TITLE: Subsets of a ball/sphere with the largest sum of distances
QUESTION [6 upvotes]: $\newcommand\R{\mathbb R}\newcommand\S{\mathbb S}$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.

REPLY [6 votes]: 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.