1. Introduction

In [Reference Peternell, Szurek, Wiśniewski, Hulek, Peternell, Schneider and Schreyer17, § 2 Theorem 2], Peternell–Szurek–Wiśniewski classified nef vector bundles on a smooth hyperquadric $\mathbb{Q}^n$ of dimension $n\geq 3$ with first Chern class $\leq 1$ over an algebraically closed field K of characteristic zero. In [Reference Ohno12, Theorem 9.3], we provided a different proof of this classification, which was based on an analysis with a full strong exceptional collection of vector bundles on $\mathbb{Q}^n$.

In this paper, we classify nef vector bundles on a smooth quadric threefold $\mathbb{Q}^3$ with first Chern class two. (In the subsequent paper [Reference Ohno14], we classify those on a smooth hyperquadric $\mathbb{Q}^n$ of dimension $n\geq 4$.) The precise statement is as follows.

Note that this list is effective: in each case exists an example. For example, if we denote by $\mathcal{N}$ a null correlation bundle on $\mathbb{P}^3$, then $\pi_p^*(\mathcal{N}(1))$ belongs to Case (9) of Theorem 1.1, where $\pi_p:\mathbb{Q}^3\to \mathbb{P}^3$ is the projection from a point $p\in \mathbb{P}^4\setminus\mathbb{Q}^3$. (Similarly, $\pi_p^*(\Omega_{\mathbb{P}^3}(2))$ belongs to Case (9) of Theorem 1.1.) Under the stronger assumption that $\mathcal{E}$ is globally generated, Ballico–Huh–Malaspina provided a classification of $\mathcal{E}$ on $\mathbb{Q}^3$ with $c_1=2$ in [Reference Ballico, Huh and Malaspina3] and [Reference Ballico, Huh and Malaspina2].

Note also that the projectivization $\mathbb{P}(\mathcal{E})$ of the bundle $\mathcal{E}$ in Theorem 1.1 is a Fano manifold of dimension r + 2, i.e. the bundle $\mathcal{E}$ in Theorem 1.1 is a Fano bundle on $\mathbb{Q}^3$ of rank r. As a related result, Langer classified smooth Fano 4-folds with adjunction theoretic scroll structure over $\mathbb{Q}^3$ in [Reference Langer10, Theorem 7.2].

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

\begin{equation*}(\mathcal{O},\mathcal{S},\mathcal{O}(1),\mathcal{O}(2))\end{equation*}

of vector bundles (see [Reference Ohno12] for more details).

The content of this paper is as follows. In § 2, we briefly recall Bondal’s theorem [Reference Alexey1, Theorem 6.2] and its related notions and results required in the proof of Theorem 1.1. In particular, we recall some finite-dimensional algebra A and fix some symbols, e.g. G, P_i and S_i, related to A and to finitely generated right A-modules. We also recall the classification [Reference Ohno13, Theorem 1.1] of nef vector bundles on a smooth quadric surface $\mathbb{Q}^2$ with Chern class $(2,2)$ in Theorem 2.3. In § 3, we recall some basic properties of the spinor bundle $\mathcal{S}$ on $\mathbb{Q}^3$. In § 4, we state Hirzebruch–Riemann–Roch formulas for vector bundles $\mathcal{E}$ on $\mathbb{Q}^3$ with $c_1=2$ and for $\mathcal{S}^\vee\otimes \mathcal{E}$. In § 5, we show some key lemmas required later in the proof of Theorem 1.1. In § 6, we provide a lower bound for the third Chern class of a nef vector bundle $\mathcal{E}$, if $h^0(\mathcal{E}(-D))\neq 0$ for some effective divisor D. In § 7, we provide the set-up for the proof of Theorem 1.1. The proof of Theorem 1.1 is carried out in § 8–19, according to which case of Theorem 2.3 $\mathcal{E}|_{\mathbb{Q}^2}$ belongs to.

2. Preliminaries

Throughout this paper, G ₀, G ₁, G ₂, G ₃ denote respectively $\mathcal{O}$, $\mathcal{S}$, $\mathcal{O}(1)$, $\mathcal{O}(2)$ on $\mathbb{Q}^3$. An important and well-known fact [Reference Kapranov9, Theorem 4.10] of the collection $(G_0,G_1,G_2,G_3)$ is that it is a full strong exceptional collection in $D^b(\mathbb{Q}^3)$, where $D^b(\mathbb{Q}^3)$ denotes the bounded derived category of (the abelian category of) coherent sheaves on $\mathbb{Q}^3$. Here we use the term ‘collection’ to mean ‘family’, not ‘set’. Thus, an exceptional collection is also called an exceptional sequence. We refer to [Reference Huybrechts7] for the definition of a full strong exceptional sequence.

Denote by G the direct sum $\bigoplus_{i=0}^3G_i$ of G ₀, G ₁, G ₂ and G ₃, and by A the endomorphism ring $\operatorname{End}(G)$ of G. The ring 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 $\mathbb{Q}^3$. 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 $e_i=\iota_i\circ p_i$. Then $e_i\in A$. Set

\begin{equation*}P_i=e_iA.\end{equation*}

Then $A\cong \oplus_i P_i$ as right A-modules, and P_i’s are projective right A-modules. We see that $P_i\otimes_A G\cong G_i$. Any finitely generated right A-module V has an ascending filtration

\begin{equation*}0=V^{\leq -1}\subset V^{\leq 0}\subset V^{\leq 1}\subset V^{\leq 2} \subset V^{\leq 3}=V\end{equation*}

by right A-submodules, where $V^{\leq i}$ is defined to be $\bigoplus_{j\leq i}Ve_j$. Set $\operatorname{Gr}^iV=V^{\leq i}/V^{\leq i-1}$ and

\begin{equation*}S_i=\operatorname{Gr}^iP_i.\end{equation*}

Then $\operatorname{Gr}^iS_i\cong K$ as K-vector spaces, $\operatorname{Gr}^jS_i=0$ for any j ≠ i, and S_i is a simple right A-module. If we set $m_i=\dim_K\operatorname{Gr}^iV$, then $\operatorname{Gr}^iV\cong S_i^{\oplus m_i}$ as right A-modules.

It follows from Bondal’s theorem [Reference Alexey1, Theorem 6.2] that

\begin{equation*}\operatorname{RHom}(G,\bullet):D^b(\mathbb{Q}^3)\to D^b (\operatorname{mod} A)\end{equation*}

is an exact equivalence, and its quasi-inverse is

\begin{equation*}\bullet\operatorname{\otimes^{L}}_AG:D^b(\operatorname{mod} A)\to D^b(\mathbb{Q}^3).\end{equation*}

For a coherent sheaf $\mathcal{F}$ on $\mathbb{Q}^3$, this fact can be rephrased in terms of a spectral sequence [Reference Ohno and Terakawa15, Theorem 1]:

(2.1)\begin{equation} 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} \end{equation}

which is called the Bondal spectral sequence. Note that $E_2^{p,q}$ is the pth cohomology sheaf $\mathcal{H}^p(\operatorname{Ext}^q(G,\mathcal{F})\operatorname{\otimes^{L}}_AG)$ of the complex $\operatorname{Ext}^q(G,\mathcal{F})\operatorname{\otimes^{L}}_AG$. When we compute the spectral sequence, we consider the ascending filtration on the right A-module $\operatorname{Ext}^q(G,\mathcal{F})$ and apply the following

Lemma 2.1. We have

(2.2)\begin{equation} S_3\operatorname{\otimes^{L}}_AG \cong \mathcal{O}(-1)[3]; \end{equation}

(2.3)\begin{equation} S_2\operatorname{\otimes^{L}}_AG\cong T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}[2]; \end{equation}

(2.4)\begin{equation} S_1\operatorname{\otimes^{L}}_AG\cong \mathcal{S}^{\vee}[1]\cong \mathcal{S}(-1)[1]; \end{equation}

(2.5)\begin{equation} S_0\operatorname{\otimes^{L}}_AG\cong \mathcal{O}, \end{equation}

where $T_{\mathbb{P}^4}$ denotes the tangent bundle of $\mathbb{P}^4$.

Proof. Since $\operatorname{RHom}(G,\mathcal{O}(-1)[3])\cong S_3$, we obtain (2.2). Note that we have an isomorphism $\operatorname{RHom}(G,\mathcal{S}^{\vee}[1])\cong S_1$ by [Reference Ohno12, Lemma 8.2 (1)]. Hence we have (2.4). It is easy to see that the last isomorphism (2.5) holds. To see (2.3), first note that we have the following exact sequence:

\begin{equation*} 0\to \mathcal{O}(-2)\to \mathcal{O}(-1)\otimes H^0(\mathcal{O}(1))^{\vee} \to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}\to 0. \end{equation*}

Serre duality shows that

\begin{equation*}H^3(\mathcal{O}(-4))\to H^3(\mathcal{O}(-3))\otimes H^0(\mathcal{O}(1))^{\vee}\end{equation*}

is dual of the canonical isomorphism

\begin{equation*} H^0(\mathcal{O})\otimes H^0(\mathcal{O}(1)) \to H^0(\mathcal{O}(1)). \end{equation*}

Hence $H^q(T_{\mathbb{P}^4}(-4)|_{\mathbb{Q}^3})=0$ for all q. Moreover, $h^q(\mathcal{S}^{\vee}(-i))=0$ for $i=0,1,2$ and all q. Therefore, we conclude that $\operatorname{RHom}(G,T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}))$ is isomorphic to $S_2[-2]$.

Remark 2.2. As the referee pointed out, Lemma 2.1 shows that

(2.6)\begin{equation} (\mathcal{O}(-1),T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}, \mathcal{S}^{\vee},\mathcal{O}) \end{equation}

is the left dual exceptional collection of $(G_0,G_1,G_2,G_3)$ $($see [Reference Alexey1] and [Reference Fonarev5] for the definition and the characterization of the left dual exceptional collection $)$. Moreover, the full exceptional collection above is strong by [Reference Bondal and Polishchuk4, Proposition 3.3] (or by showing directly that $\operatorname{Ext}^q(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}, \mathcal{S}^{\vee})=0$ for any q > 0 through the Euler exact sequence).

Our proof of Theorem 1.1 relies on the following theorem [Reference Ohno13, Theorem 1.1]:

3. Some basic properties of the spinor bundle $\mathcal{S}$ on $\mathbb{Q}^3$

We recall some basic facts and properties of the spinor bundle $\mathcal{S}$ on $\mathbb{Q}^3$ in our notation (see Ottaviani’s result [Reference Ottaviani16] and [Reference Ohno12, Theorem 8.1]). First we have an exact sequence

(3.1)\begin{equation} 0\to \mathcal{S}^{\vee}\to \mathcal{O}^{\oplus 4}\to \mathcal{S}\to 0 \end{equation}

by [Reference Ottaviani16, Theorem 2.8 (1)]. The restriction $\mathcal{S}|_{\mathbb{Q}^2}$ of $\mathcal{S}$ to a smooth hyperplane section $\mathbb{Q}^2$ of $\mathbb{Q}^3$ is isomorphic to $\mathcal{O}(1,0)\oplus \mathcal{O}(0,1)$, and $h^0(\mathcal{S})=4$. We have $\det\mathcal{S}=\mathcal{O}(1)$, and thus the canonical isomorphism

(3.2)\begin{equation} \mathcal{S}^{\vee}(1)\cong \mathcal{S}. \end{equation}

The zero locus (s)₀ of every non-zero element s of $H^0(\mathcal{S})$ is a line l in $\mathbb{Q}^3$. Thus $c_1(\mathcal{S})\cap [\mathbb{Q}^3]=h$ and $c_2(\mathcal{S})\cap [\mathbb{Q}^3]=l$. We have $h^q(\mathcal{S})=0$ for any q > 0 and $h^q(\mathcal{S}(-i))=0$ for all q if $i=1,2$ or 3.

Lemma 3.1. The natural map

\begin{equation*} H^0(\mathcal{S})\otimes H^0(\mathcal{S})\to H^0(\mathcal{O}(1)) \end{equation*}

sending $s\otimes t$ to $s\wedge t$ is surjective.

4. Hirzebruch–Riemann–Roch formulas

Let $\mathcal{E}$ be a vector bundle of rank r on $\mathbb{Q}^3$. Since the tangent bundle T of $\mathbb{Q}^3$ fits in an exact sequence

\begin{equation*}0\to T\to T_{\mathbb{P}^4}|_{\mathbb{Q}^3} \to \mathcal{O}_{\mathbb{Q}^3}(2)\to 0,\end{equation*}

the Chern polynomial $c_t(T)$ of T is

\begin{equation*}\dfrac{(1+ht)^5}{1+2ht}=1+3ht+4h^2t^2+2h^3t^3,\end{equation*}

where h denotes $c_1(\mathcal{O}_{\mathbb{Q}^3}(1))$. Then the Hirzebruch–Riemann–Roch formula implies that

\begin{equation*} \chi(\mathcal{E})=r+\dfrac{13}{12}c_1h^2 +\dfrac{3}{4}(c_1^2-2c_2)h+ \dfrac{1}{6}(c_1^3-3c_1c_2+3c_3), \end{equation*}

where we set $c_i=c_i(\mathcal{E})$. To compute $\chi(\mathcal{E}(t))$, note that

\begin{align*} c_1(\mathcal{E}(t))&=c_1+rth;\\ c_2(\mathcal{E}(t))& =c_2+(r-1)tc_1h+ \binom{r}{2} t^2h^2;\\ c_3(\mathcal{E}(t))&=c_3 +(r-2)tc_2h+ \binom{r-1}{2}t^2c_1 h^2+\binom{r}{3}t^3h^3. \end{align*}

Since $h^3=2$, we infer that

(4.1)\begin{equation} \begin{split} \chi(\mathcal{E}(t))=&\dfrac{r}{3}t^3+ \dfrac{1}{2}(c_1h^2+3r)t^2 +\dfrac{1}{2}\{3c_1h^2 +(c_1^2-2c_2)h+\dfrac{13}{3}r\}t\\ &+r+\dfrac{13}{12}c_1h^2 +\dfrac{3}{4}(c_1^2-2c_2)h +\dfrac{1}{6}(c_1^3-3c_1c_2+3c_3). \end{split} \end{equation}

Since $c_1(\mathcal{E})=dh$ for some integer d, the formula above can be written as

(4.2)\begin{equation} \begin{split} \chi(\mathcal{E}(t))=& \dfrac{r}{6}(2t+3)(t+2)(t+1)+dt^2+(d^2+3d)t-c_2ht \\ &+\dfrac{d}{6}(2d^2+9d+13) +\dfrac{1}{2}\{c_3-(d+3)c_2h\}. \end{split} \end{equation}

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

(4.3)\begin{equation} \chi(\mathcal{E}(t))= \dfrac{r}{6}(2t+3)(t+2)(t+1)+2t^2+10t+13-c_2ht +\dfrac{1}{2}\{c_3-5c_2h\}. \end{equation}

In particular,

(4.4)\begin{align} \chi(\mathcal{E}(-1))&= 5-\dfrac{3}{2}c_2h +\dfrac{1}{2}c_3; \end{align}

(4.5)\begin{align} \chi(\mathcal{E}(-2))&= 1-\dfrac{1}{2}c_2h +\dfrac{1}{2}c_3. \end{align}

Next we will compute $\chi(\mathcal{S}^{\vee}\otimes \mathcal{E}(t))$. Recall that $c_1(\mathcal{S})=h$ and that $c_1(\mathcal{S})c_2(\mathcal{S})=1$. Note also that

\begin{equation*} \begin{array}{rcl} \operatorname{rank} \mathcal{S}^{\vee}\otimes \mathcal{E}&=& 2r;\\ c_1(\mathcal{S}^{\vee}\otimes \mathcal{E})&=& 2c_1-rh;\\ c_2(\mathcal{S}^{\vee}\otimes \mathcal{E})&=& 2c_2-(2r-1)c_1h+c_1^2+\binom{r}{2}h^2+rc_2(\mathcal{S});\\ c_3(\mathcal{S}^{\vee}\otimes\mathcal{E})&=& 2c_3-2(r-1)c_2h+(r-1)^2c_1h^2+2(r-1)c_1c_2(\mathcal{S})\\ &&+2c_1c_2-(r-1)c_1^2h-\dfrac{1}{3}r(r^2-1). \end{array} \end{equation*}

The formula (4.1) together with the formulas above implies the following formula:

\begin{equation*} \begin{array}{rcl} \chi(\mathcal{S}^{\vee}\otimes \mathcal{E}(t))&=&\dfrac{2}{3}rt^3+ (c_1h^2+2r)t^2 +\{2c_1h^2+(c_1^2-2c_2)h+\dfrac{4}{3}r\}t\\ &&+\dfrac{7}{6}c_1h^2+c_1^2h-2c_2h +\dfrac{1}{3}c_1^3 +c_3-c_1c_2-c_1c_2(\mathcal{S}). \end{array} \end{equation*}

Since $c_1=dh$, the formula above becomes the following formula:

(4.6)\begin{equation} \begin{array}{rcl} \chi(\mathcal{S}^{\vee}\otimes \mathcal{E}(t))&=& \dfrac{2}{3}rt(t+1)(t+2)+2dt^2+2d(d+2)t\\ &&+\dfrac{2}{3}d(d+1)(d+2)-(2t+d+2)c_2h+c_3. \end{array} \end{equation}

For the case d = 2, we have

(4.7)\begin{equation} \chi(\mathcal{S}^{\vee}\otimes \mathcal{E}(t))= \dfrac{2}{3}rt(t+1)(t+2)+4(t+2)^2 -2(t+2)c_2h+c_3. \end{equation}

In particular,

(4.8)\begin{equation} \chi(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1))= 4-2c_2h+c_3. \end{equation}

5. Key lemmas

Lemma 5.1. We have the following exact sequence on $\mathbb{Q}^3$:

(5.1)\begin{equation} 0\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3} \to \mathcal{S}^{\vee}\otimes \operatorname{Hom}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3},\mathcal{S}^{\vee})^{\vee} \to \Omega_{\mathbb{P}^4}(1)|_{\mathbb{Q}^3}\to 0, \end{equation}

where the injection $T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3} \to \mathcal{S}^{\vee}\otimes \operatorname{Hom}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3},\mathcal{S}^{\vee})^{\vee}$ is the coevaluation morphism. Moreover, $\dim \operatorname{Hom}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3},\mathcal{S}^{\vee})^{\vee}=4$.

