TITLE: How many facets does the convex hull of all the roots of a root system have?
QUESTION [10 upvotes]: Let $V$ be an $n$-dimensional Euclidean vector space with inner product $\langle\cdot,\cdot\rangle$ and $\Phi$ an irreducible crystallographic root system in $(V,\langle\cdot,\cdot\rangle)$.
Question 1: Is there a root-theoretic formula for the number of facets of $\mathrm{ConvHull}(\Phi)$? Surely this must be known but I have not been able to find a reference.
Some thoughts about this question:
Consider instead the dual polytope $\mathcal{P}:=\{v\in V\colon \langle v,\alpha \rangle \leq 1 \textrm{ for all $\alpha \in \Phi$}\}$. Counting facets of $\mathrm{ConvHull}(\Phi)$ is the same as counting vertices of $\mathcal{P}$.
Since $\mathcal{P}$ is $W$-invariant, it is "enough" to understand the intersection of $\mathcal{P}$ with the dominant cone. This intersection is the polytope with facets:

$\langle v, \alpha_i\rangle \geq 0$ for all simple roots $\alpha_i$, $1 \leq i \leq n$;
$\langle v, \theta\rangle \leq 1$ for the highest weight $\theta$ of $\Phi$.

But the polytope given by those inequalities is well-known: it is just the fundamental alcove $A_0$ (I think technically it is the fundamental alcove for the dual root system $\Phi^\vee$? I always get tripped up by the distinction between $\Phi$ and $\Phi^\vee$). Note that $A_0$ is a simplex. Explicitly, the vertices of $A_0$ are $0$ together with $\frac{1}{a_i}\omega_i$ for $1\leq i \leq n$, where $\omega_1,\omega_2,\ldots,\omega_n$ are the fundamental coweights (i.e., the dual basis to the basis of simple roots), and $a_1,a_2,\ldots,a_n$ are the integer coefficients determined by writing $\theta = a_1\alpha_1+a_2\alpha_2+\cdots+a_n\alpha_n$. For example, the minuscule coweights (i.e., those $\omega_i$ with $a_i=1$) are a subset of the vertices of $A_0$.
So we understand the vertices of $A_0$ and $\mathcal{P}=W(A_0)$. We should be almost done.
How many $W$-orbits does $\frac{1}{a_i}\omega_i$ have? That's easy: its stabilizer is $W_i$, the Weyl group of the maximal parabolic root system obtained by removing node $i$. So $\#W(\frac{1}{a_i}\omega_i) = \#W/\#W_i$.
So the number of vertices of $\mathcal{P}$ is given by the nice root-theoretic formula $\sum_{i=1}^{n} \#W/\#W_i$, right?
Not quite. You can check that this already doesn't work for $\Phi=B_2$: $\mathrm{ConvHull}(\Phi)$ has $4$ facets, but the formula would give $8/2+8/2=8$ as an answer.
The problem: not every $\frac{1}{a_i}\omega_i$ is actually a vertex of $\mathcal{P}$.
Question 2: Which of the $\frac{1}{a_i}\omega_i$ are actually vertices of $\mathcal{P}$?
Of course an appropriate answer to Question 2 would yield an answer to Question 1 by the above discussion.

REPLY [4 votes]: These are studied under the name root polytopes by various authors in particular in the context of abelian ideal in root posets. See for example Cellini Triangulations of root polytopes and Cellini-Marietti Root polytopes and Borel subalgebras where facet ideals of abelian ideal are studied. As Sam pointed out in the comments, the latter shows that the orbits of the facets correspond to the simple roots that do not disconnect the extended Dynkin graph give a facet description of the root polytope.