Partial sum of matrix entries of representations of the symmetric group and its asymptotics

Transcription

1 Partial sum of matrix entries of representations of the symmetric group and its asymptotics Dario De Stavola 9th June 016 Abstract Many aspects of the asymptotics of Plancherel distributed partitions have been studied in the past fifty years, in particular the limit shape, the distribution of the longest rows, characters of the associated representation matrices of the symmetric group, and connections with random matrix theory. The latter prompts us to study the representation matrices themselves. We consider thus partial sums of entries of the representation matrix. These partial sums have a natural decomposition as a main term and a remainder. In this paper we describe the fluctuations of the main term. Our main tool is the expansion of symmetric functions evaluated on Jucys-Murphy elements. 1 Introduction Let λ be a partition of n, in short λ n, represented as a Young diagram in English notation. A filling of the boxes of λ with numbers from 1 to n, increasing towards right and downwards, is called a standard Young tableau. We call the number of standard Young tableaux of shape λ. We fix n and we associate to each λ the probability n!, which defines the Plancherel measure. Let us recall briefly three motivations for the study of the asymptotics of Plancherel distributed random partitions. They relate algebraic combinatorics, representation theory of the symmetric group, combinatorics of permutations, and random matrix theory. The partitions of n index the irreducible representations of the symmetric group S n. For each λ n the dimension of the corresponding irreducible representation is. A natural question concerns the asymptotics of the characters associated when λ is distributed with the Plancherel measure. A central limit theorem was proved with different techniques by Kerov, [Ker93a], [IO0], and Hora, [Hor98]: n ρ m 1 ρ ˆχ λ d ρ k mkρ/ H mk ρξ k. 1 k Here ρ is a partition of n, ˆχ λ ρ is the renormalized character associated to λ calculated on a permutation of cycle type ρ, m k ρ is the number of parts of ρ which are equal to k, H m x is the m-th Hermite polynomial, and {ξ k } k are i.i.d. standard gaussian variables. The Robinson-Schensted-Knuth algorithm allows us to interpret the longest increasing subsequence of a uniform random permutation as the first row of a Plancherel distributed partition. This motivates the study of the shape of a random partition. A limit shape result was proved independently by [KV77] and [LS77], then extended to a central limit theorem by Kerov, [IO0]. For an extensive introduction on the topic, see [Rom15]. More recently it was proved that, after rescaling, the limiting distribution of the longest k rows of a Plancherel distributed partition λ coincides with the limit distribution of the properly rescaled k largest eigenvalues of a random Hermitian matrix taken from the Gaussian Unitary Ensemble. See for example [BOO00] and references therein. Such similarities also occur for fluctuations of linear statistics, see [IO0]. dario.destavola@math.uzh.ch Institute of Mathematics, University of Zurich, Winterthurerstrasse 190, 8057, Zurich, Switzerland. Keywords: Jucys-Murphy elements, symmetric functions, symmetric group. 1

2 Among the previous three connections, the last one is the least understood and motivates the study of the asymptotics of the representation matrix itself, rather than just the trace, to confront it with similar results on random matrix theory. We consider thus, for a real valued matrix A of dimension N and u [0, 1], the partial trace and partial sum defined, respectively, as P T u A := i un A i,i N, P S ua := i,j un We stress out that these two objects are not invariant by isomorphisms of representations. We will thus use an explicit construction of irreducible representations Young seminormal representation and a specific order on the set of standard Young tableaux the last letter order, defined in Section. The partial trace and sum have been considered in the context of random orthogonal matrices by D Aristotile, Diaconis and Newman in [DDN03]. It is shown that, in this setting, both P T u A and P S u A converge, after an appropriate normalization, to Brownian motions. Our results for random representation matrices are quite different. A i,j N. We define subpartitions µ j of λ, denoted µ j λ, as partitions of n 1 obtained from λ by removing one box. Then we show in Proposition.8 that P T λ u σ := P T u π λ σ = j< j dim µ j ˆχµj σ + P T µ j ū σ, for some index j and real number ū. We call the first term of the right hand side the main term for the partial trace MT λ u σ and the second the remainder for the partial trace RT λ u σ. Our main result Theorem.10 concerns the asymptotics for the main term for the partial trace: n ρ m 1 ρ MT λ u σ d u k k m kρ/ H mk ρξ k. 3 That is, the asymptotics of the main term is the same as the asymptotics of the trace multiplied by u. Notice that this result generalizes 1, although we use Kerov s result in the proof. Here lies the difference with the random matrix theory case: the partial trace of an orthogonal random matrix converges to a Brownian motion, and thus it has a high degree of randomness; on the contrary, if we condition the partial trace of a representation matrix by the total trace, the partial trace appears to be deterministic. Informally, the main idea to prove 3 is to show that when n grows, σ ˆχµj ˆχ λ σ. j< j j< j To achieve this result we need to estimate the asymptotic of ˆχ λ σ ˆχ µ σ, for λ distributed with the Plancherel measure and µ λ. We consider power sums p ν, where ν is the cycle type of σ in which we remove 1 to each part, calculated on the multiset of contents C λ. We prove in Section 3 that p ν C λ p ν C µ o P n σ implies ˆχ λ σ ˆχ µ σ o P n σ, 4 where σ is the size of the support of the permutation σ. This will be accomplished by introducing modified power sums p ν and by analyzing its highest terms with the appropriate filtration. We cannot unfortunately prove asymptotic results on the remainder RTu λ σ, although we conjecture that n ρ m 1 ρ RTu λ σ p 0 5 for u [0, 1]. In Section 5 we describe a different conjecture, which would imply the one above, involving quotient of dimensions of irreducible representations. We give some numerical evidence. Our conjecture would imply n ρ m 1 ρ P Tu λ σ d u k mkρ/ H mk ρξ k. k

3 X Figure 1: Example of Young diagrams. Starting from the left: λ = 6, 4, 3, 3,, µ 3 = 6, 3, 3, 3, and Λ = 6, 4, 3, 3, 3. A similar decomposition as Proposition.8 can be shown for partial sums P S λ uσ := P S u π λ σ, see Proposition 4.11: if we call total sum T S λ σ = P S λ 1 σ = i π λ σ i,i, then P S λ uσ = j< j T dim µ j Sµj σ + P Sµ j ū σ, for some j and ū. Similarly to the partial trace case, we define the main term and remainder for the partial sum as, respectively, MSuσ λ := T σ, RS λ dim µ j uσ := Sµj P Sµ j ū σ. j< j We show that, for σ S r, the total sum T S λ σ can be written as linear combination of {ˆχ λ τ} τ Sr. This implies a central limit theorem for the total sum Theorem 4. and the main term of the partial sum Corollary 4.1, and a law of large numbers for the partial sum itself Theorem In section we recall results on the co-transition measure, the Young seminormal representation, partial permutations, shifted symmetric functions and we introduce the partial trace. In section 3 we prove the result concerning the asymptotics of the main term. In section 4 we study the total and partial sum, while in section 5 we describe a conjecture which would imply a convergence result on the partial trace. Preliminaries.1 Notation Set λ = λ 1, λ,... n to be a partition of n, i.e. a nonincreasing sequence of positive integers whose sum is n. We associate partitions with Young diagrams represented in English notation, as in the example below. In this setting a box in symbol is an element a, b N N, and we write = a, b λ if 1 a λ b. We say that a box λ is a corner if there exists another Young diagram with the same shape as λ without that box. Likewise, a box / λ is a bay if there exists a Young diagram with the same shape of λ together with. For a box = a, b the content is defined as c = ca, b = a b. Of particular interest are the contents of corners and bays of λ, and we call them respectively y j and x j, ordered in a way such that x 1 < y 1 < x <... < y d < x d+1, where d is the number of corners. For a corner box such that y j = c for some j, the partition of n 1 resulting from removing that box is called µ j, and we write µ j λ. We call such a µ j a subpartition of λ. Similarly, Λ j indicates the partition of n + 1 obtained from adding the bay box of content x j. Example.1. In Figure 1 we show three Young diagrams. The first is a partition λ = 6, 4, 3, 3, 18 where we stress out a bay of content and a corner of content. The second is a subpartition corresponding to removing the box X from λ, while the third has the same shape of λ with an additional box corresponding to. 3

