Proof. According to [Reference Arzhantsev, Derenthal, Hausen and Laface2, Before Proposition 2.1.2.7], there are two exact sequences

$$ \begin{align*}0 \to L \to \mathbb{Z}^{\Sigma(1)} \to N \to 0,\\ 0 \leftarrow \operatorname*{\mathrm{Cl}}(Y) \leftarrow \mathbb{Z}^{\Sigma(1)} \leftarrow M \leftarrow 0, \end{align*} $$

which are dual to each other. Here, $\mathbb {Z}^{\Sigma (1)}$ denotes the lattice with basis $\{e_\rho : \rho \in \Sigma (1)\}$ , which is assumed to be dual to itself. The top right map sends $e_\rho $ to $u_\rho $ while the lower left map sends $e_\rho $ to $[D_\rho ]$ . Since the top right map sends $e_{\rho _1}, \dots e_{\rho _k}$ to a basis of N, the lower left map sends their complement to a basis of $\operatorname *{\mathrm {Cl}}(Y)$ .

Proof. For the existence, choose an effective U-invariant $\mathbb {Q}$ -Weil divisor D on Y with $[D]=-K_X$ . Let M be the character lattice of the torus U. We write $U_\sigma \subseteq Y$ for the open subset corresponding to the cone $\sigma $ .

Choose $\chi _\sigma \in M_{\mathbb {Q}}$ such that $(\operatorname {\mathrm {div}} \chi _\sigma )_{|U_\sigma } = D_{|U_\sigma }$ . Define . Then $D(\sigma )$ is of class $-K_X$ and its support is contained in $\bigcup _{\rho \notin \sigma (1)} D_\rho $ . Moreover, a multiple of $-K_X$ being globally generated means that we have $\chi _\sigma \le \chi _{\sigma '}$ on $\sigma '$ for every $\sigma ' \in \Sigma _{\mathrm {max}}$ [Reference Cox, Little and Schenck25, Theorem 6.1.7]. Hence, $D(\sigma )$ is an effective $\mathbb {Q}$ -divisor.

Because of (2.7), there is a unique $\mathbb {Z}$ -linear combination of the $D_\rho $ with $\rho \notin \sigma (1)$ of class $-K_X$ , which must be equal to $D(\sigma )$ .

Proof. Arguing as in [Reference Salberger65, (11.5)], but using the description of $\widetilde {Z}_Y$ by the primitive collections shows

$$ \begin{align*} \widetilde{Y}_0(\mathbb{Z}) = \{\mathbf{y} \in \mathbb{Z}^{\Sigma(1)} : \gcd\{y_\rho : \rho \in S_j\}=1 \text{ for all } j=1,\dots,r\}. \end{align*} $$

Since $\widetilde {X}$ is defined by $\Phi $ in $\widetilde {Y}$ , the first result follows. The description of $\widetilde {X}(\mathbb {Z}_p)$ is obtained similarly.

By [Reference Salberger65, Lemma 11.4], $\widetilde {\pi }$ induces a $2^{\operatorname {\mathrm {rk}} \operatorname *{\mathrm {Cl}} Y} : 1$ -map $\widetilde {Y}_0(\mathbb {Z}) \to \widetilde {Y}(\mathbb {Z}) = Y(\mathbb {Q})$ . Restricting to the points where $\Phi $ vanishes gives the result.

Proof. For the first statement, note that $x^{D_0}(x^{D(\sigma )}\Phi )^{-1}$ corresponds to the divisor $D_0-D(\sigma )-X$ .

On $U^\sigma $ , we have

(3.2) $$ \begin{align} \varpi^\sigma = \frac{\pm x^{D_0}}{x^{D(\sigma)}\Phi} \bigwedge_{\rho \in \sigma(1)} \frac{\,{\mathrm d} z^\sigma_\rho}{z^\sigma_\rho} \in \Gamma(U^\sigma, \omega_Y(X)) \end{align} $$

where $\Gamma (U^\sigma , \omega _Y(X)) = \Gamma (U^\sigma , \omega _Y(D(\sigma )+X))$ since $D(\sigma )_{|U_{\sigma }} = 0$ by Lemma 2.3. With $\beta ^\sigma _\rho $ as in (2.13), let

$$ \begin{align*} \lambda = \frac{x^{D_0}}{x^{D(\sigma)}\prod_{\rho\notin \sigma(1)}x_\rho^{\beta^\sigma_\rho}}. \end{align*} $$

In view of (2.12), we obtain

$$ \begin{align*} \varpi^\sigma = \frac{\pm \lambda}{\Phi^\sigma} \bigwedge_{\rho \in \sigma(1)} \frac{\,{\mathrm d} z^\sigma_\rho}{z^\sigma_\rho} \in \Gamma(U^\sigma, \omega_Y(X)). \end{align*} $$

On $U_\sigma $ , we have

$$ \begin{align*} \operatorname{\mathrm{div}} \lambda = (\operatorname{\mathrm{div}} x^{D_0})_{|U_\sigma} - (\operatorname{\mathrm{div}} x^{D(\sigma)})_{|U_\sigma} - \sum_{\rho \notin \sigma(1)} \beta_\rho^\sigma D_\rho = (\operatorname{\mathrm{div}} x^{D_0})_{|U_\sigma} - 0 - 0 = (\operatorname{\mathrm{div}} x^{D_0})_{|U_\sigma}. \end{align*} $$

We also have $\operatorname {\mathrm {div}} \prod _{\rho \in \sigma (1)} z_\rho ^\sigma = (\operatorname {\mathrm {div}} x^{D_0})_{|U_\sigma }$ . Therefore, $\lambda = \prod _{\rho \in \sigma (1)} z_\rho ^\sigma $ on $U_\sigma $ , and we obtain the second statement.

Proof. This is similar to [Reference Blomer, Brüdern and Salberger8, Lemma 13]. Since $s_Y$ generates the $\mathscr {O}_Y$ -module $\omega _Y(D_0)$ , each

$$ \begin{align*}\varpi^\sigma = \frac{x^{D_0}}{x^{D(\sigma)} \Phi} s_Y\end{align*} $$

generates the $\mathscr {O}_Y$ -module $\omega _Y(X+D(\sigma ))$ . Since $\iota ^*\mathscr {O}_Y(D(\sigma )) = \mathscr {O}_X(D(\sigma ) \cap X)$ (using that $X \not \subseteq \operatorname *{\mathrm {supp}} D(\sigma )$ ), the isomorphism $\iota ^*\omega _Y(X) \to \omega _X$ adjoint to $\operatorname {\mathrm {Res}} \colon \omega _Y(X) \to \iota _* \omega _X$ induces an isomorphism $\iota ^*\omega _Y(X+D(\sigma )) \to \omega _X(D(\sigma ) \cap X)$ that maps $\iota ^*\varpi ^\sigma $ to $\operatorname {\mathrm {Res}} \varpi ^\sigma $ . Hence, $\operatorname {\mathrm {Res}}\varpi ^\sigma $ generates $\omega _X(D(\sigma ) \cap X)$ , that is, it is a nowhere vanishing global section.

Proof. The previous lemma shows that $\tau ^\sigma $ , as a global section of $\omega _X^{-1}$ , has corresponding divisor $D(\sigma ) \cap X$ , whose support is contained in $X \cap \bigcup _{\rho \notin \sigma } D_\rho $ , which is the complement of $X^\sigma $ (2.11).

