Enumeration on row-increasing tableaux of shape 2 n

Transcription

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

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

3 3/38 Integer partitions Let λ = (λ 1, λ 2,..., λ k ) be a partititon of n, i.e. where λ 1 λ 2 λ k > 0. λ 1 + λ λ k = n, The Ferrers diagram of λ is a left-justfied 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.

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

5 5/38 Major index and amajor index of a tablau 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 defined 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 defined 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.

6 6/38 Major index for standard Young tableaux Lemma (Stanley s q-hook length formula) For any partition λ = i λ i of n, we have Here b(λ) = i (i 1)λ i. T SYT(λ) q maj(t) = qb(λ) [n]! (1) u λ h(u). The famous RSK algorithm is a bijection between permutations of length n and pairs of SYTs of order n of the same shape. Under this bijection, the descent set of a permutation is transferred to the descent set of the corresponding recording tableau. Therefore many problems involving the statistic descent or major index of pattern-avoiding permutations can be translated to the study of descent and major index of tableaux.

7 7/38 Standard Young tableaux of shape 2 n For any positive integer n, we have C q (n) = T SYT(2 n) q maj(t) = [ ] qn 2n. (2) [n + 1] n Here [n] = 1 qn 1 q = 1 + q + q2 + + q n 1, [n]! = [n][n 1] [1] and [ n m] = [n]! [m]![n m]!. For example, whenn = 3, we have [ ] C q (3) = q3 6 = q 3 + q 5 + q 6 + q 7 + q 9. [3 + 1] 3 And there are five SYT of shape 2 3, with major index 3,6,7,5,

8 Increasing tableaux An increasing tableau is an SSYT such that both rows and columns are 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). We denote by Inc k (λ) the set of increasing tableaux of shape λ with entries are {1, 2,..., n k} Figure: There are five increasing tableaux in Inc 1(2 3). Increasing tableau is defined by O. Pechenik who studied increasing tableaux in Inc k (2 n), i.e., increasing tableaux of shape 2 n, with exactly k numbers appeared twice. O. Pechenik, Cyclic Sieving of Increasing Tableaux and Small Schröder Paths. J. Combin. Theory Ser. A, 125: , /38

9 9/38 Pechenik s result Theorem (O. Pechenik) For any positive integer n, and 0 k n we have S q (n, k) = T Inc 1(2 3) T Inc k (2 n) q maj(t) = qn+k(k+1)/2 [n + 1] For example, when n = 3, k = 1 we have [ ][ ] S q (3, 1) = q maj(t) = q [3 + 1] 1 3 [ ][ n 1 2n k k n ]. (3) = q 8 + q 7 + q 6 + q 5 + q Pechinik s proof involves the cyclic sieving of increasing tableaux. We will show a more conceptual proof using the q-hook length formula.

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

11 Schröder paths s(n, k) also counts number of small Schröder n-paths with k flat 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 is no F steps on the diagonal line, it is called a small Schröder path. 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 2n k in increasing 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 Motivation: are there any interesting result for these tableaux that correspond to all Schröder n-paths? 11/38

12 Row-increasing tableaux A row-increasing tableau is an SSYT with strictly increasing rows and weakly increasing columns, and the set of entries is a consecutive segment of positive integers. We denote by RInc m k (λ) the set of row-increasing tableaux of shape λ with set of entries {m + 1, m + 2,..., m + n k}. When m = 0, we will just denote RInc 0 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 ( )( ) 1 2n k 2n 2k r(n, k) =. (5) n k + 1 k n k r(n, k) is considered as a refinement of the large Schröder number. 12/38

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

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

15 Bijective proof of Pechenik s result Theorem (O. Pechenik) There exists a bijection γ between Inc k (2 n) and SYT(n k, n k, 1 k ) which preserves the descent set. Given T Inc k (2 n). Let A be the set of numbers that appear twice. Let B be the set of numbers that appear in the second row immediately right of an element of A. Let γ(t) be the tableau of shape (n k, n k) formed by deleting all elements of A from the first row of T and all elements of B from 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} /38

