Avoiding maximal parabolic subgroups of S k

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1 Discrete Mathematics an Theoretical Computer Science 4, 2000, Avoiing maximal paraolic sugroups of S k Toufik Mansour 1 an Alek Vainshtein 2 Department of Mathematics an Department of Computer Science, University of Haifa, Haifa, Israel tmansur@stuy.haifa.ac.il 2 alek@mathcs.haifa.ac.il receive 4 Jul 2000, accepte 16 Nov We fin an explicit expression for the generating function of the numer of permutations in S n avoiing a sugroup of S k generate y all ut one simple transpositions. The generating function turns out to e rational, an its enominator is a rook polynomial for a rectangular oar. Keywors: permutations, forien patterns, paraolic sugroups, Laguerre polynomials, rook polynomials 1 Introuction an Main Result Let [p] = {1,..., p} enote a totally orere alphaet on p letters, an let α = α 1,...,α m [p 1 ] m, β = β 1,...,β m [p 2 ] m. We say that α is orer-isomorphic to β if for all 1 i < m one has α i < α if an only if β i < β. For two permutations π S n an τ S k, an occurrence of τ in π is a susequence 1 i 1 < i 2 <... < i k n such that π i1,...,π ik is orer-isomorphic to τ; in such a context τ is usually calle the pattern. We say that π avois τ, or is τ-avoiing, if there is no occurrence of τ in π. Pattern avoiance prove to e a useful language in a variety of seemingly unrelate prolems, from stack sorting [Kn, Ch ] to singularities of Schuert varieties [LS]. A natural generalization of single pattern avoiance is suset avoiance; that is, we say that π S n avois a suset T S k if π avois any τ T. The set of all permutations in S n avoiing T S k is enote S n T. A complete stuy of suset avoiance for the case k = 3 is carrie out in [SS]. For k > 3 the situation ecomes more complicate, as the numer of possile cases grows rapily. Recently, several authors have consiere the case of general k when T has some nice algeraic properties. Paper [BDPP] treats the case when T is the centralizer of k 1 an k uner the natural action of S k on [k] see also Sec. 3 for more etail. In [AR], T is a Kazhan Lusztig cell of S k, or, equivalently, the Knuth equivalence class see [St, vol. 2, Ch. A1]. In this paper we consier the case when T is a maximal paraolic sugroup of S k. Let s i enote the simple transposition interchanging i an i + 1. Recall that a sugroup of S k is calle paraolic if it is generate y s i1,...,s ir. A paraolic sugroup of S k is calle maximal if the numer of its generators equals k 2. We enote y P l,m the maximal paraolic sugroup of S l+m generate y s 1,...,s l 1,s l+1,...,s l+m 1, an y f l,m n the numer of permutations in S n avoiing all the patterns in P l,m. In this note we fin an explicit expression for the generating function of the sequence { f l,m n}. Partially supporte y HIACS c 2000 Maison e l Informatique et es Mathématiques Discrètes MIMD, Paris, France