Proof. Let $P \in X(\mathbb {Q}_p)$ , and let $\tau $ be a local section of $\omega _X^{-1}$ not vanishing in P. Choose $\xi \in \Sigma _{\mathrm {max}}$ such that $|(\tau ^\xi /\tau )(P)|_p = \max _{\sigma \in \Sigma _{\mathrm {max}}} |(\tau ^\sigma /\tau )(P)|_p$ , which is positive by Lemma 3.3 and the fact that the sets $X^\sigma $ cover X (2.11); in particular, $\tau ^\xi $ does not vanish in P. Hence, we can compute

$$ \begin{align*} \|\tau^\xi(P)\|_p^{-1} = \max_{\sigma \in \Sigma_{\mathrm{max}}} \left|\frac{\tau^\sigma}{\tau^\xi}(P)\right|_p = \max_{\sigma \in \Sigma_{\mathrm{max}}} \frac{|(\tau^\sigma/\tau)(P)|_p}{|(\tau^\xi/\tau)(P)|_p} = 1. \end{align*} $$

On the other hand, for each $\sigma \in \Sigma _{\mathrm {max}}$ , the section $\tau ^\sigma $ extends to a global section $\widetilde \tau ^\sigma $ of $\omega _{\widetilde {X}/\mathbb {Z}_p}^{-1}$ , and $\omega _{\widetilde {X}/\mathbb {Z}_p}^{-1}$ is generated by the set of all these $\widetilde \tau ^\sigma $ as an $\mathscr {O}_{\widetilde {X}}$ -module. The computation above shows for every $\sigma \in \Sigma _{\mathrm {max}}$ that $\left|\frac {\tau ^\sigma }{\tau ^\xi }(P)\right|_p \le 1$ , hence $\tau ^\sigma (P) = a_\sigma \tau ^\xi (P)$ for some $a_\sigma \in \mathbb {Z}_p$ in the $\mathbb {Q}_p$ -module $\omega _X^{-1}(P)$ , and hence also $\widetilde \tau ^\sigma (P) = a_\sigma \widetilde \tau ^\xi (P)$ in the $\mathbb {Z}_p$ -module ${\widetilde P}^*(\omega _{\widetilde {X}/\mathbb {Z}_p}^{-1})$ . Therefore, ${\widetilde P}^*(\omega _{\widetilde {X}/\mathbb {Z}_p}^{-1})$ is generated by $\tau ^\xi (P)$ and consequently $\|\tau ^\xi (P)\|_p^*=1$ by definition of the model norm. Finally, we have

$$ \begin{align*} \|\tau(P)\|_p = |(\tau/\tau^\xi)(P)|_p \cdot \|\tau^\xi(P)\|_p = |(\tau/\tau^\xi)(P)|_p \cdot \|\tau^\xi(P)\|_p^* = \|\tau(P)\|_p^*. here \end{align*} $$

Proof. Since the sets $X^\sigma $ as in (2.11) for $\sigma \in \Sigma _{\mathrm {max}}$ cover X, our point P is contained in $X^{\xi }(\mathbb {Q})$ for some $\xi \in \Sigma _{\mathrm {max}}$ . By Lemma 3.3, we can compute $H(P)$ with as in (3.5). We have $\varpi ^\sigma = x^{-D(\sigma )}x^{D(\xi )}\varpi ^{\xi }$ by definition (3.1). Since $\operatorname {\mathrm {Res}}$ is an $\mathscr {O}_Y$ -module homomorphism (3.3), this implies $\tau ^\sigma = x^{D(\sigma )}x^{-D(\xi )}\tau ^{\xi }$ . Therefore,

(3.8) $$ \begin{align} \|\tau^{\xi}(P)\|_v^{-1} = \max_{\sigma \in \Sigma_{\mathrm{max}}} \left|\frac{\tau^\sigma}{\tau^{\xi}}(P)\right|_v = \max_{\sigma \in \Sigma_{\mathrm{max}}}\left|\frac{x^{D(\sigma)}}{x^{D(\xi)}}(P)\right|_v, \end{align} $$

hence our claim holds for . By the product formula, it follows for arbitrary $F_0$ not vanishing in P.

Proof. Let $P = \pi (P_0) \in X(\mathbb {Q})$ . For $F_0$ of degree $-K_X$ not vanishing in P and $\sigma \in \Sigma _{\mathrm {max}}$ , we can compute $(x^{D(\sigma )}/F_0)(P)$ as in Lemma 3.5, but we can also regard $x^{D(\sigma )}$ and $F_0$ as regular functions on $X_0$ that can be evaluated in $P_0$ . Here, we have $x^{D(\sigma )}(P_0)/F_0(P_0) = (x^{D(\sigma )}/F_0)(P)$ . Using Lemma 3.5, we obtain

$$ \begin{align*} H_0(P_0)=H(P) = \prod_v \max_{\sigma \in \Sigma_{\mathrm{max}}} \left|\frac{x^{D(\sigma)}}{F_0}(P)\right|_v = \prod_v \max_{\sigma \in \Sigma_{\mathrm{max}}} \left|\frac{x^{D(\sigma)}(P_0)}{F_0(P_0)}\right|, \end{align*} $$

and $\prod _v |F_0(P_0)|_v = 1$ by the product formula.

Proof. Let p be a prime and $P_0 \in \widetilde {X}_0(\mathbb {Z}_p)$ . Then $P_0\ \text {mod}\ p$ is in $\widetilde {X}_0(\mathbb {F}_p)$ . Since $\widetilde {X}_0$ is defined by the irrelevant ideal in $\widetilde {X}_1$ as in (2.15), there is a $\xi \in \Sigma _{\mathrm {max}}$ such that $x^{\underline {\xi }}(P_0\ \text {mod}\ p) \ne 0 \in \mathbb {F}_p$ . Since the support of $D(\xi )$ is as in Lemma 2.3, we have $x^{D(\xi )}(P_0\ \text {mod}\ p) \ne 0 \in \mathbb {F}_p$ , and hence $|x^{D(\xi )}(P_0)|_p=1$ . Using $x^{D(\sigma )}(P_0) \in \mathbb {Z}_p$ for all $\sigma \in \Sigma _{\mathrm {max}}$ , we conclude $\max _{\sigma \in \Sigma _{\mathrm {max}}} |x^{D(\sigma )}(P_0)|_p=1$ .

Therefore, for $P_0 \in \widetilde {X}_0(\mathbb {Z})$ , only the Archimedean factor in Lemma 3.6 remains.

Proof. We combine the $2^{\operatorname {\mathrm {rk}}\operatorname {\mathrm {Pic}} X} : 1$ -map and the description of $\widetilde {X}_0(\mathbb {Z})$ from Proposition 2.4 with the lifted height function in Corollary 3.7. The preimage of $U(\mathbb {Q})$ in $\widetilde {X}_0(\mathbb {Z})$ is the set where $x_\rho \ne 0$ for all $\rho \in \Sigma (1)$ .

Proof. In the notation of the proof of Lemma 2.3, write $D = \sum _{\rho }a_\rho D_\rho $ . Then the $-\chi _\sigma $ are the vertices, and possibly (if $-K_X$ is not ample) some other points, of the $\operatorname {\mathrm {rk}} M$ -dimensional polytope

$$ \begin{align*} P_D = \{\chi \in M_{\mathbb{Q}} : \langle n_\rho, \chi\rangle \ge -a_\rho \text{ for all}\ \rho\}\text{;} \end{align*} $$

see [Reference Cox, Little and Schenck25, §4.3 and after Lemma 9.3.9].

