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Murnaghan–Littlewood: top-degree reduced coefficients are Littlewood–Richardson coefficients

Statement

Let $α, β, γ$ be partitions with $∣ γ ∣ = ∣ α ∣ + ∣ β ∣$ . Then ${\overset{ˉ}{g}}_{α, β}^{γ} = c_{α, β}^{γ},$ the Littlewood–Richardson coefficient (Littlewood--Richardson tableaux and coefficients). In particular ${\overset{ˉ}{g}}_{α, β}^{α + β} = c_{α, β}^{α + β} = 1,$ where $α + β = (α_{1} + β_{1}, α_{2} + β_{2}, \dots)$ .

Facts & Assumptions

Given: Partitions $α, β, γ$ with $∣ γ ∣ = ∣ α ∣ + ∣ β ∣$ , and six-tuples of partitions $(μ, λ, η, ρ, θ, π)$ .

[F1]

${\overset{ˉ}{g}}_{α, β}^{γ} = \sum_{μ, λ, η, ρ, θ, π} c_{μ, θ, π}^{γ} c_{ρ, λ, θ}^{α} c_{ρ, η, π}^{β} g_{λ, η}^{μ}$ , and for every nonzero summand $∣ α ∣ = ∣ ρ ∣ + ∣ λ ∣ + ∣ θ ∣$ , $∣ β ∣ = ∣ ρ ∣ + ∣ η ∣ + ∣ π ∣$ , $∣ γ ∣ = ∣ μ ∣ + ∣ θ ∣ + ∣ π ∣$ , $∣ μ ∣ = ∣ λ ∣ = ∣ η ∣$ , with all summands nonnegative integers (The nonnegative expansion of a reduced Kronecker coefficient).

[F2]

$c_{a, b, c}^{δ} = [s_{δ}] s_{a} s_{b} s_{c}$ , symmetric in $a, b, c$ ; moreover $c_{a, b, \emptyset}^{δ} = [s_{δ}] s_{a} s_{b}$ , equal to the Littlewood–Richardson coefficient $c_{a, b}^{δ}$ (Threefold Schur coefficients $c_{a, b, c}^{δ}$ , The Littlewood–Richardson rule for products of Schur functions, Littlewood--Richardson tableaux and coefficients).

[F3]

LR tableaux have weakly increasing rows, strictly increasing columns, the prescribed content and a lattice reading word (Littlewood--Richardson tableaux and coefficients).

Proof

technique · direct

1.1F1

Degree collapse. Take a nonzero summand of the expansion in [F1]; by the degree equations of [F1], $∣ α ∣ + ∣ β ∣ - ∣ γ ∣ = (∣ ρ ∣ + ∣ λ ∣ + ∣ θ ∣) + (∣ ρ ∣ + ∣ η ∣ + ∣ π ∣) - (∣ μ ∣ + ∣ θ ∣ + ∣ π ∣) = 2 ∣ ρ ∣ + ∣ λ ∣ + ∣ η ∣ - ∣ μ ∣ = 2 ∣ ρ ∣ + ∣ μ ∣$ , using $∣ λ ∣ = ∣ η ∣ = ∣ μ ∣$ . Since $∣ γ ∣ = ∣ α ∣ + ∣ β ∣$ , this gives $2 ∣ ρ ∣ + ∣ μ ∣ = 0$ , so $ρ = μ = \emptyset$ and then $∣ λ ∣ = ∣ η ∣ = 0$ , so $λ = η = \emptyset$ .

2.1F1F2step 1.1

Collapse of the sum. With $ρ = λ = η = μ = \emptyset$ , the only possibly nonzero summands have $c_{\emptyset, \emptyset, θ}^{α} = [s_{α}] s_{θ} = δ_{α, θ}$ and $c_{\emptyset, \emptyset, π}^{β} = [s_{β}] s_{π} = δ_{β, π}$ , and $g_{\emptyset, \emptyset}^{\emptyset} = 1$ ; hence ${\overset{ˉ}{g}}_{α, β}^{γ} = c_{\emptyset, α, β}^{γ} = c_{α, β}^{γ}$ by the symmetry of the threefold coefficient in [F2], which is the first display.

3.1F2F3step 2.1

The top coefficient equals one. For $γ = α + β$ we have $c_{α, β}^{α + β} = 1$ . If $β = \emptyset$ , the skew shape is empty and its unique empty tableau proves this; this is also the induction base. Suppose $β \neq \emptyset$ . Existence: fill row $i$ of the skew shape $(α + β) / α$ , which has $β_{i}$ cells, with the letter $i$ ; the rows are constant and hence weakly increasing, the columns are strictly increasing because the letters increase down the rows, and the reading word is $1^{β_{1}} 2^{β_{2}} \dots$ , which is lattice because $β$ is weakly decreasing. Uniqueness: let $T$ be a Littlewood–Richardson tableau of shape $(α + β) / α$ and content $β$ . The first letter of its reading word is the rightmost cell of the first skew row; the lattice condition on the prefix of length one forces that letter to be $1$ , and since the row is weakly increasing every cell of the first row is $1$ . The first row is read first and consists of exactly $β_{1}$ cells, so it exhausts the $β_{1}$ letters $1$ of the content. Deleting the first row leaves the shape $(α^{'} + β^{'}) / α^{'}$ with $α^{'} = (α_{2}, α_{3}, \dots)$ , $β^{'} = (β_{2}, β_{3}, \dots)$ , carrying no letter $1$ ; its reading word, with every letter decreased by one, satisfies the lattice condition: the comparisons among letters $\geq 2$ hold because they hold for the corresponding prefixes of the original lattice word, and the comparison with letter $1$ is not required after the shift. Induction on the number of rows now shows that row $i$ consists of $β_{i}$ copies of $i$ , so $T$ is the displayed filling and $c_{α, β}^{α + β} = 1$ . With [F2] and step 2.1 this gives ${\overset{ˉ}{g}}_{α, β}^{α + β} = c_{α, β}^{α + β} = 1$ .

4.1F1F2step 2.1∎

Empty cases. If $α = \emptyset$ , step 2.1 gives ${\overset{ˉ}{g}}_{\emptyset, β}^{γ} = c_{\emptyset, β}^{γ} = [s_{γ}] s_{β} = δ_{γ, β}$ under the top-degree hypothesis. Thus equal size alone does not force $γ = β$ : the value is zero for every other partition of that size. Similarly, if $β = \emptyset$ the value is $δ_{γ, α}$ . If both inputs are empty then the top-degree hypothesis forces $γ = \emptyset$ and the coefficient is one. No choice principle is used.

Depends on

Used by

The exact global stabilization threshold for a fixed Kronecker product Theorem

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Emmanuel Briand, Rosa Orellana and Mercedes Rosas, The stability of the Kronecker product of Schur functions, arXiv:0907.4652v2 (2009); Murnaghan–Littlewood Theorem, p. 4 (standard reference, not scraped)
I. G. Macdonald, Symmetric Functions and Hall Polynomials, 2nd ed., Oxford University Press 1995; Chapter I §5 (Littlewood-Richardson rule) (standard reference, not scraped)