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Let $x_1,x_2,x_3,x_4$ be points on the unit sphere $\mathbb{S}^2$, that maximizes the quantity $$ \sum_{i < j}\| x_i - x_j \|^2, $$ where $\| x_i - x_j \|$ denotes the Euclidean distance in $\mathbb{R}^3$.

Do the $x_i$ form the shape of a Tetrahedron?

I saw this related question, which considers the total sum instead of sum of squares, but there is no proof or reference there for the case of four points.

I tried using Lagrange's multipliers but this got a bit messy; I got that $$ \sum_{i \neq j} \big(\langle v_i,x_j \rangle+\langle x_i,v_j \rangle \big)=\sum_i\lambda_i \langle x_i,v_i \rangle, $$ for every $(v_1,v_2,v_3,v_4) \in (\mathbb{R}^3)^4$, but I am not sure how to proceed from here.

Asaf Shachar
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    Yes and no, regular tetrahedron do maximize the sum but this is far from unique. A square on the equator share the same sum $16$. In fact, any 4 points which satisfies $\sum x_i = 0$ will have the same sum. – achille hui Apr 17 '22 at 07:25
  • Thanks, this is an interesting observation. Can you please elaborate on why every four points on the sphere satisfying $\sum x_i=0$, have the same sum $\sum_{i \neq j}| x_i - x_j |^2$? Using $|x_i|^2=1$ and expanding, I got that this is equivalent to $\sum x_i=0 \Rightarrow -\langle x_1,x_2 \rangle + \langle x_3,x_4 \rangle$ is constant. – Asaf Shachar Apr 17 '22 at 09:58
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    This is because $\sum\limits_{i<j}|x_i - x_j|^2 = \frac12\sum\limits_{i,j}|x_i - x_j|^2 = \frac12({\small 4\sum\limits_j |x_j|^2 + 4\sum\limits_i |x_i|^2 - 2\sum\limits_i x_i \cdot \sum\limits_j x_j}) = 16-\left|\sum_i x_i\right|^2$ – achille hui Apr 17 '22 at 10:10
  • Wow, that is brilliant, thank! I think I can make it a CW answer. This gives an if and only if solution. – Asaf Shachar Apr 17 '22 at 10:15

1 Answers1

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The complete answer is that, $x_i$ maximizes the given function, if and only if $\sum_i x_i=0$. Here is a proof base on achille hui's great comment:

$$ \sum\limits_{i<j}\|x_i - x_j\|^2 = \frac12\sum\limits_{i,j}\|x_i - x_j\|^2 = \frac12\big({\small 4\sum\limits_j \|x_j\|^2 + 4\sum\limits_i \|x_i\|^2 - 2\langle \sum\limits_i x_i , \sum\limits_j x_j}\rangle\big) $$ $$=16-\|\sum_i x_i\|^2. $$

In particular, any regular tetrahedron whose vertices lie on the sphere satisfies this condition, but there are other examples, such as a square on the equator.

Edit 1:

It is worth noting that the proof above does not use in any essential way the dimension of the sphere (here $2$), or the number of points (here $4$).

Indeed, if we want to place $N$ points $x_1,\dots,x_N$ on the $n$-dimensional sphere $\mathbb{S}^n \subseteq \mathbb{R}^{n+1}$, maximizing $\sum\limits_{i<j}\|x_i - x_j\|^2$, then the same calculation gives:

$$ \sum\limits_{i<j}\|x_i - x_j\|^2 = \frac12\sum\limits_{i,j}\|x_i - x_j\|^2 =\frac12\sum\limits_{i,j}\|x_i\|^2+\|x_j\|^2-2 \langle x_i , x_j\rangle $$ Since $$\sum\limits_{i,j}\|x_i\|^2=\sum_i \sum_j \|x_i\|^2=\sum_i N \|x_i\|^2=N \sum_i \|x_i\|^2=N^2 $$ and $$ \sum\limits_{i,j} \langle x_i , x_j\rangle=\langle \sum\limits_i x_i , \sum\limits_j x_j\rangle, $$ we get $$ \sum\limits_{i<j}\|x_i - x_j\|^2=\frac12 \big( 2N^2 - 2\langle \sum\limits_i x_i , \sum\limits_j x_j\rangle\big) =N^2-\|\sum_i x_i\|^2. $$

Edit 2: (Generalization)

Here we again focus only on four points:

$f : (0,4] \to (0,\infty)$ be any monotone increasing concave function.

We use $\sum_{i<j} |x_i - x_j|^2 =16 - \left|\sum_i x_i\right|^2.$

Since $f$ is concave and monotone increasing,

$$\sum_{i<j}f\left(|x_i - x_j|^2\right) \le 6f\left(\frac16\sum_{i<j}|x_i-x_j|^2\right) = 6f\left(\frac83-\frac16|\sum_ix_i|^2\right)\le 6f\left(\frac83\right).$$

In particular, $E_f=\sum_{i<j}f\left(|x_i - x_j|^2\right)$ is maximized whenever $|x_i-x_j|$ are all equal, i.e. $x_i$ is a tetrahedron. (This automatically implies $\sum_i x_i=0$).

When $f$ is strictly concave, the tetrahedron is the unique maximizer.

In particular this works for the map $x \to \sqrt{x}$.

Asaf Shachar
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