TERRACINI LOCI OF CURVES

EDOARDO BALLICO AND LUCA CHIANTINI

Abstract. We study subsets S of curves X whose double structure does not impose independent conditions to a linear series L, but there are divisors D ∈ |L| singular at all points of S. These subsets form the Terracini loci of X. We investigate Terracini loci, with a special look towards their non-emptiness, mainly in the case of canonical curves, and in the case of space curves.

1. Introduction

Terracini loci 𝕋(X, L, x) of a projective variety X (over an algebraically closed field of characteristic 0) are subsets of the set S(X, x) of all reduced finite subsets S ⊂ X_{reg} of cardinality x, with the property that the double scheme 2S on S does not impose independent conditions to the linear series L, and there are divisors in |L| passing through 2S. Terracini loci are certainly involved in the study of interpolation properties of the image of X in the map induced by L, but they also assume great importance in the theory of secant varieties to embedded varieties X, for they are connected with points of the abstract secant variety of X in which the differential of the map to the embedded secant variety drops rank. More details on the initial properties of Terracini loci can be found in [1].

In this paper, we focus the attention to Terracini loci of curves. Even if curves are never defective, so that their secant varieties always have the expected dimension, yet there are special points p in which the Terracini Lemma fails to provide the dimension of the Zariski tangent space to the secant variety. This happens typically when the set of points in X which generates p belongs to some Terracini locus.

We are mainly concerned with the problem of the non-emptiness of a Terracini locus 𝕋(X, L, x). Certainly 𝕋(X, L, x) is empty when X is a rational normal curve in ℙ^{r}, L is the complete hyperplane linear series and x ≤ (r + 1)/2, because every subscheme of length ≤ r + 1 in X is linearly independent. We will see that, for odd r, the emptiness of 𝕋(X, L, (r + 1)/2) characterizes rational normal curves (Proposition 5.6). This means that there is a large mass of examples of non-empty Terracini loci of curves.

Our analysis considers mainly two cases: canonically embedded curves (Section 4) and curves in ℙ^{3} (Section 5).

In the former case, the series L is the complete canonical linear series K_{X}, and one can easily expect that the existence of sets in 𝕋(X, K_{X}, x) strongly depends on the geometry of linear series on the curve. We show indeed that subsets of thetacharacteristics on X prove that 𝕋(X, K_{X}, x) is non-empty when g/2 ≤ x ≤ g − 1. For x < g/2 we see (Proposition 4.7) that the non-emptiness of 𝕋(X, K_{X}, x) is linked to the gonality of X.

2020 Mathematics Subject Classification. 14H50; 14H51.

For space curves, we use induction on the degree and genus, with the technique of smoothing reducible curves, to prove that 𝕋(X, L, x) is non-empty for some smooth curves of degree d and genus g ≤ d − 3, when 6 ≤ 2x ≤ d (Theorem 5.12). We also prove the existence of smooth special curves with non-empty Terracini loci in some components of the Hilbert scheme of space curves even outside the Brill-Noether range (Theorem 5.13 and Remark 5.14).

We would like to thank the anonymous referees for precious observations on the preliminary versions of the paper.

2. Preliminaries

We work over an algebraically closed field of characteristic 0.

2.1. Notation. For any 0-dimensional subscheme S we denote with ℓ(S) the length of S. When S is reduced, then ℓ(S) is the cardinality of S.

Let X be an integral projective variety. For any point P ∈ X_{reg}, with homogeneous maximal ideal m_{P,X}, we denote with (2P, X) the subscheme of X defined by m_{P,X}^{2}. If S = {P_{1}, …, P_{x}} ⊂ X is a finite set of points, then we denote with (2S, X) the non-reduced scheme 2S = ⋃_{i=1}^{x}(2P_{i}, X). When it is clear which X we refer to, then we will write simply 2S instead of (2S, X).

For any scheme X and for any integer x we denote with S(X, x) the set of all reduced finite subsets S ⊂ X_{reg} of length x. S(X, x) is an open subset of the symmetric product X^{(x)}.

Let X be a projective curve, and let S be a finite subset of X_{reg}. When S identifies a Cartier divisor on X, by abuse, we will continue to denote it with S. We will also denote with |S| the complete linear series associated to S, and we will denote with h^{0}(S) the dimension of the space of sections of the associated line bundle. We will use the same convention for 2S.

For any scheme Z ⊂ ℙ^{r} let ⟨Z⟩ ⊆ ℙ^{r} denote the minimal linear subspace of ℙ^{r} containing Z.

2.2. Terracini loci. We recall from [1] the definition of Terracini locus of a projective variety.

Definition 2.1. Let X be an integral projective variety, L a line bundle on X and V ⊆ H^{0}(L) a linear subspace. Set m := dim X. Fix S ∈ S(X, x). We say that S is in the Terracini locus 𝕋(X, L, V, x) if V (−(2S, X)) ≠ (0) and dim V (−(2S, X)) > dim V − x(m + 1). We say that the integer δ(S, X, L, V) := dim V (−(2S, X)) − dim V + x(m + 1) is the (Terracini) defect of S with respect to (L, V). When V = H^{0}(L), we will drop V in the notation.

We will consider, throughout the paper, mainly the case where X is a curve, i.e. m = 1. In this situation a finite set S of length x lies in the Terracini locus when dim V (−(2S, X)) > max{dim V − 2x, 0}.

We can extend the definition of Terracini loci to include some non-reduced 0dimensional subschemes.

Definition 2.2. Let C ⊂ ℙ^{r} be a smooth and connected non-degenerate curve. For each positive integer x let C^{(x)} denote the symmetric product of x copies of C. The variety C^{(x)} is a connected projective variety of dimension x parametrizing the degree x zero-dimensional schemes of X, and there is a non-empty Zariski open subset S(C, x) ⊂ C^{(x)} which parametrizes subsets of cardinality x. For each positive integer x let 𝕋̃(C, x) denote the set of all Z ∈ C^{(x)} such that dim⟨2Z⟩ ≤ 2x − 2 and ⟨2Z⟩ ≠ ∅. Obviously, 𝕋(C, 1) = 𝕋̃(C, 1) = ∅. The inclusion S(C, x) ⊂ C^{(x)} induces an inclusion 𝕋(C, x) ⊆ 𝕋̃(C, x). The semicontinuity theorem for cohomology gives that 𝕋̃(C, x) contains the closure of 𝕋(C, x) in C^{(x)}. Sometimes, 𝕋̃(C, x) is not equal to the closure of 𝕋(C, x) in C^{(x)} (Example 2.3).

Example 2.3. Fix an even integer d ≥ 4, a line D ⊂ ℙ^{2} and p ∈ D. There is a smooth degree d curve C ⊂ ℙ^{2} such that p ∈ C and D ∩ C = dp, i.e. p is a total ramification point of C with D as its tangent line. Since C has only finitely many multitangent lines, 𝕋(C, x) is finite and hence it is closed in 𝕋̃(C, x). Since d is even and d ≥ 4, \frac{d}{2}p lies in 𝕋̃(C, d/2) \ 𝕋(C, d/2).

3. Generalities

Example 3.1. Let X ⊂ ℙ^{N} be an integral curve. Here we take L = O_{X}(1) and V = the linear series of hyperplanes. For a general S = {P_{1}, …, P_{x}} ⊂ X_{reg} write 2S for (2S, X). The elements of V (−2S) correspond to hyperplanes containing the tangent lines to X at the points P_{i}’s. It follows that S lies in the Terracini locus when the span of the x tangent lines at the points of X has dimension smaller than the expected one. Then by definition 𝕋(X, L, V, 1) is always empty.

On the other hand, when the map induced by the linear series V is not birational onto the image the Terracini locus 𝕋(X, L, V, 1) may contain some points.

