ASYMPTOTIC NORMALITY OF PATTERN COUNTS IN CONJUGACY CLASSES

. We prove, under mild conditions on fixed points and two cycles, the asymptotic normality of vincular pattern counts for a permutation chosen uniformly at random in a conjugacy class. Additionally, we prove that the limiting variance is always non-degenerate for classical pattern counts. The proof uses weighted dependency graphs.

1. Introduction 1.1.Background.A natural question in the context of random combinatorial structures is the asymptotic behaviour of the number of substructures of a given type.In general, we expect such numbers to be asymptotically normal or Poisson distributed, but proving it might be difficult.For permutations, a good notion of substructures is that of permutation patterns.Unsurprisingly, the asymptotic normality of patterns counts (either classical, consecutive or vincular patters) in uniform random permutations has been extensively studied.We refer for example to [Bón10, JNZ15, Hof18, EZ20, Jan23] for such results.
More recently, there has been some interest in extending these results to other models of random permutations, and in particular to uniform random permutations in some conjugacy classes.In this spirit, asymptotic normality results have been obtained for specific patterns: see [Ful98,KL20] for descent counts and [FKL22] for the number of peaks (the latter is a sum of vincular pattern counts).These results use minimal assumptions on the conjugacy class under consideration.In another direction, the second author [Kam22] and independently Hamaker and Rhoades, [HR22, Theorem 8.8] have both proved asymptotic normality results for general patterns, but with some restrictions on the chosen conjugacy classes.In the latter reference, the problem was raised of proving the asymptotic normality of pattern counts assuming only the convergence of the proportion of fixed points and 2-cycles in our permutations [HR22, Problem 9.9].We solve this problem in the present paper; see Theorem 1 below.Additionally, we prove that the limiting Gaussian distribution is always non-degenerate for classical pattern counts (Theorem 2).
Note: In parallel to this work, the asymptotic normality of classical pattern counts in random permutations in conjugacy classes has been obtained by Dubach [Dub24] by a different method.His method only applies to classical patterns (and not to the general case of vincular patterns that we treat here), but can provide stronger results in this case (speed of convergence and large deviation estimates).It can also be used to analyze other types of statistics on random permutations in conjugacy classes, such as longest monotone subsequences, the Robinson-Schensted shape, and the number of records; all of these are out of reach with the method of weighted dependency graphs used in this paper.
Definition 1.Let π be a permutation of size k and A a subset of [k − 1].Then an occurrence of the vincular pattern (π, A) in a permutation σ is a tuple ) is an occurrence of π in σ , i.e.
• for all s in A, we have i s+1 = i s + 1.
Vincular patterns have been introduced in [BS00].They encompass the notions of classical and consecutive patterns.Indeed, when A = ∅, there are no constraints on indices being consecutive, and we recover the notion of classical patterns.On the other hand, when A = [k − 1], all indices must be consecutive, and we recover consecutive patterns.Here are some examples of occurrences of vincular patterns in the literature.
• A peak is an occurrence of (132, {1, 2}) or (231, {1, 2}).Symmetrically, a valley is an occurrence of (213, {1, 2}) or (312, {1, 2}).• The longest alternating subsequence of a permutation equals, up to an additive term equal to 1 or 2, the number of valleys, plus the number of peaks of the permutation.In other terms, it is almost a linear combination of vincular patterns (in that case, consecutive patterns).This result and some equivalent variants were obtained separately by different authors.We refer for example to Houdré and Restrepo [HR10] and Romik [Rom11].• The generating function of the vincular pattern (312, {1}) (together with other statistics) occurs as partition function of a model of statistical physics called PASEP (partially asymmetric simple exclusion process); see [JV11].
1.3.Our main result: asymptotic normality of vincular pattern counts for uniform random permutation in conjugacy classes.Throughout this article, we fix a vincular pattern (π, A) such that π is of size k.We let X (π,A) (σ ) be the number of the occurrences of (π, A) in σ and we define m by m = k − |A|, which is the number of blocks in (π, A).
For an (integer) partition λ of n, we let C λ be the conjugacy class indexed by λ in the symmetric group S n , and σ λ be a uniform random permutation in C λ .Our main result is the following statement, where m i (λ n ) denotes the number of parts of size i in λ n .
