Shuffling properties for products of random permutations Olivier - - PowerPoint PPT Presentation

Shuffling properties for products of random permutations

Halifax, June 2012 Olivier Bernardi (MIT)

Joint work with Rosena Du, Alejandro Morales, Richard Stanley

3 5 10 9 4 1 2 7 6 11 8

Warm up Question: Let π be a uniformly random permutation of {1, 2, . . . , n}. What is the probability that 1, 2, . . . , k are in distinct cycles of π?

Warm up Proof: Sampling process for π: build the cycles. Question: Let π be a uniformly random permutation of {1, 2, . . . , n}. What is the probability that 1, 2, . . . , k are in distinct cycles of π? Answer: 1 k!. 1 1 3 1 4 1 2 2 3 2

Warm up Let α = (α1, α2, . . . , αk) be a tuple of positive integers. Definition: A permutation π of {1, 2, . . . , n} is said to be α-separated if letters from different blocks B1 = {1, 2, . . . , α1}, B2 = {α1 + 1, α1 + 2, . . . , α1 + α2}, . . . Bk = {α1 + · · · + αk−1 + 1, . . . , α1 + · · · + αk}, are in different cycles of π.

Warm up Example: α = (2, 2, 1). Blocks: {1, 2} {3, 4} {5}. 3 5 10 9 4 1 2 7 6 Let α = (α1, α2, . . . , αk) be a tuple of positive integers. Definition: A permutation π of {1, 2, . . . , n} is said to be α-separated if letters from different blocks B1 = {1, 2, . . . , α1}, B2 = {α1 + 1, α1 + 2, . . . , α1 + α2}, . . . Bk = {α1 + · · · + αk−1 + 1, . . . , α1 + · · · + αk}, are in different cycles of π. 11 8

Warm up Answer: α1!α2! . . . αk! (α1 + α2 + . . . + αk)!. Let α = (α1, α2, . . . , αk) be a tuple of positive integers. Question: Let π be a uniformly random permutation of {1, 2, . . . , n}. What is the probability that π is α-separated?

Warm up Answer: α1!α2! . . . αk! (α1 + α2 + . . . + αk)!. Proof: Sampling process for π: build the cycles. 1 Let α = (α1, α2, . . . , αk) be a tuple of positive integers. Question: Let π be a uniformly random permutation of {1, 2, . . . , n}. What is the probability that π is α-separated? 1 3 1 4 1 2 2 3 2

Results

Type of question Let λ, λ′ be partitions of n. Let π, π′ be uniformly random permutations of cycle-type λ, λ′. Let α be a composition of m ≤ n. What is the probability that the product π ◦ π′ is α-separated?

Results Theorem [Du, Stanley]. Let π, π′ be uniformly random n-cycles. The probability that 1, 2, . . . , k are in different cycles of product ππ′ is 1 k! if n − k odd, 1 k! + 2 (k − 2)!(n − k + 1)(n + k)

• therwise.

Results Theorem [Du, Stanley]. Let π, π′ be uniformly random n-cycles. The probability that 1, 2, . . . , k are in different cycles of product ππ′ is 1 k! if n − k odd, 1 k! + 2 (k − 2)!(n − k + 1)(n + k)

• therwise.

+ Extension to product (n−j)-cycle × n-cycle for k = 2. Case k = 2 was conjectured by B´

• na.

More results: general composition α Theorem [BMDS]. Let π, π′ be uniformly random n-cycles. Let α = (α1, α2, . . . , αk) be a composition of size m. The probability that the product ππ′ is α-separated is (n − m)! k

i=1 αi!

(n + k)(n − 1)!

• (−1)n−mn−1

k−2

• n+m

m−k

• +

m−k

• r=0

(−1)rm−k

r

n+r+1

m

• n+k+r

r

• .

More results: general composition α Theorem [BMDS]. Let π, π′ be uniformly random n-cycles. Let α = (α1, α2, . . . , αk) be a composition of size m. The probability that the product ππ′ is α-separated is (n − m)! k

i=1 αi!

(n + k)(n − 1)!

• (−1)n−mn−1

k−2

• n+m

m−k

• +

m−k

• r=0

(−1)rm−k

r

n+r+1

m

• n+k+r

r

• .

+ Extension to π = uniformly random (n−j)-cycle.

