How to find faces of polytope defined by a Weyl orbit

Question

A few days ago I asked the following question at MSE and received no answer. I thought I would try here.

Let $\xi$ be an integral dominant weight of an irreducible root system $\Delta$, and let $\mathcal{O}_{\xi}$ be its orbit under the action of the Weyl group. The elements of the orbit are the vertices of an n-dimensional convex polytope.

Is there an efficient way to pick out all the subsets of points in $\mathcal{O}_{\xi}$ forming the faces of the convex polytope? I would like to have an algorithm that takes a Weyl orbit of an integral dominant weight as input and gives as output a list of subsets of vertices forming the faces of the polytope. By efficient, I mean that I would like something better than a brute force algorithm.

Let $G$ be the compact connected Lie group corresponding to the root system $\Delta$, and let $T\subset G$ be a maximal torus. We could also view the orbit $\mathcal{O}_{\xi}$ as the moment map image for a Hamiltonian group action of $T$ on $G/T$. In this case, the faces of the moment map image can be viewed as the image of points fixed under a one parameter subgroup of $T$.

Is there a way to understand which 1-parameter subgroups have fixed point sets that get mapped to the faces of the moment map image?

Answer 1

The faces are all of the following form: $w W_P / Stab_W(\xi)$, where $W_P$ varies over the subgroups generated by subsets of the simple reflections. In particular, for $\xi$ regular, the number of them is $\sum_P |W / W_P|$. (Note that $\mathcal O_\xi$ is only $G/T$ when $\xi$ is regular; otherwise it's $G/Stab_G(\xi)$.)

I wrote up a Hamiltonian-geometry-based proof of this, then realized the combinatorics is pretty easy too, assuming $\xi$ is regular as it seems you're doing.

The set of faces ("external" faces, to us Hamiltonians) is obviously $W$-invariant. So it's enough to find the ones nearby the basepoint, then permute them.

The polytope is "simple" -- it has the right number of edges coming out of each vertex, namely the rank of $G$. So the number of faces at the basepoint is $2^{rank(G)}$. Each $W_P$ gives one, so that's all of them. QED.

Here are some related computations in the Hamiltonian formalism. If $S \leq T$ is a connected subgroup, then $ (G/T)^S = C_G(S) N(T)/ T $. Proof: $\subseteq$ is easy. For the reverse, let $s$ be a topological generator of $S$. If $s gT = gT$, then $g^{-1} s g \in T$, so $\exists n \in N(T)$ s.t. $n g^{-1} s g n^{-1} = s$, so $g n^{-1} \in C_G(s) = C_G(S)$.

We're interested in the set $\coprod_S C_G(S) N(T)/T$ of all components, over all $S$, so we get the same set if we look at $\coprod_S N(T) C_G(S)/T$. Since the $T$-moment map is $N(T)$-equivariant, we just need to figure out which components of $\coprod_S C_G(S)/T$, i.e. which $S$, give external walls of the moment polytope.

Every Levi subgroup of $G$ (w.r.t. a fixed Weyl chamber) arises as a $C_G(S)$, where $S$ is a dominant coweight. Each of those gives a wall around the basepoint of the Weyl polyhedron, and there are the same number of each: $2^{rank}$. So all the walls near the basepoint are images of $Levi/T$, i.e. with fixed points $W_P$.

Answer 2

This is not a direct answer to your questions (which I still haven't understood completely from your formulation). But it seems important to place these questions within the extensive theoretical background, since I'm unconvinced that computing large examples (by brute force or otherwise) will provide much insight.

A convenient recent source is a short announcement by A. Khare here. He and various collaborators have posted on arXiv a number of related papers on the faces of weight polytopes. For example, a paper with Ridenour was published in 2012 in Algebras and Representation Theory; the preprint is here.

Some of this work refers back to work of Vinberg and others. Since the combinatorics of weight polytopes gets formidable even when the Weyl group is a symmetric group (presumably the best-behaved family of examples), it's a good idea to explore this literature and also to refine your own questions as far as possible.