Home Nef vector bundles on a hyperquadric with first Chern class two
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Nef vector bundles on a hyperquadric with first Chern class two

  • Masahiro Ohno EMAIL logo
Published/Copyright: April 24, 2024

Abstract

We classify nef vector bundles on a smooth hyperquadric of dimension 4 with first Chern class two over an algebraically closed field of characteristic zero.

MSC 2020: 14J60; 14J45; 14F08

1 Introduction

Peternell, Szurek and Wiśniewski classified nef vector bundles on a smooth hyperquadric n of dimension n 3 with first Chern class 1 over an algebraically closed field K of characteristic zero in [11, Section 2, Theorem 2], and we provided a different proof of this classification in [7, Theorem 9.3], proof which was based on an analysis with a full strong exceptional collection of vector bundles on n .

This paper is a continuation of the previous paper [8, Theorem 1.1], in which we have classified nef vector bundles on a smooth quadric threefold 3 with first Chern class two. In this paper, we classify those on a smooth hyperquadric n of dimension n 4 . The precise statement is as follows.

Theorem 1.1.

Let E be a nef vector bundle of rank r on a smooth hyperquadric Q n of dimension n 4 over an algebraically closed field K of characteristic zero, let S , S + , and S - be spinor bundles, and S ± denotes S + or S - . Suppose that det E O ( 2 ) . Then E is isomorphic to one of the following vector bundles or fits in one of the following exact sequences:

  1. 𝒪 ( 2 ) 𝒪 r - 1 ,

  2. 𝒪 ( 1 ) 2 𝒪 r - 2 ,

  3. n = 4 , 𝒪 ( 1 ) 𝒮 ± 𝒪 r - 3 ,

  4. 0 𝒪 ( - 1 ) 𝒪 ( 1 ) 𝒪 r 0 ,

  5. n = 4 , 0 𝒪 𝒮 ± 𝒮 ± 𝒪 r - 3 0 ,

  6. n = 5 , 0 𝒪 𝒮 𝒪 r - 3 0 ,

  7. n = 6 , 0 𝒪 𝒮 ± 𝒪 r - 3 0 ,

  8. n = 4 , 0 𝒮 ± ( - 1 ) 𝒪 ( - 1 ) 𝒪 r + 3 0 ,

  9. 0 𝒪 ( - 1 ) 2 𝒪 r + 2 0 ,

  10. 0 𝒪 ( - 2 ) 𝒪 r + 1 0 .

Note that this list is effective: in each case exists an example. Theorem 1.1 answers the question posed in [9, Question 1]. Note that Case (8) in Theorem 1.1 was missing in [9, Example 1]. Note also that the projectivization ( ) of the bundle in Theorem 1.1 is a Fano manifold of dimension n + r - 1 , i.e., the bundle in Theorem 1.1 is a Fano bundle on n of rank r.

Our basic strategy and framework for describing in Theorem 1.1 is to provide a minimal locally free resolution of in terms of some twists of the full strong exceptional collection

( 𝒪 , 𝒮 , 𝒪 ( 1 ) , 𝒪 ( 2 ) , , 𝒪 ( n - 1 ) )

of vector bundles if n is odd, and collection

( 𝒪 , 𝒮 + , 𝒮 - , 𝒪 ( 1 ) , 𝒪 ( 2 ) , , 𝒪 ( n - 1 ) )

of vector bundles if n is even (see [7] for more details).

The content of this paper is as follows. In Section 2, we briefly recall Bondal’s theorem [2, Theorem 6.2] and its related notions and results used in the proof of Theorem 1.1. In particular, we recall some finite-dimensional algebra A, and fix symbols, e.g., G and S i , related to A and finitely generated right A-modules. We also recall the classification [8, Theorem 1.1] of nef vector bundles on a smooth quadric threefold 3 with Chern class two in Theorem 2.3. In Section 3, we state Hirzebruch–Riemann–Roch formulas for vector bundles on 4 with c 1 = 2 . In Section 4, we provide the set-up for the proof of Theorem 1.1. The proof of Theorem 1.1 is carried out in Sections 5 to 13 according to which case of Theorem 2.3, | 3 belongs to.

1.1 Notation and conventions

Throughout this paper, we work over an algebraically closed field K of characteristic zero. Basically we follow the standard notation and terminology in algebraic geometry. We denote by n a smooth hyperquadric of dimension n over K, and by

  1. 𝒮 , 𝒮 + , 𝒮 - spinor bundles on n .

Note that we follow Kapranov’s convention [5, p. 499]; our spinor bundles 𝒮 , 𝒮 + , 𝒮 - are globally generated, and they are the duals of those of Ottaviani’s [10]. For a coherent sheaf , we denote by c i ( ) the i-th Chern class of and by the dual of . In particular,

  1. c i stands for c i ( ) of the nef vector bundle we are dealing with.

For a vector bundle , ( ) denotes Proj S ( ) , where S ( ) denotes the symmetric algebra of . The tautological line bundle

  1. 𝒪 ( ) ( 1 ) is also denoted by H ( ) .

Let A * 4 be the Chow ring of 4 . We denote

  1. by h the hyperplane class in A 1 4 : A 1 4 = h .

Finally, we refer to [6] for the definition and basic properties of nef vector bundles.

2 Preliminaries

Throughout this paper, G 0 , G 1 , G 2 , G 3 , G m denote respectively

𝒪 , 𝒮 , 𝒪 ( 1 ) , 𝒪 ( 2 ) , , 𝒪 ( n - 1 )

on n if n is odd, where m = n , and

𝒪 , 𝒮 + , 𝒮 - , 𝒪 ( 1 ) , , 𝒪 ( n - 1 )

on n if n is even, where m = n + 1 . An important and well-known fact [5, Theorem 4.10] of the collection ( G 0 , , G m ) is that it is a full strong exceptional collection in D b ( n ) , where D b ( n ) denotes the bounded derived category of (the abelian category of) coherent sheaves on n . An exceptional collection is also called an exceptional sequence. We refer to [4] for the definition of a full strong exceptional sequence.

Denote by G the direct sum i = 0 m G i of G 0 , , G m , and by A the endomorphism ring End ( G ) of G. Then A is a finite-dimensional K-algebra, and G is a left A-module. Denote by mod A the category of finitely generated right A-modules, and by D b ( mod A ) the bounded derived category of mod A . Let p i : G G i be the projection, and ι i : G i G the inclusion. Set e i = ι i p i . Then e i A . Set P i = e i A . Then A i P i as right A-modules, and hence P i is projective. We have P i A G G i . Any finitely generated right A-module V has an ascending filtration

0 = V - 1 V 0 V 1 V m = V

by right A-submodules, where V i = j i V e j . Set Gr i V = V i / V i - 1 and S i = Gr i P i . Then S i is a simple right A-module. If we set m i = dim K Gr i V , then Gr i V S i m i . For a coherent sheaf on n , Ext q ( G , ) is a finitely generated right A-module.

It follows from Bondal’s theorem [2, Theorem 6.2] that

RHom ( G , ) : D b ( n ) D b ( mod A )

is an exact equivalence, and its quasi-inverse is

L A G : D b ( mod A ) D b ( n ) .

For a coherent sheaf on n , this fact can be rephrased in terms of a spectral sequence [9, Theorem 1]

(2.1) E 2 p , q = 𝒯 o r - p A ( Ext q ( G , ) , G ) E p + q = { if  p + q = 0 , 0 if  p + q 0 ,

which we call the Bondal spectral sequence. Note that E 2 p , q is the p-th cohomology sheaf p ( Ext q ( G , ) L A G ) of the complex Ext q ( G , ) L A G . When we compute the spectral sequence, we consider the descending filtration on the right A-module Ext q ( G , ) , and consider distinguished triangles consisting of the pull-backs Gr i Ext q ( G , ) L A G of the graded pieces Gr i Ext q ( G , ) . Thus it is important to know what the pull-back S i L A G of the component S i of the graded piece Gr i Ext q ( G , ) is.

Here we recall the bundles Ψ i on n , which is defined by Kapranov in [5, Section 4], and which is characterized by Ancona and Ottaviani in [1, Section 5] by the non-split exact sequences

0 Ω n + 1 i ( i ) | n Ψ i Ψ i - 2 0

for i 2 with Ψ 0 = 𝒪 and Ψ 1 = Ω n + 1 1 ( 1 ) | n . Set

Ψ ~ i = Ψ n + 1 - i Ω n + 1 n + 1 ( n + 1 ) | n .

Then Ψ ~ n + 1 = 𝒪 ( - 1 ) , Ψ ~ n = T n + 1 ( - 2 ) | n Ω n + 1 n ( n ) | n , and, for 2 i n + 1 , Ψ ~ i fits in the following non-split exact sequence:

(2.2) 0 Ψ ~ i Ψ ~ i - 2 Ω n + 1 i - 2 ( i - 2 ) | n 0 .

First we note the following:

Lemma 2.1.

Let S ( ± ) denotes S + or S - if n is even and S if n is odd. Then:

  1. RHom ( 𝒪 ( k ) , 𝒮 ( ± ) ( - 1 ) ) = 0 for 0 k n - 1 .

  2. RHom ( 𝒮 ( ± ) , Ψ ~ p ) = 0 for 3 p n + 1 .

Proof.

(1) This follows from [5, Proposition 4.11].

(2) Recall that 𝒮 ( ± ) 𝒮 ( ± ) ( - 1 ) (see, e.g., [7, Theorem 8.1 (9), (10), (11)]). Hence we have RHom ( 𝒮 ( ± ) , 𝒪 ( - k ) ) = 0 for 0 k n - 1 by (1). In particular, RHom ( 𝒮 ( ± ) , Ψ ~ n + 1 ) = 0 . Set V = H 0 ( 𝒪 n + 1 ( 1 ) ) . Now Ω n + 1 p ( p ) | n is isomorphic to n + 1 - p ( T n + 1 ( - 1 ) | n ) ( - 1 ) , and we have the following exact sequence:

0 𝒪 ( - 1 ) 𝒪 V T n + 1 ( - 1 ) | n 0 .

