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SOME REMARKS ABOUT THE FM-PARTNERS OF K3 SURFACES WITH
PICARD NUMBERS 1 AND 2
PAOLO STELLARI
Abstract. In this paper we prove some results about K3 surfaces with Picard number 1 and 2.
In particular, we give a new simple proof of a theorem due to Oguiso which shows that, given an
integer N, there is a K3 surface with Picard number 2 and at least N non-isomorphic FM-partners.
We describe also the Mukai vectors of the moduli spaces associated to the Fourier-Mukai partners
of K3 surfaces with Picard number 1.
1. Introduction
In some recent papers Hosono, Lian, Oguiso and Yau (see and [13j) gave a formula that
counts the number of non-isomorphic Fourier-Mukai partners of a K3 surface. In this paper we are
interested in the case of K3 surfaces with Picard number 1 and 2.
In the second paragraph, we recall the formula for the number of the isomorphism classes of
Fourier-Mukai partners of a given K3 surface (given in |4j), which allows to count the isomorphism
classes of Fourier-Mukai partners of a K3 surface with Picard number 1 (this is also given in
[13J). As a first result, we will describe the Mukai vectors of the moduli spaces associated to the
Fourier-Mukai partners of such K3 surfaces 1 . This gives some information about the geometry of
the Fourier-Mukai partners of the given K3 surface.
In the third paragraph we prove that, given N and d positive integers, there is an elliptic K3
surface with a polarization of degree d and with at least N non-isomorphic elliptic Fourier-Mukai
partners (Theorem 13. 3|) . The most interesting consequence of this result is a new simple proof of
Theorem 1.7 in |13j (Corollary 13.41 and Remark 3.5).
We start with recalling some essential facts about lattices and K3 surfaces.
1.1. Lattices and discriminant groups. A lattice L := (L,b) is a free abelian group of finite
rank with a non-degenerate symmetric bilinear form b : L x L — > Z. Two lattices (Li,&i) and
(L 2 , b 2 ) are isometric if there is an isomorphism of abelian groups / : L\ — > L 2 such that b\(x, y) =
b 2 {f( x )> /(?/))• We write O(L) for the group of all autoisometries of the lattice L. A lattice (L,b)
is even if, for all x G L, x 2 := b(x, x) G 2Z, it is odd if there is x G L such that b(x, x) £ 2Z.
Given an integral basis for L, we can associate to the bilinear form a symmetric matrix Sl of
dimension rkL, uniquely determined up to the action of GL(rkL, Z). The integer detL:=detSL is
called discriminant and it is an invariant of the lattice. A lattice is unimodular if detL=±l. Given
(L,b) and k G Z, L(k) is the lattice (L,kb).
Given a sublattice V of L with V L, the embedding is primitive if L/V is free. In particular, a
sublattice is primitive if its embedding is primitive. Two primitive embeddings V > L and V L'
are isomorphic if there is an isometry between L and V which induces the identity on V. For a
sublattice V of L we define the orthogonal lattice V 1 - := {x G L : b(x, y) = 0, Vy G V}. Given two
lattices (Li,&i) and (£2,62)1 their orthogonal direct sum is the lattice (L,b), where L = L\ © L2
and b(xi +yi,x 2 + 2/2) = 61(^1,^2) +62(2/1,2/2), for xt,x 2 G L\ and yi,y 2 G L 2 .
2000 Mathematics Subject Classification. 14J28.
Key words and phrases. K3 surfaces, Fourier-Mukai partners.
x This result was independently proved by Hosono, Lian, Oguiso and Yau (Theorem 2.1 in 0)
1
2
PAOLO STELLARI
The dual lattice of a lattice (L,b) is L v := Hom(L,Z) = {iei® z Q: b(x,y) G Z,Vy G L}.
Given the natural inclusion L > L v , x 6(— we define the discriminant group Al:=L v / L.
