what is the 5D vector equilibrium?

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

what is the 5D vector equilibrium?

Postby Sphericality » Mon Nov 21, 2016 1:49 pm

I hope you are familiar with Bucky Fuller's term vector equilibrium?
It is his name for the cuboctahedron because of its unique property
of having the distance of its edge lengths equal to its circumradius.
This requires each of its edges to form an equilateral triangle if connected to the centerpoint.
The 2D equivalent is the hexagon, and the 3D form has 4 cross sections that are hexagons.
The VE concept also has something to do with the closest sphere packing
as the hexagon and cuboctahedron are both the maximum kissing number of their dimension.
Fuller considered the VE to be more important than the convex regular forms.

The 4D vector equilibrium, (I am fairly certain), is the 24 cell,
and it contains 8 (I believe) cuboctahedral cross sections.

My question is -
what is the name and form of the 48 cell closest sphere packing in 5D?
I guess it could be called the 48 cell and I would think it must be a CRF, but Im hoping more is known about it than that?

Other followup questions would be ...
Does it have 10 x 24cell cross sections ?
And does it fulfill the requirement to be called the 5D vector equilibrium ? (primarily that its vertexes be the same length as its circumsphere)
Does this pattern continue so that every dimension has a unique vector equilibrium ?

I look forward to your thoughts on this.
Thanks muchly ~ NDV
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Re: what is the 5D vector equilibrium?

Postby Klitzing » Mon Nov 21, 2016 8:09 pm

There is a most remarkable dimensional sequence of polytopes, each having the same circumradius to edge length ratio of unity. Those are the expanded simplices. In fact those are the Stott expansion of the regular simplex of either dimension by its dual one. Most easily those polytopes are described by their Dynkin diagram.


What truely is fascinating here, that all these (from 3D onwards) can be considered as a 3 layered stack of vertex layers. The convex hull of the top one then is the corresponding simplex (of one dimension less) itself, which thus is a true facet of that polytope, the convex hull of the equatorial one then is rght that expanded simplex (of one dimension less), which thus happens to be a pseudo facet or crosssection, and the hull of the final, opposite one then is the dual simplex (of one dimension less), again a true facet of the polytope. That is, we have

  • x3o3x = x3o || x3x || o3x
  • x3o3o3x = x3o3o || x3o3x || o3o3x
  • x3o3o3o3x = x3o3o3o || x3o3o3x || o3o3o3x
  • x3o3o3o3o3x = x3o3o3o3o || x3o3o3o3x || o3o3o3o3x
  • ...

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Re: what is the 5D vector equilibrium?

Postby Klitzing » Mon Nov 21, 2016 10:29 pm

OTOH you already mention ico, which also follows that same circumradius to edge length ratio of unity. And which likewise uses co as equatorial section. This shows that also some different solutions to that problem exist. Just to mention some of those:

  • o3x4o = co = o3x || x3x || x3o = -{3} || {6} || {3}
  • o3x3o4o = ico = o3x3o || x3o3x || o3x3o = oct || co || oct
  • o3x3o3o4o = rat = o3x3o3o || x3o3o3x || o3o3x3o = rap || spid || -rap
  • ...
as well as
  • 02,2,1 = mo = o3o3o3o3o || o3o3x3o3o || x3o3o3o3x || o3o3x3o3o || o3o3o3o3o = pt || dot || scad || dot || pt
  • 23,1 = laq = x3o3o3o3o3o || o3o3o3x3o3o || x3o3o3o3o3x || o3o3x3o3o3o || o3o3o3o3o3x = hop || -bril || staf || bril || -hop

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Re: what is the 5D vector equilibrium?

Postby wendy » Tue Nov 22, 2016 12:18 am

It' actually 1_22 = o3o3o3o3Bx. The term that Coxeter uses for these figures (or rather, the rays going to the vertex, is eutactic stars.

Some other ones are o3x5o and x3o3o5o.

All of the polytopes constructed by Wyhoff's construction have edges parallel to these vectors.
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Re: what is the 5D vector equilibrium?

Postby Klitzing » Tue Nov 22, 2016 8:44 pm

Wendy, thanks, mo clearly is 12,2, for sure. My fault.

But OTOH neither o3x5o nor x3o3o5o have a circumradius to edge length ratio of 1.
In those 2 cases the circumradius has size f, i.e. (1+sqrt(5))/2 = 1.618... times the edge length.

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Re: what is the 5D vector equilibrium?

