# Subsets of a ball/sphere with the largest sum of distances

$$\newcommand\R{\mathbb R}\newcommand\S{\mathbb S}$$Let $$B_d$$ and $$S_{d-1}$$ denote, respectively, the closed unit ball and the unit sphere in $$\R^d$$. Let us say that a finite subset $$F$$ of $$B_d$$ is maximal if the sum of all pairwise Euclidean distances between the points in the set $$F$$ is the largest possible given $$n:=\text{card}\,F$$.

In a comment to a post now apparently deleted, Noam D. Elkies noted that, by convexity, (i) any maximal set must lie on $$S_{d-1}$$ and (ii) such a set must be the set of all vertices of a regular polygon if $$d=2$$.

This suggests the following questions:

Q1. For $$d=3$$ and $$n$$ such that there exists a regular polytope with $$n$$ vertices, is it true that every maximal set must be the set of all vertices of a regular polytope?

Q2. For $$d=3$$ and any given natural $$n$$, is a maximal set of $$n$$ distinct points unique up to an orthogonal transformation?

Q3. A weaker version of Q2: For $$d=3$$ and any given natural $$n$$, are there only finitely many, up to an orthogonal transformation, maximal sets of $$n$$ distinct points?

Q4. Same questions for any $$d\ge3$$.

Q5. Same questions with the Euclidean distances $$|x-y|$$ between points $$x$$ and $$y$$ replaced by increasing functions $$f(|x-y|)$$ of the distances; e.g., here one may take $$f(r)\equiv r^2$$ or $$f(r)\equiv -1/r^{d-2}$$.

The answer to the questions about regular polytopes seem to be positive at least for $$n\le d+1$$, with the regular simplexes as the regular polytopes.

One of the correct and complete answers to any one of these questions will be accepted as an answer to this entire post.

## 1 Answer

Negative answers to some of those questions:

Q1 Not always; for $$n=8$$ a square antiprism is better than a cube. For example, in radius $$\sqrt 3$$, the cube with vertices $$(\pm 1, \pm1, \pm1)$$ gives $$16(1 + 2\sqrt{2} + \sqrt{3}) = 88.9676+$$ while changing the $$z=+1$$ vertices to $$(\pm\sqrt2, 0, 1)$$ and $$(0,\pm\sqrt2,1)$$ gives $$16\bigl(\!\sqrt{8+\sqrt{8}} + \sqrt{8-\sqrt{8}}\bigr) = 89.0362+$$. One can then optimize the $$z$$-values further. [Similarly, in ions of the form $${\rm MF}_8^{n-}$$, with eight fluorine atoms around a large central M atom (such as xenon [sic] or rhenium), the F's form a square antiprism, not a cube.]

Q5 Again not always. Indeed for $$\sum|x-y|^2$$ we have $$\mathop{\sum\!\sum}_{1 \leq i < j \leq n} |x_i - x_j|^2 = n^2 - \Bigl| \sum_i x_i \Bigr|^2$$ so the maximal configurations are precisely those whose center of mass is the origin, and once $$n$$ is at all large there is a positive-dimensional family of such configurations. There are also examples for other natural $$f$$ where the 24-cell (the $$(d,n)=(4,24)$$ regular polytope) is suboptimal; see

Henry Cohn, John H. Conway, Noam D. Elkies, and Abhinav Kumar: The $$D_4$$ root system is not universally optimal, Experimental Math. 16 (2007), 313--320, arXiv:math/0607447.

• I see now that the values $16(1 + 2\sqrt{2} + \sqrt{3}) = 88.9676+$ and $16\bigl(\!\sqrt{8+\sqrt{8}} + \sqrt{8-\sqrt{8}}\bigr) = 89.0362+$ are both too small by the same amount $16 + 8 \sqrt 2$ coming from the squares at the top and bottom of the antiprism. These squares do not change the conclusion that the antiprism is better, but must be taken into account when optimizing $z$. – Noam D. Elkies Nov 22 '20 at 15:56
• Thank you very much for your answer and, in particular, for the reference, which opens (for me) a window into the state of the art in this area. – Iosif Pinelis Nov 22 '20 at 17:27