Extending symmetric determinantal

quartic surfaces

Stephen Coughlan ∗

Abstract

We give an explicit construction for the extension of a symmetric determinantalquartic K3 surface to a Fano 6-fold. Remarkably, the moduli of the 6-fold extensionare in one-to-one correspondence with the moduli of the quartic surface. As aconsequence, we determine a 16-parameter family of surfaces of general type withpg = 1, q = 0 and K2 = 2 as weighted complete intersections inside Fano 6-folds.

1 Introduction

A symmetric determinantal quartic surface T4 ⊂ P3 is defined by the vanishing of thedeterminant of a symmetric 4 × 4 matrix with linear entries, and it is well known thatT4 has ten A1 singularities generically. There is a Weil divisor A on T4 such thatOT4(2A) = OT4(1), and by taking Proj of the graded ring

R(T4, A) =⊕n≥0

H0 (T4,OT4(nA)) ,

we obtain a model T of T4 in weighted projective space P(24, 34) defined by 14 equations,and the ten singular points are now induced by the ambient space. In this article westudy extensions of the graded ring R(T,A).

In fact, T is the “elephant” hyperplane section of the anticanonical model of a Fano3-fold W ⊂ P(1, 24, 34) with ten singular points. In other words, T is the hyperplanesection in W of weighted degree 1

T = H ∩W ⊂ P(1, 24, 34).

In terms of graded rings, there is an element a ∈ H0(W,O(−KW )) of degree 1 inR(W,−KW ) =

⊕n≥0H

0 (W,OW (−nKW )), whose vanishing defines H and so

R(T,A) = R(W,−KW )/(a).

∗Department of Mathematics and Statistics, Lederle Graduate Research Tower, University of Mas-sachusetts, Amherst, MA 01003-9305, coughlan@math.umass.edu

1

We call W an extension of T . This extension process can be iterated by incorporatingmore variables b, c, d of degree 1 into the ring. We obtain a tower of extensions

T ⊂W 3 ⊂W 4 ⊂W 5 ⊂W 6 ⊂ P(14, 24, 34),

where each Wn is a Fano n-fold with O(−KW ) = O(2− n).The 6-dimensional extension W 6 is the most important for our purposes. A priori,

there may be many different extensions of a given T . We prove that there is only one6-dimensional extension. This gives rise to a one-to-one correspondence between themoduli of the K3 surface T and the moduli of the 6-fold W 6.

For brevity, we write 12 point (or 1

2(1, . . . , 1) point) to mean the image of the originin the quotient of Cn by Z/2, acting by −1 on all the coordinates. For example, a 1

2(1, 1)point is an A1 surface singularity.

Main Theorem 1.1 Let T ⊂ P(24, 34) be a quasismooth symmetric determinantal K3surface with 10× 1

2 points. Then T determines (and is uniquely determined by) a uniqueextension to a quasismooth Fano 6-fold W ⊂ P(14, 24, 34) with 10 × 1

2 points and suchthat

T = W ∩H1 ∩H2 ∩H3 ∩H4,

where the Hi are linear hyperplanes of the projective space P(14, 24, 34).

Jan Stevens [11], first observed this extension phenomenon in 1993 when calculat-ing the deformation–extension theory for the Klein quartic curve, which has maximalsymmetry group of order 168. This extra symmetry restricts the deformation exten-sion space enough to make the computation viable. Our understanding of the extensionT ⊂ W is still somewhat limited. For instance, it should be possible to give a concretedescription of all the equations of W in terms of the symmetric matrix defining T , andthe new variables a, b, c, d.

We use this family of Fano 6-folds to give a new construction of an important familyof surfaces of general type.

Corollary 1.2 There is a 16-parameter family of topologically simply connected surfacesY of general type with pg = 1, q = 0, K2 = 2, each of which is a complete intersectionof type (1, 1, 1, 2) in a Fano 6-fold W ⊂ P(14, 24, 34) with 10× 1

2 points.

Catanese and Debarre studied surfaces of general type with pg = 1 and K2 = 2 in[3], giving an almost complete classification. The fundamental group of such surfacesis either 0 or Z/2, so the moduli space has (at least) two connected components, ofexpected dimension 16. We give a more detailed review of some of the results of [3] insection 4.1.

Todorov [14] also studied surfaces with pg = 1, K2 = 2 and π1 = Z/2 in connectionwith the Torelli problem. These surfaces have torsion group Z/2, and form a distinctirreducible connected component of the moduli space, which is quite well understood (see[3]). More recently, some examples have been constructed using Q-Gorenstein smoothingtheory in [9].

2

Our 16-parameter family lies in the irreducible component of the moduli space de-scribed in [3], but the method of construction circumvents some of the difficulties causedby the (RC) condition (see section 4.1 for details). We hope to use our family to provethat the moduli space of simply connected surfaces of general type with pg = 1 andK2 = 2 is irreducible, but so far we have not been able to do this. There is also a hyper-elliptic degeneration of this construction, which has applications to Godeaux surfaceswith fundamental group Z/2. This is treated in [7].

In Section 2 we recall some well known properties of symmetric determinantal quarticsurfaces, and explain a method for constructing such surfaces using projection from asingular point. We prove the Main Theorem in Section 3 by applying the projectionmethod to Fano 6-folds. The Corollary is proved in Section 4, and uses some beautifulresults of Takayama on Fano varieties [12], [13]. Since [3], [12], [13] use C as the basefield, we do the same. However, the Main Theorem is valid over any algebraically closedfield of characteristic zero.

Remark 1.3 The arguments in Section 3 involve some complicated calculations, be-cause we want to give explicit equations for the 6-fold W . However, a posteriori, manyof the results can be checked simply by verifying the stated formulae. Thus one canview the proofs provided as a derivation of the equations rather than essential to theargument. We remind the reader of this at the relevant points. All stated equationshave also been checked using computer algebra software.

2 Symmetric determinantal varieties

In this paragraph we review some constructions and properties of symmetric determi-nantal quartic surfaces. Section 2.1 describes symmetric determinantal quartic surfacesin terms of linear systems of quadrics in P3, following the treatment of [15]. The subse-quent section gives a model of the surface in weighted projective space, and finally wegive the projection construction in Section 2.3. This last construction is important forthe proof of the Main Theorem.

2.1 Webs of quadrics

In this section we follow [15]. Let P9 = PH0(P3,O(2)) be the space of quadrics in P3.This is the space of 4 × 4 symmetric matrices up to scalar multiplication. There is anatural stratification of this space by rank:

P9 ⊃ V 84 ⊃ V 6

10 ⊃ V 38 ,

where V nd denotes a variety of dimension n and degree d. For example, V 8

4 is a hypersur-face of degree 4 in P9, which corresponds to quadrics in P3 of rank ≤ 3, or equivalently4 × 4 symmetric matrices whose determinant vanishes. Similarly V 6

10 (respectively V 38 )

is the locus of quadrics of rank ≤ 2 (resp. ≤ 1).Now take a webM of quadrics in P3, i.e.M is a linear system of projective dimension

3 inside PH0(P3,O(2)). Choose coordinates y1, . . . , y4 for M, so that M determines a

3

4 × 4 symmetric matrix M with linear entries in yi. Let T = M ∩ V 84 be the locus

of quadrics of rank ≤ 3 in M. If M intersects V 84 transversally then T is a quartic

hypersurface in P3 defined by the vanishing of the determinant of M . The singular locusof T is the intersection of M with V 6

10, which corresponds to those quadrics in M ofrank ≤ 2. Assuming this intersection is also transverse, T has ten isolated nodes, asexpected. These conditions on M are reformulated in the following proposition, provedin [5], which enables us to define a symmetric determinantal quartic surface:

Proposition 2.1 [5, Prop. 2.1.2] Suppose that M is a basepoint free web of quadricsin P3. Moreover, assume that for any quadric q in M∩V 6

10, there is no other quadric q′

in M with sing(q) ⊂ q′. Then T ⊂ P3 is a quartic surface with ten isolated nodes. Wecall T ⊂ P3 a symmetric determinantal quartic surface.

