Computing generating functions for the number of descents in permutations which avoid a family of permutations that start with 1

Transcription

1 Computing generating functions for the number of descents in permutations which avoid a family of permutations that start with 1 Quang T. Bach Department of Mathematics University of California, San Diego Joint work with Jeffrey Remmel July 8, 2014 Quang T. Bach UCSD July 8, / 38

2 Outline 1 Symmetric Functions & Brick Tabloids 2 Reciprocity Method 3 Family of Patterns Results 4 Patterns with More Than One Descent Quang T. Bach UCSD July 8, / 38

3 Symmetric Functions & Brick Tabloids Quang T. Bach UCSD July 8, / 38

4 Permutation Statistics Let S n be the symmetric group and let σ = σ 1... σ n S n. Define Des(σ) = {i : σ i > σ i+1 } and des(σ) = Des(σ) σ j is a left-to-right minimum of σ if σ j < σ i, i < j. Let LRmin(σ) denote the number of left-to-right minima of σ For example, the left-to-right minima of σ = are 9,3,1, and LRmin(σ) = 3. Quang T. Bach UCSD July 8, / 38

5 τ-match Given a sequence σ = σ 1... σ n of distinct integers, let red(σ) be the permutation found by replacing the i th largest integer that appears in σ by i. For example, if σ = , then red(σ) = Given a permutation τ S j, define a permutation σ = σ 1... σ n S n to have a τ-match at place i provided red(σ i... σ i+j 1 ) = τ. We define τ-mch(σ) be the number of τ-matches in the permutation σ. Let N M n (τ) denote the set of permutations in S n with no τ-matches. Quang T. Bach UCSD July 8, / 38

6 Elementary & Homogeneous Symmetric Functions The n th elementary symmetric function e n is defined by E(t) = n 0 e n t n = i (1 + x i t). The n th homogeneous symmetric function h n is defined by H(t) = n 0 h n t n = i 1 1 x i t = 1 E( t). The n th power symmetric function p n is defined by p n = i x n i. If λ = (λ 1,..., λ l ) is a partition, then h λ = l i=1 h λ i, e λ = l i=1 e λ i, and p λ = l i=1 p λ i, Quang T. Bach UCSD July 8, / 38

7 Brick Tabloid Brick tabloid of of shape (n) and type λ: an 1 n rectangle chopped into bricks of lengths found in the partition λ. For example, if n = 8 and λ = (1 2, 3 2 ) then the six brick tabloids of shape (n) and type λ are: Quang T. Bach UCSD July 8, / 38

8 Brick Tabloid (cont.) Let B λ,(n) be the set of all of brick tabloids of shape (n) and type λ, and let B λ,(n) = B λ,(n). Eğecioğlu and Remmel proved that h n = λ n( 1) n l(λ) B λ,(n) e λ. Quang T. Bach UCSD July 8, / 38

9 Reciprocity Method by Jones & Remmel Quang T. Bach UCSD July 8, / 38

10 Generating Function Let NM τ,n (x, y) = σ N M n(τ) x LRmin(σ) y 1+des(σ) Once again, N M n (τ) is the set of permutations in S n with no τ-matches. We consider the generating functions of the form NM τ (t, x, y) = n 0 t n n! NM τ,n(x, y) Quang T. Bach UCSD July 8, / 38

11 Generating Function Jones and Remmel showed that if τ starts with 1 then ( ) 1 x NM τ (t, x, y) = U τ (t, y) where U τ (t, y) = 1 + n 1 U τ,n (y) tn n! Thus, U τ (t, y) = n 1 NM τ,n(1, y) tn n! We will apply the homomorphism method to give a combinatorial interpretation of the right hand side of the above equation. Quang T. Bach UCSD July 8, / 38

12 Homomorphism Method We define a homomorphism θ on the set of symmetric functions Λ as follows Then, θ(e n ) = ( 1)n NM τ,n (1, y) n! θ(e( t)) = n 0 NM τ,n (1, y) tn n! = 1 U τ (t, y) Hence, U τ (t, y) = 1 θ(e( t)) = θ(h(t)) n!θ(h n) = U τ,n (y) Quang T. Bach UCSD July 8, / 38

