Young tableaux and snakes Yuliy Baryshnikov joint work with Dan - PowerPoint PPT Presentation

Young tableaux and snakes Yuliy Baryshnikov joint work with Dan Romik (Hebrew University)

Motivation • A Young tableaux is a filling of Young diagram consisting of n boxes with numbers 1 , . . . , n increasing top-to-down and left-to-right. 1 2 6 8 9 3 5 4 n=9 7

Motivation • A Young tableaux is a filling of Young diagram consisting of n boxes with numbers 1 , . . . , n increasing top-to-down and left-to-right. 1 2 6 8 9 3 5 4 n=9 7 • The number of YTs with given shape λ has various interpretations (dimension of the representation λ of S n , for example).

Motivation • A Young tableaux is a filling of Young diagram consisting of n boxes with numbers 1 , . . . , n increasing top-to-down and left-to-right. 1 2 6 8 9 3 5 4 n=9 7 • The number of YTs with given shape λ has various interpretations (dimension of the representation λ of S n , for example). • Asymptotic regime is of interest:

Motivation (cont’d) • Consider a large shape t λ, t → ∞ :

Motivation (cont’d) • Consider a large shape t λ, t → ∞ : and a typical Young tableaux filling it: • The natural question arises:

Motivation (cont’d) • Consider a large shape t λ, t → ∞ : and a typical Young tableaux filling it: • The natural question arises: Conjecture A typical YT, considered as a function on the Young diageam t λ is close to some deterministic limiting function.

Motivation (cont’d) • Consider a large shape t λ, t → ∞ : and a typical Young tableaux filling it: • The natural question arises: Conjecture A typical YT, considered as a function on the Young diageam t λ is close to some deterministic limiting function. • How one would prove it?

Motivation (cont’d) • By finding the rate function and then solving variational problem.

Motivation (cont’d) • By finding the rate function and then solving variational problem. • Rate function will count the (normalized, per unit area) number of YT filling the shapes approximating a strip

Motivation (cont’d) • Hence we have to compute the number of Young tableaux filling the strips like

Motivation (cont’d) • Hence we have to compute the number of Young tableaux filling the strips like

Motivation (cont’d) • Hence we have to compute the number of Young tableaux filling the strips like • We start with the simplest task: finding the number of YT filling the strip of width 2 and slope 1.

Up-down permutations • A permutation σ ∈ S n is called an up-down permutation (also zig-zag permutation, alternating permutation) if it satisfies σ (1) < σ (2) > σ (3) < σ (4) > . . .

Up-down permutations • A permutation σ ∈ S n is called an up-down permutation (also zig-zag permutation, alternating permutation) if it satisfies σ (1) < σ (2) > σ (3) < σ (4) > . . . Equivalent to “2-strip” tableaux:

Up-down permutations • A permutation σ ∈ S n is called an up-down permutation (also zig-zag permutation, alternating permutation) if it satisfies σ (1) < σ (2) > σ (3) < σ (4) > . . . Equivalent to “2-strip” tableaux: 3 13 4 11 1 5 9 14 2 12 6 7 10

Up-down permutations • A permutation σ ∈ S n is called an up-down permutation (also zig-zag permutation, alternating permutation) if it satisfies σ (1) < σ (2) > σ (3) < σ (4) > . . . Equivalent to “2-strip” tableaux: 3 13 4 11 1 5 9 14 2 12 6 7 10 • Theorem (D. Andr´ e 1881): Let A n = # of n -element up-down permutations. Then ∞ A n x n � = tan x + sec x . n ! n =0

Up-down permutations • A permutation σ ∈ S n is called an up-down permutation (also zig-zag permutation, alternating permutation) if it satisfies σ (1) < σ (2) > σ (3) < σ (4) > . . . Equivalent to “2-strip” tableaux: 3 13 4 11 1 5 9 14 2 12 6 7 10 • Theorem (D. Andr´ e 1881): Let A n = # of n -element up-down permutations. Then ∞ A n x n � = tan x + sec x . n ! n =0 • Up-down permutations were named snakes and studied by V. Arnold to enumerate morsifications of real singularities .

Up-down permutations (continued) • Reminder: ∞ E n x n � sech x = n ! , ( E n ) n ≥ 0 – Euler numbers , n =0 ∞ T n x 2 n − 1 � tan x = (2 n − 1)! , ( T n ) n ≥ 0 – Tangent numbers , n =1 ∞ B n x n x � = n ! , ( B n ) n ≥ 0 – Bernoulli numbers . e x − 1 n =0

Up-down permutations (continued) • Reminder: ∞ E n x n � sech x = n ! , ( E n ) n ≥ 0 – Euler numbers , n =0 ∞ T n x 2 n − 1 � tan x = (2 n − 1)! , ( T n ) n ≥ 0 – Tangent numbers , n =1 ∞ B n x n x � = n ! , ( B n ) n ≥ 0 – Bernoulli numbers . e x − 1 n =0 A 2 n − 1 = T n = ( − 1) n − 1 4 n (4 n − 1) • In this notation: A 2 n = | E 2 n | , B 2 n . 2 n

Transfer operators Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer operators (due to N. Elkies, 2003):

Transfer operators Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer operators (due to N. Elkies, 2003): � ( x 1 , x 2 , . . . , x n ) ∈ [0 , 1] n : x 1 < x 2 > x 3 < x 4 > . . . � • Let P n = .

