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Preprint submitted on 9 Oct 2019
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A product of invariant random permutations has the same small cycle structure as uniform
Mohamed Slim Kammoun, Mylène Maïda
To cite this version:
Mohamed Slim Kammoun, Mylène Maïda. A product of invariant random permutations has the same
small cycle structure as uniform. 2019. �hal-02309521�
A product of invariant random permutations has the same small cycle structure as uniform
Mohamed Slim Kammoun
∗Mylène Maïda
†October 9, 2019
Abstract
We use moment method to understand the cycle structure of the composition of independent invariant permutations. We prove that under a good control on fixed points and cycles of length 2, the limiting joint distribution of the number of small cycles is the same as in the uniform case i.e. for any positive integerk, the number of cycles of lengthkconverges to the Poisson distribution with parameter 1k and is asymptotically independent of the number of cycles of lengthk′6=k.
1 Introduction and main results
We denote by
Snthe group of permutations of { 1, . . . , n } , by #
kσ the number of cycles of σ of length k, by # σ the total number of cycles of σ and by tr(σ) := #
1σ.
The cycle structure of a permutation chosen uniformly among the symmetric group
Snis well understood (see e.g. [Arratia, Tavaré, and Barbour,
2003] for detailed results). In particular, the following classicalresult holds:
Theorem 1.
[Arratia, Barbour, and Tavaré,
2000, Theorem 3.1] Ifσ
nfollows the uniform distribution on
Snthen for any k ≥ 1,
(#
1σ
n, . . . , #
kσ
n) −−−→
nd→∞
η
k:= (ξ
1, ξ
2, . . . , ξ
k), (1)
where −−−→
n d→∞
denotes the convergence in distribution, ξ
1, ξ
2, . . . ξ
kare independent and the distribution of ξ
dis Poisson of parameter
1d.
In this work, we question the universality class of this convergence. We show that a product of conjugation invariant permutations that do not have too many fixed points and cycles of size 2 lies within this class.
More precisely, we have the following.
Theorem 2.
Let m ≥ 2. For 1 ≤ ℓ ≤ m, let (σ
nℓ)
n≥1be a sequence of random permutations such that for any n ≥ 1, σ
ℓn∈
Sn. For any k ≥ 1, let t
nk:= #
k(
Qmℓ=1σ
ℓn). Assume that
- (H
1) For any n ≥ 1, (σ
1n, . . . , σ
nℓ) are independent.
- For any n ≥ 1 and 1 ≤ ℓ ≤ m, for any σ ∈
Sn,
σ
−1σ
nℓσ =
dσ
ℓn, (H
2)
except maybe for one ℓ ∈ { 1, . . . , m } .
∗Univ. Lille, CNRS, UMR 8524 - Laboratoire Paul Painlevé, F-59000 Lille, France. Email:
mohamed-slim.kammoun@univ-lille.fr.
†Univ. Lille, CNRS, UMR 8524 - Laboratoire Paul Painlevé, F-59000 Lille, France. Email: mylene.maida@univ-lille.fr.
- There exists 1 ≤ i < j ≤ m such that for any k ≥ 1,
n
lim
→∞EÇ
#
1σ
in√ n
åk!= 0 and lim
n→∞E
Ç
#
1σ
jn√ n
åk!= 0, (H
3)
n
lim
→∞E
(#
2σ
ni)
n = 0 and lim
n→∞
E
(#
2σ
jn)
n = 0.
(H
4)
Then for any k ≥ 1,
(t
n1, t
n2, . . . , t
nk) −−−→
nd→∞
η
k.
This convergence has also been obtained by
Mukherjee[2016] for a quite different class of permutations, namely the permutations that are equicontinuous in both coordinates and converging as a permuton (see Definitions in [Mukherjee,
2016]). Here, it is easy to check that for anyθ ∈ [0, 1], the Ewens distribution with parameter θ satisfies the convergences required in H
3and H
4. Our result tells that the product of (at least two) Ewens distributions behaves like a uniform permutation, as far as small cycles are concerned.
