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If necessary, we can restrict the following to the case where we consider only Hopf (co)operads in the category of chain complexes over fields of characteristic zero.

In case of ordinary operads, their algebras can be defined as operad maps. To be more precise if $\mathcal{O}$ is an operad and $End_V$ is the endomorphism operad of an object $V$, then an $\mathcal{O}$-algebra on $V$ can be defined as an element of

$Hom_{Operad}(\mathcal{O},End_V)$ (A)

Moreover if the operad is "Koszul", with a Koszul dual cooperad $\mathcal{O}^i$, then homotopy $\mathcal{O}$-algebras can be described as solutions to the Maurer Cartan equation

$d\phi+\phi\star\phi=0$

where $d$ and $\star$ are the differential and the convolution product on

$\Pi_{n\in\mathbb{N}} Hom_{sym.seq.}(\mathcal{O}^i(n),End_V(n))$.

Ok. Now consider $\mathcal{O}$ is moreover a Hopf operad! I wonder if there is then any additional structure on $V$ (induced or predefined), which takes this additional Hopf structure into consideration? (The same question applies to the case of the homotopy $\mathcal{O}$-algebras.)

From the question "Hopf structure on endomorphism operad" we know, that a priori $End_V$ has no Hopf structure. Therefore we can not just ask for maps in (1) to be morphisms of Hopf operads.

I know this is a vague question. That is because I can't get a grip on what to expect. On a first sight, it just seems that the Hopf structure has no reflection on the level of the $\mathcal{O}$-algebras and I wonder how to see this right.

In particular, the Maurer Cartan equation does not take the structure of a Hopf cooperad on $\mathcal{O}^i$ into conmsideration.

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    $\begingroup$ Hopf operad structure literally means that you can define a structure of an $\mathcal{O}$-algebra on the tensor product of two $\mathcal{O}$-algebras, and that it very much a "reflection on the level of $\mathcal{O}$-algebras". So you probably need to clarify your question. $\endgroup$ Commented Nov 29, 2017 at 5:17
  • $\begingroup$ I see... But I think I can not clarify the question for now. Maybe later. $\endgroup$ Commented Nov 30, 2017 at 18:25

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