The gonality of complete intersection curves
aa r X i v : . [ m a t h . AG ] M a y THE GONALITY OF COMPLETE INTERSECTION CURVES
JAMES HOTCHKISS AND BROOKE ULLERY
Abstract.
The purpose of this paper is to show that for a complete intersection curve C inprojective space (other than a few exceptions stated below), any morphism f : C → P r sat-isfying deg f ∗ O P r (1) < deg C is obtained by projection from a linear space. In particular, weobtain bounds on the gonality of such curves and compute the gonality of general completeintersection curves. We also prove a special case of one of the well-known Cayley-Bacharachconjectures posed by Eisenbud, Green, and Harris. Introduction
Let C be a projective curve. Recall that the gonality of C , gon( C ), is the minimum degreeof a surjective morphism ˜ C −→ P , where ˜ C is the normalization of C . Thus, C is rational precisely when gon( C ) = 1, and, moregenerally, gonality measures how far the curve is from being rational. Gonality is a classicalinvariant, and there has been significant interest in bounding the gonality of various classesof curves and characterizing the corresponding maps to P . Specifically, if C is embedded inprojective space, it is natural to ask whether the gonality is related to the embedding of thecurve.For example, the gonality of plane curves (i.e. complete intersection curves of codimensionone) is well understood. If C ⊂ P is a smooth curve of degree d , then the map C → P obtained by projecting from a point in C has degree d −
1, giving an upper bound on thegonality. In fact, a classical theorem of Noether states that if d ≥
3, thengon( C ) = d − , and any covering of P of degree d − C is a smooth degree d curve in projective space and V a base point freelinear system of degree at most d , one may ask when the morphism defined on C by V isobtained by projection from a linear subspace of codimension dim( V ) + 1. This question wasfirst studied for complete intersection curves in P by Ciliberto and Lazarsfeld [CL84] andlater by Basili [Bas96]. The former authors studied this question for dim( V ) = 2, the latterfor dim( V ) = 1. Specifically, Basili showed that if C ⊂ P is a smooth complete intersectioncurve, then the gonality is indeed computed by projection from a line and every minimalcovering arises in this way. Recently, Hartshorne and Schlesinger generalized Basili’s resultsto smooth ACM curves in P with the exception of a few cases [HS11]. The same resultholds for many other specific classes of curves in P (e.g. [Bal97], [EF01], [Far01], [Har02], Mar96], ). See [HS11] for a detailed review on curves in P whose gonality is computed byprojection from a line.At the other extreme, i.e. higher dimensional hypersurfaces, the recent paper [BDE + X is a smooth variety of dimension n , then the degree of irrationality of X , irr( X ),is the minimum degree of a dominant rational map X P n . Clearly, this definition agreeswith the definition of gonality in the n = 1 case, and irr( X ) = 1 if and only if X is rational.The main result of [BDE +
16] states that if X ⊂ P n +1 is a very general hypersurface of degree d ≥ n + 1, then irr( X ) = d −
1, and if d ≥ n + 2, then any dominant map X P n isobtained by projecting from a point on X .Returning to curves, the main result about gonality of complete intersection curves inhigher dimensional projective spaces to date is a lower bound due to Lazarsfeld: Theorem 1.1 (cf. [Laz97], Exercise 4.12) . Let C ⊂ P n be a smooth complete intersectioncurve of type ( a , a , . . . , a n − ) , where ≤ a ≤ · · · ≤ a n − . Then gon( C ) ≥ ( a − a · · · a n − . In light of the previous examples, one may ask whether every such map is given by projection.Our main result answers confirms this in a more general setting as long as C satisfies milddegree restrictions: Theorem 1.2.
Let C ⊂ P n be a complete intersection curve of type ( a , . . . , a n − ) , with ≤ a < a ≤ · · · ≤ a n − . Then for r < n , any morphism f : C → P r satisfying deg f ∗ O P r (1) < deg C is obtained by projecting from an ( n − r − -plane. Thus gon( C ) = a a · · · a n − − γ , where γ is the maximum number of points on C contained in an ( n − -plane.Remark . In fact, we can weaken the hypotheses of Theorem 1.2 slightly. As long as C lies on a smooth complete intersection threefold of type ( w , · · · , w n − ) and the remainingtwo degrees a and b cutting out C satisfy 4 ≤ a < b , then the conclusion of the theoremholds.By a dimension count, we are able to compute the value of γ from the previous theoremfor C general. This leads to the following corollary. Corollary 1.4.
