PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08

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Equal insertion or recording tableaux imply right or left equivalence

Facts & Assumptions

Given: $n \geq 1$ , permutations $x, y \in S_{n}$ , their insertion tableaux $P (x), P (y)$ , recording tableaux $Q (x), Q (y)$ , and the Kazhdan–Lusztig cell preorders.

[F1]

Knuth equivalence satisfies $x \sim_{K} y$ iff $P (x) = P (y)$ , and dual Knuth equivalence satisfies $x \sim_{d K} y$ iff $Q (x) = Q (y)$ (Knuth classes are the fibers of the insertion tableau).

[F2]

The star table exchanges exactly the four local triples in the two elementary Knuth relations; thus every elementary Knuth move is a right star pair, and each such pair satisfies $w^{*} \sim_{R} w$ (Star operations on strings of adjacent simple reflections, Star operations are Knuth moves and preserve the relevant cells).

[F3]

RSK is symmetric under inversion: $P (w^{- 1}) = Q (w)$ and $Q (w^{- 1}) = P (w)$ (RSK interchanges the insertion and recording tableaux under inversion).

[F4]

Inversion transports cell relations: $a \sim_{L} b$ iff $a^{- 1} \sim_{R} b^{- 1}$ ( $L$ -, $R$ - and two-sided Kazhdan–Lusztig preorders and cells).

Statement

For $x, y \in S_{n}$ : (a) $P (x) = P (y) \Rightarrow x \sim_{R} y$ ; (b) $Q (x) = Q (y) \Rightarrow x \sim_{L} y$ . Equivalently, each Knuth class lies in a single right cell and each dual Knuth class lies in a single left cell.

Proof

technique · join equal-tableau permutations by elementary Knuth moves and transport the resulting cell equivalence through inversion

1.1F1F2algebra

Insertion tableaux give right-cell equivalence. Suppose $P (x) = P (y)$ . By [F1], $x \sim_{K} y$ , so there is a finite chain $x = w_{0}, \dots, w_{m} = y$ whose successive terms differ by an elementary Knuth move. By [F2], each adjacent pair satisfies $w_{k} \sim_{R} w_{k + 1}$ ; transitivity of the right-cell equivalence gives $x \sim_{R} y$ . If the chain has length zero, reflexivity gives the same conclusion.

2.1F3F4step 1.1algebra

Recording tableaux give left-cell equivalence. Suppose $Q (x) = Q (y)$ . By [F3], $P (x^{- 1}) = Q (x) = Q (y) = P (y^{- 1})$ , so step 1.1 applied to $x^{- 1}, y^{- 1}$ gives $x^{- 1} \sim_{R} y^{- 1}$ . Inverting this cell equivalence by [F4] yields $x \sim_{L} y$ .

3.1F1step 1.1step 2.1∎

Class formulation. By [F1], every pair in a Knuth class has equal insertion tableaux, so step 1.1 puts the whole class in one right cell. Every pair in a dual Knuth class has equal recording tableaux, so step 2.1 puts that class in one left cell. These are exactly the two equivalent class statements.

Remarks

The finite Knuth chains use the locally proved star-pair right-cell equivalence; recording-tableau equality is transported through inversion of cells. Coefficientwise positivity is not required.

No choice principle is used.

Depends on

Used by

Dependency tree · two levels

19 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

Susumu Ariki, Robinson–Schensted correspondence and left cells, arXiv:math/9910117 — Proposition 3.8 and its complete proof via Knuth moves and star operations. (standard reference, not scraped)