Let $P \subset \mathbb R^n$ be a compact rational polytope (the affine space spanned by $P$ may not be $\mathbb R^n$). Let $G \subset {\bf GL}(n,\mathbb Z)$ be an arithmetic group. Let $$C = {\rm Conv}(\bigcup\_{g\in G} \ g\cdot P)$$ be the convex hull.


**Question**: does there exist a compact rational polytope $Q \subset C$ such that $\bigcup\_{g\in G}\ g\cdot Q = C$ ?



Existence of a fundamental domain for the convex hull of group action on a rational polytope
I will assume that by an "arithmetic group" you mean a subgroup $\Gamma$ commensurable to the intersection $GL(n, {\mathbb Z})\cap G$, where $G< GL(n, {\mathbb R})$ is an algebraic subgroup defined over a number field $F$; actually, $F={\mathbb Q}$ will suffice for my purposes. (I prefer to use Greek letters to denote arithmetic subgroups, as it is customary in the literature.)


I will even assume that $G$ is semisimple (which you are probably implicitly assuming). I claim that, in general, $Q$ does not exist.


The example will be given by $n=3, G=O(2,1)$, the stabilizer of the quadratic form $q=-x\_0^2+x\_1^2+x\_2^2$. I will use $\Gamma<G\cap GL(3, {\mathbb Z})$, the index two subgroup preserving the open future light cone
$$
L\_+=\{x: q(x)<0, x\_0>0\}.$$
Take
$p=(1,0,0)$ and $P=\{p\}$. Then the orbit $\Gamma \cdot p$ is an unbounded subset of $L\_+$. Moreover, for each point $x$ in the convex hull $C$ of this orbit, $C$ also contains the cone $x+ L\_+$. This is because $\Gamma$ is a lattice in $G$ and, hence, the closure of the projection of $\Gamma x$ to ${\mathbb R} P^2$ contains the quadric
$$
P(\{q(x)=0\})\subset {\mathbb R} P^2. 
$$
In particular, $q$ is unbounded from below on $C$.


Now, if $Q\subset L\_+$ is any compact subset, then its $G$-orbit (hence, $\Gamma$-orbit) is contained in a region
$$
\{x\in L\_+: a\le q(x)\le b\}
$$
where
$$
a= \max q|\_Q, b=\min q|\_Q.
$$
Hence, no bounded subset $Q$ of $L\_+$ will satisfy the containment
$$
C\subset \Gamma \cdot Q. 
$$
If $Q$ is not contained in $L\_+$, then we of course cannot have the other containment
$$
\Gamma \cdot Q\subset C. 
$$


If yo allow reductive groups $G$, then one can even get an example with $n=2$.


However, maybe by an "arithmetic group" you simply mean a finite index subgroup $\Gamma$ of $GL(n, {\mathbb Z})$. Then one can prove that for every $P$ either $C={\mathbb R}^n$ or $P=C=\{0\}$. In this case, $Q$ of course exists and for $C={\mathbb R}^n$ you can take $Q$ to be any convex polytope whose interior contains the origin.


