#  Nef vector bundles on a quadric surface with first Chern class $(2,2)$

Masahiro Ohno  Graduate School of Informatics and Engineering, The University
of Electro-Communications, Chofugaoka 1-5-1, Chofu-shi, Tokyo, 182-8585 Japan
<EMAIL_ADDRESS>
###### Abstract.

We classify nef vector bundles on a smooth quadric surface with first Chern
class $(2,2)$ over an algebraically closed field of characteristic zero.

###### Key words and phrases:

nef vector bundles, quadric surfaces, full strong exceptional collections

###### 2020 Mathematics Subject Classification:

Primary 14J60; Secondary 14F08, 14J45, 14F06

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

##  1\. Introduction

In [SW90, Sect. 3] and [PSW92, Sect. 2 Lemmas 1 and 2], Peternell-Szurek-
Wiśniewski classified nef vector bundles on a smooth quadric surface
$\mathbb{Q}^{2}$ with first Chern class $(1,1)$ over an algebraically closed
field $K$ of characteristic zero. In [Ohn22, Theorem 7.1], we provided a
different proof of this classification, which was based on an analysis with a
full strong exceptional collection of line bundles on $\mathbb{Q}^{2}$.
Following this approach, we classified those with first Chern class $(2,1)$ in
[Ohn20b, Theorem 0.1].

In this paper, we classify nef vector bundles on $\mathbb{Q}^{2}$ with first
Chern class $(2,2)$. The precise statement is as follows:

######  Theorem 1.1.

Let $\mathcal{E}$ be a nef vector bundle of rank $r$ on a smooth quadric
surface $\mathbb{Q}^{2}$ over an algebraically closed field $K$ of
characteristic zero. Suppose that $\det\mathcal{E}\cong\mathcal{O}(2,2)$. Then
$\mathcal{E}$ is isomorphic to one of the following vector bundles or fits in
one of the following exact sequences:

  1. $(1)$

$\mathcal{O}(2,2)\oplus\mathcal{O}^{\oplus r-1}$;

  2. $(2)$

$\mathcal{O}(2,1)\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus r-2}$;
$\mathcal{O}(1,2)\oplus\mathcal{O}(1,0)\oplus\mathcal{O}^{\oplus r-2}$;
$($We do not exhibit the cases obtained by replacing $(a,b)$ with $(b,a)$ in
the following:$)$

  3. $(3)$

$\mathcal{O}(1,1)^{\oplus 2}\oplus\mathcal{O}^{\oplus r-2}$;

  4. $(4)$

$0\to\mathcal{O}\xrightarrow{\iota}\mathcal{O}(1,1)\oplus\mathcal{O}(1,0)\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus
r-2}\to\mathcal{E}\to 0$;
$($More precisely, either the composite of $\iota$ and the projection

$\mathcal{O}(1,1)\oplus\mathcal{O}(1,0)\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus
r-2}\to\mathcal{O}$

is zero, or $\mathcal{E}$ is isomorphic to
$\mathcal{O}(1,1)\oplus\mathcal{O}(1,0)\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus
r-3}$.$)$

  5. $(5)$

$0\to\mathcal{O}(-1,-1)\to\mathcal{O}(1,1)\oplus\mathcal{O}^{\oplus
r}\to\mathcal{E}\to 0$;

  6. $(6)$

$0\to\mathcal{O}^{\oplus 2}\to\mathcal{O}(1,0)^{\oplus
2}\oplus\mathcal{O}(0,1)^{\oplus 2}\oplus\mathcal{O}^{\oplus
r-2}\to\mathcal{E}\to 0$;
$($More precisely,

    1. $(6$-$1)$

$0\to\mathcal{O}\to\mathcal{O}(2,0)\oplus\mathcal{O}(0,1)^{\oplus
2}\oplus\mathcal{O}^{\oplus r-2}\to\mathcal{E}\to 0$;
In this case, we have

      1. $(6$-$1$-$1)$

$\mathcal{O}(2,0)\oplus\mathcal{O}(0,2)\oplus\mathcal{O}^{\oplus r-2}$;

      2. $(6$-$1$-$2)$

$\mathcal{O}(2,0)\oplus\mathcal{O}(0,1)^{\oplus 2}\oplus\mathcal{O}^{\oplus
r-3}$;

    2. $(6$-$2)$

$\mathcal{O}(1,0)^{\oplus 2}\oplus\mathcal{O}(0,1)^{\oplus
2}\oplus\mathcal{O}^{\oplus r-4}$$)$

    3. $(6$-$3)$

$0\to\mathcal{O}(0,-1)\to\mathcal{O}(1,0)^{\oplus
2}\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus r-2}\to\mathcal{E}\to 0$;

  7. $(7)$

$0\to\mathcal{O}(-1,-1)\oplus\mathcal{O}(-1,0)\oplus\mathcal{O}(0,-1)\to\mathcal{O}^{\oplus
r+3}\to\mathcal{E}\to 0$;

  8. $(8)$

$0\to\mathcal{O}(-1,-2)\to\mathcal{O}(1,0)\oplus\mathcal{O}^{\oplus
r}\to\mathcal{E}\to 0$;

  9. $(9)$

$0\to\mathcal{O}(-1,-1)^{\oplus 2}\to\mathcal{O}^{\oplus r+2}\to\mathcal{E}\to
0$;

  10. $(10)$

$0\to\mathcal{O}(-2,-2)\to\mathcal{O}^{\oplus r+1}\to\mathcal{E}\to 0$;

  11. $(11)$

$0\to\mathcal{O}(-2,-2)\to\mathcal{O}^{\oplus r+1}\to\mathcal{E}\to k(p)\to
0$;

  12. $(12)$

$0\to\mathcal{O}(-2,-2)\to\mathcal{O}^{\oplus
r}\to\mathcal{E}\to\mathcal{O}\to 0$;

  13. $(13)$

$0\to\mathcal{O}(-1,-1)^{\oplus 4}\to\mathcal{O}^{\oplus
r}\oplus\mathcal{O}(-1,0)^{\oplus 2}\oplus\mathcal{O}(0,-1)^{\oplus
2}\to\mathcal{E}\to 0$.

Note that this list is effective: in each case exists an example. Under the
stronger assumption that $\mathcal{E}$ is globally generated, Ballico-Huh-
Malaspina gave a classification of $\mathcal{E}$ with $c_{1}=(2,2)$ in
[BHM15]. Note that the vector bundles in Case (11), (12) or (13) are nef but
not globally generated. In fact, the cokernel
$\operatorname{Coker}(\operatorname{ev})$ of the evaluation map
$\operatorname{ev}:H^{0}(\mathcal{E})\otimes\mathcal{O}\to\mathcal{E}$ is the
structure sheaf of point $p$ in Case (11) and of the entire space
$\mathbb{Q}^{2}$ in Case (12), and, in Case (13), a torsion-free sheaf of rank
one on a divisor $E$ of type $(1,1)$ (see Remark 7.2).

Note that if $\mathcal{E}$ is the same as in Theorem 1.1 then the canonical
bundle of the projective space bundle $\mathbb{P}(\mathcal{E})$ is nef.
Furthermore $\mathcal{O}_{\mathbb{P}(\mathcal{E})}(1)^{r+1}=8-c_{2}$, where
$c_{2}$ is the second Chern class of $\mathcal{E}$. Hence
$\mathbb{P}(\mathcal{E})$ is a weak Fano manifold, if $c_{2}<8$, namely except
for Cases (10), (12) and (13).

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

$(\mathcal{O},\mathcal{O}(1,0),\mathcal{O}(0,1),\mathcal{O}(1,1))$

of line bundles (see [Ohn22] for further details). In Cases (3), (4), (6),
(7), (9) and (13) of Theorem 1.1, we describe $\mathcal{E}$ in this framework;
however we do not describe every single $\mathcal{E}$ in this framework in
order to avoid unnecessarily messy descriptions. For example, we do not
describe $\mathcal{O}(2,2)$ in Case (1) of Theorem 1.1 in terms of the
aforementioned exceptional collection. Instead, we make the description neat.

We also classify nef vector bundles with a near-to-maximal degree line bundle
as a subsheaf:

######  Proposition 1.2.

Let $\mathcal{E}$ be a nef vector bundle of rank $r$ on a smooth quadric
surface $\mathbb{Q}^{2}$ with first Chern class
$c_{1}=(c_{1}^{\prime},c_{1}^{\prime\prime})$. Suppose that
$\operatorname{Hom}(\mathcal{O}(c_{1}^{\prime},b),\mathcal{E})\neq 0$ and
$\operatorname{Hom}(\mathcal{O}(c_{1}^{\prime},b+1),\mathcal{E})=0$ for some
integer $b\leq c_{1}^{\prime\prime}$. Then $\mathcal{E}$ fits in the following
exact sequence:

$0\to\mathcal{O}(c_{1}^{\prime},b)\to\mathcal{E}\to\pi_{2}^{*}\mathcal{F}_{2}\to
0,$

where $\pi_{2}:\mathcal{Q}\to\mathbb{P}^{1}$ denotes the second projection and
$\mathcal{F}_{2}$ is a nef vector bundle on $\mathbb{P}^{1}$ with degree
$c_{1}^{\prime\prime}-b$. In particular, if $b=c_{1}^{\prime\prime}$, then
$\mathcal{E}\cong\mathcal{O}(c_{1}^{\prime},c_{1}^{\prime\prime})\oplus\mathcal{O}^{\oplus
r-1}$, and if $b\leq c_{1}^{\prime\prime}-1$, then $\mathcal{E}$ fits in the
following exact sequence:

$0\to\mathcal{O}^{\oplus
c_{1}^{\prime\prime}-b-1}\to\mathcal{O}(c_{1}^{\prime},b)\oplus\mathcal{O}(0,1)^{\oplus
c_{1}^{\prime\prime}-b}\oplus\mathcal{O}^{\oplus r-2}\to\mathcal{E}\to 0.$

######  Theorem 1.3.

Let $\mathcal{E}$ be a nef vector bundle of rank $r$ on a smooth quadric
surface $\mathbb{Q}^{2}$ with first Chern class
$(c_{1}^{\prime},c_{1}^{\prime\prime})$. Suppose that
$\operatorname{Hom}(\mathcal{O}(c_{1}^{\prime}-1,c_{1}^{\prime\prime}),\mathcal{E})=\operatorname{Hom}(\mathcal{O}(c_{1}^{\prime},c_{1}^{\prime\prime}-1),\mathcal{E})=0$
and that
$\operatorname{Hom}(\mathcal{O}(c_{1}^{\prime}-1,c_{1}^{\prime\prime}-1),\mathcal{E})\neq
0$. Then $\mathcal{E}$ satisfies one of the following:

  1. $(1)$

$\mathcal{E}\cong\mathcal{O}(c_{1}^{\prime}-1,c_{1}^{\prime\prime}-1)\oplus\mathcal{O}(1,1)\oplus\mathcal{O}^{\oplus
r-2}$;

  2. $(2)$

$0\to\mathcal{O}\to\mathcal{O}(c_{1}^{\prime}-1,c_{1}^{\prime\prime}-1)\oplus\mathcal{O}(1,0)\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus
r-2}\to\mathcal{E}\to 0$;

  3. $(3)$

$0\to\mathcal{O}(-1,-1)\to\mathcal{O}(c_{1}^{\prime}-1,c_{1}^{\prime\prime}-1)\oplus\mathcal{O}^{\oplus
r}\to\mathcal{E}\to 0$.

Proposition 1.2 is a generalization of [Ohn20b, Lemma 1.2], and Theorem 1.3 is
one of [Ohn22, Theorem 7.1]. Proposition 1.2 and Theorem 1.3 will be used in
the proof of Theorem 1.1.

The content of this paper is as follows. In Section 2, we briefly recall
Bondal’s theorem [Bon89, Theorem 6.2] and its related notions and results used
in Section 7. 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. In Section 3, we collect some lemmas required in Sections
5, 6, and 7. In Section 5, we prove Proposition 1.2 and Theorem 1.3. In
Sections 6 and 7, we apply the Bondal spectral sequence [OT14, Theorem 1]. In
Section 6, we prove Theorem 1.1 when $\mathcal{E}$ contains $\mathcal{O}(D)$
as a subsheaf, where $D$ is a nonzero effective divisor. In Section 7, we
prove Theorem 1.1 in the remaining case. More precisely, in Subsection 7.1, we
first complete the classification assuming that $h^{1}(\mathcal{E})=0$, and,
in Subsection 7.2, we reduce the case $h^{1}(\mathcal{E})>0$ to the case
$h^{1}(\mathcal{E})=0$ by extending $H^{1}(\mathcal{E})\otimes\mathcal{O}$ by
$\mathcal{E}$.

