Hook Lengths and Contents. Richard P. Stanley M.I.T.

Hook Lengths and Contents p. Hook Lengths and Contents Richard P. Stanley M.I.T.

Hook Lengths and Contents p. Standard Young tableau standard Young tableau (SYT) of shape λ = (4, 4, 3, 1): < < 1 3 4 2 6 9 5 7 12 10 8 11

f λ : number of SYT of shape λ Hook Lengths and Contents p.

Hook Lengths and Contents p. f λ : number of SYT of shape λ f 3,2 = 5: 1 2 3 1 2 4 1 2 5 4 5 3 5 3 4 1 3 4 1 3 5 2 5 2 4

Hook Lengths and Contents p. Hook length formula For u λ, let h u be the hook length at u, i.e., the number of squares directly below or to the right of u (counting u once)

Hook Lengths and Contents p. Hook length formula For u λ, let h u be the hook length at u, i.e., the number of squares directly below or to the right of u (counting u once) 7 6 4 1 5 4 4 3 2 1 2 1

Hook Lengths and Contents p. Theorem (Frame-Robinson-Thrall). Let λ n. Then n! f λ = u λ h. u

Hook Lengths and Contents p. Theorem (Frame-Robinson-Thrall). Let λ n. Then n! f λ = u λ h. u f (4,4,3,1) = 12! 7 6 5 43 3 22 13 = 2970

Hook Lengths and Contents p. Nekrasov-Okounkov identity RSK algorithm (or representation theory) f 2 λ = n! λ n

Hook Lengths and Contents p. Nekrasov-Okounkov identity RSK algorithm (or representation theory) f 2 λ = n! λ n Nekrasov-Okounkov (2006), G. Han (2008): n 0 ( λ n f 2 λ ) x n (t + h 2 u ) u λ n! 2 = i 1 (1 x i ) t 1

Hook Lengths and Contents p. A corollary e k : kth elementary SF e k (x 1, x 2,... ) = x i1 x i2 x ik 1 i 1 <i 2 < <i k

Hook Lengths and Contents p. A corollary e k : kth elementary SF e k (x 1, x 2,... ) = x i1 x i2 x ik 1 i 1 <i 2 < <i k Corollary. Let g k (n) = 1 n! Then g k (n) Q[n]. f 2 λ e k(h 2 u : u λ). λ n

Hook Lengths and Contents p. Conjecture of Conjecture (Han). Let j P. Then Q[n]. 1 n! λ n f 2 λ u λ h 2j u

Hook Lengths and Contents p. Conjecture of Conjecture (Han). Let j P. Then Q[n]. 1 n! λ n f 2 λ u λ True for j = 1 by above. h 2j u

Hook Lengths and Contents p. Conjecture of Conjecture (Han). Let j P. Then Q[n]. 1 n! λ n f 2 λ u λ True for j = 1 by above. h 2j u Stronger conjecture. For any symmetric function F, f 2 λ F (h2 u : u λ) Q[n]. 1 n! λ n

Hook Lengths and Contents p. Examples Let d k (n)= 1 n! λ n f 2 λ u λ h 2k u. d 1 (n) = 1 n(3n 1) 2 d 2 (n) = 1 24 n(n 1)(27n2 67n + 74) d 3 (n) = 1 n(n 1)(n 2) 48 (27n 3 174n 2 + 511n 552).

Hook Lengths and Contents p. 1 Open variants 1 n! λ n f 2 λ u λ h u =? 1 n! λ n f λ u λ h u =? 1 n! λ n f λ u λ h 2 u =?

Hook Lengths and Contents p. 1 Contents For u λ let c(u)= i j, the content of square u = (i, j).

Hook Lengths and Contents p. 1 Contents For u λ let c(u)= i j, the content of square u = (i, j). 0 1 2 3 1 0 1 2 2 1 0 3

Hook Lengths and Contents p. 1 Semistandard tableaux semistandard Young tableau (SSYT) of shape λ = (4, 4, 3, 1): < < 1 2 2 4 2 3 6 6 4 4 7 7 x T = x 1 x 3 2 x3 4 x 5 x 2 6 x2 7

Hook Lengths and Contents p. 1 Schur functions s λ = T x T, summed over all SSYT of shape λ

Hook Lengths and Contents p. 1 Schur functions s λ = T x T, summed over all SSYT of shape λ s 3 = i j k s 1,1,1 = i<j<k x i x j x k x i x j x k = e 3.

Hook Lengths and Contents p. 1 Hook-content formula f(1 t )= f(1, 1,..., 1) (t 1 s)

Hook Lengths and Contents p. 1 Hook-content formula f(1 t )= f(1, 1,..., 1) (t 1 s) s λ (1 t ) = #{SSYT of shape λ, max t}.

