4D parallelotopes, and their coordinate systems

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

4D parallelotopes, and their coordinate systems

Postby trico » Mon Feb 10, 2020 3:31 am

Parallelotopes can be used well for Euclidean coordinate systems: They tile space, each pair of opposite facets can be associated with an axis, and specific sets of those axes can be used as a basis for a coordinate system, where integer coordinates can be mapped uniquely to any parallelotope cell in the grid.

In 2D, we have the parallelogons:
  • The square leads to the square tiling {4,4}. The 2 axes can be used directly for the Cartesian 2D coordinate system.
  • The hexagon leads to the hexagonal tiling {6,3}. Any 2 of the 3 axes can be used as a basis for a coordinate system.

In 3D, we have the parallelohedra:
  • The cube {4,3} leads to the cubic honeycomb {4,3,4}. The 3 axes can be used directly for the Cartesian 3D coordinate system.
  • The hexagonal prism (6,4,4) leads to the hexagonal prismatic honeycomb. The hexagon axis (i.e. the one orthogonal to the hexagon faces) can be used with any 2 of the 3 square axes for a coordinate system.
  • The truncated octahedron (6,6,4) leads to the bitruncated cubic honeycomb 2t{4,3,4}. Any set of 3 hexagon axes can be used for a coordinate system, as well as any set of 1 hexagon axis and 2 square axes. There's also the option to use 2 hexagon axes, then pick a square that is not directly between 2 hexagons from those axes, and use the axis orthogonal to this square, too.
  • The rhombic dodecahedron leads to the rhombic dodecahedral honeycomb. The 3 axes associated with the faces around an order-3 vertex can be used for a coordinate system. Another option is to pick 2 axes on a hexagonal equator, and add the axis from a face pair that is not on this equator.
  • The elongated dodecahedron basically leads to the same coordinate system as the one from the rhombic dodecahedra. The only difference is that one axis is stretched (assuming maximum symmetry), and there's a difference which cells share an edge or a vertex.


Now, 4D is a lot more complicated, and there is a total of 52 parallelotopes. So I was wondering how much this number can be reduced by removing parallelotopes that lead to mostly redundant coordinate systems (like the grids for the elongated dodecahedron and the rhombic dodecahedron in 3D).

For a start, we can take the parallelohedra, expand them along a 4th axis that is orthogonal to the original 3D space. If we do this with the cube, hexagonal prism, truncated octahedron, and rhombic dodecahedron, we get 4 parallelotopes that can be used for unique 4D coordinate systems:
  • Expanding the cube gives us the tesseract, which can be used for a 4D Cartesian coordinate system.
  • Expanding the hexagonal prism results in a parallelotope with 4 hexagonal prisms and 6 cubes.
  • Expanding the truncated octahedron results in a parallelotope with 2 truncated octahedra, 8 hexagonal prisms, and 6 squares.
  • Expanding the rhombic dodecahedron results in a parallelotope with 2 rhombic dodecahedra and 12 parallelepipeds.
For the coordinate systems, we can simply use the 3D coordinate systems as a basis, and add a 4th coordinate.
We could also do the same with the elongated dodecahedron, but remove the resulting parallelotope due to a (mostly) redundant grid.

To get another (parallelotope-based?) coordinate system, it should be possible to take the coordinates from two hex grids, and combine both for the coordinates of a 4D grid (think of each hex grid as representing a plane, and both planes only intersect in the origin of the coordinate system). I wouldn't be surprised if the Voronoi diagram from all points with integer coordinates in this coordinate system would be a parallelotope tiling, but my intuition about 4-dimensional space is pretty bad. :sweatdrop:

Is there a simple way to get a complete list of 4D parallelotopes with unique coordinate systems, i.e. with redundant parallelotopes removed?
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Re: 4D parallelotopes, and their coordinate systems

Postby wendy » Tue Feb 11, 2020 5:03 am

Welcome.

