I took a look at the dual polytope to \(\omega^{xx,e}_{6}\) - i.e., \(\omega^{xx,e*}_{6}\). The dual does help to understand the polytope's structure, I think.
To get the dual polytope, I derived the polytope having facet-vertex incidence matrix = the transpose of the facet-vertex incidence matrix of \(\omega^{xx,e}_{6}\). So, its vertices correspond to the facets (and facets to the vertices of) \(\omega^{xx,e}_{6}\).
The picture below is not the flux capacitor from BTTF, but a diagram of \(\omega^{xx,e*}_{6}\) that I call a tetra-oct. The diagram is a tetrahedron, but instead of edges, there are octahedra. (I made the octahedra oblong & used a Schlegel perspective to make it easier to draw.) This diagram shows all 28 vertices, 72 of the 240 edges, and all 6 of the octahedron faces of \(\omega^{xx,e*}_{6}\).
\(\omega^{xx,e*}_{6}\) is 9-dimensional, it has four 6-d tri-oct faces, and six (3-d) octahedron faces! The octahedra are regular-ish, with parallel antipodal edges.
The four (0-d) corners of the tetrahedron correspond to the four \(\omega^{xx,e}_{6}\) (8-d) facets which are three parallelograms in general position. The six (3-d) octahedra correspond to the six \(\omega^{xx,e}_{6}\) (5-d) faces which are two parallelograms in general position. The four 6-d tri-oct faces correspond to the four 2-d parallelogram faces of \(\omega^{xx,e}_{6}\). It appears \(\omega^{xx,e*}_{6}\) is equifacetted (all its facets have the same combinatorial type); intuitively appealing as the 16 facets correspond to the even permutations among the vertices of \(\Omega^{xx}_{6}\).
#octahedron