ON SECANT DEFECTIVE VARIETIES, IN PARTICULAR OF DIMENSION 4

LUCA CHIANTINI, CIRO CILIBERTO, AND FRANCESCO RUSSO

Abstract. In this paper we prove some general results on secant defective varieties. Then we focus on the 4–dimensional case and we give the full classification of secant defective 4–folds. This paper has been inspired by classical work by G. Scorza.

Introduction

Let X ⊂ ℙ^{r} be an n–dimensional variety, also called an n–fold (in this paper by variety we mean an irreducible, projective, reduced scheme). The variety X is said to be secant defective, or simply defective, if its secant variety S(X) has dimension s(X) smaller than the expected dimension, which is σ(X) = min{r, 2n + 1}. The study and classification of secant defective varieties is a classical subject which goes back to the Italian school of algebraic geometry and received in modern times a lot of attention by various authors, too many to be mentioned here. Curves are never defective. The classification of defective surfaces is classical (for a modern account, see [?]): a surface is defective if and only if it is either a cone or the Veronese surface of conics in ℙ^{5}. The classification of secant defective threefolds goes back to G. Scorza in [?], and a modern proof has been given in [?] (see also [?]). The case of 4–folds has been again considered by Scorza in the memoir [?], in which he outlines the classification of secant defective 4–folds, with the use of beautiful geometric ideas. The present paper is devoted to the full classification of secant defective 4–folds, filling up Scorza’s gaps and simplifying, in several points, his arguments, though often we follow his line of thinking. Our tools, as for Scorza, consist in delicate projective geometric analyses of the various cases which defective 4–folds fall in. These cases correspond to the different values that certain invariants of the 4–fold X acquire. The main invariants in question are basically f(X) = 9 − s(X) and γ(X), which is the dimension of the contact locus with X of a general bitangent space to X, and one has 3 ⩾ γ(X) ⩾ f(X) ⩾ 1. The difference 4 − γ(X) is dubbed by Scorza the species of X and we adopt here his terminology. Scorza’s arguments are often longwinded and, in some parts, not completely clear also due to a different and ancient terminology. Moreover, he regularly uses results, whose proofs available at that time are considered nowadays unsatisfactory (e.g. the formulas appearing in §?? have been recently revised in [?] while Scorza quotes [?]). It would be too long to precisely indicate here all the obscure parts in [?] and all the missing cases in the classification. Scorza himself states in the Introduction of [?] that his analysis is not complete for one class of fourfolds (those of second species). Our treatment is based, among other things, on deep consequences of Terracini Lemma not available to Scorza, who indeed considered 4-folds (or 3-folds) for which two general tangent spaces intersect although he suspected that this property is indeed equivalent to secant defectiveness for r ⩾ 2n + 1, see the Introduction of [?]. Some relations between the invariants of secant defective varieties defined in the paper have not been considered by Scorza and have been applied quite recently also elsewhere to obtain new contributions in the area, see for example [?, ?, ?, ?, ?].

2020 Mathematics Subject Classification. Primary 14N05; Secondary 14C20, 14M20.

Key words and phrases. Secant varieties, defective varieties.

The outcome of our analysis can be summarized in the following classification theorem:

Theorem 1. let X ⊂ ℙ^{r} be an irreducible, projective, secant defective variety of dimension 4. Then, modulo projective equivalence, the following cases occur:

(i) X is a cone; (ii) X sits in a 5 or 6–dimensional cone over a curve; (iii) X sits in a 5–dimensional cone over a surface; (iv) X is the Segre variety Seg(2, 2)= ℙ^{2} × ℙ^{2} ⊂ ℙ^{8};

(v) X is a scroll in 3–spaces as in Example ?? below; (vi) X is a scroll in 3–spaces as in Example ?? below; (vii) r = 9 and X is a scroll in 3–spaces as in Example ?? below; (viii) X is an internal projection of the Veronese 4–fold V_{2,4} of quadrics in ℙ^{14} from

finitely many points with the property that it contains at most finitely many lines through its general point;

(ix) r = 11 and X is the projection of V_{2,4} from the plane spanned by a conic on it;

(x) r = 9 and X is the projection of V_{2,4} from a 4–space spanned by a rational normal quartic curve on it;

(xi) r = 10 and X is a hyperplane section of the Segre variety Seg(2, 3)= ℙ^{2} × ℙ^{3} ⊂ ℙ^{11}; (xii) r = 9 and either X sits in a cone with vertex a line over a hyperplane section

of Seg(2, 2) in ℙ^{7} or X sits in a cone with vertex a point over the Segre variety Seg(2, 2) ⊂ ℙ^{8};

(xiii) 9 ⩽ r ⩽ 11 and X sits in a cone with vertex a line over the projection in ℙ^{r−2} of

the Veronese 3–fold V_{3,2}; (xiv) r = 9 and X sits in a 6–dimensional cone over the Veronese surface V_{2,2} in ℙ^{5}. (xv) r = 9 and X sits in a cone with vertex a line over a (defective) 3–fold in ℙ^{7} sitting

in a cone with vertex a line over the Veronese surface V_{2,2}; (xvi) X is swept out by a 3–dimensional family R of lines and it is singular along

a linear space Π of dimension ε, with 2 ⩽ ε ⩽ 3, which is intersected in one point by the general line in R, and X projects from Π to a 3–dimensional variety Y ⊂ ℙ^{r−ε−1} (with general fibres unions of lines of R), which contains a 4– dimensional family C of (generically irreducible) conics such that there is a conic in C passing through two general points of Y and the counterimage of the general conic of C via the projection from Π is a non–developable scroll spanning a 4– space with a line directrix^{1} sitting in Π;

(xvii) X is swept out by a 4–dimensional family S of surfaces spanning a 4–space, such that two general surfaces in S intersect at a point. In this case the general surface in S is rational and X itself is rational;

(xviii) the general projection of X in ℙ^{9} sits in a 6–dimensional cone with vertex a 3–

space over the Veronese surface V_{2,2}. If X is smooth, only the following cases may occur: (iv), (viii), (ix), (x), (xi), (xii) and

X sits in a cone with vertex a point over the Segre variety Seg(2, 2) ⊂ ℙ^{8}, and possibly case (xvii) if it exists.

Theorem ?? summarizes the results contained in Proposition ?? and Theorems ??, ??, ??, ??, ??, ??, ??, ??. All the cases listed in Theorem ?? do occur, except, maybe, case (xvii) for which we do not have any example. We point out that starting from dimension four there exist smooth secant defective n-folds which are of general type and have unbounded degree (for example those of the second type in case (xii)). These phenomena do not appear in dimension two or three, where all the smooth secant defective varieties are rational and belong to a small finite list.

1A directrix of a scroll surface S is an irreducible curve on S that intersects the general line of the ruling of S in one single variable point.

The paper is organized as follows. In §?? we collect some preliminaries and notation. In §?? we introduce some of the main invariants of a defective variety and we recall some important facts and notions, like Terracini’s Lemma, the second fundamental form of a variety, etc. In §?? we prove the classification Theorem ?? for n–dimensional varieties with very low dimensional secant variety. Section ?? is devoted to the classification of defective 4–folds which are scrolls in 3–spaces. Sections ??, ?? are devoted to the classification of n–folds, in particular 4–folds, of the top species. In §?? we classify 4–folds of the second species and in §?? we classify 4–folds of the first species.

As far as possible, we kept the discussion in the maximal generality, referring to an n–fold and not simply to a 4–fold.

Acknowledgements: Francesco Russo has been partially supported by the PRIN 2020 Squarefree Gröbner degenerations, special varieties and related topics and by the PRIN 2022 Birational geometry of moduli spaces and special varieties. The authors are members of the GNSAGA of the Istituto Nazionale di Alta Matematica “F. Severi”.

The authors thank Th. Dedieu for having pointed out modern references for formulae (??) and the referee for a very careful reading and for the suggestions about the exposition.

1. notation

1.1. In this paper we work over the complex field ℂ.

If X is a scheme of finite type, by dim(X) we mean the maximum of the dimensions of its irreducible components. We will denote by O_{X} the structure sheaf of X. We will denote by Reg(X) the Zariski open subset of smooth points of X and by Sing(X) the closed subset of X formed by its singular points. If x ∈ X we denote by T_{x}(X) the Zariski tangent space to x at X.

Let now X ⊆ ℙ^{r} be an irreducible projective scheme. We will denote by deg(X) the degree of X and by I_{X} the ideal sheaf of X in ℙ^{r}.

If x ∈ Reg(X) we denote by T_{X,x} ⊆ ℙ^{r} the embedded tangent space to X at x, which is a linear subspace of ℙ^{r} of dimension equal to dim(X). If either P ⊆ T_{X,x} is a linear subspace passing through x or T_{X,x} ⊆ P, then P is called a tangent subspace to X at x. If p_{0}, …, p_{m} ∈ Reg(X), then we denote by T_{X,p_{0},...,p_{m}} the join of T_{X,p_{i}} with 1 ⩽ i ⩽ m.

Let X ⊆ ℙ^{r} be a variety. For h a positive integer, an h–secant line to X is any line ℓ not contained in X such that the intersection scheme of X with ℓ has length at least h. An h–secant line is called proper if it cuts X at exactly h distinct points.

Let X ⊆ ℙ^{r} be a variety. We will denote by Sym^{2}(X) the 2–fold symmetric product of X, i.e. the variety of all unordered pairs [x, y], with x, y ∈ X.

1.2. Let v_{d,r} : ℙ^{r} → ℙ^{N(d,r)}, with N(d, r) = \binom{d+r}{r} − 1, be the d–th Veronese map of ℙ^{r}. Its image is denoted by V_{d,r} and is called the (d, r)–Veronese variety.

Let r_{1}, r_{2} be non–negative integers. One has the Segre map s_{r_{1},r_{2}} : ℙ^{r_{1}} × ℙ^{r_{2}} → ℙ^{r_{1}r_{2}+r_{1}+r_{2}}, whose image is denoted by Seg(r_{1}, r_{2}) and called the (r_{1}, r_{2})–Segre variety.

Let n, r be two non–negative integers, with r ⩾ n. Then 𝔾(n, r) denotes the Grassmann variety of type (n, r). A point of 𝔾(n, r) represents a linear subspace of dimension n in ℙ^{r}, which we also call a n–(sub)space of ℙ^{r}.

If Z ⊆ ℙ^{r}, the span of Z is denoted by ⟨Z⟩.

A 2–dimensional irreducible, projective subvariety S of 𝔾(1, n) is called a congruence of lines in ℙ^{n}. In particular, for n = 3, a congruence S of lines in ℙ^{3} is called of type (a, b) if, given a general point x ∈ ℙ^{3} and a general plane P ⊂ ℙ^{3}, there are exactly a [resp. b] lines of the congruence passing through x [resp. contained in P]. Congruences of type (1,1) consist of the set of lines intersecting two given skew lines that can be also infinitely near (this immediately follows, e.g., from [?, Thm. 2.1]). Recall that an infinitely near line s′ to a line s in ℙ^{3} is a projectivity ϕ : s → s^{∗}, where s^{∗} is the pencil of planes containing s. The congruence of lines intersecting s and s′ is the set of all lines passing through a (variable) point P of s and lying in the plane ϕ(P).

1.3. If X ⊆ ℙ^{r} is a variety and P is a linear space of dimension k in ℙ^{r}, we can consider the projection π : X −−→ ℙ^{r−k−1} of X from P. Recall that the projection is a finite morphism onto its image if P ∩ X = ∅. In this case we will say that the projection is external. If P ⊆ X we say that the projection is internal.

The variety X ⊆ ℙ^{r} is said to be non–degenerate if ⟨X⟩ = ℙ^{r}.

If X ⊆ ℙ^{r} is a variety, an extension of X is a variety X′ ⊆ ℙ^{r+k}, with k ⩾ 1, such that X is the scheme theoretical intersection of X′ with a ℙ^{r}. If X′ is a cone over X, the extension is said to be trivial. If there is a non–trivial extension of X, then X is said to be extendable.

1.4. If X is a variety, it is well known the notion of linearly equivalent divisors on X and the relation of this notion with line bundles and linear systems.

Let X be a variety and L a line bundle on X. Then we can consider the complete linear system |L| determined by L, as well as linear subsystems of it. If L is a linear system, we can consider its base locus Bsl(L), that is the scheme consisting of the intersection of all divisors in L. The divisorial part of Bsl(L) is called the divisorial base locus of L. The movable part of L is the linear system consisting of all divisors in L minus the divisorial base locus.

If X ⊆ ℙ^{r} is a variety, we can consider the hyperplane line bundle O_{X}(1) of X. The hyperplane linear system H_{X} (or simply H if there is no ambiguity) of X is the subsystem of |O_{X}(1)| which is cut out on X by the hyperplanes of ℙ^{r}. The variety X ⊆ ℙ^{r} is said to be linearly normal if there is no variety X′ ⊆ ℙ^{r+1} such that X is isomorphic to X′ via an external projection from a point. Equivalently, X ⊆ ℙ^{r} is linearly normal if h^{0}(X, O_{X}(1)) = r + 1, i.e., if H_{X} = |O_{X}(1)|.

A linear system L of dimension r determines a rational map ϕ_{L} : X −−→ ℙ^{r}. The linear system L is said to be birational if ϕ_{L} is birational onto its image.

If L_{1}, L_{2} are linear systems on the variety X, then we can consider their minimal sum L_{1} ∨ L_{2}, which consists of the minimum linear system which contains all divisors of the form D_{1} + D_{2} with D_{i} ∈ L_{i} with i = 1, 2.

If L is a linear system on the variety X and Z is a subscheme of X, we denote by L(−Z) the linear subsystem of L consisting of all divisors of L containing the scheme Z. In particular, if x ∈ X is a point, then L(−x) consists of all divisors in L containing x, whereas we denote by L(−2x) the set of all divisors in L containing the scheme defined by 𝔪_{x}^{2}, where 𝔪_{x} is the maximal ideal corresponding to the point x. In particular, if X ⊆ ℙ^{r} and x ∈ X, then H_{X}(−x) is cut out on X by all hyperplanes passing through x, whereas, if x ∈ Reg(X), then H_{X}(−2x) is cut out on X by all hyperplanes containing T_{X,x}, and it is called the linear system of hyperplanes tangent at x.

1.5. Let X ⊆ ℙ^{r}. A family F of subschemes of X is for us a closed subscheme of the Hilbert scheme of X. The family is said to be filling X if, given a general point x ∈ X, there is a scheme of F, i.e., parameterized by a point of F, passing through x.

2. Secant varieties and contact loci

2.1. Let X ⊆ ℙ^{r} be a non–degenerate irreducible projective variety of dimension n. We will denote by S(X) its secant variety, namely the Zariski closure in ℙ^{r} of the union of all proper secant lines to X, i.e. lines ⟨p_{0}, p_{1}⟩ with p_{0}, p_{1} distinct points in X. One has X = S(X) if and only if X = ℙ^{r}.

One can consider the abstract secant variety S_{X} of X, i.e. S_{X} ⊆ Sym^{2}(X) × ℙ^{r} is the Zariski closure of the set of all pairs ([p_{0}, p_{1}], x) such that p_{0}, p_{1} ∈ X are distinct points and x ∈ ⟨p_{0}, p_{1}⟩. One has the surjective morphism p_{X} : S_{X} → S(X) ⊆ ℙ^{r}, given by the restriction of the projection to the second factor. Hence:

(1)  s(X) := dim(S(X)) ⩽ σ(X) := min{r, dim(S_{X})} = min{r, 2n + 1}.

The integer σ(X) is called the expected dimension of S(X). One says that X is 1– defective, or simply defective, when strict inequality holds in (??). One defines

δ(X) := σ(X) − s(X)

to be the defect of X. A slightly different concept of defect, is the one of fibre defect defined as

f(X) := 2n + 1 − s(X)

which is the dimension of the general fibre of p_{X}, i.e., the dimension of the family of secant lines to X passing through the general point of S(X).

Remark 2.1.1. One has

f(X) = δ(X) if r ⩾ 2n + 1

whereas, if r ⩽ 2n + 1, one has

(2)  f(X) = δ(X) + 2n + 1 − r.

In any case f(X) ⩾ δ(X) and f(X) = 0 implies non–defectiveness.

2.2. Let X, Y be reduced projective schemes in ℙ^{r}. The join J(X, Y) of X and Y is the closure in ℙ^{r} of the set

{x ∈ ℙ^{r} : x ∈ ⟨p, q⟩ with p ∈ X, q ∈ Y, p ≠ q }.

If X is a variety, then S(X) = J(X, X). If X is a reducible projective scheme, then J(X, X) is in general reducible and not pure. It is convenient for us to denote this by S(X) and call it the secant variety of X. The reduced schemes X we will be considering next, even if reducible, will have the property that S(X) is pure. For them all the considerations in §?? can be repeated verbatim.

2.3. There are several notions of tangential variety to a given variety X (see [?] or [?, Lecture 15], as general references). We will adopt the following.

Let X ⊂ ℙ^{r} be an irreducible variety of dimension n. Consider the rational map

s : X × X −−→ 𝔾(1, r)

which sends the point (p_{0}, p_{1}) off the diagonal Δ to the line ⟨p_{0}, p_{1}⟩. Resolve the indeterminacies of s obtaining a morphism s̃ defined on some blow–up X of X × X.

Definition 2.3.1. A tangent line to X is any line corresponding to images via the map s̃ of points of X mapping to Δ.

We will denote by Tan(X) the locus of points of ℙ^{r} lying on some tangent line to X. This is the tangential variety to X. It does not depend on the blow–up X of X × X.

Remark 2.3.2. Tangent lines are all limits of secants ⟨p_{0}, p_{1}⟩ with p_{0}, p_{1} ∈ X distinct points, when p_{0}, p_{1} come toghether. Therefore Tan(X) ⊆ S(X). Moreover Tan(X) is a closed subvariety of ℙ^{r}, containing the variety

T (X) := \overline{⋃_{x∈Reg(X)} T_{X,x}}.

It is a well known consequence of Fulton–Hansen connectedness theorem (see [?]) that if X ⊂ ℙ^{r}, with r ⩾ 2n + 1, is smooth, then either X is not defective (in which case dim(Tan(X)) = 2n) or S(X) = Tan(X).

Note that Tan(X) = T (X) if X is smooth.

2.4. Terracini’s Lemma describes the tangent space to a secant variety at a general point of it (see [?] or, for modern versions, [?, ?, ?, ?]). In the case of interest for us, we may state it as follows.

Theorem 2.4.1 (Terracini’s Lemma). Let X ⊂ ℙ^{r} be a projective variety. If p_{0}, p_{1} ∈ X are general points and x ∈ ⟨p_{0}, p_{1}⟩ is a general point of S(X), then

T_{S(X),x} = ⟨T_{X,p_{0}}, T_{X,p_{1}}⟩ = T_{X,p_{0},p_{1}}.

Remark 2.4.2. Let us keep the notation of Terracini’s Lemma. Then

f(X) = 2n + 1 − s(X) = dim(T_{X,p_{0}} ∩ T_{X,p_{1}}) + 1.

Lemma 2.4.3. Let X ⊆ ℙ^{r} be a non–degenerate variety of dimension n ⩾ 2 and let Y be a general hyperplane section of X. Then

f(Y) = max{0, f(X) − 1}.

Proof. The assertion is trivial if X = ℙ^{r}, so we may assume that X is strictly contained in ℙ^{r} and it is not a linear subspace because it is non–degenerate.

Let Y be the section of X with a general hyperplane H and let p_{0}, p_{1} be general points of Y. Then p_{0}, p_{1} are also general points of X. One has T_{Y,p_{i}} = H ∩ T_{X,p_{i}} for i = 0, 1.

If T_{X,p_{0}} ∩ T_{X,p_{1}} = ∅, so that f(X) = 0, then also T_{Y,p_{0}} ∩ T_{Y,p_{1}} = ∅ and f(Y) = 0, and the assertion holds.

Suppose now T_{X,p_{0}} ∩T_{X,p_{1}} ≠ ∅. Note that neither p_{0} nor p_{1} lie in T_{X,p_{0}} ∩T_{X,p_{1}}, otherwise the general tangent space to X would contain the general point of X and therefore X would coincide with its general tangent space, so it would be a linear space, a contradiction. Then T_{X,p_{0}} ∩ T_{X,p_{1}} is skew with the line ⟨p_{0}, p_{1}⟩, because p_{0}, p_{1} are not in T_{X,p_{0}} ∩ T_{X,p_{1}} and if a point p ∈ ⟨p_{0}, p_{1}⟩ different from p_{0}, p_{1} lies in T_{X,p_{0}} ∩ T_{X,p_{1}}, then the whole line ⟨p_{0}, p_{1}⟩ = ⟨p, p_{1}⟩ = ⟨p_{0}, p⟩ would be contained in T_{X,p_{0}} ∩ T_{X,p_{1}}, a contradiction. Then by the generality of the hyperplane H, H ∩ T_{X,p_{0}} ∩ T_{X,p_{1}} = T_{Y,p_{0}} ∩ T_{Y,p_{1}} has dimension dim(T_{X,p_{0}} ∩ T_{X,p_{1}}) − 1 and the assertion follows. □

2.5. Given an irreducible projective variety X ⊆ ℙ^{r} of dimension n, the Gauss map of X is the rational map

g_{X} : X −−→ 𝔾(n, r)

defined at the smooth points of X by mapping x ∈ Reg(X) to the tangent space T_{X,x}. It is well known that, if x ∈ X is a general point, then the closure of the fibre of g_{X} through x is a linear subspace Γ_{X,x} of ℙ^{r} (see [?, Thm. I.2.3(c)]), which we denote by Γ_{x} if there is no danger of confusion.

Definition 2.5.1. In the above setting, Γ_{x} is called the general Gauss fibre of X and its dimension is called the tangential defect of X, denoted by t(X). We will set θ(X) = t(S(X)).

Note that, if X is smooth, then t(X) = 0 (see [?, Corollary I.2.8]).

Remark 2.5.2. In the above setting, there is a dense open set U ⊂ Reg(X), which via g_{X|U} is fibred over g_{X}(U) in open subsets of projective spaces of the same dimension t(X). This implies that if Y = X ∩ H is the section of X with a general hyperplane and x ∈ Y, then Γ_{X,x} ∩ H = Γ_{Y,x}. Hence

t(Y) = max{0, t(X) − 1}.

2.6. Given x ∈ S(X) a general point, i.e. x ∈ ⟨p_{0}, p_{1}⟩ is a general point with p_{0}, p_{1} ∈ X general points, consider the Zariski closure of the set

{p ∈ Reg(X) : T_{X,p} ⊆ T_{S(X),x} = T_{X,p_{0},p_{1}}}.

Note that this set depends only on p_{0}, p_{1} and not on x ∈ ⟨p_{0}, p_{1}⟩. We will denote by Γ_{X,p_{0},p_{1}} the union of all irreducible components of this locus containing p_{0} or p_{1}. Since p_{0}, p_{1} are general points, we can vary them and exchange them by monodromy and this monodromy action maps Γ_{X,p_{0},p_{1}} to itself. As in [?, Proposition 2.2], this implies that:

Proposition 2.6.1. Γ_{X,p_{0},p_{1}} is smooth at p_{0} and p_{1}. Hence Γ_{X,p_{0},p_{1}} is either irreducible, or it consists of only two irreducible components, one passing through p_{0} (and not through p_{1}) and the other passing through p_{1} (and not through p_{0}), both with the same dimension.

Proof. Let us prove the first assertion. Suppose, by contradiction, that Γ_{X,p_{0},p_{1}} is singular at p_{0} (and therefore, by monodromy considerations, also at p_{1}). In particular Γ_{X,p_{0},p_{1}} has positive dimension at p_{0}. Let Γ be an irreducible component of Γ_{X,p_{0},p_{1}} of positive dimension containing p_{0} and let p be a general point of Γ. We have that T_{X,p} ⊆ T_{X,p_{0},p_{1}}, hence T_{X,p,p_{1}} ⊆ T_{X,p_{0},p_{1}}. Since p is also a general point of X (because it specializes to p_{0} that is a general point of X), we have that dim(T_{X,p,p_{1}}) = dim(T_{X,p_{0},p_{1}}), hence T_{X,p,p_{1}} = T_{X,p_{0},p_{1}}. As p is a smooth point of Γ and as Γ_{X,p,p_{1}} has to be singular at p, this implies that Γ is not the only component of Γ_{X,p,p_{1}} passing through p. Since p is general in Γ, it immediately follows that Γ_{X,p_{0},p_{1}} contains a subvariety of dimension greater than Γ, which contains Γ. This contradicts the fact that Γ is a component of Γ_{X,p_{0},p_{1}}. This contradiction proves the assertion.

The fact that Γ_{X,p_{0},p_{1}} is either irreducible, or it consists of only two irreducible components, one passing through p_{0} (and not through p_{1}) and the other passing through p_{1} (and not through p_{0}) is a trivial consequence of the smoothness of Γ_{X,p_{0},p_{1}} at p_{0} and p_{1}.

Finally, if Γ_{X,p_{0},p_{1}} is reducible, the fact that its two components have the same dimension follows from the fact that they can be exchanged by monodromy. □

Let γ(X) be the dimension of Γ_{X,p_{0},p_{1}}. By the generality of p_{0}, p_{1}, it does not depend on p_{0}, p_{1}. Note that Γ_{X,p_{i}} ⊆ Γ_{X,p_{0},p_{1}} for i = 0, 1. We set

Π_{X,p_{0},p_{1}} = ⟨Γ_{X,p_{0},p_{1}}⟩.

Of course Π_{X,p_{0},p_{1}} contains ⟨p_{0}, p_{1}⟩.

