# The Distribution of Rational Points on Conics

D.R. Heath-Brown  Mathematical Institute
Radcliffe Observatory Quarter
Woodstock Road
Oxford
OX2 6GG<EMAIL_ADDRESS>
###### Abstract.

We examine the counting function for rational points on conics, and show how
the point where the asymptotic behaviour begins depends on the size of the
smallest zero.

###### 2010 Mathematics Subject Classification:

11D45 (11D09 11E08 14G05)

##  1\. Introduction

Let $q(x_{1},x_{2},x_{3})=q(\mathbf{x})\in\mathbb{Z}[x_{1},x_{2},x_{3}]$ be a
quadratic form, and write

$\mathbb{Z}_{\mathrm{prim}}^{n}=\\{(x_{1},\ldots,x_{n})\in\mathbb{Z}^{n}-\\{\mathbf{0}\\}:\,\mathrm{g.c.d.}(x_{1},\ldots,x_{n})=1\\}.$

The purpose of this paper is to investigate the behaviour as
$B\rightarrow\infty$ of the counting function

$N(B)=N(B;q)=\\#\\{\mathbf{x}\in\mathbb{Z}_{\mathrm{prim}}^{3}:\,q(\mathbf{x})=0,\,\max\\{|x_{1}|,|x_{2}|,|x_{3}|\\}\leq
B\\},$

and of its weighted form

$N(B,w)=N(B,w;q)=\sum_{\begin{subarray}{c}\mathbf{x}\in\mathbb{Z}_{\mathrm{prim}}^{3}\\\
q(\mathbf{x})=0\end{subarray}}w(B^{-1}\mathbf{x}).$

Here we take $w:\mathbb{R}^{3}\to\mathbb{R}$ to be infinitely differentiable,
with compact support. With this notation, the conic $q=0$ has
$\tfrac{1}{2}N(B)$ rational points of height at most $B$.

Provided $q$ is isotropic over $\mathbb{Q}$ (in other words, if
$q(\mathbf{x})=0$ has at least one non-zero integral solution), one has

$N(B)\sim\tfrac{1}{2}\sigma_{\infty}\mathfrak{S}(q)B\;\;\;\mbox{as}\;\;B\to\infty$

where $\sigma_{\infty}>0$ is the real density of solutions, and
$\mathfrak{S}(q)>0$ may be given explicitly in terms of the usual product of
local densities. (The factor $\tfrac{1}{2}$ is the “alpha constant” in Peyre’s
terminology [4]) Indeed one has

$N(B,w)=\tfrac{1}{2}\sigma_{\infty}(q;w)\mathfrak{S}(q)B+O_{q,w}(B\exp\\{-c\sqrt{\log
B}\\}),$ (1)

for some absolute constant $c>0$. These results follow from work of the author
[3, Corollary 2] We stress that the error term in (1) contains an unspecified
dependence on $q$. Our main aim in this paper is to obtain a good explicit
dependence, so as to show how large $B$ has to be, in terms of $q$, before one
sees the true asymptotics for $N(B)$.

In order to see the phenomena that $N(B)$ can display we present a numerical
example. Let $q_{0}$ be the form

$q_{0}(\mathbf{x})=-61x_{1}^{2}-22x_{1}x_{3}-38x_{2}^{2}+99x_{2}x_{3}+39x_{3}^{2}.$
(2)

Then the following graph shows values of $N(B;q_{0})$ for $B\leq 10000$.
10000.

$0$$2$$4$$6$$8$$10$$0$$50$$100$$150$$200$$250$$300$$B$ in 1000’s$N(B;q_{0})$

The graph appears linear from about $B=6000$ onwards, but there is a
surprising kink around $B=3500$. Indeed for $B\leq 2500$ the graph seems
linear, but with a smaller gradient than for the range $B\geq 6000$. It is
this strange behaviour that we aim to explain — see the discussion after
theorem 5.

We begin by introducing some notation and terminology. In general we will want
to allow our form $q$ to have odd cross-terms. We therefore write it in the
asymmetric shape

$q(\mathbf{x})=\sum_{1\leq i\leq j\leq 3}q_{ij}x_{i}x_{j},$

associate with $q$ the matrix

$Q=\left(\begin{array}[]{rrr}2q_{11}&q_{12}&q_{13}\\\ q_{12}&2q_{22}&q_{23}\\\
q_{13}&q_{12}&2q_{33}\end{array}\right).$ (3)

Moreover, we define the determinant, somewhat unconventionally, by

$\Delta=\Delta(q)=\tfrac{1}{2}\det(Q).$

Thus $\Delta\in\mathbb{Z}$ for any integral form, and

$\Delta\big{(}q(M\mathbf{x})\big{)}=\det(M)^{2}\Delta(q)$

for any $3\times 3$ matrix $M$. By changing the sign of $q$ if necessary we
can arrange that $\Delta(q)\geq 0$. We recall that $q$ is said to be primitive
if the coefficients $q_{ij}$ have no common factor. With this notation our
first result is the following.

######  Theorem 1.

Let $q$ be a primitive integral isotropic form with $\Delta>0$. Then there is
a positive integer $K\leq\tau(\Delta)$, and there are nonsingular $3\times 3$
integer matrices $M_{1},\ldots,M_{K}$ having the following properties.

  1. (i)

If $\Delta$ is square-free then $K=\tau(\Delta)$.

  2. (ii)

The determinant $\det(M_{k})$ is a positive divisor of $\Delta$.

  3. (iii)

For any primitive integral solution $\mathbf{x}$ of the equation
$q(\mathbf{x})=0$, there is a unique index $k$ such that $\mathbf{x}\in
M_{k}(\mathbb{Z}^{3})$.

  4. (iv)

For each $k$ there is a corresponding $D_{k}\in\mathbb{N}$ such that

$q(M_{k}\mathbf{x})=D_{k}(x_{1}x_{3}-x_{2}^{2}),$ (4)

identically in $\mathbf{x}$.

  5. (v)

We have

$\Delta\det(M_{k})^{2}=D_{k}^{3},$ (5)

so that $D_{k}\mid\Delta$ and $\det(M_{k})\mid D_{k}$. Moreover
$D_{k}\mid\det(M_{k})^{2}$.

  6. (vi)

A prime $p$ can divide $\Delta\det(M_{k})^{-1}$ only if $v_{p}(\Delta)\geq 4$.

  7. (vii)

If $\Delta$ is cube-free then for every index $k\leq\tau(\Delta)$ the set
$M_{k}(\mathbb{Z}^{3})$ contains a primitive zero of $q$.

Here $\tau(\ldots)$ is the usual divisor function, and $v_{p}(\Delta)$ is the
$p$-adic valuation. In addition to this notation we will find it convenient to
write $J(\mathbf{x})$ for the quadratic form $x_{1}x_{3}-x_{2}^{2}$, so that
$q(M_{k}\mathbf{x})=D_{k}J(\mathbf{x})$.

The theorem shows that we can partition the primitive integer zeros of $q$
into $K$ classes $\mathcal{C}_{1},\ldots,\mathcal{C}_{K}$, corresponding to
the different matrices $M_{k}$. Specifically, we define

$\mathcal{C}_{k}=\\{\mathbf{x}\in\mathbb{Z}_{\mathrm{prim}}^{3}\cap
M_{k}(\mathbb{Z}^{3}):\,q(\mathbf{x})=0\\}.$

Moreover, since the primitive integer zeros of $J$ are given exactly twice
each by $\pm(u_{1}^{2},u_{1}u_{2},u_{2}^{2})$ , the theorem shows that we can
produce the primitive integer solutions of $q(\mathbf{x})=0$ exactly twice
each as

$\mathbf{x}=\pm M_{k}(u_{1}^{2},u_{1}u_{2},u_{2}^{2}).$

Here $(u_{1},u_{2})$ must be primitive if $\mathbf{x}$ is, but unfortunately
it is not true that $\mathbf{x}$ is primitive whenever $u_{1}$ and $u_{2}$ are
coprime.

Part (vi) of the theorem shows that if $\Delta$ has no fourth-power divisors
then we have $\det(M_{k})=D_{k}=\Delta$ for every index $k$. In what follows,
it may help the reader if they first restrict attention to this simplified
case.

Part (vii) of the theorem shows that if $\Delta$ is cube-free then each of the
classes $\mathcal{C}_{k}$ is non-empty. For other values of $\Delta$ we may
discard any values of $k$ for which $\mathcal{C}_{k}$ is empty, without
affecting the claims in the theorem. Thus we will suppose in what follows that
each class $\mathcal{C}_{k}$ is non-empty.

The form $q_{0}$ given by (2) has $\Delta(q_{0})=977861=p_{0}$, say, which is
prime. Hence part (i) of the theorem shows that $K=2$, and parts (v) and (vi)
yield

$\det(M_{1})=D_{1}=\det(M_{2})=D_{2}=p_{0}.$

In fact we may take

$M_{1}=\left(\begin{array}[]{rrr}1&-45&3426\\\ 0&100&3339\\\
-1&-54&3047\end{array}\right)\;\;\mbox{and}\;\;M_{2}=\left(\begin{array}[]{rrr}39&-21&-98\\\
0&-100&-1\\\ 61&122&99\end{array}\right).$ (6)

Indeed
$q_{0}(\mathbf{x})=L_{1}(\mathbf{x})L_{2}(\mathbf{x})-p_{0}L_{3}(\mathbf{x})^{2}$
with

$L_{1}(\mathbf{x})=100x_{1}+99x_{2}+100x_{3},\;\;L_{2}(\mathbf{x})=9778x_{1}+9877x_{2}+9779x_{3},$

and

$L_{3}(\mathbf{x})=x_{1}+x_{2}+x_{3},$

and it turns out that the two classes are

$\mathcal{C}_{1}=\\{\mathbf{x}\in\mathbb{Z}_{\mathrm{prim}}^{3}:\,q(\mathbf{x})=0\;\mbox{and}\;p_{0}\mid
L_{1}(\mathbf{x})\\}$

and

$\mathcal{C}_{2}=\\{\mathbf{x}\in\mathbb{Z}_{\mathrm{prim}}^{3}:\,q(\mathbf{x})=0\;\mbox{and}\;p_{0}\mid
L_{2}(\mathbf{x})\\}.$

In order to use Theorem 1 for quantitative results we will need information on
the size of the entries in $M_{k}$. However for each $k$ there are infinitely
many choices for $M_{k}$, since the automorphism group

$\operatorname{Aut}(J)=\operatorname{Aut}_{\mathbb{Z}}(J)=\\{U\in{\rm
M}_{3}(\mathbb{Z}):J(M\mathbf{x})=J(\mathbf{x})\\}$

is infinite. (Note that $\det(U)=\pm 1$ for every $U\in\operatorname{Aut}(J)$.
Thus $U^{-1}$ will be automatically be integral.) Our next result shows that
we can always make a good choice for $M_{k}$. We will write $||\mathbf{x}||$
for the $L^{2}$-norm of the vector $\mathbf{x}$, and $||q||$ for the
$L^{2}$-norm of the coefficients of the matrix $Q$ of $q$, as given by (3).
Specifically, we have

$||q||=||Q||=\\{4q_{11}^{2}+2q_{12}^{2}+2q_{13}^{2}+4q_{22}^{2}+2q_{23}^{2}+4q_{33}^{2}\\}^{1/2}.$

######  Theorem 2.

In Theorem 1 we may choose $M_{k}$ so that if $M_{k}^{-1}$ has rows
$\mathbf{r}_{1},\mathbf{r}_{2},\mathbf{r}_{3}$, then

$||\mathbf{r}_{1}||\cdot||\mathbf{r}_{3}||\leq
9D_{k}^{-1}||q||\;\;\;\mbox{and}\;\;\;||\mathbf{r}_{2}||^{2}\leq
10D_{k}^{-1}||q||.$ (7)

Moreover if $M_{k}$ has columns $\mathbf{c}_{1},\mathbf{c}_{2},\mathbf{c}_{3}$
we will have $||\mathbf{c}_{1}||\leq||\mathbf{c}_{3}||$,

$||\mathbf{c}_{1}||\cdot||\mathbf{c}_{3}||\leq
90\det(M_{k})^{2}D_{k}^{-2}||q||^{2}\leq 90||q||^{2}$ (8)

and

$||\mathbf{c}_{2}||\leq 9\det(M_{k})D_{k}^{-1}||q||\leq 9||q||.$ (9)

The constants can certainly be improved, but for our purposes it suffices to
know that there is at least one set of numerical values that is valid. From
now on we will assume that the matrices $M_{k}$ are as described in Theorem 2.
For the form (2) we have

$||q_{0}||=\sqrt{39872}=199.679\ldots,$

and one sees that the columns of the matrices (6) amply fulfil the conditions
above.

Since $q(\mathbf{c}_{1})=q(\mathbf{c}_{3})=0$, we find that for any isotropic
form $q$ and any index $k$ there are two linearly independent zeros
$\mathbf{c}_{1},\mathbf{c}_{3}\in M_{k}(\mathbb{Z}^{3})$ with
$||\mathbf{c}_{1}||\cdot||\mathbf{c}_{3}||\leq 90||q||^{2}$; in particular
there is at least one vector in $\mathcal{C}_{k}$ of length at most $10||q||$.
(The reader should note that $\mathbf{c}_{1}$ and $\mathbf{c}_{3}$ need not be
primitive, while $\mathcal{C}_{k}$ is defined as the set of primitive zeros in
$M_{k}(\mathbb{Z}^{3})$.) Thus Theorem 2 recovers (in the case of ternary
forms) the results of Davenport [2, Theorem 1] and Cassels [1], which were
weaker in as much as they referred only to the complete set of zeros of $q$,
rather than individual classes $\mathcal{C}_{k}$.

Our next result, which is rather easy, explains how $\mathbf{c}_{1}$ and
$\mathbf{c}_{3}$ are related to the smallest and second smallest zeros of $q$
in $\mathcal{C}_{k}$. It is phrased in terms of the quantity

$\rho=\rho(q)=\frac{||q||^{3}}{\Delta},$

which we will refer to as the “aspect ratio” of $q$. We will see that
$\rho\geq 2$ in all cases, and we may expect that $\rho(q)\approx 1$ for
“typical” forms $q$. We therefore think of forms with large aspect ratio as
having untypically small determinant. For the form (2) we have
$\rho(q_{0})=8.141\ldots$.

######  Theorem 3.

We have $\rho(q)\geq 2$ for any $q$.

Let $\mathbf{z}_{1}$ be an element of $\mathcal{C}_{k}$ of minimal length, and
let $\mathbf{z}_{2}$ be an element of $\mathcal{C}_{k}$ of minimal length
subject to the condition that $\mathbf{z}_{2}\not=\pm\mathbf{z}_{1}$. Then

$||\mathbf{z}_{1}||\cdot||\mathbf{z}_{2}||\geq D_{k}/||q||.$ (10)

If

$||\mathbf{c}_{1}||<\rho^{-1/2}\det(M_{k})D_{k}^{-1}||q||,$

then $\mathbf{c}_{1}$ must be a scalar multiple of the shortest vector
$\mathbf{z}_{1}$. In general

$||\mathbf{c}_{1}||\leq
90\rho||\mathbf{z}_{1}||\;\;\;\mbox{and}\;\;\;||\mathbf{c}_{3}||\leq
90\rho||\mathbf{z}_{2}||.$

When $\rho\ll 1$, as we usually expect, we may interpret Theorem 3 as saying
that the lengths of $\mathbf{c}_{1}$ and $\mathbf{c}_{3}$ are within a
constant factor of the shortest possible lengths, namely $||\mathbf{z}_{1}||$
and $||\mathbf{z}_{2}||$. Moreover, suppose we write $c$ for the constant
$c=(91\rho)^{-1/2}$ and

$\ell=\sqrt{90}\det(M_{k})D_{k}^{-1}||q||$

for the maximum length for $\mathbf{c}_{1}$ permitted by (8). Then whenever
$\mathbf{c}_{1}$ has length at most $c\ell$, the vector $\mathbf{c}_{1}$ must
actually be the minimal zero $\mathbf{z}_{1}$, or a scalar multiple of it. For
the form $q_{0}$, the first column is a minimal zero, so that any other zero
in $\mathcal{C}_{1}$ must have length at least
$2^{-1/2}p_{0}/||q_{0}||=3462.805\ldots$ The third column of $M_{1}$ gives a
zero of length $5671.913\ldots$. In contrast, the first and third columns of
$M_{2}$ give relatively small zeros in $\mathcal{C}_{2}$.

In fact Theorem 2 follows from the following more general result.

######  Theorem 4.

Suppose $A$ is a $3\times 3$ integer matrix, and that
$J(A\mathbf{x})=q(\mathbf{x})$. Then there is a matrix
$U\in\operatorname{Aut}(J)$ such that the rows
$\mathbf{a},\mathbf{b},\mathbf{c}$ of $UA$ satisfy

$||\mathbf{a}||\cdot||\mathbf{c}||\leq
9||q||\;\;\;\mbox{and}\;\;\;||\mathbf{b}||^{2}\leq 10||q||.$

We may use the previous theorems to count primitive zeros of $q(\mathbf{x})$.
Our eventual aim is to give a sharp explicit version of the asymptotic formula
(1). We begin by estimating the number of zeros in each of the classes
$\mathcal{C}_{k}$ with height at most $B$, using the counting function

$N(B;\mathcal{C}_{k})=\sum_{\mathbf{x}\in\mathcal{C}_{k}}w(B^{-1}\mathbf{x}).$

Previously we had said that $:\mathbb{R}^{3}\to\mathbb{R}$ should be
infinitely differentiable, with compact support. We shall now be more specific
and require that $w(\mathbf{x})=0$ whenever $||\mathbf{x}||>1$. Since it is
possible that $w(\mathbf{x})$ might vanish on the zero locus of $q$ we
introduce a second weight function $w_{0}(x)$ defined as

$w_{0}(\mathbf{x})=\left\\{\begin{array}[]{cc}\exp\left\\{-\frac{1}{1-||\mathbf{x}|^{2}}\right\\},&||\mathbf{x}||<1,\\\
0,&\mbox{otherwise.}\end{array}\right.$ (11)

This has the properties required for $w$ itself, and its support includes non-
trivial points of the conic $q(\mathbf{x})=0$. We now define the real density
of points on the conic $q=0$,relative to the weight $w$ by setting

$\sigma_{\infty}(q;w)=\lim_{T\to\infty}\int_{\mathbb{R}^{3}}w(\mathbf{y})K_{T}(q(\mathbf{y}))dy_{1}dy_{2}dy_{3},$
(12)

with

$K_{T}(t)=T\max\\{1-T|t|\,,\,0\\}.$ (13)

This coincides with the constant occuring in (1), see Theorem 3 of [3] which
has a mild variant of (12). We shall show in Lemma 10 that the above limit
does indeed exist. The following asymptotic formula for $N(B;\mathcal{C}_{k})$
then holds.

