# What are some nice, high resolution tessellations of $S^3$?

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.

• The dual of the omnitruncated 600-cell has 14400 tetrahedral cells. – M. Winter Mar 21 '20 at 17:32
• I like this and I'm trying to learn more about it. Unfortunately I think it has a major drawback that while each cell is the same, the edges of a given cell have very different lengths, and some vertices have far more cells packed around than others. – A. Kriegman Apr 1 '20 at 18:18
• You will have to make some tradeoff: do you want many cells, equal edge lengths and similar vertex degrees, high degree of symmetry, only tetrahedral cells? There are examples for any of these, but not for all at the same time. All of these are probably some modifications of the 600-cell. Except if you drop symmetry, then you can get arbitrary many cells, but these polytopes do not have specific names but are just further subdivisions of the ones with names. – M. Winter Apr 1 '20 at 18:27

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.
• If we blow up that $(n,m)$-duoprism onto $S^3$ then all the vertices will land on a taurus, meaning large swathes of $S^3$ both inside and outside that taurus will have no resolution. (Note that the inside and outside of that taurus have the same shape.) Keep in mind that I'm looking for a "nice" tessellation with preferably tetrahedral cells, or at least low vertex cells. – A. Kriegman Mar 31 '20 at 16:01