PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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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.

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A differential graded algebra is an A∞-algebra with no higher products

Statement

Let $k$ be a field and let $(A, d, μ)$ be a cohomological differential graded algebra over $k$ in the convention of Cohomological differential graded algebra. Put $m_{1} = d$ , $m_{2} = μ$ , and $m_{n} = 0$ for all $n \geq 3$ . Then these maps make $A$ an uncurved cohomological A∞-algebra in the unsuspended convention of Uncurved cohomological A∞-algebra in the unsuspended convention.

Facts & Assumptions

Given: A field $k$ and a cohomological differential graded algebra $(A, d, μ)$ over $k$ .

[F1]

Uncurved cohomological A∞-algebra in the unsuspended convention: the maps $m_{n}$ must have degree $2 - n$ and satisfy the all-arity Stasheff identity.

[F2]

Cohomological differential graded algebra: the DGA multiplication is associative of degree zero, $d$ has degree $+ 1$ and $d^{2} = 0$ , and for homogeneous $a, b$ , $d (a b) = d (a) b + (- 1)^{∣ a ∣} a d (b)$ .

[F3]

Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring: the field $k$ is a commutative ring, as required by the DGA definition.

[F4]

The first three Stasheff identities: the first three A∞ equations are the square-zero, derivation, and associator-homotopy identities displayed there.

Proof

1.1givenF1F2F3

The maps have the required degrees: $m_{1} = d$ has degree $+ 1 = 2 - 1$ , $m_{2} = μ$ has degree $0 = 2 - 2$ , and each zero map $m_{n}$ for $n \geq 3$ is a homogeneous map of degree $2 - n$ . The ring and field hypotheses agree by [F3].

1.2F2F4

By [F2], $d^{2} = 0$ and $d$ satisfies the signed Leibniz rule, so the first two identities of [F4] hold for $m_{1} = d$ and $m_{2} = μ$ .

1.3F2F4algebra

Since $m_{3} = 0$ , the third identity of [F4] reduces to $μ (μ \otimes 1) - μ (1 \otimes μ) = 0$ , which is associativity from [F2].

2.1F1step 1.1algebra

For $n \geq 4$ , any nonzero summand in the Stasheff sum would require both the inner arity $s$ and the outer arity $r + 1 + t$ to lie in ${1, 2}$ . But $n = r + s + t = s + (r + 1 + t) - 1 \leq 3$ , contradicting $n \geq 4$ . Thus every term in each such sum is zero.

3.1step 1.2step 1.3step 2.1∎

Steps 1.2, 1.3, and 2.1 verify the Stasheff identity in every arity, so the specified maps define the claimed A∞-structure.

Depends on

Used by

A minimal A∞-model whose ternary operation realizes a triple Massey product Example

Dependency tree · two levels

19 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

Bernhard Keller, Introduction to A-infinity algebras and modules, §3.1 (standard reference, not scraped)
Jean-Louis Loday and Bruno Vallette, Algebraic Operads, §§9.2.1 and 9.2.5 (standard reference, not scraped)