This exact sequence induces the following exact sequence:

0 k - 1 ( T n + 1 ( - 1 ) | n ) ( - 1 ) 𝒪 k V k ( T n + 1 ( - 1 ) | n ) 0 .

Hence k ( T n + 1 ( - 1 ) | n ) ( - 1 ) has the following resolution:

0 𝒪 ( - k - 1 ) 𝒪 ( - k ) V 𝒪 ( - 1 ) k V k ( T n + 1 ( - 1 ) | n ) ( - 1 ) 0 .

Therefore

RHom ( 𝒮 ( ± ) , k ( T n + 1 ( - 1 ) | n ) ( - 1 ) ) = 0 for  0 k n - 2 .

Hence we see that RHom ( 𝒮 ( ± ) , Ψ ~ n ) = 0 and that

RHom ( 𝒮 ( ± ) , Ω n + 1 p ( p ) | n ) = 0 for  3 p n + 1 .

Now the exact sequence (2.2) inductively shows that RHom ( 𝒮 ( ± ) , Ψ ~ p ) = 0 for 3 p n + 1 . ∎

Now we can answer the question that what S i L A G is.

Lemma 2.2.

The following statements hold:

  1. If n is odd, we have

    (2.3) S p - 1 L A G Ψ ~ p [ p - 1 ] for  n + 1 p 3 ,
    (2.4) S 1 L A G 𝒮 ( - 1 ) [ 1 ] ,
    (2.5) S 0 L A G 𝒪 .

  2. If n is even, we have

    (2.6) S p L A G Ψ ~ p [ p - 1 ] for  n + 1 p 3 ,
    (2.7) S 2 L A G 𝒮 + ( - 1 ) [ 1 ] ,
    (2.8) S 1 L A G 𝒮 - ( - 1 ) [ 1 ] ,
    (2.9) S 0 L A G 𝒪 .

Proof.

It follows from [5, Proposition 4.11] that for 0 j , k n - 1 and 0 q n we have h j ( Ψ j ( - j ) ) = 1 and h q ( Ψ j ( - k ) ) = 0 unless ( q , k ) = ( j , j ) . Since

h q ( Ψ ~ p ( - k ) ) = h q ( Ψ n + 1 - p ( - 1 - k ) ) = h n - q ( Ψ n + 1 - p ( 1 + k - n ) ) ,

this implies that for 2 p n + 1 and 0 k n - 1 we have

(2.10) h p - 1 ( Ψ ~ p ( - p + 2 ) ) = 1 and h q ( Ψ ~ p ( - k ) ) = 0 unless  ( q , k ) = ( p - 1 , p - 2 ) .

Suppose that n is odd. Then (2.10) and Lemma 2.1 (2) show that RHom ( G , Ψ ~ p ) = S p - 1 [ 1 - p ] for 3 p n + 1 . Thus we obtain (2.3). Suppose that n is even. Then (2.10) and Lemma 2.1 (2) show that RHom ( G , Ψ ~ p ) = S p [ 1 - p ] for 3 p n + 1 . Thus we obtain (2.6).

Suppose that n is odd. It follows from [7, Lemma 8.2 (1)] and Lemma 2.1 (1) that RHom ( G , 𝒮 ( - 1 ) ) = S 1 [ - 1 ] . Hence we obtain (2.4). Suppose that n is even. Then it follows from [7, Lemma 8.2 (1)] and Lemma 2.1 (1) that RHom ( G , 𝒮 - ( - 1 ) ) = S 1 [ - 1 ] and RHom ( G , 𝒮 + ( - 1 ) ) = S 2 [ - 1 ] . Hence we obtain (2.8) and (2.7).

Finally, (2.5) and (2.9) hold since S 0 is nothing but a projective module P 0 = e 0 A . ∎

Theorem 1.1 will be proved, based on the following theorem [8, Theorem 1.1]:

Theorem 2.3.

Let E be a nef vector bundle of rank r on a smooth hyperquadric Q 3 of dimension 3 over an algebraically closed field K of characteristic zero, and let S be the spinor bundle on Q 3 . Suppose that det E O ( 2 ) . Then E is isomorphic to one of the following vector bundles or fits in one of the following exact sequences:

  1. 𝒪 ( 2 ) 𝒪 r - 1 ,

  2. 𝒪 ( 1 ) 2 𝒪 r - 2 ,

  3. 𝒪 ( 1 ) 𝒮 𝒪 r - 3 ,

  4. 0 𝒪 ( - 1 ) 𝒪 ( 1 ) 𝒪 r 0 ,

  5. 0 𝒪 a 𝒮 2 𝒪 r - 4 + a 0 , where a = 0 or 1 , and the composite of the injection 𝒪 a 𝒮 2 𝒪 r - 4 + a and the projection 𝒮 2 𝒪 r - 4 + a 𝒪 r - 4 + a is zero,

  6. 0 𝒮 ( - 1 ) 𝒪 ( - 1 ) 𝒪 r + 3 0 ,

  7. 0 𝒪 ( - 1 ) 2 𝒪 r + 2 0 ,

  8. 0 𝒪 ( - 2 ) 𝒪 r + 1 0 ,

  9. 0 𝒪 ( - 2 ) 𝒪 ( - 1 ) 4 𝒪 r + 3 0 .

3 Hirzebruch–Riemann–Roch formulas

Let be a vector bundle of rank r on 4 . Since the tangent bundle T of 4 fits in an exact sequence

0 T T 5 | 4 𝒪 4 ( 2 ) 0 ,

the Chern polynomial c t ( T ) of T is

( 1 + h t ) 6 1 + 2 h t = 1 + 4 h t + 7 h 2 t 2 + 6 h 3 t 3 + 3 h 4 t 4 ,

where h denotes c 1 ( 𝒪 4 ( 1 ) ) . Then the Hirzebruch–Riemann–Roch formula implies that

χ ( ) = r + 7 6 c 1 h 3 + 23 24 ( c 1 2 - 2 c 2 ) h 2 + 1 3 ( c 1 3 - 3 c 1 c 2 + 3 c 3 ) h + 1 24 ( c 1 4 - 4 c 1 2 c 2 + 4 c 1 c 3 + 2 c 2 2 - 4 c 4 ) ,

where we set c i = c i ( ) . To compute χ ( ( t ) ) , note that

c 1 ( ( t ) ) = c 1 + r h t ,
c 2 ( ( t ) ) = c 2 + ( r - 1 ) c 1 h t + ( r 2 ) h 2 t 2 ,
c 3 ( ( t ) ) = c 3 + ( r - 2 ) c 2 h t + ( r - 1 2 ) c 1 h 2 t 2 + ( r 3 ) h 3 t 3 ,
c 4 ( ( t ) ) = c 4 + ( r - 3 ) c 3 h t + ( r - 2 2 ) c 2 h 2 t 2 + ( r - 1 3 ) h 3 t 3 + ( r 4 ) h 4 t 4 .

Since h 4 = 2 , we infer that

(3.1) χ ( ( t ) ) = r 12 ( t + 1 ) ( t + 2 ) 2 ( t + 3 ) + 1 6 c 1 h 3 t 3 + 1 4 { 4 c 1 h + ( c 1 2 - 2 c 2 ) } h 2 t 2 + 1 12 { 23 c 1 h 2 + 12 ( c 1 2 - 2 c 2 ) h + 2 ( c 1 3 - 3 c 1 c 2 + 3 c 3 ) } h t + 7 6 c 1 h 3 + 23 24 ( c 1 2 - 2 c 2 ) h 2 + 1 3 ( c 1 3 - 3 c 1 c 2 + 3 c 3 ) h + 1 24 ( c 1 4 - 4 c 1 2 c 2 + 4 c 1 c 3 + 2 c 2 2 - 4 c 4 ) .

Since c 1 ( ) = d h for some integer d, the formula above can be written as

(3.2) χ ( ( t ) ) = r 12 ( t + 1 ) ( t + 2 ) 2 ( t + 3 ) + d 3 t 3 + 1 2 { 4 d + ( d 2 - c 2 h 2 ) } t 2 + 1 6 { 23 d + 12 ( d 2 - c 2 h 2 ) + ( 2 d 3 - 3 d c 2 h 2 + 3 c 3 h ) } t + 7 3 d + 23 12 ( d 2 - c 2 h 2 ) + 1 3 ( 2 d 3 - 3 d c 2 h 2 + 3 c 3 h ) + 1 12 ( d 4 - 2 d 2 c 2 h 2 + 2 d c 3 h + c 2 2 - 2 c 4 ) .

In this paper, we are dealing with the case d = 2 :

(3.3) χ ( ( t ) ) = r 12 ( t + 1 ) ( t + 2 ) 2 ( t + 3 ) + 1 3 ( 2 t 3 + 18 t 2 + 55 t + 57 ) - 1 2 c 2 h 2 t 2 + 1 2 ( c 3 h - 6 c 2 h 2 ) t + 1 12 ( 16 c 3 h - 55 c 2 h 2 ) + 1 12 ( c 2 2 - 2 c 4 ) .

In particular,

(3.4) χ ( ( - 1 ) ) = 6 - 25 12 c 2 h 2 + 5 6 c 3 h + 1 12 ( c 2 2 - 2 c 4 ) ,
(3.5) χ ( ( - 2 ) ) = 1 - 7 12 c 2 h 2 + 1 3 c 3 h + + 1 12 ( c 2 2 - 2 c 4 ) .

