Codimension two subvarieties and related questions.

Notes for the C.I.R.M. school (Trento 11-16 September 2006)

Ph. Ellia

6/09/2006

Contents

Chapter I. Introduction. 1 1. Projections and secant varieties. 1 2. Notes and complements. 4

Chapter II. Barth’s type theorems and Hartshorne’s conjecture. 7 1. Barth’s type theorems. 7 2. Hartshorne’s conjecture. 8 3. Vector bundles and codimension two subvarieties. 11 4. Stable rank two vector bundles. 17 5. Notes and complements. 20

Chapter III. Multisecants and Ran’s theorem. 23 1. Multisecants. 23 2. Ran’s theorem 24 3. Notes and complements. 28

Chapter IV. Some vanishing theorems. 29 1. Ample and k-ample vector bundles. 29 2. Linear normality. 32 3. Projective normality and Hartshorne conjecture. 34 4. Notes and complements. 37

Chapter V. Surfaces in ℙ^{4}: the theorem of Ellingsrud and Peskine. 39 1. The normal bundle: here enters the hero. 39 2. Preliminaries. 45 3. Proof of the theorem. 48 4. Notes and complements. 50

Chapter VI. Bounds for smooth divisors of hypersurfaces. 53 1. Introduction 53 2. The speciality theorem: geometrical proof. 53 3. Positivity lemma and the speciality theorem. 59 4. Notes and complements. 62

Chapter VII. Numerology. 63 1. A general expression for the u_{k}. 63 2. First consequences. 66 3. The Schwarzenberger conditions. 68 4. Bounds on e and s. 73 5. Notes and complements. 78

Chapter VIII. Conclusion. 79 1. Summarizing. 79 2. By day and by night... 79

CHAPTER I Introduction.

1. Projections and secant varieties.

Let Y be a projective variety (smooth, irreducible), of dimension r. By definition Y is endowed with a very ample line bundle, L, which embedds Y into some ℙ^{n}. In general if ϕ_{L} is the corresponding embedding, we expect ϕ_{L}(Y) := X to have a high codimension in ℙ^{n}. To remedy this situation we can try to project X, down to ℙ^{n−1} isomorphically. Here isomorphically means that the projection restricted to X is an embedding. Let O be the center of the projection and denote by π : ℙ^{n} − − > ℙ^{n−1} the projection of center O. Set π_{O} := π|X, then it is easy to see that (if O ∉ X), π_{O} is an embedding if and only if O ∉ Sec(X) where Sec(X) is the closure in ℙ^{n} of {p ∈ ℙ^{n} | p ∈< q, r >, q, r ∈ X}. Now we can estimate the dimension of Sec(X): a point on Sec(X) corresponds to a pair, (p, q), of points of X and a point on the line < p, q >. To choose a point on X consists in ∞^{r} parameters, so to choose two points consists in ∞^{2r} parameters. This being done, for (p, q) ∈ X^{2}, we have to choose a point on the line < p, q >: this is one further parameter. In conclusion this proves:

Lemma 1.1: With notations as above: dim(Sec(X)) ≤ 2r + 1.

Of course this estimate is not always sharp. For example let C ⊂ ℙ^{3} be a plane curve of degree ≥ 2. The lemma says dim(Sec(C)) ≤ 3. It is clear however that dim(Sec(X) = 2; more precisely Sec(X) =< C >, the plane spanned by C, indeed every secant to C is contained in < C > (and every line in < C > is a secant to C if d(C) ≥ 2).

A more sophisticated example is provided by the Veronese surface. Let v : ℙ^{2} → ℙ^{5} be the Veronese embedding:

[x_{0} : x_{1} : x_{2}] → [x_{0}^{2} : x_{1}^{2} : x_{2}^{2} : x_{0}x_{1} : x_{0}x_{2} : x_{1}x_{2}]

In this embedding a line of ℙ^{2} is transformed in a conic of ℙ^{5}. It turns out that given x ≠ y ∈ V = v(ℙ^{2}), there exists a conic, K ⊂ V, passing through x and y. Clearly any point on the line < x, y > will lie on the plane < K >; viceversa any point of the plane < K > lies on a secant line to K, hence to V. There are ∞^{2} such planes (as are the lines of ℙ^{2}). We conclude that dim(Sec(V) ≤ 4. It can be shown that dim(Sec(V)) = 4 (see Exercice 2.1); so dim(Sec(V)) ≠ 5, which was the expected dimension. Anyway it turns out that V can be projected isomorphically to ℙ^{4}. A celebrated theorem of Severi ([75]) states that for smooth surfaces in ℙ^{5}, this is the only exception:

Theorem 1.2: [Severi 1901] Let S ⊂ ℙ^{5} be a smooth, non degenerated surface. Then S can be projected isomorphically to ℙ^{4} if and only if S is a Veronese surface.

Another way to state this theorem is as follows:

Theorem 1.3: Let S ⊂ ℙ^{4} be a smooth surface, then h^{1}(I_{S}(1)) = 0 unless S is a projected Veronese surface.

Proof. Indeed if h^{0}(O_{S}(1)) > 4, S is the isomorphic projection of a surface S′ ⊂ ℙ^{5}.

Since a projected Veronese surface has degree 4 we obtain:

Corollary 1.4: Let S ⊂ ℙ^{4} be a smooth, non degenerated surface of degree d > 4, then S is linearly normal, i.e. h^{1}(I_{S}(1)) = 0.

One may wonder how far does this statement generalize? This is the content of Zak’s theorem on linear normality (see II.Theorem 2.2).

At this point we may summarize as follows: we can \underline{a}lways embedd a r-dimensional variety in ℙ^{2r+1}, but we must expect restrictions in order to embedd smooth r- dimensional varieties in ℙ^{n} with n < 2r + 1 (although in some cases this is possible).

An r- dimensional variety X ⊂ ℙ^{2r+1} has codimension r + 1 > dim(X). Putting it differently we could say that if codim_{ℙ^{n}}(X) ≤ dim(X), then the embedding X → ℙ^{n} is \underline{special}.

Thus the embedding of a smooth curve in ℙ^{2} is an exceptional situation, as is the embedding of a smooth surface in ℙ^{4}, the embedding of a smooth threefold in ℙ^{n}, n ≤ 6, etc...

Concerning codimension two subvarities X ⊂ ℙ^{n}, we can expect strong restrictions to their existence as soon as n ≥ 4.

Historically speaking the first results in this direction are the quoted above result of Severi, the double points formula for surfaces in ℙ^{4} and the Severi-Lefschetz theorem (nowadays the Severi-Lefschetz-Grothendieck theorem):

Theorem 1.5: [Severi-Lefschetz-Grothendieck] Let Σ ⊂ ℙ^{n} be a complete intersection with dim(Σ) ≥ 3, then P ic(Σ) ≃ ℤ.H, where H is the class of a hyperplane section.

This theorem was first proved by Severi ([76]) in the case Σ a smooth hypersurface of ℙ^{n}, n ≥ 4, then it was generalized by Lefschetz to the case of smooth complete intersections of dimension ≥ 3 and finally the smoothness assumption was relaxed by Grothendieck.

By the way we notice that this result extends the Noether-Lefschetz theorem stating that P ic(S) ≃ ℤ.H for S a general surface of degree ≥ 4 in ℙ^{3}.

Concerning codimension two subvarieties, this theorem has strong consequences. Indeed let X ⊂ Σ ⊂ ℙ^{n}, n ≥ 4 be a smooth codimension two subvariety lying on a hypersurface Σ. If X is a Cartier divisor on Σ, then by Theorem 1.5, X is linearly equivalent to mH for some m, hence, since Σ is an hypersurface, X is cut out on Σ by another hypersurface, i.e. X is the complete intersection of Σ with another hypersurface. In particular if Σ is smooth, then any smooth divisor of Σ is a complete intersection. So if X is not a complete intersection, then X cannot lie on a smooth hypersurface, in fact we have something more: if X ⊂ Σ ⊂ ℙ^{n} and if X is not a complete intersection, then dim(X ∩ Sing(Σ)) ≥ n − 4. Indeed this follows from Theorem 1.5 by cutting with a general linear space of dimension four. This again is a severe restriction.

All these results have a common denominator: the normal bundle, N_{X}, of X in ℙ^{n}, which, as we will see, plays an important role. For more details on the normal bundle, the double points formula and Severi-Lefschetz theorem, we refer to V.Section 1.

2. Notes and complements.

Exercice 2.1: Let W be a k-vector space of dimension 3 and ℙ^{2} = ℙ(W) (projective space of lines of W). We can define a map v : ℙ(W^{∗}) → ℙ(S^{2}W^{∗}) :< L >→< L^{2} > which sends a (projectivized) linear form, < L >, to its square < L^{2} >. Geometrically speaking we are doing the following: given a line R ⊂ ℙ^{2} (which is a point of ℙ_{2}^{∗}), of equation L = 0, we are sending it to the point of ℙ(S^{2}V^{∗}), the space of conics in ℙ^{2}, which corresponds to the double line L^{2} = 0. If L = ax + by + cz, then L^{2} = a^{2}x^{2} + b^{2}y^{2} + c^{2}z^{2} + 2abxy + 2acxz +2bcyz, so taking coordinates, the map v : ℙ_{2}^{∗} → ℙ(S^{2}V^{∗}) ≃ ℙ^{5} is given by (a : b : c) → (a^{2} : b^{2} : c^{2} : 2ab : 2ac : 2bc). If ch(k) ≠ 2 we recognize the Veronese embedding, so v(ℙ_{2}^{∗}) = V the Veronese surface. In other words: in the set of all conics of ℙ^{2} the Veronese surface is ℙ{double lines in ℙ^{2}}. Let M denote the space of degenerated (i.e. singular) conics. This is a degree three hypersurface. Indeed a conic z_{0}x^{2} + z_{1}y^{2} + z_{2}z^{2} + 2z_{3}xy + 2z_{4}xz + 2z_{5}yz is singular if and only if:

| \begin{matrix}z_{0} & z_{3} & z_{4} \\ z_{3} & z_{1} & z_{5} \\ z_{4} & z_{5} & z_{2} \end{matrix} | = 0

Clearly V ⊂ M.

1) Show that: Sec(V) = M.

(The key point is that a conic is reducible if and only if it can be expressed as the sum of two squares.)

2) More intrisically a point (z_{0} : ... : z_{5}) ∈ ℙ^{5} corresponds to the conic whose associated symmetric matrix is:

( \begin{matrix}z_{0} & z_{3} & z_{4} \\ z_{3} & z_{1} & z_{5} \\ z_{4} & z_{5} & z_{2} \end{matrix} )

In this setting V = ℙ{ symmetric 3 × 3 matrices of rank ≤ 1} and: M = ℙ{ symmetric matrices 3 × 3 of rank ≤ 2}.

It turns out (by general facts on determinantal varieties, or by direct computations: the equations of V are given by the 2 × 2 minors) that:

Sing(M) = V

Exercice 2.2: Generalize and look at the 2-Veronese embedding: v : ℙ(W^{∗}) → ℙ(S^{2}W^{∗}) where dim(W) = n + 1. If X is the image, what can you say about Sec(X)?

Exercice 2.3: Let C ⊂ ℙ^{3} be a smooth, non-degenerate curve. Show (in two different ways) that Sec(C) = ℙ^{3}.

Exercice 2.4: Let S ⊂ ℙ^{5} be a smooth surface, show that if dim(Sec(S)) < 4, then S is contained in a hyperplane.

Notes. For a proof of Severi’s theorem one may consult the original paper of Severi [75] and also [61]. For a generalization of Severi’s theorem by Zak, see II.Theorem 2.2.

6 I. INTRODUCTION.

CHAPTER II Barth’s type theorems and Hartshorne’s conjecture.

1. Barth’s type theorems.

For smooth complete intersections we have the following results, usually attributed to Lefschetz (at least when k = ℂ):

Theorem 1.1: Let Y ⊂ ℙ^{n} (k = ℂ) be a nonsingular complete intersection of dimension r. Then:

1. The restriction map H^{i}(ℙ^{n}, ℤ) → H^{i}(Y, ℤ) is an isomorphism for i < r, and is injective for i = r.

2. π_{1}(Y) = 1, if r ≥ 2.

3. P ic(Y) ≃ ℤ · O_{Y} (1), if r ≥ 3.

For modern proofs of these results see for example [4] for 1), and [35] for 2) and 3) (see also [42] ); in fact the smoothness assumption in 3) is not necessary.

Barth type theorems assert that, to some extend, these results hold for any smooth (not necessarily a complete intersection) variety of small codimension. Namely we have:

Theorem 1.2: Let X ⊂ ℙ^{n} (k = ℂ) be a nonsingular variety of dimension r. Then:

1. The restriction map H^{i}(ℙ^{n}, ℂ) → H^{i}(X, ℂ) is an isomorphism for i ≤ 2r − n.

2. As above but for cohomology with coefficients in ℤ.

3. π_{1}(X) = 1 if r ≥ \frac{n+1}{2}.

4. P ic(X) ≃ ℤ · O_{X}(1) if r ≥ \frac{n+2}{2}.

Part 1) was first proved by Barth ([8]), part 2) was then proved by Larsen using Barth’s techniques and Morse theory ([54]). Of course part 2) implies 1) by the formula of universal coefficients. Part 3) was proved by Barth and Larsen ([11]). Finally 4) can be deduced from 2) via the exponential sequence. Part 4) (the most relevant one in our context) has been extended to characteristic zero by Ogus ([63]), using algebraic De Rham cohomology.

For the moment let’s observe that if r < n/2, then the statement of the theorem is vacuous. On the other hand, as the codimension becomes small with respect to the dimension, one get stronger and stronger conclusions. For example in the codimension two case, if n ≥ 6, we see that X behaves, quite closely, like a complete intersection.

2. Hartshorne’s conjecture.

These Barth type theorems are certainly the main motivation for Hartshorne conjecture ([41]):

Conjecture 1: [Complete intersections] Let k be an algebraically closed field and denote by ℙ^{n} the n-dimensional projective space over k. If X ⊂ ℙ^{n} is a nonsingular subvariety of dimension r and r > \frac{2n}{3}, then X is a complete intersection.

Hartshorne himself says (cf [41]): While I am not convinced of the truth of this statement, I think it is useful to cristallize one’s ideas, and to have a particular problem in mind.

So, but what are the evidences fot this conjecture? First, if one is particularly bold, he could guess, on the basis of the Barth type theorems (Theorem 1.2), that varieties of small codimension behave like complete intersections just because they are complete intersections! Ok, but even so it seems difficult to give a precise meaning of what small means. In other words where does the fraction \frac{2}{3} come from? Actually this fraction has been found on the basis of few examples, but it seems, after Zak’s work (see Conjecture 2, Theorem 2.2), that it could well be the correct one.

Before discussing about the fraction, a few words concerning the ground field, k, are in order. It is clear that the nonsingularity assumption cannot be dropped. For example, by taking the cone in ℙ^{n} with vertex a ℙ^{n−4} over a twisted cubic in ℙ^{3}, one get a (singular) non complete intersection, codimension two subvariety of ℙ^{n}. Now smoothness behaves quite differently according to the characteristic of the ground field. There are indeed phenomena peculiar to characteristic p (existence of indecomposable rank two vector bundles on ℙ^{5}, see Section 3). They are not yet counterexamples to the conjecture but, any way, it seems safe to assume ch(k) = 0 or even k = ℂ.

From now on, unless if explicitely stated, we will assume ch(k) = 0.

To find out the fraction (if it is believed that such a fraction exists!) one tries to construct smooth X ⊂ ℙ^{n} of large dimension, which are not complete intersections. One way to do this is to project isomorphically X ⊂ ℙ^{N} into a smaller projective space.

Indeed if X ⊂ ℙ^{n} is a projected variety, then X is not linearly normal (h^{0}(O_{X}(1)) > n + 1, hence h^{1}(I_{X}(1)) ≠ 0 because X is not degenerated) and X is not a complete intersection because a complete intersection, Y, is projectively normal (in particular this implies h^{1}(I_{Y} (m)) = 0, ∀m ∈ ℤ if dim(Y) ≥ 1).

As already seen ( I) if dim(X) = r we can always embedd X in ℙ^{2r+1}. This yields non complete intersections varieties of ℙ^{n} of dimension ∼ \frac{n}{2}. To go further we need to find varieties with dim(Sec(X)) < 2r + 1. For example a projected Veronese surface S ⊂ ℙ^{4} is an example of a non complete intersection surface in ℙ^{4}. By the way let’s observe that there exist linearly normal, non complete intersections, subvarieties. For example consider the Segre embedding ℙ^{1} × ℙ^{2} → ℙ^{5}; the image is the Segre cubic threefold. It is linearly normal but it is not a complete intersection (by Bezout since the degree is three). This example is of some interest because, as we shall see, every threefold in ℙ^{5} is linearly normal (Zak’s theorem).

In [41] Hartshorne considered only a few classes of subvarieties:

1. Veronese varieties: ℙ^{n} → ℙ^{N} embedded by the complete linear system of quadrics

2. Segre varieties: ℙ^{n−1} × ℙ^{m−1} → ℙ^{nm−1}

3. Plücker embedding of Grassmannian of lines: G(1, n) → ℙ^{N}

For a smooth X ⊂ ℙ^{n} of dimension r, we define the defect of X, δ(X), by: δ(X) := 2r + 1 − dim(Sec(X)). It turns out that X can be embedded isomorphically in ℙ^{N}, N = 2r + 1 − δ(X).

Lemma 2.1: With notations as above, in case 1), δ(X) = 1, in case 2), δ(X) = 2, and δ(X) = 4 in case 3).

In particular we get:

1. The Veronese surface in ℙ^{4} (dimension 2)

2. The Segre variety ℙ^{2} × ℙ^{2} in ℙ^{7} (dimension 4)

3. The Grassmannian G(1, 5) in ℙ^{13} (dimension 8).

They are all projected varieties of dimension r in ℙ^{n} with r = \frac{2}{3}(n − 1). At this point it is more convenient to directly quote Hartshorne ([41]):

Since we do not know any examples where r is any larger with respect to n, we make the following conjecture:

Conjecture 2: [Linear normality] Let X ⊂ ℙ^{n} be a smooth variety of dimension r. If r > \frac{2}{3}(n − 1), then X is linearly normal.

This conjecture has been proved by Zak by beautiful geometrical arguments. In fact Zak did even more since he also classified the Severi varieties, i.e. the nonlinearly normal varieties in ℙ^{n} of dimension r = \frac{2}{3}(n − 1): there are exactly four of them:

Theorem 2.2: [Zak Theorem on linear normality]

(a) Conjecture 2 is true: a smooth variety X ⊂ ℙ^{n} of dimension r with r > \frac{2}{3}(n − 1) is linearly normal.

(b) There are four Severi varieties: the three varieties mentionned above and the E_{6}-variety which is a 16-dimensional variety in ℙ^{25}.

Proof. See [80] and also [56].

The E_{6} variety comes from representation theory and was not known at the time of the writing of [41]. Of course Zak’s result makes Conjecture 1 more plausible.

Coming back to our fraction, we have the Plücker embedding G(1, 4) → ℙ^{9}. The image has dimension 6 and degree 5: this is an example of a non complete intersection variety with r = \frac{2}{3}n. If we look at the standard examples of Lemma 2.1, this is the best we can achieve. Hence the fraction \frac{2}{3}...

3. Vector bundles and codimension two subvarieties.

From now on we will focus on the codimension two case of Conjecture 1. First of all we will show that this particular case of the conjecture is related to the existence of indecomposable rank two vector bundles on ℙ^{n}.

We first recall some basic facts about rank two vector bundles on ℙ^{n}. For more details see [64], [44] (and [25] if you understand italian).

Let E denote a rank two vector bundle on ℙ^{n}. The Chern classes c_{1}, c_{2} may be viewed as integers. If E(m) := E ⊗ O(m), then c_{1}(E(m)) = c_{1} + 2m and c_{2}(E(m)) = c_{1}m + c_{2} + m^{2}. In particular it is always possible to find a twist such that c_{1}(E(m)) = 0 or c_{1}(E(m)) = −1. In this case we say that E(m) is normalized.

A non-zero s ∈ H^{0}(E) induces a morphism E^{∗} →^{s^{∗}} O, let I_{Z} = Im(s^{∗}). The subscheme Z is the zero-locus of the section s (sometimes denoted (s)_{0}). There are three possibilities:

1. Z = ∅ (in this case E = O(a) ⊕ O(b))

2. Z contains a divisor

3. Z has codimension two and is locally a complete intersection.

Indeed in case 1. E is an extension of two line bundles: 0 →^{s} O → E → O(c) → 0. On ℙ^{1} every vector bundle is a direct sum of line bundles so we may assume n > 1. Then H^{1}(O(−c)) = 0 and the extension splits. In case 2. if D denotes the divisorial part of Z, the map s^{∗} factors through a map E^{∗} → O(−D) and the inclusion O(−D) → O. Hence we get a morphism: O → E(−D). Since D ∼ rH, we get a section, s′, of E(−r) which vanishes in codimension two. In particular if h^{0}(E(−1)) = 0, then every section of E vanishes in codimension two. Since E is locally free, a section s of E is locally given by two equations (f, g). If f and g have no common factor (case 3.) they define a codimension two subscheme which is (locally) a complete intersection. We may rephrase 2. in this setting: if f = hu and g = hv (u, v without common factor) then (u, v) = s′. In other words (s′)_{0} is the residual scheme of (s)_{0} with respect to the divisor D. So if s vanishes in codimension two we have an exact sequence:

0 → O(−c_{1}) → E^{∗} → I_{Z} → 0 which is nothing else than the Koszul complex of s (Z is locally a complete intersection). If we restrict to Z we get E^{∗}|Z ≃ I_{Z} ⊗ O_{Z} because the map O(−c_{1}) → E^{∗} vanishes on Z (it is locally given by t → (ft, gt) where f, g are the equations of Z). Now I_{Z} ⊗O_{Z} ≃ I_{Z}/I_{Z}^{2} ≃ N_{Z}^{∗}, the conormal bundle of Z in ℙ^{n}. Now since Z is locally a complete intersection, N_{Z} is locally free and its determinant is ω_{Z}(n + 1). Since N_{Z} ≃ E|Z, taking determinants, we conclude that ω_{Z} ≃ O_{Z}(c_{1} − n − 1) (i.e. Z is a so-called subcanonical subscheme). In conclusion if a section, s, of a rank two vector bundle vanishes in codimension two, then (s)_{0} is a l.c.i. (locally complete intersection) subscheme which is subcanonical and we have an exact sequence:

0 → O →^{s} E → I_{Z}(e + n + 1) → 0

where c_{1}(E) = e + n + 1 and ω_{Z} ≃ O_{Z}(e)

Moreover, by general properties of Chern classes, since s vanishes in the correct codimension: deg(Z) = c_{2}(E).

