The 57-cell in 5D or 4D

Question

The Hemi-dodecahedron can be nicely represented in five dimensions. Here are the six faces, where all edges have length 2.

{{{0,0,0,1,1},{1,1,0,0,0},{0,0,1,1,0},{1,0,0,0,1},{0,1,1,0,0}},
{{1,1,0,0,0},{0,0,1,1,0},{0,1,0,0,1},{1,0,0,1,0},{0,0,1,0,1}},
{{1,0,1,0,0},{0,1,0,1,0},{1,0,0,0,1},{0,0,1,1,0},{0,1,0,0,1}},
{{0,1,1,0,0},{1,0,0,1,0},{0,0,1,0,1},{0,1,0,1,0},{1,0,0,0,1}},
{{0,0,0,1,1},{0,1,1,0,0},{1,0,0,1,0},{0,1,0,0,1},{1,0,1,0,0}},
{{0,0,0,1,1},{1,1,0,0,0},{0,0,1,0,1},{0,1,0,1,0},{1,0,1,0,0}}}

Is there a nice set of coordinates in 4D or 5D (or higher) for the 57-cell? Could the above hemi-dodecahedron be used as one of the cells?

EDIT: In 4D, these coordinates can be used for the six faces of the hemi-dodecahedron / Petersen graph. All edges have length 1, and the angle between edges is always 3/4. But I'm still unclear on how to glue these cells together.

$$( ((\sqrt{5},1,1,1),(0,0,-2,0),(\sqrt{5},-1,1,-1),(0,0,0,2),(0,0,0,-2)), ((\sqrt{5},1,1,1),(0,-2,0,0),(\sqrt{5},1,-1,-1),(0,0,2,0),(0,0,-2,0)), ((\sqrt{5},1,-1,-1),(0,0,2,0),(\sqrt{5},-1,-1,1),(0,0,0,-2),(0,0,0,2)), ((\sqrt{5},-1,1,-1),(0,0,0,2),(\sqrt{5},1,-1,-1),(0,-2,0,0),(0,2,0,0)), ((\sqrt{5},-1,-1,1),(0,0,0,-2),(\sqrt{5},1,1,1),(0,-2,0,0),(0,2,0,0)), ((\sqrt{5},-1,-1,1),(0,2,0,0),(\sqrt{5},-1,1,-1),(0,0,-2,0),(0,0,2,0)))/4$$

Answer 1

(This is a bit of a reminder to OP of something we discussed in personal correspondence a few years ago, in a slightly different context.)

Let $A$ be the adjacency matrix of a combinatorial graph. A matrix whose row-vectors form an orthogonal basis of an eigenspace of $A$ has column-vectors that serve as coordinate vectors of the vertices of an harmonious geometric realization.

"Harmonious" here means that the combinatorial automorphisms of the graph induce isometries of the realization. (Oftimes, the realizations have coincident vertices, so the isometry group is usually a subgroup of the automorphism group.) By eigen-birthright, these realizations also have this property: (with the origin at the vertex centroid) moving each vertex to the sum of its neighbor's coordinate vectors gives the same result as scaling the figure by the corresponding eigenvalue.

The skeleton of the $57$-cell is the PerkelGraph, which MathWorld tells me has this spectrum: $$(-3)^{20}\;\left(\frac{1}{2}(3-\sqrt{5})\right)^{18}\;\left(\frac12(3+\sqrt{5})\right)^{18}\;6^1$$ Thus, there exists an harmonious $20$-dimensional realization, two $18$-dimensional realizations, and a $1$-dimensional realization. (This last has all of its vertices coinciding at a point.)

This is the brute-force recipe I use to generate coordinate matrices for these figures in Mathematica:

adjMat = GraphData["PerkelGraph", "AdjacencyMatrix"];

(* Get eigenvalues and eigenvectors.(Be prepared to wait!) *)

{valList, vecList} = Eigensystem[adjMat];

(* Sort values by eigendimension,and remove duplicates *)

vals = Union[Sort[valList, Count[valList, #1] < Count[valList, #2] &]];

(* Orthogonalize all eigenspaces at once.(Another lengthy process!) *)

vecList = Orthogonalize[N[vecList]];

(* Build eigenmatrices from rows corresponding to each eigenvalue *)

matList = Map[vecList[[Flatten[Position[valList, #1]]]] &, vals];

The resulting matList contains the List of matrices satisfying the description above: rows form an orthogonal basis of the eigenspace, so columns are coordinate vectors of harmonious vertices.

Unfortunately, this recipe rarely generates arithmetically "nice" coordinates. For instance, applying it to the adjacency matrix of the hemi-dodecahedron gave (a scalar multiple of) this $5$-dimensional coordinate matrix:

$$\left[ \begin{array}{cccccccccc} -\sqrt{30} & -\sqrt{30} & \sqrt{30} & -\sqrt{30} & \sqrt{30} & 0 & 0 & 0 & 0 & \sqrt{30} \\ -\sqrt{30} & \sqrt{30} & -\sqrt{30} & 0 & \sqrt{30} & -\sqrt{30} & 0 & 0 & \sqrt{30} & 0 \\ \sqrt{10} & \sqrt{10} & \sqrt{10} & -\sqrt{10} & \sqrt{10} & -2 \sqrt{10} & 0 & 2 \sqrt{10} & -2 \sqrt{10} & -\sqrt{10} \\ \sqrt{2} & \sqrt{2} & \sqrt{2} & -4 \sqrt{2} & \sqrt{2} & \sqrt{2} & 6 \sqrt{2} & -4 \sqrt{2} & \sqrt{2} & -4 \sqrt{2} \end{array} \right]$$

from this adjacency matrix (which I hope is correct):

$$\left[\begin{array}{cccccccccc} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{array}\right]$$

A far cry from the coordinates given in the question. (The jumble of numbers that result from running the recipe on the Perkel Graph itself are likewise unsightly and unhelpful, so there's no point including them here.)

The "problem" is that the recipe is at the mercy of Mathematica's eigenanalysis routines. While those routines are deterministic enough to give the same results over and over again, they don't seem to be geared toward generating the "nice" results in this context.

It may be possible to guide Mathematica through some kind of ad hoc process to produce "nice(r)" coordinates, maybe leveraging the $1$s-and-$0$s nature of the adjacency matrix, and/or the effects induced by the graph's automorphism group. I believe this pursuit ventures into Representation Theory, which isn't my forté. Nevertheless, given your own fluency in Mathematica, you might be able to make some useful refinements to the process.

(Someone should pose the problem directly to Wolfram Research and let them provide a NiceCoordinates[] function for us! :)