The order of is |detL | (see pQ, Lemma 2.1, page 12). Moreover, b induces a symmetric bilinear
form bi : Al x — > Q/Z and a corresponding quadratic form q^ : Al — > Q/Z such that, when L
is even, qi(x) = q{x) modulo 2Z, where x is the image of x G L v in A^. The elements of the triple
where is the multiplicity of positive/negative eigenvalues of the quadratic form
on L (g) R, are invariants of the lattice L.
If L is unimodular, L v = , x) : x G L}. If V is a primitive sublattice of a unimodular lattice L
such that 6 |y is non-degenerate, then there is a natural isometry of groups 7 : F v /y — > (V ± ) w /V ± .
1.2. K3 surfaces and M-polarizations. A K3 surface is a 2-dimensional complex projective
smooth variety with trivial canonical bundle and first Betti number b\ = 0. From now on, X will
be a K3 surface. The group H 2 (X,Ij) with the cup product is an even unimodular lattice and it is
isomorphic to the lattice A := C/ 3 E$(— l) 2 (for the meaning of U and E$ see page 14). The
lattice A is called K3 lattice and it is unimodular and even.
Given the lattice H 2 (X,Ij), the Neron-Severi group NS(X) is a primitive sublattice. Tx :=
NS(X) 1 - is the transcendental lattice. The rank of the Neron-Severi group p(X) :=rkNS(X) is
called the Picard number, and the signature of the Neron-Severi group is (l,p — 1), while the one of
the transcendental lattice is (2, 20 — p). If X and Y are two K3 surfaces, / : Tx — > Ty is an Hodge
isometry if it is an isometry of lattices and the complexification of / is such that /c(Cwa') = Cwy,
where H 2 >°(X) = Cuj x and H 2 '°(Y) = Cu> ¥ . We write (Tx,Cuj x ) = (T y ,C^y) to say that there
is an Hodge isometry between the two transcendental lattices.
A marking for a K3 surface X is an isometry ip : H 2 (X,Z) -> A. We write (X,ip) for a K3
surface X with a marking cp. Given Ac := A ® C and given uj G Ac we denote by [to] G P(A<c)
the corresponding line and we define the set $7 := {[uj] G P(Ac) : uj ■ uj = 0, ui ■ to > 0}. The image
in P(Ac) of the line spanned by pc(^x) belongs to O and is called period point (or period) of
the marked surface (X, ip). From now on, the period point of a marked K3 surface (X,tp) will be
indicated either by dpc(ujx) or by [ipc(uJx)]-
Given two K3 surfaces X and Y, we say that they are Fourier- Mukai-partners (or FM-partners)
if there is an equivalence between the bounded derived categories of coherent sheaves D h coh (X) and
D b coh (Y). By results due to Mukai and Orlov, this is equivalent to say that there is an Hodge
isometry (Tx,Cu>x) — ► (Ty,Cwy). We define FM(X) to be the set of the isomorphism classes of
the FM-partners of X.
Let M be a primitive sublattice of A with signature (1, t). A K3 surface X with a marking
ip : H 2 (X,Z) -> A is a marked M-polarized K3 surface if <p~~ l {M) C NS(X). A K3 surface is
M -polarizable if there is a marking ip such that (X, ip) is a marked M-polarized K3 surface. Two
marked and M-polarized surfaces (X,ip) and (X',ip') are isomorphic if there is an isomorphism
ip : X — > X' such that ip' = ip o i/j*. Form now on, we will consider the case of lattices M := (h),
with h 2 = 2d and d > 0. The pair (X,h), where X is a K3 surface and h G NS(X), with h 2 = 2d,
means a K3 surface with a polarization of degree 2d.