Theorem 1.For any vincular pattern (π, A), there exist two polynomial functions f and g such that if

Moreover, we have convergence of all moments.
We prove Theorem 1 in Section 4. A difficulty, compared to the case of uniform random permutations in the whole symmetric group S n , is that, when working in a given conjugacy class, occurrences of patterns at disjoint sets of positions are no longer independent events.Our proof thus relies then on a technique known as weighted dependency graphs introduced in [Fér18] in order to control higher cumulants for sums of random dependent variables.The main ingredient is Theorem 3 (proved in Section 3), which states the existence of a weighted dependency graph structure for the random permutation σ λ n .Both the proofs of Theorem 3 and of Theorem 1, starting from Theorem 3, require delicate combinatorial arguments, to bound complicated multi-indexed sums.
We believe that Theorem 3 is interesting in itself.It can be used to prove that other classes of permutation statistics are asymptotically normal: for example, it is immediate to adapt the proof of Theorem 1 to the number of excedances 1 , peaks and/or valleys (and thus to the length of the longest alternating subsequence).Also, the proof is readily adapted to establish joint convergence in Theorem 1, i.e. the limiting joint distribution of  is a multivariate Gaussian distribution for any family of patterns.
1.4.Non-degeneracy of the limiting Gaussian law for classical patterns.A natural problem following Theorem 1 is to determine whether the limiting law is degenerate or not, i.e. whether the limiting variance f (p 1 ) + p 2 g(p 1 ) is positive or equal to 0. In the case of a uniform permutation in S n , which corresponds 2 to p 1 = p 2 = 0, it is known that the limiting variance, f (0) in this case, is always positive; see [Hof18,Jan23].Since f and g are polynomial functions, this implies that f (p 1 ) + p 2 g(p 1 ) can only vanish when (p 1 , p 2 ) lives in a subvariety of codimension 1 of {(p 1 , p 2 ) ∈ R 2 + : p 1 + 2p 2 ≤ 1}.In other terms, we know that the limiting variance is generically positive.It would nevertheless be desirable to have a complete characterization of the degeneracy cases.
1 An excedance is a position i such that σ (i) > i. 2 Uniform random permutations in the whole symmetric group S n are of course not a special case of random permutations in conjugacy classes.However, it follows easily from the results of [HR22, Section 8] that the limiting variance for uniform random permutations is the same as for any conjugacy class with proportions of fixed points and 2 cycles tending to 0.
We could solve the question in the case of classical patterns, i.e. the case A = ∅, for which we prove that the limiting Gaussian law is always non-degenerate (except in the trivial case p 1 = 1, where most points of the permutation are fixed).For simplicity, we write π instead of (π, ∅).Our second main result is the following.
Note that in the case p 1 = 1, p 2 = 0, we can choose λ n = (1 n ), i.e. σ λ n is a.s. the identity permutation.Thus, in this case, the limiting variance f (1) is indeed 0.
The proof of this non-degeneracy statement is rather involved.It uses a method introduced in [Hof18] and [Fér20].The basic idea is to use a recursive construction of the random objects, say σ n , under consideration.Typically, such a recursive construction uses a random object of smaller size, say σ n−1 , and an extra source of randomness, say I. Then one can compute the variance of the quantity of interest, X(σ n ), by conditioning on I and using the law of total variance.If the recursive construction is well-chosen, then some terms in the law of total variance involve the variance of X(σ n−1 ), i.e., one can compute, or at least bound from below, the variance recursively.This is the basic principle, the details being quite subtle; see Section 5.

Preliminaries: weighted dependency graphs
In this section, we present the notion of weighted dependency graphs, which is the main tool in the proof of Theorem 1.The results presented in this section are taken from [Fér18].