######  Theorem 5.

For each class $\mathcal{C}_{k}$ there is a square-free divisor
$\Delta_{1}\Delta_{2}$ of $\Delta$ such that $p\mid\Delta_{1}$ for every prime
for which $p||\Delta$, and such that

$\displaystyle N(B;\mathcal{C}_{k})$ $\displaystyle=$
$\displaystyle\frac{\Delta^{1/2}}{2D_{k}^{1/2}}\sigma_{\infty}(q,w)\kappa
B\left\\{1+O_{w}\left(\psi(B)\left(\frac{||\mathbf{z}_{2}||}{B}\right)^{1/4}\right)\right\\}$
$\displaystyle\hskip 28.45274pt{}+O_{w}(1),$

with

$\kappa=\frac{6}{\pi^{2}}\prod_{p\mid\Delta_{1}}\frac{1}{1+p^{-1}}\prod_{p\mid\Delta_{2}}\frac{1-p^{-1}}{1+p^{-1}},$

and

$\psi(B)=4^{\omega(\Delta)}\rho\,\frac{\sigma_{\infty}(q,w_{0})}{\sigma_{\infty}(q,w)}\log
B.$

Here $\mathbf{z}_{2}$ is the second smallest element of $\mathcal{C}_{k}$, as
described precisely in Theorem 3.

A number of comments should be made here. Firstly, in interpreting the theorem
one should think of the factor $\psi(B)$ as being roughly of order 1, or more
generally as not being too large. We will see in Lemma 11 that
$\sigma_{\infty}(q,w)\ll\sigma_{\infty}(q,w_{0})$ when $\sup|w|\leq 1$.
However we have no estimate in the reverse direction since it is possible that
the zero locus of $q$ only just enters the support of $w$, making
$\sigma_{\infty}(q,w)$ small. None the less it is reasonable to think that
$\sigma_{\infty}(q,w_{0})\ll\sigma_{\infty}(q,w)$ in most cases of interest.

Viewing $\psi(B)$ as being small we may interpret the theorem as giving a
linear asymptotic formula for $N(B;\mathcal{C}_{k})$ which takes effect when
$B$ is not much larger than $||\mathbf{z}_{2}||$. Indeed one can easily show
that the error term $O_{w}(1)$ is insignificant when
$B\geq\rho\,||\mathbf{z}_{2}||$. Of course when $B<||\mathbf{z}_{2}||$ the
function $N(B;\mathcal{C}_{k})$ counts at most the zeros $\pm\mathbf{z}_{1}$.
Thus $N(B;\mathcal{C}_{k})$ is $O_{w}(1)$ from $B=1$ to
$B=||\mathbf{z}_{2}||$, and then begins to display its typical linear growth.

When $\Delta$ is square-free we have $D_{k}=\Delta$ for every $k$, by parts
(v) and (vi) of Theorem 1. Moreover we will have $\Delta_{1}=\Delta$ and
$\Delta_{2}=1$, so that

$\kappa=\frac{6}{\pi^{2}}\prod_{p\mid\Delta}\frac{1}{1+p^{-1}},$

for each index $k$. Thus when $\Delta$ is square-free the leading constant in
Theorem 5 is the same for each value of $k$, but the point at which linear
growth begins is potentially different.

We are now in a position to explain the observed kink in our graph of $N(B)$
for the quadratic $q_{0}$. The correspondence between $N(B)$ and the counting
functions $N(B;\mathcal{C}_{k})$ is not precise since the former is defined
using the condition $||\mathbf{x}||_{\infty}\leq B$ while the latter use
$||\mathbf{x}||=||\mathbf{x}||_{2}$. For the class $\mathcal{C}_{1}$ we may
take $\mathbf{z}_{1}=(1,0,-1)$. The zero of second smallest sup-norm in
$\mathcal{C}_{1}$ is $(3426,3339,3047)$ whence $N(B;\mathcal{C}_{1})=2$ for
$1\leq B<3426$. However as soon as $B$ is somewhat larger than $3500$ we will
have $N(B;\mathcal{C}_{1})\sim cB$, for a certain constant $c>0$. For
$\mathcal{C}_{2}$ the two zeros of smallest sup-norm are $(39,0,61)$ and
$(-98,-1,99)$ (or $(-38,-99,38)$, which has the same sup-norm) so that we will
have $N(B;\mathcal{C}_{2})\sim cB$, as soon as $B$ is somewhat larger than a
few hundred, with the same constant $c$. Thus the initial section of the graph
for $N(B,q_{0})$, up to $B=3500$ or so, reflects the range in which
$N(B;\mathcal{C}_{1})=2$ but $N(B;\mathcal{C}_{2})$ is already growing like
$cB$, and the later values of $B$ are in the range where both
$N(B;\mathcal{C}_{1})$ and $N(B;\mathcal{C}_{2})$ are growing like $cB$.

Some remarks on the shape of $\psi(B)$ are also in order. It would be
interesting to know to what extent the various factors involved could be
reduced, or indeed removed. Although this seems possible to some extent, we
hope that the present form of $\psi(B)$ will be sufficient for applications.

We can produce an asymptotic formula for $N(B,w;q)$ by summing up the formulae
for $N(B;\mathcal{C}_{k})$. Since $||\mathbf{z}_{2}||\leq||\mathbf{c}_{3}||$
and $||\mathbf{z}_{1}||\leq||\mathbf{c}_{1}||$ for each index $k$ the
inequality (8) yields

$||\mathbf{z}_{2}||\leq||\mathbf{c}_{3}||\ll||q||^{2}/||\mathbf{c}_{1}||\leq||q||^{2}/||\mathbf{z}_{1}||.$

Thus Theorem 5 has the following immediate corollary, in light of part (i) of
Theorem 1.

######  Theorem 6.

Let $\mathbf{z}_{0}$ be a non-trivial integer zero of $q$ with
$||\mathbf{z}_{0}||$ minimal. Then

$\displaystyle N(B,w;q)$ $\displaystyle=$
$\displaystyle\tfrac{1}{2}\sigma_{\infty}(q;w)\mathfrak{S}(q)B\left\\{1+O_{w}\left(\psi(B)\left(\frac{||q||^{2}}{||\mathbf{z}_{0}||B}\right)^{1/4}\right)\right\\}$
$\displaystyle\hskip 28.45274pt{}+O_{w}(\tau(\Delta)),$

with $\psi(B)$ as in Theorem 5.

This is the promised improvement of (1), with a good explicit dependence on
$q$. It produces a linear asymptotic growth as soon as $B$ is a little larger
than $||q||^{2}/||\mathbf{z}_{0}||$. Since $||\mathbf{z}_{0}||$ is typically
of order around $||q||$ this is essentially best possible. We should also
comment on the quality of the error term, which has a power saving in $B$. In
(1) there is a saving of order $\exp\\{-c\sqrt{\log B}\\}$, which has its
origins in the error term for the Prime Number Theorem. Thus one could replace
$\sqrt{\log B}$ in the exponent by some slightly larger power of $\log B$, but
one cannot hope to establish (1) with a power saving in $B$ by the methods of
[3].

The reader may compare our work with that of Sofos [5]. The latter gives an
asymptotic formula for an unweighted counting function, and has an error term
which has a better dependence on $B$ (of order $B^{1/2}\log B$) and an
explicit dependence on $q$, though a much weaker one.

In future work we plan to apply Theorem 6 to count rational points on certain
varieties that can be fibred into conics. Indeed such applications provide the
natural motivation for the present paper. In work in preparation (jointly with
Dan Loughran) we look at the counting function for Del Pezzo surfaces of
degree 5, in the case where there is a conic fibration. Another example, which
we plan to examine in due course, is the variety
$V\in\mathbb{P}^{2}\times\mathbb{P}^{2}$ cut out by the equation

$X_{0}Y_{0}^{2}+X_{1}Y_{1}^{2}+X_{2}Y_{2}^{2}=0,$

in which a rational point $P$ represented by a pair of primitive integer
vectors $(\mathbf{x},\mathbf{y})$ has height
$h(P)=||\mathbf{x}||_{\infty}^{2}||\mathbf{y}||_{\infty}$. Both these examples
require the full strength of Theorem 6.

##  2\. Proof of Theorem 1

We begin with a result that will allow us to work with matrices over
$\mathbb{Z}/mZ$, rather than $\mathbb{Z}$.

######  Lemma 1.

Let $M$ be an $n\times n$ integer matrix, with determinant coprime to some
positive integer $r$. Then there is a matrix $M^{\prime}\equiv
M\;(\text{mod}\;r)$ with prime determinant. Moreover, if $\det(M)\equiv
1\;(\text{mod}\;r)$ there is an $M^{\prime\prime}\equiv M\;(\text{mod}\;r)$ in
$\mathrm{SL}_{n}(\mathbb{Z})$.

###### Proof.

We can write $M$ in Smith Normal Form as $M=UDV$ with
$U,V\in\mathrm{SL}_{n}(\mathbb{Z})$ and $D$ diagonal. One then sees that it
suffices to prove the lemma when $M$ is diagonal, which we do by induction on
$n$. The case $n=1$ is immediate, by Dirichlet’s Theorem. If the result is
true for matrices of size $n-1$, and

$M=\mathrm{Diag}(m_{1},\ldots,m_{n})=\left(\begin{array}[]{c|c}M_{0}&0\\\
\hline\cr 0&m_{n}\end{array}\right),$

say, then $\det(M_{0})$ will be coprime to $r$ so that $M_{0}\equiv
M_{0}^{\prime}\;(\text{mod}\;r)$ with $\det(M_{0}^{\prime})$ prime. It follows
that we may write $M_{0}^{\prime}$ in Smith Normal Form as $U_{0}D_{0}V_{0}$,
with $D_{0}=\mathrm{Diag}(1,\ldots,1,p)$ say. Thus $M\equiv
M_{1}\;(\text{mod}\;r)$ with

$M_{1}=\left(\begin{array}[]{c|c}M^{\prime}_{0}&0\\\ \hline\cr
0&m_{n}\end{array}\right)=U_{1}\mathrm{Diag}(1,\ldots,1,p,m_{n})V_{1},$

where

$U_{1}=\left(\begin{array}[]{c|c}U_{0}&0\\\ \hline\cr 0&1\end{array}\right),$

and similarly for $V_{1}$. To complete the induction step it remains to show
that the lemma holds for the matrix $\mathrm{Diag}(p,m_{n})$. However

$\mathrm{Diag}(p,m_{n})\equiv\left(\begin{array}[]{cc}p&sr\\\
r&m_{n}+tr\end{array}\right)\;(\text{mod}\;r)$

and the matrix on the right has determinant $pm_{n}+tpr-sr^{2}$. Since
$pm_{n}$ will be coprime to $r$ we can make this determinant prime by taking
$t=0$ and choosing $s$ suitably. Moreover, if $pm_{n}=1+kr$, we can make the
determinant equal to 1 by choosing $s$ and $t$ so that $sr-tp=k$. This
completes the induction argument. ∎

Our next result describes the reduction of ternary forms modulo a prime $p$
and its powers. We do not assume that $p$ is odd.

######  Lemma 2.

Let $p$ be prime and let $q(\mathbf{x})$ be an integral ternary quadratic
form, not divisible by $p$ but with $p^{e}||\Delta(q)$ for some exponent
$e\geq 1$. Then there is a matrix $M\in\mathrm{SL}_{3}(\mathbb{Z})$ such that
one of the following holds.

  1. (i)

$q(M\mathbf{x})\equiv\kappa x_{3}^{2}\;(\text{mod}\;p)$, with $p\nmid\kappa$;

  2. (ii)

$q(M\mathbf{x})\equiv x_{1}x_{2}+\kappa p^{e}x_{3}^{2}\;(\text{mod}\;p^{e+1})$
for some integer $\kappa$ coprime to $p$;

  3. (iii)

$q(M\mathbf{x})\equiv q_{1}(x_{1},x_{2})+\kappa
p^{e}x_{3}^{2}\;(\text{mod}\;p^{e+1})$ for some integer $\kappa$ coprime to
$p$, with $q_{1}$ irreducible modulo $p$.

###### Proof.

In view of Lemma 1 it suffices to find a suitable $p$-adic matrix
$M\in\mathrm{SL}_{3}(\mathbb{Z}_{p})$. When $p$ is odd we can diagonalize over
$\mathbb{Z}_{p}$ to give $Ax_{1}^{2}+Bx_{2}^{2}+Cx_{3}^{2}$, say. Since $q$
has determinant divisible by $p$, but does not vanish modulo $p$ we see that
either case (i) of the lemma holds, or that we may take $p\nmid AB$ and
$p^{e}||C$. We then have case (ii) if $-AB$ is a quadratic residue of $p$, and
case (iii) otherwise.

For $p=2$ we consider the reduction of $q$ over $\mathbb{F}_{2}$. Since
$2\mid\Delta$ we find that $q(\mathbf{x})$ is equivalent to one of
$x_{3}^{2}$, or $x_{1}x_{2}$, or $x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}$ over
$\mathbb{F}_{2}$, via a matrix in $\mathrm{SL}_{3}(\mathbb{F}_{2})$. (This can
be shown by considering all possible forms $q$ modulo 2, if necessary.) The
first case leads immediately to case (i) of the lemma. In the remaining cases,
Lemma 1 shows that $q$ is equivalent to
$\tilde{q}(x_{1},x_{2})+\ell(x_{1},x_{2})x_{3}+\mu x_{3}^{2}$ over
$\mathbb{Z}_{2}$, where $\ell(x_{1},x_{2})$ is a linear form, and
$\tilde{q}(x_{1},x_{2})\equiv x_{1}x_{2}$ or
$x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}\;(\text{mod}\;2)$. Replacing $x_{1}$ and
$x_{2}$ by $x_{1}-\xi_{1}x_{3}$ and $x_{2}-\xi_{2}x_{3}$ respectively
eliminates the term $\ell(x_{1},x_{2})x_{3}$ provided that

$\ell(x_{1},x_{2})=\xi_{1}\frac{\partial\tilde{q}(x_{1},x_{2})}{\partial
x_{1}}+\xi_{2}\frac{\partial\tilde{q}(x_{1},x_{2})}{\partial x_{2}}.$

Suitable $\xi_{1},\xi_{2}\in\mathbb{Z}_{2}$ can always be found, since the
linear forms $\partial\tilde{q}/\partial x_{1}$ and
$\partial\tilde{q}/\partial x_{2}$ are congruent modulo 2 to $x_{2}$ and
$x_{1}$ respectively. We then conclude that $q$ is equivalent to
$\tilde{q}(x_{1},x_{2})+\mu^{\prime}x_{3}^{2}$ over $\mathbb{Z}_{2}$.
Computing the determinant of this we find that $2^{e}||\mu^{\prime}$. When

$\tilde{q}(x_{1},x_{2})\equiv x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}\;(\text{mod}\;2)$

we obtain case (iii) of the lemma. Finally, if $\tilde{q}(x_{1},x_{2})\equiv
x_{1}x_{2}\;(\text{mod}\;2)$ we see from Hensel’s Lemma that
$\tilde{q}(x_{1},x_{2})$ must factor over $\mathbb{Z}_{2}$, and a further
unimodular change of variables leads to case (ii) of the lemma. ∎

We next have the following lemma, which shows how we remove powers of $p$ from
$\Delta(q)$.

######  Lemma 3.

Suppose that $q(\mathbf{x})$ is an integral isotropic ternary quadratic form,
not necessarily primitive, and that $p^{e}||\Delta(q)\not=0$. Then there is a
positive integer $K\leq e+1$ such that $K=2$ when $e=1$, and there are
$3\times 3$ integer matrices $R_{1},\ldots,R_{K}$ with determinants
$\det(R_{k})=p^{\mu_{k}}$, such that the following properties hold. Firstly,
$\mu_{k}\leq e$ is a non-negative integer with $\mu_{k}\equiv
e\;(\text{mod}\;3)$ for each $k\leq K$. Secondly, the form

$p^{-(e+2\mu_{k})/3}q(R_{k}\mathbf{x})$ (14)

has integer coefficients and has determinant $p^{-e}\Delta(q)$. Thirdly, if
$q(\mathbf{x})$ vanishes for some primitive $\mathbf{x}\in\mathbb{Z}^{3}$,
then there is exactly one index $k\leq K$ for which
$R_{k}^{-1}\mathbf{x}\in\mathbb{Z}^{3}$.

