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Hook table for the shape (3,2,1)

Example

For $λ = (3, 2, 1) ⊢ 6$ the hook lengths $h (i, j) = λ_{i} - j + λ_{j}^{'} - i + 1$ are $\begin{array}{lll} 5 & 3 & 1 \\ 3 & 1 \\ 1 \end{array}$ (rows of lengths $3, 2, 1$ ), with hook product $5 \cdot 3 \cdot 1 \cdot 3 \cdot 1 \cdot 1 = 45$ . The hook length formula gives $f^{(3, 2, 1)} = 6! / 45 = 720 / 45 = 16$ . The removal recursion checks the value: $Rem ((3, 2, 1)) = {(1, 3), (2, 2), (3, 1)}$ , with removals $(2, 2, 1)$ , $(3, 1, 1)$ , $(3, 2)$ , and the same formula gives $f^{(2, 2, 1)} = 5$ , $f^{(3, 1, 1)} = 6$ , $f^{(3, 2)} = 5$ , so $f^{(3, 2, 1)} = 5 + 6 + 5 = 16$ .

Facts & Assumptions

Given: The partition $λ = (3, 2, 1) ⊢ 6$ with Young diagram $[λ]$ and conjugate $λ^{'}$ , and the removals $λ - x$ for the removable nodes $x$ .

[F1]

For $x = (i, j) \in [λ]$ one has $h (x) = λ_{i} - j + λ_{j}^{'} - i + 1$ and $P (λ) = \prod_{x \in [λ]} h (x)$ ; a node is removable exactly when it is at the end of its row and of its column (Hook, arm, leg, and hook length of a box).

[F2]

$f^{λ} = n! / P (λ)$ for $λ ⊢ n$ , with the empty product $1$ for $λ = \emptyset$ (The hook length formula).

[F3]

For $λ ⊢ n$ with $n \geq 1$ , $f^{λ} = \sum_{x \in Rem (λ)} f^{λ - x}$ , and deleting a removable node leaves the diagram of the partition $λ - x$ (The removal recursion for standard tableaux, Hook, arm, leg, and hook length of a box).

Verification

technique · direct

1.1F1given

The conjugate partition is $λ^{'} = (3, 2, 1)$ : each column of $[λ]$ has heights $3, 2, 1$ .

2.1F1step 1.1algebra

Evaluating $h (i, j) = λ_{i} - j + λ_{j}^{'} - i + 1$ : $h (1, 1) = 5$ , $h (1, 2) = 1 + 2 = 3$ , $h (1, 3) = 0 + 1 = 1$ , $h (2, 1) = 1 + 3 - 2 + 1 = 3$ , $h (2, 2) = 0 + 2 - 2 + 1 = 1$ , $h (3, 1) = 0 + 3 - 3 + 1 = 1$ ; the hook product is $5 \cdot 3 \cdot 1 \cdot 3 \cdot 1 \cdot 1 = 45$ .

2.2F1step 1.1given

The removable nodes are $(1, 3), (2, 2), (3, 1)$ : each of these is the last box of its row and of its column, while $(1, 1), (1, 2), (2, 1)$ each have a box to the right (and $(1, 2), (2, 1)$ a box below); indeed $(1, 2)$ has $(1, 3)$ to its right and $(2, 2)$ below, $(2, 1)$ has $(2, 2)$ to its right, and $(1, 1)$ has $(1, 2)$ to its right.

3.1F2step 2.1algebra

By [F2], $f^{(3, 2, 1)} = 6! / 45 = 720 / 45 = 16$ .

3.2F2step 2.2algebra

The three removals are $(2, 2, 1)$ , $(3, 1, 1)$ and $(3, 2)$ , of sizes $5$ ; by [F2] applied in size $5$ and the hook computations: $P (2, 2, 1) = 4 \cdot 2 \cdot 3 \cdot 1 \cdot 1 = 24$ so $f^{(2, 2, 1)} = 120 / 24 = 5$ ; $P (3, 1, 1) = 5 \cdot 2 \cdot 1 \cdot 2 \cdot 1 = 20$ so $f^{(3, 1, 1)} = 120 / 20 = 6$ ; $P (3, 2) = 4 \cdot 3 \cdot 1 \cdot 2 \cdot 1 = 24$ so $f^{(3, 2)} = 120 / 24 = 5$ .

4.1F3step 3.1step 3.2algebra∎

By [F3] the removal recursion predicts $f^{(3, 2, 1)} = f^{(2, 2, 1)} + f^{(3, 1, 1)} + f^{(3, 2)} = 5 + 6 + 5 = 16$ , which agrees with step 3.1.

Depends on

Used by

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Sources

David A. Craven, Groups, Geometries and Representation Theory (Spring Term 2013 lecture notes, 42 pp.) (standard reference, not scraped)
Charlotte Chan, Representation Theory of Symmetric Groups (Oxford Hilary Term 2011 lecture notes, 40 PDF pp.) (standard reference, not scraped)
C. Schensted, Longest Increasing and Decreasing Subsequences, Canadian Journal of Mathematics 13 (1961), 179-191 (13 pp.) (standard reference, not scraped)