Next we will compute χ ( 𝒮 ± ( t ) ) . Recall that c 1 ( 𝒮 ± ) = h . Note also that

rank 𝒮 ± = 2 r ,
c 1 ( 𝒮 ± ) = 2 c 1 - r h ,
c 2 ( 𝒮 ± ) = 2 c 2 - ( 2 r - 1 ) c 1 h + c 1 2 + ( r 2 ) h 2 + r c 2 ( 𝒮 ± ) ,
c 3 ( 𝒮 ± ) = 2 c 3 - 2 ( r - 1 ) c 2 h + ( r - 1 ) 2 c 1 h 2 + 2 ( r - 1 ) c 1 c 2 ( 𝒮 ± )
+ 2 c 1 c 2 - ( r - 1 ) c 1 2 h - ( r 3 ) h 3 - r ( r - 1 ) c 2 ( 𝒮 ± ) h ,
c 4 ( 𝒮 ± ) = 2 c 4 + 2 c 1 c 3 + c 2 2 - ( 2 r - 3 ) c 3 h + ( r 2 - 3 r + 3 ) c 2 h 2 + 2 ( r - 3 ) c 2 c 2 ( 𝒮 ± )
- 1 6 ( r - 1 ) ( r - 2 ) ( 2 r - 3 ) c 1 h 3 - ( r - 1 ) ( 2 r - 3 ) c 1 c 2 ( 𝒮 ) h
- ( 2 r - 3 ) c 1 c 2 h + ( r - 1 2 ) c 1 2 h 2 + ( r - 1 ) c 1 2 c 2 ( 𝒮 ) + 1 12 r 2 ( r 2 - 1 ) .

Since c 2 ( 𝒮 ± ) 2 = 1 and c 2 ( 𝒮 ± ) h 2 = 1 , formula (3.1) together with the formulas above implies the formula

χ ( 𝒮 ± ( t ) ) = r 6 t ( t + 1 ) ( t + 2 ) ( t + 3 ) + 1 3 c 1 h 3 t 3 + 1 2 ( 3 c 1 h 3 + c 1 2 h 2 - 2 c 2 h 2 ) t 2
+ 1 6 ( 14 c 1 h 3 + 9 c 1 2 h 2 - 18 c 2 h 2 + 2 c 1 3 h - 6 c 1 c 2 ( 𝒮 ± ) h - 6 c 1 c 2 h + 6 c 3 h ) t
+ 1 12 ( 15 c 1 h 3 + 14 c 1 2 h 2 - 28 c 2 h 2 + 6 c 1 3 h - 18 c 1 c 2 ( 𝒮 ± ) h - 18 c 1 c 2 h + 18 c 3 h )
+ 1 12 ( c 1 4 - 4 c 1 2 c 2 + 4 c 1 c 3 + 2 c 2 2 - 4 c 4 ) - 1 2 c 1 2 c 2 ( 𝒮 ± ) + c 2 c 2 ( 𝒮 ± ) .

Since c 1 ( ) = d h , the formula above becomes the formula

(3.6) χ ( 𝒮 ± ( t ) ) = r 6 t ( t + 1 ) ( t + 2 ) ( t + 3 ) + 2 3 d t 3 + ( d 2 + 3 d - c 2 h 2 ) t 2 + 1 3 { 2 d 3 + 9 d 2 + 11 d - 3 ( d + 3 ) c 2 h 2 + 3 c 3 h } t + 1 6 d ( d + 1 ) ( d + 2 ) ( d + 3 ) - 1 6 ( 2 d 2 + 9 d + 14 ) c 2 h 2 + 1 6 ( 2 d + 9 ) c 3 h + 1 6 ( c 2 2 - 2 c 4 ) + c 2 c 2 ( 𝒮 ± ) .

In this paper, we are dealing with the case d = 2 :

(3.7) χ ( 𝒮 ± ( t ) ) = r 6 t ( t + 1 ) ( t + 2 ) ( t + 3 ) + 2 3 ( 2 t + 5 ) ( t + 2 ) ( t + 3 ) - ( t 2 + 5 t + 20 3 ) c 2 h 2 + ( t + 13 6 ) c 3 h + 1 6 ( c 2 2 - 2 c 4 ) + c 2 c 2 ( 𝒮 ± ) .

In particular, we have the following:

(3.8) χ ( 𝒮 ± ( - 1 ) ) = 4 - 8 3 c 2 h 2 + 7 6 c 3 h + 1 6 ( c 2 2 - 2 c 4 ) + c 2 c 2 ( 𝒮 ± ) .

4 Set-up for the proof of Theorem 1.1

Let be a nef vector bundle of rank r on n of dimension n 4 with c 1 = 2 h . If h 0 ( ( - 2 ) ) 0 , then

𝒪 ( 2 ) 𝒪 r - 1

by [7, Proposition 5.1 and Remark 5.3]. This is case (1) of Theorem 1.1. Thus we will always assume that

(4.1) h 0 ( ( - 2 ) ) = 0

in the following. It follows from [7, Lemma 4.1 (1)] that

(4.2) h q ( ( t ) ) = 0 for  q > 0  and  t - 1 .

Furthermore, from (4.1) and [7, Lemma 4.1 (1)], it follows that

(4.3) h q ( ( - 2 ) ) = 0 for any  q  if  n 5 .

Denote by ( 𝒮 ( ± ) , 𝒮 ( ) ) a pair ( 𝒮 + , 𝒮 - ) or ( 𝒮 - , 𝒮 + ) of spinor bundles if n is even, and a pair ( 𝒮 , 𝒮 ) of the spinor bundle if n is odd. Ottaviani shows in [10, Theorem 2.8] that we have an exact sequence

(4.4) 0 𝒮 ( ± ) 𝒪 α 𝒮 ( ) ( 1 ) 0 ,

where α = 2 n 2 . Hence we have the following exact sequences:

(4.5) 0 𝒮 ( ± ) ( - 2 ) ( - 2 ) α 𝒮 ( ) ( - 1 ) 0

and

(4.6) 0 𝒮 ( ± ) ( - 1 ) ( - 1 ) α 𝒮 ( ) 0 .

The exact sequence (4.5) together with (4.3) implies that

(4.7) h q ( 𝒮 ( ) ( - 1 ) ) = h q + 1 ( 𝒮 ( ± ) ( - 2 ) ) for any  q  if  n 5 .

The exact sequence (4.6) together with (4.2) implies that

(4.8) h q ( 𝒮 ( ) ) = h q + 1 ( 𝒮 ( ± ) ( - 1 ) ) for any  q  if  h 0 ( ( - 1 ) ) = 0 .

In order to compute Ext q ( 𝒮 ( ) , ( - 1 ) ) , we will use the following exact sequence together with (4.5) or (4.7):

(4.9) 0 𝒮 ( ± ) ( - 2 ) 𝒮 ( ± ) ( - 1 ) ( 𝒮 ( ± ) ( - 1 ) ) | n - 1 0 .

Similarly, in order to compute Ext q ( 𝒮 ( ) , ) , we will use the following exact sequence together with (4.6) or (4.8):

(4.10) 0 𝒮 ( ± ) ( - 1 ) 𝒮 ( ± ) ( 𝒮 ( ± ) ) | n - 1 0 .

Finally, recall that 𝒮 ± | n - 1 𝒮 if n is even and that 𝒮 | n - 1 𝒮 + 𝒮 - if n is odd (see, e.g., [7, Theorem 8.1 (1) and (3)]).

4.1 The case n = 4

It follows from [7, Lemma 4.1 (2)] that, if

H ( ) r + 3 = c 2 2 - c 4 + 2 c 1 c 3 - 3 c 1 2 c 2 + c 1 4 = c 2 2 - c 4 + 4 c 3 h - 12 c 2 h 2 + 32 > 0 ,

then

h q ( ( - 2 ) ) = 0 for  q > 0 .

Let Δ λ ( ) be the Schur polynomial of (see, e.g., [3, Example 12.1.7]). Note here that

c 4 = Δ ( 4 , 0 , 0 , 0 ) ( ) 0 , c 1 c 3 - c 4 = Δ ( 3 , 1 , 0 , 0 ) ( ) 0 , c 2 2 - c 1 c 3 = Δ ( 2 , 2 , 0 , 0 ) ( ) 0 ,

since is nef (see, e.g., [6, Chapter 8]). Hence we have

(4.11) c 2 2 c 1 c 3 c 4 0 .

Therefore

H ( ) r + 3 = c 2 2 - c 4 + 4 c 3 h - 12 c 2 h 2 + 32 4 c 3 h - 12 c 2 h 2 + 32 .

Thus we see that

(4.12) h q ( ( - 2 ) ) = 0 for  q > 0  if  c 3 h - 3 c 2 h 2 + 8 > 0 .

Hence it follows from (4.5), (4.1), and (4.12) that

(4.13) h q ( 𝒮 ( ) ( - 1 ) ) = h q + 1 ( 𝒮 ( ± ) ( - 2 ) ) for any  q  if  c 3 h - 3 c 2 h 2 + 8 > 0 .

5 The case where | 3 belongs to case (1) of Theorem 2.3

This case does not arise under assumption (4.1). Indeed, if | 3 𝒪 ( 2 ) 𝒪 r - 1 and n = 4 , we have c 2 h 2 = 0 and c 3 h = 0 . Hence h q ( ( - 2 ) ) = 0 for any q by (4.12) and (4.1), and we obtain χ ( ( - 2 ) ) = 0 . From (3.5), it follows that

(5.1) 1 12 ( c 2 2 - 2 c 4 ) = - 1 .

We have h q ( 𝒮 ( - 1 ) ) = h q + 1 ( 𝒮 ± ( - 2 ) ) for any q by (4.13). In particular, h 0 ( 𝒮 ± ( - 2 ) ) = 0 . Since 𝒮 ( - 1 ) | 3 𝒮 𝒮 ( - 2 ) r - 1 , we have h 0 ( 𝒮 ( - 1 ) | 3 ) = 4 and h q ( 𝒮 ( - 1 ) | 3 ) = 0 for q > 0 . The exact sequence (4.9) then implies that h q + 1 ( 𝒮 ± ( - 2 ) ) = h q + 1 ( 𝒮 ± ( - 1 ) ) for q 1 . Hence we have

h q ( 𝒮 ( - 1 ) ) = h q + 1 ( 𝒮 ± ( - 1 ) ) for  q 1 .

Consequently, we see that h q ( 𝒮 ( - 1 ) ) = 0 for q 1 . Set a = h 0 ( 𝒮 + ( - 1 ) ) . Then

4 - a = h 1 ( 𝒮 + ( - 2 ) ) = h 0 ( 𝒮 - ( - 1 ) ) .