Now we try to go the opposite way: starting from Z we want to reconstruct E. The exact sequence above corresponds to an element of Ext^{1}(I_{Z}(e + n + 1), O). The local to global spectral sequence for Ext yields an exact sequence:

H^{1}(O(−e−n−1)) → Ext^{1}(I_{Z}(e+n+1), O) → H^{0}(Ext^{1}(I_{Z}(e+n+1), O) → H^{2}(O(−e−n−1)) → ...

If n > 2 it follows that: Ext^{1}(I_{Z}(e + n + 1), O) ≃ H^{0}(Ext^{1}(I_{Z}(e + n + 1), O). Now, by applying Hom(−, O) to the exact sequence:

0 → I_{Z} → O → O_{Z} → 0

we get:

Ext^{1}(I_{Z}(e+n+1), O) ≃ Ext^{1}(I_{Z}, O)(−e−n−1) ≃ Ext^{2}(O_{Z}, O)(−e−n−1).

But since Z has codimension two:

Ext^{2}(O_{Z}, O) ≃ Ext^{2}(O_{Z}, ω_{ℙ^{n}})(n + 1) ≃ ω_{Z}(n + 1).

Finally: Ext^{1}(I_{Z}(e + n + 1), O) ≃ ω_{Z}(−e) ≃ O_{Z} (since ω_{Z} ≃ O_{Z}(e) by assumption). In conclusion, under our assumptions: Ext^{1}(I_{Z}(e+n+1), O) ≃ H^{0}(O_{Z}). So taking 1 ∈ H^{0}(O_{Z}) we get an extension:

0 → O → F → I_{Z}(e + n + 1) → 0

It remains to show that F is locally free. This is a direct consequence of the following lemma due to Serre:

Lemma 3.1: Let A be a Noetherian local ring and I ⊂ A an ideal with a free resolution of length one:

0 → A^{q} → A^{p} →^{φ} I → 0 (∗)

Let e ∈ Ext^{1}(I, A) corresponding to the extension:

0 → A → M → I → 0 (∗∗)

Then M is a free A-module if and only if e generates the A-module Ext^{1}(I, A).

Proof. Dualizing the exact sequence (∗∗) we get:

... → Hom(A, A) →^{f} Ext^{1}(I, A) → Ext^{1}(M, A) → Ext^{1}(A, A) = 0

So Ext^{1}(M, A) = 0 if and only if f is surjcetive. Since f(1) = e, Ext^{1}(M, A) = 0 if and only if e generates Ext^{1}(I, A). It remains to show that M is free if and only if Ext^{1}(M, A) = 0. If M is free then clearly Ext^{1}(M, A) = 0. Assume Ext^{1}(M, A) = 0. Using the resolution (∗) we get a commutative diagram:

0 → A → A ⊕ A^{p} → A^{p} → 0
||  ↓ ψ  ↓ φ
0 → A →^{α}  M  →^{β} I → 0

The map ψ is defined as follows: let \overline{φ} : A^{p} → M be a lifting of φ (i.e. β ∘ \overline{φ} = φ). Such a lifting exists because, for instance, Ext^{1}(A^{p}, A) = 0. Then ψ(x, v) = α(x)+\overline{φ}(v). By the snake lemma we have coker(φ) = coker(ψ) = 0 and ker(φ) = ker(ψ). Thus we have an exact sequence:

0 → A^{q} → A ⊕ A^{p} → M → 0

Since Ext^{1}(M, A) = 0, this sequence splits and M is a direct summand of A^{p+1}. It follows that M is projective and hence (since A is local), M is a free A-module.

It follows from what has been done so far that for every section, s, of a rank two vector bundle on ℙ^{n} vanishing in codimension two, (s)_{0}, is a twocodimensional l.c.i. subscheme Z with ω_{Z} ≃ O_{Z}(c_{1} − e − n − 1); conversely given a l.c.i. subscheme, Z, of codimension two with ω_{Z} ≃ O_{Z}(e), there is (if n > 2) an exact sequence:

0 → O → E → I_{Z}(e + n + 1) → 0

This is the Serre correspondence ([73]) between codimension two l.c.i., subcanonical subschemes and rank two vector bundles. Due to its importance we rephrase this result as a theorem:

Theorem 3.2: [Serre correspondence] There is a one-to-one correspondence between

1. pairs (E, s) where E is a rank two vector bundle on ℙ^{n}, n ≥ 3, with c_{1}(E) = c_{1} and where s ∈ H^{0}(E) is a section whose zero-set has codimension two, and

2. pairs (X, ξ) where X ⊂ ℙ^{n} is a closed subcanonical subscheme of codimension two which is l.c.i. and where ξ ∈ H^{0}(ω_{X}(n + 1 − c_{1})) is a section which generates ω_{X}(n + 1 − c_{1})

Under this correspondence there is an exact sequence:

0 → O →^{s} E → I_{X}(c_{1}) → 0

Furthermore: deg(X) = c_{2}(E) and ω_{X} ≃ O_{X}(c_{1} − n − 1).

The following is some kind of Bertini’s theorem for sections of rank two vector bundles.

Proposition 3.3: Let E be a rank two vector bundle on ℙ^{n}.

1. (ch(k) arbitrary) If E(−1) is generated by global sections, then for s ∈ H^{0}(E) a sufficiently general section, the scheme of zeros (s)_{0} will be nonsingular.

2. If char(k) = 0, the same is true under the weaker hypothesis that E is generated by global sections.

Proof. 1) This follows from [51].

2) Set P := ℙ(H^{0}(E)) and define I ⊂ P × ℙ^{n} by I = {([s], x) | x ∈ (s)_{0}}. The projection induces a map f : I → ℙ^{n}. Since E is globally generated there is an exact sequence:

0 → F → H^{0}(E) ⊗ O → E → 0

where F is a vector bundle. It follows that the fibers of f are smooth, irreducible and equidimensional. Hence I is smooth. Then by the Theorem of Generic Smoothness there exists a dense open set U ⊂ P such that for every [s] ∈ U, the fiber of g : I → P (induced by projection) is smooth, but this fiber is precisely (s)_{0}.

Remark 3.4: (a) If E is a rank two vector bundle on ℙ^{n}, then for m >> 0, E(m) is generated by global sections, hence any rank two vector bundle corresponds to some smooth codimension two subvariety.

(b) Even if smooth the zero locus (s)_{0} doesn’t need to be connected (if n = 3). But if h^{1}(E(−c_{1})) = 0, (s)_{0} will be connected for every s.

(c) If n ≥ 4 and if X, Y ⊂ ℙ^{n} are two codimension-two subschemes, then X ∩ Y ≠ ∅. So if (s)_{0} is smooth, it is also irreducible.

For a rank two vector bundle we have: E^{∗} ≃ E(−c_{1}) (since E ≃ E^{∗}⊗∧^{2}E, see [40], Ch. II, Ex.5.16). This, together with Serre duality: H^{i}(E) ≃ H^{n−i}(E^{∗}(−n − 1))^{∗}, imposes strong conditions of symmetry on the cohomology. Moreover, if n ≥ 4, the existence of a smooth section imposes (at least if ch(k) = 0) the vanishing of part of the cohomology as shown in the next lemma.

Lemma 3.5: Let E be a rank two vector bundle on ℙ^{n}, n ≥ 4, admitting a section vanishing along a smooth codimension two subvariety. If char(k) = 0, then h^{i}(E(m)) = 0, for 2 ≤ i ≤ n and m ≥ −n. In particular for m ≥ −n: χ(E(m)) = h^{0}(E(m)) − h^{1}(E(m)).

Proof. We have an exact sequence:

0 → O → E → I_{X}(c_{1}) → 0 (∗)

where ω_{X} ≃ O_{X}(c_{1} − n − 1). We can write O_{X}(m) ≃ ω_{X}(m − c_{1} + n + 1). By Kodaira’s theorem, if m ≥ c_{1} − n, then h^{i}(O_{X}(m)) = 0 for 1 ≤ i ≤ n − 2. Using the exact sequence:

0 → I_{X}(m) → O(m) → O_{X}(m) → 0

this implies: h^{j}(I_{X}(m)) = 0 for 2 ≤ j ≤ n − 1 and m ≥ c_{1} − n. Now using (∗), we get that: h^{j}(E(t)) = 0 if 2 ≤ j ≤ n − 1 and t ≥ −n. Finally: h^{n}(E(l)) = h^{0}(E^{∗}(−l − n − 1)) by Serre duality. Since E has rank two: E^{∗} ≃ E(−c_{1}), so h^{n}(E(l)) = h^{0}(E(−c_{1} − l − n − 1)). Using (∗) again we get: h^{n}(E(l)) = 0 if l ≥ −n − 1.

The next Lemma shows the connection between Hartshorne conjecture and the existence of indecomposable rank two vector bundles on ℙ^{n}.

Lemma 3.6: Let E be a rank two vector bundle on ℙ^{n}, n ≥ 3. Assume E has a section vanishing along a codimension two subscheme, X ⊂ ℙ^{n}. Then E is a direct sum of line bundles if and only if X is a complete intersection.

Proof. If E ≃ O(a) ⊕ O(b), then a section is given by sections s_{a} ∈ H^{0}(O(a)), s_{b} ∈ H^{0}(O(b)) and X is the complete intersection of the two divisors (s_{a})_{0}, (s_{b})_{0}. If X is a complete intersection of type (a, b), then ω_{X} ≃ O_{X}(a + b − n − 1) and there is an exact sequence:

0 → O → O(a) ⊕ O(b) → I_{X}(a + b) → 0

Since H^{0}(O_{X}) ≃ k, the rank two vector bundle associated to X by Theorem 3.2 is uniquely determined, hence E ≃ O(a) ⊕ O(b).

Remark 3.7: Lemma 3.6 (as Theorem 3.2) is false if n = 2. For the correspondence between rank two vector bundles and codimension two subschemes on ℙ^{2} (or more generally on a surface) extra conditions are needed.

Gathering everything:

Proposition 3.8: On ℙ^{n}, n ≥ 3, the following are equivalent:

1. There exists a smooth codimension two subvariety X ⊂ ℙ^{n}, with ω_{X} ≃ O_{X}(e), which is not a complete intersection

2. There exists an indecomposable rank two vector bundle on ℙ^{n}.

Proof. 1) ⇒ 2). Follows from Theorem 3.2 and Lemma 3.6.

2) ⇒ 1). Follows from Proposition 3.3, Theorem 3.2 and Lemma 3.6.

By Theorem 1.1 every smooth codimension two subvariety X ⊂ ℙ^{n}, n ≥ 6, is subcanonical. This implies:

Proposition 3.9: Hartshorne conjecture in codimension two is equivalent to: every rank two vector on ℙ^{n}, n ≥ 7, is a dircet sum of line bundles.

This result has been one of the main motivation for a great activity in the study of rank two vector bundles on projective space. On ℙ^{3} there are plenty of (indecomposable) bundles and they have been studied also with the hope that a good understanding of the situation for n = 3 would have been of some help on ℙ^{n}, n > 3. But in spite of many efforts, in the last thirty years very few indecomposable rank two vector bundles on ℙ^{n}, n > 3, are known. More precisely:

1. If char(k) = 0, no indecomposable rank two vector bundle is known if n > 4. On ℙ^{4} there is, basically, only one known indecomposable rank two vector bundle: the Horrocks-Mumford bundle (see [49]).

2. If char(k) > 0, no indecomposable rank two vector bundle is known if n > 5. There are some examples of indecomposable rank two vector bundles on ℙ^{5} ([53] and references therein).

Due to this state of affairs one can also propose the following modified versions of Hartshorne conjecture in codimension two:

Conjecture 3: 1. Assume char(k) = 0. Every rank two vector bundle on ℙ^{5} splits. This is equivalent to require that every smooth, subcanonical, threefold X ⊂ ℙ^{5} is a complete intersection.

2. (Make your choice for char(k).) Every rank two vector bundle on ℙ^{6} splits. This is equivalent to require that every smooth, codimension two subvariety, X ⊂ ℙ^{6}, is a complete intersection.

Of course 1) implies 2) and 2) implies Hartshorne conjecture in codimension two.

Due to the correspondence with rank two vector bundle, the codimension two case of Hartshorne conjecture is particularly attractive (it should also be the easiest case of the conjecture). However it could be wrong to focus too much on this particular case; indeed it could be, as it often happens in mathematics, that it is easier or at least more natural, to prove the conjecture in its full generality. In spite of this wise advice,

from now on, we will only consider codimension two subvarieties!

4. Stable rank two vector bundles.

In this section we review, very quickly, some basic facts on stable bundles, for more details see [64], [44]. The notion of stability has been introcuded by Mumford in connection with moduli problems. For vector bundles there are different notions of stability, we will adopt the most common one (Takemoto-Mumford stability); in its simplest form the definition goes as follows:

Definition 4.1: Let E be a rank two vector bundle on ℙ^{n}. Then E is stable (resp. semi-stable) if for any non-zero morphism: O(a) → E, we have: a < \frac{c_{1}(E)}{2} (resp. a ≤ \frac{c_{1}(E)}{2}).

This can be rephrased as follows: assume E normalized (stability doesn’t depends on the twist), then: if c_{1}(E) = −1:

E is stable ⇔ E is semi-stable ⇔ h^{0}(E) = 0

if c_{1}(E) = 0:

E is stable ⇔ h^{0}(E) = 0

and:

E is semi-stable ⇔ h^{0}(E(−1)) = 0

In terms of associated subvarieties, if we have an exact sequence: 0 → O → E → I_{X}(e + n + 1) → 0 and if s := min{k | h^{0}(I_{X}(k)) ≠ 0}, then E is stable (resp. semi-stable) ⇔ 2s > e + n + 1 (resp. 2s ≥ e + n + 1). Here of course ω_{X} ≃ O_{X}(e). A stable bundle is simple (End(E) = k) and indecomposable. If a (properly) semi-stable bundle is decomposable then it is O(m) ⊕ O(m); but there exists (if n = 3) indecomposable, properly semi-stable bundles.

The discriminant of E, Δ(E) is defined by: Δ(E) := c_{1}^{2} − 4c_{2}, Δ(E) is invariant by twist.

A powerful method to study vector bundles on ℙ^{n} is to consider their restrictions to linear subvarieties, in particular to lines and to hyperplanes. Indeed by a well known theorem (sometimes attributed to Grothendieck) every rank r vector bundle on ℙ^{1} is a direct sum of line bundles. So if E is a rank two vector bundle on ℙ^{n} and if L ⊂ ℙ^{n} is a line, then: E|L ≃ O_{L}(a_{L}) ⊕ O_{L}(b_{L}). One says that (a_{L}, b_{L}) is the splitting type of E on L. We have a_{L}+b_{L} = c_{1}(E) and, in general, the splitting type varies with L. (If the splitting type is constant, the bundle is said to be uniform; it can be shown that every rank two uniform vector bundle splits if n ≥ 3.) By semi continuity there exists a non empty open subset U ⊂ G(1, n) such that the splitting type is constant on U; if L corresponds to a point of U, then (a_{L}, b_{L}) is the generic splitting type. If L ∉ U, then L is said to be a jumping line of E. A basic result is:

Theorem 4.2: [Grauert-Mülich] Let E be a semi-stable rank two vector bundle on ℙ^{n}. If c_{1} = 0, the generic splitting type of E is (0, 0); if c_{1} = −1, it is (−1, 0).

One easily deduces that the restriction of a semi-stable bundle with c_{1} = 0 to a general hyperplane is still semi-stable. In fact there is a more precise result.

Theorem 4.3: 1. (Maruyama) Let E be a semi-stable rank two vector bundle on ℙ^{n}, n ≥ 3. If H ⊂ ℙ^{n} is a general hyperplane, then E|H is still semi-stable

2. (Barth) Let E be a stable rank two vector bundle on ℙ^{n}, n ≥ 3. If H is a general hyperplane, then E|H is still stable except if n = 3, c_{1} = 0 and c_{2} = 1 (null-correlation bundles).

Now using a result of Schwarzenberger on rank two vector bundles on ℙ^{2}, we get:

Corollary 4.4: Let E be a rank two vector bundle on ℙ^{n}, n ≥ 2. If E is stable (resp. semi-stable), then Δ(E) < 0 (resp. Δ(E) ≤ 0).

5. Notes and complements.

Exercice 5.1: 1. Prove that the image of the Segre embedding ℙ^{1} ×ℙ^{2} →

ℙ^{5} has degree three.

2. Prove Lemma 2.1

3. Prove that the image of the Plücker embedding G(1, 4) → ℙ^{9} has dimension 6 and degree 5 (If you are in trouble, go to [39], p. 246,247).

Exercice 5.2: Prove that a stable rank two vector bundle on ℙ^{n} is indecomposable.

Exercice 5.3: 1. Let Y ⊂ ℙ^{3} be the union of two skew lines. Show that there is an exact sequence:

0 → O → E(1) → I_{Y} (2) → 0

where E is a stable rank two vector bundle with c_{1} = 0 and c_{2} = 1 (such a bundle is called a null-correlation bundle).

2. Show that if H ⊂ ℙ^{3} is a general plane, then E|H is properly semistable.

3. Prove that there is no stable rank two vector bundle on ℙ^{2} with c_{1} = 0 and c_{2} = 1. For this use Riemann-Roch theorem which says that if E is a rank two bundle on ℙ^{2} with c_{1}(E) = 0, then:

χ(E(k)) = (k + 1)(k + 2) − c_{2}

4. Every stable bundle with c_{1} = 0, c_{2} = 1 is a null-correlation bundle. By Maruyama’s theorem (Theorem 4.3), E|H is semi-stable for H general. Use this fact to show that h^{1}(E(−5)) = 0. Then use Riemann-Roch on ℙ^{3}:

χ(E(k)) = −c_{2}(k+2)+\frac{(k + 1)(k + 2)(k + 3)}{3} for a rank two bundle with c_{1} = 0

to conclude that E(1) has a section vanishing along two skew lines.

Exercice 5.4: A null correlation is a linear isomorphism f : ℙ^{3} → ℙ_{3}^{∗} such that x ∈ f(x) =: H_{x}, ∀x. We also say that x =: x_{H} is the null point of H = f(x).

1) Show that f induces an involution φ : G(1, 3) → G(1, 3). Set L^{∗} := φ(L). Show that if L^{∗} ∩ L ≠ ∅, then L = L^{∗}.

(It can be shown that the lines such that φ(L) = L form a general linear complex, S, in G (i.e a general hyperplane section of G in its Plücker embedding.))

2) Under suitable coordinates, f is given by the alternating matrix

A = ( \begin{matrix}0 & −1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & −1 \\ 0 & 0 & 1 & 0 \end{matrix} )

So the point p = (p_{0} : p_{1} : p_{2} : p_{3}) is sent to the hyperplane H_{p} of equation: −p_{1}X_{0} +p_{0}X_{1} −p_{3}X_{2} +p_{2}X_{3} = 0. Show that there is a commutative diagram:

0 to O(−1) → 4.O → T_{ℙ^{3}}(−1) → 0
p ↘  ↓

O(1) ↓ 0

where p is given by (x_{1}, −x_{0}, x_{3}, −x_{2}). Show that the kernel, E, of p is a rank two vector bundle. Show that E is a null correlation bundle.

[NOTE: Denote by N ⊂ ℙ^{3} × ℙ_{3}^{∗} the graph of the null correlation, then N ⊂ F where F is the incidence variety F = {(x, H) | x ∈ H} and if J denotes the ideal of N in F, then p_{∗}(J (0, 1)) is a null correlation bundle, see [9].]

3) Using Euler’s sequence show that c_{3}(Ω(2)) = 0 (twisted cotangent bundle of ℙ^{3}). Using the fact that Ω(2) is generated by global sections (could you prove it?) we conclude that a general section: O → Ω(2) doesn’t vanish (because it vanishes along a codimension three subscheme of degree c_{3} = 0). Hence the cokernel is a rank two vector bundle. Identify this rank two vector bundle.

Exercice 5.5: Consider the following result of Gruson-Peskine ([37]):

Theorem 5.1: Let C ⊂ ℙ^{3} be an integral curve of degree d. Set σ = min{k | h^{0}(I_{C∩H}(k)) ≠ 0}, where H is a general plane. If d > σ^{2} + 1, then h^{0}(I_{C}(σ)) ≠ 0.

1) Show that this result is sharp (consider sections of E(k) where E is a null-correlation bundle).

2) Use this theorem to deduce Barth’s restriction theorem (Theorem 4.3) on ℙ^{3}.

