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.

verticesof $\mathrm{ConvHull}(\Phi)$ is always the number of long roots of $\Phi$. $\endgroup$ – Sam Hopkins Jun 29 '18 at 16:41alcoved polytopein the sense of Lam and Postnikov, so perhaps the answer is known to those who study alcoved polytopes. $\endgroup$ – Sam Hopkins Jun 29 '18 at 16:53