Since the hypotheses of the proposition are open conditions onM, the parameter spacefor such webs of quadrics forms an open subset of the Grassmannian of projective 3-spaces in P(H0(P3,O(2))). Thus after dividing by coordinate changes we may identifythe moduli space of symmetric determinantal quartic surfaces with an open subset ofGr(4, 10)/PGL(4), which is 9-dimensional.

2.2 A model for T in weighted projective space

We abbreviate the weighted projective space P(2, 2, 2, 2, 3, 3, 3, 3) to P(24, 34). Similarnotation is used throughout this article. In this section we prove the following propo-sition, which is a special case of well known results on determinantal varieties. See forexample, [1] and [2].

Proposition 2.2 If T4 ⊂ P3 is a symmetric determinantal quartic, then there is anineffective Weil divisor class A on T with OT (2A) = OT (1).

Remark 2.3 This should be viewed as analogous to the theory of ineffective theta char-acteristics on curves. Let D4 ⊂ P2 be a plane quartic curve, and let A be a divisor classwith 2A = KD and h0(A) = 0. There are 36 such A and all of them are combinationsβj − βk + βl, where βi are bitangents to D. A choice of A corresponds to a symmetricdeterminantal representation of D.

Proof Let M be the 4 × 4 matrix associated to the symmetric determinantal quarticT4 ⊂ P3. Then M determines the following short exact sequence

0← A←4⊕i=1

OP3M←−−

4⊕i=1

OP3(−1)← 0, (1)

where A := cokerM . Note that A is supported on the locus where M is not invertible,which is T4. Moreover, since rankM = 3 on the nonsingular part of T4, we see that Ais invertible outside the nodes. Hence A is a Weil divisorial sheaf on T4.

4

Now applying HomOP3(−,OP3(−1)) to (1) gives

0← Ext1(A,OP3(−1))←4⊕i=1

OP3

tM←−−4⊕i=1

OP3(−1)← 0, (2)

and by Grothendieck–Serre duality, we have Ext1(A,OP3(−1)) ∼= A∨(3). Thus A =A∨(3), because M = tM . We define A to be the Weil divisor associated to A(−1) =OT (A). Clearly OT (A)[2] = OT (1) and we can calculate h0(OT (A)) = 0 from the longexact sequence associated to (1) twisted by −1. �

Remark 2.4 Let σ : T̃ → T be the resolution of singularities of T , with ten disjoint −2curves Ei. Denote by H (respectively A′) the proper transform of a hyperplane section(resp. A), on T4 ⊂ P3. Then 2A′ = H +

∑Ei, so the ten nodes form a weakly even set,

in the terminology of [2].

Now twist the short exact sequence (1) by −1 to obtain

0← OT (A)(zi)←−− 4OP2(−1)

M←− 4OP2(−2)←− 0,

where the zi are global sections ofA = OT (3A). We define a new model for the symmetricdeterminantal quartic

T = ProjR(T4, A) ⊂ P(24, 34),

with equations (z1, z2, z3, z4

)M = 0,

3∧i,j

M = zizj , (3)

where∧3i,jM = (−1)i+j detMij , the (i, j)th cofactor of M . See [2] for a proof of this.

The ten nodes of T4 become 12(1, 1) points of the weighted ambient space, defined by the

vanishing of all the zi and therefore all the 3× 3 minors of M .The four diagonal 3 × 3 minors detMii of M generate the linear system of contact

cubics to T . That is, cubic surfaces F3 which touch T along a curve C of degree 6 passingthrough all ten nodes, i.e. T ∩ F3 = 2C. Thus we can view the zi as a choice of localequation for the contact curve C. See [2], [4] for details.

Remark 2.5 By the previous remark, T admits a double cover branched over a hyper-plane and the ten nodes. This is a surface Y of general type with pg = 1, K2 = 2 of type3 in the Catanese–Debarre classification (see Section 4.1). The canonical model of Y isobtained from (3) by adjoining a new variable x of weight 1 satisfying x2 = l(y1, . . . , y4),where l is a linear form defining the branch hyperplane.

5

2.3 A projection construction for T

Let T4 ⊂ P3 be a quartic surface with at least one node. Explicitly, we can choosecoordinates so that the equation of T4 is

p2(y1, y2, y3)y24 + q3(y1, y2, y3)y4 + r4(y1, y2, y3) = 0,

with a node at P = (0, 0, 0, 1). Here p = p2(y1, y2, y3), q = q3(y1, y2, y3), r = r4(y1, y2, y3)are polynomials in y1, y2, y3 of degrees 2, 3, and 4 respectively. Then linear projectiononto the complementary plane with coordinates y1, y2, y3 gives a double covering of P2

branched in the sextic curve q2 − 4pr. The image of P under the projection is the conicp = 0, which touches the branch curve doubly in each of 6 points. We say that the conicis totally tangent to the sextic. We note for future use that the projected surface hasa model in weighted projective space T ′6 ⊂ P(1, 1, 1, 3) defined by z2 = q2 − 4pr, wherey1, y2, y3 are the coordinates of weight 1 and z is of weight 3.

If we further assume that T is a symmetric determinantal surface, then an explicitcalculation (cf. [4]) shows that the branch curve breaks into two distinct cubics. Thesetwo cubics intersect one another transversally to give 9 nodes, and the additional nodefrom the centre of projection makes 10 nodes on T4. We see in Remark 2.8 below thatthe splitting of the branch curve is equivalent to the existence of A from Proposition2.2. See [5] for another, more geometric proof of the splitting.

For the purposes of our calculations, we view the projection as an example of quasi-Gorenstein projection/unprojection. We briefly outline the ideas here, but see [8], [10]for theoretical discussion and other examples. Start with the K3 surface T ⊂ P(24, 34)with 10 × 1

2 orbifold points from Section 2.2. Choose a 12 point P , and let σ : T̃ → T

be the (1, 1)-weighted Kawamata blowup of P . The exceptional locus E ∼= P1 ⊂ T̃ isthe centre for our projection, and the projection map is determined by the linear systemσ∗A − 1

2E on T̃ , where A is the Weil divisor defined in Proposition 2.2. The image ofthis projection is T ′6,6 ⊂ P(23, 32) which is a double cover of P(2, 2, 2) branched in thetwo cubics defined by the relations of weight 6. The image of the exceptional curve E isembedded as a conic which is totally tangent to the branch sextic.

We give a normal form for the matrix defining our symmetric determinantal quartic,which is used in the proof of our main result.

Proposition 2.6 There is a Zariski dense open subset of the moduli space of symmetricdeterminantal quartics such that the matrix defining T can be written in the form

M =

b y4 g 0

a 0 fsym y1 y2

y3

, (4)

where f = y1 + α2y2 + α3y3, g = β1y1 + β2y2 + y3 and a, b are general linear forms iny1, y2, y3.

