Proof. Recall that a divisor is nef if its intersection with all negative curves is nonnegative. If E is a $(-2)$ -curve, then

$$\begin{align*}(-K-D).E = -K.E-D.E = 0 + 2\delta_{E \subset D} - \sum_{\substack{\alpha \in \mathcal{A} \\ D_\alpha \ne E }}D_\alpha. E, \end{align*}$$

and this number is nonnegative if and only if (i) and (iii) hold for E. If E is a $(-1)$ -curve, then

$$\begin{align*}(-K-D).E = -K.E-D.E = 1 + \delta_{E \in D} - \sum_{\substack{\alpha \in \mathcal{A} \\ D_\alpha \ne E }}D_\alpha. E, \end{align*}$$

and this number is nonnegative if and only if (ii) and (iii) hold for E.

Proof. For the first statement, we just have to note that $-\varepsilon \ell _0 + \sum _{1\le j \le k} a_j \ell _j$ is not effective for any $\varepsilon>0$ . Turning to the second statement, assume for contradiction that L is big and nef. Note that $\ell _0 - \ell _i$ has nonnegative intersection with all $(-1)$ -curves.

If $\ell _0-\ell _i$ has (strictly) negative intersection with a $(-2)$ -curve E, then this curve needs to have class $[E]=\ell _i - \ell _j$ for some $j\ne i$ (cf. [Reference Manin20, Theorem 25.5.3]). Writing $\ell _0 - \ell _i = L + [F]$ with effective F, we get $F.E<0$ , so $E \subset F$ , and $L \le \ell _0 - \ell _i - [E] = \ell _0 - \ell _j$ . The only negative curves that could have negative intersection with $\ell _0-\ell _j$ have class $\ell _j - \ell _k$ . As curves of classes $\ell _i - \ell _j$ , $\ell _j - \ell _k$ , etc., are contracted to a single point by $\pi $ , we can eventually find an $\ell _{i'}$ with $L\le \ell _0 - \ell _{i'}$ and such that $\ell _0-\ell _{i'}$ has nonnegative intersection with all negative curves. Then $\ell _0 -\ell _{i'}$ is nef. But $(\ell _0-\ell _{i'})^2=0$ , whence it cannot be big.

Proof. Assume for contradiction that D contains a nonnegative curve C, but that $-K-D$ is big and nef. In particular, $-K-D^\prime $ is big for all $D^\prime \subset D$ . Since C is nonnegative, it is the strict transform of a curve $C_0$ on $\mathbb {P}^2$ . Then

$$ \begin{align*} [C]=d \ell_0 - \sum_{i=1}^r a_i \ell_i, \end{align*} $$

where $d=\deg C_0$ and $a_i=C.\ell _i$ .

We first reduce to the case of $C_0$ being a line. If $d\ge 3$ , then $[-K-C] \le \sum _{1\le i \le r} a_i \ell _i$ , which is not big by Lemma 7 (i). If $C_0$ is a nondegenerate conic, then $a_1,\dots ,a_r\le 1$ since $C_0$ has multiplicity $\le 1$ in all images of the exceptional divisors. Moreover, since $C^2 \ge 0$ , at most four of the $a_i$ are nonzero. It follows that $[-K-C]\le \ell _0 - \ell _j$ , so $-K-D$ is not big or not nef by Lemma 7 (ii).

Let $C_0$ be a line. As the self-intersection of C is nonnegative, $[C]=\ell _0 - \ell _j$ for some j or $[C]=\ell _0$ . In the first case, $C_0$ contains the center $P=\pi _1 \cdots \pi _j(p_j)$ of a blowup. If $n_P>1$ , then $\pi ^{-1}(P)$ contains $(-2)$ -curves. Appealing to Lemma 5 (i), the first $(-2)$ -curve must be contained in D, as must the remaining $(-2)$ -curves by repeated applications. Let $E_0$ be the sum of these $(-2)$ -curves. Then $C'=C+E_0\subset D$ is of class $[C']=\ell _0 -\ell _{j'}$ , where $\ell _{j'}$ is the class of the final $(-1)$ -curve in the chain. If $n_P=1$ or $[C]=\ell _0$ , set $E_0=0$ and $C'=C$ ; in the first case, set $j'=j$ ; in the latter case, fix an arbitrary $j'$ and note that $[C]\le \ell _0 - \ell _{j'}$ . Then $C'$ satisfies the conditions in Lemma 5 for all negative curves in the preimage of P by this construction, and it does the same for all other curves contracted by $\pi $ as it does not meet them. For what remains, we distinguish three cases.

Case 1. The curve C does not meet any of the remaining $(-2)$ -curves, and $C.E\le 1$ for all remaining $(-1)$ -curves. Then $(-K-C')$ is nef by Lemma 5. But $(-K-C')^2\le 4 - (r - 1) \le 0$ , so it cannot be big.

Case 2. The curve C meets one of the remaining $(-2)$ -curves E. Then $E\subset D$ by Lemma 5 (i). Since E is the strict transform of a curve in $\mathbb {P}^2$ , its class satisfies $[E]= l_0 -l_{i_1} - l_{i_2} - l_{i_3}$ for some pairwise different $i_1, i_2, i_3$ or $[E] \ge 2 \ell _0 + \sum a_i\ell _i$ for some $a_i\in \mathbb {Z}$ . In the first case, $[-K-C-E]\le \ell _0 - \ell _k$ for $k\ne i_i,i_2,i_3,j$ , and in the second case, $[-K-C-E] \le \sum _{1 \le i \le r} a_i\ell _i$ . In both cases, $-K-D$ is not big or not nef by Lemma 7 (ii) or (i), respectively.

Case 3. The curve C meets a $(-1)$ -curve E with $C.E\ge 2$ . By Lemma 5 (ii), $E\subset D$ . As E is the strict transform of a curve on $\mathbb {P}^2$ , its class verifies $[E]\ge [F]$ for a $(-2)$ -class $[F]$ of the same shape as in the previous case; hence, $-K-D$ is not big or not nef.

Proof. Let D be a reduced effective divisor such that $-K-D$ is big and nef. By Proposition 8, $D=\sum E_i$ has to be supported on negative curves. Consider the complete subgraph G of the Dynkin diagram on the vertices corresponding to components of D. By Lemma 5 (iii), each of its connected components is a path or a cycle. Let $N_1$ be the number of $(-1)$ -curves in D, and $N_2$ be the number of $(-2)$ -curves. Then $v=N_1+N_2$ is the number of vertices of G, and denote by e its number of edges.

The self-intersection of the log-anticanonical divisor is

$$\begin{align*}(-K-D)^2 = K^2 + \sum_i E_i^2 + 2 \sum_i E_i.K + 2\sum_{i<j} E_i.E_j. \end{align*}$$

As $-K.E$ is zero for $(-2)$ -curves and $1$ for $(-1)$ -curves, we get

(14) $$ \begin{align} (-K-D)^2 = d + 2(e-v) - N_1. \end{align} $$

Since $-K-D$ is big and nef, this self-intersection must be positive.

If G is connected and not a cycle, then $e=v-1$ , so $d -2-N_1>0$ . In case $d=4$ , this leaves us with $N_1 \le 1$ , in case $d=3$ with $N_1 = 0$ , and in case $d\le 2$ with an immediate contradiction. In each case, the resulting divisors satisfy the asserted description using Remark 6 and that the graph is a path.

It remains to prove that G has to be connected and not a cycle. If G is not connected and does not contain a cycle, then $(-K-D)^2 = d-4-N_1 \le 0$ , so $-K-D$ cannot be big, leaving only the case of graphs G containing a cycle, in which case $d-N_1>0$ for $-K-D$ to be big.

For $N_1=0$ , we note that only Dynkin diagrams of type $\mathbf {A}$ , $\mathbf D$ , and $\mathbf E$ appear as intersection graphs of $(-2)$ -curves, and these do not contain double edges nor more general cycles.

If $N_1=1$ , then $d\ge 2$ . The sum $E_2$ of the $(-2)$ -curves in D forms a $(-2)$ -class since $E_2^2 = -2N_2 + 2(s-1)=-2$ and $-K.E_2=0$ . As the Weyl group acts transitively on $(-1)$ -curves and leaves the intersection pairing invariant, we can assume that $[E_1]=\ell _1$ . Now $\ell _1.[E_2]\ge 2$ , and so the $(-2)$ -class needs to have the form $3\ell _0 - 2\ell _1 - \ell _2 - \cdots - \ell _8$ . But such a class does not exist if $d\ge 2$ (cf. [Reference Manin20, Theorem 25.5.3]).

If $N_1 = 2$ , then $d\ge 3$ . In this case, the anticanonical model contracts $(-2)$ -curves and maps $(-1)$ -curves to lines. The resulting two lines then need to intersect with multiplicity $2$ , an impossibility.

