# The density of integral quadratic forms having a
$k$-dimensional totally isotropic subspace

Lycka Drakengren and Tom Fisher

(20th December 2022)

###### Abstract

We investigate the probability that a random quadratic form in
${\mathbb{Z}}[x_{1},...,x_{n}]$ has a totally isotropic subspace of a given
dimension. We show that this global probability is a product of local
probabilities. Our main result computes these local probabilities for
quadratic forms over the $p$-adics. The formulae we obtain are rational
functions in $p$ invariant upon substituting $p\mapsto 1/p$.

##  1 Introduction

An integral quadratic form in $n$ variables is a homogeneous polynomial of
degree $2$

$Q(x_{1},x_{2},\ldots,x_{n})=\sum_{1\leqslant i\leqslant j\leqslant
n}a_{ij}x_{i}x_{j}$ (1)

where the coefficients $a_{ij}$ belong to ${\mathbb{Z}}$. A non-zero vector
$(x_{1},x_{2},\ldots,x_{n})\in V={\mathbb{Q}}^{n}$ is called isotropic if
$Q(x_{1},x_{2},\ldots,x_{n})=0$. A subspace of $V$ is totally isotropic if all
its non-zero vectors are isotropic. We say that $Q$ is $k$-isotropic if $V$
has a $k$-dimensional totally isotropic subspace. A quadratic form that is
$1$-isotropic is simply called isotropic.

Generalising the results of [2], where only the case $k=1$ was considered, we
investigate the probability $\rho_{\rm glob}(k,n)$ that a random integral
quadratic form in $n$ variables is $k$-isotropic. More formally we define

$\rho_{\rm
glob}(k,n)=\lim_{H\to\infty}\frac{\\#\left\\{\begin{array}[]{c}\text{
quadratic forms $Q=\textstyle\sum
a_{ij}x_{i}x_{j}\in{\mathbb{Z}}[x_{1},\ldots,x_{n}]$}\\\ \text{ with
$|a_{ij}|\leqslant H$ that are $k$-isotropic over ${\mathbb{Q}}$
}\end{array}\right\\}}{(2H)^{n(n+1)/2}}$ (2)

if this limit exists.

Combining the Strong Hasse Principle [3, p. 75] and Witt’s Cancellation
Theorem (see Theorem 2.7), we know that an integral quadratic form is
$k$-isotropic over ${\mathbb{Q}}$ if and only if it is $k$-isotropic over
${\mathbb{Q}}_{p}$ for all primes $p$ and over $\mathbb{R}$. Applying a
theorem of Poonen and Stoll [7] we deduce the following result.

######  Theorem 1.1.

The probability $\rho_{\rm glob}(k,n)$ that a random integral quadratic form
in $n$ variables is $k$-isotropic exists and is given by

$\rho_{\rm glob}(k,n)=\rho_{\infty}(k,n)\prod_{p}\rho_{p}(k,n)$

where the product is over all primes $p$, and the local contributions are the
probabilities of $k$-isotropy over ${\mathbb{R}}$ and over ${\mathbb{Q}}_{p}$,
defined as in (2) but with ${\mathbb{Q}}$ replaced by ${\mathbb{R}}$ or
${\mathbb{Q}}_{p}$ as appropriate.

We fix a prime number $p$. The probability $\rho_{p}(k,n)$ may also be
interpreted as the probability that a random $p$-adic integral quadratic form
in $n$ variables is $k$-isotropic over ${\mathbb{Q}}_{p}$. Here, by a random
$p$-adic integral quadratic form, we mean a quadratic form with coefficients
in ${\mathbb{Z}}_{p}$ where the coefficients are chosen independently at
random according to Haar measure. Choosing the coefficient
$a_{ij}\in{\mathbb{Z}}_{p}$ with respect to Haar measure means that each
congruence class modulo $p$ is equally likely, and inductively for $n>1$, the
classes

$a,\>a+p^{n-1},\>a+2p^{n-1},\>\ldots,\>a+(p-1)p^{n-1}\mod p^{n}$

are equally likely where $0\leqslant a\leqslant p^{n-1}-1$ is the reduction of
$a_{ij}$ modulo $p^{n-1}$.

We now state our main theorem. It extends [2, Theorem 1.2] which treated the
case $k=1$.

