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LemmaStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedjudge pass (gpt-6.1-sol)
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The removal recursion for standard tableaux

Statement

Let n≥1 and λ⊢n. The map that sends a standard λ-tableau t to the pair (x,t−), where x is the box occupied by n and t− is the restriction of t to [λ]∖{x}, is a bijection from the set of standard λ-tableaux onto the disjoint union, over the removable nodes x∈Rem⁡(λ), of the sets of standard (λ−x)-tableaux. Consequently

fλ=∑x∈Rem⁡(λ)fλ−x(λ⊢n, n≥1),

and we adopt the convention f∅=1. For n=0 the disjoint union is empty and the recursion is not asserted: the value f∅=1 is the convention for the unique empty tableau.

Facts & Assumptions

Given: An integer n≥1, a partition λ⊢n, and the family of partitions λ−x for x∈Rem⁡(λ).

[L1]

A standard λ-tableau is a bijection t:[λ]→{1,…,n} that strictly increases along rows and down columns; the shape is determined by t (Tableaux and standard tableaux).

[L2]

The box occupied by the largest entry n of a standard λ-tableau is a removable node of λ, and deleting it leaves a standard tableau of shape λ−x (The largest standard entry lies in a removable box).

[L3]

A node x is removable exactly when [λ]∖{x} is the diagram of a partition λ−x⊢n−1; the diagram [λ−x] determines λ−x (Removable and addable nodes).

Proof

technique · direct
1.1L2given

The map is well defined: by [L2] the box x of n is removable and t− is a standard tableau of shape λ−x, and (x,t−) lies in the x-component of the displayed disjoint union.

1.2L1given

The map is injective: given its image (x,t−) one recovers t by t(x)=n and t=t− on [λ]∖{x}, so two tableaux with the same image are equal.

1.3L1L3given

The map is surjective onto the displayed union: let x∈Rem⁡(λ) and let t− be a standard tableau of shape λ−x; define t(x):=n and t(y):=t−(y) for y∈[λ−x]. Then t is a bijection [λ]→{1,…,n}, because t− is a bijection onto {1,…,n−1} and x∉[λ−x].

2.1L1L3step 1.3

The bijection t of step 1.3 is standard: adjacent pairs in [λ] not involving x are adjacent in [λ−x] and satisfy the required strict inequality by standardness of t−, while a pair involving x has its other entry in {1,…,n−1} and hence satisfies t−(⋅)≤n−1<n=t(x) in the direction of x, and the inequalities along rows and columns run into x only from the left and from above, since x is a corner.

2.2L1step 1.1step 1.3

The two constructions of steps 1.1 and 1.3 are inverse: starting from t, the tableau reconstructed from (x,t−) agrees with t because t(x)=n and t restricts to t−; starting from x,t−, the pair extracted from the reconstructed t is (x,t−) because the only entry greater than n−1 is t(x)=n.

3.1step 1.1step 1.2step 1.3step 2.1step 2.2L1∎

The components of the disjoint union are indexed by the distinct removable nodes x, and for fixed x the standard (λ−x)-tableaux number fλ−x; the bijection of steps 1.1–2.2 therefore gives the stated recursion, and for λ=∅ the union is empty while f∅=1 is the adopted convention.

Depends on

Used by

Dependency tree · two levels

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Sources