EQUIVALENCES OF DERIVED CATEGORIES AND K3 SURFACES

DMITRI ORLOV

Abstract. We consider derived categories of coherent sheaves on smooth projective varieties. We prove that any equivalence between them can be represented by an object on the product. Using this, we give a necessary and sufficient condition for equivalence of derived categories of two K3 surfaces.

§0. Introduction

Let D^{b}(X) be the bounded derived category of coherent sheaves on a smooth projective variety X . The category D^{b}(X) has the structure of a triangulated category (see [V], [GM]). We shall consider D^{b}(X) as a triangulated category.

In this paper we are concerned with the problem of description for varieties, which have equivalent derived categories of coherent sheaves.

In the paper [Mu1], Mukai showed that for an abelian variety A and its dual Â the derived categories D^{b}(A) and D^{b}(Â) are equivalent . Equivalences of another type appeared in [BO1]. They are induced by certain birational transformations which are called flops.

Further, it was proved in the paper [BO2] that if X is a smooth projective variety with either ample canonical or ample anticanonical sheaf, then any other algebraic variety X′ such that D^{b}(X′) ≃ D^{b}(X) is biregularly isomorphic to X .

The aim of this paper is to give some description for equivalences between derived categories of coherent sheaves. The main result is Theorem 2.2. of §2 . It says that any full and faithful exact functor F : D^{b}(M) ⟶ D^{b}(X) having left (or right) adjoint functor can be represented by an object E ∈ D^{b}(M × X) , i.e. F ( ) ≅ R π_{∗}(E ⊗^{L} p^{∗}( )) , where π and p are the projections on M and X respectively.

In §3 , basing on the Mukai’s results [Mu2], we show that two K3 surfaces S_{1} and S_{2} over field ℂ have equivalent derived categories of coherent sheaves iff the lattices of transcendental cycles T_{S_{1}} and T_{S_{2}} are Hodge isometric.

I would like to thank A. Polishchuk for useful notices.

§1. Preliminaries

1.1. We collect here some facts relating to triangulated categories. Recall that a triangulated category is an additive category with additional structures:

a) an additive autoequivalence T : D ⟶ D , which is called a translation functor (we usually write X[n] instead of T^{n}(X) and f[n] instead of T^{n}(f) ), b) a class of distinguished triangles:

X →^{u} Y →^{v} Z →^{w} X[1].

And these structures must satisfy the usual set of axioms (see [V]).

If X , Y are objects of a triangulated category D , then Hom_{D}^{i}(X , Y) means Hom_{D}(X , Y [i]) .

An additive functor F : D ⟶ D′ between two triangulated categories D and D′ is called exact if

a) it commutes with the translation functor, i.e there is fixed an isomorphism of functors:

t_{F} : F ∘ T ⟶^{∼} T′ ∘ F,

b) it takes every distinguished triangle to a distinguished triangle (using the isomorphism t_{F} , we replace F (X[1]) by F (X)[1] ).

The following lemma will be needed for the sequel.

1.2. Lemma [BK] If a functor G : D′ ⟶ D is a left (or right) adjoint to an exact functor F : D ⟶ D′ then functor G is also exact .

Proof. Since G is the left adjoint functor to F , there exist canonical morphisms of functors id_{D′} → F ∘ G, G ∘ F ⟶ id_{D} . Let us consider the following sequence of natural morphisms:

G ∘ T′ ⟶ G ∘ T′ ∘ F ∘ G ⟶^{∼} G ∘ F ∘ T ∘ G ⟶ T ∘ G

We obtain the natural morphism G ∘ T′ ⟶ T ∘ G . This morphism is an isomorphism. Indeed, for any two objects A ∈ D and B ∈ D′ we have isomorphisms :

Hom(G(B[1]) , A) ≅ Hom(B[1] , F (A)) ≅ Hom(B , F (A)[−1]) ≅
Hom(B , F (A[−1])) ≅ Hom(G(B) , A[−1]) ≅ Hom(G(B)[1] , A)

This implies that the natural morphism G ∘ T′ ⟶ T ∘ G is an isomorphism.

Let now A ⟶^{α} B ⟶ C ⟶ A[1] be a distinguished triangle in D′ . We have to show that G takes this triangle to a distinguished one.

Let us include the morphism G(α) : G(A) → G(B) into a distinguished triangle:

G(A) ⟶ G(B) ⟶ Z ⟶ G(A)[1].

Applying functor F to it, we obtain a distinguished triangle:

F G(A) ⟶ F G(B) ⟶ F (Z) ⟶ F G(A)[1]

(we use the commutation isomorphisms like T′ ∘ F →^{∼} F ∘ T with no mention). Using morphism id → F ∘ G , we get a commutative diagram:

A  ⟶^{α}  B  ⟶ C ⟶  A[1]
↓  ↓  ↓
F G(A) ⟶^{F G(α)} F G(B) ⟶ F (Z) ⟶ F G(A)[1]

By axioms of triangulated categories there exists a morphism μ : C → F (Z) that completes this commutative diagram. Since G is left adjoint to F , the morphism μ defines ν : G(C) → Z . It is clear that ν makes the following diagram commutative:

G(A) ⟶ G(B) ⟶ G(C) ⟶ G(A)[1]
≀↓ ≀↓ ↓ν ≀↓
G(A) ⟶ G(B) ⟶ Z ⟶ G(A)[1]

To prove the lemma, it suffices to show that ν is an isomorphism. For any object Y ∈ D let us consider the diagram for Hom :

→ Hom(G(A)[1] , Y) → Hom(Z , Y) → Hom(G(B) , Y) →
↓≀ ↓H_{Y}(ν) ↓≀
→ Hom(G(A)[1] , Y) → Hom(G(C) , Y) → Hom(G(B) , Y) →
↓≀ ↓≀ ↓≀
→ Hom(A[1] , F (Y)) → Hom(C , F (Y)) → Hom(B , F (Y)) →

Since the lower sequence is exact, the middle sequence is exact also. By the lemma about five homomorphisms, for any Y the morphism H(ν) is an isomorphism . Thus ν : G(C) → Z is an isomorphism too. This concludes the proof. □

1.3. Let X = {X^{c} →^{d^{c}} X^{c+1} →^{d^{c+1}} ⋯ → X^{0}} be a bounded complex over a triangulated category D , i.e. all compositions d^{i+1} ∘ d^{i} are equal to 0 ( c < 0 ).

A left Postnikov system, attached to X , is, by definition, a diagram

[figure: X^{c}; X^{c+1}; X^{c+2}; Y^{c} = X^{c}; Y^{c+1}; Y^{c+2}; X^{0}; Y^{0}; Y^{−1}; ⋯; i_{c} = id; j_{c}; i_{c+1}; j_{c+1}; i_{c+2}; j_{−1}; i_{0}; d^{c}; d^{c+1}; ⟲; ⋆; ⟲; ⋆; ⋆; [1]; [1]; [1]]

in which all triangles marked with ⋆ are distinguished and triangles marked with ⟲ are commutative (i.e. j_{k} ∘i_{k} = d^{k} ). An object E ∈ ObD is called a left convolution of X , if there exists a left Postnikov system, attached to X such that E = Y^{0} . The class of all convolutions of X will be denoted by Tot(X ) . Clearly the Postnikov systems and convolutions are stable under exact functors between triangulated categories.

The class Tot(X ) may contain many non-isomorphic elements and may be empty. Further we shall give a sufficient condition for Tot(X ) to be non-empty and for its objects to be isomorphic. The following lemma is needed for the sequel(see [BBD]).

1.4. Lemma Let g be a morphism between two objects Y and Y′ , which are included into two distinguished triangles:

X ⟶^{u} Y ⟶^{v} Z ⟶^{w} X[1]
[figure: f; ↓g; h; f[1]]
X′ ⟶^{u′} Y′ ⟶^{v′} Z′ ⟶^{w′} X′[1]

If v′gu = 0 , then there exist morphisms f : X → X′ and h : Z → Z′ such that the triple (f, g, h) is a morphism of triangles.

If, in addition, Hom(X[1] , Z′) = 0 then this triple is uniquely determined by g .

Now we prove two lemmas which are generalizations of the previous one for Postnikov diagrams.

1.5. Lemma Let X = {X^{c} →^{d^{c}} X^{c+1} →^{d^{c+1}} ⋯ → X^{0}} be a bounded complex over a triangulated category D . Suppose it satisfies the following condition:

Hom^{i}(X^{a} , X^{b}) = 0 for i < 0 and a < b.  (1)

Then there exists a convolution for this complex and all convolutions are isomorphic (noncanonically).

If, in addition,

Hom^{i}(X^{a} , Y^{0}) = 0 for i < 0 and for all a  (2)

for some convolution Y^{0} (and, consequently, for any one), then all convolutions are canonically isomorphic.

1.6. Lemma Let X_{1} and X_{2} be bounded complexes that satisfy (1), and let (f_{c}, ..., f_{0}) be a morphism of these complexes:

X_{1}^{c} ⟶^{d_{1}^{c}} X_{1}^{c+1} ⟶ ⋯ ⟶ X_{1}^{0}
↓f_{c}  ↓f_{c+1}  ↓f_{0}
X_{2}^{c} ⟶^{d_{2}^{c}} X_{2}^{c+1} ⟶ ⋯ ⟶ X_{2}^{0}

Suppose that

Hom^{i}(X_{1}^{a} , X_{2}^{b}) = 0 for i < 0 and a < b.  (3)

Then for any convolution Y_{1}^{0} of X_{1} and for any convolution Y_{2}^{0} of X_{2} there exists a morphism f : Y_{1}^{0} → Y_{2}^{0} that commutes with the morphism f_{0} . If, in addition,

Hom^{i}(X_{1}^{a} , Y_{2}^{0}) = 0 for i < 0 and for all a  (4)

then this morphism is unique.

Proof. We shall prove both lemmas together. Let Y^{c+1} be a cone of the morphism d^{c} :

X^{c} ⟶^{d^{c}} X^{c+1} ⟶^{α} Y^{c+1} ⟶ X^{c}[1]

By assumption d^{c+1} ∘ d^{c} = 0 and Hom(X^{c}[1] , X^{c+2}) = 0 , hence there exists a unique morphism d̄^{c+1} : Y^{c+1} → X^{c+2} such that d̄^{c+1} ∘ α = d^{c+1} .

Let us consider a composition d^{c+2}∘d̄^{c+1} : Y^{c+1} → X^{c+3} . We know that d^{c+2}∘d̄^{c+1}∘α = d^{c+2} ∘ d^{c+1} = 0 , and at the same time we have Hom(X^{c}[1] , X^{c+3}) = 0 . This implies that the composition d^{c+2} ∘ d̄^{c+1} is equal to 0 .

Moreover, consider the distinguished triangle for Y^{c+1} . It can easily be checked that Hom^{i}(Y^{c+1} , X^{b}) = 0 for i < 0 and b > c+1 . Hence the complex Y^{c+1} ⟶ X^{c+2} ⟶ ⋯ ⟶ X^{0} satisfies the condition (1). By induction, we can suppose that it has a convolution. This implies that the complex X has a convolution too. Thus, the class Tot(X ) is non-empty.

Now we shall show that under the conditions (3) any morphism of complexes can be extended to a morphism of Postnikov systems.