###  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 $\mathbb{Q}^{2}$ a smooth quadric surface
over $K$, by

  * •

$L_{1}$ a fiber $\mathbb{P}^{1}\times\\{\textrm{pt}\\}$ of the second
projection $\pi_{2}:\mathbb{P}^{1}\times\mathbb{P}^{1}\to\mathbb{P}^{1}$,

and by

  * •

$L_{2}$ a fiber $\\{\textrm{pt}\\}\times\mathbb{P}^{1}$ of the first
projection $\pi_{1}:\mathbb{P}^{1}\times\mathbb{P}^{1}\to\mathbb{P}^{1}$.

For a coherent sheaf $\mathcal{F}$, we denote by $c_{i}(\mathcal{F})$ the
$i$-th Chern class of $\mathcal{F}$ and by $\mathcal{F}^{\vee\vee}$ the double
dual of $\mathcal{F}$. In particular,

  * •

$c_{i}$ stands for $c_{i}(\mathcal{E})$ of the nef vector bundle $\mathcal{E}$
we are dealing with.

Since $\mathop{\rm Pic}\nolimits\mathbb{Q}^{2}\cong\mathbb{Z}^{\oplus 2}$, we
identify the first Chern class $c_{1}$ with its corresponding pair of
integers: $c_{1}=(c_{1}^{\prime},c_{1}^{\prime\prime})$. For any closed
subscheme $Z$ in $\mathbb{Q}^{2}$, $\mathcal{I}_{Z}$ denotes the ideal sheaf
of $Z$ in $\mathbb{Q}^{2}$; for a point $p\in\mathbb{Q}^{2}$,
$\mathcal{I}_{p}$ denotes the ideal sheaf of $p\in\mathbb{Q}^{2}$ and $k(p)$
denotes the residue field of $p\in\mathbb{Q}^{2}$. Finally we refer to [Laz04]
for the definition and basic properties of nef vector bundles.

##  2\.  Preliminaries

Throughout this paper, $G_{0}$, $G_{1}$, $G_{2}$, $G_{3}$ denote
$\mathcal{O}$, $\mathcal{O}(1,0)$, $\mathcal{O}(0,1)$, $\mathcal{O}(1,1)$,
respectively, on $\mathbb{Q}^{2}$. An important property of the collection
$(G_{0},G_{1},G_{2},G_{3})$ is that it is a full strong exceptional collection
in $D^{b}(\mathbb{Q}^{2})$, where $D^{b}(\mathbb{Q}^{2})$ denotes the bounded
derived category of (the abelian category of) coherent sheaves on
$\mathbb{Q}^{2}$. Here we use the term “collection” to mean “family”, not
“set”. Thus an exceptional collection is also called an exceptional sequence.
We refer to [Huy06] for the definition of a full strong exceptional sequence.

Denote by $G$ the direct sum $\bigoplus_{i=0}^{3}G_{i}$ of $G_{0}$, $G_{1}$,
$G_{2}$, and $G_{3}$, and by $A$ the endomorphism ring $\operatorname{End}(G)$
of $G$. Then $A$ is a finite-dimensional $K$-algebra, and $G$ is a left
$A$-module. Note that $\operatorname{Ext}^{q}(G,\mathcal{F})$ is a finitely
generated right $A$-module for a coherent sheaf $\mathcal{F}$ on
$\mathbb{Q}^{2}$. We denote by $\operatorname{mod}A$ the category of finitely
generated right $A$-modules and by $D^{b}(\operatorname{mod}A)$ the bounded
derived category of $\operatorname{mod}A$. Let $p_{i}:G\to G_{i}$ be the
projection, and $\iota_{i}:G_{i}\hookrightarrow G$ the inclusion. Set
$e_{i}=\iota_{i}\circ p_{i}$ in $A$. Any finitely generated right $A$-module
$V$ has an ascending filtration

$0=V^{\leq-1}\subset V^{\leq 0}\subset V^{\leq 1}\subset V^{\leq 2}\subset
V^{\leq 3}=V$

by right $A$-submodules, where $V^{\leq i}$ is defined to be $\bigoplus_{j\leq
i}Ve_{j}$. Set $\operatorname{Gr}^{i}V=V^{\leq i}/V^{\leq i-1}$. Let $S_{i}$
denote the right $A$-module such that $\operatorname{Gr}^{i}S_{i}=K$ and
$\operatorname{Gr}^{j}S_{i}=0$ for any $j\neq i$.

It follows from Bondal’s theorem [Bon89, Theorem 6.2] that

$\operatorname{RHom}(G,\bullet):D^{b}(\mathbb{Q}^{2})\to
D^{b}(\operatorname{mod}A)$

is an exact equivalence, and its quasi-inverse is

$\bullet\operatorname{\otimes^{L}}_{A}G:D^{b}(\operatorname{mod}A)\to
D^{b}(\mathbb{Q}^{2}).$

Since $\operatorname{RHom}(G,\mathcal{O}(-1,-1)[2])\cong S_{3}$, we obtain

$S_{3}\operatorname{\otimes^{L}}_{A}G\cong\mathcal{O}(-1,-1)[2].$

Since $\operatorname{RHom}(G,\mathcal{O}(0,-1)[1])\cong S_{2}$, we have

$S_{2}\operatorname{\otimes^{L}}_{A}G\cong\mathcal{O}(0,-1)[1].$

Similarly we have

$S_{1}\operatorname{\otimes^{L}}_{A}G\cong\mathcal{O}(-1,0)[1]\textrm{ and
}S_{0}\operatorname{\otimes^{L}}_{A}G\cong\mathcal{O}.$

We will use these facts freely in this paper. We will also apply the Bondal
spectral sequence [OT14, Theorem 1] to a coherent sheaf $\mathcal{F}$ on
$\mathbb{Q}^{2}$

(2.1)
$E_{2}^{p,q}=\mathop{{\mathcal{T}\\!or}}\nolimits_{-p}^{A}(\operatorname{Ext}^{q}(G,\mathcal{F}),G)\Rightarrow
E^{p+q}=\begin{cases}\mathcal{F}&\textrm{if}\quad p+q=0\\\ 0&\textrm{if}\quad
p+q\neq 0.\end{cases}$

Recall the following theorem [Ohn20b, Theorem 0.1], which will be applied in
Section 6.

######  Theorem 2.1.

Let $\mathcal{E}$ be a nef vector bundle of rank $r$ on a smooth quadric
surface $\mathbb{Q}^{2}$ over an algebraically closed field $K$ of
characteristic zero. If $\det\mathcal{E}\cong\mathcal{O}(2,1)$, then
$\mathcal{E}$ is isomorphic to one of the following vector bundles or fits in
one of the following exact sequences:

  1. $(1)$

$\mathcal{O}(2,1)\oplus\mathcal{O}^{\oplus r-1}$;

  2. $(2)$

$\mathcal{O}(1,1)\oplus\mathcal{O}(1,0)\oplus\mathcal{O}^{\oplus r-2}$;

  3. $(3)$

$0\to\mathcal{O}\to\mathcal{O}(1,0)^{\oplus
2}\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus r-2}\to\mathcal{E}\to 0$;

  4. $(4)$

$0\to\mathcal{O}(-1,-1)\oplus\mathcal{O}(-1,0)\to\mathcal{O}^{\oplus
r+2}\to\mathcal{E}\to 0$;

  5. $(5)$

$0\to\mathcal{O}(-2,-1)\to\mathcal{O}^{\oplus r+1}\to\mathcal{E}\to 0$.

##  3\. Lemmas

######  Lemma 3.1.

Let $\mathcal{E}$ be a nef vector bundle on a smooth complete surface $S$, and
$\mathcal{L}$ a line bundle on $S$. Suppose that there exists a non-zero
section $s\in H^{0}(\mathcal{E}\otimes\mathcal{L}^{-1})$ whose zero locus
$(s)_{0}$ has dimension $\leq 0$. Let $\mathcal{F}$ be the cokernel of the
injection $\mathcal{L}\hookrightarrow\mathcal{E}$ induced by $s$,
$\mathcal{F}^{\vee\vee}$ the double dual of $\mathcal{F}$, and $\mathcal{Q}$
the cokernel of the natural morphism $\mathcal{F}\to\mathcal{F}^{\vee\vee}$.
Then we have

$0\leq
c_{2}(\mathcal{F}^{\vee\vee})+\operatorname{length}\mathcal{Q}=c_{2}(\mathcal{F})\leq
c_{1}(\mathcal{F})^{2}.$

###### Proof.

First note that $\mathcal{F}$ is torsion-free by [Ohn22, Lemma 5.4], since
$\dim(s)_{0}\leq 0$. Hence the natural morphism
$\mathcal{F}\to\mathcal{F}^{\vee\vee}$ is injective. The exact sequence
$0\to\mathcal{L}\to\mathcal{E}\to\mathcal{F}\to 0$ shows that

(3.1) $c_{2}(\mathcal{F})+c_{1}(\mathcal{L})c_{1}(\mathcal{F})=c_{2}.$

Note here that $Q$ is a zero-dimensional coherent sheaf and that

$\operatorname{length}Q=-c_{2}(\mathcal{Q}).$

The exact sequence $0\to\mathcal{F}\to\mathcal{F}^{\vee\vee}\to\mathcal{Q}\to
0$ then implies that

$c_{2}(\mathcal{F})=c_{2}(\mathcal{F}^{\vee\vee})+\operatorname{length}\mathcal{Q}.$

Now recall that

$c_{2}\leq c_{1}(c_{1}-c_{1}(\mathcal{L}))$

by [Ohn22, Lemma 10.1]. Since $c_{1}(\mathcal{F})=c_{1}-c_{1}(\mathcal{L})$,
this inequality together with (3.1) shows that $c_{2}(\mathcal{F})\leq
c_{1}(\mathcal{F})^{2}$. Finally note that $c_{2}(\mathcal{F}^{\vee\vee})\geq
0$, since $\mathcal{F}^{\vee\vee}$ is a nef vector bundle by [Ohn20a, Lemma
9.1]. ∎

######  Lemma 3.2.

Let $Z$ be a zero-dimensional closed subscheme of length two on
$\mathbb{Q}^{2}$, and suppose that $\operatorname{length}Z\cap L_{1}\leq 1$
and $\operatorname{length}Z\cap L_{2}\leq 1$. Then $\mathcal{I}_{Z}$ fits in
an exact sequence

$0\to\mathcal{O}(-2,-2)\to\mathcal{O}(-1,-1)^{\oplus 2}\to\mathcal{I}_{Z}\to
0.$

###### Proof.

The assumption that $\operatorname{length}Z\cap L_{1}\leq 1$ and
$\operatorname{length}Z\cap L_{2}\leq 1$ shows that $Z$ is a complete
intersection of two irreducible divisors of type $(1,1)$. Therefore we obtain
the exact sequence above. ∎

######  Lemma 3.3.

Let $\mathcal{G}$ and $\mathcal{Q}$ be coherent sheaves on $\mathbb{Q}^{2}$,
and consider an exact sequence

$0\to\mathcal{G}\to\mathcal{O}(a,0)^{\oplus e}\oplus\mathcal{O}(0,b)^{\oplus
f}\oplus\mathcal{O}^{\oplus r}\to\mathcal{Q}\to 0,$

where $a$ and $b$ are positive integers, and $e$, $f$, $r$ are non-negative
integers. Suppose that $\mathcal{Q}$ is zero-dimensional and that one of the
following holds:

  1. $(1)$

$e=0$ or $f=0$;

  2. $(2)$

$e>0$, $f>0$, and $a=b=1$.

Then $\mathcal{G}$ admits a negative degree quotient on a line.

###### Proof.

We may assume that $\mathcal{Q}$ is the residue field $k(p)$ of a point
$p\in\mathbb{Q}^{2}$. Indeed, there exists a surjection $\mathcal{Q}\to k(p)$
for some $p\in\mathbb{Q}^{2}$, and let $\mathcal{K}$ be the kernel of the
composite of $\mathcal{O}(a,0)^{\oplus e}\oplus\mathcal{O}(0,b)^{\oplus
f}\oplus\mathcal{O}^{\oplus r}\to\mathcal{Q}$ and $\mathcal{Q}\to k(p)$. Then
$\mathcal{G}$ is a subsheaf of $\mathcal{K}$, and $\mathcal{K}/\mathcal{G}$
has dimension $\leq 0$. Hence $\mathcal{G}$|${}_{L_{i}}$ admits a negative
degree quotient, if $\mathcal{K}$|${}_{L_{i}}$ admits a negative degree
quotient for some line $L_{i}$ passing through $p$.

