A Labelling of the Faces in the Shi Arrangement

Transcription

1 Formal Power Seres and Algebrac Combnatorcs Séres Formelles et Combnatore Algébrque San Dego, Calforna 2006 A Labellng of the Faces n the Sh Arrangement Felpe Rncón Abstract. Let F n be the face poset of the n-dmensonal Sh arrangement, and let P n be the poset of parkng functons of length n wth the order defned by (a 1, a 2,..., a n (b 1, b 2,..., b n f a b for all. Pak and Stanley constructed a labellng of the regons n F n by elements of P n. We extend ths n a natural way to a labellng of all faces n F n by closed ntervals of P n, and explore some nterestng and unexpected propertes of ths bjecton. We gve some results that contrbute to characterze the ntervals that appear as labels and consequently to a better comprehenson of F n. Résumé. Sot F n l ensemble partellement ordonné des faces de l arrangement de Sh en dmenson n, et sot P n l ensemble partellement ordonné des fonctons de parkng de longueur n dont l ordre est défn par (a 1, a 2,..., a n (b 1, b 2,..., b n s a b pout tout. Pak et Stanley ont construt un étquetage des regons de F n avec des éléments de P n. On généralse cette étude de manère naturelle à un étquetage à toutes les faces de F n en utlsant des ntervalles fermés de P n et on éxamne quelques cureuses et nattendues proprétés de cette bjecton. On donne des resultats qu contrbuent à caractérser les ntervalles qu apparassent comme étquèttes et ans une melleure compréhenson de F n. 1. Prelmnares 1.1. The Sh arrangement. A n-dmensonal (real hyperplane arrangement s a fnte collecton of affne hyperplanes n R n. Any hyperplane arrangement A cuts R n nto open regons that are polyhedra (called the regons of A, so they have faces. More specfcally, faces of A are nonempty ntersectons between the closure of a regon and some or none hyperplanes n A. The poset consstng of all these faces ordered by ncluson s called the face poset of A. The n-dmensonal Sh arrangement S n conssts of the n(n 1 hyperplanes S n : x x j = 0, 1 for 1 < j n. Let F n be the face poset of S n, and let R n be the set of n-dmensonal faces n F n. Then R n s the set of closures of the regons of S n. However we wll dentfy the regons of S n wth ther closure, so we wll make no dstncton between the elements of R n and the regons of S n. Ths arrangement was frst consdered by Sh [4], who showed that R n = (n + 1 n 1. Faces of any hyperplane arrangement A can be descrbed by specfyng for every H A, whch sde of H contans the face. That s, for any H A defne H + and H as the two closed halfspaces determned by H (the choce of whch one s H + s arbtrary, and let H 0 = H. Then the elements n the face poset of A are precsely the nonempty ntersectons of the form F = H σh H A where σ H +,, 0}. So every face F s encoded by ts sgn sequence (σ H H A, where σ H 0 f and only f F H σh and F H Mathematcs Subject Classfcaton. Prmary 52C35; Secondary 06A07. Key words and phrases. hyperplane arrangements, Sh arrangement, parkng functons, face poset. These results wll be part of the author s undergraduate thess n mathematcs n Unversdad de los Andes under the drecton of Federco Ardla. The author s very grateful for all hs help and suggestons.

