Huybrechts, D - Fourier-Mukai Transforms in Algebraic Geometry (Cours DEA 2004)

1Cours de DEA 2004

FourierMukai Transforms in Algebraic Geometry

Daniel Huybrechts

Institut de Mathematiques Jussieu, Universite Paris 7, France

Table of Contents

1 Triangulated categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Additive categories and functors . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Triangulated categories and exact functors . . . . . . . . . . . . . . . . . 91.3 Equivalences of triangulated categories . . . . . . . . . . . . . . . . . . . . 161.4 Exceptional sequences and semi-orthogonal decompositions . . 19

2 Derived categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.1 Derived category of an abelian category . . . . . . . . . . . . . . . . . . . 232.2 Derived functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.3 Spectral sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.4 Derived category of coherent sheaves on a scheme . . . . . . . . . . 482.5 Derived functors in algebraic geometry . . . . . . . . . . . . . . . . . . . . 56

3 Derived categories and the canonical bundle I . . . . . . . . . . . . 693.1 Positive (anti-)canonical bundle . . . . . . . . . . . . . . . . . . . . . . . . . . 703.2 Autoequivalences for positive (anti-)canonical bundle . . . . . . . 773.3 Ample sequences in derived categories . . . . . . . . . . . . . . . . . . . . . 78

4 FourierMukai transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 894.1 What it is and Orlovs result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 894.2 Passage to cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

5 Derived categories and the canonical bundle II . . . . . . . . . . . 1055.1 Kodaira dimension and canonical ring . . . . . . . . . . . . . . . . . . . . . 1055.2 Nefness under derived equivalence . . . . . . . . . . . . . . . . . . . . . . . . 1075.3 Derived equivalence and birationality . . . . . . . . . . . . . . . . . . . . . 110

6 Equivalence criteria for FourierMukai transforms . . . . . . . . 1156.1 Fully faithful . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1156.2 Equivalences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

7 Spherical and exceptional objects . . . . . . . . . . . . . . . . . . . . . . . . . 1237.1 Autoequivalences induced by spherical objects . . . . . . . . . . . . . 1237.2 Braid group actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1247.3 Beilinson spectral sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

4 Table of Contents

8 Abelian varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1298.1 Basic denitions and facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1298.2 The Poincare bundle as a FourierMukai kernel . . . . . . . . . . . . 1368.3 Sl2-action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1398.4 Derived equivalences of abelian varieties . . . . . . . . . . . . . . . . . . . 1428.5 The group of autoequivalences . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

9 K3 surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1519.1 Recap: K3 surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1519.2 Equivalences of derived categories of K3 surfaces . . . . . . . . . . . 1539.3 Recap: moduli spaces of sheaves . . . . . . . . . . . . . . . . . . . . . . . . . . 157

10 Flips and flops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16110.1 Derived categories under blow-up . . . . . . . . . . . . . . . . . . . . . . . . . 16110.2 The standard ip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16610.3 The Mukai op . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16910.4 General ops in dimension three . . . . . . . . . . . . . . . . . . . . . . . . . . 172

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

1 Triangulated categories

Intro1.1 Additive categories and functors

We suppose that the reader is familiar with the notion of a category and ofa functor between two categories. Let us nevertheless recall a few things. Ifnot otherwise stated all functors are covariant.

Definition 1.1 Let A and B be two categories. A functor F : A B isfull if for any two objects A,B A the induced map F : Hom(A,B) Hom(F (A), F (B)) is surjective. The functor F is called faithful if this mapis injective for any A,B A.Definition 1.2 Two functors F, F : A B are isomorphic if there exists amorphism of functors : F F such that for any object A A the inducedmorphism A : F (A) F (A) is an isomorphism (in B). Equivalently, Fand F are isomorphic if there exist functor morphisms : F F and : F F with = id and = id.Definition 1.3 A functor F : A B is called an equivalence if there existsa functor F1 : B A such that F F1 is isomorphic to idB and F1 Fis isomorphic to idA. One calls F1 an inverse or, sometimes, quasi-inverseof F .

Two categoriesA and B are called equivalent if there exists an equivalenceF : A B.

Clearly, any equivalence is fully faithful. A partial converse is provided by

Proposition 1.4 Let F : A B be a fully faithful functor. Then F is anequivalence if and only if every object B B is isomorphic to an object ofthe form F (A) for some A A.

Proof. In order to dene the inverse functor F1, one chooses for any B Ban object AB A together with an isomorphism B : F (AB) B. Then,let F1 : B A be the functor that associates to any object B Bthis distinguished object AB A and for which F1 : Hom(B1, B2) Hom(F1(B1), F1(B2)) is given by the composition of Hom(B1, B2) =Hom(F (A1), F (A2)), f 1B2 f B1 and the inverse of the bijectionF : Hom(A1, A2) Hom(F (A1), F (A2)).

2 1 Triangulated categories

The isomorphisms F F1 = idB and F1 F = idA are given by theisomorphisms B .

The proposition immediately yields

Corollary 1.5 Any fully faithful functor F : A B defines an equivalencebetween A and the full subcategory of B of all objects B B isomorphic toF (A) for some A A.

In the following proposition we let Fun(A) be the category of all con-travariant functors, i.e. the objects are functors F : Aop Set and mor-phisms are functor morphisms. Consider the natural functor

A Fun(A)

that associates to any object A A the functor Hom( , A) : Aop Set,B Hom(B,A).Proposition 1.6 (Yoneda lemma) This functor A Fun(A) defines anequivalence of A with the subcategory of representable functors F , i.e. functorsisomorphic to some Hom( , A). In particular, A Hom( , A) is fullyfaithful.

Proof. See [18, II.3]. We will rarely work with completely arbitrary categories. All our cate-

gories will at least be additive.

Definition 1.7 A category A is an additive category if for every two objectsA,B A the set Hom(A,B) is endowed with the structure of an abeliangroup such that the following three conditions are satised:

i) The compositions Hom(A1, A2) Hom(A2, A3) Hom(A1, A3) arebilinear.

ii) There exists a zero object 0 A, i.e. an object 0 such that Hom(0, 0)is the trivial group with one element.

iii) For any two objects A1, A2 A there exists an object B A withmorphisms ji : Ai B and pi : B Ai, i = 1, 2, which make B the directsum and the direct product of A1 and A2

Exercise 1 Show that for any object A A in an additive category A thereexist unique morphisms 0 A and A 0.

A functor F : A B between additive categories A and B will usually as-sumed to be additive, i.e. the induced maps Hom(A,B) Hom(F (A), F (B))are group homomorphisms.

Everything that has been said so far carries over to additive categories.In particular, an additive functor F : A B which is an equivalence is infact an additive equivalence, i.e. the inverse functor F1 is additive as well.

1.1 Additive categories and functors 3

The Yoneda lemma is modied as follows: For an additive category A we letFun(A) be the category of all contravariant additive functors F : A Ab,where Ab is the category of abelian groups. Then the Yoneda lemma in theform of Proposition 1.6 remains true.

We will go one step further. As the categories we will eventually be interes-ted in have geometric origin, i.e. are dened in terms of certain varieties oversome base eld, we usually deal with the following special type of additivecategories. In the following we denote by k an arbitrary eld.

Definition 1.8 A k-linear category is an additive category A such that thegroups Hom(A,B) are k-vector spaces and all compositions are k-bilinear.

Additive functors between two k-linear additive categories over a commonbase eld k will assumed to be k-linear, i.e. for any two objects A,B A themap F : Hom(A,B) Hom(F (A), F (B)) is k-linear.

Once again, everything that has been mentioned before carries over lite-rally to additive categories over a eld. Usually we will state all abstractresults for additive categories, but in the applications everything will be overa base eld. In principle though, it could happen that two k-linear categoriesare equivalent as ordinary additive categories without being equivalent ask-linear categories.

The Yoneda lemma can again be adjusted to the situation: This time,one considers the category of contravariant k-linear functors from A into thecategory Vec(k) of k-vector spaces.

Definition 1.9 An additive category A is called abelian if also the followingcondition holds true:

iv) Every morphism f Hom(A,B) admits a kernel and a cokernel andthe natural map Coim(f) Im is an isomorphism.

Recall that the image Im(f) is a kernel for a cokernel B Coker(f) andthe coimage Coim(f) is a cokernel for a kernel Ker(f) A. So, condition iv)says that for any f : A B there exists the following diagram

Ker(f) i Af

B Coker(f).

Coker(i)= Ker()

In particular, the notion of exact sequences is usually only considered forabelian categories. We recall that A1

f1 A2 f2 A3 is exact if and only ifKer(f2) = Im(f1).

Examples 1.10 i) Let R be a commutative ring, then the category Mod(R)of R-modules is abelian. The full subcategory of nitely generated modulesis abelian as well.


ii) Let X be a topological space. Then the category of sheaves of abeliangroups Sh(X) is abelian. If a sheaf of commutative rings on X is xed, thenthe subcategory of sheaves of modules over this sheaf of rings is again abelian.

iii) Let X be a scheme. Then the categories Coh(X) and Qcoh(X) of allcoherent respectively quasi-coherent sheaves on X are both abelian.

Suppose F : A B is an additive functor between abelian categories. Inparticular, any sequence

A1f1 A2

f2 A3

with f2 f1 = 0 (or, in other words, Im(f1) Ker(f2)) is mapped to

F (A1)F (f1) F (A2)

F (f2) F (A3)

again with F (f2) F (f1) = F (f2 f1) = 0.Definition 1.11 The functor F is left (right) exact if any short exact se-quence

0 A1f1 A2

f2 A3 0

is mapped to a sequence

0 F (A1)F (f1) F (A2)

F (f2) F (A3) 0

which is exact except possibly in F (A3) (resp. F (A1)). The functor is exactif it is left and right exact.

Examples 1.12 i) Let A be an abelian category and A0 A. Then

Hom(A0, ) : A Ab

is a left exact functor. The contravariant functor

Hom( , A0) : A Ab

is also left exact.ii) Recall that an object P A is called projective if Hom(P, ) is right

exact (and hence exact). An object I A is called injective if Hom( , I) isright exact (and hence exact).

iii) Free modules over a ring R are projective objects in Mod(R). But(locally) free sheaves in Coh(X) are almost never projective.


Definition 1.13 Let F : A B be a functor between arbitrary categories.A functor H : B A is right adjoint to F (one writes F H) if there

exist isomorphisms

Hom(F (A), B) = Hom(A,H(B))

for any two objects A A and B B which are functorial in A and B.A functor G : B A is left adjoint to F (one writes G F ) if there exists

isomorphisms Hom(B,F (A)) = Hom(G(B), A) for any two objects A Aand B B which are functorial in A and B.

