Singular bidouble covers and the construction of interesting
algebraic surfaces.

Fabrizio Catanese

Friedrich Hirzebruch zum 70. Geburtstag gewidmet

Abstract. A bidouble cover is a finite flat Galois morphism with Galois group (Z=2)2?. The structure theorem for smooth Galois (Z=2)2? covers was given in [Cat2] [pag. 491-493] where bidouble covers of P1 ? P1 were introduced in order to find interesting properties of the moduli spaces of surfaces of general type. In this paper we develop general formulae for the case of resolutions of singular bidouble covers. P. Burniat used singular bidouble covers in order to fill out sectors of surface geography. In this paper instead, the main application is for the construction of surfaces with birational canonical map (so called simple canonical surfaces) and high K2, for instance we construct such surfaces with pg = 4; 11 <= K2 <= 28, against a prediction of F. Enriques that 24 should be the maximum allowed.
Moreover, we find, among several new examples of surfaces, some surfaces with pg = q = 1, K2 = 4; 5, and also some infinite series of surfaces whose canonical map is composed of a pencil of curves of genus 2 or 3, with non costant moduli.

1. Introduction

In this paper we present work in progress, motivated by the following problem raised by Federigo Enriques ([En], chapter VIII, page 284)

Problem 1. Try to construct simple canonical algebraic surfaces (i. e., surfaces having a birational canonical map Ö1), with geometric genus pg = 4 and K2 as high as possible.

One knows that if the canonical map Ö1 is birational, then necessarily pg >= 4 and the Castelnuovo inequality K2 >= 3pg ? 7 holds. More generally, one can ask for which values of the invariants Ø = 1 ? q + pg and K2 does there exist a simple canonical surface. Recently, while writing the survey article [Cat5], I realized that the question is quite open for small values of pg , and I noticed that one can ask a stronger question

1991 Mathematics Subject Classification. 14J29, 14J10, 14J25, 14J17.
The research of the author was performed in the realm of the Project AGE HCM, Contract ERBCHRXT 940557.

2 FABRIZIO CATANESE

Problem 2. For which values of the invariants Ø and K2 do there exist algebraic surfaces such that a general projection of the canonical map Ö1 to a 3- dimensional projective space is such that its image ?1 has ordinary singularities.

If then S is a simple canonical surface and pg attains its minimal value = 4, then K2 >= 5 and simple canonical surfaces were constructed for K2 = 6; 7 by Enriques (the case K2 = 5 being the trivial case of surfaces of degree 5 either smooth or with at worst Du Val singularities). The existence problem for 6 <= K2 <= 8 was solved by Franchetta ([Fran]), whereas other constructions for K2 = 7 were also done by Maxwell and Kodaira ([Max],[Kod] ), and finally Ciliberto gave an affirmative answer to problem 2 in the interval 6 <= K2 <= 10 ([Cil1]) showing moreover that in this range surfaces with ordinary singularities give rise to an irreducible unirational open set of the moduli space. Recently, I was informed by Ciliberto that P. Burniat had also obtained results in this direction, thus via a bibliographical search I found out that Burniat ([Bur1]) constructed simple canonical surfaces with pg = 4 and 11 <= K2 <= 16.

Enriques indeed had conjectured that K2 = 24 should be the maximumpossible value for a simple canonical surface with pg = 4, but we show in this paper that his prediction, based on the conjecture that the expected number of moduli should be strictly positive, does not hold true.

Theorem 1. There do exist algebraic surfaces with geometric genus pg = 4, and with a birational canonical map Ö1, for the following values of K2: 11 <= K2 <= 28.

The above highest value 28 for K2 is very likely to be substantially improved with the same methods of the present paper. As the reader may easily guess from the title, the above surfaces arise as minimal desingularizations of singular bidouble covers, in fact of Q = P1 ? P1. Here, a bidouble cover is a finite flat Galois morphism with Galois group (Z=2)2. The structure theorem for smooth

Galois (Z=2)2 covers was given in [Cat2] [pag. 491-493], including a calculation of their invariants, and a study of their deformation theory. There, smooth bidouble covers of P1 ? P1 were introduced in order to find interesting properties of the moduli spaces of surfaces of general type. Later, Hahn and Miranda ([H-M]) extended the structure theorem to the general case, and we present here this result in the nice formulation found by F. Schreyer. The general set up we are mostly interested in here is as follows: we have a bidouble cover f : Y ! X where X is a smooth surface, Y is singular and we let ß : S ! Y be a minimal resolution of singularities. The main problem is to calculate the invariants of S in terms of the 3 branch curves D1; D2; D3 and of the singularities of D = D1 [ D2 [ D3 (observe that Y is smooth just when the 3 branch curves are smooth and have normal crossings). Thus, one obtains formulae where there is a global contribution (yielding the value of the invariants when Y is smooth) and a local contribution, depending on the local structure of the singularities of D. We mostly analyse the case where D has ordinary singularities, and thus we can attach to any point P of X a triple [µ1(P ); µ2(P ); µ3(P )], where µi(P ) is the multiplicity of Di at P .

Once we obtain such formulae we convince ourselves that the most significant singularities are the following ones:

SINGULAR BIDOUBLE COVERS 3

Singularity type drop of K2 drop of Ø
(1, 1, 1) K2 # 1 Ø = same,
(3, 1, 0) K2 # 1 Ø # 1,
(2, 2, 0) K2 # 4 Ø # 1.

The usefulness of imposing such singularities to the branch curve was best illustrated by Burniat's beautiful construction of surfaces with pg = 0 and K2 = 6 ? m (m = 0; :::4) obtained by choosing a bidouble cover of the projective plane branched on three cubic curves having three (3; 1; 0) points and m (1; 1; 1) points (then the branch curve must be a union of lines).

Choosing smooth bidouble covers of the quadric Q = P1 ? P1 has the advantage that we have 6 discrete parameters (giving the 3 bidegrees of the branch curves), upon which the two basic surface invariants Ø and K2 depend, but as one tries to fill in sectors of the surface geography chessboard a lot of gaps are encountered: however, imposing (3; 1; 0) and (1; 1; 1) points, gives chess moves which allow to fill these gaps, and hopefully to solve in an easier and better way the geography problem attacked by Persson in [Per] (who considered surfaces with a pencil of curves of genus 2, thus with non birational canonical and bicanonical map). This will be the object of a forthcoming article, with an eye also to the construction of simple canonical surfaces. In this respect, imposing (1; 1; 1) points has the great advantage that if the most singular surface yields a simple canonical surface, then also the less singular ones achieve this goal ; whereas imposing (3; 1; 0) points has the great advantage of improving the slope K2=Ø.

These examples are particularly interesting in order to analyse the equations of canonical surfaces in Pr for r = 3; 4; 5 (cf. [C-S] [Cat5]). Another application is towards the construction of surfaces with low invariants, classically done with the technique of singular double covers. For instance, surfaces with pg = q = 1 have been classified by the author and also by E. Horikawa in the case K2 = 2 (cf. [Cat1], [Hor]), and by the author and Ciliberto in the case K2 = 3 ([Ca-Ci1], [Ca-Ci2]).

But for higher values of K2, only a sporadic example was known, due to Xiao ([Xiao1]), having K2 = 4 and Albanese map a pencil of curves of genus 2. Here we find new examples having K2 = 3; 4; 5 and Albanese map also equal to a pencil of curves of genus 2. It was surprising that, just by recalling the theory of smooth bidouble covers and giving several numerical examples, we stomped on several noteworthy examples of families of algebraic surfaces.

We want to mention for instance example 2, containing some infinite series of surfaces for which the canonical map is composed of a pencil of curves of genus 2 and 3. The principal novelty here is that, unlike the previously known examples, due to Pompilj, Beauville, Xiao and Zucconi ( cf. e.g. [Bea] , [Xiao2], [M-Y]) our examples provide fibrations with non constant moduli and with q = 0.

One interesting generalization of the theory would be to consider the case of singular Abelian covers (the smooth case having been treated by Comessatti [Com] in 1930, and later more thoroughly by Pardini [Par1]). This generalization should be viewed not only as a nice algebraic calculation, but pretty much in the spirit of Hirzebruch's beautiful constructions of algebraic surfaces as Kummer coverings ([Hir], [B-H-H]). Finally, another interesting direction of study which one may pursue is the investigation of the surgeries arising when the branch curves acquire more singularities, or become non reduced.

