Enumeration on row-increasing tableaux Rosena R. X. Du East China - PowerPoint PPT Presentation

Enumeration on row-increasing tableaux Rosena R. X. Du East China Normal University, Shanghai, China Joint work with Xiaojie Fan and Yue Zhao Shanghai Jiaotong University June 25, 2018 of shape 2 × n

Outline 1 Defjnitions and Backgrounds 2 Bijective proof of Pechenik’s result 5 Counting major index of Schröder n -paths 2/38 3 Counting major index for RInc k ( 2 × n ) 4 Counting amajor index for RInc k ( 2 × n )

Integer partitions 3/38 Let λ = ( λ 1 , λ 2 , . . . , λ k ) be a partititon of n , i.e. λ 1 + λ 2 + · · · + λ k = n , where λ 1 ≥ λ 2 ≥ · · · ≥ λ k > 0 . The Ferrers diagram of λ is a left-justfjed array of cells with λ i cells in the i -th row, for 1 ≤ i ≤ k . Figure: The Ferrers diagram of a partition λ = ( 6 , 3 , 1 ) ⊢ 10 .

Semistandard Young tableau and standard Young tableau 6 9 7 6 2 8 10 5 4 3 1 8 5 4 9 8 7 6 4 2 increasing. and every column is weakly increasing. 4/38 A semistandard Young tableau (SSYT) of shape λ is a fjlling of the Ferrers diagram of λ with positive integers such that every row is strictly increasing A standard Young tableau (SYT) of shape λ ⊢ n is a fjlling of the Ferrers diagram of λ with { 1 , 2 , . . . , n } such that every row and column is strictly Figure: A semi-standard Young tableau of shape ( 6 , 3 , 1 ) and a standard Young tableau of shape ( 6 , 3 , 1 ) .

Major index and amajor index of a tablau 7 9 7 6 8 4 3 5 10 2 1 9 6 8 4 3 5 10 2 1 5/38 A descent of an SSYT T is an integer i such that i + 1 appears in a lower row of T than i . D ( T ) : the descent set of T . The major index of T is defjned by maj ( T ) = ∑ i ∈ D ( T ) i . An ascent of T to be an integer i such that i + 1 appears in a higher row of T than i . A ( T ) : the ascent set of T . The amajor index of T is defjned by amaj ( T ) = ∑ i ∈ A ( T ) i . Figure: T ∈ SYT ( 4 , 3 , 1 , 1 , 1 ) . D ( T ) = { 2 , 5 , 6 , 8 } , maj ( T ) = 21 . A ( T ) = { 4 , 7 , 9 } , amaj ( T ) = 20 .

Major index for standard Young tableaux Lemma (Stanley’s q -hook length formula) translated to the study of descent and major index of tableaux. statistic descent or major index of pattern-avoiding permutations can be corresponding “recording tableau”. Therefore many problems involving the descent set of a permutation is transferred to the descent set of the and pairs of SYTs of order n of the same shape. Under this bijection, the The famous RSK algorithm is a bijection between permutations of length n (1) 6/38 For any partition λ = ∑ i λ i of n , we have ∑ q b ( λ ) [ n ]! q maj ( T ) = ∏ u ∈ λ h ( u ) . T ∈ SYT ( λ ) Here b ( λ ) = ∑ i ( i − 1 ) λ i .

7/38 3 4 5 6 1 2 4 3 5 6 1 2 5 4 2 6 1 3 4 2 5 6 1 3 5 2 4 6 3 1 For any positive integer n , we have q 3 q n n 3 (2) m Standard Young tableaux of shape 2 × n [ 2 n ] ∑ q maj ( T ) = C q ( n ) = . [ n + 1 ] T ∈ SYT ( 2 × n ) 1 − q = 1 + q + q 2 + · · · + q n − 1 , [ n ]! = [ n ][ n − 1 ] · · · [ 1 ] and Here [ n ] = 1 − q n [ n ] [ n ]! = [ m ]![ n − m ]! . For example, when n = 3 , we have [ 6 ] = q 3 + q 5 + q 6 + q 7 + q 9 . C q ( 3 ) = [ 3 + 1 ] And there are fjve SYT of shape 2 × 3 , with major index 3,6,7,5,9.