Let C ⊂ P n be a general complete intersection curve of type ( a , . . . , a n − ) ,with ≤ a < a ≤ · · · ≤ a n − and n ≥ . Then gon( C ) = a a · · · a n − − n + 2 . The proof of Theorem 1.2 relies on a generalization of the classical Noether-Lefschetz The-orem [Lef21], which we prove in Section 3. It states that, under certain degree restrictions,a complete intersection curve in projective space lies on a complete intersection surface with icard group generated by the hyperplane class. However, Theorem 1.2 also holds in themore general setting of arbitrary curves lying on surfaces with Picard group Z : Theorem 1.5.
Let C ⊂ P n be a smooth non-degenerate curve lying on a smooth surface S with Pic( S ) = Z · [ O S (1)] , and C ∈ |O S ( α ) | , where α ≥ . Then any morphism f : C → P r with r < n satisfying deg f ∗ O P r (1) < deg C is obtained by projecting from an ( n − r − -plane.Remark . In [Ras15], Rasmussen independently showed that, under stronger hypotheses,the fibers of a morphism C → P of degree less than the degree of C must lie in hyperplanes,which follows from our Lemma 4.4. Remark . A special case of one of the Cayley-Bacharach conjectures posed by Eisenbud,Green, and Harris (Conjecture CB12 in [EGH96]) follows easily from the proof of Theorem1.5. We discuss this in Section 5.We also obtain a bound on the gonality in this more general setting:
Corollary 1.8.
With C and S as in Theorem 1.5, deg( C ) − deg( S ) ≤ gon( C ) . If, in addition, C is linearly normal, then gon( C ) ≤ deg( C ) − n + 3 . We would like to thank Asher Auel, Ciro Ciliberto, Joe Harris, Dave Jensen, HannahLarson, Rob Lazarsfeld, Jake Levinson, Ian Shipman, David Stapleton, and Isabel Vogtfor helpful comments and conversations. The research of the second author was partiallysupported by an NSF Postdoctoral Fellowship, DMS-1502687.Regarding conventions, we will often abuse notation by using additive divisor notationwhen dealing with line bundles. Throughout, we work over the complex numbers.2.
The Cayley-Bacharach condition
Suppose Z is a set of distinct points on a smooth variety X of dimension n . Let L be aline bundle on X . The set of points Z satisfies the Cayley-Bacharach condition with respectto the complete linear system | L | if every section of L vanishing at all but one of the pointsof Z also vanishes at the remaining point. Remark . If Z satisfies the Cayley-Bacharach condition with respect to | L | , then Z failsto impose independent conditions on L . However, the converse is not true. For instance,if P , P , P ∈ P lie on a line ℓ , and P / ∈ ℓ , then the set of points { P , P , P , P } doesnot impose independent conditions on |O P (1) | , but it doesn’t satisfy the Cayley-Bacharachcondition with respect to |O P (1) | . Asher Auel and Dave Jensen showed us how to improve the upper bound from an earlier version of thispaper by applying a result of Coppens and Martens [CM91]. n the case where Z is a general fiber of a generically finite rational map X P n , byanalyzing the trace map, one obtains the following, a special case of Proposition 4.2 from[LP95]: Theorem 2.2. [LP95]
Let X be a smooth variety of dimension n , and X P n a genericallyfinite dominant rational map. Let Z ⊂ X be a finite reduced fiber. Then Z satisfies theCayley-Bacharach condition with respect to the canonical linear system | K X | . If the canonical bundle of X is sufficiently positive, this forces various geometric constraintson the fibers. For instance, if X is a hypersurface, a simple geometric argument shows thatunder certain degree hypotheses the above theorem implies that the general fiber must becollinear (cf. [BCD14]). The authors of [BDE +
16] exploited this fact to compute the degreeof irrationality of very general hypersurfaces.For our purposes, we will only be dealing with the case in which X is a curve. However,in this case, every such map is actually a morphism, and a much more general result followseasily from the Riemann-Roch Theorem: Theorem 2.3.