2 68 Toufik Mansour an Alek Vainshtein To e more precise, we prove the following more general result. Let us enote σ = s 1 s 2...s k 1, that is, σ = 2,3,...,k,1 written in one-line notation, an let a e an integer, 0 a k 1 here an in what follows k = l + m. We enote y fl,m a n the numer of permutations in S n avoiing the left coset σ a P l,m ; in particular, fl,m 0 n coincies with f l,mn. Let Fl,m a x enote the generating function of { f a l,m n}, F a l,m x = n 0 Recall that the Laguerre polynomial L α n x is given y f a l,m nxn. Ln α x = 1 n! ex α n x e x x n x n+α, an the rook polynomial of the rectangular s t oar is given y R s,t x = s!x s L t s s x 1 for s t an y R s,t x = R t,s x otherwise see [Ri, Ch. 7.4]. Main Theorem. Let λ = min{l, m}, µ = max{l, m}, then Fl,m a λ 1 xr l,m x = x r r l m r! 1 or, equivalently, F a where k = l + m = λ + µ. l,m x = k 1 x r r! 1λ x µ λ!l µ λ λ x 1 r + 1 λ x λ λ! λ 1 k + r!x r λ =r+1 µ λ 1 x r r! 1 l The proof of the Main Theorem is presente in the next section. As a corollary we immeiately get the following result see [Ma, Theorem 1]. Corollary 1 Let 0 a k 1, then { f1,k 1 a k 2!k 1 n = n+2 k for n k n! for n < k. Proof. Since R 1,k 1 = 1 + k 1x, the Main Theorem implies an the result follows. F a 1,k 1 µ r 1 m k+r 1 k 3 x = xr+1 k r 2r! = xk 2 k 3 k 2! 1 k 1x 1 k 1x + x r r!, Another immeiate corollary of the Main Theorem gives the asymptotics for fl,m a n as n. Corollary 1.2. fl,m a n cγn, where c is a constant epening on l an m, an γ is the maximal root of L µ λ λ ; in particular, γ k l 1m 1. Proof. Follows from stanar results in the theory of rational generating functions see e.g. [St, vol. 1, Ch. 4] an the fact that all the roots of Laguerre polynomials are simple see [Sz, Ch. 3.3]. The upper oun on γ is otaine in [IL]., λ,

3 Avoiing maximal paraolic sugroups of S k 69 2 Proofs First of all, we make the following simple, though useful oservation. Lemma 2 For any natural a, l, m, n such that 1 a l + m 1 one has fl,m a m+l a n = f n. Proof. Denote y ρ n an κ n the involutions S n S n that take i 1,i 2,...,i k to i k,...,i 2,i 1 reversal to n + 1 i 1,n + 1 i 2,...,n + 1 i n complement, respectively. It is easy to see that for any T S k, the involutions ρ n an κ n provie natural iections etween the sets S n T an S n ρ k T, an etween S n T an S n κ k T, respectively. It remains to note that ρ k κ k σ a P l,m = σ l+m a P m,l. From now on we assume that a 0, l 1, m 1 are fixe, an enote = n m + a. It follows from Lemma 2 that we may assume that a m, an hence n. This means, in other wors, that τ S k elongs to σ a P l,m if an only if τ 1,...,τ l is a permutation of the numers a + 1,...,a + l. In what follows we usually omit the inices a, l, m whenever appropriate; for example, instea of fl,m a n we write ust f n. For any n k an any such that 1 n, we enote y g n i 1,...,i = g a n;l,m i 1,...,i the numer of permutations π S n σ a P l,m such that π = i for = 1,...,. It is natural to exten g n to the case = 0 y setting g n /0 = f n. The following properties of the numers g n i 1,...,i can e euce easily from the efinitions. Lemma 3 m,l i Let n k an 1 i n, then g n...,i,...,i,... = 0. ii Let n k an a + 1 i for = 1,...,l, then g n i 1,...,i l = 0. iii Let n k, 1 r l, an a + 1 i for = 1,...,, r, then { gn 1 i g n i 1,...,i = 1 1,...,i r 1 1,i r+1 1,...,i 1 if 1 i r a g n 1 i 1,...,i r 1,i r+1,...,i if + 1 i r n. Proof. Property i is evient. Let us prove ii. By i, we may assume that the numers i 1,...,i l are istinct. Take an aritrary π S n such that π = i for = 1,...,l. Eviently, for any r a there exists a position r > l such that π r = r; the same is true for any r + 1. Therefore, the restriction of π to the positions 1,2,...,l, 1, 2,..., a, +1, +2,..., n in the proper orer gives an occurrence of τ σ a P l,m in π. Hence, π / S n σ a P l,m, which means that g n i 1,...,i l = 0. To prove iii, assume first that 1 i r a. Let π S n an π = i for = 1,...,. We efine π S n 1 y π 1 for 1 r 1, π = π +1 1 for r an π +1 > i r, 1 π +1 for r an π +1 < i r. We claim that π S n σ a P l,m if an only if π S n 1 σ a P l,m. Inee, the only if part is trivial, since any occurrence of τ σ a P l,m in π immeiately gives rise to an occurrence of τ in π. Conversely, any