###### Proof.

Clearly the form (14) has determinant $p^{-e}\Delta(q)$. The proof of the
lemma will be by induction on $e$. When $e=0$ we have $K=1$ and $R_{1}$ can be
taken to be the identity. To handle the induction step we assume that the
lemma holds for exponents strictly less than $e$. Suppose firstly that the
form $q$ is identically divisible by $p$, so that $e\geq 3$. Write
$q^{\prime}(\mathbf{x})=p^{-1}q(\mathbf{x})$, whence
$p^{e-3}||\Delta(q^{\prime})$. By the induction assumption we have matrices
$R^{\prime}_{1},\ldots,R^{\prime}_{J}$ with $J\leq(e-3)+1=e-2$, and exponents
$\mu^{\prime}_{k}\leq e-3\leq e$. We now claim that we can take $K=J\leq e+1$
and $R_{k}=R^{\prime}_{k}$ for every index $k$, so that
$\mu_{k}=\mu^{\prime}_{k}$. In the first place we have

$\mu_{k}=\mu^{\prime}_{k}\equiv e-3\equiv e\;(\text{mod}\;3).$

Secondly,

$p^{-(e+2\mu_{k})}q(R_{k}\mathbf{x})=p^{-\\{(e-3)+\mu^{\prime}_{k}\\}/3}q^{\prime}(R^{\prime}_{k}\mathbf{x})$

which is an integral form. Thirdly, if $q(\mathbf{x})=0$ for some primitive
$\mathbf{x}\in\mathbb{Z}^{3}$, then $q^{\prime}(\mathbf{x})=0$, whence there
is exactly one index for which ${R^{\prime}_{k}}^{-1}\mathbf{x}$ is integral.
Thus there is exactly one index for which $R_{k}^{-1}\mathbf{x}$ is integral.

When $q(\mathbf{x})$ is not identically divisible by $p$ we apply Lemma 2, and
consider separately the three possible cases. Suppose firstly that
$q(M\mathbf{x})\equiv\kappa x_{3}^{2}\;(\text{mod}\;p)$, with $p\nmid\kappa$.
In this case we must have $e\geq 2$. Then if $M^{\prime}=\mathrm{Diag}(1,1,p)$
the form $q^{\prime}(\mathbf{x})=p^{-1}q(MM^{\prime}\mathbf{x})$ will be
integral, with determinant $p^{-1}\Delta(q)$. Moreover it is still isotropic,
so that we may apply the induction hypothesis to $q^{\prime}$, with
$p^{e-1}||\Delta(q^{\prime})$. This produces matrices
$R^{\prime}_{1},\ldots,R^{\prime}_{J}$ with $J\leq e$, and exponents
$\mu^{\prime}_{k}\leq e-1$ such that
$\det(R^{\prime}_{k})=p^{\mu^{\prime}_{k}}$. We now claim that we can take
$K=J$ and $R_{k}=MM^{\prime}R^{\prime}_{k}$ in the lemma. This will have
determinant $p^{\mu_{k}}$ with $\mu_{k}=1+\mu^{\prime}_{k}\equiv
1+(e-1)=e\;(\text{mod}\;3)$, as required. Moreover

$\displaystyle p^{-(e+2\mu_{k})/3}q(R_{k}\mathbf{x})$ $\displaystyle=$
$\displaystyle p^{-(e+2\mu_{k})/3}pq^{\prime}(R^{\prime}_{k}\mathbf{x})$
$\displaystyle=$ $\displaystyle
p^{-(\\{e-1\\}+2\mu^{\prime}_{k})/3}q^{\prime}(R^{\prime}_{k}\mathbf{x})$

which is an integral form, by the induction hypothesis. Finally, when
$q(\mathbf{x})=0$ with a primitive $\mathbf{x}\in\mathbb{Z}^{3}$, we set
$\mathbf{y}=M^{-1}\mathbf{x}$, so that

$0=q(M\mathbf{y})\equiv\kappa y_{3}^{2}\;(\text{mod}\;p),$

with $p\nmid\kappa$. Then $p\mid y_{3}$, whence
${M^{\prime}}^{-1}\mathbf{y}\in\mathbb{Z}^{3}$. It follows that the vector
$\mathbf{z}=(MM^{\prime})^{-1}\mathbf{x}$ is integral, and is primitive since
$\mathbf{x}=MM^{\prime}\mathbf{z}$ is primitive. Moreover
$q^{\prime}(\mathbf{z})=p^{-1}q(\mathbf{x})=0$, whence the induction
hypothesis shows that there is exactly one $R^{\prime}_{k}$ for which
${R^{\prime}_{k}}^{-1}\mathbf{z}\in\mathbb{Z}^{3}$. Hence there is exactly one
index $k$ such that $R_{k}^{-1}\mathbf{x}\in\mathbb{Z}^{3}$. This completes
the proof of Lemma 3 when we are in case (i) of Lemma 2.

We turn next to case (ii) of Lemma 2, in which

$q(M\mathbf{x})\equiv x_{1}x_{2}+\kappa p^{e}x_{3}^{2}\;(\text{mod}\;p^{e+1})$

for some integer $\kappa$ coprime to $p$. We claim that we may take $K=e+1$ in
Lemma 3, and

$R_{k}=M\mathrm{Diag}(p^{k-1},p^{e+1-k},1)\;\;\;\mbox{for}\;\;\;1\leq k\leq
K,$

so that $\mu_{k}=e$ for every $k$. With this choice we have

$q(R_{k}\mathbf{x})\equiv p^{e}(x_{1}x_{2}+\kappa
x_{3}^{2})\;(\text{mod}\;p^{e+1}),$

so that $p^{-e}q(R_{k}\mathbf{x})$ is integral. Suppose now that
$q(\mathbf{x})=0$ with $\mathbf{x}$ primitive, and write
$\mathbf{y}=M^{-1}\mathbf{x}$, so that

$0=q(M\mathbf{y})\equiv y_{1}y_{2}+\kappa
p^{e}y_{3}^{2}\;(\text{mod}\;p^{e+1}).$ (15)

It follows that $p^{e}\mid y_{1}y_{2}$, whence there is a positive integer
$k\leq e+1$ such that $p^{k-1}\mid y_{1}$ and $p^{e+1-k}\mid y_{2}$. We then
see that $\mathbf{y}$ lies in the image of
$\mathrm{Diag}(p^{k-1},p^{e+1-k},1)$, so that $\mathbf{x}\in
R_{k}(\mathbb{Z}^{3})$. Finally if we also have $\mathbf{x}\in
R_{j}(\mathbb{Z}^{3})$ for some $j>k$ then $p^{j-1}\mid y_{1}$, whence
$p^{k}\mid y_{1}$. Since $p^{e+1-k}\mid y_{2}$ it would follow firstly that
$p^{e+1}\mid y_{1}y_{2}$, and secondly that $p\mid y_{2}$, since
$e+1-k>e+1-j\geq 0$. However when $p^{e+1}\mid y_{1}y_{2}$ the congruence (15)
shows that $p\mid y_{3}$. We then reach a contradiction, since $p$ cannot
divide $\mathbf{y}$ when $\mathbf{x}$ is primitive. This completes the proof
of Lemma 3 when we are in case (ii) of Lemma 2.

Finally we examine case (iii) of Lemma 2, in which

$q(M\mathbf{x})\equiv q_{1}(x_{1},x_{2})+\kappa
p^{e}x_{3}^{2}\;(\text{mod}\;p^{e+1})$

for some integer $\kappa$ coprime to $p$, with $q_{1}$ irreducible modulo $p$.
One sees that if $q$ is isotropic we must have $e\geq 2$. The argument is now
similar to that for case (i). Let $M^{\prime}=\mathrm{Diag}(p,p,1)$. Then the
form $p^{-2}q(MM^{\prime}\mathbf{x})$ will be integral, with determinant
$p^{-2}\Delta(q)$. This will have corresponding matrices $R^{\prime}_{k}$ with
determinant $p^{\mu^{\prime}_{k}}$, and we may take
$R_{k}=MM^{\prime}R^{\prime}_{k}$ with corresponding value
$\mu_{k}=\mu^{\prime}_{k}+2$. We leave the reader to verify that these fulfil
the conditions for Lemma 3. This completes the argument. ∎

We are now ready to prove Theorem 1

###### Proof.

We will use induction on the number of distinct prime divisors of $\Delta$. We
therefore begin by considering the case in which $\Delta=1$. Here we will have
$K=1$, and we claim that we may take $D_{1}=1$. Since $q(\mathbf{x})$ is
isotropic there is a primitive integer vector $\mathbf{z}$ such that
$q(\mathbf{z})=0$. We may then construct a unimodular integer matrix $M$ with
first column $\mathbf{z}$. This produces a form $q(M\mathbf{x})$ equivalent to
$q$ and taking the shape $x_{1}(ax_{2}+bx_{3})+q_{1}(x_{2},x_{3})$. The
coefficients $a$ and $b$ must be coprime, since $\Delta=1$. A further
unimodular transformation involving $x_{2}$ and $x_{3}$ produces
$x_{1}x_{3}+q_{2}(x_{2},x_{3})$, say. Now we replace $x_{1}$ by
$x_{1}+Ax_{2}+Bx_{3}$ for suitable integers $A,B$ to obtain a form
$x_{1}x_{3}+\lambda x_{2}^{2}$. Since the determinant is still $\Delta(q)=1$
we see that $\lambda=-1$. Thus $q$ is transformed into $x_{1}x_{3}-x_{2}^{2}$
by a unimodular integer matrix, as required.

Now suppose that $p^{e}||\Delta$. Lemma 3 produces matrices
$R_{1},\ldots,R_{K}$ with corresponding exponents $\mu_{k}$, such that the
forms

$q_{k}(\mathbf{x}):=p^{-(e+2\mu_{k})/3}q(R_{k}\mathbf{x})$

have determinant $p^{-e}\Delta$. Our induction hypothesis, applied to $q_{k}$,
now produces further matrices $M_{1,k},\ldots,M_{J,k}$ with
$J=J(k)\leq\tau(p^{-e}\Delta)$. Since the index $k$ runs up to $e+1$ at most,
there are at most

$(e+1)\tau(p^{-e}\Delta)=\tau(\Delta)$

matrices in total. Moreover if $e=1$ the index $k$ takes exactly the two
values 1 and 2. We now claim that the matrices $R_{k}M_{j,k}$ have the
required properties. Firstly, $K$ is increased by a factor 2 for each prime
factor $p||\Delta$, so that $K=\tau(\Delta)$ if $\Delta$ is square-free.
Secondly, $\det(M_{j,k})$ divides $p^{-e}\Delta$ by the induction hypothesis,
and since $\det(R_{k})\mid p^{e}$ it follows that
$\det(R_{k}M_{j,k})\mid\Delta$, as required. Thirdly, we observe that

$q(R_{k}M_{j,k}\mathbf{x})=p^{(e+2\mu_{k})/3}q_{k}(M_{j,k}\mathbf{x})=D_{j,k}J(\mathbf{x})$

for a suitable integer $D_{j,k}$.

For part (iv), let $q(\mathbf{x})=0$ for some primitive
$\mathbf{x}\in\mathbb{Z}^{3}$. Then, according to Lemma 3, there is an index
$k$ for which $R_{k}^{-1}\mathbf{x}\in\mathbb{Z}^{3}$. Moreover, if we write
$\mathbf{y}=R_{k}^{-1}\mathbf{x}$ then $\mathbf{y}$ must be a primitive
integer vector, and $q_{k}(\mathbf{y})=0$. Then, by the induction assumption
there is a choice of $j$ such that $M_{j,k}^{-1}\mathbf{y}\in\mathbb{Z}^{3}$.
Thus $(R_{k}M_{j,k})^{-1}\mathbf{x}\in\mathbb{Z}^{3}$. Finally, if we also
have $(R_{h}M_{i,h})^{-1}\mathbf{x}\in\mathbb{Z}^{3}$ we may write
$(R_{k}M_{j,k})^{-1}\mathbf{x}=\mathbf{u}\in\mathbb{Z}^{3}$ and
$(R_{h}M_{i,h})^{-1}\mathbf{x}=\mathbf{v}\in\mathbb{Z}^{3}$. Then
$R_{k}^{-1}\mathbf{x}=M_{j,k}\mathbf{u}$ and
$R_{h}^{-1}\mathbf{x}=M_{i,h}\mathbf{v}$ are both integral. According to Lemma
3 we must therefore have $k=h$. Thus
$R_{h}^{-1}\mathbf{x}=R_{k}^{-1}\mathbf{x}=\mathbf{y}$, and both
$M_{i,h}^{-1}\mathbf{y}=M_{i,k}^{-1}\mathbf{y}=\mathbf{v}$ and
$M_{j,k}^{-1}\mathbf{y}=\mathbf{u}$ are integral. Our induction hypothesis
then shows that we must have $i=j$, so that there is exactly one choice of $k$
and $j$ for which $(R_{k}M_{j,k})^{-1}\mathbf{x}$ lies in $\mathbb{Z}^{3}$.

To handle the remaining claims of the theorem we do not use the induction
argument. Given (4), we obtain the relation

$\Delta\det(M_{k})^{2}=D_{k}^{3}$

by taking determinants. Since $\det(M_{k})\mid\Delta$ we have

$\Delta\det(M_{k})^{2}\mid\Delta^{3}\;\;\;\mbox{and}\;\;\;\det(M_{k})^{3}\mid\Delta\det(M_{k})^{2},$

so that $D_{k}^{3}\mid\Delta^{3}$ and $\det(M_{k})^{3}\mid D_{k}^{3}$. Next we
write $\mathbf{x}=\operatorname{Adj}(M_{k})\mathbf{y}$ in (4) and note that
$\operatorname{Adj}(M_{k})=\det(M_{k})M_{k}^{-1}$. This yields

$\det(M_{k})^{2}q(\mathbf{y})=D_{k}J\left(\operatorname{Adj}(M_{k})\mathbf{y}\right),$

whence $D_{k}\mid\det(M_{k})^{2}$, since the form $q$ was assumed to be
primitive. This establishes part (v). For part (vi) we see that if
$p\mid\Delta$ then $p\mid D_{k}$, and hence $p\mid\det(M_{k})$. Moreover if
$p^{e}||\Delta$ and $p^{f}||\det(M_{k})$, then $3\mid e+2f$, since
$\Delta\det(M_{k})^{2}$ is a cube. We then see that we must $e=f$ whenever
$e\leq 3$. Finally, if $\Delta$ is cube-free then so is $\det(M_{k})$, whence
the entries of $M_{k}$ can have no common factor. Any vector
$M_{k}(u^{2},uv,v^{2})$ will be a zero of $q$, so we need to find integers
$u,v$ for which $M_{k}(u^{2},uv,v^{2})$ is primitive. If $p$ is a prime not
dividing $\Delta$ then $M_{k}$ is invertible modulo $p$ so that $p\nmid
M_{k}(u^{2},uv,v^{2})$ whenever $p\nmid(u,v)$. Otherwise $p$ can divide at
most two columns of $M_{k}$. If $p$ does not divide the first column then
$p\nmid M_{k}(1,0,0)$. Similarly if $p$ does not divide the third column of
$M_{k}$ then $p\nmid M_{k}(0,0,1)$. Finally, if $p$ divides the first and
third columns but not the second, then $p\nmid M_{k}(1,1,1)$. It follows, via
the Chinese Remainder Theorem, that if $(u,v)$ lies in a suitable residue
class modulo $\Delta$ then the vector $M_{k}(u^{2},uv,v^{2})$ will be coprime
to $\Delta$. One can now show via the standard arguments that the set of
integer pairs $u,v$ in such a residue class for which $u$ and $v$ are coprime,
will have positive density, given by

$\Delta^{-2}\prod_{p\nmid\Delta}(1-p^{-2}).$

We therefore obtain infinitely many pairs $u,v$ for which
$M_{k}(u^{2},uv,v^{2})$ is primitive.

This completes the proof of the theorem ∎

##  3\. Proof of Theorem 4

We begin with the following informal observation. If the coefficients of $A$
are very large compared with those of $q$, then $J(A\mathbf{x})=q(\mathbf{x})$
has coefficients which are much smaller than they might be, so that
$J(A\mathbf{x})$ “nearly vanishes”. If $A$ has rows
$\mathbf{a},\mathbf{b},\mathbf{c}$, then

$J(A\mathbf{x})=(\mathbf{a}.\mathbf{x})(\mathbf{c}.\mathbf{x})-(\mathbf{b}.\mathbf{x})^{2},$

so that $(\mathbf{a}.\mathbf{x})(\mathbf{c}.\mathbf{x})$ is approximately
equal to $(\mathbf{b}.\mathbf{x})^{2}$. If in fact they were identically
equal, the linear forms $\mathbf{a}.\mathbf{x}$, $\mathbf{b}.\mathbf{x}$ and
$\mathbf{c}.\mathbf{x}$ would have to be proportional, and so the vectors
$\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ would also be proportional.

Our next lemma confirms this, in a quantitative way.

######  Lemma 4.

Suppose $A$ has rows $\mathbf{a},\mathbf{b},\mathbf{c}$, and that
$||\mathbf{a}||\leq||\mathbf{c}||$. Write
$\mathbf{a}=\lambda\mathbf{c}+\mathbf{d}$ and
$\mathbf{b}=\mu\mathbf{c}+\mathbf{e}$, where $\mathbf{d}$ and $\mathbf{e}$ are
orthogonal to $\mathbf{c}$. Then if $J(A\mathbf{x})=q(\mathbf{x})$ we have

  1. (i)

$||\mathbf{e}||\leq 2^{-1/2}||q||^{1/2}$;

  2. (ii)

$|\lambda-\mu^{2}|\leq||q||/(2||\mathbf{c}||^{2})$;

  3. (iii)

$||\mathbf{d}-2\mu\mathbf{e}||\leq||q||/||\mathbf{c}||$; and

  4. (iv)

$|\lambda|\leq 1$.

The reader should note that the bounds (i), (ii) and (iii) above imply that

$|q(\mathbf{x})|=\left|(\mathbf{a}.\mathbf{x})(\mathbf{c}.\mathbf{x})-(\mathbf{b}.\mathbf{x})^{2}\right|\leq
2||q||\cdot||\mathbf{x}||^{2}.$

Thus they ensure that $q(\mathbf{x})$ has the expected order of magnitude,
irrespective of the size of $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$.

