Explicit Symmetric Differential Forms on Complete Intersection Varieties And

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Varieties with ample cotangent bundle have many interesting properties, however, relatively few examples of such varieties are know (see [17], [20], [7], [3]). Debarre conjectured in [7] that: if X is a general complete N intersection variety in P of multidegree high enough and such that dim X 6 codimPN X then the cotangent bundle of X should be ample. The study of this conjecture was the starting point of the present work. In [3] we were able to prove this conjecture when dim X =2, using Voisin’s variational method and inspired by the work of Siu [18] and the work of Diverio Merker and Rousseau [10]. However we were not able to make this strategy work completely in higher dimensions.

The construction of varieties with positive cotangent bundle is closely related to the construction of symmetric diﬀerential forms on it. In fact, if one wants to prove that a given variety has ample cotangent bundle, it is natural to start by producing many symmetric diﬀerential forms, to be more precise, this means proving that the cotangent bundle is big. In general, this is already a highly non-trivial question and this leads to very interesting considerations. In that direction we would like to mention the recent work of Brunebarbe, Klingler and Totaro [6] as well as the work of Roulleau and Rousseau [16].

However, ampleness is a much more restrictive condition than bigness, in some sense, bigness only requires a quantitative information on the number of symmetric differential forms whereas ampleness requires a more qualitative information on the geometry of the symmetric differential forms. The most natural way to produce symmetric differential forms is to use Riemann-Roch theorem or a variation of it. For instance under the hypothesis of Debarre’s conjecture, one can use Demailly’s holomorphic Morse inequality, in the spirit of Diverio’s work [8] and [9], to prove that the cotangent bundle is big (see [3]). Nevertheless, this approach doesn’t give much information on the constructed symmetric differential forms besides its existence. One can wonder if it is possible, given a complete intersection variety X in PN to write down explicitly the equation of a symmetric differential form on X (if such an object exists).

If X is a curve in P2 of genus greater than 1, this is a very classical exercise. In higher dimensions, very few results towards that problem are known. Brückmann [5] constructed an example of a symmetric differential form on a complete intersection in P4 given by Fermat type equations, and more recently Merker [14] was able to study examples of symmetric differential forms on complete intersection variety in P4 in the spirit of work of Siu and Yeung [19] (see also [15] for related results for higher order jet differential equations).

The aim of this paper is to develop a cohomological framework which will enable us to describe the space of holomorphic symmetric diﬀerential forms on a complete intersection variety in PN in terms of its deﬁning equations, and to give several applications. The outline of the paper is as follows.

In Section 2 we prove the main technical result of this work. In view of the possible generalizations to higher order jet diﬀerential as well as for its own sake, we will not only study the space of symmetric i ℓ1 ℓ diﬀerential forms, but also the more general spaces H (X,S ΩX ⊗···⊗ S k ΩX ) for a complete intersection variety X in PN . Recall the following vanishing result of Brückman and Rackwitz. Theorem 1.1 (Brückmann-Rackwitz [4]). Let X ⊆ PN be a complete intersection of dimension n and k codimension c. Take integers j, ℓ1,...,ℓk > 0. If j

j ℓ1 ℓk H (X,S ΩX ⊗···⊗PS ΩX )=0.

k It is natural to look at what happens in the case j = n − i=1 min{c,ℓi} in the above theorem. Our result in that direction is the following. P ∗ N Theorem A. Let c,N,e1,...,ec ∈ N , set n = N − c. Let X ∈ P be a smooth complete intersection 0 N variety of codimension c, dimension n, deﬁned by the ideal (F1,...,Fc), where Fi ∈ H (P , OPN (ei)). Take

2 c integers ℓ1,...,ℓk > c take an integer a<ℓ1 + ··· + ℓk − k, let q := n − kc and b := (k + 1) i=1 ei. Then one has a commutative diagram P

q ℓ1 ℓk / N N ℓ1−c ℓk−c H X,S ΩX ⊗···⊗ S ΩX (a) / H P ,S ΩPN ⊗···⊗ S ΩPN (a − b) ❳❳❳❳❳ ❳❳❳❳❳ ϕ E ❳❳❳❳ E ❳❳❳❳❳ ❳❳❳❳+ ϕ˜ q ℓ1 ℓk / N N ℓ1−c ℓk−c H X,S ΩX ⊗···⊗ S ΩX (a) / H P ,S ΩPN ⊗···⊗ S ΩPN (a − b) Such that all the arrowse are injectivee and such that: e e

c k {j} 1. imϕ ˜ = i=1 ker(·Fi) j=1 ker(·dFi ) . 2. im ϕ = imT ϕ ˜ ∩ im E . T Remark 1.2. The bundle Ω is described in Section 2.1 and the diﬀerent maps arising in the statement are described in Section 2.5 The important thinge to note in that result, is that it gives a way of describing the vector space

q ℓ1 ℓk N N ℓ1−c ℓk−c H X,S ΩX ⊗···⊗ S ΩX (a) as a sub-vector space of H P ,S ΩPN ⊗···⊗ S ΩPN (a − b) , that this last space is easily described, and that one can precisely determine, in terms of the deﬁning equations of X what is the relevant sub-vector space. Therefore this result (ande the more generale statements in Theorem 2.17 and Theorem 2.24) should be understood as our main theoretical tool to construct symmetric diﬀerential forms on complete intersection varieties.

In Section 3 we provide the ﬁrst applications of Theorem A. First we describe how Theorem A can be used very explicitly in Čech cohomology. Then we illustrate this by treating the case of curves in P2. After that (Proposition 3.3) we prove that the result of Brückmann and Rackwitz is optimal by providing the following non-vanishing result.

Theorem B. Let N > 2, let 0 6 c 1 and a<ℓ1 + ··· + ℓk − k. Suppose k N 0 6 q := N − c − i=1 min{c,ℓi}. Then, there exists a smooth complete intersection variety in P of codimension c, such that P q ℓ1 ℓk H (X,S ΩX ⊗···⊗ S ΩX (a)) 6=0. Then, in Corollary 3.7, we provide a new example of a family illustrating the fact that the dimension of the space of holomorphic symmetric diﬀerential forms is not deformation invariant. Theorem C. For any n > 2, for any m > 2, there is a family of varieties Y → B over a curve, of relative dimension n, and a point 0 ∈ B such that for generic t ∈ B,

0 m 0 m h (Y0,S ΩY0 ) >h (Yt,S ΩYt ).

This phenomenon has already been studied (see for instance [1], [2] and [16]) and is well known. How- ever, this example shows that invariance fails for any m > 2, whereas the other known examples (based on intersection computations) provide the result for m large enough.

In Section 4 we provide our main application, which is a special case of Debarre’s conjecture.

Theorem D. Let N,e ∈ N∗ such that e > 2N +3. If X ⊆ PN be a general complete intersection variety of multidegree (e,...,e), such that codimPN X > 3 dim X − 2, then ΩX is ample. To our knowledge, this is the ﬁrst higher dimensional result towards Debarre’s conjecture. The proof of this statement does not rely on the variational method neither does it need the Riemann-Roch theorem nor Demailly’s holomorphic Morse inequalities. The idea is to use the results of Section 2 to construct one

3 very particular example of a smooth complete intersection variety in PN (with prescribed dimension and multidegree) whose cotangent bundle is ample. Then by the openness property of ampleness, we will deduce that the result holds generically. Such an example is produced by considering intersections of deformations of Fermat type hypersurfaces.

Notation and conventions: In this paper, we will be working over he ﬁeld of complex numbers C. Given a smooth projective variety X and a vector bundle E on X, we will denote by SmE the m-th symmetric power of E, we will denote by P(E) the projectivization of rank one quotients of E, we will denote the tangent bundle of X by TX and the cotangent bundle of X by ΩX . Moreover we will denote by πX : P(ΩX ) → X the canonical projection. Given a line bundle L on X and an element σ ∈ H0(X,L) we will denote the zero

locus of σ by (σ = 0), and the base locus of L by Bs(L)= σ∈H0 (X,L)(σ = 0). Given any m ∈ N, we will denote by C[Z ,...,ZN ]m the set of homogenous polynomials of degree m in 0 T N +1 variables and by C[z1,...,zN ]6m the set of polynomials of degree less or equal to m in N variables. k Given any set E ⊆ N and any k ∈ N we will write E6= := {(i1,...,ik) such that ij 6= iℓ if j 6= ℓ} and k E< := {(i1,...,ik) such that ij

Acknowledgments. This work originated during the author’s phd thesis under the supervision of Christophe Mourougane. We thank him very warmly for his guidance, his time and all the discussions we had. We also thank Junjiro Noguchi and Yusaku Tiba for listening through many technical details. We thank Joël Merker for his many encouragements and for all the interest he showed in this work. We also thank Lionel Darondeau for motivating discussions and for his suggestions about the presentation of this paper.

2 Cohomology of symmetric powers of the cotangent bundle 2.1 The tilde cotangent bundle It will be convenient for to use the Ω bundle, studied in particular by Bogomolov and DeOliveira in [2], but also by Debarre in [7]. In some way, the bundle Ω will allow us to work naturally in homogenous coordinates. Let us recall some basic facts aboute this bundle. Consider PN = P (CN+1) with its homogenous coordinates N [Z0,...,ZN ]. Let X ⊆ P be a smooth subvariety.e We denote by γX the Gauss map N γX : X → Grass(n, P )

x 7→ TxX N N where TxX ⊂ P is the embedded tangent space of X at x, and where Grass(n, P ) denotes the grassmannian N of n-dimensional linear projective subspace of P . Let Sn+1 denote the tautological rank n +1 vector bundle on Grass(n, PN ). Then define ∗ ∨ ΩX := γX Sn+1 ⊗ OX (−1). We will refer to this bundle as the tilde cotangent bundle of X, and a holomorphic section of SmΩ will be e X called a tilde symmetric differential form. Observe that one has a natural identification

N e N+1 ΩPN =∼ C ⊗ OPN (−1) =∼ OPN (−1)dZi. i=0 M e Therefore given any homogenous degree e polynomial F ∈ C[Z0,...,ZN ] one can deﬁne a map

· dF : OPN (−e) → ΩPN (1) N ∂F g 7→ ge· dZi. ∂Zi i=0 X

4 One easily verifies that if X ⊆ PN is a smooth subvariety and if F defines a hypersurface H such that Y := X ∩ H is a smooth hypersurface in X then the above map fits into the following exact sequence,

·dF 0 → OY (−e) → ΩX |Y → ΩY → 0. (2)

N+1 We will refer to it as the tilde conormal exact sequencee . On thee other hand, let X ⊆ C \{0} be the cone ∗ ∗ above X, and let ρX = X → X be the natural projection. Observe that ρX γX Sn+1 = T X. The diﬀerential ∗ ∗ b dρX : T X → ρX TX is not invariant under the natural C action on X because for any x ∈ X, any ξ ∈ TxX ∗ 1 and any λ ∈ C , dρX,λxξb= λ dρX,xξ. We can easily compensate this by a simple twist bybOX (−1) as in the followingb b b b

∗ γX Sn+1,x → TxX ⊗ OX,x(−1) (x, ξ) 7→ (x, dρX,xξ ⊗ x).

∗ This yields an exact sequence 0 → OX (−1) → γX Sn+1 → TX(−1) → 0 which we twist and dualize to get

E 0 → ΩX → ΩX → OX → 0. (3) E Will refer to it as the Euler exact sequence. Note thate the map can be understood very explicitly. Indeed, if N N Zi E Z0dZi−ZidZ0 we consider the chart C = (Z0 6= 0) ⊂ P , with zi = for any i ∈{1,...,N}, then (dzi)= 2 . Z0 Z0 Let us mention that in our computations we will often write dzi instead of E (dzi) for simplicity. Those two exact sequences ﬁt together in the following commutative diagram 0 0

  OY (−e) OY (−e) y y

  0 −−−−→ ΩX|Y −−−−→ ΩX|Y −−−−→ OY −−−−→ 0 y y (4) e   0 −−−−→ ΩY −−−−→ ΩY −−−−→ O Y −−−−→ 0 y y e   0 0 y y

Remark 2.1. Observe that ΩX can never be ample because it has a trivial quotient. However, Debarre proved in [7] that under the hypothesis of his conjecture, the bundle ΩX (1) is ample. e 2.2 A preliminary example e The combinatorics needed in the proof of the main results of Section 2 may seem elaborate, but the idea behind it is absolutely elementary. In fact the proofs of the statements in Section 2 are only a repeated us of long exact sequences in cohomology associated to short some exact sequences which are deduced from the restriction exact sequence, the conormal exact sequence, the tilde conormal exact sequence and the Euler exact sequence. But because our purpose is to study tensor produces of symmetric powers of some vector bundle, many indices have to be taken into account, the only purpose of all the notation we will introduce is to synthesis this as smoothly as possible.

5 Let us illustrate the idea behind this by considering a basic example. Suppose that H is a smooth degree N e hypersurface in P deﬁned by some homogenous polynomial F ∈ C[Z0,...,ZN ]. Suppose that we want i m to understand the groups H (X,S ΩX (−a)) for some a ∈ N, and m > 1. To do so we look at the tilde conormal exact sequence ·dF 0e→ OX (−e) → ΩPN |X → ΩX → 0 and take the m-th symmetric power and twist it by O (−a) of it to get the exact sequence eX e m−1 ·dF m m 0 → S ΩPN |X (−e − a) → S ΩPN |X (−a) → S ΩX (−a) → 0.

i m By considering the long exact sequencee in cohomology associe ated to it, we seee that the groups H (X,S ΩX (−a)) i k can be understood from the groups H (X,S ΩPN |X (−b)) for k,b ∈ N, and from the applications appearing in the long exact sequence in cohomology. But to understand those groups, we consider the restrictione exact sequence e ·F 0 → OPN (−e) → OPN → OX → 0,

k and twist it by S ΩPN (−b) to get

k ·F k k e 0 → S ΩPN (−b − e) → S ΩPN (−b) → S ΩPN |X (−b) → 0.

i k Once again, we look at what happense in cohomology,e and we see tehat the groups H (X,S ΩPN |X (−b)) can i N ℓ be understood from the groups H (P ,S ΩPN (−c)) for ℓ,c ∈ N and the maps appearing in the long exact i N ℓ ℓ N+1 i N sequence. But observe that H (P ,S ΩPN (−c)) =∼ S C ⊗H (P , OPN (−c−ℓ))=0 for all ei < N, from this i k i m we get that H (X,S ΩPN |X (−b))=0 for alle i

The inclusions appearinge in Theorem A are of this type.e Now if one wants to describe whate is the image of this composed inclusion, one needs to look more carefully at what are exactly the maps between the cohomology groups in the diﬀerent long exact sequences. For instance the second injection comes the following exact sequence

N−1 m−1 N N m−1 ·F N N m−1 0 → H (X,S ΩPN |X (−a − e)) → H (P ,S ΩPN (−a − 2e)) → H (P ,S ΩPN (−a − e)).

