# Higher homotopy algebraic structure on the homology of an operad

Given a DGA $A$, then by standard techniques such as homological perturbation theory, the ring structure on the homology $H(A)$ extends to a minimal $A_\infty$-algebra structure such that $H(A)$ is quasi-isomorphic to $A$, and moreover, this structure on the homology is unique up to isomorphism. The $A_\infty$ algebra structure is describe explicitly, as usual, by the collection of higher multiplications $m_n: H(A)^{\otimes n} \to H(A)$ (which are closely related to Massey products).

An operad in the category of chain complexes can be thought of as a generalisation of a DGA - we now have a sequence of complexes $P(n)$, and the $\circ_i$ compositions are associative products between these. The homology $H(P(n))$ is again an operad in chain complexes (with zero differential), and the above fact for DGAs should generalise to say that $H(P)$ carries the additional structure of a strongly homotopy operad for which it is quasi-isomorphic to $P$. Has this structure been describes explicitly in terms of higher $n$-ary analogues of the $\circ_i$ compositions? I would be very grateful if someone could point me towards an appropriate reference.

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Seems to me like $H(P)$ should be an $A_\infty$-object in the category of symmetric sequences under the composition product - my guess is that you probably just replace the domain of $m_n$ with an n-fold composition product rather than a tensor. –  Tyler Lawson Jan 16 '11 at 3:13
Try looking at the work of Bruno Vallette? –  Kevin H. Lin Jan 16 '11 at 3:20
A short answer is yes, just as the higher A-infty maps are given as a twisting map from the free coalgebra on H(A) to H(A), the higher infinity operad maps are given as a twisting map from the free cooperad. I don't think I know the references well enough to give you a good long answer. There's a paper by Vallette and Merkulov which contains a lot of relevant background. Though perhaps the authors of most relevance are Chuang and Lazarev. I'm sorry that I can't be more precise. –  James Griffin Jan 17 '11 at 11:02