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

Orbits of mergings (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 sequences. Theorem. The number of S n -orbits is E n 1 . − Alternating Permutations – p. 17

Orbits of mergings (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 sequences. Theorem. The number of S n -orbits is E n 1 . − Proof. Alternating Permutations – p. 17

Orbits of mergings (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 sequences. Theorem. The number of S n -orbits is E n 1 . − Proof. Exercise. Alternating Permutations – p. 17

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. 18

Volume of a polytope (3) Let E n be the convex polytope in R n defined by x i ≥ 0 , 1 ≤ i ≤ n x i + x i +1 ≤ 1 , 1 ≤ i ≤ n − 1 . Alternating Permutations – p. 19

Volume of a polytope (3) Let E n be the convex polytope in R n defined by x i ≥ 0 , 1 ≤ i ≤ n x i + x i +1 ≤ 1 , 1 ≤ i ≤ n − 1 . Theorem. The volume of E n is E n /n ! . Alternating Permutations – p. 19

Naive proof � 1 � 1 − x 1 � 1 − x 2 � 1 − x n − 1 vol( E n ) = · · · dx 1 dx 2 · · · dx n x 1 =0 x 2 =0 x 3 =0 x n =0 Alternating Permutations – p. 20

Naive proof � 1 � 1 − x 1 � 1 − x 2 � 1 − x n − 1 vol( E n ) = · · · dx 1 dx 2 · · · dx n x 1 =0 x 2 =0 x 3 =0 x n =0 � t � 1 − x 1 � 1 − x 2 � 1 − x n − 1 f n ( t ) := · · · dx 1 dx 2 · · · dx n x 1 =0 x 2 =0 x 3 =0 x n =0 Alternating Permutations – p. 20

Naive proof � 1 � 1 − x 1 � 1 − x 2 � 1 − x n − 1 vol( E n ) = · · · dx 1 dx 2 · · · dx n x 1 =0 x 2 =0 x 3 =0 x n =0 � t � 1 − x 1 � 1 − x 2 � 1 − x n − 1 f n ( t ) := · · · dx 1 dx 2 · · · dx n x 1 =0 x 2 =0 x 3 =0 x n =0 � 1 − t � 1 − x 2 � 1 − x n − 1 f ′ n ( t ) = · · · dx 2 dx 3 · · · dx n x 2 =0 x 3 =0 x n =0 = f n − 1 (1 − t ) . Alternating Permutations – p. 20

F ( y ) f ′ n ( t ) = f n − 1 (1 − t ) , f 0 ( t ) = 1 , f n (0) = 0 ( n > 0) Alternating Permutations – p. 21

F ( y ) f ′ n ( t ) = f n − 1 (1 − t ) , f 0 ( t ) = 1 , f n (0) = 0 ( n > 0) � f n ( t ) y n F ( y ) = n ≥ 0 ⇒ ∂ 2 ∂t 2 F ( y ) = − y 2 F ( y ) , etc. Alternating Permutations – p. 21

Conclusion of proof F ( y ) = (sec y )(cos( t − 1) y + sin ty ) ⇒ F ( y ) | t =1 = sec y + tan y. Alternating Permutations – p. 22

Tridiagonal matrices An n × n matrix M = ( m ij ) is tridiagonal if m ij = 0 whenever | i − j | ≥ 2 . doubly-stochastic : m ij ≥ 0 , row and column sums equal 1 T n : set of n × n tridiagonal doubly stochastic matrices Alternating Permutations – p. 23

Polytope structure of T n Easy fact: the map T n → R n − 1 M �→ ( m 12 , m 23 , . . . , m n − 1 ,n ) is a (linear) bijection from T to E n − 1 . Alternating Permutations – p. 24

Polytope structure of T n Easy fact: the map T n → R n − 1 M �→ ( m 12 , m 23 , . . . , m n − 1 ,n ) is a (linear) bijection from T to E n − 1 . Application ( Diaconis et al.): random doubly stochastic tridiagonal matrices and random walks on T n Alternating Permutations – p. 24

A modification Let F n be the convex polytope in R n defined by x i ≥ 0 , 1 ≤ i ≤ n x i + x i +1 + x i +2 ≤ 1 , 1 ≤ i ≤ n − 2 . V n = vol( F n ) Alternating Permutations – p. 25

A modification Let F n be the convex polytope in R n defined by x i ≥ 0 , 1 ≤ i ≤ n x i + x i +1 + x i +2 ≤ 1 , 1 ≤ i ≤ n − 2 . V n = vol( F n ) n 1–3 4 5 6 7 8 9 10 n ! V n 1 2 5 14 47 182 786 3774 Alternating Permutations – p. 25