######  Theorem 1.2.

The probability $\rho_{p}(k,n)$ that a random $p$-adic integral quadratic form
in $n$ variables is $k$-isotropic over ${\mathbb{Q}}_{p}$ is given by

$\rho_{p}(k,n)=\left\\{\begin{array}[]{ll}\vspace{1ex}0&\text{ if }n\leqslant
2k-1;\\\
\vspace{1ex}\frac{1}{4}\cdot(p^{k}+1)\cdot\left(\frac{p^{k+2}-1}{(p+1)(p^{2k+1}-1)}+\prod_{i=1}^{k}\left(\frac{p^{2i-1}-1}{p^{2i}-1}\right)\right)&\text{
if }n=2k;\\\
\vspace{1ex}\frac{1}{2}+\frac{1}{2}\cdot(p^{k+1}+1)\cdot\prod_{i=1}^{k+1}\left(\frac{p^{2i-1}-1}{p^{2i}-1}\right)&\text{
if }n=2k+1;\\\
\vspace{1ex}1-\frac{1}{4}\cdot(p^{k+1}+1)\cdot\left(\frac{p^{k+3}-1}{(p+1)(p^{2k+3}-1)}-\prod_{i=1}^{k+1}\left(\frac{p^{2i-1}-1}{p^{2i}-1}\right)\right)&\text{
if }n=2k+2;\\\ 1&\text{ if }n\geqslant 2k+3.\end{array}\right.$

Combining Theorems 1.1 and 1.2 we deduce the following.

######  Corollary 1.3.

We have $\rho_{\rm glob}(k,n)=0$ for all $n\leqslant 2k+1$,

$\rho_{\rm
glob}(k,2k+2)=\rho_{\infty}(k,2k+2)\cdot\prod_{p}\left(1-\frac{p^{k+1}+1}{4}\cdot\left(\frac{p^{k+3}-1}{(p+1)(p^{2k+3}-1)}-\prod_{r=1}^{k+1}\frac{p^{2r-1}-1}{p^{2r}-1}\right)\right)$

and $\rho_{\rm glob}(k,n)=\rho_{\infty}(k,n)$ for all $n\geqslant 2k+3$.

We note two striking features of the formulae in Theorem 1.2. The first is
that they are rational functions in $p$, where the same rational function
works for all primes $p$ including $p=2$. The second is that the rational
functions are invariant upon substituting $p\mapsto 1/p$. Exactly the same two
observations were made in [1] in connection with roots of polynomials in one
variable. Moreover in that paper the substitution $p\mapsto 1/p$ also related
two auxiliary probabilities appearing in the recursion, denoted there by
$\alpha$ and $\beta$. We find that an analogous statement holds in our case;
see Corollary 5.8.

We employ two strategies for proving Theorem 1.2. The first is a direct
generalisation of the method in [2] (which only treated the case $k=1$), with
the additional idea of splitting off hyperbolic planes (see Definition 2.5).
This leads to recursive formulae that may be used to compute $\rho_{p}(k,n)$
for any given $k$ and $n$, and also show that the answer is always a rational
function in $p$. However it is still hard to write the answers uniformly in
$k$, as we do in the statement of Theorem 1.2.

The second strategy is to deduce Theorem 1.2 from a theorem of Kovaleva [6],
who computed the probability that a random $p$-adic integral quadratic form in
$n$ variables belongs to a given ${\mathbb{Q}}_{p}$-equivalence class of
quadratic forms. The answers she obtained are not rational functions in $p$,
do not exhibit the $p\leftrightarrow 1/p$ symmetries, and do not explicitly
cover the case $p=2$, where in any case it makes a difference whether we
consider random quadratic forms or random symmetric matrices. However her work
leads to a proof of Theorem 1.2 when $p$ is odd. Since we already know from
the first strategy that the answer is a rational function in $p$ it follows
that the theorem is also true when $p=2$.

Both strategies work by dividing into cases according to the
${\mathbb{F}}_{p}$-equivalence class of the quadratic form reduced mod $p$,
and from this obtaining recursive formulae for the probabilities. One
difference, not already noted above, is that in the second strategy the
quadratic form is diagonalised, whereas in the first we split off hyperbolic
planes, and so allow $2\times 2$ blocks on the diagonal.

In Section 2 we review some background on quadratic forms. In Section 3 we
discuss the global applications of our work, and in particular explain how
Theorem 1.1 and Corollary 1.3 follow from Theorem 1.2. In Section 4 we prove
some results on counting quadratic forms over finite fields, in preparation
for our first strategy for proving Theorem 1.2. The two strategies are
explained in Sections 5 and 6 respectively. Finally in Appendix A we adapt the
methods of Kovaleva to solve the recurrence relations in our first method
directly.

This work originated as a summer project carried out by the first author and
supervised by the second author. We thank the Trinity Summer Studentship
Scheme and the Research in the CMS Programme for their support.

##  2 Background on quadratic forms

We collect together some standard definitions and results on quadratic forms.
See Cassels [3] for further details. We write ${\mathbb{F}}$ for a general
field, and $V$ for a finite dimensional vector space over ${\mathbb{F}}$.

######  Definition 2.1.

A quadratic form of dimension $n$ over ${\mathbb{F}}$ is a polynomial

$Q(x_{1},x_{2},\ldots,x_{n})=\sum_{1\leqslant i\leqslant j\leqslant
n}a_{ij}x_{i}x_{j},$ (3)

where the coefficients $a_{ij}$ for $1\leqslant i\leqslant j\leqslant n$
belong to ${\mathbb{F}}$. We may also consider $Q$ as a function
$V\to{\mathbb{F}}$ where $V={\mathbb{F}}^{n}$, and refer to the pair $(V,Q)$
as a quadratic space. The corresponding symmetric bilinear form $\phi:V\times
V\to{\mathbb{F}}$ is given by

$\phi(x,y)=Q(x+y)-Q(x)-Q(y),$

where $x=(x_{1},x_{2},\ldots,x_{n})$ and $y=(y_{1},y_{2},\ldots,y_{n})$.

