EQUIVALENCES OF DERIVED CATEGORIES AND K3 SURFACES DMITRI ORLOV

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1 EQUIVALENCES OF DERIVED CATEGORIES AND K3 SURFACES DMITRI ORLOV Abstract. We consder derved categores of coherent sheaves on smooth projectve varetes. We prove that any equvalence between them can be represented by an object on the product. Usng ths, we gve a necessary and suffcent condton for equvalence of derved categores of two K3 surfaces. 0. Introducton Let D b (X) be the bounded derved category of coherent sheaves on a smooth projectve varety X. The category D b (X) has the structure of a trangulated category (see [V], [GM]). We shall consder D b (X) as a trangulated category. In ths paper we are concerned wth the problem of descrpton for varetes, whch have equvalent derved categores of coherent sheaves. In the paper [Mu1], Muka showed that for an abelan varety A and ts dual  the derved categores D b (A) and Db (Â) are equvalent. Equvalences of another type appeared n [BO1]. They are nduced by certan bratonal transformatons whch are called flops. Further, t was proved n the paper [BO2] that f X s a smooth projectve varety wth ether ample canoncal or ample antcanoncal sheaf, then any other algebrac varety X such that D b (X ) D b (X) s bregularly somorphc to X. The am of ths paper s to gve some descrpton for equvalences between derved categores of coherent sheaves. The man result s Theorem 2.2. of 2. It says that any full and fathful exact functor F : D b (M) D b (X) havng left (or rght) adjont functor can be represented by an object E D b (M X),.e. F ( ) = R π (E L p ( )), where π and p are the projectons on M and X respectvely. In 3, basng on the Muka s results [Mu2], we show that two K3 surfaces S 1 and S 2 over feld C have equvalent derved categores of coherent sheaves ff the lattces of transcendental cycles T S1 and T S2 are Hodge sometrc. I would lke to thank A. Polshchuk for useful notces. 1. Prelmnares 1.1. We collect here some facts relatng to trangulated categores. Recall that a trangulated category s an addtve category wth addtonal structures: a) an addtve autoequvalence T : D D, whch s called a translaton functor (we usually wrte X[n] nstead of T n (X) and f[n] nstead of T n (f) ), 1

2 b) a class of dstngushed trangles: X u Y v Z w X[1]. And these structures must satsfy the usual set of axoms (see [V]). If X, Y are objects of a trangulated category D, then Hom D (X, Y ) means Hom D (X, Y []). An addtve functor F : D D between two trangulated categores D and D s called exact f a) t commutes wth the translaton functor,.e there s fxed an somorphsm of functors: t F : F T T F, b) t takes every dstngushed trangle to a dstngushed trangle (usng the somorphsm t F, we replace F (X[1]) by F (X)[1] ). The followng lemma wll be needed for the sequel Lemma [BK] If a functor G : D D s a left (or rght) adjont to an exact functor F : D D then functor G s also exact. Proof. Snce G s the left adjont functor to F, there exst canoncal morphsms of functors d D F G, G F d D. Let us consder the followng sequence of natural morphsms: G T G T F G G F T G T G We obtan the natural morphsm G T T G. Ths morphsm s an somorphsm. Indeed, for any two objects A D and B D we have somorphsms : 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) Ths mples that the natural morphsm G T T G s an somorphsm. Let now A α B C A[1] be a dstngushed trangle n D. We have to show that G takes ths trangle to a dstngushed one. Let us nclude the morphsm G(α) : G(A) G(B) nto a dstngushed trangle: G(A) G(B) Z G(A)[1]. Applyng functor F to t, we obtan a dstngushed trangle: F G(A) F G(B) F (Z) F G(A)[1] (we use the commutaton somorphsms lke T F F T wth no menton). Usng morphsm d F G, we get a commutatve dagram: A F G(A) α B C A[1] F G(α) F G(B) F (Z) F G(A)[1] 2

3 By axoms of trangulated categores there exsts a morphsm µ : C F (Z) that completes ths commutatve dagram. Snce G s left adjont to F, the morphsm µ defnes ν : G(C) Z. It s clear that ν makes the followng dagram commutatve: G(A) G(B) G(C) G(A)[1] ν G(A) G(B) Z G(A)[1] To prove the lemma, t suffces to show that ν s an somorphsm. For any object Y D let us consder the dagram for Hom : Hom(G(A)[1], Y ) Hom(Z, Y ) Hom(G(B), Y ) HY (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 )) Snce the lower sequence s exact, the mddle sequence s exact also. By the lemma about fve homomorphsms, for any Y the morphsm H(ν) s an somorphsm. Thus ν : G(C) Z s an somorphsm too. Ths concludes the proof Let X = {X c dc Xc+1 dc+1 X 0 } be a bounded complex over a trangulated category D,.e. all compostons d +1 d are equal to 0 ( c < 0 ). A left Postnkov system, attached to X, s, by defnton, a dagram X c d c X c+1 d c+1 X c+2 j c j c+1 j 1 c = d c+1 c+2 0 Y c = X c [1] Y c+1 [1] Y c+2 Y 1 [1] Y 0 X 0 n whch all trangles marked wth are dstngushed and trangles marked wth are commutatve (.e. j k k = d k ). An object E ObD s called a left convoluton of X, f there exsts a left Postnkov system, attached to X such that E = Y 0. The class of all convolutons of X wll be denoted by Tot(X ). Clearly the Postnkov systems and convolutons are stable under exact functors between trangulated categores. The class Tot(X ) may contan many non-somorphc elements and may be empty. Further we shall gve a suffcent condton for Tot(X ) to be non-empty and for ts objects to be somorphc. The followng lemma s needed for the sequel(see [BBD]). 3

4 1.4. Lemma Let g be a morphsm between two objects Y and Y, whch are ncluded nto two dstngushed trangles: u v X Y Z f g h w X[1] f[1] X u Y v Z w X [1] If v gu = 0, then there exst morphsms f : X X and h : Z Z such that the trple (f, g, h) s a morphsm of trangles. If, n addton, Hom(X[1], Z ) = 0 then ths trple s unquely determned by g. Now we prove two lemmas whch are generalzatons of the prevous one for Postnkov dagrams Lemma Let X = {X c dc Xc+1 dc+1 X 0 } be a bounded complex over a trangulated category D. Suppose t satsfes the followng condton: Hom (X a, X b ) = 0 for < 0 and a < b. (1) Then there exsts a convoluton for ths complex and all convolutons are somorphc (noncanoncally). If, n addton, Hom (X a, Y 0 ) = 0 for < 0 and for all a (2) for some convoluton Y 0 (and, consequently, for any one), then all convolutons are canoncally somorphc Lemma Let X 1 and X 2 be bounded complexes that satsfy (1), and let (f c,..., f 0 ) be a morphsm of these complexes: Suppose that X c 1 d c 1 X1 c+1 X 1 0 fc fc+1 f0 X c 2 d c 2 X c+1 2 X 0 2 Hom (X1 a, X2) b = 0 for < 0 and a < b. (3) Then for any convoluton Y1 0 of X 1 and for any convoluton Y2 0 of X 2 there exsts a morphsm f : Y that commutes wth the morphsm f 0. If, n addton, then ths morphsm s unque. Hom (X a 1, Y 0 2 ) = 0 for < 0 and for all a (4) Proof. We shall prove both lemmas together. Let Y c+1 be a cone of the morphsm d c : X c d c X c+1 α Y c+1 X c [1] By assumpton d c+1 d c = 0 and Hom(X c [1], X c+2 ) = 0, hence there exsts a unque morphsm d c+1 : Y c+1 X c+2 such that dc+1 α = d c+1. 4