More results: general composition α more cycles Theorem [BMDS]. Let π be a uniformly random permutation having p cycles. Let π′ be a uniformly random n-cycle. Let α = (α1, α2, . . . , αk) be a composition of size m. The probability that the product ππ′ is α-separated is (n − m)! k

i=1 αi!

c(n, p)

n−m

• r=0

1 − k r n + k − 1 n − m − r c(n − k − r + 1, p) (n − k − r + 1)! , where c(n, p) = [xp]x(x + 1) · · · (x + n − 1) “signless Stirling numbers of the first kind”.

More results: general composition α more cycles involutions Theorem [BMDS]. Let π be a uniformly random fixed-point free involution. Let π′ be a uniformly random 2N-cycle. Let α = (α1, α2, . . . , αk) be a composition of size m. The probability that the product ππ′ is α-separated is k

i=1 αi!

(2N − 1)!(2N − 1)!! ×

min(2N−m,N−k+1)

• r=0

1 − k r 2N + k − 1 2N − m − r 2k+r−N−1(2N − k − r)! (N − k − r + 1)! .

More results: general composition α more cycles involutions symmetry Theorem [BMDS]. Let λ be a partition. Let π be a uniformly random permutation of type λ. Let π′ be a uniformly random n-cycle. Let α = (α1, α2, . . . , αk), β = (β1, β2, . . . , βk) be compositions of size m and length k. The probabilities σα

λ and σβ λ that the product ππ′ is α-separated and

β-separated are related by σα

λ

k

i=1 αi!

= σβ

λ

k

i=1 βi!

.

Strategy

Set up: Let α = (α1, . . . , αk) be a composition of m ≤ n.

• Notation. σα

λ=proba that π ◦ π′ is α-separated.

random permutation of type λ random n-cycle

Set up: Let α = (α1, . . . , αk) be a composition of m ≤ n.

• Def. For tuple A = (A1, . . . , Ak) of disjoint subsets of {1, 2, . . . , n},

we say that a permutation π is A-separated if elements in different blocks of A are in distinct cycles of π.

• Notation. σα

λ=proba that π ◦ π′ is α-separated.

3 5 10 9 4 1 2 7 6 11 8 Example: π is ({1, 3, 6}, {2, 10})-separated. random permutation of type λ random n-cycle

Set up: Let α = (α1, . . . , αk) be a composition of m ≤ n.

• Notation. σα

λ=proba that π ◦ π′ is α-separated.

• Remark. σα

λ= proba that π ◦ (1, 2, . . . , n) is A-separated.

random permutation of type λ random n-cycle random subsets (A1, . . . , Ak) with #Ai = αi random permutation of type λ

Set up: Let α = (α1, . . . , αk) be a composition of m ≤ n.

• Notation. σα

λ=proba that π ◦ π′ is α-separated.

random permutation of type λ random n-cycle random subsets (A1, . . . , Ak) with #Ai = αi

• Remark. σα

λ= proba that π−1 ◦ (1, 2, . . . , n) is A-separated.

random permutation of type λ

Set up: Let α = (α1, . . . , αk) be a composition of m ≤ n. Lemma: σα

λ =

#Sα

λ

• n

α1,α2,...,αk,n−m

• #Cλ

, where Sα

λ is set of triples (A, π, ω) such that

• A = (A1, . . . , Ak) is a tuple of disjoint subsets of [n], with #Ai = αi
• permutation π has type λ,
• permutation ω is A-separated,
• π ◦ ω = (1, 2, . . . , n).
• Notation. σα

λ=proba that π ◦ π′ is α-separated.

random permutation of type λ random n-cycle random subsets (A1, . . . , Ak) with #Ai = αi

• Remark. σα

λ= proba that π−1 ◦ (1, 2, . . . , n) is A-separated.

random permutation of type λ

Set up: We want to count set Sα

λ of triples (A, π, ω) such that

• A = (A1, . . . , Ak) is a tuple of disjoint subsets of [n], with #Ai = αi
• permutation π has type λ,
• permutation ω is A-separated,
• π ◦ ω = (1, 2, . . . , n).
• Example. Let n = 8, α = (3, 2), λ = (3, 3, 2).

Triple (A, π, ω) is in Sα

λ where

A = ({1, 4, 8}, {3, 5}), π = (1, 7, 4)(3, 2, 6)(5, 8), ω = (1, 6)(2)(3, 7, 5)(4, 8).

A formula for colored factorization of long cycle

• Def. A colored permutation is a permutation with cycles colored

in Z>0. 3 5 10 9 4 1 2 7 6 11 8 Example: Colors 1, 2, 3. It has color-type γ = (γ1, . . . , γk) if there are exactly γi elements of color i. Color-type: γ = (4, 3, 4).