If there is no danger of confusion, we write Γ_{p_{0},p_{1}} and Π_{p_{0},p_{1}} instead of Γ_{X,p_{0},p_{1}} and Π_{X,p_{0},p_{1}}.

Definition 2.6.2. In the above setting, we will call Γ_{X,p_{0},p_{1}} the tangential contact locus of X at p_{0}, p_{1}. We will call γ(X) the contact defect of X.

Proposition 2.6.3. Let X be a projective defective variety of dimension n, let Y = X ∩ H with H a general hyperplane and let p_{0}, p_{1} be general points of Y. Then

Γ_{X,p_{0},p_{1}} ∩ H ⊆ Γ_{Y,p_{0},p_{1}}.

In particular, γ(Y) ⩾ γ(X)−1 and if the equality holds then Γ_{X,p_{0},p_{1}} ∩H = Γ_{Y,p_{0},p_{1}}.

Proof. First we prove that T_{Y,p_{0},p_{1}} = T_{X,p_{0},p_{1}} ∩H. It is clear that T_{Y,p_{0},p_{1}} ⊆ T_{X,p_{0},p_{1}} ∩H, so it suffices to prove that the two spaces have the same dimension. Since T_{Y,p_{i}} = H ∩ T_{X,p_{i}} for i = 0, 1, we have that T_{Y,p_{0}} ∩ T_{Y,p_{1}} = H ∩ T_{X,p_{0}} ∩ T_{X,p_{1}} (see the proof of Lemma ??). Then dim(T_{X,p_{0},p_{1}} ∩ H) = 2n − dim(T_{X,p_{0}} ∩ T_{X,p_{1}}) − 1 =

= 2n − (dim(T_{Y,p_{0}} ∩ T_{Y,p_{1}}) + 1) − 1 = 2(n − 1) − dim(T_{Y,p_{0}} ∩ T_{Y,p_{1}}) = dim(T_{Y,p_{0},p_{1}}),

as wanted.

Now, let x ∈ Γ_{X,p_{0},p_{1}} ∩ H be a general point. Then T_{X,x} ⊆ T_{X,p_{0},p_{1}}, hence

T_{Y,x} ⊆ T_{X,x} ∩ H ⊆ T_{X,p_{0},p_{1}} ∩ H = T_{Y,p_{0},p_{1}},

proving the first assertion.

In particular, the unique component of Γ_{Y,p_{0},p_{1}} passing through p_{i}, i = 0, 1, contains the intersection of H with the unique component of Γ_{X,p_{0},p_{1}} through p_{i}. So, if γ(Y) = γ(X) − 1, then necessarily Γ_{X,p_{0},p_{1}} ∩ H = Γ_{Y,p_{0},p_{1}}. □

The following result is contained in [?, Proposition 3.9].

Proposition 2.6.4. Let X ⊂ ℙ^{r} be a non–degenerate variety such that s(X) < r. Let p_{0}, p_{1} ∈ X be general points. Then:

(i) let q_{0}, q_{1} be general points on Γ_{p_{0},p_{1}}, such that q_{i} specializes to p_{i}, for i = 0, 1 (i.e., q_{i} is a general point of a component of Γ_{p_{0},p_{1}} containing p_{i}, for i = 0, 1). Then Γ_{p_{0},p_{1}} = Γ_{q_{0},q_{1}};

(ii) f(Γ_{p_{0},p_{1}}) = f(X); (iii) Π_{p_{0},p_{1}} = S(Γ_{p_{0},p_{1}}) equals the general Gauss fibre Γ_{S(X),x} of S(X) (in particular

S(Γ_{p_{0},p_{1}}) is a linear space), whereas Γ_{p_{0},p_{1}} ≠ Π_{p_{0},p_{1}}; (iv) θ(X) = dim(Π_{p_{0},p_{1}}) = 2γ(X) + 1 − f(X).

Remark 2.6.5. As for point (ii) of Proposition ??, if Γ_{p_{0},p_{1}} is irreducible, it is clear what f(Γ_{p_{0},p_{1}}) means. If Γ_{p_{0},p_{1}} is reducible in two components (one containing p_{0} and the other containing p_{1}), then

f(Γ_{p_{0},p_{1}}) = dim(T_{Γ_{p_{0},p_{1}} ,q_{0}} ∩ T_{Γ_{p_{0},p_{1}} ,q_{1}}) + 1,

where q_{i} ∈ Γ_{p_{0},p_{1}} is a point specializing to p_{i}, for 0 ⩽ i ⩽ 1 (see Remark ??).

2.7. Let X ⊂ ℙ^{r} be a non–degenerate variety of dimension n and let p ∈ X be a general point. Consider the projection of X with centre T_{X,p} to ℙ^{r−n−1}. We call this a general tangential projection of X, and we denote it by τ_{X,p}, or by τ_{p}, or simply by τ. We denote by X_{p}, or by X_{1}, its image, and by n_{1} its dimension. As a consequence of Terracini’s Lemma, we have

(3)  n_{1} = n − f(X).

Definition 2.7.1. Let p_{0}, p_{1} ∈ X be general points. Let Ψ_{X,p_{0},p_{1}} be the component of the fibre of τ_{X,p_{1}} containing p_{0}. It is called the projection contact locus of X at p_{0}, p_{1} and, by (??), its dimension is f(X).

If there is no danger of confusion, we will write Ψ_{p_{0},p_{1}} rather than Ψ_{X,p_{0},p_{1}}.

Remark 2.7.2. One has that Γ_{p_{0},p_{1}} contains Ψ_{p_{0},p_{1}}. Indeed, if p ∈ Ψ_{X,p_{0},p_{1}} is a general point (in particular if p = p_{0}), then the projection of T_{X,p} from T_{X,p_{1}} is the tangent space to X_{1} at the point τ(p) = τ(p_{0}). In particular T_{X,p} ⊂ T_{X,p_{0},p_{1}}. Hence

(4)  γ(X) ⩾ f(X).

Actually one has γ(X) = t(X_{1})+f(X). Indeed, the component of Γ_{p_{0},p_{1}} passing through p_{1} is the counterimage, via the tangential projection τ_{p_{0}}, of the Gauss fibre Γ_{X_{p_{0}} ,τ_{p_{0}} (p_{1})}. Therefore γ(X) = f(X) holds if and only if the Gauss map of X_{1} is generically finite to its image, which is equivalent to say that it is birational to its image (see [?, Remark 3.6]). Thus, if γ(X) = f(X), then Ψ_{p_{0},p_{1}} is the irreducible component of Γ_{p_{0},p_{1}} containing p_{0}.

If X is defective, then f(X) > 0, so that γ(X) is also positive and X_{1} is strictly contained in ℙ^{r−n−1}. Otherwise n_{1} = r − n − 1 by (??) yields f(X) = 2n + 1 − r which, together with (??), implies δ(X) = 0, a contradiction.

By Lemma ??, if Y = X ∩ H with H a general hyperplane, then

f(Y) = max{0, f(X) − 1}.

This implies that, if p_{0}, p_{1} are general points of Y, then Ψ_{Y,p_{0},p_{1}} = H ∩ Ψ_{X,p_{0},p_{1}}.

Lemma 2.7.3. Let X ⊂ ℙ^{r} be a variety with f(X) ⩾ 1. Let p ∈ X be a general point and let Y by a general hyperplane section of X passing through p. Then X_{p} = Y_{p}.

Proof. We can assume that Y is a general hyperplane section of X and that p is also a general point of Y. Via the tangential projection τ_{p} we clearly have that Y_{p} ⊆ X_{p}, and X_{p}, Y_{p} are both irreducible. On the other hand dim(X_{p}) = dim(X) − f(X) = dim(Y) − f(Y) = dim(Y_{p}), so X_{p} = Y_{p} as wanted. □

Remark 2.7.4. The above lemma can be proved under milder hypotheses. For example, it suffices that X is homogeneous, Y is any hyperplane section and f(Y) = f(X) − 1. Then if p is a general point of Y one still has Y_{p} = τ_{Y,p}(Y) ⊆ τ_{X,p}(X). Moreover the tangential projection τ_{X,p} behaves as if p were a general point of X by the homogeneity hypothesis. Hence Y_{1} ⊆ X_{1}. Finally dim(X_{1}) = dim(Y_{1}) still holds and we are done.

2.8. Other relevant items related to the secant variety S(X) are the so called entry loci.

Definition 2.8.1. Let x ∈ S(X) be a point. We define the entry locus E_{x} of x with respect to X as the closure of the set

{z ∈ X : there is y ∈ X with y ≠ z and x ∈ ⟨y, z⟩}.

A useful information about the entry loci is provided by the following result (see [?, Proposition 3.13] or [?, Proposition 2.2]).

Proposition 2.8.2. Let X ⊂ ℙ^{r} be a variety with s(X) < r and let x ∈ S(X) be a general point. Then E_{x} is pure of dimension f(X).

Remark 2.8.3. Terracini’s Lemma implies that the entry locus E_{x} is contained in the tangential contact locus Γ_{p_{0},p_{1}} for all distinct pairs of points p_{0}, p_{1} ∈ X such that x ∈ ⟨p_{0}, p_{1}⟩. Therefore again f(X) ⩽ γ(X).

2.9. Let X ⊂ ℙ^{r} be a variety. Let x ∈ X be a general point and let H be a general hyperplane containing T_{X,x}. This is a general tangent hyperplane to X. Consider the Zariski closure Γ_{X,x}(H) of the set

{p ∈ Reg(X) : T_{X,p} ⊆ H}.

By [?, Thm. I.2.3, (c)] applied with m = r − 1, when H is general then Γ_{X,x}(H) is a linear subspace of ℙ^{r}. We call it the general tangent hyperplane contact locus of X at x and we will denote by d(X) its dimension, which does not depend on x and H. If there is no danger of confusion, we may write Γ_{x}(H) rather than Γ_{X,x}(H). Of course Γ_{x}(H) contains Γ_{x}, thus

d(X) ⩾ t(X).

Moreover the dual variety X^{∗} of X has dimension

dim(X^{∗}) = r − 1 − d(X)

and therefore d(X) is called the dual defect of X.

Similarly, let p_{0}, p_{1} ∈ X be general points, and let H be a general hyperplane containing T_{X,p_{0},p_{1}}, i.e. a general bitangent hyperplane to X. Consider the Zariski closure of the set

{p ∈ Reg(X) : T_{X,p} ⊆ H}.

Definition 2.9.1. The union Γ_{X,p_{0},p_{1}}(H) of all irreducible components of the aforementioned locus containing either p_{0} or p_{1} is called the general bitangent hyperplane contact locus of X at p_{0}, p_{1}. We will denote by ϵ(X) its dimension, which, by genericity, does not depend on p_{0}, p_{1} and H. We will set

Π_{X,p_{0},p_{1}}(H) := ⟨Γ_{X,p_{0},p_{1}}(H)⟩.

We will write Γ_{p_{0},p_{1}}(H) and Π_{p_{0},p_{1}}(H) instead of Γ_{X,p_{0},p_{1}}(H) and Π_{X,p_{0},p_{1}}(H) if there is no danger of confusion.

Of course Γ_{p_{0},p_{1}}(H) contains Γ_{p_{0},p_{1}}, and therefore

(5)  ϵ(X) ⩾ γ(X).

Remark 2.9.2. By (??) and (??) we see that

ϵ(X) ⩾ γ(X) ⩾ f(X).

More precisely, looking at a general tangential projection, one has d(X_{1}) = ϵ(X) − f(X) > 0. Indeed, let H be a general bitangent hyperplane to X ⊂ ℙ^{r} of dimension n, at two general points p_{0}, p_{1}. Let H′ be the hyperplane in ℙ^{r−n−1} image of H via τ_{p_{0}}, that can be considered as a general tangent hyperplane to X_{p_{0}} at τ_{p_{0}}(p_{1}). Then Γ_{X,p_{0},p_{1}}(H) is the preimage via τ_{p_{0}} of Γ_{X_{p_{0}} ,τ_{p_{0}} (p_{1})}(H′), hence ϵ(X) = d(X_{p_{0}}) + f(X) as wanted. Hence we have that ϵ(X) > f(X) if and only if d(X_{1}) = ϵ(X) − f(X) > 0, i.e. if and only if X_{1} is dual defective.

Recall the following result which parallels Proposition ?? (see [?, Theorem 1.1] and [?, Theorem 2.4]):

Theorem 2.9.3. Let X ⊂ ℙ^{r} be a defective variety. Let p_{0}, p_{1} ∈ X be general points and let H be a general hyperplane containing T_{X,p_{0},p_{1}}. One has:

(i) Γ_{p_{0},p_{1}}(H) is smooth at p_{0}, p_{1}; moreover it is either irreducible or it consists of two irreducible components of the same dimension ϵ(X) and p_{0}, p_{1} can be considered as general points each on one of these components (in accordance with Proposition ??(i)); (ii) f(Γ_{p_{0},p_{1}}(H)) = f(X); (iii) Π_{p_{0},p_{1}}(H) = S(Γ_{p_{0},p_{1}}(H)), whereas Γ_{p_{0},p_{1}}(H) ≠ Π_{p_{0},p_{1}}; (iv) dim(Π_{p_{0},p_{1}}(H)) ⩽ 2ϵ(X) + 1 − f(X).

Remark 2.9.4. As for the meaning of f(Γ_{p_{0},p_{1}}(H)) in point (ii) of Theorem ??, we adopt the same convention as in Remark ??.

Moreover we have the following:

Proposition 2.9.5. Let X ⊂ ℙ^{r} be a defective variety of dimension n. Then:

(i) r ⩾ n + 3 and f(X) ⩽ n − 1. Moreover if f(X) = n − 1 (which is the case if r = n + 3), then X is either a cone over a curve or a cone over the Veronese surface V_{2,2} in ℙ^{5};

(ii) if f(X) = γ(X) := m, then for p_{0}, p_{1} ∈ X general, Γ_{p_{0},p_{1}} is a quadric of rank at least 2 in ℙ^{m+1}. Furthermore Ψ_{p_{0},p_{1}} is an irreducible component of Γ_{p_{0},p_{1}}, hence either it coincides with Γ_{p_{0},p_{1}} or it is a linear subspace of dimension m;

(iii) in particular if f(X) = ϵ(X) := m, then for general choices of p_{0}, p_{1} and H containing T_{X,p_{0},p_{1}}, one has Γ_{p_{0},p_{1}}(H) = Γ_{p_{0},p_{1}} and (ii) holds.

Proof. Since X_{1} is non–degenerate and strictly contained in ℙ^{r−n−1} (see Remark ??), we have r − n − 1 ⩾ 2, proving the first assertion in (i). Similarly n_{1} = n − f(X) ⩾ 1, proving the second assertion in (i). Suppose that r = n + 3. Since n ⩾ 2, because curves are never defective, we have r = n + 3 ⩽ 2n + 1. Hence, by (??), we have f(X) = δ(X) + n − 2 ⩾ n − 1, so f(X) = n − 1 as stated. The rest of (i) follows from [?, Theorem 3.5].

Let us prove (ii). By Proposition ??, (iv), we have that Γ_{p_{0},p_{1}} is a hypersurface in Π_{p_{0},p_{1}}, which has dimension m + 1. It has to be a quadric otherwise the general secant to X would be a trisecant. By the Trisecant Lemma (see [?, Proposition 1.4.3]), this cannot happen unless r = n + 1, which is not the case. Then Γ_{p_{0},p_{1}} is a quadric but it cannot be a linear space, hence Γ_{p_{0},p_{1}} has rank at least 2. The rest of assertion (ii) is clear.

Part (iii) follows from (ii) and Remark ??. □

Varieties X for which part (ii) of Proposition ?? holds, i.e., for p_{0}, p_{1} ∈ X general Γ_{p_{0},p_{1}} is a quadric of rank at least 2, are locally quadratic entry locus (LQEL) varieties in the sense of [?] (see [?] for the theory of LQEL varieties and applications). If γ(X) = f(X) = 1, we will say that they present the locally conic entry locus (LCEL) case.

According to Scorza’s terminology (see [?]), we will say that a defective variety X ⊂ ℙ^{r} of dimension n is of the species k if k = n − γ(X). The top species for varieties of dimension n is n − 1. Such a variety X has γ(X) = f(X) = 1, hence it presents the LCEL case. Note that varieties X of dimension n of a given species k could be further classified according to the value of the invariant ϵ(X) ⩾ n − k = γ(X). There are however some constraints, as the following lemma indicates.

Lemma 2.9.6. Let X ⊂ ℙ^{r} be a non–degenerate variety if dimension n. Then ϵ(X) = n − 1 if and only if γ(X) = n − 1.

Proof. If γ(X) = n − 1, then ϵ(X) ⩾ n − 1 by (??), and, by Definition ??, it cannot be ϵ(X) = n so ϵ(X) = n − 1.

As for the other implication, assume ϵ(X) = n − 1. If f(X) = n − 1 then also γ(X) = n − 1 and we are done. If f(X) < n − 1, then the image X_{1} of a general tangential projection of X has a curve as a dual (see Remark ??). Then X_{1}, which is dual to a curve, is a hypersurface, and t(X_{1}) = n_{1} − 1 = n − f(X) − 1. By Remark ?? we have γ(X) = t(X_{1}) + f(X) = n − 1, as wanted. □

2.10. Let X ⊂ ℙ^{r} be a non–degenerate variety. Let x be a smooth point of X. One can consider the second fundamental form II_{X,x} of X at x (see [?, §1(b)] as a general reference or [?, Section 2.3.2] for an introduction in the same spirit of our approach). This is a linear system of quadrics in ℙ_{X,x} := ℙ(T_{x}(X)). Recall that T_{x}(X) is the Zariski tangent space to x at X, hence ℙ_{X,x}, the projective space associated to T_{x}(X), is a projective space of dimension n − 1.

Consider the general tangential projection τ_{x} : X −−→ X_{1}. This is not defined at x. Let X̃ = Bl_{x} X be the blow–up of X at x, let π_{x} : X̃ → X the induced map and consider the exceptional divisor E := E_{x}, which may be identified with ℙ_{X,x}. We have therefore the rational map

τ̃_{x} = τ_{x} ∘ π_{x} : X̃ −−→ X_{1} ⊆ ℙ^{r−n−1},

whose restriction to E is defined by a linear system of quadrics, which is precisely II_{X,x}. We will denote by B_{X,x} its base locus scheme, which consists of the points of E_{x} corresponding to directions of lines having intersection multiplicity larger than 2 with X at x (see [?]). Hence it contains the subscheme L_{X,x} of E, whose points correspond to tangent directions of the lines on X passing through x.

We may write II_{x}, L_{x}, B_{x} rather than II_{X,x}, L_{X,x}, B_{X,x} etc., if there is no danger of confusion. We will set

b(X) = dim(B_{x}) and ℓ(X) = dim(L_{x}).

Note that if ℓ(X) ⩾ 0, then ℓ(X) + n − 1 is the dimension of the union of the components of the Hilbert scheme of lines contained in X and filling X.

It is useful to collect here the main results about the second fundamental form that we will use later. First of all, it is a classical result by Terracini in [?] (see [?, (5.6), (5.7)] and also [?, Theorem 3.3.1] for a modern treatment) that

dim(T (X)) = n + 1 + dim(τ̃_{x}(E)).

Therefore, if X is smooth and defective, then by Remarks ?? and ?? one has

(6)  dim(τ̃_{x}(E)) = n − f(X) = dim(X_{1}).

Thus τ̃_{x|E} is dominant onto X_{1}, with general fibres of dimension f(X) − 1. Since X_{1} is non–degenerate in ℙ^{r−n−1}, one has

(7)  dim(II_{X,x}) = r − n − 1.

As the following result shows, (??) holds, to a certain extent, even relaxing the smoothness assumption on X.

Lemma 2.10.1. Let X ⊂ ℙ^{r} be a non–degenerate, defective variety of dimension n. Let x ∈ X be a general point and assume that the closure of the general fibre of the tangential projection τ_{x} is irreducible and contains x. Then τ̃_{x|E} is dominant onto X_{1}, with general fibres of dimension f(X) − 1. Hence (??) and (??) hold. Furthermore

(8)  r ⩽ \frac{n(n + 3)}{2}.

Proof. The assumption implies that τ̃_{x|E} : E −−→ X_{1} is dominant. Since τ̃_{x|E} is defined by a linear system of quadrics, one has

r − n − 1 ⩽ \frac{(n − 1)(n + 2)}{2}

and (??) follows. The rest of the assertion is clear. □

3. The codimension 4 case

Since defective surfaces and threefolds are classified (see [?, ?, ?, ?]), we may always assume n ⩾ 4. Let X ⊂ ℙ^{r} be an irreducible, non–degenerate, defective variety of dimension n ⩾ 4. In Proposition ?? part (i) we proved that r ⩾ n + 3 and that equality holds if and only if X is a cone over a curve or over the Veronese surface V_{2,2} ⊂ ℙ^{5}. The next theorem classifies the cases for which r ⩾ n + 4 and f(X) = n − 2 and, in particular, those for which the equality r = n + 4 holds.

Theorem 3.0.1. Let X ⊂ ℙ^{r} be a non–degenerate, defective variety of dimension n ⩾ 4, which is not a cone over a curve or over the Veronese surface V_{2,2}. Then r ⩾ n + 4 and if f(X) = n − 2, which happens in particular if r = n + 4, then X is of one of the following types:

(i) X is a cone over a non–defective surface; (ii) X sits in a (n + 1)–dimensional cone over a curve; (iii) X sits in a (n + 1)–dimensional cone over the Veronese surface V_{2,2} in ℙ^{5}; (iv) X is a cone over the Veronese threefold V_{3,2} in ℙ^{9} or over a projection of V_{3,2} in

ℙ^{8} or ℙ^{7};

(v) X is either a cone over the Segre variety Seg(2, 2) ⊂ ℙ^{8} with vertex of dimension n − 5, or a cone over a hyperplane section of Seg(2, 2).

Remark 3.0.2. Suppose X is as in one of the cases listed in Theorem ??.

If X is either a cone over a curve, or it is a cone over a Veronese surface V_{2,2}, then X is defective, with f(X) = n − 1 (see Proposition ??).

Otherwise, we claim that X is defective with f(X) = n − 2. This is immediate in cases (i), (iv), (v), since X is a cone.

Suppose we are in case (ii) so that X lies in a cone V over a curve C with vertex a (n − 1)–space Π. We may assume that C sits in a complementary space to Π. Let x ∈ X be a general point. Let π : V −−→ C be the projection of V to C from Π and let y = π(x). The subspace T_{x} := Π ∩ T_{X,x} has dimension n − 2 and

(9)  T_{X,x} = ⟨T_{x}, ℓ⟩.

with ℓ a suitable tangent line to X through x that is mapped isomorphically via π to T_{C,y} (ℓ exists because π_{|X} is dominant) so that ℓ ∩ Π = ∅. Then

(10)  T_{V,x} = ⟨Π, ℓ⟩.

The linear space T_{x} cannot stay fixed as x varies on X (otherwise X would be a cone over C). If p_{0}, p_{1} ∈ X are general points, then Π = ⟨T_{p_{0}}, T_{p_{1}}⟩, hence T_{X,p_{0},p_{1}} contains Π, it also contains P_{i} the n–spaces of the ruling of V through p_{i} for i = 0, 1, and finally T_{V,p_{0},p_{1}} because of (??) and (??). In conclusion T_{X,p_{0},p_{1}} = T_{V,p_{0},p_{1}}. Hence

dim(T_{X,p_{0},p_{1}}) = dim(T_{V,p_{0},p_{1}}) = dim(T_{V,p_{0}}) + dim(T_{V,p_{1}}) − dim(Π) =
= 2(n + 1) − (n − 1) = n + 3,

which implies f(X) = 2n + 1 − (n + 3) = n − 2 (see also Remark ??).

In case (iii) the argument is analogous to the one in case (ii). Suppose X lies in a cone V over a Veronese surface S = V_{2,2} with vertex a (n − 2)–space Π. We may assume that S sits in a complementary space to Π. Let x ∈ X be a general point. Let π : V −−→ S be the projection of V to S from Π and let y = π(x). The subspace T_{x} := Π ∩ T_{X,x} has dimension n − 3 and

(11)  T_{X,x} = ⟨T_{x}, α⟩

with α a suitable tangent plane to X through x that maps via π to T_{S,y}. Then

(12)  T_{V,x} = ⟨Π, α⟩.

As above, T_{x} cannot stay fixed as x varies on X, otherwise X would be a cone over S. Thus, for p_{0}, p_{1} ∈ X general, Π = ⟨T_{p_{0}}, T_{p_{1}}⟩, T_{X,p_{0},p_{1}} contains Π and P_{i}, the (n − 1)- spaces of the ruling of V through p_{i} for i = 0, 1, and finally T_{V,p_{0},p_{1}} because of (??) and (??). In conclusion T_{X,p_{0},p_{1}} = T_{V,p_{0},p_{1}}, yielding dim(T_{X,p_{0},p_{1}}) = n + 3 and f(X) = n − 2.

Proof of the Theorem ??. By hypothesis and by Proposition ?? part (i), we have r ⩾ n + 4 and f(X) ⩽ n−2. If r = n+4, then f(X) = n−2. Indeed, if r = n+4, then r < 2n+1 because n ⩾ 4, so that by (??) we have f(X) = δ(X) + n − 3 ⩾ n − 2, proving the claim.

Next, if f(X) = n − 2, then s(X) = 2n + 1 − f(X) = n + 3. So we may project generically to ℙ^{n+4} getting an image X′ of X. Note that the projection X → X′ is finite and bijective.

Claim 3.0.3. If X′ is as in one of the cases (i)–(v), then also X is as in the corresponding case (i)–(v).