Example 3.2. Let X be a smooth hyperelliptic curve of genus g ≥ 2. We can describe all the Terracini loci with respect to L = K_{X} and V = H^{0}(K_{X}). Let B ⊂ X be the set of the Weierstrass points of X. For any integer x let S(B, x) denote the set of all subsets of B with cardinality x. The set B is the ramification locus of the g_{2}^{1} of X. Since we work in characteristic 0, ℓ(B) = 2g+2. Fix a positive integer x and take S ∈ S(X, x). If x ≥ g, then deg(K_{X}(−(2S, X))) < 0 and hence h^{0}(K_{X}(−(2S, X))) = 0. Thus 𝕋(X, K_{X}, x) = ∅ for all x ≥ g. Assume that 1 ≤ x ≤ g − 1. Let h : X → ℙ^{1} be the morphism associated to the g_{2}^{1} of X. The linear system |K_{X}| is the minimal sum of g − 1 copies of the g_{2}^{1} of X. Hence every base-point free special line bundle on X is the sum of at most g − 1 copies of the g_{2}^{1}. Thus 𝕋(X, K_{X}, 1) = B and 𝕋(X, K_{X}, g − 1)) = S(B, g − 1). More generally, S belongs to 𝕋(X, K_{X}, x) if and only if we have that either S ∩ B ≠ ∅ or there are p, q ∈ S such that p ≠ q and h(p) = h(q) for these conditions are equivalent to h^{0}(K_{X}(−(2S, X))) > g − 2x. It is easy to realize that, for each x ∈ {2, …, g − 2}, the elements of 𝕋(X, K_{X}, x) with (maximal) defect x are the elements of S(B, x). If e := #(S ∩ B) and S has f distinct sets {p_{i}, q_{i}}, 1 ≤ i ≤ f, with p_{i} ≠ q_{i} and h(p_{i}) = h(q_{i}), then h^{0}(K_{X}(−2S)) = g − 2x + e + f.

For the rest of the section let us go back to the case in which X ⊂ ℙ^{N} is an integral curve, and we take L = O_{X}(1) and V = the linear series of hyperplanes.

Example 3.3. Let us consider what happens when x = 2 and X is a plane curve. Thus max{dim V − 2x, 0} = 0. Then S = {P, Q} ∈ S(2) belongs to the Terracini locus if and only if the tangent lines to X in P and Q coincide. Since in characteristic 0 not every tangent line is bitangent, then 𝕋(X, L, V, 2) is either empty or finite. If N > 2, the tangent lines to two general points of X span a 3-dimensional linear subspace (recall that we work in characteristic 0). The set S = {P, Q} lies in the Terracini locus 𝕋(X, L, V, 2) when the tangent lines in P, Q meet at some point P_{0}, i.e. there exists a plane containing the two tangent lines. In this case, the projection of X from P_{0} is a curve with (at least) two cusps.

4. Canonically embedded curves

Let us turn now to the case where X is a smooth curve of genus g and we consider the complete canonical linear series L = K_{X}. We will describe in several cases the locus 𝕋(X, K_{X}, x). Since we already treated the case of hyperelliptic curves in Example 3.2, we assume that g ≥ 3 and K_{X} embeds X in ℙ^{g−1}.

We start with a very easy observation, which shows that we need to distinguish two cases, depending if x is smaller than g/2 or not.

Proposition 4.1. A reduced set S of length x lies in 𝕋(X, K_{X}, x) if and only if either 2x < g and h^{0}(2S) > 1, i.e. the linear series |2S| is not a singleton, or 2x ≥ g and h^{0}(K_{X} − 2S) > 0, i.e. 2S is special.

Proof. By definition the set S lies in 𝕋(X, K_{X}, x) if and only if h^{0}(K_{X} − 2S) > min{0, h^{0}(K_{X}) − 2x}. Since h^{0}(K_{X}) = g, we distinguish between 2x < g and 2x ≥ g. In the latter case S lies in 𝕋(X, K_{X}, x) if and only if h^{0}(K_{X} − 2S) > 0. In the former case, since by Riemann-Roch h^{0}(2S) = 2x − g + 1 + h^{0}(K_{X} − 2S), S lies in 𝕋(X, K_{X}, x) if and only if h^{0}(2S) > 1. □

It follows from the previous proposition that the Terracini locus 𝕋(X, K_{X}, x) is empty if x ≥ g, because in this case the degree of K_{X} − 2S is negative. Let us consider the extremal case x = g − 1.

Example 4.2. A set S (resp. scheme) of length g − 1 belongs to 𝕋(X, K_{X}, g − 1) (resp. 𝕋̃(X, K_{X}, g − 1)) if and only if h^{0}(K_{X} − 2S) > 0 which, for degree reasons, implies that 2S is a canonical divisor. Thus subsets S ∈ 𝕋(X, K_{X}, g−1) correspond to divisors in some non empty linear series G such that 2G = K_{X}, i.e. a thetacharacteristic of X. It is well known ( [5]) that X has a finite number, exactly 2^{2g}, theta-characteristics. A theta-characteristic is odd or even, depending on the parity of h^{0}(G). The number of odd theta-characteristics is 2^{g−1}(2^{g} − 1) while there are 2^{g−1}(2^{g} + 1) even thetacharacteristics. Now assume that X is general in the moduli space Mc_{g}. In this case, by [5] Corollary 1.11, h^{0}(G) ≤ 1 for every theta-characteristic G on X, and for each odd theta-characteristic G on X the divisor D with {D} = |G| is reduced. Thus for X ∈ Mc_{g} general the Terracini locus 𝕋(X, K_{X}, g − 1) is finite, of cardinality 2^{g−1}(2^{g} − 1). There are X ∈ Mc_{g} with theta-characteristics G such that h^{0}(G) ≥ 2. For such curves 𝕋(X, K_{X}, g − 1) is infinite. On the contrary, a natural question is to ask if there are X ∈ Mc_{g} such that 𝕋(X, K_{X}, g − 1) = ∅, i.e. no reduced divisor is the zero-locus of an effective thetacharacteristic. In the case g = 3 this is equivalent to ask if there is a smooth degree 4 plane curve X with 28 flexes of higher order, i.e. 28 lines L ⊂ ℙ^{2} meeting X at a unique point. The total weight of all flexes of a smooth plane quartic is 24, because its Hessian determinant has degree 6. Thus there is no such X for g = 3. For any X we have #𝕋̃(X, K_{X}, g − 1) ≥ 2^{g−1}(2^{g} − 1) and either #𝕋̃(X, K_{X}, g − 1) = 2^{g−1}(2^{g} − 1) (case h^{0}(G) ≤ 1 for all theta-characteristic G of X) or dim 𝕋̃(X, K_{X}, g − 1) > 0.

In any case, taking subsets of a theta-characteristic, we obtain immediately the following

Proposition 4.3. Fix an integer x such that g/2 ≤ x ≤ g − 1. Then the Terracini locus 𝕋̃(X, K_{X}, x) is non empty.

Of course when X has a positive dimensional theta-characteristic, then also 𝕋(X, K_{X}, x) is infinite for all x between g/2 and g − 1.

Remark 4.4. Fix an integer x such that g/2 ≤ x ≤ g − 2. One can try to extend elements of 𝕋(X, K_{X}, x) to elements 𝕋(X, K_{X}, x + 1), with the addition of suitable points. Respect to this, we can observe:

(a) Assume that there exists S ∈ 𝕋(X, K_{X}, x) with dim⟨2S⟩ = g − 2, i.e. H := ⟨2S⟩ is a hyperplane. The point p ∈ X \ S has the property that S ∪ {p} lies in 𝕋(X, K_{X}, x + 1) if and only if p ∈ H and H is tangent to X at p. There is Z ∈ 𝕋̃(K_{X}, x + 1) containing S if and only if either H is tangent to X at a point of X \ S or there is o ∈ S such that the scheme H ∩ X contains o with multiplicity at least 4.

(b) Assume that there exists S ∈ 𝕋(X, K_{X}, x) with dim⟨2S⟩ = g−3. The differential of the rational map ϕ : X\X∩⟨2S⟩ → ℙ^{1} induced by the linear projection from ⟨2S⟩ shows that there are only finitely many p ∈ X\S such that S∪{p} ∈ 𝕋(X, K_{X}, x+1). There are at least 2 such points p, because ϕ extends to a morphism ψ : X → ℙ^{1} by the smoothness of X and X has at least 2 ramification points, because g > 0.

(c) Assume that there exists S ∈ 𝕋(X, K_{X}, x) with dim⟨2S⟩ = g − 4 − 2k − ϵ, with k non-negative integer and ϵ ∈ {0, 1}. Then S ∪ {p_{1}, …, p_{k+1}} lies in 𝕋(X, K_{X}, x + k + 1) for a general (p_{1}, …, p_{k+1}) ∈ X^{k+1}.

Let us now consider the case x < g/2.

Proposition 4.5. Assume x < (g − 1)/2, and assume that 𝕋(X, K_{X}, x) is non empty. Then

dim(𝕋(X, K_{X}, x + 1)) ≥ 1 + dim(𝕋(X, K_{X}, x)).