Now, consider the injective affine map $\phi \colon M_{\mathbb {Q}} \to \mathbb {Q}^{\Sigma (1)}$ , $\chi \mapsto \sum _{\rho } (a_\rho + \langle n_\rho , \chi \rangle )e_\rho $ as well as the linear surjective map $p\colon \mathbb {Q}^{\Sigma (1)} \to (\operatorname *{\mathrm {Cl}} Y)_{\mathbb {Q}}$ . We have $\operatorname {\mathrm {rk}} M = J - \operatorname {\mathrm {rk}} \operatorname {\mathrm {Pic}} X$ and $\operatorname {im}(p \circ \phi ) = \{-K_X\}$ . Moreover, the condition $\phi (\chi ) \in \mathbb {Q}^{\Sigma (1)}_{\ge 0}$ is equivalent to $\langle n_\rho , \chi \rangle \ge -a_\rho $ for all $\rho $ . It follows that $\phi $ restricts to a bijection $P_D \to Q = p^{-1}(-K_X) \cap \mathbb {Q}^{\Sigma (1)}_{\ge 0}$ . Hence, Q is bounded and of dimension $J - \operatorname {\mathrm {rk}} \operatorname {\mathrm {Pic}} X$ .

As we have $p(-\chi _\sigma ) = D(\sigma )$ , where $D(\sigma )$ is interpreted as an element of $\mathbb {Z}^{\Sigma (1)}$ in the obvious way, we obtain $\mathscr {V} \subseteq \phi (\{D(\sigma ) : \sigma \in \Sigma _{\mathrm {max}}\}) \subseteq Q$ . Hence, the equality

$$ \begin{align*} \max_{\sigma \in \Sigma_{\mathrm{max}}} |\mathbf{x}^{D(\sigma)}|_v = \max_{\mathbf{v} \in \mathscr{V}} |\mathbf{x}^{\mathbf{v}}|_v \end{align*} $$

holds, and, since $\phi (M) \subseteq \mathbb {Z}^{\Sigma (1)}$ , we also obtain $\mathscr {V} \subset \mathbb {Z}_{\ge 0}^{\Sigma (1)}$ .

Proof. According to Lemma 3.9, the polytope Q spans an affine subspace of dimension $J-\operatorname {\mathrm {rk}}\operatorname {\mathrm {Pic}} X$ in $\mathbb {R}^{J}$ , which does not contain $0$ since $-K_X \ne 0$ . It follows that Q spans a vector space of dimension $J-\operatorname {\mathrm {rk}}\operatorname {\mathrm {Pic}} X + 1$ in $\mathbb {R}^{J}$ . This shows $\operatorname {\mathrm {rk}} \mathscr {A}_1 = J - \operatorname {\mathrm {rk}}\operatorname {\mathrm {Pic}} X + 1$ .

Since the columns of $\mathscr {A}_1$ lie in an affine subspace of $\mathbb {R}^J$ that does not contain $0$ , a linear combination of these columns can be $0$ only if the sum of the coefficients is $0$ . It follows that we have $\operatorname {\mathrm {rk}} \left(\begin {smallmatrix}\mathscr {A}_1 \\ \mathscr {A}_3\end {smallmatrix}\right) = \operatorname {\mathrm {rk}} \mathscr {A}_1 $ . Since $\Phi $ is $\operatorname {\mathrm {Pic}} X$ -homogeneous, the first $k-1$ columns of $\mathscr {A}_2$ lie in $p^{-1}(0)$ . Moreover, note that the last column of $\mathscr {A}_2$ lies in $p^{-1}(K_X)$ since $\deg \Phi -\sum _{\rho \in \Sigma (1)} \deg (x_\rho )=K_X$ by [Reference Arzhantsev, Derenthal, Hausen and Laface2, Proposition 3.3.3.2]. Together with the fact that the columns of $\mathscr {A}_1$ lie in $p^{-1}(-K_X)$ , we obtain $ \operatorname {\mathrm {rk}} \mathscr {A} = \operatorname {\mathrm {rk}} \left(\begin {smallmatrix}\mathscr {A}_1 \\ \mathscr {A}_3\end {smallmatrix}\right)\text {.}$

Proof. According to [Reference Arzhantsev, Derenthal, Hausen and Laface2, Proposition 3.3.3.2], we have $\mathbf {\tau }_1 \in p^{-1}(-K_X)$ . It follows from $Q = p^{-1}(-K_X) \cap \mathbb {Q}_{\ge 0}^{\Sigma (1)}$ that the relative interior of Q satisfies $Q^\circ \supseteq p^{-1}(-K_X) \cap \mathbb {Q}_{> 0}^{\Sigma (1)}$ . Since all coordinates of $\mathbf {\tau }_1$ are positive, we obtain $\mathbf {\tau }_1 \in Q^\circ $ . Since the columns of $\mathscr {A}_1$ are the vertices of Q, the column $\tau _1^\top $ can be written as a linear combination of the columns of $\mathscr {A}_1$ with strictly positive coefficients whose sum is $1$ . The existence of $\mathbf {\sigma } \in \mathbb {R}^N_{> 0}$ as required follows.

Proof. As in (2.14), the implicit function theorem gives a v-adic neighborhood $U_0 \subseteq X^\xi (\mathbb {Q}_v)$ of P and an implicit function $\phi \colon V \to \mathbb {Q}_v$ for $V = \pi ^\xi _{\rho _0}(U_0) \subseteq \mathbb {Q}_v^{\xi (1)\setminus \{\rho _0\}}$ such that $\Phi ^\xi (\mathbf {z}^\xi )=0$ for all $\mathbf {z}^\xi \in X^\xi (\mathbb {Q}_v)$ with $z^\xi _{\rho _0}$ the image of $(z^\xi _\rho )_{\rho \in \xi (1)\setminus \{\rho _0\}} \in V$ under $\phi $ . We work with $\|\tau ^\xi (P)\|_v$ as in (3.5) and use $x^{D(\xi )}(\mathbf {z}^\xi )=1$ (see (2.8)) in our affine coordinates on $X^\xi (\mathbb {Q}_v)$ . Then the formulas in [Reference Peyre60, (2.2.1)] and [Reference Salberger65, Theorem 1.10] give (4.2) for $N_v \subseteq U_0$ . Indeed, our chart is

In this chart, by (3.4), the image of the local canonical section $\bigwedge _{\rho \in \xi (1) \setminus \{\rho _0\}} \,{\mathrm d} z^\xi _\rho $ under

$$ \begin{align*} \omega(\pi) \colon \pi^*\omega_{\mathbb{A}_{\mathbb{Q}}^{\xi(1) \setminus \{\rho_0\}}} \to \omega_X \end{align*} $$

is $\partial \Phi ^\xi /\partial z^\xi _{\rho _0} \cdot \operatorname {\mathrm {Res}}\varpi ^\xi $ . This implies that the image of the local anticanonical section $\bigwedge _{\rho \in \xi (1) \setminus \{\rho _0\}} \frac {\partial }{\partial z^\xi _{\rho }}$ under

$$ \begin{align*} {}^t\omega(\pi)^{-1} \colon \pi^*\omega^{-1}_{\mathbb{A}_{\mathbb{Q}}^{\xi(1) \setminus \{\rho_0\}}} \to \omega^{-1}_X \end{align*} $$

is $(\partial \Phi ^\xi /\partial z^\xi _{\rho _0})^{-1} \cdot \tau ^\xi $ . Therefore, $\mu _v(N_v)$ for $N_v \subseteq U_0$ as defined in [Peyre95, (2.2.1)] is the integral over $\pi (N_v)$ of

$$ \begin{align*} \omega_v &= \|((\partial \Phi^\xi/\partial z^\xi_{\rho_0})^{-1}\cdot\tau^\xi)(\pi^{-1}((z^\xi_\rho)_{\rho \in \xi(1)\setminus\{\rho_0\}}))\|_v \bigwedge_{\rho \in \xi(1) \setminus \{\rho_0\}} \,{\mathrm d} z^\xi_\rho\\ &=|\partial \Phi^\xi/\partial z^\xi_{\rho_0}(\mathbf{z}^\xi)|_v^{-1} \cdot \|\tau^\xi(\mathbf{z}^\xi)\|_v\bigwedge_{\rho \in \xi(1) \setminus \{\rho_0\}} \,{\mathrm d} z^\xi_\rho. \end{align*} $$

Using (3.8) together with $x^{D(\xi )}(\mathbf {z}^\xi )=1$ , we obtain (4.2).

