A NEW FAMILY OF SURFACES WITH p g =0AND K 2 =3

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1 Ann. Scient. Éc. Norm. Sup., 4 e série, t. 37, 2004, p. 507 à 531. A NEW FAMILY OF SURFACES WITH p g =0AND K 2 =3 BY MARGARIDA MENDES LOPES AND RITA PARDINI ABSTRACT. LetS be a minimal complex surface of general type with p g =0such that the bicanonical map ϕ of S is not birational and let Z be the bicanonical image. In [M. Mendes Lopes, R. Pardini, Math. Z. 241 (4) (2002) ] it is shown that either: (i) Z is a rational surface, or (ii) K 2 S =3,the map ϕ is a degree two morphism and Z is birational to an Enriques surface. Up to now no example of case (ii) was known. Here an explicit construction of all such surfaces is given. Furthermore it is shown that the corresponding subset of the moduli space of surfaces of general type is irreducible and uniruled of dimension Elsevier SAS RÉSUMÉ. SoitS une surface de type général avec p g =0dont l application bicanonique ϕ of S n est pas birationnelle, et soit Z l image bicanonique. Dans un précédent article, nous avons montré que soit Z est une surface rationnelle, soit K 2 S =3, ϕ est un morphisme de degré 2 et Z est (birationnellement) une surface d Enriques. On ne savait pas si cette seconde possibilité pouvait être réalisée. Ici nous donnons une construction explicite de toutes ces surfaces, et nous montrons aussi que la partie correspondante de l espace des modules des surfaces de type général est irréductible et unirationnelle de dimension Elsevier SAS 1. Introduction The knowledge of surfaces of general type with p g =0continues to be scarce in spite of much progress in surface theory. A minimal surface of general type with p g =0satisfies 1 K 2 9 and examples for all possible values for K 2 are known (see, e.g., [1, Chapter VII, 11]). In recent years we have undertaken to study surfaces with p g =0 by looking at their bicanonical map. When the bicanonical map is generically finite but not birational, this approach works and it has allowed us to obtain classification results and also, in some cases, information on the moduli space (see for instance [16,18,23]). The first step in describing a class of surfaces with nonbirational bicanonical map is to analyze the bicanonical image. If S is a minimal surface with p g =0, then the bicanonical image Z of S is contained in P K2, thus, in particular, for K 2 =1it is equal to P 1.ForK 2 2, Z is always a surface [27] and, if the bicanonical map ϕ is not birational, then Z is either rational or it is an Enriques surface [28]. In [17], it has been proven that the latter case can occur only if K 2 =3 and ϕ is a morphism of degree 2. No such example appears in the literature. Indeed, the known examples of minimal surfaces of general type with p g =0and K 2 =3 are the examples of Burniat and Inoue ([3,11], see also [24]), the examples due independently to J.H. Keum and D. Naie [12,21] and the recent examples due to F. Catanese [5]. The degree /04/ 2004 Elsevier SAS. All rights reserved

2 508 M. MENDES LOPES AND R. PARDINI of the bicanonical map of all these surfaces is equal to 4 (cf. [17,16]), although the Keum Naie examples are in fact double covers of nodal Enriques surfaces and their bicanonical map factorizes through the covering map. In this paper we prove that surfaces with p g =0whose bicanonical image is an Enriques surface do exist. More precisely, we give an explicit construction of all such surfaces and we prove that the corresponding subset E of the moduli space of surfaces of general type is irreducible and uniruled of dimension 6. Since the closure of E contains the Keum Naie surfaces (see Proposition 7.3), it follows that the fundamental group of our surfaces is isomorphic to Z 2 2 Z 4 (see Corollary 7.2). There is an analogy between the surfaces we study here and the so-called Todorov surfaces. These are surfaces of general type with p g =1and q =0that are mapped 2 to 1 onto a K3 surface by the bicanonical map [26,19], and they are the only surfaces whose bicanonical image is a K3 surface. This analogy is accounted for by the fact that the surfaces that we construct can also be obtained as free Z 2 -quotients of some Todorov surfaces with K 2 =6. We remark that it is known that, for fixed K 2, the Todorov surfaces form an irreducible subset of the moduli space [19]. For K 2 =2, in [6] the authors have described very precisely the irreducible component X of the moduli space containing the Todorov surfaces, showing in particular that the bicanonical map is birational for a general surface of X. A natural question, that at the moment we are not able to answer (see Remark 7.4), is whether the closure Ē of the subset of the moduli space consisting of the surfaces studied here is an irreducible component. If the answer were affirmative, then one would have a similar phenomenon to the case of surfaces with p g =0, K 2 =6and bicanonical map of degree 4, which form a component of the moduli space [16]. On the other hand, if the answer were negative, then it would follow that Todorov surfaces with K 2 =6behave similarly to those with K 2 =2. Our description of surfaces whose bicanonical image Z is an Enriques surface is based on a very detailed study of the normalization of Z. This is a polarized Enriques surface of degree 6 with 7 nodes, satisfying some additional conditions (see Proposition 2.1 and the setting of 3). The analysis and construction of these Enriques surfaces form the bulk of this paper. The main tools we use are the classification of linear systems on an Enriques surface, the analysis of the configuration of singular fibres of certain elliptic pencils, the code associated to the nodes of the surface and the corresponding Galois cover (cf. [9]). The plan of the paper is as follows: Section 2 explains the relation between the surfaces we are studying and a certain class of polarized Enriques surfaces (Σ,B) with 7 nodes; in Section 3 some properties of these Enriques surfaces are established and some examples are described; in Section 4 we make a very detailed study of the singular fibres of the elliptic pencils of Σ and we determine the code associated to the nodes of Σ; in Section 5 we describe a construction yielding pairs (Σ,B) and prove that all such pairs are obtained in that way; in Section 6 we introduce and study a quasi-projective variety parametrizing the isomorphism classes of pairs (Σ,B) and finally in Section 7 we apply the previous results to describe the family of surfaces whose bicanonical image is an Enriques surface. Notation and conventions: We work over the complex numbers. A node of a surface is an ordinary double point, namely a singularity analytically isomorphic to x 2 + y 2 + z 2 =0.The exceptional divisor of the minimal resolution of a node is a curve C P 1 such that C 2 = 2. A curve with these properties is called 2 curve, or nodal curve. We say that a projective surface Σ with canonical singularities is minimal, of general type, Enriques...iftheminimalresolutionofΣ is minimal, of general type, Enriques...Ourstandard reference for Enriques surfaces is [8], and we use freely its terminology. 4 e SÉRIE TOME N 4

