TITLE: Conceptual explanation for the sign in front of some binary operations
QUESTION [8 upvotes]: In several situations, I've seen that given a binary operation on a graded module $m:A\otimes A\to A$, a new operation $M(x,y)=(-1)^{|x|}m(x,y)$ is defined so that it satisfies some properties.
One example of this happens in Homotopy G-algebras and moduli spaces, where for a binary operation $m\in\mathcal{O}(2)$ such that $m\circ m=0$ for some operad $\mathcal{O}$, an associative product is defined by $xy=(-1)^{|x|+1}m\{x,y\}$, where the brace notation stands for the brace algebra structure on $\mathcal{O}$. In this case, the explanation I've been able to deduce is that this is necessary for the brace relation (equation (2) in the paper) to imply associativity of the product $xy$. In this case the sign $(-1)^{|x|}$ works for this purpose too.
Another more direct instance of this situations occurs in Cartan homotopy formulas and the Gauss-manian connection in cyclic homology, where given an $A_\infty$-algebra with $m_i=0$ for $i>2$, one gets a dg-algebra by defining again $xy=(-1)^{|x|}m_2(x,y)$. In this case this is because the author uses a convention for $A_\infty$-algebras in which the equations only have plus signs, so some extra sign is needed to produce the associativity relation and the Leibniz rule. So the reasons are very similar to the previous case even though the construction is simpler because there is no brace algebra here.
And another extra example for which I don't have any reference is in the case of Lie algebras. When one defines a generator of the operad of graded Lie algebras, often one takes $l(x,y)=(-1)^{|x|}[x,y]$ instead of directly defining $l$ as the bracket. If I remember correctly this was needed to obtain the Jacobi identity in purely operadic terms.

So it looks like it's very common to add that sign in order to make some relations hold. What I would like to know if there is a more conceptual explanation of why this holds systematically. Maybe it's just that it works when writing down the equations, but I'm looking for a more general intuition.
My motivation is generalizing this idea to maps of higher arity. More precisely, given an $A_\infty$-multiplication $m\in\mathcal{O}$ such that $m\circ m=0$, I want to define an $A_\infty$-structure $M$ on $\mathcal{O}$ that satisfies the sign convention
$$\sum_{n=r+s+t}(-1)^{rs+t}M_{r+1+t}(1^{\otimes r}\otimes M_s\otimes 1^{\otimes t})=0.$$
(There is also another possible convention where $rs+t$ is replaced by $r+st$)
So this is very similar to Getzler's paper where he defines $M_j(x_1,\dots, x_j)=m\{x_1,\dots x_j\}$, and this structure maps satisfy the relation $M\circ M=0$ but with all plus signs. So I need to modify these maps by some signs in a similar way as the associative case. Of course I can try to sit down and write the equations and find some necessary conditions for the signs and maybe find a pattern. But if there is a conceptual explanation for the associative case and the lie algebras, then maybe there is an easier way to find out what the signs I need are.

REPLY [4 votes]: As  Gabriel C. Drummond-Co commented, it has to do with suspensions that are implicit. I'll do it with the example of Gerstenhaber and Voronov and the others should follow similarly. Let us denote $M_2(x,y)=x\cdot y$ the product that we want to define based on the brace $m\{x,y\}$. If we define it as a map $(s\mathcal{O})^{\otimes 2}\to s\mathcal{O}$ (suspension as graded vector spaces), then the natural thing to do is using the brace $m\{-,-\}:\mathcal{O}^{\otimes 2}\to \mathcal{O}$, but to do so one has to compose with suspensions and desuspensions. Namely, $M_2(x,y)=s(m\{(s^{-1}x,s^{-1}y)\})$. And it's applying $(s^{-1})^{\otimes 2}(x,y)$ what makes the sign $(-1)^{|x|}$ appear. If we use $(s^{\otimes 2})^{-1}$ instead then we get the original sign $(-1)^{|x|+1}$.