Definition of an E-infinity algebra - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T15:00:35Z http://mathoverflow.net/feeds/question/36444 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36444/definition-of-an-e-infinity-algebra Definition of an E-infinity algebra Eric Zaslow 2010-08-23T13:20:53Z 2011-05-26T21:20:19Z <p>Hello. Can anyone give me a plain-and-simple definition of an E-infinity algebra without using the words "operad," "ring spectrum," or "stable homotopy"?</p> <p>Sorry, but I honestly couldn't find it using all on-line resources at my disposal.</p> <p>Thanks!</p> http://mathoverflow.net/questions/36444/definition-of-an-e-infinity-algebra/36478#36478 Answer by Mohammed Abouzaid for Definition of an E-infinity algebra Mohammed Abouzaid 2010-08-23T18:40:24Z 2010-08-23T19:41:09Z <p>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.</p> <p>In characteristic $p$, things are probably hopeless.</p> <p>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 <a href="http://mathoverflow.net/questions/9146/noncommutative-rational-homotopy-type" rel="nofollow">question</a>, 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).</p> http://mathoverflow.net/questions/36444/definition-of-an-e-infinity-algebra/54162#54162 Answer by Peter May for Definition of an E-infinity algebra Peter May 2011-02-03T02:16:14Z 2011-02-03T02:16:14Z <p>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.</p> http://mathoverflow.net/questions/36444/definition-of-an-e-infinity-algebra/66111#66111 Answer by Bruno V. for Definition of an E-infinity algebra Bruno V. 2011-05-26T21:20:19Z 2011-05-26T21:20:19Z <p>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 $p + q = n$. </p> <p>[See Section 13.1.13 of <a href="http://math.unice.fr/~brunov/Operads.html" rel="nofollow">http://math.unice.fr/~brunov/Operads.html</a> for more details.]</p>