Let C be a smooth curve and f : C → P r a morphism. Then any reduceddivisor Z ∈ | f ∗ O P r (1) | satisfies the Cayley-Bacharach condition with respect to | K C | .Proof. The complete linear system | Z | is base-point free, so for any P ∈ Z ,dim | Z − P | = dim | Z | − . Applying Riemann-Roch, we getdim | K C − Z | = dim | K C − ( Z − P ) | , which is equivalent to the Cayley-Bacharach property. (cid:3) A generalization of the Neother-Lefschetz theorem
In order to prove our main theorem, we need to show that a complete intersection curvelies on a complete intersection surface whose Picard group is generated by the hyperplaneclass. More precisely, the goal of this section is to prove the following theorem.
Theorem 3.1.
Let W be a smooth, complete intersection threefold in P n with n ≥ oftype ( w , . . . , w n ) , and let C be a smooth complete intersection curve in W of type ( d − e, d, w , . . . , w n ) . If ≤ d − e < d , then the very general complete intersection surface S containing C of type ( d, w , . . . , w n ) is smooth and satisfies Pic S = Z · [ O S (1)] . If the surface of type ( d − e, w , . . . , w n ) containing C is smooth, then we recover a specialcase of [Lop91, Theorem III.2.1]. Here we will deal with the case when it is possibly singular.The assumption that n ≥ P byaugmenting Proposition 3.2 below with Lemmas II.3.3’ and II.3.3” of [Lop91]. FollowingLopez’s approach, we will restrict our attention to the main technical ingredient (Corollary3.4) of Theorem 3.1 and refer to [GH78] for the remainder of the argument, which carriesthrough without incident. et T be the unique surface of type ( d − e, w , . . . , w n ) containing C , and let P be thevery general surface of type ( e, w , . . . , w n ) containing C . Let X be the total space of thepencil interpolating P ∪ T and S ; we will regard X = P ∪ T as the central fiber, which issingular at the singularities of T and along the double curve D = P ∩ T .We may choose P to meet S and T transversely, which leaves X with ordinary doublepoint singularities at the finite intersection P ∩ T ∩ S = D ∩ C = { p , . . . , p N } . Blowing up each of the p i produces a smooth model of X whose central fiber has a quadricsurface Q i over each singular point. The strict transform of T specifies a ruling of each Q i ,and blowing down along each of the specified rulings yields a family ˜ X with smooth totalspace. Away from the central fiber, X and ˜ X are isomorphic, but the central fiber of ˜ X is areducible surface ˜ X = ˜ P ∪ ˜ T , where ˜ P is the blowup of P at each of the p i , and ˜ T ∼ = T . ˜ T and ˜ P meet in a double curve ˜ D ∼ = D .Since ˜ T and ˜ P meet transversely,Pic ˜ X = Pic ˜ T × Pic ˜ D Pic ˜
P . ( ⋆ )and our goal is to compute Pic ˜ X . Consider the restrictions of Pic ˜ T and Pic ˜ P :Pic ˜ T Pic ˜ P Pic ˜ D r r The main ingredient is the following proposition:
Proposition 3.2.
