# Counting integral points on indefinite ternary quadratic equations over
number fields

Fei Xu  Fei Xu
School of Mathematical Sciences,
Capital Normal University,
105 Xisanhuanbeilu,
100048 Beijing, China<EMAIL_ADDRESS>and  Runlin Zhang  Runlin Zhang
Beijing International Center for Mathematical Research,
Peking University,
1000871 Beijing, China<EMAIL_ADDRESS>
(Date: .)

## Abstract

We study an asymptotic formula for counting integral points over an equation
defined by a non-degenerated indefinite integral ternary quadratic form $f$
representing a non-zero integer $a$ such that $-a\cdot det(f)$ is square over
a number field. In particular, we prove that the finite part of this
asymptotic formula is given by the product of local density times $1-p^{-1}$
over all finite primes $p$ over $\mathbb{Z}$.

##  1\. Introduction

Let $f(x_{1},\cdots,x_{n})$ be a non-singular indefinite quadratic form over
$\mathbb{Z}$ and $a$ be a non-zero integer. It is a classical question to
count the integral solutions

$N(f,a,T)=\\#\\{(\alpha_{1},\cdots,\alpha_{n})\in\mathbb{Z}^{n}:\
f(\alpha_{1},\cdots,\alpha_{n})=a\ \text{with}\
\left\lVert(\alpha_{1},\cdots,\alpha_{n})\right\rVert\leq T\\}$

as $T\to\infty$, where $\left\lVert\cdot\right\rVert$ is a norm on
$\mathbb{R}^{n}$. For example, when $n=2$, it is well-known that

$N(f,a,T)\sim\begin{cases}O(1)\ \ \ &\text{when
$-\det(f)\in(\mathbb{Q}^{\times})^{2}$}\\\ c\cdot
logT&\text{otherwise}\end{cases}$

as $T\to\infty$ by the Dirichlet’s unit theorem (see [PR94, Chapter 5, Theorem
5.12]). In [DRS], Duke, Rudnick and Sarnak proved that

$N(f,a,T)\sim c\cdot T^{n-2}$

as $T\to\infty$ when $n\geq 4$. Eskin and McMullen provided a much simpler
proof in [EM]. This result has been largely generalized by Eskin, Mozes and
Shah in [EMS]. For $n=3$, Duke, Rudnick and Sarnak in [DRS] also pointed out
that

$N(f,a,T)\sim\begin{cases}c\cdot T\ \ \
&\text{$-a\det(f)\not\in(\mathbb{Q}^{\times})^{2}$}\\\ c\cdot TlogT\ \ \
&\text{$-a\det(f)\in(\mathbb{Q}^{\times})^{2}$}\end{cases}$

as $T\to\infty$ for some specific quadratic forms $f$. Such general phenomenon
was proved by Oh and Shah in [OS].

In spirit of the local-global principle, one expects that the above constant
$c$ should be the contribution of numbers of local solutions over all finite
primes. Indeed, Borovoi and Rudnick in [BR95] first proved that $c$ is the
product of numbers of local solutions over all finite primes for $n\geq 4$ and
studied the case of $n=3$ and $-a\det(f)\not\in(\mathbb{Q}^{\times})^{2}$ by
introducing the density function. By applying strong approximation with
Brauer-Manin obstruction, Wei and the first author showed that $c$ is an
average of products of number of twisted local solutions by the Brauer
elements over all finite primes in [WX] (see also [CX]) for the later
situation. Since Oh and Shah in [OS] provided a different type asymptotic
formula for $n=3$ and $-a\det(f)\in(\mathbb{Q}^{\times})^{2}$, it is natural
to ask how the numbers of local solutions over finite primes contribute to the
constant $c$ in the asymptotic formula. In this paper, we will answer this
question (see Theorem 3.6).

######  Theorem 1.1.

When $n=3$ and $-a\det(f)\in(\mathbb{Q}^{\times})^{2}$, then

$N(f,a,T)\sim(\prod_{p\
\text{primes}}(1-p^{-1})\alpha_{p}(f,a))\cdot(logT\int_{B_{T}}\omega)$

as $T\to\infty$ where

$\alpha_{p}(f,a)=\lim_{k\to\infty}\frac{\\#\\{(\beta_{1},\beta_{2},\beta_{3})\in(\mathbb{Z}/(p^{k}))^{3}:\
f(\beta_{1},\beta_{2},\beta_{3})\equiv a\mod p^{k}\\}}{p^{2k}}$

for all primes $p$ and

$B_{T}=\\{(\alpha_{1},\alpha_{2},\alpha_{3})\in\mathbb{R}^{3}:\
f(\alpha_{1},\alpha_{2},\alpha_{3})=a\ \text{with}\
\left\lVert(\alpha_{1},\alpha_{2},\alpha_{3})\right\rVert\leq T\\}$

and $\omega$ is the normalized gauge form related to the equation.

We give the precise asymptotic formula for the example in [DRS] (see Example
4.12).

######  Example 1.2.

If $f(x,y,z)=x^{2}+y^{2}-h^{2}z^{2}$ with $h\in\mathbb{Z}$ and

$N(f,1,T)=\\#\\{(\alpha,\beta,\gamma)\in\mathbb{Z}^{3}:\
f(\alpha,\beta,\gamma)=1\ \text{with}\
\sqrt{\alpha^{2}+\beta^{2}+\gamma^{2}}\leq T\\}$

for $T>0$, then

$N(f,1,T)\sim\frac{8}{\pi\sqrt{1+h^{2}}}\cdot(\prod_{p\mid h,\ p\neq
2}\frac{p-(\frac{-1}{p})}{p+1})\cdot T\log T$

as $T\to\infty$.

The difficulty arising from this remaining case is that the infinity product

$\prod_{p\ \text{primes}}\alpha_{p}(f,a)$

is not convergent because $\alpha_{p}(f,a)=1+p^{-1}$ for sufficiently large
primes $p$ by [S, Hilfssatz 12]. Therefore the classical methods such as the
circle method can not be applied. Instead, we use the method in [WX] to study
this problem, which is also valid over a general number field. In order to
work over a general number field, one needs to extend the result of Oh and
Shah in [OS] which is only true over $\mathbb{Q}$, to a general number field.
Indeed, such an extension over a general number field has been given in [Zha]
up to an implicit constant. One needs to determine this implicit constant in a
precise way. Another ingredient of this paper comes from the fact that the
stabilizer of a rational point is a split torus $\mathbb{G}_{m}$ when one
considers the action of the spin group of $f$ on the equation. However, the
Tamagawa number of $\mathbb{G}_{m}$ is defined in a virtual way. One needs to
use the Tamagawa number of $\mathbb{G}_{m}$ implicitly.

Notations and terminology are standard if not explained. In particular, we fix
a number field $k$ and a non-degenerate ternary quadratic from $f(x,y,z)$ over
$k$. Let $a\in k^{\times}$ such that $-a\cdot det(f)\in(k^{\times})^{2}$ and

$X:\ \ f(x,y,z)=a$

be an affine variety defined over $k$. Assume $X(k)\neq\emptyset$. Then

$-a\cdot det(f)\in(k^{\times})^{2}\Longrightarrow f\text{ is isotropic over
$k$.}$

Fixing $v_{0}\in X(k)$, one has

$X\cong G/H\ \ \ \text{with}\ \ \
G=\operatorname{Spin}(f)\cong\operatorname{SL}_{2}\ \ \ \text{and}\ \ \
H\cong\mathbb{G}_{m}$ (1.3)

over $k$ by [CTX, §5.6], where $H$ is the stabilizer of $v_{0}$ in $G$.

######  Lemma 1.4.

For any field $F$ containing $k$, the group $G(F)$ acts on $X(F)$
transitively.

###### Proof.

Applying nonabelian Galois cohomology (see [Gi, Chapter III, Prop.3.2.2 and
§3.7] ), one has the exact sequence of pointed sets

$1\longrightarrow H(F)\longrightarrow G(F)\longrightarrow X(F)\longrightarrow
H^{1}(F,H)\longrightarrow H^{1}(F,G)$

by (1.3). Since $H^{1}(F,\mathbb{G}_{m})=0$ (Hilbert 90), one concludes that
$G(F)$ acts on $X(F)$ transitively. ∎

Since $k[X]^{\times}=k^{\times}$ by [CTX, §5.6], one obtains that $X$ admits a
unique $G$-invariant gauge form up to a scalar by [W, Chapter II, §2.2.2].
Fixing an isomorphism

$H\cong\mathbb{G}_{m}=\operatorname{Spec}[t,t^{-1}],$

we obtain the gauge form $\omega_{H}$ on $H$ by pulling back the differential
form $t^{-1}dt$ on $\mathbb{G}_{m}$ through this isomorphism. Choose both a
$G$-invariant gauge form $\omega_{G}$ on $G$ and a $G$-invariant gauge form
$\omega_{X}$ on $X$ such that $\omega_{G}$, $\omega_{H}$ and $\omega_{X}$
match together algebraically in the sense of [W, §2.4]. Let $\infty_{k}$ be
the set of all archimedean primes of $k$. We use $v\leq\infty_{k}$ to denote
all primes $v$ of $k$ and $v<\infty_{k}$ to denote all finite primes $v$ of
$k$ respectively. Moreover, the completion of $k$ with respect to
$v\leq\infty_{k}$ is denoted by $k_{v}$. Write ${\bf A}_{k}$ to be the adelic
ring of $k$. Let $\lambda_{v}$, $\mu_{v}$ and $\nu_{v}$ be the associated
measures induced by $\omega_{G}$, $\omega_{H}$ and $\omega_{X}$ on $G(k_{v})$,
$H(k_{v})$ and $X(k_{v})$ for $v\leq\infty_{k}$ respectively. Then

$\int_{G(k_{v})}\phi(g)\lambda_{v}(g)=\int_{X(k_{v})}\nu_{v}(\bar{g})\int_{H(k_{v})}\phi(gh)\mu_{v}(h)$
(1.5)

for any continuous function $\phi$ with compact support on $G(k_{v})$ for
$v\leq\infty_{k}$. Let $\parallel\cdot\parallel$ be the standard Euclidean
norm on $\mathbb{R}^{n}$ and $\left\lVert\cdot\right\rVert_{0}$ be a general
linear norm on $\mathbb{R}^{n}$ satisfying

$C_{0}^{-1}\left\lVert\cdot\right\rVert\leq\left\lVert\cdot\right\rVert_{0}\leq
C_{0}\left\lVert\cdot\right\rVert.$ (1.6)

for a fix constant $C_{0}>0$.

##  2\. Counting in a single orbit

In this section, we will provide the precise asymptotic formula for counting
the number of points in a single orbit by considering the action of an
arithmetic lattice $\Gamma$ of $G(k)$ on $X(k)$. In fact, the general
asymptotic formula has been established by the second author in [Zha] up to an
implicit constant. We will determine this implicit constant precisely.

###  2.1. A model space

Let $X_{1}$ be a variety over $k$ defined by the equation
$x_{1,1}^{2}+x_{1,2}x_{2,1}=1$ and $G_{1}=\operatorname{SL}_{2}$. Define the
action

$G_{1}\times_{k}X_{1}\longrightarrow X_{1};\ \ \
(g,\left[\begin{array}[]{cc}x_{1,1}&x_{1,2}\\\
x_{2,1}&-x_{1,1}\end{array}\right])\mapsto
g\left[\begin{array}[]{cc}x_{1,1}&x_{1,2}\\\
x_{2,1}&-x_{1,1}\end{array}\right]g^{-1}.$

Let

$M_{0}=\left[\begin{array}[]{cc}1&\\\ &-1\end{array}\right]\in X_{1}(k)\ \
\text{and $H_{1}$ be the stabilizer of $M_{0}$ in $G_{1}$. }$

Then $H_{1}$ is a maximal torus of $G_{1}$. Fix an isomorphism
$H_{1}\cong\mathbb{G}_{m}$.

