schubert polynomials, the bruhat order, and the …billey/classes/...geometryofflagmanifolds 377...

Vol. 95, No. 2 DUKE MATHEMATICAL JOURNAL (C) 1998

SCHUBERT POLYNOMIALS, THE BRUHAT ORDER,AND THE GEOMETRY OF FLAG MANIFOLDS

NANTEL BERGERON AND FRANK SOTTILE

To the memory ofMarcel Paul Schiitzenberger

CONTENTS

1. Summary 375w 3751.1. Suborders of the Bruhat order and the Cur

1.2. Substitutions and the Schubert basis 3771.3. Identities when v is a Schur polynomial 379

2. Preliminaries 3822.1. Permutations 3822.2. Schubert polynomials 3822.3. The flag manifold 383

3. Orders on 6e 3853.1. The k-Bruhat order 3853.2. A new partial order on ff’oo 3883.3. Disjoint permutations 390

w 3924. Cohomological formulas and identities for the4.1. Maps on 6eo 3924.2. An embedding of flag manifolds 3934.3. The endomorphism Xp 0 396

v when u(p) w(p) 3984.4. Identities for cuo4.5. Products of flag manifolds 3994.6. Maps Z[Xl,X2,...] Z[yl,y2,...,Zl,Z2,...] 4014.7. Products of Grassmannians 403

w 4055. Identities among the cv(X,k5.1. Proof of Theorem E(ii) 4055.2. Proof of Theorem G(ii) 4085.3. Cyclic shift 410

6. Formulas for some c(,k 4126.1. A chain-theoretic interpretation 4126.2. Skew permutations 4166.3. Further remarks 417

Received 26 February 1997. Revision received 9 June 1997.First author supported in part by a Natural Sciences and Engineering Research Council grant.Second author supported in part by Natural Sciences and Engineering Research Council grant

number OGP0170279 and National Science Foundation grant number DMS-9022140.1991 Mathematics Subject Classification. 05E15, 14M15, 05E05.

373

374 BERGERON AND SOTTILE

Extending work of Demazure [14] and of Bernstein, Gelfand, and Gelfand[7], Lascoux and Schtitzenberger [28] defined remarkable polynomial represen-tatives for Schubert classes in the cohomology of a flag manifold, called Schu-bert polynomials. For each permutation w in 6coo, there is a Schubert polynomialw [Xl,X2,...]. Schubert polynomials form an additive basis for this ring.Thus the identity

w for the ring of polynomials with respect todefines integral structure constants Curits Schubert basis. The Cur are nonnegative. They enumerate flags in a suitabletriple intersection of Schubert varieties. Evaluating a Schubert polynomial atcertain Chern classes gives a Schubert class in the cohomology of the flag mani-fold. This exhibits the cohomology of the flag manifold 10] as

;[x, x2,...]/<w w

It remains an open problem to give a bijective formula for these constants. Weexpect such a formula will have the form

cuv=w # I(saturated) chains in the Bruhat rder n ff’o frm}u to w satisfying some condition imposed by v(1)

Since every Schur symmetric polynomial Sx(xl,...,Xk) is a Schubert poly-nomial, this would generalize the Littlewood-Richardson rule [34] (cf. Section6.1), as Young tableaux are chains in Young’s lattice, a suborder of the Bruhatorder. A new proof of Pieri’s formula for Grassmannians [51] suggests a geo-metric rationale for such "chain-theoretic" formulas. Finally, known formulasfor the cur are all of this form. These include Monk’s formula [37], Pieri’s for-mulas [12], [25], [28], [39], [50], [55], and other formulas of [50].

Here, we illuminate this relation between the Bruhat order and the cuwv, refin-ing (1) and proving many new identities among the cuWo. This enables us to give a

w to compute many more, and to obtaindescription of the form (1) for some Cuonew results about the enumeration of chains in the Bruhat order. Many of theseidentities have a companion result about the Bruhat order that should imply theidentity, were such a formula as (1) known. In fact, they and the Pieri-type for-mula imply the identities (see [6]). Our combinatorial analysis leads to a newpartial order on 6aoo that contains Young’s lattice. We also compute the effect ofmany specializations of the variables in Schubert polynomials.

Algebraic structures in the cohomology of a flag manifold yield identitiesw cu wherew such as w w (commutativity) or c ,oowamong the cu cu Cu c,

:= co0wro0 (Poincar6 duality). Similar identities for the Littlewood-Richardson

GEOMETRY OF FLAG MANIFOLDS 375

coefficients have been studied combinatorially (see [1], [2], [21], [22], [56]). Weexpect the identities established here will lead to some beautiful combinatorics,

w is known. These identities imposeonce a combinatorial interpretation for the curstringent conditions on the form of any combinatorial interpretation and shouldbe useful in finding such an interpretation.

This paper is organized as follows. In Section 1 we describe our results. Sec-tion 2 contains necessary background. Section 3 contains most of our combina-torial analysis. In Section 4, we study the effect on cohomology of certain mapsbetween flag manifolds and compute specializations of the variables in a Schu-bert polynomial. In Section 5, we prove the identities when v is a Schur poly-nomial. In Section 6, we use these identities to compute many of the

1. Summary

1.1. Suborders of the Bruhat order and the cuwo. The identity

c Wo< , l (1.1.1)

defines integer constants CuWo 2,k which share many properties with the Littlewood-Richardson coefficients. They are related to chains in the k-Bruhat order, <k, asuborder of the Bruhat order. Its covers coincide with the index of summation inMonk’s formula [37]"

u" (k,k+l) u" (X1 -Jr-’’’-- Xk) u(a,b),

where the sum is over those a < k < b with (u(a, b)) (u) + 1. Young’s latticeof partitions with at most k parts is isomorphic to those permutations compara-ble to the identity in the k-Bruhat order. These are the Grassmannian permuta-tions with descent k, whose Schubert polynomials are Schur polynomials inX1,... ,Xk. If f counts the standard Young tableaux of shape 2, then [36, 1.5,Example 2]

(X1 -[’-"""-[’- Xk)m fASA(x1,... ,Xk).Abm

Considering the coefficient of w in the product u" (X1 "-’’"--Xk)m and thedefinition (1.1.1) of cuWo(,k), we obtain the following proposition.

PROPOSITION 1.1.1. The number of chains in the k-Bruhat orderfrom u to w is

2 wf Cuv(2,k)"

In particular, cW., 0 unless u k W A chain-theoretic description of theconstants CuW(,k) should provide a bijective proof of Proposition 1.1.1. By this we

mean a function z from the set of chains in [u, W]k to the set of standard Youngtableaux T whose shape is a partition A of ’(w)- ’(u) such that #z-l(T)=e , Schensted insertion [47] furnishes a proof [53] for the Littlewood-)"Richardson coefficients (cf. Section 6.1), as does Schiitzenberger’s jeu de taquin[49]. We show (Theorem 6.3.1) that if ar is a function where #z-l(T) dependsonly upon the shape of T and satisfies a condition of compatibility with the Pieriformula, then #z-t(T) *Cuo(2,k

In Section 3.1 we give a nonrecursive description of the k-Bruhat order.

TI-mORM A. Let u, w 6t’oo. Then u < k w ifand only if(i) a < k w(b);

(ii) if a < b, u(a) < u(b), and w(a) > w(b), then a < k < b.

We generalize Proposition 1.1.1 and refine (1). Let P be a parabolic suboroupof 6oo, so P is generated by some adjacent transpositions, (i, i+ 1). Define the P-Bruhat order by its covers. A cover u <v w in the P-Bruhat order is a cover inthe Bruhat order where u-lw P. When P is generated by all adjacent trans-positions except (k, k+ 1), this is the k-Bruhat order.

Let I c {1, 2,..., n-l} index the adjacent transpositions not in P. A colouredchain in the P-Bruhat order is a chain together with an element of I c {a, a+l,..., b-1} for each cover u "I, u(a, b) in the chain [30]. Iterating Monk’s rule, weobtain

(Z (,,i+) E feV(e)v’ (1.1.2)iel v:g(v)=m

where fev (P) counts the coloured chains in the P-Bruhat order from e to v. Nec-essarily, fv(P) # 0, only if v is minimal in vP. More generally, let fw(p) countthe coloured chains in the P-Bruhat order from u to w. Multiplying (1.1.2) by uand equating coefficients of w gives a generalization of Proposition 1.1.1, asfollows.

THEOREM B. Let u, w oo and P 6t’oo be a parabolic subgroup. Then

141

v 0 unless u < v w. This suggests a refine-Hence if v is minimal in vP, then Cuvment of (1). Let u, v, w 6eo, and let P be any parabolic subgroup such that v isminimal in vP. Then, for every coloured chain in the P-Bruhat order from e tov, we expect that

coloured chains in the P-Bruhat order on o from}c,, #1 u to that satisfy some condition imposed by(1.1.3)

Moreover, this rule should give a bijective proof of Theorem B.

This P-Bruhat order may be defined for parabolic subgroups of any Coxetergroup. Likewise, the problem of determining the structure constants for a Schu-bert basis also generalizes. For Weyl groups, this is the Schubert basis of coho-mology for a generalized flag manifold G/B or the analogues of Schubertpolynomials (see [8], [17], [20], [45]). For finite Coxeter groups, this is the basisAw in the coinvariant algebra [23]. Likewise, Theorem B and the expectation(1.1.3) have analogues. Of the known formulas (see [11], [24], [40], [42], [43],[44], [52] and the survey [41]), few have been expressed in a chain-theoreticmanner (see [11], [24], [40], [52]).

1.2. Substitutions and the Schubert basis. In Sections 4.3 and 4.4, we studythe cuwv when w(p)=u(p) for some p. For w6an+l and l<p<n+l, letW/p 6en be defined by deleting the pth row and w(p)th column from the permu-tation matrix of w. If y 6an and 1 < q < n + 1, then ep,q(y) if’n+1 is the permu-tation such that e.p,q(y)/p y and ep,q(y)(p) q. The index of summation in aparticular case of the Pieri formula (see [4], [28], [50]),

Cpdefines the relation v---- w. Define Wp 7Z[x1,x2,...] -- .[x1, x2,...] by

xj ifj<p,

Fp(xj)= 0 ifj=p,

xj_ ifj>p.

THEOREM C. Let u, w ff’ and p I.(i) Suppose w(p) u(p) and (w) (u) (w/p) ’(U/p). Then

(a) ep,u(p) [ulp, W/p] [u, w];(b) for every v

u/p y

Cp

(ii) For v 6eo,

cpv--,,, (y)

We prove the first assertion (Lemma 4.1.1(ii)) using combinatorial arguments.The second (in Section 4.4) and third (in Section 4.3) are proven by computing

w w c,O0u Theorem C(i)(b) givescertain maps on cohomology. Since Cur Cvuw when one of wu-1 wv-1 or coouv-1 has a fixed point and thea recursion for cur

condition on lengths is satisfied.We compute other substitutions of the variables. Let P c IN, and list the ele-

ments of P and IN P in order as

P’pl <p2 <’", IN-P.p < P <"’.

Define W/, Z[Xl, X2,...] 7Z,[yl, Y2, gl, 7,2,...] by

’tg(xp) y, ’I’e(xp;) z.In Remark 4.6.1, we define an infinite set Ie of permutations with the followingproperty.