Let us consider cones Y_{1}^{c+1} and Y_{2}^{c+1} of the morphisms d_{1}^{c} and d_{2}^{c} . There exists a morphism g_{c+1} : Y_{1}^{c+1} → Y_{2}^{c+1} such that one has the morphism of distinguished triangles:

X_{1}^{c} ⟶^{d_{1}^{c}} X_{1}^{c+1} ⟶^{α} Y_{1}^{c+1} ⟶ X_{1}^{c}[1]
↓f_{c}  ↓f_{c+1}  ↓g_{c+1}  ↓f_{c}[1]
X_{2}^{c} ⟶^{d_{2}^{c}} X_{2}^{c+1} ⟶^{β} Y_{2}^{c+1} ⟶ X_{2}^{c}[1]

As above, there exist uniquely determined morphisms d̄_{i}^{c+1} : Y_{i}^{c+1} → X_{i}^{c+2} for i = 1, 2 . Consider the following diagram:

Y_{1}^{c+1} ⟶^{d̄_{1}^{c+1}} X_{1}^{c+2}
↓g_{c+1}  ↓f_{c+2}
Y_{2}^{c+1} ⟶^{d̄_{2}^{c+1}} X_{2}^{c+2}

Let us show that this square is commutative. Denote by h the difference f_{c+2} ∘ d̄_{1}^{c+1} − d̄_{2}^{c+1} ∘ g_{c+1} . We have h ∘ α = f_{c+2} ∘ d_{1}^{c+1} − d_{2}^{c+1} ∘ f_{c+1} = 0 and, by assumption, Hom(X_{1}^{c}[1] , X_{2}^{c+2}) = 0 . It follows that h = 0 . Therefore, we obtain the morphism of new complexes:

Y_{1}^{c+1} ⟶^{d̄_{1}^{c+1}} X_{1}^{c+2} ⟶ ⋯ ⟶ X_{1}^{0}
↓g_{c+1}  ↓f_{c+2}  ↓f_{0}
Y_{2}^{c+1} ⟶^{d̄_{2}^{c+1}} X_{2}^{c+2} ⟶ ⋯ ⟶ X_{2}^{0}

It can easily be checked that these complexes satisfy the conditions (1) and (3) of the lemmas. By the induction hypothesis, this morphism can be extended to a morphism of Postnikov systems, attached to these complexes. Hence there exists a morphism of Postnikov systems, attached to X_{1} and X_{2} .

Moreover, we see that if all morphisms f_{i} are isomorphisms, then a morphism of Postnikov systems is an isomorphism too. Therefore, under the condition (1) all objects from the class Tot(X ) are isomorphic.

Now let us consider a morphism of the rightmost distinguished triangles of Postnikov systems:

Y_{1}^{−1} ⟶^{j_{1,−1}} X_{1}^{0} ⟶^{i_{1,0}} Y_{1}^{0} ⟶ Y_{1}^{−1}[1]
↓g_{−1}  ↓f_{0}  ↓g_{0}  ↓g_{−1}[1]
Y_{2}^{−1} ⟶^{j_{2,−1}} X_{2}^{0} ⟶^{i_{2,0}} Y_{2}^{0} ⟶ Y_{2}^{−1}[1]

If the complexes X_{i} satisfy the condition (4) ( i.e. Hom^{i}(X_{1}^{a} , Y_{2}^{0}) = 0 for i < 0 and all a ), then we get Hom(Y_{1}^{−1}[1] , Y_{2}^{0}) = 0 . It follows from Lemma 1.4. that g_{0} is uniquely determined. This concludes the proof of both lemmas. □

§2. Equivalences of derived categories

2.1. Let X and M be smooth projective varieties over field k . Denote by D^{b}(X) and D^{b}(M) the bounded derived categories of coherent sheaves on X and M respectively. Recall that a derived category has the structure of a triangulated category.

For every object E ∈ D^{b}(M × X) we can define an exact functor Φ_{E} from D^{b}(M) to D^{b}(X) . Denote by p and π the projections of M × X onto M and X respectively:

M × X ⟶^{π} X
p↓
M

Then Φ_{E} is defined by the following formula:

Φ_{E}( ) := π_{∗}(E ⊗ p^{∗}( ))  (5)

(we always shall write shortly f_{∗}, f^{∗}, ⊗ and etc. instead of R f_{∗}, L f^{∗}, ⊗^{L} , because we consider only derived functors).

The functor Φ_{E} has the left and the right adjoint functors Φ_{E}^{∗} and Φ_{E}^{!} respectively, defined by the following formulas:

\begin{matrix}Φ_{E}^{∗}( ) = p_{∗}(E^{∨} ⊗ π^{∗}(ω_{X}[dimX] ⊗ ( ))), \\ Φ_{E}^{!}( ) = ω_{M}[dimM] ⊗ p_{∗}(E^{∨} ⊗ ( )), \end{matrix}

where ω_{X} and ω_{M} are the canonical sheaves on X and M , and E^{∨} := R R R Hom(E, O_{M×X}) .

Let F be an exact functor from the derived category D^{b}(M) to the derived category D^{b}(X) . Denote by F^{∗} and F^{!} the left and the right adjoint functors for F respectively, when they exist. Note that if there exists the left adjoint functor F^{∗} , then the right adjoint functor F^{!} also exists and

F^{!} = S_{M} ∘ F^{∗} ∘ S_{X}^{−1},

where S_{X} and S_{M} are Serre functors on D^{b}(X) and D^{b}(M) . They are equal to ( ) ⊗ ω_{X}[dimX] and ( ) ⊗ ω_{M}[dimM] (see [BK]).

What can we say about the category of all exact functors between D^{b}(M) and D^{b}(X) ? It seems to be true that any functor can be represented by an object on the product M × X for smooth projective varieties M and X . But we are unable prove it. However, when F is full and faithfull, it can be represented. The main result of this chapter is the following theorem.

2.2. Theorem Let F be an exact functor from D^{b}(M) to D^{b}(X) , where M and X are smooth projective varieties. Suppose F is full and faithful and has the right (and,consequently, the left) adjoint functor.

Then there exists an object E ∈ D^{b}(M × X) such that F is isomorphic to the functor Φ_{E} defined by the rule (5), and this object is unique up to isomorphism.

2.3. Let F be an exact functor from a derived category D^{b}(A) to a derived category D^{b}(B) . We say that F is bounded if there exist z ∈ ℤ, n ∈ ℕ such that for any A ∈ A the cohomology objects H^{i}(F (A)) are equal to 0 for i ∉ [z, z + n] .

2.4. Lemma Let M and X be smooth projective varieties. If an exact functor F : D^{b}(M) ⟶ D^{b}(X) has a left adjoint functor then it is bounded.

Proof. Let G : D^{b}(X) ⟶ D^{b}(M) be a left adjoint functor to F . Take a very ample invertible sheaf L on X . It gives the embedding i : X ↪ ℙ^{N} . For any i < 0 we have right resolution of the sheaf O(i) on ℙ^{N} in terms of the sheaves O(j) , where j = 0, 1, .., N (see [Be]). It is easily seen that this resolution is of the form

O(i) ⟶^{∼} {V_{0} ⊗ O ⟶ V_{1} ⊗ O(1) ⟶ ⋯ ⟶ V_{N} ⊗ O(N) ⟶ 0}

where all V_{k} are vector spaces. The restriction of this resolution to X gives us the resolution of the sheaf L^{i} in terms of the sheaves L^{j} , where j = 0, 1, ..., N . Since the functor G is exact that there exist z′ and n′ such that H^{k}(G(L^{i})) are equal 0 for k ∉ [z′, z′ + n′] . This follows from the existence of the spectral sequence

E_{1}^{p,q} = V_{p} ⊗ H^{q}(G(L^{p})) ⇒ H^{p+q}(G(L^{i})).

As all nonzero terms of this spectral sequence are concentrated in some rectangle, so it follows that for all i cohomologies H (G(L^{i})) are concentrated in some segment.

Now, notice that if Hom^{j}(L^{i} , F (A)) = 0 for all i ≪ 0 , then H^{j}(F (A)) is equal to 0 . Further, by assumption, the functor G is left adjoint to F , hence

Hom^{j}(L^{i} , F (A)) ≅ Hom^{j}(G(L^{i}) , A).

If now A is a sheaf on M , then Hom^{j}(G(L^{i}) , A) = 0 for all i < 0 and j ∉ [−z′ − n′, −z′ + dimM] , and thus H^{j}(F (A)) = 0 for the same j . □

2.5. Remark We shall henceforth assume that for any sheaf F on M the cohomology objects H^{i}(F (F)) are nonzero only if i ∈ [−a, 0] .

2.6. Now we begin constructing an object E ∈ D^{b}(M × X) . Firstly, we shall consider a closed embedding j : M ↪ ℙ^{N} and shall construct an object E′ ∈ D^{b}(ℙ^{N} × X) .

Secondly, we shall show that there exists an object E ∈ D^{b}(M × X) such that E′ = (j × id)_{∗}E . After that we shall prove that functors F and Φ_{E} are isomorphic.

Let L be a very ample invertible sheaf on M such that H^{i}(L^{k}) = 0 for any k > 0 , when i ≠ 0 . By j denote the closed embedding j : M ↪ ℙ^{N} with respect to L .

Recall that there exists a resolution of the diagonal on the product ℙ^{N} × ℙ^{N} (see[Be]). Let us consider the following complex of sheaves on the product:

0 → O(−N) ⊠ Ω^{N}(N) →^{d_{−N}} O(−N + 1) ⊠ Ω^{N−1}(N − 1) → ⋯ → O(−1) ⊠ Ω^{1}(1) →^{d_{−1}} O ⊠ O (6)

This complex is a resolution of the structure sheaf O_{Δ} of the diagonal Δ .

Now by F′ denote the functor from D^{b}(ℙ^{N}) to D^{b}(X) , which is the composition F ∘ j^{∗} . Consider the product

ℙ^{N} × X ⟶^{π′} X
q↓
ℙ^{N}

Denote by

d′_{−i} ∈ Hom_{ℙ^{N} ×X}(O(−i) ⊠ F′(Ω^{i}(i)) , O(−i + 1) ⊠ F′(Ω^{i−1}(i − 1)))

the image d_{−i} under the following through map.

Hom(O(−i) ⊠ Ω^{i}(i) , O(−i + 1) ⊠ Ω^{i−1}(i − 1)) ⟶^{∼}
Hom(O ⊠ Ω^{i}(i) , O(1) ⊠ Ω^{i−1}(i − 1)) ⟶^{∼}
Hom(Ω^{i}(i) , H^{0}(O(1)) ⊗ Ω^{i−1}(i − 1)) ⟶
Hom(F′(Ω^{i}(i)) , H^{0}(O(1)) ⊗ F′(Ω^{i−1}(i − 1))) ⟶^{∼}
Hom(O ⊠ F′(Ω^{i}(i)) , O(1) ⊠ F′(Ω^{i−1}(i − 1))) ⟶^{∼}
Hom(O(−i) ⊠ F′(Ω^{i}(i)) , O(−i + 1) ⊠ F′(Ω^{i−1}(i − 1)))

It can easily be checked that the composition d_{−i+1} ∘ d_{−i} is equal to 0 . We omit the check.