Proof. The following simplified proof is due to the referee. As we have seen in Remark 2.2,

(5.2)\begin{equation} (\mathcal{O}(-1),T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}, \mathcal{S}^{\vee},\mathcal{O}) \end{equation}

is a full strong exceptional collection of $D^b(\mathbb{Q}^3)$. Since this is strong, the right mutation $\mathbf{R}_{\mathcal{S}^{\vee}}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3})$ of $T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}$ over $\mathcal{S}^{\vee}$ fits in the following distinguished triangle:

\begin{equation*} T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3} \to \mathcal{S}^{\vee}\otimes \operatorname{Hom}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3},\mathcal{S}^{\vee})^{\vee} \to \mathbf{R}_{\mathcal{S}^{\vee}}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3})\to . \end{equation*}

Now consider the mutated full exceptional collection

(5.3)\begin{equation} (\mathcal{O}(-1), \mathcal{S}^{\vee}, \mathbf{R}_{\mathcal{S}^{\vee}}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}), \mathcal{O}). \end{equation}

Note here that

(5.4)\begin{equation} \operatorname{Ext}^q(\mathcal{S}^{\vee},\mathbf{R}_{\mathcal{S}^{\vee}}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}))=0 \,\textrm{for }q\neq 0. \end{equation}

Indeed, by taking $\operatorname{RHom}(\mathcal{S}^{\vee},\bullet)$ with the triangle above, we see that $\operatorname{Hom}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3},\mathcal{S}^{\vee})^{\vee}$ is isomorphic to $\operatorname{RHom}(\mathcal{S}^{\vee}, \mathbf{R}_{\mathcal{S}^{\vee}}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}))$. On the other hand, by dualizing the collection (5.2) (and reversing the order) and then twisting it by $\mathcal{O}(-1)$ gives the following full strong exceptional collection:

(5.5)\begin{equation} (\mathcal{O}(-1), \mathcal{S}^{\vee}, \Omega_{\mathbb{P}^4}(1)|_{\mathbb{Q}^3}, \mathcal{O}). \end{equation}

Comparing two full exceptional collections (5.3) and (5.5), we infer that

\begin{equation*}\langle\mathbf{R}_{\mathcal{S}^{\vee}}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3})\rangle ={}^{\perp}\langle \mathcal{O}(-1), \mathcal{S}^{\vee}\rangle\cap \langle\mathcal{O}\rangle^{\perp}= \langle\Omega_{\mathbb{P}^4}(1)|_{\mathbb{Q}^3}\rangle.\end{equation*}

Thus, we have $\mathbf{R}_{\mathcal{S}^{\vee}}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}) \cong \Omega_{\mathbb{P}^4}(1)|_{\mathbb{Q}^3}[d]$ for some integer d, but the vanishing (5.4) implies that d = 0, namely

\begin{equation*}\mathbf{R}_{\mathcal{S}^{\vee}}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}) \cong \Omega_{\mathbb{P}^4}(1)|_{\mathbb{Q}^3}.\end{equation*}

Hence we obtain the desired exact sequence (5.1). If follows immediately from the exact sequence (5.1) that $\dim \operatorname{Hom}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3},\mathcal{S}^{\vee})^{\vee}=4$.

Proof. We have an exact sequence

\begin{equation*}0\to \Omega_{\mathbb{P}^4}(1)|_{\mathbb{Q}^3} \xrightarrow{i} H^0(\mathcal{O}(1))\otimes \mathcal{O}\to \mathcal{O}(1)\to 0,\end{equation*}

and the composite $i \circ\varphi$ can be written as

\begin{equation*}i \circ\varphi =\sum_{i=1}^r l_i\otimes s_i^{\vee}\end{equation*}

for some $l_i\in H^0(\mathcal{O}(1))$ and $s_i\in H^0(\mathcal{S})$, where $s_i^{\vee}$ denotes the dual of the morphism $\mathcal{O}\to \mathcal{S}$ determined by s_i. We may assume that $l_i\neq 0$ for all i. By replacing l_i if necessary, we may further assume that $s_1,\dots,s_r$ are linearly independent. Since $h^0(\mathcal{S})=4$, we have $r\leq 4$. Note that $\sum_{i=1}^r l_is_i^{\vee}=0$ in $\operatorname{Hom} (\mathcal{S}^{\vee},\mathcal{O}(1))$. Hence $r\geq 2$. Moreover, we have a surjective morphism

\begin{equation*}\psi:\operatorname{Coker}(i\circ \varphi)\to \mathcal{O}(1).\end{equation*}

Note that the morphism $\mathcal{O}^{\oplus r}\to \mathcal{S}$ determined by $(s_1,\dots,s_r)$ is generically surjective. Hence we see that $i\circ \varphi$ is injective. Therefore, φ is injective and

\begin{equation*}\operatorname{Coker}(\varphi)\cong \operatorname{Ker} (\psi).\end{equation*}

If r = 2, then $\operatorname{Coker}(i\circ \varphi)\cong \mathcal{T}\oplus \mathcal{O}^{\oplus 3}$ for some torsion sheaf $\mathcal{T}$ on $\mathbb{Q}^3$. Since $\mathcal{O}(1)$ is torsion-free, ψ maps $\mathcal{T}$ to zero, and we have a surjective morphism $\bar{\psi}:\mathcal{O}^{\oplus 3}\to \mathcal{O}(1)$. On the other hand, $\bar{\psi}:\mathcal{O}^{\oplus 3}\to \mathcal{O}(1)$ cannot be surjecitve since three hyperplane sections of $\mathbb{Q}^3$ always meet at a point. This is a contradiction. Hence r = 3 or 4. Suppose that r = 4. Then it follows from the exact sequence (3.1) that $\operatorname{Coker}(i\circ \varphi)\cong \mathcal{S}\oplus \mathcal{O}$. Note that ψ induces a morphism $\mathcal{S}\to \mathcal{O}(1)$, which factors through $\mathcal{I}_L(1)$ for some line L in $\mathbb{Q}^3$. Since L and a hyperplane in $\mathbb{Q}^3$ meet at a point, ψ cannot be surjective. Hence the case r = 4 does not arise, and we have r = 3.

Now it follows from the exact sequence (3.1) that the cokernel of the morphism determined by ${}^t(s_1^{\vee},s_2^{\vee},s_3^{\vee}): \mathcal{S}^{\vee}\to \mathcal{O}^{\oplus 3}$ is isomorphic to the cokernel of some non-zero morphism $\mathcal{O}\to \mathcal{S}$, and hence it is isomorphic to $\mathcal{I}_M(1)$ for some line M on $\mathbb{Q}^3$. Therefore, $\operatorname{Coker}(i\circ \varphi)\cong \mathcal{I}_M(1)\oplus \mathcal{O}^{\oplus 2}$, and we have the following exact sequence:

(5.6)\begin{equation}0\rightarrow\operatorname{Coker}(\varphi)\to {\mathcal I}_M(1)\oplus\mathcal O^{\oplus2}\xrightarrow\psi\mathcal O(1)\rightarrow0.\end{equation}

Let $\mathbb{Q}^2$ be a general hyperplane section of $\mathbb{Q}^3$ containing M. We may assume that M is a divisor of type $(1,0)$ of $\mathbb{Q}^2$. Then $\mathcal{I}_M(1)$ fits in the following exact sequence:

\begin{equation*}0\to \mathcal{O}_{\mathbb{Q}^3}\to \mathcal{I}_M(1)\to \mathcal{O}_{\mathbb{Q}^2}(0,1)\to 0.\end{equation*}

By pulling back the sequence above to a line L of type $(0,1)$ in $\mathbb{Q}^2$, we obtain the following exact sequence:

\begin{equation*} \mathcal{O}_L\to \mathcal{I}_M(1)\otimes \mathcal{O}_L\to \mathcal{O}_L\to 0. \end{equation*}

The image of $\mathcal{O}_L\to \mathcal{I}_M(1)\otimes \mathcal{O}_L$ is the torsion part of $\mathcal{I}_M(1)\otimes\mathcal{O}_L$. Therefore, $\psi\otimes 1_L$ factors through $\mathcal{O}_L^{\oplus 3}$ and induces a surjection $\mathcal{O}_L^{\oplus 3}\to \mathcal{O}_L(1)$. Hence $\operatorname{Coker}(\varphi)\otimes\mathcal{O}_L$ has $\mathcal{O}_L(-1)\oplus \mathcal{O}_L$ as a quotient.

Lemma 5.3 will be applied to ψ_a in (12.4) and (12.7) and plays a crucial role in our proof of Theorem 1.1.

Lemma 5.3. For any positive integer a and for any morphism $\psi_a:T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3} \to \mathcal{S}^{\vee \oplus a}$, there exists a line L in $\mathbb{Q}^3$ such that the cokernel $\operatorname{Coker}(\psi_a)$ of ψ_a has $\mathcal{O}_L(-1)$ as a quotient. In case a = 1, there is a one-to-one correspondence between lines L in $\mathbb{Q}^3$ and non-zero morphisms $\psi_1:T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}\to \mathcal{S}^{\vee}$ up to scalar, and the correspondence is given by the following exact sequence:

(5.7)\begin{equation} 0\to \mathcal{O}(-1)^{\oplus 2}\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}\xrightarrow{\psi_1} \mathcal{S}^{\vee} \to \mathcal{O}_L(-1)\to 0. \end{equation}

Proof. The following brilliant proof is due to the referee. This proof is much shorter than the original and enlightens the meaning of the exact sequence (5.7) more clearly.

Denote by $\operatorname{Quot}(\mathcal{S}^{\vee})$ the Quot-scheme parametrizing quotient sheaves of $\mathcal{S}^{\vee}$. Then we have a morphism

\begin{equation*}\Psi:\mathbb{P}(\operatorname{Hom}( T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}, \mathcal{S}^{\vee})^{\vee}) \to \operatorname{Quot}(\mathcal{S}^{\vee})\end{equation*}

sending $[\psi_1]$ to $\operatorname{Coker}(\psi_1)$. Note that for any line $L\subset\mathbb{Q}^3$ we have $\mathcal{S}^{\vee}|_L\cong \mathcal{O}_L(-1)\oplus \mathcal{O}_L$ so that $\mathcal{S}^{\vee}$ admits $\mathcal{O}_L(-1)$ as a quotient. Note also that the Hilbert polynomial $\chi(\mathcal{O}_L(t-1))$ of $\mathcal{O}_L(-1)$ is t. Let Z be the Hilbert scheme parametrizing lines in $\mathbb{Q}^3$. Then we have an inclusion

\begin{equation*}Z\hookrightarrow \operatorname{Quot}^t (\mathcal{S}^{\vee})\end{equation*}

sending $[L]$ to $\mathcal{O}_L(-1)$, where $\operatorname{Quot}^t (\mathcal{S}^{\vee})$ is the Quot-scheme parametrizing quotients of $\mathcal{S}^{\vee}$ with Hilbert polynomial t. It is well-known that $Z\cong \mathbb{P}^3$. Note also that

\begin{equation*} \mathbb{P}(\operatorname{Hom}( T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}, \mathcal{S}^{\vee})^{\vee}) \cong \mathbb{P}^3 \end{equation*}

by Lemma 5.1. We will show that Ψ is an isomorphism onto Z.

We first claim that the image $\mathop{\rm Im}\nolimits \Psi$ of Ψ is Z. To see this, we first apply to $\mathcal{O}_L(-1)$ for any line $L\subset\mathbb{Q}^3$ the Bondal spectral sequence (2.1). We have the following:

\begin{equation*} \operatorname{ext}^q(\mathcal{O},\mathcal{O}_L(-1)) =0 \,\textrm{for any }q;\end{equation*}

(5.8)\begin{equation} \operatorname{ext}^q(\mathcal{S},\mathcal{O}_L(-1)) =h^q(\mathcal{O}_L(-2)\oplus\mathcal{O}_L(-1)) = \begin{cases} 1&\textrm{if }q=1\\ 0&\textrm{if }q\neq 1 \end{cases}; \end{equation}

\begin{equation*} \operatorname{ext}^q(\mathcal{O}(1),\mathcal{O}_L(-1)) =h^q(\mathcal{O}_L(-2)) = \begin{cases} 1&\textrm{if }q=1\\ 0&\textrm{if }q\neq 1 \end{cases};\end{equation*}

\begin{equation*} \operatorname{ext}^q(\mathcal{O}(2),\mathcal{O}_L(-1)) =h^q(\mathcal{O}_L(-3)) = \begin{cases} 2&\textrm{if }q=1\\ 0&\textrm{if }q\neq 1 \end{cases}.\end{equation*}

Thus, $\operatorname{Ext}^3(G,\mathcal{E})=0$, $\operatorname{Ext}^2(G,\mathcal{E})=0$, $\operatorname{Hom}(G,\mathcal{E})=0$, and $\operatorname{Ext}^1(G,\mathcal{E})$ has a filtration $S_1\subset F\subset \operatorname{Ext}^1(G,\mathcal{E})$ of right A-modules such that the following sequences are exact:

\begin{equation*} 0\to F\to \operatorname{Ext}^1(G,\mathcal{E})\to S_3^{\oplus 2}\to 0;\end{equation*}

\begin{equation*} 0\to S_1\to F\to S_2\to 0.\end{equation*}

These exact sequences induce the following distinguished triangles by Lemma 2.1:

\begin{equation*}F\operatorname{\otimes^{L}}_AG\to \operatorname{Ext}^1(G,\mathcal{E})\operatorname{\otimes^{L}}_AG\to \mathcal{O}(-1)^{\oplus 2}[3]\to;\end{equation*}

\begin{equation*}\mathcal{S}^{\vee}[1]\to F\operatorname{\otimes^{L}}_AG\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}[2]\to.\end{equation*}

By taking cohomologies, we obtain the following exact sequences:

\begin{equation*}0\to E_2^{-3,1}\to \mathcal{O}(-1)\to \mathcal{H}^{-2}(F\operatorname{\otimes^{L}}_AG) \to E_2^{-2,1}\to 0;\end{equation*}

\begin{equation*} 0\to \mathcal{H}^{-2}(F\operatorname{\otimes^{L}}_AG)\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3} \xrightarrow{\psi_L} \mathcal{S}^{\vee}\to E_2^{-1,1}\to 0. \end{equation*}

Moreover, we see that $E_2^{p,q}=0$ unless q = 1 and that $E_2^{p,1}=0$ unless $p=-3, -2$ or −1. Hence we infer that $E_2^{-3,1}=0$, that $E_2^{-2,1}=0$ and that $E_2^{-1,1}\cong \mathcal{O}_L(-1)$. Therefore, $\mathcal{O}_L(-1)$ is resolved as

(5.9)\begin{equation} 0\to \mathcal{O}(-1)^{\oplus 2}\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}\xrightarrow{\psi_L} \mathcal{S}^{\vee} \to \mathcal{O}_L(-1)\to 0 \end{equation}

in terms of the full strong exceptional collection (2.6). This implies that the image $\mathop{\rm Im}\nolimits \Psi$ of Ψ contains Z. Since the source of Ψ has the same dimension as Z, we conclude that $\mathop{\rm Im}\nolimits \Psi=Z$.

