Stability of restrictions of cotangent bundles of irreducible Hermitian symmetric spaces of compact type

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Stability of restrictions of cotangent bundles of

irreducible Hermitian symmetric spaces of compact type

Indranil Biswas, Pierre-Emmanuel Chaput, Christophe Mourougane

To cite this version:

Indranil Biswas, Pierre-Emmanuel Chaput, Christophe Mourougane. Stability of restrictions of cotan- gent bundles of irreducible Hermitian symmetric spaces of compact type. Publications of the Re- search Institute for Mathematical Sciences, European Mathematical Society, 2019, 55 (2), pp.283-318.

�10.4171/PRIMS/55-2-3�. �hal-01138190�

STABILITY OF RESTRICTIONS OF COTANGENT BUNDLES OF IRREDUCIBLE HERMITIAN SYMMETRIC SPACES OF COMPACT TYPE

INDRANIL BISWAS, PIERRE-EMMANUEL CHAPUT, AND CHRISTOPHE MOUROUGANE

ABSTRACT. It is known that the cotangent bundleΩY of an irreducible Hermitian symmetric spaceY of compact type is stable. Except for a few obvious exceptions, we show that ifX ⊂Y is a complete intersection such thatPic(Y) → Pic(X)is surjective, then the restrictionΩY|X is stable. We then address some cases where the Picard group increases by restriction.

CONTENTS

1. Introduction 1

2. Restrictions with small Picard group 3

2.1. Short resolution 3

2.2. A cohomological property 4

2.3. General argument 4

3. Vanishing theorems 6

3.1. The statement 6

3.2. The case of Grassmannians (typeAn) 7

3.3. The case of quadrics (typeB_norD_n) 13

3.4. The case of Lagrangian Grassmannians (typeC_n) 14

3.5. The case of spinor Grassmannians (typeD_n) 16

3.6. The exceptional cases (typeE₆orE₇) 18

4. Restriction to a hypersurface with an increase of the Picard group 19

4.1. Another argument for general complete intersection 19

4.2. Results for all smooth divisors 20

4.3. The case ofP² 21

4.4. The case ofP³ 22

4.5. The case ofQ² 23

4.6. The case ofQ³ 23

Acknowledgements 29

References 29

1. INTRODUCTION

Stable vector bundles with zero characteristic classes on a smooth projective variety are given by the irreducible unitary representations of the fundamental group of the variety. On the other hand, it is a difficult

Date: April 1, 2015.

2000Mathematics Subject Classification. 14J60, 14M17.

1

and interesting question to produce explicit examples of stable vector bundles on algebraic varieties with non- zero characteristic classes. However, there are such vector bundles within the framework of homogeneous spaces. For example, an irreducible homogeneous bundleEon a homogeneous spaceY is stable [Ume78], [Ram66], [Bis04]. Once one has these stable vector bundles, a theorem of Mehta-Ramanathan [MR84]

and Flenner [Fle84] asserts the stability of restrictions of these bundles to general hypersurfaces Xof high enough degree with respect to the given polarisation. In this article, we address the following general question: What can be said about the stability of the restriction of an irreducible homogeneous bundleE, defined on a homogeneous spaceY, to a subvarietyX ⊂ Y ?

We give a positive answer to this question in the following setting. Recall that a Hermitian symmetric space is a Hermitian manifold in which every point is an isolated point of an isometric involution. It is homogeneous under its isometry group. Compact examples include the usual Grassmannians, the quadric hypersurfaces, the Lagrangian Grassmannians parametrising n-dimensional Lagrangian subspaces ofC²ⁿ equipped with a symplectic form, the spinor Grassmannian parametrising one family of n-dimensional isotropic subspaces ofC²ⁿ equipped with a nondegenerate quadratic form, and two exceptional manifolds.

For a compact irreducible Hermitian symmetric spacesY, the cotangent bundleΩ_Y is an irreducible homo- geneous, hence stable, vector bundle.

We assume first thatY is a compact Hermitian symmetric space, andXis a locally factorial complete in- tersection such that the restrictionPic(Y)→Pic(X)is surjective. This holds by Lefschetz theorem [Lef21]

wheneverdimX≥3(see also [Laz04, Example 3.1.25]) or wheneverXis a very general complete intersec- tion surface inPⁿexcept if it is a degreed≤3surface inP³or the intersection of two quadric threefolds in P⁴(see also [Kim91, Theorem 1]). Furthermore, we assume thatEis the cotangent (or, dually, the tangent) bundle ofY.

Theorem A(Theorem 1 and Theorem 2). LetY be a compact irreducible Hermitian symmetric space, and let X be a locally factorial positive dimensional complete intersection in Y. Assume that the restriction homomorphismPic(Y)→ Pic(X)is surjective. IfY is a projective space or a quadric, assume moreover thatXhas no linear equation. Then, the restriction ofΩY toXis stable.

In fact, our theorems are slightly more general than Theorem A since it is not necessary that X is a complete intersection : it is enough thatXhas ashort resolution, as explained in Definition 2.1. Moreover, if one is interested in general complete intersections, then using arelativeHarder-Narasimhan filtration one can get rid of the assumption on the Picard groups ; but semistability, instead of stability, is obtained (see Theorem 4).

We then deal with the case of small dimensions where the Picard group increases. Recall that irreducible Hermitian symmetric spaces of dimension2or3areP²,P³,Q²andQ³.

Theorem B(Theorem 4, Theorem 5). LetY be a compact irreducible Hermitian symmetric space of dimen- sion2or3. LetX ⊂Y be a smooth divisor, and in the case of ambient dimension3letC be a complete intersection curve.

• TakeY =P². ThenΩ_Y|X is semi-stable ifdegX≥2. IfdegX≥3, thenΩ_Y|X is stable.

• TakeY =P³. ThenΩ_Y|Xis stable ifdegX ≥2. IfCis cut-out by general non-linear hypersurfaces.

ThenΩ_Y|C is semi-stable.

• IfY =Q², thenΩ_Y|X is semi-stable but not stable.

2

• TakeY =Q³.

Assume thatdegX= 1. ThenΩ_Y|X is semi-stable.

Assume thatdegX= 2. ThenΩ_Y|X is stable.

Assume thatdegX≥9. ThenΩ_Y|X is stable.

IfdegX≥3andXis very general, thenΩ_Y|X is stable.

IfC is cut-out by general non-linear hypersurfaces, thenΩ_Y|C is semi-stable.