Consider the following complex C

C := {O(−N) ⊠ F′(Ω^{N}(N)) ⟶^{d′_{−N}} ⋯ ⟶ O(−1) ⊠ F′(Ω^{1}(1)) ⟶^{d′_{−1}} O ⊠ F′(O)}

over the derived category D^{b}(ℙ^{N} × X) . For l < 0 we have

Hom^{l}(O(−i) ⊠ F′(Ω^{i}(i)) , O(−k) ⊠ F′(Ω^{k}(k))) ≅
Hom^{l}(O ⊠ F′(Ω^{i}(i)) , H^{0}(O(i − k)) ⊗ F′(Ω^{k}(k))) ≅
Hom^{l}(j^{∗}(Ω^{i}(i)) , H^{0}(O(i − k)) ⊗ j^{∗}(Ω^{k}(k))) = 0

Hence, by Lemma 1.5. , there exists a convolution of the complex C , and all convolutions are isomorphic. By E′ denote some convolution of C and by γ_{0} denote the morphism O ⊠ F′(O) ⟶^{γ_{0}} E′ . (Further we shall see that all convolutions of C are canonically isomorphic). Now let Φ_{E′} be the functor from D^{b}(ℙ^{N}) to D^{b}(X) , defined by (5).

2.7. Lemma There exist canonically defined isomorphisms f_{k} : F′(O(k)) ⟶^{∼} Φ_{E′}(O(k)) for all k ∈ ℤ , and these isomorphisms are functorial, i.e. for any α : O(k) → O(l) the following diagram commutes

F′(O(k)) ⟶^{F′(α)} F′(O(l))
f_{k}↓  ↓f_{l}
Φ_{E′}(O(k)) ⟶^{Φ_{E′}(α)} Φ_{E′}(O(l))

Proof. At first, assume that k ≥ 0 .

Consider the resolution (6) of the diagonal Δ ⊂ ℙ^{N} × ℙ^{N} and, after tensoring it with O(k) ⊠ O , push forward onto the second component. We get the following resolution of O(k) on ℙ^{N}

{H^{0}(O(k − N)) ⊗ Ω^{N}(N)⟶ ⋯ ⟶ H^{0}(O(k − 1)) ⊗ Ω^{1}(1)⟶H^{0}(O(k)) ⊗ O} ⟶^{δ_{k}} O(k)

Consequently F′(O(k)) is a convolution of the complex D_{k} :

H^{0}(O(k − N)) ⊗ F′(Ω^{N}(N))⟶ ⋯ ⟶ H^{0}(O(k − 1)) ⊗ F′(Ω^{1}(1))⟶H^{0}(O(k)) ⊗ F′(O)

over D^{b}(X) .

On the other hand, let us consider the complex C_{k}:= q^{∗}O(k) ⊗ C on ℙ^{N} × X with the morphism γ_{k} : O(k) ⊠ F′(O) ⟶ q^{∗}O(k) ⊗ E′ , and push it forward onto the second component. It follows from the construction of the complex C that π′_{∗}(C_{k}) = D_{k} . So we see that F′(O(k)) and Φ_{E′}(O(k)) both are convolutions of the same complex D_{k} .

By assumption the functor F is full and faithful, hence, if G and H are locally free sheaves on ℙ^{N} then we have

Hom^{i}(F′(G) , F′(H)) = Hom^{i}(j^{∗}(G) , j^{∗}(H)) = 0

for i < 0 . Therefore the complex D_{k} satisfies the conditions (1) and (2) of Lemma 1.5. . Hence there exists a uniquely defined isomorphism f_{k} : F′(O(k)) ⟶^{∼} Φ_{E′}(O(k)) , completing the following commutative diagram

H^{0}(O(k)) ⊗ F′(O) ⟶^{F′(δ_{k})} F′(O(k))
id↓  ↓f_{k}
H^{0}(O(k)) ⊗ F′(O) ⟶^{π′_{∗}(γ_{k})} Φ_{E′}(O(k))

Now we have to show that these morphisms are functorial. For any α : O(k) → O(l) we have the commutative squares

H^{0}(O(k)) ⊗ F′(O) ⟶^{F′(δ_{k})} F′(O(k))
H^{0}(α) ⊗ id↓  ↓F′(α)
H^{0}(O(l)) ⊗ F′(O) ⟶^{F′(δ_{l})} F′(O(l))

and

H^{0}(O(k)) ⊗ F′(O) ⟶^{π′_{∗}(γ_{k})} Φ_{E′}(O(k))
H^{0}(α) ⊗ id↓  ↓Φ_{E′}(α)
H^{0}(O(l)) ⊗ F′(O) ⟶^{π′_{∗}(γ_{l})} Φ_{E′}(O(l))

Therefore we have the equalities:

\begin{matrix}f_{l} ∘ F′(α) ∘ F′(δ_{k}) = f_{l} ∘ F′(δ_{l}) ∘ (H^{0}(α) ⊗ id) = π′_{∗}(γ_{l}) ∘ (H^{0}(α) ⊗ id) = Φ_{E′}(α) ∘ π′_{∗}(γ_{k}) = \\ Φ_{E′}(α) ∘ f_{k} ∘ F′(δ_{k}) \end{matrix}

Since the complexes D_{k} and D_{l} satisfy the conditions of Lemma 1.6. there exists only one morphism h : F′(O(k)) → Φ_{E′}(O(l)) such that

h ∘ F′(δ_{k}) = π′_{∗}(γ_{l}) ∘ (H^{0}(α) ⊗ id)

Hence f_{l} ∘ F′(α) coincides with Φ_{E′}(α) ∘ f_{k} .

Now, consider the case k < 0 .

Let us take the following right resolution for O(k) on ℙ^{N} .

O(k) ⟶^{∼} {V_{0}^{k} ⊗ O ⟶ ⋯ ⟶ V_{N}^{k} ⊗ O(N)}

By Lemma 1.6. , the morphism of the complexes over D^{b}(X)

V_{0}^{k} ⊗ F′(O) ⟶ ⋯ ⟶ V_{N}^{k} ⊗ F′(O(N))
id ⊗ f_{0}↓≀  id ⊗ f_{N} ↓≀
V_{0}^{k} ⊗ Φ_{E′}(O) ⟶ ⋯ ⟶ V_{N}^{k} ⊗ Φ_{E′}(O(N))

gives us the uniquely determined morphism f_{k} : F′(O(k)) ⟶ Φ_{E′}(O(k)) .

It is not hard to prove that these morphisms are functorial. The proof is left to the reader. □

2.8. Now we must prove that there exists an object E ∈ D^{b}(M × X) such that j_{∗}E ≅ E′ .

Let L be a very ample invertible sheaf on M and let j : M ↪ ℙ^{N} be an embedding with respect to L . By A denote the graded algebra ⨁_{i=0}^{∞} H^{0}(M, L^{i}) .

Let B_{0} = k , and B_{1} = A_{1} . For m ≥ 2 , we define B_{m} as

B_{m} = Ker(B_{m−1} ⊗ A_{1} ⟶ B_{m−2} ⊗ A_{2})  (7)

2.9. Definition A is said to be n -Koszul if the following sequence is exact

B_{n} ⊗_{k} A ⟶ B_{n−1} ⊗_{k} A ⟶ ⋯ ⟶ B_{1} ⊗_{k} A ⟶ A ⟶ k ⟶ 0

Assume that A is n-Koszul. Let R_{0} = O_{M} . For m ≥ 1 , denote by R_{m} the kernel of the morphism B_{m} ⊗ O_{M} ⟶ B_{m−1} ⊗ L . Using (7), we obtain the canonical morphism R_{m} ⟶ A_{1} ⊗ R_{m−1} . (actually, Hom(R_{m} , R_{m−1}) ≅ A_{1}^{∗} ).

Since A is n -Koszul, we have the exact sequences

0 ⟶ R_{m} ⟶ B_{m} ⊗ O_{M} ⟶ B_{m−1} ⊗ L ⟶ ⋯ ⟶ B_{1} ⊗ L^{m−1} ⟶ L^{m} ⟶ 0

for m ≤ n .

We have the canonical morphisms f_{m} : j^{∗}Ω^{m}(m) ⟶ R_{m} , because Λ^{i}A_{1} ⊂ B_{i} and there exist the exact sequences on ℙ^{N}

0 ⟶ Ω^{m}(m) ⟶ Λ^{m}A_{1} ⊗ O ⟶ Λ^{m−1}A_{1} ⊗ O(1) ⟶ ⋯ ⟶ O(m) ⟶ 0

It is known that for any n there exists l such that the Veronese algebra A^{l} = ⨁_{i=0}^{∞} H^{0}(M, L^{il}) is n -Koszul.( Moreover, it was proved in [Ba] that A^{l} is Koszul for

l ≫ 0 ).

Using the technique of [IM] and substituting L with L^{j} , when j is sufficiently large , we can choose for any n a very ample L such that

1) algebra A is n -Koszul,

2) the complex

L^{−n} ⊠ R_{n} ⟶ ⋯ ⟶ L^{−1} ⊠ R_{1} ⟶ O_{M} ⊠ R_{0} ⟶ O_{Δ}

on M × M is exact, 3) the following sequences on M .

A_{k−n} ⊗ R_{n} ⟶ A_{k−n+1} ⊗ R_{n−1} ⟶ ⋯ ⟶ A_{k−1} ⊗ R_{1} ⟶ A_{k} ⊗ R_{0} ⟶ L^{k} ⟶ 0

are exact for any k ≥ 0 . Here, by definition, if k − i < 0 , then A_{k−i} = 0 . (see Appendix for proof).

Let us denote by T_{k} the kernel of the morphism A_{k−n} ⊗ R_{n} ⟶ A_{k−n+1} ⊗ R_{n−1} . Consider the following complex over D^{b}(M × X)

L^{−n} ⊠ F (R_{n}) ⟶ ⋯ ⟶ L^{−1} ⊠ F (R_{1}) ⟶ O_{M} ⊠ F (R_{0})  (8)

Here the morphism L^{−k} ⊠ F (R_{k}) ⟶ L^{−k+1} ⊠ F (R_{k−1}) is induced by the canonical morphism R_{k} ⟶ A_{1} ⊗ R_{k−1} with respect to the following sequence of isomorphisms

Hom(L^{−k} ⊠ F (R_{k}) , L^{−k+1} ⊠ F (R_{k−1})) ≅ Hom(F (R_{k}) , H^{0}(L) ⊗ F (R_{k−1})) ≅
≅ Hom(R_{k} , A_{1} ⊗ R_{k−1})

By Lemma 1.5. , there is a convolution of the complex (8) and all convolutions are isomorphic. Let G ∈ D^{b}(M × X) be a convolution of this complex.

For any k ≥ 0 , object π_{∗}(G ⊗ p^{∗}(L^{k})) is a convolution of the complex

A_{k−n} ⊗ F (R_{n}) ⟶ A_{k−n+1} ⊗ F (R_{n−1}) ⟶ ⋯ ⟶ A_{k} ⊗ F (R_{0}).

On the other side, we know that T_{k}[n] ⊕ L^{k} is a convolution of the complex

A_{k−n} ⊗ R_{n} ⟶ A_{k−n+1} ⊗ R_{n−1} ⟶ ⋯ ⟶ A_{k} ⊗ R_{0},

because Ext^{n+1}(L^{k} , T_{k}) = 0 for n ≫ 0 . Therefore, by Lemma 1.5. , we have π_{∗}(G ⊗ p^{∗}(L^{k})) ≅ F (T_{k}[n] ⊕ L^{k}) .

