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Kronecker Coefficients and Internal Products — Examples

1 · Prerequisites

2 · Summary

Objective. This companion computes the complete internal product table for $S_{3}$ , fixes the exact domain on which a padded family exists, and exhibits the first boundary at which eventual constancy fails to be immediacy. All three items are read through the definitions and theorems of kronecker-coefficients-and-internal-products; none of them is presented as a general rule.

The first item is the complete internal product table for $S_{3}$ (The complete internal product table for S_3): the characters of $S^{(3)}$ , $S^{(2, 1)}$ and $S^{(1, 1, 1)}$ are recomputed from the Frobenius dictionary rather than imported from another examples page, and the three-character formula gives $S^{(2, 1)} \otimes S^{(2, 1)} ≅ S^{(3)} \oplus S^{(2, 1)} \oplus S^{(1, 1, 1)}$ with all other entries trivial. The second item is the padding threshold (First-row padding below the threshold is undefined): for $α = (2, 1)$ the padded sequence is a partition exactly from $n \geq 5$ on, so the padded family, and with it any padded Kronecker coefficient with this shape, is undefined below that size.

The third item is the pre-stability failure (Row padding need not preserve a Kronecker coefficient before stability), Pak–Panova's Example 4.4 recomputed from the character tables of $S_{4}$ and $S_{5}$ : $g ((2, 2), (2, 2), (3, 1)) = 0$ while $g ((2 + t, 2), (2 + t, 2), (3 + t, 1)) = 1$ for every $t \geq 1$ . The separation hypothesis of the row-gap theorem fails at $t = 0$ , which is exactly why the padded sequence can jump at its first step; once the rows are separated, the theorem forces the value $1$ forever.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The complete internal product table for S_3

Example

For $S_{3}$ the three irreducible complex representations are $S^{(3)}$ (the trivial representation, dimension $1$ ), $S^{(2, 1)}$ (the standard representation, dimension $2$ ) and $S^{(1, 1, 1)}$ (the sign representation, dimension $1$ ). Their characters on the three conjugacy classes, indexed by cycle type in the order $(1^{3}), (2, 1), (3)$ , are

representation	$(1^{3})$	$(2, 1)$	$(3)$
$S^{(3)}$	$1$	$1$	$1$
$S^{(2, 1)}$	$2$	$0$	$- 1$
$S^{(1, 1, 1)}$	$1$	$- 1$	$1$

The complete internal (Kronecker) product table is therefore $S^{(3)} \otimes X ≅ X for every irreducible X,$ $S^{(2, 1)} \otimes S^{(2, 1)} ≅ S^{(3)} \oplus S^{(2, 1)} \oplus S^{(1, 1, 1)}, S^{(2, 1)} \otimes S^{(1, 1, 1)} ≅ S^{(2, 1)}, S^{(1, 1, 1)} \otimes S^{(1, 1, 1)} ≅ S^{(3)} .$ Equivalently, all Kronecker coefficients $g (λ, μ, ν)$ with $λ, μ, ν ⊢ 3$ are $0$ or $1$ , and the table is invariant under all permutations of the three arguments. The character table is recomputed here from the Frobenius dictionary rather than imported from an examples page.

Facts & Assumptions

Given: The partitions of $3$ : $(3)$ , $(2, 1)$ , $(1^{3})$ (Partitions, English diagrams, and conjugation); the class-function inner product $⟨ φ, ψ ⟩ = \frac{1}{6} \sum_{σ \in S_{3}} φ (σ) \overline{ψ (σ)}$ ; and the cycle-type data $z_{(1^{3})} = 6$ , $z_{(2, 1)} = 2$ , $z_{(3)} = 3$ , so the three conjugacy classes have sizes $6 / z = 1, 3, 2$ respectively (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 $∣ C_{S_{n}} (σ) ∣ = \prod_{k = 1}^{n} k^{c_{k}} c_{k}!$ ).

[F1]

$ch (χ^{λ}) = s_{λ}$ for every $λ ⊢ n$ , and $ch (f) = \sum_{ρ ⊢ n} f (ρ) p_{ρ} / z_{ρ}$ (The characteristic of a Specht character is a Schur function, The Frobenius characteristic map).

