A different way to expand an ursatope?

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

Re: A different way to expand an ursatope?

Postby quickfur » Mon Mar 14, 2016 6:39 pm

Klitzing wrote:[...]
Thus, what then would be xfox-2-oxfo-P-oooo-&#xt (P=3,4,5)? - I will have to take a look into these...

--- rk

Good news!! While the case of P=4 doesn't seem to close up in a CRF way, I have just constructed the P=3 case (with a slight modification) and verified that it's CRF!!!!!! The slight adjustment is that the last vertex layer must be x2o3x, not x2o3o, otherwise the mibdies are non-corealmar. So the corrected symbol is xfox-2-oxfo-3-ooox&#xt (note the last x in place of o).

This is a truly beautiful little CRF; it has 3 mibdies and 2 tetrahedra on one side, 1 trigonal prism, 2 octahedra(!), 3 square pyramids, and 6 tetrahedra on the other side. Even better yet, it has 17 vertices and 17 cells -- the same number of vertices as cells!

Here's a look at the 3 mibdies surrounding an edge:

Image

The 2 tetrahedra that fill in the gaps between the mibdies should be obvious.

Now here's the far side, which has a most interesting configuration of cells:

Image

I highlighted the trigonal prism which lies antipodal to the edge shared by the 3 mibdies. We see a most interesting pattern of tetrahedron-square pyramid-tetrahedron interfacing each mibdi to the trigonal prism.

Now let's take a look at the side view, which is quite interesting:

Image

Here, I turned off visibility clipping so that you can see the entire structure of the polychoron. The two octahedral cells are highlighted in green. They happen to project concentrically from this particular 4D viewpoint. The square pyramids aren't very obvious, but you can find them if you first look for the triangular prism (foreshortened to be quite squished) at the bottom of the projection.

Here's another side view, showing how two mibdies are folded together with a square pyramid sandwiched between them:

Image

I left the octahedra colored, even though they are on the far side of this projection, so that it's easier to orient yourself w.r.t. the polychoron in this 4D viewpoint.

Beautiful little thing, isn't it??! :nod:

And here are the coordinates:
Code: Select all
# x2o3o
<±1, 0, 0, -phi^2/√3>

# f2x3o
<±phi, ±1, -1/√3, 0>
<±phi, 0, 2/√3, 0>
 
# o2f3o
<0, ±phi, -phi/√3, phi/√3>
<0, 0, 2*phi/√3, phi/√3>

# x2o3x
<±1, ±1, 1/√3, phi^2/√3>
<±1, 0, -2/√3, phi^2/√3>
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Re: A different way to expand an ursatope?

Postby quickfur » Mon Mar 14, 2016 7:04 pm

The beautiful simplicity of this construction makes me wonder if a 5D equivalent might also exist, if we take 3 copies of this J62-wedge and fit them around an edge, then fill in the bottom with some other cells? Maybe a possible lace tower might look something like this?
Code: Select all
x2o3o3o
f2x3o3o
o2f3o3o
x2o3x3x

Not 100% sure about the exact symbols (my 5D visualization is rather weak, i.e., non-existent :P). Would something like this be possible? If this lace tower is CRF-able, can there be higher-dimensional analogues? Is there an infinite family of these wedge-like ursatopes? (No idea what to call them... ursawedges?)
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Re: A different way to expand an ursatope?

Postby Keiji » Mon Mar 14, 2016 8:19 pm

quickfur wrote:Good news!! While the case of P=4 doesn't seem to close up in a CRF way, I have just constructed the P=3 case (with a slight modification) and verified that it's CRF!!!!!!


This is fascinating! It has to be one of my favorite CRFs so far.

I can't tell what relation it has to the ursatopes though. Is it an analog of anything, for that matter? Or is it a new crown jewel?

Would it be a good idea to allocate D4.15 for the magnaursachoron (in case we want to rename it) and D4.16 for this one?
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Re: A different way to expand an ursatope?

Postby student91 » Mon Mar 14, 2016 8:40 pm

Really cool things you've found there!
I was just looking into the following:
quickfur wrote:In any case, looking at the cell configuration, it seems that it may not be possible to construct the analogue with octahedral symmetry after all -- because you would end up with 4 tetrids surrounding a cube, which would make a gap containing 4 pentagons surrounding a square. However, there is no Johnson solid I know of that has 4 pentagons surrounding a square. According to analogy, it would seem that we'd need 3D square teddies in order to fill in the gaps, but square teddies are non-CRF because they have √2-edges. Similarly, it doesn't seem possible to extend this construction to icosahedral symmetry either -- the corresponding gaps would require pentagonal teddies (bisected dodecahedra), and again they are non-CRF because of the phi-edges. :(

I think the icosahedral case is possible, because the "pentagonal teddies" can be completed to a dodecahedron.
So far I've got the following tower:
Code: Select all
x3o5x||x3o5f||x3f5o||F3x5o||V3x5o||F3f5o||(x3F5o+V3o5f)||f3x5f||o3x5F||(???)||(V+x)3o5x
.
I'm not sure what should be inserted into the (???), all other is just the tedrids and the completed dodecahedra.
Last edited by student91 on Tue Mar 15, 2016 11:45 pm, edited 3 times in total.
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Re: A different way to expand an ursatope?

Postby Klitzing » Mon Mar 14, 2016 9:00 pm

quickfur wrote:
Klitzing wrote:[...]
Thus, what then would be xfox-2-oxfo-P-oooo-&#xt (P=3,4,5)? - I will have to take a look into these...

--- rk

Good news!! While the case of P=4 doesn't seem to close up in a CRF way, I have just constructed the P=3 case (with a slight modification) and verified that it's CRF!!!!!! The slight adjustment is that the last vertex layer must be x2o3x, not x2o3o, otherwise the mibdies are non-corealmar. So the corrected symbol is xfox-2-oxfo-3-ooox&#xt (note the last x in place of o).

This is a truly beautiful little CRF; it has 3 mibdies and 2 tetrahedra on one side, 1 trigonal prism, 2 octahedra(!), 3 square pyramids, and 6 tetrahedra on the other side. Even better yet, it has 17 vertices and 17 cells -- the same number of vertices as cells!

...

Beautiful little thing, isn't it??! :nod:


Indeed it is! - And the pixies, as usual, too!

--- rk
Last edited by Klitzing on Mon Mar 14, 2016 9:10 pm, edited 1 time in total.
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Re: A different way to expand an ursatope?

Postby Klitzing » Mon Mar 14, 2016 9:10 pm

student91 wrote:I think the icosahedral case is possible, because the "pentagonal teddies" can be completed to a dodecahedron.

That was my feeling too ...

?mibdies?

tedrid = tridiminished rhombicosidodecahedron = J83
mibdi = metabidiminished icosahedron = J62
(some of the OBSAs, the "official Bowers style acronyms")

--- rk
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Re: A different way to expand an ursatope?

Postby student91 » Mon Mar 14, 2016 9:37 pm

Klitzing wrote:
student91 wrote:I think the icosahedral case is possible, because the "pentagonal teddies" can be completed to a dodecahedron.

That was my feeling too ...

?mibdies?

tedrid = tridiminished rhombicosidodecahedron = J83
mibdi = metabidiminished icosahedron = J62
(some of the OBSAs, the "official Bowers style acronyms")

--- rk

nvm, I just found out that the tower I found was part of a tridodecadiminished x5o3o3x
Last edited by student91 on Mon Mar 14, 2016 11:44 pm, edited 1 time in total.
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Re: A different way to expand an ursatope?