2.1.Definition.We start by recalling the notion of mixed cumulants.The mixed cumulant of a family of random variables X 1 , X 2 , . . ., X r defined on the same probability space and having finite moments is defined as If X i = X for all i, we abbreviate κ(X 1 , X 2 , . . ., X r ) as κ r (X).This is the standard cumulant of a random variable.An important property of cumulants is the following: if {X 1 , X 2 , . . ., X r } can be written as a disjoint union of two mutually independent sets of random variables, then κ(X 1 , X 2 , . . ., X r ) = 0. Hence, cumulants can be seen as a kind of measure of dependency, and the definition of weighted dependency graphs is based on these heuristics.Another important and elementary property of cumulants is that X follows the normal distribution if and only if κ r (X) = 0 for every r ≥ 3.
In the sequel, a weighted graph is a graph with weights on its edges, belonging to (0, 1].Non-edges can be interpreted as edges of weight 0, so that a weighted graph can be equivalently seen as an assignment of weights in [0, 1] to the edges of the complete graph.All our definitions are compatible with this convention. For a weighted graph H, we define M(H) to be the maximal weight of a spanning tree of H, the weight of a spanning tree being the product of the weights of its edges (if H is disconnected, there is no spanning tree and as a consequence of the above convention, we have M(H) = 0).For example, for ε < 1, the weighted graph H of Fig. 1 satisfies M(H) = ε 4 .The following definition was introduced in [Fér18].
Example of a weighted graph H with a marked spanning tree of maximum weight in red.
Definition 2. Let V be a finite set and let {Y v , v ∈ V } be a family of random variables with finite moments, defined on the same probability space.We fix a sequence C = (C 1 , C 2 , • • • ) of positive real numbers and a real-valued function Ψ defined on multisets of elements of V .A weighted graph L with vertex set V is a (Ψ , C) weighted dependency graph for Here, L[U ] is the weighted graph with vertex set U and the same weights as L. In examples of weighted dependency graphs, Ψ and C are simple or universal quantities, so that the meaningful term is M(L[U ]).Note that the smaller the weights on edges are, the smaller M(L[U ]) is, i.e. the closer to independence are the corresponding random variables {Y v , v ∈ U }. Hence, the edge weights in a weighted dependency graph quantify in some sense the dependency between the random variables {Y v , v ∈ V }.

2.2.
A useful example: uniform random permutations in S n .Let n ≥ 1 be an integer, and τ be a uniform random permutation in the symmetric group S n .For (i, j) in [n] 2 , we let A i,j = 1[τ i = j].Note that variables A i,j and A k,ℓ cannot be both equal to 1 if (i, j) (k, ℓ) but either i = k or j = ℓ.If i k and j ℓ, we have , • if two pairs v 1 = (i 1 , j 1 ) and v 2 = (i 2 , j 2 ) satisfy either i 1 = i 2 or j 1 = j 2 , then they are linked in L by an edge of weight 1. • otherwise, they are linked in L by an edge of weight 1/n.
is a sequence that does not depend on n.

Cumulant bounds.
Weighted dependency graphs are useful to bound cumulants and prove asymptotic normality results.In particular, in this paper, we will use the following lemma [Fér18, Lemma 4.10] Lemma 4. Let L be a (Ψ , C = (C r ) r≥1 ) weighted dependency graph for a family of variables {Y v , v ∈ V }.Define R and T ℓ (for ℓ ≥ 1) as follows: where w(u, v) is the weight of the edge (u, v) in L.Then, for r ≥ 1, we have 2.4.Power of weighted dependency graphs.An important property of weighted dependency graphs is the following stability property.Consider a family of random variables {Y v , v ∈ V } with a (Ψ , C) weighted dependency graph L and fix some integer d ≥ 1.We are interested in monomials Y I := v∈I Y v of degree at most d, i.e.where I ∈ V d is a list of elements of V of length d (possibly with repetitions).This defines a new family of random variables {Y I , I ∈ V d }.It turns out that this family admits a natural weighted dependency graph inherited from that of {Y v , v ∈ V }.
To state this formally, we need to introduce some more notation.Let L be a weighted graph with vertex set V .The d-th power L d of L is a weighted graph with vertex set V d .If I and J are in V d , the weight between vertices I and J is where w(u, v) is the weight between u and v in L.