Next we show that Ψ is injective. Note that the exact sequence (5.9) splits into the following two exact sequences:

(5.10)\begin{equation} 0\to \mathcal{K}\to \mathcal{S}^{\vee} \to \mathcal{O}_L(-1)\to 0; \end{equation}

(5.11)\begin{equation} 0\to \mathcal{O}(-1)^{\oplus 2}\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3} \to \mathcal{K}\to 0. \end{equation}

Since we have (5.8), the exact sequence (5.10) shows that $\mathcal{K}$ is the left mutation of $\mathcal{O}_L(-1)$ over $\mathcal{S}^{\vee}$. Moreover it follows from (5.11) that $\mathcal{O}(-1)^{\oplus 2}$ is the left mutation of $\mathcal{K}$ over $T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}$, since

\begin{equation*}K\cong \operatorname{RHom}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3},T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}) \cong \operatorname{RHom}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3},K).\end{equation*}

Therefore, ψ_L in (5.9) is uniquely determined by L up to scalar. Hence Ψ is injective.

Finally, if the composite of the morphism ψ_a and some projection $\mathcal{S}^{\vee\oplus a}\to \mathcal{S}^{\vee}$ is zero, then $\operatorname{Coker}(\psi_a)$ admits $\mathcal{S}^{\vee}$ as a quotient, and the assertion follows. Hence we may assume that the composite cannot be zero for any projection $\mathcal{S}^{\vee\oplus a}\to \mathcal{S}^{\vee}$. Then the cokernel of the composite has $\mathcal{O}_L(-1)$ as a quotient, and so does $\operatorname{Coker}(\psi_a)$.

Since the analyses of $\operatorname{Coker}(\psi_a)$ in case $a\geq 2$ in the original proof of Lemma 5.3 are indispensable for the proof of Lemma 5.4, we also provide that part of the proof as it is. Recall here that, for a coherent sheaf $\mathcal{F}$ of codimension $\geq p+1$ on a non-singular projective variety X, we have $c_i(\mathcal{F})=0$ for all $1\leq i\leq p$ (see, e.g., [Reference Fulton6, Example 15.3.6]).

Proof. The original proof of Lemma 5.3 in case $a\geq 2$

If the composite of the morphism ψ_a and some projection $\mathcal{S}^{\vee\oplus a}\to \mathcal{S}^{\vee}$ is zero, then $\operatorname{Coker}(\psi_a)$ admits $\mathcal{S}^{\vee}$ as a quotient, and the assertion follows. Hence we may assume that the composite cannot be zero for any projection $\mathcal{S}^{\vee\oplus a}\to \mathcal{S}^{\vee}$, and this implies that $a\leq 4$ by Lemma 5.1.

If a = 4, then Lemma 5.1 shows that $\operatorname{Coker}(\psi_4)\cong \Omega_{\mathbb{P}^4}(1)|_{\mathbb{Q}^3}$, and the assertion follows.

If a = 3, then ψ ₃ can be regarded as the composite of the coevaluation morphism

\begin{equation*}\psi_4:T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3} \to \mathcal{S}^{\vee}\otimes \operatorname{Hom}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3},\mathcal{S}^{\vee})^{\vee}\end{equation*}

and some projection $\mathcal{S}^{\vee}\otimes \operatorname{Hom}(T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3},\mathcal{S}^{\vee})^{\vee} \to \mathcal{S}^{\vee\oplus 3}$. Let $\mathcal{S}^{\vee}\to \mathcal{S}^{\vee\oplus 4}$ be the kernel of this projection, and let φ be the composite of the inclusion $\mathcal{S}^{\vee}\to \mathcal{S}^{\vee\oplus 4}$ and the surjection $ \mathcal{S}^{\vee\oplus 4}\to \Omega_{\mathbb{P}^4}(1)|_{\mathbb{Q}^3}$ in (5.1). Then

(5.12)\begin{equation} \operatorname{Coker}(\psi_3)\cong \operatorname{Coker}(\varphi) \end{equation}

and $\operatorname{Ker}(\psi_3)\cong \operatorname{Ker}(\varphi)$ by the snake lemma. Since $\operatorname{Hom}(\mathcal{S}^{\vee},T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3})=0$, φ cannot be zero by (5.1). Lemma 5.2 then shows that φ is injective and that the restriction $\operatorname{Coker}(\varphi)|_L$ to some line L on $\mathbb{Q}^3$ admits a negative degree quotient. Hence the assertion holds, and ψ ₃ is injective.

Suppose that a = 2. Then we can regard ψ ₂ as the composite of some $\psi_3:T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3} \to \mathcal{S}^{\vee\oplus 3}$ and some projection $\mathcal{S}^{\vee\oplus 3} \to \mathcal{S}^{\vee\oplus 2}$. Let $\mathcal{S}^{\vee}\to \mathcal{S}^{\vee\oplus 3}$ be the kernel of this projection. Note here that we have an exact sequence

\begin{equation*} 0\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3} \xrightarrow{\psi_3} \mathcal{S}^{\vee\oplus 3} \to \operatorname{Coker}(\varphi)\to 0. \end{equation*}

Denote by $\varphi_1:\mathcal{S}^{\vee}\to \operatorname{Coker}(\varphi)$ the composite of the inclusion $\mathcal{S}^{\vee}\to \mathcal{S}^{\vee\oplus 3}$ and the surjection $\mathcal{S}^{\vee\oplus 3} \to \operatorname{Coker}(\varphi)$. Then φ ₁ cannot be zero, since $\operatorname{Hom}(\mathcal{S}^{\vee},T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3})=0$. Moreover, the snake lemma implies that

\begin{equation*}\operatorname{Coker}(\psi_2)\cong \operatorname{Coker}(\varphi_1)\quad \textrm{and that }\quad \operatorname{Ker}(\psi_2)\cong \operatorname{Ker}(\varphi_1).\end{equation*}

Recall the inclusion $i:\operatorname{Coker}(\varphi)\hookrightarrow \mathcal{I}_M(1)\oplus\mathcal{O}^{\oplus 2}$ in (5.6) and consider the composite $i\circ\varphi_1$. We have the following exact sequence:

(5.13)\begin{equation} 0\to \operatorname{Coker}(\varphi_1)\to \operatorname{Coker}(i\circ \varphi_1)\to \mathcal{O}(1)\to 0. \end{equation}

Let $i\circ \varphi_1$ be equal to $(t^{\vee}, s_1^\vee, s_2^{\vee})$, where $t^{\vee}\in \operatorname{Hom}(\mathcal{S}^{\vee},\mathcal{I}_M(1))$, $t\in H^0(\mathcal{S}(1))$, $s_1^{\vee}$, $s_2^{\vee} \in \operatorname{Hom}(\mathcal{S}^{\vee},\mathcal{O})$ and s ₁, $s_2\in H^0(\mathcal{S})$. Since we have an exact sequence (5.6), we have $t^{\vee}+h_1s_1^{\vee}+h_2s_2^{\vee}=0$ for some h ₁, $h_2\in H^0(\mathcal{O}(1))$. Now we have two cases:

1. (1) s ₁ and s ₂ are linearly independent;

2. (2) s ₁ and s ₂ are linearly dependent.

(1) If s ₁ and s ₂ are linearly independent, then φ ₁ is injective, and $\operatorname{Coker}(i\circ \varphi_1)$ has rank one. Thus we see that $\operatorname{Coker}(\varphi_1)$ is a torsion sheaf. Moreover, we claim that $\operatorname{Coker}(\varphi_1)$ is pure by [Reference Huybrechts and Lehn8, Prop. 1.1.6]: first note that $\operatorname{{\mathcal Ext}}^q_{\mathbb{Q}^3}(\operatorname{Coker}(\varphi),\omega_{\mathbb{Q}^3})=0$ for all $q\geq2$; thus $\operatorname{{\mathcal Ext}}^q_{\mathbb{Q}^3}(\operatorname{Coker}(\varphi_1),\omega_{\mathbb{Q}^3})=0$ for all $q\geq2$, and hence $\operatorname{Coker}(\varphi_1)$ satisfies the generalized Serre’s condition $S_{1,1}$ in [Reference Huybrechts and Lehn8, Section 1.1]. Now we compute the Chern polynomial of $\operatorname{Coker}(\varphi_1)$. First note that $c_t(\operatorname{Coker}(\varphi))=c_t(\Omega_{\mathbb{P}^4}(1)|_{\mathbb{Q}^3})/c_t(\mathcal{S}^{\vee}) =1+lt^2-t^3$. Hence

\begin{equation*}c_t(\operatorname{Coker}(\varphi_1))=c_t(\operatorname{Coker}(\varphi))/c_t(\mathcal{S}^{\vee}) =1+ht+2lt^2.\end{equation*}

Since $\operatorname{Coker}(\varphi_1)$ is a torsion sheaf, this implies that $\operatorname{Coker}(\varphi_1)$ is supported on a hyperplane section H of $\mathbb{Q}^3$, and the length of $\operatorname{Coker}(\varphi_1)$ at the generic point of H is one. Since $\operatorname{Coker}(\varphi_1)$ is pure, this implies that $\operatorname{Coker}(\varphi_1)$ is of the form $\mathcal{I}_{Z,H}(D)$, where D is a divisor on H and $\mathcal{I}_{Z,H}$ denotes the ideal sheaf of some zero-dimensional closed subscheme Z in H. Note here that $c_t(\mathcal{O}_{H})=1+ht+2lt^2+2t^3$, that $c_t(\mathcal{O}_L)=(c_t(\mathcal{S}^{\vee})/c_t(\mathcal{O}(-1)))^{-1} =1-lt^2-t^3$ and that $c_t(k(p))=1+2t^3$, where k(p) is the residue field at a point p (see also [Reference Fulton6, Example 15.3.1] for the formula $c_t(k(p))=1+2t^3$). Hence we see that $[D]=0\cdot l$ in $A^2\mathbb{Q}^3$. Moreover, if D is of type $(d,-d)$, then $c_t(\mathcal{I}_{Z,H}(D))=1+ht+2lt^2+(2-2d^2-2\operatorname{length} Z)t^3$. Hence $(d,\operatorname{length} Z)=(0,1)$ or $(\pm 1,0)$. Therefore, $\operatorname{Coker}(\varphi_1)$ is isomorphic to either $\mathcal{I}_{p, H}$ or $\mathcal{O}_H(d,-d)$ where $d=\pm 1$. Thus the assertion holds.

(2) If s ₁ and s ₂ are linearly dependent, by replacing s_i and h_i if necessary, we may assume that $s_2=0$, and we have $t^{\vee}+h_1s_1^{\vee}=0$. Set $\varphi_1':=(t^{\vee},s_1^{\vee}): \mathcal{S}^{\vee}\to \mathcal{I}_M(1)\oplus \mathcal{O}_{\mathbb{Q}^3}$. Then $\operatorname{Coker}(i\circ\varphi_1)\cong \operatorname{Coker}(\varphi_1')\oplus \mathcal{O}_{\mathbb{Q}^3}$ and $\operatorname{Ker}(\varphi_1)\cong \operatorname{Ker}(\varphi_1')$. Note that $\varphi_1'\neq 0$ since $\varphi_1\neq 0$. Hence $s_1\neq 0$. Let L be the zero locus $(s_1)_0$ of s ₁. Then the composite of $\varphi_1'$ and the inclusion $\mathcal{I}_M(1)\oplus \mathcal{O}_{\mathbb{Q}^3}\to \mathcal{O}(1)\oplus \mathcal{O}_{\mathbb{Q}^3}$ factors through the morphism $(-h_1,1):\mathcal{O}\to \mathcal{O}(1)\oplus \mathcal{O}_{\mathbb{Q}^3}$, and we have the following commutative diagram with exact rows:

We see that $\mathop{\rm Im}\nolimits (\varphi_1')\cong \mathcal{I}_L$ and that $\operatorname{Ker}(\varphi_1')\cong \mathcal{O}(-1)$. We claim here that $\bar{h}_1\neq 0$. Assume, to the contrary, that $\bar{h}_1=0$. Then the snake lemma implies that $\operatorname{Coker}(\varphi_1')$ fits in the following exact sequence:

\begin{equation*} 0\to \mathcal{O}_L\to \operatorname{Coker}(\varphi_1')\to \mathcal{O}(1)\to \mathcal{O}_M(1)\to 0. \end{equation*}

Since $\mathcal{O}_L$ is a torsion sheaf, the surjection $\operatorname{Coker}(\varphi_1')\oplus \mathcal{O}_{\mathbb{Q}^3}\to \mathcal{O}(1)$ induces a surjection $\mathcal{I}_M(1)\oplus \mathcal{O}_{\mathbb{Q}^3}\to \mathcal{O}(1)$. On the other hand, the morphism $\mathcal{I}_M(1)\oplus \mathcal{O}_{\mathbb{Q}^3}\to \mathcal{O}(1)$ cannot be surjective since a line M and a hyperplane meets at least at one point. This is a contradiction. Hence $\bar{h}_1\neq 0$, and thus L = M. Moreover, the commutative diagram (5.14) induces the following exact sequence by the snake lemma:

\begin{equation*}0\to \operatorname{Coker}(\varphi_1')\to \mathcal{O}(1)\to k(p)\to 0,\end{equation*}

where $p=(\bar{h}_1)_0$. Therefore, $\operatorname{Coker}(\varphi_1')=\mathcal{I}_p(1)$. The exact sequence (5.13), i.e. the sequence

\begin{equation*}0\to \operatorname{Coker}(\varphi_1)\to \mathcal{I}_p(1)\oplus \mathcal{O}_{\mathbb{Q}^3} \to \mathcal{O}(1)\to 0\end{equation*}

then shows that $\operatorname{Coker}(\varphi_1)=\mathcal{I}_p$. Thus the assertion also holds if s ₁ and s ₂ are linearly dependent.

Lemma 5.4 will be applied to π in (12.8) and plays a crucial role in the proof of Theorem 1.1.

Proof. We may assume that π ≠ 0.

Suppose that $\operatorname{Coker}(\psi_a)$ admits $\mathcal{S}^{\vee}$ as a quotient; let $p:\operatorname{Coker}(\psi_a)\to \mathcal{S}^{\vee}$ be the surjection. Note that $\operatorname{Coker}(\pi)$ admits $\operatorname{Coker}(p\circ\pi)$ as a quotient. If $p\circ \pi=0$, then $\operatorname{Coker}(p\circ \pi)\cong \mathcal{S}^{\vee}$, and if $p\circ \pi\neq 0$, then $\operatorname{Coker}(p\circ \pi)\cong \mathcal{I}_L$ for some line L in $\mathbb{Q}^3$. Therefore, the restriction of $\operatorname{Coker}(\pi)$ to a line admits a negative degree quotient.

In the following, we assume that $\operatorname{Coker}(\psi_a)$ does not admit $\mathcal{S}^{\vee}$ as a quotient. Hence $a\leq 4$ by Lemma 5.1.