By (3.4), we see that the right-hand side of (4.1) coincides with (4.2) for $N_v \subseteq U_0$ . Since X is smooth, $X^\xi (\mathbb {Q}_v)$ can be covered with such $U_0$ , hence $\mu _v(N_v)$ is equal to the right-hand side for all $N_v \subseteq X^\xi (\mathbb {Q}_v)$ . Since $\varpi ^\sigma /\varpi ^\xi = x^{D(\xi )}/x^{D(\sigma )}$ by definition (3.1), we have $\tau ^\sigma \operatorname {\mathrm {Res}}\varpi ^\xi = \tau ^\sigma /\tau ^\xi = x^{D(\sigma )}/x^{D(\xi )}$ by (3.5), and hence the integrals in (4.1) are equal.

Proof. See [Reference Blomer, Brüdern and Salberger8, Lemma 16].

Proof. Our proof follows [Reference Blomer, Brüdern and Salberger8, Lemma 18]. By [Reference Salberger65, pp. 126–127], the map $\pi \colon X_0 \to X$ induces an v-adic analytic torsor $\pi _v \colon X_0(\mathbb {Q}_v) \to X(\mathbb {Q}_v)$ under $T(\mathbb {Q}_v)$ . By [Reference Salberger65, Theorem 1.22] and the previous lemma, the relative volume form $s_{X_0/X}$ defines v-adic measures on the fibers of $\pi _v$ over $X(\mathbb {Q}_v)$ . Integrating along these fibers gives a linear functional $\Lambda _v \colon C_c(X_0(\mathbb {Q}_v)) \to C_c(X(\mathbb {Q}_v))$ .

Let $\chi _p\colon X_0(\mathbb {Q}_p) \to \{0,1\}$ be the characteristic function of $\widetilde {X}_0(\mathbb {Z}_p) \subset \widetilde {X}_0(\mathbb {Q}_p) = X_0(\mathbb {Q}_p)$ . Since $\chi _p \in C_c(X_0(\mathbb {Q}_p))$ , we have $m_p(\widetilde {X}_0(\mathbb {Z}_p)) = \int _{X(\mathbb {Q}_p)} \Lambda _p(\chi _p) \mu _p$ .

We claim that $(\Lambda _p(\chi _p))(P) = (1-p^{-1})^{\operatorname {\mathrm {rk}} \operatorname {\mathrm {Pic}} X}$ for every $P \in X(\mathbb {Q}_p) = \widetilde {X}(\mathbb {Z}_p)$ . Indeed, we have $s_{\widetilde {Y}_0} = s_{\widetilde {Y}_0/\widetilde {Y}} \otimes \pi ^* s_{\widetilde {Y}}$ , where $s_{\widetilde {Y}_0/\widetilde {Y}}$ is the extension of $s_{Y_0/Y}$ to a $\widetilde {T}$ -equivariant generator of $\omega _{\widetilde {Y}_0/\widetilde {Y}}$ . Furthermore, $s_{X_0/X}$ extends to a $\widetilde {T}$ -equivariant generator $s_{\widetilde {X}_0/\widetilde {X}}$ of $\omega _{\widetilde {X}_0/\widetilde {X}}$ . For a point $P \in \widetilde {X}(\mathbb {Z}_p)$ , the torsor $\widetilde {X}_0\to \widetilde {X}$ can be pulled back to $(\widetilde {X}_0)_P \to P$ , and hence $s_{\widetilde {X}_0/\widetilde {X}}$ pulls back to a $\widetilde {T}_{\mathbb {Z}_p}$ -equivariant global section $s_{(\widetilde {X}_0)_P}$ on $\omega _{(\widetilde {X}_0)_P/\mathbb {Z}_p}$ . But the torsor over P is trivial, and $\widetilde {T} \cong \mathbb {G}_{\mathrm {m}}^r$ with $r = \operatorname {\mathrm {rk}}\operatorname {\mathrm {Pic}} X$ , hence there are affine coordinates $(t_1,\dots ,t_r)$ for the affine $\mathbb {Z}_p$ -scheme $(\widetilde {X}_0)_P$ with $s_{(\widetilde {X}_0)_P} = \,{\mathrm d} t_1/t_1 \wedge \dots \wedge \,{\mathrm d} t_r/t_r$ . Therefore,

$$ \begin{align*} (\Lambda_p(\chi_p))(P) = \int_{(\widetilde{X}_0)_P(\mathbb{Z}_p)} |s_{(\widetilde{X}_0)_P}|_p = \Big(\int_{\mathbb{Z}_p^\times} \frac{\,{\mathrm d} t}{t}\Big)^r = (1- p^{-1})^r.\\[-44pt] \end{align*} $$

Proof. Since X is smooth, $X_0$ is also smooth. Hence, for any $\mathbf {y} \in X_0(\mathbb {Q}_p)$ , we have $\partial \Phi /\partial x_\rho (\mathbf {y}) \ne 0$ for some $\rho \in \Sigma (1)$ . In particular, for any $\mathbf {y} \in \widetilde {X}_0(\mathbb {Z}_p)$ , the valuation $v_p(\partial \Phi /\partial x_\rho (\mathbf {y}))$ is finite for some $\rho $ . Hence, is finite.

There is an $\ell _1$ such that $I_p(\mathbf {y})<\ell _1$ for all $\mathbf {y} \in \widetilde {X}_0(\mathbb {Z}_p)$ . To see this, assume the contrary. Then there is a sequence $\mathbf {y}_1,\mathbf {y}_2,\ldots \in \widetilde {X}_0(\mathbb {Z}_p)$ with $I_p(\mathbf {y}_j)\ge j$ for all j. The description of $\widetilde {X}_0(\mathbb {Z}_p)$ in Proposition 2.4 shows that this sequence has an accumulation point $\mathbf {y}_0 \in \widetilde {X}_0(\mathbb {Z}_p)$ : Infinitely many $\mathbf {y}_i$ have the same first p-adic digits, infinitely many of these have the same second p-adic digits and so on; we obtain $\mathbf {y}_0$ by using these p-adic digits; $\Phi (\mathbf {y}_0)=0$ since $\Phi $ is continuous, and $\mathbf {y}_0$ satisfies the coprimality conditions since these depend only on the first p-adic digits. Passing to a subsequence, we may assume that $\mathbf {y}_0$ is the limit of the sequence $(\mathbf {y}_j)_j$ . Then $\partial \Phi /\partial x_\rho (\mathbf {y}_0) = \lim _{j \to \infty } \partial \Phi /\partial x_\rho (\mathbf {y}_j) = 0$ for all $\rho \in \Sigma (1)$ . This contradicts the smoothness of X over $\mathbb {Q}_p$ .

Let $\ell \ge \ell _1$ and $\mathbf {x} \in \widetilde {X}_0(\mathbb {Z}/p^{\ell }\mathbb {Z})$ . For any $\mathbf {y} \in \widetilde {X}_0(\mathbb {Z}_p)_{\mathbf {x}}$ , the first $\ell $ digits of $\partial \Phi /\partial x_\rho (\mathbf {y})$ depend only on $\mathbf {x}$ , and since $I_p(\mathbf {y}) < \ell _1 \le \ell $ , at least one of these digits is nonzero for some $\rho \in \Sigma (1)$ . We choose $c_{\mathbf {x}}$ and $\rho _{\mathbf {x}}$ such that digit number $c_{\mathbf {x}}$ (i. e., the coefficient of $p^{c_{\mathbf {x}}}$ in the p-adic expansion) of $\partial \Phi /\partial x_{\rho _{\mathbf {x}}}(\mathbf {y})$ is nonzero, while all lower digits of $\partial \Phi /\partial x_\rho (\mathbf {y})$ for all $\rho \in \Sigma (1)$ are zero.