THEOREM D. For every w 6oo, there exists an integer N such that if it Ip andn v, then

We prove this in Section 4.6. Theorem D gives infinitely many identities of thef,,., ,,.-(uxv). ,(uxv).a f,,.

n w ,,w It, tr Ip. Moreover, for these u, v, z with c(uxv) v O,we have [n, (u x v). z]

_[e, u] x [e, v], which suggests a chain-theoretic basis for

these identities.A combinatorial proof of Theorem D may provide insight into the problem of

determining the cuWo. In particular, it would be interesting to find a proof usingone of the combinatorial constructions of Schubert polynomials (see [3], [4],[9], [18], [27], [54]). Theorem D extends 1.5 of [29], which shows thatis a nonnegative sum of u(y)v(z). The special case of Theorem D whenP In], together with the formula -uxcWZvy CuvW. CxyZ for u,v,w n andx, y, z e 6ego, was established by Patras [38] using methods similar to ours. Also,Lascoux and Schiitzenberger [29] give the special case when P (1}.We consider more general substitutions. Let P. := (P0, P1,...) be any parti-

tion of IN. For > 0, let x (i) "-,.(i) ,.() be variables in bijection with P."--’’1 2Define Vv.: Z[xl, x2,...]-- g[x(1),_x(2),...]’ly

0 if jPo,We. (xj) x}i) if j is the Ith element of Pi.

COROLLARY 1.2.1. For every partition P. ofIN and w

re. ($w(X)) E dv"u" (P.l$u, (x(1))(u, (_x(2))Ul U2...

where each du,u (P.) is an (explicit) sum ofproducts of the cy

A ballot sequence A (al,a2,...) is a sequence of nonnegative integers where,for each i, j > 1,

#{k<jlak=i} > #{k<jlak=i+l}.

(Consider ai 0 as a vote for "none of the above.") Given a ballot sequence A,define WA" ,[Xl, x2,...] Z[Xl, x2,...] by

Wa xi0 ai 0,Xa ai # O.

COROLLARY 1.2.2. For every ballot sequence A and w if’n, there exist non-negative integers dUw (A) for u, w ff’oo such that

Moreover, each dv(A is an (explicit) sum ofproducts of the cy.Proof. If P0 :-- {il a 0} and for j > 0

P := {ilai is the jth occurrence of some integer in A},

then Pa A o I/(P0,P1 ), where A(x)i)) xj. H1.3. Identities when o is a Schur polynomial. If 2,/, and v are partitions

with at most k parts, then the Littlewood-Richardson coefficients cx are definedby the identity

They depend only on 2 and the skew partition v/In. That is, if/c and p are parti-tions with x/p v/In, then for all 2,

c c,,and the coefficient of S(Xl,...,x) in Sa(x,...,x). S(x,...,Xl) is c,. Theorder type of the interval in Young’s lattice from/t to v is determined by v/l.These facts hold also for the cuv,,t,r)"

If u <k w, let [u, W]k be the interval between u and w in the k-Bruhat order.Permutations ( and r/ are shape equivalent if there exist sets of integersP {pl <"" < pn} and Q {ql <." < qn}, where ( (respectively, //) acts asthe identity on IN- P (respectively, ]hi- Q), and

((pi) pj rl(qi) qj.

38O BERGERON AND SOTTILE

THEOREM E. Suppose U k W and x l Z where wu- is shape equivalent tozx- Then the following statements hold.

(i) We have [u, W]k --[x,z]t. When wu-= zx-1, this isomorphism is given by1) 13U- x.

(ii) For all partitions 2, Cuv(,k cxv(,!).Part (i) follows from Theorems 3.1.3 and 3.2.3, which are proven using combi-

natorial arguments. Part (ii) is proven in Section 5.1 using geometric arguments.By Theorem E, we may define the skew coefficient c for e ff’oo and 2 a partitionby c "= Cu(2,k) and also define I1 := (u)- (u) for any u e 6e with u k Cu.This leads (in Section 3.2) to a partial order on 6ao graded by I1 with the fol-lowing defining property. Let [e, ]_ be the interval in the -<_-order from the iden-tity to . If u < k u, then the map [e, ]. [u, U]k defined by

is an order isomorphism. Then Proposition 1.1.1 states that -’z fc counts thechains in [e, ].. This order is studied further in [5].We express-some of the c in terms of chains in the Bruhat order. If

u ,. k u(a, b) is a cover in the k-Bruhat order, label that edge of the Hasse dia-gram with the integer u(b). The word of a chain in the k-Bruhat order is itssequence of edge labels.

THEOREM F. Suppose U k W, and WU-1 is shape equivalent to v(lt, 1) v(v, 1)-1for some and partitions It, v. Then, for all partitions 2. and standard Youngtableaux T ofshape 2,

chains in k-Bruhat order from u to w whose wordcuWv(A,k) has recording tableau Tfor Schensted insertion

Theorem F gives a combinatorial proof of Proposition 1.1.1 for many u, w. Itis proven in Section 6.1.

If a skew partition 0 p I.I a is the union of incomparable skew partitions pand a, then

pIi-p x a

as graded posets. The skew Schur function So is defined [36, 1.5] to be ’c Szand Sp I_I a Sp. S [36, 1.5.7]. Thus

PH a 2 ac; CvC; c (1.3.1)

Permutations ( and r/ are disjoint if and / have disjoint supports and+ I 1.

THEOREM G. Let ( and I be disjoint permutations. Then(i) the map ((’, 1’) - ’1’ induces an isomorphism

(ii) for every partition , c" ,,,v cv c cvThe first statement is prown in Section 3.3 using a characterization of dis-

jointness (Lmma 3.3.1), and the second in Section 5.2 using geometry.Our last identity has no analogy with the Littlwood-Richardson coefficients.

The n-cycle (12... n) cyclicly permutes In].THEOREM H. Suppose Sn and rl ((12...n). Then, for every partition ,This is proven in Section 5.3 using geometry. Combined with Proposition

1.1.1, we obtain the following corollary.

COROLLARY 1.3.1. If U<kW and Xk z with wu-l,zx-6en, and(wu-1) (12’’’n)-- zx-, then each of the two intervals [u,w]k and [x,z]k have thesame number ofchains.These intervals [u, W]k and [x, Z]k are typically nonisomorphic. For example, in

Y’4 let u 1234, x 2134, and v 1324. If ( (1243), r/= (1423) ((1234), and(1342) t/(234), then

and v 2 v.Figure 1 shows the intervals [u, (U]2 [X,/’IX]2 and [v, V]2.

2413

/\2314 1423

\/1324

1234

3412

/2413

2314

3214

3124

2134

FIGURE 1. Effect of cyclic shift on intervals

3412

2413

/\2314 1423

\/1324

The theorems of this section, together with the "algebraic" identities cuWvw from whichc’Ua,ow v ca’i,, greatly reduce the number of distinct coefficients cu v(g,k)

all others may be determined. We indicate this for some small symmetric groupsin the table below. The first row counts the number of CuW ,,k with u <k w and121 t’(w) ’(u) in 6e,, and the second counts those c from which all the cu(X,k

may be determined using the results of this paper. For a discussion of this table,see http ://www. math. yorku, ca/bergeron/coefficient s. html.

TABLE

# c,,(,U,) 208 3600

#c 5 12

81669 2285414 79860923

62 332 3267

2. Preliminaries

2.1. Permutations. Let 6an be the group of permutations of [n] :=(1, 2,..., n}. Let (a, b) be the transposition interchanging a < b. The length Y(w)ofw e 6an counts the inversions {i <jlw(i) > w(j)} ofw. The Bruhat order < on6t’n is the partial order whose cover relation is w < w(a, b) if w(a) < w(b) and(w)+l=(w(a,b)). If u<w, let [u,w]:={vlu<v<w} be the intervalbetween u and w in 5an, a poset graded by ’(v)- Y(u). The longest elementcoo e 6an is defined by co0(j)= n + 1-j. When it is necessary to consider thelongest elements in several symmetric groups, we write con for coo 6a,.A permutation w e 6an acts on [n+l], fixing n + 1. Thus 6an c 6an+l. Define

6t’oo := ), 6an. For P {Pl < P2 < "} c IN, define qv 6a#v --0 6ao by requiringthat qv act as the identity on IN- P and qv(Q(Pi) Pc(i). This injection does notpreserve length unless P {n + 1, n + 2,...}. For this P, set In w := bv(w). Ifthere exist permutations ,,r/ and sets of positive integers P, Q such thatbv( " and q2() r/, then ff and r/are shape equivalent.

2.2. Schubert polynomials. Lascoux and Schiitzenberger invented and thendeveloped the elementary theory of Schubert polynomials in a series of papers[28]-[33]. For a self-contained exposition of some of this elegant theory see[35].

5an acts on polynomials in x1,..., Xn by permuting the variables. For a poly-nomial f,f- (i,i+l)f is antisymmetric in xi and xi+, and hence is divisible byxi Xi+l. Define the divided difference operator

ci "= (xi- Xi+l)-l(e (i, i+1)).

If al,a2,...,ae(w) is a reduced word for w, then Oal o’’’OOae(w) depends onlyupon w, defining the operator Ow. For w 6an, Lascoux and Schiitzenberger [28]defined the Schubert polynomial w by

The polynomial w is homogeneous of degree Y(w), and it is independent of thechoice of n. The set of all Schubert polynomials {wlW e Seoo } is an integral basisfor 7z[Xl, x2,...].A partition 2 is a decreasing sequence 21 > > 2k > 0 of integers. Youno’s

lattice is the set of partitions ordered by c, where/ c 2 if kt < 2i for all i. Writem for the partition with parts, each of size m. If 2k+1 0, the Schur polynomialS(Xl,... ,Xk) is

detlx-’+, ki,j=l

detlx-’l ki,j=l

which is symmetric and homogeneous of degree 121 :=A permutation w e 6ao is Grassmannian ofdescent k ifj :/: k w(j) < w(j + 1).

Then w defines and is defined by a partition 2 with 2k+1 0:

2k+-j w(j) --j j 1,...,k.

(The condition w(k+l) < w(k+2) < ..-determines the remaining values of w.) Inthis case, write w v(2, k). The raison d’etre for this definition is that v(,k)S(Xl,... ,Xk). Thus the Schubert polynomials form a basis for Z[Xl,X2,...] thatcontains all Schur symmetric polynomials S(xl,..., Xk) for all 2 and k.

2.3. The flag manifold. Let V - n. Aflag F. in V is a sequence

{0) Fo c F1 c F2 Fn-1 Fn-- V

of subspaces with dimeFi i. Flags F. and F: are opposite if Fn-j c Fj {0} forall j. The set of all flags is an ()-dimensional complex manifold IFC’V (or lF’n),called the flag manifold. There is a tautological flag .. of bundles over IF’Vwhose fibre at F. is F.. Let -xi be the first Chern class of the line bundle /_.Borel [10] showed the cohomology ring of IFV is

Z[Xl,... ,Xn]/(ei(xl,..., x.)[ i-- 1,... ,n),

where ei(xl,..., X,n) is the ith elementary symmetric polynomial in x,..., Xn.Let (S) be the linear span of S V, and U W be the set-theoretic difference

of subspaces W c U. An ordered basis fl, f2,..., fn for V determines a flagE. := ((fl,..., fn>>, where Ei <fl,..., j> for 1 < < n. A fixed flag F. gives adecomposition due to Ehresmann [16] of IFV into affine cells indexed by per-mutations w of Sn. The cell determined by w is

XvF. {E. ((fl,..., fn)) j Fn+l-w(i) gn-w(i), 1 < < n}.

Its closure is the Schubert subvariety XwF., which has codimension re(w). Also,

u < w XuF. XwF.. The Schubert class w is the cohomology class Poincar6dual to the fundamental cycle of XwF.. These classes form a basis for cohomol-ogy. Schubert polynomials were defined so that w(Xl,..., Xn) w.

