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On the number of cycles of the product of two cyclic
permutations
Michel Marcus
Labri, Université Bordeaux 1
• Joint work with Robert Cori (Labri, Bordeaux 1), Gilles Schaeffer (Lix, Ecole
Polytechnique, Palaiseau).
Product of cyclic permutations,
Lattice Paths, Siena july 2010
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Large cycles
A permutation is a large cycle if one of two conditions below holds:
it is cyclic (i.e. has a unique cycle).
Example: (1, 5, 4, 3, 2)
Product of cyclic permutations,
it has two cycles, one of which being a
fixed point. Example: (1, 3, 2, 5)(4)
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Main points
1. For a given permutation σ find the number of factorizations σ = αβ such that α is
cyclic and β is a large cycle.
2. For a cyclic permutation α find the number of factorizations α = σβ such that σ has k
cycles and β is a large cycle.
3. For a cyclic permutation α of Sn and a partition λ of the integer n find the number of
factorizations α = σβ such that σ has cycle type λ and β is a large cycle.
Product of cyclic permutations,
Lattice Paths, Siena july 2010
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The number of factorizations σ = αβ
such that α is cyclic and β is a large cycle
• We denote this number aλ , where λ is the cyclic type of σ.
• Two cases to consider: λ even or λ odd.
Product of cyclic permutations,
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Main fact
Theorem 1 (G. Boccara, 1980) Let σ be a an odd permutation with cycle type λ then
aλ = 2 · (n − 2)!
Combinatorial proofs by Machí (1992), Schaeffer (1998)
Product of cyclic permutations,
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From odd to even permutations
Theorem 2 Let σ be an even permutation with cycle type λ then
aλ
Product of cyclic permutations,
1
φ(λ)
=
n(n + 1)
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From odd to even permutations: φ(λ)
1. Let λ be an even partition of n consisting of k parts, then I := {1, 2, . . . k}
P
2. for any X ⊆ I, SX (λ) := i∈X λi
Then
φ(λ) :=
X
(−1)SX (λ)+|X| SX (λ)! (n − SX (λ))!
X⊆I
Product of cyclic permutations,
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φ(λ) :=
X
(−1)SX (λ)+|X| SX (λ)! (n − SX (λ))!
X⊆I
Example:
1. λ = [4, 3, 2], thus I := {1, 2, 3}
2. All X: ∅; {1}; {2}; {3}; {1, 2}; {1, 3}; {2, 3}; {1, 2, 3}
3. All SX : 0, 4, 3, 2, 7, 6, 5, 9
4. φ([4, 3, 2]) = +0!9! − 4!5! + 3!6! − 2!7! − 7!2! + 6!3! − 5!4! + 9!0! = 708480
Thus a[4,3,2] =
Product of cyclic permutations,
708480
9×10
= 7872
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From odd to even permutations
aλ,i denotes the number of factorizations σ = αβ such that α is cyclic, β is a (n-1)-cycle,
β(i) = i and λ is the cycle type of σ.
Proposition 1 Let λ = [λ1 , λ2 , . . .] and n − |λ| even, then
a[λ1 ,λ2 ,...] = a[λ1 ,λ2 +1,...],λ1 +1
Example :
(1, 2, 3, 4)(5, 6, 7) = (1, 5, 6, 3, 2, 4, 7) ◦ (1, 3, 2, 6, 4, 7)(5)
(1, 2, 3, 4)(6, 7) = (1, 6, 3, 2, 4, 7) ◦ (1, 3, 2, 6, 4, 7)
(1, 2, 3, 4)(5, 6) = (1, 5, 3, 2, 4, 6) ◦ (1, 3, 2, 5, 4, 6)
Product of cyclic permutations,
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From odd to even permutations
Proposition 2 Let λ be a partition of n consisting of k parts and n − k odd, then
k
aλ,λ1
X
1
=
[2(n − 2)! −
λi aλ(1,i) ,λ1 ]
λ1
i=2
Proposition 3 Let λ be a partition of n consisting of k parts and n − k even, then
k
aλ
X
1
=
[2(n − 1)! −
λi aλ(1,i) ]
λ1 + 1
i=2
where λ(1,i) := [(λ1 + 1), λ2 , . . . λi−1 , (λi − 1), . . . λk ]
Product of cyclic permutations,
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From odd to even permutations
Induction on the largest part of λ shows that formula in Theorem 2 is the solution of this
recursion.
