Surjectivity of Gaussian maps for curves on Enriques surfaces

DOI 10.1515 / ADVGEOM.2007.014 de Gruyter 2007

Surjectivity of Gaussian maps for curves on Enriques surfaces

Andreas Leopold Knutsen^∗and Angelo Felice Lopez^†

(Communicated by R. Miranda)

Abstract. Making suitable generalizations of known results we prove some general facts about Gaussian maps. These facts are then used, in the second part of the article, to give a set of conditions that insure the surjectivity of Gaussian maps for curves on Enriques surfaces. To do this we also solve a problem of independent interest: a tetragonal curve of genusg ≥7lying on an Enriques surface and general in its linear system, cannot be, in its canonical embedding, a quadric section of a surface of degreeg−1inP^g−1.

2000 Mathematics Subject Classification. Primary 14H99, 14J28. Secondary 14H51

1 Introduction

Gaussian maps have emerged in the mid 1980’s as a useful tool to study the geometry of a given varietyX ⊂P^N as soon as one has a good knowledge of the hyperplane sections Y =X∩H.

Let us briefly recall their definition and notation in the case of curves.

Notation 1.1. LetCbe a smooth irreducible curve and letL, M be two line bundles on C. We denote byµL,M : H⁰(L)⊗H⁰(M)→ H⁰(L⊗M)the multiplication map of sections and byR(L, M) = KerµL,M. The Gaussian map associated toLandM will be denoted by

ΦL,M :R(L, M)→H⁰(ωC⊗L⊗M).

This map can be defined locally byΦL,M(s⊗t) =sdt−tds(see [38]).

∗Research partially supported by a Marie Curie Intra-European Fellowship within the 6th European Com- munity Framework Programme.

†Research partially supported by the MIUR national project “Geometria delle variet`a algebriche” COFIN 2002–2004.

Perhaps the first important result, proved by Wahl, who introduced Gaussian maps, is that if a smooth curveClies on a K3 surface, then the Gaussian mapΦω_C,ω_C cannot be surjective. On the other hand, as it was proved by Ciliberto, Harris and Miranda [8], this mapΦω_C,ω_C is surjective on a curveCwith general moduli of genus10or at least12.

The link with the study of higher dimensional varieties was provided, around the same period, by Zak, who proved the following result ([39], see also [2], [28]): If Y ⊂P^ris a smooth variety of codimension at least two with normal bundleN_{Y /P}r and h⁰(N_{Y /P}r(−1)) ≤ r+ 1, then the only varietyX ⊂ P^r+1 that hasY as hyperplane section is a cone overY.

Now the point is that, ifY is a curve, we have the formula h⁰(N_{Y /P}r(−1)) =r+ 1 + cork ΦH_Y,ω_Y

whereHY is the hyperplane bundle ofY.

On the other hand, ifY is not a curve one can take successive hyperplane sections of Y. For example, whenX ⊂P^r+1is a smooth anticanonically embedded Fano threefold with general hyperplane section theK3surfaceY, in [9], Ciliberto, the second author and Miranda were able to computeh⁰(NY /P^r(−1))by calculating the coranks ofΦH_C,ω_C for the general curve sectionCofY. This then led to recover in [9] and [10], in a very simple way, a good part of the classification of smooth Fano threefolds [18], [19] and of varieties with canonical curve section [29].

To study other threefolds by means of Zak’s theorem, in many cases it is not enough to get down to curve sections and one needs to bound the cohomology of the normal bundle of surfaces. In [25] the following general result was proved:

Proposition 1.2. LetY ⊂P^rbe a smooth irreducible linearly normal surface and letH be its hyperplane bundle. Assume there is a base-point free and big line bundleD0onY withH¹(H−D0) = 0and such that the general elementD ∈ |D0|is not rational and satisfies

(i) the Gaussian mapΦ_HD,ωD(D)is surjective;

(ii) the multiplication mapsµ_VD,ωD andµ_VD,ωD(D)withV_D := Im{H⁰(H−D) → H⁰((H−D)_|D)}are surjective.

Then

h⁰(NY /P^r(−1))≤r+ 1 + cork ΦHD,ωD.

The application of the above proposition clearly points in the following direction: If one wants to study, with Gaussian maps methods, the existence of threefoldsX ⊂P^r+1 with given hyperplane sectionY, one has to know about the surjectivity of Gaussian maps of typeΦM,ω_C for curvesC ⊂Y that are general in their linear system. In the present article we study this in the case of Enriques surfaces. This is applied in [25] to prove the (sectional) genus boundg ≤17for threefoldsX ⊂P^r+1whose general hyperplane sectionY is an Enriques surface.

We prove the following result.

Theorem. LetS be an Enriques surface and letLbe a base-point free line bundle onS withL² ≥4. LetCbe a general smooth curve in|L|and letM be a line bundle onC.

Then the Gaussian mapΦM,ω_C is surjective if one of the hypotheses below is satisfied:

(i) L²= 4andh⁰(4L_|C−M) = 0;

(ii) L²= 6andh⁰((3L+KS)_|C−M) = 0;

(iii) L²≥8andh⁰(2L_|C−M) = 0;

(iv) L²≥12andh⁰(2L_|C−M) = 1;

(v) H¹(M) = 0,deg(M)≥ ¹₂L²+ 2≥6andh⁰(2L_|C−M)≤Cliff(C)−2.

The proof of this theorem will be accomplished essentially in two steps. We will first prove, in Section 2, some general facts about Gaussian maps, by generalizing some known results. Then, in the second step, in Section 5, we will deal with the specific problem of Gaussian maps for curves on Enriques surfaces. As it turns out, the most difficult point will be to show that a tetragonal curve of genusg ≥7lying on an Enriques surface and general in its linear system, in its canonical embedding, can never be a quadric section of a surface of degreeg−1inP^g−1.

Acknowledgments. The authors wish to thank Roberto Mu˜noz for several helpful dis- cussions.

2 Basic results on Gaussian maps

We briefly recall the definition, notation and some properties of gonality and Clifford index of curves.

Definition 2.1. LetX be a smooth surface. We will denote by∼(respectively≡) the linear (respectively numerical) equivalence of divisors (or line bundles) onX. We will say that a line bundleLis primitive ifL≡kL⁰for some line bundleL⁰and some integer k, impliesk=±1.

Definition 2.2. LetCbe a smooth irreducible curve of genusg ≥ 2. We denote byg^r_d a linear system of dimensionrand degreedonC and say thatC isk-gonal (and that kis its gonality) ifCpossesses ag¹_kbut nog_k−1¹ . In particular, we call a2-gonal curve hyperelliptic, a 3-gonal curve trigonal and a4-gonal curve tetragonal. We denote by gon(C)the gonality ofC.