N N m−1 ·F N N m−1 Hence, im(ϕ ˜2) = ker He (P ,S ΩPN (−a − 2e)) → He (P ,S ΩPN (−a − e)) . Toe understand similarly im(ϕ ˜2 ◦ϕ˜1) is less straightforward, combining the diﬀerent long exact sequences one obtains the following commutative diagram: e e

N−2 m ϕ˜1 / N−1 m−1 ·dF / N−1 m H (X,S ΩX (−a)) / H (X,S ΩPN |X (−a − e)) / H (X,S ΩPN |X (−a)) ❱❱❱❱ ❱❱❱❱ ϕ ❱❱❱ ϕ˜2 e ❱❱❱❱ e e ❱❱❱* N N m−1 ·dF / N N m H (P ,S ΩPN (−a − 2e)) / H (P ,S ΩPN (−a − e))

where the vertical arrows are injective. Then, by lineare algebra, we obtain that im(ϕ ˜) =e im(ϕ ˜2) ∩ ker(·dF )= ker(·F ) ∩ ker(·dF ) for suitable maps ·F and ·dF. This example already contains the main idea of the proof of the ﬁrst part of Theorem A. To study more generally tensor products of symmetric powers of the tilde cotangent bundle is done similarly by considering each factor independently, and to deduce the results concerning the cotangent bundle instead of the tilde cotangent bundle is done in a similar fashion using the Euler exact sequence.

6 2.3 An exact sequence In the rest of Section 2, the setting will be the following. Take an integer N > 2, let c ∈ {0,...,N − 1} ∗ 0 N 0 N and take e1,...,ec ∈ N . Take non-zero elements F1 ∈ H (P , OPN (e1)),...,Fc ∈ H (P , OPN (ec)), and for any i ∈{1,...,c} we set Hi = (Fi = 0). For any i ∈{1,...,c} let Xi := H1 ∩···∩ Hi. Set X := Xc and N X0 := P . We suppose that X is smooth. For simplicity, we will also suppose that Xi is smooth for each i ∈{0,...,c}. Remark 2.2. We make this additional smoothness hypothesis here so that we can work without worrying with all the conormal exact sequences between Xi and Xi+1 (and this hypothesis will be satisﬁed in all our applications). However, a more careful analysis of the proof of the main results shows that the only thing ′ we need to have is the smoothness of each of the Xis in a neighborhood of X, and this follows from the smoothness of X. To simplify our exposition, we introduce more notation. If E is a vector bundle on a variety Y , and if λ = (λ1,...,λk) is a k-uple of non-negative integers, then we set

Eλ := Sλ1 E ⊗···⊗ Sλk E.

If µ = (µ1,...,µj ) is a j-uple of non-negative integers, we set

λ ∪ µ := (λ1,...,λk,µ1,...,µj ).

The following deﬁnition gives a convenient framework for our problem. Deﬁnition 2.3. With the above notation.

0 p 1. A λ-setting is a (p + 2)-uple Σ := (Xp, λ ,...,λ ), where 0 6 p 6 c, and for any 0 6 j 6 p, j j j mj λ = (λ1,...,λmj ) ∈ N for some mj ∈ N.

0 p 2. If Σ = (Xp, λ ,...,λ ) is as above, we set: • codim Σ := p and dim Σ := N − p. 1 p j • If λ = ··· = λ = 0 set deg Σ := ep. Otherwise, let j0 := min{1 6 j 6 p such that λ 6=0} and

set deg Σ := ej0 . 3. Take Σ as above. We set:

Σ λ0 λ1 λp Σ λ0 λ1 λp Ω := ΩPN ⊗ Ω ⊗···⊗ Ω and Ω := ΩPN ⊗ Ω ⊗···⊗ Ω . |Xp X1|Xp Xp |Xp X1|Xp Xp

4. For any a ∈ Z and any j ∈ N, we set: e e e e

j Σ j Σ j Σ j Σ H Ω (a) := H Xp, Ω ⊗ OXp (a) and H Ω (a) := H Xp, Ω ⊗ OXp (a) . e e We will also need a more general deﬁnition which will allow us to work simultaneously with Ω and Ω. Deﬁnition 2.4. 1.A λ-pair (Σ, Σ) is a couple of λ-settings Σ = (X , λ0,...,λp) and Σ = (X , λ0,..., λp˜) p p˜ e such that p =p. ˜ e e e e 2. Given a λ-pair (Σ, Σ) we set:

• dim(Σ, Σ) := dime Σ = dim Σ and codim(Σ, Σ) := codim Σ = codim Σ. j j • If λ = λ = 0 for all i ∈ {1,...,p} we set deg(Σ, Σ) := ep. Otherwise, let j0 := min{j ∈ e j e j e e {1,...,p} such that λ 6=0 or λ 6=0} and set deg(Σ, Σ) = ej0 . e e e e (Σ,Σ) Σ Σ 3. With the above notation, we set Ω e := Ω ⊗ Ω . e

e 7 e e j (Σ,Σ) j (Σ,Σ) 4. For any a ∈ Z and any j ∈ N, we set H Ω (a) := H Xp, Ω ⊗ OXp (a) . To describe our fundamental exact sequence, we introduce some notion of successors. 0 p Definition 2.5. Take a λ-setting Σ = (Xp, λ ,...,λ ) with p > 1 and where for any 0 6 j 6 p one denotes j j j λ = (λ1,...,λmj ). We define λ-settings s1(Σ) and s2(Σ) as follows. j 0 p−1 • If λ =0 for all 1 6 j 6 p then set s1(Σ) := s2(Σ) := (Xp−1, λ ,...,λ ). j j • If there exists 1 6 j 6 p such that λ 6= 0, let j0 := min{j / 1 6 j 6 p and λ 6= 0} and let j0 i0 := min{i / 1 6 i 6 mj0 and λi 6=0}. Then we define

0 j0 j0 j0 j0+1 p s1(Σ) := (Xp, λ , 0,..., 0, (λ ), (λ ,...,λ ), λ ,...,λ ) i0 i0 +1 mj0

0 j0 j0 j0 j0+1 p s2(Σ) := (Xp, λ , 0,..., 0, (λ − 1), (λ ,...,λ ), λ ,...,λ ). i0 i0 +1 mj0

We will need the following generalization to λ-pairs. 0 p 0 p Deﬁnition 2.6. Take a λ-pair (Σ, Σ) where Σ = (Xp, λ ,...,λ ) and Σ = (Xp, λ ,..., λ ). Deﬁne s1(Σ, Σ) and s2(Σ, Σ) as follows. e e e e e • If λj = λj =0 for all 1 6 i 6 p set s (Σ, Σ) := s (Σ, Σ) := (s (Σ), s (Σ)) = (s (Σ), s (Σ)). e 1 2 2 2 1 1 6 6 j j > j j • If theree exists 1 j p such that λ 6=0eor λ 6=0 sete j0 := min{j e 1 / λ 6=0 or λe 6=0}.

j0 – If λ 6=0 set s1(Σ, Σ) = (Σ, s1(Σ)) ande s2(Σ, Σ) = (Σ, s2(Σ)). e j0 – If λ =0 set s1(Σ, Σ) = (s1(Σ), Σ) and s2(Σ, Σ) = (s2(Σ), Σ). e e e e e Now we come to an elementary, but important, observation. e e e e e Proposition 2.7. For any λ-pair (Σ, Σ), we have a short exact sequence e e e s (Σ,Σ) s (Σ,Σ) (Σ,Σ) 0 → Ω 2 (e− deg(Σ, Σ)) → Ω 1 → Ω → 0 (5)

0 p 0 p Proof. Take Σ = (Xp, λ ,...,λ ) and Σ = (Xp, λe ,..., λ ). We have to consider two cases.

1 p 1 p ′ ′ Case 1: λ = ··· = λ = λ = ···e= λ = 0e. Set Σe := s1(Σ) = s2(Σ) and Σ := s1(Σ) = s2(Σ) so that ′ e′ e Ω(Σ ,Σ ) = Ω(Σ,Σ). We have the restriction exact sequence |Xp e e e e e

e e 0 → OXp−1 (−ep) → OXp−1 → OXp → 0. e (Σ′,Σ′) Since ep = deg(Σ, Σ), it suﬃces to tensor this exact sequence by Ω to obtain the desired exact sequence.

j j j j Case 2: there existse j > 1 such that λ 6=0 or λ 6=0. Set j0e:= min{j > 1 / λ 6=0 or λ 6=0}. Suppose j0 j0 in a ﬁrst time that λ 6=0. Recall that ej0 = deg(Σ, Σ). Set also i0 := min{i / 1 6 i 6 mj0 and λi 6=0}. e ej0 e′ ′ 0 j0 j0 j0+1e p (Σ,Σ) Σ λe Σ Set Σ := (Xp, λ , 0,..., 0, (λ ,..., λ ), λ ,..., λ ), so that Ω = Ω ⊗ S i0 ΩX |X ⊗ Ω , i0+1 mj0 j0 p e eej0 e′ e e ej0 e′ s1(Σ,Σ) Σ λ Σ s2(Σ,Σ) Σ λ −1 Σ j0 Ω = Ω ⊗ S i0 ΩX |X ⊗ Ω and Ω = Ω ⊗ S i0 ΩX |X ⊗ Ω . By taking the λ -th e e j0−1 e p e e e j0−1 p e ie0 symmetric power of the tilde conormal exact sequence when Xj0 is seen as a hypersurface of Xj0−1 and restricting everythinge to Xp, we obtaine e e e

ej ej ej λ 0 −1 λ 0 λ 0 i0 i0 i0 0 → S ΩXj0−1 (−ej0 )|Xp → S ΩXj0−1 |Xp → S ΩXj0 |Xp → 0.

e′ It suﬃces now to tensor this exacte sequence by ΩΣ and eΩΣ to obtain the desirede result. If λj0 =0 with use the same decomposition on Σ (and the λj ’s) and we use the usual conormal exact sequence instead of the tilde conormal exact sequence. e

8 2.4 A vanishing lemma In this section, we prove a vanishing lemma which we will often use later. To be able to give the statement, we need some more notation. Given any m-uple of integers λ = (λ1,...,λm), we deﬁne nz(λ) = ♯{i such that λi 6=0} to be the number of non-zero terms in λ.

0 p j j j Deﬁnition 2.8. Take a λ-setting Σ = (Xp, λ ,...,λ ), where for all j ∈{0,...,p}, λ = (λ1,...,λmj ). Then we set

p mj j • q(Σ) := dim(Σ) − n(Σ), where n(Σ) := j=1 i=1 min{j, λi }. P pP j • i(Σ) := codim(Σ) + w(Σ), where w(Σ) := j=1 j nz(λ ).

p Pmj j p j • t(Σ) := |Σ|− nz(Σ), where |Σ| := j=0 i=1 λi and nz(Σ) := j=0 nz(λ ). We will also need the generalizationP to λ-pairs.P P

0 p 0 p Deﬁnition 2.9. Take a λ-pair (Σ, Σ) where Σ = (Xp, λ ,...,λ ), Σ = (Xp, λ ,..., λ ). Then we set

• q(Σ, Σ) := dim(Σ, Σ) − n(Σ, Σ)e , where n(Σ, Σ) := n(Σ) + n(Σ)e. e e

• i(Σ, Σ)e := codim(Σe, Σ) + w(Σe, Σ), where w(Σe, Σ) := w(Σ)+w(e Σ).

• t(Σ, Σ)e := |(Σ, Σ)|−enz(Σ), wheree |(Σ, Σ)| = |Σe| + |Σ|. e

Remark 2.10e . Let use just mention what thee purpose of thesee functions are. The integer q(Σ, Σ) will be the degree of the cohomology group on which we will get some description. The integer i(Σ, Σ) will be used as a counter in several induction arguments, and t(Σ, Σ) (as well as |Σ|) will be a bound on the twiste by OX (a) we can allow in the statements of our results. e It is straightforward but crucial to observe howe those notions behave with respect to the successors introduces in Section 2.3.

Proposition 2.11. For any λ-setting Σ0 and any λ-pair (Σ, Σ) we have:

1. q(s1(Σ0)) > q(Σ0) and q(s1(Σ, Σ)) > q(Σ, Σ). e

2. q(s2(Σ0)) = q(Σ0)+1 and q(s2e(Σ, Σ)) = qe(Σ, Σ)+1.

3. i(s1(Σ0))

4. i(s2(Σ0))

5. | s1(Σ0)| = |Σ0| and t(s1(Σ, Σ))e = t(Σ, Σ)e.

6. | s2(Σ0)| = |Σ0|− 1 and t(s2e(Σ, Σ)) > te(Σ, Σ) − 1.

We are now in position to state ande prove oure vanishing result. Lemma 2.12. Take a λ-pair (Σ, Σ). Take a ∈ Z and j ∈ N. If j < q(Σ, Σ) and a

e e Hj Ω(Σ,Σ)(a) =0. e e Remark 2.13. If we specialize this lemma to Σ = (X, 0,..., 0), Σ = (X, 0,..., 0, (ℓ1,...,ℓk)) and a =0, we j ℓ1 ℓk k obtain that H (X,S ΩX ⊗···⊗ S ΩX )=0 if j < dim X − i=1 min{c,ℓi}. This is the same conclusion as in Brückmann and Rackwitz theorem. Thereforee Lemma 2.12 can be seen as a generalization of their result. P

9 Proof of Lemma 2.12. We make an induction on i(Σ, Σ).

If i(Σ, Σ) = 0: Then Σ = (PN , λ0) and Σ = (PN , λe0), it is a straightforward induction on nz(λ0) using the symmetric powers of the Euler exact sequence. e e e If i(Σ, Σ) > 0: By Proposition 2.11 we can apply our induction hypothesis to s1(Σ, Σ) and s2(Σ, Σ). Observe also that if a

e e e 0 → Ωs2(Σ,Σ)(a − deg(Σ, Σ)) → Ωs1(Σ,Σ)(a) → Ω(Σ,Σ)(a) → 0. (8)

It then suﬃces to use (6) and (7) in the cohomologye long exact sequence associated to (8) to obtain the desired result.