Proof of the Claim ??. The assertion is trivial if X′ is as in (i), (iv) or (v).

Let us assume that X′ is as in (ii), so that X′ lies in a (n+1)–dimensional cone V′ over a curve with vertex a (n − 1)–space Π. Then X′ is swept out by a 1–dimensional family F′ of varieties of dimension n − 1, each lying in a n–space, i.e., a space of a ruling of V′.

Suppose first that the general variety in F′ spans a n–space. Since the projection X → X′ is finite and birational, then X is also swept out by the (n − 1)–dimensional varieties of a family F, which are in one–to–one correspondence with the varieties in F′. Since the projection from X to X′ is generic, also the projection of the general member of F to the general member of F′ is finite and bijective, and they both span a n–space. These n–spaces, as well as their projections to ℙ^{n+4}, meet along a (n − 1)–space, hence they form a cone V over a curve, in which X sits. So X is as in (ii).

Suppose next that the general variety in F′ does not span a n–space. This means that the general variety in F′ is a (n − 1)–space 𝔓 that intersects the vertex Π of V′ along a (n − 2)-space 𝔓′, and the same happens for the general member of F. Let 𝔓_{1}, 𝔓_{2} two general members of F′. If 𝔓′_{1} = 𝔓′_{2}, then all the (n − 1)–spaces of F′ contain the same (n − 2)–space of Π. This implies that X′ is a cone over a curve, and then also X is a cone over a curve, a contradiction. So if 𝔓_{1}, 𝔓_{2} are two general members of F′, then 𝔓′_{1} ≠ 𝔓′_{2}. Thus 𝔓_{1}, 𝔓_{2} intersect along a (n − 3)–space. Since X′ is a general projection of X, then also the members of F intersect along a (n − 3)–space. Then by [?, Lemma 4.1], one of the following cases occur:

(a) the span of the linear spaces in F is a k–space with k ⩽ n + 2;

(b) all the linear spaces in F contain the same (n − 3)–space;

(c) the general member of F cuts a fixed (n − 1)–space along a (n − 2)–space.

Case (a) cannot happen, because then X would span at most a (n + 2)–space, a contradiction. In case (b), X would be a cone, as well as X′, a contradiction again. So only case (c) happens, which implies that X also lies on a cone V over a curve, which projects to V′.

The case in which X is as in (iii) is similar, hence we will be brief. Let us assume that X′ is as in (iii), so that X′ lies in a cone V′ over a Veronese surface. Then X′ is swept out by a 2–dimensional family F′ of varieties of dimension n − 2, each lying in a (n − 1)–space, i.e. a space of the ruling of V′.

Suppose that the general variety in F′ spans a (n − 1)–space. Then X is also swept out by the (n − 2)–dimensional varieties of a family F, whose generic projections are the varieties in F′. So also the general variety of F spans a (n − 1)–space. These (n − 1)– spaces, as well as their projections to ℙ^{n+4}, meet along a (n − 2)–space, hence they form a cone V over a surface S, in which X sits. Since the cone V projects generically to the cone V′ that is a cone over a Veronese surface, we deduce that also S is a Veronese surface, hence in fact r = n + 4, X = X′ and we are in case (iii).

Suppose next that the general variety in F′ does not span a (n − 1)–space. Then the general variety in F′ is a (n − 2)–space, and the same happens for the general member of F. As above, one sees that two general members of F′ intersect along a (n − 4)– space. Since X′ is a general projection of X, then also the members of F intersect along a (n − 4)–space. Then by Lemma 4.1 in [?], one of the following cases occur:

(a’) the span of the linear spaces in F is a k–space with k ⩽ n + 1;

(b’) all the linear spaces in F contain the same (n − 4)–space; (c’) the general member of F cuts a fixed (n − 2)–space along a (n − 3)–space.

As in the previous argument, cases (a’) and (b’) cannot happen. So only case (c’) happens, which implies that X also lies on a cone V over a surface S, and we conclude as above that S is a Veronese surface, r = n + 4, X = X′ and we are in case (iii). □

By Claim ??, by substituting X with X′, we may assume r = n + 4.

Let Y = X ∩P be the section of dimension 3 of X with a general 7–dimensional linear subspace P. By Lemma ??, one has f(Y) = 1, so Y is a defective threefold which is non–degenerate in ℙ^{7}, hence Y is in the list of Theorem 1.1 of [?].

Suppose first Y is a cone with vertex a point v over a surface S. Since f(Y) = 1, this implies that S is not defective (see Remark ??). Then we claim we are in case (i). In fact, if X has degree d, also Y has degree d and the vertex v of Y is a point of Y of multiplicity d. Since Y is a general 3–fold section of X, we see that v is also a point of multiplicity d for X, and therefore X has a locus Π of points of multiplicity d which is cut out by a general ℙ^{7} in a point, so Π is a linear space of dimension n − 3, and X is a cone with vertex Π over the surface S, that is not defective.

Suppose Y is not a cone and sits in a 4–dimensional cone over a curve with vertex a plane Π. In this case we have γ(Y) = 2. Indeed, if p_{0}, p_{1} are general points in Y, then Γ_{Y,p_{0},p_{1}} consists of a pair of irreducible surfaces, each contained in one of the 3–spaces ⟨Π, p_{0}⟩ and ⟨Π, p_{1}⟩, and together spanning Π_{Y,p_{0},p_{1}} = ⟨Π, p_{0}, p_{1}⟩ which has dimension 4 (as dictated by Proposition ??, (iv)).

Let x ∈ Y be a general point, which is also a general point of X. Consider the tangential projection τ_{X,x}. The image X_{1} of X has dimension n−f(X) = 2. The restriction of τ_{X,x} to Y coincides with τ_{Y,x}, whose image Y_{1} has also dimension 3 − f(Y) = 2. This implies that X_{1} = Y_{1}. On the other hand it is immediate to see that Y_{1} is a cone over a curve. Then t(X_{1}) = t(Y_{1}) = 1. By Remark ??, we have γ(X) = t(X_{1}) + f(X) = n − 1.

Fix again two general points p_{0}, p_{1} in Y, which are also general points on X. By iterated applications of Proposition ??, we have that Γ_{Y,p_{0},p_{1}} = Γ_{X,p_{0},p_{1}}∩P. By the above description of Γ_{Y,p_{0},p_{1}} (that implies that Γ_{Y,p_{0},p_{1}} is reducible in two components containing p_{0} and p_{1}) and by iterated applications of Proposition ?? that imply that Γ_{X,p_{0},p_{1}} ∩ P ⊆ Γ_{Y,p_{0},p_{1}}, Γ_{X,p_{0},p_{1}} is the union of a pair of irreducible Weil divisors on X, each passing through one of the points p_{0}, p_{1}, each spanning at most a n–space, and together spanning Π_{X,p_{0},p_{1}} which has dimension n + 1.

Suppose first the components of Γ_{Y,p_{0},p_{1}} both span a 3–space, so that the components of Γ_{X,p_{0},p_{1}} both span a n–space. Then X is swept out by a 1–dimensional family Z of irreducible Weil divisors such that the general Z in Z spans a n–space and two general such n–spaces span a (n + 1)–space. Since X is non–degenerate in ℙ^{n+4}, this implies that all these n–spaces contain the same (n − 1)–dimensional subspace Π^{2}. By projecting X form Π we obtain a curve, so we are in case (ii).

Suppose next that the components of Γ_{Y,p_{0},p_{1}} do not span a 3–space, so they are planes. These planes cannot pass through the same line, because then Y would be a cone over a curve against the fact that f(Y) = 1. So these planes pairwise meet at a point. Then X is swept out by a 1–dimensional family Z such that the general Z in Z is a (n − 1)– space and two general such (n − 1)–spaces span a (n + 1)–space, hence they meet along a (n − 3)–space. Then the same argument, based on Lemma 4.1 in [?] we made in Claim ??, implies that: either X is a cone with vertex a (n − 3)–space, hence Y is also a cone, a contradiction; or X sits in a cone with vertex a (n − 1)–space over a curve, and we are in case (ii) again.

Suppose now that Y is not a cone and is contained in a 4–dimensional cone V over V_{2,2} with vertex a line L. Then we claim we are in case (iii). Again we have γ(Y) = 2. Indeed, if p_{0}, p_{1} are general points in Y, then Γ_{Y,p_{0},p_{1}} varies in a 2–dimensional family F of irreducible surfaces spanning linear spaces of dimension 4 and contained in 3–dimensional quadric cones with vertex L over a conic of V_{2,2}. Precisely, if 𝔣 : Y −−→ V_{2,2} is the projection from the line L, then Γ_{Y,p_{0},p_{1}} is contained in the 3–dimensional rank 3 quadric cone with vertex L over the unique conic of V_{2,2} passing through the points 𝔣(p_{0}) and 𝔣(p_{1}). Let S = Y ∩ Π be the intersection of Y with a general hyperplane Π of P ≅ ℙ^{7}, which spans the linear space Π of dimension 6 and sits in the cone over V_{2,2} with vertex v = Π ∩ L. Then S has a 2–dimensional family C of irreducible curves spanning linear spaces of dimension 3 and contained in quadric cones with vertex v over a conic of V_{2,2}: they are the hyperplane sections of the surfaces in F.

Claim 3.0.4. Let S be the surface as above. Then there is no irreducible curve Γ in Π containing v and such that the projection of S from a general point of Γ is a Veronese surface V_{2,2}.

Proof of the Claim ??. We argue by contradiction and assume there is such a curve Γ. Then for a general point x ∈ Γ, the surface S sits in the cone over V_{2,2} with vertex x, and therefore it contains a family C_{x} of curves spanning a linear space of dimension 3 and contained in a quadric cone with vertex x over a conic of V_{2,2}.

2Here (and later) we use the following trivial fact: let S be a set of n–spaces in ℙ^{r} such that any two spaces in S intersect along a (n − 1)–space (equivalently, they span a (n + 1)–space). Then either all spaces in S contain the same (n − 1)–space or they are all contained in the same (n + 1)–space.

The family C_{x} cannot be independent on x. Suppose in fact that C_{x} = C is constant. Then if C is a general curve in C, it sits in infinitely many quadric cones with vertices moving on Γ. Then C would have either degree 3 or 4. However, if deg(C) = 4 then C is the complete intersection of two quadrics in ℙ^{3}, and it cannot lie on infinitely many cones, so this case is not possible. If deg(C) = 3, then C could lie on infinitely many quadric cones, but their vertices should move on C. This implies that Γ = C, which is impossible, since C is the general curve in C.

Since the family C_{x} depends on x, this implies that S should have a 3–dimensional family D (described by all families C_{x} when x moves on Γ) of irreducible curves spanning a linear space of dimension 3. Let then p ∈ S be a general point, and consider the projection in ℙ^{5} of S from p, whose image, birational to S by the Trisecant Lemma (see [?, Proposition 1.4.3]), we denote by S′. Then S′ has a family of dimension 2 of irreducible plane curves, the images of the curves of D passing through p. By the Trisecant Lemma the curves in question are conics and S′, spanning a ℙ^{5}, is defective since it contains an irreducible conic passing through two general points of it. Then S′ is the Veronese surface V_{2,2}, but this is a contradiction, because S′ should also contain the line corresponding to the exceptional divisor of the blow–up at p. This contradiction proves the Claim. □

Next we argue by induction on the dimension n. If n = 4, let P a general pencil of hyperplanes in ℙ^{8}. Fix general P ∈ P, which cuts Y on X lying on a cone over V_{2,2} with vertex a line L. If P′ is another general element in P it cuts X along Y′ lying on a cone over V_{2,2} with vertex a line L′. The surface section S of X with P ∩ P′ := Π (which depends only on P) a priori sits inside Π in two cones over V_{2,2}, one with vertex v = L∩Π and another with vertex v′ = L′ ∩ Π. However, when P′ moves in P, by Claim ?? the point v′ cannot move, so that v = v′ and therefore L ∩ L′ = v. This implies that the lines which are vertices of the 4–dimensional cones over V_{2,2} in which the hyperplane sections of X sit, pairwise intersect each other, and this intersection is not a fixed point because the hyperplanes of ℙ^{8} have no fixed point. Then all these lines sit in a plane, which is the vertex of a cone over V_{2,2} containing X, so that we are in case (iii).

Suppose next we are in dimension n > 4 and let Z = X ∩ H be the intersection of X with a general hyperplane H. Then Z is irreducible and non-degenerate. Since X is not a cone over a curve and neither a cone over V_{2,2}, then Z cannot be a cone over a curve, neither it can be a cone over V_{2,2}. The general threefold section Y of Z, which is also a general threefold section of X, lies in a 4-dimensional cone V over V_{2,2}. In particular, two general tangent spaces to V meet along a plane, hence two general tangent spaces to Y are contained in the same ℙ^{6}. Thus Y is defective, so also Z is defective. It follows that Z satisfies all the assumption that have been assumed and discussed for X up to this point. By induction Z is in case (iii), hence it sits in a n–dimensional cone over V_{2,2} with vertex a linear space T of dimension n − 3. If p_{0}, p_{1} are general points in Z, then Γ_{Z,p_{0},p_{1}} varies in a 2–dimensional family 𝔉 of irreducible varieties of dimension n − 2 spanning linear spaces of dimension n and contained in quadric cones of dimension n − 1 and rank 3 with vertex T over a conic of V_{2,2}.

Let 𝔙 = Z ∩ Λ be the intersection of Z with a general subspace Λ of H of dimension n + 2. So that 𝔙 is contained in the cone over V_{2,2} with vertex W = Λ ∩ T of dimension n − 4. Then 𝔙 has a 2–dimensional family ℭ of irreducible varieties of dimension n − 3 spanning linear spaces of dimension n − 1 and contained in rank 3 quadric cones of dimension n − 2 with vertex W over a conic of V_{2,2}: they are the hyperplane sections of the varieties in 𝔉.

Claim 3.0.5. Let 𝔙 be the variety as above. Then there is no irreducible 1–dimensional family 𝔊 of subspaces of dimension n − 4 in Λ, containing W such that the projection of V from a general subspace of 𝔊 is a cone over the Veronese surface V_{2,2}.

Proof of the Claim ??. The proof is similar to the one of Claim ??, so we will be brief.

We argue by contradiction and assume there is such a 1–dimensional family 𝔊. Then for a general space Ξ ∈ 𝔊, 𝔙 sits in the cone over V_{2,2} with vertex Ξ, and therefore it contains a family ℭ_{Ξ} of irreducible varieties of dimension n − 3 spanning linear spaces of dimension n − 1 and contained in quadric cones of dimension n − 2 with vertex W over a conic of V_{2,2}.

As in the proof of Claim ??, one proves that ℭ_{Ξ} is not constant. This implies that 𝔙 contains a 3–dimensional family 𝔇 of varieties of dimension n − 3 spanning linear spaces of dimension n − 1. Let p ∈ 𝔙 be a general point and consider the projection in ℙ^{n+1} of 𝔙 from p, whose image, birational to 𝔙 by the Trisecant Lemma, we denote by 𝔙′. Then 𝔙′ has a family of dimension 2 of varieties of dimension n − 3 spanning linear spaces of dimension n − 2, i.e. the images of the varieties in 𝔇 containing p. The general surface section S′ of 𝔙′ spans a ℙ^{5} and it has a 2–dimensional family of irreducible plane curves, so it is the Veronese surface V_{2,2}. It is well known that V_{2,2} is not extendable (see [?, Corollary 2.4.5]), so 𝔙′ is a cone over V_{2,2} with vertex a linear space of dimension n−5. This is however impossible because 𝔙′ should contain the linear space of dimension n − 3 which is the image of the exceptional divisor corresponding to p, whereas the cone over V_{2,2} with vertex a linear space of dimension n − 5 contains no such subspace. This contradiction proves the Claim. □

At this point we can argue exactly as in the case n = 4 to see that the spaces T of dimension n − 3 vertices of the cones over V_{2,2} containing the hyperplane sections Z of X, pairwise intersect each other in a variable space of dimension n − 4, hence they lie in a space of dimension n − 2, which is the vertex of a (n + 1)–dimensional cone over V_{2,2} containing X, so we are in case (iii).

Suppose that Y is a projection in ℙ^{7} of the Veronese threefold V_{3,2} and not a hyperplane section of the Segre variety Seg(2, 2). Note that γ(Y) = 1 since the tangential contact locus to Y at two general points p_{0}, p_{1} is the unique conic in V_{3,2} passing through p_{0}, p_{1}.

By taking into account Example 2.5 of [?], we may assume that Y is not the projection of V_{3,2} from a secant (not tangent) line to it. We have to prove that we are in case (iv), and it suffices to do it when X has dimension 4. Indeed, if n > 4 then by arguing by induction we have that the general hyperplane section of X is a cone and therefore X itself is a cone.

We have f(Y) = 1 and f(X) = 2. Then γ(X) ⩾ f(X) = 2 (see Remark ??). On the other hand, by Proposition ??, we have γ(X) ⩽ γ(Y) + 1, and since γ(Y) = 1 we get γ(X) ⩽ 2 so γ(X) = 2. Then for a general choice of p_{0}, p_{1} ∈ X, one has that Γ_{p_{0},p_{1}} is a 2–dimensional quadric (see Proposition ??). Thus X is swept out by a family Q of 2–dimensional quadrics such that, given two general points p_{0}, p_{1} ∈ X, there is a quadric Q of Q containing p_{0}, p_{1}. Note that the general quadric Q is irreducible, since so is the conic passing through two general points of Y. Thus Γ_{p_{0},p_{1}} = Ψ_{p_{0},p_{1}} because Ψ_{p_{0},p_{1}} ⊆ Γ_{p_{0},p_{1}} (see Remark ??), Γ_{p_{0},p_{1}} is irreducible, and both Ψ_{p_{0},p_{1}} and Γ_{p_{0},p_{1}} have the same dimension f(X) (see Definition ?? and Proposition ?? , (ii)).

Let x be a general point of X and consider the general tangential projection τ_{x} of X and its image X_{1}, which is a non–degenerate surface in ℙ^{3}. Given a general quadric Q in Q, let us denote by Q_{1} its image via τ_{x}. There are three possibilities:

(a) Q_{1} is a linear subspace;

(b) Q_{1} is a 2–dimensional quadric;

(c) Q_{1} is a conic.

Case (a) is not possible, because it would imply that the line joining two general points of X_{1} lies in X_{1}, so X_{1} itself would be a linear space, a contradiction. Also case (b) is not possible. Indeed, this would imply that X_{1} = Q_{1} is a quadric. Since γ(X) = f(X) = 2, then Q_{1} has to be a smooth quadric (see Remark ??). On the other hand X_{1} equals Y_{1} by Lemma ??. The surface Y_{1} is in turn a projection to ℙ^{3} of the Veronese surface V_{2,2}, which is the image of the tangential projection of the Veronese threefold V_{3,2}. The only way in which a projection of the Veronese surface V_{2,2} can be a smooth quadric is if it is the projection from a secant (not tangent) line of V_{2,2}. But then Y would be the projection of V_{3,2} from a secant (not tangent) line to it, contrary to our assumptions. Thus we are left with case (c), in which Q_{1} is irreducible, otherwise Q itself would be reducible, a contradiction. So Q is a cone with vertex a point v lying in T_{X,x}. But then X is a cone, since its general tangent space contains v. Thus we have proved we are in case (iv).

Suppose finally that Y is a hyperplane section of the Segre variety Seg(2, 2), and let H be the hyperplane spanned by Y in the ℙ^{8} where Seg(2, 2) sits. Recall that Seg(2, 2) contains a 4–dimensional family F of rank 4 quadric surfaces parameterized by (ℙ^{2})^{∗} × (ℙ^{2})^{∗}: the general element Q_{L,N} of F is the image of L × N, with L, N lines in the two factors.

Claim 3.0.6. In the above setting, H is not tangent to all quadrics in F.

Proof of Claim ??. Suppose by contradiction H is tangent to all quadrics in F. Then for general lines L, N of ℙ^{2}, H would cut Q_{L,N} in a pair of lines of the type L×{x}, {y}×N, with x ∈ N, y ∈ L. Fix L and let N vary. Then for a general N ∈ (ℙ^{2})^{∗}, H contains a unique line of the form L × {x}, with x ∈ N. Let R_{L} be the set of all these points x ∈ ℙ^{2} when N varies. By the above considerations, R_{L} has to intersect the general line N in one point, so R_{L} is itself a line. So we have a rational map

𝔭 : L ∈ (ℙ^{2})^{∗} −−→ R_{L} ∈ (ℙ^{2})^{∗}.

Now we have three possibilities:

(1) the image of 𝔭 is the whole of (ℙ^{2})^{∗};

(2) the image of 𝔭 is a curve ℭ in (ℙ^{2})^{∗};

(3) 𝔭 is constant.

In case (1), H would contain the whole of Seg(2, 2), which is a contradiction. Indeed, let x, y ∈ ℙ^{2} be general points. Consider the pencil P_{x} of lines containing x, whose image via 𝔭 is a curve ℭ_{x} in (ℙ^{2})^{∗}. Hence we find a line in ℭ_{x} containing y, so H contains the point (x, y) ∈ Seg(2, 2).

One finds the same contradiction also in case (2). Indeed, let again x, y ∈ ℙ^{2} be general points. It is clear that the image of the pencil P_{x} of lines containing x via 𝔭 coincides with ℭ. So the same argument as above implies that H contains the point (x, y) ∈ Seg(2, 2).

In case (3), H would contain the image of ℙ^{2} × ℙ^{1} and, by symmetry, also of ℙ^{1} × ℙ^{2}, hence Y would be reducible, again a contradiction. This proves the claim. □

We prove now that we are in case (v), and we may assume that X has dimension 4 (recall that Seg(2, 2) is not extendable, see e.g. [?, Corollary 2.4.5]). Then X sits in ℙ^{8} and, having Y as hyperplane section, it has degree 6. We have f(X) = 2 and f(Y) = 1. Moreover the general tangential projection Y_{1} is equal to the general tangential projection of Seg(2, 2), that is a smooth quadric in ℙ^{3} (see [?, Corollary 5.4.2]). So by Remark ??, γ(Y) = f(Y) = 1. By Proposition ?? we have γ(X) ⩽ 2 and therefore γ(X) = 2 by (??). So, by Proposition ??(ii), X is swept out by a 4–dimensional family Q of 2–dimensional quadrics such that, given two general points p_{0}, p_{1} ∈ X, there is a unique quadric Q of Q containing p_{0}, p_{1}. The general quadric Q is irreducible, since so is the conic passing through two general points of Y, because of Claim ??.

Let x be a general point of Y, hence of X, and consider the general tangential projection τ_{x} of X and X_{1} its image. This coincides with the image of a general tangential projection Y_{1} of Y through x by Lemma ??. On the other hand, by Remark ??, this coincides with the image of a tangential projection of Seg(2, 2), which is a smooth quadric in ℙ^{3} (see [?, Corollary 5.4.2]). Hence X_{1} is a smooth quadric surface. For a general quadric Q in Q, let Q_{1} be its image via τ_{x}. A priori, for Q_{1} one of the possibilities (a), (b), (c) in p. ?? may happen. Again, case (a) is not possible, because it would imply that the line joining two general points of X_{1} lies in X_{1}, so X_{1} would be a linear space, a contradiction. In case (c) the conic Q_{1} is irreducible, otherwise Q would be reducible, a contradiction. So Q is a cone with vertex a point v lying in T_{X,x}. Since the image of Q is a conic under a general tangential projection of X, then the general tangent space contains the vertex v of Q, and this implies that X is a cone, a contradiction. So let us assume we are in case (b) in which Q_{1} is an irreducible quadric, thus it coincides with X_{1}, and therefore it is smooth. Moreover, since the general fibres of τ_{x} are quadrics in Q, we conclude that two general quadrics in Q intersect transversally at a point.

Let Q be the general quadric in Q and let Π be its span. Let π : X −−→ ℙ^{4} be the projection of X from Π.

Claim 3.0.7. The map π : X −−→ ℙ^{4} is birational.

Proof of Claim ??. One argues as in [?, Claim 8.13]. We reproduce here shortly the argument for the reader’s convenience.

Let X′ be the image of X via π. The restriction of π to a general quadric Q in Q is birational to its image, which is a plane of ℙ^{4} contained in X′. This yields that X′ is a subspace of ℙ^{4} and, being non–degenerate, then X′ = ℙ^{4}.

Assume by contradiction that π is not birational. Then, if x ∈ X is a general point, there is a point y ∈ X, distinct from x, such that π(x) = π(y). There is some quadric Q′ ∈ Q containing x and y. Then Q′, as the general Q ∈ Q, has non–empty intersection with Π. Since the fibre of x in π is 0–dimensional, also the fibre of x in π_{|Q′} is 0–dimensional. This implies that ⟨Q′⟩ ∩ Π is only one point z ∈ Q′, and that the line ⟨x, z⟩ is not contained in Q′, so it intersects Q′ only at x and z, contradicting π(x) = π(y). □

The restriction of π to the general hyperplane section Y of X is the projection of Y to ℙ^{4} from the plane spanned by a conic of Y.

Claim 3.0.8. The image of Y via π_{|Y} is a quadric of ℙ^{4}.

