# Curves of low positive genus and low degree on a general heptic hypersurface of [P.sup.5].

1 Introduction

H. Clemens conjectured that for any degree d > 0 a general quintic hypersurface of [P.sup.4] contains only finitely many smooth rational curves of degree d; the stronger form of the conjecture says that the same is true for singular rational curves, except rational degree 5 plane curves ([5], [6], [9], [19], [20], [21], [22], [30]). There are results on other Fano 3-folds ([21], [26, Theorem 2], [22, Theorem I. 2]) results for higher genera ([23], [24]) and results on higher dimensional hypersurfaces of general types ([32]). In particular for every integer k [greater than or equal to] 8 for a very general degree k hypersurface X [subset] [P.sup.5] and any integral curve D [subset] X, we have deg(D) [less than or equal to] 2[??] - 2, where [??] is the geometric genus of D ([31], [32, Theorem 3.9]). The case k = 7, i.e. the case of a general heptic hypersurface of [P.sup.5], was singled out as an interesting boundary case ([7], [14]). A general heptic hypersurface contains exactly 698005 lines and no rational curve (not even singular ones) with degree d if 2 [less than or equal to] d [less than or equal to] 16 ([7], [14], [28, Theorem 1.1]). There are two main tools available for rational curves, but not for smooth curves of higher genus (a very strong result on the strata by splitting type of the restricted tangent bundle and the use of semigroups related to a rational parametrization). See [6] and [7] for a full use of these tools and [8] for a full use of the latter tool in another context.

For any set S [subset or equal to] [P.sup.5] let (S) denote its linear span. In this paper we prove the following result.

Theorem 1. Let W [subset] [P.sup.5] be a general heptic hypersurface. For all integers d, g with d [less than or equal to] 16 and 1 [less than or equal to] g [less than or equal to] 3 W does not contain any smooth curve C of genus g and degree d with dim((C)) = 3.

Only numerical reasons prevent us to get degrees d a little bit higher or genera g a little bit higher or to cover the curves spanning a 3-dimensional linear subspace. We prove the following result, but several parts of its proof works for higher genera and/or higher degree.

Proposition 1. A general heptic hypersurface of [P.sup.5] contains no smooth curve C of genus 1 and degree d [less than or equal to] 14.

In section 2 (resp. section 3, resp. section 4) we prove the part of Theorem 1 concerning curves C spanning a linear subspace <C> of [P.sup.5] of dimension [less than or equal to] 2 (resp. 5, resp. 4). In section 5 we give a few results for the case dim((C)) = 3 and prove Proposition 1. Section 4 contains more lemmas than the ones needed to prove Theorem 1 and even so in many places we know how to improve the lemmas by 1. Anyway, section 3 does not work if (d, g) [member of] {(17, 0), (17, 1)}, the next interesting cases.

Since for any smooth hypersurface W [subset] [P.sup.n], n [greater than or equal to] 4, with deg(W) [greater than or equal to] 2 there is a codimension 2 subvariety Z [subset] W which is not the intersection of W and a codimension 2 subvariety of [P.sup.n] ([29], [25]), there are plenty of curves on a very general heptic hypersurface X, which are not easily described, but certainly not unexpected. A dimensional count suggests that a very general heptic hypersurface has no curve with very low genus. If we take curves with arithmetic genus q and degree d [much greater than] q, then the dimensional count is better for high q than for q = 0 (when d [much greater than] q the dimensional count is even better for degenerate curves than for non-degenerate ones).

Question 1. Let X [subset] [P.sup.5] be a very general heptic hypersurface. Is it true that X contains no elliptic curve? Is it true that for each q [member of] N there is an integer d(q) such that X has no curve of arithmetic genus q and degree d > d(q)?

Our tools cannot solve these questions (at the very least we need 7d + 1 - q < ([sup.12.sub.5])), but we made an attempt to see where not to find counterexamples to this kind of questions. A strong feature of [7] (and also of [6] and [8]) is that it works for singular rational curves and this is the best way to state problems related to Clemens' conjecture as non-existence or finiteness results for maps from moduli schemes of curves to a varying target, e.g. a very general heptic hypersurface, i.e. in the set-up of Kontsevich moduli spaces with a varying target. Unfortunately, our tools use the arithmetic genus of the image and so at most we may recover (for low degrees) singular curves with low arithmetic genus. Theorem 1 and Proposition 1 are stated only for smooth curves, because in the singular case (but with arithmetic genus q [less than or equal to] 3) we can handle only lower degrees. The first lemmas to be generalized for the interested reader are Lemmas 2 and 3, because a nondegenerate singular curve C [subset] [P.sup.5] has hyperplane sections C [intersection] H which are not curvilinear. Moreover, if q = 1 the singular curves are rational and so they do not exist for d [less than or equal to] 16 on a very general heptic by [7]. If q = 2,3 knowing by [7] the case of rational curves we may say that the non-rational ones have very mild singularities, and it helps, even for the generalizations of Lemmas 2 and 3.

Thanks are due to a referee for useful remarks.

2 Preliminaries

We work over the complex number field.

Let [M.sub.d,g] denote the set of all smooth curves C [subset] [P.sup.5] with degree d, genus g and [h.sup.1]([O.sub.C](1)) = 0. The algebraic set [M.sub.d,g] is irreducible, because we only took nonspecial line bundles for the embeddings. Fix C [member of] [M.sub.d,g] and let [N.sub.C] be the normal bundle of C in [P.sup.5]. Since dim(C) = 1, we have [h.sup.2](F) = 0 for every coherent sheaf F on C. Therefore for any surjection G [right arrow] E of coherent sheaves on C, the associated map [H.sup.1](G) [right arrow] [H.sup.1](E) is surjective. Hence [h.sup.1](E) = 0 if [h.sup.1](G) = 0. The Euler's sequence shows that [TP.sup.5] is a quotient of [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Hence [h.sup.1]([N.sub.C]) = 0. Thus [M.sub.d,g] is smooth and of dimension [h.sup.0]([N.sub.C]) = 6d + 2 - 2g.

If we drop the condition [h.sup.1]([O.sub.C](1)) = 0, then when g [less than or equal to] 3 we only need to add the case (d, g) = (4,3) (canonically embedded non-hyperelliptic curves of genus 3). See Remark 2 for a proof that a general heptic hypersurface contains no canonically embedded curve of genus 3.

For any integer r [greater than or equal to] 1 let [M.sub.d,g](r) be the set of all C [member of] [M.sub.d,g] whose linear span has dimension r. Since g > 0, we have [M.sub.d,g](1) = 0. Since 1 [less than or equal to] g < 3 and the embedding is by a non-special line bundle, [M.sub.d,g](2) = 0, unless (d, g) = (3, 1).

Let W be the set of all smooth heptic hypersurfaces W [subset] [P.sup.5] satisfying the thesis of [7], i.e. containing no rational curve of degree [less than or equal to] 16 (not even singular ones).

Let Z [subset] [P.sup.r] be any zero-dimensional scheme. For any hyperplane H [subset] [P.sup.r] let [Res.sub.H](Z) denote the residual scheme of Z with respect of H, i.e. the closed subscheme of [P.sup.r] with [I.sub.Z]: [I.sub.H] as its ideal sheaf. We have [Res.sub.H](Z) [subset or equal to] Z and deg(Z) = deg([Res.sub.H](Z)) + deg(H [intersection] Z). For every integer t we have the following residual short exact sequence of coherent sheaves on [P.sup.r] (the latter is also an OH-sheaf), which we often call the residual exact sequence of H or of the inclusion H [subset] [P.sup.r]:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1)

Remark 1. Let N [subset] [P.sup.5] be a plane and C [subset] N be a smooth cubic. The normal bundle [N.sub.C] of C in [P.sup.5] is isomorphic to [O.sub.C](3) [cross product] [O.sub.C][(1).sup.[cross product]3] and hence [h.sub.1]([N.sub.C]) = 0 and [h.sup.0]([N.sub.C]) = 18. Hence the Hilbert scheme of [P.sup.5] is smooth at [C] and of dimension 18. We have [h.sup.1]([I.sub.C](t)) = 0 for all t [member of] N and hence [h.sup.0]([I.sub.C](7)) = ([sup.12.sub.5]) - 21. Hence a general heptic hypersurface contains no plane cubic.

By Remark 1 a general heptic hypersurface contains no element of [M.sub.d,g](2).

Remark 2. Let N [subset] [P.sup.5] be a plane and C [subset] N be a smooth degree 4 curve. C has genus 3 and [O.sub.C](1) [congruent to] [[omega].sub.C]. The normal bundle [N.sub.C] of C in [P.sup.5] is isomorphic to [O.sub.C](4) [cross product] [O.sub.C][(1).sup.[cross product]3] and hence [h.sup.1]([N.sub.C]) = 3 and [h.sup.0]([N.sub.C]) = 23. [P.sup.5] has [[infinity].sup.9] planes and each plane has [[infinity].sup.14] degree 4 curves. Hence the Hilbert scheme of [P.sup.5] is smooth at [C] and of dimension 23. We have [h.sup.1]([I.sub.C](t)) = 0 for all t [member of] N and hence [h.sup.0]([I.sub.C](7)) = ([sup.12.sub.5]) - 26. Hence a general heptic hypersurface contains no degree 4 plane curves.

Lemma 1. Fix integer t [greater than or equal to] 2, r [greater than or equal to] 3 and an integral and non-degenerate curve T [subset] Pr such that [h.sup.1]([I.sub.T](t)) > 0. Fix a linear subspace [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Assume that [h.sup.1](M, [I.sub.Mp[intersection]T,M](t)) = 0 for every hyperplane M [member of] [absolute value of V]. Then [h.sup.1]([I.sub.T](t - 1)) [greater than or equal to] [h.sup.1]([I.sub.T](t))+ dim(V) - 1.

Proof. For any hyperplane M [subset] [P.sup.r] we have an exact sequence

0 [right arrow] [I.sub.T](t - 1) [right arrow] [I.sub.T](t) [right arrow] [I.sub.T[intersection]M,M](t) [right arrow] 0 (2)

Since [h.sup.1](M, [I.sub.T,M](t)) = 0, the map [H.sup.1]([I.sub.T](t - 1)) [right arrow] [H.sup.1]([I.sub.T](t)) is surjective and hence its dual [e.sub.M]: [H.sup.1][([I.sub.T](t)).sup.[disjunction]] [right arrow] [H.sup.1][([I.sub.T](t - 1)).sup.[disjunction]] is injective. Taking the equations of all hyperplanes we get a bilinear map map u: [H.sup.1][([I.sub.T](t)).sup.[disjunction]] x [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], which is injective with respect to the second variables, i.e. for every non-zero linear form [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is injective (it is [e.sub.M] with M := {l = 0}). Hence if [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] with a [not equal to] 0 and l [not equal to] 0, then u(a, l) = [e.sub.M](a) [not equal to] 0. Therefore the bilinear map u is non-degenerate in each variable. Hence [h.sup.1]([I.sub.T](t - 1)) [greater than or equal to] [h.sup.1]([I.sub.T](t)) + dim(V) - 1 by the bilinear lemma.

Lemma 2. Fix an integer a > 0 and assume d [greater than or equal to] 2g - 1 + a. Fix a zero-dimensional curvilinear scheme Z [subset] [P.sup.5] such that deg(Z) = a. Set [E.sub.Z] := {C [member of] [M.sub.d,g]: Z [subset] C}. Then every irreducible component of EZ has dimension [less than or equal to] 6d + 2 - 2g - 4a.

Proof. If [E.sub.Z] = [empty set], then the lemma is true. Hence we may assume E [not equal to] [empty set]. Fix C [member of] E. By [27, Theoreme 1.5] it is sufficient to prove that [h.sup.1]([N.sub.C](-Z)) = 0. Since C is smooth, NC is a quotient of [TP.sup.5.sub.|C]. By the Euler's sequence of [TP.sup.5] [N.sub.C] is a quotient of [O.sub.C][(1).sup.6]. Since d [greater than or equal to] 2g - 1 + a, we have [h.sup.1]([O.sub.C](1)(-Z)) = 0.

Lemma 3. Fix integers t [greater than or equal to] 1 and r [greater than or equal to] 2. Let Z [subset] [P.sup.r] denote a curvilinear zero-dimensional scheme such that c := deg(Z) [less than or equal to] 3t + r - 2, Z spans [P.sup.r] and [h.sup.1]([I.sub.Z](t)) > 0; if c = 3t + r - 2 assume that [h.sup.1](N, [I.sub.N[intersection]Z,N](t)) = 0 for every plane N [subset or equal to] [P.sup.r]. Then either there is a line L [subset] [P.sup.r] with deg(L [subset] Z) [greater than or equal to] t + 2 or there is a conic D [subset] [P.sup.r] with deg(D [intersection] Z) [greater than or equal to] 2t + 2.

Proof. The case r = 2 is true for all t by [11, Corollaire 2] (in the case c = 3t we assumed that both [h.sup.1]([I.sub.Z](t)) > 0 and [h.sup.1]([I.sub.Z](t)) = 0). Hence we may assume r [greater than or equal to] 3 and use induction on r. The case t = 1 is true (if c [less than or equal to] r, because no Z with deg(Z) [less than or equal to] r spans [P.sup.r], while if c = r + 1 because [h.sup.1]([I.sub.Z](x)) = 0 for all x > 1 if deg(Z) = r + 1 and Z spans [P.sup.r]). Hence we may assume t [greater than or equal to] 2 and use induction on t in [P.sup.r].

(a) Let M [subset] [P.sup.r] be a hyperplane such that a := deg(Z [intersection] M) is maximal. First assume [h.sup.1](M, [I.sub.Z[intersection]M,M](t)) > 0. Since Z spans M we have deg(Z [intersection] M) [less than or equal to] c - 1. The maximality property of M gives that Z [intersection] M spans M. Hence the inductive assumption gives that either there is a line L [subset] M with deg(L [intersection] Z) [greater than or equal to] t + 2 or there is a conic D [subset] M with deg(D [intersection] Z) [greater than or equal to] 2t + 2. Hence we may assume [h.sup.1] (M, [I.sub.Z[intersection]M,M](t)) = 0. The residual sequence (1) of M gives [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. We have deg([Res.sub.M](Z)) = c - a. Since M [intersection] Z spans M we have a [greater than or equal to] r [greater than or equal to] 3. Assume for the moment a = r. The maximality property of M implies that Z is in linearly general position in [P.sup.r]. Since c [less than or equal to] rt + 1, we get [h.sup.1]([I.sub.Z](t)) = 0 ([10, Theorem 3.2]). Hence c - a [less than or equal to] c - r - 1 [less than or equal to] 3t + r - 2 - r - 1 < 3(t - 1) + r - 3. By the inductive assumption either there is a line L [subset] [P.sup.r] with deg(L [intersection] Z) [greater than or equal to] t + 1 or there is a conic D [subset] [P.sup.r] with deg(D [intersection] Z) [greater than or equal to] 2t. Assume the existence of a conic D [subset] P[.bar]r with deg(D [intersection] Z) [greater than or equal to] 2t. Since r - 1 [greater than or equal to] 2, the maximality property of M gives a [greater than or equal to] 2t + r - 3. Hence c [greater than or equal to] 2t + 2t + r - 3, a contradiction. Now assume the existence of a line L [subset] [P.sup.r] with deg(L [intersection] Z) [greater than or equal to] t + 1. To prove the lemma we may assume deg(L [intersection] Z) = t + 1. Let H [subset] [P.sup.5] be a hyperplane containing L and with b := deg(H [intersection] Z) maximal among all hyperplanes containing L. Since Z spans Pr, we have b [greater than or equal to] t + r - 1. If [h.sup.1](H, [I.sub.Z[intersection]H,H](t)) > 0, then we conclude by the inductive assumption on r. Hence we may assume [h.sup.1](H, [I.sub.Z[intersections]H,H](t)) = 0. The residual sequence (1) of H gives [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. We have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Hence the inductive assumption on t gives the existence of a line L [subset] [P.sup.5] with [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. If the lemma fails, then deg(R [intersection] [Res.sub.H](Z)) = deg(R [intersection] Z) = t + 1. In this case we have c [greater than or equal to] 2t + 2. Hence the lemma is proved in degree t if c [less than or equal to] 2t + 1 (but this part is just [4, Lemma 34]).

(b) First assume R [not equal to] L and R [intersetion] L [not equal to] 0. Since deg(R [intersection] [Res.sub.H](Z)) = t + 1 and H [contains] L, we have deg((L [union] R) [intersection] Z = 2t + 2. Hence we may take D := R [union] L.

(c) Now assume R = L. We may take Z minimal with the restriction that [h.sup.1]([I.sup.Z](t)) > 0 and deg(Z [intersection] L) = t + 1. Part (a) of our proof works if instead on H we take any hyperplane U [contais] L (since as in the first part we exclude the existence of a conic D with deg([Res.sub.U](Z)) [greater than or equal to] 2t). Let Q be a quadric hypersurface containing L in its singular locus. Since deg([Res.sup.Q](Z)) [less than or equal to] 3t - 2t - 2 [less than or equal to] t - 2, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (even in the case t = 2). Therefore the residual exact sequence of Q gives [h.sup.1](Q, [I.sub.Z[intersection]Q,Q](t)) > 0 and hence [h.sup.1]([I.sub.Z[intersection]Q](t)) > 0. The minimality of Z gives Z [subset] Q. Since Z is curvilinear, taking Q = [N.sub.1] + [N.sub.2] with [N.sub.1], [N.sub.2] hyperplanes we also get that only the connected components of Z whose reduction are contained in L arise (for a minimal Z), hence we reduce to the case deg(Z) = 2t + 2. Let W [subset] Z be any degree 2t + 1 subscheme. Since deg(W [intersection] J) [less than or equal to] deg(Z [intersection] J) [less than or equal to] t + 1 for each line J, part (a) of the proof gives [h.sup.1]([I.sub.W](t)) = 0. Hence [h.sup.1](M, [I.sub.Z,M](t)) = 1. Since [h.sup.1](N, [I.sub.Z[intersection]N,N](t)) = 0 for every hyperplane N [subset] [P.sup.r], Lemma 1 gives [h.sup.1]([I.sub.Z](t - 1)) [greater than or equal to] r + [h.sup.1]([I.sub.Z](t)) = r + 1. Let N be any hyperplane plane containing L. We have [h.sup.1](N, [I.sub.Z[intersection]N](t - 1)) = 1, because deg(Z [intsection] L) = t + 1 and deg(Z [intersection] N) [less than or equal to] 2(t - 1) + 1 (use the residual exact sequence (1) of a general hyperplane M of N containing L in N). Since deg([Res.sub.N](Z)) [less than or equal to] t + 1, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (part (a) applied. Hence the residual exact sequence (1) of N gives [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], a contradiction.

(d) Now assume R [intersection] L = 0. First assume r = 3. Let Q [subset] [P.sup.3] be a general quadric containg L [union] R. Since [I.sub.L[union]R] (2) is spanned and C is curvilinear, Q [intersection] Z = Z [intersection] (R [intersection] L) (as schemes). Hence [h.sup.1](Q, [I.sub.Z[intersection]Q,Q](t)) = 0. The residual exact sequence of Q gives [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since deg([Res.sub.Q](t)) [less than or equal to] t - 2, part (a) or [4, Lemma 34] gives a contradiction. Now assume r [greater than or equal to] 4. Since L [union] R spans a 3-dimensional linear space, the maximality property of M gives b [greater than or equal to] 2t + 2 + r - 4. Hence c - b [less than or equal to] t - 1 < deg (R [intersection] [Res.sub.H](Z)), a contradiction.

3 C non-degenerate

Le t[M'.sub.d,g] be the set of all C [member of] [M.sub.d,g] spanning [P.sup.5]. [M'.sub.d,g] is smooth and irreducible and dim([M'.sub.d,g]) = 6d + 2 - 2g. A general non-special curve C [subset] [P.sup.5] of genus g and degree d [greater than or equal to] g + 5 has maximal rank ([3]). We have d [less than or equal to] 16 and 7 x 16 +1 - g < ([sup.12.sub.5]). Hence [h.sup.1]([I.sub.C](7)) = 0 for a general C [member of] [M.sub.d,g], i.e. [h.sup.0]([I.sub.C](7)) = ([sup.12.sub.5]) - 7d - 1 + g. A dimensional count shows that no C [member of] [M'.sub.d,g] is contained in a general heptic hypersurface. Hence in this section we only need to exclude all C [member of] [M'.sub.d,g] with [h.sup.1]([I.sub.C](7)) > 0. By [13, Part (ii) of Theorem at page 492] we have d [greater than or equal to] 13.

Lemma 4. No C [member of] [M'.sub.g,d], d [greater than or equal to] 11, 1 [less than or equal to] g [less than or equal to] 3, is contained in a degree 4 surface.

Proof. Fix a degree 4 surface F [subset] [P.sup.5] containing C [member of] [M'.sub.d,g]. Since C is non-degenerate, F is non-degenerate. By the classification of minimal degree surfaces either F is a cone over a rational normal curve of [P.sup.4] or it is an embedding of [F.sub.0] by the complete linear system [absolute value of h + 2f] or it is an embedding of [F.sub.2] by the complete linear system [absolute value of h + 3f].

(a) Assume that F is an embedding of F by the complete linear system [absolute value of h + 2f]. Since 1 [less than or equal to] g [less than or equal to] 3, either C [member of] [absolute value of 2h + (g + 1)f] or C [member of] [absolute value of (g + 1)h + 2f]. In the first (second) case we get d = 5 + g [less than or equal to] 8 (resp. d = 2g + 4 [less than or equal to] 10), a contradiction.

(b) Assume that F is an embedding of [F.sub.2] by the complete linear system [absolute value of h + 3f] with C [member of] [absolute value of 2h + bf]. Since g > 0, we have a [greater than or equal to] 2 and b [greater than or equal to] 2a. We have d = a + b. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], the adjunction formula gives 2g - 2 = a(-2(a - 2) + (b - 4)) + b(a - 2). If a [greater than or equal to] 3 and hence b [greater than or equal to] 6 we get b [less than or equal to] 2g - 2, a contradiction. If a = 2 and hence b = d - 2, we get 2g - 2 = 2(d - 6), a contradiction.

(c) Assume that F is a cone over a rational normal curve of [P.sup.4]. Call o its vertex and let u: G [right arrow] F be the blowing up of o. G [congruent to] [F.sub.4] and, up to this isomorphism, u is induced by the complete linear system [absolute value of h + 4f]. Let C' [subset] [F.sub.4] be the strict transform of C. Write C [member of] [absolute value of ah + bf]. Since C is smooth, C is smooth and of genus g > 0. Hence a [greater than or equal to] 2 and b [greater than or equal to] 4a. We have d = b. Hence 2 [less than or equal to] a [less than or equal to] 4 with a = 4 only if d = 16. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], the adjunction formula gives 2g - 2 = a(-4(a - 2) + (b - 6)) + b(a - 2) [greater than or equal to] b(a - 2). Hence a = 2 and 2g - 2 = 2{d - 6), a contradiction.

Lemma 5. Fix C [member of] [M'.sub.d,g] and let H [subset] [P.sup.5] be a general hyperplane.

(a) We have [h.sup.1](H, [I.sub.C[intersection]H,H](3)) [less than or equal to] max{0, d - 13}.

(b) Assume d [greater than or equal to] 14. We have [h.sup.1](H, I[Z.sub.C[intersection]H,H](3)) = d - 13 if and only if C [intersection] H is contained in a rational normal curve of H. The latter case does not occur for a curve C contained in a general heptic hypersurface if [h.sup.1]([I.sub.C](7)) [less than or equal to] 3d - 18 - 6g.

Proof. The scheme Z := C [intersection] H is a set of d points in uniform position and spanning H. Fix S [subset or equal to] Z with #(S) = min{d, 13}. Since S is in linearly general position in H = [P.sup.4], we have [h.sup.1](H, [I.sub.S,H](3)) = 0 ([10, Theorem 3.2]). Now assume d [greater than or equal to] 14. If C [intersection] H is contained in a rational normal curve D of H, then [h.sup.0](H, [I.sub.Z,H](3)) [less than or equal to] [h.sup.0](H, [I.sub.D,H](3)) = ([sup.7.sub.3]) - 13 and hence [h.sup.1](H, [I.sub.Z,H](3)) [greater than or equal to] d - 13. By part (a) the last inequality is an equality. Now assume [h.sup.1](H, [I.sub.Z,H](3)) = d - 13. First assume [h.sup.0](H, [I.sub.Z,H](2)) [greater than or equal to] 6. Since Z is uniform position, we get [h.sup.0](H, [I.sub.Z,H](2)) = 6 and that Z is contained in a rational normal curve of H ([15, Lemma 3.9]). Now assume [h.sup.0](H, [I.sub.Z,H](2)) [less than or equal to] 5. Fix any A [subset] Z with #(A) = 10. Since [h.sup.0](H, [I.sub.Z,H](2)) [less than or equal to] 5, we have [h.sup.1](H, [I.sub.A,H](2)) = 0. Fix B [subset] Z\A with #(B) = 4. Since Z is in linearly general position, B spans a hyperplane N of H and Z [intersection] N = B. We have [h.sup.1](N, [I.sub.B,N](3)) = 0. The exact sequence

0 [right arrow] [I.sub.A,H](2) [right arrow] [I.sub.A[union]B,H](3) [right arrow] [I.sub.B,N](3) [right arrow] 0

gives [h.sup.1](H, [I.sub.A[union]B,H](3)) = 0 and hence [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] d - 14, a contradiction.

Now assume that C [intersection] H is contained in a rational normal curve [T.sub.H] of H. Set a := d + 1 - 2g. Fix Z [subset] [T.sub.H] such that #(Z) = a. By Lemma 2 the set of all C [member of] [M'.sub.d,g] containing Z has codimension at least 4a in [M'.sub.d,g]. Since [P.sup.5] has [[omega].sup.5] hyperplanes each hyperplane has 15 rational normal curves and each rational normal curves has [[infinity].sup.a] subsets with cardinality a, to rule out these cases it is sufficient to test all C [member of] [M'.sub.d,g] with [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3a - 20.

Remark 3. Fix C [member of] [M'.sub.d,g] and let H [subset] [P.sup.5] be a general hyperplane. The exact sequence (2) with T := C and M := H gives [h.sup.1]([I.sub.C](t - 1)) [greater than or equal to] [h.sup.1]([I.sub.C](t)) - [h.sup.1](H, [I.sub.C[intersection]H,H](t)). Now assume d [less than or equal to] 4t + 1. Since C [intersection] H is in uniform position in H and it spans H, it is in linearly general position. Hence [h.sup.1](H, [I.sub.C[intersection]H,H](t)) = 0 ([10, Theorem 3.2]).

Lemma 6. A general heptic hypersurface contains no C [member of] [M'.sub.d,g], g > 0, with either [h.sup.0]([I.sub.C](2)) [greater than or equal to] 5, [h.sup.0]([I.sub.C](3)) [not equal to] 25 and d = 15,16, or [h.sup.0] ([I.sub.C](2)) [greater than or equal to] 6 and 13 [less than or equal to] d [less than or equal to] 16.

Proof. Fix C [member of] [M'.sub.d,g] with [h.sup.0]([I.sub.C](2)) [greater than or equal to] 5. Let K be the set-theoretic base locus of [absolute value of [I.sub.C](2)]. Since C is irreducible, we have dim(K) [less than or equal to] 3. Fix an irreducible component A of K containing C and with maximal dimension. Note that [h.sup.0]([I.sub.A](2)) = [h.sup.0]([I.sub.C](2)) and that A = C if and only if dim(A) = 1.

(a) Assume dim(A) [greater than or equal to] 3. Since dim(K) [less than or equal to] 3, we have dim(A) = 3 and deg(A) [greater than or equal to] 3. Since [h.sup.0]([I.sub.C](2)) > 2, A is not the complete intersection of 2 quadrics and hence deg(A) = 3. Since A is a minimal degree 3-fold, their classification (linearly normal rational scrolls and cones over a rational normal curve of P3), gives [h.sup.0]([I.sub.A](2)) = 3, a contradiction.

