Resolutions of unbounded complexes and homotopy (co)limits.

I want to understand once and for all what the resolution of an unbounded complex is. I've been trying to read 'Homotopy limits in triangulated categories' by Marcel Bokstedt and Amnon Neeman and can't understand at all even some of the notation. From what I've gathered, the resolution of an unbounded complex $X$ is $M$ in:

$X =$ $\underleftarrow{lim_{n}} (X_{-n}) \xrightarrow{\alpha} \underleftarrow{holim_{n}} (X_{-n}) \xrightarrow{\beta} M = \underleftarrow{holim_{n}} (I_{-n})$,

where $\alpha$ and $\beta$ are quasi-isomorphisms.

But what's the homotopy limit in this context? The thing I can relate it to is this:

A homotopy limit of a sequence of maps

$X_{0} \xrightarrow{f_0} X_{1} \xrightarrow{f_1} X_{2} \rightarrow \cdots$

is defined as the subspace of the product of $X_0$ with $X_{i}\times\hom(\Delta[1],X_{i})$, $i \geq 1$, of elements $(x_0,${$(x_i,\gamma_i)$}$_{0 \leq i \leq n}$) such that

$\gamma_i(0) = f(x_{i-1})$,

$\gamma_i(1) = x_i$.

This is completely over my head, as someone coming from Cartan-Eilenberg resolutions being the resolutions of bounded complexes I was expecting to see a bicomplex, I have no idea how to interpret that, does anyone know? Be nice please, I'm still learning.

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A resolution of complex $X$ is just a quasiisomorphism of $X$ with some other complex. The question is what properties you want to impose on this other complex. Usually people consider h-projective or h-injective resolutions and the statement is that you can construct one (sometimes). – Sasha Nov 14 '12 at 5:38
Why do you think that resolutions are related with homotopy colimits? Also, could you fix the formula? – Mikhail Bondarko Nov 14 '12 at 6:18
Notice that "holim" in the sense of Bokstedt and Neeman's paper is not the usual Bousfield-Kan notion, but a non-functorial version based on the Milnor triangle, see Definition 2.1 on p. 213 of the paper. So it is not a bicomplex, it just a cone. – Leo Alonso Nov 14 '12 at 9:48
@Leo Alonso. That's exactly what I was wondering! They define the resolution of a complex as a cone, how can a cone be a bicomplex? What do they have to do with each other? I just don't know how to reinterpret that. So the resolution of a complex is a cone? Not another complex? – Richard Jennings Nov 14 '12 at 18:13

2 Answers

In his paper "resolution of unbounded complexes" Compositio Math. (1988) N. Spaltenstein introduces the concept of K-projective and K-injective chain complexes. He proves (theorem C) that any unbounded chain complex $C.$ has a K-projective resolution or a K-injective resolution. For example if you work with chain complexes of modules over a commutative $R$. And if you take a chain complex $C.$ there exists a quasi-isomorphism from a K-projective resolution $F.$ of $C.$ into $C.$.

In his introduction N. Spalsenstein gives a very good example that motivates the introduction of such a notion (p. 124).

Let me just say that the point is that when you want to compute derived functors you do need to replace your chain complex by a "nicer" guy, this nicer guy should have nice chain homotopic properties. Because you want your computations to be chain homotopic invariant. Over a field all chain complexes are K-projective. But over a ring this is not the case even if you take a chain complexes of projective $R$-modules so you need an additional assumption (see definition p.124).

If you know closed model categories, then a K-projective chain complex is a cofibrant object for a CMC structure on chain complexes of $R$-modules.

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Thank you @David C! I actually turned to the Bokstedt-Neeman paper because I didn't find Spalstenstein's paper clear enough. What I really want to know is, according to Bokstedt-Neeman the resolution of an unbounded complex is a homotopy limit in certain diagram, which is a CONE, how can this cone be a bicomplex? How can I interpret that? – Richard Jennings Nov 14 '12 at 18:20
@Richard Jennings: a morphism of complexes can be thought of as a bicomplex with two rows --- the first row is the source complex and the second row is the target complex, and the vertical differential is just the morphism of complexes itself. – Sasha Nov 15 '12 at 5:22

Let $X$ be a complex (of $R$-modules, say). A resolution (on the right, to fix ideas) is just a quasi-isomorphism $X \to P$. If we want to derive a functor $F$ on the left, we would choose for $P$ an acyclic resolution and we have that $\mathbf{L}F X = F P$. Of course, if $P$ is homologically projective (or K-projective) then it is acyclic for any $F$. The dual thing with resolutions on the left and right derived functors. The classical construction of the unbounded resolution is a tedious step by step construction. (A totalization of a Cartan-Eilenberg bicomplex will not work except under some exactness situations). So what Bökstedt and Neeman prove is that with a simple countable homotopy colimit (a cone) you get the desired resolution for modules.

For sheaves things are more delicate. If you dualize the argument in Bökstedt and Neeman, you get K-injective resolutions for modules, but the procedure does not work for sheaves because countable products are not exact over sheaves. That's where Spaltenstein work comes into the picture.

A good discussion of these kind of questions without full proofs, but pointers to the relevant literature, are the first two chapters of Lipman's "Notes on Derived Functors and Grothendieck Duality", (in Foundations of Grothendieck Duality for Diagrams of Schemes, Lecture Notes in Mathematics, no. 1960, Springer, 2009).

Actually, what these works do is to reduce the problem of deriving additive functors to a question in homological algebra, without making recourse to model categories. The model structure for the unbounded category of complexes is a delicate thing (see the papers on this topic by Hovey and collaborators). To establish a certain model structure on the derived category of a Grothendieck Abelian category is as difficult as proving that there are K-injective resolutions by homological methods.

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Since you mentioned Hovey, let me add that if the OP decides to look more into this work then he should google "Cotorsion Pairs" and can find a lot about this. Also, I've written several MO answers and comments summarizing that work (and in one place connecting it to the Grothendieck abelian category mentioned by Lee above), so if you use google to search mathoverflow for "cotorsion pairs" or you can find some (hopefully) useful things. – David White Nov 15 '12 at 18:43