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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:

  1. 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?

  2. 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!

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2 Answers

up vote 6 down vote accepted

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.)

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I think I got it, thank you very much! By the way, is there any model category structure on the non-abelian category of chain complexes of A-algeras? –  Xiaobo Zhuang Sep 13 '12 at 11:18
    
I don't know off hand. One problem is deciding what the notion of weak equivalence should be as you cannot define a quotient of an algebra by a subalgebra, you need an ideal. There is a version of crossed differential algebra and also dg-algebras which have important applications. The Moore complex of a simplicial algebra will be a chain complex but would need more structure to get a really neat theory (see the thesis of Carassco). The starting point that I would choose would be Quillen'a paper on cohomology of algebras. It is a nice clear paper. –  Tim Porter Sep 13 '12 at 16:58
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Note that if you don't assume that your functor $F$ is additive, you can't really make a sensible left derived functor from complexes either, even if you are working in an abelian category. (For example, $F$ might not preserve the zero map!) @ZhuangXiaobo: There are almost no chain complexes of A-algebras, because maps of algebras usually have to preserve the unit. –  Tyler Lawson Sep 13 '12 at 23:38
    
Thank you very much for your warning! Things become clear to me.@Tyler Lawson –  Xiaobo Zhuang Sep 14 '12 at 0:53
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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.

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