Coefficient modules for comonad cohomology when there aren't enough abelian objects? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T16:14:50Z http://mathoverflow.net/feeds/question/43259 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43259/coefficient-modules-for-comonad-cohomology-when-there-arent-enough-abelian-objec Coefficient modules for comonad cohomology when there aren't enough abelian objects? Yemon Choi 2010-10-23T01:16:47Z 2010-10-23T01:16:47Z <p>(Please be gentle if you want to throw n-categorical stuff, or anything groupoid-esque, at me in response to this question!)</p> <p>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 <a href="http://ncatlab.org/nlab/show/semi-abelian+category" rel="nofollow">semi-abelian</a> in the sense of Borceux and Bourn: I have in mind something like the framework described in Section 5.3 of <a href="http://arxiv.org/abs/math/0607100" rel="nofollow">van der Linden's thesis.</a></p> <p>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 <a href="http://ncatlab.org/nlab/show/Moore+complex" rel="nofollow">Moore complex</a> 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 $M$.</p> <p>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:</p> <blockquote> <p>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?</p> </blockquote> <p>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.</p> <p>(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.)</p>