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In Alan Robinson's paper, Classical Obstructions and S-algebras, he provides conditions for a ring spectrum to have an $A_n$ and $\mathbb{E}_\infty$-structure.

Have the obstructions for an object of an $\infty$-operad $\mathcal{C}^\otimes$ from having an $\mathbb{E}_n$-algebra structure been studied? More precisely, what conditions must an object $\mathscr{C}$ of $\mathcal{C}^\otimes$ satisfy to be an object of $\mathrm{Alg}_{/\mathbb{E}_n}(\mathcal{C}^\otimes)$?

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  • $\begingroup$ Presumably when you say $\infty$-operad you mean (say, symmetric) monoidal $\infty$-category? $\endgroup$ Dec 11, 2014 at 0:35
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    $\begingroup$ The Goerss-Hopkins obstruction theory (developed in the papers "Moduli spaces of commutative ring spectra" and "Moduli problems for structured ring spectra," available at math.northwestern.edu/~pgoerss ) develop precisely such an obstruction theory for spectra. The general theory is more complex than the $E_1$ and $E_\infty$ version, but it does exist. $\endgroup$ Dec 11, 2014 at 2:13
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    $\begingroup$ Have you looked a tthe thesis of John Francis? He sets up the machinery of cotangent complexes in the $\infty$-categorical setting and shows that for $E_n$-algebras it takes the form of (up to some shifts) iterated $THH$. It might be worth noting that the relevant obstruction groups are very very difficult to compute. $\endgroup$ Dec 11, 2014 at 16:51
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    $\begingroup$ @Sean Thanks! I've been reading through almost all of the references, and I think I know a general outline of how the obstruction theory for E_n-algebras might look. $\endgroup$
    – user62675
    Dec 17, 2014 at 19:11
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    $\begingroup$ Yeah, there are a lot of articles by Goerss and Hopkins on this stuff. Some are more expository than others, or at least one is. You can abstract what they do to other settings. For $E_n$ obstruction theory, as I gather you are noticing, is the same except that the obstruction groups are different (maybe other things as well). $\endgroup$ Dec 18, 2014 at 18:38

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Let me expand a little on what Qiaochu and Craig mentioned.

If you want an obstruction theory for building an uber-gadget, you'll need (i) an algebraic approximation to such gadgets, and (ii) a way to resolve every uber-gadget by special uber-gadgets where the algebraic approximation remembers everything about the thing you started with.

Condition (i) usually takes the form of a functor $Uber \rightarrow Approx$ with some nice properties, and then for (ii) you hope that maybe there is an adjoint to this functor, and that $Uber$ is generated under sifted colimits by the essential image of this adjoint (I'm sweeping everything important and techncal under the rug).

Once you have this, you can start trying to formulate the problem of starting with an object in $Approx$ and adding structure until you get to an object in $Uber$ (this shouldn't be too surprising: you've essentially required that $Uber$ is monadic over $Approx$ so that elements of $Uber$ are like elements of $Approx$ with extra structure). If you do this very carefully, you'll have expressed the space of objects lifting a given one. All of this is developed for $E_\infty$-ring spectra in Goerss-Hopkins and you can find write-ups here: http://www.math.northwestern.edu/~pgoerss/ . I'm pretty sure the same proofs give you an $E_n$-obstruction theory, and this appears elsewhere as well.

Now, you probably already knew all that. But I said it to remind you that there's just no hope of proceeding if you don't have (ii) and for (ii) you need your algebraic approximation to be strong enough. So you shouldn't ask: "Given a spectrum, what's the space of $E_\infty$ ring structures on is?" You need to have a little bit to get going, some candidate for the Dyer-Lashof operations or (in the Goerss-Hopkins case) a commutative comodule algebra over $E_*E$ for a suitable cohomology theory, $E$. Otherwise your answer will probably be ridiculous and uncomputable (this is sort of what Qiaochu was pointing out.)

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    $\begingroup$ I am confused Dylan, in at least one article by Goerss and Hopkins they do not start with any candidate for the action of the Dyer-Lashof operations. It even appears in the review MR2125040. They do look at algebras in comodules though, maybe that is what you mean? $\endgroup$ Dec 11, 2014 at 16:48
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    $\begingroup$ @SeanTilson You're absolutely right, of course, they start with a commutative comodule. The Dyer-Lashof operations then appear in computing the obstructions- in fact, all of the computations take place in the category of algebras in simplicial comodules over a certain simplicial operad. I'll amend the answer accordingly. $\endgroup$ Dec 11, 2014 at 17:03
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I think this question is too general. Already the $1$-categorical version is hard: when does an object in a monoidal category have a monoid structure? When does it have a commutative monoid structure? Even in $(\text{Ab}, \otimes)$ I don't know a reasonable general statement that can be made here (that is, I already don't know good necessary and sufficient conditions for an abelian group to have a compatible ring structure, or a commutative ring structure).

Here is a nontrivial theorem in the setting of compact abelian groups: any compact topological ring is totally disconnected.

Is there a particular case you're interested in?

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  • $\begingroup$ How about when $\mathcal{C}^\otimes$ is the symmetric monoidal $\infty$-bicategory of (stable) $\infty$-categories? $\endgroup$
    – user62675
    Dec 11, 2014 at 1:03
  • $\begingroup$ I got nothing. It's already hard enough to produce examples. This problem has probably been studied for spectra and it's probably hard for spectra too. $\endgroup$ Dec 11, 2014 at 1:12
  • $\begingroup$ Do you know any references where it's been studied for spectra? I think I can use it to study the case of stable $\infty$-categories. $\endgroup$
    – user62675
    Dec 11, 2014 at 1:40
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    $\begingroup$ Well an example is Mandell and Basterra proving that $BP$ has an $E_4$-structure...and references therein $\endgroup$ Dec 11, 2014 at 1:47
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    $\begingroup$ yeah as Qiaochu said the problem seems rather general so there's a relevant example :) $\endgroup$ Dec 11, 2014 at 1:58

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