Consider the homomorphism of Lie groups

$\operatorname{Nm}:\
(k\otimes_{\mathbb{Q}}{\mathbb{R}})^{\times}\cong(\mathbb{R}^{\times})^{r}\oplus(\mathbb{C}^{\times})^{s}\longrightarrow{\mathbb{R}}_{>0};\
\ \
(x_{1},\cdots,x_{r},y_{1},\cdots,y_{s})\mapsto\prod_{i=1}^{r}|x_{i}|\cdot\prod_{j=1}^{s}|y_{j}|^{2}$

where $r$ and $s$ are the number of real and complex primes of $k$
respectively. This induces the homomorphism of Lie groups

${H_{1}(k\otimes_{\mathbb{Q}}{\mathbb{R}})\cong(\mathbb{R}^{\times})^{r}\oplus(\mathbb{C}^{\times})^{s}}$${\mathbb{R}_{>0}}$$\scriptstyle{\operatorname{Nm}}$

with the kernel $\mathrm{H}_{1}^{\operatorname{an}}$. Then

$H_{1}(k\otimes_{\mathbb{Q}}\mathbb{R})=\mathrm{H}_{1}^{\operatorname{an}}\times\mathrm{H}_{1}^{\operatorname{spl}}$

where

$\mathrm{H}_{1}^{\operatorname{spl}}=\\{\left(\left[\begin{array}[]{cc}t&\\\
&t^{-1}\end{array}\right],...,\left[\begin{array}[]{cc}t&\\\
&t^{-1}\end{array}\right]\right)\in H_{1}(k\otimes_{\mathbb{Q}}\mathbb{R}):\
t\in{\mathbb{R}}_{>0}\\}\cong{\mathbb{R}}_{>0}.$

Choose the Haar measure

$(\bigwedge_{i=1}^{r}\frac{dx_{i}}{x_{i}})\wedge(\bigwedge_{j=1}^{s}2\cdot\frac{du_{j}\wedge
dv_{j}}{u_{j}^{2}+v_{j}^{2}})\ \ \text{ on }\ \
(\mathbb{R}^{\times})^{r}\oplus(\mathbb{C}^{\times})^{s}$

where $(x_{1},\cdots,x_{r})$ and
$(u_{1}+v_{1}\sqrt{-1},\cdots,u_{s}+v_{s}\sqrt{-1})$ are the coordinates of
$(\mathbb{R}^{\times})^{r}$ and $(\mathbb{C}^{\times})^{s}$ respectively. By
the fixed isomorphism

$H_{1}(k\otimes_{\mathbb{Q}}{\mathbb{R}})\cong(\mathbb{R}^{\times})^{r}\oplus(\mathbb{C}^{\times})^{s},$

one obtains the Haar measure $\mu_{H_{1}}$ on
$H_{1}(k\otimes_{\mathbb{Q}}{\mathbb{R}})$. Choosing the Haar measure
$t^{-1}dt$ on ${\mathbb{R}}_{>0}$, one gets the Haar measure
$\mu_{\mathrm{H}_{1}^{\operatorname{spl}}}$ on
$\mathrm{H}_{1}^{\operatorname{spl}}$ by the fixed isomorphism
$\mathrm{H}_{1}^{\operatorname{spl}}\cong{\mathbb{R}}_{>0}$. There is a Haar
measure $\mu_{\mathrm{H}_{1}^{\operatorname{an}}}$ on
$\mathrm{H}_{1}^{\operatorname{an}}$ such that

$\mu_{H_{1}}\cong\mu_{\mathrm{H}_{1}^{\operatorname{an}}}\otimes\mu_{\mathrm{H}_{1}^{\operatorname{spl}}}.$

By Lemma 1.4, one can identify

$X_{1}(k\otimes_{\mathbb{Q}}{\mathbb{R}})\ \ \ \text{ with }\ \ \
G_{1}(k\otimes_{\mathbb{Q}}{\mathbb{R}})/H_{1}(k\otimes_{\mathbb{Q}}{\mathbb{R}})$

and fix Haar measures $\lambda_{G_{1}}$ on
$G_{1}(k\otimes_{\mathbb{Q}}{\mathbb{R}})$ and $\nu_{X_{1}}$ on
$X_{1}(k\otimes_{\mathbb{Q}}{\mathbb{R}})$ respectively such that
$(\lambda_{G_{1}},\mu_{H_{1}},\nu_{X_{1}})$ match up.

######  Proposition 2.1.

If $\Gamma_{1}$ is an arithmetic lattice in $G_{1}(k)$, then

$\\#\left\\{M\in\Gamma_{1}\cdot M_{0}:\ \left\lVert M\right\rVert_{0}\leq
T\right\\}\sim\frac{\mu_{\mathrm{H}_{1}^{\operatorname{an}}}{(\mathrm{H}_{1}^{\operatorname{an}}/\mathrm{H}_{1}^{\operatorname{an}}\cap\Gamma_{1}})}{\lambda_{G_{1}}{(G_{1}(k\otimes_{\mathbb{Q}}\mathbb{R})/\Gamma_{1})}}\cdot\log{T}\int_{B_{T}}\nu_{X_{1}}$

as $T\to+\infty$, where

$B_{T}=\left\\{x\in X_{1}(k\otimes_{\mathbb{Q}}\mathbb{R}):\ \left\lVert
x\right\rVert_{0}\leq T\right\\}$

and $\left\lVert\cdot\right\rVert_{0}$ is a linear norm on
$(k\otimes_{\mathbb{Q}}\mathbb{R})^{3}$ satisfying (1.6).

###### Proof.

Consider

$G_{1}(k\otimes_{\mathbb{Q}}\mathbb{R})\cong\operatorname{SL}_{2}(\mathbb{R})^{\oplus
r}\bigoplus\operatorname{SL}_{2}(\mathbb{C})^{\oplus s}\ \ \ \text{acting on}\
\ \ (k\otimes_{\mathbb{Q}}\mathbb{R})^{2}\cong(\mathbb{R}^{2})^{\oplus
r}\bigoplus(\mathbb{C}^{2})^{\oplus s}$ (2.2)

as an $\mathbb{R}$-vector space naturally. This action induces an action of
$G_{1}(k\otimes_{\mathbb{Q}}\mathbb{R})$ on

$\bigwedge^{r+2s}(k\otimes_{\mathbb{Q}}\mathbb{R})^{2}$

as an $\mathbb{R}$-vector space.

For any $v\in{\mathbb{R}}^{2}$, we define
$v^{i}\in(k\otimes_{\mathbb{Q}}\mathbb{R})^{2}$ such that the $i$-th component
of $v^{i}$ is $v$ and the rest is 0 by identification (2.2) for $1\leq i\leq
r$. Similarly for any $w\in\mathbb{C}^{2}$, we define
$w^{r+j}\in(k\otimes_{\mathbb{Q}}\mathbb{R})^{2}$ such that the $(r+j)$-th
component is $w$ and the rest is 0. Let $\\{e_{1}=(1,0),\ e_{2}=(0,1)\\}$ be
the standard basis of $\mathbb{R}^{2}$. Write

$\Lambda(e_{k}):=(\bigwedge_{i=1}^{r}e^{i}_{k})\wedge(\bigwedge_{j=1}^{s}e^{r+j}_{k})\wedge(\bigwedge_{j=1}^{s}\sqrt{-1}e^{r+j}_{k})\in\bigwedge^{r+2s}(k\otimes_{\mathbb{Q}}\mathbb{R})^{2}$

for $k=1,2$. For $g\in G_{1}(k\otimes_{\mathbb{Q}}\mathbb{R})$ and
$\varepsilon>0$, we define

$\Omega_{g,\varepsilon}:=\left\\{t\in\mathrm{H}_{1}^{\operatorname{spl}}:\
\left\lVert
gt\cdot\Lambda(e_{k})\right\rVert\geq\varepsilon,\;k=1,2\right\\}.$

By the equidistribution in [Zha, Theorem 1.2] (cf. [OS]), one has, for any
$\varepsilon>0$,

$\\#\left\\{M\in\Gamma_{1}\cdot M_{0}:\ \left\lVert M\right\rVert_{0}\leq
T\right\\}\sim\frac{\mu_{\mathrm{H}_{1}^{\operatorname{an}}}{(\mathrm{H}_{1}^{\operatorname{an}}/\mathrm{H}_{1}^{\operatorname{an}}\cap\Gamma_{1}})}{\lambda_{G_{1}}{(G_{1}(k\otimes_{\mathbb{Q}}\mathbb{R})/\Gamma_{1})}}\cdot\int_{\left\lVert
gM_{0}\right\rVert_{0}\leq
T}\mu_{\mathrm{H}_{1}^{\operatorname{spl}}}(\Omega_{g,\varepsilon})\nu_{X_{1}}$
(2.3)

as $T\to+\infty$.

Let

$\mathrm{K}:=\operatorname{SO}_{2}({\mathbb{R}})^{\oplus
r}\bigoplus\operatorname{SU}_{2}({\mathbb{R}})^{\oplus s}$

be a maximal compact subgroup of $G_{1}(k\otimes_{\mathbb{Q}}\mathbb{R})$ and

$\mathrm{U}=\left\\{\left(\left[\begin{array}[]{cc}1&x_{i}\\\
0&1\end{array}\right],\left[\begin{array}[]{cc}1&y_{j}\\\
0&1\end{array}\right]\right):\ x_{i}\in\mathbb{R},1\leq i\leq r;\
y_{j}\in\mathbb{C},1\leq j\leq s\right\\}$ (2.4)

Then the following map given by multiplication

$\mathrm{K}\times\mathrm{U}\times H_{1}(k\otimes_{\mathbb{Q}}{\mathbb{R}})\to
G_{1}(k\otimes_{\mathbb{Q}}\mathbb{R})$ (2.5)

is surjective such that all fibres are torsors under a finite group
$\mathrm{K}\cap H_{1}(k\otimes_{\mathbb{Q}}{\mathbb{R}})$ by Iwasawa
decomposition. If $g=kuh$ with $k\in\mathrm{K}$, $u\in\mathrm{U}$ and $h\in
H_{1}(k\otimes_{\mathbb{Q}}{\mathbb{R}})$ under the decomposition (2.5), then

$\mu_{\mathrm{H}_{1}^{\operatorname{spl}}}(\Omega_{g,\varepsilon})=\mu_{\mathrm{H}_{1}^{\operatorname{spl}}}(\Omega_{u,\varepsilon}).$

If

$u=\left(\left[\begin{array}[]{cc}1&x_{1}\\\
0&1\end{array}\right],\cdots,\left[\begin{array}[]{cc}1&x_{r}\\\
0&1\end{array}\right],\left[\begin{array}[]{cc}1&y_{1}\\\
0&1\end{array}\right],\cdots,\left[\begin{array}[]{cc}1&y_{s}\\\
0&1\end{array}\right]\right)\in\mathrm{U},$

then

$\log:\ \
\Omega_{u,\varepsilon}\cong\left[\frac{\log{\varepsilon}}{n},-\frac{\log{\varepsilon}}{n}+\frac{1}{2n}\left(\sum\log{(x_{i}^{2}+1)}+\sum
2\log{(|y_{j}|^{2}+1)}\right)\right]$ (2.6)

by taking logarithm with the following identities

$\displaystyle\operatorname{diag}(\exp(t),\exp(-t))\cdot\Lambda(e_{1})=\exp{((r+2s)t)}\Lambda(e_{1})$
$\displaystyle\operatorname{diag}(\exp(t),\exp(-t))\cdot\Lambda(e_{2})=\exp{(-(r+2s)t)}\Lambda(e_{2})$

for $t\in(k\otimes_{\mathbb{Q}}\mathbb{R})$,
$u\cdot\Lambda(e_{1})=\Lambda(e_{1})$ and

$\left\lVert
u\cdot\Lambda(e_{2})\right\rVert=\left\lVert\left(\begin{array}[]{c}x_{1}\\\
1\end{array}\right)^{1}\wedge...\wedge\left(\begin{array}[]{c}y_{s}\\\
1\end{array}\right)^{r+s}\right\rVert=\prod(x_{i}^{2}+1)^{1/2}\cdot\prod(|y_{j}|^{2}+1).$

Fix a Haar measures $\mu_{\mathrm{K}}$ on $\mathrm{K}$ such that
$\mu_{\mathrm{K}}(\mathrm{K})=1$. Choose a Haar measure $\mu_{\mathrm{U}}$ on
$\mathrm{U}$ such that
$\mu_{\mathrm{K}}\otimes\mu_{\mathrm{U}}\otimes\mu_{H_{1}}$ is pushed to
$\lambda_{G_{1}}$ by the surjective map (2.5). Therefore $\nu_{X_{1}}$ is
identified with (the push-forward of)
$\mu_{\mathrm{K}}\otimes\mu_{\mathrm{U}}$. Since

$\mathrm{U}\cong
k\otimes_{\mathbb{Q}}\mathbb{R}\cong\mathbb{R}^{r}\oplus\mathbb{C}^{s},$

there exists a constant $C_{U}>0$ such that $\mu_{\mathrm{U}}$ is $C_{U}$
times the standard Lebesgue measure on $\mathbb{R}^{r}\oplus\mathbb{C}^{s}$.
Moreover, we can identify $B_{T}$ as a subset of $\mathrm{K}\times\mathrm{U}$
by (2.5).