In our framework, in the case of two permutations, a weaker result can be obtained without any hypothesis on the cycles of size 2.
Proposition 3.
When m = 2, under H
1, H
2and H
3, we have convergence of the first moment i.e for any v ≥ 1,
n
lim
→∞E(t
nv) = 1 v .
Note that when one of the permutations σ
ℓnfollows the uniform distribution, under H
1, the product also follows the uniform distribution and Theorem
2is a direct consequence of Theorem
1.Our motivation to understand the cycle structure of random permutations is the relation, in the case of conjugation invariant permutations, to the longest common subsequence (LCS) of two permutations. For example, for m = 2, if σ
n−1ρ
nis conjugation invariant and
#(σ
−n1ρ
n)
√
6n
−−−→
n→∞d0.
Then for any s ∈
R,
P
Ç
LCS(σ
n, ρ
n) − 2 √ n
√
6n ≤ s
å
−−−→
n→∞
F
2(s),
where F
2is the cumulative distribution function of the GUE Tracy-Widom distribution.
Another motivation comes from traffic distributions, a non-commutative probability theory introduced by
Male[2011] to understand the moments of permutation invariant random matrices. As shown in [Male,
2011], the limit in traffic distribution of uniform permutation matrices is trivial but Theorem 1can be seen as a second-order result in this framework. It is therefore natural to ask about limiting joint fluctuations for the product of several permutation matrices, which is a really non-commutative case. To emphasize this relation, we rewrite Theorem
2as follows.
Corollary 4.
Under H
1, H
2, H
3and H
4, for any k ≥ 1,
Ätr(
Qmi=1σ
nℓ), tr((
Qmi=1σ
nℓ)
2), . . . , tr((
Qmi=1σ
ℓn)
k)
äconverges in distribution to (ξ
1, ξ
1+ 2ξ
2, . . . ,
Pd|kdξ
d), where ξ
1, ξ
2, . . . are independent and the distribution of ξ
dis Poisson of parameter
1d.
The optimality of conditions H
3and H
4will be discussed at the end of the paper.
Acknowledgements :
The first author would like to acknowledge a useful discussion with Camille Male
about traffic distributions. This work is partially supported by the Labex CEMPI (ANR-11-LABX-0007-01).
2 Proof of results
We begin with a few preliminary remarks and simplifications.
First of all, the equivalence between Theorem
2and Corollary
4is due to the following classical argument.
For any σ ∈
Sn, if c
i(σ) denotes the length of the cycle of σ containing i,
(2) tr(σ
k) =
Xn i=1
1σk(i)=i
=
Xn i=11ci(σ)|k
=
Xj|k
Xn i=1
1ci(σ)=j
=
Xj|k
j #
jσ.
In the hypothesis H
2, we assume that one of the permutations, say σ
n1, may not have a conjugation invariant distribution. In fact, it is enough to prove of Theorem
2in the case where all permutations are conjugation invariant. Indeed, if we choose τ
nuniform and independent of the σ-algebra generated by (σ
nℓ)
1≤ℓ≤m, the cycle structure of
Qmℓ=1σ
nℓis the same as
τ
n−1 Ym ℓ=1σ
nℓ!
τ
n= (τ
n−1σ
n1τ
n)
Ym ℓ=2(τ
n−1σ
ℓnτ
n) = (τ
d n−1σ
1nτ
n)
Ym ℓ=2σ
nℓand (τ
n−1σ
1nτ
n) is also conjugation invariant.
We will prove in full details the case m = 2 and indicate briefly at the end of the paper how to extend the proof to a larger number of permutations. In the sequel, σ
n1and σ
2nwill be denoted respectively by σ
nand ρ
n.