Proof. Let $\ell _1$ be as in Lemma 4.4. For $\mathbf {x} \in \widetilde {X}_0(\mathbb {Z}/p^{\ell _1}\mathbb {Z})$ and $\ell \ge \ell _1$ , let

We will see that

(4.7) $$ \begin{align}m_p(\widetilde{X}_0(\mathbb{Z}_p)_{\mathbf{x}}) = \frac{\# \widetilde{X}_0(\mathbb{Z}/p^{\ell}\mathbb{Z})_{\mathbf{x}}}{(p^{\ell})^{\#\Sigma(1)-1}} \end{align} $$

for all $\ell \ge \ell _1+c_{\mathbf {x}}$ with $c_{\mathbf {x}}<\ell _1$ as in Lemma 4.4. Since $\widetilde {X}_0(\mathbb {Z}_p)$ is the disjoint union of the sets $\widetilde {X}_0(\mathbb {Z}_p)_{\mathbf {x}}$ and $ \widetilde {X}_0(\mathbb {Z}/p^{\ell }\mathbb {Z})$ is the disjoint union of the sets $ \widetilde {X}_0(\mathbb {Z}/p^{\ell }\mathbb {Z})_{\mathbf {x}}$ for $\mathbf {x} \in \widetilde {X}_0(\mathbb {Z}/p^{\ell _1}\mathbb {Z})$ , our result follows for all .

For the proof of (4.7), we fix $\mathbf {x} \in \widetilde {X}_0(\mathbb {Z}/p^{\ell _1}\mathbb {Z})$ and let $c_{\mathbf {x}}, \rho _{\mathbf {x}}$ be as in Lemma 4.4. We claim that $\Phi (\mathbf {y}) \bmod {p^{\ell _1+c_{\mathbf {x}}}}$ is the same for all $\mathbf {y} \in \mathbb {Z}_p^{\Sigma (1)}$ with $\mathbf {y} \equiv \mathbf {x} \bmod {p^{\ell _1}}$ ; we write $\Phi ^*(\mathbf {x})$ for this value in $\mathbb {Z}/p^{\ell _1+c_{\mathbf {x}}}\mathbb {Z}$ . Indeed, for $\mathbf {y},\mathbf {y}' \in \mathbb {Z}_p^{\Sigma (1)}$ , we have

$$ \begin{align*} \Phi(\mathbf{y}') = \Phi(\mathbf{y}) + \sum_{\rho \in \Sigma(1)} (y_\rho'-y_\rho)\cdot \partial\Phi/\partial x_\rho(\mathbf{y}) + \sum_{\rho',\rho" \in \Sigma(1)} \Psi_{\rho',\rho"}(\mathbf{y},\mathbf{y}')(y_{\rho'}'-y_{\rho'})(y_{\rho"}'-y_{\rho"}) \end{align*} $$

for certain polynomials $\Psi _{\rho ',\rho "} \in \mathbb {Z}_p[X_\rho ,X^{\prime }_\rho : \rho \in \Sigma (1)]$ by Taylor expansion. If $\mathbf {y}' \equiv \mathbf {y} \bmod {p^{\ell _1}}$ , we conclude $\Phi (\mathbf {y}') \equiv \Phi (\mathbf {y}) \bmod {p^{\ell _1+c_{\mathbf {x}}}}$ .

If $\Phi ^*(\mathbf {x}) \ne 0 \in \mathbb {Z}/p^{\ell _1+c_{\mathbf {x}}}\mathbb {Z}$ , then there is no $\mathbf {y} \in \mathbb {Z}_p^{\Sigma (1)}$ with $\mathbf {y} \equiv \mathbf {x} \bmod {p^{\ell _1}}$ and $\Phi (\mathbf {y})=0$ , hence the set $\widetilde {X}_0(\mathbb {Z}_p)_{\mathbf {x}}$ is empty, and the same holds for $ \widetilde {X}_0(\mathbb {Z}/p^{\ell }\mathbb {Z})_{\mathbf {x}}$ for all $\ell \ge \ell _1+c_{\mathbf {x}}$ for similar reasons.

Now, assume $\Phi ^*(\mathbf {x}) = 0 \in \mathbb {Z}/p^{\ell _1+c_{\mathbf {x}}}\mathbb {Z}$ . By Hensel’s lemma, the map $\pi _{\rho _{\mathbf {x}}}$ that drops the $\rho _{\mathbf {x}}$ -coordinate defines an isomorphism from the integration domain $\widetilde {X}_0(\mathbb {Z}_p)_{\mathbf {x}}$ to the set

$$ \begin{align*} &\{(y_\rho)_{\rho \in \Sigma(1) \setminus \{\rho_{\mathbf{x}}\}} \in \mathbb{Z}_p^{\Sigma(1) \setminus \{\rho_{\mathbf{x}}\}} \mid y_\rho \equiv x_\rho \bmod{p^{\ell_1}}\text{ for all }\rho \in \Sigma(1) \setminus \{\rho_{\mathbf{x}}\}\}\\ ={}&\{(x_\rho+z_\rho)_{\rho \in \Sigma(1) \setminus \{\rho_{\mathbf{x}}\}} \mid z_\rho \in p^{\ell_1}\mathbb{Z}_p\} \cong (p^{\ell_1}\mathbb{Z}_p)^{\Sigma(1) \setminus \{\rho_{\mathbf{x}}\}.} \end{align*} $$

Therefore, by (4.5) and the first statement in Corollary 3.7,

$$ \begin{align*} m_p(\widetilde{X}_0(\mathbb{Z}_p)_{\mathbf{x}}) = \int_{\pi_{\rho_{\mathbf{x}}}(\widetilde{X}_0(\mathbb{Z}_p)_{\mathbf{x}})} \frac{\bigwedge_{\rho \in \Sigma(1)\setminus \{\rho_{\mathbf{x}}\}} \,{\mathrm d} y_\rho} {|\partial\Phi/\partial x_{\rho_{\mathbf{x}}}(\mathbf{y})|_p}, \end{align*} $$

where $y_{\rho _{\mathbf {x}}}$ is expressed in terms of the other coordinates using $\pi _{\rho _{\mathbf {x}}}^{-1}$ . We have $|\partial \Phi /\partial x_{\rho _{\mathbf {x}}}(\mathbf {y})|_p = p^{-c_{\mathbf {x}}}$ on the integration domain (Lemma 4.4). Thus,

$$ \begin{align*} m_p(\widetilde{X}_0(\mathbb{Z}_p)_{\mathbf{x}}) = \int_{(p^{\ell_1}\mathbb{Z}_p)^{\Sigma(1) \setminus \{\rho_{\mathbf{x}}\}} } \frac{\bigwedge_{\rho \in \Sigma(1) \setminus \{\rho_{\mathbf{x}}\}} \,{\mathrm d} z_\rho}{p^{-c_{\mathbf{x}}}} = p^{c_{\mathbf{x}}-\ell_1(\#\Sigma(1)-1)}. \end{align*} $$

On the other hand, by the discussion above, $\Phi ^*(\mathbf {x})=0 \in \mathbb {Z}/p^{\ell _1+c_{\mathbf {x}}}\mathbb {Z}$ means $\Phi (\mathbf {y})=0 \in \mathbb {Z}/p^{\ell _1+c_{\mathbf {x}}}\mathbb {Z}$ for all $\mathbf {y} \equiv \mathbf {x} \bmod {p^{\ell _1}}$ . Therefore,

$$ \begin{align*} \frac{\# \widetilde{X}_0(\mathbb{Z}/p^{\ell_1+c_{\mathbf{x}}}\mathbb{Z})_{\mathbf{x}}}{(p^{\ell_1+c_{\mathbf{x}}})^{\#\Sigma(1)-1}} = \frac{p^{c_{\mathbf{x}} \#\Sigma(1)}}{(p^{\ell_1+c_{\mathbf{x}}})^{\#\Sigma(1)-1}} = p^{c_{\mathbf{x}}-\ell_1(\#\Sigma(1)-1)}. \end{align*} $$

Using Hensel’s lemma as before, we see that $\# \widetilde {X}_0(\mathbb {Z}/p^{\ell }\mathbb {Z})_{\mathbf {x}}/(p^\ell )^{\#\Sigma (1)-1}$ has the same value for all $\ell \ge \ell _1+c_{\mathbf {x}}$ . This completes the proof of (4.7).