Finally, if $N_1=3$ , then $d = 4$ . In this case, the anticanonical model $\phi \colon \widetilde {S}\to S = Q_1 \cap Q_2 \subset \mathbb {P}^4$ is a (possibly singular) intersection of two quadrics, still contracting all $(-2)$ -curves and mapping all $(-1)$ -curves to lines. The resulting three lines need to intersect pairwise. If they were contained in a plane P, this plane would intersect $Q_1$ in three lines, an impossibility. So the three lines intersect in a point Q. The tangent space at Q needs to contain each plane containing two of these lines, whence Q is singular. Then $\widetilde {S}\to S$ factors through the blowup Y of $\mathbb {P}^4$ in Q. The strict transforms of the lines do not intersect on Y, and thus the $(-1)$ -curves on $\widetilde {S}$ do not intersect. It follows that each of them intersects a $(-2)$ -curve above Q. Hence, the $(-2)$ -curves are contained in D, and at least one of them needs to intersect three other negative curves in D. Now, Lemma 5 (iii) implies that $-K-D$ cannot be nef.

Conversely, if D is one of the divisors in the statement, then it is nef by Lemma 5, and its self-intersection is positive by equation (14); hence, it is also big.

Proof. Since $\pi $ is a $\mathbb {G}_{\mathrm {m},\mathbb {Z}}^6$ -torsor so are its restrictions to the open subschemes $\widetilde {\mathcal {U}}_i$ . Integral points $\boldsymbol {\eta }\in \operatorname {\mathrm {Spec}} R_{\mathbb {Z}}$ are lattice points $(\eta _1,\dots ,\eta _9)\in \mathbb {Z}^9$ satisfying the equation in the Cox ring. Such a point is integral on the complement of $\mathbb {V}(I_{\mathrm {irr}})=\bigcup \mathbb {V}(\eta _i,\eta _j)$ —the union running over all $i,j$ which do not share an edge in Figure 4—if it does not reduce to any of the $\mathbb {V}(\eta _i,\eta _j)$ for any prime, that is, if the gcd-condition (19) holds.

Integral points on $\mathcal {Y}_1\subset \mathcal {Y}$ are precisely those which do not reduce to $\pi ^{-1}(E_7)=\mathbb {V}(\eta _7)$ at any place; that is, they are those points satisfying $\eta _7\in \{\pm 1\}$ . Analogously, integral points on $\mathcal {Y}_2$ are those satisfying $\eta _3,\eta _4,\eta _6\in \{\pm 1\}$ , integral points on $\mathcal {Y}_3$ are those satisfying $\eta _3,\eta _4,\eta _6,\eta _7\in \{\pm 1\}$ and similarly for $\mathcal {Y}_4, \mathcal {Y}_5, \mathcal {Y}_6$ . The preimage of $\widetilde {V}$ in the universal torsor is the complement of $\eta _1\cdots \eta _7=0$ .

Proof. The first statement is a special case of Theorem 10.

For the first set, assume that $p\mid \eta _8\eta _9$ for a prime p. Then $p\nmid \eta _3\cdots \eta _6$ , since the corresponding divisors $E_3,\dots ,E_6$ share an edge with neither $E_8$ nor $E_9$ in Figure 4, while at most one of $\eta _1,\eta _2,\eta _7$ can be divisible by p. Hence, the second or third section is not divisible by p.

For the second set, assume that $p\mid \eta _2\eta _5\eta _7\eta _8$ . If $p\mid \eta _5\eta _7$ , then $p\nmid \eta _1^2\eta _2^2\eta _3^2\eta _4$ ; if $p\mid \eta _2\eta _8$ and $p\nmid \eta _7$ , then $p\nmid \eta _4\eta _5^3\eta _6^2\eta _7$ . Regarding the case $i=4$ , if $p\mid \eta _2\eta _8$ , then $p\nmid \eta _4\eta _5^3\eta _6^2\eta _7$ . For $i=5$ , assume $p\mid \eta _5\eta _7\eta _8$ . If $p\mid \eta _5\eta _7$ , then $p\nmid \eta _1^2\eta _2\eta _3^2\eta _4$ ; if $p\mid \eta _8$ and $p\nmid \eta _7$ , then $p\nmid \eta _1\eta _3\eta _4\eta _5^2\eta _6\eta _7$ .

Turning to $L_2$ , we know that the anticanonical sections in equation (17) cannot be divisible by p simultaneously, so for all sections in $L_2$ to be be divisible by p simultaneously, $p\mid \eta _1\eta _2$ . But then $p\nmid \eta _4^2 \eta _5^6 \eta _6^4 \eta _7^3$ .

Lastly, assume that the monomials in $L_1$ are divisible by p so that p in particular divides all monomials in $L_2L_3$ . Since p cannot divide all monomials in $L_2$ , it has to divide all monomials in $L_3$ . If follows that $p\mid \eta _2\eta _5\eta _8$ . If $p\mid \eta _2$ , then the last monomial in $L_3$ cannot be zero modulo p. If $p\mid \eta _8$ , then p can divide only one of $\eta _2$ or $\eta _7$ but none of the remaining variables in the latter two sections of $L_3$ , so one of those two sections is nonzero modulo p. So $p\mid \eta _5$ . Now p can divide at most one of $\eta _6$ and $\eta _7$ but none of the remaining variables, and so the last section of $L_1$ is nonzero modulo p, a contradiction.

By the same arguments (replacing vanishing modulo p by vanishing over $\overline {\mathbb {Q}}$ ), the log-anticanonical bundles $\omega _{\widetilde {S}}(D_i)^\vee $ for $i=1,2,4,5$ and the bundles $\mathcal {L}_1$ and $\mathcal {L}_2$ are base point free, whence nef. On the other hand, $\omega _{\widetilde {S}}(D_3)^\vee $ is not nef since its intersection number with $E_5$ is $-1$ .

Proof. Consider the morphism $f\colon \mathcal {Y}\to \mathbb {P}^4_{\mathbb {Z}}$ defined by $\boldsymbol {\eta }\mapsto (s_0(\boldsymbol {\eta }):\dots :s_4(\boldsymbol {\eta }))$ with the anticanonical sections $s_j$ in equation (17). We only have to show that equation (19) holds for $\boldsymbol {\eta }$ if and only if the corresponding gcd-condition in (2), resp. in (7), holds for $f(\boldsymbol {\eta })$ . To this end, we note that $\eta _1\eta _2 \eta _3\eta _4\eta _5^2\eta _6\eta _7$ is in the radical of the ideal generated by $M_2$ as in Lemma 12, so the gcd-condition (2) can be rewritten as

$$\begin{align*}\left\lvert\eta_7\right\rvert\gcd\{m(\boldsymbol{\eta}) \mid m \in M_1\} = 1 \quad \text{and}\quad \left\lvert\eta_3\eta_4\eta_6\right\rvert \gcd\{m(\boldsymbol{\eta}) \mid m \in M_2\} = 1. \end{align*}$$

By Lemma 12, these gcds are one, and so the claim follows for $i\in \{1,2\}$ . Since $\mathcal {U}_3$ and $\widetilde {\mathcal {U}}_3$ are the intersections of the respective open subschemes within the first two cases, the assertion follows for $i=3$ . The cases $i=4$ and $i=5$ can be proved using an analogous reformulation of the first two conditions in equation (7).

Proof. For $i\in \{1,2,4,5\}$ , the metrics induce height functions $\widetilde {H}_i(x)=H_{\mathbb {P}^{N_i}}(f_i(x))$ verifying $f_i(\pi (\boldsymbol {\eta })) = (m_0(\boldsymbol {\eta }): \dots : m_{N_i}(\boldsymbol {\eta }))$ for the sections $m_0,\dots ,m_{N_i}\in M_i$ constructed in Lemma 12, so

$$\begin{align*}H_i(\pi(\boldsymbol{\eta})) = \prod_v \max_{m\in M_i}\{\left\lvert m(\boldsymbol{\eta})\right\rvert_v\}. \end{align*}$$

By the same lemma, the p-adic contributions to this product are $1$ , whence $\widetilde {H}_i(\pi (\boldsymbol {\eta })) = \mathcal {H}_i(\boldsymbol {\eta })$ .

To check that these height functions coincide with the ones defined in the introduction, we note that, for example, for a point in $\mathcal {Y}_1(\mathbb {Z})$ we have $\eta _7\in \{\pm 1\}$ , and thus $|\eta _2\eta _3\eta _4\eta _5\eta _6\eta _8|=|\eta _2\eta _3\eta _4\eta _5\eta _6\eta _7\eta _8|=|x_0|$ . We get analogous identities for the other coordinates and cases. There is no section corresponding to $x_2$ in the second height function, but, for integral points on $\mathcal {Y}_2$ , we have $\left\lvert x_2\right\rvert =\sqrt {\left\lvert \eta _1^2\eta _2^2\right\rvert \left\lvert \eta _5^4\eta _7^2\right\rvert }=\sqrt {\left\lvert x_1x_3\right\rvert }$ ; hence, it can never contribute to the maximum.

