Диссертация (1136188), страница 63
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Comparison of Gelfand–Zetlin polytopes and Feigin–Fourier–Littelmann–VinbergpolytopesWe start with an elementary construction of polytopes fibered over a segment. Then we applythis construction to get the Gelfand–Zetlin and Feigin–Fourier–Littelmann–Vinberg polytopes ina uniform way.4.1. Construction with fiber polytopeLet P ⊂ Rl be a convex polytope. The set of linear functionals, whose restrictions to P attaintheir maximal values at a face F ⊂ P , form a cone CF ; the normal fan of P is defined as the setof cones CF corresponding to all faces F ⊆ Q.
We say that a polytope Q ⊂ Rl is subordinate to Pif the normal fan of P is a subdivision of the normal fan of Q. Note that the set of all polytopessubordinate to P forms a semigroup under the Minkowski sum. Denote this semigroup by SP .Let µ(t) be a piecewise-linear continuous function from a segment I ⊂ R to SP . We say thatµ(t) is convex if()t1 + t2µ(t1 ) + µ(t2 )⊂µ22NEWTON–OKOUNKOV POLYTOPES OF FLAG VARIETIESadad-1Figure 2. Newton polygons of Lλ |Yd−2 and Lµ(t) |Yd−2 for d = 3, λ = (3, 1, 0) and t = 2.for all t1 , t2 ∈ I. In other words, the set∪Pµ :=µ(t) × {t} ⊂ Rl × R = Rl+1t∈Iis a convex polytope. In this case, Pµ fibers over I and the fiber polytope is subordinate to P .Suppose now that µ′ (t) is a convex function from I to SQ for a convex polytope Q ⊂ Rl .
If thepolytopes µ(t) and µ′ (t) have the same Ehrhart polynomials for all t ∈ I then obviously so do Pµand Pµ′ . The simplest example is when P = Q and µ′ (t) is a parallel translate of µ(t). In thiscase, Pµ and Pµ′ also have the same fiber polytope but might be combinatorially different even forquite simple µ(t) and µ′ (t) (see Example 4.4).4.2.
GZ(λ) vs F F LV (λ)We now show that both GZ(λ) and F F LV (λ) can be obtained inductively from a point usingthe above construction. Recall that the Gelfand–Zetlin polytope GZ(λ) ⊂ Rd is defined by thefollowing inequalitiesλ1λ2z11z12λ3...z212zn−2.....λn......z1n−21zn−1.z2n−2z1n−1where the notationabcmeans a ≥ c ≥ b. Let Gk (λ) be the face of the Gelfand–Zetlin polytope GZ(λ) given by thel−1l= zm+1for all pairs (l, m) such that either m > j, or m = j and l ≥ i (we putequations zm0zm = λm ).VALENTINA KIRITCHENKORemark 4.1.
In [Ki, Theorem 3.4], there is an inductive construction of the Gelfand–Zetlin polytopevia convex geometric Demazure operators. The flag of facesGd (λ) ⊂ Gd−1 (λ) ⊂ Gd−2 (λ) ⊂ . . . ⊂ G1 (λ) ⊂ GZ(λ) =: G0 (λ).is exactly the flag used in this construction. In particular, by [Ki, Corollary 4.5] the numberof integer points in Gk is equal to the dimension of the Demazure module H 0 (Yk , Lλ |Yk ) for allk = 0, . .
. , d and dominant λ.Lemma 4.2. Take µ(t) as in the proof of Lemma 3.7. There exists a path z(t) ∈ Rd such thatGk−1 (λ) ∩ {zji = t} = z(t) + Gk (µ(t))for all integer t ∈ [λi+j , λj ]. In particular,Gk−1 (λ) = conv{z(t) + Gk (µ(t)) | λi+j ≤ t ≤ λj }.l(t) of z(t) ∈ Rd as follows:Proof. Define the coordinates zm(t − λi+j ) (t − λi+j−1 )..lzm(t) =.(t − λj+2 )0if m > j, l + m = i + j, λi+j ≤ tif m > j, l + m = i + j − 1, λi+j−1 ≤ t....if m > j, l + m = j + 2, λj+2 ≤ totherwiseIn particular, z(t) = 0 if i = 1. The statement of the lemma now follows by direct calculation fromthe definition of GZ(λ) and Gk (λ).Lemmas 3.6 and 4.2 together with the backward induction on k immediately yield an elementaryproof of the following theorem.Theorem 4.3. Polytopes Fk (λ) and Gk (λ) have the same Ehrhart polynomial for all k = 0,. .
. ,d. In particular, Gelfand–Zetlin polytope GZ(λ) and Feigin–Fourier–Littelmann–Vinberg polytopeF F LV (λ) have the same Ehrhart polynomial.The last statement of the theorem also follows from [FFL]. The first elementary proof of thisstatement was given in [ABS] using a different approach.Lemmas 3.6 and 4.2 imply that both F F LV (λ) and GZ(λ) can be obtained inductively froma point by iterating the construction of Section 4.1. Note that both Fk−1 (λ) and Gk−1 (λ) fiberover a segment of length λj − λi+j , and fibers are equal (up to a parallel translation) to Fk (µ(t))and Gk (µ(t)), respectively, for the same piecewise linear function µ(t) on the segment.
The onlydifference between these two cases is the presence of the shift vector z(t) in the second case.Example 4.4. cf. [Fo] For n = 3, k = 0, . . . , 3, and n = 4, k = 2, . . . , 6, there exists a unimodularchange of coordinates that maps Fk to Gk . Let n = 4, and k = 1. Then Fk provides the minimalexample when Fk is not combinatorially equivalent to Gk .We now illustrate how to obtain the inequalities defining F1 from those of F2 using Lemma 3.6or equivalently the construction of Section 4.1 (and not the definition of F1 ). For k = 2, we havei = j = 2, and{(λ1 , λ2 , λ3 , t) if λ4 ≤ t ≤ λ3µ(t) =.(λ1 , λ2 , t, t) if λ3 ≤ t ≤ λ2NEWTON–OKOUNKOV POLYTOPES OF FLAG VARIETIESBy Example 2.2 the inequalities defining F2 are0 ≤ u11 ≤ λ1 − λ2 ;0 ≤ u12 ≤ λ2 − λ3 ;u11 + u21 + u12 ≤ λ1 − λ3 ;0 ≤ u21 , u31 ;u11 + u21 + u31 ≤ λ1 − λ4 .Put u22 := t − λ4 .
Using the last statement of Lemma 3.6 as a definition of F1 , we get that F1 isdefined by inequalities:0 ≤ u11 ≤ λ1 − λ2 ;0 ≤ u12 ≤ λ2 − µ3 (u22 + λ4 );u11 + u21 + u12 ≤ λ1 − µ3 (u22 + λ4 );0 ≤ u21 , u31 ;u11 + u21 + u31 ≤ λ1 − (u22 + λ4 );0 ≤ u22 ≤ λ2 − λ4 .Using that µ3 (t) = max{λ3 , t} and eliminating redundant inequalities we get0 ≤ u11 ≤ λ1 − λ2 ;0 ≤ u12 ≤ λ2 − λ3 ;u11 + u21 + u12 ≤ λ1 − λ3 ;u12 + u22 ≤ λ2 − λ4 ;0 ≤ u21 , u31 , u22 ;u11 + u21 + u12 + u22 ≤ λ1 − λ4 ;u11 + u21 + u31 + u22 ≤ λ1 − λ4 .Similarly, one can restore G1 from G2 and check that there are only 10 inequalities for G1 .References[An] D.
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