Postby quickfur » Mon Mar 14, 2016 9:50 pm

Keiji wrote:
quickfur wrote:Good news!! While the case of P=4 doesn't seem to close up in a CRF way, I have just constructed the P=3 case (with a slight modification) and verified that it's CRF!!!!!!


This is fascinating! It has to be one of my favorite CRFs so far.

I can't tell what relation it has to the ursatopes though. Is it an analog of anything, for that matter? Or is it a new crown jewel?

It's not directly related to the ursatopes; it's more related to J62 (metabidiminished icosahedron). J62 can be thought of as two pentagons folded around an edge, with triangles inserted between them at either end of the edge, and other faces added to close up the shape. This CRF is Klitzing's generalization of J62 to 4D: fold 3 J62's around an edge, and insert tetrahedra between them, and other cells inserted to close up the shape.

Having said that, though, one could draw a potential link to the ursatopes by considering a more general scheme by which various Johnson solids may be derived. The idea is to start with some base subdimensional element, and then insert pentagons (resp. higher-dimensional pentagon analogues) around it. Say for example we start with a triangle. Then we attach 3 pentagons to it. Now, we can either proceed via the "narrow" path, that is, fold the pentagons into 3D without any intervening faces, which leads to teddi. Higher-dimensional analogues of this "narrow closure" process produce the ursatopes.

But there's another possibility, which I call the "fat" path: fold the pentagons into 3D but with intervening triangles between them. In 3D, if you start with a triangle and fold it up with intervening triangles, you get the top part of J92 (triangular hebesphenorotunda), which can be completed by adding a hexagon below the tips of the pentagons. Interestingly enough, if we started with an edge rather than a triangle, we'd have 2 pentagons sharing an edge, with triangles inserted between them, i.e., we get J62, metabidiminished icosahedron. If we start from an edge and take the narrow path, we get a pentagon (subdimensional case). So we could think of J62 as a kind of "fat pentagon" or "space-filling pentagon".

In 4D, if we fold 3 J62's around an edge and insert intervening tetrahedra, we get this new CRF. We could also fold 4 J92's around a tetrahedron (with J92's serving as a kind of "very fat pentagon"), which leads to D4.8.1 and the D4.8.x family of CRFs.

All of these CRFs have in common the process of starting with some element X, then building the next layer of vertices with X scaled by the golden ratio in some way (either a plain scaling for the "narrow" path, which corresponds with marking one of the nodes of its CD diagram with f, or a scale+expand for the "fat" path, i.e., mark a node with f and another node with x). These two layers of vertices with the golden ratio scaling produces partial pentagons, which can then be completed into full pentagons, thereby outlining the vertices of the 3rd layer. Then additional vertices can be added to close the shape up in a CRF way.

So, arguably, the ursatopes are merely one of the branches of this larger family, and this new CRF of Klitzing's is just one of the "cousins" of the ursatopes. :P

This general scheme also suggests many other CRF possibilities that so far we haven't fully explored yet, e.g., what about if we start with a tetrahedron, then insert 4 teddies, but with J62's as intervening "fat tetrahedra"? You can substitute all sorts of shapes as the analogues of the "pentagon" and the "intervening triangle", and I'd say that a good number of combinations will be CRF-able.

(Note that this scheme extends beyond CRFs too: in 3D, for example, if you start with a square, you get a non-CRF square teddy, but this square teddy is exactly the shape needed to close up the octahedral magna-ursachoron. You can also derive square analogues of J92 by choosing the "fat" path, starting with a square, which is non-CRF, but if you admit it as a cell, you can construct all sorts of (non-CRF) 4D shapes that fall under the same scheme.)

Would it be a good idea to allocate D4.15 for the magnaursachoron (in case we want to rename it) and D4.16 for this one?

Sure!
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Re: A different way to expand an ursatope?

Postby Keiji » Mon Mar 14, 2016 10:12 pm

So, in summary, what we're looking at is:

Petals around Base, Path

Pentagons around Digon, Narrow = (Degenerate)
Pentagons around Trigon, Narrow = J63

Pentagons around Digon, Fat = J62
Pentagons around Trigon, Fat = J92

J62s around Digon, Fat = D4.16
J92s around Tetrahedron, Fat = D4.8.x

Where does D4.15 fall? Would it be J83s around Tetrahedron, Fat?
Also, what is J83 itself in this scheme?

How many of these combinations are possible?
So far, we've got petals J62s, J83s and J92s (and J63s?) around base digon and tetrahedron (and going by student91's post, icosahedron?) in the fat version. That's 6 possibilities, or 12 if both the maybes are valid.
What about the narrow path, are any of those valid, or are they all degenerate or non-CRF?
Can other figures be used as the petals, J91 perhaps?
How does this generalise to 5D - can D4.15, D4.16 and D4.8.x be used as petals around a digon, a pyrochoron, a hydrochoron (presumably xylochoron would be non-CRF due to its octahedral cells)?
Can the trigon be used as the base for a 4D or 5D figure; can the tetrahedron and icosahedron be used as the base for a 5D figure?

Shooting questions at you rapid-fire I know, but hoping to provide some inspiration :D

Oh, and since I'm referring to them as petals, how about we call this larger family floratopes? :D

Happy CRMarch :] Oh, and Pi Day! It's that too!
I really should stop editing this post now...
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Re: A different way to expand an ursatope?

Postby quickfur » Mon Mar 14, 2016 10:36 pm

Will answer the rest of your questions later, but just wanted to say that putting teddies around a tetrahedron with J91's as the intervening "wedges" leads to D4.11. :D Furthermore, the "fat" path isn't necessarily unique; while the kind of wedges that can fit between petals is obviously constrained by the shape of the petals, there's still some leeway for various substitutions, as in the case of D4.11.

In the original construction that eventually led to D4.11, I only considered J62's as wedges. But, as you may already know, it's quite possible to chop up two J62's and glue them together end-to-end, and the convex hull of the result is a J91 (bilbiro). :P I decided to try out J91's just for kicks, and did produce an intermediate CRF closed off by some other configuration of cells, but then when rendering the side-views of the result, I noticed that the angle between the bilbiros appeared to be coincide with the 16-cell. That was what sparked the idea that perhaps I could close the thing up in demi-tesseractic symmetry; and the rest is history. :lol:

As you can see, this is a very fertile area for sprouting CRFs by the dozens, it's just a matter of imagination (and time)!
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Re: A different way to expand an ursatope?

Postby quickfur » Mon Mar 14, 2016 10:47 pm

I'm pretty sure the 5D 24-cell teddy exists, and is CRF. Remember that the octahedral teddy is CRF in 4D (the 3D analogue, the square teddy, is non-CRF). So you can put 24 octahedral teddies around a 24-cell and have it close up in 5D. :nod: And AFAIK, the 600-cell teddy should also exist (unless some ugly coincidence makes it co-tetrixal or hyperbolic). I'm not sure if the corresponding fat versions will exist, though. I'm pretty sure the 600-cell fat teddy wouldn't exist because the external angles of the 600-cell are too small to fit that many things in, so it can only be done in hyperbolic space. But the 24-cell fat teddy might exist. The wedges between the octahedral teddies would likely be Klitzing's D4.16, since that's the most direct 4D analogue of J62 that I know of so far.
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Re: A different way to expand an ursatope?

