$U$, and $f$ be given as in and suppose $G$ is non-empty.

Then precisely one of the following statements is true:

1.  there exists a solution $(x_1, \ldots, x_k) \in U$ to ,

2.  $G$ is disconnected: we can write $V = V_1 \cup V_2$ as a disjoint union with $V_1, V_2 \neq \emptyset$, such that for all $e \in E$ we have $e \subseteq V_1$ or $e \subseteq V_2$.

Suppose $G$ is disconnected. Without loss of generality, we can order the vertices such that $V_1 = \{1, \ldots, k'\}$ and $V_2 = \{k' + 1, \ldots, k\}$. Suppose $(x_1, \ldots, x_k) \in U$ has minimum energy. Then for all $y \in {\mathbf{R}}^d \setminus \{0\}$ we have $$\begin{gathered}
f(x_1 + y, \ldots, x_{k'} + y, x_{k'} - y, \ldots, x_k - y) \\
= f(x_1, \ldots, x_k)
+ \beta \sum_{i = 1}^{k'} \sum_{j = k' + 1}^k \left( \frac{w_i \, w_j}{\|2 \, y + x_i - x_j\|^{\delta}} - \frac{w_i \, w_j}{\|x_i - x_j\|^{\delta}} \right).\end{gathered}$$ This equation holds because all hyperedges $e_j \in E$ are either completely contained in $V_1$ or completely contained in $V_2$, such that all $x_i$ with $i \in e_j$ are either translated by $+y$ or $-y$, respectively. 