4 We study the irreducible representation π λ σ associated to λ, and we call its dimension. It is a known fact that the set SYTλ := {T : T is a standard Young tableau of shape λ} indexes a basis for π λ, so that = SYTλ. Example.. We present the set of standard Young tableaux of shape λ = 3,. T 1 = T = T 3 = T 4 = T 5 = By setting P P L λ = /n! we endow the set of partitions of n with a probability measure, called the Plancherel measure of size n. Throughout the paper we always assume that λ is distributed with the Plancherel measure.. Transition and co-transition measures Fix λ n. Two identities follow from the branching rule see for example [Sag13, section.8]: dim µ =, dim Λ = n + 1. µ λ λ Λ For v R we can define, respectively, the normalized co-transition distribution and the normalized transition distribution: F λ ctv := y j v n, F λ trv := x j v n dim Λ j n + 1. In [Ker93b] Kerov proved that the normalized transition distribution converges almost surely to the semicircular distribution, when λ is equipped with the Plancherel measure. Building on Kerov s work, we prove the same result for the normalized co-transition distribution. Lemma.3. When the set of partitions of n is equipped with the Plancherel measure, then almost surely holds for all v. lim F ctv λ = 1 n + π v 4 t dt Proof. We prove the lemma by showing that the Stieltjes transform of the normalized co-transition measure associated to the random variable λ n converges to the Stieltjes transform of the semicircular distribution. It is showed in [GH03, corollary 1] that pointwise convergence of Stieltjes transforms implies convergence in distribution, which again implies convergence of distribution functions at continuity points recall that the semicircular distribution function is continuous everywhere. We claim thus that STct λn u := 1 a.s. u u 4 =: ST n sc u. 6 u y j µ j λ In [Ker93b, sections and 4], Kerov showed that the Stieltjes transform of the transition measure converges almost surely to the Stieltjes transform of the semicircular distribution. Moreover, he also showed that, if Q λ u := j u y j and P λ u := j u x j, then STtr λn u = Qλ u P λ u, ST λ n ct u = u P λ u Q λ u. Hence STct λn u = u This proves 6 and hence the lemma. 1 STtr λn u a.s. u u u 4 = u u 4. 4

5 ũ u F λ ct u = v λ y j n v Figure : Example of graph of ũ = F λ ctv. Set F λ ct u := sup { z R s.t. F λ ct z u }. We consider F λ ct as the inverse of the step function F λ ct. We want to show that F λ ctv λ converges to u: Lemma.4. For v λ = F λ ct u for a fixed u, then F λ ctv λ = y j v λ n p u. Proof. We show in Figure an example of a normalized co-transition distribution. Since Fct λ is right continuous then u Fctv λ λ. Moreover, since the co-transition distribution is a step function, for each element of the image ũ FctR λ there exists j such that Fct λ yj n = ũ. Call j the index corresponding to F λ ctv λ, that is, F λ ct y j n = F λ ctv λ. Hence j< j / u. Thus v λ y j n, since F λ ct y j n u; v λ y j n, since for each ɛ > 0, F λ ct y j n ɛ u. Thus v λ = y j n. Therefore Equivalently j< j = F ctv λ λ dim µ j u F ctv λ λ. dim µ j u F ctv λ λ 0, dim µj p and max 0 because of the convergence of the normalized co-transition distribution towards an atom free distribution proved in the previous lemma. This proves the statement..3 Young seminormal representation and last letter order We recall the definition of Young seminormal representation, see [You77] for the original introduction, and [Gre9] for a more modern description. We need three preliminary definitions; set λ n, then 1. given a box λ recall that the content c is the difference between the row index and the column index of the box. For k n we also write c k T for the content of the box containing the number k in the tableau T. As instance, in the tableau T 1 of Example., c 4 T 1 = 1, c 1 T 1 = c 5 T 1 = 0, c T 1 = 1 and c 3 T 1 =.. For k n 1 the signed distance between k and k + 1 in the tableau T is d k T := c k T c k+1 T. For example d 1 T 1 = d T 1 = d 4 T 1 = 1; d 3 T 1 = 3. 5

6 3. If k and k + 1 are in different columns and rows we define k, k + 1T as the standard Young tableau equal to T but with the boxes containing k and k + 1 inverted. In the previous example 3, 4T 1 = T, while, 3T 1 is not defined. Definition.5. The Young seminormal representation defines the matrix associated to π λ k, k + 1 entrywise in the following way: if T and S are standard Young tableaux of shape λ, then 1/d k T if T = S; π λ k, k + 1 T,S = 1 1 d k T if k, k + 1T = S; 0 else. Notice that the adjacent transpositions k, k + 1, k = 1,..., n 1, generate the group S n, hence π λ σ is well defined for all σ S n. Example.6. Consider λ = 3,, σ =, 4, 3. We compute π λ, 4, 3 : π λ, 4, 3 = π λ 3, 4, 3 = π λ 3, 4π λ, 3 1/3 8/ /9 1/ = / 3/ /4 1/ / 3/ /4 1/ 1/3 /9 / /9 1/6 1/1 0 0 = 0 3/4 1/ / 3/ /4 1/ We define also the last letter order in SYTλ: let T, S be standard Young tableaux, then T S if the box containing n lies in a row in T which is lower that S; if n lies in the same row in both tableaux, then we look at the rows containing n 1, and so on. Notice that in the list of tableaux of shape 3, of Example. the tableaux are last letter ordered; also, the entries of the matrices in Example.6 are ordered accordingly. We write π λ σ i,j instead of π λ σ Ti,T j, where T i is the i th tableau of shape λ in the last letter order..4 The partial trace and its main term Let λ n and σ S r a permutation of the set {1,..., r} with r n. We see S r as a subgroup of S n by adding fixed points, so that π λ σ is well defined. Similarly, if ρ r n, then we define χ λ ρ = χ λ ρ 1 n r and ˆχ λ ρ = ˆχ λ ρ 1 n r. Definition.7. Let λ n, σ S r, u [0, 1]. Define the partial trace as P T λ u σ := i u π λ σ i,i. When u = 1 then the partial trace correspond to the normalized trace ˆχ λ σ = χλ σ = i π λ σ i,i. Our choice of order on the tableaux implies that, as long as σ S r and r n 1, π µ1 σ 0 0 π λ 0 π µ σ 0 σ = 0 0 π µ3 σ Hence ˆχ λ σ = j the partial trace: ˆχµj σ. This decomposition of the total trace can be easily generalized to 6

