Alternating Permutations

Transcription

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

2 Alternating Permutations p. Basic 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 >.

3 Alternating Permutations p. Euler numbers S n : symmetric group of all permutations of 1, 2,..., n

4 Alternating Permutations p. Euler numbers S n : symmetric group of all permutations of 1, 2,..., n Euler number: E n = #{w S n : w is alternating} = #{w S n : w is reverse alternating}

5 Alternating Permutations p. Euler numbers S n : symmetric group of all permutations of 1, 2,..., n Euler number: E n = #{w S n : w is alternating} = #{w S n : w is reverse alternating} E.g., E 4 = 5 : 2143, 3142, 3241, 4132, 4231

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

7 Alternating Permutations p. André s theorem Theorem (Désiré André, 1879) y:= n 0 E n x n n! = sec x + tan x E 2n is a secant number. E 2n+1 is a tangent number.

8 Alternating Permutations p. André s theorem Theorem (Désiré André, 1879) y:= n 0 E n x n n! = sec x + tan x E 2n is a secant number. E 2n+1 is a tangent number. combinatorial trigonometry

9 Alternating Permutations p. Example of combinatorial trig. sec 2 x = 1 + tan 2 x

10 Alternating Permutations p. Example of combinatorial trig. sec 2 x = 1 + tan 2 x Equate coefficients of x 2n /(2n)!: n ( ) 2n E 2k E 2(n k) 2k k=0 = n 1 k=0 ( ) 2n E 2k+1 E 2(n k) 1. 2k + 1

11 Alternating Permutations p. Example of combinatorial trig. sec 2 x = 1 + tan 2 x Equate coefficients of x 2n /(2n)!: n ( ) 2n E 2k E 2(n k) 2k k=0 = n 1 k=0 ( ) 2n E 2k+1 E 2(n k) 1. 2k + 1 Prove combinatorially (exercise).

12 Alternating Permutations p. Proof of André s theorem y := n 0 E n x n n! = sec x + tan x

13 Alternating Permutations p. Proof of André s theorem y := n 0 E n x n n! = sec x + tan x Naive proof. 2E n+1 = n k=0 ( ) n E k E n k, n 1 k (details omitted) 2y = 1 + y 2, etc.

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

15 Alternating Permutations p. Five vertices

16 Alternating Permutations p. Five vertices Slightly more complicated for E 2n

17 Alternating Permutations p. Simsun permutations (2) b 1 b 2 b k has a double descent if for some 1 < i < n, b i 1 > b i > b i+1.

18 Alternating Permutations p. Simsun permutations (2) b 1 b 2 b k has a double descent if for some 1 < i < n, b i 1 > b i > b i+1. w = a 1 a 2 a n S n is a simsun permutation if the subsequence with elements 1, 2,..., k has no double descents, 1 k n.

19 Alternating Permutations p. Simsun permutations (2) b 1 b 2 b k has a double descent if for some 1 < i < n, b i 1 > b i > b i+1. w = a 1 a 2 a n S n is a simsun permutation if the subsequence with elements 1, 2,..., k has no double descents, 1 k n. Example is not simsun: the subsequence with 1, 2, 3 is 321.

20 Alternating Permutations p. Simsun permutations (2) b 1 b 2 b k has a double descent if for some 1 < i < n, b i 1 > b i > b i+1. w = a 1 a 2 a n S n is a simsun permutation if the subsequence with elements 1, 2,..., k has no double descents, 1 k n. Example is not simsun: the subsequence with 1, 2, 3 is 321. Theorem (R. Simion & S. Sundaram) The number of simsun permutations in S n is E n+1.

21 Alternating Permutations p. 1 Orbits of mergings (3) Start with n one-element sets {1},..., {n}.

22 Alternating Permutations p. 1 Orbits of mergings (3) Start with n one-element sets {1},..., {n}. Merge together two at a time until reaching {1, 2,..., n}.

23 Alternating Permutations p. 1 Orbits of mergings (3) Start with n one-element sets {1},..., {n}. Merge together two at a time until reaching {1, 2,..., n} , , , ,

24 Alternating Permutations p. 1 Orbits of mergings (3) Start with n one-element sets {1},..., {n}. Merge together two at a time until reaching {1, 2,..., n} , , , , S n acts on these sequences.