16 Why does γ preserves the major index? For each i D(T), there are three cases: ) i / A. then in γ(t) i is still in row 1 and i + 1 is still in row 2, and 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 first 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 first column with row index j for some j 3, and i + 1 is in the first 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) /38

17 17/38 Lemma (Stanley s q-hook length formula) For any partition λ = i λ i of n, we have Here b(λ) = i (i 1)λ i. T SYT(λ) q maj(t) = Applying the above formula we have S q (n, k) = q maj(t) = = T Inc k (2 n) qb(λ) [n]! (8) u λ h(u). T SYT(n k,n k,1 k ) q maj(t) q n+k(k+1)/2 [ [2n k]! [n k]![n k 1]![n + 1][n][k]! = qn+k(k+1)/2 2n k [n + 1] n ][ n 1 k ].

18 18/38 1 Definitions and Backgrounds 2 Bijective proof of Pechenik s result 3 Counting major index for RInc k (2 n) 4 Counting amajor index for RInc k (2 n) 5 Counting major index of Schröder n-paths

19 19/38 Theorem For any positive integer n, k, we have r(n, k) = s(n, k) + s(n, k 1). (9) There is a bijection f : RInc k (2 n)\inc k (2 n) Inc k 1 (2 n). Given T RInc k (2 n)\inc k (2 n), find the minimal integer j such that T 1,j = T 2,j, i.e., the j-th column is the leftmost column of T with two identical entries. Now we first delete the entry T 2,j, then move all the entries on the right of T 2,j one box to the left and set the last entry as 2n k + 1, and define the resulting tableau to be f(t). T : f(t) : Figure: An example of f with T RInc 3(2 5)\Inc 3(2 5) and f(t) Inc 2(2 5).

20 20/38 Note that f does not preserve the major index. In fact we have Theorem For any positive integer n, k with k < n, we have R q (n, k) = S q (n, k)+s q (n, k 1)+(1 q 2n k )(S q (n 1, k 1)+S q (n 1, k 2)). (10) Proof: For all SSYTs in RInc k (2 n) \ Inc k (2 n), there are two cases. 1. If T 1,n = T 2,n, the last column of T consist of two identical entries 2n k and 2n k / D(T) We will show that the sum of q maj(t) over all these tableaux is S q (n 1, k 1) + S q (n 1, k 2).

21 21/38 1) The n-th column is the only column of T with identical entries. In this case the last column of T consist of two identical entries 2n k and 2n k / D(T). And the sum of q maj(t) over these tableaux is S q (n 1, k 1). 2) There is at least one column with identical entries in T besides the n-th column. Now let T be the tableau obtained by deleting the last column from T. Then T RInc k 1 (2 (n 1)) \ Inc k 1 (2 (n 1)). There are two cases for the last column of T. a) If T 1,n 1 T 2,n 1, then f(t ) Inc k 2 (2 (n 1)) with f(t) 1,n 1 < 2n k 1, and maj(t) = maj(t ) = maj(f(t ));

22 b) If T 1,n 1 = T 2,n 1, we have T 1,n 1 = T 2,n 1 = 2n k Since T 1,n = T 2,n = 2n k, we have 2n k 1 D(T) but 2n k 1 / D(T ), and all the other descents of T are also descents of T. Thus maj(t ) = maj(t) (2n k 1). (11) Moreover when we apply f to T we have 2n k 1 D(f(T )) but 2n k 1 / D(T ), and all the other descents of T are also descents of f(t ). Therefore we have maj(f(t )) = maj(t ) + (2n k 1). (12) Combining (11) and (12) we know that f(t ) Inc k 2 (2 (n 1)) with f(t) 1,n 1 = 2n k 1, and maj(f(t )) = maj(t). Thus the sum of q maj(t) over these tableaux of case a) and b) is S q (n 1, k 2). 22/38

23 23/38 2. If T 1,n T 2,n. In this case if we apply f to T, 2n k + 1 will be added as the last entry in row 2 of f(t) and D(f(T)) \ D(T) = {2n k} and maj(f(t)) = maj(t) + 2n k. Therefore we know that the sum of q maj(t) over these talbeaux is S q (n, k 1) q 2n k (S q (n 1, k 1) + S q (n 1, k 2)). Combining the Case 1) and Case 2) we have T RInc k (2 n)\inc k (2 n) q maj(t) = S q (n, k 1) + (1 q 2n k )(S q (n 1, k 1) + S q (n 1, k 2)) and [ R q (n, k) = qn+k(k 1)/2 n [k] k 1 ][ 2n k n ].