Proof. Pick S ∈ 𝕋(X, K_{X}, x) and p general in X. Since 2x < g then by Proposition 4.1 we have h^{0}(2S) ≥ 2. It follows h^{0}(2(S ∪ {p})) ≥ h^{0}(2S) ≥ 2. When x + 1 < g/2 , the last inequality is sufficient to conclude that S ∪ {p} sits in T (X, K_{X}, x + 1), by Proposition 4.1 again, thus the inequality on the dimensions holds. Assume x = (g − 2)/2. Since h^{0}(2(S ∪ {p})) ≥ 2, then by Riemann-Roch

h^{1}(2(S ∪ \{p\})) ≥ g − 1 − 2(\frac{g − 2}{2} + 1) + 2 = 1.

Thus 2(S ∪ {p}) is special, hence it belongs to T (X, K_{X}, x + 1), by Proposition 4.1, and we conclude as before. □

Corollary 4.6. Fix the minimal integer x such that the Terracini locus 𝕋(X, K_{X}, x) is non empty. Then for all y with x ≤ y ≤ g − 1 the Terracini locus 𝕋(X, K_{X}, y) is non-empty.

We saw in the previous section that 𝕋(X, K_{X}, 1) is non-empty if and only if X is hyperelliptic. The last 2 propositions of this section link the gonality k of X with the minimal x such that 𝕋(X, K_{X}, x) ≠ ∅.

Proposition 4.7. If 𝕋(X, K_{X}, x) ≠ ∅, then X has a linear series g_{2x}^{1}. For the converse, if X has a linear series g_{k}^{1} with k < g/2, then dim(𝕋(X, K_{X}, k)) ≥ 1.

Proof. Recall that we are assuming g ≥ 3 and X not hyperelliptic. The first assertion is trivial if x ≥ g/2, for every X satisfies 𝕋(X, K_{X}, x) ≠ ∅ and every X has a linear series g_{2x}^{1}. When x < g/2, the first assertion follows immediately by Proposition 4.1. For the second assertion, consider the degree k map f : X → ℙ^{1} associated to the linear series g_{k}^{1}. For any p ∈ ℙ^{1} let D_{p} := f^{−1}(p) denote the associated degree k divisor. Note that h^{0}(2D_{p}) ≥ h^{0}(O_{ℙ^{1}}(2p)) = 3. Then, as in Proposition 4.1, a general D_{p} belongs to 𝕋̃(X, K_{X}, k). Since we work in characteristic 0, then a general D_{p} is formed by k distinct points, hence it belongs to 𝕋(X, K_{X}, k). □

Before we can refine the previous proposition, let us see what happens for trigonal curves.

Example 4.8. Let X be a smooth trigonal curve of genus g ≥ 4, canonically embedded in ℙ^{g−1}, and let f : X → ℙ^{1} be the degree 3 morphism associated to the g_{3}^{1}. By the Castelnuovo’s inequality [8], f is unique if g ≥ 5. Let Σ ⊂ X denote the set of all ramification points of the map and let Σ′ ⊆ Σ denote the set of all p ∈ Σ which belong to fibers of cardinality 2. The points p ∈ Σ \ Σ′ are called the total ramification points of f, because the fiber containing p is supported at p. Σ′ = Σ if X is a general trigonal curve of genus g, but there are trigonal curves in which the equality fails, and also trigonal curves with Σ′ = ∅, e.g. the degree 3 Galois coverings of ℙ^{1}. Take p ∈ Σ′ and consider the point q_{p} ≠ p in the fiber of f through p. We claim that S = {p, q_{p}} belongs to 𝕋(X, K_{X}, 2), so that 𝕋(X, K_{X}, 2) is non-empty. To prove the claim, consider that h^{0}(2p + q_{p}) = 2. Then h^{0}(2p + 2q_{p}) = h^{0}(2S) ≥ 2, which proves the claim when g > 4, by Proposition 4.1. For g = 4, since h^{0}(2S) ≥ 2, the divisor 2S is special by Riemann-Roch, and the claim follows again by Proposition 4.1.

Definition 4.9. Given a linear series G on X which is a g_{d}^{1}, G is tamely ramified if there is a non-reduced divisor D in G in which all the points appear with coefficient ≤ 2.

Proposition 4.10. Let X be a smooth curve of genus g ≥ 4, canonically embedded in ℙ^{g−1}. Assume that X has a base point free pencil R ∈ Pic^{k}(X) such that h^{1}(R^{⊗2}) > 0, and let f : X → ℙ^{1} be the degree k morphism induced by |R|. Then 𝕋̃(X, K_{X}, k − 1) ≠ ∅. Assume that the f is tamely ramified. Then 𝕋(X, K_{X}, k − 1) ≠ ∅.

Proof. If k − 1 ≥ g/2 the claim follows from Proposition 4.3.

Assume k − 1 < g/2. Since g > 0 and ℙ^{1} is algebraically simply connected, there is a divisor A = m_{1}q_{1} + ⋯ + m_{s}q_{s} ∈ |R| with m_{1} ≥ 2. Since 2A is a special divisor, A ∈ 𝕋̃(X, K_{X}, k −1). Now assume that f is tamely ramified. Thus all A ∈ |R| have profiles with multiplicities 1 or 2. Since g > 0, A = 2p_{1} +⋯+2p_{i} +p_{i+1} +⋯+p_{j} ∈ |R| with p_{1}, …, p_{j} distinct points and i ≥ 1. Note that i + j = k, so that j < k. Then S = {p_{1}, …, p_{j}} sits in 𝕋(X, K_{X}, j). Namely j ≤ k − 1 < g/2 and h^{0}(R) ≥ 2 implies h^{0}(2S) ≥ 2. Then 𝕋(X, K_{X}, j) ≠ ∅, and the claim follows from Corollary 4.6. □

5. Embedded curves

In this section we consider reduced and locally complete intersection curves X embedded in a projective space X ⊂ ℙ^{r}.

Notation 5.1. We denote with N_{X} the normal bundle of X. For brevity, we will denote with 𝕋(X, x) the Terracini locus of X with respect to the (non-necessarily complete) linear series of hyperplane divisors. For all integers g ≥ 0, r ≥ 3 and d ≥ r + g, we will denote with H(d, g, r) the set of all smooth and non-degenerate curves X ⊂ ℙ^{r} of degree d and genus g such that h^{1}(O_{X}(1)) = 0. Fix X ∈ H(d, g, r). Since X is smooth, N_{X} is a quotient of O_{X}(1)^{⊕r} by the Euler’s sequence. Thus h^{1}(N_{X}) = 0. Therefore H(d, g, r) is smooth, and

dim H(d, g, r) = (r + 1)d + (r − 3)(1 − g).

Since all non-special line bundles of degree d on a curve of genus g have the same number of sections, and all smooth genus g curves are parametrized by an irreducible variety, H(d, g, r) is irreducible. See [4]. An example of J. Harris, generalized by L. Ein in [4], shows that both claims on the irreducibility and the dimension fail, for large r, if we drop the assumption on the vanishing of h^{1}(O_{X}(1)). On the other hand, we will work mainly with curves in ℙ^{3}, and in this case we can drop the assumption by [3].

It is immediate, by Bezout formula, that 𝕋(X, x) is empty if x > d/2. So we analyze the case 2x ≤ d.

Remark 5.2. Fix x with 2x < r and let X ⊂ ℙ^{r} be a non-degenerate irreducible curve such that 𝕋(X, x) ≠ ∅. Then also 𝕋(X, x + 1) ≠ ∅. Indeed if S ∈ 𝕋(X, x), then the scheme (2S, X) lies in a linear system of hyperplanes of dimension at least r − 2x + 1 ≥ 2. Thus, for p ∈ X general, the scheme (2(S ∪ {p}), X) lies in a linear system of hyperplanes of dimension at least r − 2x − 1 ≥ 0.

Notation 5.3. Take X ⊂ ℙ^{r}. An arrow in ℙ^{r} is a non-reduced scheme of length 2. The set of all arrows in ℙ^{r} supported at p is closed in the Hilbert scheme, and it has dimension r − 1. Thus the set of all arrows in ℙ^{r} has dimension 2r − 1. Note that an arrow w with support p determines a line r_{w} through p. If p ∈ X and X is smooth at p, then r_{w} is tangent to X exactly when X contains the arrow w.

In the study of Terracini loci (even not on curves) the paper [2] is very useful and we explain it in the following remark used in the proof of Proposition 5.6.

Remark 5.4. Fix an integral projective curve X in ℙ^{r}, and let W be the image of the restriction map H^{0}(O_{ℙ^{r}}(1)) → H^{0}(O_{X}(1)). For any zero-dimensional scheme Z ⊂ X set W (−Z) := H^{0}(I_{Z}(1)) ∩ W. Fix integers s > 0 and e_{i} > 0, i = 1, …, s.

