I'm looking for "nice" tessellations of $S^3$ into as many pieces as possible. Another way to think about this problem is looking for "nice" 4-polytopes with as many faces as possible, since we can project those faces onto $S^3$ to get a tessellation.
One clear candidate is the 600-cell, since it is the regular 4-polytope with the greatest number of cells. I am curious whether we can find something with more cells that is about as nice. Tetrahedral cells is a big plus.
Here's one thing I thought of trying: take each tetrahedral cell of the 600-cell, then chop off all four corners cutting through the midpoints of the edges. This gives four tetrahedra plus an octahedron. We can then cut the octahedron in half twice to get four more tetrahedra. It turns out that all eight of the tetrahedra we obtain this way have the same volume, although they are different shapes. A major disadvantage of this construction is that there is asymmetry depending on which ways we cut the octahedron in half, but the advantage is that we tessellate $S^3$ into 600 * 8 = 4800 tetrahedral cells of equal volume.
What are some other polytopes I should check out? Also, anyone who is interested in this subject should check out this project that helps visualize 4-polytopes by projecting them onto $S^3$, as I'm talking about here.
Within 3D there is the infinite set of prisms with regular bases. These somehow have the disadvantage that the 2 orthogonal components (base vs. height) are dimensionally different. But within 4D this is no longer so. There you have the Clifford symmetry allowing to have 2 fully perpendicular planes. Thus you can use the prism product of any 2 perpendicular regular polygons, thus resulting in the (n,m)-duoprism (= $\{n\}\times\{m\})$ with $n\cdot m$ vertices, … , and $n$ m-prisms plus $m$ n-prisms for cells.
You even could make this still more symmetric by choosing $n=m$. Then you will have a convex polychoron with $n^2$ vertices, …, and $2n$ facets. That one thus clearly could beat your 600-cell by far. In fact it would beat any numerical bound simply by increasing that $n$ accordingly!
But even when sticking to Wythoffian polytopes from $irreducible$ symmetry groups, the omnitruncated member of the group of the 600-cell clearly exceeds these 600 too: That one has a total cell count of 600 truncated octahedra + 1200 hexagonal prisms + 720 decagonal prisms + 120 omnitruncated icosahedra.
Here the relation to the mentioned 600-cell is easily seen: the truncated octahedra are situated at the body centers of the former tetrahedra, the hexagonal prisms are situated at the face centers of the former triangles, the decagonal prisms are situated perpendicular to the former edges, and the omnitruncated icosahedra are situated at the former vertices.
--- rk