Proof. Let us write $c_1(V)=\frac {d}{k}Q+\alpha $ , where $\alpha \in \langle Q\rangle ^\perp _{\mathbb {Q}}\subset \operatorname {Pic}(X)_{\mathbb {Q}}$ . The desired inequality is obtained by plugging this into (2.1) and using the fact that $\alpha ^2\le 0$ by the Hodge index theorem.

Proof. In characteristic zero this was proved by Kuleshov-Orlov [Reference Kuleshov and Orlov15, Prop. 2.9, 2.10]. Below we review the steps of the proof to check that the argument carries over to arbitrary characteristic. For coherent sheaves $F,G$ we set $h^i(F,G):=\dim \operatorname {Ext}^i(F,G)$ ; $TF$ denotes the maximal torsion subsheaf in F; $T^0F\subset TF$ is the maximal torsion subsheaf with $0$ -dimensional support.

Step 1. For a torsion-free coherent sheaf F on a smooth projective surface S one has

(2.4) $$ \begin{align} h^1(F,F)\ge h^1(F^{\vee\vee},F^{\vee\vee})+h^1(F^{\vee\vee}/F,F^{\vee\vee}/F). \end{align} $$

For a nonzero coherent sheaf M with $0$ -dimensional support (such as $F^{\vee \vee }/F$ above) on S, one has

(2.5) $$ \begin{align} h^1(M,M)>0. \end{align} $$

These statements are proved in [Reference Mukai20] for sheaves on a symplectic surface over $\mathbb {C}$ , however, the arguments work for arbitrary S and in any characteristic. Namely, the first assertion follows from the spectral sequence converging to $R\operatorname {End}(F)$ coming from the resolution $[F^{\vee \vee }\to F^{\vee \vee }/F]$ of F. Since $F^{\vee \vee }$ is locally free and $F^{\vee \vee }/F$ has $0$ -dimensional support, we have

$$ \begin{align*}\operatorname{Ext}^1(F^{\vee\vee},F^{\vee\vee}/F)=\operatorname{Ext}^1(F^{\vee\vee}/F,F^{\vee\vee})=0,\end{align*} $$

and thus the associated filtration on $\operatorname {Ext}^1(F,F)$ has middle subquotient

$$ \begin{align*}\operatorname{Ext}^1(F^{\vee\vee},F^{\vee\vee})\oplus \operatorname{Ext}^1(F^{\vee\vee}/F,F^{\vee\vee}/F),\end{align*} $$

giving inequality (2.4).

The inequality (2.5) is derived as follows: first, we note that $\chi (M,M)=0$ and by Serre duality, $h^0(M,M)=h^2(M,M)$ . Hence, $h^1(M,M)=2h^0(M,M)$ , which is positive for nonzero M.

Step 2. For an exact sequence $0\to G_2\to E\to G_1\to 0$ of coherent sheaves on a smooth projective surface, such that $h^0(G_2,G_1)=h^2(G_1,G_2)=0$ , one has

$$ \begin{align*}h^1(E,E)\ge h^1(G_1,G_1)+h^1(G_2,G_2).\end{align*} $$

If in addition $h^1(E,E)=0$ then one has

$$ \begin{align*}h^0(E,E)=h^0(G_1,G_1)+h^0(G_2,G_2)+\chi(G_1,G_2), \ \ h^2(E,E)=h^2(G_1,G_1)+h^2(G_2,G_2)+\chi(G_2,G_1).\end{align*} $$

This is [Reference Kuleshov and Orlov15, Lem. 2.2(2)]. The proof is a homological algebra argument valid in any characteristic.

Step 3. Let G be a coherent sheaf on X with $h^1(G,G)=0$ . Then

1. (1) $T^0G=0$ , $h^1(TG,TG)=h^1(G/TG,G/TG)=0$ ;

2. (2) $h^0(G,G)>h^2(G,G)$ . Furthermore, for any coherent sheaf F on X one has $h^0(F,G)\ge h^2(G,F)$ .

In the case $T^0G=0$ , the proof of (2) is given in [Reference Kuleshov and Orlov15, Lem. 2.4, Rem. 2.5]. The proof of (1) is given in [Reference Kuleshov and Orlov15, Cor. 2.7], based on Step 2 and the case $T^0G=0$ of (2). Note that to check that $T^0G=0$ , one applies Step 2 to the sequence $0\to T^0G\to TG\to TG/T^0G\to 0$ and deduces that $h^1(T^0G,T^0G)=0$ , which implies the vanishing of $T^0G$ by Step 1 (see (2.5)).Footnote ² All these proofs work in arbitrary characteristic.

Step 4. Let F be an exceptional coherent sheaf on X. Then F is either locally free or $F\simeq {\mathcal {O}}_C(d)$ , where C is a $(-1)$ -curve, $d\in {\mathbb {Z}}$ .

The proofs given in [Reference Kuleshov and Orlov15, Lem. 2.8, Prop. 2.9], based on Steps 1, 2 and 3, are valid in arbitrary characteristic.

Step 5. Let A be an exceptional object in $D^b(\operatorname {Coh} X)$ . Then $A\simeq F[n]$ , where F is a coherent sheaf.

The proof in [Reference Kuleshov and Orlov15, Prop. 2.10], based on Step 3(2), is valid in arbitrary characteristic.

Proof. (i) The fact that $V|_C$ is simple follows immediately by applying the functors $\operatorname {Ext}^*(V,?)$ to the exact sequence $0\to V(-C)\to V\to V|_C\to 0$ and observing that $\operatorname {Ext}^1(V,V(-C))\simeq \operatorname {Ext}^1(V,V)^*=0$ by Serre duality (see [Reference Kuleshov and Orlov15, Lem. 3.6]).

To prove the second statement we note that a smooth anticanonical divisor $C\subset X$ always exists (see [Reference Demazure7, Thm. 1]). Then $\deg (V)=\deg (V|_C)$ and $\operatorname {rk}(V)=\operatorname {rk}(V|_C)$ , and we use the well known fact that the degree and rank of a simple vector bundle $V|_C$ on an elliptic curve are relatively prime.

(ii) Let us choose a smooth anticanonical divisor $C\subset X$ (it exists by [Reference Demazure7, Thm. 1]). By (i), $V|_C$ is a simple bundle on an elliptic curve. It is well known that such a bundle is stable. Now suppose $0\to V_1\to V\to V_2\to 0$ is a destabilising sequence with $V_2$ torsion free. Then the sequence $0\to V_1|_C\to V|_C\to V_2|_C\to 0$ is exact, so it destablises $V|_C$ , a contradiction. Thus, V is semistable. Since the degree and rank of V are relatively prime (by part (i)), we deduce that V is stable. Note that in characteristic zero, stability of V was proved in [Reference Gorodentsev12].