4 FABRIZIO CATANESE

2. The basic model of a bidouble cover and the structure theorem

The simplest example of a bidouble cover is the mapping f : P2 ! P2 which is given by squaring the coordinates, i. e. f(y1; y2; y3) = (x21; x22; x23).

The most convenient way is to think of it as a projection of the embedding of P2 to the Veronese surface V . The cone over the Veronese surface consists in the algebraic set of rank 1 (3 ? 3) matrices, thus V is defined by the determinantal equation

rk

0

@

x1 w3 w2
w3 x2 w1
w2 w1 x3

1
A = 1: (?)

The isomorphism of P2 with V is given by setting

xi = y2i
(2.1)

wk = yi ? yj

(fi; j; kg being a permutation of f1; 2; 3g).
The mapping f is then the projection f : V ! P2 given by the diagonal entries of the symmetric matrix.

f : P2 ! P2 is branched on the three coordinate lines Dj = div(xj), whereas the ramification divisor R on the source is given by the three coordinate lines Rj = div(yj ), which represent the respective Fix point sets of the three non trivial elements öj in the Galois group.

The above toy model contains all the basic features of the general case, as illustrated by the following structure theorem which is proven in [Cat2] [pag. 491- 493] and [H-M][pag. 27-29] (except for the last statements, not explicitly stated in these references, and for which we give hints of proof).

Especially, the toy model illustrates the profound difference between the birational and the biregular point of view: birationally, in the toy model we take the two square roots of the rational functions x1x2=x2, x1x3=x2 (x being a general linear form). But, biregularly, f is not a fibre product of two double coverings, only the normalization of such a fibre product given by equations
u2 = x1x2; v2 = x1x3.

Theorem 2. Let f : Y ! X be a finite flat (Z=2)2- Galois cover with X an algebraic variety in char (k) 6= 2. Then there exist 3 Cartier divisors Dj = div(xj) on X and 3 line bundles Li, with fibre coordinate wi, such that Y is embedded in the total space of the vector bundle L1 ? L2 ? L3, direct sum of the line bundles Li. Moreover Y is defined there by the determinantal equation (?). Conversely, assume that the above data satisfy the following linear equivalences on X,

ffl Li + Di ? Lj + Lk.
for each permutation (i; j; k) of (1; 2; 3):

then they determine a bidouble cover defined by the above determinantal equations (?). One has a decomposition according to the characters of (Z=2)2:

ffl f?OY ?= OX ? L
i OX (?Li):
Assume in addition that X is a smooth variety, then:

ffl Y is smooth if and only if the divisors Dj are smooth, they do not have a common intersection and have pairwise transversal intersections.

SINGULAR BIDOUBLE COVERS 5

ffl Y is Cohen - Macaulay (this follows from the determinantal equations) and for its dualizing sheaf !Y (which, if Y is normal, equals the sheaf of Zariski differentials) we have (cf. [H], ex. III. 6. 10 and III. 7. 2)
f?!Y ?= Hom(f?OY ; !X) ?= !X ? L
i !X (Li).
ffl Y is normal if and only if the divisors Dj are reduced and have no common components (since by the Cohen - Macaulay property, normality is equivalent to non singularity in codimension 1).

Remark 1. The determinantal equation (?) obviously implies some linear equiv-
alences on X, namely (using the additive notation for Cartier divisors) that

ffl 2Li ? Dj + Dk
ffl Li + Di ? Lj + Lk.

But, (cf. the statement of the structure theorem) the first linear equivalences follows from the second ones: this shows that one can choose the Li 's with the sole restriction that the linear system jLj + Lk ? Lij is non empty, and then pick up a divisor Di in the system jLj + Lk ? Lij.

Whereas the first equations show that the linear equivalence classes of the branch divisors cannot be chosen freely, being subject to (mod2) congruences.

Definition 1. A bidouble cover is said to be of simple type if there is a permutation of the indices (1, 2, 3) such that the divisor Dk is trivial (equivalently, Lk ? Li + Lj).

Remark 2. Simple bidouble covers are useful for many applications, but especially in dimension higher than 2, where otherwise the condition that the three branch divisors have an empty intersection may be hard to fulfill.

From now on, we shall constantly assume that X is smooth.

Remark 3. Assume that Y is not normal, i. e., that either the divisors Dj are not reduced, or that they have a common component. Then the recipe for finding the normalization of Y is quite easy:

ffl if Di = D0i + 2A, replace Di by D0i and Lj by Lj ? A for j 6= i
ffl if there is a divisor A with Di >= A for all i, replace each Di by Di ? A and

each Lj by Lj ? A
ffl if Di = D0i + A, Dj = D0j + A, but Dk ? A is not effective, replace Di by D0i, Dj by D0j , and Dk by Dk + A, accordingly replace Lk by Lk ? A.

We then obtain a bidouble cover which is nonsingular in codimension 1, whence normal (cf. theorem 2).

Remark 4. The structure of bidouble covers is easier to understand locally, where all line bundles are trivial, and therefore one can take three square roots y21 = x1; y22 = x2; y23 = x3, which define a Galois (Z=2)3-cover Z.

Y is the quotient of Z by the involution multiplying all the three functions yi by ?1.

In fact, the functions wi = yjyk are just needed to obtain a set of generators as OX-module of the ring OY of functions on Z invariant under the above involution. These observations show that if X is Gorenstein (e. g. smooth, as in our case), then also Z is Gorenstein: therefore Y is then 1=2-Gorenstein.

6 FABRIZIO CATANESE

In particular, if we set N = 2KX + D = 2KX + P3j=1 Lj , 2KY is the Cartier

divisor f?(N ).

Observe moreover that in this context one verifies immediately the determinantal equations
w2i = xjxk, wiwj = wkxk
which can be interpreted as giving the ring structure to the direct sum
f?OY = OX ? L
i OX (?Li).

Assume now that our bidouble cover is smooth, or just Gorenstein and locally factorial. Then the above functions y1; y2; y3 globalize to sections of line bundles on Y , since the Weil divisors
Ri = div(yi)
are by hypothesis Cartier divisors.
We should point out the importance of these functions, since they are precisely the eigenfunctions for the action of the Galois group, indeed they satisfy

ffl öiyi = ?yi
ffl ökyi = yi for k 6= i.

Using those, we can write more precisely:
f?!Y = y1y2y3!X ? L
i yi!X(Li).
This means that, if u1; ::un are local coordinates on X, then the sections of !Y
can be written, locally on X, uniquely as
(y1y2y3f0(u1; ::un) + P3j=1 yjfj(u1; ::un))?,

? = (du1 ^ :: ^ dun)(y1y2y3)?1 being a local generator of !Y .
The above way of writing sections of the canonical sheaf is the most appropriate to analyse the behaviour of the canonical map, and in particular its base points.

In fact, one can properly understand it by analogy with the absolute case of direct image, that is, global sections: if F = div(y) is the fixed part of a linear system jDj = F + jM j on a variety Y , then H0(OY (D) = yH0(OY (M ).

An equivalent formula, which is used to prove the former one, and does not require the assumption that the ramification divisors Ri be Cartier, is:
f0(u1; ::un)(du1 ^ :: ^ dun) + P3j=1 fj(u1; ::un))(du1 ^ :: ^ dun)(wj)?1 ).

On the other hand, since ?2 = (du1 ^ :: ^ dun)2(y1y2y3)?2 = (du1 ^ :: ^ dun)2(x1x2x3)?1 = (du1 ^ :: ^ dun)2(w1w2w3)?1 , setting as above

ffl N = 2KX + D = 2KX + P3j=1 Lj
we obtain also the important formulae

ffl H0(OY ((2m + 1)KY ) =
= y1y2y3H0(OX (KX + mN ) ? L
i yiH0(OX (KX + mN + Li),
and
ffl H0(OY ((2m)KY ) = H0(OX(mN ) ? L
i wiH0(OX(mN ? Li).
Please note also that, since f?(N ) = 2KY , we have
ffl K2Y = N2,

whereas
ffl pg(Y ) = pg(X) + P3i=1 h0(OX(KX + Li).

Example 1. Let X = Q = P1 ? P1: then we have three pairs of integers, the bidegrees (ni; mi) of Di, or equivalently
the bidegrees (ai; bi) of Li.