5 Let us consder a composton d c+2 d c+1 : Y c+1 X c+3. We know that d c+2 d c+1 α = d c+2 d c+1 = 0, and at the same tme we have Hom(X c [1], X c+3 ) = 0. Ths mples that the composton d c+2 d c+1 s equal to 0. Moreover, consder the dstngushed trangle for Y c+1. It can easly be checked that Hom (Y c+1, X b ) = 0 for < 0 and b > c+1. Hence the complex Y c+1 X c+2 X 0 satsfes the condton (1). By nducton, we can suppose that t has a convoluton. Ths mples that the complex X has a convoluton too. Thus, the class Tot(X ) s non-empty. Now we shall show that under the condtons (3) any morphsm of complexes can be extended to a morphsm of Postnkov systems. Let us consder cones Y1 c+1 and Y2 c+1 of the morphsms d c 1 and d c 2. There exsts a morphsm g c+1 : Y1 c+1 Y2 c+1 such that one has the morphsm of dstngushed trangles: X c 1 d c 1 α X1 c+1 Y1 c+1 X 1 c[1] fc fc+1 gc+1 fc [1] X c 2 d c 2 X c+1 2 β Y c+1 2 X c 2 [1] As above, there exst unquely determned morphsms Consder the followng dagram: dc+1 : Y c+1 X c+2 for = 1, 2. Y1 c+1 gc+1 Y c+1 2 d c+1 1 X c+2 1 fc+2 d c+1 2 X c+2 2 c+1 Let us show that ths square s commutatve. Denote by h the dfference f c+2 d 1 d c+1 2 g c+1. We have h α = f c+2 d c+1 1 d c+1 2 f c+1 = 0 and, by assumpton, Hom(X1 c [1], Xc+2 2 ) = 0. It follows that h = 0. Therefore, we obtan the morphsm of new complexes: Y1 c+1 d c+1 1 X1 c+2 X 1 0 gc+1 fc+2 f0 Y c+1 2 d c+1 2 X c+2 2 X 0 2 It can easly be checked that these complexes satsfy the condtons (1) and (3) of the lemmas. By the nducton hypothess, ths morphsm can be extended to a morphsm of Postnkov systems, attached to these complexes. Hence there exsts a morphsm of Postnkov systems, attached to X 1 and X 2. Moreover, we see that f all morphsms f are somorphsms, then a morphsm of Postnkov systems s an somorphsm too. Therefore, under the condton (1) all objects from the class Tot(X ) are somorphc. 5

6 Now let us consder a morphsm of the rghtmost dstngushed trangles of Postnkov systems: Y1 1 j 1, 1 1,0 X1 0 Y1 0 Y1 1 [1] g 1 f0 g0 g 1 [1] Y 1 2 j 2, 1 X 0 2 2,0 Y 0 2 Y 1 2 [1] If the complexes X satsfy the condton (4) (.e. Hom (X1 a, Y 2 0 ) = 0 for < 0 and all a ), then we get Hom(Y1 1 [1], Y2 0) = 0. It follows from Lemma 1.4. that g 0 s unquely determned. Ths concludes the proof of both lemmas. 2. Equvalences of derved categores 2.1. Let X and M be smooth projectve varetes over feld k. Denote by D b (X) and D b (M) the bounded derved categores of coherent sheaves on X and M respectvely. Recall that a derved category has the structure of a trangulated category. For every object E D b (M X) we can defne an exact functor Φ E from D b (M) to D b (X). Denote by p and π the projectons of M X onto M and X respectvely: M X Then Φ E s defned by the followng formula: p M π X Φ E ( ) := π (E p ( )) (5) (we always shall wrte shortly f, f, and etc. nstead of R f, L f, L, because we consder only derved functors). The functor Φ E has the left and the rght adjont functors Φ E and Φ! E respectvely, defned by the followng formulas: Φ E ( ) = p (E π (ω X [dmx] ( ))), Φ! E ( ) = ω M [dmm] p (E ( )), where ω X and ω M are the canoncal sheaves on X and M, and E := R Hom(E, O M X ). Let F be an exact functor from the derved category D b (M) to the derved category D b (X). Denote by F and F! the left and the rght adjont functors for F respectvely, when they exst. Note that f there exsts the left adjont functor F, then the rght adjont functor F! also exsts and F! = S M F S 1 X, where S X and S M are Serre functors on D b (X) and D b (M). They are equal to ( ) ω X [dmx] and ( ) ω M [dmm] (see [BK]). 6

7 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 projectve varetes M and X. But we are unable prove t. However, when F s full and fathfull, t can be represented. The man result of ths chapter s the followng theorem Theorem Let F be an exact functor from D b (M) to D b (X), where M and X are smooth projectve varetes. Suppose F s full and fathful and has the rght (and,consequently, the left) adjont functor. Then there exsts an object E D b (M X) such that F s somorphc to the functor Φ E defned by the rule (5), and ths object s unque up to somorphsm Let F be an exact functor from a derved category D b (A) to a derved category D b (B). We say that F s bounded f there exst z Z, n N such that for any A A the cohomology objects H (F (A)) are equal to 0 for [z, z + n] Lemma Let M and X be smooth projectve varetes. If an exact functor F : D b (M) D b (X) has a left adjont functor then t s bounded. Proof. Let G : D b (X) D b (M) be a left adjont functor to F. Take a very ample nvertble sheaf L on X. It gves the embeddng : X P N. For any < 0 we have rght resoluton of the sheaf O() on P N n terms of the sheaves O(j), where j = 0, 1,.., N (see [Be]). It s easly seen that ths resoluton s of the form { } O() V 0 O V 1 O(1) V N O(N) 0 where all V k are vector spaces. The restrcton of ths resoluton to X gves us the resoluton of the sheaf L n terms of the sheaves L j, where j = 0, 1,..., N. Snce the functor G s exact that there exst z and n such that H k (G(L )) are equal 0 for k [z, z + n ]. Ths follows from the exstence of the spectral sequence E p,q 1 = V p H q (G(L p )) H p+q (G(L )). As all nonzero terms of ths spectral sequence are concentrated n some rectangle, so t follows that for all cohomologes H (G(L )) are concentrated n some segment. Now, notce that f Hom j (L, F (A)) = 0 for all 0, then H j (F (A)) s equal to 0. Further, by assumpton, the functor G s left adjont to F, hence Hom j (L, F (A)) = Hom j (G(L ), A). If now A s a sheaf on M, then Hom j (G(L ), A) = 0 for all < 0 and j [ z n, z + dmm], and thus H j (F (A)) = 0 for the same j Remark We shall henceforth assume that for any sheaf F on M the cohomology objects H (F (F)) are nonzero only f [ a, 0] Now we begn constructng an object E D b (M X). Frstly, we shall consder a closed embeddng j : M P N and shall construct an object E D b (P N X). 7