We refer to properties of a quadratic space $(V,Q)$ and properties of $V$ or
$Q$ interchangeably.

######  Definition 2.2.

Let $(V,Q)$ be a quadratic space and let $\phi$ be its associated symmetric
bilinear form. The radical of $(V,Q)$, when not over a field of characteristic
$2$, is the vector space consisting of vectors $x\in V$ such that
$\phi(x,y)=0$ for all $y\in V$. In characteristic $2$, we further require that
$Q(x)=0$. A quadratic space is regular if its radical is zero-dimensional and
singular otherwise. The rank of the quadratic form $Q$ is $n-r$, where $r$ is
the dimension of the radical.

######  Definition 2.3.

Quadratic spaces $(V_{1},Q_{1})$ and $(V_{2},Q_{2})$ over the same field
${\mathbb{F}}$ are isometric if there is a linear isomorphism
$T:V_{1}\rightarrow V_{2}$ such that $Q_{1}(x)=Q_{2}(Tx)$ for all $x\in
V_{1}$. In this situation, the forms $Q_{1}$ and $Q_{2}$ are said to be
equivalent over ${\mathbb{F}}$. In other words, quadratic forms over
${\mathbb{F}}$ are equivalent if they are related by a linear substitution
given by a matrix $P\in\operatorname{GL}_{n}({\mathbb{F}})$. This defines an
equivalence relation on the set of quadratic forms with coefficients in
${\mathbb{F}}$. More generally, if $R\subset{\mathbb{F}}$ is a subring, then
we say that quadratic forms are equivalent over $R$ (or $R$-equivalent) if
they are related by a matrix $P\in\operatorname{GL}_{n}(R)$.

The next two definitions are closely related. The first naturally extends the
definitions we already gave in the introduction in the case
${\mathbb{F}}={\mathbb{Q}}$.

######  Definition 2.4.

Let $(V,Q)$ be a quadratic space. A non-zero vector $x\in V$ is called
isotropic if $Q(x)=0$. A quadratic space $(V,Q)$ is isotropic if $V$ contains
an isotropic vector, and totally isotropic if all its non-zero vectors are
isotropic. If $V$ has a subspace $V_{0}$ of dimension $k$ such that the
quadratic space $(V_{0},Q)$ is totally isotropic, then we say that the
quadratic space $(V,Q)$ is $k$-isotropic. In particular, a quadratic space is
$1$-isotropic if and only if it is isotropic.

######  Definition 2.5.

A hyperbolic plane is a quadratic space $(V,Q)$ of dimension $2$ where $Q$ is
equivalent over ${\mathbb{F}}$ to the form $q(x_{1},x_{2})=x_{1}x_{2}$.

######  Lemma 2.6.

A regular quadratic space $(V,Q)$ is isotropic if and only if $V$ has a
subspace $V_{0}$ such that the quadratic space $(V_{0},Q)$ is a hyperbolic
plane.

###### Proof.

See [3, p. 15]. ∎

We now introduce some theorems that will be useful for finding isotropic
spaces.

######  Theorem 2.7 (Witt’s Cancellation Theorem).

Let $(V,Q)$ be a quadratic space. Let $V_{1}$, $V_{2}$ be subspaces of $V$.
Denote by $V_{1}^{\perp}$ and $V_{2}^{\perp}$ the orthogonal complements of
$V_{1}$ resp. $V_{2}$ in $V$. If $(V_{1},Q)$ and $(V_{2},Q)$ are regular and
isometric, then $(V_{1}^{\perp},Q)$ and $(V_{2}^{\perp},Q)$ are also
isometric.

###### Proof.

See [5, pp. 89–92] for quadratic forms over a field of characteristic not $2$,
and [5, p. 118] for the case of characteristic $2$. ∎

Let $p$ be a prime. We write ${\mathbb{F}}_{p}$ for the finite field with $p$
elements, ${\mathbb{Q}}_{p}$ for the field of $p$-adic numbers, and
${\mathbb{Z}}_{p}$ for the ring of $p$-adic integers. The following argument
is one we will revisit in the proof of Lemma 5.2.

######  Lemma 2.8.

Let $Q$ be a quadratic form with coefficients in ${\mathbb{Z}}_{p}$ that
reduces over ${\mathbb{F}}_{p}$ to a form that is both $k$-isotropic and
regular. Then $Q$ is $k$-isotropic (and regular) over ${\mathbb{Q}}_{p}$.