Postby Keiji » Tue Mar 15, 2016 12:45 am

I've finally worked out the incidence matrix for the D4.16, it's on its wiki page. That was some hard work but satisfying as always. Now it's time for bed :)
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Re: A different way to expand an ursatope?

Postby Klitzing » Tue Mar 15, 2016 12:55 pm

Dear Keiji, not that it wouldn't be true, the superdiagonal part of the incidence matrix indeed is derivable from the supradiagonal one together with the diagonal elements. But, like the supradiagonal ones provide an easy access to the respective subelement counts of the elements of the considered polytope, the superdiagonal ones too provide valuable informations. E.g. the superdiagonal parts of the rows of the vertex block provide the subelement counts of the respective vertex figures. Those of the edge block provide the subelement counts of the respective edge figures etc. Thus, when an incidence matrix is provided in full matrix form, then those informations can be read off very quickly. Whereas, when you only provide the supradiagonal part (and provide the diagonal elements as a further separate row below), then this kind of Information is only contained implicite. It can not be read off directly.

Further, the dynkin symbols have lately been extended widely, and also are available in a typewriter friendly notation. Thus the respective subelements of the rows of an incidence matrix most often can be directly related to their dynkin subsymbols. This additional Information is much more informative than just a numerical or alphabetical distinction of the equivalence classes of symmetry. Moreover they generally allow for a subpression of most of the wordy explanations of respective interrelations or positionings. This is why I generally add those as an extra column too. (For more ease these dynkin symbols are surely allowed to be complemented with their readable names too, e.g. as shown below.)

E.g. the just provided matrix (in the wiki) then could "better" read as follows:
Code: Select all
xfox oxfo3ooox&#xt   → height(1,2) = sqrt[(7+3*sqrt(5))/24] = 0.755761
                       height(2,3) = sqrt[(3+sqrt(5))/24] = 0.467086
                       height(3,4) = sqrt(3)/6 = 1/sqrt(12) = 0.288675

o... o...3o...     | 2 * * * | 1 3 0 0  0  0 0 0 | 3 3 0  0 0 0 0 0 0 0 | 3 1 0 0 0 0
.o.. .o..3.o..     | * 6 * * | 0 1 2 1  2  0 0 0 | 1 2 1  2 2 1 0 0 0 0 | 2 1 1 1 0 0
..o. ..o.3..o.     | * * 3 * | 0 0 0 2  0  4 0 0 | 1 0 0  4 0 0 2 2 0 0 | 2 0 0 2 1 0
...o ...o3...o     | * * * 6 | 0 0 0 0  2  2 1 2 | 0 0 0  2 1 2 2 2 2 1 | 1 0 1 2 2 1
-------------------+---------+-------------------+----------------------+------------
x... .... ....     | 2 0 0 0 | 1 * * *  *  * * * | 3 0 0  0 0 0 0 0 0 0 | 3 0 0 0 0 0
oo.. oo..3oo..&#x  | 1 1 0 0 | * 6 * *  *  * * * | 1 2 0  0 0 0 0 0 0 0 | 2 1 0 0 0 0
.... .x.. ....     | 0 2 0 0 | * * 6 *  *  * * * | 0 1 1  0 1 0 0 0 0 0 | 1 1 1 0 0 0
.oo. .oo.3.oo.&#x  | 0 1 1 0 | * * * 6  *  * * * | 1 0 0  2 0 0 0 0 0 0 | 2 0 0 1 0 0
.o.o .o.o3.o.o&#x  | 0 1 0 1 | * * * * 12  * * * | 0 0 0  1 1 1 0 0 0 0 | 1 0 1 1 0 0
..oo ..oo3..oo&#x  | 0 0 1 1 | * * * *  * 12 * * | 0 0 0  1 0 0 1 1 0 0 | 1 0 0 1 1 0
...x .... ....     | 0 0 0 2 | * * * *  *  * 3 * | 0 0 0  0 0 0 2 0 2 0 | 1 0 0 0 2 1
.... .... ...x     | 0 0 0 2 | * * * *  *  * * 6 | 0 0 0  0 0 1 0 1 1 1 | 0 0 1 1 1 1
-------------------+---------+-------------------+----------------------+------------
xfo. .... ....&#xt | 2 2 1 0 | 1 2 0 2  0  0 0 0 | 3 * *  * * * * * * * | 2 0 0 0 0 0
.... ox.. ....&#x  | 1 2 0 0 | 0 2 1 0  0  0 0 0 | * 6 *  * * * * * * * | 1 1 0 0 0 0
.... .x..3.o..     | 0 3 0 0 | 0 0 3 0  0  0 0 0 | * * 2  * * * * * * * | 0 1 1 0 0 0
.ooo .ooo3.ooo&#x  | 0 1 1 1 | 0 0 0 1  1  1 0 0 | * * * 12 * * * * * * | 1 0 0 1 0 0
.... .x.o ....&#x  | 0 2 0 1 | 0 0 1 0  2  0 0 0 | * * *  * 6 * * * * * | 1 0 1 0 0 0
.... .... .o.x&#x  | 0 1 0 2 | 0 0 0 0  2  0 0 1 | * * *  * * 6 * * * * | 0 0 1 1 0 0
..ox .... ....&#x  | 0 0 1 2 | 0 0 0 0  0  2 1 0 | * * *  * * * 6 * * * | 1 0 0 0 1 0
.... .... ..ox&#x  | 0 0 1 2 | 0 0 0 0  0  2 0 1 | * * *  * * * * 6 * * | 0 0 0 1 1 0
...x .... ...x     | 0 0 0 4 | 0 0 0 0  0  0 2 2 | * * *  * * * * * 3 * | 0 0 0 0 1 1
.... ...o3...x     | 0 0 0 3 | 0 0 0 0  0  0 0 3 | * * *  * * * * * * 2 | 0 0 1 0 0 1
-------------------+---------+-------------------+----------------------+------------
xfox oxfo ....&#xt | 2 4 2 2 | 1 4 2 4  4  4 1 0 | 2 2 0  4 2 0 2 0 0 0 | 3 * * * * *  mibdi
.... ox..3oo..&#x  | 1 3 0 0 | 0 3 3 0  0  0 0 0 | 0 3 1  0 0 0 0 0 0 0 | * 2 * * * *  tet
.... .x.o3.o.x&#x  | 0 3 0 3 | 0 0 3 0  6  0 0 3 | 0 0 1  0 3 3 0 0 0 1 | * * 2 * * *  oct
.... .... .oox&#x  | 0 1 1 2 | 0 0 0 1  2  2 0 1 | 0 0 0  2 0 1 0 1 0 0 | * * * 6 * *  tet
..ox .... ..ox&#x  | 0 0 1 4 | 0 0 0 0  0  4 2 2 | 0 0 0  0 0 0 2 2 1 0 | * * * * 3 *  squippy
...x ...o3...x     | 0 0 0 6 | 0 0 0 0  0  0 3 6 | 0 0 0  0 0 0 0 0 3 2 | * * * * * 1  trip

--- rk
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Re: A different way to expand an ursatope?

Postby quickfur » Tue Mar 15, 2016 7:22 pm

Klitzing wrote:[...]
Got a promising idea! What then about teddi || ... || tedrid?
[...]
Code: Select all
               x3o       f3o   o3x                  teddi
                                                
       o3x       F3o             f3x   x3o       
                                                       F=ff=x+f=2x+v,
                                                       V=F+v=2f=2x+2v
x3o x3f   F3x       V3x       F3f  Vx3oF  f3x o3x   tedrid

[...]
--- rk

Finally got around to this today. Very nice!!!! It has a beautiful arrangement of pentagonal rotundae connecting the teddi to the tetrid, and a fascinating cluster of octahedra around the bottom.