Proof. Let 𝔔 be a 2–dimensional quadric in Seg(2, 2) and consider the projection 𝔤 : Seg(2, 2) −−→ ℙ^{4} from the 3–space ⟨𝔔⟩. By the same argument as in Claim ??, we see that 𝔤 is birational and maps a general quadric in Seg(2, 2) to a plane. This yields that 𝔤^{−1} : ℙ^{4} −−→ Seg(2, 2) is defined by the linear system of quadrics containing two skew lines a, b in ℙ^{4} and the planes images of the quadrics on Seg(2, 2) are the planes intersecting in one point each of the lines a, b. Then, via 𝔤, the hyperplane section Y of Seg(2, 2) is mapped to a quadric (containing a and b), and at the same time it is projected to this quadric from the plane of a conic contained in it. □

Hence the birational map π^{−1} : ℙ^{4} −−→ X is defined by a linear system L of quadrics. Since the general hyperplane through Π is mapped to the general hyperplane of ℙ^{4}, the

hyperplanes of ℙ^{4}, plus a fixed hyperplane P′, form a sublinear system of L. This implies that 2P′ ∈ L, hence the base locus B of L sits in P′. Moreover P′ has to be the image of the exceptional divisor of the blow–up of X along Q, and therefore P′ is the image of the hyperplane P tangent to X along Q. Note that P cuts X along a threefold Z of degree 6 singular along Q.

The general quadric of Q maps via π to a plane which intersects P′ along a line. The planes thus obtained vary in a 4–dimensional family Q′ with the property that given two general points x, y of ℙ^{4} there is a unique plane of the family Q′ containing x, y. The line ⟨x, y⟩ cuts P′ in a point z. There is then a unique secant line ℓ to the base locus B of L passing through z (because of the uniqueness of the quadric in Q containing two general points of X), so that the unique plane of Q′ through x, y is ⟨x, y, ℓ⟩. This implies that the intersection lines of the planes in Q′ with P′ vary in a congruence C of secants to B such that given a general point z ∈ P′, there is a unique line in C containing z.

We claim that C is a congruence of type (1, 1). Let in fact S be a general plane in P′. We have to show that in S there is a unique line of the congruence C. Note that S is the projection via π of a general subspace T of dimension 6 contained in P and containing Π. The intersection of T with X is a surface (of degree 6) consisting of Q counted twice, plus another quadric Q′. The lines of C in S are just the images of Q′ via the projection from Π. This implies that Q and Q′ intersect along a line and the projection of Q′ from Π is the unique line of the congruence C sitting on S. This proves our claim, hence C is given by all lines in P′ which meet two skew lines L_{1}, L_{2} of P′ that, a priori, can be also infinitely near (see §??) which form the base locus B of L.

In conclusion L is the 8–dimensional linear system of quadrics in ℙ^{4} passing through two skew lines. The skew lines cannot be infinitely near because, as we said, the image X_{1} of a general tangential projection of X is a smooth quadric, whereas, if the two lines were infinitely near, it would be a cone. Finally the image of ℙ^{4} via this linear system of quadrics is Seg(2, 2) (see Claim ??). □

Remark 3.0.9. In the hypoteses of Theorem ??, if in addition we assume that X is smooth, then X = Seg(2, 2).

Indeed, since s(X) = 2n + 1 − f(X) = n + 3, if X is smooth, it can be isomorphically projected to a ℙ^{n+3} and Zak’s theorem on linear normality (see Chapter I, Corollary 2.11 of [?]) implies 3n ⩽ 2(n + 2) and hence n = 4. Then the generic projection of X to ℙ^{8} is a Severi variety (see again Definition 1.2 and Theorem 4.7, Chapter IV, [?]; see also [?]), whence the assertion.

4. Scrolls

In this section we will dispose of the classification of defective, non–degenerate 4–folds X in ℙ^{r}, which are scrolls in 3–spaces over a curve. The cases f(X) ⩾ 2 (and r = 8) have been dealt with in Proposition ?? and in Theorem ??. So we will assume here that f(X) = 1 and r ⩾ 9.

Before stating and proving the classification theorem, let us construct some examples.

Example 4.0.1. Let us consider a plane Π in ℙ^{r}, with r ⩾ 9. Consider an irreducible surface scroll Σ sitting in a (r − 3)–space skew with Π. Let C be the curve parameterizing the lines of Σ. Let C ⊂ Π^{∗} be an irreducible curve. Suppose there is a rational dominant map ϕ : C −−→ C. Then we can consider the 4–fold scroll X which is the Zariski closure of the union of the 3–spaces joining a line of Σ, corresponding to a point c of C where ϕ is defined, to the line corresponding to the point ϕ(c). Assume that C ⊂ Π^{∗} is not a line, so that the the 3–spaces of the ruling of X do not pass through the same point, hence X is not a cone. Since the 3–spaces of the ruling of X pairwise meet at a point, then also two general tangent spaces to X meet so that X is defective.

Note that X is singular at all points where two 3–spaces of the ruling meet.

Example 4.0.2. Consider an irreducible 1–dimensional family P of planes in ℙ^{r}, with r ⩾ 9, such that two general planes of the family intersect at a point, and it does not happen that either all the planes of P pass through the same point or all the planes of P intersect the same plane along a line. Then by Lemma 4.1 from [?] (Morin’s theorem), the planes of P span at most a 5–space.

Consider an irreducible curve C ⊂ ℙ^{r} which has a dominant rational map ϕ : C −−→ P, and is in a sufficiently general position with respect to the planes of the family P. Then consider the 4–dimensional scroll X which is the Zariski closure of the union of all 3– spaces joining a point x ∈ C where ϕ is defined, with the plane corresponding to ϕ(c). Then X is not a cone and it is defective. As in Example ??, X is singular.

Example 4.0.3. Let S ⊂ ℙ^{9} be an irreducible, projective surface, and let C be an irreducible 1–dimensional family of curves on S not passing through the same point, such that the general curve in C is irreducible, spans a 3–dimensional space and two general curves of C intersect at a point. Let X be the 4–fold scroll which is swept out by the 3– dimensional spaces spanned by the curves of C. Then X is not a cone and it is defective. As in Example ??, X is singular.

For a specific example of this type, consider the Veronese surface V_{2,3} ⊂ ℙ^{9}. It contains the images of the lines of ℙ^{2} which are rational normal cubics, each spanning a 3–dimensional linear space. Let C be an irreducible 1–dimensional family of these curves corresponding to a family of lines which do not pass through the same point.

Other infinitely many examples are obtained in the following way. Consider the Veronese 3–fold V_{3,2} ⊂ ℙ^{9}, which is defective, any two of its tangent spaces meeting at a point. We can view V_{2,3} as the image of the dual Veronese map which sends a point H of the dual (ℙ^{3})^{∗} of ℙ^{3} (i.e., H is a plane of ℙ^{3}) to the quadric 2H in the ℙ^{9} parameterizing the quadrics of ℙ^{3}. Then the tangent space to V_{2,3} at the point 2H consists of all the points of ℙ^{9} corresponding to reducible quadrics of the form H + H′ with H′ varying in (ℙ^{3})^{∗}. Consider any irreducible curve Γ ⊂ (ℙ^{3})^{∗} and consider the 4–fold X which is swept out by the tangent spaces to V_{3,2} at the points 2H corresponding to points H of Γ. We claim that X is of the type described above. Indeed, consider the surface S ⊂ ℙ^{9}, birational to the symmetric product Γ(2) of the curve Γ, consisting of all points H + H′ with H, H′ ∈ Γ. Then the tangent space to V_{3,2} at the point 2H, where H ∈ Γ, cuts S in the curve C_{H}, isomorphic to Γ, described by the reducible quadrics H + H′, with H′ varying in Γ. When H varies in Γ the curve C_{H} varies in a 1–dimensional family C of curves on S, enjoying the required property.

Theorem 4.0.4. Let X ⊂ ℙ^{r}, with r ⩾ 9, be a non–degenerate, linearly normal, defective 4–fold scroll in 3–spaces over a curve with f(X) = 1. Then one of the following cases occurs:

(i) X is a cone with vertex a point over a non–defective threefold; (ii) X is as in Example ??; (iii) X is as in Example ??; (iv) r = 9 and X is as in Example ??.

In all these cases X is singular.

Proof. If X is a cone, the assumption f(X) = 1 implies that X has only one point as vertex and that its general hyperplane section is non–defective (see Lemma ??). Assume next that X is not a cone.

Claim 4.0.5. Two general 3–spaces of the ruling of X meet at a single point.

Proof of the Claim ??. Let p_{0}, p_{1} ∈ X be general points, and let P_{0}, P_{1} be the 3–spaces of the ruling of X through p_{0}, p_{1} respectively. Consider the tangential projection τ_{p_{0}} : X −−→ X_{1}. Since X is defective with f(X) = 1, then X_{1} ⊂ ℙ^{r−5} is a non–degenerate 3–fold by (??). Since X is a scroll and X_{1} is non–degenerate, the 3–spaces of the ruling of X are mapped by τ_{p_{0}} to planes, hence P_{1} ( that is contained in T_{X,p_{1}}) intersects T_{X,p_{0}} at a point p which is the unique intersection point of T_{X,p_{0}} and T_{X,p_{1}} because f(X) = 1 (see Remark ??). Similarly, P_{0} has to intersect T_{X,p_{1}} at p, hence P_{0} and P_{1} have to intersect at p, and only at p. □

Let P be a general 3–space of the ruling of X. The intersection point of P with another general 3–space P′ of the ruling cannot stay fixed as P′ varies, because we are assuming X is not a cone. Then the Zariski closure of the union of these intersections is an irreducible curve C_{P} in P. Moreover if P and P′ are general 3–spaces of the ruling of X, then C_{P} and C_{P′} intersect at one point, i.e., the intersection point of P and P′.

If C_{P} is a line, then these lines pairwise meet at a point. Since we are assuming X is not a cone, then they lie in one and the same plane. Then we are in case (ii).

If C_{P} is not a line, but spans a plane, then these planes pairwise meet at a point but cannot pass through the same point because X is not a cone. So, by applying Lemma 4.1 from [?], either they intersect the same plane along a line, or they span at most a 5–space. In the former case we are again in case (ii) and the curves C_{P} are lines, contrary to our assumption. So we are in the latter case and in case (iii).

Finally, assume C_{P} spans P. Then take four general 3–spaces P_{i} of the ruling, for 1 ⩽ i ⩽ 4. If p_{i} = P ∩ P_{i}, for 1 ⩽ i ⩽ 4, then the points p_{1}, …, p_{4} are independent, so they span P. This implies that X is contained in ⟨P_{1}, P_{2}, P_{3}, P_{4}⟩ which has dimension at most 9. On the other hand, we have r ⩾ 9, hence r = 9, and we are in case (iv). □

5. Top species: the irreducible case

5.1. Here we will consider the case of an irreducible, non–degenerate, defective variety X ⊂ ℙ^{r} of dimension n ⩾ 4, with r ⩾ 2n + 1. Then we have δ(X) = f(X) (see Remark ??). The case δ(X) = n − 2 is covered by Theorem ??. So we may assume 1 ⩽ δ(X) ⩽ n − 3.

In this section we will assume f(X) = δ(X) = γ(X). Hence, for p_{0}, p_{1} ∈ X general points, one has that Γ_{p_{0},p_{1}} = Ψ_{p_{0},p_{1}} equals the general entry locus and it is a quadric of rank at least 2 in ℙ^{δ(X)+1} (see Proposition ??, (ii)). Therefore X is a, perhaps singular, LQEL variety (see §??). If δ(X) = 1, then X presents the LCEL case. This happens when X is a variety of the top species.

By taking a general linear section Y of X of dimension m = n − δ(X) + 1 we have Y ⊂ ℙ^{s}, with s = r − δ(X) + 1 ⩾ 2m + 1, and δ(Y) = 1 by Lemma ??. Hence the varieties X in question are extensions of varieties of the top species, thus presenting the LCEL case. Therefore, as a first instance, we will limit ourselves to consider only the top species case. In this section we will also stick to the irreducible case, i.e. the case in which the general entry locus is an irreducible conic.

5.2. From now on in this section we let X ⊂ ℙ^{r} be a non-degenerate, defective variety of dimension n ⩾ 4 with r ⩾ 2n + 1. We will assume that X is of the top species, so that δ(X) = f(X) = γ(X) = 1, and presents the irreducible case. By definition of LCEL variety (see p. ??), there is an irreducible family C of dimension 2n − 2 of generically irreducible conics such that, given two general points p_{0}, p_{1} ∈ X, there is a unique conic of C containing p_{0}, p_{1}.

Lemma 5.2.1. In the above setting, let x ∈ X be a general point and let t be a general tangent direction to X at x. Then there is a unique irreducible conic of the family C passing through x and tangent to t.

Proof. Let X[2] be the Hilbert scheme of X parameterizing the 0–dimensional subschemes of X of length 2. Consider the set

Ω = {(Γ, ξ) ∈ C × X[2] : ξ ⊂ Γ}

The second projection p_{2} : Ω → X[2] is birational, by definition of LCEL variety, and it is surjective by the projectivity of C. Suppose, by contradiction, that, given x and t general, there is more than one conic through x tangent to t. Then, by Zariski’s Main Theorem, given a general non–reduced length 2 subscheme ξ of X (which is a smooth point of X[2]), there would be a positive dimensional family of conics in C containing ξ. This would imply that the subset Ω′ of Ω consisting of pairs (Γ, ξ) such that ξ is non–reduced has a component of dimension at least 2n, since the locus of non reduced schemes of length 2 of X has dimension 2n − 1. The first projection p_{1} : Ω′ → C has general fibres of dimension 1, so p_{1}(Ω′) would have dimension at least 2n − 1, a contradiction. The general conic in C passing through x is smooth at x by hypothesis and its tangent direction at x is general by the unicity property proved above. In conclusion, there is a unique irreducible conic in C passing through a general point x and having a general tangent direction t at x. □

If x ∈ X is a general point, define the Scorza map

S_{x} : X −−→ T_{X,x}

sending a general point y ∈ X to the point T_{X,x} ∩ T_{X,y}. If C_{x,y} is the unique conic in X passing through two general points x and y of X, then one has

S_{x}(y) = T_{C_{x,y},x} ∩ T_{C_{x,y},y}.

Indeed, T_{C_{x,y},x} ∩ T_{C_{x,y},y} is a point, contained in T_{X,x} ∩ T_{X,y} and therefore coinciding with T_{X,x} ∩ T_{X,y}, which is also a point. If there is no danger of confusion, we write S instead of S_{x}.

Proposition 5.2.2. In the above setting, the Scorza’s map S_{x} is birational so that X is rational. Moreover its inverse S_{x}^{−1} is defined by a linear system of hypersurfaces of degree d having a point of multiplicity d − 2 at x.

Proof. We prove that S_{x} is birational by constructing its inverse S_{x}^{−1}. Indeed, let y ∈ T_{X,x} be a general point. Let r = ⟨x, y⟩. By Lemma ??, there is a unique irreducible conic C ∈ C passing through x and tangent to r at x. Let s be the tangent line to C through y other than r. It is tangent to C at a point z. Then z = S_{x}^{−1}(y). This proves that S_{x} is birational.

As for the rest of the assertion, it follows from the fact that S_{x}^{−1} maps a general line through x to a conic on X. □

The main point of our analysis is to study the Scorza map. Specifically, following Scorza’s ideas in [?], we will prove that, under suitable hypotheses, its inverse is defined by a linear system of quadrics.

5.3. We will study now the family of conics C. In doing so, we will use some ideas contained in [?], where however X is assumed to be smooth.

Let x ∈ X be a general point. We will denote by p_{x} : F_{x} → C_{x} the universal family of conics in C passing through x. One has F_{x} ⊆ C_{x} × X. Let ϕ_{x} : F_{x} → X be the projection to the second factor, which is birational. Then F_{x} is irreducible, so also C_{x} is irreducible with dim(C_{x}) = n − 1 and dim(F_{x}) = n.

The map p_{x} has a natural section σ_{x} mapping a point of C_{x} corresponding to a conic C, to the point x ∈ C ⊂ F_{x}. The image E_{x} of σ_{x} is contracted by ϕ_{x} to the point x. Consider the blow–up π_{x} : X̃ → X of X at x. So we have the following commutative diagram

\xymatrix{F_{x} \ar[d]^{p_{x}} \ar[dr]^{ϕ_{x}} \ar@{-->}[r]^{ψ_{x}} & X̃ \ar[d]^{π_{x}} \\ C_{x} \ar[u]^{σ_{x}} & X}

where ψ_{x} is a birational map whose indeterminacy locus is contained in E_{x}. Let S_{x} be the closed subscheme of C_{x} parameterizing those conics in C_{x} which are singular at x. We will denote by D_{x} the image of S_{x} via the section σ_{x}. Then we claim that D_{x} is exactly the indeterminacy locus of ψ_{x}. In fact D_{x} is clearly contained in the indeterminacy locus of ψ_{x}. To prove that ψ_{x} is defined off D_{x}, one first remarks that E_{x} is the preimage of x via ϕ_{x}. Then [?, Lemma 4.3] assures that E_{x} is a Cartier divisor at points off D_{x}. So we restrict ψ_{x} to the open subset U_{x} = F_{x} \ D_{x} and we have that the schematic preimage of x via ϕ_{x} is a Cartier divisor on U_{x}. Then we can apply the universal property of the blow–up to conclude that ψ_{x} is defined on U_{x}.

Thus ψ_{x} is defined at the general point of E_{x} and its inverse is defined at the general point of the exceptional divisor E_{x} of the blow–up π_{x} : X̃ → X. Hence ψ_{x} induces a birational map ψ_{x,0} : E_{x} −−→ E_{x} (see also Lemma ??). The geometric meaning of the map ψ_{x,0} is clear: it associates to a point (c, x) ∈ E_{x} \ D_{x} the point of E_{x} corresponding to the direction of the tangent line at x to the conic C corresponding to the point c ∈ C_{x}.

Let D_{x} be the closure in E_{x} of the points corresponding to all directions at x of irreducible components of conics in S_{x}. We set

χ(X) = dim(D_{x}),

which is independent on x ∈ X. Note that D_{x} ⊆ L_{x}, where L_{x} was introduced in §??, thus χ(X) ⩽ ℓ(X). Given a point in S_{x} corresponding to a conic C singular at x the plane Π_{C} spanned by C is tangent to X at x and the points corresponding to the directions of the lines in Π_{C} through x describe a line r_{C} in E_{x}. It may either happen that C has rank 2 or rank 1. In the former case we get two distinct points p, q ∈ D_{x} and the line r_{C} joins the points p and q. In the latter case C consists of a line Z counted with multiplicity 2, whose direction corresponds to a unique point z ∈ E_{x}. The plane Π_{C} still gives rise to a line r_{C} in E_{x} through z. The following lemma shows that something similar to the rank 2 case for C happens in this situation (recall the definitions and notation from §??).

Lemma 5.3.1. In the above setting, i.e., when C consists of a line Z counted with multiplicity 2, the line r_{C} belongs to the Zariski tangent space to the scheme L_{x} at the point z. Every point in L_{x} is a base point for II_{x} and all quadrics in II_{x} are tangent to r_{C} at z.

Proof. The plane Π_{C} is tangent to X at the general point of the line Z. Therefore we have an inclusion of normal sheaves N_{Z,Π_{C}} → N_{Z,X}, hence an inclusion N_{Z,Π_{C}} (−x) → N_{Z,X}(−x), which induces an inclusion H^{0}(Z, N_{Z,Π_{C}} (−x)) ⊆ H^{0}(Z, N_{Z,X}(−x)). The 2–dimensional space H^{0}(Z, N_{Z,Π_{C}} (−x)) is the tangent space to the pencil of lines through the point x in the plane Π_{C}, whereas H^{0}(Z, N_{Z,X}(−x)) is the Zariski tangent space to L_{x} at z. So this proves the first assertion. Every point in L_{x} is a base point for II_{x} (see §??). As for the final assertion, consider any quadric Q in II_{x}. Then there is a hyperplane section H of X tangent at x, having a double point at x with the projectified tangent cone at x coinciding with Q. The hyperplane cutting out H contains the plane Π_{C} which is tangent also to H along Z. This implies that the quadric Q has intersection multiplicity at least 2 with r_{C} at z, as wanted. □

Let us denote by J_{X,x}, or simply by J_{x}, the Zariski closure of the union of all tangent spaces to L_{x} in E_{x} at the points of D_{x}.

Lemma 5.3.2. In the above setting, let D ⊆ E_{x} \ D_{x} be an irreducible complete curve contracted to a point by ψ_{x,0}. Then the general point (c, x) ∈ D corresponds to a rank 2 conic C = N + M, where N is a fixed line through x and M decribes a cone not containing x, with vertex a point of N.

If q ∈ E_{x}\D_{x} and if there is an irreducible curve D in E_{x}\D_{x} such that ψ_{x,0}(D) = {q}, then D is not complete and q ∈ J(D_{x}, D_{x}) ∪ J_{x}.

Proof. We shall adapt to our situation an argument of Kebekus (see [?, Proof of Theorem 3.4]). Let ψ_{x,0}(D) = p. Take D̃, the normalization of D, and consider the scheme Y over D̃ which is gotten by base–change from our family.

Suppose first the general fiber of π_{Y} : Y → D̃ is irreducible, so that Y is an irreducible surface. Let D_{0} ⊂ Y be the section of π_{Y} : Y → D̃ induced by σ_{x}. Note that Y is smooth along D_{0} by the hypothesis that D ∩ D_{x} is empty. Since the induced morphism contracts D_{0} to the point x ∈ X, the restriction of the differential of ϕ_{Y} : Y → X induces a morphism T_{ϕ_{Y}} : N_{D_{0},Y} → l_{p} ≃ ℂ, where l_{p} is the line in T_{x}(X) corresponding to p ∈ E_{x}. Since N_{D_{0},Y} is not trivial (ϕ_{Y} (D_{0}) = x implies D_{0}^{2} < 0), the map T_{ϕ_{Y}} is not constant, so it has a zero. The corresponding curve of the family is singular at x. This is a contradiction (here we used Kebekus’ argument).

Suppose now that every fiber of π_{Y} : Y → D̃ is reducible, consisting of two irreducible components each one isomorphically mapping to different lines contained in X. Only one of these two lines, call it N, contains x and has the direction corresponding to p. So N does not move. Let M be the other line, which varies in a 1–dimensional family. By the assumption D ∩ D_{x} = ∅, the intersection point y of N with M has to stay fixed when M varies. Then Y has an irreducible component mapping to a cone with vertex y not passing through x.

Let q ∈ E_{x} be an indeterminacy point of ψ_{x,0}^{−1}, and let D ⊆ E_{x} \ D_{x} be an irreducible curve such that ψ_{0}(D) = q. Assume that q ∉ J_{x}, in particular q ∉ D_{x}. Then the general point (c, x) ∈ D corresponds to an irreducible conic C containing x, whose tangent line has direction corresponding to q. By Kebekus’ argument that we already used above, D cannot be complete, hence there is some point (c_{0}, x) ∈ \overline{D} ∩ D_{x}. Let C_{0} be the conic corresponding to c_{0}, which is singular at x. Then the direction corresponding to q has to belong to the Zariski tangent space to C_{0} at x. If C_{0} has rank 1, then q ∈ J_{x} by Lemma ??, a contradiction. Then C_{0} has rank 2, and, by the above argument, q ∈ J(D_{x}, D_{x}). □

Next we will restrict our attention to the case χ(X) ⩽ 0, i.e. dim(D_{x}) ⩽ 0.

Lemma 5.3.3. Suppose dim(D_{x}) ⩽ 0. Then ψ_{x,0}^{−1} induces an isomorphism between E_{x} \ (J(D_{x}, D_{x}) ∪ J_{x}) and its image, so that for every point q ∈ E_{x} \ (J(D_{x}, D_{x}) ∪ J_{x}) there is a unique conic in C_{x} with tangent line at x having direction corresponding to q.

Proof. Let

\xymatrix{Ẽ_{x} \ar[d]_{α} \ar[dr]^{ψ̃_{x,0}} \\ E_{x} \ar@{-->}[r] & E_{x},}

be a resolution of ψ_{x,0} : E_{x} −−→ E_{x}.

Consider a point q ∈ E_{x} \ (J(D_{x}, D_{x}) ∪ J_{x}), and suppose that ψ_{x,0}^{−1} is not defined at q. Then also ψ̃_{x,0}^{−1} is not defined at q. By Zariski’s Main Theorem, there is an irreducible, complete curve C ⊆ ψ̃_{x,0}^{−1}(q). Let D = α(C), which we may assume to be a curve by the assumption that ψ_{x,0}^{−1} is not defined at q. Clearly D cannot be contained in D_{x}, which is a finite set, since by assumption D_{x} is finite. This gives a contradiction to Lemma ??.

Thus ψ̃_{x,0}^{−1} is defined at q, hence also ψ_{x,0}^{−1} is defined at q. If (c, x) = ψ_{x,0}^{−1}(q), then the conic C corresponding to c is smooth at x, with tangent line having the direction corresponding to q. □

If χ(X) ⩽ 0 one has

(13)  dim(J(D_{x}, D_{x})) ⩽ 1.

We also need an estimate on the dimension of J_{x}.

Lemma 5.3.4. If χ(X) ⩽ 0, one has dim(J_{x}) ⩽ n − 3.

Proof. Since dim(D_{x}) ⩽ 0, J_{x} is a union of finitely many linear spaces T_{L_{x},z} with z ∈ D_{x}. We have to prove that none of these spaces has dimension m ⩾ n − 2.

Suppose first there is a point z ∈ D_{x} such that T_{L_{x},z} has dimension n − 1, i.e. it coincides with E_{x}. Since T_{L_{x},z} ⊆ T_{B_{x},z} ⊆ T_{Q,z} for all quadrics Q in II_{X,x}, we deduce that all quadrics in II_{X,x} are cones whose vertex contains z and hence that the map defined by II_{X,x} on E_{x} has fibres of positive dimension. Since this map coincides with τ̃_{x|E_{x}} : E_{x} −−→ X_{1} (see Lemma ?? and remember that f(X) = 1 so that the general fibre of τ_{x} is an irreducible conic), which is generically finite, we get a contradiction.

