# How to divide a $3$ D-sphere in “equivalent” parts?

My goal is to put $n$ points on a sphere in $\mathbb{R}^3$ to divide it in $n$ parts, so that their disposition would be as "equivalent" as possible. I don't exactly know what "equivalent" mathematically means, perhaps that the min distance between two points maximal is.

Anyway in $2$ dimensions it is simple to divide a circle in $n$ parts. In $3$ dimensions I can figure out some good repartitions for particular $n$ values but I lack a more general approach.

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–  Rahul Aug 17 '11 at 17:10
Given $n$ points on a circle $S^1$ these points divide $S^1$ in $n$ parts in an obvious way, and it is easy to describe a configuration where these points are "equivalent". On a $2$-sphere this is another matter. If your $n$ points are the vertices of a regular $n$-gon on the equator they are certainly "equivalent" insofar as there is a group of isometries of $S^2$ permuting these points transitively. But perhaps you have something else in mind. In any case, $n$ points on $S^2$ do not divide $S^2$ into $n$ parts without further ado. –  Christian Blatter Aug 17 '11 at 19:56
The ones cited by Rahul Narain may be better, but you could also look at math.stackexchange.com/questions/11499/… –  Ross Millikan Apr 14 '12 at 2:58

A nontrivial problem, I think. You might find some links into the research literature on Ed Saff's homepage.

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You can try the physical method: treat each point as an electron constrained in a sphere, and randomly distribute these point particles in the sphere, then you can solve the equations of motion to reach a stable (minimum energy) state, where each particle maximally separated from its closest neighbours (electric repulsive forces). Here is how to apply this method on a sphere surface.

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