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The vertices of a Platonic solid are equally distributed on its circumscribing sphere in a very strong sense: each of them has the same number of nearest neighbours and all distances between nearest neighbours are the same.

It seems clear to me that the Platonic solids also provide the only examples for such equidistributed arrangements of points on the sphere, i.e.

(∗) One can equidistribute (in the strong sense above) only 4, 6, 8, 12, or 20 points on the 2-sphere.

The latter is - thus - a consequence from the fact that there are only five Platonic solids, which in turn is a consequence of Eulers theorem, for which in turn there is a whole bunch of proofs.

My questions are:

(1) Are there other "independent" and maybe more "direct" proofs of (∗)?

(2) Are there corresponding results for general $n$-spheres?

I assume that one can equidistribute on every $n$-sphere $n+2$ points (the generalized tetrahedron) and $2^{n+1}$ points (the generalized cube).

(3) Are there other such (maybe partial) functions $f_3(n), f_4(n), f_5(n),\dots$ with e.g. $f_3(2) = 6, f_4(2) = 12, f_5(2) = 20$?

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  • $\begingroup$ Nice question! I don't understand why you ask (3), though -- you've already said how these points can be equidistributed, namely as specified by a simplex and a hypercube; why is this an "assumption"? $\endgroup$
    – joriki
    Jun 7 '13 at 7:10
  • $\begingroup$ I am often unsure what to be sure of and what not. And what seemed clear to me (∗) might turn out to be wrong (see Jyrki's answer below). $\endgroup$ Jun 7 '13 at 7:26
  • $\begingroup$ I edited my question. $\endgroup$ Jun 7 '13 at 11:05
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Unless I misunderstood your definition of strong equidistribution I think that your claim that 20 is the maximum cardinality is wrong.


Edit: A trivial example is that $n$ equally spaced points along the equator (with their longitudes differing by $2\pi/n$) are strongly equidistributed according to this definition. You can choose $n$ to be as large as you wish.


For a more interesting example consider the Icosahedral group $G$ of order 120 acting faithfully on $S^2$. If you pick any point $x$ on the surface of the sphere and consider the orbit $C=Gx$, it will have the strong equidistribution. This is because $G\le O_3(\mathbb{R})$ and thus you can turn any point of the orbit $C$ to another by rotating and reflecting the sphere (both of these operations preserve all distances, so strong equidistribution follows).

The same principle can be extended to other dimensions by using a finite subgroup of $O_n(\mathbb{R})$.

Related problems are studied under the umbrella of telcomm applications. I would suspect that a lot of examples of sets with strong equidistribution are to be found within the class of spherical codes. Check out the link to N. J. A. Sloane's database of good spherical codes.

Group codes (orbits of finite groups of orthogonal transformations) are an important subclass.


A few formalities: Let $G\le O_n(\mathbb{R})$ be a finite group, $x\in\mathbb{R}^n$ an arbitrary initial point, then the associated group code is $$ C=Gx=\{gx\mid g\in G\}. $$ Because $G$ is a group we have $Gy=C$ for all $y\in C$. Because $G\le O_n(\mathbb{R})$ we have $$ d(x,y)=d(gx,gy) $$ for all the vectors $x,y\in\mathbb{R}^n$ and all the elements $g\in G$.

Claim 1. Let $y\in C$ be arbitrary. The distances between closest neighbor $z\in C$ to $y$ is the same for all $y$.

Proof. If $y'\in C$ is the closest neighbor of $y$, and $g\in G$ is such that $gy\in z$ (such an element exists by construction of $C$), then $$ d(z,gy')=d(gy,gy')=d(y,y'), $$ so $z'=gy'$ is at the same distance from $z$ as $y'$ from $y$. There cannot be points of $C$ any closer to $z$, because then we could switch the roles of $z$ and $y$ and contradict ourselves.

Claim 2. Let $x_1,x_2,\ldots,x_n$ be all the points of $C$ at minimum distance $d$ from $x$. Let $y\in C$ be arbitrary. Then there are exactly $n$ points of $C$ at distance $d$ from $y$ also.

Proof. Again we select an orthogonal transformation $g\in G$ such that $gx=y$. Then all the points $y_i=gx_i,i=1,2,\ldots,n$ are at distance $d$ from $y$. They are also all distinct. There cannot be more than $n$ points at distance $d$ from $y$, because then we, again, contradict ourselves by reversing the roles of $x$ and $y$.

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  • $\begingroup$ If you select the initial point $x$ carefully so that it is equidistant from the three sides of the spherical triangle forming a fundamental domain of $G$ then you also maximize the minimum distance between two points of $C$. The same construction was generalized to higher dimensional finite reflection groups by Thomas Mittelholzer and yours truly in '96. $\endgroup$ Jun 7 '13 at 7:21
  • $\begingroup$ Would you say what Patrick Boucher claims here softimageblog.com/archives/115 (under "Defining Evenly Distributed") is false, or did I interpret him wrongly? $\endgroup$ Jun 7 '13 at 7:30
  • $\begingroup$ Will the 120 points of the orbit be the vertices of a convex polyhedron? And if so: why would this not be a regular polyhedron? $\endgroup$ Jun 7 '13 at 7:40
  • $\begingroup$ @Hans: You do get a convex polyhedron with 120 vertices. But it won't be regular. Look at that picture of a triangulation of the sphere by 120 triangles. Pick $x$ equidistant from the sides of the triangle. The resulting polyhedron will have faces that squares, hexagons and 10-gons. All corresponding to points where 4, 6 or 10 triangles meet (respectively). $\endgroup$ Jun 7 '13 at 9:18
  • $\begingroup$ AFAICT Boucher does not define strong equidistribution at all. His evenly distributed requires even less than you do, and he specifically seeks to have a more general concept than that of Platonic solids. It is somewhat telling that he does not mention the concept of a spherical code. Nor does he mention Sloane. I would have thought that anyone playing seriously with these things would study at least Conway & Sloane. $\endgroup$ Jun 7 '13 at 9:26

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