Here's a projection centered on the top teddi, with other teddies highlighted in yellow and midbies in magenta:

Image

The red edges outline the tetrid at the bottom of the polychoron.

Here are the pentagonal rotundae wrapping around the teddi:

Image

Each pair of rotundae shares a triangular face.

Now let's look at the octahedra:

Image

There are 10 octahedra here, with the ones at the bottom forming a partial network of face-sharing octahedra filling in the gaps between the teddies and mibdies. They are complemented by the square pyramids:

Image

It's not so easy to tell from these images, but if you consider the arrangement of octahedra and square pyramids together, you'll realize that they are actually a fragment of the octahedra on the surface of the rectified 600-cell o5o3x3o, and the teddies and mibdies are various truncations of the o5o3x3o's icosahedra. In other words, this CRF is a diminishing of the o5o3x3o.

The interesting thing about this CRF is that it's a diminishing of a bistratic stacking of two segmentochora (o5o3x || o5x3o || x5o3x), but it is itself not decomposable into two CRF segmentochora.

Just for fun, here's a side view of this CRF, showing its shape as a truncated cap:

Image

You can vaguely see the outline of a mibdi teddi filling up the wedge-like gap between the two rotundae.

P.S., here are the full coordinates, for posterity: :P
Code: Select all
#
# J62
#

# x3o
<±1, -1/√3, -phi^2/√3, -1/phi>
<0, 2/√3, -phi^2/√3, -1/phi>

# f3o
<±phi, -phi/√3, 1/(phi*√3), -1/phi>
<0, 2*phi/√3, 1/(phi*√3), -1/phi>

# o3x
<±1, 1/√3, phi^2/√3, -1/phi>
<0, -2/√3, phi^2/√3, -1/phi>


#
# Tridiminished o5x3o
#

# o3x:
<0, -2/√3, -2*phi^2/√3, 0>
<±1, 1/√3, -2*phi^2/√3, 0>

# (x3f deleted)

# F3o:
<±phi^2, -phi^2/√3, -2/√3, 0>
<0, 2*phi^2/√3, -2/√3, 0>

# (f3f deleted)

# (o3F deleted)

# f3x:
<±1, phi^3/√3, 2*phi/√3, 0>
<±phi^2, -1/(phi*√3), 2*phi/√3, 0>
<±phi, -(phi+2)/√3, 2*phi/√3, 0>

# x3o:
<0, 2/√3, 2*phi^2/√3, 0>
<±1, -1/√3, 2*phi^2/√3, 0>


#
# J83
#

# x3o
<±1, -1/√3, -(4*phi+1)/√3, 1>
<0, 2/√3, -(4*phi+1)/√3, 1>

# x3f
<±phi^2, 1/(phi*√3), -(2*phi+3)/√3, 1>
<±phi, (phi+2)/√3, -(2*phi+3)/√3, 1>
<±1, -phi^3/√3, -(2*phi+3)/√3, 1>

# F3x
<±1, (2*phi+3)/√3, -(2*phi+1)/√3, 1>
<±(phi+2), -phi/√3, -(2*phi+1)/√3, 1>
<±phi^2, -(phi+3)/√3, -(2*phi+1)/√3, 1>

# V3x
<±(2*phi+1), -(2*phi-1)/√3, -1/√3, 1>
<±1, (4*phi+1)/√3, -1/√3, 1>
<±2*phi, -2*phi^2/√3, -1/√3, 1>

# F3f
<±phi^3, -1/√3, (2*phi-1)/√3, 1>
<±phi, (3*phi+2)/√3, (2*phi-1)/√3, 1>
<±phi^2, -(3*phi+1)/√3, (2*phi-1)/√3, 1>

# x3F
<±1, -(2*phi+3)/√3, phi^3/√3, 1>
<±(phi+2), phi/√3, phi^3/√3, 1>
<±phi^2, (phi+3)/√3, phi^3/√3, 1>

# V3o
<±2*phi, -2*phi/√3, phi^3/√3, 1>
<0, 4*phi/√3, phi^3/√3, 1>

# f3x
<±phi^2, -1/(phi*√3), (2*phi+3)/√3, 1>
<±phi, -(phi+2)/√3, (2*phi+3)/√3, 1>
<±1, phi^3/√3, (2*phi+3)/√3, 1>

# o3x
<±1, 1/√3, (4*phi+1)/√3, 1>
<0, -2/√3, (4*phi+1)/√3, 1>

The distance between the teddi layer and the tridiminished o5x3o is 1/phi, and the distance between the tridiminished o5x3o and the tetrid is exactly 1. (This is for edge length 2; for edge length 1 obviously the values would be halved.)
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Re: A different way to expand an ursatope?

Postby Keiji » Tue Mar 15, 2016 7:30 pm

Klitzing wrote:Dear Keiji, not that it wouldn't be true, the superdiagonal part of the incidence matrix indeed is derivable from the supradiagonal one together with the diagonal elements. But, like the supradiagonal ones provide an easy access to the respective subelement counts of the elements of the considered polytope, the superdiagonal ones too provide valuable informations. E.g. the superdiagonal parts of the rows of the vertex block provide the subelement counts of the respective vertex figures. Those of the edge block provide the subelement counts of the respective edge figures etc. Thus, when an incidence matrix is provided in full matrix form, then those informations can be read off very quickly. Whereas, when you only provide the supradiagonal part (and provide the diagonal elements as a further separate row below), then this kind of Information is only contained implicite. It can not be read off directly.


Displaying the imat in the way I do was a conscious and deliberate decision. :) I personally find it easier to read off the element counts from the bottom row where as you find it easier to read them off from the diagonal. I also find that if the asterisks and the top-right half of the matrix are included it makes it more confusing. Perhaps it's because I'm more of a programmer/engineer than a mathematician. You're free to display it in the more conventional format on your website of course, and maybe I'll make an optional setting to display it like this but for now it will stay as it is.

Klitzing wrote:Further, the dynkin symbols have lately been extended widely, and also are available in a typewriter friendly notation. Thus the respective subelements of the rows of an incidence matrix most often can be directly related to their dynkin subsymbols. This additional Information is much more informative than just a numerical or alphabetical distinction of the equivalence classes of symmetry. Moreover they generally allow for a subpression of most of the wordy explanations of respective interrelations or positionings. This is why I generally add those as an extra column too. (For more ease these dynkin symbols are surely allowed to be complemented with their readable names too, e.g. as shown below.)


This I would like to do, however I only have a basic understanding of these symbols and so I'm not confident I will get them right. If there's a way to automate it (programmatically determine the correct symbol for each row based on the imat and the symbol for the whole figure) then I'd like to do this. I can see the general pattern in your matrix, but I'm not going to write these myself unless I can fully understand what's going on.
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Re: A different way to expand an ursatope?

Postby quickfur » Tue Mar 15, 2016 9:14 pm

I'd highly recommend getting familiar with the Dynkin symbols, since they are extremely useful not only in notating shapes precisely, they also help a great deal in computations of coordinates. They are especially handy in dealing with CRFs, because the regular polygons in CRFs have well-known symbols and can be manipulated easily that way.

