Does anyone know any way or any algorithm that can exactly and/or numerically find the coordinates of $n_{k}+2$ equally spaced points on the $(n_{k}-1)$-dimensional unit sphere $S^{n_{k}-1}$ for some sequence of integers $1<n_{1}<n_{2}\cdots<n_{k}<\cdots$? Many thanks.
1 Answer
Depends on what you mean by equally spaced.
If you want that all points have pairwise identical distances, then this is not possible. This is only possible for at most $n+1$ points, but not more.
If you want to maximize the minimal distance of $n+2$ points in the $(n-1)$-sphere, you have to pick some of the vertices of the $n$-dimensional crosspolytope, that is $(\pm1,...,\pm 1)$.
Here is a reference to that:
- Conway, John Horton, and Neil James Alexander Sloane. Sphere packings, lattices and groups.
In Chapter 1, subsection 2.6 they introduce the function $A(n,\theta)$ to denote the maximal number of points on an $(n-1)$-dimensional sphere (in $\Bbb R^n$) so that any two points have an angle of at least $\theta$ between them. They state
- $A(n, \pi/2)=2n$ (for the vertices of the crosspolytope), and
- $A(n,\theta)\le n+1$ for $\theta>\pi/2$ (for the vertices of a simplex).
Now, let's say you have an arrangement of $n+2$ points that maximizes the minimal distance between any two points. Let $\theta$ be this (angular) distance. By 2., we know that $\theta \le\pi/2$, and by 1. we know that $\pi/2$ can be achieved (and we actually have space to add more points).
-
$\begingroup$ Thank you for your answer, but why does one have to pick some of the vertices of the n-dimensional crosspolytope? Thanks. $\endgroup$– nancyDApr 2, 2020 at 9:21
-
$\begingroup$ Since you are saying "this is only possible for at most n+1 points, but not more", how about 6 and 20 points on $S^2$? Thanks. $\endgroup$– nancyDApr 2, 2020 at 9:21
-
$\begingroup$ @nancyD I will edit the answer later, to add some further information (right now, I can't). For now: for 6 and 20 points the pairwise distance is not identical, e.g. for 6 points there are antipodal pairs of vertices, and non-antipodal ones. $\endgroup$ Apr 2, 2020 at 9:45
-
$\begingroup$ @nancyD I updates the answer with a reference. $\endgroup$ Apr 2, 2020 at 10:54
-
$\begingroup$ $6$ and $20$ points on $S^2$ would seem to be irrelevant, nancy, since your question specifically asks for the number of points to be greater than the dimension by exactly three. $\endgroup$ Apr 2, 2020 at 11:17