John Carlos Baez on Nostr: Hmm, I think Wikipedia's article on the 600-cell explains how to get *one* of these ...

Hmm, I think Wikipedia's article on the 600-cell explains how to get *one* of these 10×10 sheets. If so, the question is just how to partition the 600 tetrahedra into 6 of these 10×10 sheets:

"The 120-cell can be decomposed into two disjoint tori. Since it is the dual of the 600-cell, this same dual tori structure exists in the 600-cell, although it is somewhat more complex. The 10-cell geodesic path in the 120-cell corresponds to the 10-vertex decagon path in the 600-cell.

Start by assembling five tetrahedra around a common edge. This structure looks somewhat like an angular "flying saucer". Stack ten of these, vertex to vertex, "pancake" style. Fill in the annular ring between each pair of "flying saucers" with 10 tetrahedra to form an icosahedron. You can view this as five vertex-stacked icosahedral pyramids, with the five extra annular ring gaps also filled in. The surface is the same as that of ten stacked pentagonal antiprisms: a triangular-faced column with a pentagonal cross-section. Bent into a columnar ring this is a torus consisting of 150 cells, ten edges long, with 100 exposed triangular faces, 150 exposed edges, and 50 exposed vertices. Stack another tetrahedron on each exposed face. This will give you a somewhat bumpy torus of 250 cells with 50 raised vertices, 50 valley vertices, and 100 valley edges. The valleys are 10 edge long closed paths and correspond to other instances of the 10-vertex decagon path mentioned above (great circle decagons). These decagons spiral around the center core decagon, but mathematically they are all equivalent (they all lie in central planes)."

It would take me some time to absorb this. But to continue....

(2/n)