The fact that V is determined by its class in $\mathrm {K}_0$ follows from stability (see [Reference Gorodentsev12, Cor. 2.5]). Indeed, if $[V_1]=[V_2]$ then $\chi (V_1,V_2)=\chi (V_1,V_1)=1$ . By stability, $h^2(V_1,V_2)=h^0(V_2,V_1(-Q))=0$ . Therefore, $h^0(V_1,V_2)>0$ , so there is a nonzero morphism $V_1\to V_2$ which must be an isomorphism by stability.

The class in $\mathrm {K}_0$ is determined by $\operatorname {rk}(V)$ , $c_1(V)$ and $\chi (V)$ . If $H^*(X,V)=0$ then $\chi (V)=0$ , so V is determined by $\operatorname {rk}(V)$ and $c_1(V)$ .

(iii) See [Reference Kuleshov and Orlov15, Cor. 3.2]. The proof is valid in arbitrary characteristic.

Proof. Let $\alpha :\operatorname {Pic}(X)\to \operatorname {Pic}(X)$ be such an isometry. Since $\alpha $ preserves the intersection number with Q, we see that $E^{\prime }_1=\alpha (E_1),\ldots ,E^{\prime }_4=\alpha (E_4)$ are classes of disjoint $(-1)$ -curves (see [Reference Dolgachev8, Sec. 8.2.6]), therefore, they can be contracted. Let $X'$ denote the corresponding blown down surface. Then $X'$ is a del Pezzo surface of degree $8$ , hence, either $X'\simeq {\mathbb P}^1\times {\mathbb P}^1$ or $X'\simeq F_1$ , the first Hirzebruch surface. The classes $s'=\alpha (s)$ and $f'=\alpha (f)$ form a basis of $\operatorname {Pic}(X')$ , with $(s')^2=(f')^2=0$ , $s'\cdot f'=1$ . Thus, the intersection form on $\operatorname {Pic}(X')$ is even, so $X'\simeq {\mathbb P}^1\times {\mathbb P}^1$ , and $s'$ and $f'$ are the classes of the rulings (since we know their (self-)intersections as well as their intersections with Q).

Proof. This follows from the above calculation of the endomorphism algebra of (3.5) and from Bondal’s theorem that relates the category generated by a strong exceptional collections with the derived category of modules over its endomorphism algebra (see [Reference Bondal1]).

Proof. (i) Since the classes of objects of the exceptional collection (3.5) freely generate $\mathrm {K}_0({\mathcal C})$ , the Euler form $\chi $ is determined by the condition that $\chi (E,E)=1$ for each object E of (3.5) and by

(3.12) $$ \begin{align} \chi({\mathcal{O}}_{\mathcal C}(-1),{\mathcal{O}}_{\mathcal C})=2, \ \chi({\mathcal{O}}_{\mathcal C}(-1),S_i)=\chi({\mathcal{O}}_{\mathcal C},S_i)=1, \ \chi(S_i,S_j)=0 \end{align} $$

with other values on objects of (3.5) vanishing by semiorthogonality. Note that equalities (3.12) follow from [Reference Meltzer19, Prop. 3.2.5(i)]).

Thus, for $\chi ^-(A,B)=\chi (A,B)-\chi (B,A)$ we have

$$ \begin{align*} &\chi^-(a_1[{\mathcal{O}}_{\mathcal C}(-1)]+a_0[{\mathcal{O}}_{\mathcal C}]+\sum_i b_i [S_i],a^{\prime}_1[{\mathcal{O}}_{\mathcal C}(-1)]+a^{\prime}_0[{\mathcal{O}}_{\mathcal C}]+\sum_i b^{\prime}_i [S_i])\\ &=2(a_1a^{\prime}_0-a^{\prime}_1a_0)+\sum_i(a_0+a_1)b^{\prime}_i-\sum_i (a^{\prime}_0+a^{\prime}_1)b_i. \end{align*} $$

On the other hand,

$$ \begin{align*}\operatorname{rk}(a_1[{\mathcal{O}}_{\mathcal C}(-1)]+a_0[{\mathcal{O}}_{\mathcal C}]+\sum_i b_i [S_i])=a_0+a_1, \ \ \deg(a_1[{\mathcal{O}}_{\mathcal C}(-1)]+a_0[{\mathcal{O}}_{\mathcal C}]+\sum_i b_i [S_i])=-2a_1+\sum_i b_i.\end{align*} $$

This immediately shows that the kernel $\Lambda _n$ of $\chi ^-(\cdot ,\cdot )$ coincides with $\ker (\operatorname {rk},\deg )\subset \mathrm {K}_0({\mathcal C})$ . Formulas (3.12) also show that the classes $\varepsilon _i\in (\Lambda _n)_{\mathbb {Q}}$ given by (3.6) are orthonormal with respect to $\chi (\cdot ,\cdot )$ .

Furthermore, $\Lambda _n=\ker (\operatorname {rk},\deg )$ is generated by the classes $[S_i]-[S_j]$ and $[{\mathcal {O}}_{\mathcal C}(-1)]-[{\mathcal {O}}_{\mathcal C}]+2[S_i]$ , which are given in terms of the basis $(\varepsilon _i)$ as

$$ \begin{align*}[S_i]-[S_j]=\varepsilon_i-\varepsilon_j, \ \ [{\mathcal{O}}_{\mathcal C}(-1)]-[{\mathcal{O}}_{\mathcal C}]+2[S_i]=2\varepsilon_i,\end{align*} $$

so $\Lambda _n$ coincides with the root lattice $\mathrm {Q}(D_n)$ .

(ii) Using the exceptional collection (3.5), we see that the classes (3.7) form a basis of $\mathrm {K}_0({\mathcal C})$ . Also, one immediately checks that the first two classes are orthogonal to $\Lambda _n$ with respect to $\chi (\cdot ,\cdot )$ . Since the restriction of $\chi (\cdot ,\cdot )$ to $\Lambda _n$ is nondegenerate, this gives the required decomposition $\mathrm {K}_0({\mathcal C})_{\mathbb {Q}}=M\oplus (\Lambda _n)_{\mathbb {Q}}$ . The formulas for $\pi [{\mathcal {O}}_{\mathcal C}]$ , $\pi [{\mathcal {O}}_{\mathcal C}(-1)]$ and $\pi [S_i]$ are straightforward. The formula for $\pi [S^{\prime }_i]$ follows from the equality $[S_i]+[S^{\prime }_i]=[{\mathcal {O}}_{\mathcal C}]-[{\mathcal {O}}_{\mathcal C}(-1)]$ that follows from (3.4).