3 A NEW FAMILY OF SURFACES WITH p g =0AND K 2 =3 509 Given an automorphism σ of a variety X, we say that a map f : X Y is composed with σ if f σ = f. IfG is a finite group, a G-cover is a finite map f : X Y of normal varieties together with a faithful G-action on X such that f is isomorphic to the quotient map X X/G. If G = Z 2, then we say that f is a double cover. Contrary to what is often done (cf. for instance [22]), we do not require that f be flat. We denote linear equivalence by and numerical equivalence by num. The group of line bundles modulo numerical equivalence on a variety Y is denoted by Num(Y ). 2. Surfaces with p g =0and K 2 =3and Enriques surfaces As explained in the introduction, the bulk of this paper is a very detailed study of a class of polarized Enriques surfaces with 7 nodes. In this section we explain the relation between such Enriques surfaces and a class of minimal surfaces of general type with nonbirational bicanonical map. Let S be a minimal surface of general type with p g (S) =0and KS 2 =3. We denote by ϕ : S P 3 the bicanonical map of S and we assume that S has an involution σ such that: (a) ϕ is composed with σ; (b) the quotient surface T := S/σ is birational to an Enriques surface. We denote by X the canonical model of S. Abusing notation, we denote by the same letter the involution induced by σ on X. PROPOSITION 2.1. In the above setting: (i) the quotient surface Σ:=X/σ is an Enriques surface with 7 nodes; (ii) the quotient map π : X Σ is branched on the nodes of Σ and on a divisor B with negligible singularities, contained in the smooth part of Σ; (iii) B is ample and B 2 =6; (iv) the bicanonical system 2K X = π B is base point free; (v) the degree of the bicanonical map of S (and X) is either 2 or 4. Proof. Since the bicanonical map ϕ of S factorizes through σ, by [18, Proposition 2.1] the isolated fixed points of σ are 7. The quotient surface T := S/σ has 7 nodes, which are the images of the isolated fixed points of σ. The quotient map S T is branched on the nodes and on a smooth divisor B 0 contained in the smooth part of T. By Lemma 7 of [28] and the remark following it, there exists a birational morphism r : T Σ,whereΣ is an Enriques surface with 7 nodes, such that the exceptional curves of r are contained in the smooth part of T and the divisor B := r(b 0 ) has negligible singularities. Let S X π Σ be the Stein factorization of the induced map S Σ.Themapπ : X Σ is a double cover branched on the nodes of Σ and on the divisor B. The singularities of X occur above the singularities of B, hence they are canonical and there is a birational morphism from X to the canonical model X of S. More precisely, there is a commutative diagram: X π Xπ Σ Σ where the horizontal arrows represent birational morphisms and π, π are the quotient maps for the involutions induced by σ on X and X. By adjunction we have 2K X = π (2K Σ + B)=π B, hence B is nef and B 2 =6.Since the bicanonical map of X factorizes through π, we actually have 2K X = π B. Thesame

4 510 M. MENDES LOPES AND R. PARDINI argument as in the proof of [17, Theorem 5.1] shows that the system B is free. Thus (Σ,B) is a pair as in the setting of 3 and we can apply Corollary 3.3, which is proven in 3, showing that B is ample. It follows that the horizontal maps in the above diagram are isomorphisms and we can identify X with X and Σ with Σ. The system B, being free, is not hyperelliptic, therefore by [7, Proposition 5.2.1] either it is birational or it has degree 2. Hence the degree of the bicanonical map of X and S has degree either 2 or 4. This completes the proof. The previous proposition has a converse: PROPOSITION 2.2. Let Σ be an Enriques surface with 7 nodes and let π : X Σ be a double cover branched on the nodes of Σ and on a divisor B such that: (a) B is ample and B 2 =6; (b) B is contained in the smooth part of Σ and it has negligible singularities. Then X is the canonical model of a minimal surface S of general type with p g (S)=0and KS 2 =3and the bicanonical map of S factorizes through the map S Σ. Condition (a) can be replaced by: (a ) B is free and B 2 =6. Proof. Assume that conditions (a) and (b) are satisfied. The singular points of X lie above the singularities of B.SinceB has negligible singularities, the singularities of X are canonical and one has 2K X = π B.SinceB is ample, K X is also ample and X is the canonical model of a surface S of general type. One has K 2 S = K 2 X = 1 2 B2 =3. To compute the birational invariants χ(s) and p g (S), we consider the minimal resolution of singularities η : Y Σ and the flat double cover π : X Y obtained from π by taking base change with η. The surface X has canonical singularities and it is birational to X and S. The branch locus of π consists of the inverse image B of B and of the 2-curves N 1,...,N 7 that are exceptional for η.so π is given by a relation 2L B +N 1 + +N 7,whereL is a line bundle on Y. A standard computation gives χ(s)=χ( X)=χ(Y )+χ(l 1 )=1. By Kawamata Viehweg vanishing one has h i (K Y + L)=0for i>0, hence p g (S)=p g ( X)=h 0 (K Y + L)=χ(K Y + L)=0. Since the singularities of the branch divisor B of π are negligible, it is well known that the smooth minimal model S of X can be obtained by repeatedly blowing up Σ at the singular points of B and taking base change and normalization. Hence the involution of S induced by σ has the same number of isolated base points as σ, that is 7, and the bicanonical map factorizes through it by [16, Proposition 2.1]. If (a ) holds, then B is ample by Corollary 3.3, and so condition (a ) implies condition (a). 3. Enriques surfaces with 7 nodes: examples Recall that for Enriques surfaces we adopt the notation and the terminology of [8]. Our notation for the singular fibres of an elliptic pencil is the same as in [1], Chapter V, 7. In this section we consider the following situation: Set-up: Σ is a nodal Enriques surface with 7 nodes, B is a base point free linear system of Σ such that B 2 =6. We denote by η : Y Σ the minimal desingularization and by N 1,...,N 7 the 4 e SÉRIE TOME N 4