2. FM-PARTNERS OF A K3 SURFACE WITH p = 1 AND ASSOCIATED MUKAI VECTORS
In this section we want to describe the Mukai vectors of the moduli spaces associated to the
M-polarized FM-partners of a K3 surface X with Picard number 1, By Orlov's results ({Hp,
q = \FM(X)\ is the same as the number of non-isomorphic compact 2-dimensional fine moduli
spaces of stable sheaves on X. Obviously, on a K3 surface with Picard number 1 and NS(X) = (h)
there is only one (/i)-polarization of degree h 2 = 2d. So the concept of FM-partner and the
concept of M-polarized FM-partner coincide. If M = (h) we are sure, by Orlov, that if we find
q non-isomorphic moduli spaces, then these are representatives of all the isomorphism classes of
M-polarized FM-partners of X.
FM-PARTNERS OF K3 SURFACES WITH PICARD NUMBERS 1 AND 2
3
We recall briefly the counting formula for the isomorphism classes of FM-partners of a given K3
surface. Given a lattice S, the genus of S is the set G(S) of all the isometry classes of lattices S'
such that As = Ag> and the signature of S' is equal to the one of S.
Let Tx be the transcendental lattice of an abelian surface or of a K3 surface X with period Clux-
We can define the group
G := Hodge (T x ,Cux) = {ge 0(T X ) : g(Coj x ) = Cto x }.
We know (see j2] Theorem 1.1, page 128), that the genus of a lattice, with fixed rank and dis-
criminant, is finite. The map O(S) — > O(As) defines an action of 0(5*) on 0(A$). On the other
hand, taken g G G, and given a marking ip for X, ip o g o ip~ 1 induces an isometry on the lattice
T := (p(T x ), thus tp defines a homomorphism G » 0(T). The composition of this map and the
map 0(T) -^0(A T ) gives an action of G on 0(i r ) = 0(i s ).
Theorem 2.1. [4, Theorem 2.3]. Let X be a K3 surface and let £(NS(X)) = = {Si, ■ ■ ■ ,S m }
Then
m
\FM(X)\=J2\0(Sj)\0(A Sj )/G\,
where the actions of the groups G and O(Sj) are defined as before.
The following corollary (which is Theorem 1.10 in j 1 3) i determines the number q of FM-partners
of a surface with Picard number 1.
Corollary 2.2. Let X be a K3 surface with p(X) = 1 and such that NS(X) = (h), with h? = 2d.
(i) The group 0(A S ) is trivial if d = 1 while, if d > 1, 0(A S ) = (Z/2Z)^ d ), where p{d) is the
number of distinct primes q such that q\d. In particular, if d > 2, then \0(As)\ = 2 p<yd \
(ii) For all markings ip of X, the image of Hx,<p '■= {<P ° g ° <P~ l '■ g € G} C O(T) in O(At) by
the map O(T) -> 0(A T ) is {±id}.
In particular, \FM(X)\ = 2 P ^~ 1 , where now we set p(l) = 1.
Assertion (i) is known and it can also be found in ^H] (Lemma 3.6.1).
Using the notation of p], we put H*(X,Z) := H°(X,Z) H 2 (X,Z) H 4 (X,Z). Given a :=
(011,012,013) and (3 := (Pi, 02, $3) in H*(X,Z), using the cup product we define the bilinear form
a ■ (3 := —cei U ^3 + 02 U /?2 — «3 U /3i .
From now on, depending on the context, a ■ (3 will mean the bilinear form denned above or the cup
product on H 2 (X,Z).
We give to H*(X,Z) an Hodge structure considering
H*(X,C) 2 >° := H 2 '°(X),
H*(X,C) ' 2 := H°' 2 (X),
H*(X,C)^ := H°(X,C)®H 1 ' 1 (X)®H 4 (X,C).
H(X,Z) is the group H*(X,Z) with the bilinear form and the Hodge structure denned before.
For v = (r,h,s) G H(X,Z) with r G H°(X,Z) = Z, s £ H 4 (X,Z) = Z and /i G H 2 (X,Z),
M(v) is the moduli space of stable sheaves £ on I such that ikE = r, ci(E) = h and s =
ci(E) 2 /2 - 02(E) + r. If the stability is denned with respect to A G H 2 (X,Z) we write M A (v).
