# Explicit computations for derived functors

Let $F$ be a left exact functor from the category of sheaves of abelian groups to the category of abelian groups, $\mathscr{F}$ a sheaf of abelian groups on a topological space $X$. Since injective resolutions always exist, and acyclic ones are sent to acyclic ones, we may define the right derived functor $RF$ by $RF(\mathscr{F}):=F(I^\cdot)$ for any injective resolutions $\mathscr{F}\to I^\cdot$.

However, though they exist, an injective resolution is usually messy, so we often do not use it for computations. An example is we use $\check{C}$ech resolution to compute for $F=(f:X\to\{pt\})_*$; note that $R^iF(\mathscr{F})=H^i(X,\mathscr{F})$. More precisely, we pick a good open cover $\mathcal{U}$ for $X$, and then we have $$H^i(X,\mathscr{F}) = \check{H}^{\,i}(\mathcal{U},\mathscr{F}) \mbox{ [Harshorne A.G. ex.III.4.11].}$$ For good enough spaces, we can pick such a good cover and compute the right-hand side precisely.

Questions:

1. In general is there a way to compute a derived functor first by resolving by a Cech complex with a good cover?

2. If we cannot expect my first question to be true, is it at least possible for some specific functors?

3. When $F=(f:X\to Y)_*$, we have a clearer description: $R^iF(\mathscr{F})$ is the sheaf that associates to the presheaf $$V\mapsto H^i(f^{^1}(V), \mathscr{F}|_{f^{-1}(V)}) \mbox{ [Harshorne A.G. III.8.1].}.$$ Good! this makes things more explicit! I notice that this result can be obtained by my first question (if it is true). However, Hartshorne uses a quite complicated proof that refers to other concepts such as "effacable", "universal $\delta$-functors". I also found that if I prove it directly by definition, it will be a mess (Homology sheaf is a quotient by the image sheaf.. so you have to take two sheafifications!). Is there a plain explanation?

4. For the third question, is it possible to get an explicit result just for $RF(\mathscr{F})$ but not $R^iF(\mathscr{F})$? I would like to know since $RF(\mathscr{F})$ contains more information.

1) : This is true and is a consequence of the Čech-to-derived functor spectral sequence (see the wikipedia page, where they mention that we can use this to compute derived functors).

2) : see 1)

3) I'm not sure I really understand this part, but taking a (or two) sheafification is not really a big deal, and indeed this description is really useful in particular when you are taking the stalks.

4) You can't obtain $RF(\mathcal F)$ unless some really special case, since you would need an explicit resolution which is hard to produce. However there is a case when you can compute it : Deligne's highly non-trivial theorem tells you that $$Rf_* \underline{\Bbb Q}_X = \bigoplus_i R^if_*\underline{\Bbb Q}_X[-i]$$ when $f : X \to Y$ is a proper morphism between projective smooth varieties. Concretely this means that $Rf_* \underline{\Bbb Q}_X$ is isomorphic as a complex to its cohomology in the right degree, with zero differential. The proof uses tools from Hodge theory.

a) Let me add that projective (i.e by locally free sheaves) resolutions are more easy to find, and you can compute the full complex $R\text{Hom}$ or the derived tensor product very explicitely. For example if $X \subset \Bbb P^n$ is an hypersurface, the exact sequence $$0 \to \mathcal{O}_{\Bbb P^n}(-X) \to \mathcal{O}_{\Bbb P^n} \to \mathcal{O}_X \to 0$$ can be interpreted as a $2$-step projective resolution of the coherent sheaf $\mathcal O_X$. Since $\text{Hom}$ is a bifunctor, you can choose to take a projective resolution, rather than an injective one :-)
• Thanks a lot for your thorough explanation! I took some time to digest and read more about this. After learning more stuff in this field, I realized why I asked this question a month ago: I thought we NEED injective resolution in order to define derived functors. Now I understand that usually we do not (it depends on the functors of interest). For example, to derive $\Gamma$ we only need a soft resolution! – Student Oct 7 '18 at 11:46