Therefore it follows from (3.8) and (5.1) that

a = χ ( 𝒮 + ( - 1 ) ) = 2 + c 2 c 2 ( 𝒮 + )

and that

4 - a = χ ( 𝒮 - ( - 1 ) ) = 2 + c 2 c 2 ( 𝒮 - ) .

Since c 2 ( 𝒮 ± ) [ 4 ] is represented by a 2-plane 2 , the nefness of | 2 implies that c 2 c 2 ( 𝒮 ± ) 0 by [6, Theorem 8.2.1]. Hence we conclude that a = 2 and c 2 c 2 ( 𝒮 ± ) = 0 . Set σ = c 2 ( 𝒮 + ) [ 4 ] and τ = c 2 ( 𝒮 - ) [ 4 ] . Then A 2 4 is generated by σ and τ. Since c 2 σ = 0 and c 2 τ = 0 , it follows from c 2 A 2 4 that c 2 2 = 0 . Hence we have c 4 = 0 by (4.11). This however contradicts (5.1). Therefore this case does not arise under assumption (4.1).

6 The case where | 3 belongs to case (2) of Theorem 2.3

6.1 The case n = 4

Suppose that | 3 𝒪 ( 1 ) 2 𝒪 r - 2 . Then c 2 h 2 = 2 and c 3 h = 0 . Hence h q ( ( - 2 ) ) = 0 for any q by (4.12) and (4.1). Since h 0 ( ( - 1 ) | 3 ) = 2 and h q ( ( - 1 ) | 3 ) = 0 for q > 0 , this implies that h 0 ( ( - 1 ) ) = 2 and that h q ( ( - 1 ) ) = 0 for q > 0 . Since h q ( ( - 2 ) | 3 ) = 0 for any q, we infer that h q ( ( - 3 ) ) = 0 for any q. Since h 3 ( ( - 3 ) | 3 ) = r - 2 and h q ( ( - 3 ) | 3 ) = 0 unless q = 3 , we infer that h 4 ( ( - 4 ) ) = r - 2 and that h q ( ( - 4 ) ) = 0 unless q = 4 .

We have

h q ( 𝒮 ( - 1 ) ) = h q + 1 ( 𝒮 ± ( - 2 ) ) for any  q

by (4.13). Since ( 𝒮 ± ( - 1 ) ) | 3 𝒮 ( - 1 ) 2 𝒮 ( - 2 ) r - 2 , we see that h q ( ( 𝒮 ± ( - 1 ) ) | 3 ) = 0 for any q. The exact sequence (4.9) then implies that

h q + 1 ( 𝒮 ± ( - 2 ) ) = h q + 1 ( 𝒮 ± ( - 1 ) ) for any  q .

Hence h q ( 𝒮 ( - 1 ) ) = h q + 1 ( 𝒮 ± ( - 1 ) ) for any q. Thus h q ( 𝒮 ( - 1 ) ) = 0 for any q.

Apply to ( - 1 ) the Bondal spectral sequence (2.1). We see that Hom ( G , ( - 1 ) ) = S 0 2 , that Ext q ( G , ( - 1 ) ) = 0 for 1 q 3 , and that Ext 4 ( G , ( - 1 ) ) = S 5 r - 2 . Hence we have E 2 p , q = 0 unless q = 0 or q = 4 , E 2 p , 0 = 0 unless p = 0 , E 2 0 , 0 = 𝒪 2 , E 2 p , 4 = 0 unless p = - 4 , and E 2 - 4 , 4 = 𝒪 ( - 1 ) r - 2 by Lemma 2.2 (2). Therefore ( - 1 ) fits in an exact sequence

0 𝒪 2 ( - 1 ) 𝒪 ( - 1 ) r - 2 0 .

Hence 𝒪 ( 1 ) 2 𝒪 r - 2 . This is case (2) of Theorem 1.1.

6.2 The case n 5

Suppose that | n - 1 𝒪 ( 1 ) 2 𝒪 r - 2 . We will show that is isomorphic to 𝒪 ( 1 ) 2 𝒪 r - 2 by applying to ( - 1 ) the Bondal spectral sequence (2.1).

First, h q ( ( - 2 ) ) = 0 for any q by (4.3). Hence we see that h 0 ( ( - 1 ) ) = 2 . We have h q ( ( - 1 ) ) = 0 for q > 0 by (4.2). Note that h q ( ( - t ) | n - 1 ) = 0 for any q and any t ( 2 t n - 2 ) . Since h q ( ( - 2 ) ) = 0 for any q, this implies that h q ( ( - t ) ) = 0 for any q and any t ( 2 t n - 1 ) . As h n - 1 ( ( - n + 1 ) | n - 1 ) = r - 2 and h q ( ( - n + 1 ) | n - 1 ) = 0 unless q = n - 1 , we infer that h n ( ( - n ) ) = r - 2 and that h q ( ( - n ) ) = 0 unless q = n .

Note that h q ( ( 𝒮 ( ± ) ( - 1 ) ) | n - 1 ) = 0 for any q. Therefore the exact sequence (4.9) implies that

h q + 1 ( 𝒮 ( ± ) ( - 2 ) ) = h q + 1 ( 𝒮 ( ± ) ( - 1 ) ) for any  q .

This implies that h q ( 𝒮 ( ) ( - 1 ) ) = h q + 1 ( 𝒮 ( ± ) ( - 1 ) ) for any q by (4.7). Hence Ext q ( 𝒮 ( ) , ( - 1 ) ) = 0 for any q.

Lemma 2.2 shows that E 2 p , q = 0 unless ( p , q ) = ( 0 , 0 ) or ( - n , n ) , that E 2 0 , 0 = 𝒪 2 , and that E 2 - n , n = 𝒪 ( - 1 ) r - 2 . Therefore ( - 1 ) fits in an exact sequence

0 𝒪 2 ( - 1 ) 𝒪 ( - 1 ) r - 2 0 .

Hence 𝒪 ( 1 ) 2 𝒪 r - 2 . This is case (2) of Theorem 1.1.

7 The case where | 3 belongs to case (3) of Theorem 2.3

7.1 The case n = 4

Suppose that | 3 𝒪 ( 1 ) 𝒮 𝒪 r - 3 . Then c 2 h 2 = 3 and c 3 h = 1 . We will show that 𝒪 ( 1 ) 𝒮 ± 𝒪 r - 3 by applying to ( - 1 ) the Bondal spectral sequence (2.1).

We have h q ( ( - 1 ) ) = 0 for q > 0 by (4.2). Note that h q ( ( - 1 ) | 3 ) = 0 for q > 0 and that h 0 ( ( - 1 ) | 3 ) = 1 . Hence h q ( ( - 2 ) ) = 0 for q 2 . Assumption (4.1) together with (3.5) and (4.11) shows that

- h 1 ( ( - 2 ) ) = χ ( ( - 2 ) ) = - 5 12 + 1 12 ( c 2 2 - 2 c 4 ) - 5 + c 4 12 .

Note here that c 1 c 3 = 2 since c 3 h = 1 . Hence c 4 2 by (4.11). Therefore h 1 ( ( - 2 ) ) 7 12 , and thus h 1 ( ( - 2 ) ) = 0 . Since h 0 ( ( - 1 ) | 3 ) = 1 , this implies that h 0 ( ( - 1 ) ) = 1 . Now that h q ( ( - 2 ) ) = 0 for any q, we have

h q ( ( - 3 ) ) = h q - 1 ( ( - 2 ) | 3 ) = 0 for any  q .

Hence h q ( ( - 4 ) ) = h q - 1 ( ( - 3 ) | 3 ) for any q. Note that h q ( ( - 3 ) | 3 ) = 0 for q < 3 and that h 3 ( ( - 3 ) | 3 ) = r - 3 . Thus h q ( ( - 4 ) ) = 0 for q < 4 and h 4 ( ( - 4 ) ) = r - 3 .

As 𝒮 ( - 1 ) | 3 𝒮 ( 𝒮 𝒮 ( - 1 ) ) 𝒮 ( - 1 ) r - 3 , it follows from [7, Lemma 8.2 (1)] that h 1 ( 𝒮 ( - 1 ) | 3 ) = 1 and that h q ( 𝒮 ( - 1 ) | 3 ) = 0 unless q = 1 . Since h q ( ( - 2 ) ) = 0 for any q, the exact sequence (4.5) implies that h q ( 𝒮 ( - 1 ) ) = h q + 1 ( 𝒮 ± ( - 2 ) ) for any q. In particular, we have

h 4 ( 𝒮 ( - 1 ) ) = 0 and h 0 ( 𝒮 ± ( - 2 ) ) = 0 .

Since we have the exact sequence (4.9) and h q ( 𝒮 ( - 1 ) | 3 ) = 0 unless q = 1 , we infer that h 0 ( 𝒮 ± ( - 2 ) ) = h 0 ( 𝒮 ± ( - 1 ) ) and that h q + 1 ( 𝒮 ± ( - 2 ) ) = h q + 1 ( 𝒮 ± ( - 1 ) ) for q 2 . Hence we have h 0 ( 𝒮 ± ( - 1 ) ) = 0 and h q ( 𝒮 ( - 1 ) ) = h q + 1 ( 𝒮 ± ( - 1 ) ) for q 2 . Consequently, h q ( 𝒮 ( - 1 ) ) = 0 for q 2 . Note also that h 1 ( 𝒮 ± ( - 2 ) ) = 0 and that h q ( 𝒮 ± ( - 2 ) ) = 0 for q 3 . Set a = h 1 ( 𝒮 ± ( - 1 ) ) . Since h 1 ( 𝒮 ( - 1 ) | 3 ) = 1 , we see that a = 0 or 1 and that 1 - a = h 2 ( 𝒮 ± ( - 2 ) ) = h 1 ( 𝒮 ( - 1 ) ) . Thus ( h 1 ( 𝒮 + ( - 1 ) ) , h 1 ( 𝒮 - ( - 1 ) ) ) is either ( 0 , 1 ) or ( 1 , 0 ) .

