Maria Valtorta

6
THE GEOMETRY OF K3 SURFACES
LECTURES DELIVERED AT THE
SCUOLA MATEMATICA INTERUNIVERSITARIA
CORTONA, ITALY
JULY 31|AUGUST 27, 1988
DAVID R. MORRISON
1. Introduction
This is a course about K3 surfaces and several related topics. I want to begin by
working through an example which will illustrate some of the techniques and results
we will encounter during the course. So consider the following problem.
3Problem. Find an example of C XP ,where C is a smooth curve of genus 3
and degree 8 andX is a smooth surface of degree 4.
Of course, smooth surfaces of degree 4 are one type of K3 surface. (For those who
don’t know, a K3 surface is a (smooth) surfaceX which is simply connected and has
trivial canonical bundle. Such surfaces satisfy (O ) = 1, and for every divisor DX
on X, DD is an even integer.)
We rst try a very straightforward approach to this problem. Let C be any smooth
curve of genus 3, and let Z be any divisor on C of degree 8. (For example, we may
takeZ to be the sum of any 8 points onC.) I claim that the linear systemjZj de nes
an embedding of C. This follows from a more general fact, which I hope you have
seen before.
Theorem. LetC be a smooth curve of genusg, and letZ be a divisor onC of degree
d> 2g. Then the linear systemjZj is base-point-free, and de nes an embedding of
C.
Proof. First suppose that P is a base point of Z. We have an exact sequence
0!O (Z−P)!O (Z)!O (Z)! 0C C P
0 0and the assumption that P is a base point implies H (O (Z−P)) = H (O (Z)).C C
But deg(Z−P)=d−1> 2g−2 and deg(Z)=d> 2g−2 so that both of these divisors
0Z and Z−P are non-special. By Riemann-Roch, h (O (Z−P)) =d− 1−g +1 =C
0d−g +1 =h (O (Z)), a contradiction.C
1P
P
P
P

6
P
2 DAVID R. MORRISON
Similarly, suppose that P and Q are not separated by the linear systemjZj.(We
include the case P =Q, where we suppose that the maximal ideal of C at P is not
embedded.) Then in the long exact cohomology sequence associated to
0!O (Z−P−Q)!O (Z)!O (Z)! 0C C P+Q
0 0we must have that the map H (O (Z))! H (O (Z)) is not surjective. ThisC P+Q
1implies that H (O (Z−P−Q))=(0). ButC
1 0 H (O (Z−P−Q)) H (O (K −Z +P +Q))=C C C
and deg(K −Z +P +Q)= 2g− 2−d +2 < 0 so this divisor cannot be eectiveC
(again a contradiction). Q.E.D.
I have included this proof because later we will study the question: on a surface,
when does a linear system have base points, and when does it give an embedding?
To return to our problem, we have a curveC of genus 3 and a divisorZ of degree
8. This divisor is non-special and gives an embedding (since 8> 2 3). By Riemann-
Roch,
0h (O (Z)) = 8−3+1 = 6C
5sojZj maps C intoP .
3The theory of generic projections guarantees that we can project C intoP from
5 3P in such a way as to still embedC. So we assume from now on: CP is a smooth
curve of degree 8 and genus 3. (The linear systemjO (1)j is not complete.)C
0 0We now consider the map H (O 3(k))! H (O (k)) for various degrees k.WeC
have
(k +1)(k +2)(k +3)0h (O 3(k)) =
6
0h (O (k)) = 8k−3+1 = 8k− 2C
(by Riemann-Roch).
0 0It is easy to see that for k 3we have h (O 3(k))<h (O (k)) while for k 4C
0 0we have h (O 3(k))>h (O (k)). This means that for k 3, the linear system cutC
out on C by hypersurfaces of degree k is incomplete, while for k 4there must be
a hypersurface of degree k containing C.Infact,for k =4 we have
0
3h (O (4)) = 35
0h (O (4)) = 30C
4so that there is at least aP of quartic hypersurfaces X containing C. By Bertini’s
4theorem, the generic X is smooth away from C (the base locus of thisP ). So we
3have almost solved our problem: we have constructed CXP with C smooth,
but X is only known to be smooth away from C.
Where do we go from here? We could continue to pursue these methods of projec-
3 4tive geometry inP . For example, we might consider a generic pencil inside ourP