7 Proposition.8 Decomposition of the partial trace. Fix u [0, 1], λ n and σ S r with r n 1. Set Fct λ u = y j n = v λ for some j as in Lemma.4. Define ū = u < 1, dim µ j j< j then P T λ u σ = j< j dim µ j ˆχµj σ + P T µ j ū σ. 8 Proof. We can decompose the partial trace as P T λ u σ = i u π λ σ i,i = πλ σ i,i i + j< j <i u j< j π λ σ i,i. The first sum in the RHS is i j< j πλ σ i,i = j< j ˆχµj σ. We consider now the second sum of the RHS: we have u j j by the definition of j, thus < i u. j< j j j Hence π λ σ i,i = π µ j σĩ,ĩ, where ĩ = i, so that j< j 0 < ĩ u j< j = ū dim µ j. Therefore <i u j< j and the proposition is proved. π λ σ i,i = ĩ ū dim µ j π λ σĩ,ĩ dim µ j = P T µ j ū σ, We define y j v λ n σ to be the main term for the partial trace denoted MT λ ˆχµj u σ, while dim µ j P T µ j ū σ is called the remainder denoted Ruσ. λ The main goal of this paper is to establish the asymptotic behavior of the main term for the partial trace. We settle some notation: if a permutation σ has cycle type ρ we write m k σ = m k ρ for the number of parts of ρ which are equal to k or, equivalently, for the number of cycles of σ which are of length k. We define the weights of σ and ρ as wtσ := wtρ := Suppσ = ρ m 1 ρ. We recall now a result due to [Ker93a], with a complete proof given by [IO0] and, independently, by [Hor98]: Theorem.9. Fix ρ r n. The asymptotic behavior of the irreducible character ˆχ λ is given by: n wtρ ˆχ λ d ρ k mkρ/ H mk ρξ k, k where {ξ k } k is a sequence of independent standard Gaussian random variables, and H m x, m 1, is the Hermite polynomial of degree m defined by the recurrence relation xh m x = H m+1 x + mh m 1 x and initial data H 0 x = 1 and H 1 x = x. Our result is the following: 7

8 Theorem.10. Let v λ = F λ ct u, σ of cycle type ρ. Then n wtσ MTu λ σ = n wtσ y j v λ n ˆχµj σ d u k mkρ/ H mk ρξ k. k Notice that the procedure of Proposition.8 can be iterated: consider σ S r and s N such that n r > s. Set µ s := λ and µ s 1 := µ, where µ is defined according to Proposition.8. Similarly set u s := u and u s 1 := ū, then we can rewrite 8 as P T λ u σ = MT µs u s σ + dim µs 1 P T µs 1 σ. u s 1 Applying again Proposition.8 to the second term of the right hand side, we see that there exists µ s µ s 1 and u s [0, 1] such that P T λ u σ = MT µs u s σ + By iterating we obtain the following proposition: dim µs 1 MT µs 1 σ + u s 1 dim µs P T µs σ. u s Proposition.11. Let σ S r. Set s such that n s > r, then there exists a sequence of partitions µ 0 µ 1... µ s = λ and a sequence of real numbers 0 u 0,..., u s < 1 such that P T λ u σ =.5 Partial permutations s dim µ i MT µ i σ + u i i=1 dim µ0 0 P T µ σ. u 0 In this section we recall some results in the theory of partial permutations, introduced in [IK99]. All the definitions and results in this section can be found in [Fér1]. Definition.1. A partial permutation is a pair σ, d, where d N is finite and σ is a bijection d d. We call P n the set of partial permutations σ, d such that d {1,..., n}. The set P n is endowed with the operation σ, d σ, d = σ σ, d d, where σ is the bijection from d d to itself defined by σ d = σ and σ d \d = Id; the same holds for σ. Define the algebra B n = C[P n ]. There is an action of the symmetric group S n on P n defined by τ σ, d := τστ 1, τd, and we call A n the subalgebra of B n of the invariant elements under this action. Set α ρ;n := σ, d. d {1,...,n}, d = ρ σ S d of type ρ Proposition.13. The family α ρ;n ρ n forms a linear basis for A n. There is a natural projection B n+1 B n which sends to 0 the partial permutations whose support contains n + 1. The projective limit through this projection is called B := lim B n, and similarly A := lim A n. The family α ρ := α ρ;n n 1 forms a linear basis of A. We recall the definition of Jucys-Murphy element, described for example in [Juc66] and [Mur81], and its generalization as a partial permutation. Definition.14. The i-th Jucys-Murphy element is the element of the group algebra C[S n ] defined by J i := 1, i +, i i 1, i, J 1 = 0. The partial Jucys-Murphy element ξ i is the element of B n defined by ξ i := j<ij, i, {j, i}, ξ 1 := 0. Proposition.15. If f is a symmetric function, fξ 1,..., ξ n A n. The sequence f n = fξ 1,..., ξ n is an element of the projective limit A, that we denote Γ f ; moreover, f Γ f is an algebra morphism. With the formalism of the virtual alphabet, we define Ξ as the projective limit of the sequence ξ 1,..., ξ n for n ; equivalently, Ξ is the unique element of n=1 C[S n] such that fξ := Γ f. 8

9 For a partition ν = ν 1,..., ν q and n ν we consider p ν ξ 1,..., ξ n = q i=1 ξνi ξνi n. Then one can show that p ν Ξ = m i ν! α ν+1 + c ρ α ρ, i ρ < ν +q with ν + 1 = ν 1 + 1,..., ν q + 1 and c ρ non-negative integers. Let us fix some notation: we write n k for the k-th falling factorial, that is, nn 1... n k + 1, and z ρ = i ρ i i m iρ! for the order of the centralizer of a permutation of type ρ. We call K ρ1 n ρ = 1 σ, #{σ of cycle type ρ1 n ρ } where the sum runs over the permutations σ S n of cycle type ρ1 n ρ. We consider a morphism of algebras ϕ n : A n ZC[S n ] which sends σ, d to σ. On the basis α ρ;n it acts as follows: ϕ n α ρ;n = n ρ z ρ K ρ1 n ρ. 9 In particular ˆχ λ ϕ n α ν;n = 1 z ν λ ν ˆχ λ ν. It follows from a result of Jucys [Juc74, Equation 1] that ˆχ λ ϕ n fξ 1,..., ξ n = fc λ, 10 where f is a symmetric function and C λ is the multiset of contents of the diagram λ, C λ = {c, λ}..6 Shifted symmetric functions We recall some results on the algebra of shifted symmetric functions Λ and their relations with A. The results and definitions in this section follow [IK99]. Definition.16. Let Λ n be the algebra of polynomials with complex coefficients in x 1,..., x n that become symmetric in the new variables x i = x i i. There is a natural projection Λ n + 1 Λ n which sends x n+1 to 0. The algebra of shifted symmetric functions Λ is defined as the projective limit of Λ n according to this projection. We can apply shifted symmetric functions to partitions in the following way: if f is a shifted symmetric function and λ = λ 1, λ,..., λ l a partition, then we set fλ = fλ 1, λ,..., λ l. As in the symmetric functions algebra, there are several interesting basis for the algebra of shifted symmetric functions. We present one of them: Proposition.17. There exists a family of shifted symmetric functions {p ρ} ρ r λ n, { n p r ˆχ λ ρλ = ρ1, if n r; n r 0, otherwise. Moreover, the family {p ρ}, where ρ runs over all partitions, forms a basis for Λ. such that, for each These functions are called shifted power sums; they were introduced in [OO98, Section 15], where it was proved that p ρλ 1 1, λ,..., λ l l = p ρ λ. In order to simplify the notation we write p k, where k is a positive integer, instead of p k. Similarly, p 1 k replaces p 1 k. As polynomials in the variables x 1,..., x n,... the function p ρ have degree deg p ρx 1,..., x n,... = ρ. We are interested in two filtrations of the algebra Λ, which are described in [IK99, section 10]: degp ρ 1 := ρ 1 := ρ + m 1 ρ. This filtration is usually referred as the Kerov filtration; degp ρ N := ρ N = ρ + lρ, where lρ = i N m iρ is the number of parts of ρ. Lemma.18. There is an isomorphism F : A Λ that sends α ρ to p ρ/z ρ. 9