The case $i=3$ is more complicated. The log-anticanonical height function $\widetilde {H}_3$ induced by the metrics on $\mathcal {L}_1$ and $\mathcal {L}_2$ satisfies

$$ \begin{align*} \widetilde{H}_3(\pi(\boldsymbol{\eta}))=\frac{\max_{s\in L_1} \left\lvert s(\boldsymbol{\eta})\right\rvert}{\max_{s\in L_2} \left\lvert s(\boldsymbol{\eta})\right\rvert} = \max\left\{ \max_{s \in L_3} \left\lvert s(\boldsymbol{\eta})\right\rvert, \frac{\left\lvert\eta_1^3 \eta_2^3 \eta_3^2 \eta_4 \eta_8 \eta_9\right\rvert}{\max_{s \in L_2} \left\lvert s(\boldsymbol{\eta})\right\rvert} \right\} \end{align*} $$

by an analogous argument as in the previous cases. Now note that, for $\boldsymbol {\eta } \in \mathcal {Y}_3(\mathbb {Z}) \cap \pi ^{-1}(\widetilde {V})(\mathbb {Q})$ , we can simplify this to

(21) $$ \begin{align} \widetilde H_3(\pi(\boldsymbol{\eta}))= \max\{\left\lvert\eta_2\eta_5\eta_8\right\rvert, \left\lvert\eta_1\eta_2\eta_5^2\right\rvert, \left\lvert\eta_5^4\right\rvert, \min\{\left\lvert\eta_1^2\eta_2^2\right\rvert, \left\lvert\eta_8\eta_9\right\rvert\}\}. \end{align} $$

Indeed, $\eta _3,\eta _4,\eta _6,\eta _7$ have absolute value $1$ and the remaining variables absolute value at least one. Then

$$\begin{align*}\frac{\left\lvert\eta_1^3 \eta_2^3 \eta_3^2 \eta_4 \eta_8 \eta_9\right\rvert}{\max_{s \in L_2} \left\lvert s(\boldsymbol{\eta})\right\rvert} \le \min\{\left\lvert\eta_1^2\eta_2^2\right\rvert, \left\lvert\eta_8\eta_9\right\rvert\} \end{align*}$$

follows from $\eta _1^3\eta _2^3\eta _3^4\eta _4^2\eta _6, \eta _1\eta _2\eta _7\eta _8\eta _9\in L_2$ . Now assume that $\eta _1^2\eta _2^2$ is larger than the other three terms in the maximum in equation (21); then the maximum over $L_2$ can only be attained by $\eta _1^3\eta _2^3\eta _3^4\eta _4^2\eta _6$ or $\eta _1\eta _2\eta _7\eta _8\eta _9$ , and the inverse inequality follows.

Finally, we note that $\left\lvert \eta _3\right\rvert =\left\lvert \eta _4\right\rvert =\left\lvert \eta _6\right\rvert =\left\lvert \eta _7\right\rvert =1$ implies $\left\lvert \eta _1^2\eta _2^2\right\rvert =\left\lvert x_1\right\rvert $ and $\left\lvert \eta _8\eta _9\right\rvert =\left\lvert x_4\right\rvert $ .

Proof. We combine Lemma 11 and Lemma 14. The $\#D_i$ coordinates $\eta _j$ belonging to irreducible components $E_j\subset D_i$ satisfy $\left\lvert \eta _j\right\rvert =1$ . By symmetry, we can further assume that $\eta _j=1$ , making the $2^6$ -to- $1$ -correspondence from Lemma 11 a $2^{6-\#D_i}$ -to- $1$ -correspondence.

Proof. The proof is as in [Reference Derenthal11, Lemma 8.4], with slightly different height functions and some $\eta _i=1$ , which leads to different error terms. In the first case, using the second height condition, the error term is

$$ \begin{align*} \ll \sum_{\eta_1, \dots, \eta_6} 2^{\omega(\eta_3)+\omega(\eta_3\eta_4\eta_5\eta_6)} \ll \sum_{\eta_2, \dots, \eta_6} \frac{2^{\omega(\eta_3)+\omega(\eta_3\eta_4\eta_5\eta_6)}B}{|\eta_2\eta_3^2\eta_4^2\eta_5^2\eta_6^2|} \ll B\log B. \end{align*} $$

In the second case, using the second and the third height conditions, it is

$$ \begin{align*} \ll \sum_{\eta_1, \eta_2, \eta_5, \eta_7} 2^{\omega(\eta_5)} \ll \sum_{\eta_1, \eta_5} \frac{2^{\omega(\eta_5)}B}{|\eta_1\eta_5^2|} \ll B\log B. \end{align*} $$

In the third case, using the second height condition, it is

$$ \begin{align*} \ll \sum_{\eta_1, \eta_2, \eta_5} 2^{\omega(\eta_5)} \ll \sum_{\eta_1, \eta_5} \frac{2^{\omega(\eta_5)}B}{|\eta_1\eta_5^2|} \ll B\log B. \end{align*} $$

The remaining cases are very similar.

Proof. In the first case, by equation (18), the last height condition is

(23) $$ \begin{align} |(\eta_2\eta_8^2+\eta_4\eta_5^3\eta_6^2\eta_8)/\eta_1| \le B. \end{align} $$

Hence, by [Reference Derenthal11, Lemma 5.1(4)],

$$ \begin{align*} V_{1,1}(\eta_1, \dots, \eta_6;B) \ll \frac{B^{1/2}}{|\eta_1\eta_2|^{1/2}} = \frac{B}{|\eta_1\eta_2\eta_3\eta_4\eta_5\eta_6|} \left(\frac{B}{|\eta_1\eta_2\eta_3^2\eta_4^2\eta_5^2\eta_6^2|}\right)^{-1/2}. \end{align*} $$

In the second case, we use

$$ \begin{align*} V_{2,1}(\eta_1, \eta_2, \eta_5, \eta_7; B) \ll \frac{B}{|\eta_1\eta_2\eta_5\eta_7|}. \end{align*} $$

In the third case, we use

$$ \begin{align*} V_{3,1}(\eta_1, \eta_2, \eta_5; B) \ll \frac{B}{|\eta_1\eta_2\eta_5|}. \end{align*} $$

Therefore, [Reference Derenthal11, Proposition 4.3, Corollary 7.10] gives the result in the first three cases. The final cases are similar to the second and third cases.

Proof. We must show that adding the condition $|\eta _2\eta _3^2\eta _4^2\eta _5^2\eta _6^2| \le B$ , removing the condition $|\eta _1| \ge 1$ , and replacing equation (23) by $|\eta _2\eta _8^2/\eta _1| \le B$ in the integration domain changes the integral by $\ll B(\log B)^4$ .

Adding the condition $|\eta _2\eta _3^2\eta _4^2\eta _5^2\eta _6^2| \le B$ does not change $V_{1,0}(B)$ since this inequality follows from $|\eta _1| \ge B$ and the second height condition. Afterwards, we can remove the condition $|\eta _1| \ge 1$ from $V_{1,0}(B)$ since this changes the integral by

$$ \begin{align*} \int_{\substack{|\eta_1|\le 1,\ |\eta_2|,\dots,|\eta_6|\ge 1\\ \mathcal{H}_1(\eta_1,\dots,\eta_6,\eta_8)\le B\\ |\eta_2\eta_3^2\eta_4^2\eta_5^2\eta_6^2| \le B}} \frac{\mathop{}\!\mathrm{d} \eta_1 \dots \mathop{}\!\mathrm{d} \eta_6 \mathop{}\!\mathrm{d} \eta_8}{|\eta_1|} \ll \int_{\substack{|\eta_1| \le 1,\ |\eta_2|,\dots,|\eta_6|\ge 1\\ |\eta_2\eta_3^2\eta_4^2\eta_5^2\eta_6^2| \le B}} \frac{B^{1/2} \mathop{}\!\mathrm{d} \eta_1 \dots \mathop{}\!\mathrm{d} \eta_6}{|\eta_1\eta_2|^{1/2}}, \end{align*} $$

where we estimate the integral over $\eta _8$ as in the proof of Lemma 17. Now we observe that the new condition $|\eta _2\eta _3^2\eta _4^2\eta _5^2\eta _6^2| \le B$ together with $|\eta _2|,\dots ,|\eta _6| \ge 1$ implies $|\eta _2|,\dots ,|\eta _6|\le B$ ; all these conditions and $|\eta _1|\le 1$ allow us to bound the error as required.

Finally, we must replace equation (23) by $|\eta _2\eta _8^2/\eta _1| \le B$ . A comparison of $\mathcal {H}_1$ in Cox coordinates (Lemma 14) with $H_1$ as in equation (3) motivates the transformation