Our arguments are very different in the two situations. In the case of complete intersections with no increase of Picard groups, we use a new vanishing theorem (Theorem 3) that may be of independent interest (see Section 3.1). In fact, by standard cohomological arguments, a subbundle F ⊂ Ω_Y|X of the restriction ofΩY contradicting stability yields the non-vanishing of some cohomology group H^q(Y,Ω^p(l)), whereq is related to the codimension ofX and lto the degree of F. Our vanishing theorem implies the desired stability inequality (see Section 2). In small dimensions, we use tools from projective geometry to make explicit the new line bundles that appear on the subvarietyXand some arguments of representation theory (see Section 4).

2. RESTRICTIONS WITH SMALL PICARD GROUP

LetY be a compact irreducible Hermitian symmetric space, but not isomorphic to Q². We denote the ample generator of the Picard group of Y by O_Y(1). For a sheaf F on Y and an integer l, we denote the tensor product F ⊗ O_Y(1)^⊗lby F(l). We denote bydegY the top degree self-intersection degY :=

O_Y(1)^dimY and byc₁(Y)the index of the Fano manifoldY. We recall thatc₁(Y)is defined by the equality

−K_Y = c₁(Y)O_Y(1)as elements of the N´eron-Severi group ofY.

2.1. Short resolution. We will prove stability of the restriction ofΩ_Y to subschemes whose structure sheaf has a short resolution in the following sense :

Definition 2.1. A subschemeX⊂Y is said to have ashort resolutionif there is a resolution 0→ F_k→ F_k−1→ · · · → F1 → F0→ OX →0,

whereF₀ = O_Y,F_i := ⊕_jO_Y(−d_ij)withd_ij ≥ ifori > 0, and the lengthkof the resolution satisfies k <dim(Y).

Example 2.2. The Koszul resolution of complete intersections is a short resolution for a positive-dimensional complete intersection inY. If, moreover, none of the equations is linear, then the integersd_ijin Definition 2.1 satisfyd_ij ≥i+ 1.

Our second class of examples are some arithmetically Cohen-Macaulay subschemes. LetX ⊂P^N be a subscheme defined by a homogeneous idealJ in the homogeneous coordinate ringA := C[X₀, . . . , X_N].

Recall that X is calledarithmetically Cohen-Macaulay if the depth of A/J is equal to the dimension of A/J, namelydimX+ 1. Let, moreover, I ⊂Adenote the homogeneous ideal ofY. The reason why we will consider arithmetically Cohen-Macaulay subschemes is the following :

Fact 2.3. LetX ⊂Y ⊂P^N be an arithmetically Cohen-Macaulay subscheme, and assume thatA/J has finite projective dimension overA/I. Then, the structure sheafO_X has a resolution by split vector bundles overY of lengthk= dimY −dimX

0→ F_k→ F_k−1→ · · · → F₁ → F₀→ O_X →0,

3

whereF0 =OY andFi :=⊕jOY(−dij)withdij ≥ifori >0. In particular, ifdim(X)>0, thenXhas a short resolution.

If, moreover,Xis not linearly degenerate inP^N, then we haved_ij ≥i+ 1.

Proof. We have I ⊂ J ⊂ A. By Auslander-Buchsbaum formula [Mat89, Theorem 19.1], we have the equality

pd_A/I(A/J) = depth(A/I)−depth(A/J).

Moreover, any homogeneous space embedded by a homogeneous ample line bundle is arithmetically Cohen- Macaulay (see for example [BK05, Corollary 3.4.4]). Thus,

depth(A/I) = dim(A/I) = dimY + 1.

Therefore, pd_A/I(A/J) = dimY −dimX. Hence a minimal free resolution ofA/J overA/Ihas length k= dimY −dimX. Moreover, since for such a resolution the differentials have positive degree, we have d_ij ≥ifor all∀i, j. IfXis not included in any hyperplane, it has no equation of degree1, sod_1j ≥2for

all∀j, and we deduce thatd_ij ≥i+ 1.

2.2. A cohomological property. We will need the following nonvanishing property to prove our stability results. Its proof is postponed to Section 3.

Proposition 2.4. LetY be a compact irreducible Hermitian symmetric space, but not a projective space.

Letl, p, qbe integers, withq <dimY, such thatH^q(Y,Ω^p_Y(l))6= 0. Then, l+q ≥ p c₁(Y)

dim(Y), with equality holding if and only if

• p= dim(Y), q = 0andl=c₁(Y),

• orp=q = 0andl≥0,

• orY is a quadric andl= 0,

• orY ≃Q⁴, l= 2, p = 3, q = 1.

Of course, ifq = dim(Y)were allowed, then the above inequality would fail, since we may havelvery negative andp= dim(Y)(then apply Serre duality).

2.3. General argument. We can now prove the

Theorem 1. LetY be any compact irreducible Hermitian symmetric space excluding a projective space and a quadric. LetXbe a locally factorial positive dimensional subvariety ofY having a short resolution and such thatPic(X) = Z· O_Y(1)_|X. Then the restriction ofΩ_Y toXis stable.

Note that if dimX ≥ 3, the constraint on the Picard group of the complete intersection is ensured by Lefschetz theorem.

Proof. We will later prove a slightly weaker result for quadrics and projective spaces (Theorem 2), thus, for the momentY is any compact irreducible Hermitian symmetric space.

LetF be a coherent subsheaf of Ω_Y|X of rank 0 < p < dimY. Since X is assumed to be locally factorial, the rank one reflexive subsheafdetF := (VpF)^⋆⋆ofVpΩ_Y|X is invertible [Har80, Proposition

4

1.9] and hence isomorphic toOX(−d)for some integerd. We have µ(Ω_Y) = O_X(1)^dim(X)−1·K_Y

rank Ω_Y|X =− c₁(Y)

dim(Y) ·degX·degY

µ(F) = O_X(1)^dim(X)−1·detF rankF =−d

p ·degX·degY . The inclusionF ⊂ Ω_Y|X yields the non-vanishing of

H⁰(X, Hom(detF,Ω^p_Y|X)) = H⁰(X,Ω^p_Y(d)_|X), from which we have to deduce the stability inequality

µ(F) < µ(Ω_Y|X), equivalently, d > p c₁(Y) dim(Y). Consider a resolution ofO_X as in Definition 2.1. The resolution

0→ F_k⊗Ω^p_Y(d)→ · · · → F₁⊗Ω^p_Y(d)→ F₀⊗Ω^p_Y(d)→Ω^p_Y(d)_|X →0

translates the non-vanishing ofH⁰(X,Ω^p_Y(d)_|X)into the nonvanishing of one of the cohomology groups in the decomposition

Hⁱ(Y,F_i⊗Ω^p_Y(d)) =⊕_jHⁱ(Y,Ω^p_Y(d−d_ij)), say ofHⁱ(Y,Ω^p_Y(d−d_ij)).

