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The hook length formula

Statement

For every n≥0 and every λ⊢n, the number fλ of standard λ-tableaux is fλ=n!∏x∈[λ]h(x), the empty product for λ=∅ being 1, so that f∅=1 and, for n≥1, fλ=1 for λ=(n) and λ=(1n). In particular, over C, dim⁡CSλ=n!∏x∈[λ]h(x) for the Specht module Sλ, including dim⁡CS∅=1.

Facts & Assumptions

Given: An integer n≥0 and a partition λ⊢n, with fλ the number of standard λ-tableaux and P(λ)=∏x∈[λ]h(x) the hook product.

[F1]

fλ=∑x∈Rem⁡(λ)fλ−x for n≥1, and f∅=1; the boxes of [λ] are the boxes of [λ−x] together with x for x∈Rem⁡(λ) (The removal recursion for standard tableaux).

[F2]

For x∈Rem⁡(λ) with λ⊢n≥1: P(λ)/P(λ−x)=R(x):=∏y∈Rxhλ(y)/(hλ(y)−1) (Removing a corner changes hooks in its row and column).

[F3]

∑x∈Rem⁡(λ)R(x)=n, the empty sum being 0 (The hook-product ratios sum to the size).

[F4]

The family of standard polytabloids is a C-basis of the Specht module Sλ, so dim⁡CSλ=fλ for every λ⊢n, including n=0 (Standard polytabloids form a basis of a complex Specht module).

[F5]

P(λ) is the product of the n positive integers h(x), one for each box of [λ]; for λ=∅ it is the empty product 1, for λ=(n) the product is n!, and for λ=(1n) the conjugate diagram gives the same multiset of hooks (Hook, arm, leg, and hook length of a box).

Proof

technique · strong induction on $n$
1.1baseF1F5given

Base cases: for n=0 the only partition is ∅, whose set of standard tableaux is the singleton consisting of the empty tableau, so f∅=1=0!/1 with empty product 1; for n=1 the only partition is (1), whose single box has h=1 and exactly one standard tableau, so f(1)=1=1!/1.

1.2ihgiven

Induction hypothesis: for every m with 0≤m<n and every partition μ⊢m, fμ=m!/P(μ).

1.3F4given

The dimension clause: by [F4], dim⁡CSλ=fλ for every λ⊢n, including λ=∅ where both sides are 1; this holds for all n because [F4] covers every n≥0.

2.1step 1.2F1F2F3algebra

For n≥2 and λ⊢n, [F1] gives fλ=∑x∈Rem⁡(λ)fλ−x; each λ−x is a partition of n−1<n, so step 1.2 gives fλ−x=(n−1)!/P(λ−x), and [F2] turns this into (n−1)!R(x)/P(λ). Summing over the removable nodes and using [F3], fλ=(n−1)!P(λ)∑xR(x)=(n−1)! nP(λ)=n!P(λ).

3.1step 2.1F5given

The two identities f(n)=1=f(1n) follow because the hook multiset of (n) and of (1n) is {1,2,…,n} by [F5], so the formula gives n!/n!=1 in both cases.

4.1step 1.1step 1.2step 2.1step 1.3step 3.1discharge-induction∎

Strong induction on n: the base cases are step 1.1, the inductive step is step 2.1 with the hypothesis step 1.2, and steps 1.3 and 3.1 record the dimension and endpoint clauses; hence fλ=n!/∏x∈[λ]h(x) and dim⁡CSλ=fλ hold for every n≥0 and every λ⊢n.

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Sources