Let Z be a general subscheme of X_{reg} with s connected components of length e_{1}, …, e_{s}. Then we claim that the main result of [2] and its proof imply that:

dim W (−Z) = max{0, dim W − e_{1} − ⋯ − e_{s}}.

Namely the proof there yields that a general scheme in X (hence obviously curvilinear) which sits in no hypersurfaces of degree d imposes independent conditions to the linear system of hypersurfaces of degree d. Notice that the scheme Z is a Cartier divisor of X, and the integer dim W (−Z) is the codimension of the linear space ⟨Z⟩ in ℙ^{r}.

For rational curves we can easily show the following.

Proposition 5.5. Fix integers d ≥ r ≥ 3 and x with r ≤ 2x ≤ d. Then there is a smooth and non-degenerate rational curve X ⊂ ℙ^{r} such that 𝕋(X, x) ≠ ∅.

Proof. Fix a hyperplane H ⊂ ℙ^{r}. Let Z ⊂ H be a general union of x arrows and let Z′ be a general set of d − 2x points. By [9, Theorem 1.6] there is a smooth and non-degenerate rational curve X ⊂ ℙ^{r} such that (Z ∪ Z′) = X ∩ H. Let S ⊂ H be the reduction of Z. Since X is smooth and 2x ≥ r, then S ∈ 𝕋(X, x). □

Indeed, we have a characterization of rational normal curves in terms of Terracini loci.

Proposition 5.6. Let r ≥ 3 be an odd integer and X ⊂ ℙ^{r} a smooth, connected and non-degenerate curve. Then X is a rational normal curve if and only if 𝕋(X, (r + 1)/2)) = ∅.

Proof. Set x := (r + 1)/2. The “ if ” part is true, because if X is a rational normal curve each zero-dimensional scheme Z ⊂ X of degree r + 1 is linearly independent. Now assume 𝕋(X, x) = ∅. Set d := deg(X) and fix a general S ⊂ X of cardinality x−1, say S = {p_{1}, …, p_{x−1}}. Let V be the linear span of the double scheme (2S, X). For all positive integers a_{1}, …, a_{x−1} set V (a_{1}, …, a_{x−1}) := ⟨a_{1}p_{1} +⋯+a_{x−1}p_{x−1}⟩. Note that V = V (2, …, 2). Since S is general in X, Remark 5.4 applied to the curve X gives

dim V (a_{1}, …, a_{x−1}) = min{r, x − 2 + ∑_{i=1}^{x−1}(a_{i} − 1)}.

Thus dim V = r − 2. Since dim V = r − 2 and dim V (a_{1}, …, a_{x−1}) = r − 1 if there is j ∈ {1, …, x − 1} such that a_{j} = 3 and a_{i} = 2 for all i ≠ j, the scheme V ∩ X contains each p_{i} with multiplicity 2. First assume (V ∩ X)_{red} ≠ S and take o ∈ (V ∩ X)_{red} \ S. Since dim⟨2o ∪ V ⟩ ≤ dim V + 1, S ∪ {o} ∈ 𝕋(X, x), a contradiction. Thus (V ∩ X)_{red} = S. Since we proved that each p_{i} appear with multiplicy 2 in the scheme-theoretic intersection V ∩ X, we have V ∩ X = (2S, X), which has degree 2x − 2 = r − 1. Let u denote the linear projection from V to ℙ^{1}. Since X is smooth, u_{|X\S} extends to a morphism u′ : X → ℙ^{1}, and the degree of u′ is d − r + 1. The assumption 𝕋(X, x) = ∅ implies that u′ has no ramification point, except possibly at the points of S. Fix a ∈ S, say a = p_{1}. The point p_{1} is a ramification point of u′ only if V (4, 2, …, 2) is a hyperplane. This is false, because dim V (4, 2, …, 2) = r. Hence u′ has no ramification points. This is possible only if d = r, hence X is a rational normal curve. □

Since in the rest of the paper we will often argue by induction, taking the smoothing of nodal, reducible curves, we need some preliminary results on normal bundles of reducible curves.

Remark 5.7. Let X ⊂ ℙ^{r} be a reduced curve with only locally complete intersection singularities. N_{X} is a vector bundle of rank (r − 1) on X and

deg(N_{X}) = (r + 1) deg(X) + (r − 1)(1 − p_{a}(X)).

There is a map ϕ : T ℙ_{|X}^{r} → N_{X} which is surjective outside Sing(X). Consider the restriction to X of the Euler’s sequence of T ℙ^{r}:

(1)  0 → O_{X} → O_{X}(1)^{⊕(r+1)} → T ℙ_{|X}^{r} → 0

Now assume h^{1}(O_{X}(1)) = 0. Since dim X = 1, then h^{2}(O_{X}) = 0. Thus (1) gives h^{1}(T ℙ_{|X}^{r}) = 0. The map ϕ induces the following exact sequence of coherent sheaves on X:

(2)  0 → Ker(ϕ) → T ℙ_{|X}^{r} → Im(ϕ) → 0

Since dim X = 1, h^{2}(Ker(ϕ)) = 0. Thus (2) gives h^{1}(Im(ϕ)) = 0. If X is smooth, then N_{X} = Im(ϕ) and hence h^{1}(N_{X}) = 0. Now assume X singular. Consider the exact sequence

(3)  0 → Im(ϕ) → N_{X} → N_{X}/Im(ϕ) → 0

Since N_{X}/Im(ϕ) is supported by the finite set Sing(X), then h^{1}(N_{X}/Im(ϕ)) = 0. Thus (3) gives h^{1}(N_{X}) = 0 even if the non-special curve is singular. If X is smooth and rational, then h^{1}(O_{X}) = 0. As above we obtain h^{1}(T ℙ^{r}(−1)_{|X}) = 0 and h^{1}(N_{X}(−1)) = 0.

In particular we study the case of curves in ℙ^{3}.

Lemma 5.8. Let C ⊂ ℙ^{3} be a reduced curve and let C′ ⊂ ℙ^{3} be a smooth conic that meets C at i ≤ 3 points p_{1}, …, p_{i} ∈ C_{reg} with C ∪ C′ nodal at each p_{i}. Let H be the plane spanned by C′. If i > 1, assume also that T_{p_{1}}C ⊈ H, T_{p_{2}}C ⊈ H and T_{p_{1}}C ∩ T_{p_{2}}C = ∅.

1) If i = 1 then N_{C∪C′|C′} is the direct sum of two line bundles, one of degree 3 and one of degree 4.

2) If i = 2 then N_{C∪C′|C′} is the direct sum of two line bundles, both of degree 4.

3) If i = 3, then N_{C∪C′|C′} is the direct sum of two line bundles, one of degree 4 and one of degree 5. In any case h^{1}(N_{C∪C′|C′}(−1)) = 0.

Proof. Note that N_{C′} ≅ O_{C′}(2) ⊕ O_{C′}(1), i.e. N_{C′} is the direct sum of a degree 4 line bundle and a degree 2 line bundle. We prove 2) first. Set Y := T_{p_{1}}C ∪ T_{p_{2}}C ∪ C′. By [7, Corollary 3.2] the bundle N_{C∪C′|C′} is obtained from N_{C′} by making two positive elementary transformations (in the sense of [7, §2]), which depend uniquely on C′, p_{1}, p_{2}, and the lines T_{p_{1}}C, T_{p_{2}}C. Thus N_{C∪C′|C′} ≃ N_{Y |C′}. Since Y is connected and deg(Y) = 4, h^{0}(O_{Y} (2)) < 10. Thus h^{0}(I_{Y} (2)) > 0. Since T_{p_{1}}C ∩ T_{p_{2}}C = ∅, Y is contained neither in a reducible quadric, nor in a quadric cone. Thus Y lies in a smooth quadric Q with, say, T_{p_{1}}C ∈ |O_{Q}(1, 0)|. Then also T_{p_{2}}C, which does not meet T_{p_{1}}C, lies in |O_{Q}(1, 0)|. Obviously C′ ∈ |O_{Q}(1, 1)|, and we conclude that Y ∈ |O_{Q}(3, 1)|. Thus N_{Y} is an extension of O_{Y} (2) by the restriction to Y of the line bundle O_{Q}(3, 1).

Hence N_{Y |C′} is an extension of two line bundles of degree 4.

