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Since this old question of mine from late 2010 recently got an upvote, I thought it might be worth unearthing in case new readers have new contributions. I hope this isn't an abuse of community norms.


Please be gentle if you want to throw n-categorical stuff, or anything groupoid-esque, at me in response to this question!

During my PhD I did a fair bit of superficial reading around so-called "cotriple homology" or "comonad homology" - the idea being that on a nice category $\cal C$ carrying a comonad $(\bot,\varepsilon,\delta)$, for each object $A\in\cal C$ one can build a simplicial resolution of $A$ with the face and degeneracy maps built out of the data of the comonad. Here I am not assuming that $\cal C$ is abelian, but the question I will get round to is motivated by examples which are semi-abelian in the sense of Borceux and Bourn: I have in mind something like the framework described in Section 5.3 of van der Linden's thesis.

Anyway: given any functor $F$ from $\cal C$ to an abelian category $\cal M$, we can hit one of these simplicial resolutions with $F$ to get a simplicial object in $\cal M$, and then take its Moore complex to get a chain complex and thence homology groups (or, if $F$ is contravariant, we get a cosimplicial object and cohomology groups.) One classical example of this machinery in action is given by taking $\cal C$ to be the category of groups, fixing an abelian object $M\in \cal C$ - i.e. an abelian group - and then taking $F(G) = \operatorname{Hom}_{\cal C}(G,M)$ when we recover the usual group cohomology of $G$ with coefficients in $M$.

Now in the examples I've seen discussed, the category $\cal C$ has a good supply of abelian objects, or at least one can fix $A\in\cal C$ and then look for abelian objects in the slice category ${\cal C}/A$. My question is the following, which is unfortunately vague but which hopefully admits a sensible answer from someone who understands all this better than I do:

If neither $\cal C$ or ${\cal C}/A$ have non-zero abelian objects, what is the next sensible or natural candidate for the functor $F$, so as to get a worthwhile (co)homology theory?

Note that at the moment I still want my target category $\cal M$ to be abelian, which may be misguided -- if so, clarification of why would be most welcome. I am aware that one consider nonabelian coefficients, but would like to satisfy myself that there are no good choices of abelian coefficients before moving on.

(I do have a particular $\cal C$ in mind, which I think satisfies the conditions in my question; but I thought I'd phrase the question more generally in case there is some broader principle that can be applied.)

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How about taking F to be some Hom-functor anyway, and then taking the homotopy groups of the simplicial set that you get? –  Peter Arndt Oct 23 '10 at 1:48
    
Peter: that had occurred to me, but for reasons I'm leaving in the background for the moment, I was hoping to get a set-up closer to that in group cohomology or Andre-Quillen cohomology. However, it may be that the simplicial homotopy groups are the "only reasonable game in town" - the question is really asking if this is indeed the case. –  Yemon Choi Oct 23 '10 at 20:25

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