Alternating Permutations Richard P. Stanley M.I.T. Alternating - PowerPoint PPT Presentation

Alternating Permutations Richard P. Stanley M.I.T. Alternating Permutations – p. 1

Definitions A sequence a 1 , a 2 , . . . , a k of distinct integers is alternating if a 1 > a 2 < a 3 > a 4 < · · · , and reverse alternating if a 1 < a 2 > a 3 < a 4 > · · · . Alternating Permutations – p. 2

Definitions A sequence a 1 , a 2 , . . . , a k of distinct integers is alternating if a 1 > a 2 < a 3 > a 4 < · · · , and reverse alternating if a 1 < a 2 > a 3 < a 4 > · · · . S n : symmetric group of all permutations of 1 , 2 , . . . , n Alternating Permutations – p. 2

E n E n = # { w ∈ S n : w is alternating } = # { w ∈ S n : w is reverse alternating } via a 1 a 2 · · · a n �→ n + 1 − a 1 , n + 1 − a 2 , . . . , n + 1 − a n Alternating Permutations – p. 3

E n E n = # { w ∈ S n : w is alternating } = # { w ∈ S n : w is reverse alternating } via a 1 a 2 · · · a n �→ n + 1 − a 1 , n + 1 − a 2 , . . . , n + 1 − a n E.g., E 4 = 5 : 2143, 3142, 3241, 4132, 4231 Alternating Permutations – p. 3

André’s theorem Theorem (Désiré André, 1879) x n � y := E n n ! = sec x + tan x n ≥ 0 Alternating Permutations – p. 4

André’s theorem Theorem (Désiré André, 1879) x n � y := E n n ! = sec x + tan x n ≥ 0 E n is an Euler number , E 2 n a secant number , E 2 n − 1 a tangent number . Alternating Permutations – p. 4

Naive proof 52439 1 826 � �� � ���� ← − − → alternating alternating Alternating Permutations – p. 5

Naive proof 52439 1 826 � �� � ���� ← − − → alternating alternating Obtain w that is either alternating or reverse alternating. Alternating Permutations – p. 5

Naive proof 52439 1 826 � �� � ���� ← − − → alternating alternating Obtain w that is either alternating or reverse alternating. n � n � � ⇒ 2 E n +1 = E k E n − k , n ≥ 1 k k =0 Alternating Permutations – p. 5

Completion of proof n � n � � E k E n − k , n ≥ 1 2 E n +1 = k k =0 Alternating Permutations – p. 6

Completion of proof n � n � � E k E n − k , n ≥ 1 2 E n +1 = k k =0 x n � = 2 y ′ 2 E n +1 n ! n ≥ 0 � n � � n � x n � � = 1 + y 2 E k E n − k k n ! n ≥ 0 k =0 Alternating Permutations – p. 6

Completion of proof n � n � � E k E n − k , n ≥ 1 2 E n +1 = k k =0 x n � = 2 y ′ 2 E n +1 n ! n ≥ 0 � n � � n � x n � � = 1 + y 2 E k E n − k k n ! n ≥ 0 k =0 ⇒ 2 y ′ = 1 + y 2 , etc . Alternating Permutations – p. 6

A generalization ∃ more sophisticated approaches. E.g., let f k ( n ) = # { w ∈ S n : w ( r ) < w ( r + 1) ⇔ k | r } f 2 ( n ) = E n Alternating Permutations – p. 7

A generalization ∃ more sophisticated approaches. E.g., let f k ( n ) = # { w ∈ S n : w ( r ) < w ( r + 1) ⇔ k | r } f 2 ( n ) = E n Theorem. �� � − 1 f k ( kn ) x kn ( − 1) n x kn � ( kn )! = . ( kn )! n ≥ 0 k ≥ 0 Alternating Permutations – p. 7

Combinatorial trigonometry Define sec x and tan x by x n � sec x + tan x = E n n ! . n ≥ 0 Alternating Permutations – p. 8

Combinatorial trigonometry Define sec x and tan x by x n � sec x + tan x = E n n ! . n ≥ 0 Exercise. sec 2 x = 1 + tan 2 x Alternating Permutations – p. 8

Combinatorial trigonometry Define sec x and tan x by x n � sec x + tan x = E n n ! . n ≥ 0 Exercise. sec 2 x = 1 + tan 2 x tan x + tan y Exercise. tan( x + y ) = 1 − (tan x )(tan y ) Alternating Permutations – p. 8

Some occurences of Euler numbers (1) E 2 n − 1 is the number of complete increasing binary trees on the vertex set [2 n + 1] = { 1 , 2 , . . . , 2 n + 1 } . Alternating Permutations – p. 9

Five vertices 1 1 1 1 1 1 2 2 2 2 2 2 5 3 5 5 4 5 3 4 3 3 3 4 4 3 4 5 5 4 1 1 1 1 1 1 3 3 2 2 2 2 2 2 4 4 5 5 4 5 5 4 3 5 5 3 3 4 4 3 1 1 1 1 3 3 2 2 2 4 2 3 3 5 5 4 4 5 5 4 Alternating Permutations – p. 10

Five vertices 1 1 1 1 1 1 2 2 2 2 2 2 5 3 5 5 4 5 3 4 3 3 3 4 4 3 4 5 5 4 1 1 1 1 1 1 3 3 2 2 2 2 2 2 4 4 5 5 4 5 5 4 3 5 5 3 3 4 4 3 1 1 1 1 3 3 2 2 2 4 2 3 3 5 5 4 4 5 5 4 Slightly more complicated for E 2 n Alternating Permutations – p. 10

Simsun permutations Definition. A permutation w = a 1 · · · a n ∈ S n is simsun if for all 1 ≤ k ≤ n , the restriction b 1 · · · b k of w to 1 , 2 , . . . , k has no double descents b i − 1 > b i > b i +1 . Alternating Permutations – p. 11