Clearly, if H is right adjoint to F then F is left adjoint to H .

Remarks 1.14 i) Suppose F H . Then the identity morphism idF (A) Hom(F (A), F (A)) induces a morphism A H(F (A)). The naturality ofisomorphisms in the denition of the adjoint functor ensures that these mor-phisms dene a functor morphism

h : idA H F.

In the same vein, inserting A = H(B) yields a canonical morphism F (H(B)) B and, therefore, a functor morphism

g : F H idB.

ii) Using the Yoneda lemma 1.6, one veries that a left (or right) adjointfunctor, if it exists at all, is unique up to isomorphism.

iii) If F is an additive functor (in particular, A and B are additive),then one requires the isomorphisms Hom(F (A), B) = Hom(A,H(B)) to beisomorphisms of abelian groups. Similar, if everything is k-linear, then alsothese isomorphisms are required to be k-linear. A priori, an adjoint functorof an additive functor need not be additive, but only those will usually occur.

iv) If A and B are abelian categories and F : A B is left adjoint toH : B A then F is right exact and H is left exact.Exercise 2 Suppose F H . Show that

f H(f) hA : A H(F (A)) H(B)

describes the adjunction morphism Hom(F (A), B) = Hom(A,H(B)).

Exercise 3 Prove the assertion iv).

Exercise 4 Suppose F H . Show that for the induced morphisms g : F H id and h : id H F the composition H (H F ) H = H (F H) Hof hH( ) and H(g) is the identity. See [30, IV.1] and [18, II.3] for a converse.


Example 1.15 Let f : X Y be a morphism between two schemes X andY . Then f : Qcoh(Y ) Qcoh(X) is right exact and f : Qcoh(X) Qcoh(Y ) is left exact. Moreover, f f. If f is proper, the same holds forthe categories of coherent sheaves on X and Y .

Lemma 1.16 Let F : A B be a functor and G F . Then the inducedfunctor morphism g : G F idA induces for any A,B A the followingcommutative diagram

Hom(A,B)

gA

F

Hom(G(F (A)), B) Hom(F (A), F (B))

Here, the isomorphism is given by adjunction.Similarly, if F H then the natural functor morphism h : idA H F

induces for all A,B A the following commutative diagram:

Hom(A,B)hB

F

Hom(A,H(F (A)))

Hom(F (A), F (B))

Again, the isomorphism is given by adjunction.

Proof. As G F , the following diagram commutes for all : A B and allC B :

Hom(G(C), A)

Hom(C,F (A))

F ()

Hom(G(C), B) Hom(C,F (B))

Applied to C = F (A) it yields

Hom(G(F (A)), A)

Hom(F (A), F (A))


Clearly, under the second vertical homomorphism the identity idF (A) Hom(F (A), F (A)) is mapped to F () Hom(F (A), F (B)). On the otherhand, its image under

Hom(F (A), F (A)) = Hom(G(F (A)), A) Hom(G(F (A)), B)


is just gA.This proves the commutativity of the lower triangle. The commutativity

of the upper one is proved similarly.

Corollary 1.17 Suppose a fully faithful functor F : A B admits a leftadjoint G F . Then the natural functor morphism G F idA is anisomorphism.

Similarly, if a fully faithful functor F : A B admits a right adjointF H, then the natural functor morphism idA H F is an isomorphism.

Proof. Since F : Hom(A,B) Hom(F (A), F (B)) is bijective, the commu-tativity of the diagram above proves that G F idA induces bijectionsHom(A,B) Hom((G F )(A), B) for all A and B. By the Yoneda lemma1.6, this shows that G F idA is an isomorphism.

The proof of the second statement is similar. The same arguments also show the converse:

Corollary 1.18 Let F : A B and G : B A be two functors such thatG F . If the induced functor morphism G F idA is an isomorphism,then F is fully faithful.

Similarly, if F H such that idA H F is an isomorphism, then F isfully faithful.

Remark 1.19 In short, if F H , then:F is fully faithful h : idA H F

and if G F , then:F fully faithful g : G F idA.

Exercise 5 Suppose G F H and F fully faithful. Construct a canonicalhomomorphism H G.

In many cases, adjoint functors exist. The case that interests us most,is the case of equivalences. Here, the existence of left and right adjoints isgranted by the following general result.

Proposition 1.20 Let F : A B be an equivalence of categories. Then Fadmits a left adjoint and a right adjoint. More precisely, if F : B A is aninverse functor of F then F F F .

Proof. Very roughly, this is due to the following sequence of canonical isomor-phisms Hom(F (A), B) = Hom(F (F (A)), F (B)) = Hom(A,F (B)), wherewe use F (F (A)) = A.


Remark 1.21 These results justify the approach that is usually followed whenproving the equivalence of certain categories: One rst tries to construct anatural functor F and a left adjoint G F (or right adjoint F H). Thenone veries that the adjunction morphisms are bijective, which proves thatthe functor F is fully faithful. Eventually, one has to ensure that any objectin the target category is isomorphic to an object in the image of F .

Definition 1.22 Let A be an k-linear category. A Serre functor is a k-linearequivalence S : A A such that for any two objects A,B A there existsan isomorphism A,B : Hom(A,B)

Hom(B,S(A)) (of k-vector spaces)which is functorial in A and B.

If needed, the induced pairing Hom(A,B) Hom(B,S(A)) k will bewritten f g.Remark 1.23 In the original paper by Bondal and Kapranov [6] an additionalcondition was required, namely that for any two objects A,B A the fol-lowing diagram is commutative:

Hom(A,B)A,B

S

Hom(B,S(A))

Hom(S(A), S(B))S(A),S(B) Hom(S(B), S2(A))

S

In fact, it turns out that this is automatic1. Indeed, inserting the additionaldiagonal arrow B,S(A) : Hom(S(A), S(B)) Hom(B,S(A)) induced bythe dening property of a Serre functor, one reduces to the commutativity ofthe two triangles. Thus one has to show that

Hom(A,B)

S

A,B

Hom(B,S(A))

Hom(S(A), S(B))B,S(A)

is commutative or, in other words that for f Hom(B,S(A)) and g Hom(A,B) one has f g = S(g) f, where the bracket indicates thepairing dened by . Since is functorial in the second component, we havethe commutative diagram1 Thanks to Raphael Rouquier for explaining this to me.

1.2 Triangulated categories and exact functors 9

Hom(A,B)A,B Hom(B,S(A))

Hom(B,B)

g

B,B Hom(B,S(B))

(S(g))

which is just saying f g = S(g) f (start with the identity idB in thelower left corner).

In order to avoid any trouble with the dual, one usually assumes that allHoms in A are nite dimensional. Under this hypothesis it is easy to see thata Serre functor, if it exists, is unique up to isomorphism. More generally onehas the following

Lemma 1.24 Let A and B be k-linear categories over a field k with finitedimensional Homs. If A and B endowed with a Serre functor SA respectivelySB. Then any k-linear equivalence F : A B commutes with Serre duality,i.e. there exists an isomorphism SB F = F SA.

Proof. This is an application of the Yoneda lemma 1.6: Since F is fullyfaithful, one has Hom(A,S(B)) = Hom(F (A), F (S(B))) and Hom(B,A) =Hom(F (B), F (A)) for any two objects A,B A.

Together with the two isomorphisms Hom(A,S(B)) = Hom(B,A) andHom(F (B), F (A)) = Hom(F (A), S(F (B)), this yields a functorial (in Aand B) isomorphism Hom(F (A), F (S(B))) = Hom(F (A), S(F (B))). Usingthe hypothesis that F is an equivalence and, in particular, that any objectin B is isomorphic to some F (A), one concludes that there exists a functorisomorphism F SA = SB F .

Remark 1.25 Let F : A B be a functor between categories A and Bendowed with Serre functors SA respectively SB. If G F then F SA G S1B . (As before we assume that all Homs are nite dimensional.)Indeed, under the given assumptions we have the following functorial iso-morphisms Hom(A1, (SA G S1B )(A2)) = Hom((G S1B )(A2), A1) =Hom(SB(A2), F (A1)) = Hom(F (A1), SB(S1B (A2))) = Hom(F (A1), A2).

A similar argument allows the construction of a left adjoint if a rightadjoint F H is given. In particular, for functors between categories withSerre functors the existence of the left respectively right adjoint implies theexistence of the other one.

1.2 Triangulated categories and exact functors

Let us right away start with the denition of a triangulated category.


Definition 1.26 Let D be an additive category. The structure of a triangu-lated category on D is given by an additive equivalence T : D D, the shiftfunctor, and a set of distinguished triangles A B C T (A) subject tothe axioms TR1 TR4.

Before actually explaining the four axioms TR1-4, let us introduce the no-tation A[1] := T (A) for any object A D and f [1] := T (f) Hom(A[1], B[1])for any morphism f Hom(A,B). Similarly, one writes A[n] := T n(A) andf [n] := T n(f) for n Z. Thus, a triangle will be denoted by A B C A[1]. A morphism between two triangles A B C A[1] andA B C A[1] is given by a commutative diagram

A

f

B

g

C

h

A[1]

f [1]

A B C A[1]

It is an isomorphism if f, g, and h are isomorphisms.Here are the axioms for a triangulated category:

TR1 i) Any triangle of the form

Aid A 0 A[1]

is distinguished.ii) Any triangle isomorphic to a distinguished triangle is distinguished.iii) Any morphism f : A B can be completed to a distinguished triangle

Af B C A[1] .

TR2 The triangle

Af B

g Ch A[1]

is a distinguished triangle if and only if

Bg C

h A[1]f [1] B[1]

is a distinguished triangle.

TR3 Suppose there exists a commutative diagram of distinguished triangleswith vertical arrows f and g:

A

f

B

g

C

h

A[1]

f [1]

A B C A[1]


Then the diagram can be completed to a commutative diagram, i.e. to amorphism of triangles, by (a not necessarily unique) morphism h.

TR4to be completed

The rst two axioms TR1 and TR2 seem very natural. Essentially, theyare saying that the set of distinguished triangle is preserved under shift andisomorphism and that there are enough distinguished triangles available. Thethird one, TR3, seems a little less so, due to the non-uniqueness of the com-pleting morphism. It is in fact this axiom that makes people believe that thenotion of a triangulated category is still not optimal and should be replacedby something more canonical. The last axiom, TR4, is crucial but not usedvery often in these notes (an exception is the proof of the existence result ofOrlov, Theorem 4.11).

