Shuffling properties for products of
random permutations
Olivier Bernardi (MIT)
Joint work with Rosena Du, Alejandro Morales, Richard Stanley
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Halifax, June 2012
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
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?
1
Answer:
.
k!
Proof: Sampling process for π: build the cycles.
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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
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.
Example: α = (2, 2, 1). Blocks: {1, 2} {3, 4} {5}.
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Warm up
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?
α1 !α2 ! . . . αk !
.
Answer:
(α1 + α2 + . . . + αk )!
Warm up
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?
α1 !α2 ! . . . αk !
.
Answer:
(α1 + α2 + . . . + αk )!
Proof: Sampling process for π: build the cycles.
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Results
Type of question
Let λ, λ0 be partitions of n.
Let π, π 0 be uniformly random permutations of cycle-type λ, λ0 .
Let α be a composition of m ≤ n.
What is the probability that the product π ◦ π 0 is α-separated?
Results
Theorem [Du, Stanley]. Let π, π 0 be uniformly random n-cycles.
The probability that 1, 2, . . . , k are in different cycles of product ππ 0 is
1
k!
1
2
+
k! (k − 2)!(n − k + 1)(n + k)
if n − k odd,
otherwise.
Results
Theorem [Du, Stanley]. Let π, π 0 be uniformly random n-cycles.
The probability that 1, 2, . . . , k are in different cycles of product ππ 0 is
1
k!
1
2
+
k! (k − 2)!(n − k + 1)(n + k)
if n − k odd,
otherwise.
+ Extension to product (n−j)-cycle × n-cycle for k = 2.
Case k = 2 was conjectured by Bóna.
More results:
general composition α
Theorem [BMDS]. Let π, π 0 be uniformly random n-cycles.
Let α = (α1 , α2 , . . . , αk ) be a composition of size m.
The probability that the product ππ 0 is α-separated is
Qk
(n − m)! i=1 αi !
(n + k)(n − 1)!
(−1)n−m n−1
k−2
n+m
m−k
+
m−k
X
r=0
(−1)r
m−k
!
n+r+1
r
m
n+k+r
r
.
More results:
general composition α
Theorem [BMDS]. Let π, π 0 be uniformly random n-cycles.
Let α = (α1 , α2 , . . . , αk ) be a composition of size m.
The probability that the product ππ 0 is α-separated is
Qk
(n − m)! i=1 αi !
(n + k)(n − 1)!
(−1)n−m n−1
k−2
n+m
m−k
+
m−k
X
r=0
(−1)r
m−k
!
n+r+1
r
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 0 be a uniformly random n-cycle.
Let α = (α1 , α2 , . . . , αk ) be a composition of size m.
The probability that the product ππ 0 is α-separated is
Qk
(n − m)! i=1 αi !
c(n, p)
n−m
X
r=0
1−k
r
n + k − 1 c(n − k − r + 1, p)
,
n − m − r (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 0 be a uniformly random 2N -cycle.
Let α = (α1 , α2 , . . . , αk ) be a composition of size m.
The probability that the product ππ 0 is α-separated is
Qk
αi !
×
(2N − 1)!(2N − 1)!!
k+r−N −1
min(2N −m,N −k+1)
X
1−k
2N + k − 1 2
(2N − k − r)!
.
(N − k − r + 1)!
r
2N − m − r
r=0
i=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 0 be a uniformly random n-cycle.
Let α = (α1 , α2 , . . . , αk ), β = (β1 , β2 , . . . , βk ) be compositions of size
m and length k.
β
α
The probabilities σλ and σλ that the product ππ 0 is α-separated and
β-separated are related by
σλβ
σλα
Qk
i=1
αi !
= Qk
i=1
βi !
.
Strategy
Set up:
Let α = (α1 , . . . , αk ) be a composition of m ≤ n.
Notation. σλα =proba that π ◦ π 0 is α-separated.
random permutation of type λ
random n-cycle
Set up:
Let α = (α1 , . . . , αk ) be a composition of m ≤ n.
Notation. σλα =proba that π ◦ π 0 is α-separated.
random permutation of type λ
random n-cycle
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.
