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Row padding need not preserve a Kronecker coefficient before stability

Statement refuted

Refuted claim: adding $t^{k}$ to the first $k$ rows of the first two arguments and $t k$ to the first part of the third preserves the Kronecker coefficient for every $t \geq 0$ as soon as the three first parts are ordered $λ_{1} + t \leq μ_{1} + t \leq ν_{1} + t k$ (Row-gap reduction and conditional k-stability of Kronecker coefficients).

The witness is $λ = μ = (2, 2)$ , $ν = (3, 1)$ , $k = 1$ ( $n = 4$ , $s = n - ν_{1} = 1$ ): $g ((2, 2), (2, 2), (3, 1)) = 0, g ((2 + t, 2), (2 + t, 2), (3 + t, 1)) = 1 for every t \geq 1.$ Thus adding $(t)$ to the first rows does not preserve the coefficient from $t = 0$ on: the row-separation hypothesis of the theorem fails at $t = 0$ , since $\min {λ_{1}, μ_{1}} = 2 < 3 = \max {λ_{2}, μ_{2}} + (n - ν_{1})$ . Eventual constancy is not immediacy.

Facts & Assumptions

Given: The partitions $λ = μ = (2, 2)$ and $ν = (3, 1)$ of $n = 4$ , with $k = 1$ and $s = 1$ ; the shifted triples $λ + (t), μ + (t), ν + (t)$ of $n + t$ (First-row padding of a partition); the conjugacy classes of $S_{4}$ and $S_{5}$ with sizes $1, 6, 3, 8, 6$ on $(1^{4}), (2, 1^{2}), (2^{2}), (3, 1), (4)$ and $1, 10, 15, 20, 20, 30, 24$ on $(1^{5}), (2, 1^{3}), (2^{2}, 1), (3, 1^{2}), (3, 2), (4, 1), (5)$ .

[F1]

For $ρ, σ, τ ⊢ m$ , $g (ρ, σ, τ) = \frac{1}{m!} \sum_{υ \in S_{m}} χ^{ρ} (υ) χ^{σ} (υ) χ^{τ} (υ)$ (The Kronecker coefficient is a three-character inner product).

[F2]

Cycle type indexes the conjugacy class, and the class of cycle type $υ$ has size $m! / z_{υ}$ with $z_{υ} = \prod_{i} i^{m_{i} (υ)} m_{i} (υ)!$ (The conjugacy classes of $S_{n}$ are indexed by the tuples $(c_{1}, \dots, c_{n})$ with $\sum k c_{k} = n$ , If $σ \in S_{n}$ has $c_{k}$ cycles of length $k$ , then $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}!$ ).

[F3]

$M^{ρ}$ is the permutation module on the $ρ$ -tabloids; its character is $χ_{M^{ρ}} (υ) = # {ρ -tabloids fixed by υ}$ , and Young's rule $M^{ρ} = ⨁_{λ} K_{λ ρ} S^{λ}$ gives the irreducible multiplicities (Young subgroups, tabloids, and permutation modules, Young's rule for complex permutation modules, Characters add on direct sums, multiply on tensor products, and conjugate on duals).

[F5]

Kostka numbers count fillings with prescribed content, weak rows and strict columns (Semistandard tableaux and Kostka numbers).

[F4]

Row-gap theorem, part (i) (Row-gap reduction and conditional k-stability of Kronecker coefficients).

Counterexample

technique · direct

1.1F2F3F5algebra

( $S_{4}$ characters.) For $m = 4, 5$ and content $(m - j, j)$ with $j = 1, 2$ , entries are $1, 2$ . Strict columns force at most two rows, and every lower-row entry is $2$ , so the possible shapes are $(m - a, a)$ with $0 \leq a \leq j$ . For each such shape, the lower row is $2^{a}$ and the top row is $1^{m - j} 2^{j - a}$ ; it is semistandard since $a \leq j \leq m - j$ . Thus each of these Kostka numbers is $1$ and all others vanish by [F5]. Also $M^{(m)}$ is trivial and its sole Specht constituent is $S^{(m)}$ by [F3]. For $ρ = (3, 1)$ the tabloids are the ordered $(3, 1)$ -set pairs of $[4]$ , equivalently the $3$ -subsets, and a $3$ -subset is fixed by $υ$ exactly when its complementary point is fixed; hence $χ_{M^{(3, 1)}} = (4, 2, 0, 1, 0)$ on the five classes in the order of the Given. Subtracting the trivial character and using [F3] gives $χ^{(3, 1)} = (3, 1, - 1, 0, - 1)$ , whose norm $\frac{1}{24} [9 + 6 + 3 + 0 + 6] = 1$ confirms it is the standard character. For $ρ = (2, 2)$ the tabloids are the ordered pairs $(A, A^{c})$ of $2$ -subsets, so $χ_{M^{(2, 2)}} (υ) = # {A : ∣ A ∣ = 2, υ A = A}$ , giving $(6, 2, 2, 0, 0)$ ; by [F3] $χ^{(2, 2)} = χ_{M^{(2, 2)}} - χ^{(4)} - χ^{(3, 1)} = (2, 0, 2, - 1, 0)$ , with norm $\frac{1}{24} [4 + 0 + 12 + 8 + 0] = 1$ .