With notation as above,(i)
Pic P = Z · [ O P (1)] (ii) Ker r = Z · [ O ˜ P ( P p i )( − dH )] (iii) Im r ∩ Im r = Z · [ O ˜ D (1)] (iv) r is injective.Proof. By our assumptions on the degrees, (i) follows from the classical Noether-Lefschetztheorem, and (ii) follows from (i). Moreover, (iii) follows from a standard monodromyargument which is given in [Lop91, Lemma II.3.3], and it remains to show (iv).First, note that the singularities of T are isolated, since C is smooth and moves in T .Furthermore, T is a complete intersection and hence has, for instance, Cohen-Macaulaysingularities, so T is normal. Notationally, since ˜ T ∼ = T and ˜ D ∼ = D , we will work with T and D for simplicity.Lemma II.2.4 of [Lop91] shows that unless e = 1 and T is either ruled by lines, the Veronesesurface, or its general projection to P or P , then there exists a pencil | V | of irreduciblecurves within |O T ( e ) | . If T is the Veronese surface or its general projection to P , r is clearlyinjective. The general projection of the Veronese surface to P (the Steiner surface) is notnormal (and we have excluded n = 3). herefore, if we assume that T is not ruled, then we may assume that T possesses a pencil | V | of irreducible curves within |O T ( e ) | . Let f : T ′ → P be a resolution of | V | : T ′ T P ǫ f Let L be an element of Pic T , and assume L that restricts trivially to a general member of | V | . Then ǫ ∗ L restricts trivially to every fiber of f over an open subscheme U of P , andcohomology and base change implies that ǫ ∗ L | f − ( U ) is actually the pullback of an invertiblesheaf on P | U . Since there are finitely many fibers in the complement of f − ( U ), all of whichare irreducible, ǫ ∗ L is globally the pullback of a line bundle O P ( ℓ ) on P .But f ∗ O P ( ℓ ) ∼ = ℓ ( eH − E ), where E is supported on the exceptional divisor Exc( ǫ ) of ǫ . Onthe complement of Exc( ǫ ), ǫ ∗ L ∼ = L ∼ = ℓeH , and since T is normal, the isomorphism L ∼ = ℓeH extends across all of T . It follows that L is trivial, and the restriction map Pic T → Pic C isinjective for the very general curve C in | V | .Next, assume that T is ruled by lines and e = 1. If T is smooth, then it cannot be ruled—itwould be of general type. So T possesses some singularities, and the upshot is that T mustbe a cone: Lemma 3.3.
Let T be a singular, normal, irreducible surface in P n which is ruled by lines.Then T is a cone over a smooth, degenerate curve.Proof. Let Z be a connected component of the curve in G (1 , n ) which sweeps out T , and let Z ′ be its normalization. Note that Z is irreducible, as otherwise T would be singular alonga line, and likewise the normalization Z ′ → Z is bijective. The universal line Φ → Z pullsback to a family of lines Φ ′ → Z ′ , and there is a natural map π : Φ ′ → T .First, we claim that π is birational. Assume, for the sake of contradiction, that deg π ≥ Z . For every point p along Λ,there is an additional line on T meeting Λ at π , and since Z is irreducible, it follows thatevery line which comes from Z meets Λ. Applying the same argument to three distinct Λyields that T is a plane, which contradicts various of the hypotheses on T .Second, we claim that π contracts a curve. This follows from the birationality of π ,the normality of T , and Zariski’s Main Theorem, as well as the fact that π cannot be anisomorphism. Let E denote a curve contracted by π . By definition of Φ ′ , E cannot besupported on the fibers of Φ ′ → Z ′ , so E must meet every fiber. Then π ( E ) lies on everyline which comes from Z , and by taking a general hyperplane H ∩ T section of T , we seethat T may be regarded as the cone over H ∩ T with vertex π ( E ). (cid:3) To conclude the proof of Proposition 3.2, note that Cl T is isomorphic to Pic D by [Har77,Exercise II.6.3]. Since T is normal, Pic T embeds into Cl T , and thus the restriction map r : Pic T → Pic D is injective. (cid:3) orollary 3.4. Pic ˜ X = Z · [ O ˜ X (1)] ⊕ Z · [ M ] , where M | ˜ T = O ˜ T M | ˜ P = O ˜ P ( e ) ⊗ O ˜ P ( ˜ D ))Note that by our description of the Picard group in ( ⋆ ), it suffice to specify M by specifyingits restrictions to the components of ˜ X . The remainder of the proof of Theorem 3.1 followsthe argument given in [GH78, pg. 37-39], which carries through in this setting withoutrevision. 4. Proofs of the main theorems
The structure of the proof of Theorem 1.5 is as follows: we take a surface of Picard rankone containing the curve, and apply the following theorem of Griffiths and Harris to constructa vector bundle on the surface.
