Grid on a sphere

For which n can we divide the surface of a sphere into n equal parts, such that each part has k neighboring parts, and the total grid has k-fold rotational symmetry around an axis from any of the parts center to the center of the sphere? And no part can be distinguished from another other then fixing a coordinate system.

Is it the same as for the plane ?

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How's the center of a part defined? – anon Oct 30 '11 at 7:22
The point around which each part has k-fold rotational symmetry, if we rotate it without intersecting the tangent plane of the sphere at this point – user1708 Oct 30 '11 at 7:25
If you relax the constraints a bit and only require a transitive symmetry group, then you get more than the Platonic solids. For example the 3D finite reflections groups give rise to subdivisions of the sphere into 48 or 120 congruent parts (with the possibility of pairing up adjacent parts to halve their number). – Jyrki Lahtonen Oct 30 '11 at 8:41

Your choices are either $n\ge 2$ identical wedges each stretching from north pole to south pole (with $k=2$), or one of the five Platonic solids projected radially on their circumscribed sphere:

• $k=3, n=4$ (tetrahedron)
• $k=3, n=8$ (octahedron)
• $k=3, n=20$ (icosahedron)
• $k=4, n=6$ (cube)
• $k=5, n=12$ (dodecahedron)

There is no solution where the parts have smaller angular diameter than for the icosahedron.

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Would any square grid on a sphere then be a cube with each side subdivided into squares ? How many ways is there to square grid a sphere? – user1708 Oct 30 '11 at 7:39
The only square grid is the cube itself. A subdivided cube does not have global rotational symmetry around the centers of each subdivision. – Henning Makholm Oct 30 '11 at 7:47
I know, but if we disregard that, and consider the cube and it subdivided equal, then there is the way its done on globuses, is there any more ? – user1708 Oct 30 '11 at 7:50
If you disregard that requirement, then it is a completely different question. I'm not sure you get more solutions that way -- the remaining requirement that all faces must be identical and that each face must itself be rotationally symmetric will rule out many tilings. The subdivided cube is still out (different small squares on the cube project to different shapes on the sphere), as is a geographic grid. – Henning Makholm Oct 30 '11 at 8:00
In fact, just requiring rotational symmetry for each face alone will restrict (with the same $k$ for every face) will imply that all corner angles (measured between the great-circles that connect neighboring corners) are the same everywhere, because all side lengths must then also be the same. Therefore this alone will force us to use a Platonic solid. – Henning Makholm Oct 30 '11 at 8:29