4 70 Toufik Mansour an Alek Vainshtein occurrence of τ in π that oes not inclue i r gives rise to an occurrence of τ in π. Assume that there exists an occurrence of τ in π that inclues i r. Since r l, this occurrence of τ contains a entries that are situate to the right of i r an are strictly less than i r. However, the whole π contains only a 1 such entries, a contraiction. It now follows from 1 that property iii hols for 1 i r a. The case + 1 i r n is treate similarly with the help of the transformation π S n π S n 1 given y π = π for 1 r 1, π +1 1 for r an π +1 > i r, π +1 for r an π +1 < i r. Now we introuce the quantity that plays the crucial role in the proof of the Main Theorem. For n k an 1 l we put An, = A a l,m n, = i 1,...,i =a+1 As efore, this efinition is extene to the case = 0 y setting An,0 = g n /0 = f n. g n i 1,...,i. Theorem 4 Let n k + 1 an 1 l 1, then An, + 1 = An, m An 1, An 1, 1. 2 Proof. First of all, we introuce two auxiliary sums: Bn, = B a +1 l,m n, = g n i 1,...,i, i 1,...,i =a+1 Cn, = Cl,m a n, = g n i 1,...,i, i 1,...,i =a where = n m + a; once again, Bn,0 = Cn,0 = f n. Let us prove three simple ientities relating together the sequences {An,}, {Bn,}, {Cn,}. Lemma 5 Let n k an 1 l, then: An, = An, 1 m abn 1, 1 acn 1, 1, m aan, = m abn, m abn 1, 1, aan, = acn, acn 1, 1. Proof. To prove the first ientity, oserve that y efinitions an Lemma 3iii for the case r =, one has An, 1 An, = n i 1,...,i 1 =a+1 i =1 g n i 1,...,i An,

5 Avoiing maximal paraolic sugroups of S k 71 = = i 1,...,i 1 =a+1 a i =1 g n i 1,...,i + n i =+1 g n i 1,...,i agn 1 i 1 1,...,i m ag n 1 i 1,...,i 1 i 1,...,i 1 =a+1 = acn 1, 1 + m abn 1, 1, an the result follows. The secon ientity is trivial for a = m, so assume that 0 a m 1 an oserve that y efinitions an Lemma 3ii an iii, one has Bn, = i 1,...,i =a+1 = An, + g n i 1,...,i + g n i 1,...,i 1, + 1,i +1...,i =1 i 1,...,î,...,i =a+1 g n 1 i 1,...,i 1,i +1,...,i =1 i 1,...,î,...,i =a+1 = An, + Bn 1, 1, an the result follows. Finally, the thir ientity is trivial for a = 0, so assume that 1 a m an oserve that y efinitions an Lemma 3ii an iii, one has Cn, = an the result follows. i 1,...,i =a+1 = An, + g n i 1,...,i + =1 = An, + Cn 1, 1, =1 g n i 1,...,i 1,a,i +1...,i i 1,...,î,...,i =a+1 g n 1 i 1 1,...,i 1 1,i +1 1,...,i 1 i 1,...,î,...,i =a+1 Now we can complete the proof of Theorem 4. Inee, using twice the first ientity of Lemma 5, one gets An, + 1 = An, m abn 1, acn 1, 1, An 1, = An 1, 1 m abn 2, 1 acn 2, 1. Next, the other two ientities of Lemma 5 imply An, + 1 An 1, = An, An 1, 1 m abn 1, m abn 2, 1 acn 1, acn 2, 1 = An, An 1, 1 m aan 1, aan 1,, an the result follows. The next result relates the sequence {An,} to the sequence { f n}.