Note that a priori we have not required that in a triangle A B C A[1] the composition A C is zero. But this can be easily deduced fromTR3.

Exercise 6 Prove the last statement.

Proposition 1.27 Let A B C A[1] be a distinguished triangle in atriangulated category D. Then for any object A0 D the induced sequencesare exact:

Hom(A0, A) Hom(A0, B) Hom(A0, C)

Hom(C,A0) Hom(B,A0) Hom(A,A0)

Due to TR2, also Hom(A0, B) Hom(A0, C) Hom(A0, A[1]) is exactand similarly for Hom( , A0). Thus, one obtains in fact long exact sequences.

Exercise 7 Consider a commutative diagram of distinguished triangle

A

f

B

g

C

h

A[1]

f [1]

A B C A[1]

Show that if two of the vertical morphisms are isomorphisms then so is thethird. Also note that f and g might be zero without h being so.

Exercise 8 Let A B C A[1] be a distinguished triangle in a trian-gulated category D. Suppose that C A[1] is trivial. Show that then thetriangle is split, i.e. is given by a direct sum decomposition B = A C.


Definition 1.28 An additive functor

F : D D

between triangulated categories D and D is called exact ifi) There exists a functor isomorphism

F TD TD F

ii) Any distinguished triangle

A B C A[1]

in D is mapped to a distinguished triangle

F (A) F (B) F (C) F (A)[1] ,

where F (A[1]) is identied with F (A)[1] via the functor isomorphism in i).

Remark 1.29 Once again, the notions of a triangulated category and of anexact functor have to be adjusted when one is interested in additive categoriesover a eld k. In this case, the shift functor should be k-linear and one usuallyconsiders only k-linear exact functors.

Also note, that in this case the two long exact cohomology sequencesin Proposition 1.27 associated to a distinguished triangle are long exact se-quences of k-vector spaces.

Proposition 1.30 Let F : D D be an exact functor between triangulatedcategories. If F H, then H : D D is exact.

Similarly, if G F then G : D D is exact.See [6, 36]

Proof. Let us rst show that the adjoint functor H commutes with the shiftfunctors T on D and T on D. Since F is exact, one has an isomorphismF T = T F and also F T1 = T 1 F .

This yields the following functorial isomorphisms

Hom(A,H(T (B))) = Hom(F (A), T (B)) = Hom(T 1(F (A)), B)= Hom(F (T1(A)), B) = Hom(T1(A), H(B))= Hom(A, T (H(B))).

As everything is functorial, the Yoneda lemma yields an isomorphism

H T T H .

Next, we have to show that H maps a distinguished triangle D to a distin-guished triangle in D. Let A B C A[1] be a distinguished triangle


in D. The induced morphism H(A) H(B) can be completed to a dis-tinguished triangle H(A) H(B) C0 H(A)[1]. Here we tacitly useH(A[1]) = H(A)[1] given by the above isomorphism H T = T H .

Using the assumption that F is exact and the adjunction morphismsF (H(A)) A and F (H(B)) B, one obtains a commutative diagram ofdistinguished triangles

F (H(A))

F (H(B))

F (C0)

F (H(A))[1]

A B C A[1]

which can be completed by the dotted arrow according to the axiom TR3.Applying H to the whole diagram and using the adjunction id H F ,

one gets

H(A)

H(B)

C0

hC0

H(A)[1]

H(F (H(A)))

H(F (H(B)))

H(F (C0))

H()

H(F (H(A)))[1]

H(A) H(B) H(C) H(A)[1]

Here, the curved vertical arrows are in both cases the identity morphisms(see Exercise 4). To conclude one would like to apply Exercise 7, but we arenot allowed to use that H(A) H(B) H(C) H(A)[1] is distinguished.So, we have to use once more that F is exact in order to make sure that

Hom(A0, H(B)) Hom(A0, H(C)) Hom(A0, H(A)[1])= Hom(F (A0), B) = Hom(F (A0), C) = Hom(F (A0), A[1])is exact. Then we obtain Hom(A0, C0) = Hom(A0, H(C)) for all A0 andhence H() hC0 : C0 H(C). Thus, H(A) H(B) H(C) H(A)[1]is isomorphic to the distinguished triangle H(A) H(B) C0 H(A)[1],so it is itself distinguished.

A subcategory D D of a triangulated category is a triangulated sub-category if D admits the structure of a triangulated category such that theinclusion is exact. If D D is a full subcategory, then D is a triangulatedsubcategory if and only if D is invariant under shift and for any distinguishedtriangle A B C A[1] in D with A,B D also C D.Definition 1.31 A full triangulated subcategory D D is called admissibleif the inclusion has a right adjoint : D D, i.e. there exists functorialisomorphisms HomD(A,B) = HomD(A, (B)) for all A D and B D.


The orthogonal complement of a(n admissible) subcategory D D is thefull subcategory D of all objects A D such that Hom(B,A) = 0 for allB D.Remarks 1.32 i) The right adjoint functor : D D for an admissible fulltriangulated subcategory D D is exact.

ii) The orthogonal complement of an admissible subcategory is a triangu-lated subcategory. Indeed, the condition Hom(B,C) = 0 yields Hom(B,C[i]) =Hom(B[i], C) = 0, as D is invariant under shift. If C1 C2 C3 C1[1]is a distinguished triangle in D with C1, C2 D then the long exact se-quence obtained from applying Hom(B, ) shows that also C3 D.

iii) More explicitly, one shows that a full triangulated subcategoryD Dis admissible if and only if for all A D there exists a distinguished triangleB A C B[1] with B D and C D.

SupposeD is admissible. By adjunction, the identity in HomD((A), (A))denes a morphism B := (A) A which can be completed to a distin-guished triangle B A C B[1]. In order to see that indeed C D,one applies Hom(B, ) to it and uses Hom(B, B) = Hom(B, (A)) =Hom(B, A) for all B D.

Conversely, if such a distinguished triangle is given for any A, then onedenes the functor : D D by (A) = B. One uses Hom(B,C) = 0 forB D and C D to show that B does not depend (up to isomorphism)on the choice of the triangle.

iv) Admissible subcategories occur whenever there is a fully faithful exactfunctor F : D D that admits a right adjoint. Indeed, in this case thefunctor F denes an equivalence between D and an admissible subcategoryof D.

The notion of equivalence that will be important for us is the following.

Definition 1.33 Two triangulated categoriesD andD are equivalent if thereexists an exact equivalence F : D D.

So usually one rst tries to nd an exact functor F . In order to show thatF denes an equivalence of triangulated categories, one has to verify that F ,as an additive functor, is fully faithful and any object in D is isomorphic toone in the image.

We conclude this section by a discussion of Serre functors in the contextof triangulated categories. The following result was proved by Bondal andKapranov [6]:

Proposition 1.34 Any Serre functor on a triangulated category over a fieldk is exact.


Proof. By Lemma 1.24, a Serre functor S commutes with the shift functor T ,as T is an autoequivalence. It remains to show that under S a distinguishedtriangle A

f B g C h A[1] is mapped to a distinguished triangle. In a rststep, one completes S(A) S(B) to a distinguished triangle

S(A)S(f) S(B)

C0 S(A)[1]

Next, one tries to construct a commutative diagram

S(A)

=

S(B)

=

C0

S(A)[1]

=

S(A) S(B)

S(g) S(C)S(h) S(A[1]).

The compatible long exact sequences, induced by applying Hom(D, ) to thehorizontal sequences, and the Yoneda lemma would then show that : C0 S(C) must be an isomorphism. (Note that the long sequence induced bythe bottom sequence is dual to the long exact sequence induced by applyingHom( , D) to the distinguished triangle A B C A[1] and hence itselfexact.)

It remains to prove the existence of . Via Serre duality can be consid-ered as a linear form on Hom(C,C0). The two commutativity conditionsi) = S(g) and ii) S(h) = can be expressed as follows: i) isequivalent to saying that the linear form : Hom(C,C0) k composedwith : Hom(C, S(B)) Hom(C,C0) coincides with the linear formy : Hom(C, S(B)) Hom(B,S(B)) k, where the last map is the pairingidB . Similarly, ii) is equivalent to saying that ( h) : Hom(A[1], C0) Hom(C,C0) k equals x : Hom(A[1], C0) Hom(A[1], SA[1]) k givenby idA .

In order to ensure the existence of the linear form : Hom(C,C0) k, itsuces to show that for any Hom(C, S(B)) and Hom(S(A)[1], C0)with = h one has y() = x(). First, TR3 shows that there existsa commutative diagram

C

h A[1]

f [1] B[1]

[1]

g[1] C[1]

[1]

S(B)

C0 S(A)[1]

S(f)[1] S(B)[1]

But then y() = idB ( g) = idB (S(f) ) = f , where forthe last equality we used the functoriality of the Serre pairing Hom( , B) =Hom(B,S( )). Similarly, one nds x() = idA ( ) = f .


1.3 Equivalences of triangulated categories

In this section we discuss criteria that allow to decide whether a given exactfunctor is fully faithful or even an equivalence. This continues the discussionof Remark 1.21 in the context of triangulated categories.

Let us begin with the denition of a spanning class. In many geometricsituations, a spanning class can be found (cf. Proposition 2.54 or Corollary2.59).

Definition 1.35 Let D be a triangulated category. A collection Ob(D)spans D if for all B D the following two conditions hold:

i) If Hom(A,B[i]) = 0 for all A and all i Z then B = 0.ii) If Hom(B[i], A) = 0 for all A and all i Z then B = 0.

Proposition 1.36 Let F : D D be an exact functor between triangulatedcategories. Assume F admits a left adjoint G and a right adjoint H. If Ob(D) spans D and for all objects A,B the natural homomorphismsF : Hom(A,B[i]) Hom(F (A), F (B[i])) are bijective for all i Z, then Fis fully faithful. See [8, 36]

Proof. First recall that H and G are both exact due to Proposition 1.30. Thiswill be used throughout.