A formula for colored factorization of long cycle

• Def. A colored permutation is a permutation with cycles colored

in Z>0. It has color-type γ = (γ1, . . . , γk) if there are exactly γi elements of color i. Thm [Schaeffer Vassilieva 08, Vassilieva Morales 09]. Let γ, γ′ be compositions of size n of length k, k′ The number of colored permutations π, π′ of color-types γ, γ′ such that ππ′ = (1, 2, . . . , n) is Bγ,γ′ = n(n − k)!(n − k′)! (n − k − k′ + 1)! .

Navigating between cycle-type and color-type

• Def. Symmetric functions in x = x1, x2, x3 . . .

Bases indexed by partitions λ = (λ1, . . . , λℓ):

• Power basis:

pλ(x) = ℓ

i=1 pλi(x) where pk(x) = i≥1 xk i .

• Monomial basis: mλ(x) =

γ1,γ2,...∼λ

• i≥1 xγi

i .

Navigating between cycle-type and color-type

• Def. Symmetric functions in x = x1, x2, x3 . . .

Bases indexed by partitions λ = (λ1, . . . , λℓ):

• Power basis:

pλ(x) = ℓ

i=1 pλi(x) where pk(x) = i≥1 xk i .

• Monomial basis: mλ(x) =

γ1,γ2,...∼λ

• i≥1 xγi

i .

Rk. If π is a permutation of type λ, then pλ(x) is the generating function of the colorings of π. 3 5 8 9 4 1 2 7 6 p4(x) p2(x) p2(x)p1(x)

pλ(x) =

• coloring of π
• i≥0

x#colored i

i

=

• µ⊢n

mµ(x) #colorings of π of color-type µ

Strategy Write the generating function of the numbers #Sα

λ in one

symmetric function basis and compute the coefficients in another:

• λ⊢n

pλ(x) #Sα

λ =

• µ⊢n

mµ(x) computableµ.

Generating function We want to count set Sα

λ of triples (A, π, ω) such that

• A = (A1, . . . , Ak) is a tuple of disjoint subsets of [n], with #Ai = αi
• permutation π has type λ,
• permutation ω is A-separated,
• π ◦ ω = (1, 2, . . . , n).

Generating function We want to count set Sα

λ of triples (A, π, ω) such that

• A = (A1, . . . , Ak) is a tuple of disjoint subsets of [n], with #Ai = αi
• permutation π has type λ,
• permutation ω is A-separated,
• π ◦ ω = (1, 2, . . . , n).

Generating function: Gα

n(x, t) :=

• λ⊢n

pλ(x)

• (A,π,ω)∈Sα

λ

tc(ω,A), where c(ω, A) = number of cycles of ω containing no element in A. Hence #Sα

λ = [pλ(x)]Gα n(x, 1).

Generating function We want to count set Sα

λ of triples (A, π, ω) such that

• A = (A1, . . . , Ak) is a tuple of disjoint subsets of [n], with #Ai = αi
• permutation π has type λ,
• permutation ω is A-separated,
• π ◦ ω = (1, 2, . . . , n).

Generating function: Gα

n(x, t) :=

• λ⊢n

pλ(x)

• (A,π,ω)∈Sα

λ

tc(ω,A), where c(ω, A) = number of cycles of ω containing no element in A.

• Example. n = 8, α = (3, 2).

A = ({1, 4, 8}, {3, 5}), π = (1, 7, 4)(3, 2, 6)(5, 8), ω = (1, 6)(2)(3, 7, 5)(4, 8). Triple (A, π, ω) contributes p3(x)2p2(x) t1. Hence #Sα

λ = [pλ(x)]Gα n(x, 1).

Generating function We want to count set Sα

λ of triples (A, π, ω) such that

• A = (A1, . . . , Ak) is a tuple of disjoint subsets of [n], with #Ai = αi
• permutation π has type λ,
• permutation ω is A-separated,
• π ◦ ω = (1, 2, . . . , n).

where T α

λ,r is set of triples (A, π, ω) such that

• A = (A1, . . . , Ak) is a tuple of disjoint subsets of [n], with #Ai = αi
• colored permutation π has color-type λ,
• colored permutation ω uses k + r colors, and elements in Ai have

color i,

• π ◦ ω = (1, 2, . . . , n).