13 Homomorphism Method Therefore, to compute the generating function NM τ (t, x, y) we first need to compute n!θ(h n ) to obtain the polynomials U τ,n (y) Then U τ (t, y) = 1 + n 1 U τ,n (y) tn n! and ultimately, ( 1 NM τ (t, x, y) = U τ (t, y) ) x Quang T. Bach UCSD July 8, / 38

14 Previous Results from Jones & Remmel Jones and Remmel showed that one can compute n!θ(h n ) in several cases where des(τ) = 1 and τ starts with 1. Using the recursion formula from Eğecioğlu and Remmel Jones and Remmel showed that n!θ(h n ) = λ n h n = λ n( 1) n l(λ) B λ,(n) e λ. ( 1) l(λ) B B λ,(n) ( n b 1,..., b l(λ) ) l(λ) NM τ,bi (1, y) i=1 where B = (b 1,..., b l(λ) ) B λ,(n) Quang T. Bach UCSD July 8, / 38

15 Previous Results from Jones & Remmel Combinatorial interpretation for the right hand side of the previous equation in terms of brick tabloids: Fix a brick B = (b 1,..., b l(λ) ) Fill the cells of B with a permutation σ = σ 1... σ n such that within each brick b i, the entries in the cells reduce to permutation in N M bi (τ) Label each descents of σ that occurs within a brick by y Label y at the end of each bricks b i For example, when n = 10 and τ = 1324, a possible object created this way is -y y y -y y -y Quang T. Bach UCSD July 8, / 38

16 Previous Results from Jones & Remmel - Involution Next, Jones and Remmel defined an involution on B λ,(n) as follows Scan the cells from left to right, looking for the first cell c such that either c is is labeled with a y or c is at the end of brick b i, σ c > σ c+1, and there is no τ-match that lies entirely with in the cells of b i and b i+1 -y y y -y y -y y y y -y y y 7 3 The involution is sign-reversing, weight-preserving so the fixed-points of this involution give us n!θ(h n ) = U τ,n (y). Quang T. Bach UCSD July 8, / 38

17 Fixed Points of the Involution Lemma Let O be a fixed point of the involution. Then O satisfies the following conditions The elements in each brick of O are increasing The first element of each brick of O form an increasing sequence, reading from left to right For any two consecutive bricks b i and b i+1 in O, either (a.) There is an increase between b i and b i+1, or (b.) There is a decrease between b i and b i+1 but there is a τ-match that lies entirely with in the cells of b i and b i+1 For example, when τ = 1423, a typical fixed point of this involution is -y y -y y 10 Quang T. Bach UCSD July 8, / 38

18 U τ,n (y) for τ = 1324 Jones and Remmel showed that for certain τ, the fixed points of the involutions satisfied certain natural recurrences which allowed them to show that the U τ,n (y) satisfied simple recurrence relations. Theorem For τ = 1324, U τ,1 (y) = y, and for n 2, n/2 U τ,n (y) = (1 y)u τ,n 1 (y) + ( y) k 1 C k 1 U τ,n 2k+1 (y) k=2 where C k = 1 ( 2k ) k+1 k is the k-th Catalan number. Quang T. Bach UCSD July 8, / 38

19 U τ,n (y) for τ = p with p 5 Theorem For τ = p where p 5, U τ,1 (y) = y, and for n 2, U τ,n (y) = (1 y)u τ,n 1 (y) + n 2 +1 p 2 k=2 ( y) k 1 U τ,n [(k 1)(p 2)+1] (y) Please refer to their previous papers for further results. Quang T. Bach UCSD July 8, / 38

20 Extension to Family of Patterns Quang T. Bach UCSD July 8, / 38

21 Family of Patterns We will extend Jones and Remmel s method to permutations which avoid certain family of patterns. If Γ is a set of permutations, then we let N M n (Γ) be the set of permutations σ S n which have no consecutive occurrences of any of the permutations in Γ. Quang T. Bach UCSD July 8, / 38