2.1 Preliminary results
To prove Theorem
2, we will use the same objects introduced in [Kammoun,2019, pages 12-13] where onecan get further details and examples. To a couple of permutations and a subset of p indices, we will associate a set of 2p graphs. For technical reasons, we prefer working with σ
n−1ρ
nrather than σ
nρ
n: for any k ≥ 1, we define ˜ t
nk:= #
k(σ
−n1ρ
n). Under H
2, σ
n=
dσ
n−1and consequently under H
1and H
2, ∀ k ≥ 1 (t
n1, t
n2, . . . , t
nk) and (˜ t
n1, ˜ t
n2, . . . , ˜ t
nk) have the same distribution.
Let us now recall the combinatorial objects we will use.
• We denote by
Gnk
the set of oriented simple graphs with vertices { 1, 2, . . . , n } and having exactly k edges. Given g ∈
Gnk
, we denote by E
gthe set of its edges and by A
g:= [
1(i,j)∈Eg]
1≤i,j≤nits adjacency matrix.
• A connected component of g is called trivial if it does not have any edge and a vertex i of g is called isolated if E
gdoes not contain any edge of the form (i, j) or (j, i) nor a loop (i, i). Let g ∈
Gnk
, we denote by g ˜ the graph obtained from g after removing isolated vertices.
• We say that two oriented simple graphs g
1and g
2are isomorphic if one can obtain g
2by changing the labels of the vertices of g
1. In particular, if g
1, g
2∈
Gnkthen g
1, g
2are isomorphic if and only if there exists a permutation matrix σ such that A
g1σ = σA
g2.
• Let R be the equivalence relation such that g
1R g
2if g ˜
1and g ˜
2are isomorphic. We denote by
Gˆ
k:=
∪
n≥1Gnk
/ R the set of equivalence classes of ∪
n≥1Gnkfor the relation R . Let n ∈
N∗and σ, ρ ∈
Sn. Let m ∈ { 1, . . . , n } be fixed.
• We denote by (i
m1(σ, ρ) = m, i
m2(σ, ρ), . . . , i
mkm(σ,ρ)
(σ, ρ)) the cycle of σ
−1◦ ρ containing m, so that k
m(σ, ρ) := c
m(σ
−1◦ ρ) is the length of this cycle. For i ≤ k
m(σ, ρ), we define j
lm(σ, ρ) := ρ(i
ml(σ, ρ)). In particular, i
m1(σ, ρ), i
m2(σ, ρ), . . . , i
mkm(σ,ρ)
(σ, ρ) are pairwise distinct and j
1m(σ, ρ), j
2m(σ, ρ), . . . , j
kmm(σ,ρ)
(σ, ρ) are pairwise distinct. For sake of simplicity, when it is clear, we will use the notations
k
m, i
mland j
lminstead of k
m(σ, ρ), i
ml(σ, ρ) and j
lm(σ, ρ).
• We denote by G
1m(σ, ρ) ∈
Gnkm
and G
2m(σ, ρ) ∈
Gnkm
the graphs with vertices { 1, . . . , n } such that E
Gm1 (σ,ρ)
= { (i
m1, j
mkm) }
[km[−1 l=1
{ (i
ml+1, j
lm) }
!
and E
Gm2 (σ,ρ)
=
km
[
l=1
{ (i
ml, j
lm) }
and by g
σthe graph such that A
gσ= σ. By construction, for any positive integer m ≤ n, G
1m(σ, ρ) (resp. G
2m(σ, ρ)) is a sub-graph of g
σ(resp. g
ρ). Moreover, we want to emphasize that G
1m(σ, ρ) and G
2m(σ, ρ) have the same set of non-isolated vertices.
For i ∈ { 1, 2 } , let G ˆ
im(σ, ρ) be the equivalence class of G
im(σ, ρ).
• Let I = (s
1, s
2, . . . , s
l) a set of distinct indices of { 1, . . . , n } . We denote by
G
I(σ, ρ) = ( G
1s1(σ, ρ), G
2s1(σ, ρ), G
1s2(σ, ρ), . . . , G
1sl(σ, ρ), G
2sl(σ, ρ)) and
G ˆ
I(σ, ρ) = ( ˆ G
s11(σ, ρ), G ˆ
2s1(σ, ρ), G ˆ
1s2(σ, ρ), . . . , G ˆ
s1l(σ, ρ), G ˆ
2sl(σ, ρ)).