5 A NEW FAMILY OF SURFACES WITH p g =0AND K 2 =3 511 disjoint nodal curves contracted by η, andweset B := η B. Furthermore, we assume that there exists L Pic(Y ) such that B + N N 7 =2L. Remark 3.1. The choice of this set-up is suggested by the results of the previous section. Indeed, the condition that the class B + N N 7 be divisible by 2 in Pic(Y ) means that, given a curve B B, there exists a double cover π : X Y branched on the union of B and N 1,...,N 7.If B is disjoint from N 1,...,N 7 (recall that BN i =0) and it has at most negligible singularities, then the surface X has canonical singularities, occurring above the singularities of B. Fori =1,...,7, one has π N i =2C i,wherec i is a 1-curve contained in the smooth part of X. If we denote by X the surface obtained by contracting the C i,then π induces a double cover π : X Σ branched over the image B of B and over the nodes of Σ. By Proposition 2.2, X is the canonical model of a minimal surface of general type with p g =0and K 2 =3and the bicanonical map of X factorizes through π. Notice that, since Y is an Enriques surface, the line bundle L + K Y also satisfies the relation 2(L + K Y ) B + N N 7, so that a pair (Σ,B) as in the set-up determines two nonisomorphic double covers of Σ with the same branch locus. In order to describe some examples, we need to prove first some general facts. PROPOSITION 3.2. There exist three elliptic half-pencils Ẽ1, Ẽ2, Ẽ3 on Y such that: (i) Ẽ i Ẽ j =1for i j; (ii) B = Ẽ1 + Ẽ2 + Ẽ3. Proof. Notice that the system B, being base point free, is not hyperelliptic. Hence, if B is not as stated, then by Proposition and Theorem of [7] there are the following possibilities: (1) B = 2Ẽ0 + Ẽ1 + θ 2, Ẽ0Ẽ1 = Ẽ0θ 2 =1, Ẽ1θ 2 =0; (2) B = 3Ẽ0 +2θ 0 + θ 1, Ẽ0θ 0 = θ 0 θ 1 =1,Ẽ0θ 1 =0, where Ẽm are elliptic half-pencils and θ m are nodal curves, m =0, 1, 2. Consider the nodal curves N 1,...,N 7 and recall that B + N N 7 is divisible by 2 in Pic(Y ). In case (1), suppose first that θ 2 is not one of the curves N i.sincen i B =0, necessarily the curves θ 2,N 1,...,N 7 are disjoint and so by Lemma 4.2 of [17] the divisor θ 2 + N N 7 is divisible by 2 in Pic(Y ). Hence also Ẽ1 is divisible by 2 in Pic(Y ), a contradiction. So θ 2 is one of the curves N i,sayθ 2 = N 7. Then there exists L Pic(Y ) such that D = Ẽ 1 + N N 6 =2L and we get D 2 = 12 and L 2 = 3, contradicting the fact that the intersection form on an Enriques surface is even. Consider now case (2). As in case (1), if neither of the curves θ 0,θ 1 is one of the nodal curves N 1,...,N 7, we conclude that the half-pencil Ẽ0 is divisible by 2 in Pic(Y ), a contradiction. If θ 1 is one of the curves N i, we also arrive at a contradiction as in case (1). Finally suppose that θ 0 is one of the curves N i,sayn 7.ThenẼ0(N N 7 )=1.SinceẼ0 B =2, B + N N 7 is not divisible by 2 in Pic(Y ), a contradiction. So B is as stated. COROLLARY 3.3. The divisor B is ample on Σ. Proof. Denote by the set of irreducible curves C of Y with R B BC =0. By [8, Corollary 4.1.1], is a finite set and the corresponding classes are independent in Num(Y ). R B By Proposition 3.2, we can write B = Ẽ1 + Ẽ2 + Ẽ3,wheretheẼi are elliptic half-pencils. If we denote by V the subspace of Num(Y ) Q spanned by the classes of Ẽ1, Ẽ2, Ẽ3, then the classes of the curves of belong to V R B, which has dimension 7. So we have = {N R B 1,...,N 7 } and B = η B is ample on Σ.

6 512 M. MENDES LOPES AND R. PARDINI The following is a partial converse to Proposition 3.2: LEMMA 3.4. Let Y be a smooth Enriques surface containing 7 disjoint nodal curves N 1,...,N 7. Assume that Ẽ1, Ẽ2, Ẽ3 are elliptic half pencils on Y such that ẼiẼj =1if i j and ẼiN j =0for every i, j.ifweset B := Ẽ1 + Ẽ2 + Ẽ3, then B + N N 7 is divisible by 2 in Num(Y ). Proof. Recall that Num(Y ) is an even unimodular lattice of rank 10. Let M be the sublattice of Num(Y ) spanned by the classes of Ẽ1 Ẽ2 and Ẽ1 Ẽ3. The discriminant of M is equal to 3, hence M is primitive. Let M be the sublattice spanned by the classes of the N i and by the class of B. The primitive closure of M is M and the code W associated to the set of classes B,N 1,...,N 7 is naturally isomorphic to the quotient group M /M. Computing discriminants one gets: 2 8 3=disc(M )=disc(m )2 2dimW =disc(m) 2 2dimW =3 2 2dimW, namely dim W =4. Using the fact that the intersection form on Y is even, it is easy to check that the elements of W have weight divisible by 4. Since W has length 8, this implies that W is the extended Hamming code (see, e.g., [13]), and in particular it contains the vector of weight 8, i.e., B + N N 7 is divisible by 2 in Num(Y ). For i =1, 2, 3, we denote by Ẽ i the unique effective divisor in Ẽi + K Y and we write F i = 2Ẽi = 2Ẽ i. Thus F i is an elliptic pencil with double fibres 2Ẽi and 2Ẽ i.the classes Ẽi are nef, hence BN j =0implies ẼiN j =0for every i, j and therefore for i =1, 2, 3 F i induces an elliptic pencil F i = 2E i = 2E i on Σ, wheree i = η Ẽ i, E i = η Ẽ i and B = E 1 + E 2 + E 3. Example 3.5. This example appears in [21] and in an unpublished paper by J. Keum [12]. One considers an Enriques surface Σ with 8 nodes as in Example 1 of [17]. The surface Σ has two isotrivial elliptic pencils F 1 and F 2 with F 1 F 2 =4. The system F 1 + F 2 gives a degree 2 morphism onto a Del Pezzo quartic in P 4 such that the nodes of Σ are mapped to smooth points (cf. [21, 2]). We take Σ to be the surface obtained by resolving one of the nodes of Σ and we denote by C the corresponding nodal curve. We denote by the same letter the pull-backs of F 1, F 2 on Σ and we set B := F 1 + F 2 C. By the above discussion, the system B is free and it gives a degree 2 map onto a Del Pezzo cubic in P 3. By [17, Lemma 4.2], the class of N N 7 + C is divisible by 2 in Pic(Y ), hence the class of B + N N 7 is also divisible by 2. Let 2E i be a double fibre of F i, i =1, 2. By Riemann Roch there exists an effective divisor E 3 E 1 + E 2 C. WehaveE 3 2 =0and E 1 E 3 = E 2 E 3 =1. We claim that E 3 is an elliptic half-pencil, so that B E 1 + E 2 + E 3 as predicted by Proposition 3.2. We now work on the nonsingular surface Y and, as usual, we denote by D the pull-back on Y of a divisor D of Σ. LetG i F i, i =1, 2, be the fibre containing C. By the description of Σ given in [17], G 1 and G 2 are fibres of type I0 and the divisor G 3 := G 1 + G 2 2C is an elliptic configuration of type I2. It follows that G 3 is an elliptic pencil and that 2Ẽ3 is a double fibre of G 3. Example 3.6. Let C P 3 be the Cayley cubic, defined by x 1 x 2 x 3 + x 0 x 2 x 3 + x 0 x 1 x 3 + x 0 x 1 x 2 =0. The singularities of C are 4 nodes, that occur at the coordinate points and form an even set. The 6 lines joining the nodes are of course contained in C. We label these lines by e 1,e 1,e 2,e 2,e 3,e 3 in such a way that, for i =1, 2, 3, e i,e i is a pair of skew lines and e 1,e 2,e 3 are coplanar. The surface C contains 3 more lines l 1,l 2,l 3, contained in the plane x 0 + x 1 + x 2 + x 3 =0. An elementary geometric argument shows that, up to a permutation 4 e SÉRIE TOME N 4