22 Generating Function We are interested in generating functions of the form M Γ (x, y, t) = n 0 M Γ,n (x, y) tn n! = t n n! n 0 σ N M n(γ) x LRmin(σ) y 1+des(σ) If all the permutations ( in Γ start with 1, we can show that M Γ (x, y, t) is always of the form 1 x, U τ (t,y)) where U Γ (t, y) = 1 + n 1 U Γ,n (y) tn n! Quang T. Bach UCSD July 8, / 38

23 Homomorphism Method We show that the machinery of Jones and Remmel can easily be extended to handle the case of families of permutations that start with 1. That is, we can define a homomorphism Θ on Λ by setting Then n!θ n (h n ) = U Γ,n (y). Θ(e n ) = ( 1)n M Γ,n (1, y). n! One can then give a combinatorial interpretation to n!θ n (h n ) in terms of filled brick tabloids and define an involution on such filled brick tabloids so that the following lemma holds. Quang T. Bach UCSD July 8, / 38

24 Fixed Points of the Involution Lemma Let O be a fixed point of the involution. Then O satisfies the following conditions The elements in each brick of O are increasing The first element of each brick of O form an increasing sequence, reading from left to right For any two consecutive bricks b i and b i+1 in O, either (a.) There is an increase between b i and b i+1, or (b.) There is a decrease between b i and b i+1 but there is a τ-match that lies entirely with in the cells of b i and b i+1, for any permutation τ Γ For example, when Γ = {1324, 1423}, a typical fixed point is -y y -y y -y Quang T. Bach UCSD July 8, / 38

25 U Γ,n (y) for Γ = {1324, 1423} Theorem For Γ = {1324, 1423}, U Γ,1 (y) = y, and for n 2, U Γ,n (y) = (1 y)u Γ,n 1 (y) y(n 2) [U Γ,n 2 (y) + yu Γ,n 3 (y)] Quang T. Bach UCSD July 8, / 38

26 U Γ,n (y) for Γ = {1324, 1423} U Γ,n (y) = (1 y)u Γ,n 1 (y) y(n 2) [U Γ,n 2 (y) + yu Γ,n 3 (y)] Let O be a fixed point of the involution The first elements of each brick in O are increasing 1 is in the first cell of O If 2 is in the second cell of O, we have two possibilities depending whether 2 starts the second brick or not Remove the first cell to obtain a fixed point for n or -y 1 2 Quang T. Bach UCSD July 8, / 38

27 U Γ,n (y) for Γ = {1324, 1423} U Γ,n (y) = (1 y)u Γ,n 1 (y) y(n 2) [U Γ,n 2 (y) + yu Γ,n 3 (y)] If 2 is not in the second cell then Brick b 1 has two cells, 2 is in the first cell of brick b 2 There is a Γ match among the cells of bricks b 1, b 2 that involves 1 and 2. Choose a number to fill in the second cell of O, remove the first two cells to obtain a fixed point for n 2 that has at least two cells in the first brick. -y 1 2 Γ-match Quang T. Bach UCSD July 8, / 38

28 U Γ,n (y) for Γ = {1324, 1423} Mathematica allows us to compute the polynomials U Γ,n (y) for the first few values of n. n U Γ,n (y) 6 y + 14y 2 36y y 4 5y 5 + y 6 7 y + 20y 2 90y y 4 29y 5 + 6y 6 y 7 8 y + 27y 2 188y y 4 162y y 6 7y 7 + y 8 9 y + 35y 2 348y y 4 842y y 6 55y 7 +8y 8 y 9 10 y + 44y 2 591y y y y 6 424y 7 +71y 8 9y 9 + y y + 54y 2 941y y y y y y 8 89y y 10 y 11 Observation: U Γ,n (y) are unimodal with peaks occur in pairs. Quang T. Bach UCSD July 8, / 38