We shall use the following commutative diagram (see Lemma 1.16):

Hom(A,B)hB

gA

F

Hom(A,H(F (B)))


(1.1)

for all A,B D.We rst show that for any A the natural homomorphism gA :

G(F (A)) A is an isomorphism. In order to see this, one chooses a dis-tinguished triangle

G(F (A))gA A C G(F (A))[1]

Applying Hom( , B) for an arbitrary B D induces a long exact sequencewhich combined with the commutative lower triangle yields

Hom(C,B[i]) Hom(A,B[i])

F

Hom(G(F (A)), B[i])

Hom(FA,FB[i])

1.3 Equivalences of triangulated categories 17

If B , the maps F : Hom(A,B[i]) Hom(FA,FB[i]) are bijective byassumption. Hence, Hom(C,B[i]) = 0 for all i Z and all B . Since spans D, one nds C = 0 and, therefore, gA : G(F (A)) = A.

Note that this immediately implies that for A and any B Din fact all homomorphisms in the commutative diagram (1.1) are bijective.In particular, hB : Hom(A,B[i]) Hom(A,H(F (B[i]))). This applied toB D and using a distinguished triangle of the form (again use TR1 for itsexistence)

BhB H(F (B)) C B[1]

shows that Hom(A,C[i]) = 0 for all i Z and all A . Hence, C = 0and, thus, hB : B

H(F (B)). In particular, hB : Hom(A,B) Hom(A,H(F (B))) is bijective for any A D. Using the commutativity ofthe upper triangle, this proves that F : Hom(A,B) Hom(F (A), F (B)) isbijective for all A,B D, i.e. F is fully faithful. (The last argument was alsostated as Corollary 1.18.)

Suppose we know already that the functor is fully faithful. What do weneed to know in order to be able to decide whether it is in fact an equivalence?The following lemma provides a rst criterion, which however is often not easyto check. But we will build upon the arguments used in its proof to deduceProposition 1.39, which is extremely useful.

Lemma 1.37 Let F : D D be a fully faithful exact functor between trian-gulated categories and suppose that F has a right adjoint F H. Then F isan equivalence if and only if H(C) = 0 implies C = 0 for any C D.

Proof. By Corollary 1.17 one knows that for any A the natural morphismhA : A H(F (A)) is an isomorphism.

In order to prove the assertion, one has to verify that also the adjunctionmorphism gB : F (H(B)) B is an isomorphism for any B D. Indeed,H would be an inverse of F in this case. Note that Corollary 1.17 does notapply, because we dont know whether H is fully faithful.

For any B D the morphism gB : F (H(B)) B can be completed to adistinguished triangle F (H(B)) B C F (H(B))[1]. Since H is exactby Proposition 1.30, we obtain a distinguished triangle

H(F (H(B)))H(gB ) H(B) H(C) H(F (H(B)))[1]

in D.Since H(gB) hH(B) = idH(B) (see Exercise 4) and, therefore, H(gB) is

an isomorphism, one nds H(C) = 0. Hence, C = 0 by assumption. Thisshows that gB is an isomorphism.

The analogous statement for a left adjoint functor G F also holds true.


Definition 1.38 A triangulated category D is decomposed into triangulatedsubcategories D1,D2 D if

i) Both categories D1 and D2 contain objects non-isomorphic to 0.ii) For all A D there exists a distinguished triangle

B1 A B2 B1[1]

with Bi Di, i = 1, 2.iii) Hom(B1, B2) = Hom(B2, B1) = 0 for all B1 D1 and B2 D2.A triangulated category that cannot be decomposed is called indecom-

posable.

Later, we will see that the derived category of an integral scheme is inde-composable (see Lemma 2.51).

Exercise 9 Show that condition ii) in the presence of iii) just says that A isthe direct sum of B1 and B2. In particular, the denition is symmetric in D1and D2 despite the chosen order in ii).Proposition 1.39 Let F : D D be a fully faithful exact functor betweentriangulated categories. Suppose that D contains objects not isomorphic to 0and that D is indecomposable.

Then F is an equivalence of categories if and only if F has a left adjointG F and a right adjoint F H such that for any object B D one has:H(B) = 0 implies G(B) = 0. See [8].

Proof. In order to prove the proposition, one introduces two full triangu-lated subcategories D1,D2 D whose objects consists of all B D withF (H(B)) = B (induced by adjunction) respectively of all C D withH(C) = 0. Clearly, both are triangulated subcategories. (Note that the cate-gory D1 is the image of F , i.e. the full subcategory of all objects isomorphicto an element of the form F (A).)

The arguments in the proof of the previous lemma show that any B Dcan be decomposed by a distinguished triangle B1 B B2 B[1] withBi Di.

Furthermore, for all B1 D1 and B2 D2 we haveHom(B1, B2) = Hom(F (H(B1)), B2) = Hom(H(B1), H(B2)) = 0

and

Hom(B2, B1) = Hom(B2, F (H(B1)) = Hom(G(B2), H(B1)) = 0,for H(B2) = 0 by assumption implies G(B2) = 0.

Since D is indecomposable, either D1 or D2 is trivial, i.e. every element inone of the two subcategories is isomorphic to 0. Suppose D2 D is an equiv-alence. Thus, for any B D one would have H(B) = 0. Using that F is fully

1.4 Exceptional sequences and semi-orthogonal decompositions 19

faithful then yields Hom(A,A) = Hom(F (A), F (A)) = Hom(A,H(F (A))) =0. In particular, for any A D one has idA = 0 and, therefore, A = 0. Thiswould contradict the non-triviality of D, as D contains at least one objectnot isomorphic to 0.

Hence, D1 D is an equivalence, i.e. every object B D satisesF (H(B)) = B, which yields the assertion.

Remark 1.40 The proposition can be best applied when G = H . This parti-cular case will in fact occur in the applications. So, if F is fully faithful andH F H then F is an equivalence whenever D is indecomposable.

The following is a combination of the two propositions in the presence ofSerre functors

Corollary 1.41 Let F : D D be an exact functor between triangulatedcategories D and D with left adjoint G F and right adjoint F H. Fur-thermore assume that Ob(D) is a spanning class satisfying the followingconditions

i) For all A,B the natural morphisms

Hom(A,B[i]) Hom(F (A), F (B)[i])

are bijective for all i Z.ii) The categories D and D admit Serre functors SD respectively SD such

that for all A one has F (SD(A)) = SD(F (A)).iii) The category D is indecomposable and D is non-trivial.Then F is an equivalence. See [11].

Proof. One has to verify the condition that H(B) = 0 implies G(B) = 0needed for Proposition 1.39.

Suppose H(B) = 0. On the other hand, using adjunction and the com-patibility of the Serre functors with F , one nds for A :

Hom(A,H(B)) = Hom(F (A), B) = Hom(B,SD(F (A)))= Hom(B,F (SD(A))) = Hom(G(B), SD(A))= Hom(A,G(B)).Hence, Hom(A,G(B)) = 0 for all A and, therefore, G(B) = 0.

1.4 Exceptional sequences and semi-orthogonaldecompositions

In the geometric context, the derived categories in question will usually beindecomposable (see Proposition 2.51). However, a weaker notion of a decom-position of a triangulated category, so called semi-orthogonal decompositions,


turns out to be very useful. This is the topic of this section. We will rst in-troduce a special version of a semi-orthogonal decomposition that is inducedby an exceptional sequence.

Definition 1.42 An object E D in a k-linear triangulated category D iscalled exceptional if

Hom(E,E[i]) ={

k if i = 00 else

An exceptional sequence is a sequence E1, . . . , En of exceptional objectssuch that Hom(Ei, Ej [k]) = 0 for all i > j and all k. An exceptional se-quence is full if D is generated by {Ei}, i.e. any full triangulated subcategorycontaining all objects Ei is equivalent to D (via the inclusion).Lemma 1.43 Let D be a k-linear triangulated category such that for anyA,B D the vector space i Hom(A,B[i]) is finite-dimensional.

If E D is exceptional, then objects of the form E[i]ji form anadmissible triangulated subcategory E of D.

Proof. We leave it to the reader to check that E is indeed triangulated.In order to see that it is admissible one considers for any object A D thecanonical morphism

Hom(E,A[i]) E[i] A,which can be completed to a distinguished triangle Hom(E,A[i])E[i]A B. Using that E is exceptional, one nds Hom(E,B[i]) = 0. Hence,B E.

The concept of a (full) exceptional sequence is generalized by the following

Definition 1.44 A sequence of full admissible triangulated subcategoriesD1, . . . ,Dn D is semi-orthogonal if for all i > j one has Dj Di . Such asequence denes a semi-orthogonal decomposition of D if D is generated bythe Di, i.e. D is the smallest triangulated subcategory of D containing all ofthem.

Examples 1.45 i) Let D D be an admissible full triangulated subcategory.Then D1 := D,D2 := D D is a semi-orthogonal decomposition of D.

ii) Let E1, . . . , En be an exceptional sequence in D. Then the admissibletriangulated subcategories (see Lemma 1.43) D1 := E1, . . . ,Dn := Enis a semi-orthogonal sequence. If the exceptional sequence is full, thenD1, . . . ,Dn D is a semi-orthogonal decomposition.Lemma 1.46 A semi-orthogonal sequence of full admissible triangulated sub-categories D1, . . . ,Dn D generates D, i.e. defines a semi-orthogonal decom-position, if and only if any object A D with A Di for i = 1, . . . , n istrivial.

1.4 Exceptional sequences and semi-orthogonal decompositions 21

Proof. Suppose D1, . . . ,Dn D is a semi-orthogonal decomposition. Forany A0 D one denes the full triangulated subcategory A0 of all ob-jects A D with Hom(A,A0[i]) = 0 for all i Z. If A0

Di , thenD1, . . . ,Dn A0. Hence, A0 = D and, in particular, A0 A0. Thelatter yields Hom(A0, A0) = 0 and thus A0 = 0.

Let us now assume thatDi = {0}. For simplicity we assume n = 2 and

leave the general case to the reader. We have to show that any A0 D iscontained in the triangulated subcategory generated by D1 and D2. Since D2is admissible, one nds a distinguished triangle A A0 B A[1] withA D2 andB D2 . The latter can be decomposed further by a distinguishedtriangle C B C C[1] with C D1 and C D1 (use that D1 isadmissible). As C D1 D2 and B D2 , one nds C D2 . Hence,C D1 D2 , which implies C = 0 by assumption. Thus, B = C D1.But then A0 sits in a distinguished triangle with the other two objects in D1respectively D2.

Exercise 10 Let D1,D2 D be a semi-orthogonal decomposition of lengthtwo. Show that the inclusion D1 D2 is an equivalence.

2 Derived categories

This chapter is meant as a reminder. The material is essentially standard(derived categories, Serre duality, etc.) but a few more recent results areblend in to prepare the stage for the later sections. Many arguments are onlysketched, if at all, and the reader not familiar with the material or feelinguncomfortable with it should go back to the literature.

