I am a number theory graduate student learning a bit of homological algebra, and I am curious about higher complexes in abelian categories. I apologize if my post is slightly vague as I am not an expert in this area. I will use capital roman letters to denote objects or complexes, but the usage is clearly stated.


To motivate my question, let us start from a single object $M$ in an Abelian category $\mathcal{C}$. If $\mathcal{C}$ has enough projectives, we can form a projective resolution $P\to M$ of $M$, and apply a right exact additive functor $F:\mathcal{C}\to\mathcal{D}$ to $P$, and calculuate homology. Here $\mathcal{D}$ is just some other abelian category. This will give us the derived functors of $F$, and is a standard and well-known construction.

Of course, it doesn't stop there. In the same abelian category $\mathcal{C}$ with enough projectives, any chain complex $M$ also has a (left) Cartan-Eilenberg resolution $P\to M$. Recall $P$ is an upper half plane double complex and the map $P\to M$ is just a chain map $P_{\bullet,0}\to M_\bullet$. Finally, $P$ is required to satisfy some axioms making it into a sort of 2-dimensional version of a projective resolution. I won't go into detail because this is also fairly standard.

The point is that we can also apply a right-exact functor $F$ to this double complex $P$ and take the the homology of the total direct-sum complex of $P$ (if it exists!); that is $H_i(Tot^\oplus(FP))$, to get the hyperderived functors of $F$.

The Question

It seems as though there is a natural generalization. One can easily define an $n$-complex in an analogous fashion to $2$-complexes. Higher dimensional complexes don't really show up much as far as I can tell, although I believe in Cartan-Eilenberg a $4$-complex is used somewhere (sorry, I don't have the book with me!).

So I suppose my question is:

Suppose $\mathcal{C}$ is an abelian category with enough projectives. Is it true that for any $n$, an $n$-complex $M$ has some appropriate higher Cartan-Eilenberg resolution (which would be an $n+1$-complex)?

Appropriate means that if $P\to M$ is this hypothetical higher Cartan-Eilenberg resolution, then applying a right exact additive functor $F$ to $P$ and taking the homology of the total direct-sum (if it exists) complex gives the "correct" notion of $n$-hyperderived functors.


I have searched the literature for this concept but I could not find anything relevant. I am thinking that there are two possibilities (a) yes, higher Cartan-Eilenberg resolutions exists and are interesting, or (b) yes, higher Cartan-Eilenberg resolutions exist but don't capture any new information and so are not that interesting. I'd be a bit surprised if they don't exist but I do not have enough experience in homological algebra to understand the bigger picture here.

Also, we could have phrased this question in terms of injectives and (right) Cartan-Eilenberg resolutions.


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    $\begingroup$ At least if you only consider non-negative chain complexes, this will be true because if $A$ is abelian with enough projectives, then $Ch_{\geq 0}(A)$ is again abelian with enough projectives (by the Dold-Kan correspondence), and hence also $Ch_{\geq 0}(Ch_{\geq 0}(A))$, etc. $\endgroup$ Commented Feb 26, 2012 at 21:34
  • $\begingroup$ have looked up huybrechts book on fourier-mukai transforms? that and the two books by kashiwara-schapira are my favorite places to learn homloogical algebra (and perhaps gelfand-manin, but only during the weekends) $\endgroup$ Commented Feb 27, 2012 at 23:36

1 Answer 1


The process can certainly be iterated as explained by Marc (see also Weibel, Homological Algebra, 1.2.5. Moreover cf. 1.2.3, 2.2.2 for the fact that the category of chain complexes over an abelian category with enough projectives is again an abelian category with enough projectives).

However, it seems to me that it isn't often used. A reason might be, that in many (most ?) cases one isn't interested in a double complex (or higher dimensional analogs) itself but in the (co)homology of of its total complex (the definition of hyperderived functors in your question is an example for this point of view).

However, in this case there is no need to jump into a higher dimension to define a projective resolution. For, there is an alternative definition for the projective resolution of a chain complex that is - in my opinion - much more elegant and easier to work with than with Cartan-Eilenberg's definition:

A projective resolution of the chain complex $C$ is a complex $P$ of projectives together with a quasi-isomorphism $f: P \to C$ (i.e. $f$ is a chain map such that $H_n(f): H_n(P) \to H_n(C)$ is an isomorphism for all $n$).

Note that such a $P$ is in general no projective object in the category of chain complexes, but it yields the same hyperderived functors, hypercohomology spectral sequences, etc. For a textbook reference of this definition see for example

  • McCleary, A User's Guide to Spectral Sequences (before Theorem 12.12)
  • Benson, Representations and Cohomology I, Definition 2.7.4
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    $\begingroup$ In Dave Benson's errata to his book (abdn.ac.uk/~mth192/papers/b/benson/comments.dvi) he comments "2.7.4 claims to define a projective resolution P of a chain complex C. The definition is incorrect; the Cartan-Eilenberg theory of proper resolutions cannot be avoided here." $\endgroup$
    – M T
    Commented Feb 27, 2012 at 8:25
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    $\begingroup$ The error arises from the fact that "complexes of projectives" are not necessarily "projectively cofibrant" in the projective model structure for chain complexes. I believe that this is discussed in Hovey's book on model categories. $\endgroup$ Commented Feb 27, 2012 at 8:34
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    $\begingroup$ There is no problem as long as the complex is bounded above. If you want unbounded complexes (which does not seem the case) you need to use K-projective resolutions. See Spaltenstein, 1988 or Bökstedt-Neeman. $\endgroup$
    – Leo Alonso
    Commented Feb 27, 2012 at 11:29
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    $\begingroup$ @mt: But it works in the bounded case and thus applies in many situations. I guess that the OP who says he starts to learn homological algebra won't start with the "pathological" cases. $\endgroup$
    – Ralph
    Commented Feb 27, 2012 at 12:04
  • $\begingroup$ Also we should remark that it is not immediate that a projective object in the category of chain complexes is actually a CE resolution, and perhaps there are counterexamples to this. @Ralph: I am interested in pathological cases. I finished reading in detail the first five chapters of Weibel and this is one of the things that confused me a bit, so actually this kind of pathology is kind of enlightening. $\endgroup$
    – user1437
    Commented Feb 27, 2012 at 20:39

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