(24) $$ \begin{align} \eta_8=\frac{B}{\eta_2\eta_3\eta_4\eta_5\eta_6}x_0,\quad \eta_1=\frac{B}{\eta_2\eta_3^2\eta_4^2\eta_5^2\eta_6^2} x_2, \end{align} $$

which turns $\frac {\mathop {}\!\mathrm {d} \eta _1 \mathop {}\!\mathrm {d} \eta _8}{|\eta _1|}$ into $\frac {B\mathop {}\!\mathrm {d} x_0 \mathop {}\!\mathrm {d} x_2}{|x_2\eta _2\eta _3\eta _4\eta _5\eta _6|}$ , and the transformation $\eta _3 =\frac {B}{\eta _4^2\eta _5^4\eta _6^3} x_3$ , which turns $\frac {\mathop {}\!\mathrm {d} \eta _3}{|\eta _3|}$ into $\frac {\mathop {}\!\mathrm {d} x_3}{|x_3|}$ . These transformations turn $\mathcal {H}_1(\eta _1,\dots ,\eta _6,\eta _8) \le B$ into

$$ \begin{align*} |x_0|,|x_2|,|x_3|,|x_0(x_0+x_3)/x_2| \le 1. \end{align*} $$

Furthermore, they turn $|\eta _3| \ge 1$ and $|\eta _2\eta _3^2\eta _4^2\eta _5^2\eta _6^2| \le B$ (which imply $|\eta _2\eta _4^2\eta _5^2\eta _6^2| \le B$ ) into a condition $X_3 \le |x_3| \le X_3'$ for certain $X_3$ and $X_3'$ , whose values (depending on $\eta _2,\eta _4,\eta _5,\eta _5,B$ ) will not matter to us. Altogether, this shows that

$$ \begin{align*} V_{1,0}(B) = \int_{\substack{|\eta_2|, |\eta_4|, |\eta_5|, |\eta_6| \ge 1\\|\eta_2 \eta_4^2 \eta_5^2 \eta_6^2|\le B}} W(\eta_2,\eta_4,\eta_5,\eta_6,B) \frac{B\mathop{}\!\mathrm{d} \eta_2 \mathop{}\!\mathrm{d} \eta_4 \mathop{}\!\mathrm{d} \eta_5 \mathop{}\!\mathrm{d} \eta_6}{|\eta_2\eta_4\eta_5\eta_6|} + O(B(\log B)^4) \end{align*} $$

with

$$ \begin{align*} W(\eta_2,\eta_4,\eta_5,\eta_6,B) = \int_{\substack{ |x_0|,|x_2|,|x_3|,|x_0(x_0+x_3)/x_2| \le 1\\ X_3 \le |x_3| \le X_3' }} \frac{\mathop{}\!\mathrm{d} x_0 \mathop{}\!\mathrm{d} x_2 \mathop{}\!\mathrm{d} x_3}{|x_2 x_3|}\text{.} \end{align*} $$

Now the following Lemma 19 shows that we can replace $|x_0(x_0+x_3)/x_2|\le 1$ by $|x_0^2/x_2| \le 1$ with an error of $O(1)$ . We plug this back into $V_{1,0}(B)$ and observe that the integral of $O(1)\cdot B/|\eta _2\eta _4\eta _5\eta _6|$ is $\ll B(\log B)^4$ , while the inverse of our previous transformations turn the main term into $V_{1,0}'(B)$ since they turn $|x_0^2/x_2| \le 1$ into $|\eta _2\eta _8^2/\eta _1| \le B$ and since we can remove the condition $|\eta _2 \eta _4^2 \eta _5^2 \eta _6^2|\le B$ , which is implied by the others.

Proof. As a first step, we integrate over $x_2$ to get

(25) $$ \begin{align} W(\eta_2,\eta_4,\eta_5,\eta_6,B) = \int_{\substack{ |x_0(x_0+x_3)| \le 1 \\ |x_0|,|x_3| \le 1 \\ X_3 \le |x_3| \le X_3' }} (-2\log |x_0| - 2 \log |x_0+x_3|) \frac{\mathop{}\!\mathrm{d} x_0 \mathop{}\!\mathrm{d} x_3}{|x_3|} \end{align} $$

and shall integrate the two terms individually.

To determine the integral over the first one, we remove the condition $|x_0(x_0+x_3)| \le 1$ , introducing an error of at most

$$\begin{align*}|R_1(\eta_2,\eta_4,\eta_5,\eta_6,B)| \le 4 \int_{\substack{ x_0,|x_3| \le 1, x_0\ge 0\\ |x_0(x_0+x_3)| \ge 1 \\ }} -\log x_0 \frac{\mathop{}\!\mathrm{d} x_0 \mathop{}\!\mathrm{d} x_3}{|x_3|} \end{align*}$$

by using the symmetry in the signs of $x_0$ and $x_3$ . The last inequality implies that $x_3$ has a distance of at least $1/|x_0|$ (which is $\geq 1$ ) from $-x_0$ . Since $x_0>0$ and $x_3>-1$ , it cannot be smaller, and thus $-x_0+1/x_0 \le x_3 \le 1$ holds. We thus get

$$ \begin{align*} |R_1(\eta_2,\eta_4,\eta_5,\eta_6,B)| &\ll \int_{0\leq x_0 \leq 1} -\log x_0 \left(\int_{-x_0+\frac{1}{x_0}\le x_3 \le 1} \frac{\mathop{}\!\mathrm{d} x_3}{|x_3|} \right) \mathop{}\!\mathrm{d} x_0 \\ & \ll \int_{0\leq x_0 \leq \frac{\sqrt{5}-1}{2}} \left| \log x_0 \log \left(-x_0 + \frac{1}{x_0}\right)\right| \mathop{}\!\mathrm{d} x_0 \ll 1\text{.} \end{align*} $$

We can now integrate the first term in equation (25) over $x_0$ and get

(26) $$ \begin{align} \int_{\substack{ |x_0|,|x_3|, |x_0(x_0+x_3)| \le 1 \\ X_3 \le |x_3| \le X_3' }} -2\log |x_0| \frac{\mathop{}\!\mathrm{d} x_0 \mathop{}\!\mathrm{d} x_3}{|x_3|} = \int_{\substack{ |x_3| \le 1 \\ X_3\le |x_3| \le X_3' }} 4 \frac{\mathop{}\!\mathrm{d} x_3}{|x_3|} + O(1)\text{.} \end{align} $$

To treat the second term, we begin with a change of variables $x_0'=x_0+x_3$ and add the condition $|x_0'|\le 1$ , introducing an error of at most

$$ \begin{align*} |R_2(\eta_2,\eta_4,\eta_5,\eta_6,B)| \le \int_{\substack{ |x_0'-x_3|,|x_3|, |x_0'(x_0'-x_3)| \le 1 \\ x_0'> 1 }} 4 \log x_0' \frac{\mathop{}\!\mathrm{d} x_0' \mathop{}\!\mathrm{d} x_3}{|x_3|}\text{,} \end{align*} $$

again using the symmetry of the integral. The third condition implies $|x_3-x_0'| \leq 1/|x_0'| < 1$ —that is, $x_0' - 1/x_0' < x_3$ —and thus we get

$$ \begin{align*} |R_2(\eta_2,\eta_4,\eta_5,\eta_6,B)| & \ll \int_{x_0'> 1} \log x_0' \left(\int_{x_0'-\frac{1}{x_0'}}^{1} \frac{\mathop{}\!\mathrm{d} x_3}{|x_3|} \right) \mathop{}\!\mathrm{d} x_0' \\ & \ll \int_{1< x_0' \le 2} \log x_0' \left\lvert\log\left(x_0'-\frac{1}{x_0'}\right)\right\rvert \mathop{}\!\mathrm{d} x_0' \ll 1\text{.} \end{align*} $$

(For the second inequality, note that $x_0' - 1/x_0'\le 1$ implies $x_0'\le 2$ .) Thus, the second term of equation (25) is

$$\begin{align*}\int_{\substack{ |x_0'(x_0'-x_3)| , |x_0'|,|x_0'-x_3|,|x_3| \le 1 \\ X_3 \le |x_3| \le X_3' }} - 2 \log |x_0'| \frac{\mathop{}\!\mathrm{d} x_0' \mathop{}\!\mathrm{d} x_3}{|x_3|} +O(1)\text{.} \end{align*}$$

The condition $|x_0'(x_0'-x_3)|\le 1$ is implied by the second and third condition, so we can remove it. Removing $|x_0'-x_3|\le 1$ introduces an error of at most

$$\begin{align*}|R_3(\eta_2,\eta_4,\eta_5,\eta_6,B)| \le \int_{\substack{ x_0',|x_3| \le 1\\ |x_0'-x_3|> 1\\ x_0'\ge 0 }} - 2 \log x_0' \frac{\mathop{}\!\mathrm{d} x_0' \mathop{}\!\mathrm{d} x_3}{|x_3|} \end{align*}$$

by the symmetry of the integral. The conditions imply $-1 \le x_3 \le x_0'-1$ and thus

$$ \begin{align*} |R_3(\eta_2,\eta_4,\eta_5,\eta_6,B)| &\ll \int_{0 \le x_0' \le 1} - \log x_0' \left(\int_{-1}^{x_0'-1} \frac{\mathop{}\!\mathrm{d} x_3}{|x_3|} \right) \mathop{}\!\mathrm{d} x_0'\\ & \ll \int_{0 \le x_0' \le 1} \log x_0' \log|x_0'-1|\mathop{}\!\mathrm{d} x_0' \ll 1\text{.} \end{align*} $$

Thus, the integral of the second summand of equation (25) is

(27) $$ \begin{align} \int_{\substack{ |x_0'|,|x_3| \le 1 \\ X_3 \le |x_3| \le X_3' }} - 2 \log |x_0'| \frac{\mathop{}\!\mathrm{d} x_0' \mathop{}\!\mathrm{d} x_3}{|x_3|} +O(1) = \int_{\substack{ |x_3| \le 1 \\ X_3 \le |x_3| \le X_3'}} 4 \frac{\mathop{}\!\mathrm{d} x_3}{|x_3|} +O(1)\text{.} \end{align} $$

Since

(28) $$ \begin{align} \int_{|x_0|,|x_2|,|x_0^2/x_2| \le 1} \frac{\mathop{}\!\mathrm{d} x_0 \mathop{}\!\mathrm{d} x_2}{|x_2|} = 8, \end{align} $$

adding equations (26) and (27) yields the desired result.

