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I'm interested in the subject of maximally separated points (e.g., the minimum of the distances between any two of the points is maximal) in various spaces, and I've been trying to think about how this works on a spherical surface.

For the simpler case of a circle, you can easily separate any number of points maximally simply by dividing the circle into the appropriate number of equal arcs and taking the points to be the vertices between the arcs.

On a spherical surface things are more complicated. For 0, 1, 2, or 3 points, the maximal spacing is on any great circle as above. For 4, 6, 8, 12, or 20 points the maximal spacing must be on the vertices of the relevant platonic solid inscribed in the sphere, otherwise there must be some points spaced more closely than this and you can separate them further. For the case of five points, the solution seems to be two points on opposite poles of a diameter plus three equally-separated points on a great circle lying in a plane perpendicular to that diameter. For seven or more points, other than the platonic cases, I have great difficulty determining if there's any solution better than the poles-plus-great-circle solution given for five points. Is there any better solution I'm missing?

Tseug
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  • For a large number of points, i.e. $1000$ or so, this seems sub-optimal to me because of the mostly unused upper and lower hemisphere. – Ragnar Aug 16 '14 at 19:59
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    The umbrella concept here is Spherical Codes. IIRC at least Thomas Ericsson of Linköping Technical University, Sweden, maintains a database of best known constellations. On other manifolds the question is also of interest to telecommunication engineers. At Steifel manifolds and Grassmannians. – Jyrki Lahtonen Aug 16 '14 at 20:02
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    Ah! According to Wikipedia the good ol' N.J.A. Sloane is the one with an on-line database. – Jyrki Lahtonen Aug 16 '14 at 20:05
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    http://en.wikipedia.org/wiki/Tammes_problem – Will Jagy Aug 16 '14 at 20:15
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    journal page 212 at http://www.scipress.org/journals/forma/pdf/2103/21030197.pdf Figure 6(b) – Will Jagy Aug 16 '14 at 20:17
  • For 8 points, the vertices of a square-based antiprism beats the vertices of a cube. For 20 points, the vertices of a regular dodecahedron is not the best solution. This answer shows a better one. – Rosie F May 26 '20 at 09:11

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