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In the definition of operads, if we restrict our attention to $\mathbb{S}$-modules where the action by the symmetric groups is free, then the free operads have still an underlying free $\mathbb{S}$-module? Even the colimits over this kind of operads have still an underling free $\mathbb{S}$-module?

And finally, in this kind of symmetric operads, the free operad construction (using trees) will be much simpler right? I mean if it is similar to the one for the non-symmetric case (where essentially we only need to label every vertex with the elements of the $\mathbb{S}$-module)? In the sense that we don't need all the technical combinatorial details about the behavior of $\mathbb{S}$-modules.

I'm not an expert in operads but these questions came to me when i was reading the book Algebraic Operads of Bruno Vallette and Jean-Louis Loday. In Section 5.5, the free operad construction is described.

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  • $\begingroup$ What do you mean by "having an underling free S-module"? $\endgroup$
    – Ryan Budney
    Commented May 25, 2016 at 18:28
  • $\begingroup$ By "underlying S-module", the S-module obtained after apply to an operad the forgetful functor. And by "free" that the action of symmetric groups in this S-module is free. $\endgroup$
    – emmagvr
    Commented May 25, 2016 at 18:45
  • $\begingroup$ To clarify, I think that S-module in this context means what is also known as "symmetric sequences" (a.k.a. functor on finite sets and bijections) $\endgroup$
    – Denis Nardin
    Commented May 25, 2016 at 22:15
  • $\begingroup$ What you say about colimits is not true because there are non-free maps between free S-modules. For instance, over Q, you can get the free operad on any quotient S-module of a free S-module by applying the free functor to the quotient map followed by a splitting back to the free S-module and then pushing out along the zero map. $\endgroup$
    – Gabriel C. Drummond-Cole
    Commented May 26, 2016 at 2:37
  • $\begingroup$ It is useful to note that if $X$ is binary, for example, then the free operad $F$ on $X$ has $F(2) = X(2)$, so this is a free $S_2$-module iff $X(2)$ is. It follows that you can pick counterexamples by hand. Similarly, $F(3)$ has dimension $3$, and so cannot be free as an $S_3$-module, $F(4)$ has dimension strictly less than $24$, etc: the answer follows purely by linear algebraic considerations. :) $\endgroup$
    – Pedro
    Commented Dec 10, 2021 at 15:04

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  1. What you say about free operads on free $\mathbb{S}$-modules being themselves free $\mathbb{S}$-modules and being describable in terms of the free non-symmetric operad is correct.

    In section 5.9.11 of Loday–Vallette, they construct adjoint functors between ns operads and (symmetric) operads. The free ns operad and free symmetric operad functors also admit right adjoint forgetful functors. Finally, the a free $S$-module functor on arity graded modules also has a forgetful adjoint.

The forgetful functors commute by inspection: $$ \require{AMScd} \begin{CD} Op@>>>ns\ Op\\ @VVV @VVV\\ \mathbb{S}\text{-}mod@>>>\mathbb{N}\text{-}mod \end{CD} $$ and so the left adjoint free functors commute $$ \require{AMScd} \begin{CD} Op@<{\otimes \mathbb{K}[\mathbb{S}_n]}<<ns\ Op\\ @A{F}AA @A{F_{ns}}AA\\ \mathbb{S}\text{-}mod@<{\otimes\mathbb{K}[\mathbb{S}_n]}<<\mathbb{N}\text{-}mod \end{CD} $$ and thus the free operad $F(M)$ for $M\cong N\otimes \mathbb{K}[\mathbb{S}_n]$ is canonically isomorphic to $F_{ns}(N)\otimes \mathbb{K}[\mathbb{S}_n]$.

  1. As explained in my comment, this is no longer true for colimits of free operads on free $\mathbb{S}$-modules. Take a free $\mathbb{S}$-module $M$ and an endomorphism $f$ of $M$ whose cokernel are not free $\mathbb{S}$-modules. For instance, let $M$ be a free $\mathbb{S}_2$-module and let $f$ be symmetrization or skew-symmetrization.

Then the pushout of the diagram $$ \require{AMScd} \begin{CD} F(M)@>{F(f)}>>F(M)\\ @VVV \\ 0 \end{CD} $$ is the free operad on the cokernel of $f$ (in our example, the trivial or sign representation of $\mathbb{S}_2$), and in particular is not free as an $\mathbb{S}$-module.

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