If F. and F: are opposite flags, then XuF. n XvF: is an irreducible, genericallytransverse intersection, a consequence of [15] (cf. [50, Section 5]). Thus its codi-mension is ’(u)+ ’(v), and the fundamental cycle of XuF. c XvF: is Poincar6dual to u" . Since

7Z[X 1,""", Xn ]-----rTZ[X1,..., Xn+m]/(ei(x Xn+m)

is an isomorphism on Z(x’... x" ai < m), identities of Schubert polynomialsfollow from product formulas for Schubert classes. The Schubert basis is self-dual. If Y(w) + ?(v) (), then

if v o0w, (2.3.1)w" 0 otherwise.

Let GrasskV be the Grassmannian of k-dimensional subspaces of V, a k(n-k)-dimensional manifold. A flag F. induces a cellular decomposition indexed bypartitions 2 c (n-k)k. The closure of the ceilindexed by 2 is the Schubert vari-ety fF.:

fLF. "= {H GrasskVl dimH c Fn+j_k_, >j, j 1,...,k}.

The cohomology class Poincar dual to the fundamental cycle of fxF. isSx(xl,... ,Xk), where Xl,... ,xk are negative Chern roots of the tautological k-plane bundle on GrasskV. Write S;t for S(x,..., xk), if k is understood. TheseSchubert classes form a basis for cohomology, p c 2 fuF. = fxF., and if F., F:are opposite flags, then

[ta F. Z C/t2=(n-k)

where the Cuv are the Littlewood-Richardson coefficients [21]This Schubert basis is self-dual. If 2 c (n- k)k, then let 2c, the complement of

2, be the partition (n- k- 2k,..., n- k- 21). Suppose I,1 / I1 k(n- k), then

S(n-k)’S2(X1,...,Xk)’SIt(XI,...,Xk) 0if/t 2c,otherwise.

We suppress the dependence of 2 on n and k.A map f:X Y between manifolds induces a group homomorphism

f." H*X H*Y via Poincar6 duality and the functorial map on homology.

This map satisfies the projection formula (cf. [19, 8.1.7]). Let e H*X andfl e H* Y, then

f.(f*= c fl) (2.3.2)

For an (oriented) manifold X of dimension d, Has Z. [pt] is generated by theclass of a point. Let deg H*X be the map that selects the coefficient of [pt].Then deg(f.fl) deg(fl).

Let nk" IFV GrasskV be defined by Ztk(E.) Ek. Then rt-lfF. Xv(,k)F.and nk Xoov(C,k)F. F. is generically one-to-one. Thus

S;t v(2,k)

@w [ S if w 090v(2c, k),0 othcrwisc.

The cohomology of IFCV x IFgW (dimW m) has an integral basis of classes@u (R) x for u e 6en and x e 6am. Likewise the cohomology of GrasskV x GrasslWhas a basis S (R) S for 2 (n- k)k and/ (m- l) t.

While we use the cohomology rings of complex varieties, our results andmethods are valid for the Chow rings [19] and/-adic (6tale) cohomology [13] ofthese same varieties over any field.

3. Orders on 9(R)

3.1. The k-Bruhat order. The k-Bruhat order <k is a suborder of the Bruhatorder on 6 related to the coefficients cw It was called the k-coloureduEhresmanoSdre in [30]. Its covers are given by the index of summation inMonk’s formula [37]

u <kw

Thus w covers u in the k-Bruhat order (u <k W) if ’(w)= ’(u)+ 1 andw u(a, b) where a < k < b. The k-Bruhat order has a nonrecursive character-ization.

THEOREM A. Let u, w e 6"o. Then u < k W if and only if(i) a < k w(b),

(ii) a < b, u(a) < u(b), and w(a) > w(b), together imply a < k < b.

Proof. We show the k-Bruhat order is the transitive relation uk w definedby (i) and (ii). If u < k u(a, b) is a cover, then uk u(a, b). Thus u <k w implies

---k W. Algorithm 3.1.1 completes the proof. [-1

ALGORITHM 3.1.1 (Produces a chain in the k-Bruhat order).nput Permutations u, w oo with uk w.output A chain in the k-Bruhat orderfrom w to u.

Output w. While u v w, do(1) Choose a < k with u(a) minimal subject to u(a) < w(a).(2) Choose k < b with u(b) maximal subject to w(b) < w(a) < u(b).(3) w :-- w(a, b), output w.At every iteration of 1, U<kW. Moreover, this algorithm terminates in

e(w) -e(u) iterations and the sequence ofpermutations produced is a chain in thek-Bruhat orderfrom w to u.

Proof. It suffices to consider a single iteration. We show it is possible tochoose a and b, then uk w(a, b), and finally w(a, b) ".k w.

In (1), u v w, so one may always choose a. Suppose Uk w 6t’n and it isnot possible to choose b. In that case, if j > k and w(j)< w(a), then alsou(j) < w(a). Similarly, ifj < k and w(j) < w(a), then u(j) < w(j) < w(a). Thus< w(a) uw-l() < w(a), which contradicts uw-l(w(a)) u(a) < w(a).Let w’ "= w(a, b). Since w(b) > u(a) implies (i) for (u, w’), suppose w(b) < u(a).

Set bl := u-lw(b). Then w(bl) u(bl) and the minimality of u(a) shows thatbl > k and w(bl) < u(bl). Similarly, if b2 "= u-lw(bl), then b2 > k and w(b2) <u(b2). Continuing, we obtain a sequence bl,b2,.., with u(a)> u(bl)>u(b2) > ..., a contradiction.

(u, w’) satisfies (ii). Suppose < j and u(i) < u(j). If j < k, then w(i) < w(j).To show w’(i) < w’(j), it suffices to consider the case j a. But then u(i) < u(a),and thus u(i)- w(i)- w’(i), by the minimality of u(a). Then w’(i)< u(a)<w(b) w’(a). Similarly, if k < i, then w’(i) < w’(j).

Finally, suppose w does not cover w’ in the k-Bruhat order. Since w(a) > w(b),there exists a c with a < c w(c) > w(b). If k < c, then (ii) impliesu(c) > u(b) and the maximality of u(b) implies w(a) < w(c), a contradiction. Thecase c < k similarly leads to a contradiction.

Remark 3.1.2. Algorithm 3.1.1 depends only upon ( wu-.+/-nput A permutation ( 6oo.output: Permutations (,(1,... ,fro e such that if u <k(U, then

is a saturated chain in the k-Bruhat order.Output . While # e, do(1) Choose a minimal subject to a < ((a).(2) Choose fl maximal subject to ((fl) < ((a) < ft.(3) := ((a, fl), output .To see this is equivalent to Algorithm 3.1.1, set u(a) and fl u(b) so that

w(a) () and w(b) ((fl). Thus w(a, b) (, fl)u.More is true, the full interval [u, W]k depends only upon wu-.

THEOREM 3.1.3. If UkW and X <ky with wu-1 --7,X-1, then the mapH vu-lx induces an isomorphism [u, W]k - Ix, Z]k.This is a consequence of the following lemma.

LEMMA 3.1.4. Let u k W and x k Z with wu-1 ZX-1. Then< (,)u < w= x (,)x < z.

Proof. Let wu- zx-1. The position of y in u is u-l(y).Suppose (a, fl)x does not cover x in the k-Bruhat order, so there is a y with< y < fl and x-1() < x-l(y) < x-l(fl). Then we have

()... ()... (fl)

Since u k (O,)U, either k < u-l(fl) < U-I() or else uillustrate u, (g, fl)u, and w for each possibility:

-l(y) < u-1 (a) < k. We

k < u-l(fl) < U-I(’) U-l(y) < U-1(0) k

w ()... (fl)... (,) ()... ()... (fl)

Assume k < U-1 (fl) < lg -1 () Then Theorem A and (, fl)u < k w imply y > ((y)and (() < (), since < y and both have positions greater than k in (oqfl)u.Let c "= x-1 (y). If c < k, then x < k z implies y < ((y) so y ((y). Also, 0 < yimplies ()< ((y) and thus ((y)= < fl < (), a contradiction. Similarly,c > k or u-1() < u-1(0), which leads to a contradiction. Thus x "k (o, fl)x.To show y := (o,fl)x <k z, first note that (y,z) satisfies (i) of Theorem A,

because (, fl)u < k w. For (ii), we need only show the following:(a) If o < y < fl and x-1 (y) (x-l(t) SO that y yx-l(y) < yx-l(o) fl, then

gX-1 () () ((fl) ZX-1

(b) If 0 < r < fl and x-1 (fl) < x-l(?) so that a yx-l(fl) < yx-l() ?, then() < ().

If a < y <fl, then one of these does occur, as x,.k (a, fl)x=y. Supposex-1 (y) < x-l(a), as the other case is similar.

Since x-l(y) < k and x < z, we have y < (y), by condition (i). If u-l(y) <U-I(0), then (a, fl)u k W = ((y) < ((t). If u-l(fl) < u-l(y), then y ((y), and so((y) y < fl < ’(a). Since u < k (a, fl)u, we cannot have u-1 (a) < u-1 (y) <u- (fl).

Define up "= {a 10 < (()} and down (fl [fl > ((fl)}.

TrmOREM 3.1.5. Let ( 6oo.(i) For u 6aoo, u < k U if and only if the following conditions are satisfied:

(a) u-lup {1,..., k};(b) u-ldown( {k + 1, k + 2,...};(c) for all z, fl up (respectively, , fldown(), < fl and u-l()<

u-l(fl) together imply (() < ((fl).(ii) If #up < k, then there is a permutation u such that u

Proof. Statement (i) follows from Theorem A. For (ii), let {a,..., ak} containup and possibly some fixed points of (, and let {ak+l,ak+2,...} be its comple-ment in IN. Index these sets so that ((ai) < ((ai+) for # k. Define u ff’oo byu(i) ai. Then (u is Grassmannian with descent k, and Theorem A implies

3.2. A new partial order on 6oo. For (6 oco, define I’1 to be the difference of# { (g, fl) ((up) x ((down) g > fl) and

#{a,b up( or a,bdownla > b and ((a) < ((b)}

+ # {(a, b) up x down a > b}.

LEMMA 3.2.1. Ifu <k (u, then (u) + (u).

Proof. By Theorem 3.1.3, d((u)- ’(u) depends only upon (. Using the per-mutation u in the proof of Theorem 3.1.5 shows that it equals I1. If c ((c),then the number of inversions involving c is the same for both u and (u. The firstterm above counts the remaining inversions in (u and the last two terms theremaining inversions in u. E]

By Theorem 3.1.3, [u, (t/]k depends only upon ( if u k (u. In fact, it is inde-pendent of k as well. That is, if x <(x, then the map v xu-lv defines an iso-morphism [u, (U]k

Definition 3.2.2. For , r/ 6coo, let r/5 if there exist u e 6coo and a positiveinteger k such that u < k r/u < k (U. If U is chosen as in the proof of Theorem 3.1.5,then we see that /__. ( if the following occur:

(1) if < then r/() < (();(2) if > r/(), then r/()(3) if , e up (respectively, o,fl down) with 0 < and ((o)< ((fl), then<

Figure 2 illustrates

_on 6e4. For ( if’,, define ( := co0(co0.

THEOREM 3.2.3. Suppose u,(i) (6coo, _) is a graded poset with rankfunction I 1.(ii) The map 2 v(A,k) exhibits Young’s lattice of partitions with at most k

parts as an induced suborder of (6coo, 5).

