There are two constructions that look quite similar to me: the derived category of an abelian category, and the homotopy category of a model category. Is there any explicit relationship between these two constructions? (This question is related to, and indeed the inspiration for, one of my previous questions.)
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asked Oct 12, 2009 at 19:15
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5 Answers
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6
Yes. The former is a special case of the latter. There is a model category structure on the category of (say bounded) chain complexes of objects in your given abelian category. The weak equivalences are the quasi-isomorphisms, and the homotopy category is the derived category.
In the case of R-modules, for a ring R, this is explained in detail in this paper by Dwyer-Spalinski.
answered Oct 12, 2009 at 19:46
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$\begingroup$ Do you know a reference for the general case? $\endgroup$ Oct 12, 2009 at 19:51
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$\begingroup$ I think this can be found in the standard texts "Model Categories" by Mark Hovey or "Model Categories and Their Localizations" by Philip Hirschhorn. It is also probably in Quillen's original papers. I would look at Quillen's stuff first. $\endgroup$ Oct 12, 2009 at 20:06
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$\begingroup$ I wasn't able to find it in Hovey's book. I'll try Hirschhorn's book, and Quillen's stuff. Thanks! $\endgroup$ Oct 12, 2009 at 20:10
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2$\begingroup$ Weak equivalences may be quasi-isomorphisms, but what are fibrations and cofibrations, or at least the fibrant and cofibrant objects? Since this question does not seem to have a good answer in general, what you say is unlikely to be true. $\endgroup$ Dec 2, 2009 at 16:56
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$\begingroup$ This is an exercise in Gelfand-Manin, right? Section V.2, maybe? $\endgroup$ Sep 8, 2011 at 23:10
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Unfortunately, it isn't quite right to say that derived categories of abelian categories are a special case of model categories. Morally this might be true, but for a general abelian category there is no known model category structure on its (unbounded) category of chain complexes whose weak equivalences are the quasi-isomorphisms. There is such a model structure when the abelian category is a Grothendieck category; this is shown in
- Mark Hovey -- Model category structures on chain complexes of sheaves (2001).
Quillen originally gave the example of a model structure on the category of non-negatively bounded complexes of R-modules, but the case of unbounded complexes of R-modules seems not to have appeared in print until the publication of Hovey's book
- Mark Hovey -- Model categories (1999).
As far as I'm aware, none of the standard references on model categories talk about unbounded derived categories of abelian categories---probably because in general they don't arise as the homotopy category of any known model structure on the category of chain complexes!
answered Sep 8, 2011 at 21:02
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6$\begingroup$ @Beren, in those cases when there is no known model category structure on the category of unbounded complexes on an abelian category, the derived category is not know to exist either. $\endgroup$ Sep 9, 2011 at 7:49
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1$\begingroup$ @Fernando: Thanks for pointing that out. Is there a feeling among experts that all derived categories of abelian categories which exist should have models (even if we aren't currently aware of such models)? Secondly, are there any abelian categories for which we have proofs that their derived categories don't exist? $\endgroup$ Sep 9, 2011 at 12:45
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4$\begingroup$ @Beren: 1) As far as I remember, I don't know any abelian category with a derived category which does not come from a sort of model category. 2) Look at Freyd's Abelian Categories Chapter 6, Exercise A, pp. 131–132. $\endgroup$ Sep 10, 2011 at 8:07
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Both give rise to derivators, and indeed thinking about homotopy theories as non-abelian derived categories is what led Grothendieck to introduce then (note that Heller and Franke independently came up with derivators, but I'm not sure they had the same motivation)
answered Sep 8, 2011 at 23:33
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Some information may be found at nLab: homotopy category. Following the links there you also find information on all the other keywords mentioned above.
Urs Schreiber
answered Oct 15, 2009 at 8:20
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I think you don't want any bounded condition. I don't see how the category of chain complexes with bounded cohomology could be a model category. It doesn't have all small colimits; just take longer and longer chain complexes with trivial differentials, and you get something with unbounded cohomology.
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1$\begingroup$ Bounded can mean that they are universally bounded (ie, consider only chain complexes that are positively graded). $\endgroup$ Oct 15, 2009 at 17:29
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$\begingroup$ This is what most people call "concentrated in positive degree". I've never heard anyone refer to it as being bounded. $\endgroup$– user332Oct 16, 2009 at 16:07
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1$\begingroup$ @Rex, I think that Quillen model categories are not required to be cocomplete. $\endgroup$ Sep 9, 2011 at 7:51
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1$\begingroup$ That's not Quillen's, he only requires finite (co)limits $\endgroup$ Sep 9, 2011 at 8:26
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1$\begingroup$ Of course, if you adapt the definition of the notion somebody else is asking about, you'll probably always be true, correct and precise! ;-) $\endgroup$ Sep 10, 2011 at 8:11