2 For the Sh arrangement t s useful to represent ths sequence as a matrx. We wll assume as conventon that for < j f H : x = x j then H : x x j, and f H : x = x j + 1 then H : x x j + 1. Frst defne M n as the set of all n n matrces whose entres belong to +,, 0}. Then for any F F n consder ts sgn sequence (σ H H Sn, and defne ts assocated matrx M F M n as follows: σ H f j <, where H : x j = x (M F,j = σ H f < j, where H : x = x j f = j. For example, the matrx assocated to the regon defned by x n x n 1... x 1 x n + 1 has all entres equal to, except for dagonal ones whch are 0. In general, f F F n then F s a regon f and only f all non-dagonal entres of M F are dfferent from zero. And f F, G F n then F G f and only f M G has the same entres as M F except for some non-dagonal zero entres of M F whch become or + n M G. However, there s another way of representng a face that wll be very useful for us. For notaton smplcty, f n s a postve nteger let [n] = 1, 2,...,n}. Now, f F F n, we wll say a functon X : [n] R s an nterval representaton of F f the pont ( X(1, X(2,..., X(n R n belongs to F and not to any other face properly contaned n F. We wll denote by X n the set of all functons from [n] to R. Two nterval representatons X, X X n wll be called equvalent f they represent the same face. We can magne these nterval representatons as ways n whch n numbered ntervals of length 1 can be placed on the real lne: any X X n can be thought as the collecton of the n ntervals [X(, X( + 1] for [n]. Interval [X(, X( + 1] wll be refered as the -th nterval of X. So the face represented by X s determned only by the relatve poston of the endponts of the ntervals of X Parkng functons. A parkng functon of length n s a sequence P = (P 1, P 2,..., P n [n] n such that f Q 1 Q 2... Q n s the ncreasng rearrangement of the terms of P, then Q. Parkng functons were frst consdered by Konhem and Wess [3] under a slghtly dfferent defnton, but equvalent to ours. Let P n be the poset of the parkng functons of length n wth the order defned by (P 1, P 2,..., P n (Q 1, Q 2,..., Q n f P Q for all [n]. Pak and Stanley constructed a bjecton between R n and the parkng functons of length n as follows [5]: Let R 0 R n be the regon defned by x n x n 1... x 1 x n + 1, and defne ts label λ(r 0 = (1, 1,...,1 Z n. Suppose that R, R R n, R s labelled and R s unlabelled, R and R are only separated by the hyperplane H : x = x j ( < j, and R 0 and R are on the same sde of H; then defne λ(r = λ(r+e (e Z n s the -th vector of the canoncal bass. If under the same hypothess R and R are only separated by the hyperplane H : x = x j +1 ( < j and R 0 and R are on the same sde of H; then defne λ(r = λ(r+e j. Fgure 1 shows the projecton of the arrangement S 3 on the plane defned by x + y + z = 0, and the labellng of the regons n a smplfed notaton x y=1 221 x y= y z= y z=1 x z=1 x z=0 Fgure 1. Arrangement S 3 and the labellng λ

3 Notce that n our conventon, f R R n then we have that λ(r = (a 1 + 1, a 2 + 1,...,a n + 1 where a s the number of + entres n the -th column of M R. Stanley showed that ths labellng was n fact a bjecton between R n and the elements of P n, that s, he showed that all these labels were parkng functons, each appearng once. 2. The labellng of F n We wll now extend ths labellng to all faces n F n. Frst we prove a lemma that allows us to defne the labellng. Lemma 2.1. Let F F n. Then there exst two unque regons F, F + R n such that F F, F F + and for any regon R R n, f F R then λ(f λ(r λ(f + n P n. Moreover, F F + = F. Proof. Consder an nterval representaton X X n of F. Clearly the lemma s true f F R n, that s, f there are no equaltes n X of the form X( = X(j or X( = X(j + 1 wth < j, because n ths case F = F + = F. In other case, let r be the maxmum X( for whch there