Simsun permutations Definition. A permutation w = a 1 · · · a n ∈ S n is simsun if for all 1 ≤ k ≤ n , the restriction b 1 · · · b k of w to 1 , 2 , . . . , k has no double descents b i − 1 > b i > b i +1 . Example. 346251 is not simsun: restriction to 1 , 2 , 3 , 4 is 3 421 . Alternating Permutations – p. 11

Simsun permutations Definition. A permutation w = a 1 · · · a n ∈ S n is simsun if for all 1 ≤ k ≤ n , the restriction b 1 · · · b k of w to 1 , 2 , . . . , k has no double descents b i − 1 > b i > b i +1 . Example. 346251 is not simsun: restriction to 1 , 2 , 3 , 4 is 3 421 . Theorem (R. Sim ion and S. Sun daram) The number of simsun permutations w ∈ S n is E n . Alternating Permutations – p. 11

Chains of partitions (2) Start with n one-element sets { 1 } , . . . , { n } . Alternating Permutations – p. 12

Chains of partitions (2) Start with n one-element sets { 1 } , . . . , { n } . Merge together two at a time until reaching { 1 , 2 , . . . , n } . Alternating Permutations – p. 12

Chains of partitions (2) Start with n one-element sets { 1 } , . . . , { n } . Merge together two at a time until reaching { 1 , 2 , . . . , n } . 1 − 2 − 3 − 4 − 5 − 6 , 12 − 3 − 4 − 5 − 6 , 12 − 34 − 5 − 6 125 − 34 − 6 , 125 − 346 , 123456 Alternating Permutations – p. 12

Chains of partitions (2) Start with n one-element sets { 1 } , . . . , { n } . Merge together two at a time until reaching { 1 , 2 , . . . , n } . 1 − 2 − 3 − 4 − 5 − 6 , 12 − 3 − 4 − 5 − 6 , 12 − 34 − 5 − 6 125 − 34 − 6 , 125 − 346 , 123456 S n acts on these chains. Alternating Permutations – p. 12

Chains of partitions (2) Start with n one-element sets { 1 } , . . . , { n } . Merge together two at a time until reaching { 1 , 2 , . . . , n } . 1 − 2 − 3 − 4 − 5 − 6 , 12 − 3 − 4 − 5 − 6 , 12 − 34 − 5 − 6 125 − 34 − 6 , 125 − 346 , 123456 S n acts on these chains. Theorem. The number of S n -orbits is E n 1 . − Alternating Permutations – p. 12

Orbit representatives for n = 5 12 − 3 − 4 − 5 123 − 4 − 5 1234 − 5 12 − 3 − 4 − 5 123 − 4 − 5 123 − 45 12 − 3 − 4 − 5 12 − 34 − 5 125 − 34 12 − 3 − 4 − 5 12 − 34 − 5 12 − 345 12 − 3 − 4 − 5 12 − 34 − 5 1234 − 5 Alternating Permutations – p. 13

A polytope volume Let P n be the convex polytope in R n with equations x i ≥ 0 , 1 ≤ i ≤ n x i + x i +1 ≤ 1 , 1 ≤ i ≤ n − 1 . Alternating Permutations – p. 14

A polytope volume Let P n be the convex polytope in R n with equations x i ≥ 0 , 1 ≤ i ≤ n x i + x i +1 ≤ 1 , 1 ≤ i ≤ n − 1 . Theorem. vol ( P n ) = 1 n ! E n Alternating Permutations – p. 14

Ulam’s problem Let E ( n ) be the expected length is ( w ) of the longest increasing subsequence of w . Alternating Permutations – p. 15

Ulam’s problem Let E ( n ) be the expected length is ( w ) of the longest increasing subsequence of w . w = 5 24 1 7 36 ⇒ is( w ) = 3 Alternating Permutations – p. 15

Two highlights Long story . . . Alternating Permutations – p. 16

Two highlights Long story . . . Logan-Shepp, Vershik-Kerov: E ( n ) ∼ 2 √ n Alternating Permutations – p. 16

Two highlights Long story . . . Logan-Shepp, Vershik-Kerov: E ( n ) ∼ 2 √ n Write is n ( w ) for is( w ) when w ∈ S n . Baik-Deift-Johansson: � is n ( w ) − 2 √ n � ≤ t n →∞ Prob lim = F ( t ) , n 1 / 6 the Tracy-Widom distribution . Alternating Permutations – p. 16

Alternating subsequences? as( w ) = length of longest alternating subseq. of w w = 5 6 2 1 834 7 ⇒ as( w ) = 5 Alternating Permutations – p. 17

The main lemma MAIN LEMMA. ∀ w ∈ S n ∃ alternating subsequence of maximal length that contains n . Alternating Permutations – p. 18

The main lemma MAIN LEMMA. ∀ w ∈ S n ∃ alternating subsequence of maximal length that contains n . a k ( n ) = # { w ∈ S n : as( w ) = k } b k ( n ) = a 1 ( n ) + a 2 ( n ) + · · · + a k ( n ) = # { w ∈ S n : as( w ) ≤ k } . Alternating Permutations – p. 18

Recurrence for a k ( n ) n � n − 1 � � ⇒ a k ( n ) = j − 1 j =1 � ( a 2 r ( j − 1) + a 2 r +1 ( j − 1)) a s ( n − j ) 2 r + s = k − 1 Alternating Permutations – p. 19

The main generating function b k ( n ) t k x n � Define B ( x, t ) = n ! k,n ≥ 0 Theorem. 2 /ρ t e ρx − 1 B ( x, t ) = ρ, 1 − 1 − ρ √ 1 − t 2 . where ρ = Alternating Permutations – p. 20