# Repertory of the different sorts of operads

Many different types of operads have emerged in recent years (symmetric, shuffle, cyclic, anticyclic, coloured, etc.).

I would like, for any of these, list the following data:

1. Description of the free object (combinatorial objects involved, expression for the composition maps, expression for auxiliary map (as e.g., the action of the symmetric group or the cyclic action);
2. Given a collection $G$ of generators and its generating series $G(t)$, give an expression for the Hilbert series $H_G(t)$ of the operad generated by $G$;
3. Give some examples of worthwile and well-studied operads that fit into this kind of operads.

References are naturally welcome but I would like to create here a big list (with one kind of operad by answer if possible).

-
A big list is usually Community Wiki. – Todd Trimble Feb 26 '13 at 11:36
A list would be huge. But a few comments on the question. I wouldn't have invented the word operad'' without having symmetric in mind: to me, that word is redundant. A generic answer to (1) is built into the word: it is a portmanteau (Humpty Dumpty: Lewis Carroll) of the words "operation" and "monad". So "operads" in the original spirit must have accompanying monads with isomorphic categories of algebras, hence the free objects must be given by that monad. Very often (2) is quite meaningless, for example for the original operads in spaces. – Peter May Feb 26 '13 at 14:13
Ryan, I think what he means is that for each flavor of operad (symmetric, cyclic, etc. etc.), i.e., for each item in the list, give a few good representative examples of that flavor. – Todd Trimble Feb 26 '13 at 19:12
Andrew, you are the only one, I sincerely hope. Samuele, sure there are interesting examples, but then you are only talking about one object multicategories, not quite the same flavor. Your distinction between free and algebraic clearly needs a better understanding of monads, which prescribe free objects quite generally. – Peter May Feb 26 '13 at 22:17

One operad that's popped up in my work is the kind of operad that has the same formal properties as the endomorphism operad of a space $X$, provided $X$ is equipped with an action of a topological group $G$. Level $n$ of the endomorphism operad is just the space of maps $Map(X^n, X)$, but we think of this space as one with an action of the group
$$G \times (\Sigma_n \ltimes G^n) \equiv G \times (\Sigma_n \wr G).$$
I sometimes denote this group $\Sigma^*_n \wr G$, and the family of groups $\Sigma^* \wr G$. I've been calling these operads $\Sigma^* \wr G$-operads as I thought the name was kind of harmless and more or less descriptive.
The free objects over this operad look like a disjoint union of rooted trees, where the vertices of the trees are decorated by points in the generating $\Sigma^*_k \wr G$-spaces. There is an equivalence relation on these decorated trees, generated by one relation for each edge of the trees -- corresponding to the endomorphism operad's equivariance.
These types of operads come up (and are useful) with $G$ being various orthogonal groups, or diffeomorphism groups of balls (or other automorphism groups of balls) for certain embedding spaces, like spaces of knots. For classical knots, like knots in the 3-sphere, there is an almost free operad of this type that describes the homotopy-type of the space of knots completely, up to the computation of certain (finite) symmetry groups of some hyperbolic links in $S^3$. The operads I'm talking about I call splicing operads. These are subspaces of $Map(X^n,X)$ but in this case, $X$ is a ball, and the maps are smooth maps with various restrictions on them (to get knots of various types).
Another way to think about these operads is that they are a particular case of a colored operad, where there is only one color but this color has nontrivial automorphism group $G$. So we need to allow the collection of colors to form a category, not just a set. A shameless self-promotion: this point of view is explained in my paper "On the operad structure of admissible $G$-covers". – Dan Petersen Feb 27 '13 at 9:13