Timeline for Definition of an E-infinity algebra
Current License: CC BY-SA 2.5
12 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 26, 2011 at 21:10 | comment | added | Bruno V. | @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". | |
| Aug 24, 2010 at 3:11 | comment | added | Agustí Roig | @Eric. If I remember correctly (with the help of Markl, Shnider, Stasheff, "Operads in Algebra...", page 19, :-) ), $C_\infty$ algebras are $A_\infty$ algebras which are strictly commutative. Whereas, of course, $E_\infty$ algebras are not. | |
| Aug 24, 2010 at 2:19 | comment | added | Micah Miller | 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. | |
| Aug 23, 2010 at 21:49 | comment | added | Mohammed Abouzaid | My understanding is that if one cares about quasi-isomorphism types of commutative dga's then the two theories ($E_\infty$ or $C_{\infty}$) give the same answer rationally. It would be great if we had a comment from experts confirming or denying this. | |
| Aug 23, 2010 at 20:48 | comment | added | Eric Zaslow | 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? | |
| Aug 23, 2010 at 19:41 | history | edited | Mohammed Abouzaid | CC BY-SA 2.5 |
fixed link
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| Aug 23, 2010 at 19:29 | comment | added | Agustí Roig | @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. | |
| Aug 23, 2010 at 19:12 | history | edited | Mohammed Abouzaid | CC BY-SA 2.5 |
added 688 characters in body
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| Aug 23, 2010 at 18:56 | comment | added | Eric Zaslow | 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! | |
| Aug 23, 2010 at 18:50 | vote | accept | Eric Zaslow | ||
| Aug 23, 2010 at 18:40 | history | answered | Mohammed Abouzaid | CC BY-SA 2.5 |