Theorem 4.1. [GH78]
Let S be a smooth projective surface, L a line bundle on S , and Z ⊂ S a reduced set of points. Then there exists a rank two vector bundle E with det E = L along with a section s ∈ H ( E ) with Z ( s ) = Z if and only if Z satisfies the Cayley-Bacharachproperty with respect to | K S + L | . We then determine that the vector bundle is Bogomolov unstable, which gives a lower boundon the degree of the fiber and forces the fiber to be contained in a hyperplane. Analyzingthe geometry, we conclude that each hyperplane must contain a single fiber, and that thehyperplanes lie in a linear pencil.Combining the Griffiths-Harris theorem with Theorem 2.3, we easily obtain the following.
Lemma 4.2.
Let C be a smooth curve satisfying the assumptions of Theorem 1.5. Let f : C → P r be a morphism, and let Γ ∈ | f ∗ O P r (1) | be general (and thus reduced). Then thereis a rank two vector bundle E on S sitting in the short exact sequence → O S → E → I Γ ,S ( α ) → . (4.1) Proof.
By Theorem 2.3, Γ satisfies the Cayley-Bacharach condition with respect to the canon-ical linear series | K C | . In particular, by adjunction, it satisfies the Cayley-Bacharach condi-tion with respect to | K S + αH | where H is a hyperplane section on S . The statement thenfollows from Theorem 4.1. (cid:3) Using (4.1), we can check that E is Bogomolov unstable, producing a second representationof E as extension. The plan is to compare the two. We first recall Bogomolov’s InstabilityTheorem. Theorem 4.3. [Bog78]
Let F be a rank two vector bundle on a smooth projective surface X . If c ( F ) − c ( F ) > , (4.2) hen F is Bogomolov unstable . That is, there exists a finite subscheme Z ⊂ X (possiblyempty), plus line bundles L and M on X sitting in an exact sequence → L → F → M ⊗ I Z → where ( L − M ) > and ( L − M ) A > for all ample divisors A . We can now use this along with 4.1 to prove our key lemma. The technique of the proof issimilar to that of Reider’s Theorem (cf. [Laz97] Theorem 2.1, or [Rei88] for Reider’s originalproof).
Lemma 4.4.
Let C be a smooth curve satisfying the assumptions of Theorem 1.5. Let f : C → P r be a morphism with r < n . Suppose deg f ∗ O P r (1) < d C := deg( C ) , and let Γ ∈ | f ∗ O P r (1) | be a divisor. Then(1) Γ lies in a hyperplane, and(2) deg f ∗ O P r (1) ≥ deg( S ) · ( α − .Proof. Without loss of generality, we can take Γ to be general in | f ∗ O P r (1) | . Let E be thevector bundle on S obtained in Lemma 4.2. First, we show that E is Bogomolov unstable.By (4.1), the Chern classes of E are given by c ( E ) = αH,c ( E ) = d Γ , where d Γ is the length of Γ, i.e. deg f ∗ O P r (1). Let d S be the degree of S . Then c ( E ) − c ( E ) = α d S − d Γ , which greater than zero since d Γ < d C = d S α , and α ≥
4. Thus, E sits in the short exactsequence 0 → L → E → M ⊗ I Z → M is effective. Since Pic( S ) = Z , we can write L = O S ( λ ). By (4.1)and (4.4), c ( E ) = c ( L ) + c ( M ) = c ( αH ) . Thus, M ≡ αH − L . By the instability of E , we have(2 L − αH ) · H = (2 λ − α ) d S > . So 2 λ > α ≥ . (4.5)In particular, λ is positive. Thus, the composite map L → E → I Γ ,S ⊗ O S ( α )is nonzero, as otherwise L would map to the kernel, O S , of the right-hand map. Twistingdown by λ , we obtain a nonzero map O S → I Γ ,S ⊗ O S ( α − λ ) = I Γ ,S ⊗ M. This implies that h ( M ) ≥ h ( I Γ ,S ⊗ M ) > . herefore, there is an effective curve C ∈ | M | which contains Γ. Also, α > λ (since Γ is nonempty).Now we approximate the intersection pairing L · M . Let d Z denote the length of Z . Thenby (4.1) and (4.4), we obtain d Γ = c ( E ) = L · M + d Z . Thus d Γ ≥ L · M. Collecting inequalities, we havedeg C = αd S > d Γ ≥ L · M = λ ( α − λ ) d S . Combining this inequality with (4.5), we get 0 < α − λ <
2. Thus, λ = α −
1, which proves(2). For (1), notice L = O S ( α − M = O S (1) . In particular, Γ ⊂ C lies in a hyperplane, as desired. (cid:3) Now that we know a general divisor in | f ∗ O P r (1) | will lie in a hyperplane, to prove Theorem1.5, it only remains to show that the corresponding pencil of hyperplanes forms a linear penciland that a member of the pencil contains only one fiber. First we’ll prove this in the casewhen r = 1, and then reduce the general case to this one. Lemma 4.5.
