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Derived functors

We have a way to derive an additive functor F:ABF : \mathcal A \to \mathcal B between abelian categories. If we have enough K-injectives, then the right derived functor is RF(X):=F(IX)\mathbf{R}F(X) := F(I_X) where XIXX \to I_X is a quasi-isomorphism with IXI_X K-injective.

Similarly, if we have enough K-projectives, then the left derived functor is LF(X):=F(PX)\mathbf{L}F(X) := F(P_X) where PXXP_X \to X is a quasi-isomorphism with PXP_X K-projective.

These provide method, at least theoretically, to compute the derived functors:

  • Figure out an explicit resolution
  • Apply FF
  • Compute homology.

However, as with most mathematical objects, it is also desirable to have an abstract characterization of derived functors.

Definition. An additive functor F:TSF: \mathcal T \to \mathcal S is called exact if there is a natural isomorphism F[1][1]FF \circ [1] \cong [1] \circ F and for any triangle XYZX[1]X \to Y \to Z \to X[1] in T\mathcal T we have a triangle FXFYFZFX[1]FX \to FY \to FZ \to FX[1] in S\mathcal S.

Any addivite functor F:ABF: \mathcal A \to \mathcal B extends to an exact functor F:K(A)K(B)F : K(\mathcal A) \to K(\mathcal B) since it takes cone complexes to cone complexes.

We would like to descend FF to an exact functor between the derived categories but there is only possible if FF was already exact as functor between the abelian categories.

Instead, we take the “closest” exact functor to FF in a precise way.

Historically, there exists (at least) two specifications for derived functors.

Definition. (Verdier) The right derived functor of an additive functor F:ABF: \mathcal A \to \mathcal B is an exact functor RF:D(A)D(B)\mathbf{R}F: D(\mathcal A) \to D(\mathcal B) together with a natural transformation jBFRFjBj_{\mathcal B} \circ F \to \mathbf{R}F \circ j_{\mathcal B} such that for any other exact functor G:D(A)D(B)G : D(\mathcal A) \to D(\mathcal B) and natural transformation jBFGjBj_{\mathcal B} \circ F \to G \circ j_{\mathcal B} there exists a unique natural transformation RFG\mathbf{R}F \to G making the diagram

There is also a corresponding left derived functor. Here jA:AD(A)AA[0]\begin{aligned} j_{\mathcal A} : \mathcal A & \to D(\mathcal A) \\ A & \mapsto A[0] \end{aligned} is the fully-faithful embedding.

Thus, we can view Verdier’s definition of RF\mathbf{R}F as the “closest” extension of FF to a quasi-isomorphism preserving functor.

Another definition, due to Deligne, has the advantage of formulating the existence RF(X)\mathbf{R}F(X) as a representability problem. Like most representability problems, if we make our category bigger we can solve it.

Fix an object XX and note that the class of quasi-isomorphism XXX \to X^\prime is filtered since we can complete

Definition. (Deligne) Let XX be an object of D(A)D(\mathcal A) then RF(X)\mathbf{R}F(X) is the object of D(B)D(\mathcal B) that represents YcolimXq.iXHomD(B)(Y,FX)Y \mapsto \operatorname{colim}_{X \underset{q.i}{\to} X^\prime} \operatorname{Hom}_{\mathcal D(\mathcal B)}(Y, FX^\prime )

If Deligne’s RF\mathbf{R}F exists, it satisfies Verdier’s specification also.

The motivation for this definition is the following fact which we won’t prove.

Proposition. We have HomD(A)(Y,X)=colimXq.iXHomK(A)(Y,X)\operatorname{Hom}_{D(\mathcal A)}(Y,X) = \operatorname{colim}_{X \underset{q.i}{\to} X^\prime} \operatorname{Hom}_{K(\mathcal A)}(Y,X^\prime)

Another fact:

Propsition. The full subcategory of objects of D(A)D(\mathcal A) where RF\mathbf{R}F is exists is a triangulated subcategory and is closed under retracts.

Let’s see that under our common assumptions that things agree.

Proposition. Assume that A\mathcal A has enough injectives and countable products. Then RF(X)F(IX)\mathbf{R}F(X) \cong F(I_X) for a quasi-isomorphism XIXX \to I_X with IXI_X K-injective.