Suppose that $\mathcal{Q}=k(p)$. If the composite

$\mathcal{O}(a,0)^{\oplus e}\oplus\mathcal{O}(0,b)^{\oplus
f}\hookrightarrow\mathcal{O}(a,0)^{\oplus e}\oplus\mathcal{O}(0,b)^{\oplus
f}\oplus\mathcal{O}^{\oplus r}\to k(p)$

is zero, then we have a surjection $\mathcal{O}^{\oplus r}\to k(p)$, and its
kernel is isomorphic to $\mathcal{I}_{p}\oplus\mathcal{O}^{\oplus r-1}$ and
thus $\mathcal{G}$ admits $\mathcal{I}_{p}\oplus\mathcal{O}^{\oplus r-1}$ as a
quotient. Hence $\mathcal{G}$|${}_{L_{i}}$ admits a negative degree quotient.
If the composite

$\mathcal{O}(a,0)^{\oplus e}\oplus\mathcal{O}(0,b)^{\oplus
f}\hookrightarrow\mathcal{O}(a,0)^{\oplus e}\oplus\mathcal{O}(0,b)^{\oplus
f}\oplus\mathcal{O}^{\oplus r}\to k(p)$

is surjective, let $\mathcal{K}$ be its kernel. Then $h^{1}(\mathcal{K})=0$
and $\mathcal{G}$ is isomorphic to $\mathcal{K}\oplus\mathcal{O}^{\oplus r}$.
Thus the problem is reduced to the case $r=0$, and we assume that $r=0$. Then
we may assume that some direct summand, say, $\mathcal{O}(a,0)$ yields a
surjection $\mathcal{O}(a,0)\to k(p)$. If $f=0$, then $\mathcal{G}$ is
isomorphic to $\mathcal{I}_{p}(a,0)\oplus\mathcal{O}(a,0)^{\oplus e-1}$, and
thus $\mathcal{G}\mid_{L_{2}}$ admits $\mathcal{O}_{L_{2}}(-1)$ as a quotient
for the line $L_{2}$ passing through $p$. If $f>0$ and $a=b=1$, then
$\mathcal{G}$ is isomorphic to
$\mathcal{I}_{p}(1,0)\oplus\mathcal{O}(1,0)^{\oplus
e-1}\oplus\mathcal{O}(0,1)^{\oplus f}$, since
$\operatorname{Ext}^{1}(\mathcal{O}(1,0)^{\oplus
e-1}\oplus\mathcal{O}(0,1)^{\oplus f},\mathcal{O}(1,0))=0$. Therefore
$\mathcal{G}$|${}_{L_{2}}$ admits $\mathcal{O}_{L_{2}}(-1)$ as a quotient. ∎

######  Lemma 3.4.

Let $\mathcal{E}$ be a vector bundle of rank $r$ on $\mathbb{Q}^{2}$ with
$c_{1}=(2,2)$. Then

$\chi(\mathcal{E}(p,q))=8-c_{2}+2p+2q+r(p+1)(q+1)$

for any integers $p$, $q$. In particular,
$\chi(\mathcal{E}(-1,0))=\chi(\mathcal{E}(0,-1))=6-c_{2}$.

###### Proof.

The Riemann-Roch formula for a vector bundle $\mathcal{E}(p,q)$ of rank $r$ on
$\mathbb{Q}^{2}$ with
$c_{1}(\mathcal{E})=(c_{1}^{\prime},c_{1}^{\prime\prime})$ is

$\chi(\mathcal{E}(p,q))=c_{1}^{\prime}c_{1}^{\prime\prime}-c_{2}+(q+1)c_{1}^{\prime}+(p+1)c_{1}^{\prime\prime}+r(p+1)(q+1).$

Since $(c_{1}^{\prime},c_{1}^{\prime\prime})=(2,2)$ by assumption, this yields
the desired formula. ∎

######  Lemma 3.5.

Let $\mathcal{E}$ be a nef vector bundle of rank $r$ on $\mathbb{Q}^{2}$ with
$c_{1}=(2,2)$. Then

$\operatorname{Ext}^{2}(G,\mathcal{E})=0.$

###### Proof.

It follows from [Ohn22, Lemma 4.5. (1)] that $H^{q}(\mathcal{E}(1,1))=0$ for
$q>0$. Note here that we have the following exact sequences for integers $a$,
$b$:

$\begin{split}H^{1}(\mathcal{E}|_{L_{1}}(a))\to H^{2}(\mathcal{E}(a,b-1))\to
H^{2}(\mathcal{E}(a,b));\\\ H^{1}(\mathcal{E}|_{L_{2}}(b))\to
H^{2}(\mathcal{E}(a-1,b))\to H^{2}(\mathcal{E}(a,b)).\end{split}$

Since $\mathcal{E}$ is nef, both $H^{1}(\mathcal{E}|_{L_{1}}(a))$ and
$H^{1}(\mathcal{E}|_{L_{2}}(b))$ vanish if $a,b\geq-1$. Hence we see that
$\operatorname{Ext}^{2}(G,\mathcal{E})=0$. ∎

######  Lemma 3.6.

Let $\mathcal{E}$ be a nef vector bundle of rank $r$ on $\mathbb{Q}^{2}$ with
$c_{1}=(2,2)$. Suppose that $H^{1}(\mathcal{E})=0$ and that
$\operatorname{Hom}(\mathcal{O}(1,0),\mathcal{E})=\operatorname{Hom}(\mathcal{O}(0,1),\mathcal{E})=0$.
Then $c_{2}\geq 6$, and $\operatorname{Ext}^{1}(G,\mathcal{E})$ fits in the
following exact sequence of right $A$-modules:

$0\to S_{1}^{\oplus c_{2}-6}\oplus S_{2}^{\oplus
c_{2}-6}\to\operatorname{Ext}^{1}(G,\mathcal{E})\to S_{3}^{\oplus c_{2}-4}\to
0.$

Moreover we have an isomorphism of right $A$-modules:

$\operatorname{Hom}(G,\mathcal{E})\cong S_{0}^{\oplus r+8-c_{2}}.$

###### Proof.

Since both $\operatorname{Hom}(\mathcal{O}(1,0),\mathcal{E})$ and
$\operatorname{Hom}(\mathcal{O}(0,1),\mathcal{E})$ vanish by assumption,
Lemmas 3.4 and 3.5 imply that

$\begin{split}-h^{1}(\mathcal{E}(-1,0))&=\chi(\mathcal{E}(-1,0))=6-c_{2};\\\
-h^{1}(\mathcal{E}(0,-1))&=\chi(\mathcal{E}(0,-1))=6-c_{2};\\\
-h^{1}(\mathcal{E}(-1,-1))&=\chi(\mathcal{E}(-1,-1))=4-c_{2}.\end{split}$

Thus $c_{2}-6=h^{1}(\mathcal{E}(-1,0))\geq 0$. Moreover the above formulas
together with the assumption $h^{1}(\mathcal{E})=0$ show the structure of the
right $A$-module $\operatorname{Ext}^{1}(G,\mathcal{E})$, i.e., the desired
exact sequence of right $A$-modules. Furthermore we see by Lemmas 3.4, 3.5,
and the assumption $h^{1}(\mathcal{E})=0$ that
$h^{0}(\mathcal{E})=\chi(\mathcal{E})=r+8-c_{2}$, and thus we obtain
$\operatorname{Hom}(G,\mathcal{E})\cong S_{0}^{\oplus r+8-c_{2}}$. ∎

######  Lemma 3.7.

Let $\mathcal{E}$ be a nef vector bundle of rank $r$ on $\mathbb{Q}^{2}$ with
$c_{1}=(2,2)$. If $c_{2}\leq 7$, then $H^{1}(\mathcal{E})=0$.

###### Proof.

This follows from [Ohn22, Lemma 4.5. (2)]. ∎

##  4\. Set-up for Sections 5 and 6

Let $\mathcal{E}$ be a nef vector bundle of rank $r$ on $\mathbb{Q}^{2}$ with
$c_{1}=(c_{1}^{\prime},c_{1}^{\prime\prime})$. Suppose that

$\operatorname{Hom}(\mathcal{O}(a+1,b),\mathcal{E})=\operatorname{Hom}(\mathcal{O}(a,b+1),\mathcal{E})=0\textrm{
and }\operatorname{Hom}(\mathcal{O}(a,b),\mathcal{E})\neq 0$

for some integers $a$, $b$. We have an inclusion
$\mathcal{O}(a,b)\hookrightarrow\mathcal{E}$, and let $\mathcal{F}$ be its
cokernel. Set $\mathcal{L}:=\mathcal{O}(a,b)$, and we will apply Lemma 3.1.
Let $\mathcal{Q}$ be the cokernel of the natural inclusion
$\mathcal{F}\hookrightarrow\mathcal{F}^{\vee\vee}$. Then we have

(4.1) $0\leq
c_{2}(\mathcal{F}^{\vee\vee})+\operatorname{length}\mathcal{Q}=c_{2}(\mathcal{F})\leq
2(c_{1}^{\prime}-a)(c_{1}^{\prime\prime}-b).$

Throughout Sections 5 and 6, we will also use the symbols $\mathcal{F}$ and
$\mathcal{Q}$.

##  5\. Nef vector bundles having a near-to-maximal degree line bundle

######  Proposition 5.1 (=Proposition 1.2).

Let $\mathcal{E}$ be a nef vector bundle of rank $r$ on a smooth quadric
surface $\mathbb{Q}^{2}$ with first Chern class
$c_{1}=(c_{1}^{\prime},c_{1}^{\prime\prime})$. Suppose that
$\operatorname{Hom}(\mathcal{O}(c_{1}^{\prime},b),\mathcal{E})\neq 0$ and
$\operatorname{Hom}(\mathcal{O}(c_{1}^{\prime},b+1),\mathcal{E})=0$ for some
integer $b\leq c_{1}^{\prime\prime}$. Then $\mathcal{E}$ fits in the following
exact sequence:

$0\to\mathcal{O}(c_{1}^{\prime},b)\to\mathcal{E}\to\pi_{2}^{*}\mathcal{F}_{2}\to
0,$

where $\pi_{2}:\mathcal{Q}\to\mathbb{P}^{1}$ denotes the second projection and
$\mathcal{F}_{2}$ is a nef vector bundle on $\mathbb{P}^{1}$ with degree
$c_{1}^{\prime\prime}-b$. In particular, if $b=c_{1}^{\prime\prime}$, then
$\mathcal{E}\cong\mathcal{O}(c_{1}^{\prime},c_{1}^{\prime\prime})\oplus\mathcal{O}^{\oplus
r-1}$, and if $b\leq c_{1}^{\prime\prime}-1$, then $\mathcal{E}$ fits in the
following exact sequence:

$0\to\mathcal{O}^{\oplus
c_{1}^{\prime\prime}-b-1}\to\mathcal{O}(c_{1}^{\prime},b)\oplus\mathcal{O}(0,1)^{\oplus
c_{1}^{\prime\prime}-b}\oplus\mathcal{O}^{\oplus r-2}\to\mathcal{E}\to 0.$

###### Proof.

Suppose that
$\operatorname{Hom}(\mathcal{O}(c_{1}^{\prime},b),\mathcal{E})\neq 0$. Since
$\mathcal{E}$ is nef, we have
$\operatorname{Hom}(\mathcal{O}(c_{1}^{\prime}+1,b),\mathcal{E})=0$. Moreover
$\operatorname{Hom}(\mathcal{O}(c_{1}^{\prime},b+1),\mathcal{E})=0$ by
assumption. It then follows from (4.1) that that $\mathcal{Q}=0$. Hence
$\mathcal{F}\cong\mathcal{F}^{\vee\vee}$, and $\mathcal{F}$ is a nef vector
bundle with first Chern class $(0,c_{1}^{\prime\prime}-b)$. Thus
$\mathcal{F}\cong\pi_{2}^{*}\mathcal{F}_{2}$ for some nef vector bundle
$\mathcal{F}_{2}$ on $\mathbb{P}^{1}$ of degree $c_{1}^{\prime\prime}-b$. If
$b=c_{1}^{\prime\prime}$, then
$\mathcal{E}\cong\mathcal{O}(c_{1}^{\prime},c_{1}^{\prime\prime})\oplus\mathcal{O}^{\oplus
r-1}$. If $b\leq c_{1}^{\prime\prime}-1$, we obtain the result by [Ohn20b,
Lemma 1.1]. This completes the proof of Proposition 1.2. ∎

######  Theorem 5.2 (=Theorem 1.3).

Let $\mathcal{E}$ be a nef vector bundle of rank $r$ on a smooth quadric
surface $\mathbb{Q}^{2}$ with first Chern class
$(c_{1}^{\prime},c_{1}^{\prime\prime})$. Suppose that
$\operatorname{Hom}(\mathcal{O}(c_{1}^{\prime}-1,c_{1}^{\prime\prime}),\mathcal{E})=\operatorname{Hom}(\mathcal{O}(c_{1}^{\prime},c_{1}^{\prime\prime}-1),\mathcal{E})=0$
and that
$\operatorname{Hom}(\mathcal{O}(c_{1}^{\prime}-1,c_{1}^{\prime\prime}-1),\mathcal{E})\neq
0$. Then $\mathcal{E}$ satisfies one of the following:

  1. $(1)$

$\mathcal{E}\cong\mathcal{O}(c_{1}^{\prime}-1,c_{1}^{\prime\prime}-1)\oplus\mathcal{O}(1,1)\oplus\mathcal{O}^{\oplus
r-2}$;

  2. $(2)$

$0\to\mathcal{O}\to\mathcal{O}(c_{1}^{\prime}-1,c_{1}^{\prime\prime}-1)\oplus\mathcal{O}(1,0)\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus
r-2}\to\mathcal{E}\to 0$;

  3. $(3)$

$0\to\mathcal{O}(-1,-1)\to\mathcal{O}(c_{1}^{\prime}-1,c_{1}^{\prime\prime}-1)\oplus\mathcal{O}^{\oplus
r}\to\mathcal{E}\to 0$.