Transfer operators Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer operators (due to N. Elkies, 2003): � ( x 1 , x 2 , . . . , x n ) ∈ [0 , 1] n : x 1 < x 2 > x 3 < x 4 > . . . � • Let P n = . • Compute vol( P n ) in two ways:

Transfer operators Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer operators (due to N. Elkies, 2003): � ( x 1 , x 2 , . . . , x n ) ∈ [0 , 1] n : x 1 < x 2 > x 3 < x 4 > . . . � • Let P n = . • Compute vol( P n ) in two ways: First, vol( P n ) = A n n ! ;

Transfer operators Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer operators (due to N. Elkies, 2003): � ( x 1 , x 2 , . . . , x n ) ∈ [0 , 1] n : x 1 < x 2 > x 3 < x 4 > . . . � • Let P n = . • Compute vol( P n ) in two ways: First, vol( P n ) = A n n ! ; • Second, vol( P n ) =

Transfer operators Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer operators (due to N. Elkies, 2003): � ( x 1 , x 2 , . . . , x n ) ∈ [0 , 1] n : x 1 < x 2 > x 3 < x 4 > . . . � • Let P n = . • Compute vol( P n ) in two ways: First, vol( P n ) = A n n ! ; • Second, � 1 vol( P n ) = dx 1 0 x 1

Transfer operators Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer operators (due to N. Elkies, 2003): � ( x 1 , x 2 , . . . , x n ) ∈ [0 , 1] n : x 1 < x 2 > x 3 < x 4 > . . . � • Let P n = . • Compute vol( P n ) in two ways: First, vol( P n ) = A n n ! ; • Second, � 1 � 1 vol( P n ) = dx 1 dx 2 x 2 0 x 1 x 1

Transfer operators Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer operators (due to N. Elkies, 2003): � ( x 1 , x 2 , . . . , x n ) ∈ [0 , 1] n : x 1 < x 2 > x 3 < x 4 > . . . � • Let P n = . • Compute vol( P n ) in two ways: First, vol( P n ) = A n n ! ; • Second, � 1 � 1 � x 2 vol( P n ) = dx 1 dx 2 dx 3 x 2 x 3 0 x 1 0 x 1

Transfer operators Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer operators (due to N. Elkies, 2003): � ( x 1 , x 2 , . . . , x n ) ∈ [0 , 1] n : x 1 < x 2 > x 3 < x 4 > . . . � • Let P n = . • Compute vol( P n ) in two ways: First, vol( P n ) = A n n ! ; • Second, � 1 � 1 � 1 � x 2 x 4 vol( P n ) = dx 1 dx 2 dx 3 dx 4 x 2 x 3 0 x 1 0 x 3 x 1

Transfer operators Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer operators (due to N. Elkies, 2003): � ( x 1 , x 2 , . . . , x n ) ∈ [0 , 1] n : x 1 < x 2 > x 3 < x 4 > . . . � • Let P n = . • Compute vol( P n ) in two ways: First, vol( P n ) = A n n ! ; • Second, � 1 � 1 � 1 � x 2 x 4 vol( P n ) = dx 4 . . . dx 1 dx 2 dx 3 x 2 x 3 0 x 1 0 x 3 = x 1

Transfer operators Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer operators (due to N. Elkies, 2003): � ( x 1 , x 2 , . . . , x n ) ∈ [0 , 1] n : x 1 < x 2 > x 3 < x 4 > . . . � • Let P n = . • Compute vol( P n ) in two ways: First, vol( P n ) = A n n ! ; • Second, � 1 � 1 � 1 � x 2 x 4 vol( P n ) = dx 4 . . . dx 1 dx 2 dx 3 x 2 x 3 0 x 1 0 x 3 = � . . . ◦ T ◦ S ◦ T ◦ S 1 , 1 � L 2 [ 0 , 1 ] , x 1

Transfer operators Many proofs of Andr´ e’s theorem, mostly algebraic. Proof using transfer operators (due to N. Elkies, 2003): � ( x 1 , x 2 , . . . , x n ) ∈ [0 , 1] n : x 1 < x 2 > x 3 < x 4 > . . . � • Let P n = . • Compute vol( P n ) in two ways: First, vol( P n ) = A n n ! ; • Second, � 1 � 1 � 1 � x 2 x 4 vol( P n ) = dx 4 . . . dx 1 dx 2 dx 3 x 2 x 3 0 x 1 0 x 3 = � . . . ◦ T ◦ S ◦ T ◦ S 1 , 1 � L 2 [ 0 , 1 ] , x 1 where � 1 � x ( Tf )( x ) = f ( y ) dy , ( Sg )( x ) = g ( y ) dy 0 x