Increasing tableaux 1 5 1 2 4 3 4 5 3 An increasing tableau is an SSYT such that both rows and columns are 4 2 4 5 Increasing tableau is defjned by O. Pechenik who studied increasing tableaux numbers appeared twice. O. Pechenik, Cyclic Sieving of Increasing Tableaux and Small Schröder Paths. J. Combin. Theory Ser. A , 125: 357–378, 2014. 4 3 3 4 strictly increasing, and the set of entries is an initial segment of positive integers (if an integer i appears, positive integers less than i all appear). 1 2 3 2 2 5 1 2 4 2 3 5 1 8/38 We denote by Inc k ( λ ) the set of increasing tableaux of shape λ with entries are { 1 , 2 , . . . , n − k } . Figure: There are fjve increasing tableaux in Inc 1 ( 2 × 3 ) . in Inc k ( 2 × n ) , i.e., increasing tableaux of shape 2 × n , with exactly k

Pechenik’s result 5 5 1 2 4 2 3 5 1 2 3 3 4 1 2 2 4 3 4 5 1 3 4 2 4 5 Pechinik’s proof involves the cyclic sieving of increasing tableaux. We will show a more conceptual proof using the q -hook length formula. 4 3 Theorem (O. Pechenik) q 4 k n (3) 2 9/38 1 3 1 For any positive integer n , and 0 ≤ k ≤ n we have [ n − 1 ][ 2 n − k ] ∑ q maj ( T ) = q n + k ( k + 1 ) / 2 S q ( n , k ) = . [ n + 1 ] T ∈ Inc k ( 2 × n ) For example, when n = 3 , k = 1 we have [ 3 − 1 ][ 5 ] ∑ q maj ( T ) = = q 8 + q 7 + q 6 + q 5 + q 4 . S q ( 3 , 1 ) = [ 3 + 1 ] T ∈ Inc 1 ( 2 × 3 )

A refjnement of small Shröder number n Theory Ser. A , 76: 175–177, 1996. R. P. Stanley, Polygon dissections and standard Young tableaux. J. Combin. In 1996 Stanley gave a bijection between the above two sets. counts the following sets: (4) 1 k 10/38 Setting q = 1 , we get the cardinality of Inc k ( 2 × n ) : ( n − 1 )( 2 n − k ) s ( n , k ) = . n + 1 s ( n , k ) is considered as a refjnement of the small Schröder number which 1. Dissections of a convex ( n + 2 ) -gon into n − k regions; 2. SYTs of shape ( n − k , n − k , 1 k ) ; Pechenik gave a nice bijection between SYTs of shape ( n − k , n − k , 1 k ) and increasing tableaux in Inc k ( 2 × n ) .

Schröder paths 1 1 3 4 2 3 4 1 2 4 3 2 4 1 2 3 2 3 4 Motivation: are there any interesting result for these tableaux that correspond to all Schröder n -paths? 4 1 11/38 3 is no F steps on the diagonal line, it is called a small Schröder path. order, if i appears only in row 1 (2), it corresponds to a U ( D ) step, if i appears in both rows, it corresponds to an F step. 1 2 3 3 1 4 1 2 4 2 3 4 1 2 s ( n , k ) also counts number of small Schröder n -paths with k fmat steps. A Schröder n -path is a lattice path goes from ( 0 , 0 ) to ( n , n ) with steps ( 0 , 1 ) , ( 1 , 0 ) and ( 1 , 1 ) and never goes below the diagonal line y = x . If there There is an obvious bijection between SSYTs in RInc k ( 2 × n ) and Schröder n -paths with k steps: read the numbers i from 1 to 2 n − k in increasing