Proof. The difference that we must estimate is the integral over

$$\begin{align*}\left\lvert\eta_2\eta_5\eta_8\right\rvert, \left\lvert\eta_1\eta_2\eta_5^2\right\rvert, \left\lvert\eta_5^4\right\rvert, \left\lvert(\eta_2\eta_8^2+\eta_5^3\eta_8)/\eta_1\right\rvert \le B \le \left\lvert\eta_1^2\eta_2^2\right\rvert. \end{align*}$$

Using the condition $\left\lvert (\eta _2\eta _8^2+\eta _5^3\eta _8)/\eta _1\right\rvert \le B$ in [Reference Derenthal11, Lemma 5.1(4)], we have

$$ \begin{align*} |V_{3,0}'(B)-V_{3,0}(B)| \ll \int_{\substack{|\eta_1|,|\eta_2|,|\eta_5|\ge 1\\|\eta_1\eta_2\eta_5^2| \le B}} \frac{B^{1/2} \mathop{}\!\mathrm{d}\eta_1 \mathop{}\!\mathrm{d}\eta_2 \mathop{}\!\mathrm{d}\eta_5}{|\eta_1\eta_2|^{1/2}}. \end{align*} $$

The remaining conditions imply $|\eta _1|,|\eta _2|\le B$ . Now the result follows by integrating first over $|\eta _5| \le (B/|\eta _1\eta _2|)^{1/2}$ and then over $1 \le |\eta _1|,|\eta _2| \le B$ .

Proof. Again, we apply the coordinate change (24), which shows that

$$ \begin{align*} V_{1,0}'(B)= \int_{\substack{ |\eta_2|, |\eta_3|, |\eta_4|, |\eta_5|, |\eta_6| \ge 1\\ |\eta_3 \eta_4^2 \eta_5^4 \eta_6^3 |, |\eta_2 \eta_3^2 \eta_4^2 \eta_5^2\eta_6^2|\le B }} \frac{B\mathop{}\!\mathrm{d} \eta_2 \mathop{}\!\mathrm{d} \eta_3 \mathop{}\!\mathrm{d} \eta_4 \mathop{}\!\mathrm{d} \eta_5 \mathop{}\!\mathrm{d} \eta_6}{|\eta_2\eta_3\eta_4\eta_5\eta_6|}\cdot \int_{|x_0|,|x_2|,|x_0^2/x_2| \le 1} \frac{\mathop{}\!\mathrm{d} x_0 \mathop{}\!\mathrm{d} x_2}{|x_2|} \text{.} \end{align*} $$

The integral over $x_0,x_2$ is $8$ by equation (28). Restricting to positive $\eta _i$ introduces a factor of $2^5$ . Substituting $\eta _i = B^{t_i}$ turns $\mathop {}\!\mathrm {d}\eta _i/\eta _i$ into $\log B \mathop {}\!\mathrm {d} t_i$ , and we thus arrive at

$$ \begin{align*} V_{1,0}'(B) =2^5 \int_{\substack{ t_2,t_3,t_4,t_5,t_6 \ge 0\\ t_3 + 2 t_4 + 4 t_5 + 3 t_6 \le 1\\ t_2 + 2 t_3 + 2 t_4 + 2 t_5 + 2 t_6 \le 1 }} 8 B(\log B)^5\mathop{}\!\mathrm{d} t_2 \mathop{}\!\mathrm{d} t_3 \mathop{}\!\mathrm{d} t_4 \mathop{}\!\mathrm{d} t_5 \mathop{}\!\mathrm{d} t_6. \end{align*} $$

This integral can be interpreted as the volume of a polytope, which we compute using Magma.

For the second case, using the first height condition yields

$$ \begin{align*} V_{2,0}(B) &= 2\int_{|\eta_1|,|\eta_2|,|\eta_5|,|\eta_7| \ge 1,\ |\eta_1^2\eta_2^2|,|\eta_5^4\eta_7^2|\le B} \frac{B \mathop{}\!\mathrm{d}\eta_1 \mathop{}\!\mathrm{d}\eta_2 \mathop{}\!\mathrm{d}\eta_5 \mathop{}\!\mathrm{d}\eta_7}{|\eta_1\eta_2\eta_5\eta_7|}. \end{align*} $$

We proceed as in the first case; here, we can compute the volume by hand. The final two cases are analogous.

For the third case, we observe that $|\eta _1\eta _2\eta _5^2|$ can be ignored in the definition of $\mathcal {H}_3'$ since it is the geometric average of $|\eta _1^2\eta _2^2|$ and $|\eta _5^4|$ . Now the computation is very similar to the second case.

Proof. Consider the image $(x,y)$ of an integral point $\pi (\eta _1,\dots ,\eta _9) \in \widetilde {\mathcal {U}}_1(\mathbb {Z}_p)$ . Assume ${|x|< 1}$ . Then $\eta _5 \not \in \mathbb {Z}_p^\times $ or $\eta _6 \not \in \mathbb {Z}_p^\times $ (since $\eta _7\in \mathbb {Z}_p^\times $ ). In both cases, the coprimality conditions imply $\eta _8\in \mathbb {Z}_p^\times $ , and thus $|xy^2|=|\eta _7\eta _8^2/ \eta _1^3 \eta _2 \eta _3^3 \eta _4^2 \eta _6 | \ge 1$ .

On the other hand, let us consider a point $(x,y)$ in the above set and construct an integral point $(\eta _1,\dots ,\eta _9)$ on the torsor with $f(\pi (\eta _1,\dots ,\eta _9))=(x,y)$ . If $|x| < 1$ , we distinguish two cases for $|y|$ :

1. (i) If $1/|x|^{1/2} \le |y| < 1/|x|$ , let $\eta _5=xy$ , $\eta _6=1/xy^2$ , $\eta _9=-1-x/y$ and the remaining coordinates be $1$ . Then $\eta _9 \in -1 + p\mathbb {Z}_p\subset \mathbb {Z}_p^\times $ since $|x/y|\le |x|^{1/2} < 1$ , and thus the coprimality conditions are satisfied.

2. (ii) If $1/|x|\le |y|$ , let $\eta _4=1/xy$ , $\eta _6=x$ , $\eta _9=-1-x/y$ , and let all the other coordinates be $1$ . Since $|x/y| \le |x|^2 < 1$ , we again have $\eta _9 \in -1+p\mathbb {Z}_p\subset \mathbb {Z}_p^\times $ , and thus the coprimality conditions hold.

If $|x| \ge 1$ , we distinguish three cases for $|y|$ .

1. (i) If $|y| < 1$ , let $\eta _2=1/x$ , $\eta _8=y$ , $\eta _9=-1-y/x$ and the remaining coordinates be $1$ . Then $\eta _9\in -1+p\mathbb {Z}_p \subset \mathbb {Z}_p^\times $ since $|y/x| < 1$ .

2. (ii) If $1 \le |y| < |x|$ , let $\eta _3=1/y$ , $\eta _2=y/x$ , $\eta _9=-1-y/x$ and the remaining coordinates be $1$ . Again, we have $|y/x| < 1$ so that $\eta _9\in -1+p\mathbb {Z}_p\subset \mathbb {Z}_p^\times $ .

3. (iii) Finally, if $|x|\le |y|$ , let $\eta _3=1/x$ , $\eta _4=x/y$ , $\eta _1=-1-x/y$ and the remaining coordinates be $1$ . If $|y|> |x|$ , we have $\eta _1\in -1+p\mathbb {Z}_p\subset \mathbb {Z}_p^\times $ ; if $|x|=|y|$ , we have $\eta _4\in \mathbb {Z}_p^\times $ . In both cases, the coprimality conditions on the torsor are satisfied.

We now turn to $\widetilde {\mathcal {U}}_2$ . Let $(x,y)$ be in the image of the set of integral points. If $|y|>1$ , we have either $|\eta _5|<1$ or $|\eta _1|<1$ . In the first case, we get $|xy^2|=|\eta _7\eta _8^2/\eta _1^3\eta _2|=|\eta _7|\le 1$ (since all other variables have to be units); for the second case, we note that

$$\begin{align*}x+y = \frac{\eta_4\eta_5^3\eta_6^2\eta_7 + \eta_2\eta_8}{\eta_1\eta_2\eta_3\eta_4\eta_5\eta_6} = - \frac{\eta_1\eta_9}{\eta_1\eta_2\eta_3\eta_4\eta_6}, \end{align*}$$

and thus $|x+y|=|\eta _9|\le 1$ (since all other variables have to be units).