6

Before proving the proposition, we first derive equations for T ′6,6 ⊂ P(2, 2, 2, 3, 3)starting from the K3 surface T defined by equations (3), assuming that M is in thenormal form (4). Now T has a 1

2 point at P = (0, 0, 0, 1), with local coordinates near Pgiven by z3, z4. Thus if we project away from P we expect to eliminate y4, z3, z4, (cf. [10],example 9.13). Calculating cofactors detM11, detM22 of M we obtain equations:

z21 = F6(y1, y2, y3) = a(y1y3 − y2

2)− y1f2,

z22 = G6(y1, y2, y3) = b(y1y3 − y2

2)− y3g2,

(5)

defining T ′6,6. These are the only equations remaining from (3) that do not involve y4,z3, z4. In particular the cofactor −detM12 for z1z2 involves y4 and so does not survivethe projection. The product F6G6 = q2 − 4pr defines the branch sextic, and the totallytangent conic is defined by y1y3 − y2

2 = 0.

Proof (of proposition 3) The statement of the proposition is equivalent to exhibitinga Zariski dense open subset of the moduli space of projected K3 surfaces T ′ via explicitequations of the form dictated by (5). This effectively reverses the projection mapdescribed in the preceding discussion. Start from

P1 ϕ−→ T ′6,6 ⊂ P(2, 2, 2, 3, 3)

with 9 × 12 points. Write yi, zi for the coordinates on P(2, 2, 2, 3, 3) of weight 2, 3

respectively, and u, v for the coordinates on P1. The image of ϕ in the plane P(2, 2, 2) isa conic which touches the branch curve at exactly six points, three on each component.

After coordinate changes, the embedding of P1 is

ϕ : P1 → P(2, 2, 2, 3, 3)

(u, v) 7→ (u2, uv, v2, u3 + α2u2v + α3uv

2, β1u2v + β2uv

2 + v3). (6)

We note that u is a factor of ϕ∗(z1) and likewise v divides ϕ∗(z2). This assumes thatthere are at least two distinct points of tangency, one on each component of the branchcurve.

The coordinate ring k[P1] is a graded module over k[P2] via ϕ∗, so referring toequation (6), we can write ϕ∗(zi) as:

ϕ∗(z1) = (y1 + α2y2 + α3y3)u = fu

ϕ∗(z2) = (β1y1 + β2y2 + y3)v = gv.

Then by using the module structure to render ϕ∗(z21), ϕ∗(z1z2) and ϕ∗(z2

2) as polynomialsin yi, we see that the image of ϕ is given by the equations

C1 : z21 = y1f

2, (7)

C2 : z1z2 = y2fg, (8)

C3 : z22 = y3g

2, (9)

Q : y1y3 = y22. (10)

7

Note that the choice of representation for the first three equations is only unique modulothe conic Q of equation (10); for example we could have written z2

2 = β21y

21y3 +2β1β2y

32 +

(β22 + 2β1)y2

2y3 + y33 instead.

The projected surface T ′ is given by taking two combinations

C1 + a(y1, y2, y3)Q

C3 + b(y1, y2, y3)Q,(11)

where a, b are linear. Comparing these equations with (5), we see that the correspondingmatrix has normal form (4). There are 9 moduli for this construction: 3 from varyingthe six points of tangency and a further 3 for each of the linear forms a, b. The onlyassumption we have made is that there are at least two distinct points of tangency, oneon each branch curve. Since this is an open condition, the proposition is proved. �

Remark 2.7 1. The divisibility assumptions we have made on ϕ∗(zi) are somewhatartificial, and not strictly necessary. One can obtain similar normal forms withweaker assumptions, for example using only row and column operations. Ourapproach enables us to give a clean statement of the Main Theorem, but we donot believe it is the optimal result of this type.

2. If T is quasismooth then it has normal form (4), but the converse is not true. Thuswe obtain symmetric determinantal quartic surfaces with singularities which areworse than 10×A1. Such surfaces still have extensions as in the conclusion of ourMain Theorem.

Remark 2.8 Suppose T ⊂ P3 is a quartic surface with a node P , and let T ′6 ⊂P(1, 1, 1, 3) be the image of T under the projection from P . Then the branch sextic splitsinto two cubics if and only if there is a Weil divisor class A on T with O(A)[2] = O(1)as in the conclusion of Proposition 2.2. Indeed, suppose the equation of the branchcurve splits as F3(y1, y2, y3)G3(y1, y2, y3). Then we may define T ′6,6 ⊂ P(2, 2, 2, 3, 3) by

equations z21 = F , z2

2 = G, where z1, z2 are the generators of degree 3, and F , G are nowof weighted degree 6.

3 Extending determinantal formats

In this section we treat extensions of symmetric determinantal quartic surfaces, culmi-nating in the proof of the Main Theorem. We use the projection construction for theK3 surface T , which is not as symmetric as the determinantal representation but is verybeautiful in its own way. Unfortunately a consequence of this approach is that we arenot able to completely understand how the symmetric matrix is involved in the exten-sion. The proof involves some complicated calculations, but we only use elementarypolynomial manipulations and linear algebra.

We first give a formal definition of our notion of an extension:

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Definition 3.1 Let X ⊂ PM be a variety whose homogeneous ideal I(X) is generated byfi(xj) for i = 1, . . . ,m, where x0, . . . , xM are coordinates on PM . We say Y ⊂ PM+N isan extension of X if the ideal of Y is generated by some Fl(xj , yk) for l = 1, . . . , n, withthe property that I(X) is the saturation of the ideal generated by Fl(xj , 0). We assumethat each extension variable yi occurs in some Fl, to avoid degenerate situations such aswhen the extension is a cone.

Remark 3.2 Note that when the coordinate ring ofX is Cohen–Macaulay, this is knownas the hyperplane section principle; we may choose m extended equations Fi(xj) withFi(xj , 0) = fi(xj), and the saturation step is not necessary. More generally, the extendedideal has a larger number of generators than the original ideal. We see instances of thisin the calculations which follow.

Given this definition, we are now able to state the Main Theorem.

Main Theorem 3.3 Let T ⊂ P(24, 34) be a quasismooth symmetric determinantal K3surface with 10 × 1

2 points, which has normal form (4). Then T determines (and isuniquely determined by) a unique quasismooth Fano 6-fold W ⊂ P(14, 24, 34) with 10× 1

2points and such that

T = W ∩H1 ∩H2 ∩H3 ∩H4,

where the Hi are hyperplanes of the projective space P(14, 24, 34).

Strategy of proof We start from a putative quasismooth Fano 6-foldW ⊂ P(14, 24, 34)with 10× 1

2 points, which is an extension of a symmetric determinantal quartic surfaceT in normal form as described in Proposition 2.6. Then T has a preferred node P , whichcorresponds to a 1

2 point P̃ in W . The projection from P yields T ′6,6 ⊂ P(23, 32) withequations (11), together with the totally tangent conic, which is the image of ϕ definedin (6) with equations (7)–(10). This is encoded by the bottom row of the diagram.

Wvvn n n

P5 Φ // W ′⋃

⋃ ⋃T

wwo o o o

P1ϕ // T ′

Next consider the projection from P̃ , which has image W ′6,6 ⊂ P(14, 23, 32) and also gives

a map Φ: P5 → W ′. Note that W ′ is an extension of T ′, and since Φ|P1 = ϕ, Im(Φ)is an extension of Im(ϕ). This gives the upper row of the diagram. We prove that Φand W ′ exist and are actually uniquely determined by ϕ and T ′. Thus by reversing theprojection, W exists and is unique. The projection method is what makes the proofpossible, because we can work explicitly with the equations of T ′ and W ′, whereas thoseof W are very complicated.