We now assume thatY is not a projective space. In our setting Proposition 2.4 reads (d−d_ij) +i≥p c₁(Y)

dim(Y). (2.1)

It follows thatd≥p_dim(Y^c1(Y)₎.

In case of the equalityd =p_dim(Y^c1(Y)₎, we get thatd_ij = i, and the equality in (2.1) holds. Now assume that Y is not a quadric. Therefore, Proposition 2.4 gives thatp = 0 orp = dim(Y), equivalently, either

F ={0}orF = Ω_Y. Thus,Ω_Y is stable.

Some remarks regarding the two excluded cases in Theorem 1.

Remark2.5. IfX ⊂ Y is a linear section of the quadricY, then, as the above proof shows, the restriction ofΩ_Y toXis semi-stable.

Remark2.6. IfX⊂Y is a smooth quadric inside the quadricY, then we have an exact sequence 0→N_X^∗_|Y →Ω_Y|X →Ω_X →0.

All these vector bundles have equal slope, soΩ_Y|X is not stable. We believe that in this situation the only destabilizing subsheaf isN_X|Y^∗ . The above argument shows that a destabilizing sheaf must have rank equal to the codimension ofX.

Remark2.7. Similarly, ifX ⊂Y is contained in a linear subspaceHin a projective spaceY, then for the exact sequence,

0→N_H|Y^∗

|X →Ω_Y|X →Ω_H|X →0, the slope ofN_H|Y^∗

|X is strictly bigger than the slope ofΩ_Y|X. Thus,Ω_Y|X is not even semi-stable.

Thus, to get a result similar to Theorem 1 in these two cases, we exclude the case whereX has a linear equation :

5

Theorem 2. Let Y be a smooth quadric or a projective space. Then ΩY is stable. Let X be a locally factorial subvariety inY having a short resolution and such thatPic(X) = Z· O_Y(1)_|X. Assume thatX is contained in no hyperplane section ofY. Then the restriction ofΩ_Y toXis stable.

Note that by Lefschetz theorem, this theorem applies to very general cubic hypersurfaces ofQ³.

Proof. We continue with the notation of Theorem 1. IfY is a quadric, by the above proof of Theorem 1, we have the non-vanishing of someHⁱ(Y,F_i⊗Ω^p_Y(d)). Ifi >0, by (2.1) we have for somejthe inequality

d−d_ij +i≥p c₁(Y)

dim(Y) (=p).

Since, by Fact 2.3, d_ij > i, we get thatd > p. Ifi = 0, thenH⁰(Y,Ω^p_Y(d)) 6= 0, and by a result due to Snow (see Section 3.3), we get that eitherp = 0orp = dim(Y). This implies stability as in the proof of Theorem 1.

Assume now thatY is the projective spacePⁿ. We may assume that0< p < n. We wish to prove that d

p > n+ 1 n .

Since ^p+1_p > ⁿ⁺¹_n , it is enough to prove thatd ≥ p+ 1. For integersp, q, l, we haveH^q(Pⁿ,Ω^p_Pn(l))6= 0 if and only if one of the following hold :

(1) l >0, p < landq = 0, (2) l= 0andp=q,

(3) l <0, n−p <−landq=n.

Once again, we get thatHⁱ(Y,F_i⊗Ω^p_Y(d)) 6= 0for somei. Ifi = 0, sinceF₀ = O_Y, this implies that p= 0. Ifi >0, sincei < n, this implies thati=pandd=d_ij for somej. Thus we haved≥i+ 1 =p+ 1,

as claimed.

3. VANISHING THEOREMS

3.1. The statement. We will now explain the proof of Proposition 2.4. Actually, we will prove a stronger vanishing theorem that may be useful for other purposes.

Theorem 3. LetY be a compact irreducible Hermitian symmetric space, but not a projective space. Let l, p, qbe integers, withl >0,p >0, such thatH^q(Y,Ω^p_Y(l))6= 0.

(1) Then,

l+q ≥p c₁(Y) dim(Y). Furthermore, equality holds if and only if

• p= dim(Y), q = 0andl=c₁(Y),

• orY is a quadric andl= 0,

• orY ≃Q⁴, l= 2, p= 3, q = 1.

(2) If moreoverq >0, then

l+q ≤p.

The proof of Theorem 3 is given in the next subsections. Surprisingly enough, the first item is very intricate, and the proof of the vanishing theorem in this case entails involved combinatorial arguments. The second item is much easier.

6

Let us explain that Theorem 3 implies Proposition 2.4.

Assume thatH^q(Y,Ω^p_Y(l))6= 0. Ifl = 0, then by Hodge theory we havep=q. Sincec₁(Y)≤dim(Y) with equality holding if and only if Y is a quadric [KO73, page 37], this case is settled. Ifp = 0, then q = 0ifl > 0, whileq = dim(Y)ifl < 0. Thus this case is also settled.

In the second case of the theorem, let us prove that actually l+q < p c₁(Y)

dim(Y) + dim(Y)−c₁(Y). (3.1) Firstly, the theorem states thatl+q ≤ p, and we havep≤dim(Y). Since

(1− c₁(Y)

dim(Y))(l+q)≤dim(Y)−c₁(Y), we have

l+q≤ c₁(Y)

dim(Y)(l+q) + dim(Y)−c1(Y)≤p c₁(Y)

dim(Y)+ dim(Y)−c1(Y).

Note that the equality here would imply thatp= dim(Y), in which case we cannot haveq >0. Thus, the inequality (3.1) is proved. Now, coming back to the proof of the theorem, ifl < 0, then by Serre duality we have

H^dim(Y)−q(Y,Ω^dim(Y_Y ^)−p(−l)) 6= 0.

The relation (3.1) gives that(−l) + (dim(Y)−q)<(dim(Y)−p)_dim(Y^c1(Y)₎ + dim(Y)−c₁(Y), or in other words,

l+q > p c1(Y) dim(Y).

