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THE ROOK MONOID IS LEXICOGRAPHICALLY SHELLABLE MAHIRBILENCAN 0 1 0 2 1. INTRODUCTION. n a Let n be a positive integer and let [n] = {1,....,n}. Let P be a finite graded poset of rank J nwithminimumand maximumelementsdenotedbyˆ0andˆ1,respectively. LetC(P)bethe 8 2 setofpairs(x,y) ∈ P ×P suchthaty coversx. P iscalledlexicographicallyshellable(see Section 2 for a slightly more general definition), if there exists a map f : C(P) → [n] such ] O that C (1) in every interval [x,y] of P there is a unique unrefinable chain c : x = x < x < . 0 1 h ··· < x = y suchthatf(x ,x ) < f(x ,x )foralli = 0,...,k −1, t k+1 i i+1 i+1 i+2 a (2) the sequence f(c) := (f(x,x ),...,f(x ,y)) of the unique chain c from (1) is lexi- m 1 k cographicallyfirstin{f(d)|disanunrefinablechainin[x,y]}. [ 1 TheconceptoflexicographicallyshellabilityisintroducedbyBjo¨rnerin[3]andisshownto v implytheweakerpropertyofshellabilityof∆(P),thesimplicialcomplexofallchainsofP. 4 0 Shellability of a simplicial complex is a combinatorial property with important topological 1 andalgebraicconsequences. Formoresee[5],[17],or[19]. 5 . In this paper we are concerned with the question of shellability for R , the rook monoid 1 n 0 of 0/1 matrices of size n with at most one 1 in each row and each column. One can view 0 elementsofR asnon-attackingrookplacementsonann×nchessboard. Thisexplainsthe 1 n : nomenclature. ThepartialorderingonR thatweareinterestedincomesfromthetopology v n i ofthe“matrixSchubertvarieties.” Letusexplain. X Let K be an algebraically closed field, and let G = GL be the general linear group over r n a K. Denote by B ⊂ G the subgroup of invertible upper triangular matrices. Let M be the n setofalln×nmatricesoverK. NotethatM isamonoidundermatrixmultiplication. The n ZariskiclosureinM ofanorbitoftheaction n (1.1) (x,y)·g = xgy−1, g ∈ G, x,y ∈ B ofB×B onGiscalledamatrixSchubertvariety. ItiswellknownthatthematrixSchubert varieties are parametrized by the symmetric group S , and G has the “Bruhat-Chevalley n Date:January28,2010. 1 2 MAHIRBILENCAN decomposition” (cid:71) (1.2) G = BwB. w∈Sn Clearly, the action (1.1) on G extends to an action on M . In [13], Renner shows that n the orbits of the extended action are parametrized by the rook monoid R , furthermore, the n analogueof(1.2)holds: (cid:71) (1.3) M = BrB. n r∈Rn TheBruhat-Chevalley-RennerorderingonR isdefinedby n r ≤ t ⇐⇒ BrB ⊆ BtB. Here,thebarontheorbitBtB denotestheZariskiclosureinM . n ThemainresultofthispaperisthattherookmonoidR withrespecttoBruhat-Chevalley- n Renner ordering is a lexicographically shellable poset. Consequently, we know that for any interval I in R , the simplicial complex ∆(I) has the homotopy type of a wedge of spheres n orballs. Reductive monoids. The monoid of n×n matrices is an important member of the family ofvarietiescalledalgebraicmonoids. Toplaceourworkappropriatelyinthisgeneralsetting and to help the reader unfamiliar with the theory of algebraic monoids let us briefly recall thedefinitionsandrelevantcombinatorialresultswithoutdetail. Seeoneof[14],[11]or[16] formore. Let G be a reductive group. Fix a maximal torus T and a Borel subgroup B such that T ⊂ B ⊂ G. The Weyl group W associated with (G,T) is