Take for example the octahedral teddy, which can be expressed as the symbol ooo4oox3xfo&#xt. This is just a compact notation for the following lace tower:
Code: Select all
o4o3x
o4o3f
o4x3o

The "ooo" in "ooo4oox3xfo..." appears as the first column of the tower; the "oox" appears as the second column, and the "xfo" appears as the 3rd column. So basically the linear symbol is just a compression of this lace tower (since all elements are of the form .4.3. so we just write the 4 and 3 once, and transpose the columns into rows between them. The "&#xt" just means "the vertex layers are to be laced with unit edges". That is, the exact distance between, say, o4o3x and o4xo3f is not specified, but can be computed using basic trigonometry, since we know what the final edge length (i.e. the hypotenuse) must be.

Now let's look at each individual layer of this tower. While the "standard" interpretation of the Dynkin symbol involves confusing (to me!) explanations of mirrors and angles and what-not, I personally find it easiest to think of it as a coordinate system (IIRC this was Wendy's idea) along the edges of a regular polytope.

Let's take o4o3x to start. Whenever you see 'o' in the symbol, it means the corresponding coordinate in the coordinate system is zero; whenever you see 'x' the corresponding coordinate is 1. But what are these coordinates referring to? I find it most helpful to think in terms of folding polygons (resp. polytopes) around a ridge (i.e., (n-2)-dimensional element). The beautiful thing about Dynkin symbols is that substrings of the symbol exactly describe sub-dimensional elements of the polytope. So in o4o3x, for example, the substring o3x describes a triangle. Why? Because the 3 means that 'x', which is a unit edge (remember, 'x' means '1'), is folded around a 3-fold axis. What does that make? A triangle, of course! The 'o' means that the 'x' is folded around the 3-fold axis without any intervening edges. If you were to write x3x, for example, that would mean that it's a hexagon, because you're folding 'x' around a 3-fold axis, with the other 'x' indicating a unit displacement around the joint. So the 'x' describes every other edge of the hexagon, and the other 'x' describes the other 3 edges. Notice that the symbol x3x is palindromic, which means you can read it backwards and it still means the same thing, which makes sense 'cos it doesn't matter which set of 3 edges you start with the make the hexagon. You could start with folding edges 1, 3, 5, say, with a unit distance between them (i.e., interpret the first 'x' as the starting shape and the other 'x' as displacement), or fold edges 2, 4, 6 with unit distance between them (i.e., interpret the second 'x' as the starting shape and the first 'x' as displacement), and the result would be the same. The Dynkin symbol neatly captures this equivalence by being equivalent to its mirror image.

OK, so now we've determined that o3x is a triangle. What about x3o? Remember that 'o' means 'zero', so it's just a single point. We're folding a single point around a 3-fold axis (as indicated by the '3'), with a displacement of 1 (as indicated by the 'x'). So again, we have a triangle. You could equivalently interpret this by reading from the other direction, i.e., 'x' means an edge, '3' means fold it around a 3-fold axis, and o means zero displacement. Either interpretation gives you a triangle. However, there's a difference between o3x and x3o: remember that this is a coordinate system, so while (1,0) in the Cartesian coordinate system has the same length as (0,1), it's pointing in a different direction. Similarly, x3o describes a triangle that's pointing in a different direction from o3x. Which direction? Exactly the direction perpendicular to the "displacements" we were dealing with in the x3x example. Or, more generally speaking, in the dual orientation. So o3x is a triangle, and x3o is the dual triangle. The fact that the Dynkin symbol is not palindromic indicates a distinction between the two. In the case of x3x, we're talking about the hexagon inverted along a 3-fold (not 6-fold) axis; so in the triangular coordinate system .3., the hexagon is in some sense "self-dual". Or, to be more precise, orienting the hexagon in the orientation of the dual triangle does not change it.

OK, enough of boring old 2D shapes. Now let's go back to the original symbol o4o3x. There are various ways of reading this, but let's build upon what we've learned above, that o3x describes a triangle. So we can read this symbol as o4(o3x), that is, fold triangles around a 4-fold axis. What do we get? We get a square pyramid. But it doesn't end there; the Dynkin symbol is transitive across symmetries, so we're talking about something that contains triangles meeting 4 to a vertex. That is, of course, an octahedron. So that's how o4o3x describes an octahedron. Now, we could equivalently read the symbol as (o4o)3x, that is o4o describes points folded around a 4-fold axis (it's still just a point -- it's a kind of degenerate square, you can think of it), and then folding these degenerate squares (i.e. points) in 3D around 3-fold axes. Remember that .4. is a coordinate system for squares, so o4o can be thought of as a point in the middle of a square. So folding o4o around a 3-fold axis is equivalent to folding squares around a 3-fold axis, that is, you get a cube, and then placing points in the middle of each square face. I.e., we get an octahedron, as expected.

(This, of course, suggests that if ithe squares weren't degenerate, we'd actually get a cube, which is precisely what happens when we write x4o3o. That is, x4o describes a square (fold 4 edges around a 4-fold axis), and x4o3o = (x4o)3o means fold squares around a 3-fold axis, so you get a cube.)

OK, so o4o3x means an octahedron. But what does o4o3f mean??! Well, f is shorthand for the Golden Ratio phi=(1+√5)/2. It means we start not with a unit edge, but with an edge of length phi, and fold it along a 3-fold axis, producing a triangle scaled by phi, and then fold that around a 4-fold axis, that is, an octahedron scaled by phi. So we see that the second layer of vertices in the octahedral teddy is an octahedron scaled by phi.

If we were to connect these two layers together with unit edges, we'd get an octahedron of unit edge length on top, an octahedron of edge length phi on the bottom, and in the middle, the lacing edges, also of unit length, would trace out trapeziums with 3 unit edges and 1 phi edge. If you think about this carefully, you'd realize that these trapeziums are nothing other than incomplete pentagons. So the first two layers of our tower o4o3x || o4o3f describes a partial shape that has incomplete pentagons. These pentagons are completed by the final layer, o4x3o.

What is o4x3o? Note that, while in 2D it didn't matter which way you read the symbols, in 3D and above it does matter which side of the symbol is connected to the rest of it, that is, o4x3o is not the same as o4o3x. This is because o3x describes a triangle in "normal" orientation, such that folding it around a 4-fold axis would eventually produce an octahedron. x3o, however, describes triangles in the same positions as the o3x triangles, but in dual orientation. So imagine if you were to take an octahedron and spin its faces in-place so that they are now in dual orientation to the original triangles. What you get are 8 triangles that are in the same orientation as a cuboctahedron. And in fact, that's what o4x3o describes: a cuboctahedron. Equivalently, you can read it is (o4x)3o, i.e., o4x is a square in dual orientation to the squares in a cube (a cube is x4o3o, remember), so if you fold dual squares around 3-fold axes, you get a cuboctahedron.