7 A NEW FAMILY OF SURFACES WITH p g =0AND K 2 =3 513 N 6 N 1 N 5 A 3 A 2 N 2 A 1 N 3 N 4 Fig. 1. of the indices, we may assume that the line l i meets e i,e i and it does not meet e j,e j for i j. For i =1, 2, 3 we denote by f i the moving part of the linear system cut out on C by the planes containing the line l i. The general f i is a smooth conic, hence one has K C f i = 2, fi 2 =0and f i f j =2for i j. The singular fibres of f i are 2e i, 2e i and l j +l k,wherei, j, k is a permutation of 1, 2, 3. Consider a curve D O C (2) such that D is contained in the smooth part of C and it has at most simple singularities. Let Σ be the double cover of C branched on D and on the four nodes of C. The surface Σ has canonical singularities, occurring over the singular points of D. Standard computations (cf. the proof of Proposition 2.2) show that Σ is an Enriques surface and that the pull-back of the system of hyperplanes of P 3 is a complete system B on Σ with B 2 =6.For i =1, 2, 3, we consider on Σ the system F i obtained by pulling-back f i. The system F i is an elliptic pencil, with double fibres 2E i and 2E i,wheree i, E i are the pull-backs of e i,ande i, respectively. For i j one has E i E j =1.Furthermore,B E 1 + E 2 + E 3. We consider now a special case of the above construction: we take D to be the union of the section H 0 of C with the plane x 0 + x 1 + x 2 + x 3 =0and of the section H with a general hyperplane tangent to C.SoH has an ordinary double point at the tangency point and is smooth elsewhere and H and H 0 intersect transversely at 3 points. The surface Σ thus obtained has 7 nodes, occurring above the singularities of H 0 + H. As usual we denote by η : Y Σ the minimal resolution and by N 1,...,N 7 the exceptional curves of η. The strict transform on Y of the line l i is a nodal curve A i. It is not difficult to check that one can relabel the curves N 1,...,N 7 in such a way that N 7 corresponds to the singularity of Σ above the double point of H and the incidence relations of the set of curves A 1,A 2,A 3,N 1,...,N 6 are as shown in the dual graph in Fig. 1. As usual we denote by Ẽi, Ẽ i, Fi, B the pull-backs on Y of E i, E i, F i, B. The singular fibre G i of F i corresponding to the fibre l j + l k of f i is of type I2. More precisely, we have G 1 = N 2 + N 3 + N 5 + N 6 +2(A 3 + A 2 + N 1 ), G 2 = N 1 + N 3 + N 4 + N 6 +2(A 3 + A 1 + N 2 ), G 3 = N 1 + N 2 + N 4 + N 5 +2(A 1 + A 2 + N 3 ).

8 514 M. MENDES LOPES AND R. PARDINI By Lemma 3.4, one of the classes B + N N 7 and B + K Y + N N 7 is divisible by 2 in Num(Y ). We will show later (Corollary 4.5) that the second case actually occurs. Hence we set B := B + K Σ. Examples 3.5 and 3.6 share the common feature that either the system B or the system B + K Σ is not birational. The next example shows that this does not happen in general. Example 3.7. By deforming Example 1 we show the existence of a pair (Σ,B) such that both B and B + K Σ are birational. We start with a pair (Σ 0,B 0 ) as in Example 3.5. The Kuranishi family p : Y U of the minimal resolution Y 0 of Σ 0 is smooth of dimension 10 by [1, Theorem VIII.19.3]. We may assume that U is contractible and that the family Y is differentiablytrivial. Hence for every fibre Y t := p 1 (t) the inclusion Y t Yinduces an isomorphism H 2 (Y t, Z) H 2 (Y, Z).TheLeray spectral sequence gives h 1 (Y, O Y )=h 2 (Y, O Y )=0, hence by the exponential sequence every integral cohomology class of Y comes from a unique holomorphic line bundle. In particular, there exist line bundles Ẽ1, Ẽ2, Ẽ3, N 1,...,N 7, C that restrict on the central fibre Y 0 to Ẽ1, Ẽ2, Ẽ3, N 1,...,N 7,C, respectively. We set B := Ẽ1 + Ẽ2 + Ẽ3 and for t U we denote by B t, Ẽi,t, N i,t, C t the restrictions to Y t of the above bundles. Obviously, the class of B t + N 1,t + + N 7,t is divisible by 2 in H 2 (Y t, Z) for every t U. By [4, Theorem 3.7], the subset U 1 of U where the classes N i,t are effective and irreducible is smooth of dimension 3, while the subset U 2 of U 1 where also C t is effective is smooth of dimension 2. SinceC t Ẽ1,t + Ẽ2,t Ẽ3,t, by [8, Theorem 4.7.2] the system B t is birational for t U 1 \ U 2. On the other hand, by semicontinuity we may assume that B t + K Yt is birational for every t U, since it is birational on the central fibre Y 0. So the required example can be obtained by taking Y t with t U 1 \ U 2 and by blowing down the nodal curves N 1,t,...,N 7,t. 4. Enriques surfaces with 7 nodes: codes and singular fibres We keep the set-up and the notation of the previous section. Here we make a detailed study of the code associated to the nodal curves N 1,...,N 7 and of the singular fibres of the pencils F i. These results are needed in the following section, where we give a construction of all the pairs (Σ,B) as in the set-up of 3. We denote by V and V num, respectively, the code and the numerical code associated to N 1,...,N 7 (cf. [17, 2], [2]). Namely, V is the kernel of the map Z 7 2 Pic(Y )/2Pic(Y ) that maps (x 1,...,x 7 ) to the class of x 1 N x 7 N 7. The code V num is defined in analogous way, replacing Pic(Y ) by Num(Y ). Clearly, V is a subcode of V num of codimension 1. We say that a divisor D is even if it is divisible by 2 in Pic(Y ). In particular, if D = x i N i then D is even if and only if (x 1,...,x 7 ) V (we denote by the same letter the integer x i and its class in Z 2 ). LEMMA 4.1. dim V num =3. Proof. Since the determinant of the matrix (ẼiẼj) i,j=1,2,3 is equal to 2, the classes Ẽ 1, Ẽ2, Ẽ3 span a primitive sublattice L of rank 3 of Num(Y ).IfL is the sublattice spanned by the classes N 1,...N 7,thenV num is isomorphic to the quotient group L /L.Sowehave namely dim V num = =disc(l )=2 2dimVnum disc(l )=2 2dimVnum disc(l), LEMMA 4.2. The linear system B separatesthe nodes of Σ. 4 e SÉRIE TOME N 4