3.2. The case of Grassmannians (typeA_n). Fix positive integersa, b ≥ 2. Let G_ab := G(a, a+b) be the Grassmannian that parametrisesa-dimensional linear subspaces of a fixed(a+b)-dimensionalk-vector spaceV. It is the homogeneous spaceG(a, a+b) = SU(a+b)/[SU(a+b)∩U(a)×U(b)]. LetO_ab(1)be the Pl¨ucker polarisation on G_ab, which is also the positive generator ofPic(G_ab). The cotangent bundleΩ_ab of G_abbeing homogeneous and irreducible isµ-stable. We assumedimG_ab =ab≥4so that by Lefschetz hyperplane theorem, the restriction O_ab(1)_|X generates the Picard group of every smooth hypersurface X of G_ab. Theorem 1 reads in this case as follows :

Proposition 3.1 (Theorem 1 for Grassmannians). Assume that ab ≥ 4. Let X be a smooth complete intersection in G_absuch thatPic(X) = Z· O_ab(1)_|X. Then the restriction ofΩ_abtoXis semi-stable. It is stable except exactly whena = b = 2withXbeing a hyperplane section.

In caseY is a Grassmannian, by [Sno86], the non-vanishing ofH^q(Y,Ω^p_Y(l))implies the existence of a partition ofp, which isl-admissible with cohomological degreeqin the following sense.

Definition 3.2([Sno86]). Letλbe a partition and lan integer. We say that λisl-admissible if no hook- number of λ is equal to l. The (l-)cohomological degree of an l-admissible partition is the number of hook-numbers which are greater thanl.

Hence, as we have seen in the proof of Theorem 1, the above Proposition 3.1 is a consequence of the following combinatorial result :

7

Proposition 3.3(First part of Theorem 3 for Grassmannians). Letlbe a non-negative integer. Letλbe a l-admissible partition ofpin a rectanglea×b, with cohomological degreeq. Then, we have the inequality

l+q ≥ pa+b ab ,

with equality holding if and only if either λ = (b^a), l = a+band q = 0, orλ = (2,1), l = 2 and q = 1.

Proof. The case wherel = 0is easy as then we have p = q, and ifa , b ≥ 2, the inequalityab ≥ a+b holds.

8

Pn i=1

xi

+n−1

Pn i=2

xi

+n Pn i=2

xi

+n−2

Pn i=3

xi

+n−1 Pn i=3

xi

+n−3

· · ·

Pn n−k+1

xi

+k+1 Pn n−k+1

xi

+k−1 Pn n−3

xi

+3

Pn n−2

xi

+4 Pn n−2

xi

+2

xn−1 +xn

+3

xn−1 +xn +1

xn

+2 x_n 1

n−P1 i=1

xi

+n−2

n−P1 i=2

xi

+n−1

n−P1 i=2

xi

+n−3

n−P1 i=3

xi

+n−2

n−P1 i=3

xi

+n−4

· · · ·

n−1P

n−3

xi

+2

xn−1 +xn−2

+3

xn−1 +xn−2

+1

xn−1

+2 xn−1 1

n−2P

i=1

xi

+n−3

n−2P

i=2

xi

+n−2

n−2P

i=2

xi

+n−4

n−2P

i=3

xi

+n−3

n−2P

i=3

xi

+n−5

· · · ^{+xxn−n−23}

+1

xn−2

+2 x_n−2 1

n−3P

i=1

xi

+n−4

n−3P

i=2

xi

+n−3

n−3P

i=2

xi

+n−5

n−3P

i=3

xi

+n−4

n−3P

i=3

xi

+n−6

· · · x_n−3 1

... ... ... ... ...

k+2P

j=1

xj

+k+1

k+2P

j=2

xj

+k+2

k+2P

j=2

xj

+k

k+2P

j=3

xj

+k+1

k+2P

j=3

xj

+k−1

k+1P

j=1

xj

+k

k+1P

j=2

xj

+k+1

k+1P

j=2

xj

+k−1

Pk j=1

xj

+k−1

k+1P

j=2

xj

+k+1

k+1P

j=2

xj

+k−1

x_k 1

... ... ...

x1 +x2 +x3 +2

x2 +x3

+3

x2 +x3

+1

x3

+2 x3 1

x1 +x2

+1

x2

+2 x₂ 1

x₁ 1

We choose to denote partitions by non-decreasing sequencesλ = (λ1 ≤λ2 ≤ · · · ≤λ_b^′). Letx1 = λ1

be the first part; for all2 ≤ i ≤ n, letx_i denote the increaseλ_i −λ_i−1 between consecutive parts. We denote byq_i the number of boxes in thei^throw strictly bigger thanl. From the reading of the top row, we obtain that the gaplhas the expression

l = Xn

j=i

xj+ (n+ 1−i)

for some indexiwith2≤i≤n. Ifx_n−2 = 0for example, then the whole block

xn−2

+xn−1

+xn+2

x_n−1 +xn

+3 x_n−1

+x_n−2 +1

x_n−1 +2

x_n−2 1

gets removed, and we get two consecutive valuesxn−1+xn+ 2and x_n−2+x_n−1+x_n+ 3 = x_n−1+x_n+ 3

as candidates for gaps on the top row. For example, withx₁= 2, x₂ = 0, x₃ = 2, x₄= 2, we get6and7as candidates for gaps on the top row.

9 8 5 4 2 1

6 5 2 1

3 2

2 1

The number of boxes strictly bigger thanlon the top row isq_n=Pi−1

j=1x_j. We have l+q=

Xn j=1

x_j+ (n+ 1−i) +

n−1X

j=1

q_j =b^′+a^′+ 1−i+

n−1X

j=1

q_j.

On the other hand, counting boxes in the complementary partition, we have a^′− p

b^′ = = a^′b^′−p b^′ =

Pn

j=2(j−1)x_j Pn

j=1x_j .

Up to changing the fixed vector spaceV by its dual, we may assume thata^′ ≤b^′, so thatb^′− _a^p′ ≥a^′− _b^p′. We have to show thatl+q ≥ p

a^′ + p

b^′, that is(a^′− p

b^′) + (b^′− p

a^′)≥i−1−

n−1X

j=1

q_j. It is enough to prove that

a^′− p

b^′ ≥ i−1−Pn−1 j=1 q_j

2 . (3.2)

10

Letk = #{i|ci,1 < l}, whereci,1denotes the hook number in the(i,1)-box. We have

( Xk j=1

x_j) +k−1< l≤(

k+1X

j=1

x_j) +k

(see the red boxes). Asl ≤ (Pk+1

j=1xj) +k, the numberlis one of the gaps in the(k+ 1)^throw, and hence, l ≤ (Pk+1

j=2x_j) +k(see the green boxes). Asl ≤ (Pk+2

j=2x_j) +k(see the blue box), it is a gap in the (k+ 2)^throw withl ≤(Pk+2

j=3xj) +k(see the green boxes). Arguing further, we get that on then^throw, l≤(Pn

j=n−k+1x_j) +kforcing the indexito satisfy the inequality i ≥ n−k+ 1.