2.5 Statements for the tilde cotangent bundle In this section we will prove the ﬁrst half of Theorem A, to be precise, we will describe the map ϕ˜ and give the announced description of im(ϕ ˜). In fact, to prove this result we will need to prove a more general statement. Before we state our results we need more notation and some more deﬁnitions. If ℓ ∈ N and if λ := (λ1,...,λk) is a k-uple of integers such that λi > ℓ for any i ∈{1,...,k} then we set

λ − ℓI := (λ1 − ℓ,...,λk − ℓ).

0 p 0 p If we have two λ-settings of the same dimension Σ1 = (Xp, λ ,...,λ ) and Σ2 = (Xp,µ ,...,µ ), we set

0 0 p p Σ1 ∪ Σ2 := (Xp, λ ∪ µ ,...,λ ∪ µ ).

Deﬁnition 0 p 6 6 j j j 2.14. Take a λ-setting Σ = (Xp, λ ,...,λ ) where for all 0 j p we denote λ = (λ1,...,λmj ). j We say that Σ is simple if for any 1 6 j 6 p and for any 1 6 i 6 mj, λi > j. More generally, we say that a λ-pair (Σ, Σ) is simple if Σ and Σ are simple.

It is easye to observe how simplicitye behaves with respect to the successors s1 and s2.

Proposition 2.15. If Σ is simple, then s1(Σ) and s2(Σ) are simple and moreover

q(s1(Σ)) = q(s2(Σ)) = q(Σ)+1.

Deﬁnition 2.16. If Σ is simple, we deﬁne

N 0 1 2 p Σlim := (P , λ ∪ (λ − I) ∪ (λ − 2I) ∪···∪ (λ − pI)), and p p p p j

b := ei 1+ mj = ei + mj ei. Σ   i=1 j=i i=1 j=1 i=1 X X X X X  

10 0 p Fix a simple setting Σ = (Xp, λ ,...,λ ) as above. Note that for any i ∈ {1,...,p} one has a natural morphism ·Fi OPN −→ OPN (ei) which induces a morphism Σlim ·Fi Σlim Ω (a) −→ Ω (a + ei). This induces an application in cohomology, e e

N Σlim ·Fi N Σlim H (Ω (a)) −→ H (Ω (a + ei)), which we will still denote ·Fi. This shoulde not lead to anye confusion.

Similarly, taking the symmetric powers of the application (2), for any i ∈{1,...,p} and for any m ∈ N one has a natural morphism m ·dFi m+1 S ΩPN −→ S ΩPN (ei). Now ﬁx j ∈{1,...,p} and k ∈{1,...,m }. Set j e e 1 0 1 j−1 j j η := λ ∪ (λ − I) ∪···∪ (λ − (j − 1)I) ∪ (λ1 − j,...,λk−1 − j) 2 j j j+1 p η := (λk+1 − j,...,λmj − j) ∪ (λ − (j + 1)I) ∪···∪ (λ − pI).

η1 j η2 Σlim λk−j So that Ω = ΩPN ⊗ S ΩPN ⊗ ΩPN . As above, we get a morphism

j j λ −j ·dFi λ −j+1 e e e eS k ΩPN −→ S k ΩPN (ei).

η1 η2 If we tensor this morphism by ΩPN , ΩPN ande OPN (a) we get ae morphism

{j,k} 1 j 2 ·dFi η η Σelim e λk −j+1 Ω (a) −→ ΩPN ⊗ S ΩPN ⊗ ΩPN (a + ei), which yields a morphism e e e e

{j,k} 1 j 2 ·dFi η η N Σlim N N λk−j+1 H Ω (a) −→ H P , ΩPN ⊗ S ΩPN ⊗ ΩPN (a + ei) . We are now in position toe state our result. e e e

0 p Theorem 2.17. Take a simple λ-setting Σ = (Xp, λ ,...,λ ) as above. Take an integer a < |Σ|. Then, there exists an injection ϕ˜ q(Σ) Σ N Σlim . (H Ω (a) ֒→ H Ω (a − bΣ Moreover, e e p p mj q(Σ) Σ {j,k} ϕ˜ H Ω (a) = ker(·Fi) ker ·dF .  i  i=1 j=i k \ \ \=1 e   To simplify our presentation, we will decompose the proof of Theorem 2.17 in a couple of propositions. The following proposition describes the map ϕ.˜

11 Proposition 2.18. With the notation of Theorem 2.17. Let q := q(Σ). There is a chain of inclusions:

(Hq(ΩΣ(a)) ֒→ Hq+1 Ωs2(Σ)(a − deg Σ

2 q+2 s2(Σ) ((e ֒→ H Ωe (a − degΣ − degs2(Σ ··· e k−1 k q+k s2 (Σ) i (H Ω a − deg s2(Σ →֒ i=0 !! X ··· e N−q−1 N−q N s2 (Σ) i (H Ω a − deg s2(Σ →֒ i=0 !! X e And moreover

N−q 1. s2 (Σ) = Σlim.

N−q−1 i 2. i=0 deg s2(Σ) = bΣ. Proof.PThis is an induction on i(Σ).

N 0 If i(Σ) = 0, the result is clear since Σ = (P , λ ) and therefore q = N, bΣ =0 and Σ=Σlim.

Now suppose i(Σ) > 0. Then we can suppose, by induction, that the result holds for s2(Σ) which is simple and satisﬁes q(s2(Σ)) = q(Σ)+1. Consider the exact sequence

0 → Ωs2(Σ)(a − deg Σ) → Ωs1(Σ)(a) → ΩΣ(a) → 0.

By Proposition 2.15 and Lemma 2.12e we obtain e e

Hj Ωs2(Σ)(a − deg Σ) = Hj Ωs1(Σ)(a) = 0 if j 6 q. By looking at the long exact sequencee induced in cohomologye we obtain an exact sequence

0 → Hq ΩΣ(a) → Hq+1 Ωs2(Σ)(a − deg Σ) → Hq+1 Ωs1(Σ)(a) . (9) Applying the induction hypothesise to s2(Σ) wee get the desired chain of inclusions.e N−q Now observe that s2(Σ)lim =Σlim, and therefore, by induction, we obtain s2 (Σ) = Σlim. 1 p To see the last point, we have to prove that bΣ = bs2(Σ) + degΣ. If λ = ··· = λ = 0, then degΣ = ep, p p−1 6 6 bΣ = i=1 ei and bs2(Σ) = i=1 ei, and therefore bΣ = bs2(Σ) + degΣ. Now if there exists 1 j p such j j that λ 6=0. Set, as usual, j0 := min{j > 1 / λ 6=0}, so that degΣ = ej0 . Then mj =0 for all 1 6 j < j0 and thereforeP P

p p j j0−1 j0 p p j bs2(Σ) = ei + mj ei + ei + (mj0 − 1) ei = ei + mj ei − ej0 = bΣ − degΣ. i=1 j=j +1 i=1 i=1 i=1 i=1 j=j +1 i=1 X X0 X X X X X0 X

The map q Σ N Σlim ϕ˜ : H Ω (a) → H Ω (a − bΣ) in Theorem 2.17 is just the composition ofe all the injectionse in Proposition 2.18. We will need an easy linear algebra lemma.

12 Lemma 2.19. Take a commutative diagram of vector spaces

g f1 A / B / C ❅❅ ❅❅ϕ ❅ h1 h2 ❅❅ f2 D / E

such that g,h1 and h2 are injective, such that f1 ◦ g =0 and such that g(A) = ker f1, then

ϕ(A)= h1(B) ∩ ker f2. The key observation is the following proposition. Proposition 2.20. With the notation of Theorem 2.17. Let q := q(Σ). There is a commutative diagram

ι ·Aq Hq ΩΣ(a) / Hq+1 Ωs2(Σ)(a − deg Σ) / Hq+1 Ωs1(Σ)(a) e e e . . . .

k k−1 ·Aq+k k−1 k−1 q+k s2 (Σ) i / q+k s2 s1(Σ) i H Ω a − i=0 deg s2(Σ) / H Ω a − i=1 degs2(Σ) e P e P k+1 k ·Aq+k+1 k k q+k+1 s2 (Σ) i / q+k+1 s2 s1(Σ) i ϕ˜ H Ω a − i=0 degs2(Σ) / H Ω a − i=1 deg s2(Σ) P P e e . . . .

( N−q N−q−1 ·AN N−q−1 N−q−1 N s2 (Σ) i / N s2 s1(Σ) i H Ω a − i=0 deg s2(Σ) / H Ω a − i=1 degs2(Σ) P P where all the vertical arrowse and the map ι are injective. The maps ·Aqe+k are described as follows. Let 0 p 1 p Σ = (Xp, λ ,...,λ ). If λ = ··· = λ =0 the map ·Aq+k is just the map ·Fp induced by OPN → OPN (ep). j j If there exists 1 6 j 6 p such that λ 6= 0, set j0 := min{j > 1/λ 6= 0}, then the maps ·Aq+k are maps induced by ·dFj0 so that ·Aq is the one induced by the map appearing in the exact sequence (9) and the map {j0,1} ·AN is just ·dFj0 . The proof of Theorem 2.17 now easily follows by induction from Proposition 2.20 and Lemma 2.19. Proof of Theorem 2.17. We proceed by induction on i(Σ). If i(Σ) = 0, there is nothing to prove. Now suppose i(Σ) > 0. With the same notation as above. Thanks to Proposition 2.20 we obtain the following commutative diagram, whose vertical arrows are injective.

ι ·Aq Hq ΩΣ(a) / Hq+1 Ωs2(Σ)(a − deg Σ) / Hq+1 Ωs1(Σ)(a) ◗◗◗ ◗◗◗ ϕ˜ ◗◗ ϕ˜2 e ◗◗◗ e e ◗◗( N−q ·AN N−q−1 N s (Σ) / N s s1(Σ) H Ω 2 (a − bΣ) / H Ω 2 (a − bΣ + deg(Σ)) e e 13 By Lemma 2.19 we obtain imϕ ˜ = imϕ ˜2 ∩ ker(·AN ). Now it suﬃces to apply the induction hypothesis to s2(Σ) to obtain the announced description of imϕ ˜2, which induces the announced description for imϕ. ˜ We now give the proof of the proposition.

0 p Proof of Proposition 2.20. Let Σ = (Xp, λ ,...,λ ). We need to treat two cases.

1 p Case 1: λ = ··· = λ =0, then s1(Σ) = s2(Σ). Therefore one has for any k > 0 a commutative diagram

0 0

k+1 k ·Fp k k s2 (Σ) i / s2 s1(Σ) i Ω a − i=0 degs2(Σ) / Ω a − i=1 degs2(Σ) e P e P k k−1 ·Fp k−1 k−1 s1 s2 (Σ) i / s1 s2 s1(Σ) i Ω a − i=0 deg s2(Σ) / Ω a − i=1 deg s2(Σ) e P e P k k−1 ·Fp k−1 k−1 s2 (Σ) i / s2 s1(Σ) i Ω a − i=0 deg s2(Σ) / Ω a − i=1 deg s2(Σ) e P e P 0 0

where the horizontal maps are just multiplication by Fp. Here the vertical exact sequences come from the k k k−1 k+1 k exact sequence (9) applied to s2(Σ). Since Σ is simple, we see that s2 (Σ), s2 s1(Σ), s2 (Σ), s2 s1(Σ), k k−1 s1 s2(Σ) and s1 s2 s1(Σ) are simple, and that

k k−1 k+1 k k k−1 q(s2 (Σ)) = q(s2 s1(Σ)) = q(Σ)+k and q(s2 (Σ)) = q(s2 s1(Σ)) = q(s1 s2 (Σ)) = q(s1 s2 s1(Σ)) = q(Σ)+k+1.

By considering the diagram in cohomology associated to the above diagram and by applying Lemma 2.12, we obtain the following commutative diagram

0 0

k k−1 ·Fp k−1 k−1 q+k s2 (Σ) i / q+k s2 s1(Σ) i H Ω a − i=0 degs2(Σ) / H Ω a − i=1 deg s2(Σ) e P e P k+1 k ·Fp k k q+k+1 s2 (Σ) i / q+k+1 s2 s1(Σ) i H Ω a − i=0 deg s2(Σ) / H Ω a − i=1 degs2(Σ) P P now we put all thosee squares together to obtain our claim. e

j j Case 2: there exists 1 6 j 6 p such that λ 6= 0. Set j0 := min{j > /λ 6= 0}. First let us explain what k k−1 s (Σ) s s1(Σ) the maps ·Aq+k are. Observe that for any 1 6 k 6 N − q, the bundles Ω 2 and Ω 2 are of the form

k s2 (Σ) Σ1 ℓ Σ2 Ω = Ω ⊗ S ΩXi ⊗ Ω e e k−1 s s1(Σ) Σ1 ℓ+1 Σ2 Ω 2 = Ω ⊗ S ΩX ⊗ Ω e e e i e e e e e 14 where Σ1 and Σ2 are simple. The map ·dFj0 is then the map induced by IdΣ1 ⊗ ·dFj0 ⊗ IdΣ2 . As before we have a commutative diagram

0 0

k+1 k ·dFj0 k k s2 (Σ) i / s2 s1(Σ) i Ω a − i=0 degs2(Σ) / Ω a − i=1 degs2(Σ) P P e ·B e ·B k k−1 ·dFj0 k−1 k−1 s1 s2 (Σ) i / s1 s2 s1(Σ) i Ω a − i=0 deg s2(Σ) / Ω a − i=1 deg s2(Σ) e P e P k k−1 ·dFj0 k−1 k−1 s2 (Σ) i / s2 s1(Σ) i Ω a − i=0 deg s2(Σ) / Ω a − i=1 deg s2(Σ) e P e P 0 0

k where the map ·B comes from the short exact sequence (9) applied to s2(Σ). More precisely, the map ·B will be either induced by ·Fj for some j, either induced by ·dFj for some j. If ·B = ·Fj then the commutativity is clear. If ·B is induced by some map ·dFj then there are two cases to consider. Recall that by construction each of those maps will be the identity on all except one term in the tensor product. So either the maps ·B and ·A act on the same factor of the tensor product, either they act on different factors. If they act on different factors, the commutativity is clear. If they act on the same factor it suffices to use the commutativity of the following diagram:

/ ℓ−1 ·dFi / ℓ / ℓ / 0 / S ΩXi−1| (−2ei) / S ΩXi−1| (−ei) / S ΩXi (−ei) / 0 Xi Xi

dF dF dF e · j0 e · j0 e · j0 / ℓ ·dFi / ℓ+1 / ℓ+1 / 0 S ΩXi−1| (−ei) S ΩXi−1| S ΩXi 0. Xi Xi

Now the rest of the proof followse as in the ﬁrst case bye looking at what happense in cohomology.