9 A NEW FAMILY OF SURFACES WITH p g =0AND K 2 =3 515 Proof. This follows by [8, Lemma 4.6.3]. LEMMA 4.3. Denote by P 1,...,P 7 the image points of N 1,...,N 7 via the system B. If N 1 + N 2 + N 3 + N 4 is an even divisor, then P 5,P 6,P 7 are collinear. Proof. Notice first of all that the points P 1,...,P 7 are distinct by Lemma 4.2. Since B + N N 7 is even by assumption, there exists M Pic(Y ) such that 2M B N 5 N 6 N 7.SetM := M + K Y.SinceM 2 = M 2 =0, there exist effective divisors D M and D M. So the linear system B N 5 N 6 N 7 contains two distinct divisors 2D and 2D, hence it has positive dimension. This means that P 5,P 6,P 7 lie on a line. PROPOSITION 4.4. The code V has dimension 2. Proof. By Lemma 4.1, to show that dim V =2it is enough to show that V V num.so assume by contradiction that V = V num.sincedim V =3 and all the elements of V have weight 4, V is isomorphic to the Hamming code (see, e.g., [13]). By the definition of the Hamming code, the set of indices {1,...,7} is in one-to-one correspondence with the nonzero vectors of Z 3 2. The vectors corresponding to distinct indices i 1, i 2, i 3 span a plane of Z 3 2 if and only if there is v =(x 1,...,x 7 ) V \{0} such that x i1 = x i2 = x i3 =0. By Lemma 4.3, this happens if and only if the points P i1,p i2,p i3 lie on a line in P 3. Hence the points P 1,...,P 7 form a configuration isomorphic to the finite plane P 2 (Z 2 ). Since the line through two of the P i contains a third point of the set, it is easy to check that P 1,...,P 7 lie in a plane. On the other hand, it is well known that the plane P 2 (Z 2 ) cannot be embedded in P 2 (C). Sowehavea contradiction and the proof is complete. We are now able to complete the description of Example 3.6: COROLLARY 4.5. Let Σ be the surface of Example 3.6.Then B + K Y + N N 7 is divisible by 2 in Pic(Y ). Proof. In the notation of Example 3.6 we have G 1 =2(A 2 + A 3 + N 1 )+N 2 + N 3 + N 5 + N 6 2Ẽ1, hence N 2 + N 3 + N 5 + N 6 is an even divisor. The same argument shows that the divisors N 1 + N 2 + N 4 + N 5 and N 1 + N 3 + N 4 + N 6 are also even. By Proposition 4.4 these are the only nonzero elements of V. By Lemma 3.4 we know that one of the classes B + N N 7 and B + K Y + N N 7 is even. Assume by contradiction that B + N N 7 is even. Pulling back to Y the section of C with H 0 we get B 2(A 1 + A 2 + A 3 + N 1 + N 2 + N 3 )+N 4 + N 5 + N 6. Hence it follows that N 1 + N 2 + N 3 + N 7 is also an even divisor, a contradiction. The next result describes the possible configurations of singular fibres of the pencils F i and relates them to the properties of the systems B and B + K Σ. THEOREM 4.6. The possible configurations of fibres with singular support of the pencils F i are the following: (1) up to a permutation of the indices, the pencils F 1 and F 2 are isotrivial with 2 fibres of type I 0, while F 3 has a fibre of type I 2 and two fibres of type I 2 or 2 I 2. In this case the system B has degree 2 and the system B + K Σ is birational;

10 516 M. MENDES LOPES AND R. PARDINI (2) each of the pencils F i has a fibre of type I2 and two fibres of type I 2 or 2 I 2. The dual graph of the set of nodal curves that form the I2 fibres is the same as in Fig. 1. In this case B is birational and B + K Σ has degree 2; (3) each of the pencils F i has a fibre of type I0 and three fibres of type I 2 or 2 I 2. In this case the systems B and B + K Σ are both birational. Remark 4.7. The proof of Theorem 4.6 below actually shows more, namely that case (1) of Theorem 4.6 corresponds exactly to Example 3.5 (cf. Lemma 4.10) and that case (2) corresponds exactly to Example 3.6. The proof of Theorem 4.6 is somewhat involved and requires some auxiliary lemmas. LEMMA 4.8. Assume that for i j, the pencils F i and F j on Y have singular fibres of type I0 or I 2, G i =2C i + N i1 + + N i4, respectively, G j =2C j + N j1 + + N j4. Then C i C j =0and the set {i 1,...,i 4 } {j 1,...,j 4 } consists of two elements. Proof. The curve C i is irreducible if G i is of type I0 and it is a chain of 3 nodal curves if G i is of type I2. One has: 4= F i Fj = G i G j =2C i G j, namely C i (2C j + N j1 + + N j4 )=2. We remark that C i N jt is equal to 1 if j t {i 1,...,i 4 }, and it is equal to 0 otherwise. Since there are 7 of the N i, one has C i (N j1 + + N j4 ) > 0. So either we have C i C j =0and {i 1,...,i 4 } {j 1,...,j 4 } consists of two elements, or C i C j = 1 and {i 1,...,i 4 } = {j 1,...,j 4 }. Assume by contradiction that we are in the second case. This implies in particular that C i and C j are not both irreducible. Assume that C i is irreducible. Then C j is a chain of 3 nodal curves C j = A 1 + N + A 2 such that each of the end curves A 1, A 2 meets exactly two of the curves N i1,...,n i4 and the central curve N is one of the N i.infact,ifn were not one of the N i,then the classes of N 1,...,N 7,N,A 1,A 2, being independent, would be a basis of H 2 (Y,Q), against the Index Theorem. Furthermore, from C i C j = 1 it follows that C i is equal to A 1 or A 2, a contradiction, since C i meets all the curves N i1,...,n i4.sowehavec j = A 1 + N + A 2 as above and, with an analogous notation, C i = B 1 + N + B 2,whereN is again one of the N i. Observe that θc i 1 for every irreducible curve θ. SinceN and N are different from N i1,...,n i4, the relations N Fj =0, N F i =0give: N C j = NC i =0and 1=C i C j =(A 1 + A 2 )C i, hence, say, A 1 C i = 1, A 2 C i =0. So we can assume that A 1 = B 1, while A 2 is disjoint from C i and B 2 is disjoint from C j. Say that A 1 = B 1 meets the curves N i1 and N i2. Then the connected divisor =N i1 + N i2 + A 1 + N + N is orthogonal to both F i and F j, so its support is contained in both G i and G j.ifn N, then the intersection form on the components of is semidefinite, hence by Zariski s Lemma is the support of both G i and G j, but this is impossible. Hence N = N, but this contradicts the fact that A 2 and C i are disjoint. LEMMA 4.9. (i) The fibres with reducible support that occur in the pencils F i can be of the following types: I 2, 2 I 2, I0, I2 ; (ii) each pencil F i has at least a fibre of type I0 or I 2 ; (iii) a fibre of type I0 of F i contains 3 or 4 of the N i, each with multiplicity 1, and a fibre of type I2 contains 4 of the N i with multiplicity 1 and one with multiplicity 2. Proof. We recall first of all that the multiple fibres of an elliptic pencil are of type mi k, k 0 ([1], Chapter V, 7) and that the multiple fibres of an elliptic pencil on an Enriques 4 e SÉRIE TOME N 4

