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Of course, before group completion, topological commutative monoids do not model all connected $E_\infty$ spaces -- among the grouplike ones, they model only products of Eilenberg-Mac Lane spaces. But after group completion, perhaps they do. Similarly, I wonder: Do strictly symmetric monoidal (i.e. the symmetry isomorphisms $a \otimes b \cong b \otimes a$ are identities) topologically-enriched categories model all connective spectra?

Maybe something even stronger is true: Do strictly symmetric monoidal groupoids (or equivalently, 1-types with a commutative monoid structure) model all connective spectra after group completion? Do strictly symmetric monoidal groups (equivalently, $K(G,1)$'s with a commutative monoid structure) model all 0-connective spectra after group completion?

If the answer to some version of this question is affirmative, I'd liken it to the fact that every space is the realization of a category -- you can get some extra strictness in a model at the expense of some noninvertibility.

This question obviously has some affinities to another question I asked a few days ago.

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    $\begingroup$ I am skeptical about this: every strictly commutative monoid has a natural structure of $\mathbb{N}$-module. After group-completion this should give them the structure of $H\mathbb{Z}$-modules, which of course not all connective spectra have (since all $H\mathbb{Z}$-modules are generalized Eilenberg-MacLane spectra). $\endgroup$
    – Denis Nardin
    Commented Jan 2, 2017 at 8:53
  • $\begingroup$ That's a good point! I think it can be made precise: group completion is $\Omega B$; $B$ preserves finite products, and $\Omega$ is lax monoidal, so the composite is lax monoidal and so preserves module structures. Although does $B$ really preserve finite products in the category of $E_\infty$ spaces? $\endgroup$
    – Tim Campion
    Commented Jan 2, 2017 at 15:40
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    $\begingroup$ Yes it does, since $E_\infty$-spaces are an additive category (so finite products and finite coproducts are the same) $\endgroup$
    – Denis Nardin
    Commented Jan 2, 2017 at 15:50

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Do strictly symmetric monoidal topologically-enriched categories model all connective spectra?

If your modeling functor is some reasonable variant of the ordinary classifying space functor, I think the answer must be No.

What I know for sure about this question is written (with Angélica Osorno) in this paper: Modeling stable one-types. In that paper we study how to model the Postnikov data of spectra with just two nonzero homotopy groups (in dimension 0 and 1). The Postnikov data in such a low-dimensional case consists only of the two (abelian) homotopy groups $\pi_0$ and $\pi_1$, together with a Postnikov invariant.

Therefore we restrict not only to symmetric monoidal categories, but to those symmetric monoidal categories where all morphisms are invertible and all objects are invertible up to isomorphism. (The standard term for these is "Picard categories", but a more descriptive name would be "grouplike symmetric monoidal groupoids".)

The sloganized summary of the paper is that the symmetry of the Picard category is the data which corresponds to the Postnikov invariant of the stable 1-type. (There's a neat connection with the theory of quadratic maps, and some calculations of Eilenberg and Mac Lane from the 1950's, which partially explain why you might expect this.) I'll skip the precise details here -- you can try the paper, or just ask me later.

For a specific example, we write down a symmetric monoidal category which models the Postnikov truncation of the zeroth space of the sphere spectrum. It has a strictly associative and unital monoidal structure, but the symmetry is nontrivial.

Now, maybe there's some other way of modeling connective spectra with strictly symmetric monoidal but topologically enriched categories. But I think our work proves it can't be done in a way that generalizes the ordinary classifying space functor when the enrichment is discrete.

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