• Aucun résultat trouvé

A product of invariant random permutations has the same small cycle structure as uniform

N/A
N/A
Protected

Academic year: 2021

Partager "A product of invariant random permutations has the same small cycle structure as uniform"

Copied!
13
0
0

Texte intégral

(1)

HAL Id: hal-02309521

https://hal.archives-ouvertes.fr/hal-02309521

Preprint submitted on 9 Oct 2019

HAL

is a multi-disciplinary open access archive for the deposit and dissemination of sci- entific research documents, whether they are pub- lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.

L’archive ouverte pluridisciplinaire

HAL, est

destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

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�

(2)

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 lengthk6=k.

1 Introduction and main results

We denote by

Sn

the group of permutations of { 1, . . . , n } , by #

k

σ the number of cycles of σ of length k, by # σ the total number of cycles of σ and by tr(σ) := #

1

σ.

The cycle structure of a permutation chosen uniformly among the symmetric group

Sn

is well understood (see e.g. [Arratia, Tavaré, and Barbour,

2003] for detailed results). In particular, the following classical

result holds:

Theorem 1.

[Arratia, Barbour, and Tavaré,

2000, Theorem 3.1] If

σ

n

follows the uniform distribution on

Sn

then for any k ≥ 1,

(#

1

σ

n

, . . . , #

k

σ

n

) −−−→

nd

→∞

η

k

:= (ξ

1

, ξ

2

, . . . , ξ

k

), (1)

where −−−→

n d

→∞

denotes the convergence in distribution, ξ

1

, ξ

2

, . . . ξ

k

are independent and the distribution of ξ

d

is 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

n

)

n1

be a sequence of random permutations such that for any n ≥ 1, σ

n

Sn

. For any k ≥ 1, let t

nk

:= #

k

(

Qmℓ=1

σ

n

). Assume that

- (H

1

) For any n ≥ 1, (σ

1n

, . . . , σ

n

) are independent.

- For any n ≥ 1 and 1 ≤ ℓ ≤ m, for any σ ∈

Sn

,

σ

1

σ

n

σ =

d

σ

n

, (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.

(3)

- There exists 1 ≤ i < j ≤ m such that for any k ≥ 1,

n

lim

→∞E

Ç

#

1

σ

in

√ n

åk!

= 0 and lim

n→∞E

Ç

#

1

σ

jn

√ n

åk!

= 0, (H

3

)

n

lim

→∞

E

(#

2

σ

ni

)

n = 0 and lim

n→∞

E

(#

2

σ

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 any

θ ∈ [0, 1], the Ewens distribution with parameter θ satisfies the convergences required in H

3

and 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

2

and 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

n

follows the uniform distribution, under H

1

, the product also follows the uniform distribution and Theorem

2

is 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

n1

ρ

n

is conjugation invariant and

#(σ

n1

ρ

n

)

6

n

−−−→

n→∞d

0.

Then for any s ∈

R

,

P

Ç

LCS(σ

n

, ρ

n

) − 2 √ n

6

n ≤ s

å

−−−→

n

→∞

F

2

(s),

where F

2

is 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 1

can 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

2

as follows.

Corollary 4.

Under H

1

, H

2

, H

3

and H

4

, for any k ≥ 1,

Ä

tr(

Qmi=1

σ

n

), tr((

Qmi=1

σ

n

)

2

), . . . , tr((

Qmi=1

σ

n

)

k

)

ä

converges in distribution to (ξ

1

, ξ

1

+ 2ξ

2

, . . . ,

Pd|k

d

), where ξ

1

, ξ

2

, . . . are independent and the distribution of ξ

d

is Poisson of parameter

1d

.

The optimality of conditions H

3

and H

4

will 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).

(4)

2 Proof of results

We begin with a few preliminary remarks and simplifications.

First of all, the equivalence between Theorem

2

and Corollary

4

is 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=1

1ci(σ)|k

=

X

j|k

Xn i=1

1ci(σ)=j

=

X

j|k

j #

j

σ.

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

2

in the case where all permutations are conjugation invariant. Indeed, if we choose τ

n

uniform and independent of the σ-algebra generated by (σ

n

)

1m

, the cycle structure of

Qmℓ=1

σ

n

is the same as

τ

n1 Ym ℓ=1

σ

n

!