11 A NEW FAMILY OF SURFACES WITH p g =0AND K 2 =3 517 surface are precisely two double fibres ([1], Chapter VIII). The nodal curves N 1,...,N 7 are contained in fibres of F i for i =1, 2, 3. For every singular fibre F s of F i, we denote by r(f s ) the number of irreducible curves contained in F s and different from N 1,...,N 7. Since the subspace orthogonal to the class of F i in H 2 (Y,Q) has dimension 9, Zariski s Lemma implies that 8+ F s (r(f s ) 1) 9, namely r(f s ) 2 for every singular fibre F s of F i and there is at most one singular fibre F s with r(f s )=2. This shows that the possible types are m I 2, m I 3, mi 4, I0, I 1, I 2, III, IV and that, except possibly one, the fibres with reducible support are of type m I 2, III or I0. On the other hand, we have 12 = c 2(Y )= F s singular e(f s), hence the quantity F s reducible e(f s) is 12. Using this remark and the fact that the 7 curves N 1,...,N 7 are contained in fibres of F i it is easy to show that types m I 3, I1, III and IV cannot occur and that the fibres cannot all be of type m I 2 or m I 4. This proves (ii). Now assume that, say, F 1 has a fibre G 1 of type I 4 (or 2 I 4 ). Then G 1 contains two of the N i, hence we can write the support of G 1 as C 1 + C 2 + N 1 + N 2,whereC 1 C 2 =0, C i N j =1. Assume that C 1 Ẽ 2 =0. Then the connected fundamental cycle C 1 + N 1 + N 2 is contained in a reducible fibre G 2 of F 2.SinceC 1 N i =0for i>2, thefibreg 2 is necessarily of type I 4 (or 2 I 4 ). Since r(g 1 )=r(g 2 )=2by the above discussion and by (ii) it follows that both F 1 and F 2 have fibres G 1, respectively G 2, of type I 0 and that the nodal curves appearing with multiplicity 1 in G 1 and G 2 are a subset of {N 3,...,N 7 }, contradicting Lemma 4.8. This shows that the intersection numbers C 1 Ẽ 2,C 1 Ẽ 3,C 2 Ẽ 2,C 2 Ẽ 3 are all strictly positive. Since 4=(Ẽ2 + Ẽ3) F 1 =(Ẽ2 + Ẽ3)G 1 C 1 Ẽ 2 + C 1 Ẽ 3 + C 2 Ẽ 2 + C 2 Ẽ 3, these numbers are all equal to 1. So the class of C 1 C 2 is orthogonal to N 1,...,N 7, Ẽ1, Ẽ2, Ẽ3. Since the classes of N 1,...,N 7, Ẽ1, Ẽ2, Ẽ3 are a basis of H 2 (Y,Q), the class C 1 C 2 is numerically equivalent to 0. On the other hand, we have (C 1 C 2 ) 2 = 4, a contradiction. This finishes the proof of (i). Statement (iii) follows by examining the admissible types of fibres, recalling that r(f s ) 2 for every singular fibre F s. LEMMA Assume that there exists a nodal curve C Y such that Then we have case (1) of Theorem 4.6. Ẽ 1 C = N 1 C = = N 7 C =0. Proof. By [17, Lemma 4.2], the divisor C + N N 7 is divisible by 2 in Pic(Y ). Hence CẼi is even for i =1, 2, 3.ThecurveC is contained in a fibre of F 1, hence C(Ẽ2 + Ẽ3)=C B B F 1 =4. On the other hand, since B + N N 7 is also even, the divisor B + C is even and so ( B + C) 2 =4+2 BC is divisible by 8. Hence we have BC =2.From2= BC = CẼ2 + CẼ3, it follows, say, CẼ2 =0, CẼ3 =2. Now, as in the proof of Lemma 4.9, we consider the contributions to c 2 (Y ) and to the Picard number of Y of the various types of singular fibres. Since there are 8 disjoint nodal curves contained in the fibres of F 1 and F 2, one sees that the only possibility is that the fibres with singular support of both pencils are two fibres of type I0 and that each fibre of type I0 contains four of the curves N 1,...,N 7,C, each with multiplicity 1. Recall that an elliptic pencil with 2 fibres of type I0 on an Enriques surface is isotrivial. By Lemma 4.8 we can label the curves N i in such a way that the singular fibres of F 1 are N 1 + N 2 + N 3 + N 4 +2A 1 and N 5 + N 6 + N 7 + C +2A 2 and the singular fibres of F 2 are N 1 + N 2 + N 5 + N 6 +2B 1 and N 3 + N 4 + N 7 + C +2B 2.

12 518 M. MENDES LOPES AND R. PARDINI Computing intersection numbers, one sees that Ẽ3 num Ẽ 1 + Ẽ2 C, namely Ẽ 3 Ẽ1 + Ẽ2 C or Ẽ 3 Ẽ1 + Ẽ2 C + K Y. Since B + C is even, we conclude that Ẽ3 Ẽ1 + Ẽ2 C. Hence the system B is equal to F 1 + F 2 C and it has degree 2 by [8, Theorem 4.7.2]. Looking at the adjunction sequence for C, one gets 0=h 0 (C + K Y )=h 0 (Ẽ1 + Ẽ2 Ẽ 3)=0, hence B + K Y is birational again by [8, Theorem 4.7.2]. Now F 3 F 1 + F 2 2C N 5 + N 6 + N 7 +2A 2 + N 3 + N 4 + N 7 +2B 2, hence N 3 + N 4 + N 5 + N 6 +2(A 2 + N 7 + B 2 ) is a fibre of F 3 of type I2. Thus Lemma 4.9 and the formula 12 = c 2 (Y )= F s e(f s ) imply that the remaining fibres with singular support are two fibres of type I 2 or 2 I 2. Proof of Theorem 4.6. By Lemma 4.10, we may assume that every nodal curve disjoint from N 1,...,N 7 satisfies CẼi > 0 for i =1, 2, 3. Hence none of the pencils F i can have two fibres of type I 0. By Lemma 4.9 and its proof one sees that in principle the possible configurations of fibres with reducible support are: (i) one fibre of type I 0 and three fibres of type I 2 or 2 I 2 ; (ii) one fibre of type I 2 and two fibres of type I 2 or 2 I 2. Notice that in both cases all the fibres with singular support are reducible. Assume that, say, F 1 has a fibre G 1 of type I 2. By Lemma 4.9, (iii), G 1 contains five of the N i and two more components A 2 and A 3. Each of the curves A 2 and A 3 meets 3 of the N i and there is only one of the N i that intersects both. We set λ 2 =2A 2 + i (A 2N i )N i and λ 3 =2A 3 + i (A 3N i )N i. One has λ 2 2 = λ2 3 = 2 and G 1 = λ 2 + λ 3.Since 2=Ẽ2G 1 =2Ẽ2(A 2 + A 3 ), we may assume Ẽ2A 2 =1, Ẽ 2 A 3 =0. Since by Corollary 3.3 the curves N 1,...,N 7 are the only nodal curves orthogonal to Ẽ1, Ẽ2, Ẽ3, wehavealsoẽ3a 3 =1, Ẽ3A 2 =0. The support of λ 3, being connected and orthogonal to Ẽ2, is contained in a fibre G 2 of F 2.SinceA 3 meets precisely three of the N i, by Lemma 4.9 the fibre G 2 is also of type I2 and we can write as above G 2 = λ 1 + λ 3,whereλ 1 =2A 1 + i (A 1N i )N i, with A 1 a nodal curve different from the N i andsuchthata 1 Ẽ 3 =0, A 1 Ẽ 1 =1. Notice that the three nodal curves A 1, A 2 and A 3 are distinct. The same argument shows that λ 1 and λ 2 are contained in fibres of F 3 of type I2. By the proof of Lemma 4.9 each pencil F i has at most one fibre of type I2, hence λ 1 and λ 2 are contained in the same fibre G 3 and G 3 = λ 1 + λ 2. Assume that the curve N i that appears with multiplicity 2 in G 1 and G 2 is the same, say N 7.ThenN 7 is a component of λ 1, λ 2 and λ 3 and Lemma 4.8 implies that, up to a permutation of 1,...,6 the incidence relations between the curves N 1,...,N 7,A 1,A 2,A 3 are given by the dual graph of Fig. 2. The divisor A 1 + A 2 + A 3 + N N 7 is simply connected, hence its inverse image in the K3 cover of Y consists of two disjoint divisors isomorphic to it. It is easy to check that the intersection matrix of the components of these divisors is nondegenerate of type (2, 18),but this contradicts the Index Theorem. So, up to a permutation of the indices, G 1 contains N 1 with multiplicity 2, G 2 contains N 2 with multiplicity 2 and G 3 contains N 3 with multiplicity 2. Using Lemma 4.8 again, one shows that, up to a permutation of the indices, N 7 is not contained in G 1, G 2, G 3 and the incidence relations between A 1,A 2,A 3,N 1,...,N 6 are given by the dual graph in Fig. 1. Computing intersection numbers, one checks that B and the divisor λ1 + λ 2 + λ 3 are numerically equivalent. The argument used to prove Corollary 4.5 shows that they are not 4 e SÉRIE TOME N 4