(b) Assume dim(A) = 2. Let T be the intersection of 3 general elements of [absolute value of [I.sub.C](2)]. Since A is non-degenerate, we have 4 [less than or equal to] deg(A) [less than or equal to] 7 (even if T has a 3-dimensional component) by Bezout's theorem ([12, Theorem 2.2.5]). By Lemma 4 we may assume deg(A) [greater than or equal to] 5. The exact sequence

0 [right arrow] [I.sub.A](1) [right arrow] [I.sub.A](2) [right arrow] [I.sub.A[intersection]H,H](2) [right arrow] 0 (3)

gives [h.sup.0]([I.sub.C](2)) = [h.sup.0]([I.sub.A](2)) [less than or equal to] [h.sup.0](H, [I.sub.A[intersection]H,H](2)). The integral curve A [intersection] H spans H. Set q := [p.sub.a](A [intersection] H). We have q [less than or equal to] deg(A [intersection] C) - 4 for all deg(A) [member of] {5, 6, 7} by an elementary case of the Castelnuovo's bound for the arithmetic genus of a curve in [P.sup.r]. By [13, part (ii) of Theorem at page 492] if deg(A) [less than or equal to] 6, then [h.sup.1](H, [I.sub.A[intersection]H,H](2)) = 0, except the case deg(A) = 6, A [intersection] H a smooth rational curve having a quadrisecant line. Assume for the moment deg(A) [less than or equal to] 6 and that we are not in this exceptional case. We get [h.sup.0]([I.sub.A[intersection]H,H](2)) = 14 - 2 deg(A) + q [less than or equal to] 4, unless deg(A) = 5 and q = 1.

(b1) Assume deg(A) = 5 and q = 1. We may assume [h.sup.0]([I.sub.A](2)) = 5. So we do need to check this case if d = 13, 14 and hence here we assume d [member of] {15, 16}. In this case A [intersection] H is linearly normal and arithmetically Cohen-Macaulay with [h.sup.0](H, [I.sub.A[intersection]H,H](3)) = 35 - 15 = 20. Since [h.sup.0]([I.sub.A](2)) = [h.sup.0](H, [I.sub.A[intersection]H,H](2)), then A is linearly normal by (3). By (3) we get that A is arithmetically Cohen-Macaulay and in particular [h.sup.1]([I.sub.A](2)) = 0. From (3) we get [h.sup.0]([I.sub.A](3)) = [h.sup.0]([I.sub.A](2)) + [h.sup.0](H, [I.sub.A[intersection]H,H](3)) = 25. Since C [subset] A, we get [h.sup.0]([I.sub.C](3)) [greater than or equal to] 25. By assumption [h.sup.0]([I.sub.C](2)) [not equal to] 25 and so [h.sup.0]([I.sub.C](3)) [greater than or equal to] 26. Therefore there is a cubic hypersurface T [subset] [P.sup.5] with C [subset] T and A [subset or not equal to] T, so that C is contained in the locally Cohen-Macaulay curve A [intersection] T. Since deg(A [intersection] T) = 15, we get d [not equal to] 16 and C [not equal to] A [intersection] T. Since [[omega].sub.A[intersection]T] [congruent to] [O.sub.A[intersection]T](2), we get g > 3, a contradiction.

(b2) Now assume deg(A) = 6 and that D := A [intersection] H is a smooth rational curve spanning H and with a line L [subset] H with deg(L [intersection] D) = 4. We have L [subset] K, [p.sub.a](L [union] D) = 3 and L [union] D is a linearly normal curve of H = [P.sup.4]. Let N [subset] H be a general hyperplane. Since [h.sup.i](H, [I.sub.D[union]L](1)) = 0, i = 0, 1, a standard exact sequence gives [h.sup.0](H, [I.sub.D[union]L](2)) = [h.sup.0](N, [I.sub.N[intersection](D[union]L),N](1)) and hence [h.sup.0](H, [I.sub.D,H](2)) = [h.sup.0](H, [I.sub.D[union]L,H](2)) [less than or equal to] [h.sup.0](N, [I.sub.N[intersection]D,N](2)). Since N [intersection] D is in uniform position in N and it spans N, we get [h.sup.1](N, [I.sub.N[intersection]D,N](2)) = 0 and hence [h.sup.0](N, [I.sub.D[intersection]N,N](2)) = 4. Hence [h.sup.0]([I.sub.C](2)) [less than or equal to] [greater than or equal to] 4, a contradiction.

(b3) Now assume deg(A) = 7. By Bezout's theorem ([12, Theorem 2.2.5]) we have T [intersection] H = (A [intersection] H) [union] [L.sub.H] with [L.sub.H] a linear space of dimension [greater than or equal to] 1. If [L.sub.H] is a line, then the complete intersection T [intersection] H links A [intersection] H to a line and hence A [intersection] H is arithmetically Cohen-Macaulay. We get [h.sup.0]([I.sub.C](2)) = [h.sup.0]([I.sub.C[intersection]H](2)) = 1 + q [less than or equal to] 4. Now assume dim([L.sub.H]) > 1. Since T is a general intersection of 3 elements of [absolute value of [I.sub.C](2)] and H is general, we first get that K has a 3-dimensional component, B, which is a linear space, then we get T = A [union] B and then we get A [union] B = K. Hence [h.sup.0]([I.sub.C](2)) = 3, a contradiction.

(c) Assume dim(A) = 1, i.e. A = C, and d = 15, 16. Let T be the intersection of 3 general elements of [absolute value of [I.sub.C](2)]. Let B be an irreducible component of [T.sub.red] containing C. Since we may apply steps (a) and (b) to every irreducible component of K containing C, we have B [subset or not equal to] K and hence there is a quadric containing C, but not containing B. By Bezout we have deg(B) [less than or equal to] 8 with equality if and only if dim(T) = 2 and B = T. Intersecting B with a general element of [absolute value of [I.sub.C](2)] we get a locally Cohen-Macaulay scheme E of pure dimension 1 with deg(E) [less than or equal to] 16 and E [contains or equal to] C. We exclude the case d = 16, because C has not the genus, 17, of an intersection of 4 quadrics. Now assume d = 15. Since B = T, T has dimension 2 and hence it is a complete intersection. Hence the complete intersection E links C to a line. Therefore C is arithmetically Cohen-Macaulay and in particular [h.sup.1]([I.sub.C](2)) = 0, contradicting the inequality 2d + 1 - g > 21.

(d) Assume d = 13, 14. Take K, A as in the previous steps. Fix a general hyperplane H [subset] [P.sup.5]. Since C is non-degenerate, we have [h.sup.0]([I.sub.C](1)) = 0. By (3) with C instead of A we have [h.sup.0](C (2)) [less than or equal to] [h.sup.0](H, [I.sub.C[intersection]H,H](2)). Assume [h.sup.0](H, [I.sub.C[intersection]H,H](2)) > 6. By [15, Lemma 3.9] C [intersection] H is contained in a rational normal curve D [subset] H. Since d > 8, we have D [subset] K. By step (a) we get the existence of A [subset or equal to] K with C [subset] A and A a degree 4 surface, contradicting Lemma 4.

Lemma 7. Fix integers d, g such that 6 [less than or equal to] g + 5 [less than or equal to] d < 15. Let r be the set of all C [member of] [M'.sub.d,g] contained in the smooth locus a quadric. For any v [member of] {0, 1, 2} and integer x > 0 let [[GAMMA].sub.v,x] be the set of all C [member of] [M'.sub.d,g] contained in a quadric with singular locus V of dimension v and x = deg(V [intersection] C). Then dim([GAMMA]) [less than or equal to] 4d + 21 + g, [[GAMMA].sub.0,x] = [empty set] for all x [greater than or equal to] 2, dim([[GAMMA].sub.0,1]) [less than or equal to] 4d + 22 + g, dim([[GAMMA].sub.1,x]) [less than or equal to] 4d + g + x + 18, and dim([[GAMMA].sub.2,x]) [less than or equal to] 4d + 14 + g.

Proof. Fix C [member of] [M'.sub.d,g] with [h.sup.0]([I.sub.C](2)) > 0 and let Q [subset] [P.sup.5] be any quadric containing C. Since C is non-degenerate, Q is irreducible. The Hilbert scheme Hilb(Q) of Q has [H.sup.0]([N.sub.C,Q]) as its tangent space at [C]. Let V be the singular locus of Q. Let [tau] be the tangent sheaf of Q. Since the algebraic group Aut(Q) acts transitively on Q\V and [H.sup.0]([tau]) is the tangent space to Aut(Q) at the identity, [H.sup.0]([tau]) spans t at each point of Q\V.

(i) First assume that either Q is smooth or C [intersection] V = [empty set]. Since dim [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], to handle these curves C it is sufficient to prove that the set of all C [member of] [M.sub.d,g] contained in Q and with V [intersection] Q = 0 has dimension [less than or equal to] 4d + 1 + g. Since C [intersection] V = [empty set], the normal sheaf [N.sub.C,Q] is a rank 3 vector bundle on C with degree 4d + 2g - 2. Since [N.sub.C,Q] is a quotient of [[tau].sub.|C], it is spanned. Take 2 general sections of [H.sup.0]([N.sub.C,Q]). These sections induces a rank 2 subsheaf G' of [N.sub.C,Q] isomorphic to [O.sup.[cross product]2.sub.c]. Let G be the saturation of G' in [N.sub.C,Q], i.e. the only rank 2 subsheaf of [N.sub.C,Q] containing G' and with [N.sub.C,Q]/G a line bundle. Since [h.sup.1](G') = 2g, we have [h.sup.1](G) [less than or equal to] 2g. First assume deg(G) [less than or equal to] 3d - 1, i.e. deg([N.sub.C,Q]/G) > 2g - 2. We get [h.sup.1]([N.sub.C,Q]/G) = 0 and so [h.sup.1]([N.sub.C,Q]) [less than or equal to] 2g. Riemann Roch gives [h.sup.0]([N.sub.C,Q]) [less than or equal to] 4d + g + 1. Now assume deg(G) [greater than or equal to] 3d. Let [G.sub.1] be the saturation of a general section of G. So G is the extension of two line bundles, G/[G.sub.1] and [G.sub.1], with deg(G/[G.sub.1]) + deg([G.sub.1]) = 3d [greater than or equal to] 2(2g - 1) and both with a non-zero section. We get that at least one of the line bundles G/[G.sub.1] and [G.sub.1] is non-special and the other one, [Laplace], has [h.sup.1]([Laplace]) [less than or equal to] g . Hence [h.sup.1](G) [less than or equal to] g. Since the line bundle [N.sub.C,Q]/G has a non-zero section, we have [h.sup.1]([N.sub.C,Q]/G) [less than or equal to] g and hence [h.sup.1]([N.sub.C,Q]) [less than or equal to] 2g even in this case.

(ii) Now assume that C [intersection] V [not equal to] [empty set]. The set of all quadrics of rank 5 (resp. 4, resp. 3) has dimension 19 (resp. 17, resp. 14). Since C is not a plane curve and dim(V) [less than or equal to] 2, the scheme C [intersection] V is finite. Let u: [??] [right arrow] Q be the blowing up of V, E := [u.sup.-1](V) the exceptional divisor, and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] the strict transform of C. Since C is smooth, u maps isomorphically [??] onto C. Let [[PSI].sub.v,x] be closure in Hilb([??]) of the strict transforms of all C [subset] Q with deg(C [intersection] V) = x. Take a general D [member of] [[PSI].sub.v,x]. Since Aut([??]) acts transitively of [??]\E, step (a) of the proof gives [h.sup.1]([N.sub.D,[??]]) [less than or equal to] 2g. Hence it is sufficient to give a winning upper bound for deg([N.sub.D,[??]]), i.e. a winning lower bound for [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. The group Pic([??]) is freely generated by E and the pull-back H of [O.sub.Q](1). We have D x H = d and D x E = x. We claim that [[omega].sub.[??]] [congruent to] [O.sub.[??]](-3H + cE) with c = -2 + v. First assume v > 0. Let M [subset] Q be a general hyperplane. M [intersection] Q is a 3-dimensional quadric with vertex of dimension v - 1. We apply [18], Example 8.5 (2), if v = 1 and [18], Example 8.5 (3) if v = 2. Now assume v = 0. To apply the previous formulas we need to take a hyperplane M [contains] V whose pull-back has E as a component.

Lemma 8. A general heptic hypersurface contains no C [member of] [M'.sub.d,g], 13 [less than or equal to] d [less than or equal to] 16, 1 [less than or equal to] g [less than or equal to] 3, with no line R [subset] [P.sup.5] with deg(R [intersection] C) [greater than or equal to] 7 and no conic D [subset] [P.sup.5] with deg(C [intersection] D) [greater than or equal to] 12.

Proof. Recall that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and that we are assuming [h.sup.1]([I.sub.C](7)) > 0. By the cases r = 4, c [less than or equal to] 3t + 1 and t = 5, 6, 7 of Lemma 3 we have [h.sup.1](M, [I.sub.C[intersection]M,M](t)) = 0 for all hyperplanes M [subset] [P.sup.5]. Lemma 1 gives [h.sup.1]([I.sub.C](4)) [greater than or equal to] [h.sup.1]([I.sub.C](5)) + 5 [greater than or equal to] [h.sup.1]([I.sub.C](6)) + 10 [greater than or equal to] [h.sup.1]([I.sub.C](7)) + 15. Fix a general hyperplane H [subset] [P.sup.5]. Lemma 5 and Remark 3 give [h.sup.1]([I.sub.C](3)) [greater than or equal to] [h.sup.1]([I.sub.C](4)) and that either [h.sup.1]([I.sub.C](2)) [greater than or equal to] [h.sup.1]([I.sub.C](3)) - max{0, d - 14} or d [greater than or equal to] 14, [h.sup.1]([I.sub.C](2)) [greater than or equal to] [h.sup.1]([I.sub.C](3)) + 13 - d and [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3d - 18 - g. First assume d [greater than or equal to] 14, [h.sup.1]([I.sub.C](2)) [greater than or equal to] [h.sup.1]([I.sub.C](3)) + 13 - d and [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3d - 18 - g. We first get [h.sup.1]([I.sub.C](3)) [greater than or equal to] [h.sup.1] ([I.sub.C](4)) [greater than or equal to] 3d - 3 - g and then [h.sup.1]([I.sub.C](2)) [greater than or equal to] 2d + 10 - g, i.e. [h.sup.0]([I.sub.C](2)) [greater than or equal to] 31, contradicting Lemma 6.

Now assume [h.sup.1]([I.sub.C](2)) [greater than or equal to] [h.sup.1](C(3)) - max{0, d - 14} and hence [h.sup.1]([I.sub.C](2)) [greater than or equal to] 16 - max{0, d - 14}, i.e. [h.sup.0]([I.sub.C](2)) [greater than or equal to] 36 + g - 2d - max{0, d - 14}. We get [h.sup.0]([I.sub.C](2)) [greater than or equal to] 21 + 16 - 1 - 26 + g = 11 + g [greater than or equal to] 7 if d = 13 and [h.sup.0]([I.sub.C](2) [greater than or equal to] 21 + 16 - d + 14 - 2d - 1 + g = 50 - 3d + g if 14 [less than or equal to] d [less than or equal to] 16. If either 13 [less than or equal to] d [less than or equal to] 15 or d = 16 and g = 3 we use Lemma 6, except that if d = 16 and g = 3 we also need to exclude that [h.sup.0]([I.sub.C](3)) = 25. Assume d = 16 and g = 3. Since [h.sup.1]([I.sub.C](3)) [greater than or equal to] 16 and ([sup.8.sub.3]) = 56, we have [h.sup.0]([I.sub.C](3)) [greater than or equal to] 56 + 16 - 1 - 48 + 3 = 26.

Now assume d = 16 and g = 1,2. We proved that [h.sup.0]([I.sub.C](2)) [greater than or equal to] 2 + g. To repeat the same proof it would be sufficient to have [h.sup.1]([I.sub.C](7)) [greater than or equal to] 4. If C is contained in the smooth locus of at least one quadric, then by Lemma 7 we may assume [h.sup.1]([I.sub.C](7)) [greater than or equal to] 6d + 2 - 2g - (4d - 21 + g) = 2d - 19 - 3g [greater than or equal to] 17. Now assume that a general quadric hypersurface containing c has singular locus V of dimension v and with x := deg(C [intersection] V) > 0. Since x [less than or equal to] 6 if v = 1, it is sufficient to use Lemma 7.

Notation 1. For each integer a > 0 let [A'.sub.a] denote the set of all C [member of] [M'.sub.d,g] such that there is a line L [subset] [P.sup.5] with deg(L [intersection] C) = a and [A".sub.a] := [U.sub.b[greater than or equal to]a][A'.sub.a]. For each integer a > 0 let [B'.sub.a] denote the set of all C [member of] [M'.sub.d,g] such that there is a conic D [subset] [P.sup.5] with deg(L [intersection] D) = a and [B".sub.a] := [U.sub.b[greater than or equal to]a][B'.sub.a].

Lemma 9. A general W [member of] W contains no C [member of] [M'.sub.d,g], 13 [less than or equal to] d [less than or equal to] 16, 1 [less than or equal to] g [less than or equal to] 3, with [A".sub.7] [not equal to] [empty set].

Proof. (a) In this step we prove that W contains no element of [A".sub.9]. Fix C [member of] [A".sub.9] and take a line L [subset] [P.sup.5] such that deg(L [intersection] C) [greater than or equal to] 9. Set b := 8 if (d, g) = (13, 3) and b := 9 if (d, g) [not equal to] (13,3). Fix Z [subset or equal to] C [intersection] L such that deg(Z) = 9. Set Z := {A [member of] [M.sub.d,g]: A [contains] Z}. Lemma 2 gives that dim([E.sub.Z]) [less than or equal to] 6d + 2 - 2g - 4b. Since L has [[infinity].sup.b] degree b subschemes and [P.sup.5] has [[infinity].sup.8] lines, to prove that no element of [A".sub.9] is contained in W it is sufficient to test the ones, C, with the additional condition that either [h.sup.1]([I.sub.C](7)) [greater than or equal to] 19 (case (d, g) [not equal to] (13, 3)) or [h.sup.1]([I.sub.C](7)) [greater than or equal to] 16 (case (d, g) = (13, 3)). By the cases t = 3, 4, 5, 6, 7 of Remark 3 we get [h.sup.1]([I.sub.C](3)) [greater than or equal to] [h.sup.1]([I.sub.C](7)). Therefore [h.sup.1]([I.sub.C](2)) [greater than or equal to] 17 if (d, g) [not equal to] (13, 3) and [h.sup.1]([I.sub.C](2)) [greater than or equal to] 16 if (d, g) = (13,3) (Lemma 5). We get [h.sup.0]([I.sub.C](2)) [greater than or equal to] 37 + g - 2d [greater than or equal to] 6 if (d, g) [not equal to] (13, 3) and [h.sup.0]([I.sub.C](3)) [greater than or equal to] 13 if (d, g) = (13, 3). Apply Lemma 6.

(b) In this step we prove that a general heptic contains no element of [A'.sub.8]. By part (a) it is sufficient to exclude all elements of [A'.sub.8]\[A'.sub.9],. By Lemma 2 it is sufficient to exclude all C [member of] [A'.sub.8]\[A'.sub.9] with [h.sup.1]([I.sub.C](7)) [greater than or equal to] 16. Fix any such C. By Lemma 1 and the case r = 4, t = 7 and c = d [less than or equal to] 3t + 1 of Lemma 3 we have [h.sup.1]([I.sub.C](6)) [greater than or equal to] 21. By the cases t = 3, 4, 5, 6 of Remark 3 we get [h.sup.1]([I.sub.C](3)) [greater than or equal to] 21. Lemma 3 gives [h.sup.1]([I.sub.C](2)) [greater than or equal to] 19 and hence [h.sup.0]([I.sub.C](2)) [greater than or equal to] 39 - 2d + g [greater than or equal to] 8. Use Lemma 6.

(c) Fix C [member of] [A'.sub.7]\[A'.sub.9]. By Lemma 2 it is sufficient to exclude the curves C with [h.sup.1]([I.sub.C](7)) [greater than or equal to] 13. Since C [not member of] [A".sub.9], we have [h.sup.1]([I.sub.C](6)) [greater than or equal to] 18 (Lemma 1). The case t = 4, 5, 6 of Remark 3 gives [h.sup.1]([I.sub.C](3)) [greater than or equal to] 18. Lemma 5 gives [h.sup.1]([I.sub.C](2)) [greater than or equal to] 16 and hence [h.sup.0]([I.sub.C](2)) [greater than or equal to] 36 + g - 2d. Apply Lemma 6, except that if d = 16 and g = 1 we also need to prove that [h.sup.0]([I.sub.C](3)) = 25. Since [h.sup.1]([I.sub.C](3)) > 18, we have [h.sup.0]([I.sub.C](3)) [greater than or equal to] 56 + 18 - 48 = 26.

Lemma 10. A general W [member of] W contains no C [member of] [M'.sub.d,g] such that either there is a conic D [subset] [P.sup.5] with deg(D [intersection] C) [greater than or equal to] 12 or there is a line L [subset] [P.sup.5] with deg(L [intersection] C) [greater than or equal to] 6.

Proof. By Lemma 9 it is sufficient to exclude all C [member of] [B'.sub.12]\[A".sub.7] and all C [member of] [A'.sub.6]\[A'.sub.7].

(a) We first exclude all C [member of] [B".sub.14]\[A.sub.7]. In this case we have d = 16 and C [member of] [B'.sub.14]. Take a conic D [member of] [P.sup.5] such that deg(D [intersection] C) = 14. Since C [not member of] [A".sub.7], D is a smooth conic. Set x(1) := 14, x(2) := 13 and x(3) := 11. Fix any Z [subset] D such that deg(Z) = x(g). Since x(g) + 2g - 1 [less than or equal to] 16, the set [E.sub.Z] of all non-degenerate A [member of] [M.sub.16,g] with Z [subset] A has dimension 6d + 2 - 2g - 4x(g) (Lemma 2). Since D is a smooth curve, it has [[infinity].sup.x(g)] degree x(g) subschemes. [P.sup.5] has [[infinity].sup.9] planes and each plane has [[infinity].sup.5] conics. Hence it is sufficient to exclude all non-degenerate C [member of] [B'.sub.14]\[A.sub.7] with [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3x(g) - 14. By Lemma 1 and the case r = 4, t = 7, c = d [less than or equal to] 3t + 1 of Lemma 3 (note that deg(C [intersection] D') < 16 for each conic D' and that [A".sub.9] = 0) we have [h.sup.1]([I.sub.C](6)) [greater than or equal to] 3x(g) - 9 [greater than or equal to] 22. Then we continue as in the last two lines of step (a) of the proof of Lemma 9.

(b) Now we exclude all C [member of] [A'.sub.6]\[A".sub.7]. Since 13 [greater than or equal to] 6 + 2g - 1, by Lemma 2 we may assume [h.sup.1]([I.sub.C](7)) [greater than or equal to] 10. By step (a) we may assume that C [not member of] [B".sub.12]. Lemma 1 and the cases t = 6, 7, r = 4, of Lemma 3 give [h.sup.1]([I.sub.C](5)) [greater than or equal to] 5 + [h.sup.1]([I.sub.C](6)) [greater than or equal to] 10 + [h.sup.1]([I.sub.C](7)) [greater than or equal to] 20. Then we continue as in the proof of Lemma 9.

(c) Now we exclude all C [member of] [B'.sub.12] [union] [B'.sub.13]\[A.sub.7]. By step (b) we may assume C [not member of] [A'.sub.6]. Take a conic D such that deg(D [intersection] C) [member of] {12, 13}. Note that if [B".sub.12] = 0, then d [not equal to] 13. Since C [not member of] [A".sub.6], the conic D is smooth. Set [y.sub.d,g] := min{12, d + 1 - 2g}. Fix any zero-dimensional scheme Z [subset] D with deg(Z) = [y.sub.d,g]. By Lemma 2 the set of all non-degenerate C containing Z has dimension [less than or equal to] 6d + 2 - 2g - 4[y.sub.d,g]. Every smooth conic has [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] zero-dimensional schemes of degree [y.sub.d,g]. [P.sup.5] has [[infinity].sup.9] planes and each plane has [[infinity].sup.5] conics. Hence it is sufficient to check the curves C with the additional condition [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3[y.sub.d,g] - 14. Since 13 [less than or equal to] d [less than or equal to] 16 and g [less than or equal to] 3, we have [y.sub.d,g] [greater than or equal to] 8 and so 3[y.sub.d,g] - 14 > 10. We conclude as in step (b).

By Lemmas 8 and 10 we proved the part of Theorem 1 concerning non-degenerate C [member of] [M.sub.d,g].

4 Curves spanning a hyperplane

In this section we consider curves C [member of] [M.sub.d,g] spanning a hyperplane. For any hyperplane M [subset] [P.sup.5] let [M'.sub.d,g](M) be the set of all curves C [member of] [M.sub.d,g] spanning M. [M'.sub.d,g] (M) is a smooth and irreducible variety of dimension 5d + 1 - g. A general element of [M'.sub.d,g](M) has maximal rank ([1]). Since ([sup.11].sub.4]) > 7d + 1 - g, we have [h.sup.1](M, [I.sub.C,M](7)) = 0 for a general C [member of] [M'.sub.d,g](M). Since [P.sup.5] has [[infinity].sup.5] hyperplanes to prove that a general W [member of] W contains no curve in [M.sub.d,g] spanning a hyperplane it is sufficient to fix a hyperplane M [subset] [P.sup.5] and exclude all C [member of] [M'.sub.d,g] (M) such that [h.sup.1](M, [I.sub.C,M](7)) [greater than or equal to] d - 4 - g.

Since g > 0, by [13, part (ii) of Theorem on page 492] we have [h.sup.1](M, [I.sub.C,M](7)) = 0 if d [less than or equal to] 11. Hence in this section we assume 12 [less than or equal to] d [less than or equal to] 16 and we fix the hyperplane M [subset] [P.sup.5].

Remark 4. For any C [member of] [M'.sub.d,g](M) let [alpha](C) (or just [alpha]) denote the minimal integer t such that [h.sup.0](M, [I.sub.C,M](t)) > 0. Since C spans M, we have [alpha] [greater than or equal to] 2. Since d [less than or equal to] 16, we have 4d + 1 - g < 70 = ([sup.8.sub.4]) and so [alpha] [less than or equal to] 4. Since ([sup.6.sub.2]) = 15 and ([sup.7.sub.3]) = 35, we have [alpha] = 2 if and only if [h.sup.1](M, [I.sub.C,M](2)) [greater than or equal to] 2d - g - 13 and [alpha] [less than or equal to] 3 if and only if [h.sup.1](M, [I.sub.C,M](3)) [greater than or equal to] 3d - g - 33.

Lemma 11. Fix C [member of] [M'.sub.d,g](M), d [less than or equal to] 16, g > 0. There is no plane N [subset] M with deg(N [intersection] C) [greater than or equal to] 15.

Proof. Assume the existence of a plane N [subset] M such that deg(N [intersection] C) [greater than or equal to] 15. Fix a hyperplane H [subset] M with H [contains] N. Since the scheme C [intersection] H spans H, we get d = 16 and deg(C [intersection] N) = 15. Since C is smooth, for a general H [contains] N, H contains a tangent line of C if and only if this tangent line is contained in N. Hence for a general H we have C [intersection] H = (C [intersection] N) [union] {[p.sub.H]} with [p.sub.H] [member of] C\C [intersection] N. The pencil of all hyperplanes H [contains] N shows that C is rational, a contradiction.

Lemma 12. A general W [member of] W contains no C [member of] [M.sub.d,g] such that C [member of] [M'.sub.d,g] (M) for some hyperplane M and C is contained in a degree 3 surface A of M.

Proof. Since C spans M, then A spans M. The classification of minimal degree surfaces gives that either A is the Hirzebruch surface F1 embedded by the complete linear system [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] or it is a cone over a rational normal curve of [P.sup.3].

(a) Assume that A is the Hirzebruch surface [F.sub.1] embedded by the complete linear system [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and take a, b such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since C has genus g > 0, we have b [greater than or equal to] a [greater than or equal to] 2. We have d = (ah + bf) x (h + 2f) = a + b. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], the adjunction formula gives 2g - 2 = (ah + bf). ((a - 2)h + (b - 3)f) = (b - a)(a - 2) + a(b - 3). If a = 2 and hence b = d - 2 [greater than or equal to] 10, we get 2g - 2 [greater than or equal to] 7, a contradiction. If a [greater than or equal to] 3, we have 2g - 2 > 4, a contradiction.

(b) Assume that A is a cone with vertex o over a rational normal curve of [P.sup.3]. Let u: [F.sub.3] [right arrow] A be a minimal desingularization of A. [F.sub.3] is the Hirzebruch surface with the same name with h = [u.sup.-1](o) and u induced by the complete linear system [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Let C [subset] [F.sub.3] be the strict transform of C. Take a, b such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] with b [greater than or equal to] 3a. We have d = b. We have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] Since C is smooth, u induces an isomorphism of C; onto C. The adjunction formula gives 2g - 2 = (ah + bf) x ((a - 2)h + (d - 5)f) = (a - 2)(d - 3a) + a(d - 5). Since g > 0, we have a [greater than or equal to] 2 and hence 2 [less than or equal to] a [less than or equal to] [d/3]. In all cases we get 2g - 2 > 4, a contradiction.