It follows immediately from Remark 2.5. that the cohomology sheaves H^{i}(π_{∗}(G ⊗ p^{∗}(L^{k}))) = H^{i}(F (T_{k})[n]) ⊕ H^{i}(F (L^{k})) concentrate on the union [−n − a, −n] ∪ [−a, 0] for any k > 0 ( a was defined in 2.5. ). Therefore the cohomology sheaves H^{i}(G) also concentrate on [−n − a, −n] ∪ [−a, 0] . We can assume that n > dimM + dimX + a .

This implies that G ≅ C ⊕ E , where E, C are objects of D^{b}(M × X) such that H^{i}(E) = 0 for i ∉ [−a, 0] and H^{i}(C) = 0 for i ∉ [−n − a, −n] . Moreover, we have π_{∗}(E ⊗ p^{∗}(L^{k})) ≅ F (L^{k}) .

Now we show that j_{∗}(E) ≅ E′ . Let us consider the morphism of the complexes over D^{b}(ℙ^{N} × X) .

O(−n) ⊠ F′(Ω^{n}(n)) ⟶ ⋯ ⟶  O ⊠ F′(O)
↓can ⊠ F (f_{n})  ↓can ⊠ F (f_{0})
j_{∗}(L^{−n}) ⊠ F (R_{n}) ⟶ ⋯ ⟶ j_{∗}(O_{M}) ⊠ F (R_{0})

By Lemma 1.6. , there exists a morphism of convolutions ϕ : K ⟶ j_{∗}(G) . If N > n , then K is not isomorphic to E′ , but there is a distinguished triangle

S ⟶ K ⟶ E′ ⟶ S[1]

and the cohomology sheaves H^{i}(S) ≠ 0 only if i ∈ [−n − a, −n] . Now, since Hom(S , j_{∗}(E)) = 0 and Hom(S[1] , j_{∗}(E)) = 0 , we have a uniquely determined morphism ψ : E′ ⟶ j_{∗}(E) such that the following diagram commutes

K ⟶^{ϕ} j_{∗}(G)
↓ ↓
E′ ⟶^{ψ} j_{∗}(E)

We know that π′_{∗}(E′ ⊗ q^{∗}(O(k))) ≅ F (L^{k}) ≅ π_{∗}(E ⊗ p^{∗}(L^{k})) . Let ψ_{k} be the morphism π′_{∗}(E′⊗q^{∗}(O(k))) ⟶ π_{∗}(E ⊗p^{∗}(L^{k})) induced by ψ . The morphism ψ_{k} can be included in the following commutative diagram:

S^{k}A_{1} ⊗ F (O) ⟶^{can} F (L^{k}) ⟶^{∼} π′_{∗}(E′ ⊗ q^{∗}(O(k)))
can↓  ↓ψ_{k}
A_{k} ⊗ F (O) ⟶^{can} F (L^{k}) ⟶^{∼} π_{∗}(E ⊗ p^{∗}(L^{k}))

Thus we see that ψ_{k} is an isomorphism for any k ≥ 0 . Hence ψ is an isomorphism too. This proves the following:

2.10. Lemma There exists an object E ∈ D^{b}(M × X) such that j_{∗}(E) ≅ E′ , where E′ is the object from D^{b}(ℙ^{N} × X) , constructed in 2.6. .

2.11. Now, we prove some statements relating to abelian categories. they are needed for the sequel.

Let A be a k -linear abelian category (henceforth we shall consider only k -linear abelian categories). Let {P_{i}}_{i∈ℤ} be a sequence of objects from A .

2.12. Definition We say that this sequence is ample if for every object X ∈ A there exists N such that for all i < N the following conditions hold:

a) the canonical morphism Hom(P_{i} , X) ⊗ P_{i} ⟶ X is surjective,

b) Ext^{j}(P_{i} , X) = 0 for any j ≠ 0 ,

c) Hom(X , P_{i}) = 0 .

It is clear that if L is an ample invertible sheaf on a projective variety in usual sense, then the sequence {L^{i}}_{i∈ℤ} in the abelian category of coherent sheaves is ample.

2.13. Lemma Let {P_{i}} be an ample sequence in an abelian category A . If X is an object in D^{b}(A) such that Hom (P_{i} , X) = 0 for all i ≪ 0 , then X is the zero object.

Proof. If i ≪ 0 then

Hom(P_{i} , H^{k}(X)) ≅ Hom^{k}(P_{i} , X) = 0

The morphism Hom(P_{i} , H^{k}(X)) ⊗ P_{i} ⟶ H^{k}(X) must be surjective for i ≪ 0 , hence H^{k}(X) = 0 for all k . Thus X is the zero object. □

2.14. Lemma Let {P_{i}} be an ample sequence in an abelian category A of finite homological dimension. If X is an object in D^{b}(A) such that Hom (X , P_{i}) = 0 for all i ≪ 0 . Then X is the zero object.

Proof. Assume that the cohomology objects of X are concentrated in a segment [a, 0] . There exists the canonical morphism X ⟶ H^{0}(X) . Consider a surjective morphism P_{i_{1}}^{⊕k_{1}} ⟶ H^{0}(X) . By Y_{1} denote the kernel of this morphism. Since Hom (X , P_{i_{1}}) = 0 we have Hom^{1}(X , Y_{1}) ≠ 0 . Further take a surjective morphism P_{i_{2}}^{⊕k_{2}} ⟶ Y_{1} . By Y_{2} denote the kernel of this morphism. Again, since Hom (X , P_{i_{2}}) = 0 , we obtain Hom^{2}(X , Y_{2}) ≠ 0 . Iterating this procedure as needed, we get contradiction with the assumption that A is of finite homological dimension. □

2.15. Lemma Let B be an abelian category, A an abelian category of finite homological dimension, and {P_{i}} an ample sequence in A . Suppose F is an exact functor from D^{b}(A) to D^{b}(B) such that it has right and left adjoint functors F^{!} and F^{∗} respectively. If the maps

Hom^{k}(P_{i} , P_{j}) ⟶^{∼} Hom^{k}(F (P_{i}) , F (P_{j}))

are isomorphisms for i < j and all k . Then F is full and faithful.

Proof. Let us take the canonical morphism f_{i} : P_{i} ⟶ F^{!}F (P_{i}) and consider a distinguished triangle

P_{i} ⟶^{f_{i}} F^{!}F (P_{i}) ⟶ C_{i} ⟶ P_{i}[1].

Since for j ≪ 0 we have isomorphisms:

Hom^{k}(P_{j} , P_{i}) ⟶^{∼} Hom^{k}(F (P_{j}) , F (P_{i})) ≅ Hom^{k}(P_{j} , F^{!}F (P_{i})).

We see that Hom (P_{j} , C_{i}) = 0 for j ≪ 0 . It follows from Lemma 2.13. that C_{i} = 0 . Hence f_{i} is an isomorphism.

Now, take the canonical morphism g_{X} : F^{∗}F (X) ⟶ X and consider a distinguished triangle

F^{∗}F (X) ⟶^{g_{X}} X ⟶ C_{X} ⟶ F^{∗}F (X)[1]

We have the following sequence of isomorphisms

Hom^{k}(X , P_{i}) ⟶^{∼} Hom^{k}(X , F^{!}F (P_{i})) ≅ Hom^{k}(F^{∗}F (X) , P_{i})

This implies that Hom (C_{X} , P_{i}) = 0 for all i . By Lemma 2.14. , we obtain C_{X} = 0 . Hence g_{X} is an isomorphism. It follows that F is full and faithful. □

Let A be an abelian category possessing an ample sequence {P_{i}} . Denote by D^{b}(A) the bounded derived category of A . Let us consider the full subcategory j : C ↪ D^{b}(A) such that ObC := {P_{i} | i ∈ ℤ} . Now we would like to show that if there exists an isomorphism of a functor F : D^{b}(A) ⟶ D^{b}(A) to identity functor on the subcategory C , then it can be extended to the whole D^{b}(A) .

2.16. Proposition Let F : D^{b}(A) ⟶ D^{b}(A) be an autoequivalence. Suppose there exists an isomorphism f : j ⟶^{∼} F |_{C} ( where j : C ↪ D^{b}(A) is a natural embedding). Then it can be extended to an isomorphism id ⟶^{∼} F on the whole D^{b}(A) .

Proof. First, we can extend the transformation f to all direct sums of objects C componentwise , because F takes direct sums to direct sums.

Note that X ∈ D^{b}(A) is isomorphic to an object in A iff Hom^{j}(P_{i} , X) = 0 for j ≠ 0 and i ≪ 0 . It follows that F (X) is isomorphic to an object in A , because

Hom^{j}(P_{i} , F (X)) ≅ Hom^{j}(F (P_{i}) , F (X)) ≅ Hom^{j}(P_{i} , X) = 0

for j ≠ 0 and i ≪ 0 .

2.16.1 At first, let X be an object from A . Take a surjective morphism v : P_{i}^{⊕k} ⟶ X . We have the morphism f_{i} : P_{i}^{⊕k} ⟶ F (P_{i}^{⊕k}) and two distinguished triangles:

Y ⟶^{u} P_{i}^{⊕k}  ⟶^{v} X ⟶ Y [1]
↓f_{i}
F (Y) ⟶^{F (u)} F (P_{i}^{⊕k}) ⟶^{F (v)} F (X) ⟶ F (Y)[1]

Now we show that F (v)∘f_{i} ∘u = 0 . Consider any surjective morphism w : P_{j}^{⊕l} ⟶ Y . It is sufficient to check that F (v)∘f_{i}∘u∘w = 0 . Let f_{j} : P_{j}^{⊕l} ⟶ F (P_{j}^{⊕l}) be the canonical morphism. Using the commutation relations for f_{i} and f_{j} , we obtain

F (v) ∘ f_{i} ∘ u ∘ w = F (v) ∘ F (u ∘ w) ∘ f_{j} = F (v ∘ u ∘ w) ∘ f_{j} = 0

because v ∘ u = 0 .

Since Hom(Y [1] , F (X)) = 0 , by Lemma 1.4. , there exists a unique morphism f_{X} : X ⟶ F (X) that commutes with f_{i} .

2.16.2 Let us show that f_{X} does not depend from morphism v : P_{i}^{⊕k} ⟶ X . Consider two surjective morphisms v_{1} : P_{i_{1}}^{⊕k_{1}} ⟶ X and v_{2} : P_{i_{2}}^{⊕k_{2}} ⟶ X . We can take two surjective morphisms w_{1} : P_{j}^{⊕l} ⟶ P_{i_{1}}^{⊕k_{1}} and w_{2} : P_{j}^{⊕l} ⟶ P_{i_{2}}^{⊕k_{2}} such that the following diagram is commutative:

P_{j}^{⊕l} ⟶^{w_{2}} P_{i_{2}}^{⊕k_{2}}
↓w_{1}  ↓v_{2}
P_{i_{1}}^{⊕k_{1}} ⟶^{v_{1}} X

Clearly, it is sufficient to check the coincidence of the morphisms, constructed by v_{1} and v_{1} ∘ w_{1} . Now, let us consider the following commutative diagram:

P_{j}^{⊕l}  ⟶^{w_{1}} P_{i_{1}}^{⊕k_{1}}  ⟶^{v_{1}} X
↓f_{j}  ↓v_{2}  ↓f_{X}
F (P_{j}^{⊕l}) ⟶^{F (w_{1})} F (P_{i_{1}}^{⊕k_{1}}) ⟶^{F (v_{1})} F (X)

Here the morphism f_{X} is constructed by v_{1} . Both squares of this diagram are commutative. Since there exists only one morphism from X to F (X) that commutes with f_{j} , we see that the f_{X} , constructed by v_{1} , coincides with the morphism, constructed by v_{1} ∘ w_{1} .

