MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anyone give me a plain-and-simple definition of an E-infinity algebra without using the words "operad," "ring spectrum," or "stable homotopy"?

Sorry, but I honestly couldn't find it using all on-line resources at my disposal.


share|cite|improve this question
This can be done without the second two phrases, but trying to define it without the concept of an operad might be very difficult. – Tyler Lawson Aug 23 '10 at 14:12
In characteristic zero, you can obtain an equivalent category to E-infinity algebras by taking commutative differential graded algebras (unbounded) and inverting the quasi-isomorphisms. Then you have commutativity on the nose, and don't have to worry about all the higher stuff. It's of course a cheat, but it's without operads, ring spectra and stable homotopy theory. – Timo Schürg Aug 23 '10 at 14:26
If you want a formula, then perhaps you should look at the operad defined in the papers of McClure-Smith and Berger-Fresse . Of course, these involve the word "operad", but they give quite explicit example(s) of a chain-level E-infinity operad. – Charles Rezk Aug 23 '10 at 15:54
Charles and Tim, shame on you both! (Smily icon here.) I understand that the algebraic structure is just some superficial indication of a deeper structure, but I do not want this question hijacked by you structuralists! (Tim, where would symplectic geometry be if Fukaya hadn't written down formulas?) That said, I'm sure that there's a smart someone out there who can turn Charles's references into a formula. (But I ain't him!) Curmudgeonly yours, – Eric Zaslow Aug 23 '10 at 16:34
@Eric: Yes, there is a very short and powerful way to describe A-infinity algebras in terms of a very mild amount of chain-level data, which comes from one very nice A-infinity operad. I guess I feel like the problem is the E-infinity operads, in terms of generators and relations, tend to be much larger and more difficult to work with on a practical basis; one needs infinitely many cup-i products at the very least before one has even left the binary operations. – Tyler Lawson Aug 23 '10 at 16:45
up vote 13 down vote accepted

In characteristic 0, Kadeishvili has a notion of $C_{\infty}$ algebra which models rational homotopy theory. See the last paragraph of the introduction of his paper arXiv:0811.1655. His point of view is to simply consider $A_{\infty}$ algebras whose operations satisfy a certain property with respect to shuffle maps. So your computer doesn't have to remember any new operations, just check that the old ones are right.

In characteristic $p$, things are probably hopeless.

Added Remark: I just want to make clear that this does not give a "trivial proof" that a commutative dga is formal as a commutative dga if the underlying dga is formal in the "non-commutative" sense. The reason is that when you transfer from cochains from cohomology, you are restricted in the kind of morphisms allowed if you are interested in the commutative theory. So, just as in the answers to this question, there is some work to be done if you want results like that (to be completely honest, there is not yet a proof that I completely understand, so declare myself agnostic).

share|cite|improve this answer
Thanks, Mohammed. This is a very simple definition. The "certain property" above is just that the m_k vanish when you replace any d<k inputs by the signed sum of their d! permutations. My computer thanks you, too! – Eric Zaslow Aug 23 '10 at 18:56
@Eric. Beware, Eric, that Mohammed is talking about $C_\infty$ algebras: they are not the same as $E_\infty$ algebras. I think there is no simple description of $E_\infty$ algebras in terms of generators (operations) and relations like the ones avaliable for $A_\infty$ or $C_\infty$ ones. McClure-Smith's paper cited by Charles Rezk seems to provide the simplest one, but still is far more complicated than those for $A_\infty$ or $C_\infty$ cases. – a.r. Aug 23 '10 at 19:29
Grrr! Why doesn't life get simpler? So then, how to understand Kadeishvili's comment, "In rational case E-infinity o-- can be replaced by commutative o-- C acting on appropriate cochains. And in order to step from cochains to cohomology we replace C be the o-- C-infinity," where "o--" is a forbidden word. I gather that E-infinity involves the structure of permuting inputs, and that there are many higher products coming from ways of organizing said inputs. This is intuitive, but also suggests that there is some algebraic simplification by capping the number of inputs at say 3. Or, why not? – Eric Zaslow Aug 23 '10 at 20:48
not an expert, but here's something using one of the forbidden words. Both $C_\infty$ and $E_\infty$ operads are cofibrant resolutions of the commutative operad, but $E_\infty$ has the added condition that $E_\infty(n)$ is contractible. Working rationally, there are maps between $E_\infty$ and $C_\infty$ since they are resolutions of the same object. – Micah Miller Aug 24 '10 at 2:19
@Agusti: this statement is not true. A $C_\infty$ algebra is a $A\infty$ algebra which vanishes on the image of non-trivial shuffles, not that they are "strictly symmetric". – Bruno V. May 26 '11 at 21:10

In characteristic 0, one can define an $E_\infty$-algebra simply by an $A_\infty$-algebra $(A, d, \lbrace \mu_n\rbrace_{n\ge 2})$ such that each operations $\mu_n$ vanishes on the sum of all $(p, q)$-shuffles for $p + q = n$.

[See Section 13.1.13 of for more details.]

share|cite|improve this answer

Drinfeld once remarked to me something to the effect that he likes the definition of an operad because it is so simple. One doesn't have to be a Drinfeld to appreciate the truth of that statement. It is the simplicity of the notion that led me to search for a name with a nice ring to it, that people would remember. Steenrod operations were originally defined using operads implicitly. For odd primes, I believe there is still no ``simple'', by which I understand combinatorially explicit, construction of the operations.

share|cite|improve this answer
Dear Professor May, is there still no way to do it even if one uses the simplicial approach of Hovey-Shipley-Smith to obtain a symmetric monoidal product of (symmetric) spectra? From what I understand (perhaps quite wrongly!), is that their approach avoids any explicit mention of operads, although it does still depend on doing a variant of stable homotopy theory. – Harry Gindi Feb 3 '11 at 6:23
You are mixing contexts and definitions in a fairly confused way. $E_{\infty}$ algebras make sense in many categories. There are many symmetric monoidal categories of spectra (historically, the first, by a nose, was constructed in EKMM (Elmendorf-Kriz-Mandell-May). In any such category, commutative monoids, alias commutative ring spectra, are equivalent to $E_{\infty}$ algebras in the relevant category of spectra. – Peter May Feb 4 '11 at 2:13
Ah, thanks for the clarification! – Harry Gindi Feb 4 '11 at 6:26

A completely explicit definition that works over any ring is given as Proposition 18 in the following preprint of Malte Dehling and Bruno Vallette that was posted on the arXiv today.

share|cite|improve this answer
That's a very nice paper. – Mark.Neuhaus Aug 4 '15 at 2:28
Thank you! Feel free to ask if you have any question. – Bruno V. Aug 4 '15 at 7:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.