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Frobenius sewing and exact two-dimensional topological amplitudes — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Nonsemisimple Frobenius sewing is also possible

Example

Let $A = C [x] / (x^{2})$ with basis $(1, x)$ , unit $1$ , multiplication inherited from the quotient, and Frobenius functional $ϵ (1) = 0$ , $ϵ (x) = 1$ . The bilinear pairing $b (a, b) = ϵ (a b)$ has matrix $(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ in that basis, hence is nondegenerate. With the conjugate-linear involution fixing $x$ , the form $h (a, b) = ϵ (a^{*} b)$ is Hermitian but indefinite. Here a finite-dimensional algebra is semisimple when its Jacobson radical (the intersection of its maximal left ideals) is zero; $A$ is not semisimple because $x$ is a nonzero nilpotent in that radical. The closed oriented classification therefore gives a closed two-dimensional TQFT from a nonsemisimple Frobenius algebra without asserting positive reflection form data.

Example references

Closed oriented two-dimensional TQFTs are commutative Frobenius algebras

Facts & Assumptions

Given: The quotient algebra $A = C [x] / (x^{2})$ , its basis $(1, x)$ , the stated functional, and the conjugate-linear involution $1^{*} = 1$ , $x^{*} = x$ .

[F1]

Closed oriented two-dimensional TQFTs are commutative Frobenius algebras: over any field, every finite-dimensional commutative Frobenius algebra supplies a strong symmetric monoidal functor on the closed oriented two-dimensional cobordism category, whose morphisms have no free boundary.

Verification

1.1givenalgebra

Every class in $A$ is uniquely $a + b x$ with $a, b \in C$ , and $(a + b x) (c + d x) = a c + (a d + b c) x$ because $x^{2} = 0$ . This is the quotient of the associative commutative algebra $C [x]$ , so $A$ is a two-dimensional commutative unital associative algebra with nonzero nilpotent $x$ .

1.2givenalgebra

The pairing values are $b (1, 1) = 0$ , $b (1, x) = 1$ , $b (x, 1) = 1$ , and $b (x, x) = 0$ , so its matrix is $(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ and determinant $- 1 \neq 0$ . Hence $b$ is nondegenerate.

2.1givenstep 1.1step 1.2algebra

Define $Δ (1) = 1 \otimes x + x \otimes 1$ and $Δ (x) = x \otimes x$ . On both basis elements, $(ϵ \otimes 1) Δ = (1 \otimes ϵ) Δ = {id}_{A}$ as counit identities; the two coassociative composites on $1$ both equal $1 \otimes x \otimes x + x \otimes 1 \otimes x + x \otimes x \otimes 1$ , and on $x$ both equal $x \otimes x \otimes x$ . For the two Frobenius identities, bilinearity reduces to the four basis pairs $(1, 1), (1, x), (x, 1), (x, x)$ ; on these pairs all three composites $Δ μ$ , $(1 \otimes μ) (Δ \otimes 1)$ , and $(μ \otimes 1) (1 \otimes Δ)$ respectively equal $Δ (1), Δ (x), Δ (x), 0$ . Thus the specified maps make $A$ a commutative Frobenius algebra.

2.2givenstep 1.1algebra

The ideal $(x)$ is maximal because $A / (x) ≅ C$ . For every maximal ideal $M$ , the quotient $A / M$ is a field and the image of $x$ there is nilpotent because $x^{2} = 0$ ; a field has no nonzero nilpotents, so $x \in M$ . Hence $0 \neq x \in J (A)$ , the intersection of all maximal left ideals, and $A$ is not semisimple under the stated definition.

2.3givenstep 1.2algebra

For $a = a_{0} + a_{1} x$ and $b = b_{0} + b_{1} x$ , $h (a, b) = \overline{a_{0}} b_{1} + \overline{a_{1}} b_{0}$ , so $h (a, b) = \overline{h (b, a)}$ and the form is Hermitian. Also $h (1 + x, 1 + x) = ϵ ((1 + x)^{2}) = 2$ while $h (1 - x, 1 - x) = ϵ ((1 - x)^{2}) = - 2$ , so it takes both signs and is not positive definite.

3.1givenF1step 2.1∎

By step 2.1 the algebra is finite-dimensional, commutative, and Frobenius, so [F1] applies over $k = C$ and gives the closed oriented two-dimensional TQFT. This conclusion uses the closed category only and adds no open/closed or positivity claim.