Next, suppose that there is a point z ∈ D_{x} such that T_{L_{x},z} is a hyperplane in E_{x}. Then if a quadric in II_{X,x} has a double point at a general point y ∈ E_{x}, then it is singular along the whole line ⟨y, z⟩. Since the quadrics in II_{X,x} singular at y correspond to hyperplanes tangent to the image of E_{x} at the image of y, the general Gauss fibre of the image of II_{X,x} has positive dimension. Since the image of E via the map determined by II_{X,x} is X_{1}, we find a contradiction because, being γ(X) = f(X), the Gauss map of X_{1} is generically finite (see Remarks ?? and ??)). □

5.4. We go back now to the study of Scorza’s map S_{x}. If χ(X) ⩽ 0, by (??) and Lemma ??, a general line M ⊂ E_{x} = ℙ^{n−1} does not intersect J(D_{x}, D_{x}) ∪ J_{x}. Moreover, the line M does not intersect the base locus B_{x} of II_{x} because dim(B_{x}) ⩽ n − 3. Indeed, the linear system of quadrics II_{x} cannot have a fixed component because this would be a hyperplane in E_{x}. Then the map induced by II_{x} would be a linear projection being given by a linear system of hyperplanes and X_{1}, coinciding with the image of E_{x} via a linear projection, would be a linear space. This is not possible, because X_{1} is non–degenerate and strictly contained in ℙ^{r−n−1} (see Remark ??). In particular, M does not intersect L_{x} that is contained in B_{x}.

Hence, by Lemma ??, L = ψ_{x,0}^{−1}(M) is a smooth irreducible rational curve such that for every point (c, x) ∈ L the conic C corresponding to c is smooth. By taking the universal family over L we obtain a smooth surface S with a morphism β : S → L = ℙ^{1}, such that every fiber F of β is irreducible and maps to a conic in X through x via the tautological morphism ϕ_{S} : S → X, which is clearly birational onto its image Y = ϕ_{S}(S). The surface S is isomorphic to a surface 𝔽_{e}, with e ⩾ 0. The section E_{0} with E_{0}^{2} = −e is contracted to x on Y, hence e ⩾ 1.

The next lemma is the main tool for the study of Scorza’s map.

Lemma 5.4.1. In the above setting, suppose again χ(X) ⩽ 0. Then Y is a Veronese surface V_{2,2}.

Proof. First of all we claim that T_{X,x} ∩ Y = {x}. Indeed, suppose z ∈ Y is different from x. Then there is a smooth conic C on Y containing x and z. The line ⟨x, z⟩ intersects transversally C at x. If z ∈ T_{X,x}, then the plane ⟨C⟩ would be contained in T_{X,x} because it containes two distinct tangent lines to X at x, namely ⟨x, z⟩ and the tangent line to C at x. Moreover the tangent line to C at x has intersection multiplicity larger than 2 with X at x (because it is the limit of lines tangent at x and secant to X at a general point of C) and therefore the point in E_{x} corresponding to its direction sits in the base locus B_{x} of II_{x}, a contradiction.

Consider now the tangential projection τ_{x} : X −−→ X_{1}. The restriction of τ_{x} to Y is thus defined on Y \ {x} and τ_{x}(Y), which is a curve because the irreducible conics through x contained in Y are contracted to points by τ_{x}, coincides with τ̃_{x}(M), which, by the genericity of M, is a conic Γ. This yields dim(⟨Y ⟩) ⩾ 5, because Y is projected from the span of its tangent cone at x, which is at least a plane.

Let E_{1} ⊂ S be the general curve in the linear system |E_{0} + eF |, where F is a fibre of the ruling of S ≅ 𝔽_{e} and E_{0} is the negative section. Then E_{1} does not intersect E_{0}, therefore ϕ_{S}(E_{1}) does not pass through x so that it is projected via τ_{x} onto Γ. Moreover the restriction of τ_{x} to ϕ_{S}(E_{1}) is birational, because the pull–back to S of the fibres of the restriction of τ_{x} to Y are the curves in |F | and F · E_{1} = 1. This yields deg(ϕ_{S}(E_{1})) = 2. Let H be a general divisor in the linear system |ϕ_{S}^{∗}(O_{Y} (1))|. Then H ≡ 2E_{0} + bF (remember that the curves in |F | are mapped to conics on Y, so H · F = 2). Since 0 = H · E_{0} = −2e + b, we have H ≡ 2(E_{0} + eF). Moreover, as we saw, 2 = H · E_{1} = 2E_{1}^{2} = 2e, hence e = 1. This implies the assertion because dim(⟨Y ⟩) ⩾ 5. □

We can now prove our main result about Scorza’s map which can be seen as a strong improvement of Lemma ??.

Theorem 5.4.2. Let X ⊂ ℙ^{r}, with r ⩾ 2n + 1, be a non-degenerate, defective variety, of dimension n ⩾ 4. Assume that X is of the top species, presents the irreducible case and χ(X) ⩽ 0. Then S_{x}^{−1} is defined by a linear system of quadrics, r ⩽ \frac{n(n+3)}{2} and X is a projection to ℙ^{r} of the Veronese variety V_{n,2}.

Proof. By Lemma ??, the restriction of S_{x}^{−1} to a general plane Π through x maps Π to a Veronese surface V_{2,2}. This implies that S_{x}^{−1} is defined by a linear system of quadrics (see Proposition ??), whence the claim follows. □

5.5. The previous results suffice for the classification of defective 4–folds of the top species presenting the irreducible case. Recall the definition of b(X) from §??.

Proposition 5.5.1. Let X ⊂ ℙ^{r}, with r ⩾ 9, be a non–degenerate, defective variety, of dimension 4. Assume that X is of the top species and presents the irreducible case. Then b(X) ⩽ 1 and if the equality holds then, for a general point x ∈ X, one has:

(i) either B_{x} is a smooth conic and r = 9; (ii) or B_{x} is a line and r ⩽ 11; (iii) or B_{x} is the union of a line L and of a point p ∉ L, and r ⩽ 10.

In cases (ii) and (iii), one has ϵ(X) ⩾ 2.

Proof. By hypothesis, dim(X_{1}) = 3 and the general fiber of the general tangential projection τ_{x} : X −−→ X_{1} ⊂ ℙ^{r−5} is an irreducible conic. Then τ̃_{x}(E_{x}) = X_{1} ⊂ ℙ^{r−5} (see Lemma ??) so that dim(II_{x}) = r − 5 ⩾ 4. Therefore the positive dimensional irreducible components of B_{x} ⊂ E_{x} = ℙ^{3}, which are one dimensional because II_{x} has no fixed components (see §??), are necessarily lines or conics. Indeed, an irreducible non–degenerate curve A ⊂ ℙ^{3} has degree at least three and h^{0}(I_{A}(2)) ⩽ 3 (see also [?, Theorem 6.1]). If the irreducible component is not a line, then it is a plane curve defined by quadric surfaces in ℙ^{3} and hence a conic.

If B_{x} contains a conic C, then II_{x} ⊆ |H^{0}(I_{C,E_{x}}(2))| = ℙ^{4} together with dim(II_{x}) = r − 5 ⩾ 4 implies that equality holds and hence that B_{x} = C. Therefore X_{1} is a quadric hypersurface in ℙ^{4}, being the image of the second fundamental form of X at x by Lemma ??. Since X is of the top species, i.e., γ(X) = f(X) = 1, then X_{1} is smooth (recall that γ(X) = t(X_{1}) + f(X), see Remark ??) and hence C is smooth.

If B_{x} contains a line L, then it cannot contain any other line. Indeed, as we have shown above, B_{x} cannot contain any other line intersecting L because in this case it would be a reducible conic C and X_{1} would be an irreducible quadric cone, yielding γ(X) = f(X)+ 1 > f(X) contrary to our assumption. Since the linear system of quadrics containing two skew lines has only dimension 3, then B_{x} cannot contain any line skew with L. Suppose there is a point p ∈ L such that all quadrics in II_{x} are tangent to a given plane Π at p. Then if q ∈ E_{x} is a general point, the quadrics in II_{x} singular at q are singular at p (the tangent space at p contains ⟨Π, q⟩ = E_{x}) and hence they must be singular along the whole line ⟨p, q⟩. This would imply that X_{1} has general Gauss fiber of positive dimension (by the argument used in the second part of the proof of Lemma ??), against the hypothesis that X is of the top species, i.e., γ(X) = 1. If there are two distinct points p, q ∈ B_{x} off L, again X_{1} would be a quadric cone in ℙ^{4}, which leads again to a contradiction. In fact, the line L′ = ⟨p, q⟩ has to be skew with L, otherwise it is contained in B_{x}, a contradiction. Moreover the map determined by the linear system of quadrics II_{x}, contracts L′ to a point v ∈ X_{1} ⊂ ℙ^{4}, and the projection of X_{1} from v to ℙ^{3} is the image of E_{x} = ℙ^{3} via the linear system of quadrics containing L and L′, that is well known to be a smooth quadric in ℙ^{3}. This proves that only cases (i), (ii), (iii) are possible. In case (i), one has dim(II_{x}) = 4 and r = 9. In case (ii), one has dim(II_{x}) ⩽ 6 and r ⩽ 11. In case (iii), one has dim(II_{x}) ⩽ 5 and r ⩽ 10.

In cases (ii), (iii), X_{1} turns out to be a scroll in planes, and therefore d(X_{1}) > 0 (see §??). This proves the final assertion (see Remark ??). □

We can now prove the classification theorem for defective 4–folds of the top species presenting the irreducible case. It suffices to assume the 4–fold to be linearly normal.

Theorem 5.5.2. Let X ⊂ ℙ^{r}, with r ⩾ 9, be a non–degenerate, linearly normal, defective variety, of dimension 4. Assume that X is of the top species and presents the irreducible case. Then X is one of the following:

(i) an internal projection of the Veronese 4–fold V_{4,2} ⊂ ℙ^{14} from finitely many points with the property that ℓ(X) ⩽ 0;

(ii) the projection of V_{4,2} ⊂ ℙ^{14} from the plane spanned by a conic on it; (iii) the projection of V_{4,2} ⊂ ℙ^{14} from a 4–space spanned by a rational normal quartic

curve on it; (iv) a hyperplane section of Seg(2, 3) ⊂ ℙ^{11}.

Proof. If ℓ(X) ⩽ 0, then also χ(X) ⩽ 0. Then, as we saw in Theorem ??, S_{x}^{−1} is defined by a linear system Λ of quadrics in ℙ^{4} and the base locus B of Λ is clearly finite, because ℓ(X) = 0. Thus we are in case (i).

Assume that ℓ(X) > 0. Since L_{x} ⊆ B_{x}, we have b(X) ⩾ ℓ(X), Proposition ?? yields ℓ(X) = 1.

Suppose S_{x}^{−1} is still defined by a linear system Λ of quadrics (with a base locus B), which is the case if χ(X) ⩽ 0 (see Theorem ??). If z ∈ ℙ^{4} is a general point, mapped by S_{x}^{−1} to a general point y ∈ X, then II_{y} is projectively equivalent to the projection in ℙ^{3} from z of the linear system of cones in Λ with vertex in z. By Proposition ?? and since ℓ(X) = 1, one has b(X) = 1 and:

(a) either B is a smooth conic;

(b) or B is a line;

(c) or B is the union of a line L and of a point p ∉ L.

Cases (a), (b) lead to cases (iii) and (ii) respectively because S_{x}^{−1} : ℙ^{4} −−→ X ⊂ ℙ^{r} is given by the complete linear system of quadrics vanishing on a conic in ℙ^{4} in case (a), respectively on a line in ℙ^{4} in case (b) (the linear systems are complete because X is linearly normal). In case (c), X is the image of ℙ^{4} by the complete linear system of quadrics vanishing on a line and at a point so that X is an internal projection of the 4–fold appearing in case (ii) and it sits in ℙ^{10} (again the linear systems are complete because X is linearly normal). It is easy to compute deg(X) = 10. Indeed, this degree is the number of intersection points of four general quadrics in ℙ^{4} containing a line L and of a point p ∉ L, off p and L. This is the same as computing the self intersection minus 1 of the linear system cut out on a smooth Del Pezzo surface S of degree 4 in ℙ^{4} (intersection of two general quadrics in ℙ^{4} through L) by the quadrics containing the line L of S: one easily checks that this self–intersection is 11. On the other hand X also sits on Seg(2, 3), since Λ is the sum of two linear systems of hyperplanes: Λ_{1} given by the hyperplanes containing the line L; Λ_{2} given by the hyperplanes containing the point p, with dimension 2 and 3 respectively. Indeed, Λ_{1} yields a rational map ϕ_{1} : ℙ^{4} −−→ ℙ^{2} while Λ_{2} defines a rational map ϕ_{2} : ℙ^{4} −−→ ℙ^{3}. Hence ϕ_{1} × ϕ_{2} : ℙ^{4} −−→ ℙ^{2} × ℙ^{3} ⊂ ℙ^{11} is given by a linear system of quadrics vanishing on L and p, that is in Λ, proving the claim. Hence X is contained in a hyperplane section of Seg(2, 3). As we saw, deg(X) = 10. Since also deg(Seg(2, 3)) = 10, we see that X coincides with a hyperplane section of Seg(2, 3), so we are in case (iv).

Assume now that S_{x}^{−1} is not defined by a linear system of quadrics, so that χ(X) = ℓ(X) = 1 and b(X) = 1.

To treat this case we need a preliminary remark. Since ℓ(X) = 1, there is an irreducible component R of the Hilbert scheme of lines in X of dimension 4, thus, by [?, Theorem 2], we have the following possibilities:

(1) either X is swept out by a 1–dimensional family Q of 3–dimensional quadrics generically of rank 5;

(2) or X is swept out by a 2–dimensional family P of planes; and the lines of the family R are the lines contained in the quadrics of Q in case (1), the lines contained in the planes of P in case (2).

By Proposition ??, in case (1) we have only one quadric Q of the family Q passing through the general point x ∈ X (because the lines contained in such a Q through x are parametrized by a conic C_{x} ⊆ L_{x} ⊆ B_{x}) and in case (2) we have only one plane P of the family P passing through x (because the lines contained in such a P through x are parametrized by a line ℓ_{x} ⊆ L_{x} ⊆ B_{x}).

Let us now set B := B_{x}. Then by Proposition ?? we have for B the three cases (a), (b) and (c).

Case (a) corresponds to case (1) above, i.e. X ⊂ ℙ^{9} is ruled by a 1–dimensional family Q of generically smooth 3–dimensional quadrics. Let ψ : X̃ → X be the tautological morphism associated to the universal family α : X̃ → Q. Then ψ is birational since, as we saw, there is only one quadric of the family Q passing through the general point x ∈ X. Let x ∈ X be a general point, which can be seen as a general point of X̃. Recall that X possesses a family C of (generically irreducible) conics such that given two general points of X there is a conic of C containing them. Consider the strict transform C̃_{x} on X̃ of the family C_{x} of conics of C through x. Let D ⊂ X̃ be a Cartier divisor which is the pull-back via α of a divisor in a base point free linear system on Q. If C corresponds to the general point of C̃_{x}, then C ·D > 0. On the other hand the proper transform C′ on X̃ of a reducible conic of C with a double point at x ∈ X is not contained in the exceptional locus of ψ by the generality of x and it is contained in the unique element Q_{x} of Q̃ (that is the family of fibres of α) passing through x (because Q_{x} contains all the lines in X passing through x). Hence C′ · D = 0. Therefore these reducible conics are not contained in C_{x}, contrary to the assumption χ(X) = 1. Case (a) is thus excluded.

Cases (b) and (c) correspond to case (2) above, i.e., X is ruled by a 2–dimensional family P of planes. In case (b) one comes to a contradiction with an argument similar to the one we made in case (a). Indeed, let ψ : X̃ → X be the birational tautological morphism over the universal family α : X̃ → P. The proper transform C′ on X̃ of a reducible conic with a double point at the general point x lies in the plane of the family through x. Let D ⊂ X̃ be a Cartier divisor which is the pull-back via α of a divisor in a base point free linear system on P. Then D · C′ = 0, while C · D > 0 for the strict transform of a general conic in C_{x} through x, a contradiction.

In case (c) we have X ⊂ ℙ^{10} and the image of the general tangential projection τ_{x} is X_{1} = Seg(1, 2) ⊂ ℙ^{5}, since it is the image of E_{x} = ℙ^{3} via the second fundamental form II_{x}, that is the linear system of quadrics with base locus a line and a point. Thus we have two rational maps ϕ_{1} : X −−→ ℙ^{1} and ϕ_{2} : X −−→ ℙ^{2}. Set L_{1} = ϕ_{1}^{∗}(|H^{0}(ℙ^{1}, O_{ℙ^{1}}(1))|) [resp. L_{2} = ϕ_{2}^{∗}(|H^{0}(ℙ^{2}, O_{ℙ^{2}}(1))|)]. Then L_{1} ∨ L_{2} coincides τ_{x}^{∗}(|H^{0}(Seg(1, 2), (1))|), which is H(−2x) (the linear system of hyperplanes with a double point at x) and both linear systems L_{1} and L_{2} have a base point at x. Indeed L_{1} [resp. L_{2}] induces on E_{x} the linear system of planes passing through the line L [resp. containing the point p] and therefore x must be a base point for L_{1} [resp. L_{2}]. By moving x we see that L_{1} and L_{2} vary in continuous families, and therefore in larger linear systems, since X is rational (see Proposition ??). If N_{1} and N_{2} are the complete linear systems containing L_{1} and L_{2}, we have dim(N_{1}) = dim(L_{1}) + 1 = 2 and dim(N_{2}) = dim(L_{2}) + 1 = 3, because L_{1} and L_{2} have the base point x and, by their very definition, are complete under this constraint, because their minimal sum gives the rational map τ_{x} from X to Seg(1, 2) which is linearly normal. Finally, since the minimal sum L_{1} ∨ L_{2} coincides with H(−2x), we have N_{1} ∨ N_{2} = H. By using N_{1} and N_{2} we define a morphism ϕ : X → Seg(2, 3), which is an embedding because N_{1} ∨ N_{2} = H. Then X is realized as a hyperplane section of Seg(2, 3) ⊂ ℙ^{11} because dim(H) = 10. In conclusion, we are in case (iv). □

Remark 5.5.3. Let X be as in Theorem ?? and assume in addition that X is smooth. Then, either we are in cases (ii), (iii) or (iv), or we are in case (i) and either X is the Veronese variety V_{4,2} in ℙ^{14}, or X is the projection to ℙ^{13} of V_{4,2} from a point on it. Indeed, any internal projection of V_{4,2} from more than one point is singular because this projection contracts the unique conic through two (distinct or infinitely near) points to a singular point of the image.

6. Top species: the reducible case

6.1. Next we continue the analysis of varieties X ⊂ ℙ^{r} of dimension n ⩾ 4, with r ⩾ 2n + 1, of the top species, but we turn to the reducible case in which the general entry locus is a reducible conic. Remember that by definition of top species we have γ(X) = 1, which implies f(X) = 1 (and δ(X) = 1 because r ⩾ 2n + 1).

Let p_{0}, p_{1} be general points of X. Then Γ_{p_{0},p_{1}} consists of two incident lines R_{0}, R_{1}, with p_{i} ∈ R_{i} and p_{i} ∉ R_{1−i}, i = 0, 1. In this case Ψ_{p_{0},p_{1}} = R_{0} (see Definition ?? and Proposition ??). This implies, by monodromy, that R_{0}, R_{1} move in one and the same irreducible family R of lines filling X.

Lemma 6.1.1. In the above setting, either X is a cone with vertex a point over a non– defective variety of dimension n − 1 or dim(R) ⩾ n.

Proof. Since there is some line in R through the general point of X, one has dim(R) ⩾ n − 1. Suppose the equality holds. Then that there is a finite number t of lines of R passing through the general point of X, and, by the irreducibility of R, they are exchanged by monodromy.

Let p_{0}, p_{1} be general points of X and let, as above, Γ_{p_{0},p_{1}} = R_{0} ∪ R_{1} be the unique tangential contact locus containing p_{0}, p_{1}, with p_{i} ∈ R_{i} and p_{i} ∉ R_{1−i}, i = 0, 1. Since, given p_{1}, the line R_{0} ∈ R through p_{0} is uniquely determined, this is the unique line in R containing p_{0} and t = 1.

If we let p_{0} stay fixed and move p_{1}, the line R_{0} stays also fixed, whereas R_{1}, the unique line in R through p_{1}, always intersects R_{0}. This proves that the lines in R pairwise meet at a point, and therefore they all pass through a fixed point, thus X is a cone. □

The previous lemma suggests that our analysis would take advantage from some classification result for varieties of dimension n with a family of lines of dimension at least n. Unfortunately such a classification is missing in general, but it is luckily available for n = 4 (see [?]). Therefore we turn our attention to the 4–dimensional case.

Theorem 6.1.2. Let X ⊂ ℙ^{r}, r ⩾ 9, be a non–degenerate, linearly normal, defective variety of dimension 4 of the top species, not a cone, presenting the reducible case. Then X is a scroll over a curve and therefore one of the cases listed in Theorem ?? occurs.

Proof. By Lemma ??, X is covered by lines moving in a family R of dimension at least 4. Let S be an irreducible component of the Hilbert scheme of lines contained in X and containing R. If dim(S) ⩾ 6, then through two general points of X there would pass a line contained in X and X ⊂ ℙ^{r} would be a linear subspace. Hence dim(S) ⩽ 5 and if the equality holds, then X is swept out by a 1–dimensional family F of 3–spaces by a classical result of B. Segre (see [?] or the modern reference [?, Theorem 1]). Thus, if dim(S) = 5, then X is a scroll and we are done.

Suppose now that S = R has dimension 4. By [?, Theorem 2], as in the proof of Theorem ??, from which we keep the notation, we have the following alternatives (1) and (2):

(1) either X is swept out by a 1–dimensional family Q of 3–dimensional quadrics generically of rank 5;

(2) or X is swept out by a 2–dimensional family P of planes;

Suppose we are in case (1). Let x, y ∈ X be general points and let Q_{x}, Q_{y} be 3– dimensional quadrics in Q containing r_{x} and r_{y} respectively. Let p be the intersection point of r_{x} and r_{y}, and let Z_{x,y} be the union of the irreducible components of Q_{x} ∩ Q_{y} containing p, that has pure dimension 2 by Krull Theorem. We claim that Z_{x,y} is an irreducible quadric surface. Indeed, Z_{x,y} is a surface whose linear span has dimension at most 3 (the linear spans of Q_{x} and Q_{y} are distinct by genericity) and it is defined by quadrics. Moreover Z_{x,y} is irreducible because Q_{x}, Q_{y} have rank 5 and cannot contain planes.

Consider now the 1–dimensional family K of 4–spaces spanned by the quadrics of Q. By keeping, in the above argument, x fixed and letting y vary, we see that two general 4– spaces in K intersect in a 3–space. Hence the 4–spaces in K all contain the same 3–space Π. Thus X is contained in a 5–dimensional cone over a curve C with vertex Π and S(X) would be the cone over S(C) from Π, yielding s(X) = 7, which contradicts δ(X) = 1.

Suppose we are in case (2). Then, with the same argument as above, we see that the planes in P pairwise meet. By [?, Lemma 4.1] and taking into account that X is of the top species and is not a cone, we see that X should be contained in a 5–dimensional cone with vertex a plane Π over a non–defective surface Y and S(X) would be the cone with vertex Π over S(Y). Let π be the projection of X to Y from Π. Given two general points p_{0}, p_{1} ∈ X mapping via π to general points q_{0}, q_{1} ∈ Y, then Γ_{X,p_{0},p_{1}} is the pull–back via π of Γ_{Y,q_{0},q_{1}} = {q_{0}, q_{1}}, which contradicts γ(X) = 1. □

7. Second species

7.1. From now on we will concentrate on the 4–fold case. Let X ⊂ ℙ^{r}, r ⩾ 9, be a non– degenerate, defective, projective variety of dimension 4. By Proposition ?? and Theorem ?? we may assume f(X) = δ(X) = 1. By Theorem ?? we may also assume X is neither a cone nor a scroll. It suffices to assume the 4–fold to be linearly normal.

We will consider here the case in which X is of the second species, i.e. γ(X) = 2. Therefore, given p_{0}, p_{1} ∈ X general points, the tangential contact locus Γ_{p_{0},p_{1}} is a surface, spanning a 4–space Π_{p_{0},p_{1}} by Proposition ??, part (v). Again we have two cases: the irreducible and the reducible case, according to the possibilities that Γ_{p_{0},p_{1}} is irreducible or it consists of two irreducible components each passing through one of the two points p_{0}, p_{1}.

7.2. We first examine the irreducible case. In this case the tangential contact surfaces move in a 4–dimensional irreducible family S, such that given two general points p_{0}, p_{1} ∈ X there is a unique irreducible surface Γ = Γ_{p_{0},p_{1}} in S containing p_{0} and p_{1} (see §??). A priori the surfaces in S may have a base locus scheme B which is the largest subscheme B of X such that B is contained in all surfaces in S.

There are two subcases to be considered:

(i) two general surfaces in S have no isolated point in common off the base locus scheme B; (ii) two general surfaces in S have some isolated point in common off the base locus

scheme B. Let x ∈ X be a general point. Then we have a subfamily S_{x} of surfaces in S passing through x. Each irreducible component of S_{x} has dimension 2. Actually S_{x} is irreducible, because otherwise if y ∈ X is another general point, there would be more than one surface in S containing x and y.