###### Proof.

The case $k=1$ is a consequence of Hensel’s lemma. In fact if we go via Lemma
2.6 then we only need Hensel’s lemma for a quadratic polynomial in one
variable. For general $k>1$ we proceed by induction on $k$. Once we know that
$Q$ is isotropic, we may assume via a ${\mathbb{Z}}_{p}$-equivalence that

$Q(x_{1},\ldots,x_{n})=x_{1}x_{2}+Q^{\prime}(x_{3},\ldots,x_{n})$

for some quadratic form $Q^{\prime}$ with coefficients in ${\mathbb{Z}}_{p}$.
The reduction of $Q^{\prime}$ mod $p$ is then $(k-1)$-isotropic (and regular)
over ${\mathbb{F}}_{p}$ by Theorem 2.7, and the induction hypothesis applies.
∎

######  Theorem 2.9.

  1. (i).

A quadratic form over ${\mathbb{F}}_{p}$ in $3$ or more variables is always
isotropic.

  2. (ii).

A quadratic form over ${\mathbb{Q}}_{p}$ in $5$ or more variables is always
isotropic.

###### Proof.

(i) This is a consequence of the Chevalley-Warning theorem. See for example
[8, p. 5].
(ii) This is Meyer’s theorem. See for example [3, p. 41]. ∎

We make the following definition concerning quadratic forms over
${\mathbb{F}}_{p}$.

######  Definition 2.10.

A quadratic form over ${\mathbb{F}}_{p}$ belongs to the class $[l,m,n]$ if it
is equivalent to a form

$Q(x_{1},x_{2},\ldots,x_{n})=\sum_{i=1}^{l}x_{r+2i-1}x_{r+2i}+f(x_{r+2l+1},\ldots,x_{n})$
(4)

where $f$ is a regular anisotropic form of dimension $m$. Note that $l$ is the
number of hyperbolic planes in the orthogonal decomposition, $m\in\\{0,1,2\\}$
by Theorem 2.9(i), $n$ is the dimension of the form, and $r=n-2l-m$ is the
dimension of the radical.

Repeated application of Lemma 2.6 shows that every quadratic form over
${\mathbb{F}}_{p}$ belongs to the class $[l,m,n]$ for some $l,m,n$, and
Theorem 2.7 shows that $l,m,n$ are uniquely determined by the quadratic form.

##  3 Global densities

In this section we explain how Theorem 1.1 and Corollary 1.3 follow from
Theorem 1.2. First we read off from Theorem 1.2 the asymptotic behaviour of
$\rho_{p}(k,n)$ as $p\to\infty$.

######  Corollary 3.1.

Let $k\geqslant 1$. As $p\to\infty$, we have the following approximations.

$\rho_{p}(k,n)=\left\\{\begin{array}[]{ll}\vspace{1ex}\frac{1}{2}+O(\frac{1}{p})&\text{
if }n=2k;\\\ \vspace{1ex}1-\frac{1}{2p}+O(\frac{1}{p^{2}})&\text{ if
}n=2k+1;\\\ 1-\frac{1}{4p^{3}}+O(\frac{1}{p^{4}})&\text{ if
}n=2k+2.\end{array}\right.$

###### Proof.

In the case $n=2k+2$ we consider the Taylor series expansions

$\frac{p^{k+3}-1}{(p+1)(p^{2k+3}-1)}=\frac{1}{p^{k+1}}\left(1-\frac{1}{p}+\frac{1}{p^{2}}-\frac{1}{p^{3}}+O\left(\frac{1}{p^{4}}\right)\right),$

and

$\prod_{i=1}^{k+1}\frac{p^{2i-1}-1}{p^{2i}-1}=\frac{1}{p^{k+1}}\left(1-\frac{1}{p}+\frac{1}{p^{2}}-\frac{2}{p^{3}}+O\left(\frac{1}{p^{4}}\right)\right),$

using the big $O$ notation. Substituting these expansions into the formula for
$\rho_{p}(k,n)$ in Theorem 1.2 gives the approximation for $\rho_{p}(k,n)$ as
claimed. The other cases are similar but easier. ∎

Fix values of $k$ and $n$. We write $U_{\infty}$ for the set of quadratic
forms in $\mathbb{R}[x_{1},\ldots,x_{n}]$ that are not $k$-isotropic over
$\mathbb{R}$. Likewise we write $U_{p}$ for the set of quadratic forms in
${\mathbb{Z}}_{p}[x_{1},\ldots,x_{n}]$ that are not $k$-isotropic over
${\mathbb{Q}}_{p}$. Let $\mu_{\infty}$ denote the standard Lebesgue measure on
$\mathbb{R}^{d}$, and let $\mu_{p}$ denote the Haar measure on
${\mathbb{Z}}_{p}^{d}$ normalised to have total volume $1$.

######  Lemma 3.2.

Let $1\leqslant k\leqslant n$ and $d=\binom{n+1}{2}$. Suppose that the
following condition holds for all sufficiently large primes $p$:

> Every quadratic form in ${\mathbb{Z}}_{p}[x_{1},\ldots,x_{n}]$ whose
> reduction mod $p$ has rank at least $n-1$ is $k$-isotropic over
> ${\mathbb{Q}}_{p}$.