[F2]

For partitions $ρ$ and $λ$ of the same integer, $h_{λ} = \sum_{ρ} N (λ, ρ) p_{ρ} / z_{ρ}$ in $Λ_{Q}$ , in particular $h_{1} = p_{1}$ , $h_{2} = \frac{1}{2} p_{1}^{2} + \frac{1}{2} p_{2}$ and $h_{3} = \frac{1}{6} p_{1}^{3} + \frac{1}{2} p_{1} p_{2} + \frac{1}{3} p_{3}$ (Complete homogeneous functions expand in power sums with cycle-distribution coefficients).

[F3]

The family ${p_{ρ} / z_{ρ} : ρ ⊢ n}$ is a $Q$ -basis of $Λ_{Q}^{n}$ , so coefficients in that basis are unique (Power sums form a rational but not integral stable basis).

[F4]

$g (λ, μ, ν) = \frac{1}{6} \sum_{σ \in S_{3}} χ^{λ} (σ) χ^{μ} (σ) χ^{ν} (σ)$ (The Kronecker coefficient is a three-character inner product), and $g$ is symmetric in the three arguments (Kronecker coefficients are nonnegative integers with full S_3 and conjugation symmetries).

[F5]

The tabloid module $M^{μ}$ is the permutation representation on the $μ$ -tabloids (Young subgroups, tabloids, and permutation modules), and Young's rule gives $M^{μ} ≅ ⨁_{λ ⊢ 3} (S^{λ})^{\oplus K_{λ μ}}$ , so the Young permutation character is $φ^{μ} = \sum_{λ} K_{λ μ} χ^{λ}$ by additivity of characters (Young's rule for complex permutation modules, Characters add on direct sums, multiply on tensor products, and conjugate on duals); in particular $φ^{(3)} = χ^{(3)}$ and $φ^{(2, 1)} = χ^{(3)} + χ^{(2, 1)}$ because content $(2, 1)$ consists of two $1$ 's and one $2$ : its only semistandard shapes are $(3)$ and $(2, 1)$ , with a unique filling for each. A three-cell column is impossible (Semistandard tableaux and Kostka numbers).

[F6]

$ch (φ^{μ}) = h_{μ}$ for every $μ ⊢ 3$ (The characteristic of a Young permutation character is complete homogeneous).

[F7]

$s_{λ} = \det (h_{λ_{i} - i + j})$ for every partition $λ$ , so $s_{(3)} = h_{3}$ (Jacobi–Trudi and dual Jacobi–Trudi identities).

[F8]

The sign representation is one-dimensional with $sgn (σ) = (- 1)^{3 - c (σ)}$ , where $c (σ)$ is the number of cycles of $σ$ counted with fixed points; the sign twist of $S^{(3)}$ is $S^{(1, 1, 1)}$ (Sign twist corresponds to the omega involution); hence, once step 1.1 identifies $S^{(3)}$ as trivial, $χ^{(1, 1, 1)}$ has values $1, - 1, 1$ on $(1^{3}), (2, 1), (3)$ (The sign representation of $S_{n}$ and the restriction ${Res}_{H}^{G} (V)$ of a representation to a subgroup, A $k$ -cycle has sign $(- 1)^{k - 1}$ , and $sgn (σ) = (- 1)^{n - c (σ)}$ when fixed points are counted as cycles).

Verification

technique · direct

1.1F1F2F3F7

Character of $S^{(3)}$ : by [F7] $s_{(3)} = h_{3}$ , and [F2] expands $h_{3} = \frac{1}{6} p_{1}^{3} + \frac{1}{2} p_{1} p_{2} + \frac{1}{3} p_{3}$ in the basis ${p_{ρ} / z_{ρ}}$ . By [F1] the coefficient of $p_{ρ} / z_{ρ}$ in $s_{(3)} = ch (χ^{(3)})$ is $χ^{(3)} (ρ)$ , so uniqueness [F3] gives the values $χ^{(3)} (1^{3}) = 1$ , $χ^{(3)} (2, 1) = 1$ , $χ^{(3)} (3) = 1$ .