The hope is that by passing quickly to the results by Bondal, Bridgeland,Orlov, et al. on derived categories of coherent sheaves, the topic of this course,the reader will get to know derived categories from a more geometric pointof view.

2.1 Derived category of an abelian category

In this section we shall recall the fundamental aspects of derived categories.We begin by stating the existence of the derived category as a theorem, andexplain the technical features, necessary for any calculation, later on. Derivedfunctors will only be discussed in Section 2.2.

In the sequel, we will mostly be interested in the derived category ofthe abelian category of (coherent) sheaves or of modules over a ring. Werecommend the textbooks [18, 27] for other examples.

Remark 2.1 Often, an object in a given abelian categoryA is studied in termsof its resolutions. To be more specic, recall that by denition a coherentsheaf F on a scheme X can locally be given by nitely many generators andnitely many relations. In other words, at least locally there exists an exactsequence Om1X Om2X F 0. On a smooth projective variety X anycoherent sheaf F admits a locally free resolution of length n = dim(X), i.e.there exists an exact sequence of the form

0 En . . . E1 E0 F 0with Ei locally free. Thus, in order to study arbitrary coherent sheaves on Xone can restrict to locally free sheaves and complexes of those.

More generally, when working with an abelian category, it is often neces-sary and natural to allow also complexes of objects in A. This leads to thenotion of the category of complexes.

24 2 Derived categories

Let us briey recall the denition of the category of complexes Kom(A)of an abelian category A. A complex in A consists of a diagram of objectsand morphisms in A of the form

. . . Ai1di1 Ai

di Ai+1di+1 . . .

satisfying di di1 = 0, or equivalently Im(di1) Ker(di), for all i Z.A morphism f : A B between two complexes A and B is a com-

mutative diagram

. . . Ai1di1A

fi1

AidiA

fi

Ai+1di+1A

fi+1

. . .

. . . Bi1di1B Bi

diB Bi+1di+1B . . .

Definition 2.2 The category of complexes Kom(A) of an abelian category Ais the category whose objects are complexes A in A and whose morphismsare morphisms of complexes.

Proposition 2.3 The category of complexes Kom(A) of an abelian categoryis again abelian.

Proof. The proof is straightforward. E.g. the zero object is the complex . . .0 0 0 . . . and the kernel of a morphism f : A B is the complexof the kernels Ker(f i), i Z.

Also note that mapping an object A A to the complex A with A0 = Aand Ai = 0 for i = 0 identies A with a full subcategory of Kom(A).

The complex category Kom(A) has two more features: cohomology andshifts.

Recall that the cohomology Hi(A) of a complex A is the quotient

Hi(A) :=Ker(di)Im(di1)

A,

i.e. Hi(A) = Coker(Im(di1) Ker(di)). A complex A is acyclic ifHi(A) = 0 for all i Z. Any complex morphism f : A B inducesnatural homomorphisms

Hi(f) : Hi(A) Hi(B).

The induced map for the cohomology objects is used to dene quasi-isomorphisms, which play a central role in the passage to the derived category.

Definition 2.4 A morphism of complexes f : A B is a quasi-iso-morphism (or qis, for short) if for any i Z the induced map Hi(f) :Hi(A) Hi(B) is an isomorphism.

2.1 Derived category of an abelian category 25

Let us now turn to the shift functor.

Definition 2.5 Let A Kom(A) be a given complex. Then A[1] is thecomplex with (A[1])i := Ai+1 and dierential diA[1] := di+1A .

The shift f [1] of a morphism of complexes f : A B is the complexmorphism A[1] B[1] given by f [1]i := f i+1.Corollary 2.6 The shift functor T : Kom(A) Kom(A), A A[1] de-fines an equivalence of abelian categories.

More precisely, the inverse functor T1 is given by A A[1], where,more generally, A[k]i = Ak+i and diA[k] = (1)kdi+kA for any k Z.

Note however, that Kom(A) endowed with the shift functor T does notdene a triangulated category. (In fact, we would also have to give the class ofdistinguished triangles and the canonical choices, like short exact sequencesor mapping cones, do not work.)

Maybe Exercise here.The basic idea that leads to the denition of the derived category is toinvert all quasi-isomorphisms in Kom(A). We will give further explanationswhy this is desirable later on. Let us begin with the following existence the-orem.

Theorem 2.7 Let A be an abelian category and Kom(A) be its category ofcomplexes. Then there exists a category D(A), the derived category of A, anda functor

Q : Kom(A) D(A)such that

i) If f : A B is a quasi-isomorphism, then D(f) is an isomorphismin D(A).

ii) (Universality) Any functor F : Kom(A) D satisfying property i)factorizes uniquely over Q : Kom(A) D(A), i.e. there exists a uniquefunctor (up to isomorphism) G : D(A) D with F = G Q.

As stated, the theorem is a pure existence result. In order to be ableto work with the derived category, we have to understand which objectsbecome isomorphic under Q : Kom(A) D(A) and, more complicated, howto represent morphisms in the derived category. Explaining this, will at thesame time provide a proof for the above theorem. Moreover, we will see thefollowing facts:

Corollary 2.8 i) The objects of Kom(A) and D(A) are identified under Q :Kom(A) D(A).

ii) The cohomology objects Hi(A) of an object A D(A) are well-definedobjects of the abelian category A.

iii) Viewing any object in A as a complex concentrated in zero yields anequivalence between A and the full subcategory of D(A) that consists of allcomplexes A with Hi(A) = 0 for i = 0.


Contrary to the category of complexes Kom(A), the derived categoryD(A) is in general not abelian, but it always is triangulated. The shift functorindeed descends to D(A) and a natural class of distinguished triangles canbe found as will be explained shortly.

In the following, we shall rst explain why dividing out by quasi-iso-morphisms is a natural thing to do. At the same time, we will introduce thehomotopy category of complexes. This will be an intermediate step in passingfrom Kom(A) to D(A):

Kom(A)

D(A)

K(A)

Recall that a projective resolution of an object A A consists of a mor-phism of complexes P A, where the object A A is considered as acomplex concentrated in zero, P is a complex with P i = 0 for i > 0, and P i

projective (see example 1.12, ii)) for all i such that:

. . . P2 P1 P 0 A 0

is exact. Clearly, a projective resolution P A is a quasi-isomorphism.Similarly, an injective resolution of A A is an exact sequence 0

A I0 I1 . . . with all Ii injective. This is usually viewed as a quasi-isomorphism A I.

Often, a given object admits, if at all, various dierent projective (in-jective) resolutions. Thus, replacing an object in an abelian category e.g.by a projective resolution is not a unique procedure. It turns out, however,that projective resolutions are unique up to a certain equivalence relation.This will lead us to the concept of homotopically equivalent morphisms and,eventually, to the homotopy category of complexes K(A).Definition 2.9 Two morphisms of complexes f, g : A B are calledhomotopically equivalent, f g, if there exists a collection of homomorphismshi : Ai Bi1, i Z, such that f i gi = hi+1 diA + di1B hi.

The homotopy category of complexes K(A) is the category with objectsOb(Kom(A)) and morphisms HomK(A)(A, B) := HomKom(A)(A, B)/ .

That the denition makes sense, e.g. that the composition is well-denedin K(A), follows from the following assertions which are all easily veried.Proposition 2.10 i) Homotopy equivalence between complex morphisms A B is an equivalence relation.

ii) Homotopically trivial morphisms form an ideal in the morphisms ofKom(A).


iii) If f g : A B, then Hi(f) = Hi(g) for all i.iv) If f : A B and g : B A are given such that f g idB

and g f idA, then f and g are quasi-isomorphisms and, more precisely,Hi(f)1 = Hi(g).

Remark 2.11 Note that the denition of K(A) makes sense for any additivecategory. This will be needed later when we consider the full subcategory ofall injective objects in a given abelian category (cf. Proposition 2.26).

Why considering morphisms of complexes only up to homotopy curesthe ambiguity problem, e.g. for projective resolutions, is explained by thefollowing fact.

Proposition 2.12 Let A,B A and let P A be a projective resolution.Furthermore, assume that B B is a resolution but not necessarily pro-jective. Then any morphism A B can be extended to a complex morphismf : P B, i.e. there exists a commutative diagram

. . . P2

f2

P1

f1

P 0

f0

A

. . . B2 B1 B0 B

.

Moreover, f : P B is unique up to homotopy.

Proof. See e.g. [18]. In fact, one only needs a complex morphism P A andnot necessarily a resolution.

Similar arguments also prove

Proposition 2.13 Let A B be a quasi-isomorphism. Then for any com-plex P of projective objects P i with P i = 0 for i 0 the induced mapHomK(A)(P , A) HomK(A)(P , B) is bijective. Exercise 11 Find the opposite statements of Propositions 2.12 and 2.13 forinjective resolutions.

In general, for arbitrary resolutions A A and B B a given mapf : A B cannot be extended. If, however, a projective resolution P Aexists, then one nds morphisms to both resolutions:

P

A

B

A B


Here, in A P B the morphism P A is a quasi-isomorphism,whereas P B is an ordinary morphism of complexes.

We shall see that a morphism in the derived category is, in general, nolonger represented by an honest morphism between complexes but by a dia-gram

P

qis

A B

(without the complex in the middle necessarily being projective).All this motivates our wish to invert arbitrary quasi-isomorphisms and

we will now explain the construction of the derived category.The rst step is to give the objects of D(A). This is easy, we set

Ob(D(A)) = Ob(Kom(A)). The set of morphism HomD(A) between two com-plexes A and B viewed as objects in D(A) is the set of all equivalence classesof diagrams of the form

Cqis

A B

where C A is a quasi-isomorphism. Two such diagrams are equivalent ifthey are dominated (in the homotopy category K(A)!) by a third one of thesame sort, i.e.

C

C1

C

2

A B

(In particular, the compositions C C1 A and C C2 A arehomotopy equivalent and both are quasi-isomorphisms.)

In this way, we have dened objects and morphisms of our category D(A),but we still have to check a number of properties. In particular, we have todene the composition of two morphisms. If two morphisms

C1qis

A B

and C2qis

B C


are given, we want the composition of both to be given by a commutative (inthe homotopy category K(A)!) diagram of the form

C

qis

C1qis

C2

qis

A B C

There are two obvious problems: One has to ensure that such a diagramexists and that it is unique up to equivalence.

Both things are true, but we need to introduce the mapping cone in orderto explain why.