SINGULAR BIDOUBLE COVERS 7

Notice that the ni 's are either all even or all odd, and likewise the mi's, and non negative, and (ai; bi) = 1=2[(nj; mj) + (nk; mk)].
Let Y be a smooth bidouble cover of type (n1; m1); (n2; m2); (n3; m3)):
then

setting n =
3

X

j=1
nj ; m =
3
X

j=1
mj ;(2.2)

Ø(Y ) = 1

4 ((n ? 4)(m ? 4) +

3
X

j=1
njmj)(2.3)

pg(Y ) =
3
X

j=1
max(0; (aj ? 1))max(0; (bj ? 1))(2.4)

K2Y = 2(n ? 4)(m ? 4):

We are now interested to see which surfaces of general type with low pg can we obtain. We see immediately that no surface with pg = 0 can be thus obtained, whereas for pg = 1 we get the following list

Li = (1, 2) (2, 1) (2, 2) K2 = 2; pg = 1; Ø = 2 Di = (3, 1) (1, 3) (1, 1)
Li = (1, 2) (3, 1) (2, 2) K2 = 4; pg = 1; Ø = 2 Di = (4, 1) (0, 3) (2, 1),
Li = (1, 3) (3, 1) (2, 2) K2 = 8; pg = 1; Ø = 2 Di = (4, 0) (0, 4) (2, 2).

where the first surfaces have been considered in [C-D].
For pg = 2 there is already some overlapping for the values of K2 and Ø.

Li = (1, 2) (a, 1) (3, 2) K2 = 2a; pg = 2; Ø = 3 Di = (a+2, 1) (4-a, 3) (a-2, 1) where 2 <= a <= 4, so K2 = 4; 6; 8
Li = (0, 2) (3, 1) (3, 2) K2 = 4; pg = 2; Ø = 2 Di = (6, 1) (0, 3) (0, 1),
Li = (0, 3) (3, 1) (3, 2) K2 = 8; pg = 2; Ø = 1 Di = (6, 0) (0, 4) (0, 2),
Li = (1, 3) (a, 1) (3, 2) K2 = 4a; pg = 2; Ø = 3 Di = (a+2, 0) (4-a, 4) (a-2, 2) where 2 <= a <= 4, so K2 = 8; 12; 16
Li = (1, t) (2, 2) (2, 2) K2 = 2t; pg = 2; Ø = 3 Di = (3, 4-t) (1, t) (1, t)
where 1 <= t <= 4, so K2 = 2; 4; 6; 8

Remark 5. Observe that in the last case the canonical pencil has no fixed part, but it has 2t base points. Moreover let us point out that, in the fourth case, where the three branch divisors are divisible by 2, then (cf. prop. 2. 7 of [Cat2]) the fundamental group of Y is equal to (Z=2), while in the other cases with Ø = 3 the surface can be shown to be simply connected.

Example 2. We give here series of surfaces of general type with canonical map composed of a pencil of genus a = 2 or a = 3. To this purpose, choose

Li = (m+1, 2) (1, a) (m+k, 1) Di = (k, a-1) (2m+k, 3-a) (2-k, 1+a) whence in particular 2 <= a <= 3; 0 <= k <= 2, K2 = 2(a ? 1)(2m + k ? 2), pg = m, Ø = m + 1.

Notice that the fibres of the canonical pencil Ö1 : Y ! P1 have non constant moduli.

Next, we proceed to search for canonical surfaces: in order to do this, we need to make a preliminary observation

Lemma 3. Let f : Y ! X be a smooth bidouble cover, where H0(!X ) = 0. Assume moreover that the canonical map Ö1 of Y is birational: then for each i = 1; 2; 3, H0(!X (Li)) 6= 0:

8 FABRIZIO CATANESE

Proof. Let y be a point of Y which does not map to the branch divisor D. If, say, H0(!X(L1)) = 0 we take the element ö1 in the Galois group, such that it takes the same value (?1) on the two characters Ø2; Ø3.

Then the image point Ö1(ö1(y)) is the same as Ö1(y) (since in projective space v = ?v), contradicting the birationality of Ö1.

Example 3. The first possibility for the canonical map Ö1 of Y to be birational is then the case where pg = 4, and where moreover
Li = (3, 2) (2, 2) (2, 2), K2 = 12; pg = 4; Ø = 5, Di = (1, 2) (3, 2) (3, 2).
We shall verify that for a general choice of the branch divisors, Ö1 is indeed birational. It is also interesting to look at the other cases where pg = 4 and the canonical map has a 2-dimensional image

Li = (3, 2) (3, 2) (n, 1) K2 = 2(n + 2); Ø = 5 Di = (n, 1) (n, 1) (6-n, 3)
where 0 <= n <= 6
Li = (3, 2) (3, 2) (1, m) K2 = 6m; Ø = 5 Di = (1, m) (1, m) (5, 4-m),
where 1 <= m <= 4
Li = (3, 2) (2, 3) (1, a) K2 = 4(1 + a); Ø = 5 Di = (0, 1+a) (2, a-1) (4, 5-a), where 1 <= a <= 5
Li = (4, 2) (2, 2) (n, 1) K2 = 2(n + 2); Ø = 5 Di = (n-2, 1) (n+2, 1) (6-n, 3) where 2 <= n <= 6
Li = (3, 3) (1, n) (m, 1) K2 = 2mn; Ø = 5 Di = (m-2, n-2) (m+2, 4-n) (4-m, n+2)
where 2 <= n; m <= 4.

In the first two cases the image ?1 of the canonical map is a smooth quadric, Ö1 has 2n, respectively 2m base points, and therefore degree 2, respectively 2m.

In the third case Ö1 has 2 + 2a base points, and degree 2 onto a ruled surface ?1 of degree 1 + a whose equation can be easily written, as follows.

If one takes coordinates (u0; u1)(t0; t1) on X = Q = P1 ? P1 and takes into account that the canonical map is given by setting
(z0; z1; z2; z3) = (y1u0; y1u1; y2t0; y2t1),
one obtains the following equations for ?1:
x1(z2; z3) = x2(z0; z1; z2; z3).
In fact, x1(z2; z3) = (ya+1
2 )x1(t0; t1) = (ya+1
2 )y21 =

x2(u0; u1; t0; t1)(ya?1
2 )y21 = x2(z0; z1; z2; z3).
In the fourth case, the canonical image ?1 is a quadric cone, Ö1 has 2n base points, and degree 2.

Finally, in the fifth case, the canonical system has a fixed part, equal to R1, and the canonical map coincides with the bidouble cover f followed by the standard embedding of Q as a quadric.

We prove now the assertion about birationality of Ö1 made in the previous example. In doing so, we shall establish a quite general method and criterion, valid for all bidouble covers with pg = 4.

Proposition 4. Let Y be a general bidouble cover of X = Q = P1 ? P1, of type (1, 2) (3, 2) (3, 2): then the canonical map Ö1 of Y is birational. The canonical image ?1 does not have ordinary singularities.

Proof. Observe that the canonical map is given by setting
(z0; z1; z2; z3) = (y1u0; y1u1; y2; y3). Therefore there are no base points and it suffices, to show the second assertion, to show that the map Ö : R1 ! P1 = fz0 =

SINGULAR BIDOUBLE COVERS 9

z1 = 0g) has degree 8. To show this, we change coordinates on Q = P1 ? P1 so that D1 = fu1t20 = u0t21g, and we compose Ö with the map which squares the coordinates on P1.

It follows that, since we have a degree 2 cover R1 ! D1, the degree of Ö equals the degree of : D1 ! P1, where in affine coordinates (t2; t) = (x2(t2; t); x3(t2; t)). It is then clear that, varying x2; x3, we obtain all possible polynomials of degree 8.

To verify the birationality of the canonical map, let us consider two general points y; y0 of Y . There are two cases to consider:
Case I: f(y) = f(y0).
Then there is an involution ö in the Galois group such that y0 = ö(y) ; moreover, since there are three non trivial eigenspaces in H0(!Y ), we can choose two corresponding characters Ø; Ø0 such that Ø(ö) = 1; Ø0(ö) = ?1. Letting s; s0 be non zero sections in the corresponding eigenspaces, we obtain (s(y0); s0(y0)) = (s(y); ?s0(y)): if y; y0 are general, then s; s0 do not vanish at y and Ö1(y0) 6= Ö1(y). Thus we are reduced to consider
Case II: f(y) = x 6= x0 = f(y0).
The case where x; x0 do not belong to the same curve of the pencil Cu = fu1=u0 = ug is immediate, therefore it suffices to show the birationality of the restriction of Ö1 to Cu. We have the following diagram:

(Cu) Ö! P2
# #fi
(Lu) ! P2

where Ö is given by (y1u0; y2; y3), fi is the standard bidouble cover obtained by squaring the coordinates, Lu is the curve image of Cu on X, and finally, then, is given by (x1u20; x2; x3).