• For i ∈ { 1, 2 } , let G
i{1,2,...,k}(σ, ρ) be the graph such that E
G{1,2,...,k}
i (σ,ρ)
= ∪
kl=1E
Gℓi(σ,ρ)
and G ˆ
i{1,2,...,k}(σ, ρ) be the equivalence class of G
i{1,2,...,k}(σ, ρ).
Using the conjugation invariance and the relation (2), Theorem
2is equivalent to the following: under the same hypotheses, for any v
1, v
2, v
3, . . . , v
k≥ 1,
n
lim
→∞X
ˆ
gi,ˆg′i∈Gˆvi,1≤i≤k
n
kPÄG ˆ
{1,2,...,k}(σ
n, ρ
n) = (ˆ g
1, g ˆ
′1, g ˆ
2, . . . g ˆ
′k)
ä= C
v1,v2,...,vk, (*)
where C
v1,v2,...,vkis a constant independent of the laws of the permutations. Note that, for any v
i≥ 1,
Gˆ
vi
and therefore the number of terms of the sum is finite.
For example, if we take P (x) = x
2, we have
EÄP
Ĉ t
n1ää=
EÑ Xn i=1
1ci(σ−1◦ρ)=1
!2é
=
Xn i=1EÄ1ci(σ−1◦ρ)=1
ä
+
Xn i6=jEÄ1ci(σ−1◦ρ)=11cj(σ−1◦ρ)=1
ä
= n
EÄ1c1(σ−1◦ρ)=1ä
+ (n
2− n)
EÄ1c1(σ−1◦ρ)=11c2(σ−1◦ρ)=1ä
−−−→
n→∞C
1+ C
1,1= 1 + 1 = 2
Similarly, if we take P (x, y) = xy, we obtain
E(P (ˆ t
n1, ˆ t
n2)) −−−→
n→∞dC
1,2= C
2,1= 1.
Before getting into the proof of (*), let us gather some useful combinatorial and then probabilistic results.
Lemma 5.
[Kammoun,
2019, Lemma 15] Ifm
1∈ { i
ml 2, 1 ≤ l ≤ k
m2} , then G
1m1(σ, ρ) = G
1m2(σ, ρ) and G
2m1(σ, ρ) = G
2m2(σ, ρ).
Lemma 6.
For any m ≤ n, for any permutation σ, ρ ∈
Sn, k
m(ρ, σ) = k
m(σ, ρ),
j
ℓm(ρ, σ) = j
kmm(σ,ρ)−ℓ+1
(σ, ρ), ∀ 1 ≤ ℓ ≤ k
m(σ, ρ), i
mℓ(ρ, σ) = i
mkm(σ,ρ)−ℓ+2(σ, ρ), ∀ 2 ≤ ℓ ≤ k
m(σ, ρ), i
m1(ρ, σ) = i
m1(σ, ρ) = m,
A
Gm1 (σ,ρ)
= A
TG2ρ(m)(ρ−1,σ−1)
.
Lemma 7.
If all non trivial connected components of G
1m1(σ, ρ) and G
2m1(σ, ρ) have 2 vertices then both G
1m1(σ, ρ) and G
2m1(σ, ρ) have no 2-cycles .
Proof. Using the symmetries of the problem (Lemmas
5and
6), it suffices to prove that if all non trivialconnected components of G
11(σ, ρ) and G
21(σ, ρ) have 2 vertices then it is impossible to have at the same time (1, 2) ∈ G
21(σ, ρ) and (2, 1) ∈ G
12(σ, ρ). To simplify notations, let k
1:= k
1(σ, ρ) = c
1(σ
−1◦ ρ), i
1o:= i
1o(σ, ρ) and j
o1:= j
o1(σ, ρ).
