4.4 Symmetric Groups Farid Aliniaeifard York University - - PowerPoint PPT Presentation

4.4 Symmetric Groups

Farid Aliniaeifard

York University <http://math.yorku.ca/~faridanf/>

June 18, 2015

Overview

3.3 Λ is the unique indecomposable PSH 4.4 Symmetric Groups

3.3 Λ is the unique indecomposable PSH

◮ Goal: If A has only one primitive elements in its PSH-basis Σ,

then A must be isomorphic as a PSH to the ring of Symmetric functions Λ, after rescale the grading of A. PSH-isomorphism: A PSH-morphism A

φ

→ A

′ betweeen two PSH’s A and A ′ having

PSH-bases Σ and Σ

′ is a graded Hopf Algebra morphism for which

φ(N)Σ ⊂ NΣ

′. If A = A ′ and Σ = Σ ′ it will be called a

PSH-endomorphism. If φ is an isomorphism and restrict to a bijection Σ → Σ

′, it will be called a PSH-isomorphism; if it is both

a PSH-isomorphism and an endomorphism, it is a PSH-automorphism.

Theorem: Let A be a PSH with a PSH-basis Σ containing only one primitive ρ, and assume that the grading has been rescaled so that ρ has degree 1. Then, after renaming ρ = e1 = h1, one can find unique sequences {hn}n=0,1,2,... and {en}n=0,1,2,... of elements of Σ having the following properties:

◮ (a) h0 = e0 = 1, and h1 = e1 := ρ has ρ2 a sum of two

elements of Σ, namely ρ2 = h2 + e2.

◮ (b) For all n = 0, 1, 2, . . . , there exit unique elements hn, en in

An ∩ Σ that satisfy h⊥

2 en = 0

and e⊥

2 hn = 0

with h2, e2 being the two elements of Σ introduced in (a).

◮ (c) For k = 0, 1, 2, . . . , n one has

h⊥

k hn = hn−k and σ⊥hn = 0 for σ ∈ Σ \ {h0, h1, . . . , hn}

e⊥

k en = en−k and σ⊥en = 0 for σ ∈ Σ \ {e0, e1, . . . , en}.

In particular, e⊥

k hn = 0 = e⊥ k hn for k ≥ 2. ◮ (d) Their coproducts are

∆(hn) =

• i+j=n

hi ⊗ hj ∆(en) =

• i+j=n

ei ⊗ ej.

◮ (e) The elements hn, en in A satisfy the same relation

• i+j=n

(−1)ieihj = δo,n as their coproduct in Λ, along with the property that A = Z[h1, h2, . . .] = Z[e1, e2, . . .]

◮ (f) There is exactly one nontrivial automorphism A ω

→ A as a PSH, swapping hn ↔ en.

◮ (g) There are exactly two PSH-isomorphisms A → Λ.

γ : A → Λ hn → hn(X) en → en(X) and γω : A → Λ en → hn(X) hn → en(X)

⊥

Recall: Given a Hopf algebra A of finite type, and its (graded) dual A◦, let (., .) = (., .)A be the paring (f , a) = f (a) for f ∈ A◦ and a ∈ A. Then define for each f in A◦ an operator A

f ⊥

→ A as follows: for ∈ A with ∆(a) = a1 ⊗ a2, let f ⊥(a) =

• (f , a1)a2.

4.4 Symmetric Groups

Notations:

◮ Consider the tower of symmetric groups Gn = Sn and

A = A(G∗) =: A(S).

◮ Denote by 1Sn, sgnSn the trivial character and sign character

• f Sn

◮ For a partition λ of n, denote by 1Sλ, sgnSλ the trivial and sign

character restrict to Younge subgroup Sλ = Sλ1 × Sλ2 × . . .

◮ denote by 1λ the class function which is the characteristic

function for the Sn-conjugacy class of permutation of cycle type λ

◮ zλ := m1!.m2! . . . if λ = (1m1!, 2m2, . . .) with multiplicity mi

for the part i

Theorem

Irreducible complex characters {χλ} of Sn are indexed by partition λ in Parn, and one has a PSH-isomorphism, the Frobenius characteristic map, A = A(Sn) ch → Λ that for n ≥ 0 and λ ∈ Parn sends 1Sn → hn sgnSn → en χλ → sλ IndSn

Sλ1Sλ → hλ

IndSn

SλsgnSλ → eλ

1λ → pλ

zλ

(where ch is extended to a C-linear map AC → ΛC) and for n ≥ 1 sends 1(n) → pn n

Proof.

◮ A with

m := Indi+j

i,j

: Ai ⊗ Aj → Ai+j ∆ := ⊕i+j=nResi+j

i,j

: An → ⊕i+j=nAi ⊗ Aj is a PSH with PSH-basis Σ = ⊔n≥0Irr(Sn) .

◮ The unique irreducible character ρ = 1S1 of S1 is the only

element of C = Σ ∩ P, P is the set of primitive elements.

◮ Thus Theorem (g) tells us that there are two

PSH-isomorphisms A → Λ, each of which sends Σ to the PSH-basis of Schur functions {sλ} for Λ.

◮ we can pin down one of the two isomorphisms to call ch, by

insisting that it map the two characters 1S2 and sgnS2 in Irr(S2) to h2 and e2.

◮ 1⊥ S2 annihilates sgnSn and sgnS2 annihilates 1Sn ◮ Theorem (b) ⇒ ch(1Sn) = hn and ch(sgnSn) = en. ◮ By induction products

IndSn

Sλ1Sλ → hλ

IndSn

SλsgnSλ → eλ ◮ AC is a Hopf algebra and has the C-bilinear form (., .)C. ◮ 1(n) is a primitive element in AC ⇒ ch(1(n)) is a scalar

multiple of pn (Corollary 3.9).

◮ To find the scalar:

pn = mn ⇒ (hn, pn)Λ = (hn, mn)Λ = 1, while ch−1(hn) = 1Sn, we have (1Sn, 1(n)) = 1 n!(n − 1)! = 1 n ⇒ ch(1(n)) = pn

n . ◮ Exercise 4.28(d) ⇒ ch(1λ) = pλ zλ

References

Grinberg and Reiner Hopf Algerba in Combinatorics Bruce Sagan (2000) The Symmetric Group

The End