alternating permutations - mit mathematics rstan/transparencies/altperm-ams.pdf · simsun...

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  • Alternating PermutationsRichard P. Stanley

    M.I.T.

    Alternating Permutations p. 1

  • Basic definitions

    A sequence a1, a2, . . . , ak of distinct integers isalternating if

    a1 > a2 < a3 > a4 < ,and reverse alternating if

    a1 < a2 > a3 < a4 > .

    Alternating Permutations p. 2

  • Euler numbers

    Sn : symmetric group of all permutations of 1, 2, . . . , n

    Euler number:

    En = #{w Sn : w is alternating}= #{w Sn : w is reverse alternating}

    E.g., E4 = 5 : 2143, 3142, 3241, 4132, 4231

    Alternating Permutations p. 3

  • Euler numbers

    Sn : symmetric group of all permutations of 1, 2, . . . , n

    Euler number:

    En = #{w Sn : w is alternating}= #{w Sn : w is reverse alternating}

    E.g., E4 = 5 : 2143, 3142, 3241, 4132, 4231

    Alternating Permutations p. 3

  • Euler numbers

    Sn : symmetric group of all permutations of 1, 2, . . . , n

    Euler number:

    En = #{w Sn : w is alternating}= #{w Sn : w is reverse alternating}

    E.g., E4 = 5 : 2143, 3142, 3241, 4132, 4231

    Alternating Permutations p. 3

  • Andrs theorem

    Theorem (Dsir Andr, 1879)

    y:=

    n0En

    xn

    n!= sec x + tan x

    E2n is a secant number.

    E2n+1 is a tangent number.

    combinatorial trigonometry

    Alternating Permutations p. 4

  • Andrs theorem

    Theorem (Dsir Andr, 1879)

    y:=

    n0En

    xn

    n!= sec x + tan x

    E2n is a secant number.

    E2n+1 is a tangent number.

    combinatorial trigonometry

    Alternating Permutations p. 4

  • Andrs theorem

    Theorem (Dsir Andr, 1879)

    y:=

    n0En

    xn

    n!= sec x + tan x

    E2n is a secant number.

    E2n+1 is a tangent number.

    combinatorial trigonometry

    Alternating Permutations p. 4

  • Example of combinatorial trig.

    sec2 x = 1 + tan2 x

    Equate coefficients of x2n/(2n)!:

    n

    k=0

    (

    2n

    2k

    )

    E2kE2(nk)

    =n1

    k=0

    (

    2n

    2k + 1

    )

    E2k+1E2(nk)1.

    Prove combinatorially (exercise).

    Alternating Permutations p. 5

  • Example of combinatorial trig.

    sec2 x = 1 + tan2 x

    Equate coefficients of x2n/(2n)!:

    n

    k=0

    (

    2n

    2k

    )

    E2kE2(nk)

    =n1

    k=0

    (

    2n

    2k + 1

    )

    E2k+1E2(nk)1.

    Prove combinatorially (exercise).

    Alternating Permutations p. 5

  • Example of combinatorial trig.

    sec2 x = 1 + tan2 x

    Equate coefficients of x2n/(2n)!:

    n

    k=0

    (

    2n

    2k

    )

    E2kE2(nk)

    =n1

    k=0

    (

    2n

    2k + 1

    )

    E2k+1E2(nk)1.

    Prove combinatorially (exercise).

    Alternating Permutations p. 5

  • Proof of Andrs theorem

    y :=

    n0En

    xn

    n!= sec x + tan x

    Naive proof.

    2En+1 =n

    k=0

    (

    n

    k

    )

    EkEnk, n 1

    2y = 1 + y2, etc.(details omitted)

    Alternating Permutations p. 6

  • Proof of Andrs theorem

    y :=

    n0En

    xn

    n!= sec x + tan x

    Naive proof.

    2En+1 =n

    k=0

    (

    n

    k

    )

    EkEnk, n 1

    2y = 1 + y2, etc.(details omitted)

    Alternating Permutations p. 6

  • Some occurences of Euler numbers

    (1) E2n1 is the number of complete increasingbinary trees on the vertex set[2n + 1] = {1, 2, . . . , 2n + 1}.

    Alternating Permutations p. 7

  • Five vertices

    1 1

    3 34554

    2 2

    1 1

    2 24554

    3 3

    1

    5

    1

    34435

    11

    354

    534

    2 2 2 2

    1111

    33224 5 5 4 4 5 5 4

    2233

    1

    5

    1

    3 4 4 35

    1 1

    3 54

    5 34

    2222

    Slightly more complicated for E2n

    Alternating Permutations p. 8

  • Five vertices

    1 1

    3 34554

    2 2

    1 1

    2 24554

    3 3

    1

    5

    1

    34435

    11

    354

    534

    2 2 2 2

    1111

    33224 5 5 4 4 5 5 4

    2233

    1

    5

    1

    3 4 4 35

    1 1

    3 54

    5 34

    2222

    Slightly more complicated for E2n

    Alternating Permutations p. 8

  • Simsun permutations

    (2) b1b2 bk has a double descent if for some1 < i < n,

    bi1 > bi > bi+1.

    w = a1a2 an Sn is a simsun permutation ifthe subsequence with elements 1, 2, . . . , k has nodouble descents, 1 k n.Example. 3241 is not simsun: the subsequencewith 1, 2, 3 is 321.

    Theorem (R. Simion & S. Sundaram) Thenumber of simsun permutations in Sn is En+1.