Lemma 13. Fix C [member of] [M'.sub.d,g] (M) and let H [subset] M be a general hyperplane. Fix an integer t [greater than or equal to] 1.

(a) If 3t + 1 [greater than or equal to] d, then [h.sup.1](H, [I.sub.C[intersection]H,H](t)) = 0.

(b) Assume d [greater than or equal to] 3t + 2. Then [h.sup.1]([I.sub.C[intersection]H,H](t)) [less than or equal to] d - 3t - 1 and equality holds only if C [intersection] H is contained in a rational normal curve of H.

(c) If C is contained in a general heptic hypersurface and C [intersection] H is contained in a rational normal curve of M, then [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3d - 23 (case g = 1) or [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3d - 16 - 6g (case g = 2, 3).

Proof. The scheme C [intersection] H is a set of d points of H spanning H = [P.sup.3] and in uniform position. Part (a) follows from [10, Theorem 3.2] and part (b) is proved as in Lemma 8.

Now we prove part (c). Let T [subset] H be a rational normal curve of H containing C [intersection] H.

(i) First assume g = 1 and set Z := C [intersection] H. Since T has rod subsets of cardinality d, [P.sup.5] has [[infinity].sup.8] 3-dimensional linear spaces, each 3-dimensional H [subset] [P.sup.5] is contained in [[infinity].sup.1] hyperplanes of [P.sup.5] and each 3-dimensional linear space has [[infinity].sup.12] rational normal curves we get part (c) for g = 1 if we prove that [h.sup.1]([N.sub.C,M](-Z)) = 0, i.e. [h.sup.1]([N.sub.C,M](-1)) = 0. Since C is a curve, [h.sup.2](F) = 0 for all coherent sheaves F on C. Hence it is sufficient to prove that [h.sup.1](C, [TM.sub.|C] (-1)) = 0. Assume [h.sup.1](C, [TM.sub.|C](-1)) > 0. By duality there is a non-zero map [TM.sub.|C] [right arrow] [O.sub.C](1). Fix homogeneous coordinates [z.sub.0], [z.sub.1], [z.sub.2], [z.sub.3], [z.sub.4] of M. The Euler's sequence gives a nonzero map [O.sub.C][(1).sup.[cross product]5] [right arrow] [O.sub.C](1), i.e. ([a.sub.0], [a.sub.1], [a.sub.2], [a.sub.5]) [member of] [C.sup.5]\{0}. Hence C is contained in the hyperplane {[[summation].sub.i][a.sub.i][z.sub.i] = 0}, a contradiction.

(ii) Now assume g [greater than or equal to] 2. Set b := d + 2 - 2g and take Z [subset] C [intersection] H with deg(Z) = b. Since T has [[infinity].sup.b] subsets of cardinality b, [P.sup.5] has [[infinity].sup.8] 3-dimensional linear spaces, each 3-dimensional H [subset] P[.bar]5 is contained in [[infinity].sup.1] hyperplanes of [P.sup.5] and each 3-dimensional linear space has [[infinity].sup.12] rational normal curves, we get part (c) for g = 2, 3 if we prove that [h.sup.1]([N.sub.C,M](-Z)) = 0. Assume [h.sup.1]([N.sub.C,M](-Z)) > 0. Since [N.sub.C,M] is a quotient of [O.sub.C][(1).sup.[cross product]5], we get [h.sup.1]([O.sub.C](1)(-Z)) > 0. Since deg([O.sub.C](1)(-Z)) = 2g - 2, we get [O.sub.C](1)(-Z) [congruent to] [[omega].sub.Z]. Since b < d, there is Z' [subset] C [intersection] H with #(Z [intersection] Z') = b - 1. Set {p} := Z\Z [intersection] Z'. By monodromy the value of [h.sup.1]([O.sub.C](1)(-A)) is the same for all A [subset] C [intersection] H with cardinality b. Hence [O.sub.C](1)(-Z') [congruent to] [O.sub.C](1)(-Z), i.e. p and q are linearly equivalent, contradicting the assumption g > 0.

Lemma 14. A general W [member of] W contains no C [member of] [M.sub.d,g] such that C [member of] [M'.sub.d,g](M) for some hyperplane M and [h.sup.0](M, [I.sub.C,M](2)) [greater than or equal to] 3.

Proof. Fix C [member of] [M'.sub.d,g](M) such that [h.sup.0](M, [I.sub.C,M](2)) [greater than or equal to] 3. Let K [subset] M be the settheoretic base locus of [absolute value of [I.sub.C,M](2)]. Since C spans M, every quadric hypersurface of M containing C is irreducible. Hence dim(K) [less than or equal to] 2. Let A [subset or equal to] K be any irreducible component of K containing C. First assume dim(A) = 2. Since C spans M, then A spans M and hence deg(A) [greater than or equal to] 3. Fix 2 general [Q.sub.1], [Q.sub.2] elements of [absolute value of [I.sub.C,M](2)]. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], Bezout gives A [??] [Q.sub.1] [intersection] [Q.sub.2] and hence deg(A) = 3. Use Lemma 12. Now assume dim(A) = 1, i.e. A = C. Fix general [Q.sub.1], [Q.sub.2], [Q.sub.3] and set T := [Q.sub.1] [intersection] [Q.sub.2] [intersection] [Q.sub.3]. Since d > 8, and C is an irreducible component of T, the contradiction comes from Bezout ([12, Theorem 2.2.5]).

Lemma 15. Fix C [member of] [M'.sub.d,g](M) and let H [subset] M be a general hyperplane. Set Z := C [intersection] H.

(i) If [h.sup.0](H, [I.sub.Z,H](2)) = 0, then we have [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] max{0, d - 13} and [h.sup.1](H, [I.sub.Z,H](4)) = 0.

(ii) If [h.sup.0](H, [I.sub.Z,H](2)) = 1, then we have [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] max{0,d - 12}, [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] 2 if d [member of] {15, 16} and [h.sup.1](H, [I.sub.Z,H](4)) = 0.

(iii) If [h.sup.0](H, [I.sub.Z,H](2)) = 2, then [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] d - 11, [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] d - 12 for all d [greater than or equal to] 13, [h.sup.1](H, [I.sub.Z,H](4)) [less than or equal to] max{0, d - 15} and Z is contained in a complete intersection T [subset] H of 2 quadrics; if d = 16 and [h.sup.1](H, [I.sub.Z,H](4)) > 0, then C is not contained in a general heptic hypersurface; if d = 12 and [h.sup.1](H, [I.sub.Z,H](3)) > 0, then [h.sup.0]( M, [I.sub.C,M](2)) > 0.

(iv) To rule out from a general heptic hypersurface all C [member of] [M.sub.d,g] such that there is a hyperplane M [subset] [P.sup.5] with C [member of] [M'.sub.d,g](M) and [h.sup.0](H, [I.sub.C[intersection]H,H](2)) = 2 for a general hyperplane H of M it is sufficient to rule out all C [member of] [M'.sub.d,g](M) with [h.sup.1](M, [I.sub.C,M](7)) [greater than or equal to] 3d - 6g - 21.

Proof. The set Z is in uniform position in H and it spans H. First assume [h.sup.0](H, [I.sub.Z,H](2)) = 0. Fix any S [subset] C [intersection] H with #(S) = 9. S is contained in a unique quadric surface QS and QS [intersection] C [intersection] H = S. Fix p [member of] C [intersection] H and any S' [subset] Z\(S [union] {p}) with #(S') [less than or equal to] 3. Since Z is in uniform position, there is a plane N [subset] H with N [intersection] Z = S'. The cubic [Q.sub.S] [union] N shows that [h.sup.1](H, [I.sub.SUS'[union]{p}](3)) = [h.sup.1](H, [I.sub.SUS'](3)). Hence [h.sup.1] (H, [I.sub.Z,H](3)) [less than or equal to] max{0, d - 13}. Since d - 10 [less than or equal to] 9, we have [h.sup.0](H, [I.sub.Z\(S[union]{p})](2)) > 0. Take a general Q' [member of] [absolute value of [I.sub.Z\(S[union]{p}),H](2)]. Since Z is in uniform position and #(Z\(S [union] {p})) [less than or equal to] 8, we have Q' [intersection] Z = Z\(S [union] {p}). Hence (Q [union] Q') [intersection] Z = Z\{p}. Hence [h.sup.1](H, [I.sub.Z\{p},H](4)) = [h.sup.1](H, [I.sub.Z,H](4)). Taking smaller subsets of Z\(S [union] {p}) we get in finitely steps that [h.sup.1](H, [I.sub.Z,H](4)) = 0.

Now assume [h.sup.0](H, [I.sub.Z,H](2)) = 1. Take [S.sub.1] [subset] Z with #([S.sub.1]) = 8 and let [Q.sub.1] [subset] H be a general element of [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since Z is in uniform position, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Taking planes we get [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] max{0, d - 12}. Since d [less than or equal to] 17, any [S.sub.2] [subset] Z\[S.sub.1] with #([S.sub.2]) [less than or equal to] 8 has [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for a general [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], we get [h.sup.1](H, [I.sub.Z,H](4)) = 0. Now assume d [member of] {15, 16}. Let Q [subset] H be the quadric surface containing Z. Since Z is in uniform position, Q is irreducible. Since [h.sup.0](Q, [O.sub.Q](3)) = 16, it is sufficient to prove that [h.sup.0](Q, [I.sub.Z,Q](2)) [less than or equal to] 18 - d. In particular we may assume [h.sup.0](Q, [I.sub.Z,Q](3)) > 0. Fix a general D [member of] [absolute value of [I.sub.Z,Q](3)]. Since Z is in uniform position in Q (or by monodromy), D is an integral curve and hence it is a canonically embedded integral curve, which is the complete intersection of Q and a cubic surface. Riemann-Roch gives [h.sup.0](D, [O.sub.D](3)) = 15. Since D is arithmetically Cohen-Macaulay, to prove that [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] 2 it is sufficient to prove that [h.sup.1](D, [I.sub.Z,D](3)) [less than or equal to] 2. Take any zero-dimensional scheme Z' [subset] D such that deg(Z') = 16 and Z' [contains or equal to] Z. It is sufficient to prove that [h.sup.1](D, [I.sub.Z',D](3)) [less than or equal to] 2. This is true (by duality), because the very ampleness of [[omega].sub.D] [congruent to] [O.sub.D](1) implies [h.sup.0](D, [Laplace]) [less than or equal to] 1 for every rank 1 torsion free sheaf [Laplace] on D with deg([Laplace]) = 2.

Now assume [h.sup.0]([I.sub.Z,H](2)) = 2. Since Z is in uniform position, for all [S.sub.3] [subset] Z with #([S.sub.3]) [less than or equal to] 7 we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for a general [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] As above we get [h.sup.1](H, [I.sub.Z,H](4)) [less than or equal to] max{0, d - 15} and [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] d - 11. Let T be the intersection of all elements of [absolute value of [I.sub.Z,H](2)]. Since Z is in uniform position, all elements of [absolute value of [I.sub.Z,H](2)] are irreducible. Hence dim(T) = 1. Assume for the moment that T is either not reduced or reducible. Since Z is in uniform position and it is a set, there is an irreducible curve [T.sub.1] [subset] T with deg([T.sub.1]) [less than or equal to] 3 and Z [subset] [T.sub.1]. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] we get a contradiction. Hence T is an integral degree 4 curve with [p.sub.a](T) = 1. By monodromy for a general H we have Z [subset] [T.sub.reg]. Set b := d + 1 - 2g. Since [T.sub.reg] has [[infinity].sup.b] subsets with cardinality b, [P.sup.5] has [[infinity].sup.8] 3-dimensional linear spaces and each H [member of] G(3, 5) has [[infinity].sup.16] complete intersections of 2 quadrics, to rule out all C [member of] [M.sub.d,g] for which there is M and H with C [member of] [M'.sub.d,g](M) and C [intersection] H contained in an integral degree 4 complete intersection, it is sufficient to rule out all C [member of] [M'.sub.d,g](M) with [h.sup.1](M, [I.sub.C,M](7)) [greater than or equal to] 3d - 6g - 21. If [h.sup.0](M, [I.sub.C,M](2)) = 0, then C is arithmetically Gorenstein ([17]) and hence [h.sup.1](M, [I.sub.C,M](7)) = 0, a contradiction. Since [p.sub.a](T) = 1 and [h.sup.1]( H, [I.sub.T,H](3)) = 0, we have [h.sup.1](H, [I.sub.A,H](3)) = 0 for every zero-dimensional scheme A [subset] T with either deg(A) [less than or equal to] 11 or deg(A) = 12 and A [not member of] [absolute value of [O.sub.T](3)]. Hence [h.sup.1](H, [I.sub.Z,H](3)) = 0 if d [less than or equal to] 11. Assume d = 12 and [h.sup.1](H, [I.sub.Z,H](3)) > 0. We get Z [member of] [O.sub.T](3) I and we conclude quoting [17]. Assume d [greater than or equal to] 13. Take [A.sub.1], [A.sub.2] [subset] C [inset] H such that #([A.sub.1]) = #(A2) = 12 and #([A.sub.1] [intersection] [A.sub.2]) = 11. Set {p} := [A.sub.1]\[A.sub.1] [intersection] [A.sub.2] and {q} = [A.sub.2]\[A.sub.1]. By monodromy we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], i.e. [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Assume [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], i.e. [A.sub.i] [member of] [absolute value of [O.sub.T](3)]. We get that p and q are linear equivalent, contradicting the inequality [p.sub.a](T) > 0.

Now assume d = 16 and [h.sup.1](H, [I.sub.C[intersection]H,T](4)) > 0. Since [[omega].sub.T] [congruent to] [O.sub.T], we have C [intersection] H [member of] [absolute value of [O.sub.T](4)], i.e. (since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is the complete intersection of a quartic and 2 quadrics. By [33, Theorem 0.1] we have d = 2 (mod 3) and that C [intersection] H is contained in rational normal curve; both statements are false.

Lemma 16. Fix C [member of] [M'.sub.d,g](M) d [greater than or equal to] 13, such that [alpha](C) = 3 and [h.sup.0](M, [I.sub.C,M](3)) [greater than or equal to] 2. Let H [subset] M be a general hyperplane. Then

(a) C [intersection] H is not contained in a curve of degree [less than or equal to] 3 or degree 4 and arithmetic genus 1.

(b) [h.sup.1](H, [I.sub.C[intersection]H,H](4)) = 0, [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] max{0, d - 12}, [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] 2 if d = 15 and [h.sup.1] (H, [I.sub.Z,H](3)) [less than or equal to] 3 if d = 16.

Proof. Let K [subset] M be the set-theoretic base locus of [absolute value of [I.sub.C,M](3)]. Since [alpha](C) = 3 and [h.sup.0](M, [I.sub.C,M](3)) [greater than or equal to] 2, we have dim(K) [less than or equal to] 2. Assume that for a general H [subset] M, the set C [intersection] H is contained in a curve [T.sub.H] with c := deg([T.sub.H]) [less than or equal to] 4 and c minimal. Since H is general, C [intersection] H is a set of d points of H spanning H and in uniform position. Since d > c, monodromy and the minimality of c gives that [T.sub.H] is irreducible. Since d > 12, Bezout's theorem gives [T.sub.H] [subset] K [intersection] H. Hence there is an irreducible component A of K with dim(A) = 2 and [T.sub.H] [subset] A [intersection] H for a general hyperplane H. For a general H, A [intersection] H is irreducible. Since [T.sub.H] [subset or equal to] A [intersection] H, we get [T.sub.H] = A [intersection] H and hence deg(A) = c. Lemma 12 gives c = 4. Assume [p.sub.a]([T.sub.H]) = 1, i.e. assume that [T.sub.H] is the complete intersection of 2 quadrics. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], the exact sequence (3) gives that A is a complete intersection of 2 quadric hypersurfaces of M. Since C [subset] A, we get [alpha](C) = 2, a contradiction. Now we prove part (b). By part (a) we have [h.sup.0](H, [I.sub.C[intersection]H,H](2)) [less than or equal to] 1. Use parts (i) and (ii) of Lemma 17.

Lemma 17. Let [GAMMA] be the set of all C [member of] [M.sub.d,g] such that there is a hyperplane M [subset] [P.sup.5] with C [member of] [M'.sub.d,g] (M) and [h.sup.0](M, [I.sub.C,M](2)) > 0. Then dim([GAMMA]) [less than or equal to] 3d + 19 + g.

Proof. Since [P.sup.5] has [[infinity].sup.5] hyperplanes, it is sufficient to prove that for each hyperplane M the set [PSI] of all C [member of] [M'.sub.d,g](M) and [h.sup.0](M, [I.sub.C,M](2)) > 0 has dimension [less than or equal to] 3d + 14 + g. Fix C [member of] [PSI] and let Q [subset] M be any quadric containing C. Since C spans M, Q is irreducible. The Hilbert scheme Hilb(Q) of Q has [H.sup.0]([N.sub.C,Q]) as its tangent space at C. Let [tau] be the tangent sheaf of Q.

(a) First assume that either Q is smooth or C does not intersect the singular locus V of C. Since dim [absolute value of [O.sub.M] (2)] = 14, to handle this case it is sufficient to prove that [h.sup.0]([N.sub.C,Q]) [less than or equal to] 3d + g. Since the algebraic group Aut(Q) acts transitively on Q\V and [H.sup.0]([tau]) is the tangent space to Aut(Q) at the identity, [H.sup.0]([tau]) spans [tau] at each point of Q\V. Since C [subset] Q\V, [N.sub.C,Q] is a quotient of [tau[].sub.|C] and hence it is spanned. Since [[omega].sub.Q] [congruent to] [O.sub.Q](-3), [N.sub.C,Q] is a rank 2 vector bundle with degree 3d + 2g - 2. Hence [h.sup.1] (det([N.sub.C,Q])) = 0. Since [N.sub.C,Q] is spanned, it is an extension of det([N.sub.C,Q]) by [O.sub.C] and hence [h.sup.1] ([N.sub.C,Q]) [less than or equal to] g. Riemann-Roch gives [h.sup.0]([N.sub.C,Q]) [less than or equal to] 3d + g.

(b) Now assume C [intersection] V [not equal to] 0 and set x := deg(C [intersection] V). Since C is smooth, x = 1 if dim(V) = 0. By step (a) and the fact that M has [[infinity].sup.13] singular quadrics it is sufficient to prove that [h.sup.0]([N.sub.C,Q]) [less than or equal to] 3d + 1 + g. The vector space [H.sup.0]([tau]) is the tangent space at the identity map of the automorphism group Aut(Q). Since Q\V is homogeneous, [[tau].sub.|Q\V] is a spanned vector bundle. Since C is not a line and dim(V) [less than or equal to] 1, the set V [intersection] C is finite. Dualizing the natural map from the conormal sheaf of C in Q to [[OMEGA].sup.1.sub.Q] we get a map w: [[tau].sub.|C] [right arrow] [N.sub.C,Q] which is surjective outside the finite set C [intersection] V. Let u: [??] [right arrow] Q be the blowing up of V, E := [u.sup.-1] (V) the exceptional divisor, and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] the strict transform of C. Since C is smooth, u maps isomorphically [??] onto C. Let Y be closure in Hilb([??]) of the strict transforms of all C [subset] Q with deg(C [intersection] V) = x. It is sufficient to prove that dim([PSI]) [less than or equal to] 3d + 1 + g. Take a general D [member of] [PSI]. Since Aut([??]) acts transitively of [??]\E, step (a) of the proof gives [h.sup.1]([N.sub.D/[??]]) [less than or equal to] g. Hence it is sufficient to prove that deg([N.sub.D,[??]]) [less than or equal to] 3d + 2g - 1, i.e. deg([[tau].sub.[??]|D]) [less than or equal to] 3d + 1, i.e. deg([[omega].sub.[??]]|D) [greater than or equal to] -3d - 1. The group Pic([??]) is freely generated by E and the pull-back H of [O.sub.Q](1). We have D x H = d and D x E = x. We have [[omega].sub.[??]] [congruent to] [O.sub.[??]](-3H + cE) with c = -1 if dim(H) = 0 ([18], Example 8.5 (2)) and c = 0 if dim(H) = 1 ([18], Example 8.5 (3)). Hence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and the proof is complete.

Lemma 18. A general heptic hypersurface contains no C [member of] [M.sub.d,g] such that there is a hyperplane M [subset] [P.sup.5] with C [member of] [M'.sub.d,g] (M), there is no line L [subset] M with deg(L [intersection] C) > 6, no conic D [subset] M with deg(D [intersection] C) [greater than or equal to] 12.

Proof. By [11, Corollaire 2], Lemma 11 and the assumptions on the lines and the conics of C, for any integer t > 5 and any plane N c M we have [h.sup.1](N, [I.sub.C[intersection]N,N](t)) = 0. Lemma 1 and the case r = 4 and t = 5,6,7 of Lemma 3 give [h.sup.1] (M, [I.sub.C,M](4)) [greater than or equal to] [h.sup.1] (M, [I.sub.C,M](7)) + 12 [greater than or equal to] d + 7 - g. Let H [subset] M be a general hyperplane.

(a) Assume that the set H [intersection] C is not contained in a rational normal curve of H. By Lemma 15 (assuming that C is contained in a general heptic hypersurface) [h.sup.1](H, [I.sub.C[intersection]H,H](4)) = 0. Set [beta] = 0 if [less than or equal to]< 15 and [beta] = 1 if d = 16. By Lemma 15 we have [h.sup.1] (H, [I.sub.C,H](3)) [less than or equal to] 1 if d = 12 and [h.sup.1](H,[I.sub.C[intersection]H,H](3)) [less than or equal to] d - 12 if d [greater than or equal to] 13. Hence [h.sup.1](H, [I.sub.C[intersection]H,H](4)) + [h.sup.1](H, [I.sub.C[intersection]H,H(3)]) [less than or equal to] 3 + [beta]. By (1) for a general hyperplane H [subset] M we have [h.sup.1] (M, [I.sub.C,M](2)) [greater than or equal to] [h.sup.1] (M,[I.sub.C,M](4)) - [beta] - [beta] [greater than or equal to] d + 4 - g - f, i.e. [h.sup.0] (M, [I.sub.C,M](2)) [greater than or equal to] 18 - d - [beta]. if d [less than or equal to] 15, then we conclude by Lemma 14. Now assume d = 16. We got [h.sup.0] (M, [I.sub.C,M](2)) > 0. By Lemma 17 we may assume [h.sup.1](M,[I.sub.C,M](7)) [greater than or equal to] 3d - 17 - 3g = 21 - 4g. We first get [h.sup.1](M,[I.sub.C,M](4)) [greater than or equal to] 33 - 3g and then [h.sup.1] (M, [I.sub.C,M] (2)) [greater than or equal to] 29 - 3g, i.e. [h.sup.0](M, [I.sub.C,M] (2)) > 10 - 2g [greater than or equal to] 4, contradicting Lemma 14.

(b) Now assume that C [intersection] H is contained in a rational normal curve [E.sub.H] of H. By part (c) of Lemma 13 we may assume that [h.sup.1] (M, [I.sub.C,M](7)) [greater than or equal to] 3d - 16 - 6g - [epsilon] with [epsilon] = 1 if g = 1 and [epsilon] = 0 if g [epsilon] {2,3}. We first get [h.sup.1] (M,[I.sub.C,M](4)) > 3d - 4 - 6g - [epsilon] and then [h.sup.1](M,[I.sub.C,M](2)) [greater than or equal to] 3d - 4 - 6g - [epsilon] - min{d - 13,0} - d + 10, i.e. [h.sup.0](M,[I.sub.C,M](2)) > 20 - 5g - [epsilon] - min{d - 13,0}. Lemma 14 concludes unless (d, g) = (16,3). Assume (d, g) = (16,3). We just proved that [h.sup.0](M, [I.sub.C,M](2)) [greater than or equal to] 2. By Lemma 14 we may assume that [h.sup.0] (M, [I.sub.C,M](2)) = 2. Let [summation] the intersection of two different quadric hypersurfaces of M containing C. Since C is integral and C is not contained in a degree 3 surface (Lemma 12), [summation] is an integral complete intersection surface. Hence for a general hyperplane H [subset] M, the scheme [F.sub.H] := H [intersection] [summation] is an integral complete intersection of two quadrics. By Bezout we have #([E.sub.H] [intersection] [F.sub.H]) [less than or equal to] 6, contradicting the inclusion of C [intersection] H in [E.sub.H] [intersection] [F.sub.H].

Notation 2. Fix d, g. For any integer a [greater than or equal to] 0 let [F.sub.a] (resp. [G.sub.a]) denote the set of all C [member of] [M.sub.d,g] such that there is a hyperplane M [subset] [P.sup.5] and a line L [subset] M (resp. a smooth conic L [subset] M) with C [member of] [M'.sub.d,g] and deg(L [intersection] C) = a. Set F[".sub.a] := [U.sub.b[greater than or equal to]a] [F.sub.b] and [G".sub.a] := [[union].sub.b[greater than or equal to]a] [G.sub.b].

The proof of Lemma 2 gives the following result.

Lemma 19. Fix a hyperplane M [subset] [P.sup.5] and an integer b [greater than or equal to] 4 such that d [greater than or equal to] b + 2g - 1. Let Z [subset] M be a zero-dimensional scheme with deg(Z) = b. Then the set of all C [member of] [M'.sub.d,g](M) containing Z has codimension at least 3b in [M'.sub.d,g](M).

Lemma 20. Fix d, g with 12 [less than or equal to] d [less than or equal to] 16 and 1 [less than or equal to] g [less than or equal to] 3. For any integer a [greater than or equal to] 6 set [u.sub.a] := min{a, d + 1 - 2g}.

(i) Every irreducible component of [F".sub.a], a [greater than or equal to] 6, has dimension [less than or equal to] 5d + 12 - g - 2[u.sub.a].

(ii) Every irreducible component of [G".sub.a], a [greater than or equal to] 12, has dimension [less than or equal to] 5d + 17 - g - 2[u.sub.a].

Proof. Fix a hyperplane M [subset] [P.sup.5]. Set [F.sub.a](M) := {C [member of] [F.sub.a]: C [subset] M}, [G.sub.a](M) := {C [member of] [G.sub.a]: C [subset] m}, [F".sub.a] (M) := [U.sub.b[greater than or equal to]a] F(M) and G!J(M) := [[union].sub.b[greater than or equal to]a] [G.sub.b](M). Since in the definitions of Fa and Ga we prescribed that C [member of] [M'.sub.d,g](M), every C [member of] [G".sub.a] [union] [F".sub.a] is contained in a unique hyperplane. Hence to prove the lemma we may fix a hyperplane M, give an upper bound for dim([F".sub.a](M)) and dim([G".sub.a](M)) and then add +5 to that upper bound. Since the upper bounds in (i) and (ii) are non-decreasing functions of a to prove part (i) (resp. part (ii)) it is sufficient to give a suitable upper bound for all [F.sub.a](M), a [greater than or equal to] 6, and all [G.sub.a](M), a [greater than or equal to] 12. Fix C [member of] [F.sub.a](M) (resp. C [member of] [G.sub.a](M)) and a line L [subset] M (resp. a smooth conic L [subset] M) with deg(L [intersection] C) = a. Fix a zero-dimensional scheme Z [subset or equal to] C [intersection] Z with deg(Z) = [u.sub.a]. Use Lemma 19, that the smooth curve L has [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] subschemes of degree [u.sub.a], that M has [[infinity].sup.6] lines and that M has [[infinity].sup.11] conics.

Lemma 21. A general heptic hypersurface contain no C [member of] [M.sub.d,g] such that there is a hyperplane M with C [member of] [M'.sub.d,g] (M) and either there is a line L [subset] M with deg(C [intersection] L) [greater than or equal to] 6 or there is conic D [subset] M with deg(D [intersection] C) [greater than or equal to] 12.

Proof. By Lemma 11 and [11, Corollaire 2] we have [h.sup.1] (N, [I.sub.C[intersection]N,N](t)) = 0 if t [greater than or equal to] 5 and the plane N contains neither a line R with deg(R [intersection] C) [greater than or equal to] t + 2 nor a conic D with deg(D [intersection] C) [greater than or equal to] 2t + 2. Until step (e) we assume that either [alpha](C) [not equal to] 2 or d [less than or equal to] 12.

(a) We first exclude (with the silent assumption [alpha](C) [not equal to] 2) all C [member of] [M.sub.d,g] such that there is a hyperplane M and a line L [subset] M with C [member of] [M'.sub.d,g](M) and deg(C [intersection] L) [greater than or equal to] 9. Fix C, L such that deg(C [intersection] L) [greater than or equal to] 9. Set [u.sub.9] := min{9, d + 1 - 2g}. By Lemma 20 we may assume [h.sup.1] (M,[I.sub.C,M](7)) [greater than or equal to] d - 10 - g + 2[u.sub.9]. Part (a) of Lemma 13 for t = 5,6,7 gives [h.sup.1] (M, [I.sub.C,M](4)) [greater than or equal to] [h.sup.1] (M,[I.sub.C,M](7)).

