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Theorem 4.4.4.7 in Lurie's Higher Algebra (or Theorem 4.3.22 in DAG III) states (roughly speaking) that under certain conditions the ∞-category of commutative ∞-monoids in a given symmetric monoidal ∞-category C is equivalent to the underyling ∞-category of the model category of strictly commutative monoids in a symmetric monoidal model category that presents C. In other words, E_∞-monoids can be strictified to strictly commutative monoids, provided that the relevant symmetric monoidal model category satisfies certain (rather strong) conditions, like being freely powered.

In Example 4.3.25 in DAG III Lurie verifies the conditions of the above theorem for the symmetric monoidal model category of symmetric simplicial spectra equipped with the S-model structure. In particular, he obtains that E_∞-ring spectra can be strictified to (strictly) commutative symmetric ring spectra.

I wonder if a similar result is true for motivic symmetric spectra. Some of the arguments in Lurie's writeup and references therein seem to be specific to simplicial sets, so it's not completely obvious whether all arguments carry through to the motivic case.

More generally, the same question can be asked for symmetric spectra in an arbitrary model category M. Obviously, some additional assumptions must be imposed on M, so I wonder what the full list of conditions might be.

Can motivic E_∞-ring spectra be strictified to commutative motivic symmetric ring spectra?

References on this matter will be appreciated.

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If I'm not wrong, usually you need to change the tensor product in order to have an equivalence between the category of E-ininity monoids with respect to the old tensor product and the category of strict commutative monoid with respect to a new tensor product (operadic one). here is a ref (subsection 4.2) in the case of topological spaces arxiv.org/pdf/0811.0553v1.pdf –  Fedotov Nov 5 '13 at 19:49
    
@Fedotov: Symmetric spectra (as opposed to ordinary spectra) already provide for this. Strictly commutative monoids in ordinary spectra don't model all E_∞-ring spectra, but the opposite is true for symmetric spectra. –  Dmitri Pavlov Nov 5 '13 at 20:33
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The example that you cite is a mistake: the category of symmetric spectra is not freely powered in the sense of DAG III. While it is possible to prove strictification results in the setting of symmetric spectra, one cannot do so simply by applying the results of DAG III. –  Jacob Lurie Nov 6 '13 at 11:37
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Let me remark that in my thesis I come up with a hypothesis on a model category which is a bit weaker than the one in DAG III, but which still guarantees that commutative monoids inherit a model structure. Furthermore, symmetric spectra with the positive model structure do satisfy this hypothesis. Details can be found in my research statement: dwhite03.web.wesleyan.edu/research.html. I hope to have a preprint on my website soon about this. Another useful point is that it's sufficient to check both my hypothesis and Professor Lurie's on generating (trivial) cofibrations –  David White Nov 20 '13 at 15:05

1 Answer 1

This is very related to some work in progress of mine, which began as part of a larger project of Aaron Mazel-Gee and Markus Spitzweck. Details can be found in the long version of my research statement, section 5.1. If you want more details, please email me separately, as there are still things I want to work out before making a draft public.

I am much more accustomed to model categories so let me phrase my answer in that context. Basically, one way to get a model structure for stable motivic homotopy theory is to use Mark Hovey's stabilization machine from Spectra and Symmetric Spectra in General Model Categories. This gives you a model structure on motivic symmetric spectra which generalizes the usual one on symmetric spectra. Classically, in order to get the strictification you desire (which operad people might call rectification), you would need to pass to the positive stable model structure on symmetric spectra, introduced in Brooke Shipley's A Convenient Model Category for Commutative Ring Spectra. The basic point of this model structure is to force the unit to no longer be cofibrant, removing Gaunce Lewis's obstacle to having a good model category spectra. In this model category, commutative monoids do inherit a model structure (with weak equivalences and fibrations maps which are such as maps of symmetric spectra). Furthermore, a cofibrant commutative monoid must be cofibrant as a symmetric spectrum (this is what "convenient" means). Finally, rectification holds as shown for example in Theorem 1.4 in Elmendorf-Mandell Rings, Modules, and Algebras in Infinite Loop Space Theory.

The theorem in my research statement shows that Hovey's machine can be tweaked to output a positive stable model structure, at least when the input model category is combinatorial. In particular, we checked that this applies to motivic symmetric spectra. To apply the Elmendorf-Mandell result it seems you must also assume the model category is simplicial, i.e. satisfies the SM7 axiom. Thankfully, motivic symmetric spectra does satisfy SM7. Determining in what generality these rectification results hold is still future work for me, so I don't want to say anything too definitively on this yet. For sure both $E_\infty$ ring spectra and strict commutative ring spectra inherit model structures, and it seems highly likely they are Quillen equivalent, so you have a good hope for rectification coming from these considerations.