10 Our goal in this section is to develop [IO0, Proposition 4.1], which gives information about the top degree term of p ρ p θ for partitions ρ and θ, where top degree refers to the Kerov filtration. Consequently, we obtain results on the product α ρ α θ Corollary.3. We will write V k < for the algebra whose basis is {p ρ such that ρ 1 < k}. Notice that since deg 1 is a filtration, p ρ V k < V < ρ k By abuse of notation we will often write p ρ + V k < to indicate p ρ plus a linear combination of elements with deg 1 < k. We recall a result of Ivanov and Olshanski [IO0, Proposition 4.1]: Lemma.19. For any partition ρ and k, { k p ρp k = p ρ k + mk ρ p + V < ρ\k 1 k ρ, if m 1+k kρ 1 V < ρ, if m 1+k kρ = 0, 1 where the partition ρ \ k is obtained by removing one part equal to k. Lemma.0. For a partition ρ, p ρp 1 k = p ρ 1 k + V < ρ 1+k. Proof. We prove the statement by induction on k, the case k = 1 being proved in [IO0, Proposition 4.11]. This implies that for any k p 1 k p 1 = p 1 k+1 + V < k+. Suppose the lemma true for k > 1, then where we repetitively used property 11. p ρp 1 k+1 = p ρp 1 k p 1 + p ρv < k+ = p ρ 1 k p 1 + V < ρ 1+k+ = p ρ 1 k 1 + V < ρ 1+k+, Define Ṽ k = as the algebra with basis {p θ such that θ 1 = k and m 1 θ > 0}, a slightly different statement as property 11 holds: Lemma.1. For a positive integer k and a partition ρ p ρ Ṽ = k Ṽ = ρ 1+k + V < ρ 1+k. 13 Proof. It is enough to show that, for a partition θ such that θ 1 = k and m 1 θ > 0, Set θ = θ 1. By the previous lemma p θ = p θ p 1 + V < θ 1+ and p ρ p θ Ṽ = ρ 1+ θ 1 + V < ρ 1+ θ p ρ p θ = p ρ p θ p 1 + V < ρ 1+ θ 1+. Set C τ ρ, θ to be the structure constants in the basis {p τ } of the product p ρ p θ, that is, p ρ p θ = τ 1 ρ 1+ θ 1 C τ ρ, θ p τ, where the restriction τ 1 ρ 1 + θ 1 is a consequence of 11. Thus p ρ p θ = C τ ρ, θ p τ p 1 + V < ρ 1+ θ 1+ τ 1 ρ 1+ θ 1 = p τ 1 + V < τ 1+ + V < ρ 1+ θ 1+ This proves 14 and hence the lemma. τ 1 ρ 1+ θ 1 C τ ρ, θ Ṽ = ρ 1+ θ 1+ + V < ρ 1+ θ 1+ = Ṽ = ρ 1+ θ 1 + V < ρ 1+ θ 1. 10

11 Proposition.. Let ρ, θ be partitions. Then p ρ p θ = p ρ θ + Ṽ = ρ 1+ θ 1 + V < ρ 1+ θ 1. Proof. We prove the statement by induction on the number lθ m 1 θ of non-one parts of θ, with the base case being θ = 1 k, shown in the previous lemma. Consider now the claim true for p ρ p θ and set θ = θ k for k. Then by Lemma.19 p θ = p θ k = p θ p k + { k mk θ p θ\k 1 k + V < θ 1, if m k θ 1 V < θ 1, if m k θ = 0, = p θ p k + Ṽ = θ 1 + V < θ 1. Because of Property 11 and Lemma.1, By the inductive step Hence, by applying again Lemma.19 We substitute the previous expression in 15 p ρ p θ = p ρ p θ p k + Ṽ = ρ 1+ θ 1 + V < ρ 1+ θ p ρ p θ = p ρ θ + Ṽ = ρ 1+ θ 1 + V < ρ 1+ θ 1. p ρ p θ p k = p ρ θ p k + Ṽ = ρ 1+ θ 1+k + V < ρ 1+ θ 1+k = p ρ θ k + Ṽ = ρ 1+ θ 1+k + V < ρ 1+ θ 1+k. which concludes the proof. p ρ p θ = p ρ θ k + Ṽ = ρ 1+ θ 1+k + V < ρ 1+ θ 1+k, We can obtain a similar result in the algebra A by applying the isomorphism F 1 defined in.18. It is easy to see that F 1 Ṽ k = is the space of linear combinations of α ρ such that ρ 1 = k and m 1 ρ > 0. Similarly F 1 V k < is the space of linear combinations of α ρ such that ρ 1 < k. Corollary.3. Let ρ, θ be partitions. Then α ρ α θ = α ρ θ + F 1 Ṽ = k + F 1 V < k. 3 Asymptotic of the main term for the partial trace In this section we prove Theorem.10. The main step is to prove that ˆχ λ ρ ˆχ µ ρ o P n wtρ, where ρ is a fixed partition, wtρ = ρ m 1 ρ, λ is a partition of n and µ λ. Fro a partition ν = ν 1,..., ν q we define slightly modified power sums p ν ; we will show Lemma 3.5 and Equation 10 that ˆχ λ ϕ n p ν ξ 1,..., ξ n ˆχ µ ϕ n 1 p ν ξ 1,..., ξ n 1 = p ν C λ p ν C µ o P n ν +q. 16 In order to translate this result on a bound on ˆχ λ ρ ˆχ µ ρ we need to study the expansion of the modified power sums evaluated on Ξ, the infinite set of Jucys-Murphy elements in the theory of partial permutations: p ν ξ 1,..., ξ n = c ς α ς;n. ς With the right choice of ν and the right filtration we will prove Proposition 3.6 that p ν C λ p ν C µ = ˆχ λ ϕ n c ς α ς;n ˆχ µ ϕ n 1 c ς α ς;n 1 = ˆχ λ ρ ˆχ µ ρ + o P n wtρ. 17 ς ς Comparing 16 and 17, this gives ˆχ λ ρ ˆχ µ ρ o P n wtρ. 11

