CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Untwisted conjugation changes the module

Statement refuted

Refuted claim: the same-shape duality $C_{M} (λ) ≅ P (λ)^{\circ}$ of Murphy cells are the same-shape contragredient Specht duals may be applied without a sign twist, i.e. one may conclude $C_{M} (λ^{'}) ≅ P (λ)^{\circ}$ directly.

This is false. Over $Q$ at $q = 1$ and $n = 3$ the Murphy cell module $C_{M} ((1^{3}))$ is the sign module, in which $T_{i}$ acts by $- 1$ , while the ordinary polytabloid module $P ((3))$ and its contragredient dual are trivial, in which $T_{i}$ acts by $+ 1$ . Hence $C_{M} (λ^{'}) ≅ P (λ)^{\circ}$ without $θ$ fails in general. The correct bridge is $C_{M} (λ^{'}) ≅ P (λ)^{θ}$ of Ordinary Specht modules are the signed conjugate Murphy cells; combining it with same-shape duality gives $P (λ^{'}) ≅ P (λ)^{\circ, θ}$ , which is a different formula. No choice principle is used.

Facts & Assumptions

Given: The field $Q$ , the specialization $q = 1$ of the type- $A$ Hecke algebra $H$ , the polytabloid module $P ((3)) = z_{(3)} H$ with its DJ form, the Murphy cell module $C_{M} ((1^{3}))$ with its form, and the comparison theorem Ordinary Specht modules are the signed conjugate Murphy cells together with its corollary Polytabloid specialization and the exact signed multiplicity dictionary.

[F1]

At $q = 1$ the specialization of $H$ is the group algebra $Q [S_{3}]$ and the sign automorphism $θ$ is the pullback along $w \mapsto sgn (w) w$ ; the partitions $(3)$ and $(1^{3})$ are conjugate (Polytabloid specialization and the exact signed multiplicity dictionary).

[F2]

The comparison theorem gives $P ((3)) ≅ C_{M} ((1^{3}))^{θ}$ with the exact unit $u_{(3)} = q^{ℓ (w_{0} (S_{(1^{3})})) - ℓ (w_{(3)})}$ (Ordinary Specht modules are the signed conjugate Murphy cells).

[F3]

$P ((3)) = x_{(3)} H$ is spanned by $x_{(3)} = \sum_{w \in S_{3}} T_{w}$ , and $x_{(3)} T_{i} = q x_{(3)}$ for $i = 1, 2$ ; the contragredient dual of a one-dimensional module with trivial action is trivial (Ordinary Specht modules are the signed conjugate Murphy cells).

[F4]

In $C_{M} ((1^{3}))$ one has $f^{(1^{3})} = 1$ , the higher ideal $I^{⊳ (1^{3})}$ contains the elements $1 + T_{i}$ , and the cell form has value $G_{(1^{3})} (m_{0}, m_{0}) = 1$ (Polytabloid specialization and the exact signed multiplicity dictionary, Ordinary Specht modules are the signed conjugate Murphy cells).

Counterexample

technique · direct

1.1F3algebra

The ordinary module is trivial. At $q = 1$ one has $x_{(3)} = \sum_{w \in S_{3}} T_{w}$ and $x_{(3)} T_{i} = \sum_{w} T_{w} T_{i} = \sum_{w} T_{w s_{i}} = x_{(3)}$ for each generator, because right multiplication by $s_{i}$ permutes the group and every $T_{w s_{i}}$ is again one of the summed basis elements; with the scalar carried along this reads $x_{(3)} T_{i} = q x_{(3)} = x_{(3)}$ at $q = 1$ , and $P ((3)) = x_{(3)} H = Q x_{(3)}$ is one-dimensional. Its dual is therefore also one-dimensional with trivial action.

1.2F4algebra

The cell is sign. At $q = 1$ the element $1 + T_{i}$ lies in the higher ideal $I^{⊳ (1^{3})}$ , because $1 + T_{1} = x_{(2, 1)}$ belongs to that two-sided ideal and conjugation by the invertible braid word $T_{1} T_{2}$ sends $T_{1}$ to $T_{2}$ , so $1 + T_{2}$ also belongs to it; hence in the one-dimensional cell $m_{0} (1 + T_{i}) = 0$ , that is, $m_{0} T_{i} = - m_{0}$ . The cell form value is $1$ , and the module is the sign module.

2.1step 1.1step 1.2F1F2algebra∎

No untwisted isomorphism. A right $H$ -module isomorphism $f ⁣ : C_{M} ((1^{3})) \to P ((3))^{\circ}$ would satisfy $f (m h) = f (m) h$ for every $h$ ; taking $h = T_{i}$ and a nonzero $m$ gives $f (- m) = f (m)$ because $C_{M} ((1^{3}))$ has $T_{i}$ acting by $- 1$ and $P ((3))^{\circ}$ has $T_{i}$ acting by $+ 1$ , while $f (- m) = - f (m)$ in characteristic zero, so $- f (m) = f (m)$ forces $f (m) = 0$ , contradicting injectivity over $Q$ . Therefore no such isomorphism exists and the untwisted claim is false. The twisted statement is consistent: on $C_{M} ((1^{3}))^{θ}$ the generator $T_{i}$ acts by the ordinary action of $θ (T_{i}) = - T_{i}$ , that is by $+ 1$ , matching $P ((3))$ , and the unit is $u_{(3)} = 1$ by [F2].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

17 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; author-hosted scan (standard reference, not scraped)
Andrew Mathas, Iwahori-Hecke algebras and Schur algebras of the symmetric group, author-hosted prepublication manuscript, 79 PDF pages (standard reference, not scraped)