Then $\rho_{\rm glob}(k,n)$ exists and is given by

$\rho_{\rm glob}(k,n)=\frac{\mu_{\infty}([-1,1]^{d}\setminus
U_{\infty})}{2^{d}}\cdot\prod_{p}(1-\mu_{p}(U_{p})).$ (5)

###### Proof.

As noted in the introduction, a quadratic form is $k$-isotropic over
${\mathbb{Q}}$ if and only if it is $k$-isotropic over ${\mathbb{Q}}_{p}$ for
all primes $p$ and over $\mathbb{R}$. We then apply [7, Lemmas 20 and 21] with
$U_{\infty}$ and $U_{p}$ as defined above, $S=\emptyset$ and
$f,g\in{\mathbb{Z}}[a_{11},a_{12},\ldots,a_{nn}]$ two distinct
$(n-1)\times(n-1)$ minors of the generic $n\times n$ symmetric matrix of
coefficients. ∎

In our earlier notation the factors on the right hand side of (5) may be
written

$\rho_{\infty}(k,n)=\frac{\mu_{\infty}([-1,1]^{d}\setminus
U_{\infty})}{2^{d}}\qquad\text{ and }\qquad\rho_{p}(k,n)=1-\mu_{p}(U_{p}).$
(6)

######  Lemma 3.3.

If $n\geqslant 2k+2$ then the condition in Lemma 3.2 is satisfied.

###### Proof.

Let $Q\in{\mathbb{Z}}_{p}[x_{1},\ldots,x_{n}]$ be a quadratic form whose
reduction mod $p$ has rank at least $n-1$. In the terminology of Definition
2.10, the reduction of $Q$ mod $p$ belongs to the class $[l,m,n]$ for some
$l,m,n$ with $m\in\\{0,1,2\\}$. Our assumptions then give $2l+m\geqslant
n-1\geqslant 2k+1$. Since $k$ and $l$ are integers it follows that $l\geqslant
k$. Then $Q$ is $k$-isotropic over ${\mathbb{Q}}_{p}$ by Lemma 2.8. ∎

Proof of Theorem 1.1. Let $\rho_{\rm glob}(k,n)$ be as defined in (2), and let
$\overline{\rho}_{\rm glob}(k,n)$ be the same quantity with the limit replaced
by lim sup. We write $\rho_{p}(k,n)$ for the probabilities computed in Theorem
1.2. A standard argument (see for example [4, Proposition 3.2]) uses the local
conditions at finitely many primes to show that

$\overline{\rho}_{\rm glob}(k,n)\leqslant\prod_{p<M}\rho_{p}(k,n).$ (7)

If $n\leqslant 2k+1$ then by Corollary 3.1 the right hand side of (7) tends to
$0$ as $M\to\infty$. Therefore $\rho_{\rm glob}(k,n)=0$ and the equality
claimed in Theorem 1.1 holds since both sides are zero.

If $n\geqslant 2k+2$ then Lemmas 3.2 and 3.3 apply. Combining (5) and (6)
gives

$\rho_{\rm glob}(k,n)=\rho_{\infty}(k,n)\prod_{p}\rho_{p}(k,n)$

as required. $\Box$

Corollary 1.3 follows immediately from Theorem 1.1, Theorem 1.2 and the
observation in the last proof that the local product is zero for $n\leqslant
2k+1$.

######  Remark 3.4.

We do not know an accurate method for computing the probabilities
$\rho_{\infty}(k,n)$, but we can estimate them using a Monte Carlo simulation.
On this basis we record the following numerical values that are likely to be
accurate to the number of decimal places recorded.

$\begin{array}[]{c|ccc}k&\prod_{p}\rho_{p}(k,2k+2)&\rho_{\infty}(k,2k+2)&\rho_{\rm
glob}(k,2k+2)\\\ \hline\cr 1&0.98743625&0.9823&0.9699\\\
2&0.98229463&0.9705&0.9533\\\ 3&0.98007620&0.9623&0.9431\\\
4&0.97906880&0.9561&0.9361\\\ 5&0.97859528&0.9512&0.9309\end{array}$

######  Remark 3.5.

In [2] (which only treats the case $k=1$) some alternatives to the definition
(2) were considered. The global densities so defined may still be computed as
a product over all places, and the local contributions at the finite places
are the same as before. However the local contributions at infinity can
change, and for one natural choice of distribution these were computed
exactly. It is possible that something similar could be done for $k>1$, but we
did not pursue this.

##  4 Counting quadratic forms over ${\mathbb{F}}_{p}$

In this section we prove some formulae counting quadratic forms over
${\mathbb{F}}_{p}$. We consider quadratic forms over ${\mathbb{F}}_{p}$
according to their class $[l,m,n]$ as defined in Definition 2.10.

