# How universal is operadic approach to studying algebras?

I have just started to read about operads, so this question might be silly.

So it seems to me that any "reasonable" class of algebras can actually be defined as a class of all algebras over a certain operad. For example, associative algebras are algebras over the associative operad $\mathcal{As}$, commutative algebras - over the commutative operad $\mathcal{Com}$. The same is true for Lie, Poisson, Gerstenhaber, $A_{\infty}$-algebras.

So my question is: is there a (interesting) class of algebras that can't be viewed as a class of algebras over a certain operad? I mean, how universal is the operadic approach to studying algebras?

If there are algebras that are not algebras over an operad, then what are the obstructions?

Thank you very much!

• I believe the "correct" analog of Lie algebras in characteristic $2$ requires $[x,x]=0$ (which is equivalent to the more customary $[x,y]=-[y,x]$ only in characteristics $\neq2$), and it seems that, because $x$ occurs twice in the term $[x,x]$, this is not operadic. – Andreas Blass Aug 31 '13 at 14:47
• That's a good point, Andreas, and one that's bothered me in the past. But as far as I know the question is open: when $k$ is a field of characteristic 2, is there an operad of $k$-vector spaces whose algebras are exactly the Lie algebras over $k$? (It's not obvious that there is, nor that there isn't.) – Tom Leinster Aug 31 '13 at 16:53
• The original definition was designedly not all embracing; algebras defined by identities with repeated variables are common and not covered. Some verbiage explaining the motivation is in math.uchicago.edu/~may/PAPERS/mayi.pdf – Peter May Aug 31 '13 at 21:07
• "Cartesian operads are equivalent to Lawvere theories" see ncatlab.org/toddtrimble/published/Towards+a+doctrine+of+operads – Buschi Sergio Sep 1 '13 at 5:54
• Sergio: "cartesian operads" are more general than operads, so I don't think they're directly relevant to the original question (although they provide a useful perspective). – Tom Leinster Sep 1 '13 at 13:43

## 5 Answers

One example is a bialgebra. An operad only has operations taking several inputs and one output, so there is no room for comultiplication. On the other hand this is not a serious obstacle. A small modification of the defining axioms will accommodate also this example: bialgebras are algebras over a properad, rather than an operad.

A better example is maybe then a field. There is no possibility of encoding the existence of the inverse in the language of operads. More generally, operads and their relatives are all about encoding operations and relations which are multilinear, and inversion is just the simplest example of something nonlinear.

The answer rather depends on what you mean by an "algebra". For instance, universal algebraists would certainly call groups a kind of algebra, and there's no operad for which the algebras are groups.

For operads of sets (i.e. operads where the operations of each arity form a set rather than, say, a vector space or topological space), it's well-understood which algebraic theories can be obtained. I'll split the answer into two, according to whether or not you wish your operads to come equipped with a symmetric group action.

• The algebraic theories that arise from non-symmetric operads are exactly the so-called strongly regular theories. These are the finitary algebraic theories that can be presented using only equations in which the same variables appear in the same order, and only once each, on each side. For example, the associativity equation $(xy)z = x(yz)$ is fine, but the inverse equation $x x^{-1} = 1$ is not (for two reasons: $x$ appears twice on the left-hand side, and $x$ appears on the left-hand side but not the right-hand side).

• The algebraic theories that arise from symmetric operads can be characterized similarly, just dropping the "in the same order" requirement.

For operads of vector spaces, I guess there are similar theorems. E.g. the Jacobi identity $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0$ is OK because although you've got variables appearing several times on one side of the equation, each variable appears exactly once in each summand. As far as I know, though, no one's actually proved a theorem like this. Someone should!

(Note that although the equation $x x^{-1} = 1$ doesn't satisfy the requirements above, it doesn't follow from this that there is no operad whose algebras are exactly groups. Conceivably there is some other, sneaky presentation of the theory of groups that uses only equations satisfying the requirements. In fact, there isn't — it can be proved that there really is no operad for groups. But that requires a different theorem.)

• What is the statement of the theorem that someone should prove? Besides, do you have an example of an algebraic structure which, like groups, cannot be straightforwardly captured by operads, but, unlike groups, a "sneaky" presentation of it makes that it can be? – Samuele Giraudo Sep 4 '13 at 21:16
• Samuele: I haven't tried to formulate a precise statement. That's part of the challenge, though it shouldn't be too hard. The kind of statement I have in mind would be similar to the statements implicit in my two bullet points (for which see Theorem C.1.1 of my book Higher Operads, Higher Categories and Theorem 2.8.5 of Miles Gould's thesis, respectively). As for the second question, nothing springs to mind -- I guess we human beings like operadic-type equations enough that we tend to use them whenever we can. – Tom Leinster Sep 8 '13 at 21:51

Self-distributive laws, defined by $x(yz)=(xy)(xz)$ are an example of such a kind of algebras. They are useful in the study of braids and large cardinals.

Here the obstruction is that one need to duplicate the variable $x$ to write the axiom.

Lie algebras in characteristic $2$ are not algebras over any operad, the relation $[x,x]=0$ cannot be encoded by an operad since it is quadratic, not multilinear. I therefore don't agree with your statement that any reasonable kind of algebras are algebras over an operad.

• Fernando, can you prove that there is no linear operad over a field of characteristic 2 whose algebras are the Lie algebras over that field? I agree, of course, that the equation $[x, x] = 0$ makes it hard to construct one. But I don't know a proof that it's impossible. – Tom Leinster Aug 31 '13 at 22:51
• @Fernando I am sorry for the statement about reasonble algebras. I usually deal with characteristic $0$ case, so I didn't think about char=$2$. – Sasha Patotski Aug 31 '13 at 22:54
• Tom, that's essentially a study of free algebras, isn't it? Determine the degree $n$ part of the free algebra on $n$ generators, for all $n\geq 0$, and check they don't assemble to an operad defining Lie algebras. This gives a definite answer, for sure. – Fernando Muro Sep 1 '13 at 6:52

I put in a remark about this question back in 2013. I'm going to expand on that for emphasis. Let me shamelessly quote myself, from the expository paper http://www.math.uchicago.edu/~may/PAPERS/mayi.pdf where I describe the coining of the word operad.
$$$$To these ends, I consciously sacrificed all-embracing generality: many types of algebras defined by identities are deliberately excluded". In particular, algebras defined by identities with repeated variables are deliberately excluded despite their evident interest. I knew about alternating algebras and Jordan algebras from a course at Yale with Nathan Jacobson and about graded Lie algebras (char 2 or 3) from my thesis, and they are deliberately excluded.