On the other hand, let $(x,y)$ be in the set on the right-hand side in the statement of the lemma. We want to construct an integral point on the torsor lying above $(x,y)$ . If $|y|\le 1$ and $|x|\le 1$ , let $\eta _8=y$ , $\eta _7=x$ , $\eta _9=-x-y$ and the remaining variables be $1$ , which satisfies the coprimality conditions. If $|y|\le 1$ and $|x|>1$ , let $\eta _8=y$ , $\eta _2=1/x$ , $\eta _9=-1-y/x$ and the remaining variables be $1$ . Then $\eta _9\in -1-p\mathbb {Z}_p\subset \mathbb {Z}_p^\times $ , so $(\eta _1,\dots ,\eta _9)$ is integral. Let now $|y|>1$ . If $|xy^2|\le 1$ , let $\eta _5=1/y$ , $\eta _7=xy^2$ , $\eta _9=-1-xy$ and the remaining variables be $1$ ; again, $\eta _9\in \mathbb {Z}_p^\times $ . Finally, if $|x+y|\le 1$ , let $\eta _1=1/x$ , $\eta _9=-x-y$ , $\eta _8=-\eta _1\eta _9-1$ and the remaining variables be $1$ . Then $\eta _8\in \mathbb {Z}_p^\times $ , so $(\eta _1,\dots ,\eta _9)$ is integral, and, since $\eta _8/\eta _1=(-\eta _1\eta _9-1)/\eta _1 = x+y-x=y$ , it indeed lies above $(x,y)$ . For the disjoint union description of $\widetilde {\mathcal {U}}_2$ , we just have to observe that $|y|>1$ and $|xy^2| \le 1$ implies $|x|=|y|^{-2}<1$ , while $|y|>1$ and $|x+y|\le 1$ implies $|x|=|y|>1$ .

The third set consists of points that are integral with respect to both $Q_1$ and $Q_2$ . Therefore, we obtain it as the intersection of the previous two sets. For the description of $\widetilde {\mathcal {U}}_3$ as a disjoint union, we start with the one of $\widetilde {\mathcal {U}}_2$ and intersect each set with $\widetilde {\mathcal {U}}_1$ . Here, $|y| \le 1$ implies $|x|\ge 1$ since otherwise $|xy^2|<1$ . Furthermore, $|y|>1$ and $|xy^2|\le 1$ implies $|x|<1$ ; hence, $|xy^2|\ge 1$ must hold. Finally, $|y|>1$ and $|x+y|\le 1$ implies $|x|=|y|>1$ .

The final two cases are analogous.

Proof. In the first case, we have

(29) $$ \begin{align} \mathop{}\!\mathrm{d} f_*\tau_{(\widetilde{S},D_1),v} = \|(\mathop{}\!\mathrm{d} x \wedge \mathop{}\!\mathrm{d} y) \otimes 1_{E_7} \|_{\omega_{\widetilde{S}}(D),v}^{-1} \mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y. \end{align} $$

To make sense of this, we need a metric on the log-canonical bundle, not just on a line bundle isomorphic to it. To this end, we consider the isomorphism between the canonical bundle $\omega _{\widetilde {S}}$ and the line bundle whose meromorphic sections are elements of degree $\omega _{\widetilde {S}}$ of the field of fractions of the Cox ring that maps $\mathop {}\!\mathrm {d} x \wedge \mathop {}\!\mathrm {d} y$ to $1/\eta _1^2\eta _2^2\eta _3^3\eta _4^2\eta _6$ ; in addition, we consider the isomorphisms between $\mathcal {O}(E_i)$ and the line bundles whose sections are elements of the Cox ring mapping $1_{E_i}$ to $\eta _i$ . Together, these induce an isomorphism from each $\omega _{\widetilde {S}}(D_1)$ to the line bundle whose sections are functions of the Cox ring of degree $\omega _{\widetilde {S}}(D_1)$ , and we can pull back the adelic metric we constructed along this isomorphism (and similarly for the log-canonical bundles in the remaining cases). In Cox coordinates, the norm in equation (29) at a point $\boldsymbol {\eta }$ is

(30) $$ \begin{align} \frac{|\eta_1^2\eta_2^2\eta_3^3\eta_4^2\eta_6|}{|\eta_7| \max\{|\eta_2\eta_3\eta_4\eta_5\eta_6\eta_8|,|\eta_1\eta_2\eta_3^2\eta_4^2\eta_5^2\eta_6^2|,|\eta_3\eta_4^2\eta_5^4\eta_6^3\eta_7|,|\eta_8\eta_9|\}}. \end{align} $$

In the second case, we can analogously determine the norm

$$\begin{align*}\|(\mathop{}\!\mathrm{d} x \wedge \mathop{}\!\mathrm{d} y) \otimes 1_{E_3} \otimes 1_{E_4} \otimes 1_{E_6}\|_{\omega_{\widetilde{S}}(D_2),v}^{-1} \end{align*}$$

in Cox coordinates:

(31) $$ \begin{align} \frac{1}{|\eta_3\eta_4\eta_6|} \frac{|\eta_1^2\eta_2^2\eta_3^3\eta_4^2\eta_6|}{ \max\{|\eta_2\eta_5\eta_7\eta_8|,|\eta_1^2\eta_2^2\eta_3^2\eta_4|,|\eta_4\eta_5^4\eta_6^2\eta_7^2|\}}. \end{align} $$

In the third case, the norm

$$\begin{align*}\|(\mathop{}\!\mathrm{d} x \wedge \mathop{}\!\mathrm{d} y) \otimes 1_{E_3} \otimes 1_{E_4} \otimes 1_{E_6} \otimes 1_{E_7}\|_{\omega_{\widetilde{S}}(D_3),v}^{-1} \end{align*}$$

at a point $\boldsymbol {\eta }$ in Cox coordinates is

(32) $$ \begin{align} \frac{1}{|\eta_3\eta_4\eta_6\eta_7|} \frac{|\eta_1^2\eta_2^2\eta_3^3\eta_4^2\eta_6|}{ \max\{|\eta_2\eta_5\eta_8|,|\eta_1\eta_2\eta_3\eta_4\eta_5^2\eta_6|,|\eta_4\eta_5^4\eta_6^2\eta_7|, M_0(\boldsymbol{\eta})\}} \end{align} $$

with

$$ \begin{align*} M_0(\boldsymbol{\eta}) = \frac{\left\lvert\eta_1^3 \eta_2^3 \eta_3^2 \eta_4 \eta_8 \eta_9\right\rvert}{\max_{s \in B} \{\left\lvert s(\boldsymbol{\eta})\right\rvert\}}. \end{align*} $$

Then $M_0(g(x,y))=M(x,y)$ as above after removing terms that can never contribute to the minimum.

In the remaining two cases, the norms of interest are

$$ \begin{align*} &\frac{ 1 }{ \left\lvert\eta_3\eta_4\eta_5\eta_6\eta_7\right\rvert } \frac{|\eta_1^2\eta_2^2\eta_3^3\eta_4^2\eta_6|}{\max\{\eta_2\eta_8, \eta_1\eta_2\eta_3\eta_4\eta_5\eta_6, \eta_4\eta_5^3\eta_6^2\eta_7\}} \quad \text{and}\\ & \frac{1}{\left\lvert\eta_2\eta_3\eta_4\eta_6\right\rvert} \frac{|\eta_1^2\eta_2^2\eta_3^3\eta_4^2\eta_6|}{\max\{\left\lvert\eta_5\eta_7\eta_8\right\rvert,\left\lvert\eta_1^2\eta_2\eta_3^2\eta_4\right\rvert,\left\lvert\eta_1\eta_3\eta_4\eta_5^2\eta_6\eta_7\right\rvert\}}, \end{align*} $$

respectively.

Proof. We compute

(33) $$ \begin{align} \tau_{(\widetilde{S},D_i),p}(\widetilde{\mathcal{U}}_i(\mathbb{Z}_p))=\int_{f(\widetilde{\mathcal{U}}_i(\mathbb{Z}_p)\cap V'(\mathbb{Q}_p))} \mathop{}\!\mathrm{d} f_*\tau_{(\widetilde{S},D_i),p} \end{align} $$

for $i\in \{1,\dots ,5\}$ . For $i=1$ , the previous two lemmas transform this into

$$\begin{align*}\int_{\substack{x,y\in\mathbb{Q}_p\\|x|\ge 1 \text{ or }|xy^2| \ge 1}} \frac{1}{|x|\max\{|y|,1,|x|,|y(y+x)|\}} \mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y. \end{align*}$$

Subdividing the domain of integration into the regions with $|x|>|y|$ , $|x|=|y|$ , and $|x|<|y|$ in order to simplify the denominator, we get

(34) $$ \begin{align} \begin{aligned} &\int_{\substack{|y|<|x|\\|x|\ge 1}} \frac{1}{|x|\max\{|x|,|xy|\}}\mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y + \int_{\substack{|y|=|x|\\|x|\ge 1}} \frac{1}{|x|\max\{|x|,|y(y+x)|\}} \mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y \\ &\qquad + \int_{\substack{|x| < |y|\\ |xy^2| \ge 1}} \frac{1}{|xy^2|} \mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y \end{aligned} \end{align} $$

after simplifying the description of the domains ( $|x|<1$ would imply $|xy^2| \le |x|^3<1$ in the first two cases; $|y|^2<1/|x|$ would imply $|y|^2 <1/|x| \le 1 \le |x|^2 < |y|^2 $ in the third case).