If we are in case (i), two general surfaces in S_{x} have a positive dimensional intersection off B, hence they intersect off B in a curve C passing through x. Since the surfaces in S_{x} span in general a 4–space which varies with the surface, there are three subcases to be considered:

(i1) the curve C spans a 3–space; (i2) the curve C spans a plane; (i3) the curve C is a line.

7.3. In this section we discuss the subcase (i1). We prove that:

Lemma 7.3.1. The subcase (i1) does not occur.

Proof. In subcase (i1) the spans of two general surfaces in S_{x} intersect along a 3–space containing the point x. Since they fill up X they cannot lie in a 5–space, so they pass through the same 3–space P_{x} containing all curves which are the intersections of two general surfaces in S_{x}.

We claim that then all surfaces in S_{x} must contain the same curve C_{x}, passing through x, which spans P_{x}. In fact, let Γ be the general surface in S_{x}. If the other surfaces in S_{x} cut out on Γ different curves C passing through x, then these curves would, on one side, sweep out Γ, on the other all lie in P_{x}, hence Γ would lie in P_{x} which is a 3–space, a contradiction. This proves our claim.

When x moves, the curve C_{x} moves in a family C of dimension 3, which fills up X. Let y ∈ X be another general point. Then the surface Γ of S which contains x and y, contains also C_{x} and C_{y} (actually, for a general point z ∈ Γ, it contains C_{z}), and this implies that P_{x} and P_{y}, that are contained in the 4–space spanned by Γ, intersect in a plane. This yields that there is a plane Π such that for x ∈ X general, the 3–space P_{x} contains Π. Let π : X −−→ ℙ^{r−3} be the projection from Π and let Y be the image of X, which has to span the ℙ^{r−3}. Moreover dim(Y) ⩽ 3 because for x ∈ X general the curve C_{x} is contracted to a point by π. Let q_{0}, q_{1} ∈ Y be general points and let p_{0}, p_{1} be points in the counterimages of q_{0}, q_{1} respectively. The surface Γ_{p_{0},p_{1}}, which spans a 4–space contaning Π, is mapped by π to the line ⟨q_{0}, q_{1}⟩ ⊂ Y. This implies that Y is a linear space, a contradiction. □

7.4. In this section we discuss the subcase (i2). We keep the notation introduced above. We prove that:

Theorem 7.4.1. In subcase (i2) then either:

(a) r = 9 and X sits in a cone with vertex a line over a hyperplane section of Seg(2, 2) in ℙ^{7};

(b) or 9 ⩽ r ⩽ 11 and X sits in a cone with vertex a line over a projection in ℙ^{r−2} of the Veronese 3–fold V_{3,2} in ℙ^{9} from a linear space of dimension r−10, intersecting V_{3,2} in a 0–dimensional scheme of lenght 11 − r.

Proof. Let x ∈ X be a general point and let Γ, Γ′ be general elements in S_{x}. Then they intersect in a curve C which spans a plane P and passes through x.

Claim 7.4.2. Keeping Γ fixed and letting Γ′ vary, the curve C stays fixed.

Proof of the Claim ??. Suppose the assertion is false. Consider the family F_{x} of plane curves cut out on Γ by the other surfaces of S_{x}, and recall that S_{x} has dimension 2.

If F_{x} has dimension 2, then, since x is a general point of Γ, the surface Γ possesses a 3–dimensional family F of plane curves. This implies that Γ is either a plane or spans a 3–space, a contradiction since Γ spans a 4–space. In fact, since F has dimension 3, two general curves of F intersect in more than one point (through two general points of Γ there passes a 1-dimensional family of curves in F). If these intersection points span a plane, then two general curves lie in the same plane and Γ would be the plane containing all the curve in F. If they span a line, the planes spanning two general curves in F intersect along a line, so the planes spanning curves in F pairwise intersect along a moving line, hence they span a 3–space.

Hence F_{x} has dimension 1, i.e., a general curve in F_{x} belongs to a 1–dimensional family of surfaces in S_{x}. Then Γ has a 2–dimensional family F_{Γ} of plane curves and therefore there is some curve of F_{Γ} which passes through two general points of Γ. By moving Γ we see that X possesses an irreducible family F of plane curves such that there is some curve of F passing through two general points of X: indeed, if Γ is general in S, then two general points of Γ are also general on X. This yields that F has dimension at least 6.

Consider now the set I of all pairs (C, Γ), with C in F and Γ in S such that C ⊂ Γ. Since on the general surface Γ ∈ S there is the general curve C ∈ F, we can find an irreducible component J of I such that the two projections π_{1} : J → F and π_{2} : J → S are both surjective. Looking at π_{2}, since S has dimension 4 and since for Γ general in S the family F_{Γ} has dimension 2, we see that dim(J) = 6. On the other hand, by what we saw above, the general fibre of π_{1} : J → F has dimension at least 1, which implies that F has dimension at most 5, a contradiction. □

By Claim ??, all surfaces in S_{x} contain a fixed plane curve C passing through x. Let us denote again by F the family of these curves C, which has the property that, given a general point x ∈ X there is a unique curve of F containing x. Hence F has dimension 3. Moreover the general surface Γ in S contains a 1–dimensional family F_{Γ} of curves in F. We denote by P the family of planes spanned by the curves in F.

Let p_{0}, p_{1} ∈ X be general points. The surface Γ_{p_{0},p_{1}} which contains p_{0} and p_{1} contains also the curves C_{0}, C_{1} of F which pass through p_{0}, p_{1} respectively. Hence the planes P_{0}, P_{1} in P spanned by C_{0}, C_{1}, lying in the 4–space spanned by Γ_{p_{0},p_{1}}, intersect in at least one point.

Claim 7.4.3. Two general planes in P do not intersect in a single point.

Proof of the Claim ??. Suppose by contradiction that two general planes in P intersect in a single point. Then by [?, Lemma 4.1], the only possibilities are:

(1) there is a point p such that all the planes in P contain p;

(2) there is a plane Π such that all the planes in P intersect Π along a line.

Assume (1) happens. Let π : X −−→ ℙ^{r−1} be the projection of X from p. The image Y of X is a 4–fold, otherwise X would be a cone with vertex p, a contradiction. Hence π : X −−→ Y is a generically finite rational map. If Γ is a general surface in S, then the 4–dimensional span of Γ contains p, so Γ is projected from p to a surface in ℙ^{3}. Hence Y contains a 4–dimensional family Q of irreducible surfaces in 3–space such that there is a surface Q of Q through two general points of Y. Hence a general secant line to Y lies in the 3-space generated by Q. By the Trisecant Lemma (see [?, Proposition 1.4.3]), a general secant line intersects Y (and a fortiori Q) transversally only in the two points. This implies that the surfaces in Q are quadrics and that through the general point of S(Y) there passes at least a 2–dimensional family of secant lines to Y. Then γ(Y) = f(Y) = 2 and the general contact locus of Y (see Definition ??) is an irreducible quadric. So we can apply Theorem ?? to Y. Cases (i) and (ii) of Theorem, ?? do not apply to Y, because in those cases the contact surfaces are not irreducible. In case (iii) of Theorem ?? the tangential contact loci are not surfaces, so also this case does not apply. In case (iv) of Theorem ??, Y is a cone with vertex a point v and the images of the planes of P via π are the lines generating the cones. This implies that the planes of P all intersect along the fixed line ⟨p, v⟩, contrary to the assumption. Suppose we are in case (v) of Theorem ??. If Y is a cone over a hyperplane section of Seg(2, 2), the same argument as above proves that the planes of P all intersect along a fixed line, a contradiction. It remains to discuss the case in which Y = Seg(2, 2). Then X sits in a cone with vertex p over Seg(2, 2). Now the quadrics in Q are just all the quadrics in Seg(2, 2), because they form a family of dimension 4. Two general of them pairwise intersect at a single point. On the other hand, if y ∈ Y is a general point, two general quadrics in Q containing y intersect along the image via π of a curve in F, hence they must intersect along a line, a contradiction again.

We claim that also case (2) cannot happen. Indeed, let π : X −−→ ℙ^{r−3} be the projection of X from the plane Π. Then the general curve of F is contracted to a point by π, so the image of π is a 3–fold Y. Let q_{0}, q_{1} ∈ Y be general points and let p_{0}, p_{1} ∈ X be such that π(p_{i}) = q_{i}, for i = 1, 2. We may assume that p_{0}, p_{1} ∈ X are also general points. Consider the surface Γ_{p_{0},p_{1}}. It contains the curves C_{0}, C_{1} of F which pass through p_{0}, p_{1} respectively. The planes P_{0}, P_{1}, spanned by C_{0}, C_{1} respectively, cut Π in two distinct lines. This implies that the span of Γ_{p_{0},p_{1}} contains Π, hence the image of Γ_{p_{0},p_{1}} via π is a line, precisely the line joining q_{0} and q_{1}. This yields that Y ⊂ ℙ^{r−3} is a linear subspace because the line through two general points of Y is contained in Y. This contradicts the fact that Y has to span ℙ^{r−3} because r − 3 ⩾ 6 > 3. □

Finally we have to examine the case in which the planes spanned by the curves of F meet pairwise along a line, and then they all pass through the same line s. Let π : X −−→ ℙ^{r−2} be the projection of X from s. The general curve in F is contracted to a point by π, so the image of π is a 3–fold Y.

Claim 7.4.4. Y is a defective 3–fold with f(Y) = 1, possessing a 4–dimensional family C of generically irreducible conics such that if q_{0}, q_{1} ∈ Y are general points there is a unique conic in C containing q_{0} and q_{1}.

Proof. Let q_{0}, q_{1} ∈ Y be general points and let p_{0}, p_{1} ∈ X be such that π(p_{i}) = q_{i}, for i = 0, 1. Then we may assume that p_{0}, p_{1} ∈ X are also general points. Consider the surface Γ_{p_{0},p_{1}} spanning a 4–space Π_{p_{0},p_{1}}, which is swept out by a family F_{Γ} of plane curves whose planes all contain the line s. Then the projection of Γ_{p_{0},p_{1}} from s is a curve spanning the plane image of Π_{p_{0},p_{1}} and γ(Y) = 1. Hence Y has a 4–dimensional family C of generically irreducible plane curves such that if q_{0}, q_{1} ∈ Y are general points there is a unique curve in C containing q_{0} and q_{1}. By the Trisecant Lemma, this implies that the curves in C are conics. Hence 1 ⩽ f(Y) ⩽ γ(Y) = 1, yielding f(Y) = 1. □

In conclusion, in this case X, that is linearly normal, sits in the cone with vertex s over Y, that is also linearly normal and is in the list of defective threefolds in [?, Theorem 1.1]. Actually in that list the only 3–folds Y possessing a 4–dimensional family C of generically irreducible conics such that if q_{0}, q_{1} ∈ Y are general points there is a unique conic in C containing q_{0} and q_{1}, are the hyperplane section of Seg(2, 2) in ℙ^{7} and the Veronese 3–fold V_{3,2} and its projections, that must be internal projections because of the linear normality. This leads to cases (a) and (b) of Theorem ??. □

Remark 7.4.5. If X is a 4–fold as in Theorem ??, then X is in fact defective, with δ(X) = f(X) = 1 and irreducible tangential contact loci of dimension 2.

First, suppose X sits in a cone with vertex a line s over a hyperplane section Y of Seg(2, 2) in ℙ^{7}. Hence we have a projection π : X −−→ Y ⊂ ℙ^{7} from s. If p_{0}, p_{1} ∈ X are general points and q_{i} = π(p_{i}) with i = 0, 1, then from ⟨s, T_{X,p_{i}}⟩ ∩ ℙ^{7} = T_{Y,q_{i}}, we get ⟨s, T_{Y,q_{0}}, T_{Y,p_{1}}⟩ = ⟨T_{X,p_{0}}, T_{X,p_{1}}⟩ = T_{X,p_{0},p_{1}}. Since T_{Y,q_{0},q_{1}} = ℙ^{6} (recall that γ(Y) = f(Y) = 1 = δ(Y)), we get T_{X,p_{0},p_{1}} = ℙ^{8} and f(X) = 1. From T_{X,p_{0},p_{1}} = ⟨s, T_{Y,q_{0},q_{1}}⟩ we deduce that the general tangential contact locus is the pull back of a general conic in Y via π so that it has dimension 2, yielding γ(X) = 2 (see Remark ?? for the relations between the invariant of X and Y).

The other case of Theorem ?? is similar because a projection Y ⊂ ℙ^{r−2} of V_{3,2} ⊂ ℙ^{9} has δ(Y) = f(Y) = γ(Y) = 1 (as well as V_{3,2}, its projection Y is a LCEL variety).

Remark 7.4.6. Suppose X is a 4–fold as in Theorem ??. Then we claim X ⊂ ℙ^{9} is never smooth. The 4-fold X projects from the line s onto the 3-fold Y. If X is smooth, then s ⊂ X. Indeed, if this is not the case, s ∩ X = {p_{1}, …, p_{h}} with h ⩾ 1. If m_{i} = length_{p_{i}}(s ∩ X) ⩾ 1, we can resolve the projection from s by blowing-up a finite number of points above each p_{i} because the strict transform of s will cut the exceptional divisor in a finite number of points. The last exceptional divisor is a smooth ℙ^{3}, which would be sent by the projection onto a linear space of dimension 3 (the linear system of hyperplanes through s induces on this ℙ^{3} a base point free linear system in |O(1)| because we have blown-up a point and the strict transform of a general hyperplane through s has now multiplicity one at this point). Hence Y would be a linear space of dimension 3, which is impossible since Y is non-degenerate (see the proof of [?, Lemma 3.19] for a similar argument).

Let us assume that s ⊂ X, let E ⊂ X̃ be the exceptional divisor of the blowing-up of X along s and let π̃ : X̃ → Y be the induced morphism. Then E, which is a ℙ^{2}-bundle over s, is mapped onto Y in such a way that each ℙ^{2} in E is a linear space in Y. In particular Y is ruled by planes.

Suppose that Y ⊂ ℙ^{7} is a hyperplane section of Seg(2, 2) ⊂ ℙ^{8}, that is a divisor of type (1, 1) on ℙ^{2} × ℙ^{2}. Recall that a plane on Seg(2, 2) = ℙ^{2} × ℙ^{2} is either of the form ℙ^{2} × {p} or of the form {q} × ℙ^{2}, with p, q points in ℙ^{2}. Since Y contains infinitely many planes, there would exist a curve Γ ⊂ ℙ^{2} such that either Y = ℙ^{2} × Γ or Y = Γ × ℙ^{2}, contradicting the fact that Y is a divisor of type (1, 1).

Suppose now that Y is a projection of V_{3,2} ⊂ ℙ^{9}. If r = 11, then Y = V_{3,2} that does not contain planes. If r = 10, then Y is the internal projection of V_{3,2} from a point, that contains a unique plane, i.e., the image of the exceptional divisor of the blow–up of V_{3,2} at the centre of projection. If r = 9, then Y is the projection of V_{3,2} from a secant or tangent line. This is also easily seen not to contain infinitely many planes, concluding the proof of our claim. Indeed, if the threefold Y contains infinitely many planes, it would be a scroll in planes and its general hyperplane section Z would be a surface scroll. However the general hyperplane section Z of Y in this case is the image of a smooth quadric surface Q in ℙ^{3} birationaly mapped in ℙ^{6} via the linear system |O_{Q}(2) ⊗ I_{ζ}|, where ζ is a lenght two 0–dimensional subscheme of Q. Then the irregularity of Z is 0. On the other hand, the general hyperplane section of Z is a smooth elliptic curve, because so is the general curve in |O_{Q}(2) ⊗ I_{ζ}|. If Z would be a scroll it would have irregularity 1, a contradiction.

7.5. In this section we discuss the subcase (i3). We keep notation and conventions introduced above, in particular we assume that X is not a cone.

Lemma 7.5.1. In the subcase (i3), if x ∈ X is a general point, then all surfaces in S_{x} contain the same line s_{x} passing through x. Hence X contains a 3–dimensional family R of lines such that there is a unique line in R passing through the general point of X. Moreover the surfaces Γ in S are ruled, swept out by a 1–dimensional family of lines in R.

Proof. Fix Γ a general surface in S_{x}. If the other surfaces in S_{x} intersect Γ in a moving line, then Γ would possess a positive dimensional family of lines through its general point. Then Γ would be a plane, which is impossible. So the intersection of all surfaces in S_{x} is a unique line through x and the assertion follows. □

We will prove the following:

Theorem 7.5.2. In the subcase (i3) then either:

(1) r = 9 and X sits in a cone with vertex a line over a hyperplane section of Seg(2, 2) in ℙ^{7};

(2) or 9 ⩽ r ⩽ 11 and X sits in a cone with vertex a line over a projection in ℙ^{r−2} of the Veronese 3–fold V_{3,2} in ℙ^{9} from a linear space of dimension r−10, intersecting V_{3,2} in a 0–dimensional scheme of lenght 11 − r;

(3) or r = 9 and X sits in a cone with vertex a line over a linearly normal (defective) 3–fold in ℙ^{7} sitting in a cone with vertex a line over the Veronese surface V_{2,2}, in particular X sits in a 6–dimensional cone with vertex a 3–space over the Veronese surface V_{2,2};

(4) or X is swept out by a 3–dimensional family R of lines and it is singular along a linear space Π of dimension ε, with 2 ⩽ ε ⩽ 3, which is cut out in one point by the general line in R, and X projects from Π to a 3–dimensional variety Y ⊂ ℙ^{r−ε−1} (with general fibres unions of lines of R), which contains a 4–dimensional family C of (generically irreducible) conics such that there is a conic in C passing through two general points of Y and the counterimage of the general conic of C via the projection from Π is a non–developable scroll spanning a 4–space with a line directrix sitting in Π. In any event r ⩽ 10 + ε.

Before proving Theorem ??, we need a few preliminary lemmas which are due to Scorza (see [?], §11).

First of all we consider Σ ⊂ ℙ^{r}, with r ⩾ 3, an irreducible, non–degenerate, projective surface, which is a non–developable scroll. Let s be a line of the ruling of Σ such that the general point of s is smooth for Σ and the tangent plane to Σ at the smooth points of s is not constant. We will say that s is a non–developable line of the ruling, otherwise, if the tangent plane at the smooth points of s is constant, the line is called developable. We recall the following standard fact in basic projective–differential geometry (see, e.g., [?, §1]):

Proposition 7.5.3 (Chasles’ Theorem). In the above setting, if s is a non–developable line of the ruling of Σ, there is a 3–space P_{s} containing s and there is a projectivity ω_{s} : s → s^{∗} (where s^{∗} is the pencil of planes containing s in P_{s}), such that for all points x ∈ s smooth for Σ, one has T_{Σ,x} = ω_{s}(x).

The 3–space P_{s} is called the tangent space to Σ along s. If s is instead a developable line of the ruling of Σ, we denote by P_{s} the plane tangent to Σ at a general point of s.

One has the possibility of defining a notion of tangent plane to Σ even at points that are not smooth for Σ. We proceed in this way. We consider a pair (s, x), where s is a line of the ruling whose general point is smooth for Σ and x ∈ s. Then we define the tangent plane T_{Σ,s,x} to Σ at (s, x) in the following way:

• T_{Σ,s,x} = ω_{s}(x) if s is a non–developable line of the ruling;

• T_{Σ,s,x} = T_{Σ,y} where y is a general point of s, if s is a developable line of the

ruling. Sometimes one says that T_{Σ,s,x} is a tangent plane to Σ at x on s and note that it is a

limit of tangent planes at smooth points of Σ on s.

Remark 7.5.4. Let C ⊂ Σ be a directrix of Σ, let x ∈ C be a general point and let s be a (general) line of the ruling containing x. Then T_{Σ,s,x} contains the tangent line to C at x. This follows from the discussion in [?, §1].

Note that given x ∈ Σ there are at most finitely many lines of the ruling of Σ containing x, because Σ is not a cone. So there are at most finitely many tangent planes to Σ at x. Moreover it is clear that if x is smooth for Σ there is a unique tangent plane to Σ at x that coincides with T_{Σ,x}.

Lemma 7.5.5. Let Σ ⊂ ℙ^{4} be an irreducible, non–degenerate, projective surface, which is a non–developable scroll. Let s be a general line of the ruling of Σ. Then there is a curve ℭ_{s} on Σ, union of irreducible components such that some tangent plane to Σ at the general points of each of these components intersect s at a point. More precisely, ℭ_{s} consists of an irreducible component ℭ′_{s} which is unisecant to the lines of the ruling of Σ, plus finitely many lines of the ruling, whose set we denote by 𝔖_{s}. If the curve ℭ′_{s} = ℭ′ is independent on s, then Σ has a directrix line R and in this case ℭ′ = R.

Proof. Let t be a further general line of the ruling of Σ and let P_{t} be the tangent 3–space to Σ along t. Then P_{t} cannot contain s which is a general line of the ruling, because Σ is non–degenerate in ℙ^{4}. Hence P_{t} cuts s at a point q_{t}. Then the plane ⟨q_{t}, t⟩ is tangent to Σ at a point p_{s} ∈ t. The curve ℭ′_{s}, which is unisecant to the lines of the ruling, whose general point is p_{s}, is a component of the desired curve ℭ_{s}. The remaining components of ℭ_{s} are those non–developable lines u of the ruling such that P_{u} contains s plus the developable lines u of the ruling such that P_{u} intersects s, that fill up the set 𝔖_{s}.

Suppose now ℭ′_{s} = ℭ′ does not depend on s. Keeping the above notation, take t general but fixed and let s vary. Then the point p := p_{s} = ℭ′∩t does not depend on s, and therefore the general line of the ruling intersects in one point the tangent plane T_{Σ,t,p}, that contains t. Similarly, the general line of the ruling intersects in one point the tangent plane T_{Σ,s,q}, where q = ℭ′ ∩ s, that contains s. Note that T_{Σ,t,p} and T_{Σ,s,q} are distinct (because t and s being general lines of the ruling of Σ are not coplanar), and intersect along a line, precisely the line R joining the two points T_{Σ,t,p} ∩ s and T_{Σ,s,q} ∩ t. So ⟨T_{Σ,t,p}, T_{Σ,s,q}⟩ is a 3–space. Since all lines of the ruling intersect both T_{Σ,t,p} and T_{Σ,s,q} and since Σ is non–degenerate in ℙ^{4}, then the general line of the ruling intersects R in a variable point (because Σ is not a cone), i.e., Σ has the line directrix R.

Finally, assume that Σ has a line directrix R. Then ℭ′ = R, because if p ∈ R is a general point, and t a (general) line of the ruling containing p, then T_{Σ,t,p} contains R (see Remark ??), hence it meets the general line of the ruling. □

To state the second lemma, recall that an irreducible, projective surface S ⊂ 𝔾(1, r) is called a congruence of lines in ℙ^{r}. Given an irreducible projective curve C ⊂ S it corresponds to a scroll in ℙ^{r}, and it is well known that the scroll is developable if and only if for the general point x ∈ C the tangent line to C at x is contained in 𝔾(1, r) (see for example [?, Proposition 12.4.1]).

Lemma 7.5.6. Let S be a congruence of lines in ℙ^{r}, with r ⩾ 3. If all curves contained in S correspond to developable scrolls, then one of the following occurs:

(α) there is a plane Π ⊂ ℙ^{r} such that S consists of all the lines in Π; (β) there is a point p ∈ ℙ^{r} such that S consists of lines containing p.

Proof. Consider the variety W ⊂ ℙ^{r}, r ⩾ 3, described by the lines in S. By projecting W down generically in ℙ^{3} it suffices to prove the assertion for r = 3. So we assume r = 3, hence S ⊂ ℚ ⊂ ℙ^{5}, where ℚ = 𝔾(1, 3) is a smooth quadric.

Let x ∈ S be a general point. Every line passing through x in the tangent plane T_{S,x} is tangent to some curve on S, and therefore this line is contained in ℚ. Hence T_{S,x} ⊂ ℚ and therefore T (S) ⊆ ℚ. This implies that all tangent planes to S in smooth points of S belong to one and the same system of planes of ℚ, hence they pairwise intersect at a point, thus either S is defective or it spans a 4–space.

If S is defective, it could either be a cone or the Veronese surface V_{2,2}. The latter case is impossible, because T (V_{2,2}) is a cubic hypersurface in ℙ^{5} and not a quadric.

Claim 7.5.7. If the surface S is a cone, then either case (α) or (β) occurs.

Proof of Claim ??. We want to show that the lines of S pairwise intersect, because this implies that we are in case (α) or (β). So we argue by contradiction and assume that two general lines in S are skew. Since S is a cone, its vertex corresponds to a line s in ℙ^{3} such that all lines corresponding to the points of S intersect s. Now take a curve C on S such that two general points on C correspond to skew lines. Then C corresponds to a developable surface Σ in ℙ^{3} that is not a cone since it contains skew lines of the ruling. So Σ is the tangent developable surface to a curve Z that is not a plane curve, because two general tangent lines to Z are skew. However, the general tangent line to Z has to intersect s, which is not possible. This contradiction proves the assertion. □

Suppose now that S spans a 4–space Π, which therefore contains also T (S). This means that the quadric ℚ′ cut out by Π on ℚ is singular, because it contains planes. Hence ℚ′ is a cone with vertex a point corresponding to a line s, and the lines corresponding to the points of S all intersect the line s. Even in this case, arguing as before, we can conclude that we are either in case (α) or (β). □

As a consequence we have the:

Lemma 7.5.8. Let V ⊂ 𝔾(1, r) be a projective variety of dimension n ⩾ 2 possessing an irreducible family of dimension 2(n − 1) of (generically) irreducible curves C such that:

(ι) all curves in C correspond to developable scrolls; (ιι) given two general points of V there is a curve in C containing them.

