An Asymptotic Version of a Theorem of Knuth Jonathan Novak MSRI - PowerPoint PPT Presentation

An Asymptotic Version of a Theorem of Knuth Jonathan Novak MSRI & Waterloo Permutation Patterns 2010 August 10, 2010

Symmetry

Schensted pairs s ( d , N ) = no. of Schensted pairs on partitions λ ⊢ N , ℓ ( λ ) ≤ d s (3 , 9) = 94 359 1 2 3 4 5 1 3 5 7 9 6 7 8 2 4 6 9 8 s ( d , N ) = ?

Knuth’s formula Theorem s (2 , N ) = dim R (2 , N ) Proof. ♣ ♣ ♣ ♣ ♥ ♥ ♥ ♥ � ♣ ♣ ♣ ♣ ♥ ♥ ♣ ♣ ♥ ♥ ♣ ♣ ♥ ♥ ♥ ♥ Corollary � 2 N � 1 s (2 , N ) = . N + 1 N

A general formula � (dim λ ) 2 , s ( d , N ) = λ ⊢ N ℓ ( λ ) ≤ d where N ! � ( λ i − λ j + j − i ) . dim λ = � d i =1 ( λ i − i + d )! 1 ≤ i < j ≤ d Challenge: use this formula to estimate s (3 , 10 10 ) .

Regev’s formula Theorem For any fixed d ≥ 1 , � d − 1 � d 2 N + d 2 1 − d 2 � 1 − d 2 (2 N ) s ( d , N ) ∼ (2 π ) i ! 2 2 i =0 � d − 1 � � � � � d 2 1 − d 2 1 − d 2 � 1 − d d 2 N N 2 2 = (2 π ) i ! d 2 2 2 i =0 � �� � � �� � S ( d , N ) growth rate GUE partition function as N → ∞ .

Regev’s formula Proof. “Continuous” Schensted pairs: ♣ ♣ ♣ ♣ ♥ ♥ ♥ ♥ . . . ♣ ♣ ♥ ♥ � �� � β times � (dim λ ) β s ( d , N ; β ) := λ ⊢ N ℓ ( λ ) ≤ d � C β e − β W ( y 1 ,..., y d − 1 ) d y d , N s ( d , N ; β ) → Ω d − 1 � �� � Mehta-Dyson-Selberg

Asymmetry

Asymmetry ♠ ♠ ♠ ♠ ♥ ♥ ♥ ♥ ♠ ♥ � ? ♠ ♥ ♠ ♠ ♠ ♠ ? � ♠ ♥ ♥ ♥ ♠ ♥ ♥ ♥

Symmetry

Symmetry

Symmetry

Symmetry

Asymptotic symmetry ♠ ♠ ♠ ♠ ♠ ♥ ♥ ♥ ♥ ♥ ♣ ♣ ♣ ♣ ♣ ♣ ♠ ♠ ♠ ♥ ♥ ♥ ♣ ♣ ♣ ♣ ♣ ♣ � ♠ ♥ ♣ ♣ ♣ ♣ ♣ ♣ Conjecture: s ( d , N ) ∼ dim R ( d , 2 N / d )

Verification of asymptotic symmetry Exact formula: ( dq )! dim R ( d , q ) = � d − 1 ( q + i )! i =0 i ! Dimension of a d × ∞ strip: � d − 1 � � 1 − d 2 1 − d d dq + 1 2 q dim R ( d , q ) ∼ (2 π ) i ! 2 2 i =0 Scaling dictated by symmetry: q � 2 N / d Reproduces Regev’s formula: � d − 1 � � d 2 N + d 2 1 − d 2 1 − d 2 (2 N ) dim R ( d , 2 N / d ) ∼ (2 π ) i ! 2 2 i =0

Asymptotic Knuth theorem Theorem For any fixed d ≥ 1 , s ( d , dn ) ∼ dim R ( d , 2 n ) as n → ∞ . Corollary The number of permutations in S ( dn ) with no decreasing subsequence of length d + 1 is asymptotically equal to the number of involutions in S (2 dn ) with longest decreasing subsequence of length exactly d and longest increasing subsequence of length exactly 2 n .

