Caractères irréductibles du groupe symétrique et grands

Transcription

Caractères irréductibles du groupe symétrique et grands
diagrammes de Young
Valentin Féray
collaborateurs : Piotr Śniady (Wroclaw), Pierre-Loïc Méliot (MLV)
LaBRI
GT CÉA, 9 avril 2010
Valentin Féray (LaBRI)
Caractères de Sn
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Outline of the talk
1
Rappel de la première séance
2
Asymptotique des caractères
Equivalents classiques
Forme limite de diagrammes de Young
Caractères de Sn
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Représentation de Sn
Exact formula for character values
We want to compute the value of the irreducible characters of Sn :
indexed by λ ⊢ n ;
ι
on a small permutation π ∈ Sk ֒→ Sn .
Caractères de Sn
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Représentation de Sn
Exact formula for character values
We want to compute the value of the irreducible characters of Sn :
indexed by λ ⊢ n ;
ι
on a small permutation π ∈ Sk ֒→ Sn .
Definition
Nσ1 ,σ2 (T ) is the number of functions 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).
Σπ (λ) :=
n · (n − 1) . . . (n − k + 1)χλ (ι(π))
=
dim λ
Caractères de Sn
X
(−1)σ2 Nσ1 ,σ2 (λ).
σ1 ,σ2 ∈Sk
σ2 σ1 =π
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Asymptotics is easy to read on this formula
Model : fix a permutation π0 ∈ Sk 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 by c)
7→
Question : asymptotics of
χλ (π)
dim λ
?
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7→
λ
Question : asymptotics of χdim(π)
λ ?
Easy on the N’s : Nσ1 ,σ2 (c · λ) = c |C (σ2 )| Nσ1 ,σ2 (λ)
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7→
λ
λ ?
Dominant term of Σπ (λ) ∼ nk χλ (π)/ dim(λ) :


X µ
Y

λj i 
Nπ,Idk (λ) =
µi ∈type(π)
j
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7→
λ
λ ?


X µ
Y
Y

λj i  =
Nπ,Idk (λ) =
pµi (λ)
µi ∈type(π)
j
Caractères de Sn
µi ∈type(π)
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Consider π = ι(π0 ) (i.e. we just add fixpoints) and λ = c • λ0 = λ dilated
by c (i.e. horizontal and vertical lengths are multiplied by c)
7→
Question : asymptotics of
χλ (π)
dim λ
?
Caractères de Sn
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7→
λ
λ ?
Easy on the N’s : Nσ1 ,σ2 (c • λ) = c |C (σ1 )|+|C (σ2 )| Nσ1 ,σ2 (λ)
Caractères de Sn
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7→
λ
λ ?
Easy on the N’s : Nσ1 ,σ2 (c • λ) = c |C (σ1 )|+|C (σ2 )| Nσ1 ,σ2 (λ)
X
±Nσ1 ,σ2 (λ)
σ1 σ2 =π
|C (σ1 )|+|C (σ2 )| maximal
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Free cumulants
Σπ (c • λ) =
But,
X
σ1 σ2 =π
|C (σ1 )|+|C (σ2 ) maximal
σ1 σ2 = π
|C (σ1 )| + |C (σ2 )| maximal
≃
Q
Caractères de Sn
±Nσ1 ,σ2 (λ)
NCµi ≃
Q
Trees(µi ).
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Free cumulants
Σπ (c • λ) =
X
σ1 σ2 =π
|C (σ1 )|+|C (σ2 ) maximal
±Nσ1 ,σ2 (λ)
Q
Q
σ1 σ2 = π
≃ NCµi ≃ Trees(µi ).
|C (σ1 )| + |C (σ2 )| maximal
With generating series, one can prove (Rattan, 2006)
Y
rhs =
Rµi +1 (λ)
But,
Rk : free cumulants defined from the shape ωλ 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 diagrams has a limit)
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.
Caractères de Sn
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Description of the problem
Consider Plancherel’s probability measure on Young diagrams of size n
P(λ) =
(dim λ)2
n!
∃ equivalent description as a Markov process
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 ∈ Sn . Let us consider the random variable :
tr ρλ (π)
.
Xπ (λ) = χ (π) =
dim Vλ
λ
Let us compute its expectation :
E(Xπ ) =
1 X
(dim Vλ ) · tr ρλ (π)
n!
λ⊢n
=
1
1 “
tr L V dim Vλ ” (π) =
tr
(π)
n!
n! C[Sn ]
λ⊢n λ
Last expression is easy to evaluate :
E(Xπ ) = δπ,Idn
Caractères de Sn
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Convergence of cumulants
Recall : we proved that
Thus
Q
i
Rki +1 ≈ Σ(1
... k1 )(k1 +1 ... k1 +k2 )...
E(R2 ) ≈ E n · χ· (Id) = n
E(Ri ) ≈ E ni · χ· ((1 . . . i)) = 0 if i > 2
Var(Ri ) = E(Ri2 ) − E(Ri )2 ≈ E n2i · χ· ((1 . . . i)(i + 1 . . . 2i)
2
− E ni · χ· ((1 . . . i)) = 0 if i ≥ 2
Caractères de Sn
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Convergence of cumulants
Recall : we proved that
Thus
Q
i
Rki +1 ≈ Σ(1
... k1 )(k1 +1 ... k1 +k2 )...
lim E (R2 /n) = 1
√ i
lim E Ri / n = 0 if i > 2
√ i
lim Var Ri / n = 0 if i ≥ 2
Easy to make it formal because Rk ∈ Vect(Σπ ).
√ i
⇒ Random variables Ri / n converge in probability towards the sequence
(0, 1, 0, 0, . . . )
General lemma from Kerov :
convergence of cumulants ⇒ convergence of Young diagrams
Caractères de Sn
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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◦ (h1/√n (λn )) → δΩ ,
where Ω is an explicit function drawn here :
Caractères de Sn
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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.
Q
we use the first approximation for characters Σπ (λ) ≈ pµi (λ).
Caractères de Sn
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Q
Example : q−Plancherel measure (q < 1)
defined using representation of Hecke algebras
one can prove
Thus
(1 − q)|µ|
n(n − 1) . . . (n − |µ| + 1)
Eq (Σπ ) = Q
µi
i 1−q
(1 − q)k k
n
Eq (pk ) ≈ Q
k
i 1−q
Caractères de Sn
Varq (pk ) ≈ 0
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Q
Example : q−Plancherel measure (q < 1)
defined using representation of Hecke algebras
one can prove
Thus
(1 − q)|µ|
n(n − 1) . . . (n − |µ| + 1)
Eq (Σπ ) = Q
µi
i 1−q
(1 − q)k
lim Eq (pk /nk ) = Q
k
i1−q
Caractères de Sn
lim Varq (pk /nk ) = 0
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Theorem (F., Méliot, 2010)
Let q < 1. In probability, under q-Plancherel measure,
∀k ≥ 1,
pk (λ)
(1 − q)k
−→
.
M
n,q
|λ|k
1 − qk
∀i ≥ 1,
λi
−→Mn,q (1 − q) q i −1 ;
n
Moreover,
We also obtained the second-order asymptotics.
Caractères de Sn
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End of the talk
Thank you for listening
Questions ?
Caractères de Sn
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Caractères irréductibles du groupe symétrique et grands

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