######  Definition 4.1.

Consider a quadratic form

$Q(x_{1},x_{2},\ldots,x_{n})=\sum_{1\leqslant i\leqslant j\leqslant
n}a_{ij}x_{i}x_{j}$

over ${\mathbb{F}}_{p}$ where the coefficients $a_{ij}$ are chosen
independently at random according to counting measure.

Let $\pi_{0}(l,m,n)$ be the probability that $Q$ belongs to the class
$[l,m,n]$.

Let $\pi_{1}(l,m,n)$ be the probability that $Q$ belongs to the class
$[l,m,n]$ given that $a_{11}\not=0$.

Let $\pi_{2}(l,m,n)$ be the probability that $Q$ belongs to the class
$[l,m,n]$ given that $a_{11}x_{1}^{2}+a_{12}x_{1}x_{2}+a_{22}x_{2}^{2}$ is a
regular anisotropic form.

Note that Theorem 2.9(i) implies that $\pi_{i}(l,m,n)=0$ if $m\geqslant 3$.

The group $\operatorname{GL}_{n}({\mathbb{F}}_{p})$ acts on the set of
$n$-dimensional quadratic forms over ${\mathbb{F}}_{p}$ by linear
substitutions. The class $[l,m,n]$ is the union of either one or two orbits,
and so its size may be computed as a sum of orbit sizes. To begin with we only
consider forms that are regular. In these cases the orbit sizes can be
computed using the orbit-stabiliser theorem and the following theorem.

######  Lemma 4.2.

Let $Q_{0}$ be a quadratic form over ${\mathbb{F}}_{p}$ belonging to the class
$[l,m,n]$. Suppose that $Q_{0}$ is regular, equivalently $n=2l+m$. Then the
stabiliser in $\operatorname{GL}_{n}({\mathbb{F}}_{p})$ of $Q_{0}$ is an
orthogonal group of order $S(m,n)$ where

$\displaystyle S(0,2k)$
$\displaystyle=2p^{k(k-1)}(p^{k}-1)\prod_{i=1}^{k-1}(p^{2i}-1);$
$\displaystyle S(1,2k+1)$
$\displaystyle=\left\\{\begin{array}[]{ll}2p^{k^{2}}\prod_{i=1}^{k}(p^{2i}-1)&\text{
if }p\neq 2;\\\ p^{k^{2}}\prod_{i=1}^{k}(p^{2i}-1)&\text{ if }p=2;\\\
\end{array}\right.$ $\displaystyle S(2,2k)$
$\displaystyle=2p^{k(k-1)}(p^{k}+1)\prod_{i=1}^{k-1}(p^{2i}-1).$

###### Proof.

See [5, pp. 81–82] for $p$ an odd prime, and [5, pp. 147–150] for the case
$p=2$. ∎

To find the orbit size when the radical has dimension $r=n-2l-m$, we multiply
the orbit size of the regular part under the action of
$\operatorname{GL}_{n-r}({\mathbb{F}}_{p})$ by the number

$N(r,n)=\prod_{i=0}^{r-1}\frac{p^{n}-p^{i}}{p^{r}-p^{i}}$

of $r$-dimensional subspaces of ${\mathbb{F}}_{p}^{n}$. The orbit size
$O(l,m,n)$ of a form in $[l,m,n]$ under the action of
$\operatorname{GL}_{n}({\mathbb{F}}_{p})$ is therefore given by

$O(l,m,n)=N(n-2l-m,n)\cdot\frac{|\operatorname{GL}_{2l+m}({\mathbb{F}}_{p})|}{S(m,2l+m)}.$

If $m=1$ and $p$ is odd, there are two orbits belonging to $[l,m,n]$; other
values of $m$ give a unique orbit (see [5, p. 79]). Hence, using that
$|\operatorname{GL}_{n}({\mathbb{F}}_{p})|=\prod_{i=0}^{n-1}(p^{n}-p^{i})$,
and dividing by the total number of quadratic forms of dimension $n$, we
obtain the values of $\pi_{0}(l,m,n)$ recorded in the next lemma. Note that
the only form in the class $[0,0,n]$ is the form where all the coefficients
are zero.

######  Lemma 4.3.

For $l+m>0$ and $n=2l+m+r$ we have

$\pi_{0}(l,m,n)=\left\\{\begin{array}[]{ll}\vspace{1ex}\frac{1}{p^{n(n+1)/2}}\cdot\prod_{i=0}^{r-1}\frac{p^{n}-p^{i}}{p^{r}-p^{i}}\cdot\frac{\prod_{i=0}^{2l-1}(p^{2l}-p^{i})}{2p^{l(l-1)}(p^{l}-1)\prod_{i=1}^{l-1}(p^{2i}-1)}&\text{
if }m=0;\\\
\vspace{1ex}\frac{1}{p^{n(n+1)/2}}\cdot\prod_{i=0}^{r-1}\frac{p^{n}-p^{i}}{p^{r}-p^{i}}\cdot\frac{\prod_{i=0}^{2l}(p^{2l+1}-p^{i})}{p^{l^{2}}\prod_{i=1}^{l}(p^{2i}-1)}&\text{
if }m=1;\\\
\frac{1}{p^{n(n+1)/2}}\cdot\prod_{i=0}^{r-1}\frac{p^{n}-p^{i}}{p^{r}-p^{i}}\cdot\frac{\prod_{i=0}^{2l+1}(p^{2l+2}-p^{i})}{2p^{l(l+1)}(p^{l+1}+1)\prod_{i=1}^{l}(p^{2i}-1)}&\text{
if }m=2.\end{array}\right.$