Hook Lengths and Contents p. 1 Hook-content formula f(1 t )= f(1, 1,..., 1) (t 1 s) s λ (1 t ) = #{SSYT of shape λ, max t}. Theorem. s λ (1 t ) = u λ t + c u h u.

Hook Lengths and Contents p. 1 A curiosity Let κ(w) denote the number of cycles of w S n. Then 1 n! u,v S n q κ(uvu 1v 1) = λ n (q + c u ). u λ

Hook Lengths and Contents p. 1 A curiosity Let κ(w) denote the number of cycles of w S n. Then 1 n! u,v S n q κ(uvu 1v 1) = Restatement: n! λ n e k (c u : u λ) λ n (q + c u ). u λ = #{(u, v) S n S n : κ(uvu 1 v 1 ) = n k}

Hook Lengths and Contents p. 1 Another curiosity w S n q κ(w2) = λ n f λ (q + c u ) u λ

Hook Lengths and Contents p. 1 Han s conjecture for contents Theorem. For any symmetric function F, 1 n! f 2 λ F (c u : u λ) Q[n]. λ n Idea of proof. By linearity, suffices to take F = e µ.

Hook Lengths and Contents p. 1 Power sum symmetric functions x = (x 1, x 2,... ) p m = p m (x) = i xm i p λ = p λ1 p λ2

Hook Lengths and Contents p. 1 Power sum symmetric functions x = (x 1, x 2,... ) p m = p m (x) = i xm i p λ = p λ1 p λ2 Let l(λ) be the length (number of parts) of λ p m (1 t ) = t p λ (1 t ) = t l(λ)

Hook Lengths and Contents p. 1 Some notation x (1),..., x (k) : disjoint sets of variables

Hook Lengths and Contents p. 1 Some notation x (1),..., x (k) : disjoint sets of variables ρ(w): cycle type of w S n, e.g., w = (1, 3, 4)(2)(5) p ρ(w) = p (3,1,1) = p 3 p 2 1

Hook Lengths and Contents p. 1 Some notation x (1),..., x (k) : disjoint sets of variables ρ(w): cycle type of w S n, e.g., w = (1, 3, 4)(2)(5) p ρ(w) = p (3,1,1) = p 3 p 2 1 H λ = u λ h u

Hook Lengths and Contents p. 2 The fundamental tool Theorem. λ n H k 2 λ s λ (x (1) ) s λ (x (k) ) = 1 n! w 1 w k =1 in S n p ρ(w1)(x (1) ) p ρ(wk )(x (k) )

Hook Lengths and Contents p. 2 The fundamental tool Theorem. λ n H k 2 λ s λ (x (1) ) s λ (x (k) ) = 1 n! w 1 w k =1 in S n p ρ(w1)(x (1) ) p ρ(wk )(x (k) ) Proof. Follows from representation theory: the relation between multiplying conjugacy class sums in Z(CG) (the center of the group algebra of the finite group G) and the irreducible characters of G.

Hook Lengths and Contents p. 2 Set x (i) = 1 t i, so s λ(x (i) ) t i +c u h u p ρ(wi )(x (i) ) t κ(w i) i. Get λ n H 2 λ 1 n! (t 1 + c u ) (t k + c u ) = u λ w 1 w k =1 in S n κ(w t 1 ) 1 t κ(w k) k.

Hook Lengths and Contents p. 2 Completion of proof Take coefficient of t n µ 1 1 t n µ k 1 n! λ n k : f 2 λ e µ(c u : u λ) = #{(w 1,..., w k ) S k n : w 1 w k = 1, c(w i ) = n µ i }.

Hook Lengths and Contents p. 2 Completion of proof Take coefficient of t n µ 1 1 t n µ k 1 n! λ n k : f 2 λ e µ(c u : u λ) = #{(w 1,..., w k ) S k n : w 1 w k = 1, c(w i ) = n µ i }. Elementary combinatorial argument shows this is a polynomial in n.

Hook Lengths and Contents p. 2 Shifted parts Let λ = (λ 1,..., λ n ) n. partitions of 3: 300, 210, 111

Hook Lengths and Contents p. 2 Shifted parts Let λ = (λ 1,..., λ n ) n. partitions of 3: 300, 210, 111 shifted parts of λ: λ i + n i, 1 i n

Hook Lengths and Contents p. 2 Shifted parts Let λ = (λ 1,..., λ n ) n. partitions of 3: 300, 210, 111 shifted parts of λ: λ i + n i, 1 i n shifted parts of (3, 1, 1, 0, 0): 7, 4, 3, 1, 0

Hook Lengths and Contents p. 2 Contents and shifted parts Let F (x; y) be symmetric in x and y variables separately, e.g., p 1 (x)p 2 (y) = (x 1 + x 2 + )(y 2 1 + y2 2 + ).