6 72 Toufik Mansour an Alek Vainshtein Theorem 6 Let n k an 1 l, then An, = m 1! f n. Proof. Let Dn, = D a l,m n, enote the right han sie of the aove ientity. We claim that for n k+1 an 1 l 1, Dn, satisfies the same relation 2 as An, oes. Inee, Dn 1, = = m 1! m 1 1! 1 =1 f n 1 m f n + 1! 1 f n 1, an Dn 1, 1 = 1 = m 1 1! m 1 1! 1 =1 f n 1 1 f n, 1 an hence m Dn, m Dn 1, Dn 1, 1 = f n + m 1 +1! f n 1 m + 1! + m m + m 1 f n = m + 1 m = f n + 1! f n ! f n 1 = Dn, =1 It follows that Dn, as well as An, are efine uniquely for n k an 1 l y initial values Dk,, Dn,0, an Dn,1 Ak,, An,0, an An,1, respectively. It is easy to see that for n k one has An,0 = Dn,0 = f n. Next, the first ientity of Lemma 5 for = 1 gives An,1 = An,0 m abn 1,0 acn 1,0 = f n m f n 1 for n k. On the other han, y efinition, Dn,1 = f n m f n 1 for n k, an hence An,1 = Dn,1. Finally, a simple cominatorial argument shows that l Ak, =! k! l!m! for 1 l.

7 Avoiing maximal paraolic sugroups of S k 73 On the other han, Dk, = m 1! k! l!m!, since f r = r! for 1 r k 1 an f k = k! l!m!. To prove Ak, = Dk, it remains to check that m l 1! k! =! k!, which follows from Lemma 7 elow. Finally, we are reay to prove the Main Theorem state in Sec. 1. First of all, y Lemma 3ii, An,l = 0 for n k. Hence, y Theorem 6, l m l 1! f n = 0 for n k, or, equivalently, l m 1! l x f n x n = 0 for n k. As was alreay mentione, f r = r! for 1 r k 1, therefore, summing over n k yiels l m l 1! x F a k 1 l,m x x i i! = 0. 3 Recall that the rook polynomial of the rectangular s t oar, s t, satisfies relation s t s R s,t x =! x see [Ri, Ch. 7.4]. Hence, 3 is equivalent to Fl,m a l m xr λ,µ x = 1! l x k 1 i=0 i=0 x i k 1 i! = x r r! r m l 1 r. Let us ivie the external sum in the aove expression into three parts: the sum from r = 0 to λ 1, the sum from r = λ to µ 1, an the sum from r = µ to k 1. By Lemma 7 elow, the thir sum vanishes, while the secon sum is equal to µ 1 x r λ r λ!k r 1! r! 1, µ r 1!r! r=λ an the first expression of the Main Theorem follows. The secon expression is otaine easily from 3 an relation etween rook polynomials an Laguerre polynomials given in Sec. 1. It remains to prove the following technical result, which is apparently known; however, we faile to fin a reference to its proof, an ecie to present a short proof inspire y the rilliant ook [PWZ].

8 74 Toufik Mansour an Alek Vainshtein Lemma 7 Let 1 s t an let Then: Ms,t = Ms,t = s i=0 s t 1 i i i n. i n t s n if n s +t, s 0 if t n s +t 1, 1 s s+t n 1 s n if s n t 1. s Proof. Direct check reveals that Ms,t is a hypergeometric series; to e more precise, [ ] Ms,t = 2 F t, s 1 n ;1. Since s is a nonpositive integer, the Gauss formula applies see [PWZ, Ch. 3.5], an we get Recall that Γ z +t + sγ z Ms,t = lim. z n Γt zγs z ΓxΓ1 x = π sinπx. 4 If n s +t, we apply 4 for x = z +t + s, x = t z, x = s z, x = z, an get Γn t + 1Γn s + 1 Ms,t = Γn t s + 1Γn + 1 lim sinπt zsinπs z z n sinπzsinπt + s z = If t n s +t 1, we apply 4 for x = t z, x = s z, x = z, an get Ms,t = Γn t + 1Γn s + 1Γs +t n Γn + 1 lim z n Finally, if s n t 1, we apply 4 for x = s z, x = z, an get Ms,t = Γs +t nγn s + 1 Γt nγn + 1 lim z n sinπs z sinπz n t s n s. sinπt zsinπs z sinπz s+t n 1 = 1 s s n. s = 0. 3 Concluing remarks Oserve first, that accoring to the Main Theorem, F a l,m x oes not epen on a; in other wors, S np l,m = S n σ a P l,m for any a. We otaine this fact as a consequence of lengthy computations. A natural question woul e to fin a iection etween S n P l,m an S n σ a P l,m that explains this phenomenon.