Definition 2.3. LetCbe a smooth irreducible curve of genusg≥4and letAbe a line bundle onC. The Clifford index ofAis the integer

Cliff(A) := degA−2(h⁰(A)−1).

The Clifford index ofCis

Cliff(C) := min

Cliff(A) :h⁰(A)≥2, h¹(A)≥2 .

We say that a line bundleAonC contributes to the Clifford index ofC ifh⁰(A) ≥ 2, h¹(A)≥2.

2.1 Preliminaries on Gaussian maps. We recall some well-known facts about Gauss- ian maps.

Proposition 2.4. [38, Proposition 1.10] LetCbe a smooth irreducible nonhyperelliptic curve of genusg ≥ 3, letC ⊂ P^g−1 be its canonical embedding and letM be a line bundle onC. We have two exact sequences

0−→Cokerµ_M,ωC −→H¹(Ω¹

P^g−1|C⊗ω_C⊗M)−→H¹(M)^⊕g

−→H¹(ωC⊗M)−→0 (1) and

0−→Coker ΦM,ω_C −→H¹(N_C/^∗ _Pg−1⊗ωC⊗M)−→H¹(Ω¹_Pg−1|C⊗ωC⊗M)

−→H¹(ω²_C⊗M)−→0. (2) In particular

(a) ifH⁰(N_C/P^g−1⊗M⁻¹) = 0thenΦM,ω_C is surjective;

(b) ifH¹(M) = 0andµM,ωC is surjective thencork ΦM,ωC =h⁰(NC/P^g−1⊗M⁻¹).

In the sequel we will collect some results about Gaussian maps of type Φ_M,ωC for curvesC of low genus or low gonality or with Clifford index higher thanh⁰(2K_C − M) + 2.

We start with an elementary but useful fact.

Lemma 2.5. Fora≥2, letQ1, . . . , Qabe linearly independent homogeneous polynomi- als of degree 2 inX0, . . . , Xr. Suppose that the relations amongQ1, . . . , Qaare gener- ated byRi= [Ri1, . . . , Ria], for1≤i≤b. If(c1, . . . , ca)∈C^a− {0}, then there exists anisuch that

a

X

j=1

cjRij6= 0.

Proof. Suppose to the contrary thatPa

j=1cjRij = 0for everyiwith1≤i≤b. Without loss of generality assume thatc16= 0, so that

Ri1=−

a

X

j=2

c⁻¹₁ cjRij, 1≤i≤b. (3)

Claim 2.6. SetQ⁰₁=Q1, Q⁰_j =Qj−c⁻¹₁ cjQ1for2≤j≤a. Then (i) Q⁰₁, . . . , Q⁰_aare linearly independent;

(ii) the relations amongQ⁰₁, . . . , Q⁰_aare generated bySi = [0, Ri2, . . . , Ria], for1 ≤ i≤b.

Proof. Consider a relation Pa

j=1R⁰_jQ⁰_j = 0, where theR⁰_j’s are polynomials. Then R⁰₁Q1+Pa

j=2R⁰_j(Qj−c⁻¹₁ cjQ1) = 0, whence

R⁰₁−

a

X

j=2

c⁻¹₁ c_jR⁰_j

Q₁+

a

X

j=2

R⁰_jQ_j= 0. (4)

If allR⁰_j’s are complex numbers we getR⁰_j= 0for allj, proving (i).

To see (ii), by (4) and the hypothesis of the lemma we deduce that there are polyno- mialsdjsuch that

"

R⁰₁−

a

X

j=2

c⁻¹₁ c_jR⁰_j, R₂⁰, . . . , R⁰_a

#

=

b

X

i=1

d_iR_i=

" _b X

i=1

d_iR_i1,

b

X

i=1

d_iR_i2, . . . ,

b

X

i=1

d_iR_ia

#

whenceR⁰_j=Pb

i=1d_iR_ijfor2≤j≤aand R₁⁰ =

a

X

j=2

c⁻¹₁ c_jR⁰_j+

b

X

i=1

d_iR_i1=

b

X

i=1

d_i

a

X

j=2

c⁻¹₁ c_jR_ij+R_i1

!

= 0

by (3). Now

b

X

i=1

diSi =

"

0,

b

X

i=1

diRi2, . . . ,

b

X

i=1

diRia

#

= [R⁰₁, R⁰₂, . . . , R⁰_a]. 2

Conclusion of the proof of Lemma 2.5. Consider the Koszul relation[Q⁰₂,−Q⁰₁,0, . . . ,0]

among Q⁰₁, . . . , Q⁰_a. By the claim there are polynomials di such that Pb

i=1diSi = [Q⁰₂,−Q⁰₁,0, . . . ,0], giving the contradictionQ⁰₂= 0. 2 In many cases, to compute the corank of Gaussian maps, or, as in Proposition 2.4, to compute a suitable cohomology group involving the normal bundle, it is quite convenient to know some surface containing the given curve. The result below will help to compute the cohomology of the normal bundle with the help of the surface.

Lemma 2.7. LetY ⊂P^rbe an integral subvariety that is scheme-theoretically an inter- section of quadrics and let X ⊂ Y be a smooth irreducible nondegenerate subvariety.

LetL=OY(1)andM a line bundle onX. Suppose that either (i) h⁰(2L_|X−M) = 0or

(ii) h⁰(2L_|X−M) = 1and the relations among the quadrics cutting outY are gener- ated by linear ones.

LetFX,Y = HomO_Pr(JY /P^r,OX). ThenH⁰(FX,Y ⊗M⁻¹) = 0.

Remark 2.8. WhenY is smooth we have thatFX,Y =N_{Y /P}r

|X. The fact thatY ⊂P^r is scheme-theoretically an intersection of quadrics certainly holds ifY satisfies property N1, that isY is projectively normal and its homogeneous ideal is generated by quadrics

([26, Def.1.2.5], [16]). Also the fact that the relations among the quadrics cutting outY are generated by linear ones certainly holds ifY satisfies propertyN2, that isY satisfies propertyN1and the relations among the quadrics generating its homogeneous ideal are generated by linear ones ([26, Def.1.2.5], [16]). The difference, in our case, is that we do not assumeY to be linearly normal.

Proof of Lemma 2.7. Let{Q1, . . . , Q_a} be linearly independent quadrics cutting outY scheme-theoretically and consider the corresponding beginning of the minimal free reso- lution ofJ_{Y /P}r:

M

i≥0

O_P^r(−3−i)^⊕bi −→ O_P^r(−2)^⊕a−→ JY /P^r−→0.