We see that Ext 4 ( G , ( - 1 ) ) = S 5 r - 3 , that Ext q ( G , ( - 1 ) ) = 0 for q = 3 , 2 , and that Hom ( G , ( - 1 ) ) = S 0 . Then Lemma 2.2 shows that E 2 p , 4 = 0 unless p = - 4 , that E 2 - 4 , 4 = 𝒪 ( - 1 ) r - 3 , that E 2 p , q = 0 for any p if q = 3 , 2 , that E 2 p , 0 = 0 unless p = 0 , and that E 2 0 , 0 = 𝒪 .

If ( h 1 ( 𝒮 + ( - 1 ) ) , h 1 ( 𝒮 - ( - 1 ) ) ) = ( 0 , 1 ) , then Ext 1 ( G , ( - 1 ) ) = S 2 , and Lemma 2.2 shows that E 2 p , 1 = 0 unless p = - 1 , and that E 2 - 1 , 1 = 𝒮 + ( - 1 ) .

If ( h 1 ( 𝒮 + ( - 1 ) ) , h 1 ( 𝒮 - ( - 1 ) ) ) = ( 1 , 0 ) , then Ext 1 ( G , ( - 1 ) ) = S 1 , and Lemma 2.2 shows that E 2 p , 1 = 0 unless p = - 1 , and that E 2 - 1 , 1 = 𝒮 - ( - 1 ) .

Hence ( - 1 ) has a filtration 𝒪 F 1 ( ( - 1 ) ) ( - 1 ) such that F 1 ( ( - 1 ) ) fits in the following exact sequences:

0 𝒪 F 1 ( ( - 1 ) ) 𝒮 ± ( - 1 ) 0 ,
0 F 1 ( ( - 1 ) ) ( - 1 ) 𝒪 ( - 1 ) r - 3 0 .

Therefore F 1 ( ( - 1 ) ) 𝒪 𝒮 ± ( - 1 ) , and thus 𝒪 ( 1 ) 𝒮 ± 𝒪 r - 3 . This is case (3) of Theorem 1.1.

7.2 The case n 5

We will show that this case does not arise. Suppose, to the contrary, that there exists a nef vector bundle on 5 such that | 4 𝒪 ( 1 ) 𝒮 ± 𝒪 r - 3 . We have h q ( 𝒮 ( - 1 ) ) = h q + 1 ( 𝒮 ( - 2 ) ) for any q by (4.7). In particular, we have h 0 ( 𝒮 ( - 2 ) ) = 0 . It follows from [7, Lemma 8.2 (2)] that h q ( ( 𝒮 + 𝒮 - ) ( - 1 ) | 4 ) = 0 unless q = 1 and that h 1 ( ( 𝒮 + 𝒮 - ) ( - 1 ) | 4 ) = 1 . The exact sequence (4.9) implies that h 0 ( 𝒮 ( - 2 ) ) = h 0 ( 𝒮 ( - 1 ) ) and that h q + 1 ( 𝒮 ( - 2 ) ) = h q + 1 ( 𝒮 ( - 1 ) ) for q 2 . Hence we have h 0 ( 𝒮 ( - 1 ) ) = 0 and h q ( 𝒮 ( - 1 ) ) = 0 for q 2 . Note also that h 1 ( 𝒮 ( - 2 ) ) = 0 . Hence we have the following exact sequence:

0 H 1 ( 𝒮 ( - 1 ) ) H 1 ( ( 𝒮 + 𝒮 - ) ( - 1 ) | 4 ) H 2 ( 𝒮 ( - 2 ) ) 0 .

This is, however, a contradiction, because h 1 ( 𝒮 ( - 1 ) ) = h 2 ( 𝒮 ( - 2 ) ) and h 1 ( ( 𝒮 + 𝒮 - ) ( - 1 ) | 4 ) = 1 . Therefore the case n 5 does not arise.

8 The case where | 3 belongs to case (4) of Theorem 2.3

8.1 The case n 4

Suppose that | n - 1 fits in an exact sequence

0 𝒪 ( - 1 ) 𝒪 ( 1 ) 𝒪 r | n - 1 0 .

Then h 0 ( ( - 1 ) | n - 1 ) = 1 and h q ( ( - 1 ) | n - 1 ) = 0 for q > 0 .

We claim here that h q ( ( - 2 ) ) = 0 for any q. If n 5 , this follows from (4.3). Suppose that n = 4 . Then c 2 h 2 = 4 and c 3 h = 4 . Note that h q ( ( - 1 ) ) = 0 for q > 0 by (4.2). Hence h q ( ( - 2 ) ) = 0 for q 2 . Assumption (4.1) together with (3.5) and (4.11) shows that

- h 1 ( ( - 2 ) ) = χ ( ( - 2 ) ) = 1 12 ( c 2 2 - 2 c 4 ) - c 4 12 .

Since c 3 h = 4 , we have c 1 c 3 = 8 . Hence c 4 8 by (4.11). Therefore h 1 ( ( - 2 ) ) 2 3 , and thus h 1 ( ( - 2 ) ) = 0 . Hence the claim holds.

Since h 0 ( ( - 1 ) | n - 1 ) = 1 , the claim above implies that h 0 ( ( - 1 ) ) = 1 . Hence we have an injection 𝒪 ( 1 ) . Let be its cokernel. Then is torsion-free by (4.1), and we have the following exact sequence:

0 𝒪 ( 1 ) 0 .

Since 𝒪 n - 1 ( 1 ) | n - 1 is injective, | n - 1 fits in the following exact sequences:

0 ( - 1 ) | n - 1 0

and

0 𝒪 ( - 1 ) 𝒪 r | n - 1 0 .

We will apply to the Bondal spectral sequence (2.1). We have h q ( ) = 0 for q > 0 , since h q ( ) = 0 for q > 0 by (4.2). Moreover, h q ( ( - 1 ) ) = 0 for any q, since h q ( ( - 1 ) ) = 0 for q > 0 by (4.2). Thus h 0 ( ) = h 0 ( | n - 1 ) = r . Note that h q ( ( - k ) | n - 1 ) = 0 for any q and any k ( 1 k n - 3 ) , that h n - 2 ( ( - n + 2 ) | n - 1 ) = 1 , and that h q ( ( - n + 2 ) ) = 0 unless q = n - 2 . Hence we see that h q ( ( - k ) ) = 0 for any q and any k ( 2 k n - 2 ) , that h n - 1 ( ( - n + 1 ) ) = 1 , and that h q ( ( - n + 1 ) ) = 0 unless q = n - 1 . Since is torsion-free, it follows from (4.4) that there is an exact sequence

0 𝒮 ( ± ) ( - 1 ) ( - 1 ) α 𝒮 ( ) 0 ,

where α = 2 n 2 . Hence h q ( 𝒮 ( ) ) = h q + 1 ( 𝒮 ( ± ) ( - 1 ) ) for any q. Note that h q ( 𝒮 ( ± ) | n - 1 ) = 0 for any q. Since we have an exact sequence

0 𝒮 ( ± ) ( - 1 ) 𝒮 ( ± ) ( 𝒮 ( ± ) ) | 3 0 ,

we infer that h q + 1 ( 𝒮 ( ± ) ( - 1 ) ) = h q + 1 ( 𝒮 ( ± ) ) for any q. Hence we see that h q ( 𝒮 ( ± ) ) = 0 for any q. Therefore Ext q ( G , ) = 0 unless q = n - 1 or 0, Ext n - 1 ( G , ) = S n if n is odd, Ext n - 1 ( G , ) = S n + 1 if n is even, and Hom ( G , ) = S 0 r . Hence E 2 p , q = 0 unless ( p , q ) = ( - n , n - 1 ) or ( 0 , 0 ) , E 2 - n , n - 1 = 𝒪 ( - 1 ) , and E 2 0 , 0 = 𝒪 r by Lemma 2.2. Thus we have an exact sequence

0 𝒪 ( - 1 ) 𝒪 r 0 .

Since h 1 ( 𝒪 ( 1 ) ) = 0 , we can lift up the surjection 𝒪 r to a morphism 𝒪 r , and we have the following commutative diagram with exact rows:

Since left and right vertical arrows are surjective, so is the middle arrow, and the snake lemma yields

Ker ( 𝒪 ( 1 ) 𝒪 r ) 𝒪 ( - 1 ) .

Therefore belongs to case (4) of Theorem 1.1.

9 The case where | 3 belongs to case (5) of Theorem 2.3

9.1 The case n = 4

Suppose that | 3 fits in the following exact sequence:

0 𝒪 a 𝒮 2 𝒪 r - 4 + a | 3 0 ,

where a = 0 or 1, and the composite of the injection

𝒪 a 𝒮 2 𝒪 r - 4 + a

and the projection

𝒮 2 𝒪 r - 4 + a 𝒪 r - 4 + a

is zero. We will apply to ( - 1 ) the Bondal spectral sequence (2.1). First note that h q ( ( - 1 ) | 3 ) = 0 for any q and that h q ( ( - 1 ) ) = 0 for q > 0 by (4.2). Hence we see that h q ( ( - 2 ) ) = 0 for any q by (4.1). Thus we infer that h 0 ( ( - 1 ) ) = 0 . Since h q ( ( - 2 ) | 3 ) = 0 for any q, we infer that h q ( ( - 3 ) ) = 0 for any q. Note here that h q ( ( - 3 ) | 3 ) = 0 unless q = 2 or 3, and that h 2 ( ( - 3 ) | 3 ) a 1 . Set b = h 2 ( ( - 3 ) | 3 ) . Then we see that h 3 ( ( - 3 ) | 3 ) = r - 4 + b . Moreover, h 3 ( ( - 3 ) ) = b and h 4 ( ( - 4 ) ) = r - 4 + b .

The exact sequence above induces the following exact sequence:

0 𝒮 ( - 1 ) a ( 𝒮 𝒮 ( - 1 ) ) 2 𝒮 ( - 1 ) r - 4 + a 𝒮 ( - 1 ) | 3 0 .