25 Alternating Permutations p. 1 Orbits of mergings (3) Start with n one-element sets {1},..., {n}. Merge together two at a time until reaching {1, 2,..., n} , , , , S n acts on these sequences. Theorem. The number of S n -orbits is E n 1.

26 Alternating Permutations p. 1 Orbit representatives for n =

27 Alternating Permutations p. 1 Volume of a polytope (4) 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.

28 Alternating Permutations p. 1 Volume of a polytope (4) 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!.

29 Alternating Permutations p. 1 The nicest proof Triangulate E n so that the maximal simplices σ w are indexed by alternating permutations w S n.

30 Alternating Permutations p. 1 The nicest proof Triangulate E n so that the maximal simplices σ w are indexed by alternating permutations w S n. Show Vol(σ w ) = 1/n!.

31 Alternating Permutations p. 1 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

32 Alternating Permutations p. 1 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 n to E n.

33 Alternating Permutations p. 1 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 n to E n. Application (Diaconis et al.): random doubly stochastic tridiagonal matrices and random walks on T n

34 Alternating Permutations p. 1 Distribution of is(w) Yesterday: is(w) = length of longest increasing subsequence of w S n E(n) 2 n

35 Alternating Permutations p. 1 Distribution of is(w) Yesterday: is(w) = length of longest increasing subsequence of w S n E(n) 2 n For fixed t R, ( lim Prob isn (w) 2 n n n 1/6 ) t = F (t), the Tracy-Widom distribution.

36 Alternating Permutations p. 1 Analogues of distribution of is(w) Length of longest alternating subsequence of w S n

37 Alternating Permutations p. 1 Analogues of distribution of is(w) Length of longest alternating subsequence of w S n Length of longest increasing subsequence of an alternating permutation w S n.

38 Alternating Permutations p. 1 Analogues of distribution of is(w) 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!

39 Alternating Permutations p. 1 Longest alternating subsequences as(w)= length longest alt. subseq. of w D(n)= 1 as(w)? n! w S n w = as(w) = 5

40 Alternating Permutations p. 1 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}.

41 Alternating Permutations p. 2 The case n = 3 w as(w)

42 Alternating Permutations p. 2 The case n = 3 w as(w) a 1 (3) = 1, a 2 (3) = 3, a 3 (3) = 2 b 1 (3) = 1, b 2 (3) = 4, b 3 (3) = 6

43 Alternating Permutations p. 2 The main lemma Lemma. w S n alternating subsequence of maximal length that contains n.

44 Alternating Permutations p. 2 The main lemma Lemma. w S n alternating subsequence of maximal length that contains n. Corollary. a k (n) = 2r+s=k 1 n j=1 ( ) n 1 j 1 (a 2r (j 1) + a 2r+1 (j 1)) a s (n j)

45 Alternating Permutations p. 2 The main generating function Theorem. B(x, t)= B(x, t) = k,n 0 b k (n)t k xn n! 2/ρ 1 1 ρ t e ρx 1 ρ, where ρ= 1 t 2.

46 Alternating Permutations p. 2 Formulas for b k (n) Corollary. b 1 (n) = 1 b 2 (n) = n b 3 (n) = 1 4 (3n 2n + 3) b 4 (n) = 1 8 (4n (2n 4)2 n ).

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

48 Alternating Permutations p. 2 Mean (expectation) of as(w) D(n) = 1 n! w S n as(w) = 1 n! n k=1 k a k (n), the expectation of as(w) for w S n

49 Alternating Permutations p. 2 Mean (expectation) of as(w) D(n) = 1 n! w S n as(w) = 1 n! n k=1 k a k (n), the expectation of as(w) for w S n Let A(x, t) = a k (n)t k xn = (1 t)b(x, t) n! k,n 0 ( ) 2/ρ = (1 t) 1 1 ρ t e 1. ρx ρ

50 Alternating Permutations p. 2 Formula for D(n) n 0 D(n)x n = A(x, 1) t = 6x 3x2 + x 3 6(1 x) 2 = x + n 2 4n x n.

51 Alternating Permutations p. 2 Formula for D(n) n 0 D(n)x n = A(x, 1) t = 6x 3x2 + x 3 6(1 x) 2 = x + n 2 4n x n. D(n) = 4n + 1 6, n 2 (why?)