2.6 Twisting the Euler exact sequence

From the previous section, we have a good understanding of the groups Hq(Σ) ΩΣ(a) when Σ is simple. Now we want to use this to deduce a similar description for the groups Hq(Σ) ΩΣ(a) . To do this, we will use cohomological technics similar to the ones we used in the previous sections,e but instead of building everything on the restriction exact sequence and the conormal exact sequence, we will use the Euler exact sequence. Again, everything will be based on a suitable exact sequence, and to deﬁne it, we need another way of taking successors for simple pairs.

0 p 0 p Deﬁnition 2.21. Take a simple λ-pair (Σ, Σ) such that Σ := (Xp, λ ,...,λ ), Σ := (Xp, λ ,..., λ ) and j j0 such that nz(Σ) 6=0. Set j0 := min{j > 0 / λ 6=0}, i0 := min{i > 1 / λi 6=0} and let e e e e ′ j0 j0 j0+1 p Σ := (Xp, 0,..., 0, (λ ,...,λ ), λ ,...,λ ) i0+1 mj0

′ 0 j0−1 j0 j0 j0+1 p Σ := (Xp, λ ,..., λ , λ ∪ (λi0 ), λ ,..., λ )

′′ 0 j0−1 j0 j0 j0+1 p Σ = (Xp, λ ,..., λ , λ ∪ (λ − 1), λ ,..., λ ), e e e e i0 e e e e e e e e 15 and set ′ ′ ′ ′′ h1(Σ, Σ) := (Σ , Σ ) and h2(Σ, Σ) := (Σ , Σ ). We make an elementary observation. e e e e

Proposition 2.22. If (Σ, Σ) is a simple pair such that nz(Σ) 6= 0, then h1(Σ, Σ) is simple, q(h1(Σ, Σ)) = q(Σ, Σ), and q(h2(Σ, Σ)) > q(Σ, Σ). e e e The following proposition is crucial to us. e e e Proposition 2.23. With the above notation. 1. The Euler exact sequence (3) yields an exact sequence

e E e e 0 → Ω(Σ,Σ) → Ωh1(Σ,Σ) → Ωh2(Σ,Σ) → 0.

2. Suppose codim(Σ, Σ) > 0. For any 1 6 i, j 6 2, there is an isomorphism

e e e Ωhi sj (Σ,Σ) =∼ Ωsj hi(Σ,Σ).

3. Suppose codim(Σ, Σ) > 0. Set b := deg(Σ, Σ). There is a commutative diagram 0 0 0 e e

e e e s (Σ,Σ) s h (Σ,Σ) s h (Σ,Σ) 0 −−−−→ Ω 2  (−b) −−−−→ Ω 2 1  (−b) −−−−→ Ω 2 2  (−b) −−−−→ 0 y y y

e e e s (Σ,Σ) s h(Σ,Σ) s h(Σ,Σ) 0 −−−−→ Ω 1 −−−−→ Ω 1 1 −−−−→ Ω 1 2 −−−−→ 0 y y y (10)

e e e (Σ,Σ) h (Σ,Σ) h (Σ,Σ) 0 −−−−→ Ω  −−−−→ Ω 1 −−−−→ Ω 2 −−−−→ 0 y y y

   0 0 0 y y y

Proof. The proof is very similar to what was done in the previous sections, and we give only a rough outline of it.

j0 1. With the notation of the deﬁnition of h1 and h2, it suﬃces to take the λi0 ’s power of the Euler exact sequence to get j j j λ 0 λ 0 λ 0 −1 i0 i0 i0 0 → S ΩXj0 → S ΩXj0 → S ΩXj0 → 0.

Then one just has to tensor this by a suitable tensore product of symmetrice powers of Ω’s and Ω’s. 2. One just has to check the diﬀerent cases. Most of the time the announced isomorphism is juste the identity, but in some cases, the isomorphism is a reordering of some factors in the tensor product under consideration. 3. One takes the symmetric powers of Diagram (4) for suitable X and Y , and twist it by a suitable tensor product of symmetric powers of Ω′s and Ω′s.

16 2.7 Statements for the cotangent bundle We are now in position to state and prove the second half of Theorem A. Observe that for any simple λ-setting e E e e (Σ, Σ), the Euler exact sequence (3) yields a morphism Ω(Σ,Σ) → ΩΣ∪Σ. Observe also that ΩΣlim∪Σlim =∼ e Ω(Σ∪Σ)lim (just as in Proposition 2.23, this isomorphism is just a reordering of the diﬀerent factors of the e e tensore product). Therefore, this induces a map E :ΩΣlim ⊗ ΩΣlim → Ωe(Σ∪Σ)lim . As usual we wille also denote bye E the map induced between the diﬀerent cohomology groups. We have the following. e Theorem 2.24. If (Σ, Σ) is a simple pair and a

e e q(Σ,Σ) (Σ,Σ) N Σ e e / e lim Σlim e H Ω (a) H Ω ⊗ Ω (a − bΣ − bΣ) ❯❯❯ ❯❯❯ ϕ E ❯❯❯ E ❯❯❯❯ e ❯❯❯* q ,e e ϕ˜ N e (Σ Σ) Σ∪Σ / (Σ∪Σ)lim e H Ω (a) H Ω (a − bΣ − bΣ) where all the arrows are injective ande where ϕ˜ is the map frome Theorem 2.17. Moreover, im ϕ = imϕ ˜∩im E . We would like to point out a special case of particular interest (which was denoted by Theorem A in the introduction).

Corollary 2.25. Take integers ℓ1,...,ℓk > c take an integer a<ℓ1 + ··· + ℓk − k, let q := n − kc and c b := (k + 1) i=1 ei. Then one has a commutative diagram P q (ℓ1,...,ℓk) / N N (ℓ1−c,...,ℓk−c) H X, ΩX (a) / H P , ΩPN (a − b) ❯❯ ❯❯❯❯ ϕ E ❯❯❯ E ❯❯❯❯ ❯❯❯* ϕ˜ q (ℓ1,...,ℓk) / N N (ℓ1−c,...,ℓk−c) H X, ΩX (a) / H P , ΩPN (a − b) Such that: e e

c k {j} 1. imϕ ˜ = i=1 ker(·Fi) j=1 ker ·dFi . 2. im ϕ = imT ϕ ˜ ∩ im E . T

{j} In the corollary, the map ·dFi is just the map

dF {j} N N (ℓ1−c,...,ℓk−c) · i N N (ℓ1−c,...ℓj −c+1,ℓk−c) H P , ΩPN (a − b) → H P , ΩPN (a − b + ei) induced by the map e e ℓj −c ·dFi ℓ−c+1 S ΩPN → S ΩPN (ei). Remark 2.26. Let us consider more precisely the case k = 1 in Corollary 2.25, suppose for simplicity that e e N =2N0 is even. At ﬁrst sight, it might seem that if c>n this result doesn’t tell us anything, but in fact, we can still get some information by using a simple trick. Indeed, suppose we want to construct symmetric

diﬀerential forms on X, then it suﬃces to write Y = H1 ∩···∩ HN0 and X = Y ∩ HN0+1 ∩···∩ Hc. 0 m Corollary 2.25 then gives us information on H (Y,S ΩY ). And if we are able to construct an element 0 m 0 m ω ∈ H (Y,S ΩY ) one can consider the induced restriction ωX ∈ H (X,S ΩX ). Similar arguments can by done when N is odd. We would like to mention that in the proof of our main application in Section 4, we will in fact use a similar trick with Theorem 2.17 and not only Corollary 2.25.

17 0 p 0 p Proof of Theorem 2.24. Take a simple λ-pair (Σ, Σ) where Σ = (Xp, λ ,...,λ ) and Σ = (Xp, λ ,..., λ ). We make an induction on nz(Σ). Let q := q(Σ, Σ). If nz(Σ) = 0 there’s nothing to prove. We now suppose that nz(Σ) 6=0. Diagram (10) yields a commutativee square e e e e e e Hq Ω(Σ,Σ) −−−−→ Hq Ωh1(Σ,Σ)

e e Hq+1 Ωs2(Σ,Σ)(− deg(Σ, Σ)) −−−−→ Hq+1 Ωs2 h1(Σ,Σ) (− deg(Σ, Σ)) y y where, by applying Lemma 2.12, all the arrowse are injective. Now set e

j−1 j k e q+i q+i s2h1 (Σ,Σ) ℓ Hj,k := H Ω a − deg s2(Σ, Σ) . ℓ !! X=0 e Putting all the above cartesian squares together, we obtain the following commutative diagram whose arrows are all injective.

q q q q q H0,0 −−−−→ H0,1 −−−−→ · · · −−−−→ H0,k −−−−→ H0,k+1 −−−−→ · · · −−−−→ H0,nz(Σ)

  . . . q+1 q+1 . . . H1y,0 −−−−→ H1y,1 −−−−→ · · · y. y. y.

. . . . . . . . y. y. y. y.

q+j q+j q+j H −−−−→ · · · · · · −−−−→ H −−−−→ H  −−−−→ · · · (11) j,y0 j,ky j,ky+1

q+j+1 q+j+1 q+j+1 H  −−−−→ · · · · · · −−−−→ H  −−−−→ H  −−−−→ · · · j+1y ,0 j+1y ,k j+1y,k+1

. . . . . . . . y. y. y. .

N N  H  −−−−→ · · · · · · · · · −−−−→ H  Ny−q,0 N−yq,nz(Σ) Observe that e q ∼ q Σ∪Σ H0,nz(Σ) = H (Ω (a)) N N Σ Σ H =∼ H (Ω lim ⊗ Ω lim (a − b − be )) N−q,0 e Σ Σ e N N Σ ∪Σ H =∼ H (Ω lim lim (a − b − be )). N−q,nz(Σ) e Σ Σ We would like to point out that to make the proofe completely precise, one should take into account the diﬀerent isomorphisms coming from Proposition 2.23.2. But for simplicity we neglect those details here. Set q N q N ϕ to be the map H0,0 → HN−q,nz(Σ). Observe also that the composed map H0,nz(Σ) → HN−q,nz(Σ) is just the map ϕ˜ from Theorem 2.17. This gives us the commutative diagram of Theorem 2.24. The proof of the

18 rest of the statement is similar to the proof of Theorem 2.17. When one looks at the long exact sequence associated to Diagram 10, we get a commutative diagram

e e e 0 −−−−→ Hq Ω(Σ,Σ)(a) −−−−→ Hq Ωh1(Σ,Σ)(a) −−−−→ Hq Ωh2(Σ,Σ)(a)

 e  e  e 0 −−−−→ Hq+1 Ωs2(Σ,Σ)(a − b) −−−−→ Hq+1 Ωs2 ◦h1(Σ,Σ)(a − b) −−−−→ Hq+1 Ωs2 ◦h2(Σ,Σ)(a − b) y y y where b := deg(Σ, Σ). A quick induction yields a commutative diagram

e Eq e e 0 / Hq eΩ(Σ,Σ)(a) / Hq Ωh1(Σ,Σ)(a) / Hq Ωh2(Σ,Σ)(a) ❯❯❯❯ ❯❯❯ ϕ1 ❯❯❯ ϕ2 ❯❯❯❯ ❯❯❯* E N N−q ,e N N N−q h ,e τ N N−q h ,e / s2 (Σ Σ) e / s2 1(Σ Σ) e / s2 2(Σ Σ) e 0 H Ω (a − bΣ,Σ) H Ω (a − bΣ,Σ) H Ω (a − bΣ,Σ) e e Where bΣ,Σ = bΣ + bΣ. By Lemma 2.19 we obtain

im ϕ1 = im ϕ2 ∩ ker(τ) = im ϕ2 ∩ im EN .

From Diagram 11 one can extract the commutative diagram

ϕ q / q / q H0,0 / H0,1 / H ■ ❑❑ 0,nz(Σ) ■■ ❑❑ ′ ■■ϕ1 ❑❑ϕ ■■ ϕ2 ❑❑ ϕ˜ ■■$ ❑❑❑% E $ E ′ % N N / N / N HN−q,0 HN−q,1 HN−q,nz(Σ) 8 E

′ ′ Our induction hypothesis is that im ϕ = im E ∩ imϕ. ˜ Using the fact that im ϕ1 = im ϕ2 ∩ im EN , we get

′ ′ ′ ′ ′ im ϕ = E (im ϕ1)= E (im EN ) ∩ E (im ϕ2) = im E ∩ im ϕ = im E ∩ imϕ ˜ ∩ im E = im E ∩ imϕ. ˜

3 Applications

We give different applications of theorems 2.17 and 2.24. These two statements basically give us a way of computing different cohomology groups on complete intersection varieties by reducing the problem to the computation of the kernel of a linear map depending (in some explicit way) on the defining equations of our complete intersection. In general computing this kernel is a difficult question because of the dimension of the spaces that are involved. However, in some special cases, one can make those computations, and this gives us some noteworthy conclusions.

3.1 Explicit computation in Čech cohomology In this section, we explain how to make theorems 2.17 and 2.24 explicit via the use of Čech cohomology. N We will use the standard homogenous coordinates [Z0 : ... : ZN ] on P and the standard aﬃne subsets N Ui := (Zi 6= 0) ⊂ P . Let U := (Ui)06i6N .