24 24/38 1 Definitions and Backgrounds 2 Bijective proof of Pechenik s result 3 Counting major index for RInc k (2 n) 4 Counting amajor index for RInc k (2 n) 5 Counting major index of Schröder n-paths

25 25/38 Counting amajor index for RInc k (2 n) R q (n, k) = T RInc k (2 n) q amaj(t) = qk(k 1)/2 [n k + 1] We will prove the above formula by showing that T RInc k (2 n) q maj(t) = q n k T RInc k (2 n) [ 2n k k ][ 2n 2k n k q amaj(t). For example, there are 6 row-increasing tableaux in RInc 2 (2 3), with the (maj, amaj) pairs (4,3),(4,5),(2,4),(5,1),(3,3),(6,2). ] We want to establish a bijection Φ : RInc k (2 n) RInc k (2 n) such that maj(φ(t)) = amaj(t) + n k.

26 26/38 The general result Theorem There is a bijection Φ : RInc k (2 n) RInc k (2 n) that preserves the second row, and maj(φ(t)) = amaj(t) + n k. T : Φ(T) : Figure: An example of the map Φ with n = 13, k = 6, l = 3, amaj(t) = 95 and maj(φ(t)) = 102

27 27/38 The prime case A row-increasing tableau T is prime if for each integer j satisfies T 1,j+1 = T 2,j + 1, T 2,j+1 also appears in row 1 in T. princ m k (λ): prime row-increasing tableaux of shape λ with set of entries {m + 1, m + 2,..., m + n k}. For each T princ m k (2 n), let A be the set of numbers that appear twice, and B be the set of numbers that appear in the second row immediately left of an element of A in cyclic order. Let g(t) be the tableau of shape 2 n obtained by first deleting all elements in A from the first row and then inserting all elements in B into the first row and list them in increasing order, and keep the entries in row 2 unchanged. In the following example, we have A = {2, 6, 9} and B = {3, 8, 9}. T : g g(t) :

28 28/38 Lemma The map g is an injection from princ m k (2 n) to RInc m k (2 n) which satisfies the following: 1) If T 2,1 appears only once in T, then g(t) 1,i+1 g(t) 2,i for each i, 1 i n 1; 2) T 2,1 appears twice in T if and only if g(t) 1,n = g(t) 2,n. Sketch of Proof: there are two cases: T 2,1 appears only once in T; T : g g(t) :

29 29/38 Lemma The map g is an injection from princ m k (2 n) to RInc m k (2 n) which satisfies the following: 1) If T 2,1 appears only once in T, then g(t) 1,i+1 g(t) 2,i for each i, 1 i n 1; 2) T 2,1 appears twice in T if and only if g(t) 1,n = g(t) 2,n. Sketch of Proof: there are two cases: T 2,1 appears twice in T; T : g g(t) : T :

30 Lemma For each T princ m k (2 n) we have maj(g(t)) = { amaj(t) + n k, if T1,1 = T 2,1 ; amaj(t) + m + n k, if T 1,1 T 2,1. (13) Proof: let T 0 be the skew shape tableau obtained by deleting the numbers in A from row 1 of T and push all the remaining numbers to the right, and keep the second row unchanged. T : g g(t) : T 0 : D(g(T)) \ D(T 0 ) = A(T) \ A(T 0 ). and therefore maj(g(t)) maj(t 0 ) = amaj(t) amaj(t 0 ); 30/38