By [7, Lemma 8.2 (1)], we see that h 1 ( 𝒮 ( - 1 ) | 3 ) = 2 and h q ( 𝒮 ( - 1 ) | 3 ) = 0 unless q = 1 . As h q ( ( - 2 ) ) = 0 for any q, the exact sequence (4.5) implies that

h q ( 𝒮 ( - 1 ) ) = h q + 1 ( 𝒮 ± ( - 2 ) ) for any  q .

Since h q ( 𝒮 ( - 1 ) | 3 ) = 0 unless q = 1 , the exact sequence (4.9) implies that h 0 ( 𝒮 ± ( - 2 ) ) = h 0 ( 𝒮 ± ( - 1 ) ) and that h q + 1 ( 𝒮 ± ( - 2 ) ) = h q + 1 ( 𝒮 ± ( - 1 ) ) for q 2 . Hence h 0 ( 𝒮 ± ( - 1 ) ) = 0 and h q ( 𝒮 ( - 1 ) ) = 0 for q 2 . Note also that h 1 ( 𝒮 ± ( - 2 ) ) = 0 . Set c = h 1 ( 𝒮 ± ( - 1 ) ) . Since h 1 ( 𝒮 ( - 1 ) | 3 ) = 2 , we see that 0 c 2 and that 2 - c = h 2 ( 𝒮 ± ( - 2 ) ) = h 1 ( 𝒮 ( - 1 ) ) . Thus ( h 1 ( 𝒮 + ( - 1 ) ) , h 1 ( 𝒮 - ( - 1 ) ) ) = ( 0 , 2 ) , ( 1 , 1 ) , or ( 2 , 0 ) . We may assume that ( h 1 ( 𝒮 + ( - 1 ) ) , h 1 ( 𝒮 - ( - 1 ) ) ) = ( c , 2 - c ) .

Now we have the following equalities:

Ext 4 ( G , ( - 1 ) ) = S 5 r - 4 + b , Ext 3 ( G , ( - 1 ) ) = S 5 b , Ext 1 ( G , ( - 1 ) ) = S 1 c S 2 2 - c , Ext q ( G , ( - 1 ) ) = 0 for  q = 2 , 0 .

Lemma 2.2 then shows that E 2 - 4 , 4 = 𝒪 ( - 1 ) r - 4 + b , that E 2 - 4 , 3 = 𝒪 ( - 1 ) b , that E 2 - 1 , 1 = 𝒮 + ( - 1 ) 2 - c 𝒮 - ( - 1 ) c , and that E 2 p , q = 0 unless ( p , q ) = ( - 4 , 4 ) , ( - 4 , 3 ) , or ( - 1 , 1 ) . Hence we infer that ( - 1 ) fits in the following exact sequence:

0 𝒪 ( - 1 ) b 𝒮 + ( - 1 ) 2 - c 𝒮 - ( - 1 ) c ( - 1 ) 𝒪 ( - 1 ) r - 4 + b 0 .

This yields case (5) of Theorem 1.1. More precisely, if b = 0 , then is isomorphic to 𝒮 + 2 - c 𝒮 - c 𝒪 r - 4 . If b = 1 , note that, for non-zero elements s 1 , s 2 of H 0 ( 𝒮 ± ) , the intersection ( s 1 ) 0 ( s 2 ) 0 can be empty if and only if ( s 1 , s 2 )  or  ( s 2 , s 1 ) H 0 ( 𝒮 + ) × H 0 ( 𝒮 - ) . Since is a vector bundle, this implies that if b = 1 , then c must be equal to 1. Now the last long exact sequence is a splice of two short exact sequences

0 𝒪 ( - 1 ) 𝒮 + ( - 1 ) 𝒮 - ( - 1 ) 𝒦 0

and

0 𝒦 ( - 1 ) 𝒪 ( - 1 ) r - 4 0 .

The first exact sequence shows that Ext 1 ( 𝒪 ( - 1 ) , 𝒦 ) = 0 , which implies that the second sequence splits and ( - 1 ) 𝒦 𝒪 ( - 1 ) r - 4 . Thus taking a direct sum of the first exact sequence and the trivial sequence

0 𝒪 ( - 1 ) r - 4 𝒪 ( - 1 ) r - 4 0

and then twisting by 𝒪 ( 1 ) give the desired resolution of .

9.2 The case n = 5

Suppose that | 4 fits in the following exact sequence:

0 𝒪 𝒮 ± 𝒮 ± 𝒪 r - 3 | 4 0 .

We will apply to ( - 1 ) the Bondal spectral sequence (2.1). First note that h q ( ( - 1 ) | 4 ) = 0 for any q and that h q ( ( - 2 ) ) = 0 for any q by (4.3). Hence we see that h q ( ( - 1 ) ) = 0 for any q. Since h q ( ( - 2 ) | 4 ) = 0 for any q, we infer that h q ( ( - 3 ) ) = 0 for any q. Furthermore, h q ( ( - 4 ) ) = 0 for any q, since h q ( ( - 3 ) | 4 ) = 0 for any q. Note here that h q ( ( - 4 ) | 4 ) = 0 unless q = 3 or 4, and that h 3 ( ( - 4 ) | 4 ) 1 . Set a = h 3 ( ( - 4 ) | 4 ) . Then we see that h 4 ( ( - 4 ) | 4 ) = r - 4 + a . Moreover, h 4 ( ( - 5 ) ) = a and h 5 ( ( - 5 ) ) = r - 4 + a .

The exact sequence above induces the following exact sequence:

0 𝒮 ( - 1 ) | 4 ( 𝒮 + 𝒮 - ) ( 𝒮 ± 𝒮 ± 𝒪 r - 3 ) ( - 1 ) ( 𝒮 ( - 1 ) ) | 4 0 .

By [7, Lemma 8.2 (2)], we see that h 1 ( ( 𝒮 ( - 1 ) ) | 4 ) = 2 and that h q ( ( 𝒮 ( - 1 ) ) | 4 ) = 0 unless q = 1 . By (4.7), we have h q ( 𝒮 ( - 1 ) ) = h q + 1 ( 𝒮 ( - 2 ) ) for any q. Since h q ( ( 𝒮 ( - 1 ) ) | 4 ) = 0 unless q = 1 , the exact sequence (4.9) implies that h 0 ( 𝒮 ( - 2 ) ) = h 0 ( 𝒮 ( - 1 ) ) and that h q + 1 ( 𝒮 ( - 2 ) ) = h q + 1 ( 𝒮 ( - 1 ) ) for q 2 . Hence we have h 0 ( 𝒮 ( - 1 ) ) = 0 and h q ( 𝒮 ( - 1 ) ) = 0 for q 2 . Note also that h 1 ( 𝒮 ( - 2 ) ) = 0 . Since h 1 ( ( 𝒮 ( - 1 ) ) | 4 ) = 2 and h 1 ( 𝒮 ( - 1 ) ) = h 2 ( 𝒮 ( - 2 ) ) , we see that h 1 ( 𝒮 ( - 1 ) ) = 1 .

Now we have the following equalities:

Ext 5 ( G , ( - 1 ) ) = S 5 r - 4 + a , Ext 4 ( G , ( - 1 ) ) = S 5 a , Ext 1 ( G , ( - 1 ) ) = S 1 , Ext q ( G , ( - 1 ) ) = 0 for  q = 3 , 2 , 0 .

Lemma 2.2 then shows that E 2 - 5 , 5 = 𝒪 ( - 1 ) r - 4 + a , that E 2 - 5 , 4 = 𝒪 ( - 1 ) a , that E 2 - 1 , 1 = 𝒮 ( - 1 ) , and that E 2 p , q = 0 unless ( p , q ) = ( - 5 , 5 ) , ( - 5 , 4 ) , or ( - 1 , 1 ) . Hence we infer that ( - 1 ) fits in the following exact sequence:

0 𝒪 ( - 1 ) a 𝒮 ( - 1 ) ( - 1 ) 𝒪 ( - 1 ) r - 4 + a 0 .

This yields Case (6) of Theorem 1.1. Note that c 4 ( 𝒮 ) = 0 since c 4 ( 𝒮 ) h = 0 .

9.3 The case n = 6

Suppose that | 5 fits in the following exact sequence:

0 𝒪 𝒮 𝒪 r - 3 | 5 0 .

We will apply to ( - 1 ) the Bondal spectral sequence (2.1). First note that h q ( ( - k ) | 5 ) = 0 for any q if 1 k 4 and that h q ( ( - k ) ) = 0 for any q if 1 k 2 by (4.1), (4.2), and (4.3). Hence we also see that h q ( ( - k ) ) = 0 for any q if 3 k 5 . Note here that h q ( ( - 5 ) | 5 ) = 0 unless q = 4 or 5, and that h 4 ( ( - 5 ) | 5 ) 1 . Set a = h 4 ( ( - 5 ) | 5 ) . Then we see that h 5 ( ( - 5 ) | 5 ) = r - 4 + a . Moreover, h 5 ( ( - 6 ) ) = a and h 6 ( ( - 6 ) ) = r - 4 + a .

The exact sequence above induces the following exact sequence:

0 𝒮 ( - 1 ) 𝒮 𝒮 ( - 1 ) 𝒮 ( - 1 ) r - 3 𝒮 ( - 1 ) | 5 0 .