We also infer that on each row above thek^th, there are boxes strictly bigger thanl, so that considering the number of boxes strictly bigger thanlon non-top rows we havePn−1

j=1qj ≥ n−k−1, in particular

i−1−

n−1X

j=1

q_j ≤ k .

From the choice of the levelkwe have(Pk

j=1x_j) +k≤l = Pn

j=ix_j+ (n+ 1−i). Now cancelling the common terms in these sums, we infer the following bound on the first increases in terms of the last ones

min(k,i−1)X

j=1

xj ≤

Xn j=max(k+1,i)

xj+ (n−k+ 1−i)≤

Xn j=max(k+1,i)

xj. (3.3)

Therefore

a^′− p b^′ =

Pn

j=2(j−1)x_j Pn

j=1x_j =

Pn

j=2(j−1)x_j Pmin(k,i−1)

j=1 xj+Pn

min(k+1,i)xj

≥

Pn

j=2(j−1)xj

Pn

j=max(k+1,i)x_j+Pn

min(k+1,i)x_j (3.4)

=

Pn

j=2(j−1)x_j Pmax(k+1,i)−1

j=min(k+1,i) xj+ 2Pn

max(k+1,i)xj

≥ min(k, i−1)Pn

j=min(k+1,i)x_j 2Pn

j=min(k+1,i)x_j (3.5)

= min(k, i−1)

2 ≥ i−1−Pn−1 j=1q_j

2 .

This proves the inequality for semi-stability.

If the inequality in (3.5) is not strict, then comparing numerators, we have

x₂ =x₃ =· · ·=x_min(k,i−1) = 0 and xmin(k+2,i+1)=· · ·=x_n= 0.

The partition is not a rectangle, so thatx_min(k+1,i)6= 0.

11

x1

+xk+1

+n−1 x1

+xk+1

+n−2 · · · ^x_+n^k+1 +n−k^xk+1

−1

xk+1

+n−k−2 · · · n−k

... ... ... ... ... ...

x1

+xk+1

+k+1 x1

+xk+1

+k · · · _+k+2^{xk+1 x}₊₁^{k+1 xk+1} · · · 2

x1

+xk+1

+k x1

+xk+1

+k−1 · · · _+k+1^{xk+1 xk+1 x}₋₁^k+1 · · · 1

x1

+k−1 x1

+k−2 · · · k

... ... ...

x₁+1 ^x1 · · · 2

x₁ x₁−1 · · · 1

Comparing denominators, we find thatmin(k+ 1, i) = max(k+ 1, i). So the levelksatisfies the equation k=i−1anda^′ −_b^p′ ≥ ⁱ⁻¹₂ . If the inequality in (3.2) is not strict, thenPn−1

j=1q_j = 0, so the levelkmust also be maximalk=n−1.

x1

+xn

+n−1 · · · _+n^xn x_n ^x₋₁ⁿ · · · 2 1

x1

+n−2 · · · n−1

... ...

x1

+1 · · · 2

x1 · · · 1

If the first inequality in (3.4) is not strict, then recalling (3.3) we find a third constraint on the level k, namelyn−k+ 1−i= 0. This gives usk=i−1 =n−1 = 1. This proves the requirement (3.2) with strict inequality except in the following case:

3 1 1

We now prove the second part of Theorem 3.

12

Proposition 3.4 (Second part of Theorem 3 for Grassmannians). Letl be a positive integer. Let λ be a l-admissible partition ofpwithl-cohomological degreeq >0. Then, we have the inequality

l+q ≤p.

Moreover, if the equalityp=l+qholds, thenλis a hook (i.e., it’s shape is(a,1^b)).

Proof. LetY(λ) := {(i, j) | j ≤λ_i} ⊂ N² be the Young diagram ofλ. Forx ∈ Y(λ), we denote by h(x) the hook number atx. We havep = #Y(λ)and q = #{x ∈ Y(λ) |h(x) > l}. Since q > 0, let x ∈ Y(λ)such that h(x) > l. Moreover, we can assume thatx is minimal for this property, namely that h(y) < l if yis south-east from x. By definition ofh(x), there areh(x)−1elements z ∈ Y(λ) which are either on the same row asxon its right, or underxin the same column. For these elements, we have h(z)< l. This implies thatp−q= #{y ∈Y(λ)|h(y)< l} ≥h(x)−1≥l.

We now deal with the case of equality (that will not be used in the sequel). If the equalityp = l+q occurs, withq >0as above, then we first show thatxis on the first row.

x^′′

x x^′

x^′′

x x^′

Caseλ_i−1=λ_i Caseλ_i−1 > λ_i

Ifxis not on the first row, then the hook number of the boxx^′′, very right on the row over that ofxnot in the same row ofxneither on the same column, is2or1. Hence this box contribute to{y ∈Y(λ)|h(y)< l}

and therefore,p−q > l. In the same way, we can show thatxis on the first column and that all the boxes with hook number smaller thanlare on the hook ofx. Thereforeλis a hook.

3.3. The case of quadrics (typeBnorDn). LetY be a non singular quadric hypersurface of dimensionn with its natural polarisationO_Y(1) =O_Pⁿ⁺¹(1)_|Y. It is the homogeneous space

Y = SO(n+ 2)/(SO(n)×SO(2)).

From the adjunction formula,c₁(Y) =n+ 2−2 = dimY. Recall a theorem of Snow.

Theorem ([Sno86, page 174]). LetY be a nonsingular quadric hypersurface of dimensionn. IfH^q(Y,Ω^p_Y(l))6= 0, then

• eitherp=qandl= 0,

• orq =n−pandl=−n+ 2p,

• orq = 0andl > p,

• orq =nandl <−n+p.

13

As whenq = 0and 0 < p < n, the inequality l > pholds, the cotangent bundleΩY of the quadric is stable as soon as the Picard group of the quadric is the restriction of that ofPⁿ⁺¹, for example whenn≥3.

Theorem 3 follows for quadrics by checking the above cases.

3.4. The case of Lagrangian Grassmannians (typeC_n). In this case,Y is the so-called Lagrangian Grass- mannian, parametrisingn-dimensional Lagrangian subspaces ofC²ⁿequipped with the standard symplectic form. It is the homogeneous spaceY = Sp(2n,C)/U(n). By [Sno88], the non-vanishing ofH^q(Y, Ω^p_Y(l)) amounts to the existence of an l-admissibleC_n-sequence of weightpand cohomological degree q, in the following sense :

Definition 3.5. Fixl, n ∈ Nwith l > 0. A n-uple of integers (x_i)_1≤i≤n will be called anl-admissible C_n-sequenceif

• ∀1≤i≤n,|x_i|=i

• ∀i≤j, xi+xj 6= 2l.

