Isolated maxima for sum of distances of points on a manifold

Question

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

Are the maxima of $f_n$ on $X^n / (G \times S_n)$ isolated?

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 as an older paper of Berman and Hanes with some estimates of the maximum value in this case which have subsequently been improved upon).

Answer 1

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.