19 Remark 3.1. In some situations it is more natural to use other open coverings of PN more suitable to the geometric problem under consideration. See for instance Section 3.2 for an illustration of this. Recall, see for example [11], that if a > N +1 then

N N N 1 H (P , OPN (−a)) =∼ Hˇ (U, OPN (−a)) =∼ · C (12) Zi0 ··· ZiN i0+···+iN =a 0 N i0,...,iMN >1

Recall also that N

ΩPN = OPN (−1)dZi = V ⊗ OPN (−1), i=0 M N e where V = i=0 C · dZi. Now take integers ℓ1,...,ℓk > 0 and a<ℓ1 + ··· + ℓk − N − 1. We have

ℓ ,...,ℓ L N N ( 1 k) ∼ (ℓ1,...,ℓk) N N H P , ΩPN (a) = V ⊗ H P , OPN (a − ℓ1 −···− ℓk) (ℓ1,...,ℓ ) 1 e =∼ V k ⊗ · C. Zi0 ··· ZiN i0,...,iN >1 0 N i0+···+iN =Mℓ1+···+ℓk−a

N N (ℓ1,...,ℓk) Therefore an element of H P , ΩPN (a) can be thought of as an element of the form e dZJ1 ⊗···⊗ dZJk ω = ωI . J1,...,Jk ZI NN+1 J1,...,JXk,I∈ |J1|=ℓ1,...,|Jk|=ℓk |I|=ℓ1+···+ℓk −a I>I

I N+1 Where ωJ1,...,Jk ∈ C and where I := (1,..., 1) ∈ N . If K = (k0,...,kN ) and J = (j0,...,jN ) are both N+1 in N we write K > J if ki > ji for any 0 6 i 6 N. We also use the standard multi-index notation: if N+1 I i0 iN I i0 iN I = (i0,...,iN ) ∈ N then Z := Z0 ··· ZN and dZ := dZ0 ··· dZN .

M N N We now describe explicitly the maps ·F and ·dF . Start with a monomial Z ∈ H (P , OPN (e)) of N+1 M degree e, where M ∈ N . Take λ = (ℓ1,...,ℓk) as above. The multiplication by Z induces the map

N N λ N N λ H P , ΩPN (a) → H P , ΩPN (a + e) J1 Jk J J dZ ⊗···⊗dZ 1 k I M if I + > M ω = dZ ⊗···⊗dZ 7→ ω · ZM = Z − I eZI 0e if I + I M. 0 N If we take any F ∈ H (P , OPN (e)), it suﬃces to decompose F as a sum of monomials and extend the above description by linearity.

M 0 N Now consider an element ξ = Z dZi ∈ H P , ΩPN (e) . Fix 1 6 j 6 k. This induces a map N N (ℓ1,...,ℓk) N N (ℓ1,...,ℓj +1,...,ℓk) H P , ΩPN (a) → He P , ΩPN (a + e) J1 Jj J dZ ⊗···⊗dZidZ ⊗···⊗dZ k J1 Jk > dZ ⊗···⊗dZ {j} ZI−M if I + I M ω = e I 7→ ω · ξ = e Z 0 if I + I M. {j} N N again, it suﬃces to extend by linearity to describe the maps ·dF for any j and any F ∈ H (P , OPN (e)). The maps E are understood very similarly and we don’t provide all the details here. However, because in our computations we will have to use the coboundary map in the long exact sequence in cohomology associated to a short exact sequence, we would like to recall how such a coboundary map can be understood. Suppose

20 that X ⊆ PN is a projective variety, that U is an open covering of X and that one has the following exact sequence of sheaves on X ϕ ρ 0 →F →E → G → 0 This gives us the following maps between the Čech complexes:

......

dˇ dˇ dˇ ϕ ρ Ck(U, F) k / Ck(U, E) k / Ck(U, G)

dˇ dˇ dˇ ϕk+1 ρk+1 Ck+1(U, F) / Ck+1(U, E) / Ck+1(U, G)

ˇ ˇ ˇ d d d ......

k k+1 Here dˇ denotes the Čech differential. The coboundary map δk : Hˇ (U, G) → Hˇ (U, F) is obtained by applying the snake lemma in the above diagram (we suppose that U is sufficiently refined so that one can make this work). The problem we will be facing is the following: suppose that δk is injective and that we k+1 U ˇ k+1 U have (σi0 ,...,ik ) ∈ Z ( , F) representing a class σ ∈ H ( , F) such that σ ∈ im(δk), how to compute a −1 k U Čech representative for δk (σ) in C ( , G)? This is just a diagram chase, and goes as follows: first compute k U ˇ ϕk+1(σi0 ,...,ik ), by hypothesis, there exists (τi0 ,...,ik ) ∈ C ( , E) such that ϕk+1(σi0,...,ik ) = d(τi0 ,...,ik ). And −1 then δk (σ) is just represented by the cocycle ρk(τi0 ,...,ik ). For us the maps ϕk will be either multiplication by F or multiplication by dF for some polynomial F and ρk will just by a restriction map.

3.2 The case of plane curves Let us just explain how we can use this strategy to construct the “classical” diﬀerential form on smooth 2 curves in P of genus greater than 1. Let F ∈ C[Z0,Z1,Z2] be a homogenous polynomial of degree e > 3. 2 ∂F F Such that the curve C := (F = 0) ⊆ is smooth. For any i ∈{0, 1, 2}, set Fi := and U := (Fi 6= 0) P ∂Zi i UF F UF 2 and := (Ui )06i62. Because C is smooth, is an open covering of P . From Proposition 2.18 we get a diagram whose arrows are all injective

0 ϕ˜0 / 1 H (C, ΩC ) / H (C, OC (−e)) ◗◗◗ ◗◗◗ ◗◗◗ ϕ˜1 e ϕ˜ ◗◗ ◗◗( 2 2 H (P , OP2 (−2e))

By Corollary 2.25 we get that im(ϕ ˜) = ker(·F ) ∩ ker(·dF ). We will work in Čech cohomology with respect to F P,2 2 2 U . For any degree e − 3 polynomial P ∈ C[Z0,Z1,Z2] consider the element ω ∈ H (P , OP2 (−2e)) given by the following cocycle: P P ˇ 2 UF ω012 := ∈ H ( , OP2 (−2e)). e F0F1F2 By Euler’s formula we have eF = Fe0Z0 + F1Z1 + F2Z2. Hence, we obtain

P P 1 P Z0 Z1 Z2 2 F ω · F = · (F Z + F Z + F Z )= + + =0 ∈ Hˇ (U , OP2 (−e)) 012 e F F F 0 0 1 1 2 2 e F F F F F F 0 1 2 1 2 0 2 0 1 e

21 And because dF = F0dZ0 + F1dZ1 + F2dZ2 we obtain similarly

P dZ0 dZ1 dZ2 2 F ω · dF = P + + =0 ∈ Hˇ (U , ΩP2 (−e)) 012 F F F F F F 1 2 0 2 0 1 hence, we see that eωP,2 ∈ im(ϕ ˜). This already proves the existence of ae tilde differential form on C and in fact, Corollary 2.25 even proves that this tilde differential form will yield a true differential form simply E 2 2 −1 P,2 because, with the notatione of Corollary 2.25, im( ) = H (P , OP2 (−e)). It remains to compute ϕ˜ (ω ), −1 P,2 to do so, we first compute ϕ˜1 (ω ). This is done as follows: e P P Z0 Z1 Z2 P P 2 P ω · F = + + = ω − ω + ω = dˇ (ω ) 6 6 012 e F Fe F F F F 12 02 01 ij 0 i

P k P ZkdZi ZkdZj (ZiFi + Zj Fj )dZk k P 1 Zk Zi 1 Zk Zj ωij · dF = (−1) + − = (−1) + e F F F F e F dZk dZi F dZk dZj j i i j j i

For any i ∈{0, 1, 2}, take j, k ∈{0, 1}\{i} such that j < k, and set e P i P Zj Zk ωi := (−1) . eF dZj dZk i

P ˇ P P,0 With this formula, it is straighforwarde to check that (ωij · F )06i

Z0dZi − ZidZ0 1 Z0 Zi dzi = 2 = 2 . Z Z dZ0 dZi 0 0

Hence if i ∈{1, 2} and j ∈{1, 2}\{i} we obtain

e−3 P i P Z0 Zj i Z0 Q 2 i 1 Q ωi = (−1) = (−1) e−1 Z0 dzi = (−1) dzj . eF dZ0 dZj e f i eZ0 fi i

P,0 0 From this we obtaine that ω |U 0 ∈ H (U0 ∩ C, ΩC ) is represented in Čech cohomology with respect to (U ′ ∩ C,U ′ ∩ C) by the cocycle (−Q dz2 ,Q dz1 ), as expected. 1 2 f1 f2 Remark 3.2. It is true that this computation is longer than the usual one, however the strategy is a bit different. Indeed, the classical approach consists in finding a differential form locally (on U0 ∩ C) which is done by a “guessing” process, after what one checks that the constructed differential form extends on C. Here we somehow approach the problem the other way around, using Corollary 2.25, once we have found the P element ω012, we already know that we have a global differential form on C, and the entire computation just comes down to compute a local representative of it, this process is long but of a mechanical nature. In our opinion this second strategy reduces the importance of the “guessing” part and is therefore more suitable for higher dimensional generalizations.

22 3.3 Optimality in Brückmann and Rackwitz theorem We are now in position to prove optimality in Theorem 1.1. By applying theorems 2.17 and 2.24 to a very particular complete intersection, we will prove the following (so called Theorem B in the introduction).

Proposition 3.3. Let N > 2, let 0 6 c 1 and a<ℓ1 + ··· + ℓk − k. k N Suppose 0 6 q := N − c − i=1 min{c,ℓi}. Then, there exists a smooth complete intersection variety in P of codimension c, such that P q ℓ1 ℓk H (X,S ΩX ⊗···⊗ S ΩX (a)) 6=0. We will construct an example as follows. Take a c × (N + 1) matrix

a10 ··· a1N . . . A =  . .. .  , a ··· a  c0 cN    where the aij ∈ C are such that for any p ∈{1,...,c}, the p × p minors of the matrix A are non zero. Fix an integer e > 1. For each p ∈{1,...,c}, set

N e Fp := apj Xj and Xp := (F1 = 0) ∩···∩ (Fp = 0). j=0 X

One easily check that Xi is smooth for any i ∈{1,...,c}. We will show that if we take e ≫ 1 this example is suﬃcient to prove Proposition 3.3

3.3.1 The simple case

0 c We ﬁrst treat the simple case. Take a simple setting Σ = (Xc, λ ,...,λ ), if for any 0 6 j 6 c one denotes j j j j > 6 6 c λ = (λ1,...,λmj ) where λk j for any 1 k mj. In our situation, one obtains: n(Σ) = j=1 jmj , c c mj j nz(Σ) = j=1 mj , |Σlim| = j=0 k=1(λk − j) and bΣ = e(c + n(Σ)). The statement is the following.P

PropositionP 3.4. Take X =PXc andP Σ as above. Take a

N +1+2|Σ |− a e > lim then Hq(Σ)(ΩΣ(a)) 6=0. q(Σ)+1

Proof. It suﬃces to apply theorems 2.17 and 2.24 to a non-zero element of the form

c mj P j λk −j N N Σlim ω := e−1 e−1 (Z0dZ1 − Z1dZ0) ∈ H P , ΩPN (a − bΣ) . Z0 ··· ZN j=0 k O O=1 e Where P ∈ C[Z0,...,ZN ] is a homogenous polynomial of degree (q(Σ)+1)e+a−N −1−2|Σlim|. The condition on the degree just insures that such an element exists. Our statement follows at once using theorems 2.17 N e N e−1 and 2.24 and the description of Section 3.1 with Fp = j=0 apj Zj and dFp = e j=0 apj Zj dZj . P P 3.3.2 The general case To complete the proof of Proposition 3.3 we also have to treat the non-simple case. If Σ is not simple, we can not apply directly theorems 2.17 and 2.24. We need another type of successors. Take any setting 0 p j j j 6 6 Σ = (Xp, λ ,...,λ ) where λ = (λ1,...,λmj ). Suppose, without loss of generality, that for all 1 j p j and for all 1 6 i 6 mj we have λi > 1. And suppose also that Σ is not simple. We deﬁne the successors

23 j c1 and c2 as follows: consider j0 := max 1 6 j 6 p such that ∃ 1 6 i 6 mj with λi < j and i0 := j0 n o min 1 6 i 6 mj0 such that λi < j0 . Set n o 0 j0−2 j0 −1 j0 j0+1 p c1(Σ) := Xp, λ ,...,λ , λ ∪ (λi0 ),µ,λ ,...,λ

0 j0−2 j0 −1 j0 j0+1 p c2(Σ) := Xp, λ ,...,λ , λ ∪ (λi0 − 1),µ,λ ,...,λ j0 j0 j0 j0 ′ where µ := λ ,...,λ , λ ,...,λ . With those notation set also deg Σ := ej . As in Proposition 1 i0−1 i0+1 mj0 0 2.7 we obtain an exact sequence 0 → Ωc2(Σ)(− deg′ Σ) → Ωc1(Σ) → ΩΣ → 0.

We can now prove the following. Proposition 3.5. Let Σ be any setting, and denote q := q(Σ). For any integer a

′ . (Hq ΩΣ (a) ֒→ Hq ΩΣ(a Proof. Of course, we can suppose that Σ is not simple. Observe that q(c2(Σ)) = q(Σ)+1 = q +1. Therefore, applying Lemma 2.12, we obtain Hq Ωc2(Σ)(a − deg′ Σ) =0. This yields an injection . (Hq Ωc1(Σ)(a) ֒→ Hq ΩΣ(a

′ r It suﬃces now to observe that after a suﬃcient number r of iterations, we obtain a simple setting Σ = c1(Σ) such that q(Σ′)= q. By induction we therefore get a chain of injections

′ r . (Hq ΩΣ (a) = Hq Ωc1(Σ)(a) ֒→ · · · ֒→ Hq Ωc1(Σ)(a) ֒→ Hq ΩΣ(a

The proof of Proposition 3.3 now follows directly from propositions 3.4 and 3.5.

3.4 Examples for the non-invariance under deformation Applying theorems 2.17 and 2.24 to a very particular family of complete intersection surfaces we will prove 0 m that the numbers h (X,S ΩX ) are not deformation invariant as soon as m > 2. Our example is the 2 2 following. Take α := (α1, α2) ∈ C and β := (β1,β2) ∈ C . Take a0,...,a4 ∈ C such that ai 6= aj if i 6= j. > e e Take an integer e 5 and set e1 = ⌊ 2 ⌋ and e2 = ⌈ 2 ⌉. Set

e e e e e e1 e2 e1 e2 Fα := Z0 + Z1 + Z2 + Z3 + Z4 + α1Z0 Z1 + α2Z2 Z3 e e e e e e1 e2 e1 e2 Gβ := a0Z0 + a1Z1 + a2Z2 + a3Z3 + a4Z4 + β1Z0 Z1 + β2Z2 Z3 . And set Xα,β = (Fα = 0) ∩ (Gβ = 0). Proposition 3.6. With the above notation, we have:

0 2 1. h (X0,0,S ΩX0,0 ) 6=0.