Moreover, $\pi_{0}(0,0,n)=1/p^{n(n+1)/2}$.

Next we compute the probabilities $\pi_{1}(l,m,n)$ in terms of the
probabilities $\pi_{0}(l,m,n)$.

######  Lemma 4.4.

We have

$\pi_{1}(l,m,n)=\left\\{\begin{array}[]{ll}\pi_{0}(l-1,1,n-1)/2&\text{ if
}m=0\text{ and }l\geqslant 1;\\\ \pi_{0}(l,0,n-1)+\pi_{0}(l-1,2,n-1)&\text{ if
}m=1\text{ and }l\geqslant 1;\\\ \pi_{0}(l,1,n-1)/2&\text{ if }m=2.\\\
\end{array}\right.$

Moreover, $\pi_{1}(0,0,n)=0$ and $\pi_{1}(0,1,n)=1/p^{n(n-1)/2}$.

###### Proof.

We first suppose that $p$ is an odd prime. Let $Q$ be a quadratic form in $n$
variables over ${\mathbb{F}}_{p}$ with first coefficient $a_{11}\not=0$. We
must compute the probability $\pi_{1}(l,m,n)$ that $Q$ belongs to the class
$[l,m,n]$. By a linear substitution to eliminate the cross-terms containing
$x_{1}$, we may assume that
$Q(x_{1},x_{2},\ldots,x_{n})=a_{11}x_{1}^{2}+F(x_{2},x_{3},\ldots,x_{n})$ for
some $F\in{\mathbb{F}}_{p}[x_{2},\ldots,x_{n}]$. The class of $Q$ is
determined by the class of $F$ and the value of $a_{11}$. Since the
coefficients of $F$, like those of $Q$, are randomised according to counting
measure, we can compute $\pi_{1}(l,m,n)$ in terms of the
$\pi_{0}(l^{\prime},m^{\prime},n-1)$ for suitable $l^{\prime}$ and
$m^{\prime}$. More precisely, we note that if $F$ belongs to the class
$[l,0,n-1]$ or $[l-1,2,n-1]$ then $Q$ belongs to the class $[l,1,n]$, whereas
if $F$ belongs to the class $[l-1,1,n-1]$ then it is equally likely that $Q$
belongs to the class $[l,0,n]$ or $[l-1,2,n]$. The stated formulae follow.

To prove the lemma when $p=2$ we outline an alternative method for computing
$\pi_{1}(l,m,n)$ that gives the answer as a rational function in $p$. In this
alternative method we compute $\pi_{1}(l,m,n)$ by finding the probability that
a form in $[l,m,n]$ satisfies $a_{11}\not=0$, and then multiply by
$\pi_{0}(l,m,n)\cdot\frac{p}{p-1}$ according to Bayes’ formula. The second
factor comes from the fact that $a_{11}\not=0$ with probability
$\frac{p-1}{p}$. Since $a_{11}=Q(1,0,\ldots,0)$ and
$\operatorname{GL}_{n}({\mathbb{F}}_{p})$ acts transitively on
${\mathbb{F}}_{p}^{n}\setminus\\{0\\}$ it suffices to show that

$N(Q)=\\#\\{x\in{\mathbb{F}}_{p}^{n}\mid Q(x)=0\\}$

is a polynomial in $p$, where the polynomial depends only on $l,m,n$. We prove
this claim by induction on $l$, noting that if
$Q(x_{1},\ldots,x_{n})=x_{1}x_{2}+Q^{\prime}(x_{3},\ldots,x_{n})$ then
$N(Q)=(2p-1)N(Q^{\prime})+(p-1)(p^{n-2}-N(Q^{\prime}))$, whereas if $l=0$ then
$N(Q)=p^{n-m}$. ∎

To determine the values of $\pi_{2}(l,m,n)$, we use a method similar to the
one we used for calculating $\pi_{1}(l,m,n)$ for $p$ an odd prime. However,
this proof also includes the case $p=2$.

######  Lemma 4.5.

We have

$\pi_{2}(l,m,n)=\left\\{\begin{array}[]{ll}\pi_{0}(l-2,2,n-2)&\text{ if
}m=0\text{ and }l\geqslant 2;\\\ \pi_{0}(l-1,1,n-2)&\text{ if }m=1\text{ and
}l\geqslant 1;\\\ \pi_{0}(l,0,n-2)&\text{ if }m=2.\end{array}\right.$

Moreover, $\pi_{2}(0,0,n)=\pi_{2}(0,1,n)=\pi_{2}(1,0,n)=0$.