Notes. As shown in [41] the Barth-Lefschetz theorem (i.e. 1) of Theorem 1.2) can be derived from the Hard-Lefschetz theorem. It is also possible (see [72]) to derive the Barth-Lefschetz theorem from the vanishing of le Potier (see IV.Theorem 1.6). Along the same lines, see also the recent work of Badescu ([6]). Another approach to the Barth-type theorems is via Fulton-Hansen connectedness theorem ([34]); this result is also related to Theorem 2.2. The correspondance of Serre (Theorem 3.2) can be extended, under suitable assumptions, to wider classes of codimension two, closed subschemes (see for example [45]); but in higher codimension there is no such general statement. It is known that, if r ≥ n, there exist families of rank r indecomposable vector bundles on ℙ^{n}, but for r < n, little is known. An example of indecomposable rank n − 1 vector bundle on ℙ^{n}, n ≥ 3, was given by Tango ([79]). Then there are some example of rank three bundles on ℙ^{5}, ℙ^{4} and, in finite characteristic, a rank two bundle on ℙ^{5} and (many) rank two bundles on ℙ^{4}. We refer to [53] for a recent account on this topic. In characteristic zero there is only the Horrocks-Mumford bundle on ℙ^{4} (and the derived bundles, for instance pull-backs by finite coverings, [19]). The theorem of Grauert-Mülich (Theorem 4.2) has been extended to arbitrary rank (Grauert-Mülich-Spindler theorem, see [64]). Barth’s restriction theorem (Theorem 4.3) is proved in [9], (see also [24] for the rank three case and a good description of the method). All these results are proved via the standard construction, a basic technique for vector bundles. As we have seen (Exercice 5.5) the restriction theorem is related to the so-called lifting problem, cf [16] for the relation between these two points of view. For a neat, algebraic approach to the lifting problem, see [50].

CHAPTER III Multisecants and Ran’s theorem.

Ran’s result ([67]) asserts that rank two vector bundles having a section and whose Chern classes satisfy a certain numerical inequality have to split. It has many interesting consequences. The starting point of Ran’s proof is the following obvious fact: if there is a (k + 1)-secant line to X through a general point of ℙ^{n}, then h^{0}(I_{X}(k)) = 0. So if for some other reason (instability of the associated bundle) one knows that h^{0}(I_{X}(k)) ≠ 0, then there is no (k+1)- secant line to X through a general point of ℙ^{n}. If one has a formula giving the degree of Σ_{k+1}^{p} = {L ∈ G(1, n) | p ∈ L and length(L ∩ X) ≥ k + 1}, then this formula must yield deg(Σ_{k+1}^{p}) = 0. In our case this gives informations on the Chern classes of the associated bundle.

1. Multisecants.

Let X ⊂ ℙ^{n} be a closed subscheme of codimension two. We would like to estimate the dimension of k-secants to X (or k-secants through a (general) point of ℙ^{n}).

Definition 1.1: We denote by Sec^{k}(X) ⊂ G(1, n) the locus of k-secants to X: Sec^{k}(X) = {L ∈ G(1, n) | length(X ∩ L) ≥ k}. Furthermore if p ∉ X, we define Σ_{k}^{p}(X) = {L ∈ G(1, n) | p ∈ L and length(X ∩ L) ≥ k}.

Let’s denote by H^{k}(X) = {p_{1} + ... + p_{k} | p_{i} ∈ X}; it has dimension k.dim(X). Now let’s denote by Al^{k} = {p_{1}+...+p_{k} | p_{i} ∈ ℙ^{n} and < p_{1}, ..., p_{k} >= L, a line }. We have a map: p : Al^{k} → G(1, n) : p_{1} + ... + p_{k} →< p_{1}, ..., p_{k} >. If L ∈ G(1, n), then p^{−1}(L) ≃ Hilb^{k}(L) ≃ ℙ^{k}. We conclude that Al^{k} has dimension 2n + k − 2. Considering the intersection Al^{k} ∩ H^{k}(X) in H^{k}(ℙ^{n}), we may expect that this intersection has dimension: (2n+k−2)+k(n−2)−kn = 2n−k−2. So we conclude that the expected dimension of Sec^{k}(X) is 2n−k−2.

Since the lines passing through a point p ∈ ℙ^{n} form a ℙ^{n−1}, the expected dimension of Σ_{k}^{p}(X) (for p general) is: (2n−2−k)+(n−1)−(2n−2) = n−k−1. So we expect a finite number of (n − 1)-secants to X through a general point of ℙ^{n} and no k-secants if k > n − 1. Also we expect Σ_{k+1}^{p}(X) to be of codimension one in Σ_{k}^{p}(X) for k ≤ n − 1. If C ⊂ ℙ^{3} is a smooth, irreducible, non plane curve, then there are \frac{(d−1)(d−2)}{2} − g 2-secants to C through a general point (d = deg(C), g = g(C)). In particular if ω_{C} ≃ O_{C}(e), we find \frac{d(d−e−3)}{2} 2-secant lines to C through a general point. If E is the rank two vector bundle associated to C (through 0 → O → E → I_{C}(e+4) → 0) and if e(k) := c_{2}(E(−k)), a short computation shows that: \frac{d(d−e−3)}{2} = \frac{e(0).e(1)}{2}. As we will see this is a special case of a more general result.

Of course it is not difficult to give examples where the expected situation is not satisfied. For instance there is no bisecant line to a plane curve through a general point of ℙ^{3}. Also let C ⊂ Q where Q ⊂ ℙ^{3} is a smooth quadric surface. If C has bidegree (a, b), then C has ∞^{1} a-secant lines and ∞^{1} bsecant lines. In general we expect Sec^{k}(C) = ∅ if k > 4 for a curve in ℙ^{3}. In a similar way if S ⊂ ℙ^{4} is a smooth surface lying on a quadric hypersurface, then Σ_{3}^{p}(S) = ∅ if p ∈ ℙ^{4} is general, this is not against the general prediction (a finite number of trisecant lines through a general point), but of course we don’t expect that number to be always zero!

These few examples suggest that exceptional (non predicted) behaviour of secant lines should have some geometrical significance.

2. Ran’s theorem

In this section X ⊂ ℙ^{n} is a closed subscheme of codimension two, which is the zero-locus of a section of a rank two vector bundle:

0 → O → E → I_{X}(e + n + 1) → 0 (∗)

In particular: ω_{X} ≃ O_{X}(e). We define e(k) := c_{2}(E(−k)) = −k(e + n + 1) + d + k^{2}. If p ∈ ℙ^{n} \ X, then Σ_{k}^{p}(X) := {L ∈ G(1, n) | p ∈ L and length(X ∩ L) ≥ k} (we do not assume that p is a general point). The key lemma of the whole story is:

Lemma 2.1: With notations as above, if the codimension of Σ_{k+1}^{p}(X) in Σ_{k}^{p}(X) is ≥ 2, then e(k) = 0.

Proof. Clearly we have dim(Σ_{0}^{p}(X)) = n − 1 and dim(Σ_{1}^{p}(X)) = n − 2. So we may proceed by induction. So we are reduced to prove that if there exists an irreducible curve C ⊂ Σ_{k}^{p}(X) such that C ∩ Σ_{k+1}^{p}(X) = ∅ and min{length(L ∩ X) | L ∈ C} = k, then e(k) = 0. Let I ⊂ G(1, n) × ℙ^{n} denote the incidence variety: I = {(L, x) | x ∈ L}. We have natural maps (induced by projections): \underline{p} : I → ℙ^{n}, q : I → G(1, n). If C̃ → C is a normalization of C, then we have a ℙ^{1}-bundle induced by q: π : S → C̃. Also let us denote by g : S → ℙ^{n} the map induced by \underline{p}. (The image is the cone of k-secants to X passing through the point p, parametrized by C.) Put D := g^{−1}(X) (as a scheme). By assumption the fibers of D over C̃ all have length k. Hence D is flat over C̃, i.e. D is a divisor on S. Let f denote a fiber of π and let Z denote the section of π contracted by g. If H = g^{−1}(H) (H ⊂ ℙ^{n} an hyperplane), then: D ≡ kH − mf for some m. Since D.Z = 0 (p ∉ X) and H.Z = 0 (Z is contracted), it follows that m = 0 and D ≡ kH. Now let’s pull-back on S the section of the associated bundle E defining X, we get: O → g^{∗}(E). This map vanishes precisely on the divisor D, i.e. the image of g^{∗}(E^{∗}) → O is O(−D). We have an exact sequence:

0 → L^{∗} → g^{∗}(E^{∗}) → O(−D) → 0

Then L^{∗} is reflexive (because g^{∗}(E^{∗}) is locally free and O(−D) is torsion-free, see [45] Prop. 1.1), hence locally free ([45] Prop. 1.9). In conclusion, we have:

0 → O(D) → g^{∗}(E) → L → 0

where L is a line bundle on S. Computing Chern classes we get: c_{1}(L) = (c_{1} − k)H and c_{2}(g^{∗}(E)) = k(c_{1} − k)H^{2}. So d = k(c_{1} − k), i.e. e(k) = d − k(c_{1} − k) = 0.

The original statement in [67] is:

Corollary 2.2: Let X ⊂ ℙ^{n} be a codimension two closed subscheme which is the zero locus of a section of a rank two vector bundle. If e(0)e(1)...e(k) ≠ 0 with k ≤ n − 2, then: Σ_{k+1}^{p}(X) ≠ ∅.

Proof. Observe that dim(Σ_{0}^{p}(X)) = n − 1 and apply Lemma 2.1.

Corollary 2.3: With notations as above, assume e(0)...e(k) ≠ 0 and k ≤ n − 2, then: h^{0}(I_{X}(k)) = 0.

Proof. Let p ∈ ℙ^{n} be a general point. By Lemma 2.1, Σ_{i+1}^{p}(X) has codimension at most one in Σ_{i}^{p}(X) for i ≤ k. Since dim(Σ_{0}^{p}(X)) = n − 1, dim(Σ_{k+1}^{p}(X)) ≥ n − 2 − k. It follows that there exists a (k + 1)-secant line to X through p, hence h^{0}(I_{X}(k)) = 0.

Theorem 2.4: Let E be a rank two vector bundle on ℙ^{n}, n ≥ 3, with Chern classes c_{1}, c_{2}. Assume E has a section vanishing in codimension two.

(i) If c_{1} ≥ \frac{c_{2}}{α} + α, for some α, 0 < α ≤ n − 2, or (ii) if c_{2} ≤ n − 2 then E splits.

Proof. By assumption we have an exact sequence:

0 → O → E → I_{X}(c_{1}) → 0

where X has codimension two.

(i) Since e(α) = c_{2} + α(α − c_{1}), the assumption implies e(α) ≤ 0. Now, e(0) = c_{2} = d ≥ 0 and it follows that the equation e(t) = t^{2} − c_{1}t + c_{2} = 0 has a real root, hence its discriminant Δ = c_{1}^{2} − 4c_{2} ≥ 0. This implies, by Maruyama’s theorem ( II.Theorem 4.3) that E is not stable. So if r := max\{m | h^{0}(E(−m)) ≠ 0\}, then c_{1}(E(−r)) = c_{1} − 2r ≤ 0, i.e. r ≥ \frac{c_{1}}{2} (∗). Observe also that c_{1} ≥ \frac{c_{2}}{α} +α implies c_{1} > 0, hence r > 0. A section of E(−r) vanishes in codimension two:

0 → O → E(−r) → I_{Z}(c_{1} − 2r) → 0

and d(Z) = c_{2}(E(−r)) = e(r) ≥ 0 (∗∗) Also the exact sequence:

0 → O(−r) → E(−r) → I_{X}(c_{1} − r) → 0

shows h^{0}(I_{X}(c_{1} − r)) ≠ 0 (+). Now we have: e(r) = e(c_{1} − r) ≥ 0 (∗∗) and r ≥ \frac{c_{1}}{2} (∗), this implies, since e(α) ≤ 0 and r > 0, that: c_{1}−r ≤ α ≤ n−2. By (+) and Corollary 2.3, ∃i, i ≤ c_{1} − r, such that e(i) = 0. Since e(t) is stricly decreasing for t ≤ \frac{c_{1}}{2} , hence for t ≤ c_{1} − r, the only possibility is i = c_{1} − r. So e(c_{1} − r) = e(r) = d(Z) = 0. It follows that Z = ∅ and E splits. (ii) Assume now c_{2} = d(X) =: d ≤ n − 2. Any (d + 1)-secant to X is contained in X, so Σ_{d+1}^{p} = ∅. By Corollary 2.2, since d ≤ n − 2, there exists i, 0 < i ≤ n − 2 such that e(i) = c_{2} + i(i − c_{1}) = 0. We conclude with (i) since the assumption is satisfied with α = i.

A first important consequence of this theorem:

Corollary 2.5: Let Z ⊂ ℙ^{n}, n ≥ 3, be a closed subscheme of codimension two, section of a rank two vector bundle. If Z is not a complete intersection, then d(Z) ≥ n − 1.

Proof. By assumption we have an exact sequence:

0 → O → E → I_{Z}(c_{1}) → 0

Since d(Z) = c_{2}, by Theorem 2.4 (ii), if d(Z) ≤ n − 2, then E splits and Z is a complete intersection ( II.Section 3.Lemma 3.6 ).

Remark 2.6: Notice that Corollary 2.5 implies that any smooth, codimension two subvariety which is indefinitely extendable is a complete intersection. Indeed let be given for every n a linear embedding: i_{n} : ℙ^{n} → ℙ^{n+1} and a smooth codimension two subvariety X_{n} ⊂ ℙ^{n} such that X_{n+1} ∩ i_{n}(ℙ^{n}) ≃ X_{n}. Then X_{n} is a complete intersection for every n. It just suffices to take n such that n − 2 ≥ d, where d = d(X_{n}) (which is independent of n). This ”Babylonian” problem must not be confused with the analogous problem for rank two vector bundle: for every n a rank two vector bundle, E_{n}, is given on ℙ^{n} such that E_{n+1}|i_{n}(ℙ^{n}) ≃ E_{n}. As shown in [12], one still has that E_{n} splits for every n, but the proof is quite different.

Another consequence:

Proposition 2.7: Let Z ⊂ ℙ^{n}, n ≥ 3, be a closed subscheme of codimension two, section of a rank two vector bundle. If h^{0}(I_{Z}(n − 2)) ≠ 0, then Z is a complete intersection.

Proof. The assumption implies Σ_{n−1}^{p}(Z) = ∅. By Corollary 2.2 there exists i, 1 < i ≤ n − 2 such that e(i) = 0. The Chern classes of the associated bundle, E, verify: c_{1} = \frac{c_{2}}{i} + i. By Theorem 2.4, E splits and Z is a complete intersection.

3. Notes and complements.

Exercice 3.1: Let E be a normalized rank two vector bundle on ℙ^{n}, n ≥ 3. If E is not stable we define the index of unstability, r(E) (or just r if no confusion can arise) by: r(E) := max{k | h^{0}(E(−k)) ≠ 0}. Of course: r(E) ≥ 0.

(i) Show that r(E) is the only integer t ≥ 0 such that E(−t)) has a section vanishing in codimension two. (ii) If H ⊂ ℙ^{n} is an hyperplane, show that r(E) = r(E_{H}). (iii) Deduce that an indefinitely extendable unstable rank two vector bundle is a direct sum of line bundles. Which is the problem with stable bundles?

Notes. For variations on Ran’s theorem see [47], [48]. Proposition 2.7 was first noticed by Ballico and Chiantini in [7] where they derived other interesting consequences for codimension two subvarieties, then improved in [47]. We will take this again in Chapter VII. The result of Barth and Van de Ven about indefinitely extendable rank two vector bundles has been extended to higher rank by Sato (see [17] and bibliography therein).

CHAPTER IV Some vanishing theorems.

In this chapter we will give some vanishing theorems and in particular a (partial) proof of Zak’s theorem on linear normality. Before to proceed we recall some basic facts about ample and k-ample vector bundles.

1. Ample and k-ample vector bundles.

Definition 1.1: Let X be a projective scheme. A line bundle, L, on X is said to be ample if for some m ≥ 0, L^{m} is very ample. A vector bundle, E, on X is said to be ample if O_{Q}(1) is ample (Q = ℙ(E) := P roj(E)).

Proposition 1.2: Let X be a projective scheme.

1. Let E be an ample vector bundle on X and let Y ⊂ X be a closed subscheme. Then E|Y is ample.

2. Let f : X → Y be a finite surjective morphism. A vector bundle E on Y is ample if and only if f^{∗}(E) is ample on X.

3. Let 0 → F → E → G → 0 be an exact sequence where E, F, G are vector bundles on X. If E is ample, then G is ample. If F and G are ample then E is ample. In particular if E = ⨁ E_{i}, then E is ample if and only if E_{i} is ample ∀i.

4. If E is generated by global sections and if F is ample, then E ⊗ F is ample. The same conclusion holds if E is assumed to be ample.

Proof. See [42], [43].

For examples of ample vector bundles we have:

Lemma 1.3: The tangent bundle T_{ℙ^{n}} is ample for every n ≥ 1. If X ⊂ ℙ^{n} is a smooth subvariety then N_{X} is ample.

Proof. By the Euler sequence T_{ℙ^{n}} is a quotient of (n+1).O_{ℙ^{n}}(1) which is ample (by 3. of Proposition 1.2). We conclude again by 3. of Proposition 1.2.

Always by Euler’s sequence we have (n + 1).O_{X} → N_{X}(−1) → 0 and we conclude with 4. of Proposition 1.2.

We recall Mori’s theorem (conjectured by Hartshorne): if X is a smooth projective variety with T_{X} ample, then X ≃ ℙ^{n} for some n.

As we have seen in the above proof, if X ⊂ ℙ^{n} is a smooth subvariety, then N_{X}(−1) is generated by global sections and N_{X} is ample. We may wonder if N_{X}(−1) is ample. This is not true in general.

Lemma 1.4: Let X ⊂ ℙ^{n} be a smooth subvariety. The following are equivalent:

1. N_{X}(−1) is not ample

2. there exists an integral curve C ⊂ X and an hyperplane H ⊂ ℙ^{n} such that H ∩ X is singular along C.

Proof. Combining the Euler sequence and the exact sequence: 0 → T_{X}(−1) → T_{ℙ^{n}}(−1)|X → N_{X}(−1) → 0, we get a surjection: V^{∗} ⊗ O_{X} → N_{X}(−1) → 0 (V = H^{0}(O_{ℙ^{n}}(1))), which yields: ℙ(N_{X}(−1)) ⊂ ℙ_{n}^{∗} × X. In this inclusion ℙ(N_{X}(−1)) is identified with the conormal variety C_{X} = {(x, H) ∈ X × ℙ_{n}^{∗} | T_{x}X ⊂ H}. The projection induces f : C_{X} → ℙ_{n}^{∗} and the image is X^{∗} the dual variety of X. We have f^{∗}(O_{X^{∗}}(1)) = O_{Q}(1) (Q = ℙ(N_{X}(−1))). Now f is finite if and only if for every H, f^{−1}(h) = Sing(X ∩ H) is finite. Moreover if f is finite then O_{Q}(1) is ample by 2. of Proposition 1.2. This proves 1. ⇒ 2. Conversely if C is contained in a fibre of f then D.C = 0 for D ∈ |O_{Q}(1)| and this shows (1. of Proposition 1.2) that O_{Q}(1) is not ample.

Remark 1.5: 1. Observe that if X ⊂ ℙ^{n} is linearly normal, 2. of Lemma 1.4

is equivalent to: h^{0}(O_{X}(H − 2C)) ≠ 0.

2. Let X be a variety of dimension ≥ 2 and let L be an ample line bundle on X with h^{0}(L) ≠ 0. Consider a very ample multiple of L: L^{r} (r ≥ 2). Let φ_{L^{r}}(X) = Y ⊂ ℙ(H^{0}(L^{r}), then N_{Y} (−1) is not ample.

3. For examples of smooth surfaces S ⊂ ℙ^{4} with N_{S}(−1) not ample, see [13].

Ampleness is one of the many notions of ”positivity” and as such yields vanishing theorems in cohomology. One of the most useful vanishing is:

1. AMPLE AND k-AMPLE VECTOR BUNDLES. 31

Theorem 1.6: [le Potier vanishing] Let E be an ample vector bundle of rank r on the smooth variety X. Then h^{i}(∧^{j}E ⊗ ω_{X}) = 0 if i + j ≥ r + 1.

We will need the following ([23]):

Proposition 1.7: Let E be a rank r ample vector bundle on the smooth variety X. Let O_{X}(1) be a very ample line bundle on X. Then: h^{i}(∧^{j}E⊗ω_{X}(t)) = 0 for i + j ≥ r + 1 and t ≥ 0.

Proof. Observe that if i = 0, then j ≥ r + 1 and ∧^{j}E = 0, so we have only to care to the cases i > 0. Let H be a general divisor in |O_{X}(1)|, then we have:

0 → ∧^{j}E ⊗ ω_{X}(t) → ∧^{j}E ⊗ ω_{X}(t + 1) → ∧^{j}E ⊗ ω_{H}(t) → 0

(by adjunction ω_{X}|H ≃ ω_{H}(−1)). Assume dim(X) = 1. We have h^{i}(∧^{j}E ⊗ ω_{H}(t)) = 0, ∀t if i > 0, for dimesional reasons. Now make induction on t (the case t = 0 follows from Theorem 1.6). This proves the result if dim(X) = 1. Now make induction on dim(X) and on t.

The notion of k-ampleness due to Sommese ([77]) extends ampleness:

Definition 1.8: A vector bundle, E, on the smooth variety X is said to be k-ample if there exists a morphism f : X → Y with fibers of dimension ≤ k and an ample vector bundle, F, on Y such that f^{∗}(F) = E.

Stated otherwise: a line bundle L on X is k-ample if L^{⊗m} is globally generated for some m ≥ 0 and if the corresponding map to projective space has fibers of dimension ≤ k. A vector bundle E is k-ample if O_{Q}(1) is k-ample on Q = ℙ(E). The interest of this notion is due to:

Theorem 1.9: Let E be a k-ample vector bundle of rank r on X. Then h^{i}(X, E^{∗}) = 0 if i ≤ dim(X) − k − r.

Proof. See [77] Prop. 1.14.

