## March 19, 2005

### Derived Categories for Dummies, Part II

#### Posted by urs I spent yesterday sitting on the beach at Vietri and reading Weibel, ‘An introduction to homological algebra’, trying to understand the details of derived functors. It takes many, many pages to really define them in detail, but here I try to summarize the key steps. Last time I reviewed the definition of the derived category $D\left(C\right)$ of any abelian category $C$. $D\left(C\right)$ is like the category $K\left(C\right)$ of (cochain) complexes in $C$ modulo chain homotopy but modded out by

(1)$q:K\left(C\right)\to D\left(C\right)$

(cf. Weibel 10.3.1)

which sends quasi-isomrphisms to true isomorphisms.

(2)$F:C\to D$

it lifts in the obvious way to a functor

(3)$KF:K\left(C\right)\to KD\phantom{\rule{thinmathspace}{0ex}}.$

The problem is to turn this into a functor

(4)$RF:D\left(C\right)\to D\left(D\right)$

in a way that respects $q$, i.e. which preserves quasi-isomorphisms.

So what one wants is to have an $RF$ which makes it true that there is a natural transformation from

(5)$F\circ q$

to

(6)$q\circ RF\phantom{\rule{thinmathspace}{0ex}}.$

If such an $RF$ exists it is called the total derived functor (10.5.1).

The idea here is simple enough. Things become more involved when this implicit definition of $RF$ is to be made more explicit. In order to do that the rest of the theory revolves around determining the cohomology of $RF$.

So if $RF$ is applied to some cochain complex $X$, what is the $n$-th cohomology of the complex $RF\left(X\right)$, i.e. what is the right hand side of

(7)${ℝ}^{n}F\left(x\right):={H}^{n}RF\left(X\right)$

?

Here the ${ℝ}^{n}F$ are called the hyper-derived functors, by definition (5.7.4).

In order to understand hyper-derived functors it is finally necessary to first understand ordinary derived functors (to be distinguished from the ‘total’ derived functors that I started with).

So the idea is this:

Given any object $A$ in an abelian catgeory we can find resolutions of this object. A resolution is an exact cochain complex

(8)$0\to A\to {I}^{0}\to {I}^{1}\to \cdots$

which hence has ${H}^{0}=A$.

(This is familiar in physics from BRST quantization, where $A$ would be the space of physical states at ghost number 0.)

There are certain such resolutions which enjoy a property called injectivity. This is the case when each of the ${I}^{n}$ is injective which in turn means that $\mathrm{Hom}\left(\cdot ,I\right)$ is an exact functor.

So here is what an ordinary right derived functor ${R}^{n}F$ does to an object $A$:

(9)${R}^{n}F\left(A\right):={H}^{n}\left(F\left({I}_{A}\right)\right)$

it assigns to it the $n$th-cohomology of the result of applying $F$ to any (fixed but arbitrary) injective resolution ${I}_{A}$ of $A$.

Fine, but that’s not enough. One needs to boost this construction by one dimension to get hyper-derived functors.

Given any chain complex $A$, we can find its right Cartan-Eilenberg resolution, which is a (cochain) complex of cochain complexes, i.e. a double complex, with injective objects everywhere, being a resolution of the complex $A$ in analogy to the above construction in an appropriate sense. The $n$th hypercohomology of this (the cohomology of the total complex) is the $n$th hyper-derived functor of $F$ applied to ${R}^{n}F\left(A\right)$.

Using all this the implicitly defined total right derived functor $RF$ can be described finally as indicated at the beginning:

Its $n$th cohomology is given by

(10)${H}^{n}RF\left(X\right)={ℝ}^{n}F\left(X\right)\phantom{\rule{thinmathspace}{0ex}}.$

Furthermore, if $X$ is the cochain representing an object $x$ then hypercohomology on $X$ coincides with cohomology

(11)${ℝ}^{n}F\left(X\right)={R}^{n}F\left(X\right)$

and at 0th position it reproduces $F$ applied to the original object

(12)${ℝ}^{0}F\left(X\right)=F\left(x\right)\phantom{\rule{thinmathspace}{0ex}}.$

This gives the relation of the total derived functor $RF$ to the original functor $F$.

Posted at March 19, 2005 11:46 AM UTC

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