Product of cyclic permutations,
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A result of D. Zagier (1995)
Theorem 3 Given a cyclic permutation α ∈ Sn , let bn,k denote the number of
factorizations α = σβ such that σ has k cycles and β is a large cycle. If n − k is even,
then β is cyclic and
2 · s(n + 1, k)
bn,k =
n(n + 1)
where s(n + 1, k) is the (n + 1, k) unsigned Stirling number of the first kind.
Product of cyclic permutations,
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Once more from odd to even
In Sn+1 , consider all the products of the following form :
σ = αβ
where σ has k cycles, n + 1 − k is odd, α is cyclic and β is a large cycle.
Product of cyclic permutations,
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Once more from odd to even
In Sn+1 , consider all the products of the following form :
σ
= α β
s(n + 1, k)
where σ has k cycles, n + 1 − k is odd, α is cyclic and β is a large cycle.
Product of cyclic permutations,
Lattice Paths, Siena july 2010
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Once more from odd to even
In Sn+1 , consider all the products of the following form :
σ
= α β
s(n + 1, k) · 2 · (n − 1)!
where σ has k cycles, n + 1 − k is odd, α is cyclic and β is a large cycle.
Product of cyclic permutations,
Lattice Paths, Siena july 2010
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Once more from odd to even
In Sn+1 , consider all the products of the following form :
σ = α
β
n!
where σ has k cycles, n + 1 − k is odd, α is cyclic and β is a large cycle.
Product of cyclic permutations,
Lattice Paths, Siena july 2010
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Once more from odd to even
In Sn+1 , consider all the products of the following form :
σ = α
β
n! bn+1,k
where σ has k cycles, n + 1 − k is odd, α is cyclic and β is a large cycle.
Product of cyclic permutations,
Lattice Paths, Siena july 2010
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Once more from odd to even
Hence, if n + 1 − k is odd :
σ
= α
β
s(n + 1, k) · 2 · (n − 1)! = n! bn+1,k
thus
bn+1,k =
Product of cyclic permutations,
2 · s(n + 1, k)
n
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Back to the result of D. Zagier (1995)
From odd to even, and from even to odd...
Now, build Bn+1,k (odd) from Bn,k (even).
(1, 2, 3, 4, 5) ◦ (1, 3, 4, 5, 2) = (1, 4)(2)(3, 5)
(1, 2, 3, 4, 5, 6) ◦ (1, 3, 4, 5, 2)(6) = (1, 4, 6)(2)(3, 5)
Then, consider conjugation by powers of α
(1, 2, 3, 4, 5, 6) ◦ (1, 4, 3, 5, 6)(2) = (1, 5)(2, 3, 6)(4)
etc.
Product of cyclic permutations,
( α2 )
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Back to the result of D. Zagier (1995)
From odd to even, and from even to odd...
Now, build Bn+1,k (odd) from Bn,k (even).
(1, 2, 3, 4, 5) ◦ (1, 3, 4, 5, 2) = (1, 4)(2)(3, 5)
←β
(1, 2, 3, 4, 5, 6) ◦ (1, 3, 4, 5, 2)(6) = (1, 4, 6)(2)(3, 5) ← β 0
Then, consider conjugation by powers of α
(1, 2, 3, 4, 5, 6) ◦ (1, 4, 3, 5, 6)(2) = (1, 5)(2, 3, 6)(4) ← β20
(α2 )
etc.
So, from an element of Bn,k (even), one can build n + 1 elements of Bn+1,k .
Product of cyclic permutations,
Lattice Paths, Siena july 2010
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Back to the result of D. Zagier (1995)
From odd to even, and from even to odd...
