The structure of an algebra $A$ over a operad $O$ is encoded by an operad morphisms from $O$ to $\{Hom(A^{\otimes k},\, A)\}_{k}$. The same structure can be stored using the structure $M_OA\to A$ of an algebra over the monad $M_O$ which is defined on an object $A$ by $\bigoplus_j O(j)\otimes_{\Sigma_j} A^{\otimes j}$ and its structure maps $M_O\circ M_O\to M_O$ and $1\to M_O$ are induced from composition and unit respectively.

The structure of an coalgebra $C$ over a operad $O$ is encoded by an operad morphisms from $O$ to $\{Hom(C,C^{\otimes k})\}_{k}$.

Is there a monadic interpretation of coalgebras over an operad?

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    $\begingroup$ A coalgebra is just an algebra in the opposite category, so the category of coalgebras over an operad is comonadic rather than monadic in general. $\endgroup$ Oct 3, 2017 at 16:58
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    $\begingroup$ @DenisNardin However, I believe that the encoding of an operad action as a monad action uses the fact that $\otimes$ distributes over the infinite coproduct $\bigoplus_j$. Since $\otimes$ doesn't usually distribute over infinite products, it's not clear to me that coalgebras over an operad can usually be encoded as a coalgebra over a comonad. $\endgroup$ Oct 3, 2017 at 19:15
  • $\begingroup$ MO's related links points me to mathoverflow.net/q/152785/49 $\endgroup$ Oct 3, 2017 at 19:17

1 Answer 1


What works is a comonadic interpretation of coalgebras over cooperads. Coalgebras over cooperads are like conilpotent coalgebras over operads.


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