defined to be the quotient group W = N (T)/T, where N (T) is the normalizer of T in G. In the case of G = GL the G G n Weyl group is isomorphic to the symmetric group S . The Bruhat-Chevalley order on the n Weyl group W is defined by w ≤ v ⇐⇒ BwB ⊆ BvB. It is shown by different authors that the Bruhat-Chevalley orders are lexicographically shellable (see [10], [4], and in the specialcaseofthesymmetricgroup,see[7]). The generalization of the Bruhat-Chevalley ordering in the realm of algebraic monoids is due to Renner, [13]. An algebraic monoid is an algebraic variety M together with an associativebinaryoperationm : M ×M → M whichisamorphismofvarieties. An interesting class of algebraic monoids can be described as follows. Let ρ : G → 0 GL(V) be a rational representation of a semisimple algebraic group G . By abuse of nota- 0 tion,letK∗ denotethescalarmatricesinthe(affine)spaceEnd(V)oflineartransformations onV. Then,theZariskiclosureM = K∗ ·ρ(G )inEnd(V)isareductivemonoid. 0 Let G be the (reductive) group of invertible elements of a reductive monoid M, and let T ⊂ B ⊂ G be a maximal torus and a Borel subgroup. It is shown in [13] that reductive THEROOKMONOIDISLEXICOGRAPHICALLYSHELLABLE 3 monoidshavedecompositionsintodoublecosetsofB (cid:71) M = Br˙B, r˙ ∈ N (T)/T, G r∈R indexed by a finite monoid R, now called the Renner monoid of M. Here N (T) is the G Zariski closure in M of the normalizer in G of T. The Bruhat-Renner ordering on R is defined as before. In the special case of the defining representation ρ : G → GL(Kn) of 0 G = SL ,theRennermonoidR isisomorphictotherookmonoidR . TheWeylgroupW 0 n n of (G,T) forms the group of invertible elements in the Renner monoid R, and the Bruhat- ChevalleyorderingonW extendstotheBruhat-RennerorderingonR. ThereisacrosssectionlatticeΛ ⊂ Rofidempotents,parametrizingtheG×G−orbitsin M (cid:71) M = GeG. e∈Λ Furthermore, (cid:71) R = WeW. e∈Λ Let e ∈ Λ. In [12], Putcha shows that the subposets WeW ⊆ R of W × W-orbits in R are lexicographically shellable posets. It is also known that the cross section lattice Λ ⊆ R is an(upper) semimodularlattice, henceshellable. However, showing that a Rennermonoid isshellableseemstobeadifficultproblem. 2. BACKGROUND. 2.0.1. Lexicographic shellability. Let P be a finite poset with a maximum and a minimum element, denoted by ˆ1 and ˆ0 respectively. We assume that P is graded of rank n. In other words, all maximal chains of P have equal length n. Denote by C(P) the set of covering relations C(P) = {(x,y) ∈ P ×P : y coversx}. An edge-labeling on P is a map of the form f = f : C(P) → Γ for some poset Γ. The P,Γ Jordan-Ho¨ldersequence(withrespecttof)ofamaximalchainc : x < x < ··· < x < 0 1 n−1 x ofP isthen-tuple n f(c) = (f((x ,x )),f((x ,x )),...,f((x ,x ))) ∈ Γn. 0 1 1 2 n−1 n Fix an edge labeling f, and a maximal chain c : x < x < ··· < x . We call both of the 0 1 n maximalchaincanditsimagef(c)increasing,if f((x ,x )) ≤ f((x ,x )) ≤ ··· ≤ f((x ,x )) 0 1 1 2 n−1 n holdsinΓ. 