Definition of an E-infinity algebra 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.
Thanks!
 A: 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. http://arxiv.org/abs/1503.02701
A: 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 http://math.unice.fr/~brunov/Operads.html for more details.]
A: 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.
A: 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).