###### Proof.

First we have $0\leq
c_{2}(\mathcal{F}^{\vee\vee})+\operatorname{length}\mathcal{Q}=c_{2}(\mathcal{F})\leq
2$ by (4.1).

For the case $c_{2}(\mathcal{F})=0$, we have
$\operatorname{length}\mathcal{Q}=0$ and
$\mathcal{F}\cong\mathcal{F}^{\vee\vee}$. Moreover
$\mathcal{F}\cong\mathcal{O}(1,1)\oplus\mathcal{O}^{\oplus r-2}$ by [Ohn22,
Theorem 7.1] since $c_{2}(\mathcal{F})=0$. Hence
$\mathcal{E}\cong\mathcal{O}(c_{1}^{\prime}-1,c_{1}^{\prime\prime}-1)\oplus\mathcal{O}(1,1)\oplus\mathcal{O}^{\oplus
r-2}$. This is Case (1) of Theorem 1.3.

For the case $c_{2}(\mathcal{F})=1$, if $\operatorname{length}\mathcal{Q}=0$,
then $\mathcal{F}\cong\mathcal{F}^{\vee\vee}$, and
$\mathcal{F}\cong\mathcal{O}(1,0)\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus
r-3}$ by [Ohn22, Theorem 7.1]. If $\operatorname{length}\mathcal{Q}=1$, then
$\mathcal{Q}\cong k(p)$ for some point $p\in\mathbb{Q}^{2}$, and
$c_{2}(\mathcal{F}^{\vee\vee})=0$. Thus we infer that
$\mathcal{F}^{\vee\vee}\cong\mathcal{O}(1,1)\oplus\mathcal{O}^{\oplus r-2}$ by
[Ohn22, Theorem 7.1]. Hence
$\mathcal{F}\cong\mathcal{I}_{p}(1,1)\oplus\mathcal{O}^{\oplus r-2}$. In both
cases, $\mathcal{F}$ fits in the following exact sequence:

$0\to\mathcal{O}\to\mathcal{O}(1,0)\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{r-2}\to\mathcal{F}\to
0.$

Therefore $\mathcal{E}$ belongs to Case (2) of Theorem 1.3.

Suppose that $c_{2}(\mathcal{F})=2$. If $\operatorname{length}\mathcal{Q}=0$,
then $\mathcal{F}\cong\mathcal{F}^{\vee\vee}$, and $\mathcal{F}$ fits in the
following exact sequence:

$0\to\mathcal{O}(-1,-1)\to\mathcal{O}^{\oplus r}\to\mathcal{F}\to 0$

by [Ohn22, Theorem 7.1]. Hence $\mathcal{E}$ belongs to Case (3) of Theorem
1.3. If $\operatorname{length}\mathcal{Q}=1$, then
$c_{2}(\mathcal{F}^{\vee\vee})=1$, and thus
$\mathcal{F}^{\vee\vee}\cong\mathcal{O}(1,0)\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus
r-3}$ by [Ohn22, Theorem 7.1]. Lemma 3.3 then implies that $\mathcal{F}$
admits a negative degree quotient on a line. This however contradicts that
$\mathcal{E}$ is nef. Therefore this case does not arise. If
$\operatorname{length}\mathcal{Q}=2$, then $c_{2}(\mathcal{F}^{\vee\vee})=0$,
and thus $\mathcal{F}^{\vee\vee}\cong\mathcal{O}(1,1)\oplus\mathcal{O}^{\oplus
r-2}$ by [Ohn22, Theorem 7.1]. Hence there exists a surjection
$\mathcal{O}(1,1)\to\mathcal{Q}$, since $\mathcal{F}$ cannot admit a negative
degree quotient. Therefore $\mathcal{Q}\cong\mathcal{O}_{Z}$ for some zero-
dimensional closed subscheme $Z$ of length two. Moreover
$\operatorname{length}Z\cap L_{1}\leq 1$ and $\operatorname{length}Z\cap
L_{2}\leq 1$, since $\mathcal{F}$ cannot admit a negative degree quotient.
Lemma 3.2 then implies that $\mathcal{F}$ fits in the following exact
sequence:

$0\to\mathcal{O}(-1,-1)\to\mathcal{O}^{\oplus r}\to\mathcal{F}\to 0.$

Hence $\mathcal{E}$ belongs to Case (3) of Theorem 1.3. This completes the
proof of Theorem 1.3. ∎

##  6\. Proof of Theorem 1.1 when $\mathcal{E}$ contains an effective divisor

In this section, we prove Theorem 1.1 when
$\operatorname{Hom}(\mathcal{O}(a,b),\mathcal{E})\neq 0$, where $a$ and $b$
are non-negative integers such that $a+b\geq 1$.

Suppose that $a+b\geq 2$. Then we obtain Cases (1), (2), (3), (4), (5), (6-1)
of Theorem 1.1 using Proposition 1.2 and Theorem 1.3.

In the following in this section, we consider the case $a+b=1$. Moreover we
may assume, without loss of generality, that $(a,b)=(0,1)$. Since we have
completed the case $a+b\geq 2$, we may assume that

(6.1) $\operatorname{Hom}(\mathcal{O}(p,q),\mathcal{E})=0\textrm{ for
}(p,q)=(2,0),(1,1),(0,2).$

We then apply the set-up in Section 4 with
$(c_{1}^{\prime},c_{1}^{\prime\prime})=(2,2)$. We have the following exact
sequence

$0\to\mathcal{F}\to\mathcal{F}^{\vee\vee}\to\mathcal{Q}\to 0,$

and we see that $\mathcal{F}^{\vee\vee}$ is a nef vector bundle of rank $r-1$
with first Chern class $(2,1)$. It follows from (4.1) that

$0\leq
c_{2}(\mathcal{F}^{\vee\vee})+\operatorname{length}\mathcal{Q}=c_{2}(\mathcal{F})\leq
4.$

Since we have an exact sequence

$0\to\mathcal{O}(-1,0)\to\mathcal{E}(-1,-1)\to\mathcal{F}(-1,-1)\to 0,$

it follows from $\operatorname{Hom}(\mathcal{O}(1,1),\mathcal{E})=0$ that
$H^{0}(\mathcal{F}(-1,-1))=0$. Thus

(6.2)
$h^{0}(\mathcal{F}^{\vee\vee}(-1,-1))\leq\operatorname{length}\mathcal{Q}\leq
4-c_{2}(\mathcal{F}^{\vee\vee}).$

The basic strategy for dealing with the case
$\operatorname{Hom}(\mathcal{O}(0,1),\mathcal{E})\neq 0$ under the assumption
(6.1) is as follows: first apply Theorem 2.1 to $\mathcal{F}^{\vee\vee}$ in
order to determine the structure of $\mathcal{F}^{\vee\vee}$ based on the
value of $c_{2}(\mathcal{F}^{\vee\vee})$; then analyze the structure of
$\mathcal{F}$ based on the length of $\mathcal{Q}$, with range of (6.2). This
will be carried out in Subsections 6.1, 6.2, 6.3, 6.4, and 6.5.

###  6.1.  The case $c_{2}(\mathcal{F}^{\vee\vee})=0$

We have

$\mathcal{F}^{\vee\vee}\cong\mathcal{O}(2,1)\oplus\mathcal{O}^{\oplus r-2}$

by Theorem 2.1. Hence $h^{0}(\mathcal{F}^{\vee\vee}(-1,-1))=2$, and thus
$2\leq\operatorname{length}\mathcal{Q}\leq 4$ by (6.2). Since $\mathcal{F}$
cannot admit negative degree quotients, we have a surjection
$\mathcal{O}(2,1)\to\mathcal{Q}$, and thus $\mathcal{Q}\cong\mathcal{O}_{Z}$
for some zero-dimensional closed subscheme $Z$. Note that

$2\leq\operatorname{length}Z\leq 4.$

Moreover $\mathcal{F}$ fits in an exact sequence

$0\to\mathcal{I}_{Z}(2,1)\to\mathcal{F}\to\mathcal{O}^{\oplus r-2}\to 0,$

and we have $\operatorname{length}Z\cap L_{1}\leq 2$ and
$\operatorname{length}Z\cap L_{2}\leq 1$.

####  6.1.1.  Suppose that $\operatorname{length}Z=2$.

If $\operatorname{length}Z\cap L_{1}=2$ for some line $L_{1}$, then $Z$ lies
on the line $L_{1}$, and $\mathcal{I}_{Z}$ fits in an exact sequence

$0\to\mathcal{O}(-2+a,-1)\to\mathcal{O}(-2,0)\oplus\mathcal{O}(a,-1)\to\mathcal{I}_{Z}\to
0$

for any $a\leq 0$. Here we take $a$ to be $-1$. Then $\mathcal{F}$ fits in an
exact sequence

$0\to\mathcal{O}(-1,0)\to\mathcal{O}(0,1)\oplus\mathcal{O}(1,0)\oplus\mathcal{O}^{\oplus
r-2}\to\mathcal{F}\to 0.$

Therefore $\mathcal{E}$ belongs to Case $(6$-$3)$ of Theorem 1.1. If
$\operatorname{length}Z\cap L_{1}\leq 1$ for any line $L_{1}$, then Lemma 3.2
shows that $\mathcal{F}$ fits in the following exact sequence

$0\to\mathcal{O}(0,-1)\to\mathcal{O}(1,0)^{\oplus 2}\oplus\mathcal{O}^{\oplus
r-2}\to\mathcal{F}\to 0.$

Therefore $\mathcal{E}$ belongs to Case $(6$-$3)$ of Theorem 1.1.

####  6.1.2.  Suppose that $\operatorname{length}Z=3$.

Let $p$ be an associated point of $Z$; we have an exact sequence

$0\to k(p)\to\mathcal{O}_{Z}\to\mathcal{O}_{Z^{\prime}}\to 0,$

where $Z^{\prime}$ is a zero-dimensional closed subscheme of $\mathbb{Q}^{2}$
of length $2$. This exact sequence induces an exact sequence

$0\to\mathcal{I}_{Z}\to\mathcal{I}_{Z^{\prime}}\to k(p)\to 0.$

By taking $p\in Z$ suitably, we may assume that
$\operatorname{length}Z^{\prime}\cap L_{i}\leq 1$ for $i=1,2$. Lemma 3.2 then
implies that $h^{0}(\mathcal{I}_{Z^{\prime}}(1,1))=2$ and that
$\mathcal{I}_{Z^{\prime}}(1,1)$ is globally generated. Hence we have a
surjection $H^{0}(\mathcal{I}_{Z^{\prime}}(1,1))\to H^{0}(k(p))$, and thus
$h^{0}(\mathcal{I}_{Z}(1,1))=1$. Let $D_{1}$ be the divisor of type $(1,1)$
containing $Z$. Since $h^{0}(\mathcal{O}_{Z})=3$, we see that
$h^{0}(\mathcal{I}_{Z}(2,1))\geq
h^{0}(\mathcal{O}(2,1))-h^{0}(\mathcal{O}_{Z})=3$. Note here that reducible
divisors of type $(2,1)$ containing $Z$ correspond to a linear subspace of
$H^{0}(\mathcal{I}_{Z}(2,1))$ of dimension two. Therefore $Z$ is contained in
an irreducible divisor $D_{2}$ of type $(2,1)$. Then $\dim D_{1}\cap D_{2}=0$
and $Z\subset D_{1}\cap D_{2}$. Note that the irreducibility of $D_{2}$
implies its smoothness since its arithmetic genus is zero. Since the
intersection number $D_{1}D_{2}=3$, we have
$\operatorname{length}Z=3=\operatorname{length}D_{1}\cap D_{2}$. Hence
$Z=D_{1}\cap D_{2}$. Therefore $\mathcal{I}_{Z}$ fits in the following exact
sequence:

(6.3)
$0\to\mathcal{O}(-3,-2)\to\mathcal{O}(-2,-1)\oplus\mathcal{O}(-1,-1)\to\mathcal{I}_{Z}\to
0.$

Hence $\mathcal{F}$ fits in the following exact sequence:

$0\to\mathcal{O}(-1,-1)\to\mathcal{O}(1,0)\oplus\mathcal{O}^{\oplus
r-1}\to\mathcal{F}\to 0,$

and thus $\mathcal{E}$ fits in the following exact sequence:

$0\to\mathcal{O}(-1,-1)\to\mathcal{O}(1,0)\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus
r-1}\to\mathcal{E}\to 0.$

Therefore $\mathcal{E}$ belongs to Case $(7)$ of Theorem 1.1.