Let f : C → P be a morphism with deg f < deg C (again, with C satisfyingthe assumptions of Theorem 1.5). Then f is the projection from an ( n − -plane in P n .Proof. Let d C = deg( C ) and d S = deg( S ). Let Γ ⊂ C be a fiber of f , and suppose it doesn’tspan a hyperplane. That is, assume Γ ⊂ G , where G ⊂ P n is an ( n − G determines a morphism C → P of degree at most d C − deg f ≤ αd S − ( α − d S = d S < ( α − d S ≤ gon( C ) , which is a contradiction, since the degree cannot be smaller than the gonality. Thus, eachfiber of f spans a hyperplane.For the sake of contradiction, suppose two distinct fibers lie in H ⊂ P n , a hyperplane.Then by Lemma 4.4, 2 d S ( α − ≤ f < d C = d S α. But α ≥
4. Thus, no two fibers span the same hyperplane.Let Γ and Γ be fibers of f contained in hyperplane sections h and h , respectively.There is a linear equivalence between h − Γ and h − Γ . If h − Γ is not equal to h − Γ on the level of cycles, then, possibly removing base points, there exists a base-point freepencil on C of degree length ( h − Γ ) = d C − deg f ≤ d S , but, by Lemma 4.4, such a pencil cannot exist. herefore, h − Γ = h − Γ . Thus, since Γ and Γ are disjoint, h − Γ = h ∩ h . Sincewe chose Γ and Γ arbitrarily, these equalities hold for every pair of fibers.Let Λ ⊂ Gr ( n − , n ) be the pencil of hyperplanes that are spanned by fibers of f , and K = \ H ∈ Λ H ⊂ P n . Assume dim
K < n − . Then for each pair H , H ∈ Λ, H ∩ H ∩ C = h ∩ h . Thus, { H ∩ H | H , H ∈ Λ } is a nonconstant family of n − P n hasdimension at least n −
1, and intersects C in h ∩ h . But length ( h ∩ h ) < length h = d C ,whereas every hypersurface intersects C in at least d C points, a contradiction.Thus, dim K = n −
2, and Λ is the unique linear pencil corresponding to the projectionfrom K . (cid:3) Proof of Theorem 1.5.
By similar reasoning as in the beginning of the proof of Lemma 4.5,each divisor in the linear system W ⊆ H ( f ∗ O P r (1)) corresponding to f spans a hyperplane,and no two divisors lie in the same hyperplane. That is, there is a natural injection π : P ( W ∨ ) → P ( V ∨ ) , where V = H ( O P n (1)).It remains to show that the image of π is a linear subvariety. Note that it suffices to showthat the image of a general line P ( W ∨ ) is a line in P ( V ∨ ). A general line in P ( W ∨ ) is a linearsubsystem of dimension one, corresponding to the composition C f / / Im f pr / / P where pr is a projection from an ( r − f . Thus the degree of pr ◦ f isdeg[(pr ◦ f ) ∗ O P (1)] = deg[ f ∗ (pr ∗ O P (1))] = deg( f ∗ O P r (1)) . So pr ◦ f satisfies the hypotheses of Lemma 4.5, which guarantees that the image of thelinear system associated to pr ◦ f under π is a line, as needed. (cid:3) Corollary 1.8 follows quickly from Lemma 4.4 (2) and a theorem of Coppens and Martens.
Proof of Corollary 1.8.