The vector v is isotropic ii v ■ v = Q. The vector u G Z) is primitive if H(X, Z)/Zv is free.
As we have observed, the results of Orlov in ^1] imply that each FM-partner of X is isomorphic
to an Mh(v). We determine a set of Mukai vectors which corresponds bijectively to the isomorphism
classes of the FM-partners of X in FM(X), First of all, we recall the following theorem due to
Mukai (PU).
4
PAOLO STELLARI
Theorem 2.3. [10, Theorem 1.5 3]. If X is a K3 surface, v = (r,h,s) is an isotropic vector
in H l,1 {X, Z) = H*(X, C) ' H H*(X,Z) and Ma(v) is non-empty and compact, then there is an
isometry ip : v^/Zv — > H 2 (Ma(v),Z) which respects the Hodge structure.
If NS(X) = Z/t with h 2 = 2d = 2pf . . .p^, where > 0, > 1 and pi odd primes with pi ^ pj
Hi ^ j, then we consider the Mukai vectors
v\ = v ju '" ' js ■ = (v ejl v" 3 " h n js+1 ■ ■ ■ v 3m \
where / = . . . ,j s } and J = {j s +i, ■ ■ ■ ,j m } are a partition of {1, ... , m} such that I II J =
{1, . . . ,m}. The following theorem shows how to determine |-FM(X)| of them corresponding to
non-isomorphic moduli spaces of stable sheaves.
Theorem 2.4. Let X be a K3 surface with NS(X) = Zh such that h 2 = 2d = 2pf ...p e ™. Then,
for all Vj as above, Mh(vj) is a 2-dimensional compact fine moduli space of stable sheaves on X .
Moreover, if M^Vj 1 ) = M^v 1 ^), then v 1 ^ = Vj 2 2 or vf 2 = (si,h,ri), with v 1 ^ = (ri,h,si), where
the multindexes Ik and Jk vary over all the partitions of {1, ... , m).
Proof. The vectors Vj are all isotropic and they are primitive in H(X,Z), so, by Theorem 5.4 in
[TT)] M^Vj) is non-empty. Moreover the hypothesis of Theorem 4.1 in are satisfied and so the
moduli spaces are compact. By Corollary 0.2 in they are 2-dimensional, while they are fine by
the results in the appendix of |1()| .
If m = or m = 1 then, by Corollary 12.21 we have only one moduli space with respectively
v = (1, h, 1) in the first case and v = (1, h,p e ) in the second case.
Otherwise we must prove that if
vx = (ri, h, si) = Vj\ v 1 ^ = (r 2 , h, s 2 ) = v 2 ,
with t>2 7^ (si,h, n), then
M h (vi) ¥ M h (v 2 )
But by Theorem 12.31 and Torelli theorem, if we put
Mi := v^/Zv! and M 2 := v£ /Zv 2
then it suffices to show that there are no Hodge isometries between Mi and M 2 . Obviously, it
suffices to show that there are no Hodge isometries between the transcendental lattices which lifts
to an isometry of the second cohomology groups.
By definition, a representative of a class in Mj (i = 1, 2) is a vector (a, b, c) such that bh = asj+crj,
hence
bh = asi (mod r^),
for i = 1,2. Prom now on we will write (a, b, c) for the equivalence class or for a representative of
the class. In fact, all the arguments we are going to propose are independent from the choice of a
representative.
The Hodge structures on Mi and M 2 are induced by the ones defined on H(X,Z), so, up to an
isometry, we identify NS(M/j(ui)) and NS(M/ l (v 2 )) with
Si := ((0, h, 2a x )> C Mi and S 2 := ((0, h, 2s 2 )) C M 2
respectively.
Now, we can describe the transcendental lattices T\ := and T 2 := S% of M/ l (vi) and Mh(v 2 )
respectively.