12 Remark 3.1. It is easy to see that ˆχ λ ρ ˆχ µ ρ o P n ρ lρ, where lρ is the number of parts of ρ; for example, considering balanced diagrams, in [FŚ11] the authors prove an explicit formula for the normalized character, which implies ˆχ λ ρ ˆχ µ ρ o P n ρ lρ. Alternatively one can look at the descriptions of normalized characters expressed as polynomials in terms of free cumulants, studied by Biane in [Bia03]. The action of removing a box from a random partition λ affects free cumulants in a sense described in [DFŚ10], which furnish the aforementioned bound. On the other hand we needed a stronger result, namely Proposition 3.6, and for this reason we introduce the modified power sums. Through the section, σ will be a fixed permutation, ρ its cycle type, and we consider ν = ν 1,..., ν q, such that ρ = ν + 1 = ν 1 + 1,..., ν q + 1, 1..., Modified power sums Definition 3.. For a partition ν = ν 1,..., ν q the modified power sum is where Catk = k 1 k k+1 p ν x 1, x,... = q i=1 Lemma 3.3. Set ν = ν 1,..., ν q, then p νi x 1, x,... Cat νi νi! α ν i 1 +1, is the k-th Catalan number if k is an integer, and 0 otherwise. p ν Ξ = ς c ς α ς, 18 for some non-negative integers c ς. We have 1. the sum runs over the partitions ς such that ς 1 ν N ;. c ν+1 = i m iν!; 3. if c ς 0, ς 1 = ν N and ς ν + 1 then m 1 ς > 0. Proof. Part 1. Since degα ς 1 = ς 1 is a filtration, it is enough to prove the statement for q = 1, that is, ν = k for some positive integer k. We consider p k Ξ = j 1, h, {j 1, h}... j k, h, {j k, h} j 1,...,j k <h Let σ, d = j 1, h... j k, h, {j 1,..., j k, h} be a term of the previous sum and let ς d be the cycle type of σ. This is the outline of the proof: firstly, we show an upper bound for ς 1, that is, ς 1 k +. Then we check some conditions that σ must satisfy in order to have ς 1 = k + if such a σ exists. We prove that ς 1 = k + if and only if σ, d = Id, d with d = k/ + 1 and k even. Finally, we calculate how many times the partial permutation Id, d, with d = k/ + 1, appears in the modified power sum p k Ξ. This number is the coefficient of α in p k 1 +1 kξ, and we show that it is equal to Cat k k!. Since we defined p k Ξ = p k Ξ Cat k k! α1 this shows that all the terms in p k kξ satisfy +1 ς 1 k + 1 = ν N when ν = k. We first estimate ς 1 = {j 1,..., j k, h} + m 1 ς = #{j i, s.t. σj i = j i } + #{j i, s.t. σj i j i } + δh is fixed by σ + δh is not fixed by σ, where δ is the Kronecker delta. We stress out that, when counting the set cardinalities above, we do not count multiplicities; for example, if σ = 1, 5, 5, 51, 53, 5 = 13, 5, then #{j i, s.t. σj i = j i } = #{1, } =, #{j i, s.t. σj i j i } = #{3} = 1, ς =, 1, 1. 1

13 Notice that in the sequence j 1,..., j k all fixed points must appear at least twice, while non fixed points must appear at least once, hence while obviously, Thus ς 1 k +. #{j i, s.t. σj i = j i } + #{j i, s.t. σj i j i } k, δh is fixed by σ + δh is not fixed by σ. Suppose there exists σ such that ς 1 = k +, then from the proof of the inequality ς 1 k + we know that σ satisfies for each i, j i appears at most twice; 0 j i is fixed by σ iff it appears exactly twice in the multiset {j 1,..., j k }; 1 h is a fixed point. We prove by induction on k that if ς 1 = k + and α ς appears in the sum 19 then k is even and ς = 1 k +1. If k = 1 then ς = and ς 1 = < 3 = k +. If k = then ς can be either ς = 3 or ς = 1, 1. By our request that ς 1 = 4 = k + we see that we must have ς = 1, 1. Consider now the statement true up to k 1 and σ = j 1, h... j k, h. By property h is fixed, so that j k must appear at least twice, and by 0 j k appears exactly twice. Hence, by 1 j k is fixed, thus it exists a unique l < k 1 such that j l+1 = j k and where σ = j 1, h... j l, h j k, h j l+, h... j k 1, hj k, h = τ 1 j k, hτ j k, h = τ 1 τ, τ 1 = { Id if l = 0 j 1, h... j l, h if l > 0, τ = { Id if l = k j l+, j k... j k 1, j k if l < k. If l = 0 or l = k we can apply induction on either τ 1 or τ and the result follows, so consider 0 < l < k. We claim that the sets {j 1,... j l } and {j l+,..., j k } are disjoint. Notice first that j k does not appear in {j 1,... j l }. Suppose j a = j b for some a l < b, and choose b minimal with this property. Then j b is not fixed by τ, that is, τ j b = j b with either b < b or b = k. In the first case b < b, by minimality of b, j b does not appear in τ 1 ; in the second case b = k we know that j k does not appear in {j 1,... j l }. Hence in either case σj b = j b. This is a contradiction, since j b appears twice in σ and therefore must be fixed property 1. This proves that {j 1,... j l } {j l+,..., j k } =. Therefore τ 1 and τ respect properties 0,1 and, and we can apply the inductive hypothesis to obtain that τ 1 = Id = τ and both l and k l are even. We conclude hence that if ς 1 = k + and α ς appears in the sum 18 then ς = 1 k +1 and k even. We assume now that k is even and we calculate the coefficient of α 1 k +1 in p kξ, which is equal to the coefficient of Id, d = j 1, h... j k, h, {j 1,..., j k, h} in the sum 19, for a fixed d of cardinality k/ + 1. In order to do this we count the number of lists L = j 1, h,..., j k, h such that {j 1,..., j k, h} = d and j 1, h... j k, h = Id. We call L d the set of these lists. Define a pair set partition of the set [k] = {1,..., k} as a set partition A = {{r 1, s 1 },..., {r k, s k }} of [k] into pairs. Such a set partition is said to be crossing if r a < r b < s a < s b for some a, b k/, otherwise the set partition is said to be non crossing. Calling S k the set of non crossing pair set partitions, it is known that S k = Catk/, see [Hor98]. We build a map ψ k : L d S k and we prove that this map is k!-to-one, which implies that Ld = c 1 k +1 = Cat k k!. Let L L d, L = L 1,..., L k = j 1, h,..., j k, h; we have proven that, since j 1, h... j k, h = Id and d = {j 1,..., j k, h} = k/ + 1, then each element j i, h must appear exactly twice in L. We construct a pair partition ψ k L such that a pair {r, s} ψ k L iff j r = j s. By [Hor98, Lemma ] this set partition is non crossing. This map is clearly surjective, although not injective: every permutation γ acting on d = {j 1,..., j k, h} which fixes h acts also on L d : γl = γj 1, h,..., γj k, h. Notice that h > j 1,..., j k and this is why γh = h in order to have an action on L d. Moreover ψ k L = ψ k L if and only if L = γl for some γ in S d\h. Thus ψ k is a k!-to-one map, hence c1 = Cat k k +1 k! and the first part of the proof is concluded. Part. The second statement is shown in [Fér1, section ]. 13

