ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

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.

A nonabelian group algebra is symmetric Frobenius but not a closed sector

Example

For every field $k$ , the group algebra $k [S_{3}]$ is symmetric Frobenius under the coefficient-of-identity form. It is noncommutative, so it cannot itself be the commutative algebra assigned to a circle by a closed oriented two-dimensional TQFT. This says nothing about whether it can be the open sector of a separate knowledgeable open/closed theory.

Facts & Assumptions

Given: A field $k$ and the symmetric group $S_{3}$ , with permutation products composed right to left.

[F1]

The group ring has basis ${[g] : g \in S_{3}}$ and multiplication $[g] [h] = [g h]$ (The group ring $R [G]$ of finitely supported formal $R$ -linear combinations of group elements, The group ring $R [G]$ is a unital $R$ -algebra with basis $G$ , and each $g \in G$ is a unit of $R [G]$ ).

[F2]

The coefficient-of-identity form on $k [G]$ is associative, symmetric, and nondegenerate for every finite group $G$ and every field $k$ (For a finite group and a field of characteristic p, the group algebra is a symmetric Frobenius algebra via the coefficient of the identity). The item's title is narrower than its actual statement.

[F3]

The algebra assigned to a circle by a closed oriented two-dimensional TQFT is commutative (Closed oriented two-dimensional TQFTs are commutative Frobenius algebras).

Verification

1.1givenF1F2algebra

Apply [F2] to $G = S_{3}$ . It gives the symmetric nondegenerate associative form $⟨ x, y ⟩ =$ coefficient of $[e]$ in $x y$ ; by the Frobenius definition this makes $k [S_{3}]$ a symmetric Frobenius algebra over any field. In $S_{3}$ , $(12) (23) = (123)$ and $(23) (12) = (132)$ under the stated composition convention. These permutations are distinct, and their corresponding group elements are distinct basis vectors by [F1]. Therefore $[(12)] [(23)] = [(123)] \neq [(132)] = [(23)] [(12)],$ so the multiplication is not commutative.

2.1step 1.1F3contradiction∎

If this algebra were itself assigned to the circle by a closed oriented two-dimensional TQFT, [F3] would make its multiplication commutative, contradicting step 1.1. Hence it is not a closed sector of that form. The argument concerns only the closed circle object; it neither constructs nor rules out a separate open/closed extension with a different commutative closed algebra.

Depends on

Used by

Nothing in the library uses this result yet.

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

Peter Webb, A Course in Finite Group Representation Theory, §§1.1 and 8.5 (standard reference, not scraped)
Aaron D. Lauda and Hendryk Pfeiffer, Open-closed strings: Two-dimensional extended TQFTs and Frobenius algebras, arXiv:math/0510664v3 (standard reference, not scraped)