exsts a j > such that X( = X(j or X( = X(j + 1. Take k as the maxmum such that X( = r. Defne a new nterval representaton X X n by X X( f k ( = X( + ɛ f = k where ɛ s a suffcently small postve real number so that for all j, f X(k < X(j then X(k + ɛ < X(j, and f X(k < X(j + 1 then X(k + ɛ < X(j + 1. So X s the same nterval representaton as X, but ts k-th nterval s moved a lttle bt to the rght. Let F F n be the face represented by X. By the defnton of X t s clear that nequaltes n X reman unchanged n X, and also equaltes that do not nvolve X(k. That s, f X( < X(j then X ( < X (j, f X( < X(j + 1 wth < j then X ( < X (j + 1, and f X( > X(j + 1 wth < j then X ( > X (j + 1. Also f X( = X(j and, j k then X ( = X (j, and f X( = X(j + 1 wth < j and, j k then X ( = X (j + 1. Notce as well that there are no equaltes n X of the form X( = X(k + 1 wth < k because t mply be a contradcton wth the maxmalty of r, nether equaltes of the form X(k = X( wth k < because they contradct the choce of k. So all equaltes n X nvolvng X(k must be of the form X(k = X( + 1 wth k <, or X( = X(k wth < k. In the frst case we have that X (k > X( + 1 = X ( + 1, so (M F k, = +. In the second case X ( = X( < X (k, so (M F k, = +. All ths shows that M F has the same entres as M F except for the non-dagonal zero entres n the k-th row and k-th column of M F, whch become + n M F. If we repeat ths constructon startng wth the face F we obtan a face F, satsfyng that M F has the same entres as M F except for some non-dagonal zero entres n M F that become + n M F. And contnung wth ths process we fnally get a face F +, such that M F + s the same matrx as M F but replacng all ts non-dagonal zero entres by +. Consder now the same constructon, but defnng X by movng the k-th nterval of X a lttle bt to the left. The non-dagonal zero entres n the k-th row and k-th column of M F become now n M F, so repeatng the process we fnally get a face F such that M F s the same matrx as M F but replacng all ts non-dagonal zero entres by. By ths descrpton of ther assocated matrces, t s easy to see that F + F = F. Now, let R R n be any regon contanng F. Remember that M R must be the same matrx as M F, but changng the nondagonal zero entres n M F by or +. Then for every [n] the number of + entres n the -th column of M R must be at least the number of + entres n the -th column of M F, and at most the number of + entres n the -th column of M F +. Hence λ(r P n must satsfy the relaton ( λ(f ( λ(r ( λ(f + for all, that s, λ(f λ(r λ(f + n P n. Ths property mples easly the unqueness of F and F +, so the proof s complete. Ths lemma s nterestng by tself, as the followng result shows. Corollary 2.2. Let R 1, R 2,...,R k R n, and defne P = (P1, P2,..., Pn = λ(r for 1 k. If ( Q = max P1, max P2,...,max Pn s not a parkng functon then k =1 R =.

4 Proof. If F = k =1 R then F F n. Hence by the Lemma we have that P λ(f + for all, but ths mples that Q s a parkng functon. We now defne the labellng of the faces n F n. Denote by Int(P n the set of all closed ntervals of P n. Defnton 2.3. The labellng λ : F n Int(P n s defned by λ(f = [ λ(f, λ(f + ]. We wll use λ also for ths labellng because t can be consdered as an extenson of the labellng we had for regons (by dentfyng λ(r wth λ(r }. Notce that dfferent faces have dfferent labels, because F F + = F for all F F n. Unfortunately, not all closed ntervals of P n appear as labels of some face. 3. Propertes of the labellng Clearly the man property of ths labellng s stated n the followng surprsng theorem. Theorem 3.1. Let F F n. Then λ(f = λ(r R R n and F R }. Proof. Let I(F = λ(r R R n and F R }. Lemma 2.1 tells us that I(F λ(f. Now, notce that n ( (λ(f λ(f = + ( λ(f + 1 =1 because P = (P 1, P 2,..., P n s a parkng functon n λ(f f and only f ( λ(f P ( λ(f + for