14 Part 3. This claim is proven by induction on q, the number of parts of ν. The base case is q = 1 and p k Ξ = c ς α ς. ς 1 k+1 We consider the α ς s appearing in the sum such that ς 1 = k + 1 and m 1 ς = 0, so that ς = k + 1. Expanding the sum in a similar way as Equation 19, we see that ς = k + 1 = #{j 1,..., j k, h}, and all the elements in this set must be pairwise different. Hence σ = j 1, h... j k, h = h, j k,..., j 1 and ς = k + 1. Thus the statement for q = 1 is proved. Let ν = ν 1,..., ν q with q. Set ν = ν 1,..., ν q 1 and suppose, by inductive hypothesis, that the assertion is true for p ν Ξ. Then p ν Ξ = p ν Ξ p νq Ξ = = We apply now Corollary.3 and obtain p ν Ξ = ς 1 ν N θ 1 ν q+1 ς 1 ν N c ς α ς ς 1 ν N θ 1 ν q+1 c ς c θ α ς α θ. θ 1 ν q+1 c θ α θ c ς c θ α ς θ + F 1 Ṽ = ν N + F 1 V < ν N. Hence there exists only one term in the previous sum such that ς 1 = ς θ 1 = ν N and m 1 ς = 0, that is, ς = ν + 1, and the proof is completed. Let now λ n be a random partition distributed with the Plancherel measure. We say that a function X on the set of partitions is X o P n β if n β Xλ p 0. Similarly, X O P n β is stochastically bounded by n β if for any ɛ > 0 there exists M > 0 such that P P L Xλn β > M ɛ. For example, as a consequence of Kerov s result on the convergence of characters Theorem.9, ˆχ λ ρ O P n wtρ/. If a function Xλ, µ j depends also on a subpartition µ j λ, by Xλ, µ j o P n β we mean that max j Xλ, µ j o P n β, and similarly for the notion of stochastical boundedness. Lemma 3.4. For each partition ρ we have that ˆχ λ ϕ n α ρ;n O P n ρ 1, where ρ 1 = ρ + m 1 ρ. Proof. From Equation 9, ˆχ λ ϕ n α ρ;n = cρ z ρ n ρ ˆχ λ ρ. Since ˆχ λ ρ O P n wtρ/ the lemma follows. Lemma 3.5. Let µ j λ. In probability ˆχ λ ϕ n p ν ξ 1,..., ξ n ˆχ µj ϕ n 1 p ν ξ 1,..., ξ n 1 = p ν C λ p ν C µj o P n ν +q. Proof. The first equality come from the fact that ˆχ λ ϕ n applied to a symmetric function of partial Jucys-Murphy elements equals the same symmetric function evaluated on the contents of λ. We prove now the asymptotic bound. Consider first the case q = 1, i.e., ν = k for some positive integer k, in which case p k C λ = p k C µj y j = p k C µj + y k j + Cat k k + 1 n k +1 n 1 k +1 = p k C µj + y k j + Cat k n 1 k. Notice that y j < max{λ 1, λ 1} since y j is the content of a box of λ. It is shown in [Rom15, Lemma 1.5] that, with probability that goes to 1, both λ 1 and λ 1 are smaller than 3 n, hence for each subpartition µ j of λ, one has y j < 3 n with probability 1 for n big enough. Thus p k C λ p k C µj O P n k o P n k+1. In the general case q > 1 we expand the product, obtaining p ν C λ p ν C µj = p νi C µj νi y νi j + Cat n 1 ν i. A {1,...,q} i A i / A 14

15 We use now Lemma 3.3 and 3.4, which show that the factor i A p ν i C µj is in O P n 1 i A νi+ A and i/ A Therefore p ν C λ p ν C µj O P n l, with νi y νi j + Cat n 1 ν i O P n 1 i / A νi. l = 1 max A {1,...,q} Proposition 3.6. Let λ n, µ λ. Let ρ r < n then ν i + A < ν + q. i ˆχ λ ρ ˆχ µ ρ o P n wtρ. Proof. Set ν = ν 1,..., ν q such that ρ = ν 1 + 1,..., ν q + 1, 1,..., 1. obviously, ˆχ λ ν+1 = ˆχ λ ρ and ˆχ µ ν+1 = ˆχµ ρ; moreover, ν + q = wtρ. We want to prove that ˆχ λ ν+1 ˆχ µ ν+1 o P n ν +q. We prove the statement by induction on ν = ν + q. The base case is ν + 1 = 1 and ˆχ λ 1 = 1 = ˆχµ 1, so the proposition is trivially true. Consider n ν +q p ν C λ p ν C µ o P 1 because of the previous lemma. It can be rewritten as n ν +q ς s.t. ς 1 ν +q where the coefficients c ς are described in Lemma 3.3. Equivalently n ν +q p ν C λ p ν C µ = n ν +q = n ν +q ˆχ λ ϕ n ˆχ µ ϕ n 1 cς α ς;n p 0, ς 1 ν +q ς 1 ν +q We split the previous sum and notice that n ν +q c ς n ς ˆχ λ ς n 1 ς ˆχ µ ς z ς c ς ˆχ λ ς n ς n 1 ς + n 1 ς ˆχ λ ς ˆχ µ ς. z ς c ς ς z ς ς n 1 ς 1 ˆχ λ ς p 0. Indeed n 1 ς 1 n ν +q n ν +q 1 and n ν +q 1 ˆχ λ ς p 0 since ς 1 ν + q. We deal with the sum n ν +q c ς ς z ς n 1 ς ˆχ λ ς ˆχ µ ς. We separate in this sum the terms with ς 1 = ν + q and ς ν + 1, the terms with ς 1 < ν + q, and the term corresponding to ς = ν + 1. Case ς 1 = ν + q and ς ν + 1: we want to estimate n ν +q ς 1= ν +q m 1ς>0 c ς n 1 ς ˆχ λ ς ˆχ µ ς z ς where the restriction m 1 ς > 0 is a consequence of Lemma 3.3, part 3. We consider one term of the previous sum and we write ν := ς 1, removed of the 0 parts. Notice as before that ˆχ λ ς = ˆχ λ ν+1, and ˆχ µ ς = ˆχ µ ν+1. Thus ν + 1 < ς 1 = ν + q and we can apply the induction hypothesis. Therefore ˆχ λ ν+1 ˆχµ ν+1 o P n ν+1 and, n ς 1 n 1 ς ˆχ λ ς ˆχ µ ς o P n ς 1 m 1 ς ν+1 = o P 1. Case ς 1 < ν + q. We can apply induction again, to have ˆχ λ ς ˆχ µ ς o P n ς m 1 ς. Therefore n ν +q n 1 ς ˆχ λ ς ˆχ µ ς o P n ν +q + ς 1 o P 1. 15