τ

n

= (τ

n1

σ

n1

τ

n

)

Ym ℓ=2

n1

σ

n

τ

n

) = (τ

d n1

σ

1n

τ

n

)

Ym ℓ=2

σ

n

and (τ

n1

σ

1n

τ

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

n1

and σ

2n

will be denoted respectively by σ

n

and ρ

n

.

2.1 Preliminary results

To prove Theorem

2, we will use the same objects introduced in [Kammoun,2019, pages 12-13] where one

can 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

n1

ρ

n

rather than σ

n

ρ

n

: for any k ≥ 1, we define ˜ t

nk

:= #

k

n1

ρ

n

). Under H

2

, σ

n

=

d

σ

n1

and consequently under H

1

and 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

Gn

k

the set of oriented simple graphs with vertices { 1, 2, . . . , n } and having exactly k edges. Given g ∈

Gn

k

, we denote by E

g

the set of its edges and by A

g

:= [

1(i,j)Eg

]

1i,jn

its 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

g

does not contain any edge of the form (i, j) or (j, i) nor a loop (i, i). Let g ∈

Gn

k

, we denote by g ˜ the graph obtained from g after removing isolated vertices.

• We say that two oriented simple graphs g

1

and g

2

are isomorphic if one can obtain g

2

by changing the labels of the vertices of g

1

. In particular, if g

1

, g

2

Gnk

then g

1

, g

2

are isomorphic if and only if there exists a permutation matrix σ such that A

g1

σ = σA

g2

.

• Let R be the equivalence relation such that g

1

R g

2

if g ˜

1

and g ˜

2

are isomorphic. We denote by

G

ˆ

k

:=

n1Gn

k

/ R the set of equivalence classes of ∪

n1Gnk

for 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

mk

m(σ,ρ)

(σ, ρ)) the cycle of σ

1

◦ ρ containing m, so that k

m

(σ, ρ) := c

m

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

mk

m(σ,ρ)

(σ, ρ) are pairwise distinct and j

1m

(σ, ρ), j

2m

(σ, ρ), . . . , j

km

m(σ,ρ)

(σ, ρ) are pairwise distinct. For sake of simplicity, when it is clear, we will use the notations

k

m

, i

ml

and j

lm

instead of k

m

(σ, ρ), i

ml

(σ, ρ) and j

lm

(σ, ρ).

(5)

• We denote by G

1m

(σ, ρ) ∈

Gnk

m

and G

2m

(σ, ρ) ∈

Gnk

m

the graphs with vertices { 1, . . . , n } such that E

Gm

1 (σ,ρ)

= { (i

m1

, j

mkm

) }

[

km[1 l=1

{ (i

ml+1

, j

lm

) }

!

and E

Gm

2 (σ,ρ)

=

km

[

l=1

{ (i

ml

, j

lm

) }

and by g

σ

the 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=1

E

G

i(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

2

is equivalent to the following: under the same hypotheses, for any v

1

, v

2

, v

3

, . . . , v

k

≥ 1,

n

lim

→∞

X

ˆ

gigiGˆvi,1ik

n

kPÄ

G ˆ

{1,2,...,k}

n

, ρ

n

) = (ˆ g

1

, g ˆ

1

, g ˆ

2

, . . . g ˆ

k

)

ä

= C

v1,v2,...,vk

, (*)

where C

v1,v2,...,vk

is a constant independent of the laws of the permutations. Note that, for any v

i

≥ 1,

G

ˆ

v

i

and therefore the number of terms of the sum is finite.

For example, if we take P (x) = x

2

, we have

EÄ

P

Ä

ˆ t

n1ää

=

E

Ñ Xn i=1

1ci−1ρ)=1

!2é

=

Xn i=1

EÄ1ci−1ρ)=1

ä

+

Xn i6=j

EÄ1ci−1ρ)=11cj−1ρ)=1

ä

= n

EÄ1c1−1ρ)=1

ä

+ (n

2

− n)

EÄ1c1−1ρ)=11c2−1ρ)=1

ä

−−−→

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→∞d

C

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] If

m

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

m

(ρ, σ) = j

km

m(σ,ρ)ℓ+1

(σ, ρ), ∀ 1 ≤ ℓ ≤ k

m

(σ, ρ), i

m

(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

Gm

1 (σ,ρ)

= A

T

G2ρ(m)−1−1)

.