Applying the left exact functorHomO_Pr(−,OX)we get an exact sequence 0−→ FX,Y −→ OX(2)^⊕a−→M

i≥0

OX(3 +i)^⊕bi

whence an exact sequence

0−→H⁰(FX,Y ⊗M⁻¹)−→H⁰(2L_|X−M)^⊕a−→^ϕ M

i≥0

H⁰ (3 +i)L_|X−M⊕bi

.

ThenH⁰(F_X,Y ⊗M⁻¹) = Kerϕ.

If we are under hypothesis (i), then obviouslyH⁰(F_X,Y ⊗M⁻¹) = 0.

If we are under hypothesis (ii), thenb_i = 0fori≥1and we will prove thatKerϕ= 0.

To this end letσbe a generator ofH⁰(2L_|X−M). For1≤i≤b0letRi= [Ri1, . . . , Ria] be the linear relations generating all relations amongQ1, . . . , Qa, so that the mapϕis given by the matrix(Rij|X). If0 6= (c1σ, . . . , caσ)∈Kerϕthen, for everyisuch that 1 ≤ i ≤ b0, we have Pa

j=1Rij|Xcjσ = 0 whence(Pa

j=1cjRij)_|X = 0. As X is nondegenerate andPa

j=1c_jR_ij is a linear polynomial, we deduce thatPa

j=1c_jR_ij = 0

for alliwith1≤i≤b₀, contradicting Lemma 2.5. 2

Now the first general result about Gaussian maps.

Proposition 2.9. LetC be a smooth irreducible nonhyperelliptic curve of genusg ≥ 3 and letM be a line bundle onC. We have

(a) Ifg= 3thencork Φ_M,ωC ≥h⁰(4K_C−M)−corkµ_M,ωC −3h¹(M), with equality ifH⁰(−M) = 0.

(b) Ifg= 4thencork ΦM,ω_C ≥h⁰(2KC−M)+h⁰(3KC−M)−corkµM,ω_C−4h¹(M), with equality ifH⁰(−M) = 0.

(c) Ifg= 5andCis nontrigonal thencork ΦM,ω_C ≥3h⁰(2KC−M)−corkµM,ω_C − 5h¹(M), with equality ifH⁰(−M) = 0.

(d) Suppose thatC is a plane quintic andA is the very ampleg²₅ onC. IfH⁰(5A− M) = 0then ΦM,ω_C is surjective. If H¹(M) = 0 andµM,ω_C is surjective then cork ΦM,ω_C ≥h⁰(5A−M), with equality holding if in additionh⁰(4A−M)≤1.

(e) Suppose thatC is trigonal,g ≥ 5andAis ag₃¹onC. Ifh⁰(2KC−M)≤1and H⁰(3KC−(g−4)A−M) = 0thenΦM,ω_C is surjective. IfH¹(M) = 0andµM,ω_C

is surjective thencork ΦM,ω_C ≥h⁰(3KC−(g−4)A−M), with equality holding if in additionh⁰(2KC−M)≤1.

Proof. Assertions (a), (b) and (c) follow easily from Proposition 2.4.

Let us prove (d). In the canonical embeddingC ⊂P⁵we have thatCis contained in the Veronese surfaceY and we have an exact sequence

0−→N_C/Y ⊗M⁻¹−→N_C/P5⊗M⁻¹−→N_{Y /P}5

|C⊗M⁻¹−→0. (5)

Observe thath⁰(N_C/Y ⊗M⁻¹) = h⁰(5A−M). Now ifh⁰(5A−M) = 0then also h⁰(2K_C−M) = h⁰(4A−M) = 0and from (5) and Proposition 2.4 (a), we see that to prove (d) we just need to show thatH⁰(N_{Y /P}5

|C⊗M⁻¹) = 0. The latter follows by Lemma 2.7 and Remark 2.8 since, as is well-known,Y satisfies propertyN3.

Now ifH¹(M) = 0andµ_M,ωC is surjective, we have thatcork Φ_M,ωC =h⁰(N_C/P5

⊗M⁻¹)≥h⁰(5A−M)by Proposition 2.4 (b) and (5). If we also assume thath⁰(4A− M) =h⁰(2K_C−M)≤1then we can apply again Lemma 2.7 and Remark 2.8. We get thath⁰(N_{Y /P}5

|C⊗M⁻¹) = 0, whence, from (5), thath⁰(N_C/P5⊗M⁻¹) =h⁰(5A−M).

To see (e) recall that, in the canonical embeddingC ⊂P^g−1, we have [33, 6.1] that C∈ |3H−(g−4)R|on a rational normal surfaceY ⊂P^g−1, whereHis its hyperplane bundle andRits ruling. Since, as is well-known,Y satisfies propertyN_g−3, applying, as

in Case (d), Lemma 2.7 and Proposition 2.4 we get (e). 2

Note that the Cases (d), (e) of the above proposition and the corollary below are a slight improvement of [35, Theorem 2.4] (because we also consider the caseh⁰(2KC− M) = 1).

Corollary 2.10. LetCbe a smooth irreducible curve of genusg≥5and letMbe a line bundle onC. Then the Gaussian mapΦ_M,ωC is surjective if one of the hypotheses below is satisfied:

(a) Cis a plane quintic anddegM ≥25,M 6= 5Aif equality holds, whereAis the very ampleg²₅onC;

(b) Cis trigonal anddegM ≥max{4g−6,3g+ 6},M 6= 3K_C−(g−4)Aifg≤12 anddegM = 3g+ 6, whereAis ag¹₃onC.

Proof. (a) follows immediately from 2.9(d) while (b) is a consequence of 2.9(e) since, if h⁰(2K_C−M)≥2, thendeg(2K_C−M)≥3, a contradiction. 2

Another easy but useful consequence of the proof of Lemma 2.7 is the following.

Proposition 2.11. LetCbe a smooth irreducible curve of genusg ≥5and letM be a line bundle onC. Suppose that either

(i) Cliff(C) = 2andh⁰(2KC−M) = 0or (ii) Cliff(C)≥3andh⁰(2KC−M)≤1.

ThenΦM,ω_C is surjective.

Proof. SinceCliff(C)≥2, by [37], [34], the resolution of the ideal sheaf of the canonical embeddingC⊂P^g−1starts as

M

i≥0

O_P^g−1(−3−i)^⊕bi −→ O_P^g−1(−2)^⊕a −→ J_C/P^g−1 −→0

withbi= 0fori≥1whenCliff(C)≥3. Restricting toCand dualizing we get an exact sequence

0−→N_C/Pg−1 −→ OC(2)^⊕a −→M

i≥0

OC(3 +i)^⊕bi

whence an exact sequence

0−→H⁰(N_C/Pg−1⊗M⁻¹)−→H⁰(2KC−M)^⊕a−→^ϕ M

i≥0

H⁰ (3 +i)KC−M⊕bi

.

