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This is a follow-up question to another question I asked last month. In MMSS's "Model Categories of Diagram Spectra," the authors consider many different models for spectra and prove monoidal Quillen equivalences between them. One of the models is $W$-spaces, i.e. diagrams of shape $W$ in $Top$ where $W$ is the category of based spaces which are homeomorphic to finite CW complexes, and $Top$ means compactly generated spaces.

Does the category of commutative monoids in $W$-spaces inherit a model structure?

By commutative here I mean strictly commutative, not $E_\infty$. My interest is for the background section of a paper I'm working on now about when commutative monoids form a model category. If it's not known one way or the other I can try my theorem on the category of $W$-spaces, but I suspect that category won't satisfy my hypotheses. If it's known to be true, maybe I can recover it as a special case of my theorem. If it's known to be false, that would be good to know too.

Note that this question comes up on page 5 of MMSS and the authors say they don't know. On page 47 they point out the obstacle which prevented them from proving CommMon($W$-spaces) inherits a model structure, and it's the same obstacle Peter May pointed out in his answer to my previous question. They couldn't prove that every relative $Sym(K^+)$-cell complex was a stable equivalence (where $K^+$ are the positive stable generating trivial cofibrations) because they couldn't prove $W$-spaces have the property that for every cell $S$-module $M$, $(E\Sigma_j)_+ \wedge_{\Sigma_j} M^j \to M^j/\Sigma_j$ is a weak equivalence. If someone had proven this in the past decade I'm sure Peter May would know. What I'm wondering is if someone found a different way to prove CommMon($W$-spaces) forms a model category or if someone found evidence that it can't, similar to the evidence in this answer that CDGA can't be a model category, or on page 13 of this paper of Schwede's that CommMon($\Gamma$-spaces) can't be a model category. Even if no hard evidence exists, I'd like to hear some soft evidence, i.e. why one would or would not believe this to be true.

The reason one might hope the answer is yes is that Theorems 0.1 and 0.4 prove that $W$-spaces are Quillen equivalent to Orthogonal Spectra and Symmetric Spectra, and that this Quillen equivalence carries over when we pass to the model categories of monoids (the fact that such model structures exist follows from the fact that all these categories satisfy the pushout product and monoid axioms). Both Symmetric Spectra and Orthogonal Spectra have a notion of positive model structures, which makes the category of commutative monoids inherit a model structure. It's natural to wonder if $W$-spaces has something like that too.

MMSS also raise another natural question (on page 6):

Is $Ho(E_\infty$ $W$-spaces$) \cong Ho($Commutative $W$-spaces$)$?

It seems like with all the work that has been done on operads in the last decade this should be known. Can someone confirm that and give a reference?

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As a side note, even in the presence of a negative answer to the second question, the first question might be interesting. For instance, in topological spaces one does not have an equivalence between $E_\infty$ monoids and topological abelian monoids; the latter are interesting (e.g. Dold-Kan and Dold-Thom). In equivariant spectra the "commutative" rings (that appear in Hill-Hopkins-Ravenel) are strictly stronger than the most "homotopical" notion of a commutative ring object, because they possess equivariant norms; they're still very interesting. – Tyler Lawson Aug 15 '12 at 5:10

There is a very illuminating paper I did not know about when I last answered a similar question: Tyler Lawson. ``Commutative $\Gamma$ rings do not model all commutative ring spectra". Homology, Homotopy and Applications Vol 11, 2009, 189-194. The title says it all. Together with the monoidal functor from $\Gamma$-spaces to $W$-spaces of MMSS, Tyler's result makes clear that commutative $W$-rings cannot give a homotopy category equivalent to the categories of orthogonal, symmetric, or EKMM commutative ring spectra. Incidentally, the question has a minor slip. David writes: "By commutative here I mean strictly commutatative, not $E_{\infty}$". As explained in EKMM, strictly commutative EKMM ring spectra are Quillen equivalent to $E_{\infty}$ ring spectra as defined way back in the 1970's. Those still have the nicest connection to spaces, namely $E_{\infty}$ ring spaces. See my paper: What precisely are $E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra. Geometry & Topology Monographs 16(2009), 215--282 (and on my web page) for a modern overview.

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Thanks for the answer, I will definitely check that paper out. This completely answers the second question. Does it also mean you can't have a model structure on commutative $W$-rings induced from the one on $W$-spaces? I.E. if there was such a model structure, would it have to give a homotopy category equivalent to that of commutative symmetric ring spectra? – David White Aug 11 '12 at 16:02
I hadn't thought about that, since to me the answer to the second question makes the first question uninteresting. With the obvious definitions, if you had a model structure then there would be a Quillen adjunction to commutative symmetric or orthogonal spectra that is not a Quillen equivalence. It seems more likely to me that you just don't get a model structure. But it is murky. Unclear what kind of prolongation you get from commutative orthogonal spectra when you know they can't give you what they ought to give you. – Peter May Aug 11 '12 at 16:46
Thanks for your speedy reply. I agree that this information makes the question less interesting. You mentioned in your answer to my previous question that you didn't think commutative $W$-rings would have a model structure (this opinion also comes up in MMSS, but without attribution). Since the Lawson paper is relatively new, could you sketch why you originally didn't think commutative $W$-rings would have a model structure? I was hoping to draw you out on this with my comment in the question above about "soft evidence." I would have asked on the other thread, but it seemed off topic. – David White Aug 15 '12 at 0:48
Gut feelings of my collaborators and myself. But also the technical point that we doubted that the map $$(E\Sigma_j)_+ \wedge_{\Sigma_j} X^{(j)}\longrightarrow X^{(j)}/\Sigma_j$$ is an equivalence for cofibrant $W$-spaces $X$. – Peter May Aug 21 '12 at 15:49

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