Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The recent article found here revisits Thomason's proof that symmetric monoidal categories model all connective spectra, but stops short of showing that there is a full closed model structure on this category (as does, it seems, Thomason's original paper.) Is there such a thing?

My guess is some lifting similar to how the model structure on small categories is derived would work, but I'm not sure if there are any complications.

share|improve this question
    
Elmendorf's conjecture below has a corresponding result in terms of dendroidal sets which is proved here: arxiv.org/abs/1203.6891 –  Peter Arndt Apr 2 '12 at 13:23

1 Answer 1

up vote 7 down vote accepted

One basic problem is that the category of symmetric monoidal categories isn't complete. Its completion, in a basic sense, is the category of multicategories, on which it seems reasonable to conjecture there is a model category structure whose homotopy category "is" the connective part of stable homotopy -- we hope to prove this soon. See Elmendorf and Mandell, "Permutative categories, multicategories, and algebraic K-theory", which just appeared in Algebraic and Geometric Topology.

share|improve this answer
    
Yes, shortly after writing this it had occurred to me that I wasn't sure if the the category had all the required limits and colimits and this might be an obstruction. The idea of using multicategories is intriguing. I'll be sure to have a look at your paper. –  Eric Finster Mar 20 '10 at 23:45
    
What's the state of this work in progress? Isn't the category of multicategories just the category of (non-symmetric) colored operads? Does the recent work of Marcy Robertson get the required model structure? –  David White Aug 10 '12 at 18:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.