Suppose that a = 4. Then $\operatorname{Coker}(\psi_4)\cong \Omega_{\mathbb{P}^4}(1)|_{\mathbb{Q}^3}$ by Lemma 5.1. Since $\Omega_{\mathbb{P}^4}(1)|_{L}\cong \mathcal{O}_L(-1)\oplus\mathcal{O}_L^{\oplus 3}$ for any line L in $\mathbb{Q}^3$, if $\operatorname{Coker}(\pi)|_L$ does not admit a negative degree quotient for any line L in $\mathbb{Q}^3$, we see that $\operatorname{Coker}(\pi)|_L\cong \mathcal{O}_L^{\oplus 3}$ for any line L in $\mathbb{Q}^3$. This implies that $\operatorname{Coker}(\pi)\cong \mathcal{O}_{\mathbb{Q}^3}^{\oplus 3}$ by [Reference Wiśniewski18, (3.6.1) Lemma]. Thus $\Omega_{\mathbb{P}^4}(1)|_{\mathbb{Q}^3} \cong \mathcal{O}(-1)\oplus \mathcal{O}^{\oplus 3}$, which contradicts $H^0(\Omega_{\mathbb{P}^4}(1)|_{\mathbb{Q}^3})=0$. Therefore, $\operatorname{Coker}(\pi)|_L$ admits a negative degree quotient for some line L in $\mathbb{Q}^3$.

Suppose that a = 3. Recall that $\operatorname{Coker}(\psi_3)\cong \operatorname{Coker}(\varphi)$ in (5.12). Recall also the inclusion $i:\operatorname{Coker}(\varphi)\hookrightarrow \mathcal{I}_M(1)\oplus\mathcal{O}^{\oplus 2}$ in (5.6) and consider the composite $i\circ\pi$. We have the following exact sequence:

(5.15)\begin{equation} 0\to \operatorname{Coker}(\pi)\to \operatorname{Coker}(i\circ \pi) \xrightarrow{\rho} \mathcal{O}(1)\to 0. \end{equation}

Let $i\circ \pi$ be equal to $(t, g_1, g_2)$, where $t\in \operatorname{Hom}(\mathcal{O}(-1),\mathcal{I}_M(1))\cong H^0(\mathcal{I}_M(2))$, g ₁, $g_2 \in \operatorname{Hom}(\mathcal{O}(-1),\mathcal{O})\cong H^0(\mathcal{O}(1))$. Since we have an exact sequence (5.6), we have $t+h_1g_1+h_2g_2=0$ for some h ₁, $h_2\in H^0(\mathcal{O}(1))$. Now we have two cases:

1. (1) g ₁ and g ₂ are linearly independent;

2. (2) g ₁ and g ₂ are linearly dependent.

(1) If g ₁ and g ₂ are linearly independent, then the cokernel of the morphism $(g_1,g_2):\mathcal{O}(-1)\to \mathcal{O}^{\oplus 2}$ is of the form $\mathcal{I}_C(1)$, where C is the conic defined by g ₁ and g ₂. Hence $\operatorname{Coker}(i\circ \pi)$ fits in the following exact sequence:

\begin{equation*}0\to \mathcal{I}_M(1)\to \operatorname{Coker}(i\circ \pi)\to \mathcal{I}_C(1)\to 0.\end{equation*}

Now consider the composite of the injection $\mathcal{I}_M\to \operatorname{Coker}(i\circ\pi)(-1)$ and the surjection $\rho (-1):\operatorname{Coker}(i\circ\pi)(-1)\to \mathcal{O}$. The composite is nothing but the inclusion $\mathcal{I}_M\hookrightarrow \mathcal{O}$ and its cokernel is $\mathcal{O}_M$. Thus the surjection $\rho (-1)$ induces a surjection $\bar{\rho}(-1):\mathcal{I}_C\to \mathcal{O}_M$. This implies that $C\cap M=\emptyset$. Moreover $\operatorname{Coker}(\pi)(-1)\cong \operatorname{Ker} (\bar{\rho}(-1)) \cong \mathcal{I}_{C\sqcup M}$. Hence $\operatorname{Coker}(\pi)\cong \mathcal{I}_{C\sqcup M}(1)$. Note that the conic C and the line M can be joined by a line L in $\mathbb{Q}^3$. Indeed, any hyperplane section H containing M intersects C at some point p, and the point p and M can be joined by a line L in H. Now we see that $\operatorname{Coker}(\pi)|_L$ admits a negative degree quotient.

(2) If g ₁ and g ₂ are linearly dependent, by replacing g_i and h_i if necessary, we may assume that $g_2=0$, and we have $t+h_1g_1=0$. Set $\pi_1':=(t,g_1): \mathcal{O}(-1)\to \mathcal{I}_M(1)\oplus \mathcal{O}_{\mathbb{Q}^3}$. Then $\operatorname{Coker}(i\circ\pi)\cong \operatorname{Coker}(\pi')\oplus \mathcal{O}_{\mathbb{Q}^3}$. Note that $\pi'\neq 0$ since π ≠ 0. Hence $g_1\neq 0$. Let H be the hyperplane defined by g ₁. Then we have the following commutative diagram with exact rows:

We claim here that $\bar{h}_1\neq 0$. Assume, to the contrary, that $\bar{h}_1=0$. Then the snake lemma shows that we have the following exact sequence:

\begin{equation*} 0\to \mathcal{O}_H\to \operatorname{Coker}(\pi')\to \mathcal{O}(1)\to \mathcal{O}_M(1)\to 0. \end{equation*}

Since $\mathcal{O}_H$ is a torsion sheaf, the surjection $\rho:\operatorname{Coker}(\pi')\oplus \mathcal{O}_{\mathbb{Q}^3}\to \mathcal{O}(1)$ sends $\mathcal{O}_H$ to zero, and thus ρ induces a surjection $\mathcal{I}_M(1)\oplus \mathcal{O}_{\mathbb{Q}^3}\to \mathcal{O}(1)$. On the other hand, the morphism $\mathcal{I}_M(1)\oplus \mathcal{O}_{\mathbb{Q}^3}\to \mathcal{O}(1)$ cannot be surjective since a line M and a hyperplane meets at least at one point. This is a contradiction. Hence $\bar{h}_1\neq 0$. Then the kernel of the morphism $-\bar{h}_1:\mathcal{O}_H\to \mathcal{O}_M(1)$ is $\mathcal{O}_H(-M)$ and the cokernel of $-\bar{h}_1$ is k(p) for some point $p\in M$. Hence the commutative diagram (5.16) induces the following exact sequence by the snake lemma:

\begin{equation*}0\to \mathcal{O}_H(-M)\to \operatorname{Coker}(\pi')\to \mathcal{O}(1)\to k(p)\to 0.\end{equation*}

Since $\mathcal{O}_H(-M)$ is a torsion sheaf, the surjection $\rho:\operatorname{Coker}(\pi')\oplus \mathcal{O}_{\mathbb{Q}^3}\to \mathcal{O}(1)$ sends $\mathcal{O}_H(-M)$ to zero, and thus the inclusion $\mathcal{O}_H(-M)\hookrightarrow \operatorname{Coker}(\pi')\oplus\mathcal{O}_{\mathbb{Q}^3}$ induces an inclusion $\mathcal{O}_H(-M)\hookrightarrow \operatorname{Coker}(\pi)$. The exact sequence (5.15) induces the following exact sequence:

\begin{equation*}0\to \operatorname{Coker}(\pi)/ \mathcal{O}_H(-M) \to \mathcal{I}_p(1)\oplus \mathcal{O}_{\mathbb{Q}^3} \to \mathcal{O}(1)\to 0.\end{equation*}

This shows that $\operatorname{Coker}(\pi)/\mathcal{O}_H(-M)=\mathcal{I}_p$.

Suppose that a = 2. As we have seen in the original proof of Lemma 5.3, $\operatorname{Coker}(\psi_2)$ is isomorphic to $\operatorname{Coker}(\varphi_1)$, and $\operatorname{Coker}(\varphi_1)$ is one of the following: $\mathcal{I}_{p,H}$; $\mathcal{O}_H(d,-d)$ where $d=\pm 1$; $\mathcal{I}_p$. If $\operatorname{Coker}(\varphi_1)=\mathcal{I}_{p,H}$, then $\operatorname{Coker}(\pi)$ admits $\mathcal{O}_C(-p)$ as a quotient, where C is a conic on H. If $\operatorname{Coker}(\varphi_1)=\mathcal{O}_{H}(d,-d)$ with $d=\pm 1$, then $\operatorname{Coker}(\pi)$ admits $\mathcal{O}_L(-1)$ as a quotient, where L is a line on H. If $\operatorname{Coker}(\varphi_1)=\mathcal{I}_{p}$, then $\operatorname{Coker}(\pi)$ admits $\mathcal{I}_{p,H}$ as a quotient. Hence the assertion follows if a = 2.

Suppose that a = 1. Then $\operatorname{Coker}(\psi_1)\cong \mathcal{O}_L(-1)$ by Lemma 5.3. Since π ≠ 0, the morphism $\pi:\mathcal{O}(-1)\to \mathcal{O}_L(-1) $ is surjective, and $\operatorname{Ker}(\pi)\cong \mathcal{I}_L(-1)$. This completes the proof.

6. A lower bound for the third Chern class

Note that

(6.1)\begin{equation} c_3\geq 2c_1c_2-c_1^3 \end{equation}

for a nef vector bundle $\mathcal{E}$ on a complete threefold X, since $H(\mathcal{E})^{r+2} =c_3-2c_1c_2+c_1^3\geq 0$ for a nef line bundle $H(\mathcal{E})$. If there exists an injection $\mathcal{L} \to \mathcal{E}$ from a line bundle $\mathcal{L}$, then we have a lower bound, which is better if $\mathcal{L}\cong \mathcal{O}(D)$ for some effective divisor D, as the following lemma shows:

Lemma 6.1. Let $\mathcal{E}$ be a nef vector bundle of rank r on a complete variety X of dimension three. Let $\mathcal{L}$ be a line bundle on X such that $H^0(\mathcal{E}\otimes \mathcal{L}^{-1})\neq 0$. Then we have the following inequality:

\begin{equation*}c_3\geq 2c_1c_2-c_1^3+ (c_1^2-c_2)c_1(\mathcal{L}).\end{equation*}

Lemma 6.1 will be applied to $\mathcal{E}$ in § 12.1.

7. Set-up for the proof of Theorem 1.1

Let $\mathcal{E}$ be a nef vector bundle of rank r on $\mathbb{Q}^3$ with $c_1=2h$. It follows from [Reference Ohno12, Lemma 4.1 (1)] that

(7.1)\begin{equation} h^q(\mathcal{E}(t))=0 \,\textrm{for } q \gt 0 \,\textrm{and } t\geq 0. \end{equation}

Moreover, if $H(\mathcal{E})^{r+2} =c_3-2c_1c_2+c_1^3 =c_3-4c_2h+16 \gt 0$, then

(7.2)\begin{equation} h^q(\mathcal{E}(-1))=0 \,\textrm{for } q \gt 0 \end{equation}

by [Reference Ohno12, Lemma 4.1 (2)]. Note here that

(7.3)\begin{equation} c_3\geq 0 \end{equation}

by [Reference Lazarsfeld11, Theorem 8.2.1], since $\mathcal{E}$ is nef. Hence we see that

(7.4)\begin{equation} h^q(\mathcal{E}(-1))=0 \,\textrm{for } q \gt 0 \,\textrm{if } c_2h\leq 3. \end{equation}

It follows from [Reference Ohno12, Lemma 4.3] that

(7.5)\begin{equation} \operatorname{Ext}^q(\mathcal{S},\mathcal{E}(2))=0 \,\textrm{for } q \gt 0. \end{equation}

The exact sequence (3.1) together with the isomorphism (3.2) implies that $\mathcal{S}^{\vee}\otimes \mathcal{E}(2)$ fits in an exact sequence

\begin{equation*} 0\to \mathcal{S}^{\vee}\otimes \mathcal{E}(1)\to \mathcal{E}(1)^{\oplus 4}\to \mathcal{S}^{\vee}\otimes\mathcal{E}(2)\to 0. \end{equation*}

It then follows from (7.1) and (7.5) that

(7.6)\begin{equation} \operatorname{Ext}^q(\mathcal{S},\mathcal{E}(1))=0 \,\textrm{for } q\geq 2. \end{equation}

If $h^0(\mathcal{E}(-2))\neq 0$, then $\mathcal{E}\cong \mathcal{O}(2)\oplus \mathcal{O}^{\oplus r-1}$ by [Reference Ohno12, Proposition 5.1 and Remark 5.3]. Thus, we will always assume that

(7.7)\begin{equation} h^0(\mathcal{E}(-2))=0 \end{equation}

in the following. It follows from Theorem 2.3 that

(7.8)\begin{equation} h^q(\mathcal{E}|_{\mathbb{Q}^2})=0 \,\textrm{for } q\geq 2. \end{equation}

Moreover

(7.9)\begin{equation} h^1(\mathcal{E}|_{\mathbb{Q}^2})= \begin{cases} 1 &\textrm{if }\mathcal{E}|_{\mathbb{Q}^2}\,\textrm{belongs to Case (11) of Theorem~2.3;}\\ 0 &\textrm{otherwise.} \end{cases} \end{equation}

The vanishing (7.1) then shows that

(7.10)\begin{equation} h^3(\mathcal{E}(-1))=0. \end{equation}

Moreover

(7.11)\begin{equation} h^2(\mathcal{E}(-1))=0 \,\textrm{unless }\mathcal{E}|_{\mathbb{Q}^2}\,\textrm{belongs to Case }(11)\ \textrm{of Theorem~2.3.} \end{equation}

It follows from Theorem 2.3 that

(7.12)\begin{equation} h^q(\mathcal{E}(-1)|_{\mathbb{Q}^2})=0 \,\textrm{for } q\geq 2. \end{equation}

The vanishing (7.10) then shows that

(7.13)\begin{equation} h^3(\mathcal{E}(-2))=0. \end{equation}

The exact sequence (3.1) together with (3.2) also induces the following exact sequence

(7.14)\begin{equation} 0\to \mathcal{S}^{\vee}\otimes\mathcal{E}(-1) \to \mathcal{E}(-1)^{\oplus 4}\to \mathcal{S}^{\vee}\otimes\mathcal{E} \to 0. \end{equation}

This exact sequence (7.14) and an exact sequence

(7.15)\begin{equation} 0\to \mathcal{S}^{\vee}\otimes \mathcal{E}(-1) \to \mathcal{S}^{\vee}\otimes \mathcal{E} \to \mathcal{S}^{\vee}\otimes \mathcal{E}|_{\mathbb{Q}^2} \to 0 \end{equation}

will be used to compute $\operatorname{Ext}^q(\mathcal{S},\mathcal{E})$.

8. The case where $\mathcal{E}|_{\mathbb{Q}^2}$ belongs to Case (1) of Theorem 2.3

The assumption (7.7) implies that this case does not arise. Indeed, if $\mathcal{E}|_{\mathbb{Q}^2}\cong \mathcal{O}(2,2) \oplus \mathcal{O}^{\oplus r-1}$, then $h^q(\mathcal{E}(-1)|_{\mathbb{Q}^2})=0$ for q > 0. Moreover $c_2h=0$. Hence $h^q(\mathcal{E}(-1))=0$ for q > 0 by (7.4). This implies that $h^q(\mathcal{E}(-2))=0$ for $q\geq 2$. The assumption (7.7) then shows that

\begin{equation*}0\geq -h^1(\mathcal{E}(-2))=\chi(\mathcal{E}(-2))= 1+\dfrac{1}{2}c_3\end{equation*}

by (4.5). This contradicts (7.3). Hence this case does not arise.