(iii) Formula (3.9) can be checked on the generating classes $[{\mathcal {O}}_{\mathcal C}(-1)]$ , $[{\mathcal {O}}_{\mathcal C}]$ , $([S_i])$ .

Proof. Let $\pi (v)=\sum _{i=1}^4 y_i\varepsilon _i$ . From formula (3.10) with $n=4$ , we immediately see that $\chi (v,v)=1$ if and only if $\sum _i y_i^2=1$ . Furthermore, by (3.11), one has $\chi (v,[S_i])=r/2+y_i$ , which implies that $y_i\equiv r/2\ \mod {\mathbb {Z}}$ . On the other hand, this formula gives $\chi ([{\mathcal {O}}_{\mathcal C}],v)=d/2+\sum _i y_i/2$ . Hence, $\sum _i y_i\equiv d\ \mod 2{\mathbb {Z}}$ , which shows that $\pi (v)$ is in $B(r,d)$ .

Since $\pi $ restricts to an isomorphism on $\ker (\operatorname {rk},\deg )=\Lambda _n$ (see Lemma 3.3(i),(ii)), a class $v\in \mathrm {K}_0({\mathcal C})$ is determined by $\operatorname {rk}(v)$ , $\deg (v)$ and $\pi (v)$ . Hence, the map (3.13) is injective. Since $|B(r,d)|=8$ for coprime $(r,d)$ , it remains to show that for every coprime $(r,d)$ with $r>0$ , there exist exactly $8$ isomorphism classes of exceptional sheaves V on ${\mathcal C}$ (with distinct classes in $\mathrm {K}_0({\mathcal C})$ ) such that $\operatorname {rk}(V)=r$ and $\deg (V)=d$ .

To this end we use the fact that for any slope $\mu $ there exists an autoequivalence $\Phi $ of $D^b(\operatorname {Coh} {\mathcal C}_4)$ taking indecomposable bundles of slope $\mu $ to indecomposable bundles of slope $0$ (see [Reference Lenzing and Meltzer18, Thm. 6.3]). Therefore, it is enough to check our assertion for slope $0$ , i.e., to show that there are $8$ isomorphism classes of line bundles of degree $0$ on ${\mathcal C}$ . But this is clear from the identification of $\operatorname {Pic}({\mathcal C})$ with the abelian group L (see (3.1)): the subgroup $\ker (\deg )\subset L$ is isomorphic to $({\mathbb {Z}}/2)^3$ .

Proof. See Appendix A.

Proof. Let $v:=\pi [V]\in \mathrm {P}(D_5)$ . Using formula (3.9), the condition $\chi (V,V)=1$ can be rewritten as

$$ \begin{align*}(v,v)=r^2/4+1.\end{align*} $$

On the other hand, stability of V (see [Reference Meltzer19, Prop. 3.2.7]) implies that for any line bundle L of degree 0 one has $\operatorname {Hom}(L,V)=0$ , hence, $\chi (L,V)\le 0$ . Taking L to be either ${\mathcal {O}}$ or of the form ${\mathcal {O}}(x_i-x_j)$ or ${\mathcal {O}}(x_i+x_j-x_k-x_l)$ and applying formula (3.9) (taking into account (3.8)), we get the inequality

(3.14) $$ \begin{align} (\omega, v)\le \frac{r}{4}-\frac{d}{2} \end{align} $$

for every $\omega $ in the $W(D_5)$ -orbit of $\omega _0:=\sum _{i=1}^5 \varepsilon _i/2$ . Let $w\in W(D_5)$ be an element such that $v_0:=w\cdot v$ is in the closure of the fundamental chamber of $D_5$ . Since our system of inequalities (3.14) is $W(D_5)$ -invariant, the same system of inequalities holds for $v_0=\sum _i y_i\varepsilon _i$ . Using the obtained inequality for $(\omega _0,v_0)=\sum y_i/2$ , we get that

(3.15) $$ \begin{align} \sum_{i=1}^5 y_i\le \frac{r}{2}-d<\frac{r}{2}+\sqrt{r}-1, \end{align} $$

where the second inequality follows from the assumption that $r>(-d+1)^2$ . Thus, the assumptions of Lemma 3.7 hold for $v_0$ . Since the inequality here is strict, cases (3) and (4) of this lemma do not arise, and thus we are left with cases (1), (2), and (5). In the first two cases, the fact that $y_1=r/2$ implies that one of the coordinates of $v=\pi (V)$ is equal to $\pm r/2$ . Then, by formula (3.11), for some i either $\chi (V,S_i)=0$ of $\chi (V,S^{\prime }_i)=0$ .

In case (5) we find that $r=11$ and inequalities (3.15) become

$$\begin{align*}y_1+y_2+y_3+y_4+y_5=15/2\le 11/2-d<9/2+\sqrt{11}, \end{align*}$$

forcing $-d=2$ .

Proof. Since X is an intersection of two quadrics in ${\mathbb P}^4$ , in characteristic $\neq 2$ the existence of an equivalence $\Phi $ of $\langle {\mathcal {O}}_X\rangle ^\perp $ with the derived category of a weighted projective line ${\mathcal C}_5$ follows from [Reference Kuznetsov16, Cor. 5.7]. We give a different proof below which works in any characteristic and constructs an equivalence $\Phi $ matching specific exceptional collections in both categories.

Step 1. The full exceptional collection on X (see [Reference Orlov22]),

$$ \begin{align*}({\mathcal{O}}_{e_i}(-1)[-1])_{i=1,\ldots,5},{\mathcal{O}}_X(-2h),{\mathcal{O}}_X(-h),{\mathcal{O}}_X\end{align*} $$

gives a full exceptional collection in $\langle {\mathcal {O}}_X\rangle ^\perp $ (by deleting the last object). Since $\operatorname {Hom}^*({\mathcal {O}}_{e_i}(-1), {\mathcal {O}}_X(-2h+\sum _i e_i))=0$ (e.g., by Serre duality), the exact sequence

(4.2) $$ \begin{align} 0\to {\mathcal{O}}_X(-2h)\to {\mathcal{O}}_X(-2h+\sum_i e_i)\to \bigoplus_i {\mathcal{O}}_{e_i}(-1)\to 0 \end{align} $$

shows that the left mutation of ${\mathcal {O}}_X(-2h)$ through the subcategory $\langle {\mathcal {O}}_{e_i}(-1)[-1], i=1,\ldots ,5\rangle $ is ${\mathcal {O}}_X(-2h+\sum _i e_i)[-1]$ . Hence, we get a full exceptional collection

$$ \begin{align*}{\mathcal{O}}_X(-2h+\sum_i e_i)[-1], ({\mathcal{O}}_{e_i}(-1)[-1])_{i=1,\ldots,5}, {\mathcal{O}}_X(-h)\end{align*} $$

in $\langle {\mathcal {O}}_X\rangle ^\perp $ . Next, since $\operatorname {Hom}^*({\mathcal {O}}_X(e_i-h),{\mathcal {O}}_X(-h))=0$ , the exact sequence

$$ \begin{align*}0\to {\mathcal{O}}_X(-h)\to {\mathcal{O}}_X(e_i-h)\to {\mathcal{O}}_{e_i}(-1)\to 0\end{align*} $$

shows that the right mutation of ${\mathcal {O}}_{e_i}(-1)[-1]$ through ${\mathcal {O}}_X(-h)$ is ${\mathcal {O}}_X(e_i-h)$ . Thus, we get an exceptional collection