Postby wendy » Fri Nov 25, 2016 12:59 am

OTH o3v5o and v3o3o5o do. The other figure that makes a eutactic star is a3o4o3a, ie xo3oo4oo3ox (the unmodified compound).
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Re: what is the 5D vector equilibrium?

Postby Klitzing » Fri Nov 25, 2016 10:57 am

Well, o3v5o and v3o3o5o would clearly have unit circumradius, but then with a v=1/f= 0.618.. sized edge length. So the ratio still isn't the searched one.

xo3oo4oo3ox is just the compound of x3o4o3o (24-cell) and its dual o3o4o3x (again a 24-cell, arranged such, that the vertices of the second are situated just atop the body centers of the cells of the first, and vice versa). As the 24-cell itself bows to that special ratio property, this compound clearly does too.

One might try to build the convex hull of that compound, in order to look for a convex polychoron. Then the circumradius obviously would not change. But the new edge size (bridgeing the respective tips) would happen to fall below unity. In fact it would have size sqrt(2-sqrt(2)).

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Re: what is the 5D vector equilibrium?

Postby quickfur » Sun Dec 11, 2016 3:38 pm

Klitzing wrote:[...]
What truely is fascinating here, that all these (from 3D onwards) can be considered as a 3 layered stack of vertex layers.

This is actually true in 2D too: the hexagon can be considered a tristratic polygon of a unit edge, a double-length edge, and another unit edge. Though of course, in this case the middle layer happens to be no longer unit-edged. :P

Edit: Albeit, the middle layer can be considered an expanded 1-simplex, which is a special case in that no other node exists in the CD diagram so what is expanded is the only node, hence x -> u, forming the decomposition of the hexagon as: x || u || x.
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Re: what is the 5D vector equilibrium?

Postby Sphericality » Thu May 25, 2017 6:34 am

Thank you very much everyone for your replies -
I am in awe of the expertise here.
A beginner's glossary of all the terms used
would make everything so much more useful to the uninitiated non-mathematician. :oops:
I have been trying to find a good primer on Dynkin diagram basics.
In fact I would love to master the different notations used here on the Hi.gher space forums.
Googling them has so far produced only examples of complex Lie groups and worse.
All I want is an explanation for how these notations work with the basic polygons, polyhedrons and polytopes.

Did I miss one here on the forums ?
I have searched the Wiki and the Intro, and even made it to Bendwavy in my Googling.
Despite having what I think is a good conceptual grasp of higher space
I failed to find something about the notations basic enough for me. :mrgreen:

I am still trying to understand how the triangular/tetrahedral honeycombs progress into higher dimensions.
I realise that in 3D the spaces between the tetra are octa, but what happens in 4D and beyond ?

I note that in my original post that started this thread I made the grossly mistaken assumption that the
progression of kissing numbers from 6 to 12 to 24 would continue in some logically predictable way lol.

It seems the answer to my first question: what is the name and form of the 48 cell closest sphere packing in 5D ...
is that my assumption was incorrect and it is actually proven that the closest sphere packing in 5D has between 40 and 44 4D spheres,
and that figure has no name and is not a CRF ? Yes ?

Do I understand correctly that @Klitzings answer to my second question is that the 5D VE is this form here
https://bendwavy.org/klitzing/incmats/spid.htm - called a Spid ?
And that the pattern does indeed continue to infinite dimensions ?

Klitzing wrote:There is a most remarkable dimensional sequence of polytopes, each having the same circumradius to edge length ratio of unity. Those are the expanded simplices. In fact those are the Stott expansion of the regular simplex of either dimension by its dual one. Most easily those polytopes are described by their Dynkin diagram.
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Re: what is the 5D vector equilibrium?

Postby student91 » Thu May 25, 2017 8:40 am

I have once written this down, I think it is quite understandable, though far from complete. The last chapters are highly conceptual and thus not very understandable, but I hope you can understand the rest.
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Re: what is the 5D vector equilibrium?

Postby Klitzing » Fri May 26, 2017 10:06 pm

Sphericality wrote:Thank you very much everyone for your replies -
I am in awe of the expertise here.
:)
Sphericality wrote:A beginner's glossary of all the terms used
would make everything so much more useful to the uninitiated non-mathematician. :oops:
I have been trying to find a good primer on Dynkin diagram basics.
In fact I would love to master the different notations used here on the Hi.gher space forums.
Googling them has so far produced only examples of complex Lie groups and worse.
All I want is an explanation for how these notations work with the basic polygons, polyhedrons and polytopes.

