TITLE: Right actions of operads and monads
QUESTION [10 upvotes]: Given an operad $A$, there is an associated monad $M_A$ given by $M_A(X) = \coprod_n A(n) \otimes X^{\otimes n}$ such that being an $A$-algebra and being an algebra over the monad $M_A$ is the same thing (the categories are equivalent). This is very classical. However, there is also the notion of a right action of an operad on $X$, given by compatible maps
$$
X^{\otimes n} \otimes A(k_1) \otimes \cdots \otimes A(k_n) \to X^{\otimes \sum k_i}
$$
(one could also replace $X^{\otimes n}$ by a more general sequence of spaces $X(n)$).
Also, a right module over a monad $M \in End(C)$ is a functor $F\colon C \to D$ to some other category with a right action map $F \circ M \to F$ which is unital and associative.
Is there any relationship between these two concepts? Is there a purely monadic way of describing a right action of a monad $M$, which specializes to the right action of an operad $A$ if $M=M_A$?

REPLY [3 votes]: I think I found a construction that looks right in many circumstances. In the simplest case of a monad $M$ on sets, define a right action of $M$ on an object $X$ in a concrete category C as a natural transformation of functors in sets $Y$,
$$
X \times M(Y) \to M(X \times Y)
$$
where $X \times S$ for a set $S$ denotes the $S$-fold coproduct of $X$ in the category C. There is a unitality and an associativity condition.
For example, if $C$ is the category of commutative algebras and $M$ is the free commutative monoid monad, then a right $M$-action on $X$ is precisely a bialgebra structure on $X$. This isn't entirely obvious. You can enrich this definition and do it over categories other than sets.
I know this is very brief, but I don't know if there are people still following this... So if you want to know more, have comments, or have seen something like that before, let me know.