Definition 2.14 Let f : A B be a complex morphism. Its mapping coneis the complex C(f) with

C(f)i = Ai+1 Bi and diC(f) :=(di+1A 0

f i+1 diB

).

Note that in the literature one nds dierent conventions for the denitionof the dierential dC(f), e.g. f i+1 with an extra sign. The reader easily checksthat with this denition the mapping cone is a complex. Moreover, there existtwo natural complex morphisms

: B C(f) and : C(f) A[1]

given by the natural injection Bi Ai+1 Bi and the natural projectionAi+1 Bi A[1]i = Ai+1, respectively.

Clearly, the two compositions A B C(f) and B C(f) A[1]are both trivial. In fact, B C(f) A[1] is a short exact sequences ofcomplexes. In particular, we obtain the long exact cohomology sequence

Hi(A) Hi(B) Hi(C(f)) Hi+1(A) . . .

Also, by construction any commutative diagram can be completed as fol-lows

A1f1

B1

C(f1)

A1[1]

A2

f2 B2 C(f2) A2[1]


This probably reminds the reader of the axiom TR3. The following propo-sition should be viewed in light of the axiomTR2. In fact, the triangles denedby the mapping cone will form the distinguished triangles in the homotopyand in the derived category (cf. Proposition 2.19). The next proposition willat the same time help us to solve our problem with the denition of thecomposition of two morphisms in the derived category.

Proposition 2.15 Let f : A B be a morphism of complexes and C(f)its mapping cone with the two natural morphisms : B C(f) and :C(f) A[1]. Then there exists a complex morphism g : A[1] C()which is an isomorphism in K(A) and such that the following diagram iscommutative in K(A):

B

id

C(f)

id

A[1]f

g

B[1]

id

B

C(f) C()

B[1]

Proof. The morphism g : A[1] C() is easy to dene: We let

A[1]i = Ai+1 C()i = Bi+1 C(f)i = Bi+1 Ai+1 Bi

be the map (f i+1, id, 0). We leave it to the reader to verify that this isindeed a complex morphism.

The inverse g1 in K(A) can be given as the projection onto the middlefactor.

The commutativity (in Kom(A)!) of the diagram

A[1]f

g

B[1]

id

C()

B[1]

is straightforward. (But note the annoying sign, which is in fact responsiblefor the sign in TR2.)

The diagramC(f)

id

A[1]

g

C(f)

C()

does not commute in Kom(A), but it does commute up to homotopy. Toprove this, one rst checks that g g1 is indeed homotopic to the identityand then uses g1 = .


For the details see [27, 1.4]. Let us rst show, how to use the construction of the mapping cone in order

to compose two maps in the derived category. In order to do this, we considera quasi-isomorphism A

f B and an arbitrary morphism g : C B.Proposition 2.16 Then there exists a commutative diagram in K(A) (!):

C0

qis C

g

A

qis

f B

Proof. The idea is to construct a commutative diagram of the form

C( g)[1]

C

g

C(f)

=

C( g)

A

f B C(f) A[1]

Due to the previous proposition we know that B C(f) A[1] in K(A)is isomorphic to B C(f) C(). But there exists a natural morphismC( g) C().

Using the long exact cohomology sequences, one proves that the morphismC0 := C( g)[1] C is a quasi-isomorphism.

It should be clear how to use the proposition in order to show that twomorphisms in the derived category can be composed. Similar arguments showthat the equivalence class of the composition is well-dened.

Exercise 12 One might be tempted to dene C0 more directly as the bredsum of A and C over B. Find an example that shows that this in generaldoes not work. (E.g. try a surjection for B with kernel A.)

Exercise 13 Show that a complex A is isomorphic to 0 in D(A) if and onlyif Hi(A) = 0 for all i. On the other hand, nd an example of a complexmorphism f : A B such that Hi(f) = 0 for all i, but without f beingtrivial in D(A). See [18, 22]

In fact, f is zero in D(A) if and only if there exists a qis g : C Asuch that f g is homotopically zero.Exercise 14 Let A be a complex with m := max{i | Hi(A) = 0} < .Show that there exists a morphism : A Hm(A)[m] in the derivedcategory such that Hm() : Hm(A) Hm(A) equals the identity.

Similarly, if m := min{i | Hi(A) = 0} > , then there exists :Hm(A)[m] A with Hm() = id.


Remark 2.17 Behind the construction of the derived category there is a gene-ral procedure, called localization. Roughly, one constructs the localization of acategory with respect to a localizing class of morphisms. In our case, these arethe quasi-isomorphisms. It turns out, that quasi-isomorphisms form indeed alocalizing class in K(A) (but not in Kom(A) !). For details see [18, 27].Definition 2.18 A triangle A1 A2 A3 A1[1] in K(A) (respectivelyin D(A)) is called distinguished if it is isomorphic (in K(A) respectively inD(A)) to one of the form A f B C(f) A[1] with f a complexmorphism.

Proposition 2.19 With the above choices the homotopy category of com-plexes K(A) and the derived category D(A) of an abelian category have boththe structure of a triangulated category.

Moreover, the natural functor QA : K(A) D(A) is exact.

Proof. Again we refer to the literature, e.g. [18, IV.2.]. The main techniqueis the mapping cone. Note that there is the additional diculty in the de-rived category that an isomorphism of two triangles is not given by honestmorphisms.

Exercise 15 Let A := Vecf (k) be the abelian category of nite-dimensionalvector spaces over a eld k. Show that D(A) is equivalent to iZA. Moreprecisely, any complex A D(A) is isomorphic to its cohomology complexH(A) (with trivial dierentials).

Exercise 16 Show more generally, that the assertion in the last exercise holdstrue, whenever the abelian category A is semi-simple, i.e. such that any shortexact sequence in A splits. See [18, III.2.3].

By denition, objects in the categories K(A) and D(A) are unbounded.It often is more convenient to work with bounded complexes.

Definition 2.20 Let Kom(A), with = +,, or b, be the category ofcomplexes A with Ai = 0 for i 0, i 0, respectively |i| 0.

By dividing out rst by homotopy equivalence and then by quasi-isomor-phisms one obtains the categories K(A) and D(A) with = +,, or b. Letus consider the natural functors D(A) D(A) given by just forgetting theboundedness condition.

Proposition 2.21 The natural functors D(A) D(A), where = +,, orb, define equivalences of D(A) with the full triangulated subcategories of allcomplexes A D(A) with Hi(A) = 0 for i 0, i 0, respectively |i| 0.


Proof. The idea is the following: Suppose Hi(A) = 0 for i > i0. Then thecommutative diagram

. . . Ai02

=

Ai01

=

Ker(di0A )

0

. . .

. . . Ai02 Ai01 Ai0 Ai0+1 . . .

denes a quasi-isomorphism between a complex in K(A) and A.For details see [27]. Note that the statement is about the derived and not

about the homotopy category. The same arguments prove iii) in Corollary 2.8 saying thatA is canonically

equivalent to the full subcategory of all objects A D(A) with Hi(A) = 0for i = 0 (cf. [18, III.5]).Remark 2.22 Eventually, we will be interested in the bounded derived cate-gory of coherent sheaves Db(Coh(X)). However, there are reasons that willoblige us to work with a bigger abelian category, e.g. Qcoh(X) or Sh(X),and/or with the unbounded derived categories.

As in the process of deriving functors one has to work with injectiveresolutions and those are almost never bounded, derived functors are oftendened in the bigger categories of unbounded (or only partially bounded)complexes. Only a posteriori they are then restricted to the smaller categories.

Not only, that we would like to stay in the bounded derived category, butare sometimes forced to work with unbounded complexes, we also have toleave the abelian categoryCoh(X) we are primarily interested in and work inthe bigger onesQcoh or even Sh. The reason is essentially the same, we wantto replace coherent sheaves by their injective resolutions, but there almost nocoherent injective sheaves. We will come back to this question, but the readermight keep the following inclusions of abelian categories Coh Qcoh Shin mind.

The general context can be set as follows: Consider a full abelian subcat-egory A B of an abelian category B. Then there are two derived categoriesD(A) and D(B) with an obvious functor D(A) D(B) between them. Ingeneral, this functor has no reason to be nice, e.g. it need not be faithful orfull.

One might wonder whether this functor denes an equivalence betweenD(A) and the full subcategory of D(B) containing those complexes whosecohomology is in A. In general, this does not hold, but at least the passagefrom Db(Sh(X)) to Db(Qcoh(X)) can be explained in this way due to thefollowing

Proposition 2.23 Let A B be a thick subcategory and suppose that anyA A can be embedded in an object A A which is injective as an objectof B.


Then the natural functor defines an equivalence

D+(A) = D+A(B)

of D+(A) and the full triangulated subcategory D+A(B) D+(B) of complexeswith cohomology in A.

Analogously, Db(A) = DbA(B).

Proof. Recall that A B is thick if any extension in B of objects in A is againin A. This is clearly necessary in order to ensure that D+A(B) is a triangulatedsubcategory.

The idea of the proof is to replace any complex bounded below in B withcohomology in A by an injective resolution contained in A. For the proof werefer to the literature [18, 27]. Compare also the proof of the next proposition,which explains the main techniques in a slightly simpler case.

Example 2.24 Prove that under the assumption of the proposition D+A(B) isa triangulated subcategory of D+(B).

Suppose I A is the full additive subcategory containing all injectives ofan abelian categoryA. The inclusion I A and QA : K(A) D(A) inducea natural exact functor : K(I) D(A). (As for an abelian category, K(I)is triangulated.) In many situation, this functor denes an equivalence. Inorder to prove this, we will need the following

Lemma 2.25 Let A, I Kom+(A) such that all Ii are injective. ThenHomK(A)(A, I) = HomD(A)(A, I).

Proof. Clearly, there is a natural map HomK(A)(A, I) HomD(A)(A, I)and we have to show that for any morphism

B

qis

A I

in D(A) there exists a unique morphism of complexes A I making thediagram commutative up to homotopy. In other words, one has to show thatfor any qis B A in Kom+(A) the induced map HomK(A)(A, I) HomK(A)(B, I) is bijective.

Completing B A to a distinguished triangle in K+(A) and usingthe long exact Hom( , I)-sequence reduces the claim to the assertion thatHomK(A)(C, I) = 0 for any acyclic complex C, i.e. a complex qis to 0. Ahomotopy between any complex morphism C I and the zero map can be


explicitly constructed by standard algebraic homology methods, see [26, I.6].In fact, the arguments are quite similar to the ones used to prove Proposition2.13 using injective objects instead of projective ones.