###### Proof.

We begin by observing that in general one has

$||M\mathbf{x}||\leq||M||\cdot||\mathbf{x}||$

for any $3\times 3$ matrix $M$, whence

$|q(\mathbf{x})|\leq\tfrac{1}{2}||\mathbf{x}||\cdot||Q\mathbf{x}||\leq\tfrac{1}{2}||Q||\cdot||\mathbf{x}||^{2}=\tfrac{1}{2}||q||\cdot||\mathbf{x}||^{2}.$
(16)

Taking $\mathbf{x}=\mathbf{e}$ we have
$|q(\mathbf{e})|\leq\tfrac{1}{2}||q||\cdot||\mathbf{e}||^{2}$. Moreover

$q(\mathbf{e})=(\mathbf{a}.\mathbf{e})(\mathbf{c}.\mathbf{e})-(\mathbf{b}.\mathbf{e})^{2}=-(\mathbf{b}.\mathbf{e})^{2},$

since $\mathbf{e}$ and $\mathbf{c}$ are orthogonal. However
$\mathbf{b}.\mathbf{e}=||\mathbf{e}||^{2}$, again since $\mathbf{e}$ and
$\mathbf{c}$ are orthogonal. It follows that

$||\mathbf{e}||^{4}=|q(\mathbf{e})|\leq\tfrac{1}{2}||q||\cdot||\mathbf{e}||^{2},$

and hence that $||\mathbf{e}||^{2}\leq\tfrac{1}{2}||q||$. The first claim of
the lemma then follows.

Alternatively we may take $\mathbf{x}=\mathbf{c}$ in (16). Here we have

$q(\mathbf{c})=(\mathbf{a}.\mathbf{c})(\mathbf{c}.\mathbf{c})-(\mathbf{b}.\mathbf{c})^{2}=\lambda||\mathbf{c}||^{4}-\mu^{2}||\mathbf{c}||^{4},$

whence (16) yields

$|\lambda-\mu^{2}|\cdot||\mathbf{c}||^{4}\leq\tfrac{1}{2}||q||\cdot||\mathbf{c}||^{2}.$

This gives us the second assertion of the lemma.

Thirdly we consider $\mathbf{c}Q\mathbf{f}^{T}$, where
$\mathbf{f}=\mathbf{d}-2\mu\mathbf{e}$. Recalling that $\mathbf{a}$ etc. are
row vectors, we have

$\mathbf{c}Q\mathbf{f}^{T}=\mathbf{c}A^{T}\left(\begin{array}[]{rrr}0&0&1\\\
0&-2&0\\\
1&0&0\end{array}\right)A\mathbf{f}^{T}=(\mathbf{c}.\mathbf{c})(\mathbf{a}.\mathbf{f})-2(\mathbf{b}.\mathbf{c})(\mathbf{b}.\mathbf{f})+(\mathbf{a}.\mathbf{c})(\mathbf{c}.\mathbf{f}).$

However $\mathbf{b}.\mathbf{c}=\mu||\mathbf{c}||^{2}$, and since $\mathbf{f}$
is orthogonal to $\mathbf{c}$ we have
$\mathbf{a}.\mathbf{f}=\mathbf{d}.\mathbf{f}$,
$\mathbf{b}.\mathbf{f}=\mathbf{e}.\mathbf{f}$ and $\mathbf{c}.\mathbf{f}=0$,
so that

$\mathbf{c}Q\mathbf{f}^{T}=\\{(\mathbf{d}.\mathbf{f})-2\mu(\mathbf{e}.\mathbf{f})\\}||\mathbf{c}||^{2}=||\mathbf{f}||^{2}||\mathbf{c}||^{2}.$

On the other hand

$|\mathbf{c}Q\mathbf{f}^{T}|\leq||\mathbf{c}||\cdot||Q\mathbf{f}^{T}||\leq||\mathbf{c}||\cdot||\mathbf{f}||\cdot||Q||=||\mathbf{c}||\cdot||\mathbf{f}||\cdot||q||.$

Thus $||\mathbf{f}||\leq||q||/||\mathbf{c}||$ as in the third claim of the
lemma.

The final part is merely a trivial consequence of our initial assumption that
$||\mathbf{a}||\leq||\mathbf{c}||$. ∎

We are now ready to prove Theorem 4. Suppose we have found a matrix $UA$ with
rows $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$, such that
$||\mathbf{a}||\cdot||\mathbf{c}||$ is minimal. Since
$J(UA\mathbf{x})=J(A\mathbf{x})=q(\mathbf{x})$ we may apply Lemma 4 to $UA$.
Premultiplying $UA$ by

$U_{2}=\left(\begin{array}[]{rrr}0&0&1\\\ 0&1&0\\\
1&0&0\end{array}\right)\in\operatorname{Aut}(J)$ (17)

if necessary we may assume that $||\mathbf{a}||\leq||\mathbf{c}||$. Similarly,
premultiplying by

$\left(\begin{array}[]{rrr}1&0&0\\\ 0&-1&0\\\
0&0&1\end{array}\right)\in\operatorname{Aut}(J)$

if necessary, we may assume that $\mu\geq 0$, in the notation of Lemma 4.

We begin the proof by observing that it suffices to show that we have
$||\mathbf{a}||\cdot||\mathbf{c}||\leq 9||q||$. To see this we note that the
choice $\mathbf{x}=\mathbf{b}$ in (16) yields

$\left|(\mathbf{a}.\mathbf{b})(\mathbf{c}.\mathbf{b})-||\mathbf{b}||^{4}\right|\leq\tfrac{1}{2}||q||\cdot||\mathbf{b}||^{2},$

whence

$\displaystyle||\mathbf{b}||^{4}$ $\displaystyle\leq$
$\displaystyle\tfrac{1}{2}||q||\cdot||\mathbf{b}||^{2}+|(\mathbf{a}.\mathbf{b})(\mathbf{c}.\mathbf{b})|$
$\displaystyle\leq$
$\displaystyle\\{\tfrac{1}{2}||q||+||\mathbf{a}||\cdot||\mathbf{c}||\\}||\mathbf{b}||^{2}$
$\displaystyle\leq$ $\displaystyle 10||q||\cdot||\mathbf{b}||^{2},$

given that $||\mathbf{a}||\cdot||\mathbf{c}||\leq 9||q||$. This gives us the
required second bound $||\mathbf{b}||^{2}\leq 10||q||$.

We also note that if $||\mathbf{c}||\leq 3\sqrt{||q||}$ then

$||\mathbf{a}||\cdot||\mathbf{c}||\leq||\mathbf{c}||^{2}\leq 9||q||,$

since we are assuming that $||\mathbf{a}||\leq||\mathbf{c}||$. Thus we may
suppose that

$||\mathbf{c}||\geq 3\sqrt{||q||}$ (18)

for the remainder of the proof.

We now consider $U_{1}UA$ where

$U_{1}=\left(\begin{array}[]{rrr}1&0&0\\\ -1&1&0\\\ 1&-2&1\end{array}\right).$

Then $U_{1}\in\operatorname{Aut}(J)$ and $U_{1}UA$ has rows $\mathbf{a}$,
$-\mathbf{a}+\mathbf{b},\mathbf{a}-2\mathbf{b}+\mathbf{c}$. Since $UA$ was
chosen with $||\mathbf{a}||\cdot||\mathbf{c}||$ minimal, we conclude that

$||\mathbf{a}||\cdot||\mathbf{a}-2\mathbf{b}+\mathbf{c}||\geq||\mathbf{a}||\cdot||\mathbf{c}||,$

and hence that

$||\mathbf{a}-2\mathbf{b}+\mathbf{c}||\geq||\mathbf{c}||.$

We now substitute $\mathbf{a}=\lambda\mathbf{c}+\mathbf{d}$ and
$\mathbf{b}=\mu\mathbf{c}+\mathbf{e}$, yielding

$||(\lambda-2\mu+1)\mathbf{c}+\mathbf{d}-2\mathbf{e}||\geq||\mathbf{c}||.$

Thus parts (ii) and (iii) of Lemma 4 yield

$\displaystyle||\mathbf{c}||$ $\displaystyle\leq$
$\displaystyle|\lambda-2\mu+1|\cdot||\mathbf{c}||+||\mathbf{d}-2\mathbf{e}||$
$\displaystyle\leq$
$\displaystyle|\mu^{2}-2\mu+1|\cdot||\mathbf{c}||+\frac{||q||}{2||\mathbf{c}||}+\frac{||q||}{||\mathbf{c}||}+2|\mu-1|\cdot||\mathbf{e}||.$

However parts (ii) and (iv) of the lemma, along with our assumption (18), show
that

$\mu^{2}\leq|\lambda|+||q||/(2||\mathbf{c}||^{2})\leq\frac{19}{18}.$

Since we are assuming that $\mu\geq 0$ we conclude that $0\leq\mu\leq 37/36$.
In particular $|1-\mu|\leq 1$ so that (18) yields

$\displaystyle||\mathbf{c}||$ $\displaystyle\leq$
$\displaystyle(1-\mu)^{2}||\mathbf{c}||+\frac{3||q||}{2||\mathbf{c}||}+2||\mathbf{e}||$
$\displaystyle\leq$
$\displaystyle(1-\mu)^{2}||\mathbf{c}||+\frac{\sqrt{||q||}}{2}+\sqrt{2||q||}$
$\displaystyle\leq$ $\displaystyle(1-\mu)^{2}||\mathbf{c}||+2\sqrt{||q||},$

by part (i) of Lemma 4. It now follows that

$\mu(2-\mu)\leq\frac{2\sqrt{||q||}}{||\mathbf{c}||},$ (19)

and since $0\leq\mu\leq{37/36}$ we deduce that

$\frac{35}{36}\mu\leq\frac{2\sqrt{||q||}}{||\mathbf{c}||},$

whence

$0\leq\mu\leq\frac{72}{35}\frac{\sqrt{||q||}}{||\mathbf{c}||}.$

From part (ii) of Lemma 4 we now have

$|\lambda|\leq\frac{||q||}{2||\mathbf{c}||^{2}}+\left(\frac{72}{35}\right)^{2}\frac{||q||}{||\mathbf{c}||^{2}}\leq
5\frac{||q||}{||\mathbf{c}||^{2}}.$

Moreover, parts (i) and (iii) yield

$||\mathbf{d}||\leq||q||/||\mathbf{c}||+\sqrt{2}|\mu|\sqrt{||q||}\leq\left(1+\frac{72\sqrt{2}}{35}\right)\frac{||q||}{||\mathbf{c}||}\leq
4\frac{||q||}{||\mathbf{c}||}.$

We therefore conclude that

$||\mathbf{a}||\leq|\lambda|\cdot||\mathbf{c}||+||\mathbf{d}||\leq
9\frac{||q||}{||\mathbf{c}||},$

which suffices for the theorem.

##  4\. Deduction of Theorem 2

Theorem 2 will follow from Theorem 4. We have

$M_{k}^{T}QM_{k}=D_{k}\left(\begin{array}[]{rrr}0&0&1\\\ 0&-2&0\\\
1&0&0\end{array}\right).$

In general one has $M\operatorname{Adj}(M)=\det(M)I$, so that

$\det(M_{k})^{2}Q=D_{k}A^{T}\left(\begin{array}[]{rrr}0&0&1\\\ 0&-2&0\\\
1&0&0\end{array}\right)A,$

with $A=\operatorname{Adj}(M_{k})$. It follows that

$J(A\mathbf{x})=\det(M_{k})^{2}D_{k}^{-1}q(\mathbf{x}),$

where

$AM_{k}=\det(M_{k})I.$ (20)

We may now apply Theorem 4, which provides a matrix
$U\in\operatorname{Aut}(J)$ such that the rows
$\mathbf{a},\mathbf{b},\mathbf{c}$ of $UA$ satisfy

$||\mathbf{a}||\cdot||\mathbf{c}||\leq 9\det(M_{k})^{2}D_{k}^{-1}||q||$ (21)

and

$||\mathbf{b}||^{2}\leq 10\det(M_{k})^{2}D_{k}^{-1}||q||.$ (22)

If the columns of $M_{k}U^{-1}$ are
$\mathbf{c}_{1},\mathbf{c}_{2},\mathbf{c}_{3}$, and $U_{2}$ is given by (17),
then the columns of $M_{k}U^{-1}U_{2}^{-1}$ are
$\mathbf{c}_{3},-\mathbf{c}_{2},\mathbf{c}_{1}$, while the rows of $U_{2}UA$
are $\mathbf{c},-\mathbf{b},\mathbf{a}$. Thus we are free to replace $U$ by
$U_{2}U$ if we wish. We may therefore suppose without loss of generality that
the columns of $M_{k}U^{-1}$ have $||\mathbf{c}_{1}||\leq||\mathbf{c}_{3}||$.
We may also replace $U$ by $-U$, which will not affect the properties (21) and
(22) or the lengths $||\mathbf{c}_{1}||$ and $||\mathbf{c}_{3}||$. Thus we may
also suppose without loss of generality that $\det(U)=+1$.

Having suitably modified $U$ we still have

$J(UA\mathbf{x})=\det(M_{k})^{2}D_{k}^{-1}q(\mathbf{x}),$

so that

$(UA)^{T}\left(\begin{array}[]{rrr}0&0&1\\\ 0&-2&0\\\
1&0&0\end{array}\right)UA=\det(M_{k})^{2}D_{k}^{-1}Q.$

We now claim that we may replace $M_{k}$ by $N_{k}=M_{k}U^{-1}$ in Theorem 1.
Part (i) of the theorem obviously remains true. Since
$\det(N_{k})=\det(M_{k})$ the second and sixth assertions of Theorem 1 are
immediate. Moreover

$q(N_{k}\mathbf{x})=q\big{(}M_{k}(U^{-1}\mathbf{x})\big{)}=D_{k}J(U^{-1}\mathbf{x})=D_{k}J(\mathbf{x}),$

giving us the fourth assertion, and also the fifth since the value of $D_{k}$
is the same for $N_{k}$ as it was for $M_{k}$. Finally,
$U^{-1}(\mathbb{Z}^{3})=\mathbb{Z}^{3}$ since $\det(U)=1$, whence
$N_{k}(\mathbb{Z}^{3})=M_{k}(\mathbb{Z}^{3})$. This suffices for the third
assertion of the theorem.

We proceed to consider the rows of $N_{k}^{-1}$. Since $N_{k}=M_{k}U^{-1}$, we
have

$\operatorname{Adj}(N_{k})=\operatorname{Adj}(U^{-1})\operatorname{Adj}(M_{k})=UA=B.$

Thus

$N_{k}^{-1}=\det(N_{k})^{-1}\operatorname{Adj}(N_{k})=\det(N_{k})^{-1}B.$

Since $\det(N_{k})=\det(M_{k})$ the first pair of inequalities in Theorem 2
now follow from (21) and (22).

To handle the columns $\mathbf{c}_{i}$ of $N_{k}$ we begin with the
observation that

$N_{k}=\det(N_{k})\operatorname{Adj}(N_{k}^{-1}).$

If the rows of $N_{k}^{-1}$ are $\mathbf{r}_{1}=\mathbf{u}$,
$\mathbf{r}_{2}=\mathbf{v}$ and $\mathbf{r}_{3}=\mathbf{w}$, then the first
column of $\operatorname{Adj}(N_{k}^{-1})$ will be

$\big{(}v_{2}w_{3}-v_{3}w_{2},v_{3}w_{1}-v_{1}w_{3},v_{1}w_{2}-v_{2}w_{1}\big{)}^{T},$

and hence will have Euclidean length at most
$||\mathbf{v}||\cdot||\mathbf{w}||$. It follows that

$||\mathbf{c}_{1}||\leq\det(N_{k})||\mathbf{r}_{2}||\cdot||\mathbf{r}_{3}||,$
(23)

and similarly that

$||\mathbf{c}_{2}||\leq\det(N_{k})||\mathbf{r}_{1}||\cdot||\mathbf{r}_{3}||\;\;\;\mbox{and}\;\;\;||\mathbf{c}_{3}||\leq\det(N_{k})||\mathbf{r}_{1}||\cdot||\mathbf{r}_{2}||.$
(24)

Thus (23) and the second part of (24) yield

$||\mathbf{c}_{1}||\cdot||\mathbf{c}_{3}||\leq\det(N_{k})^{2}||\mathbf{r}_{1}||\cdot||\mathbf{r}_{2}||^{2}||\mathbf{r}_{3}||,$

so that the first inequality of (8) follows from (7). The second part of (8)
is then a consequence of the fact that $\det(N_{k})\mid D_{k}$, as noted in
Theorem 1. To establish (9) we merely combine the first part of (24) with (7),
and again use the fact that $\det(N_{k})\mid D_{k}$. This completes the proof
of Theorem 2.

##  5\. Proof of Theorem 3

The matrix (3) has three real eigenvalues, whose product is $\det(Q)=2\Delta$.
For any vector $\mathbf{x}$ we have
$||Q\mathbf{x}||\leq||Q||\cdot||\mathbf{x}||$, so that if $\lambda$ is an
eigenvalue we must have $|\lambda|\leq||Q||$. Since $||q||$ is defined to be
$||Q||$ we conclude that $2\Delta\leq||q||^{3}$, giving us the required bound
$\rho(q)\geq 2$.

If $\mathbf{z}_{1},\mathbf{z}_{2}\in\mathbb{Z}^{3}$ are linearly independent
zeros of $q$ from the same class class $\mathcal{C}_{k}$, then
$q(\mathbf{z}_{1}+\mathbf{z}_{2})$ cannot vanish, since a non-singular conic
cannot have three collinear zeros. However $\mathbf{z}_{1}+\mathbf{z}_{2}$
will be in $M_{k}(\mathbb{Z}^{3})$ so that we must have $D_{k}\mid
q(\mathbf{z}_{1}+\mathbf{z}_{2})$ by (4). It follows that
$q(\mathbf{z}_{1}+\mathbf{z}_{2})-q(\mathbf{z}_{1})-q(\mathbf{z}_{2})$ is a
non-zero multiple of $D_{k}$. Recalling the definition (3) of the matrix $Q$
of $q$ we see that
$q(\mathbf{z}_{1}+\mathbf{z}_{2})-q(\mathbf{z}_{1})-q(\mathbf{z}_{2})=\mathbf{z}_{1}^{T}Q\mathbf{z}_{2}$.
We therefore find that

$D_{k}\leq|q(\mathbf{z}_{1}+\mathbf{z}_{2})-q(\mathbf{z}_{1})-q(\mathbf{z}_{2})|=|\mathbf{z}_{1}^{T}Q\mathbf{z}_{2}|\leq||\mathbf{z}_{1}||\cdot||Q||\cdot||\mathbf{z}_{2}||,$

and hence that $||\mathbf{z}_{1}||\cdot||\mathbf{z}_{2}||\geq D_{k}/||q||$.
This gives us the second assertion of the theorem.