First assume d [less than or equal to] 12 and so [u.sub.9] = d + 1 - 2g. Lemma 13 gives [h.sup.1] (M, [I.sub.C,M] (3)) [greater than or equal to] [h.sup.1] (M, [I.sub.C,M] (4)) and [h.sup.1] (M, [I.sub.C,M] (2)) [greater than or equal to] [h.sup.1] (M, [I.sub.C,M] (3)) - d + 10. Hence we have [h.sup.0] (M, [I.sub.C,M] (2)) [greater than or equal to] 14 - 2d + g + 3d + 3 - 6g - 2 = 15 + d - 5g > 0, a contradiction.

Now assume d [greater than or equal to] 13. Note that [u.sub.9] [greater than or equal to] 8 and hence [h.sup.1] (M, [I.sub.C,M] (4)) [greater than or equal to] d + 6 - g. We get [h.sup.1] (M, [I.sub.C,M] (3)) [greater than or equal to] d + 3 - g (Lemma 12) and hence [h.sup.0] (M, [I.sub.C,M] (3)) > 39 - 2d - 1 [greater than or equal to] 2, then [alpha](C) = 3 and then (by Lemma 16) [h.sup.1] (M, [I.sub.C,M](3)) > [h.sup.1] (M, [I.sub.C,M](4)) > d + 6 - gand [h.sup.1] (M, [I.sub.C,M](2)) [greater than or equal to] d + 3 - g,i.e. [h.sup.0](M, [I.sub.C,M](2)) > 17 - d > 0, a contradiction.

(b) Now we exclude all C [member of] [M.sub.d,g] such that there is a hyperplane M and a line L [subset] M with C [member of] [M.sub.d,g] (M) and deg(C [intersection] L) [greater than or equal to] 6. Fix C, L such that deg(C [intersection] L) [greater than or equal to] 6. Since d [greater than or equal to] 2g - 1 + 6, by Lemma 20 we may assume [h.sup.1] (M, [I.sub.C,M](7)) [greater than or equal to] d + 2 - g. By step (a) we may assume that deg(R [intersection] C) [less than or equal to] 8 for all lines R. Lemma 1 and the case t = 7 of Lemma 13 give [h.sup.1] (M, [I.sub.C,M](4)) [greater than or equal to] [h.sup.1](M, [I.sub.C,M](6)) [greater than or equal to] d + 6 - g. We solved this case in step (a).

(c) Now we exclude all C such that there is a conic D with deg(D [intersection] C) [greater than or equal to] 12 (they have d [greater than or equal to] 13). By step (b) we may assume deg(R [intersection] C) [less than or equal to] 5 for all lines R. Thus D is smooth. Set b := min{12,d + 1 - 2g}. By Lemma 20 we may assume that [h.sup.1] (M, [I.sub.C,M] (7)) [greater than or equal to] d - 15 - g + 2b. By Lemma 1 and the case t = 7 of Lemma 13 we may assume [h.sup.1] (M, [I.sub.C,M] (6)) [greater than or equal to] d - 11 - g + 2b. Since 2b [greater than or equal to] 17, we conclude as in steps (a) and (b).

(d) Now we assume d [greater than or equal to] 13 and [alpha](C) = 2. By Lemma 17 we may assume [h.sup.1] (M, [I.sub.C,M](7)) [greater than or equal to] 3d - 17 - 3g.

(d1) First assume that there is no line L with deg(L [intersection] C) [greater than or equal to] 9. By Lemmas 1 and 13 (case t = 7), we have [h.sup.1] (M, [I.sub.C,M](6)) > [h.sup.1] ([I.sub.C,M](7)) + 4 [greater than or equal to] 3d - 13 - 3g. Part (a) of Lemma 13 for t = 5,6 gives [h.sup.1] (M, [I.sub.C,M](4)) [greater than or equal to] 3d - 13 - 3g. The cases t = 3,4 of Lemma 13 gives [h.sup.1] (M, [I.sub.C,M] (2)) [greater than or equal to] 3d - 13 - 3g + 10 - d - min{0,d - 13}, i.e. [h.sup.0](M,Ic,m(2)) [greater than or equal to] 11 - 2g - min{0,d - 13}. Lemma 14 gives d = 16, g = 3 and [h.sup.0] (M, [I.sub.C,M](2)) = 2. Let K be the intersection of 2 different elements of [absolute value of [I.sub.C,M](2)]. K is a degree 4 surface and for a general hyperplane H of M K [intersection] H is an integral degree 4 curve, which is the complete intersection of 2 quadrics. Hence for a general H the set C [intersection] H is not contained in a rational normal curve of H. Hence [h.sup.1] (M, [I.sub.C,M](3)) [greater than or equal to] [h.sup.1] (M, [I.sub.C,M](4)) - 2 and [h.sup.1](M, [I.sub.C,M](2)) [greater than or equal to] [h.sup.1] (M, [I.sub.C,M](3)) - 4 (Lemma 13). Hence [h.sup.1](M, [I.sub.C,M](2)) [greater than or equal to] 20, i.e. [h.sup.0] (M, [I.sub.C,M](2)) [greater than or equal to] 5, a contradiction.

(d2) Now assume the existence of a line L [subset] M such that e := deg(L [intersection] C) [greater than or equal to] 9. Let U be the set of all lines R [subset] M such that deg(R [intersection] C) [greater than or equal to] 7 and let V be the set of all planes spanned by conics D such that deg(D [intersection] C) [greater than or equal to] 12.

(d2.1) Assume for the moment that U [union] V is finite. Let N [subset] M be a general plane. We have N [intersection] R = [empty set] for all R [member of] U and N [intersection] A is a single point for every A [member of] V. Let V [subset] [H.sup.0]([O.sub.M](1)) be the 2-dimensional linear subspace parametrizing all hyperplanes U of M containing N. For any such U we have R [??] U for all R [member of] U and dim(U [intersection] A) = 1 for each A [member of] V. Hence U contains no line R with deg(R [intersection] C) [greater than or equal to] 7 and no conic D with deg(C [intersection] D) [greater than or equal to] 12. By Lemma 11 and [11, Corollaire 2] we have [h.sup.1] (A, [I.sub.C[intersection]A,A](5)) = 0 for every plane A [subset] U and then Lemma 3 for r = 3, t = 5 and c < 16 gives [h.sup.1](U, [I.sub.C[intersection]U,U] (5)) = 0. Lemma 1 for t = 5,6,7 gives [h.sup.1] (M, [I.sub.C,M] (4)) [greater than or equal to] [h.sup.1](M, [I.sub.C,M](7)) + 3. As in step (d1) we get 15 [less than or equal to] d [less than or equal to] 16 and g = 3. As in step (d1) we exclude the case [h.sup.0](M, [I.sub.C,M] (2)) = 2 (and in particular the case d = 15), i.e. we may assume d = 16 and [h.sup.0] (M, [I.sub.C,M] (2)) = 1. Let Q C M denote the quadric containing C. We have [h.sup.1](M, [I.sub.C,M](3)) [greater than or equal to] [h.sup.1] (M,[I.sub.C,M](4)) - 3 [greater than or equal to] 22, i.e. h0(M,Icm(3)) > 35 - 46 + 22 = 11 [greater than or equal to] 7. Hence [h.sup.0](Q, [I.sub.C,Q](3)) [greater than or equal to] 2. Fix two general [U.sub.1], [U.sub.2] [member of] [absolute value of [I.sub.C,Q](3)] and set K := [U.sub.1] [intersection] [U.sub.2]. K is the complete intersection in M of a quadric and 2 cubics, because any element of [absolute value of [I.sub.C,Q](3)] is an irreducible surface, since [h.sup.0](Q, [I.sub.C,Q](2)) = [h.sup.0](M, [I.sub.C,M](2)) - 1 = 0. The complete intersection K links C to a degree 2 locally Cohen-Macaulay curve E. By Bezout we have L [subset] K and so either E is a double structure on L (a rope) or it is the union of L and another line R. Set R := L if E is a rope. The complete intersection K links R to the curve c [union] L. Since R is arithmetically Cohen-Macaulay, C [union] L is arithmetically Cohen-Macaulay. In particular [h.sup.1](M, [I.sub.C[union]L,M](7)) = 0. Since [h.sup.1](M, [I.sub.C,M](7)) [greater than or equal to] 3d - 14 - 3g > d [greater than or equal to] deg(C [intersection] L), we got a contradiction.

(d2.2) Now assume that U is infinite. Take R [member of] U with R [not equal to] L. Assume R [intersection] L [not equal to] [empty set]. Since deg(L [intersection] R) = 1, we get deg(L [intersection] C) = 9, deg(R [intersection] C) = 7 L [intersection] R [subset or equal to] C. The plane A spanned by R [union] L has deg(A [intersection] C) [greater than or equal to] 15, contradicting Lemma 11. Now assume R [intersection] L = [empty set]. The hyperplane H spanned by R [intersection] L intersects C in a scheme of degree [greater than or equal to] 16. We get d = 16, deg(L [intersection] C) = 9 and deg(R [intersection] C) = 7. Since C is smooth, the linear projection M\L [right arrow] [P.sup.2] from L induces a morphism u: C [right arrow] [P.sup.2] such that deg(u) x deg(u(C) = 7. Since there are infinitely many R [member of] U, we get deg(u) [greater than or equal to] 7 and so u(C) is a line, contradicting the assumption C [member of] [M'.sub.d,g](M).

(d2.3) Now assume that V is infinite. Let V' be an irreducible component of V with dim(V') > 0. Assume the existence of a conic D [subset] M such that deg(D [intersection] C) [greater than or equal to] 12 and call A the plane spanned by D. If A [intersection] L is a point, then the 3-dimensional linear space spanned by A [union] L intersects C in a scheme of degree [greater than or equal to] 12 + 9 - 1 > d, a contradiction.

Now assume L [subset] A for infinitely many planes A [member of] V'. For any plane N [subset] M with N [intersection] L = [empty set], the set N [intersection] A is a single point. Hence as in step (d2.1) we may ignore all these planes A [member of] V', even if infinite.

Now take a general A [member of] V' and assume A [intersection] L = [empty set] and f := deg(D [intersection] C) with D [subset] A a conic and f [greater than or equal to] 12. Take general [A.sub.1], [A.sub.2] [member of] V' and let [D.sub.i] [susbet] [A.sub.i], i = 1, 2, be the conic with deg([D.sub.i] [intersection] C) = f. First assume that [A.sub.1] [intersection] [A.sub.2] is a line (this is the case if either [D.sub.1] and [D.sub.2] have a common component or [D.sub.1] [intersection] [D.sub.2] is a zero-dimensional scheme of degree [greater than or equal to] 2). We get that either all A [member of] V' are contained in a fixed hyperplane of M (absurd, because C spans M) or there is a line R [subset] M contained in all A [member of] V'. If N [subset] M is a plane with R [intersection] N = [empty set], then no conic of some A [member of] V' is contained in a hyperplane of M containing N.

(d2.3.1) Assume [D.sub.1] [intersection] [D.sub.2] = [empty set]. We have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and therefore we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Fix any [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since deg(C [intersection] Q) [greater than or equal to] 2f + e [greater than or equal to] 33, Bezout gives C [subset] Q'. Hence [h.sup.0](M, [I.sub.C,M](2)) [greater than or equal to] 2. By Lemma 14 we may assume [h.sup.0](M, [I.sub.C,M](2)) = 2 and call K' the degree 4 irreducible surface which is the base locus of [absolute value of [I.sub.C,M](2)]. For a general hyperplane H [subset] M the scheme K' [intersection] H is an integral curve complete intersection of 2 quadrics and containing C [intersection] H. By Bezout C [intersection] H is not contained in a rational normal curve of H and so [h.sup.0](H, [I.sub.C[intersection]H,H](2)) = 2. Lemma 15 gives [h.sup.1](M, [I.sub.C,M](3)) [greater than or equal to] [h.sup.1](M, [I.sup.C,M](4)) [greater than or equal to] 3d - 17 - 3g, i.e. [h.sup.0](M, [I.sub.C,M](3)) [greater than or equal to] 17 - 2g > 10. Hence the map [H.sup.0](M, [I.sub.C,M](2)) [cross product] [H.sup.0](M, [O.sub.M](1)) [right arrow] [H.sup.0](M, [I.sub.C,M](3)) is not surjective. Therefore C is contained in the intersection of K' with a cubic hypersurface, contradicting Bezout and the inequality d > 12.

(d2.3.2) Now assume that the scheme [D.sub.1] [intersection] [D.sub.2] is a single point. First assume that either e [greater than or equal to] 10 or f [greater than or equal to] 13 or d [less than or equal to] 15 or [D.sub.1] [intersection] [D.sub.2] [not member of] C. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Fix any [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since deg(C [intersection] Q') [greater than or equal to] 2f + e - deg([D.sub.1] [intersection] [D.sub.2] [intersection] C) [greater than or equal to] 2d + 1, Bezout gives Q' [contains] C, contradicting Lemma 14. Now assume e = 9, f = 12, d = 16 and that [D.sub.1] [intersection] [D.sub.2] is a point p [subset or equal to] C. Take a general q [member of] C. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Fix any [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since deg(C [intersection] Q') [greater than or equal to] 2f - 1 + e - deg([D.sub.1] [intersection] [D.sub.2] [intersection] C) + 1 [greater than or equal to] 2d + 1, then Q' [contains] C. Hence [h.sup.0](M, [I.sub.C,M](2)) [greater than or equal to] 2. By Lemma 14 we may assume [h.sup.0](M, [I.sub.C,M](2)) = 2 and call K' the degree 4 irreducible surface which is the base locus of [absolute value of [I.sub.C,M](2)]. We conclude as in step (d2.3.1).

Proof of Theorem 1: We just proved Theorem 1 for all C [member of] [M.sub.d,g] spanning a hyperplane of [P.sup.5], and so we completed the proof of Theorem 1

5 Curves in a 3-space

In this section we assume that C spans a 3-dimensional linear subspace U [subset] [P.sub.5]. We prove Proposition 1 and give a few results useful to extend it to higher degrees and higher genera. The Hilbert scheme [M'.sub.d,g](U) of all non-special curves of degree d and genus g of U is smooth and irreducible of dimension 4d. Since the Grassmannian G(3,5) of all 3-dimensional linear subspaces of [P.sup.5] has dimension 8, to exclude these curves C it is sufficient to exclude the ones with [h.sup.1] ([I.sub.C](7)) [greater than or equal to] 2d + 2 - 2g - 8. However, for a general W [member of] W and a general U [member of] G(3,5) the surface W [intersection] U is a general surface of degree 7 and hence any curve of degree [less than or equal to] 16 on it is a complete intersection by a theorem of Max Noether. Hence to exclude all C [member of] [M.sub.d,g] spanning some U [member of] G(3,5) it is sufficient to exclude the ones with [h.sup.1] ([I.sub.C](7)) [greater than or equal to] 2d - 2g - 5. Fix a general X [member of] [M'.sub.d,g](U). Since X has maximal rank ([2]) and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], we have [h.sup.1] ([I.sub.x](7)) = [h.sup.1](U, [I.sub.X,U](7)) = 0 and so [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since dim([M'.sub.d,g](U)) = 4d, [P.sup.5] has [[infinity].sup.8] 3-dimensional linear spaces and 7d + 1 - g > 4d + 8, a general heptic hypersurface contains no curve C [member of] [M.sub.d,g] with the Hilbert function of X in degree 1 and degree 7. Hence we may increase by one these bounds and so we may assume [h.sup.1] (U, [I.sub.C,U] (7)) [greater than or equal to] 2d - 2g - 4, i.e. for any d,g to rule out the existence of any C [member of] [M.sub.d,g] with dim(<C>) = 3 for a general heptic hypersurface it is sufficient to exclude all C [member of] [M'.sub.d,g](U) with [h.sup.1] (U, [I.sub.C,U](7)) [greater than or equal to] 2d - 2g - 4. By [13] we may assume d [greater than or equal to] 11. Hence 11 [less than or equal to] d [less than or equal to] 16 and 1 [less than or equal to] g [less than or equal to] 3. Let [alpha] or [alpha](C) be the minimal degree of a surface of U containing C. Since C is non-degenerate, we have [alpha] [greater than or equal to] 2. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], we have [alpha] [less than or equal to] 7.

Lemma 22. We have [alpha] [not equal to] 2.

Proof. Assume [alpha] = 2. Let Q [subset] U be a quadric containing C. If Q is a quadric cone, then C is arithmetically Cohen-Macaulay ([16, V, Ex. 2.9]), because C is smooth, and in particular [h.sup.1] ([I.sub.C](7)) = 0, a contradiction. Hence Q is a smooth quadric. Write C [member of] [absolute value of [O.sub.Q] (u, [upsilon])] with, say, u [less than or equal to] [upsilon]. Since g > 0, we have u > 1. Since [h.sup.1] ([O.sub.C] (1)) = 0, we have u [less than or equal to] 2. Since 1 [less than or equal to] g [less than or equal to] 3, we get u = 2 and [upsilon] = g + 1. In all cases we have [h.sup.1] ([I.sub.C](7) ) = 0, a contradiction.

The following lemma is a particular case of step (b2) of Remark 5 proved below.

Lemma 23. Fix C [member of] [M'.sub.d,g](u), d [less than or equal to] 16. Let H [subset] U be a general hyperplane. We have [h.sup.1] (H, [I.sub.C[intersection]H,H](t)) = 0 for all t [greater than or equal to] 5, [h.sup.1] (H, [I.sub.C[intersection]H,H](4)) = 0 if d [less than or equal to] 12 and [h.sup.1] (H, [I.sub.C[intersection]H,H](4)) [less than or equal to] 1 if d [member of] {13,14}.

Remark 5. Let H c U be a general plane. The set Z := C [intersection] H is formed by d points of H in uniform position and spanning H.

(a) In particular Z is in linearly general position and hence [h.sup.1] (H, [I.sub.Z,H](t)) = 0 for all t with 2t [greater than or equal to] d - 1 ([10, Theorem 3.2]). If d [greater than or equal to] 2t + 2 we get [h.sup.1] (H, [I.sub.Z,H](t)) < d - 2t - 1.

(b) Now assume 2t [less than or equal to] d - 2, but [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], i.e., if d = 16 assume 5 [less than or equal to] t [less than or equal to] 7, while if 11 [less than or equal to] d [less than or equal to] 15 assume 4 [less than or equal to] t [less than or equal to] [(d - 2)/2]. Assume [h.sup.1](H, [I.sub.Z,H](t)) > 0. We get [h.sup.0] (H, [I.sub.Z,H](t)) > 0. Since C [intersection] H is in uniform position (or by a monodromy argument) and d > t, either a general D [member of] [absolute value of [I.sub.Z,H](t)] is irreducible or the base locus of [absolute value of [I.sub.Z,H](t)] contains an irreducible curve T [contains] Z and D = cT for some integer c > 2. The latter case does not occur if t is a prime, because Z spans H.

(b1) Assume D = cT with c [greater than or equal to] 2. We get [absolute value of [I.sub.Z,H](t)] = {cT} and hence [h.sup.0](H, [I.sub.Z,H](t)) = 1. Therefore [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], we get [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since Z spans H, t is not a prime. Since d [less than or equal to] 16, we get t < 6. Hence t = 4, c = 2, deg(T) = 2 and d = 15. Since T is an integral conic (i.e. a smooth conic), we get [h.sup.1] (H, [I.sub.Z,H](y)) = 0 for all y [greater than or equal to] 7. Let K [subset] U be a degree a surface containing C. Recall that [alpha] [less than or equal to] 7. Since d = 15 > 2[alpha], Bezout theorem gives T [subset] K. Varying H we see that K contains a 3-dimensional family of conics and hence it is a projection of a Veronese surface. Hence [alpha] [less than or equal to] 4. Since C has odd degree K must be an inner projection of the Veronese surface and hence [alpha] = 3. Since C is smooth, we see that C is isomorphic to a plane curve of degree [greater than or equal to] 8, contradicting the assumption g [less than or equal to] 3.

(b2) Assume that D is irreducible. Since [h.sup.i] ([O.sub.h]) = 0, i = 1,2, then we have [h.sup.1] (H, [I.sub.Z,H](t)) = [h.sup.1] (D, [I.sub.Z,D](t)). Since [congruent to] [O.sub.D](t - 3), we get d [greater than or equal to] 3t. If d = 3t we also get that Z is the complete intersection of D with a plane cubic; since [h.sup.1] (U, [I.sub.C,U](7)) > 0, [33, Theorem 0.1] gives that Z is contained in a smooth conic, contradicting the inequality d > 2t and the irreducibility of D. Now assume d [greater than or equal to] 3t + 1. Since [h.sup.1](D, [I.sub.Z,D](t)) > 0, [I.sub.Z,D](t) is a subsheaf of [[omega.sub.]D] [congruent to] [O.sub.D] (t - 3) and hence there is Z' [subset] Z such that deg(Z') = 3t and Z' is the complete intersection of D and and a cubic curve. We have h1 (H,IZ,H(t)) = [h.sub.1] (D, [I.sub.Z,D](t)) and [h.sup.1] (H, [I.sub.Z,H](t)) = [h.sup.1] (D, [I.sub.Z',D](t)) = [h.sup.1](D, [O.sub.D](t - 3)) = 1. Since [O.sub.D] (t - 3) is very ample, we have [h.sup.0] (D, [I.sub.E,D] (t)) = [h.sup.0] (D, [I.sub.Z',D](t)) - deg(E) (i.e. [h.sup.1] (D, [I.sub.E,D](t)) = [h.sup.1](D, [I.sub.Z',D](t)) for every E [contains] Z' with deg(E) - deg(Z') [less than or equal to] 2. Hence if [h.sup.1] (H, [I.sub.Z,H](t)) > 0, then d [greater than or equal to] 3t + 1 and if d [less than or equal to] 3t + 2, then [h.sup.1] (H, [I.sub.Z,H](t)) [less than or equal to] 1.

5.1 Proof of Proposition 1

In this subsection we take g = 1 and d [less than or equal to] 14.

Lemma 24. Fix a zero-dimensional scheme Z [subset] U and set b := deg(Z). Let [E.sub.Z] be the set of all C [member of] [M'.sub.d,1](U) such that C [contains] Z. If d [greater than or equal to] b, then every irreducible component of [E.sub.Z] has dimension 4d - 2b.

Proof. Fix C [member of] [E.sub.Z]. It is sufficient to prove that [h.sup.1] ([N.sub.C,U](-Z)) = 0 ([27, Theorem 1.5]). Since [N.sub.C] is a quotient of [O.sub.C][(1).sup.[direct sum]4], we are done if [h.sup.1](OC(1)(-Z)) = 0. Since d [greater than or equal to] b and [[omega].sub.C] [congruent to] [O.sub.Z], this is the case, unless d = b and [O.sub.C](1) [congruent to] [O.sub.C](Z). By duality we only need to exclude the existence of a non-zero map [N.sub.C] [right arrow] [O.sub.C] (1). Assume that this is the case. The restriction to C of Euler's sequence of T[P.sup.3] gives a non-zero map [O.sub.C][(1).sup.[direct sum]4] [right arrow] [O.sub.C](1). This map gives the equation of a hyperplane of U containing C, a contradiction.

Lemma 25. A general W [member of] W contains no C [member of] [M.sub.d,1], d [less than or equal to] 14, such that there is U [member of] G(3,5) with C [member of] [M'.sub.d,1] (U), deg(R n[intersection]C) [less than or equal to] 7 for each line R and deg(C [intersection] D) [less than or equal to] 13 for each conic D.

Proof. Fix C [member of] [M'.sub.d,1](U) such that deg(R [intersection] C) [less than or equal to] 7 for each line R and deg(R [intersection] D) [less than or equal to] 13 for each conic D. By Lemmas 1 and 23 we have [h.sup.1] (U, [I.sub.C,U](5)) > [h.sup.1](U, [I.sub.C,U](6)) + 3 [greater than or equal to] [h.sup.1] (U, [I.sub.C,U](7)) + 6 [greater than or equal to] 2d. By Lemma 23 we have [h.sup.1](U, [I.sub.C,U](3)) [greater than or equal to] [h.sup.1] (U, [I.sub.C,U](7)) - [epsilon] [greater than or equal to] 2d - [epsilon] with [epsilon] = 0 if d [less than or equal to] 12 and [epsilon] = 1 if d [member of] {13,14}, i.e. [h.sup.0](U, [I.sub.C,U](3)) [greater than or equal to] 20 - d - [epsilon] > 2. Since [alpha] > 2 (Lemma 22), C is contained in the intersection of 2 integral cubic surfaces and so d [less than or equal to] 9, a contradiction.

Lemma 26. A general W [member of] W contains no C [member of] [M.sub.d,1], d [less than or equal to] 14, such that there is U [member of] G(3,5) with C [member of] [M'.sub.d,1] (U) and a conic D with deg(D [intersection] C) [greater than or equal to] 14.

Proof. Take C [member of] [M.sub.d,1], d [less than or equal to] 14, such that there is U [member of] G(3,5) with C [member of] [M'.sub.d,1](U) and a conic D with deg(D [intersection] C) > 14. We have d = 14 and deg(D [intersection] C) = 14. By Lemma 25 we may assume that deg(R [intersection] C) [less than or equal to] 6 for every line R. Hence D is a smooth conic and so it has [[infinity].sup.14] degree 14 subschemes. Since dim(G(3,5)) = 8, U has [[infinity].sup.3] planes and each plane has [[infinity].sup.5] conics, to prove the lemma (quoting Lemma 24) it is sufficient to exclude all C [member of] [M.sub.14,4] spanning a 3-space U and with [h.sup.1] (U, [I.sub.C,U](7)) [greater than or equal to] 6d + 2 - 2g - 4d - 8 + 28 - 14 - 8 = 2d - 2. By Lemma 23 we have [h.sup.1] (U, [I.sub.C,U](3)) [greater than or equal to] h1 (U, [I.sub.C,U](7)) - [epsilon] [greater than or equal to] 2d - 2 - [epsilon] with [epsilon] = 0 if d [less than or equal to] 12 and [epsilon] = 1 if d [member of] {13,14}, i.e. [h.sup.0](U, [I.sub.C,U](3)) [greater than or equal to] 18 - d - [epsilon] [greater than or equal to] 2 and so Bezout gives d [less than or equal to] 9, a contradiction.

Lemma 27. A general W [member of] W contains no C [member of] [M.sub.d,1], d [less than or equal to] 14, such that there is U [member of] G(3,5) with C [member of] [M'.sub.d,1] (U) and a line R with deg(R [intersection] C) [greater than or equal to] 8 for some line R.

Proof. Fix U and C [member of] [M'.sub.d,1](U). By Lemma 26 we may assume deg(D [intersection] C) [less than or equal to] 13 for each conic D. Let R [subset] U be a line such that b := deg(C [intersection] R) is maximal. First assume b [greater than or equal to] 9. Fix Z [subset] R [intersection] C with deg(Z) = 9. Since dim(G(3,5)) = 8, U has [[infinity].sup.4] lines and each line has [[infinity].sup.9] subschemes of degree 9, Lemma 24 shows that it is sufficient to exclude the curves C with [h.sup.1] (U, [I.sub.C,U] (7)) [greater than or equal to] 2d - 8 + 18 - 9 - 4 = 2d - 3. By Lemma 23 we have [h.sup.1] (U, [I.sub.C,U](3)) [greater than or equal to] 2d - 3 - [epsilon] with [epsilon] = 0 if d [less than or equal to] 12 and [epsilon] = 1 if d [member of] {13, 14}, i.e. [h.sup.0](U, [I.sub.C,U](3)) [greater than or equal to] 17 - d - [epsilon] [greater than or equal to] 2 and so Bezout gives d [less than or equal to] 9, a contradiction. Now assume b = 8. As above we see that it is sufficient to exclude all C [member of] [M'.sub.d,1](U) with [h.sup.1] (U, [I.sub.C,U](7)) [greater than or equal to] 2d - 4. By Lemmas 1 and 23 we have [h.sup.1](U, [I.sub.C,U](6)) [greater than or equal to] 2d - 1 and hence [h.sup.0](U, [I.sub.C,U](3)) [greater than or equal to] 4.

Proof of Proposition!. Lemmas 25, 26 and 27 prove Proposition 1.

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E. Ballico *

* The author was partially supported by MIUR and GNSAGA of Indam (Italy)

Received by the editors in December 2015--In revised form in March 2016.

Communicated by J. Fine.

