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It is well know that the $\infty$-category of group-like $E_\infty$-spaces and the $\infty$-category of connective spectra are equivalent, see e.g. May - "$E_\infty$-spaces, group completions and permutative categories" or Lurie - "Higher Algebra", Remark 5.1.3.17

Now the category of $E_\infty$-spaces (here space means simplicial set) carries a model structure as well as the category of spectra. Is there a direct (left) Quillen functor

$E_\infty$-space $\to$ Spectra

whose derived functor restricts to such an equivalence? I have been unable to find a discussion of this in the litertatur. The only thing I can find are indirect functors going through $\Gamma$-spaces or related categories. The Bar-construction which is usually used is not left Quillen (!?).

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What exactly do you mean by an $E_\infty$ space? If you have a particular $E_\infty$ operad in mind, then it depends which one you are using. If you want a version of the category of $E_\infty$ spaces in which the operad is allowed to vary, you should state that as well. Similarly, the precise answer will depend on your version of the category of spectra. –  Neil Strickland Mar 6 '12 at 16:50
    
Take for example the Barrat-Eccles Operad and the classical (Bousfield-Frielander) category of spectra in simplicial sets. I was hoping that there might be an answer which is independet of the specific choice of $E_\infty$-algebra. But I want the Operad to remain fix. –  Thomas Nikolaus Mar 6 '12 at 16:53
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I would have expected that a suitable version of the bar construction would provide a left Quillen functor. Could you clarify exactly which version you are considering, and why it is not left Quillen? (I would be inclined to use the operad from Steiner's paper "A canonical operad pair" to construct orthogonal spectra of topological spaces, but no doubt there are other possibilities, including some that are more simplicial.) –  Neil Strickland Mar 7 '12 at 8:41
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One way to make sense of 'restricting to an equivalence' is to consider the group-like $E_\infty$ spaces as a left Bousfield localization of the category of $E_\infty$ spaces. I think you just need to invert the map from the free $E_\infty$ space on $S^0$ to $QS^0$. The fibrant objects in this category will be group-like and fibrant replacement will be group completion. To do such a construction I would want the category of $E_\infty$ spaces to be left proper. This should follow from the $E_\infty$ operad being cofibrant, by Spitzweck's thesis. –  Justin Noel Mar 7 '12 at 9:07
    
@Justin: This is actually exactly what I did. Just that it turned out that its easier to localize at a diffrent map than the inlcusion $S^0 \to QS^0$. @Neil: I am actually in a hurry, I will comment on that later. But what makes you believe that the bar construction is even left adjoint? –  Thomas Nikolaus Mar 7 '12 at 13:37

1 Answer 1

up vote 6 down vote accepted

Of course, as several people have noted, the answer depends on the choice of details. There is a variant of my original passage from $E_{\infty}$ spaces to spectra that certainly works, as was noted in ``Units of ring spectra and Thom spectra'' by Ando, Blumberg, Gepner, Hopkins, and Rezk (arXiv: 0810.4535v3).

Take the Steiner $E_{\infty}$ operad for definiteness and denote the monad on based spaces associated to it by $\mathbf{C}$. Take spectra to mean Lewis-May spectra since it is very convenient to have the $(\Sigma^{\infty},\Omega^{\infty})$ adjunction for the question at hand, and that is incompatible with symmetric monoidal categories of spectra. Of course, that means I'm not using simplicial sets, but I don't suffer from a prejudice in their favor: when I write space I prefer to actually mean space.

Then, as discussed in modern terms in my paper "What precisely are $E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra?'' Geometry & Topology Monographs 16(2009), 215--282, the spectrum associated to a $\mathbf{C}$-space $X$ is the two-sided bar construction $B(\Sigma^{\infty},\mathbf{C},X)$. For cofibrant $X$, this is equivalent to the ``tensor product''
$\Sigma^{\infty}\otimes_{\mathbf{C}}X$, which is defined by an obvious coequalizer. This functor from $\mathbf{C}$-spaces to spectra is left adjoint to $\Omega^{\infty}$. Further details are as one would expect.

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Thank you Peter, thats exactly the kind of thing I was looking for. Actually I do not see a reason why this should not work for simplicial $E_\infty$-spaces and simplicial spectra.... –  Thomas Nikolaus Mar 8 '12 at 14:03
    
Think about what $\Omega^{\infty}$ would mean in the simplicial world. There are serious advantages to being eclectic. For a different, Segalic answer to your original question see Mandell, May, Schwede, Shipley ``Model categories of diagram spectra'', especially \S 18. –  Peter May Mar 9 '12 at 0:23
    
I agree 100% with you comment about being eclectic. Its just that I got a situation where I naturally obtain simplicial $E_\infty$-spaces and thats why I was interested in them... –  Thomas Nikolaus Mar 9 '12 at 19:58

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