For $T>0$ and $\eta\in(0,1)$, we define another subset $B_{T,\eta}$ of
$\mathrm{K}\times\mathrm{U}$ by

$B_{T,\eta}:=\left\\{(k,u)\in B_{T}:\ |x_{i}|\geq\eta T,\ 1\leq i\leq r;\
|y_{j}|\geq\eta T,\ 1\leq j\leq s\right\\}$

by using the coordinates (2.4). When
$\left\lVert\cdot\right\rVert_{0}=\left\lVert\cdot\right\rVert$, we write
$B_{T}^{\operatorname{st}}$ and $B^{\operatorname{st}}_{T,\eta}$ for the
corresponding sets respectively. Then

$B^{\operatorname{st}}_{T/C_{0}}\subset B_{T}\subset
B^{\operatorname{st}}_{C_{0}T}$

by (1.6).

Since

$\left[\begin{array}[]{cc}1&x\\\
0&1\end{array}\right]\left[\begin{array}[]{cc}1&0\\\
0&-1\end{array}\right]\left[\begin{array}[]{cc}1&x\\\
0&1\end{array}\right]^{-1}=\left[\begin{array}[]{cc}1&-2x\\\
0&-1\end{array}\right]$

and $B_{T}^{\operatorname{st}}$ is left $\mathrm{K}$-invariant, we can write

$\displaystyle B_{T}^{\operatorname{st}}$
$\displaystyle\cong\mathrm{K}\times\left\\{(x_{i},y_{j}):\
2(r+s)+4\sum|x_{i}|^{2}+4\sum|y_{j}|^{2}\leq T^{2}\right\\}$
$\displaystyle=\mathrm{K}\times\left\\{(x_{i},y_{j}):\
\sum|x_{i}|^{2}+\sum|y_{j}|^{2}\leq\frac{1}{4}(T^{2}-2r-2s)\right\\}.$

By (2.6), we find that

$\log(\Omega_{g,\varepsilon})\subset\left[\frac{\log{\varepsilon}}{n},-\frac{\log{\varepsilon}}{n}+\log{T}\right]$

for $[g]\in B_{T}^{\operatorname{st}}$. On the other hand if $[g]$ is in
$B_{T,\eta}$, then

$\log(\Omega_{g,\varepsilon})\supset\left[\frac{\log{\varepsilon}}{n},-\frac{\log{\varepsilon}}{n}+\log{T}+\log{\eta}\right].$

Consequently, one has

$\mu_{\mathrm{H}_{1}^{\operatorname{spl}}}(\Omega_{g,\varepsilon})\leq\log{T}+\log{C_{0}}+\frac{2\log{\varepsilon}}{n}$

for $[g]\in B_{T}\subset B^{\operatorname{st}}_{C_{0}T}$ and

$\log{T}+\frac{2\log{\varepsilon}}{n}+\log{\eta}\leq\mu_{\mathrm{H}_{1}^{\operatorname{spl}}}(\Omega_{g,\varepsilon})\leq\log{T}+\log{C_{0}}+\frac{2\log{\varepsilon}}{n}$

for $[g]\in B_{T,\eta}$. Asymptotically, this is just $\log{T}$.

It remains to show that the complement of $B_{T,\eta}$ is negligible in
$B_{T}$. Indeed, it is not hard to see from the definition that there exists
$C_{1}>0$ (independent of $\eta$) such that

$\nu_{X_{1}}(B^{\operatorname{st}}_{T}\setminus
B^{\operatorname{st}}_{T,\eta})\leq(C_{1}\eta)\cdot\nu_{X_{1}}(B^{\operatorname{st}}_{T}).$

We may also require $C_{1}$ to satisfy

$\nu_{X_{1}}(B^{\operatorname{st}}_{C_{0}T})\leq
C_{1}\cdot\nu_{X_{1}}(B^{\operatorname{st}}_{C_{0}^{-1}T}).$

Since

$(B_{T}\setminus B_{T,\eta})\subset(B_{C_{0}T}\setminus
B_{C_{0}T,C_{0}^{-1}\eta}),$

we have

$\nu_{X_{1}}(B_{T}\setminus
B_{T,\eta})\leq(C_{1}^{2}C_{0}^{-1}\eta)\cdot\nu_{X_{1}}(B_{T}).$

By firstly letting $T\to+\infty$ and then letting $\eta\to 0$, we conclude
that for any $\varepsilon>0$

$\int_{B_{T}}\mu_{\mathrm{H}_{1}^{\operatorname{spl}}}(\Omega_{g,\varepsilon})\nu_{X_{1}}\sim\log{T}\int_{B_{T}}\nu_{X_{1}}$

as $T\to+\infty$. The result follows from combining (2.3). ∎

###  2.2. The general case

For general case (1.3), we fix an isomorphism $H\cong\mathbb{G}_{m}$ which
induces the homomorphism of Lie groups

${H(k\otimes_{\mathbb{Q}}{\mathbb{R}})\cong(\mathbb{R}^{\times})^{r}\oplus(\mathbb{C}^{\times})^{s}}$${\mathbb{R}_{>0};}$$\scriptstyle{\operatorname{Nm}}$
(2.7)
$\displaystyle(x_{1},\cdots,x_{r},y_{1},\cdots,y_{s})\mapsto\prod_{i=1}^{r}|x_{i}|\cdot\prod_{j=1}^{s}|y_{j}|^{2}$

with the kernel $\mathrm{H}^{\operatorname{an}}$ as before. Similarly, one has

$H(k\otimes_{\mathbb{Q}}\mathbb{R})=\mathrm{H}^{\operatorname{an}}\times\mathrm{H}^{\operatorname{spl}}\
\ \ \text{ with }\ \ \ \mathrm{H}^{\operatorname{spl}}\cong{\mathbb{R}}_{>0}.$

Choosing the Haar measure $\mu_{\mathrm{H}^{\operatorname{spl}}}$ on
$\mathrm{H}^{\operatorname{spl}}$ induced by $t^{-1}dt$ on
${\mathbb{R}}_{>0}$, one obtains a Haar measure
$\mu_{\mathrm{H}^{\operatorname{an}}}$ on $\mathrm{H}^{\operatorname{an}}$
such that
$\mu_{\infty_{k}}\cong\mu_{\mathrm{H}^{\operatorname{an}}}\otimes\mu_{\mathrm{H}^{\operatorname{spl}}}$,
where $\mu_{\infty_{k}}$ is given by (1.5). Now we can extend [OS, Theorem
1.2] over $\mathbb{Q}$ to a general number field.

######  Theorem 2.8.

If $\Gamma$ is an arithmetic lattice in $G(k)$, then

$\\#\left\\{v\in\Gamma\cdot v_{0}:\ \left\lVert v\right\rVert_{0}\leq
T\right\\}\sim\frac{\mu_{\mathrm{H}^{\operatorname{an}}}{(\mathrm{H}^{\operatorname{an}}/\mathrm{H}^{\operatorname{an}}\cap\Gamma})}{\lambda_{\infty_{k}}{(G(k\otimes_{\mathbb{Q}}\mathbb{R})/\Gamma)}}\cdot\log{T}\int_{B_{T}}\nu_{\infty_{k}}$

as $T\to+\infty$, where $\lambda_{\infty_{k}}$ and $\nu_{\infty_{k}}$ are
given by (1.5),

$B_{T}=\left\\{x\in X(k\otimes_{\mathbb{Q}}\mathbb{R}):\ \left\lVert
x\right\rVert_{0}\leq T\right\\}$

and $\left\lVert\cdot\right\rVert_{0}$ is a linear norm on
$(k\otimes_{\mathbb{Q}}\mathbb{R})^{3}$ satisfying (1.6).

###### Proof.

Let $det(f)=\delta$. Then $-\delta f$ is also an isotropic quadratic form over
$k$ with the same discriminant as that of

$f_{1}(x,y,z):=x^{2}+yz\ \ \ \text{ up to $(k^{\times})^{2}$}.$

This implies that the quadratic space $(k^{3},-\delta f)$ is isometric to the
quadratic space $(k^{3},f_{1})$ over $k$ by [OM, 58:4 Theorem and 58:6]. There
is an invertible $k$-linear map $\sigma$ from $k^{3}$ to $k^{3}$ such that
$\sigma$ induces an isomorphism of algebraic groups

$\Phi:G=\operatorname{Spin}(f)=\operatorname{Spin}(-\delta f)\longrightarrow
G_{1}$

over $k$. Since any two maximal split tori of $G_{1}$ over $k$ are conjugated
by an element in $G_{1}(k)$, there is $g\in G_{1}(k)$ such that the following
diagram of algebraic groups

$\textstyle{H\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g\circ\Phi|_{H}}$$\scriptstyle{\cong}$$\textstyle{H_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g\circ\Phi}$$\scriptstyle{\cong}$$\textstyle{G_{1}}$

commutes, where the vertical arrows are the inclusion maps. Therefore
$g\circ\Phi$ induces

$\overline{g\circ\Phi}:\ X\longrightarrow X_{1}$

an isomorphism of homogeneous spaces over $k$ by sending $v_{0}$ to $M_{0}$.
The conjugation of $g$ induces an invertible $k$-linear map on $k^{3}$ given
by

$g:\ k^{3}\longrightarrow k^{3};\ \ \ \left[\begin{array}[]{cc}x&y\\\
z&-x\end{array}\right]\mapsto g\left[\begin{array}[]{cc}x&y\\\
z&-x\end{array}\right]g^{-1}$

for any $(x,y,z)\in k^{3}$. By considering $k$-points, one has the following
commutative diagram

$\textstyle{X(k)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\overline{g\circ\Phi}}$$\scriptstyle{\cong}$$\textstyle{X_{1}(k)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{k^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g\circ\sigma}$$\scriptstyle{\cong}$$\textstyle{k^{3}}$

where the vertical arrows are the inclusion maps of the chosen coordinates.

Let $\Gamma_{1}$ be the image of $\Gamma$ under the map $g\circ\Phi$ and
$\left\lVert\cdot\right\rVert_{1}$ be the linear norm on
$(k\otimes_{\mathbb{Q}}\mathbb{R})^{3}$ defined by
$\left\lVert\cdot\right\rVert_{1}:=\left\lVert
g\circ\sigma(\cdot)\right\rVert_{0}$. Since $\overline{g\circ\Phi}$ induces
the bijection between the following sets

$\displaystyle\left\\{v\in\Gamma\cdot v_{0}:\ \left\lVert
v\right\rVert_{0}\leq T\right\\}$
$\displaystyle\longleftrightarrow\left\\{M\in\Gamma_{1}\cdot M_{0}:\
\left\lVert M\right\rVert_{1}\leq T\right\\}$

and the Haar measures are preserved, one concludes that Theorem 2.8 follows
from Proposition 2.1. ∎

##  3\. Main result

Let $\mathfrak{o}_{k}$ be the ring of integers of $k$ and $\mathcal{X}$ be a
separated scheme of finite type over $\mathfrak{o}_{k}$ such that
$\mathcal{X}\times_{\mathfrak{o}_{k}}k=X$. Since $H\cong\mathbb{G}_{m}$ over
$k$, there is the unique maximal open compact subgroup $N_{v}$ of $H(k_{v})$
for all $v<\infty_{k}$. Write $N_{v}=H(k_{v})$ for $v\in\infty_{k}$.