Let A = { η > 1; j
η1∈ { i
11, i
12, . . . , i
1η−1} or i
1η∈ { j
11, j
21, . . . , j
1η−1}} . Suppose that (1, 2) ∈ G
21(σ, ρ) and (2, 1) ∈ G
21(σ, ρ) then k
1≥ 2 and there exists a unique 1 < l ≤ k
1such that i
1l= 2 and j
l1= 1 so that A is non-empty. Let ℓ
′:= inf (A) ≥ 2. Assume that ℓ
′> 2. If j
ℓ1′∈ { i
11, i
12, . . . , i
1ℓ′−1} , then there exists ℓ
′′< ℓ
′such that j
ℓ1′= i
1ℓ′′and since the component of G
21(σ, ρ) containing i
1ℓ′has two vertices and by definition (i
1ℓ′, j
ℓ1′) and (i
1ℓ′′, j
ℓ1′′) are two edges of G
21(σ, ρ), then j
ℓ1′′= i
1ℓ′. Since (i
1ℓ′, j
ℓ1′−1) = (j
1ℓ′′, j
ℓ1′−1) and (i
1ℓ′′+1, j
1ℓ′′) are edges of G
11(σ, ρ) and since G
11(σ, ρ) has only connected components of size 2, we have necessarily i
1ℓ′′+1= j
ℓ1′−1. One can check easily that ℓ
′′< ℓ
′− 2 otherwise either G
11(σ, ρ) or G
21(σ, ρ) has a loop. Indeed, if ℓ
′′= ℓ
′− 2, then (i
1ℓ′′+1, j
ℓ1′′+1) = (j
ℓ1′−1, j
ℓ1′′+1) = (j
ℓ1′−1, j
ℓ1′−1) is an edge of G
21(σ, ρ) and if ℓ
′′= ℓ
′− 1, then (i
1ℓ′′+1, j
1ℓ′′) = (j
ℓ1′−1, j
ℓ1′′) = (j
ℓ1′−1, j
ℓ1′−1) is an edge of G
11(σ, ρ). This implies that ℓ
′− 1 ∈ A, which is absurd. i
1ℓ′∈ { j
11, j
21, . . . , j
ℓ1′−1} can be treated using the same techniques and one can extend easily to ℓ
′= 2.
We now introduce the following notation : given g ∈
Gnk
, we denote by
Sn,g:= { σ ∈
Sn; ∀ (i, j) ∈ E
g, σ(i) = j } .
In other words,
Sn,gis the set of permutations σ such that g is a sub-graph of g
σ. It is not difficult to prove the two following lemmas.
Lemma 8.
Let g
1, g
1′, g
2, . . . , g
′k∈ ∪
ℓGnℓ
and let g, g
′be such that E
g= ∪
kℓ=1E
giand E
g′= ∪
kℓ=1E
g′i
. Assume that there exists ρ, σ such that
G
{1,2,...,k}(σ, ρ) = (g
1, g
1′, g
2, . . . , g
′k).
Then for any random permutation ρ
n, σ
n,
P
\k i=1
{ σ
n∈
Sn,gi
, ρ
n∈
Sn,g′ i}
!
=
PÄG
{1,2,...,k}(σ
n, ρ
n) = (g
1, g
′1, g
2, . . . , g
k′)
ä=
PG
1{1,2,...,k}(σ
n, ρ
n) = g, G
2{1,2,...,k}(σ
n, ρ
n) = g
′.
Proof. We will only prove the first equality. The second one can be obtained using the same argument.
Let σ
′, ρ
′be two permutations. We have seen that G
2m(σ
′, ρ
′) is a subset of g
ρ′, so that G
2m(σ
′, ρ
′) = g
m′⇒ ρ
′∈
Sn,g′m
, and that G
1m(σ
′, ρ
′) is a subset of g
σ′, so that
G
1m(σ
′, ρ
′) = g
m⇒ σ
′∈
Sn,gm
. Consequently,
PÄ
G
{1,2,...,k}(σ
n, ρ
n) = (g
1, g
′1, g
2, . . . , g
k′)
ä≤
P\k i=1
{ σ
n∈
Sn,gi
, ρ
n∈
Sn,g′ i}
!