0 2 2. For generic α and β, h (Xα,β,S ΩXα,β )=0.

24 0 2 Proof. The fact that h (X0,0,S ΩX0,0 ) 6=0 is a very particular case of Proposition 3.4. The rest of the proof is a straightforward computation, but we give it for the sake of completeness. Observe ﬁrst that

0 2 ∼ 0 2 H Xα,β,S ΩXα,β = H Xα,β,S ΩXα,β . ϕ 0 2 ˜e 4 N N By Theorem 2.17 and we obtain an injection H Xα,β,S ΩXα,β ֒→ H P , OP (−4e) such that 0 2 e ϕ˜ H Xα,β,S ΩXα,β = ker(·Fα) ∩ ker(·Gβ) ∩ ker(·dFα) ∩ ker(·dGβ ). 4 N e 4 ∂Fα Let ξ ∈ H , OPN (−4e) , observe that ξ ∈ ker (·dF ) if and only if ξ ∈ ker · . Moreover, P α i=0 ∂Zi

4 ∂Fα 4 ∂Fα since eF = Zi , we see that if ξ ∈ ker · then ξ ∈ ker(·Fα). A similar argument for Gβ i=0 ∂Zi i=0 ∂Zi T shows that P 4 T ∂F ∂G imϕ ˜ = ker · α ∩ ker · β . ∂Zi ∂Zi i=0 \ Now we proceed to a standart Gauss algorithm

I 1 ∂Fα ∂G ξ I · = 0 ξ ∈ ker · ∂Fα ∩ ker · β ⇔ I Z ∂Z0 ∂Z0 ∂Z0 I 1 ∂Gβ ξ I · = 0 ( I Z ∂Z0 P

P I 1 e−1 e1−1 e2 I ξ ZI · (eZ0 + e1α1Z0 Z1 ) =0 ⇔ I 1 e−1 e1−1 e2 ξ I · (ea Z + e β Z Z ) = 0 PI Z 0 0 1 1 0 1 P I 1 e−1 I ξ ZI · Z0 = 0 ⇔ I 1 e1−1 e2 ξ I · Z Z = 0 PI Z 0 1 ξPI =0 ∀I = (i ,...,i ) / i >e − 1 ⇔ 0 4 0 ξI =0 ∀I = (i ,...,i ) / i >e − 1 and i > e . 0 4 0 1 1 2

Note that to do this, we suppose β1 6= a0α1. Similarly, we ﬁnd

I ∂Fα ∂Gβ ξ =0 ∀I = (i0,...,i4) / i1 >e − 1 ξ ∈ ker · ∂Z ∩ ker · ∂Z ⇔ I 1 1 ξ =0 ∀I = (i0,...,i4) / i0 >e1 and i1 > e2 − 1 I ∂Fα ∂Gβ ξ =0 ∀I = (i0,...,i4) / i2 >e − 1 ξ ∈ ker · ∂Z ∩ ker · ∂Z ⇔ I 2 2 ξ =0 ∀I = (i0,...,i4) / i2 >e1 − 1 and i3 > e2 I ∂Fα ∂Gβ ξ =0 ∀I = (i0,...,i4) / i3 >e − 1 ξ ∈ ker · ∂Z ∩ ker · ∂Z ⇔ I 3 3 ξ =0 ∀I = (i0,...,i4) / i2 >e1 and i3 > e2 − 1 and ξ ∈ ker · ∂Fα ∩ ker · ∂Gβ ⇔ ξI =0 ∀I = (i , ··· ,i ) / i >e − 1. ∂Z4 ∂Z4 0 4 4 Now we just have to observe that any multi-index I = (i0,...,i4) with |I| = 4a satisﬁes one of those conditions, because otherwise we would get

4e = i0 + ··· + i4 6 2(e2 + e − 2) + e − 1 6 4e − 4,

which is a contradiction. And therefore, if ξ ∈ imϕ, ˜ then ξI =0 for all I. Hence, imϕ ˜ = {0}. Observe that from this simple example, one can easily generate other families for which the dimension of the space of holomorphic diﬀerential forms jumps.

25 Corollary 3.7. For any n > 2, for any m > 2, there is a family of varieties Y → B over a curve, of relative dimension n, and a point 0 ∈ B such that for generic t ∈ B,

0 m 0 m h (Y0,S ΩY0 ) >h (Yt,S ΩYt ). Proof. Let X be the family of surfaces constructed in Proposition 3.6. Then it suﬃces to consider the family X × A for any (n − 2)−dimensional abelian variety A.

4 Varieties with ample cotangent bundle 4.1 Statements In this section we prove our main application of the results of Section 2. Our statement is the following partial result towards Debarre’s conjecture (denoted by Theorem D in the introduction). > > 3N−2 > N Theorem 4.1. Let N,c,e ∈ N such that N 2, c 4 and e 2N +3. If X ⊆ P be a general complete intersection of codimension c and multidegree (e,...,e), then ΩX is ample. 3N−2 > N > Remark 4.2. Observe that the condition c 4 can be rephrased by codimP (X) 3 dim(X) − 2. Since ampleness is an open condition (see for instance Theorem 1.2.17 in [12] and Proposition 6.1.9 in [13]), it suffices to construct one example of a smooth complete intersection variety with ample cotangent bundle satisfying the hypothesis of the theorem to prove that the result holds for a general complete intersection variety. We will construct such an example by considering deformations of Fermat type complete intersection varieties. To do so, let us introduce some notation. ∗ ⊕N+1 Fix N > 2, ε ∈ N and e ∈ N . Set Aε := C[Z0,...,ZN ]ε. For any s = (s0,...,sN ) ∈ Aε we set: N e Fs(Z)= F (s,Z) := siZi . (13) i=0 X This is a homogeneous equation of degree e0 := ε + e and for a general choice of s, it defines a smooth N N hypersurface in P which we will denote by Xs. For any m,a ∈ N, and any smooth variety X ⊆ P , we will m ⊗m ∗ write LX := OP(ΩX )(1) and LX (−a) := LX ⊗ πX OX (−a). With those notation we have the following. > > 3N−2 > Theorem 4.3. Let N,c,e,ε,a ∈ N such that N 2, c 4 and set n := N − c. Suppose that ε 1 and j ⊕N+1 that e > N +1+ a + Nε. Then for any 0 6 j 6 N there exists s ∈ Aε such that X := Xs1 ∩···∩ Xsc n is a smooth complete intersection variety such that LX (−a) is nef. Let us first prove that this result implies Theorem 4.1. Proof of Theorem 4.3 ⇒ Theorem 4.1. Take a = ε = 1, e > N(1 + ε)+ a +1=2N +2 (this is equivalent > > 3N−2 to e0 2N +3) and c 4 . Then Theorem 4.3 implies that there exists a smooth complete intersection N n variety X ⊂ P of codimension c and multidegree (e,...,e) such that LX (−1) is nef. But this implies n that LX is ample and therefore that LX is ample, and thus, ΩX is ample. But since ampleness is an open condition in families, we deduce that a general complete intersection variety in PN of codimension c and multidegree (e,...,e) has ample cotangent bundle. The proof of Theorem 4.3 is an induction based on the following technical lemma. > > 3N−2 > > Lemma 4.4. Let N,c,e,ε,a ∈ N such that N 2, c 4 , ε 1 and e N +1+a+Nε. Set n := N −c. j ⊕N+1 For any 0 6 j 6 N take s ∈ Aε such that X := Xs1 ∩···∩ Xsc is a smooth complete intersection variety. For any i ∈{0,...,N}, set Hi := (Zi = 0) and Wi := X ∩ Hi. We look at P(ΩWi ) as a subvariety 1 c of P(ΩX ). Then, for a general choice of s ,...,s , there exists E ⊆ P(ΩX ), such that dim E = 0 and such that N n Bs(LX (−a)) ⊆ P(ΩWi ) ∪ E i=0 [

26 The proof of Lemma 4.4 is the content of Section 4.2 and Section 4.3. But for now let us explain how to obtain Theorem 4.3 from Lemma 4.4. > 3N−2 Proof of Theorem 4.3. Let H be a hyperplan section of X. Fixing c 4 , the proof is an induction on n = dim X (or equivalently on N). If n = 1, then X is a curve. By the adjunction formula, KX = OX (c(e + ε) − N − 1). Therefore ΩX (−a)= KX (−a) is nef. n Now suppose that n > 2. We have to prove that for any irreducible curve C ⊆ P(ΩX ), C · LX (−a) > 0. n n n Certainly, if C 6⊂ Bs(LX (−a)), C · LX (−a) > 0. Now suppose that C ⊆ Bs(LX (−a)), from Lemma 4.4 n N we know that Bs(LX (−a)) ⊆ i=0 P(ΩWi ) ∪ E for some zero-dimensional set E. Therefore, there exists i ∈{0,...,N} such that C ⊆ P(ΩWi ). But on can view Wi as a codimension c complete intersection variety N−1 S in Hi =∼ P deﬁned by equations of the same type than X (with one less variable). Observe moreover that 3N−2 3(N−1)−2 P > > LX (−a)| (Wi) = LWi (−a) and that if c 4 , one has c 4 . From our induction hypothesis we n−1 n−1 > > ∗ therefore know that LWi (−a) is nef. Thus, LWi (−a) · C 0 and in particular (n − 1)LWi · C aπX H · C. Hence, n n ∗ ∗ nLX · C = (n − 1)LW · C > aπ H · C > aπ H · C. n − 1 i n − 1 X n And ﬁnally LX (−a) · C > 0.

4.2 Constructing symmetric differential forms To prove Lemma 4.4 we first need to construct sufficiently many symmetric differential forms on complete intersection varieties of the above type, this is the purpose of this section. The setting is the following. We fix > > > N > > > N,c,ε,e,a,r,n ∈ N such that N 2, N c 2 , a 0, ε 0, e a+N(ε+1)+1, r = e−1 and n = N −c. 6 6 j N+1 6 6 p For any 1 j c, we take s ∈ Aε such that for any 1 p c and for any I := (i1,...,ip) ∈{1,...,c}6=, I the set X := Xsi1 ∩···∩ Xsip is a smooth complete intersection variety. Of course, this last condition holds ij (1,...,c) if the s ’s are general. We also denote X := X . For any 0 6 i 6 N, any 1 6 j 6 c and any v ∈ Aε, we set ai(v) := Ziv and αi(v) := Zidv + evdZi. So that :

N N j j r j j r F (s ,Z)= ai(si )Zi and dF (s ,Z) := dFsj (Z)= αi(si )Zi (14) i=0 i=0 X X

For any i ∈{0,...,n} set Ui := (Zi 6= 0), U = (Ui)06i6N and UX := (Ui ∩ X)06i6N , UX is an open covering of X. Our ﬁrst result is the following.

n Lemma 4.5. With the above notations, for any I = (i1,...,in) ∈{1,...,c}6=, and for any degree e−a−Nε− I,P 0 n N − 1 homogeneous polynomial P ∈ C[Z0,...,ZN ], there is a non-zero element ω ∈ H (X,S ΩX (−a)) I,P I,P I,P ˇ 0 U n such that, when computed in Čech cohomology one has ω = (ω0 ,..., ωN ) ∈ H ( X ,S ΩX (−a)) with e e 1 1 1 1 a0(s0) ··· aj−1(sje−1) aje+1(sj+1) e ··· aN (sN ) e ...... j c c c c I,P (−1) P a0(s0) ··· aj−1(sj−1) aj+1(sj+1) ··· aN (sN ) ωj = r i1 i1 i1 i1 . Z α0(s ) ··· αj−1(s ) αj+1(s ) ··· αN (s ) i 0 j−1 j+1 N ...... e in in in in α0(s0 ) ··· αj−1(sj−1) αj+1(sj+1) ··· αN (sN )

j j j j Proof. For simplicity, for any i ∈{ 1,...,N} and any j ∈{1,...,c}, we write ai := ai(si ), αi := αi(si ) and n c−n Fj = Fsj . As in the statement, take I = (i1,...,in) ∈ {1,...,c}6=, and take (in+1,...,ic) ∈ {1,...,c}6= such that {i1,...,ic} = {1,...,c}. For any j ∈{0,...,c} write Ij = (i1,...,ij). We can apply Theorem 2.17 and Theorem 2.24 to the simple λ-setting Σ = (X, λ0,...,λc) where λi = ∅ if i 6= n and λn = n (note that

27 q(Σ) = 0). Here the order in which we intersect the diﬀerent hypersurfaces to obtain X is crucial. We obtain an injection 0 n ϕ˜ N N H X,S ΩXIn |X ⊗ OX (−a) → H P , OPN (−a − Ne0) such that e c n im(ϕ ˜)= ker(·Fi ) ker(·Fi ) ∩ ker(·dFi ) . (15)  j  j j j=n+1 i=1 ! \ \ \   Observe that our hypothesis on the degree ensures that e−a−Nε−N −1 > 0. Fix any degree e−a−Nε−N −1 polynomial P . Then we get a well deﬁned element

I,P P ˇ N U ω0,...,N := r r ∈ H ( , OPN (−a − Ne0)) . Z0 ··· ZN

e I,P −1 I,P From (14) and (15) we see that ω0,...,N ∈ im(ϕ ˜). To obtain the explicit description of ϕ˜ (ω0···N ) ∈ ˇ 0 n H UX ,S ΩXIn |X (−a) we have to describe explicitly the inclusion ϕ˜, to do so, we have to unravel the e e proof of Theorem 2.17 in our situation. This inclusion is described in Proposition 2.18, applying this proposition ine our situation, we see that ϕ˜ is obtained as the composition of the following chain of inclusions:

0 n 1 n−1 (H (X,S ΩXIn |X (−a)) ֒→ H X,S ΩXIn−1 |X (−a − e0

2 n−2 (e ֒→ H X,S ΩeXIn−2 |X (−a − 2e0 . . . . e n−1 ((H (X, ΩXI1 |X (−a − (n − 1)e0 →֒ n ((H (X, OX (−a − ne0 →֒ n+1 Ice−1 ((H (X , OXIc−1 (−a − (n + 1)e0 →֒ . . . . N N ((H (P , OPN (−a − Ne0 →֒

For any 0 6 ℓ 6 N, we denote by ωℓ element in the Hℓ group in the above chain of inclusion whose N N I,P image in H (P , OPN (−a − N(e + ε))) under the above inclusions is ω . Moreover, for any 0 6 ℓ 6 N ℓ+1 ◦ N−ℓ and for any K = (k0,...,kℓ) ∈ {0,...,Ne }< if we let K = (kℓ+1,...,kN ) ∈ {0,...,N}< such that {k0,...,kN } = {0,...,N}, we set ε(K) to be the signature of the followinge permutation 0 1 ··· N σK := , k k ··· kN 0 1 m m m m m m and for any m ∈ {1,...,c}, we set LK := (akℓ+1 ,...,akN ) and ΛK := (αkℓ+1 ,...,αkN ). We also set 1 XK := Zr ···Zr . k0 kℓ Let ℓ ∈{0,...,N}, we are going to prove by induction that ωℓ can be represented in Čech cohomology ℓ by the following cocycle (ω ) ℓ+1 : K K∈{0,...,N}< e i1 e LK 6 6 ℓ N(N−ℓ+1) . 1. If n ℓ N, write m := N − ℓ, then ωK = (−1) ε(K)P . XK .