Then there is a point p ∈ ℙ^{r} such that any point of V corresponds to a line containing p (hence r ⩾ n + 1).

Proof. The assumption (ιι) implies that given a general point x ∈ V and a general tangent line t to V at x, there is a curve in C which passes through x, is smooth at x and whose tangent line at x is t. This in turn implies that T (V) ⊆ 𝔾(1, r) and all curves in V correspond to developable scrolls and the same happens for every surface S ⊂ V. By Lemma ?? this implies that the points of V correspond to lines which pairwise intersect at a point. They cannot correspond to the lines in a plane, hence they correspond to lines passing through a fixed point. □

Now we are ready for the:

Proof of Theorem ??. By Lemma ??, X contains a 3–dimensional family R of lines such that there is a unique line in R passing through the general point of X. Let p_{0}, p_{1} ∈ X be general points. Let s_{0}, s_{1} be the unique lines in R containing p_{0}, p_{1} respectively. Then Γ := Γ_{p_{0},p_{1}} (that moves in the family S of dimension 4) is a scroll of which s_{0}, s_{1} are general lines of the ruling. Then we may apply Lemma ?? to R and conclude that the general surface in S is not a developable scroll, otherwise X would be a cone.

By Lemma ??, there is a point x ∈ s_{1} such that a tangent plane to Γ at x intersects s_{0}. We need to consider two cases:

a) for general p_{0}, either the point x ∈ s_{1} moves (and so is a smooth point of X) or it is fixed but it is still a regular point of X;

b) for general p_{0}, the point x is in Sing(X) (in particular the point x does not move

on s_{1} when p_{0} moves). We first assume that we are in case a). In particular, the point x is smooth for X.

Claim 7.5.9. Given a general line s in R consider the linear space

Π_{s} = ⟨ \overline{⋃_{y∈s∩Reg(X)} T_{X,y}} ⟩.

Then:

(1) the general line s′ ∈ R intersects Π_{s};

(2) given two general lines s, s′ ∈ R, one has dim(⟨Π_{s}, Π_{s′}⟩) ⩽ 8.

Proof of the Claim ??. Given two general lines s, s′ ∈ R, there is a unique surface Γ containing both of them. Morover, there is a point x ∈ s such that a tangent plane to Γ at x on s intersects s′. This implies that s′ intersects Π_{s} and it proves (1) by the generality of s and s′.

Both Π_{s_{0}}, Π_{s_{1}} are contained in the 8–space T_{X,p_{0},p_{1}} which is tangent to X along Γ, proving (2). □

Set now

Π := ⋂_{s∈R} Π_{s}

and ε = dim(Π).

Claim 7.5.10. If s ∈ R is a general line, then s intersects Π in a point. Hence ε ⩾ 1.

Proof of the Claim ??. Let s, s′, t ∈ R be general lines. Then t has non–empty intersection with both Π_{s} and Π_{s′} by part (1) of Claim ??. If t intersects Π_{s} and Π_{s′} in distinct points, then t lies in ⟨Π_{s}, Π_{s′}⟩. This is not possible, because t ∈ R is general, and then all of X would lie in ⟨Π_{s}, Π_{s′}⟩ which has dimension at most 8 by part (2) of Claim ??. Hence t intersects ⟨Π_{s}, Π_{s′}⟩ in a point of their intersection. By the genericity of s and s′ we see that t intersects Π. It is not possible that Π is a point, because then X would be a cone. Hence ε ⩾ 1. □

Consider now the irreducible closed set ℑ ⊂ S × (ℙ^{r})^{∗} whose general member is a pair (Γ, H) with Γ ∈ S general and H a hyperplane tangent to X along Γ. We have the two projections p_{1} : ℑ → S and p_{2} : ℑ → (ℙ^{r})^{∗}. The map p_{1} is surjective, with general fibre of dimension r − 9, hence dim(ℑ) = dim(S) + (r − 9) = r − 5. Let 𝔍 be the image of p_{2}. One has dim(𝔍) ⩾ 1, because otherwise there would be a unique hyperplane tangent to X along the general surface Γ ∈ S, that is impossible, because X is non–degenerate.

We denote by h the dimension of the general fibre of p_{2} so that dim(𝔍) = r − 5 − h. If h ⩾ 1 and if H ∈ 𝔍 is general, then H is tangent to X along a h–dimensional family of surfaces Γ ∈ S, that fills up a 3–dimensional variety X_{H}.

Claim 7.5.11. If h ⩾ 1 and if x ∈ X is general, the general 3–fold X_{H} containing x contains the unique line of R containing x.

Proof of the Claim ??. Since h ⩾ 1, the general 3–fold X_{H} is swept out by surfaces in S, so there is some surface Γ ∈ S contained in X_{H} and containing x. Then Γ contains the unique line in R containing x as wanted. □

Claim 7.5.12. The map p_{2} is generically finite to 𝔍, i.e., h = 0, unless, may be, r = 9 and h = 2.

Proof of the Claim ??. We argue by contradiction and assume h ⩾ 1. We have a rational map η : X −−→ R that sends a general point x ∈ X to the unique line in R containing x. Claim ?? implies that if h ⩾ 1 and H is general in 𝔍, there is a surface Σ_{H} contained in R, such that X_{H} = η^{∗}(Σ_{H}). We can assume Σ_{H} is smooth, otherwise we work on a desingularization. Given the general surface Γ ∈ S along which H is tangent to X, there is an irreducible curve C_{H,Γ} on Σ_{H} such that Γ = η^{∗}(C_{H,Γ}). The curves of the type C_{H,Γ} form a family of dimension h on Σ_{H}. Since two general surfaces in S intersect along a line, we see that C_{H,Γ}^{2} = 1 on Σ_{H}. Then, by [?, Prop. 0.18], one has h ⩽ 2. Hence dim(𝔍) = r − 5 − h ⩾ r − 7 ⩾ 2.

Let now p_{0}, p_{1} be general points of X. We have a family 𝔊 of dimension r − 7 − h (which is positive, unless r = 9 and h = 2) of H ∈ 𝔍 such that p_{0}, p_{1} ∈ X_{H}. For all H ∈ 𝔊, X_{H} has to contain the unique surface Γ containing p_{0}, p_{1} and H has to be tangent to X along Γ. Note that 𝔊 is not a linear system, because otherwise it would cut out a linear system on X and the general divisor of this linear system would have singular points filling up X, against Bertini’s theorem. Hence 𝔊 has dimension strictly smaller than the linear system of hyperplanes tangent to X along Γ. Hence we have dim(𝔊) = r −7−h < r −9, so h > 2 a contradiction. □

Claim 7.5.13. If h = 0 then ε ⩽ 3. If r = 9 and h = 2, then ε ⩽ 5.

Proof of the Claim ??. Assume first h = 0. The linear space Π is contained in all hyperplanes of the family 𝔍. If h = 0, we have dim(𝔍) = r − 5. Moreover, 𝔍 is not a linear system, because otherwise it would cut out a linear system on X and the general divisor of this linear system would have singular points filling up X, against Bertini’s theorem. Hence 𝔍 has dimension strictly smaller that the family of hyperplanes containing Π, thus r − 5 < r − (ϵ + 1), i.e., ε ⩽ 3.

Consider next the case r = 9 and h = 2. In this case dim(𝔍) = 2. As above, 𝔍 is not a linear system. Hence 𝔍 has dimension strictly smaller that the family of hyperplanes containing Π, thus 2 < 9 − (ϵ + 1) i.e., ε ⩽ 5. □

Next we consider the projection π : X −−→ ℙ^{r−ε−1} from Π and we denote by Y its image.

Claim 7.5.14. The case r = 9, h = 2 and ε ⩾ 4 does not happen. Hence one always has ε ⩽ 3.

Proof of the Claim ??. We have dim(𝔍) = 2. Let H be general in 𝔍 and consider again the threefold X_{H} along which H is tangent to X. By the Infinitesimal Bertini’s Theorem (see [?, Thm. 2.2]), X_{H} spans a linear space of dimension at most 6 containing Π.

Suppose first ε = 4. If H and H′ are general in 𝔍, they do intersect at some general point of X, because dim(𝔍) = 2 and if x ∈ X is a general point, there is a 1–dimensional family of hyperplanes H in 𝔍 such that x ∈ X_{H}.

The 3–fold X_{H} (for H ∈ 𝔍 general) is mapped by π either to a point p_{H} or to a line ℓ_{H}. In either case, both p_{H} or ℓ_{H} cannot stay fixed when H varies, because X is non– degenerate in ℙ^{9}. On the other hand, if the image of X_{H} is a point p_{H}, since general H and H′ in 𝔍 intersect at some general point of X, then p_{H} = p_{H′}, a contradiction. By the same argument, if the image of X_{H} is a line ℓ_{H}, then for general H and H′ in 𝔍 one has that ℓ_{H} and ℓ_{H′} intersect at general points of both ℓ_{H} and ℓ_{H′}. Since Y is swept out by lines of the type ℓ_{H}, we see that Y should be a plane, and this is a contradiction since Y has to be non degenerate in ℙ^{4}.

The case ε = 5 can be excluded in a similar way. In this case in the map π the 3–fold X_{H} is mapped to a point p_{H} that cannot stay fixed when H varies, because X is non– degenerate in ℙ^{9}. On the other hand, since general H and H′ in 𝔍 intersect at some general point of X, then p_{H} = p_{H′}, a contradiction. □

So, from now on we may assume that ε ⩽ 3.

Claim 7.5.15. One has 2 ⩽ dim(Y) ⩽ 3.

Proof of the Claim ??. The projection π contracts all lines of R to points in Y, hence dim(Y) ⩽ 3. It cannot be the case that dim(Y) = 0 because Y has to span ℙ^{r−ε−1} and r − ε − 1 ⩾ 8 − ε ⩾ 5. To finish we have to exclude that dim(Y) = 1. Assume, by contradiction, that this is the case. Then X is swept out by a 1–dimensional family of 3–folds (the fibres of π), each lying in a ℙ^{ε+1}. This implies ε = 3, otherwise X would be a cone with vertex Π over a curve. Then the general tangent space to X intersects Π in at least a plane, two general tangent spaces to X intersect at least along a line and δ(X) ⩾ 2 by Terracini Lemma, a contradiction. □

Claim 7.5.16. The case dim(Y) = 2 is not possible.

Proof of the Claim ??. Assume by contradiction that dim(Y) = 2. Then X sits in the (ε + 3)–dimensional cone V with vertex Π over Y. Since X is not a cone, we have ε ⩾ 2.

Let x ∈ X be general. Then T_{X,x} ⊂ T_{V,x}, and T_{V,x} is a (ε + 3)-space which contains Π. Since the projection of T_{X,x} from Π is the general tangent space to Y, we see that T_{X,x} intersects Π in a line.

Let s_{x} be the line which T_{X,x} cuts out on Π. Since X is not a cone, the line s_{x} moves when x varies and describes a family L of lines in Π.

We claim that two general lines in L intersect each other. Otherwise, two general lines in L are skew, and then ε = 3. Let p_{0}, p_{1} ∈ X be general points. Then T_{X,p_{0},p_{1}}, contains Π, T_{X,p_{0}} and T_{X,p_{1}}, hence it coincides with T_{V,p_{0},p_{1}}, which is tangent to X along the union of the two surfaces which are the fibres of π passing through p_{0} and p_{1}. This is a contradiction, because the surface Γ_{p_{0},p_{1}} is irreducible by hypothesis. We have thus proved that two general lines in L meet.

Since the lines in L pairwise meet, then they all belong to the same plane P ⊆ Π, because they cannot pass through the same point otherwise X would be a cone. Consider the projection π′ : X −−→ ℙ^{r−3} from P. The image of π′ is a surface Y′ because the general tangent space to X intersects the centre of projection along a line. Moreover X sits in the 5–dimensional cone W over Y′ with vertex P. Now we repeat the same argument as above. Namely, let p_{0}, p_{1} ∈ X be general points. Then T_{X,p_{0},p_{1}}, containing P, T_{X,p_{0}} and T_{X,p_{1}}, coincides with T_{W,p_{0},p_{1}}, which is tangent to X along the union of two surfaces, i.e., the fibres of π′ passing through p_{0} and p_{1}, leading again to a contradiction. □

Hence one has dim(Y) = 3. Note now that, by the proof of (2) of Claim ??, all tangent 8–spaces to X along a surface Γ in S contain Π. Under the projection π from Π all these 8–spaces are mapped to a family of dimension 4 (=dim(S)) F of (7 − ε)–spaces, which are tangent to Y along the projection of the surfaces in S from Π. If q_{0}, q_{1} ∈ Y are general points, then through them there passes a projection of a surface in S and T_{Y,q_{0},q_{1}} = ⟨T_{Y,q_{0}}, T_{Y,q_{1}}⟩ is contained in a (7−ε)–space in F. This implies that the two tangent spaces to Y at the general points q_{0}, q_{1} ∈ Y intersect in dimension λ with

6 − λ = dim(T_{Y,q_{0}}) + dim(T_{Y,q_{1}}) − λ = dim(T_{Y,q_{0},q_{1}}) ⩽ 7 − ε

which implies λ ⩾ ε − 1.

If ε = 3, then two general tangent spaces to Y ⊂ ℙ^{r−4} intersect in dimension 2. Then a general curve section of Y has tangent lines which pairwise meet, hence it would be a plane curve, whereas it spans a ℙ^{r−6} and r − 6 ⩾ 3, a contradiction.

If ε = 2, then two general tangent spaces to Y ⊂ ℙ^{r−3} intersect in dimension 1. Then the general hyperplane section of Y, which spans a ℙ^{r−4}, is a defective surface, hence either it is a cone or a Veronese surface V_{2,2}. Hence Y itself is either a cone over a curve or a cone over the Veronese surface V_{2,2}. In either case, X would be contained in a cone over a curve or in a cone over a Veronese surface and hence it would be of the first species by Remark ??, contradiction.

Finally, if ε = 1, then two general tangent spaces to Y ⊂ ℙ^{r−2} intersect in dimension 0, i.e., Y is a defective 3–fold and X sits in a cone with vertex a line over Y. By [?, Theorem 1.1], then Y is either a hyperplane section of Seg(2, 2) (and then r = 9), or the Veronese 3–fold V_{3,2} in ℙ^{9} or a projection of it in ℙ^{8} or ℙ^{7} (and then 9 ⩽ r ⩽ 11), or it sits in a 4–dimensional cone over the Veronese surface V_{2,2} (and again r = 9) or in a 4-dimensional cone over a curve. However the last case is not possible. Indeed, if Y sits in a 4-dimensional cone over a curve, then the general contact locus of X would be reducible. Hence we are in one of the cases (1), (2), (3) of the statement of Theorem ??.

Now we consider case b). So let us assume that x ∈ Sing(X). Note that, by Lemma ??, this is the case only if Γ_{p_{0},p_{1}} has a line directrix which passes through x and, by the generality of s_{1}, it consists of singular points of X. We have the:

Claim 7.5.17. Suppose that when p_{0} moves, the point x does not move on s_{1} and x ∈ Sing(X). Then there is a subspace Π ⊆ Sing(X) of dimension ε, with 1 ⩽ ε ⩽ 3, such that if ℓ ∈ R is a general line, then ℓ intersects Π in a point. Moreover if Γ is a general surface in S, then Γ is a (non–developable) scroll with line directrix contained in Π.

Proof of the Claim ??. In this situation, as p_{0}, p_{1} vary, the point x describes a variety V ⊆ Sing(X), of dimension ε, with 1 ⩽ ε ⩽ 3. So V is described by the line directrices of the scrolls Γ_{p_{0},p_{1}} as p_{0}, p_{1} vary. Hence the general line in R intersects V in its general point because the intersection point is on a general directrix. Let x, y ∈ V be general points. Take lines a, b ∈ R, with x ∈ a and y ∈ b, which are general lines in R, and consider the surface Γ in S containing a, b (we can suppose that a passes through a general point p_{0}, that b passes through a general point p_{1} and take Γ = Γ_{p_{0},p_{1}}). By construction and by definition of x and y, the line joining x and y is exactly the line directrix of Γ, and it lies on V. In conclusion, V is a linear space Π proving the assertion. □

Now again we consider the projection π : X −−→ ℙ^{r−ε−1} from Π and let Y be its image. Exactly as in Claims ?? and ??, we see that Y has to be a 3–fold. If Γ is a general surface in S, its image via π is a plane curve, because Γ spans a 4–space and we are projecting it from the line directrix which sits in this 4–space. Hence Y contains a family C of generically irreducible plane curves (the images of the surfaces Γ in S), such that there is a curve of C passing through two general points of Y. Then, by the Trisecant Lemma, the curves of C are conics. The counterimage via π of the general conic in C is a non–developable scroll with a line directrix contained in Π. Note that either r − ε − 1 ⩽ 6, or Y ⊂ ℙ^{r−ε−1} is defective. Suppose that r − ε − 1 > 6 and Y ⊂ ℙ^{r−ε−1} is defective. Then, by [?, Theorem 1.1], either r − ε − 1 ⩽ 9, or Y is a cone, or Y sits in a 4–dimensional cone over a curve. We claim that these two latter cases do not occur. Suppose first that Y is a cone. It could be a cone with vertex a line over a curve C or a cone with vertex a point p over a surface S. In the former case, since Y contains the family C of conics, then C is a conic, hence Y (that is linearly normal) sits in ℙ^{4}, so that r − ε − 1 = 4, a contradiction, since we are assuming r − ε − 1 > 6. In the latter case, since Y contains the family C of conics, the surface S has also a family C′ of conics (the projections of the conics in C form p), such that there is a conic in C′ passing through two general points of S. Then, by the classification of defective surfaces, S is either the Veronese surface V_{2,2} or it is contained in a ℙ^{n}, with n ⩽ 4. Hence r − ε − 1 ⩽ 6, again a contradiction. In conclusion, we have r − ε − 1 ⩽ 9, i.e., r ⩽ 10 + ε. If ε = 1 and r − 2 ⩾ 7, again by [?, Theorem 1.1], we are in cases (1), (2) or (3). Otherwise we are in case (4) of the statement of Theorem ??. □

Remark 7.5.18. If X is as in cases (1), (2), (3) of Theorem ??, then X is indeed defective, f(X) = 1 and the general contact loci of X are irreducible surfaces. In case (4), which has been overlooked by Scorza in [?, §11], X is still defective with f(X) = 1, because X is covered by a 4–dimensional family of scrolls Γ each spanning a 4–space and such that through two general points of X there is a unique scroll Γ containing them. Then the general secant line s to X is also a secant to a scroll Γ and therefore there is a 1– dimensional family of secants to Γ, hence to X, containing the general point of s. It is however not clear if the surfaces Γ are contact loci, neither we have examples of such 4–folds.

We can repeat verbatim the argument of Remark ?? and conclude that all 4–folds as in Theorem ?? are singular.

7.6. In this section we discuss case (ii), in which two general surfaces in S have some isolated point in common off the base locus scheme B. We keep all notation and conventions introduced above, in particular we assume X is not a cone. We will denote by j the dimension of the intersection of the spans of two general surfaces in S, which are ℙ^{4}. Thus 0 ⩽ j ⩽ 3.

We will prove the:

Theorem 7.6.1. In case (ii) we have the following possibilities:

(∗) X sits in a cone with vertex a point over Seg(2, 2) in ℙ^{8}; (∗∗) r = 9 and X sits in a cone with vertex a line over a hyperplane section of

Seg(2, 2) in ℙ^{7}; (∗ ∗ ∗) or 9 ⩽ r ⩽ 11 and X sits in a cone with vertex a line over the Veronese 3–fold

V_{3,2} in ℙ^{9} or the projection of it in ℙ^{8} or ℙ^{7}; (∗ ∗ ∗∗) j = 0, hence given two general surfaces Γ, Γ′ in S, the 4–spaces spanned by

them intersect in one point which is also the intersection of Γ and Γ′. In this case the general surface in S is rational and X itself is rational.

Proof. We start with the following:

Claim 7.6.2. There is no subspace Π of ℙ^{r} with dim(Π) = k < r − 1 such that for the general surface Γ in S the 4–space spanned by Γ intersects Π in a subspace of dimension Proof of the Claim ??. Suppose by contradiction that there is such a subspace Π. Let x ∈ X be a general point. Then for the general surface Γ in S_{x} the span of such a surface is contained in ⟨Π, x⟩, which has dimension k + 1 < r. Hence the whole of X would be contained in ⟨Π, x⟩, a contradiction because X is non–degenerate in ℙ^{r}. □

Claim 7.6.3. There is no plane P such that for the general surface Γ in S the 4–space spanned by Γ contains P.

Proof of the Claim ??. Suppose by contradiction that there is such a plane P. Let π : X −−→ ℙ^{r−3} be the projection from P, and Y the image of π. If Γ is the general surface in S, then Γ is mapped to a line by π. Thus Y is a 3–fold such that if x, y ∈ Y are general points, there is a line in Y containing x and y. Hence Y would be a 3–space, a contradiction, because Y has to span the ℙ^{r−3} and r − 3 ⩾ 6. □

Let first be j = 3. Then, since the 4–spaces spanned by the surfaces in S cannot lie in a 5–space, they have to pass through the same 3–space, and this contradicts Claim ??.

Let j = 2. Consider two general surfaces Γ_{1}, Γ_{2} in S, and their spans Π_{1}, Π_{2}. Then Π′ = ⟨Π_{1}, Π_{2}⟩ has dimension 6. Let Γ be another general surface in S and Π be its span. Of course Π cannot be contained in Π′ and by Claim ?? it cannot intersect it in a 3–space. This implies that Π intersects Π′ in the intersection plane of Π_{1} and Π_{2}, which contradicts Claim ??.

Let j = 1. Let us argue as in the case j = 2. Consider two general surfaces Γ_{1}, Γ_{2} in S, and their spans Π_{1}, Π_{2}. Then Π′ = ⟨Π_{1}, Π_{2}⟩ has dimension 7. Let Γ be another general surface in S and Π be its span, which cannot be contained in Π′ and by Claim ?? it cannot intersect it in a 3–space. Then either Π intersects Π′ in the intersection line of Π_{1} and Π_{2} (hence all the spans of the surfaces of S pass through the same line ℓ), or it intersects Π′ in a plane which contains the two distinct intersection lines of Π with Π_{1} and Π_{2}.

In the former case let π : X −−→ ℙ^{r−2} be the projection of X from the line ℓ and Y be its image. We claim that Y is a defective 3–fold such that there is an irreducible conic in Y containing two general points of Y. To prove the claim, note first that if Γ is a general surface in S, then its projection from ℓ is not a plane. Otherwise Y would be swept out by a family of planes such that through two general points of Y there is a plane of the family passing through them. This implies that Y ⊂ ℙ^{r−2} would be a linear subspace of dimension at most 4, which is a contradiction because r − 2 ⩾ 7 and Y has to be non– degenerate. Then the general Γ in S is projected from ℓ to a plane curve, which implies that dim(Y) = 3. Moreover Y is covered by a family of generically irreducible plane curves such that through two general points of Y there is a curve of the family passing through them. The Trisecant Lemma implies that the plane curves are conics and Y is in particular defective, because r − 2 ⩾ 7.

We claim that in this situation we are in cases (∗∗) or (∗ ∗ ∗). Indeed, let us look at the list of defective 3–folds in [?, Theorem 1.1]. To prove the claim it suffices to prove that:

• Y cannot be a cone;

• Y cannot be contained in a 4–dimensional cone over a curve;

• Y cannot be contained in a 4–dimensional cone over the Veronese surface V_{2,2}. To prove the first and second items, we argue as at the end of the proof of Theorem ??.

Indeed, suppose Y is a cone. If it is a cone with vertex a line over a curve then this curve is a conic, hence Y sits in ℙ^{4}, a contradiction, because Y is linearly normal in ℙ^{r−2} and r − 2 ⩾ 7. If Y is a cone with vertex a point over a surface S, then S has also a family of conics such that there is a conic passing through two general points of S. Then S is either the Veronese surface V_{2,2} or it sits in a ℙ^{n}, with n ⩽ 4. So Y would be contained in a ℙ^{m}, with m ⩽ 6, whereas Y is linearly normal in ℙ^{r−2} and r − 2 ⩾ 7. To prove the last item, we notice that in this case we have r = 9. Moreover, by [?, Example 2.4], a threefold Y contained in a 4–dimensional cone over the Veronese surface V_{2,2} has f(Y) = 1 and γ(Y) = 2. Let p_{0}, p_{1} ∈ X general points and set q_{0} = π(p_{0}), q_{1} = π(q_{1}). The surface Γ_{X,p_{0},p_{1}} spans a 4–space that contains the line ℓ, hence the hyperplane T_{X,p_{0},p_{1}} in ℙ^{9} also contains ℓ and it is projected from ℓ to the hyperplane T_{Y,q_{0},q_{1}} in ℙ^{7}. Since γ(Y) = 2, we have that T_{Y,q_{0},q_{1}} is tangent to Y along a surface Γ_{Y,q_{0},q_{1}}, and this would imply that T_{X,p_{0},p_{1}} is tangent to X along π^{−1}(Γ_{Y,q_{0},q_{1}}) that is a 3–fold, a contradiction since γ(X) = 2.

