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Hooks are bead-gap pairs on the abacus

Statement

Let e≥2, let λ⊢n and let X={h1>h2>⋯>hm} be a β-set for λ, so that P(X)=λ (Beta-sets, the shift, and the e-runner abacus of a partition). Write Hi(λ) for the set of hook lengths in row i of [λ]: thus h∈Hi(λ) means h=hi,j(λ) for some node (i,j)∈[λ], and Hi(λ)=∅ for i>ℓ(λ) (Hook, arm, leg, and hook length of a box).

(i) For every h≥0 and every i∈[m], h∈Hi(λ) if and only if h≤hi and hi−h∉X. Equivalently, the hooks of λ correspond bijectively to the bead-gap pairs (b,g) with b∈X, g∉X and 0≤g<b; the hook Hi,j(λ) corresponds to (b,g)=(hi,  hi−hi,j(λ)), and then ∣Hi,j(λ)∣=b−g.

(ii) Fix a node (i,j)∈[λ] and let (b,g) be its bead-gap pair as in (i). Then X′:=(X∖{b})∪{g} is again a β-set, now for the partition λ∖Ri,j obtained from λ by removing the removable border strip Ri,j that corresponds to the hook of (i,j), of size hi,j(λ) and height ℓ(i,j)=λj′−i (Removable border strips and their height, Hooks of a partition correspond bijectively to removable border strips); in terms of parts, with b0:=ℓ(i,j)=λj′−i, λ∖Ri,j=(λ1,…,λi−1, λi+1−1,…,λi+b0−1, j−1, λi+b0+1,… ), so that ∣λ∖Ri,j∣=n−hi,j(λ). Finally X′ is obtained from the configuration AX by moving the bead from position b to the gap g.

(iii) In particular e∣hi,j(λ) if and only if b≡g(mode), that is, if and only if the bead b and the gap g lie on the same runner of the e-abacus; then g=b−e⋅(hi,j(λ)/e) and the move in (ii) is the slide of the bead b up hi,j(λ)/e rows on its own runner. A hook of length e in row i of λ exists if and only if hi≥e and hi−e∉X; Thus λ has a hook of length e if and only if some bead of X has a gap in the position e labels above it on its own runner.

No choice principle is used.

Facts & Assumptions

Given: An integer e≥2, a partition λ⊢n with t:=ℓ(λ) parts, and a β-set X={h1>⋯>hm} for λ, so that P(X)=λ (Beta-sets, the shift, and the e-runner abacus of a partition).

[F1]

Hook lengths: hi,j(λ)=λi−j+λj′−i+1 for (i,j)∈[λ], and for fixed i the map j↦hi,j(λ) is strictly decreasing on 1≤j≤λi, because hi,j(λ)−hi,j+1(λ)=1+λj′−λj+1′≥1; hence the hook lengths in a fixed row are pairwise distinct and hi,j(λ)≥1 (Hook, arm, leg, and hook length of a box).

[F2]

For a β-set Y={y1>⋯>yr} one has P(Y)j=yj−(r−j) for 1≤j≤r, after which trailing zeros are deleted: indeed there are yj positions 0,1,…,yj−1 below the bead yj, of which exactly r−j are occupied by the beads yj+1,…,yr, so yj−(r−j) of them are gaps, and the definition of P takes exactly this difference (Beta-sets, the shift, and the e-runner abacus of a partition).

[F3]

The β-sets of λ are exactly the sets Xλ+l, l∈N0, where Xλ={b1(λ)>⋯>bt(λ)} with bi(λ)=hi,1(λ)=λi+t−i, and ∣Xλ+l∣=t+l (Beta-sets, the shift, and the e-runner abacus of a partition).

[F4]

A removable border strip of λ is a nonempty edge-connected set of nodes containing no 2×2 block whose complement in [λ] is again a Young diagram; its height is its number of occupied rows minus one, and for a strip that is the hook of a node it equals the leg length of that node (Removable border strips and their height, Hooks of a partition correspond bijectively to removable border strips).

Proof

technique · direct
1.1F1F2F3algebra

Pad λ with zeros to length m. By [F3], hi=λi+m−i. Below hi there are exactly m−i beads, hence hi−(m−i)=λi gaps. For a node (i,j) put gj=hi−hi,j(λ)=m+j−1−λj′. These values are distinct by [F1], satisfy 0≤gj<hi, and are gaps: equality gj=hx=λx+m−x would give λx−j+λj′−x+1=0. If x≤λj′, both differences are nonnegative; otherwise both are at most −1, since λx<j. Neither case gives zero. Thus the λi numbers gj exhaust the gaps below hi, including the empty set when λi=0.

1.2F1F4constructalgebra

Fix (i,j) and put r=λj′ and g=m+j−1−r. The rim strip of [F4] has complement with rows μp=λp outside i≤p≤r, μp=λp+1−1 for i≤p<r, and μr=j−1: in row p the strip occupies the rim nodes with column at least j, whose complement ends at max⁡(j−1,λp+1−1). For p<r one has λp+1≥j, while λr+1<j. These formulas give the displayed partition in (ii), with trailing zeros deleted, and its size is n−hi,j(λ).

2.1F1step 1.1algebra

Consequently, for h≥0, h∈Hi(λ) exactly when h≤hi and hi−h∉X: at h=0 the latter fails because hi∈X, and for h>hi its first inequality fails. For 1≤h≤hi this is the gap correspondence in step 1.1. The bead hi determines i, and the distinct row hook lengths determine j, so (i,j)↦(hi,gj) is the asserted bijection with length hi−gj.

2.2F2step 1.1step 1.2algebra

The new value g is below hr because hr−g=λr−j+1≥1, and, if r<m, above hr+1 because g−hr+1=j−λr+1≥1. Hence the decreasing list of X′=(X∖{hi})∪{g} is h1,…,hi−1,hi+1,…,hr,g,hr+1,…,hm. Applying [F2] gives precisely the parts μp in step 1.2. Thus P(X′)=λ∖Ri,j, and the abacus change is the relocation of the bead hi to g.

3.1step 2.1step 2.2algebra∎

The length is b−g, so e∣hi,j(λ) exactly when b≡g(mode). In that case the bead remains on its runner, moves up hi,j(λ)/e rows, and all runner bead counts are unchanged. Without this divisibility assumption the relocation may change runners. In row i a hook of length e exists exactly when hi≥e and hi−e∉X, by step 2.1; existential quantification over i gives the final bead criterion in (iii).

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