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CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
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The horizontal Pieri rule

Statement

Assume the Axiom of Choice. Let V=Cr, r≥1, let λ be a partition with ℓ(λ)≤r and let d≥0. Then Sλ(V)⊗Sym⁡d(V)≅⨁νSν(V), the sum over those partitions ν of ∣λ∣+d with ℓ(ν)≤r and [λ]⊆[ν] for which the skew diagram ν/λ is a horizontal strip (at most one box in each column, Skew diagrams and semistandard skew tableaux), each summand occurring with multiplicity one; equivalently sλhd=∑ν/λ horizontalsν in the rank-r Schur basis, where Sym⁡d(V) is the d-th symmetric power (Symmetric and exterior powers over an arbitrary field).

Facts & Assumptions

Given: AC, V=Cr, a partition λ with ℓ(λ)≤r and an integer d≥0.

[F1]

For d>0, the one-row Specht module S(d) is trivial: its tabloid module has one basis element and all column stabilizers are trivial. Thus S(d)(V)≅(V⊗d)Sd. This is isomorphic to the quotient symmetric power of Symmetric and exterior powers over an arbitrary field: the averaging operator P=d!−1∑σ∈Sdσ annihilates every coinvariance relation and induces the inverse to the quotient map restricted to invariants, because q(Pt)=q(t) and P fixes invariant tensors. These maps commute with GL⁡(V) (Schur modules and their characters, Column antisymmetrizers, polytabloids, and Specht modules). For d=0, use the empty partition in place of (d); its Schur module, V⊗0 and Sym⁡0V are all C.

[F2]

Littlewood--Richardson rule: for partitions λ,μ with ℓ(λ),ℓ(μ)≤r, Sλ(V)⊗Sμ(V)≅⨁ν: ℓ(ν)≤rSν(V)⊕cλμν, with cλμν the number of LR tableaux of shape ν/λ and content μ, and cλμν=0 unless λ⊆ν and ∣ν∣=∣λ∣+∣μ∣ (The Littlewood--Richardson tensor-product rule, Littlewood--Richardson tableaux and coefficients).

[F3]

A semistandard skew tableau of shape ν/λ and content (d) has all its entries equal to 1; weak increase along rows is automatic, and strict increase down columns forces every column of ν/λ to contain at most one box, so such a tableau exists if and only if ν/λ is a horizontal strip, and then it is unique; its reading word is the constant word 1 1⋯1, a lattice word (Skew diagrams and semistandard skew tableaux, Semistandard tableaux and Kostka numbers, Littlewood--Richardson tableaux and coefficients).

[F4]

The complete symmetric polynomial hd=∑1≤i1≤⋯≤id≤rxi1⋯xid equals s(d) by the one-row tableau expansion (and h0=s∅=1), and the Schur polynomials sν, ℓ(ν)≤r, are linearly independent: after multiplying a finite relation by aδr, the coefficient of the strictly decreasing exponent vector ν+δr is exactly that relation’s coefficient of sν (Semistandard tableaux expand Schur characters, Stable Schur functions from bialternants, The Littlewood--Richardson tensor-product rule).

Proof

1.1F1F2givenalgebra

Apply the Littlewood--Richardson rule [F2] with μ=(d) for d>0 and μ=∅ for d=0, and use S(d)(V)=Sym⁡d(V) from [F1]: Sλ(V)⊗Sym⁡d(V)≅⨁ν: ℓ(ν)≤rSν(V)⊕cλ,(d)ν.

2.1F2F3step 1.1algebra

For d=0, the unique empty tableau gives the single summand ν=λ. For d>0, the coefficient cλ,(d)ν counts LR tableaux of shape ν/λ and content (d); by [F3] this number is 1 when ν/λ is a horizontal strip and 0 otherwise, and it vanishes unless λ⊆ν and ∣ν∣=∣λ∣+d by [F2]. Substituting into step 1.1 gives the direct-sum decomposition, the sum being over precisely those horizontal strips.

3.1F1F4step 1.1step 2.1algebra∎

Taking characters in step 2.1 and using ch⁡Sν(V)=sν(x1,…,xr) and ch⁡Sym⁡d(V)=hd [F4] gives sλhd=∑ν/λ horizontalsν in the rank-r Schur basis; the two displayed statements are equivalent by the linear independence of the Schur characters [F4].

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Sources