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.



How many facets does the convex hull of all the roots of a root system have?
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](http://arxiv.org/abs/1612.06143) and Cellini-Marietti [Root polytopes and Borel subalgebras](https://arxiv.org/abs/1203.0756) 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.