Proof. (Expand to view)

Given any quasi-isomorphism XXX \to X^\prime, we can find a quasi-isomorphism XIX^\prime \to I^\prime with II^\prime K-injective. Thus, colimXq.iXHom(Y,FX)colimXq.iIHom(Y,FI) \operatorname{colim}_{X \underset{q.i}{\to} X^\prime} \operatorname{Hom}(Y, FX^\prime ) \cong \operatorname{colim}_{X \underset{q.i}{\to} I^\prime} \operatorname{Hom}(Y, FI^\prime ) where the later is colimit is take over all quasi-isomorphism XIX \to I^\prime with II^\prime K-injective. Given two such XIX \to I and XIX \to I^\prime, we have seen that II and II^\prime are homotopy equivalent. Thus, FIFI and FIFI^\prime are homotopy equivalent. In particular, they are quasi-isomorphic. So the colimit is constant and isomorphic to to any FIFI.

We get also have the analogous statement for left derived functors.

Definition. Given an object AAA \in \mathcal A, the i-th derived functor RiF(A)\mathbf{R}^iF(A) is RiF(A):=Hi(RF(A))\mathbf{R}^iF(A) := H^i(\mathbf{R}F(A))

So, as an example, ExtAi(A,B):=Hi(RHom(A,B))\operatorname{Ext}^i_{\mathcal A}(A,B) := H^i(\mathbf{R}\operatorname{Hom}(A,B)) Given that Hom is a bifunctor, there is possibly some ambiguity in the notation RHom(A,B)\mathbf{R}\operatorname{Hom}(A,B).

Lemma. Assume that A\mathcal A has enough projectives and enough injectives. Then, RHom(A,B)Hom(P,B)Hom(A,I)\mathbf{R}\operatorname{Hom}(A,B) \cong \operatorname{Hom}(P,B) \cong \operatorname{Hom}(A,I) where PAP \to A is a projective resolution and BIB \to I is an injective resolution.

Proof. (Expand to view)

We just need to exhibit a quasi-isomorphism. We have

with the both being quasi-isomorphisms because PP is K-projective and II is K-injective.

One of the most often used properties of RiF\mathbf{R}^iF or LiF\mathbf{L}^iF is the following.

Proposition. Let 0ABC00 \to A \to B \to C \to 0 be an exact sequence in A\mathcal A. Then we have a long exact sequence RiFARiFBRiFCRi+1FA\cdots \to \mathbf{R}^iFA \to \mathbf{R}^iFB \to \mathbf{R}^iFC \to \mathbf{R}^{i+1}FA \to \cdots

Proof. (Expand to view)

Let ϕ:AB\phi : A \to B denote the map. We have a map C(ϕ)CC(\phi) \to C which is a quasi-isomorphism. Thus, ABC A \to B \to C is isomorphic in D(A)D(\mathcal A) to a triangle and hence is one. Since RF\mathbf{R}F is exact, we have a triangle RFARFBRFCRFA[1] \mathbf{R}FA \to \mathbf{R}FB \to \mathbf{R}F C \to \mathbf{R}FA[1] Taking the long exact sequence of homology gives the conclusion.

We don’t necessarily need to use injectives or projectives to compute derived functors.

Definition. An object AA of F\mathcal F is called (right) adapted to FF if RiFA=0  for i0.\mathbf{R}^iFA = 0 \ \text{ for } i \neq 0.

Lemma. Let AAA \to A^\prime be a quasi-isomorphism of bounded below complexes whose components are adapted to FF. Then FAFAFA \to FA^\prime is a quasi-isomorphism.

Proof. (Expand to view)

[Homework](/homework/02/)

Of course any injective object is adapted for any FF. But, in general, that can be more adapted objects and it may be easier to build resolutions out of them.

Definition. Given a left RR-module MM and a right RR-module NN, the i-th Tor modules is ToriR(M,N):=Hi(MLN)\operatorname{Tor}^R_i(M,N) := H^{-i}(M \overset{\mathbf{L}}{\otimes} N)

Note that since R\otimes_R is right exact we have Tor0R(M,N)MRN.\operatorname{Tor}_0^R(M,N) \cong M \otimes_R N.

Similar to the above, we have that Tor is symmetric in the two variables.

