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.
Motivation
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.
Comments
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.
Thanks
Best Answer
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:
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