In the case in which a general 4–space Π intersects Π′ in a plane which contains the two distinct intersection lines of Π with Π_{1} and Π_{2}, the intersection lines of the spans of two general surfaces of S are generically distinct but they pairwise intersect in one point. Then either all these lines lie in the same plane or they pass through the same point p but do not lie in the same plane. The former case is impossible by Claim ??. In the latter case consider the projection π : X −−→ ℙ^{r−1} from p and let Y be its image. Since X is not a cone, then Y is a 4–fold. The general surface in S, which is a tangential contact surface on X, maps to a surface in Y, which is a tangential contact surface spanning a 3–space. Hence Y has an irreducible, 4–dimensional family Q of generically irreducible tangential contact surfaces spanning a 3–space, such that given two general points of Y there is a surface in Q containing the two points. By the Trisecant Lemma, the general surfaces in Q are irreducible quadrics and therefore f(Y) ⩾ 2.

If f(Y) = 3, then by Proposition ??, (i), Y would either be a cone over a curve or a cone over V_{2,2}. Both cases are impossible because in the former case Y would not contain the family Q of contact quadrics, in the latter Y would lie in a 7–space, whereas Y spans a ℙ^{r−1} and r − 1 ⩾ 8.

So we have that f(Y) = 2, and Y is in the list of Theorem ??. The only cases in this list which are compatible with the present situation are the cases (iv) and (v) of Theorem ?? (in the other cases the contact loci are not surfaces), and then we are in cases (∗), (∗∗) or (∗ ∗ ∗).

Finally suppose we are in case j = 0. Then we are in case (∗ ∗ ∗∗) and we will prove that the general surface in S is rational and X itself is rational. To see this, we need the following:

Claim 7.6.4. Let x ∈ X be a general point and let t be a general tangent direction of X at x. Then there is a unique surface in S_{x} tangent to t.

Proof of the Claim ??. We assume S_{x}, that has dimension 2, to be irreducible, otherwise we substitute S_{x} with one of its components. Let π_{x} : F_{x} → S_{x} be the universal family over S_{x}, that has dimension 4. There is the obvious rational map f_{x} : F_{x} −−→ X that is birational because through two general points of X there passes a unique surface in S. The map f_{x} induces another obvious birational map f̃_{x} : F_{x} −−→ X̃, where X̃ → X is the blow–up of X at x, with exceptional divisor E. Look at the birational map f̃_{x}^{−1}. Its indeterminacy locus on X̃ has codimension at least 2, so f̃_{x}^{−1} is defined at the general point of E. If ξ is a general point of E, that corresponds to a pair (x, t), with t general tangent direction at x, then f̃_{x}^{−1}(ξ) is the unique required surface. □

Claim ?? implies that S_{x}, which has dimension 2, is unirational, because there is a dominant map of ℙ(T_{x}(X)) ≅ ℙ^{3} to S_{x} (recall that T_{x}(X) is the Zariski tangent space to X at x), so S_{x} it is rational. Let now Γ be a general surface in S. We have a natural map ρ : S_{x} −−→ Γ, which takes the general surface Γ′ in S_{x} and maps it to the unique intersection point of Γ′ and Γ, which is also the unique intersection point of their spans. This map is dominant by the generality of Γ, hence Γ is unirational, so it is rational. Finally let x, y ∈ X be general points. There is a natural map σ : S_{x} × S_{y} −−→ X which maps the pair (Γ, Γ′) of general surfaces in S_{x} × S_{y} to their unique intersection point. This map is dominant because through a general point z ∈ X there passes a unique surface in S_{x} and a unique surface in S_{y}, intersecting at z. Moreover, it is also generically injective, because given z ∈ X general, there is a unique surface in S_{x} [resp. in S_{y}] containing z. This proves that X is rational. □

Remark 7.6.5. In cases (∗), (∗∗), (∗∗∗) listed in Theorem ??, it is clear that X is defective and verifies (ii). In case (∗ ∗ ∗∗) the 4–fold X is still defective, since the surfaces Γ on S span 4–spaces, hence f(X) = 1. However we do not have examples of such 4–folds, hence we do not know if this case really occurs.

Moreover, if X is smooth, as we have seen many times, cases (∗∗) and (∗ ∗ ∗) are not possible.

7.7. Here we discuss the reducible case, in which given two general points p_{0}, p_{1} ∈ X there is a unique tangential contact surface Γ_{p_{0},p_{1}} containing p_{0}, p_{1}, which consists of two irreducible components each passing through one of the two points p_{0}, p_{1}. We will denote by T the family of surfaces of X such that any tangential contact surface Γ is the sum of two components each in T . By the uniqueness of the tangential contact locus, given a general point x ∈ X there is a unique surface in T containing x, hence T is irreducible (see §??) and clearly dim(T ) = 2. We will denote by i the dimension of the span of the general surface in T . Recall that, since we are studying the case γ(X) = 2 and f(X) = δ(X) = 1, the general tangential contact surface Γ spans a 4–space by Proposition ??. Hence the union of two general surfaces in T spans a 4–space. Therefore 2 ⩽ i ⩽ 3. Indeed, clearly i ⩾ 2 and moreover i ⩽ 4 because, as we said, the union of two general surfaces in T spans a 4–space. Moreover it cannot be the case that i = 4, otherwise the span of a general surface in T would contain any other surface in T , hence it would contain X.

Proposition 7.7.1. In the above setting, if X is not a cone, then X sits in a cone with vertex a plane over a surface.

Proof. With the above notation, assume that i = 3. Then if S, S′ are two general surfaces in T , one has dim(⟨S⟩ ∩ ⟨S′⟩) = 2. Then either the spans of the surfaces in T lie in a projective space of dimension 3, which is clearly not possible, or there is a plane P which is contained in ⟨S⟩ for all S in T . By projecting from P we see X sits in a cone with vertex the plane P over a surface.

Assume i = 2, then T is a family of planes. These planes pairwise span a 4–space, hence they pairwise intersect at a point. By [?, Lemma 4.1] there is a plane P such that all planes in T intersect P along a line. By projecting from P we see again that X sits in a cone with vertex the plane P over a surface. □

Remark 7.7.2. If X is a 4–fold as in Proposition ??, then X is defective. In fact if P is the plane vertex of the 5–dimensional cone in which X sits, and if x ∈ X is a general point, then T_{X,x} intersects P along a line, and therefore if y ∈ X is another general point, T_{X,x} and T_{X,y} do intersect at a point.

A 4–fold X as in Proposition ?? is never smooth, because it is singular at points of the plane P vertex of the 5–dimensional cone over a surface S in which it sits. To see this we proceed in this way.

First, assume that X does not contain P, so that it intersects P along a curve C. Take a general hyperplane H in ℙ^{r} and consider the section Y of X with H, that sits in a cone with vertex the line s = P ∩ H over the surface S. Then Y intersects s at the intersection points p_{1}, …, p_{h} of H with C. We show that Y is singular at one of these points, thus showing, by the genericity of H, that X is singular. To see this, argue as in Remark ??. Namely, resolve the projection of Y from s by blowing-up a finite number of points above each p_{i}. If Y is smooth at one of the points p_{i}, the last exceptional divisor is a ℙ^{2}, which would be sent by the projection onto a plane. Hence S would be a plane, which is impossible since Y is non-degenerate.

Alternatively, X could contain P. In this case, again proceed as in Remark ?? by blowing up P in X. Then the exceptional divisor E of the blow–up is a ℙ^{2}-bundle over P, and it is mapped under the projection from P onto S in such a way that each ℙ^{2} in E is mapped to a plane. In particular S would be a plane, which leads to the same contradiction as above.

8. First species

8.1. Next we consider an irreducible, non–degenerate 4–fold, not a scroll, X ⊂ ℙ^{r}, r ⩾ 9, of the first species, i.e., γ(X) = 3. Taking into account Proposition ?? and Theorem ?? we may assume f(X) = δ(X) = 1.

Given p_{0}, p_{1} ∈ X general points, the tangent spaces to X at p_{0} and p_{1} intersect at one point and the tangential contact locus Γ_{p_{0},p_{1}} is a 3–fold, spanning a 6–space Π_{p_{0},p_{1}} (see Proposition ??). We have two cases: the irreducible and the reducible case, according to the possibilities that Γ_{p_{0},p_{1}} is irreducible or it consists of two irreducible components each passing through one of the two point p_{0}, p_{1}.

Since all the above properties are preserved by a general projection to ℙ^{9} because dim(S(X)) = 8, we will assume that r = 9.

Lemma 8.1.1. In the above setting we have the following cases:

(i) in the irreducible case the 3–folds Γ_{p_{0},p_{1}} move in a 2–dimensional family which is a linear system;

(ii) in the reducible case, the two components of Γ_{p_{0},p_{1}} move in a 1–dimensional (not necessarily rational) pencil.

Proof. Since we are assuming r = 9 and f(X) = 1, given p_{0}, p_{1} ∈ X general points, there is a unique hyperplane bitangent to X at p_{0} and p_{1}, hence there is a unique 3–fold Γ_{p_{0},p_{1}} as above, which is smooth at p_{0} and p_{1}. In the irreducible case the family of 3-folds Γ_{p_{0},p_{1}} has dimension two since through a general point there passes a one dimensional family of such 3-folds. It follows from [?, Theorem 5.10] that we are in case (i). Case (ii) follows from the fact that given x ∈ X general there is a unique component of a contact variety passing through x (hence the family has dimension 1). □

Let x ∈ X be a general point and consider the tangential projection τ_{x} : X −−→ X_{1} ⊂ ℙ^{4}. Since X is defective and f(X) = 1, then X_{1} is a 3–fold.

Lemma 8.1.2. In the above setting, we have t(X_{1}) = d(X_{1}) = 2, hence X_{1} is swept out by a 1–dimensional family P of planes along which the tangent space to X_{1} is constant, so that the dual variety X_{1}^{∗} of X_{1} is a curve. Each plane of the family P is the image via τ_{x} of a 3–fold Γ_{x,y}.

Proof. The first assertion follows by Remark ?? and by the fact that for a hypersurface Z in projective space one has t(Z) = d(Z). The second assertion follows from the fact that the general contact locus of X projects onto the general tangential locus of X_{1}, which is a plane of the family P. □

8.2. In this section we examine the irreducible case. We keep all notation and convention we introduced above.

Lemma 8.2.1. In the above setting, if we are in the irreducible case, then the curve X_{1}^{∗} is rational.

Proof. The varieties Γ_{x,y} passing through x form a rational pencil, by Lemma ??,(i), so that the family P is also rational. The rationality of X_{1}^{∗} follows. □

Next we want to study more closely the general tangential projection τ_{x} : X −−→ X_{1} ⊂ ℙ^{4} and the 3–fold X_{1}. By Lemma ?? and by [?, Theorem 2.20], we deduce that X_{1} is of one of the following types:

(i) there is a non–degenerate, irreducible, projective curve C ⊂ ℙ^{4} such that X_{1} is the closure of the union of the osculating planes to C at all non–flex points of C; the tangent space to X_{1} along a general osculating plane to C is the corresponding osculating 3–space;

(ii) X_{1} is the cone with vertex a point over a tangent developable scroll Σ in ℙ^{3}; (iii) X_{1} is the cone with vertex a line over a plane curve.

Lemma 8.2.2. In the above setting only case (iii) occurs.

Proof. We follow Scorza’s argument in §6 of [?]. Suppose first we are in case (i). There are formulae (see [?], p. 490, and, for modern versions, [?], [?, p. 39] and [?, Proposition 10.4.13]) that give the degree n_{1} of the tangent developable to C, the degree n_{2} of X_{1}, i.e., the locus of osculating planes to C, the class n_{3} of C, i.e., the degree of X_{1}^{∗}. For any branch ξ of C, let (α, α_{1}, α_{2}) be the (truncated) rank sequence of ξ:

• α is the degree of the branch, i.e. the intersection multiplicity of the branch with the general hyperplane through the origin of the branch;

• α + α_{1} is the intersection multiplicity of ξ with a general hyperplane containing the tangent line to ξ at the origin (the tangent line to a branch singular at the origin is the reduced tangent cone to the branch at the origin);

• α + α_{1} + α_{2} is the intersection multiplicity of ξ with a general tangent hyperplane

passing through the osculating plane to the branch. Note that for a general branch one has α = α_{1} = α_{2} = 1. Let d be the degree of C, while, as we know, its genus is 0. The formulae in question give

\begin{aligned}n_{1} & = 2(d − 1) − ∑(α − 1) \\ n_{2} & = 3(d − 2) − ∑(2α + α_{1} − 3) \\ n_{3} & = 4(d − 3) − ∑(3α + 2α_{1} + α_{2} − 6) \end{aligned} ; (14)

where the sums are taken over all branches of C.

Consider again the general tangential projection τ_{x} : X −−→ X_{1} ⊂ ℙ^{4}. Let Y be a general hyperplane section of X passing through x. Note that Y is not defective because f(X) = δ(X) = 1 (recall Lemma ??). However if p_{0}, p_{1} are general points of Y, the 7–space T_{Y,p_{0},p_{1}} is tangent to Y along a surface Γ_{Y,p_{0},p_{1}}, i.e. it is tangent along the intersection of the 3–fold Γ_{X,p_{0},p_{1}} with the hyperplane spanned by Y.

We can then consider the tangential projection τ_{Y,x} : Y −−→ Y_{1} ⊂ ℙ^{4}. Since Y is not defective, then Y_{1} has dimension 3 and therefore Y_{1} = X_{1} (because f(X) = 1, f(Y) = 0 and (??) and Lemma ?? hold). We let s be the degree of the map τ_{Y,x}.

Take a general plane Π in ℙ^{4} which cuts out a plane curve Z of degree n_{2} on X_{1}. By Bertini’s Theorem the curve Z is irreducible. By Lemma ?? the curve Z is rational. We let D be the normalization of the preimage Z′ of Z via the generically finite map τ_{Y,x}, so that D is a smooth irreducible curve.

Note that, since Z is rational, on D there is the linear series g_{s}^{1} consisting of the fibres of the map of D to Z (that is rational). This is the pull–back on D of the movable part of the linear series cut out on Z′ by the rational pencil described by the surfaces Γ_{Y,x,y} on Y passing through x. The hyperplanes of ℙ^{4} (which come via the tangential projections τ_{X,x} and τ_{Y,x} from hyperplanes containing T_{X,x} or T_{Y,x} respectively) cut out on D the pull– back to D of the linear series cut out on Z by the lines of the plane Π, hence the pull–back on D of the linear series cut out on Z′ by these hyperplanes has dimension 2 and degree sn_{2}, i.e., it is a g_{sn_{2}}^{2}, which is composite with the above g_{s}^{1}. Hence, a hyperplane passing through T_{Y,x} and tangent to Y along a surface Γ_{Y,x,y} cuts out on Z′ a divisor whose pull– back on D consists of a divisor of the g_{s}^{1}, counted with multiplicity 2, plus n_{2} − 2 more divisors of the g_{s}^{1}.

Let p ∈ D be a general point, which by abusing notation, we may identify with a general point of Z′ hence with a general point of X and let q be its image on Z. Note that q is also a general point of X_{1}.

The number of hyperplanes passing through T_{Y,x} and p and tangent to Y along surfaces of the type Γ_{Y,x,y} which do not pass through p equals the number of divisors of the g_{sn_{2}}^{2} which contain the divisor of the g_{s}^{1} containing p and then another divisor of the g_{s}^{1} counted with multiplicity 2. This number coincides with the number of lines of the plane Π passing through q and tangent to Z at a point different from q. By Hurwitz formula, and taking into account that Z is rational, we have that this number is 2(n_{2} − 2).

On the other hand, we can compute this number in another way. Indeed, this number is nothing but the number of tangent hyperplanes to X_{1} passing through q but not tangent to X_{1} in q. By duality this equals the number of intersections of a general tangent hyperplane to X_{1}^{∗} with X_{1}^{∗} off the contact point, i.e., n_{3} − 2.

In conclusion we have n_{3} − 2 = 2(n_{2} − 2), which, by using (??), reads

∑(α − 1) = 2(d − 1) + ∑(α_{2} + 5)

and implies n_{1} = 2(d−1)−∑(α−1) < 0, a contradiction. Hence case (i) cannot happen.

Assume next we are in case (ii). Consider the tangent developable scroll Σ ⊂ ℙ^{3} and let ℭ be the rational curve of degree d whose tangent lines fill up Σ. For any branch ξ of ℭ, let (α, α_{1}) be the (truncated) rank sequence of ξ, with the same meaning as above. Then the degree of Σ, hence of X_{1}, is

n_{2} = 2(d − 1) − ∑(α − 1)

whereas the class of ℭ, which coincides with the degree of X_{1}^{∗}, is

n_{3} = 3(d − 2) − ∑(2α + α_{1} − 3).

Then the same argument as above leads to the equality

d + ∑(α_{1} − 1) = 0

which is also impossible. Then also case (ii) cannot happen, and only case (iii) can occur. □

Theorem 8.2.3. Let X ⊂ ℙ^{9} be a non–degenerate variety of dimension 4, not a scroll, of the first species, which is not a cone. If X presents the irreducible case, then X sits in a 6–dimensional cone with vertex a 3–space over a Veronese surface V_{2,2} in a ℙ^{5} skew with the vertex.

Proof. By Lemma ??, given a general point x ∈ X, all hyperplanes containing T_{X,x} and tangent to a 3–fold Γ_{x,y} also pass through a fixed ℙ^{6}, i.e., the join of T_{X,x} with the line vertex of X_{1} (which is a cone with vertex a line by Lemma ??). By Lemma ??, these hyperplanes form, in the dual plane of this ℙ^{6} an irreducible rational curve.

Let us consider the 2–dimensional family K of hyperplanes tangent to X along a 3–fold of the type Γ_{p_{0},p_{1}}. By duality, this family describes an irreducible surface S ⊂ (ℙ^{9})^{∗} which possesses a family of dimension at least 2 of irreducible, rational, plane curves.

Claim 8.2.4. The surface S spans a ℙ^{n}, with 2 ⩽ n ⩽ 5.

Proof of the Claim ??. The bound n ⩾ 2 is obvious. Let us prove that n ⩽ 5. By [?, Lemma 4.1], the elements of the family P of the planes spanned by the plane curves on S either pass through the same point p, or intersect a given plane P along lines, or they span at most a 5–space. In the first case, by projecting S from p we find a surface which possesses a 2–dimensional family of lines, hence it is a plane, thus n = 3. In the second case, by projecting S from P, we see that each plane curve of S is contracted to one point and, since these curves intersect each other, S itself is contracted to a point, which again implies n = 3. The Claim is thus proved. □

More precisely:

Claim 8.2.5. One has n = 5. Moreover all tangent spaces to X intersect a fixed 3–space Π along a line.

Proof of the Claim ??. First of all n = 2 is impossible, because then the hyperplanes of K vary in a 2–dimensional linear system cutting on X a linear system of surfaces which are all singular with the singular points filling up X, contrary to Bertini’s theorem.

Assume n ⩾ 3. In this case all the hyperplanes of K contain one and the same subspace Π of dimension 8 − n. Let x ∈ X be a general point. The hyperplanes in K containing T_{X,x} all contain the same ℙ^{6} (see the beginning of the proof), which contains Π and also T_{X,x}. Hence T_{X,x} cuts Π along a subspace of dimension at least 6 − n.

If n = 3, two general tangent spaces to X intersect in at least a line, a contradiction, because f(X) = 1. Assume n = 4. All tangent spaces to X intersect Π along a plane. Consider the projection π : X −−→ ℙ^{4}, of X from Π. The image of X is a curve Λ. The general hyperplane of K maps via π to a general bitangent hyperplane to Λ and the 3–folds Γ_{p_{0},p_{1}} would then be reducible in pairs of fibres of π, a contradiction since we are in the irreducible case.

In conclusion this proves that n = 5. Then all the hyperplanes of K contain one and the same 3–space Π and we can conclude again that all tangent spaces to X intersect Π along a line. □

Consider the projection π : X −−→ ℙ^{5} of X from Π. The image of X is a surface S. The general hyperplane of K maps via π to a general bitangent hyperplane to S, which is also tangent along an irreducible curve, the image of Γ_{p_{0},p_{1}}. To finish our proof we need to prove the:

Claim 8.2.6. One has S = V_{2,2}

Proof of the Claim ??. If x, y ∈ S are general points, there is a bitangent hyperplane to S at x and y, hence the tangent spaces to S at x and y intersect at a point, thus S is defective with irreducible contact curve. Hence S = V_{2,2} (see [?, Theorem 1.3]). □

This ends our proof. □

Remark 8.2.7. Let X ⊂ ℙ^{9} be a 4–fold as in the statement of Theorem ??, sitting in a 6–dimensional cone V with vertex a 3–space Π over a Veronese surface V_{2,2}. We claim that X is in general defective, of the first species, presenting the irreducible case.

Indeed, let p_{0}, p_{1} ∈ X be general points. The space T_{V,p_{0},p_{1}} has dimension 8 and it contains T_{X,p_{0}} and T_{X,p_{1}}, which therefore in general intersect at a point, proving that X is defective with f(X) = 1. Moreover V is swept out by a 2–dimensional family of quadric cones of rank 3, which project from Π the conics of the Veronese surface V_{2,2}. For each of these quadric cones Q, there is a hyperplane which is tangent to V along Q. In general the intersection of Q with X is an irreducible 3–fold of the type Γ_{p_{0},p_{1}}, proving our claim.

Also in this case X is never smooth.

8.3. In this section we examine the reducible case. We consider irreducible, defective, projective threefolds X ⊂ ℙ^{r}, of the first species, with r ⩾ 9, not a scroll, with f(X) = 1. By projecting generically to ℙ^{9} we may assume that r = 9. If p_{0}, p_{1} ∈ X are general points, the tangent spaces to X at p_{0} and p_{1} span a hyperplane which is tangent to X along a contact 3–fold Γ_{p_{0},p_{1}} which is reducible in two irreducible components each containing p_{0} and p_{1} respectively. By Lemma ??, these two components vary in a pencil which we denote by G, and we denote by Φ its general element.

Theorem 8.3.1. Let X ⊂ ℙ^{r} be a non–degenerate variety of dimension 4, not a scroll, of the first species with f(X) = 1, which is not a cone. If X presents the reducible case, then one of the following occurs:

(i) X lies in a 6–dimensional cone with vertex a ℙ^{4} over a curve; (ii) X lies in a 5–dimensional cone with vertex a plane over a scroll surface.

Proof. Let us denote by m the dimension of ⟨Φ⟩, with Φ ∈ G a general element. Since we are assuming X not a scroll, we have m ⩾ 4. On the other hand, if Φ and Φ′ are two general elements of G, then ⟨Φ, Φ′⟩ has dimension 6 (see Proposition ??,(iv)). This implies that m ⩽ 5.

If m = 4 and if Φ and Φ′ are two general elements of G, then ⟨Φ⟩ ∩ ⟨Φ′⟩ has dimension 2. By [?, Lemma 4.1], we have only the following two possibilities:

(a) there is a fixed plane Π such that, for a general element Φ of G, one has Π ⊂ ⟨Φ⟩;

(b) there is a fixed 4–space Π such that, for a general element Φ of G, one has that

Π ∩ ⟨Φ⟩ has dimension 3. If (a) holds, we are in case (ii) and if (b) holds we are in case (i).

If m = 5 and if Φ and Φ′ are two general elements of G, then ⟨Φ⟩ ∩ ⟨Φ′⟩ has dimension 4. This implies that there is a 4–space Π such that if Γ is a general element of G, one has that Π ⊂ ⟨Φ⟩. Then we are again in case (i). □

Remark 8.3.2. Note that if (i) of Theorem ?? occurs, then X ⊂ ℙ^{r} is actually defective and in general it has f(X) = 1 and presents the reducible case. Indeed, if p_{0}, p_{1} ∈ X are general points, and V is the 6–dimensional cone containing X, then T_{V,p_{0},p_{1}} has dimension 8 and it contains T_{X,p_{0}} and T_{X,p_{1}}, which therefore in general intersect at a point.

If we are in case (ii) of Theorem ??, X is still defective with, in general, f(X) = 1. In fact, let V be the 5–dimensional cone on which X sits, with vertex the plane Π. Then V is swept out by a 1–dimensional family K of 4–spaces, i.e., the spans of Π with the lines of the scroll Σ which is the base of the cone. The general such 4–space P intersects X along a 3–fold X_{P}. If x ∈ X is a general point and P is the unique element of K containing x, then T_{X,x} intersects P in the 3–space T_{X_{P} ,x} and therefore it intersects Π along a line. Therefore two general tangent spaces to X intersect in general at a point in Π, proving that X is defective.

It is interesting to note that, in case (ii) of Theorem ??, X is not in general of the first species. In fact, let p_{0}, p_{1} ∈ X be general points. Then T_{V,p_{0},p_{1}} has dimension 8 and, containing T_{X,p_{0},p_{1}}, it coincides with it. Now in general the bitangent hyperplane to V at p_{0}, p_{1} is only tangent along the 3–spaces generators of V passing through p_{0}, p_{1}, hence in a general situation it is bitangent to X only along the surfaces in which these generators cut X.

However, it may happen that, in some particular cases, X is of the top species. For example this is the case if the scroll Σ is developable. In this case in fact the tangent space to V is constant along the 4–spaces of the family K so that in this case a general bitangent hyperplane to X is tangent along two threefolds of the form X_{P} with P ∈ K.

In any event, X is never smooth in these cases.

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Dipartimento di Ingegneria dell’Informazione e Scienze Matematiche, Università di Siena, Via Roma, 56, 53100 Siena, Italia

   Email address: chiantini@unisi.it

Dipartimento di Matematica, Università di Roma Tor Vergata, Via O. Raimondo, 00173 Roma, Italia

   Email address: cilibert@axp.mat.uniroma2.it

Dipartimento di Matematica e Informatica, Università di Catania, Città Universitaria, Viale A. Doria 6, 95125 Catania, Italia

   Email address: francesco.russo@unict.it