(4.3) $$ \begin{align} {\mathcal{O}}_X(-2h+\sum_i e_i)[-1], {\mathcal{O}}_X(-h), ({\mathcal{O}}_X(e_i-h))_{i=1,\ldots,5}. \end{align} $$

Step 2. Let us show that collection (4.3) is strong. Let us set $L:={\mathcal {O}}_X(-h+\sum _i e_i)$ . It is convenient to tensor our collection with ${\mathcal {O}}_X(h)$ to get the collection

(4.4) $$ \begin{align} L[-1], {\mathcal{O}}_X, ({\mathcal{O}}_X(e_i))_{i=1,\ldots,5}. \end{align} $$

Since $\deg (L)=2$ , $\deg ({\mathcal {O}}_X)=0$ and $\deg ({\mathcal {O}}_X(e_i))=1$ , we have

$$ \begin{align*}\operatorname{Hom}(L,{\mathcal{O}}_X)=\operatorname{Hom}(L,{\mathcal{O}}_X(e_i))=\operatorname{Hom}({\mathcal{O}}_X,L(-Q))=\operatorname{Hom}({\mathcal{O}}_X(e_i),L(-Q))=0,\end{align*} $$

which by Serre duality implies that

$$ \begin{align*}\operatorname{Ext}^2(L,{\mathcal{O}}_X)=\operatorname{Ext}^2(L,{\mathcal{O}}_X(e_i))=0.\end{align*} $$

We also have $\operatorname {Ext}^{>0}({\mathcal {O}}_X,{\mathcal {O}}_X(e_i))=H^{>0}(X,{\mathcal {O}}_X(e_i))=0$ (this follows from the vanishing of cohomology of ${\mathcal {O}}_X(e_i)|_{e_i}$ ).

Step 3. Next, let us compute the endomorphism algebra of collection (4.3), or equivalently of (4.4). Since $\operatorname {Hom}({\mathcal {O}}_X,{\mathcal {O}}_X(e_i))$ is $1$ -dimensional, we just need to calculate the natural maps

(4.5) $$ \begin{align} H^1(X,L^{-1})\to H^1(X,L^{-1}(e_i)). \end{align} $$

Let $C\subset X$ denote the strict transform of the unique conic D passing through the point $p_1,\ldots ,p_5$ . Consider the commutative diagram

where the left horizontal arrows are given by C and the vertical arrows are given by $e_i$ . It follows that the restriction maps $H^1(X,L^{-1})\to H^1(C,{\mathcal {O}}_C(-3))$ and $H^1(X,L^{-1}(e_i))\to H^1(C,{\mathcal {O}}_C(-2))$ are isomorphisms, so the map (4.5) coincides with the morphism $H^1(C,{\mathcal {O}}_C(-3))\to H^1(C,{\mathcal {O}}_C(-2))$ given by the point $C\cap e_i$ .

Step 4. Since $H^1(C,{\mathcal {O}}_C(-3))$ is 2-dimensional, and the points $C\cap e_i$ on C map to the corresponding points $p_i$ on $D\simeq {\mathbb P}^1$ , by Lemma 3.2, there exists an equivalence

where ${\mathcal C}={\mathcal C}_5$ is associated with the configuration $(p_i)$ on D, identifying the exceptional collection (4.3) with the image of

$$ \begin{align*}{\mathcal{O}}_{\mathcal C}(-1), {\mathcal{O}}_{\mathcal C}, (S_i)_{i=1,\ldots,5}.\end{align*} $$

We claim that

$$ \begin{align*}\widetilde{\Phi}(S^{\prime}_i)\simeq {\mathcal{O}}_X(-Q+h-e_i).\end{align*} $$

First, we can check that the left mutation of $S_i$ through $\langle {\mathcal {O}}_{\mathcal C}(-1),{\mathcal {O}}_{\mathcal C}\rangle $ is isomorphic to $S^{\prime }_i[-1]$ . Namely, combining the exact sequence (3.2) with (3.3) tensored by ${\mathcal {O}}_{{\mathcal C}}(-1)$ , we get an exact triangle

$$ \begin{align*}S^{\prime}_i[1]\to F\to S_i\to\ldots\end{align*} $$

with $F\in \langle {\mathcal {O}}_{{\mathcal C}}(-1),{\mathcal {O}}_{{\mathcal C}}\rangle $ . On the other hand, the left mutation of ${\mathcal {O}}_X(e_i-h)$ through ${\mathcal {O}}_X(-h)$ is ${\mathcal {O}}_{e_i}(-1)[-1]$ (see Step 1), and the exact sequence

$$ \begin{align*}0\to {\mathcal{O}}_X(-Q+h-e_i)\to {\mathcal{O}}_X(-2h+\sum_j e_j)\to {\mathcal{O}}_{e_i}(-1)\to 0\end{align*} $$

together with the vanishings $\operatorname {Hom}^*({\mathcal {O}}_X(-2h+\sum _j e_j),{\mathcal {O}}_X(-Q+h-e_i))= H^*(X,{\mathcal {O}}_X(-e_i))=0$ show that the left mutation of ${\mathcal {O}}_{e_i}(-1)[-1]$ through ${\mathcal {O}}_X(-2h+\sum _j e_j)[-1]$ is ${\mathcal {O}}_X(-Q+h-e_i)[-1]$ . Since $\widetilde {\Phi }$ is compatible with mutations, this implies our claim.

Step 5. To obtain the equivalence $\Phi $ we compose $\widetilde {\Phi }$ with the autoequivalence

$$ \begin{align*}F\mapsto F\otimes {\mathcal{O}}_{\mathcal C}(2)(-\sum x_i)[1]\end{align*} $$

of $D^b(\operatorname {Coh} {\mathcal C})$ . Then $\Phi $ identifies the exceptional collection (4.3) in $\langle {\mathcal {O}}_X\rangle ^\perp $ with the collection

$$ \begin{align*}{\mathcal{O}}_{\mathcal C}(-3)(\sum x_i), {\mathcal{O}}_{\mathcal C}(-2)(\sum x_i), (S^{\prime}_i)_{i=1,\ldots,5}\end{align*} $$

on ${\mathcal C}$ , and sends $S_i$ to ${\mathcal {O}}_X(-Q+h-e_i)[1]$ .