29 U Γ,n (y) for Γ = {1324, 123} This method also applies even when the permutations of the family Γ all have different length, and different number of descents. Theorem For Γ = {1324, 123}, U Γ,1 (y) = y, and for n 2, U Γ,n (y) = yu Γ,n 1 (y) yu Γ,n 2 (y) n/2 + k=2 ( y) k C k 1 U Γ,n 2k (y) where C k = 1 ( 2k ) k+1 k is the k-th Catalan number. Quang T. Bach UCSD July 8, / 38

30 For general Γ p k 1,k 2 Let p 4 and k 1, k 2 2 with k 1 + k 2 = p. We consider the family of permutations Γ p k 1,k 2 in S p defined as Γ p k 1,k 2 = {σ S p : σ 1 = 1, σ k1 +1 = 2, σ 1 < σ 2 <... < σ k1 & σ k1 +1 < σ k1 +2 <... < σ p } Then the permutations in Γ p k 1,k 2 all start with 1 and have one descent. Therefore, the previous method still applies to these patterns. Quang T. Bach UCSD July 8, / 38

31 For general Γ p k 1,k 2 For any Γ = Γ p k 1,k 2 defined before, let m = min(k 1, k 2 ) and M = max(k 1, k 2 ) Then we obtain the recursion for U Γ,n (y) as follows: Theorem For Γ p k 1,k 2, U Γ,1 (y) = y, and for n 2, U Γ,n (y) = (1 y)u Γ,n 1 (y) ( ) [ ] m 1 n 1 y U Γ,n M(y) + y U Γ,n M i (y) k 1 1 i=1 Quang T. Bach UCSD July 8, / 38

32 Unimodality of U Γ,n (y) for Γ p k 1,k 2 Interestingly, when the permutations of Γ p k 1,k 2 polynomials U Γ,n (y) are no longer unimodal. is long enough, the Let k 1 = 6 and k 2 = 4, Mathematica allows us to compute the polynomials U Γ,n (y) for some small values of n n U Γ,n (y) 10 y + 65y 2 36y y 4 126y y 6 84y 7 +36y 8 9y 9 + y y + 192y 2 227y y 4 210y y 6 210y y 8 45y y 10 y y + 445y 2 923y y 4 330y y 6 462y y 8 165y y 10 11y 11 + y 12 Quang T. Bach UCSD July 8, / 38

33 Permutations with More Than One Desents Quang T. Bach UCSD July 8, / 38

34 Permutations that start with 1 and have more than one descents Suppose τ is a permutation that starts with 1 and have more than one descents. Then The current involution no longer applies We need to put more conditions on our involution to ensure that we are breaking and combining the bricks at the same position For some simple permutations τ of this type, we can still compute the recursion for U τ,n (y) Quang T. Bach UCSD July 8, / 38

35 U τ,n (y) for τ = Theorem For τ = , U τ,1 (y) = y, and for n 2, U τ,n (y) = (1 y)u τ,n 1 (y) + where C k = 1 k+1 n 4 2 ( 1) k+1 2 y k+1 C k [U τ,n 2k 2 (y) + yu τ,n 2k 3 (y)] k=2 ( 2k k ) is the k-th Catalan number. Quang T. Bach UCSD July 8, / 38

36 U τ,n (y) for τ = Theorem For τ = , U τ,1 (y) = y, and for n 2, U τ,n (y) = (1 y)u τ,n 1 (y) + n 4 2 ( ) ( 1) k+1 n k 3 2 y k+1 U τ,n 2k 3 (y) k + 1 k=2 Quang T. Bach UCSD July 8, / 38

37 U Γ,n (y) for Γ = {14253, 15243} Miles Jones obtained a recursion for U τ,n (y) where τ = We now want to extend his result to the family Γ = {14253, 15243}. The same method also applies for family of permutations that start with 1 and have more than one descents. Theorem For Γ = {14253, 15243}, U Γ,1 (y) = y, and for n 2, U Γ,n (y) = (1 y)u Γ,n 1 (y) + n 3 2 k ( 1) k+1 2 y k+1 (n 2k 2) (n 2i 1) U τ,n 2k 3 (y) k=2 i=1 Quang T. Bach UCSD July 8, / 38

38 The End Quang T. Bach UCSD July 8, / 38