As in the proof of Lemma 2.7 we have thatH⁰(NC/P^g−1⊗M⁻¹) = 0under hypothesis (i) andH⁰(N_C/Pg−1 ⊗M⁻¹) = Kerϕ = 0under hypothesis (ii). Therefore we conclude

by Proposition 2.4 (a). 2

Using an appropriate generalization of the methods of [5, Proof of Theorem 2] we can also get surjectivity whenh⁰(2KC−M)≥2.

Proposition 2.12. LetC be a smooth irreducible curve of genus g ≥ 4and letM be a line bundle on C. Suppose there exists an integerm ≥ 1 and an effective divisor D=P1+· · ·+Pmsuch that

(i) H¹(M −2Pi) = 0for1≤i≤m;

(ii) h⁰(D) = 1andh⁰(2KC−M−D) = 0;

(iii) m≤Cliff(C)−2.

ThenΦM,ω_C is surjective.

Proof. As is well-known we haveCliff(C)≤ b^g−1₂ c, whencem≤Cliff(C)−2≤g−4.

We start by observing thatK_C−Dis very ample. In fact, ifK_C−Dis not very ample, there are two pointsQ₁, Q₂∈Csuch that

h⁰(KC−D−Q1−Q2) =h⁰(KC−D)−1 =g−2−m+h⁰(D) =g−1−m

whenceh¹(D+Q1+Q2) =g−1−m≥3andh⁰(D+Q1+Q2) = 2by Riemann–

Roch. Therefore D +Q1+Q2 contributes to the Clifford index of C and we have Cliff(C)≤Cliff(D+Q1+Q2) =m, contradicting (iii).

Consider the embeddingC⊂PH⁰(KC−D) =P^r, wherer=g−1−m. We claim that, in the latter embedding,Chas no trisecant lines. As a matter of fact if there exist three pointsQ1, Q2, Q3 ∈ Csuch that their linear spanhQ1, Q2, Q3iis a line, we have that

1 = dimhQ1, Q2, Q3i=h⁰(KC−D)−1−h⁰(KC−D−Q1−Q2−Q3)

=g−1−m−h⁰(K_C−D−Q₁−Q₂−Q₃)

whenceh¹(D+Q1+Q2+Q3) =g−2−m≥2and againh⁰(D+Q1+Q2+Q3) = 2.

ThereforeD+Q1+Q2+Q3contributes to the Clifford index ofCand we getCliff(C)≤ Cliff(D+Q1+Q2+Q3) =m+ 1, contradicting (iii).

Note further that by (ii) and (iii) we have

deg(K_C−D) = 2g−2−m≥2g+ 2−2h¹(K_C−D)−Cliff(C)

therefore Green–Lazarsfeld’s theorem [26, Proposition 2.4.2] gives that C is scheme- theoretically cut out by quadrics inP^r. Hence we have a surjection

OC(2D−2K_C)^⊕α→N_C/^∗

P^r →0.

Setting, as in [5],RL =N_C/^∗

PH⁰(L)⊗Lfor any very ample line bundleL, we deduce a surjection

OC(M−KC+D)^⊕α→RK_C−D⊗M →0.

By (ii), we have thatH¹(M −KC+D) =H⁰(2KC−M−D)^∗= 0whence H¹(R_KC−D⊗M) = 0.

Now there is an exact sequence [5, 2.7], [13, Proof of Theorem 5]

0−→RK_C−D⊗M −→RK_C⊗M −→

m

M

i=1

OC(M−2Pi)−→0

and therefore by (i) we deduce that

H⁰(N_C/Pg−1⊗M⁻¹)∼=H¹(N_C/^∗ _Pg−1⊗ωC⊗M)^∗∼=H¹(RK_C ⊗M)^∗= 0.

Hence we get the surjectivity ofΦM,ω_C by Proposition 2.4 (a). 2 We will often use the above result in the following simplified version.

Corollary 2.13. LetCbe a smooth irreducible curve of genusg≥4and letMbe a line bundle onCsuch thatH¹(M) = 0anddeg(M)≥g+ 1. Suppose that

h⁰(2K_C−M)≤Cliff(C)−2.

ThenΦM,ω_C is surjective.

Proof. Letm= Cliff(C)−2. Thenm≥0by hypothesis and whenm= 0the surjectiv- ity ofΦM,ω_C holds by Proposition 2.11. Whenm≥1choose general pointsP1, . . . , Pm

ofCand apply Proposition 2.12. 2

Corollary 2.14. [35, Corollary 1.7] LetCbe a smooth irreducible curve of genusg≥5 nontrigonal and not isomorphic to a plane quintic. LetM be a line bundle onC.

Then the Gaussian mapΦ_M,ωC is surjective ifdegM ≥ 4g−4andM 6= 2KCif equality holds.

Proof. Immediate consequence of Corollary 2.13 or of Proposition 2.11. 2

2.2 Gaussian maps on tetragonal curves. In this subsection we improve Tendian’s results [35] about Gaussian maps on tetragonal curves. Moreover note that, even though the statement in [35, Theorem 2.10] is almost correct, the proof certainly contains a gap (see Remark 2.17).

We start with some generalities on tetragonal curves following again [33, 6.2].

Definition-Notation 2.15. LetCbe a smooth irreducible tetragonal curve of genusg≥6 not isomorphic to a plane quintic. LetAbe ag₄¹onCand letV_A⊂P^g−1=PH⁰(ω_C)be the rational normal scroll spanned by the divisors in|A|,H_Athe hyperplane bundle and R_Aa ruling ofV_A. LetE_Abe the rank 3 vector bundle onP¹so thatV_Ais the image of PEAunder the morphism given by|O_PEA(1)|. LetHeAandReAbe the pull-backs, under this morphism, ofHA andRArespectively. Then there are two integersb1,A, b2,Asuch thatb_1,A≥b_2,A≥0,b_1,A+b_2,A=g−5and there are two surfacesYe_A∼2He_A−b1,ARe_A, ZeA∼2HeA−b2,AReAsuch that, ifYA, ZAare their images inP^g−1thenC=YA∩ZA. We also define

b2(C) = min

b2,A, Aag₄¹onC . We have

Lemma 2.16. The surfaceY_A⊂P^g−1has degreeg−1 +b_2,Aand satisfies propertyN₂. Proof. We set for simplicityY =YA,Ye =YeA,V =VA,E =EA,H =HA,R=RA, He = HeA,Re = ReA,bi = bi,A, i = 1,2. Note thatHe³ = degV = g−3,Re² = 0 andHe².Re= 1. LetXe ∈ |O

Ye(He)|be a general curve. Since|O

Ye(He)|is not composed with a pencil we have thatXeis irreducible. MoreoverXeis smooth outsideHe∩Sing(Ye), whenceXe is also reduced.