Besides, we say that a function Ψ defined on multisets of elements of a set ) for all multisets U 1 and U 2 (here and throughout the paper, ⊎ is the disjoint union of multisets, i.e. if an element belongs to both U 1 and U 2 with multiplicity m 1 and m 2 , it belongs to 1 ⊎ U 2 with multiplicity m 1 + m 2 ).Finally, a function Ψ defined on lists of V induces a function on lists of multisets of elements of V by setting The

A weighted dependency graph structure for uniform random permutation in conjugacy classes
Fix an integer partition λ n of size n, for (i, j) in [n] 2 , we introduce the random variables Note that the law of B i,j depends on λ n .Given a subset α = {(i t , j t ), t ≤ T } of [n] 2 , we write S(α) = {i 1 , . . ., i T , j 1 , . . ., j T } for the support of α.Furthermore, we denote by cc(V , E) the number of connected components of the graph (V , E).We also write CC(α) instead of cc(S(α), α); this is the number of components of the graph with edge set α and no isolated vertices.We now let ( 7) We define similarly Ψ on lists of elements of [n] 2 , and it is insensitive to the order and repetitions of arguments.
The choice of Ψ may seem surprising, but in fact, it is quite natural.Indeed, the joint moments M α := E( (i,j)∈α B i,j ) are nonzero if and only if (S(α), α) is a disjoint union of directed paths and cycles.In this case, M α is O(Ψ (α)), and this is optimal (in the sense that M α = Θ(Ψ (α)) if m k (λ) = Θ(n) for any fixed k).Furthermore, as proved in the next lemma, Ψ has been chosen to be super-multiplicative, so that we can use Proposition 5. Lemma 6.The function Ψ defined by (7) is super-multiplicative.
When α contains a cycle C of length at least 2, removing an edge of C from α does not change any quantity of (8).By symmetry, we may assume that neither α nor α ′ contain a cycle, except possibly loops.Then, writing loops(α) for the number of loops in α (counted with multiplicity), one has The latter holds true since for any multigraph G = (V , E), one has The goal of this section is to prove the following proposition.
Theorem 3.There is a universal sequence C = (C r ) r≥1 such that the following holds.For any n ≥ 1 and any λ ⊢ n, the complete graph L on [n] 2 with weights if {i, j} and {k, ℓ} are disjoint; 1 otherwise, is a (Ψ , C) weighted dependency graph for the family {B i,j , (i, j) ∈ [n] 2 }.
Fix r ≥ 1 and let α = {{(i t , j t ), 1 ≤ t ≤ r}}, be a multiset of size r of elements of [n] 2 .We want to show that (10) where M(L[α]) is defined in Section 2.1.We start with a lemma.
Lemma 7.For any α = {(i 1 , j 1 ), . . ., (i r , j r )}, we have Proof.By definition, the weighted graph L, and therefore also L[α], have only edges with weight 1 or n −1 .To form a spanning tree of maximal weight, we need to use as many edges of weight 1 as possible.It follows that obtained by keeping only edges of weight 1.It remains to relate connected components of L 1 [α] to those of (S(α), α).By construction, L 1 [α] has r vertices decorated with pairs (i t , j t ) and edges between pairs with a non-empty intersection.Consider a variant, denoted K 1 [α], where we have 2r vertices with decorations i t and j t (t running from 1 to r) and with two kinds of edges • for each t, the vertex decorated with i t is connected to that decorated with j t ; • vertices with equal decorations are connected.See Fig. 2 for an example.Contracting the first type of edges in Contracting the second type of edges instead gives the graph (S(α), α) considered above.Hence, both have the same number of connected components and the lemma follows.
We now prove this inequality.The idea is to represent σ λ as τ −1 • ρ • τ, where ρ is a fixed permutation in C λ and τ a uniform random permutation in S n .As in Section 2.2,  we write A i,j = 1[τ i = j] and we have By multilinearity of cumulants, one has Combining Proposition 3 together with Proposition 5 for m = 2, we know that L 2 is a (Ψ , D) weighted dependency graph for the family {A i,k A j,k ′ , (i, k, j, k ′ ) ∈ [n] 4 }, where • L 2 is the second power of the graph L defined in Proposition 3; namely, in the graph L 2 , the vertex associated with A i s ,k s A j s ,k ′ s is linked to that of A i t ,k t A j t ,k ′ t by an edge of weight 1 if {i s , j s } ∩ {i t , j t } ∅ or {k s , k ′ s } ∩ {k t , k ′ t } ∅, and by an edge of weight 1/n otherwise.