If (a,b,c) ■ (0, h, 2s\) = then bh = (mod r%). Indeed, let us suppose that bh = K (mod ri)
where K ^ (mod r\). Then, by simple calculations, we obtain
/ , v / s / n ■ h — K s
(a, b, c = L, 0, H + 0, n,
FM-PARTNERS OF K3 SURFACES WITH PICARD NUMBERS 1 AND 2
5
as equivalence classes. Here n = b — kh, for a particular k G Z, bh = Ls% (mod ri) and H is an
integer. But now
= (a, b, c) • (0, h, 2si) = -2Lsi + nh = -2Ls x + (b - kh)h =
= -2bh + 2wn +bh- kh 2 ,
with w G Z. So
bh = (mod ri).
This is a contradiction. By these remarks and simple calculations, a class y in Ti, as an element
of the quotient Mi, has representative (0,n, nh/r{). But (0,n,nh/ri) ■ (0,h, 2si) = nh = 0. So
y = (0, n, 0) and
Ti = {(0,n,0) :nGT x }.
Analogously we have
T 2 = {(0,n,0):iieT x }.
By Lemma 4.1 in (see also point (ii) of Corollary I2.2|) . if / : (Ti,Ca>i) — > (T2,Ca;2) is a Hodge
isometry, then all the Hodge isometries from T\ into T2 are / and — /. But in this case Mi and M2
inherit their Hodge structure from H(X,Z). Hence the two Hodge isometries /, g : T\ — ► T2 are
(0, n, 0) J-+ (0, n, 0) or (0, n, 0) 1-^ (0, -n, 0).
Let us show that / cannot be lifted to an isometry from Mi into Mi- Equivalently, this means
that there are no isomorphisms between Mh{v\) and Mh{v2) which induces /.
We start by observing that, if (a, b, c) € Mj with i = 1,2, then
(a, b, c) • (0, /i, 2si) = —bh (mod rj).
Indeed, if bh = K (mod r«) then a = L (mod rj) and so (a, c) = (L, 0, if) + (0, n, n ' h ~ K ) with n
and if as before. So, (a, 6, c) • (0, h, 2si) = —2bh + 2wri + bh — kh? = —bh (mod r^) and n/i = bh
(mod rj).
Now let us suppose that there is an isometry <p : Mi — > M2 which induces /. We can prove that
there is (a, 6, c) G Mi, with bh = (mod ri), such that ip(a,b,c) = (d, e, /) G M2 with eh ^
(mod r2). First of all, by our hypotheses about r± and r2, we can suppose that there is a prime p
which divides T2 but which does not divide r\ (otherwise we can change the roles of Mi and M2 in
the following argument). By Theorem 1.14.4 in ^21, there is an isometry
V? : H 2 (X, Z) — ► U 3 E 8 {-1) 2 = A
such that k\ := ip(h) = (1, d, 0, . . . , 0), where h 2 = 2d. Let k2 := (0, r±, 0, . . . , 0). Now k\ ■ k2 = r\
and we can take n := ip~ 1 (k2). Obviously, the vector (0, n,n ■ hjr\) G Mi is such that n ■ h =
(mod r\). Let us suppose that ^((0, n, n-h/r\)) = (d, e, f) withe-/i = (mod r2). By the previous
remark, this is equivalent to say that
/ / n ■ h\\ ( m ■ h s
for a given m G H 2 (X, Z).
Since rkSi =rkS" 2 = 1, either <p((0, h, 2s x )) = (0,h,2s 2 ) or cp((Q, h, 2si)) = -(0,h,2s 2 ). In
particular, if 93 correspond to case (1) (the same argument holds if ip is as in case (2)), then
(ft ' h\ ( ( ti ' h\
0, n, —^—\ ■ (0, h, 2si) = ip M 0, n, — — J • (0, h, 2si)
m ■ h\
0, m, • (0, h, 2s2) = m ■ h.
T2 J
In particular, m ■ h = n ■ h = r\ which is not divisible by T2- This gives a contradiction and thus
eh ^ (mod r2).