LetL = O

Xe(H), Xe = Y ∩H, so thatϕ_L(Xe) = X. We will first prove thatX satisfies propertyN2. To this end by [3, Theorem A] it is enough to show that

degX ≥2p_a(X) + 3. (6)

Taking intersections inPEwe have

degX= degL=He²·Ye =He²·(2He−b₁R) = 2ge −6−b₁=g−1 +b₂. (7) On the other hand, using the cohomology of the scroll, we get

pa(X) = 1−χ(OX) = 1−χ(OY) +χ(OY(−1))

= 1−χ(OV) +χ(OV(−2H+b1R)) +χ(OV(−1))−χ(OV(−3H+b1R))

=g−4−b₁.

Now2pa(X) + 3 = 2g−5−2b1 ≤ 2g−6−b1if and only ifb1 ≥1. The latter holds becauseb1≥b2≥0andg≥6. Therefore (6) is proved.

Again using the cohomology of the scroll it is easy to prove thatH¹(J_{Y /P}^g−1(j)) = 0 for everyj∈Zand thatH¹(OY(j)) = 0for everyj≥0. Applying [16, Theorem 2.a.15 and Theorem 3.b.7] (that hold for any scheme) we deduce thatY satisfies propertyN2

sinceY ∩Hdoes. 2

Remark 2.17. In Tendian’s paper it is assumed that a general hyperplane sectionY ∩H is smooth, but in fact it can be singular [33, 6.5] when theg₄¹exhibitsCas a double cover of an elliptic or hyperelliptic curve.

Proposition 2.18. LetC be a smooth irreducible tetragonal curve of genusg ≥ 6 not isomorphic to a plane quintic. LetAbe ag₄¹, setb2=b2,Aand letMbe a line bundle on C. We have

(i) Ifh⁰(2KC−M)≤1andh⁰(2KC−M−b2A) = 0, thenΦM,ω_C is surjective;

(ii) IfH¹(M) = 0andµM,ω_C is surjective, thencork ΦM,ω_C ≥h⁰(2KC−M−b2A), with equality holding ifh⁰(2KC−M)≤1.

Proof. LetY be the surface arising in the scroll defined byAand set, as in Lemma 2.7, FC,Y = Hom_O

Pg−1(J_{Y /P}^g−1,OC). Applying the left exact functorHom_O

Pg−1(−,OC) to the exact sequence

0−→ J_{Y /P}^g−1−→ J_C/P^g−1 −→ J_C/Y −→0

we get an exact sequence

0−→N_C/Y ⊗M⁻¹−→N_C/Pg−1⊗M⁻¹−→ FC,Y ⊗M⁻¹. (8) Observe thath⁰(N_C/Y⊗M⁻¹) =h⁰(2K_C−M−b₂A). Now ifh⁰(2K_C−M−b₂A) = 0, from (8) and Proposition 2.4 (a), we see that to prove (i) we just need to show that

ifh⁰(2K_C−M)≤1thenH⁰(F_C,Y ⊗M⁻¹) = 0. (9) On the other hand, under the hypotheses in (ii), we have thatcork Φ_M,ωC =h⁰(N_C/Pg−1

⊗M⁻¹)by Proposition 2.4 (b). Now from (8) we get that h⁰(N_C/Pg−1 ⊗M⁻¹) ≥ h⁰(2K_C−M−b₂A)and to prove equality we need again to prove (9).

To conclude we just note that (9) holds by Lemmas 2.7 and 2.16. 2

3 Linear series on quadric sections of surfaces of degreeg−1inP^g−1 In this section we will use some well-known vector bundle methods ([26], [36]) to study linear series on curves of genusgthat are, in their canonical embedding, a quadric section of a surface of degreeg−1inP^g−1. We recall that when the surface is a smooth Del Pezzo the gonality and Clifford index of such curves are known by [31], [21]. Most of the results we prove are probably known, at least in the smooth case, but we include them anyway for completeness’ sake.

Lemma 3.1. Let X be a smooth surface with−KX ≥ 0. Let C ⊂ X be a smooth irreducible curve of genusgand letAbe a base-point freeg_k¹onC. Suppose that2g− 2−KX.C−4k≥max{0,3−4χ(OX)}and, ifh¹(OX)≥1, thath⁰(N_C/X⊗A⁻¹)≥ 2h¹(OX) + 1. Then there exist two line bundles L, M on X and a zero-dimensional subschemeZ⊂Xsuch that the following hold:

(i) C∼M+L;

(ii) k=M.L+ length(Z)≥M.L≥L²≥0;

(iii) there exists an effective divisor DonC of degreeM.L+L²−k ≥ 0 such that A∼=L_|C(−D);

(iv) ifL²= 0thenM.L=kandA∼=L_|C; (v) Lis base-component free and nontrivial;

(vi) ifC∼ −2KXthen3L²+M.L∈4Z.

Proof. LetF = Ker{H⁰(A)⊗ OX →A}andE =F^∗. As is well-known ([26])Eis a rank two vector bundle sitting in an exact sequence

0−→H⁰(A)^∗⊗ OX −→ E −→N_C/X⊗A⁻¹−→0 (10) and moreoverc₁(E) =Candc₂(E) =k, so that∆(E) :=c₁(E)²−4c₂(E) =C²−4k= 2g−2−K_X.C−4k≥0. LetHbe an ample line bundle onXand suppose thatEisH- stable. Thenh⁰(E ⊗E^∗) = 1by [14, Corollary 4.8] andh²(E ⊗E^∗) =h⁰(E ⊗E^∗(K_X))≤ h⁰(E ⊗ E^∗) = 1, therefore2≥h⁰(E ⊗ E^∗) +h²(E ⊗ E^∗) =h¹(E ⊗ E^∗) +χ(E ⊗ E^∗)≥ 4χ(OX) + ∆(E)≥3, a contradiction. HenceEis notH-stable and ifM is the maximal destabilizing subbundle we have an exact sequence

0−→M −→ E −→ JZ/X⊗L−→0 (11) whereLis another line bundle onXandZis a zero-dimensional subscheme ofX. Com- puting Chern classes in (11) we get (i) and the equality in (ii). Since the destabilizing condition reads(M −L).H ≥0and since(M −L)² = ∆(E) + 4 length(Z)≥0, we see thatM −Lbelongs to the closure of the positive cone ofX. We now claim thatE is globally generated off a finite set. In fact ifh¹(OX)≥1we have by hypothesis that h⁰(N_C/X ⊗A⁻¹) ≥ 2h¹(OX) + 1and the claim follows by (10) since the mapψ : H⁰(E)→H⁰(N_C/X⊗A⁻¹)is nonzero. On the other hand ifh¹(OX) = 0we have that ψis surjective, whence, again by (10), we just need to prove thath⁰(N_C/X⊗A⁻¹)≥1.