(iii) If u k (u, then/] - /]u induces an isomorphism [e, ]. -- [u, (U]k.(iv) If _-< (, then -+ /]-1 induces an isomorphism [q, (]. --+ [e, (/]-1].<.(v) For every infinite set P c IN, q, /’oo ---+ ff’oo is an injection ofgraded posets.

Thus, if (,/] 5/’o are shape equivalent, then [e, (]. - [e,(vi) The map/] -+ /](-1 induces an order-reversing isom-orphism between [e, (]_

and [e, (-1]..(vii) The homomorphism ( ( on n is an automorphism of (n, 5).Theorem E(i) follows from the definition of 5 and Theorem 3.2.3(v).

Proof. Statements (i)-(v) follow from the definitions. Suppose uwith u,/]u, (u ’n. If w := (u, then wcoo < n_k/](-lwco0 < n_k(-1W(.O0, whichproves (vi). Similarly, u k w :> .1 n_k implies (vii).

(13)(24)

(1234) (1324) (1342) (1243) (1423) (1432)

(124) (234) (134) (243) (12)(34) (14)(23) (123) (142) (132) (143)

(24). (34) (23) (14) (12) (13)

FIGURE 2.

_on

Example 3.2.4. Let (=(24)(153) and /] (35) (174) {1,3,4,5,7} ((). Then21345<2(.21345 and 3215764<3/].3215764. Figure 3 shows [21342,(.21345]2 [3215764,/]. 3215764]3 and [e, (]..

45123 5273461 (24)(153)

35124 43125 4273561 5243761 (15423) (13)(24)

24135 32145 41235 3254761 4235761 5214763 (143) (123)(243)

23145 31245 3245761 4215763 (13) (23)

21345 3215764 e

FIGURE 3. Isomorphic intervals in 42, 43, and

_

3.3. Disjoint permutations. Let ( S’n and 1,..., n be the vertices of a con-vex, planar n-gon numbered consecutively. Define the directed geometric graphF to be the union of directed chords (, (()) for in the support supp of (.

Permutations ( and are disjoint if the edge sets of F and F (drawn on thesame n-gon) are disjoint as subsets of the plane. This implies (but is not equiv-alent to) supp6 c supp .

In Figure 4, the pair of cycles on the left is disjoint and the other pair is not.We relate this definition to that given in Section 1.3.

84

FIGURE 4. Graphs of the permutations (1782)(345) and (13)(24)

LEMMA 3.3.1. Let (,1 ff’oo. Then the edges ofF are disjoint from the edgesofr if and only if supp( c supp J25 and I(I + Ill loll.

Proof. Suppose and r/have disjoint support, and let (a, ((a)) be an edge ofF and (b, r/(b)) be an edge of F. The contribution of the endpoints of theseedges to I(r/[ I(I is zero if the edges do not cross, which proves the forwardimplication.For the reverse, suppose they cross. The contribution is 1 if a < ((a) and

b > r/(b) (or vice versa), and zero otherwise. Since each edge is part of a directedcycle, there are at least four crossings, one of each type shown in Figure 5. There,the numbers increase in a clockwise direction, with the least number in thenortheast (/z). Thus [(gl > I1 / [l. t2

r/(b) a b a r/(b) ((a) b ((a)

Fmum 5. Crossings

LEMMA 3.3.2. Let < fl, ( 6aoo, and suppose (-<. (, fl)(. Then(i) and fl are connected in F(,#)(;

(ii) if (c, d) is any chord meeting F(, then (c, d) meets(iii) ifp and q are connnected in F(, then they are connected in(iv) if( and r are disjoint and (’ -< (, then (’ and rl are disjoint.

Proof. Suppose uand (u(j)=fl, and set a= u(i) and b u(j). Since (u ".k (,fl)fu is a cover,< k < j, and thus a < < fl < b, as u < k (u. Thus the edges (a, fl) and (b, ) of

F(,fl) meet, proving (i).For (ii), note that F(,fl) differs from F only by the (possible) deletion of edges

(a, ) and (b, fl) and the addition of the edges (a, fl) and (b, ). Checking allpossibilities for (c,d), (a, ), and (b, fl) shows (ii). Statement (iii) follows from(ii) by considering F- (a, )- (b, fl). The contrapositive of (iv) is also a con-sequence of (ii). If (’ and 0 are not disjoint and (’ 5 (, then ( and 0 are not dis-joint.

LEMMA 3.3.3. Suppose u, (, 0 6a with (, 0 disjoint. Then

U k 0u tl . k (U and u k 0u.

Proof. Suppose u k 0u. Let < k so that u(i) < (0u(i). Sincesupp supp, , u(i) < (u(i). Similarly, if k < j, then u(j) > (u(j), showingthat (i) of Theorem A holds for the pair (u, (u).For (ii), suppose i< j, u(i)<u(j), and (u(i)>(u(j). If j<k, then

u(i) suppI. Since u <k(OU, and (,0 have disjoint supports, (u(i)= O(u(i)<q(u(j), thus u(j) supp,, and so

u(i) < u(j) < (u(i) < ou(j).

But then the edge (u(i), (u(i)) of Fc meets the edge (u(j), ou(j)) of Fq, a con-tradiction. The assumption that k < leads similarly to a contradiction. Thusu < k (u, and similarly u < k 0u.

Suppose now that u <k (u and u <k 0u. Condition (i) of Theorem A holds for(u, (0u) as ( and 0 have disjoint support. For (ii), let < j with u(i) < u(j), andsuppose j < k. If the set (u(i),u(j)} meets at most one of supp or suppq, say,supp, then u <k (t/ implies (ou(i) < (ou(j). Suppose now that u(i) supp andu(j) suppq. Since u <k(U, we have (u(i) < (u(j) u(j). But u <kOU impliesu(j) < ou(j). Thus Ofu(i) (u(i) < u(j) < ou(j) O(u(j). Similar argumentssuffice when k < i. [-]

Proof of Theorem G(i). Suppose ( and 0 are disjoint. By Lemmas 3.3.2 and3.3.3, the map [e, (]_ x [e, 0]- - [e, (0]-, defined by ((’, 0’) H ’0’, is an injection.For surjectivity, let

_0- By Lemma 3.3.2(iii) and downward induction from

(0 to , F has no edges connecting supp to supp,. Set := [supp and

" := lsupp,. Then ’", and r and " are disjoint. Surjectivity follows byshowing ’

_( and "

_0.

It suffices to consider the case -<. (, fl) (0 of a cover. By Lemma 3.3.2(i),and fl are connected in F, so we may assume that , fl suppI. Then " 0 and(,fl)’ (. We show that ’ (,fl)’ ( is a cover, which completes theproof.

Choose u 6coo with u < k u < k //U. Let a "= (’u)-1(0) and b := (’u)-1 (fl).Since ’ and r/ are disjoint, g, fl supp, and so a, b supp,. Thus (z, fl)’rlu’rlu(a,b), showing a < k < b, as ’rlu ".k (o,fl)’rlu.

Since ’ and /are disjoint and ’r/, Lemma 3.3.3 implies u <k ’u. Thus

I’1 + (u)= ’(’u). But since ’ and are disjoint and ’/-<. r/is a cover, wehave

so ’(’u)4-1 Y(’u(a,b)). Since a < k < b and (u ’u(a,b), this implies’u <k ’u. UIExample 3.3.4.

u 2316745. ThenLet (=(2354) and r/= (176), which are disjoint. Let

u <3 (r/u 3571624, u <3 flu 3516724, u <3r/u 2371645.

The intervals [u, u]3, [u, /u]3, and [u, ff/u]3 are illustrated in Figure 6.

3516724

/ \2371645 2516734 3416725

2361745 2416735

2316745 2316745

3571624

2571634 3471625 3561724

2471635 2561734 3461725 3516724

2371645 2461735 2516734 3416725

2361745 2416735

2316745

FIGURE 6. Intervals of disjoint permutations

4. Cohomological formulas and identities for the

4.1. Maps on 6"00. For p, q IN and w oo, define ep,q(W) oq’oo"

w(j) j q,

ep,q(W)(j) q j p,w(j- 1) j > p and w(j) < q,

w(j-1) + l j>pandw(j)>q.

Note that ep,p #q_{p}. If p # q, then ep,q’oo - oo is not a group homo-morphism. The map ep,q has a left inverse/p, defined by

u/p (j)u(j) 1

u(j + 1)u(j / 1)- 1

j u(p),j > p and u(j) < u(p),j > p and u(j) > u(p).

Representing permutations as matrices, u/p erases the pth row and u(p)thcolumn of u, and eva adds a new pth row and qth column consisting mostly ofzeroes, but with a 1 in the (p, q)th position. For example,

e3,3(23154) 243165; 264351/3 25341.

LEMMA 4.1.1. Suppose u < w and p, q are positive integers. Then we have thefollowing:

(i) ep,q(u) < ep,q(w).(ii) If(w) :(u) :(ep,q(W)) (ep,q(U)), then

(iii) If u, w :n and either of p or q is equal to either 1 or n + 1, then:(w) :(u)

(iv) If u <k w and u(p) w(p), then U/p < k,W/p and [u, Wig -- [u/p, w/pig,, wherek’ is equal to k if k < p, and k-1 otherwise. Furthermore, wu-l=

Proof. Suppose u .6. u(a, b) is a cover. Then 8p,q(U) < e,p,q(tl(a, b)) is a cover ifeither p < a or b < p, or else a < p < b and either q < u(a) or u(b) < q. If, how-ever, a < p < b and u(a) < q < u(b), then there is a chain of length 3 from ep,q(u)to ep,q(u(a,b)) ep,q(u)(a,b+l):

ep,q(U) ,6. ep,q(U)(a, p) .6. ep,q(U)(a, b+l, p) .6. ep,q(U)(a, b+l).

The lemma follows from this. For example, under the hypothesis of (ii), eva(wand ep,q(u) each have the same number of inversions involving q. Thus, ifp,q(U) V ep,q(W), then v(p) q.

4.2. An embeddin# of fla# manifolds. Let W V with W-n andV- n+l. Suppose f e V-W so that V (W, f). For p [n+l], define theinjection p" IF:W IFV by

<Ej-l, f>if j<p,if j>p.

PROPOSITION 4.2.1 [50, Lemma 12].eoery p, q e In+ 1],

Let E. e IFW and w e fin. Then, for

!pXwE. c Xep,(w)ln+2_qE..

Recall that e is the identity permutation.

COROLLARY 4.2.2. Let w e n and E., E e IFveW be opposite fla#s. Then IE.and n+iE are opposite fla#s in IFfV, and

@pXE. Xep,, ()@n+ E. c Xe..+, (e) %k E Xep,, (e),+ E c Xep,.+, (w) E..

Proof. Since XeE[ ]Fd’W, Proposition 4.2.1 with q 1 or n + 1 implies thatpXwE. is a subset of either intersection

X8,1(w)n+1E. Xep,n+(e)1E

or

Xs,, (e) @n+ E X,,n+l (w) 1E..

Since E. and E are opposite flags, @n+E. and klE are opposite flags, so bothintersections are generically transverse and irreducible. Since

’($p,1 (W)) ’(W) - p- 1, (ep,n+ (w)) (w) + n + 1 p,

both intersections have the same dimension as IpXwE., which proves equality.

Since e,n+ (e) v(n+l-p, p), where n + 1 p is the partition of n + 1 pinto a single part, we see that ,,n+l(e)= hn+l-p(Xl,...,Xp), the complete sym-metric polynomial of degree n+ 1 p in x,...,xp. Similarly, ,l(e)en_(x, ,xp_) Xl xp_, as ep,1 v(1p-,p- 1), where 1 p-1 is the parti-tion of p-1 into p-1 equal parts, each of size 1.