Example: π is ({1, 3, 6}, {2, 10})-separated.
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Set up:
Let α = (α1 , . . . , αk ) be a composition of m ≤ n.
Notation. σλα =proba that π ◦ π 0 is α-separated.
random permutation of type λ
random n-cycle
Remark. σλα = proba that π ◦ (1, 2, . . . , n) is A-separated.
random permutation of type λ
random subsets (A1 , . . . , Ak )
with #Ai = αi
Set up:
Let α = (α1 , . . . , αk ) be a composition of m ≤ n.
Notation. σλα =proba that π ◦ π 0 is α-separated.
random permutation of type λ
random n-cycle
Remark. σλα = proba that π −1 ◦ (1, 2, . . . , n) is A-separated.
random permutation of type λ
random subsets (A1 , . . . , Ak )
with #Ai = αi
Set up:
Let α = (α1 , . . . , αk ) be a composition of m ≤ n.
Notation. σλα =proba that π ◦ π 0 is α-separated.
random permutation of type λ
random n-cycle
Remark. σλα = proba that π −1 ◦ (1, 2, . . . , n) is A-separated.
random permutation of type λ
Lemma:
σλα =
random subsets (A1 , . . . , Ak )
with #Ai = αi
#Sλα
n
α1 ,α2 ,...,αk ,n−m
#Cλ
,
where Sλα 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).
Set up:
We want to count set Sλα 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 Sλα 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 .
It has color-type γ = (γ1 , . . . , γk ) if there are exactly γi elements of
color i.
Example: Colors 1, 2, 3.
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4
Color-type: γ = (4, 3, 4).
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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 0 be compositions of size n of length k, k 0
The number of colored permutations π, π 0 of color-types γ, γ 0 such
that ππ 0 = (1, 2, . . . , n) is
Bγ,γ 0
n(n − k)!(n − k 0 )!
=
.
0
(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 , . . . , λ` ):
Q`
P
• Power basis:
pλ (x) = i=1 pλi (x) where pk (x) = i≥1 xki .
P
Q
γi
• Monomial basis: mλ (x) = γ1 ,γ2 ,...∼λ
x
i≥1 i .
Navigating between cycle-type and color-type
Def. Symmetric functions in x = x1 , x2 , x3 . . .
Bases indexed by partitions λ = (λ1 , . . . , λ` ):
Q`
P
• Power basis:
pλ (x) = i=1 pλi (x) where pk (x) = i≥1 xki .
P
Q
γi
• Monomial basis: mλ (x) = γ1 ,γ2 ,...∼λ
x
i≥1 i .
Rk. If π is a permutation of type λ, then pλ (x) is the generating
function of the colorings of π.
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pλ (x) =
X
coloring of
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Y
colored
x#
i
π i≥0
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2
p4 (x) p2 (x) p2 (x)p1 (x)
i = X m (x) #colorings of π of color-type µ
µ
µ`n
Strategy
Write the generating function of the numbers #Sλα in one
symmetric function basis and compute the coefficients in another:
X
X
pλ (x) #Sλα =
mµ (x) computableµ .
λ`n
µ`n
Generating function
We want to count set Sλα 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 Sλα 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)
:=
X
λ`n
pλ (x)
X
tc(ω,A) ,
α
(A,π,ω)∈Sλ
where c(ω, A) = number of cycles of ω containing no element in A.
Hence #Sλα = [pλ (x)]Gα
n (x, 1).
Generating function
We want to count set Sλα 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)
:=
X
λ`n
pλ (x)
X
tc(ω,A) ,
α
(A,π,ω)∈Sλ
where c(ω, A) = number of cycles of ω containing no element in A.
Hence #Sλα = [pλ (x)]Gα
n (x, 1).
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)2 p2 (x) t1 .
Generating function
We want to count set Sλα 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).
Proposition.
Gα
n (x, t + k) =
X
λ`n
pλ (x)
X
(t + k)c(ω,A) =
α
(A,π,ω)∈Sλ
X
t
α
#Tλ,r mλ (x)
.
r
λ`n,r≥0
α
where Tλ,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).