Next, if $\mathbf{c}_{1}$ is not a scalar multiple of $\mathbf{z}_{1}$ we will
have

$||\mathbf{c}_{1}||\geq||\mathbf{z}_{2}||\geq||\mathbf{z}_{1}||.$

This would lead to the inequalities

$D_{k}/||q||\leq||\mathbf{z}_{1}||\cdot||\mathbf{z}_{2}||\leq||\mathbf{c}_{1}||^{2}<\rho^{-1}\det(M_{k})^{2}D_{k}^{-2}||q||^{2}.$

We then have a contradiction, by virtue of (5). This establishes the third
claim of the theorem.

Finally, since $||\mathbf{c}_{3}||\geq||\mathbf{z}_{2}||$ we have

$\displaystyle||\mathbf{c}_{1}||$ $\displaystyle\leq$ $\displaystyle
90\frac{\det(M_{k})^{2}D_{k}^{-2}||q||^{2}}{||\mathbf{c}_{3}||}$
$\displaystyle\leq$ $\displaystyle
90\frac{\det(M_{k})^{2}D_{k}^{-2}||q||^{2}}{||\mathbf{z}_{2}||}$
$\displaystyle\leq$ $\displaystyle
90\frac{\det(M_{k})^{2}D_{k}^{-2}||q||^{2}}{D_{k}/||q||}||\mathbf{z}_{1}||$
$\displaystyle=$ $\displaystyle 90\frac{||q||^{3}}{\Delta}||\mathbf{z}_{1}||,$

by (8), (10) and (5). Similarly, since
$||\mathbf{c}_{1}||\geq||\mathbf{z}_{1}||$ we have

$\displaystyle||\mathbf{c}_{3}||$ $\displaystyle\leq$ $\displaystyle
90\frac{\det(M_{k})^{2}D_{k}^{-2}||q||^{2}}{||\mathbf{c}_{1}||}$
$\displaystyle\leq$ $\displaystyle
90\frac{\det(M_{k})^{2}D_{k}^{-2}||q||^{2}}{||\mathbf{z}_{1}||}$
$\displaystyle\leq$ $\displaystyle
90\frac{\det(M_{k})^{2}D_{k}^{-2}||q||^{2}}{D_{k}/||q||}||\mathbf{z}_{2}||$
$\displaystyle=$ $\displaystyle 90\frac{||q||^{3}}{\Delta}||\mathbf{z}_{2}||.$

This completes our proof of Theorem 3.

##  6\. Preliminaries for the proof of Theorem 5

If $M=M_{k}$ we see from Theorem 1 that

$N(B;\mathcal{C}_{k})=\frac{1}{2}\sum_{\begin{subarray}{c}\mathbf{u}\in\mathbb{Z}^{2}\\\
M\mathbf{u}^{2}\mathrm{primitive}\end{subarray}}\left\\{w(B^{-1}M\mathbf{u}^{2})+w(-B^{-1}M\mathbf{u}^{2})\right\\},$

where we write

$M\mathbf{u}^{2}=M\left(\begin{array}[]{c}u_{1}^{2}\\\ u_{1}u_{2}\\\
u_{2}^{2}\end{array}\right)\;\;\;\mbox{when}\;\;\mathbf{u}=(u_{1},u_{2}),$

for notational convenience. It is thus also convenient to set
$w_{+}(\mathbf{x})=w(\mathbf{x})+w(-\mathbf{x})$ so that $w_{+}$ is an even
function, supported on the set $||\mathbf{x}||\leq 1$. We then have

$N(B;\mathcal{C}_{k})=\frac{1}{2}\sum_{\begin{subarray}{c}\mathbf{u}\in\mathbb{Z}^{2}\\\
M\mathbf{u}^{2}\mathrm{primitive}\end{subarray}}w_{+}(B^{-1}M\mathbf{u}^{2}).$

We begin by considering the condition that $M\mathbf{u}^{2}$ should be
primitive. Our goal is the following result.

######  Lemma 5.

The set of primes may be partitioned into sets
$\mathcal{P}_{0},\mathcal{P}_{1},\mathcal{P}_{2}$ with the following
properties. Firstly, if $p\nmid\Delta$ then $p\in\mathcal{P}_{0}$. Secondly,
if $p\in\mathcal{P}_{0}$ then $p\mid M\mathbf{u}^{2}$ if and only if
$p\mid\mathbf{u}$. Thirdly, if $p\in\mathcal{P}_{n}$ for $n=1$ or 2 then there
are distinct lattices $\Lambda_{i}(p)\subseteq\mathbb{Z}^{2}$ for $1\leq i\leq
n$ having determinant $p$, and such that $p\mid M\mathbf{u}^{2}$ if and only
if $\mathbf{u}$ lies in one of the lattices $\Lambda_{i}(p)$. Finally, if
$p||\Delta$ then $p\in\mathcal{P}_{1}$.

###### Proof.

If $p$ is a prime not dividing $\det(M)$ (and in particular for any prime not
dividing $\Delta$) the matrix $M$ will be invertible modulo $p$, so that
$p\mid M\mathbf{u}^{2}$ if and only if $p\mid\mathbf{u}^{2}$. In this case the
condition that $p\mid M\mathbf{u}^{2}$ is equivalent to $p\mid\mathbf{u}$, and
$p$ will be in $\mathcal{P}_{0}$. On the other hand, if $p\mid\det(M)$ then
$M$ is singular modulo $p$. It cannot vanish modulo $p$, since the class
$\mathcal{C}_{k}$ corresponding to $M=M_{k}$ is assumed to be non-empty,
whence $M$ has rank 1 or 2 modulo $p$. Suppose firstly that $M$ has rank 1
over $\mathbb{F}_{p}$, with a non-zero row $(A,B,C)$ say. If the quadratic
form $Au^{2}+Buv+Cv^{2}$ is irreducible modulo $p$ then $p\mid
M\mathbf{u}^{2}$ implies $p\mid\mathbf{u}$. In this case $p$ will be in
$\mathcal{P}_{0}$. If the form $Au^{2}+Buv+Cv^{2}$ splits into distinct
factors as $L_{1}(u,v)L_{2}(u,v)$ then $p\mid M\mathbf{u}^{2}$ if and only if
$\mathbf{u}$ lies in one or both of the lattices $\Lambda_{1},\Lambda_{2}$
given by $p\mid L_{i}(u,v)$. In this case $p$ will be in $\mathcal{P}_{2}$. On
the other hand, if $Au^{2}+Buv+Cv^{2}$ has a repeated factor $L(u,v)^{2}$,
then one has $p\mid M\mathbf{u}^{2}$ if and only if $\mathbf{u}$ lies in the
lattice $\Lambda$ given by $p\mid L(u,v)$, so that $p\in\mathcal{P}_{1}$. A
similar analysis applies when $M$ has rank 2 over $\mathbb{F}_{p}$, showing in
this case that the condition $p\mid M\mathbf{u}^{2}$ is either equivalent to
$\mathbf{u}\in\Lambda$ for some lattice $\Lambda\subset\mathbb{Z}^{2}$ of
determinant $p$, or is equivalent to $p\mid\mathbf{u}$.

Finally, suppose that $p||\Delta$. Then $p||\det(M)$, by part (iii) of Theorem
1, so that $M$ has rank 2 over $\mathbb{F}_{p}$. Using row operations one sees
that there is a matrix $U\in\mathrm{GL}_{3}(\mathbb{Z}_{p})$ such that $UM=R$
takes one of the forms

$R_{1}=\left(\begin{array}[]{ccc}1&0&a\\\ 0&1&b\\\
0&0&p\end{array}\right)\;\;\mbox{or}\;\;R_{2}=\left(\begin{array}[]{ccc}1&a&0\\\
0&p&0\\\
0&0&1\end{array}\right)\;\;\mbox{or}\;\;R_{3}=\left(\begin{array}[]{ccc}p&0&0\\\
0&1&0\\\ 0&0&1\end{array}\right).$

Then if $M=M_{k}$ and $D=D_{k}$ we have $p||D$ by (5), and

$\displaystyle J(\operatorname{Adj}(R)\mathbf{x})$ $\displaystyle=$
$\displaystyle\det(M)^{2}J(R^{-1}\mathbf{x})$ $\displaystyle=$
$\displaystyle\det(M)^{2}D^{-1}q(MR^{-1}\mathbf{x})$ $\displaystyle=$
$\displaystyle\det(M)^{2}D^{-1}q(U^{-1}\mathbf{x}).$

Since $U$ is invertible modulo $p$ we conclude that
$J(\operatorname{Adj}(R)\mathbf{x})$ vanishes modulo $p$. When $R=R_{1}$ we
have

$\operatorname{Adj}(R_{1})=\left(\begin{array}[]{ccc}p&0&-a\\\ 0&p&-b\\\
0&0&1\end{array}\right)$

so that

$J(\operatorname{Adj}(R_{1})\mathbf{x})=(px_{1}-ax_{3})x_{3}-(px_{2}-bx_{3})^{2}\equiv-(a+b^{2})x_{3}^{2}\;(\text{mod}\;p).$

In this case we conclude that $p\mid a+b^{2}$. Since $U$ is invertible modulo
$p$ the condition $p\mid M\mathbf{u}^{2}$ is equivalent to $p\mid
R\mathbf{u}^{2}$, and for $R=R_{1}$ this becomes

$u_{1}^{2}+au_{2}^{2}\equiv u_{1}u_{2}+bu_{2}^{2}\equiv 0\;(\text{mod}\;p).$

Since $a\equiv-b^{2}\;(\text{mod}\;p)$ this holds precisely when $p\mid
u_{1}+bu_{2}$. Thus for $R=R_{1}$ there is a single lattice condition.

For $R=R_{2}$ we calculate that

$J(\operatorname{Adj}(R_{2})\mathbf{x})=(px_{1}-ax_{2})px_{3}-x_{2}^{2},$

which cannot vanish identically modulo $p$. This case is therefore forbidden.
When $R=R_{3}$ we see that $p\mid R\mathbf{u}^{2}$ if and only if $p\mid
u_{2}$, which again gives us a single lattice condition. Thus whenever
$p||\Delta$ the condition $p\mid M\mathbf{u}^{2}$ gives us a single lattice
condition with determinant $p$. ∎

Lemma 5 allows us to handle the primitiveness condition in the definition of
the sum $N(B;\mathcal{C}_{k})$ as follows.

######  Lemma 6.

Suppose that $w(\mathbf{x})$ is supported on the disc $||\mathbf{x}||\leq 1$.
Then there is a square-free divisor $\Delta_{1}\Delta_{2}$ of $\Delta$, and
lattices $\Lambda^{(1)},\ldots,\Lambda^{(J)}$, where

$J=2^{\omega(\Delta_{1})}3^{\omega(\Delta_{2})},$

with the following properties. Firstly if $p||\Delta$ then $p\mid\Delta_{1}$;
secondly the determinant $\mathrm{d}(\Lambda^{(j)})$ divides
$\Delta_{1}\Delta_{2}^{2}$ for every index $j$; thirdly

$N(B;\mathcal{C}_{k})=\frac{1}{2}\sum_{j=1}^{J}\lambda\left(\mathrm{d}(\Lambda^{(j)})\right)\sum_{\begin{subarray}{c}d=1\\\
(d,\Delta_{1}\Delta_{2})=1\end{subarray}}^{\infty}\mu(d)\sum_{\mathbf{u}\in\Lambda^{(j)}-\\{\mathbf{0}\\}}w_{+}(d^{2}B^{-1}M\mathbf{u}^{2}),$

where $\lambda(n)=(-1)^{\Omega(n)}$ is the Liouville function; and fourthly

$\sum_{j=1}^{J}\frac{\lambda\left(\mathrm{d}(\Lambda^{(j)})\right)}{\mathrm{d}(\Lambda^{(j)})}\sum_{\begin{subarray}{c}d=1\\\
(d,\Delta_{1}\Delta_{2})=1\end{subarray}}^{\infty}\frac{\mu(d)}{d^{2}}=\frac{6}{\pi^{2}}\prod_{p\mid\Delta_{1}}\frac{1}{1+p^{-1}}\prod_{p\mid\Delta_{2}}\frac{1-p^{-1}}{1+p^{-1}}.$

###### Proof.

For the proof we use the notation $\mathbb{1}(A)$ for the characteristic
function for the property $A$. We begin by observing that

$\mathbb{1}(p\nmid
M\mathbf{u}^{2})=1-\mathbb{1}(p\mid\mathbf{u}),\;\;\;p\in\mathcal{P}_{0},$
$\mathbb{1}(p\nmid
M\mathbf{u}^{2})=1-\mathbb{1}(\mathbf{u}\in\Lambda_{1}(p)),\;\;\;p\in\mathcal{P}_{1},$

and finally,

$\mathbb{1}(p\nmid
M\mathbf{u}^{2})=1-\mathbb{1}(\mathbf{u}\in\Lambda_{1}(p))-\mathbb{1}(\mathbf{u}\in\Lambda_{2}(p))+\mathbb{1}(\mathbf{u}\in\Lambda_{1}(p)\cap\Lambda_{2}(p)),$

when $p\in\mathcal{P}_{2}$. We now take $\Delta_{1}$ to be the product of the
primes in $\mathcal{P}_{1}$ and $\Delta_{2}$ to be the product of the primes
in $\mathcal{P}_{2}$, so that $\Delta_{1}\Delta_{2}\mid\Delta$. Let
$\Lambda^{(j)}$ (for $1\leq j\leq J$) run over all lattices formed by the
intersection of none, some, or all, of the lattices $\Lambda_{i}(p)$ (for
$i=1$ or 2 and $p\in\mathcal{P}_{1}\cup\mathcal{P}_{2}$). Then
$J=2^{\omega(\Delta_{1})}3^{\omega(\Delta_{2})}$, and

$\mathbb{1}\left((\Delta_{1}\Delta_{2},M\mathbf{u}^{2})=1\right)=\sum_{j=1}^{J}\lambda\left(\mathrm{d}(\Lambda^{(j)})\right)\mathbb{1}(\mathbf{u}\in\Lambda^{(j)}),$

since $\Lambda_{1}(p)\cap\Lambda_{2}(p)$ will have determinant $p^{2}$ when
$p\mid\Delta_{2}$. The conditions $p\nmid M\mathbf{u}^{2}$ for primes
$p\in\mathcal{P}_{0}$ are produced by

$\sum_{\begin{subarray}{c}d\mid\mathbf{u}\\\
(d,\Delta_{1}\Delta_{2})=1\end{subarray}}\mu(d),$

so that

$\displaystyle N(B;\mathcal{C}_{k})$ $\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{\begin{subarray}{c}\mathbf{u}\in\mathbb{Z}^{2}\\\
M\mathbf{u}^{2}\mathrm{primitive}\end{subarray}}w_{+}\left(\frac{M}{B}\mathbf{u}^{2}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{\begin{subarray}{c}\mathbf{u}\in\mathbb{Z}^{2}-\\{\mathbf{0}\\}\\\
M\mathbf{u}^{2}\mathrm{primitive}\end{subarray}}w_{+}\left(\frac{M}{B}\mathbf{u}^{2}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{j=1}^{J}\lambda\left(\mathrm{d}(\Lambda^{(j)})\right)\sum_{\begin{subarray}{c}d=1\\\
(d,\Delta_{1}\Delta_{2})=1\end{subarray}}^{\infty}\mu(d)\sum_{\begin{subarray}{c}\mathbf{u}\in\Lambda^{(j)}-\\{\mathbf{0}\\}\\\
d\mid\mathbf{u}\end{subarray}}w_{+}\left(\frac{M}{B}\mathbf{u}^{2}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{j=1}^{J}\lambda\left(\mathrm{d}(\Lambda^{(j)})\right)\sum_{\begin{subarray}{c}d=1\\\
(d,\Delta_{1}\Delta_{2})=1\end{subarray}}^{\infty}\mu(d)\sum_{\mathbf{u}\in\Lambda^{(j)}-\\{\mathbf{0}\\}}w_{+}\left(d^{2}\frac{M}{B}\mathbf{u}^{2}\right)$

as required. Here we should note that the $d$-summation is finite for all
relevant $\mathbf{u}$.

The final part is clear, by multiplicativity. ∎

In light of Lemma 6 our focus moves to sums of the shape

$S(\Lambda,B,M_{k})=\sum_{\mathbf{x}\in\Lambda-\\{\mathbf{0}\\}}w_{+}(B^{-1}M_{k}\mathbf{x}^{2}),$

where $\Lambda$ is an integer lattice, $w_{+}$ is an even weight supported in
the disc $||\mathbf{x}||\leq 1$, and $M_{k}$ is an integer matrix of the shape
described in Theorems 1 and 2. We first need to understand the range of
summation in $S(\Lambda,B,M_{k})$.

######  Lemma 7.

Let $M$ be one of the matrices $M_{k}$, as described in Theorem 2. Let
$X_{1}=\sqrt{B||\mathbf{r}_{1}||}$ and $X_{2}=\sqrt{B||\mathbf{r}_{3}||}$, so
that

$X_{1}X_{2}\leq 3BD_{k}^{-1/2}||q||^{1/2}.$

Then if $w_{+}(B^{-1}M\mathbf{x}^{2})\not=0$ with
$\mathbf{x}\in\mathbb{R}^{2}$ we have both $|x_{1}|\leq X_{1}$ and
$|x_{2}|\leq X_{2}$. Moreover if $w_{+}(B^{-1}M\mathbf{x}^{2})\not=0$ with
$\mathbf{x}\in\mathbb{Z}^{2}$ we have $B\geq 1$.