52 Alternating Permutations p. 2 Formula for D(n) n 0 D(n)x n = A(x, 1) t = 6x 3x2 + x 3 6(1 x) 2 = x + n 2 4n x n. D(n) = 4n + 1, n 2 (why?) 6 Compare E(n) 2 n.

53 Alternating Permutations p. 2 Variance of as(w) V (n)= 1 n! w S n ( as(w) 4n + 1 ) 2, n 2 6 the variance of as(n) for w S n

54 Alternating Permutations p. 2 Variance of as(w) V (n)= 1 n! w S n ( as(w) 4n + 1 ) 2, n 2 6 the variance of as(n) for w S n Corollary. V (n) = 8 45 n , n 4

55 Alternating Permutations p. 2 Variance of as(w) V (n)= 1 n! w S n ( as(w) 4n + 1 ) 2, n 2 6 the variance of as(n) for w S n Corollary. V (n) = 8 45 n , n 4 similar results for higher moments

56 Alternating Permutations p. 2 A new distribution? P (t) = lim n Prob w Sn ( as(w) 2n/3 n ) t

57 Alternating Permutations p. 2 A new distribution? P (t) = lim n Prob w Sn Stanley distribution? ( as(w) 2n/3 n ) t

58 Alternating Permutations p. 2 Limiting distribution Theorem (Pemantle, Widom, (Wilf)). ( as(w) 2n/3 lim Prob w S n n n ) t (Gaussian distribution) = 1 t 45/4 π e s2 ds

59 Alternating Permutations p. 2 Limiting distribution Theorem (Pemantle, Widom, (Wilf)). ( as(w) 2n/3 lim Prob w S n n n ) t (Gaussian distribution) = 1 t 45/4 π e s2 ds

60 Alternating Permutations p. 2 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.)

61 Alternating Permutations p. 2 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 + 3E 2 + 3E 4 + E 6 = 1 + 3E 2 + 3E 4 + E 6 = = 80

62 Alternating Permutations p. 3 Another example (1 + t) E = 1 + Et + ( ) E t ( ) E t = 1 + Et E(E 1)t2 + = 1 + E 1 t (E 2 E 1 ))t 2 + = 1 + t (1 1)t2 + = 1 + t + O(t 3 ).

63 Alternating Permutations p. 3 Alt. fixed-point free involutions fixed-point free involution w S 2n : all cycles of length two

64 Alternating Permutations p. 3 Alt. fixed-point free involutions fixed-point free involution w S 2n : all cycles of length two Let f(n) be the number of alternating fixed-point free involutions in S 2n.

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

66 Alternating Permutations p. 3 An umbral theorem Theorem. F (x) := n 0 f(n)x n

67 Alternating Permutations p. 3 An umbral theorem Theorem. F (x) := n 0 f(n)x n = ( ) (E 1 + x 2 +1)/4 1 x

68 Alternating Permutations p. 3 Proof idea Proof. Uses representation theory of the symmetric group S n.

69 Alternating Permutations p. 3 Proof idea Proof. Uses representation theory of the symmetric group S n. There is a character χ of S n (due to H. O. Foulkes) such that for all w S n, χ(w) = 0 or ± E k.

70 Alternating Permutations p. 3 Proof idea Proof. Uses representation theory of the symmetric group S n. There is a character χ of S n (due to H. O. Foulkes) such that for all w S n, χ(w) = 0 or ± E k. Now use known results on combinatorial properties of characters of S n.

71 Alternating Permutations p. 3 Ramanujan s Lost Notebook Theorem (Ramanujan, Berndt, implicitly) As x 0+, 2 n 0 ( ) n(n+1) 1 x 1 + x k 0 f(k)x k = F (x), an analytic (non-formal) identity.

72 Alternating Permutations p. 3 A formal identity Corollary (via Ramanujan, Andrews). F (x) = 2 n 0 n q n j=1 (1 q2j 1 ) 2n+1 j=1 (1 + qj ), where q = ( 1 x 1+x) 2/3, a formal identity.