Proof. We combine Lemma 4.3 and Proposition 4.5 with (4.6).

Proof. As in (3.10), let $\mathscr {A}_3 = (1,\dots ,1) \in \mathbb {R}^{1\times \Sigma _{\mathrm {max}}} = \mathbb {R}^{1 \times N}$ . Let $\{e_\rho : \rho \in \Sigma (1)\} \cup \{e_0\}$ be the standard basis of $\mathbb {R}^{\Sigma (1)} \times \mathbb {R}$ . We define $\deg (e_\rho ) = \deg (x_\rho )$ for $\rho \in \Sigma (1)$ and $\deg (e_0) = K_X$ . Consider the sequence of linear maps

The second map is surjective, and the image of the first is contained in the kernel of the second. Since we have $\operatorname {\mathrm {rk}} \mathscr {A}_1 = \#\Sigma (1) + 1 - \operatorname {\mathrm {rk}} \operatorname {\mathrm {Pic}} X$ by Lemma 3.10, this sequence is exact. It follows that the dual sequence

is exact as well. Let $\{d^\vee _\rho : \rho \notin \sigma (1)'\} \cup \{K_X^\vee \}$ be the $\mathbb {R}$ -basis of $(\operatorname {\mathrm {Pic}} X)_{\mathbb {R}}^\vee $ dual to the $\mathbb {R}$ -basis of $(\operatorname {\mathrm {Pic}} X)_{\mathbb {R}}$ given above. We have

$$ \begin{align*} \deg^\vee(d_\rho^\vee) = e_\rho - \sum_{\rho' \in \sigma(1)'} b_{\rho,\rho'} e_{\rho'} \quad \text{and} \quad \deg^\vee(K_X^\vee) = e_0 - \sum_{\rho' \in \sigma(1)'} b_{\rho'} e_{\rho'} \end{align*} $$

for all $\rho \notin \sigma (1)'$ . Since these elements lie in the kernel of the leftmost map in the dual exact sequence, this gives the required relations between the rows of the matrix $\mathscr {A}_1$ and the row $\mathscr {A}_3$ .

Proof. Let $\operatorname {\mathrm {vol}}_{\mathbb {Z}}$ be the volume on $(\operatorname {\mathrm {Pic}} X)_{\mathbb {R}}$ defined by the lattice $\operatorname {\mathrm {Pic}} X$ , and let $\operatorname {\mathrm {vol}}_{\mathbb {R}}$ be the volume on $(\operatorname {\mathrm {Pic}} X)_{\mathbb {R}}$ defined by the basis $\{K_X\} \cup \{\deg (x_\rho ) : \rho \notin \sigma (1)'\}$ . Since the determinant of the transformation matrix is $-\alpha ^\sigma _{\rho _1}$ , we have $\operatorname {\mathrm {vol}}_{\mathbb {Z}} = |\alpha ^\sigma _{\rho _1}|\operatorname {\mathrm {vol}}_{\mathbb {R}}$ . For the corresponding dual volumes on $(\operatorname {\mathrm {Pic}} X)^\vee _{\mathbb {R}}$ , we have $\operatorname {\mathrm {vol}}^\vee _{\mathbb {Z}} = |\alpha ^\sigma _{\rho _1}|^{-1}\operatorname {\mathrm {vol}}^\vee _{\mathbb {R}}$ .

Peyre considers the unique $(\operatorname {\mathrm {rk}} \operatorname {\mathrm {Pic}} X-1)$ -form $\operatorname {\mathrm {vol}}_{\mathrm {P}}$ on $(\operatorname {\mathrm {Pic}} X)_{\mathbb {R}}^\vee $ such that $\operatorname {\mathrm {vol}}_{\mathrm {P}} \wedge K_X = \operatorname {\mathrm {vol}}^\vee _{\mathbb {Z}}$ . We also consider the form $\operatorname {\mathrm {vol}}_V = \bigwedge _{\rho \notin \sigma (1)'} \deg (x_\rho )$ . Note that we have $\operatorname {\mathrm {vol}}_V \wedge K_X = \operatorname {\mathrm {vol}}_{\mathbb {R}}^\vee $ . It follows that we have $\operatorname {\mathrm {vol}}_{\mathrm P} = |\alpha ^\sigma _{\rho _1}|^{-1} \operatorname {\mathrm {vol}}_V$ . These forms can be restricted to volumes on any affine subspace parallel to the subspace $V = \{\phi \in (\operatorname {\mathrm {Pic}} X)_{\mathbb {R}}^\vee : \langle \phi , K_X\rangle = 0\}$ . Hence,

$$ \begin{align*} \alpha(X) &= \operatorname{\mathrm{vol}}_P{}\{r \in (\operatorname{\mathrm{Eff}} X)^\vee : \langle r, K_X\rangle = -1\}\\ &= |\alpha^\sigma_{\rho_1}|^{-1}\operatorname{\mathrm{vol}}_V{}\{r \in (\operatorname{\mathrm{Pic}} X)_{\mathbb{R}}^\vee : \langle r, K_X \rangle = -1, \langle r, \deg x_\rho\rangle \ge 0 \text{ for all}\ \rho \in \Sigma(1)\}\\ &= |\alpha^\sigma_{\rho_1}|^{-1}\operatorname{\mathrm{vol}}_V{} \left\{r_0 K_X^\vee + \sum_{\rho \notin \sigma(1)'} r_\rho d^\vee_\rho : \begin{aligned} &r_0 = -1, r_\rho \ge 0\text{ for all }\rho \notin \sigma(1)',\\ & b_{\rho'}-\textstyle\sum_{\rho \notin \sigma(1)'} r_{\rho}b_{\rho,\rho'} \ge 0 \text{ for all } \rho' \in \sigma(1)' \end{aligned} \right\}, \end{align*} $$

and the claim follows.

Proof. Consider the case $\Psi =x_\rho $ first. For $\rho \in \sigma (1)\setminus \{\rho _0\}$ , the claim holds by definition of the substitution. For $\rho = \rho _1$ , we have $\Psi (f(\mathbf {z}))=1=t_{\rho _1}^{-1}\cdot t_{\rho _1} = t_{\rho _1}^{-\alpha ^\sigma _{\Psi ,\rho _1}}\Psi (g(\mathbf {t}))$ . For $\rho \notin \sigma (1)'$ , we have $\Psi (f(\mathbf {z}))=1\cdot 1=t_{\rho _1}^{-\alpha ^\sigma _{\Psi ,\rho _1}}\Psi (g(\mathbf {t}))$ . Therefore, the claim holds for all monomials and hence also for all homogeneous polynomials and all homogeneous rational functions in $\{x_\rho : \rho \in \Sigma (1) \setminus \{\rho _0\}\}$ . In particular, in the case $\Psi =x_{\rho _0}$ , since $\phi $ is such a rational function of degree $\deg (x_{\rho _0})$ , the substitution gives $\Psi (f(\mathbf {z}))=\phi (\mathbf {z},\mathbf {1}) = t_{\rho _1}^{-\alpha ^\sigma _{\rho _0,\rho _1}}\phi (\mathbf {t},\mathbf {1}) = t_{\rho _1}^{-\alpha ^\sigma _{\Psi ,\rho _1}}\Psi (g(\mathbf {t}))$ . Now, the claim follows for all monomials, homogeneous polynomials and finally all homogeneous rational functions in $\{x_\rho : \rho \in \Sigma (1)\}$ .