Its weight is defined to be

p := X

xi>0

x_i

and its cohomological degree isq := #{(i, j) | i≤jandx_i+x_j >2l}.

Notation 3.6. Given an integerx, we denotex⁺ := M ax(0, x). Therefore, we havep = P

ix⁺_i . The set of all(u, v) ∈ N² such thatu ≤v andx_u +x_v >2l will be denoted byQ. The cardinality#Qwill be denoted by q. Moreover we adopt the following convention: ifC is a condition on(u, v), thenQ(C) will denote the subset ofQconsisting of pairs satisfying C. For example, given an integerv0, the setQ(v=v0) consists of all pairs(u, v)inQsuch thatv=v₀.

Since this excludes the case ofY being a projective space or a quadric, which occurs whenn ≤ 2, the first part of Theorem 3 amounts to the following proposition is this case :

Proposition 3.7(First part of Theorem 3 for Lagrangian Grassmannians). Let(x_i)be anl-admissibleC_n- sequence of weightpand cohomological degreeq, withn≥3. Then

l+q ≥ 2p n , with equality occurring if and only ifx_i =i, l=n+ 1andq= 0.

Proof. Lett= #{i|x_i > l}. Snow classified the cases wherel= 1[Sno88, Theorem 2.2]. We then have p= ^t(t+1)₂ andq=t². Sincet≤ ⁿ₂, we have

2p

n = 2t(t+ 1)

n ≤t+ 1≤t²+ 1 =q+l .

Moreover, if the equality holds, thent= 1and thusn= 2, but this value ofnhas been excluded. Therefore, the proposition is true in this case.

We now assume thatl≥2. Giveni, j, ifx_i > landx_j > l, then evidentlyx_i+x_j >2l. Therefore, q ≥ t(t+ 1)

2 ≥ 2t−1. On the other hand, we havep≤ ^l(l−1)₂ +tn. If2p≥(l+q)n, then

l(l−1) + 2tn ≥ (2t+l−1)n .

14

Sincel >1, this impliesn≤l, and soq = 0.

Ifl =n, thenx_n =−n, so we have2p≤ n(n−1), therefore, ^2p_n ≤ n−1 < l, and the proposition is true.

Ifl > n, since2p≤n(n+ 1), we get that ^2p_n ≤n+ 1≤l, and if the equality holds thenl=n+ 1and

p= ⁿ⁽ⁿ⁺¹⁾₂ .

We now prove the second part of Theorem 3 :

Proposition 3.8 (Second part of Theorem 3 for Lagrangian Grassmannians). Let(x_i) be anl-admissible C_n-sequence of weightpand cohomological degreeq, withn≥3andq >0. Then

l+q ≤p . Moreover, the equalityp=q+lholds if and only if

x= (−1,−2, . . . ,−l, l+ 1,−(l+ 2),−(l+ 3), . . . ,−n)

withp=l+ 1andq= 1.

Proof. The proof is similar to that of Proposition 3.4. Letj be the minimal integer such that there exists i ≤ j withx_i +x_j > 2l. We have x_j = j ≥ l+ 1. We want to boundq = #Q. We observe that if (u, j) ∈ Qwithj−l ≤ u < j, then x_u > 0. Otherwise, x_u = −uand 0 < x_u+x_j = −u+j ≤ l,

contradicting the assumption thatx_u+x_j >2l. Hence1≤x⁺_u, and therefore

#Q(v=j, j−l≤u < j) ≤ X

j−l≤u<j

x⁺_u .

Actually, a similar inequality holds withj−lreplaced byj−2l, but in the sequel we will use the inequality j−l≥1. In fact, we have

#Q(v=j, u < j−l) ≤ j−l−1. Finally,#Q(v > j)≤P

v>jx⁺_v. Therefore, by minimality ofj, we have the inequality:

q = = #Q(v=j=u) + #Q(v=j, j−l≤u < j) + #Q(v=j, u < j−l) + #Q(v > j)

≤ 1 +P

j−l≤u<jx⁺_u + (j−l−1) +P

v>jx⁺_v

≤ P

u<jx⁺_u +x_j+P

v>jx⁺_v −l=p−l . This proves the inequality.

We now deal with the case of equality. Assume thatq=p−l. Then, asking for equalities in the previous estimates, we find that for j −l ≤ u < j, if x_u > 0, then x_u = 1, and for u < j−l, x_u +j > 2l by the first inequality and x_u < 0by the second. In particular, j −l−1 ≤ 0and hencej = l+ 1 for otherwisex_j−l−1+j=−(j−l−1) +j=l+ 1>2l. Forv₀such thatj=l+ 1< v₀, from the equality Q(v =v₀) = x⁺_v0, we infer that ifx_v0 >0then for allu ≤v₀,x_u+x_v0 >2l. In particular, x_v0−1 >0 andxv0−2 > 0. By decreasing induction, we find thatx_l = l, contradicting thel-admissibility. Hence, for j < v <2l, we getx_v <0. Finally,xis of the form(−1,−2,−3,· · · ,−l, l+1,−(l+2),−(l+3), . . . ,−n) or(1,−2,−3,· · · ,−l, l+ 1,−(l+ 2),−(l+ 3), . . . ,−n). In the second case, one hasp=l+ 2thusq= 2 thusx₁+x_l+1 >2lthusl= 1. But thenxis not1-admissible sincex₁ = 1.

15

3.5. The case of spinor Grassmannians (typeDn). In this caseY is the so-called spinor Grassmannian, that parametrises a family of n-dimensional isotropic subspaces of C²ⁿ equipped with a non-degenerate quadratic form. It is the homogeneous space Y = SO(2n)/U(n). By [Sno88], the non-vanishing of H^q(Y, Ω^p_Y(l))amounts to the existence of an l-admissible D_n-sequence of weight pand cohomological degreeqin the following sense :

Definition 3.9. Fixn, l ∈ Nwithl > 0. A n-uple of integers(x_i)_{0≤i≤n−1} will be called anl-admissible D_n-sequenceif

• |x_i|=ifor all0≤i≤n−1,

• x_i+x_j 6=lfor alli < j.

Its weight is defined to be

p := X

xi>0

xi

and its cohomological degreeq := #{(i, j) | i < jandx_i+x_j > l}.