73 Alternating Permutations p. 3 Simple result, hard proof Recall: number of n-cycles in S n is (n 1)!.

74 Alternating Permutations p. 3 Simple result, hard proof Recall: number of n-cycles in S n is (n 1)!. Theorem. Let b(n) be the number of alternating n-cycles in S n. Then if n is odd, b(n) = 1 n d n E n /n. µ(d)( 1) (d 1)/2 E n/d

75 Alternating Permutations p. 3 Special case Corollary. Let p be an odd prime. Then b(p) = 1 ) (E p ( 1) (p 1)/2. p

76 Alternating Permutations p. 3 Special case Corollary. Let p be an odd prime. Then b(p) = 1 ) (E p ( 1) (p 1)/2. p Combinatorial proof?

77 Alternating Permutations p. 3 Inc. subsequences of alt. perms. Recall: is(w) = length of longest increasing subsequence of w S n. Define C(n) = 1 E n w is(w), where w ranges over all E n alternating permutations in S n.

78 Alternating Permutations p. 3 β Little is known, e.g., what is I.e., C(n) = n β+o(1). β = lim n log C(n) log n? Compare lim n log E(n) log n = 1/2.

79 Alternating Permutations p. 3 β Little is known, e.g., what is I.e., C(n) = n β+o(1). β = lim n log C(n) log n? Compare lim n log E(n) log n = 1/2. Easy: β 1 2.

80 Alternating Permutations p. 4 Limiting distribution? What is the (suitably scaled) limiting distribution of is(w), where w ranges over all alternating permutations in S n?

81 Alternating Permutations p. 4 Limiting distribution? What is the (suitably scaled) limiting distribution of is(w), where w ranges over all alternating permutations in S n? Is it the Tracy-Widom distribution?

82 Alternating Permutations p. 4 Limiting distribution? What is the (suitably scaled) limiting distribution of is(w), where w ranges over all alternating permutations in S n? Is it the Tracy-Widom distribution? Possible tool: umbral analogue of Gessel s determinantal formula.

83 Alternating Permutations p. 4 Descent sets Let w = a 1 a 2 a n S n. Descent set of w: D(w) = {i : a i > a i+1 } {1,..., n 1}

84 Alternating Permutations p. 4 Descent sets Let w = a 1 a 2 a n S n. Descent set of w: D(w) = {i : a i > a i+1 } {1,..., n 1} D( ) = {1, 4, 5} D( ) = {1, 3, 5} (alternating) D( ) = {2, 4, 6} (reverse alternating)

85 Alternating Permutations p. 4 β n (S) β n (S) = #{w S n : D(w) = S}

86 Alternating Permutations p. 4 β n (S) β n (S) = #{w S n : D(w) = S} w D(w) {1} 312 {1} 132 {2} 231 {2} 321 {1, 2} β 3 ( ) = 1, β 3 (1) = 2 β 3 (2) = 2, β 3 (1, 2) = 1

87 Alternating Permutations p. 4 u S Fix n. Let S {1,, n 1}. Let u S = t 1 t n 1, where { a, i S t i = b, i S.

88 Alternating Permutations p. 4 u S Fix n. Let S {1,, n 1}. Let u S = t 1 t n 1, where { a, i S t i = b, i S. Example. n = 8, S = {2, 5, 6} {1,..., 7} u S = abaabba

89 Alternating Permutations p. 4 A noncommutative gen. function Ψ n (a, b) = S {1,...,n 1} β n (S)u S.

90 Alternating Permutations p. 4 A noncommutative gen. function Ψ n (a, b) = Example. Recall S {1,...,n 1} β n (S)u S. β 3 ( ) = 1, β 3 (1) = 2, β 3 (2) = 2, β 3 (1, 2) = 1 Thus Ψ 3 (a, b) = aa + 2ab + 2ba + bb

91 Alternating Permutations p. 4 A noncommutative gen. function Ψ n (a, b) = Example. Recall S {1,...,n 1} β n (S)u S. β 3 ( ) = 1, β 3 (1) = 2, β 3 (2) = 2, β 3 (1, 2) = 1 Thus Ψ 3 (a, b) = aa + 2ab + 2ba + bb = (a + b) 2 + (ab + ba)

92 Alternating Permutations p. 4 The cd-index Theorem. There exists a noncommutative polynomial Φ n (c, d), called the cd-index of S n, with nonnegative integer coefficients, such that Ψ n (a, b) = Φ n (a + b, ab + ba).