Hook Lengths and Contents p. 2 Contents and shifted parts Let F (x; y) be symmetric in x and y variables separately, e.g., p 1 (x)p 2 (y) = (x 1 + x 2 + )(y 2 1 + y2 2 + ). Theorem. Let r(n)= 1 f 2 λ n! F (c u, u λ; λ i + n i, 1 i n λ n Then r(n) Z (n fixed), and r(n) Q[n].

Hook Lengths and Contents p. 2 Similar results are false Note. Let P (n)= 1 n! f 2 λ ( λ 2 i ). λ n i

Hook Lengths and Contents p. 2 Similar results are false Note. Let P (n)= 1 n! f 2 λ ( λ 2 i ). λ n i Then P (3) = 16 3 Z P (n) Q[n]

Hook Lengths and Contents p. 2 ϕ Λ Q = {symmetric functions over Q} Define a linear transformation ϕ : Λ Q Q[t] by ϕ(s λ ) = n (t + λ i + n i) i=1 H λ.

Hook Lengths and Contents p. 2 Key lemma for shifted parts Lemma. Let µ n, l = l(µ). Then ϕ(p µ ) = ( 1) n l m i=0 ( m i ) t(t + 1) (t + i 1), where m= m 1 (µ), the number of parts of µ equal to 1.

Hook Lengths and Contents p. 2 Proof of main result Theorem. For any symmetric function F, f 2 λ F (h2 u : u λ) Q[n]. 1 n! λ n

Hook Lengths and Contents p. 2 Proof of main result Theorem. For any symmetric function F, f 2 λ F (h2 u : u λ) Q[n]. 1 n! λ n Proof based on the multiset identity {h u : u λ} {λ i λ j i + j : 1 i < j n} = {n + c u : u λ} {1 n 1, 2 n 2,..., n 1}.

Hook Lengths and Contents p. 2 Example. λ = (3, 1, 0, 0), λ (1, 2, 3, 4) = (2, 1, 3, 4): {4, 2, 1, 1} {3, 5, 6, 2, 3, 1} = {3, 4, 5, 6} {1, 1, 1, 2, 2, 3}

Hook Lengths and Contents p. 3 {h u : u λ} {λ i λ j i + j : 1 i < j n} = {n + c u : u λ} {1 n 1, 2 n 2,..., n 1}.

Hook Lengths and Contents p. 3 {h u : u λ} {λ i λ j i + j : 1 i < j n} = {n + c u : u λ} {1 n 1, 2 n 2,..., n 1}. Note. λ i λ j i + j = (λ i + n i) (λ j + n j) Not symmetric in λ i + n i and λ j + n j, but (λ i λ j i + j) 2 is symmetric.

Hook Lengths and Contents p. 3 Further results Recall: Nekrasov-Okounkov (2006), G. Han (2008): n 0 ( λ n f 2 λ ) x n (t + h 2 u ) u λ n! 2 = i 1 (1 x i ) t 1

Hook Lengths and Contents p. 3 Content analogue Content Nekrasov-Okounkov identity n 0 ( λ n f 2 λ ) x n (t + c 2 u ) u λ n! 2 = (1 x) t. Proof: simple consequence of dual Cauchy identity" and the hook-content formula.

Hook Lengths and Contents p. 3 Fujii et al. variant 1 n! λ n f 2 λ u λ (c 2 u i2 ) k 1 i=0 = (2k)! (k + 1)! 2(n) k+1. where (n) k+1 = n(n 1) (n k).

Hook Lengths and Contents p. 3 Okada s conjecture 1 n! λ n f 2 λ u λ k (h 2 u i2 ) i=1 = 1 2(k + 1) 2 ( )( ) 2k 2k + 2 (n) k+1 k k + 1

Hook Lengths and Contents p. 3 Proof of Okada s conjecture Proof (Greta Panova).

Hook Lengths and Contents p. 3 Proof of Okada s conjecture Proof (Greta Panova). Both sides are polynomials in n.

Hook Lengths and Contents p. 3 Proof of Okada s conjecture Proof (Greta Panova). Both sides are polynomials in n. The two sides agree for 1 n k + 2.

Hook Lengths and Contents p. 3 Proof of Okada s conjecture Proof (Greta Panova). Both sides are polynomials in n. The two sides agree for 1 n k + 2. Key step: both sides have degree k + 1.

Hook Lengths and Contents p. 3 Central factorial numbers x n = k T (n, k)x(x 1 2 )(x 2 2 ) (x k 2 )

Hook Lengths and Contents p. 3 Central factorial numbers x n = k T (n, k)x(x 1 2 )(x 2 2 ) (x k 2 ) n\k 1 2 3 4 5 1 1 2 1 1 3 1 5 1 4 1 21 14 1 5 1 85 147 30 1

Hook Lengths and Contents p. 3 Okada-Panova redux Corollary. = k+1 j=1 1 n! λ n f 2 λ u λ h 2k u T (k + 1, j) 1 2j 2 ( )( 2(j 1) 2j j 1 j ) (n) j.