The first of the integrals in equation (34) is

(35) $$ \begin{align} &\int_{|x|\ge 1} \frac{1}{|x|^2} \int_{|y|<|x|} \frac{1}{\max\{1,|y|\}} \mathop{}\!\mathrm{d} y \mathop{}\!\mathrm{d} x = \int_{|x|\ge 1} \frac{1}{|x|^2} \left(\frac{1}{p} + \int_{1\le |y| < |x|} \frac{1}{|y|} \mathop{}\!\mathrm{d} y\right) \mathop{}\!\mathrm{d} x \nonumber\\ & \quad = \int_{|x|\ge 1} \frac{1}{|x|^2}\left(\frac{1}{p} + \left(1-\frac{1}{p}\right)|v(x)| \right) \mathop{}\!\mathrm{d} x = \frac{1}{p} + \sum_{\delta\ge 0} \left(1-\frac{1}{p}\right)^2\frac{\delta}{p^\delta} = \frac{2}{p} \text{,} \end{align} $$

while the second integral is

(36) $$ \begin{align} \int_{\substack{|y+x|\le \frac{1}{p} \\ |x|\ge 1}} \frac{1}{|x|^2} + \int_{\substack{|y+x| \ge 1 \\ |x|\ge 1, |y| = |x|}} \frac{1}{|xy(x+y)|}. \end{align} $$

The first integral in equation (36) is $\frac {1}{p}\int _{|x|\ge 1}\frac {1}{|x|^2}\mathop {}\!\mathrm {d} x = \frac {1}{p}$ . Turning to the second one, we note that $|x| = |y|$ is implied by the ultrametric triangle inequality if $|x+y|<|x|$ . The set of $y\in \mathbb {Q}_p$ with $|x+y|=|x|$ and $|y|=|x|$ has volume $|x|-2|x|/p$ since the two sets $\{y \mid |y-0|< |x|\}$ and $\{y \mid |y+x|< |x|\}$ have volume $|x|/p$ and are disjoint (because $|y|<|x|$ implies $|y+x|=|x|$ ). We thus get

$$ \begin{align*} &\int_{|x| \ge 1} \frac{1}{|x|^2} \left(\sum_{0\leq \delta< |v(x)|} \left(1-\frac{1}{p}\right)\frac{p^\delta}{p^\delta} + \left(1-\frac{2}{p}\right)\frac{|x|}{|x|}\right) \mathop{}\!\mathrm{d} x \\ & \quad = \int_{|x| \ge 1} \frac{1}{|x|^2} \left( \left(1-\frac{1}{p}\right)|v(x)| + \left(1-\frac{2}{p}\right) \right)\mathop{}\!\mathrm{d} x = \frac{1}{p} + 1-\frac{2}{p} = 1 - \frac{1}{p}\text{,} \end{align*} $$

computing the integral over x similarly as in equation (35). The second integral in equation (34) thus evaluates to $1$ . Finally, the third integral in equation (34) is

$$ \begin{align*} &\int \frac{1}{|y|^2} \int_{1/|y|^2 \le |x| < |y|} \frac{1}{|x|} \mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y = \int_{|y|\ge 1} \frac{1}{|y|^2} \sum_{-2|v(y)|\le \delta < |v(y)|} \left(1-\frac{1}{p}\right) \mathop{}\!\mathrm{d} y \\ & \quad = \int_{|y| \ge 1} \left(1-\frac{1}{p} \right) \frac{3|v(y)|}{|y|^2} = \frac{3}{p}\text{,} \end{align*} $$

again computed analogously to the previous ones. Adding the three terms in equation (34), we arrive at our claim for $i=1$ .

For $i=2$ , we get

$$ \begin{align*} &\int_{|y|\le 1\text{, } |xy^2| \le 1 \text {, or } |x+y|\le 1} \frac{1}{|x|\max\{|y|,|x^{-1}|,|x|\}} \mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y \\ & \quad = \int_{|y|\le 1} \frac{1}{\max\{1,|x^2|\}} \mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y + \int_{\substack{|y|> 1 \\ |x+y|\le 1}} \frac{1}{|y^2|} \mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y + \int_{\substack{|y|> 1 \\ |x|\le 1/|y|^2}} \mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y \end{align*} $$

for the integral (33) (since $|x|=|y|$ in the second case). The first integral is then

$$\begin{align*}1 + \int_{|x|>1}\frac{1}{|x^2|} \mathop{}\!\mathrm{d} x = 1+\frac{1}{p}, \end{align*}$$

while the other two integrals are

$$\begin{align*}\int_{|y|>1}\frac{1}{|y^2|} \mathop{}\!\mathrm{d} y = \frac{1}{p}. \end{align*}$$

For $i=3$ , we compute the integral on the right hand side of equation (33) to be

$$ \begin{align*} &\int_{f(\widetilde{\mathcal{U}}_3(\mathbb{Z}_p)\cap V'(\mathbb{Q}_p))} \frac{1}{|x|\max\{|y|,1,|x|,M(g(x,y))\}} \mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y \\ & \quad = \int_{|y|\le 1,\ |x| \ge 1} \frac{1}{|x^2|} \mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y + \int_{\substack{|y|> 1 \\ |x+y|\le 1}} \frac{1}{|y^2|} \mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y + \int_{\substack{|y|> 1 \\ |x|= 1/|y|^2}} \mathop{}\!\mathrm{d} x \mathop{}\!\mathrm{d} y \end{align*} $$

(again using that $|x|=|y|$ in the second case). The first integral is then

$$\begin{align*}\int_{|x| \ge 1}\frac{1}{|x^2|} \mathop{}\!\mathrm{d} x = 1, \end{align*}$$

while the second one is

$$\begin{align*}\int_{|y|>1}\frac{1}{|y^2|} \mathop{}\!\mathrm{d} y = \frac{1}{p}, \end{align*}$$

and the third one is

$$ \begin{align*} \left(1-\frac 1 p\right)\int_{|y|>1} \frac{1}{|y^2|} \mathop{}\!\mathrm{d} y = \frac{1}{p} - \frac{1}{p^2}. \end{align*} $$

The final two cases are similar: for $i=4$ , the integrals over the two disjoint sets in Lemma 22 are $1$ and $p^{-1}$ , respectively, while, for $i=5$ , the integrals over the three disjoint sets are $1$ , $p^{-1}$ and $p^{-1}$ , respectively.

Proof. Following [Reference Chambert-Loir and Tschinkel7, 2.1.12], we can compute the unnormalized Tamagawa volume of $E_7$ by integrating

$$\begin{align*}\|\mathop{}\!\mathrm{d} y \|_{\omega_{E_7},\infty}^{-1}= \lim_{x\to 0} \left(|x| \| (\mathop{}\!\mathrm{d} x \wedge \mathop{}\!\mathrm{d} y) \otimes 1_{E_7}\|_{\omega_{\widetilde{S}}(E_7), \infty}^{-1} \right)\text{.} \end{align*}$$

Again evaluating equation (30) in the image of $(x,y)$ , we get the volume

$$ \begin{align*} \tau^\prime_{1,E_7,\infty}(E_7(\mathbb{R}))&=\int_{\mathbb{R}} \lim_{x\to 0}\frac{|x|}{|x|\max\{ 1,|x|,|y|,|y(y+x)|\}}\mathop{}\!\mathrm{d} y \\ &= \int_{\mathbb{R}} \frac{1}{\max\{1,|y^2|\}} \mathop{}\!\mathrm{d} y =4, \end{align*} $$

which we normalize by multiplying with $c_{\mathbb {R}}=2$ .

For the second and third cases, we work in neighborhoods of the two intersection points $D_{A_1}=E_3\cap E_4$ and $D_{A_2}=E_4 \cap E_6$ . The Tamagawa measures on these points are simply real numbers. In order to compute them, we consider the charts

$$ \begin{align*} g' &\colon \mathbb{A}^2_{\mathbb{Q}} \to \widetilde{S},\ (a,b) \mapsto (1:1:a:b:1:1:1:1:-1-b)\ \quad \text{and}\\ g" &\colon \mathbb{A}^2_{\mathbb{Q}} \to \widetilde{S},\ (c,d) \mapsto (1:1:1:c:1:d:1:1:-1-cd). \end{align*} $$

We have $x=1/a=d$ , $y=1/ab=1/cd$ for these charts. Since

$$ \begin{align*} \|\mathop {}\!\mathrm {d} x \wedge \mathop {}\!\mathrm {d} y \|= |\det (J_{f\circ g'})|\| \mathop {}\!\mathrm {d} a\wedge \mathop {}\!\mathrm {d} b\|, \end{align*} $$

we can use equation (31) to compute the norms

$$ \begin{align*} \|(\mathop{}\!\mathrm{d} a \wedge \mathop{}\!\mathrm{d} b) \otimes 1_{E_3} \otimes 1_{E_4} \otimes 1_{E_6}\|_{\omega_{\widetilde{S}}(D_2), \infty} &= \max\{|a^3b^2|,|ab|,|ab^2|\} \quad \text{and}\\ \|(\mathop{}\!\mathrm{d} c \wedge \mathop{}\!\mathrm{d} d) \otimes 1_{E_3} \otimes 1_{E_4} \otimes 1_{E_6}\|_{\omega_{\widetilde{S}}(D_2), \infty} & = \max\{|c^2d|,|cd|,|c^2d^3|\}\text{.} \end{align*} $$

Analogously to the first case, we now arrive at

$$\begin{align*}\tau^\prime_{2,D_{A_1},\infty} = \lim_{(a,b)\to (0,0)} \frac{|ab|}{\max\{|a^3b^2|,|ab|,|ab^2|\}} = 1, \end{align*}$$

and, similarly, $\tau ^\prime _{2,D_{A_2},\infty }=1$ for the unnormalized measures on the points $D_{A_i}(\mathbb {R})$ , which we multiply with $c_{\mathbb {R}}^2=4$ .