9. The case where $\mathcal{E}|_{\mathbb{Q}^2}$ belongs to Case (2) of Theorem 2.3

Suppose that

\begin{equation*}\mathcal{E}|_{\mathbb{Q}^2}\cong \mathcal{O}(2,1)\oplus\mathcal{O}(0,1) \oplus \mathcal{O}^{\oplus r-2}.\end{equation*}

Then $h^0(\mathcal{E}(-1)|_{\mathbb{Q}^2})=2$ and $h^q(\mathcal{E}(-1)|_{\mathbb{Q}^2})=0$ for q > 0. Moreover $c_2h=2$. Hence

\begin{equation*}h^q(\mathcal{E}(-1))=0\,\textrm{for }q \gt 0\end{equation*}

by (7.4). It then follows from (4.4) and (7.3) that $h^0(\mathcal{E}(-1))=\chi(\mathcal{E}(-1))=2+\dfrac{1}{2}c_3\geq 2$. On the other hand, we have $h^0(\mathcal{E}(-1))\leq h^0(\mathcal{E}(-1)|_{\mathbb{Q}^2})=2$ by (7.7). Therefore, the restriction map $H^0(\mathcal{E}(-1))\to H^0(\mathcal{E}(-1)|_{\mathbb{Q}^2})$ is an isomorphism,

\begin{equation*}h^0(\mathcal{E}(-1))=2 \,\textrm{and }c_3=0.\end{equation*}

Hence we see that

\begin{equation*}h^q(\mathcal{E}(-2))=0 \,\textrm{for all } q.\end{equation*}

Since $\mathcal{E}(-2)|_{\mathbb{Q}^2}\cong \mathcal{O}(0,-1)\oplus\mathcal{O}(-2,-1) \oplus \mathcal{O}(-2,-2)^{\oplus r-2}$, we have $h^q(\mathcal{E}(-2)|_{\mathbb{Q}^2})=0$ for q < 2 and $h^2(\mathcal{E}(-2)|_{\mathbb{Q}^2})=r-2$. Therefore

\begin{equation*}h^q(\mathcal{E}(-3))=0 \,\textrm{for } q \lt 3\, \textrm{and }h^3(\mathcal{E}(-3))=r-2.\end{equation*}

Next we will compute $\operatorname{Ext}^q(\mathcal{S},\mathcal{E}(-1))$. Since

\begin{equation*}\mathcal{S}^{\vee}\otimes\mathcal{E}(t)|_{\mathbb{Q}^2} \cong (\mathcal{O}(-1,0)\oplus\mathcal{O}(0,-1)) \otimes (\mathcal{O}(2+t,1+t)\oplus\mathcal{O}(t,1+t)\oplus\mathcal{O}(t,t)^{\oplus r-2}), \end{equation*}

we see that $h^q(\mathcal{S}^{\vee}\otimes\mathcal{E}(t)|_{\mathbb{Q}^2})=0$ for q > 0 and $t\geq 0$. Hence it follows from (7.6) that

\begin{equation*}\operatorname{Ext}^q(\mathcal{S},\mathcal{E}(-1))=0 \,\textrm{for } q\geq 2.\end{equation*}

Since $c_2h=2$ and $c_3=0$, the formula (4.8) shows that

\begin{equation*}h^0(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1)) =h^1(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1)).\end{equation*}

Set $a=h^0(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1))$. Note that $\mathcal{S}^{\vee}\otimes \mathcal{E}(-1)$ fits in an exact sequence

\begin{equation*} 0\to \mathcal{S}^{\vee}\otimes \mathcal{E}(-2)\to \mathcal{E}(-2)^{\oplus 4}\to \mathcal{S}^{\vee}\otimes\mathcal{E}(-1)\to 0 \end{equation*}

by (3.1) and (3.2). Since $h^q(\mathcal{E}(-2))=0$ for all q, this exact sequence shows that

\begin{equation*}h^q(\mathcal{S}^{\vee}\otimes \mathcal{E}(-2))= \begin{cases} 0& \textrm{if }q=0, 3\\ a& \textrm{otherwise.} \end{cases} \end{equation*}

On the other hand, we have an exact sequence

(9.1)\begin{equation} 0\to \mathcal{S}^{\vee}\otimes \mathcal{E}(-2)\to \mathcal{S}^{\vee}\otimes\mathcal{E}(-1) \to (\mathcal{S}^{\vee}\otimes\mathcal{E}(-1))|_{\mathbb{Q}^2} \to 0. \end{equation}

Since

\begin{equation*}\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)|_{\mathbb{Q}^2} \cong (\mathcal{O}(-1,0)\oplus\mathcal{O}(0,-1)) \otimes (\mathcal{O}(1,0)\oplus\mathcal{O}(-1,0)\oplus\mathcal{O}(-1,-1)^{\oplus r-2}), \end{equation*}

we see that

\begin{equation*}h^q(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1)|_{\mathbb{Q}^2})= \begin{cases} 1& \textrm{if }q=0, 1\\ 0& \textrm{if }q=2, 3. \end{cases} \end{equation*}

Hence the exact sequence (9.1) implies that a = 1.

We apply to $\mathcal{E}(-1)$ the Bondal spectral sequence (2.1). We have $\operatorname{Ext}^3(G,\mathcal{E}(-1)) \cong S_3^{\oplus r-2}$, $\operatorname{Ext}^2(G,\mathcal{E}(-1)) =0$ and $\operatorname{Ext}^1(G,\mathcal{E}(-1)) \cong S_1$. Moreover, $\operatorname{Hom}(G,\mathcal{E}(-1))$ fits in an exact sequence

\begin{equation*} 0\to S_0^{\oplus 2}\to \operatorname{Hom}(G,\mathcal{E}(-1))\to S_1\to 0. \end{equation*}

Now Lemma 2.1 shows that $E_2^{p,3}=0$ unless $p=-3$, that $E_2^{-3,3}\cong \mathcal{O}(-1)^{\oplus r-2}$, that $E_2^{p,2}=0$ for all p, that $E_2^{p,1}=0$ unless $p=-1$, that $E_2^{-1,1}\cong \mathcal{S}(-1)$ and that a distinguished triangle

\begin{equation*}\mathcal{O}^{\oplus 2}\to \operatorname{Hom}(G,\mathcal{E}(-1))\operatorname{\otimes^{L}}_A G \to \mathcal{S}(-1)[1]\to \end{equation*}

exists. Hence we have the following exact sequence:

(9.2)\begin{equation} 0\to E_2^{-1,0}\to \mathcal{S}(-1)\to \mathcal{O}^{\oplus 2}\to E_2^{0,0}\to 0. \end{equation}

Note here that $E_2^{-1,0}\cong E_{\infty}^{-1,0}=0$. Hence we see that $E_2^{0,0}$ is a non-zero torsion sheaf. On the other hand, $\mathcal{E}(-1)$ has $E_2^{0,0}$ as a subsheaf, so that $E_2^{0,0}$ must be torsion-free. This is a contradiction. Therefore, this case does not arise.

10. The case where $\mathcal{E}|_{\mathbb{Q}^2}$ belongs to Case (3) of Theorem 2.3

Suppose that $\mathcal{E}|_{\mathbb{Q}^2}\cong \mathcal{O}(1,1)^{\oplus 2} \oplus \mathcal{O}^{\oplus r-2}$. Then $c_2\cdot h=2$. Hence $h^q(\mathcal{E}(-1))=0$ for q > 0 by (7.4). Since $h^q(\mathcal{E}(-1)|_{\mathbb{Q}^2})=0$ for q > 0, this implies that $h^q(\mathcal{E}(-2))=0$ for $q\geq 2$. The assumption (7.7) together with (4.5) and (7.3) shows that

\begin{equation*}0\geq -h^1(\mathcal{E}(-2))=\chi(\mathcal{E}(-2)) =\frac{1}{2}c_3\geq 0.\end{equation*}

Hence $h^1(\mathcal{E}(-2))=0$ and $c_3=0$. Thus $h^0(\mathcal{E}(-1))=h^0(\mathcal{E}(-1)|_{\mathbb{Q}^2})=2$. Since $h^q(\mathcal{E}(-2))=0$ for any q, we see that $h^q(\mathcal{E}(-3))=h^{q-1}(\mathcal{E}(-2)|_{\mathbb{Q}^2})$ for all q. Hence $h^q(\mathcal{E}(-3))=0$ unless q = 3 and $h^3(\mathcal{E}(-3))=r-2$. Since

\begin{equation*}\mathcal{S}^{\vee}\otimes\mathcal{E}(t)|_{\mathbb{Q}^2} \cong (\mathcal{O}(-1,0)\oplus\mathcal{O}(0,-1)) \otimes (\mathcal{O}(1+t,1+t)^{\oplus 2}\oplus\mathcal{O}(t,t)^{\oplus r-2}), \end{equation*}

we see that $h^q(\mathcal{S}^{\vee}\otimes\mathcal{E}(t)|_{\mathbb{Q}^2})=0$ for q > 0 and $t\geq -1$. Hence it follows from (7.6) that $\operatorname{Ext}^q(\mathcal{S},\mathcal{E}(-t))=0$ for $q\geq 2$ and $t=0,1,2$. Since the exact sequence (3.1) together with (3.2) induces an exact sequence

\begin{equation*} 0\to \mathcal{S}^{\vee}\otimes \mathcal{E}(-2)\to \mathcal{E}(-2)^{\oplus 4}\to \mathcal{S}^{\vee}\otimes\mathcal{E}(-1)\to 0, \end{equation*}

the vanishing $h^1(\mathcal{E}(-2))=0$ implies that $h^1(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1))=0$. Since $h^0(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1)|_{\mathbb{Q}^2})=0$, this implies that $h^1(\mathcal{S}^{\vee}\otimes \mathcal{E}(-2))=0$. Hence $h^0(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1))= h^0(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1)|_{\mathbb{Q}^2})=0$. We apply to $\mathcal{E}(-1)$ the Bondal spectral sequence (2.1). We see that $\operatorname{Hom}(G,\mathcal{E}(-1))\cong S_0^{\oplus 2}$, that $\operatorname{Ext}^q(G,\mathcal{E}(-1))=0$ for $q=1,2$ and that $\operatorname{Ext}^3(G,\mathcal{E}(-1)) \cong S_3^{\oplus r-2}$. Hence $E_2^{p,q}=0$ unless q = 0 or q = 3, $E_2^{p,0}=0$ unless p = 0, $E_2^{0,0}=\mathcal{O}^{\oplus 2}$, $E_2^{p,3}=0$ unless $p=-3$ and $E_2^{-3,3}=\mathcal{O}(-1)^{\oplus r-2}$ by Lemma 2.1. Therefore, $\mathcal{E}(-1)$ fits in an exact sequence

\begin{equation*} 0\to \mathcal{O}^{\oplus 2}\to \mathcal{E}(-1)\to \mathcal{O}(-1)^{\oplus r-2} \to 0. \end{equation*}

Hence $\mathcal{E}\cong \mathcal{O}(1)^{\oplus 2}\oplus\mathcal{O}^{\oplus r-2}$. This is Case (2) of Theorem 1.1.

11. The case where $\mathcal{E}|_{\mathbb{Q}^2}$ belongs to Case (4) of Theorem 2.3

Suppose that $\mathcal{E}|_{\mathbb{Q}^2}$ fits in an exact sequence

\begin{equation*} 0\to \mathcal{O} \to \mathcal{O}(1,1)\oplus \mathcal{O}(1,0)\oplus \mathcal{O}(0,1)\oplus \mathcal{O}^{\oplus r-2}\to \mathcal{E}|_{\mathbb{Q}^2}\to 0. \end{equation*}

Then $c_2h=3$. Hence $h^q(\mathcal{E}(-1))=0$ for q > 0 by (7.4). Note that $h^q(\mathcal{E}(-1)|_{\mathbb{Q}^2})=0$ for q > 0 and that $h^0(\mathcal{E}(-1)|_{\mathbb{Q}^2})=1$. Hence $h^q(\mathcal{E}(-2))=0$ for $q\geq 2$. The assumption (7.7) together with (4.5) and (7.3) shows that

\begin{equation*}0\geq -h^1(\mathcal{E}(-2))=\chi(\mathcal{E}(-2)) =-\frac{1}{2}+\frac{1}{2}c_3\geq -\frac{1}{2}.\end{equation*}

Hence $h^1(\mathcal{E}(-2))=0$ and $c_3=1$. Now that $h^q(\mathcal{E}(-2))=0$ for any q, we have $h^q(\mathcal{E}(-3))=h^{q-1}(\mathcal{E}(-2)|_{\mathbb{Q}^2})$ for any q. Set $a=h^1(\mathcal{E}(-2)|_{\mathbb{Q}^2})$. Then a = 0 or 1, and $h^2(\mathcal{E}(-2)|_{\mathbb{Q}^2})=r-3+a$. Hence we see that $h^q(\mathcal{E}(-3))=0$ for $q\leq 1$, that $h^2(\mathcal{E}(-3))=a$ and that $h^3(\mathcal{E}(-3))=r-3+a$. Moreover, the assumption (7.7) implies that $h^0(\mathcal{E}(-1))=h^0(\mathcal{E}(-1)|_{\mathbb{Q}^2})=1$. Since $\mathcal{E}|_{\mathbb{Q}^2}(-2,-1)$ fits in an exact sequence

\begin{equation*} \begin{array}{ll} 0\to \mathcal{O}(-2,-1) \to \mathcal{O}(-1,0)\oplus \mathcal{O}(-1,-1)\oplus \mathcal{O}(-2,0)\oplus &\mathcal{O}(-2,-1)^{\oplus r-2}\\ &\to \mathcal{E}|_{\mathbb{Q}^2}(-2,-1)\to 0, \end{array} \end{equation*}

we see that $h^q(\mathcal{E}|_{\mathbb{Q}^2}(-2,-1))=0$ unless q = 1. Hence $h^q(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)|_{\mathbb{Q}^2})=0$ unless q = 1. Note that $h^q(\mathcal{S}^{\vee}\otimes\mathcal{E}(t)|_{\mathbb{Q}^2})=0$ for $t\geq 0$ and $q\geq 1$. Hence it follows from (7.6) that $\operatorname{Ext}^q(\mathcal{S},\mathcal{E}(-t))=0$ for $q\geq 2$ and $t=0,1$. Note that $\mathcal{S}^{\vee}\otimes \mathcal{E}(-2)$ is a subbundle of $\mathcal{E}(-2)^{\oplus 4}$ by (3.1). Since $h^0(\mathcal{E}(-2))=0$, this implies that $h^0(\mathcal{S}^{\vee}\otimes\mathcal{E}(-2))=0$. Since we have an exact sequence

\begin{equation*} 0\to \mathcal{S}^{\vee}\otimes \mathcal{E}(-2) \to \mathcal{S}^{\vee}\otimes\mathcal{E}(-1) \to \mathcal{S}^{\vee}\otimes\mathcal{E}(-1)|_{\mathbb{Q}^2}\to 0 \end{equation*}

and $h^0(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)|_{\mathbb{Q}^2})=0$, we infer that $h^0(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1))=0$. Now, from (4.8), it follows that

\begin{equation*} -h^1(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =\chi(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =4-2\cdot 3+1=-1. \end{equation*}

We apply to $\mathcal{E}(-1)$ the Bondal spectral sequence (2.1). We have the following isomorphisms: $\operatorname{Ext}^3(G,\mathcal{E}(-1))\cong S_3^{\oplus r-3+a}$; $\operatorname{Ext}^2(G,\mathcal{E}(-1))\cong S_3^{\oplus a}$; $\operatorname{Ext}^1(G,\mathcal{E}(-1))\cong S_1$; $\operatorname{Hom}(G,\mathcal{E}(-1))\cong S_0$. Lemma 2.1 then shows that $E_2^{p,q}=0$ unless $(p,q)=(-3,3)$, $(-3,2)$, $(-1,1)$ or $(0,0)$, that $E_2^{-3,3}=\mathcal{O}(-1)^{\oplus r-3+a}$, that $E_2^{-3,2}=\mathcal{O}(-1)^{\oplus a}$, that $E_2^{-1,1}=\mathcal{S}(-1)$ and that $E_2^{0,0}=\mathcal{O}$. Hence $E_3^{-3,2}=0$ and $E_3^{-1,1}$ fits in the following exact sequence:

\begin{equation*}0\to \mathcal{O}(-1)^{\oplus a}\to \mathcal{S}(-1)\to E_3^{-1,1}\to 0.\end{equation*}

Moreover $\mathcal{E}(-1)$ has a filtration $\mathcal{O}\subset F(\mathcal{E}(-1))\subset \mathcal{E}(-1)$ such that $F(\mathcal{E}(-1))$ fits in the following exact sequences:

\begin{equation*}0\to F(\mathcal{E}(-1))\to \mathcal{E}(-1)\to \mathcal{O}(-1)^{\oplus r-3}\to 0;\end{equation*}

\begin{equation*}0\to \mathcal{O}\to F(\mathcal{E}(-1))\to E_3^{-1,1}\to 0.\end{equation*}

In particular, we see that $F(\mathcal{E}(-1))$ is a vector bundle, since so is $\mathcal{E}(-1)$. On the other hand, since $\operatorname{Ext}^1(\mathcal{S}(-1),\mathcal{O})=0$, $F(\mathcal{E}(-1))$ fits in the following exact sequence:

\begin{equation*} 0\to \mathcal{O}(-1)^{\oplus a}\to \mathcal{O}\oplus\mathcal{S}(-1)\to F(\mathcal{E}(-1))\to 0. \end{equation*}

This implies that a = 0. Indeed, if a = 1, then $F(\mathcal{E}(-1))$ cannot be a vector bundle, since the intersection of a line and a hyperplane section cannot be empty. Therefore $F(\mathcal{E}(-1))\cong \mathcal{O}\oplus \mathcal{S}(-1)$, and thus $\mathcal{E}\cong \mathcal{O}(1)\oplus \mathcal{S}\oplus \mathcal{O}^{\oplus r-3}$. This is Case (3) of Theorem 1.1.