Did I miss one here on the forums ?
I have searched the Wiki and the Intro, and even made it to Bendwavy in my Googling.
Despite having what I think is a good conceptual grasp of higher space
I failed to find something about the notations basic enough for me. :mrgreen:

I could recommend you as a quick notational reference:
https://bendwavy.org/klitzing/explain/dynkin-notation.htm
and for a more detailed intro:
https://bendwavy.org/klitzing/explain/dynkin.htm

Sphericality wrote:I am still trying to understand how the triangular/tetrahedral honeycombs progress into higher dimensions.
I realise that in 3D the spaces between the tetra are octa, but what happens in 4D and beyond ?

Well the general situation here is a circular Dynkin diagram having just one node being marked. Within my notation that one is represented by x3o3...3o3*a, where the "x" represents the marked node (graphically that would be a ringed dot, but within the typewriter friendly representation thereof it is represented by an "x"), the "o"s represent the unmarked nodes (graphically that would be an unringed dot), and the final "*a" just reminds that the typewriter friendly linear representation has to be closed back to the a-th (= 1st) node again. The cells of that highdimensional honeycomb then are obtained by deleting any node, provided the remainder still has a ringed node (= "x") within every connected sub-diagram. Here we would just get the followings: x3o3o3...3o, o3x3o3...3o, ..., o3o3o3...3x. The first one then is the simplex. The last one is ist dual, again a simplex. The second one is the rectified simplex (having its vertices at the center positions of the edges of a simplex). The third then would be a birectified simplex (having its vertices at the center positions of the triangles of a simplex). Etc. - So, for a 3D honeycomb x3o3o3o3*a we have: x3o3o (tetrahedra), o3x3o (rectified tetrahedra = octahedra), and o3o3x (birectified tetrahedra = the dual tetrahedra). - For the 4D case x3o3o3o3o3*a we would get for 4D cells: x3o3o3o (pentachora = 5-cells), o3x3o3o (rectified pentachora), o3o3x3o (birectified pentachora = same as the rectified pentachora, just in inverted relative orientation), and o3o3o3x (trirectified pentachora = dual pentachora). Etc.

Sphericality wrote:I note that in my original post that started this thread I made the grossly mistaken assumption that the
progression of kissing numbers from 6 to 12 to 24 would continue in some logically predictable way lol.

It seems the answer to my first question: what is the name and form of the 48 cell closest sphere packing in 5D ...
is that my assumption was incorrect and it is actually proven that the closest sphere packing in 5D has between 40 and 44 4D spheres,
and that figure has no name and is not a CRF ? Yes ?

Wrt. to sphere packings and kissing numbers I would you highly recommend the standard literature for that purpose:
"Sphere Packings, Lattices and Groups" by John Conway and Neil J. A. Sloane, cf. http://www.springer.com/de/book/9780387985855.

Sphericality wrote:Do I understand correctly that @Klitzings answer to my second question is that the 5D VE is this form here
https://bendwavy.org/klitzing/incmats/spid.htm - called a Spid ?
And that the pattern does indeed continue to infinite dimensions ?

Klitzing wrote:There is a most remarkable dimensional sequence of polytopes, each having the same circumradius to edge length ratio of unity. Those are the expanded simplices. In fact those are the Stott expansion of the regular simplex of either dimension by its dual one. Most easily those polytopes are described by their Dynkin diagram.

Yes you do. - Within 3D you have x3o3x, i.e the cuboctahedron (which is nothing but the expanded tetrahedron). The faces here are x3o . = triangle, x . x = square, and . o3x = dual triangle. It can be placed as axial stack of a top triangle x3o, an equatorial hexagon x3x, and a bottom triangle in dual orientation, i.e. o3x. The equatorial section then shows that it has a circumradius to edge ratio of 1. - The 4D one then is x3o3o3x, the expanded pentachoron, also known as runcinated (i.e. 4th node being ringed in addition to the 1st one) pentachoron, or alternatively as small prismated decachoron. Its cells then are x3o3o . = tetrahedra, x3o . x = triangular prisms, x . o3x = further triangular prisms (in different orientation), and . o3o3x = further tetrahedra (in dual orientation). This one then too can be given as axial stack: x3o3o (tetrahedron) atop equatorial x3o3x (cuboctahedron) atop o3o3x (dual tetrahedron). Thus this figure too has the same circumradius to edge ratio of 1. - The 5D case then would be x3o3o3o3x, the expanded hexteron = stericated (= 1st and 5th nodes being ringed only) hexateron = small cellated dodecateron. Its 4D boundaries are: x3o3o3o . = pentachoron, x3o3o . x = tetrahedron-prism, x3o . o3x = triangular duoprism, x . o3o3x = further tetrahedron-prism in different orientation), and . o3o3o3x = further (dual) pentachoron. Again it can be given in axial stack orientation. Then it is x3o3o3o (pentachoron) atop equatorial x3o3o3x (small prismated decachoron) atop o3o3o3x (dual pentachoron). Thus this figure again has the same circumradius to edge ratio of 1. Etc.