8 Secondly, we shall show that there exsts an object E D b (M X) such that E = (j d) E. After that we shall prove that functors F and Φ E are somorphc. Let L be a very ample nvertble sheaf on M such that H (L k ) = 0 for any k > 0, when 0. By j denote the closed embeddng j : M P N wth respect to L. Recall that there exsts a resoluton of the dagonal on the product P N P N (see[be]). Let us consder the followng 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) Ths complex s a resoluton of the structure sheaf O of the dagonal. Now by F denote the functor from D b (P N ) to D b (X), whch s the composton F j. Consder the product Denote by P N X π q P N d Hom P N X (O( ) F (Ω ()), O( + 1) F (Ω 1 ( 1))) the mage d under the followng through map. Hom(O( ) Ω (), O( + 1) Ω 1 ( 1)) X Hom(O Ω (), O(1) Ω 1 ( 1)) Hom(Ω (), H 0 (O(1)) Ω 1 ( 1)) Hom(F (Ω ()), H 0 (O(1)) F (Ω 1 ( 1))) Hom(O F (Ω ()), O(1) F (Ω 1 ( 1))) Hom(O( ) F (Ω ()), O( + 1) F (Ω 1 ( 1))) It can easly be checked that the composton d +1 d s equal to 0. We omt the check. Consder the followng complex C C := {O( N) F (Ω N (N)) d N O( 1) F (Ω 1 (1)) d 1 O F (O)} over the derved category D b (P N X). For l < 0 we have Hom l (O( ) F (Ω ()), O( k) F (Ω k (k))) = Hom l (O F (Ω ()), H 0 (O( k)) F (Ω k (k))) = Hom l (j (Ω ()), H 0 (O( k)) j (Ω k (k))) = 0 8

9 Hence, by Lemma 1.5., there exsts a convoluton of the complex C, and all convolutons are somorphc. By E denote some convoluton of C and by γ 0 denote the morphsm O F γ 0 (O) E. (Further we shall see that all convolutons of C are canoncally somorphc). Now let Φ E be the functor from D b (P N ) to D b (X), defned by (5) Lemma There exst canoncally defned somorphsms f k : F (O(k)) Φ E (O(k)) for all k Z, and these somorphsms are functoral,.e. for any α : O(k) O(l) the followng dagram commutes F (O(k)) f k Φ E (O(k)) F (α) Φ E (α) F (O(l)) fl Φ E (O(l)) Proof. At frst, assume that k 0. Consder the resoluton (6) of the dagonal P N P N and, after tensorng t wth O(k) O, push forward onto the second component. We get the followng resoluton 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)) s a convoluton 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 consder the complex C k := q O(k) C on P N X wth the morphsm γ k : O(k) F (O) q O(k) E, and push t forward onto the second component. It follows from the constructon of the complex C that π (C k ) = D k. So we see that F (O(k)) and Φ E (O(k)) both are convolutons of the same complex D k. By assumpton the functor F s full and fathful, hence, f G and H are locally free sheaves on P N then we have Hom (F (G), F (H)) = Hom (j (G), j (H)) = 0 for < 0. Therefore the complex D k satsfes the condtons (1) and (2) of Lemma Hence there exsts a unquely defned somorphsm f k : F (O(k)) Φ E (O(k)), completng the followng commutatve dagram H 0 (O(k)) F (O) d H 0 (O(k)) F (O) F (δ k ) π (γ k ) F (O(k)) fk Φ E (O(k)) Now we have to show that these morphsms are functoral. For any α : O(k) O(l) we have the commutatve squares H 0 (O(k)) F (O) H 0 (α) d H 0 (O(l)) F (O) 9 F (δ k ) F (δ l ) F (O(k)) F (α) F (O(l))

10 and H 0 (O(k)) F (O) H 0 (α) d Therefore we have the equaltes: H 0 (O(l)) F (O) π (γ k ) π (γ l) Φ E (O(k)) ΦE (α) Φ E (O(l)) f l F (α) F (δ k ) = f l F (δ l ) (H 0 (α) d) = π (γ l ) (H 0 (α) d) = Φ E (α) π (γ k ) = Φ E (α) f k F (δ k ) Snce the complexes D k and D l satsfy the condtons of Lemma 1.6. there exsts only one morphsm h : F (O(k)) Φ E (O(l)) such that h F (δ k ) = π (γ l ) (H 0 (α) d) Hence f l F (α) concdes wth Φ E (α) f k. Now, consder the case k < 0. Let us take the followng rght resoluton for O(k) on P N. O(k) { V0 k O VN k O(N) } By Lemma 1.6., the morphsm of the complexes over V k D b (X) 0 F (O) VN k F (O(N)) d f 0 d fn V k 0 Φ E (O) V k N Φ E (O(N)) gves us the unquely determned morphsm f k : F (O(k)) Φ E (O(k)). It s not hard to prove that these morphsms are functoral. The proof s left to the reader Now we must prove that there exsts an object E D b (M X) such that j E = E. Let L be a very ample nvertble sheaf on M and let j : M P N be an embeddng wth respect to L. By A denote the graded algebra H 0 (M, L ). Let B 0 = k, and B 1 = A 1. For m 2, we defne B m as =0 B m = Ker(B m 1 A 1 B m 2 A 2 ) (7) 2.9. Defnton A s sad to be n -Koszul f the followng sequence s exact B n k A B n 1 k A B 1 k A A k 0 Assume that A s n-koszul. Let R 0 = O M. For m 1, denote by R m the kernel of the morphsm B m O M B m 1 L. Usng (7), we obtan the canoncal morphsm R m A 1 R m 1. (actually, Hom(R m, R m 1 ) = A 1 ). Snce A s 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 10

11 for m n. We have the canoncal morphsms f m : j Ω m (m) R m, because Λ A 1 B and there exst 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 s known that for any n there exsts l such that the Veronese algebra A l = H 0 (M, L l ) s n -Koszul.( Moreover, t was proved n [Ba] that A l s Koszul for =0 l 0 ). Usng the technque of [IM] and substtutng L wth L j, when j s suffcently large, we can choose for any n a very ample L such that 1) algebra A s n -Koszul, 2) the complex L n R n L 1 R 1 O M R 0 O on M M s exact, 3) the followng 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 defnton, f k < 0, then A k = 0. (see Appendx for proof). Let us denote by T k the kernel of the morphsm A k n R n A k n+1 R n 1. Consder the followng complex over D b (M X) L n F (R n ) L 1 F (R 1 ) O M F (R 0 ) (8) Here the morphsm L k F (R k ) L k+1 F (R k 1 ) s nduced by the canoncal morphsm R k A 1 R k 1 wth respect to the followng sequence of somorphsms 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 s a convoluton of the complex (8) and all convolutons are somorphc. Let G D b (M X) be a convoluton of ths complex. For any k 0, object π (G p (L k )) s a convoluton 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 sde, we know that T k [n] L k s a convoluton 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 mmedately from Remark 2.5. that the cohomology sheaves H (π (G p (L k ))) = H (F (T k )[n]) H (F (L k )) concentrate on the unon [ n a, n] [ a, 0] for any k > 0 ( a was defned n 2.5. ). Therefore the cohomology sheaves H (G) also concentrate on [ n a, n] [ a, 0]. We can assume that n > dmm + dmx + a. 11