Let $\mathcal{G}$ and $\mathcal{H}$ be smooth connected group schemes over
$\mathfrak{o}_{k}$ such that

$\mathcal{G}\times_{\mathfrak{o}_{k}}k=G\ \ \ \text{and}\ \ \
\mathcal{H}\times_{\mathfrak{o}_{k}}k=H$

respectively. There is a finite set $S$ of primes of $k$ containing
$\infty_{k}$ such that

$\mathcal{X}\times_{\mathfrak{o}_{k}}\mathfrak{o}_{k_{v}}\cong(\mathcal{G}\times_{\mathfrak{o}_{k}}\mathfrak{o}_{k_{v}})/(\mathcal{H}\times_{\mathfrak{o}_{k}}\mathfrak{o}_{k_{v}})\
\ \ \text{and}\ \ \ \mathcal{H}(\mathfrak{o}_{k_{v}})=N_{v}$ (3.1)

for all primes $v\not\in S$ induced by (1.3), where $\mathfrak{o}_{k_{v}}$ is
the integral ring of $k_{v}$.

$h_{k}=[H({\bf A}_{k}):H(k)\cdot\prod_{v\leq\infty_{k}}N_{v}]\ \ \ \text{and}\
\ \ H(k)\cap\prod_{v\leq\infty_{k}}N_{v}\cong\mathfrak{o}_{k}^{\times}$ (3.2)

where $h_{k}$ is the class number of $k$.

Fix a maximal open compact subgroup $M_{v}$ of $G(k_{v})$ containing $N_{v}$
for $v\in(S\setminus\infty_{k})$. Define

$\Gamma_{v}=\begin{cases}\mathcal{G}(\mathfrak{o}_{k_{v}})\ \ \
&\text{$v\not\in S$}\\\ \\{g\in M_{v}:\
g\cdot\mathcal{X}(\mathfrak{o}_{k_{v}})=\mathcal{X}(\mathfrak{o}_{k_{v}})\\}\
\ \ &\text{$v\in(S\setminus\infty_{k})$}\\\ G(k_{v})\ \ \
&\text{$v\in\infty_{k}$}.\end{cases}$

Then $\Gamma_{v}$ acts on $\mathcal{X}(\mathfrak{o}_{k_{v}})$ such that the
number of orbits $[\mathcal{X}(\mathfrak{o}_{k_{v}}):\Gamma_{v}]$ is finite
for all $v\leq\infty_{k}$. Moreover, one has that $N_{v}\subset\Gamma_{v}$ for
all $v\leq\infty_{k}$.

######  Lemma 3.3.

There are finitely many $x_{1},\cdots,x_{l}\in\mathcal{X}(\mathfrak{o}_{k})$
such that

$\mathcal{X}(\mathfrak{o}_{k})=\bigcup_{i=1}^{l}[(\prod_{v\leq\infty_{k}}\Gamma_{v})\cdot
x_{i}\cap\mathcal{X}(\mathfrak{o}_{k})]\ \ \ \text{ with }\ \ \
l=\prod_{v\in(S\setminus\infty_{k})}[\mathcal{X}(\mathfrak{o}_{k_{v}}):\Gamma_{v}].$

###### Proof.

By (3.1) and [Gi, Chapter III, Proposition 3.2.2], the following sequence of
pointed sets

$1\rightarrow{\mathcal{H}}(\mathfrak{o}_{k_{v}})\rightarrow{\mathcal{G}}(\mathfrak{o}_{k_{v}})\rightarrow{\mathcal{X}}(\mathfrak{o}_{k_{v}})\rightarrow\check{H}^{1}(\mathfrak{o}_{k_{v}},{\mathcal{H}})$

is exact for $v\not\in S$. Since $\mathcal{H}$ is connected and smooth over
$\mathfrak{o}_{k_{v}}$, every torsor over $\mathfrak{o}_{k_{v}}$ under
$\mathcal{H}$ is trivial by Hensel’s Lemma (see [PR94, Chapter 3, Thm.3.11])
and Lang’s Theorem (see [PR94, Chapter 6, Thm.6.1]). This implies
$\check{H}^{1}(\mathfrak{o}_{k_{v}},{\mathcal{H}})$ is trivial. Therefore

$[\mathcal{X}(\mathfrak{o}_{k_{v}}):\Gamma_{v}]=1$

for all primes $v\not\in S$.

Since $X$ satisfies strong approximation off $\infty_{k}$ by [CTX, Theorem 3.7
and §5.6], one obtains

$(\prod_{v\leq\infty_{k}}\Gamma_{v})\cdot\mathcal{X}(\mathfrak{o}_{k})=\prod_{v\leq\infty_{k}}\mathcal{X}(\mathfrak{o}_{k_{v}})$

by Lemma 1.4. This implies that

$l=\prod_{v\leq\infty_{k}}[\mathcal{X}(\mathfrak{o}_{k_{v}}):\Gamma_{v}]=\prod_{v\in(S\setminus\infty_{k})}[\mathcal{X}(\mathfrak{o}_{k_{v}}):\Gamma_{v}]$

as desired. ∎

Let

$\Gamma=G(k)\cap(\prod_{v\leq\infty_{k}}\Gamma_{v}).$

Then $\Gamma$ is an arithmetic lattice in $G(k)$ and acts on
$\mathcal{X}(\mathfrak{o}_{k})$.

######  Remark 3.4.

By [BHC, 6.9 Theorem], the number of $\Gamma$-orbits of
$\mathcal{X}(\mathfrak{o}_{k})$ is finite when $k=\mathbb{Q}$. It should be
pointed out that a simple proof of this finiteness in [WX, Corollary 2.5] is
not correct because the natural map in the proof is not injective. However,
this finiteness result plays no role in the proof of the main results in [WX].
In fact, the precise descriptions of number of orbits are given by [WX, Prop.
2.9, Prop.2.11 and Prop.2.12] and in particular the finiteness of orbits
follows in those cases over a general number field.

Since $\Gamma$ acts on $[((\prod_{v\leq\infty_{k}}\Gamma_{v})\cdot
x)\cap\mathcal{X}(\mathfrak{o}_{k})]$ for any
$x\in\mathcal{X}(\mathfrak{o}_{k})$, one can give the precise description of
number of orbits.

######  Lemma 3.5.

If $x\in\mathcal{X}(\mathfrak{o}_{k})$, then there are
$y_{1},\cdots,y_{h_{x}}$ in $\mathcal{X}(\mathfrak{o}_{k})$ such that

$((\prod_{v\leq\infty_{k}}\Gamma_{v})\cdot
x)\cap\mathcal{X}(\mathfrak{o}_{k})=\bigcup_{i=1}^{h_{x}}\Gamma y_{i}$

with

$h_{x}=[H_{x}({\bf
A}_{k}):H_{x}(k)\cdot\prod_{v\leq\infty_{k}}(H_{x}(k_{v})\cap\Gamma_{v})]<\infty$

where $H_{x}$ is the stabilizer of $x$ in $G$.

###### Proof.

For any

$y\in((\prod_{v\leq\infty_{k}}\Gamma_{v})\cdot
x)\cap\mathcal{X}(\mathfrak{o}_{k}),$

there is $\sigma\in\prod_{v\leq\infty_{k}}\Gamma_{v}$ such that $y=\sigma x$.
By Lemma 1.4, there is $\tau\in G(k)$ such that $y=\tau x$. This implies that
$\tau^{-1}\sigma\in H_{x}({\bf A}_{k})$. Define a map

$\Phi:\ ((\prod_{v\leq\infty_{k}}\Gamma_{v})\cdot
x)\cap\mathcal{X}(\mathfrak{o}_{k})\longrightarrow H_{x}({\bf
A}_{k})/[H_{x}(k)\cdot\prod_{v\leq\infty_{k}}(H_{x}(k_{v})\cap\Gamma_{v})]$

by sending $y$ to $\tau^{-1}\sigma$. Suppose there are

$\sigma^{\prime}\in\prod_{v\leq\infty_{k}}\Gamma_{v}\ \ \ \text{ and }\ \ \
\tau^{\prime}\in G(k)$

such that $y=\sigma^{\prime}x=\tau^{\prime}x$. Then

$\sigma^{-1}\sigma^{\prime}\in\prod_{v\leq\infty_{k}}(H_{x}(k_{v})\cap\Gamma_{v})\
\ \ \text{and}\ \ \ \tau^{-1}\tau^{\prime}\in H_{x}(k).$

This implies that the map $\Phi$ is well-defined. For any $\xi\in H_{x}({\bf
A}_{k})\subset G({\bf A}_{k})$, there are

$\gamma\in\prod_{v\leq\infty_{k}}\Gamma_{v}\ \ \ \text{ and }\ \ \ \eta\in
G(k)$

such that $\xi=\eta\cdot\gamma$ by strong approximation for $G$ (see [PR94,
Theorem 7.12]). Therefore

$\eta^{-1}x=\gamma x\in((\prod_{v\leq\infty_{k}}\Gamma_{v})\cdot x)\cap
X(k)=((\prod_{v\leq\infty_{k}}\Gamma_{v})\cdot
x)\cap\mathcal{X}(\mathfrak{o}_{k})$

and $\Phi$ is surjective. Suppose

$\Phi(y)=\Phi(y_{1})\ \ \ \text{ with }\ \
y,y_{1}\in((\prod_{v\leq\infty_{k}}\Gamma_{v})\cdot
x)\cap\mathcal{X}(\mathfrak{o}_{k}).$

There are

$\sigma,\sigma_{1}\in\prod_{v\leq\infty_{k}}\Gamma_{v}\ \ \ \text{ and }\ \ \
\tau,\tau_{1}\in G(k)$

with $y=\sigma x=\tau x$ and $y_{1}=\sigma_{1}x=\tau_{1}x$ respectively such
that

$(\tau^{-1}\sigma)^{-1}(\tau_{1}^{-1}\sigma_{1})\in
H_{x}(k)\cdot\prod_{v\leq\infty_{k}}(H_{x}(k_{v})\cap\Gamma_{v}).$

This implies that there are $h\in H_{x}(k)$ and
$\delta\in\prod_{v\leq\infty_{k}}(H_{x}(k_{v})\cap\Gamma_{v})$ such that

$\tau_{1}\tau^{-1}h=\sigma_{1}\sigma^{-1}\delta\in(G(k)\cap\prod_{v\leq\infty_{k}}\Gamma_{v})=\Gamma.$

Then $y_{1}\in\Gamma y$. Therefore the map induced by $\Phi$

$[(\prod_{v\leq\infty_{k}}\Gamma_{v})\cdot
x\cap\mathcal{X}(\mathfrak{o}_{k})]/\Gamma\xrightarrow{\cong}H_{x}({\bf
A}_{k})/[H_{x}(k)\cdot\prod_{v\leq\infty_{k}}(H_{x}(k_{v})\cap\Gamma_{v})]$

is a bijection. The finiteness of $h_{x}$ follows from [PR94, Chapter 5,
Thm.5.1]. ∎

Define

$N({\mathcal{X}},T)=\\#\\{(\alpha,\beta,\gamma)\in{\mathcal{X}}(\mathfrak{o}_{k})\subset
X(k):\ \ \left\lVert(\alpha,\beta,\gamma)\right\rVert_{0}\leq T\\}$

for $T>0$, where $\left\lVert\cdot\right\rVert_{0}$ is a norm on
$(k\otimes_{\mathbb{Q}}\mathbb{R})^{3}$ satisfying (1.6). Let $r$ and $s$ be
the numbers of real and complex primes of $k$ respectively, $w_{k}$ be the
number of roots of unity in $k$, $d_{k}$ be the discriminant of $k$ and
$R_{k}$ be the regulator of $k$. The main result of this section is the
following theorem.