9 Avoiing maximal paraolic sugroups of S k 75 Secon, it is well known that rook polynomials or the corresponing Laguerre polynomials are relate to permutations with restricte positions, see [Ri, Ch.7,8]. Laguerre polynomials also arise in a natural way in the stuy of generalize erangements see [FZ] an references therein. It is tempting to fin a cominatorial relation etween permutations with restricte positions an permutations avoiing maximal paraolic sugroups, which coul explain the occurrence of Laguerre polynomials in the latter context. Finally, one can consier permutations avoiing nonmaximal paraolic sugroups of S k. The first natural step woul e to treat the case of sugroups generate y k 3 simple transpositions. It is convenient to enote y P l1,l 2,l 3 with l 1 + l 2 + l 3 = k the sugroup of S k generate y all the simple transpositions except for s l1 an s l1 +l 2 ; further on, we set f l1,l 2,l 3 n = S n P l1,l 2,l 3, an F l1,l 2,l 3 x = n 0 f l1,l 2,l 3 nx n. It is easy to see that F l1,l 2,l 3 x = F l3,l 2,l 1 x, so one can assume that l 1 l 3. This sai, the main result of [BDPP] can e formulate as follows: let k 3, then k 3 F 1,1,k 2 x = x r r! + r=1 k 3!xk k 1x 1 2k 1x + k 3 2 x 2. To the est of our knowlege, this is the only known instance of F l1,l 2,l 3 x. It is worth noting that even in this, simplest case of nonmaximal paraolic sugroup, the generating function is no longer rational. References [AR] R. Ain an Yu. Roichman, Shape avoiing permutations, 1999, preprint math.co/ [BDPP] E. Barcucci, A. Del Lungo, E. Pergola, an R. Pinzani, Permutations avoiing an increasing numer of length-increasing forien susequences, Discrete Math. Theor. Comput. Sci , [FZ] D. Foata an D. Zeilerger, Laguerre polynomials, weighte erangements, an positivity, SIAM J. Discr. Math , [IL] M. Ismail an X. Li, Bouns on the extreme zeros of orthogonal polynomials, Proc. AMS , [Kn] D. Knuth, The Art of Computer Programming, vol. 1, Aison Wesley, Reaing, MA, [LS] [Ma] V. Lakshmiai an B. Sanhya, Criterion for smoothness of Schuert varieties in Sln/B, Proc. Inian Aca. Sci , T. Mansour, Permutations containing an avoiing certain patterns, Proc. 12th Conference on Formal Power Series an Algeraic Cominatorics Moscow, 2000, [PWZ] M. Petkovšek, H. Wilf, an D. Zeilerger, A=B, A. K. Peters, Wellesley, MA, [Ri] J. Rioran, An Introuction to Cominatorial Analysis, Wiley, New York, [SS] R. Simion an F. Schmit, Restricte permutations, Europ. J. Com , [St] R. Stanley, Enumerative Cominatorics, vols 1, 2, Camrige University Press, Camrige, 1997, [Sz] G. Szego, Orthogonal Polynomials, AMS, Provience, RI, 1967.

10 76 Toufik Mansour an Alek Vainshtein