We begin with some preparatory results on the general form that Φ must take. Defineϕ as in (6) and factor ϕ via ϕ0 : P1 → P(2, 2, 2), the standard parametrisation of theplane conic:

9

Lemma 3.4 Let u, v, a, b, c, d be the coordinates of P5. Then up to automorphisms ofP5 and P(14, 23) and excluding cones, the only extension Φ0 : P5 → P(14, 23) of ϕ0, is

Φ∗0(a) = a, Φ∗0(b) = b, Φ∗0(c) = c, Φ∗0(d) = d,

Φ∗0(y1) = u2 −dv +bd− c2,

Φ∗0(y2) = uv + bu +cv −ad+ bc,

Φ∗0(y3) = v2 − au +ac− b2.(12)

Proof Let f : (C2u,v, 0)→ (C3

y1,y2,y3 , 0) be the map germ (u, v) 7→ (u2, uv, v2). We havesyzygies vf(y1) = uf(y2) and vf(y2) = uf(y3), which can be written in matrix form(

0 −v uv −u 0

),

where the 2× 2 cofactors recover the images f(yi). We deform the entries of this matrixto make it generic: (

a b− v c+ ub+ v c− u d

).

This gives a versal unfolding of f , and the 2 × 2 cofactors give the definition of Φ∗0(yi)from (12). The uniqueness follows from versality and the weighting on the ambientspace. �

Lemma 3.5 The image of Φ0 in P(14, 23) is given by the vanishing of the determinantof the symmetric matrix A

A =

y1 −y2 L1

−y2 y3 L2

L1 L2 L3

, (13)

where the outsized entries are

L1 = by1 + cy2 + dy3

L2 = ay1 + by2 + cy3

L3 = y1y3 − y22 + b2y1 + (2bc− ad)y2 + c2y3.

Proof Write M , R for the coordinate rings of P5, P(14, 23) respectively. Then byequation (12), M is a graded R-module via Φ∗0 generated by 1, u, v with presentation

0←M(v,u,1)←−−−− 2R(−1)⊕R A←− 2R(−3)⊕R(−4).

The map A is given by the matrix (13), which can be chosen to be symmetric by orderingthe basis in a nonstandard way. The image of Φ0 is supported on the locus where A isnot an isomorphism. �

10

Remark 3.6 In fact, A satisfies the rank condition (RC): remove the last row of A,and denote the ij-th minor of A by Aij . Then

A11 = (y3 − ac+ b2)A33 + a(−A23)

−A12 = (y2 + ad− bc)A33 − b(−A23)− cA13

A22 = (y1 − bd+ c2)A33 + dA13.

Hence following [3], we recover the algebra structure on M , which in turn is equivalentto the definition of Φ0:

v2 = (y3 − ac+ b2) · 1 +a · uuv = (y2 + ad− bc) · 1 −b · u −c · vu2 = (y1 − bd+ c2) · 1 +d · v.

Let S be the coordinate ring of P(14, 23, 32) and let π∗ : R → S be the inclusionhomomorphism. By Lemma 3.4, Φ: P5 → P(14, 23, 32) must satisfy π ◦ Φ = Φ0 so thatthe following diagram commutes:

P(14, 23, 32) = ProjS

π

������

ProjM = P5

Φ

::uuuuuuuuuuuuuuΦ0 // P(14, 23) = ProjR

Moreover, M is a graded module over S via Φ∗, with the same generators 1, u, v butwith more relations. We will not insist on writing ϕ∗, Φ∗0, Φ∗ when the context is clear.

Lemma 3.7 The map Φ: P5 → P(14, 23, 32) is determined by Φ∗0 as defined in (12),together with Φ∗(zi), which must take the form

Φ∗(z1) = Φ∗(f + s4)u+ s5v,

Φ∗(z2) = t4u+ Φ∗(g + t5)v(14)

wheref = y1 + α2y2 + α3y3, g = β1y1 + β2y2 + y3,

and si(a, b, c, d), ti(a, b, c, d), i = 4, 5 are homogeneous polynomials of degree 2, yet to bedetermined.

Proof Since Φ is a lift of Φ0 and Φ|P1 = ϕ, the general forms for Φ∗(zi) are

Φ∗(z1) = u3 + α2u2v + α3uv

2 + s1u2 + s2uv + s3v

2 + s4u+ s5v,

Φ∗(z2) = β1u2v + β2uv

2 + v3 + t1u2 + t2uv + t3v

2 + t4u+ t5v

11

where si(a, b, c, d), ti(a, b, c, d) are homogeneous polynomials of degree 1 or 2 as appro-priate. Now using the R-module structure on M determined by Φ∗0, we can write

u3 + α2u2v + α3uv

2 = Φ∗0(y1 + α2y2 + α3y3)u+ l.o.t.

Then by making coordinate changes z1 7→ z1 + s1y1 and similar, we may remove anyterms which are quadratic in u, v, including those involving si, ti for i = 1, 2, 3. Puttingthis all together gives the form for Φ∗(z1) in (14). The calculation for Φ∗(z2) is similar.We note that all the coordinate changes used in this proof do not affect the map ϕ onthe level of K3 surfaces. �

Given the general form for Φ, it remains to find suitable values of s4, s5, t4, t5 sothat W ′6,6 ⊂ P(14, 23, 32) containing Im(Φ) exists. From the proof of Proposition 2.6, wesee that this is equivalent to the kernel of Φ∗ containing equations extending (7), (9)and (10). In order to find these extended equations, we define

M ′ = R+Rz1 +Rz2 ⊂M = R+Ru+Rv.

This is the R-submodule consisting of all elements of degree 0 or 1 in z1, z2. We give aconvenient representation for M ′.

Lemma 3.8 The submodule M ′ ⊂M is the image of the R-module map(v, u, 1

)B,

where B is the matrix 0 s5 g + t5 y1 −y2 L1

0 f + s4 t4 −y2 y3 L2

1 0 0 L1 L2 L3

.

Proof This is a combination of Lemmas 3.5 and 3.7. The submodule M ′ = R+Rz1 +Rz2 is the image of the composite map

M(v,u,1)←−−−− 2R(−1)⊕R B←− R⊕ 4R(−3)⊕R(−4).

After multiplication by(v, u, 1

), we see from (14) that the first three columns of B give

the module generators 1, z1, z2 respectively. The last three columns are the matrix A of(13), which maps to 0 under the composite. �

Theorem 3.9 (I) The kernel of Φ∗ : S →M contains equations extending (7), (9) ofthe form

z21 − y1f

2 ∈M ′,z2

2 − y3g2 ∈M ′

if and only if s4 = s5 = t4 = t5 = 0.

(II) Given part (I), the equations are

C̃1 : z21 − y1f

2 = (c2 − bd)f2 − (δ1L1 − δ2L2)df + (δ1y2 + δ2y3)dz1 + α3dfz2 (15)

C̃3 : z22 − y3g

2 = (b2 − ac)g2 − (−δ3L1 + δ1L2)ag + β1agz1 + (δ3y1 + δ1y2)az2, (16)

where δ1 = 1− α3β1, δ2 = α2 − α3β2, δ3 = β2 − α2β1.

12

Corollary 3.10 The kernel of Φ∗ contains the following equation extending (8)

C̃2 : z1z2 − fgy2 = fg(ad− bc)− bgz1 − cfz2,

and three (nontrivial) equations extending multiples of (10), of the form

Q̃i : yi(y1y3 − y22) ∈M ′

for i = 1, 2, 3.

