I know from [this paper by Ward](https://arxiv.org/abs/1208.5543) that one can obtain the (signs of) the Gerstenhaber bracket using operadic suspension on any operad $\mathcal{O}$. More precisely, the insertion $\tilde{\circ}$ of the new operad has the appropriate sign when compared to the original insertion $\circ$ on $\mathcal{O}$. Applying this to elements $m\_i\in\mathcal{O}(i)$ of degree $2-i$ one obtains the expression defining $A\_\infty$-algebras, so if $m=\sum\_i m\_i$, $m\tilde{\circ} m=0$ is the same as an $A\_\infty$-multiplication.


I would like to know if there is a bigraded analogue of this construction that could give me the expression defining *derived* $A\_\infty$-algebras (look [here](https://arxiv.org/abs/0909.3222) for a definition). I have tried several approaches that did not work, such as trying to mimique Ward's definition for the bigraded case. There is also a use of operadic suspension in a bigraded context in [this paper by Livernet et al](https://arxiv.org/abs/1401.5251) which is applied only to the vertical component of the bigraded modules, so it doesn't really do the job.


I also know from [Roitzheim and Whitehouse](https://arxiv.org/abs/0909.3222) that there is a way to obtain the derived $A\_\infty$-equation by aplying the same insertion obtained from operadic suspension in the graded case but twisting the second argument. More precisely, if $f$ is of bidegree $(i,j)$ then $f^\sharp=(-1)^i f$. If $m=\sum\_{ij}m\_{ij}$ with $m\_{ij}\in \mathcal{O}(j)$ of bidigree $(i,2-(i+j))$, then $m\circ m^\sharp=0$ is equivalent to having a derived $A\_\infty$-multiplication. But it seems to me that no operadic insertion can be obtained from this twist (to be an operadic it should satisfy certain associativity relations and the twist destroys associativity).


So to make it clear, given a bigraded operad $\mathcal{O}$ with given insertion map $\circ$ and elements $m\_{ij}\in\mathcal{O}(j)$ of bidegree $(i,2-(i+j))$, I would like to find some new operad $\tilde{\mathcal{O}}$ with insertion map $\tilde{\circ}$ such that $m=\sum\_{ij}m\_{ij}\in\tilde{\mathcal{O}}$ (under some canonical identification) and such that $m\tilde{\circ} m=0$ is equivalent to $(-1)^{rq+t+pj}m\circ m=0$, the equation deifining derived $A\_\infty$-algebras. Is this even possible?



Bigraded operadic suspension
In the paper [Derived A-infinity algebras in an operadic context](https://arxiv.org/abs/1110.5167), Section 5.1, there is a star operation which is the convolution operation on $\mathrm{Hom}(C,P)$ where $C$ is a cooperad and $P$ is an operad. In the case of $C=dAs^¡$ and $P=\mathrm{End}\_A$, this complex is in bijection with the Hochschild complex of $A$ and we get an operation $\star$ such that if $m=\sum\_{ij}m\_{ij}$ with $m\_{ij}$ of arity $j$ and bidigree $(i,2-i-j)$, then $m\star m=0$ if and only if $m$ is a derived $A\_\infty$-multiplication on $A$.


Thaking the explicit definition of $\star$ we can define a similar operation for any operad $\mathcal{O}$ with infinitesimal composition $\circ$, so that $\star$ is $\circ$ up to sign.The sign can be computed as a sequence of (vertical) operadic suspension and totalization.


