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 underling 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.

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.

In the definition of operads, if we restrict our attention to S-modules where the action by the symmetric groups is free, then the free operads have still an underling "free S-module"? Even the colimits over this kind of operads have still an underling "free 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 S-module)?. In the sense that we don't need all the technical combinatorial details about the behavior of 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.

Thanks!

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 underling 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.