Since the classes of the objects of our exceptional collections generate the corresponding Grothendieck groups, the assertions concerning the induced map on $\mathrm {K}_0$ follow, including the relation between the ranks and degrees.

Proof. (i) Viewing X as the blow up of ${\mathbb P}^1\times {\mathbb P}^1$ at $4$ points $q_1,\ldots ,q_4$ , gives the standard full exceptional collection in $D^b(\operatorname {Coh} X)$ ,

$$ \begin{align*}({\mathcal{O}}_{E_i}(-1)[-1])_{i=1,\ldots,4}, {\mathcal{O}}_X(-s-f), {\mathcal{O}}_X(-s), {\mathcal{O}}_X(-f), {\mathcal{O}}_X).\end{align*} $$

Deleting the last two objects we get a full exceptional collection in $\langle {\mathcal {O}}_X(-f),{\mathcal {O}}_X\rangle ^{\perp }$ .

Let $F_i$ denote the exceptional curve with the class $f-E_i$ (the proper transform of the ruling f through $q_i$ ). We claim that the right mutation of ${\mathcal {O}}_{E_i}(-1)[-1]$ through $\langle {\mathcal {O}}_X(-s-f),{\mathcal {O}}_X(-s)\rangle $ is ${\mathcal {O}}_X(-s)|_{F_i}$ . Indeed, this follows from the exact sequences

$$ \begin{align*}0\to {\mathcal{O}}_X(-s-f)\to {\mathcal{O}}_X(-s-f+E_i)\to {\mathcal{O}}_{E_i}(-1)\to 0,\end{align*} $$

$$ \begin{align*}0\to {\mathcal{O}}_X(-s-f+E_i)\to {\mathcal{O}}_X(-s)\to {\mathcal{O}}_X(-s)|_{F_i}\to 0\end{align*} $$

and the vanishings $\operatorname {Hom}^*({\mathcal {O}}_{E_i},{\mathcal {O}}_X(-s-f))=\operatorname {Hom}^*({\mathcal {O}}_{F_i},{\mathcal {O}}_X)=0$ , obtained using Serre duality. Namely, the first sequence shows that the right mutation of ${\mathcal {O}}_{E_i}(-1)[-1]$ through $\langle {\mathcal {O}}_X(-s-f)\rangle $ is ${\mathcal {O}}_X(-s-f+E_i)$ , while the second sequence shows that the right mutation of ${\mathcal {O}}_X(-s-f+E_i)$ through $\langle {\mathcal {O}}_X(-s)\rangle $ is ${\mathcal {O}}_X(-s)|_{F_i}$ . Thus, we get a full exceptional collection

(4.6) $$ \begin{align} {\mathcal{O}}_X(-s-f), {\mathcal{O}}_X(-s), ({\mathcal{O}}_X(-s)|_{F_i})_{i=1,\ldots,4} \end{align} $$

in $\langle {\mathcal {O}}_X(-f),{\mathcal {O}}_X\rangle ^{\perp }$ . This collection is strong and its endomorphism algebra is determined by the compositions

$$ \begin{align*}\operatorname{Hom}({\mathcal{O}}_X(-s),{\mathcal{O}}_X(-s)|_{F_i})\otimes \operatorname{Hom}({\mathcal{O}}_X(-s-f),{\mathcal{O}}_X(-s))\to \operatorname{Hom}({\mathcal{O}}_X(-s-f),{\mathcal{O}}_X(-s)|_{F_i}).\end{align*} $$

These compositions can be identified with the restriction maps

$$ \begin{align*}H^0(X,{\mathcal{O}}(f))\to H^0(X,{\mathcal{O}}_X(f)|_{F_i})\simeq H^0(F_i,{\mathcal{O}}).\end{align*} $$

Since $F_i$ lies in a fiber of the projection , the above maps are identified with the restriction maps $H^0({\mathbb P}^1,{\mathcal {O}}(1))\to {\mathcal {O}}(1)|_{p_f(q_i)}$ .

By Lemma 3.2, we get an equivalence of $ \langle {\mathcal {O}}_X(-f),{\mathcal {O}}_X\rangle ^{\perp }$ with $D^b(\operatorname {Coh} {\mathcal C})$ identifying the collection (4.6) with (3.5), where ${\mathcal C}={\mathcal C}_4$ is associated with the points $p_f(q_1),\ldots ,p_f(q_4)$ in ${\mathbb P}^1$ . To get the equivalence $\Phi _f$ we use in addition tensoring by ${\mathcal {O}}_{{\mathcal C}}(-1)$ which takes the collection (3.5) to

$$ \begin{align*}{\mathcal{O}}_{\mathcal C}(-2), {\mathcal{O}}_{\mathcal C}(-1), (S_i)_{i=1,\ldots,4}.\end{align*} $$

Since the left mutation of $S_i$ through $\langle {\mathcal {O}}_{\mathcal C}(-2),{\mathcal {O}}_{\mathcal C}(-1)\rangle $ is $S^{\prime }_i[-1]$ (see Step 4 in the proof of Prop. 4.1), we have $\Phi _f(S^{\prime }_i)={\mathcal {O}}_{E_i}(-1)$ .

(ii) The formulas for the rank and the first Chern class (and hence, for degree) are checked on our exceptional collections. To get the required formulas for the linear map A we compute using (2.9) and (2.11):

$$ \begin{align*} & \frac{f}{2}-E_i=\frac{1}{2}h-e_i-\frac{1}{2}e_5=\frac{1}{2}(\varepsilon_1+\varepsilon_2+\varepsilon_3+\varepsilon_4)-\varepsilon_i, \ \text{for } i=1,2,3,\\ & \frac{f}{2}-E_4=-\frac{1}{2}h+e_4+\frac{1}{2}e_5=\frac{1}{2}(-\varepsilon_1-\varepsilon_2-\varepsilon_3+\varepsilon_4). \end{align*} $$

It is easy to check that A sends $B(0,1)$ to $B(1,1)$ , $B(1,1)$ to $B(1,0)$ , and $B(1,0)$ to $B(0,1)$ .

Proof. Let $V'$ be an exceptional object in $D^b(\operatorname {Coh} {\mathcal C})$ such that $\Phi _f(V')\simeq V$ . Then by Proposition 4.2(ii), $\operatorname {rk}(V')=r$ , $\deg (V')=d$ and

$$ \begin{align*}c_1(V)=\frac{r+d}{2}f-\frac{r}{2}(Q-f)+A(y'),\end{align*} $$

where $y'=\pi ([V])$ . It remains to observe that by Lemma 3.5, $y'\in B(r,d)$ , and so by Proposition 4.2(ii), $y=A(y')\in B(d,r+d)$ . The last assertion follows from Lemma 3.5.