12 Ths mples that G = C E, where E, C are objects of D b (M X) such that H (E) = 0 for [ a, 0] and H (C) = 0 for [ n a, n]. Moreover, we have π (E p (L k )) = F (L k ). Now we show that j (E) = E. Let us consder the morphsm of the complexes over D b (P N X). O( n) F (Ω n (n)) O F (O) can F (fn) can F (f0 ) j (L n ) F (R n ) j (O M ) F (R 0 ) By Lemma 1.6., there exsts a morphsm of convolutons φ : K j (G). If N > n, then K s not somorphc to E, but there s a dstngushed trangle S K E S[1] and the cohomology sheaves H (S) 0 only f [ n a, n]. Now, snce Hom(S, j (E)) = 0 and Hom(S[1], j (E)) = 0, we have a unquely determned morphsm ψ : E j (E) such that the followng dagram commutes K E φ ψ j (G) j (E) We know that π (E q (O(k))) = F (L k ) = π (E p (L k )). Let ψ k be the morphsm π (E q (O(k))) π (E p (L k )) nduced by ψ. The morphsm ψ k can be ncluded n the followng commutatve dagram: S k A 1 F (O) can A k F (O) can F (L k ) can F (L k ) π (E q (O(k))) ψk π (E p (L k )) Thus we see that ψ k s an somorphsm for any k 0. Hence ψ s an somorphsm too. Ths proves the followng: Lemma There exsts an object E D b (M X) such that j (E) = E, where E s the object from D b (P N X), constructed n Now, we prove some statements relatng to abelan categores. they are needed for the sequel. Let A be a k -lnear abelan category (henceforth we shall consder only k -lnear abelan categores). Let {P } Z be a sequence of objects from A Defnton We say that ths sequence s ample f for every object X A there exsts N such that for all < N the followng condtons hold: a) the canoncal morphsm Hom(P, X) P X s surjectve, b) Ext j (P, X) = 0 for any j 0, c) Hom(X, P ) = 0. 12

13 It s clear that f L s an ample nvertble sheaf on a projectve varety n usual sense, then the sequence {L } Z n the abelan category of coherent sheaves s ample Lemma Let {P } be an ample sequence n an abelan category A. If X s an object n D b (A) such that Hom (P, X) = 0 for all 0, then X s the zero object. Proof. If 0 then Hom(P, H k (X)) = Hom k (P, X) = 0 The morphsm Hom(P, H k (X)) P H k (X) must be surjectve for 0, hence H k (X) = 0 for all k. Thus X s the zero object Lemma Let {P } be an ample sequence n an abelan category A of fnte homologcal dmenson. If X s an object n D b (A) such that Hom (X, P ) = 0 for all 0. Then X s the zero object. Proof. Assume that the cohomology objects of X are concentrated n a segment [a, 0]. There exsts the canoncal morphsm X H 0 (X). Consder a surjectve morphsm P k 1 1 H 0 (X). By Y 1 denote the kernel of ths morphsm. Snce Hom (X, P 1 ) = 0 we have Hom 1 (X, Y 1 ) 0. Further take a surjectve morphsm P k 2 2 Y 1. By Y 2 denote the kernel of ths morphsm. Agan, snce Hom (X, P 2 ) = 0, we obtan Hom 2 (X, Y 2 ) 0. Iteratng ths procedure as needed, we get contradcton wth the assumpton that A s of fnte homologcal dmenson Lemma Let B be an abelan category, A an abelan category of fnte homologcal dmenson, and {P } an ample sequence n A. Suppose F s an exact functor from D b (A) to D b (B) such that t has rght and left adjont functors F! and F respectvely. If the maps Hom k (P, P j ) Hom k (F (P ), F (P j )) are somorphsms for < j and all k. Then F s full and fathful. Proof. Let us take the canoncal morphsm f : P F! F (P ) and consder a dstngushed trangle P Snce for j 0 we have somorphsms: Hom k (P j, P ) f F! F (P ) C P [1]. Hom k (F (P j ), F (P )) = Hom k (P j, F! F (P )). We see that Hom (P j, C ) = 0 for j 0. It follows from Lemma that C = 0. Hence f s an somorphsm. Now, take the canoncal morphsm g X : F F (X) X and consder a dstngushed trangle F F (X) g X X C X F F (X)[1] 13

14 We have the followng sequence of somorphsms Hom k (X, P ) Hom k (X, F! F (P )) = Hom k (F F (X), P ) Ths mples that Hom (C X, P ) = 0 for all. By Lemma 2.14., we obtan C X = 0. Hence g X s an somorphsm. It follows that F s full and fathful. Let A be an abelan category possessng an ample sequence {P }. Denote by D b (A) the bounded derved category of A. Let us consder the full subcategory j : C D b (A) such that ObC := {P Z}. Now we would lke to show that f there exsts an somorphsm of a functor F : D b (A) D b (A) to dentty functor on the subcategory C, then t can be extended to the whole D b (A) Proposton Let F : D b (A) D b (A) be an autoequvalence. Suppose there exsts an somorphsm f : j F C ( where j : C D b (A) s a natural embeddng). Then t can be extended to an somorphsm d F on the whole D b (A). Proof. Frst, we can extend the transformaton f to all drect sums of objects C componentwse, because F takes drect sums to drect sums. Note that X D b (A) s somorphc to an object n A ff Hom j (P, X) = 0 for j 0 and 0. It follows that F (X) s somorphc to an object n A, because for j 0 and 0. Hom j (P, F (X)) = Hom j (F (P ), F (X)) = Hom j (P, X) = At frst, let X be an object from A. Take a surjectve morphsm v : P k X. We have the morphsm f : P k F (P k ) and two dstngushed trangles: Y F (Y ) u F (u) P k f F (P k ) v X Y [1] F (v) F (X) F (Y )[1] Now we show that F (v) f u = 0. Consder any surjectve morphsm w : Pj l Y. It s suffcent to check that F (v) f u w = 0. Let f j : Pj l F (Pj l ) be the canoncal morphsm. Usng the commutaton relatons for f and f j, we obtan F (v) f u w = F (v) F (u w) f j = F (v u w) f j = 0 because v u = 0. Snce Hom(Y [1], F (X)) = 0, by Lemma 1.4., there exsts a unque morphsm f X : X F (X) that commutes wth f Let us show that f X does not depend from morphsm v : P k two surjectve morphsms v 1 : P k 1 1 X and v 2 : P k 2 2 surjectve morphsms w 1 : Pj l P k 1 1 and w 2 : Pj l P k 2 2 dagram s commutatve: Pj l w1 w 2 P k 1 1 v 1 P k v2 X X. Consder X. We can take two such that the followng

15 Clearly, t s suffcent to check the concdence of the morphsms, constructed by v 1 and v 1 w 1. Now, let us consder the followng commutatve dagram: Pj l w 1 P k 1 v 1 1 X fj v2 fx F (P l j ) F (w 1 ) F (P k 1 1 ) F (v 1 ) F (X) Here the morphsm f X s constructed by v 1. Both squares of ths dagram are commutatve. Snce there exsts only one morphsm from X to F (X) that commutes wth f j, we see that the f X, constructed by v 1, concdes wth the morphsm, constructed by v 1 w Now we must show that for any morphsm X φ Y P l j f Y φ = F (φ) f X we have equalty: v Y. Choose a surjectve morphsm P k Take a surjectve morphsm that the composton φ u lfts to a morphsm ψ : P k 0 the map Hom(P k square: P l j u X such. We can do t, because for, Pj l ) Hom(P k, Y ) s surjectve. We get the commutatve P k ψ Pj l u v By h 1 and h 2 denote f Y φ and F (φ) f X respectvely. We have the followng sequence of equaltes: and h 1 u = f Y φ u = f Y v ψ = F (v) f j ψ = F (v) F (ψ) f h 2 u = F (φ) f X u = F (φ) F (u) f = F (φ u) f = F (v ψ) f = F (v) F (ψ) f Consequently, the followng square s commutatve for t = 1, 2. Z P k F (ψ) f F (W ) F (P l j ) u X φ Y X Z[1] ht F (v) F (Y ) F (W )[1] By Lemma 1.4., as Hom(Z[1], F (Y )) = 0, we obtan h 1 = h 2. Thus, f Y φ = F (φ) f X. Now take a cone C X of the morphsm f X. Usng the followng somorphsms Hom(P, X) = Hom(F (P ), F (X)) = Hom(P, F (X)), we obtan Hom j (P, C X ) = 0 for all j, when 0. Hence, by Lemma 2.13., C X = 0 and f X s an somorphsm Let us defne f X[n] : X[n] F (X[n]) = F (X)[n] for any X A by f X[n] = f X [n]. 15