######  Theorem 3.6.

If $q_{v}$ is the number of elements in the residue field of $k$ at
$v<\infty_{k}$, then

$N({\mathcal{X}},T)\sim\frac{2^{r}(2\pi)^{s}}{w_{k}\sqrt{|d_{k}|}}R_{k}h_{k}\cdot(\prod_{v<\infty_{k}}(1-q_{v}^{-1})\int_{\mathcal{X}(\mathfrak{o}_{k_{v}})}\nu_{v})\cdot(\log{T}\int_{B_{T}}\nu_{\infty_{k}})$

as $T\to\infty$ where

$B_{T}=\\{(\alpha,\beta,\gamma)\in X(k\otimes_{\mathbb{Q}}\mathbb{R}):\ \
\left\lVert(\alpha,\beta,\gamma)\right\rVert_{0}\leq T\\}.$

###### Proof.

Since $\mathcal{X}(\mathfrak{o}_{k})$ can be decomposed into the following
disjoint union

$\mathcal{X}(\mathfrak{o}_{k})=\bigcup_{i=1}^{l}[(\prod_{v\leq\infty_{k}}\Gamma_{v})\cdot
x_{i}\cap\mathcal{X}(\mathfrak{o}_{k})]=\bigcup_{i=1}^{l}\bigcup_{j=1}^{h_{x_{i}}}\Gamma
y_{i,j}$

with

$(\prod_{v\leq\infty_{k}}\Gamma_{v})\cdot
x_{i}\cap\mathcal{X}(\mathfrak{o}_{k})=\bigcup_{j=1}^{h_{x_{i}}}\Gamma
y_{i,j}\ \ \ \text{and}\ \ \ h_{x_{i}}=[H_{x_{i}}({\bf
A}_{k}):H_{x_{i}}(k)\cdot\prod_{v\leq\infty_{k}}(H_{x_{i}}(k_{v})\cap\Gamma_{v})]$

for $1\leq i\leq l$ by Lemma 3.3 and Lemma 3.5, one obtains that

$N({\mathcal{X}},T)=\sum_{i=1}^{l}\sum_{j=1}^{h_{x_{i}}}N(y_{i,j},T)$

where

$N(y_{i,j},T)=\\{(\alpha,\beta,\gamma)\in\Gamma y_{i,j}:\ \
\left\lVert(\alpha,\beta,\gamma)\right\rVert_{0}\leq T\\}$

for $1\leq j\leq h_{x_{i}}$ and $1\leq i\leq l$. By Theorem 2.8, one has

$N(y_{i,j},T)\sim\frac{\mu_{\mathrm{H}_{y_{i,j}}^{\operatorname{an}}}{(\mathrm{H}_{y_{i,j}}^{\operatorname{an}}/\mathrm{H}_{y_{i,j}}^{\operatorname{an}}\cap\Gamma})}{\lambda_{\infty_{k}}{(G(k\otimes_{\mathbb{Q}}\mathbb{R})/\Gamma)}}\cdot\log{T}\int_{B_{i,j}(T)}\nu_{\infty_{k}}$

with

$B_{i,j}(T)=\\{z\in G(k\otimes_{\mathbb{Q}}\mathbb{R})y_{i,j}\subseteq
X(k\otimes_{\mathbb{Q}}\mathbb{R}):\ \left\lVert z\right\rVert_{0}\leq T\\}$

as $T\to\infty$ for $1\leq j\leq h_{x_{i}}$ and $1\leq i\leq l$.

By Lemma 1.4, there is $g_{i,j}\in G(k)$ such that

$g_{i,j}v_{0}=y_{i,j}\ \ \ \text{ and }\ \ \ H_{y_{i,j}}=g_{i,j}Hg_{i,j}^{-1}$
(3.7)

for $1\leq j\leq h_{x_{i}}$ and $1\leq i\leq l$. Since $H(k_{v})\cap
g_{i,j}^{-1}\Gamma_{v}g_{i,j}$ is an open and compact subgroup of $H(k_{v})$
for all $v<\infty_{k}$, one has

$(H(k_{v})\cap g_{i,j}^{-1}\Gamma_{v}g_{i,j})\subseteq N_{v}\ \ \ \text{and}\
\ \ H(k_{v})\cap g_{i,j}^{-1}\Gamma_{v}g_{i,j}=N_{v}\cap
g_{i,j}^{-1}\Gamma_{v}g_{i,j}$ (3.8)

for all $v<\infty_{k}$, $1\leq j\leq h_{x_{i}}$ and $1\leq i\leq l$. Then

$\displaystyle[(H(k)\cap\prod_{v\leq\infty_{k}}N_{v}):(H(k)\cap\prod_{v\leq\infty_{k}}(H(k_{v})\cap
g_{i,j}^{-1}\Gamma_{v}g_{i,j}))]^{-1}\mu_{\mathrm{H}_{y_{i,j}}^{\operatorname{an}}}{(\mathrm{H}_{y_{i,j}}^{\operatorname{an}}/\mathrm{H}_{y_{i,j}}^{\operatorname{an}}\cap\Gamma})$
(3.9) $\displaystyle=$
$\displaystyle[(H(k)\cap\prod_{v\leq\infty_{k}}N_{v}):(H(k)\cap\prod_{v\leq\infty_{k}}(H(k_{v})\cap
g_{i,j}^{-1}\Gamma_{v}g_{i,j}))]^{-1}\mu_{\mathrm{H}^{\operatorname{an}}}{(\mathrm{H}^{\operatorname{an}}/\mathrm{H}^{\operatorname{an}}\cap
g_{i,j}^{-1}\Gamma g_{i,j}})$ $\displaystyle=$
$\displaystyle\mu_{\mathrm{H}^{\operatorname{an}}}{(\mathrm{H}^{\operatorname{an}}/\mathrm{H}^{\operatorname{an}}\cap(H(k)\cap\prod_{v\leq\infty_{k}}N_{v})})=\frac{2^{r}(2\pi)^{s}}{w_{k}}R_{k}$

by the product formula (see [N, Chapter III, (1.3) Proposition]) and (2.7) and
the computation in [Vo, P.140].

By Lemma 1.4, there is $g_{i}\in G(k)$ such that

$g_{i}v_{0}=x_{i}\ \ \ \text{ and }\ \ \ H_{x_{i}}=g_{i}Hg_{i}^{-1}$ (3.10)

for $1\leq i\leq l$. By the same argument as (3.8) and (3.9), one obtains that

$\displaystyle[(H(k)\cap\prod_{v\leq\infty_{k}}N_{v}):(H(k)\cap\prod_{v\leq\infty_{k}}(H(k_{v})\cap
g_{i}^{-1}\Gamma_{v}g_{i}))]^{-1}\mu_{\mathrm{H}_{x_{i}}^{\operatorname{an}}}{(\mathrm{H}_{x_{i}}^{\operatorname{an}}/\mathrm{H}_{x_{i}}^{\operatorname{an}}\cap\Gamma})$
(3.11) $\displaystyle=$
$\displaystyle\mu_{\mathrm{H}^{\operatorname{an}}}{(\mathrm{H}^{\operatorname{an}}/\mathrm{H}^{\operatorname{an}}\cap(H(k)\cap\prod_{v\leq\infty_{k}}N_{v})})=\frac{2^{r}(2\pi)^{s}}{w_{k}}R_{k}$

for $1\leq i\leq l$.

For each $1\leq j\leq h_{x_{i}}$, there is

$(\gamma_{v})_{v\leq\infty_{k}}\in\prod_{v\leq\infty_{k}}\Gamma_{v}\ \ \
\text{such that}\ \ \ (\gamma_{v})_{v\leq\infty_{k}}\cdot x_{i}=y_{i,j}$

by the partition of $\mathcal{X}(\mathfrak{o}_{k})$. One can write

$h_{v}=g_{i,j}^{-1}\gamma_{v}g_{i}\in H(k_{v})$

for all $v\leq\infty_{k}$ by (3.7) and (3.10). Since $H(k_{v})$ is
commutative, one obtains that

$H(k_{v})\cap g_{i,j}^{-1}\Gamma_{v}g_{i,j}=H(k_{v})\cap
h_{v}g_{i}^{-1}\gamma_{v}^{-1}\Gamma_{v}\gamma_{v}g_{i}h_{v}^{-1}=H(k_{v})\cap
h_{v}g_{i}^{-1}\Gamma_{v}g_{i}h_{v}^{-1}=H(k_{v})\cap
g_{i}^{-1}\Gamma_{v}g_{i}$

for all $v\leq\infty_{k}$. This implies that

$\mu_{\mathrm{H}_{y_{i,j}}^{\operatorname{an}}}{(\mathrm{H}_{y_{i,j}}^{\operatorname{an}}/\mathrm{H}_{y_{i,j}}^{\operatorname{an}}\cap\Gamma})=\mu_{\mathrm{H}_{x_{i}}^{\operatorname{an}}}{(\mathrm{H}_{x_{i}}^{\operatorname{an}}/\mathrm{H}_{x_{i}}^{\operatorname{an}}\cap\Gamma})$
(3.12)

for all $1\leq j\leq l_{x_{i}}$ by (3.9) and (3.11).

Since

$B_{T}=B_{i,j}(T)\ \ \text{ for $1\leq j\leq h_{x_{i}}$ and $1\leq i\leq l$ }$

by Lemma 1.4, one concludes that

$N({\mathcal{X}},T)\sim\sum_{i=1}^{l}\frac{h_{x_{i}}\cdot\mu_{\mathrm{H}_{x_{i}}^{\operatorname{an}}}{(\mathrm{H}_{x_{i}}^{\operatorname{an}}/\mathrm{H}_{x_{i}}^{\operatorname{an}}\cap\Gamma})}{\lambda_{\infty_{k}}{(G(k\otimes_{\mathbb{Q}}\mathbb{R})/\Gamma)}}\cdot\log{T}\int_{B_{T}}\nu_{\infty_{k}}$
(3.13)

as $T\to\infty$ by (3.12). Since $h_{x_{i}}$ is equal to

$\displaystyle[H_{x_{i}}({\bf
A}_{k}):H_{x_{i}}(k)\prod_{v\leq\infty_{k}}(H_{x_{i}}(k_{v})\cap\Gamma_{v})])=[H({\bf
A}_{k}):H(k)\prod_{v\leq\infty_{k}}(H(k_{v})\cap g_{i}^{-1}\Gamma_{v}g_{i})])$
$\displaystyle=$ $\displaystyle
h_{k}[H(k)\prod_{v\leq\infty_{k}}N_{v}:H(k)\prod_{v\leq\infty_{k}}(H(k_{v})\cap
g_{i}^{-1}\Gamma_{v}g_{i})]$ $\displaystyle=$ $\displaystyle
h_{k}[(H(k)\cap\prod_{v<\infty_{k}}N_{v}):(H(k)\cap\prod_{v<\infty_{k}}(H(k_{v})\cap
g_{i}^{-1}\Gamma_{v}g_{i}))]^{-1}\prod_{v<\infty_{k}}[N_{v}:(N_{v}\cap
g_{i}^{-1}\Gamma_{v}g_{i})]$

for $1\leq i\leq l$ by (3.2) and (3.10), one obtains that

$h_{x_{i}}\cdot\mu_{\mathrm{H}_{x_{i}}^{\operatorname{an}}}{(\mathrm{H}_{x_{i}}^{\operatorname{an}}/\mathrm{H}_{x_{i}}^{\operatorname{an}}\cap\Gamma})=\frac{2^{r}(2\pi)^{s}}{w_{k}}R_{k}h_{k}\prod_{v<\infty_{k}}[N_{v}:(N_{v}\cap
g_{i}^{-1}\Gamma_{v}g_{i})]$ (3.14)

for $1\leq i\leq l$ by (3.11). Since the Tamagawa number of $G$ is 1 by [W,
Chapter IV; Thm.4.4.1], one obtains

$1=\lambda_{G}(G(k)\backslash G({\bf
A}_{k}))=\lambda_{G}(G(k)\backslash(G(k)\cdot\prod_{v\leq\infty_{k}}\Gamma_{v}))$

by strong approximation for $G$ (see [PR94, Theorem 7.12]), where
$\lambda_{G}=\prod_{v\leq\infty_{k}}\lambda_{v}$. This implies that

