Polarization Problem on a Higher-Dimensional Sphere for a Simplex

Abstract

We study the problem of maximizing the minimal value over the sphere $S^{d-1}\subset {\mathbb {R}}^d$ of the potential generated by a configuration of $d+1$ points on $S^{d-1}$ (the maximal discrete polarization problem). The points interact via the potential given by a function f of the Euclidean distance squared, where $f:[0,4]\rightarrow (-\infty ,\infty ]$ is continuous (in the extended sense), decreasing on [0, 4], and finite and convex on (0, 4], with a concave or convex derivative $f'$. We prove that the configuration of the vertices of a regular d-simplex inscribed in $S^{d-1}$ is optimal. This result is new for $d>3$ (certain special cases for $d=2$ and $d=3$ are also new). As a byproduct, we find a simpler proof for the known optimal covering property of the vertices of a regular d-simplex inscribed in $S^{d-1}$.

Appendix A

In Sect. 4 we use the following known result, see [23, (5.10)] or [18]. For completeness, we give its proof.

Proposition A.1

Let $\omega _d^*=\{\mathbf{x}_0,\mathbf{x}_1,\ldots ,\mathbf{x}_d\}$ be the set of vertices of a regular d-simplex inscribed in $S^{d-1}$, $d\ge 2$. Then for any point $\mathbf{x}\in {\mathbb {R}}^d$,

$$\begin{aligned} \sum _{i=0}^{d}(\mathbf{x}\cdot \mathbf{x}_i)^2=\frac{d+1}{d}|\mathbf{x}|^2. \end{aligned}$$

(35)

Proof

This proposition can be proven in different ways. We will prove it using induction on dimension. For $d=2$, without loss of generality, we can let $\mathbf{x}_0=(1,0)$, $\mathbf{x}_1=(-1/2,\sqrt{3}/2)$, and $\mathbf{x}_2=(-1/2,-\sqrt{3}/2)$. Then for every $\mathbf{x}=(x_1,x_2)\in {\mathbb {R}}^2$, we have

$$\begin{aligned} \sum _{i=0}^{2}(\mathbf{x}\cdot \mathbf{x}_i)^2=x_1^2+\frac{(x_1-\sqrt{3} x_2)^2}{4}+\frac{(x_1+\sqrt{3} x_2)^2}{4}=\frac{3}{2} x_1^2+\frac{3}{2} x_2^2=\frac{3}{2}\quad |\mathbf{x}|^2. \end{aligned}$$

Assume now that $d>2$ and that equality (35) holds for every $\mathbf{x}\in {\mathbb {R}}^{d-1}$, and prove it for any $\mathbf{x}\in {\mathbb {R}}^d$. Without loss of generality, we can denote $\mathbf{x}_0=(0,\ldots ,0,1)\in {\mathbb {R}}^{d}$ and let $H:=\{\mathbf{x}\in {\mathbb {R}}^d:\mathbf{x}\,{\bot }\,\mathbf{x}_0\}$. Given $\mathbf{x}\in {\mathbb {R}}^d$, let $a\in {\mathbb {R}}$ and $\mathbf{b}\in H$ be such that $\mathbf{x}=a\mathbf{x}_0+\mathbf{b}$. Then $|\mathbf{x}|^2=a^2+|\mathbf{b}|^2$. Let also $\mathbf{z}_i\in H$ be such that $\mathbf{x}_i=-{\mathbf{x}_0}/{d}+\mathbf{z}_i$, $i=1,\ldots ,d$. Observe that $\mathbf{z}_1,\ldots ,\mathbf{z}_d$ are the vertices of a regular simplex in H inscribed in the sphere of radius $R={\sqrt{d^2-1}}/{d}$ centered at $\mathbf{0}$. Then $\mathbf{x}\cdot \mathbf{x}_i=(a\mathbf{x}_0+\mathbf{b})\cdot (-{\mathbf{x}_0}/{d}+\mathbf{z}_i)=-{a}/{d}+\mathbf{b}\cdot \mathbf{z}_i$, $i=1,\ldots ,d$. Using the induction assumption and the fact that $\sum _{i=1}^{d}{} \mathbf{z}_i=\mathbf{0}$ as the vertices of a regular simplex with the center of mass at $\mathbf{0}$, we obtain

$$\begin{aligned} \sum _{i=0}^{d}(\mathbf{x}\cdot \mathbf{x}_i)^2&=a^2+\sum _{i=1}^{d}\biggl (\!{-}\frac{a}{d}+\mathbf{b}\cdot \mathbf{z}_i\biggr )^{\!2}=\biggl (\!1+\frac{1}{d}\biggr )a^2-\frac{2a}{d}\quad \mathbf{b}\cdot \sum _{i=1}^{d}{} \mathbf{z}_i\\&\quad +\sum _{i=1}^{d}(\mathbf{b}\cdot \mathbf{z}_i)^2=\biggl (\!1+\frac{1}{d}\biggr )a^2+R^2\sum _{i=1}^{d}\left( \mathbf{b}\cdot \frac{\mathbf{z}_i}{R}\right) ^2\\&=\frac{d+1}{d}a^2+\frac{dR^2}{d-1}|\mathbf{b}|^2=\frac{d+1}{d}(a^2+|\mathbf{b}|^2)=\frac{d+1}{d}|\mathbf{x}|^2, \end{aligned}$$

and (35) follows by induction. $\square$