COROLLARY 4.2.3. Let w e fin. In H*IFYV,

ep,(w) hn+l-p(X,1, Xp) e,n+l(w) Xl Xp-1,

and these products are equal to (p),@w.We compute . The Pieri formulas of [50] show that if u e Sen and k, m < n

are positive integers, then

1u toow em(Xl Xk) 0

Ck,U

otherwise,(4.2.1)

1u" toow hn+l-m(X1,... ,Xk) 0

k,U ----+Wotherwise,

(4.2.2)

Ck,where u w if there is a chain in the k-Bruhat order,

U’<k ((ZI,fll)U <,k <k (Om,m)’’" (l,fll)u W,

such that fll >’">tim" When k=m, it follows that {tl,...,tk}--Cp

{u(1),...,u(k)}. When k m p-1, write for this relation. Similarly,rk,

u w if there is a chain in the k-Bruhat order,

U <k (l,fll)u <k <k (n+l-m,n+l-m)’’" (l,fll)u W

such that fll ( f12 < < fin+l-m" Recall that (Dn is the longest element in 5e,.

THEOREM 4.2.4. Let v 5fn+l. In H*IFdn,

--,(y)

{ Xi(ii) p(Xi)= 0

Xi-1

rl,,n+l-p, ep,.+l (y)

i<p,

i=p,i>p.

Proof. In H*lFd.,

$’@ E deg(@o.y $, @o) @y.y9n

By the projection formula (2.3.2) and Corollary 4.2.3, we have

deg(o.y, ff@v) deg(v. (p)[email protected]) deg(@o "ep,.+l(to.y)’Xl"’’ Xp-1).

Note that $p,n+l (O)ny) O)n+lep,1 (y). By (4.2.1), the triple product

v" e,,n+l(ony) XI’’’Xp-1

Cpis zero unless v ----. 8.p,1 (y), and in this case it equals o+1. This establishes thefirst equality of (i). For the second, use the other formula for (g,).y from Cor-ollary 4.2.3 and (4.2.2).For (ii), let -. be the tautological flag on lFg+, o. the tautological flag on

lFgn, and 1 the trivial line bundle. Then

if i<p,if i--p,

ifip.

But -xi is the Chern class of both /-i_ and oi/#i_.

4.3. The endomorphism xp O. For p IN and v oo, define

cpAp(v) "-- {y . .oo v ,p,l(Y) ).

rp,n+l-pLEMMA 4.3.1. Ifv , and p < n, then Ap(v) {y Iv ep,+l(y)).ep

Proof. If v n, P < n, and v-,w, then w n+l, so Ap(v) c n. But thenrp,n+l-p}Av(v and {ye n[V ep,n+(y)} index the two equal sums in Theorem

4.2.4(i). [:]

Let Wp [x,x2,...] -* ,[x1,x2,...] be defined by

x if < p,

(x) 0 if i= p,x_l if > p.

TxamOWM C(ii). For v oo and p IN, Fvo ’yz,4v(o) y-Proof. For p < n+ 1, the homomorphism Fv induces the map

H*IF’.+ H*IF’n by Theorem 4.2.4(ii). Choosing n large enough completesthe proof. V]

COROLLARY 4.3.2. For w, x, y oo and p IN,

uAp(x) vAv(y wAv(z

Proof. Apply t’v to the identity . y -Cy to obtain

ueAp(x) vAp(y) z wAp(z)

Expanding the product u" v and equating the coefficients of w proves theidentity.

Example 4.3.3. We compute I/3(413652). The polynomial 413652 is

-- XX22X3X4 + XlX2X3X5 -- XlX2X3X4 q’- -- XlX2X3X42 xxaxE4x5 2 XXEXaX4xs.-+- X Xa X4X5 "]- "-I-

Thus I/3(Z413652) XX2X3X4 dr" XX2X3X4 -4t- X31X2XX4 However,

and

52341 XX2X3X4

3 2 3 242531 X1X2X3X4 -Jr-X1X2X3X4

which shows I3(413652)--52341 -42531- To see that this agrees with Theo-c3rem C, we compute the permutations w such that x ---, w:

623451 631452 531642 523641

\/613452 513462

413652

FIGUR 7

Of these, only the two underlined permutations are of the form 83,1 (U)"

and

631452 e3,1(52341)

531642 e3,1(42531).

LEMMA 4.3.4. Let 2 be a partition and p,k be positive integers. ThenAp(v(2, k)) ((v(2, k’)}, where k’ k- 1 ifp < k, and k otherwise.

Proof. By the combinatorial definition of Schur functions [46, Section 4.4],.rp(v(2,k) v(2,k’). [--]

Lemma 4.3.4 implies that v(2, k’) is the only solution x to v(;t, k) ’-- tp,X (X), astatement about chains in the Bruhat order.

4.4. Identitiesfor Cu when u(p) w(p)

LEMMA 4.4.1. Let u, w 5an+l with u(p) w(p) for some p [m + 1], and sup-pose e(w) -e(u) e(Wll,) -e(uD,). Then in H*IFn+I,

Proof. Let E., E be opposite flags in W. By Proposition 4.2.1,

(4.4.1)

Note that COn(W/p) (Ogn+W)/. Since u(p) w(p), i]Sw(p)E, and n+2_u(p)E areopposite in V. Also, as :(w) ---P ’(u) :(w/p)- :(u/p), both sides of (4.4.1) havethe same dimension, so they are equal. [2]

Proof of Theorem C (i) (b). It suffices to compute this in H*IF’n+I for n suchthat p < n, v e 6en+l, and Ap(v) c ff’n. By Lemma 4.4.1,

Since -xenYco.z CxZ y for x, y, z 5an and/p,1 (OOny) fDn+l Bp,1

’,’ulv Y(lp), (ony)

Cu/vyaln+le.v,(y X.1 Xp-1,yeS/’.

by Corollary 4.2.3. Thus

w =deg(@u @o.+,w @)Ct O

E "ul’’wlljy deg (f;aln+lev,l(Y) (Xl X,_l)"

t;u/n y"yeln(o)

When p 1, this has the following consequence.w 0 unless v 1 x y. In that case,COROLLARY 4.4.2. If U(1) W(1), then cxv

WCu xy Cu/ y"

4.5. Products offlag manifolds. Let P, Q e ([n+nml); that is, P, Q c [n + m] andeach has order n. List P, Q, and their complements pc, QC in order

P=pl <’" <Pn, pc := [n + m]- P p <... < p,,Q ql < < q,, QC := In+ m]- Q q <... < q.

Define a function ep,Q ,an oq’m oq’m+n by

ep,Q(V, w)(pi) qv(i)

wl(p;) qw(j)

Here ee,Q(V, w) is the permutation matrix obtained by placing the entries of v inthe blocks P x Q and those of w in the blocks pc x QC. If P [n + 1] {p} andQ In + 1]- {q}, then ep,Q(v, e) ep,q(V)..

Suppose V n, W m, and P ctn+ml Define a maptl J"

e ]FdV x IFdW ]Fd(V W)

by e(E., F.)j (Ei, Fe Pi, P, < j). Equivalently, if el,..., e is a basis for V andf,... ,fm a basis for W, then

Ce(((e, e,,)), ((f, ,fro))) ((gl,..., gn+m)),

where gp, ei and gp? =jS. From this, it follows that if E.,E[ IFgV andF., F e IFdW are pairs of opposite flags, then ,e(E., F.) and %.+.e(E, F) areopposite flags in V W.

LEMMA 4.5.1.F. IFdW,

Let P, Q ([n+m]), I) . ,-n, and w ,P’m Then, for E. IFdV and

p(Xm,,vE. X Xo mwF.) c Xo.+,,,,,q.(v,w),Q(E.,F.),

qtl,(XvE, x XwF.) = Xee,o.(v,w)lPrOn+mQ(E.,F.).

Proof. For a flag G., define G =,oGj Gj-1. By the definition of ffQ, we have

E IpQ(E.,F.),t, and F Q(E.,1-,.)q. Since

COn+mQ n + m + 1 qn < < n + m + 1 qx,

En+l-j o.+,,,9_,(E., F.)n+m+l_,Fn+l-j o.+,,Q(E., F.)n+m+l_,l,

the lemma follows from the definitions of Schubert varieties and ,p.


COROLLARY 4.5.2. Let E.,E IF#V and F.,F IFW be pairs of oppositeflags and let P (In+m]) Set Q {m + 1 m + n}. Then, for every v 5 and

Proof. Since Ogn+m[n] Q, XeE. ]FV, and XeF. ]F’W, Lemma 4.5.1shows that pl,(XvE. XwF.) is a subset of any of the four intersections. Equalityfollows, as they have the same dimension. Indeed, for x, z 6en and y, u 6em,

:(ep,[.](x, y)) :(x) + :(y) + #{i In], j [m] lPi >

:(ee,Q(Z, u)) :(z) + :(u) + #{i [n], j e [m] p > Pi}.

Thus, :(ep,[n](x, y)) + :(ee,Q(Z, u)) :(x) + :(y) + :(z) + :(u) + n m, and so

If (x, y, z, u) is one of (v, w, e, e), (v, e, e, w), (e, w, v, e), and (e, e, v, w), then the left-hand side is the dimension of the corresponding intersection, and the right-handside is the dimension of XoE. x XwF..COROLLARY 4.5.3. Let Q {m + 1,..., m + n} fOn+m[rl]. For every v 6en,

w m, and P ([+m]), the identities hold in I-I*lF’n+m:

ee,[n](v,w) ee,Q(e,e) ee,[n](v,e) et,,Q(e,w)

ee,[n](e,w) e,,O.(v,e) ee,[n](e,e) ee,Q(V,W),

and this common cohomolooy class is (ffp).(@v (R) @w).THEOtEM 4.5.4. Let x 6an+m and P ([n+m]). Then we have the followin9.(i) Let

(v,w)px E C’,[[:ll(e,e)xV ( w

Ve,We g’I

(1)(DmceP’t"l x@o (R) @wee,[nl(e,OmW)

v,wem

(ii) Let Q {m + 1,..., m + n}. For every v n and w m, we have

+P,t.I (e,e) x ep.t, (e,o,w) x ep,t, (o+v,e) x *P.I+I (o,o,orew) x

II II II IICee,e(v,w) r,e(V,m) ee,e(,w) C",e(,m)ee,o (e,e) x ee,(e,mW) x ee,a(nv,e) x ee,(nV,mW) X"

Remark 4.5.5. Each structure constant in (ii) is of the form cCy where is oneof v x w,v x ,-.,f)-I x w, or -1 x -1. Each interval [y,(y] is isomorphic to[e, v] x [e, w]. This is consistent with the expectation that the Cx should onlydepend upon [y, z] and x.

z #o.+.y forProof. In (ii), the second row is a consequence of the first, as cyx, y, z Y’n+m, and the first row is a consequence of the identities in (i). For (i),there exist constants dVxw defined by

(R)

Since the Schubert basis is self-dual (2.3.1), we have

dxw dcg($,x. (o,,o (R) OmW))dcg(x. (ke),(o.v (R)

Each expression for ($p),(o.v (R) o.w) of Corollary 4.5.3 yields one of thesums in (i). For example, the last expression yields

dVxW deg (x e.,.t,l(e,e) ro,+,,,e.,,,.t,,l(t,,w) )..ee,tnl(V,w)t:ee,tnl (e,e) x

since O.)n+meP,[n] (V, W) ep,,,+m[n] ((.OnV, (.OmW).4.6. Maps Z[xl,x2,...] Z[y,yE,...,z.t,z2,...].

