Uncurved cohomological A∞-algebra in the unsuspended convention

Let $k$ be a field (Field) and let $A={⨁}_{j\in \mathbb{Z}}{A}^j$ be a $\mathbb{Z}$-graded $k$-vector space; its tensor maps use the convention in Koszul sign rule for graded tensor products and permutations. An uncurved cohomological A∞-algebra in the unsuspended convention is a sequence of homogeneous $k$-linear maps

$m_n:A^{{\otimes }_{k}n}⟶A,\mathrm{\mid }{m}_n\mathrm{\mid }=2-n,n\ge 1,$

such that for every $n\ge 1$ the following equality of maps $A^{{\otimes }_{k}n}\to A$ holds:

${\sum }_{\begin{array}{c}r,t\ge 0,s\ge 1\\ r+s+t=n\end{array}}(-1{)}^{r+st}{m}_{r+1+t}(1^{\otimes r}\otimes m_s\otimes 1^{\otimes t})=0.$

Here $1$ denotes the identity map of $A$, the tensor maps use the Koszul rule, and the displayed sign $(-1{)}^{r+st}$ is part of the fixed convention. The degree condition means that for homogeneous inputs of degrees $j_1,\dots ,j_n$, $m_n$ has output degree $j_1+\cdots +j_n+2-n$.

“Uncurved” means no arity-zero operation $m_0$ is included. “Nonunital” means that no unit is part of this definition. A unit or a strict-unit condition on higher operations is additional data and must be stated separately. No graded commutativity is assumed for $m_2$.

The arity-two identity is the chain-map condition for $m_2$. Its commutative square is shown below; the downward map on $A^{\otimes 2}$ is the tensor differential with the Koszul sign rule. This diagram records the equation only; it is not a proof of any Stasheff identity.