Remark 3.11 1. Part (I) of the theorem uniquely determines Φ up to automorphism.The coordinate changes used in Lemma 3.7 and the theorem do not affect the orig-inal map ϕ so Φ is completely determined by ϕ.

2. Since the coordinate ring of Im(ϕ) is not Cohen–Macaulay, the standard hyper-plane section principle goes awry (cf. Remark 3.2). Equation (10) does not extenddirectly, and we replace it in the kernel of Φ∗ with three equations Q̃i extendingyi(y1y3 − y2

2) for i = 1, 2, 3.

Remark 3.12 One consequence of Corollary 3.10 is that Q̃i for i = 1, 2, 3 correspondto generators of weighted degree 6 for the kernel of the map B defined in Lemma 3.8.Now, some of the arguments which follow amount to finding preimages under B, andthe kernel obscures the calculations. Therefore when necessary, we work with B′, thesubmatrix of B formed by the first five columns. Note that this does not affect anyassertions about uniqueness, because adding multiples of Q̃i is not permitted, becauseit changes the equation we are extending.

The “if” part of Theorem 3.9 is a straightforward verification that when s4 = s5 =t4 = t5 = 0, equations (15), (16) are in the kernel of Φ∗ by direct substitution. Thederivation of the equations is outlined in the proof of Proposition 3.13 below. Theremainder of this section proves the “only if” part.

We attempt the same elimination calculation used in the proof of Proposition 2.6,but with Φ∗ instead of ϕ∗. Since Φ∗ is more complicated than ϕ∗, we need severaladditional steps. Observe that by definition of Φ∗ (or from Remark 3.6), we can writeu2, uv, v2 as

u2 = Φ∗(y1 − bd+ c2) + dv

uv = Φ∗(y2 + ad− bc)− bu− cvv2 = Φ∗(y3 − ac+ b2) + au.

Thus by squaring Φ∗(zi) defined in Lemma 3.7 and rendering u2, uv, v2 as above, wearrive at

Φ∗(z2

1 − f21 (y1 − bd+ c2)− 2f1s5(y2 + ad− bc)− s2

5(y3 − ac+ b2))≡ 0

Φ∗(z2

2 − t24(y1 − bd+ c2)− 2g1t4(y2 + ad− bc)− g21(y3 − ac+ b2)

)≡ 0

13

modulo (a, b, c, d)M , where f1 = f + s4 and g1 = g + t5. The residual parts to thesecongruences are

J : f21dv − 2f1s5(bu+ cv) + s2

5au

K : t24dv − 2g1t4(bu+ cv) + g21au,

(17)

which are homogeneous expressions of degree 6 in (a, b, c, d)M . We prove that the tworesidual terms J , K are contained in the submodule M ′ if and only if s4 = s5 = t4 =t5 = 0

We first derive equations (15) and (16) assuming that s4 = s5 = t4 = t5 = 0. Underthese assumptions, J = df2v and K = ag2u.

Proposition 3.13 Suppose s4 = s5 = t4 = t5 = 0. Then the preimage of J (respectivelyK) under the map

(v, u, 1

)B is unique, and given by η (resp. ξ) where

η1 = (−δ1L1 + δ2L2)df η2 = (δ1y2 + δ2y3)d η3 = α3df

η4 = δ1df η5 = −δ2df η6 = 0

and

ξ1 = (δ3L1 − δ1L2)ag ξ2 = β1ag ξ3 = (δ3y1 + δ1y2)a

ξ4 = −δ3ag ξ5 = δ1ag ξ6 = 0.

Proof We calculate the preimage η of J = df2v under B. We drop the multiplicationby(v, u, 1

)and write J = t

(df2, 0, 0

). To avoid complications arising from the

kernel of B (cf. Remark 3.12), we define B′ to be the submatrix

B′ =

0 0 g y1 −y2

0 f 0 −y2 y3

1 0 0 L1 L2

.

We solve for η using the top row first. Since the first row of B′ only involves y3 as partof g, we must write

df2 = df(y1 + α2y2 + α3y3)

= df (y1 + α2y2 + α3(g − β1y1 − β2y2))

or as an expression in the top row of B′,

df2 =(0, 0, g, y1, −y2

)

∗∗

α3df

(1− α3β1)df

(−α2 + α3β2)df

,

14

where the starred entries are not yet determined. Thus we have

η3 = α3df, η4 = (1− α3β1)df, η5 = (−α2 + α3β2)df.

Now the second row of B′ must satisfy

0 =(0, f, 0, −y2, y3

)∗∗η3

η4

η5

.

Thus the starred entry in the second row is in fact determined by the rest of η:

η2 =1

f(y2η4 − y3η5).

Finally, we use the first column of B′ to remove any accidental terms arising when wemultiply the third row by η. Hence

η1 = −L1η4 − L2η5, η6 = 0.

Unravelling these recursive expressions gives η as appearing in the statement of theproposition. The calculation of ξ, the preimage of K = ag2u, is similar and is omitted.�

Next we prove that si = ti = 0. Reinstate si, ti to the matrix B. The obstructionsto J and K being in the image of B are then

J ′ := J −(v, u, 1

)Bη = J −

(v, u, 1

) s5 g + t5f + s4 t4

0 0

(η2

η3

)

K ′ := K −(v, u, 1

)Bξ = K −

(v, u, 1

) s5 g + t5f + s4 t4

0 0

(ξ2

ξ3

).

Here the last two equalities are because si, ti only appear in the second and third columnsof B. A careful computation using equations (17) and Proposition 3.13 gives

J ′ =(v, u, 1

)s4(2f + s4)d− 2(f + s4)s5c− η2s5 − η3t5−2(f + s4)s5b+ s2

5a− η2s4 − η3t40

K ′ =

(v, u, 1

) −2(g + t5)t4c+ t24d− ξ2s5 − ξ3t5−2(g + t5)t4b+ t5(2g + t5)a− ξ2s4 − ξ3t4

0

.

These expressions are of weighted degree 6 in (a, b, c, d)3M , and linear in y1, y2, y3.

Proposition 3.14 The image of B contains J ′ and K ′ if and only if s4 = s5 = t4 =t5 = 0. Therefore J ′ = K ′ = 0.

15

Proof The “if” part follows from the previous proposition. In particular all termsinvolve some si or ti by construction, and by degree considerations, all terms involvingyi must also be linear in si, ti.

For the converse, we attempt to write J ′, K ′ as expressions in M ′ = R+Rz1 +Rz2.First we observe that J ′ and K ′ are of degree 3 in u, v because of terms involving yiuand yiv. Thus the important parts of J ′, K ′ are the terms involving yi:

J ′ =(v, u, 1

)2(s4d− s5c)f − η2s5 − η3t5−2fs5b− η2s4 − η3t4

0

+ l.o.t.

=(v, u, 1

)(2(s4d− s5c)− α3t5d)f − (δ1y2 + δ2y3)s5d(−2s5b− α3t4d)f − (δ1y2 + δ2y3)s4d

0

+ l.o.t.,

K ′ =(v, u, 1

) −2gt4c− ξ2s5 − ξ3t5−2(t4b− t5a)g − ξ2s4 − ξ3t4

0

+ l.o.t.

=(v, u, 1

) (−2t4c− β1s5a)g − (δ3y1 + δ1y2)t5a(−2(t4b− t5a)− β1s4a)g − (δ3y1 + δ1y2)t4a

0

+ l.o.t.

