Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In checking the details of the correspondence between operads over a symmetric monoidal category and monads on some associated endofunctor of the category, I cannot make the obvious proof work without assuming that the monoidal product distributes over coproducts. But no such assumption is mentioned in my sources (for example Operads, Algebras, Modules by May (PDF).) Am I missing something in my argument?

Let $\mathcal{C}$ (mathcal C) be an operad (for simplicity take it non-symmetric) over a symmetric monoidal category $\mathcal{V},$ with composition $\gamma\colon \mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes\dotsb\otimes\mathcal{C}(m_n)\to\mathcal{C}(m_1+\dotsb+m_n).$ We define a functor $C\colon \mathcal{V}\to\mathcal{V}$ (Roman C) by $CX = \coprod_i \mathcal{C}(i)\otimes X^{\otimes i}$.

Then one wants to verify that the operad structure $\gamma$ gives a monad on $C$. That is, we need a natural morphism $C^2X\to CX,$ or $\coprod_i \mathcal{C}(i)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes i}\to \coprod_k \mathcal{C}(k)\otimes X^{\otimes k}$. By universal prop of coproducts, it will suffice to exhibit an arrow $\mathcal{C}(n)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}\to \coprod_k \mathcal{C}(k)\otimes X^{\otimes k}$ for all $n$.

Clearly we have $\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n}\to \mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes\dotsb\otimes \mathcal{C}(m_n) X^{\otimes m_1+\dotsb+m_n}\to \\ \mathcal{C}(m_1+\dotsb+m_n)\otimes X^{\otimes m_1+\dotsb+m_n}\to\coprod_k\mathcal{C}(k)\otimes X^{\otimes k},$ where the first arrow is by symmetry of the monoidal structure, the second arrow is the operad composition $\gamma,$ and the third arrow is the canonical inclusion into the coproduct.

Therefore by universal prop of coproducts, we have $\coprod_{m_1,\dotsc,m_n}\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n}\to\coprod_k\mathcal{C}(k)\otimes X^{\otimes k}.$

In general, again using inclusion morphisms of coproducts, we have arrows $\mathcal{C}(m_\ell)\otimes X^{\otimes m_\ell}\to \coprod_j \mathcal{C}(j)\otimes X^{\otimes j}.$ Then by functorality of the monoidal product, we have $\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n}\to \mathcal{C}(n)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}$. By universal property of coproducts, we therefore have an arrow from the coproduct $\coprod_{m_1,\dotsc,m_n}\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n}\to \mathcal{C}(n)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}$.

To summarize, we have the obvious maps $\mathcal{C}(n)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}\leftarrow \coprod_{m_1,\dotsc,m_n}\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n} \to \\ \coprod_k \mathcal{C}(k)\otimes X^{\otimes k}.$ Unless we know that the arrow on the left is an isomorphism, we do not get the structure map for a monad on $C$. And that arrow on the left will generally not be an isomorphism if the monoidal product in $\mathcal{V}$ does not distribute over the coproduct. For example, if the monoidal product is the coproduct itself.

share|improve this question
4  
You're completely correct. I've also found such things. The point is that people often consider operads in closed symmetric monoidal categories, hence $\otimes$ preserves $\coprod$. I had wondered about the insistence on closedness. First I thought it was only to define endomorphism operads, but it also solves more subtle problems like the one you point out. –  Fernando Muro Dec 25 '13 at 9:52
1  
I've certainly always assumed this distributivity, and I'm pretty sure it (or something close to it) is needed. If it's not mentioned in that paper of Peter May's, it must just have been a slip on his part. –  Tom Leinster Dec 25 '13 at 17:09
3  
Quite right, Tom. I did mention taking V to be closed on page 3 of the paper Joe cites and Fernando rightly points out that that fixes everything. I should have assumed that or the distributivity explicitly (as I'm sure I did elsewhere). In all of my applications, V has been closed. –  Peter May Dec 25 '13 at 20:59
    
Thank you for clarifying, gentlemen. I had thought requiring monoidal product to preserve coproducts would be an unusual requirement, which is why I had to ask. I somehow forgot that this was implied by closedness of the monoidal product, which is a natural for enrichment categories. So obvious in hindsight... thanks again! –  Joe Hannon Dec 26 '13 at 2:13

1 Answer 1

up vote 4 down vote accepted

I also noticed this at some point. I think you are right. One reference where this assumption is explicitly spelled out is the paper of Getzler and Jones.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.