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I am not a geometer, so I might be misusing some terms. So let me try to be more explicit regarding what I mean.

"Subdivide a sphere into cells" means to partition the set of all points on a sphere into finitely many non-overlapping (except at the boundary), contiguous regions (I'll call them cells). Note, that this doesn't exclude regions with arbitrarily complex boundaries (even infinite ones).

"Uniformly" means that all cells in the subdivision should be the same shape and size. The shape doesn't need to be in any way regular or symmetrical, but you are not allowed to flip/mirror the shape, only translate and rotate it.

"Arbitrarily small" means that for any given $\varepsilon > 0$, you should be able to generate such a subdivision, so that each cell fits into a ball of radius $\varepsilon$.


Some examples:

You can split a sphere into 8 equal "triangle" cells with angles 90-90-90 by halving it 3 times. But further subdividing those "triangles" is tricky (for example, you can't use edgewise subdivision to further refine them, because you will end up with "triangles" with different angles).

You can split a sphere into an arbitrary amount of equal "orange slices", but they won't be "arbitrarily small" because they only get "thinner", not "shorter".

Spherical polyhedra or spherical tilings satisfy the "arbitrarily small" requirement, but there are only finitely many uniform polyhedra, so this seems like a dead end. On the other hand, uniform spherical tilings assume that the cells themselves are also polyhedral, that they are vertex-transitive and so on. So it's not clear if relaxing these assumption allows for more tilings.


My intuition is telling me, that the answer to the question in the title is "No" (such a subdivision is not possible), but I wasn't able to come up with a proof. Thoughts?

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    $\begingroup$ en.wikipedia.org/wiki/… should help $\endgroup$
    – student91
    Commented Mar 2, 2023 at 6:44
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    $\begingroup$ In particular, when you do not do your orange slices, you have to end up with one of the other symmetry groups, and there's only finitely many such symmetry groups so no "new" symmetry groups to go even smaller. $\endgroup$
    – student91
    Commented Mar 2, 2023 at 6:49
  • $\begingroup$ @student91 I might be a bit sleep deprived, but it's not immediately obvious, why do different subdivisions have to "correspond" to different symmetry groups. At first glance it seems to me, that these symmetry groups correspond to the spherical polyhedra, not arbitrary subdivisions. $\endgroup$
    – RuRo
    Commented Mar 2, 2023 at 7:47
  • $\begingroup$ If I would write that out it would be more of an answer than a comment, I don't have time right now to write a full answer (feel free to do so yourself). You can look at $\{f\in SO(3)\colon\forall\mathrm{region},\ f(\mathrm{region})=\text{other region}\}$, which is a finite subgroup of $SO(3)$ (bc. it is strict subgroup of all permutations of regions, so finite), which therefore should be one of the groups on that list. $\endgroup$
    – student91
    Commented Mar 2, 2023 at 8:27
  • $\begingroup$ @student91 yes, I understand that part. Let's call this subgroup $s(D)$ for each subdivision $D$. What I don't understand is why we can't have $\|D_1\| < \|D_2\|$ even if $s(D_1) = s(D_2)$. Why does increasing the number of regions automatically entail changing the symmetry. You can subdivide a region of the plane aperiodically with something like the sphinx tiling, so why can't you subdivide a region of the sphere aperiodically? P.S. you don't have to answer if you don't have the time. $\endgroup$
    – RuRo
    Commented Mar 2, 2023 at 11:57

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I don't think this is possible either. I offer a very handwavy argument showing why.

You are trying find a monohedral tiling of the surface of the sphere. I think it is safe to assume w.l.o.g. that all the edges are straightened out, so that the tile shape is a spherical polygon, say an $n$-gon.

In the flat plane the angles of a triangle add to $\pi$ radians, but on a spherical triangle its angles add up to more than that. In general, the smaller the spherical triangle's area (relative to sphere), the closer its angle sum approaches $\pi$. Basically, smaller triangles are less curved. The same thing happens with spherical $n$-gons.

The problem is when you are tiling the spherical surface, at any point where three or more tiles meet the angles will have to add up to exactly $2\pi$. The surplus angle has to go somewhere. If you impose a fixed limit on how many tiles meet at a vertex, then with an $n$-gon there there is a fixed limit on how many ways you can combine its vertex angles to $2\pi$. This in turn means that there is a limit on how many surpluses it can cope with, i.e. how many tiles you can fit together before it has to enclose the sphere. In other words, as you increase the total number of tiles, you will also have to increase the number of tiles that meet at some of the vertices.

So the tile will have to become thinner and thinner in order to fit around such a vertex. It seems that the tile's length needs to be at least about one sixth of the sphere's circumference, but I can't quite find a good argument for that. Here is a nice example:
Triangle tiling
I found this in the paper "Some New Tilings of the Sphere with Congruent Triangles" by Robert J. MacG. Dawson

Note that there are some interesting variations of the orange-slice dissection. If you have an even number of slices you can rotate the east/west hemispheres by any amount to break up most of the symmetry. Or you can halve the pieces, to make a spherical bipyramid. You can then offset the north/south hemispheres and tweak the edges along the equator to make a spherical trapezohedron or anything in between. With a spherical bipyramid that has an order that is a multiple of 4, then you can turn any octant by a third of a turn to make it fully asymmetric.

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