###### Proof.

We set $B^{-1}M\mathbf{x}^{2}=\mathbf{y}$, so that $||\mathbf{y}||\leq 1$ if
$w_{+}(B^{-1}M\mathbf{x}^{2})\not=0$. If $M^{-1}$ has rows
$\mathbf{r}_{1},\mathbf{r}_{2},\mathbf{r}_{3}$, as in Theorem 2, then

$|(M^{-1}\mathbf{y})_{1}|=|\mathbf{r}_{1}^{T}\mathbf{y}|\leq||\mathbf{r}_{1}||\cdot||\mathbf{y}||\leq||\mathbf{r}_{1}||,$

so that $x_{1}^{2}\leq B||\mathbf{r}_{1}||$. Similarly $x_{2}^{2}\leq
B||\mathbf{r}_{3}||$, and the first result follows. If
$\mathbf{x}\in\mathbb{Z}^{2}-\\{\mathbf{0}\\}$ with
$w_{+}(B^{-1}M\mathbf{x}^{2})\not=0$ we have $B^{-1}||M\mathbf{x}^{2}||\leq
1$. Since $\mathbf{x}^{2}$ does not vanish we see that $M\mathbf{x}^{2}$ must
be a non-zero integer vector, since $M$ is nonsingular. It follows that
$||M\mathbf{x}^{2}||\geq 1$, whence $B\geq 1$ as claimed. ∎

We now give a crude bound for $S(\Lambda,B,M_{k})$.

######  Lemma 8.

We have $S(\Lambda,B,M_{k})=0$ if $B<1$, and otherwise

$S(\Lambda,B,M_{k})\ll_{w}D_{k}^{-1/2}||q||^{1/2}\left\\{\frac{B}{\mathrm{d}(\Lambda)}+B^{1/2}||\mathbf{c}_{3}||^{1/2}\right\\}.$

###### Proof.

The first claim is obvious, given Lemma 7. Generally

$S(\Lambda,B,M_{k})\ll_{w}\\#\\{\mathbf{x}\in\Lambda:\,|x_{1}|\leq
X_{1},\,|x_{2}|\leq X_{2}\\}.$

If we set

$\Lambda_{0}=\\{(x_{1}/X_{1},x_{2}/X_{2}):\,(x_{1},x_{2})\in\Lambda\\},$

then $\mathrm{d}(\Lambda_{0})=\mathrm{d}(\Lambda)/X_{1}X_{2}$, and

$S(\Lambda,B,M_{k})\ll_{w}\\#\\{\mathbf{y}\in\Lambda_{0}:\,||\mathbf{y}||_{\infty}\leq
1\\}.$

Thus

$S(\Lambda,B,M_{k})\ll_{w}\mathrm{d}(\Lambda_{0})^{-1}+\lambda_{1}^{-1}+1\ll
X_{1}X_{2}\mathrm{d}(\Lambda)^{-1}+\lambda_{1}^{-1}+1,$

where $\lambda_{1}$ is the length of the shortest non-zero vector in
$\Lambda_{0}$. However one has $||\mathbf{x}||\geq 1$ for every non-zero
vector in $\Lambda$, and hence

$\lambda_{1}\geq\max(X_{1},X_{2})^{-1}.$

We therefore obtain the bound

$S(\Lambda,B,M_{k})\ll_{w}\frac{X_{1}X_{2}}{\mathrm{d}(\Lambda)}+\max(X_{1},X_{2})+1.$

If $\max(X_{1},X_{2})\leq\tfrac{1}{2}$ and $\mathbf{x}$ is an integer vector
for which $w_{+}(B^{-1}M\mathbf{x}^{2})$ is non-zero, then we must have
$\mathbf{x}=\mathbf{0}$, which is excluded from the sum $S(\Lambda,B,M_{k})$.
Thus $S(\Lambda,B,M_{k})=0$ when $\max(X_{1},X_{2})\leq\tfrac{1}{2}$. It
therefore follows that

$\displaystyle S(\Lambda,B,M_{k})$ $\displaystyle\ll_{w}$
$\displaystyle\frac{X_{1}X_{2}}{\mathrm{d}(\Lambda)}+\max(X_{1},X_{2})$
$\displaystyle\ll_{w}$ $\displaystyle
D_{k}^{-1/2}||q||^{1/2}\frac{B}{\mathrm{d}(\Lambda)}+B^{1/2}\max\\{||\mathbf{r}_{1}||\,,\,||\mathbf{r}_{3}||\\}^{1/2}.$

We now claim that

$||\mathbf{r}_{1}||\ll\frac{||q||}{D_{k}}||\mathbf{c}_{3}||\;\;\;\mbox{and}\;\;\;||\mathbf{r}_{3}||\ll\frac{||q||}{D_{k}}||\mathbf{c}_{1}||.$
(25)

Clearly this will suffice for the lemma, since we have chosen $M_{k}$ so that
$||\mathbf{c}_{1}||\leq||\mathbf{c}_{3}||$.

To prove (25) we begin with the observation that the scalar product
$\mathbf{r}_{1}.\mathbf{c}_{1}$ takes the value $1$, since $M^{-1}M=I$.
Similarly we have $\mathbf{r}_{3}.\mathbf{c}_{3}=1$. It follows that

$||\mathbf{r}_{1}||\cdot||\mathbf{c}_{1}||\geq
1\;\;\;\mbox{and}\;\;\;||\mathbf{r}_{3}||\cdot||\mathbf{c}_{3}||\geq 1.$ (26)

Thus

$||\mathbf{r}_{1}||\leq||\mathbf{r}_{1}||\cdot||\mathbf{r}_{3}||\cdot||\mathbf{c}_{3}||,$

so that the first part of (25) follows from (7). The second part may be proved
entirely analogously. ∎

##  7\. Theorem 5 — The leading term

To estimate $S(\Lambda,B,M_{k})$ we will use the following form of the Poisson
summation formula.

######  Lemma 9.

Let $N\in$GL${}_{2}(\mathbb{R})$, so that $\Lambda=N(\mathbb{Z}^{2})$ is a
two-dimensional lattice. Then

$S(\Lambda,B,M_{k})+w_{+}(\mathbf{0})=\mathrm{d}(\Lambda)^{-1}\sum_{\mathbf{a}\in\mathbb{Z}^{2}}I(\mathbf{a},M,N),$

with

$I(\mathbf{a},M,N)=\int_{\mathbb{R}^{2}}w_{+}(B^{-1}M\mathbf{x}^{2})e(-\mathbf{a}^{T}N^{-T}\mathbf{x})dx_{1}dx_{2}.$

###### Proof.

Writing $\varpi(\mathbf{x})=w_{+}(B^{-1}M\mathbf{x}^{2})$, the Poisson
summation formula yields

$S(\Lambda,B,M_{k})+w_{+}(\mathbf{0})=\sum_{\mathbf{y}\in\mathbb{Z}^{2}}\varpi(N\mathbf{y})=\sum_{\mathbf{a}\in\mathbb{Z}^{2}}\int_{R^{2}}\varpi(N\mathbf{z})e(-\mathbf{a}^{T}\mathbf{z})dz_{1}dz_{2}.$

If we substitute $\mathbf{x}=N\mathbf{z}$ we have
$\mathbf{a}^{T}\mathbf{z}=\mathbf{a}^{T}N^{-T}\mathbf{x}$, and the result
follows since $|\det(N)|=\mathrm{d}(\Lambda)$. ∎

The main term in Theorem 5 will come from the integral with
$\mathbf{a}=\mathbf{0}$.

######  Lemma 10.

Define $K_{T}(t)$ as in (13). Then if $q(M\mathbf{x})=DJ(\mathbf{x})$ as in
Theorem 1 we have

$\int_{\mathbb{R}^{3}}w(\mathbf{y})K_{T}(q(\mathbf{y}))dy_{1}dy_{2}dy_{3}\to\frac{D^{1/2}}{\Delta^{1/2}}\int_{\mathbb{R}^{2}}w_{+}(M\mathbf{x}^{2})dx_{1}dx_{2},$

as $T\to\infty$, and hence

$\int_{\mathbb{R}^{2}}w_{+}(B^{-1}M\mathbf{x}^{2})dx_{1}dx_{2}=\sigma_{\infty}(q;w)\frac{\Delta^{1/2}}{D^{1/2}}B,$

where $\sigma_{\infty}(q;w)$ is given by (12).

###### Proof.

Since

$\int_{\mathbb{R}^{3}}w(\mathbf{y})K_{T}(q(\mathbf{y}))dy_{1}dy_{2}dy_{3}=\int_{\mathbb{R}^{3}}w(-\mathbf{y})K_{T}(q(\mathbf{y}))dy_{1}dy_{2}dy_{3}$

we have

$\sigma_{\infty}(q;w)=\frac{1}{2}\lim_{T\to\infty}\int_{\mathbb{R}^{3}}w_{+}(\mathbf{y})K_{T}(q(\mathbf{y}))dy_{1}dy_{2}dy_{3}.$

Then, writing $\tilde{w}(\mathbf{x})=w_{+}(D^{-1/2}M\mathbf{x})$ we have

$\displaystyle\int_{\mathbb{R}^{3}}w_{+}(\mathbf{y})K_{T}(q(\mathbf{y}))dy_{1}dy_{2}dy_{3}$
(27) $\displaystyle=$
$\displaystyle\det(M)\int_{\mathbb{R}^{3}}w_{+}(M\mathbf{z})K_{T}(q(M\mathbf{z}))dz_{1}dz_{2}dz_{3}$
$\displaystyle=$
$\displaystyle\det(M)\int_{\mathbb{R}^{3}}w_{+}(M\mathbf{z})K_{T}(DJ(\mathbf{z}))dz_{1}dz_{2}dz_{3}$
$\displaystyle=$
$\displaystyle\frac{\det(M)}{D^{3/2}}\int_{\mathbb{R}^{3}}\tilde{w}(\mathbf{x})K_{T}(J(\mathbf{x}))dx_{1}dx_{2}dx_{3}$
$\displaystyle=$
$\displaystyle\Delta^{-1/2}\int_{\mathbb{R}^{3}}\tilde{w}(\mathbf{x})K_{T}(J(\mathbf{x}))dx_{1}dx_{2}dx_{3}.$

The function $\tilde{w}$ will be even, so that the above becomes

$\frac{2}{\Delta^{1/2}}\int_{0}^{\infty}\int_{\mathbb{R}^{2}}\tilde{w}(\mathbf{x})K_{T}(J(\mathbf{x}))dx_{1}dx_{2}dx_{3}.$

Since $w_{+}(\mathbf{y})$ is supported on the set $||\mathbf{y}||\leq 1$ we
see that $\tilde{w}(\mathbf{z})$ is supported on a subset of $[-C,C]^{3}$ for
some $C=C(M,D)>0$. If we write $x_{0}=x_{2}^{2}/x_{3}$ and $x_{1}=x_{0}+u$,
then both $\tilde{w}(\mathbf{x})$ and $\tilde{w}(x_{0},x_{2},x_{3})$ vanish
unless $|x_{2}|\leq C$ and $x_{3}\leq C$. If $\phi$ is the function

$\phi(u,x_{2},x_{3})=\tilde{w}(x_{0}+u,x_{2},x_{3})-\tilde{w}(x_{0},x_{2},x_{3})$

it follows that

$\displaystyle\int_{0}^{\infty}\int_{\mathbb{R}^{2}}\\{\tilde{w}(\mathbf{x})-\tilde{w}(x_{0},x_{2},x_{3})\\}K_{T}(J(\mathbf{x}))dx_{1}dx_{2}dx_{3}$
(28) $\displaystyle=$
$\displaystyle\int_{0}^{C}\int_{-C}^{C}\int_{-\infty}^{\infty}\phi(u,x_{2},x_{3})K_{T}(x_{3}u)dudx_{2}dx_{3}$
$\displaystyle\ll$ $\displaystyle T\int_{0}^{C}\int_{-C}^{C}\int_{|u|\leq
1/Tx_{3}}|\phi(u,x_{2},x_{3})|dudx_{2}dx_{3}.$

We may assume that $\tilde{w}(x_{0}+u,x_{2},x_{3})$ and
$\tilde{w}(x_{0},x_{2},x_{3})$ do not both vanish, and hence that either
$|x_{0}|\leq C$ or $|x_{0}+u|\leq C$, (or both). In this case

$\phi(u,x_{2},x_{3})\ll_{w,M}\min(1,|u|).$ (29)

We proceed to consider separately the ranges $|x_{0}|\geq 2C$ and $|x_{0}|\leq
2C$. When $|x_{0}|\geq 2C$ the bound $|x_{0}+u|\leq C$ implies that $|u|\geq
x_{0}/2$, and since $|u|\leq 1/Tx_{3}$ we conclude that $x_{2}^{2}\leq 2/T$.
The bound $|x_{0}|\geq 2C$ then shows that $x_{3}\leq x_{2}^{2}/2C\leq 1/CT$.
Moreover $u$ is restricted to a range $|x_{0}+u|\leq C$ of length $O_{M}(1)$.
The estimate in (29) is $O_{w,M}(1)$, so that the corresponding contribution
to (28) is

$\ll_{w,m}T\int_{0}^{1/CT}\int_{x_{2}^{2}\leq
2/T}dx_{2}\,dx_{3}\ll_{w,M}T^{-1/2}.$

On the other hand, when $|x_{0}|\leq 2C$ we have $x_{2}^{2}\leq 2Cx_{3}$ and

$\int_{|u|\leq 1/Tx_{3}}\min(1,|u|)du\ll(Tx_{3})^{-1}\min(1\,,\,1/Tx_{3}),$

so that the corresponding contribution to (28) is

$\ll_{w,M}\int_{0}^{C}x_{3}^{-1/2}\min(1\,,\,1/Tx_{3})dx_{3}\ll_{w,M}T^{-1/2}.$

We therefore conclude that

$\displaystyle\int_{\mathbb{R}^{3}}w_{+}(\mathbf{y})K_{T}(q(\mathbf{y}))dy_{1}dy_{2}dy_{3}$
$\displaystyle=$
$\displaystyle\frac{2}{\Delta^{1/2}}\int_{0}^{\infty}\int_{\mathbb{R}^{2}}\tilde{w}(x_{0},x_{2},x_{3})K_{T}(J(\mathbf{x}))dx_{1}dx_{2}dx_{3}+O_{w,M}(T^{-1/2}).$

The function $\tilde{w}(x_{0},x_{2},x_{3})$ is independent of $x_{1}$ and

$\int_{\mathbb{R}}K_{T}(x_{1}x_{3}-x_{2}^{2})dx_{1}=x_{3}^{-1}.$

Moreover with the substitutions $x_{3}=u_{2}^{2}$ and $x_{2}=u_{1}u_{2}$ we
have

$\displaystyle\int_{0}^{\infty}\int_{\mathbb{R}}\tilde{w}(x_{2}^{2}/x_{3},x_{2},x_{3})x_{3}^{-1}dx_{2}dx_{3}$
$\displaystyle=$ $\displaystyle
2\int_{0}^{\infty}\int_{\mathbb{R}}\tilde{w}(u_{1}^{2},u_{1}u_{2},u_{2}^{2})du_{1}\,du_{2}$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{2}}\tilde{w}(u_{1}^{2},u_{1}u_{2},u_{2}^{2})du_{1}\,du_{2}$
$\displaystyle=$ $\displaystyle
D^{1/2}\int_{\mathbb{R}^{2}}w_{+}(M\mathbf{u}^{2})du_{1}\,du_{2},$

and the lemma follows. ∎

Our next result tells us about the size of $\sigma_{\infty}(q;w)$. Recall that
$q$ is isotropic, and hence indefinite, with $\Delta>0$, so that the matrix
$Q$ of $q$ has one positive eigenvalue $\lambda$ say, and two negative ones
$-\mu$ and $-\nu$ say. We may assume that $\mu\geq\nu(>0)$. With this notation
we have $\Delta=\tfrac{1}{2}\lambda\mu\nu$. We remind the reader of the
notation $f\asymp g$, meaning that both $f\ll g$ and $g\ll f$ hold. in our
context the two implied constants will be absolute. Thus we will have

$\max(\lambda,\mu,\nu)\asymp||q||,$

for example.

######  Lemma 11.

If $\sup|w(\mathbf{x})|\leq 1$ we have

$|\sigma_{\infty}(q;w)|\leq 2e^{4/3}\sigma_{\infty}(q;w_{0}),$

where the weight $w_{0}$ is given by (11). Moreover

$\sigma_{\infty}(q;w_{0})\asymp\frac{\min(\lambda,\mu,\nu)^{1/2}}{\Delta^{1/2}}\log(2\mu/\nu)$

when $\lambda\geq\mu\geq\nu$,

$\sigma_{\infty}(q;w_{0})\asymp\frac{\min(\lambda,\mu,\nu)^{1/2}}{\Delta^{1/2}}\log(2\lambda/\nu)$

when $\mu\geq\lambda\geq\nu$, and

$\sigma_{\infty}(q;w_{0})\asymp\frac{\min(\lambda,\mu,\nu)^{1/2}}{\Delta^{1/2}}$

when $\mu\geq\nu\geq\lambda$.

Thus

$\frac{1}{||q||}\ll\sigma_{\infty}(q;w_{0})\ll\frac{\rho^{1/4}}{||q||}$

in every case.

###### Proof.

According to Lemma 10 we have

$\displaystyle|\sigma_{\infty}(q;w)|$ $\displaystyle=$
$\displaystyle\frac{\Delta^{1/2}}{D^{1/2}}\left|\int_{\mathbb{R}^{2}}w_{+}(M\mathbf{x}^{2})dx_{1}dx_{2}\right|$
$\displaystyle\leq$ $\displaystyle
2\frac{\Delta^{1/2}}{D^{1/2}}\int_{|M\mathbf{x}^{2}|\leq 1}dx_{1}dx_{2}$
$\displaystyle\leq$ $\displaystyle
2e^{4/3}\frac{\Delta^{1/2}}{D^{1/2}}\int_{|M\mathbf{x}^{2}|\leq
1}\exp\\{-1/(1-||M\mathbf{x}||^{2}/4)\\}dx_{1}dx_{2}$ $\displaystyle\leq$
$\displaystyle
2e^{4/3}\frac{\Delta^{1/2}}{D^{1/2}}\int_{\mathbb{R}^{2}}w_{0}(\tfrac{1}{2}M\mathbf{x}^{2})dx_{1}dx_{2}$
$\displaystyle=$ $\displaystyle
4e^{4/3}\frac{\Delta^{1/2}}{D^{1/2}}\int_{\mathbb{R}^{2}}w_{0}(M\mathbf{x}^{2})dx_{1}dx_{2}.$

On the other hand, applying Lemma 10 to the weight $w_{0}$ we find that

$\sigma_{\infty}(q;w_{0})=2\frac{\Delta^{1/2}}{D^{1/2}}\int_{\mathbb{R}^{2}}w_{0}(M\mathbf{x}^{2})dx_{1}dx_{2}.$

The first claim of the lemma then follows.