2. Linear normality.

Following [22] we are going to give a (partial) proof of Zak’s theorem on linear normality, the same method will provide some other interesting vanishings. The ingredients needed are Zak’s lemma on tangencies and bundles of principal parts.

Lemma 2.1: Let X ⊂ ℙ^{n} be a smooth non degenerate subvariety of dimension t. Let L be a k-dimensional linear subspace (k ≥ t). Then:

dim(Sing(L ∩ X)) ≤ k − t

.

This is Zak’s lemma on tangencies which is a key point in the proof of II.Theorem 2.2.

Lemma 2.2: Let X ⊂ ℙ^{n} be a smooth subvariety of dimension t. Then N_{X}(−1) is (n − t − 1)-ample. In particular h^{i}(N_{X}^{∗}(1)) = 0, 0 ≤ i ≤ 3t − 2n + 1.

Proof. Consider again the morphism p : C_{X} = ℙ(N_{X}(−1)) → X^{∗} ⊂ ℙ_{n}^{∗} (see proof of Lemma 1.4). If h ∈ X^{∗}, then p^{−1}(h) ≃ Sing(H ∩ X). By Lemma 2.1, dim(p^{−1}(h) ≤ n − 1 − t. This proves that N_{X}(−1) is n − 1 − tample (observe that p^{∗}(O_{ℙ_{n}^{∗}}(1)) = O_{Q}(1), where Q = ℙ(N_{X}(−1))). We conclude with Theorem 1.9

Before to go on we need to recall some facts on bundles of principal parts. Let X be a smooth variety and denote by Δ ⊂ X × X the diagonal. Let J ⊂ O_{X×X} denote the ideal sheaf of Δ. Moreover let Δ_{k+1} denote the k-th infinitesimal neighbourhood of Δ in X ×X (Δ_{k+1} is defined by J^{k+1}). Finally denote by p, q the projections X × X → X.

If E is a vector bundle on X, set: P^{k}(E) := q_{∗}(p^{∗}(E)|Δ_{k+1}). Recall that J /J^{2} ≃ Ω_{X}^{1} and S^{a}(Ω_{X}^{1}) ≃ J^{a}/J^{a+1}, a ≥ 1. In particular from the exact sequence:

0 → J^{k+1}/J^{k+2} → O_{X×X}/J^{k+2} → O_{X×X}/J^{k+1} → 0

by twisting by p^{∗}(E) and applying q_{∗}, we get:

0 → S^{k+1}(Ω_{X}^{1}) ⊗ E → P^{k+1}(E) → P^{k}(E) → 0

Observe that R^{i}(q_{∗}(p^{∗}(E) ⊗ J^{k+1}/J^{k+2})) = 0 by Grauert’s theorem ([40] III.Cor. 12.9). For k = 0 we get:

0 → Ω_{X}^{1} ⊗ E → P^{1}(E) → E → 0

which shows that P^{1}(E) is a vector bundle. Then, by induction, we see that P^{k}(E) is a vector bundle for every k ≥ 1. Finally ∀k ≥ 1 there is a surjection P^{k}(E) → E → 0. We will use the following fact:

(Fact +) The surjection P^{1}(E) → E → 0 induces, a surjective map H^{0}(P^{1}(E)) → H^{0}(E) → 0. Indeed we have H^{0}(E) ≃ H^{0}(p^{∗}(E)) and:

H^{0}(E) ≃ H^{0}(p^{∗}(E)) →^{g} H^{0}(p^{∗}E|Δ_{2}) →^{f} H^{0}(p^{∗}E|Δ) ≃ H^{0}(E)

since f ∘ g is an isomorphism, f is surjective. Applying q_{∗} we get the desired surjection.

We are ready to prove:

Theorem 2.3: Let X ⊂ ℙ^{n} be a smooth subvariety of dimension t. If t ≥ \frac{2n}{3}, then h^{1}(I_{X}(1)) = 0, i.e. X is linearly normal in ℙ^{n}.

Proof. We have a comutative diagram:

0  0
↓  ↓
N_{X}^{∗}(1) =  N_{X}^{∗}(1)
↓  ↓
0 → Ω_{ℙ^{n}}^{1}|X → V ⊗ O_{X} → O_{X}(1) → 0
↓ ↓ ||
0 → Ω_{X}^{1}(1) → P^{1}(O_{X}(1)) → O_{X}(1) → 0
↓  ↓
0  0

By Lemma 2.2, h^{1}(N_{X}^{∗}(1)) = 0. This implies V → H^{0}(P^{1}(O_{X}(1))) is surjective. Since H^{0}(P^{1}(O_{X}(1))) → H^{0}(O_{X}(1)) is surjective (Fact (+) above), it follows that V = H^{0}(O_{ℙ^{n}}(1)) → H^{0}(O_{X}(1)) is surjective, hence h^{1}(I_{X}(1)) = 0.

Remark 2.4: Zak’s theorem states that if dim(X) > \frac{2(n−1)}{3}, then X ⊂ ℙ^{n} is linearly normal. In particular if X ⊂ ℙ^{n} has codimension two, then X is linearly normal as soon as n ≥ 5. The result above shows linear normality in the codimension two case only for n ≥ 6. If V ⊂ ℙ^{5} is a smooth threefold such that N_{V} (−1) is ample, then h^{2}(N_{V} (−1)⊗ ω_{V}) = 0 by Theorem 1.6. By duality: h^{1}(N_{V}^{∗} (1)) = 0 and the above proof applies. For another cohomological proof (using the formal neighborhood of X) of Theorem 2.3 see [65].

3. Projective normality and Hartshorne conjecture.

Let’s recall the following:

Definition 3.1: A smooth subvariety X ⊂ ℙ^{n} is said to be projectively normal if h^{1}(I_{X}(m)) = 0, ∀m ∈ ℤ. Instead X is said to be k-normal if h^{1}(I_{X}(m)) = 0 for m ≤ k.

In our context the importance of the notion of projective normality is due to the following result of Evans-Griffith:

Theorem 3.2: Let E be a rank r vector bundle on ℙ^{n}. Set H_{∗}^{i}(E) = ⨁_{m∈ℤ} H^{i}(E(m)).

If H_{∗}^{i}(E) = 0 for 1 ≤ i ≤ r − 1, then E splits as a direct sum of line bundles.

In fact this theorem is one of the many consequences of a deep result in commutative algebra (”syzygies theorem”) but it can be proved with standard techniques of algebraic geometry. We begin with a simple lemma ([23])

Lemma 3.3: Let E be a rank r vector bundle on the reduced irreducible variety X. Assume that E is generated by its sections.

1. If h^{0}(E^{∗}) ≠ 0, then E = O_{X} ⊕ F for some rank r − 1 vector bundle F on X.

2. If X = ℙ^{n} and H^{n}(E(−n − 1)) ≠ 0, then E = O_{ℙ^{n}} ⊕ F.

Proof. First we observe that 2. is a direct consequence of 1. Indeed: h^{n}(E(−n − 1)) = h^{0}(E^{∗}) (Serre duality). So H^{0}(E^{∗}) ≠ 0 and, by 1., E = O_{X} ⊕ F. For 1., a section, s, of E^{∗} yields a morphism: E →^{s^{∗}} O. Let I_{Y} := Im(s^{∗}) so that E → I_{Y} → 0. Since E is generated by global sections, the same occurs for I_{Y}, it follows that Y = ∅ and we have an exact sequence 0 → F → E →^{s^{∗}} O → 0. Since E is globally generated we can find a section O →^{t} E such that s^{∗} ∘ t ≠ 0:

O
↓ t ↘ s^{∗} ∘ t
E →^{s^{∗}} O → 0

Now g = s^{∗} ∘ t is an isomorphism and s^{∗} ∘ (t ∘ g^{−1}) = 1, so the sequence splits and E ≃ F ⊕ O.

Before to proceed to the proof of Theorem 3.2 we recall some facts about regularity: a coherent sheaf, F, on ℙ^{n} is said to be m-regular if h^{i}(F(m−i)) = 0 for i > 0. If F is m-regular then F(m) is generated by global sections, the natural map H^{0}(F(m)) ⊗ H^{0}(O(t)) → H^{0}(F(m + t)) is surjective and F is (m + t)-regular for t ≥ 0 (see [62], p.100).

Proof of Theorem 3.2. By twisting the bundle we may assume that E is (−1)-regular but not (−2)-regular. Since E(−1) is generated by global sections, E is ample (4. of Proposition 1.2). By Proposition 1.7 h^{j}(E⊗O(−n− 1) ⊗ O(t)) = 0 if t ≥ 0. It turns out (t = n − 1 − j) that: h^{j}(E(−2 − j)) = 0 if r ≤ j ≤ n − 1. By assumption h^{j}(E(−2 − j)) = 0 if 1 ≤ j < r. Since E is not (−2)-regular it follows that h^{n}(E(−2 − n)) ≠ 0. Since E(−1) is generated by global sections, we conclude (by Lemma 3.3) that: E(−1) = O ⊕ F where F is a rank r − 1 vector bundle with H_{∗}^{i}(F) = 0 if 1 ≤ i ≤ r. Going on this way we get that E is a direct sum of line bundles.

We have the following generalization of the classical Gherardelli’s theorem:

Corollary 3.4: Let X ⊂ ℙ^{n}, n ≥ 3, be a smooth variety of codimension two, with ω_{X} ≃ O_{X}(e). The following are equivalent:

1. X is projectively normal

2. X is a complete intersection.

Proof. By Serre’s correspondance (see Theorem ??) a nonzero section of ω_{X}(−e) ≃ O_{X} yields an exact sequence:

0 → O → E → I_{X}(e + n + 1) → 0

where E is a rank two vector bundle. We have H_{∗}^{1}(E) = H_{∗}^{1}(I_{X}) and we conclude by Theorem 3.2.

Since by Barth’s theorem any smooth codimension two subvariety of ℙ^{n}, n ≥ 6, is subcanonical we may rephrase this corollary as:

Corollary 3.5: A smooth codimension two subvariety of ℙ^{n}, n ≥ 6, is a complete intersection if and only if it is projectively normal.

So one possible way to approach Hartshorne conjecture in codimension two is to prove: h^{1}(I_{X}(m)) = 0, ∀m > 0, i.e. that X is k-normal for every k (notice that Zak’s theorem corresponds to 1-normality). This motivates the search of vanishing theorems for I_{X}. For lack of time and space we won’t develop here this topic (see the Notes for some informations).

4. Notes and complements.

Exercice 4.1: Lemma 2.1 has some striking consequences. Let X ⊂ ℙ^{n} be a smooth projective variety of dimension r.

1) Show that the Gauss map: g : X → G(r, n) : x → T_{x}X is finite.

2) Let Z be an arbitrary hyperplane section of X. Prove that Z is nonsingular in codimension 2r − n − 1.

3) The dual variety X^{∗} ⊂ ℙ_{n}^{∗} is defined as the projection of the conormal variety C_{X} ⊂ ℙ_{n}^{∗} × X (see proof of Lemma 1.4). Show that: dim(X^{∗}) ≥ r.

Exercice 4.2: Horrocks theorem states that a vector bundle E on ℙ^{n} is a direct sum of lines bundles if and only if the whole intermediate cohomology vanishes (H_{∗}^{i}(E) = 0 for 1 ≤ i < n). See [64] for a proof. Use this result to prove the classical statement of Gherardelli:

Theorem: Let C ⊂ ℙ^{3} be a smooth, irreducible curve. The following are equivalent:

1) C is a complete intersection.

2) C is subcanonical and projectively normal.

Notes. Arguing along the lines of Theorem 2.3, Ein obtained ([22]) some higher vanishing, for instance: h^{1}(I_{X}(2)) = 0 if n ≥ 8. Similar results have been obtained in [65] by essentially the same techniques except that the ”princiapl parts bundles” argument is replaced by the study of the formal neighborhood. We should also mention [68] where quadratic normality is proved (by a different method) for n ≥ 16. Another very interesting approach is due to Alzati and Ottaviani: using a result of Mather they first show ([2]):

Theorem: Let X ⊂ ℙ^{n} be a smooth subvariety of codimension d. For p ∈ ℙ^{n} let X^{2} := {x ∈ X | p ∈ T_{x}X}. Then when X^{i} is smooth define X^{i+1} in a similar way. Then for p general each X^{i} is smooth and (when non empty) of codimension d in X^{i−1}.

In the codimension two case dim(X^{i}) = n − 2i and X^{i} is the smooth zero locus of a suitable section of N_{X_{i−1}|X_{i−2}} (look at the sections coming from the natural map H^{0}(O_{ℙ^{n}}(1)) → H^{0}(N_{X}(−1))). Then applying the results of Ein ([22]) to the X^{i} they get various vanishings; for instance: X ⊂ ℙ^{n} of codimension two is t-normal if n ≥ 6t − 2 (t ≥ 2) and h^{r}(O_{X}(t)) = 0 if r ≥ 1 and n ≥ 6t + r (t ≥ 1) ([3]). See also [1] and references therein.

38 IV. SOME VANISHING THEOREMS.

CHAPTER V Surfaces in ℙ^{4}: the theorem of Ellingsrud and Peskine.

1. The normal bundle: here enters the hero.

In this chapter we will prove Ellingsrud-Peskine theorem stating that the degree of a smooth surface, not of general type, in ℙ^{4} cannot be arbitrarily large. So ”special” (i.e. of non general type, for instance rational) surfaces in ℙ^{4} are rare: this supports the philosophy that, non complete intersections, codimension two subvariety of ℙ^{n} get more and more difficult to find as n increases.

Actually there is another reason to look at this result. As we will see, the key point of the proof is a lemma giving a bound on c_{2}(N_{S}^{∗}(s)) when S lies on a hypersurface of degree s. In the next chapters we will consider generalizations of this lemma which will allow us to bound the degree of a smooth codimension two subvariety lying on a hypersurface of degree s by a function of s.

As we already saw in IV and it will become the leitmotiv of the next chapters, the normal bundle of X ⊂ ℙ^{n} carries a tremendous amount of informations (for instance in the codimension two case if X is associated to a rank two vector bundle, E, then N_{X} ≃ E|X). So, before to plunge into the proof of Ellingsrud-Peskine theorem, we take the opportunity to make a little general digression on the normal bundle.

In this section X will denote a smooth, non degenerated, codimension two subvariety of ℙ^{n}, n ≥ 4.

The normal bundle N_{X} of X in ℙ^{n} is defined as usual by the exact sequence:

0 → T_{X} → T_{ℙ^{n}}|X → N_{X} → 0 (∗)

40 V. SURFACES IN ℙ^{4}: THE THEOREM OF ELLINGSRUD AND PESKINE.

It is also important to consider the dual, the conormal bundle, N_{X}^{∗}, defined by the exact sequence:

0 → I_{X}^{2} → I_{X} → N_{X}^{∗} → 0 (∗∗)

It turns out that: N_{X}^{∗} ≃ I_{X}/I_{X}^{2} ≃ I_{X} ⊗ O_{X}.

Lemma 1.1: 1. det(N_{X}) ≃ ω_{X}(n + 1), N_{X}^{∗} ⊗ ω_{X}(n + 1) ≃ N_{X} 2. N_{X}(m) is generated by global sections for m ≥ −1

Proof. (1) To compute the determinant just use (∗). The isomorphism N_{X}^{∗} ⊗ ω_{X}(n + 1) ≃ N_{X} follows from the fact that N_{X} has rank two (= codim(X)).

(2) This follows from (∗) taking into account that, from the Euler sequence, T_{ℙ^{n}}(m) and hence T_{ℙ^{n}}(m)|X, is globally generated for m ≥ −1.

As we will see later (2) reflects a strong positivity property of ℙ^{n}. Another important fact is that for n ≥ 4 (hence dim(X) ≥ 2), N_{X} has a second Chern class (the normal bundle of a curve in ℙ^{3} has no c_{2}), concerning this second Chern class we have:

Lemma 1.2: [”formule clef”] Let i : X → ℙ^{n} denote the inclusion. Then i_{∗}(c_{2}(N_{X})) = X · X

Proof. See [40], Appendix A.

This formula generalizes the self-intersection formula for a curve on a surface. We can now give a quick proof of the double points formula for smooth surfaces in ℙ^{4} (see [40] Appendix A, Example 4.1.3).

Proposition 1.3: [Double points formula] Let S ⊂ ℙ^{4} be a smooth surface, then:

d(d − 5) − 10(π − 1) + 12χ = 2K^{2}

here d is the degree of S, π is the sectional genus, χ = 1 − q + p_{g} (q = h^{1}(O_{S}) the irregularity, p_{g} = h^{0}(ω_{S}) the geometric genus) and K^{2} is the self-intersection of a canonical divisor.

Proof. In the Chow ring of ℙ^{4} S is equivalent to d times a hyperplane. It follows that X ·X = d^{2}. By Lemma 1.2, the degree of c_{2}(N_{X}) = d^{2}. Now we compute this degree using the exact sequence (∗). From the Euler sequence:

0 → O → 5.O(1) → T_{ℙ^{n}} → 0, we get that the Chern polynomial of T_{ℙ^{n}} is given by: C_{t}(T_{ℙ^{n}}) = (1 + ht)^{5}, where h is the class of a hyperplane. It follows that c_{1}(T_{ℙ^{n}}) = 5h, c_{2}(T_{ℙ^{n}}) = 10h^{2}. Using (∗) we get: (1−Kt+c_{2}t^{2})(1+c_{1}(N_{X})t+c_{2}(N_{X})t^{2}) = 1+5Ht+10H^{2}t^{2} (here c_{2} := c_{2}(T_{S})). It follows that:

c_{1}(N_{X}) = 5H + K, c_{2}(N_{X}) = 10H^{2} − c_{2} + 5HK + K^{2}

Since 1 + p_{a} = \frac{1}{12}(K^{2} + c_{2}), we have that c_{2} = 12χ − K^{2}. Of course H^{2} = d and by the adjunction formula: HK = 2π − 2 − d (by abuse of notations we sistematically identify a codimension two cycle with its degree). Putting everything together we get the desired formula.

This formula is similar to the formula which gives the genus of a smooth plane curve and imposes very strong restrictions. For example assume that the generic hyperplane section of S is a (smooth) rational curve. Then S is rational hence q = p_{g} = 0 and K^{2} ≤ 9, since π = 0 by assumption, we get: d(d−5)+22 ≤ 18, so d ≤ 4. It follows that a smooth rational curve C ⊂ ℙ^{3}, of degree d > 4, cannot be realized as a hyperplane section of a smooth surface S ⊂ ℙ^{4}. Of course, there are many more interesting applications of the double points formula.

There is another way to use the second Chern class of N_{X}: assume X ⊂ Σ ⊂ ℙ^{n}, where Σ is a degree s hypersurface. The inclusion X ⊂ Σ corresponds to 0 → I_{Σ} ≃ O(−s) → I_{X}, restricting to X (i.e. applying − ⊗ O_{X}), we get O_{X}(−s) → N_{X}^{∗}. If X is not contained in the singular locus of Σ (which will be the case for example if s is the minimal degree of a hypersurface containing X), then we get a non-zero section of N_{X}^{∗}(s): O_{X} →^{σ} N_{X}^{∗}(s). This section vanishes precisely on the intersection of X with the singular locus of Σ (recall: N_{X}^{∗} ≃ I_{X}/I_{X}^{2}); more precisally: (σ)_{0} = X ∩ Jac(Σ), where Jac(Σ) is the Jacobian scheme of Σ (defined by the partials of an equation of Σ) and where the intersection is to be taken in the scheme-theoretic way. As for any section of a rank two vector bundle, the zero-locus (σ)_{0} can contain a divisor or it can be of pure codimension two (maybe empty). In the latter case the degree of (σ)_{0} is given by (the degree of) c_{2}(N_{X}^{∗}(s)) = c_{2}(N_{X}(−s)) (see [40], Appendix A). For instance we have:

Proposition 1.4: [Jacobi formula] Let S ⊂ ℙ^{4} be a smooth surface lying on a degree s hypersurface Σ. Assume dim(S ∩ Sing(Σ)) = 0. Then:

π = 1 + \frac{d(d + s^{2} − 4s)}{2s} − \frac{μ}{2s}

where μ denotes the degree of S ∩ Jac(Σ).

Proof. As above let O_{S} →^{σ} N_{X}^{∗}(s) denote the section corresponding to the inclusion S ⊂ Σ. Since (σ)_{0} = S ∩ Jac(Σ), our assumption implies that σ vanishes in codimension two. It follows that μ = c_{2}(N_{X}^{∗}(s)) = c_{2}(N_{X}(−s)). Now if E is a rank two vector bundle and if L is a line bundle, then c_{2}(E⊗L) = c_{1}(E)c_{1}(L)+c_{2}(E)+c_{1}(L)^{2}. It turns out that: μ = (5H+K)(−sH)+d^{2}+s^{2}H^{2} and the desired formula follows.

We observe that 1+ \frac{d(d+s^{2}−4s)}{2s} = 1+ \frac{d}{2}(\frac{d}{s} +s−4) is the genus of a curve in ℙ^{3} which is the complete intersection of a surface of degree s and a surface of degree d/s. Now if an integral curve C ⊂ ℙ^{3} has the degree and the genus of a complete intersection of type (a, b) and lies on an integral surface of degree a, then C is a complete intersection of type (a, b).

Proposition 1.5: Let C ⊂ ℙ^{3} be an integral curve of degree d. Assume that C lies on an integral surface of degree s and that p_{a}(C) = 1 + \frac{d}{2}(\frac{d}{s} + s − 4), then C is a complete intersection of type (s, d/s).

This proposition will be a consequence of a general result of Gruson-Peskine ([36], [37]) about the genus of space curves. To state the theorem let’s denote by G(d, s) the maximal (arithmetic) genus of integral curves of degree d not lying on a surface of degree < s:

G(d, s) := max{g(C)|∃C ⊂ ℙ^{3} of degree d with h^{0}(I_{C}(s − 1)) = 0}.

