Symmetric and non-symmetric operads

Question

I have trouble understanding why they are distinct notions of symmetric operad and non-symmetric operads : are they really both needed ? It seems like symmetric operads are more general, but I do not understand why we just could not forget about the action of the permutations, for example for the topological $\mathbb{E}_k$ operads.

Answer 1

Symmetric operads are more general than non-symmetric operads in the sense that given a non-symmetric operad $P$, you can always build a symmetric operad $P_\Sigma$ by defining $P_\Sigma(n) = P(n) \times \Sigma_n$ and compositions are induced from $P$ and block composition of permutations. Then an algebra over the symmetric operad $P_\Sigma$ is exactly the same thing as an algebra over the non-symmetric operad $P$. The converse is not true: there are symmetric operads whose category of algebras are not equivalent to the category of algebras over a non-symmetric operad.

However, the symmetric group is relatively complicated, especially in positive characteristic. It is much easier to study the homotopy theory of non-symmetric operads than the homotopy theory of symmetric operads. So if you know that the operad you study comes from a non-symmetric operad, it can be nice.

I do not understand why we just could not forget about the action of the permutations, for example for the topological $E_k$ operads.

Now that is a different question. If you do this then you get a non-symmetric operad which is definitely not what you want.

Maybe it's easier to see on simpler operads. If you take the operad of commutative algebras and you forget the symmetric group action, then it's a non-symmetric algebra: the operad encoding associative (but not necessarily commutative) algebras! That's clearly not what you want. Another example: consider the operad governing Lie algebras. If you forget the action of the symmetric groups, then you get a non-symmetric operad which is free (see Salvatore–Tauraso), which is rather strange!

In the case of the $E_k$-operad, consider e.g. the little $2$-cubes operads. You have an operation of arity $2$ obtained by splitting the unit square in two. There is a path from this operation to the operation where you exchange the two inputs, which is rather nice, because it tells you that this operation is homotopy commutative. But if you forget about the symmetric group action, then "exchanging the inputs" makes no sense, and the two operations that you get by deciding whether $1$ is on the left or on the right are basically unrelated.