For the remainder of the proof it will be convenient to write

$I(T)=\int_{\mathbb{R}^{3}}w_{0}(\mathbf{y})K_{T}(q(\mathbf{y}))dy_{1}dy_{2}dy_{3}.$

Now let $U$ be a real orthogonal matrix diagonalising $q(\mathbf{x})$, so that
$q(U\mathbf{x})=\mathrm{Diag}(\lambda,-\mu,-\nu)$ say. Substituting
$U\mathbf{y}$ in place of $\mathbf{y}$, and noting that the weight $w_{0}$ is
invariant under rotations, we deduce that

$I(T)=\int_{\mathbb{R}^{3}}w_{0}(\mathbf{y})K_{T}(\lambda y_{1}^{2}-\mu
y_{2}^{3}-\nu y_{3}^{2})dy_{1}dy_{2}dy_{3},$

whence

$I(T)\leq\int_{[-1,1]^{3}}K_{T}(\lambda y_{1}^{2}-\mu y_{2}^{3}-\nu
y_{3}^{2})dy_{1}dy_{2}dy_{3}$

and

$I(T)\gg\int_{[-1/2,1/2]^{3}}K_{T}(\lambda y_{1}^{2}-\mu y_{2}^{3}-\nu
y_{3}^{2})dy_{1}dy_{2}dy_{3}$

We now consider three cases. Firstly, suppose that $\lambda\geq\mu$. Then if
$\mathbf{y}\in[-1,1]^{3}$ and $K_{T}(\lambda y_{1}^{2}-\mu y_{2}^{3}-\nu
y_{3}^{2})\not=0$ we have

$\lambda y_{1}^{2}\leq T^{-1}+\mu y_{2}^{2}+\nu y_{3}^{2}\leq T^{-1}+2\mu.$

It follows that $|y_{1}|\leq 2\sqrt{\mu/\lambda}$ as soon as
$T\geq(2\mu)^{-1}$. Writing $\xi=\min(1\,,\,2\sqrt{\mu/\lambda}$ we then
deduce that

$\displaystyle I(T)$ $\displaystyle\leq$
$\displaystyle\int_{[-1,1]^{2}}\int_{-\xi}^{\xi}K_{T}(\lambda y_{1}^{2}-\mu
y_{2}^{3}-\nu y_{3}^{2})dy_{1}dy_{2}dy_{3}$ $\displaystyle=$
$\displaystyle\sqrt{\mu/\lambda}\int_{[-1,1]^{2}}\int_{-\xi\sqrt{\lambda/\mu}}^{\xi\sqrt{\lambda/\mu}}K_{T}(\mu
y^{2}-\mu y_{2}^{3}-\nu y_{3}^{2})dydy_{2}dy_{3},$

on substituting $y_{1}=\sqrt{\mu/\lambda}y$. Since $\xi\sqrt{\lambda/\mu}\leq
2$ we obtain

$\displaystyle I(T)$ $\displaystyle\leq$
$\displaystyle\sqrt{\mu/\lambda}\int_{[-2,2]^{3}}K_{T}(\mu y^{2}-\mu
y_{2}^{3}-\nu y_{3}^{2})dydy_{2}dy_{3}$ $\displaystyle=$ $\displaystyle
8\sqrt{\mu/\lambda}\int_{[-1,1]^{3}}K_{T}(4\mu z_{1}^{2}-4\mu z_{2}^{3}-4\nu
z_{3}^{2})dz_{1}dz_{2}dz_{3}$ $\displaystyle=$ $\displaystyle
8\sqrt{\mu/\lambda}.(4\mu)^{-1}\int_{[-1,1]^{3}}K_{4\mu
T}(z_{1}^{2}-z_{2}^{3}-\nu\mu^{-1}z_{3}^{2})dz_{1}dz_{2}dz_{3},$

for $T$ sufficiently large. Similarly when $\lambda\geq\mu$ we have

$\displaystyle I(T)$ $\displaystyle\gg$
$\displaystyle\int_{[-1/2,1/2]^{2}}\int_{-\xi}^{\xi}K_{T}(\lambda
y_{1}^{2}-\mu y_{2}^{3}-\nu y_{3}^{2})dy_{1}dy_{2}dy_{3}$ $\displaystyle=$
$\displaystyle\sqrt{\mu/\lambda}\int_{[-1/2,1/2]^{2}}\int_{-\xi\sqrt{\lambda/\mu}}^{\xi\sqrt{\lambda/\mu}}K_{T}(\mu
y^{2}-\mu y_{2}^{3}-\nu y_{3}^{2})dydy_{2}dy_{3}.$

Since $\xi\sqrt{\lambda/\mu}\geq 1$ we obtain

$\displaystyle I(T)$ $\displaystyle\gg$
$\displaystyle\sqrt{\mu/\lambda}\int_{[-1/2,1/2]^{3}}K_{T}(\mu y^{2}-\mu
y_{2}^{3}-\nu y_{3}^{2})dydy_{2}dy_{3}$ $\displaystyle=$
$\displaystyle\tfrac{1}{8}\sqrt{\mu/\lambda}\int_{[-1,1]^{3}}K_{T}(\tfrac{1}{4}\mu
z_{1}^{2}-\tfrac{1}{4}\mu z_{2}^{3}-\tfrac{1}{4}\nu
z_{3}^{2})dz_{1}dz_{2}dz_{3}$ $\displaystyle=$ $\displaystyle
8\sqrt{\mu/\lambda}.(\tfrac{1}{4}\mu)^{-1}\int_{[-1,1]^{3}}K_{\tfrac{1}{4}\mu
T}(z_{1}^{2}-z_{2}^{3}-\nu\mu^{-1}z_{3}^{2})dz_{1}dz_{2}dz_{3}.$

Hence if we write

$J(T;\delta)=\int_{[-1,1]^{3}}K_{T}(z_{1}^{2}-z_{2}^{2}-\delta
z_{3}^{2})dz_{1}dz_{2}dz_{3}$

we have

$(\lambda\mu)^{-1/2}J(\tfrac{1}{4}\mu T;\nu\mu^{-1})\ll
I(T)\ll(\lambda\mu)^{-1/2}J(4\mu T;\nu\mu^{-1}).$

On taking the $\limsup$ as $T\to\infty$ this yields

$\lim_{T\to\infty}I(T)\asymp(\lambda\mu)^{-1/2}\limsup_{T\to\infty}J(T;\nu\mu^{-1})$
(30)

when $\lambda\geq\mu$.

Suppose next that $\mu\geq\lambda\geq\nu$. Then if $K_{T}(\lambda
y_{1}^{2}-\mu y_{2}^{3}-\nu y_{3}^{2})\not=0$ with $\mathbf{y}\in[-1,1]^{3}$
we have

$\mu y_{2}^{2}\leq\mu y_{2}^{2}+\nu y_{3}^{2}\leq T^{-1}+\lambda y_{1}^{2}\leq
T^{-1}+\lambda.$

It follows that $|y_{2}|\leq 2\sqrt{\lambda/\mu}$ as soon as
$T\geq\lambda^{-1}$. We may then replace the range $[-1,1]$ for $y_{2}$ by
$[-\xi,\xi]$ where $\xi=\min(1\,,\,2\sqrt{\lambda/\mu})$ this time. Proceeding
much as before we find that

$\displaystyle I(T)$ $\displaystyle\leq$
$\displaystyle\sqrt{\lambda/\mu}\int_{[-2,2]^{3}}K_{T}(\lambda
y_{1}^{2}-\lambda y^{3}-\nu y_{3}^{2})dy_{1}dydy_{3}$ $\displaystyle=$
$\displaystyle 8\sqrt{\lambda/\mu}.(4\lambda)^{-1}\int_{[-1,1]^{3}}K_{4\lambda
T}(z_{1}^{2}-z_{2}^{3}-\nu\lambda^{-1}z_{3}^{2})dz_{1}dz_{2}dz_{3}$

and

$I(T)\gg(\lambda\mu)^{-1/2}\int_{[-1,1]^{3}}K_{4\lambda
T}(z_{1}^{2}-z_{2}^{3}-\nu\lambda^{-1}z_{3}^{2})dz_{1}dz_{2}dz_{3}.$

Thus

$\lim_{T\to\infty}I(T)\asymp(\lambda\mu)^{-1/2}\limsup_{T\to\infty}J(T;\nu\lambda^{-1})$
(31)

when $\mu\geq\lambda\geq\nu$.

Thirdly we suppose that $\lambda\leq\nu$. In this case if
$\mathbf{y}\in[-1,1]^{3}$ and $K_{T}(\lambda y_{1}^{2}-\mu y_{2}^{3}-\nu
y_{3}^{2})\not=0$ we have $|y_{2}|\leq 2\sqrt{\lambda/\mu}$ and $|y_{3}|\leq
2\sqrt{\lambda/\nu}$ when $T\geq\lambda^{-1}$. We then replace the ranges for
$y_{2}$ and $y_{3}$ by by $[-\xi_{2},\xi_{2}]$ and $[-\xi_{3},\xi_{3}]$
respectively, where $\xi_{2}=\min(1\,,\,2\sqrt{\lambda/\mu})$ and
$\xi_{3}=\min(1\,,\,2\sqrt{\lambda/\nu})$. A similar argument to before then
shows that

$\lim_{T\to\infty}I(T)\asymp(\mu\nu)^{-1/2}\limsup_{T\to\infty}J(T;1)$ (32)

when $\nu\geq\lambda$.

It remains to consider $J(T;\delta)$, where $0<\delta\leq 1$. To obtain a
lower bound we restrict the variables $z_{1},z_{2}$ to the square given by
$|z_{1}+z_{2}|\leq 1$ and $|z_{1}-z_{2}|\leq 1$, which lies inside the region
$[-1,1]^{2}$. A suitable change of variable then shows that

$J(T;\delta)\geq\frac{1}{2}\int_{[-1,1]^{3}}K_{T}(u_{1}u_{2}-\delta
z_{3}^{2})du_{1}du_{2}dz_{3}.$

We now restrict $u_{1},u_{2}$ further, so that $u_{1},u_{2}\geq 0$ and
$\delta/4\leq u_{1}u_{2}\leq\delta/2$. For any such $u_{1},u_{2}$ and any
$T\geq 4\delta^{-1}$ the inequality

$|u_{1}u_{2}-\delta z^{2}|\leq 1/(2T)$

implies

$\delta z^{2}\leq u_{1}u_{2}+(2T)^{-1}\leq\delta,$

whence one automatically has $z\in[-1,1]$. Moreover it also implies

$\delta z^{2}\geq u_{1}u_{2}-(2T)^{-1}\geq\delta/8,$

whence $|z|\geq\tfrac{1}{3}$, say. It follows that one has

$|u_{1}u_{2}-\delta z_{3}^{2}|\leq 1/(2T)$

for an admissible set of values for $z_{3}$ of measure $\gg(\delta T)^{-1}$.
We therefore conclude that

$\displaystyle J(T;\delta)$ $\displaystyle\gg$
$\displaystyle\delta^{-1}\mathrm{Meas}\\{(u_{1},u_{2})\in[0,1]^{2}:\,\delta/4\leq
u_{1}u_{2}\leq\delta/2\\}$ (33) $\displaystyle\gg$
$\displaystyle\delta^{-1}\int_{2/\delta}^{1}\frac{\delta}{4u_{1}}du_{1}$
$\displaystyle\gg$ $\displaystyle\log(2/\delta).$

To obtain an upper bound for $J(T;\delta)$ we extend the range of
$z_{1},z_{2}$ to the square given by $|z_{1}+z_{2}|\leq 2$ and
$|z_{1}-z_{2}|\leq 2$. A suitable change of variable now shows that

$J(T;\delta)\leq\frac{1}{2}\int_{[-2,2]^{3}}K_{T}(u_{1}u_{2}-\delta
z_{3}^{2})du_{1}du_{2}dz_{3}.$

When $|u_{1}u_{2}|\leq 2/T$ the integrand is only non-zero when $z_{3}^{2}\leq
3/(\delta T)$, so that this range contributes

$\ll T\int_{-2}^{2}\min\left(1\,,\,\frac{1}{|u_{2}|T}\right)(\delta
T)^{-1}du_{2}\ll\delta^{-1/2}\frac{\log T}{\sqrt{T}}.$

The range $|u_{1}u_{2}|\leq 2/T$ therefore makes no contribution when we let
$T$ go to infinity. When $|u_{1}u_{2}|\geq 2/T$ the integrand is only non-zero
when $\sqrt{|u_{1}u_{2}|/(2\delta)}\leq|z_{3}|\leq 2$. Thus the set of values
for $z_{3}$ for which $|u_{1}u_{2}-\delta z_{3}^{2}|\leq T^{-1}$ consists of
at most two intervals, having total length
$O(T^{-1}(|u_{1}u_{2}|\delta)^{-1/2})$. Moreover the set is empty unless
$|u_{1}u_{2}|\leq 8\delta$.It follows that the corresponding contribution to
$J(T;\delta)$ is

$\displaystyle\ll$ $\displaystyle
T\int_{-2}^{2}\int_{|u_{1}|\leq\min(2\,,\,8\delta/|u_{2}|)}T^{-1}(|u_{1}u_{2}|\delta)^{-1/2}du_{1}du_{2}$
$\displaystyle\ll$
$\displaystyle\delta^{-1/2}\int_{-2}^{2}\frac{\min(2\,,\,8\delta/|u_{2}|)^{1/2}}{|u_{2}|^{1/2}}du_{2}$
$\displaystyle\ll$ $\displaystyle\log(2/\delta).$

In view of the lower bound (33) we therefore have
$J(T;\delta)\asymp\log(2/\delta)$, and the second claim of Lemma 11 then
follows from (30), (31) and (32).

For the third claim of the lemma we note that

$\min(\lambda,\mu,\nu)\geq\frac{\lambda\mu\nu)}{\max(\lambda,\mu,\nu)}\gg\frac{\Delta}{||q||^{2}}.$

This produces the lower bound for $\sigma_{\infty}(q,w_{0})$. For the upper
bound we observe for example that when $\lambda\geq\mu\geq\nu$ we have

$\min(\lambda,\mu,\nu)^{1/2}\log(2\mu/\nu)=\nu^{1/2}\log(2\mu/\nu)\ll\nu^{1/2}(\mu\nu)^{1/4}$

and that

$(\mu\nu)^{1/4}\ll\Delta^{1/4}\lambda^{-1/4}\ll\Delta^{1/4}||q||^{-1/4}.$

Thus when $\lambda\geq\mu\geq\nu$ we have

$\sigma_{\infty}(q;w_{0})\ll\Delta^{-1/4}||q||^{-1/4}=\rho^{1/4}||q||^{-1}.$

When $\mu\geq\lambda\nu$ or $\mu\geq\nu\geq\lambda$ we may argue similarly.
This suffices to complete the proof of the lemma. ∎

##  8\. Theorem 5 — The error term

In Lemma 9 the contribution from $\mathbf{a}\not=\mathbf{0}$ will produce an
error term, as we now show.

######  Lemma 12.

Let $\Lambda\subseteq\mathbb{Z}^{2}$ be a 2-dimensional lattice, and let
$\mathbf{n}_{1},\mathbf{n}_{2}$ be a basis for $\Lambda$ chosen so that
$||\mathbf{n}_{1}||\cdot||\mathbf{n}_{2}||\ll\mathrm{d}(\Lambda)$. Then if
$N=(\mathbf{n}_{1}|\mathbf{n}_{2})$ we have $\Lambda=N(\mathbb{Z}^{2})$.
Moreover for any integer $K\geq 0$ we have

$I(\mathbf{a},M_{k},N)\ll_{w,K}BD_{k}^{-1/2}||q||^{1/2}\left(\frac{B_{0}}{B}\right)^{K/2}||\mathbf{a}||^{-K},$

with

$B_{0}=\rho\,\mathrm{d}(\Lambda)^{2}||\mathbf{c}_{3}||.$

###### Proof.

Let $N^{-1}\mathbf{a}=(b_{1},b_{2})^{T}$ and suppose that
$|b_{1}|\geq|b_{2}|$, say. If we integrate by parts $K$ times with respect to
$x_{1}$ we find that

$I(\mathbf{a},M_{k},N)\ll_{K}|b_{1}|^{-K}\int_{\mathbb{R}^{2}}\left|\frac{\partial^{K}}{\partial
x_{1}^{K}}w(B^{-1}M_{k}\mathbf{x}^{2})\right|dx_{1}dx_{2}.$

Since $\mathbf{a}=N(b_{1},b_{2})^{T}$ we have

$||\mathbf{a}||\ll\max(||\mathbf{n}_{1}||,||\mathbf{n}_{2}||)\max(|b_{1}|,|b_{2}|)\ll\mathrm{d}(\Lambda)|b_{1}|,$

whence

$I(\mathbf{a},M_{k},N)\ll_{K}||\mathbf{a}||^{-K}\mathrm{d}(\Lambda)^{K}\int_{\mathbb{R}^{2}}\left|\frac{\partial^{K}}{\partial
x_{1}^{K}}w(B^{-1}M_{k}\mathbf{x}^{2})\right|dx_{1}dx_{2}.$

We will write the components of $B^{-1}M_{k}\mathbf{x}^{2}$ as
$f_{1}(x_{1}),f_{2}(x_{1}),f_{3}(x_{1})$, where the $f_{i}$ are quadratic
polynomials which also involve $x_{2}$. Then the $K$-th order partial
derivatives of $w(B^{-1}M_{k}\mathbf{x}^{2})$ with respect to $x_{1}$ will be
sums of various terms $T_{n}$. Each $T_{n}$ will be a product containing a
single partial derivative of $w$, of order at most $K$, along with various
first and second derivatives of the $f_{i}$. If there are $r$ first
derivatives and $s$ second derivatives then $r+2s=K$. It therefore follows
that

$\frac{\partial^{K}}{\partial
x_{1}^{K}}w(B^{-1}M_{k}\mathbf{x}^{2})\ll_{w,K}F_{1}^{r}F_{2}^{s}$

for some exponents with $r+2s=K$, where

$F_{1}=\sup\\{|f_{i}^{\prime}(x_{1})|:\,w(B^{-1}M_{k}\mathbf{x}^{2})\not=0\\},$

and $F_{2}$ is the maximum of $|f_{1}^{\prime\prime}|,|f_{2}^{\prime\prime}|$
and $|f_{3}^{\prime\prime}|$.