1.2F1F2F3F5F6algebra

Character of $S^{(2, 1)}$ : the Young permutation character of the tabloid module $M^{(2, 1)}$ has characteristic $ch (φ^{(2, 1)}) = h_{2} h_{1}$ by [F6], and $φ^{(2, 1)} = χ^{(3)} + χ^{(2, 1)}$ by Young's rule [F5]. Applying [F1] therefore gives $ch (χ^{(2, 1)}) = h_{2} h_{1} - h_{3}$ ; by the expansions [F2], $h_{2} h_{1} - h_{3} = (\frac{1}{2} p_{1}^{2} + \frac{1}{2} p_{2}) p_{1} - (\frac{1}{6} p_{1}^{3} + \frac{1}{2} p_{1} p_{2} + \frac{1}{3} p_{3}) = \frac{1}{3} p_{1}^{3} - \frac{1}{3} p_{3}$ . By [F1] and uniqueness [F3], the values are $χ^{(2, 1)} (1^{3}) = 6 \cdot \frac{1}{3} = 2$ , $χ^{(2, 1)} (2, 1) = 0$ and $χ^{(2, 1)} (3) = 3 \cdot (- \frac{1}{3}) = - 1$ .

2.1F1F8step 1.1

Character of $S^{(1, 1, 1)}$ : step 1.1 gives the trivial character of $S^{(3)}$ , so by [F8] the sign character has values $1, - 1, 1$ on $(1^{3}), (2, 1), (3)$ , so its characteristic is $ch (χ^{(1, 1, 1)}) = \frac{1}{6} p_{1}^{3} - \frac{1}{2} p_{1} p_{2} + \frac{1}{3} p_{3}$ by [F1]; this is consistent with $ch (χ^{(1, 1, 1)}) = s_{(1, 1, 1)} = e_{3}$ from [F1].

3.1F4step 1.1step 2.1step 1.2algebra

Triple products. Write $T = (3)$ , $V = (2, 1)$ and $E = (1^{3})$ . Their character rows from steps 1.1, 1.2 and 2.1 are $(1, 1, 1)$ , $(2, 0, - 1)$ and $(1, - 1, 1)$ on classes of sizes $1, 3, 2$ . Formula [F4] gives $g (T, T, T) = (1 + 3 + 2) / 6 = 1$ , $g (T, E, E) = (1 + 3 + 2) / 6 = 1$ , $g (T, V, V) = g (E, V, V) = (4 + 0 + 2) / 6 = 1$ and $g (V, V, V) = (8 + 0 - 2) / 6 = 1$ . The other unordered triples give $g (T, T, E) = g (E, E, E) = (1 - 3 + 2) / 6 = 0$ and $g (T, T, V) = g (T, E, V) = g (E, E, V) = (2 + 0 - 2) / 6 = 0$ . These are all ten unordered triples; [F4] supplies every ordering.

4.1F4step 3.1∎

Since the multiplicity of $S^{ν}$ in $S^{λ} \otimes S^{μ}$ is $g (λ, μ, ν)$ , step 3.1 gives exactly the displayed table: $S^{(2, 1)} \otimes S^{(2, 1)} ≅ S^{(3)} \oplus S^{(2, 1)} \oplus S^{(1, 1, 1)}$ , $S^{(2, 1)} \otimes S^{(1, 1, 1)} ≅ S^{(2, 1)}$ , $S^{(1, 1, 1)} \otimes S^{(1, 1, 1)} ≅ S^{(3)}$ and $S^{(3)} \otimes X ≅ X$ for every irreducible $X$ . All coefficients are $0$ or $1$ , and the table equals each of its permutations by [F4].

Remarks

The example gives the complete internal product table for $S_{3}$ and exhibits the multiplicities of the tensor square of the standard representation, which decomposes as the direct sum of all three irreducibles.
The three characters were recomputed from the Frobenius dictionary rather than cited from a character table on another page, and the count of elements in each class used $1, 3, 2 = 6 / z_{ρ}$ .

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedOpen item page →

First-row padding below the threshold is undefined

Statement refuted

Refuted claim: for every partition $α$ and every integer $n \geq ∣ α ∣$ the displayed sequence $(n - ∣ α ∣, α_{1}, α_{2}, \dots)$ is a partition of $n$ , so that the padded family and the padded Kronecker coefficient $g_{n} (α, β, γ)$ would be defined as soon as the size condition $n \geq ∣ α ∣$ holds (First-row padding of a partition).