From an element of Bn,k (even), one can build n + 1 elements of Bn+1,k .
Conversely, one finds β 0 (thus β) from βi0 :
α−i βi0 αi = β 0 (fixes n + 1)
Hence
bn,k =
Product of cyclic permutations,
2 · s(n + 1, k)
n(n + 1)
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A generalisation of Zagier’s result : bλ
Given a cyclic permutation α of Sn and a partition λ of the integer n,
bλ denotes the number of factorizations α = σβ
such that σ has cycle type λ and β is a large cycle.
Product of cyclic permutations,
Lattice Paths, Siena july 2010
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A generalisation of Zagier’s result : bλ
Theorem 4 (V. Feray & E. Vassilieva, 2010)
Let λ be an odd partition of n, then
n X
(p − 1) · |λ(p) |p−1 · bλ(p)
cλ =
2
p∈P (λ)
where
• cλ denotes the cardinality of the conjugacy class defined by λ
• P (λ) is the set of all parts of λ; e.g. P ([5, 5, 3, 1, 1]) = {1, 3, 5}
• λ(p) is obtained from λ by replacing one occurrence of p by p − 1; e.g.
[5, 5, 3, 1, 1](5) = [5, 4, 3, 1, 1]
• |λ|x denotes the number of occurrences of the integer x in λ
Product of cyclic permutations,
Lattice Paths, Siena july 2010
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A generalisation of Zagier’s result : bλ
Once more, we start from theorem 1 and we consider all the products of the following shape:
σ
= α
β
cλ · 2 · (n − 2)! = (n − 1)! bλ
where σ has cycle type λ, n − k is odd, α is cyclic and β is a large cycle.
Hence
bλ =
Product of cyclic permutations,
2
cλ
n−1
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A generalisation of Zagier’s result : bλ
Now consider bλ,j̄ , like bλ , but β(j) 6= j
Clearly, this number does not depend on the choice of j (consider conjugation), hence
n
∀j ∈ {1, 2, . . . n}, bλ =
b
n − 1 λ,j̄
Product of cyclic permutations,
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A generalisation of Zagier’s result : bλ
Finally,
bλ,n̄ =
X
(p − 1) · |λ(p) |p−1 · bλ(p)
p∈P (λ)
Consider
• α = (1, 2, 3, 4, 5, 6) and β = (1, 2, 4, 6, 5)(3)
• thus σ = (1, 6, 5)(2)(3, 4) ( λ = [3, 2, 1])
• Now delete 3 and rename all elements greater than 3 :
– σ 0 = (1, 5, 4)(2)(3) and β 0 = (1, 2, 3, 5, 4); σ 0 is of type λ(2)
– conversely, from σ 0 , σ = (1, 6, 5)(2, 3)(4) or σ = (1, 6, 5)(2)(3, 4)
Product of cyclic permutations,
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A generalisation of Zagier’s result : bλ
Hence
X
n
bλ =
(p − 1) · |λ(p) |p−1 · bλ(p)
n−1
p∈P (λ)
which is also Theorem 4, since cλ = n−1
bλ
2
n X
cλ =
(p − 1) · |λ(p) |p−1 · bλ(p)
2
p∈P (λ)
Product of cyclic permutations,
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Open problems
Find combinatorial proofs of the following results (D. Zagier) :
• Let λ be an even partition of n with f parts equal to 1, then
2 · (n − 1)!
2 · (n − 1)!
≤ aλ ≤
n−f +2
n − f + 19
29
• Given a n − cycle α and a partition λ of n, let pk (λ) denote the number of
permutations σ having cycle type λ and such that the product σα has k cycles; then
n
X
pk (λ) · Ek (x) · (1 − x)n+1−|λ|−k
k=1
|λ|
Y
= cλ (1 + x + x2 . . . + xλi −1 )
i=1
where Ek (x) is Euler polynomial
Product of cyclic permutations,
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Grazie mille !
Product of cyclic permutations,
Lattice Paths, Siena july 2010