Let $\psi $ be the numerator of $\phi $ . Because of (4.8), $t_{\rho _1}$ appears in at most one monomial of $\psi (\mathbf {t},\mathbf {1})$ ; we assume that it appears in the first monomial with odd exponent. Therefore, either the exponent of $t_{\rho _1}$ in the first monomial of $t_{\rho _1}^{-\alpha ^\sigma _{\psi ,\rho _1}}\psi (\mathbf {t},\mathbf {1})$ is odd, or the exponents of $t_{\rho _1}$ in all other monomials of this expression are odd. But since our substitution gives $\psi (\mathbf {z},\mathbf {1})=t_{\rho _1}^{-\alpha ^\sigma _{\psi ,\rho _1}}\psi (\mathbf {t},\mathbf {1})$ , the exponent of $t_{\rho _1}$ in a certain monomial of $t_{\rho _1}^{-\alpha ^\sigma _{\psi ,\rho _1}}\psi (\mathbf {t},\mathbf {1})$ can only be odd if there is a $z_\rho $ with odd exponent in the corresponding monomial of $\psi (\mathbf {z},\mathbf {1})$ , and then the exponent of $t_\rho $ in this monomial of $\psi (\mathbf {t},\mathbf {1})$ is also odd.

Proof. Our starting point is (4.11). We use the identity (for positive real s)

$$ \begin{align*} \frac{1}{s} = \int_{z_{\rho_1}> 0,\ sz_{\rho_1}\le 1} \,{\mathrm d} z_{\rho_1} \end{align*} $$

to deduce

$$ \begin{align*} \omega_\infty = \int_{(\mathbf{z},z_{\rho_1}) \in \mathbb{R}^{\sigma(1) \setminus \{\rho_0\}} \times \mathbb{R}_{>0},\ H_\infty(f(\mathbf{z}))\cdot z_{\rho_1} \le 1} \frac{\,{\mathrm d}\mathbf{z}\,{\mathrm d} z_{\rho_1}}{|\partial\Phi/\partial x_{\rho_0}(f(\mathbf{z}))|}. \end{align*} $$

We use the transformation $z_{\rho _1} = t_{\rho _1}^{\alpha ^\sigma _{\rho _1}}$ (with positive $t_{\rho _1}$ ) and the transformations from Lemma 4.9. The latter give $H_\infty (f(\mathbf {z})) = t_{\rho _1}^{-\alpha ^\sigma _{\rho _1}}H_\infty (g(\mathbf {t}))$ since all monomials appearing in the definition of the anticanonical height function $H_\infty $ have degree $-K_X$ ; therefore, $H_\infty (f(\mathbf {z}))\cdot z_{\rho _1} = H_\infty (g(\mathbf {t}))$ . Furthermore, $|\partial \Phi /\partial x_{\rho _0}(f(\mathbf {z}))| = |t_{\rho _1}^{-\alpha ^\sigma _{\partial \Phi /\partial x_{\rho _0},\rho _1}}\partial \Phi /\partial x_{\rho _0}(g(\mathbf {t}))|$ (even without using the observation that these are the same constants by (4.8)). We obtain $\,{\mathrm d} z_{\rho _1} = |\alpha ^\sigma _{\rho _1} t_{\rho _1}^{\alpha ^\sigma _{\rho _1}-1}| \,{\mathrm d} t_{\rho _1}$ and

$$ \begin{align*} \,{\mathrm d} \mathbf{z} = |t_{\rho_1}^{-\sum_{\rho' \in \sigma(1) \setminus \{\rho_0\}}\alpha^\sigma_{\rho',\rho_1}}|\bigwedge_{\rho' \in \sigma(1) \setminus \{\rho_0\}} \,{\mathrm d} t_{\rho'}. \end{align*} $$

The integration domain is unchanged.

We have $-K_X=\sum _{\rho ' \in \Sigma (1)} \deg (x_{\rho '}) - \deg (\Phi )$ by [Reference Arzhantsev, Derenthal, Hausen and Laface2, Proposition 3.3.3.2], and $\deg (\partial \Phi /\partial x_{\rho _0})=\deg (\Phi )-\deg (x_{\rho _0})$ . Therefore, $\alpha ^\sigma _{\rho _1}=\sum _{\rho ' \in \Sigma (1)} \alpha ^\sigma _{\rho ',\rho _1}-\alpha ^\sigma _{\Phi ,\rho _1}$ and $\alpha ^\sigma _{\partial \Phi /\partial x_{\rho _0},\rho _1}=\alpha ^\sigma _{\Phi ,\rho _1}-\alpha ^\sigma _{\rho _0,\rho _1}$ . Since $\alpha ^\sigma _{\rho ',\rho }=\delta _{\rho '=\rho }$ for all $\rho ',\rho \notin \sigma (1)$ , we conclude that

$$ \begin{align*}\alpha^\sigma_{\rho_1}=\sum_{\rho' \in \sigma(1) \setminus \{\rho_0\}}\alpha^\sigma_{\rho',\rho_1}+1-\alpha^\sigma_{\partial\Phi/\partial x_{\rho_0},\rho_1}.\end{align*} $$

This shows that the powers of $t_{\rho _1}$ cancel out so that $\,{\mathrm d}\mathbf {z}\,{\mathrm d} z_{\rho _1}/|\partial \Phi /\partial x_{\rho _0}(f(\mathbf {z}))| = \,{\mathrm d} \mathbf {t}/|\partial \Phi /\partial x_{\rho _0}(g(\mathbf {t}))|$ . Therefore,

$$ \begin{align*} \omega_\infty = |\alpha^\sigma_{\rho_1}| \int_{\mathbf{t} \in \mathbb{R}^{\sigma(1) \setminus \{\rho_0\}} \times \mathbb{R}_{>0},\ H_\infty(g(\mathbf{t}))\le 1} \frac{\,{\mathrm d} \mathbf{t}}{|\partial\Phi/\partial x_{\rho_0}(g(\mathbf{t}))|}. \end{align*} $$

We claim that

has the same value as $\omega _\infty $ . Indeed, $\phi (\mathbf {t},\mathbf {1})$ (the $\rho _0$ -component of $g(\mathbf {t})$ ) is the only place where the sign of $t_{\rho _1}$ might matter. Our claim is clearly true if $t_{\rho _1}$ does not appear in $\phi (\mathbf {t},\mathbf {1})$ or if $t_{\rho _1}$ has an even exponent in $\phi (\mathbf {t},\mathbf {1})$ . If $t_{\rho _1}$ appears in $\phi (\mathbf {t},\mathbf {1})$ with odd exponent, then the change of variables and for all $t_\rho $ appearing in the final statement of Lemma 4.9 in $\omega _\infty ^-$ shows that $\omega _\infty ^-=\omega _\infty $ . Therefore,

$$ \begin{align*} \mu_\infty(X(\mathbb{R})) = \omega_\infty = \frac{1}{2}(\omega_\infty+\omega_\infty^-) = \frac{|\alpha^\sigma_{\rho_1}|}{2} \int_{\mathbf{t} \in \mathbb{R}^{\sigma(1) \setminus \{\rho_0\}} \times \mathbb{R}_{\ne 0},\ H_\infty(g(\mathbf{t}))\le 1} \frac{\,{\mathrm d} \mathbf{t}}{|\partial\Phi/\partial x_{\rho_0}(g(\mathbf{t}))|}. \end{align*} $$

Since $\operatorname {\mathrm {rk}} \operatorname {\mathrm {Pic}} X = \#\Sigma (1)-\#\sigma (1)$ and replacing $\mathbb {R}^{\sigma (1) \setminus \{\rho _0\}} \times \mathbb {R}_{\ne 0}$ by $\mathbb {R}^{\sigma (1)' \setminus \{\rho _0\}}$ does not change the integral, this completes the proof.