Sλα :“Separated factorizations of (1, .., n)”
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1
6
4
5
2
×
2
3
5
6
1
4
A-separated
type λ
3
2
=
4
1
5
6
pλ (x) t
t
0
mµ (x)
1
coloring respecting A
color-type γ
3
6
4
5
2
×
2
3
6
4
3
2
5
1
=
α
Tγ,r
:“Separated colored factorizations of (1, .., n)”
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1
6
5
Counting separated colored factorization
Let α = (α1 , . . . , αk ) be a partition of size m.
Prop. For any partition λ of size n and length `,
n(n − `)!(n − k − r)! n + k − 1
α
#Tλ,r =
.
(n − ` − k − r + 1)! n − m − r
α
Tλ,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
Let α = (α1 , . . . , αk ) be a partition of size m.
Prop. For any partition λ of size n and length `,
n(n − `)!(n − k − r)! n + k − 1
α
#Tλ,r =
.
(n − ` − k − r + 1)! n − m − r
Proof.
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3
1
6
4
5
2
×
2
3
6
4
3
5
1
=
4
1
6
First choose colored factorization of color-type λ, λ0 with λ0 of
n(n − `)!(n − k − r)!
length k + r: Bλ,λ0 =
.
(n − ` − k − r + 1)!
5
Counting separated colored factorization
Let α = (α1 , . . . , αk ) be a partition of size m.
Prop. For any partition λ of size n and length `,
n(n − `)!(n − k − r)! n + k − 1
α
#Tλ,r =
.
(n − ` − k − r + 1)! n − m − r
Proof.
2
3
1
6
4
5
2
×
2
3
6
4
3
5
1
=
4
1
5
6
First choose colored factorization of color-type λ, λ0 with λ0 of
n(n − `)!(n − k − r)!
length k + r: Bλ,λ0 =
.
(n − ` − k − r + 1)!
Qk
λi
Choose the elements of Ai among those colored i:
i=1 αi .
Summation simplifies because Bλ,λ0 is independent of λ0 .
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 0 be a random n-cycle.
Probability of α-separation for π ◦ π 0 is given by
σλα =
#Sλα
n
α1 ,α2 ,...,αk ,n−m
#Cλ
,
where
#Sλα = [pλ (x)]Gα
n (x, 1)
and
Gα
n (x, t)
=
X
n(n − `λ )!(n − k − r)!
mλ (x)
(n − k − `λ − r + 1)!
λ`n,r≥0
n+k−1
n−m−r
t−k
.
r
Harvesting the generating function
Symmetry
For α = (α1 , . . . , αk ) partition of size m,
α
#S
λ
σλα =
n
α1 ,α2 ,...,αk ,n−m
where
#Cλ
,
#Sλα = [pλ (x)]Gα
n (x, 1)
and
X
t−k
n(n − `λ )!(n − k − r)! n + k − 1
α
.
Gn (x, t) =
mλ (x)
(n − k − `λ − r + 1)! n − m − r
r
λ`n,r≥0
Symmetry
For α = (α1 , . . . , αk ) partition of size m,
α
#S
λ
σλα =
n
α1 ,α2 ,...,αk ,n−m
#Cλ
,
where
#Sλα = [pλ (x)]Gα
n (x, 1)
and
X
t−k
n(n − `λ )!(n − k − r)! n + k − 1
α
.
Gn (x, t) =
mλ (x)
(n − k − `λ − r + 1)! n − m − r
r
λ`n,r≥0
Cor. If α and β are compositions of same size and same length then
β
Gα
(x,
t)
=
G
n
n (x, t).
β
α
σ
σ
β
α
In particular, for all λ, #Sλ = #Sλ and Qk λ
= Qk λ .
i=1 αi !
i=1 βi !
Symmetry
For α = (α1 , . . . , αk ) partition of size m,
α
#S
λ
σλα =
n
α1 ,α2 ,...,αk ,n−m
#Cλ
,
where
#Sλα = [pλ (x)]Gα
n (x, 1)
and
X
t−k
n(n − `λ )!(n − k − r)! n + k − 1
α
.