Row-increasing tableaux 1 1 3 A row-increasing tableau is an SSYT with strictly increasing rows and weakly 2 3 4 1 2 4 3 2 4 1 2 3 2 3 4 1 k (5) 4 4 1 3 increasing columns, and the set of entries is a consecutive segment of positive integers. 3 1 2 3 1 denote RInc 0 4 4 2 1 1 12/38 2 4 2 3 k ( λ ) the set of row-increasing tableaux of shape λ with We denote by RInc m set of entries { m + 1 , m + 2 , . . . , m + n − k } . When m = 0 , we will just k ( λ ) as RInc k ( λ ) . It is obvious that Inc k ( λ ) ⊆ RInc k ( λ ) . Figure: There are 6 row-increasing tableaux in RInc 2 ( 2 × 3 ) . It is not hard to show that RInc k ( 2 × n ) is counted by ( 2 n − k )( 2 n − 2 k ) r ( n , k ) = . n − k + 1 n − k r ( n , k ) is considered as a refjnement of the large Schröder number.

Our Main Results k (7) k Theorem (6) 13/38 following results. Theorem We study the statistics maj and amaj of SSYTs in RInc k ( 2 × n ) and get the For any positive integer n , and 0 ≤ k ≤ n we have [ 2 n − k ][ 2 n − 2 k ] ∑ q maj ( T ) = q n + k ( k − 3 ) / 2 R q ( n , k ) = . [ n − k + 1 ] n − k T ∈ RInc k ( 2 × n ) For any positive integer n , and 0 ≤ k ≤ n we have [ 2 n − k ][ 2 n − 2 k ] ∑ q amaj ( T ) = q k ( k − 1 ) / 2 � R q ( n , k ) = . [ n − k + 1 ] n − k T ∈ RInc k ( 2 × n )

1 Defjnitions and Backgrounds 2 Bijective proof of Pechenik’s result 5 Counting major index of Schröder n -paths 14/38 3 Counting major index for RInc k ( 2 × n ) 4 Counting amajor index for RInc k ( 2 × n )

Bijective proof of Pechenik’s result 4 9 7 6 8 4 3 5 2 1 9 8 7 6 3 Theorem (O. Pechenik) 8 6 5 4 2 1 deleting all elements of A from the fjrst row of T and all elements of B from B be the set of numbers that appear in the second row immediately right of which preserves the descent set. 15/38 There exists a bijection γ between Inc k ( 2 × n ) and SYT ( n − k , n − k , 1 k ) Given T ∈ Inc k ( 2 × n ) . Let A be the set of numbers that appear twice. Let an element of A . Let γ ( T ) be the tableau of shape ( n − k , n − k ) formed by the second row of T . It is not hard to prove that γ is a bijection. E.g., in the following example, we have A = { 4 , 6 , 8 } and B = { 6 , 7 , 9 } .

16/38 9 9 7 6 8 1 3 5 2 1 4 8 2 7 6 4 3 8 6 5 4 Why does γ preserves the major index? �→ For each i ∈ D ( T ) , there are three cases: ∈ A . then in γ ( T ) i is still in row 1 and i + 1 is still in row 2 , and 1) i / therefore i ∈ D ( γ ( T )) ; 2) i ∈ A and i / ∈ B . In this case i + 1 ∈ B . And in γ ( T ) , i is in row 2 and i + 1 is in the fjrst column with row index j for some j ≥ 3 , thus i ∈ D ( γ ( T )) ; 3) i ∈ A and i ∈ B . Since we also have i + 1 ∈ B , then in S , i is in the fjrst column with row index j for some j ≥ 3 , and i + 1 is in the fjrst column with row index j + 1 , thus i ∈ D ( γ ( T )) ; Combining the above three cases, we have D ( T ) ⊆ D ( γ ( T )) . Similarly we can show that D ( γ ( T )) ⊆ D ( T ) .