4
$\begingroup$

Let $\Phi$ be a (reduced, crystallographic) root system with Weyl group $\mathcal{W}$, and $p$ a (nonzero) minuscule weight for $\Phi$: its orbit $\mathcal{W}p$ is the set of vertices of a convex polytope which, given together with the affine lattice $p+L$ generated by these vertices (a translate of the root lattice inside the weight lattice), defines a projective toric variety $X$.

Question: Is there a more geometric description of this toric variety $X$?

Specifically, can we relate it to the partial flag variety $G/P$ quotient of the semisimple algebraic group $G$ associated to $\Phi$ by the parabolic subgroup associated to $p$? Can we ($T$-equivariently? canonically?) embed $X$ in $G/P$?

Note: A presumably closely related toric variety is the one associated with the fan of Weyl chambers and the weight lattice: see here for a discussion and the paper by de Mari, Procesi & Shayman, "Hessenberg Varieties" (Trans. Amer. Math. Soc. 332 (1992), 529–534); this is the closure of a general orbit of a maximal torus $T \subset G$ acting on the flag variety $G/B$: see Batyrev & Blume, "The Functor of Toric Varieties Associated with Weyl Chambers", Tohoku Math. J. 63 (2011), 581–604 and the refernces to Klyachko found there. But to be honest, while I write "closely related" because intuitively it should be, I don't really see the details of this close relation.

$\endgroup$

1 Answer 1

3
$\begingroup$

In general if $T$ acts on a projective variety $X$ with moment polytope $\Phi(X)$, then a general point $x\in X$ will have $\Phi(\overline{T\cdot x}) = \Phi(X)$ i.e. be an abnormal toric variety with this same polytope. In the case of Grassmannians, Fink and Speyer prove that the variety is in fact normal, and I expect this works in other minuscule cases.

The close relation you claim is a map: take $y\in G/B$, $\pi:G/B\to G/P$, and look at $\overline{T\cdot y} \twoheadrightarrow \overline{T\cdot \pi(y)}$, inducing maps between their moment polytopes and normalizations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.