One of the interesting things you can do with coordinate systems, is to use the 'matrix-dot' to find the dot-product or angles between vectors. It is widely but unknownely used here.

If \(v_i\) is a set of vectors, and \(z_iv_i\) is a span of vectors, then there is a matrix \(M_{ij} = [v_i\cdot v_j]\) such that the dot product of two vectors \(x\) and \(y\) is \( M_{ij} x_i y_j\).

The common use here is that the vectors run from the centre of polytopes to their vertex, when the edge is taken as 2, and the polytopes are arranged in the same symmetry. From this, all polytopes derived by mirror-edge, effectively correspond to 'cartesian coordinates' in that system, and the matrix-dot of the vertex-vector is its circumradius. Likewise, the matrix-dot of the difference of vertices indicate the base of a triangle, the height being that of the matching segmentatope (or lace prism).
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the dream we dream together is reality.

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Re: 4D parallelotopes, and their coordinate systems

Postby trico » Sat Feb 15, 2020 4:14 pm

Okay, I asked for help to remove "redundant" 4D parallelotopes. But it turns out that I have more trouble figuring out what the 52 4D parallelotopes are than I thought.

The (zonotopal) ones I found so far by combining lower-dimensional parallelotopes:
  • tesseract (Cartesian product of 4 line segments)
  • 4-6 duoprism (Cartesian product of 1 hexagon and 2 line segments)
  • 6-6 duoprism (Cartesian product of 2 hexagons)
  • truncated octahedral prism (Cartesian product of 1 truncated octahedron and 1 line segment)
  • rhombic dodecahedral prism (Cartesian product of 1 rhombic dodecahedron and 1 line segment)
    rhombic_dodecahedral_prism_net.png
    (12.38 KiB) Not downloaded yet
  • elongated dodecahedral prism (Cartesian product of 1 elongated dodecahedron and 1 line segment)
The only other zonotopal 4D parallelotope that I know of is the omnitruncated 5-cell, aka the order-5 permutohedron.
And the only non-zonotopal 4D parallelotope that I know of is the 24-cell.

I found a link to a paper on the wikipedia article for parallelohedra (Once more about the 52 four-dimensional parallelotopes). But there are a few concepts which I'm not familiar with, which seem to be necessary to figure out what each listed parallelotope is: Root systems and Dynkin diagrams, and matroids based on graphs. And Minkowski sums are mentioned as a way to construct the non-zonohedral parallelotopes.
I saw a pinned thread in this subforum (Notions and Notations.) where the Dynkin notation is explained, I guess this would be a good place to start...
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Re: 4D parallelotopes, and their coordinate systems

Postby trico » Thu Apr 16, 2020 7:40 pm

I found a non-zonotopal parallelotope other than the 24-cell: It can be derived from a tesseract, where 2 polar opposite cubes are fixed, and the 6 other cubes are cut at the equator plane between those 2 fixed cubes. Then each of the 12 resulting square frusta is turned into a square bifrustum. And for each edge from a pole cube, and the corresponding closest edge from the other pole cube, an orthobifastigium (like the gyrobifastigium Johnson solid, but non-rotated), is added between those edges, but doesn't contain those edges.

Coordinates are specified depending on a parameter 0 < a < 1:
(a=0 results in a degenerate cube, and a=1 results in a degenerate 24-cell)

Vertices:
  • pole cube vertices: (+-2a,+-(1-a²),+-(1-a²),+-(1-a²))
  • equator plane vertices: y/z/w permutations of (0,+-(1+a²),+-(1-a²),+-(1-a²))
  • remaining orthobifastigium vertices: (+-a,+-1,+-1,+-1)

Cell centers:
  • pole cubes: (+-2a,0,0,0)
  • square bifrusta: y/z/w permutations of (+-a,+-1,0,0)
  • orthobifastigia: y/z/w permutations of (0,+-1,+-1,0)