Error term Complements: µ ⊂ R ( d , q ) , µ ∗ R ( d , q ) = ( q − µ d , q − µ d − 1 , . . . , q − µ 1 ) Theorem s ( d , dn ) = dim R ( d , 2 n ) + E ( d , dn ) , where E ( d , dn ) = 1 � � (dim µ − dim µ ∗ ) 2 (dim ν ) 2 + . 2 µ ⊢ dn ν ⊢ dn ν 1 > 2 n µ ⊂ R ( d , 2 n ) � �� � � �� � large deviation asymmetry

Laplace method If you want to understand a sum/integral where the integrand contains a large parameter, the maximum of the integrand is the centre of the universe. √ n , . . . , n + y d √ n ) ∼ ? dim( n + y 1

Laplace method Theorem For any fixed y 1 > · · · > y d , y 1 + · · · + y d = 0 , √ n , . . . , n + y d √ n ) = e − W ( y 1 ,..., y d ) , n →∞ C d , dn dim( n + y 1 lim where d W ( y 1 , . . . , y d ) = 1 � � y 2 i − log( y i − y j ) . 2 i =1 1 ≤ i < j ≤ d Proof. √ n , . . . , n + y d √ n ) dim( n + y 1 (( y i − y j ) √ n + j − i ) . Γ( dn + 1) � = √ n + i + d + 1) � d i =1 Γ( n + y i 1 ≤ i < j ≤ d

Laplace method � (dim λ ) β s ( d , N ; β ) = λ ⊢ N ℓ ( λ ) ≤ d � n →∞ C β e − β W ( y 1 ,..., y d ) d y lim d , dn s ( d , dn ; β ) = Ω d − 1 Ω d − 1 = { y 1 > · · · > y d , y 1 + · · · + y d = 0 } ⊂ R d − 1 Regev: evaluate this (difficult) integral.

Laplace method � (dim µ )(dim µ ∗ ) . dim R ( d , 2 n ) = µ ⊢ dn µ ⊂ R ( d , 2 n ) � (dim µ ) γ (dim µ ∗ ) δ . t ( d , dn ; γ, δ ) = µ ⊢ dn µ ⊂ R ( d , 2 n ) Exactly the same argument: � n →∞ C γ + δ e − γ W ( y 1 ,..., y d ) e − δ W ( − y d ,..., − y 1 ) d y . lim d , dn t ( d , dn ; γ, δ ) = Ω d − 1

Symmetry returns s ( d , dn ) ∼ dim R ( d , 2 n ) � � � e − 2 W ( y 1 ,..., y d ) d y = e − W ( y 1 ,..., y d ) e − W ( − y d ,..., − y 1 ) d y Ω d − 1 Ω d − 1 ⇑ W ( y 1 , . . . , y d ) = W ( − y d , . . . , − y 1 )

Symmetry returns Energy: d W ( y 1 , . . . , y d ) = 1 � � y 2 i − log( y i − y j ) . 2 i =1 1 ≤ i < j ≤ d Symmetry: W ( y 1 , . . . , y d ) = W ( − y d , . . . , − y 1 )

Symmetry returns W ( y 1 , . . . , y d ) = W ( − y d , . . . , − y 1 ) Theorem For any 0 ≤ γ < β, � � (dim λ ) β ∼ (dim µ ) γ (dim µ ∗ ) β − γ λ ⊢ dn µ ⊢ dn ℓ ( λ ) ≤ d µ ⊂ R ( d , 2 n ) Corollary � � (dim λ ) 2 ∼ (dim µ )(dim µ ∗ ) = dim R ( d , 2 n ) λ ⊢ dn µ ⊢ dn ℓ ( λ ) ≤ d µ ⊂ R ( d , 2 n )

Mehta-Dyson integral Energy: d W ( t 1 , . . . , t d ) = 1 � � t 2 i − log( t i − t j ) . 2 i =1 1 ≤ i < j ≤ d Partition function (Mehta-Dyson integral): � e − β W ( t 1 ,..., t d ) d t Ψ( d ; β ) = W d W d = { t 1 > · · · > t d } ⊂ R d

Mehta-Dyson conjecture: d Γ(1 + i β 2 ) Ψ( d ; β ) = 1 � 2 − β d ( d − 1) 2 β − d d d !(2 π ) . 4 Γ(1 + β 2 ) i =1 • Bombieri: Selberg = ⇒ Mehta-Dyson • Symmetry = ⇒ Ψ( d ; 2) • Dyson: Ψ( d ; 2 k ) = ⇒ Ψ( d ; β ) . • Symmetry = ⇒ Ψ( d ; 2 k )???

Double-Scaling limit Baik-Deift-Johansson, Okounkov, Borodin-Okounkov-Olshanski, Johansson: Theorem For d , N → ∞ at the rate d ∼ 2 N 1 / 2 + tN 1 / 6 , s ( d , N ) ∼ F ( t ) N ! , where F ( t ) = Tracy-Widom distribution function. s ( d , dn ) = dim R ( d , 2 n )+1 � � (dim µ − dim µ ∗ ) 2 + (dim ν ) 2 . 2 µ ⊢ dn ν ⊢ dn ν 1 > 2 n µ ⊂ R ( d , 2 n ) Asymptotics of E ( d , dn ) in double scaling limit???

Acknowledgements Many thanks to Michael Albert and Andrei Okounkov.