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Re: what is the 5D vector equilibrium?

Postby Klitzing » Fri May 26, 2017 11:04 pm

student91 wrote:I have once written this down.
I think we 3 should continue somehow on that article, ain't we?
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Re: what is the 5D vector equilibrium?

Postby student91 » Sat May 27, 2017 4:33 pm

Klitzing wrote:I think we 3 should continue somehow on that article, ain't we?
--- rk
Yes, I guess so, My summer holiday starts in july, I could start then doing something about it. Last time we tried to collaborately make something of it I had the impression everybody just added everything they wanted to include about the subject. However, I think we should only include the fundamental concepts in order to make it as comprehensible as possible. I am far from happy with the thing I posted there, but it is a start. Considering this, Are you happy if I just try to do it on my own? :D Of course you get to review it several times after I'm finished.
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Re: what is the 5D vector equilibrium?

Postby student5 » Sat May 27, 2017 7:24 pm

student91 wrote:
Klitzing wrote:I think we 3 should continue somehow on that article, ain't we?
--- rk
Yes, I guess so, My summer holiday starts in july, I could start then doing something about it. Last time we tried to collaborately make something of it I had the impression everybody just added everything they wanted to include about the subject. However, I think we should only include the fundamental concepts in order to make it as comprehensible as possible. I am far from happy with the thing I posted there, but it is a start. Considering this, Are you happy if I just try to do it on my own? :D Of course you get to review it several times after I'm finished.

Can I help as well? or would that just increase the clutter even more? :XD:
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Re: what is the 5D vector equilibrium?

Postby Klitzing » Sat May 27, 2017 9:06 pm

Last time I retarded the process, cause we already have lots of EKFs, even way too much to enlist them all in a single article. But on the other hand those all had been selected manually. We well could have overlooked some. So it might be good not only to outline the process, but also to have some rigour in the to be presented part of the enlisting. E.g. something like completeness thereof. My last proposal of those days was to complement it possibly by some computer aided research. What would you think, could that be set up somehow? Or should we stick with the process only and presenting just some few examples?
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Re: what is the 5D vector equilibrium?

Postby student91 » Sat May 27, 2017 9:31 pm

Klitzing wrote:Last time I retarded the process, cause we already have lots of EKFs, even way too much to enlist them all in a single article.
hmmm, I see your point. I think our main focus should be outlining the principles and ideas that underly the construction of the EKFs. Only when this is achieved people will be interested in articles full of lists (I think). Furthermore in my opinion the public domain does not yet have an comprehensible, complete and short explanation about the Coxeter-Dynkin notation.(It seems everyone here has his own explanation, I'm not sure my explanation is useful to people who don't understand it already :\) I tried to achieve both of these in one article, but I guess by this I kinda lost my focus. I would be happy to discuss this after my exams.
But on the other hand those all had been selected manually. We well could have overlooked some. So it might be good not only to outline the process, but also to have some rigour in the to be presented part of the enlisting. E.g. something like completeness thereof. My last proposal of those days was to complement it possibly by some computer aided research. What would you think, could that be set up somehow? Or should we stick with the process only and presenting just some few examples?
--- rk
I think our work back then was fairly thorough, I think we will be able to proof completeness without a computer, given we restrict ourselves to the (in my opinion most interesting) .5.3.3.-symmetry. But again, my exams have my priority now. Feel free to write suggestions.
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Re: what is the 5D vector equilibrium?

Postby Sphericality » Wed May 31, 2017 11:58 pm

Thank you all. The paper by student91 above even in its current form was a revelation for me :-)
https://drive.google.com/file/d/0B3tiav ... sp=sharing
I think Im actually beginning to understand this stuff hehehe 8) :mrgreen: 8) :o_o: 8) :nod: :arrow: :arrow: :arrow: :idea:
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Re: what is the 5D vector equilibrium?

Postby szluk » Sun Jan 31, 2021 1:26 pm

Klitzing wrote:There is a most remarkable dimensional sequence of polytopes, each having the same circumradius to edge length ratio of unity. Those are the expanded simplices. In fact those are the Stott expansion of the regular simplex of either dimension by its dual one. Most easily those polytopes are described by their Dynkin diagram.