2.16.3 Now we must show that for any morphism X ⟶^{ϕ} Y we have equality:

f_{Y} ∘ ϕ = F (ϕ) ∘ f_{X}

Take a surjective morphism P_{j}^{⊕l} ⟶^{v} Y . Choose a surjective morphism P_{i}^{⊕k} ⟶^{u} X such that the composition ϕ∘u lifts to a morphism ψ : P_{i}^{⊕k}⟶P_{j}^{⊕l} . We can do it, because for i ≪ 0 the map Hom(P_{i}^{⊕k} , P_{j}^{⊕l}) → Hom(P_{i}^{⊕k} , Y) is surjective. We get the commutative square:

P_{i}^{⊕k} ⟶^{u} X
↓ψ  ↓ϕ
P_{j}^{⊕l} ⟶^{v} Y

By h_{1} and h_{2} denote f_{Y} ∘ ϕ and F (ϕ) ∘ f_{X} respectively. We have the following sequence of equalities:

h_{1} ∘ u = f_{Y} ∘ ϕ ∘ u = f_{Y} ∘ v ∘ ψ = F (v) ∘ f_{j} ∘ ψ = F (v) ∘ F (ψ) ∘ f_{i}

and

h_{2} ∘ u = F (ϕ) ∘ f_{X} ∘ u = F (ϕ) ∘ F (u) ∘ f_{i} = F (ϕ ∘ u) ∘ f_{i} = F (v ∘ ψ) ∘ f_{i} = F (v) ∘ F (ψ) ∘ f_{i}

Consequently, the following square is commutative for t = 1, 2 .

Z ⟶ P_{i}^{⊕k}  ⟶^{u} X ⟶ Z[1]
F (ψ) ∘ f_{i}↓  ↓h_{t}
F (W) ⟶ F (P_{j}^{⊕l}) ⟶^{F (v)} F (Y) ⟶ F (W)[1]

By Lemma 1.4. , as Hom(Z[1] , F (Y)) = 0 , we obtain h_{1} = h_{2} . Thus, f_{Y} ∘ ϕ = F (ϕ) ∘ f_{X} .

Now take a cone C_{X} of the morphism f_{X} . Using the following isomorphisms

Hom(P_{i} , X) ≅ Hom(F (P_{i}) , F (X)) ≅ Hom(P_{i} , F (X)),

we obtain Hom^{j}(P_{i} , C_{X}) = 0 for all j , when i ≪ 0 . Hence, by Lemma 2.13. , C_{X} = 0 and f_{X} is an isomorphism.

2.16.4 Let us define f_{X[n]} : X[n] ⟶ F (X[n]) ≅ F (X)[n] for any X ∈ A by

f_{X[n]} = f_{X}[n].

It is easily shown that these transformations commute with any u ∈ Ext^{k}(X , Y) . Indeed, since any element u ∈ Ext^{k}(X , Y) can be represented as a composition u = u_{0}u_{1} ⋯ u_{k} of some elements u_{i} ∈ Ext^{1}(Z_{i} , Z_{i+1}) and Z_{0} = X, Z_{k} = Y , we have only to check it for u ∈ Ext^{1}(X , Y) . Consider the following diagram:

Y ⟶ Z ⟶ X ⟶^{u} Y [1]
f_{Y} ↓  ↓f_{Z}  ↓f_{Y} [1]
F (Y) ⟶ F (Z) ⟶ F (X) ⟶^{F (u)} F (Y)[1]

By an axiom of triangulated categories there exists a morphism h : X → F (X) such that (f_{Y} , f_{Z}, h) is a morphism of triangles. On the other hand, since Hom(Y [1] , F (X)) = 0 , by Lemma 1.4. , h is a unique morphism that commutes with f_{Z} . But f_{X} also commutes with f_{Z} . Hence we have h = f_{X} . This implies that

f_{Y} [1] ∘ u = F (u) ∘ f_{X}

2.16.5 The rest of the proof is by induction over the length of a segment, in which the cohomology objects of X are concentrated. Let X be an object from D^{b}(A) and suppose that its cohomology objects H^{p}(X) are concentrated in a segment [a, 0] . Take v : P_{i}^{⊕k} ⟶ X such that

a) Hom^{j}(P_{i} , H^{p}(X)) = 0 for all p and for j ≠ 0,
b) u : P_{i}^{⊕k} ⟶ H^{0}(X) is the surjective morphism,  (9)
c) Hom(H^{0}(X) , P_{i}) = 0.

Here u is the composition v with the canonical morphism X ⟶ H^{0}(X) . Consider a distinguished triangle:

Y [−1] ⟶ P_{i}^{⊕k} ⟶^{v} X ⟶ Y

By the induction hypothesis, there exists the isomorphism f_{Y} and it commutes with f_{i} . So we have the commutative diagram:

Y [−1] ⟶ P_{i}^{⊕k}  ⟶^{v} X ⟶ Y
f_{Y} [−1]↓  ↓f_{i}  ↓f_{Y}
F (Y)[−1] ⟶ F (P_{i}^{⊕k}) ⟶^{F (v)} F (X) ⟶ F (Y)

Moreover we have the following sequence of equalities

Hom(X , F (P_{i}^{⊕k})) ≅ Hom(X , P_{i}^{⊕k}) ≅ Hom(H^{0}(X) , P_{i}^{⊕k}) = 0

Hence, by Lemma 1.4. , there exists a unique morphism f_{X} : X ⟶ F (X) that commutes with f_{Y} .

2.16.6 We must first show that f_{X} is correctly defined. Suppose we have two morphisms v_{1} : P_{i_{1}}^{⊕k_{1}} ⟶ X and v_{2} : P_{i_{2}}^{⊕k_{2}} ⟶ X . As above, we can find P_{j} and surjective morphisms w_{1}, w_{2} such that the following diagram is commutative

P_{j}^{⊕l} ⟶^{w_{2}} P_{i_{2}}^{⊕k_{2}}
↓w_{1}  ↓u_{2}
P_{i_{1}}^{⊕k_{1}} ⟶^{u_{1}} H^{0}(X)

We can find a morphism ϕ : Y_{j} ⟶ Y_{i_{1}} such that the triple (w_{1}, id, ϕ) is a morphism of distinguished triangles.

P_{j}^{⊕l} ⟶^{v_{1}∘w_{1}} X ⟶ Y_{j} ⟶ P_{j}^{⊕l}[1]
w_{1}↓  ↓id  ↓ϕ  ↓w_{1}[1]
P_{i_{1}}^{⊕k_{1}}  ⟶^{v_{1}} X ⟶ Y_{i_{1}} ⟶ P_{i_{1}}^{⊕k_{1}}[1]

By the induction hypothesis, the following square is commutative.

Y_{j}  ⟶^{ϕ} Y_{i_{1}}
f_{Y_{j}} ↓  ↓f_{Y_{i}}
F (Y_{j}) ⟶^{F (ϕ)} F (Y_{i_{1}})

Hence, we see that the f_{X} , constructed by v_{1} ∘ w_{1} , commutes with f_{Y_{i_{1}}} and, consequently, coincides with the f_{X} , constructed by v_{1} ; because such morphism is unique by Lemma 1.4. . Therefore morphism f_{X} does not depend on a choice of morphism v : P_{i}^{⊕k} ⟶ X .

2.16.7 Finally, let us prove that for any morphism ϕ : X ⟶ Y the following diagram commutes

\begin{matrix}X & ⟶^{ϕ} & Y \\ f_{X} ↓ & & ↓f_{Y} \\ F (X) & ⟶^{F (ϕ)} & F (Y) \end{matrix}  (10)

Suppose the cohomology objects of X are concentrated on a segment [a, 0] and the cohomology objects of Y are concentrated on [b, c] . Case 1. If c < 0 , we take a morphism v : P_{i}^{⊕k} ⟶ X that satisfies conditions (9) and Hom(P_{i}^{⊕k} , Y) = 0 . We have a distinguished triangle:

P_{i}^{⊕k} ⟶^{v_{1}} X ⟶^{α} Z ⟶ P_{i}^{⊕k}[1]

Applying the functor Hom(− , Y) to this triangle we found that there exists a morphism ψ : Z ⟶ Y such that ϕ = ψ ∘ α . We know that f_{X} , defined above, satisfy

F (α) ∘ f_{X} = f_{Z} ∘ α

If we assume that the diagram

Z ⟶^{ψ} Y
f_{Z} ↓  ↓f_{Y}
F (Z) ⟶^{F (ψ)} F (Y)

commutes, then diagram (10) commutes too.

This means that for verifying the commutativity of (10) we can substitute X by an object Z . And the cohomology objects of Z are concentrated on the segment [a, −1] . Case 2. If c ≥ 0 , we take a surjective morphism v : P_{i}^{⊕k} ⟶ Y [c] that satisfies conditions (9) and Hom(H^{c}(X) , P_{i}^{⊕k}) = 0 . Consider a distinguished triangle

P_{i}^{⊕k}[−c] ⟶^{v[−c]} Y ⟶^{β} W ⟶ P_{i}^{⊕k}[−c + 1]

Note that the cohomology objects of W are concentrated on [b, c − 1] .

By ψ denote the composition β ∘ ϕ . If we assume that the following square

X ⟶^{ψ} W
f_{X} ↓  ↓f_{W}
F (X) ⟶^{F (ψ)} F (W)

commutes, then, since F (β) ∘ f_{Y} = f_{W} ∘ β ,

F (β) ∘ (f_{Y} ∘ ϕ − F (ϕ) ∘ f_{X}) = f_{W} ∘ ψ − F (ψ) ∘ f_{X} = 0.

We chose P_{i} such that Hom(X , P_{i}^{⊕k}[−c]) = 0 . As F (P_{i}^{⊕k}) is isomorphic to P_{i}^{⊕k} , then Hom(X , F (P_{i}^{⊕k}[−c])) = 0 . Applying the functor Hom(X , F (−)) to the above triangle we found that the composition with F (β) gives an inclusion of Hom(X , F (Y)) into Hom(X , F (W)) . This follows that f_{Y} ∘ ϕ = F (ϕ) ∘ f_{X} , i.e. our diagram (10) commutes.

Combining case 1 and case 2, we can reduce the checking of commutativity of diagram (10) to the case when X and Y are objects from the abelian category A . But for those it has already been done. Thus the proposition is proved. □

2.17. Proof of theorem. 1) Existence. Using Lemma 2.10. and Lemma 2.7. , we can construct an object E ∈ D^{b}(M × X) such that there exists an isomorphism of the functors f̄: F |_{C}⟶^{∼} Φ_{E}|_{C} on full subcategory C ⊂ D^{b}(M) , where ObC = {L^{i} | i ∈ ℤ} and L is a very ample invertible sheaf on M such that for any k > 0 H^{i}(M, L^{k}) = 0 , when i ≠ 0 .

By Lemma 2.15. the functor Φ_{E} is full and faithfull. Moreover, the functors F^{!} ∘ Φ_{E} and Φ_{E}^{∗} ∘ F are full and faithful too, because we have the isomorphisms:

F^{!}(f̄) : F^{!} ∘ F |_{C}≅ id_{C} ⟶^{∼} F^{!} ∘ Φ_{E}|_{C}

Φ_{E}^{∗}(f̄) : Φ_{E}^{∗} ∘ F |_{C}⟶^{∼} Φ_{E}^{∗} ∘ Φ_{E}|_{C}≅ id_{C}

and conditions of Lemma 2.15. is fulfilled.

