3
$\begingroup$

Let $G$ be a semisimple algebraic group over $\mathbb{C}$ and for a highest weight $\lambda$, denote by $V_{\lambda}^w$ the Demazure module associated with $\lambda$ and $w$. More precisely, $V_{\lambda}^w=U(\mathfrak{b}).v_{w \lambda}$. There are many nice descriptions of these modules (sections of line bundles restricted to Schubert varieties, and more concretely the Demazure character formula).

If $P(V_{\lambda}^w)$ denotes the set of weights which appear in the Demazure module, I can consider the convex hull in $X^*(T) \otimes_{\mathbb{Z}} \mathbb{R}$. The extremal vertices of the resulting polytope will be $w'(\lambda)$ for $w' \leq w$ via an argument of Bernstein, Gelfand and Gelfand.

My questions generally run as follows: Is there a nice description of the facets of the resulting polytope? Is it true that if $F$ is a facet of this polytope that there exists a coweight $\check{\Lambda_i}$ such that for all $v \in F$ we have $\langle v, \check{\Lambda_i} \rangle =n$ for $n \in \mathbb{Z}$? Are they pseudo-Weyl polytopes in the sense of Kamnitzer (MV Cycles and Polytopes)?

I would be grateful for any knowledge in this direction.

$\endgroup$
3
$\begingroup$

If I understand the question correctly, there is a nice description of the faces. They show up as so called reduced Kogan faces. However, taking the convex hull is not the right operation, as Demazure characters (or key polynomials) are given as the integer point transform of a union of reduced Kogan faces. In particular, the set is not always convex.

This paper deals with a closely related question on key polynomials and convexity., See also the sequel

Furthermore, Postnikov and Stanley have a paper on 312-avoiding permutations, for which the key polynomials are a bit special, as they are given as lattice points in a single Kogan face. See Alexander Postnikov and Richard P. Stanley. Chains in the Bruhat order. Journal of Algebraic Combinatorics, 29(2):133–174, March 2008.

I have some references and formulas here, but it is work in progress..

$\endgroup$
  • $\begingroup$ Thank you so much! This is precisely what I was looking for. $\endgroup$ – Marc Besson May 4 at 13:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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