The content of Case I is that the map fi, which has degree 4, remains also of degree 4 when restricted to the image of Cu, since indeed the Galois orbit of a general point has cardinality 4. The conclusion is that degÖ = deg , whence it is equivalent to show that the map is birational onto its image.

Now, in this particular case, is given, on Lu = P1, by three polynomials of degree 2, so the only possibility is that the image of Lu should be a line for each u. (In this case the image of Q = P1 ? P1 under the map 1 given by (x1u20; x1u21; x2; x3) would be a sextic surface ruled by lines and with a 4-ple line).

We can directly show that this case does not occur for general choice of the branch loci, since on a general curve Lu we can fix the three divisors Di \ Lu as we want, therefore the three sections (x1; x2; x3) do not map Lu to a line.

Example 4. We restrict now to smooth covers Y of Q which have pg = 5; 6 and fulfill the necessary condition for the birationality of Ö1 given in lemma 3.

Li = (4, 2) (2, 2) (2, 2) K2 = 16; pg = 5; Ø = 6 Di = (0, 2) (4, 2) (4, 2) Li = (3, 2) (3, 2) (2, 2) K2 = 16; pg = 5; Ø = 6 Di = (2, 2) (2, 2) (4, 2), Li = (2, 3) (3, 2) (2, 2) K2 = 18; pg = 5; Ø = 6 Di = (3, 1) (1, 3) (3, 3).

10 FABRIZIO CATANESE

Li = (3, 3) (2, 2) (2, 2) K2 = 18; pg = 6; Ø = 7 Di = (1, 1) (3, 3) (3, 3)
Li = (3, 2) (3, 2) (3, 2) K2 = 20; pg = 6; Ø = 7 Di = (3, 2) (3, 2) (3, 2),
Li = (4, 2) (3, 2) (2, 2) K2 = 20; pg = 6; Ø = 7 Di = (1, 2) (3, 2) (5, 2),
Li = (3, 2) (3, 2) (2, 3) K2 = 24; pg = 6; Ø = 7 Di = (2, 3) (2, 3) (4, 1)
Li = (4, 2) (2, 3) (2, 2) K2 = 24; pg = 6; Ø = 7 Di = (0, 3) (4, 1) (4, 3).

The two surfaces with branch divisors of respective types ((1, 2) (3, 2) (5, 2)) and ((3, 2), (3, 2), (3, 2)) are homeomorphic but they belong to different irreducible components of the moduli space (cf. [Cat2] [p. 500] where it is shown that for the first surfaces the local dimension of the moduli space is at least 39, whereas for the second ones the local moduli space is smooth of dimension equal to 38).

The last calculations about local moduli spaces are based on the concept of natural deformations of bidouble covers which consists in modifying the determinantal equations to require the following matrix to have rank equal to 1:

0

@

x1 + fl1w1 w3 w2
w3 x2 + fl2w2 w1
w2 w1 x3 + fl3w3

1
A = 1 (?)

where fli is a section of H0(OX (Di ? Li)), for i = 1; 2; 3.

Example 5. Example (a, b, c) consists of two smooth simple covers Y , Y 0 of respective types ((2a, 2b), (2c, 2b)), and ((2a + 2, 2b), (2c - 2, 2b)).

By the previous formulae, these two surfaces have the same invariants Ø(S) = 2(a + c ? 2)(b ? 1) + 4b(a + c), K2S = 16(a + c ? 2)(b ? 1).

If one moreover assumes that a >= 2c + 1, a >= b + 2, c >= b + 2, and that a, b, c are even and >= 3, one gets different irreducible components of the moduli space (this again follows directly from the results of [Cat2]).

3. Resolutions and Adjunction ideals

Throughout this section we assume that ß : S ! Y is the minimal resolution of a normal bidouble cover f : Y ! X.

Moreover, here we assume throughout that S; Y; X are surfaces, in order to simplify the treatment.

We recall first of all that, in view of the characterization of !Y as the sheaf of 2? forms regular on the smooth locus, one has immediately:
ß?!S æ !Y
(in the Gorenstein case, !Y is invertible, therefore we can write
ß?!S = A !Y , and A is called the adjunction ideal).
Recalling moreover that
f?!Y = y1y2y3!X ? L
i yi!X(Li),
we can write
f?ß?!S = y1y2y3A0!X ? L
i Aiyi!X(Li).
In order to obtain the explicit determination of the "adjunction" ideals Ai, observe that, if u; v are local coordinates on X, then the sections of !S can be written, locally on X, uniquely as
f0(u; v)(du ^ dv) + P3j=1 fj(u; v))(du ^ dv)(wj)?1 ).

Therefore the ideals Aj are precisely the ideals of the functions fj(u; v) for which we do not get on S a 2?form with poles on the exceptional divisors.

SINGULAR BIDOUBLE COVERS 11

It is therefore straightforward to observe that
A0 = OX .
In general, though, it happens that Aj is a proper ideal, and since Y is normal, it is an ideal of finite colength. What we shall now do will be to determine the ideals Aj for the singular points of D, and in particular calculate the colength
l(OX=Ai).
The interpretation of this calculation will be that the imposition of certain singularities on the branch curve D will produce a "drop" of the value of Ø equal to r = P
i l(OX=Ai). Moreover, calculating the dimension of the vector spaces H0(Ai !X (Li)) one can compute pg, whence also the irregularity q of S.

To calculate the contribution to the drop of K2, since we are not sure whether the surface S is minimal, it is better to calculate the bigenus P2(S) = H0(!2S): in fact, we have
K2 + Ø = Ø(!2S)
and for minimal surfaces of general type, indeed holds
K2 + Ø = P2.
To this purpose, we need to calculate similar formulae (i. e., ideals of "biadjunction") for f?ß?!2S .

We remind the reader that the local generator of (!2Y ) (in the sense of sheaves associated to Weil divisors) was found to be ?2 = (du ^ dv)2(y1y2y3)?2 = (du ^ dv)2(x1x2x3)?1.
Whence we obtained
H0(!2Y ) = H0(OX(2KX + D)) ? L
i wiH0(OX(2KX + D ? Li)).
Therefore we obtain
f?ß?!2S = B0(2KX + D) ? L
i wiBi(2KX + D ? Li),
and a rather immediate description of the ideals Bj , as follows:

f0 2 B0 iff div(f0) >= div(x1x2x3) ? 2div(du ^ dv)
fi 2 Bi iff div(fi) >= div(x1x2x3) ? 2div(du ^ dv) ? div(wi), i. e.,
fi 2 Bi iff div(fi) >= 2div(yi) + div(yjyk) ? 2div(du ^ dv).

Also in this case the conclusion is that the sum of the colengths of the ideals Bj contributes to the drop of Ø(!2S), whence, subtracting the drop of Ø, we get the drop for K2 of the minimal model (in case S is of general type); whereas, the biadjunction ideals do not necessarily impose independent conditions when S is not minimal.

We shall analyse now the case where D has ordinary singularities, therefore we can attach to any point P of X a triple [µ1(P ); µ2(P ); µ3(P )], where µi(P ) is the multiplicity of the curve Di in P .
It will be convenient to write µi = µi(P ) = 2mi + ?i, where 0 <= ?i <= 1.
We have four possible cases:

[?1; ?2:?3] = (0; 0; 0)
[?1; ?2:?3] = (1; 1; 1)
[?1; ?2:?3] = (0; 1; 1)
[?1; ?2:?3] = (1; 0; 0).

On the blow up X0 of X at P the full transforms of the Di 's would not give a normal cover, therefore we apply the recipe indicated in remark 3, by which the multiplicities of E in the four cases can be respectively reduced to be (0, 0, 0), (0, 0, 0), (1, 0, 0), (1, 0, 0).