Gn (x, t) =
mλ (x)
(n − k − `λ − r + 1)! n − m − r
r
λ`n,r≥0
Cor. If α and β are compositions of same size and same length then
β
Gα
(x,
t)
=
G
n
n (x, t).
β
α
σ
σ
β
α
In particular, for all λ, #Sλ = #Sλ and Qk λ
= Qk λ .
i=1 αi !
i=1 βi !
α
Remark. This result comes from“symmetry” in formula for Tλ,r
,
which has a bijective proof [B., Morales 12].
Involution × n-cycle
involution
2N -cycle
size m length k
Theorem [BMDS].
The probability that the product π ◦ π 0 is α-separated is
Qk
i=1 αi !
×
(2N − 1)!(2N − 1)!!
k+r−N −1
min(2N −m,N −k+1)
X
1−k
2N + k − 1 2
(2N − k − r)!
.
(N − k − r + 1)!
r
2N − m − r
r=0
Involution × n-cycle
involution
2N -cycle
size m length k
Theorem [BMDS].
The probability that the product π ◦ π 0 is α-separated is
Qk
i=1 αi !
×
(2N − 1)!(2N − 1)!!
k+r−N −1
min(2N −m,N −k+1)
X
1−k
2N + k − 1 2
(2N − k − r)!
.
(N − k − r + 1)!
r
2N − m − r
r=0
Proof. Use
#S2αN
n(n − `λ )!(n − k − r)! n + k − 1 1 − k
.
= [p2N (x)]
mλ (x)
r
(n − k − `λ − r + 1)! n − m − r
λ`n,r≥0
X
and [p2N ]mλ = 0 unless λ = 2N −s , 1s for some s.
Permutation with p cycles × long cycle
Permutation with long cycle
p cycles
size m, length k
Theorem [BMDS].
The probability that the product π ◦ π 0 is α-separated is
Qk
n−m
X
(n − m)!
αi !
1−k
n + k − 1 c(n − k − r + 1, p)
i=1
c(n, p)
r=0
r
n−m−r
where c(n, p) = [xp ]x(x + 1) · · · (x + n − 1).
(n − k − r + 1)!
.
Permutation with p cycles × long cycle
Permutation with long cycle
p cycles
size m, length k
Theorem [BMDS].
The probability that the product π ◦ π 0 is α-separated is
Qk
n−m
X
(n − m)!
αi !
1−k
n + k − 1 c(n − k − r + 1, p)
i=1
c(n, p)
n−m−r
r
r=0
(n − k − r + 1)!
.
where c(n, p) = [xp ]x(x + 1) · · · (x + n − 1).
Proof. One needs to compute
X
µ
n(n − `λ )!(n − k − r)! n + k − 1 1 − k
.
[pµ ]
mλ (x)
r
(n − k − `λ − r + 1)! n − m − r
λ`n,r≥0
X
of length
p
Moreover, using the principal specialization gives
X
µ
of length
X
[pµ ]
p
λ
of length
mλ (x) =
`
n − 1 (−1)`−p c(`, p)
`−1
`!
.
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.
type λ∗
We compute
in terms of
long cycle
size m, length k
σλα∗ = proba that π ∗ ◦ π 0∗ is α-separated
α0
σλ = proba that π ◦ π 0 is α0 -separated.
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.
type λ∗
We compute
in terms of
long cycle
size m, length k
σλα∗ = proba that π ∗ ◦ π 0∗ is α-separated
α0
σλ = proba that π ◦ π 0 is α0 -separated.
type λ
long cycle
Theorem.
α
σλ∗
×
=
n!
n+r
n+r
α1,...,αk ,n+r−m
r
m−k
m−p
n+p
n+m+r−p
n+m+r−p−1
m−k
+ n
X
n
n+m
n+m
p
p=0
(n − m + p)!(m − k − p + 1)!
(m−k−p+1,1k−1 )
σλ
.
Thanks.
3
2
3
1
6
4
5
2
×
2
3
6
4
5
1
=
4
1
6
5