In particular, for any (i 1 , . . ., i r ), (j 1 , . . ., j r ) and (k 1 , . . ., k r ), we have The weighted graph L 2 has only edges with weight 1 and 1/n.Therefore for any subgraph formed by its edges of weight 1.
In the case we are interested in, we get where Let us illustrate this definition with an example.We take λ = (6, 2, 1) and choose ρ to be (1, 2, 3, 4, 5, 6)(7, 8)(9) (in the product of cycle notation).As in Fig. 2, we let r = 3, i 1 = j 2 = 7, j 1 = 4, i 2 = i 3 = j 3 = 9.We consider the term indexed by (k 1 , k 2 , k 3 ) = (9, 5, 4), yielding (ρ (in fact, only one of these intersections being nonempty would suffice for 2 to be linked to 3).One can check however that 1 is not linked to 3 since To help us in our analysis, we now introduce, for each r-tuple k , all on vertex set [r]×{1, 2}.For all these graphs, the idea is to label the vertex (s, 1) with the pair (i s , k s ) and (s, 2) with the pair (j s , ρ k s ).Then we put an edge between two vertices labelled by (x, y) and (x ′ , y ′ ) if the first coordinates in the labels coincide; , we add edges between (s, 1) and (s, 2) for any s ≤ r (for clarity, these extra edges will be drawn in blue).We continue our example above by representing the four associated graphs in Fig. 3.We note that a coincidence between a value in {i 1 , . . ., i r , j 1 , . . ., j r } and a value in the set {k 1 , . . ., k r , ρ k 1 , . . ., ρ k r } has no influence on any of the graphs considered here.
Comparing their definitions, we immediately see that G A k 1 ,...,k r can be obtained from G ∨ k 1 ,...,k r by merging, for each s ≤ r, the vertices (s, 1) and (s, 2) into a single vertex s.Since G ∨ k 1 ,...,k r has edges between (s, 1) and (s, 2) for any s ≤ r, this merge operation does not From left to right: the four graphs G (1) associated with the particular values of i 1 , . . ., i r , j 1 , . . ., j r , k 1 , . . ., k r given on page 12. change the number of connected components.Besides, again directly from definition, we see that connected components of G ∧ k 1 ,...,k r correspond to distinct pairs in the set With these observations, Eqs. ( 13) and ( 14) imply We now need the following easy, possibly well-known, graph theoretical lemma.
Lemma 8. Let V be a vertex set and E 1 , E 2 be two sets of edges on V .Then Proof The union E 1 ∪ E 2 may however contain cycles, so we only have an inequality Together with the standard inclusion-exclusion principle ; this is for example the case of the bottom edge in Fig. 3).Consequently, G ∧ k 1 ,...,k r has more connected components than this intersection.Combining these observations with Lemma 8, for any k 1 , . . ., k r , we have ).
We need a final lemma.
Lemma 9. Fix ρ in S n .For any graph G on vertex set [r]×{1, 2}, the number of lists ) .Thus the latter sum is bounded by a constant depending only on r, namely 2 ( 2r 2 ) .Inserting this inequality in (18), this proves (11) and thus Theorem 3.

Proof of asymptotic normality of vincular pattern counts
By the method of moments, convergence (1), together with moment convergence, hold if • and higher cumulants tend to 0 after normalization, i.e. κ r (X π (σ The existence of polynomial functions f and g such that the first item holds is proven in [HR22] (see Proposition 7.2 and Theorem 8.14 there).We will bound the cumulants using the weighted dependency graph structure of Section 3.
Note that we have We denote by I (π,A) the set of tuples (i 1 , . . ., i k , j 1 , . . ., j k ) as in the above sum.To bound the cumulant of X (π,A) (σ λ n ), we need a weighted dependency graph for the family of products . This is given by applying the stability property of weighted dependency graphs of Proposition 5 to the specific dependency graph of Theorem 3, as we now explain.