.
Now suppose that there exists ρ
′, σ
′such that
G
{1,2,...,k}(σ
′, ρ
′) = (g
1, g
′1, g
2, . . . , g
k′).
Let σ, ρ such that σ ∈ ∩
ki=1Sn,gi
and ρ ∈ ∩
ki=1Sn,g′i
. By definition and by iteration on ℓ, one can check that for any ℓ
′≤ k, i
ℓℓ′(σ
′, ρ
′) = i
ℓℓ′(σ, ρ) and j
ℓℓ′(σ
′, ρ
′) = j
ℓℓ′(σ, ρ). Consequently,
G
{1,2,...,k}(σ, ρ) = (g
1, g
1′, g
2, . . . , g
′k).
Finally we obtain
PÄ
G
{1,2,...,k}(σ
n, ρ
n) = (g
1, g
′1, g
2, . . . , g
k′)
ä≥
P\k i=1
{ σ
n∈
Sn,gi
, ρ
n∈
Sn,g′ i}
!
.
Lemma 9.
[Kammoun,
2019, Lemma 16] Letg
1, g
2∈
Gnk
. Assume that there exists ρ ∈
Snsuch that A
g2ρ = ρA
g1. If ρ has a fixed point on any non-trivial connected component of g
1, then
Sn,g1
∩
Sn,g2
= ∅ or A
g1= A
g2.
Lemma 10.
For any graph g ∈
Gnk
having f loops, p non-trivial connected components and v non-isolated vertices, for any random permutation σ
nwith conjugation invariant distribution on
Sn,
P
(σ
n∈
Sn,g) ≤
P(σ
n(1) = 1, . . . , σ
n(f ) = f )
n−p v−p
(v − p)! ≤ 1
n−p v−p
(v − p)! .
Proof. It is an adaptation of the proof of [Kammoun,
2019, Corollary 17]. By conjugation invariance, onecan suppose without loss of generality that the loops of g are (1, 1), (2, 2), . . . (f, f ) and the set of non isolated vertices of g are { 1, 2, . . . , v } .
If there exist i, j, l, with j 6 = l such that { (i, j) ∪ (i, l) } ⊂ E
gor { (j, i) ∪ (l, i) } ⊂ E
gthen
Sn,g= ∅ . Therefore, if
Sn,g6 = ∅ , then non-trivial connected components of g having w vertices are either cycles of length w or isomorphic to g
w, where A
gw= [
1j=i+1]
1≤i,j≤w.
Let g ∈
Gnk
such that
Sn,g6 = ∅ . Fix p vertices x
1= 1, x
2= 2, . . . , x
f= f, x
f+1, . . . , x
peach belonging to a different non-trivial connected components of g. Let x
p+1< x
p+2< · · · < x
vbe such that { x
p+1, . . . , x
v} = { 1, 2, . . . , v } \ { x
1, . . . x
p} be the other non-isolated vertices. Let
F = { (y
i)
p+1≤i≤v; y
i∈ { 1, 2, . . . , n } \ { x
1, . . . x
p} pairwise distinct } . Given y = (y
i)
p+1≤i≤v∈ F , we denote by g
y∈
Gnk
the graph isomorphic to g obtained by fixing the labels of x
1, x
2, . . . , x
pand by changing the labels of x
iby y
ifor p + 1 ≤ i ≤ v. Since non trivial connected components of g of length w are either cycles or isomorphic to g ¯
w, if y 6 = y
′∈ F , then g
y6 = g
y′and by Lemma
9, Sn,gy
∩
Sn,gy′
= ∅ . Since σ
nis conjugation invariant, we have
P(σ
n∈
Sn,gy
) =
P(σ
n∈
Sn,gy′
) =
P(σ
n∈
Sn,g). Remark also that for any y ∈ F and any i ≤ f , (i, i) is a loop of g
y. Thus,
Sn,gy
⊂ { σ ∈
Sn; ∀ i ≤ f, σ
n(i) = i } and thus
P(σ
n∈
Sn,g) =
P
y∈FP
(σ
n∈
Sn,gy
)
card(F ) =
P(σ
n∈ ∪
y∈FSn,gy
)
card(F ) ≤
P(σ
n(1) = 1, . . . , σ
n(f ) = f )
n−p v−p
(v − p)!