Sinceg ≥2k+ 1 + ¹₂KX.Cwe getdeg(N_C/X⊗A⁻¹) = 2g−2−KX.C−k ≥g.

Thereforeh⁰(N_C/X⊗A⁻¹)≥1by Riemann–Roch and the claim is proved.

SinceE is globally generated off a finite set then so isL. It follows thatL ≥ 0,L is base-component free andL² ≥0. Now the signature theorem [4, VIII.1] implies that (M −L).L ≥ 0thus proving (ii). To see (iii) and (iv) note that ifM.L > 0 then the nefness ofLimplies thatH⁰(−M) = 0. On the other hand ifM.L = 0thenL² = C.L = 0whenceL ≡ 0 by the Hodge index theorem and thereforeC ≡ M. Then M.H = C.H > 0whence again H⁰(−M) = 0. Twisting (10) and (11) by−M we deduce thath⁰(L_|C⊗A⁻¹)≥h⁰(E(−M))≥1. This proves (iii) and also (v). Moreover it givesdeg(L_|C⊗A⁻¹)≥0, whence, ifL²= 0, we get thatM.L≥k. By (ii) it follows thatM.L=kand thereforedeg(L_|C⊗A⁻¹) = 0, whenceL_|C∼=A. This proves (iv).

Finally suppose thatC∼ −2KX. We haveχ(L) =χ(OX) +¹₂L.(L−KX)whence 2L.(L−K_X)is divisible by4. But2L.(L−K_X) = 2L²+L.C= 3L²+M.L, giving

(vi). 2

We now analyze linear series on curves on surfaces of degreerinP^r. We will use the following

Definition-Notation 3.2. For1≤n ≤9we denote byΣnthe blow-up ofP² atnpos- sibly infinitely near points, byHe the strict transform of a line and byGithe total inverse image of the blown-up points. LetQ⊂P³be a quadric cone with vertexV. We denote by BlV Qthe blow-up ofQalongV and byHe the strict transform of a plane. LetCn ⊂Pⁿ be the cone over a smooth elliptic curve inPⁿ⁻¹and letV be the vertex. We denote by Bl_V C_nthe blow-up ofC_nalongV, byC₀the inverse image ofV and byfthe numerical class of a fiber.

Remark 3.3. We recall that by [30, Theorem 8] a linearly normal integral surfaceY ⊂P^r of degreeris either the anticanonical image ofΣ_9−rorC_ror the 2-Veronese embedding inP⁸of an irreducible quadric inP³or the 3-Veronese embedding inP⁹ofP².

Proposition 3.4. LetX be a surface amongΣn,BlV QorBlV Cn as in Definition 3.2 and letCbe a smooth irreducible curve such that, ifX = Σ_norBl_VQthenC∼ −2KX, while ifX = Bl_V C_nthenC≡ −2KX−2C₀. We have:

(a) ifX = Σ1thenChas no complete base-point freeg₆¹;

(b) ifX = Σ2then every complete base-point freeg₄¹onCis(He−Gi)_|C,i= 1,2;

(c) ifX = Σ2then every complete base-point freeg₆¹onCis(2He−G1−G2)_|C−P1−P2, whereP1, P2are two points ofC;

(d) ifX = Σ3then every complete base-point freeg₄¹onCis(He−Gi)_|C,i= 1,2,3;

(e) ifX = Σ₃then every complete base-point freeg¹₅onCis eitherHe_|C−P or(2He− G₁−G₂−G₃)_|C−P, for some pointP ∈C;

(f) ifX = Σ3andAis a complete base-point freeg₆¹onCthen eitherA∼= (2He−Gi− Gj)_|C−P1−P2, for1≤i < j≤3andP1, P2are two points ofCor(−KX)_|C−Ais another complete base-point freeg₆¹onCdifferent from(2He−Gi−Gj)_|C−P1−P2; (g) ifX = BlV C6thenChas no complete base-point freeg₅¹and every complete base-

point freeg₄¹onCis(f1+f2)_|C, wheref1, f2are two fibers;

(h) ifX = BlV QthenChas a unique complete base-point freeg₄¹, namelyf_|C, where f is the pull-back of a line of the coneQ;

(i) ifX = BlV Qthen every complete base-point freeg¹₆onCisHe_|C−P1−P2, where P1, P2are two points ofC;

(j) ifX= Bl_V Qthen there is no effective divisorZ⊂Csuch thatf_|C+Zis a complete base-point freeg²₈onC.

Proof. We record, for later use, the following fact onX = Σn. LetLbe a nef line bundle onXwithL ∼aHe −Pn

i=1biGi. Then a=L.He ≥0, bi=L.Gi≥0, L²=a²−

n

X

i=1

b²_i, L.(−KX) = 3a−

n

X

i=1

bi (12) and the Cauchy–Schwartz inequality(Pn

i=1bi)²≤nPn

i=1b²_i implies that

(3a+L.KX)²≤n(a²− L²). (13) We will now apply Lemma 3.1 to a base-point freeg_k¹indicated in (a)-(i) and we will set z= length(Z).

(a) We haveK_X² = 8whenceC²= 32, k= 6and from (ii) of Lemma 3.1 we deduce that6 =M.L+z≥M.L≥L²≥0. Now if3≤L²≤6we have a contradiction by the Hodge index theorem applied toCandL. The same theorem implies, forL² = 2, that C ≡4L. ButC ∼6He −2G1whence the contradiction4L.He =C.He = 6. IfL²= 1 writeL ∼aHe −b1G1. Thena² =b²₁+ 1thereforea= 1, b1 = 0andL ∼H. Thene deg(L_|C⊗A⁻¹) = H.Ce −6 = 0, whenceA∼=He_|Cby (iii) of Lemma 3.1. Therefore we have the contradictionh⁰(A) = 3. If L² = 0by (iv) of Lemma 3.1 we have that M.L= 6whence3L²+M.L= 6, contradicting (vi) of Lemma 3.1. This proves (a).

(b) We have K_X² = 7, C² = 28andk = 4. By (ii) of Lemma 3.1 and the Hodge index theorem applied toCandLwe see that we are left with the caseL² = 0whence A∼=L_|C. By (12), (13) we deduce thatL∼He −G_ifori= 1,2. This proves (b).

(c) We haveK_X² = 7whenceC² = 28andk= 6. From (ii) of Lemma 3.1 and the Hodge index theorem applied toCandLwe get0≤L²≤2. The same theorem implies, forL²= 2, thatz= 0, M.L= 6. By (iii) of Lemma 3.1 there are two pointsP₁, P₂∈C such thatA ∼=L_|C−P₁−P₂. By (12), (13) we deduce thatL ∼ 2He −G₁−G₂. If L²= 1again by (ii) of Lemma 3.1 and the Hodge index theorem applied toCandLwe get that0 ≤z ≤1and5 ≤M.L≤ 6. By (vi) of Lemma 3.1 we have thatM.L = 5 whencedeg(L_|C⊗A⁻¹) = 0, so thatA ∼= L_|C by (iii) of Lemma 3.1. By (12), (13) we deduce thatL ∼ He, giving the contradictionh⁰(A) = 3. IfL² = 0we have that M.L= 6by (iv) of Lemma 3.1 contradicting (vi) of Lemma 3.1. This proves (c).

(d) We haveK_X² = 6whenceC²= 24andk= 4. From (ii) of Lemma 3.1 and the Hodge index theorem applied toCandLwe get0≤L²≤1. The same theorem implies, forL² = 1, thatz = 0, M.L = 4, contradicting (vi) of Lemma 3.1. ThereforeL² = 0 and (iv) of Lemma 3.1 implies thatM.L= 4andA∼=L_|C. By (12), (13) we deduce that L∼He−Gi. This proves (d).

(e) We haveK_X² = 6whenceC² = 24andk= 5. From (ii) of Lemma 3.1 and the Hodge index theorem applied toC andLwe getL² ≤2 with equality only whenz = 0, M.L= 5, contradicting (vi) of Lemma 3.1. WhenL²= 1, the same theorem together with (vi) of Lemma 3.1 implies thatz = 0, M.L= 5, whenceA ∼=L_|C−P by (iii) of Lemma 3.1. By (12), (13) we deduce that eitherL∼He orL∼2He−G1−G2−G3. If L² = 0then (iv) of Lemma 3.1 implies thatM.L= 5, contradicting (vi) of Lemma 3.1.

This proves (e).

(f) We haveK_X² = 6whenceC² = 24andk= 6. From (ii) of Lemma 3.1 and the Hodge index theorem applied toCandLwe see, for3≤L² ≤5, thatz= 0, M.L= 6, contradicting (vi) of Lemma 3.1. IfL² = 2by the Hodge index theorem and (vi) of Lemma 3.1 we have thatz= 0, M.L= 6. By (12), (13) we deduce thatL∼2He−Gi− Gjfori6=jand by (iii) of Lemma 3.1 we have that there are two pointsP1, P2∈Csuch thatA∼=L_|C−P1−P2. IfL²= 1by the Hodge index theorem and (vi) of Lemma 3.1 we have thatz= 1, M.L= 5. By (12), (13) we deduce that eitherL∼HeorL∼2He−G1− G2−G3. By (iii) of Lemma 3.1 we have thatA∼=L_|C, giving the contradictionh⁰(A) = 3. IfL²= 0by (iv) of Lemma 3.1 we have thatM.L= 6contradicting (vi) of Lemma 3.1.

Finally whenL²= 6the Hodge index theorem applied toCandLimplies thatC ≡2L andz = 0. Therefore L ∼ M ∼ −KX whence the exact sequence (11) splits since Ext¹(OX(−KX),OX(−KX)) = 0and we getE ∼=OX(−KX)^⊕2. ThereforeE(KX)

is globally generated and so is (−KX)_|C ⊗A⁻¹ by (10). Moreover again by (10) we get that(−KX)_|C⊗A⁻¹ is ag₆¹. Also such ag₆¹ cannot coincide with the other type (2He −Gi−Gj)_|C−P1−P2, for otherwise we would have that(−KX)_|C⊗A⁻¹ ∼ (2He −G_i−G_j)_|C−P₁−P₂, whenceA ∼= (He −G_k)_|C+P₁+P₂would have two base points. This proves (f).

(g) We have thatX ∼=P(OE⊕ OE(−1))whereE⊂P⁵is a smooth elliptic normal curve. LetC₀be a section andf be a fiber so thatC₀²=−6and the intersection form is even. MoreoverC ≡2C₀+ 12f,C² = 24andk = 4,5. From (ii) of Lemma 3.1 and the Hodge index theorem applied toCandLwe deduce, ifL²≥2, thatk= 5,L²= 2, z = 0andM.L= 5. On the other hand ifL²= 0we have thatM.L=kandA∼=L_|C by (iv) of Lemma 3.1. Let L ≡ aC0+bf so thatM ≡ (2−a)C0+ (12−b)f and L²= 2a(b−3a). Moreover, by (v) of Lemma 3.1 we havea=f.L≥0. Now ifL²= 2 we geta= 1,b= 4giving the contradictionM.L= 6. ThereforeL²= 0whence either a= 0orb= 3a. In the second case we getk =M.L= 6a, a contradiction. Therefore a= 0andk=M.L= 2b, that isk= 4,b= 2andL≡2f as desired. This proves (g).

(h) We have thatX ∼=P(O_P1⊕ O_P1(−2)). LetC0be a section andf be a fiber so thatC₀²=−2and the intersection form is even. MoreoverC∼4C0+ 8f,C²= 32and k= 4. From (ii) of Lemma 3.1 and the Hodge index theorem applied toCandLwe have a contradiction ifL²≥2. HenceL²= 0, M.L= 4andA∼=L_|Cby (iv) of Lemma 3.1.

Then we get that eitherL∼f orL∼C₀+f. SinceC₀.C= 0, this proves (h).

(i) We retain the notation used in (h) except that nowk= 6. From (ii) of Lemma 3.1 and the Hodge index theorem applied toCandLwe deduce, ifL² ≥ 2, thatL² = 2, z = 0, M.L= 6andC ≡4L, whenceL ∼C₀+ 2f ∼He. By (iii) of Lemma 3.1 we have that there are two pointsP1, P2∈Csuch thatA∼=He_|C−P1−P2. WhenL²= 0 we getM.L= 6by (iv) of Lemma 3.1, contradicting (vi) of Lemma 3.1. This proves (i).