16 It s easly shown that these transformatons commute wth any u Ext k (X, Y ). Indeed, snce any element u Ext k (X, Y ) can be represented as a composton u = u 0 u 1 u k of some elements u Ext 1 (Z, Z +1 ) and Z 0 = X, Z k = Y, we have only to check t for u Ext 1 (X, Y ). Consder the followng dagram: Y Z X u Y [1] f Y fz fy [1] F (Y ) F (Z) F (X) F (u) F (Y )[1] By an axom of trangulated categores there exsts a morphsm h : X F (X) such that (f Y, f Z, h) s a morphsm of trangles. On the other hand, snce Hom(Y [1], F (X)) = 0, by Lemma 1.4., h s a unque morphsm that commutes wth f Z. But f X also commutes wth f Z. Hence we have h = f X. Ths mples that f Y [1] u = F (u) f X The rest of the proof s by nducton over the length of a segment, n whch the cohomology objects of X are concentrated. Let X be an object from D b (A) and suppose that ts cohomology objects H p (X) are concentrated n a segment [a, 0]. Take v : P k X such that a) Hom j (P, H p (X)) = 0 for all p and for j 0, b) u : P k H 0 (X) s the surjectve morphsm, (9) c) Hom(H 0 (X), P ) = 0. Here u s the composton v wth the canoncal morphsm X H 0 (X). Consder a dstngushed trangle: Y [ 1] P k v X Y By the nducton hypothess, there exsts the somorphsm f Y and t commutes wth f. So we have the commutatve dagram: f Y [ 1] Y [ 1] P k f F (Y )[ 1] F (P k ) v X Y fy F (v) F (X) F (Y ) Moreover we have the followng sequence of equaltes Hom(X, F (P k )) = Hom(X, P k ) = Hom(H 0 (X), P k ) = 0 Hence, by Lemma 1.4., there exsts a unque morphsm f X : X F (X) that commutes wth f Y We must frst show that f X s correctly defned. Suppose we have two morphsms v 1 : P k 1 1 X and v 2 : P k 2 2 X. As above, we can fnd P j and surjectve 16

17 morphsms w 1, w 2 such that the followng dagram s commutatve Pj l w1 w 2 P k 1 u 1 1 P k 2 2 u2 H 0 (X) We can fnd a morphsm φ : Y j Y 1 such that the trple (w 1, d, φ) s a morphsm of dstngushed trangles. P l j v 1 w 1 X Yj P l j [1] w 1 d φ w1 [1] P k 1 v 1 1 X Y1 P k 1 1 [1] By the nducton hypothess, the followng square s commutatve. Y j f Yj F (Y j ) φ Y 1 fy F (φ) F (Y 1 ) Hence, we see that the f X, constructed by v 1 w 1, commutes wth f Y1 and, consequently, concdes wth the f X, constructed by v 1 ; because such morphsm s unque by Lemma Therefore morphsm f X does not depend on a choce of morphsm v : P k X Fnally, let us prove that for any morphsm φ : X Y the followng dagram commutes X f X F (X) φ Y fy F (φ) F (Y ) 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 morphsm v : P k X that satsfes condtons (9) and Hom(P k, Y ) = 0. We have a dstngushed trangle: P k v 1 X α Z P k [1] Applyng the functor Hom(, Y ) to ths trangle we found that there exsts a morphsm ψ : Z Y such that φ = ψ α. We know that f X, defned above, satsfy If we assume that the dagram F (α) f X = f Z α Z f Z F (Z) ψ Y fy F (ψ) F (Y ) commutes, then dagram (10) commutes too. Ths means that for verfyng the commutatvty of (10) we can substtute X by an object Z. And the cohomology objects of Z are concentrated on the segment [a, 1]. 17 (10)

18 Case 2. If c 0, we take a surjectve morphsm v : P k Y [c] that satsfes condtons (9) and Hom(H c (X), P k ) = 0. Consder a dstngushed trangle P k [ c] v[ c] Y β W P k [ c + 1] Note that the cohomology objects of W are concentrated on [b, c 1]. By ψ denote the composton β φ. If we assume that the followng square X f X F (X) ψ W fw F (ψ) F (W ) commutes, then, snce 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 such that Hom(X, P k [ c]) = 0. As F (P k ) s somorphc to P k, then Hom(X, F (P k [ c])) = 0. Applyng the functor Hom(X, F ( )) to the above trangle we found that the composton wth F (β) gves an ncluson of Hom(X, F (Y )) nto Hom(X, F (W )). Ths follows that f Y φ = F (φ) f X,.e. our dagram (10) commutes. Combnng case 1 and case 2, we can reduce the checkng of commutatvty of dagram (10) to the case when X and Y are objects from the abelan category A. But for those t has already been done. Thus the proposton s proved Proof of theorem. 1) Exstence. Usng Lemma and Lemma 2.7., we can construct an object E D b (M X) such that there exsts an somorphsm of the functors f : F C ΦE C on full subcategory C D b (M), where ObC = {L Z} and L s a very ample nvertble sheaf on M such that for any k > 0 H (M, L k ) = 0, when 0. By Lemma the functor Φ E s full and fathfull. Moreover, the functors F! Φ E and Φ E F are full and fathful too, because we have the somorphsms: F! ( f) : F! F C = dc F! Φ E C Φ E( f) : Φ E F C Φ E Φ C E = dc and condtons of Lemma s fulflled. Further, the functors F! Φ E and Φ E F are equvalences, because they are adjont each other. Consder the somorphsm F! ( f) : F! F C= dc F! Φ C E on the subcategory C. By Proposton we can extend t onto the whole D b (M), so d F! Φ E. Snce F! s the rght adjont to F, we get the morphsm of the functors f : F Φ E such that f C = f. It can easly be checked that f s an somorphsm. Indeed, let C Z be a cone of the morphsm f Z : F (Z) Φ E (Z). Snce F! (f Z ) s an somorphsm, we 18

