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 $M=M_A$?

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    $\begingroup$ Operads aren't really algebraic topology. They properly belong in the classification of "universal algebra". $\endgroup$ Mar 22, 2010 at 17:25
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    $\begingroup$ They're not category theory or universal algebra, either. It's not clear the retagging is helpful. $\endgroup$ Mar 22, 2010 at 17:34
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    $\begingroup$ What kind of symmetric monoidal category are you working in? You at least need the ⊗ to distribute over coproducts to get this equivalence with monads. (And since I don't know what the appropriate distributivity should be in the case of right actions, I don't know how to answer the question.) $\endgroup$ Mar 22, 2010 at 17:50
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    $\begingroup$ Thanks Tyler for drawing the discussion away from the subject of how to tag it... Good point, yes, my monoidal product distributes over coproducts, or else what I said wouldn't be true. The monad itself, however, will usually not commute with coproducts. I don't quite understand your parenthesized sentence -- any definition, however crazy, of a "right module" for a monad that generalizes the right action of an operad will make me happy. If you like, work in vector spaces. $\endgroup$
    – Tilman
    Mar 22, 2010 at 19:40
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    $\begingroup$ I mean the distributivity with respect to the right action. Unless I am mistaken, a right action is a family of maps $X \otimes A(n) \to X^{\otimes n}$ and the natural "comonadic" thing would be a map $X \to \prod Hom(A(n), X^{\otimes n})$. However, this formula doesn't actually define a comonad in e.g. vector spaces and so I'm not sure how to connect a right action to something that looks monadic. $\endgroup$ Mar 25, 2010 at 19:56

2 Answers 2


I hadn't seen this question before. Better late than never, perhaps. I'll give some categorical background first, which may well be familiar, but to get everything sorted out we should probably recall it anyway.

A more than sufficient set of hypotheses is that we're working in a cocomplete closed symmetric monoidal category $V$. Let $\mathbb{P}$ be the permutation category. There is a well-known monoidal product $\boxtimes$ on $V^{\mathbb{P}^{op}}$ called "substitution product", given objectwise by the formula

$$(F \boxtimes G)(n) = \sum_k F(k) \otimes_{S_k} G^{\otimes_{Day} k}$$

where $\otimes_{Day}$ is the Day convolution product. (The more standard notation is $\circ$ instead of $\boxtimes$, but that is potentially confusing here.) Monoids in the monoidal category $(V^{\mathbb{P}^{op}}, \boxtimes)$ are the same thing as operads. The nLab article on operads provides plenty of explanation and background for this view on operads.

And, as in any monoidal category, a monoid $A$ induces a monad structure on the monoidal category, in fact two monad structures: a left one $A \boxtimes -$ whose algebras are left $A$-modules, and a right one $- \boxtimes A$ whose algebras are right $A$-modules. This applies in particular to operads $A$, so we have two monads, $L_A$ and $R_A$ respectively, both acting on $V^{\mathbb{P}^{op}}$.

Then a right action of an operad on a functor $F: \mathbb{P}^{op} \to V$ is the same thing as an (ordinary, left-sided) algebra/module of $R_A = - \boxtimes A$. But you can also consider a right action as inducing a right module $F \boxtimes -$ over the monad $L_A = A \boxtimes -$. So the answer to your question is certainly 'yes', but you'll have to decide for yourself how interesting the answer is.

Perhaps I could say a few more things. In your notion of right action on an object $X$ of $V$, what you essentially did is embed $V$ in $V^{\mathbb{P}^{op}}$ by sending an object $X$ to the evident $\mathbb{P}$-representation $X^{\otimes \bullet}$, so that your notion of right action on $X$ takes the form of a right $A$-module

$$(X^{\otimes \bullet}) \boxtimes A \to X^{\otimes \bullet}$$

in the category $V^{\mathbb{P}^{op}}$. That's fine, but I'll note that that choice of embedding doesn't quite match the one used for the usual notion of (left) algebra over an operad. For this, we embed $V$ in $V^{\mathbb{P}^{op}}$ by mapping an object $X$ of $V$ to the functor $\hat{X}: \mathbb{P}^{op} \to V$ where $\hat{X}(0) = X$ and otherwise $\hat{X}(n)$ is the initial object $0$. Let $i: V \to V^{\mathbb{P}^{op}}$ denote this embedding. Then, it is easy to see that the composite

$$V^{\mathbb{P}^{op}} \times V \stackrel{id \times i}{\to} V^{\mathbb{P}^{op}} \times V^{\mathbb{P}^{op}} \stackrel{\circ}{\to} V^{\mathbb{P}^{op}}$$

factors up to isomorphism through the embedding $i: V \to V^{\mathbb{P}^{op}}$, thus giving a functor

$$V^{\mathbb{P}^{op}} \times V \stackrel{\bullet}{\to} V$$

where the monoidal category $V^{\mathbb{P}^{op}}$ acts on $V$, in such a way that there is a coherent natural isomorphism $(F \circ G) \bullet X \cong F \bullet (G \bullet X)$. If you work through the details, you find that

$$F \bullet X = \sum_k F(k) \otimes_{S_k} X^{\otimes k}$$

This type of structure, where a monoidal category $M$ acts on a category $C$ in this coherent categorified fashion, is called an actegory, and the general nonsense in this case is that a monoid $A$ in $M$ induces a monad on $C$, given objectwise by $X \mapsto A \bullet X$. In particular, an operad as monoid in $V^{\mathbb{P}^{op}}$ induces a monad on $V$, and it's the usual monad on $V$ attached to an operad $A$ with components valued in $V$.

  • $\begingroup$ Thanks! However, my question was aimed in a different direction: Suppose we have a monad F on some category C, is there a way of defining a "right action" of the same monad F on C such that if F comes from an operad (via the usual, left action construction) then a right F-object is the same as a right algebra over the operad. You're constructing a different monad from an operad whose algebras are right algebras over the operad. I think I actually know a partial answer to my original question (I didn't when I posted), see below. $\endgroup$
    – Tilman
    May 30, 2010 at 8:15
  • $\begingroup$ I think I did correctly transcribe your notion of right action of an $V$-valued operad $A$ on a $V$-object $X$, namely as a right module structure of the P-rep $X^{\otimes *}$ over the monoid $A$. But never mind: it might be more fruitful to pursue the thread below your own answer. If this is coming from things related to plethysm, I'd be interested. $\endgroup$
    – Todd Trimble
    May 30, 2010 at 16:15

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.

  • $\begingroup$ Yes, more please. :) $\endgroup$
    – JBorger
    May 30, 2010 at 13:41
  • $\begingroup$ Jim, I already sent you some notes! And the formal plethories are coming up pretty soon, too... $\endgroup$
    – Tilman
    May 30, 2010 at 15:25
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    $\begingroup$ Well, that's very close to things I've seen before, namely strengths on functors. However, before trying to say more in these damned boxes, I'm having trouble parsing something: a priori, the right side of your arrow lives in sets, and the left-hand side in C. Do you mean the left hand side is the underlying set of the M(Y)-fold coproduct of X? And the argument inside the M on the right is similarly the underlying set of the Y-fold coproduct of X? Anyway, it's good to get more of an idea what's motivating your original question. $\endgroup$
    – Todd Trimble
    May 30, 2010 at 16:09

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