Do solutions to the Thomson problem maximize volume for the particular polyhedra formed?

Question

There's nothing I really need to add to the question: Is it possible to prove that the solutions to the Thomson problem for $N>3$ have the largest volume for polyhedra with all vertices 1 unit from the origin or do there exist counterexamples to this?

Answer 1

Nope.

For $N = 8$, the best known configuration that minimize the energy is a square anti-prism.

Let $\theta \approx 55.91399164489755^\circ$, $c = \cos\theta$ and $s = \sin\theta$. One way to realize the square anti-prism is place the vertices at $$(\pm s,0,c),\quad(0,\pm s,c),\quad\left(\pm \frac{s}{\sqrt{2}}, \pm \frac{s}{\sqrt{2}}, -c\right) $$

The base of the anti-prism is a square with side $2s \approx 1.65639443625$. The height of the anti-prism will be $2c \approx 1.12087353058$.

Using this dimensions, one can reproduce the energy $19.675287861$ on wiki page of Thomson problem. The volume of this anti-prism is $\frac23(5+\sqrt{2})cs^2 \approx 1.643789905948143$.

In contrast, the largest $8$-vertex polyhedron inside a sphere has volume $$\sqrt{\frac{475+29\sqrt{145}}{250}} \approx 1.815716104224$$ Aside from it isn't a square anti-prism, I don't really know how to describe this polyhedron. Look at this answer for coordinates and a picture for that polyhedron. Follow references there for what is know about largest $n$-vertex polyhedron inscribed inside a sphere.