12 FABRIZIO CATANESE

In the first two cases (since locally the blow up and the reduction process have brought us to the normal crossings condition) the resolution of singularities is obtained by taking the corresponding cover of X0, which introduces a single smooth curve C, obtained as the preimage of the exceptional curve E.
The curve C has self intersection = ?4, and its genus g satisfies
2g ? 2 = ?8 + 4 P3j=1 mj in the first case,
respectively 2g ? 2 = ?8 + 4 P3j=1 mj + 6 in the second case.
Thus g = 2 P3j=1 mj ? 3, resp. g = 2 P3j=1 mj .
We calculate:

(du ^ dv) has order of zero on C = 1
xi has order of zero on C = µi
(du ^ dv)=wi has order on C = 1 ? (µj + µk)=2
fi 2 Ai iff ord fi >= (µj + µk ? 2)=2
fi 2 Bi, i >= 1 iff ord fi >= µi ? 2 + (µj + µk)=2
f0 2 B0 iff ord f0 >= µi + µj + µk ? 2

Therefore in all these cases the adjoint and biadjoint ideals are just powers of the maximal ideal MP of the point P . Since the colength of the ideal MrP equals r(r + 1)=2, which fortunately vanishes for r = 0; ?1, it follows that Ø drops by the following amount:
P3i=1(1=8)(µj + µk ? 2)(µj + µk) = (1=8)[P3i=1(µj + µk)2 ? 2(µj + µk)] =

(1=4) P3i=1(µi)2 ? (1=2) P3i=1 µi + (1=4)ö2(µ) =

= ((1=2) P3i=1 µi)2 ? (1=2) P3i=1 µi ? (1=4)ö2(µ)
where ö2 is the second elementary symmetric function.
For the calculation of how much K2 drops, recall the formula
r(r + 1)=2 ? s(s + 1)=2 = (1=2)(r ? s)(r + s + 1).
Whence (apply the formula to r = l(OX=Bi); l = l(OX=Ai)) we obtain that K2 drops by:
(1=2)((P3i=1 µi) ? 2)((P3i=1 µi) ? 1) + (1=2)(P3i=1(µi ? 1)((P3i=1 µi) ? 2)) =

= ((P3i=1 µi) ? 2)2.

There remains now to treat the third and fourth cases. Here, after we perform a blow up of the point P , and the reduction process, it happens that the exceptional curve E is part of the branch locus, and that the branch locus does not consist of three smooth curves having normal crossings.

Indeed, by our choice, the first branching divisor D01 consists of the union of E and of the strict transform of D1, while, for i = 2; 3, D0i equals the strict transform of Di.

Therefore the bidouble cover Z of X0 corresponding to the branch divisors D0i is normal with only 2µ1 nodes as singularities, corresponding to the inverse images of the intersection points of E and of the strict transform of D1.

But we can observe that Z is Gorenstein, and !Z is the direct image of !S under the minimal resolution of singularities of Z. Therefore everything works similarly to the first two cases: the major difference being that, since C is part of the ramification locus, then

SINGULAR BIDOUBLE COVERS 13

(du ^ dv) has order on C = 3
xi has order of zero on C = 2µi
(du ^ dv)=wi has order on C = 3 ? µj ? µk
fi 2 Ai iff ord fi >= (µj + µk ? 3)=2
fi 2 Bi, i >= 1 iff ord fi >= µi ? 3 + (µj + µk)=2
f0 2 B0 iff ord f0 >= µi + µj + µk ? 3

Moreover, the numbers (µj + µk)=2 can just be half integers.
This remark gives rise to a simplification of the situation, in the sense that the third and fourth cases give rise (with the exception of the third case with µ1 = 0) to singularities which are not worth their price!

Proposition 5. Given an ordinary singularity of the branch curve D of type (2m1+1; 2m2; 2m3), the adjoint and biadjoint ideals are the same as for a singularity of type (2m1; 2m2; 2m3). In particular, Ø and K2 drop exactly by the same amounts as for a singularity of type (2m1; 2m2; 2m3). Similarly an ordinary singularity of the branch curve D of type (2m1; 2m2 + 1; 2m3 + 1) has the same adjoint and biadjoint ideals as a singularity of type (2m1 ? 1; 2m2 + 1; 2m3 + 1). In the latter case, when m1 = 0, this means that the drops of Ø and K2 are given by the same formula as in the first two cases, but with multiplicities (?1; µ2; µ3), i. e.,
Ø drops by (1=4)[(µ2 + µ3 ? 1)2 ? (µ2 + µ3) + 2 ? µ2µ3],
while K2 drops by [(µ2 + µ3 ? 3)2.

Proof. Note that in both cases we replace a triple (µ1; µ2; µ3) where the µi's do not have the same parity by a new triple where the µi's have the same parity, but the sum (µ1 + µ2 + µ3) goes down by 1.

Therefore, the ideal B0 does not change. We also notice that r = (µ1 + µ2 + µ3) ? 3 is always at least ?1, since we can disregard the case (1; 0; 0) (it yields no singular point) ; so the calculation that r(r +1)=2 conditions are imposed is correct (for r = ?2 we would obtain a positive number instead of 0).

The same reasoning yields that the ideals Ai, for i = 2; 3, and the ideal B1 do not change. In the case of A1, we may observe that (µ2 + µ3) is an even integer, whence an integer d is at least (µ2 + µ3 ? 3)=2 iff it is at least (µ2 + µ3 ? 2)=2. A completely similar argument works for Bi when i = 2; 3.

We can summarize our results as follows:
µ1; µ2; µ3 of the same parity

K2 drops by ((P3i=1 µi) ? 2)2

Ø drops by ((1=2)P3i=1 µi)2 ? (1=2) P3i=1 µi ? (1=4)ö2(µ)
µ1; µ2; µ3 of different parity

K2 drops by ((P3i=1 µi) ? 3)2

Ø drops by (1=4)[(P3i=1 µi)2 ? 3(P3i=1 µi) ? ö2(µ) ? µ1 + 3].

For the effective construction of algebraic surfaces, it is convenient to have at hand a numerical table of the numerical changes of the invariants Ø and K2 for singularities of small multiplicities ; clearly, in view of proposition 5, we will not consider the triples of three strictly positive multiplicities having different parities, and also those (like (0; 1; 2)) which do not affect any of the two invariants. We will moreover restrict ourselves to the cases where the drop of K2 is at most 25 (recall that this drop is a perfect square): this means, in terms of the sum of the

multiplicities, (P3i=1 µi) <= 7, resp. <= 8 in the case where µ1 = 0; µ2; µ3 are odd.

14 FABRIZIO CATANESE

Singularity type drop of K2 drop of Ø
(1, 1, 1) K2 # 1 Ø = same,
(0, 1, 3) K2 # 1 Ø # 1, (2, 2, 0) K2 # 4 Ø # 1, (0, 0, 4) K2 # 4 Ø # 2, (3, 1, 1) K2 # 9 Ø # 2, (3, 3, 0) K2 # 9 Ø # 3, (0, 1, 5) K2 # 9 Ø # 4, (2, 2, 2) K2 # 16 Ø # 3, (4, 2, 0) K2 # 16 Ø # 4, (0, 0, 6) K2 # 16 Ø # 6, (3, 3, 1) K2 # 25 Ø # 5, (0, 3, 5) K2 # 25 Ø # 7, (0, 1, 7) K2 # 25 Ø # 9,

Remark 6. Of course, one can also search for non ordinary singularities. For instance, if we have that the multiplicities of the Di's are (1; 1; 1) and they have a common tangent which gives an ordinary singularity of type (1; 1; 1) on the blow up, then K2 # 2, Ø = same.

If instead the multiplicities are (2; 1; 1) and D2; D3 have a common tangent while D has at least two distinct tangents, then K2 # 1, Ø = same.

Finally, if the multiplicities are (2; 1; 1), all the tangents are equal, then K2 # 8 Ø # 2.

This means that the first singularity is a superposition of two ordinary singularities of type (1; 1; 1), the second is a useless degeneration of this singularity, the third could be convenient.

We end this section with a useful

Lemma 6. Assume that we have a singular bidouble cover f : Y ! X where pg(X) = 0. Let P be a point of type (1; 0; 3) for D: then, on the desingularization S of Y , Ö1 has no base points on the elliptic curve E which is the inverse image of P in S, if the following conditions are satisfied

ffl there is a section f1 of A1!X(L1) not vanishing at P , and
ffl there is a section f2 of A2!X(L2) vanishing simply at P but with tangent direction different from the one of the divisor D1 = div(x1).