(6)

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

5

and

6), it suffices to prove that if all non trivial

connected 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

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

1η1

} or i

1η

∈ { j

11

, j

21

, . . . , j

1η1

}} . 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

1

such 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

1

∈ { i

11

, i

12

, . . . , i

11

} , then there exists ℓ

′′

< ℓ

such that j

1

= i

1′′

and since the component of G

21

(σ, ρ) containing i

1

has two vertices and by definition (i

1

, j

1

) and (i

1′′

, j

1′′

) are two edges of G

21

(σ, ρ), then j

1′′

= i

1

. Since (i

1

, j

11

) = (j

1′′

, j

11

) and (i

1′′+1

, j

1′′

) are edges of G

11

(σ, ρ) and since G

11

(σ, ρ) has only connected components of size 2, we have necessarily i

1′′+1

= j

11

. 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

1′′+1

, j

1′′+1

) = (j

11

, j

1′′+1

) = (j

11

, j

11

) is an edge of G

21

(σ, ρ) and if ℓ

′′

= ℓ

− 1, then (i

1′′+1

, j

1′′

) = (j

11

, j

1′′

) = (j

11

, j

11

) is an edge of G

11

(σ, ρ). This implies that ℓ

− 1 ∈ A, which is absurd. i

1

∈ { j

11

, j

21

, . . . , j

11

} can be treated using the same techniques and one can extend easily to ℓ

= 2.

We now introduce the following notation : given g ∈

Gn

k

, we denote by

Sn,g

:= { σ ∈

Sn

; ∀ (i, j) ∈ E

g

, σ(i) = j } .

In other words,

Sn,g

is 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

∈ ∪

Gn

and let g, g

be such that E

g

= ∪

kℓ=1

E

gi

and E

g

= ∪

kℓ=1

E

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,g

i

, ρ

n

Sn,g i

}

!

=

PÄ

G

{1,2,...,k}

n

, ρ

n

) = (g

1

, g

1

, g

2

, . . . , g

k

)

ä

=

P

G

1{1,2,...,k}

n

, ρ

n

) = g, G

2{1,2,...,k}

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

, ρ

) is a subset of g

ρ

, so that G

2m

, ρ

) = g

m

⇒ ρ

Sn,g

m

, and that G

1m

, ρ

) is a subset of g

σ

, so that

G

1m

, ρ

) = g

m

⇒ σ

Sn,g

m

. Consequently,

PÄ

G

{1,2,...,k}

n

, ρ

n

) = (g

1

, g

1

, g

2

, . . . , g

k

)

ä

P

\k i=1

{ σ

n

Sn,g

i

, ρ

n

Sn,g i

}

!

.

Now suppose that there exists ρ

, σ

such that

G

{1,2,...,k}

, ρ

) = (g

1

, g

1

, g

2

, . . . , g

k

).

(7)

Let σ, ρ such that σ ∈ ∩

ki=1Sn,g

i

and ρ ∈ ∩

ki=1Sn,g

i

. By definition and by iteration on ℓ, one can check that for any ℓ

≤ k, i

, ρ

) = i

(σ, ρ) and j

, ρ

) = j

(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}

n

, ρ

n

) = (g

1

, g

1

, g

2

, . . . , g

k

)

ä

P

\k i=1

{ σ

n

Sn,g

i

, ρ

n

Sn,g i

}

!

.

Lemma 9.

[Kammoun,

2019, Lemma 16] Let

g

1

, g

2

Gn

k

. Assume that there exists ρ ∈

Sn

such that A

g2

ρ = ρA

g1

. If ρ has a fixed point on any non-trivial connected component of g

1

, then

Sn,g

1

Sn,g

2

= ∅ or A

g1

= A

g2

.

Lemma 10.

For any graph g ∈

Gn

k

having f loops, p non-trivial connected components and v non-isolated vertices, for any random permutation σ

n

with conjugation invariant distribution on

Sn

,

P

n

Sn,g

) ≤

P

n

(1) = 1, . . . , σ

n

(f ) = f )

np vp

(v − p)! ≤ 1

np vp

(v − p)! .

Proof. It is an adaptation of the proof of [Kammoun,

2019, Corollary 17]. By conjugation invariance, one

can 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

g

or { (j, i) ∪ (l, i) } ⊂ E

g

then

Sn,g

= ∅ . Therefore, if

Sn,g

6 = ∅ , 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

]

1i,jw

.