Lim K e

28 i1 LK . . ic ℓ N(N−ℓ+1) LK 2. If 0 6 ℓ 6 n, write m := n − ℓ, then ω = (−1) ε(K)P XK . K Λi1 K . . e im ΛK

Of course, if we look at ℓ =0 in the last case, up to a sign, we get the description announced in the statement I,P 0 n of the lemma, and it suﬃces to look at ω as an element in H (X,S ΩX (−a)) by considering its image 0 n 0 n under the natural restriction map H (X,S ΩXIn |X (−a)) → H (X,S ΩX (−a)). e e We start by computing a Čech cocycle fore ωN−1. Recall that one hase the twisted restriction short exact sequence ·Fi1 PN PN e 0 → O (−a − Ne0) → O (−a − (N − 1)e0) → OXi1 (−a − (N − 1)e0) → 0, which yields the following in cohomology:

δ·F F N−1 i1 N N · i1 N N PN PN 0 → H (Xi1 , OXi1 (−a − (N − 1)e0)) → H (P , O (−a − Ne0)) → H (P , O (−a − (N − 1)e0)).

N−1 N−1 N I,P By the very deﬁnition of ω , we have δ·Fi1 (ω )= ω = ω . As explained in Section 3.1, to compute N−1 N a cocycle for ω , we compute ω0,...,N · Fi1 and write it as the Čech diﬀerential of the desired cocycle. We have e e e e

e e N N ij r ω0,...,N · Fi1 = P · X0,...,N · aj Zj j=0 X e N N ij j j N(N−1+1) ij = P · aj X0,...,ˆj,...,N = P · (−1) (−1) (−1) aj X0,...,ˆj,...,N j=0 j=0 X X ˆ j i1 i1 If we write K = (0,..., j,...,N), we have ε(K) = (−1) and LK = (aj ) and XK = X0,...,ˆj,...,N . Therefore, N N−1 we see that ω · F = dˇ (ω ) N where as announced 0,...,N i1 K K∈{0,...,N}< N−1 N(N−1+1) i1 e e ωK = (−1) ε(K)P |LK|XK . We will only treat one case of the induction, the other case is treated the exact same way. Let n<ℓ 6 N, e ℓ ℓ let m = N − ℓ and suppose that we know that ω can be represented by a cocycle (ω ) ℓ+1 as in K K∈{0,...,N}< the case 1. We have to prove that ωℓ−1 can be represented by a cocycle of the form described in case 1. We have e e ℓ ℓ Im e ω ∈ H X , OXIm (−a − ℓe0) , and we have the following short exact sequence e ·Fim+1 0 → OXIm (−a − ℓe0) → OXIm (−a − (ℓ − 1)e0) → OXIm+1 (−a − (ℓ − 1)e0) → 0. In cohomology, it yields an injection (which is precisely the one appearing in the above chain of inclusions)

δ·F im+1 ℓ−1 Im+1 ℓ Im (H X , OXIm+1 (−a − (ℓ − 1)e0) ֒→ H X , OXIm (−a − ℓe0

ℓ−1 ℓ+1 To compute a Čech cohomology cocycle of ω , we proceed as before. For any K ∈ {0,...,N}< we are ℓ ℓ+1 going to compute ωK · Fim+1 modulo Fi1 ,...,Fim . Fix K = (k0,...,kℓ) ∈{0,...,N}< and as usual, take

29 ◦ N−ℓ ◦ K = (kℓ+1,...,kN ) ∈ {0,...,N}< such that K ∪ K = {0,...,N}. Because we are working modulo

Fi1 ,...,Fim , we have N ai1 Zr = − ℓ ai1 Zr Fi1 = 0 j=ℓ+1 kj kj j=0 kj kj ......  . .  . . ⇔ P . P .  N im r ℓ im r  Fi = 0  a Z = − a Z  m  j=ℓ+1 kj kj j=0 kj kj N im+1 r ℓ im+1 r Fim+1 − Fim+1 = 0 j=ℓ+1 ak Zk − Fim+1 = − j=0 ak Zk   P j j P j j   We may write this as follows:  P P

i1 i1 r ℓ i1 r ak ··· ak 0 − j ak Zk ℓ+1 N Zkℓ+1 =0 j j . . . . .  . . .  .  .  . . .  .  = P . . im im r ℓ im r ak ··· ak 0 Z − a Z  ℓ+1 N   kN   j=0 kj kj   im+1 im+1   ℓ im+1 r   a ··· a −1   Fim+1   − a Z   kℓ+1 kN     Pj=0 kj kj        By a simple application of Cramer’s rule we obtain P

ai1 ··· ai1 0 ai1 ··· ai1 − ℓ ai1 Zr kℓ+1 kN kℓ+1 kN j=0 kj kj ...... P . Fim = aim ··· aim 0 aim ··· aim − ℓ aim Zr kℓ+1 kN kℓ+1 kN j=0 kj kj im+1 im im+1 im+1 ℓ im+1 a ··· a −1 a ··· a − a Zr kℓ+1 kN kℓ+1 kN Pj=0 kj kj which yields P i i i i1 a 1 ··· a 1 a 1 LK ℓ kℓ+1 kN kj . . . . r . Fi = . . . Z . . m+1 . . . kj im j=0 im+1 im+1 im+1 LK X ak ··· ak ak ℓ+1 N j

We just have to be a bit careful with the signs. For any j ∈{0,...,ℓ} we set Kj = K \{kj}. We also take

νj ∈{ℓ+1,...,N −1} such that kνj < kj < kνj +1 (if kj < kℓ+1, we just take νj = ℓ). With this notation, we ◦ ◦ have Kj = (kℓ+1,...,kνj , kj , kνj +1,...,kN ) (or Kj = (kj , kνj +1,...,kN ) if kj < kℓ+1). A straightforward −1 νj −j computation shows that σKj = (kj , kj+1,...,kνj ) ◦ σK , hence ε(Kj ) = (−1) ε(K). On the other hand, observe that ai1 ··· ai1 ai1 i1 kℓ+1 kN kj LKj ...... = (−1)N−νj ...... im+1 im+1 im+1 im+1 ak ··· ak ak LK ℓ+1 N j j

Now we are ready for our computation:

i i1 L 1 LK ℓ Kj ℓ N(N−ℓ) . N(N−ℓ) N−νj . r ω · Fi = (−1) ε(K)P XK . Fi = (−1) ε(K)P XK (−1) . Z K m+1 . m+1 . kj im j=0 im+1 L X L K Kj

i1 L ℓ Kj N(N−ℓ) N j j−νj . = (−1) P (−1) ε(K)(−1) (−1) . XK . j j=0 im+1 X LK j i1 L ℓ Kj N(N−ℓ+1) j . = (−1) ε(Kj )(−1) P . XK . . j j=0 im+1 X LK j

30 ℓ ˇ ℓ−1 ′ ℓ Hence ω · Fim+1 = d ωK′ ′ ℓ , where for any K ∈{0,...,N}<, K K ∈{0,...,N}< i1 e e LK′ ℓ−1 N(N−ℓ+1) ′ . . ′ ωK′ = (−1) ε(K )P . XK im+1 L ′ K

as announced. The last step is to use the tilde-symmetric diﬀerential forms constructed in Lemma 4.5 to get usual symmetric diﬀerential forms. Recall that in our situation, from the exact sequence

n n n−1 0 → S ΩX (−a) → S ΩX (−a) → S ΩX (−a) → 0

0 n−1 and the vanishing H X,S ΩX (−a) (deducede from Lemma 2.12),e we get an isomorphism e 0 n 0 n H (X,S ΩX (−a)) =∼ H X,S ΩX (−a) . Therefore we just have to compute the image ωI,P of ωI,P under thise isomorphism. To do this we just have to I,P dehomogenize ω . For simplicity, we only work in the U0 = (Z0 6= 0) chart. We use the following notation: for any i ∈{1,...,N}, z = Z1 , so that dZ = Z dz + z dZ . Let us make an elementary observation. Take i Z0 i 0 i e i 0 ℓ > 1, and let Ge ∈ C[Z0,...,ZN ] be a homogeneous degree ℓ polynomial. Let g be the dehomogenized of G with respect to Z0. Then we have ℓ−1 ℓ dG = ℓZ0 gdZ0 + Z0dg. Indeed,

N N ∂G ∂G ∂G 2 Z0dG = Z0 dZi = Z0 dZ0 + (Z0 dzi + ZidZ0) ∂Zi ∂Z ∂Zi i=0 0 i=1 X X N N N ∂G 2 ∂G 2 ℓ−1 ∂g ℓ ℓ+1 = ZidZ0 + Z0 dzi = ℓGdZ0 + Z0 Z0 dzi = ℓZ0gdZ0 + Z0 dg. ∂Zi ∂Zi ∂zi i=0 i=1 i=1 X X X We introduce some more notation. For any u ∈ Aε =∼ C[z1,...,zN ]6ε and any q ∈{1,...,N} we set

bq(u) := zqu and βq(u) := zqdu + eudzq.

j j For any 1 6 j 6 c and for any 0 6 i 6 N we let ti (resp. fj ) to be the dehomogenized of si (resp. Fj ) with j j ε+1 j ε+1 j respect to Z0. Therefore we have ai(si )= Zisi = Z0 ziti = Z0 bi(ti ) and

j j j ε−1 j ε j ε j αi(si ) = Zidsi + esi dZi = Z0zi(εZ0 ti dZ0 + Z0 dti )+ eZ0 ti (zidZ0 + Z0dzi) ε j ε+1 j j ε j ε+1 j = e0Z0 ziti dZ0 + Z0 (zidti + eti dzi)= e0Z0 bi(ti )dZ0 + Z0 βi(ti ). Therefore, by the elementary properties of the determinant we get

1 1 1 1 a1(s1) ··· aN (sN ) b1(t1) ··· bN (tN ) ...... c c c c I,P P a1(s1) ··· aN (sN ) N(ε+1)−r b1(t1) ··· bN (tN ) ω0 = r i1 i1 = PZ0 i1 i1 . Z α1(s ) ··· αN (s ) β1(t ) ··· βN (t ) 0 1 N 1 N ...... e α (sin ) ··· α (sin ) β (tin ) ··· β (tin ) 1 1 N N 1 1 N N

This completes the proof of the following.

31 Proposition 4.6. With the above notation. Take a homogenous polynomial P ∈ C[Z0,...,ZN ] of degree e − a − Nε − N − 1 and let Q ∈ C[z1,...,zN ] be the dehomogenization of P with respect to Z0. Then, for n any I = (i1,...,in) ∈{1,...,c}6=, the symmetric diﬀerential form

1 1 b1(t1) ··· bN (tN ) . . . . c c I,Q b1(t1) ··· bN (tN ) 0 n ω0 (t) := Q i1 i1 ∈ H (U0 ∩ X,S ΩX ) β1(t ) ··· βN (t ) 1 N . . . .

β (tin ) ··· β (tin ) 1 1 N N

I,P 0 n extends to a twisted symmetric diﬀerential form ω ∈ H (X,S ΩX (−a)).

4.3 Estimating the base locus 4.3.1 Dimension count lemmas We will need some elementary results. The following is classical, we provide a proof because we will use the idea of it again afterwards.

∗ Lemma 4.7. Let p, q ∈ N such that n 6 p. Fix a rank n matrix A ∈ Matn,p(C). For any r ∈ N with r

−1 ∆Γ = {B ∈ Matp,q(C) such that im(AB) ⊆ Γ} = B ∈ Matp,q(C) such that im(B) ⊆ A (Γ)

−1 q r+p−n But since rk A = n, A is surjective and therefore dim(A (Γ)) = r+p−n. Therefore, ∆Γ =∼ Hom(C , C ) and thus dim(∆Γ)= q(r + p − n). In particular,

dim ∆ = q(r + p − n) + dim Grass(r, Cn)= q(r + p − n)+ r(n − r)= qp − (q − r)(n − r).

p,q Since Σr = pr1(∆), the result follows. From the previous lemma one can easily deduce the following.

∗ M ∨ Corollary 4.8. Let M,N,c ∈ N such that c 6 N. Let ℓ1,...,ℓN ∈ (C ) be non-zero linear forms. For any r ∈ N∗ such that r < c, let

1 1 ℓ1(x1) ··· ℓN (xN ) j 16j6c M Nc . . 6 Σr = (xi )16i6N ∈ (C ) such that rk  . .  r .  ℓ (xc ) ··· ℓ (xc )   1 1 N N      Then, dim(Σr) 6 MNc − (N − r)(c − r).  The following lemma will be crucial to us.

32 ∗ Lemma 4.9. Take c,n,M ∈ N such that n 6 c and let N := n+c. Take linear forms ℓ1,...,ℓn, λ1,...,λn ∈ M ∨ (C ) such that for any i ∈{1,...,c}, λi 6=0 and ker ℓi 6= ker λi. Fix A ∈ Glc(C) and B ∈ Matc(C). For j 16j6c M cn any x = (xi )16i6n ∈ (C ) consider

1 1 ℓ1(x1) ··· ℓn(xn) . .  A . .  c c ℓ1(x1) ··· ℓn(xn) S (x)=  1 1  .  λ (x ) ··· λn(x )   1 1 n   . .   B . .     λ (xc ) ··· λ (xc )   1 1 n n    For any r ∈{c,...,N − 1}, let

M cn M cn Σr := x ∈ (C ) such that rk S (x) 6 r ⊆ (C ) .

Then, dim(Σr) 6 Mnc − (N − r)(2c − r).