What truely is fascinating here, that all these (from 3D onwards) can be considered as a 3 layered stack of vertex layers. The convex hull of the top one then is the corresponding simplex (of one dimension less) itself, which thus is a true facet of that polytope, the convex hull of the equatorial one then is rght that expanded simplex (of one dimension less), which thus happens to be a pseudo facet or crosssection, and the hull of the final, opposite one then is the dual simplex (of one dimension less), again a true facet of the polytope.

--- rk


Thank you for this very helpful explanation.
It follows that the number of the external vertices (other than the origin) of n-VE is given by
|v(n)|=n*(n+1)
This is 0, 0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210,… A279019 OEIS sequence; least possible number of diagonals of simple convex polyhedron with n faces.

I was thinking about orienting n-VE in Cartesian coordinate system, as in the attached images for n=2 and n=3. In these cases it contains 2*(n-1) Cartesian vertices defining n-1 Cartesian coordinate system with the origin at [0...00] and having addresses
a_1=[1,0,...,0,0] x 2
a_2=[0,1,...,0,0] x 2
...
a_(n-1)=[0,0,...,1,0] x 2
(negative coordinates can be encoded in a vertex number).
For 1 ≤ n ≤ 3 n-VE contains 2^n supplementary vertices having addresses
a_j=[c,c,...,c,e] x 2^n
which form is dictated by radial symmetry condition and the form of Cartesian vertices. But this orientation cannot be provided for n>3, as in general any n-VE provides addressability for only
|v(n)| - 2*(n-1)
or
n^2-n+2
(A014206 OEIS sequence) supplementary vertices, while 2^n is required.
Solving c and e for a_j we have
c=1/2
e=√(5-n)/2

Does that mean that n-VE for n>3 cannot be oriented like that in Cartesian coordinates?
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Re: what is the 5D vector equilibrium?

Postby Klitzing » Mon Feb 01, 2021 5:36 pm

The x3x hexagon itself is to be aligned within the x1, x2 plane like a 2-unit edge along x1 (its diagonal) plus two opposite unit edges in parallel somewhere with positive/negative x2 coordinate.
Next place a x3o triangle in parallel somewhere at positive x3 coordinate above and a further o3x (dual) triangle at negative x3 coordinate below.
Then place a x3o3o tetrahedron in parallel somewhere at positive x4 coordinate above and a further o3o3x (dual) tetrahedron at negative x4 coordinate below.
Further place a x3o3o3o pentachoron in parallel somewhere at positive x5 coordinate above and a further o3o3o3x (dual) pentachoron at negative x5 coordinate below.
Go on like that forever. This construction in fact is exactly what I already had described in my last post.

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Re: what is the 5D vector equilibrium?

Postby szluk » Mon Feb 01, 2021 7:08 pm

Thank you. I do not question that n-VE can be constructed in n-dimensional space.

I would simply like to have an (n-1)-space of each n-VE spanned by its 2(n-1) vertices and defining Cartesian coordinate system. That seems possible only for n=2,3.
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Re: what is the 5D vector equilibrium?

Postby Klitzing » Thu Feb 04, 2021 8:45 pm

You probably would be interested to know about:
Theorem 4.5. A regular simplex of dimension n can be inscribed in a hypercube
of dimension n if and only if n = 2^m − 1 for some m.

and the correction thereof:

Theorem 5.3. It is possible to inscribe a regular simplex of dimension n in a
hypercube of dimension n if and only if a Hadamard matrix of order n + 1 exists.

Cf.

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Re: what is the 5D vector equilibrium?

Postby szluk » Mon Feb 08, 2021 10:00 am

Thank you very much.
This is certainly related to the fact that Boolean {0, 1}^n address space ({n}-cube) is isomorphic to (2^n-1)-simplex (https://arxiv.org/abs/2007.03782).
In this paper I considered (2^n)-cube defined by 2^n n-cubes sharing common origin (00…0). Later on when I learned about Buckminster Fuller research I realized that (2^n)-cube is an n-dimensional companion of his n-vector equilibrium (they are the same for n ≤ 1). But it turns out that it does not preserve orthogonality so nicely as cubes.
I guess it must have something to do with the kissing number which equals |v| (# of external vertices of ve) for 1 ≤ n ≤ 3 but for n=4 it is 24 (not 20 as in the case of runcinated 4-simplex) and for n > 4 only lower and upper bounds are known (with the exception of n = 8 and n = 24).

--SŁ
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