11-cell

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11-cell

Postby Seldon » Fri Mar 16, 2007 12:09 pm

In the science magazine DISCOVER, there was an article about the fourth dimension. The article said that a regular 11-cell is possible in tetraspace. What do you guys make of this?

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Postby Keiji » Fri Mar 16, 2007 1:04 pm

It's not possible to have an 11-cell anywhere. The only n-dimensional regular shape (n > 2) that can possibly have an odd number of (n-1)-hypercells is a simplex. The 4D simplex has 5 cells.
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Postby Nick » Fri Mar 16, 2007 8:29 pm

What's an 11-cell?
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Postby pat » Fri Mar 16, 2007 8:51 pm

Certainly, one can make an 11-cell, just not a regular 11-cell.

Take a 10-faced polyhedron (pentagonal bipyramid or octahedron with two corners lopped off or a cube with four corners lopped off or whatever you like). Draw a point in the center. Connect every vertex to that point. (Assuming the polyhedron was convex enough that all of the connections are internal to the polyhedron...) You've just drawn the surface of an 11-celled pyramid. One cell for each face, plus the original polyhedron.

If you don't like pyramids, then just lop the corners off of a 4-simplex and then lop off one more corner. Or, lop three corners off a 4-cube. Or, just pick 11 half-hyperspaces that all include the origin, bound a finite region, and such that none of the half-hyperspaces is superfluous (ie. that for each half-hyperspace X, the corresponding negative hyperspace -X intersects the interior of the intersection of the remaining 10 half-hyperspaces).
Last edited by pat on Fri Mar 16, 2007 8:59 pm, edited 2 times in total.
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Postby pat » Fri Mar 16, 2007 8:57 pm

The 10-dimensional simplex has 11 10-dimensional cells. But, that's probably not what the Discover article was trying to say.

Which issue of Discover is this?
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Postby wendy » Sat Mar 17, 2007 7:13 am

The 11cell as reported in Discover is a {3,5,3}, having icosahedral cells and dodecahedral vertices. These are not the euclidean versions of these, but rather the elliptic version, where opposite sides are identified.

You can see from its symbol that it lives as a tiling in H3 space.

In practice, the thing can not really be rendered in Euclidean geometry, but is something like a repeated pattern in H3. It is something like colouring the chessboard in black and white, but here colouring the {3,5,3} in eleven different colours such that the opposite walls of each cell, and the opposite corners at a vertex, were always the same.

There is a matching {5,3,5}, a tiling of dodecahedra, five at an edge, with 57 different colours.

See, eg the wikipedia entries for '11-cell' and '57-cell'.
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Postby papernuke » Sun Mar 25, 2007 1:05 am

I still dont get what an 11 cell is.

[edit] although i didnt ask, i looked at the stuff awnsering nick's question, but i dont get it.
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Postby papernuke » Sun Mar 25, 2007 1:06 am

what do the {#,#,#}s mean? is it length breadth and depth?
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Postby pat » Sun Mar 25, 2007 2:31 am

No, Nuke... they refer to the types of facets. It is the Schlafli symbol.

{ p } refers to a regular, two-dimensional polytope with p sides. A p-polygon. So, { 3 } is an equilateral triangle, { 5 } is a regular pentagon, etc.

{ p, q } refers to a regular, three-dimensional polytope with facets that are { p }'s where q of them meet at each vertex. So, for example, the tetrahedron is { 3, 3 } because each facet is an equilateral triangle and three of them meet at each vertex. The cube is { 4, 3 } because each facet is a { 4 } (a square), and 3 of them meet at each vertex. The octahedron is { 3, 4 } because each facet is an equilateral triangle { 3 }, and 4 of them meet at each vertex.

{ p, q, r } refers to a regular, four-dimensional polytope with facets that are { p, q }'s where r of them meet at each vertex. So, the hypercube is { 4, 3, 3 } because three facets meet at each vertex and each of those facets is a cube { 4, 3 }.

The n-dimensional simplex is { 3, 3, 3, ..., 3 } with (n-1) 3's.
The n-dimensional cube is { 4, 3, 3, 3, ..., 3 } with (n-2) 3's.
The n-dimensional cross-polytope is { 3, 3, 3, ..., 3, 4 } with (n-2) 3's.

There's a nice chart on the Schlafli Symbol page
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Postby bo198214 » Sun Mar 25, 2007 2:57 pm

pat wrote:{ p, q, r } refers to a regular, four-dimensional polytope with facets that are { p, q }'s where r of them meet at each vertex. So, the hypercube is { 4, 3, 3 } because three facets meet at each vertex and each of those facets is a cube { 4, 3 }.