Remark3.10. Observe that the only1-admissibleDn-sequence is the sequence (0,−1, . . . ,−n)withp = q = 0. In fact, the1-admissibility condition leads to the implication(x_v >0 =⇒ x_v−1 > 0), and thus to x₁ = 1andx₀+x₁ = 1.

We continue to use Notation 3.6 except that now Q = {(i, j) | i < j andx_i +x_j > l}. SinceY is not a projective space or a quadric, we haven ≥ 5. The first part of theorem 3 amounts to the following proposition in this case :

Proposition 3.11 (First part of Theorem 3 for spinor Grassmannians). Let (x_i) be an l-admissible D_n- sequence of weightpand cohomological degreeq, withn≥5. Then

l+q ≥ 4p n , with equality occurring if and only ifxi =iandl= 2(n−1).

Proof. First of all, ifx_n−1 =−(n−1), letx^′ be the sequence of lengthn−1withx^′_i =x_i fori≤n−2.

Thenx^′ is evidentlyl-admissible. It has weightpand cohomological degree q. By induction on n, we get thatl+q ≥ _n−1^4p > ^4p_n. Thus, in the rest of the proof, we assume thatx_n−1 =n−1.

Let us first assume thatl >2(n−1). In this case, we haveq= 0. Since in any casep≤ ⁿ⁽ⁿ⁻¹⁾₂ , we get that ^4p_n ≤2(n−1)< l+q.

Let us now assume thatn ≤ l ≤ 2(n−1). Then, we denote by u the unique integer that satisfies the following condition

#{i | x_i >0, l−n+ 1≤i≤n−1} = u+ 1.

Since the sequence (x_i)isl-admissible, if forl−n+ 1 ≤i ≤n−1we havex_i > 0, thenl−n+ 1 ≤ l−i≤n−1andx_l−i<0. This implies that

n−1−(l−n+ 1)≥2u ,

that is,l≤2(n−u−1). Sincen≤l≤2n−2u−2, we haven≥2u+ 2. The sum of positivexi’s with l−n+ 1≤i≤n−1can be at most(u+ 1)(n−1)−^u(u+1)₂ . Therefore, we have

4p ≤ 4(u+ 1)(n−1)−2u(u+ 1) + 2(l−n+ 1)(l−n).

16

On the other handq≥u, so introducing

∆ := (u+l)n−2(l−n+ 1)(l−n)−4(u+ 1)(n−1) + 2u(u+ 1), the proposition amounts to the positivity of∆whenevern≤l≤2(n−u−1).

After fixingu and n, the above defined∆is a concave function on l, so we only need to consider the values of∆whenl=nand whenl= 2(n−u−1). Whenl=n, we get that

∆ = n²−(4 + 3u)n+ 2u²+ 6u+ 4.

Fixingu, the two roots of this polynomial aren=u+ 2andn= 2u+ 2. Since we know thatn≥2u+ 2, we have∆≥0forl=n≤2(n−u−1). Forl= 2(n−1−u), we have

∆ = 3un−6u(u+ 1)

Since once againn ≥ 2u+ 2, we get that ∆≥ 0, and hence the inequality in the proposition follows for anylsuch thatn≤l≤2(n−u−1).

Moreover, we show that the equalityl+q = ^4p_n can only occur ifxi = iandl = 2(n−1). Indeed, let us assume that∆ = 0. By the concavity argument, we have eitherl=norl= 2(n−u−1). Ifl=n, we also get by the above argument thatn= 2u+ 2. Since the inequality

4p ≤ 4(u+ 1)(n−1)−2u(u+ 1)

becomes an equality, we conclude thatxis of the form(−0,−1,· · · ,−u, u+ 1, u+ 2,· · ·,2u+ 1). This implies thatq = ^u(u+1)₂ , and sinceq =u, we haveu = 1andn= 4, and the last equality contradicts the hypothesis of the proposition. Ifl= 2(n−u−1), since∆ = 3un−6u(u+ 1) = 0, we haven= 2(u+ 1) oru = 0. The case of n = 2(u+ 1), n = l, was already dealt with earlier. Thus we haveu = 0and l= 2(n−1). The equality

4p= 4(u+ 1)(n−1)−2u(u+ 1) + 2(l−n+ 1)(l−n) amounts top= ⁽ⁿ⁻¹⁾ⁿ₂ , so thatx_i =ifor alli, and we are in the case of the proposition.

Let us now assume that l < n. We consider the sequence (x^′_i) with x^′_i = xi for i < n− 1 and x^′_n−1 =−(n−1). We observe that(x^′_i)isl-admissible with weightp^′ =p−(n−1)and cohomological degreeq^′ satisfyingq^′ ≤q−(n−l). In fact,x_i+x_n−1> lfori < n−1−l, andx_n−1−l=n−1−lby l-admissibility, so thatx_n−1−l+x_n−1 > l.

By our very first argument, we have _n−1^4p′ < d+q^′, so that^4p_n < d+q^′+ 4. Therefore, ifq^′ ≤q−4, then we are done. This is indeed the case ifn−l≥4. Thus, we assume thatq^′ ≥q−3, and son≤l+ 3. We now consider these cases.

Ifn=l+ 3, we havexn−1 =l+ 2, and sox2 = 2. Sinceq^′ ≥q−3, we get thatxi <0for3≤i≤n−2.

Thus we havep≤l+ 5. The inequality in the proposition is implied by the inequalityl+ 3 > ^4(l+5)_l+3 , which in turn is true forl ≥ 3. Observe that the valuel = 2is excluded because we would then havex₄ = 4.

Therefore, eitherx₂+x₀ = 2(ifx₂= 2) orx₂+x₄= 2(ifx₂=−2).

Ifn =l+ 2, then there is at most one integerisuch that2≤ i≤n−2andx_i > 0. By admissibility, x₁ = 1, and thereforex_l−1 =−l+ 1. Moreover, we havex_l =−l. This implies thatp≤2l. The inequality of the proposition is implied by the inequalityl+ 2> _l+2^8l , which in turn is true forl≥3.

The valuen=l+ 1would contradictl-admissibility, since we would then havex_l=l.

We now prove the second part :

17

Proposition 3.12(Second part of Theorem 3 for spinor Grassmannians). Let(xi) be anl-admissibleDn- sequence of weightpand cohomological degreeq, withn≥5andq >0. Then

l+q ≤p .

Moreover, the equality p = q +l holds if and only if there are exactly two indices i, j such that x_i >

0, x_j > 0and they satisfy the condition that either x_i+x_j =l+ 1(in this caseq = 1) orxis equal to (0,1,−2,−3, . . . ,−l, l+ 1,−(l+ 2),−(l+ 3), . . . ,−n)(thenq= 2).