Proof. According to [Reference Peyre61, 5.1], Peyre’s constant for X is $c = \alpha (X) \beta (X) \tau _H(X)$ . Here, the cohomological constant is

$$ \begin{align*} \beta(X)=\#H^1(\operatorname{\mathrm{Gal}}(\overline{\mathbb{Q}}/\mathbb{Q}), \operatorname{\mathrm{Pic}}(X \otimes_{\mathbb{Q}} \overline{\mathbb{Q}})) = 1 \end{align*} $$

since X is split. Recall (4.3) for $\tau _H(X)$ . By Lemma 4.8 and Proposition 4.10, $\alpha (X)\mu _\infty (X(\mathbb {R}))= c^\ast c_\infty $ . Furthermore, we use Proposition 4.6 for the p-adic densities.

Proof. If $n=1$ this is the familiar form of Weyl’s inequality. If $n\ge 2$ , then we apply repeated Weyl differencing. Let $h\in \mathbb N$ . On combining [Reference Vaughan72, Lemma 2.4] with [Reference Vaughan72, Exercise 2.8.1], one has

$$ \begin{align*}\Big| \sum_{X<x\le 2X} e(\beta x^h)\Big|^{2^{h-1}} \le (2X)^{2^{h-1}-h} \sum_{\substack{|u_j|\le X\\ 1\le j < h}} \sum_{x\in I(\mathbf u)} e\big(h!\beta u_1u_2\dots u_{h-1}(x + {\textstyle \frac12}|\mathbf u|_1)\big),\end{align*} $$

where the $I(\mathbf u)$ are certain subintervals of $[X,2X]$ . Note here that the sum on the right is real and nonnegative. One trivially has

$$ \begin{align*}\Big| \sum_{-2X\leq x< -X} e(\beta x^h)\Big|= \Big| \sum_{X<x\le 2X} e(\beta x^h)\Big|,\end{align*} $$

and hence it follows that

(6.2) $$ \begin{align} \Big| \sum_{X<|x|\le 2X} e(\beta x^h)\Big|^{2^{h-1}} \ll X^{2^{h-1}-h} \sum_{\substack{|u_j|\le X\\ 1\le j < h}} \sum_{x\in I(\mathbf u)} e\big(h!\beta u_1u_2\dots u_{h-1} (x + {\textstyle \frac12}|\mathbf u|_1)\big). \end{align} $$

By Hölder’s inequality,

$$ \begin{align*}|W(\alpha,\mathbf Y;\mathbf{h})|^{2^{h_1 -1}} \le (Y_2\cdots Y_n)^{2^{h_1-1}-1} \sum_{\substack{\frac12 Y_\nu< |y_\nu|\le Y_\nu \\ 2\le \nu\le n}} \Big|\sum_{\frac12 Y_1<|y_1|\le Y_1} e(\alpha y_1^{h_1}y_2^{h_2}\cdots y_n^{h_n})\Big|^{ 2^{h_1 -1}}.\end{align*} $$

We apply (6.2) with $\beta =\alpha y_2^{h_2}\cdots y_n^{h_n}$ to the sum over $y _1$ . We write $\mathbf {h}'=(h_2,h_3,\ldots , h_n)$ , $\mathbf Y'=(Y_2,Y_3,\ldots , Y_n)$ and then find that

$$ \begin{align*}|W(\alpha,\mathbf Y;\mathbf{h})|^{2^{h_1 -1}} \ll Y_1^{2^{h_1-1}-h_1} \langle \mathbf Y'\rangle^{2^{h_1 -1}-1} \sum_{\substack{|u_j|\le Y_1\\ 1\le j < h_1}} \sum_{y\in I_1(\mathbf u)} W(h_1! \alpha u_1u_2\cdots u_{h_1-1}(y+ {\textstyle \frac12}|\mathbf u|_1) , \mathbf Y'; \mathbf{h}'),\end{align*} $$

where $I_1(\mathbf u)$ are certain subintervals of $[\frac 12 Y_1, Y_1]$ . Now, we apply Hölder’s inequality again to bring in $|W(\beta ,\mathbf Y'; \mathbf {h}')|^{2^{h_2-1}}$ . We may then estimate the sum over $y_2$ by (6.2). Repeated use of this process produces the inequality

(6.3) $$ \begin{align} |W(\alpha,\mathbf Y;\mathbf{h})|^{2^{h_1-1}\cdots 2^{h_n-1}} \ll \langle \mathbf Y\rangle^{2^{h_1+\dots+h_n-n}} {\mathbf Y}^{-\mathbf{h}} \sum_{\mathbf u_1,\ldots, \mathbf u_n} \sum_{\substack{y_\nu\in I_\nu(\mathbf u_\nu) \\ 1\le \nu <n}} \Big| \sum_{y_n\in I_n(\mathbf u_n)} e(\alpha vy_n)\Big| \end{align} $$

in which $\mathbf u_\nu \in \mathbb {Z}^{h_{\nu } - 1}$ runs over integer vectors with $|\mathbf u_\nu |\le Y_\nu $ for $1\le \nu \le n$ , the $I_\nu (\mathbf u_\nu )$ are certain subintervals of $[\frac 12 Y_\nu , Y_\nu ]$ and

$$ \begin{align*}v= h_1! h_2!\cdots h_n!\langle \mathbf u_1\rangle\cdots \langle \mathbf u_n\rangle y_1y_2\cdots y_{n-1}.\end{align*} $$

Note that $v=0$ will occur in (6.3) only when one of the $\mathbf u_\nu $ has a zero entry so that the total contribution to (6.3) from summands with $v=0$ does not exceed $ \langle \mathbf Y\rangle ^{2^{h_1+\dots +h_n-n}} Y_n^{-1}$ , which is acceptable. For nonzero v, the innermost sum in (6.3) does not exceed $\min (Y_n, \|\alpha v\|^{-1})$ . Further, we have $v\ll \mathbf Y^{\mathbf {h}}Y_n^{-1}$ , and a divisor function estimate shows that there are no more than $O(|v|^\varepsilon )$ choices for $\mathbf u_\nu $ , $y_\nu $ that correspond to the same v. This shows that

$$ \begin{align*}|W(\alpha,\mathbf Y;\mathbf{h})|^{2^{h_1-1}\cdots 2^{h_n-1}} \ll \langle \mathbf Y\rangle^{2^{|\mathbf{h}|_1-n}} Y_n^{-1} + \langle \mathbf Y\rangle^{2^{|\mathbf{h}|_1-n}+\varepsilon} {\mathbf Y}^{-\mathbf h}\! \sum_{1\le v \ll \mathbf Y^{\mathbf{h}}Y_n^{-1}}\! \min(Y_n, \|\alpha v\|^{-1}).\end{align*} $$

Reference to [Reference Vaughan72, Lemma 2.1] completes the proof.