There's another way to look at the Dynkin symbol, and that is by taking a regular polytope of the same symmetry as the basis for interpreting it. For example, let's take the cube as the "standard" shape for symbols of the form a4b3c. Then the first element a refers to the vertices of the cube, the second element b refers to the edges of the cube, and the last element c refers to the faces of the cube. By "refer" I mean draw vectors from the origin (assuming the cube is origin-centered) to the indicated element. For the first element, you have 8 vectors (there are 8 vertices in the cube), that is, <±1,±1,±1>, or apacs<1,1,1>. The second element corresponds to the vectors apacs<0, √2, √2> (why √2 will be explained in a bit), and the 3rd element to the vectors apacs<0, 0, √2>. Given a symbol a4b3c, the resulting polyhedron is the convex hull of a*apacs<1,1,1> + b*apacs<0, √2, √2> + c*apacs<0, 0, √2>. So when you write x4o3o, that means a=1, b=0, c=0, so you get a cube. When you write o4x3o, that means a=0, b=1, c=0, so you get a cuboctahedron (of the same edge length as x4o3o, and that is why we use <0, √2, √2> rather than <0, 1, 1>). If you write x4x3o, for example, your polyhedron would be apacs<1,1,1> + apacs<0, √2, √2>, which is equivalently apacs<1, 1+√2, 1+√2>, which, if you graph it out, is a truncated cube (of the right edge lengths!!). Similarly, o4x3x describes a truncated octahedron, x4o3x describes a (small) rhombicuboctahedron, etc.. So once you know the basis vectors for a particular edge labelling of the Dynkin diagram, the Dynkin symbol basically tells you how to compute its coordinates, in addition to being an intuitive description of how to obtain the shape by folding lower-dimensional elements into higher dimensions(!).

Now coming back to our tower...

The bottom symbol is o4x3o. Note that the orientation of the triangles is in dual orientation to the top symbol o4o3x. So in terms of the partial pentagons produced by the first two layers, the vertices of o4x3o are precisely in the right place to complete the pentagons and close up the shape, as they would if we were to lace the last layer to the 2nd by unit edges, as the suffix &#x indicates.

This isn't all there is to the lace tower, though. Remember what I said about substrings of the Dynkin symbol? That principle can also be applied to lace towers! For example, we can read off the 1st two nodes of our tower, and we'd get o4o || o4o || o4x. That is, a point, another point, then a square. Well, point || point is obviously an edge, so that means that in the position of symmetry described by the first 2 nodes of the symbol, there are edges, not full-dimensioned elements. Then we have point || square, which is obviously a square pyramid. This tells us that there are square pyramids in this position of symmetry. But what is this position of symmetry? Remember the symbol for the cube: x4o3o. The first two positions is (x4o), i.e., where the faces of the cube lie. So that tells us that the square pyramids are found where the faces of a cube, oriented in the same way as the octahedral teddy, would lie. Therefore, there are exactly 6 square pyramids. (See? We can know this just by reading the Dynkin symbol, without even "visualizing" the polychoron!).

Similarly, if we read the last 2 symbols of each layer, we get o3x || o3f || x3o. That is, triangle || phi-scaled triangle || dual triangle. This describes a teddi. And since these last 2 symbols correspond with the faces of an octahedron, this tells us that there are exactly 8 teddies in this polychoron. And so, we learn that the polychoron has 6 square pyramids, 8 teddies, 1 octahedron, and 1 cuboctahedron. :D

There's still more to this, of course. E.g., you can read the first and last symbols (deleting the middle node, which is equivalent to writing '2', and that lets you read off prism symmetries, i.e., elements that correspond to the edges of the cube). For example, if we delete the middle column, we get o2x || o2f || o2o. The 2 here is actually an abuse of notation, because in the "real" Dynkin diagram there's actually no edge between the two nodes, but it serves as a nice mnemonic for "put two copies of the base shape on parallel (hyper)planes to make a prism". Here, o2x is a unit edge (i.e., a point prism), o2f is a phi-scaled edge, and o2o is a point. This, in fact, describes a pentagon (the o2f is just a short-chord of the pentagon, you see). Since the first node is consistently o, the "prism" here is degenerate; it just flattens into a polygon (the height of the prism is 0). If we had x2x || x2f || x2o instead, we'd have a pentagonal prism, which can be interpreted as square (x2x) || 1*phi rectangle (x2f) || line (x2o).

Just for kicks, let's read off the element types of a 4D uniform polychoron, say o5x3o3x. The first 3 nodes are o5x3o, so that means we have icosidodecahedra (o5(x3o) == dual triangles folded around 5-fold axes (thus attached by their vertices), or equivalently, (o5x)3o == dual pentagons folded around 3-fold axes, attached by their vertices) in the position corresponding to the cells of the 120-cell, so 120 icosidodecahedra. The last 3 nodes are x3o3x, i.e., cuboctahedron (as rhombi-tetrahedron: (x3o)3x = 4 dual triangles folded around 3-fold axes a unit displacement apart, this causes the original tetrahedron edges to expand into squares and 4 more triangles from the dual tetrahedron to appear), in the position corresponding to the cells of the 600-cell, so there are 600 cuboctahedra. Deleting the 2nd node gives us o2o3x, i.e., triangles (triangular prisms of height 0), and deleting the 3rd node gives o5x2x, i.e., pentagonal prisms. So we see that o5x3o3x consists of icosidodecahedra, cuboctahedra, and pentagonal prisms (and triangles, but they are subdimensional).

OK this post has become far longer than I anticipated. I'll shut up now. :oops:

Edit: Corrected some silly errors in the o5x3o3x example.

Edit 2: Corrected yet more silly errors.
Last edited by quickfur on Wed Mar 16, 2016 6:02 pm, edited 1 time in total.
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Re: A different way to expand an ursatope?

Postby Klitzing » Wed Mar 16, 2016 4:05 pm

quickfur wrote:E.g., you can read the first and last symbols (deleting the middle node, which is equivalent to writing '2', and that lets you read off prism symmetries, i.e., elements that correspond to the edges of the cube). For example, if we delete the middle column, we get o2x || o2f || o2o. The 2 here is actually an abuse of notation, because in the "real" Dynkin diagram there's actually no edge between the two nodes, but it serves as a nice mnemonic for "put two copies of the base shape on parallel (hyper)planes to make a prism".

No, the other way round. This introduced 2 tells about a digonal symmetry. And indeed, links of Dynkin symbols, which ought bear a mark 2, simply are neglected. Infact, in the wording of mirrors, these link marks just tell that the mirrors are to be placed at an angle of pi/k (where k is the link mark). x3o is a triangle and its mirrors of symmetry are inclined by pi/3. Same for the square x4o: these are inclined in pi/4. And again so for x2o, that is we have a very skinny rectangle, on side has "edges" of length x (aka 1) and the other "edges" are of size o (aka 0). But right angles (pi/2) are quite often being used, e.g. in prisms etc. This is why these links in Dynkin symbols usually are omitted. Even a symbol of type x4o3o (the cube) ought to be read with all not explicitely connected nodes being additionally linked with such 2's. This best can be seen in x4o3x, the rhombicuboctahedron. here the first node (x) and the last node (again x) are to be thought of as being linked by a further link marked 2. Thus the according subsymbol is truely x2x. There is no fake here. And what is that x2x? It is just a rectangle where the 2 side lengths happen to be both x, ie. have the same length. Thus it describes squares. And which squares are those? Those are the ones in the 12 "rhombical" orientations.

Just for kicks, let's read off the element types of a 4D uniform polychoron, say o5x3o3x. The first 3 nodes are o5x3o, so that means we have icosidodecahedra (o5(x3o) == dual triangles folded around 5-fold axes (thus attached by their vertices), or equivalently, (o5x)3o == dual pentagons folded around 3-fold axes, attached by their vertices) in the position corresponding to the cells of the 120-cell, so 120 icosidodecahedra. The last 3 nodes are x3o3x, i.e., cuboctahedron (as rhombi-tetrahedron: (x3o)3x = 4 dual triangles folded around 3-fold axes a unit displacement apart, this causes the original tetrahedron edges to expand into squares and 4 more triangles from the dual tetrahedron to appear), in the position corresponding to the cells of the 600-cell, so there are 600 cuboctahedra. Deleting the 2nd node gives us o2x3x, i.e., hexagons (hexagonal prisms of height 0), and deleting the 3rd node gives o5x2x, i.e., pentagonal prisms. So we see that o5x3o3x consists of icosidodecahedra, cuboctahedra, and pentagonal prisms (and hexagons, but they are subdimensional).