By [7, Lemma 8.2 (1)], we see that h 1 ( ( 𝒮 ( - 1 ) ) | 5 ) = 1 and that h q ( ( 𝒮 ( - 1 ) ) | 4 ) = 0 unless q = 1 . By (4.7), we have h q ( 𝒮 ( - 1 ) ) = h q + 1 ( 𝒮 ± ( - 2 ) ) for all q. Since h q ( 𝒮 ( - 1 ) | 5 ) = 0 unless q = 1 , the exact sequence (4.9) implies that h 0 ( 𝒮 ± ( - 2 ) ) = h 0 ( 𝒮 ± ( - 1 ) ) and that h q + 1 ( 𝒮 ± ( - 2 ) ) = h q + 1 ( 𝒮 ± ( - 1 ) ) for q 2 . Hence we have h 0 ( 𝒮 ± ( - 1 ) ) = 0 and h q ( 𝒮 ( - 1 ) ) = 0 for q 2 . Note that h 1 ( 𝒮 ( - 2 ) ) = 0 . Set b = h 1 ( 𝒮 + ( - 1 ) ) . Since h 1 ( 𝒮 ( - 1 ) | 5 ) = 1 , we see that b = 0 or 1 and that 1 - b = h 2 ( 𝒮 + ( - 2 ) ) = h 1 ( 𝒮 - ( - 1 ) ) . Hence we see that ( h 1 ( 𝒮 + ( - 1 ) ) , h 1 ( 𝒮 - ( - 1 ) ) ) = ( 0 , 1 ) , or ( 1 , 0 ) .

Now we have the following equalities:

Ext 6 ( G , ( - 1 ) ) = S 7 r - 4 + a , Ext 5 ( G , ( - 1 ) ) = S 7 a , Ext 1 ( G , ( - 1 ) ) = S 2  if  b = 0 , Ext 1 ( G , ( - 1 ) ) = S 1  if  b = 1 , Ext q ( G , ( - 1 ) ) = 0  unless  q = 6 , 5 , 1 .

Lemma 2.2 then shows that E 2 - 6 , 6 = 𝒪 ( - 1 ) r - 4 + a , that E 2 - 6 , 5 = 𝒪 ( - 1 ) a , that E 2 - 1 , 1 = 𝒮 ± ( - 1 ) , and that E 2 p , q = 0 unless ( p , q ) = ( - 6 , 6 ) , ( - 6 , 5 ) , or ( - 1 , 1 ) . Hence we infer that ( - 1 ) fits in the following exact sequence:

0 𝒪 ( - 1 ) a 𝒮 ± ( - 1 ) ( - 1 ) 𝒪 ( - 1 ) r - 4 + a 0 .

This yields case (7) of Theorem 1.1. Note that c 4 ( 𝒮 ± ) = 0 since c 4 ( 𝒮 ± ) h 2 = 0 .

9.4 The case n 7

We will show that this case does not arise. Suppose, to the contrary, that there exists a nef vector bundle on 7 such that | 6 fits in the following exact sequence:

0 𝒪 𝒮 ± 𝒪 r - 3 | 6 0 .

We have h q ( 𝒮 ( - 1 ) ) = h q + 1 ( 𝒮 ( - 2 ) ) for any q by (4.7). By [7, Lemma 8.2 (2)], we see that

h 1 ( ( 𝒮 + 𝒮 - ) ( - 1 ) | 4 ) = 1

and that

h q ( ( 𝒮 + 𝒮 - ) ( - 1 ) | 3 ) = 0 unless  q = 1 .

The exact sequence (4.9) implies h 0 ( 𝒮 ( - 2 ) ) = h 0 ( 𝒮 ( - 1 ) ) and h q + 1 ( 𝒮 ( - 2 ) ) = h q + 1 ( 𝒮 ( - 1 ) ) for q 2 . Hence we have h 0 ( 𝒮 ( - 1 ) ) = 0 and h q ( 𝒮 ( - 1 ) ) = 0 for q 2 . Note also that h 1 ( 𝒮 ( - 2 ) ) = 0 . Hence we have the following exact sequence:

0 H 1 ( 𝒮 ( - 1 ) ) H 1 ( ( 𝒮 + 𝒮 - ) ( - 1 ) | 6 ) H 2 ( 𝒮 ( - 2 ) ) 0 .

This is, however, a contradiction, because h 1 ( 𝒮 ( - 1 ) ) = h 2 ( 𝒮 ( - 2 ) ) and h 1 ( ( 𝒮 + 𝒮 - ) ( - 1 ) | 4 ) = 1 . Therefore the case n 7 does not arise.

10 The case where | 3 belongs to case (6) of Theorem 2.3

10.1 The case n = 4

Suppose that | 3 fits in the following exact sequence:

0 𝒮 ( - 1 ) 𝒪 ( - 1 ) 𝒪 r + 3 | 3 0 .

We will apply to the Bondal spectral sequence (2.1). First note that h q ( ( - 1 ) | 3 ) = 0 for all q. Hence we have h q ( ( - 2 ) ) = h q ( ( - 1 ) ) for any q. Since we have (4.1) and (4.2), this implies that h q ( ( - k ) ) = 0 for k = 1 , 2 and any q. Thus we see that h 0 ( ) = h 0 ( | 3 ) = r + 3 . Note here that h 2 ( ( - 2 ) | 3 ) = 1 and that h q ( ( - 2 ) | 3 ) = 0 unless q = 2 . Hence we see that h 3 ( ( - 3 ) ) = 1 and that h q ( ( - 3 ) ) = 0 unless q = 3 .

The exact sequence above induces the following exact sequence:

0 ( 𝒮 𝒮 ( - 1 ) ) 𝒮 ( - 1 ) 𝒮 r + 3 𝒮 | 3 0 .

By [7, Lemma 8.2 (1)], we see that h 0 ( 𝒮 | 3 ) = 1 and h q ( 𝒮 | 3 ) = 0 unless q = 1 . We have h q ( 𝒮 ) = h q + 1 ( 𝒮 ± ( - 1 ) ) for any q by (4.8). In particular, we have h 0 ( 𝒮 ± ( - 1 ) ) = 0 . Since h q ( 𝒮 | 3 ) = 0 unless q = 0 , the exact sequence (4.10) implies that h q + 1 ( 𝒮 ± ( - 1 ) ) = h q + 1 ( 𝒮 ± ) for q 1 . Hence we have h q ( 𝒮 ) = 0 for q 1 . Set a = h 0 ( 𝒮 + ) . Since h 0 ( 𝒮 | 3 ) = 1 , we see that a = 0 or 1 and that 1 - a = h 1 ( 𝒮 + ( - 1 ) ) = h 0 ( 𝒮 - ) . Therefore ( h 0 ( 𝒮 + ) , h 0 ( 𝒮 - ) ) = ( 0 , 1 ) , or ( 1 , 0 ) .

Now we see that Ext 3 ( G , ) = S 5 , that Ext q ( G , ) = 0 for q = 2 , 1 , and that Hom ( G , ) fits in the following exact sequence:

0 S 0 r + 3 Hom ( G , ) S 2 0 if  a = 0 ,
0 S 0 r + 3 Hom ( G , ) S 1 0 if  a = 1 .

Lemma 2.2 then shows that E 2 - 4 , 3 = 𝒪 ( - 1 ) , that E 2 p , q = 0 unless ( p , q ) = ( - 4 , 3 ) , or ( 0 , 0 ) , and that E 2 0 , 0 fits in the following exact sequence:

0 𝒮 ± ( - 1 ) 𝒪 r + 3 E 2 0 , 0 0 .

Hence we infer that fits in the following exact sequence:

0 𝒮 ± ( - 1 ) 𝒪 ( - 1 ) 𝒪 r + 3 0 .

This is case (8) of Theorem 1.1.

10.2 The case n 5

We will show that this case does not arise. Suppose, to the contrary, that there exists a nef vector bundle on 5 such that | 4 fits in the following exact sequence:

0 𝒮 ± ( - 1 ) 𝒪 ( - 1 ) 𝒪 r + 3 | 4 0 .

Since h 0 ( ( - 1 ) | 4 ) = 0 , it follows from (4.1) that h 0 ( ( - 1 ) ) = 0 . Hence we have h q ( 𝒮 ) = h q + 1 ( 𝒮 ( - 1 ) ) for any q by (4.8). In particular, h 0 ( 𝒮 ( - 1 ) ) = 0 . Note that h q ( ( 𝒮 + 𝒮 - ) | 4 ) = 0 unless q = 0 , and that h 0 ( ( 𝒮 + 𝒮 - ) | 4 ) = 1 by [7, Lemma 8.2 (2)]. Hence the exact sequence (4.10) implies that h q + 1 ( 𝒮 ( - 1 ) ) = h q + 1 ( 𝒮 ) for q 1 . Hence we have h q ( 𝒮 ) = 0 for q 1 . Now we have the following exact sequence:

0 H 0 ( 𝒮 ) H 0 ( ( 𝒮 + 𝒮 - ) | 4 ) H 1 ( 𝒮 ( - 1 ) ) 0 .

This is, however, a contradiction, because h 0 ( 𝒮 ) = h 1 ( 𝒮 ( - 1 ) ) and h 0 ( ( 𝒮 + 𝒮 - ) | 4 ) = 1 . Therefore the case n 5 does not arise.

11 The case where | 3 belongs to case (7) of Theorem 2.3

11.1 The case n 4

Suppose that | n - 1 fits in an exact sequence

0 𝒪 ( - 1 ) 2 𝒪 r + 2 | n - 1 0 .

Then we see that h 0 ( | n - 1 ) = r + 2 , that h q ( | n - 1 ) = 0 for q > 0 , that h q ( ( - k ) | n - 1 ) = 0 for any q if 1 k n - 3 , that h n - 2 ( ( - n + 2 ) | n - 1 ) = 2 , and that h q ( ( - n + 2 ) | n - 1 ) = 0 unless q = n - 2 .

We will apply to the Bondal spectral sequence (2.1). Since we have (4.1), we see that h 0 ( ( - 1 ) ) = 0 . Hence h q ( ( - 1 ) ) = 0 for any q by (4.2). Then we see that h q ( ( - k ) ) = 0 for any q if 1 k n - 2 , that h 0 ( ) = r + 2 , that h q ( ) = 0 for q > 0 , that h n - 1 ( ( - n + 1 ) ) = 2 , and that h q ( ( - n + 1 ) ) = 0 unless q = n - 1 .

We have h q ( 𝒮 ( ) ) = h q + 1 ( 𝒮 ( ± ) ( - 1 ) ) for any q by (4.8). Note that h q ( ( 𝒮 ( ± ) ) | n - 1 ) = 0 for any q. The exact sequence (4.10) then implies that h q + 1 ( 𝒮 ( ± ) ( - 1 ) ) = h q + 1 ( 𝒮 ( ± ) ) for any q. Hence we see that h q ( 𝒮 ( ± ) ) = 0 for any q.

