Asymptotic of characters of symmetric groups and limit shape of Young diagrams

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Asymptotic of characters of symmetric groups and limit shape of Young diagrams

Valentin Féray

coworkers : Piotr Śniady (Wroclaw), Pierre-Loïc Méliot (Marne-La-Vallée)

Laboratoire Bordelais de Recherche en Informatique CNRS

Séminaire Lotharingien de Combinatoire (64), Lyon, France

Valentin Féray (LaBRI, CNRS) Characters ofS_n 2010-03-29 1 / 16

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Outline of the talk

1 Character values of symmetric groups An exact formula

Asymptotic behaviours

2 Application : limit shape of Young diagrams

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Character values of symmetric groups An exact formula

Young symmetrizer

Let T be a ﬁlling of λ= (3,2,2).

2 3 6 4 1 7 5 Consider :

row-stabilizer RS(T) =S_{_2,3,6_}×S_{_1,4_}×S_{_5,7_}. column-stabilizer CS(T) =S_{_2,4,7_}×S_{_1,3,5_}.

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Young symmetrizer

Let T be a ﬁlling of λ= (3,2,2).

2 3 6 4 1 7 5

n!s_λ

dimλ = X

σ1∈RS(T) σ2∈CS(T)

(−1)^σ²p_type(σ₂_σ₁₎

Work in progress with P. Śniady : analog for zonal polynomials

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An equivalent formulation

Recall : character value χ^λ(µ) fulﬁllss_λ=P

µ χ^λ(µ)

zµ p_µ. n!χ^λ(π)

dimλ = X

σ₂σ₁=π

(−1)^σ²N_σ^′₁_,σ₂(λ), where

Deﬁnition

N_σ^′₁_,σ₂(λ) is the number of bijectionsf :{1, . . . ,n} ≃λsuch that for all i,f(i) and f(σ1(i)) (resp.f(σ2(i))) are in the same row(resp.

column).

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Nice behaviour on short permutations

If π ∈Sk

֒→ι Sn, (π(i) =i ∀i >k),

thenN_σ^′₁_,σ₂(λ) =0 unlessσ1(i) =σ2(i) =π(i)∀ i >k. In this case the formula becomes :

n!χ^λ(ι(π))

dimλ = X

σ1,σ2∈Sk σ2σ1=π

(−1)^σ²N_ι(σ^′ ₁_),ι(σ₂₎(λ)

But N_ι(σ^′ ₁_),ι(σ₂₎= #{f :{1, . . . ,k}֒→λwith usual conditions}·

(n−k)!

| {z }

choices of the places ofk+1,...,n

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Nice behaviour on short permutations

Deﬁnition

N_σ^′₁_,σ₂(λ) is the number of injections f :{1, . . . ,k}֒→λ such that, for all i,f(i) and f(σ1(i)) (resp.f(σ2(i))) are in the same row(resp.

column).

Σ_π(λ) := n·(n−1). . .(n−k+1)χ^λ(ι(π))

dimλ = X

(−1)^σ²N_σ^′₁_,σ₂(λ).

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Forgetting injectivity

Deﬁnition

N_σ₁_,σ₂(T) is the number of functions f :{1, . . . ,k}→λsuch that, for all i,f(i) and f(σ1(i)) (resp.f(σ2(i))) are in the same row(resp.

column).

Σ_π(λ) := n·(n−1). . .(n−k+1)χ^λ(ι(π))

dimλ = X

(−1)^σ²N_σ₁_,σ₂(λ).

Idea of proof : the total contribution of a non-injective function in rhs is easily seen to be 0.

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Character values of symmetric groups Asymptotic behaviours

Asymptotics is easy to read on this formula

Model : ﬁx a permutation π0 ∈S_k and a partition λ0⊢k

Consider π=ι(π0) (i.e. we just add ﬁxpoints) and λ=c·λ0=λ multiplied by c (i.e. horizontal lengths are multiplied byc)

7→

Question : asymptotics of ^χ_dim^λ^(π)_λ ?

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Asymptotics is easy to read on this formula

7→

Easy on the N’s :Nσ1,σ2(c·λ) =c^|^C^(σ²⁾^|Nσ1,σ2(λ)

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Asymptotics is easy to read on this formula

7→

Easy on the N’s :Nσ1,σ2(c·λ) =c^|^C^(σ²⁾^|Nσ1,σ2(λ) Dominant term of Σ_π(λ) :

N_π,Id_k(λ) = Y

µi∈type(π)



X

j

λ^µ_jⁱ





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Asymptotics is easy to read on this formula

7→

Easy on the N’s :Nσ1,σ2(c·λ) =c^|^C^(σ²⁾^|Nσ1,σ2(λ) Dominant term of Σ_π(λ) :

N_π,Id_k(λ) = Y



X λ^µ_jⁱ



 = Y

pµi(λ)

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Asymptotics is easy to read on this formula

Consider π=ι(π0) (i.e. we just add ﬁxpoints) and λ=c•λ0=λdilated by c (i.e. horizontal and verticallengths are multiplied byc)