...
Edit: Corrected some silly errors in the o5x3o3x example.

Still one left. When you omit the second node here, you'll get o2o3x, not o2x3x. Thence there are no degenerate hexagon "cells", but rather just degenerate triangle "cells". (You ought think of them as triangular prisms of zero height. The zero height here clearly is given by the first node o.)

But I clearly second quickfur's aim. One truely gains a lot additional understanding about polytopes when one is able to read Dynkin symbols! :nod:

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Re: A different way to expand an ursatope?

Postby Klitzing » Wed Mar 16, 2016 4:29 pm

Klitzing wrote:And, for sure, the same diminishing concept would hold too, when considering any other srid-diminishing as a base of this bistratic rox cap. I.e. we also could derive some
  • ike || id || srid   =   bistratic rox cap
  • gyepip (J11) || ... || dirid (J76)
  • pap || ... || pabidrid (J80)
  • mibdi (J62) || ... || mabidrid (J81)
  • teddi (J63) || ... || tedrid (J83)   (the one described above & within the last post)

Aggregated now also the cell count for pap || ... || pabidrid:
  • pap: 1
  • oct: 10+10 = 20
  • mibdi (J62): 10
  • pero (J6): 2
  • squippy (J1): 10+10 = 20
  • pabidrid (J80): 1
And its lace city can be given as
Code: Select all
            x5x
              
              
              
              
    o5f       
            x5f
              
x5o           
              
            F5o
    x5x       
            o5F
              
o5x           
              
            f5x
    f5o       
              
              
              
              
            x5x

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Re: A different way to expand an ursatope?

Postby Klitzing » Wed Mar 16, 2016 5:14 pm

Still am wrangling with the namings for these diminishings of the bistratic cap of rox. - According to Wendy, I'd prefer kind a wording like "around-symmetrically N-diminished bistratic cap of rectified hexacosachoron". As the diminishing does not take place in axial direction, rather within the common subsymmetry which gets stacked. In our current example that one is the o3o5o subsymmetry which gets stacked resp. diminished. And diminishing then takes place simultaneously to all parallel layers, i.e. throughout. Esp. when considering such bidiminishings (the para and the meta version this is) then an unspecified usage else would suggest an axial application, which here is NOT being meant!

Any agreements? / supports? / or rather further opinions?

What then could be according OBSAs?   => @Hedrondude?
Perhaps something like "arsted biscrox" for the one of this first post and "arspabd biscrox" for the today one?

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Re: A different way to expand an ursatope?

Postby Klitzing » Thu Mar 17, 2016 1:10 pm

Klitzing wrote:And, for sure, the same diminishing concept would hold too, when considering any other srid-diminishing as a base of this bistratic rox cap. I.e. we also could derive some
  • ike || id || srid   =   bistratic rox cap
  • gyepip (J11) || ... || dirid (J76)
  • pap || ... || pabidrid (J80)
  • mibdi (J62) || ... || mabidrid (J81)
  • teddi (J63) || ... || tedrid (J83)   (the one described above & within the last post)

In this context we surely would have to consider the respective chopped off bits themselves too. Here we then have some
  • peppy (J2) || ... || pecu (J5)
The according lace city then is
Code: Select all
    x5o
       
       
o5o   
    o5f
       
       
x5o   
       
       
    x5x
       
       
o5x   

which shows that this fellow then happens to be a known segmentochoron (in a different orientation). In fact it is nothing but gyepip (J11) || pero (J6). Thus, after all no need for further investigations. Just for completeness to remind.

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Re: A different way to expand an ursatope?

Postby Klitzing » Fri Mar 18, 2016 12:32 pm

quickfur wrote:... So the corrected symbol is xfox-2-oxfo-3-ooox&#xt (note the last x in place of o).

This is a truly beautiful little CRF; it has 3 mibdies and 2 tetrahedra on one side, 1 trigonal prism, 2 octahedra(!), 3 square pyramids, and 6 tetrahedra on the other side. Even better yet, it has 17 vertices and 17 cells -- the same number of vertices as cells!

Here's a look at the 3 mibdies surrounding an edge:
Image
The 2 tetrahedra that fill in the gaps between the mibdies should be obvious.

Now here's the far side, which has a most interesting configuration of cells:
Image
I highlighted the trigonal prism which lies antipodal to the edge shared by the 3 mibdies. We see a most interesting pattern of tetrahedron-square pyramid-tetrahedron interfacing each mibdi to the trigonal prism.
...

Dreamed up, based on that, what about replacing these 3 (orange) mibdies (J62) by 3 mabidrids (J81)?

These then pairwise connect by their decagons. The central edge (with the former 3 incident pentagons) here would become a trip. And the innermost tips of the 2 tets (formerly tip-connected to that edge) there get replaced by 2 teddies. - Not too clear ad hoc what comes next. Would have to think a bit about that great relative...

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Re: A different way to expand an ursatope?

Postby Klitzing » Fri Mar 18, 2016 11:55 pm

... by means of direct neighbourhood all adjacent cells to the mabidrids (J81) then are more or less clear: Squares of mabdirid then all become adjoined by trips, pentagons by some ike diminishing. In fact those within the triangular cone clearly are teddies (J63), those within the slit atop the decagons are mibdies (J62), and the remaining ones are either gyepips (J11) or paps (so far not yet settled on those). The triangles of mabidrid are to be adjoined to coes or tricues. And at least those atop the teddies ought be full coes.

Still not closed so far. The whole far end remains to be constructed. Esp how to "translate" the squippies, octs and the single (green) trip of the small relative...

--- rk
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Re: A different way to expand an ursatope?

Postby quickfur » Sat Mar 19, 2016 12:19 am

Very promising idea indeed. Unfortunately I'll be busy this weekend so will have to wait till next week to investigate how to close up this figure. I'm quite certain it should be possible to close up, probably just needs a bigger corona on the far end.
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Re: A different way to expand an ursatope?

Postby student91 » Sat Mar 19, 2016 8:06 pm

Klitzing wrote:[...]
Dreamed up, based on that, what about replacing these 3 (orange) mibdies (J62) by 3 mabidrids (J81)?

These then pairwise connect by their decagons. The central edge (with the former 3 incident pentagons) here would become a trip. And the innermost tips of the 2 tets (formerly tip-connected to that edge) there get replaced by 2 teddies.[...]
Klitzing wrote:... by means of direct neighbourhood all adjacent cells to the mabidrids (J81) then are more or less clear: Squares of mabdirid then all become adjoined by trips, pentagons by some ike diminishing. In fact those within the triangular cone clearly are teddies (J63), those within the slit atop the decagons are mibdies (J62), and the remaining ones are either gyepips (J11) or paps (so far not yet settled on those). The triangles of mabidrid are to be adjoined to coes or tricues. And at least those atop the teddies ought be full coes.

Still not closed so far. The whole far end remains to be constructed. Esp how to "translate" the squippies, octs and the single (green) trip of the small relative...

