7

2

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?

flag

1 Answer

7

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.)

link|flag
I think that "has a set of generators" is implied by "presentable", right? – Reid Barton Nov 9 2009 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 2009 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 2009 at 0:44
1 
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 2009 at 0:51

Your Answer

Get an OpenID
or

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