13 A NEW FAMILY OF SURFACES WITH p g =0AND K 2 =3 519 N 4 N 3 A 3 N 7 N 5 A 1 A N 2 2 N 6 N 1 Fig. 2. linearly equivalent, hence λ 1 + λ 2 + λ 3 B + K Y. Now the system λ 1 + λ 2 + λ 3 has degree 2 by Theorem 7.2 of [7] and B is birational. This settles case (2). We are left with the case in which each of the pencils F i has a fibre of type I0 and 3 fibres of type I 2 or 2 I 2. We recall that by [7, Proposition 5.2.1] and [8, Theorem 4.7.2] B has degree 2 if Ẽ1 + Ẽ2 Ẽ3 is nonempty and it is birational otherwise. Assume that there is Ẽ1 + Ẽ2 Ẽ3. SinceẼ1 =Ẽ2 =0and Ẽ3 =2, all the components of are nodal curves contained in fibres of F 1 and F 2 and there is a component θ of with θẽ3 > 0, but this cannot happen because of the configuration of reducible fibres of the pencils F i. Since this argument is purely numerical it shows also that B + K Y is birational. 5. Enriques surfaces with 7 nodes: a general construction Here we describe a construction giving pairs (Σ,B) as in the set-up of 3 and we prove that all such pairs can be obtained that way. Construction 5.1. Consider the following automorphisms of the projective line P 1 : (x 0,x 1 ) (x e1 0, x 1 ); (x 0,x 1 ) (x e2 1,x 0 ). The subgroup Γ generated by e 1 and e 2 is isomorphic to Z 2 2;wesete 3 := e 1 + e 2. The action of Γ can be lifted to the line bundle O P 1(2). Two such liftings differ by a character of Γ, hence for any chosen lifting a basis of eigenvectors of H 0 (P 1, O P 1(2)) is given by: s(x 0,x 1 ):=x x 2 1, d(x 0,x 1 ):=x 2 0 x 2 1, p(x 0,x 1 )=x 0 x 1. Denote by G the subgroup of automorphisms of P 1 P 1 P 1 generated by the elements: (e 1,e 1, 1), (e 1, 1,e 1 ), (e 2,e 2,e 2 ).

14 520 M. MENDES LOPES AND R. PARDINI The group G is isomorphic to Z 3 2. We denote by G 0 the subgroup of index 2 generated by (e 1,e 1, 1) and (e 1, 1,e 1 ). The fixed locus of the nonzero elements of G 0 has dimension 1, while the fixed locus of the elements of G \ G 0 has dimension 0. Notice that, although the action of Γ on P 1 does not lift to a linear representation on the space H 0 (O P 1(1)), the action of G on P 1 P 1 P 1 is induced by a linear representation on H 0 (O P 1 P 1 P1(1, 1, 1)). Hence it is possible to lift the G-action to the line bundle O P 1 P 1 P1(1, 1, 1) and, compatibly, to all its multiples. Notice also that the possible G-actions on O P 1 P 1 P1(1, 1, 1) differ by a character of G, and thus they all induce the same action on O P 1 P 1 P 1(2, 2, 2). Denoting the homogeneous coordinates on P1 P 1 P 1 by x =(x 0,x 1 ), y =(y 0,y 1 ), z =(z 0,z 1 ), under this action the space H 0 (O P 1 P 1 P1(2, 2, 2)) decomposes into eigenspaces as follows: T 0 := s(x)s(y)s(z),s(x)d(y)d(z),d(x)s(y)d(z),d(x)d(y)s(z),p(x)p(y)p(z) ; T 1 := s(x)s(y)d(z),s(x)d(y)s(z),d(x)s(y)s(z),d(x)d(y)d(z) ; T 2 := s(x)s(y)p(z),p(x)p(y)s(z),d(x)d(y)p(z) ; T 3 := s(x)p(y)p(z),p(x)s(y)s(z),p(x)d(y)d(z) ; T 4 := p(x)s(y)d(z),p(x)d(y)s(z),d(x)p(y)p(z) ; T 5 := s(x)p(y)s(z),p(x)s(y)p(z),d(x)p(y)d(z) ; T 6 := s(x)d(y)p(z),p(x)p(y)d(z),d(x)s(y)p(z) ; T 7 := s(x)p(y)d(z),p(x)d(y)p(z),d(x)p(y)s(z). The subspace T 0 corresponds to the trivial character and the subspace T 1 corresponds to the character orthogonal to G 0. The system T 0 is base point free, while for i>0 the base locus of the system T i is nonempty and it contains the fixed locus of some element of G \ G 0. Let Z T 0 be general. By Bertini s theorem Z is a smooth surface. The elements (e 1, 1,e 1 ), (1,e 1,e 1 ) and (e 1,e 1, 1) act on Z fixing 8 points each and the remaining nonzero elements of G act freely on Z. By the adjunction formula, Z is a K3 surface, hence the quotient surface Z/G is a nodal Enriques surface with 6 nodes. For i =1, 2, 3, the projection on the ith factor P 1 P 1 P 1 P 1 induces an elliptic pencil on Z, which in turn gives an elliptic pencil F i = 2E i on Z/G. A standard argument shows that for i j one has E i E j =1. Assume now that Z has 8 nodes that form a G-orbit and no other singularities. Then the quotient surface Z/G has an extra node, which is the image of the 8 nodes of Z. By Lemma 3.4 either E 1 + E 2 + E 3 or E 1 + E 2 + E 3 + K Z/G is a divisor B as in the set-up of 3. Set Σ:=Z/G and denote by π : Z Σ the quotient map. Notice that π B = π (B + K Σ ) is isomorphic to O Z (2, 2, 2) and π H 0 (Σ,B) and π H 0 (Σ,B + K Σ ) are eigenspaces of H 0 (Z, O Z (2, 2, 2)). So, considering the dimensions, they correspond to the restrictions to Z of T 0 and T 1. We will show later (Lemma 6.3) that the restriction of T 0 is equal to π H 0 (Σ,B+ K Σ ) and the restriction of T 1 is equal to π H 0 (Σ,B). The fact that Construction 5.1 can actually be performed, namely that there exists Z as required, is a consequence of the following theorem and of the examples given in 3. THEOREM 5.2. Let (Σ,B) be a pair as in the set-up of 3. Then(Σ,B) can be obtained from Construction 5.1. Proof. Let V be the code associated with the nodes of Σ, which is isomorphic to Z 2 2 by Proposition 4.4. By [9], Proposition 2.1 and Remark 2, there is a Galois cover π 0 : Z 0 Σ with Galois group Hom(V,C ) = Z 2 2 branched on the 6 nodes of Σ that appear in V.ThemapZ 0 Σ 4 e SÉRIE TOME N 4