16 We obtain thus that which proves the statement. n ν +q n 1 ν +q i m iν! z ν+1 ˆχ λ ν+1 ˆχ µ ν+1 + o P 1 p 0, Proof of Theorem.10. Fix σ S r and let ρ be its cycle type. We have MT λ u σ = y j v λ n ˆχµj ρ = ˆχλ ρ + o P n ν +q, y j v λ n since / 1, and the previous proposition. Hence n ν +q MTu λ σ = n ν +q y j v λ n ˆχλ ρ + o P 1. Finally, by Lemma.4 and Theorem.9, we obtain n ν +q MT λ u σ d u k km kρ/ H mk ρξ k. 4 Sum of the entries of an irreducible representation In this chapter our goal is to describe the sum of the entries of the matrix associated to a Young seminormal representation, up to a certain index depending on the dimension of the representation. We stress out that the objects we study really depend on the representation matrix, and change, for example, under isomorphisms of the representation. Some calculations are similar to those in the previous chapter: first we consider the sum of all the entries in the matrix before this role was played by the trace, and then we study the sum of the entries whose indices i, j satisfy i u, j u, while before we were considering the partial trace. 4.1 Total sum Definition 4.1. Define the normalized total sum associated to an irreducible representation and a permutation σ S r as T S λ σ := π λ σ i,j The following is the main result of the section: i,j Theorem 4.. Fix σ S r and let λ n. Define the two real numbers r [ ] m := E r P L [T S ν σ] and v := E r P L ˆχ ν,1,...,1 T Sν σ, Where E r P L [X] is the average of the random variable X considered with the Plancherel measure dim ν /r! for ν r. Then n T S λ σ m d N 0, v, a normal random variable of variance v. Note that the limit N 0, v is degenerate in the case v = 0. It is nontrivial to see when v is nonzero. At the end of the section we study the values of v and m when σ is an adjacent transposition, and we write the explicit values of v and m for permutations of S 4. In order to prove the Theorem, we need some preliminary results. Proposition 4.3. Let λ n and r n. There exists a bijection φ r between SYTλ φr ν r SYTν SYTλ/ν Proof. Let T be a standard Young tableau of shape λ, the image φ r T = U, V of T is defined as follows: the boxes of T whose entries are smaller or equal than r identify a tableau U of shape ν λ. Define now V as a standard Young tableau of skew shape λ/ν, and in each box write a r, where a is the value inside the corresponding box in T. 16

17 Example 4.4. Here is an example for λ = 6, 4, 3, 3, 3, 1 and r = 8: , Lemma 4.5. Let λ n, σ S r with r n. For two standard Young tableaux T, T SY T λ, set φ r T = U, V and φ r T = U, V, then { π λ 0 if V V σ T,T = π ν σ U,U if V = V, where ν = shu = shu is the shape of the tableau U in the second case. Proof. We will prove the lemma by induction on the number of factors in the minimal decomposition of σ into adjacent transpositions. Recall that if σ = k, k + 1, k < r, by Definition.5 1/d k T if T = T ; π λ k, k + 1 T,T = 1 1 d k T if k, k + 1T = T. 0 else. Hence if V V then π λ k, k + 1 T,T = 0, otherwise π λ k, k + 1 T,T = π ν k, k + 1 U,U since d k T = d k U. Consider a general σ S r and write it as σ = k, k + 1 σ. Then π λ σ T,T = π λ k, k + 1 T,S π λ σ S,T S SYTλ = 1 1 d k T πλ σ T,T + 1 d k T πλ σ k,k+1t,t δ {k,k+1t SYTλ}. We apply the inductive hypothesis on σ, obtaining { 1 0 if V V d k T πλ σ T,T = 1 d k U πν σ U,U if V = V, { 0 if V V 1 1 d k T π λ σ k,k+1t,t = 1 1 d k U π λ σ k,k+1u,u if V = V, since if k, k + 1T is a standard Young tableau, then φ r k, k + 1T = k, k + 1U, V. δ {k,k+1t SYTλ} = δ {k,k+1u SYTν}. To conclude 1 1 d k U πλ σ U,U + 1 d k U πλ σ k,k+1u,u δ {k,k+1u SYTν} = π λ σ U,U Similarly, Corollary 4.6. Let λ and σ be as before, then Proof. By the previous lemma: T S λ σ = ν r T S ν σ dim ν /ν. T S λ σ := T,T SYTλ π λ σ i,j = ν r U,U SYTν π ν σ U,U /ν, and the conclusion is immediate. 17

18 Lemma 4.7. Let ν = r n = λ, then /ν = 1 χ ν τχ λ τ. r! Proof. Fix ν, we prove the statement by induction on n. If n = r then the LHS is equal to δ λ,ν, the Kronecker delta, and same for the RHS by the character orthogonality relation of the first kind. Suppose r = n 1, then by 7 we have χ λ τ = µ j λ χµj τ for each τ S n 1. Hence the RHS of the statement can be written as 1 χ ν τχ λ 1 τ = χ ν τχ µj τ = δ µj=ν = δ ν λ, r! r! which is equal to the LHS. µ j λ µ j λ By the inductive hypothesis we consider the statement true for each λ n 1. Hence /ν = / λ dim λ/ν Proposition 4.8. Let σ S r, λ n, then T S λ σ = λ n 1 = λ λ 1 χ ν τχ λ τ r! = 1 χ ν τχ λ τ. r! E r P L [ˆχ ν τt S ν σ] ˆχ λ τ. Proof. We apply Corollary 4.6 and Lemma 4.7: T S λ σ = T S ν 1 σ dim ν χ ν τˆχ λ τ r! ν r = dim ν ˆχ ν τt S ν σ ˆχ λ τ r! ν r and we recognize inside the parenthesis the average E r P L [ˆχν τt S ν σ] taken with the Plancherel measure of partitions of r. Proof of Theorem 4.. We write the previous proposition as T S λ σ = m + v ˆχ λ 1, + E r P L [ˆχ ν τt S ν σ] ˆχ λ τ, wtτ> where the first two terms correspond respectively to τ = Id and the sum of all transpositions. By Kerov s Theorem.9, we have that n E r P L [ˆχ ν τt S ν σ] ˆχ λ τ p 0; wtτ> on the other hand n v ˆχ λ 1, d vn 0,, therefore n T S λ σ m d N 0, v. 18

19 Example 4.9. Let σ = r 1, r, r >, be an adjacent transposition. In this example we will show that m = E r P L [T Sν σ] is strictly positive and that v := [ ] r E r P L ˆχ ν,1,...,1 T Sν σ = 1. First, notice that, for all ν r and for all T, S standard Young tableaux of shape ν, d r 1 T = d r 1 T, where T is a tableau of shape ν conjugated to T, which implies that and if T S then π ν σ T,S 0. Hence Since T S λ σ = ˆχ λ σ + T,S SYTν T S π ν σ T,S = { π ν σ T,S if T = S π ν σ T,S if T S, m = dim ν T S ν σ = 1 dim ν T S ν σ + T S ν σ. r! r! ν r ν r m = 1 ν r dim ν r! π ν σ T,S dim ν, then T,S SYTν T S For r >, at least one summand is nonzero. Consider now v and decompose T S ν σ as above: r [ ] v = Er P L ˆχ ν,1,...,1 ˆχν σ + E r P L By the character relations of the second kind we get E r P L π ν σ T,S + π ν σ T,S > 0. ˆχν,1,...,1 T,S SYTν T S π ν σ T,S dim ν [ ] ˆχ ν,1,...,1 ˆχν σ = dim ν ˆχ ν,1,...,1 r! ˆχν σ = rr 1. ν r On the other hand, using T for the conjugate of T as above, 1 1 r! ν r E r P L ˆχν,1,...,1 T,S SYTν T S T,S SYTν T S π ν σ T,S dim ν = χ ν,1,...,1 πν σ T,S + χ ν,1,...,1 πν σ T,S = 0 since π ν σ T,S = π ν σ T,S if T S and χν,1,...,1 = χν,1,...,1. Finally, v = 1. Here we write the values of m = E r P L [T Sν σ] and v = [ ] r E r P L ˆχ ν,1,...,1 T Sν σ. when σ S 4. σ Id 3,4,3,3,4,4,3,4 1, 1,3,4 1,,3 1,,3,4 1,,4,3 1,,4 1,3, m 1 1/ /3 5/1 1/6-1/ /3 1/3 1/3 0-1/3 v / 4/3 13/ /3 1/3 0 σ 1,3,4, 1,3 1,3,4 1,3,4 1,3,,4 1,4,3, 1,4, 1,4,3 1,4 1,4,,3 1,4,3 m -1/1 -/3-1/6 7/1 1/6-1/1 0-5/1-1/4 -/3-7/1 v 1/ 1 0-7/6-1/3-7/6-4/3-5/6-1/6 1/3 1/6 4. Partial sum of the entries of an irreducible representation Definition Let λ n, σ S n, u R. Then P S λ uσ := i,j u π λ σ i,j. 19