$\lambda_{\infty_{k}}(G(k\otimes_{\mathbb{Q}}\mathbb{R})/\Gamma)^{-1}=\prod_{v<\infty_{k}}\int_{\Gamma_{v}}\lambda_{v}$

by reduction theory (see [PR94, Chapter 5, §5.2]). Partitioning the set
$\\{x_{i}\\}_{i=1}^{l}$ into the following disjoint union

$\\{x_{i}\\}_{i=1}^{l}=\prod_{v\in(S\setminus\infty_{k})}\\{x_{v,j}:\ 1\leq
j\leq l_{v}\\}\ \ \ \text{such that}\ \ \
\prod_{v\in(S\setminus\infty_{k})}l_{v}=l\ \ \ \text{and}\ \ \
\mathcal{X}(\mathfrak{o}_{k_{v}})=\bigcup_{j=1}^{l_{v}}\Gamma_{v}x_{v,j}$

for $v\in(S\setminus\infty_{k})$ by Lemma 3.3 and writing $g_{v,j}$ for
$g_{i}$ in (3.10) when we use $x_{v,j}$ for $x_{i}$, one obtains

$\displaystyle\int_{\mathcal{X}(\mathfrak{o}_{k_{v}})}\nu_{v}=\sum_{j=1}^{l_{v}}(\int_{\Gamma_{v}}\lambda_{v})(\int_{H(k_{v})\cap
g_{v,j}^{-1}\Gamma_{v}g_{v,j}}\mu_{v})^{-1}$ (3.15) $\displaystyle=$
$\displaystyle(\sum_{j=1}^{l_{v}}[N_{v}:N_{v}\cap
g_{v,j}^{-1}\Gamma_{v}g_{v,j}])(\int_{\Gamma_{v}}\lambda_{v})(\int_{N_{v}}\mu_{v})^{-1}$

for all primes $v\in(S\setminus\infty_{k})$. For any primes $v\not\in S$, one
simply has

$\int_{\mathcal{X}(\mathfrak{o}_{k_{v}})}\nu_{v}\cdot\int_{N_{v}}\mu_{v}=\int_{\Gamma_{v}}\lambda_{v}$
(3.16)

by (3.1). Combining (3.13), (3.14), (3.15) and (3.16), one concludes that

$\displaystyle
N({\mathcal{X}},T)\sim\frac{2^{r}(2\pi)^{s}}{w_{k}}R_{k}h_{k}\cdot(\sum_{i=1}^{l}\prod_{v<\infty_{k}}[N_{v}:(N_{v}\cap
g_{i}^{-1}\Gamma_{v}g_{i})]\int_{\Gamma_{v}}\lambda_{v})\cdot(\log{T}\int_{B_{T}}\nu_{\infty_{k}})$
$\displaystyle=\frac{2^{r}(2\pi)^{s}}{w_{k}}R_{k}h_{k}\cdot((\prod_{v\not\in
S}\int_{\Gamma_{v}}\lambda_{v})(\sum_{i=1}^{l}\prod_{v\in(S\setminus\infty_{k})}[N_{v}:(N_{v}\cap
g_{i}^{-1}\Gamma_{v}g_{i})]\int_{\Gamma_{v}}\lambda_{v}))\cdot(\log{T}\int_{B_{T}}\nu_{\infty_{k}})$
$\displaystyle=\frac{2^{r}(2\pi)^{s}}{w_{k}}R_{k}h_{k}\cdot((\prod_{v\not\in
S}\int_{\Gamma_{v}}\lambda_{v})(\prod_{v\in(S\setminus\infty_{k})}\sum_{j=1}^{l_{v}}[N_{v}:(N_{v}\cap
g_{v,j}^{-1}\Gamma_{v}g_{v,j})]\int_{\Gamma_{v}}\lambda_{v}))\cdot(\log{T}\int_{B_{T}}\nu_{\infty_{k}})$
$\displaystyle=\frac{2^{r}(2\pi)^{s}}{w_{k}}R_{k}h_{k}\cdot((\prod_{v\not\in
S}\int_{\Gamma_{v}}\lambda_{v})(\prod_{v\in(S\setminus\infty_{k})}\int_{\mathcal{X}(\mathfrak{o}_{k_{v}})}\nu_{v}\cdot\int_{N_{v}}\mu_{v}))\cdot(\log{T}\int_{B_{T}}\nu_{\infty_{k}})$
$\displaystyle=\frac{2^{r}(2\pi)^{s}}{w_{k}}R_{k}h_{k}\cdot(\prod_{v<\infty_{k}}\int_{\mathcal{X}(\mathfrak{o}_{k_{v}})}\nu_{v}\cdot\int_{N_{v}}\mu_{v})\cdot(\log{T}\int_{B_{T}}\nu_{\infty_{k}})$

as $T\to\infty$. Since

$\int_{N_{v}}\mu_{v}=(1-q_{v}^{-1})[\mathfrak{o}_{k_{v}}:D_{k_{v}}]^{-\frac{1}{2}}\
\ \ \text{and}\ \ \
\prod_{v<\infty_{k}}[\mathfrak{o}_{k_{v}}:D_{k_{v}}]=|d_{k}|$

by [Tate, Lemma 2.3.3] and [N, Chapter III, (2.9) Theorem and (2.11)
Corollary] where $D_{k_{v}}$ is the absolute different of $k_{v}$ for all
$v<\infty_{k}$, one obtains the desired result. ∎

If $k=\mathbb{Q}$, then $s=0$, $w_{k}=2$ and $r=d_{k}=R_{k}=h_{k}=1$ in
Theorem.3.6. One obtains Theorem 1.1 by [BR95, (1.8.0) and Lemma 1.8.1].

##  4\. An example

As an application of Theorem 3.6, we will give a precise asymptotic formula
for counting the number of integral points of the following equation

$\mathcal{X}:\ \ \ x^{2}+y^{2}-\delta z^{2}=1$ (4.1)

with $\delta\in\mathbb{Z}$ and $\delta>0$, which has been studied in [DRS]. We
keep the same notations as those in the previous sections.

######  Lemma 4.2.

Let $X=\mathcal{X}\times_{\mathbb{Z}}\mathbb{Q}$. Then

$\omega_{X}=\frac{dx\wedge dy}{-2\delta z}=\frac{dz\wedge
dy}{-2x}=\frac{dx\wedge dz}{-2y}$

is a gauge form on $X$ such that the induced Tamagawa measure $\nu_{p}$ on
$X(\mathbb{Q}_{p})$ satisfies

$\int_{\mathcal{X}(\mathbb{Z}_{p})}\nu_{p}=\lim_{k\to\infty}\frac{\\#\\{(\beta_{1},\beta_{2},\beta_{3})\in(\mathbb{Z}/(p^{k}))^{3}:\
f(\beta_{1},\beta_{2},\beta_{3})\equiv 1\mod p^{k}\\}}{p^{2k}}$ (4.3)

for all primes $p$.

###### Proof.

The equality of $\omega_{X}$ follows from

$x\cdot dx+y\cdot dy-\delta z\cdot dz=0$

by differentiating the given equation (4.1).

Let $SO(f)$ be the special orthogonal group defined by $f=x^{2}+y^{2}-\delta
z^{2}$. Since $X$ is also a homogeneous space of $SO(f)$ by [CTX, §5.6], it is
enough to show that $\omega_{X}$ is $SO(f)$-invariant by [W, Theorem 2.4.1].
Let

$(x^{\prime},y^{\prime},z^{\prime})=(x,y,z)\cdot A\ \ \ \text{with}\ \ \
A=\left[\begin{array}[c]{llll}a_{11}\ \ a_{12}\ \ a_{13}\\\ a_{21}\ \ a_{22}\
\ a_{23}\\\ a_{31}\ \ a_{32}\ \ a_{33}\end{array}\right]\in SO(f).$

Since

$A\cdot\left[\begin{array}[c]{llll}1&0&\ \ 0\\\ 0&1&\ \ 0\\\
0&0&-\delta\end{array}\right]=\left[\begin{array}[c]{llll}1&0&\ \ 0\\\ 0&1&\ \
0\\\ 0&0&-\delta\end{array}\right]\cdot(A^{\prime})^{-1},$

one obtains

$\left[\begin{array}[c]{llll}a_{11}&a_{12}&-\delta\cdot a_{13}\\\
a_{21}&a_{22}&-\delta\cdot a_{23}\\\ a_{31}&a_{32}&-\delta\cdot
a_{33}\end{array}\right]=\left[\begin{array}[c]{llll}\ \ A_{11}&\ \ A_{12}&\ \
A_{13}\\\ \ \ A_{21}&\ \ A_{22}&\ \ A_{23}\\\ -\delta\cdot A_{31}&-\delta\cdot
A_{32}&-\delta\cdot A_{33}\end{array}\right]$ (4.4)

where $A_{ij}$ is the cofactor of $a_{ij}$ in $A$ for $1\leq i,j\leq 3$.
Therefore

$dx^{\prime}\wedge dy^{\prime}=A_{33}\cdot dx\wedge dy+A_{13}\cdot dy\wedge
dz+A_{23}\cdot dz\wedge dx$ $=(A_{33}-A_{13}\frac{x}{\delta
z}-A_{23}\frac{y}{\delta z})\cdot dx\wedge dy=\frac{z^{\prime}}{z}\cdot
dx\wedge dy$

by the equality of $\omega_{X}$ and (4.4).

In order to show (4.3), one considers a morphism

$\psi:\operatorname{Spec}(k[x,y,z])\longrightarrow\operatorname{Spec}(k[t]);\
(x,y,z)\mapsto f(x,y,z)-1$

induced by $f$. Since $\omega_{X}$ satisfies

$\omega_{X}\wedge\psi^{*}(dt)=(\frac{dx\wedge dy}{-2\delta z})\wedge(2x\cdot
dx+2y\cdot dy-2\delta z\cdot dz)=dx\wedge dy\wedge dz,$

one concludes that (4.3) holds by [BR95, Lemma 1.8.1]. ∎

The local densities of $f$ can be computed as follows.

######  Lemma 4.5.

If $f(x,y,z)=x^{2}+y^{2}-\delta z^{2}$ with $\delta\in\mathbb{Z}$ and
$\delta>0$ and

$\alpha_{p}(f,1)=\lim_{k\to\infty}\frac{\\#\\{(\beta_{1},\beta_{2},\beta_{3})\in(\mathbb{Z}/(p^{k}))^{3}:\
f(\beta_{1},\beta_{2},\beta_{3})\equiv 1\mod p^{k}\\}}{p^{2k}}$

for all primes $p$, then

$\alpha_{p}(f,1)=\begin{cases}1+(\frac{\delta}{p})p^{-1}\ \ \ &\text{$p$ is
odd with $(p,\delta)=1$}\\\ 1-(\frac{-1}{p})p^{-1}\ \ \ &\text{$p$ is odd with
$p|\delta$}\\\ 1\ \ \ &\text{$p=2$ with $4\mid\delta$ or $\delta\equiv 1\mod
8$}\\\ 3\cdot 2^{-2}\ \ \ &\text{$p=2$ with $\delta\equiv 3,7\mod 8$}\\\
2^{-1}\ \ \ &\text{$p=2$ with $2|\delta$ but $4\nmid\delta$ or $\delta\equiv
5\mod 8$}\end{cases}$

where $(\frac{*}{p})$ are the Legendre symbols.

###### Proof.

When $p$ is odd and $(p,\delta)=1$, then

$\alpha_{p}(f,1)=1+(\frac{\delta}{p})p^{-1}$

by [S, Hilfssatz 12].

When $p$ is odd with $p|\delta$ and $(\frac{-1}{p})=1$, then $f$ is equivalent
to $xy-\delta z^{2}$ over $\mathbb{Z}_{p}$. Since

$\\#\\{(\beta_{1},\beta_{2},\beta_{3})\in(\mathbb{Z}/(p^{k}))^{3}:\
\beta_{1}\beta_{2}-\delta\beta_{3}^{2}\equiv 1\mod
p^{k}\\}=(p^{k}-p^{k-1})p^{k}$

for sufficiently large $k$, one obtains that $\alpha_{p}(f,1)=1-p^{-1}$.