--- rk

I have been calculating dynkin-coordinates for this thing. here are the coordinates for the three roxes:
Code: Select all
xx3o||Ff3o||Ao3x||VF3o||Bx3x||FA3o||(xB3o+Bx3f)||VF3x||(fV3x+Ao3F)||Ff3u||oF3f||xx3F
Where: F=f+x, A=f+2x, B=2f+x, V=2f, u=2x.
I'm quite sure these are correct, though there might be a little mistake somewhere.
I'm now working on closing it, though I won't promise a closed thing, I find it quite difficult doing this without renderings.
esp. the u in the third layer at the right came up quite unexpectedly.
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Re: A different way to expand an ursatope?

Postby quickfur » Sun Mar 20, 2016 5:32 am

I've been finding that lace towers are (much!) easier to compute coordinates for than &#zx coordinates. Probably an artifact of me working in Cartesian coordinates for my polytope viewer.
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Re: A different way to expand an ursatope?

Postby student91 » Sun Mar 20, 2016 12:27 pm

quickfur wrote:I've been finding that lace towers are (much!) easier to compute coordinates for than &#zx coordinates. Probably an artifact of me working in Cartesian coordinates for my polytope viewer.

I think this is because lace-towers all are build according to a 3-dimensional symmetry. Luckily, all 3-dimensional symmetries have a quite direct link to the [2,2]-symmetry (the symmetry of cartesian apacs-coordinates). However, the only link [3,3,3]-symmetry has to [2,2,2]-symmetry, is AFAIK via [5,3,3]-symmetry. So it ought be possible, by selecting some vertices from o5o3o3x, to obtain relatively convenient coordinates for x3o3o3o up to o3o3o3x, and then be able to easilly compute coordinates for [3,3,3]-symmetry &#zx-lacings.

EDIT: coordinates for x3o3o3o should be findable in x5o3o3o (when the edge-length of x5o3o3o is 2/phi^2, the x3o3o3o has an edge-length of 2sqrt(5)), and coordinates for o3x3o3o should be findable in o5x3o3o, again with an edge-length of sqrt(20) when rahi has an edge-length of 4-2f.
Last edited by student91 on Sun Mar 20, 2016 11:40 pm, edited 1 time in total.
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Re: A different way to expand an ursatope?

Postby Klitzing » Sun Mar 20, 2016 12:46 pm

quickfur wrote:
Klitzing wrote:[...]
Thus, what then would be xfox-2-oxfo-P-oooo-&#xt (P=3,4,5)? - I will have to take a look into these...

--- rk

Good news!! While the case of P=4 doesn't seem to close up in a CRF way, I have just constructed the P=3 case (with a slight modification) and verified that it's CRF!!!!!! The slight adjustment is that the last vertex layer must be x2o3x, not x2o3o, otherwise the mibdies are non-corealmar. So the corrected symbol is xfox-2-oxfo-3-ooox&#xt (note the last x in place of o).

... Here's a look at the 3 mibdies surrounding an edge:

Image

Thought about these again. Found that you'll need xfox...2oxfo...Poooa...&#xt where a=a(P) and, as you already pointed out a(3)=x, but now found out that you likewise would get a(4)=q and a(5)=f. Thus, in order to close that thing with edges only, which won't exceed unity (x), we'd have to add a further layer at least when P>3. - Thus, what would be appropriate there?

Case P=4 has some x4o at the second layer and some o4q at the fourth. accordingly this kind of asks for some further x4o layer somewhere beyond. Then this open gap could become a co, each. So what then would be required for the remaining node? I.e. we'll have solve for xfoxb2oxfox4oooqo&#xt, getting b=1/F, a not feasible length (for CRF). Thus this looks not to become a possible ansatz.

Now, what about case P=5? Again we have some x5o in layer 2 and then some o5f in layer 4. Thus this is the start for some pero, each. So the required next layer would be some x5x. Therefore we try to solve correspondingly for xfoxb2oxfox5ooofx&#xt. Here we derive simply b=o, a quite feasible length!

So, what then ought be that xfoxo2oxfox5ooofx&#xt ?
First we have 5 mibdies around one edge. Then we add, similar to the 3fold case, 2 peppies into the dimples. Next we attach 2 peroes to these base pentagons, which in turn connect by their base decagons mirror symmetrically. The remaining gaps then seem to be fillable by 2*5 further peppies (1 vertex of layer 2 and 3 each, 2 of layer 4 and 5 each) and 3*5 equatorial tets (1*5 of type 2 vertices of layer 4 and 5 each; 2*5 of type 1 vertex of layer 3 and 5 each, 2 of layer 4). The later peppies then ought connect at one of their triangles pairwise. - After all a promising candidate for endorsement! :)

--- rk
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Re: A different way to expand an ursatope?

Postby student91 » Sun Mar 20, 2016 12:54 pm

Klitzing wrote:So, what then ought be that xfoxo2oxfox5ooofx&#xt ?
This is just a tripentadiminishing of o5o3o3x, just as the expanded form is a tripentadiminishing of x5o3o3x
Last edited by student91 on Sun Mar 20, 2016 7:10 pm, edited 1 time in total.
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Re: A different way to expand an ursatope?

Postby Klitzing » Sun Mar 20, 2016 2:47 pm

student91 wrote:
Klitzing wrote:So, what then ought be that xfoxo2oxfox5ooofx&#xt ?
This is just a tridiminishing of o5o3o3x, just as the expanded form is a tridiminishing of x5o3o3x

What? Cannot follow how that one ought be some tri-diminishing.

But indeed, it closely is related to ex. In fact, we already did consider its 2/10-luna:
Image
(cf. here) or alternatively as a lace city (cf. here):
Code: Select all
            o5o       
                   o5x
                      
       x5o           
o5o                   
                   f5o
            o5f       
                      
o5x                   
       f5o           
                      
                   x5x


In order to introduce those mibdies around the "central" edge (which is displayed top left in the lace city as o5o-o5o) we just have to omit the 5 neighbouring vertices (which then are the equatorial ones of the red pentagonal dipyramid, as shown in quickfur's pic):
Code: Select all
            o5o       
                   o5x
                      
                      
o5o                   
                   f5o
            o5f       
                      
o5x                   
       f5o           
                      
                   x5x

and we are done. Thus it rather comes out to be a penta-diminishing of the 2/10-lune of ex! (Thus, at best we could speak of some hepta-diminishing of ex.)

But then it also proves thereby its convexity! Moreover my construction was completely correct. :]

--- rk
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Re: A different way to expand an ursatope?

Postby Klitzing » Tue Mar 22, 2016 1:02 pm

student91 wrote:nvm, I just found out that the tower I found was part of a tridodecadiminished x5o3o3x

(Then, in turn, you were refering to that post. )

But then, as the srids (x3o5x) are bases of doe||srid = ox3oo5xx&#x caps, those or their interlocking counterparts, the tedrids, come in underneath the doe positions of sidpith (x3o3o5x) or, equivalently, the vertex positions of ex (x3o3o5o). Moreover, as the tedrids and the base srid in that 5fold version will connect at triangles (not at pentagons), we clearly have 20 tedrids and 1 srid so far. Therefore this ought be rather a 21-diminishing (at least).

In fact, we have ex as a vertex layer: o3o5o || x3o5o || o3o5x || f3o5o || ... And we thus would diminish sidpith at those does, which correspond to all vertices of the here given first and third layer of ex. o3o5o clearly makes 1 diminishing. And o3o5x clearly has 20 vertices, i.e. gives rise to 20 further diminishings.

--- rk
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