24-cell construction, from parallelepipeds?

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

24-cell construction, from parallelepipeds?

Postby ae » Fri Aug 11, 2006 8:43 pm

The way the vertices of 24-cell are two sets of axis-aligned vertices
which are permutations of (\pm1,0,0,0) and diagonal vertices of
(\pm 1/2, \pm 1/2, \pm 1/2, \pm 1/2) is very similar to the structure
of rhombic dodecahedron in 3D. The vertices of rhombic dodecahedron
are at permutations of (\pm 1, 0, 0) and (\pm 1/2, \pm 1/2, \pm 1/2).

One way of constructing a rhombic dodecahedron is by putting together
four parallelepipeds:

http://www.cs.sfu.ca/gruvi/Reza/researc ... rd_dec.pdf


I am wondering if there is any way of constructing a 24-cell with some finite
number of four dimensional parallelepipeds.

Any ideas?![/code]
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Re: 24-cell construction, from parallelepipeds?

Postby quickfur » Fri Aug 11, 2006 10:19 pm

ae wrote:The way the vertices of 24-cell are two sets of axis-aligned vertices which are permutations of (\pm1,0,0,0) and diagonal vertices of (\pm 1/2, \pm 1/2, \pm 1/2, \pm 1/2) is very similar to the structure of rhombic dodecahedron in 3D. The vertices of rhombic dodecahedron are at permutations of (\pm 1, 0, 0) and (\pm 1/2, \pm 1/2, \pm 1/2).

Correct. In fact, the 24-cell is in many ways the 4D analogue of the rhombic dodecahedron (see below). It is a happy coincidence that in 4D, it also happens to have regular facets.

One way of constructing a rhombic dodecahedron is by putting together four parallelepipeds:

http://www.cs.sfu.ca/gruvi/Reza/researc ... rd_dec.pdf

Correct. It might interest you to know that there are exactly two ways of doing this. This actually corresponds with the vertex-first projection of a 4D cube into 3D: four of the cells project onto one possible arrangement of the four parallelopipeds, and the other four cells project onto the other possible arrangement.

Now, the other way of constructing a rhombic dodecahedron is to take two 3D cubes, cut one of them into 6 square pyramids, and stick the pyramids onto the faces of the other cube. If you do the analogous thing in 4D: cut a 4D cube into 8 cubical pyramids and stick them onto the facets of another 4D cube, you end up with the 24-cell.

I am wondering if there is any way of constructing a 24-cell with some finite number of four dimensional parallelepipeds.

Any ideas?![/code]

I'm not familiar enough with 5D to be able to answer this either way, but my intuition suggests that this doesn't work. The main reason is that in 3D, the faces of the rhombic dodecahedron are trapezoids---distorted squares, if you will, which correspond with projections of the square ridges of a 4D cube. Unfortunately, in 4D, the 24-cell has octahedral facets, which are not parallelopipeds, and so cannot correspond with projected 5D cube ridges.

I know Wendy has mentioned before that in 4D and above, there are actually (at least) 2 possible analogues of the rhombic dodecahedron: one is shape resulting from cutting an n-cube into pyramids and attaching them to another n-cube, and the other is the maximal projection of an (n+1)-cube into n-space. I'm not sure what the second shape in 4D is called, though. Its cells would have to be parallelopipeds, so it would be neither regular nor uniform (Archimedean). Maybe Wendy can help us with this one. :-)
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Re: 24-cell construction, from parallelepipeds?

Postby ae » Fri Aug 11, 2006 10:48 pm

I know Wendy has mentioned before that in 4D and above, there are actually (at least) 2 possible analogues of the rhombic dodecahedron: one is shape resulting from cutting an n-cube into pyramids and attaching them to another n-cube, and the other is the maximal projection of an (n+1)-cube into n-space. I'm not sure what the second shape in 4D is called, though. Its cells would have to be parallelopipeds, so it would be neither regular nor uniform (Archimedean). Maybe Wendy can help us with this one. :-)


Right, this maximal projection of 5D cube onto 4D would have 32-2 = 30 vertices,
and I am also having a difficulty finding a 4D polytope with 30 vertices.
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Postby wendy » Sat Aug 12, 2006 8:01 am

The m-notation is used to construct the dual of the corresponding x notation.

The two kinds of 4D rhombic dodecahedra, are m3o[3o]3m and o3m[3o]4o. In general, one can replace [3o] by any repeation of this eg [3o3o3o3o3o] to get the corresponding m+3-dimensional version. When m=0, one gets the figures m3o3m = o3m4o = rhombic dodecahedron.

In higher dimensions, there are two distinct figures, as follows:

o3m[3o]4o. This produces the "double-cube", formed by placing a pyramid on each face of the measure polytope. In 3d, 4d, and 5d, this leads to the most efficient packing of spheres. This figure can be thought of as adding to a unit-edge cube, a cross-polytope (octahedron, 16ch), of edge sqrt(2). The vertices of this fall at the centres of adjacent cubes in the cubic tiling.

m3o[3o]3o. This produces a rather interesting figure with in general, 2^(n+1)-2 vertices, and n(n+1) faces. The thing also tiles, and is also has equal margin-angle (of 120 degrees).

You can make this figure by taking (n+1) rhombic prisms, such that when the obtuse vertex is at the centre of a simplex (eg pentachoron), the body of the thing fills all space. You can also make this, by projecting a (n+1) cube down the long axis. (n+1) faces remain visible, and gives the construction shown above.

If one considers the tiling of cubes, and restricts the condition that no part of the cube shall pass w+x+y+z... = (say) 30, when this tiling is projected downwards, you get each cube becoming an m3o[3o]3m.

If you just leave the exposed faces, the thing behaves as a cats-eye mirror, because light falling on it from any angle shall be reflected back to where it comes from. This tiling is designated b3b3o[3o]3o3z, and is a tiling of parallopieds that has no "through-planes" (ie is not a slice-and-dice job).

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Postby wendy » Sat Aug 12, 2006 8:05 am

These figures all have 30 vertices in 4d.

m3o3o3m has 20 faces and 30 vertices
o3x3o3x has 30 vertices.
o3x3x3o has 10 alike faces and 30 vertices

Only the first is a zonohedron (ie can be made of rhombs), the other two feature triangles. This is also why the 24-choron can not be built of rhombs: all hedra need to be even-edged and opposite edges parallel for it to be a zonotope.
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