Is it possible to prove that for every $n > 3$ the maximal possible volume of a convex polyhedron having $n$ vertices inscribed in a sphere of unit radius is an algebraic number?
Update: What can be said about degrees of these algebraic numbers for different $n$?
Yes. Let $\mathbb{R}^{alg}$ be the field of real algebraic numbers. This is a real closed field which means that, for any statement of first order logic, using the symbols $0$, $1$, $+$, $\times$, $=$, $<$, that statement is true in $\mathbb{R}^{alg}$ if and only if it is true in $\mathbb{R}$.
The statement "The volume of $\mathrm{Hull}(x_1, x_2, \ldots, x_n)$ is $V$" is a first order statement in the coordinates of the $x_i$ and $V$. Proof sketch There are finitely many simplicial complexes on the abstract vertex set $1$, $2$, ..., $n$. For each of these simplicial complexes, the statement that this simplicial complex is a triangulation of $\mathrm{Hull}(x_1, x_2, \ldots, x_n)$ is finitely many polynomial inequalities and, if the simplicial complex is such a triangulation, then the volume of $\mathrm{Hull}(x_1, x_2, \ldots, x_n)$ is a polynomial in the coordinates of the $x$'s. So we can encode $\mathrm{Vol}(\mathrm{Hull}(x_1, \ldots, x_n)) = V$ as
$\Delta_1$ encodes a triangulation of the convex hull and $V=\cdots$ OR $\Delta_2$ encodes a triangulation of the convex hull and $V=\cdots$ OR ...
So I can talk about $\mathrm{Vol}(\mathrm{Hull}(x_1, \ldots, x_n))$ in the first order theory of ordered fields. Consider the statement:
There exists an $A$ such that there are $n$ points on the unit sphere with $A = \mathrm{Vol}(\mathrm{Hull}(x_1, \ldots, x_n))$ and such that, for any points $y_1$, \dots, $y_n$ on the unit sphere, we have $A \geq \mathrm{Vol}(\mathrm{Hull}(x_1, \ldots, x_n))$.
This is a first order statement by the above discussion. It is true when all the variables range over $\mathbb{R}$, because $(S^2)^n$ is compact. So it is also true when all variables range over $\mathbb{R}^{alg}$. Let $A^{alg}$ be the maximum volume of a sphere with real algebraic coordinates; we have just shown that this number exists. Let $A$ be the maximum volume of a sphere with real coordinates.
It remains to check that $A = A^{alg}$. Let $f(t)$ be the minimal polynomial of $A^{alg}$ over $\mathbb{Q}$. Let the real roots of $f$ be $x_1$, $x_2$, ...., $x_n$ with $A^{alg} = x_r$. Consider the statement
The number $A$ in the previous paragraph obeys $f(A) = 0$. There are $n-1$ other numbers $x_1$, $x_2$, ..., $x_{r-1}$, $x_{r+1}$, ..., $x_n$ with $f(x_1)=f(x_2) = \cdots = f(x_n)=0$ and $x_1 < x_2 < \cdots > < x_{r-1} < A < x_{r+1} < \cdots < x_n$.
This is a first order statement which has been constructed to be true in $\mathbb{R}^{alg}$. So it is also true in $\mathbb{R}$, and we see that $A$ is the $r$-th root of $f$ in $\mathbb{R}$. Thus, $A = A^{alg}$, as claimed.
This is a hard question. First part (probably the easiest here) is to check that the polyhedron is actually a critical point of volume (with respect to, say, the coordinates of the points). This is actually not obvious (even for the cube). If it is, then you are asking for the gradient of volume to vanish. The gradient is a collection of polynomial functions, so if the variety is zero-dimensional (over $\mathbb{C}$) then the answer to your question is, of course, yes, by standard elimination. That, however, is far from obvious even if there are no infinitesimal real deformations.
Interestingly, in hyperbolic space life is comparatively easier, since volume of ideal polyhedra (which are in precise correspondence to inscribed euclidean polyhedra) is a convex function of the dihedral angles (an old result of mine), so one has all sorts of uniqueness results. None of which seem to translate to Euclidean space.