Further, the functors F^{!} ∘ Φ_{E} and Φ_{E}^{∗} ∘ F are equivalences, because they are adjoint each other.

Consider the isomorphism F^{!}(f̄) : F^{!} ∘ F |_{C}≅ id_{C} ⟶^{∼} F^{!} ∘ Φ_{E}|_{C} on the subcategory C . By Proposition 2.16. we can extend it onto the whole D^{b}(M) , so id ⟶^{∼} F^{!} ∘ Φ_{E} .

Since F^{!} is the right adjoint to F , we get the morphism of the functors f : F ⟶ Φ_{E} such that f|_{C} = f̄ . It can easily be checked that f is an isomorphism. Indeed, let C_{Z} be a cone of the morphism f_{Z} : F (Z) ⟶ Φ_{E}(Z) . Since F^{!}(f_{Z}) is an isomorphism, we obtain F^{!}(Z) = 0 . Therefore, this implies that Hom(F (Y) , C_{Z}) = 0 for any object Y . Further, there are isomorphisms F (L^{k}) ≅ Φ_{E}(L^{k}) for any k . Hence, we have

Hom^{i}(L^{k} , Φ_{E}^{!}(C_{Z})) = Hom^{i}(Φ_{E}(L^{k}) , C_{Z})) = Hom^{i}(F (L^{k}) , C_{Z})) = 0

for all k and i .

Thus, we obtain Φ_{E}^{!}(C_{Z}) = 0 . This implies that Hom(Φ_{E}(Z) , C_{Z}) = 0 . Finally, we get F (Z) = C_{Z}[−1] ⊕ Φ_{E}(Z) . But we know that Hom(F (Z)[1] , C_{Z}) = 0 . Thus, C_{Z} = 0 and f is an isomorphism.

2) Uniqueness. Suppose there exist two objects E and E_{1} of D^{b}(M × X) such that Φ_{E_{1}} ≅ F ≅ Φ_{E_{2}} . Let us consider the complex (8) over D^{b}(M × X) (see the proof Lemma 2.10. ).

L^{−n} ⊠ F (R_{n}) ⟶ ⋯ ⟶ L^{−1} ⊠ F (R_{1}) ⟶ O_{M} ⊠ F (R_{0})

By Lemma 1.5. , there exists a convolution of this complex and all convolutions are isomorphic. Let G ∈ D^{b}(M × X) be a convolution of the complex (8). Now consider the following complexes

L^{−n} ⊠ F (R_{n}) ⟶ ⋯ ⟶ L^{−1} ⊠ F (R_{1}) ⟶ O_{M} ⊠ F (R_{0}) ⟶ E_{k}

Again by Lemma 1.5. , there exists a unique up to isomorphism convolutions of these complexes.

Hence we have the canonical morphisms G ⟶ E_{1} and G ⟶ E_{2} . Moreover, it has been proved above (see the proof of Lemma 2.10. ) that C_{1} ⊕ E_{1} ≅ G ≅ C_{2} ⊕ E_{2} for large n , where E_{k}, C_{k} are objects of D^{b}(M × X) such that H^{i}(E_{k}) = 0 for i ∉ [−a, 0] and H^{i}(C_{k}) = 0 for i ∉ [−n − a, −n] ( a was defined in 2.5. ). Thus E_{1} and E_{2} are isomorphic.

This completes the proof of Theorem 2.2. □

2.18. Theorem Let M and X be smooth projective varieties. Suppose F : D^{b}(M) ⟶ D^{b}(X) is an equivalence. Then there exists a unique up to isomorphism object E ∈ D^{b}(M × X) such that the functors F and Φ_{E} are isomorphic.

It follows immediately from Theorem 2.2.

§3. Derived categories of K3 surfaces

3.1. In this chapter we are trying to answer the following question: When are derived categories of coherent sheaves on two different K3 surfaces over field ℂ equivalent?

This question is interesting, because there exists a procedure for recovering a variety from its derived category of coherent sheaves if the canonical (or anticanonical) sheaf is ample. Besides, if D^{b}(X) ≃ D^{b}(Y) and X is a smooth projective K3 surface, then Y is also a smooth projective K3 surface. This is true, because the dimension of a variety and Serre functor are invariants of a derived category.

The following theorem is proved in [BO2].

3.2. Theorem (see [BO2]) Let X be smooth irreducible projective variety with either ample canonical or ample anticanonical sheaf. If D = D^{b}(X) is equivalent to D^{b}(X′) for some other smooth algebraic variety, then X is isomorphic to X′ .

However, there exist examples of varieties that have equivalent derived categories, if the canonical sheaf is not ample. Here we give an explicit description for K3 surfaces with equivalent derived categories.

3.3. Theorem Let S_{1} and S_{2} be smooth projective K3 surfaces over field ℂ . Then the derived categories D^{b}(S_{1}) and D^{b}(S_{2}) are equivalent as triangulated categories iff there exists a Hodge isometry f_{τ} : T_{S_{1}} ⟶^{∼} T_{S_{2}} between the lattices of transcendental cycles of S_{1} and S_{2} .

Recall that the lattice of transcendental cycles T_{S} is the orthogonal complement to Neron-Severi lattice N_{S} in H^{2}(S, ℤ) . Hodge isometry means that the one-dimensional subspace H^{2,0}(S_{1}) ⊂ T_{S_{1}} ⊗ ℂ goes to H^{2,0}(S_{2}) ⊂ T_{S_{2}} ⊗ ℂ .

Now we need some basic facts about K3 surfaces (see [Mu2]). If S is a K3 surface, then the Todd class td_{S} of S is equal to 1 + 2w , where 1 ∈ H^{0}(S, ℤ) is the unit element of the cohomology ring H^{∗}(S, ℤ) and w ∈ H^{4}(S, ℤ) is the fundamental cocycle of S . The positive square root \sqrt{td_{S}} = 1 + w lies in H^{∗}(S, ℤ) too.

Let E be an object of D^{b}(S) then the Chern character

ch(E) = r(E) + c_{1}(E) + \frac{1}{2}(c_{1}^{2} − 2c_{2})

belongs to integral cohomology H^{∗}(S, ℤ) .

For an object E , we put v(E) = ch(E)\sqrt{td_{S}} ∈ H^{∗}(S, ℤ) and call it the vector associated to E (or Mukai vector).

We can define a symmetric integral bilinear form (,) on H^{∗}(S, ℤ) by the rule

(u, u′) = rs′ + sr′ − αα′ ∈ H^{4}(S, ℤ) ≅ ℤ

for every pair u = (r, α, s), u′ = (r′, α′, s′) ∈ H^{0}(S, ℤ) ⊕ H^{2}(S, ℤ) ⊕ H^{4}(S, ℤ) . By H̃(S, ℤ) denote H^{∗}(S, ℤ) with this inner product (,) and call it Mukai lattice.

For any objects E and F , inner product (v(E), v(F)) is equal to the H^{4} component of ch(E)^{∨} · ch(F) · td_{S} . Hence, by Riemann-Roch- Grothendieck theorem, we have

(v(E), v(F)) = χ(E, F) := ∑_{i}(−1)^{i}dimExt^{i}(E , F)

Let us suppose that D^{b}(S_{1}) and D^{b}(S_{2}) are equivalent. By Theorem 2.2. there exists an object E ∈ D^{b}(S_{1} × S_{2}) such that the functor Φ_{E} gives this equivalence.

Now consider the algebraic cycle Z := p^{∗}\sqrt{td_{S_{1}}} · ch(E) · π^{∗}\sqrt{td_{S_{2}}} on the product S_{1} × S_{2} , where p and π are the projections

S_{1} × S_{2} ⟶^{π} S_{2}
p↓
S_{1}

It follows from the following lemma that the cycle Z belongs to integral cohomology H^{∗}(S_{1} × S_{2}, ℤ) .

3.4. Lemma [Mu2] For any object E ∈ D^{b}(S_{1} × S_{2}) the Chern character ch(E) is integral, which means that it belongs to H^{∗}(S_{1} × S_{2}, ℤ)

The cycle Z defines a homomorphism from integral cohomology of S_{1} to integral cohomology of S_{2} :

f : H^{∗}(S_{1}, ℤ) ⟶ H^{∗}(S_{2}, ℤ)
∪  ∪
α  ↦ π_{∗}(Z · p^{∗}(α))

The following proposition is similar to Theorem 4.9 from [Mu2].

3.5. Proposition If Φ_{E} is full and faithful functor from D^{b}(S_{1}) to D^{b}(S_{2}) then: 1) f is an isometry between H̃(S_{1}, ℤ) and H̃(S_{2}, ℤ) ,

2) the inverse of f is equal to the homomorphism

f′ : H^{∗}(S_{2}, ℤ) ⟶ H^{∗}(S_{1}, ℤ)
∪  ∪
β  ↦ p_{∗}(Z^{∨} · π^{∗}(β))

defined by Z^{∨} = p^{∗}\sqrt{td_{S_{1}}} · ch(E^{∨}) · π^{∗}\sqrt{td_{S_{2}}} , where E^{∨} := R R R Hom(E, O_{S_{1}×S_{2}}) .

Proof. The left and right adjoint functors to Φ_{E} are:

Φ_{E}^{∗} = Φ_{E}^{!} = p_{∗}(E^{∨} ⊗ π^{∗}( ))[2]

Since Φ_{E} is full and faithful, the composition Φ_{E}^{∗} ∘ Φ_{E} is isomorphic to id_{D^{b}(S_{1})} . Functor id_{D^{b}(S_{1})} is given by the structure sheaf O_{Δ} of the diagonal Δ ⊂ S_{1} × S_{1} . Using the projection formula and Grothendieck-Riemann-Roch theorem, it can easily be

shown that the composition f′ ∘ f is given by the cycle p_{1}^{∗}\sqrt{td_{S_{1}}} · ch(O_{Δ}) · p_{2}^{∗}\sqrt{td_{S_{1}}} , where p_{1}, p_{2} are the projections of S_{1} × S_{1} to the summands. But this cycle is equal to Δ .

Therefore, f′ ∘ f is the identity, and, hence, f is an isomorphism of the lattices, because these lattices are free abelian groups of the same rank.

Let ν_{S} : S ⟶ Specℂ be the structure morphism of S . Then our inner product (α, α′) on H̃(S, ℤ) is equal to ν_{∗}(α^{∨} · α′) . Hence, by the projection formula, we have

(α, f(β)) = ν_{S_{2},∗}(α^{∨} · π_{∗}(π^{∗}\sqrt{td_{S_{2}}} · ch(E) · p^{∗}\sqrt{td_{S_{1}}} · p^{∗}(β))) =
= ν_{S_{2},∗}π_{∗}(π^{∗}(α^{∨}) · p^{∗}(β) · ch(E) · \sqrt{td_{S_{1}×S_{2}}}) =
= ν_{S_{1}×S_{2},∗}(π^{∗}(α^{∨}) · p^{∗}(β) · ch(E) · \sqrt{td_{S_{1}×S_{2}}})

for every α ∈ H^{∗}(S_{2}, ℤ), β ∈ H^{∗}(S_{1}, ℤ) . In a similar way, we have

(β, f′(α)) = ν_{S_{1}×S_{2},∗}(p^{∗}(β^{∨}) · π^{∗}(α) · ch(E)^{∨} · \sqrt{td_{S_{1}×S_{2}}})

Therefore, (α, f(β)) = (f′(α), β) . Since f′ ∘ f is the identity, we obtain

(f(α), f(α′)) = (f′f(α), α′) = (α, α′)

Thus, f is an isometry. □

3.6. Consider the isometry f . Since the cycle Z is algebraic, we obtain two isometries

f_{alg} : −N_{S_{1}}⊥U ⟶^{∼} −N_{S_{2}}⊥U and f_{τ} : T_{S_{1}} ⟶^{∼} T_{S_{2}} , where N_{S_{1}}, N_{S_{2}} are Neron-Severi

lattices, and T_{S_{1}}, T_{S_{2}} are the lattices of transcendental cycles. It is clear f_{τ} is a Hodge isometry.

