Tessellating "steradian"?

Question

A solid angle is called a steradian, which is essentially a cone with origin at the centre of the sphere.

One neat thing about angles in 2D is that they perfectly tessellate the circle. i.e. given an angle $\theta = 2\pi / n$ you can cut the circle into $n$ identical pizza slices.

Steradians do not tessellate the surface of the sphere the same way, it is not possible to fully cover the surface without intersections.

I am wondering if there is a mathematical object that can tessellate the surface of a sphere (e.g. as a series of spherical quads) such that if the set of slices is called $S = \{s_i\}$ then $\cap\{s_i\} = edges$ and $\cup\{s_i\} = S^2$ the unit sphere.

An additional property of the cut pieces should be that they all have the same surface area.

Answer 1

Yes. Take any regular polyhedron with all of its vertices on the sphere. Join them up according to the edges of the polyhedron. The pieces that you've cut the sphere into satisfy your condition (or rather, they fail to satisfy it in the same way that angles in 2D fail it - the intersections are the edges, rather than the empty set).

Note, in particular, that this exact construction also gives your 2D version - your tessellating angles are precisely the result of taking a regular polygon with its vertices on the circle and cutting appropriately. However, in 3 or higher dimensions, there are only finitely many such (as there are only finitely many regular n-polytopes for n > 2).

Indeed, these are the only solutions: if you had some other solution, then we could construct a regular polytope by connecting the points at which three or more components meet with straight lines, which would then give the same decomposition when passed through the above procedure.

Answer 2

The regular polyhedra in another answer can be subdivided. Think of a cube projected onto a sphere. Draw the diagonals and medians of each square face dividing the face into eight identical pieces. Done to the entire cube this gives 48 identical pieces. A regular octahedron to which a similar process is applied gives the same 48-piece division. A regular dodecahedron or icosahedron gives 120, the maximum possible number by this method; a regular tetrahedron gives 24.

Answer 3

An elegant approach to generating solutions of congruent pieces is by the use of kaleidoscopes. See for example Ball and Coxeter; Mathematical Recreations and Essays, Chapter V on Polyhedra.

A solution set not described in the above reference is the set of regular hosohedra. Here the pieces are "lunes" like slices of melon rind, and the set corresponds to the set of polygons.

If "the same surface area" allows pieces of different shapes, then the problem becomes much more complicated.