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 $P_{*,0}\to A$. 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 $H_i(Tot^\oplus(P))$, 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