Now, the only elements of M ′ which are of degree 3 in u, v are those elements of Rz1+Rz2

which do not involve yi. So we use z1 (respectively z2) to remove terms involving y1u(resp. y3v). Recalling that z1, z2 correspond to the second and third rows of B, weobtain J ′′, K ′′:

J ′′ = J ′ − (−2s5b− α3t4d)

s5

f + s4

0

− (2α3(s4d− s5c)− α23t5d− δ2ds5

)g + t5t40

K ′′ = K ′ −

(−2β1(t4b− t5a)− β2

1s4a− δ3t4a) s5

f + s4

0

− (−2t4c− β1s5a)

g + t5t40

.

Thus

J ′′ =(v, u, 1

)2(s4d− s5c)(f − α3g)− (η2 − δ2dg)s5 − (η3 − α23dg)t5

−η2s4

0

+ l.o.t.

K ′′ =(v, u, 1

) −ξ3t5−2(t4b− t5a)(g − β1f)− (ξ2 − β2

1af)s4 − (ξ3 − δ3af)t40

+ l.o.t.

We claim that J ′′, K ′′ are elements of R.1, of degree ≤ 2 in u, v. On the other hand,since

u2v = y1v + l.o.t. and uv2 = y2v + l.o.t.

= y2u+ l.o.t. = y3u+ l.o.t.,

16

we can read off the four coefficients of u2v and uv2 occurring in J ′′,K ′′

J ′′

{u2v : 2(s4d− s5c)(1− α3β1) + δ2β1s5d− α3(1− α3β1)t5d− δ1s4d

uv2 : 2(s4d− s5c)(α2 − α3β2)− (δ1 − δ2β2)s5d− α3(α2 − α3β2)t5d− δ2s4d,

K ′′

{u2v : −2(t4b− t5a)(β2 − α2β1)− β1(β2 − α2β1)s4a− (δ1 − δ3α2)t4a− δ3t5a

uv2 : −2(t4b− t5a)(β3 − α3β1)− β1(1− α3β1)s4a+ δ3α3t4a− δ3t5a.

By our claim, all four of these coefficients must vanish. Writing this condition as simul-taneous linear equations in s4, s5, t4, t5 we obtain

C

s4

s5

t4t5

= 0, (18)

where C is the coefficient matrixδ1d −2δ1c+ β1δ2d 0 −α2δ1dδ2d −2δ2c+ (β2δ2 − δ1)d 0 −α2δ2d−β1δ3a 0 (α1δ3 − δ1)a− 2δ3b δ3a−β1δ1a 0 α2δ3a− 2δ1b δ1a

.

The proposition follows from:

Claim 3.15 The linear equations (18) have a unique solution s4 = s5 = t4 = t5 = 0corresponding to a Fano 6-fold W ′6,6 ⊂ P(14, 23, 32) containing the image of Φ.

Proof of claim We define ∆ = δ21 − δ2δ3. This is the resultant of Φ∗(z1) and Φ∗(z2),

which appears later as the determinant of matrix (19). If ∆ = 0, then Φ∗(zi) have acommon root, and so both components of the branch curve have a common point oftangency with the conic. This contradicts the hypothesis that T is quasismooth.

Hence we may assume that ∆ 6= 0, and there are two cases. The easy case is whenδ1 6= 0. Then by performing row and column operations on C, we find that the uniquesolution to (18) is s4 = s5 = t4 = t5 = 0 and the claim is proved.

If δ1 = 0, then since ∆ 6= 0 we may assume that δ2δ3 6= 0. The first and last rows ofC then force s5 = t4 = 0. However, the remaining two rows of C reduce to s4 = α3t5.It appears that we do not have a unique solution! We suppose that t5 6= 0 and carefullyexamine the consequences. Recall that by equation (14), we have

Φ∗(z1) = Φ∗(y1 + α2y2 + α3y3 + α3t5)u

Φ∗(z2) = Φ∗(β1y1 + β2y2 + y3 + t5)v.

Thus by making a coordinate change y3 7→ y3 + t5 (or equivalently y1 7→ y1 + s4), wemay assume that Φ∗(z1) = Φ∗(f)u and Φ∗(z2) = Φ∗(g)v as before, but now Φ∗(y3) =v2 − au+ ac− b2 + t5.

17

The obvious amendments to matrix B allow us to recover extended equations C̃1, C̃2

and C̃3, using the same method as in the proof of Proposition 3.13. However, Corollary3.10 no longer holds if t5 is nonzero. That is, there are no extended equations Q̃i. This isshown using similar methods to the proof of Corollary 3.10 below, so we omit the details.Finally, we note that if the defining equations of the K3 surface T ′ do not involve Q,then the symmetric determinantal surface T is not quasismooth. Therefore, we musthave t5 ≡ 0.

� Theorem 3.9 is proved by observing that si = ti = 0 by Proposition 3.14, so that

z21 − (y1 − bd+ c2)f2 =

(v, u, 1

)Bη

by Proposition 3.13. Now by Lemma 3.8, this is equivalent to

z21 = (y1 − bd+ c2)f2 + η1 + η2z1 + η3z2.

Likewise using ξ, the equation for z22 is

z22 = (y3 − ac+ b2)g2 + ξ1 + ξ2z1 + ξ3z2.

Written out in full, these are equations (15), (16) in the statement of the theorem. Thisconcludes the proof of Theorem 3.9, and its Corollary is proved in Section 3.1. �

Proof of main theorem Given the existence of equations extending (7–10), we cannow prove the Main Theorem. Indeed, given a symmetric determinantal quartic K3surface in normal form, projection from the distinguished node P gives the K3 surfaceT ′6,6 ⊂ P(23, 32) defined by equations

C1 + a(y1, y2, y3)Q, C3 + b(y1, y2, y3)Q,

where a =∑3

i=1 aiyi, b =∑3

i=1 biyi are linear as in (11). Then the Fano 6-fold W ′6,6 ⊂P(14, 23, 32) extending T ′ is defined by equations

C̃1 +3∑i=1

aiQ̃i, C̃3 +3∑i=1

biQ̃i,

where C̃i and Q̃i are constructed by Theorem 3.9 and its Corollary. Clearly W ′ containsthe image of Φ and is unique. Hence W ′ is the image, under projection from a 1

2 point,of the unique Fano 6-fold W extending T . �

3.1 Proof of Corollary 3.10

To prove the corollary we calculate the equations extending (8) and multiples of (10).First we have the following easy Proposition.

Proposition 3.16 The equation extending (8) is

C̃2 : z1z2 = fg(y2 + ad− bc)− bgz1 − cfz2.

18

Proof First note that

z1z2 − fg(y2 + ad− bc) = −fg(bu+ cv),

so we show that fg(bu+ cv) is in the image of B. We must find some λ such that

fg

cb0

= Bλ.

Simply choose λ2 = bg, λ3 = cf and the other λi = 0. This gives C̃2 as above. �The equations Q̃i extending Q : (y1y3 − y2

2) are much more complicated. First notethat by Lemma 3.5,

L3 + L2u+ L1v = 0,

and L3 = y1y3 − y22 + b2y1 + (2bc− ad)y2 + c2y3. Thus the three extensions Q̃i of Q are

equivalent to finding µ(i) such that

Q̃i : yiL3 = −(v, u, 1

)Bµ(i),

where µ(i) is the preimage of yi(L2u+L1v) under B. We now see that each Q̃i correspondsto a degree 6 generator of the kernel of B, as claimed in Remark 3.12.

