I recently proved something for the operad $Comm$ valued in a model category $M$ and am trying to generalize it to other symmetric operads. I'm very new to operads, so please forgive me if there are errors in this question. I'll edit it as people point out in the comments. This result I proved happens to be known for cofibrant operads, so I'm hoping to get it for some class of operads weaker than that. I was hoping the following hypothesis would do the trick, since this was a key property of $Comm$ I was using: "$O(n)$ is cofibrant in $M$ for all $n$.'' If I had to, I was willing to go up to the hypothesis: "$O(n)$ is cofibrant in $M^{\Sigma_n}$ for all $n$.'' So feel free to assume this is the situation if it helps.

Unfortunately, the proof for $Comm$ secretly relied on the fact that $Comm(n)$ splits as $Comm(s)\times Comm(n-s)$, so when I had an action of $\sigma \in \Sigma_n$ on some $A^s\otimes B^{n-s}$ which didn't do any mixing I was able to break it down into $\sigma|_A$ and $\sigma|_B$ and pass to $\Sigma_n$-coinvariants in a nice way. To make my proof work for a more general operad $O$ I must assume $O(n)=O(s)\times O(n-s)$ for all $1 < s < n$.

(1) How strong is this condition? Is it often satisfied for operads which arise in practice?

I can see immediately that this condition fails for the operad $Assoc$, since $\Sigma_n \neq \Sigma_s \times \Sigma_{n-s}$. However, my intuition is saying it may hold for the "little $n$-somethings" operads. The condition seems to imply that the whole operad can be recovered from $O(1)$, which seems pretty strong.

(2) Does this condition imply the operad is freely generated in some way over $O(1)$?

The nLab article on free operads is a bit dense for my current level of Operad-foo, but I think I understand it well enough to see that the notion of free in (2) is going to have to be different from the nLab notion.

(3) Does this condition imply any kind of cofibrancy for the operad $O$?

By cofibrant I mean either in the model category of operads, in the projective model structure on symmetric sequences (if you view operads as monoids in that category with respect to the circle product), aritywise in $M$ or $M^{\Sigma_n}$, or in some other way that I'm not familiar with.

I'm not sure if it's helpful or not, but I did some musing about this in the context of thinking about operads as parameterizing $n$-ary operations. In that context, the condition seems to be saying that an $n$-ary operation breaks up as a product of $n$ operations of arity $1$, which is quite strong. On the other hand, it is true that understanding all $1$-ary operations lets me understand all $n$-ary operations by Hom-Tensor duality (in a closed symmetric monoidal category), so maybe this isn't so crazy.

  • $\begingroup$ I think that your comments around questions (1) and (2) very nearly give a proof that $O$ is the direct product of the operad $Comm$ with the monoid $O(1)$ (so $O(n) = Comm(n) \times O(1)^{\times n}$). $\endgroup$ – Craig Westerland Jun 7 '12 at 22:49
  • $\begingroup$ I don't know that that condition does hold for the little n-somethings operad. $E\Sigma_i \times E\Sigma_j$ is not going to have a free $\Sigma_{i+j}$, this would be implied by such a splitting, I think. All of the spaces are contractible, but there is the free symmetric group action that you need in order for it to be an $E_\infty$ operad. I could be wrong though. Your last comment also leads me to believe that this shouldn't happen in general. How can there be an interesting product on an $E_\infty$ algebra if such a thing is true? $\endgroup$ – Sean Tilson Jun 8 '12 at 3:07
  • $\begingroup$ @Craig Westerland: Why is $O(1)$ a monoid? Since $O(1)\times O(1)=O(2)$ I would need a map $O(2)\rightarrow O(1)$, which I can't see at this moment. Also, why do you need the $Comm(n)$ part? Isn't it just true that $O(n)=O(1)^n$? $\endgroup$ – David White Jun 8 '12 at 13:24
  • $\begingroup$ @Sean: You could well be right. I'm learning more and more not to trust my gut about these things. In particular, I think you are right about how splitting would require a free $\Sigma_{i+j}$ action which isn't there. I'm curious to hear more about the connection to the product on $E_\infty$ algebras if you have time. $\endgroup$ – David White Jun 8 '12 at 13:28
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    $\begingroup$ @David White: Operadic composition always gives me a map $O(1) \times O(1) \to O(1)$ which is associative (since it's an operad) and unital (if its a unital operad), making $O(1)$ into a monoid. As for the $Comm(n)$ factor, you certainly don't need it, in the sense that the object $O(n)$ is isomorphic to both $O(1)^{\times n}$ and $Comm(n) \times O(1)^{\times n}$. But I'd like a description of $O(n)$ that indicates $O$'s operadic structure. I was hoping that it might be a (semi-)direct product (see Salvatore-Wahl), which would yield $O(n) = Comm(n) \times O(1)^{\times n}$ $\endgroup$ – Craig Westerland Jun 8 '12 at 23:27

I'm rushed, so this might not be right, but it feels right. Consider reduced operads in a cartesian monoidal category, so that $C(0) = \ast$. A monoid $M$ in our category gives a reduced operad $R(M)$ with $j$th object $M^j$. The structure map $$M^k\times M^{j_1}\times\cdots M^{j_k}\to M^{j_1+\cdots + j_k}$$
has coordinates given by multiplying the $h$th coordinate of $M^k$ with the coordinates of $M^{j_h}$. This is a right adjoint to the functor $L$ that sends a reduced operad $C$ to $C(1)$ with the evident product and unit. The isomorphism of hom sets uses the degeneracy operators that are there because we are working with reduced operads. Obviously $LR= id$. Very often $id\to RL$ is an inclusion. That happens for little whatnots. Very rarely, only in uninteresting cases, is $id\to RL$ an isomorphism. That doesn't happen for little whatnots.

  • $\begingroup$ Thanks, this is just the sort of answer I was looking for. $\endgroup$ – David White Jun 9 '12 at 12:50

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