all. Now let X X n be an nterval representaton of F, and defne A(F, = j [n] j > and X( = X(j } and B(F, = j [n] j < and X(j = X(+1 }. Then c(f, = A + B s the number of non-dagonal zero entres n the -th column of M F. So c(f, s the dfference between the number of + entres n the -th column of M F + and the number of + entres n the -th column of M F. Hence c(f, = ( λ(f + ( λ(f, and n ( λ(f = c(f, + 1. =1 We wll then prove that I(F n =1( c(f, + 1, whch s equvalent to the equalty between I(F and λ(f by a cardnalty argument. Notce that I(F s the number of regons that contan F as a face. Then I(F s the number of ways (up to equvalence n whch the ntervals of X can be moved a lttle bt, changng all equaltes n X of the form X( = X(j or X( = X(j + 1 (1 < j n to nequaltes. So we wll prove there are at least n =1( c(f, + 1 dfferent ways of dong ths. The proof s by nducton on n. If n = 2 there are 5 faces n F 2, and t s easy to check that for each one of them the equalty holds. Now assume the asserton s true for n 1. Consder F F n and let X X n be an nterval representaton of F. Let r be the mnmum X(, and let k be the mnmum such that X( = r. By the choce of k there s no such that < k and X( = X(k, or > k and X(k = X( + 1. That s, for all k we have that k / A(F, and k / B(F,. Then, gnorng the k-th nterval, by nducton hypothess there are at least k( c(f, + 1 dfferent ways of movng (as explaned before all ntervals of X except the k-th nterval. Consder one of these ways n whch these ntervals can be moved, and for k let X ( be the new poston of the -th nterval. We can assume wthout loss of generalty that the ntervals were moved very lttle, so that there exsts an open nterval U around X(k + 1 such that X ( + 1 U f and only f X(k + 1 = X( + 1, and X ( U f and only f X( = X(k + 1. Then the c(f, k ponts of X (+1 A(F, k } X ( B(F, k } separate the nterval U n c(f, k+1 dsjont open ntervals U 0, U 1,...,U c(f,k. For every j such that 0 j c(f, k let z j be some pont nsde nterval U j, and defne Y j X n as follows: Y j ( = X ( k z j f = k. So Y j s an nterval representaton obtaned by movng all ntervals of X a lttle bt (as explaned before. Because U was chosen suffcently small, Y j represents a regon n R n that contans F. Moreover, f j then Y and Y j represent dfferent regons, because Y (k U and Y j (k U j. So we have proved that for

5 every way of movng all ntervals of X except the k-th nterval there are at least c(f, k + 1 dfferent regons n R n that contan F. Hence ( ( n ( I(F c(f, k + 1 c(f, + 1 = c(f, + 1 as we wanted, so the proof s complete. Ths Theorem can also be stated as follows. k Corollary 3.2. Let R 1, R 2,..., R k R n such that F k =1 R, and defne P = (P1, P2,..., Pn = λ(r for 1 k. If R R n s such that then F R. ( mn P 1, mn P2,..., mn Pn ( λ(r max Another mportant consequence s stated n the next corollary. =1 P 1, max Corollary 3.3. Let F, G F n. Then F G f and only f λ(f λ(g. P2,..., maxpn So f we defne I n as the poset of all ntervals of P n appearng as labels, ordered by reverse ncluson, then λ s an somorphsm between F n and I n. Ths means that the characterzaton of all ntervals n I n wll gve us a complete combnatoral descrpton of F n. We already know that all ntervals of P n consstng of exactly one element appear n I n as labels of some regon. Now, every F F n has a dmenson, whch determnes the rank of F n the poset F n. To see how ths dmenson s represented n I n we need the followng defnton. Let X X n. A chan of X s a k-tuple (a 1, a 2,..., a k [n] k constructed as follows: Choose a 1 so that there s no < a 1 such that X( = X(a 1 + 1, nether > a 1 such that X(a 1 = X(. Once a j has been chosen, f there exsts some < a j such that X( = X(a j then a j+1 = max < a j X( = X(a j }. If ths does not exst, but there exsts some l > a j such that