####  6.1.3.  Suppose that $\operatorname{length}Z=4$.

Then $h^{0}(\mathcal{O}_{Z})=4$, and thus $h^{0}(\mathcal{I}_{Z}(2,1))\geq 2$.
Note here that every closed subscheme $Z_{1}\subset Z$ of length $3$
determines uniquely an effective divisor $D_{1}$ containing $Z_{1}$ of type
$(1,1)$ since $\operatorname{length}Z\cap L_{1}\leq 2$ and
$\operatorname{length}Z\cap L_{2}\leq 1$. Moreover $D_{1}$ does not contain
$Z$; otherwise $\mathcal{I}_{Z}(2,1)|_{D_{1}}$ and thus $\mathcal{F}|_{D_{1}}$
would admit a negative degree quotient, which is a contradiction. Hence we see
that the number of reducible divisors containing $Z$ of type $(2,1)$ is
finite. Therefore we can take two irreducible divisors $D_{1}$ and $D_{2}$
containing $Z$ of type $(2,1)$. Then $\dim D_{1}\cap D_{2}=0$ and $Z\subset
D_{1}\cap D_{2}$. Since the intersection number $D_{1}D_{2}=4$, we have
$\operatorname{length}Z=4=\operatorname{length}D_{1}\cap D_{2}$. Thus
$Z=D_{1}\cap D_{2}$, and we have an exact sequence

$0\to\mathcal{O}(-2,-1)\to\mathcal{O}^{\oplus 2}\to\mathcal{I}_{Z}(2,1)\to 0.$

Hence $\mathcal{F}$ fits in the following exact sequence:

$0\to\mathcal{O}(-2,-1)\to\mathcal{O}^{\oplus r}\to\mathcal{F}\to 0.$

Therefore $\mathcal{E}$ belongs to Case $(8)$ of Theorem 1.1.

###  6.2.  The case $c_{2}(\mathcal{F}^{\vee\vee})=1$

We have
$\mathcal{F}^{\vee\vee}\cong\mathcal{O}(1,1)\oplus\mathcal{O}(1,0)\oplus\mathcal{O}^{r-3}$
by Theorem 2.1. Hence $h^{0}(\mathcal{F}^{\vee\vee}(-1,-1))=1$, and thus
$1\leq\operatorname{length}\mathcal{Q}\leq 3$ by (6.2). Since $\mathcal{F}$
cannot admit negative degree quotients, we see that the composite

$\mathcal{O}(1,1)\oplus\mathcal{O}(1,0)\hookrightarrow\mathcal{F}^{\vee\vee}\to\mathcal{Q}$

is surjective. Let $\mathcal{K}$ be the kernel of the surjection
$\mathcal{O}(1,1)\oplus\mathcal{O}(1,0)\to\mathcal{Q}$. Then $\mathcal{F}$
fits in an exact sequence

$0\to\mathcal{K}\to\mathcal{F}\to\mathcal{O}^{\oplus r-3}\to 0.$

Let $\mathcal{Q}_{1}$ and $\mathcal{Q}_{2}$ denote the image and the cokernel
of the composite

$\mathcal{O}(1,1)\hookrightarrow\mathcal{O}(1,1)\oplus\mathcal{O}(1,0)\to\mathcal{Q}$

respectively. Note that we have a surjection
$\mathcal{O}(1,0)\to\mathcal{Q}_{2}$. Hence
$\mathcal{Q}_{2}\cong\mathcal{O}_{Z_{2}}$ for some closed subscheme of
dimension $\leq 0$. If $Z_{2}\neq\emptyset$, then $\mathcal{K}$ admits
$\mathcal{I}_{Z_{2}}(1,0)$ as a quotient, and thus $\mathcal{F}|_{L_{2}}$
admits a negative degree quotient for some line $L_{2}$. This is a
contradiction. Hence $Z_{2}=\emptyset$, and thus the composite
$\mathcal{O}(1,1)\hookrightarrow\mathcal{O}(1,1)\oplus\mathcal{O}(1,0)\to\mathcal{Q}$
is surjective. Thus $\mathcal{Q}\cong\mathcal{O}_{Z}$ for some zero-
dimensional closed subscheme $Z$ with

$1\leq\operatorname{length}Z\leq 3,$

and $\mathcal{K}$ fits in an exact sequence

(6.4) $0\to\mathcal{I}_{Z}(1,1)\to\mathcal{K}\to\mathcal{O}(1,0)\to 0$

with $\operatorname{length}Z\cap L_{1}\leq 2$ and $\operatorname{length}Z\cap
L_{2}\leq 1$.

####  6.2.1. Suppose that $\operatorname{length}Z=1$.

Then $Z$ is a point $p$, and $\mathcal{K}$ fits in the following exact
sequence:

$0\to\mathcal{O}\to\mathcal{O}(1,0)^{\oplus
2}\oplus\mathcal{O}(0,1)\to\mathcal{K}\to 0.$

Thus $\mathcal{E}$ fits in an exact sequence

$0\to\mathcal{O}\to\mathcal{O}(1,0)^{\oplus 2}\oplus\mathcal{O}(0,1)^{\oplus
2}\oplus\mathcal{O}^{\oplus r-3}\to\mathcal{E}\to 0.$

Since $\mathcal{E}$ is a vector bundle and
$h^{0}(\mathcal{E}(-2,0))=h^{0}(\mathcal{E}(0,-2))=0$ by our assumption (6.1),
we see that $\mathcal{E}$ belongs to Case $(6\textrm{-}2)$ of Theorem 1.1.

####  6.2.2.  Suppose that $\operatorname{length}Z=2$.

If $\operatorname{length}Z\cap L_{1}\leq 1$ for any line $L_{1}$, then Lemma
3.2 shows that $\mathcal{K}$ fits in the following exact sequence:

$0\to\mathcal{O}(-1,-1)\to\mathcal{O}(1,0)\oplus\mathcal{O}^{\oplus
2}\to\mathcal{K}\to 0.$

Therefore $\mathcal{E}$ belongs to Case $(7)$ of Theorem 1.1. If
$\operatorname{length}Z\cap L_{1}=2$ for some line $L_{1}$, then $Z$ lies on
the line $L_{1}$, and $\mathcal{I}_{Z}$ fits in the following exact sequence:

$0\to\mathcal{O}(-2,-1)\to\mathcal{O}(-2,0)\oplus\mathcal{O}(0,-1)\to\mathcal{I}_{Z}\to
0.$

Note here that the exact sequence (6.4) does not split, since
$\mathcal{I}_{Z}(1,1)|_{L_{1}}$ admits a negative degree quotients. We will
apply the Bondal spectral sequence (2.1) to $\mathcal{K}$. Note that
$E_{2}^{p,q}$ is the $p$-th cohomology
$\mathcal{H}^{p}(\operatorname{Ext}^{q}(G,\mathcal{K})\operatorname{\otimes^{L}}_{A}G)$
of the complex
$\operatorname{Ext}^{q}(G,\mathcal{K})\operatorname{\otimes^{L}}_{A}G$. Since
$h^{1}(\mathcal{I}_{Z})=1$ and $h^{q}(\mathcal{I}_{Z})=0$ for $q=0,2$, we
infer that $h^{1}(\mathcal{K}(-1,-1))=1$ and $h^{q}(\mathcal{K}(-1,-1))=0$ for
$q=0,2$. Since $h^{q}(\mathcal{I}_{Z}(1,0))=0$ for all $q$, we see that
$h^{q}(\mathcal{K}(0,-1))=0$ for all $q$. Since
$h^{q}(\mathcal{I}_{Z}(0,1))=1$ for $q=0,1$ and
$h^{2}(\mathcal{I}_{Z}(0,1))=0$, the non-splitting of (6.4) implies that
$h^{q}(\mathcal{K}(-1,0))=0$ for $q=1,2$ and $h^{0}(\mathcal{K}(-1,0))=1$.
Since $h^{0}(\mathcal{I}_{Z}(1,1))=2$ and $h^{q}(\mathcal{I}_{Z}(1,1))=0$ for
$q=1,2$, we see that $h^{0}(\mathcal{K})=4$ and $h^{q}(\mathcal{K})=0$ for
$q=1,2$. Hence $\operatorname{Ext}^{2}(G,\mathcal{K})=0$,
$\operatorname{Ext}^{1}(G,\mathcal{K})\cong S_{3}$, and
$\operatorname{Hom}(G,\mathcal{K})$ fits in an exact sequence of right
$A$-modules

$0\to S_{0}^{\oplus 4}\to\operatorname{Hom}(G,\mathcal{K})\to S_{1}\to 0,$

which yields the following distinguished triangle

$\mathcal{O}^{\oplus
4}\to\operatorname{Hom}(G,\mathcal{K})\operatorname{\otimes^{L}}_{A}G\to\mathcal{O}(-1,0)[1]\to.$

By taking cohomologies of the triangle above, we have the following exact
sequence:

$0\to E_{2}^{-1,0}\to\mathcal{O}(-1,0)\to\mathcal{O}^{\oplus 4}\to
E_{2}^{0,0}\to 0.$

We also see that $E_{2}^{-2,1}\cong\mathcal{O}(-1,-1)$ and that
$E_{2}^{p,q}=0$ unless $(p,q)=(-2,1)$ $(-1,0)$ or $(0,0)$. Since $E^{-1}=0$,
this implies that $E_{2}^{-1,0}=E_{\infty}^{-1,0}=0$. Since $\mathcal{K}$ fits
in the following exact sequence:

$0\to E_{2}^{-2,1}\to E_{2}^{0,0}\to\mathcal{K}\to 0,$

the vanishing $\operatorname{Ext}^{1}(\mathcal{O}(-1,-1),\mathcal{O}(-1,0))=0$
shows that $\mathcal{K}$ fits in an exact sequence

$0\to\mathcal{O}(-1,-1)\oplus\mathcal{O}(-1,0)\to\mathcal{O}^{\oplus
4}\to\mathcal{K}\to 0.$

Thus $\mathcal{E}$ fits in the following exact sequence:

$0\to\mathcal{O}(-1,-1)\oplus\mathcal{O}(-1,0)\to\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus
r+1}\to\mathcal{E}\to 0.$

Therefore $\mathcal{E}$ belongs to Case $(7)$ of Theorem 1.1.

####  6.2.3.  Suppose that $\operatorname{length}Z=3$.

As we have observed in the case $\operatorname{length}Z=3$ in Subsection 6.1,
$Z$ is the intersection $D_{1}\cap D_{2}$ of an effective divisor $D_{1}$ of
type $(1,1)$ and an irreducible divisor $D_{2}$ of type $(2,1)$, and
$\mathcal{I}_{Z}$ fits in the exact sequence (6.3). Note here that the exact
sequence (6.4) does not split since $\mathcal{I}_{Z}(1,1)|_{D_{1}}$ admits a
negative degree quotient. Again we will apply the Bondal spectral sequence
(2.1) to $\mathcal{K}$. Since $h^{1}(\mathcal{I}_{Z})=2$ and
$h^{q}(\mathcal{I}_{Z})=0$ for $q=0,2$, we infer that
$h^{1}(\mathcal{K}(-1,-1))=2$ and $h^{q}(\mathcal{K}(-1,-1))=0$ for $q=0,2$.
Since $h^{1}(\mathcal{I}_{Z}(1,0))=1$ and $h^{q}(\mathcal{I}_{Z}(1,0))=0$ for
$q=0,2$, we see that $h^{1}(\mathcal{K}(0,-1))=1$ and
$h^{q}(\mathcal{K}(0,-1))=0$ for $q=0,2$. Since
$h^{1}(\mathcal{I}_{Z}(0,1))=1$ and $h^{q}(\mathcal{I}_{Z}(0,1))=0$ for
$q=0,2$, the non-splitting of (6.4) implies that $h^{q}(\mathcal{K}(-1,0))=0$
for all $q$. Since $h^{0}(\mathcal{I}_{Z}(1,1))=1$ and
$h^{q}(\mathcal{I}_{Z}(1,1))=0$ for $q=1,2$, we see that
$h^{0}(\mathcal{K})=3$ and $h^{q}(\mathcal{K})=0$ for $q=1,2$. Hence
$\operatorname{Ext}^{2}(G,\mathcal{K})=0$,
$\operatorname{Hom}(G,\mathcal{K})\cong S_{0}^{\oplus 3}$, and
$\operatorname{Ext}^{1}(G,\mathcal{K})$ fits in an exact sequence of right
$A$-modules

$0\to S_{2}\to\operatorname{Ext}^{1}(G,\mathcal{K})\to S_{3}^{\oplus 2}\to 0,$

which yields the following distinguished triangle:

$\mathcal{O}(-1,-1)^{\oplus
2}[1]\to\mathcal{O}(0,-1)[1]\to\operatorname{Ext}^{1}(G,\mathcal{K})\operatorname{\otimes^{L}}_{A}G\to.$

Thus $E_{2}^{-2,1}$ and $E_{2}^{-1,1}$ lie in the following exact sequence:

$0\to E_{2}^{-2,1}\to\mathcal{O}(-1,-1)^{\oplus
2}\xrightarrow{\pi}\mathcal{O}(0,-1)\to E_{2}^{-1,1}\to 0.$

Moreover $E_{2}^{0,0}\cong\mathcal{O}^{\oplus 3}$, and $E_{2}^{p,q}=0$ unless
$(p,q)=(-2,1)$, $(-1,1)$ or $(0,0)$. Since $\mathcal{K}$ fits in the exact
sequence $0\to E_{3}^{0,0}\to\mathcal{K}\to E_{2}^{-1,1}\to 0$,
$E_{2}^{-1,1}|_{L_{2}}$ cannot admit a negative degree quotient, and thus the
restriction $\pi|_{L_{2}}$ of the morphism $\pi$ above to any line $L_{2}$ is
surjective. Hence $\pi$ itself is surjective, i.e., $E_{2}^{-1,1}=0$ and
$E_{2}^{-2,1}\cong\mathcal{O}(-2,-1)$. Therefore $\mathcal{K}\cong
E_{3}^{0,0}$ and $\mathcal{K}$ fits in an exact sequence

$0\to\mathcal{O}(-2,-1)\to\mathcal{O}^{\oplus 3}\to\mathcal{K}\to 0.$

Therefore $\mathcal{E}$ belongs to Case $(8)$ of Theorem 1.1.