6
PAOLO STELLARI
The previous remarks show that if
nh
(a,b,c) = 0,n,
V n
then
(777 . h — /V"
0,m,
with L (mod r 2 ). Let us take (0, N, 0) € T x and
<p(0,N,0) = f(0,N,0) = {0,N,0)eT 2 .
Then
/ re/A
(*) nN = (0,re,— J • (0,iV,0) =
/ m ■ h — K
(L,0,H) + 0,m,
(0, iV, 0) = miV.
Because H 2 (X, Z) is unimodular and (*) is true for every iV € Tx, we have m — n = kh € NS(X),
where k G Z. But now n/i = (0,re,^) • (0,/i,2si) = [(L, 0, H) + (0,m,^^^)] • (0,/i,2s 2 ) =
mh — 2Ls2 = nh + kh 2 — 2Ls 2 = nh + 2/cr 2 S2 — 2Ls 2 - So L = (mod r 2 ) which is contradictory.
Repeating the same arguments for g, we see that neither / nor g lifts to an isometry of the
second cohomology groups. So, by Torelli Theorem, Mh,(v\) ^ M/ l (u 2 )- □
3. Genus and polarizations when p = 2
In this paragraph we are interested in the number of non isomorphic FM-partners of K3 surfaces
with a given polarization and Picard number 2.
Our main result is Theorem 13.31 First of all, we recall the following lemma which is an easy
corollary of Nikulin's Theorem 1.14.2 in |12j and whose hypotheses are trivially verified if p = 2.
Lemma 3.1. Let L be an even unimodular lattice and let T\ and T 2 be two even sublattice with
the same signature (i( + ),t(_)) ; where > and t(_) > 0. Let the corresponding discriminant
groups (A^jqTi) and (AT 2 ,qT 2 ) be isometric andletrkTi > 2+£(At 1 ), where £(At 1 ) is the minimal
number of generators of At x ■ Then T\ = T 2 .
We prove the following lemma.
Lemma 3.2. Let Ld in be the lattice (Z 2 ,Md in ), where
M,
2d n
n
with d and n positive integers such that (2d, n) = 1. Then
(i) the discriminant group Ai dn is cyclic;
(ii) if d\, di, n\ and n 2 are positive integers such that (2di,n\) = (2d 2 ,n 2 ) = 1 then Az dini =
Al 4 n if and only if
(a.l) ni = n 2 ;
(b.l) there is an integer a such that (a,n) = 1 and d±a 2 = d 2 (mod n 2 );
(iii) if Ld 1)n — Ld 2 ,n then one of the following conditions holds
(a.2) d\ = <i 2 (mod re);
(b.2) did 2 = 1 (mod re).
Proof. Let ed,n = (1)0)* and fd, n = (0, 1)* be generators of the lattice Ld, n - Under the hypothesis
(2d,n) = 1, (i) follows immediately because
FM-PARTNERS OF K3 SURFACES WITH PICARD NUMBERS 1 AND 2
7
has order jdetM^I = n 2 and it is cyclic with generator
ne d , n ~ ^df d ,r
Indeed,
L
■2
f d,n • >
I Tl&d,n 2dfd,n fd,r
* n z n
and f dn has order n in Ai dn .
First we prove that the conditions (a.l) and (b.l) are necessary. The orders of Ai d ^ ni and
AL d2 n are n\ and n 2 respectively with ni, n 2 > 0, so n := n\ = n 2 (which is (a.l)). If Ai, d n and
Ai, d n are isomorphic as groups, there is an integer a prime with n such that the isomorphism is
determined by
fdx,n ^ a fd 2 ,n-
But now
?2 _ -2di ?2 _ -2d 2
dl,n _ n 2 d2,n _ n 2 '
and if we want Ai d ^ n and A^^ n to be isometric as lattices, we must require
-24 _ 2 -2d 2
— 2~ = a — — ( m °d 2).