Proof. Consider the element $w\in W(D_5)$ sending $\varepsilon _1$ to $-\varepsilon _1$ , $\varepsilon _5$ to $-\varepsilon _5$ and fixing $\varepsilon _i$ for $i=2,3,4$ . For the first item it is enough to prove that for any $y\in B(a+b,a)$ , one has $wD_{ab}(y)=D_{ba}(w(y))$ where $w(y)\in B(a+b,b)$ . Indeed, we have

$$ \begin{align*}wD_{ab}(y)=(a/2)(Q/2+\varepsilon_5)+(b/2)(Q/2-\varepsilon_5)+w(y).\end{align*} $$

Note that w preserves $B(0,1)$ and swaps $B(1,0)$ with $B(1,1)$ , hence, $wB(a+b,a)=B(a+b,b)$ , and our claim follows.

The second item follows from

$$ \begin{align*}-D_{-a,-b}(y)=(a/2)f+(b/2)(Q-f)-y=D_{a,b}(-y)\end{align*} $$

(note that the longest element of $W(D_4)$ takes y to $-y$ ).

Proof. Uniqueness follows from the fact that an exceptional object $E\in \langle {\mathcal {O}}_X\rangle ^\perp $ is determined by its rank and $c_1$ (see Lemma 2.3(ii)). By Lemma 4.5 and duality, it is enough to prove existence of an exceptional pair $(V,{\mathcal {O}}_X)$ with V of rank r and $c_1(V)=D_{d+r,-r}(y)$ . But this immediately follows from Corollary 4.4.

Proof. A class descends to a del Pezzo surface of degree $5$ if and only if it is orthogonal to the class of an exceptional curve. In terms of the basis $(Q,\varepsilon _1,\ldots ,\varepsilon _5)$ of $\operatorname {Pic}(X)_{\mathbb {Q}}$ , the set of classes of 16 exceptional curves is the $W(D_5)$ -orbit of

$$ \begin{align*}2h-\sum_{i=1}^5 e_i=Q/4+\sum_{i=1}^5 \varepsilon_i/2.\end{align*} $$

In other words, all exceptional classes have form $e=Q/4+x$ , where $x=\sum _{i=1}^5 x_i\varepsilon _i$ has $x_i=\pm 1/2$ , with even number of $x_i$ ’s negative. Recall that the basis $(\varepsilon _i)$ is orthonormal with respect to the negative of the intersection form. Hence, the intersection number of the class e with $Q/2-\varepsilon _5$ (resp., $Q/2+\varepsilon _5$ ) is $1/2+x_5$ (resp., $1/2-x_5$ ). Thus, we have

$$ \begin{align*}e\cdot D_{a,b}(y)=\begin{cases} a/2+x\cdot y, & x_5=1/2, \\ b/2+ x\cdot y, & x_5=-1/2.\end{cases}\end{align*} $$

Recall that $y=\sum _{i=1}^4 y_i\varepsilon _i\in B(a+b,a)$ (and so $y_5=0$ ). By the definition of $B(a+b,a)$ , either $y_i=\pm 1/2$ for $i=1,\ldots ,4$ , or $y_{i_0}=\pm 1$ for some $i_0\in [1,4]$ and $y_i=0$ for $i\neq i_0$ . Since $x_i=\pm 1/2$ , this implies that $|x\cdot y|\le 1$ . Therefore, $e\cdot D_{a,b}(y)$ can be zero only if either $|a|\le 2$ or $|b|\le 2$ . It follows that either a or b belongs to the set $\{\pm 1, \pm 2\}$ (since a and b are relatively prime). By Lemma 4.5, using an action of $W(D_5)$ on geometric markings of X, we can replace the pair $(a,b)$ by $(b,a)$ and by $(-a,-b)$ . Thus, it is enough to consider the cases $b=2$ and $b=1$ , with some choice of y.

In the case $b=2$ , a is odd, so $y\in B(1,1)$ , i.e., $y_i=\pm 1/2$ , with odd number of negative $y_i$ ’s. Hence, $e=Q/4+y-\varepsilon _5/2$ is an exceptional class with $e\cdot D_{a,b}(y)=0$ (recall that $y\cdot y=-(y,y)=-1$ ).

In the case where $b=1$ and a is odd, we can take $y=-\varepsilon _4$ . Then

$$ \begin{align*}D_{a,1}(\varepsilon_4)=\frac{a}{2}(Q/2-\varepsilon_5)+\frac{1}{2}(Q/2+\varepsilon_5)-\varepsilon_4=\frac{a-1}{2}(Q/2-\varepsilon_5)+Q/2-\varepsilon_4=\frac{a-1}{2}f+s\end{align*} $$

which comes from ${\mathbb P}^1\times {\mathbb P}^1$ .

In the case where $b=1$ and a is even, we can take $y=-\sum _{i=1}^4 \varepsilon _i/2$ . Then

$$ \begin{align*}D_{a,1}(-\sum_{i=1}^4\varepsilon_i/2)=\frac{a}{2}(3Q/4-\sum_{i=1}^4\varepsilon_i/2-\varepsilon_5/2)-(\frac{a}{2}-1)(Q/4-\sum_{i=1}^4\varepsilon_i/2+\varepsilon_5/2)=\frac{a}{2}h-(\frac{a}{2}-1)e_5,\end{align*} $$

which comes from $F_1$ , and even from ${\mathbb P}^2$ in the case $a=2$ .

Proof. By Proposition 4.8, if $r=2$ or $d-r=\pm 2$ then the divisor class $c_1(V)$ descends to a del Pezzo surface of degree $5$ . Similarly, if $r=1$ or $d-r=\pm 1$ then this class descends to a del Pezzo surface of degree $8$ (to ${\mathbb P}^1\times {\mathbb P}^1$ if d is even, and to $F_1$ if d is odd). Hence, by Lemma 2.3(iii), V itself also descends (the descended bundle still forms an exceptional pair with ${\mathcal {O}}_X$ ).

Proof. The formula for the rank and $c_1$ of $M(V)$ follow from additivity of these invariants in exact triangles and the fact that for an exceptional pair $(V,{\mathcal {O}})$ one has $\chi (V^\vee )=-\deg (V)$ as follows from (2.2).