I count 4 (not 3) cubes at each vertex of a tesseract.

In wikipedia (http://en.wikipedia.org/wiki/Schläfli_symbol, but because the umlaut it can not be made a link in phpBB) the explanation is that { p, q, r} indeed means that it has {p,q} facets, but not that r of them meet at each vertex, but that the vertex figure is {q,r} (the vertex figure is the figure made of the neighboured vertices of vertex). Only in the 3dim case { p, q} the vertex figure must be {q} (a q-gon) which means that each vertex has q neighbors and hence also q meeting facets.

However in 4d it seems to be equivalent to that r facets meet at each *edge* (in the tesseract indeed 3 cubes meet at each edge), though they spoke merely of a not further defined "edge figure" {r} in wikipedia.

I dont know whether this can be generalized that in {q<sub>1</sub>,...,q<sub>d-1</sub>} always q<sub>d-1</sub> facets meet at each d-3 dimensional subface? Wendy?
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Postby wendy » Tue Mar 27, 2007 7:51 am

The Schläfli symbol works like this. S is the dimension of the surface of the thing, or the solid dimension of a tiling.

{p} is a polygon.

{p,q} is a polyhedron, being p, q at a vertex (S-2)

{p,q,r} is polychoron, being {p,q}, r at an edge (S-2)

Basically, a face (N-1 d) is bounded by margins (N-2 d). When you pack polytopes together, faces go together by pairs, eg cube-cube join by a square.

The numbers are actually a count around the margin. [I am using dashes to show where numbers are removed: these are purely to set context.]

So, {4,3,4} is the ordinary packing of cubes in 3d.

It has four-sided hedra {4,-,-}

These are arranged around a sphere-like thing, three at a corner. (evidently you can only have two at an edge) This gives {4,3,-}

You then pack cubes, with two at a face, and 4 at an edge. This gives {4,3,4}.

You see eight cubes at a corner, because the {-,3,4} is {3,4} has eight faces. Each triangle of the octahedron {3,4} becomes a cube {4,3} in the cubic tiling {4,3,4}.

So the 11 cell {3,5,3}, you are packing icosashedra {3,5,-} into a vertex figure that is a dodecahedron {-, 5, 3} Of course, there are three {-,3} pentagons {5,-} at a vertex, and thence three {-,-,3} icoashedra {3,5,-} at an edge of {3,5,3}.
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Postby bo198214 » Tue Mar 27, 2007 9:54 am

but thats not exactly an answer to my question, wether for example {p,q,r,s} would imply that s {p,q,r} objects meet at each 2 dimensional subface.
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Postby wendy » Wed Mar 28, 2007 7:56 am

{p,q,r,s}

This can be either a 4d tiling, or a 5d polytope. The surface is however always 4d.

The names cell, wall, sill relate to solid tilings, of N, N-1, N-2 dimensions. The values in brackets evaluate N as 4.

The cells (4d) are solid in this space, are {p,q,r}.

The walls (3d) divide cells, are {p,q}. Two cells meet at a wall.

The sills (2d) bound the walls. These are polygons, and the space around (ie, orthogonal to) the individual sills, is a space divided into rays (representing walls) and sectors (representing the cells). The number of these sectors is the number {r}.
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Postby bo198214 » Wed Mar 28, 2007 11:58 am

1. Are you sure that r is the number of sectors, not s? What otherwise does s denote?
2. You make a pseudo-generalization to N dimensions because you still consider {p,q,r,s} instead of {q<sub>1</sub>,...,q<sub>N-1</sub>}. So how would the proper generalization look? I would guess simply by replacing p by q<sub>1</sub>,...,q<sub>N-4</sub>. For example {p,q,r} by {q<sub>1</sub>,...,q<sub>N-4</sub>,q<sub>N-3</sub>,q<sub>N-2</sub>}. Right?

@moderators: This thread can be moved to the geometry section.
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Postby wendy » Thu Mar 29, 2007 7:45 am

The numbers {p2, p3, p4, ...} will complete a pN regular polytope.

The dual of {p2, p3, ..., pN-1, pN} is {pN, pN-1, ..., p3, p2}.

If one is using Schläfli symbols, then the numbers stand for the running margin-count.

