3x *b3o3o" a further true sentence - via correspondingly applied Stott expansion on either side - i.e. something along the lines "oox3xux3oox3xxx&#xt is(?) a bistratic cap of o3x3x *b3o3x"?

Afterthought:
at least the circumradii of either of these therein used segmentotera oo3xu3oo3xx&#x resp. ox3ux3ox3xx&#x
and that of pithin = o3x3x *b3o3x all evaluate to be = sqrt(45/8).
--- rk

[Klitzing]

What I'm not fully clear here is whether the Stott expansions of across-symmetry would translate into those of the truncate.
E.g. would it be correct that we obtain from the true sentence "oox3xux3oox3ooo&#xt is a bistratic cap of o3x3x *b3o3o" a further true sentence - via correspondingly applied Stott expansion on either side - i.e. something along the lines "oox3xux3oox3xxx&#xt is(?) a bistratic cap of o3x3x *b3o3x"?

Seems to be right indeed:
for pithin can be given as oox(uo)(ox)xx-3-xux(xu)(xu)xo-3-oox(xo)(oo)ox-3-xxx(ou)(dx)ux-&#xt, where u=2x, d=3x and the "layers" within parantheses are meant to be co-realmic (i.e. the here to be used vertex set would be that of the according compound). The strictly positive heights there would be all = sqrt(2/5) = 0.632456.
Thus its bistratic cap thus indeed is oox3xux3oox3xxx&#xt.

Thus Stott expansions of across-symmetry of the D-dim tower here seems to translate directly (via the (D-1)-dim tutsatope-base resp. its facial usage within the D-dim truncate into (full-dim) Stott expansion of the D-dim truncate.
Moreover, as the latter clearly is orbiform (moreover: uniform), the former thus is forced to be orbiform as well.

This then in turn seems to support that Stott expansions of across-symmetry of D-dim towers generally will respect orbiformity, i.e. whenever the pre-image was, then the result will be too. - Not a clear proof, but a further support on the conjecture.

--- rk

[Klitzing]

Dear Wendy, Jonathan, et al.,

thought a bit about that recent urs(ul)a- and tutsa- stuff. These describe genuinely bistratic axial polytopes with some common symmetry, where a link maked 3 gets some special role. In the one case one errects a teddi thereon in the other a tut. The remainder of the across diagram could be anything (within the restrictions provided by Wendy for the teddies, resp. by me for the tuts). Off-special nodes could be ringed or un-ringed by choice, as long those apply through all 3 layers in the same pattern.

Came now up with the bistratic caps of rahi. These can be given as ooo3xox5ofx&#xt. Thus here we have kind a perotope. I.e. the special link now is marked 5, again we have a bistratic polytope, and obviously the name here is borrowed from the 3D representant xox5ofx&#xt = pero = J6.

Sure, links marked 5 are kind of restricted to lower dimensions only (as long as we consider spherical symmetry only and the across diagram remains connected), but we might consider o5o, o3o5o, and o3o3o5o here.

So, could anyone come up with similar restrictions (as refered to above) to the applicability of perotopes? (Either by means of some general result or at least by means of the rather few individual checkings?)

 

Btw. Polyhedron Dude coined general acronyms for ursatopes and tutsatopes:
un-expanded ones, i.e. those which are based on a quasiregular figure, easily can adopt their name therefrom: X-u (X-ursulate), resp. X-tum (X-tutism).
For the expanded ones one has to consider both parallel bases: X-a-Y-u (X atop Y -ursulate) or X-al-Y-u (X atop inverted (=alternate) Y -ursulate), similarily X-a-Y-tum resp. X-al-Y-tum.
Thus, it seems we need for further acronyms here again...

--- rk

[Mercurial, the Spectre]

It's only limited to four dimensions. To see: look at the segments centered icosahedrally. It can be represented as rectified {5,3^n} || {5,3^n} || truncated {5,3^n}. In 2D, you get a pentagon. In 3D, you get the pentagonal rotunda. In 4D, you have the id || f-doe || tid, where the {5,3^n} part is phi times the rectified{5,3^n}'s edge length. Since these involve rectified {5,3^n} cells you have the fact that it is a part of rectified {5,3^n}. This is what Klitzing told earlier. {5,3,3,3} is hyperbolic.

In fact, perotopes can be applied to base polytopes related to the pentagon. In 3D this is trivial, and that is the pentagonal rotunda. The family of perotopes derived from above is one example. The 4D variant could be called "peroid". In 4D, one has these possible bases: dodecahedron, icosidodecahedron (covered), and rhombicosidodecahedron. Now look at the f-part of pero. They are antialigned with the base, so the edges of the original are necessarily vertices of the f-part. This implies that the f-part is the rectification of base part. This excludes the dodecahedron as a base. Then the third layer is simply the truncate of the f-part, creating decagonal faces.

Let's look at the one remaining case: srid || f-id || grid. I don't know if that can be realized as CRF, but it's the only one in 4D other than above.

[Klitzing]

xox5ofx&#xt = pero

xox5ofx3ooo&#xt does not work because of imaginary height between leayer 1 and 2

ooo3xox5ofx&#xt = id-first bistratic cap of o3o3x5o
xxx3xox5ofx&#xt = id-first bistratic subsegment of x3o3x5o

ooo3ooo3xox5ofx&#xt does not work because of imaginary height between leayer 1 and 2

--- rk

[mr_e_man]

Because of the properties of prisms, any vertex figure of a CRF n-polytope is also an edge figure of a CRF (n+1)-polytope, and any (n-1)-dyad (with its ditopal angle) is also an n-dyad.

Denote by A_n the set of all ditopal angles of CRF n-polytopes. Then A_n ⊆ A_n+1. Also denote by A_∞ the union of all A_n; thus A_∞ is the set of all ditopal angles in any number of dimensions.

A_-1 = A₀ = A₁ = {}. The line segment is a CRF 1-polytope; it doesn't have any (2D) faces, so all faces are regular, vacuously. It does have a ditopal measure, but it's a length, not an angle; so A₁ is empty.

The regular polygon angles are A₂ = {180° - 360°/k | k∈ℕ, k≥3}; the smallest one is 60°, and the largest ones approach 180°.

The CRF polyhedron dihedral angles are A₃ (shown in my list of verfs); the smallest one is 31.7175°, and the largest ones approach 180°, and there's also an infinite subset approaching 90° (from antiprisms).

The CRF polychoron dichoral angles are A₄ (not fully known yet); the smallest known one is 10.8123°, in the pentagonal magnabicupolic ring.

Is there any lower limit to A_∞ (other than 0°)? Or do the angles get arbitrarily small?

In other words, is inf(A_∞) = 0° ?

If so, do we need to go to A_∞, or is there some fixed dimension n such that inf(A_n) = 0° ?