# How many facets does the convex hull of all the roots of a root system have?

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.

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.

• I should say, in case it was not clear, that the number of vertices of $\mathrm{ConvHull}(\Phi)$ is always the number of long roots of $\Phi$. – Sam Hopkins Jun 29 '18 at 16:41
• Another remark: $\mathcal{P}$ is an alcoved polytope in the sense of Lam and Postnikov, so perhaps the answer is known to those who study alcoved polytopes. – Sam Hopkins Jun 29 '18 at 16:53
• You have probably seen this, but let me record it anyways: these are studied under the name "root polytopes" by various authors in particular in the context of abelian ideal in root posets. In particular, Cellini (arxiv.org/abs/1612.06143) and others studied facet ideals of abelian ideal. I will try to get back to this later... – Christian Stump Jun 29 '18 at 17:21
• @ChristianStump: Thanks very much for the pointer! I think this paper of Cellini and Marietti (arxiv.org/abs/1203.0756) might contain the answer to Question 2 I'm looking for: on pg. 4 it states "In particular, the orbits of the facets correspond to the simple roots of Φ that do not disconnect the extended Dynkin graph." – Sam Hopkins Jun 29 '18 at 18:01
• @ChristianStump: if you post this as an answer I will accept it! – Sam Hopkins Jun 29 '18 at 18:07

• Note there is a way to describe "simple roots corresponding to nodes that do not disconnect the extended Dynkin diagram" in a way that avoids discussion of the extended Dynkin diagram. Namely, let $\Phi_i$ denote the root system generated by $\{\alpha_1,\ldots,\hat{\alpha}_i,\ldots,\alpha_n,-\theta\}$ (where the hat denotes omission). Then the node $i$ does not disconnect the extended Dynkin diagram iff $\Phi_i$ is irreducible. – Sam Hopkins Jul 2 '18 at 11:54