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Murnaghan–Nakayama values of the character of (5,4,4)

Example

The partition $λ = (5, 4, 4)$ of $13$ has hook lengths, read row by row, $7, 6, 5, 4, 1$ ; $5, 4, 3, 2$ ; $4, 3, 2, 1$ , so its removable border strips have sizes ${1, 2, 3, 4, 5, 6, 7}$ and there is no removable strip of size $8, 9, 10, 11, 12$ or $13$ (Hooks of a partition correspond bijectively to removable border strips). The Murnaghan–Nakayama rule (The Murnaghan–Nakayama character rule) gives:

$χ^{(5, 4, 4)} = - 1$ on the class of cycle type $(5, 4, 3, 1)$ of $S_{13}$ : the two legal removals of a $5$ -strip have signs $+$ and $-$ , and the recursion yields $χ^{(3, 3, 2)} (4, 3, 1) - χ^{(5, 3)} (4, 3, 1) = 0 - 1 = - 1$ ;
$χ^{(5, 4, 4)} = 2$ on the class of cycle type $(7, 3, 3)$ : the unique legal removal of a $7$ -strip has sign $+$ and yields $χ^{(3, 3)} (3, 3) = 1 - (- 1) = 2$ ;
$χ^{(5, 4, 4)} = 0$ on every class whose cycle type contains a part in ${8, 9, 10, 11, 12, 13}$ , since $λ$ has no removable border strip of those sizes (A missing removable strip of the cycle length forces a zero character value).

Facts & Assumptions

Given: The partition $λ = (5, 4, 4) ⊢ 13$ and the character $χ^{(5, 4, 4)}$ of the complex Specht module (Column antisymmetrizers, polytabloids, and Specht modules).

[F1]

Murnaghan–Nakayama rule: $χ^{λ} (σ c) = \sum_{λ / μ} (- 1)^{ht (λ / μ)} χ^{μ} (σ)$ for a cycle $c$ on a specified set of points fixed by $σ$ (a specified fixed point for length $1$ ), summed over the removable border strips of $λ$ of size the length of $c$ ; the empty sum is $0$ (The Murnaghan–Nakayama character rule).

[F2]

Hook/rim-hook correspondence: the removable border strips of $λ$ correspond bijectively to the nodes of $[λ]$ , the strip of a node having size the hook length of the node; hence the sizes of the removable strips are the hook lengths, and a strip of size $r$ occupying $c$ columns has height $r - c$ (Hooks of a partition correspond bijectively to removable border strips, Removable border strips and their height, Hook, arm, leg, and hook length of a box).

[F3]

Vanishing criterion: if no hook length of $λ$ equals $r$ , then $χ^{λ}$ vanishes on every class of $S_{13}$ containing an $r$ -cycle (A missing removable strip of the cycle length forces a zero character value).

[F4]

Full-cycle/hook values in the recursive steps: on a $3$ -cycle class of $S_{3}$ , $χ^{(3)} = 1$ and $χ^{(2, 1)} = - 1$ (The value of an irreducible character on a full cycle is supported on hooks).

Verification

technique · direct

1.1F2F3algebra

For a node in row $i$ , column $j$ the hook length is the arm length $5 - j$ or $4 - j$ or $4 - j$ (number of nodes to its right in its row) plus the leg length (number of nodes below it in its column) plus $1$ ; computing this row by row gives $7, 6, 5, 4, 1$ ; $5, 4, 3, 2$ ; $4, 3, 2, 1$ , whose multiset is ${7, 6, 5, 5, 4, 4, 4, 3, 3, 2, 2, 1, 1}$ . By [F2] the removable strip sizes are exactly ${1, 2, 3, 4, 5, 6, 7}$ , so no removable strip has size $8, 9, 10, 11, 12$ or $13$ , and [F3] gives $χ^{(5, 4, 4)} = 0$ on every class whose cycle type contains a part in ${8, 9, 10, 11, 12, 13}$ .

1.2F1F2algebra

On the class $(5, 4, 3, 1)$ the part $5$ is stripped first; the hooks of length $5$ sit at the nodes $(1, 3)$ and $(2, 1)$ , with rim strips ${(1, 4), (1, 5), (2, 4), (3, 3), (3, 4)}$ leaving $(3, 3, 2)$ (three columns, height $2$ , sign $+$ ) and ${(2, 4), (3, 1), (3, 2), (3, 3), (3, 4)}$ leaving $(5, 3)$ (four columns, height $1$ , sign $-$ ). The child values are $χ^{(3, 3, 2)} (4, 3, 1) = 0$ , since the $4$ -strips of $(3, 3, 2)$ lead to $(3, 1)$ (height $1$ ) and $(2, 1, 1)$ (height $2$ ) and both $(3, 1)$ and $(2, 1, 1)$ have no $3$ -strip, and $χ^{(5, 3)} (4, 3, 1) = (- 1) \cdot χ^{(2, 2)} (3, 1) = (- 1) \cdot (- 1) = 1$ , the unique $4$ -strip of $(5, 3)$ leading to $(2, 2)$ with height $1$ and the unique $3$ -strip of $(2, 2)$ leading to a single node with height $1$ . Hence $χ^{(5, 4, 4)} (5, 4, 3, 1) = + 0 - 1 = - 1$ .

1.3F1F2F4algebra

On the class $(7, 3, 3)$ the unique hook of length $7$ is the one at $(1, 1)$ , whose rim strip occupies all seven rim nodes from $(1, 5)$ to $(3, 1)$ and leaves $(3, 3)$ with height $2$ , so $χ^{(5, 4, 4)} (7, 3, 3) = + χ^{(3, 3)} (3, 3)$ ; stripping a $3$ -strip from $(3, 3)$ gives the strips to $(3)$ (height $0$ ) and to $(2, 1)$ (height $1$ ), so $χ^{(3, 3)} (3, 3) = χ^{(3)} (3) - χ^{(2, 1)} (3) = 1 - (- 1) = 2$ by [F4]. Hence the value on the class $(7, 3, 3)$ is $2$ .

2.1step 1.1step 1.2step 1.3F1algebra∎

Steps 1.1, 1.2 and 1.3 give the three displayed values: $χ^{(5, 4, 4)} = - 1$ on $(5, 4, 3, 1)$ from the two sign-opposite $5$ -strip removals, $χ^{(5, 4, 4)} = 2$ on $(7, 3, 3)$ from the unique $7$ -strip removal and the two-term recursion for $(3, 3)$ , and $χ^{(5, 4, 4)} = 0$ on every class containing a cycle of length in ${8, 9, 10, 11, 12, 13}$ because the removable strip sizes are ${1, \dots, 7}$ . This settles all three claims of the statement.

Remarks

Consistency with the hook lengths. The two $5$ -strips stripped in step 1.2 correspond to the two hooks of length $5$ of $(5, 4, 4)$ , at the nodes $(1, 3)$ and $(2, 1)$ , and the unique $7$ -strip stripped in step 1.3 to the hook of length $7$ at $(1, 1)$ ; the remaining hook lengths $6, 4, 4, 4, 3, 3, 2, 2, 1, 1$ are not used in the first removals; the later $4$ , $3$ and $1$ cycles are evaluated in the resulting smaller shapes, and the absent hook lengths $8, \dots, 13$ are exactly the forced zeros of step 1.1.

Depends on

Used by

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Sources

G. D. James, The Representation Theory of the Symmetric Groups, Lecture Notes in Mathematics 682, Springer 1978 (complete author-hosted scan, 161 pp.) (standard reference, not scraped)
William Woods, Representation theory of the symmetric groups (Cambridge Part III essay, 33 pp.) (standard reference, not scraped)