2010 Mathematics Subject Classification: 14J32; 14M10; 14H50.

Dept. of Mathematics University of Trento 38123 Povo (TN), Italy email: ballico@science.unitn.it

H. Clemens conjectured that for any degree d > 0 a general quintic hypersurface of [P.sup.4] contains only finitely many smooth rational curves of degree d; the stronger form of the conjecture says that the same is true for singular rational curves, except rational degree 5 plane curves ([5], [6], [9], [19], [20], [21], [22], [30]). There are results on other Fano 3-folds ([21], [26, Theorem 2], [22, Theorem I. 2]) results for higher genera ([23], [24]) and results on higher dimensional hypersurfaces of general types ([32]). In particular for every integer k [greater than or equal to] 8 for a very general degree k hypersurface X [subset] [P.sup.5] and any integral curve D [subset] X, we have deg(D) [less than or equal to] 2[??] - 2, where [??] is the geometric genus of D ([31], [32, Theorem 3.9]). The case k = 7, i.e. the case of a general heptic hypersurface of [P.sup.5], was singled out as an interesting boundary case ([7], [14]). A general heptic hypersurface contains exactly 698005 lines and no rational curve (not even singular ones) with degree d if 2 [less than or equal to] d [less than or equal to] 16 ([7], [14], [28, Theorem 1.1]). There are two main tools available for rational curves, but not for smooth curves of higher genus (a very strong result on the strata by splitting type of the restricted tangent bundle and the use of semigroups related to a rational parametrization). See [6] and [7] for a full use of these tools and [8] for a full use of the latter tool in another context.

For any set S [subset or equal to] [P.sup.5] let (S) denote its linear span. In this paper we prove the following result.

Theorem 1. Let W [subset] [P.sup.5] be a general heptic hypersurface. For all integers d, g with d [less than or equal to] 16 and 1 [less than or equal to] g [less than or equal to] 3 W does not contain any smooth curve C of genus g and degree d with dim((C)) = 3.

Only numerical reasons prevent us to get degrees d a little bit higher or genera g a little bit higher or to cover the curves spanning a 3-dimensional linear subspace. We prove the following result, but several parts of its proof works for higher genera and/or higher degree.

Proposition 1. A general heptic hypersurface of [P.sup.5] contains no smooth curve C of genus 1 and degree d [less than or equal to] 14.

In section 2 (resp. section 3, resp. section 4) we prove the part of Theorem 1 concerning curves C spanning a linear subspace <C> of [P.sup.5] of dimension [less than or equal to] 2 (resp. 5, resp. 4). In section 5 we give a few results for the case dim((C)) = 3 and prove Proposition 1. Section 4 contains more lemmas than the ones needed to prove Theorem 1 and even so in many places we know how to improve the lemmas by 1. Anyway, section 3 does not work if (d, g) [member of] {(17, 0), (17, 1)}, the next interesting cases.

Since for any smooth hypersurface W [subset] [P.sup.n], n [greater than or equal to] 4, with deg(W) [greater than or equal to] 2 there is a codimension 2 subvariety Z [subset] W which is not the intersection of W and a codimension 2 subvariety of [P.sup.n] ([29], [25]), there are plenty of curves on a very general heptic hypersurface X, which are not easily described, but certainly not unexpected. A dimensional count suggests that a very general heptic hypersurface has no curve with very low genus. If we take curves with arithmetic genus q and degree d [much greater than] q, then the dimensional count is better for high q than for q = 0 (when d [much greater than] q the dimensional count is even better for degenerate curves than for non-degenerate ones).

Question 1. Let X [subset] [P.sup.5] be a very general heptic hypersurface. Is it true that X contains no elliptic curve? Is it true that for each q [member of] N there is an integer d(q) such that X has no curve of arithmetic genus q and degree d > d(q)?

Our tools cannot solve these questions (at the very least we need 7d + 1 - q < ([sup.12.sub.5])), but we made an attempt to see where not to find counterexamples to this kind of questions. A strong feature of [7] (and also of [6] and [8]) is that it works for singular rational curves and this is the best way to state problems related to Clemens' conjecture as non-existence or finiteness results for maps from moduli schemes of curves to a varying target, e.g. a very general heptic hypersurface, i.e. in the set-up of Kontsevich moduli spaces with a varying target. Unfortunately, our tools use the arithmetic genus of the image and so at most we may recover (for low degrees) singular curves with low arithmetic genus. Theorem 1 and Proposition 1 are stated only for smooth curves, because in the singular case (but with arithmetic genus q [less than or equal to] 3) we can handle only lower degrees. The first lemmas to be generalized for the interested reader are Lemmas 2 and 3, because a nondegenerate singular curve C [subset] [P.sup.5] has hyperplane sections C [intersection] H which are not curvilinear. Moreover, if q = 1 the singular curves are rational and so they do not exist for d [less than or equal to] 16 on a very general heptic by [7]. If q = 2,3 knowing by [7] the case of rational curves we may say that the non-rational ones have very mild singularities, and it helps, even for the generalizations of Lemmas 2 and 3.

Thanks are due to a referee for useful remarks.

2 Preliminaries

We work over the complex number field.

Let [M.sub.d,g] denote the set of all smooth curves C [subset] [P.sup.5] with degree d, genus g and [h.sup.1]([O.sub.C](1)) = 0. The algebraic set [M.sub.d,g] is irreducible, because we only took nonspecial line bundles for the embeddings. Fix C [member of] [M.sub.d,g] and let [N.sub.C] be the normal bundle of C in [P.sup.5]. Since dim(C) = 1, we have [h.sup.2](F) = 0 for every coherent sheaf F on C. Therefore for any surjection G [right arrow] E of coherent sheaves on C, the associated map [H.sup.1](G) [right arrow] [H.sup.1](E) is surjective. Hence [h.sup.1](E) = 0 if [h.sup.1](G) = 0. The Euler's sequence shows that [TP.sup.5] is a quotient of [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Hence [h.sup.1]([N.sub.C]) = 0. Thus [M.sub.d,g] is smooth and of dimension [h.sup.0]([N.sub.C]) = 6d + 2 - 2g.

If we drop the condition [h.sup.1]([O.sub.C](1)) = 0, then when g [less than or equal to] 3 we only need to add the case (d, g) = (4,3) (canonically embedded non-hyperelliptic curves of genus 3). See Remark 2 for a proof that a general heptic hypersurface contains no canonically embedded curve of genus 3.

For any integer r [greater than or equal to] 1 let [M.sub.d,g](r) be the set of all C [member of] [M.sub.d,g] whose linear span has dimension r. Since g > 0, we have [M.sub.d,g](1) = 0. Since 1 [less than or equal to] g < 3 and the embedding is by a non-special line bundle, [M.sub.d,g](2) = 0, unless (d, g) = (3, 1).

Let W be the set of all smooth heptic hypersurfaces W [subset] [P.sup.5] satisfying the thesis of [7], i.e. containing no rational curve of degree [less than or equal to] 16 (not even singular ones).

Let Z [subset] [P.sup.r] be any zero-dimensional scheme. For any hyperplane H [subset] [P.sup.r] let [Res.sub.H](Z) denote the residual scheme of Z with respect of H, i.e. the closed subscheme of [P.sup.r] with [I.sub.Z]: [I.sub.H] as its ideal sheaf. We have [Res.sub.H](Z) [subset or equal to] Z and deg(Z) = deg([Res.sub.H](Z)) + deg(H [intersection] Z). For every integer t we have the following residual short exact sequence of coherent sheaves on [P.sup.r] (the latter is also an OH-sheaf), which we often call the residual exact sequence of H or of the inclusion H [subset] [P.sup.r]:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1)

Remark 1. Let N [subset] [P.sup.5] be a plane and C [subset] N be a smooth cubic. The normal bundle [N.sub.C] of C in [P.sup.5] is isomorphic to [O.sub.C](3) [cross product] [O.sub.C][(1).sup.[cross product]3] and hence [h.sub.1]([N.sub.C]) = 0 and [h.sup.0]([N.sub.C]) = 18. Hence the Hilbert scheme of [P.sup.5] is smooth at [C] and of dimension 18. We have [h.sup.1]([I.sub.C](t)) = 0 for all t [member of] N and hence [h.sup.0]([I.sub.C](7)) = ([sup.12.sub.5]) - 21. Hence a general heptic hypersurface contains no plane cubic.

By Remark 1 a general heptic hypersurface contains no element of [M.sub.d,g](2).

Remark 2. Let N [subset] [P.sup.5] be a plane and C [subset] N be a smooth degree 4 curve. C has genus 3 and [O.sub.C](1) [congruent to] [[omega].sub.C]. The normal bundle [N.sub.C] of C in [P.sup.5] is isomorphic to [O.sub.C](4) [cross product] [O.sub.C][(1).sup.[cross product]3] and hence [h.sup.1]([N.sub.C]) = 3 and [h.sup.0]([N.sub.C]) = 23. [P.sup.5] has [[infinity].sup.9] planes and each plane has [[infinity].sup.14] degree 4 curves. Hence the Hilbert scheme of [P.sup.5] is smooth at [C] and of dimension 23. We have [h.sup.1]([I.sub.C](t)) = 0 for all t [member of] N and hence [h.sup.0]([I.sub.C](7)) = ([sup.12.sub.5]) - 26. Hence a general heptic hypersurface contains no degree 4 plane curves.

Lemma 1. Fix integer t [greater than or equal to] 2, r [greater than or equal to] 3 and an integral and non-degenerate curve T [subset] Pr such that [h.sup.1]([I.sub.T](t)) > 0. Fix a linear subspace [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Assume that [h.sup.1](M, [I.sub.Mp[intersection]T,M](t)) = 0 for every hyperplane M [member of] [absolute value of V]. Then [h.sup.1]([I.sub.T](t - 1)) [greater than or equal to] [h.sup.1]([I.sub.T](t))+ dim(V) - 1.

Proof. For any hyperplane M [subset] [P.sup.r] we have an exact sequence

0 [right arrow] [I.sub.T](t - 1) [right arrow] [I.sub.T](t) [right arrow] [I.sub.T[intersection]M,M](t) [right arrow] 0 (2)

Since [h.sup.1](M, [I.sub.T,M](t)) = 0, the map [H.sup.1]([I.sub.T](t - 1)) [right arrow] [H.sup.1]([I.sub.T](t)) is surjective and hence its dual [e.sub.M]: [H.sup.1][([I.sub.T](t)).sup.[disjunction]] [right arrow] [H.sup.1][([I.sub.T](t - 1)).sup.[disjunction]] is injective. Taking the equations of all hyperplanes we get a bilinear map map u: [H.sup.1][([I.sub.T](t)).sup.[disjunction]] x [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], which is injective with respect to the second variables, i.e. for every non-zero linear form [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is injective (it is [e.sub.M] with M := {l = 0}). Hence if [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] with a [not equal to] 0 and l [not equal to] 0, then u(a, l) = [e.sub.M](a) [not equal to] 0. Therefore the bilinear map u is non-degenerate in each variable. Hence [h.sup.1]([I.sub.T](t - 1)) [greater than or equal to] [h.sup.1]([I.sub.T](t)) + dim(V) - 1 by the bilinear lemma.

Lemma 2. Fix an integer a > 0 and assume d [greater than or equal to] 2g - 1 + a. Fix a zero-dimensional curvilinear scheme Z [subset] [P.sup.5] such that deg(Z) = a. Set [E.sub.Z] := {C [member of] [M.sub.d,g]: Z [subset] C}. Then every irreducible component of EZ has dimension [less than or equal to] 6d + 2 - 2g - 4a.

Proof. If [E.sub.Z] = [empty set], then the lemma is true. Hence we may assume E [not equal to] [empty set]. Fix C [member of] E. By [27, Theoreme 1.5] it is sufficient to prove that [h.sup.1]([N.sub.C](-Z)) = 0. Since C is smooth, NC is a quotient of [TP.sup.5.sub.|C]. By the Euler's sequence of [TP.sup.5] [N.sub.C] is a quotient of [O.sub.C][(1).sup.6]. Since d [greater than or equal to] 2g - 1 + a, we have [h.sup.1]([O.sub.C](1)(-Z)) = 0.

Lemma 3. Fix integers t [greater than or equal to] 1 and r [greater than or equal to] 2. Let Z [subset] [P.sup.r] denote a curvilinear zero-dimensional scheme such that c := deg(Z) [less than or equal to] 3t + r - 2, Z spans [P.sup.r] and [h.sup.1]([I.sub.Z](t)) > 0; if c = 3t + r - 2 assume that [h.sup.1](N, [I.sub.N[intersection]Z,N](t)) = 0 for every plane N [subset or equal to] [P.sup.r]. Then either there is a line L [subset] [P.sup.r] with deg(L [subset] Z) [greater than or equal to] t + 2 or there is a conic D [subset] [P.sup.r] with deg(D [intersection] Z) [greater than or equal to] 2t + 2.

Proof. The case r = 2 is true for all t by [11, Corollaire 2] (in the case c = 3t we assumed that both [h.sup.1]([I.sub.Z](t)) > 0 and [h.sup.1]([I.sub.Z](t)) = 0). Hence we may assume r [greater than or equal to] 3 and use induction on r. The case t = 1 is true (if c [less than or equal to] r, because no Z with deg(Z) [less than or equal to] r spans [P.sup.r], while if c = r + 1 because [h.sup.1]([I.sub.Z](x)) = 0 for all x > 1 if deg(Z) = r + 1 and Z spans [P.sup.r]). Hence we may assume t [greater than or equal to] 2 and use induction on t in [P.sup.r].

(a) Let M [subset] [P.sup.r] be a hyperplane such that a := deg(Z [intersection] M) is maximal. First assume [h.sup.1](M, [I.sub.Z[intersection]M,M](t)) > 0. Since Z spans M we have deg(Z [intersection] M) [less than or equal to] c - 1. The maximality property of M gives that Z [intersection] M spans M. Hence the inductive assumption gives that either there is a line L [subset] M with deg(L [intersection] Z) [greater than or equal to] t + 2 or there is a conic D [subset] M with deg(D [intersection] Z) [greater than or equal to] 2t + 2. Hence we may assume [h.sup.1] (M, [I.sub.Z[intersection]M,M](t)) = 0. The residual sequence (1) of M gives [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. We have deg([Res.sub.M](Z)) = c - a. Since M [intersection] Z spans M we have a [greater than or equal to] r [greater than or equal to] 3. Assume for the moment a = r. The maximality property of M implies that Z is in linearly general position in [P.sup.r]. Since c [less than or equal to] rt + 1, we get [h.sup.1]([I.sub.Z](t)) = 0 ([10, Theorem 3.2]). Hence c - a [less than or equal to] c - r - 1 [less than or equal to] 3t + r - 2 - r - 1 < 3(t - 1) + r - 3. By the inductive assumption either there is a line L [subset] [P.sup.r] with deg(L [intersection] Z) [greater than or equal to] t + 1 or there is a conic D [subset] [P.sup.r] with deg(D [intersection] Z) [greater than or equal to] 2t. Assume the existence of a conic D [subset] P[.bar]r with deg(D [intersection] Z) [greater than or equal to] 2t. Since r - 1 [greater than or equal to] 2, the maximality property of M gives a [greater than or equal to] 2t + r - 3. Hence c [greater than or equal to] 2t + 2t + r - 3, a contradiction. Now assume the existence of a line L [subset] [P.sup.r] with deg(L [intersection] Z) [greater than or equal to] t + 1. To prove the lemma we may assume deg(L [intersection] Z) = t + 1. Let H [subset] [P.sup.5] be a hyperplane containing L and with b := deg(H [intersection] Z) maximal among all hyperplanes containing L. Since Z spans Pr, we have b [greater than or equal to] t + r - 1. If [h.sup.1](H, [I.sub.Z[intersection]H,H](t)) > 0, then we conclude by the inductive assumption on r. Hence we may assume [h.sup.1](H, [I.sub.Z[intersections]H,H](t)) = 0. The residual sequence (1) of H gives [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. We have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Hence the inductive assumption on t gives the existence of a line L [subset] [P.sup.5] with [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. If the lemma fails, then deg(R [intersection] [Res.sub.H](Z)) = deg(R [intersection] Z) = t + 1. In this case we have c [greater than or equal to] 2t + 2. Hence the lemma is proved in degree t if c [less than or equal to] 2t + 1 (but this part is just [4, Lemma 34]).

(b) First assume R [not equal to] L and R [intersetion] L [not equal to] 0. Since deg(R [intersection] [Res.sub.H](Z)) = t + 1 and H [contains] L, we have deg((L [union] R) [intersection] Z = 2t + 2. Hence we may take D := R [union] L.

(c) Now assume R = L. We may take Z minimal with the restriction that [h.sup.1]([I.sup.Z](t)) > 0 and deg(Z [intersection] L) = t + 1. Part (a) of our proof works if instead on H we take any hyperplane U [contais] L (since as in the first part we exclude the existence of a conic D with deg([Res.sub.U](Z)) [greater than or equal to] 2t). Let Q be a quadric hypersurface containing L in its singular locus. Since deg([Res.sup.Q](Z)) [less than or equal to] 3t - 2t - 2 [less than or equal to] t - 2, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (even in the case t = 2). Therefore the residual exact sequence of Q gives [h.sup.1](Q, [I.sub.Z[intersection]Q,Q](t)) > 0 and hence [h.sup.1]([I.sub.Z[intersection]Q](t)) > 0. The minimality of Z gives Z [subset] Q. Since Z is curvilinear, taking Q = [N.sub.1] + [N.sub.2] with [N.sub.1], [N.sub.2] hyperplanes we also get that only the connected components of Z whose reduction are contained in L arise (for a minimal Z), hence we reduce to the case deg(Z) = 2t + 2. Let W [subset] Z be any degree 2t + 1 subscheme. Since deg(W [intersection] J) [less than or equal to] deg(Z [intersection] J) [less than or equal to] t + 1 for each line J, part (a) of the proof gives [h.sup.1]([I.sub.W](t)) = 0. Hence [h.sup.1](M, [I.sub.Z,M](t)) = 1. Since [h.sup.1](N, [I.sub.Z[intersection]N,N](t)) = 0 for every hyperplane N [subset] [P.sup.r], Lemma 1 gives [h.sup.1]([I.sub.Z](t - 1)) [greater than or equal to] r + [h.sup.1]([I.sub.Z](t)) = r + 1. Let N be any hyperplane plane containing L. We have [h.sup.1](N, [I.sub.Z[intersection]N](t - 1)) = 1, because deg(Z [intsection] L) = t + 1 and deg(Z [intersection] N) [less than or equal to] 2(t - 1) + 1 (use the residual exact sequence (1) of a general hyperplane M of N containing L in N). Since deg([Res.sub.N](Z)) [less than or equal to] t + 1, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (part (a) applied. Hence the residual exact sequence (1) of N gives [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], a contradiction.

(d) Now assume R [intersection] L = 0. First assume r = 3. Let Q [subset] [P.sup.3] be a general quadric containg L [union] R. Since [I.sub.L[union]R] (2) is spanned and C is curvilinear, Q [intersection] Z = Z [intersection] (R [intersection] L) (as schemes). Hence [h.sup.1](Q, [I.sub.Z[intersection]Q,Q](t)) = 0. The residual exact sequence of Q gives [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since deg([Res.sub.Q](t)) [less than or equal to] t - 2, part (a) or [4, Lemma 34] gives a contradiction. Now assume r [greater than or equal to] 4. Since L [union] R spans a 3-dimensional linear space, the maximality property of M gives b [greater than or equal to] 2t + 2 + r - 4. Hence c - b [less than or equal to] t - 1 < deg (R [intersection] [Res.sub.H](Z)), a contradiction.

3 C non-degenerate

Le t[M'.sub.d,g] be the set of all C [member of] [M.sub.d,g] spanning [P.sup.5]. [M'.sub.d,g] is smooth and irreducible and dim([M'.sub.d,g]) = 6d + 2 - 2g. A general non-special curve C [subset] [P.sup.5] of genus g and degree d [greater than or equal to] g + 5 has maximal rank ([3]). We have d [less than or equal to] 16 and 7 x 16 +1 - g < ([sup.12.sub.5]). Hence [h.sup.1]([I.sub.C](7)) = 0 for a general C [member of] [M.sub.d,g], i.e. [h.sup.0]([I.sub.C](7)) = ([sup.12.sub.5]) - 7d - 1 + g. A dimensional count shows that no C [member of] [M'.sub.d,g] is contained in a general heptic hypersurface. Hence in this section we only need to exclude all C [member of] [M'.sub.d,g] with [h.sup.1]([I.sub.C](7)) > 0. By [13, Part (ii) of Theorem at page 492] we have d [greater than or equal to] 13.

Lemma 4. No C [member of] [M'.sub.g,d], d [greater than or equal to] 11, 1 [less than or equal to] g [less than or equal to] 3, is contained in a degree 4 surface.

Proof. Fix a degree 4 surface F [subset] [P.sup.5] containing C [member of] [M'.sub.d,g]. Since C is non-degenerate, F is non-degenerate. By the classification of minimal degree surfaces either F is a cone over a rational normal curve of [P.sup.4] or it is an embedding of [F.sub.0] by the complete linear system [absolute value of h + 2f] or it is an embedding of [F.sub.2] by the complete linear system [absolute value of h + 3f].

(a) Assume that F is an embedding of F by the complete linear system [absolute value of h + 2f]. Since 1 [less than or equal to] g [less than or equal to] 3, either C [member of] [absolute value of 2h + (g + 1)f] or C [member of] [absolute value of (g + 1)h + 2f]. In the first (second) case we get d = 5 + g [less than or equal to] 8 (resp. d = 2g + 4 [less than or equal to] 10), a contradiction.

(b) Assume that F is an embedding of [F.sub.2] by the complete linear system [absolute value of h + 3f] with C [member of] [absolute value of 2h + bf]. Since g > 0, we have a [greater than or equal to] 2 and b [greater than or equal to] 2a. We have d = a + b. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], the adjunction formula gives 2g - 2 = a(-2(a - 2) + (b - 4)) + b(a - 2). If a [greater than or equal to] 3 and hence b [greater than or equal to] 6 we get b [less than or equal to] 2g - 2, a contradiction. If a = 2 and hence b = d - 2, we get 2g - 2 = 2(d - 6), a contradiction.

(c) Assume that F is a cone over a rational normal curve of [P.sup.4]. Call o its vertex and let u: G [right arrow] F be the blowing up of o. G [congruent to] [F.sub.4] and, up to this isomorphism, u is induced by the complete linear system [absolute value of h + 4f]. Let C' [subset] [F.sub.4] be the strict transform of C. Write C [member of] [absolute value of ah + bf]. Since C is smooth, C is smooth and of genus g > 0. Hence a [greater than or equal to] 2 and b [greater than or equal to] 4a. We have d = b. Hence 2 [less than or equal to] a [less than or equal to] 4 with a = 4 only if d = 16. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], the adjunction formula gives 2g - 2 = a(-4(a - 2) + (b - 6)) + b(a - 2) [greater than or equal to] b(a - 2). Hence a = 2 and 2g - 2 = 2{d - 6), a contradiction.

Lemma 5. Fix C [member of] [M'.sub.d,g] and let H [subset] [P.sup.5] be a general hyperplane.

(a) We have [h.sup.1](H, [I.sub.C[intersection]H,H](3)) [less than or equal to] max{0, d - 13}.

(b) Assume d [greater than or equal to] 14. We have [h.sup.1](H, I[Z.sub.C[intersection]H,H](3)) = d - 13 if and only if C [intersection] H is contained in a rational normal curve of H. The latter case does not occur for a curve C contained in a general heptic hypersurface if [h.sup.1]([I.sub.C](7)) [less than or equal to] 3d - 18 - 6g.

Proof. The scheme Z := C [intersection] H is a set of d points in uniform position and spanning H. Fix S [subset or equal to] Z with #(S) = min{d, 13}. Since S is in linearly general position in H = [P.sup.4], we have [h.sup.1](H, [I.sub.S,H](3)) = 0 ([10, Theorem 3.2]). Now assume d [greater than or equal to] 14. If C [intersection] H is contained in a rational normal curve D of H, then [h.sup.0](H, [I.sub.Z,H](3)) [less than or equal to] [h.sup.0](H, [I.sub.D,H](3)) = ([sup.7.sub.3]) - 13 and hence [h.sup.1](H, [I.sub.Z,H](3)) [greater than or equal to] d - 13. By part (a) the last inequality is an equality. Now assume [h.sup.1](H, [I.sub.Z,H](3)) = d - 13. First assume [h.sup.0](H, [I.sub.Z,H](2)) [greater than or equal to] 6. Since Z is uniform position, we get [h.sup.0](H, [I.sub.Z,H](2)) = 6 and that Z is contained in a rational normal curve of H ([15, Lemma 3.9]). Now assume [h.sup.0](H, [I.sub.Z,H](2)) [less than or equal to] 5. Fix any A [subset] Z with #(A) = 10. Since [h.sup.0](H, [I.sub.Z,H](2)) [less than or equal to] 5, we have [h.sup.1](H, [I.sub.A,H](2)) = 0. Fix B [subset] Z\A with #(B) = 4. Since Z is in linearly general position, B spans a hyperplane N of H and Z [intersection] N = B. We have [h.sup.1](N, [I.sub.B,N](3)) = 0. The exact sequence

0 [right arrow] [I.sub.A,H](2) [right arrow] [I.sub.A[union]B,H](3) [right arrow] [I.sub.B,N](3) [right arrow] 0

gives [h.sup.1](H, [I.sub.A[union]B,H](3)) = 0 and hence [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] d - 14, a contradiction.

Now assume that C [intersection] H is contained in a rational normal curve [T.sub.H] of H. Set a := d + 1 - 2g. Fix Z [subset] [T.sub.H] such that #(Z) = a. By Lemma 2 the set of all C [member of] [M'.sub.d,g] containing Z has codimension at least 4a in [M'.sub.d,g]. Since [P.sup.5] has [[omega].sup.5] hyperplanes each hyperplane has 15 rational normal curves and each rational normal curves has [[infinity].sup.a] subsets with cardinality a, to rule out these cases it is sufficient to test all C [member of] [M'.sub.d,g] with [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3a - 20.

Remark 3. Fix C [member of] [M'.sub.d,g] and let H [subset] [P.sup.5] be a general hyperplane. The exact sequence (2) with T := C and M := H gives [h.sup.1]([I.sub.C](t - 1)) [greater than or equal to] [h.sup.1]([I.sub.C](t)) - [h.sup.1](H, [I.sub.C[intersection]H,H](t)). Now assume d [less than or equal to] 4t + 1. Since C [intersection] H is in uniform position in H and it spans H, it is in linearly general position. Hence [h.sup.1](H, [I.sub.C[intersection]H,H](t)) = 0 ([10, Theorem 3.2]).

Lemma 6. A general heptic hypersurface contains no C [member of] [M'.sub.d,g], g > 0, with either [h.sup.0]([I.sub.C](2)) [greater than or equal to] 5, [h.sup.0]([I.sub.C](3)) [not equal to] 25 and d = 15,16, or [h.sup.0] ([I.sub.C](2)) [greater than or equal to] 6 and 13 [less than or equal to] d [less than or equal to] 16.

Proof. Fix C [member of] [M'.sub.d,g] with [h.sup.0]([I.sub.C](2)) [greater than or equal to] 5. Let K be the set-theoretic base locus of [absolute value of [I.sub.C](2)]. Since C is irreducible, we have dim(K) [less than or equal to] 3. Fix an irreducible component A of K containing C and with maximal dimension. Note that [h.sup.0]([I.sub.A](2)) = [h.sup.0]([I.sub.C](2)) and that A = C if and only if dim(A) = 1.

(a) Assume dim(A) [greater than or equal to] 3. Since dim(K) [less than or equal to] 3, we have dim(A) = 3 and deg(A) [greater than or equal to] 3. Since [h.sup.0]([I.sub.C](2)) > 2, A is not the complete intersection of 2 quadrics and hence deg(A) = 3. Since A is a minimal degree 3-fold, their classification (linearly normal rational scrolls and cones over a rational normal curve of P3), gives [h.sup.0]([I.sub.A](2)) = 3, a contradiction.

(b) Assume dim(A) = 2. Let T be the intersection of 3 general elements of [absolute value of [I.sub.C](2)]. Since A is non-degenerate, we have 4 [less than or equal to] deg(A) [less than or equal to] 7 (even if T has a 3-dimensional component) by Bezout's theorem ([12, Theorem 2.2.5]). By Lemma 4 we may assume deg(A) [greater than or equal to] 5. The exact sequence

0 [right arrow] [I.sub.A](1) [right arrow] [I.sub.A](2) [right arrow] [I.sub.A[intersection]H,H](2) [right arrow] 0 (3)

gives [h.sup.0]([I.sub.C](2)) = [h.sup.0]([I.sub.A](2)) [less than or equal to] [h.sup.0](H, [I.sub.A[intersection]H,H](2)). The integral curve A [intersection] H spans H. Set q := [p.sub.a](A [intersection] H). We have q [less than or equal to] deg(A [intersection] C) - 4 for all deg(A) [member of] {5, 6, 7} by an elementary case of the Castelnuovo's bound for the arithmetic genus of a curve in [P.sup.r]. By [13, part (ii) of Theorem at page 492] if deg(A) [less than or equal to] 6, then [h.sup.1](H, [I.sub.A[intersection]H,H](2)) = 0, except the case deg(A) = 6, A [intersection] H a smooth rational curve having a quadrisecant line. Assume for the moment deg(A) [less than or equal to] 6 and that we are not in this exceptional case. We get [h.sup.0]([I.sub.A[intersection]H,H](2)) = 14 - 2 deg(A) + q [less than or equal to] 4, unless deg(A) = 5 and q = 1.