Let g ∈

Gn

k

such that

Sn,g

6 = ∅ . Fix p vertices x

1

= 1, x

2

= 2, . . . , x

f

= f, x

f+1

, . . . , x

p

each belonging to a different non-trivial connected components of g. Let x

p+1

< x

p+2

< · · · < x

v

be 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+1iv

; y

i

∈ { 1, 2, . . . , n } \ { x

1

, . . . x

p

} pairwise distinct } . Given y = (y

i

)

p+1iv

∈ F , we denote by g

y

Gn

k

the graph isomorphic to g obtained by fixing the labels of x

1

, x

2

, . . . , x

p

and by changing the labels of x

i

by y

i

for 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

y

6 = g

y

and by Lemma

9, Sn,g

y

Sn,g

y

= ∅ . Since σ

n

is conjugation invariant, we have

P

n

Sn,g

y

) =

P

n

Sn,g

y

) =

P

n

Sn,g

). Remark also that for any y ∈ F and any i ≤ f , (i, i) is a loop of g

y

. Thus,

Sn,g

y

⊂ { σ ∈

Sn

; ∀ i ≤ f, σ

n

(i) = i } and thus

P

n

Sn,g

) =

P

yFP

n

Sn,g

y

)

card(F ) =

P

n

∈ ∪

yFSn,g

y

)

card(F ) ≤

P

n

(1) = 1, . . . , σ

n

(f ) = f )

np vp

(v − p)!

≤ 1

np vp

(v − p)! .

Lemma 11.

Let σ

n

be a random permutation with conjugation invariant distribution on

Sn

such that, for any k ≥ 1, lim

n→∞E

Å

#1nσn

kã

= 0. Then, for any f ≥ 1,

P

n1

(1) = 1, . . . , σ

n1

(f ) = f ) = o(n

f2

).

(8)

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

n

with conjugation invariant distribution on

Sn

,

P

n

Sn,g

) ≤

P

(c

1

n

) = 2)

np 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,g

y

⊂ { σ ∈

Sn

; c

1

(σ) = 2 } . Thus,

P

n

Sn,g

) =

P

yFP

n

Sn,g

y

)

card(F ) =

P

n

∈ ∪

yFSn,g

y

)

card(F ) ≤

P

(c

1

n

) = 2)

card(F ) =

P

(c

1

n

) = 2)

np p

p! .

By the previous combinatorial lemmas, we get that the main contribution will come from the following subset of graphs. Let T

kn

Gn

k

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

k

be the equivalence class of the graphs of ∪

n

T

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

n

with conjugation invariant distribution on

Sn

,

1

np p

p!

Ç

1 − p

2

− p

n − 1 − p

P

n

(1) = 1)

å

P

n

Sn,g

) ≤ 1

np p

p! .

Proof. The upper bound is due to Lemma

10

with 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

n

Sn,g

| σ

n

Sn

\

Spn

) = 0. If

P

n

Spn

) = 0, then necessarily by conjugation invariance, 1 −

pn21p

− p

P

n

(1) = 1) ≤ 0.

Suppose now that

P

n

Spn

) 6 = 0. We obtain

P

n

Sn,g

) =

P

n

Sn,g

| σ

n

Spn

)

P

n

Spn

). Using again the conjugation invariance, we obtain

P

n

Sn,g

| σ

n

Spn

) = 1

np p

p!

and

P

n

Spn

) = 1 −

P

n

Sn

\

Spn

)

≥ 1 −

Xp i=1

P

n

(i) ≤ p)

= 1 − p

Ç

P

n

(1) = 1) + (1 −

P

n

(1) = 1))(p − 1) n

å

≥ 1 − p

2

− p

n − 1 − p

P

n

(1) = 1).

(9)

2.2 Proof of Proposition 3

Proof. We will adapt the proof of [Kammoun,

2019, Lemma 14]. Let

v

1

≥ 1 be fixed. In the case k = 1, since C

1

= 1, (*) holds if we have:

∀ ˆ g

1

, ˆ g

2

G

ˆ

v

1

,

P

(( ˆ G

11

n

, ρ

n

), G ˆ

21

n

, ρ

n

)) = (ˆ g

1

, ˆ g

2

))) = C

ˆg1g2

n + o

Å

1 n

ã

and

X

ˆ g1g2Gˆv

1

C

gˆ1g2

= C

1

= 1.

