ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The Young permutation characteristic for shape $(2, 1)$

Example

For $S_{3}$ , the Young permutation module $M^{(2, 1)}$ has $ch (M^{(2, 1)}) = h_{2} h_{1} = s_{(3)} + s_{(2, 1)}$ , and its character takes the values $(3, 1, 0)$ on cycle types $(1^{3}), (2, 1), (3)$ , decomposing as $χ^{(3)} + χ^{(2, 1)}$ ; this agrees with Young's rule $M^{(2, 1)} ≅ S^{(3)} \oplus S^{(2, 1)}$ and with the dictionary table.

Facts & Assumptions

Given: The partition $(2, 1)$ of $3$ , the cycle types $(1^{3}), (2, 1), (3)$ of $S_{3}$ , and a permutation $w \in S_{3}$ .

[F1]

$ch (φ^{(2, 1)}) = h_{(2, 1)} = h_{2} h_{1}$ , where $φ^{(2, 1)}$ is the character of $M^{(2, 1)}$ , and $h_{λ} = \prod_{i} h_{λ_{i}}$ (The characteristic of a Young permutation character is complete homogeneous, Elementary and complete families freely generate the stable ring).

[F2]

$φ^{(2, 1)} (w)$ is the number of $(2, 1)$ -tabloids fixed by $w$ (The character of a permutation representation counts fixed points).

[F3]

Young's rule: $M^{(2, 1)} ≅ (S^{(3)})^{\oplus K_{(3), (2, 1)}} \oplus (S^{(2, 1)})^{\oplus K_{(2, 1), (2, 1)}} \oplus (S^{(1, 1, 1)})^{\oplus K_{(1, 1, 1), (2, 1)}}$ , with $K_{λ μ}$ the number of semistandard $λ$ -tableaux of content $μ$ (Young's rule for complex permutation modules, Semistandard tableaux and Kostka numbers).

[F4]

For partitions of the same integer, $h_{μ} = \sum_{λ} K_{λ μ} s_{λ}$ with $K_{λ μ} = 0$ unless $λ ⊵ μ$ and $K_{μ μ} = 1$ (The Kostka change of basis is dominance-unitriangular).

[F5]

In the dictionary example for $S_{3}$ , $s_{(3)} = \frac{p_{1}^{3} + 3 p_{1} p_{2} + 2 p_{3}}{6}$ , $s_{(2, 1)} = \frac{p_{1}^{3} - p_{3}}{3}$ , with $χ^{(3)} = (1, 1, 1)$ and $χ^{(2, 1)} = (2, 0, - 1)$ on the cycle types $(1^{3}), (2, 1), (3)$ (The Frobenius characteristic dictionary for $S_{3}$ ).

Verification

technique · direct

1.1F2given

Listing the three $(2, 1)$ -tabloids by their two-element row: ${12} ∣ {3}$ , ${13} ∣ {2}$ , ${23} ∣ {1}$ . The identity fixes all three; the transposition $(12)$ fixes exactly ${12} ∣ {3}$ , because a fixed tabloid must have both its row sets $w$ -invariant, and $(12)$ preserves ${1, 2}$ and ${3}$ but moves ${1, 3}$ to ${2, 3}$ and ${2, 3}$ to ${1, 3}$ ; the $3$ -cycle $(123)$ fixes none, since a row set of size $2$ is never invariant under a $3$ -cycle. Hence $φ^{(2, 1)} = (3, 1, 0)$ on the cycle types $(1^{3}), (2, 1), (3)$ .

1.2F1

By [F1], $ch (φ^{(2, 1)}) = h_{2} h_{1}$ .

1.3F3F4givenalgebra

The Kostka numbers for $μ = (2, 1)$ are $K_{(3), (2, 1)} = 1$ (the single semistandard tableau $1 1 2$ of shape $(3)$ ) and $K_{(2, 1), (2, 1)} = 1$ (the tableau with first row $1 1$ and second row $2$ ), while $K_{(1, 1, 1), (2, 1)} = 0$ both because two entries equal to $1$ would have to occur in the same column and because $(1, 1, 1)$ does not dominate $(2, 1)$ [F4]; so Young's rule [F3] gives $M^{(2, 1)} ≅ S^{(3)} \oplus S^{(2, 1)}$ .

2.1F1F5step 1.2algebra

By [F5] the dictionary example gives $s_{(3)} + s_{(2, 1)} = \frac{p_{1}^{3} + 3 p_{1} p_{2} + 2 p_{3}}{6} + \frac{p_{1}^{3} - p_{3}}{3} = \frac{3 p_{1}^{3} + 3 p_{1} p_{2}}{6} = \frac{p_{1}^{3} + p_{1} p_{2}}{2}$ ; on the other hand $h_{2} = \frac{p_{1}^{2} + p_{2}}{2}$ , so $h_{2} h_{1} = \frac{(p_{1}^{2} + p_{2}) p_{1}}{2} = \frac{p_{1}^{3} + p_{1} p_{2}}{2}$ . Therefore $h_{2} h_{1} = s_{(3)} + s_{(2, 1)}$ , and step 1.2 identifies this with $ch (φ^{(2, 1)})$ .

2.2F5step 1.1algebra

Adding the values of $χ^{(3)}$ and $χ^{(2, 1)}$ from [F5] gives $(1, 1, 1) + (2, 0, - 1) = (3, 1, 0)$ on the cycle types $(1^{3}), (2, 1), (3)$ , exactly the values computed for $φ^{(2, 1)}$ in step 1.1.

3.1F5step 1.3step 2.1step 2.2∎

The two routes agree: directly, $φ^{(2, 1)} = (3, 1, 0) = 1 \cdot χ^{(3)} + 1 \cdot χ^{(2, 1)}$ by step 2.2, matching the Young's-rule decomposition $M^{(2, 1)} ≅ S^{(3)} \oplus S^{(2, 1)}$ of step 1.3; symmetrically, $ch (φ^{(2, 1)}) = h_{2} h_{1} = s_{(3)} + s_{(2, 1)}$ by steps 1.2 and 2.1, matching the dictionary table of [F5].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

44 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

I. G. Macdonald, Symmetric Functions and Hall Polynomials, 2nd ed., Chapter I §7 (standard reference, not scraped)
G. D. James, The Representation Theory of the Symmetric Groups, §6 (standard reference, not scraped)

The Young permutation characteristic for shape (2,1)(2,1)