Proposition. Gα

n(x, t + k) =

• λ⊢n

pλ(x)

• (A,π,ω)∈Sα

λ

(t + k)c(ω,A) =

• λ⊢n,r≥0

#T α

λ,r mλ(x)

t r

• .

1 2 3 4 5 6

× =

3 4 6 1 5 2 1 4 6 3 2 5 pλ(x) t type λ A-separated 1 2 3 4 5 6

× =

3 4 6 1 5 2 1 4 6 3 2 5 coloring respecting A Sα

λ :“Separated factorizations of (1, .., n)”

color-type γ mµ(x) t

• T α

γ,r:“Separated colored factorizations of (1, .., n)”

Counting separated colored factorization Let α = (α1, . . . , αk) be a partition of size m.

• Prop. For any partition λ of size n and length ℓ,

#T α

λ,r = n(n − ℓ)!(n − k − r)!

(n − ℓ − k − r + 1)! n + k − 1 n − m − r

• .

T α

λ,r is set of triples (A, π, ω) such that

• A = (A1, . . . , Ak) is a tuple of disjoint subsets of [n], with #Ai = αi
• colored permutation π has color-type λ,
• colored permutation ω uses k + r colors, and elements in Ai have

color i,

• π ◦ ω = (1, 2, . . . , n).

Counting separated colored factorization 1 2 3 4 5 6

× =

3 4 6 1 5 2 1 4 6 3 2 5 Let α = (α1, . . . , αk) be a partition of size m.

• Prop. For any partition λ of size n and length ℓ,

#T α

λ,r = n(n − ℓ)!(n − k − r)!

(n − ℓ − k − r + 1)! n + k − 1 n − m − r

• .

Proof. First choose colored factorization of color-type λ, λ′ with λ′ of length k + r: Bλ,λ′ = n(n − ℓ)!(n − k − r)! (n − ℓ − k − r + 1)! .

Counting separated colored factorization 1 2 3 4 5 6

× =

3 4 6 1 5 2 1 4 6 3 2 5 Let α = (α1, . . . , αk) be a partition of size m.

• Prop. For any partition λ of size n and length ℓ,

#T α

λ,r = n(n − ℓ)!(n − k − r)!

(n − ℓ − k − r + 1)! n + k − 1 n − m − r

• .

Proof. First choose colored factorization of color-type λ, λ′ with λ′ of length k + r: Bλ,λ′ = n(n − ℓ)!(n − k − r)! (n − ℓ − k − r + 1)! . Choose the elements of Ai among those colored i: k

i=1

λi

αi

• .

Summation simplifies because Bλ,λ′ is independent of λ′.

Summary Let α = (α1, . . . , αk) be a partition of size m. Let λ be a partition of n ≥ m. Let π be a random permutation of type λ. Let π′ be a random n-cycle. Probability of α-separation for π ◦ π′ is given by σα

λ =

#Sα

λ

• n

α1,α2,...,αk,n−m

• #Cλ

, where #Sα

λ = [pλ(x)]Gα n(x, 1)

and Gα

n(x, t) =

• λ⊢n,r≥0

mλ(x)n(n − ℓλ)!(n − k − r)! (n − k − ℓλ − r + 1)! n + k − 1 n − m − r t − k r

• .

Harvesting the generating function

Symmetry For α = (α1, . . . , αk) partition of size m, σα

λ =

#Sα

λ

• n

α1,α2,...,αk,n−m

• #Cλ

, where #Sα

λ = [pλ(x)]Gα n(x, 1)

and Gα

n(x, t) =

• λ⊢n,r≥0

mλ(x)n(n − ℓλ)!(n − k − r)! (n − k − ℓλ − r + 1)! n + k − 1 n − m − r t − k r

• .

Symmetry For α = (α1, . . . , αk) partition of size m, σα

λ =

#Sα

λ

• n

α1,α2,...,αk,n−m

• #Cλ

, where #Sα

λ = [pλ(x)]Gα n(x, 1)

and Gα

n(x, t) =

• λ⊢n,r≥0

mλ(x)n(n − ℓλ)!(n − k − r)! (n − k − ℓλ − r + 1)! n + k − 1 n − m − r t − k r

• .
• Cor. If α and β are compositions of same size and same length then

Gα

n(x, t) = Gβ n(x, t).

In particular, for all λ, #Sα

λ = #Sβ λ and

σα

λ

k

i=1 αi!

= σβ

λ

k

i=1 βi!

.

Symmetry For α = (α1, . . . , αk) partition of size m, σα

λ =

#Sα

λ

• n

α1,α2,...,αk,n−m

• #Cλ

, where #Sα

λ = [pλ(x)]Gα n(x, 1)

and Gα

n(x, t) =

• λ⊢n,r≥0

mλ(x)n(n − ℓλ)!(n − k − r)! (n − k − ℓλ − r + 1)! n + k − 1 n − m − r t − k r

• .
• Remark. This result comes from“symmetry” in formula for T α

λ,r,

which has a bijective proof [B., Morales 12].

• Cor. If α and β are compositions of same size and same length then

Gα

n(x, t) = Gβ n(x, t).

In particular, for all λ, #Sα

λ = #Sβ λ and

σα

λ

k

i=1 αi!

= σβ

λ

k

i=1 βi!

.

Involution × n-cycle Theorem [BMDS]. The probability that the product π ◦ π′ is α-separated is k

i=1 αi!

(2N − 1)!(2N − 1)!! ×

min(2N−m,N−k+1)

• r=0

1 − k r 2N + k − 1 2N − m − r 2k+r−N−1(2N − k − r)! (N − k − r + 1)! . 2N-cycle involution size m length k

Involution × n-cycle Theorem [BMDS]. The probability that the product π ◦ π′ is α-separated is k

i=1 αi!

(2N − 1)!(2N − 1)!! ×

min(2N−m,N−k+1)

• r=0

1 − k r 2N + k − 1 2N − m − r 2k+r−N−1(2N − k − r)! (N − k − r + 1)! . 2N-cycle involution size m length k

• Proof. Use

#Sα

2N = [p2N (x)]

• λ⊢n,r≥0

mλ(x) n(n − ℓλ)!(n − k − r)! (n − k − ℓλ − r + 1)! n + k − 1 n − m − r 1 − k r

• .

and [p2N ]mλ = 0 unless λ = 2N−s, 1s for some s.

Permutation with p cycles × long cycle Theorem [BMDS]. The probability that the product π ◦ π′ is α-separated is (n − m)! k

i=1 αi!

c(n, p)

n−m

• r=0

1 − k r n + k − 1 n − m − r c(n − k − r + 1, p) (n − k − r + 1)! . where c(n, p) = [xp]x(x + 1) · · · (x + n − 1). Permutation with p cycles long cycle size m, length k

Permutation with p cycles × long cycle Theorem [BMDS]. The probability that the product π ◦ π′ is α-separated is (n − m)! k

i=1 αi!

c(n, p)

n−m

• r=0

1 − k r n + k − 1 n − m − r c(n − k − r + 1, p) (n − k − r + 1)! . where c(n, p) = [xp]x(x + 1) · · · (x + n − 1). Permutation with p cycles long cycle size m, length k

• Proof. One needs to compute
• µ of length p

[pµ]

• λ⊢n,r≥0

mλ(x) n(n − ℓλ)!(n − k − r)! (n − k − ℓλ − r + 1)! n + k − 1 n − m − r 1 − k r

• .

Moreover, using the principal specialization gives

• µ of length p

[pµ]

• λ of length ℓ

mλ(x) = n − 1 ℓ − 1 (−1)ℓ−pc(ℓ, p) ℓ! .

Adding fixed-points to π Let λ be a partition of n without part of size 1. Let λ∗ be partition of n + r obtained by adding r parts of size 1. We compute σα

λ∗= proba that π∗ ◦ π′∗ is α-separated

in terms of σα′

λ = proba that π ◦ π′ is α′-separated.

type λ∗ long cycle size m, length k type λ long cycle

Adding fixed-points to π Let λ be a partition of n without part of size 1. Let λ∗ be partition of n + r obtained by adding r parts of size 1. We compute σα

λ∗= proba that π∗ ◦ π′∗ is α-separated

in terms of σα′

λ = proba that π ◦ π′ is α′-separated.

type λ∗ long cycle size m, length k Theorem.

σα

λ∗ =

n!

• n+r

α1,...,αk,n+r−m

n+r

r

• ×

m−k

• p=0
• n+p

n

n+m+r−p

n+m

• + m−p

n

n+m+r−p−1

n+m

m−k

p

• (n − m + p)!(m − k − p + 1)!

σ(m−k−p+1,1k−1)

λ

.

type λ long cycle

Thanks.

1 2 3 4 5 6

× =

3 4 6 1 5 2 1 4 6 3 2 5