7→

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Asymptotics is easy to read on this formula

7→

Easy on the N’s :N_σ₁_,σ₂(c•λ) =c^|^C^(σ¹⁾^|⁺^|^C(σ²⁾^|N_σ₁_,σ₂(λ)

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Asymptotics is easy to read on this formula

7→

Easy on the N’s :N_σ₁_,σ₂(c•λ) =c^|^C^(σ¹⁾^|⁺^|^C(σ²⁾^|N_σ₁_,σ₂(λ) Dominant term of Σ_π(λ) :

X

σ1σ2=π

|C(σ1)|+|C(σ2)|maximal

±N_σ₁_,σ₂(λ)

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Free cumulants

Σ_π(c•λ) = X

σ1σ2=π

|C(σ1)|+|C(σ2)maximal

±N_σ₁_,σ₂(λ)

But,

σ₁σ₂ =π

|C(σ1)|+|C(σ2)|maximal

≃Q

NCµi ≃Q

Trees(µi).

With generating series, one can prove (Rattan, 2006) rhs =Y

R_µ_i₊₁(λ)

Rk : free cumulants deﬁned from the shapeω_λ by Biane (1998).

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Remarks

Works in more general context than sequences c·λ₀ and c •λ₀ (in fact, works as soon as a sequence of Young diagram has alimit)

These results were already known (Vershik & Kerov 81, Biane 98), but : we provide uniﬁed approach of both cases ;

our bound for error terms are better.

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Application Limit shape of Young diagrams

Description of the problem

Consider Plancherel’s probability measure on Young diagrams of size n P(λ) =(dimλ)²

n!

Question : is there a limit shape for (renormalized rotated) Young diagram taken randomly with Plancherel’s measure when n→ ∞?

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Normalized character values have simple expectations !

Fix π ∈S_n. Let us consider the random variable : X_π(λ) =χ^λ(π) = tr ρ_λ(π)

dimV_λ . Let us compute its expectation :

E(X_π) = 1 n!

X

λ⊢n

(dimV_λ)·tr ρ_λ(π)

= 1 n!tr^“L

λ⊢nV_λ^dim^V^λ”(π) = 1

n!trC[S_n](π) Last expression is easy to evaluate :

E(X_π) =δ_π,Id_n

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Convergence of cumulants

Recall : we proved that Q

iR_k_i₊₁≈Σ_k₁_,...,k_r. Thus

E(R2)≈n

E(Ri)≈0 if i >2 Var(Ri)≈0 if i ≥2

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Convergence of cumulants

Recall : we proved that Q

iR_k_i₊₁≈Σ_k₁_,...,k_r. Thus

limE(R2/n) =1 limE

Ri/√ nⁱ

=0 if i >2 lim Var

R_i/√ nⁱ

=0 if i ≥2

Easy to make it formal because R_k ∈Vect(Σ_π).

⇒ Random variables Ri/√

nⁱ converge in probability towards the sequence (0,1,0,0, . . .)

General lemma from Kerov :

convergence of cumulants⇒convergence of Young diagrams

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Existence of a limiting curve

Theorem (Logan and Shepp 77, Kerov and Vershik 77)

Let us take randomly (with Plancherel measure) a sequence of Young diagram λn of size n. Then, in probability, for the uniform convergence topology on continuous functions, one has :

r45^◦(h_1/^√_n(λ_n))→δ_Ω, where Ωis an explicit function drawn here :

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Convergence of q-Plancherel measure

Case where expectation of character values are big : there can not be a limit shape after dilatation.

we use the ﬁrst approximation for charactersΣ_π(λ)≈Q

p_µ_i(λ).

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Convergence of q-Plancherel measure

p_µ_i(λ).

Example : q−Plancherel measure (q<1)

deﬁned using representation of Hecke algebras one can prove

Eq(Σ_π) = (1−q)^|^µ^| Q

i1−q^µⁱ n(n−1). . .(n− |µ|+1) Thus

Eq(p_k)≈ (1−q)^k

Q 1−q^kn^k Varq(p_k)≈0

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Convergence of q-Plancherel measure

p_µ_i(λ).

Example : q−Plancherel measure (q<1)

deﬁned using representation of Hecke algebras one can prove

Eq(Σ_π) = (1−q)^|^µ^| Q

i1−q^µⁱ n(n−1). . .(n− |µ|+1) Thus

limEq(p_k/n^k) = (1−q)^k Q

i1−q^k lim Varq(p_k/n^k) =0

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Convergence of q-Plancherel measure

Theorem (F., Méliot, 2010)

Let q<1. In probability, under q-Plancherel measure,

∀k ≥1, p_k(λ)

|λ|^k −→Mn,q

(1−q)^k 1−q^k . Moreover,

∀i ≥1, λi

n −→Mn,q (1−q)qⁱ⁻¹; We also obtained the second-order asymptotics.

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End of the talk

Thank you for listening Questions ?