≤ 1
n−p v−p
(v − p)! .
Lemma 11.
Let σ
nbe a random permutation with conjugation invariant distribution on
Snsuch that, for any k ≥ 1, lim
n→∞EÅ
#√1nσn
kã
= 0. Then, for any f ≥ 1,
P
(σ
n1(1) = 1, . . . , σ
n1(f ) = f ) = o(n
−f2).
Lemma 12.
For any p ≥ 1, let g be a graph with p non trivial components each having 2 vertices. Assume that at least one of these components is a cycle. Then for any random permutation σ
nwith conjugation invariant distribution on
Sn,
P
(σ
n∈
Sn,g) ≤
P(c
1(σ
n) = 2)
n−p p
p! .
Proof. Remark that by conjugation invariance, one can suppose without loss of generality that the set of non isolated vertices of g are { 1, 2, . . . , 2p } and that (1, 2), (2, 1) ∈ E
g. Using the same definitions as the previous proof with f = 0 and v = 2p and by choosing x
1= 1, we have
Sn,gy
⊂ { σ ∈
Sn; c
1(σ) = 2 } . Thus,
P(σ
n∈
Sn,g) =
P
y∈FP
(σ
n∈
Sn,gy
)
card(F ) =
P(σ
n∈ ∪
y∈FSn,gy
)
card(F ) ≤
P(c
1(σ
n) = 2)
card(F ) =
P(c
1(σ
n) = 2)
n−p p
p! .
By the previous combinatorial lemmas, we get that the main contribution will come from the following subset of graphs. Let T
kn⊂
Gnk
be the set of graphs g having exactly k non trivial component each having one edge and two vertices.
For example, T
13=
1 2
,
2 1,
1 3,
3 1,
2 3,
3 2
. Let T
“kbe the equivalence class of the graphs of ∪
nT
kn.
Their contribution is as follows.
Lemma 13.
For any p ≥ 1, n ≥ 2p and any graph g ∈ T
pn, for any random permutation σ
nwith conjugation invariant distribution on
Sn,
1
n−p p
p!
Ç
1 − p
2− p
n − 1 − p
P(σ
n(1) = 1)
å≤
P(σ
n∈
Sn,g) ≤ 1
n−p p
p! .
Proof. The upper bound is due to Lemma
10with v = 2p. Using the conjugation invariance, one can suppose without loss of generality that E
g= { (1, i
1), (2, i
2), . . . , (p, i
p) } where i
j> p are all distinct. Let
Spn
= { σ ∈
Sn, ∀ i ≤ p, σ(i) > p } .
Remark that
P(σ
n∈
Sn,g| σ
n∈
Sn\
Spn) = 0. If
P(σ
n∈
Spn) = 0, then necessarily by conjugation invariance, 1 −
pn2−−1p− p
P(σ
n(1) = 1) ≤ 0.
Suppose now that
P(σ
n∈
Spn) 6 = 0. We obtain
P(σ
n∈
Sn,g) =
P(σ
n∈
Sn,g| σ
n∈
Spn)
P(σ
n∈
Spn). Using again the conjugation invariance, we obtain
P
(σ
n∈
Sn,g| σ
n∈
Spn) = 1
n−p p
p!
and
P
(σ
n∈
Spn) = 1 −
P(σ
n∈
Sn\
Spn)
≥ 1 −
Xp i=1P
(σ
n(i) ≤ p)
= 1 − p
ÇP
(σ
n(1) = 1) + (1 −
P(σ
n(1) = 1))(p − 1) n
å
≥ 1 − p
2− p
n − 1 − p
P(σ
n(1) = 1).