X(a j = X(l+1, then a j+1 = max l > a j X(a j = X(l + 1 }. The chan ends when there are no such nor l as n the last step. X can have several dfferent chans, but the defnton mples that all of them must be dsjont, and every [n] must belong to some chan of X. It s easy to see that chans represent sets of ntervals that are bnded one to another n X. That s, f we move a lttle bt the j-th nterval to obtan a new nterval representaton X X n, then for all n the same chan as j we must also move the -th nterval n the same way f we want X to represent the same face as X. Hence, the number of chans of X s the dmenson of the face represented by X. Proposton 3.1. Let F F n, and λ(f = [P, Q]. Then dm(f = [n] P = Q }. Proof. Let X X n be an nterval representaton of F. Remember the defntons of A(F,, B(F, and c(f, gven n the proof of Theorem 3.1. Notce that f H = (a 1, a 2,...,a k [n] k s a chan of X then c(f, a j = 0 f and only f j = 1, because a j A(F, a j+1 B(F, a j+1. So the number of chans of X s equal to the number of [n] such that c(f, = 0. But we had seen that c(f, = Q P, so the proof s complete. Contnung wth the same deas we can prove the followng proposton. Proposton 3.2. Let F F n, and λ(f = [P, Q]. Then for some m N. Q 1 P 1, Q 2 P 2,..., Q n P n } = 0, 1, 2,..., m} Proof. Let X X n be an nterval representaton of F. Notce that f H = (a 1, a 2,...,a k [n] k s a chan of X then A(F, a j+1 B(F, a j+1 A(F, a j B(F, a j a j }for all j, so c(f, a j+1 c(f, a j + 1. Then Q aj+1 P aj+1 Q aj P aj + 1 for all j. Rememberng that Q a1 P a1 = 0 we have that Q a1 P a1, Q a2 P a2,..., Q ak P ak } = 0, 1,...,m H } for some m H N. Therefore, by takng the unon over all chans of X, the proof s fnshed.

6 Last proposton restrcts a lot the possble ntervals appearng as labels, and makes a step toward the characterzaton of the elements of I n. We now characterze the possble szes of the ntervals that appear as labels of faces of a fxed dmenson. Proposton 3.3. The set λ(f F Fn and dm(f = k } s the set of all postve numbers d such that d = 2 a1 3 a2...(m + 1 am for some m N, where a > 0 for all m, and a 1 + a a m = n k. Proof. Let F F n be a face such that dm(f = k, and let λ(f = [P, Q]. Defne a = j Qj P j = }. Proposton 3.2 tells us there exsts m N such that a > 0 f and only f m. Then n λ(f = [P, Q] = (Q P + 1 = 2 a1 3 a2... (m + 1 am. =1 It s clear that a 0 + a a m = n, so by Proposton 3.1 we have that a 1 + a a m = n k. On the other hand, f we take a 0, a 1,...,a m such that a > 0 for all m and a 0 + a a m = n then t s easy to construct an nterval representaton X of a face F F n satsfyng a = j c(f, j = }. Therefore, rememberng that f λ(f = [P, Q] then c(f, j = Q j P j, the proposton follows. Remember that f F F n then λ(f = R Rn F R}, so ths proposton s also gvng some geometrcal nformaton about the Sh arrangement. Fnally, we characterze the ntervals appearng as labels of 1-dmensonal faces. Proposton 3.4. Let I = [P, Q] be an nterval of P n. Then I s the label of a 1-dmensonal face f and only f the followng statements hold: Q s a permutaton of [n]. P s determned by Q n the followng way. Denote ( a 1, a 2,..., a n = ( Q 1 (1, Q 1 (2,..., Q 1 (n, and let 0 = 0 < 1 < 2 <... < k = n be the numbers such that 1, 2,..., k 1 } = j [n] a j < a j+1 }. Then for all r [n], f j s such that j < r j+1 we have that where 1 = 0. P ar = j 1 + l [n] j 1 < l j and a l > a r } + 1, Proof. To see that the condtons are necessary, let F F n be a 1-dmensonal face such that λ(f = [P, Q], and let X X n be an nterval representaton of F. Then X conssts only of one chan H = (b 1, b 2,..., b n. Remember that Q 1 s the number of non-dagonal + or 0 entres n the -th column of M F, that s, Q = j [n] j > and X(j X(} + j [n] j < and X(j X( + 1} + 1. But all ntervals of X are on the same