IN- P, and suppose pc is infinite. List P and pc.Let P c ]N, define pc .=

fP’Pl<P2 < "" < Ps

pC.p<p<....

if #P=s,

otherwise,


Define We" Z[xl, x2,...]---, Z[Yl, Y2,... ,z,z2,...] by

Then there exist dv(P) 7z for u, v, w 5aoo defined by

Fp(@w(X)) dvv(P) @u(y)@o(z).

For l,dIN and Rc{d+l,...,d+21} with #R=I, define /(l, d, R) :=(P [d]) w R.

THEOREM D’. Let P IN and w 5aoo. For any integers I> v(w) and dexceeding the last descent of w and any subset R of {d + 1,..., d + 2/} of car-dinality l, set n := # P(l, d, R), m := d + 21 n, and "= e(t,a,R),[n](e, e). Thendw (P) O, unless u 5an and v 5am, and in that case,

dv(P) Cnw

Moreover, dvv(P) = 0 implies that a #P c [d] exceeds the last descent ofu, andd- a exceeds the last descent of v.

Remark 4.6.1. Theorem D’ generalizes [29, 1.5] (see also [35, 4.19]) where itis shown that dO([a]) > 0. Define Ii, to be

{e(l,l,R),[n](e e) e IN, n + # (P c [/]), R {l+ 1,...,3/}, #R- l}.

For w if’n, choose "N IN so that N/3 exceeds both the re(w) and the lastdescent of w. If z e Ie with z 5av, then z e(l,d,R),[n](e e) for l, d, R satisfying

the conditions of Theorem D’. So dv(P) cn w for r Ip 5aN, which estab-lishes Theorem D.

Apply the ring homomorphism Wp to both sides of the product

Expand this in terms of q(y)(z), and equate coefficients to obtain the fol-lowing corollary.

COROLLARY 4.6.2. Let w, , rl, 5ao, and P IN. Then there exists an integerN IN such that if Ip 61, then, ...(,x),,,..,, c., , c.-("x")"c(’x’a)".,_,, , du , c,a.

u,v,,#

Proof of Theorem D’. First, n(x) e E[Xl,..., xs] whenever s exceeds the lastdescent of n (see [28]; see also [35, 4.13]). Thus, @w(X) Z[xl,...,xa]. So ifdvV(P) v 0, then u(y) Z[yl,..., Ya] and v(z) e .[Zl,... ,Zb]. Hence a, respec-tively, b, exceeds the last descent of u, respectively, v. Since deg w(x) < l, bothdeg u(y) and deg o(z) are at most I. Consider the commutative diagram

Z[*l,...,Xd]C- ,7z[y,..., y, z,..., zs]

H*IF:.+m H’F:. (R) H’IF:re.

Here, Wp is the restriction of W to g[xl,...,Xn+m]. The vertical arrows areinjective on the module g(x1... x ai < l) and its image

Moreover, as P c [d] P [d], the composition o of the top row coincideswith Wp o l. Since w(X) (Xl... x] i < l), the formula for ff,(w) in The-orem 4.5.4 computes Pp(w(X)). !-]

4.7. Products of Grassmannians. Let k < n and 1< m be integers, V- tEnand W - m. Define k,l GrasskV x GrasslW Grassk+l(V W) by

k,l (H,K) H K.

THEOREM 4.7.1(i) For every Schubert class S, H*Grassk+lV W,

k,l(S2) 2c;S,(R)S.lUcY

(ii) IfSc (R) S: e H*GrasskV (R) H* GrasstW, then

(k,l),(Stzc ( Svc) ACluvS2

where 2c, luc, and vc are defined by lu n- k- k+l-i, V m- 1-Vl+l-i, and2 m + n k l- ik+l+l_i.

Remark 4.7.2. Suppose -x1,... ,--Xk are Chern roots of the tautologicalbundle over GrasskV, -Yl,...,-Y those of the tautological bundle overGrassW, and f H*Grassk+lV 3 W (which is a symmetric polynomial in the


negative Chern roots of the tautological bundle over Grassk+lV W). Then

9, f f(x Xk yl Yl).

Let A A(z) be the ring of symmetric functions, which is the inverse limit (inthe category of graded rings) of the rings of symmetric polynomials in the vari-ables z,... ,Zn. Fixing 2 and choosing k, l,n, and rn large enough gives a newproof of the following proposition.

PROPOSITION 4.7.3 [36, 1.5.9].variables. Then

Let 2 be a partition and x, y be infinite sets of

S (x, y) s,(y),

where S# are Schurfunctions in the ring A ofsymmetric functions.If we define a linear map A A(z) A(x) (R). A(y) by A(f(z)) =f(x, y), then

A is induced by the maps 0,. Moreover, the obvious commutative diagrams ofspaces give a new proof of [36, 1.5.25]" A is a cocommutative Hopf algebra withcomultiplication A.

Proof of Theorem 4.7.1. The first statement is a consequence of the second.Schubert classes form a basis for the cohomology ring, so there exist integralconstants dv such that

a vs. (R) s..

Since the Schubert basis diagonalizes the intersection pairing,

dv deg((,/(S). (S, (R) Sv)).

Apply (k,l), and use the second assertion to obtain

d deg(S;t. (kd),(Sc (R) Svc))

The second assertion is a consequence of the following lemma.

LEMMA 4.7.4. Suppose lu, v are partitions with lu c (n- k)k and v (m- l).Let E. lFveV and F. lFveW, and let G be any flag opposite to [n](E.,F.) with

G W. Then

k,l(lucEo X flvcF.) fpC[n](E.,F.) C f(n_k), (4.7.1)

where p is the partition

We finish the proof of Theorem 4.7.1. Lemma 4.7.4 implies

(gk,l), (Size ( Svc) [p[n](E., F.) c (n_k)’G]

Since deg(S. S St) cr, we see that

C/(n-k) C2

Here,/z LI v is a skew partition with two components/t and v, and the last equal-ity is a special case of (1.3.1) in Section 1.3.

ProofofLemma 4.7.4. Since

"](n_k)t {M Grassk+lV W ldim M c Gm l}

and G’m W, we see that qk,l(GrasSkV X GrasslW) c D(n_k),G. The inclusion in(4.7.1) follows, as the definitions imply

tPk,l(gE. X vcF.) p[n](E.,F.).

Equality follows, as the cycles have the same dimension. The intersection hasdimension IP[- I(n- k)ll- I/z[ + Ivl- dimfl,E, x fvF.. [-1

5. Identities among the w

5.1. Proof of Theorem E (ii). Combining Lemma 4.3.4 with Theorem C(i)(b),we deduce the following lemma.

LEMMA 5.1.1. Suppose x <k z and x(p) z(p). Let k’ k- 1 if p < k andk’ k, otherwise. Thenfor all partitions 2, we have

Z CZ/pCx v(2,k) x/ v(2,k’)"

By Lemma 4.1.1(iv), 7,X-1 and zip(x/p)-1 are shape-equivalent.

LEMMA 5.1.2. Let x,z,u,wAan. Suppose XkZ tlkW and ZX-1--WU-1.Further suppose that w is Grassmannian with descent k, the permutation wu- hasno fixed points, and, for k < < n, u(i) x(i). Then, for all partitions 2 with atmost k parts,

CuWv(2,k) zCxv(2,k)"

Proof of Theorem E(ii) usin# Lemma 5.1.2. We reduce Theorem E(ii) toLemma 5.1.2. First, by Lemma 5.1.1, it suffices to prove Theorem E(ii) whenx,z,u, w 6an, k l, with wu-l= zx- and the permutation wu- has no fixedpoints.

Define s 6an by

u(i) l <i<k,:=

x(i) k<i<n,

and set := wu-s. Then s k and

w(i) l<i<k,t(i)=

z(i) k<i<n.

w zIt suffices to show that Cu,,(;,k and cs,,(,k) each equal ct,(;,k). Thus we may furtherassume u(i) x(i) for 1 < < k or u(i) x(i) for k < < n.

Suppose that u(i) x(i) for 1 < < k. If for v 6en, := O9ovo9o,

z w CCxv(A,k) Cuv(A,k) Cycv-(,k tv(2,k)"

Set l--n-k and ,t the partition conjugate to 2. Then . < k,, 1 < k,,e(-1) 1/-1, /3(2, k) v(2 t, l), and (i) fi(i) for < < n. Thus we mayassume x(i) u(i) for 1 < < k.

Finally, there is a permutation s 6en such that wu-ls is Grassmannian ofdescent k. Thus it suffices to further assume that w is Grassmannian with descentk, the situation of Lemma 5.1.2. [-]

We prove Lemma 5.1.2 by studying two intersections of Schubert varieties andtheir image under the projection IF#’V GrasskV. Let el,..., en be a basis for V,and set F. <(el,..., en>>. Let M(w) c Mnntl?. be the set of matrices satisfyingthe conditions

(a) M(w)i,w(i 1,(b) M(w)i,j 0 if either w(i) < j or else w- (j) < i.

Then M(w) e(w) because the only unconstrained entries of M(w) are M(w)i,jwhen j < w(i) and < w-l(j), and there are e(w) such entries.

Example 5.1.3. M(25134) is the set of matrices

a 1 0 0 0b 0 c d 1

01 0 0 00 0 10000 1

(a, b, c, d)

Fix a basis el,...,en for V. For M(w) and 1 <i< n, define the vectorjS(t) "= Y’ ,e. Thenf(),... ,f,() are the "row vectors" of the matrix , andthey form a basis for V as has determinant (-1)e(). Set E.()=((fl(),... ,fn())). Since jS()e Fw(O- Fw(0-1, we see that E.() XoowF.. Infact, M(w) parameterizes the Schubert cell XoowF.. When w is Grassmannianwith descent k, matrices in M(w) have a simple form: if k < i, then f() ew(0.For opposite flags F., F[, o0w" u is the class Poinear6 dual to the funda-

mental cycle of XoowF. X,,F. We use the projection formula (2.3.2) to computethe coefficient cuv(a,k as

Cv(2,k deg(S(xl,..., xk). @oow @u)

deg(nk),(S(x,... ,Xk). @oow" @u)deg(Sx. (nk),(oow" @u)).

Thus Lemma 5.1.2 is a consequence of Lemma 5.1.4, which shows

LEMMA 5.1.4 Let u, w, x, z satisfy the hypotheses ofLemma 5.1.2. Then, ifF. andF are opposite flags in V,

nk (XoowF. r XuF) nk (XoozF. c XxF)

and the projections 7tk onto the image cycle have the same degree.

Proof. Let el,...,e, be a basis for V such that F. ((e,...,en)) andF ((e,,... ,el)), and define M(w) as before. Let A c M(w) consist of thosematrices such that E.(0) X,F[. If j > k, set Oj() =Jj() ew(j). For j < kconstruct Oj() inductively, setting Oj() to be the intersection of F’n+_,(j) andthe affine space () + (Oi()li < j and u(i) < u(j)). Since E.() e X,F. anddimEj()cFtn+l_u(j)= #{i<jlu(i)>u(j)}, this intersection consists of asingle, nonzero vector Oj().The algebraic map A -- (0(),...,On()) V" parameterizes a basis of

V. Moreover, for A, E.()= ((01(),...,0,())), and if 1 <j < k, then

9j() Fn+l_u() Fw(j). For A,

G.(tx) :-- <<gu-,x(1)(oO, Ou-,x(n)(O0>> . XoozF. c XxF, (5.1.1)

and thus A parameterizes a subset of XoozF. cXxF’. Indeed, for 1 <j < k,gu-’x(j) e Fn+l_x(j)c Fy(j). Also for j > k, we have u-ix(j)=j w-ly(j); thusgj(o) =j() ey(j) and Gj(a) Ej(a). Then the definition of Schubert varietiesin Section 2.3 implies (5.1.1).