The lower bound is immediate from Lemma 4.4 (2).For the upper bound, since S is non-degenerate, deg( S ) ≥ n −
1. Thus, deg( C ) ≥ n − C ) ≥ n −
7, then C has a 2 n − n − (cid:3) Theorem 1.2 follows easily from Theorem 1.5 and Theorem 3.1. roof of Theorem 1.2. If n = 2, the Theorem follows from Noether’s Theorem, so we assume n ≥
3. Consider the linear subsystem D of |O P n ( a n − ) | consisting of sections vanishing on C .Since C is a complete intersection, it is generated in its highest degree a n − , so the base locusof D is C . Thus, by the strong Bertini Theorem (see e.g. [EH16, Proposition 5.6]), a generalmember of D is smooth. Choosing such a hypersurface, and proceeding by induction, wecan find a smooth complete intersection cubic of type ( a , a , . . . , a n − ) containing C that issmooth. By Theorem 3.1, we can then find a smooth complete intersection surface S of type( a , . . . , a n − ) satisfying the hypotheses of Theorem 1.5. (See the comment below Theorem3.1 for the n = 3 case.) The conclusion follows by applying Theorem 1.5. (cid:3) We conclude this section with a proof of Corollary 1.4.
Proof of Corollary 1.4.
In light of Theorem 1.2, we need to show that a general such curvehas a (2 n − n − n − n − G be the Grassmannian of ( n − P n and P n [2 n − the curvilinear locus ofthe Hilbert scheme of 2 n − P n . Define P ⊂ G × P n [2 n − to be the incidencecorrespondence P = { ( P, Z ) | Z ⊂ P } . Let Ψ = |O P n ( a ) | × · · · × |O P n ( a n − ) | . Define the locus Y ⊂ Ψ × P to be Y = { (( S , . . . , S n − ) , ( P, Z )) | Z ⊂ S ∩ · · · ∩ S n − ∩ P } . Since a i ≥ n ≥
3, we have (cid:18) a i + nn (cid:19) > n − . Thus, a general Z ∈ P n [2 n − will impose independent conditions on hypersurfaces of degree a i , so the fibers of the projection Y → P have dimension X i (cid:0) h ( O P n ( a i )) − − (2 n − (cid:1) = − n + 2 n + X i (cid:18) a i + nn (cid:19) . Thus,dim( Y ) = 2( n −
1) + (2 n − n − − n + 2 n + X i (cid:18) a i + nn (cid:19) = X i (cid:18) a i + nn (cid:19) − n, while dim(Ψ) = X i (cid:18) a i + nn (cid:19) − n + 1so Y can’t dominate Ψ.Now we show that a general such curve has a (2 n − n − P ′ ⊂ G × P n [2 n − to be the same incidence correspondence as above, only with the length of the ubschemes being one less. We similarly define the incidence correspondence Y ′ ⊂ Ψ × P ′ . Then dim( Y ′ ) = X i (cid:18) a i + nn (cid:19) − n + 1 = dim(Ψ) . Thus, we need to show that a general fiber of the projection Y ′ → Ψ is zero-dimensional. Infact, it suffices to show there exists a fiber of dimension zero. That is, we want to show thatthere is a curve with a positive number of finitely many (2 n − n − P be an ( n − Z a set of 2 n − P . At each point of Z ,choose a general tangent direction in P n . That is, choose a general 0-dimensional subschemesupported at Z , with length 2 at each point. Call this length 4 n − Z ′ . Since (cid:0) a i + nn (cid:1) ≥ n − i , there will be some complete intersection curve C of the requireddegrees containing Z ′ .It now suffices to show that the dimension of the family of ( n −
2) planes meeting the2 n − Z ′ is zero at P . The locus of ( n − n − G , which completes the proof. (cid:3) A Cayley-Bacharach Conjecture
By modifying the proof of Lemma 4.4, we are able to prove the following special case ofone of the Cayley-Bacharach conjectures (Conjecture CB12 of [EGH96]).
Theorem 5.1.