chan, so we have that for all j [n] j > b and X(j X(b } j [n] j < b and X(j X(b + 1} = b 1, b 2,...,b 1 }, hence Q b =. Ths shows that Q s a permutaton of [n], and that a = b for all. Notce that the numbers 0, 1,..., k satsfy that for all m, j < m j+1 f and only f X(a m = X(a 1 j. Then j = l [n] X(l > X(a1 j}. Remember also that P 1 s the number of + entres n the -th column of M F, that s, P = j [n] j > and X(j > X(} + j [n] j < and X(j > X( + 1} + 1. Let r [n] and j such that j < r j+1, so X(a r = X(a 1 j. Therefore, because X conssts only of the chan H, as we wanted. P ar = l l > a r and X(l > X(a r } + l l < a r and X(l > X(a r + 1 } + 1 = l l > a r and X(l = X(a r + 1 } + l X(l > X(ar + 1 } + 1 = l l > a r and X(l = X(a 1 (j 1 } + l X(l > X(a1 (j 1 } + 1 = m a m > a r and j 1 < m j } + j 1 + 1,

7 On the other hand, t s easy to see that f [P, Q] s an nterval of P n satsfyng the prevous condtons, then t appears as the label of a 1-dmensonal face. In fact, the functon X X n defned by X ( Q 1 ( = l [n] l < and Q 1 (l < Q 1 (l + 1 } represents a 1-dmensonal face F such that λ(f = [P, Q]. Ths characterzaton has an nterestng corollary. Corollary 3.4. Each regon R R n such that λ(r s a permutaton of [n] contans a unque 1- dmensonal face F F n. Moreover, each 1-dmensonal face F F n s contaned n a unque regon R R n such that λ(r s a permutaton of [n]. Proof. Suppose R s a regon such that λ(r s a permutaton of [n]. By the last characterzaton we know that there exsts a unque P P n such that [P, Q] s the label of a 1-dmensonal face F. Theorem 3.1 mples that F R. Moreover, f F s a one dmensonal face contaned n R then Q λ(f, and because Q s a maxmal element of P n we have that λ(f = [P, Q] for some P P n. Therefore P = P and F = F, so the face F s unque. Now, f F s a 1-dmensonal face then by the characterzaton λ(f = [P, Q], wth Q a permutaton of [n]. By Theorem 3.1, f R s the regon such that λ(r = Q then F R. Moreover, f R s a regon that contans F then Q = λ(r [P, Q]. Therefore, f Q s a permutaton of [n] then Q = Q, because Q s a maxmal element of P n. So R = R, provng that the regon R s unque. Corollary 3.5. The number of 1-dmensonal faces of S n s n!. Ths s a partcular example of a general result frst stated by Athanasads [1]. However, ths bjectve proof allows a better comprehenson of the geometrcal organzaton of these faces. 4. Perspectves After developng these results, t seems clear that there are stll many aspects to understand about ths labellng. We are now workng on three man problems. In frst place, we are tryng to acheve a total and smple characterzaton of the ntervals of I n. Ths would gve a complete combnatoral descrpton of the poset F n, thus a better comprehenson of the geometry of the Sh arrangement. We are also tryng to generalze to hgher dmensons the way n whch 1-dmensonal faces were counted, obtanng ths way a smlar result to the one gven by Athanasads [1]. Fnally, we want to apply all these results to the theory of random walks on hyperplane arrangements, as defned by Brown and Dacons n [2]. References [1] C.A. Athanasads, Characterstc polynomals of subspace arrangements and fnte felds, Adv. Math. 122 (1996, [2] K. S. Brown and P. Dacons, Random walks and hyperplane arrangements, The Annals of Probablty 26 (1998, no. 4, [3] A. G. Konhem and B. Wess, An occupancy dscplne and applcatons, SIAM J. Appled Math. 14 (1966, [4] J.-Y. Sh, The Kazhdan-Lusztg Cells n Certan Affne Weyl Groups, Lecture Notes n Mathematcs, no. 1179, Sprnger, Berln/Hedelberg/New York, [5] R. Stanley, Hyperplane arrangements, parkng functons and tree nversons, Mathematcal Essays n Honor of Gan-Carlo Rota, Brkhäuser, Boston/Basel/Berln, 1998, Departamento de Matemátcas, Unversdad de los Andes, Bogotá, Colomba E-mal address: ed-rnco@unandes.edu.co