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Sep 21, 2017 at 20:23 vote accept 54321user
Sep 21, 2017 at 20:15 comment added 54321user @DenisNardin That's okay! In this case the most natural constraint is to have the $E_\infty$-structure life the cup product structure on cohomology. Also, I know very little about obstruction theory other than the basics. Any suggestions/references are welcome!
Sep 21, 2017 at 17:46 answer added Tyler Lawson timeline score: 9
Sep 20, 2017 at 7:41 comment added Denis Nardin @54321user Sorry for the late reply. It is roughly as asking how many different ring structures exist on a given abelian group. A reasonable, but in general hard with very few applications. If you want to put some more constraints on the $E_\infty$-structure (for example you fix a given product on the homotopy groups) it becomes one of the most powerful recent tools in modern homotopy theory: obstruction theory. It is still very hard though.
Sep 18, 2017 at 1:37 comment added 54321user @DenisNardin Is it a bad question to ask then how many different $E_\infty$-structures exist?
Sep 18, 2017 at 1:35 comment added 54321user @DylanWilson Writing down the structure maps for an operad interaction would certainly be interesting. Really my goal is to bring some of this $E_\infty$-language down to earth. Thanks for the suggestion by specializing in the case $G=\mathbb{Z}/p$!
Sep 17, 2017 at 1:23 comment added Dylan Wilson But maybe you mean something different, like just for us to write down like the structure maps for an operad action?
Sep 17, 2017 at 1:22 comment added Dylan Wilson @54321user when G=Z/p and you take coefficients in a ring of characteristic p, there is actually a description by a single generator and single relation (described at the linked "Related question" mathoverflow.net/questions/48138/…) due to Mandell. You might be able to bootstrap that up to understanding other abelian p-groups when the coefficients are still in a ring (or preferably field) of char p. But Z coefficients will probably be sorta ugly, and so will more complicated G
Sep 16, 2017 at 10:23 comment added Denis Nardin @54321user It's not that there are competing definitions for $E_∞$-algebras, is that there are competing ways in which an $E_∞$-algebra structure can be given (but they have all been proven to be equivalent). I believe Tyler was wondering exactly what you mean by "explicit" (generators-and-relations? that's going to be awful).
Sep 12, 2017 at 22:28 comment added 54321user @TylerLawson I'm not sure how explicity of a description you want for $C^*(K(G,n);\mathbb{Z})$. For example, I could take the singular cochains for the symmetric space of a moore space, $Sym^\infty(M(G,n))$. Also, I was not aware that there were competing definitions for $E_\infty$-algebras. I was using the definition in the Lurie-Gaitsgory paper on the Weil-conjectures for function fields.
Sep 8, 2017 at 1:21 comment added Tyler Lawson I guess this depends on what you're looking for. An answer to this question probably involves three things: an explicit description of a model for $C^*(K(G,n);\Bbb Z)$, an explicit model for what an $E_\infty$ structure on a cochain complex is, and an explicit way that $C^*(K(G,n);\Bbb Z)$ has this structure. It would be easier to provide a helpful answer knowing if you have specific models in mind or if describing those would be part of what you need.
Sep 7, 2017 at 22:44 comment added Saal Hardali On second thought I think this statement is true only over fields. As for a reference I'd like to know of one too sadly I have no idea! (It's just something I have been told several times).
Sep 7, 2017 at 22:35 comment added 54321user @SaalHardali do you have a reference for this claim in the $A_\infty$ case?
Sep 7, 2017 at 22:22 comment added Saal Hardali Perhaps this is only enough for the $A_\infty$-structure though...
Sep 7, 2017 at 22:15 comment added Saal Hardali I'm far from an expert but I think if you know all steenrod operations and all massey products on the cohomology then you know everything.
Sep 7, 2017 at 21:48 history asked 54321user CC BY-SA 3.0