Let us recall the notation from Section 2.4, adapted to the present situation.If each of α 1 , . . ., α ℓ is a list of k elements (or k-list for short) of [n] 2 , we define where in the right-hand side, lists are seen as sets, and we recall that Ψ is defined in (7).We also consider the k-th power L k of the weighted graph L from Theorem 3. Concretely, L k has vertices indexed by k-lists of [n] 2 and its edge weights are given by 1 otherwise.
Using Proposition 5, there is a universal sequence D k = (C k r ) r≥1 such that L k is a (Ψ , D k ) weighted dependency graph for the set of random variables k t=1 B i t ,j t , indexed by klists α = ((i 1 , j 1 ), . . ., (i k , j k )) of elements of [n] 2 .We shall consider the restriction of this weighted graph to k-lists in I (π,A) : this is a (Ψ , D k ) weighted dependency graph for the set of random variables k t=1 B i t ,j t , indexed by I (π,A) .
To use Lemma 4, we want to bound the quantities R and (T ℓ ) ℓ≥1 , whose definitions are now recalled for this specific weighted dependency graph.
We start by bounding R.
Edges are chosen as follows: • in G (I) , we connect (s, 1) and (s, 2) for all s ≤ k; • in G (A) , we connect (s, 1) and (s + 1, 1) for s ∈ A; • in both graphs, we additionally connect vertices with equal labels.
Furthermore, let G (∩) be the intersection of G (I) and G (A) ; it only has edges connecting vertices with equal labels, and hence its number of components is the number of distinct values in {i 1 , . . ., i k , j 1 , . . ., j k }.Thus (19) can be rewritten as .
We now introduce the union G (∪) of G (I) and G (A) (union in terms of edge-set).Since it contains the edges {((s, 1), (s, 2)), s ≤ k} ∪ {((s, 1), (s + 1, 1)), s ∈ A}, it has at most m = k − #A connected components.Using Eq. ( 22) and Lemma 8, we get But, for a given graph G, there are at most n cc(G) lists (i 1 , . . ., i k , j 1 , . . ., j k ) in I (π,A) such that G (A) = G.Indeed, choosing a value i s or j s corresponding to a vertex of G forces the values of variables i t or j t in the same component.We finally get R ≤ 2 ( 2k 2 ) n m .□ The next step is to bound T ℓ for all ℓ ≥ 1.
Lemma 11.For any ℓ ≥ 1, there exists C ′′ k,ℓ (depending on the size k of the pattern (π, A) and on ℓ, but not on n) such that , the following holds true To prove this lemma, we need the following simple combinatorial fact to control the various terms in (20).
We write as usual β = ((i 1 , j 1 ), • • • , (i k , j k )).First we recall that the constraints i s+1 = i s +1 for s ∈ A splits the sequence (i 1 , • • • , i k ) into m blocks of necessary consecutive values.This allows to define F α (β) (F stands for forced) as the number of indices s such that i s is not in S(α), but either i s−1 is in the same block as i s (i.e.i s − 1 ∈ A), or one i t in the same block is in S(α).Then, for fixed α, i, j, the number of lists β in . Indeed, to construct β, one needs i new values, but j of them are forced by the constraints i s+1 = i s for s ∈ A.
On the other hand, in each block of indices, all but possibly one i s in the block are in F α (β) ⊎ S(α).Using the definition of Ψ (Eq.( 7)), Lemma 12 and the inequality above, we have This yields the following bound: Recalling that there are at most C ′ k,ℓ,i,j n i−j terms in the latter sum, we conclude that Bringing everything together, we get that Since this holds for any tuple (α 1 , . . ., α ℓ ), we have T ℓ ≤ C ′′ k,ℓ n m−1 , as wanted.□ Back to the proof of Theorem 1, Lemma 4 gives us for some constants C ′′′ k,r .Consequently, if r > 2, we get: This concludes the proof of Theorem 1.

Non-degeneracy of the limiting law
In this section, we prove Theorem 2. Recall that this theorem only considers classical patterns, i.e. the case A = ∅, and that we write π instead of (π, ∅) for simplicity.Note that in this case, the parameter m in Theorem 1 is simply equal to the size k of the pattern.