2.1F2F3step 1.1algebra

( $S_{5}$ characters.) The tabloids of shape $(3, 2)$ are the $3$ -subsets of $[5]$ , and a $3$ -subset is invariant precisely when it is a union of cycles; counting over the seven classes gives $χ_{M^{(3, 2)}} = (10, 4, 2, 1, 1, 0, 0)$ . The tabloids of shape $(4, 1)$ are the $4$ -subsets, whose complements are points, so $χ_{M^{(4, 1)}} (υ) = # f i x (υ) = (5, 3, 1, 2, 0, 1, 0)$ and $χ^{(4, 1)} = (4, 2, 0, 1, - 1, 0, - 1)$ . Subtracting the two characters already found from $χ_{M^{(3, 2)}}$ by [F3] gives $χ^{(3, 2)} = (10, 4, 2, 1, 1, 0, 0) - (1, 1, 1, 1, 1, 1, 1) - (4, 2, 0, 1, - 1, 0, - 1) = (5, 1, 1, - 1, 1, - 1, 0)$ , of norm $\frac{1}{120} [25 + 10 + 15 + 20 + 20 + 30 + 0] = 1$ .

2.2F1F2step 1.1algebra

( $t = 0$ .) Applying [F1] with $m = 4$ and the class sizes of [F2] gives $g ((2, 2), (2, 2), (3, 1)) = \frac{1}{24} (1 \cdot 2^{2} \cdot 3 + 6 \cdot 0^{2} \cdot 1 + 3 \cdot 2^{2} \cdot (- 1) + 8 \cdot (- 1)^{2} \cdot 0 + 6 \cdot 0^{2} \cdot (- 1)) = \frac{12 - 12}{24} = 0$ .

3.1F1F2step 2.1algebra

( $t = 1$ .) Applying [F1] with $m = 5$ and the class sizes of [F2] gives $g ((3, 2), (3, 2), (4, 1)) = \frac{1}{120} (1 \cdot 5^{2} \cdot 4 + 10 \cdot 1^{2} \cdot 2 + 15 \cdot 1^{2} \cdot 0 + 20 \cdot 1^{2} \cdot 1 + 20 \cdot 1^{2} \cdot (- 1) + 30 \cdot (- 1)^{2} \cdot 0 + 24 \cdot 0^{2} \cdot (- 1)) = \frac{100 + 20 + 20 - 20}{120} = \frac{120}{120} = 1$ .

4.1F4step 3.1

(All $t \geq 1$ .) The triple $((3, 2), (3, 2), (4, 1))$ satisfies the hypotheses of [F4] with $k = 1$ : $3 = λ_{1} \leq μ_{1} = 3 \leq ν_{1} = 4$ and $\min {3, 3} = 3 \geq \max {2, 2} + (5 - 4) = 3$ . Hence $g ((3 + u, 2), (3 + u, 2), (4 + u, 1)) = g ((3, 2), (3, 2), (4, 1)) = 1$ for every $u \geq 0$ , and substituting $u = t - 1$ gives $g ((2 + t, 2), (2 + t, 2), (3 + t, 1)) = 1$ for every $t \geq 1$ .

5.1step 2.2step 4.1F4∎

Consequently $t \mapsto g ((2 + t, 2), (2 + t, 2), (3 + t, 1))$ takes the value $0$ at $t = 0$ and the value $1$ for every $t \geq 1$ : the padded family is not constant from $t = 0$ on, and the separation hypothesis of [F4] is exactly what fails at $t = 0$ .

Remarks

The example is Pak-Panova's Example 4.4, recomputed here from the character tables of $S_{4}$ and $S_{5}$ rather than cited. Together with Row-gap reduction and conditional k-stability of Kronecker coefficients it shows that eventual constancy is not immediate constancy: the padded coefficient can jump at the first value of $t$ for which the rows $k$ and $k + 1$ are separated.
The all- $t$ conclusion in step 4.1 uses the proved row-gap theorem at the separated $t = 1$ triple; its proof supplies the single-cut Littlewood–Richardson/Jacobi–Trudi argument locally. The $t = 0$ and $t = 1$ values remain independent finite character computations.

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Sources

Igor Pak and Greta Panova, Bounds on the Kronecker coefficients, arXiv:1406.2988v2 (2014) (standard reference, not scraped)
Charlotte Chan, Representation Theory of Symmetric Groups (complete lecture notes) (standard reference, not scraped)