If instead Q is a point of type (1; 1; 1) for D, and A is the rational curve lying over Q in the resolution S of Y , Ö1 has no base points on A if there are at least two sections fi; fj of Ai!X(Li), resp. Aj!X(Lj) not vanishing at Q.

Proof. Note that E2 = ?1 and KSE = 1; observe moreover that we are looking on S at the linear system generated by the divisors of the 2?forms fi(du ^ dv)=wi.

In the case of the point P of type (1; 0; 3), f1(du^dv)=w1, if f1 does not vanish at P , yields a divisor which does not contain E , but contains the divisor R1, thus this divisor meets E in the two points corresponding to the inverse image of the tangent direction of D1 = div(x1). Whereas f2(du ^ dv)=w2 if f2 vanishes simply at P , yields also a divisor which does not contain E , but contains the divisor R2, and the inverse image F2 of the proper transform of div(f2). We are done since R1\R2\E = ;, and moreover R1\F2\E = ; under our assumption on the tangent of div(f2).

SINGULAR BIDOUBLE COVERS 15

In the case of the point Q of type (1; 1; 1), we use the fact that Ri \ Rj \ A = ;.

4. Simple canonical surfaces

In this section we shall give some existence results for simple canonical surfaces with pg = 4 and K2 >= 11. Our results are not yet extensive, but we present the basic methods in order to obtain a large segment of values for high K2. There is the following strategy principle which is best illustrated in the next theorem. Since the (1; 1; 1) points do not give adjoint conditions, assume that we get a connected family of singular bidouble covers yielding equisingular deformations of all the singularities of the branch curves, with the exception of a certain number m of (1; 1; 1) points each deforming to a triple of double points: if the canonical map is birational for the resolution of a singular cover Y o with m (1; 1; 1) points, then it is still birational for all the surfaces which occur as resolutions of some sufficiently small deformation of Y o.

Theorem 7. There do exist algebraic surfaces with geometric genus pg = 4, K2 = 11; 12:::17, q = 0, and with birational canonical map Ö1.

Proof. Let, once again, X = Q = P1 ? P1: and take three divisors D1; D2; D3 of respective bidegrees (1; 3)(3; 1)(3; 3). Then the respective bidegrees of L1; L2; L3 are (3; 2)(2; 3)(2; 2), thus for a smooth such bidouble cover one would have K2 = 18, pg = 5 and q = 0. We consider the case where D3 has a triple point P , P 2 D1, (whence P gives then a (1; 0; 3) point of D) and moreover we have 0 <= m <= 6 (1; 1; 1) points of D. If we take coordinates (u0; u1)(t0; t1) on X = Q = P1 ? P1, such that P is the point u1 = t1 = 0, we have that the canonical map of the minimal resolution S of Y is given by (y1u0; y1u1; y2t1; y3), since the adjoint ideal A2 is the maximal ideal of the point P . We may also rewrite (y1u0; y1u1; y2t1; y3) = (w3u0; w3u1; x2t1; w1) and thus notice that Ö1 is induced by a family of rational maps defined on the vector bundle where Y is embedded. Clearly these rational maps have as base points on Y exactly the point P 0 corresponding to the (1; 0; 3) point P (y1 = t1 = y3 = 0) and the (1; 1; 1) points where y1 = y2 = y3 = 0.

We observe preliminarly that Ö1 has no base points on S. This follows in fact from lemma 6, which ensures that there are no base points on the exceptional curves (for the other points, it is evident).

Let us then consider the following family of branch divisors, given, in affine coordinates t = t1=t0; u = u1=u0 as follows:
D3 = div[u3(1 ? t2) + u2(b1t + b2t2 + b3t3) + u(c2t2 + c3t3) ? (b3 + c3)t3],
so that D3 has an ordinary triple point at P with tangent distinct from t = 0, then
D2 = div(t1u0(u1 ? u0)(u1 + u0)
so that D2 intersects D3 at P and at 9 points
(u = 1; t = 1), (u = 1; t = 1), (u = 1; t = ?1), (u = 1; t = 1), (u = 1; t = ff), (u = 1; t = fi),

(u = ?1; t = fl), (u = ?1; t = ffi), (u = ?1; t = ?), such that (for general choice of the parameters, as it is easy to verify) the seven numbers
1; ?1; ff; fi; fl; ffi; ? are all distinct.
We consider now the divisors D2; D3 as fixed, and we consider a family of divisors D1 given as follows
D1 = div[(t0 ? >=t1)(µt0 ? t1)f(u; t)]

16 FABRIZIO CATANESE

where f(u; t) is a general equation of bidegree (1; 1) (corresponding to the choice of a plane in 3-space).

Clearly, for >= = 0; µ = ff there is an appropriate choice fo of f such that we get 6 points of type (1; 1; 1) (i. e., we take a plane passing through three of the remaining six points). Moreover, in a neighbourhood of this point in the parameter space for (>=; µ; f) we obtain bidouble covers with any number 0 <= m <= 6 of (1; 1; 1) points. There remains therefore, since birationality is an open property, to show that the canonical map for the minimal resolution So of Y o is birational.

This last fact is indeed obvious since we have seen that there are no base points and K2So = 11, a prime number.

Remark 7. The reader might have noticed that indeed the previous construction could easily be pushed to yield also canonical surfaces with 8 <= K2 <= 10. We did not treat this case since 8; 9; 10 are not prime numbers, thus the proof would have been more complicated (but it can be done along the lines of proposition 4 and Theorem 9). A natural question, however, would be whether these surfaces belong to the previously known families of canonical surfaces in those degrees, for instance to the closure of the ones considered by Ciliberto in [Cil1].

The next theorem goes entirely along the lines of the foregoing one, by a specialization argument and by the fact that 17 is again a prime number.

Theorem 8. There do exist algebraic surfaces with geometric genus pg = 4; q = 0 and 17 <= K2 <= 22, having a birational canonical map Ö1.

Proof. We consider as base surface X = Q = P1 ? P1 and bidouble covers Y with the Dj's of respective bidegrees (2; 3); (2; 3); (4; 1). Thus the Li's have bidegrees (3; 2); (3; 2); (2; 3) and if Y were smooth, it would have K2Y = 24 and pg = 6, by the formulae we have previously given.

We shall impose two points of respective types (3; 1; 0), (1; 3; 0) (thus they both lie in D1 \ D2, and moreover 0 <= m <= 5 points of type (1; 1; 1) on D.
We first fix the choice of D1; D2.
We set D1 to be the sum of the line ft1 = 0g and of two irreducible plane sections ß1; ß2 of the quadric Q. ß1 is chosen to pass through the points (0; 0); (1; 1), ß2 is chosen to pass through the points (0; 0); (u2; 1).

Similarly, we let D2 be the sum of the line ft0 = 0g and of two irreducible plane sections ß3; ß4 of Q, where ß3 passes through the points (0; 0); (1; 1), ß4 passes through (1; 1); (u1; 0).

It is easy to verify that, for the 6 points given by (u1; 0); (u2; 1) and by the four intersection points ßi \ ßj, the values of the first, resp. second, coordinate are all distinct.

It suffices therefore to take a divisor D3 of type (4; 1) consisting of 5 lines passing through any required number 0 <= m <= 5 of these points.

For m = 5 we get a resolution S which has K2S = 17, and for which, by lemma 6 again, the canonical map Ö1 is base point free. Therefore Ö1 is then birational, and the theorem follows by the usual specialization argument.

SINGULAR BIDOUBLE COVERS 17

We continue by producing an example of a canonical surface in P3 with very high degree, K2 = 28, for which we give a direct proof of birationality of the canonical map.

Theorem 9. There do exist algebraic surfaces with geometric genus pg = 4 and K2 = 28; q = 0, having a birational canonical map Ö1.

Proof. We consider as base surface X = P1 ?P1 and bidouble covers Y with the Dj's of respective bidegrees (4; 2); (2; 4); (2; 2). Thus the Li's have bidegrees (2; 3); (3; 2); (3; 3) and if Y were smooth, it would have K2Y = 32 and pg = 8. We shall impose four points of type (0; 1; 3), one on D1 \ D3, one on D2 \ D3, two on D1 \ D2: thus we get surfaces with K2 = 28 and pg = 4.

Choose coordinates (x0; x1); (y0; y1) on X = P1 ? P1, and let z1; z2; z3 be then the equations of the ramification divisors on Y (so, div(z2i ) = Di).

