resolution by simplicial objects versus resolution by chain complex - MathOverflow

resolution by simplicial objects versus resolution by chain complex

2012-09-13T03:39:49Z

I'm reading Illusie's book "complexe de cotangent et Deformations I". And I'm puzzled on the definition of cotangent complex.

I formulate my question as follows:

Suppose $C$ and $D$ are abelian categories, and $F:C \to D$ is a functor. I would like to consider its left-derived functor $LF$. There are two ways:

(1) $C$ embedded into $K(C)$ ,the homotopy category of complexes in $C$, and denote the left-derived functor of F as $L^1F$, so you calculate $L^1F$ by resolution of chain complexes.

(2) $C$ embedded into $sC$, the category of simplicial objects in C, and denote the left-derived functor of F as $L^2F$, so you calculate $L^2F$ by resolution of simplicial objects.

then, my question is:

for any object $X$ in $C$, what is the relation ship with $L^1F(X)$ and $L^2F(X)$? Is there any relation like Dold-Kan?
then what is the reason for the defition of cotangent complex to use simplicial resolution? Is it simply because there is no left-derived functor for the Kahler differential functor if I embed the category $C$ into $K(C)$?

Appreciate very much!

Answer by David Roberts for resolution by simplicial objects versus resolution by chain complex

2012-09-13T05:03:33Z

There is an equivalence between the category of simplicial objects in $A$ and connective chain complexes in $A$ (see this nLab page). This is exactly the same as Dold-Kan. There is also a Quillen equivalence between these categories (see previous link). Since derived functors between abelian categories is secretly about working with Quillen model structures on categories of (connective) chain complexes, you can calculate the derived functor using either.

I don't know the particular benefits of working with the cotangent complex, rather than with a resolution involving chain complexes.

Answer by Tim Porter for resolution by simplicial objects versus resolution by chain complex

2012-09-13T10:47:03Z

The point in Illusie's (and earlier Quillen's) work is that you are not really working in an abelian category! Look at Quillen's paper, his homology is the homology of commutative algebras (and the abelian objects there are not that interesting as they have trivial multiplication). The homology is the derived functor of the (relative) abelianisation in a non-abelian setting. Chain complexes are not even available.

For abelian categories yes, David's comment is correct, but in Illusie you start with a non-abelian setting, therefore need simplicial resolutions, then use (relative) abelianisation /Kahler module of differentials, to get to simplicial modules, and finally can use Dold-Kan to get to the chain complexes and a form of homology.

(By the way, I would be careful in how you think of 'embedding' $C$ into $K(C)$, as that is where the subtleties start.)