We will say that an integral curve, C, realizes G(d, s), if it has degree d, genus G(d, s) and satisfies h^{0}(I_{C}(s − 1)) = 0.

Theorem 1.6: 1. If d > s(s − 1) then:

G(d, s) = 1 + \frac{d(d + s^{2} − 4s) − r(s − 1)(s − r)}{2s} (∗)

where d + r ≡ 0 (mod s), 0 ≤ r < s. Furthermore C realizes G(d, s) if and only if C is linked to a plane curve of degree r by a complete intersection of surfaces of degree s and (d + r)/s.

2. If s(s − 1) ≥ d > s^{2} − 2s + 2 then:

G(d, s) = s^{3} − 5s^{2} + 9s − 6 + ν(ν + 2s − 3)/2

where ν = d − (s^{2} − 2s + 3). Moreover if C realizes G(d, s), then C is arithmetically Cohen-Macaulay.

This theorem, especially 1), will play an important role in the sequel.

Proof of Proposition 1.5. 1) Assume first d ≥ s(s − 1). If h^{0}(I_{C}(s − 1)) ≠ 0, then C is contained in a complete intersection (s − 1, s), it follows that d = s(s − 1) and that C is a complete intersection of type (s − 1, s). So we may assume h^{0}(I_{C}(s−1)) = 0. If d > s(s−1), the conclusion follows from 1) of Theorem 1.6. By 2) of Theorem 1.6, the case d = s(s − 1) cannot occur.

2) Assume d < s(s − 1). Let t be the smallest integer ≥ d/s. We have t < s (because d/s < s − 1) and d > t(t − 1) (otherwise d/s < d/t ≤ t − 1). Finally we also have h^{0}(I_{C}(t−1)) = 0 (otherwise d ≤ s(t−1)). By 1) of Theorem 1.6, p_{a}(C) ≤ 1 + \frac{d}{2}(\frac{d}{t} + t − 4). Since \frac{d}{t} + t ≤ \frac{d}{s} + s (because ts ≥ d), it follows that p_{a}(C) = 1+ \frac{d}{2}(\frac{d}{t} +t−4) and t = d/s. We conclude with 1) of Theorem 1.6.

By combining the previous results we can give a quick proof of some particular (although quite general!) cases of the Severi-Lefschetz-Grothendieck theorem ( I.Theorem 1.5). But before to proceed we would like to point out the following facts which will be used in the sequel without any further mention:

Lemma 1.7: (i) If X ⊂ ℙ^{n}, n ≥ 4, is smooth of codimension two, then X is irreducible. (ii) let X as above, then X is a complete intersection if and only if its general hyperplane section is a complete intersection.

Theorem 1.8: Let X ⊂ ℙ^{n}, n ≥ 4, be a smooth codimension two subvariety. Assume that X is a Cartier divisor on the integral hypersurface Σ. Then X is the complete intersection of Σ and another hypersurface.

Proof. We see, by taking a section with a general 4-dimensional linear subspace, that it is enough to prove the theorem for n = 4. Since X is a Cartier divisor on Σ, X ∩ Sing(Σ) = ∅. Indeed if X is locally defined on Σ by the equation f = 0, and if g = 0 is an equation of Σ, then X is locally (in ℙ^{n}) defined by (f, g); X being smooth, the hypersurface g = 0 can’t have singularities along X. By Proposition 1.4, a general hyperplane section, C, of X has the genus of a complete intersection (s, d/s). Since C lies on an integral surface of degree s, by Proposition 1.5, C is a complete intersection (s, d/s). This implies that X also is a complete intersection (s, d/s) and the conclusion follows.

Theorem 1.9: Let Σ ⊂ ℙ^{n}, n ≥ 4, be a smooth hypersurface. Then P ic(Σ) ≃ ℤ.H where H is an hyperplane section.

Proof. The theorem amounts to show that every Cartier divisor on Σ is the complete intersection of Σ with another hypersurface. It is enough to prove the theorem for n = 4. Let X ⊂ Σ be a Cartier divisor. If X is smooth, the conclusion follows from Theorem 1.8. Assume X not necessarily smooth. We claim that I_{X,Σ}(d), where d = deg(X), is globally generated. Assume this claim for a while. Then, by Bertini’s theorem, a general element, Y, in the linear system |dH − X| is a smooth divisor on Σ. By the first part of the proof Y is a complete intersection of Σ and another hypersurface: Y ∼ mH. It follows that: X ∼ (d − m)H and we are done. To prove the claim we will show that X is the schematical intersection of Σ and of the cones of vertex a point x ∉ X, base X. For simplicity we will denote by C_{x} the cone of vertex x, base X and by R_{x} the residual to X in the complete intersection C_{x} ∩ Σ. We have to show that ∀z ∈ Σ, ∃x such that z ∉ R_{x}. First assume z ∉ X. If x ∉ C_{z}, then z ∉ C_{x} and, a fortiori, z ∉ R_{x}. Now assume z ∈ X. Let L be a line through z such that: (i) L is transversal to Σ at z (i.e. L ⊄ T_{z}Σ) (ii) L ∩ Supp(X) = {z}. Since Σ is smooth such an L exists. Let x be a point of L \ (L ∩ Σ), then the projection of center x, π_{x} : X → ℙ^{n−1} induces an isomorphism between X (locally at z) and π_{x}(X) (locally at π_{x}(z)) and z ∉ R_{x}.

Remark 1.10: a) Theorem 1.9 is the original statement of Severi; the proof given above is essentially Severi’s proof. By the way in the proof of the claim it is necessary to assume L ∩ Supp(X) = {z}. For example let X ⊂ ℙ^{3} be a smooth twisted cubic and let Σ be a smooth surface containing X. Let L be a bisecant to X, L ⊄ Σ. Take x ∈ L \ (L ∩ Σ), the projection π_{x}(X) is, say, a nodal cubic (L ∩ X = {z, z′}, z ≠ z′), so the cone C_{x} is singular along L. It follows that C_{x} ∩ Σ is singular at z and z′. This shows that the linked curve, R_{x}, has to pass through z and z′.

b) The claim used in the proof is proved (for divisors on a smooth surface in ℙ^{3}) in [62], Lecture 16.

c) If X ⊂ ℙ^{n}, n ≥ 3, is a smooth codimension two subvariety of degree d, then I_{X}(d) is globally generated (X is the intersection of the cones of vertex p ∉ X, base X). If X is not smooth, this is no longer true: for example if X ⊂ ℙ^{3} is a double line with p_{a}(X) ≤ −2, then I_{X}(2) is not globally generated. On the other hand the claim proves that I_{X}(d) is globally generated if X lies on a smooth hypersurface of degree s ≤ d (use 0 → O(−s) → I_{X} → I_{X,Σ} → 0).

d) There are several difficulties to extend this proof to arbitrary hypersurfaces: first of all it may be no longer true that |dH − X| is base-point free; secondly, even if the claim holds, a general element, Y ∈ |dH − X|, won’t be in general a smooth Cartier divisor. Indeed if Σ has a curve in its singular locus, then Y (which at end should be ∼ mH) will meet that curve and at a point of the intersection, Y won’t be a smooth Cartier divisor and the argument breaks down (cp [18]).

2. Preliminaries.

Now we turn to surfaces in ℙ^{4} and fix some notations: let S ⊂ ℙ^{4} be a smooth surface of degree d and let s = s(S) be the smallest degree of an hypersurface containing S. We set G = G(d, s) (see Theorem 1.6). We will use the following classical result:

Proposition 2.1: (Roth’s lemma) Let C be a general hyperplane section of S ⊂ ℙ^{4}, a smooth, non degenerated surface of degree d. If h^{0}(I_{C}(σ)) ≠ 0 and if d > σ^{2}, then h^{0}(I_{S}(σ)) ≠ 0.

Although we won’t use it let us point out a more precise result:

Theorem 2.2: Let X ⊂ ℙ^{4} be an integral, linearly normal, locally Cohen-Macaulay surface of degree d. Let C denote a general hyperplane section of X. If d > σ^{2} − σ + 2 and if h^{0}(I_{C}(σ)) ≠ 0, then h^{0}(I_{X}(σ)) ≠ 0.

Proof. See [59].

By Proposition 2.1 we conclude that, if d > (s − 1)^{2}, then: π = G − δ with δ ≥ 0.

Lemma 2.3: (Main lemma) With notations as above, let S ⊂ ℙ^{4} be a smooth surface of degree d = sk + r and sectional genus π = G−δ. Set c_{2}(N^{∗}(s)) =: μ = d(d+s^{2}−4s)−s(2π−2). Then: 0 ≤ μ ≤ (s − 1)^{2}d, or, in other words: δ ≤ \frac{ks(s−1)^{2}+r(r−1)(s−1)}{2s}.

Proof. By definition of s, S ⊂ Σ, where Σ is an integral hypersurface of degree s; moreover S is not contained in the singular locus of Σ. The inclusion S ⊂ Σ induces an inclusion: I_{Σ} = O(−s) → I_{S}, restricting to S and twisting by s, we get: O_{S} →^{σ} N_{S}^{∗}(s). The zero-locus of σ is precisely S ∩ Jac(Σ), where Jac(Σ) is the subscheme of Σ defined by J, the Jacobian ideal of Σ (generated by the partials of an equation of Σ). On the other hand: (σ)_{0} = D + Z where D is a divisor (empty if σ vanishes in codimension two) and where Z is a codimension two subscheme of S. We have that Z is the zero-locus of O_{S} → N_{S}^{∗}(sH − D); hence:

deg(Z) = c_{2}(N_{S}^{∗}(sH − D)) = −D.c_{1}(N_{S}^{∗}(s)) + μ + D^{2} (∗)

Since (σ)_{0} = S ∩ Jac(Σ), I_{Z}(−D) = J ⊗ O_{S}. Since J (s − 1) is globally generated (look at the partials of an equation of Σ), we conclude that: I_{Z}((s − 1)H − D) is globally generated, hence:

deg(Z) ≤ ((s − 1)H − D)^{2} = (s − 1)^{2}d − 2(s − 1)HD + D^{2}

Combining with (∗) and taking into account that c_{1}(N_{S}^{∗}(s)) = (2s − 5)H − K, we get:

μ ≤ (s − 1)^{2}d − (K + 3H)D

Since ω_{S}(3) is globally generated (Kodaira), we have: (K + 3H)D ≥ 0 and the result follows.

Remark 2.4: The conclusion c_{2}(N_{S}^{∗}(t)) ≤ d(d + t^{2} − 4t) − t(2π − 2) of Lemma 2.3 holds as soon as S is not contained in Sing(Σ), where Σ is an hypersurface of degree t (t not necessarily equal to s).

The following is a classical argument ”à la Castelnuovo”:

Lemma 2.5: Let S ⊂ ℙ^{4} be a smooth surface with d > s^{2}. Then, with notations as above: h^{2}(I_{S}(t)) ≤ δ(d − t) ∀ t.

Proof. Let C be the general hyperplane section of S and set a = max{c, e} where c = max{t/h^{1}(I_{C}(t)) ≠ 0}, e = max{t/h^{2}(I_{C}(t)) ≠ 0}. Consider the exact sequence:

0 → I_{S}(t) → I_{S}(t + 1) → I_{C}(t + 1) → 0

Clearly h^{2}(I_{S}(a)) = 0 and h^{2}(I_{S}(t)) ≤ h^{1}(I_{C}(t + 1)) + h^{2}(I_{S}(t + 1)) for every

t, then by descending induction: h^{2}(I_{S}(t)) ≤ ∑_{m≥t+1}^{a} h^{1}(I_{C}(m)).

Now we set Γ = C ∩ H the general plane section of C, again:

0 → I_{C}(t) → I_{C}(t + 1) → I_{Γ}(t + 1) → 0

Let ρ_{m} : H^{0}(I_{C}(m)) → H^{0}(I_{Γ}(m)) denote the natural map of restriction and set R_{m} = Coker(ρ_{m}), r_{m} = dim(R_{m}). Then we set Q_{m} = Ker(f_{m}), where f_{m} : H^{1}(I_{Γ}(m)) → H^{2}(I_{C}(m − 1)) and define q_{m} = dim(Q_{m}). We have the following two exact sequences:

0 → R_{m} → H^{1}(I_{C}(m − 1)) → H^{1}(I_{C}(m)) → Q_{m} → 0
0 → Q_{m} → H^{1}(I_{Γ}(m)) → H^{2}(I_{C}(m − 1)) → H^{2}(I_{C}(m)) → 0

From the first one we obtain: ∑_{m≥1} r_{m} = ∑_{m≥1} q_{m} and h^{1}(I_{C}(m)) ≤ ∑_{t≥m+1} r_{t}

for every m. The latter is clear by descending induction starting from: h^{1}(I_{C}(c + 1)) = 0, h^{1}(I_{C}(c)) = r_{c+1}. Then looking at the second sequence we get:

∑_{m≥1} q_{m} = ∑_{m≥1} h^{1}(I_{Γ}(m)) − h^{2}(I_{C}) = g(χ(C)) − π

here χ(C) denotes the numerical character of C. Now, since d > s^{2}, we have g(χ(C)) ≤ G (see [36]). Finally, as well known: a ≤ d (in fact there are better bounds). Combining with π = G − δ yields the result.

Lemma 2.6: Let S ⊂ ℙ^{4} be a smooth surface with d = deg(S) > s^{2}, set d = sk + r, then:

h^{2}(I_{S}(k)) ≥ B_{k} − \frac{dk(k + 1)}{2} + k(π − 1) − χ(S)

where B_{k} = h^{0}(O(k)) − h^{0}(O(k − s)) − 1.

Proof. We have χ(I_{S}(k)) = χ(O_{P^{4}}(k)) − χ(O_{S}(k)), so h^{2}(I_{S}(k)) ≥ h^{0}(O_{P^{4}}(k)) − h^{0}(I_{S}(k)) − χ(O_{S}(k)). Since S lies on an irreducible surface of degree s, and d > s^{2}, then h^{0}(I_{S}(k)) = h^{0}(O_{P^{4}}(k − s)), unless S is a complete intersection (s, k). In that case h^{0}(I_{S}(k)) = h^{0}(O_{P^{4}}(k−s))+1. By Riemann-Roch: χ(O_{S}(k)) = \frac{kH(kH−K)}{2} + χ(S) = \frac{dk(k+1)}{2} −k(π−1)+χ(S). Setting B_{k} = h^{0}(O_{P^{4}}(k))−h^{0}(O_{P^{4}}(k−s))−1 we get the desired formula.

3. Proof of the theorem.

We are ready to prove the theorem of Ellingsrud-Peskine. In fact we will prove a slightly more general result.

Theorem 3.1: There exists an integer M such that deg(S) ≤ M for every smooth surface S ⊂ ℙ^{4} with p_{g} ≤ π.

Proof. We recall the double points formula: 10(π − 1) = d(d − 5) + 12χ − 2K^{2}. First suppose S is of general type, then we can use Miyaoka’s inequality ([60]): K^{2} ≤ 9χ. This, together with the hypothesis p_{g} ≤ π, implies that: π − 1 ≥ \frac{d(d−5)−12}{16}. Now if d is sufficiently large, \frac{d(d−5)−12}{16} > \frac{d^{2}+45d}{18} ≥ G(d, 9) − 1. It follows that h^{0}(I_{C}(8)) ≠ 0. Thanks to Roth’s lemma (Proposition 2.1) we can lift this property to the surface S, since we can assume d > 64, so h^{0}(I_{S}(8)) ≠ 0. Now if S is not of general type and if d > 90, then h^{0}(I_{S}(5)) ≠ 0 (cf [32], Prop. 1). To conclude we have to show that surfaces with p_{g} ≤ π, lying on an irreducible hypersurface of degree s ≤ 8, have bounded degree. Of course we may assume that deg(S) = d > s^{2}. By Lemma 2.6:

h^{2}(I_{S}(k)) ≥ B_{k}−\frac{dk(k + 1)}{2}+k(π−1)−1−p_{g} ≥ B_{k}−\frac{dk(k + 1)}{2}+(π−1)(k−1)−2

(Use: p_{g} ≤ π). Since π = G − δ, we can write:

h^{2}(I_{S}(k)) ≥ B_{k} − \frac{dk(k + 1)}{2} +(k−1)(\frac{d(d + s^{2} − 4s) − r(s − 1)(s − r)}{2s} −δ)−2.

On the other hand, by Lemma 2.5: h^{2}(I_{S}(k)) ≤ δ(d − k), so we get:

(∗) δ(d − 1) ≥ B_{k} − \frac{dk(k + 1)}{2} + (k − 1)(\frac{d(d + s^{2} − 4s) − r(s − 1)(s − r)}{2s}) − 2

Let us look at the second part of the inequality (∗), it is a degree three polynomial, whose leading term is \frac{s}{6}k^{3}. Then a miracle occurs: thanks to Lemma 2.3, δ is bounded by a linear expression in k, so (∗) cannot hold for k >> 0 and we are done.

Remark 3.2: It is possible to compute an actual (though, certainly, not sharp at all) bound for d, in fact inequality (*) gives: M ≈ 10^{3}.

We immediately get:

Corollary 3.3: There exists an integer M such that deg(S) ≤ M for every smooth surface S ⊂ ℙ^{4} with h^{0}(ω_{S}(−1)) = 0.

Proof. From the exact sequence: 0 → ω_{S}(−1) → ω_{S} → ω_{C}(−1) → 0 if h^{0}(ω_{S}(−1)) = 0, it follows that p_{g} ≤ h^{0}(ω_{C}(−1)). Since ω_{C}(−1)) is special, Clifford’s theorem does apply and we get: h^{0}(ω_{C}(−1)) ≤ π − \frac{d}{2} and then also that p_{g} ≤ π − \frac{d}{2}. We conclude with Theorem 3.1.

Remark 3.4: Since q ≤ π and since π is bounded if so is d, the irregularity of surfaces with p_{g} ≤ π is bounded.

Finally:

Theorem 3.5: There exists an integer M such that if S ⊂ ℙ^{4} is a smooth surface with deg(S) > M, then the canonical linear system |K_{S}| induces a birational map. In particular every surface of non-general type has degree at most M.

Proof. By Corollary 3.3 if d > M, then h^{0}(ω_{S}(−1)) ≠ 0 and it follows that |K| = |K−H|+|H| is very ample outside of the base locus of |K−H|.

4. Notes and complements.

Exercice 4.1: Prove Lemma 1.7. Notice that in (ii) the assumption n ≥ 4 is necessary: if C ⊂ Q ⊂ ℙ^{3} is a smooth curve of degree 2d on a smooth quadric surface, then C ∩ H is a complete intersection (2, d) (prove it!). By the way, there is a beautiful theorem of Strano ([78]) stating that these are the only exceptions.

Exercice 4.2: Let S ⊂ ℙ^{4} be a smooth surface of degree d. Assume S is contained in a quadric hypersurface. Then: if d is even, S is a complete intersection (2, \frac{d}{2}), if d is odd, S is linked to a plane by a complete intersection (2, \frac{d+1}{2}). So in any case S is arithmetically Cohen-Macaulay.

Exercice 4.3: Let Σ ⊂ ℙ^{4} be a cubic hypersurface with isolated singularities.

a) Show that the scheme Jac(Σ) has length ≤ 16.

b) If S ⊂ Σ is a smooth surface, then:

π = \frac{d(d − 3)}{6} − \frac{μ}{6}

where μ = #(S∩Sing(Σ)). Moreover: if d ≡ 0 (mod 3), then: μ ∈ {0, 6, 12}, otherwise μ ∈ {4, 10, 16}.

c) Show that if μ = 16, then Σ is a cone.

Exercice 4.4: Let C ⊂ S ⊂ ℙ^{3}, where C is a smooth, irreducible curve of degree d and where S is a smooth surface of degree s. Set e := max{t | h^{1}(O_{C}(t)) ≠ 0}, c := max{t | h^{1}(I_{C}(t)) ≠ 0}. Show that:

c ≤ d + e(1 − s) + s^{2} − 4s

(if L^{∗} := I_{C,S}, show that L(s − 4 − e) has a section vanishing along a curve, B. Then use the fact that I_{B,S}(δ) is globally generated (δ := deg(B)).

NOTE: For s = 2, we obtain: c ≤ d − e − 4. A result of Gruson-Peskine (completed by Han) says that this inequality holds for every curve in ℙ^{3}.

Notes. For smooth surfaces on cubic hypersurfaces in ℙ^{4} with isolated singularities, see [5]. The classification of smooth surfaces on cubic hypersurfaces has been completed by Koelblen ([52]): if S lies on an irreducible cubic hypersurface Σ, then one of the following occurs:

• S is a.C.M. and is linked on Σ to an a.C.M. surface of degree ≤ 3 which is a cone if Σ is a cone and which is smooth otherwise

• S is linked on Σ to a Veronese surface

• S is linked on Σ to a quintic elliptic scroll. The proof is rather technical and difficult. It would be nice to have a simpler proof. For surfaces on hyperquartics with isolated singularities, see [27]. The statement and the proof of Theorem 3.5 are taken from [26]; the proof is a simplified version of the original proof of [32]. This approach has been generalized in []. It is conjectured that the degree of a smooth surface of non general type should be ≤ 15. In [14] it is shown that such a degree is ≤ 105; then many papers have appeared claiming for better bounds, but they all contain gaps, the situation is very confused and I don’t feel comfortable enough to give further references. For the construction of smooth surfaces of non general type see [20]. The analogous of Ellingsrud-Peskine theorem for threefold in ℙ^{5} is proved in [15] (see also [31] for a shorter proof). Since the general hyperplane section of a non general type threefold in ℙ^{5} is, in general, a surface of general type, this is not a direct consequence of Theorem 3.5. No effective bound is known nor conjectured. For higher codimensional subvarieties see [71]. Another astonishing conjecture is about the irregularity of smooth surfaces in ℙ^{4}: there exists an integer B such that if S ⊂ ℙ^{4} is a smooth surface, then q(S) ≤ B (some people believe that B = 2 should work). Little is known concerning this problem.