12. The case where $\mathcal{E}|_{\mathbb{Q}^2}$ belongs to Case (5) of Theorem 2.3

Suppose that $\mathcal{E}|_{\mathbb{Q}^2}$ fits in an exact sequence

\begin{equation*} 0\to \mathcal{O}(-1,-1) \to \mathcal{O}(1,1)\oplus \mathcal{O}^{\oplus r}\to \mathcal{E}|_{\mathbb{Q}^2}\to 0. \end{equation*}

Then $c_2h=4$. Note that

(12.1)\begin{equation} h^q(\mathcal{S}^{\vee}\otimes \mathcal{E}|_{\mathbb{Q}^2}) = \begin{cases} 4& \textrm{if}\quad q= 0\\ 0& \textrm{if}\quad q\neq 0, \end{cases} \end{equation}

and that

(12.2)\begin{equation} h^q(\mathcal{E}(-1)|_{\mathbb{Q}^2}) = \begin{cases} 1& \textrm{if}\quad q= 0, 1\\ 0& \textrm{if}\quad q\neq 0, 1. \end{cases} \end{equation}

Hence we have

\begin{equation*}h^0(\mathcal{E}(-1))\leq 1\end{equation*}

by (7.7).

12.1. Suppose that $h^0(\mathcal{E}(-1))=1$

Lemma 6.1 then shows that $c_3\geq 4$. Hence $H^q(\mathcal{E}(-1))$ vanishes for q > 0 by (7.2). The formula (4.4) then shows that

\begin{equation*}h^0(\mathcal{E}(-1))=-1+\dfrac{1}{2}c_3.\end{equation*}

Thus we have $c_3=4$. We also see that $h^q(\mathcal{E}(-2))=0$ unless q = 2 and that $h^2(\mathcal{E}(-2))=1$ by (12.2) and (7.7). We have $h^0(\mathcal{E})=r+5$. Since we have an exact sequence

\begin{equation*} 0\to \mathcal{S}^{\vee}\otimes \mathcal{E}(-2) \to \mathcal{E}(-2)^{\oplus 4} \to \mathcal{S}^{\vee}\otimes \mathcal{E}(-1) \to 0, \end{equation*}

we see that $h^0(\mathcal{S}^{\vee}\otimes \mathcal{E}(-2))=0$ and that $h^3(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1))=0$. Note that $h^0(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1)|_{\mathbb{Q}^2})=0$. Since we have an exact sequence

\begin{equation*} 0\to \mathcal{S}^{\vee}\otimes \mathcal{E}(-2) \to \mathcal{S}^{\vee}\otimes \mathcal{E}(-1) \to \mathcal{S}^{\vee}\otimes \mathcal{E}(-1)|_{\mathbb{Q}^2} \to 0, \end{equation*}

we infer that $h^0(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1))=0$. Since we have an exact sequence (7.14), we see that $h^q(\mathcal{S}^{\vee}\otimes \mathcal{E})=0$ for $q\geq 2$. The exact sequence (7.15) together with (12.1) shows that $h^2(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1))=0$. Now the formula (4.8) shows that

\begin{equation*}-h^1(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =\chi(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1))=0,\end{equation*}

since $c_3=4$ and $c_2h=4$. The exact sequence (7.14) then implies that $h^q(\mathcal{S}^{\vee}\otimes\mathcal{E})=0$ unless q = 0 and that $h^0(\mathcal{S}^{\vee}\otimes\mathcal{E})=4$. Since $h^0(\mathcal{E}(-1))=1$, we have an injection $\mathcal{O}(1)\to \mathcal{E}$. Let $\mathcal{F}$ be its cokernel: we have the following exact sequence:

\begin{equation*} 0\to \mathcal{O}(1)\to \mathcal{E}\to \mathcal{F}\to 0. \end{equation*}

We apply to $\mathcal{F}$ the Bondal spectral sequence (2.1). We see that $h^q(\mathcal{F})=0$ unless q = 0 and that $h^0(\mathcal{F})=r$. Moreover $h^q(\mathcal{F}(-1))=0$ for any q, $h^q(\mathcal{F}(-2))=0$ unless q = 2 and $h^2(\mathcal{F}(-2))=1$. Finally, we have $h^q(\mathcal{S}^{\vee}\otimes \mathcal{F})=0$ for all q. Therefore $\operatorname{Ext}^q(G,\mathcal{F})=0$ for q = 3 and 1, $\operatorname{Ext}^2(G,\mathcal{F})\cong S_3$ and $\operatorname{Hom}(G,\mathcal{F})\cong S_0^{\oplus r}$. Hence $E_2^{p,q}=0$ unless $(p\cdot q)=(-3,2)$ or $(0,0)$, $E_2^{-3,2}=\mathcal{O}(-1)$ and $E_2^{0,0}=\mathcal{O}^{\oplus r}$ by Lemma 2.1. Thus, we have an exact sequence

\begin{equation*} 0\to \mathcal{O}(-1)\to \mathcal{O}^{\oplus r}\to \mathcal{F}\to 0. \end{equation*}

Therefore $\mathcal{E}$ belongs to Case (4) of Theorem 1.1.

12.2. Suppose that $h^0(\mathcal{E}(-1))=0$

Then $h^0(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1))=0$ by (7.14). Note that $H^q(\mathcal{E}|_{\mathbb{Q}^2})$ vanishes for all q > 0. Since $h^q(\mathcal{E})=0$ for all q > 0 by (7.1), we have $h^q(\mathcal{E}(-1))=0$ for all $q\geq 2$. Hence (4.4) and (7.3) imply that

\begin{equation*}0\geq -h^1(\mathcal{E}(-1))=\chi(\mathcal{E}(-1))=-1+\dfrac{1}{2}c_3\geq -1.\end{equation*}

Therefore, $(h^1(\mathcal{E}(-1)),c_3)$ is either $(0,2)$ or $(1,0)$. Since $h^3(\mathcal{E}(-1))=0$, we first have $h^3(\mathcal{S}^{\vee}\otimes \mathcal{E})=0$ by (7.14). Secondly, we have $h^3(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1))=0$ by (12.1) and (7.15). Thirdly, we have $h^2(\mathcal{S}^{\vee}\otimes \mathcal{E})=0$ by (7.14) since $h^2(\mathcal{E}(-1))=0$. Finally, we have $h^2(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1))=0$ by (12.1) and (7.15). Hence

(12.3)\begin{equation} -h^1(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =\chi(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =-4+c_3 \end{equation}

by (4.8). We apply to $\mathcal{E}$ the Bondal spectral sequence (2.1).

12.2.1. Suppose that $(h^1(\mathcal{E}(-1)),c_3)=(0,2).$

Then $h^1(\mathcal{S}^{\vee}\otimes \mathcal{E})=0$ by (7.14). Moreover $h^1(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1))=2$ by (12.3). Hence we have $h^0(\mathcal{S}^{\vee}\otimes \mathcal{E})=2$ by (7.14). Since $h^q(\mathcal{E}(-1)|_{\mathbb{Q}^2})=1$ for q = 0, 1 and $h^q(\mathcal{E}(-1)|_{\mathbb{Q}^2})=0$ for q = 2, 3, we infer that $h^q(\mathcal{E}(-2))=1$ for q = 1, 2, and that $h^q(\mathcal{E}(-2))=0$ unless q = 1 or 2. Since $h^0(\mathcal{E}|_{\mathbb{Q}^2})=r+4$, we see that $h^0(\mathcal{E})=r+4$. Therefore, we have an exact sequence

\begin{equation*} 0\to S_0^{\oplus r+4}\to \operatorname{Hom}(G,\mathcal{E})\to S_1^{\oplus 2}\to 0 \end{equation*}

and the following: $\operatorname{Ext}^1(G,\mathcal{E})\cong S_3$; $\operatorname{Ext}^2(G,\mathcal{E})\cong S_3$ and $\operatorname{Ext}^3(G,\mathcal{E})=0$. Therefore, Lemma 2.1 implies that $E_2^{p,q}=0$ unless $(p,q)=(-3,1)$, $(-3,2)$, $(-1,0)$ or $(0,0)$, that $E_2^{-3,1}\cong \mathcal{O}(-1)$, that $E_2^{-3,2}\cong \mathcal{O}(-1)$ and that there is an exact sequence

\begin{equation*} 0\to E_2^{-1,0}\to \mathcal{S}(-1)^{\oplus 2}\to \mathcal{O}^{\oplus r+4} \to E_2^{0,0}\to 0.\end{equation*}

It follows from the Bondal spectral sequence (2.1) that $E_2^{-3,1}\cong E_2^{-1,0}$, that $E_2^{-3,2}\cong E_3^{-3,2}$, that $E_2^{0,0}\cong E_3^{0,0}$ and that there is an exact sequence

\begin{equation*}0\to E_3^{-3,2}\to E_3^{0,0}\to \mathcal{E}\to 0.\end{equation*}

Hence we obtain the following exact sequences:

\begin{equation*} 0\to \mathcal{O}(-1)\to \mathcal{S}(-1)^{\oplus 2}\to \mathcal{O}^{\oplus r+4} \to E_3^{0,0}\to 0;\end{equation*}

\begin{equation*}0\to \mathcal{O}(-1)\to E_3^{0,0}\to \mathcal{E}\to 0. \end{equation*}

The latter exact sequence shows that $E_3^{0,0}$ is a vector bundle since so is $\mathcal{E}$. The former exact sequence then splits into the following two exact sequences with $\mathcal{G}$ a vector bundle of rank three:

\begin{equation*}0\to \mathcal{O}(-1)\to \mathcal{S}(-1)^{\oplus 2}\to \mathcal{G}\to 0;\end{equation*}

\begin{equation*}0\to \mathcal{G}\to \mathcal{O}^{\oplus r+4}\to E_3^{0,0}\to 0.\end{equation*}

The latter exact sequence shows that the dual $\mathcal{G}^{\vee}$ of $\mathcal{G}$ is globally generated. The injection $\mathcal{O}(-1)\to \mathcal{S}(-1)^{\oplus 2}$ in the former exact sequence gives rise to two global sections s ₀, s ₁ of $\mathcal{S}$, and we infer that $(s_0)_0\cap (s_1)_0=\emptyset$ since $\mathcal{G}$ is a vector bundle. Hence s ₀ and s ₁ are linearly independent. We also see that $\mathcal{G}^{\vee}$ fits in the following exact sequence:

\begin{equation*}0\to \mathcal{G}^{\vee}\to \mathcal{S}^{\oplus 2}\to \mathcal{O}(1)\to 0.\end{equation*}

Note that the induced map $H^0(\mathcal{S})^{\oplus 2}\to H^0(\mathcal{O}(1))$ sends $(t_0,t_1)$ to $s_0\wedge t_0+s_1\wedge t_1$, and Lemma 3.1 implies that it is surjective. Therefore $h^0(\mathcal{G}^{\vee})=3$. Since $\mathcal{G}^{\vee}$ is a globally generated vector bundle of rank three, this implies that $\mathcal{G}^{\vee}\cong \mathcal{O}^{\oplus 3}$. On the other hand, the exact sequence above shows that $c_1(\mathcal{G}^{\vee})=1$. This is a contradiction. Hence the case $(h^1(\mathcal{E}(-1)),c_3)=(0,2)$ does not arise.

12.2.2. Suppose that $(h^1(\mathcal{E}(-1)),c_3)=(1,0)$

Then $h^1(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1))=4$ by (12.3). Set $a:=h^0(\mathcal{S}^{\vee}\otimes \mathcal{E})$. Then $h^1(\mathcal{S}^{\vee}\otimes \mathcal{E})=a$ by (7.14). From (12.2), it follows that $h^q(\mathcal{E}(-2))=0$ unless q = 1 or 2 and that $(h^1(\mathcal{E}(-2)),h^2(\mathcal{E}(-2)))=(1,0)$ or $(2,1)$. Note also that $h^0(\mathcal{E})=r+3$.

12.2.2.1. Suppose that $(h^1(\mathcal{E}(-2)),h^2(\mathcal{E}(-2)))=(1,0).$ Then we see that $\operatorname{Ext}^3(G,\mathcal{E})=0$, that $\operatorname{Ext}^2(G,\mathcal{E})=0$, that $\operatorname{Ext}^1(G,\mathcal{E})$ has a filtration $S_1^{\oplus a}\subset F\subset \operatorname{Ext}^1(G,\mathcal{E})$ of right A-modules such that the following sequences are exact:

\begin{equation*} 0\to F\to \operatorname{Ext}^1(G,\mathcal{E})\to S_3\to 0;\end{equation*}

\begin{equation*} 0\to S_1^{\oplus a}\to F\to S_2\to 0,\end{equation*}

and that $\operatorname{Hom}(G,\mathcal{E})$ fits in the following exact sequence of right A-modules:

\begin{equation*} 0\to S_0^{\oplus r+3}\to \operatorname{Hom}(G,\mathcal{E})\to S_1^{\oplus a}\to 0. \end{equation*}

These exact sequences induce the following distinguished triangles by Lemma 2.1:

\begin{equation*}F\operatorname{\otimes^{L}}_AG\to \operatorname{Ext}^1(G,\mathcal{E})\operatorname{\otimes^{L}}_AG\to \mathcal{O}(-1)[3]\to;\end{equation*}

\begin{equation*}\mathcal{S}(-1)[1]^{\oplus a}\to F\operatorname{\otimes^{L}}_AG\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}[2]\to;\end{equation*}

\begin{equation*}\mathcal{O}^{\oplus r+3}\to \operatorname{Hom}(G,\mathcal{E})\operatorname{\otimes^{L}}_AG\to \mathcal{S}(-1)[1]^{\oplus a}\to.\end{equation*}

By taking cohomologies, we obtain the following exact sequences by (3.2):

\begin{equation*}0\to E_2^{-3,1}\to \mathcal{O}(-1)\to \mathcal{H}^{-2}(F\operatorname{\otimes^{L}}_AG) \to E_2^{-2,1}\to 0;\end{equation*}

(12.4)\begin{equation} 0\to \mathcal{H}^{-2}(F\operatorname{\otimes^{L}}_AG)\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3} \xrightarrow{\psi_a} \mathcal{S}^{\vee\oplus a}\to E_2^{-1,1}\to 0; \end{equation}

\begin{equation*}0\to E_2^{-1,0}\to \mathcal{S}^{\vee\oplus a} \to\mathcal{O}^{\oplus r+3}\to E_2^{0,0}\to 0.\end{equation*}

Moreover, we have the following exact sequences:

\begin{equation*}0\to E_2^{-2,1}\to E_2^{0,0}\to E_3^{0,0}\to 0;\end{equation*}

\begin{equation*}0\to E_2^{-3,1}\to E_2^{-1,0}\to 0;\end{equation*}

\begin{equation*}0\to E_3^{0,0}\to \mathcal{E}\to E_2^{-1,1}\to 0.\end{equation*}

Since $\mathcal{E}$ is nef, $E_2^{-1,1}$ cannot admit negative degree quotients. Hence it follows from Lemma 5.3 that a = 0. Then $E_2^{-1,1}=0$, $E_2^{-3,1}=E_2^{-1,0}=0$, $E_2^{0,0}=\mathcal{O}^{\oplus r+3}$, and we have the following exact sequence:

\begin{equation*}0\to \mathcal{O}(-1)\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}\to E_2^{-2,1}\to 0.\end{equation*}

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

(12.5)\begin{equation} 0\to \mathcal{O}(-1)\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3} \to \mathcal{O}^{\oplus r+3}\to \mathcal{E}\to 0. \end{equation}

Since $T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}$ fits in an exact sequence

\begin{equation*} 0\to \mathcal{O}(-2)\to \mathcal{O}(-1)^{\oplus 5}\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}\to 0, \end{equation*}

the exact sequence (12.5) induces the following exact sequence:

\begin{equation*} 0\to \mathcal{O}(-2)\to \mathcal{O}(-1)^{\oplus 4}\to \mathcal{O}^{\oplus r+3}\to \mathcal{E}\to 0. \end{equation*}

This is Case (9) of Theorem 1.1.