Lemma 13.Assume π has size at least 2 and p 1 < 1.Then the function is non constant.
In both cases, we can conclude that the function ϕ π p 1 (x, y) + ϕ π p 1 (y, x) is not constant.□ We also note that |ϕ π p 1 (x, y) − ϕ π p 1 (x ′ , y ′ )| is bounded from above by the probability that at least one of P 1 , . . ., P k−1 has its x-coordinate between x and x ′ or its y-coordinate between y and y ′ .Using a union bound, this implies i.e. ϕ π p 1 is (k − 1)-Lipschitz when using the L 1 norm on [0, 1].

5.2.
A recursive inequality on the variance.Let λ be a partition with at least one part equal to 2 and denote by λ ′ the partition obtained by removing a part of length 2 to λ.Then a uniform random permutation σ = σ λ in C λ can be constructed as follows.
• Choose two distinct indices i and j uniformly at random between 1 and n, and set σ i = j and σ j = i • Take a uniform random permutation σ ′ of [n]\{i, j} of cycle type λ ′ , independent of {i, j}, and set σ k = σ ′ k for k {i, j}.Using the law of total variance, we have Moreover, the quantity X π (σ ) can be decomposed as X π (σ ) = X π (σ ′ ) + C, where C is the number of occurrences of π in σ , using position i or j (or both).Note that X π (σ ′ ) is independent of i, j, so that E(X π (σ ′ )|i, j) is the constant random variable a.s.equal to In particular, we get an initial bound (25) v(λ) ≥ Var E(C|i, j) .

5.3.
Analysing the initial bound (25).Recall that C counts the number of occurrences of π in σ that use position i or j (or both).We first consider the number C 1 of those using i but not j.
Lemma 14.Let x, y be in [0, 1] and i = i(n) and j = j(n) be two sequences chosen such that i = xn + o(n) and j = yn + o(n).Then we have where ϕ π p 1 is defined in Eq. (24) and the error term is uniform on i and j.
We start with some estimates on the distribution of finitely many values (σ (i 1 ), . . ., σ (i m )) of a uniform random permutation σ of cycle type λ.

Lemma 15.
i) Fix m ≥ 1 and let i 1 , . . ., i m be distinct indices in {1, . . ., n}.Then P(∃s t : ii) Let j 1 , . . ., j m also be distinct indices in {1, . . ., n}.We assume that there is no s t such that i s = j t and we let r = |{t : i t = j t }|.Then In both estimates, the error terms depend on m, but are uniform in (i 1 , . . ., i m , j 1 , . . ., j m ) and in λ.Moreover, the same estimates hold conditionally to i, j in the construction of Section 5.2, provided that the tuple (i 1 , . . ., i m , j 1 , . . ., j m ) does not contain i or j.
Proof.For the first item, let us note that a given index i s is a fixed point of σ with probability m 1 (λ)/n.Conditionally of i s not being a fixed point, by conjugation invariance, its image σ i s is uniformly distributed on [n] \ {i s }, and thus for fixed s t, we have The first item in the lemma follows by a simple union bound.