Proof. Recall that we have the operation $(V,{\mathcal {O}}_X)\mapsto (R(V)=V^\vee (-Q),{\mathcal {O}}_X)$ on exceptional pairs. Since $\mu (R(V))=-\mu (V)-4$ and R preserves the rank (see Lemma 5.1), for each fixed rank r, passing from V to $R(V)$ transforms the condition $\mu <-r-1/r-4$ into the condition $\mu>r+1/4$ . Thus, it is enough to prove the assertion assuming that $\mu>r+1/r$ . Since an exceptional bundle in $\langle {\mathcal {O}}_X\rangle ^\perp $ is determined by its rank and $c_1$ (see Lemma 2.3(ii)), it is enough to check that for this range either $c_1(V)=D_{d+r,-r}(y)$ or $(r,d)=(2,7)$ and $c_1(V)=5h-2Q$ for some geometric marking.

Recall that by Proposition 4.1, we have an equivalence

where ${\mathcal C}={\mathcal C}_5$ is a weighted projective line. Let $V'=\Phi ^{-1}(V)\in D^b(\operatorname {Coh}{\mathcal C})$ . Then by (4.1), one has $r':=\operatorname {rk}_{\mathcal C}(V')=2r+d>0$ and $d':=\deg _{\mathcal C}(V')=-r$ , so changing $\Phi $ by an even shift we can assume $V'$ to be a vector bundle on ${\mathcal C}$ . Our assumption $d/r>r+1/r$ is equivalent to $r'>(1-d')^2$ .

Applying Proposition 3.8 we deduce that either $V'$ is right orthogonal to one of the objects $S_i$ or $S^{\prime }_i$ , or $(r,d)=(2,7)$ and $\pi [V']$ is in the $W(D_5)$ -orbit of $-\frac {5}{2}\sum _{i=1}^5 \varepsilon _i$ . In the latter case, we get that with respect to an appropriate geometric marking, $\pi [V']=-\frac {5}{2}\sum _{i=1}^5 \varepsilon _i$ . Since $\deg (V)=7$ , we have $c_1(V)=\frac {7}{4}Q+\alpha $ , where $\alpha \in \langle Q\rangle ^\perp _{\mathbb {Q}}$ . By Proposition 4.1, we get $\alpha =\pi [V']=-\frac {5}{2}\sum _{i=1}^5 \varepsilon _i$ , with respect to the identification $\langle Q\rangle ^\perp =\mathrm {Q}(D_5)$ . Hence,

$$ \begin{align*}c_1(V)=\frac{7}{4}Q-\frac{5}{2}\sum_{i=1}^5 \varepsilon_i=5h-2Q.\end{align*} $$

In the former case we get that either $\operatorname {Hom}^*(S_i,V')=0$ or $\operatorname {Hom}^*(S^{\prime }_i,V')=0$ for some i. Applying the equivalence $\Phi $ above, we get that either $\operatorname {Hom}^*({\mathcal {O}}_X(e_i-h),V)=0$ or $\operatorname {Hom}^*({\mathcal {O}}_X(Q-e_i+h),V)=0$ for some i. But the classes $e_i-h=-Q/2+\varepsilon _i$ and $-Q-e_i+h=-Q/2-\varepsilon _i$ , where $i=1,\ldots ,5$ , form a single $W(D_5)$ -orbit. Thus, choosing an appropriate geometric marking, we can assume that $\operatorname {Hom}^*({\mathcal {O}}_X(-f),V)=0$ (recall that $f=h-e_5$ ). But then V is in the subcategory $\langle {\mathcal {O}}_X(-f),{\mathcal {O}}_X\rangle ^\perp $ , so by Corollary 4.4, $c_1(V)$ is of the form $D_{d+r,-r}(y)$ .

Proof. For each k, $5\le k\le 9$ , let us consider the quadratic order of discriminant $k(k-4)$ ,

$$ \begin{align*}O_k:={\mathbb{Z}}+{\mathbb{Z}}\frac{-k+\sqrt{k(k-4})}{2}\subset O\subset \mathbb{Q}(\sqrt{k(k-4)}),\end{align*} $$

where O is the ring of integers in the quadratic extension, and let $\sigma :O_k\to O_k$ denote the Galois conjugation. Note that $O_k\simeq {\mathbb {Z}}[x]/(x^2+kx+k)$ . For every V with $(\deg (V),\operatorname {rk}(V))=(d,r)$ , we set

$$ \begin{align*}\xi(V)=\xi(d,r):=-d+r\frac{-k+\sqrt{k(k-4)}}{2}\in O_k.\end{align*} $$

Note that

$$ \begin{align*}\operatorname{Nm}(\xi(d,r))=d^2+krd+kr^2.\end{align*} $$

In particular, the norm is always a square modulo k.

Let us consider two operations (5.1) on exceptional pairs $(V,{\mathcal {O}})$ (where we allow V to be any exceptional object of the derived category). It is easy to check that

(5.5) $$ \begin{align} \xi(R(V))=-\sigma\xi(V), \ \ \xi(RM(V))=u\cdot \xi(V), \end{align} $$

where $u=-1+k/2+\sqrt {k(k-4)}/2$ is a unit in $O_k$ .

The following elementary observation will be used implicitly in the arguments below. Let ${\mathbb {Z}}\subset A={\mathbb {Z}}[x]/(x^2+mx+n)$ be a quadratic extension, and let p be a prime number. If an element $a\in A$ satisfies $\operatorname {Nm}(a)=\pm p$ then the ideal $(a)$ in A is prime. Indeed, in this case it is easy to see that $A/(a)\simeq {\mathbb {Z}}/(p)$ .

Now we consider the case of each k separately.

Case $k=9$ . In this case $O_9={\mathbb {Z}}+{\mathbb {Z}}\cdot 3\frac {-1+\sqrt {5}}{2}$ has conductor $3$ in the ring of integers $O\subset \mathbb {Q}(\sqrt {5})$ . Note that $u=u_f^4$ , where $u_f=(1+\sqrt {5})/2$ is the fundamental unit in O. We are looking for elements of $O_9$ of norm $-9$ . Since $3$ generates a prime ideal in O, they are of the form $\pm 3u_f^{2n+1}$ . Thus, up to the action of the operations (5.5), we can reduce to the case $\xi (V)=\pm 3u_f$ , in which case V is a shift of ${\mathcal {O}}(-2)$ .

Case $k=8$ . In this case $O_8={\mathbb {Z}}+{\mathbb {Z}}\cdot 2\sqrt {2}$ has conductor $2$ in $O={\mathbb {Z}}[\sqrt {2}]$ , and $u=u_f^2$ , where $u_f=1+\sqrt {2}$ is the fundamental unit in O. We are looking for elements of $O_8$ with norm in $[-8,-1]$ . Since a norm of an element in $O_8$ is a square modulo $8$ , the norm is actually either $-8$ , $-7$ , or $-4$ .