Both cycles XoowF. c XuF and XoozF. XxF are irreducible and have thesame dimension, (w)- E(u) ]wu-l[. Since G.(a) G.(fl) if and only if a fl,the loci of flags {G.(a) a e A} is dense in XF XooyF.. Also, for a A, wehave Gk(a)= Ek(a), as u-ix permutes {1,...,k}, showing the equality of theimage cycles. Since the association E.(a) G.(a) induces a rational map

XoowF. c XuF XooF. c XxF

covering the projections k, these projections have the same degree. This com-pletes the proof. I--1

5.2. Proof of Theorem G (ii).tions and 2 is any partition, then

We show that if and q are disjoint permuta-

c cc y c,lU,

LEMMA 5.2.1. Let (,gl e qfn+m be disjoint permutations. Suppose k > #up,l> #upq, n > #supp(, and m >/ #suppq. Let u 6en+m be a permutation such

that u < k+l(rlu. Let Q be any element of ([n+m]supp,) that contains supp( forwhich k # u-(Q) c [k + 1]. Set Qe := In + m] Q.

Define (’e 5fn and rl’e ,m by (kQ((’)=( and QC(l’lt) ---q. Set P= u-l(Q),pc u-(QC), and define v e 5n and w e 5m by u(pi) qv(o and u(p) qv(i),where

P Pl < P2 < < Pn, pe p < p <... < Pn,

Q=ql < q2 <... < qn, QC= q < q <... < qn"

Then we have the followino:(i) v < k (’v and w < ll’ltW;

(ii) u ee,Q(V, w) and (rlu= ee,Q((’v, q’w);(iii) for all pairs ofopposite flaos E., E[ lFEn and F., F ]FYm,

!PP (Xco.(’vE, xvg) x (Xtoml,wF. ( XwF)


Proof. Since U k+l (U, (i) follows from Theorem A. Statement (ii) is alsoimmediate. For (iii), Lemma 4.5.1 shows the inclusion c. Since ’ is shape-equivalent to (, as is ’ to r/, and and /are disjoint, I1 I’1 / I’1, This showsthat both cycles have the same dimension, and hence are equal, as @Q(E., F.) and. (,+)(E F) are opposite.0

Note that if u k (U, then

c deg(S,. (Zr,k),(@ooCu @u)).

Thus the skew coefficients c are defined by the identity in H*Grassk V,

(Zk),(oo," u) C S. (5.2.1)2=(n-k)

Proof of Theorem G (ii). We use the notation of Lemma 5.2.1. The followingdiagram commutes, since [k +/] {pl,..., Pk, P,..., Pf}:

IF/’, x ]F/’m IFd,,+m

Grasskffn x Grasslmk,t GraSSk+ln+m"

From this and Lemma 5.2.1, we see that

rk+l(Xon/,OuQ(E., F.) c Xuoo,/n,Q(E, F))is equal to

qk,l (nk (Xo,’oE. XoE) x nl (Xo.,rwF. c XwF)).

Thus (nk+l), (mn+m(u" u) is equal to

(,),((),(,.) (),(.,. )).

This, together with (5.2.1) and Theorem 4.7.1(ii), gives

C(A’lSA (k+/), ((Dn+ml’lll"

Clu Cv (k,l), (Su ( Svc)It - -v vS2",v

We are done, as (’, and ’, are shape-equivalent pairs.

5.3. Cyclic shiftTHOM H’ (Cyclic shift). Let u, w, x, z with u k w and x z. Suppose

wu- n and zx- is shape-equivalent to (wu-) (12"’’n)’, for some t. Then, forevery partition 2,

c,k zCx u(,l)"

Proof. It suffices to prove a restricted case. Suppose u, w e if’n, u < k w, and wis Grassmannian with descent k. We construct permutations x,z 6fn withx <k z and zx-1 (wu-1) (12"’’n) for which

nk (XoowF. XuF) rk (XoozG. c XxG). (5.3.1)

Here el,..., en is a basis for V, and the flags F., F, G., and G are

F. ((el,...,

G. ((e,, el,..., en-1

F ((en,...,el)),

G ((en-1,..., el,

Then (5.3.1) implies CuWv(2,k) Cv(2,k), which completes the proof.If wu-1 (n) n, then zx-1 1 x wu-1, which is shape-equivalent to wu-1, and

the result follows by Theorem E(ii). Assume wu-l(n) n. Then w(k)= n andu(k) < n, as w is Grassmannian with descent k. Set m := u(k), p := u-l(n)(> k),and 1"= w(p). Define x 6en by

u(j) + 1

x(j)1

m+lu(j- 1)+ 1

1 < j < k or p < j,

j=k,j=k+l,k+l<j<p.

Then x < k ,7,’= (WU-1) (12 n) X, where

w(j)-+- 1

z(j)1+ 11

w(j- 1)+ 1

l<j<korp<j,j=k,j=k+l,k+l<j<p.

To show (5.3.1), let gl(),... ,g,() for A be the parameterized basis forflags E.() XuF XoowF. constructed in the proof of Lemma 5.1.4. Sincegk() F’n+l_,,(k Fw(k), u(k) m, and w(k) n, there exist regular functionsfl() on A such that

n-1

g(o0 e,, + ’,8)(oOe.j=m

Since F <e.> Ep(o0 Ep-1 (0) and gp() el, there exist regular functionsg() on A with gp(o0 nowhere vanishing such that

p

j=l

k-1 p

j=l j=k+l

as gk() is the only gj(a) whose en-coefficient is nonzero. Thus

p k-1

e,,- gi(a)ew gk(a) + Y6(a)g(a) Ek(a) Ek-1 (00.j=k+l j=l

Define a basis hi (a),..., hn(a) for V by

h(a) e,- (Ef-k+iengj-i (00

1 <j < k or p <j,

j=k,j=k+l,k+l<j<p.

We claim E(a) <<hl(),..., hn(a)>> is a flag in XoozG. XxG, which implies(5.3.1). Since hk(a) e Ek(a) Ek-l(a) and hi(a) gi(a) for j < k, we have

E() <hi (a),..., hk(X)> Ek(a).

Thus if a #- a’, then E(a) # E(a’) and so (E(a) a e A} is a subset of the inter-section XoozG. XxG of dimension equal to dim A d(w) d(u) (z) d(x),the dimension of Xa,ozG.XxG. Thus (E(a) laeA} is dense, and soE’k(a) Ek(a) implies (5.3.1).For notational convenience, set G := G -Gi_1 and similarly for F. To

establish this claim, we first show that hi(a e Gz(i) for j 1,..., n, which shows

hi(a),... ,hn(a) is a parameterized basis for V and E()e XoozG.. Then, for a

fixed a eA, we construct h’,...,h’n that satisfy E(a)= ((h’,...,h’)) andh) e Gn+ x j for j 1,. n, showing E (=) e XxG_()Note that if < n, then Gi+l (en, Fi) Thus hi( e F .) G . for 1 j < k

and p< j, and f k+ 1 <j p, then hj()eFw( Gzf. Then, since

G (en), we see that hk+()= en e G G+I_:S. FlnallLslnce w s Grass-mannian of descent k, if k + 1 p, then ,,k,)--)w(p) l, which showshk() ll (k)" Thus () Xoz..

XWe now show that E(e) s xG.. Note that if ab<n, thenF+_, Fb G_, Gb+. Thus if 1 j < k, h(e) (e) e F+_u() F() c

G+_x(l. Since x(k)= 1, we see that h(e)e G+_x(l V. Fix e e A and set

n-1

hk+, :-- gk(a) en i(x)ej Gtn+l_(m+l) Gtn+l_x(k+l).j=m

Since hk+l + hk+l (o0 gk(), we see that E+ (0) (E(), n+1).Finally, since E.() XuF, if k < j, there exists a vector

such that <Ej-1 (0), gj> Ej(o). For k + 1 < j < p, set

lj-1 k,j-len e (en-1..., en+l-u(j-1)} Gn+l_x(j),

as gk() is the only vector among {gl(0),...,gn(0)} that is not in the span ofEel,..., en-1. If p < j, set h; g yk,len e Gn+_x(i). Then ((h,., h’n)) (a)

completing the proof. [3

6. Formulas for some

6.1. A chain-theoretic interpretation. We give a chain-theoretic interpretationfor some coefficients c which is similar to the results of [50]. If eitheru <k (, fl)u or -<. (, fl)( is a cover, label that edge in the Hasse diagram with theinteger fl max{,fl}. Given a saturated chain in the k-Bruhat order from u tou (or, equivalently, a saturated 5-chain from e to ), the word of that chain is itssequence of edge labels. Given a word 09 a.a2...am, Schensted insertion(see [47] or [46, Section 3.3]) of o into the empty tableau gives a pair (S, T) ofYoung tableaux, where S is the insertion tableau and T the recording tableau of

Let/z c 2 be partitions. A permutation is shape-equivalent to a skew Youngdiagram 2//z if there is a k such that is shape-equivalent to v(2, k) v(/z, k) -1. Itfollows that ( is shape-equivalent to some skew partition 2//z if and only if

(whenever , fl up or , fl e down),

< <

Tn.OREU F’. Let be partitions, and suppose is shape-equivalentto /. Thenfor every partition v,

(i) c[ c’, and(ii) for every standard Young tableau T ofshape v,

-<-chains from e to ( whose )word has recording tableau T

Equivalently, if u < k W and WU-1 then

( chains in k-Bruhat order from u to 1CuW(v,k) # w whose word has recording tableau T

Remark 6.1.1. Theorem F’(ii) gives a combinatorial proof of Proposition 1.1,when wu- is shape-equivalent to a skew partition. Theorem F’(ii) is deceptivelysimilar to Theorem 8 of [50].

THEOREM 8 [50]. Suppose v (p, 1q-l). Then for every u, w 6aoo and k IN,the constant Cuv(v,k counts either

(i) word a <... < ap > ap + 1 >... > a + q- 1

or

(ii) { chains in k’Bruhat orderfrom u to w with }word a >... > aq < aq + 1 <... < ap + q- 1

The recording tableaux of words in (i) have 1,2,...,p in the first row and1,p/ 1,...,p+q-1 in the first column. These are the only words with thisrecording tableau. Similarly, the recording tableaux of words in (ii) have1, 2,..., q in the first column and 1, q+l,..., p/q-1 in the first row. Theorem F,however, is not a generalization of this result: The permutation := (143652) isnot shape-equivalent to any skew partition as 4, 5 down but ((4) > (5). Never-theless, c[41 1. Interestingly, does satisfy the conclusions of Theorem F.While tle hypothesis of Theorem F is not necessary for the conclusion to

hold, some hypotheses are necessary. Let (162)(354), a product of two dis-joint 3-cycles. Then (L..6) (132)(465) v(,2). V(m, 2)- Hence, by Theo-rem H, we have

c__ =c =c =1.

(This is also a consequence of Theorem G and the form of the Pieri formula in[28] or of [50, Theorem 5].) If u 312645, then (u 561234 and the labeledHasse diagram of [u, u]2 is

561234

461235 521634

361245 421635 512634

321645 412635

312645

Fotr 8

The labels of the six chains are

2456, 2465, 2645, 4526, 4256, 4265,

and these have (respective) recording tableaux

This list omits , and the third and fourth tableaux are identical.