52 V. SURFACES IN ℙ^{4}: THE THEOREM OF ELLINGSRUD AND PESKINE.

CHAPTER VI Bounds for smooth divisors of hypersurfaces.

1. Introduction

In this chapter we will see how generalizations of the ”main lemma” of Ellingsrud- Peskine ( V.Section 2.Lemma 2.3) yield bounds (by functions of s) on the invariants (d, e) of a smooth X ⊂ ℙ^{n}, n ≥ 5, with ω_{X} ≃ O_{X}(e). We recall that the subcanonical condition, ω_{X} ≃ O_{X}(e), is automatically fullfilled if n ≥ 6 ( II.Theorem 1.2).

The first result (speciality theorem) gives such bounds (although not the better ones) and we will give two proofs of this result. The first, more geometrical, is really a direct generalization of the ”main lemma”; the second, more general, is based on a simple argument of positivity (positivity lemma). In fact the positivity lemma yields a series of n−2 inequalities on ℙ^{n} which can be used to get more and more precise bounds. These inequalities will be used also to show that smooth codimension two subvarieties lying on hypersurfaces of degree ≤ n are complete intersections.

2. The speciality theorem: geometrical proof.

Let’s start with some general facts.

2.1. Some basic facts. We will denote by X ⊂ ℙ^{n}, n ≥ 4, a smooth, codimension two, subvariety of degree d with ω_{X} ≃ O_{X}(e); as usual we set s := min{t | h^{0}(I_{X}(t)) ≠ 0}. Let Σ be an hypersurface of degree s containing X. Clearly Σ is integral and X is not contained in the singular locus of Σ. As already seen the inclusion X ⊂ Σ induces a non-zero section: O_{X} →^{σ} N_{X}^{∗}(s) such that (σ)_{0} = X ∩ Jac(Σ). Now as any section of a rank two bundle there are three possibilities for σ:

1. (σ)_{0} = ∅

2. (σ)_{0} contains a divisor of X

3. (σ)_{0} has codimension two (and is l.c.i.). In the first case by Severi-Lefschetz-Grothendieck theorem, X is a complete intersection: X = Σ ∩ Σ_{1}. This proves:

Lemma 2.1: Let X ⊂ ℙ^{n}, n ≥ 4, be a smooth codimension two subvariety. Assume X lies on the hypersurface Σ. If X is not a complete intersection, then dim(X ∩ Sing(Σ)) ≥ n − 4. Moreover if deg(Σ) = s, then n − 4 ≤ dim(Sing(Σ) ∩ X) ≤ n − 3.

The following lemma (except point 2.b) is a direct consequence of the ”main lemma”:

Lemma 2.2: 1. Let X ⊂ ℙ^{n}, n ≥ 4, be a smooth codimension two subvariety of degree d with ω_{X} ≃ O_{X}(e), then:

s(n + 1 + e) − s^{2} ≤ d ≤ s(n − 1 + e) + 1.

2. Assume moreover P ic(X) ≃ ℤ.H. Let Σ be an hypersurface of degree s containing X and suppose dim(Sing(Σ) ∩ X) = n − 3. Let T ∼ lH denote the divisorial part of Sing(Σ) ∩ X, then: a)

(s − l)(n + 1 + e) − (s − l)^{2} ≤ d ≤ (s − l)(n − 1 + e) + 1

b) (ch(k) = 0)

h^{1}(I_{X}(l)) ≠ 0.

Proof. 1) Let S denote the intersection of X with a general ℙ^{4} ⊂ ℙ^{n}. By the ”main lemma” ( V.Section 2.Lemma 2.3, see also V.Remark 2.4), 0 ≤ μ ≤ (s − 1)^{2}d (∗), where μ = d(d + s^{2} − 4s) − s(2π − 2). If C is a general hyperplane section of S, then, by adjunction, ω_{C} ≃ O_{C}(n + e − 3). It follows that: 2π − 2 = d(n + e − 3). Inserting in (∗) we get the result.

2)(a) This is just a slight variation on the first point. For the convenience of the reader we will sketch it briefly. The inclusion X ⊂ Σ induces σ : O_{X} → N_{X}^{∗}(s). The zero locus (σ)_{0} is the scheme theoretical intersection of the jacobian of Σ with X. By hypotesis, (σ)_{0} has codimension one in X and we can divide (σ)_{0} by the codimension one part getting a section σ′ ∈ H^{0}(N_{X}^{∗}(s − l)) thus (σ′)_{0} = c_{2}(N_{X}^{∗}(s − l)) in A^{2}(X). A short calculation, using c_{1}(N_{X}) = (n + 1)H + K = (n + 1 + e)H and c_{2}(N_{X}) = dH^{2} (”formule clef”), gives: c_{2}(N_{X}^{∗}(s−l)) = c_{2}(N_{X}(−s+l)) = [−(s−l)(n+1+e)+d+(s−l)^{2}].H^{2}. Taking degree: deg(σ′)_{0} = [−(s−l)(n+1+e)+d+(s−l)^{2}].H^{n−2} = [−(s−l)(n+1+e)+ d+(s−l)^{2}]d. It follows that d ≥ (s−l)(n+e+1)−(s−l)^{2}. On the other hand, if F = 0 is an equation of Σ, the jacobian is contained in V(F′_{i}) ∩ V(F′_{k}) ∩ X, for suitable indices, and the linear system cut out by the partials of F has a degree l fixed part and a degree s−l−1 moving one, hence (σ′)_{0} is contained in the intersection of two divisors of degree s−l−1 without common components. This implies that (s − l − 1)^{2}.H^{2} − [−(s − l)(n + 1 + e) + d + (s − l)^{2}].H^{2} is effective. Taking degrees again yields: d ≤ (s − l)(n − 1 + e) + 1.

(b) Suppose by contradiction that X is l-normal. Since P ic(X) ≃ Z.H, T is cut out on X by an hypersurface: T = X ∩ V(P), where P is an homogeneous polynomial. Let F = 0 be an equation of Σ and set F′_{i} = ∂F/∂x_{i}. Since T ⊂ (Sing(Σ) ∩ X), we have X ∩ V(P) ⊂ X ∩ V(F′_{i}). This implies that P divides F′_{i}, mod(I(X). So: P G_{i} − F′_{i} ∈ I(X). By minimality of s, this implies P G_{i} = F′_{i}. By Euler’s relation (ch(k) = 0), P divides F, but this is impossible, since by minimality of s again, Σ is integral.

Remark 2.3: The left-hand side inequality in 1) can be written as:

e ≤ \frac{d}{s} + s − n − 1

This is a special case of the Speciality Lemma of Gruson-Peskine ([38]) which states that if X ⊂ ℙ^{n} is an integral, codimension two subvariety, then s ≤ [P (t)]_{+} + t for every positive integer t, where P (x) = x^{2} − (e + n + 1)x + d; moreover if P (t) < 0 equality holds if and only if X is a complete intersection (s, d/s). As noticed in [38] this result is closely connected with Barth’s restriction theorem ( II.Theorem 4.3 and Corollary 4.4), see also Exercice 4.1. This has been reconsidered in [69].

By the way this lemma has some interesting consequences:

Corollary 2.4: Let X ⊂ ℙ^{n}, n ≥ 4, be a smooth, codimension two subvariety. Assume X is numerically a complete intersection (a, b), a ≤ b (i.e. d = ab, ω_{X} ≃ O_{X}(a + b − n − 1)).Then:

1. If b > a(a − 3) + 3 then X is a complete intersection.

2. If a ≤ n − 1 then X is a complete intersection.

Proof. Exercice 4.2 Corollary 2.5: Let X ⊂ ℙ^{n}, n ≥ 4, be a smooth codimension two subvariety with ω_{X} ≃ O_{X}(e). If X is not a complete intersection, then: e ≥ −n + 2.

Proof. By Lemma 2.2: d(X) = d ≤ s(n − 1 + e) + 1. If e ≤ −n + 1, then d ≤ 1 and X is a complete intersection.

2.2. The speciality theorem: weak version. We are now ready to prove a weak version of the speciality theorem. It is possible, with similar geometrical arguments, to prove the strong form (i.e. without assuming P ic(X) ≃ ℤ.H if n = 5) of the theorem (see [28]) but the proof gets more intricated. Since we will give later a simple and general proof (via the positivity lemma), we prefer for the moment, to keep things simple, to present this weak version where the geometrical arguments are transparent.

Theorem 2.6: [Speciality theorem, weak version]

Let X ⊂ ℙ^{n}, n ≥ 5 be a smooth codimension two subvariety. If n = 5 assume P ic(X) ≃ ℤ.H. Let Σ be an hypersurface of degree s containing X. If X is not a complete intersection, then:

e ≤ \frac{(s − 1)[(s − 1)^{2} − n + 1]}{n − 1} − n + 1

where ω_{X} ≃ O_{X}(e).

Of course the strength of this statement is that the bound of e doesn’t depend on d.

Notations 2.7: From now on, X ⊂ ℙ^{n}, n ≥ 5, will denote a smooth, non degenerate, codimension two subvariety which is not a complete intersection. Since P ic(X) ≃ ℤ.H (by assumption if n = 5 and by Barth’s theorem if n ≥ 6), X is subcanonical: ω_{X} ≃ O_{X}(e). By Serre’s construction we may associate to X a rank two vector bundle:

0 → O → E → I_{X}(e + n + 1) → 0

The Chern classes of E are: c_{1}(E) = e + n + 1, c_{2}(E) = d(X) =: d. Let Σ be an hypersurface of degree s := min{t | h^{0}(I_{X}(t)) ≠ 0} containing X. Then Σ gives a section of I_{X}(s) which lifts to a section σ_{Σ} ∈ H^{0}(E(−e − n − 1 + s)). Since X is not a complete intersection, Z, the zero-locus of σ_{Σ}, has codimension two and:

z := d(Z) = c_{2}(E(−e−n−1+s)) = d−s(e+n+1−s), ω_{Z} ≃ O_{Z}(−e−2n−2+2s).

The following lemma is very useful:

Lemma 2.8: With notations as above, the scheme X ∩ Z is the base locus of the jacobian system of Σ on X: X ∩ Z = X ∩ Jac(Σ) as schemes. So, the fundamental cycle ([33] 1.5) of Z in A_{∗}(X) is c_{2}(N_{X}^{∗}(s)) as soon as X and Z intersect in the expected codimension.

Proof. As already said, the inclusion X ⊂ Σ yields a non-zero section: O_{X} →^{t} N_{X}^{∗}(s) such that (t)_{0} = X ∩ Jac(Σ). Now observe that E(−e − n − 1) |_{X}≃ N_{X}^{∗} and consider the section: O →^{σ} E(−e−n−1+s) whose zero-locus is Z. Since σ, by construction, is obtained from a lifting of Σ, σ|X ∈ H^{0}(N_{X}^{∗}(s)) is nothing but t. On the other hand it is clear that (σ|X)_{0} = X ∩Z. It follows that: X ∩ Z = X ∩ Jac(Σ).

For the proof of Theorem 2.6 we will first assume that X ∩Z has the right dimension.

Lemma 2.9: With notations as in 2.7 assume dim(X ∩ Z) = n − 4. Then Y := X ∩ Z is a subcanonical, l.c.i. scheme with ω_{Y} ≃ O_{Y} (2s − n − 1). Moreover Y is the base locus of the jacobian system of Σ in X.

Proof. We are assuming that Y is the proper intersection of X and Z hence we have:

0 → O_{X} → E |_{X} (−e − n − 1 + s) → I_{Y |X}(−e − n − 1 + 2s) → 0

so N_{Y,X}^{∗} ≃ E |_{X} (−s) and the first statement follows by adjunction. For the last statement use Lemma 2.8.

Notations 2.10: Keep the assumptions of Lemma Lemma 2.9 and denote by Σ_{1} and Σ_{2} two general partials of Σ. Since dim(X ∩ Z) = n − 4, C := X ∩ Σ_{1} ∩ Σ_{2} is a subcanonical, l.c.i. scheme containing Y such that N_{C,X} ≃ O_{C}(s − 1) ⊕ O_{C}(s − 1). We have ω_{C} ≃ O_{C}(e + 2s − 2). The scheme C is a complete intersection in X which links Y to another subscheme. By abuse of notations, for any A ∈ A_{i}(X) ⊂ A_{∗}(X) we denote by d(A) ∈ ℤ the degree of A: d(A) := d(A ∩ H^{i}), A ∩ H^{i} ∈ A_{0}(ℙ^{n}) ≃ ℤ.

For any closed subscheme Γ ⊂ X we still denote by Γ ∈ A_{∗}(X) the fundamental cycle of Γ ([33] 1.5).

Lemma 2.11: With notations as in 2.10, denote by R the residual to Y with respect to C. Then C = Y ∪ R is a geometric linkage and Δ := R ∩ Y is a Cartier divisor of Y such that: I_{Δ|Y} ≃ O_{Y} (−e − n + 1). Furthermore: d(Δ) ≤ (s − 1)d((s − 1)^{2} − z) and: z(e + n − 1) ≤ (s − 1)[(s − 1)^{2} − z].

Proof. Denote by Y_{red} the support of Y and set Y_{red} = Y_{1}∪⋯∪Y_{r} where Y_{i}, 1 ≤ i ≤ r, are the irreducible components of Y_{red}. Furthermore, denote by P_{i} the general point of Y_{i}. Since Y is l.c.i. in X and since I_{Y |X}(s − 1) is globally generated by the partials of Σ, we can find two general elements in Jac(Σ) generating the fibers of N_{Y |X}^{∗}(s − 1) at each P_{i}, 1 ≤ i ≤ r. This implies that R ∪ Y is a geometric linkage.

Now consider the local Noether sequence (exact sequence of liaison):

0 → I_{C} → I_{R} → ω_{Y} ⊗ ω_{C}^{−1} → 0.

we get

ω_{Y} ⊗ ω_{C}^{−1} ≃ \frac{I_{R}}{I_{C}} ≃ \frac{I_{R} + I_{Y}}{I_{C} + I_{Y}} ≃ \frac{I_{Δ}}{I_{y}} ≃ I_{Δ|Y}

(the second isomorphism follow by geometric linkage, since I_{R} ∩ I_{Y} = I_{C}) hence ω_{Y} ⊗ ω_{C}^{−1} ≃ O_{Y} (−e − n + 1) ≃ I_{Δ|Y} and we are done. For the last statement, the scheme Δ ⊂ R is the base locus of the jacobian system of Σ in R, hence Δ ⊂ Σ̃ ∩ R with Σ̃ a general element of Jac(Σ) and d(Δ) ≤ d(R) · (s − 1). We conclude since d(R) · (s − 1) = (d(C) − z) · (s − 1) = ((s − 1)^{2}d − zd) · (s − 1). The last inequality follows from d(Δ) = d(Y) · (e + n − 1) = d · z · (e + n − 1).

Now we can conclude the proof of Theorem 2.6.

Proof of Theorem 2.6. It is enough to prove the theorem for s minimal. Let Σ be an hypersurface of minimal degree containing X, we set s := d(Σ) and d := d(X). According to Lemma 2.2 we distinguish two cases.

1) dim(X ∩ Sing(Σ)) = n − 3. In this case, by Lemma 2.2, we have d ≤ (s − 2)(n − 1 + e) + 1. On the other hand z = d − s(e + n + 1 − s) (see 2.7). It follows that: z ≤ (s−1)^{2} −2(n−1+e). Since z ≥ n−1 ( III.Corollary 2.5), we get: \frac{(s−1)^{2}−n+2}{2}−n+1 ≥ e. One checks (using s ≥ n−1, cf III.Proposition 2.7) that this implies the bound of Theorem 2.6.

2) dim(X ∩ Sing(Σ)) = n − 4. By the last inequality of Lemma 2.11, e ≤ (s − 1)[\frac{(s−1)^{2}}{z} − 1] − n + 1. Since z ≥ n − 1, we get the result.

As a consequence we obtain:

Theorem 2.12: Let Σ ⊂ ℙ^{n}, n ≥ 5, be an integral hypersurface of degree s. Let X ⊂ Σ be a smooth subvariety with dim(X) = n − 2. If n = 5 assume P ic(X) ≃ ℤ.H. If X is not a complete intersection, then

d(X) < \frac{s(s − 1)[(s − 1)^{2} − n + 1]}{n − 1} + 1.

Proof. Just combine Theorem 2.6 with the inequality d ≤ s(n−1+e)+1 of Lemma 2.2.

3. Positivity lemma and the speciality theorem.

The most general form of the positivity lemma is:

Lemma 3.1: [positivity lemma] Let F be a rank two vector bundle on a smooth connected variety X of dimension m and let L be an invertible sheaf such that h^{0}(E ⊗ L) ≠ 0. Put

\frac{1 + c_{1}(L) t}{1 − c_{1}(F) t + c_{2}(F) t^{2}} = ∑ u_{i}t^{i}

in A_{∗}(X)[[t]], where A_{∗}(X) is the Chow ring of X and t is an indeterminate. Assume F is globally generated. Then the u_{i}’s can be represented by pseudoeffective cycles (see [55], 2.2.B), in particular u_{m} has non-negative degree.

Proof. Set Q := P (F) (in Grothendieck notation Proj(SymF)) and denote by p : Q → X the projection. The Chow ring of Q is

\frac{A_{∗}(X)[x]}{(x^{2} − c_{1}(F) x + c_{2}(F))}

(where the indeterminate x corresponds to the tautological quotient of p^{∗}(F)) and the Gysin map p_{∗} : A_{∗}(Q) → A_{∗}(X) sends α + β x to β. By hypothesis, there is an effective divisor D of first Chern class x + c_{1}(L). Since F is globally generated x is nef and D.x^{i} is pseudo-effective. Then p_{∗}(D.x^{i}) = p_{∗}(x^{i+1} + c_{1}(L)x^{i}) = u_{i} (by the formula giving the Gysin map), so u_{i} is pseudo-effective.

We will apply this lemma to the following situation: X is a subvariety of codimension two of ℙ^{n}, n ≥ 4, with ω_{X} ≃ O_{X}(e) and F = N_{X}(−1). One knows that F is globally generated because it is a quotient of T_{ℙ^{n}}(−1), which is globally generated on ℙ^{n} (Euler sequence). By the subcanonical assumption (automatically fullfilled if n ≥ 6), we have an exact sequence:

0 → O → E → I_{X}(e + n + 1) → 0

where E is a rank 2 vector bundle on ℙ^{n} with Chern classes c_{1}(E) = e+n+1, c_{2}(E) = deg(X) =: d, moreover: N_{X} = E ⊗ O_{X}.

Let s = min{t : h^{0}(I_{X}(t) ≠ 0} and q = min(s, e + n) and notice that h^{0}(N_{X}^{∗}(q)) > 0. Apply the positivity lemma with L = O_{X}(q−e−n). Then the u_{i} can be computed in A_{∗}(ℙ^{n}) = \frac{ℤ[t]}{t^{n+1}} (by abuse of notation we consider u_{i} as an integer) and the positivity lemma applied to X ∩ ℙ^{i+2} says that u_{i} ≥ 0 for i ≤ n − 2. Let s_{i} be the Segre classes of E(−1). One has u_{i} = c_{1}(L).s_{i−1} + s_{i}. If s ≥ e + n one has L ≃ O, u_{i} = s_{i}; this case is not new ([47]), so we focus on the other case (q = s).

Lemma 3.2: The u_{i} are computed by induction on i by u_{0} = 1, u_{1} = s − 1, u_{i} = (e + n − 1)u_{i−1} + (d − e − n)u_{i−2}. Set z := d − s(e + n + 1) + s^{2}, then the first u_{i}’s are:

u_{2} = (s − 1)^{2} − z
u_{3} = (s − 1)^{3} − z(e + n + s − 2) and:
u_{4} = u_{2}^{2} − z(e + n − 1)^{2}.

Proof. Left to the reader.

As an immediate consequence we have:

Theorem 3.3: [Speciality theorem] Let X ⊂ ℙ^{n}, n ≥ 5, be a smooth subvariety of codimension two with ω_{X} ≃ O_{X}(e). Let Σ ⊂ ℙ^{n} denote an hypersurface of degree s containing X. If X is not a complete intersection then:

(i) If n = 5 : e ≤ \frac{(s − 1)^{3}}{4} − 3 − s and: d ≤ \frac{s(s − 1)[(s − 1)^{2} − 4]}{4} + 1

(ii) If n ≥ 6 : e ≤ \frac{(s − 1)^{2} − n + 1}{\sqrt{n − 1}} − n + 1 and: d ≤ \frac{s[(s − 1)^{2} − n + 1]}{\sqrt{n − 1}} + 1.

Proof. (i) By u_{3} ≥ 0: (s − 1)^{3} ≥ z(e + n + s − 2). Observe that, since X is not a complete intersection, z = c_{2}(E(−e − n − 1 + s)) is the degree of a codimension two subscheme which is not a complete intersection. By III.Corollary 2.5, z ≥ n − 1. It follows that: (s − 1)^{3} ≥ (n − 1)(e + n + s − 2) which gives the bound on e. By u_{2} ≥ 0: d ≤ s(n − 1 + e) + 1 and this gives the bound on d. (ii) The proof is similar using u_{4} ≥ 0 instead of u_{3} ≥ 0.

Remark 3.4: • The condition u_{2} ≥ 0 (which gives us: d ≤ s(n − 1 + e) + 1) is nothing but the ”main lemma” of Ellingsrud-Peskine (cp. Lemma 2.2), whereas the condition: u_{3} ≥ 0 is equivalent to z(e + n − 1) ≤ (s − 1)[(s − 1)^{2} − z], the second inequality of Lemma 2.11. So we may say that the ”geometrical” proof of the Speciality Theorem is worked out on ℙ^{4}, ℙ^{5}; the Positivity Lemma allows us to work also on ℙ^{6} (u_{4} ≥ 0) and so to get a better bound if n ≥ 6. In the next chapter we will try to use the full strengh of the Positivity Lemma.

