How to compute derived functors

Quick description

Computations with derived functors are used in many contexts in algebra, number theory, geometry (especially complex analytic or algebraic geometry), and topology. Often one needs to compute the result of some derived functor applied to a particular object, as an end in itself, or in order to substitute the answer into another calculation that one is making. In this article we discuss some of the basic techniques for computing derived functors.

It should be noted that often one just wants to compute the order (if it turns out to be a finite group) or the dimension (if it turns out to be a vector space) of the particular value of the derived functor in question. We also discuss techniques for doing this here.

This article is incomplete. Add more examples. Discuss Euler characteristics, Herbrand quotients, etc.

Prerequisites

A knowledge of basic homological algebra, together with other relevant fields (typically at the graduate level) depending on the example.

Basic tools

Suppose that $F$ is a left exact functor (to fix ideas; but one could equally well apply this discussion to a right exact functor) on some abelian category (e.g. a category of sheaves or modules).

If $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is an exact sequence of objects, then applying $F$ and its derived functors gives a long exact sequence

$0 \rightarrow F(A) \rightarrow F(B) \rightarrow F(C) \rightarrow R^1F(A) \rightarrow R^1F(B) \rightarrow R^1F(C) \rightarrow \cdots$

$\quad \cdots \rightarrow R^iF(A) \rightarrow R^iF(B) \rightarrow R^iF(C) \rightarrow \cdots,$

where $R^i F$ denote the $i$ th right derived functor of $F$ . So if we know a lot about the values of $R^iF$ for two of the three objects in the original exact sequence, we can hope to obtain information about the values of $R^iF$ applied to the remaining object.

Example 1

The simplest situation occurs when all higher derived functors $B$ and $C$ (say) vanish. For example, suppose we wish to compute $\mathrm{Ext}^1(G,\Z/m\Z)$ in the category of abelian groups, where $G$ is a finite abelian group and $m$ is a positive integer. We put $\Z/m\Z$ in the short exact sequence

$0 \rightarrow \Z/m\Z \rightarrow \Q/\Z \rightarrow \Q/\Z \rightarrow 0,$

in which the second arrow is multiplication by $1/m$ , and the third arrow is multiplication by $m$ .

Applying $\mathrm{Hom}(G,\,)$ to this short exact sequence, we get a long exact sequence of $\mathrm{Ext}$ s. Since $\Q/\Z$ is divisible, and thus injective, we see that $\mathrm{Ext}^1(G,\Q/\Z)$ vanishes. Thus this long exact sequence of $\mathrm{Ext}$ s simplifies somewhat to the four term exact sequence

$0 \rightarrow \mathrm{Hom}(G,\Z/m\Z) \rightarrow \mathrm{Hom}(G,\Q/\Z) \rightarrow \mathrm{Hom}(G,\Q/\Z) \rightarrow \mathrm{Ext}^1(G,\Z/m\Z) \rightarrow 0,$

where in particular the third arrow is multiplication by $m$ . Now $\mathrm{Hom}(G,\Q/\Z)$ is the Pontrjagin dual $G^*$ of the finite abelian group $G$ . Thus we obtain a natural isomorphism

$\mathrm{Ext}^1(G,\Z/m\Z) \cong G^*/m G^* \cong (G[m])^*,$

where the first isomorphism comes from the long exact sequence, and the second isomorphism is a general property of Pontrjagin duality. (Here $G[m]$ denotes the $m$ -torsion subgroup of $G$ .) In general, if $G \rightarrow H$ is a homomorphism of finite abelian groups, it induces a transpose map $H^* \rightarrow G^*$ , via the formula $f^*(\chi) = \chi\circ f,$ for any $= \mathrm{Hom}(H,\Q/\Z).$ One then checks that the kernel of $f$ and the cokernel of $f^*$ are in natural duality. If $f$ is multiplication by $m$ thought of as a map from $G$ to itself, then $f^*$ coincides with multiplication by $m$ thought of as a map from $G^*$ to itself (as one verifies from the definition: $f^*(\chi)(g) = \chi(f(g)) = \chi(m g) = m\chi(g)$ ). Thus the $G[m]$ and $G^*/m G^*$ are in duality, as claimed. This gives about as an explicit solution to our computation as we could hope for. (Note that is makes sense that the answer is given in terms of the dual of $G$ , rather than directly in terms of $G$ , since $\mathrm{Ext}^1$ is a contravariant functor of $G$ .)

Example 2

The preceding example demonstrates a general principle, namely, that it is useful to have a good supply of objects some or all of the derived functors of which are known to vanish. This is one reason that vanishing theorems for derived functors are given so much prominence in the theory of various derived functors.

Here are some basic examples of vanishing results:

All higher right derived functors vanish when applied to an injective object (essentially by definition).

All higher left derived functors vanish when applied to a projective object (again, essentially by definition).

In particular, if $A$ and $B$ are modules over a ring $R$ , then $\mathrm{Ext^i}(A,B)$ vanishes if $i >0$ and either $A$ is projective (e.g. free) or $B$ is injective.

All higher $\mathrm{Tor}$ s vanish when one of the arguments is flat. (Again, essentially by definition of flat.) One gets a good supply of flat modules over any commutative ring $R$ by noting that any localization of $R$ is flat over $R$ . If $R$ is furthermore Noetherian, then the Artin-Rees theorem shows that the completion of $R$ at any ideal is flat over $R$ .

If $L/K$ is a Galois extension of fields, with Galois group $G$ , then the group cohomology $H^i(G,L)$ vanishes if $i > 0$ , and also $H^1(G,L^{\times})$ vanishes. (The latter result is Hilbert's celebrated Theorem 90.)

If $\mathcal F$ is a sheaf of abelian groups on a topological space $X$ which is supported on a finite set of closed points, then $H^i(X,\mathcal F)$ ( $i$ th sheaf cohomology) vanishes if $i > 0$ .

Example 3

The reader can examine the links given above for further examples of methods for computatings of derived functors.