20 We can now argue in a similar way as we did in Section.4: summing entries of π λ σ up to a certain index is equivalent to sum all the entries of the submatrices π µj σ for j < j and the right choice of j, plus a remainder which is again a partial sum. We present the analogous of Proposition.8; we omit the proof, since it follows the same argument of the partial trace version: Proposition Fix u [0, 1], λ n and σ S r with r n 1. Set Fct λ u = y j n = v λ. Define ū = u < 1, dim µ j j< j then P S λ uσ = y j<v λ n T dim µ j Sµj σ + P Sµ j ū σ. 3 As before, we call the main term for the partial sum MSuσ λ := y j<v λ n T Sµj σ and remainder for the partial sum RSuσ λ dim µ j := P Sµ j ū σ. The connection between the main term for the partial trace and the main term for the partial sum is easily described by applying Proposition 4.8: MS λ uσ = y j<v λ n T Sµj σ = E r P L [ˆχ ν τt S ν σ] ˆχ µj τ y j<v λ n = E r P L [ˆχ ν τt S ν σ] ˆχµj τ y j<v λ n = E r P L [ˆχ ν τt S ν σ] MTu λ τ, 4 We can thus apply Theorem.10 on the convergence of the main term for the partial trace to describe the asymptotic of MS λ uσ; notice that the only term in the previous sum which does not disappear when λ increases is the one in which τ is the identity; moreover, the second highest degree term correspond to τ being a transposition, and thus of order O P n 1. We get: Corollary 4.1. Set σ S r and let λ n. Consider as before r m := E r P L [T S ν σ] and v := E r P L [ ] ˆχ ν,1,...,1 T Sν σ. Then n MS λ u σ u m d u N 0, v. Proof. We rewrite 4 as: MS λ uσ = m MT λ u Id + E r P L [ˆχ ν τt S ν σ] MT λ u τ + E r P L [ˆχ ν τt S ν σ] MT λ u τ. wtτ= wtτ> By Lemma.4, MT λ u Id p u. By Theorem.10 n 1 wtτ= E r P L [ˆχ ν τt S ν σ] MT λ u τ d v u N 0, ; on the other hand E r P L [ˆχ ν τt S ν σ] MT λ u τ o P n 1. The corollary is proved. wtτ> 0

21 Although we cannot show a satisfying result on the convergence of the partial trace because we cannot prove that n wtρ dim µ j P T µ j ū σ 0, we are more lucky with the partial sum: Theorem Let u R, λ n and σ S r, then We will prove the theorem after three lemmas. P S λ uσ p ue r P L [T S ν σ]. Lemma Let as usual σ S r and λ n, with n > r. Then π λ σ i,j = 0 for all i, j such that i j > r! Proof. We iteratively use the Formula 7: π λ σ = π µn 1 σ = µ n 1 λ µ n 1 λ µ n µ n 1 π µn σ = µ r λ π µr σ. Therefore π λ σ is a block matrix such that the only nonzero blocks are those on the diagonal. conclude, notice that dim µ r r! To Lemma With the usual setting of λ, σ and u we have P S λ uσ u r! lσ, where lσ is the length of the reduced word of σ, i.e. the minimal number of adjacent transpositions occurring in the decomposition of σ. Proof. We first prove a bound on the absolute value of an entry of the matrix π λ σ T,S lσ for each T, S 5 by induction on lσ. The base case is σ = k, k + 1, for which π λ k, k + 1 T,S 1. Suppose σ = k, k + 1 σ, then π λ 1 1 σ T,S = d k T πλ σ T,S + 1 d k T πλ σ k,k+1t,s δ k,k+1t SY T λ l σ + l σ = lσ. We see that P Suσ λ = i,j u π λ σ i,j u r! max T,S SYTλ π λ σ T,S lσ+1 u r!, which allows us to conclude. Notice that we used the previous lemma in the first inequality, which shows that the number of nonzero terms appearing in the sum is bounded by u r! Lemma Let σ Id, then P T λ u σ p 0. Proof. Recall the decomposition of the partial trace into main term and remainder Proposition.8: P T λ u σ = MT λ u σ + RT λ u σ. By Theorem.10, if σ Id then MT λ u σ o P n wtσ o P 1. On the other hand we can estimate the remainder through the bound on a singular entry calculated in 5: RT λ u σ lσ dim µ j p 0, where lσ is the length of the reduced word of σ, µ j is the subpartition corresponding to the renormalized content y j n = Fct λ u, and ū = u. dim µ j j< j 1

22 Proof of Proposition We claim that the partial sum P S λ uσ and the following quantity are asymptotically close E r P L [ˆχ ν τt S ν σ] P T λ u τ, that is, P Sλ uσ E r P L [ˆχ ν τt S ν σ] P Tu λ τ p 0. We substitute in the previous expression the decomposition formulas for the partial sum and partial trace, respectively Propositions 4.11 and.8, and we simplify according to the equality of Equation 4, so that it remains: dim µ j P Sµ j ū σ E r P L [ˆχ ν τt S ν σ] P T µ j ū τ. We recall from 5 that π λ τ T,S lτ for each T, S, which implies that P T µ j ū τ lτ ū hence dim µ j dim µ j P Sµ j ū σ E r P L [ˆχ ν τt S ν σ] P T µ j ū τ ū r! lσ + E r P L [ˆχ ν τt S ν σ] lτ ū since the expression inside the parenthesis is bounded and dim µ j/ p 0. On the other hand E r P L [ˆχ ν τt S ν σ] P T λ u τ = p 0 u Er P L [T S ν τ] + o1 d ue r P L [T S ν τ]. For the same reasons for which we cannot prove convergence of the partial trace, here we cannot show dim µ j a second order asymptotic, indeed we know that the term P Sµ j ū σ disappear when λ grows, but we do not know how fast. This will be discussed more deeply in the next section. We can now generalize the concept of partial sum, and Lemma 4.15 allows us to describe its asymptotics. Definition Let λ n, σ S n, u 1, u R. Define the partial sum of the entries of the irreducible representation matrix associated to π λ σ stopped at u 1, u as Corollary Let u 1, u [0, 1], then P S λ u 1,u σ := i u 1 j u π λ σ i,j. P S λ u 1,u σ p min{u 1, u }E r P L [T S ν σ] Proof. Suppose u 1 < u, we only need to prove that P Su λ 1,u σ P Su λ 1 σ = i u 1 u 1 <j u π λ σ i,j p 0. The number of nonzero terms in the above sum is bounded by r! Equation 5 we have π λ σ i,j lσ. Therefore by Lemma 4.14; moreover by P S λ u 1,u σ P S λ u 1 σ p 0, and the corollary follows.