Lemma. If PMP \to M and QNQ \to N are projective resolutions, then PRNP \otimes_R N and MRQM \otimes_R Q are quasi-isomorphic.

Proof. (Expand to view)

We have two maps PRNPRQMRQ P \otimes_R N \leftarrow P \otimes_R Q \to M \otimes_R Q which we want to check are quasi-isomorphisms. Here PRQP \otimes_R Q is a complex with (PRQ)i=a+b=iPaRQb (P \otimes_R Q)^i = \bigoplus_{a+b=i} P^a \otimes_R Q^b and δPRQ(pq)=δP(p)q+(1)apδQ(q) \delta_{P \otimes_R Q}(p \otimes q) = \delta_P(p) \otimes q + (-1)^a p \otimes \delta_Q(q) where pqPaRQbp \otimes q \in P^a \otimes_R Q^b. Consider the subcategory of chain complexes satisfying the condition that XRQXRN X \otimes_R Q \to X \otimes_R N is a quasi-isomorphism. This contains any projective module and is closed under shifts. It is also closed under taking cones. This tells us that any bounded complex of projectives lives in this category. Now let XX be a bounded above complex of projectives. Then, to consider any particular homology map, we only need to use τjX\tau_{\geq j} X for some jj. Thus, XX can also be a bounded above complex of projectives and we see the two maps are isomorphisms.

Recall that FF is flat if FRF \otimes_R - is exact.

Proposition. The following are equivalent:

  • FF is flat.
  • ToriR(F,)=0\operatorname{Tor}_i^R(F,-) = 0 for i>0i > 0.
  • Tor1R(F,)=0\operatorname{Tor}_1^R(F,-) = 0.
Proof. (Expand to view)

Assume that FF is flat. And take PMP \twoheadrightarrow M with PP projective. Then, we have a short exact sequence 0Tor1R(F,M)FRKFPFM0 0 \to \operatorname{Tor}_1^R(F,M) \to F \otimes_R K \to F \otimes P \to F \otimes M \to 0 Since FF is flat, we know that 0FRKFPFM0 0 \to F \otimes_R K \to F \otimes P \to F \otimes M \to 0 is exact also. Thus Tor1R(F,M)=0\operatorname{Tor}_1^R(F,M) = 0. We also have Tor2R(F,M)Tor1R(F,K) \operatorname{Tor}_2^R(F,M) \cong \operatorname{Tor}_1^R(F,K) As we have Tor1\operatorname{Tor}_1 vanishing for any MM, we get Tor2R(F,M)=0\operatorname{Tor}_2^R(F,M) = 0 for any MM also. Moving up the chain we get ToriR(F,)=0\operatorname{Tor}_i^R(F,-) = 0. If ToriR(F,)=0\operatorname{Tor}_i^R(F,-) = 0 for i>0i > 0, then certainly Tor1R(F,)=0\operatorname{Tor}_1^R(F,-) = 0. Assume that Tor1R(F,)=0\operatorname{Tor}_1^R(F,-) = 0. Take any exact sequence 0ABC0 0 \to A \to B \to C \to 0 and we get an exact sequence Tor1R(F,C)FRCFRBFA0 \cdots \to \operatorname{Tor}_1^R(F,C) \to F \otimes_R C \to F \otimes_R B \to F \otimes_A \to 0 Since Tor1R(F,C)=0\operatorname{Tor}_1^R(F,C) = 0, we have that 0FRCFRBFRA0 0 \to F \otimes_R C \to F \otimes_R B \to F \otimes_R A \to 0 is exact.

Let’s compute an example.

Example. We really only have one projective resolution so far: the Koszul resolution for k[x1,,xn]/(x1,,xn)k[x_1,\ldots,x_n]/(x_1,\ldots,x_n). In the tensor product, K(x1,,xn)k[x1,,xn]kK(x_1,\ldots,x_n) \otimes_{k[x_1,\ldots,x_n]} k and the maps reduce to 00 since their images land in the submodule generated by (x1,,xn)(x_1,\ldots,x_n). So Hi(K(x1,,xn)k[x1,,xn]k)=Kik=k(ni)H^{-i}\left(K(x_1,\ldots,x_n) \otimes_{k[x_1,\ldots,x_n]} k \right) = K^i \otimes k = k^{n \choose i}