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?




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*Update:* What can be said about degrees of these algebraic numbers for different $n$?



Do maximal polyhedra have algebraic volume?
Yes. Let $\mathbb{R}^{alg}$ be the field of real algebraic numbers. This is a [real closed field](http://en.wikipedia.org/wiki/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.