93 Alternating Permutations p. 4 The cd-index Theorem. There exists a noncommutative polynomial Φ n (c, d), called the cd-index of S n, with nonnegative integer coefficients, such that Example. Recall Ψ n (a, b) = Φ n (a + b, ab + ba). Ψ 3 (a, b) = aa + 2ab + 2ba + b 2 = (a + b) 2 + (ab + ba). Therefore Φ 3 (c, d) = c 2 + d.

94 Alternating Permutations p. 4 Small values of Φ n (c, d) Φ 1 = 1 Φ 2 = c Φ 3 = c 2 + d Φ 4 = c 3 + 2cd + 2dc Φ 5 = c 4 + 3c 2 d + 5cdc + 3dc 2 + 4d 2 Φ 6 = c 5 + 4c 3 d + 9c 2 dc + 9cdc 2 + 4dc 3 +12cd dcd + 12d 2 c.

95 Alternating Permutations p. 4 S µ Let deg c = 1, deg d = 2.

96 Alternating Permutations p. 4 S µ Let deg c = 1, deg d = 2. µ: cd-monomial of degree n 1

97 Alternating Permutations p. 4 S µ Let deg c = 1, deg d = 2. µ: cd-monomial of degree n 1 Replace each c in µ with 0, each d with 10, and remove final 0. Get the characteristic vector of a set S µ [n 2].

98 Alternating Permutations p. 4 S µ Let deg c = 1, deg d = 2. µ: cd-monomial of degree n 1 Replace each c in µ with 0, each d with 10, and remove final 0. Get the characteristic vector of a set S µ [n 2]. Example. n = 10: µ = cd 2 c 2 d = , the characteristic vector of S µ = {2, 4, 8} [8]

99 Alternating Permutations p. 4 cd-index coefficients Recall: w = a 1 a 2 a n S n is a simsun permutation if the subsequence with elements 1, 2,..., k has no double descents, 1 k n.

100 Alternating Permutations p. 4 cd-index coefficients Recall: w = a 1 a 2 a n S n is a simsun permutation if the subsequence with elements 1, 2,..., k has no double descents, 1 k n. Theorem (Simion-Sundaram, variant of Foata-Schützenberger) The coefficient of µ in Φ(c, d) is equal to the number of simsun permutations in S n 1 with descent set S µ.

101 Alternating Permutations p. 4 An example Example. Φ 6 = c 5 +4c 3 d+9c 2 dc+9cdc 2 +4dc 3 +12cd 2 +10dcd+12d 2 c, dcd S dcd = {1, 4}

102 Alternating Permutations p. 4 An example Example. Φ 6 = c 5 +4c 3 d+9c 2 dc+9cdc 2 +4dc 3 +12cd 2 +10dcd+12d 2 c, dcd S dcd = {1, 4} The ten simsun permutations w S 5 with D(w) = {1, 4}: 21354, 21453, 31254, 31452, , 42351, 51243, 51342, 52341,

103 Alternating Permutations p. 4 An example Example. Φ 6 = c 5 +4c 3 d+9c 2 dc+9cdc 2 +4dc 3 +12cd 2 +10dcd+12d 2 c, dcd S dcd = {1, 4} The ten simsun permutations w S 5 with D(w) = {1, 4}: but not , 21453, 31254, 31452, , 42351, 51243, 51342, 52341,

104 Alternating Permutations p. 5 Two consequences Theorem. (a) Φ n (1, 1) = E n (the number of simsum permutations w S n ).

105 Alternating Permutations p. 5 Two consequences Theorem. (a) Φ n (1, 1) = E n (the number of simsum permutations w S n ). (b) (Niven, de Bruijn) For all S {1,..., n 1}, β n (S) E n, with equality if and only if S = {1, 3, 5,... } or S = {2, 4, 6... }

106 Alternating Permutations p. 5 An example Φ 5 = 1c 4 + 3c 2 d + 5cdc + 3dc 2 + 4d = 16 = E 5

107 Alternating Permutations p. 5