When $p$ is odd with $p|\delta$ and $(\frac{-1}{p})=-1$, then

$\\#\\{(\bar{\xi},\bar{\eta})\in(\mathbb{Z}/p\mathbb{Z})^{2}:\ \ \
\bar{\xi}^{2}+\bar{\eta}^{2}=\bar{1}\mod p\\}=p+1$

and one of $\xi$ and $\eta$ is not divisible by $p$. For such a pair
$(\bar{\xi},\bar{\eta})$, one can make a substitution

$\begin{cases}x=\xi+px_{1}\\\ y=\eta+py_{1}\end{cases}$

and obtain the equation

$2\xi x_{1}+px_{1}^{2}+2\eta y_{1}+py_{1}^{2}-(\delta
p^{-1})z^{2}=(1-\xi^{2}-\eta^{2})p^{-1}$ (4.6)

over $\mathbb{Z}$ by (4.1). Without loss of generality, we assume that $\xi$
is not divisible by $p$. By Hensel’s Lemma, the equation (4.6) has a unique
solution for $x_{1}$ over $\mathbb{Z}_{p}$ for $y_{1}$ and $z$ taking any
values in $\mathbb{Z}$. This implies that

$\\#\\{(\beta_{1},\beta_{2},\beta_{3})\in(\mathbb{Z}/(p^{k}))^{3}:\
f(\beta_{1},\beta_{2},\beta_{3})\equiv 1\mod p^{k}\\}=(p+1)p^{k-1}p^{k}$

for sufficiently large $k$ and $\alpha_{p}(f,1)=1+p^{-1}$.

When $p=2$ and $(\delta,2)=1$, then

$f\cong\begin{cases}2xy+\delta z^{2}\ \ \ &\text{if
$(\frac{-1,-\delta}{\mathbb{Q}_{2}})=-1$}\\\ 2(x^{2}+xy+y^{2})-3\delta z^{2}\
\ \ &\text{if $(\frac{-1,-\delta}{\mathbb{Q}_{2}})=1$}\end{cases}$ (4.7)

over $\mathbb{Z}_{2}$ by [OM, 93:18. Example (iv) and 58:6].

$\bullet$ Case $\delta\equiv 1\mod 8$. Then $f\cong 2xy+z^{2}$ by (4.7). Since
an odd integer $m$ is a square in $\mathbb{Z}_{2}$ if and only if $m\equiv
1\mod 8$ by [OM, 63:1. Local Square Theorem], one concludes that

$\\#\\{(\beta_{1},\beta_{2},\beta_{3})\in(\mathbb{Z}/(2^{k}))^{3}:\
2\beta_{1}\beta_{2}+\beta_{3}^{2}\equiv 1\mod
2^{k}\\}=2(2^{2k}-2^{2(k-1)}-2\cdot 2^{k-2}\cdot 2^{k-1})$

for sufficiently large $k$ and $\alpha_{2}(f,1)=1$.

$\bullet$ Case $\delta\equiv 3\mod 8$. Then $f\cong 2(x^{2}+xy+y^{2})-z^{2}$
by (4.7). Since $z$ and at least one of $x$ and $y$ take odd integers for the
equation $2(x^{2}+xy+y^{2})-z^{2}=1$, one can write $z^{2}=1+8t$.

Suppose both $x$ and $y$ are odd. Write $x=1+2x_{1}$ and $y=1+2y_{1}$. Then

$\frac{1}{2}(3x_{1}+3y_{1}+1)+x_{1}^{2}+x_{1}y_{1}+y_{1}^{2}=t$ (4.8)

and $x_{1}$ and $y_{1}$ have a different parity. Moreover, for any even (resp.
odd) $x_{1}$, there is a unique odd (resp. even) $y_{1}\in\mathbb{Z}_{2}$
satisfying (4.8) by Hensel’s Lemma.

Suppose $x$ is odd and $y$ is even. Write $x=1+2x_{1}$ and $y=2y_{1}$. Then

$2x_{1}(x_{1}+1)+y_{1}(1+2x_{1})+2y_{1}^{2}=2t.$ (4.9)

For any $x_{1}\in\mathbb{Z}_{2}$, there is a unique $y_{1}\in\mathbb{Z}_{2}$
satisfying (4.9) by Hensel’s Lemma.

Summarizing the above arguments with symmetry of $x$ and $y$, one concludes
that

$\\#\\{(\beta_{1},\beta_{2},\beta_{3})\in(\mathbb{Z}/(2^{k}))^{3}:\
2(\beta_{1}^{2}+\beta_{1}\beta_{2}+\beta_{2}^{2})-\beta_{3}^{2}\equiv 1\mod
2^{k}\\}=2^{k-1}(2\cdot 2^{k-2}+2\cdot 2^{k-1})$

for sufficiently large $k$ and $\alpha_{2}(f,1)=3\cdot 2^{-2}$.

$\bullet$ Case $\delta\equiv 5\mod 8$. Then $f\cong 2xy+5z^{2}$ by (4.7). By
[OM, 63:1 Local Square Theorem], one concludes

$\\#\\{(\beta_{1},\beta_{2},\beta_{3})\in(\mathbb{Z}/(2^{k}))^{3}:\
2\beta_{1}\beta_{2}+5\beta_{3}^{2}\equiv 1\mod 2^{k}\\}=2\cdot(2\cdot
2^{k-1}\cdot 2^{k-2})$

for sufficiently large $k$ and $\alpha_{2}(f,1)=2^{-1}$.

$\bullet$ Case $\delta\equiv 7\mod 8$. Then $f\cong 2(x^{2}+xy+y^{2})-5z^{2}$
by (4.7). By [OM, 63:1 Local Square Theorem], the equation
$2(x^{2}+xy+y^{2})-5z^{2}=1$ is solvable over $\mathbb{Z}_{2}$ if and only if

$(x^{2}+xy+y^{2}-1)\in 2\mathbb{Z}_{2}^{\times}.$ (4.10)

Suppose both $x$ and $y$ are odd. Write $x=1+2x_{1}$ and $y=1+2y_{1}$. Then
$(x_{1}+y_{1})\in 2\mathbb{Z}_{2}$ by (4.10). Suppose $x$ is even and write
$x=2x_{1}$. Then both $x_{1}$ and $y$ have to be odd by (4.10). The similar
result is true for even $y$. Therefore

$\displaystyle\\#\\{(\beta_{1},\beta_{2},\beta_{3})\in(\mathbb{Z}/(2^{k}))^{3}:\
2(\beta_{1}^{2}+\beta_{1}\beta_{2}+\beta_{2}^{2})-5\beta_{3}^{2}\equiv 1\mod
2^{k}\\}$ $\displaystyle=2\cdot(2^{2(k-1)}-2\cdot 2^{2(k-2)}+2\cdot
2^{k-2}\cdot 2^{k-1})=3\cdot 2^{2(k-1)}$

for sufficiently large $k$ and $\alpha_{2}(f,1)=3\cdot 2^{-2}$.

When $p=2$ and $2|\delta$ but $4\nmid\delta$, then $f$ is equivalent to

$(\mathbb{Z}_{2}e_{1}+\mathbb{Z}_{2}e_{2})\perp\mathbb{Z}_{2}e_{3}\ \ \
\text{with}\ \ Q(e_{1})=1,B(e_{1},e_{2})=1,Q(e_{2})=2\ \ \text{and}\
Q(e_{3})=\delta$

by [OM, 93:17. Example]. Since $2|\delta$ but $4\nmid\delta$, there is
$\kappa\in\mathbb{Z}_{2}$ such that

$Q(e_{2}+e_{3})=2+\delta=4\kappa.$

Since $\mathbb{Z}_{2}e_{1}+\mathbb{Z}_{2}(e_{2}+e_{3})$ splits
$(\mathbb{Z}_{2}e_{1}+\mathbb{Z}_{2}e_{2})\perp\mathbb{Z}_{2}e_{3}$ by [OM,
82:15], one obtains that

$(\mathbb{Z}_{2}e_{1}+\mathbb{Z}_{2}e_{2})\perp\mathbb{Z}_{2}e_{3}=(\mathbb{Z}_{2}e_{1}+\mathbb{Z}_{2}(e_{2}+e_{3}))\perp\mathbb{Z}_{2}e_{3}^{\prime}$

with $Q(e_{3}^{\prime})=(1-4\kappa)^{-1}\delta$ by comparing the discriminants
of lattices. One concludes that

$f\cong x^{2}+2xy+4\kappa y^{2}+(1-4\kappa)^{-1}\delta z^{2}$

over $\mathbb{Z}_{2}$. Therefore

$\\#\\{(\beta_{1},\beta_{2},\beta_{3})\in(\mathbb{Z}/(2^{k}))^{3}:\
\beta_{1}^{2}+2\beta_{1}\beta_{2}+4\kappa\beta_{2}^{2}+(1-4\kappa)^{-1}\delta\beta_{3}^{2}\equiv
1\mod 2^{k}\\}=2^{k-1}\cdot 2^{k}$

for sufficiently large $k$ and $\alpha_{2}(f,1)=2^{-1}$.

When $p=2$ and $4\mid\delta$, then $f\cong x^{2}+2xy+2y^{2}-\delta z^{2}$ over
$\mathbb{Z}_{2}$ by [OM, 93:17. Example]. Therefore

$\\#\\{(\beta_{1},\beta_{2},\beta_{3})\in(\mathbb{Z}/(2^{k}))^{3}:\
\beta_{1}^{2}+2\beta_{1}\beta_{2}+2\beta_{2}^{2}-\delta\beta_{3}^{2}\equiv
1\mod 2^{k}\\}=2\cdot 2^{k-1}\cdot 2^{k}$

for sufficiently large $k$ and $\alpha_{2}(f,1)=1$. ∎

The singular integral of $\mathcal{X}$ can be computed as follows.

######  Lemma 4.11.

If $\delta>0$ and

$B_{T}=\\{(x,y,z)\in\mathbb{R}^{3}:\ x^{2}+y^{2}-\delta z^{2}=1\ \text{with}\
\sqrt{x^{2}+y^{2}+z^{2}}\leq T\\}$

for $T\geq 1$, then

$\int_{B_{T}}\frac{1}{2\delta z}dxdy=2\pi\sqrt{\frac{T^{2}-1}{1+\delta}}.$

###### Proof.

Since $z^{2}=\delta^{-1}(x^{2}+y^{2}-1)\geq 0$, one obtains that

$\int_{B_{T}}\frac{1}{2\delta z}dxdy=\int_{D_{T}}\frac{1}{\delta z}dxdy$

where $D_{T}$ can be described as

$\\{(x,y)\in\mathbb{R}^{2}:\
\sqrt{\frac{\delta}{1+\delta}}\leq\sqrt{x^{2}+y^{2}-(\delta+1)^{-1}}\leq\sqrt{\frac{\delta}{1+\delta}}\cdot
T\\}.$

By using the polar coordinates $x=\rho\cos\theta$ and $y=\rho\sin\theta$, one
obtains

$\int_{B_{T}}\frac{1}{2\delta
z}dxdy=\frac{2\pi}{\sqrt{\delta}}\int_{1}^{\sqrt{\frac{\delta
T^{2}+1}{1+\delta}}}\frac{\rho}{\sqrt{\rho^{2}-1}}d\rho=2\pi\sqrt{\frac{T^{2}-1}{1+\delta}}$

as desired. ∎

Write $\delta=\delta_{0}\delta_{1}^{2}$ where both $\delta_{0}$ and
$\delta_{1}$ are positive integers and $\delta_{0}$ is square free. Let

$S=\\{p\ \text{primes}:\ p\mid\delta_{1}\ \text{and}\ (p,2\delta_{0})=1\\}.$

######  Example 4.12.

Suppose that $f(x,y,z)=x^{2}+y^{2}-\delta z^{2}$ and

$N(f,1,T)=\\#\\{(\alpha,\beta,\gamma)\in\mathbb{Z}^{3}:\
f(\alpha,\beta,\gamma)=1\ \text{with}\
\sqrt{\alpha^{2}+\beta^{2}+\gamma^{2}}\leq T\\}$

for $T>0$.

