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It's known that from every operad arises a cartesian monad whose algebras are the algebras for the operad. Leinster proved that there are different operads from which arise the same monad, in this way he proved that operads cannot be identified with monads. Now I'm wondering: is it true that every cartesian monad arises from an operad, i.e. every such a monad is the monad associated to an operad?

I little specification, what I mean here by operad is a generalized operad i.e. a $T$-operad for some cartesian monad $(T,\mu,\eta)$ in a cartesian category $\mathcal C$.

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3 Answers 3

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The answer is no, and it is explained in Appendix C of Leinster's book Higher Operads, Higher Categories.

Briefly, a 'plain operad', for Leinster, is a non-symmetric operad in Set; a T-operad for T the free-monoid monad. Leinster shows that if T is a cartesian monad then a T-operad is the same thing as a cartesian monad S equipped with a cartesian transformation $S \to T$. So a plain operad is the same thing as a cartesian monad on Set that is 'augmented over' the free-monoid monad. But not every cartesian monad on Set is so augmented.

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What category are you working in? Operads are definable in any symmetric monoidal category. They always have associated monads in that category such that the categories of algebras over the operad and monad are isomorphic. That relationship between operads and monads motivated my coining of the word "operad'', back in 1971. The original definition was for spaces, but Max Kelly and I understood the general situation right away. It is rare and special that the monad arising from an operad is cartesian, so it is not true that ``from every operad arises a cartesian monad''. Incidentally, it is not only true that different operads can have the same associated monad, it is also true that the same operad can have different associated monads in different categories. For example, an operad $\mathcal{C}$ in spaces with $\mathcal C(0)$ a point has a monad in unbased spaces and a quite different monad in based spaces with isomorphic categories of algebras.

With my original definition of an operad, not every cartesian monad arises from an operad. For a perhaps esoteric example, consider the monad in the category of globular sets that defines strict $\omega$-categories.

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Thanks for the answer, I've just edited the question adding the specification about the category I'm working in. –  Giorgio Mossa May 29 '11 at 16:22
    
Is that true? The usual construction of a monad from an operad requires that the symmetric monoidal structure commutes with colimits. –  Tilman May 29 '11 at 23:07
    
Could you show an example of operad whose associated monad isn't cartesian? I tried to search but I didn't find any such operad. –  Giorgio Mossa May 30 '11 at 13:44
2  
Peter's original usage of "operad" includes symmetries. In that case it is easy to find operads whose monad isn't cartesian, like the terminal (symmetric) operad, whose algebras are commutative monoids. But for non-symmetric operads, it is always true (at least in a suitable generality) that the associated monad is cartesian. –  Mike Shulman Jun 3 '11 at 4:18
    
So symmetric operads aren't special case of generalized operads, are they? –  Giorgio Mossa Jun 8 '11 at 11:50

Let $T$ be a Cartesian monad on a Cartesian category $E.$ Let $1$ denote the terminal object of $E$ and view it as an object in the bicategory of spans $E_T$. Then the identity morphism for $1,$ which is the span $$T\left(1\right) \stackrel{id_T(1)}{\longleftarrow} T\left(1\right) \to 1,$$ is an internal monad, which I will denote by $C$. (The monad structure comes from the $T$-algebra structure on $T\left(1\right)$). So, in the terminology of Leinster, $C$ is a $T$-operad. The algebras for $C$ are the algebras for the induced monad $T_C$ on $E/C_0=E/1 \simeq E,$ which can easily be checked to be $T$ itself.

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For clarity, I should mention that my answer is to the edited question. Before the edit, the question asked about when a Cartesian monad arised from an operad. After the edit, it asked when it arised from a generalized operad in the sense of Leinster, which, as I argued above, is always. –  David Carchedi May 30 '11 at 12:36

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