###  6.3.  The case $c_{2}(\mathcal{F}^{\vee\vee})=2$

We see that $\mathcal{F}^{\vee\vee}$ fits in the following exact sequence:

$0\to\mathcal{O}\to\mathcal{O}(1,0)^{\oplus
2}\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus
r-3}\to\mathcal{F}^{\vee\vee}\to 0$

by Theorem 2.1.

Suppose that $\mathcal{Q}=0$. Then $\mathcal{F}\cong\mathcal{F}^{\vee\vee}$,
and the assumption (6.1) implies that $\mathcal{E}$ belongs to Case
$(6\textrm{-}2)$ of Theorem 1.1.

Suppose that $\mathcal{Q}\neq 0$. Let $\mathcal{G}$ be the kernel of the
composite

$\mathcal{O}(1,0)^{\oplus 2}\oplus\mathcal{O}(0,1)\oplus\mathcal{O}^{\oplus
r-3}\to\mathcal{F}^{\vee\vee}\to\mathcal{Q}$

of the surjections. Then $\mathcal{G}$ fits in an exact sequence

$0\to\mathcal{O}\to\mathcal{G}\to\mathcal{F}\to 0.$

Since $\mathcal{G}$ admits a negative degree quotient on a line by Lemma 3.3,
this implies that $\mathcal{F}$ also admits a negative degree quotient on a
line. This is a contradiction. Hence this case does not arise.

###  6.4.  The case $c_{2}(\mathcal{F}^{\vee\vee})=3$

We see that $\mathcal{F}^{\vee\vee}$ fits in the following exact sequence:

$0\to\mathcal{O}(-1,-1)\oplus\mathcal{O}(-1,0)\to\mathcal{O}^{\oplus
r+1}\to\mathcal{F}^{\vee\vee}\to 0$

by Theorem 2.1. We have $\operatorname{length}\mathcal{Q}=0$ or $1$.

If $\mathcal{Q}=0$, then $\mathcal{E}$ belongs to Case $(7)$ of Theorem 1.1.

If $\mathcal{Q}=k(p)$ for some point $p\in\mathbb{Q}$, $\mathcal{F}$ fits in
an exact sequence

$0\to\mathcal{O}(-1,-1)\oplus\mathcal{O}(-1,0)\to\mathcal{I}_{p}\oplus\mathcal{O}^{\oplus
r}\to\mathcal{F}\to 0.$

Hence $\mathcal{F}$ fits in an exact sequence

$0\to\mathcal{O}(-1,-1)^{\oplus
2}\oplus\mathcal{O}(-1,0)\xrightarrow{\varphi}\mathcal{O}(-1,0)\oplus\mathcal{O}(0,-1)\oplus\mathcal{O}^{\oplus
r}\to\mathcal{F}\to 0.$

We claim here that the composite of the inclusion
$\mathcal{O}(-1,0)\to\mathcal{O}(-1,-1)^{\oplus 2}\oplus\mathcal{O}(-1,0)$,
the morphism $\varphi$ above and the projection
$\mathcal{O}(-1,0)\oplus\mathcal{O}(0,-1)\oplus\mathcal{O}^{\oplus
r}\to\mathcal{O}(-1,0)$ is non-zero. Assume, to the contrary, that it is zero.
Then $\varphi$ induces a morphism $\mathcal{O}(-1,-1)^{\oplus
2}\to\mathcal{O}(-1,0)$, whose cokernel is a quotient of $\mathcal{F}$. Since
$\mathcal{F}|_{L_{1}}$ cannot admit a negative degree quotient for any line
$L_{1}$, the morphism $\mathcal{O}(-1,-1)^{\oplus 2}\to\mathcal{O}(-1,0)$ is
surjective, and thus its kernel is $\mathcal{O}(-1,-2)$. Hence the kernel of
the composite of $\varphi$ and the projection
$\mathcal{O}(-1,0)\oplus\mathcal{O}(0,-1)\oplus\mathcal{O}^{\oplus
r}\to\mathcal{O}(-1,0)$ is isomorphic to
$\mathcal{O}(-1,-2)\oplus\mathcal{O}(-1,0)$. Therefore the exact sequence
above induces the following exact sequence:

$0\to\mathcal{O}(-1,-2)\oplus\mathcal{O}(-1,0)\xrightarrow{\psi}\mathcal{O}(0,-1)\oplus\mathcal{O}^{\oplus
r}\to\mathcal{F}\to 0.$

Since the composite of the inclusion
$\mathcal{O}(-1,0)\to\mathcal{O}(-1,-2)\oplus\mathcal{O}(-1,0)$, the morphism
$\psi$ above and the projection $\mathcal{O}(0,-1)\oplus\mathcal{O}^{\oplus
r}\to\mathcal{O}(0,-1)$ is zero, $\psi$ induces a morphism
$\mathcal{O}(-1,-2)\to\mathcal{O}(0,-1)$, whose cokernel is a quotient of
$\mathcal{F}$ and is isomorphic to either $\mathcal{O}(0,-1)$ or
$\mathcal{O}_{D_{1}}(-p)$, where $D_{1}$ is an effective divisor of type
$(1,1)$ and $p$ is a point on $D_{1}$. This is a contradiction. Hence the
claim holds.

The claim above implies that the composite

$\mathcal{O}(-1,0)\to\mathcal{O}(-1,-1)^{\oplus
2}\oplus\mathcal{O}(-1,0)\xrightarrow{\varphi}\mathcal{O}(-1,0)\oplus\mathcal{O}(0,-1)\oplus\mathcal{O}^{\oplus
r}\to\mathcal{O}(-1,0)$

is an isomorphism, and thus the kernel of the morphism

$\mathcal{O}(-1,-1)^{\oplus
2}\oplus\mathcal{O}(-1,0)\xrightarrow{\varphi}\mathcal{O}(-1,0)\oplus\mathcal{O}(0,-1)\oplus\mathcal{O}^{\oplus
r}\to\mathcal{O}(-1,0)$

is isomorphic to $\mathcal{O}(-1,-1)^{\oplus 2}$. Therefore $\mathcal{F}$ fits
in an exact sequence

$0\to\mathcal{O}(-1,-1)^{\oplus
2}\xrightarrow{\iota}\mathcal{O}(0,-1)\oplus\mathcal{O}^{\oplus
r}\to\mathcal{F}\to 0.$

Since the composite of the injection $\iota$ above and the projection
$\mathcal{O}(0,-1)\oplus\mathcal{O}^{\oplus r}\to\mathcal{O}(0,-1)$ is
surjective, the exact sequence above induces the following exact sequence:

$0\to\mathcal{O}(-2,-1)\to\mathcal{O}^{\oplus r}\to\mathcal{F}\to 0.$

Hence $\mathcal{E}$ belongs to Case $(8)$ of Theorem 1.1.

###  6.5.  The case $c_{2}(\mathcal{F}^{\vee\vee})=4$

We see that $\mathcal{Q}=0$, that $\mathcal{F}\cong\mathcal{F}^{\vee\vee}$,
and that $\mathcal{F}$ fits in the following exact sequence:

$0\to\mathcal{O}(-2,-1)\to\mathcal{O}^{\oplus r}\to\mathcal{F}\to 0$

by Theorem 2.1. Therefore $\mathcal{E}$ belongs to Case $(8)$ of Theorem 1.1.

##  7\. Proof of Theorem 1.1 when $\mathcal{E}$ contains no effective divisors

In the following, we assume that
$\operatorname{Hom}(\mathcal{O}(0,1),\mathcal{E})=\operatorname{Hom}(\mathcal{O}(1,0),\mathcal{E})=0$.

###  7.1. The case $h^{1}(\mathcal{E})=0$

We will apply the Bondal spectral sequence (2.1) to $\mathcal{E}$. Note that
$E_{2}^{p,q}=\mathcal{H}^{p}(\operatorname{Ext}^{q}(G,\mathcal{E})\operatorname{\otimes^{L}}_{A}G)$.
Lemma 3.5 implies that $E_{2}^{p,2}=0$, and Lemma 3.6 shows that

$c_{2}\geq 6.$

Before going into the details of the proof of Subsection 7.1, we describe
herein the structure of Subsection 7.1 and provide a roadmap to the
classification of Theorem 1.1. The proof of Subsection 7.1 is divided into
three parts: in 7.1.1, we will deal with the case $c_{2}=6$ to obtain Case
$(9)$ of Theorem 1.1; in 7.1.2, we will describe the common set-up for the
case $c_{2}\geq 7$ and deal with the case $c_{2}=7$ to obtain Case $(11)$ of
Theorem 1.1; in 7.1.3, we will show that $c_{2}=8$ and deal with the case
$c_{2}=8$ to obtain Cases $(12)$ and $(13)$ of Theorem 1.1.

From Lemma 3.6, we obtain a distinguished triangle

$\begin{split}(S_{1}\operatorname{\otimes^{L}}_{A}G)^{\oplus
c_{2}-6}\oplus(S_{2}\operatorname{\otimes^{L}}_{A}G)^{\oplus
c_{2}-6}&\to\operatorname{Ext}^{1}(G,\mathcal{E})\operatorname{\otimes^{L}}_{A}G\\\
&\to(S_{3}\operatorname{\otimes^{L}}_{A}G)^{\oplus c_{2}-4}\to.\end{split}$

Hence we have the following distinguished triangle:

$\begin{split}(\mathcal{O}(-1,0)^{\oplus
c_{2}-6}\oplus\mathcal{O}(0,-1)^{\oplus
c_{2}-6})[1]&\to\operatorname{Ext}^{1}(G,\mathcal{E})\operatorname{\otimes^{L}}_{A}G\\\
&\to\mathcal{O}(-1,-1)^{\oplus c_{2}-4}[2]\to.\end{split}$

Taking cohomologies of the triangle above yields the following exact sequence:

(7.1) $\begin{split}0\to E_{2}^{-2,1}\to\mathcal{O}(-1,-1)^{\oplus
c_{2}-4}&\xrightarrow{\varphi}\mathcal{O}(-1,0)^{\oplus
c_{2}-6}\oplus\mathcal{O}(0,-1)^{\oplus c_{2}-6}\\\ &\to E_{2}^{-1,1}\to
0.\end{split}$

We also see that $E_{2}^{p,1}=0$ unless $p=-2$ or $-1$. Lemma 3.6 also shows
that

$\operatorname{Hom}(G,\mathcal{E})\operatorname{\otimes^{L}}_{A}G\cong(S_{0}\operatorname{\otimes^{L}}_{A}G)^{\oplus
r+8-c_{2}}\cong\mathcal{O}^{\oplus r+8-c_{2}}.$

Taking cohomologies, we see that $E_{2}^{p,0}=0$ unless $p=0$ and that
$E_{2}^{0,0}\cong\mathcal{O}^{\oplus r+8-c_{2}}$. Therefore we have an exact
sequence

(7.2) $0\to E_{2}^{-2,1}\to\mathcal{O}^{\oplus r+8-c_{2}}\to E_{3}^{0,0}\to
0,$

and $\mathcal{E}$ fits in an exact sequence

(7.3) $0\to E_{3}^{0,0}\to\mathcal{E}\to E_{2}^{-1,1}\to 0.$

Note here that $E_{2}^{-1,1}$ cannot admit a negative degree quotient since
$\mathcal{E}$ is nef.