We let D1 = div(y0x1(x1 ? x0)[x1y1(x1 ? x0) + y0x0b1(x)]), where b1(x) is a linear function in (x0; x1).

Similarly, we let D2 = div(x0y1(y1 ? y0)[x1y1(y1 ? y0) + x0y0n1(y)], where n1(y) is a linear function in (y0; y1), finally we let D3 = div(G(x; y)) where the polynomial G of bidegree (2; 2) vanishes on the point P1;1 which is a triple point for D1, and on the point P1;1 which is a triple point for D2. In particular, the equation of G can be written as follows:

G(x; y) = x0y0ff(x; y) + (x1 ? x0)(y1 ? y0)fl(x; y), where ff; fl are homogeneous of bidegree (2; 2).

The reader will notice that P1;0 and P0;1 are points of respective multiplicities (1; 3; 0); (3; 1; 0).
Then the canonical map of the resolution S of Y is given by:
(z1(y1 ? y0); z2(x1 ? x0); z3x0y0; z3x1y1).
The proof proceeds now as for proposition 4: we consider the pencil L>= on X = P1 ? P1 given by L>= = div(>=0x0y0 + >=1x1y1).

Let moreover C>= be the inverse image of L>= on S. The pencil L>= is a pencil of conics on the quadric, with base points P1;0 and P0;1. The divisors cut by the branch loci on the general L>= are, respectively

3P0;1 + P1;0 + B0, P0;1 + 3P1;0 + N 0, G0, where B0; N 0 have degree 2, but G0 has degree 4.
The conclusion is that C>= has genus = 5, indeed C2>= = 0, C>=KS = 8.
Since the canonical system has no fixed part, there are a priori only three possibilities:

i) the restriction Ö of Ö1 to a general C>= is birational, whence Ö1 is birational, or
ii) Ö has degree 2, whence deg Ö1 = 2
iii) Ö is a degree 4 covering of a conic Q>=, whence deg Ö1 = 4.
Case ii) leads to a contradiction as follows.
There is then an involution ö of which the canonical map is composed, thus ö is central in Aut(S).

Since Ö1 separates the four points of a general orbit of (Z=2)2, we get an action

of (Z=2)3 on S, whence an action of (Z=2) on X = P1 ? P1. Such involution ø on X must leave the three branch divisors invariant, since they are the images of the fixed point sets of the non trivial involutions in (Z=2)2. Moreover, ø is biregular on the blow up of X in the four essential singularities of the branch divisor (since it comes from an involution which is biregular on S). Let us denote by Z such a blow

18 FABRIZIO CATANESE

up: it is a del Pezzo surface, and the anticanonical system, which has projective dimension equal to 4, is free from base points and contracts exactly two divisors, ?x and ?y, ?x being the proper transform of the line through P1;1 and P1;0, while ?y is the proper transform of the line through P1;1 and P0;1.
We want to derive a contradiction by showing that ø is already biregular on X. To this purpose we look then at the lines on Z (for the anticanonical mapping), i. e., the irreducible curves C with CKZ = ?1 = C2 (since these are preserved under any automorphism). Among the lines are the exceptional divisors Ei of the blow up; the other lines (and also the curves contracted by j ? KZ j) have class

a?x + b?y + P4i=1(2 ? ri)Ei, and since they are contained in a two dimensional system of hyperplanes, it follows that a; b <= 1, and clearly 0 <= ri.
The diophantine equations
P4i=1 r2i = 1 + 2ab <= 3
P4i=1 ri = 2(a + b) ? 1 <= 3

show that ri <= 1 and that the only other lines on Z, beyond the four we already considered, are the proper transforms of the four lines on X passing exactly through one of the blown up points. In fact, the solutions with a = b = 1 and three ri = 1 should be discarded, since they correspond to planes on the quadric X passing through three of the blown up points, whence they give reducible curves.

The conclusion is now that ø preserves the four lines which meet ?x; ?y, that is, the four exceptional curves Ei of the blow up. After composing with an automorphism of X, we get a ø 0 with ø 0(?x) = ?x, ø 0(?y) = ?y and ø 0(Ei) = Ei.
But then it follows that ø is an automorphism of X.
Since the branch divisors are left invariant, it follows easily that ø does not exchange the two factors, leaves the four essential singularities fixed, whence it is the identity, a contradiction.
Let us now consider case iii), where Ö is a degree 4 covering of a conic Q>=.
Then Q>= is irreducible, and its equation is an eigenvector for the Galois group, therefore it must have the form
P3i=1 ai(>=)t2i

where Ö is given by (t1; t2; t3) with the ti's eigenvectors for the Galois group.
In fact, if the equation were not invariant, it would have the form titj , against the irreducibility of Q>=.
We obtain therefore a rational map
>= ! (a1(>=); a2(>=); a3)(>=) such that
a1(>=)G1(y1 ? y0)2 + a2(>=)G2(x1 ? x0)2 + a3(>=)G3 = 0,
where Di = div(Gi), and G3 equals our previous G.
Since (>=0; >=1) = (x1y1; x0y0),
we may get rid of denominators in the rational functions and substitute. We obtain an equation of the form

A1(x1y1; x0y0)G1(y1 ? y0)2 + A2(x1y1; x0y0)G2(x1 ? x0)2
+A3(x1y1; x0y0)G3 = 0,
where degA1 = degA2 = m; degA3 = m+2 and m is minimalwith this property. We observe now that x0 divides G2, therefore we obtain:
A1G1(y1 ? y0)2 = ?A3G3 (mod(x0)).
Now, we can vary G = G3 in a linear system so that the intersection with div(x0) moves: we get a contradiction unless x0 divides both A1; A3.

SINGULAR BIDOUBLE COVERS 19

By the special form of these polynomials, it follows that indeed x0y0 divides both A1; A3.

But then y0 divides A2, and by the same principle x0y0 divides A2. The conclusion is that x0y0 divides A1; A2; A3 against the minimality of m, and we are done.

We have given above a geometric proof, but one can indeed mimic the idea of proof of Theorem 6 (replacing the prime number 11 by 23) and obtain also the following result:

Theorem 10. There do exist algebraic surfaces with geometric genus pg = 4, 23 <= K2 <= 28, having a birational canonical map Ö1.

Proof. We consider everything, X = P1 ? P1, the Dj's of bidegrees
(4; 2); (2; 4); (2; 2), the Li's of bidegrees (2; 3); (3; 2); (3; 3), as in the previous theorem, except that we specialize the branch curves so as to obtain, beyond the given four points of type (0; 1; 3), also r points of type (1; 1; 1).

We choose coordinates (x0; x1); (y0; y1) on X = P1 ? P1 exactly as before, and calculate the further intersection points of
D1 = div(y0x1(x1 ? x0)[x1y1(x1 ? x0) + y0x0b1(x)] with
D2 = div(x0y1(y1 ? y0)[x1y1(y1 ? y0) + x0y0n1(y)].
Note that n1(y), b1(x) are linear functions: therefore our points are, beyond
P1;1, P1;1, P0;1, P1;0, P0;0, the points P0;1, Po;n, P1;?1, P1;?2, P1;0, Pb;0, Pµ1;1, Pµ2;1,
and the two points which are solutions of the system
n1(y)(x1 ? x0) = b1(x)(y1 ? y0),
[x1y1(x1 ? x0) + y0x0b1(x)] = [x1y1(y1 ? y0) + x0y0n1(y)] = 0.
Our notation above is aimed to be almost self explanatory: the point b is the root of b1(x), n is the root of n1(y), ?1; ?2 are the roots of y21 ? y1y0 + y0n1(y), while µ1; µ2 are the roots of x1(x1 ? x0) + x0b1(x).
We choose D3 = div(G(x; y)), where
G(x; y) = x0y0ff(x; y) + (x1 ? x0)(y1 ? y0)fl(x; y), and in order to let it pass through

0 <= r <= 5 of these points, we impose one after the other the following specialization conditions:
I) P1;1 2 D3:
this amounts to fl(x; y) = x0fl2(y) + y0fl1(x)
II) P0;1 2 D3:
this amounts to ff(x; y) = x1ff2(y) + (y1 ? y0)ff1(x)
III) P1;0 2 D3:
this amounts to ff(x; y) = a1x1y1 + a2(y1 ? y0)(x1 ? x0), since P0;1 and P1;0 form the complete intersection x1y1 = (y1 ? y0)(x1 ? x0) = 0.
IV) Pb;0 2 D3:
this amounts to b being a root of (set y0 = 1; y1 = 0) the polynomial
(a2 + fl2(1; 0))x0 + fl1(x). This condition is easily solved by setting
b1(x) = (a2 + fl2(1; 0))x0 + fl1(x).
V) P0;n 2 D3:
this amounts to n being a root of (set x0 = 1; x1 = 0) the polynomial
(ff1(1; 0) ? fl1(1; 0))y0 ? fl2(y). This condition is easily solved by setting
n1(y) = (ff1(1; 0) ? fl1(1; 0))y0 ? fl2(y).