We outline the derivation of µ(1). For clarity, we use a dot to denote a zero entry ina matrix. Suppose we have two 6× 3 matrices D and E, such that

y1

· · ·y1 y2 y3

· · ·

= BD, y1

y1 y2 y3

· · ·· · ·

= BE.

Proposition 3.17 Given D, E as above, the extended relation Q̃1 is

y1L3 = −(µ(1)1 + µ

(1)2 z1 + µ

(1)3 z2),

where

µ(1) = D

abc

+ E

bcd

.

Proof Since L2 = ay1 + by2 + cy3, multiplying D on the right by the column vectort(a, b, c

)gives ·

y1L2

·

= BD

abc

.

Similarly L1 = by1 + cy2 + dy3, soy1L1

··

= BE

bcd

.

19

Taking the sum of these gives

y1L3 = −(v, u, 1

)y1L1

y1L2

·

= −(v, u, 1

)Bµ(1),

and so by Lemma 3.8, the equation extending Q is

Q̃1 : y1L3 = −(µ(1)1 + µ

(1)2 z1 + µ

(1)3 z2).

�Thus it remains to find suitable matrices D and E. In order to write down D, E, we

must first define the (transpose) resultant matrix of Φ∗(z1) and Φ∗(z2):

T =

1

α2 1 β1

α3 α2 1 β2 β1

α3 α2 1 β2 β1

α3 1 β2

1

. (19)

We split T and its inverse into 3× 3 blocks

T =

(V1 W1

V2 W2

), T−1 =

(v1 w1

v2 w2

),

so that in particular,

V1v1 +W1v2 = I3 V1w1 +W1w2 = 0 (20)

V2v1 +W2v2 = 0 V2w1 +W2w2 = I3. (21)

We may now state:

Lemma 3.18 There is a 6× 3 matrix D satisfying

y1

· · ·y1 y2 y3

· · ·

= BD,

where D = tD1v1 + tD2v2 and

D1 =

· y1 · · · ·

−α2(L1y2 + L2y1) y2 · α2y2 α2y1 ·· y3 · · · ·

,

D2 =

(β1L1 − β2L2)y1 + L1y3 · y1 −β1y1 − y3 β2y1 ·

−L2g · y2 · g ·(β1L1 − β2L2)y3 · y3 −β1y3 β2y3 ·

.

20

Proof One proof is to simply evaluate the matrix product and check that it gives thecorrect answer, but this is not very enlightening. Instead, we sketch the derivation of D.For convenience, we recall that

B =

0 0 g y1 −y2 L1

0 f 0 −y2 y3 L2

1 0 0 L1 L2 L3

.

Now D1 and D2 are chosen so that

B · tD1 =

· · ·y1f y2f + α2Q y3f· · ·

, B · tD2 =

· · y23

y2g + β2Q y3g β1y2y3 + β2y23

· · ·

,

where as usual, Q = y1y3 − y22. Determining such D1 and D2 is not difficult, via the

same procedure used in the proof of Proposition 3.13.Next consider the following matrix product, which can be easily verified from the

definition of T from (19):

(y2

1, y1y2, y1y3, y2y3, y23, 0

)(V1

V2

)=(y1f, y2f + α2Q, y3f

).

We multiply on the right by the block matrix v1

y1

(y1, y2, y3

)V1v1 + y3

(y2, y3, 0

)V2v1 =

(y1f, y2f + α2Q, y3f

)v1,

and use identities (20) to obtain

y1

(y1, y2, y3

)(I3 −W1v2)− y3

(y2, y3, 0

)W2v2 =

(y1f, y2f + α2Q, y3f

)v1.

Thus rearranging this expression, we get

y1

(y1, y2, y3

)=(y1f, y2f + α2Q, y3f

)v1+

y1

(y1, y2, y3

)W1v2 + y3

(y2, y3, 0

)W2v2. (22)

In the same way, consider

(0, y2

1, y1y2, y1y3, y2y3, y23

)(W1

W2

)=(y1g, y2g + β2Q, y3g

).

Rearranging into block matrices and multiplying on the right by v2 leads to(0, 0, 0

)=(y1g, y2g + β2Q, y3g

)v2−

y1

(0, y1, y2

)W1v2 − y3

(y1, y2, ∗

)W2v2. (23)

21

This is almost trivial, except that we have replaced y3 with a ∗ in the last product. Thatis because W2v2 = −V2v1 via (21), but the third row of V2 is empty, so the starred entryis not involved in the overall matrix product.

Now writing equations (23) and (22) as rows of a single matrix equation togetherwith an empty third row, and collecting coefficients of v1 and v2, we obtain

y1

· · ·y1 y2 y3

· · ·

=

· · ·y1f y2f + α2Q y3f· · ·

v1+

· · y23

y2g + β2Q y3g β1y2y3 + β2y23

· · ·

v2.

Thus on comparing the above equation with B · tD1 and B · tD2 calculated above, we seethat D gives the required preimage. �

The second lemma gives E. Let τ be the cyclic permutation(1, 3, 5

) (2, 4, 6

)of order 3 acting on the rows of T , so that τ−1 acts on the columns of T−1 to giveτ(T )−1 = τ−1(T−1). We write τ(T ) and τ(T )−1 in block form as

τ(T ) =

(V τ

1 V τ2

W τ1 W τ

2

), τ(T )−1 =

(vτ1 vτ2wτ1 wτ2

).

Lemma 3.19 There is a 6× 3 matrix E satisfying

y1

y1 y2 y3

· · ·· · ·

= BE,

where E = tE1wτ1 + tE2w

τ2 and

E1 =

(−α2L1 + α3L2)y1 y1 · α2y1 −α3y1 ·

−L1f y2 · f · ·(−α2L1 + α3L2)y3 + L2y1 y3 · α2y3 −y1 − α3y3 ·

,

E2 =

· · y1 · · ·

−β2(L1y3 + L2y2) · y2 β2y3 β2y2 ·· · y3 · · ·

.

�

The proof is the same as for Lemma 3.18, so we omit it. In any case, one can evaluatethe matrix product to check the result.

The derivation of Q̃2 and Q̃3 is similar, requiring further permutations of T . We skipto the end result, and write

µ(i) = D(i)

bcd

+ E(i)

abc

for i = 2, 3.

22

Here D(i), E(i) are composed of

D(i) = tD(i)1 vσi1 + tD

(i)2 vσi2 , E(i) = tE

(i)1 wτi1 + tE

(i)2 wτi2 ,

where vσi1 denotes the first 3 × 3 block of σi(T−1), etc. For i = 2, the permutations to

be applied to T are

σ2 =(1, 6, 5, 4, 3, 2

), τ2 =

(1, 2, 3, 4, 5, 6

).

Then

D(2)1 =

α3(L1y2 + L2y1) y1 · −α3y2 −α3y1 ·

· y2 · · · ·L1y2 + L2y1 y3 · −y2 −y1 ·

D(2)2 =

L1g · y1 −g · ·

(β1L1 − β2L2)y2 − y3L2 · y2 −β1y2 β2y2 + y3 ·(β1L1 − β2L2)y3 · y3 −β1y3 β2y3 ·

,

E(2)1 =

(−α2L1 + α3L2)y1 y1 · α2y1 −α3y1 ·

(−α2L1 + α3L2)y2 − y1L1 y2 · y1 + α2y2 −α3y2 ·L2f y3 · · −f ·

E(2)2 =

L1y3 + L2y2 · y1 −y3 −y2 ·

· · y2 · · ·β1(L1y3 + L2y2) · y3 −β1y3 −β1y2 ·

.