Thus we have proved that if the derived categories of two K3 surfaces are equivalent, then there exists a Hodge isometry between the lattices of transcendental cycles.

3.7. Let us begin to prove the converse. Suppose we have a Hodge isometry

f_{τ} : T_{S_{2}} ⟶^{∼} T_{S_{1}}

It implies from the following proposition that we can extend this isometry to Mukai lattices.

3.8. Proposition [Ni] Let ϕ_{1}, ϕ_{2} : T ⟶ H be two primitive embedding of a lattice T in an even unimodular lattice H . Assume that the orthogonal complement N := ϕ_{1}(T)^{⊥}

in H contains the hyperbolic lattice U = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) as a sublattice.

Then ϕ_{1} and ϕ_{2} are equivalent, that means there exists an isometry γ of H such that ϕ_{1} = γϕ_{2} .

We know that the orthogonal complement of T_{S} in Mukai lattice H̃(S, ℤ) is isomorphic to N_{S} ⊥ U . By Proposition 3.8. , there exists an isometry

f : H̃(S_{2}, ℤ) ⟶^{∼} H̃(S_{1}, ℤ)

such that f|_{T_{S_{2}}} = f_{τ} .

Put v = f(0, 0, 1) = (r, l, s) and u = f(1, 0, 0) = (p, k, q) .

We may assume that r > 1 . One may do this, because there are two types of isometries on Mukai lattice that are identity on the lattice of transcendental cycles. First type is multiplication by Chern character e^{m} of a line bundle:

ϕ_{m}(r, l, s) = (r, l + rm, s + (m, l) + \frac{r}{2}m^{2})

Second type is the change r and s (see [Mu2]). So we can change f in such a way that r > 1 and f|_{T_{S_{2}}} = f_{τ} .

First, note that vector v ∈ U ⊥ −N_{S_{1}} is isotropic, i.e (v, v) = 0 . It was proved by Mukai in his brilliant paper [Mu2] that there exists a polarization A on S_{1} such that the moduli space M_{A}(v) of stable bundles with respect to A , for which vector Mukai is equal to v , is projective smooth K3 surface. Moreover, this moduli space is fine, because there exists the vector u ∈ U ⊥ −N_{S_{1}} such that (v, u) = 1 . Therefore we have a universal vector bundle E on the product S_{1} × M_{A}(v) .

The universal bundle E gives the functor Φ_{E} : D^{b}(M_{A}(v)) ⟶ D^{b}(S_{1}) .

Let us assume that Φ_{E} is an equivalence of derived categories. In this case, the cycle Z = π_{S_{1}}^{∗}\sqrt{td_{S_{1}}} · ch(E) · p^{∗}\sqrt{td_{M}} induces the Hodge isometry

g : H̃(M_{A}(v), ℤ) ⟶ H̃(S_{1}, ℤ),

such that g(0, 0, 1) = v = (r, l, s) . Hence, f^{−1} ∘ g is an isometry too, and it sends (0, 0, 1) to (0, 0, 1) . Therefore f^{−1} · g gives the Hodge isometry between the second cohomologies, because for a K3 surface S

H^{2}(S, ℤ) = (0, 0, 1)^{⊥}/ℤ(0, 0, 1).

Consequently, by the strong Torelli theorem (see [LP]), the surfaces S_{2} and M_{A}(v) are isomorphic. Hence the derived categories of S_{1} and S_{2} are equivalent.

3.9. Thus, to conclude the proof of Theorem 3.3. , it remains to show that the functor Φ_{E} is an equivalence.

First, we show that the functor Φ_{E} is full and faithful. This is a special case of the following more general statement, proved in [BO1].

3.10. Theorem [BO1] Let M and X be smooth algebraic varieties and E ∈ D^{b}(M × X) . Then Φ_{E} is fully faithful functor, iff the following orthogonality conditions are verified:

i) Hom_{X}^{i}(Φ_{E}(O_{t_{1}}) , Φ_{E}(O_{t_{2}})) = 0  for every i and t_{1} ≠ t_{2}.
ii) Hom_{X}^{0}(Φ_{E}(O_{t}) , Φ_{E}(O_{t})) = k,
Hom_{X}^{i}(Φ_{E}(O_{t}) , Φ_{E}(O_{t})) = 0,  for i ∉ [0, dimM].

Here t , t_{1} , t_{2} are points of M , O_{t_{i}} are corresponding skyscraper sheaves.

In our case, Φ_{E}(O_{t}) = E_{t} , where E_{t} is stable sheaf with respect to the polarization A on S_{1} for which v(E_{t}) = v . All these sheaves are simple and Ext^{i}(E_{t} , E_{t}) = 0 for i ∉ [0, 2] . This implies that condition 2) of Theorem 3.10. is fulfilled.

All E_{t} are stable sheaves, hence Hom(E_{t_{1}} , E_{t_{2}}) = 0 . Further, by Serre duality Ext^{2}(E_{t_{1}} , E_{t_{2}}) = 0 . Finally, since the vector v is isotropic, we obtain Ext^{1}(E_{t_{1}} , E_{t_{2}}) = 0 .

This yields that Φ_{E} is full and faithful. As our situation is not symmetric (a priori), it is not clear whether the adjoint functor to Φ_{E} is also full and faithful. Some additional reasoning is needed.

3.11. Theorem In the above notations, the functor Φ_{E} : D^{b}(M_{A}(v)) ⟶ D^{b}(S_{1}) is an equivalence.

Proof. Assume the converse, i.e. Φ_{E} is not an equivalence, then, since the functor Φ_{E} is full and faithful, there exists an object C ∈ D^{b}(S_{1}) such that Φ_{E}^{∗}(C) = 0 . By Proposition 3.5. , the functor Φ_{E} induces the isometry f on the Mukai lattices, hence the Mukai vector v(C) is equal to 0 .

Object C satisfies the conditions Hom^{i}(C , E_{t}) = 0 for every i and all t ∈ M_{A}(v) , where E_{t} are stable bundles on S_{1} with the Mukai vector v .

Denote by H^{i}(C) the cohomology sheaves of the object C . There is a spectral sequence which converges to Hom^{i}(C , E_{t})

E_{2}^{p,q} = Ext^{p}(H^{−q}(C) , E_{t}) ⟹ Hom^{p+q}(C , E_{t})  (11)

It is depicted in the following diagram

[figure: q; p; ⋮; ⋮; d_{2}; d_{2}; d_{2}]

We can see that Ext^{1}(H^{q}(C) , E_{t}) = 0 for every q and all t , and every morphism d_{2} is an isomorphism.

To prove the theorem, we need the following lemma.

3.12. Lemma Let G be a sheaf on K3 surface S_{1} such that Ext^{1}(G , E_{t}) = 0 for all t . Then there exists an exact sequence

0 ⟶ G_{1} ⟶ G ⟶ G_{2} ⟶ 0

that satisfies the following conditions:

1) Ext^{i}(G_{1} , E_{t}) = 0 for every i ≠ 2, and Ext^{2}(G_{1} , E_{t}) ≅ Ext^{2}(G , E_{t})
2) Ext^{i}(G_{2} , E_{t}) = 0 for every i ≠ 0, and Hom(G_{2} , E_{t}) ≅ Hom(G , E_{t})

and μ_{A}(G_{2}) < μ_{A}(G) < μ_{A}(G_{1}) .

Proof. Firstly, there is a short exact sequence

0 ⟶ T ⟶ G ⟶ G̃ ⟶ 0,

where T is a torsion sheaf, and G̃ is torsion free.

Secondly, there is a Harder-Narasimhan filtration 0 = I_{0} ⊂ ... ⊂ I_{n} = G̃ for G̃ such that the successive quotients I_{i}/I_{i−1} are A -semistable, and μ_{A}(I_{i}/I_{i−1}) > μ_{A}(I_{j}/I_{j−1}) for i < j .

Now, combining T and the members of the filtration for which μ_{A}(I_{i}/I_{i−1}) > μ_{A}(E_{t}) (resp. = , < ) to one, we obtain the 3-member filtration on G

0 = J_{0} ⊂ J_{1} ⊂ J_{2} ⊂ J_{3} = G.

Let K_{i} be the quotients sheaves J_{i}/J_{i−1} . We have

μ_{A}(K_{1}) > μ_{A}(K_{2}) = μ_{A}(E_{t}) > μ_{A}(K_{3})

(we suppose, if needed, μ_{A}(T) = +∞ ).

Moreover, it follows from stability of E_{t} that

Hom(K_{1} , E_{t}) = 0  and  Ext^{2}(K_{3} , E_{t}) = 0

Combining this with the assumption that Ext^{1}(G , E_{t}) = 0 , we get Ext^{1}(K_{2} , E_{t}) = 0 . To prove the lemma it remains to show that K_{2} = 0 . Note that K_{2} is A -semistable. Hence there is a Jordan-Hölder filtration for K_{2} such

that the successive quotients are A -stable. The number of the quotients is finite. Therefore we can take t_{0} such that

Hom(K_{2} , E_{t_{0}}) = 0  and  Ext^{2}(K_{2} , E_{t_{0}}) = 0

Consequently, χ(v(K_{2}), v(E_{t})) = 0 . Thus, as Ext^{1}(K_{2} , E_{t}) = 0 for all t , we obtain Ext^{i}(K_{2} , E_{t}) = 0 for every i and all t .

Further, let us consider Φ_{E}^{∗}(K_{2}) . We have

Hom (Φ_{E}^{∗}(K_{2}) , O_{t}) ≅ Hom (K_{2} , E_{t}) = 0,

This implies Φ_{E}^{∗}(K_{2}) = 0 . Hence v(K_{2}) = 0 , because f is an isometry. And, finally, K_{2} = 0 . The lemma is proved. □

Let us return to the theorem. The object C possesses at least two non-zero consequent cohomology sheaves H^{p}(C) and H^{p+1}(C) . They satisfy the condition of Lemma 3.12. Hence there exist decompositions with conditions 1),2):

0 ⟶ H_{1}^{p} ⟶ H^{p}(C) ⟶ H_{2}^{p} ⟶ 0 and 0 ⟶ H_{1}^{p+1} ⟶ H^{p+1}(C) ⟶ H_{2}^{p+1} ⟶ 0

Now consider the canonical morphism H^{p+1}(C) ⟶ H^{p}(C)[2] . It induces the morphism s : H_{1}^{p+1} ⟶ H_{2}^{p}[2] . By Z denote a cone of s .

Since d_{2} of the spectral sequence (11) is an isomorphism, we obtain

Hom (Z , E_{t}) = 0 for all t.