20 FABRIZIO CATANESE

At this point it is easy to verify that once we have imposed further r = 5 points of type (1; 1; 1), namely (P1;1; P0;1; P1;0; Pb;0; P0;n), for general choice of the remaining parameters we get a divisor D with ordinary singularities.

It follows as in theorem 6 that we get a surface So with jKSo j free from base points, and with K2 = 23.

Therefore our surface So has a birational canonical morphism, and the same holds for all the surfaces S with K2 = 24; :::28 which are the minimal resolutions of singular bidouble covers in a suitable neighbourhood of Y o.

5. Burniat's surfaces and Burniat type surfaces

This short section is dedicated to some constructions of surfaces of general type with pg = 0, or, when the singular points do not impose independent conditions, to surfaces with pg = q = 1 with K2 = 3; 4; 5 and Albanese map equal to a pencil of genus 2 curves. We call all these Burniat type surfaces, since in the book by Barth Peters and van de Ven [B-P-V] the name of Burniat surfaces refers precisely to certain surfaces S of general type with pg = 0 and K2 = 6 ? m; 0 <= m <= 4 constructed by Paul Burniat as resolutions of singular bidouble covers of the plane (cf. [Bur2]). A more modern exposition was later given by Peters in [Pet], but unfortunately only based on the theory of double coverings. Let us see then how the Burniat surfaces occur rather canonically in the framework of bidouble covers.

Example 6. Consider X = P2 and consider a bidouble cover Y of type (3; 3; 3) (in this cases the degrees of the Li's and Dj's are all equal to 3). If Y is smooth, then it has K2Y = 9 and pg = 3, by our standard formulae.

We would like to make pg drop by 3, but K2 by at most 8. A quick inspection of the table shows that this is only possible if we impose exactly three (0; 1; 3) points and then we can try to further impose m (1; 1; 1) points, where m should be at most 5. But if a cubic curve C has a triple point, then it is a union of three lines. Therefore must be D1 = A1 + A2 + A3, where the three lines go through a point a, and similarly D2 = B1+B2+B3 consists of three lines through b, D3 = C1+C2+C3 consists of three lines through c.

To have the desired three (0; 1; 3) points we must moreover assume that A1 joins a and b, and similarly B1 joins b and c, C1 joins c and a. For special positions of the configuration one can obtain that the two lines B2; B3 go through any number m of the four points of intersection Ai \ Cj ; i; j = 2; 3.

Therefore Burniat obtains in this way surfaces S with pg = 0 and K2 = 6 ? m; 0 <= m <= 4.

Example 7. Consider X = P2 and consider a bidouble cover Y with the Li's of degrees (3; 2; 4) and the Dj's of degrees (3; 5; 1). If Y is smooth, then it has K2Y = 9 and pg = 4, by our standard formulae.

We make pg and K2 drop by 4 by imposing four (0; 1; 3) points. This is only possible if the quintic curve D2 = B has three triple points (which are then non collinear, since B is reduced), and D1 = A consists of three lines A = A1 + A2 + A3 through a point P . But then B splits into a triangle B1 + B2 + B3 and a conic B0 through the vertices of the triangle. We let the point P belong to B0, the lines A2; A3 go through two of the vertices of the triangle, and D3 go through the third vertex. According to whether D3 does or does not go through the intersection point B0\A1 we get a surface S with S with K2 = 4, respectively K2 = 5. We get pg = 0

SINGULAR BIDOUBLE COVERS 21

(thus also q = 0) since the canonical system is given by the lines through the points of type (0; 1; 3) lying in B \ A and the constants vanishing on the (0; 3; 1) point B \ D3.

K2 = 4; pg = 0:

The following examples yield surfaces with pg = q = 1, K2 = 3; 4; 5 and genus g = 2 of the fibres of the Albanese pencil, thus answering Problem 5. 10 raised in [Ca-Ci1]. In fact, surfaces with pg = q = 1, were classified in [Cat1], [Ca-Ci1], [Ca-Ci2] for the values K2 <= 3, and only one example, due to Xiao ([Xiao1]), was known for higher K2: namely, K2 = 4 and genus g = 2 of the Albanese fibres (notice that under the last assumption holds: K2 <= 6).

Example 8. Here again we consider X = P2 and a bidouble cover Y of the same degrees as in the previous example, namely (3; 2; 4) for the Li's, and (3; 5; 1) for the Dj 's.

Again we impose four points of type (0; 1; 3), B splits into a triangle B1+B2+B3 and a conic B0 through the vertices of the triangle, D1 = A consists of three lines A = A1 + A2 + A3 through a point P .
We have two cases:
Case I): P does not belong to B0 and the three lines A1; A2; A3 pass through the three respective vertices of the triangle; finally, D3 is a general line through P . The difference from the previous example is that there is no essential singularity in B \ D3, thus Ø drops to 1, but also pg = 1.

Here, K2 = 5. Moreover, consider the pencil of lines through P : we can decompose the Galois covering as first taking a double cover branched on four lines through P (thus we obtain a ruled surface of irregularity 1), and then taking a double cover branched on the 5? section given by B and on the section coming from the blow up of the point P . Therefore, the above pencil lifts to the Albanese pencil, which is a pencil of genus 2 curves.

22 FABRIZIO CATANESE

Case I: K2 = 5; pg = q = 1:

Case II): P belongs to B0, and the three lines A1; A2; A3 pass through the three respective vertices of the triangle; the line D3 can be chosen to go through zero, one or two of the points of transversal intersection of B with A. Accordingly, we get a surface S with K2 = 5; 4; or3, and again pg = q = 1. The pencil of conics through the four points of type (0; 1; 3) induces as before the Albanese pencil, and the Albanese fibres have genus 2, since they are double covers of those conics branched on 6 points (2 come from the intersection with D3, the other 4 from the four sections coming from the blow ups of the four points).

Case II: K2 = 3; pg = q = 1:

SINGULAR BIDOUBLE COVERS 23

Example 9. Here we consider instead as base surface X = P1 ? P1 and consider a bidouble cover Y with the Li's and the Dj's of bidegrees (2; 2). If Y is smooth, then it has K2Y = 8 and pg = 3, by our standard formulae.

We make pg and K2 drop by 3 by imposing three points of type (0; 1; 3). We are moreover able to impose up to three points of type (1; 1; 1).

The way to do so is to let any divisor Di be a union of a horizontal line Hi, a vertical line Vi, and a curve Ci of type (1; 1) passing through Pi = Vi \ Hi; notice that Ci is a plane section of the quadric surface Q in P3, hence we can let these curves pass through any three non collinear points (that is, lying neither on a vertical line V nor on a horizontal line H).

Case I): let Ci go through Pi; Pi+1 (here 1; 2; 3 are viewed as rest classes modulo 3), and possibly through a fourth point P . Accordingly, we get three points of type (0; 1; 3) and zero or one point of type (1; 1; 1).
Thus K2 = 5 or 4, pg = 0.
Case II): C1 goes through P1; P2; P3, C2 goes through P1; P2, C3 goes through P3 and the two points V1 \ H2, V2 \ H1. Here, we get three points of type (0; 1; 3) and two points of type (1; 1; 1).
Thus K2 = 3, pg = 0.

There remains to calculate the fundamental groups for the above surfaces (these are known for the classical Burniat surfaces, cf. [B-P-V] or [Pet]). In particular, is the last example essentially different from the one given by Burniat?

Acknowledgements. I would like to thank C. Ciliberto for pointing out Burniat's work to my attention, and M. Lehn for helping with the pictures of the Burniat type surfaces.

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Mathematisches Institut der Georg-August-Universität Göttingen, Bunsenstr. 3/5, D-37073 Göttingen, Germany
E-mail address: catanese@cfgauss.uni-math.gwdg.de