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Is it possible to evenly distribute N=5 points on a 3-sphere such that:

  • the distance between any two points is equal.
  • the points are all distinct.

in complete analogy to the even distribution of 4 points on a 2-sphere such that they correspond to the vertices of its inscribed tetrahedron?

If so, can this be generalized to higher dimensional spheres as N increases? This seems to be suggested here:

If you are working with spheres of a higher dimension, there are more finite groups of rotations; the orbits of a point under these groups will give sets of points which could reasonably be called "uniform".

seraph
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  • "Uniform" in the sense of your boxed quote means (among other things) that the multiset of distances from any fixed point of the set is the same, so that is different from your question as it does not require the distance between any pair of points to be the same. – Jyrki Lahtonen Aug 18 '14 at 04:18
  • In this related answer I try to describe what the use of groups means in the construction of such sets of points. – Jyrki Lahtonen Aug 18 '14 at 04:42

1 Answers1

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The $N+1$ vertices of an $N$-simplex give a generalization of a regular tetrahedron (the case $N=3$). One way of seeing this is the following. The construction is the $(N+1)$-dimensional analogue of the observation that in $\Bbb{R}^3$ the coordinate axes intersect the plane $x+y+z=1$ at the vertices of an equilateral triangle.

Consider the points $P_0=(1,0,\ldots,0)$, $P_1=(0,1,0,\ldots,0)$, $\ldots$, $P_N=(0,0,\ldots,1)$. The distance between any pair of them is equal to $\sqrt2$. But they all reside in $\Bbb{R}^{N+1}$, so we need to fix that. Consider the orthogonal projection $\pi:\Bbb{R}^{N+1}\to V$, where $V$ is the zero-sum subspace $$ V=\{(x_0,x_1,\ldots,x_N)\in\Bbb{R}^{N+1}\mid \sum_{i=0}^Nx_i=0\}. $$ We see that $V$ consists of vectors orthogonal to $\vec{n}=(1,1,1,\ldots,1)$. The formula for the projection is thus: $$ \pi(\vec{x})=\vec{x}-\frac{\vec{x}\cdot\vec{n}}{\Vert\vec{n}\Vert^2}\vec{n}. $$ Anyway, because $\vec{OP_i}\cdot\vec{n}=1$ for all $i$, the points $P_i$ all lie on the same coset of $V$. Thus their projections $Q_i=\pi(P_i)$ are all still equidistant from each other, and all belong to $V$. Because $V$ is $N$-dimensional we have thus constructed a set of $N+1$ points of an $N$-dimensional space equidistant from each other and also at the same distance from the origin of that space.

I chose to do it this way, because the coordinates of the points $Q_i,i=0,1,\ldots,N,$ w.r.t. to any orthonormal basis of $V$ are bit more cumbersome.

Jyrki Lahtonen
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