4 MAHIRBILENCAN Let k > 0 be a positive integer. We consider the lexicographic (total) ordering on the k−fold cartesian product Γk = Γ ×···×Γ. An edge labeling f : C(P) → Γ is called an EL−labeling,if (1) ineveryinterval[x,y] ⊆ P ofrankk > 0thereexistsauniquemaximalchaincsuch thatf(c) ∈ Γk isincreasing, (2) theJordan-Ho¨ldersequencef(c) ∈ Γk oftheuniquechaincfrom(1)isthesmallest among the Jordan-Ho¨lder sequences of maximal chains x = x < x < ··· < x = 0 1 k y. AposetP iscalledEL-shellable,ifithasanEL−labeling. Remark 2.1. There are various lexicographic shellability conditions in the literature and the EL−shellability defined here is among the stronger ones. A deep relationship between EL−shellabilityofaCoxetergroupW andtheKazhdan-LusztigtheoryoftheHeckealgebra associatedwithW isfoundbyDyerin[6]. 2.0.2. The symmetric group. S is the set of all permutations of [n]. Let us represent the n elements of S in one line notation w = (w ,...,w ) ∈ S so that w(i) = w . It is well n 1 n n i knownthattheS isagradedposetwithrespecttoBruhat-Chevalleyordering. LetB bethe n invertibleuppertriangularmatricesinSL . GradingonS isgivenbythelengthfunction n n (2.1) (cid:96)(w) = dim(BwB)−dim(B) = inv(w), where (2.2) inv(w) = |{(i,j) : 1 ≤ i < j ≤ n, w > w }|. i j NotethatdimB = (cid:0)n+1(cid:1). 2 The Bruhat-Chevalley ordering on S is the smallest partial order generated by the tran- n sitive closure of the following (covering) relations. The permutation x = (a ,....,a ) is 1 n coveredbythepermutationy = (b ,...,b ),if(cid:96)(y) = (cid:96)(x)+1and 1 n (1) ak = bk fork ∈ {1,...,(cid:98)i,...,(cid:98)j,...,n}(hatmeansomitthosenumbers), (2) a = b ,a = b ,anda < a . i j j i i j An EL−labeling for S is constructed by Edelman [7] as follows. Let Γ = [n]×[n] be the n poset of pairs, ordered lexicographically: (i,j) ≤ (r,s) if i < r, or i = r and j < s. Define f((x,y)) = (a ,a ),ify = (b ,...,b )coversx = (a ,...,a )suchthat i j 1 n 1 n (1) ak = bk fork ∈ {1,...,(cid:98)i,...,(cid:98)j,...,n}, (2) a = b ,a = b ,anda < a . i j j i i j Forn = 3,theEL−labelingofS isasdepictedintheFigure1. 3 Theorem2.2. ([7])ThesymmetricgroupS withBruhat-Chevalleyorderingislexicograph- n icallyshellable. THEROOKMONOIDISLEXICOGRAPHICALLYSHELLABLE 5 (3,2,1) (2,3) (1,2) (2,3,1) (3,1,2) (1,3) (1,3) (1,2) (2,3) (1,3,2) (2,1,3) (2,3) (1,2) (1,2,3) FIGURE 1. EL−labelingofS3 2.0.3. Therookmonoid. Recallfrom[13]thattherankfunctiononR isgivenby n (2.3) (cid:96)(x) = dim(BxB), x ∈ R . n Thereisacombinatorialformulafor(cid:96)(x),x ∈ R similarto(2.1). Toexplainletusrepresent n the elements of R by n-tuples, as we did implicitly for S in the previous subsection. Let n n x = (x ) ∈ R anddefinethesequence(a ,...,a )by ij n 1 n (cid:40) 0 ifthej’thcolumnconsistsofzeros, (2.4) a = j i ifx = 1. ij By abuse of notation, we denote both the matrix and the sequence (a ,...,a ) by x. For 1 n example,theassociatedsequenceofthepartialpermutationmatrix   0 0 0 0 0 0 0 0 x =   1 0 0 0 0 0 1 0 isx = (3,0,4,0). Let x = (a ,....,a ) ∈ R . A pair (i,j) of indices 1 ≤ i < j ≤ n is called a coinversion 1 n n pairforx,if0 < a < a . Byabuseofnotation,weusecoinvforboththesetofcoinversion i j pairsofx,aswellasitscardinality. Example2.3. Letx = (4,0,2,3). Then,theonlycoinversionpairforxis(3,4). Therefore, coinv(x) = 1. 