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Feb 14, 2019 at 18:52 vote accept David White
Aug 21, 2012 at 15:49 comment added Peter May 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 $X$.
Aug 15, 2012 at 0:48 comment added David White 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.
Aug 11, 2012 at 16:46 comment added Peter May 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.
Aug 11, 2012 at 16:02 comment added David White 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?
Aug 11, 2012 at 13:58 history answered Peter May CC BY-SA 3.0