Let $X$ be a closed Riemannian manifold and consider the function $f\_n : X \times \cdots \times X \to \mathbb{R}$ where the domain of $f\_n$ is the $n$-fold cartesian product of $X$ and where $f\_n(p\_1,...,p\_n) = \sum\_{i \neq j} d(p\_i, p\_j)$ where $d$ is the distance function on $X$. The function $f\_n$ is invariant under the action of the isometry group $G = \text{Isom}(X)$ on the $n$-fold product $X^n$ as well as the action of the symmetric group on $n$ letters $S\_n$ acting on $X^n$ by permuting the factors



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> Are the maxima of $f\_n$ on $X^n / (G \times S\_n)$ isolated?
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As an example, this is the case for $X = S^1$ where the maxima is unique and achieved by the regular $n$-gon. My initial inspiration was that I was considering this question for $S^2$ with the round metric (where I have a feeling the answer is well known and I suspect the maximum is unique, but I do not know and would appreciate a reference if anyone knows -- [here](https://www.jstor.org/stable/2006132) as an older paper of Berman and Hanes with some estimates of the maximum value in this case which have subsequently been improved upon).



Isolated maxima for sum of distances of points on a manifold
Not necessarily. Consider the sphere $(\mathbb{S}^2,g)$ with its usual metric and give it a new metric $hg$, where $h\leq1$, $h$ has three local minima $h(p\_1)=0.7,h(p\_2)=0.8,h(p\_3)=0.9$ (where $p,q,r$ are close to one another) and $h=1$ outside very small neighborhoods of $p\_1,p\_2,p\_3$. Then Isom$(\mathbb{S}^2,hg)=\{\text{Id}\}$, but $f\_2$ has the same maximum as in the usual metric, and this maximum is achieved in uncountably many points of $(\mathbb{S}^2)^2$, so they are not isolated.


