CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6.1-sol)

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A semistandard tableau with non-lattice reading word is not a Littlewood--Richardson tableau

Statement refuted

For partitions $λ \subseteq ν$ and $μ$ with $∣ ν ∣ = ∣ λ ∣ + ∣ μ ∣$ , every semistandard skew tableau of shape $ν / λ$ and content $μ$ is a Littlewood--Richardson tableau, i.e. its reading word is automatically a lattice word (Littlewood--Richardson tableaux and coefficients).

Facts & Assumptions

[F1]

Reading words read the rows from right to left beginning with the top row and are lattice words when every prefix contains at least as many letters $i$ as letters $i + 1$ for every $i \geq 1$ (Littlewood--Richardson tableaux and coefficients).

[F2]

A semistandard skew tableau of shape $ν / λ$ weakly increases along rows, strictly increases down columns, and has content $μ$ when the entry $i$ occurs $μ_{i}$ times (Semistandard tableaux and Kostka numbers, Skew diagrams and semistandard skew tableaux).

Counterexample

Given: $ν = (2, 1)$ , $λ = \emptyset$ in English coordinates (Partitions, English diagrams, and conjugation), $μ = (1, 1, 1)$ , and the tableau $T$ of shape $(2, 1)$ with first row $1 3$ and second row $2$ , $\begin{array}{cc} 1 & 3 \\ 2 \end{array} .$

1.1F2given

The tableau $T$ is a semistandard skew tableau of shape $ν / λ = (2, 1)$ and content $μ = (1, 1, 1)$ : its first row is weakly increasing ( $1 \leq 3$ ), its first column is strictly increasing ( $1 < 2$ ) and its second column is a single cell, and its entries are exactly one $1$ , one $2$ and one $3$ .

1.2F1given

The reading word of $T$ is obtained by reading each row from right to left starting with the top row: the first row contributes $3, 1$ and the second row contributes $2$ , so $w (T) = 3 1 2$ (Littlewood--Richardson tableaux and coefficients).

2.1F1step 1.1step 1.2algebra

The first prefix of $w (T)$ is the single letter $3$ ; it contains zero copies of the letter $2$ and one copy of the letter $3$ , so the prefix condition of a lattice word fails for $i = 2$ . Hence $T$ is a semistandard skew tableau of shape $(2, 1)$ and content $(1, 1, 1)$ whose reading word is not a lattice word, and by definition $T$ is not a Littlewood--Richardson tableau; this refutes the statement.

3.1F2step 1.2step 2.1algebra∎

The instance is sharp: the semistandard tableaux of shape $(2, 1)$ and content $(1, 1, 1)$ are exactly $T$ and the tableau with first row $1 2$ and second row $3$ . Indeed the top-left entry must be $1$ : were it $2$ , the cell below it in the first column would carry the only remaining letter larger than $2$ , namely $3$ , leaving $1$ in the top-right cell in violation of weak row increase; were it $3$ , no letter would remain below it in strict column increase. With $1$ in the top-left cell, the remaining letters $2$ and $3$ may be placed in the other two cells in either order, since $2, 3 > 1$ and each row and column condition involves at most those two cells, giving exactly the two tableaux. The second of these has reading word $2 1 3$ , which also fails the lattice condition at its first letter $2$ ; consistently, $c_{\emptyset, (1, 1, 1)}^{(2, 1)} = 0$ by the defining count of LR tableaux, while $s_{\emptyset} = 1$ and $s_{(2, 1)} (x_{1}, x_{2}, x_{3}) = x_{1}^{2} x_{2} + x_{1}^{2} x_{3} + x_{1} x_{2}^{2} + 2 x_{1} x_{2} x_{3} + x_{1} x_{3}^{2} + x_{2}^{2} x_{3} + x_{2} x_{3}^{2}$ is not the squarefree polynomial $s_{(1, 1, 1)} = e_{3}$ (Littlewood--Richardson tableaux and coefficients, Skew Jacobi–Trudi and tableau expansion, Stable Schur functions from bialternants).

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Sources

J. R. Stembridge, A Concise Proof of the Littlewood--Richardson Rule, Electronic Journal of Combinatorics 9 (2002), #N5, 4 pp. (standard reference, not scraped)