(j) Again we use the notation in (i). Suppose there is an effective divisorZ⊂Csuch thatf_|C+Zis a complete base-point freeg₈²onC. By Riemann–Roch we get that

h⁰((2C0+ 3f)_|C−Z) =h⁰(KC−f_|C−Z) = 3

and the exact sequence

0−→ OX(−2C0−5f)−→ J_Z/X(2C0+ 3f)−→ J_Z/C(2C0+ 3f)−→0

gives that alsoh⁰(J_Z/X(2C0+ 3f)) = 3, whence, sinceh⁰(2C0+ 3f) = 6, thatZdoes not impose independent conditions to|2C0+ 3f|. Now letZ⁰⊂Zbe an effective divisor of degree3and setZ⁰+P =Z. By the exact sequence

0−→ OX(−2C0−5f)−→ JZ⁰/X(2C0+ 3f)−→ JZ⁰/C(2C0+ 3f)−→0

and Riemann–Roch we have

h⁰(JZ⁰/X(2C0+ 3f)) =h⁰(JZ⁰/C(2C0+ 3f)) =h¹(f_|C+Z−P)

=h⁰(f_|C+Z−P) + 1 = 3.

ThereforeZ is in special position with respect to 2C0+ 3f ∼ L+KX, whereL ∼ 4C0+ 7f. By [32], [17], [7], [27] there is a rank2vector bundleE onX sitting in an

exact sequence

0−→ OX−→ E −→ J_Z/X⊗ L −→0 (14) withc1(E) = L andc2(E) = 4so that∆(E) = L²−16 = 8 > 0. ThereforeE is Bogomolov unstable and ([6], [32]) there are two line bundlesA, B onX and a zero- dimensional subschemeW ⊂X sitting in an exact sequence

0−→A−→ E −→ J_W/X⊗B−→0. (15) MoreoverL ∼A+B,A.B+ length(W) = 4,(A−B)²= 8 + 4 length(W)andA−B lies in the positive cone ofX.

We record for later use two extra properties ofAandB. For every nef line bundleM such thatM²≥0we have:

(A−B).M ≥0; (16)

A.M ≥0. (17)

To prove (16) and (17) letM be a nef line bundle such thatM² ≥0. ThenM.H ≥0 for every ampleH, whenceM lies in the closure of the positive cone ofX, therefore (A−B).M ≥0by [4, VIII.1]. Now ifA.M <0then alsoB.M <0by (16), whence h⁰(A) =h⁰(B) = 0, asM is nef. But this and (15) giveh⁰(E) = 0, contradicting (14).

Now(A−B)²≥8and(A+B)²=L²= 24therefore

A²+B²≥16. (18)

MoreoverLlies in the positive cone ofX, whence, by [4, VIII.1],(A−B).L>0, that is

A²> B². (19)

Now ifA²≤8we deduce by (19) thatB²≤6, contradicting (18). Therefore

A²≥10. (20)

Suppose thatA∼aC0+a1f so thatB ∼(4−a)C0+ (7−a1)f. IntersectingAwith the nef divisorsf, C0+ 2fand using (17), we see thata≥0, a1≥0, whenceA≥0and in factA >0by (20). Alsoa >0, for otherwiseA²= 0. Now the exact sequences (14) and (15) twisted by−Agive

h⁰(J_Z/X(B))≥h⁰(E(−A))≥1 (21) whence alsoB >0. The nefness ofC₀+ 2f then implies7−a₁ =B.(C₀+ 2f)≥0, whencea1 ≤7, while the nefness off implies that4−a =B.f ≥0, whencea≤4.

By (16) withM =C0+ 2fwe get2a1−7 = (A−B).(C0+ 2f)≥0, whencea1≥4.

Finally by (20) we have thata(a1−a)≥5. Therefore we have proved that

1≤a≤4, 4≤a₁≤7, a(a₁−a)≥5. (22) Ifa= 1,2we get thatA²+B²≤12, contradicting (18). Recall now thatC0∩C=∅ sinceC0.C = 0. Whena= 3we haveA²+B²= 4a1−6whencea1= 6,7by (18).

Whena1 = 7we haveB ∼C0, whenceB =C0. By (21) we deduce the contradiction Z ⊂C0∩C=∅. Whena1 = 6we haveB ∼C0+f, whenceB =C0∪F for some rulingF. As above we have thatZ∩C0 =∅, whenceZ ⊂F ∩C. SinceF.C = 4we have thatZ = F ∩C, whenceZ ∼ f_|C and thereforef_|C+Z ∼ 2f_|C is a complete base-point freeg²₈onC. This is of course a contradiction since onXwe have that2f_|Cis a complete base-point freeg³₈onC. Finally whena= 4we haveB∼(7−a₁)fwhence a₁ ≤6asB > 0. By (22) we geta₁ = 6whenceB ∼ f, therefore againB =F for some rulingF. HenceZ ⊂F∩C, giving the same contradiction above. This proves (j).

2 Remark 3.5. LetCbe a smooth tetragonal curve of genus7such thatdimW₄¹(C) = 0 anddimW₅¹(C) = 1(as in the caseC ∼ −2K_X onX = Σ₃). By [1]W₆¹(C)has an irreducible component of dimension at least3and whose general elementAis a complete g₆¹onC. MoreoverAis base-point free sincedimW₄¹(C) = 0anddimW₅¹(C) = 1.

Also the same holds forKC−Athus proving that, for these curves, there is a family of dimension at least 3of complete base-point freeg¹₆’s whose residual is also base-point free.

4 Some results on Enriques surfaces

We will use the following well-known

Definition 4.1. LetLbe a line bundle on an Enriques surfaceS such thatL² >0. Fol- lowing [12] we define

φ(L) = inf

|F.L|:F∈PicS, F²= 0, F 6≡0 .

This function has two important properties:

(i) φ(L)²≤L²([12, Corollary 2.7.1]);

(ii) IfLis nef, then there exists a genus one pencil|2E|such thatE.L=φ(L)(by [11, 2.11] or by [12, Corollary 2.7.1, Proposition 2.7.1 and Theorem 3.2.1]).

We will often use the

Definition 4.2. LetSbe an Enriques surface. A nodal curve onS is a smooth rational curve contained inS.

We will now briefly recall some results on line bundles on Enriques surfaces, proved in [24] and [22], that we will often use.

Lemma 4.3. [22, Lemma 2.2] LetL >0and∆>0be divisors on an Enriques surface S withL² ≥0,∆² =−2andk :=−∆.L > 0. Then there exists anA > 0such that A²=L²,A.∆ =kandL∼A+k∆. Moreover ifLis primitive then so isA.