A “naive” recurrence V n = f n (1 , 1) , where f 0 ( a, b ) = 1 , f n (0 , b ) = 0 for n > 0 ∂ ∂af n ( a, b ) = f n − 1 ( b − a, 1 − a ) . Alternating Permutations – p. 26

f n ( a, b ) for n ≤ 3 f 1 ( a, b ) = a f 2 ( a, b ) = 1 2(2 ab − a 2 ) f 3 ( a, b ) = 1 6( a 3 − 3 a 2 − 3 ab 2 + 6 ab ) Alternating Permutations – p. 27

f n ( a, b ) for n ≤ 3 f 1 ( a, b ) = a f 2 ( a, b ) = 1 2(2 ab − a 2 ) f 3 ( a, b ) = 1 6( a 3 − 3 a 2 − 3 ab 2 + 6 ab ) Is there a “nice” generating function for f n ( a, b ) or V n = f n (1 , 1) ? Alternating Permutations – p. 27

Distribution of is ( w ) is( w ) = length of longest increasing subsequence of w ∈ S n Alternating Permutations – p. 28

Distribution of is ( w ) is( w ) = length of longest increasing subsequence of w ∈ S n is(48361572) = 3 Alternating Permutations – p. 28

Distribution of is ( w ) is( w ) = length of longest increasing subsequence of w ∈ S n is( 4 83 6 15 7 2) = 3 Alternating Permutations – p. 28

Distribution of is ( w ) is( w ) = length of longest increasing subsequence of w ∈ S n is( 4 83 6 15 7 2) = 3 Vershik-Kerov, Logan-Shepp: 1 � E ( n ) := is( w ) n ! w ∈ S n 2 √ n ∼ Alternating Permutations – p. 28

Limiting distribution of is ( w ) Baik-Deift-Johansson: For fixed t ∈ R , � is n ( w ) − 2 √ n � n →∞ Prob lim ≤ t = F ( t ) , n 1 / 6 the Tracy-Widom distribution . Alternating Permutations – p. 29

Alternating analogues Length of longest alternating subsequence of w ∈ S n Alternating Permutations – p. 30

Alternating analogues Length of longest alternating subsequence of w ∈ S n Length of longest increasing subsequence of an alternating permutation w ∈ S n . Alternating Permutations – p. 30

Alternating analogues Length of longest alternating subsequence of w ∈ S n Length of longest increasing subsequence of an alternating permutation w ∈ S n . The first is much easier! Alternating Permutations – p. 30

Longest alternating subsequences as( w ) = length of longest alt. subseq. of w w = 5 6 2 1 834 7 ⇒ as( w ) = 5 Alternating Permutations – p. 31

Longest alternating subsequences as( w ) = length of longest alt. subseq. of w w = 5 6 2 1 834 7 ⇒ as( w ) = 5 D ( n ) = 1 � as( w ) ∼ ? n ! w ∈ S n Alternating Permutations – p. 31

Definitions of a k ( n ) and b k ( 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. 32

The case n = 3 w as( w ) 1 23 1 1 32 2 213 3 2 3 1 2 312 3 3 2 1 2 Alternating Permutations – p. 33

The case n = 3 w as( w ) 1 23 1 1 32 2 213 3 2 3 1 2 312 3 3 2 1 2 a 1 (3) = 1 , a 2 (3) = 3 , a 3 (3) = 2 b 1 (3) = 1 , b 2 (3) = 4 , b 3 (3) = 6 Alternating Permutations – p. 33

The main lemma Lemma. ∀ w ∈ S n ∃ alternating subsequence of maximal length that contains n . Alternating Permutations – p. 34

The main lemma Lemma. ∀ w ∈ S n ∃ alternating subsequence of maximal length that contains n . Corollary. 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. 34

The main generating function b k ( n ) t k x n � 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. 35

Formulas for b k ( n ) Corollary. ⇒ b 1 ( n ) = 1 b 2 ( n ) = n 4 (3 n − 2 n + 3) 1 b 3 ( n ) = 8 (4 n − (2 n − 4)2 n ) 1 b 4 ( n ) = . . . Alternating Permutations – p. 36

Formulas for b k ( n ) Corollary. ⇒ b 1 ( n ) = 1 b 2 ( n ) = n 4 (3 n − 2 n + 3) 1 b 3 ( n ) = 8 (4 n − (2 n − 4)2 n ) 1 b 4 ( n ) = . . . no such formulas for longest increasing subsequences Alternating Permutations – p. 36

Mean (expectation) of as( w ) n D ( n ) = 1 as( w ) = 1 � � k · a k ( n ) , n ! n ! w ∈ S n k =1 the expectation of as( w ) for w ∈ S n Alternating Permutations – p. 37

Mean (expectation) of as( w ) n D ( n ) = 1 as( w ) = 1 � � k · a k ( n ) , n ! n ! w ∈ S n k =1 the expectation of as( w ) for w ∈ S n Let a k ( n ) t k x n � A ( x, t ) = n ! = (1 − t ) B ( x, t ) k,n ≥ 0 � � 2 /ρ t e ρx − 1 = (1 − t ) . 1 − 1 − ρ ρ Alternating Permutations – p. 37

Formula for D ( n ) ∂ D ( n ) x n = � ∂tA ( x, 1) n ≥ 0 = 6 x − 3 x 2 + x 3 6(1 − x ) 2 4 n + 1 � x n . = x + 6 n ≥ 2 Alternating Permutations – p. 38