Proposition 2.26 Suppose that A contains enough injectives, i.e. any objectin A can be embedded into an injective object. Then the natural functor :K+(I) D+(A) induced by QA is an equivalence.

Proof. Let us rst check that even without the hypothesis the functorK+(I) D+(A) is fully faithful. We have to verify two things. Firstly, ifa complex morphism f : I J of complexes of injectives Ii, Jj is zero inD(A) then f is homotopic to zero. Secondly, any morphism in D(A) can becompleted to a commutative diagram in K(A)

A

qis

I J.

Both assertion follow from the above lemma. (So, it is the injectivity of J

that matters for both parts.)Hence, K+(I) is equivalent to the full subcategory of complexes A

D+(A) isomorphic (in D+(A)) to a complex of injectives I K+(I).By assumption, any object A A admits an injective resolution. In other

words, any A A is quasi-isomorphic to a complex I K+(A). Thus, itsuces to show that this implies that any complex A K+(A) is quasi-isomorphic to a bounded below complex of injectives. Or, in other words,a quasi-inverse 1 can be given by choosing an injective resolution qis toany given complex. This is done by induction. We only indicate the rststep. For the complete proof see [27, I.1.1.7]. Suppose A is of the form0 A0 A1 A2 . . .. By assumption, there exists an injective objectI0 and a monomorphism A0 I0. The induced morphism A I0 := (I0 0 . . .) has the property that Hi(A) Hi(I0 ) is an isomorphism for i < 0and injective for i = 0. In the next step one constructs I1 = I

0 I1 0 . . . such that the induced cohomology map is injective for i = 1 andan isomorphism for i < 1. The denition of I1 and the morphism I0 I1is easy: Consider the object (I0 A1)/A0 and choose an injective object I1containing it. The morphism I0 I1 is the composition of the inclusion andthe natural morphism I0 (I0 A1)/A0 and the cohomological propertiesare readily veried.

In the course of the proof we have indeed used the assumption that allcomplexes are bounded below. For the analogous statement replacing injec-tive by projective, one would have to work in D(A) for the same reason.

The proposition will be essential for the construction of right derivedfunctors.


2.2 Derived functors

Let F : A B be an additive functor between abelian categories. In general,F will not preserve exactness. In particular, the image under F of an exactcomplex in A, which is automatically quasi-isomorphic to the zero complex,will not necessarily be exact. In other words, the naive extension of F to afunctor between the derived categories D(A) D(B) does not make sense,except when F is an exact functor. More generally, one has the following easy

Lemma 2.27 Let F : K(A) K(B) be an exact (in the triangulated sense)functor. Then F naturally induces a commutative diagram

K(A)

K(B)

D(A) D(B)

if one of the following conditions holds true:i) Under F a quasi-isomorphism is mapped to a quasi-isomorphism.ii) The image of an acyclic complex is again acyclic.

Derived functors are designed to replace a given functor F on the level ofderived categories. For exact functors this is straightforward, but in generalthe construction is rather involved. In any case, one always has to assumesome kind of exactness. For a left exact functor F : A B (see Denition1.11) one constructs the right derived functor

RF : D+(A) D+(B)

and for a right exact functor F : A B one constructs the left derivedfunctor

LF : D(A) D(B).Both constructions are completely analogous and we shall discuss only RF . Inthe application, however, there is a dierence in that Qcoh(X) for a smoothprojective variety always contains enough injectives, but never enough pro-jectives (well, except if X is a point).

Before explaining the general construction, we list all left or right exactfunctors that will be used in the geometric context. We will come back tothose at the beginning of the next section. In this section, we shall developonly the abstract machinery.

Examples 2.28 i) Let X be a topological space. Then the global section func-tor

(X, ) : Sh(X) Ab

2.2 Derived functors 37

is a left exact functor. Similarly, for a scheme X it denes left exact functors (X, ) : Qcoh(X) Ab and (X, ) : Coh(X) Ab. If X is a projectivevariety over a eld k, then one has the left exact functor

(X, ) : Coh(X) Vecf (k)

into the category of nite-dimensional vector spaces.ii) Let f : X Y be a continuous map. Then the direct image functor

f : Sh(X) Sh(Y ) is left exact. Similarly, if f : X Y is a morphism ofschemes, then

f : Qcoh(X) Qcoh(Y )

is left exact. If f : X Y is a proper morphism of noetherian schemes, thenthe direct image denes a left exact functor

f : Coh(X) Coh(Y ) .

This in particular applies to morphisms of projective varieties.Note that in general, (Y, ) f = (X, ).iii) Suppose X is a scheme and F Qcoh(X). Then

Hom(F , ) : Qcoh(X) Ab

is left exact. For a coherent sheaf on a projective variety over a eld k onehas the left exact functor

Hom(F , ) : Coh(X) Vecf (k) .

iv) Consider, as before, a quasi-coherent sheaf F on a scheme X . Thenthe sheaf hom denes a left exact functor

Hom(F , ) : Qcoh(X) Qcoh(X) .

If F is coherent, then Hom(F , ) takes coherent sheaves to coherent ones.Also note that Hom(F , ) = Hom(F , ).

v) Let X be a topological space endowed with a sheaf of commutativerings R. Consider the abelian category of sheaves of R-modules ShR(X). IfF ShR(X), then

(F R ) : ShR(X) ShR(X)

is a right exact functor.vi) Let f : X Y be a continuous map. Then the inverse image denes

an exact functorf1 : Sh(Y ) Sh(X) .


If f : X Y is a morphism of schemes, one denes f := (OXf1OY )f1.Thus,

f : Qcoh(Y ) Qcoh(X)

is a right exact functor, as it is the composition of the exact functor f1 :Sh(Y ) Sh(X) and the right exact functor (OXf1OY ) : Shf1OY (X)ShOX (X).

Clearly, the inverse image in this sense sends a coherent sheaf to a coherentsheaf, i.e.

f : Coh(Y ) Coh(X) .

Now, back to the abstract setting: We let F : A B be a left exactfunctor of abelian categories. Furthermore, we assume that A admits enoughinjectives. In particular, we will use the equivalence : K+(IA) D+(A)naturally induced by the functor QA : K+(A) D+(A) (see Proposition2.26). By 1 we denote a quasi-inverse of given by choosing a complex ofinjective objects qis to any given bounded below complex. Thus, we have thediagram

K+(IA)

K+(A) K(F )

QA

K+(B)QB

D+(A)1

D+(B)Here, K(F ) is the functor (. . . Ai1 Ai Ai+1 . . .) (. . .

F (Ai1) F (Ai) F (Ai+1) . . .) which is well-dened for the homotopycategories.

Definition 2.29 The right derived functor is the functor

RF := QB K(F ) 1 : D+(A) D+(B).

Here is the list of the main properties of the right derived functor.

Proposition 2.30 i) There exists a natural morphism of functors QB K(F ) RF QA.

ii) The right derived functor RF : D+(A) D+(B) is an exact functorof triangulated categories.

iii) Suppose G : D+(A) D+(B) is an exact functor. Then any functormorphism QB K(F ) G QA factorizes over a unique functor morphismRF G.

Proof. i) Let A D+(A) and I := 1(A). The natural transformationid 1 yields a functorial morphism A I in D+(A), which itselfis given by some A

qis C I. Using the injectivity of Ii yields a unique


morphism A I in K(A) (see the explanations in the proof of Proposition2.26), which, moreover, is independent of the choice of C. Altogether, thisyields a functorial morphism K(F )(A) K(F )(I) = RF (A).

ii) The category K(IA) is triangulated and : K+(IA) D+(A) is clearlyexact. Thus, also the inverse functor is exact (cf. Proposition 1.30). Hence,RF is the composition of three exact functors and, therefore, itself exact.

iii) See [18, III.6.11]. These properties determine the right derived functor RF of a left exact

functor F up to unique isomorphism.

Definition 2.31 Let F : A B be a left exact functor and RF : D+(A) D+(B) its right derived functor. Then for any complex A D+(A) onedenes:

RiF (A) := Hi(RF (A)) B.Remark 2.32 The induced additive functors RF i : A B are the higherderived functors of F . Note that RF iF (A) = 0 for i < 0 and R0F (A) = Afor any A A. Indeed, if A I0 I1 . . . is an injective resolution,then RF i(A) = Hi(. . . F (I0) F (I1) . . .) and, therefore, R0F (A) =Ker(F (I0) F (I1)) = F (A), as F is left exact. An object A A is F -acyclicif RiF (A) = 0 for i = 0.

Sometimes, in order to emphasize the presence of a complex, one writesRiF (A) instead of RF i(A).

Remark 2.33 Going through the above arguments, one nds that our hypoth-esis can be weakened. In this remark we will explain two generalizations ofthe above arguments. Firstly, the functor might only be given on the homo-topy categories (and not on the level of the abelian categories) and, secondly,we might have to work with abelian categories which do not contain enoughinjectives.

Let us give the most general statement right away (see [22, Thm.II.5.1]):Suppose A and B are abelian categories and

F : K+(A) K(B)

is an exact functor (recall that both categories are triangulated). Then theright derived functor

RF : D+(A) D(B)

satisfying i)-iii) of the above proposition exists whenever one can nd a trian-gulated subcategory KF K+(A) which is adapted to F , i.e. which satisesthe two conditions:

i) If A KF is acyclic, i.e. Hi(A) = 0 for any i, then F (A) is acyclic.ii) Any A K+(A) is quasi-isomorphic to a complex in KF .


We will need this more general statement, e.g. when working with thesheaf-hom Hom(F, E) of complexes of sheaves and as well when deningthe left derived functor of the tensor product of sheaves. In the former, weobviously start out with a functor that only lives on the level of complexesand in the latter we have to deal with the fact that the category of sheavesover a ring does not contain enough projectives.

Let us spell out the second case in more details. The following explainshow to dene the derived functor of a left exact functor F : A B in thecase that A does not contain enough injectives. See [18, III.6].

Suppose F : A B is a left exact functor. In this situation one denesadapted already on the level of the abelian categories. A class of objectsIF Ob(A) stable under nite sums is F -adapted if the following conditionshold true:

i) Suppose A Kom+(A) is acyclic, i.e. Hi(A) = 0 for all i, and Ai IFfor all i. Then F (A) is again acyclic.

ii) Any object in A can be embedded into an object of IF .Under these conditions, the localization of K+(IF ) by quasi-isomorphisms

between complexes in IF is equivalent to D+(A). This is due to ii). Conditioni) ensures that the image under F of a quasi-isomorphism of complexes in IFis again a quasi-isomorphism. Hence, we may dene K(F ) on the localizationof K+(IF ). The right derived functor of F is then dened in the same wayby using these two facts.