Proof. Letjbe the smallest integer such that there existsi < j withxi+xj > l. Observe thatxj >0, and byD_n-admissibility, x₀+x_j 6=lso thatj 6=l. We first deal with the casej > l. The argument in this case is similar to the case of typeC_n:

q= #Q = #Q(v=j, j−l≤u < j) + #Q(v=j, u < j−l) + #Q(v > j)

≤ P

j−l≤u<jx⁺_u + (j−l) +P

v>jx⁺_v

≤ P

u<jx⁺_u +x⁺_j −l+P

v>jx⁺_v =p−l .

If under the assumptionj > lthe equalityp=q+lholds, then by the second inequality we havex_u≤0for u < j−l, and by the first inequality we havex⁺_u ≤1forj−l≤u < j. Ifx_j−l=−(j−l), thenx_j−l+x_j =l, contradictingl-admissibility. Therefore,x_j−l=j−l≤1, sol−l= 1andx₁ = 1. Whenv > j, the first inequality leads to the implication(xv >0 =⇒ ∀u < v, xu+xv > l). Assuming the existence of av > j such thatx_v > 0, we get thatx_v−1 > 0and x_v−2 > 0because l > 1(see Remark 3.10). By descending induction, this would lead tox_j−1>0. Thenj−1 = 1, hencej= 2,l= 1, which is a contradiction. Thus, ifv > j, thenx_v <0. Hencex= (0,1,−2,−3,· · · ,−l, l+ 1,−(l+ 2),−(l+ 3), . . . ,−n)(withp=l+ 2 andq= 2).

Let us now assume thatj < l, and letibe the largest integer such thati < j andx_i+x_j > l. Note that x_i >0andx_j >0, and by maximality ofiwe havex_k<0fori < k < j. We have

q = #Q = 1 + #Q(v=j, u < i) + #Q(v > j)

≤ 1 +P

u<ix⁺_u +P

v>jx⁺_v

≤ (x_i+x_j−l) +P

u<ix⁺_u +P

v>jx⁺_v =p−l .

The inequality is proved. In the case of q = p−l, for1 ≤u < i we havex⁺_u ≤ 1by the first inequality.

By the second inequality we havex_i+x_j =l+ 1. Using the descending induction argument, if there is a v > j such that x_v > 0, then x_l > 0, andx₀ +x_l = lcontradicting the admissibility. The only positive entries are amongx₁, x_i, x_j. In this case, sincei+j =l+ 1, we haveQ={(i, j)}, soq = 1,p=l+ 1

andx₁ <0.

3.6. The exceptional cases (type E₆ or E₇). Now Y is homogeneous under a group of type E₆ (case EIII) orE₇ (caseEV II). In the first case, we havedim(Y) = 16andc₁(Y) = 12. In the second case, we havedim(Y) = 27andc₁(Y) = 18. These values are well-known to the specialists; several arguments for the computation ofc1can be found at the end of Section 2.1 in [CMP08].

From Tables 4.4 and 4.5 in [Sno88] we conclude that the inequalities we are looking for hold :

Proposition 3.13(Theorem 3 for the exceptional cases). LetY be a Hermitian symmetric space of type EIIIorEV II. Letl, p, q be integers withl >0, p >0, and such that

H^q(Y,Ω_Y(l))6= 0.

Then,p_dim(Y^c1(Y)₎ ≤l+q. Equality implies thatp= dim(Y), l=c₁(Y)andq= 0.

Assume moreover thatq >0. Thenl+q≤p.

18

4. RESTRICTION TO A HYPERSURFACE WITH AN INCREASE OF THEPICARD GROUP

In this section, we assume that we are given a compact irreducible Hermitian symmetric space Y as in Section 2. We assumeX ⊂Y is a smooth divisor but we do not make any assumption onPic(X).

4.1. Another argument for general complete intersection. In this section, we want to get rid of the assumption on the Picard group. This can be done at the cost of considering only general complete intersec- tions.

LetV = Γ(Y,OY(1))^∗ be the minimal homogeneous embedding ofY, so thatY ⊂PV. Let S =PS^h1V^∗×. . .×PS^hcV^∗,

and letZ ⊂Y ×S be the universal family of complete intersections defined by (x,([H₁], . . . ,[H_c]))∈Z ⇐⇒ ∀i, H_i(x) = 0.

Thus we have morphisms p : Z → Y and q : Z → S such that for generic s ∈ S, the inverse image Y_s:=q⁻¹(s)is a complete intersection inY.

To prove by contradiction, assume that semistability does not hold. We will use the relative Harder- Narasimhan filtration relative toq : Z → S [HL10, Theorem 2.3.2] (the idea of using this relative version appears e.g. in the proof of [HL10, Theorem 7.1.1]). There is a birational projective morphismf :T →S which induces a commutative diagram

g^∗p^∗ΩY

p^∗ΩY

Z_T

g //Z

q

p //Y

T f

//S

and there is a filtration ofg^∗p^∗Ω_Y, which induce for a generic points∈Sthe Harder-Narasimhan filtration ofΩ_Y|Ys. We denote byF the first term of this filtration and bykits rank. The rank one reflexive subsheaf detF := (VkF)^⋆⋆ofp^∗VkΩ_Y is invertible. SinceSis smooth, f is an isomorphism in codimension 1, so alsog, and sinceZis also smooth,detFdefines a line bundle onZand it is a subsheaf ofVk

p^∗ΩY. We will denotedetFbyL. Now,pis a locally trivial morphism with fibres isomorphic to products of projective spaces, soPic(Z) ≃ Pic(Y)×Pic(S), and Lcan be expressed as p^∗L_Y ⊗q^∗L_S, for some line bundles L_Y ∈Pic(Y)andL_S ∈Pic(S).

Letdbe the integer such thatL_Y ≃ O_Y(−d), and letX=Y_s. Givens∈S, we haveL_|X×{s} ≃ L_Y|X, and hence for generics∈S, this yields an injection of sheavesOY(−d)_|X ⊂p^∗ΩY.

Leth=h₁. . . h_c, we have :

µ(Ω_Y|X) = O_X(1)^dim(X)−1·K_Y

rank Ω_Y|X =− c₁(Y)

dim(Y) ·h·degY µ(F_|X) = O_X(1)^dim(X)−1·detF

rankF =−d

p ·h·degY . SinceO_Y(−d)_|X ⊂ Ω_Y|X, it follows that

H⁰(X, Hom(O_Y(−d)_|X,Ω^k_Y|X)) = H⁰(X,Ω^k_Y|X(d))

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