EDIT: It has been pointed out to me that this answer was slightly lacking in a couple of places. For one thing, the idea of the positive model structure originally goes back to Jeff Smith, so I should have included his name above. Secondly, the real feature of importance in the positive model structure is that for any cofibrant spectrum $X$, the natural map $(E\Sigma_n)_+ \wedge_{\Sigma_n} X^{\wedge n} \to X^{\wedge n}/\Sigma_n$ is a weak equivalence, i.e. the map from the the homotopy colimit (which is the extended powers) to the colimit (symmetric powers). I believe I can prove this for motivic symmetric spectra, so the rest of Shipley's paper should go through (I'm still writing this up, though, so for now you should treat this as hearsay). If you're interested in learning more about this property, I asked a couple of questions about it on MO and got some very informative answers from Peter May.

I should also mention that classically if you instead work with orthogonal spectra rather than symmetric spectra then you also need to pass to a positive variant in order to get the property above, and hence rectification. See Mandell-May-Schwede-Shipley Model categories of diagram spectra. This has also been done for equivariant orthogonal spectra in a paper of Mandell-May. I don't know anything about orthogonal motivic spectra and I don't know how that would be done, because motivic spaces are built on simplicial sets rather than topological spaces, and I don't know how to represent $O(n)$ in that setting. Classically, if you use S-modules then rectification comes for free because commutative spectra and $E_\infty$ spectra are the same thing (see EKMM, the point is that the monoidal product already builds in higher homotopies). Po Hu has developed motivic S-modules but does not seem to mention commutative monoids or $E_\infty$ at all in her paper. I don't know that side of the story well enough to know if it still comes for free or not.

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How do your results relate to those of Jens Hornbostel, as found in Section 3 of his paper “Preorientations of the derived motivic multiplicative group” (msp.warwick.ac.uk/agt/2013/13-05/p083.xhtml)? In particular, in Theorem 3.4 he constructs a version of the positive flat stable model structure on motivic spectra and in Theorem 3.10 he proves a generalization of the results by Elmendorf and Mandell that you cited and Harper for the motivic case, which gives a Quillen equivalence between algebras over the Barratt-Eccles operad and commutative monoids in motivic symmetric spectra. –  Dmitri Pavlov Nov 14 '13 at 17:58
    
@DmitriPavlov: Yes, someone else emailed me with that paper yesterday. I wasn't aware of it, but there seems to be substantial overlap. I've been reading it in my spare time (there's not much; many job applications are due tomorrow!) and was going to post another answer to your question once I finished. I do think it answers your question in the affirmative (there's a remark before Theorem 3.10 about $E_\infty$ and $Com$), and nothing feels at all fishy about Hosnbostel's approach. I'm surprised no one pointed me to this paper before, because it seems very good. –  David White Nov 14 '13 at 20:02
    
To answer your other question, it seems my results are more general because they hold for general categories of symmetric spectra. Also, my work might generalize to other operads via section 3 of my research statement and a different generalization of Harper's result that all operads are admissible in the positive model structure (i.e. that algebras over any operad inherit a model structure). I do plan to see if that holds in the generality of my setup, and if so then it'll resolve Hornbostel's conjecture 3.7. I'm taking a break from this project till job stuff is over, so it might be a while –  David White Nov 14 '13 at 20:04
    
Hornbostel's and Harper's results are about algebras over simplicial operads, whereas Lurie's result is about algebras over ∞-preoperads (Definition 2.1.4.2 in Lurie's Higher Algebra). So one still has to pass between (model categories of) algebras over simplicial operads and ∞-preoperads. Section 6.2 in the paper by Heuts, Hinich, and Moerdijk constructs a left Quillen equivalence from ∞-preoperads to simplicial operads by passing through marked dendroidal sets and dendroidal sets, which might be useful here, but it's unclear to me how to proceed. –  Dmitri Pavlov Nov 20 '13 at 14:46
    
@DmitriPavlov. I'm really not familiar with the language of $\infty$-preoperads at all. What kind of algebraic structure are you trying to encode? If the $\infty$-preoperad structure can be strictified to $P$-alg structure over some simplicial operad $P$ then you'd be all set, right? –  David White Nov 20 '13 at 15:00

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