1) By [7, Corollary 3.2] the bundle N_{C∪C′|C′} is obtained from N_{C′} by making one general positive elementary transformation, and as above one general positive elementary transformation of N_{C∪C′|C′} is a direct sum of a line bundle of degree 4 and a line bundle of degree 3.

3) Here N_{C∪C′|C′} is obtained from N_{C′} by making 3 positive elementary transformations, one at p_{1}, one at p_{2}, and one at p_{3}. Hence N_{C∪C′|C′} is obtained from one positive elementary transformation on the vector bundle E on C′ obtained by making positive elementary transformations only at p_{1} and p_{2}. By 1) E is isomorphic to O_{C′}(2) ⊕ O_{C′}(2). Since any vector bundle on C′ splits in a direct sum of line bundles, the claim follows. □

Remark 5.9. We will use repeatedly a result of J. Kleppe, who proved that the functor of deformations inside ℙ^{r} of a (possibly reducible) curve T ⊂ ℙ^{r} with fixed hyperplane section H ∩ T has tangent space H^{0}(T, N_{T} (−1)) and obstruction space H^{1}(N_{T} (−1)). See e.g. [9, Th. 1.5].

Remark 5.10. Let C and D be locally complete intersection space curves. Assume S := C∩D = C_{reg}∩D_{reg} and that at each p ∈ S the tangent lines of C and D at p are different. Set Y := C ∪D. Call N_{C}^{+} (resp. N_{D}^{+}) the rank 2 vector bundle on C (resp. D) obtained from N_{C} (resp. N_{D}) making a positive elementary transformation in the direction of T_{p}D (resp. T_{p}C) at all p ∈ S. We have deg(N_{C}^{+}) = deg(N_{C}) + #S and deg(N_{D}^{+}) = deg(N_{D}) + #S. By [7, Cor. 3.2] we have N_{C}^{+} ≅ N_{Y |D} and N_{D}^{+} ≅ N_{Y |D}. Thus the Mayer-Vietoris exact sequence of the vector bundle N_{Y} gives the following exact sequence

(4)  0 → N_{C∪D} → N_{C}^{+} ⊕ N_{D}^{+} → N_{Y |S} → 0.

Thus to prove that h^{1}(N_{Y} (−1)) = 0 it is sufficient to prove that h^{1}(N_{C}^{+}(−1)) = h^{1}(N_{D}^{+}(−1)) = 0 and that the restriction map H^{0}(D, N_{D}^{+}(−1)) → H^{0}(N_{Y |S}) is surjective.

A similar statement holds for the proof that h^{1}(N_{C∪D}) = 0.

Lemma 5.11. Let H ⊂ ℙ^{3} be a plane and let C′ ⊂ H be a smooth conic. Fix 3 distinct points p_{1}, p_{2}, p_{3} of C′ and lines L_{1}, L_{2}, L_{3} such that H ∩ L_{i} = {p_{i}} for all i. Let E (resp. F) be the vector bundle obtained from N_{C′} making positive elementary transformations at p_{1} and p_{2} (resp. p_{1}, p_{2} and p_{3}) with respect to L_{1} and L_{2} (resp. L_{1}, L_{2} and L_{3}). Then E is a direct sum of 2 line bundles of degree 4 and F is a direct sum of a line bundle of degree 5 and a line bundle of degree 4.

Proof. Set X := C′ ∪ L_{1} ∪ L_{2} and Y := X ∪ L_{3}. Since H ∩ L_{i} = {p_{i}}, then L_{i} is transversal to H. Thus X and Y are nodal at p_{1}, p_{2} and p_{3}. Note that E ≅ N_{X|C′} and F ≅ N_{Y |C′}.

First assume L_{1} ∩ L_{2} ≠ ∅. Thus X is nodal with 3 nodes and arithmetic genus 1. Call M the plane containing L_{1} ∪ L_{2}. To prove that E ≅ O_{C′}(2) ⊕ O_{C′}(2) it is sufficient to prove that X is the complete intersection of M ∪ H and a quadric. This is true, because h^{0}(O_{X}(2)) = 8, h^{0}(O_{ℙ^{3}}(2)) = 10 and hence h^{0}(I_{X}(2)) ≥ 2.

Now assume L_{1} ∩ L_{2} = ∅. In this case X is contained neither in a reducible quadric nor in a quadric cone. Since h^{0}(O_{X}(2)) = 9, X is contained in a smooth quadric, Q. Call |O_{Q}(1, 0)| the ruling of Q containing L_{1}, and hence also containing L_{2}. Since C′ ∈ |O_{Q}(1, 1)|, then X ∈ |O_{Q}(3, 1)|. Since N_{Q} ≅ O_{Q}(2), we have an exact sequence

(5)  0 → N_{X,Q} → N_{X} → N_{X}(2) → 0

We have N_{X.Q} ≅ O_{X}(3, 1) so that its restriction to C′ has degree 4. Since (5) is an exact sequence of vector bundles, its restriction to C′ is an exact sequence of vector bundles on C′ ≅ ℙ^{1} in which the leftmost and the rightmost terms are line bundles of degree 4. Since C′ ≅ ℙ^{1}, E is a direct sum of two line bundles of degree 4.

The bundle F is obtained from E making a positive elementary transformation, and all rank 2 vector bundles on ℙ^{1} split. Hence F is a direct sum of a line bundle of degree 5 and a line bundle of degree 4. □

Now we are ready to prove some non-emptiness results in ℙ^{3}. Recall that H(d, g, 3) denotes the set of smooth and non-degenerate curves space curves X of degree d and genus g, such that h^{1}(O_{X}(1)) = 0.

Theorem 5.12. Fix integers g ≥ 0, d ≥ g + 3 and x such that 6 ≤ 2x ≤ d. Then there is X ∈ H(d, g, 3) such that 𝕋(X, x) ≠ ∅ and h^{1}(N_{X}(−1)) = 0.

Proof. Fix a plane H ⊂ ℙ^{3}. We will find X ∈ H(d, g, 3) such that X is tangent to H at x points of H spanning H. We first dispose of the case x = ⌊d/2⌋ and d = g + 3 in steps (a) and (b), leaving the case x = ⌊d/2⌋ and d > g + 3 to step (c) and the case 3 ≤ x < ⌊d/2⌋ to step (d).

(a) Assume d even and x = d/2.

(a1) Assume d = 6. By Proposition 5.6 𝕋(C, 2) ≠ ∅ for each smooth curve C of genus 1 and degree 4. Hence by a change of coordinates we find a smooth curve C ⊂ ℙ^{3} of degree 4 and genus 1 which is tangent to H at 2 distinct points, say q_{1} and q_{2}. Since C is the complete intersection of 2 quadric surfaces, the normal bundle of C splits as N_{C} ≅ O_{C}(2) ⊕ O_{C}(2). Since C has genus 1, we get h^{1}(N_{C}(−1)) = 0. Fix a general q_{3} ∈ H and let M ⊂ ℙ^{3} be a general plane containing q_{3}. Since q_{3} and M are general, we may assume that M is transversal to C, C ∩ M ∩ H = ∅, and C ∩ M spans M. Fix three distinct points {p_{1}, p_{2}, p_{3}} ∈ C ∩ M which span M. There is a smooth conic C′ containing {p_{1}, p_{2}, p_{3}, q_{3}}, tangent in q_{3} to the line H ∩ M but not containing the fourth point p_{4} of C ∩ M. Indeed, q_{3} is general in H ∩ M, while there are only at most two conics passing through p_{1}, p_{2}, p_{3}, p_{4} and tangent to H ∩ M. Since M is transversal to C, the curve Y := C ∪ C′ is nodal. By Lemma 5.8 the vector bundle N_{C′}^{+}(−1) is the direct sum of a line bundle of degree 2 and a line bundle of degree 3. Since line bundles of degree 2, 3 separate any set of three points in the smooth conic C′, the restriction map H^{0}(C′, N_{C′}^{+}(−1)) → H^{0}(N_{C′}^{+}(−1)_{|{p_{1},p_{2},p_{3}}}) is surjective. Remark 5.10 gives h^{1}(N_{Y} (−1)) = 0, so that Y is smoothable by [7, Theorem 4.1]. By semicontinuity, a general member X_{0} of a smoothing family of Y satisfies h^{1}(N_{X_{0}}(−1)) = 0. By construction 𝕋(Y, 3) ≠ ∅. Yet, in order to conclude, we need more: we need to smooth Y in a family of space curves whose elements Y_{λ} satisfies 𝕋(Y_{λ}, 3) ≠ ∅. In other words, we need:

Claim 1: There are an affine smooth and connected curve Δ, o ∈ Δ and a flat family {Y_{λ}}_{λ∈Δ} of space curves such that Y_{0} = Y, the general element of the family is smooth, and 𝕋(Y_{λ}, 3) ≠ ∅ for all λ ∈ Δ.