Grid coordinates:
2(xa,y,z,w), where y,z,w are integer, and x has the same parity as y+z+w.

a-values for which 2 edge types have the same length:
  • a = sqrt2-1
  • a = 1/2
  • a = sqrt((9 - sqrt65)/2)

Other interesting a-values, which affect the angles between the cell centers (2a,0,0,0), (a,1,0,0), (-a,1,0,0):
  • a = 1/sqrt2: angles correspond to the ones between the face centers (2,0,0), (1,1,1), (-1,1,1) on a truncated octahedron
  • a = 1/sqrt3: the cell centers lie on a regular hexagon
  • a = 1/sqrt5: angles correspond to the ones between the cell centers (4,-1,-1,-1,-1,), (2,2,2,-3,-3), (2,-3,-3,2,2) on an omnitruncated 5-cell
    (coordinates are 5D, but specify a 4D subspace)

I like a = 1/sqrt3 for the simple angles. I also like a = 1/sqrt5, since the grid *should* be a subgrid of the one for omnitruncated 5-cells. I also want to explore the a = 1/sqrt2 option, and see if there is an interesting relation to the truncated octahedron in 3-space.


I hope I didn't make any mistakes. And I hope that I can figure out how to generate other non-zonotopal parallelotopes like this one...

P.S.: Atm. I can't work well at the PC for technical reasons. Apologies if it takes longer for me to respond...
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Re: 4D parallelotopes, and their coordinate systems

Postby trico » Thu Apr 23, 2020 3:24 pm

I'm learning more and more about 4-space and 4-parallelotopes, but the biggest barrier for me atm. seems to be the mathematical language barrier. It'd help me a lot if someone could confirm whether I understand the following correctly :nod:, or not. :glare:

I'm working with this article to try to identify as many 4-parallelotopes as possible, and see whether there are "redundant" ones which have nearly the same grid (similar to the rhombic and elongated dodecahedron in 3D):
Once more about the 52 four-dimensional parallelotopes

Table 2 on page 13 (chapter 6) is very useful: It lists each non-zonotopal parallelotope as a Minkowski sum of the 24-cell P_V(D4) with a zonotopal parallelotope Z(U).
Note that the zonotopes with dimension 3 or less are specified above the table.

From what I understand, the roots for each zonotope are specified as facet vectors of the 24-cell, which can be put into 3 tesseractic subgroups of mutually orthogonal vectors (named 'quadruple' in the article):
[list=][*]quadruple #1: 12⁺ = +-(1,1,0,0), 12⁻ = +-(1,-1,0,0), 34⁺ = +-(0,0,1,1), 34⁻ = +-(0,0,1,-1)
[*]quadruple #2: 13⁺ = +-(1,0,1,0), 13⁻ = +-(1,0,-1,0), 24⁺ = +-(0,1,0,1), 24⁻ = +-(0,1,0,-1)
[*]quadruple #3: 14⁺ = +-(1,0,0,1), 14⁻ = +-(1,0,0,-1), 23⁺ = +-(0,1,1,0), 23⁻ = +-(0,1,-1,0)[/list]
Multiple vectors from the same quadruple are enclosed in () in table 2, and from what I understand, the resulting non-zonotopal parallelotope has 24 facets + 2 * (number of represented quadruples).
(which is a very important piece of information for me, since I only consider parallelotopes "redundant" if they have the same number of facets, and topologically the same grid if adjacency is defined through shared facets)

Now, I think those roots are a list of edges, correct? And only a small number of edges required to define the polytope's symmetry is listed, since an (irregular) cube has 12 edges, and not just 3? Or are these intended to be facet vectors that define the zonotope?

Is it difficult to calculate the Minkowski sums with the 24-cell? Is there a good tutorial for Minkowski sums? Also, about the parallelotope that I found in my previous post: What is the generating zonotope? I can imagine it as a modification of the 24-cell with either a line segment or a cube, but these are just guesses.
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