(b1) Assume deg(A) = 5 and q = 1. We may assume [h.sup.0]([I.sub.A](2)) = 5. So we do need to check this case if d = 13, 14 and hence here we assume d [member of] {15, 16}. In this case A [intersection] H is linearly normal and arithmetically Cohen-Macaulay with [h.sup.0](H, [I.sub.A[intersection]H,H](3)) = 35 - 15 = 20. Since [h.sup.0]([I.sub.A](2)) = [h.sup.0](H, [I.sub.A[intersection]H,H](2)), then A is linearly normal by (3). By (3) we get that A is arithmetically Cohen-Macaulay and in particular [h.sup.1]([I.sub.A](2)) = 0. From (3) we get [h.sup.0]([I.sub.A](3)) = [h.sup.0]([I.sub.A](2)) + [h.sup.0](H, [I.sub.A[intersection]H,H](3)) = 25. Since C [subset] A, we get [h.sup.0]([I.sub.C](3)) [greater than or equal to] 25. By assumption [h.sup.0]([I.sub.C](2)) [not equal to] 25 and so [h.sup.0]([I.sub.C](3)) [greater than or equal to] 26. Therefore there is a cubic hypersurface T [subset] [P.sup.5] with C [subset] T and A [subset or not equal to] T, so that C is contained in the locally Cohen-Macaulay curve A [intersection] T. Since deg(A [intersection] T) = 15, we get d [not equal to] 16 and C [not equal to] A [intersection] T. Since [[omega].sub.A[intersection]T] [congruent to] [O.sub.A[intersection]T](2), we get g > 3, a contradiction.

(b2) Now assume deg(A) = 6 and that D := A [intersection] H is a smooth rational curve spanning H and with a line L [subset] H with deg(L [intersection] D) = 4. We have L [subset] K, [p.sub.a](L [union] D) = 3 and L [union] D is a linearly normal curve of H = [P.sup.4]. Let N [subset] H be a general hyperplane. Since [h.sup.i](H, [I.sub.D[union]L](1)) = 0, i = 0, 1, a standard exact sequence gives [h.sup.0](H, [I.sub.D[union]L](2)) = [h.sup.0](N, [I.sub.N[intersection](D[union]L),N](1)) and hence [h.sup.0](H, [I.sub.D,H](2)) = [h.sup.0](H, [I.sub.D[union]L,H](2)) [less than or equal to] [h.sup.0](N, [I.sub.N[intersection]D,N](2)). Since N [intersection] D is in uniform position in N and it spans N, we get [h.sup.1](N, [I.sub.N[intersection]D,N](2)) = 0 and hence [h.sup.0](N, [I.sub.D[intersection]N,N](2)) = 4. Hence [h.sup.0]([I.sub.C](2)) [less than or equal to] [greater than or equal to] 4, a contradiction.

(b3) Now assume deg(A) = 7. By Bezout's theorem ([12, Theorem 2.2.5]) we have T [intersection] H = (A [intersection] H) [union] [L.sub.H] with [L.sub.H] a linear space of dimension [greater than or equal to] 1. If [L.sub.H] is a line, then the complete intersection T [intersection] H links A [intersection] H to a line and hence A [intersection] H is arithmetically Cohen-Macaulay. We get [h.sup.0]([I.sub.C](2)) = [h.sup.0]([I.sub.C[intersection]H](2)) = 1 + q [less than or equal to] 4. Now assume dim([L.sub.H]) > 1. Since T is a general intersection of 3 elements of [absolute value of [I.sub.C](2)] and H is general, we first get that K has a 3-dimensional component, B, which is a linear space, then we get T = A [union] B and then we get A [union] B = K. Hence [h.sup.0]([I.sub.C](2)) = 3, a contradiction.

(c) Assume dim(A) = 1, i.e. A = C, and d = 15, 16. Let T be the intersection of 3 general elements of [absolute value of [I.sub.C](2)]. Let B be an irreducible component of [T.sub.red] containing C. Since we may apply steps (a) and (b) to every irreducible component of K containing C, we have B [subset or not equal to] K and hence there is a quadric containing C, but not containing B. By Bezout we have deg(B) [less than or equal to] 8 with equality if and only if dim(T) = 2 and B = T. Intersecting B with a general element of [absolute value of [I.sub.C](2)] we get a locally Cohen-Macaulay scheme E of pure dimension 1 with deg(E) [less than or equal to] 16 and E [contains or equal to] C. We exclude the case d = 16, because C has not the genus, 17, of an intersection of 4 quadrics. Now assume d = 15. Since B = T, T has dimension 2 and hence it is a complete intersection. Hence the complete intersection E links C to a line. Therefore C is arithmetically Cohen-Macaulay and in particular [h.sup.1]([I.sub.C](2)) = 0, contradicting the inequality 2d + 1 - g > 21.

(d) Assume d = 13, 14. Take K, A as in the previous steps. Fix a general hyperplane H [subset] [P.sup.5]. Since C is non-degenerate, we have [h.sup.0]([I.sub.C](1)) = 0. By (3) with C instead of A we have [h.sup.0](C (2)) [less than or equal to] [h.sup.0](H, [I.sub.C[intersection]H,H](2)). Assume [h.sup.0](H, [I.sub.C[intersection]H,H](2)) > 6. By [15, Lemma 3.9] C [intersection] H is contained in a rational normal curve D [subset] H. Since d > 8, we have D [subset] K. By step (a) we get the existence of A [subset or equal to] K with C [subset] A and A a degree 4 surface, contradicting Lemma 4.

Lemma 7. Fix integers d, g such that 6 [less than or equal to] g + 5 [less than or equal to] d < 15. Let r be the set of all C [member of] [M'.sub.d,g] contained in the smooth locus a quadric. For any v [member of] {0, 1, 2} and integer x > 0 let [[GAMMA].sub.v,x] be the set of all C [member of] [M'.sub.d,g] contained in a quadric with singular locus V of dimension v and x = deg(V [intersection] C). Then dim([GAMMA]) [less than or equal to] 4d + 21 + g, [[GAMMA].sub.0,x] = [empty set] for all x [greater than or equal to] 2, dim([[GAMMA].sub.0,1]) [less than or equal to] 4d + 22 + g, dim([[GAMMA].sub.1,x]) [less than or equal to] 4d + g + x + 18, and dim([[GAMMA].sub.2,x]) [less than or equal to] 4d + 14 + g.

Proof. Fix C [member of] [M'.sub.d,g] with [h.sup.0]([I.sub.C](2)) > 0 and let Q [subset] [P.sup.5] be any quadric containing C. Since C is non-degenerate, Q is irreducible. The Hilbert scheme Hilb(Q) of Q has [H.sup.0]([N.sub.C,Q]) as its tangent space at [C]. Let V be the singular locus of Q. Let [tau] be the tangent sheaf of Q. Since the algebraic group Aut(Q) acts transitively on Q\V and [H.sup.0]([tau]) is the tangent space to Aut(Q) at the identity, [H.sup.0]([tau]) spans t at each point of Q\V.

(i) First assume that either Q is smooth or C [intersection] V = [empty set]. Since dim [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], to handle these curves C it is sufficient to prove that the set of all C [member of] [M.sub.d,g] contained in Q and with V [intersection] Q = 0 has dimension [less than or equal to] 4d + 1 + g. Since C [intersection] V = [empty set], the normal sheaf [N.sub.C,Q] is a rank 3 vector bundle on C with degree 4d + 2g - 2. Since [N.sub.C,Q] is a quotient of [[tau].sub.|C], it is spanned. Take 2 general sections of [H.sup.0]([N.sub.C,Q]). These sections induces a rank 2 subsheaf G' of [N.sub.C,Q] isomorphic to [O.sup.[cross product]2.sub.c]. Let G be the saturation of G' in [N.sub.C,Q], i.e. the only rank 2 subsheaf of [N.sub.C,Q] containing G' and with [N.sub.C,Q]/G a line bundle. Since [h.sup.1](G') = 2g, we have [h.sup.1](G) [less than or equal to] 2g. First assume deg(G) [less than or equal to] 3d - 1, i.e. deg([N.sub.C,Q]/G) > 2g - 2. We get [h.sup.1]([N.sub.C,Q]/G) = 0 and so [h.sup.1]([N.sub.C,Q]) [less than or equal to] 2g. Riemann Roch gives [h.sup.0]([N.sub.C,Q]) [less than or equal to] 4d + g + 1. Now assume deg(G) [greater than or equal to] 3d. Let [G.sub.1] be the saturation of a general section of G. So G is the extension of two line bundles, G/[G.sub.1] and [G.sub.1], with deg(G/[G.sub.1]) + deg([G.sub.1]) = 3d [greater than or equal to] 2(2g - 1) and both with a non-zero section. We get that at least one of the line bundles G/[G.sub.1] and [G.sub.1] is non-special and the other one, [Laplace], has [h.sup.1]([Laplace]) [less than or equal to] g . Hence [h.sup.1](G) [less than or equal to] g. Since the line bundle [N.sub.C,Q]/G has a non-zero section, we have [h.sup.1]([N.sub.C,Q]/G) [less than or equal to] g and hence [h.sup.1]([N.sub.C,Q]) [less than or equal to] 2g even in this case.

(ii) Now assume that C [intersection] V [not equal to] [empty set]. The set of all quadrics of rank 5 (resp. 4, resp. 3) has dimension 19 (resp. 17, resp. 14). Since C is not a plane curve and dim(V) [less than or equal to] 2, the scheme C [intersection] V is finite. Let u: [??] [right arrow] Q be the blowing up of V, E := [u.sup.-1](V) the exceptional divisor, and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] the strict transform of C. Since C is smooth, u maps isomorphically [??] onto C. Let [[PSI].sub.v,x] be closure in Hilb([??]) of the strict transforms of all C [subset] Q with deg(C [intersection] V) = x. Take a general D [member of] [[PSI].sub.v,x]. Since Aut([??]) acts transitively of [??]\E, step (a) of the proof gives [h.sup.1]([N.sub.D,[??]]) [less than or equal to] 2g. Hence it is sufficient to give a winning upper bound for deg([N.sub.D,[??]]), i.e. a winning lower bound for [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. The group Pic([??]) is freely generated by E and the pull-back H of [O.sub.Q](1). We have D x H = d and D x E = x. We claim that [[omega].sub.[??]] [congruent to] [O.sub.[??]](-3H + cE) with c = -2 + v. First assume v > 0. Let M [subset] Q be a general hyperplane. M [intersection] Q is a 3-dimensional quadric with vertex of dimension v - 1. We apply [18], Example 8.5 (2), if v = 1 and [18], Example 8.5 (3) if v = 2. Now assume v = 0. To apply the previous formulas we need to take a hyperplane M [contains] V whose pull-back has E as a component.

Lemma 8. A general heptic hypersurface contains no C [member of] [M'.sub.d,g], 13 [less than or equal to] d [less than or equal to] 16, 1 [less than or equal to] g [less than or equal to] 3, with no line R [subset] [P.sup.5] with deg(R [intersection] C) [greater than or equal to] 7 and no conic D [subset] [P.sup.5] with deg(C [intersection] D) [greater than or equal to] 12.

Proof. Recall that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and that we are assuming [h.sup.1]([I.sub.C](7)) > 0. By the cases r = 4, c [less than or equal to] 3t + 1 and t = 5, 6, 7 of Lemma 3 we have [h.sup.1](M, [I.sub.C[intersection]M,M](t)) = 0 for all hyperplanes M [subset] [P.sup.5]. Lemma 1 gives [h.sup.1]([I.sub.C](4)) [greater than or equal to] [h.sup.1]([I.sub.C](5)) + 5 [greater than or equal to] [h.sup.1]([I.sub.C](6)) + 10 [greater than or equal to] [h.sup.1]([I.sub.C](7)) + 15. Fix a general hyperplane H [subset] [P.sup.5]. Lemma 5 and Remark 3 give [h.sup.1]([I.sub.C](3)) [greater than or equal to] [h.sup.1]([I.sub.C](4)) and that either [h.sup.1]([I.sub.C](2)) [greater than or equal to] [h.sup.1]([I.sub.C](3)) - max{0, d - 14} or d [greater than or equal to] 14, [h.sup.1]([I.sub.C](2)) [greater than or equal to] [h.sup.1]([I.sub.C](3)) + 13 - d and [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3d - 18 - g. First assume d [greater than or equal to] 14, [h.sup.1]([I.sub.C](2)) [greater than or equal to] [h.sup.1]([I.sub.C](3)) + 13 - d and [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3d - 18 - g. We first get [h.sup.1]([I.sub.C](3)) [greater than or equal to] [h.sup.1] ([I.sub.C](4)) [greater than or equal to] 3d - 3 - g and then [h.sup.1]([I.sub.C](2)) [greater than or equal to] 2d + 10 - g, i.e. [h.sup.0]([I.sub.C](2)) [greater than or equal to] 31, contradicting Lemma 6.

Now assume [h.sup.1]([I.sub.C](2)) [greater than or equal to] [h.sup.1](C(3)) - max{0, d - 14} and hence [h.sup.1]([I.sub.C](2)) [greater than or equal to] 16 - max{0, d - 14}, i.e. [h.sup.0]([I.sub.C](2)) [greater than or equal to] 36 + g - 2d - max{0, d - 14}. We get [h.sup.0]([I.sub.C](2)) [greater than or equal to] 21 + 16 - 1 - 26 + g = 11 + g [greater than or equal to] 7 if d = 13 and [h.sup.0]([I.sub.C](2) [greater than or equal to] 21 + 16 - d + 14 - 2d - 1 + g = 50 - 3d + g if 14 [less than or equal to] d [less than or equal to] 16. If either 13 [less than or equal to] d [less than or equal to] 15 or d = 16 and g = 3 we use Lemma 6, except that if d = 16 and g = 3 we also need to exclude that [h.sup.0]([I.sub.C](3)) = 25. Assume d = 16 and g = 3. Since [h.sup.1]([I.sub.C](3)) [greater than or equal to] 16 and ([sup.8.sub.3]) = 56, we have [h.sup.0]([I.sub.C](3)) [greater than or equal to] 56 + 16 - 1 - 48 + 3 = 26.

Now assume d = 16 and g = 1,2. We proved that [h.sup.0]([I.sub.C](2)) [greater than or equal to] 2 + g. To repeat the same proof it would be sufficient to have [h.sup.1]([I.sub.C](7)) [greater than or equal to] 4. If C is contained in the smooth locus of at least one quadric, then by Lemma 7 we may assume [h.sup.1]([I.sub.C](7)) [greater than or equal to] 6d + 2 - 2g - (4d - 21 + g) = 2d - 19 - 3g [greater than or equal to] 17. Now assume that a general quadric hypersurface containing c has singular locus V of dimension v and with x := deg(C [intersection] V) > 0. Since x [less than or equal to] 6 if v = 1, it is sufficient to use Lemma 7.

Notation 1. For each integer a > 0 let [A'.sub.a] denote the set of all C [member of] [M'.sub.d,g] such that there is a line L [subset] [P.sup.5] with deg(L [intersection] C) = a and [A".sub.a] := [U.sub.b[greater than or equal to]a][A'.sub.a]. For each integer a > 0 let [B'.sub.a] denote the set of all C [member of] [M'.sub.d,g] such that there is a conic D [subset] [P.sup.5] with deg(L [intersection] D) = a and [B".sub.a] := [U.sub.b[greater than or equal to]a][B'.sub.a].

Lemma 9. A general W [member of] W contains no C [member of] [M'.sub.d,g], 13 [less than or equal to] d [less than or equal to] 16, 1 [less than or equal to] g [less than or equal to] 3, with [A".sub.7] [not equal to] [empty set].

Proof. (a) In this step we prove that W contains no element of [A".sub.9]. Fix C [member of] [A".sub.9] and take a line L [subset] [P.sup.5] such that deg(L [intersection] C) [greater than or equal to] 9. Set b := 8 if (d, g) = (13, 3) and b := 9 if (d, g) [not equal to] (13,3). Fix Z [subset or equal to] C [intersection] L such that deg(Z) = 9. Set Z := {A [member of] [M.sub.d,g]: A [contains] Z}. Lemma 2 gives that dim([E.sub.Z]) [less than or equal to] 6d + 2 - 2g - 4b. Since L has [[infinity].sup.b] degree b subschemes and [P.sup.5] has [[infinity].sup.8] lines, to prove that no element of [A".sub.9] is contained in W it is sufficient to test the ones, C, with the additional condition that either [h.sup.1]([I.sub.C](7)) [greater than or equal to] 19 (case (d, g) [not equal to] (13, 3)) or [h.sup.1]([I.sub.C](7)) [greater than or equal to] 16 (case (d, g) = (13, 3)). By the cases t = 3, 4, 5, 6, 7 of Remark 3 we get [h.sup.1]([I.sub.C](3)) [greater than or equal to] [h.sup.1]([I.sub.C](7)). Therefore [h.sup.1]([I.sub.C](2)) [greater than or equal to] 17 if (d, g) [not equal to] (13, 3) and [h.sup.1]([I.sub.C](2)) [greater than or equal to] 16 if (d, g) = (13,3) (Lemma 5). We get [h.sup.0]([I.sub.C](2)) [greater than or equal to] 37 + g - 2d [greater than or equal to] 6 if (d, g) [not equal to] (13, 3) and [h.sup.0]([I.sub.C](3)) [greater than or equal to] 13 if (d, g) = (13, 3). Apply Lemma 6.

(b) In this step we prove that a general heptic contains no element of [A'.sub.8]. By part (a) it is sufficient to exclude all elements of [A'.sub.8]\[A'.sub.9],. By Lemma 2 it is sufficient to exclude all C [member of] [A'.sub.8]\[A'.sub.9] with [h.sup.1]([I.sub.C](7)) [greater than or equal to] 16. Fix any such C. By Lemma 1 and the case r = 4, t = 7 and c = d [less than or equal to] 3t + 1 of Lemma 3 we have [h.sup.1]([I.sub.C](6)) [greater than or equal to] 21. By the cases t = 3, 4, 5, 6 of Remark 3 we get [h.sup.1]([I.sub.C](3)) [greater than or equal to] 21. Lemma 3 gives [h.sup.1]([I.sub.C](2)) [greater than or equal to] 19 and hence [h.sup.0]([I.sub.C](2)) [greater than or equal to] 39 - 2d + g [greater than or equal to] 8. Use Lemma 6.

(c) Fix C [member of] [A'.sub.7]\[A'.sub.9]. By Lemma 2 it is sufficient to exclude the curves C with [h.sup.1]([I.sub.C](7)) [greater than or equal to] 13. Since C [not member of] [A".sub.9], we have [h.sup.1]([I.sub.C](6)) [greater than or equal to] 18 (Lemma 1). The case t = 4, 5, 6 of Remark 3 gives [h.sup.1]([I.sub.C](3)) [greater than or equal to] 18. Lemma 5 gives [h.sup.1]([I.sub.C](2)) [greater than or equal to] 16 and hence [h.sup.0]([I.sub.C](2)) [greater than or equal to] 36 + g - 2d. Apply Lemma 6, except that if d = 16 and g = 1 we also need to prove that [h.sup.0]([I.sub.C](3)) = 25. Since [h.sup.1]([I.sub.C](3)) > 18, we have [h.sup.0]([I.sub.C](3)) [greater than or equal to] 56 + 18 - 48 = 26.

Lemma 10. A general W [member of] W contains no C [member of] [M'.sub.d,g] such that either there is a conic D [subset] [P.sup.5] with deg(D [intersection] C) [greater than or equal to] 12 or there is a line L [subset] [P.sup.5] with deg(L [intersection] C) [greater than or equal to] 6.

Proof. By Lemma 9 it is sufficient to exclude all C [member of] [B'.sub.12]\[A".sub.7] and all C [member of] [A'.sub.6]\[A'.sub.7].

(a) We first exclude all C [member of] [B".sub.14]\[A.sub.7]. In this case we have d = 16 and C [member of] [B'.sub.14]. Take a conic D [member of] [P.sup.5] such that deg(D [intersection] C) = 14. Since C [not member of] [A".sub.7], D is a smooth conic. Set x(1) := 14, x(2) := 13 and x(3) := 11. Fix any Z [subset] D such that deg(Z) = x(g). Since x(g) + 2g - 1 [less than or equal to] 16, the set [E.sub.Z] of all non-degenerate A [member of] [M.sub.16,g] with Z [subset] A has dimension 6d + 2 - 2g - 4x(g) (Lemma 2). Since D is a smooth curve, it has [[infinity].sup.x(g)] degree x(g) subschemes. [P.sup.5] has [[infinity].sup.9] planes and each plane has [[infinity].sup.5] conics. Hence it is sufficient to exclude all non-degenerate C [member of] [B'.sub.14]\[A.sub.7] with [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3x(g) - 14. By Lemma 1 and the case r = 4, t = 7, c = d [less than or equal to] 3t + 1 of Lemma 3 (note that deg(C [intersection] D') < 16 for each conic D' and that [A".sub.9] = 0) we have [h.sup.1]([I.sub.C](6)) [greater than or equal to] 3x(g) - 9 [greater than or equal to] 22. Then we continue as in the last two lines of step (a) of the proof of Lemma 9.

(b) Now we exclude all C [member of] [A'.sub.6]\[A".sub.7]. Since 13 [greater than or equal to] 6 + 2g - 1, by Lemma 2 we may assume [h.sup.1]([I.sub.C](7)) [greater than or equal to] 10. By step (a) we may assume that C [not member of] [B".sub.12]. Lemma 1 and the cases t = 6, 7, r = 4, of Lemma 3 give [h.sup.1]([I.sub.C](5)) [greater than or equal to] 5 + [h.sup.1]([I.sub.C](6)) [greater than or equal to] 10 + [h.sup.1]([I.sub.C](7)) [greater than or equal to] 20. Then we continue as in the proof of Lemma 9.

(c) Now we exclude all C [member of] [B'.sub.12] [union] [B'.sub.13]\[A.sub.7]. By step (b) we may assume C [not member of] [A'.sub.6]. Take a conic D such that deg(D [intersection] C) [member of] {12, 13}. Note that if [B".sub.12] = 0, then d [not equal to] 13. Since C [not member of] [A".sub.6], the conic D is smooth. Set [y.sub.d,g] := min{12, d + 1 - 2g}. Fix any zero-dimensional scheme Z [subset] D with deg(Z) = [y.sub.d,g]. By Lemma 2 the set of all non-degenerate C containing Z has dimension [less than or equal to] 6d + 2 - 2g - 4[y.sub.d,g]. Every smooth conic has [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] zero-dimensional schemes of degree [y.sub.d,g]. [P.sup.5] has [[infinity].sup.9] planes and each plane has [[infinity].sup.5] conics. Hence it is sufficient to check the curves C with the additional condition [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3[y.sub.d,g] - 14. Since 13 [less than or equal to] d [less than or equal to] 16 and g [less than or equal to] 3, we have [y.sub.d,g] [greater than or equal to] 8 and so 3[y.sub.d,g] - 14 > 10. We conclude as in step (b).

By Lemmas 8 and 10 we proved the part of Theorem 1 concerning non-degenerate C [member of] [M.sub.d,g].

4 Curves spanning a hyperplane

In this section we consider curves C [member of] [M.sub.d,g] spanning a hyperplane. For any hyperplane M [subset] [P.sup.5] let [M'.sub.d,g](M) be the set of all curves C [member of] [M.sub.d,g] spanning M. [M'.sub.d,g] (M) is a smooth and irreducible variety of dimension 5d + 1 - g. A general element of [M'.sub.d,g](M) has maximal rank ([1]). Since ([sup.11].sub.4]) > 7d + 1 - g, we have [h.sup.1](M, [I.sub.C,M](7)) = 0 for a general C [member of] [M'.sub.d,g](M). Since [P.sup.5] has [[infinity].sup.5] hyperplanes to prove that a general W [member of] W contains no curve in [M.sub.d,g] spanning a hyperplane it is sufficient to fix a hyperplane M [subset] [P.sup.5] and exclude all C [member of] [M'.sub.d,g] (M) such that [h.sup.1](M, [I.sub.C,M](7)) [greater than or equal to] d - 4 - g.

Since g > 0, by [13, part (ii) of Theorem on page 492] we have [h.sup.1](M, [I.sub.C,M](7)) = 0 if d [less than or equal to] 11. Hence in this section we assume 12 [less than or equal to] d [less than or equal to] 16 and we fix the hyperplane M [subset] [P.sup.5].

Remark 4. For any C [member of] [M'.sub.d,g](M) let [alpha](C) (or just [alpha]) denote the minimal integer t such that [h.sup.0](M, [I.sub.C,M](t)) > 0. Since C spans M, we have [alpha] [greater than or equal to] 2. Since d [less than or equal to] 16, we have 4d + 1 - g < 70 = ([sup.8.sub.4]) and so [alpha] [less than or equal to] 4. Since ([sup.6.sub.2]) = 15 and ([sup.7.sub.3]) = 35, we have [alpha] = 2 if and only if [h.sup.1](M, [I.sub.C,M](2)) [greater than or equal to] 2d - g - 13 and [alpha] [less than or equal to] 3 if and only if [h.sup.1](M, [I.sub.C,M](3)) [greater than or equal to] 3d - g - 33.

Lemma 11. Fix C [member of] [M'.sub.d,g](M), d [less than or equal to] 16, g > 0. There is no plane N [subset] M with deg(N [intersection] C) [greater than or equal to] 15.

Proof. Assume the existence of a plane N [subset] M such that deg(N [intersection] C) [greater than or equal to] 15. Fix a hyperplane H [subset] M with H [contains] N. Since the scheme C [intersection] H spans H, we get d = 16 and deg(C [intersection] N) = 15. Since C is smooth, for a general H [contains] N, H contains a tangent line of C if and only if this tangent line is contained in N. Hence for a general H we have C [intersection] H = (C [intersection] N) [union] {[p.sub.H]} with [p.sub.H] [member of] C\C [intersection] N. The pencil of all hyperplanes H [contains] N shows that C is rational, a contradiction.

Lemma 12. A general W [member of] W contains no C [member of] [M.sub.d,g] such that C [member of] [M'.sub.d,g] (M) for some hyperplane M and C is contained in a degree 3 surface A of M.

Proof. Since C spans M, then A spans M. The classification of minimal degree surfaces gives that either A is the Hirzebruch surface F1 embedded by the complete linear system [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] or it is a cone over a rational normal curve of [P.sup.3].

(a) Assume that A is the Hirzebruch surface [F.sub.1] embedded by the complete linear system [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and take a, b such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since C has genus g > 0, we have b [greater than or equal to] a [greater than or equal to] 2. We have d = (ah + bf) x (h + 2f) = a + b. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], the adjunction formula gives 2g - 2 = (ah + bf). ((a - 2)h + (b - 3)f) = (b - a)(a - 2) + a(b - 3). If a = 2 and hence b = d - 2 [greater than or equal to] 10, we get 2g - 2 [greater than or equal to] 7, a contradiction. If a [greater than or equal to] 3, we have 2g - 2 > 4, a contradiction.

(b) Assume that A is a cone with vertex o over a rational normal curve of [P.sup.3]. Let u: [F.sub.3] [right arrow] A be a minimal desingularization of A. [F.sub.3] is the Hirzebruch surface with the same name with h = [u.sup.-1](o) and u induced by the complete linear system [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Let C [subset] [F.sub.3] be the strict transform of C. Take a, b such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] with b [greater than or equal to] 3a. We have d = b. We have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] Since C is smooth, u induces an isomorphism of C; onto C. The adjunction formula gives 2g - 2 = (ah + bf) x ((a - 2)h + (d - 5)f) = (a - 2)(d - 3a) + a(d - 5). Since g > 0, we have a [greater than or equal to] 2 and hence 2 [less than or equal to] a [less than or equal to] [d/3]. In all cases we get 2g - 2 > 4, a contradiction.

Lemma 13. Fix C [member of] [M'.sub.d,g] (M) and let H [subset] M be a general hyperplane. Fix an integer t [greater than or equal to] 1.