Formula for D ( n ) ∂ D ( n ) x n = � ∂tA ( x, 1) n ≥ 0 = 6 x − 3 x 2 + x 3 6(1 − x ) 2 4 n + 1 � x n . = x + 6 n ≥ 2 ⇒ D ( n ) = 4 n + 1 , n ≥ 2 6 Alternating Permutations – p. 38

Formula for D ( n ) ∂ D ( n ) x n = � ∂tA ( x, 1) n ≥ 0 = 6 x − 3 x 2 + x 3 6(1 − x ) 2 4 n + 1 � x n . = x + 6 n ≥ 2 ⇒ D ( n ) = 4 n + 1 , n ≥ 2 6 Compare E ( n ) ∼ 2 √ n . Alternating Permutations – p. 38

Variance of as( w ) � 2 � V ( n ) = 1 as( w ) − 4 n + 1 � , n ≥ 2 n ! 6 w ∈ S n the variance of as( w ) for w ∈ S n Alternating Permutations – p. 39

Variance of as( w ) � 2 � V ( n ) = 1 as( w ) − 4 n + 1 � , n ≥ 2 n ! 6 w ∈ S n the variance of as( w ) for w ∈ S n Corollary. V ( n ) = 8 45 n − 13 180 , n ≥ 4 Alternating Permutations – p. 39

Variance of as( w ) � 2 � V ( n ) = 1 as( w ) − 4 n + 1 � , n ≥ 2 n ! 6 w ∈ S n the variance of as( w ) for w ∈ S n Corollary. V ( n ) = 8 45 n − 13 180 , n ≥ 4 similar results for higher moments Alternating Permutations – p. 39

A new distribution? � as( w ) − 2 n/ 3 � √ n P ( t ) = lim n →∞ Prob w ∈ S n ≤ t Alternating Permutations – p. 40

A new distribution? � as( w ) − 2 n/ 3 � √ n P ( t ) = lim n →∞ Prob w ∈ S n ≤ t Stanley distribution? Alternating Permutations – p. 40

Limiting distribution Theorem (Pemantle, Widom, (Wilf)). � as( w ) − 2 n/ 3 � √ n n →∞ Prob w ∈ S n lim ≤ t √ � t 45 / 4 1 e − s 2 ds √ π = −∞ (Gaussian distribution) Alternating Permutations – p. 41

Limiting distribution Theorem (Pemantle, Widom, (Wilf)). � as( w ) − 2 n/ 3 � √ n n →∞ Prob w ∈ S n lim ≤ t √ � t 45 / 4 1 e − s 2 ds √ π = −∞ (Gaussian distribution) Alternating Permutations – p. 41

Umbral enumeration Umbral formula: involves E k , where E is an indeterminate (the umbra ). Replace E k with the Euler number E k . (Technique from 19th century, modernized by Rota et al.) Alternating Permutations – p. 42

Umbral enumeration Umbral formula: involves E k , where E is an indeterminate (the umbra ). Replace E k with the Euler number E k . (Technique from 19th century, modernized by Rota et al.) Example. (1 + E 2 ) 3 = 1 + 3 E 2 + 3 E 4 + E 6 = 1 + 3 E 2 + 3 E 4 + E 6 = 1 + 3 · 1 + 3 · 5 + 61 = 80 Alternating Permutations – p. 42

Another example � E � � E � (1 + t ) E = 1 + Et + t 2 + t 3 + · · · 2 3 = 1 + Et + 1 2 E ( E − 1) t 2 + · · · = 1 + E 1 t + 1 2( E 2 − E 1 )) t 2 + · · · = 1 + t + 1 2(1 − 1) t 2 + · · · = 1 + t + O ( t 3 ) . Alternating Permutations – p. 43

An umbral quiz Let B be the Bell number umbra. Then Alternating Permutations – p. 44

An umbral quiz Let B be the Bell number umbra. Then (1 + t ) B = ?? Alternating Permutations – p. 44

An umbral quiz Let B be the Bell number umbra. Then (1 + t ) B = e t Alternating Permutations – p. 44

Alt. fixed-point free involutions fixed point free involution w ∈ S 2 n : all cycles of length two (number = 1 · 3 · 5 · · · (2 n − 1) ) Alternating Permutations – p. 45

Alt. fixed-point free involutions fixed point free involution w ∈ S 2 n : all cycles of length two (number = 1 · 3 · 5 · · · (2 n − 1) ) Let f ( n ) be the number of alternating fixed-point free involutions in S 2 n . Alternating Permutations – p. 45

Alt. fixed-point free involutions fixed point free involution w ∈ S 2 n : all cycles of length two (number = 1 · 3 · 5 · · · (2 n − 1) ) Let f ( n ) be the number of alternating fixed-point free involutions in S 2 n . n = 3 : 214365 = (1 , 2)(3 , 4)(5 , 6) 645231 = (1 , 6)(2 , 4)(3 , 5) f (3) = 2 Alternating Permutations – p. 45