A few days ago I asked the [following question](https://math.stackexchange.com/questions/610929/weyl-orbits-of-integral-dominant-weights-and-convex-polytopes) 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?



How to find faces of polytope defined by a Weyl orbit
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 $W\_P$. 


