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Suppose that $K$ is an $E_\infty$-algebra on a space $X$ (more generally, any ringed topos; also, feel free to assume that $X$ is a point). That is, $K$ is a cochain complex of sheaves on $X$, endowed with a multiplication which is commutative and associative `up to coherent homotopy' (=an algebra structure over an $E_\infty$-operad). You can assume for simplicity that the sheaves $K^i$ and the cohomology sheaves $H^i(K)$ are flat (so we don't have to worry about derived vs usual tensor product). Over $\mathbb{Q}$, we know that every such $K$ is quasi-isomorphic to a commutative differential graded algebra (cdga), see e.g. 1, Part II, Corollary 1.5, but I'm interested in the integral case.

Question 1. Is there a known criterion for $K$ being quasi-isomorphic to a cdga?

For each prime $p$, $H^*(K)\otimes \mathbb{F}_p$ has a natural action of the mod $p$ Steenrod algebra (1, I 7). I think I might have read somewhere that if $K$ is quasi-isomorphic to a cdga, this action has to be trivial. Is that true?

Question 2. Is the vanishing of the Steenrod operations on $H^*(K)\otimes \mathbb{F}_p$ necessary and/or sufficient for the existence of a cdga quasi-isomorphic to $K$?

1 I. Kriz, J. P. May Operads, Algebras, Modules, and Motives, Asterisque 233, available online here.

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  • $\begingroup$ I might be confused but I thought that it was $H^*(K\otimes \mathbb{F}_p)$ that had the action of the Steenrod algebra. $\endgroup$ – Denis Nardin Mar 5 '16 at 14:38
  • $\begingroup$ @DenisNardin Above I assume that the $K^i$ are flat, so $H^*(K\otimes \mathbb{F}_p) = H^*(K)\otimes \mathbb{F}_p$. $\endgroup$ – Piotr Achinger Mar 5 '16 at 15:23
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    $\begingroup$ Yes, there is a known condition. It appears in my paper Commutative Monoids in General Model Categories, Definition 4.5, where I call the condition the "rectification axiom": personal.denison.edu/~whiteda/files/ResearchPapers/…. I didn't cover your particular example in that paper, but I'd love to think more about it. In my most recent paper with Donald Yau we have some purely algebraic examples and the work we did there might turn out to be relevant. I'm about to get on an airplane, so can't write more now. $\endgroup$ – David White Mar 5 '16 at 21:04
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    $\begingroup$ A related question to the one you're asking is about commutative differential graded $\mathbb{Z}$-algebras. In this setting, the transferred model structure I referred to earlier is not known to exist, so that approach to rectification is unlikely to work. Now that I've had a chance to think about it, I think an approach that avoids model categories, more in line with your Question 2, is better suited to answer this question, since you're interested in the integral case. Sorry I can't be of more help. $\endgroup$ – David White Mar 6 '16 at 16:09
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    $\begingroup$ Certainly it is a necessary condition, but I don't know if it is sufficient. I would look at the work of Mike Mandell. To see that it is necessary, note that the operations can be constructed as null-homotopies of $\mu-\tau\circ\mu$ where $\mu$ is the product and $\tau$ is the twist map. If this algebra is commutative, this difference is identically 0 and so you can take the constant null-homotopy. This forces the operations to vanish, except for when they agree with the Frobenius, these are nonzero. $\endgroup$ – Sean Tilson Mar 8 '16 at 10:22

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