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Is there a category of operads which allows morphisms which take operations to operations with more or fewer arguments? One example should be when you fix arguments to obtain maps with fewer inputs. I'm grateful for any comments or references.

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  • $\begingroup$ Well, if you think of the category of operads as a subcategory of the category of Collections then the answer is no. If $C$ is your base category then the category of Collections is $\prod_n{C^{\Sigma_n}}$ and the morphisms in this category are levelwise. There are good reasons to think of Operads inside $Coll(C)$. In particular, it gives you the notion of $\Sigma$-cofibrant operad (i.e. cofibrant in the projective model structure on $Coll(C)$) which is very useful for doing homotopy theory $\endgroup$
    – David White
    Commented Dec 20, 2012 at 15:35
  • $\begingroup$ Here's another reason why we shouldn't allow such maps. An operad is a special case of a colored operad, which is the same thing as a multicategory. Maps of colored operads are functors of multicategories. They are defined in Section 1 of "Localization of Algebras over Coloured Operads" by Casacuberta et al. In that definition, the maps are levelwise, and the authors remark that this matches with the notion of a functor of multicategories. Many who study operads think this approach via enriched categories and multicategories is the right one, so "morphism of operads" should match it $\endgroup$
    – David White
    Commented Dec 20, 2012 at 15:46
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    $\begingroup$ David, while your comments are otherwise reasonable, the idea that there is an approach that deserves to be called the right one'' to operads is anathema to me. It is like saying, for example, that groupoids give the right approach'' to groups. For some purposes yes, in general, ex-cathedra, obviously no. Operads, like groups, give a fundamental algebraic structure that appears in a variety of contexts and admits a variety of generalizations and elaborations. $\endgroup$
    – Peter May
    Commented Dec 20, 2012 at 15:56
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    $\begingroup$ Mathematically, the question implicitly highlights one designed in limitation to the generality of the kinds of algebraic structure that operads define: they are not designed (or rather designed not) to deal with identities with repeated variables, such as x^2 = 1 or even x x^{-1} = 1. There are several reasons for that choice. $\endgroup$
    – Peter May
    Commented Dec 20, 2012 at 15:59
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    $\begingroup$ David, the category of operads isn't a subcategory of the category of collections (at least, not in the obvious way), since one collection can carry many different operad structures. $\endgroup$
    – Tom Leinster
    Commented Dec 20, 2012 at 17:00

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I'm not certain I understand the intent of the question, but perhaps the following is the kind of thing Poisson is looking for.

A non-symmetric operad is a sequence $(P_n)_{n \geq 0}$ of sets together with an identity element and maps defining composition, all satisfying some axioms.

A symmetric operad is the same, but also comes with a map $\theta_*\colon P_n \to P_n$ for each bijection $\theta\colon \mathbf{n} \to \mathbf{n}$, again satisfying axioms. (Here I use $\mathbf{n}$ to denote a fixed $n$-element set, say $\{1, \ldots, n\}$.)

A finitary algebraic theory is the same, but also comes with a map $\theta_* \colon P_n \to P_m$ for each function $\theta\colon \mathbf{n} \to \mathbf{m}$, again satisfying axioms.

The three types of structure have successively greater expressive power. For example, there is no non-symmetric operad encoding the theory of commutative monoids (because expressing the equation $xy = yx$ requires the nontrivial bijection $\mathbf{2} \to \mathbf{2}$). There is a symmetric operad encoding the theory of commutative monoids, but there is none encoding the theory of commutative monoids in which every element is idempotent (because expressing the term $x^2$ requires the surjection $\mathbf{2} \to \mathbf{1}$).

This doesn't mean that finitary algebraic theories are 'better' than operads, because there's a trade-off: the contexts in which it's possible to talk about algebras are successively narrower. That is, given an operad $P$, you can talk about algebras for $P$ in an arbitrary monoidal category $\mathcal{E}$; but if $P$ has the structure of a symmetric operad, you need $\mathcal{E}$ to be symmetric monoidal in order to talk about $P$-algebras in $\mathcal{E}$; and if $P$ is a finitary algebraic theory, you need the monoidal structure on $\mathcal{E}$ to be actual categorical product $\times$.

There are many well-known equivalent definitions of "finitary algebraic theory": clone, Lawvere theory, finitary monad on $\mathbf{Set}$, etc, or the classical definition from universal algebra (modulo choice of presentation). The definition I'm implicitly referring to above is perhaps not so well-known, though it's simple enough. You can find the details in Definitions 2.3.1 and 2.3.2 of Miles Gould's thesis, and the equivalence to the other definitions is Theorem 2.3.12.

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  • $\begingroup$ Thanks, I will have a look! The situation I was looking at is the following. Consider the operad of BV-algebras (commutative multiplication m and unary operation d). We can obtain a Gerstenhaber structure by considering [a,b]= d(ab)-d(a)b-ad(b) (ignoring Koszul signs). What does this correspond to at the operad level? $\endgroup$
    – Poisson
    Commented Dec 20, 2012 at 17:42
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    $\begingroup$ Ah, OK. I was talking about operads of sets, whereas here you probably want operads of vector spaces, also known as linear operads. (Thus, each $P_n$ is a vector space rather than a set.) An equation like the one you mention can be expressed by a linear operad; you don't even need symmetry. $\endgroup$
    – Tom Leinster
    Commented Dec 20, 2012 at 17:56
  • $\begingroup$ Yes, perhaps I should have been more clear about that. The equation I can express just fine, the problem is that I turn a unary operation into a binary operation, thus it is not a map of operads. $\endgroup$
    – Poisson
    Commented Dec 20, 2012 at 18:05
  • $\begingroup$ The specific example in Poisson's comment here doesn't seem to match the original question. $\endgroup$
    – Andreas Blass
    Commented Dec 20, 2012 at 18:09
  • $\begingroup$ Yes, I took another example here, where we map an operation to an operation with more inputs in contrast to the example with fixing inputs where we reduce the number of inputs. Perhaps it was a bit naive to think that a theory existed to take care of these two examples easily. $\endgroup$
    – Poisson
    Commented Dec 20, 2012 at 18:18

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