The witness is $α = (2, 1)$ , of size $∣ α ∣ = 3$ and first part $α_{1} = 2$ . For $n = 3$ and $n = 4$ the size condition $n \geq ∣ α ∣$ holds, but the displayed sequences are $(3 - ∣ α ∣, 2, 1) = (0, 2, 1), (4 - ∣ α ∣, 2, 1) = (1, 2, 1),$ and neither is a partition: the first has a nonpositive leading entry, and the second is not weakly decreasing. Only from the threshold $∣ α ∣ + α_{1} = 3 + 2 = 5$ on does the family become a family of partitions, starting with $α [5] = (2, 2, 1)$ . Hence the padded sequence is undefined below the threshold, and for this shape the stable family begins at $n = 5$ .

Facts & Assumptions

Given: The partition $α = (2, 1)$ and the integers $n = 3, 4, 5$ (Partitions, English diagrams, and conjugation).

[F1]

Read $∣ \emptyset ∣ = \emptyset_{1} = 0$ . For a partition $α$ and an integer $n \geq ∣ α ∣ + α_{1}$ one sets $α [n] = (n - ∣ α ∣, α_{1}, α_{2}, \dots)$ , except that $\emptyset [0] = \emptyset$ ; this padding is a partition of $n$ . Below the threshold $α [n]$ is undefined (First-row padding of a partition).

[F2]

A partition of $n$ is a finite weakly decreasing sequence of positive integers with sum $n$ ; trailing zeros are not parts, so a displayed sequence that is not weakly decreasing, or that contains a nonpositive entry, is not a partition (Partitions, English diagrams, and conjugation).

[F3]

$α = (2, 1)$ has $∣ α ∣ = 2 + 1 = 3$ and $α_{1} = 2$ , so its padding threshold is $∣ α ∣ + α_{1} = 5$ . [F1, given]

[F4]

The size-indexed padded coefficient $g_{n} (α, β, γ) = g (α [n], β [n], γ [n])$ is defined exactly for $n \geq \max {∣ α ∣ + α_{1}, ∣ β ∣ + β_{1}, ∣ γ ∣ + γ_{1}}$ ; below this common threshold it is undefined (Padded Kronecker coefficient sequences).

Counterexample

technique · direct

1.1F2F3given

For $n = 3$ the displayed sequence is $(3 - 3, 2, 1) = (0, 2, 1)$ . Its leading entry $0$ is not a positive integer and $0 < 2$ shows that the sequence is not weakly decreasing, so by [F2] it is not a partition of $3$ , although $3 = ∣ α ∣$ and hence $n \geq ∣ α ∣$ holds.

1.2F2F3given

For $n = 4$ the displayed sequence is $(4 - 3, 2, 1) = (1, 2, 1)$ . Its entries are positive and sum to $4$ , but $1 < 2$ , so it is not weakly decreasing and by [F2] it is not a partition of $4$ , although $4 \geq ∣ α ∣$ holds.

2.1F1F2F3step 1.1step 1.2

For $n = 5$ , which is the threshold $∣ α ∣ + α_{1}$ computed in [F3], the displayed sequence is $(5 - 3, 2, 1) = (2, 2, 1)$ : the entries are positive, weakly decreasing and sum to $5$ , so it is a partition of $5$ . By [F1] the padding $α [5] = (2, 2, 1)$ is defined, and by step 1.1 and step 1.2 it is the first member of the family; the threshold is therefore exact.

3.1step 1.1step 1.2step 2.1F1F4∎

Consequently the refuted claim fails at $n = 3$ and $n = 4$ : the size condition $n \geq ∣ α ∣$ is not enough, and the padded family of $α = (2, 1)$ is a family of partitions only from $n = 5$ on. Any padded Kronecker coefficient $g_{n} (α, β, γ)$ with $α = (2, 1)$ and $n < 5$ is likewise undefined, because one of its three entries is not a partition.

Remarks

The failed threshold $∣ α ∣$ is exactly weaker than the true one $∣ α ∣ + α_{1}$ ; the extra term $α_{1}$ is what keeps the padded partition weakly decreasing, since the new first part $n - ∣ α ∣$ must be at least the old first part $α_{1}$ .
The example also shows why the size-indexed padded Kronecker coefficients of Padded Kronecker coefficient sequences are defined only from the maximum of the three thresholds: a single unpadded entry makes the whole expression undefined.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

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.

Kronecker Coefficients and Internal Products — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The complete internal product table for S_3

Example

Facts & Assumptions

Verification

Remarks

First-row padding below the threshold is undefined

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Row padding need not preserve a Kronecker coefficient before stability

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources