Диссертация (1136188), страница 21

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A simple vertex u is then labeled by wv if p(u) = wp(v) forsome element w from the Weyl group. Put s1 = sα1 and s2 = sα2 . We denote by [u1 , u2 ] theedge of the Gelfand–Zetlin polytope connecting vertices u1 and u2 .All faces of Qλ except for a unique non-simple vertex can be represented as(v, B) for some choice of a simple vertex v and a Borel subgroup B, for example, ifB = B + , then (v, B + ) = v, (s1 v, B + ) is the edge [s1 v, v], and (s2 s1 v, B + ) is the face{y = b} ∩ Qλ . If B = B − , then (v, B − ) = Qλ , (s1 v, B − ) is the face {x = 0} ∩ Qλ , and(s2 s1 v, B − ) is the edge [s2 s1 v, s1 s2 s1 v].All faces of Qλ that do not contain the non-simple vertex are admissible.

Inparticular, there are two two-dimensional admissible faces 1 = (s2 s1 v, B + ) and 2 =(s1 v, B − ) corresponding to the cells O(s2 s1 v, B + ) and O(s1 v, B − ). It is easy to checkthat these two cells represent different Schubert cycles.

Denote these cycles by Z 21 andZ 12 , respectively (that is, we label the cohomology class of O(wv, B + ) by Z w and encodew = s1 s2 by 12, etc). There are also six admissible edges that connect simple vertices ofQλ . These correspond to two Schubert cycles of dimension 1. Namely, the edges [v, s1 v],[s1 s2 v, s1 s2 s1 v], and [s2 s1 v, s2 v] correspond to Z 1 and the other three edges correspond toZ 2 .

Then Theorem 3.3 together with Example 2.2 applied to the two-dimensional admissible faces tells thatHλ Z 12 = bZ 1 + (a + b)Z 2 ;Hλ Z 21 = (a + b)Z 1 + aZ 2 .Downloaded from http://imrn.oxfordjournals.org/ at Higher School of Economics on February 29, 2012E3E12524V. KiritchenkoRemark 4.1.

Note that if we only considered faces (u, B − ) for the lower triangularBorel subgroup B − (that is, proceeded as in [13, 14]), then we would not be able to represent the Schubert cycle Z 21 by a single admissible face. Instead, we would get the unionof two faces: the rectangular one {x = z} and the triangular one {y = a}. The union ofthese two faces looks like the admissible face 1 (corresponding to Z 21 by my construction) broken into two pieces.This example extends to arbitrary n. Namely, the Schubert divisor with the Schubert polynomial x1 + .

. . + xn−1 corresponds to the union of n − 1 facets of type (v, B − )type (v, B + ).We now describe heuristic Schubert calculus on the faces of Qλ . We can represent Schubert cycle Z 21 by faces in two different ways: as 1 and as F1 + F2 , where F1and F2 denote the faces given by the equations y = a and x = z, respectively. The latterrepresentation comes from [14]. We also represent Z 12 by 2 .

Finally, we represent theone-dimensional Schubert cycle Z 1 in two ways, by the edge E 1 = [s1 s2 v, s1 s2 s1 v] and theedge E 3 = [s2 v, s2 s1 v], and represent Z 2 by the edge E 2 = [s2 s1 v, s1 s2 s1 v] (see Figure 1). We2 by intersecting the corresponding faces:can now compute Z 21 Z 12 and Z 12(F1 + F2 ) ∩ 2 = E 1 + E 2 ,which is exactly the identity Z 21 Z 12 = Z 1 + Z 2 . Similarly,(F1 + F2 ) ∩ 1 = E 32 = Z . We can also get the identities Z Zgives the identity Z 2111 12 = Z 2 Z 21 = [ pt] andZ 1 Z 21 = Z 2 Z 12 = 0 by choosing the edges representing Z 1 and Z 2 so that they have transverse intersection with 1 or 2 , for example, to find Z 1 Z 12 , we represent Z 1 by E 3 andZ 12 by 2 and get that 2 ∩ E 3 = pt.

Similarly, to find Z 1 Z 21 , we represent Z 1 by E 1 andZ 21 by 1 , which yields 1 ∩ E 1 = ∅.An analogous Schubert calculus on the Gelfand–Zetlin polytope can be done forarbitrary n [11]. It can be rigourously justified using the concept of the polytope ringassociated with the volume polynomial of the Gelfand–Zetlin polytope. The elements ofthe polytope ring can be naturally identified with linear combinations of faces.Downloaded from http://imrn.oxfordjournals.org/ at Higher School of Economics on February 29, 2012according to [14]. On the other hand, it can be represented by a single admissible face ofGelfand–Zetlin Polytopes and Flag VarietiesD(v)D(s2v)The diagrams for the vertices v, s1 v, and s2 v of the Gelfand–Zetlin polytope for SL 3 .5 Geometry and Combinatorics of the Gelfand–Zetlin PolytopeTo prove Proposition 3.2, we have to study the faces of the Gelfand–Zetlin polytopeQλ .

