CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30

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A reducible Specht module in characteristic two

Statement refuted

If the signed column-antisymmetrizer construction is made over any field, then every resulting Specht module is irreducible.

Facts & Assumptions

Given: Let $K = F_{2}$ and $λ = (3, 1)$ . Let $v_{i}$ be the $(3, 1)$ -tabloid whose singleton second row is $i$ , for $1 \leq i \leq 4$ . Put $V = ⨁_{i = 1}^{4} K v_{i}$ , with $S_{4}$ acting by $σ v_{i} = v_{σ (i)}$ . Define the modular polytabloid directly by reducing each coefficient $sgn (γ) \in {1, - 1}$ to $K$ in $e_{t} = \sum_{γ \in C_{t}} \overline{sgn (γ)} γ \cdot {t}$ , and let $S_{K}^{(3, 1)}$ be the span of these vectors over all tableaux $t$ .

[F1]

A tabloid is a row-equivalence class, tabloids form the permutation-module basis, and $S_{n}$ acts by relabelling entries (Young subgroups, tabloids, and permutation modules).

[F2]

The column stabilizer consists of permutations preserving each column set (Row and column stabilizers).

[F3]

The signed column sum is $κ_{t} = \sum_{γ \in C_{t}} sgn (γ) γ$ and the polytabloid is $e_{t} = κ_{t} \cdot {t}$ (Column antisymmetrizers, polytabloids, and Specht modules).

[F4]

The sign takes values in ${1, - 1}$ (The sign is a homomorphism $S_{n} \to {+ 1, - 1}$ , surjective exactly when $n \geq 2$ ).

[F5]

A finite-dimensional representation is a finite-dimensional vector space with a group homomorphism to its group of invertible linear maps (A finite-dimensional representation $ρ : G \to GL (V)$ over a field, and its degree).

[F6]

A subrepresentation is an invariant linear subspace, and an irreducible representation has no proper nonzero subrepresentation (Subrepresentations, direct sums of representations, and irreducibility).

Counterexample

technique · direct

1.1givenF1F2F3F4algebra

Every tableau of shape $(3, 1)$ has a first column of size two and two singleton columns; if its bottom entry is $i$ and the entry above it is $j$ , then [F2] gives $C_{t} = {1, (i j)}$ . Its tabloid is $v_{i}$ , and $(i j) \cdot v_{i} = v_{j}$ . By [F3] and [F4], both signs reduce to $1$ in $K$ because $1 = - 1$ in $F_{2}$ , so $e_{t} = v_{i} + v_{j}$ . Thus all polytabloids lie in $W : = \ker ϵ$ , where $ϵ (\sum_{i} a_{i} v_{i}) = \sum_{i} a_{i}$ .

2.1givenF1F2F3F4step 1.1algebra

The tableaux with top rows $[4, 2, 3]$ , $[4, 1, 3]$ , $[4, 1, 2]$ and respective bottom entries $1, 2, 3$ give $b_{1} = v_{1} + v_{4}$ , $b_{2} = v_{2} + v_{4}$ , $b_{3} = v_{3} + v_{4}$ . These vectors are independent by their first three coordinates. If $x = \sum_{i} a_{i} v_{i} \in W$ , then $a_{4} = a_{1} + a_{2} + a_{3}$ , so $x = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$ . Therefore $W$ has basis $b_{1}, b_{2}, b_{3}$ , and since each is a polytabloid while every polytabloid lies in $W$ , $S_{K}^{(3, 1)} = W$ .

3.1givenF1F5F6step 2.1algebra∎

The vector $w = v_{1} + v_{2} + v_{3} + v_{4}$ is nonzero, has $ϵ (w) = 4 = 0$ in $K$ , and is fixed by every permutation in $S_{4}$ . Hence $K w$ is a nonzero subrepresentation of $S_{K}^{(3, 1)}$ by [F5] and [F6]. It is proper because $S_{K}^{(3, 1)} = W$ has the three-element basis from step 2.1, whereas $K w$ has dimension one. Thus this Specht module is reducible, refuting the claimed field-independent irreducibility.

Depends on

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Sources

Charlotte Chan, Representation Theory of Symmetric Groups, Remark 4.6, printed p. 16 (modular warning; the concrete F_2 example is computed here) (standard reference, not scraped)
Mark Wildon, Representation Theory of the Symmetric Group, Examples 2.3(2) and 2.6(B), printed pp. 5-6 (hook-shape tabloids and polytabloids over a field; the reducibility witness is computed here) (standard reference, not scraped)