12.2.2.2. Suppose that $(h^1(\mathcal{E}(-2)),h^2(\mathcal{E}(-2)))=(2,1).$ Then we see that $\operatorname{Ext}^3(G,\mathcal{E})=0$, that $\operatorname{Ext}^2(G,\mathcal{E})\cong S_3$, that $\operatorname{Ext}^1(G,\mathcal{E})$ has a filtration $S_1^{\oplus a}\subset F\subset \operatorname{Ext}^1(G,\mathcal{E})$ of right A-modules such that the following sequences are exact:

\begin{equation*} 0\to F\to \operatorname{Ext}^1(G,\mathcal{E})\to S_3^{\oplus 2}\to 0;\end{equation*}

\begin{equation*} 0\to S_1^{\oplus a}\to F\to S_2\to 0,\end{equation*}

and that $\operatorname{Hom}(G,\mathcal{E})$ fits in the following exact sequence of right A-modules:

\begin{equation*} 0\to S_0^{\oplus r+3}\to \operatorname{Hom}(G,\mathcal{E})\to S_1^{\oplus a}\to 0. \end{equation*}

Lemma 2.1 implies that $\operatorname{Ext}^2(G,\mathcal{E})\operatorname{\otimes^{L}}_AG\cong \mathcal{O}(-1)[3]$ and that the three exact sequences above induce the following distinguished triangles:

\begin{equation*}F\operatorname{\otimes^{L}}_AG\to \operatorname{Ext}^1(G,\mathcal{E})\operatorname{\otimes^{L}}_AG\to \mathcal{O}(-1)^{\oplus 2}[3]\to;\end{equation*}

\begin{equation*}\mathcal{S}(-1)[1]^{\oplus a}\to F\operatorname{\otimes^{L}}_AG\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}[2]\to;\end{equation*}

\begin{equation*}\mathcal{O}^{\oplus r+3}\to \operatorname{Hom}(G,\mathcal{E})\operatorname{\otimes^{L}}_AG\to \mathcal{S}(-1)[1]^{\oplus a}\to.\end{equation*}

By taking cohomologies, we see that $E_2^{p,2}=0$ unless $p=-3$, that $E_2^{-3,2}\cong \mathcal{O}(-1)$, and that we have the following exact sequences by (3.2):

(12.6)\begin{equation} 0\to E_2^{-3,1}\to \mathcal{O}(-1)^{\oplus 2}\to \mathcal{H}^{-2}(F\operatorname{\otimes^{L}}_AG) \to E_2^{-2,1}\to 0; \end{equation}

(12.7)\begin{equation} 0\to \mathcal{H}^{-2}(F\operatorname{\otimes^{L}}_AG)\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3} \xrightarrow{\psi_a} \mathcal{S}^{\vee\oplus a}\to E_2^{-1,1}\to 0; \end{equation}

\begin{equation*}0\to E_2^{-1,0}\to \mathcal{S}^{\vee\oplus a} \to\mathcal{O}^{\oplus r+3}\to E_2^{0,0}\to 0.\end{equation*}

Moreover, we have the following exact sequences:

(12.8)\begin{equation} 0\to E_3^{-3,2}\to E_2^{-3,2}\xrightarrow{\pi} E_2^{-1,1}\to E_3^{-1,1}\to 0; \end{equation}

\begin{equation*}0\to E_2^{-2,1}\to E_2^{0,0}\to E_3^{0,0}\to 0;\end{equation*}

\begin{equation*}0\to E_2^{-3,1}\to E_2^{-1,0}\to 0;\end{equation*}

\begin{equation*}0\to E_3^{-3,2}\to E_3^{0,0}\to E_4^{0,0}\to 0;\end{equation*}

\begin{equation*}0\to E_4^{0,0}\to \mathcal{E}\to E_3^{-1,1}\to 0.\end{equation*}

Since $\mathcal{E}$ is nef, $E_3^{-1,1}$ cannot admit negative degree quotients. If a > 0, it follows from Lemmas 5.4 and 5.3 that a = 1, that $E_3^{-1,1}=0$, that $E_3^{-3,2}\cong \mathcal{I}_L(-1)$ for some line $L\subset\mathbb{Q}^3$, that $E_2^{-1,1}\cong \mathcal{O}_L(-1)$ and that $ \mathcal{H}^{-2}(F\operatorname{\otimes^{L}}_AG)\cong \mathcal{O}(-1)^{\oplus 2}$. Therefore, $\mathcal{E}\cong E_4^{0,0}$ and the exact sequence (12.6) becomes the following exact sequence:

\begin{equation*}0\to E_2^{-3,1}\to \mathcal{O}(-1)^{\oplus 2}\to \mathcal{O}(-1)^{\oplus 2} \to E_2^{-2,1}\to 0.\end{equation*}

Set $\mathcal{O}(-1)^{\oplus b}\cong E_2^{-3,1}$ for some non-negative integer $b\leq 2$. Then $E_2^{-2,1}\cong \mathcal{O}(-1)^{\oplus b}$ and we have the following exact sequences:

\begin{equation*}0\to \mathcal{O}(-1)^{\oplus b}\to \mathcal{S}^{\vee} \to\mathcal{O}^{\oplus r+3}\to E_2^{0,0}\to 0;\end{equation*}

\begin{equation*}0\to\mathcal{O}(-1)^{\oplus b} \to E_2^{0,0}\to E_3^{0,0}\to 0;\end{equation*}

\begin{equation*}0\to \mathcal{I}_L(-1)\to E_3^{0,0}\to \mathcal{E}\to 0.\end{equation*}

Since $\mathcal{O}^{\oplus r+3}$ is torsion-free and $\mathcal{S}^{\vee}$ is not isomorphic to $\mathcal{O}^{\oplus 2}$, we see that $b\leq 1$. Note here that $E_3^{0,0}$ is torsion-free, and so is $E_2^{0,0}$. If b = 1, we get an exact sequence

\begin{equation*}0\to \mathcal{I}_M\to \mathcal{O}^{\oplus r+3}\to E_2^{0,0}\to 0\end{equation*}

for some line M in $\mathbb{Q}^3$. Since we can extend $\mathcal{I}_M\to \mathcal{O}^{\oplus r+3}$ to an injection $\mathcal{O}\to \mathcal{O}^{\oplus r+3}$ by taking double duals, we infer that $E_2^{0,0}$ contains a torsion sheaf $\mathcal{O}_M$. This is a contradiction. Hence b = 0, and $E_2^{0,0}$ fits in the following exact sequences:

\begin{equation*}0\to \mathcal{S}^{\vee} \to\mathcal{O}^{\oplus r+3}\to E_2^{0,0}\to 0;\end{equation*}

\begin{equation*}0\to \mathcal{I}_L(-1)\to E_2^{0,0}\to \mathcal{E}\to 0.\end{equation*}

Since $\mathcal{I}_L(-1)$ is torsion-free but not locally free, so is $E_2^{0,0}$. Hence the former exact sequence together with (3.1) implies that $E_2^{0,0}\cong \mathcal{I}_M(1)\oplus \mathcal{O}^{\oplus r}$ for some line M in $\mathbb{Q}^3$. This can be shown by the similar argument as in the proof of Lemma 5.2. Indeed, by taking a free basis of $\mathcal{O}^{\oplus r+3}$ suitably, we may assume that the injection $\mathcal{S}^{\vee}\to \mathcal{O}^{\oplus r+3}$ is written as ${}^t(s_1^{\vee},\dots,s_{m}^{\vee},0,\dots,0)$ for some linearly independent elements $s_1,\dots,s_m$ of $H^0(\mathcal{S})$, where $s_i^{\vee}$ denotes the dual of the morphism $\mathcal{O}\to \mathcal{S}$ defined by s_i. We have $2=\operatorname{rank} \mathcal{S}^{\vee}\leq m\leq h^0(\mathcal{S})=4$. Since $E_2^{0,0}$ is torsion-free, we have $3\leq m$. Since $E_2^{0,0}$ is not locally free, it follows from the exact sequence (3.1) that m ≠ 4. Hence m = 3. Moreover, the exact sequence (3.1) shows that if we extend $(s_1,s_2,s_3)$ to a basis $(s_1,s_2,s_3,s_4)$ of $H^0(\mathcal{S})$ then there exists a basis $(t_1,t_2,t_3,t_4)$ of $H^0(\mathcal{S})$ such that $\sum_{i=1}^4t_is_i^{\vee}=0$ and that the cokernel of the morphism ${}^t(s_1^{\vee},s_2^{\vee},s_3^{\vee})$ is isomorphic to the cokernel of the morphism $t_4:\mathcal{O}\to \mathcal{S}$. Hence the cokernel of ${}^t(s_1^{\vee},s_2^{\vee},s_3^{\vee})$ is isomorphic to $\mathcal{I}_M(1)$ for some line M on $\mathbb{Q}^3$. Therefore $E_2^{0,0}\cong \mathcal{I}_M(1)\oplus \mathcal{O}^{\oplus r}$. By taking the double dual of the injection $\mathcal{I}_L(-1)\to E_2^{0,0}$ in the latter exact sequence, we obtain a commutative diagram with exact rows

for some coherent sheaf $\mathcal{F}$. Note that $\operatorname{Tor}_q^{\mathcal{O}_p}(\mathcal{F}_p,k(p))=0$ for $q\geq 2$ and any point p. Since $\mathcal{E}$ is torsion-free, the snake lemma implies that L = M and that we have an exact sequence

\begin{equation*}0\to \mathcal{O}_L(-1)\to \mathcal{O}_M(1) \to \mathcal{O}_Z\to 0\end{equation*}

for some closed subscheme Z of length two. Moreover, $\mathcal{E}$ fits in the following exact sequence:

\begin{equation*}0\to \mathcal{E}\to \mathcal{F}\to \mathcal{O}_Z\to 0.\end{equation*}

For an associated point p of Z, the exact sequence above induces a coherent sheaf $\mathcal{G}$ and the following exact sequence:

\begin{equation*} 0\to \mathcal{E}\to \mathcal{G}\to k(p)\to 0. \end{equation*}

Since $\operatorname{Tor}_3^{\mathcal{O}_p}(\mathcal{F}_p,k(p))=0$, we have $\operatorname{Tor}_3^{\mathcal{O}_p}(\mathcal{G}_p,k(p))=0$. Note that $\operatorname{Tor}_q^{\mathcal{O}_p}(\mathcal{E}_p,k(p))=0$ for $q\geq 1$. Hence $\operatorname{Tor}_3^{\mathcal{O}_p}(k(p),k(p))=0$, which contradicts the fact that $\operatorname{Tor}_3^{\mathcal{O}_p}(k(p),k(p))=1$. Therefore, a cannot be positive: a = 0. Thus $0=E_2^{-1,1}=E_3^{-1,1}$, $0=E_2^{-1,0} = E_2^{-3,1}$, $\mathcal{O}^{\oplus r+3}\cong E_2^{0,0}$, $E_3^{-3,2}\cong E_2^{-3,2}\cong \mathcal{O}(-1)$, $E_4^{0,0}\cong \mathcal{E}$, and we have the following exact sequences:

\begin{equation*}0\to\mathcal{O}(-1)^{\oplus 2}\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3} \to E_2^{-2,1}\to 0;\end{equation*}

\begin{equation*}0\to E_2^{-2,1}\to \mathcal{O}^{\oplus r+3}\to E_3^{0,0}\to 0;\end{equation*}

\begin{equation*}0\to \mathcal{O}(-1)\to E_3^{0,0}\to \mathcal{E}\to 0.\end{equation*}

Since $T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}$ fits in an exact sequence

\begin{equation*} 0\to \mathcal{O}(-2)\to \mathcal{O}(-1)^{\oplus 5}\to T_{\mathbb{P}^4}(-2)|_{\mathbb{Q}^3}\to 0, \end{equation*}

$E_2^{-2,1}$ has a resolution of the following form:

\begin{equation*} 0\to \mathcal{O}(-2)\to \mathcal{O}(-1)^{\oplus 3}\to E_2^{-2,1}\to 0. \end{equation*}

Therefore, we see that $\mathcal{E}$ belongs to Case (9) of Theorem 1.1.

13. The case where $\mathcal{E}|_{\mathbb{Q}^2}$ belongs to Case (6) of Theorem 2.3

Suppose that $\mathcal{E}|_{\mathbb{Q}^2}$ fits in the following exact sequence:

\begin{equation*}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}|_{\mathbb{Q}^2}\to 0.\end{equation*}

Then $h^q(\mathcal{E}(-1)|_{\mathbb{Q}^2})=0$ for any q, and $c_2h=4$. Since $h^q(\mathcal{E}|_{\mathbb{Q}^2})=0$ for any q > 0, the vanishing (7.1) shows that $h^q(\mathcal{E}(-t))=0$ for $q\geq 2$ and $t=1,2$. The assumption (7.7) together with (4.5) and (7.3) shows that

\begin{equation*}0\geq -h^1(\mathcal{E}(-2))=\chi(\mathcal{E}(-2))=-1+\frac{1}{2}c_3\geq -1.\end{equation*}

Therefore we have two cases: $(h^1(\mathcal{E}(-2)),c_3)=(0,2)$ or $(1,0)$. Note here that $h^q(\mathcal{E}(-1))=h^q(\mathcal{E}(-2))$ for any q. In particular, $h^0(\mathcal{E}(-1))=h^0(\mathcal{E}(-2))=0$ by (7.7).

We claim here that $h^q(\mathcal{S}^{\vee}\otimes \mathcal{E}(t)|_{\mathbb{Q}^2})=0$ for q > 0 and $t\geq 0$. Indeed, we see that

\begin{equation*}h^q((\mathcal{O}(-1,0)\oplus\mathcal{O}(0,-1))\otimes (\mathcal{O}(1+t,t)^{\oplus 2}\oplus\mathcal{O}(t,1+t)^{\oplus 2} \oplus \mathcal{O}(t,t)^{\oplus r-3}))=0\end{equation*}

for q > 0 and $t\geq 0$. Hence we obtain the claim. Then it follows from (7.6) that

(13.1)\begin{equation} h^q(\mathcal{S}^{\vee}\otimes\mathcal{E}(t))=0\,\textrm{for }q\geq 2 \,\textrm{and }t=0, -1. \end{equation}

Since $h^0(\mathcal{E}(-1))=0$, the exact sequence (7.14) together with (13.1) shows that $h^q(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1))=0$ unless q = 1. Hence

(13.2)\begin{equation} -h^1(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =\chi(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =-4+c_3 \end{equation}

by (4.8).