Proof of Claim 1: Set Z′ := (2q_{1}, C) ∪ (2q_{2}, C), Z′′ := (2q_{3}, C′) and Z := Z′ ∪ Z′′. Note that Z ∩ C = Z′ and Z ∩ C′ = Z′′. Since q_{1}, q_{2} and q_{3} are smooth points of Y, Z is a degree 6 Cartier divisor of Y. Thus N_{Y} (−Z) is a rank 2 vector bundle on Y with deg(N_{Y} (−Z)) = deg(N_{Y})−12. The vector space H^{0}(N_{Y} (−Z)) is the tangent space to the functor of deformations of Y inside ℙ^{3} in families of curves containing Z, while H^{1}(N_{Y} (−Z)) is an obstruction space of this functor [9, Th. 1.5]. To prove Claim 1 it is sufficient to find a smoothing family {Y_{λ}}_{λ∈Δ} of Y = Y_{o} such that Z ⊂ Y_{λ} for all λ ∈ Δ. Note that N_{C′}(−Z) = N_{C′}(−Z′′) is a direct sum of degree 2 line bundles on C′ ≅ ℙ^{1}. Thus h^{1}(N_{C′}(−Z−p_{i})) = 0 for all i. In particular (with the terminology of [7]) h^{1}(N_{C}(−Z)^{−}) = 0. Since N_{C} ≅ O_{C}(2) ⊕ O_{C}(2), and deg(Z ∩ C) = 4, h^{1}(N_{C}(−Z − p_{1} − p_{2} − p_{3})) = 0. Thus h^{1}(F) = 0 for every vector bundle F on C obtained from N_{C}(−Z − p_{1} − p_{2} − p_{3}) making finitely many positive elementary transformations. Consider the Mayer-Vietoris exact sequence

(6) 0 → N_{Y} (−Z) → N_{Y} (−Z)_{|C} ⊕ N_{Y} (−Z)_{|C′} → N_{Y} (−Z)_{|{p_{1},p_{2},p_{3}}} → 0

Since N_{Y} (−Z)_{|C}(−p_{1} − p_{2} − p_{3}) is obtained from N_{C}(−Z − p_{1} − p_{2} − p_{3}) by making positive elementary transformations, h^{1}(N_{Y} (−Z)_{|C}(−p_{1} − p_{2} − p_{3})) = 0 (here we consider p_{1}, p_{2}, p_{3} as points of the smooth curve C). The sequence

0 → N_{Y} (−Z)_{|C}(−p_{1} − p_{2} − p_{3}) → N_{Y} (−Z)_{|C} → N_{Y} (−Z)_{|{p_{1},p_{2},p_{3}}} → 0

shows that the map ϕ : H^{0}(N_{Y} (−Z)_{|C}) → H^{0}(N_{Y} (−Z)_{|{p_{1},p_{2},p_{3}}}) is surjective, and h^{1}(N_{Y} (−Z)_{|C}) = 0. We also know that N_{Y} (−Z)_{|C′} is a direct sum of 2 line bundles of non-negative degree on the smooth conic C′, so that h^{1}(N_{Y} (−Z)_{|C′}) = 0. Hence the surjectivity of ϕ gives a fortiori, in sequence (6), that h^{1}(N_{Y} (−Z)) = 0. Thus we may apply the proof of [7, Th. 4.1] since the deformation functor of Y which maintains Z fixed is unobstructed. We obtain a family {Y_{λ}}_{λ∈Δ} as in the statement, in which the general element contains Z, hence it is tangent to H at three points. □

(a2) Assume d ≥ 8 and that the theorem is true for the triples (d′, g′, x′) such that x = d′/2, d′ = g′ + 3 and d′ ≤ d − 2. Take a solution C for (d′, g′, x′) = (d − 2, g − 2, x − 1) and use the proof of step (a1) with this C instead of an elliptic curve, as follows. By induction, there is a plane H which contains a set Z′ of x − 1 arrows in C. Then take a general plane M and three points o_{1}, o_{2}, o_{3} of M ∩ C. Fix a conic C′ ⊂ M passing through o_{1}, o_{2}, o_{3} and tangent to H at a point p. Since the base locus of the family of conics in M passing through o_{1}, o_{2}, o_{3} and tangent to H is exactly given by o_{1}, o_{2}, o_{3}, we may assume that C′ misses any other point of M ∩ C. Set Y = C ∪ C′, Z′′ = (2p, C′), and Z = Z′ ∪ Z′′.

Claim 2: h^{1}(N_{Y} (−Z)) = 0.

Proof of Claim 2: The bundle N_{C′}^{+}(−Z′′ − o_{1} − o_{2} − o_{3}) is a direct sum of a line bundle of degree 0 and a line bundle of degree −1. Thus h^{1}(C′, N_{Y |C′}(−Z − o_{1} −o_{2} −o_{3})) = 0. As in Claim 1, just working on C′ instead of C and with o_{1}, o_{2}, o_{3} instead of p_{1}, p_{2}, p_{3}, this implies that the restriction map

H^{0}(N_{Y} (−Z)_{|C′}) → H^{0}(N_{Y} (−Z)_{|{o_{1},o_{2},o_{3}}})

is surjective. Then the analogue of (6) gives h^{1}(N_{Y} (−Z)) = 0. □

So, we can continue the induction by taking a general element X in a family {Y_{λ}}_{λ∈Δ} giving a smoothing of Y and fixing Z. Notice that at any step X satisfies h^{1}(N_{X}(−Z)) = 0 by semicontinuity, since h^{1}(N_{Y} (−Z)) = 0.

(b) Assume d odd and x = (d − 1)/2. Since x ≥ 3, d ≥ 7. By Proposition 5.6 each C ∈ H(5, 2, 3) has 𝕋(C, 2) ≠ ∅. To adapt the proof of step (a) we first find C ∈ H(5, 2, 3) with h^{1}(N_{C}(−1)) = 0. Let Y′ ⊂ ℙ^{3} be a smooth rational cubic. Take a plane M ⊂ ℙ^{3} transversal to Y′ and let C′ ⊂ M be a smooth conic containing Y′ ∩ M. Set T := Y′ ∪ C′. Lemma 5.8 gives h^{1}(N_{T} (−1)) = 0. By [7, Theorem 4.1] and Lemma 5.8, T is smoothable in a family preserving the scheme-theoretic intersection (Y′ ∪ C′) ∩ H as in step (a1). By semicontinuity, a general element C of the family has h^{1}(N_{C}(−1)) = 0. Next we proceed by induction as in step (a), adding suitable conics C′, constructed as in step (a), to curves C which are solutions for the triple (d′, g′, x′) = (d − 2, g − 2, x − 1), and taking a general smoothing X of C ∪ C′ . To see that we get h^{1}(N_{X}(−1)) = 0 we apply Lemma 5.11 to C′.

(c) Assume d > g + 3. If d − g − 3 is even, start with a curve of genus g and degree d′ = g + 3, constructed as in step (a2) when d′ is even, or constructed as in step (b) if d′ is odd. Continue for (d − g − 3)/2 steps, by adding to the previously constructed curve C a smooth conic C′, tangent to H, with #(C′ ∩ C) = 1 and C′ ∪C nodal. By Lemma 5.8 we always get h^{1}(N_{C∪C′}(−1)) = 0. After (d−g −3)/2 steps we get the claimed curve. Assume that d − g − 3 is odd. When g = 0, start with a rational quartic, by using Proposition 5.6. When g ≥ 1, start with a curve C of genus g − 1 and degree d′ = g + 2, constructed as in step (a2) when d′ is even, or constructed as in step (b) if d′ is odd. In the first step add to C a smooth conic C′, tangent to H, which meets the curve C at two points whose tangent lines t_{1}, t_{2} to C are disjoint and different from the tangent lines to C′. We may take such a conic C′ in a plane which does not contain t_{1}, t_{2}. Then continue for (d − g − 4)/2 steps, by adding to the previously constructed curve C a smooth conic C′, tangent to H, which intersects C at a unique point and with C ∪C′ nodal. In any case, the assumptions of Lemma 5.8 hold for C ∪ C′. Thus by Lemma 5.8 we always get h^{1}(N_{C∪C′}(−1)) = 0, and we can continue the induction, by taking a smoothing of C ∪ C′ which preserves the intersection with H.