If $\operatorname {Nm}(a)=-8=-(\sqrt {2})^6$ for $a\in O$ , then using the fact that the ideal $(\sqrt {2})\subset O$ is prime, we get

$$ \begin{align*}a=(\sqrt{2})^3v,\end{align*} $$

where $v\in O$ has $\operatorname {Nm}(v)=1$ , so $v=\pm u_f^{2n}$ . Hence, our operations (5.5) (together with the shift) reduce to the case $\xi (V)=2\sqrt {2}$ , i.e., $\operatorname {rk}(V)=1$ , $\deg (V)=-4$ .

Similarly, if $\operatorname {Nm}(a)=-7$ , then we use the prime decomposition $(7)=(-1+2\sqrt {2})(1+2\sqrt {2})$ . Hence, replacing a by $-\sigma (a)$ if needed, we get $a=(-1+2\sqrt {2})v$ , where $\operatorname {Nm}(v)=1$ , so we can reduce to $\xi (V)=-1+2\sqrt {2}$ , i.e., $\operatorname {rk}(V)=1$ , $\deg (V)=-3$ .

Finally, if $\operatorname {Nm}(a)=-4$ then $a=2v$ (again using primality of $(\sqrt {2})$ ), where $v=\pm u_f^{2n+1}$ , so we can reduce to $\xi (V)=2(-1+\sqrt {2})$ , i.e., $\operatorname {rk}(V)=1$ , $\deg (V)=-2$ .

Thus, the question reduces to which of these degrees can occur for a line bundle L forming an exceptional pair $(L,{\mathcal {O}})$ . For $X_8={\mathbb P}^1\times {\mathbb P}^1$ , the degree is always even, and both degrees $-2$ and $-4$ occur (for $L={\mathcal {O}}(-1,0)$ and $L={\mathcal {O}}(-1,-1)$ ).

On the other hand, it is easy to see that for $X_8=F_1$ , the exceptional pair $(L,{\mathcal {O}})$ should have $L={\mathcal {O}}(-s+nf)$ or $L={\mathcal {O}}(-f)$ or $L={\mathcal {O}}(-2s-2f)$ , where f is the class of a fiber of $F_1\to {\mathbb P}^1$ , and s is the class of a section with $s^2=-1$ . Since $Q=2s+3f$ ,

$$ \begin{align*}\deg({\mathcal{O}}(-s+nf))=2n-1, \ \ \deg({\mathcal{O}}(-f))=-2, \ \ \deg({\mathcal{O}}(-2s-2f)=-6.\end{align*} $$

Thus, for $X_8=F_1$ , the cases $\operatorname {Nm}(\xi (d,r))=-4$ and $\operatorname {Nm}(\xi (d,r))=-7$ occur, while the case $\operatorname {Nm}(\xi (d,r))=-8$ does not occur. Since $\operatorname {Nm}(\xi (d,r))\equiv d^2\ \mod (8)$ , we have $\operatorname {Nm}(\xi (d,r))=-7$ if and only if d is odd, and $\operatorname {Nm}(\xi (d,r))=-4$ if and only if $d\equiv 2\ \mod (4)$ .

Case $k=7$ . In this case $O_8=O={\mathbb {Z}}+{\mathbb {Z}}\frac {\sqrt {21}-1}{2}$ is the ring of integers in $\mathbb {Q}(\sqrt {21})$ , and we are looking for elements of O with the norm in $[-7,-1]$ . We have $u=u_f=\frac {5+\sqrt {21}}{2}$ .

Since the norm is a square modulo $7$ , the norm has to be in $\{-7,-6,-5,-3\}$ . It is easy to rule out $-7$ . Also, $d^2+7rd+7r^2$ can be even only if both d and r are even, which rules out $-6$ .

Using the fact that $\operatorname {Nm}(\frac {-1+\sqrt {21}}{2})=-5$ , we see that if $\operatorname {Nm}(a)=-5$ then replacing a by $-\sigma (a)$ if needed, we get $a=\pm \frac {-1+\sqrt {21}}{2}\cdot u_f^n$ , so we can reduce the problem to $\xi (V)=\frac {-1+\sqrt {21}}{2}$ , i.e., $\operatorname {rk}(V)=1$ , $\deg (V)=-3$ .

Similarly, since $\operatorname {Nm}(\frac {-3+\sqrt {21}}{2})=-3$ , if $\operatorname {Nm}(a)=-3$ then replacing a by $-\sigma (a)$ if needed, we get $a=\pm \frac {-3+\sqrt {21}}{2}\cdot u_f^n$ , so we can reduce to $\xi (V)=\frac {-3+\sqrt {21}}{2}$ , i.e., $\operatorname {rk}(V)=1$ , $\deg (V)=-2$ .

In both cases we can take $(V,{\mathcal {O}})$ to be the pull-back of the similar pair on $F_1$ under the blow up map $X_7\to X_8\simeq F_1$ .

Case $k=6$ . In this case $O_6={\mathbb {Z}}[\sqrt {3}]$ , we have $u=u_f=2+\sqrt {3}$ , and we are looking for elements with the norm in $[-6,-1]$ . Since the norm is a square modulo $6$ , it has to be in $\{-6,-5,-3,-2\}$ . It is easy to rule out $-6$ . If $d^2+6rd+6r^2=-5$ then $d^2\equiv -1\ \mod (4)$ which is impossible. Thus, the only two possibilities are $\operatorname {Nm}(a)=-3$ and $\operatorname {Nm}(a)=-2$ . Arguing as before, we reduce to solutions of the form $\pm \sqrt {3}u_f^n$ and $\pm (-1+\sqrt {3})u_f^n$ , so we reduce to the cases $(r,d)=(1,-3)$ and $(r,d)=(1,-2)$ . Both cases are realised as pull-backs under the blow up map $X_6\to X_7$ .

Case $k=5$ . In this case $O_5=O={\mathbb {Z}}+{\mathbb {Z}}\frac {-1+\sqrt {5}}{2}$ , and $u=u_f^2$ , where $u_f=(1+\sqrt {5})/2$ , and we are looking for elements with the norms in $[-5,-1]$ (with $(r,d)$ relatively prime). Since the norm is a square modulo $5$ , it has to be in $\{-5,-4,-1\}$ . If $d^2+5rd+5r^2=-4$ then $d^2+rd+r^2$ is even which is possible only if both r and d are even, which is a contradiction. Hence, the only possible norms are $-5$ and $-1$ . The norm $-5$ is realised by elements of the form $\pm \sqrt {5}u_f^{2n}$ , while the norm $-1$ is realised by elements of the form $\pm u_f^{2n+1}$ . Thus, we reduce to the cases $(r,d)=(2,-5)$ and $(r,d)=(1,-2)$ . The latter case is realised as the pull-back under the blow up map $X_5\to X_6$ .