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Restriction to the two-block Young subgroup has LR multiplicities

Statement

Let 0≤r≤n and let K≤Sn be the subgroup preserving each of the two blocks B1={0,…,n−r−1} and B2={n−r,…,n−1} of the underlying set {0,…,n−1} (The finite symmetric group Sn, one-line notation, and cycle notation); under the order-preserving relabelling i↦i−1, these are the two blocks of sizes n−r and r in the outer-product definition (The outer induction product of symmetric-group characters); it is the image of the external direct product Sn−r×Sr acting on B1 and B2 (The external direct product G×H with componentwise multiplication). Identify the complex character ring R(K) with R(Sn−r)⊗ZR(Sr) by external tensor products χ⊠ψ of characters (The tensor product of two complex representations, The character ring of a direct product is the tensor product of the factor character rings). For a partition λ⊢n let χλ be the character of the complex Specht module Sλ, an irreducible character of Sn (Column antisymmetrizers, polytabloids, and Specht modules, Complex Specht modules are irreducible), and let cνμλ be the Littlewood–Richardson coefficient (Littlewood--Richardson tableaux and coefficients). Then, in R(K), χλ∣K=∑ν⊢n−r ∑μ⊢rcνμλ (χν⊠χμ).

In particular the multiplicity of the irreducible character χν⊠χμ in the restricted character χλ∣K equals cνμλ; for r=n this says that the multiplicity is 1 if μ=λ and 0 otherwise. No choice principle is used.

Facts & Assumptions

Given: Integers 0≤r≤n, the symmetric group Sn=Sym⁡({0,…,n−1}) with the composition and cycle conventions of The finite symmetric group Sn, one-line notation, and cycle notation, the two blocks B1={0,…,n−r−1} and B2={n−r,…,n−1} (so B1=∅ when r=n), and the subgroup K of Sn preserving each block.

[F1]

Concretely K=Sym⁡(B1)×Sym⁡(B2): the restrictions to its two blocks determine each element uniquely. It is the image of the external direct product Sn−r×Sr under the two-block embedding of The outer induction product of symmetric-group characters, transported by i↦i−1 (The external direct product G×H with componentwise multiplication). This uses block sizes, not a partition (n−r,r); either block may be empty.

[F2]

The Specht characters exhaust the irreducible characters of each symmetric group (Specht modules classify the complex irreducibles of Sn). External tensor products: for characters χ′ of Sn−r and ψ′ of Sr the assignment (σ,τ)↦χ′(σ)ψ′(τ) defines the character χ′⊠ψ′ of Sn−r×Sr on the tensor product of the underlying representations (The tensor product of two complex representations); the map R(Sn−r)⊗ZR(Sr)→R(K), χ′⊗ψ′↦χ′⊠ψ′, is an isomorphism of rings, and as χ′,ψ′ range over irreducible characters the products χ′⊠ψ′ are pairwise distinct, orthonormal, and form a Z-basis of R(K) (The character ring of a direct product is the tensor product of the factor character rings).

[F3]

For a finite group G the irreducible complex characters are orthonormal, ⟨χi,χj⟩=δij (The first orthogonality relation for irreducible complex characters), and they form an orthonormal basis of the space of class functions (The irreducible complex characters form an orthonormal basis of cf(G)); every irreducible character lies in R(G), the Z-span of the irreducible characters (Virtual characters and the character ring R(G) of a finite group); and ⟨χV,χW⟩=dim⁡Hom⁡G(W,V) for finite-dimensional complex representations V,W (The class-function inner product ⟨χV,χW⟩ equals dim⁡Hom⁡G(W,V)).

[F4]

Frobenius reciprocity: for H≤G, a complex character θ of H and a complex character ψ of G, ⟨Ind⁡HGθ,ψ⟩G=⟨θ,Res⁡HGψ⟩H (Frobenius reciprocity for complex characters, The induced character Ind⁡HGχ of a complex character).

[F5]

For ν⊢n−r and μ⊢r, the outer induction product χν∘χμ is Ind⁡Sn−r×SrSn(χν⊠χμ) under the identification of [F1], and the outer Littlewood–Richardson rule gives χν∘χμ=∑λ⊢ncνμλ χλ in R(Sn) (The outer induction product of symmetric-group characters, The outer Littlewood–Richardson rule, Littlewood--Richardson tableaux and coefficients).

Proof

technique · direct
1.1F2F5given

For ν⊢n−r and μ⊢r put θνμ:=χν⊠χμ∈R(K); by [F2] it is an honest character of K, and the family (θνμ)ν,μ is orthonormal and Z-spans R(K). By [F5] its induction to Sn is the outer product χν∘χμ=∑λ⊢ncνμλχλ.

2.1step 1.1F1F3F4F5algebra

Fix λ⊢n. Applying [F1] and [F4] to the character θνμ of K and the character χλ of Sn, and then using [F5] and the orthonormality of the irreducible characters χλ′ from [F3], gives ⟨θνμ,χλ∣K⟩K=⟨Ind⁡Sn−r×SrSnθνμ,χλ⟩Sn=∑λ′⊢ncνμλ′⟨χλ′,χλ⟩=cνμλ.

2.2step 1.1F2F3algebra

The characters θνμ are exactly the irreducible characters of K. Indeed, every irreducible character χi of K lies in R(K) by [F3] and hence is a finite integral combination χi=∑ν,μnνμθνμ because the θνμ form a Z-basis of R(K) by [F2]. Computing the norm with [F3] and the orthonormality of the θνμ gives 1=⟨χi,χi⟩=∑ν,μnνμ2, so exactly one coefficient equals ±1 and the others are 0, that is, χi=±θνμ for a single pair (ν,μ); a character has positive degree, so χi=θνμ. Hence the irreducible characters of K form a subfamily of the orthonormal family (θνμ); the irreducibles are an orthonormal basis of the class functions by [F3], so their number is dim⁡cf(K), while an orthonormal family in cf(K) has at most dim⁡cf(K) elements; the numbers are therefore equal and the two families coincide.

3.1step 2.1step 2.2F2F3

The restricted character χλ∣K is a class function on K, and by [F3] it expands in the orthonormal basis of irreducible characters, which is the family (θνμ) by step 2.2, with coefficient ⟨χλ∣K,θνμ⟩K. This is the conjugate of ⟨θνμ,χλ∣K⟩K=cνμλ from step 2.1 and hence equals that same real integer: χλ∣K=∑ν⊢n−r∑μ⊢r⟨θνμ,χλ∣K⟩K θνμ. Substituting the value cνμλ from step 2.1 into each coefficient gives the displayed identity of the statement.

4.1step 3.1F2F3algebra∎

The coefficient of the irreducible character θνμ=χν⊠χμ in the expansion of step 3.1 is cνμλ, so this is the multiplicity of χν⊠χμ in χλ∣K. If r=n then K=Sn and the only partition of n−r=0 is ν=∅, so the identity reads χλ=∑μ⊢nc∅μλχμ; comparing with the orthonormal-basis expansion χλ=∑μδλμχμ from [F3] gives c∅μλ=δλμ, so the multiplicity is 1 for μ=λ and 0 otherwise. Similarly r=0 gives μ=∅ and cν∅λ=δλν; when n=0 both partitions are empty and the coefficient is 1.

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