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What is the standard reference for the fact that the classifying space of a strict monoidal category is a topological monoid with respect to the operation induced by the tensor product?

EDIT: The first version of the question was stated for strict symmetric monoidal categories, but as was pointed out in the comments, a symmetry is of course not necessary to just get a monoid structure on the classifying space.

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Of course, you can drop the word "symmetric": a strict monoidal category is the same as a monoid in $(Cat, \times)$, and since the classifying space functor $Cat \to kSpace$ preserves products, it takes monoids to monoids. This isn't covered in Segal's Categories and Cohomology Theories? (Maybe it is; I don't recall.) –  Todd Trimble Nov 27 '12 at 19:26
    
@Todd: Thank you. I will have a look. I know that the proof of this is really just the fact that geometric realizations are product preserving. But I happen to write a paper, which uses this and has a main audience outside of topology. So I need to back things up with as many references as possible. –  Ulrich Pennig Nov 27 '12 at 19:49
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Theo, if you're asking Todd, then he'd say "topological monoid" here is an abuse of language where strictly speaking we are taking $kSpace$ as our "convenient" category of topological spaces. I thought that was a pretty standard maneuver; it's well known that $Top$ has some undesirable properties (such as not being cartesian closed, when we'd really like function spaces with all our hearts). –  Todd Trimble Nov 27 '12 at 23:12
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As Todd already pointed out, this is special case of the observation that every lax monoidal functor $C \to D$ extends to a functor $\mathsf{Mon}(C) \to \mathsf{Mon}(D)$ (applied to $C=(\mathsf{Cat},\times)$ and $D=(\mathsf{CGHaus},\times)$). Since this is trivial, it is always just mentioned in the literature (for example Saavedra Rivano, Categories Tannakiennes, I.6.1.4.). –  Martin Brandenburg Nov 28 '12 at 0:25
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You not only can but should drop the word ``symmetric''. Otherwise someone in a naive audience may ask whether your topological monoid is commutative, and of course it is not. With the standard notion of a strict symmetric monoidal category (aka a permutative category), the classifying space gives rise to a spectrum whose zeroth space is a group completion of your monoid. –  Peter May Dec 2 '12 at 23:47
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up vote 3 down vote accepted

Well, if you are going to reference me somewhere, I can give you something more explicit. The cited Corollary 11.7 is only about topological monoids. However Theorem 4.10 of "$E_{\infty}$ spaces, group completions, and permutative categories" ( http://www.math.uchicago.edu/~may/PAPERS/13.pdf ) has the precise statement requested: "If $(\mathcal{A},\Box,\ast)$ is a strict monoidal category, then $B\mathcal{A}$ is a topological monoid with product $B\Box$." The result goes on to say precisely what holds with respect to commutativity when $\mathcal{A}$ is permutative (= strict symmetric monoidal).

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Sorry to be pedantic, but I can't resist: permutative is symmetric strict monoidal, not strict symmetric monoidal. That is, the monoidal structure is strict, but in general the symmetry isn't. –  Tom Leinster Dec 4 '12 at 14:55
    
Right, that is pedantic: mathematics over euphony. But you can argue the other way: whatever the word order, this is as strict as you can get while still being equivalent to symmetric monoidal, therefore strict symmetric monoidal. Your pedantic comment is one reason to prefer "permutative". –  Peter May Dec 4 '12 at 15:03
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There is an S missing in the link. It should be PAPERS, not PAPER. This link works: math.uchicago.edu/~may/PAPERS/13.pdf –  Martin Dec 4 '12 at 15:24
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My apologies, and thanks whoever you are. –  Peter May Dec 4 '12 at 15:45
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Thank you! I can cite this for permutative categories --> infinite loop spaces as well. –  Ulrich Pennig Dec 4 '12 at 16:14
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My reference for this is now Corollary 11.7 in J. P. May. The geometry of iterated loop spaces. Springer-Verlag, Berlin, 1972. Lectures Notes in Mathematics, Vol. 271.

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