(a) If 3t + 1 [greater than or equal to] d, then [h.sup.1](H, [I.sub.C[intersection]H,H](t)) = 0.

(b) Assume d [greater than or equal to] 3t + 2. Then [h.sup.1]([I.sub.C[intersection]H,H](t)) [less than or equal to] d - 3t - 1 and equality holds only if C [intersection] H is contained in a rational normal curve of H.

(c) If C is contained in a general heptic hypersurface and C [intersection] H is contained in a rational normal curve of M, then [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3d - 23 (case g = 1) or [h.sup.1]([I.sub.C](7)) [greater than or equal to] 3d - 16 - 6g (case g = 2, 3).

Proof. The scheme C [intersection] H is a set of d points of H spanning H = [P.sup.3] and in uniform position. Part (a) follows from [10, Theorem 3.2] and part (b) is proved as in Lemma 8.

Now we prove part (c). Let T [subset] H be a rational normal curve of H containing C [intersection] H.

(i) First assume g = 1 and set Z := C [intersection] H. Since T has rod subsets of cardinality d, [P.sup.5] has [[infinity].sup.8] 3-dimensional linear spaces, each 3-dimensional H [subset] [P.sup.5] is contained in [[infinity].sup.1] hyperplanes of [P.sup.5] and each 3-dimensional linear space has [[infinity].sup.12] rational normal curves we get part (c) for g = 1 if we prove that [h.sup.1]([N.sub.C,M](-Z)) = 0, i.e. [h.sup.1]([N.sub.C,M](-1)) = 0. Since C is a curve, [h.sup.2](F) = 0 for all coherent sheaves F on C. Hence it is sufficient to prove that [h.sup.1](C, [TM.sub.|C] (-1)) = 0. Assume [h.sup.1](C, [TM.sub.|C](-1)) > 0. By duality there is a non-zero map [TM.sub.|C] [right arrow] [O.sub.C](1). Fix homogeneous coordinates [z.sub.0], [z.sub.1], [z.sub.2], [z.sub.3], [z.sub.4] of M. The Euler's sequence gives a nonzero map [O.sub.C][(1).sup.[cross product]5] [right arrow] [O.sub.C](1), i.e. ([a.sub.0], [a.sub.1], [a.sub.2], [a.sub.5]) [member of] [C.sup.5]\{0}. Hence C is contained in the hyperplane {[[summation].sub.i][a.sub.i][z.sub.i] = 0}, a contradiction.

(ii) Now assume g [greater than or equal to] 2. Set b := d + 2 - 2g and take Z [subset] C [intersection] H with deg(Z) = b. Since T has [[infinity].sup.b] subsets of cardinality b, [P.sup.5] has [[infinity].sup.8] 3-dimensional linear spaces, each 3-dimensional H [subset] P[.bar]5 is contained in [[infinity].sup.1] hyperplanes of [P.sup.5] and each 3-dimensional linear space has [[infinity].sup.12] rational normal curves, we get part (c) for g = 2, 3 if we prove that [h.sup.1]([N.sub.C,M](-Z)) = 0. Assume [h.sup.1]([N.sub.C,M](-Z)) > 0. Since [N.sub.C,M] is a quotient of [O.sub.C][(1).sup.[cross product]5], we get [h.sup.1]([O.sub.C](1)(-Z)) > 0. Since deg([O.sub.C](1)(-Z)) = 2g - 2, we get [O.sub.C](1)(-Z) [congruent to] [[omega].sub.Z]. Since b < d, there is Z' [subset] C [intersection] H with #(Z [intersection] Z') = b - 1. Set {p} := Z\Z [intersection] Z'. By monodromy the value of [h.sup.1]([O.sub.C](1)(-A)) is the same for all A [subset] C [intersection] H with cardinality b. Hence [O.sub.C](1)(-Z') [congruent to] [O.sub.C](1)(-Z), i.e. p and q are linearly equivalent, contradicting the assumption g > 0.

Lemma 14. A general W [member of] W contains no C [member of] [M.sub.d,g] such that C [member of] [M'.sub.d,g](M) for some hyperplane M and [h.sup.0](M, [I.sub.C,M](2)) [greater than or equal to] 3.

Proof. Fix C [member of] [M'.sub.d,g](M) such that [h.sup.0](M, [I.sub.C,M](2)) [greater than or equal to] 3. Let K [subset] M be the settheoretic base locus of [absolute value of [I.sub.C,M](2)]. Since C spans M, every quadric hypersurface of M containing C is irreducible. Hence dim(K) [less than or equal to] 2. Let A [subset or equal to] K be any irreducible component of K containing C. First assume dim(A) = 2. Since C spans M, then A spans M and hence deg(A) [greater than or equal to] 3. Fix 2 general [Q.sub.1], [Q.sub.2] elements of [absolute value of [I.sub.C,M](2)]. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], Bezout gives A [??] [Q.sub.1] [intersection] [Q.sub.2] and hence deg(A) = 3. Use Lemma 12. Now assume dim(A) = 1, i.e. A = C. Fix general [Q.sub.1], [Q.sub.2], [Q.sub.3] and set T := [Q.sub.1] [intersection] [Q.sub.2] [intersection] [Q.sub.3]. Since d > 8, and C is an irreducible component of T, the contradiction comes from Bezout ([12, Theorem 2.2.5]).

Lemma 15. Fix C [member of] [M'.sub.d,g](M) and let H [subset] M be a general hyperplane. Set Z := C [intersection] H.

(i) If [h.sup.0](H, [I.sub.Z,H](2)) = 0, then we have [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] max{0, d - 13} and [h.sup.1](H, [I.sub.Z,H](4)) = 0.

(ii) If [h.sup.0](H, [I.sub.Z,H](2)) = 1, then we have [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] max{0,d - 12}, [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] 2 if d [member of] {15, 16} and [h.sup.1](H, [I.sub.Z,H](4)) = 0.

(iii) If [h.sup.0](H, [I.sub.Z,H](2)) = 2, then [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] d - 11, [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] d - 12 for all d [greater than or equal to] 13, [h.sup.1](H, [I.sub.Z,H](4)) [less than or equal to] max{0, d - 15} and Z is contained in a complete intersection T [subset] H of 2 quadrics; if d = 16 and [h.sup.1](H, [I.sub.Z,H](4)) > 0, then C is not contained in a general heptic hypersurface; if d = 12 and [h.sup.1](H, [I.sub.Z,H](3)) > 0, then [h.sup.0]( M, [I.sub.C,M](2)) > 0.

(iv) To rule out from a general heptic hypersurface all C [member of] [M.sub.d,g] such that there is a hyperplane M [subset] [P.sup.5] with C [member of] [M'.sub.d,g](M) and [h.sup.0](H, [I.sub.C[intersection]H,H](2)) = 2 for a general hyperplane H of M it is sufficient to rule out all C [member of] [M'.sub.d,g](M) with [h.sup.1](M, [I.sub.C,M](7)) [greater than or equal to] 3d - 6g - 21.

Proof. The set Z is in uniform position in H and it spans H. First assume [h.sup.0](H, [I.sub.Z,H](2)) = 0. Fix any S [subset] C [intersection] H with #(S) = 9. S is contained in a unique quadric surface QS and QS [intersection] C [intersection] H = S. Fix p [member of] C [intersection] H and any S' [subset] Z\(S [union] {p}) with #(S') [less than or equal to] 3. Since Z is in uniform position, there is a plane N [subset] H with N [intersection] Z = S'. The cubic [Q.sub.S] [union] N shows that [h.sup.1](H, [I.sub.SUS'[union]{p}](3)) = [h.sup.1](H, [I.sub.SUS'](3)). Hence [h.sup.1] (H, [I.sub.Z,H](3)) [less than or equal to] max{0, d - 13}. Since d - 10 [less than or equal to] 9, we have [h.sup.0](H, [I.sub.Z\(S[union]{p})](2)) > 0. Take a general Q' [member of] [absolute value of [I.sub.Z\(S[union]{p}),H](2)]. Since Z is in uniform position and #(Z\(S [union] {p})) [less than or equal to] 8, we have Q' [intersection] Z = Z\(S [union] {p}). Hence (Q [union] Q') [intersection] Z = Z\{p}. Hence [h.sup.1](H, [I.sub.Z\{p},H](4)) = [h.sup.1](H, [I.sub.Z,H](4)). Taking smaller subsets of Z\(S [union] {p}) we get in finitely steps that [h.sup.1](H, [I.sub.Z,H](4)) = 0.

Now assume [h.sup.0](H, [I.sub.Z,H](2)) = 1. Take [S.sub.1] [subset] Z with #([S.sub.1]) = 8 and let [Q.sub.1] [subset] H be a general element of [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since Z is in uniform position, we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Taking planes we get [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] max{0, d - 12}. Since d [less than or equal to] 17, any [S.sub.2] [subset] Z\[S.sub.1] with #([S.sub.2]) [less than or equal to] 8 has [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for a general [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], we get [h.sup.1](H, [I.sub.Z,H](4)) = 0. Now assume d [member of] {15, 16}. Let Q [subset] H be the quadric surface containing Z. Since Z is in uniform position, Q is irreducible. Since [h.sup.0](Q, [O.sub.Q](3)) = 16, it is sufficient to prove that [h.sup.0](Q, [I.sub.Z,Q](2)) [less than or equal to] 18 - d. In particular we may assume [h.sup.0](Q, [I.sub.Z,Q](3)) > 0. Fix a general D [member of] [absolute value of [I.sub.Z,Q](3)]. Since Z is in uniform position in Q (or by monodromy), D is an integral curve and hence it is a canonically embedded integral curve, which is the complete intersection of Q and a cubic surface. Riemann-Roch gives [h.sup.0](D, [O.sub.D](3)) = 15. Since D is arithmetically Cohen-Macaulay, to prove that [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] 2 it is sufficient to prove that [h.sup.1](D, [I.sub.Z,D](3)) [less than or equal to] 2. Take any zero-dimensional scheme Z' [subset] D such that deg(Z') = 16 and Z' [contains or equal to] Z. It is sufficient to prove that [h.sup.1](D, [I.sub.Z',D](3)) [less than or equal to] 2. This is true (by duality), because the very ampleness of [[omega].sub.D] [congruent to] [O.sub.D](1) implies [h.sup.0](D, [Laplace]) [less than or equal to] 1 for every rank 1 torsion free sheaf [Laplace] on D with deg([Laplace]) = 2.

Now assume [h.sup.0]([I.sub.Z,H](2)) = 2. Since Z is in uniform position, for all [S.sub.3] [subset] Z with #([S.sub.3]) [less than or equal to] 7 we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for a general [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] As above we get [h.sup.1](H, [I.sub.Z,H](4)) [less than or equal to] max{0, d - 15} and [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] d - 11. Let T be the intersection of all elements of [absolute value of [I.sub.Z,H](2)]. Since Z is in uniform position, all elements of [absolute value of [I.sub.Z,H](2)] are irreducible. Hence dim(T) = 1. Assume for the moment that T is either not reduced or reducible. Since Z is in uniform position and it is a set, there is an irreducible curve [T.sub.1] [subset] T with deg([T.sub.1]) [less than or equal to] 3 and Z [subset] [T.sub.1]. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] we get a contradiction. Hence T is an integral degree 4 curve with [p.sub.a](T) = 1. By monodromy for a general H we have Z [subset] [T.sub.reg]. Set b := d + 1 - 2g. Since [T.sub.reg] has [[infinity].sup.b] subsets with cardinality b, [P.sup.5] has [[infinity].sup.8] 3-dimensional linear spaces and each H [member of] G(3, 5) has [[infinity].sup.16] complete intersections of 2 quadrics, to rule out all C [member of] [M.sub.d,g] for which there is M and H with C [member of] [M'.sub.d,g](M) and C [intersection] H contained in an integral degree 4 complete intersection, it is sufficient to rule out all C [member of] [M'.sub.d,g](M) with [h.sup.1](M, [I.sub.C,M](7)) [greater than or equal to] 3d - 6g - 21. If [h.sup.0](M, [I.sub.C,M](2)) = 0, then C is arithmetically Gorenstein ([17]) and hence [h.sup.1](M, [I.sub.C,M](7)) = 0, a contradiction. Since [p.sub.a](T) = 1 and [h.sup.1]( H, [I.sub.T,H](3)) = 0, we have [h.sup.1](H, [I.sub.A,H](3)) = 0 for every zero-dimensional scheme A [subset] T with either deg(A) [less than or equal to] 11 or deg(A) = 12 and A [not member of] [absolute value of [O.sub.T](3)]. Hence [h.sup.1](H, [I.sub.Z,H](3)) = 0 if d [less than or equal to] 11. Assume d = 12 and [h.sup.1](H, [I.sub.Z,H](3)) > 0. We get Z [member of] [O.sub.T](3) I and we conclude quoting [17]. Assume d [greater than or equal to] 13. Take [A.sub.1], [A.sub.2] [subset] C [inset] H such that #([A.sub.1]) = #(A2) = 12 and #([A.sub.1] [intersection] [A.sub.2]) = 11. Set {p} := [A.sub.1]\[A.sub.1] [intersection] [A.sub.2] and {q} = [A.sub.2]\[A.sub.1]. By monodromy we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], i.e. [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Assume [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], i.e. [A.sub.i] [member of] [absolute value of [O.sub.T](3)]. We get that p and q are linear equivalent, contradicting the inequality [p.sub.a](T) > 0.

Now assume d = 16 and [h.sup.1](H, [I.sub.C[intersection]H,T](4)) > 0. Since [[omega].sub.T] [congruent to] [O.sub.T], we have C [intersection] H [member of] [absolute value of [O.sub.T](4)], i.e. (since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is the complete intersection of a quartic and 2 quadrics. By [33, Theorem 0.1] we have d = 2 (mod 3) and that C [intersection] H is contained in rational normal curve; both statements are false.

Lemma 16. Fix C [member of] [M'.sub.d,g](M) d [greater than or equal to] 13, such that [alpha](C) = 3 and [h.sup.0](M, [I.sub.C,M](3)) [greater than or equal to] 2. Let H [subset] M be a general hyperplane. Then

(a) C [intersection] H is not contained in a curve of degree [less than or equal to] 3 or degree 4 and arithmetic genus 1.

(b) [h.sup.1](H, [I.sub.C[intersection]H,H](4)) = 0, [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] max{0, d - 12}, [h.sup.1](H, [I.sub.Z,H](3)) [less than or equal to] 2 if d = 15 and [h.sup.1] (H, [I.sub.Z,H](3)) [less than or equal to] 3 if d = 16.

Proof. Let K [subset] M be the set-theoretic base locus of [absolute value of [I.sub.C,M](3)]. Since [alpha](C) = 3 and [h.sup.0](M, [I.sub.C,M](3)) [greater than or equal to] 2, we have dim(K) [less than or equal to] 2. Assume that for a general H [subset] M, the set C [intersection] H is contained in a curve [T.sub.H] with c := deg([T.sub.H]) [less than or equal to] 4 and c minimal. Since H is general, C [intersection] H is a set of d points of H spanning H and in uniform position. Since d > c, monodromy and the minimality of c gives that [T.sub.H] is irreducible. Since d > 12, Bezout's theorem gives [T.sub.H] [subset] K [intersection] H. Hence there is an irreducible component A of K with dim(A) = 2 and [T.sub.H] [subset] A [intersection] H for a general hyperplane H. For a general H, A [intersection] H is irreducible. Since [T.sub.H] [subset or equal to] A [intersection] H, we get [T.sub.H] = A [intersection] H and hence deg(A) = c. Lemma 12 gives c = 4. Assume [p.sub.a]([T.sub.H]) = 1, i.e. assume that [T.sub.H] is the complete intersection of 2 quadrics. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], the exact sequence (3) gives that A is a complete intersection of 2 quadric hypersurfaces of M. Since C [subset] A, we get [alpha](C) = 2, a contradiction. Now we prove part (b). By part (a) we have [h.sup.0](H, [I.sub.C[intersection]H,H](2)) [less than or equal to] 1. Use parts (i) and (ii) of Lemma 17.

Lemma 17. Let [GAMMA] be the set of all C [member of] [M.sub.d,g] such that there is a hyperplane M [subset] [P.sup.5] with C [member of] [M'.sub.d,g] (M) and [h.sup.0](M, [I.sub.C,M](2)) > 0. Then dim([GAMMA]) [less than or equal to] 3d + 19 + g.

Proof. Since [P.sup.5] has [[infinity].sup.5] hyperplanes, it is sufficient to prove that for each hyperplane M the set [PSI] of all C [member of] [M'.sub.d,g](M) and [h.sup.0](M, [I.sub.C,M](2)) > 0 has dimension [less than or equal to] 3d + 14 + g. Fix C [member of] [PSI] and let Q [subset] M be any quadric containing C. Since C spans M, Q is irreducible. The Hilbert scheme Hilb(Q) of Q has [H.sup.0]([N.sub.C,Q]) as its tangent space at C. Let [tau] be the tangent sheaf of Q.

(a) First assume that either Q is smooth or C does not intersect the singular locus V of C. Since dim [absolute value of [O.sub.M] (2)] = 14, to handle this case it is sufficient to prove that [h.sup.0]([N.sub.C,Q]) [less than or equal to] 3d + g. Since the algebraic group Aut(Q) acts transitively on Q\V and [H.sup.0]([tau]) is the tangent space to Aut(Q) at the identity, [H.sup.0]([tau]) spans [tau] at each point of Q\V. Since C [subset] Q\V, [N.sub.C,Q] is a quotient of [tau[].sub.|C] and hence it is spanned. Since [[omega].sub.Q] [congruent to] [O.sub.Q](-3), [N.sub.C,Q] is a rank 2 vector bundle with degree 3d + 2g - 2. Hence [h.sup.1] (det([N.sub.C,Q])) = 0. Since [N.sub.C,Q] is spanned, it is an extension of det([N.sub.C,Q]) by [O.sub.C] and hence [h.sup.1] ([N.sub.C,Q]) [less than or equal to] g. Riemann-Roch gives [h.sup.0]([N.sub.C,Q]) [less than or equal to] 3d + g.

(b) Now assume C [intersection] V [not equal to] 0 and set x := deg(C [intersection] V). Since C is smooth, x = 1 if dim(V) = 0. By step (a) and the fact that M has [[infinity].sup.13] singular quadrics it is sufficient to prove that [h.sup.0]([N.sub.C,Q]) [less than or equal to] 3d + 1 + g. The vector space [H.sup.0]([tau]) is the tangent space at the identity map of the automorphism group Aut(Q). Since Q\V is homogeneous, [[tau].sub.|Q\V] is a spanned vector bundle. Since C is not a line and dim(V) [less than or equal to] 1, the set V [intersection] C is finite. Dualizing the natural map from the conormal sheaf of C in Q to [[OMEGA].sup.1.sub.Q] we get a map w: [[tau].sub.|C] [right arrow] [N.sub.C,Q] which is surjective outside the finite set C [intersection] V. Let u: [??] [right arrow] Q be the blowing up of V, E := [u.sup.-1] (V) the exceptional divisor, and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] the strict transform of C. Since C is smooth, u maps isomorphically [??] onto C. Let Y be closure in Hilb([??]) of the strict transforms of all C [subset] Q with deg(C [intersection] V) = x. It is sufficient to prove that dim([PSI]) [less than or equal to] 3d + 1 + g. Take a general D [member of] [PSI]. Since Aut([??]) acts transitively of [??]\E, step (a) of the proof gives [h.sup.1]([N.sub.D/[??]]) [less than or equal to] g. Hence it is sufficient to prove that deg([N.sub.D,[??]]) [less than or equal to] 3d + 2g - 1, i.e. deg([[tau].sub.[??]|D]) [less than or equal to] 3d + 1, i.e. deg([[omega].sub.[??]]|D) [greater than or equal to] -3d - 1. The group Pic([??]) is freely generated by E and the pull-back H of [O.sub.Q](1). We have D x H = d and D x E = x. We have [[omega].sub.[??]] [congruent to] [O.sub.[??]](-3H + cE) with c = -1 if dim(H) = 0 ([18], Example 8.5 (2)) and c = 0 if dim(H) = 1 ([18], Example 8.5 (3)). Hence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and the proof is complete.

Lemma 18. A general heptic hypersurface contains no C [member of] [M.sub.d,g] such that there is a hyperplane M [subset] [P.sup.5] with C [member of] [M'.sub.d,g] (M), there is no line L [subset] M with deg(L [intersection] C) > 6, no conic D [subset] M with deg(D [intersection] C) [greater than or equal to] 12.

Proof. By [11, Corollaire 2], Lemma 11 and the assumptions on the lines and the conics of C, for any integer t > 5 and any plane N c M we have [h.sup.1](N, [I.sub.C[intersection]N,N](t)) = 0. Lemma 1 and the case r = 4 and t = 5,6,7 of Lemma 3 give [h.sup.1] (M, [I.sub.C,M](4)) [greater than or equal to] [h.sup.1] (M, [I.sub.C,M](7)) + 12 [greater than or equal to] d + 7 - g. Let H [subset] M be a general hyperplane.

(a) Assume that the set H [intersection] C is not contained in a rational normal curve of H. By Lemma 15 (assuming that C is contained in a general heptic hypersurface) [h.sup.1](H, [I.sub.C[intersection]H,H](4)) = 0. Set [beta] = 0 if [less than or equal to]< 15 and [beta] = 1 if d = 16. By Lemma 15 we have [h.sup.1] (H, [I.sub.C,H](3)) [less than or equal to] 1 if d = 12 and [h.sup.1](H,[I.sub.C[intersection]H,H](3)) [less than or equal to] d - 12 if d [greater than or equal to] 13. Hence [h.sup.1](H, [I.sub.C[intersection]H,H](4)) + [h.sup.1](H, [I.sub.C[intersection]H,H(3)]) [less than or equal to] 3 + [beta]. By (1) for a general hyperplane H [subset] M we have [h.sup.1] (M, [I.sub.C,M](2)) [greater than or equal to] [h.sup.1] (M,[I.sub.C,M](4)) - [beta] - [beta] [greater than or equal to] d + 4 - g - f, i.e. [h.sup.0] (M, [I.sub.C,M](2)) [greater than or equal to] 18 - d - [beta]. if d [less than or equal to] 15, then we conclude by Lemma 14. Now assume d = 16. We got [h.sup.0] (M, [I.sub.C,M](2)) > 0. By Lemma 17 we may assume [h.sup.1](M,[I.sub.C,M](7)) [greater than or equal to] 3d - 17 - 3g = 21 - 4g. We first get [h.sup.1](M,[I.sub.C,M](4)) [greater than or equal to] 33 - 3g and then [h.sup.1] (M, [I.sub.C,M] (2)) [greater than or equal to] 29 - 3g, i.e. [h.sup.0](M, [I.sub.C,M] (2)) > 10 - 2g [greater than or equal to] 4, contradicting Lemma 14.

(b) Now assume that C [intersection] H is contained in a rational normal curve [E.sub.H] of H. By part (c) of Lemma 13 we may assume that [h.sup.1] (M, [I.sub.C,M](7)) [greater than or equal to] 3d - 16 - 6g - [epsilon] with [epsilon] = 1 if g = 1 and [epsilon] = 0 if g [epsilon] {2,3}. We first get [h.sup.1] (M,[I.sub.C,M](4)) > 3d - 4 - 6g - [epsilon] and then [h.sup.1](M,[I.sub.C,M](2)) [greater than or equal to] 3d - 4 - 6g - [epsilon] - min{d - 13,0} - d + 10, i.e. [h.sup.0](M,[I.sub.C,M](2)) > 20 - 5g - [epsilon] - min{d - 13,0}. Lemma 14 concludes unless (d, g) = (16,3). Assume (d, g) = (16,3). We just proved that [h.sup.0](M, [I.sub.C,M](2)) [greater than or equal to] 2. By Lemma 14 we may assume that [h.sup.0] (M, [I.sub.C,M](2)) = 2. Let [summation] the intersection of two different quadric hypersurfaces of M containing C. Since C is integral and C is not contained in a degree 3 surface (Lemma 12), [summation] is an integral complete intersection surface. Hence for a general hyperplane H [subset] M, the scheme [F.sub.H] := H [intersection] [summation] is an integral complete intersection of two quadrics. By Bezout we have #([E.sub.H] [intersection] [F.sub.H]) [less than or equal to] 6, contradicting the inclusion of C [intersection] H in [E.sub.H] [intersection] [F.sub.H].

Notation 2. Fix d, g. For any integer a [greater than or equal to] 0 let [F.sub.a] (resp. [G.sub.a]) denote the set of all C [member of] [M.sub.d,g] such that there is a hyperplane M [subset] [P.sup.5] and a line L [subset] M (resp. a smooth conic L [subset] M) with C [member of] [M'.sub.d,g] and deg(L [intersection] C) = a. Set F[".sub.a] := [U.sub.b[greater than or equal to]a] [F.sub.b] and [G".sub.a] := [[union].sub.b[greater than or equal to]a] [G.sub.b].

The proof of Lemma 2 gives the following result.

Lemma 19. Fix a hyperplane M [subset] [P.sup.5] and an integer b [greater than or equal to] 4 such that d [greater than or equal to] b + 2g - 1. Let Z [subset] M be a zero-dimensional scheme with deg(Z) = b. Then the set of all C [member of] [M'.sub.d,g](M) containing Z has codimension at least 3b in [M'.sub.d,g](M).

Lemma 20. Fix d, g with 12 [less than or equal to] d [less than or equal to] 16 and 1 [less than or equal to] g [less than or equal to] 3. For any integer a [greater than or equal to] 6 set [u.sub.a] := min{a, d + 1 - 2g}.

(i) Every irreducible component of [F".sub.a], a [greater than or equal to] 6, has dimension [less than or equal to] 5d + 12 - g - 2[u.sub.a].

(ii) Every irreducible component of [G".sub.a], a [greater than or equal to] 12, has dimension [less than or equal to] 5d + 17 - g - 2[u.sub.a].

Proof. Fix a hyperplane M [subset] [P.sup.5]. Set [F.sub.a](M) := {C [member of] [F.sub.a]: C [subset] M}, [G.sub.a](M) := {C [member of] [G.sub.a]: C [subset] m}, [F".sub.a] (M) := [U.sub.b[greater than or equal to]a] F(M) and G!J(M) := [[union].sub.b[greater than or equal to]a] [G.sub.b](M). Since in the definitions of Fa and Ga we prescribed that C [member of] [M'.sub.d,g](M), every C [member of] [G".sub.a] [union] [F".sub.a] is contained in a unique hyperplane. Hence to prove the lemma we may fix a hyperplane M, give an upper bound for dim([F".sub.a](M)) and dim([G".sub.a](M)) and then add +5 to that upper bound. Since the upper bounds in (i) and (ii) are non-decreasing functions of a to prove part (i) (resp. part (ii)) it is sufficient to give a suitable upper bound for all [F.sub.a](M), a [greater than or equal to] 6, and all [G.sub.a](M), a [greater than or equal to] 12. Fix C [member of] [F.sub.a](M) (resp. C [member of] [G.sub.a](M)) and a line L [subset] M (resp. a smooth conic L [subset] M) with deg(L [intersection] C) = a. Fix a zero-dimensional scheme Z [subset or equal to] C [intersection] Z with deg(Z) = [u.sub.a]. Use Lemma 19, that the smooth curve L has [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] subschemes of degree [u.sub.a], that M has [[infinity].sup.6] lines and that M has [[infinity].sup.11] conics.

Lemma 21. A general heptic hypersurface contain no C [member of] [M.sub.d,g] such that there is a hyperplane M with C [member of] [M'.sub.d,g] (M) and either there is a line L [subset] M with deg(C [intersection] L) [greater than or equal to] 6 or there is conic D [subset] M with deg(D [intersection] C) [greater than or equal to] 12.

Proof. By Lemma 11 and [11, Corollaire 2] we have [h.sup.1] (N, [I.sub.C[intersection]N,N](t)) = 0 if t [greater than or equal to] 5 and the plane N contains neither a line R with deg(R [intersection] C) [greater than or equal to] t + 2 nor a conic D with deg(D [intersection] C) [greater than or equal to] 2t + 2. Until step (e) we assume that either [alpha](C) [not equal to] 2 or d [less than or equal to] 12.

(a) We first exclude (with the silent assumption [alpha](C) [not equal to] 2) all C [member of] [M.sub.d,g] such that there is a hyperplane M and a line L [subset] M with C [member of] [M'.sub.d,g](M) and deg(C [intersection] L) [greater than or equal to] 9. Fix C, L such that deg(C [intersection] L) [greater than or equal to] 9. Set [u.sub.9] := min{9, d + 1 - 2g}. By Lemma 20 we may assume [h.sup.1] (M,[I.sub.C,M](7)) [greater than or equal to] d - 10 - g + 2[u.sub.9]. Part (a) of Lemma 13 for t = 5,6,7 gives [h.sup.1] (M, [I.sub.C,M](4)) [greater than or equal to] [h.sup.1] (M,[I.sub.C,M](7)).