Proofof Theorem F’. Suppose ( v(2, k). v(/z, k) -1. Then

[e, (].< - [v(/t, k), v(2, k)] k [/z, 2]=.

The first isomorphism preserves the edge labels; in the second, these labels cor-respond to diagonals in a Young diagram. If v = v’ is a cover in Young’s lattice,

In that case, vi + 1 and the label ofthere is a unique row such that vi : vi. vthe diagonal on whichthe corresponding edge in the k-Bruhat order is k- + vi,

the new box of v’ lies.A chain in Young’s lattice from/z to 2 is a standard skew tableau R of shape

2//z. Consider its word al...am as a two-rowed array

1 2 m)W---al a2 am

Then the entry of R is in the ath diagonal.

Let S and T be, respectively, the insertion and recording tableaux of w. Con-sider the two-rowed array consisting of the columns (’) arranged in lexico-graphic order. Then the insertion and recording tableaux of this new array are Tand S, respectively (see [26], [48]).The second row of this new array, the word inserted to obtain T, is the

"diagonal" word of R; the entries of R are ordered lexicographically by diagonal.By Lemma 6.1.2, the diagonal word is Knuth-equivalent to the original word.Thus Tis the unique tableau of partition shape Knuth-equivalent to R. Thisgives a combinatorial bijection

( _-chainsfrometo(whose I / skewtableauxRofshape 1word has recording tableau T == 2/t Knuth-equivalent to T

proving the theorem in this case, as it is well-known that

skew tableaux R of shape2// Knuth-equivalent to T

Now suppose ( is shape-equivalent to v(2, k). v(p,k)-1. By Theorem E(ii),’/ proving (i). Assume 2,/t, and k have been chosen so that (Cv

p(V(2, k). v(/t,k) -1), for some P. By Theorem 3.2.3(iii), p induces an iso-

morphism

p’[e, v(2, k). v(/t,k) -11 [e, (]

If /-<. (a, fl) is a cover in [e, v(2, k) v(fl, k)-l]., then qpr -<. qp((a, fl)rl) is a coverin [e,(]_ with label p, where P =pl < p2 <’". Thus, if is a chain in[e, v(2, k-). v(p, k)-l]. whose word al,..., am has recording tableau T, then p(,)is a chain in [e, (]. with word Pal,..., Pare, which also has recording tableau T.

Order the diagonals of a skew Young tableau R beginning with the diagonalincident to the end of the first column of R. The diagonal word of R is the entriesof R listed in lexicographic order by diagonal, with magnitude-breaking ties. Thetableau on the left below has diagonal word 7 58 379 148262658. Schenstedinsertion of the initial segment 7 58 379 148 (those diagonals incident upon thefirst column) gives the tableau on the right, whose row word is the initialsegment.

5 7 8

3 4 6122

3 7 9

4 8

This observation is the key to the proof of the following lemma.

LEMMA 6.1.2.column word.

The diagonal word of a skew tableau is Knuth-equivalent to its

Proof. Let d(R) be the diagonal word of a skew tableau R. We show thatd(R) is Knuth-equivalent to the word c.d(R’), where c is the first column of Rand R’ is R with c removed. An induction completes the proof.

Suppose the first column of R has length b and R has r diagonals. For1 < j < b, let w, "= a. aJ be the subword of d(R) consisting of the jth diagonal." 1’’" sThen af <... < a, sl < s2 < < Sb, and f k < sj, then a > "k"J+ >’" > a, asthese are consecutive entries in the kth column of R.For 1 < < b, let 7 be the insertion tableau of the word w.w2.., we. Then the

kth column of 7 is a >... > a/, where sy_l < k<sy. Hence c.d(R’)=c.row(T’).Wb+l... Wr, where row(T’) is the row word of Tb with its first columnc removed. Since the column word of a tableau is Knuth-equivalent to itsrow word, we have the Knuth-equivalences c.row(T’)=r c.col(T’)= col(Tb)--r row(Tb), and this completes the proof.

6.2. Skew permutations. Define the set of skew permutations to be the small-est set of permutations containing all skew partitions v(2, k). v(/z, k)-1 which isclosed under the following conditions.

(1) Shape equivalence. If g is shape-equivalent to a skew permutation (, then /is skew.

(2) Cyclic shift. If ( 6en is skew, then so is ((12...n).(3) Products of disjoint permutations. If (, r/are disjoint and skew, then (r/is

skew.A shape of a skew permutation is a (nonunique) skew partition O, which is

defined inductively. If ( is shape-equivalent to 2/lu, then ( has shape 2//z. If( e ff’n is a skew permutation with shape 0, then ((12...n) has shape 0. If ( andare disjoint skew permutations with respective shapes p and a, then 0/has skewshape pH a.

THEOREM 6.2.1. Let be a skew permutation with shape 0.(i) For all partitions v,

(ii) The number of chains in the interval [e, (].< is equal to the number of stan-dard Young tableaux ofshape O.

Proof. The number of standard skew tableaux of shape 0 is 0-f c, hence(ii) is consequence of (i) and Proposition 1.1. To show (i), we need only considerthe last part (3) of the recursive definition of skew permutations, by TheoremsE(ii) and H. Suppose ( and / are disjoint skew permutations with respectiveshapes p and a, and for all partitions v, cv cP and Cv c. Then by TheoremG(ii),

..pLItr [--]1)

Example 6.2.2. Consider the graph of (1978)(26354)"

9

6 ,r’54

Fmum 9

Thus the two cycles (1978) and /= (26354) are disjoint. Note that ( is shape-equivalent to (1423) and (1423) (1234) (1342). Similarly, r/is shape-equivalentto (15243) and (15243)(12345)= (13542). Both of these cycles, (1423) and(15243), are skew partitions. Let 2 t and / q, and v [. Then

v(2,2) 13245, v(/,,2)= 34125, v(v,2)= 35124

and

v(2, 2) <2 (1342). v(2, 2) v(/,, 2),

v(2, 2) <2 (13542). v(2, 2) v(v, 2).

Hence, for every partition x, c / and c cV,/. Thus it follows thatc" c, where p is any of the four skew partitions:

6.3. Further remarks. For small symmetric groups, it is instructive to ex-amine all permutations and determine to which class they belong. Here, weenumerate each class in 6e4, if’s, and ,_a6.

skewpartitions

TABLE 2

shape equivalent toa skew partition

skewpermutation

4 14

5aS 42

132

21

79

311

24

120

678

If ( is one of the forty-two permutations in 6e6 that is not a skew permutation,and ( is not one of

(125634), (145236), (143652), (163254), (153)(246), or (135)(264), (6.3.1)

0 for all partitions v. It would bethen there is a skew partition 0 such that coy cinteresting to understand why this occurs for all but these six permutations. Canone characterize those permutations ( such that there exists a skew partition 0

0 for all partitions v?with c cFor each of these six "exceptional" permutations in (6.3.1), there is a skew

partition 0 for which c c for all v c ab, where a #up and b # down.For these we have 0 ab. For example, let (153)(246). If u 214365, thenu < 3 (u and there are forty-two chains in [u, (u]3. Also

c =1, c =2, c =1,

which verifies Proposition 1.1 as f 5, f 16, and fma 5. In this case,0= fl. Since upc {1,2,4} and down {6,5,3}, we see that a b 3.However, 0 ab.A bijective interpretation of the CuW(,,k) should also give a bijective proof of

Proposition 1.1. We show that a function z from chains to standard tableauxsatisfying some extra conditions provides a bijective interpretation of the Cu(,,k).Let ch[u, W]k denote the set of (saturated) chains in the interval [u, W]k. For a par-tition/ and integer m, let/ m be the set of partitions 2 with 2 -/ a horizontalstrip of length m. These arise in the classical Pieri formula

S/(Xl,...,Xk)’hm(X1,...,Xk)-- E S2(X1,...,Xk).;t/,m

If T is a standard tableau of shape/t and m any integer, let T m be the set oftableaux U that contain T as an initial segment such that U- T is a horizontalstrip whose entries increase from left to right.

THEOREM 6.3.1. Suppose that for every u < w, there is a map

standard Young tableau T whosech[u, W]k --.r

shape is a partition of d(w)- #(u)

such that we have the followin#.(1) duW,v(Lk)----- # {), ch[u, W]k z(), T} depends only upon the shape 2 of the

standard tableau T.(2) If), t.e is the concatenation oftwo chains and e, then z(t) is a subtableau

of z(7). (This means that z(7) is a recording tableau.)(3) Suppose 7=.e with tch[u,x]k and hence ech[x,w]k. Then

rk,z(tLe) () m only if x----w; e(c) := e ch[x, W]k must be unique for this tooccur.

Then, for every standard tableau T ofshape 2 and u < k w,

CVo(2,k) duW,v(Lk).Such a map z is a generalization of Schensted insertion. In that respect, the

existence of such a map would generalize Theorem F’.

Proof. We induct on 2. Assume the theorem holds for all u, w, and partitions7 with fewer rows than 2, or if 2 and 7 have the same number of rows, then thelast row of 2 exceeds the last row of 7t.

The form of the Pieri formulas expressed in [50], [55] (see also Section 4.2),and condition (3) prove the theorem when 2 consists of a single row. Assumethat 2 has more than one row, and set/z to be 2 with its last row removed. Let mbe the length of the last row of 2, and let T be any tableau of shape/z. Recall thatU shape(U) gives a one-to-one correspondence between T m and :z m.

yBy the definition of cu v(/,k), we have

u’S/(X1,’’’,Xk) E Cuv(l,k)u <ty

By the Pieri formula for Schubert polynomials,

Y w.u Sl(X1,...,Xk) hm(X1,...,Xk) E E Cuv(Iz,k)W U<ky

rk,y"--’- W

By the classical Pieri formula, this also equals

u E S(X Xk E E CWu v(n,k) w"n lz m w rt lz , m

Hence

w yE Cur(n,k) E Cur(It,k)"telt,m u <k Y

rk,y’---+ W

We exhibit a bijection between the two sets

gT,k,m H { ch[u, y] [z() T}U <ky

y.--.-- w

and Hrtela,m L, where

Ln {), e ch[u, W]k z(,) e T m and z() has shape

This completes the proof. Indeed, by the induction hypothesis

MT,k,m E YCu v(l.t,k)U <kyrk,

y’-- W

and for n /t rn with n : 2,

# Ln Cuv(,k).

Thus the bijection shows

CuWv(2,k) E Y E wCuv(g,k Cuv(n,k # L,

U<ky nlz*m

y----- w

which is #r-(U), for any U of shape 2.To construct the desired bijection, consider first the map

MT,k,m H Ln

defined by t e ch[u, y]k--.e(t). By property (3), z(t.e(t)) e T m, so this injectivemap has the stated range. To see that it is surjective, let rr e t m and e L. Letbe the first Igl steps in the chain 7, so that &e, and suppose s ch[u, Y]k"

rk,Then z()= T, so z(.e) z(),m. By (3), this implies y---w, and hence- MT,k,m. I--]


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BERGERON: DEPARTMENT OF MATHEMATICS AND STATISTICS, YORK UNIVERSITY, TORONTO; ONTAJOM3J 1P3, CANADA; [email protected]; http ://www.math. yorku, ca/Who/Faculty/Bezgezon

SOTTILE: DEPARTMENT OF MATHEMATICS, UNIVERSrrY OF WISCONSIN, MADISON, WISCONSIN 53706,USA; [email protected]; http ://www. math. wisc. edu/" sott ile

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