6
THE GEOMETRY OF K3 SURFACES 3
0 0of quartics: the base locus of this pencil is C[C with C another curve of degree
08. (C = C since the arithmetic genus would be wrong). We could try varying the
0pencil and showing that the induced family of divisorsC\C onC has no base point.
But the arguments are very intricate, and I’m not sure if they work! So we will try
a dierent approach.
In this second approach, we construct X rst instead of C. Whatwe wantis
aK3 surface X together with two curves H and C on X.(H is the hyperplane
3section from the embeddingXP ). We want C to have genus 3,jHj to de ne an
3embeddingXP ,andC should have degree 8 under this embedding. To translate
these properties into numerical properties of H and C on X, notice that for any
curveDX we have
deg(K )=deg((K +D)j )= deg(Dj )=DDD X D D
(using the adjunction formula, and the fact that K is trivial). Thus, g(D)=X
1(DD)+1.2
The numerical versions of our properties are: C C =4 (so that g(C)= 3),
3HH =4 (so thatwe get a quartic XP )and HC =8 (so that C will have
3degree 8 inP ). In addition, we wantC to be smooth andjHj to de ne an embedding
3inP .
The rst step is to check that the topology of a K3 surface permits curves with these
numerical properties to exist. The topological properties I have in mind concern the
2intersection pairing onH (X;Z). Each curveD on X has a cohomology class [D]2
2H (X;Z), and the intersection number for curves coincides with the cup product
pairing
2 2 4 H (X;Z)⊗H (X;Z)!H (X;Z) Z:=
Poincar e duality guarantees that this is a unimodular pairing (i. e. in a basis, the
matrix for the pairing has determinant1). In addition, the signature of the pairing
(the number of +1 and−1 eigenvalues) can be computed as (3,19) for a K3 surface.
2The pairing is also even: this means xx2 2Z for all x2 H (X;Z). These three
properties together imply that the isomorphism type of this bilinear form overZ is
2 3(−E ) U whereE is the unimodular even positive de nite form of rank 8, and8 8
U is the hyperbolic plane. Let denote this bilinear form; is called the K3 lattice.
We will return to study these topological properties in more detail (and de ne the
terms!) in section 11.
2 In our example, we need a submodule of H (X;Z) generated by 2 elements h=
and c such that the matrix of the pairing on these 2 elements is
!
48
:
84
The fact that such a submodule exists is a consequence of



4 DAVID R. MORRISON
Theorem (James [12]). Given an even symmetric bilinear formL over the integers
such that L has signature (1;r− 1) and the rank r of L is 10, there exists a
submodule of isomorphic to L.
1(Later we will study re nements of this theorem in which the rank is allowed to
be larger.)
2So there is no topological obstruction in our case. Moreover, later in the course
we will partially prove the following:
Fact. Given a submodule L of on which the form has signature (1 ;r− 1) with
r 20, there exists a (20−r)-dimensional family of K3 surfacesfXg,eachequippedt
2 with an isomorphism H (X;Z) in such a way that elements of L correspond=t
to cohomology classes of line bundles on X.Moreover,for t generic, these are thet
only line bundles on X,that is, the Neron-Severi group NS(X ) [together with itst t
intersection form] is isomorphic toL.
I said we would partially prove this: what we will not prove (for lack of time) is the
global Torelli theorem and the surjectivity of the period map for K3 surfaces. This
3fact depends on those theorems.
To return to our problem: we now have a K3 surface X and two line bundles
O (H),O (C) with the correct numerical properties, which generate NS(X). ForX X
2 0 any line bundleL onX,wehaveH (L) H (L ) (sinceK is trivial), from which= X
follows the Riemann-Roch inequality:
L L0 0 h (L)+h (L ) +2
2
(since (O )=2). Thus, ifL L− 2eitherL orL is e ective.X
In our situation we conclude:H is e ective and C is e ective. Replacing H by
−H if necessary we may assume H is e ective. (The choice of C is then determined
by HC =8.)
4To nish our construction, we need another fact which will be proved later.
2Fact. Let H be an e ective divisor on a K3 surface X with H 4. IfjHj has
base points or does not de ne an embedding, then there is a curve E on X with
2 2 1E =−2or E = 0. Moreover, when it does de ne an embedding H (O (H)) = 0X
2and H (O (H)) = 0.X
5In our situation, we wish to rule out the existence of such an E.Wehave E
2 2 2mH +nC sinceH andC generate NS(X). Thus, E =4m +16mn+4n is always
1Need cross-reference.
2Need cross-reference.
3For statements of the theorems and further discussion, see section 12.
4Need cross-reference: section 5?
5The symbol denotes linear equivalence.


THE GEOMETRY OF K3 SURFACES 5
2 abdivisible by 4, soE =−2 is impossible. Moreover, if a rank 2 quadratic form ( )bc
2ab 48represents 0 then− det ( )= b −ac is a square.Inourcase,− det ( )= 48 isbc 84
2not asquare, so E = 0 is impossible.
We conclude thatjHj is very ample. Then HC = 8 implies−C is not e ective,
so that C must be e ective. Furthermore, jCj is then very ample, so this linear
system contains a smooth curve (which we denote again by C). By Riemann-Roch,
0 HH 3h (O (H)) = +2 = 4, sojHj maps X intoP as a smooth quartic surface, andX 2
3we have CXP as desired.
2. K3 surfaces and Fano threefolds
We will use in this course a de nition of K3 surfaces