Proof. Using elementary operations on the c ﬁrst columns, we see that we may suppose that A = Ic is the identity matrix of size c. We will write everything in matricial notation. For any i ∈ {1,...,M} any j 16j6c M cn j ∈{1,...,c} and any x = (xi )16i6n ∈ (C ) we let

j xi,1 M ∨ M ∨ j . M Li = (ℓi,1,...,ℓi,M ) ∈ (C ) , Λi = (λi,1,...,λi,M ) ∈ (C ) and Xi =  .  ∈ C xj  i,M    j j j j t such that ℓi(xi ) = LiXi and λi(xi )=ΛiXi . For any matrix Q we denote the transposed of Q by Q. Set moreover

t t t 1 t 1 L1 ··· 0 Λ1 ··· 0 X1 ··· Xn ...... L :=  . .. .  , Λ :=  . .. .  and X :=  . .  . 0 ··· tL 0 ··· tΛ tXc ··· tXc  n   n   1 n        I XL With those notations, S(x)= c , and by elementary operations on the lines we get that B XΛ I XL rk(S(x)) = rk c = c + rk(XΛ − BXL). 0 XΛ − BXL

Write B = (bi,j )16i,j6c. By a straightforward computation we obtain that for any j ∈ {1,...,n} the j-th column of the matrix XΛ − BXL is exactly Kj Xj where

1 Xj Λj ··· 0 b11Lj ··· b1cLj ...... Xj =  .  and Kj =  . .. .  −  . .  Xc 0 ··· Λ b L ··· b L  j   j   c1 j cc j        In particular, we have XΛ − BXL = K1X1 ··· KnXn . An easy computation shows that under our ′ ′ hypothesis on the ℓis and the λis, rk Kj = c for all j ∈{1,...,n}. The end of the proof goes as in Lemma 4.8. Let ∆ := (X, Γ) ∈ (CM )nc × Grass(r − c, Cc) such that im(XΛ − BXL) ⊆ Γ

33 M cn c c Let pr1 : ∆ → (C ) and pr2 : ∆ → Grass(r−c, C ) be the natural projections. For any Γ ∈ Grass(r−c, C ) −1 let ∆Γ = pr1(pr2 (Γ)). Then

M nc ∆Γ = X ∈ (C ) such that im(XΛ − BXL) ⊆ Γ M nc = X ∈ (C ) such that im K1X1 ··· Kn Xn ⊆ Γ M nc −1 −1 = X ∈ (C ) such that im(X1) ⊆ K1 (Γ),..., im(Xn) ⊆ Kn (Γ) −1 Since for each j ∈{1,...,n},Kj is surjective, we obtain that dim(Kj (Γ)) = r − c + Mc − c = Mc + r − 2c. Mc+r−2c ⊕n In particular ∆Γ =∼ Hom(C, C ) , which implies that dim(∆Γ)= n(Mc + r − 2c). And therefore,

dim(∆) = n(Mc+r −2c)+dimGrass(r −c, Cc)= n(Mc+r −2c)+(r −c)(2c−r)= Mnc−(n+c−r)(2c−r).

But since Σr = pr1(∆), this concludes the proof of the lemma. We are also going to need the following elementary algebraic geometry lemma. Lemma 4.10. Let X and Y be two algebraic varieties (not necessarily irreducible). Let f : X → Y be a regular map. Suppose that Y = Y0 ∪···∪ Yr such that Yi ∩ Yj = ∅ if i 6= j. For each i ∈ {0,...,r} take ni ∈ N. Suppose that for each y ∈ Yi and for each i ∈{0,...,r} , dim Xy 6 ni. Then

dim X 6 max (dim(Yi)+ ni). 06i6r

4.3.2 Proof of Lemma 4.4 We are now in position to prove Lemma 4.4. It will be a immediate consequence of the following more precise statement. > > 3N−2 > > Lemma 4.11. Let N,c,e,ε,a ∈ N such that N 2, c 4 , ε 1 and e N +1+ a + Nε, set j ⊕N+1 q := e−(N +1+a+Nε). Set n := N −c. For any 0 6 j 6 N take s ∈ Aε such that X := Xs1 ∩···∩Xsc is a smooth complete intersection variety. For any i ∈{0,...,N}, set Hi := (Zi = 0) and Wi := X ∩Hi. We 1 c look at P(ΩWi ) as a subvariety of P(ΩX ). Then, for a general choice of s ,...,s , there exists E ⊆ P(ΩX ) such that dim E 6 0 and

N I,P Bs(LX (−a)) ⊆ (ω =0)= P(ΩWi ) ∪ E. I∈{1,...,c}n i=0 \ 6= [ P ∈C[Z0,...,ZN ]q

I,P 0 n ∼ 0 n Where ω ∈ H (P(ΩX ), LX (−a)) = H (X,S ΩX (−a)) is the symmetric diﬀerential form constructed in n Proposition 4.6 viewed as a global section of LX (−a).

Proof. For simplicity, we only treat the chart U0, the other charts are dealt with in the exact same way. Let N N−1 us precise the notation of Proposition 4.6: for any (u,z,ξ) ∈ Aε × C × P we set

bq(u,z) := zqu(z), and βq(u,z,ξ) := zqduz(ξ)+ eu(z)ξq.

1 c j 16j6c N c N N−1 Moreover, for any t = (t ,...,t ) = (ti )16i6N ∈ (Aε ) and for any (z, ξ) ∈ C × P we set

1 1 1 1 b1(t1,z) ··· bN (tN ,z) β1(t1,z,ξ) ··· βN (tN ,z,ξ) . . ′ . . B(t,z) :=  . .  and B (t,z,ξ) :=  . .  . b (tc,z) ··· b (tc ,z) β (tc ,z,ξ) ··· β (tc ,z,ξ)  1 1 N N   1 1 N N      ∼ N N−1 From now on we make the identiﬁcation P(ΩPN |U0 ) = C × P . So that if for any j ∈{1,...,c} we take j N+1 N s ∈ Aε such that Xs = (Fs1 = 0) ∩···∩ (Fsc = 0) is a smooth complete intersection in P and if for any

34 j j j ∈{1,...,c}, we let t be the dehomogeneization on s , ftj be the dehomogeneization of Fsj and Xt = Xs, then, one naturally has

N N−1 ft1 (z)= ··· = ftc (z)=0 N N−1 ∼ P(ΩXt∩U0 )= (x, ξ) ∈ C × P such that ⊆ C × P = P(ΩU0 ) df 1 (ξ)= ··· = dftc,z(ξ)=0 t ,z N ∼ N Moreover, set W := i=1 ((zi = 0) ∩ (ξi = 0)) = i=1 P(ΩHi∩U0 ), where Hi = (Zi = 0). Our aim is to prove that, for a general t ∈ (N+1)c, there exists E ⊆ (Ω ) such that dim E 6 0 and SAε SP Xt∩U0 I,Q P(ΩXt∩U0 ) ∩ ω0 (t,z,ξ)=0 = (P(ΩXt∩U0 ) ∩ W ) ∪ E. I,Q \ n The first thing to observe is that if (x, ξ) ∈ W then for any I ∈{1,...,c}6= and any Q ∈ C[z1,...,zN ]6q one I,Q I,Q has ω (t,z,ξ)=0, and therefore W ⊆ I,Q ω0 (t,z,ξ)=0 . The more difficult part is to prove that, generically, this is actually an equality up to a finite number of points. T I,Q Observe that if rk(B(t,z)) < c, then ω0 (t,z,ξ)= 0 for all I and all Q. On the other hand, if rk(B(t,z)) = c then B(t,z) ωI,Q(t,z,ξ)=0 ∀I, ∀Q ⇐⇒ rk < N. 0 B′(t,z,ξ) (N+1)c I,Q Therefore, for any t ∈ Aε , the locus I,Q ω0 (t,z,ξ)=0 is precisely T B(t,z) (z, ξ) ∈ CN × PN−1 such that rk B(t,z)

(N+1)c N (N+1)c (N+1)c N N (N+1)c N N−1 (N+1)c ρ1 : Aε × C → Aε ρ2 : Aε × C → C p1 : Aε × C × P → Aε (N+1)c N N−1 (N+1)c N (N+1)c N N−1 N N−1 p12 : Aε × C × P → Aε × C p23 : Aε × C × P → C × P .

N k For any k ∈ {0,...,N}, set Yk = (z1,...,zN ) ∈ C / ∃I ∈{1,...,N}6= such that zi =0 ⇐⇒ i ∈ I . Observe that dim Yk = N − k. Moreover,n set o

X 1 c (N+1) c N c := (t ,...,t ),z ∈ (Aε ) × C such that ft1 (z)= ··· = ftc (z)=0 n o X ′ 1 c (N+1) c N N−1 ft1 (z)= ··· = ftc (z)=0 c := (t ,...,t ),z,ξ ∈ (Aε ) × C × P such that df 1 (ξ)= ··· = dftc,z(ξ)=0 t ,z ∆c := {(t,z) ∈ Xc such that rk B(t,z)

ρ1 (N+1)c Of course, Xc → Aε is just the universal family of complete intersection of Fermat type we are con- X ′ sidering (or to be precise, the dehomogeneization of it), and c is just the projectivization of the relative X ρ1 (N+1)c (N+1)c X −1 X ′ cotangent bundle of c → Aε . For any t ∈ Aε , we write c,t = ρ1 (t) (and similarly for c,t, (N+1)c ∆c,t, etc.). By the above discussion, our aim is to prove that for a general t ∈ Aε , Γc,t is ﬁnite. To (N+1)c ′ 0 do so, we will prove that for a general t ∈ Aε , ∆c,t is empty and dim Γc,t 6 0. This will follow at once from the two following claims. (N+1)c Claim 1. dim ∆c < dim Aε

35 0 (N+1)c Claim 2. dim Γc 6 dim Aε

Proof of Claim 1. Take k ∈ {0,...,N}. Fix (z1,...,zN ) ∈ Yk. Without loss of generality, we may assume that z1 = ··· = zk =0 and that zk+1 ··· zN 6=0. In particular,

1 1 tk+1(z) ··· tN (z) . . rk B(t,z)=rk  . .  . tc (z) ··· tc (z)  k+1 N    z −1 Set ∆c := ρ1 ρ2 ({z}) ∩ ∆c . Of course, 1 1 tk+1(z) ··· tN (z) z j 16j6c (N+1)c . . . . 1 c ∆c =  (ti )16i6N ∈ Aε such that rk  . .  < c, and ft (z)= ··· = ft (z)=0 .  tc (z) ··· tc (z)   k+1 N     1 1   tk+1(z) ··· tN (z)  . . (N+1)c By Lemma 4.8 the set deﬁned by rk  . . 

z (N+1)c dim ∆c 6 dim Aε − max{0,N − k − c +1}− c. j j e j e j Indeed, for any j ∈ {1,...,c}, the equation ftj (z) = t0(z)+ t1(z)z1 + ··· + tN (z)zN = 0 is aﬃne (in ti ) j z and involves the term t0(z) which appears in none of the other equations deﬁning ∆c . So that each of the equations ftj (z)=0 increases the codimension by one, and we get the announced dimension. Therefore by Lemma 4.10 we obtain that

(N+1)c dim ∆c 6 max dim Yk + dim Aε − max{0,N − k − c +1}− c 06k6N (N+1)c (N+1)c (N+1)c = max dim Aε − max{0,N − k − c +1} + N − k − c 6 dim Aε − 1 < dim Aε . 06k6N

To prove Claim 2 we will need the following observation. N N−1 Claim 3. Let (x, ξ) ∈ C × P \ W . For any q ∈{1,...,N}, βq(·,z,ξ) 6=0 and ker bq(·,z) 6= ker βq(·,z,ξ).

N Proof of Claim 3. For k ∈{1,...,N}, we set δk = (0,..., 0, 1, 0,..., 0) ∈ N be the multi-index whose only N N−1 non-zero term is in the k-th slot. For any u ∈ Aε any (z, ξ) ∈ C × P and any q ∈{1,...,N} we have by deﬁnition,

I I+δq bq(u,z) = zq uI z = zqu + z zquδ + ··· + zN zquδ + z uI   0 1 1 N |XI|6ε 26X|I|6ε  N  I−δk I βq(u,z,ξ) = zq ikuI z ξk + eξq uI z     k |XI|6ε X=1 |XI|6ε     N I−δk I = eξqu0 + (zqξ1 + ez1ξq)uδ1 + ··· + (zqξN + ezN ξq)uδN + zq ikz ξk + ez ξq uI . k ! 26X|I|6ε X=1 N−1 N−1 Take (x, ξ) ∈ C × P \ W. Let q ∈ {1,...,N}. In view of the above expression, if βq = 0 then in particular eξq = zqξ1 + ez1ξq = ··· = zqξN + ezN ξq =0,

36 and because (z, ξ) ∈/ W , we get zq 6= 0, so that ξq = ξ1 = ··· = ξN = 0, which is not possible because N−1 ξ ∈ P . It remains to prove that ker bq(·,z) 6= ker βq(·,z,ξ). If bq(·,z)=0, this is obvious. Suppose bq(·,z) 6= 0 and therefore zq 6= 0. If ker bq(·,z) = ker βq(·,z,ξ), then from the above expression we get in particular that 1 z 1 z 1 = ··· = N =0, eξ z ξ + ez ξ eξ z ξ + ez ξ q q 1 1 q q q N N q

From which we deduce once again that ξ1 = ··· = ξN =0 .

c Proof of Claim 2. For any K = (k1,...,kc) ∈{1,...,N}6= let

1 1 bk1 (tk1 ,z) ··· bkc (tkc ,z) K (N+1)c N N−1 . . U = (t,z,ξ) ∈ Aε × C × P such that det  . .  6=0 . c c  bk (t ,z) ··· bk (t ,z)   1 k1 c kc    K K 0 0 0 K  For any K, UΓ := U ∩Γc is an open subset of Γc and Γc = K UΓ , therefore for our dimension estimation, we might as well restrict ourselves to U := U (1,...,c) and U := U (1,...,c). Fix (z, ξ) ∈ N × N−1 and set ΓS Γ C P z,ξ −1 (N+1)c UΓ := p1(p23 (z, ξ) ∩ UΓ). Using Claim 3 and Lemma 4.9 we see that the set deﬁned in Aε by

B(t,z) rk

0 (N+1)c dim Γc = dim Aε − 4c +3N − 2.

0 (N+1)c In particular, dim Γc 6 dim Aε as soon as −4c +3N − 2 6 0. This last condition is equivalent to > 3N−2 c 4 , which is exactly our hypothesis.

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