• We stress the fact that Lemma 3.1 and hence Theorem 3.3 are valid over a field of arbitrary characteristic.

The speciality theorem implies that the family of smooth, non Cartier, divisors on a hypersurface Σ ⊂ ℙ^{n}, n ≥ 6, is limited.

Theorem 3.3 also says that on any fixed hypersurface of ℙ^{n}, n ≥ 5, Hartshorne’s conjecture is true, except maybe for finitely many degrees. We can also rephrase it as stating that a smooth codimension two subvariety of ℙ^{n}, n ≥ 5 (if n = 5 we must assume X subcanonical) which is not a complete intersection cannot lie on a hypersurface of too low degree (too low with respect to its own degree). Morally speaking one could think that this implies that ”very” unstable bundles have to split, but this is just ”moral suasion”. Indeed given a rank two vector bundle E on ℙ^{n}, one, in general, is unable to estimate the first twist of E having a smooth section. So there are no new results on the vector bundle side.

As we have seen Theorem 3.3 is a direct consequence of Lemma 3.1 which in turn reflects the positivity of the tangent bundle of ℙ^{n}: the basic fact is that if X ⊂ ℙ^{n} is smooth, then N_{X}(−1) being a quotient of T_{ℙ^{n}}(−1), is globally generated. If X is no longer smooth, then N_{X}(−1) is no longer a quotient of T_{ℙ^{n}}(−1) and N_{X}(−1) is not necessarily generated by global sections. So we see that the smoothness assumption cannot be dropped.

The speciality theorem uses only the few first inequalities of Lemma 3.1, what happens if n > 6? Do we get stronger and stronger inequalities, do these inequalities prove the conjecture for n > N_{0}? That’s what we are going to investigate in the next chapter.

4. Notes and complements.

Exercice 4.1: Let X ⊂ ℙ^{n} of codimension two, with ω_{X} ≃ O_{X}(e) and let E be the associated bundle:

0 → O → E → I_{X}(e + n + 1) → 0

As usual let s := min{k | h^{0}(I_{X}(k)) ≠ 0}. Assume X is not a complete intersection.

(i) Show that E(−e − n − 1 + s) has a section vanishing in codimension two. Deduce that: e + n + 1 < \frac{d}{s} + s (compute c_{2}(E(−s))). (ii) If E is not stable show that s ≤ a where a is the smallest root of P (x) = x^{2} − (e + n + 1)x + d (use II.Corollary 4.4).

Exercice 4.2: A codimension two subvariety X ⊂ ℙ^{n} is said to be a numerically complete intersection (n.c.i.) of type (a, b), a ≤ b, if d = ab, ω_{X} ≃ O_{X}(a + b − n − 1) (so the associated bundle is numerically split, i.e. it has the same Chern classes as O(a) ⊕ O(b)).

(i) Show that if s ≥ a, then X is a complete intersection (use V.Theorem 1.6; as usual s denotes the minimal degree of an hypersurface containing X). (ii) Prove Corollary 2.4.

Notes. The Speciality Theorem was first proved in [29]; the improved version and the Positivity Lemma are proved in [31].

CHAPTER VII Numerology.

1. A general expression for the u_{k}.

Notations 1.1: Let’s summarize the situation and take some notations: X ⊂ ℙ^{n} is a smooth, codimension two subvariety of degree d with ω_{X} ≃ O_{X}(e). We will assume that X is not a complete intersection. Associated to X there is a rank two vector bundle:

0 → O → E → I_{X}(e + n + 1) → 0

the Chern classes of E are c_{1} = e + n + 1, c_{2} = d; so the Chern classes, C_{i}, of E(−1) are given by: C_{1} = e + n − 1, C_{2} = d − e − n. The Segre classes, s_{k}, of E(−1) are defined, for k ≤ n − 2, by:

\frac{1}{1 − C_{1}X + C_{2}X^{2}} = ∑_{k≥0} s_{k}X^{k} (∗)

Since E|X ≃ N_{X}, we know ([47]) that s_{k} ≥ 0. The classes u_{k} are defined, for k ≤ n − 2, by:

u_{k} = (s − e − n)s_{k−1} + s_{k}

By the Positivity Lemma we have: u_{k} ≥ 0 for k ≤ n − 2. As already observed, this follows from (∗) if s ≥ e + n. We recall that E is stable if and only if: 2s > e + n + 1; so the Positivity Lemma gives new informations also for some stable bundles. By the results of Maruyama and Barth, if E is stable (resp. semi-stable), then Δ < 0 (resp. Δ ≤ 0), where Δ := c_{1}^{2} − 4c_{2}.

To obtain the sharpest results one should distinguish two cases: (I) s ≤ e + n, (II) s > e + n. In the first case the u_{k} give better bounds than the s_{k}, whereas in case (II) it is the contrary. Of course inequalities deriving from the u_{k} (resp. the s_{k}) are valid in any case (independently of s, e).

There are several ways to derive a general expression for the s_{k}, u_{k} (see Exercice 5.1), here we choose a uniform approach. With notations as above we first observe that C_{1} = e + n − 1 and C_{2} = d − e − n are positive ( for C_{1} this follows from VI. Corollary 2.5, for C_{2} from the fact that N_{X}(−1) is globally generated). We set: e(X) = X^{2} − C_{1}X + C_{2}.

• If Δ := C_{1}^{2}−4C_{2} > 0, then C_{1} > 2ρ, where ρ := \sqrt{C_{2}}. Since C_{1}/2ρ > 1, there exists a unique t ∈ ℝ such that: ch t = C_{1}/2ρ and we have:

1 − C_{1}X + C_{2}X^{2} = 1 − 2ρch tX + ρ^{2}X^{2}

Here ch x and sh x are the hyperbolic trigonometric functions defined by:

ch x = \frac{e^{x} + e^{−x}}{2} , sh x = \frac{e^{x} − e^{−x}}{2}

• If Δ < 0, then C_{1}/2ρ < 1 and there exists a unique θ, 0 < θ < π such that: cos θ = C_{1}/2ρ. If we set: t := iθ, then: ch t = cos θ, sh t = isin θ and we may still write:

1 − C_{1}X + C_{2}X^{2} = 1 − 2ρch tX + ρ^{2}X^{2}

In any case (i.e. if Δ ≠ 0), the two roots, a, b, of e(X) = 0 are: a = ρe^{−t}, b = ρe^{t}. The case Δ = 0 is considered as a limit case.

Proposition 1.2: With notations as above:

1. If Δ ≠ 0, then:

s_{k} = ρ^{k} · \frac{sh (k + 1)t}{sh t}

and

u_{k} = ρ^{k}[\frac{s − 1}{ρ} \frac{sh kt}{sh t} − \frac{sh (k − 1)t}{sh t}]

2. If Δ = 0, then:

s_{k} = ρ^{k}(k + 1)

and

u_{k} = ρ^{k}[k \frac{s − 1}{ρ} − (k − 1)]

Proof. Assume Δ ≠ 0. Since 1 − C_{1}X + C_{2}X^{2} = (1 − aX)(1 − bX) = (1 − ρe^{−t})(1 − ρe^{t}), after a partial decomposition, we get: s_{k} = ρ^{k} · \frac{sh (k+1)t}{sh t} (to be replaced by ρ^{k}(k + 1) if Δ = 0). Since u_{k} = (s − e − n)s_{k−1} + s_{k}, we get the expression for u_{k} (and replacing sh x by x the corresponding expression in case Δ = 0).

Remark 1.3: We observe that the invariant s doesn’t appear in s_{k}.

1. A GENERAL EXPRESSION FOR THE u_{k}. 65

Now we introduce a new invariant:

Definition 1.4: Let Δ > 0 (resp. Δ < 0), the function f(x) = \frac{sh (x+1)t}{sh xt} (resp. g(x) = \frac{sin(x+1)θ}{sinxθ}) is strictly decreasing on ]0, +∞[ (resp. ]0, \frac{π}{θ} − 1[). Moreover, if Δ > 0, lim_{x→+∞}f(x) = e^{t}. Since Δ > 0, E is not stable and 2s < e+n+1, since 0 < z = e(s−1), we have s−1 < a (see also VI.refESpec), hence s − 1 < a = ρe^{−t}, e^{t} < \frac{ρ}{s−1}. We conclude that there exists an unique α such that f(α) = ρ/(s − 1) (resp. such that g(α) = ρ/(s − 1)). Notice that in the case Δ < 0, (α + 1)θ < π. Also the condition s_{n−2} ≥ 0, implies sin (n − 1)θ ≥ 0, i.e. θ ≤ \frac{π}{n−1}. Similarly considering the function v(x) = (x − 1)/x we define, in case Δ = 0, α such that: \frac{α}{α+1} = \frac{s−1}{ρ}. In any case, by the Positivity Lemma: α ≥ n − 3.

Remark 1.5: Let’s stress the following point: when Δ > 0, the assumption that X is not a complete intersection implies that z > 0. This in turn implies s−1 < a and leads to the existence of α. If X is a complete intersection (p, q), p ≤ q, then z = 0 and the inequality s−1 < a doesn’t hold (s−1 = a = p−1).

The following lemma (”la panoplie du parfait potache”) translates the inequalities u_{k} ≥ 0 into equalities involving the invariants d, s, e, n (and also α, t,...):

Lemma 1.6: With notations as above we have: if Δ > 0:

\frac{σ}{sh t} = \frac{s − 1}{sh αt} = \frac{ρ}{sh (α + 1)t} = \frac{e + n − s}{sh (α + 2)t}

if Δ < 0:

\frac{σ}{sin θ} = \frac{s − 1}{sin αθ} = \frac{ρ}{sin (α + 1)θ} = \frac{e + n − s}{sin (α + 2)θ}

if Δ = 0:

α + 2 = \frac{e + n − s}{σ} = \frac{ρ}{σ} + 1 = \frac{s − 1}{σ} + 2

Proof. First assume Δ > 0. By definition z = e(s − 1). Inserting s − 1 = \frac{ρsh αt}{sh (α+1)t}, we get

z = ρ^{2}[\frac{sh^{2}(αt)}{sh^{2}(α + 1)t} − 2ch t.\frac{sh(αt)}{sh(α + 1)t} + 1] = ρ^{2}[\frac{sh^{2}(αt)}{sh^{2}(α + 1)t} − \frac{sh(α − 1)t}{sh(α + 1)t}].

For the last equality check that sh (α + 1)t + sh (α − 1)t = 2ch t.sh (αt). Finally: ρ^{2}[\frac{sh^{2}(αt)}{sh^{2}(α+1)t} − \frac{sh(α−1)t}{sh(α+1)t}] = [\frac{ρsh t}{sh (α+1)t}]^{2}. For this check that sh^{2}(αt) − sh(α−1)t.sh(α+1)t = sh^{2}(t). We conclude that \frac{σ}{sh t} = \frac{ρ}{sh(α+1)t}. This proves the first three equalities. For the last one:

\frac{ρ}{sh (α + 1)t} = \frac{s − 1}{sh αt} = \frac{2ρch t − (s − 1)}{2sh (α + 1)t ch t − sh αt}

To conclude observe that: 2ρch t = e + n − 1 and 2sh (α + 1)t ch t − sh αt = sh (α + 2)t. The proof in case Δ < 0 is similar. If Δ = 0, observe that z = e(s) = (s − ρ − 1)^{2}, hence σ = ρ − s + 1.

Remark 1.7: Observe that when Δ < 0 and s = e + n, then sin(α + 2)θ = 0.

2. First consequences.

As a first consequence we get, for n ≥ 5, bounds on e and d by functions of s, n:

Proposition 2.1: Let X ⊂ ℙ^{n}, n ≥ 5, be a smooth codimension two subvariety of degree d and with ω_{X} ≃ O_{X}(e). As usual let s := min{k | h^{0}(I_{X}(k)) ≠ 0}. Then, if X is not a complete intersection:

e + n − s ≤ (n − 1)^{− \frac{1}{n−4}} (s − 1)^{\frac{n−2}{n−4}}

and

d ≤ s[1 + (n − 1)^{− \frac{1}{n−4}} (s − 1)^{\frac{n−2}{n−4}} ]

Proof. First of all we assume Δ > 0 and we observe that f(t) = logsh t is concave. Since αt = \frac{1}{α}t+ \frac{α−1}{α}(α+1)t we have f(αt) = f(\frac{1}{α}t+ \frac{α−1}{α}(α+1)t) ≥ \frac{1}{α}f(t) + \frac{α−1}{α}f((α + 1)t). Taking the exponentials we find

sh αt ≥ (sh t)^{\frac{1}{α}} (sh(α + 1)t)^{\frac{α−1}{α}} (+)

Similarly, writing αt = \frac{2}{α+1}t + \frac{α−1}{α+1}(α + 2)t and exponentiating the inequality coming from the concavity of f(t) we get

sh αt ≥ (sh t)^{\frac{2}{α+1}} (sh(α + 2)t)^{\frac{α−1}{α+1}} (++)

By Lemma 1.6 \frac{σ}{sh t} = \frac{h−1}{sh αt} = \frac{ρ}{sh (α+1)t} = \frac{e+n−h}{sh (α+2)t}, hence (+) gives

s − 1 ≥ (σ)^{\frac{1}{α}} (ρ)^{\frac{α−1}{α}}

and (++) gives

s − 1 ≥ (σ)^{\frac{2}{α+1}} (e + n − s)^{\frac{α−1}{α+1}}

from which follow

ρ ≤ σ^{\frac{−1}{α−1}} (s − 1)^{\frac{α}{α−1}} , e + n − s ≤ σ^{\frac{−2}{α−1}} (s − 1)^{\frac{α+1}{α−1}} .

and finally d = ρ^{2} + e + n ≤ s(1 + σ^{\frac{−2}{α−1}} (s − 1)^{\frac{α+1}{α−1}} ).

In order to conclude the case Δ > 0 it suffices to show that

σ^{\frac{−2}{α−1}} (s − 1)^{\frac{α+1}{α−1}} ≤ (n − 1)^{\frac{−1}{n−4}} (s − 1)^{\frac{n−2}{n−4}} . Since z = σ^{2} ≥ n − 1 we have \frac{(s−1)^{2}}{σ^{2}} ≤ \frac{(s−1)^{2}}{n−1} so σ^{\frac{−2}{α−1}} (s − 1)^{\frac{α+1}{α−1}} = (\frac{(s−1)^{2}}{σ^{2}} )^{\frac{1}{α−1}} (s − 1) ≤ (\frac{(s−1)^{2}}{n−1} )^{\frac{1}{α−1}} (s − 1) and we are done because α ≥ n − 3.

The case Δ < 0 (Δ = 0) can be proved the same way by using f(t) = logsin t (f(t) = log t) which is concave as well for t ∈]0, π[.

We observe that, as n increases, these inequalities are better than those given by the Speciality Theorem ( VI.Theorem 3.3).

In some sense the next proposition is a further improvment, except in the case Δ > 0 where the bound depends on Δ, hence on e.

Proposition 2.2: Let X ⊂ ℙ^{n}, n ≥ 4 be a smooth codimension two subvariety with ω_{X} ≃ O_{X}(e). If X is not a complete intersection, then:

1. If Δ > 0, then: d < M^{2}s^{2} + sM\sqrt{Δ}, where M = \frac{n−2}{n−3}.

2. If Δ ≤ 0, then: d < M^{2}s(s − 1) + s

Proof. 1) By Lemma 1.6

\frac{ρ}{s − 1} = \frac{sh(α + 1)t}{shαt}

and

\frac{sh(α + 1)t}{shαt} ≤ \frac{sh(n − 2)t}{sh(n − 3)t}

since α ≥ n − 3 (Definition 1.4). One can check that

\frac{sh(n − 2)t}{sh(n − 3)t} ≤ e^{t} \frac{n − 2}{n − 3}

so

\frac{ρ}{s − 1} ≤ e^{t} \frac{n − 2}{n − 3}

and

a = \frac{ρ}{e^{t}} ≤ (s − 1)\frac{n − 2}{n − 3}

Then we have

w := a − (s − 1) ≤ (s − 1)[\frac{n − 2}{n − 3} − 1] = \frac{s − 1}{n − 3} < \frac{s}{n − 3}

and

d = (a + 1)(b + 1) = (s + w)^{2} + \sqrt{Δ}(s + w).

Finally we get d < (\frac{n−2}{n−3})^{2}s^{2} + s\sqrt{Δ}\frac{n−2}{n−3}.

2) Assume first Δ < 0. By Lemma 1.6 we have: \frac{ρ}{s−1} = \frac{sin((α+1)θ)}{sin((α)θ)} ≤ \frac{α+1}{α}, indeed \frac{sinx}{x} is decreasing on 0 < x ≤ π. It follows that ρ ≤ M(s − 1). Since ρ = \sqrt{ab} = \sqrt{d − e − n}, we get the result taking into account the inequality: s(e + n + 1 − s) ≤ d (z ≥ 0). The case Δ = 0 follows directly from u_{k} ≥ 0, taking into account the inequality s(e + n + 1 − s) ≤ d.

Remark 2.3: We always have: s(e + n + 1 − s) ≤ d ( VI.Lemma 2.2). If we know d ≤ s^{2} when X is not a complete intersection, then e + n + 1 − 2s ≤ 0 and the associated bundle has to be stable. In conclusion a bound d ≤ Cs^{2} with C ≤ 1 would prove the non existence of indecomposable unstable bundles (”Grauert-Schneider conjecture”). In Proposition 2.2 we have such a bound (by taking a high twist of the bundle, Δ may be considered as a constant) but with C = M^{2} > 1 (and lim_{n→∞} M^{2} = 1).

3. The Schwarzenberger conditions.

The Chern classes of a vector bundle on ℙ^{n} have to satisfy some numerical conditions (this follows from the Atiyah-Singer index theorem, roughly speaking the expression of χ(E(k)) given by Riemann-Roch has to be an integer). These conditions, known as the Scharzenberger conditions ([46]), are as follows (in the rank two case):

Let E be a rank two bundle on ℙ^{n} with Chern classes c_{1}, c_{2}. Write c_{1} = a + b, c_{2} = ab, then:

\binom{a − 1 + m}{m} + \binom{b − 1 + m}{m} ∈ ℤ for m = 2, ..., n (Sm)

Let’s start with some preliminaries computations: Writing: (a + b)^{2} = a^{2} + b^{2} + 2ab, we get: a^{2} + b^{2} = (a + b)^{2} − 2ab = c_{1}^{2} − 2c_{2}.

a^{2} + b^{2} = c_{1}^{2} − 2c_{2} (P 2)

Going on this way expressing (a+b)^{i} in function of the elementary symmetric functions, we get:

a^{3} + b^{3} = c_{1}^{3} − 3c_{2}c_{1} (P 3)
a^{4} + b^{4} = c_{1}^{4} − 4c_{2}c_{1}^{2} + 2c_{2}^{2} (P 4)
a^{5} + b^{5} = c_{1}^{5} − 5c_{2}c_{1}^{3} + 5c_{2}^{2}c_{1} (P 5)
a^{6} + b^{6} = c_{1}^{6} − 6c_{2}c_{1}^{4} + 9c_{2}^{2}c_{1}^{2} − 2c_{2}^{3} (P 6)

3.1. The condition S2. The condition is:

\binom{a + 1}{2} + \binom{b + 1}{2} = \frac{(a + b) + a^{2} + b^{2}}{2} ∈ ℤ

i.e. using a + b = c_{1} and (P 2):

\frac{c_{1} + c_{1}^{2} − 2c_{2}}{2} = \frac{c_{1}(c_{1} + 1)}{2} − c_{2} ∈ ℤ

Of course this condition is always satisfied.

3.2. The condition S3. The condition is:

\binom{a + 2}{3} + \binom{b + 2}{3} ∈ ℤ

One computes:

\binom{a + 2}{3} = \frac{a^{3} + 3a^{2} + 2a}{6}

So:

\binom{a + 2}{3} + \binom{b + 2}{3} = \frac{(a^{3} + b^{3}) + 3(a^{2} + b^{2}) + 2(a + b)}{6}

Then, using (P 2), (P 3), we get:

\frac{c_{1}^{3} − 3c_{2}c_{1} + 3c_{1}^{2} − 6c_{2} + 2c_{1}}{6} ∈ ℤ (S3)

Now using the fact that the Schwarzenberger conditions are independant of the twist, we may assume c_{1} = 0 or c_{1} = −1.

Lemma 3.1: The condition (S3) is as follows:

1. If c_{1} = 0: no condition

2. If c_{1} = −1: c_{2} ≡ 0 (mod2) (c_{2} is even).

3.3. The condition S4. Arguing as above we get that S4 is equivalent to:

\frac{(a^{4} + b^{4}) + 6(a^{3} + b^{3}) + 11(a^{2} + b^{2}) + 6(a + b)}{24} ∈ ℤ

Using (P 4), ..., (P 2) we get that S4 is equivalent to:

c_{1}^{4} − 4c_{2}c_{1}^{2} + 2c_{2}^{2} + 6c_{1}^{3} − 18c_{2}c_{1} + 11c_{1}^{2} − 22c_{2} + 6c_{1} ≡ 0 (mod24)

Lemma 3.2: The condition S4 is as follows:

1. If c_{1} = 0: c_{2}(c_{2} + 1) ≡ 0 (mod12)

2. If c_{1} = −1: c_{2}(c_{2} − 4) ≡ 0 (mod12)

In particular the first possible cases for c_{1} = 0 and c_{2} > 0 are: (0, 3), (0, 8). It is still unknown if there are unstable vector bundle with c_{1} = 0, c_{2} = 3.