19 obtan F! (Z) = 0. Therefore, ths mples that Hom(F (Y ), C Z ) = 0 for any object Y. Further, there are somorphsms F (L k ) = Φ E (L k ) for any k. Hence, we have Hom (L k, Φ! E(C Z )) = Hom (Φ E (L k ), C Z )) = Hom (F (L k ), C Z )) = 0 for all k and. Thus, we obtan Φ! E (C Z) = 0. Ths mples that Hom(Φ E (Z), C Z ) = 0. Fnally, 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 s an somorphsm. 2) Unqueness. Suppose there exst two objects E and E 1 of D b (M X) such that Φ E1 = F = ΦE2. Let us consder the complex (8) over D b (M X) (see the proof Lemma ). L n F (R n ) L 1 F (R 1 ) O M F (R 0 ) By Lemma 1.5., there exsts a convoluton of ths complex and all convolutons are somorphc. Let G D b (M X) be a convoluton of the complex (8). Now consder the followng complexes L n F (R n ) L 1 F (R 1 ) O M F (R 0 ) E k Agan by Lemma 1.5., there exsts a unque up to somorphsm convolutons of these complexes. Hence we have the canoncal morphsms G E 1 and G E 2. Moreover, t has been proved above (see the proof of Lemma ) that C 1 E 1 = G = C2 E 2 for large n, where E k, C k are objects of D b (M X) such that H (E k ) = 0 for [ a, 0] and H (C k ) = 0 for [ n a, n] ( a was defned n 2.5. ). Thus E 1 and E 2 are somorphc. Ths completes the proof of Theorem Theorem Let M and X be smooth projectve varetes. Suppose F : D b (M) D b (X) s an equvalence. Then there exsts a unque up to somorphsm object E D b (M X) such that the functors F and Φ E are somorphc. It follows mmedately from Theorem Derved categores of K3 surfaces 3.1. In ths chapter we are tryng to answer the followng queston: When are derved categores of coherent sheaves on two dfferent K3 surfaces over feld C equvalent? Ths queston s nterestng, because there exsts a procedure for recoverng a varety from ts derved category of coherent sheaves f the canoncal (or antcanoncal) sheaf s ample. Besdes, f D b (X) D b (Y ) and X s a smooth projectve K3 surface, then Y s also a smooth projectve K3 surface. Ths s true, because the dmenson of a varety and Serre functor are nvarants of a derved category. The followng theorem s proved n [BO2]. 19

20 S Theorem (see [BO2]) Let X be smooth rreducble projectve varety wth ether ample canoncal or ample antcanoncal sheaf. If D = D b (X) s equvalent to D b (X ) for some other smooth algebrac varety, then X s somorphc to X. However, there exst examples of varetes that have equvalent derved categores, f the canoncal sheaf s not ample. Here we gve an explct descrpton for K3 surfaces wth equvalent derved categores Theorem Let S 1 and S 2 be smooth projectve K3 surfaces over feld C. Then the derved categores D b (S 1 ) and D b (S 2 ) are equvalent as trangulated categores ff there exsts a Hodge sometry f τ : T S1 TS2 between the lattces of transcendental cycles of S 1 and S 2. Recall that the lattce of transcendental cycles T S s the orthogonal complement to Neron-Sever lattce N S n H 2 (S, Z). Hodge sometry means that the one-dmensonal subspace H 2,0 (S 1 ) T S1 C goes to H 2,0 (S 2 ) T S2 C. Now we need some basc facts about K3 surfaces (see [Mu2]). If S s a K3 surface, then the Todd class td S of S s equal to 1 + 2w, where 1 H 0 (S, Z) s the unt element of the cohomology rng H (S, Z) and w H 4 (S, Z) s the fundamental cocycle of S. The postve square root tds = 1 + w les n H (S, Z) too. Let E be an object of D b (S) then the Chern character ch(e) = r(e) + c 1 (E) (c2 1 2c 2 ) belongs to ntegral cohomology H (S, Z). For an object E, we put v(e) = ch(e) td S H (S, Z) and call t the vector assocated to E (or Muka vector). We can defne a symmetrc ntegral blnear form (, ) on H (S, Z) by the rule (u, u ) = rs + sr αα H 4 (S, Z) = Z for every par u = (r, α, s), u = (r, α, s ) H 0 (S, Z) H 2 (S, Z) H 4 (S, Z). By H(S, Z) denote H (S, Z) wth ths nner product (, ) and call t Muka lattce. For any objects E and F, nner product (v(e), v(f )) s equal to the H 4 component of ch(e) ch(f ) td S. Hence, by Remann-Roch- Grothendeck theorem, we have (v(e), v(f )) = χ(e, F ) := ( 1) dmext (E, F ) Let us suppose that D b (S 1 ) and D b (S 2 ) are equvalent. By Theorem 2.2. there exsts an object E D b (S 1 S 2 ) such that the functor Φ E gves ths equvalence. Now consder the algebrac cycle Z := p td S1 ch(e) π td S2 on the product S 1 S 2, where p and π are the projectons S 1 S 2 p π S 2

21 It follows from the followng lemma that the cycle Z belongs to ntegral cohomology H (S 1 S 2, Z) Lemma [Mu2] For any object E D b (S 1 S 2 ) the Chern character ch(e) s ntegral, whch means that t belongs to H (S 1 S 2, Z) The cycle Z defnes a homomorphsm from ntegral cohomology of S 1 to ntegral cohomology of S 2 : f : H (S 1, Z) H (S 2, Z) α π (Z p (α)) The followng proposton s smlar to Theorem 4.9 from [Mu2] Proposton If Φ E s full and fathful functor from D b (S 1 ) to D b (S 2 ) then: 1) f s an sometry between H(S1, Z) and H(S2, Z), 2) the nverse of f s equal to the homomorphsm f : H (S 2, Z) H (S 1, Z) β p (Z π (β)) defned by Z = p td S1 ch(e ) π td S2, where E := R Hom(E, O S1 S 2 ). Proof. The left and rght adjont functors to Φ E are: Φ E = Φ! E = p (E π ( ))[2] Snce Φ E s full and fathful, the composton Φ E Φ E s somorphc to d D b (S 1 ). Functor d D b (S 1 ) s gven by the structure sheaf O of the dagonal S 1 S 1. Usng the projecton formula and Grothendeck-Remann-Roch theorem, t can easly be shown that the composton f f s gven by the cycle p1 tds1 ch(o ) p2 tds1, where p 1, p 2 are the projectons of S 1 S 1 to the summands. But ths cycle s equal to. Therefore, f f s the dentty, and, hence, f s an somorphsm of the lattces, because these lattces are free abelan groups of the same rank. Let ν S : S SpecC be the structure morphsm of S. Then our nner product (α, α ) on H(S, Z) s equal to ν (α α ). Hence, by the projecton formula, we have (α, f(β)) = ν S2, (α π (π td S2 ch(e) p td S1 p (β))) = = ν S2, π (π (α ) p (β) ch(e) td S1 S 2 ) = = ν S1 S 2, (π (α ) p (β) ch(e) td S1 S 2 ) for every α H (S 2, Z), β H (S 1, Z). In a smlar way, we have (β, f (α)) = ν S1 S 2, (p (β ) π (α) ch(e) td S1 S 2 ) Therefore, (α, f(β)) = (f (α), β). Snce f f s the dentty, we obtan (f(α), f(α )) = (f f(α), α ) = (α, α ) 21