Note that if A contains enough injectives, then IA is F -adapted for anyleft exact functor F . In this case, we may enlarge IA by all F -acyclic objects,i.e. by those objects A A with RiF (A) = 0 for i = 0. This yields a largeradapted class for F .

Exercise 17 Show that enlarging a given F -adapted class IF , e.g. IA of anabelian category with enough injectives, by all F -acyclic ones yields again anF -adapted class.

Exercise 18 Let IF be an adapted class for a left exact functor F : A B.Show that RiF (A) = 0 for all i = 0 and A IF .Exercise 19 Suppose we know that the right derived functor RF exists. LetA be a complex of F -acyclic objects. Show that RF (A) = K(F )(A).

In other words, in order to compute RF (A) of an arbitrary complex A

it suces to nd a qis A = I K+(A) such that all Ii are F -acyclic. ThenRF (A) = K(F )(I).Exercise 20 Write down the conditions for a class of objects to be adaptedto a right exact functor. (This actually is the situation one needs to considerfor the tensor product.)

In the next section we will study in detail a number of derived functors forthe derived category of coherent sheaves. Here, we shall stay in the general


situation and consider Hom(A, ) : A Ab for an arbitrary object A A.Clearly, Hom(A, ) is left exact and if A contains enough injectives, onedenes

Exti(A, ) := Hi RHom(A, ).It turns out, that these Ext-groups can in fact be interpreted internally inthe derived category:

Proposition 2.34 Suppose A,B A are objects of an abelian category con-taining enough injectives. Then there are natural isomorphisms

ExtiA(A,B) = HomD(A)(A,B[i]),where A,B are considered as complexes concentrated in degree zero.

Proof. Suppose B I is an injective resolution of B. By construction,RHom(A,B) as an object of the derived category D(Ab) is isomorphic tothe complex (Hom(A, Ii))iN. Therefore, Exti(A,B) is the cohomology ofthis complex.

A morphism f Hom(A, Ii) is a cycle, i.e. in the kernel of Hom(A, Ii) Hom(A, Ii+1), if and only if f denes a morphism of complexes f : A I[i].This morphism of complexes is homotopically trivial if and only if f is aboundary, i.e. in the image of Hom(A, Ii1) Hom(A, Ii).

Hence, Exti(A,B) = HomK(A)(A, I[i]). Since I is a complex of injec-tives, we have HomK(A)(A, I[i]) = HomD(A)(A, I[i]) (see Lemma 2.25).Using B = I, this yields the assertion.

Remarks 2.35 i) Again, the above arguments can easily be generalized to adescription of ExtA(A, B). Suppose A Kom(A). Then

HomK(A)(A, ) : K+(A) K(Ab)

is an exact functor. More explicitly, a complex B = (Bi)iZ is mapped tothe complex (HomK(A)(A, Bi))iZ.

The full triangulated subcategory of complexes of injectives is adapted tothis functor (always under the assumption that A has enough injectives) andwe may dene RHom(A, B) D(Ab) for any B D+(A) and

Exti(A, B) := Hi(RHom(A, B)).

ThenExti(A, B) = HomD(A)(A, B[i]).

Copying the above proof, one encounters the inner hom of two complexesA and B. Namely, Hom(A, B) with Homi(A, B) =

Hom(Ak, Bk+i)

and dierential diA (1)idiB. The i-th cohomology of Hom(A, B) com-putes HomK(A)(A, B[i]).


ii) It is noteworthy, that these Ext-groups only depend on A as an ele-ment of the derived category. Hence, we have dened a bifunctor

D(A)opp D+(A) D(Ab) ,

which is exact in each variable. This can also be seen as a consequence of thegeneral principle (see Lemma 2.27), as Proposition 2.34 shows that under theinner Hom( , B) an acyclic complex is mapped to an acyclic complex.

iii) Suppose that A has enough projectives. Then one can dene for anycomplex B Kom(A) the right derived functor of the left exact functorHom( , B) : K(A)opp K(Ab). This yields the exact contravariant func-tor

RHom( , B) : D(A)opp D(Ab)Analogously to ii), it only depends on B as an object in the derived categoryand one thus denes again a bifunctor exact in the two variables

D(A)opp D(A) D(Ab) .

If A has enough projectives and enough injectives, then the two bifunctorsin ii) and iii) give rise to the same bifunctor (cf. [22, I.6.3])

D(A)opp D+(A) D(Ab) .

That the cohomology of both yields the same, follows from the above remarks.Also note that in a category A with enough injectives, but possibly not

enough projectives, the derived functor

RHom( , B) : D(A)opp D(Ab)

nevertheless exists for any B K+(A) due to ii)iv) Composition in the derived category naturally leads to composition

for Ext-groups:

Exti(A, B) Extj(B, C) Exti+j(A, C),

where we assume A, B, C D+(A).Indeed, elements in

Exti(A, B) = HomD(A)(A, B[i])

and

Extj(B, C) = HomD(A)(B, C[j]) = HomD(A)(B[i], C[i+ j])

can be composed to an element in Exti+j(A, C) = HomD(A)(A, C[i+ j]).

2.3 Spectral sequences 43

Proposition 2.36 Let F1 : A B and F2 : B C be two left exact functorsof abelian categories. Assume that there exist adapted classes IF1 A andIF2 B for F1 respectively F2 such that F1(IF1) IF2 .

Then the derived functors RF1 : D+(A) D+(B), RF2 : D+(B) D+(C), and R(F2 F1) : D+(A) D+(C) exist and there is a natural iso-morphism

R(F2 F1) = RF2 RF1.

Proof. The existence ofRF1 and RF2 follows from the assumptions. Moreover,since F1(IF1) IF2 , the class IF1 is also adapted to the composition F2 F1and, hence, R(F2 F1) exists as well.

A natural morphism R(F2 F1) RF2 RF1 is given by the universalityproperty of the derived functor R(F2 F1).

If A D+(A) is isomorphic to a complex I K+(IF1), then thismorphism R(F2 F1)(A) = (K(F2) K(F1))(I) R(F2)((RF1)(A)) =R(F2)(K(F1)(I)) = K(F2)(K(F1)(I)) is an isomorphism.

Remarks 2.37 i) Suppose that A and B both contain enough injectives. Thenthe assumption of the proposition is satised if F1(IA) IB, but this mightbe dicult to verify. We might, however, enlarge IB by all F2-acyclic objectsin B, i.e. by objects B with RiF2(B) = 0 for i = 0. This yields a newadapted class IF2 and we then only have to show that for any injectiveobject I IA A the image F1(I) B is F2-acyclic. The proposition isoften applied in this form.

ii) The proposition also holds true for the derived functors of exact func-tors on the homotopy categories. In this case, one starts with exact functorsF1 : K+(A) K+(B) and F2 : K+(B) K(C). In this situation, one hasto assume that there exist adapted triangulated subcategories KF1 K+(A)and KF2 K+(B) such that F1(KF1) KF2 .

2.3 Spectral sequences

In this short section we will explain how spectral sequences occur whenevertwo derived functors are composed. We will not enter the very technicaldetails of this machinery, but hope to provide at least the amount necessaryto follow the applications in the later sections.

The main data of a spectral sequence in an abelian category A is acollection of objects (Ep,qr , E

n), n, p, q, r Z, r 1, and morphismsdp,qr : E

p,qr Ep+r,qr+1r subject to the following conditions:

i) dp+r,qr+1r dp,qr = 0 for all p, q, r, which thus yields a complexEp+r,qr+r .

ii) Ep,qr+1 = H0(Ep+r,qr+r ). These isomorphisms are actually part ofthe data.


iii) For any (p, q) there exists an r0 such that dp,qr = dpr,q+r1r = 0 for

r r0. In particular, Ep,qr = Ep,qr0 for all r r0. This object is called Ep,q .iv) There is a decreasing ltration . . . F p+1En F pEn . . . En,

such that

F pEn = 0 and

F pEn = En, and isomorphisms Ep,q =F pEp+q/F p+1Ep+q.

So, in some sense the objects Ep,qr converge towards En. Usually, all the

objects of one layer, say Ep,qr with r xed, are explicitly given. Then onewrites

Ep,qr Ep+q.Let us try to visualize these data as follows: For r = 1 one has the following

system of horizontal complexes:

q

Ep2,q+11 Ep1,q+11

Ep,q+11 Ep+1,q+11

. . .

Ep2,q1 Ep1,q1 E

p,q1

Ep+1,q1 . . .

Ep2,q11 Ep1,q11

Ep,q11 Ep+1,q11

. . .

p

For r = 2 it looks like this:

q

Ep2,q+12

E

p1,q+12

Ep,q+12

E

p+1,q+12

. . .

Ep2,q2

E

p1,q2

Ep,q2

E

p+1,q2

. . .

Ep2,q12 Ep1,q12 E

p,q12 E

p+1,q12

. . .

p

2.3 Spectral sequences 45

Remark 2.38 One almost never goes beyond E2. Usually the argument will golike this: For some reason one knows that all dierentials are trivial. Hence,En admits a ltration the subquotients of which are isomorphic to Ep,np2 .E.g. if the objects are all vector spaces, then this yields a non-canonicalisomorphism En =

Ep,np2 .

The standard source for spectral sequences are nice ltrations on com-plexes. Such nice ltrations occur naturally on the total complex of a doublecomplex concentrated in e.g. the rst quadrant. We briey sketch this partof the theory.

A double complex K, consists of objects Ki,j for i, j Z and morphismsdi,jI : K

i,j Ki+1,j and di,jII : Ki,j Ki,j+1 satisfying d2I = d2II = 0, i.e. Ki,and K,j form complexes for all i, j, and dIdII +dIIdI = 0. The total complexK := tot(K,) of a double complex K, is the complex Kn =

i+j=n K

i,j

with d = dI + dII .On the total complex K of a double complex K, there exists a natural

decreasing ltration (in fact, there are two natural ones due to the symmetryof the situation):

F Kn :=

i+j=n

j

Ki,j ,

which satises dI(F Kn) F (Kn+1). Clearly, the graded objects grKn =F Kn/(F +1Kn) = Kn, form the complex K,[] (up to the globalsign (1)). Hence, Hk(gr(K)) = Hk(K,) and the coho

Huybrechts, D - Fourier-Mukai Transforms in Algebraic Geometry (Cours DEA 2004)

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