Let ˆ g

1

, g ˆ

2

G

ˆ

v

1

be two unlabeled graphs having respectively p

1

and p

2

connected components and v ≤ 2v

1

vertices. We denote by

p

n

(ˆ g

1

, ˆ g

2

) :=

P

(( ˆ G

11

n

, ρ

n

), G ˆ

21

n

, ρ

n

)) = (ˆ g

1

, g ˆ

2

)).

Let B

ˆgn1g2

be the set of couples (g

1

, g

2

) ∈ (

Gnv

1

)

2

having 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

i

is g ˆ

i

and there exists σ, ρ such that G

11

(σ, ρ) = g

1

and G

21

(σ, ρ) = g

2

. By Lemma

8

and H

1

, we have

p

n

(ˆ g

1

, ˆ g

2

) =

X

(g1,g2)Bnˆg

1g2

P

(( G

11

n

, ρ

n

), G

21

n

, ρ

n

)) = (g

1

, g

2

))

=

X

(g1,g2)Bnˆg

1g2

P

n

Sn,g1

, ρ

n

Sn,g2

) =

X

(g1,g2)Bˆgn

1g2

P

n

Sn,g1

)

P

n

Sn,g2

) (3)

Starting from (3), we now distinguish different cases, depending on the structure of g ˆ

1

and g ˆ

2

.

• Case 1: ˆ g

1

and ˆ g

2

have respectively f

1

and f

2

loops 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

10

and

11,

p

n

(ˆ g

1

, g ˆ

2

) = o

Å

n

−f12−f2

ã X

(g1,g2)Bgnˆ

1g2

1

np1 vp1

(v − p

1

)!

1

np2 vp2

(v − p

2

)!

= card(B

ˆgn1g2

)

np1

vp1

(v − p

1

)!

nvpp2

2

(v − p

2

)! o

Å

n

−f12−f2 ã

n1 v1

v!

2

o

Å

n

−f12−f2 ã

np1

vp1

(v − p

1

)!

nvpp2

2

(v − p

2

)! = n

v1(vp1+vp2)

o

Å

n

−f12−f2 ã

= o(n

1

).

• Case 2: g ˆ

1

and g ˆ

2

do not contain any loop, so that p

1

v2

and p

2

v2

. Then, again by Lemma

10,

p

n

(ˆ g

1

, ˆ g

2

) ≤

X

(g1,g2)Bnˆg

1g2

1

np1

vp1

(v − p

1

)!

1

np2

vp2

(v − p

2

)!

= card(B

ˆgn1g2

)

np1

vp1

(v − p

1

)!

nvpp22

(v − p

2

)!

n1 v1

v!

2

np1

vp1

(v − p

1

)!

nvpp2

2

(v − p

2

)! = O

Ä

n

v1(vp1+vp2)ä

.

Therefore, if p

1

<

v2

, as p

1

v21

we have

p

n

(ˆ g

1

, ˆ g

2

) = O(n

32

).

Références

Documents relatifs

Prove that the gamma distribution is actually a continuous probability distributions that is it is non-negative over x &gt; 0 and its integral over R + is equal to 1.. Exercise 8

One idea is to compare how rare primes are with how rare cycles are: We know that roughly one out of every log x integers up to x is prime, and that exactly one in every N

Minimum percentage targets (with standard deviations) for representative reserve networks within eight mammal provinces in Canada using three sample plot sizes (diamonds: 13,000 km

L’accès aux archives de la revue « Rendiconti del Seminario Matematico della Università di Padova » ( http://rendiconti.math.unipd.it/ ) implique l’accord avec les

We then extend the domain to the infinite line limit, remove the regularization, and construct a traveling wave solution for the original set of equations satisfying the

The purpose of this work is to provide more concrete formulas (in terms of integrals of certain derivatives) for the invariant inner products, in the special cases

First introduced by Faddeev and Kashaev [7, 9], the quantum dilogarithm G b (x) and its variants S b (x) and g b (x) play a crucial role in the study of positive representations

This paper is organized as follows: Section 2 describes time and space discretiza- tions using gauge formulation, Section 3 provides error analysis and estimates for the