6 MAHIRBILENCAN In[2],weshowthatthedimension,(cid:96)(x) = dim(BxB)oftheorbitBxB,x ∈ R isgiven n by (cid:40) n (cid:88) a +n−i, ifa (cid:54)= 0 (2.5) (cid:96)(x) = ( a∗)−coinv(x), wherea∗ = i i i i 0, ifa = 0 i=1 i Ifx = (a ,...,a ) ∈ S beapermutation. Then 1 n n n (cid:88) (cid:96)(x) = ( a +n−i)−coinv(x) i i=1 (cid:18) (cid:19) (cid:18) (cid:19) n+1 n = + −coinv(x) 2 2 (cid:18) (cid:19) n+1 = +inv(x), 2 whichagreeswiththeformula(2.1). Infact,using(2.5)itiseasytoseethatifx ∈ R ,then n (cid:88) (2.6) (cid:96)(x) = a +inv(x), i where inv(x) = |{(i,j) : 1 ≤ i < j ≤ n, a > a }|. i j In[9],acharacterizationoftheBruhat-ChevalleyorderingontherookmonoidR isgiven. n Theorem 2.4. [9] Let x = (a ,...,a ), y = (b ,...,b ) ∈ R . The Bruhat-Chevalley order 1 n 1 n n onR isthesmallestpartialorderonR generatedbydeclaringx ≤ y ifeither n n (1) thereexistsan1 ≤ i ≤ nsuchthatb > a andb = a forallj (cid:54)= i,or i i j j (2) there exist 1 ≤ i < j ≤ n such that b = a , b = a with b > b , and for all i j j i i j k ∈/ {i,j},b = a . k k The following two Lemmas proved in [2] are critical for deciding whether x ≤ y is a coveringrelationornot. Lemma 2.5. Let x = (a ,...,a ) and y = (b ,...,b ) be elements of R . Suppose that 1 n 1 n n ak = bk forallk = {1,...,(cid:98)i,...,n}andai < bi. Then,(cid:96)(y) = (cid:96)(x)+1ifandonlyifeither (1) b = a +1,or i i (2) there exists a sequence of indices 1 ≤ j < ··· < j < i such that the set 1 s {a ,...,a }isequalto{a +1,...,a +s},andb = a +s+1. j1 js i i i i Example 2.6. Let x = (4,0,5,0,3,1), and let y = (4,0,5,0,6,1). Then (cid:96)(x) = 21, and (cid:96)(y) = 22. Letz = (6,0,5,0,3,1). Then(cid:96)(z) = 23. THEROOKMONOIDISLEXICOGRAPHICALLYSHELLABLE 7 Lemma 2.7. Let x = (a ,...,a ) and y = (b ,...,b ) be two elements of R . Suppose 1 n 1 n n that a = b , a = b and b < b where i < j. Furthermore, suppose that for all k ∈ j i i j j i {1,...(cid:98)i,...,(cid:98)j,...,n}, ak = bk. Then, (cid:96)(y) = (cid:96)(x) + 1 if and only if for s = i + 1,...,j − 1, eithera < a ,ora < a . j s s i Example 2.8. Let x = (2,6,5,0,4,1,7), and let y = (4,6,5,0,2,1,7). Then (cid:96)(x) = 35, and(cid:96)(y) = 36. Letz = (7,6,5,0,4,1,2). Then(cid:96)(z) = 42. 3. AN EL−LABELING OF Rn. Recall that covering relations of the Bruhat-Renner ordering on R are characterized by n the Lemma 2.5, and 2.7. For simplicity, a covering relation is called type 1 if it is as in Lemma2.5,anditiscalledtype2ifitisasinLemma2.7. Usingthesetwolemmas,wedefineanEL−labelingonR n F : C(R ) −→ Γ, n whereΓ istheposetΓ = {0,1,...,n}×{0,1,...,n}withrespecttolexicographicordering. Let(x,y) ∈ C(R ). Wedefine n (cid:40) (a ,b ), ify coversxbytype1 i i (3.1) F((x,y)) = (a ,a ), ify coversxbytype2. i j Forn = 3,theEL−labelingisasdepictedintheFigure2below. Remark3.1. (1) LetF((x,y)) = (a,b)forsome(x,y) ∈ C(R ). Then,bisnever0. n (2) If y covers x by type 2, then the set of nonzero entries of y is the same as the set of nonzeroentriesofx. Ify coversxbytype1,thenthesymmetricdifferenceoftheset of nonzero entries of y and the set of nonzero entries of x has at most 2, and at least 1elements. Theorem 3.2. Let Γ = {0,1,...,n}×{0,1,...,n}, and let F : C(R ) −→ Γ be the edge- n labeling,definedasin(3.1). ThenF isanEL−labelingforR . n WeprovethistheoreminthenextSection. ThecompletelabelingofR isshowninFigure 3 2. 