####  7.1.1.  The case $c_{2}=6$

We have $E_{2}^{-1,1}=0$ and $E_{2}^{-2,1}\cong\mathcal{O}(-1,-1)^{\oplus 2}$
by (7.1). Thus $E_{3}^{0,0}\cong\mathcal{E}$ by (7.3) and the exact sequence
(7.2) gives Case $(9)$ of Theorem 1.1.

####  7.1.2.  The case $c_{2}\geq 7$

Consider the composite

$\psi:\mathcal{O}(-1,-1)^{\oplus
c_{2}-4}\xrightarrow{\varphi}\mathcal{O}(-1,0)^{\oplus
c_{2}-6}\oplus\mathcal{O}(0,-1)^{\oplus c_{2}-6}\to\mathcal{O}(0,-1)$

of the morphism $\varphi$ in the exact sequence (7.1) and the projection
$\mathcal{O}(-1,0)^{\oplus c_{2}-6}\oplus\mathcal{O}(0,-1)^{\oplus
c_{2}-6}\to\mathcal{O}(0,-1)$. Since $E_{2}^{-1,1}|_{L_{2}}$ cannot admit a
negative degree quotient for any line $L_{2}$, the composite $\psi$ is
surjective, and thus the exact sequence (7.1) induces the following exact
sequence:

(7.4) $\begin{split}0\to
E_{2}^{-2,1}&\to\mathcal{O}(-2,-1)\oplus\mathcal{O}(-1,-1)^{\oplus c_{2}-6}\\\
&\xrightarrow{\alpha}\mathcal{O}(-1,0)^{\oplus
c_{2}-6}\oplus\mathcal{O}(0,-1)^{\oplus c_{2}-7}\to E_{2}^{-1,1}\to
0.\end{split}$

Note that the morphism $\alpha$ above is not zero since $E_{2}^{-1,1}$ cannot
admit a negative degree quotient.

Now suppose that $c_{2}=7$. Then the exact sequence (7.4) becomes the
following exact sequence:

$0\to
E_{2}^{-2,1}\to\mathcal{O}(-2,-1)\oplus\mathcal{O}(-1,-1)\xrightarrow{\alpha}\mathcal{O}(-1,0)\to
E_{2}^{-1,1}\to 0.$

If the composite

$\mathcal{O}(-1,-1)\to\mathcal{O}(-2,-1)\oplus\mathcal{O}(-1,-1)\xrightarrow{\alpha}\mathcal{O}(-1,0)$

of the inclusion
$\mathcal{O}(-1,-1)\to\mathcal{O}(-2,-1)\oplus\mathcal{O}(-1,-1)$ and the
morphism $\alpha$ is zero, then we get an exact sequence
$0\to\mathcal{O}(-2,-1)\to\mathcal{O}(-1,0)\to E_{2}^{-1,1}\to 0$, and thus
$E_{2}^{-1,1}\cong\mathcal{O}_{D_{1}}(-p)$ for some effective divisor $D_{1}$
of type $(1,1)$ and some point $p$ on $D_{1}$. This is a contradiction. Hence
the composite above is non-zero and injective, and thus we obtain the
following exact sequence:

$0\to E_{2}^{-2,1}\to\mathcal{O}(-2,-1)\to\mathcal{O}_{L_{1}}(-1)\to
E_{2}^{-1,1}\to 0.$

Since $E_{2}^{-1,1}$ cannot admit a negative degree quotient, the morphism
$\mathcal{O}(-2,-1)\to\mathcal{O}_{L_{1}}(-1)$ is non-zero, and its image is
$\mathcal{O}_{L_{1}}(-2)$. Hence

$E_{2}^{-1,1}\cong k(p)$

for some point $p\in L_{1}$ and $E_{2}^{-2,1}\cong\mathcal{O}(-2,-2)$. The
exact sequences (7.2) and (7.3) then yield Case $(11)$ of Theorem 1.1.

####  7.1.3.  The case $c_{2}\geq 8$

Since $\mathcal{E}$ is nef, we have $c_{2}\leq c_{1}^{2}=8$. Hence $c_{2}=8$.
The exact sequence (7.4) then becomes the following exact sequence:

$\begin{split}0\to
E_{2}^{-2,1}\to\mathcal{O}(-2,-1)\oplus\mathcal{O}(-1,-1)^{\oplus
2}&\xrightarrow{\alpha}\mathcal{O}(-1,0)^{\oplus 2}\oplus\mathcal{O}(0,-1)\\\
&\to E_{2}^{-1,1}\to 0.\end{split}$

Since $E_{2}^{-1,1}$ cannot admit a negative degree quotient, the composite

$\mathcal{O}(-2,-1)\oplus\mathcal{O}(-1,-1)^{\oplus
2}\xrightarrow{\alpha}\mathcal{O}(-1,0)^{\oplus
2}\oplus\mathcal{O}(0,-1)\to\mathcal{O}(0,-1)$

of $\alpha$ and the projection $\mathcal{O}(-1,0)^{\oplus
2}\oplus\mathcal{O}(0,-1)\to\mathcal{O}(0,-1)$ is surjective. Let
$\mathcal{K}$ be its kernel. Then $\mathcal{K}$ fits in the following exact
sequence:

(7.5) $0\to
E_{2}^{-2,1}\to\mathcal{K}\xrightarrow{\mu}\mathcal{O}(-1,0)^{\oplus 2}\to
E_{2}^{-1,1}\to 0.$

Note that the morphism $\mu$ in (7.5) cannot be zero since $E_{2}^{-1,1}$
cannot admit a negative degree quotient. Moreover $\mathcal{K}$ is either
$\mathcal{O}(-3,-1)\oplus\mathcal{O}(-1,-1)$ or $\mathcal{O}(-2,-1)^{\oplus
2}$. Therefore we have two cases:

  1. $(a)$

$\mathcal{K}\cong\mathcal{O}(-3,-1)\oplus\mathcal{O}(-1,-1);$

  2. $(b)$

$\mathcal{K}\cong\mathcal{O}(-2,-1)^{\oplus 2}$.

We consider Case (a) in 7.1.3.1 and address Case (b) in 7.1.3.2.

#####  7.1.3.1 Case $(a)$

If the composite

$\nu_{a}:\mathcal{O}(-1,-1)\to\mathcal{K}\xrightarrow{\mu}\mathcal{O}(-1,0)^{\oplus
2}$

of the inclusion $\mathcal{O}(-1,-1)\to\mathcal{K}$ and $\mu$ is zero, then
$\mu$ induces an exact sequence

$\mathcal{O}(-3,-1)\xrightarrow{\bar{\mu}}\mathcal{O}(-1,0)^{\oplus 2}\to
E_{2}^{-1,1}\to 0.$

Since $\bar{\mu}\neq 0$, the composite

$\mathcal{O}(-3,-1)\xrightarrow{\bar{\mu}}\mathcal{O}(-1,0)^{\oplus
2}\to\mathcal{O}(-1,0)$

of $\bar{\mu}$ and some projection $\mathcal{O}(-1,0)^{\oplus
2}\to\mathcal{O}(-1,0)$ is nonzero, and thus its cokernel is isomorphic to
$\mathcal{O}_{D}(-p)$ for some divisor $D$ of type $(2,1)$ and a point $p\in
D$. Since this cokernel $\mathcal{O}_{D}(-p)$ is also a quotient of
$E_{2}^{-1,1}$, this contradicts the property of $E_{2}^{-1,1}$. Hence the
composite $\nu_{a}$ above is nonzero. Let $\mathcal{C}_{a}$ be the cokernel of
$\nu_{a}$. Since $\nu_{a}$ gives rise to a section $(s_{1},s_{2})\in
H^{0}(\mathcal{O}(0,1))^{\oplus 2}$, we see that

$\mathcal{C}_{a}\cong\begin{cases}\mathcal{O}(-1,1)\quad\textrm{if
}(s_{1})_{0}\cap(s_{2})_{0}=\emptyset;\\\
\mathcal{O}(-1,0)\oplus\mathcal{O}_{L_{1}}(-1)\quad\textrm{if
}(s_{1})_{0}\cap(s_{2})_{0}=L_{1}.\end{cases}$

Note that $\mathcal{C}_{a}$ also fits in an exact sequence

$0\to
E_{2}^{-2,1}\to\mathcal{O}(-3,-1)\xrightarrow{\bar{\mu}_{a}}\mathcal{C}_{a}\to
E_{2}^{-1,1}\to 0.$

We claim here that $\bar{\mu}_{a}$ is injective. To see this, first note that
$\bar{\mu}_{a}$ is nonzero since $E_{2}^{-1,1}$ cannot admit a negative degree
quotient on a line $L_{1}$. If $\bar{\mu}_{a}$ is not injective, then
$\mathcal{C}_{a}\cong\mathcal{O}(-1,0)\oplus\mathcal{O}_{L_{1}}(-1)$ and the
composite of $\bar{\mu}_{a}$ and the projection
$\mathcal{O}(-1,0)\oplus\mathcal{O}_{L_{1}}(-1)\to\mathcal{O}(-1,0)$ is zero,
and thus $E_{2}^{-1,1}$ admits $\mathcal{O}(-1,0)$ as a quotient. This is a
contradiction. Hence the claim holds: $\bar{\mu}_{a}$ is injective, and thus

$E_{2}^{-2,1}=0.$

The exact sequences (7.1), (7.2), and (7.3) then become the following exact
sequences:

(7.6) $\displaystyle 0\to\mathcal{O}(-1,-1)^{\oplus
4}\to\mathcal{O}(-1,0)^{\oplus 2}\oplus\mathcal{O}(0,-1)^{\oplus 2}\to
E_{2}^{-1,1}\to 0;$ (7.7) $\displaystyle 0\to\mathcal{O}^{\oplus
r}\to\mathcal{E}\to E_{2}^{-1,1}\to 0.$

These two exact sequences result in Case $(13)$ of Theorem 1.1. Here we
address the structure of $E_{2}^{-1,1}$ some more; if $\mathcal{C}_{a}$ is
isomorphic to $\mathcal{O}(-1,0)\oplus\mathcal{O}_{L_{1}}(-1)$, then the
composite of $\bar{\mu}_{a}$ and the projection
$\mathcal{O}(-1,0)\oplus\mathcal{O}_{L_{1}}(-1)\to\mathcal{O}(-1,0)$ is
nonzero, and its cokernel is a quotient of $E_{2}^{-1,1}$ and isomorphic to
$\mathcal{O}_{D}(-p)$ for some point $p$ on a divisor $D$ of type $(2,1)$.
This is a contradiction. Hence $\mathcal{C}_{a}\cong\mathcal{O}(-1,1)$, and we
see that

$E_{2}^{-1,1}\cong\mathcal{O}_{E}(\mathfrak{d})$

for some divisor $\mathfrak{d}$ of degree $0$ on a divisor $E$ of type
$(2,2)$.

#####  7.1.3.2 Case $(b)$

Since $\mu\neq 0$, the composite

$\nu_{b}:\mathcal{O}(-2,-1)\to\mathcal{K}\xrightarrow{\mu}\mathcal{O}(-1,0)^{\oplus
2}$

of a suitable inclusion $\mathcal{O}(-2,-1)\to\mathcal{K}$ and $\mu$ is
nonzero. Let $\mathcal{C}_{b}$ be the cokernel of $\nu_{b}$. Then
$\mathcal{C}_{b}$ fits in an exact sequence

$0\to
E_{2}^{-2,1}\to\mathcal{O}(-2,-1)\xrightarrow{\bar{\mu}_{b}}\mathcal{C}_{b}\to
E_{2}^{-1,1}\to 0.$

Since $\nu_{b}$ gives rise to a section $(t_{1},t_{2})\in
H^{0}(\mathcal{O}(1,1))^{\oplus 2}$, $\mathcal{C}_{b}$ satisfies one of the
following:

  1. (i)

If $(t_{1})_{0}\cap(t_{2})_{0}=C$ for some divisor $C$ of type $(1,1)$, then

$\mathcal{C}_{b}\cong\mathcal{O}(-1,0)\oplus\mathcal{O}_{C}(-p)$

for some point $p\in C$;

  2. (ii)

If $(t_{1})_{0}\cap(t_{2})_{0}=L_{2}$, then $\mathcal{C}_{b}$ fits in an exact
sequence

$0\to\mathcal{O}_{L_{2}}(-1)\to\mathcal{C}_{b}\to\mathcal{O}(-1,1)\to 0;$

  3. (iii)

If $(t_{1})_{0}\cap(t_{2})_{0}=L_{1}$, then $\mathcal{C}_{b}$ fits in an exact
sequence

$0\to\mathcal{O}_{L_{1}}(-2)\to\mathcal{C}_{b}\to\mathcal{O}\to 0;$

  4. (iv)

If $(t_{1})_{0}\cap(t_{2})_{0}=Z$ for some zero-dimensional subscheme of
length $2$, then

$\mathcal{C}_{b}\cong\mathcal{I}_{Z}(0,1).$

We will divide 7.1.3.2 into two cases: in 7.1.3.2.1, we consider the case
where $\bar{\mu}_{b}$ is not injective; in 7.1.3.2.2, we address the case
where $\bar{\mu}_{b}$ is injective.

