A codimension $d$ face of a polytope is called *rationally smooth* if it lies on only $d$ facets, because this is exactly the condition for the corresponding toric variety to have only orbifold singularities (not worse) there.

Is there some good reason that the moment polytope of a full flag manifold $G/B$ should have only smooth faces? It's easy to prove, and it doesn't hold for partial flag manifolds like $Gr(2,4)$ (whose moment polytope is an octahedron). Both of these varieties are smooth, of course, and neither is a toric variety; what the rational smoothness of the faces tells you is that the normalization of a generic $T$-orbit closure in the full flag manifold is orbifold, and in a Grassmannian it's not.

[Added: in general, if $X$ carries an *algebraic* action of a torus $T$, then $\overline{T\cdot x}$ for a generic $x\in X$ will be a variety with the same moment polytope as $X$. That doesn't make it a toric variety under Fulton's book's definition, as it may not be normal.]

Is there any other reason to predict that a given Hamiltonian space $X$ should have a moment polytope with this property?

Motivation: I have another such $X$, that isn't smooth actually, and its *nonabelian* moment polytope has this property inside the positive Weyl chamber. I would like to know "why".