First assume d [less than or equal to] 12 and so [u.sub.9] = d + 1 - 2g. Lemma 13 gives [h.sup.1] (M, [I.sub.C,M] (3)) [greater than or equal to] [h.sup.1] (M, [I.sub.C,M] (4)) and [h.sup.1] (M, [I.sub.C,M] (2)) [greater than or equal to] [h.sup.1] (M, [I.sub.C,M] (3)) - d + 10. Hence we have [h.sup.0] (M, [I.sub.C,M] (2)) [greater than or equal to] 14 - 2d + g + 3d + 3 - 6g - 2 = 15 + d - 5g > 0, a contradiction.

Now assume d [greater than or equal to] 13. Note that [u.sub.9] [greater than or equal to] 8 and hence [h.sup.1] (M, [I.sub.C,M] (4)) [greater than or equal to] d + 6 - g. We get [h.sup.1] (M, [I.sub.C,M] (3)) [greater than or equal to] d + 3 - g (Lemma 12) and hence [h.sup.0] (M, [I.sub.C,M] (3)) > 39 - 2d - 1 [greater than or equal to] 2, then [alpha](C) = 3 and then (by Lemma 16) [h.sup.1] (M, [I.sub.C,M](3)) > [h.sup.1] (M, [I.sub.C,M](4)) > d + 6 - gand [h.sup.1] (M, [I.sub.C,M](2)) [greater than or equal to] d + 3 - g,i.e. [h.sup.0](M, [I.sub.C,M](2)) > 17 - d > 0, a contradiction.

(b) Now we exclude all C [member of] [M.sub.d,g] such that there is a hyperplane M and a line L [subset] M with C [member of] [M.sub.d,g] (M) and deg(C [intersection] L) [greater than or equal to] 6. Fix C, L such that deg(C [intersection] L) [greater than or equal to] 6. Since d [greater than or equal to] 2g - 1 + 6, by Lemma 20 we may assume [h.sup.1] (M, [I.sub.C,M](7)) [greater than or equal to] d + 2 - g. By step (a) we may assume that deg(R [intersection] C) [less than or equal to] 8 for all lines R. Lemma 1 and the case t = 7 of Lemma 13 give [h.sup.1] (M, [I.sub.C,M](4)) [greater than or equal to] [h.sup.1](M, [I.sub.C,M](6)) [greater than or equal to] d + 6 - g. We solved this case in step (a).

(c) Now we exclude all C such that there is a conic D with deg(D [intersection] C) [greater than or equal to] 12 (they have d [greater than or equal to] 13). By step (b) we may assume deg(R [intersection] C) [less than or equal to] 5 for all lines R. Thus D is smooth. Set b := min{12,d + 1 - 2g}. By Lemma 20 we may assume that [h.sup.1] (M, [I.sub.C,M] (7)) [greater than or equal to] d - 15 - g + 2b. By Lemma 1 and the case t = 7 of Lemma 13 we may assume [h.sup.1] (M, [I.sub.C,M] (6)) [greater than or equal to] d - 11 - g + 2b. Since 2b [greater than or equal to] 17, we conclude as in steps (a) and (b).

(d) Now we assume d [greater than or equal to] 13 and [alpha](C) = 2. By Lemma 17 we may assume [h.sup.1] (M, [I.sub.C,M](7)) [greater than or equal to] 3d - 17 - 3g.

(d1) First assume that there is no line L with deg(L [intersection] C) [greater than or equal to] 9. By Lemmas 1 and 13 (case t = 7), we have [h.sup.1] (M, [I.sub.C,M](6)) > [h.sup.1] ([I.sub.C,M](7)) + 4 [greater than or equal to] 3d - 13 - 3g. Part (a) of Lemma 13 for t = 5,6 gives [h.sup.1] (M, [I.sub.C,M](4)) [greater than or equal to] 3d - 13 - 3g. The cases t = 3,4 of Lemma 13 gives [h.sup.1] (M, [I.sub.C,M] (2)) [greater than or equal to] 3d - 13 - 3g + 10 - d - min{0,d - 13}, i.e. [h.sup.0](M,Ic,m(2)) [greater than or equal to] 11 - 2g - min{0,d - 13}. Lemma 14 gives d = 16, g = 3 and [h.sup.0] (M, [I.sub.C,M](2)) = 2. Let K be the intersection of 2 different elements of [absolute value of [I.sub.C,M](2)]. K is a degree 4 surface and for a general hyperplane H of M K [intersection] H is an integral degree 4 curve, which is the complete intersection of 2 quadrics. Hence for a general H the set C [intersection] H is not contained in a rational normal curve of H. Hence [h.sup.1] (M, [I.sub.C,M](3)) [greater than or equal to] [h.sup.1] (M, [I.sub.C,M](4)) - 2 and [h.sup.1](M, [I.sub.C,M](2)) [greater than or equal to] [h.sup.1] (M, [I.sub.C,M](3)) - 4 (Lemma 13). Hence [h.sup.1](M, [I.sub.C,M](2)) [greater than or equal to] 20, i.e. [h.sup.0] (M, [I.sub.C,M](2)) [greater than or equal to] 5, a contradiction.

(d2) Now assume the existence of a line L [subset] M such that e := deg(L [intersection] C) [greater than or equal to] 9. Let U be the set of all lines R [subset] M such that deg(R [intersection] C) [greater than or equal to] 7 and let V be the set of all planes spanned by conics D such that deg(D [intersection] C) [greater than or equal to] 12.

(d2.1) Assume for the moment that U [union] V is finite. Let N [subset] M be a general plane. We have N [intersection] R = [empty set] for all R [member of] U and N [intersection] A is a single point for every A [member of] V. Let V [subset] [H.sup.0]([O.sub.M](1)) be the 2-dimensional linear subspace parametrizing all hyperplanes U of M containing N. For any such U we have R [??] U for all R [member of] U and dim(U [intersection] A) = 1 for each A [member of] V. Hence U contains no line R with deg(R [intersection] C) [greater than or equal to] 7 and no conic D with deg(C [intersection] D) [greater than or equal to] 12. By Lemma 11 and [11, Corollaire 2] we have [h.sup.1] (A, [I.sub.C[intersection]A,A](5)) = 0 for every plane A [subset] U and then Lemma 3 for r = 3, t = 5 and c < 16 gives [h.sup.1](U, [I.sub.C[intersection]U,U] (5)) = 0. Lemma 1 for t = 5,6,7 gives [h.sup.1] (M, [I.sub.C,M] (4)) [greater than or equal to] [h.sup.1](M, [I.sub.C,M](7)) + 3. As in step (d1) we get 15 [less than or equal to] d [less than or equal to] 16 and g = 3. As in step (d1) we exclude the case [h.sup.0](M, [I.sub.C,M] (2)) = 2 (and in particular the case d = 15), i.e. we may assume d = 16 and [h.sup.0] (M, [I.sub.C,M] (2)) = 1. Let Q C M denote the quadric containing C. We have [h.sup.1](M, [I.sub.C,M](3)) [greater than or equal to] [h.sup.1] (M,[I.sub.C,M](4)) - 3 [greater than or equal to] 22, i.e. h0(M,Icm(3)) > 35 - 46 + 22 = 11 [greater than or equal to] 7. Hence [h.sup.0](Q, [I.sub.C,Q](3)) [greater than or equal to] 2. Fix two general [U.sub.1], [U.sub.2] [member of] [absolute value of [I.sub.C,Q](3)] and set K := [U.sub.1] [intersection] [U.sub.2]. K is the complete intersection in M of a quadric and 2 cubics, because any element of [absolute value of [I.sub.C,Q](3)] is an irreducible surface, since [h.sup.0](Q, [I.sub.C,Q](2)) = [h.sup.0](M, [I.sub.C,M](2)) - 1 = 0. The complete intersection K links C to a degree 2 locally Cohen-Macaulay curve E. By Bezout we have L [subset] K and so either E is a double structure on L (a rope) or it is the union of L and another line R. Set R := L if E is a rope. The complete intersection K links R to the curve c [union] L. Since R is arithmetically Cohen-Macaulay, C [union] L is arithmetically Cohen-Macaulay. In particular [h.sup.1](M, [I.sub.C[union]L,M](7)) = 0. Since [h.sup.1](M, [I.sub.C,M](7)) [greater than or equal to] 3d - 14 - 3g > d [greater than or equal to] deg(C [intersection] L), we got a contradiction.

(d2.2) Now assume that U is infinite. Take R [member of] U with R [not equal to] L. Assume R [intersection] L [not equal to] [empty set]. Since deg(L [intersection] R) = 1, we get deg(L [intersection] C) = 9, deg(R [intersection] C) = 7 L [intersection] R [subset or equal to] C. The plane A spanned by R [union] L has deg(A [intersection] C) [greater than or equal to] 15, contradicting Lemma 11. Now assume R [intersection] L = [empty set]. The hyperplane H spanned by R [intersection] L intersects C in a scheme of degree [greater than or equal to] 16. We get d = 16, deg(L [intersection] C) = 9 and deg(R [intersection] C) = 7. Since C is smooth, the linear projection M\L [right arrow] [P.sup.2] from L induces a morphism u: C [right arrow] [P.sup.2] such that deg(u) x deg(u(C) = 7. Since there are infinitely many R [member of] U, we get deg(u) [greater than or equal to] 7 and so u(C) is a line, contradicting the assumption C [member of] [M'.sub.d,g](M).

(d2.3) Now assume that V is infinite. Let V' be an irreducible component of V with dim(V') > 0. Assume the existence of a conic D [subset] M such that deg(D [intersection] C) [greater than or equal to] 12 and call A the plane spanned by D. If A [intersection] L is a point, then the 3-dimensional linear space spanned by A [union] L intersects C in a scheme of degree [greater than or equal to] 12 + 9 - 1 > d, a contradiction.

Now assume L [subset] A for infinitely many planes A [member of] V'. For any plane N [subset] M with N [intersection] L = [empty set], the set N [intersection] A is a single point. Hence as in step (d2.1) we may ignore all these planes A [member of] V', even if infinite.

Now take a general A [member of] V' and assume A [intersection] L = [empty set] and f := deg(D [intersection] C) with D [subset] A a conic and f [greater than or equal to] 12. Take general [A.sub.1], [A.sub.2] [member of] V' and let [D.sub.i] [susbet] [A.sub.i], i = 1, 2, be the conic with deg([D.sub.i] [intersection] C) = f. First assume that [A.sub.1] [intersection] [A.sub.2] is a line (this is the case if either [D.sub.1] and [D.sub.2] have a common component or [D.sub.1] [intersection] [D.sub.2] is a zero-dimensional scheme of degree [greater than or equal to] 2). We get that either all A [member of] V' are contained in a fixed hyperplane of M (absurd, because C spans M) or there is a line R [subset] M contained in all A [member of] V'. If N [subset] M is a plane with R [intersection] N = [empty set], then no conic of some A [member of] V' is contained in a hyperplane of M containing N.

(d2.3.1) Assume [D.sub.1] [intersection] [D.sub.2] = [empty set]. We have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and therefore we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Fix any [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since deg(C [intersection] Q) [greater than or equal to] 2f + e [greater than or equal to] 33, Bezout gives C [subset] Q'. Hence [h.sup.0](M, [I.sub.C,M](2)) [greater than or equal to] 2. By Lemma 14 we may assume [h.sup.0](M, [I.sub.C,M](2)) = 2 and call K' the degree 4 irreducible surface which is the base locus of [absolute value of [I.sub.C,M](2)]. For a general hyperplane H [subset] M the scheme K' [intersection] H is an integral curve complete intersection of 2 quadrics and containing C [intersection] H. By Bezout C [intersection] H is not contained in a rational normal curve of H and so [h.sup.0](H, [I.sub.C[intersection]H,H](2)) = 2. Lemma 15 gives [h.sup.1](M, [I.sub.C,M](3)) [greater than or equal to] [h.sup.1](M, [I.sup.C,M](4)) [greater than or equal to] 3d - 17 - 3g, i.e. [h.sup.0](M, [I.sub.C,M](3)) [greater than or equal to] 17 - 2g > 10. Hence the map [H.sup.0](M, [I.sub.C,M](2)) [cross product] [H.sup.0](M, [O.sub.M](1)) [right arrow] [H.sup.0](M, [I.sub.C,M](3)) is not surjective. Therefore C is contained in the intersection of K' with a cubic hypersurface, contradicting Bezout and the inequality d > 12.

(d2.3.2) Now assume that the scheme [D.sub.1] [intersection] [D.sub.2] is a single point. First assume that either e [greater than or equal to] 10 or f [greater than or equal to] 13 or d [less than or equal to] 15 or [D.sub.1] [intersection] [D.sub.2] [not member of] C. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Fix any [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since deg(C [intersection] Q') [greater than or equal to] 2f + e - deg([D.sub.1] [intersection] [D.sub.2] [intersection] C) [greater than or equal to] 2d + 1, Bezout gives Q' [contains] C, contradicting Lemma 14. Now assume e = 9, f = 12, d = 16 and that [D.sub.1] [intersection] [D.sub.2] is a point p [subset or equal to] C. Take a general q [member of] C. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], we have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Fix any [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since deg(C [intersection] Q') [greater than or equal to] 2f - 1 + e - deg([D.sub.1] [intersection] [D.sub.2] [intersection] C) + 1 [greater than or equal to] 2d + 1, then Q' [contains] C. Hence [h.sup.0](M, [I.sub.C,M](2)) [greater than or equal to] 2. By Lemma 14 we may assume [h.sup.0](M, [I.sub.C,M](2)) = 2 and call K' the degree 4 irreducible surface which is the base locus of [absolute value of [I.sub.C,M](2)]. We conclude as in step (d2.3.1).

Proof of Theorem 1: We just proved Theorem 1 for all C [member of] [M.sub.d,g] spanning a hyperplane of [P.sup.5], and so we completed the proof of Theorem 1

5 Curves in a 3-space

In this section we assume that C spans a 3-dimensional linear subspace U [subset] [P.sub.5]. We prove Proposition 1 and give a few results useful to extend it to higher degrees and higher genera. The Hilbert scheme [M'.sub.d,g](U) of all non-special curves of degree d and genus g of U is smooth and irreducible of dimension 4d. Since the Grassmannian G(3,5) of all 3-dimensional linear subspaces of [P.sup.5] has dimension 8, to exclude these curves C it is sufficient to exclude the ones with [h.sup.1] ([I.sub.C](7)) [greater than or equal to] 2d + 2 - 2g - 8. However, for a general W [member of] W and a general U [member of] G(3,5) the surface W [intersection] U is a general surface of degree 7 and hence any curve of degree [less than or equal to] 16 on it is a complete intersection by a theorem of Max Noether. Hence to exclude all C [member of] [M.sub.d,g] spanning some U [member of] G(3,5) it is sufficient to exclude the ones with [h.sup.1] ([I.sub.C](7)) [greater than or equal to] 2d - 2g - 5. Fix a general X [member of] [M'.sub.d,g](U). Since X has maximal rank ([2]) and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], we have [h.sup.1] ([I.sub.x](7)) = [h.sup.1](U, [I.sub.X,U](7)) = 0 and so [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since dim([M'.sub.d,g](U)) = 4d, [P.sup.5] has [[infinity].sup.8] 3-dimensional linear spaces and 7d + 1 - g > 4d + 8, a general heptic hypersurface contains no curve C [member of] [M.sub.d,g] with the Hilbert function of X in degree 1 and degree 7. Hence we may increase by one these bounds and so we may assume [h.sup.1] (U, [I.sub.C,U] (7)) [greater than or equal to] 2d - 2g - 4, i.e. for any d,g to rule out the existence of any C [member of] [M.sub.d,g] with dim(<C>) = 3 for a general heptic hypersurface it is sufficient to exclude all C [member of] [M'.sub.d,g](U) with [h.sup.1] (U, [I.sub.C,U](7)) [greater than or equal to] 2d - 2g - 4. By [13] we may assume d [greater than or equal to] 11. Hence 11 [less than or equal to] d [less than or equal to] 16 and 1 [less than or equal to] g [less than or equal to] 3. Let [alpha] or [alpha](C) be the minimal degree of a surface of U containing C. Since C is non-degenerate, we have [alpha] [greater than or equal to] 2. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], we have [alpha] [less than or equal to] 7.

Lemma 22. We have [alpha] [not equal to] 2.

Proof. Assume [alpha] = 2. Let Q [subset] U be a quadric containing C. If Q is a quadric cone, then C is arithmetically Cohen-Macaulay ([16, V, Ex. 2.9]), because C is smooth, and in particular [h.sup.1] ([I.sub.C](7)) = 0, a contradiction. Hence Q is a smooth quadric. Write C [member of] [absolute value of [O.sub.Q] (u, [upsilon])] with, say, u [less than or equal to] [upsilon]. Since g > 0, we have u > 1. Since [h.sup.1] ([O.sub.C] (1)) = 0, we have u [less than or equal to] 2. Since 1 [less than or equal to] g [less than or equal to] 3, we get u = 2 and [upsilon] = g + 1. In all cases we have [h.sup.1] ([I.sub.C](7) ) = 0, a contradiction.

The following lemma is a particular case of step (b2) of Remark 5 proved below.

Lemma 23. Fix C [member of] [M'.sub.d,g](u), d [less than or equal to] 16. Let H [subset] U be a general hyperplane. We have [h.sup.1] (H, [I.sub.C[intersection]H,H](t)) = 0 for all t [greater than or equal to] 5, [h.sup.1] (H, [I.sub.C[intersection]H,H](4)) = 0 if d [less than or equal to] 12 and [h.sup.1] (H, [I.sub.C[intersection]H,H](4)) [less than or equal to] 1 if d [member of] {13,14}.

Remark 5. Let H c U be a general plane. The set Z := C [intersection] H is formed by d points of H in uniform position and spanning H.

(a) In particular Z is in linearly general position and hence [h.sup.1] (H, [I.sub.Z,H](t)) = 0 for all t with 2t [greater than or equal to] d - 1 ([10, Theorem 3.2]). If d [greater than or equal to] 2t + 2 we get [h.sup.1] (H, [I.sub.Z,H](t)) < d - 2t - 1.

(b) Now assume 2t [less than or equal to] d - 2, but [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], i.e., if d = 16 assume 5 [less than or equal to] t [less than or equal to] 7, while if 11 [less than or equal to] d [less than or equal to] 15 assume 4 [less than or equal to] t [less than or equal to] [(d - 2)/2]. Assume [h.sup.1](H, [I.sub.Z,H](t)) > 0. We get [h.sup.0] (H, [I.sub.Z,H](t)) > 0. Since C [intersection] H is in uniform position (or by a monodromy argument) and d > t, either a general D [member of] [absolute value of [I.sub.Z,H](t)] is irreducible or the base locus of [absolute value of [I.sub.Z,H](t)] contains an irreducible curve T [contains] Z and D = cT for some integer c > 2. The latter case does not occur if t is a prime, because Z spans H.

(b1) Assume D = cT with c [greater than or equal to] 2. We get [absolute value of [I.sub.Z,H](t)] = {cT} and hence [h.sup.0](H, [I.sub.Z,H](t)) = 1. Therefore [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], we get [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since Z spans H, t is not a prime. Since d [less than or equal to] 16, we get t < 6. Hence t = 4, c = 2, deg(T) = 2 and d = 15. Since T is an integral conic (i.e. a smooth conic), we get [h.sup.1] (H, [I.sub.Z,H](y)) = 0 for all y [greater than or equal to] 7. Let K [subset] U be a degree a surface containing C. Recall that [alpha] [less than or equal to] 7. Since d = 15 > 2[alpha], Bezout theorem gives T [subset] K. Varying H we see that K contains a 3-dimensional family of conics and hence it is a projection of a Veronese surface. Hence [alpha] [less than or equal to] 4. Since C has odd degree K must be an inner projection of the Veronese surface and hence [alpha] = 3. Since C is smooth, we see that C is isomorphic to a plane curve of degree [greater than or equal to] 8, contradicting the assumption g [less than or equal to] 3.

(b2) Assume that D is irreducible. Since [h.sup.i] ([O.sub.h]) = 0, i = 1,2, then we have [h.sup.1] (H, [I.sub.Z,H](t)) = [h.sup.1] (D, [I.sub.Z,D](t)). Since [congruent to] [O.sub.D](t - 3), we get d [greater than or equal to] 3t. If d = 3t we also get that Z is the complete intersection of D with a plane cubic; since [h.sup.1] (U, [I.sub.C,U](7)) > 0, [33, Theorem 0.1] gives that Z is contained in a smooth conic, contradicting the inequality d > 2t and the irreducibility of D. Now assume d [greater than or equal to] 3t + 1. Since [h.sup.1](D, [I.sub.Z,D](t)) > 0, [I.sub.Z,D](t) is a subsheaf of [[omega.sub.]D] [congruent to] [O.sub.D] (t - 3) and hence there is Z' [subset] Z such that deg(Z') = 3t and Z' is the complete intersection of D and and a cubic curve. We have h1 (H,IZ,H(t)) = [h.sub.1] (D, [I.sub.Z,D](t)) and [h.sup.1] (H, [I.sub.Z,H](t)) = [h.sup.1] (D, [I.sub.Z',D](t)) = [h.sup.1](D, [O.sub.D](t - 3)) = 1. Since [O.sub.D] (t - 3) is very ample, we have [h.sup.0] (D, [I.sub.E,D] (t)) = [h.sup.0] (D, [I.sub.Z',D](t)) - deg(E) (i.e. [h.sup.1] (D, [I.sub.E,D](t)) = [h.sup.1](D, [I.sub.Z',D](t)) for every E [contains] Z' with deg(E) - deg(Z') [less than or equal to] 2. Hence if [h.sup.1] (H, [I.sub.Z,H](t)) > 0, then d [greater than or equal to] 3t + 1 and if d [less than or equal to] 3t + 2, then [h.sup.1] (H, [I.sub.Z,H](t)) [less than or equal to] 1.

5.1 Proof of Proposition 1

In this subsection we take g = 1 and d [less than or equal to] 14.

Lemma 24. Fix a zero-dimensional scheme Z [subset] U and set b := deg(Z). Let [E.sub.Z] be the set of all C [member of] [M'.sub.d,1](U) such that C [contains] Z. If d [greater than or equal to] b, then every irreducible component of [E.sub.Z] has dimension 4d - 2b.

Proof. Fix C [member of] [E.sub.Z]. It is sufficient to prove that [h.sup.1] ([N.sub.C,U](-Z)) = 0 ([27, Theorem 1.5]). Since [N.sub.C] is a quotient of [O.sub.C][(1).sup.[direct sum]4], we are done if [h.sup.1](OC(1)(-Z)) = 0. Since d [greater than or equal to] b and [[omega].sub.C] [congruent to] [O.sub.Z], this is the case, unless d = b and [O.sub.C](1) [congruent to] [O.sub.C](Z). By duality we only need to exclude the existence of a non-zero map [N.sub.C] [right arrow] [O.sub.C] (1). Assume that this is the case. The restriction to C of Euler's sequence of T[P.sup.3] gives a non-zero map [O.sub.C][(1).sup.[direct sum]4] [right arrow] [O.sub.C](1). This map gives the equation of a hyperplane of U containing C, a contradiction.

Lemma 25. A general W [member of] W contains no C [member of] [M.sub.d,1], d [less than or equal to] 14, such that there is U [member of] G(3,5) with C [member of] [M'.sub.d,1] (U), deg(R n[intersection]C) [less than or equal to] 7 for each line R and deg(C [intersection] D) [less than or equal to] 13 for each conic D.

Proof. Fix C [member of] [M'.sub.d,1](U) such that deg(R [intersection] C) [less than or equal to] 7 for each line R and deg(R [intersection] D) [less than or equal to] 13 for each conic D. By Lemmas 1 and 23 we have [h.sup.1] (U, [I.sub.C,U](5)) > [h.sup.1](U, [I.sub.C,U](6)) + 3 [greater than or equal to] [h.sup.1] (U, [I.sub.C,U](7)) + 6 [greater than or equal to] 2d. By Lemma 23 we have [h.sup.1](U, [I.sub.C,U](3)) [greater than or equal to] [h.sup.1] (U, [I.sub.C,U](7)) - [epsilon] [greater than or equal to] 2d - [epsilon] with [epsilon] = 0 if d [less than or equal to] 12 and [epsilon] = 1 if d [member of] {13,14}, i.e. [h.sup.0](U, [I.sub.C,U](3)) [greater than or equal to] 20 - d - [epsilon] > 2. Since [alpha] > 2 (Lemma 22), C is contained in the intersection of 2 integral cubic surfaces and so d [less than or equal to] 9, a contradiction.

Lemma 26. A general W [member of] W contains no C [member of] [M.sub.d,1], d [less than or equal to] 14, such that there is U [member of] G(3,5) with C [member of] [M'.sub.d,1] (U) and a conic D with deg(D [intersection] C) [greater than or equal to] 14.

Proof. Take C [member of] [M.sub.d,1], d [less than or equal to] 14, such that there is U [member of] G(3,5) with C [member of] [M'.sub.d,1](U) and a conic D with deg(D [intersection] C) > 14. We have d = 14 and deg(D [intersection] C) = 14. By Lemma 25 we may assume that deg(R [intersection] C) [less than or equal to] 6 for every line R. Hence D is a smooth conic and so it has [[infinity].sup.14] degree 14 subschemes. Since dim(G(3,5)) = 8, U has [[infinity].sup.3] planes and each plane has [[infinity].sup.5] conics, to prove the lemma (quoting Lemma 24) it is sufficient to exclude all C [member of] [M.sub.14,4] spanning a 3-space U and with [h.sup.1] (U, [I.sub.C,U](7)) [greater than or equal to] 6d + 2 - 2g - 4d - 8 + 28 - 14 - 8 = 2d - 2. By Lemma 23 we have [h.sup.1] (U, [I.sub.C,U](3)) [greater than or equal to] h1 (U, [I.sub.C,U](7)) - [epsilon] [greater than or equal to] 2d - 2 - [epsilon] with [epsilon] = 0 if d [less than or equal to] 12 and [epsilon] = 1 if d [member of] {13,14}, i.e. [h.sup.0](U, [I.sub.C,U](3)) [greater than or equal to] 18 - d - [epsilon] [greater than or equal to] 2 and so Bezout gives d [less than or equal to] 9, a contradiction.

Lemma 27. A general W [member of] W contains no C [member of] [M.sub.d,1], d [less than or equal to] 14, such that there is U [member of] G(3,5) with C [member of] [M'.sub.d,1] (U) and a line R with deg(R [intersection] C) [greater than or equal to] 8 for some line R.

Proof. Fix U and C [member of] [M'.sub.d,1](U). By Lemma 26 we may assume deg(D [intersection] C) [less than or equal to] 13 for each conic D. Let R [subset] U be a line such that b := deg(C [intersection] R) is maximal. First assume b [greater than or equal to] 9. Fix Z [subset] R [intersection] C with deg(Z) = 9. Since dim(G(3,5)) = 8, U has [[infinity].sup.4] lines and each line has [[infinity].sup.9] subschemes of degree 9, Lemma 24 shows that it is sufficient to exclude the curves C with [h.sup.1] (U, [I.sub.C,U] (7)) [greater than or equal to] 2d - 8 + 18 - 9 - 4 = 2d - 3. By Lemma 23 we have [h.sup.1] (U, [I.sub.C,U](3)) [greater than or equal to] 2d - 3 - [epsilon] with [epsilon] = 0 if d [less than or equal to] 12 and [epsilon] = 1 if d [member of] {13, 14}, i.e. [h.sup.0](U, [I.sub.C,U](3)) [greater than or equal to] 17 - d - [epsilon] [greater than or equal to] 2 and so Bezout gives d [less than or equal to] 9, a contradiction. Now assume b = 8. As above we see that it is sufficient to exclude all C [member of] [M'.sub.d,1](U) with [h.sup.1] (U, [I.sub.C,U](7)) [greater than or equal to] 2d - 4. By Lemmas 1 and 23 we have [h.sup.1](U, [I.sub.C,U](6)) [greater than or equal to] 2d - 1 and hence [h.sup.0](U, [I.sub.C,U](3)) [greater than or equal to] 4.

Proof of Proposition!. Lemmas 25, 26 and 27 prove Proposition 1.

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E. Ballico *

* The author was partially supported by MIUR and GNSAGA of Indam (Italy)

Received by the editors in December 2015--In revised form in March 2016.

Communicated by J. Fine.

2010 Mathematics Subject Classification: 14J32; 14M10; 14H50.

Dept. of Mathematics University of Trento 38123 Povo (TN), Italy email: ballico@science.unitn.it

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