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A missing removable strip of the cycle length forces a zero character value

Statement

Let $λ ⊢ n$ and let $ρ \in S_{n}$ be a permutation containing a cycle of length $1 \leq r \leq n$ (equivalently, $c_{r} (ρ) > 0$ in the cycle-count notation of Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type, with fixed points counted as cycles of length $1$ ). If $λ$ has no removable border strip of size $r$ (Removable border strips and their height), then $χ^{λ} (ρ) = 0.$ By the hook/rim-hook correspondence this is equivalent to: if no hook length of $λ$ equals $r$ (Hook, arm, leg, and hook length of a box), then $χ^{λ} (ρ) = 0$ (Hooks of a partition correspond bijectively to removable border strips).

Facts & Assumptions

Given: A partition $λ ⊢ n$ , an integer $r \geq 1$ , and a permutation $ρ \in S_{n}$ containing a cycle of length $r$ , with $χ^{λ}$ the character of the complex Specht module $S^{λ}$ (Column antisymmetrizers, polytabloids, and Specht modules).

[F1]

Murnaghan–Nakayama rule: for a cyclic permutation $c$ on a specified $r$ -point set $D$ and a permutation $σ \in S_{n}$ fixing $D$ pointwise, $χ^{λ} (σ c) = \sum_{λ / ν} (- 1)^{ht (λ / ν)} χ^{ν} (σ)$ , summed over the removable border strips of $λ$ of size $r$ , the empty sum being $0$ (The Murnaghan–Nakayama character rule).

[F2]

Hook/rim-hook correspondence: for every node $x \in [λ]$ the rim hook $ξ (x)$ is a removable border strip of $λ$ of size the hook length $h (x)$ , and every removable border strip of $λ$ is $ξ (x)$ for exactly one node $x$ ; hence the sizes of the removable border strips of $λ$ are exactly the hook lengths of $λ$ (Hooks of a partition correspond bijectively to removable border strips, Hook, arm, leg, and hook length of a box, Removable border strips and their height).

[F3]

Cycle decomposition: the disjoint cycles of $ρ$ have pairwise disjoint supports, so selecting a cycle on $r$ points (including a specified fixed point when $r = 1$ ) leaves a permutation $σ$ fixing those points. For $1 \leq r \leq n$ , the cycle-count entry $c_{r} (ρ)$ is the number of $r$ -element orbits, so such a cycle exists exactly when $c_{r} (ρ) > 0$ (Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type, The finite symmetric group $S_{n}$ , one-line notation, and cycle notation).

Proof

technique · direct

1.1givenF3algebra

Select a cycle of length $r$ , with its set $D$ of $r$ points; for $r = 1$ select a fixed point and put $c = 1$ , otherwise let $c$ be its cyclic permutation. Put $σ : = ρ c^{- 1}$ , the product of the remaining disjoint cycles. Then $ρ = σ c$ , and $σ$ fixes $D$ pointwise, because the supports of the cycles of $ρ$ are pairwise disjoint.

1.2F2algebra

By the hook/rim-hook correspondence [F2], the removable border strips of $λ$ are exactly the rim hooks $ξ (x)$ indexed by the nodes of $[λ]$ , and the size of $ξ (x)$ is the hook length $h (x)$ . Therefore $λ$ has a removable border strip of size $r$ if and only if some hook length of $λ$ equals $r$ , so the two hypotheses of the statement are equivalent.

2.1step 1.1step 1.2F1givenalgebra∎

Apply the Murnaghan–Nakayama rule [F1] to the specified cycle $(D, c)$ and the compatible permutation $σ$ of step 1.1: $χ^{λ} (ρ) = χ^{λ} (σ c) = \sum_{λ / ν} (- 1)^{ht (λ / ν)} χ^{ν} (σ)$ , the sum running over the removable border strips $λ / ν$ of size $r$ . By hypothesis $λ$ has no such strip, so the index set is empty and the empty sum is $0$ ; thus $χ^{λ} (ρ) = 0$ . By the equivalence of step 1.2 the same conclusion holds under the hook-length hypothesis.

Remarks

Emptiness is the only mechanism used. No cancellation takes place: the value is zero because every term of the recursion is absent. A permutation of $S_{n}$ cannot contain a cycle of length $r > n$ . Within $1 \leq r \leq n$ , the criterion applies whenever $r$ lies in a gap of the hook-length multiset, as for $λ = (2, 2, 1, 1)$ and $r = 3$ , whose hook lengths $5, 4, 2, 2, 1, 1$ avoid $3$ .

Depends on

Used by

The hook-divisibility vanishing criterion Corollary
A disconnected removable rim set is not a border strip Counterexample
Murnaghan–Nakayama values of the character of (5,4,4) Example

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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)