Let us consider the second item.By symmetry, we can assume that i s = j s for s ≤ r and i s j s for s > r.Recalling that the set F(σ ) of fixed points of σ is a uniform random subset of [n] of size m 1 (λ), we have Now, conditionally on i 1 , . . ., i r being fixed points, the probability that i r+1 , . . ., i m are not fixed points is Combining this computation with item i), the probability of i 1 , . . ., i r being fixed points, of i r+1 , . . ., i m being not fixed points and of (σ i r+1 , . . ., σ i m ) being disjoint from (i r+1 , . . ., i m ) is p r 1 (1 − p 1 ) m−r + O(n −1 ).But under this event, by conjugation invariance, the tuple (σ i r+1 , . . ., σ i m ) is a uniform random tuple of distinct values in [n] \ {i 1 , . . ., i m }.Hence, the conditional probability that is equal to (j r+1 , • • • , j m ) is 1/n m−r (1 + O(n −1 )).We finally get P(∀s ≤ m, σ (i s ) = j s ) = p r 1 (1 − p 1 ) m−r + O(n −1 ) n m−r (1 + O(n −1 )) , which proves the second item.The conditional statement in the lemma follows easily from the unconditional one since for h {i, j}, we have σ h = σ ′ h , where σ ′ is a uniform random permutation of cycletype λ ′ of the set [n] \ {i, j}.□ Proof of Lemma 14. Recall that C 1 is the number of occurrences of π in σ using position i but not j.Denoting by i 1 , . . ., i k−1 the other positions of such an occurrence of π (in any order, hence the symmetry factor in the next formula) and j 1 , . . ., j k−1 their images by σ , we can write C 1 as follows: (27 By Lemma 15, item i), for each i 1 , . . ., i k−1 , the total contribution of terms with some j t i t , but j t ∈ {i 1 , . . ., i k−1 , i}, is O(n −1 ) in expectation.Hence, the total contribution of such terms to the double sum in (27) is O(n k−2 ) in expectation.Furthermore, from Lemma 15, item ii), if (j 1 , . . ., j k−1 ) is such that for each t, either j t = i t or j t {i 1 , . . ., i k−1 , i}, we have, conditionally to i, j, This gives We let (P (n) i ) i≤k−1 be a vector of independent random variables with the following distribution, conditioned not to share coordinates: • with probability p 1 , we set P Lemma 16.There exist K 3 , K 4 > 0 such that for any n > K 3 , E Var(C|i, j) ≤ K 4 n 2k−3 .
Proof.Conditionally on i, j, we have j r 1 =j,j r 2 =i B i r 1 ,j r 1 B i r 2 ,j r 2 ℓ {r 1 ,r 2 } B i ℓ ,j ℓ .
Conditionally on i, j, we have, in the two first term B i r ,j r = B i,j = 1, and in the third one B i r 1 ,j r 1 = B i,j = 1 = B j,i = B i r 2 ,j r 2 The other B factors have indices different from i and j.Since σ ′ is a uniform permutation on the conjugacy class indexed by λ ′ we have (Ψ , D k−1 ) weighted dependency graph for the random variables k−1 t=1 B i t ,j t .One can easily adapt the proof of Section 4 to obtain that the quantities R and T ℓ associated with the above dependency graph satisfy R ≤ 32 ( 2(k−1) 2 ) (n − 2) k−1 and T ℓ ≤ K 7,k n k−2 (note that the product k−1 t=1 B i t ,j t under consideration have degree k − 1).From Lemma 4, this implies Var(C|i, j) ≤ K 4 n 2k−3 .One can conclude since K 4 is independent of the choice of i and j. □ Plugging in Eq. (30) and Lemma 16 in Eq. ( 26), we get that, for some constant K 5 > 0, For, i ≤ m 2 (λ), we denote by λ (i) = ((λ ′ ) • • • ) ′ the partition obtained by removing i blocks of size 2 from λ.For i ≤ K 6 n 1/2 (where K 6 is a positive constant), the inequality (31) holds substituting λ by λ (i) and we have following proposition was proved in [Fér18, Proposition 5.11].Proposition 5. Consider random variables {Y v , v ∈ V } with a (Ψ , C) weighted dependency graph L and assume that Ψ is super-multiplicative.We fix some integer d ≥ 1.Then L d is a (Ψ , D d ) weighted dependency graph for the family {Y I , I ∈ V d }, where the constants D d = (D d,r ) r≥1 depend only on d, r and C.
Hence, variables A i,j and A k,ℓ are weakly dependent.It turns out that this family of random variables admits a nice weighted dependency graph.This was shown in [Fér18, Proposition 8.1], which we copy here.(We use the symbol # for the number of distinct elements in a set or multiset. )Proposition 3. Consider the weighted graph L on vertex set [n] 2 defined as follows: k r is the graph with vertex set [r] and, for each s, t in[r], it has an edge between s and t if and only if {i s , j s } ∩ {i t , j t } ∅ or {k s , ρ k s }∩{k t , ρ k t } ∅.We note that G A k 1 ,...,k r depends on i 1 , . . ., i r , j 1 , . . ., j r , but since these indices are fixed throughout the proof, we keep this dependence implicit in the notation.Only the dependence in k 1 , . . ., k r is made explicit.