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There is a theorem of Schwede and Shipley which classifies categories of modules over an A ring spectrum as those stable presentable (∞,1)-categories with a compact generator. Suppose I allow my A rings to "have many objects", that is, I consider categories of the form FunSp(Iop, Sp) where Sp is the category of spectra, I is a small Sp-enriched category (in some appropriate sense) and Funsp denotes the category of Sp-enriched functors. Is there a classification of which stable presentable categories can be obtained in this way? Is it possible that all stable presentable categories are of this form?

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up vote 7 down vote accepted

According to the abstract of http://arxiv.org/abs/math/0108143 (Schwede & Shipley, Classification of Stable Model Categories), they deal with the case of stable model categories (=stable presentable (∞,1)-categories, I suppose) which have a set of compact generators, and show they are the same as model categories of functors from spectral-enriched categories.

(Edited, in the light of Reid's comment, to include the hypothesis of compact generators.)

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I think that "has a set of generators" is implied by "presentable", right? –  Reid Barton Nov 9 '09 at 0:24
    
Ah. I should have said "compact generators", as that appears the condition Schwede-Shipley require. So "presentable" implies "set of generators", but "compact generators" is much more special. –  Charles Rezk Nov 9 '09 at 0:33
    
I don't have a ready example in the case where you drop the hypothesis of compactness. If you form the localization of spectra with respect to a homology theory E, the localization L_E(S) of the sphere spectrum is a generator for this stable homotopy theory, but in general L_E(S) is not compact. But this doesn't preclude the existence of some other compact generator, and that actually happens in some cases, e.g., E= a Morava K-theory. –  Charles Rezk Nov 9 '09 at 0:44
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Corollary B.13 of the reference [HS99] gives examples of what I presume are stable presentable categories without any nonzero compact objects. –  Reid Barton Nov 9 '09 at 0:51
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