2.2 Proof of Proposition 3
Proof. We will adapt the proof of [Kammoun,
2019, Lemma 14]. Letv
1≥ 1 be fixed. In the case k = 1, since C
1= 1, (*) holds if we have:
∀ ˆ g
1, ˆ g
2∈
Gˆ
v1
,
P(( ˆ G
11(σ
n, ρ
n), G ˆ
21(σ
n, ρ
n)) = (ˆ g
1, ˆ g
2))) = C
ˆg1,ˆg2n + o
Å
1 n
ã
and
Xˆ g1,ˆg2∈Gˆv
1
C
gˆ1,ˆg2= C
1= 1.
Let ˆ g
1, g ˆ
2∈
Gˆ
v1
be two unlabeled graphs having respectively p
1and p
2connected components and v ≤ 2v
1vertices. We denote by
p
n(ˆ g
1, ˆ g
2) :=
P(( ˆ G
11(σ
n, ρ
n), G ˆ
21(σ
n, ρ
n)) = (ˆ g
1, g ˆ
2)).
Let B
ˆgn1,ˆg2be the set of couples (g
1, g
2) ∈ (
Gnv1
)
2having the same non-isolated vertices such that 1 is a non-isolated vertex of both graphs and, for i ∈ { 1, 2 } , the equivalence class of g
iis g ˆ
iand there exists σ, ρ such that G
11(σ, ρ) = g
1and G
21(σ, ρ) = g
2. By Lemma
8and H
1, we have
p
n(ˆ g
1, ˆ g
2) =
X(g1,g2)∈Bnˆg
1,ˆg2
P
(( G
11(σ
n, ρ
n), G
21(σ
n, ρ
n)) = (g
1, g
2))
=
X(g1,g2)∈Bnˆg
1,ˆg2
P
(σ
n∈
Sn,g1, ρ
n∈
Sn,g2) =
X(g1,g2)∈Bˆgn
1,ˆg2
P
(σ
n∈
Sn,g1)
P(ρ
n∈
Sn,g2) (3)
Starting from (3), we now distinguish different cases, depending on the structure of g ˆ
1and g ˆ
2.
• Case 1: ˆ g
1and ˆ g
2have respectively f
1and f
2loops i.e edges of type (i, i) with f
1+ f
2> 0. Then 2p
1− f
1≤ v and 2p
2− f
2≤ v. Consequently, by Lemmas
10and
11,p
n(ˆ g
1, g ˆ
2) = o
Ån
−f12−f2ã X
(g1,g2)∈Bgnˆ
1,ˆg2
1
n−p1 v−p1
(v − p
1)!
1
n−p2 v−p2
(v − p
2)!
= card(B
ˆgn1,ˆg2)
n−p1
v−p1
(v − p
1)!
nv−−pp22
(v − p
2)! o
Ån
−f12−f2 ã≤
n−1 v−1
v!
2o
Ån
−f12−f2 ãn−p1
v−p1
(v − p
1)!
nv−−pp22
(v − p
2)! = n
v−1−(v−p1+v−p2)o
Ån
−f12−f2 ã= o(n
−1).
• Case 2: g ˆ
1and g ˆ
2do not contain any loop, so that p
1≤
v2and p
2≤
v2. Then, again by Lemma
10,p
n(ˆ g
1, ˆ g
2) ≤
X(g1,g2)∈Bnˆg
1,ˆg2
1
n−p1
v−p1
(v − p
1)!
1
n−p2
v−p2
(v − p
2)!
= card(B
ˆgn1,ˆg2)
n−p1
v−p1
(v − p
1)!
nv−−pp22(v − p
2)!
≤
n−1 v−1
v!
2n−p1
v−p1
(v − p
1)!
nv−−pp22