22 Thus, f s an sometry Consder the sometry f. Snce the cycle Z s algebrac, we obtan two sometres f alg : N S1 U N S2 U and f τ : T S1 TS2, where N S1, N S2 are Neron-Sever lattces, and T S1, T S2 are the lattces of transcendental cycles. It s clear f τ s a Hodge sometry. Thus we have proved that f the derved categores of two K3 surfaces are equvalent, then there exsts a Hodge sometry between the lattces of transcendental cycles Let us begn to prove the converse. Suppose we have a Hodge sometry f τ : T S2 TS1 It mples from the followng proposton that we can extend ths sometry to Muka lattces Proposton [N] Let φ 1, φ 2 : T H be two prmtve embeddng of a lattce T n an even unmodular lattce H. Assume( that the) orthogonal complement N := φ 1 (T ) 0 1 n H contans the hyperbolc lattce U = as a sublattce. 1 0 Then φ 1 and φ 2 are equvalent, that means there exsts an sometry γ of H such that φ 1 = γφ 2. We know that the orthogonal complement of T S n Muka lattce H(S, Z) s somorphc to N S U. By Proposton 3.8., there exsts an sometry such that f TS2 = f τ. f : H(S 2, Z) H(S 1, Z) 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 ths, because there are two types of sometres on Muka lattce that are dentty on the lattce of transcendental cycles. Frst type s multplcaton by Chern character e m of a lne bundle: φ m (r, l, s) = (r, l + rm, s + (m, l) + r 2 m2 ) Second type s the change r and s (see [Mu2]). So we can change f n such a way that r > 1 and f TS2 = f τ. Frst, note that vector v U N S1 s sotropc,.e (v, v) = 0. It was proved by Muka n hs brllant paper [Mu2] that there exsts a polarzaton A on S 1 such that the modul space M A (v) of stable bundles wth respect to A, for whch vector Muka s equal to v, s projectve smooth K3 surface. Moreover, ths modul space s fne, because there exsts the vector u U N S1 such that (v, u) = 1. Therefore we have a unversal vector bundle E on the product S 1 M A (v). The unversal bundle E gves the functor Φ E : D b (M A (v)) D b (S 1 ). Let us assume that Φ E s an equvalence of derved categores. In ths case, the cycle Z = πs tds1 1 ch(e) p td M nduces the Hodge sometry g : H(M A (v), Z) H(S 1, Z), 22

23 such that g(0, 0, 1) = v = (r, l, s). Hence, f 1 g s an sometry too, and t sends (0, 0, 1) to (0, 0, 1). Therefore f 1 g gves the Hodge sometry between the second cohomologes, because for a K3 surface S H 2 (S, Z) = (0, 0, 1) /Z(0, 0, 1). Consequently, by the strong Torell theorem (see [LP]), the surfaces S 2 and M A (v) are somorphc. Hence the derved categores of S 1 and S 2 are equvalent Thus, to conclude the proof of Theorem 3.3., t remans to show that the functor Φ E s an equvalence. Frst, we show that the functor Φ E s full and fathful. Ths s a specal case of the followng more general statement, proved n [BO1] Theorem [BO1] Let M and X be smooth algebrac varetes and E D b (M X). Then Φ E s fully fathful functor, ff the followng orthogonalty condtons are verfed: ) Hom X (Φ E(O t1 ), Φ E (O t2 )) = 0 for every and t 1 t 2. ) Hom 0 X (Φ E(O t ), Φ E (O t )) = k, Hom X (Φ E(O t ), Φ E (O t )) = 0, for / [0, dmm]. Here t, t 1, t 2 are ponts of M, O t are correspondng skyscraper sheaves. In our case, Φ E (O t ) = E t, where E t s stable sheaf wth respect to the polarzaton A on S 1 for whch v(e t ) = v. All these sheaves are smple and Ext (E t, E t ) = 0 for [0, 2]. Ths mples that condton 2) of Theorem s fulflled. All E t are stable sheaves, hence Hom(E t1, E t2 ) = 0. Further, by Serre dualty Ext 2 (E t1, E t2 ) = 0. Fnally, snce the vector v s sotropc, we obtan Ext 1 (E t1, E t2 ) = 0. Ths yelds that Φ E s full and fathful. As our stuaton s not symmetrc (a pror), t s not clear whether the adjont functor to Φ E s also full and fathful. Some addtonal reasonng s needed Theorem In the above notatons, the functor Φ E : D b (M A (v)) D b (S 1 ) s an equvalence. Proof. Assume the converse,.e. Φ E s not an equvalence, then, snce the functor Φ E s full and fathful, there exsts an object C D b (S 1 ) such that Φ E (C) = 0. By Proposton 3.5., the functor Φ E nduces the sometry f on the Muka lattces, hence the Muka vector v(c) s equal to 0. Object C satsfes the condtons Hom (C, E t ) = 0 for every and all t M A (v), where E t are stable bundles on S 1 wth the Muka vector v. 23

24 Denote by H (C) the cohomology sheaves of the object C. There s a spectral sequence whch converges to Hom (C, E t ) It s depcted n the followng dagram E p,q 2 = Ext p (H q (C), E t ) = Hom p+q (C, E t ) (11) q. d 2 d 2 d 2. p d 2 We can see that Ext 1 (H q (C), E t ) = 0 for every q and all t, and every morphsm s an somorphsm. To prove the theorem, we need the followng lemma Lemma Let G be a sheaf on K3 surface S 1 such that Ext 1 (G, E t ) = 0 for all t. Then there exsts an exact sequence that satsfes the followng condtons: 0 G 1 G G 2 0 1) Ext (G 1, E t ) = 0 for every 2, and Ext 2 (G 1, E t ) = Ext 2 (G, E t ) 2) Ext (G 2, E t ) = 0 for every 0, and Hom(G 2, E t ) = Hom(G, E t ) and µ A (G 2 ) < µ A (G) < µ A (G 1 ). Proof. Frstly, there s a short exact sequence 0 T G G 0, where T s a torson sheaf, and G s torson free. Secondly, there s a Harder-Narasmhan fltraton 0 = I 0... I n = G for G such that the successve quotents I /I 1 are A -semstable, and µ A (I /I 1 ) > µ A (I j /I j 1 ) for < j. Now, combnng T and the members of the fltraton for whch µ A (I /I 1 ) > µ A (E t ) (resp. =, < ) to one, we obtan the 3-member fltraton on G 0 = J 0 J 1 J 2 J 3 = G. Let K be the quotents sheaves J /J 1. We have µ A (K 1 ) > µ A (K 2 ) = µ A (E t ) > µ A (K 3 ) (we suppose, f needed, µ A (T ) = + ). 24

25 Moreover, t follows from stablty of E t that Hom(K 1, E t ) = 0 and Ext 2 (K 3, E t ) = 0 Combnng ths wth the assumpton that Ext 1 (G, E t ) = 0, we get Ext 1 (K 2, E t ) = 0. To prove the lemma t remans to show that K 2 = 0. Note that K 2 s A -semstable. Hence there s a Jordan-Hölder fltraton for K 2 such that the successve quotents are A -stable. The number of the quotents s fnte. Therefore we can take t 0 such that Hom(K 2, E t0 ) = 0 and Ext 2 (K 2, E t0 ) = 0 Consequently, χ(v(k 2 ), v(e t )) = 0. Thus, as Ext 1 (K 2, E t ) = 0 for all t, we obtan Ext (K 2, E t ) = 0 for every and all t. Further, let us consder Φ E (K 2). We have Hom (Φ E(K 2 ), O t ) = Hom (K 2, E t ) = 0, Ths mples Φ E (K 2) = 0. Hence v(k 2 ) = 0, because f s an sometry. And, fnally, K 2 = 0. The lemma s 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 satsfy the condton of Lemma Hence there exst decompostons wth condtons 1),2): 0 H p 1 Hp (C) H p 2 0 and 0 Hp+1 1 H p+1 (C) H p Now consder the canoncal morphsm H p+1 (C) H p (C)[2]. It nduces the morphsm s : H p+1 1 H p 2 [2]. By Z denote a cone of s. Snce d 2 of the spectral sequence (11) s an somorphsm, we obtan 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 p+1 1 ) > µ A (E t ) > µ A (H p 2 ). Therefore v(z) 0. Ths contradcton proves the theorem. There exsts the another verson of Theorem Theorem Let S 1 and S 2 be smooth projectve K3 surfaces over feld C. Then the derved categores D b (S 1 ) and D b (S 2 ) are equvalent as trangulated categores ff there exsts a Hodge sometry f : H(S 1, Z) H(S 2, Z) between the Muka lattces of S 1 and S 2. Here the Hodge sometry means that the one-dmensonal subspace H 2,0 (S 1 ) H(S 1, Z) C goes to H 2,0 (S 2 ) H(S 2, Z) C. Appendx. The facts, collected n ths appendx, are not new; they are known. However, not havng a good reference, we regard t necessary to gve a proof for the statement, whch s used n the man text. We explot the technque from [IM]. 25