    Alternating Permutations p. 9

  • Simsun permutations

    (2) b1b2 bk has a double descent if for some1 < i < n,

    bi1 > bi > bi+1.

    w = a1a2 an Sn is a simsun permutation ifthe subsequence with elements 1, 2, . . . , k has nodouble descents, 1 k n.

    Example. 3241 is not simsun: the subsequencewith 1, 2, 3 is 321.

    Theorem (R. Simion & S. Sundaram) Thenumber of simsun permutations in Sn is En+1.

    Alternating Permutations p. 9

  • Simsun permutations

    (2) b1b2 bk has a double descent if for some1 < i < n,

    bi1 > bi > bi+1.

    w = a1a2 an Sn is a simsun permutation ifthe subsequence with elements 1, 2, . . . , k has nodouble descents, 1 k n.Example. 3241 is not simsun: the subsequencewith 1, 2, 3 is 321.

    Theorem (R. Simion & S. Sundaram) Thenumber of simsun permutations in Sn is En+1.

    Alternating Permutations p. 9

  • Simsun permutations

    (2) b1b2 bk has a double descent if for some1 < i < n,

    bi1 > bi > bi+1.

    w = a1a2 an Sn is a simsun permutation ifthe subsequence with elements 1, 2, . . . , k has nodouble descents, 1 k n.Example. 3241 is not simsun: the subsequencewith 1, 2, 3 is 321.

    Theorem (R. Simion & S. Sundaram) Thenumber of simsun permutations in Sn is En+1.

    Alternating Permutations p. 9

  • Orbits of mergings

    (3) Start with n one-element sets {1}, . . . , {n}.

    Merge together two at a time until reaching{1, 2, . . . , n}.

    123456, 123456, 123456125346, 125346, 123456

    Sn acts on these sequences.

    Theorem. The number of Sn-orbits is En1.

    Alternating Permutations p. 10

  • Orbits of mergings

    (3) Start with n one-element sets {1}, . . . , {n}.Merge together two at a time until reaching{1, 2, . . . , n}.

    123456, 123456, 123456125346, 125346, 123456

    Sn acts on these sequences.

    Theorem. The number of Sn-orbits is En1.

    Alternating Permutations p. 10

  • Orbits of mergings

    (3) Start with n one-element sets {1}, . . . , {n}.Merge together two at a time until reaching{1, 2, . . . , n}.

    123456, 123456, 123456125346, 125346, 123456

    Sn acts on these sequences.

    Theorem. The number of Sn-orbits is En1.

    Alternating Permutations p. 10

  • Orbits of mergings

    (3) Start with n one-element sets {1}, . . . , {n}.Merge together two at a time until reaching{1, 2, . . . , n}.

    123456, 123456, 123456125346, 125346, 123456

    Sn acts on these sequences.

    Theorem. The number of Sn-orbits is En1.

    Alternating Permutations p. 10

  • Orbits of mergings

    (3) Start with n one-element sets {1}, . . . , {n}.Merge together two at a time until reaching{1, 2, . . . , n}.

    123456, 123456, 123456125346, 125346, 123456

    Sn acts on these sequences.

    Theorem. The number of Sn-orbits is En1.

    Alternating Permutations p. 10

  • Orbit representatives for n = 5

    12345 12345 1234512345 12345 1234512345 12345 1253412345 12345 1234512345 12345 12345

    Alternating Permutations p. 11

  • Volume of a polytope

    (4) Let En be the convex polytope in Rn definedby

    xi 0, 1 i nxi + xi+1 1, 1 i n 1.

    Theorem. The volume of En is En/n!.

    Alternating Permutations p. 12

  • Volume of a polytope

    (4) Let En be the convex polytope in Rn definedby

    xi 0, 1 i nxi + xi+1 1, 1 i n 1.

    Theorem. The volume of En is En/n!.

    Alternating Permutations p. 12

  • The nicest proof

    Triangulate En so that the maximal simplicesw are indexed by alternating permutationsw Sn.

    Show Vol(w) = 1/n!.

    Alternating Permutations p. 13

  • The nicest proof

    Triangulate En so that the maximal simplicesw are indexed by alternating permutationsw Sn.

    Show Vol(w) = 1/n!.

    Alternating Permutations p. 13

  • Tridiagonal matrices

    An n n matrix M = (mij) is tridiagonal ifmij = 0 whenever |i j| 2.

    doubly-stochastic: mij 0, row and columnsums equal 1

    Tn: set of n n tridiagonal doubly stochasticmatrices

    Alternating Permutations p. 14

  • Polytope structure of Tn

    Easy fact: the map

    Tn Rn1M 7 (m12,m23, . . . ,mn1,n)

    is a (linear) bijection from Tn to En.

    Application (Diaconis et al.): random doublystochastic tridiagonal matrices and random walkson Tn

    Alternating Permutations p. 15

  • Polytope structure of Tn

    Easy fact: the map

    Tn Rn1M 7 (m12,m23, . . . ,mn1,n)

    is a (linear) bijection from Tn to En.

    Application (Diaconis et al.): random doublystochastic tridiagonal matrices and random walkson Tn

    Alternating Permutations p. 15

  • Distribution of is(w)

    Yesterday: is(w) = length of longest increasingsubsequence of w Sn

    E(n) 2n

    For fixed t R,

    limn

    Prob

    (

    isn(w) 2

    n

    n1/6 t)

    = F (t),

    the Trac