In the third case, using the same change of variables and the description (32) of the metric in Cox coordinates, we get

$$\begin{align*}\left\lVert(\mathop{}\!\mathrm{d} a \wedge \mathop{}\!\mathrm{d} b) \otimes 1_{E_3} \otimes 1_{E_4} \otimes 1_{E_6} \otimes 1_{E_7}\right\rVert_{\omega_{\widetilde{S}}(D_3), \infty} = \left\lvert ab\right\rvert\max\{1,\left\lvert ab\right\rvert, \left\lvert b\right\rvert, M_0(g'(a,b))\} \end{align*}$$

with $M_0(g'(ab))\to 0$ , as $(a,b)\to (0,0)$ , whence $\tau ^\prime _{3,D_{A_1},\infty }=1$ for the unnormalized measure. Finally,

$$\begin{align*}\left\lVert(\mathop{}\!\mathrm{d} c \wedge \mathop{}\!\mathrm{d} d) \otimes 1_{E_3} \otimes 1_{E_4} \otimes 1_{E_6} \otimes 1_{E_7}\right\rVert_{\omega_{\widetilde{S}}(D_3), \infty} = \left\lvert cd\right\rvert\max\{1,\left\lvert cd\right\rvert, \left\lvert cd^2\right\rvert, M_0(g"(c,d))\}, \end{align*}$$

where again $M_0(g"(c,d))\to 0$ , and we get $\tau ^\prime _{3,D_{A_2},\infty }=1$ for the unnormalized measure. Again, we multiply both measures with $c_{\mathbb {R}}^2=4$ .

The computations in the cases $i=4,5$ are analogous.

Proof. To compute $\alpha _{i,A}$ , we choose $j_0 \in \{1,\dots ,7\}$ such that $E_{j_0} \in A$ and such that the classes of $E_j$ for $j \in \{1,\dots ,7\} \setminus \{j_0\}$ form a basis of $\operatorname {\mathrm {Pic}} \widetilde {S}$ . The latter holds for $j_0 \in \{1,2,3,6,7\}$ since the data in [Reference Derenthal12] show that $\operatorname {\mathrm {Pic}} \widetilde {S}$ has rank $6$ and is generated by the classes of the negative curves $E_1, \dots , E_7$ , where

(38) $$ \begin{align} E_1+E_2+E_3-2E_5-E_6-E_7 \end{align} $$

is a principal divisor. An inspection of Figure 4 shows that $E_{j_0} \in A$ for some $j_0 \in \{3,6,7\}$ .

Hence, there are unique linear combinations $\sum _{j \ne j_0} a_j E_j$ of class $\omega _{\widetilde {S}}(D_i)^\vee $ and $\sum _{j \ne j_0} b_j E_j$ of the same class as $E_{j_0}$ ; the coefficients $a_j,b_j \in \mathbb {Z}$ can be computed using equation (38) and the fact that $2E_1+2E_2+3E_3+2E_4+E_6$ has anticanonical class by equation (17). For the following computations, it is useful to know that

(39) $$ \begin{align} 2E_4+4E_5+3E_6+2E_7,\quad 2E_1+2E_2+3E_3+2E_4,\quad E_1+E_2+2E_3+2E_4+2E_5+2E_6 \end{align} $$

have class $\omega _{\widetilde {S}}(E_{j_0})^\vee $ for $j_0=3,6,7$ , respectively (expressed without using $E_{j_0}$ ).

Let $J = \{j \in \{1,\dots ,7\} \mid E_j \subset D_i,\ E_j \notin A\}$ and $J' = \{1,\dots ,7\} \setminus (J \cup \{j_0\})$ . By definition (37),

$$ \begin{align*} \operatorname{\mathrm{Pic}} \widetilde{U}_{i,A} = (\operatorname{\mathrm{Pic}} \widetilde{S})/\langle E_j \mid j \in J\rangle; \end{align*} $$

hence, a basis is given by the classes of $E_j$ for $j \in J'$ modulo the classes of $E_j$ for $j \in J$ , and its effective cone is generated by the classes of $E_j$ for $j \in J' \cup \{j_0\}$ modulo the classes of $E_j$ for $j \in J$ . Working with the dual basis, we obtain

$$ \begin{align*} \alpha_{i,A} = \operatorname{\mathrm{vol}}\left\{(t_j) \in \mathbb{R}_{\ge{}0}^{J'} \mid \sum_{j \in J'} a_j t_j =1,\ \sum_{j \in J'} b_j t_j\ge 0\right\}. \end{align*} $$

If $A=\{E_{j_0},E_{j_1}\}$ is a $1$ -simplex, then $j_1 \in J$ , and the next step is to eliminate the variable $t_{j_1}$ using the equation, which gives a description of $\alpha _{i,A}$ as the volume of a polytope in $\mathbb {R}_{\ge {}0}^{J_i}$ with $J_i$ as in equation (22) defined by two inequalities.

In the first case, we have $\widetilde {U}_{1,E_7}=\widetilde {S}$ and corresponding effective cone $\operatorname {\mathrm {Eff}}\widetilde {S}$ , whose dual is the nef cone of $\widetilde {S}$ . Working with the dual basis of the classes of $E_1,\dots ,E_6$ and using descriptions (38) and (39) for $E_7$ and $\omega _{\widetilde {S}}(D_1)^\vee $ , we obtain

$$ \begin{align*} \alpha_1=\operatorname{\mathrm{vol}}\left\{(t_1,\dots, t_6) \in \mathbb{R}_{\ge{}0}^6\ \left|\ \begin{aligned} &t_1+t_2+t_3-2t_5-t_6 \ge 0\\ &t_1+t_2+2t_3+2t_4+2t_5+2t_6 = 1\end{aligned}\right.\right\} \end{align*} $$

and eliminate $t_1$ .

In the second case, there are two constants $\alpha _{2,A_i}$ associated with the maximal faces $A_1=\{E_3,E_4\}$ and $A_2=\{E_4,E_6\}$ of the Clemens complex. The subvarieties used in their definition are $\widetilde {U}_{2,A_1}=\widetilde {S} \setminus E_6$ and $\widetilde {U}_{2,A_2}=\widetilde {S} \setminus E_3$ . In the first case, we have $J=\{6\}$ , choose $j_0=3$ and obtain $J'=\{1,2,4,5,7\}$ . Therefore, the Picard group of $\widetilde {U}_{A_1}$ is $(\operatorname {\mathrm {Pic}}\widetilde {S})/\langle E_6 \rangle $ with a basis is given by the classes of $E_1,E_2,E_4,E_5,E_7$ modulo $E_6$ , and its effective cone is generated by the classes of $E_1,\dots ,E_5,E_7$ modulo $E_6$ . Since $E_3$ has the same class as $-E_1-E_2+2E_5+E_6+E_7$ in $\operatorname {\mathrm {Pic}} \widetilde {S}$ by equation (38), while $E_4+4E_5+2E_6+2E_7$ has class $\omega _{\widetilde {S}}(D_2)^\vee $ by equation (39), we obtain (working modulo $E_6$ )

$$ \begin{align*} \alpha_{2,A_1} = \operatorname{\mathrm{vol}}\left\{(t_1,t_2,t_4,t_5,t_7) \in \mathbb{R}_{\ge{}0}^5\ \left|\ \begin{aligned} &-t_1-t_2+2t_5+t_7 \ge 0\\ &t_4+4t_5+2t_7 = 1\end{aligned}\right.\right\} \end{align*} $$

and eliminate $t_4$ .

The computation of $\alpha _{2,A_2}$ is similar. Here, we choose $j_0=6$ , and our basis is given by the classes of $E_1,E_2,E_4,E_5,E_7$ modulo $E_3$ . The divisor $E_6$ has the same class as $E_1+E_2+E_3-2E_5-E_7$ , while $2E_1+2E_2+2E_3+E_4$ has class $\omega _{\widetilde {S}}(D_2)^\vee $ . Therefore,

$$ \begin{align*} \alpha_{2,A_2} = \operatorname{\mathrm{vol}}\left\{(t_1,t_2,t_4,t_5,t_7) \in \mathbb{R}_{\ge{}0}^5\ \left|\ \begin{aligned} &t_1+t_2-2t_5-t_7 \ge 0\\ &2t_1+2t_2+t_4 = 1\end{aligned}\right.\right\}; \end{align*} $$

again, we eliminate $t_4$ .

The further cases are analogous. The only exceptional case is the computation of $\alpha _{4,A_1}$ . Working with $J=\{5,6,7\}$ , $j_0=3$ and $J'=\{1,2,4\}$ , a similar computation as for $\alpha _{2,A_1}$ shows

$$ \begin{align*} \alpha_{4,A_1} = \operatorname{\mathrm{vol}}\left\{(t_1,t_2,t_4) \in \mathbb{R}_{\ge{}0}^5\ \left|\ \begin{aligned} &-t_1-t_2 \ge 0,\\ &t_4 = 1\end{aligned}\right.\right\}, \end{align*} $$

which clearly has volume $0$ in the hyperplane $t_4=1$ .

Proof. This follows from Lemma 25 and Lemma 26. For $i=2,\dots ,6$ , we observe that the polytopes of volumes $\alpha _{i,A}$ in Lemma 26 fit together to the one appearing in the description of $C_i$ in Lemma 21.