4. PROOFS. Let R be the rook monoid. Let Γ = {0,...,n}×{0,...,n}. Then, for any k > 0, Γk = n Γ×···×Γ istotallyorderedwithrespecttothelexicographicordering. LetF bethelabeling on R , as defined in (3.1). Let [x,y] ⊆ R be an interval, and c : x = x < ··· < x = y be n n 0 k amaximalchainin[x,y]. LetF(c)theJordan-Ho¨ldersequenceoflabelsofc: (4.1) F(c) = (F((x ,x )),...,F((x ,x ))) ∈ Γk. 0 1 k−1 k 8 MAHIRBILENCAN (3,2,1) (2,3) (1,2) (0,1) (2,3,1) (3,1,2) (3,2,0) (0,1) (1,3) (0,2) (2,3) (1,3) (2,3) (0,1) (0,2) (1,2) (1,2) (1,3,2) (2,1,3) (2,3,0) (3,0,2) (3,1,0) (2,3) (0,1) (0,2) (1,2) (0(,01,)3) (0,2()1,2)(0,(31),3) (0,3)(2,3) (1,2) (1,3) (2,3) (0,1) (1,2,3) (0,3,2) (1,3,0) (2,0,3) (2,1,0) (3,0,1) (0,2()0,3) (1,2) (0,1) (0,3)(2,3) (0,2()1,2) (1,3) (1,2) (0(1,3,3)) (0,1) (2,3) (0,1) (2,3) (0,1) (0,2,3) (0,3,1) (1,0,3) (1,2,0) (2,0,1) (3,0,0) (0,1) (0,1) (2,3) (0,1) (0,2) (0,2) (0,1) (0,3) (1,3) (1,3) (2,3) (1,2) (2,3) (1,2) (0,1,3) (0,2,1) (0,3,0) (1,0,2) (2,0,0) (2,3) (0,1) (1,2) (0,1) (0,3) (2,3) (0,1) (0,2) (0,2) (1,2) (0,1,2) (0,0,3) (0,2,0) (1,0,0) (0,2) (2,3) (0,2) (1,2) (0,1) (0,1) (0,0,2) (0,1,0) (1,2) (0,1) (0,0,1) (0,1) (0,0,0) FIGURE 2. EL−labelingoftherookmonoidR3 THEROOKMONOIDISLEXICOGRAPHICALLYSHELLABLE 9 Proposition 4.1. Let c : x = x < ··· < x = y be a maximal chain in [x,y] such 0 k thatitsJordan-Ho¨ldersequenceF(c)islexicographicallysmallestamongallJordan-Ho¨lder sequences(ofchainsfrom[x,y])inΓk. Then, (4.2) F((x ,x )) ≤ F((x ,x )) ≤ ··· ≤ F((x ,x )). 0 1 1 2 k−1 k Beforewestartourproof,letusgiveanexampleinthecaseofn = 3. Example4.2. Letx = (0,1,0)andy = (3,1,2)inR . ItiseasytocheckfromFigure2that 3 in[x,y]themaximalchain c : x < (1,0,0) < (1,0,2) < (1,2,0) < (1,2,3) < (2,1,3) < y hasthe(lexicographically)smallestJordan-Ho¨ldersequence. Obviously, F(c) = ((0,1),(0,2),(0,2),(0,3),(1,2),(2,3)) isanon-decreasingsequence. Proof. Assumethat(4.2)isnottrue. Then,thereexistthreeconsecutiveterms x < x < x t−1 t t+1 inc,suchthat (4.3) F((x ,x )) > F((x ,x )). t−1 t t t+1 Obviously,wehavethefollowing4casestoconsider. Case1: type(x ,x ) = 1,andtype(x ,x ) = 1. t t+1 t−1 t Case2: type(x ,x ) = 1,andtype(x ,x ) = 2. t t+1 t−1 t Case3: type(x ,x ) = 2,andtype(x ,x ) = 1. t t+1 t−1 t Case4: type(x ,x ) = 2,andtype(x ,x ) = 2. t t+1 t−1 t In each of these cases we construct an element z ∈ [x,y] which covers x , and such t−1 that F((x ,z)) < F((x ,x )). Since we assume that F(c) is the lexicographically first t−1 t−1 t Jordan-Ho¨lder sequence, this provides us with the contradictions we seek. To this end, let x = (a ,...,a ),x = (b ,...,b )andx = (c ,...,c ). t−1 1 n t 1 n t+1 1 n Case 1: Since type(x ,x ) = 1, there exists an index 1 ≤ r ≤ n such that b = a for t−1 t k k all k (cid:54)= r and a < b . Likewise, there exists 1 ≤ s ≤ n such that c = b for all k (cid:54)= s, r r k k and b < c . Therefore, F((x ,x )) = (a ,b ) and F((x ,x )) = (b ,c ). Furthermore, s s t−1 t r r t t+1 s s by the assumption, (a ,b ) > (b ,c ). Since a < b , r cannot be equal to s. Otherwise, r r s s r r F((x ,x )) = (a ,b ) > F((x ,x )) = (b ,c ), which is absurd. Therefore, either t−1 t r r t t+1 r r r > s,orr < s. Hence,b = a . s s Suppose first that r > s. Define z = (d ,...,d ) ∈ R by d = a for k (cid:54)= s, and 1 n n k k d = c . It is easy to check that z covers x , and that F((x ,z)) = (a ,c ). Since s s t−1 t−1 s s F((x ,x )) = (a ,b ) > (b ,c ) = (a ,c ) = F((x ,z)),wefindacontradiction. t−1 t r r s s s s t−1 10 MAHIRBILENCAN Next,supposethatr < s. Observethata = a = 0isnotpossible(because,type(x ,x ) = r s t−1 t 1). Similar to the previous case, define z = (d ,...,d ) by d = a for k (cid:54)= s, and 1 n k k d = b . It is easy to check that z covers x and that F((x ,z)) = (a ,b ) is less s r t−1 t−1 s r than F((x ,x )) = (a ,b ). This, too, contradicts the hypotheses (on F(c)). Therefore, t−1 t r r Case1isfinished. Case 2: Since type(x ,x ) = 1, there exists r ∈ [n] such that b = c for k (cid:54)= r, and t t+1 k k b < c , and since type(x ,x ) = 2 there exist i < j such that b = a for k ∈/ {i,j}, and k k t−1 t k k b = a , b = a witha < a . Then(a ,a ) > (b ,c )by(4.3). i j j i i j i j r r Suppose first that either r < i or r > j is true. Let d = a for k (cid:54)= r and let d = c . Put k k r r z = (d ,...,d ) ∈ R . Then, z covers x and F((x ,z)) = (a ,c ) < F((x ,x )) = 1 n n t−1 t−1 r r t−1 t (a ,a ). Thisisacontradiction,asbefore. i j Next, suppose that i ≤ r ≤ j. Then, the first case is i < r < j. Since type(x ,x ) = 2, t−1 t either a > a , or a < a . If a > a , then a > a . This contradicts F((x ,x )) = r j r i r j r i t t+1 (b ,c ) = (a ,c ) < F((x ,x )) = (a ,a ). Therefore, a ≤ a . If a = a , then it is r r r r t−1 t i j r i r i easy to see that a = a = 0. But r > i, and type(x ,x ) = 2. Therefore, a = a = 0 r i t−1 t r i is not possible. So, we conclude that a > a . Since type(x ,x ) = 2, c < a . Note that i r t t+1 r i a < a . Finally, define z = (d ,...,d ) ∈ R by letting d = a for k (cid:54)= r, and d = c . It i j 1 n n k k r r iseasytoseethatz coversx ,andthatF((x ,z)) = (a ,c ) < F((x ,x )) = (a ,a ). t−1 t−1 r r t−1 t i j Thisisacontradiction,asbefore. The remaining cases are r = i and r = j. If r = j, then F((x ,x )) = (a ,a ), and t−1 t i j F((x ,x )) = (a ,c ). Therefore c < a . Define z = (d ,...,d ) ∈ R by d = a for t t+1 i j j j 1 n n k k k (cid:54)= i,andd = c . Itiseasytoseethatz coversx ,andthatF((x ,z)) < F((x ,x )). i j t−1 t−1 t−1 t This is a contradiction. Finally, if r = i, then F((x ,x )) = (a ,a ) < F((x ,x )) = t−1 t i j t t+1 (a ,c ),contradicting(4.3). ThisfinishesCase2. j i TheCase3issimilartotheCase2,soweomittheproof. Case 4: Since type(x ,x ) = 2, there exist 1 ≤ i < j ≤ n such that a < a , b = a , t−1 t i j i j b = a , and since type(x ,x ) = 2, there exist 1 ≤ r < s ≤ n such that b < b , c = b j i t t+1 r s r s andc = b . s r If j < r or s < i, define z = (d ,...,d ) by d = a for k (cid:54)= {r,s}, and d = a , d = a . 1 n k k s r r s It is easy to check that z covers x , and that F((x ,z)) = (a ,a ) < F((x ,x )) = t−1 t−1 r s t−1 t (a ,a ). Thiscontradictsthehypotheses. Therefore,oneofthefollowingholds: i j (a) i ≤ r ≤ j ≤ s,or (b) r ≤ i ≤ s ≤ j We proceed with (a). If i < r < j < s, we see that a > a . Since type(x ,x ) = 2, we i r t t+1 seefurtherthata > a > a . Definez = (d ,...,d )byd = a fork ∈/ {r,s}andd = a , j i s 1 n k k r s d = a . It is easy to see that z covers x , and that F((x ,z)) = (a ,a ) < (a ,a ) = s r t−1 t−1 r s i j F((x ,x ))). Acontradiction,asbefore. Ifi = r < j < s,thenF((x ,x )) = (a ,a ) < t−1 t t−1 t i j F((x ,x )) = (a ,a ). This contradicts (4.3). The case i < r < j = s is similar, so, we t t+1 j s