The leading coefficient of $Bf_{i}$ will be the $i$-th entry in the first
column of $M_{k}$, so that its modulus will be at most $||\mathbf{c}_{1}||$,
in the notation of Theorem 2. It follows that $F_{2}\leq
2||\mathbf{c}_{1}||/B$. Similarly the coefficient of $x_{1}$ in $Bf_{i}$ will
have modulus at most $||\mathbf{c}_{2}||$, so that

$F_{1}\leq X_{1}||\mathbf{c}_{1}||/B+X_{2}||\mathbf{c}_{2}||/B.$

It follows via Lemma 7 that

$\displaystyle\max_{r+2s=K}F_{1}^{r}F_{2}^{s}$ $\displaystyle\leq$
$\displaystyle F_{1}^{K}+F_{2}^{K/2}$ $\displaystyle\ll_{K}$
$\displaystyle(X_{1}||\mathbf{c}_{1}||/B)^{K}+(X_{2}||\mathbf{c}_{2}||/B)^{K}+(||\mathbf{c}_{1}||/B)^{K/2}$
$\displaystyle\ll_{K}$ $\displaystyle
B^{-K/2}\\{||\mathbf{r}_{1}||^{K/2}||\mathbf{c}_{1}||^{K}+||\mathbf{r}_{3}||^{K/2}||\mathbf{c}_{2}||^{K}+||\mathbf{c}_{1}||^{K/2}\\}$
$\displaystyle\ll_{K}$ $\displaystyle B^{-K/2}E^{K/2},$

with

$E=||\mathbf{r}_{1}||\cdot||\mathbf{c}_{1}||^{2}+||\mathbf{r}_{3}||\cdot||\mathbf{c}_{2}||^{2}+||\mathbf{c}_{1}||.$

Lemma 7 shows that the support of $w(B^{-1}M\mathbf{x}^{2})$ is included in a
rectangle of area $O(BD_{k}^{-1/2}||q||^{1/2})$, and we therefore conclude
that

$I(\mathbf{a},M_{k},N)\ll_{w,K}\frac{B||q||^{1/2}}{D_{k}^{1/2}}\left\\{\mathrm{d}(\Lambda)^{2}B^{-1}E\right\\}^{K/2}||\mathbf{a}||^{-K}.$
(34)

We now claim that

$E\ll\rho\,||\mathbf{c}_{1}||\ll\rho\,||\mathbf{c}_{3}||$ (35)

when $|b_{1}|\geq|b_{2}|$, as we are currently supposing. In the alternative
case the argument is completely analogous, leading to exactly the same bound.
To establish our claim we use (23), (7) and (5) to show that

$\displaystyle||\mathbf{r}_{1}||\cdot||\mathbf{c}_{1}||^{2}$
$\displaystyle\leq$
$\displaystyle\det(M_{k})||\mathbf{r}_{1}||\cdot||\mathbf{r}_{2}||\cdot||\mathbf{r_{3}}||\cdot||\mathbf{c}_{1}||$
$\displaystyle\leq$ $\displaystyle
9\sqrt{10}\det(M_{k})||q||^{3/2}D_{k}^{-3/2}||\mathbf{c}_{1}||$
$\displaystyle\ll$ $\displaystyle\rho^{1/2}||\mathbf{c}_{1}||.$

This is sufficient for the term
$||\mathbf{r}_{1}||\cdot||\mathbf{c}_{1}||^{2}$, since $\rho\geq 2$ by Theorem
3. Secondly, the bounds (26), (7), (9) and (5) yield

$\displaystyle||\mathbf{r}_{3}||\cdot||\mathbf{c}_{2}||^{2}$
$\displaystyle\leq$
$\displaystyle||\mathbf{r}_{1}||\cdot||\mathbf{c}_{1}||\cdot||\mathbf{r}_{3}||\cdot||\mathbf{c}_{2}||^{2}$
$\displaystyle\ll$ $\displaystyle
D_{k}^{-1}||q||\cdot||\mathbf{c}_{1}||\cdot||\mathbf{c}_{2}||^{2}$
$\displaystyle\ll$
$\displaystyle\det(M_{k})^{2}D_{k}^{-3}||q||^{3}||\mathbf{c}_{1}||$
$\displaystyle\ll$ $\displaystyle\rho||\mathbf{c}_{1}||,$

which is sufficient for the term
$||\mathbf{r}_{3}||\cdot||\mathbf{c}_{2}||^{2}$. Finally, since $\rho\geq 2$
we have $||\mathbf{c}_{1}||\leq\rho||\mathbf{c}_{1}||$. This give us the
required estimate (35) for $E$, whence (34) produces the bound

$I(\mathbf{a},M_{k},N)\ll_{w,K}||\mathbf{a}||^{-K}\frac{B||q||^{1/2}}{D_{k}^{1/2}}\left\\{\frac{\mathrm{d}(\Lambda)^{2}}{B}\rho\,||\mathbf{c}_{3}||\right\\}^{K/2}.$

The lemma now follows. ∎

We can now summarize the results of our analysis of $S(\Lambda,B,M_{k})$.

######  Lemma 13.

We have $S(\Lambda,B,M_{k})=0$ if $B<1$, and otherwise

$S(\Lambda,B,M_{k})\ll_{w}D_{k}^{-1/2}||q||^{1/2}\left\\{\frac{B}{\mathrm{d}(\Lambda)}+B^{1/2}||\mathbf{c}_{3}||^{1/2}\right\\}.$

Moreover

$\displaystyle S(\Lambda,B,M_{k})$ $\displaystyle=$
$\displaystyle\sigma_{\infty}(q;w)\frac{\Delta^{1/2}}{D_{k}^{1/2}\mathrm{d}(\Lambda)}B+O_{w}(1)$
(36) $\displaystyle\hskip
28.45274pt{}+O_{w,K}(D_{k}^{-1/2}||q||^{1/2}B(B_{0}/B)^{K}),$

for any integer $K\geq 2$, with

$B_{0}=\rho\,\mathrm{d}(\Lambda)^{2}||\mathbf{c}_{3}||.$

###### Proof.

The first half is the content of Lemma 8, while the second follows from Lemma
9, together with Lemma 10 for the term $\mathbf{a}=\mathbf{0}$, and Lemma 12
for $\mathbf{a}\not=\mathbf{0}$. Here we replace $K$ by $2K$ and observe that

$\sum_{\begin{subarray}{c}\mathbf{a}\in\mathbb{Z}^{2}\\\
\mathbf{a}\not=0\end{subarray}}||\mathbf{a}||^{-2K}\ll_{K}1$

for $K\geq 2$. ∎

##  9\. Completing the proof of Theorem 5

To prove Theorem 5 we will apply Lemma 6, using the crude upper bound from
Lemma 13 when $\mathrm{d}(\Lambda^{(j)})$ or $d$ is large, and the asymptotic
estimate (36) otherwise. We therefore begin by choosing a real parameter
$d_{0}\geq 1$, which we will specify later, and noting that we can restrict
attention to the range $d\leq\sqrt{B}$, by virtue of the first clause of Lemma
13.

The contribution to $N(B;\mathcal{C}_{k})$ from terms with $d_{0}\leq
d\leq\sqrt{B}$, summed over all the lattices $\Lambda^{(j)}$, will be

$\displaystyle\ll_{w}$ $\displaystyle
3^{\omega(\Delta)}\frac{||q||^{1/2}}{D_{k}^{1/2}}\sum_{d_{0}\leq
d\leq\sqrt{B}}\left\\{Bd^{-2}+B^{1/2}d^{-1}||\mathbf{c}_{3}||^{1/2}\right\\}$
$\displaystyle\ll_{w}$ $\displaystyle
3^{\omega(\Delta)}\frac{||q||^{1/2}}{D_{k}^{1/2}}\left\\{Bd_{0}^{-1}+B^{1/2}(\log
B)||\mathbf{c}_{3}||^{1/2}\right\\}.$

Similarly, the contribution from terms with $\mathrm{d}(\Lambda^{(j)})\geq
d_{0}$ and $d\leq d_{0}$ will be

$\displaystyle\ll_{w}$
$\displaystyle\frac{||q||^{1/2}}{D_{k}^{1/2}}\sum_{\begin{subarray}{c}j=1\\\
\mathrm{d}(\Lambda^{(j)})\geq d_{0}\end{subarray}}^{J}\sum_{d\leq
d_{0}}\left\\{\frac{Bd^{-2}}{\mathrm{d}(\Lambda^{(j)})}+B^{1/2}d^{-1}||\mathbf{c}_{3}||^{1/2}\right\\}$
$\displaystyle\ll_{w}$ $\displaystyle
3^{\omega(\Delta)}\frac{||q||^{1/2}}{D_{k}^{1/2}}\left\\{Bd_{0}^{-1}+B^{1/2}(\log
B)||\mathbf{c}_{3}||^{1/2}\right\\}.$

We now examine the terms for which both $d\leq d_{0}$ and
$\mathrm{d}(\Lambda^{(j)})\leq d_{0}$. We have

$\sum_{\begin{subarray}{c}j=1\\\ \mathrm{d}(\Lambda^{(j)})\geq
d_{0}\end{subarray}}^{J}\frac{1}{\mathrm{d}(\Lambda^{(j)})}\,\sum_{d=1}^{\infty}\frac{1}{d^{2}}\ll
3^{\omega(\Delta)}d_{0}^{-1},$

and

$\sum_{j=1}^{J}\frac{1}{\mathrm{d}(\Lambda^{(j)})}\,\sum_{d\geq
d_{0}}\frac{1}{d^{2}}\ll 3^{\omega(\Delta)}d_{0}^{-1}.$

Hence, when we use the asymptotic formula (36) for terms in which both $d$ and
$\mathrm{d}(\Lambda^{(j)})$ are at most $d_{0}$, the main term contributes

$\tfrac{1}{2}\sigma_{\infty}(q;w)\frac{\Delta^{1/2}}{D_{k}^{1/2}}B\left\\{\kappa+O(3^{\omega(\Delta)}d_{0}^{-1})\right\\},$

where

$\kappa=\sum_{j=1}^{J}\lambda\left(\mathrm{d}(\Lambda^{(j)})\right)\sum_{\begin{subarray}{c}d=1\\\
(d,\Delta_{1}\Delta_{2})=1\end{subarray}}^{\infty}\mu(d)d^{-2}=\frac{6}{\pi^{2}}\prod_{p\mid\Delta_{1}}\frac{1}{1+p^{-1}}\prod_{p\mid\Delta_{2}}\frac{1-p^{-1}}{1+p^{-1}},$

by Lemma 6. Using the estimate from Lemma 11 we see that the $O$-term above
contributes

$\ll_{w}3^{\omega(\Delta)}\frac{\rho^{1/4}}{||q||}\frac{\Delta^{1/2}}{D_{k}^{1/2}}\frac{B}{d_{0}}=3^{\omega(\Delta)}\rho^{-1/4}\frac{||q||^{1/2}}{D_{k}^{1/2}}\frac{B}{d_{0}}\leq
3^{\omega(\Delta)}\frac{||q||^{1/2}}{D_{k}^{1/2}}\frac{B}{d_{0}}.$

On the other hand, the error term $O_{w}(1)$ in (36) contributes
$O_{w}(d_{0}^{2})$ while the second error term contributes

$\displaystyle\ll_{w,K}$
$\displaystyle\frac{||q||^{1/2}}{D_{k}^{1/2}}B(B_{1}/B)^{K}\sum_{\begin{subarray}{c}j=1\\\
\mathrm{d}(\Lambda^{(j)})\leq d_{0}\end{subarray}}^{J}\sum_{d\leq
d_{0}}\mathrm{d}(\Lambda^{(j)})^{2K}d^{2K-2}$ $\displaystyle\ll_{w,K}$
$\displaystyle\frac{||q||^{1/2}}{D_{k}^{1/2}}B(B_{1}/B)^{K}3^{\omega(\Delta)}d_{0}^{4K-1},$

with

$B_{1}=\rho\,||\mathbf{c}_{3}||.$

Thus, if we assume that $B\geq B_{1}$ we may take $d_{0}=B^{1/4}B_{1}^{-1/4}$
and $K=2$ so that the total of all the above error terms is

$\ll_{w}d_{0}^{2}+3^{\omega(\Delta)}\frac{||q||^{1/2}}{D_{k}^{1/2}}\left\\{\frac{B}{d_{0}}+B^{1/2}(\log
B)||\mathbf{c}_{3}||^{1/2}\right\\}.$

In the notation of Theorem 3 we have

$||\mathbf{c}_{3}||^{2}\geq||\mathbf{z}_{1}||\cdot||\mathbf{z}_{2}||\geq
D_{k}/||q||,$ (37)

whence

$d_{0}^{2}=\frac{B^{1/2}}{B_{1}^{1/2}}\ll\frac{B^{1/2}}{||\mathbf{c}_{3}||^{1/2}}\ll\frac{||q||^{1/2}}{D_{k}^{1/2}}B^{1/2}||\mathbf{c}_{3}||^{1/2}.$

Thus the error term $d_{0}^{2}$ above is dominated by the final term. Moreover

$\frac{B}{d_{0}}=B^{3/4}B_{1}^{1/4}\gg
B^{1/2}\max\\{||\mathbf{c}_{1}||\,,\,||\mathbf{c}_{3}||\\}^{1/2}$

when $B\geq B_{1}$. We therefore deduce that

$N(B;\mathcal{C}_{k})=\tfrac{1}{2}\sigma_{\infty}(q,w)\kappa\frac{\Delta^{1/2}}{D_{k}^{1/2}}B+O_{w}\left(3^{\omega(\Delta)}\frac{||q||^{1/2}}{D_{k}^{1/2}}B_{1}^{1/4}B^{3/4}\log
B\right)$

for $B\geq B_{1}$.

When $\rho^{-1}||\mathbf{c}_{3}||\ll B\leq B_{1}$ we argue as above with
$d_{0}=1$, showing that

$N(B;\mathcal{C}_{k})\ll_{w}1+3^{\omega(\Delta)}\frac{||q||^{1/2}}{D_{k}^{1/2}}\left\\{B+B^{1/2}(\log
B)||\mathbf{c}_{3}||^{1/2}\right\\}.$

However for $B\leq B_{1}$ we have

$\sigma_{\infty}(q,w)\kappa\frac{\Delta^{1/2}}{D_{k}^{1/2}}B\ll
3^{\omega(\Delta)}\frac{||q||^{1/2}}{D_{k}^{1/2}}B_{1}^{1/4}B^{3/4}$

by Lemma 11, and $B\ll B_{1}^{1/4}B^{3/4}$. Moreover
$B^{1/2}||\mathbf{c}_{3}||^{1/2}\ll B_{1}^{1/4}B^{3/4}$ when
$B\gg\rho^{-1}||\mathbf{c}_{3}||$. It follows that

$\displaystyle N(B;\mathcal{C}_{k})$ $\displaystyle=$
$\displaystyle\tfrac{1}{2}\sigma_{\infty}(q,w)\kappa\frac{\Delta^{1/2}}{D_{k}^{1/2}}B+O_{w}(1)$
(38) $\displaystyle\hskip
14.22636pt{}+O_{w}\left(3^{\omega(\Delta)}\frac{||q||^{1/2}}{D_{k}^{1/2}}B_{1}^{1/4}B^{3/4}\log
B\right)$

in the range $\rho^{-1}||\mathbf{c}_{3}||\ll B\leq B_{1}$. On the other hand,
if we have $B<(90\rho)^{-1}||\mathbf{c}_{3}||$ then $||\mathbf{x}||\leq B$
implies $||\mathbf{x}||<||\mathbf{z}_{2}||$, in the notation of Theorem 3, so
that $N(B;\mathcal{C}_{k})$ counts at most the points $\pm\mathbf{z}_{1}$. In
this case we will have $N(B;\mathcal{C}_{k})\ll_{w}1$, and hence (38) holds
for $B<(90\rho)^{-1}||\mathbf{c}_{3}||$ too. To complete the proof of Theorem
5 it remains to observe that $\kappa\gg(3/4)^{\omega(\Delta)}$, that
$\sigma_{\infty}(q,w)\gg||q||^{-1}$, by Lemma 11, and that
$||\mathbf{c}_{3}||\ll\rho\,||\mathbf{z}_{2}||$, by Theorem3.

## References

  * [1] J.W.S. Cassels, Bounds for the least solutions of homogeneous quadratic equations, Proc. Cambridge Philos. Soc., 51 (1955), 262–264.
  * [2] H. Davenport, Homogeneous quadratic equations, Mathematika, 18 (1971), 1–4.
  * [3] D.R. Heath-Brown, A new form of the circle method, and its application to quadratic forms, J. reine angew. Math., 481 (1996), 149–206.
  * [4] E. Peyre, Hauteurs et mesures de Tamagawa sur les variétés de Fano, Duke Math. J., 79 (1995), 101–218.
  * [5] E. Sofos, Uniformly counting rational points on conics, Acta Arith., 166 (2014), 1–14.