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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".

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Do I understand correctly that the moment polytope of $G/B$ is the convex hull of a $W$-orbit of a regular weight $\lambda?$ –  Victor Protsak Aug 11 '10 at 22:47
    
Yeah, and for $G/P$ it's an appropriately nonregular weight. –  Allen Knutson Aug 11 '10 at 23:43

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