If $\delta_{0}>1$, then

$N(f,1,T)\sim\frac{c}{\pi\sqrt{1+\delta}}\cdot L(1,\chi)\cdot\frac{\prod_{p\in
S}(1-(\frac{-1}{p})p^{-1})(1+(\frac{\delta_{0}}{p})p^{-1})^{-1}}{\prod_{p\mid\delta_{0},\
p\neq 2}(1+(\frac{-1}{p})p^{-1})}\cdot T$

as $T\to\infty$ where

$c=\begin{cases}8\ \ \ &\text{$\delta_{0}\equiv 1\mod 8$}\\\ 12\ \ \
&\text{$\delta_{0}\equiv 5\mod 8$ and $2\nmid\delta$}\\\ 24\ \ \
&\text{$\delta_{0}\equiv 5\mod 8$ and $2\mid\delta$}\\\ 12\ \ \
&\text{$\delta_{0}\equiv 3,7\mod 8$ and $2\nmid\delta$}\\\ 16\ \ \
&\text{$\delta_{0}\equiv 3,7\mod 8$ and $2\mid\delta$}\\\ 8\ \ \
&\text{$2\mid\delta_{0}$ and $2\nmid\delta_{1}$}\\\ 16\ \ \
&\text{$2\mid\delta_{0}$ and $2\mid\delta_{1}$}\end{cases}$

and $L(\chi,s)$ is the Dirichlet $L$-function associated to the non-trivial
quadratic character $\chi$ of ${\rm
Gal}(\mathbb{Q}(\sqrt{\delta_{0}})/\mathbb{Q})$.

If $\delta_{0}=1$, then

$N(f,1,T)\sim\frac{8}{\pi\sqrt{1+\delta}}\cdot(\prod_{p\mid\delta,\ p\neq
2}\frac{p-(\frac{-1}{p})}{p+1})\cdot T\log T$

as $T\to\infty$.

###### Proof.

By [OM, 102:7 and 104:5 Theorem], the number $h(f)$ of classes in $gen(f)$ is
given by the formula

$h(f)=[\mathbb{G}_{m}({\bf
A}_{\mathbb{Q}}):\theta(SO(f_{\mathbb{Q}}))\cdot(\theta(SO(f_{\mathbb{R}})\times\prod_{p\
\text{primes}}SO(f_{\mathbb{Z}_{p}})))]$

where $\theta$ is the spinor norm map. Since

$\theta(SO(f_{\mathbb{Q}}))=\mathbb{Q}^{\times},\ \ \
\theta(SO(f_{\mathbb{R}}))=\mathbb{R}^{\times}\ \ \ \text{and}\ \ \
\theta(SO(f_{\mathbb{Z}_{p}}))\supseteq\mathbb{Z}_{p}^{\times}$

for odd primes $p$ by [OM, 101:8, 55:2a and 92:5], then
$-1\in\theta(SO(f_{\mathbb{Q}}))$ implies that

$(i_{p})_{p\leq\infty_{\mathbb{Q}}}\in\theta(SO(f_{\mathbb{Q}}))\cdot(\theta(SO(f_{\mathbb{R}})\times\prod_{p\
\text{primes}}SO(f_{\mathbb{Z}_{p}})))$

where $i_{p}=-1$ for $p=2$ and $i_{p}=1$ otherwise. Since $5$ is represented
by $f$ over $\mathbb{Z}_{2}$, one has $5\in\theta(SO(f_{\mathbb{Z}_{2}}))$.
Therefore

$\theta(SO(f_{\mathbb{Q}}))\cdot(\theta(SO(f_{\mathbb{R}})\times\prod_{p\
\text{primes}}SO(f_{\mathbb{Z}_{p}})))\supseteq\mathbb{Q}^{\times}\cdot(\mathbb{R}^{\times}\times(\prod_{p\
\text{primes}}\mathbb{Z}_{p}^{\times}))$

and $h(f)=1$.

If $\delta_{0}>1$, then

$N(f,1,T)\sim(\prod_{p\
\text{primes}}\alpha_{p}(f,1))\cdot\int_{B_{T}}\frac{1}{2\delta z}dxdy$

as $T\to\infty$ by [CX, §5] and Lemma 4.2, where $\alpha_{p}(f,1)$ and $B_{T}$
are defined in Lemma 4.5 and Lemma 4.11 respectively.

$\bullet$ Case $\delta_{0}\equiv 1\mod 8$. Then $2$ splits completely in
$\mathbb{Q}(\sqrt{\delta_{0}})$ and

$L(\chi,1)=2\cdot\prod_{p\nmid
2\delta_{0}}(1-(\frac{\delta_{0}}{p})p^{-1})^{-1}.$

Therefore

$\displaystyle N(f,1,T)\sim\prod_{p\mid\delta,\ p\neq
2}(1-(\frac{-1}{p})p^{-1})\cdot\prod_{p\nmid
2\delta}(1+(\frac{\delta_{0}}{p})p^{-1})\cdot\frac{2\pi}{\sqrt{1+\delta}}T$
$\displaystyle=$
$\displaystyle\frac{2\pi}{\sqrt{1+\delta}}\cdot\frac{\prod_{p\mid\delta,\
p\neq 2}(1-(\frac{-1}{p})p^{-1})}{\prod_{p\in
S}(1+(\frac{\delta_{0}}{p})p^{-1})}\cdot\prod_{p\nmid
2\delta_{0}}\frac{1-p^{-2}}{1-(\frac{\delta_{0}}{p})p^{-1}}\cdot T$
$\displaystyle=$
$\displaystyle\frac{2\pi}{\sqrt{1+\delta}}\cdot\frac{\prod_{p\mid\delta,\
p\neq 2}(1-(\frac{-1}{p})p^{-1})}{\prod_{p\in
S}(1+(\frac{\delta_{0}}{p})p^{-1})}\cdot
2^{-1}L(1,\chi)\cdot\frac{4}{3}\prod_{p\mid\delta_{0}}(1-p^{-2})^{-1}\cdot\zeta(2)^{-1}\cdot
T$ $\displaystyle=$ $\displaystyle\frac{8}{\pi\sqrt{1+\delta}}\cdot
L(1,\chi)\cdot\frac{\prod_{p\mid\delta,\ p\neq
2}(1-(\frac{-1}{p})p^{-1})}{\prod_{p\in
S}(1+(\frac{\delta_{0}}{p})p^{-1})}\cdot\prod_{p\mid\delta_{0}}(1-p^{-2})^{-1}\cdot
T$ $\displaystyle=$ $\displaystyle\frac{8}{\pi\sqrt{1+\delta}}\cdot
L(1,\chi)\cdot\frac{\prod_{p\in
S}(1-(\frac{-1}{p})p^{-1})(1+(\frac{\delta_{0}}{p})p^{-1})^{-1}}{\prod_{p\mid\delta_{0}}(1+(\frac{-1}{p})p^{-1})}\cdot
T$

as $T\to\infty$ by Lemma 4.5 and Lemma 4.11 and $\zeta(2)=\frac{\pi^{2}}{6}$.

$\bullet$ Case $\delta_{0}\equiv 5\mod 8$. Then $2$ is inert in
$\mathbb{Q}(\sqrt{\delta_{0}})$ and

$L(\chi,1)=\frac{2}{3}\cdot\prod_{p\nmid
2\delta_{0}}(1-(\frac{\delta_{0}}{p})p^{-1})^{-1}.$

Therefore

$\displaystyle
N(f,1,T)\sim\begin{cases}\frac{1}{2}\cdot\prod_{p\mid\delta}(1-(\frac{-1}{p})p^{-1})\cdot\prod_{p\nmid
2\delta}(1+(\frac{\delta_{0}}{p})p^{-1})\cdot\frac{2\pi}{\sqrt{1+\delta}}T\ \
\ &2\nmid\delta\\\ \prod_{p\mid\delta,\ p\neq
2}(1-(\frac{-1}{p})p^{-1})\cdot\prod_{p\nmid\delta}(1+(\frac{\delta_{0}}{p})p^{-1})\cdot\frac{2\pi}{\sqrt{1+\delta}}T\
\ \ &2\mid\delta\end{cases}$

as $T\to\infty$ by Lemma 4.5 and Lemma 4.11 and the result follows.

$\bullet$ Case $\delta_{0}\equiv 3,7\mod 8$. Then $2$ is ramified in
$\mathbb{Q}(\sqrt{\delta_{0}})$ and

$L(\chi,1)=\prod_{p\nmid 2\delta_{0}}(1-(\frac{\delta_{0}}{p})p^{-1})^{-1}.$

Therefore

$\displaystyle
N(f,1,T)\sim\begin{cases}\frac{3}{4}\cdot\prod_{p\mid\delta}(1-(\frac{-1}{p})p^{-1})\cdot\prod_{p\nmid
2\delta}(1+(\frac{\delta_{0}}{p})p^{-1})\cdot\frac{2\pi}{\sqrt{1+\delta}}T\ \
\ &2\nmid\delta\\\ \prod_{p\mid\delta,\ p\neq
2}(1-(\frac{-1}{p})p^{-1})\cdot\prod_{p\nmid\delta}(1+(\frac{\delta_{0}}{p})p^{-1})\cdot\frac{2\pi}{\sqrt{1+\delta}}T\
\ \ &2\mid\delta\end{cases}$

as $T\to\infty$ by Lemma 4.5 and Lemma 4.11 and the result follows.

$\bullet$ Case $2\mid\delta_{0}$. Then $2$ is ramified in
$\mathbb{Q}(\sqrt{\delta_{0}})$ and

$L(\chi,1)=\prod_{p\nmid\delta_{0}}(1-(\frac{\delta_{0}}{p})p^{-1})^{-1}.$

Therefore

$\displaystyle N(f,1,T)\sim\begin{cases}\frac{1}{2}\cdot\prod_{p\mid\delta,\
p\neq
2}(1-(\frac{-1}{p})p^{-1})\cdot\prod_{p\nmid\delta}(1+(\frac{\delta_{0}}{p})p^{-1})\cdot\frac{2\pi}{\sqrt{1+\delta}}T\
\ \ &2\nmid\delta_{1}\\\ \prod_{p\mid\delta,\ p\neq
2}(1-(\frac{-1}{p})p^{-1})\cdot\prod_{p\nmid\delta}(1+(\frac{\delta_{0}}{p})p^{-1})\cdot\frac{2\pi}{\sqrt{1+\delta}}T\
\ \ &2\mid\delta_{1}\end{cases}$

as $T\to\infty$ by Lemma 4.5 and Lemma 4.11 and the result follows.

Otherwise $\delta_{0}=1$. By Theorem 3.6, Lemma 4.2, Lemma 4.5 and Lemma 4.11,
one obtains

$\displaystyle N(f,1,T)\sim(\prod_{p\
\text{primes}}(1-p^{-1})\alpha_{p}(f,1))\cdot\log
T\int_{B_{T}}\frac{1}{2\delta z}dxdy$ $\displaystyle=$
$\displaystyle\frac{1}{2}\prod_{p\mid\delta,\ p\neq
2}(1-p^{-1})(1-(\frac{-1}{p})p^{-1})\cdot\prod_{p\nmid\delta,\ p\neq
2}(1-p^{-2})\cdot\frac{2\pi}{\sqrt{1+\delta}}T\log T$ $\displaystyle=$
$\displaystyle\frac{1}{2}\cdot\frac{\prod_{p\mid\delta,\ p\neq
2}(1-p^{-1})(1-(\frac{-1}{p})p^{-1})}{\prod_{p\mid\delta,\ p\neq
2}(1-p^{-2})}\cdot\frac{4}{3}\zeta(2)^{-1}\cdot\frac{2\pi}{\sqrt{1+\delta}}T\log
T$ $\displaystyle=$
$\displaystyle\frac{8}{\pi\sqrt{1+\delta}}\cdot(\prod_{p\mid\delta,\ p\neq
2}\frac{1-(\frac{-1}{p})p^{-1}}{1+p^{-1}})\cdot T\log T$

as $T\to\infty$. ∎

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