26 Let X be a smooth projectve varety and L be a very ample nvertble sheaf on X such that H (X, L k ) = 0 for any k > 0, when 0. Denote by A the coordnate algebra for X wth respect to L,.e. A = H 0 (X, L k ). Now consder the varety X n. Frst, we ntroduce some notatons. Defne subvaretes (n) ( 1,..., k )( k+1,..., m ) Xn by the followng rule: By (n) ( 1,..., k )( k+1,..., m) := {(x 1,..., x n ) x 1 = = x k ; x k+1 = = x m } S (n) denote (n) (n,...,) Further, let let Σ (n) := n k=1 T (n) (n) (k,k 1) := 1 k=1 k=0. It s clear that S(n) = X. (n) (n,...,)(k,k 1) (note that T (n) 1 and T (n) 2 are empty) and. We see that T (n) S (n). Denote by I (n) the kernel of the restrcton map O (n) S O (n) T 0. Usng nducton by n, t can easly be checked that the followng complex on X n s exact. (Note that P n : 0 J Σ (n) I n (n) I (n) n 1 I(n) 2 I (n) 1 0 I (n) 1 = O (n) n,...,1 and I (n) 2 = O (n) n,...,2 complex s a short exact sequence on X X : P 2 : 0 J O X X O 0 ). For example, for n = 2 ths Denote by π (n) the projecton of X n onto th component, and by π (n) j denote the projecton of X n onto the product of 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 (n) 1 (J Σ (n) (O L L)). Proposton A.1 Let L be a very ample nvertble sheaf on X as above. Suppose that for any m such that 1 < m n + dmx + 2 the followng condtons hold: a) H (X m, J Σ (m) (L L)) = 0 for 0 b) R π (m) 1 (J Σ (m) (O L L)) = 0 for 0 c) R π (m) 1m (J Σ (m) (O L L O)) = 0 for 0 Then we have: 1) algebra A s n-koszul,.e the sequence B n k A B n 1 k A B 1 k A A k 0 s exact; 2) the followng 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 (f k < 0, then A k = 0 by defnton); 3) the complex L n R n L 1 R 1 O M R 0 O gves n-resoluton of the dagonal on X X,.e. t s exact. 26

27 Proof. 1) Frst, note that H (X m, I (m) k (L L)) = H (X k 1, J Σ (k 1) (L L)) A m k+1 By condton a), they are trval for 0. Consder the complexes P m (L L) for m n + dmx + 1. Applyng the functor H 0 to these complexes and usng condton 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 + dmx + 1. Now put m = n + dmx + 1. Denote by W m the complex I (m) m I (m) m 1 I(m) 2 I (m) 1 0 Take the complex W m (L L L ) and apply functor H 0 to t. We obtan the followng sequence: B m 1 k A B m 2 k A +1 B 1 k A m 1 A m 0 The cohomologes of ths sequence are condton b) that H j (X m, J Σ (m) (L L L )). It follows from H j (X m, J Σ (m) (L L L )) = H j (X, R 0 π (m) m (J Σ (m) (L L O)) L ) Hence they are trval for j > dmx. Consequently, we have the exact sequences: B n k A m n+ 1 B n 1 k A m n+ B 1 k A m+ 2 A m+ 1 for 1. And for 1 the exactness was proved above. Thus, algebra A s n-koszul. 2) The proof s the same as for 1). We have somorphsms R π (m) 1 (I(m) k (O L L)) = R π (k 1) 1 (J Σ (k 1) (O L L)) A m k+1 Applyng functor R 0 π (m) 1 to the complexes Pm (O L L)) for m n+dmx+2, we obtan 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 + dmx + 2. Put m = n + dmx + 2. Applyng functor R 0 π (m) 1 to the complex W m (O L L L )), we get the complex A R m 2 A m+ 3 R 1 A m+ 2 R 0 L m+ 2 0 The cohomologes of ths complex are R j π (m) 1 (J Σ (m) (O L L L )) = R j p 1 (R 0 π (m) 1m (J Σ (m) (O L L O)) (O L )) They are trval for j > dmx. 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 27

28 are exact for all k 0. 3) Consder the complex W n+2 (O L L L ). Applyng the functor R 0 π (n+2) 1(n+2) to t, we obtan the followng complex on X X : L n R n L 1 R 1 O M R 0 O By condton c), t s exact. Ths fnshes the proof. Note that for any ample nvertble sheaf L we can fnd j such that for the sheaf L j the condtons a),b),c) are fulflled. [Ba] References J. Backeln, On the rates of growth of the homologes of Veronese subrng, Lecture Notes n Math., vol.1183, Sprnger-Verlag, [Be] A. A. Belnson, Coherent sheaves on P n and problems of lnear algebra, Funkconalny analz ego prl., v.12, p (1978) (n Russan). [BBD] A. A. Belnson, J. Bernsten, P. Delgne, Fasceaux pervers, Astérsque 100 (1982). [BK] A. I. Bondal, M. M. Kapranov, Representable functors, Serre functors, and mutatons, Izv. Akad. Nauk SSSR Ser.Mat. 53 (1989), pp ; Englsh transl. n Math. USSR Izv. 35 (1990), pp [BO1] A. I. Bondal, D. O. Orlov, Semorthogonal decomposton for algebrac varetes, Preprnt MPI/95-15 (see also alg-geom ). [BO2] A. I. Bondal, D. O. Orlov, Reconstructon of a varety from the derved category and groups of autoequvalences, Preprnt MPI/ [GM] S. I. Gelfand, Yu. I. Mann, Introducton to Cohomology Theory and Derved Categores, Nauka, 1988 [IM] (n Russan). S. P. Inamdar, V. B. Mehta, Frobenus splttng of Schubert varetes and lnear syzyges, Amercan Journal of Math., vol.116, N6 (1994), pp [LP] E. Loojenga, C. Peters, Torell theorem for Kähler K3 surfaces, Composto Math., v.58 (1981), pp [Mu1] S. Muka, Dualty between D(X) and D( ˆX) wth ts applcaton to Pcard sheaves, Nagoya Math. J. 81 (1981), pp [Mu2] S. Muka, On the modul space of bundles on a K3 surface I, Vector bundles on algebrac varetes, [N] Tata Insttute of Fundamental Research, Oxford Unversty Press, Bombay and London, V. V. Nkuln, Integral symmetrc blnear forms and some of ther applcatons, Englsh trans., Math. USSR Izvesta, v.14 (1980), pp [V] J. L. Verder, Categores dervées, n SGA 4 1/2, Lecture Notes n Math., v.569, Sprnger-Verlag, Algebra secton, Steklov Math. Insttute, Vavlova 42, Moscow, RUSSIA E-mal address: orlov@geness.m.ras.ru 28