On the other hand, when i = 3, we permute T by

σ3 =(1, 5, 3

) (2, 6, 4

), τ3 = id,

but this time the required matrices are the same as for Q̃1. In other words, D(3)1 = D1,

D(3)2 = D2 from Lemma 3.18 and E

(3)1 = E1, E

(3)2 = E2 from Lemma 3.19.

4 Surfaces with pg = 1 and K2 = 2

In this section, we review some of the results of [3] on surfaces with pg = 1 and K2 = 2,before using our 6-fold extensions to give a new construction of a family of such surfaces.

4.1 Connection with work of Catanese and Debarre

In this section, Y is a surface of general type with pg = 1 and K2 = 2. We outline themethods used in [3] and some of the results. Following ideas of Enriques, Catanese andDebarre considered the bicanonical map ϕ2K of Y , which is actually a morphism onto a

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surface Σ ⊂ P(1, 2, 2, 2) ∼= P3. The various possibilities are classified in detail, using thefact that deg(ϕ) · deg(Σ) = 8. The two cases (numbered according to [3]) that we needare

3. degϕ = 2 and Σ is a symmetric determinantal quartic K3 surface.

4. ϕ : Y → Σ is birational,

In both of these cases, Y is simply connected. The birational case is the most important,because these surfaces are the least well understood and yet they form a Zariski denseopen subset of the moduli space.

The canonical ring R = R(Y,KY ) is a module over S = R(Y, 2KY ), the bicanonicalring. The free resolution of R is given by

R← S ⊕ 4S(−3)A←− S(−8)⊕ 4S(−5)← 0,

where A is a symmetric 5× 5 matrix with entries of the appropriate degree (we assumehere that the canonical curve is not hyperelliptic, but see [7] for a new approach to thehyperelliptic case). Since R is a commutative S-algebra, A satisfies the (RC) condition:

Let A′ be the n × (n + 1) matrix obtained by removing the first row of A.Then the ideal generated by the n×n minors of A′ contains the n×n minorsof A.

If Y is of type 3 or 4, then the ring structure on R can actually be recovered from thematrix A.

The form of A is given in [3] for all cases, including hyperelliptic ones. Unfortunately,when ϕ is birational, it is difficult to write down an explicit general solution to the (RC)condition. One approach, originally due to Enriques, is to project a del Pezzo surface ofdegree 7 into P3. The image of this projection is classically called the biadjoint surfaceto Y . It is a hypersurface defined by the vanishing of the determinant of a certain5 × 5 matrix. After a change of variables, one can derive a matrix A satisfying the(RC) condition, and corresponding to a surface of general type Y (cf. [3]). This gives adense open subset of an irreducible component of the moduli space, which is the uniqueirreducible component containing those surfaces of type 3. Such surfaces are simplyconnected. It is conjectured (but still not proven) that this family forms an irreducibleconnected component of the moduli space.

4.2 A new construction

Using Theorem 1.1, we obtain our family of surfaces.

Theorem 4.1 There is a 16 parameter family of topologically simply connected surfacesY of general type with pg = 1, q = 0, K2 = 2 and no torsion, each of which is a completeintersection of type (1, 1, 1, 2) in a Fano 6-fold W ⊂ P(14, 24, 34) with 10× 1

2 points.

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Proof To obtain Y from W take 3 transverse hyperplane sections of weight 1 andone hypersurface section of weight 2, avoiding the isolated orbifold 1

2 points. SinceW is quasismooth, by the adjunction formula the surface Y has ωY = OY (1) and isnonsingular. Furthermore it is clear from the construction of Y that

pg(Y ) = h0(Y,KY ) = h0(W,−KW )− 3 = 1.

Consider Y as a quadric section of a Fano 3-fold W 3. Then the standard short exactsequence

0→ OW 3(−2)→ OW 3 → OY → 0,

implies that H1(OY ) = 0 by Kodaira vanishing, so Y is regular, and the Riemann–Rochformula gives K2

Y = 2.To prove that Y is topologically simply connected, we use the following result of

Takayama, which we state in a much weaker formulation than that contained in [12],[13]:

Theorem 4.2 Let W be a normal projective variety with terminal singularities, suchthat −KW is ample. Then W is topologically simply connected, and so is any resolutionof singularities of W .

Thus by the Lefschetz hyperplane theorem, Y is simply connected.Finally, the Main Theorem says that the family of Fano 6-folds depends on the

same number of moduli as the family of symmetric determinantal quartics, which is 9.Furthermore, counting the number of choices for linear and quadric sections of W showsthat we have a 9 + 3 + 4 = 16 parameter family of surfaces Y . �

Remark 4.3 The expected dimension of the moduli space of surfaces with pg = 1, q = 0and K2 = 2 is 16. We have not proved that our family has dimension 16, although ithas the right number of parameters. It would be interesting to know if our family maybe used to construct an irreducible component of the moduli space.

Acknowledgements I would like to thank Miles Reid for introducing me to thisproblem, which forms part of my University of Warwick PhD thesis [6]; in addition Ithank Yongnam Lee, Jan Stevens and the anonymous referees for their useful commentsand suggestions. This research was partially supported by the World Class Universityprogram through the National Research Foundation of Korea funded by the Ministry ofEducation, Science and Technology (R33-2008-000-10101-0).

References

[1] A. Beauville, Determinantal hypersurfaces, Michigan Math. J. 48, 39–64 (2000)

[2] F. Catanese, Babbage’s conjecture, contact of surfaces, symmetric determinantalvarieties and applications, Invent. Math. 63, 433–465 (1981)

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[3] F. Catanese, O. Debarre, Surfaces with K2 = 2, pg = 1, q = 0, Jour. Reine. Angew.Math. 395, 1–55 (1989)

[4] A. Cayley, A memoir on quartic surfaces, Proc. London Math. Soc. s1–3, 19 (1869),A second memoir on quartic surfaces, Proc. London Math. Soc. s1–3, 198 (1869),A third memoir on quartic surfaces, Proc. London Math. Soc. s1–3, 233 (1869)

[5] F. Cossec, Reye congruences, Trans. Amer. Math. Soc. 280, no. 2, 737–751 (1983)

[6] S. Coughlan, Key varieties for surfaces of general type, University of Warwick PhDthesis, vi + 81 pp. (2009)

[7] S. Coughlan, Extending hyperelliptic K3 surfaces, and Godeaux surfaces with tor-sion Z/2, arXiv:0909.5548, (2009)

[8] S. Papadakis, M. Reid, Kustin–Miller unprojection with complexes, J. AlgebraicGeom. 13 (2004) 249–268

[9] H. Park, J. Park, D. Shin, Surfaces of general type with pg = 1 and q = 0, J. KoreanMath. Soc. 50 (3), 493–507, (2013)

[10] M. Reid, Graded rings and birational geometry, Proc. of algebraic geometry sym-posium (Kinosaki, Oct 2000), K. Ohno (Ed.), 1–72

[11] J. Stevens, Unpublished calculations, (1993)

[12] S. Takayama, Simple connectedness of weak Fano varieties, J. Algebraic Geom. 9(2000), no. 2, 403–407

[13] S. Takayama, Local simple connectedness of resolutions of log-terminal singularities,Internat. J. Math. 14 (2003), no. 8, 825–836

[14] A. Todorov, Counterexamples of the global Torelli theorem, Invent. Math. 63(1981), no. 2, 287–304

[15] A. Tyurin, The intersection of quadrics, Uspehi Mat. Nauk 30 (1975), no 6 (186),51–99

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