31 31/38 2. when T 1,1 T 2,1, maj(t 0 ) = amaj(t 0 ) + m + n k. Suppose the descents of (T 0 ) appears in columns k + x 1, k + x 2,..., k + x d in row 1, and the ascents of (T 0 ) appears in columns y 1, y 2,..., y d 1 in row 2. Here d, x 1,..., x d, y 1,... y d 1 are all positive integers and x d = n k. It is not hard to check that T 0 is determined by the set X = {x 1, x 2,..., x d } and Y = {y 1, y 2,..., y d 1 }. And we have D(T 0 ) = {m + x 1, m + x 2 + y 1,..., m + x d + y d 1 }; A(T 0 ) = {m + x 1 + y 1, m + x 2 + y 2,..., m + x d 1 + y d 1 }. Therefore we have maj(t 0 ) amaj(t 0 ) = m + x d = m + n k. An example for m = 4, n = 6, k = 2, d = 2, X = {2, 4} and Y = {2}. T 0 : Similarly we can prove that when T 1,1 = T 2,1, maj(t 0 ) = amaj(t 0 ) + n k.

32 32/38 The general case Given T RInc k (2 n), we can uniquely decompose T into prime row-increasing tableaux T 1 T 2 T l, and set Φ(T) = g(t 1 )g(t 2 ) g(t l ). An example with n = 13, k = 6, and l = 3. Here A(T) = {3, 8, 9, 11, 13, 15, 17, 19}, A(T 0 1 ) = {3}, A(T0 2 ) = {11, 13}, A(T 0 3 ) =, D(T0 1 ) = {1, 5}, D(T0 2 ) = {10, 12, 14}, D(T0 3 ) = {18}. D(Φ(T)) = {1, 5, 8, 10, 12, 14, 15, 18, 19}. amaj(t) = 95 and maj(φ(t)) = 102. T : Φ(T) :

33 33/38 Proof for the general case (sketch) Suppose T j princ mj k j (2 n j ) for integers m j, k j, n j with m j, k j 0, and n j > 0. We have n 1 + n n l = n, k 1 + k k l = k. The smallest entry of T i is m i + 1 with m 1 = 0, and m j = 2(n 1 + n n j 1 ) (k 1 + k k j 1 ), 2 j l, And the the largest entry of T j is m j+1 for each j, 1 j l 1, It is easy to check that A(T) = A(T 1 ) A(T 2 ) A(T l ) {m 2, m 3,..., m l } and l l amaj(t) = amaj(t j ) + m j. j=1 j=2

34 34/38 Proof for the general case (sketch) Moreover we have D(Φ(T)) = D(g(T 1 )) D(g(T 2 )) D(g(T l )). Since for each j, 1 j l, maj(g(t j )) = amaj(t j ) + m j + n j k j, Therefore we have maj(φ(t)) = = l l maj(g(t j )) = (amaj(t j ) + m j + n j k j ) j=1 j=1 l l amaj(t j ) + m j + n k j=1 j=1 = amaj(t) + n k.

35 35/38 1 Definitions and Backgrounds 2 Bijective proof of Pechenik s result 3 Counting major index for RInc k (2 n) 4 Counting amajor index for RInc k (2 n) 5 Counting major index of Schröder n-paths

36 36/38 Counting major index of Schröder n-paths Let P be a Schröder n-path that goes from the origin (0, 0) to (n, n) with k F steps, we can associate with P a word w = w(p) = w 1 w 2 w 2n k over the alphabet {0, 1, 2} with exactly k 1 s. (8, 8) (0, 0) Figure: A Schröder P with ω(p) =

37 Counting major index of Schröder n-paths The descent set of w is the set of all positions of the descents of w, D(w) = {i : 1 i n, w i > w i+1 }. The major index of w is defined as maj(w) = i D(w) i. And define maj(p) = maj(w(p)). In 1993, Bonin, Shapiro and Simion study the major index for Schöder paths and gave the following result: [ ][ ] q maj(p) 1 2n k 2n 2k =. (14) [n k + 1] k n k Here the sum is over all Schröder n-paths with exactly k F steps. An obvious bijection between SSYTs in RInc k (2 n) and Schröder n-paths with k steps: read the numbers i from 1 to 2n k in increasing 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. A naive thinking is that if the i-th step corresponds to a descent in P, then i is an ascent of T, i.e., D(P)=A(T) and maj(p) = amaj(t). But this is NOT true. 37/38

38 Thank you! 38/38