Consequently, we have Φ_{E}^{∗}(Z) = 0 . On the other hand, we know that μ_{A}(H_{1}^{p+1}) > μ_{A}(E_{t}) > μ_{A}(H_{2}^{p}) . Therefore v(Z) ≠ 0 . This contradiction proves the theorem. □

There exists the another version of Theorem 3.3.

3.13. Theorem Let S_{1} and S_{2} be smooth projective K3 surfaces over field ℂ . Then the derived categories D^{b}(S_{1}) and D^{b}(S_{2}) are equivalent as triangulated categories iff there exists a Hodge isometry f : H̃(S_{1}, ℤ) ⟶^{∼} H̃(S_{2}, ℤ) between the Mukai lattices of S_{1} and S_{2} .

Here the ‘Hodge isometry’ means that the one-dimensional subspace H^{2,0}(S_{1}) ⊂ H̃(S_{1}, ℤ) ⊗ ℂ goes to H^{2,0}(S_{2}) ⊂ H̃(S_{2}, ℤ) ⊗ ℂ .

Appendix.

The facts, collected in this appendix, are not new; they are known. However, not having a good reference, we regard it necessary to give a proof for the statement, which is used in the main text. We exploit the technique from [IM].

Let X be a smooth projective variety and L be a very ample invertible sheaf on X such that H^{i}(X, L^{k}) = 0 for any k > 0 , when i ≠ 0 . Denote by A the coordinate algebra for X with respect to L , i.e. A = ⨁_{k=0}^{∞} H^{0}(X, L^{k}) .

Now consider the variety X^{n} . First, we introduce some notations. Define subvarieties Δ_{(i_{1},...,i_{k})(i_{k+1},...,i_{m})}^{(n)} ⊂ X^{n} by the following rule:

Δ_{(i_{1},...,i_{k})(i_{k+1},...,i_{m})}^{(n)} := {(x_{1}, ..., x_{n})|x_{i_{1}} = ⋯ = x_{i_{k}}; x_{i_{k+1}} = ⋯ = x_{m}}

By S_{i}^{(n)} denote Δ_{(n,...,i)}^{(n)} . It is clear that S_{i}^{(n)} ≅ X^{i} .

Further, let T_{i}^{(n)} := ⋃_{k=1}^{i−1} Δ_{(n,...,i)(k,k−1)}^{(n)} (note that T_{1}^{(n)} and T_{2}^{(n)} are empty) and

let Σ^{(n)} := ⋃_{k=1}^{n} Δ_{(k,k−1)}^{(n)} . We see that T_{i}^{(n)} ⊂ S_{i}^{(n)} . Denote by I_{i}^{(n)} the kernel of the restriction map O_{S_{i}^{(n)}} ⟶ O_{T_{i}^{(n)}} ⟶ 0 .

Using induction by n , it can easily be checked that the following complex on X^{n}

P_{n}: 0 ⟶ J_{Σ^{(n)}} ⟶ I_{n}^{(n)} ⟶ I_{n−1}^{(n)} ⟶ ⋯ ⟶ I_{2}^{(n)} ⟶ I_{1}^{(n)} ⟶ 0

is exact. (Note that I_{1}^{(n)} = O_{Δ_{n,...,1}^{(n)}} and I_{2}^{(n)} = O_{Δ_{n,...,2}^{(n)}} ). For example, for n = 2 this complex is a short exact sequence on X × X :

P_{2}: 0 ⟶ J_{Δ} ⟶ O_{X×X} ⟶ O_{Δ} ⟶ 0

Denote by π_{i}^{(n)} the projection of X^{n} onto i^{th} component, and by π_{ij}^{(n)} denote the projection of X^{n} onto the product of i^{th} and j^{th} components.

Let B_{n} := H^{0}(X^{n}, J_{Σ^{(n)}}⊗(L⊠⋯⊠L)) and let R_{n−1} := R^{0}π_{1∗}^{(n)}(J_{Σ^{(n)}}⊗(O⊠L⊠⋯⊠L)) .

Proposition A.1 Let L be a very ample invertible sheaf on X as above. Suppose that for any m such that 1 < m ≤ n + dimX + 2 the following conditions hold:

a) H^{i}(X^{m}, J_{Σ^{(m)}} ⊗ (L ⊠ ⋯ ⊠ L)) = 0  for i ≠ 0
b) R^{i}π_{1∗}^{(m)}(J_{Σ^{(m)}} ⊗ (O ⊠ L ⊠ ⋯ ⊠ L)) = 0  for i ≠ 0
c) R^{i}π_{1m∗}^{(m)}(J_{Σ^{(m)}} ⊗ (O ⊠ L ⊠ ⋯ ⊠ L ⊠ O)) = 0 for i ≠ 0

Then we have:

1) algebra A is n-Koszul, i.e the sequence

B_{n} ⊗_{k} A ⟶ B_{n−1} ⊗_{k} A ⟶ ⋯ ⟶ B_{1} ⊗_{k} A ⟶ A ⟶ k ⟶ 0

is exact; 2) the following complexes on X :

A_{k−n} ⊗ R_{n} ⟶ A_{k−n+1} ⊗ R_{n−1} ⟶ ⋯ ⟶ A_{k−1} ⊗ R_{1} ⟶ A_{k} ⊗ R_{0} ⟶ L^{k} ⟶ 0

are exact for any k ≥ 0 (if k − i < 0 , then A_{k−i} = 0 by definition); 3) the complex

L^{−n} ⊠ R_{n} ⟶ ⋯ ⟶ L^{−1} ⊠ R_{1} ⟶ O_{M} ⊠ R_{0} ⟶ O_{Δ}

gives n-resolution of the diagonal on X × X , i.e. it is exact.

Proof.

1) First, note that

H^{i}(X^{m}, I_{k}^{(m)} ⊗ (L ⊠ ⋯ ⊠ L)) = H^{i}(X^{k−1}, J_{Σ^{(k−1)}} ⊗ (L ⊠ ⋯ ⊠ L)) ⊗ A_{m−k+1}

By condition a), they are trivial for i ≠ 0 .

Consider the complexes P_{m} ⊗ (L ⊠ ⋯ ⊠ L) for m ≤ n + dimX + 1 . Applying the functor H^{0} to these complexes and using condition a), we get the exact sequences:

0 ⟶ B_{m} ⟶ B_{m−1} ⊗_{k} A_{1} ⟶ ⋯ ⟶ B_{1} ⊗_{k} A_{m−1} ⟶ A_{m} ⟶ 0

for m ≤ n + dimX + 1 .

Now put m = n + dimX + 1 . Denote by W_{m} the complex

I_{m}^{(m)} ⟶ I_{m−1}^{(m)} ⟶ ⋯ ⟶ I_{2}^{(m)} ⟶ I_{1}^{(m)} ⟶ 0

Take the complex W_{m} ⊗ (L ⊠ ⋯ ⊠ L ⊠ L^{i}) and apply functor H^{0} to it. We obtain the following sequence:

B_{m−1} ⊗_{k} A_{i} ⟶ B_{m−2} ⊗_{k} A_{i+1} ⟶ ⋯ ⟶ B_{1} ⊗_{k} A_{m−1} ⟶ A_{m} ⟶ 0

The cohomologies of this sequence are H^{j}(X^{m}, J_{Σ^{(m)}} ⊗ (L ⊠ ⋯ ⊠ L ⊠ L^{i})) . It follows from condition b) that

H^{j}(X^{m}, J_{Σ^{(m)}} ⊗ (L ⊠ ⋯ ⊠ L ⊠ L^{i})) = H^{j}(X, R^{0}π_{m∗}^{(m)}(J_{Σ^{(m)}} ⊗ (L ⊠ ⋯ ⊠ L ⊠ O)) ⊗ L^{i})

Hence they are trivial for j > dimX . Consequently, we have the exact sequences:

B_{n} ⊗_{k} A_{m−n+i−1} ⟶ B_{n−1} ⊗_{k} A_{m−n+i} ⟶ ⋯ ⟶ B_{1} ⊗_{k} A_{m+i−2} ⟶ A_{m+i−1}

for i ≥ 1 . And for i ≤ 1 the exactness was proved above.

Thus, algebra A is n-Koszul.

2) The proof is the same as for 1). We have isomorphisms

R^{i}π_{1∗}^{(m)}(I_{k}^{(m)} ⊗ (O ⊠ L ⊠ ⋯ ⊠ L)) ≅ R^{i}π_{1∗}^{(k−1)}(J_{Σ^{(k−1)}} ⊗ (O ⊠ L ⊠ ⋯ ⊠ L)) ⊗ A_{m−k+1}

Applying functor R^{0}π_{1∗}^{(m)} to the complexes P_{m}⊗(O⊠L⊠⋯⊠L)) for m ≤ n+dimX+2 , we obtain the exact complexes on X

0 ⟶ R_{m−1} ⟶ A_{1} ⊗ R_{m−2} ⟶ ⋯ ⟶ A_{m−2} ⊗ R_{1} ⟶ A_{m−1} ⊗ R_{0} ⟶ L^{m−1} ⟶ 0

for m ≤ n + dimX + 2 .

Put m = n + dimX + 2 . Applying functor R^{0}π_{1∗}^{(m)} to the complex W_{m} ⊗ (O ⊠ L ⊠ ⋯ ⊠ L ⊠ L^{i})) , we get the complex

A_{i} ⊗ R_{m−2} ⟶ ⋯ ⟶ A_{m+i−3} ⊗ R_{1} ⟶ A_{m+i−2} ⊗ R_{0} ⟶ L^{m+i−2} ⟶ 0

The cohomologies of this complex are

R^{j}π_{1∗}^{(m)}(J_{Σ^{(m)}} ⊗(O⊠L⊠⋯⊠L⊠L^{i})) ≅ R^{j}p_{1∗}(R^{0}π_{1m∗}^{(m)}(J_{Σ^{(m)}} ⊗(O⊠L⊠⋯⊠L⊠O))⊗(O⊠L^{i}))

They are trivial for j > dimX . Thus, the sequences

A_{k−n} ⊗ R_{n} ⟶ A_{k−n+1} ⊗ R_{n−1} ⟶ ⋯ ⟶ A_{k−1} ⊗ R_{1} ⟶ A_{k} ⊗ R_{0} ⟶ L^{k} ⟶ 0

are exact for all k ≥ 0 .

3) Consider the complex W_{n+2}⊗(O⊠L⊠⋯⊠L⊠L^{−i}) . Applying the functor R^{0}π_{1(n+2)∗}^{(n+2)} to it, we obtain the following complex on X × X :

L^{−n} ⊠ R_{n} ⟶ ⋯ ⟶ L^{−1} ⊠ R_{1} ⟶ O_{M} ⊠ R_{0} ⟶ O_{Δ}

By condition c), it is exact.

This finishes the proof.

Note that for any ample invertible sheaf L we can find j such that for the sheaf L^{j} the conditions a),b),c) are fulfilled.

References

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[Mu2] S. Mukai, On the moduli space of bundles on a K3 surface I, Vector bundles on algebraic varieties, Tata Institute of Fundamental Research, Oxford University Press, Bombay and London, 1987.

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[V] J. L. Verdier, Categories derivées, in SGA 4 1/2, Lecture Notes in Math., v.569, Springer-Verlag, 1977.

Algebra section, Steklov Math. Institute, Vavilova 42, 117333 Moscow, RUSSIA

   E-mail address: orlov@class.mi.ras.ru