3.4. The condition S5. We have:

\binom{a − 1 + 5}{5}+\binom{b − 1 + 5}{5} = \frac{(a^{5} + b^{5}) − 10(a^{4} + b^{4}) + 35(a^{3} + b^{3}) + 50(a^{2} + b^{2}) + 24(a + b)}{5! = 120}

Using the previous relations (P_{i}) we get:

(c_{1}^{5}−5c_{2}c_{1}^{3}+5c_{2}^{2}c_{1})+10(c_{1}^{4}−4c_{2}c_{1}^{2}+2c_{2}^{2})+35(c_{1}^{3}−3c_{2}c_{1})+50(c_{1}^{2}−2c_{2})+24c_{1} ≡ 0 mod(120)

If c_{1} = 0 this yields:

\frac{c_{2}(c_{2} − 5)}{6} ∈ ℤ

This gives nothing more than (S4): if c_{2}(c_{2}−11) = 12m, then c_{2}^{2}−11c_{2}+6c_{2} = 12m + 6c_{2} and c_{2}(c_{2} − 5) = 6(2m + c_{2}).

If c_{1} = −1 this yields:

\frac{c_{2}(c_{2} − 2)}{8} ∈ ℤ

Since c_{2} = 2k (cf S3), c_{2}(c_{2} − 2) = 2k(2k − 2) = 4k(k − 1); since k(k − 1) is even, this is ≡ 0 mod(8) and the condition follows from (S3). In conclusion, condition (S5) doesn’t bring any new information.

3.5. The condition S6. We have

\binom{a + 5}{6} = (a^{6} + 15a^{5} + 85a^{4} + 225a^{3} + 274a^{2} + 120a)/720

Using the relations (P_{i}) it follows that:

(c_{1}^{6} − 6c_{2}c_{1}^{4} + 9c_{2}^{2}c_{1}^{2} − 2c_{2}^{3}) + 15(c_{1}^{5} − 5c_{2}c_{1}^{3} + 5c_{2}^{2}c_{1}) + 85(c_{1}^{4} − 4c_{2}c_{1}^{2} + 2c_{2}^{2})+

+225(c_{1}^{3} − 3c_{2}c_{1}) + 274(c_{1}^{2} − 2c_{2}) + 120c_{1} ≡ 0 mod (720)

So we get the ”normalized” conditions:

Lemma 3.3: The normalized Shwarzenberger conditions on ℙ^{6} are: 1. If c_{1} = 0: \frac{−c_{2}(c_{2}^{2}−85c_{2}+274)}{360} ∈ ℤ

2. If c_{1} = −1: \frac{−c_{2}(c_{2}^{2}−52c_{2}+72)}{360} ∈ ℤ

In conclusion:

Proposition 3.4: Let E be a rank two vector bundle on ℙ^{6} with Chern classes c_{1}, c_{2}, then:

1. If c_{1} = 0 then:

c_{2}(c_{2} − 11) ≡ 0 (mod 12) and \frac{−c_{2}(c_{2}^{2} − 85c_{2} + 274)}{360} ∈ ℤ

2. If c_{1} = −1 then:

c_{2} is even, c_{2}(c_{2} − 4) ≡ 0 (mod 12) and \frac{−c_{2}(c_{2}^{2} − 52c_{2} + 72)}{360} ∈ ℤ

3.6. The invariant Δ_{min}(n). Following Holme and Schneider ([47]), we define, in case Δ = c_{1}^{2} − 4c_{2} < 0, −Δ_{min}(n) as the minimal value of 4c_{2} − c_{1}^{2} > 0 for c_{1}, c_{2} satisfying the Schwarzenberger conditions on ℙ^{n}. Since we may assume c_{1} = 0 or c_{1} = −1, it is not difficult (with a computer) to compute Δ_{min}(n) for the first values of n:

Lemma 3.5: We have: −Δ_{min}(6) = −Δ_{min}(7) = 71 and −Δ_{min}(8) = 119.

Proof. Left to the computer.

It is clear that −Δ_{min}(n) is an increasing function of n. We try now to investigate this aspect of the question.

3.7. Some arithmetic. Let E be a rank two vector bundle on ℙ^{n} with Chern classes c_{1}, c_{2} (and Δ = c_{1}^{2} − 4c_{2} not a square). Let R = ℤ[X]/(X^{2} − c_{1}X + c_{2}). The Schwarzenberger condition says that: T r_{ℚR/ℚ}\binom{ξ+k}{n} ∈ ℤ for ξ= class of X, k ∈ ℤ. Let p be a prime number, then we have three cases:

1. inert (pR is prime)

2. decomposable (R/pR ≃ 𝔽_{p} × 𝔽_{p})

3. ramified (p | Δ)

Lemma 3.6: If there exists a rank 2 vector bundle E of Chern classes (c_{1}, c_{2}) on ℙ^{n}, then for each prime p < n, the discriminant Δ = c_{1}^{2} − 4c_{2} is a square mod p (possibly 0).

Proof. Assuming the contrary, we may suppose that n−1 = p is a prime such that Δ is not a square mod p. Let ξ be a root of X^{2} − c_{1}X + c_{2} in 𝔽_{p^{2}}. Then T r_{𝔽_{p^{2}} /𝔽_{p}}(ξ(ξ+1) … (ξ+p)) ≡ −Δ mod p, since (ξ+1) … (ξ+p) = F (ξ)− ξ where F is the Frobenius automorphism of 𝔽_{p^{2}} and T r_{𝔽_{p^{2}} /𝔽_{p}}(ξ(F (ξ) − ξ)) ≡ −Δ. So, if x is the image of X in R, one has T r_{R/ℤ}(x(x + 1) … (x + p)) is not divisible by p and T r_{ℚR/ℚ}\binom{x+p}{p+1} = χ(E) is not an integer. Contradiction.

By [74] pp. 134-135 one knows that there exists some prime p < n such that Δ is not a square mod p, when n ≥ c(log|Δ|)^{2} under Generalized Riemann Hypothesis or when n ≥ 2(|Δ|)^{A} without restrictions (c and A effective constants). This means that |Δ_{min}(n)| ≥ e^{\sqrt{\frac{n}{c}}} under GRH or |Δ_{min}(n)| ≥ (\frac{n}{2})^{\frac{1}{A}}.

In particular it follows that the Chern classes of an indefinitely extendable rank two vector bundle are the Chern classes of a split bundle (i.e. Δ is a square).

4. BOUNDS ON e AND s. 73

4. Bounds on e and s.

Aim of this section is to give (lower) bounds (by functions of n) on the invariants e, s of a smooth codimension two subvariety X ⊂ ℙ^{n} which is a not a complete intersection. We will distinguish several cases according to the stability (or non-stability) of the associated bundle.

The next proposition deals with the non stable case:

Proposition 4.1: Let X ⊂ ℙ^{n}, n ≥ 4, be a smooth subvariety of codimension two. Assume X is not a complete intersection.

1. if Δ ≥ 0, then s − 1 ≥ (n − 3)\sqrt{n − 1} and e ≥ (2n − 4)\sqrt{n − 1} − n.

2. if Δ < 0 and e + n + 1 ≥ 2s, then s − 1 ≥ \frac{2}{π} (n − 3)\sqrt{n − 1} and

e ≥ \frac{2}{π} (2n − 4)\sqrt{n − 1} − n.

Proof. 1.) Assume first Δ > 0. By Lemma 1.6: \frac{s−1}{σ} = \frac{sh(αt)}{sh t} ≥ α. Since σ ≥ \sqrt{n − 1} and α ≥ n − 3, we get the result. In the same way, from Lemma 1.6: \frac{e+n−s}{σ} = \frac{sh(α+2)}{sh t} ≥ α+2 ≥ n−1, hence: e+n−s ≥ (n−1)\sqrt{n − 1} and the result follows.

Assume now Δ = 0. We have z = (s − a − 1)^{2}. Since z ≥ n − 1, it follows that a + 1 − s ≥ \sqrt{n − 1} (note that a + 1 > s if X is not a c.i.), so a ≥ s − 1 + \sqrt{n − 1} and we get s − 1 ≥ \frac{(n−3)}{(n−2)}(s − 1 + \sqrt{n − 1}) hence s − 1 ≥ (n − 3)\sqrt{n − 1}. We conclude as above since \frac{e+n−s}{σ} = α + 2 ≥ n − 1. 2.) By Lemma 1.6 \frac{e+n−s}{s−1} = \frac{sin(α+2)θ}{sinαθ}. The assumption e + n + 1 ≥ 2s implies that sin(α + 2)θ ≥ sin αθ. This in turn, implies αθ < \frac{π}{2} (we have (α + 2)θ < \frac{3π}{2} , cf Definition 1.4). By Lemma 1.6 \frac{s−1}{σ} = \frac{sin(αθ)}{sin θ}. Since sinαx/sin x is decreasing, we get: \frac{s−1}{σ} > \frac{1}{sin(π/2α)}, hence s−1 > \frac{2}{π} (n−3)\sqrt{n − 1}. The proof for e is similar using \frac{e+n−s}{σ} = \frac{sin(α+2)θ}{sin θ} of Lemma 1.6.

For the stable case we need the following:

Lemma 4.2: Let E be a rank two vector bundle admitting a section with smooth codimension two zero locus. Assume Δ = c_{1}^{2} − 4c_{2} < 0 and write: c_{1} = δ + \overline{δ}, c_{2} = δ\overline{δ}, a = δ − 1 = ρe^{iθ}, ρ > 0, arg(θ) < π. Then:

c_{2} > \frac{−Δ_{min}(n)}{4sin^{2}(θ)} + c_{1} − 1

Proof. We have −Δ = 4C_{2} − C_{1}^{2} where C_{i} = c_{i}(E(−1)). Moreover C_{1} = a + \overline{a}, C_{2} = a\overline{a}. It follows that: C_{1}^{2} = (a + \overline{a})^{2} = 4C_{2}cos^{2}(θ). Now 4C_{2} − C_{1}^{2} = 4C_{2}sin^{2}(θ). By definition: −Δ > −Δ_{min}(n), so 4C_{2}sin^{2}(θ) > −Δ_{min}(n). Since C_{2} = c_{2} + 1 − c_{1}, we conclude.

Corollary 4.3: Let X ⊂ ℙ^{n} be a smooth codimension two subvariety. Assume Δ < 0.

1. If n = 6, then: e ≥ 5

2. If n = 7, then: e ≥ 8

3. If n ≥ 8, then: e ≥ 2n − 1.

Proof. Denote by C_{i} the Chern classes of E(−1) where E is the rank two bundle associated to X. We recall that we may assume C_{i} ≥ 0. By Lemma 4.2:

C_{2} > \frac{−Δ_{min}(n)}{4sin^{2}(θ)} ≥ \frac{−Δ_{min}(n) · (n − 1)^{2}}{4π^{2}}

the last inequality coming from the fact that θ ≤ \frac{π}{n−1} (see Definition 1.4).

(i) If n = 6, then s_{3} = C_{1}(C_{1}^{2} − 2C_{2}) and s_{3} ≥ 0 implies: (e + n − 1)^{2} = C_{1}^{2} ≥ 2C_{2}. It follows that:

(e + n − 1)^{2} = C_{1}^{2} > \frac{−Δ_{min}(n) · (n − 1)^{2}}{2π^{2}}

inserting n = 6, −Δ_{min}(6) = 71 the result follows. (ii), (iii) For n ≥ 7 we can use s_{5} ≥ 0, where s_{5} = C_{1}(C_{1}^{2} − C_{2})(C_{1}^{2} − 3C_{2}), i.e. C_{1}^{2} ≥ 3C_{2}. As above we get:

(e + n − 1)^{2} > \frac{−Δ_{min}(n)3(n − 1)^{2}}{4π^{2}}

If n = 7, −Δ_{min}(7) = 71 and we conclude as above. If n ≥ 8, −Δ_{min}(n) ≥ −Δ_{min}(8) = 119 and this yields: (e+n−1)^{2} > 9(n−1)^{2}, hence the result.

This proposition is of some interest only in the stable case (2s > e+n+1) since in the other cases we have better results (cf Proposition 4.1). By the way it is still possible to get a little improvment:

Lemma 4.4: Let X ⊂ ℙ^{n} be a smooth codimension two subvariety with Δ < 0, then:

1. If n = 7, then: e ≥ 9

2. If n = 6, then: e ≥ 8.

Proof. (i) According to Corollary 4.3 we have only to consider the case e = 8. If E is the associated bundle, the Chern classes of E(−1) are: C_{1} = 14, C_{2} = d − 15. The condition s_{5} ≥ 0 implies C_{1}^{2} ≥ 3C_{2}, so d ≤ 80. It follows that the normalized twist has Chern classes: c_{1} = 0 and c_{2} = d − 64 ≤ 16. But there is no value of c_{2}, 1 ≤ c_{2} ≤ 16, which satisfies the Schwarzenberger conditions on ℙ^{6}. (ii) The proof is similar (cf [47]).

Gathering everything together (cp [47]):

Theorem 4.5: Let X ⊂ ℙ^{n}, n ≥ 6, be a smooth codimension two subvariety with ω_{X} ≃ O_{X}(e). Then: e ≥ n + 2. More precisely:

• If Δ ≥ 0, then: e ≥ −n + (2n − 4)\sqrt{n − 1}

• if Δ < 0 and e + n + 1 ≥ 2s (non-stable case), then: e ≥ −n + \frac{2}{π} (2n − 4)\sqrt{n − 1}

• if Δ < 0 then: e ≥ n + 2 if n < 8 and e ≥ 2n − 1 if n ≥ 8.

Now we turn to the invariant s.

Lemma 4.6: Let X ⊂ ℙ^{n}, n ≥ 6, be a smooth subvariety of codimension two. Assume Δ < 0 and e + n + 1 − 2s ≤ 0.

(i) If n ≥ 6, then s ≥ n + 2 (ii) If n ≥ 8, then s ≥ 3n/2

Proof. (i) If n ≥ 6, then e ≥ n+2 (Theorem 4.5), hence (n+2)+n+1 ≤ e + n + 1 ≤ 2s, thus s ≥ n + 2. (ii) As above using e ≥ 2n − 1 (Theorem 4.5).

Remark 4.7: In case Δ < 0, we may proceed as follows: by Lemma 1.6: s − 1 = ρ\frac{sin(αθ)}{sin(α+1)θ}, so s − 1 = δ \frac{sin(αθ)}{sin θ.sin(α+1)θ}, where δ = \frac{1}{2} \sqrt{−Δ} = ρsin θ, θ(α + 1) < π. Let’s denote by m(α) the minimum of the function φ(θ) = \frac{sin(αθ)}{sin θ.sin(α+1)θ} on ]0, \frac{π}{α+1}[. This minimum is reached for the solution, β, of \frac{sin(α+1)β}{sin β} = \sqrt{α + 1} and is an increasing function of α. So we have: s − 1 ≥

δm(α) ≥ δm(n − 3) ≥ ^{\frac{\sqrt{−Δ_{min}(n)}}{2}}.m(n − 3). It is possible to compute an approximated value of m_{n} := m(n − 3). For instance we have: m_{5} ≃ 1, 6949, m_{6} ≃ 2, 2845, m_{7} ≃ 2, 8203, m_{8} ≃ 3, 3233 (and m_{40} ≃ 16, 1647). Since −Δ_{min}(8) = 119, we get: s − 1 ≥ 19 if n = 8, which is better than 12 = \frac{3.8}{2}.

We immediately get:

Theorem 4.8: Let X ⊂ ℙ^{n}, n ≥ 11, be a smooth codimension two subvariety. If X is not a complete intersection, then s ≥ \frac{3n}{2}.

Proof. Assume first Δ < 0. If e+n+1 > 2s, then, by Proposition 4.1 it is enough to check that 1 + \frac{2}{π} (n − 3)\sqrt{n − 1} ≥ \frac{3n}{2} if n ≥ 11. If e + n + 1 ≤ 2s, then s ≥ \frac{3n}{2} by Lemma 4.6. If Δ ≥ 0, by Proposition 4.1, it is enough to check that 1+(n−3)\sqrt{n − 1} ≥ \frac{3n}{2} if n ≥ 11 (actually this is satisfied for n ≥ 8).

The following improves Ran’s result (s ≥ n − 1 if X is not a complete intersection, see III.Proposition 2.7):

Theorem 4.9: Let X ⊂ ℙ^{n}, n ≥ 4, be a smooth codimension two subvariety. Assume ch(k) = 0. If n < 6 assume X subcanonical. If h^{0}(I_{X}(n)) ≠ 0, then X is a complete intersection.

Proof. For the case n = 4 we refer to [30]. If n = 5 by [28] we may assume s = 5 and by [7] e ≥ 3. From u_{3} ≥ 0 (see VI.Lemma 3.2) we get: z ≤ \frac{(s−1)^{3}}{e+n+s−2}, i.e. z ≤ 5. In fact 4 ≤ z ≤ 5, since z ≥ n−1 ( III.Corollary 2.5). Arguing as in [28] Lemma 2.6, every irreducible component of Z_{red} appears with multiplicity, hence Z is either a multiplicity z structure on a linear subspace or is contained in a cubic hypersurface. The last case is not possible ( III.Proposition 2.7). In the first case by [57] (or also [58] observing that the proof of the main theorem works in the case of a codimension two linear subspace of ℙ^{5}), Z is a complete intersection. If 6 ≤ n ≤ 7. By u_{4} ≥ 0: f(z) = \frac{[(s−1)^{2}−z]^{2}}{z} ≥ (e + n − 1)^{2}. Since f(z) is decreasing and z ≥ n − 1, f(n − 1) ≥ (e + n − 1)^{2}, i.e. (s − 1)^{2} ≥ \sqrt{n − 1}(e + n−1)+n−1. By Theorem 4.5: e ≥ n+2, so (s−1)^{2} ≥ \sqrt{n − 1}(2n+1)+n−1, but this inequality is not satisfied if s ≤ n, 6 ≤ n ≤ 7. Now assume 8 ≤ n ≤ 10. If Δ ≥ 0, we conclude with Proposition 4.1. If Δ < 0 we conclude by Remark 4.7 (s ≥ 20 if n = 8). For n ≥ 11, we conclude with Theorem 4.8.

Remark 4.10: 1. Of course the case n = 4 requires a quite different approach (on ℙ^{4} we only have u_{2} ≥ 0, the ”main lemma” of Ellingsrud-Peskine). Also recall that there exist smooth subcanonical surfaces, which are not complete intersections, lying on quintic hypersurfaces: these are the abelian surfaces of degree 10 corresponding to the Horrocks-Mumford bundle.

2. It should be possible, with some effort, to improve a little bit Theorem 4.9, at least for n ≥ 6.

5. Notes and complements.

Exercice 5.1: Sequences (v_{n})_{n≥0} inductively defined by v_{n} = αv_{n−1}+βv_{n−2}, n ≥ 2 (α, β fixed) form a two dimensional vector space. A sequence (q^{n}) belongs to this vector space if and only if: q^{2} = αq + β. If the discriminant, Δ, of this equation is ≠ 0, the two roots, a, b, yield a basis and expressing (u_{k}) in this basis (by looking to u_{0}, u_{1}) we get:

u_{k} = (\frac{a}{b})^{k} − \frac{a − s + 1}{b − s + 1} if Δ > 0

u_{k} = (s − 1)sin(kθ) − ρsin(k − 1)θ if Δ < 0

If Δ = 0, then −β = a^{2} and one sees, by induction, that: u_{k} = a^{k−1}[k(s − 1) − (k − 1)a]. Check all this and recover Proposition 1.2.

Exercice 5.2: Check all the computations of this chapter.

Notes. This chapter is taken from [31]. The inequality e ≥ n + 2, if n ≥ 6 was first proved in [47] improving a former result of [7] (but for n = 5 the result of [7] is still the best). It is possible to obtain a bound on d by a function of n ([7], [47]). For n = 6 one can prove that X is a complete intersection if d ≤ 73 (see [28]). For subcanonical surfaces in ℙ^{4}, see [7], [21], [30] and the huge literature on the Horrocks-Mumford bundle.

CHAPTER VIII Conclusion.

1. Summarizing.

The results obtained so far show that there are strong constraints to the existence of smooth codimension two X ⊂ ℙ^{n}, n ≥ 6, which are not complete intersections:

• X is necessarily subcanonical (ω_{X} ≃ O_{X}(e))

• if s := min{k | h^{0}(I_{X}(k)) ≠ 0}, then the degree, d := d(X), is bounded by a function of s

• no such X exists on a hypersurface of degree ≤ n (i.e. s ≥ n + 1); also e ≥ n + 2 (and there are better bounds under suitable assumptions on Δ := (e + n + 1)^{2} − 4d)

• there are strong restrictions (mainly due to the Schwarzenberger conditions) to extend X to ℙ^{n+t}, t big enough (on the other hand if X exists on ℙ^{n}, then, by hyperplane sections, we get X_{k} ⊂ ℙ^{k} for all k < n and X is an extension of X_{k}).

We may conclude that there is indeed a general phenomenon which forces many of the smooth codimension two X ⊂ ℙ^{n} (n big enough) to be a complete intersection.

2. By day and by night...

By day try to prove the conjecture, by night try to find a counter-example!

(Or the contrary, depending on your personal convictions.)

All the results so far obtained do not exclude the existence of some sporadic X ⊂ ℙ^{n} which are not complete intersections. Also from the ”vector bundle side” we may conclude that any (c_{1}, c_{2}) allowed by the Schwarzenberger conditions on ℙ^{n}, n ≥ 4 is still a possible candidate for the Chern classes of an indecomposable bundle. Even the most investigated case (c_{1} = 0, c_{2} = 3 on ℙ^{4}) is still possible: there is no semi-stable bundle with these Chern classes ([10], [7]), but the non-stable case is not excluded. For instance are there such bundles with an arbitrarily large index of unstability?

Also we have to point out that, if char(k) > 0, there are examples of indecomposable rank two vector bundles on ℙ^{4}, ℙ^{5} (see [53] and references therein), at the moment these examples are peculiar to characteristic p... Also it seems incredible that the Horrocks-Mumford bundle and its derived bundles are the unique indecomposable bundles on ℙ^{4}.

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