This is true if and only if
d\ = a 2 d,2 (mod n 2 ).
So the necessity of condition (b.l) is proved. In the same way it follows that (a.l) and (b.l) are
also sufficient.
Let us consider point (iii). The lattices L di n and L d2jn are isometric if and only if there is a
matrix A G GL(2, Z) such that
(*) A t M dl>n A = M d2jU .
Let L dl ^ n and L d2 ^ n be isometric and let
A
x y
z t
Then from (*) we obtain the two relations
(1) (fa = x 2 d\ + xzn;
(2) 2y(yd 1 + tn) = 0.
By (2) we have only two possibilities: either y = or yd\ = —tn. Let y = 0. From the relation
1 = |det(A)| = \xt - yz\ = \xt\
it follows that x = ±1 and so, from (1), we have d\ = d 2 (mod n), which is condition (a. 2).
Let us consider the case yd\ = —tn. We know that (d±,n) = 1 and hence y = cn and t = —cd\,
with c 6 Z. From
1 = |det(>l)| = | — cxdi — cnz|
it follows that c = ±1. We suppose c = 1 (if c = — 1 then the same arguments work by simple
changes of signs). Multiplying both members of relation (1) by d\ we have
d 2 c?i = x 2 d\ (mod n).
But we know that ±1 = det(A) = —xd\ — nz and so — xd\ = ±1 (mod n). Thus
1 = x 2 d\ (mod n)
and from this we obtain (b.2). □
Now we can prove the following theorem (note that point (iii) and (v) are exactly Theorem 1.7
in[ni).
8
PAOLO STELLARI
Theorem 3.3. Let N and d be positive integers. Then there are N K3 surfaces X\,. . . ,Xjy with
Picard number p = 2 such that
(i) Xi is elliptic, for every i G {1, . . . , N};
(ii) there is i G {1, ... , iV} such that Xi has a polarization of degree 2d;
(iii) NSpQ) ? NS(Xj) ifi + j, where i,j G {1,. . . ,N};
(iv) \detNS(Xi)\ is a square, for every i G {1, . . . , N};
(v) there is an Hodge isometry between (T Xi ,Cu Xi ) and (T x .,Cu) X .), for all i,j G {1,. . . , N}.
In particular, Xi and Xj are non-isomorphic FM-partners, for all i,j G {1,. . . , N}.
Proof. The surjectivity of the period map for K3 surfaces implies that, given a sublattice S of A
with rank 2 and signature (1,1), there is at least one K3 surface X such that its transcendental
lattice Tx is isometric to T := S- 1 .
So the theorem follows if we can show that, for an arbitrary integer N, there are at least A?"
sublattices of A with rank 2, signature (1,1) and representing zero which are non-isometric but
whose orthogonal lattices are isometric in A.
Let us consider in U © U <^-> A the following sublattices
Sd,n '■-
/1\
Q
n
1
w
V /
with (2(7, n) = 1 and n > 0.
We can observe that, when n and d vary, the lattices S qt7l are primitive in A and the matrices
associated to their quadratic forms are exactly the M q ^ n . Since M q ^ n has negative determinant, the
lattice has signature (1,1). Moreover, 5 ?jn represents zero
Let n > 2 be a prime number such that n > d 2 N . We choose d\ := d, di := d2 2 ,. . . , d^ = dN 2 .
By definition, there is an integer such that (ai,n) = 1 and
a 2 di = di (mod n 2 ),
for every i G {1, . . . , A^}. Thus the hypotheses (b.l) of Lemma l3~2l are satisfied and by point (ii) of
the same lemma,
where i,j G {1, A^}. By Lemma l3~2l there are isometries
with i G {2, N}. Now let (Xi, (fx) be a marked K3 surface associated to the lattice Sd lt n- By the
surjectivity of the period map we can consider the marked K3 surfaces (Xi, cpi), with i G {2, A^},
such that
(1) <Pi,c(<CuXi) = ^dv^cOCWJ);
(2) ^(NS(Xi)) = S^n;
(3) <Pi(T Xi ) = Sl n .
