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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