Upper-bound on volume of polytope inscribed in the sphere

Question

To my great surprise, I was unable to find any general reference on the volume of (symmetric) polytopes inscribed in, say, the unit sphere.

Let $P = absconv(v_1,\dots,v_k) $ where the $v_i$'s are unit vectors. Then, I would like to have an upper bound on the quantity :

$$\frac{vol(P)}{vol(B^2)} $$ where $B^2$ is the euclidean unit ball. Intuitively the maximal volume polytope with k vertices should be made of vertices $v_i$'s forming an $\epsilon$-net of the sphere for some $\epsilon$ (that is being "uniformly" distributed) but that's beyond the point.

In particular, I'm expecting that if $k$ is sub-exponential in $n$ then this volume ratio should go to zero. That is if $k_n = o(e^{cn})$ for every $c>0$ and $P_n$ is a polytope with $k_n$ vertices on the unit sphere of $R^n$, then $$vol(P_n) = o(vol(B^2_n))$$

Is this true ? Is there any quantitative statement about it ?

Answer 1

The following is a direct consequence of Theorem 1 in

• Elekes. "A geometric inequality and the complexity of computing volume." Discrete & Computational Geometry (1986).

If $k_n\in o(2^n)$ then $\mathrm{vol}(P_n)/\mathrm{vol}(B_n^2)\to 0$.

I recommend to take a look at the proof, it is short ($\sim$ 10 lines) and very elegant.

Likely one can say much more, but I have no appropriate source at hand: I suspect that you need super-exponentially many vertices to have a positive volume relative to the sphere. This is true for randomly chosen vertices.