Obviuosly, the surfaces Xi are FM-partners of X\.
Now we show that, when i j,
First of all we know that, obviously, dj ^ di (mod n) if i ^ j. On the other hand,
didj < d 2 N 4 < n,
so
1 ^ didj (mod n).
FM-PARTNERS OF K3 SURFACES WITH PICARD NUMBERS 1 AND 2
9
Hence, by point (iii) of Lemma, l3~2l the lattices can not be isometric. The K3 surfaces X\,... ,Xn
are obviously elliptic and the discriminant of their Neron-Severi group is a square. Moreover X\
has a polarization of degree 2d.
This shows that it is possible to find N K3 surfaces which satisfy the hypotheses of the theorem.
□
The previous theorem gives a new proof of the following result due to Oguiso (|13J).
Corollary 3.4. [13, Theorem 1.7]. Let N be a natural number. Then there are N K3 surfaces
X\,. . . , Xn with Picard number p = 2 such that
(i) NSpQ) £ NS(X,-) ifi + j, where i,j G {1,. . .,N};
(ii) there is an Hodge isometry between (Tx^CuiXi) and (Tx^CuJXj), for all i,j G {1,. . . ,N}.
Remark 3.1. The proof proposed by Oguiso in ^H] is based on deep results in number theory.
In particular, it uses a result of Iwaniec [Jj about the existence of infinitely many integers of type
4n 2 + 1 which are product of two not necessarily distinct primes. Theorem 13.31 gives an elementary
proof of Theorem 1.7 in [13] entirely based on simple remarks about lattices and quadratic forms.
Lemma l3~Tl is true also when L = U @U @JJ ' . The period map is onto also for abelian surfaces
(see [16J). Thus, using the lattices S^ n described before, the following proposition (similar to a
result given in |6J) can be proved with the same techniques.
Proposition 3.5. Let N and d be positive integers. Then there are N abelian surfaces X\,. . . ,Xn
with Picard number p = 2 such that
(i) NS(JQ) $ NSpT,-) ifiji j, with i,j G {1,. . .,N};
(ii) there is i G {1, . . . , N} such that Xj has a polarization of degree 2d;
(iii) there is an Hodge isometry between (Tx^CuxJ and (Tx^CuXj), for all i,j G {1,. . . ,N}.
The following easy remark shows that it is possible to obtain an arbitrarily large number of
M-polarizations on a K3 surface, for certain M.
Remark 3.2. Let iV be a natural number. Then there are a primitive sublattice M of A with
signature (1,0) and a K3 surface X with p(X) = 2 such that X has at least N non-isomorphic
M-polarizations. In particular X has at least iV non-isomorphic M-polarized FM-partners.
In fact, let S = U, where U is, as usual, the hyperbolic lattice. Then, by the surjectivity of the
period map, there is a K3 surface X such that NS(X) = S.
Let d be a natural number with d = p^ 1 . . .p e ™ . In S there are 2 P ^~ 1 primitive vectors with
autointersection 2d. Indeed they are all the vectors of type
Jj — {P h ,P js+1 ■■■P jn ),
for / and J that vary in all possible partitions III J = {1, . . . , re}.
The group 0(U) has only four elements (i.e. ±id, the exchange of the vectors of the base and
the composition of this map with —id). So it is easy to verify that all these polarizations are not
isomorphic. Choosing d to be divisible by a sufficiently large number of distinct primes, we can
find at least iV non-isomorphic M-polarizations. The last assertion follows from Lemma l3~Tl
Acknowledgements. The author would like to express his thanks to Professor Bert van Geemen for his
suggestions and helpful discussions.
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10
PAOLO STELLARI
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Dipartimento di Matematica "F. Enriques", Universita degli Studi di Milano, Via Cesare Saldini
50, 20133 Milano, Italy
E-mail address: [email protected]