First, we describe explicitly the simple vertices of Qλ and the edges going out ofsimple vertices. We mostly follow [13]. Brief explanations are provided for the reader’sconvenience, for more details see [13, 2.1–2.3]. Next, we will find out under whichconditions two simple vertices are connected by the edge (see Lemma 5.2). Finally, weprove Proposition 3.2 and formulate and prove a Chevalley formula for arbitrary faces(v, B) (see Theorem 5.5).We describe the faces of Qλ by triangular diagrams following [13]. Put x0,i := λifor i = 1, . . . , n.

It is easy to see that each face of Qλ is defined by the equations of theform xi, j = xi−1, j or xi, j = xi−1, j+1 for some i = 1, . . . , n − 1, j = 1, . . . , n − i. For a face ,encode all the equations defining by the following graph D(). Draw n rows indexedby 1, . . . , n with n − i + 1 points pi,1 , . . . , pi,n−i+1 in the ith row. These are the vertices ofthe graph D() (each vertex pi, j corresponds to the coordinate xi−1, j ). For each equalityLxi, j = xi−1, j and xi, j = xi−1, j+1 defining the face , we draw the edge ei+1,j of type L beRtween the vertices pi+1, j and pi, j and the edge ei+1,j of type R between pi+1, j and pi, j+1 ,respectively. The resulting graph is the diagram of the face . Figure 2 shows the diagrams for the vertices v, s1 v, and s2 v of the Gelfand–Zetlin polytope for SL 3 consideredin Section 4.5.1 Simple verticesIt is easy to show that v is a simple vertex of the Gelfand–Zetlin polytope if and only if thecorresponding diagram D(v) has exactly n − i edges starting at the ith row and ending atthe (i + 1)-st row (for all positive i < n) and two such edges never start or end at the sameDownloaded from http://imrn.oxfordjournals.org/ at Higher School of Economics on February 29, 2012Fig.

2.D(s1v)25252526V. Kiritchenkopoint. In other words, the graph D(v) is the disjoint union of n simple trees T1 (v), . . . ,Tn(v). Each tree Ti (v) starts at the first row of D(v) and ends at the ith row (that is, eachTi looks like the Dynkin diagram Ai ). The vertex of Ti (v) in the first row will be called thestarting point of Ti (v). Note that the coordinates xi, j and xk,l of the vertex v are equal ifand only if the vertices pi+1, j and pk+1,l belong to the same tree. The diagram D(v) canalso be thought of as an RC-graph or a pipe dream (see [13, 14] for details on connectionbetween pipe dreams and faces of the Gelfand–Zetlin polytope).

Let us call the diagramof a simple vertex also simple. There is a different way to characterize simple diagramsedges end at the ith row of D(v), and the edges ei,L j are strictly to the left of the edges ei,Rj .Each simple diagram D(v) defines a permutation σv of elements 1,. . . ,n as follows: the vertex p1,i is the starting point of the tree Tσv (i) . It is easy to check that thisgives a bijective correspondence between simple vertices of Qλ and elements of the symmetric group Sn, which is isomorphic to the Weyl group of G (we choose the isomorphism which sends the elementary transposition (i (i + 1)) to the simple reflection sαi ).This bijection is compatible with the bijection between the vertices of the weight polytope Pλ and elements of the Weyl group, that is, p(v) = σv λ. Indeed, using the formulafor the projection p : Qλ → Pλ from Section 2.1, we get that if p(u) = sαi p(v) (and thusn− jp(u) = p(v) − ( p(v), αi )αi ), then the sums of coordinates k=1 x j,k for the vertices v andu only differ for j = i.

This is only possible if the trees T j (v) and T j (u) have the samestarting points for all j = i, (i + 1), which implies σv = sαi σu.5.2 EdgesWe now describe the edges of the Gelfand–Zetlin polytope. Let u and v be two vertices ofthe Gelfand–Zetlin polytope. We say that the diagram D(u) is obtained from the diagramD(v) by switching the edge ei,L j ,if the diagrams have the same set of edges with oneexception: instead of the edge ei,L j , the diagram D(v) has the edge ei,Rj . Switching of ei,Rjis defined in the same way, for example, the diagrams D(s1 v) and D(s2 v) on Figure 2 areR and e R , respectively. It isobtained from the diagram D(v) by switching the edges e2,13,1easy to see that two vertices u and v are connected by an edge of the Gelfand–Zetlinpolytope if and only if their diagrams can be obtained from each other by switching theedge ei,L j or ei,Rj for some i and j.

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