######  7.1.3.2.1  Suppose that $\bar{\mu}_{b}$ is not injective.

First consider Case (i). We see that the composite of $\bar{\mu}_{b}$ and the
projection $\mathcal{C}_{b}\to\mathcal{O}(-1,0)$ is zero, and thus
$E_{2}^{-1,1}$ admits $\mathcal{O}(-1,0)$ as a quotient. This is a
contradiction. Hence Case (i) does not occur. Next consider Case (ii). Then
the composite of $\bar{\mu}_{b}$ and the projection
$\mathcal{C}_{b}\to\mathcal{O}(-1,1)$ is zero, and $E_{2}^{-1,1}$ admits
$\mathcal{O}(-1,1)$ as a quotient, which is a contradiction. Hence Case (ii)
does not occur. For Case (iii), we first observe that
$\bar{\mu}_{b}:\mathcal{O}(-2,-1)\to\mathcal{C}_{b}$ factors through
$\mathcal{O}_{L_{1}}(-2)$. If the induced morphism
$\mathcal{O}(-2,-1)\to\mathcal{O}_{L_{1}}(-2)$ is zero, then $\bar{\mu}_{b}=0$
and $E_{2}^{-1,1}\cong\mathcal{C}_{b}$. We claim here that the exact sequence
in Case (iii) splits. Assume, to the contrary, that it does not split. We will
apply the Bondal spectral sequence (2.1) to $\mathcal{C}_{b}$. Set
$E_{2}^{p,q}(\mathcal{C}_{b}):=\mathcal{H}^{p}(\operatorname{Ext}^{q}(G,\mathcal{C}_{b})\operatorname{\otimes^{L}}_{A}G)$
in order to avoid confusion with $E_{2}^{p,q}$ for $\mathcal{E}$. Non-
splitting of the exact sequence in (iii) implies that
$h^{q}(\mathcal{C}_{b})=0$ for all $q$. We see that
$h^{1}(\mathcal{C}_{b}(-1,0))=2$, that $h^{q}(\mathcal{C}_{b}(-1,0))=0$ for
$q=0,2$, that $h^{1}(\mathcal{C}_{b}(0,-1))=1$, that
$h^{q}(\mathcal{C}_{b}(0,-1))=0$ for $q=0,2$, that
$h^{1}(\mathcal{C}_{b}(-1,-1))=2$, and that $h^{q}(\mathcal{C}_{b}(-1,-1))=0$
for $q=0,2$. Hence $\operatorname{Ext}^{2}(G,\mathcal{C}_{b})=0$,
$\operatorname{Hom}(G,\mathcal{C}_{b})=0$, and
$\operatorname{Ext}^{1}(G,\mathcal{C}_{b})$ fits in the following exact
sequence of right $A$-modules:

$0\to S_{1}^{\oplus 2}\oplus
S_{2}\to\operatorname{Ext}^{1}(G,\mathcal{C}_{b})\to S_{3}^{\oplus 2}\to 0,$

which induces the following distinguished triangle:

$\mathcal{O}(-1,-1)^{\oplus 2}[1]\to(\mathcal{O}(-1,0)^{\oplus
2}\oplus\mathcal{O}(0,-1))[1]\to\operatorname{Ext}^{1}(G,\mathcal{C}_{b})\operatorname{\otimes^{L}}_{A}G\to.$

Therefore we have the following exact sequence

$0\to E_{2}^{-2,1}(\mathcal{C}_{b})\to\mathcal{O}(-1,-1)^{\oplus
2}\to\mathcal{O}(-1,0)^{\oplus 2}\oplus\mathcal{O}(0,-1)\to
E_{2}^{-1,1}(\mathcal{C}_{b})\to 0.$

Moreover we see that $E_{2}^{p,q}=0$ unless $(p,q)=(-2,1)$ or $(-1,1)$. Thus
we have $E_{2}^{-2,1}(\mathcal{C}_{b})=E^{-1}(\mathcal{C}_{b})=0$, where
$E^{-1}(\mathcal{C}_{b})$ denotes the $E^{-1}$ for $\mathcal{C}_{b}$. Now we
see that $E_{2}^{-1,1}(\mathcal{C}_{b})\cong\mathcal{C}_{b}\cong
E_{2}^{-1,1}$. Since $E_{2}^{-1,1}|_{L_{2}}$ cannot admit a negative degree
quotient, the exact sequence above induces the following exact sequence:

$0\to\mathcal{O}(-2,-1)\to\mathcal{O}(-1,0)^{\oplus 2}\to E_{2}^{-1,1}\to 0,$

which implies that $E_{2}^{-1,1}$ admits a negative degree quotient on a
divisor of type $(1,1)$. This is a contradiction. Hence the exact sequence in
Case (iii) splits if $\bar{\mu}_{b}=0$. Then $E_{2}^{-1,1}$ admits
$\mathcal{O}_{L_{1}}(-2)$ as a quotient, which is a contradiction. Hence the
induced morphism $\mathcal{O}(-2,-1)\to\mathcal{O}_{L_{1}}(-2)$ is nonzero and
thus surjective, and the following conclusions are drawn:

$\displaystyle E_{2}^{-2,1}$ $\displaystyle\cong\mathcal{O}(-2,-2);$
$\displaystyle E_{2}^{-1,1}$ $\displaystyle\cong\mathcal{O}.$

The exact sequences (7.2) and (7.3) then become the following exact sequences:

(7.8) $\displaystyle 0\to\mathcal{O}(-2,-2)\to$
$\displaystyle\mathcal{O}^{\oplus r}\to E_{3}^{0,0}\to 0;$ (7.9)
$\displaystyle 0\to E_{3}^{0,0}\to$ $\displaystyle\mathcal{E}\to\mathcal{O}\to
0.$

These two exact sequences yield Case $(12)$ of Theorem 1.1. Finally consider
Case (iv). In this case, $\bar{\mu}_{b}=0$, and thus $\mathcal{C}_{b}\cong
E_{2}^{-1,1}$. Then $E_{2}^{-1,1}|_{C}$ admits $\mathcal{O}_{C}(-1)$ as a
quotient for some divisor $C$ of type $(1,1)$ passing through $Z$. This is a
contradiction. Hence Case (iv) does not occur.

######  Remark 7.1.

If $\mathcal{E}$ fits in the exact sequences (7.8) and (7.9), then the
cokernel of the evaluation map
$H^{0}(\mathcal{E})\otimes\mathcal{O}\to\mathcal{E}$ is $\mathcal{O}$.

######  7.1.3.2.2  Suppose that $\bar{\mu}_{b}$ is injective.

Then

$E_{2}^{-2,1}=0.$

As in Case $(a)$, this implies that $E_{2}^{-1,1}$ fits in the exact sequences
(7.6) and (7.7) and we obtain Case $(13)$ of Theorem 1.1. Here we address the
structure of $E_{2}^{-1,1}$ some more; if $\mathcal{C}_{b}$ belongs to Case
(i), then $E_{2}^{-1,1}$ admits $\mathcal{O}_{D_{1}}(-1)$ as a quotient for
some divisor $D_{1}$ of type $(1,1)$, which is a contradiction. Hence Case (i)
does not occur. If $\mathcal{C}_{b}$ belongs to Case (ii), then $E_{2}^{-1,1}$
admits $\mathcal{O}_{D_{2}}(-1)$ as a quotient for some divisor $D_{2}$ of
type $(1,2)$, which is a contradiction. Hence Case (ii) does not occur. If
$\mathcal{C}_{b}$ belongs to Case (iii), then $E_{2}^{-1,1}$ fits in the
following exact sequence:

$0\to\mathcal{O}_{L_{1}}(-2)\to E_{2}^{-1,1}\to\mathcal{O}_{D_{3}}\to 0,$

where $D_{3}$ is a divisor of type $(2,1)$. In this case, $E_{2}^{-1,1}$ lies
on the divisor $L_{1}+D_{3}$ of type $(2,2)$. Finally if $\mathcal{C}_{b}$
belongs to Case (iv), then $E_{2}^{-1,1}$ fits in the following exact
sequence:

$0\to\mathcal{O}(-2,-1)\to\mathcal{I}_{Z}(0,1)\to E_{2}^{-1,1}\to 0,$

and thus

$E_{2}^{-1,1}\cong\mathcal{O}_{E}(\mathfrak{d})$

for some divisor $\mathfrak{d}$ of degree $0$ on an effective divisor $E$ of
type $(2,2)$.

######  Remark 7.2.

If $\mathcal{E}$ fits in the exact sequences $(\ref{deg0Elliptic})$ and
$(\ref{deg0EllipticQuot})$, then the cokernel of the evaluation map
$H^{0}(\mathcal{E})\otimes\mathcal{O}\to\mathcal{E}$ is supported on a divisor
$E$ of type $(2,2)$.

###  7.2. The case $h^{1}(\mathcal{E})>0$

Recall that $c_{2}\leq c_{1}^{2}=8$ since $\mathcal{E}$ is nef. The assumption
$h^{1}(\mathcal{E})>0$ then implies that $c_{2}=c_{1}^{2}=8$ by Lemma 3.7. The
natural isomorphism
$H^{1}(\mathcal{E})\cong\operatorname{Ext}^{1}(\mathcal{O},\mathcal{E})$
induces the following exact sequence:

$0\to\mathcal{E}\to\mathcal{E}_{0}\to H^{1}(\mathcal{E})\otimes\mathcal{O}\to
0.$

It follows from [Laz04, Theorem 6.2.12 (ii)] that $\mathcal{E}_{0}$ is a nef
vector bundle of rank $r+h^{1}(\mathcal{E})$ with
$c_{1}(\mathcal{E}_{0})=(2,2)$, $c_{2}(\mathcal{E}_{0})=8$, and
$h^{1}(\mathcal{E}_{0})=0$. Note here the evaluation map
$\operatorname{ev}:H^{0}(\mathcal{E}_{0})\otimes\mathcal{O}\to\mathcal{E}_{0}$
factors through $\mathcal{E}$ since $H^{0}(\mathcal{E})\cong
H^{0}(\mathcal{E}_{0})$. Hence we have a surjection
$\operatorname{Coker}(\operatorname{ev})\to
H^{1}(\mathcal{E})\otimes\mathcal{O}$. From Remarks 7.1 and 7.2, it follows
that $\operatorname{Coker}(\operatorname{ev})\cong\mathcal{O}$, that
$h^{1}(\mathcal{E})=1$, and that $\mathcal{E}$ is globally generated. Thus
$\mathcal{E}$ belongs to Case $(12)$ of Theorem 1.1. This completes the proof
of Theorem 1.1.

## References

  * [BHM15] Edoardo Ballico, Sukmoon Huh, and Francesco Malaspina.  Globally generated vector bundles on a smooth quadric surface.  Sci. China Math., 58(3):633–652, 2015.
  * [Bon89] Alexey I. Bondal.  Representations of associative algebras and coherent sheaves.  Izv. Akad. Nauk SSSR Ser. Mat., 53(1):25–44, 1989.
  * [Huy06] Daniel Huybrechts.  Fourier-Mukai transforms in algebraic geometry.  Oxford Mathematical Monographs. The Clarendon Press Oxford University Press, Oxford, 2006.
  * [Laz04] Robert Lazarsfeld.  Positivity in algebraic geometry. II. Positivity for vector bundles, and multiplier ideals., volume 49 of Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. A Series of Modern Surveys in Mathematics.  Springer-Verlag, Berlin, 2004.
  * [Ohn20a] Masahiro Ohno.  Nef vector bundles on a projective space with first Chern class three.  Rend. Circ. Mat. Palermo Series 2, 69(2):425–458, 2020.
  * [Ohn20b] Masahiro Ohno.  Nef vector bundles on a quadric surface with the first Chern class (2,1).  Adv. Geom, 20(1):109–116, 2020.
  * [Ohn22] Masahiro Ohno.  Nef vector bundles on a projective space or a hyperquadric with the first Chern class small.  Rend. Circ. Mat. Palermo Series 2, 71(2):755–781, 2022.
  * [OT14] Masahiro Ohno and Hiroyuki Terakawa.  A spectral sequence and nef vector bundles of the first Chern class two on hyperquadrics.  Ann. Univ. Ferrara Sez. VII Sci. Mat., 60(2):397–406, 2014.
  * [PSW92] Thomas Peternell, Michał Szurek, and Jarosław A. Wiśniewski.  Numerically effective vector bundles with small Chern classes.  In K. Hulek, T. Peternell, M. Schneider, and F.-O. Schreyer, editors, Complex Algebraic varieties, Proceedinigs, Bayreuth, 1990, number 1507 in Lecture Notes in Math., pages 145–156, Berlin, 1992. Springer.
  * [SW90] Michał Szurek and Jarosław A. Wiśniewski.  Fano bundles of rank 2 on surfaces.  Compositio Math., 76:295–305, 1990.