Let Γ be any subscheme of a zero-dimensional complete intersection of hy-persurfaces of degrees d ≤ · · · ≤ d n in P n , so that Γ is contained in a complete intersectionsurface S of type ( d , . . . , d n ) with Pic( S ) generated by the hyperplane class. Set k := d + · · · + d n − n − . If Γ fails to impose independent conditions on hypersurfaces of degree k + e + 2 , where ≤ e ≤ d − , then deg(Γ) ≥ ( e + 1) · d · d · · · · · d n . Remark . In the notation of the original statement of the conjecture, we are consideringthe case s = 3, and setting m = k + e + 2. Notice that in this case, we are able to obtain astronger bound than the one given in the conjecture.Before the proof, we need a slightly more general formulation of Theorem 4.1 in order todeal with non-reduced 0-cycles. Theorem 5.3 (cf. [Laz97], Prop 3.9) . Let S be a smooth projective surface, Z ⊂ S a zero-dimensional subscheme, and L a line bundle on S . Given an element η ∈ Ext ( L ⊗I Z,S , O S ) ,denote by F η the sheaf arising from the extension: → O S → F η → L ⊗ I Z,S → . Then F η fails to be locally free if and only if there exists a proper (possibly empty) subscheme Z ′ ( Z such that η ∈ Im (cid:8) Ext ( L ⊗ I Z ′ ,S , O S ) → Ext ( L ⊗ I Z,S , O S ) (cid:9) . roof of Theorem 5.1. Set m := k + e .First we show that there is some non-empty subscheme Z ⊆ Γ such that h ( I Z, P n ( m )) = 0but h ( I Z ′ , P n ( m )) = 0 for all proper subschemes Z ′ ( Z by induction on the length of Γ.Notice that since h ( O P n ( m )) = 0, the condition h ( I Z, P n ( m )) = 0 is equivalent to Z failingto impose independent conditions on O P n ( m ), so if the proper subschemes of Γ all satisfythe desired cohomological property, we are done. Otherwise, there is some nonempty Z ( Γsuch that h ( I Z, P n ( m )) = 0, and we are done by the induction hypothesis. For the base case,assume Γ consists of a single point. Then the only proper subscheme is the empty set, whichtrivially imposes independent conditions on O P n ( m ), as desired. Since deg(Γ) ≥ deg( Z ), wecan replace Γ with Z for the remainder of the proof.Since S is a complete intersection with embedding line bundle O S (1), we know h ( O S ( m )) =0. Thus, we obtain the following commutative diagram with exact rows: h ( O P n ( m )) / / (cid:15) (cid:15) (cid:15) (cid:15) h ( O Z ( m )) / / ∼ = (cid:15) (cid:15) h ( I Z, P n ( m )) / / (cid:15) (cid:15) h ( O S ( m )) / / h ( O Z ( m )) / / h ( I Z,S ( m )) / / h ( I Z, P n ( m )) = 0 if and only if h ( I Z,S ( m )) = 0.Thus, since ω S = O S ( k ), Serre duality yieldsExt ( I Z,S ( e + 2) , O S ) ∼ = h ( I Z,S ( m )) ∨ = 0and similarly Ext ( I Z ′ ,S ( e + 2) , O S ) = 0 . Therefore, by Theorem 5.3, we can find a nontrivial extension0 → O S → E → I Z,S ( e + 2) → E locally free.For the sake of contradiction, assumedeg( Z ) < ( e + 1) · d · d · · · · · d n = e · deg( S ) . Then c ( E ) − c ( E ) = ( e + 2) deg( S ) − Z ) > (( e + 2) − e + 1)) deg( S ) ≥ . Thus, by Theorem 4.3, E is Bogomolov unstable, and we can write it as an extension0 → L → E → M ⊗ I W → , where L and M are line bundles satisfying the conditions from Theorem 4.3 and W ⊂ S is afinite subscheme. Since the Picard group of S is generated by the hyperplane class, we canset L = O S ( λ ).Following the proof of Lemma 4.4 ( e + 2 taking the place of α ), we conclude that2 λ > e + 2 ≥ ,e + 2 > λ, and λ ( e + 2 − λ ) · deg( S ) ≤ deg( Z ) < ( e + 2) · deg( S ) . ombining these inequalities, we obtain e + 2 − λ = 1 . This implies ( e + 1) · deg( S ) ≤ deg( Z ) , which gives us a contradiction. (cid:3) References [Bal97] E. Ballico. On the gonality of curves in P n . Comment. Math. Univ. Carolin. , 38(1):177–186, 1997.[Bas96] B´en´edicte Basili. Indice de Clifford des intersections compl`etes de l’espace.
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