Thus Ext q ( G , ) = 0 unless q = n - 1 or 0, Ext n - 1 ( G , ) = S n 2 if n is odd, Ext n - 1 ( G , ) = S n + 1 2 if n is even, and Hom ( G , ) = S 0 r + 2 . Hence E 2 p , q = 0 unless ( p , q ) = ( - n , n - 1 ) or ( 0 , 0 ) , E 2 - n , n - 1 = 𝒪 ( - 1 ) 2 , and E 2 0 , 0 = 𝒪 r by Lemma 2.2. Therefore we have the following exact sequence:

0 𝒪 ( - 1 ) 2 𝒪 r + 2 0 .

This is case (9) of Theorem 1.1.

12 The case where | 3 belongs to case (8) of Theorem 2.3

12.1 The case n 4

Suppose that | n - 1 fits in an exact sequence

0 𝒪 ( - 2 ) 𝒪 r + 1 | n - 1 0 .

Then we see that h 0 ( | n - 1 ) = r + 1 , that h q ( | n - 1 ) = 0 for q > 0 , that h q ( ( - k ) | n - 1 ) = 0 for any q if 1 k n - 4 , that h n - 2 ( ( - n + 3 ) | n - 1 ) = 1 , that h q ( ( - n + 3 ) | n - 1 ) = 0 unless q = n - 2 , that h n - 2 ( ( - n + 2 ) | n - 1 ) = n + 1 , and that h q ( ( - n + 2 ) | n - 1 ) = 0 unless q = n - 2 .

We will apply to the Bondal spectral sequence (2.1). We have h 0 ( ( - 1 ) ) = 0 by (4.1). It follows from (4.2) that h 0 ( ) = r + 1 , that h q ( ) = 0 for q > 0 , that h q ( ( - k ) ) = 0 for any q if 1 k n - 3 , that h n - 1 ( ( - n + 2 ) ) = 1 , that h q ( ( - n + 2 ) ) = 0 unless q = n - 1 , that h n - 1 ( ( - n + 1 ) ) = n + 2 , and that h q ( ( - n + 1 ) ) = 0 unless q = n - 1 .

We have h q ( 𝒮 ( ) ) = h q + 1 ( 𝒮 ( ± ) ( - 1 ) ) for any q by (4.8). Note that h q ( ( 𝒮 ( ± ) ) | n - 1 ) = 0 for any q, since n - 1 3 . The exact sequence (4.10) then implies that h q + 1 ( 𝒮 ( ± ) ( - 1 ) ) = h q + 1 ( 𝒮 ( ± ) ) for any q. Hence we see that h q ( 𝒮 ( ± ) ) = 0 for any q.

Thus Ext q ( G , ) = 0 unless q = n - 1 or 0, Hom ( G , ) = S 0 r + 1 , and Ext n - 1 ( G , ) fits in the following exact sequence:

0 S n - 1 Ext n - 1 ( G , ) S n n + 2 0 if  n  is odd ,
0 S n Ext n - 1 ( G , ) S n + 1 n + 2 0 if  n  is even .

Therefore Lemma 2.2 implies that E 2 p , q = 0 unless ( p , q ) = ( - n , n - 1 ) , ( - n + 1 , n - 1 ) or ( 0 , 0 ) , that E 2 0 , 0 𝒪 r + 1 , and that E 2 - n , n - 1 and E 2 - n + 1 , n - 1 fit in the following exact sequence:

(12.1) 0 E 2 - n , n - 1 𝒪 ( - 1 ) n + 2 T n + 1 ( - 2 ) | n E 2 - n + 1 , n - 1 0 .

The Bondal spectral sequence induces the following isomorphisms and exact sequences:

E 2 - n , n - 1 E n - n , n - 1 , E 2 0 , 0 E n 0 , 0 , 0 E n - n , n - 1 E n 0 , 0 E n + 1 0 , 0 0 , 0 E n + 1 0 , 0 E 2 - n + 1 , n - 1 0 .

Note here that E 2 - n + 1 , n - 1 | L cannot admit a negative degree quotient for any line L in n since is nef. We show that E 2 - n + 1 , n - 1 = 0 . First note that the exact sequence (12.1) induces the following exact sequence:

0 E 2 - n , n - 1 𝒪 ( - 1 ) n + 2 𝒪 ( - 2 ) 𝑝 𝒪 ( - 1 ) n + 2 E 2 - n + 1 , n - 1 0 .

Consider the composite of the inclusion 𝒪 ( - 1 ) n + 2 𝒪 ( - 1 ) n + 2 𝒪 ( - 2 ) and the morphism p above, and let 𝒪 ( - 1 ) a be the cokernel of this composite. Then we have the following exact sequence:

𝒪 ( - 2 ) 𝜋 𝒪 ( - 1 ) a E 2 - n + 1 , n - 1 0 .

We claim here that a = 0 . Suppose, to the contrary, that a > 0 . Since E 2 - n + 1 , n - 1 cannot be isomorphic to 𝒪 ( - 1 ) a , the morphism π above is not zero. Therefore the composite of π and some projection 𝒪 ( - 1 ) a 𝒪 ( - 1 ) is not zero, whose quotient is of the form 𝒪 H ( - 1 ) for some hyperplane H in n . Hence E 2 - n + 1 , n - 1 admits 𝒪 H ( - 1 ) as a quotient. This is a contradiction. Thus a = 0 and E 2 - n + 1 , n - 1 = 0 . Now the exact sequence (12.1) is the restriction of the Euler sequence, and thus E 2 - n , n - 1 𝒪 ( - 2 ) . Therefore fits in the following exact sequence:

0 𝒪 ( - 2 ) 𝒪 r + 1 0 .

This is case (10) of Theorem 1.1.

13 The case where | 3 belongs to case (9) of Theorem 2.3

13.1 The case n 4

We will show that this case does not arise. Suppose, to the contrary, that there exists a nef vector bundle on 4 such that | 3 fits in the following exact sequence:

0 𝒪 ( - 2 ) 𝛼 𝒪 ( - 1 ) 4 𝒪 r + 3 | 3 0 .

We see that h 0 ( | 3 ) = r + 3 , that h q ( | 3 ) = 0 for q > 0 , that h 1 ( ( - 1 ) | 3 ) = 1 , and that h q ( ( - 1 ) | 3 ) = 0 for q 1 . Note that α ( - 1 ) is surjective and thus the image of H 0 ( α ( - 1 ) ) has dimension at least four. Hence H 0 ( α ( - 1 ) ) is injective, and its dual H 3 ( α ( - 2 ) ) : H 3 ( 𝒪 ( - 4 ) ) H 3 ( 𝒪 ( - 3 ) 4 ) is surjective. Therefore we infer that h 1 ( ( - 2 ) | 3 ) = 1 and that h q ( ( - 2 ) | 3 ) = 0 for q 1 .

We will apply to the Bondal spectral sequence (2.1). Since we have (4.1) and h q ( ( - 1 ) ) = 0 for q > 0 by (4.2), we see that h 0 ( ( - 1 ) ) = 0 , that h 2 ( ( - 2 ) ) = 1 , and that h q ( ( - 2 ) ) = 0 unless q = 2 . Hence h 0 ( ) = r + 3 and h q ( ) = 0 for q > 0 by (4.2). Moreover, we see that h 2 ( ( - 3 ) ) = 2 and that h q ( ( - 3 ) ) = 0 unless q = 2 .

We have h q ( 𝒮 ) = h q + 1 ( 𝒮 ± ( - 1 ) ) for any q by (4.8). Note that h q ( 𝒮 | 3 ) = 0 for any q. Hence h q + 1 ( 𝒮 ± ( - 1 ) ) = h q + 1 ( 𝒮 ± ) for any q by (4.10), and thus h q ( 𝒮 ± ) = 0 for any q.

Now we see that Hom ( G , ) S 0 r + 3 , that Ext q ( G , ) = 0 unless q = 0 or 2, and that Ext 2 ( G , ) fits in the following exact sequence:

0 S 4 Ext 2 ( G , ) S 5 2 0 .

Therefore Lemma 2.2 shows that E 2 p , q = 0 unless ( p , q ) = ( - 3 , 2 ) or ( 0 , 0 ) , that E 2 0 , 0 𝒪 r + 3 , and that E 2 - 3 , 2 fits in the following exact sequence:

0 𝒪 ( - 1 ) 2 T 5 ( - 2 ) | 4 E 2 - 3 , 2 0 .

Hence fits in the following exact sequence:

0 𝒪 ( - 1 ) 2 T 5 ( - 2 ) | 4 𝒪 r + 3 0 .

Since T 5 ( - 2 ) | 4 fits in an exact sequence

0 𝒪 ( - 2 ) 𝒪 ( - 1 ) 6 T 5 ( - 2 ) | 4 0 ,

we infer that fits in the following exact sequence:

0 𝒪 ( - 1 ) 2 𝒪 ( - 2 ) 𝒪 ( - 1 ) 6 𝒪 r + 3 0 .

Hence fits in the following exact sequence:

0 𝒪 ( - 2 ) 𝒪 ( - 1 ) 4 𝒪 r + 3 0 .

Since is locally free, the injection 𝒪 ( - 2 ) 𝒪 ( - 1 ) 4 has locally free quotient, and thus this map is determined by four hyperplanes whose intersection with 4 is empty. However the intersection of four hyperplanes in 5 and 4 cannot be empty. Hence this case does not arise.


Communicated by Shigeharu Takayama


Award Identifier / Grant number: 21K03158

Funding statement: This work was partially supported by JSPS KAKENHI (C) Grant Number 21K03158.

Acknowledgements

Deep appreciation goes to the referee for his careful reading the manuscript and kind suggestions and comments.

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Received: 2023-12-13
Revised: 2024-04-02
Published Online: 2024-04-24
Published in Print: 2025-02-01

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