(d) Assume 3 ≤ x ≤ ⌊d/2⌋. We start with some curve Y such that 𝕋(Y, 2) ≠ ∅ and h^{1}(N_{Y} (−1)) = 0. We take Y of genus 1 and degree 4 if d is even or genus 2 and degree 5 if g is odd. Then we continue as in steps (a), (b) and (c) above, except that in ⌊d/2⌋ − x steps we add a smooth conic C′ not tangent to H. □

For space curves X ⊂ ℙ^{3} with h^{1}(O_{X}(1)) ≠ 0 we prove the following result.

Theorem 5.13. Fix integers x ≥ 3, g ≥ 0 and d such that g ≥ 2x + 2. Write g = 2x + 1 + 5s − q, 0 ≤ q ≤ 4, and assume d ≥ 2x + 4 + 3s. Then there is a smooth and connected curve X ⊂ ℙ^{3} such that deg(X) = d, p_{a}(X) = g, h^{1}(N_{X}) = 0, h^{1}(O_{X}(2)) = 0 and 𝕋(X, x) ≠ ∅.

Remark 5.14. In Theorem 5.13, for each fixed x we find g_{0} such that for all g ≥ g_{0} and all d ≥ \frac{3}{4}g + 3 there is a smooth curve X ⊂ ℙ^{3} of genus g and degree d with 𝕋(X, x) ≠ ∅. The same is true for a slowing increasing function x(g) of g. Note that for a fixed x and for g ≫ x these curves X cover a range of degrees and genera larger that the Brill-Noether range d ≥ \frac{3}{4}g + 3.

For the proof of Theorem 5.13, we need a series of preliminary lemmas.

Lemma 5.15. Fix an integer e ∈ {1, 2, 3, 4}. Let C ⊂ ℙ^{3} be an integral and nondegenerate curve of degree d. If e = 4 assume d ≥ 4. Take a union Y ⊂ ℙ^{3} of finitely many curves such that C ⊈ Y. Then there is a smooth conic C′ such that #(C ∩ C′) = e, C ∪ C′ is nodal at each point of C ∩ C′, and C′ ∩ Y = ∅.

Proof. Take a general plane M ⊂ ℙ^{3}. The plane M is transversal to C, Y ∩ M is finite and Y ∩C ∩M = ∅. By the trisecant lemma no 3 of the d points of C ∩M are collinear. Fix S ⊆ C ∩ M such #S = e. Since no 3 of the points of S are collinear, S is the scheme-theoretic base locus of the (5 − e)-dimensional linear space |I_{S}(2)|. A general element of |I_{S}(2)| is smooth. Thus there is a smooth C′ ∈ |I_{S}(2)| such that C′ ∩ Y = ∅ and C′ ∩ C = S. Since M is transversal to C and C′ ⊂ M, then C ∪ C′ is nodal at each point of S. □

Lemma 5.16. Let C ⊂ ℙ^{3} be an integral and non-degenerate curve of degree 4. For a general S ⊂ C of length 6 there is a rational normal curve T_{S} ⊂ ℙ^{3} such that S = C ∩ T_{S} and T_{S} ∪ C is nodal. Moreover, if p_{a}(C) = 1 then T_{S_{1}} ∩ T_{S_{2}} = ∅ for a general S_{1} × S_{2} ⊂ C × C such that both S_{1} and S_{2} have length 6.

Proof. Let U denote the set of all A ⊂ ℙ^{3} such that #A = 6 and A is in linear general position. For every A ∈ U there is a unique rational normal curve T_{A} containing A. Set U(C) := {S ∈ U | S ⊂ C}. Since C is integral and non-degenerate, U(C) is an integral quasi-projective variety of dimension 6. The set S is a general element of U(C). Since deg(C) ≠ 3, then C ≠ T_{S}. We need to prove that S is equal to C ∩ T_{S} (set-theoretically), that T ∪ C is nodal, and the last assertion, concerning a general S_{1} × S_{2} ⊂ C × C, when p_{a}(C) = 1. If p_{a}(C) = 1, the curve C is the complete intersection of 2 quadrics and (since it has at most one singular point and with embedding dimension 2) it is contained in a smooth quadric Q, say C ∈ |O_{Q}(2, 2)|. Fix a general S ⊂ C of length 6. By the generality of S, since h^{0}(O_{Q}(1, 2)) = h^{0}(O_{Q}(2, 1)) = 6, we get h^{0}(I_{S,Q}(2, 1)) = h^{0}(I_{S,Q}(1, 2)) = 0. Thus T_{S} ⊈ Q. Bezout theorem gives S = T_{S} ∩ Q as schemes. Thus T_{S} ∩ C = S and T_{S} ∪ C is nodal. If p_{a}(C) = 0 then h^{0}(I_{C}(2)) = 1 and the unique quadric surface Q containing C is smooth [6, Ex. V.2.9], with either C ∈ |O_{Q}(1, 3)| or C ∈ |O_{Q}(3, 1)|. We conclude as in the case p_{a}(C) = 1. Now we prove the last claim, for p_{a}(C) = 1. It is sufficient to find S_{1}, S_{2} ∈ U(C) such that T_{S_{1}} ∩ T_{S_{2}} = ∅. The pencil |I_{C}(2)| has only finitely many (i.e. 4) singular elements. Take smooth quadrics Q_{1}, Q_{2} ∈ |I_{S}(2)| such that Q_{1} ≠ Q_{2}. Note that C ∈ |O_{Q_{i}}(2, 2)|, i = 1, 2. Take a general T_{i} ∈ |O_{Q_{i}}(2, 1)|. Each T_{i} is a rational normal curve and deg(T_{i} ∩ C) = 6. Bertini Theorem gives that S_{i} := T_{i} ∩ C has cardinality 6. Since S_{i} ⊂ T_{i} and T_{i} is a rational normal curve, S_{i} is in linear general position. Thus T_{i} = T_{S_{i}}. Since T_{S_{i}} ≠ C, we have T_{S_{1}} ⊈ Q_{2} and T_{S_{2}} ⊈ Q_{1}. Since T_{i} ⊂ Q_{i}, i = 1, 2, we get T_{S_{1}} ∩Q_{2} = T_{S_{1}} ∩Q_{1} ∩Q_{2} = T_{S_{1}} ∩C = S_{1}. Since T_{S_{2}} ⊂ Q_{2}, then T_{S_{1}} ∩ T_{S_{2}} ⊆ T_{S_{1}} ∩ Q_{2} = S_{1}. Since S_{1} ∩ S_{2} = ∅ and T_{S_{2}} ∩ C = S_{2}, we get T_{S_{1}} ∩ T_{S_{2}} = ∅. □

Lemma 5.17. For q = 0, …, 4 and x ≥ 2 the space H(2x+4, 2x+1−q, 3) contains smooth curves X_{0} which satisfy the following conditions.

(1) 𝕋(X_{0}, x) is non-empty, and there exists a plane H which contains a union of x arrows Z ⊂ X_{0} with h^{1}(N_{X_{0}}(−Z)) = 0.

(2) For a general choice of s subsets S_{1}, …, S_{s} ⊂ X_{0} of cardinality 6, there are disjoint rational normal curves T_{1}, …, T_{s} such that for all i the union X_{0} ∪ T_{i} is nodal, and T_{i} meets X_{0} exactly at S_{i}.