This means, that if you have polytopes of the type {p2,p3,...pM-1,pM}, and you try to put these together, then obviously, these go face-to-face, as {p2,p3,...,pM-1}. Perpendicular to a face, you have a ray, with one point inside, and the second end outside. It's a line.

To add a number, you can specify how many faces share a common margin {p2,p3,...pM-2}. Around (perpendicular to) the margin one has a 2d space. In the final construction, there are going to be {pM+1} faces around this margin, and therefore each face appears as a sector perpendicular to {p2,p3,...,pM-2} element.

So when we tile space with {p2,,,pM} into a {p2...pM+1}, we end up with having p(M+1) cells around an element {p2...pM-2}.

For example, in the case of a cube {4,3}. We can pack three or four cubes around {} [ie a line], to get {4,3,3} or {4,3,4}. Of a tesseract {4,3,3} we pack 3 or 4 around {4,-,-} = {4} = square, to get a {4,3,3,3} or {4,3,3,4}.

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Postby Keiji » Thu Mar 29, 2007 9:41 am

wendy wrote:{4,3,4}


What is {4,3,4}? Is it simply the tiling of cubes over a realm?
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Postby bo198214 » Thu Mar 29, 2007 10:03 am

@Wendy
So is this a "yes" to
bo198214 wrote:in {q<sub>1</sub>,...,q<sub>d-1</sub>} always q<sub>d-1</sub> facets meet at each d-3 dimensional subface?

Or in your terminology: in a polytope {p<sub>2</sub>,...,p<sub>N</sub>} always meet p<sub>N</sub> facets at an N-3 dimensional subface (which is of type {p<sub>2</sub>,...,p<sub>N-3</sub>})?

And is it "yes" to the question, whether using the r instead of the s, was a typo in your previous post?

@Keiji
{4,3,4} is the tiling of E3 by cubes.
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Postby wendy » Sat Mar 31, 2007 7:50 am

The notation {p1,p2,p3...} is called the Schläfli symbol. He was the first of many to find it.

It can be read in two ways.

1. As a series of "margin figures". A margin is a thing that divides faces, in the terminology of George Olshevsky (dinogeorge) a 'ridge'. Tilings don't have ridges, though.

2. As a series of vertex-figures, where each edge of {p2,p3,...} becomes a p1. This is how Gosset read them.

In the first form, we have

N is the dimension of a polytope, while M is the dimension of the surface or tiling. So always, M=N-1.

{4,3,4} as squares {4}, three at an (N-3) = cubes, four at an edge (N-2 or N-3) gives a tiling of cubes.

In the second reading, we have {4} square, gives {3,4} when each edge of the square becomes a triangle, becomes {4,3,4} cubic, when each edge of the octahedron becomes a square.
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Postby bo198214 » Sat Mar 31, 2007 8:52 am

*wonders whether Wendy can answer with yes or no*
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Postby wendy » Sun Apr 01, 2007 7:13 am

@Wendy
So is this a "yes" to bo198214 wrote:
in {q1,...,qd-1} always qd-1 facets meet at each d-3 dimensional subface?

Or in your terminology: in a polytope {p2,...,pN} always meet pN facets at an N-3 dimensional subface (which is of type {p2,...,pN-3})?


yes

It's {s} as ye observe.

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Postby bo198214 » Sun Apr 01, 2007 3:28 pm

Wow, thanks. :)
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Re: 11-cell

Postby headcircus » Mon May 14, 2007 5:00 am

Seldon wrote:In the science magazine DISCOVER, there was an article about the fourth dimension. The article said that a regular 11-cell is possible in tetraspace. What do you guys make of this?

Seldon


The 11-cell.. is my current obsession. Can you tell? To me it is more than just a shape, a tiling or the like. It's more like a symbol for unlimited possibilities... simply because it is so absract there is nowhere to put it and certainly nothing anyone can do with it.

But the kicker is, instead of this meaning that it is completely valueless, I believe what Freeman Dyson is quoted as saying in that article with "the 11-cell might turn out to be important."

And where I believe it will be important is not in ways (cant type t e r ms on here) of practical applications necessarily or even in ways of science and math (although this may be true), I believe it is more that its discovery and illustration bring to light a reminder yet again that what we perceive to be the limitations of our knowledge and experience always need to be expanded and are currently very narrow.

In essence, it is a symbol (at least for me) of completely unlimited possibilities. The scientist in me is not happy about that, but the artist in me loves it.
The avatar is an 11-cell (hendecachoron) (Computer model courtesy of Carlo Sequin, UC Berkeley, styled by Jaron Lanier)
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