15 A NEW FAMILY OF SURFACES WITH p g =0AND K 2 =3 521 can be factorized as Z 0 Z 1 Σ, where both maps are double covers branched on a set of 4 nodes. By [17, Proposition 3.1], Z 1 is a nodal Enriques surface with 6 nodes, hence, by ib., Z 0 is an Enriques surface with 4 nodes. Let K Σ be the K3-cover of Σ and consider the following cartesian diagram: (5.1) p Z π 0 Z 0 K The surface Z is a K3 surface with 8 nodes and the map Z Z 0 is the K3-cover of Z 0.The composite map π : Z Σ is a Galois cover with Galois group isomorphic to Z 3 2. Notice (cf. [9], proof of Proposition 2.1 and Remark 2) that, although the cover π 0 : Z 0 Σ is not uniquely determined (in fact there are four different possibilities), the cover π : Z Σ does not depend on the choice of Z 0. For the reader s convenience the proof is broken into steps. Step 1: For i =1, 2, 3 there exist elliptic pencils C i on Z such that π F i 4C i. Since π is unramified in codimension 1, if F i F i is general then π F i is a disjoint union of linearly equivalent elliptic curves. Hence, to prove the statement it is enough to show that π F i has 4 connected components. Let π : Z Y the Galois cover obtained from π by taking base change with the minimal desingularization Y Σ and, as usual, denote by F i the elliptic pencil of Y induced by F i. By Theorem 4.6 the pencil F i has a fibre G i of type I0 or I 2.We write G i =2A i + N1 i + + N4. i By the results of 4, the nonzero elements of V correspond to the even sets N1 i + + N4, i i =1, 2, 3. So by the definition of Z and Z (cf. also [9, 2]) we have the following formula: ( π = O O Z Y K Y i=1,2,3 Σ ) ( O Y (A i Ẽi) i=1,2,3 ) O Y (A i Ẽ i). The restriction of the line bundles O Y ( K Y ), O Y (A i Ẽi) and O Y (A i Ẽ i ) to a general F i is trivial, hence π Fi has at least 4 connected components. So for i =1, 2, 3 we can write π F i m i Ci where C i is a smooth connected elliptic curve and m i =4or m i =8. Notice that C i Cj 2 for i j, since otherwise the product of the pencils C i and C j would give a birational map Z P 1 P 1. On the other hand, for i j we have 32 = π Fi π Fj = m i m j Ci Cj 16 C i Cj, hence m i = m j =4and C i Cj =2. Finally, the pencils C i induce pencils C i on Z such that π F i 4C i. Step 2: The product of the pencils C 1, C 2 and C 3 defines an embedding ψ : Z P 1 P 1 P 1 such that ψ(z) is a divisor of type (2, 2, 2). We remark that D = C 1 + C 2 + C 3 is ample on Z, since by Step 1 2D = π B and B is ample on Σ by Corollary 3.3. We let ɛ : W Z be the minimal desingularization, we denote by N 1,...,N 8 the exceptional curves of ɛ and we set D := ɛ D, Ci := ɛ C i. The divisor D is nef and big and the N i are the only irreducible curves that have intersection equal to 0 with D. Since D 2 =12,byReider s

16 522 M. MENDES LOPES AND R. PARDINI Theorem and by the fact that the intersection form on a K3 surface is even, if two points x, y W are not separated by D, then there exists an effective connected divisor A x, y such that either: (a) A 2 = 2, A D =0,or(b)A 2 =0, A D =1or (c) A 2 =0, A D =2. Possibility (a). Recall that the corresponds to the case when both x and y belong to one of the curves N i pencils C i have no multiple fibres, because the double fibres of the pencils F i disappear when one takes the K3 cover (actually, it is not hard to prove that any elliptic fibration on a K3 has no multiple fibres). In case (b) one would have, say, A C 1 = A C 2 =0, namely A would be a fibre of both C 1 and C 2, which is impossible. If A 2 =0and A D =2,thenwehave,say,A C 1 =0 and A is a fibre of C 1. But in this case A D =4, a contradiction. The above discussion shows that the map ψ is one-to-one onto its image and that the differential of ψ at every smooth point of Σ is nonsingular. In particular the image of ψ is an hypersurface with at most isolated singularities, hence it is normal. It follows that ψ is an isomorphism. The fact that the image is a divisor of type (2, 2, 2) is a consequence of the fact that C i C j =2if i j. Step 3: There exist coordinates on P 1 P 1 P 1 such that the surface ψ(z) is an element of the linear system T 0 defined in Construction 5.1 and the action of the Galois group of Z Σ coincides with the group action defined there. Denote by G the Galois group of π. By the definition of the map ψ, the three copies of P 1 in P 1 P 1 P 1 can be naturally identified with (the dual of) the linear systems C 1, C 2 and C 3, hence G acts on P 1 P 1 P 1 and the embedding ψ : Z P 1 P 1 P 1 is G-equivariant with respect to the given actions. We have seen in Step 1 that for every i =1, 2, 3 there is a nonzero g i G such that g i acts trivially on C i. Since the fixed locus of g i on C 1 C 2 C 3 has positive dimension and Z is ample, g i has fixed points on Z. Since by construction the cover π : Z Σ factorizes through the K3 cover K Σ, it follows that g 1, g 2, g 3 do not generate G. On the other hand g i must act nontrivially on C j for j i, since otherwise the fixed locus of g i on P 1 P 1 P 1 would be a divisor and g i would fix a curve of Z pointwise. Hence G 0 := {1,g 1,g 2,g 3 } is a subgroup of G isomorphic to Z 2 2.Fixh G \ G 0.Foreveryi we can choose homogeneous coordinates on P 1 = C i such that, using the notation of Construction 5.1, the nonzero element of G 0 /g i acts as e 1 and h acts as e 2. With respect to these coordinates we have: g 1 =(1,e 1,e 1 ), g 2 =(e 1, 1,e 1 ), g 3 =(e 1,e 1, 1), h =(e 2,e 2,e 2 ), namely the G-action on P 1 P 1 P 1 is the same as in Construction 5.1 and the surface Z,beingG-invariant, belongs to one of the linear systems T i, i =0,...,7. In addition, each of the nonzero elements of G 0 fixes 8 points of Z and the elements of G \ G 0 act freely on Z. This is the same as saying that Z is in general position with respect to the fixed loci of all the elements, hence, as we have remarked in Construction 5.1, Z must be an element of T Enriques surfaces with 7 nodes: a parametrization The aim of this section is to construct a quasi-projective variety parametrizing the isomorphism classes of pairs (Σ,B) as in the set up of 3 and to study the geometry of this space. In addition, we show the existence of a tautological family on a finite Galois cover of the parametre space. This tautological family admits a section and a simultaneous resolution. These results are used in the next section to describe the subset of the moduli space of surfaces with p g =0and K 2 =3consisting of the surfaces S that have an involution σ such that: (1) the quotient surface S/σ is birational to an Enriques surface; (2) the bicanonical map ϕ of S is composed with σ. We use all the notation from the previous sections. For a pair (Σ,B) as in the set-up of 3 we denote as usual by η : Y Σ the minimal desingularization and by N 1,...,N 7 the exceptional 4 e SÉRIE TOME N 4