Proof. We make induction on x ≥ 2. To deal with the case x = 2 (i.e. d = 8 and g = 5 − q) we start with a smooth elliptic quartic curve C. We know that H^{1}(N_{C}(−1)) = 0 and 𝕋(C, 2) is non-empty. Since 𝕋(C, 2) ≠ ∅, there is a subset Z ⊂ C of cardinality 2 such that the tangent lines to C at the points of Z lie in a plane H. By Lemma 5.16, for a general choice of s subsets S_{1}, …, S_{s} ⊂ C of cardinality 6 we find disjoint rational normal curves T_{1}, …, T_{s} such that C ∪ T_{i} is nodal and T_{i} ∩ C = S_{i} for all i. Take a general plane H′ ⊂ ℙ^{3}. Thus H′ is transversal to C and Z ∩ H′ = ∅. Fix e ∈ {0, 1, 2, 3} and W ⊂ H′∩C such that #W = 4−e. Take a general smooth conic C′ ⊂ H′ containing W. Since #W ≤ 4 and C′ is general, T_{i} ∩ C′ = ∅ for all i. Set Y := C ∪ C′. The normal bundle N_{Y} (−Z)_{|C} is obtained from N_{C}(−Z) by making positive elementary transformations, so that h^{1}(N_{Y} (−Z)_{|C}) = 0. Recall that 1 ≤ #W ≤ 4. For each o ∈ W the tangent line T_{o}C of C at o is transversal to H′. If #W = 4 Lemma 5.11 shows that the restriction map ρ : H^{0}(N_{Y} (−Z)_{|C′}) → H^{0}(N_{Y} (−Z)_{|W}) surjects, so that the exact sequence

(7)  0 → N_{Y} (−Z) → N_{Y} (−Z)_{|C} ⊕ N_{Y} (−Z)_{|C′} → N_{Y} (−Z)_{|W} → 0

proves that h^{1}(N_{Y} (−Z)) = 0. If #W ≤ 3, then the surjectivity of ρ only use that N_{C′} is a direct sum of 2 line bundles of degree ≥ 2 and that N_{Y} (−Z)_{|C′} is obtained from N_{C′} making positive elementary transformations. By semicontinuity, we can find a smoothing D of C ∪ C′ which preserves Z. Thus we get curves D in H(6, g, 3), for g = #W ∈ {1, 2, 3, 4}, such that 𝕋(D, 2) ≠ ∅, h^{1}(N_{D}(−Z)) = 0, and such that for general subsets S_{1}, …, S_{s} ⊂ X_{0} of cardinality 6 we have the disjoint rational normal curves T_{1}, …, T_{s}, as in the statement. Then consider a conic C′′ which meets D in one or in two general points. As above, in both cases Y′ = C′′ ∪ D satisfies h^{1}(N_{Y′}(−Z)) = 0, and we can find a general smoothing X_{0} of Y′ which preserves Z. X_{0} has degree 8, and for its genus g we can obtain any number between 1 and 5, because #W is any integer between 1 and 4. Moreover, h^{1}(N_{X_{0}}(−Z)) = 0, 𝕋(X_{0}, 2) ≠ ∅, and by semicontinuity X_{0} satisfies condition 2 of the statement. This concludes the case x = 2. Assume we constructed the required curve X ∈ H(2(x − 1) + 4, 2(x − 1) + 1 − q, 3). There exists a subscheme Z ⊂ X formed by x − 1 arrows which is contained in a plane H, and moreover h^{1}(N_{X}(−Z)) = 0. Take a general plane M which is transversal to both H and X, and misses Z. As in the proof of (a1) of Theorem 5.12, for p_{1}, p_{2}, p_{3} ∈ M ∩X there exists a smooth conic C′ passing through p_{1}, p_{2}, p_{3}, tangent to H at p_{4} ∉ X, which misses the remaining points of M ∩ X. Take Y = X ∪ C′, and define Z_{0} as the union of Z and the arrow at p_{4} tangent to C′. Notice that Z_{0} lies in H. As in the proof of Claim 1, the vanishing of H^{1}(N_{X}(−Z)) implies the surjectivity of the map H^{0}(N_{Y} (−Z_{0})_{|X}) → H^{0}(N_{Y} (−Z_{0})_{|{p_{1},p_{2},p_{3}}}), and the analogue of sequence (6) shows that h^{1}(N_{Y} (−Z_{0})) = 0. Thus there exists a smoothing X_{0} ∈ H(2x + 4, 2x + 1 − q, 3) of Y which preserves Z_{0}. The existence of Z_{0} ⊂ X_{0} implies 𝕋(X_{0}, x) ≠ ∅. By semicontinuity h^{1}(N_{X_{0}}(−Z_{0})) = 0, and X_{0} satisfies condition 2 of the statement. □

We notice that, in the previous lemma, the condition h^{1}(N_{X_{0}}(−Z)) = 0 implies h^{1}(N_{X_{0}}) = 0.

Lemma 5.18. Take a reduced curve C ⊂ ℙ^{3} such that h^{1}(O_{C}(2)) = 0. Let T ⊂ ℙ^{3} be a smooth rational curve which is not an irreducible component of C. Set f := deg(T) and e := deg(C ∩ T). If e ≤ 2f + 1, then h^{1}(O_{C∪T} (2)) = 0.

Proof. Consider the Mayer-Vietoris exact sequence

(8)  0 → O_{C∪T} (2) → O_{C}(2) ⊕ O_{T} (2) → O_{T ∩C}(2) → 0

By assumption h^{1}(O_{C}(2)) = 0. Obviously, h^{1}(O_{T} (2)) = 0. Since e ≤ 2f + 1, T ≅ ℙ^{1} and deg(O_{T} (2)) = 2f, the restriction map H^{0}(O_{T} (2)) → H^{0}(O_{C∩T} (2)) is surjective. Hence (8) gives h^{1}(O_{C∪T} (2)) = 0. □

Lemma 5.19. Let C ⊂ ℙ^{3} be an integral locally complete intersection curve and Z ⊂ C_{reg} a zero-dimensional scheme. Assume h^{1}(N_{C}(−Z)) = 0. Take S ⊂ C_{reg} such that #S = 6, S ∩ Z = ∅, S is in linearly general position and the only rational normal curve T containing S meets C only at S and with C ∪ T nodal at each point of S. Set Y := C ∪ T. Then h^{1}(N_{Y} (−Z)) = 0.

Proof. By assumption S is the scheme-theoretic intersection of C and T. Thus we have the following Mayer-Vietoris exact sequence

(9)  0 → N_{Y} (−Z) → N_{Y} (−Z)_{|C} ⊕ N_{Y} (−Z)_{|T} → N_{Y} (−Z)_{|S} → 0

Since Z ∩ T = ∅, N_{Y} (−Z)_{|T} ≅ N_{Y |T}. Since N_{Y} (−Z)_{|C} is obtained from N_{C}(−Z) making positive elementary transformations and h^{1}(N_{C}(−Z)) = 0, h^{1}(N_{Y} (−Z)_{|C}) = 0. Since N_{T} is a direct sum of 2 line bundles of degree 5, #S = 6 and N_{Y} (−Z)_{|T} is obtained from N_{T} making positive elementary transformations, h^{1}(N_{Y} (−Z)_{|T}) = 0 and the restriction map H^{0}(N_{Y} (−Z)_{|T}) → N_{Y} (−Z)_{|S} is surjective. Thus (9) gives h^{1}(N_{Y} (−Z)) = 0. □

Now we are ready for the proof of Theorem 5.13.

Proof of Theorem 5.13. Recall that g = 2x + 1 + 5s − q, where q ∈ {0, …, 4}. By assumption d ≥ 2x + 4 + 3s. We first dispose of the case d = 2x + 4 + 3s. The case s = 0 is covered by Lemma 5.17. Consider a curve X_{0} ∈ H(4x + 2, 2x + 1, 3) as in the statement of Lemma 5.17. Since 𝕋(X_{0}, x) is non-empty, we can take a set Z of x coplanar arrows in X_{0} supported at x points, with h^{1}(N_{X_{0}}(−Z)) = 0. Moreover, for a choice of s general subsets S_{1}, …, S_{s} ⊂ X_{0} of cardinality 6 there are disjoint rational normal curves T_{1}, …, T_{s} such that for all i the union X_{0} ∪ T_{i} is nodal, and T_{i} meets X_{0} exactly at S_{i}. Define Y = X_{0} ∪ T_{1} ⋯ ∪ T_{s}. Then h^{1}(N_{Y} (−Z)) = 0. Arguing by induction on s one finds a smoothing X_{s} of Y which preserves Z. The curve X_{s} belongs to H(2x + 4 + 3s, 1 + 2x + 5s − q, 3). The existence of Z provides that 𝕋(X_{s}, x) ≠ ∅. By semicontinuity we also know that h^{1}(N_{X_{s}}(−Z)) = 0. Finally assume d = (2x + 4 + 3s) + t, for some t > 0. Then we obtain the required curve in H(d, g, 3) by induction on t. We start with X_{s,0} = X_{s}. Then we construct X_{s,t+1} from X_{s,t} by adding a line ℓ which meets X_{s,t} at one general point, so that X_{s,t} ∪ ℓ is nodal, and taking a smooth deformation which fixes Z. The condition h^{1}(O_{X}(2)) = 0 follows by applying several times Lemma 5.18 applied to smooth rational curves of degree ≤ 3. □

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E. Ballico, Dipartimento di Matematica, Università di Trento, Italy

   Email address: edoardo.ballico@unitn.it

L. Chiantini, Dipartimento di Ingegneria dell’Informazione e Scienze Matematiche, Università di Siena, Italy

   Email address: luca.chiantini@unisi.it