13.1. Suppose that $(h^1(\mathcal{E}(-2)),c_3)=(0,2)$

Then $h^q(\mathcal{E}(-2))=0$ for any q. Hence $h^q(\mathcal{E}(-1))=0$ for any q. Set $a=h^1(\mathcal{E}(-2)|_{\mathbb{Q}^2})$. Then $a\leq 2$ and $h^2(\mathcal{E}(-2)|_{\mathbb{Q}^2})=r-4+a$. Thus $h^2(\mathcal{E}(-3))=a$, $h^3(\mathcal{E}(-3))=r-4+a$ and $h^q(\mathcal{E}(-3))=0$ unless q = 2 or 3. It follows from (13.2) that $h^1(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1))=2$. We apply to $\mathcal{E}(-1)$ the Bondal spectral sequence (2.1). We have $\operatorname{Ext}^3(G,\mathcal{E}(-1))\cong S_3^{\oplus r-4+a}$, $\operatorname{Ext}^2(G,\mathcal{E}(-1))\cong S_3^{\oplus a}$, $\operatorname{Ext}^1(G,\mathcal{E}(-1))\cong S_1^{\oplus 2}$ and $\operatorname{Hom}(G,\mathcal{E}(-1))=0$. Lemma 2.1 then shows that $E_2^{-3,3}\cong \mathcal{O}(-1)^{\oplus r-4+a}$, that $E_2^{-3,2}\cong \mathcal{O}(-1)^{\oplus a}$, that $E_2^{-1,1}\cong \mathcal{S}(-1)^{\oplus 2}$ and that $E_2^{p,q}=0$ unless $(p,q)=(-3,3)$, $(-3,2)$ or $(-1,1)$. Then $\mathcal{E}(-1)$ fits in the $(-1)$-twist of the following exact sequence:

(13.3)\begin{equation} 0\to \mathcal{O}^{\oplus a} \to \mathcal{S}^{\oplus 2}\to \mathcal{E}\to \mathcal{O}^{\oplus r-4+a} \to 0. \end{equation}

This sequence splits into the following two exact sequences:

\begin{equation*} 0\to \mathcal{O}^{\oplus a} \to \mathcal{S}^{\oplus 2}\to \mathcal{F}\to 0;\end{equation*}

\begin{equation*}0\to \mathcal{F}\to \mathcal{E}\to \mathcal{O}^{\oplus r-4+a} \to 0,\end{equation*}

where $\mathcal{F}$ is a globally generated vector bundle of rank $4-a$. We claim here that $a\leq 1$. Indeed, if a = 2, then we have the following exact sequences:

\begin{equation*}0\to \mathcal{O} \to \mathcal{S}^{\oplus 2}\to \mathcal{G}\to 0;\end{equation*}

\begin{equation*}0\to \mathcal{O}\to \mathcal{G}\to \mathcal{F}\to 0,\end{equation*}

where $\mathcal{G}$ is a globally generated vector bundle of rank 3. Since $\mathcal{F}$ is a vector bundle, $\mathcal{G}$ must have a nowhere vanishing global section, and thus $c_3(\mathcal{G})=0$. On the other hand, $c_3(\mathcal{G})=c_3(\mathcal{S}^{\oplus 2})=2c_2(\mathcal{S})h=2$. This is a contradiction. Hence the case a = 2 does not arise. Now note that $\mathcal{E}$ is isomorphic to $\mathcal{F}\oplus \mathcal{O}^{\oplus r-4+a}$ since $h^1(\mathcal{F})=0$. Therefore, $\mathcal{E}$ fits in an exact sequence

\begin{equation*} 0\to \mathcal{O}^{\oplus a}\to \mathcal{S}^{\oplus 2}\oplus \mathcal{O}^{\oplus r-4+a} \to \mathcal{E}\to 0, \end{equation*}

where the composite of the inclusion $\mathcal{O}^{\oplus a}\to \mathcal{S}^{\oplus 2}\oplus \mathcal{O}^{\oplus r-4+a}$ and the projection $\mathcal{S}^{\oplus 2}\oplus \mathcal{O}^{\oplus r-4+a} \to \mathcal{O}^{\oplus r-4+a}$ is zero. This is Case (5) of Theorem 1.1.

13.2. Suppose that $(h^1(\mathcal{E}(-2)),c_3)=(1,0)$

Then $h^1(\mathcal{E}(-1))=1$. Hence $h^0(\mathcal{E})=h^0(\mathcal{E}|_{\mathbb{Q}^2})-1=r+3$. It follows from (13.2) that $h^1(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1))=4$. Set $a=h^0(\mathcal{S}^{\vee}\otimes \mathcal{E})$. Then the exact sequence (7.14) shows that $a\leq 4$, that $h^q(\mathcal{S}^{\vee}\otimes \mathcal{E})=0$ unless q = 0 or 1 and that $h^1(\mathcal{S}^{\vee}\otimes\mathcal{E})=a$. Hence we have $\operatorname{Ext}^q(G,\mathcal{E})=0$ for q = 2 and 3, and $\operatorname{Hom}(G,\mathcal{E})$ fits in an exact sequence

\begin{equation*}0\to S_0^{\oplus r+3}\to \operatorname{Hom}(G,\mathcal{E})\to S_1^{\oplus a}\to 0.\end{equation*}

Moreover, $\operatorname{Ext}^1(G,\mathcal{E})$ has a filtration $S_1^{\oplus a}\subset F\subset \operatorname{Ext}^1(G,\mathcal{E})$ of right A-modules such that the following sequences are exact:

\begin{equation*} 0\to F\to \operatorname{Ext}^1(G,\mathcal{E})\to S_3\to 0;\end{equation*}

\begin{equation*} 0\to S_1^{\oplus a}\to F\to S_2\to 0.\end{equation*}

Now the structures of right A-modules $\operatorname{Ext}^q(G,\mathcal{E})$’s are the same as those of $\operatorname{Ext}^q(G,\mathcal{E})$’s in § 12.2.2.1, and we conclude that $\mathcal{E}$ belongs to Case (9) of Theorem 1.1.

14. The case where $\mathcal{E}|_{\mathbb{Q}^2}$ belongs to Case (7) of Theorem 2.3

Suppose that $\mathcal{E}|_{\mathbb{Q}^2}$ fits in the following exact sequence:

\begin{equation*} 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}|_{\mathbb{Q}^2}\to 0. \end{equation*}

Then $c_2h=5$. It then follows from (6.1) that $c_3\geq 4$. Note that

(14.1)\begin{equation} h^q(\mathcal{S}^{\vee}\otimes \mathcal{E}|_{\mathbb{Q}^2}) = \begin{cases} 2& \textrm{if}\quad q= 0\\ 0& \textrm{if}\quad q\neq 0, \end{cases} \end{equation}

and that

(14.2)\begin{equation} h^q(\mathcal{E}(-1)|_{\mathbb{Q}^2}) = \begin{cases} 1& \textrm{if}\quad q= 1\\ 0& \textrm{if}\quad q\neq 1. \end{cases} \end{equation}

Hence we have

\begin{equation*}h^0(\mathcal{E}(-1))=0\end{equation*}

by (7.7). Note that $H^q(\mathcal{E}|_{\mathbb{Q}^2})$ vanishes for any q > 0. Since $h^q(\mathcal{E})=0$ for any q > 0 by (7.1), we have $h^q(\mathcal{E}(-1))=0$ for any $q\geq 2$. Hence it follows from (4.4) that

\begin{equation*}0\geq -h^1(\mathcal{E}(-1))=\chi(\mathcal{E}(-1))=-\dfrac{5}{2}+\dfrac{1}{2}c_3 \geq -\dfrac{1}{2}.\end{equation*}

Therefore $c_3=5$ and $h^1(\mathcal{E}(-1))=0$. Now it follows from (7.14) that $h^q(\mathcal{S}^{\vee}\otimes\mathcal{E})=h^{q+1}(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1))$ for any q. In particular, $h^0(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1))=0$. Moreover $h^{q+1}(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =h^{q+1}(\mathcal{S}^{\vee}\otimes\mathcal{E})$ for $q\geq 1$ by (14.1) and (7.15). Hence $h^q(\mathcal{S}^{\vee}\otimes\mathcal{E})=0$ for $q\geq 1$ and $h^{q}(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1))=0$ for $q\geq 2$. Therefore

\begin{equation*} -h^1(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =\chi(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =-6+c_3=-1 \end{equation*}

by (4.8). Thus $h^0(\mathcal{S}^{\vee}\otimes \mathcal{E})=1$. We apply to $\mathcal{E}$ the Bondal spectral sequence (2.1). From (14.2), it follows that $h^q(\mathcal{E}(-2))=0$ unless q = 2 and that $h^2(\mathcal{E}(-2))=1$. Since $h^0(\mathcal{E}|_{\mathbb{Q}^2})=r+3$, we see that $h^0(\mathcal{E})=r+3$. Hence we have an exact sequence

\begin{equation*} 0\to S_0^{\oplus r+3}\to \operatorname{Hom}(G,\mathcal{E})\to S_1\to 0, \end{equation*}

and the following: $\operatorname{Ext}^q(G,\mathcal{E}) = 0$ for $q=1,3$; $\operatorname{Ext}^2(G,\mathcal{E})\cong S_3$. Therefore, Lemma 2.1 implies that $E_2^{p,q}=0$ unless $(p,q)=(-3,2)$ or $(0,0)$, that $E_2^{-3,2} \cong \mathcal{O}(-1)$, and that there is the following exact sequence:

\begin{equation*} 0\to \mathcal{S}(-1)\to \mathcal{O}^{\oplus r+3} \to E_2^{0,0}\to 0.\end{equation*}

Note that we have the following exact sequence:

\begin{equation*}0\to E_2^{-3,2}\to E_2^{0,0}\to \mathcal{E}\to 0.\end{equation*}

Since $\operatorname{Ext}^1(\mathcal{O}(-1),\mathcal{S}(-1))=0$, this implies that $\mathcal{E}$ fits in the following exact sequence:

\begin{equation*} 0\to \mathcal{S}(-1)\oplus \mathcal{O}(-1) \to \mathcal{O}^{\oplus r+3}\to \mathcal{E}\to 0. \end{equation*}

This is Case (6) of Theorem 1.1.

15. The case where $\mathcal{E}|_{\mathbb{Q}^2}$ belongs to Case (8) of Theorem 2.3

Suppose that $\mathcal{E}|_{\mathbb{Q}^2}$ fits in the following exact sequence:

\begin{equation*} 0\to \mathcal{O}(-1,-2) \to \mathcal{O}(1,0)\oplus \mathcal{O}^{\oplus r}\to \mathcal{E}|_{\mathbb{Q}^2}\to 0. \end{equation*}

Then $c_2h=6$. It then follows from (6.1) that $c_3\geq 8$. Note that

(15.1)\begin{equation} h^q(\mathcal{E}(-1)|_{\mathbb{Q}^2}) = \begin{cases} 2& \textrm{if}\quad q= 1\\ 0& \textrm{if}\quad q\neq 1, \end{cases} \end{equation}

and that

(15.2)\begin{equation} h^q(\mathcal{S}^{\vee}\otimes \mathcal{E}|_{\mathbb{Q}^2}) = \begin{cases} 1& \textrm{if}\quad q=0, 1\\ 0& \textrm{if}\quad q\neq 0, 1. \end{cases} \end{equation}

Hence we have

\begin{equation*}h^0(\mathcal{E}(-1))=0\end{equation*}

by (7.7). Note that $H^q(\mathcal{E}|_{\mathbb{Q}^2})$ vanishes for all q > 0. Since $h^q(\mathcal{E})=0$ for all q > 0 by (7.1), we have $h^q(\mathcal{E}(-1))=0$ for all $q\geq 2$. It follows from (4.4) that

\begin{equation*}0\geq -h^1(\mathcal{E}(-1))=\chi(\mathcal{E}(-1))=-4+\dfrac{1}{2}c_3 \geq 0.\end{equation*}

Therefore $c_3=8$ and $h^1(\mathcal{E}(-1))=0$. Now it follows from (7.14) that $h^q(\mathcal{S}^{\vee}\otimes\mathcal{E})=h^{q+1}(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1))$ for any q. In particular, $h^0(\mathcal{S}^{\vee}\otimes \mathcal{E}(-1))=0$. Moreover $h^{q+1}(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =h^{q+1}(\mathcal{S}^{\vee}\otimes\mathcal{E})$ for $q\geq 2$ by (7.15) and (15.2). Hence $h^q(\mathcal{S}^{\vee}\otimes\mathcal{E})=0$ for $q\geq 2$ and $h^{3}(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1))=0$. Hence

\begin{equation*} -h^1(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1))+ h^2(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =\chi(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =-8+c_3=0 \end{equation*}

by (4.8). Set $a=h^0(\mathcal{S}^{\vee}\otimes\mathcal{E})$. Then $a=h^1(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =h^2(\mathcal{S}^{\vee}\otimes\mathcal{E}(-1)) =h^1(\mathcal{S}^{\vee}\otimes\mathcal{E})$. We see that a = 1 by (7.15) and (15.2). We apply to $\mathcal{E}$ the Bondal spectral sequence (2.1). It follows from (15.1) that $h^q(\mathcal{E}(-2))$ vanishes unless q = 2 and that $h^2(\mathcal{E}(-2))=2$. Since $h^0(\mathcal{E}|_{\mathbb{Q}^2})=r+2$, we see that $h^0(\mathcal{E})=r+2$. Therefore, $\operatorname{Ext}^3(G,\mathcal{E})=0$, $\operatorname{Ext}^2(G,\mathcal{E})\cong S_3^{\oplus 2}$, $\operatorname{Ext}^1(G,\mathcal{E}) \cong S_1$ and $\operatorname{Hom}(G,\mathcal{E})$ fits in the following exact sequence:

\begin{equation*} 0\to S_0^{\oplus r+2}\to \operatorname{Hom}(G,\mathcal{E})\to S_1\to 0. \end{equation*}

Therefore, Lemma 2.1 implies that $E_2^{p,q}=0$ unless $(p,q)=(-3,2)$, $(-1,1)$, $(-1,0)$ or $(0,0)$, that $E_2^{-3,2}\cong \mathcal{O}(-1)^{\oplus 2}$, that $E_2^{-1,1}\cong \mathcal{S}(-1)$ and that there exists the following exact sequence:

\begin{equation*}0\to E_2^{-1,0}\to \mathcal{S}(-1)\to \mathcal{O}^{\oplus r+2}\to E_2^{0,0}\to 0.\end{equation*}

The Bondal spectral sequence implies that $E_2^{-1,0}=0$, that $E_2^{0,0}\cong E_3^{0,0}$ and that we have the following exact sequences:

\begin{equation*} 0\to E_3^{-3,2}\to \mathcal{O}(-1)^{\oplus 2}\xrightarrow{\varphi} \mathcal{S}(-1)\to E_3^{-1,1}\to 0; \end{equation*}

\begin{equation*}0\to E_3^{-3,2}\to E_3^{0,0}\to E_4^{0,0}\to 0;\end{equation*}

\begin{equation*} 0\to E_4^{0,0}\to \mathcal{E}\to E_3^{-1,1}\to 0. \end{equation*}

Since $\mathcal{E}$ is nef, $E_3^{-1,1}$ cannot admit a negative degree quotient. Hence φ ≠ 0. Thus, there exists an inclusion $\iota:\mathcal{O}(-1)\to \mathcal{O}(-1)^{\oplus 2}$ such that $\varphi\circ \iota\neq 0$. Now we have a morphism $\bar{\varphi}:\mathcal{O}(-1)\cong\operatorname{Coker}(\iota)\to \operatorname{Coker}(\varphi\circ\iota)\cong \mathcal{I}_L$ for some line L in $\mathbb{Q}^3$ and $\bar{\varphi}$ fits in the following exact sequence:

\begin{equation*} 0\to E_3^{-3,2}\to \mathcal{O}(-1)\xrightarrow{\bar{\varphi}} \mathcal{I}_L\to E_3^{-1,1}\to 0. \end{equation*}

This shows that $E_3^{-1,1}|_M$ admits a negative degree quotient for some line M in $\mathbb{Q}^3$. This is a contradiction. Therefore, $\mathcal{E}|_{\mathbb{Q}^2}$ cannot belong to Case (8) of Theorem 2.3.

16. The case where $\mathcal{E}|_{\mathbb{Q}^2}$ belongs to Case (9) of Theorem 2.3

Suppose that $\mathcal{E}|_{\mathbb{Q}^2}$ fits in the following exact sequence:

\begin{equation*} 0\to \mathcal{O}(-1,-1)^{\oplus 2} \to \mathcal{O}^{\oplus r+2}\to \mathcal{E}|_{\mathbb{Q}^2}\to 0. \end{equation*}

Then $c_2h=6$. It then follows from (6.1) that $c_3\geq 8$. Note that

(16.1)\begin{equation} h^q(\mathcal{E}(-1)|_{\mathbb{Q}^2}) = \begin{cases} 2& \textrm{if}\quad q= 1\\ 0& \textrm{if}\quad q\neq 1, \end{cases} \end{equation}

and that

(16.2)\begin{equation} h^q(\mathcal{S}^{\vee}\otimes \mathcal{E}|_{\mathbb{Q}^2}) =0 \,\textrm{for all } q. \end{equation}

Hence we have

\begin{equation*}h^0(\mathcal{E}(-1))=0\end{equation*}

by (7.7). Note that $H^q(\mathcal{E}|_{\mathbb{Q}^2})$ vanishes for all q > 0. Since $h^q(\mathcal{E})=0$ for all q > 0 by (7.1), we have $h^q(\mathcal{E}(-1))=0$ for all $q\geq 2$. It follows from (4.4) that