###### Proof.

It suffices to consider
$Q(x_{1},x_{2},\ldots,x_{n})=f(x_{1},x_{2})+F(x_{3},\ldots,x_{n})$ for
$f\in{\mathbb{F}}_{p}[x_{1},x_{2}]$ regular anisotropic and
$F\in{\mathbb{F}}_{p}[x_{3},\ldots,x_{n}]$. The class of $F$ then determines
the class of $Q$, and this gives the formulae as stated. ∎

##  5 First method: Reduction modulo $p$ and recursion

In this section we give our first method for computing the probability
$\rho_{p}(k,n)$ that a random $p$-adic integral quadratic form in $n$
variables is $k$-isotropic.

######  Definition 5.1.

Let $Q$ be a random $p$-adic integral quadratic form in $n$ variables.

Let $\delta_{0}(k;l,m,n)$ be the probability that $Q$ is $k$-isotropic given
that its reduction mod $p$ belongs to the class $[l,m,n]$.

Let $\delta_{1}(k;l,m,n)$ be the probability that $Q$ is $k$-isotropic given
that its reduction mod $p$ belongs to the class $[l,m,n]$, the coefficients
$a_{11},a_{12},\ldots,a_{1n}$ are all divisible by $p$, but $p^{2}$ does not
divide $a_{11}$.

Let $\delta_{2}(k;l,m,n)$ be the probability that $Q$ is $k$-isotropic given
that its reduction mod $p$ belongs to the class $[l,m,n]$, the coefficients
$a_{11},a_{12},\ldots,a_{1n}$ and $a_{22},a_{23},\ldots,a_{2n}$ are all
divisible by $p$, but the reduction of
$\frac{1}{p}(a_{11}x_{1}^{2}+a_{12}x_{1}x_{2}+a_{22}x_{2}^{2})$ mod $p$ is a
regular anisotropic form.

By definition $\delta_{0}(k;0,0,n)$ is the probability of $k$-isotropy given
that $Q$ vanishes mod $p$. This is the same as $\rho_{p}(k,n)$. Our next two
results establish recursive relations for computing the $\delta_{i}(k;l,m,n)$.

######  Lemma 5.2.

For $i\in\\{0,1,2\\}$ we have

$\delta_{i}(k;l,m,n)=\left\\{\begin{array}[]{ll}\delta_{i}(k-l;0,m,n-2l)&\text{
if }k>l;\\\ 1&\text{ if }k\leqslant l.\end{array}\right.$

###### Proof.

A quadratic form whose reduction modulo $p$ belongs to the class $[l,m,n]$ is
equivalent over ${\mathbb{Z}}_{p}$ to a form which satisfies

$Q(x_{1},\ldots,x_{n})\equiv\sum_{i=1}^{l}x_{r+2i-1}x_{r+2i}+f(x_{r+2l+1},\ldots,x_{n})\mod
p,$ (8)

for $f$ a regular anisotropic form over ${\mathbb{F}}_{p}$ of dimension
$m\in\\{0,1,2\\}$. We claim that $Q$ is equivalent over ${\mathbb{Z}}_{p}$ to
a form

$Q^{\prime}(x_{1},\ldots,x_{n})=\sum_{i=1}^{l}x_{r+2i-1}x_{r+2i}+Q^{\prime\prime}(x_{1},\ldots,x_{r},x_{r+2l+1},\ldots,x_{n}),$
(9)

where $Q^{\prime\prime}(x_{1},\ldots,x_{r},x_{r+2l+1},\ldots,x_{n})\equiv
f(x_{r+2l+1},\ldots,x_{n})\mod p$. If $k\leqslant l$ it follows immediately
that $Q$ is $k$-isotropic. If $k>l$ then by Witt’s Cancellation Theorem, the
form $Q$ is $k$-isotropic if and only if the form $Q^{\prime\prime}$ is
$(k-l)$-isotropic. So it only remains to prove the claim, and at the same time
convince ourselves that $Q^{\prime\prime}$ is suitably randomised.

For simplicity let us suppose that $l=1$. Then the
${\mathbb{Z}}_{p}$-equivalence taking (8) to (9) is built out of two sorts of
transformations. First we let $\operatorname{GL}_{2}({\mathbb{Z}}_{p})$ act on
the variables $x_{r+1}$ and $x_{r+2}$ by linear substitution. (As in the proof
of Lemma 2.8 the required substitution exists by Hensel’s lemma.) Then we make
substitutions for $x_{r+1}$ and $x_{r+2}$ where we add to each $p$ times a
linear combination of the other variables. Thus
$Q^{\prime\prime}(x_{1},\ldots,x_{r},x_{r+3},\ldots,x_{n})$ is obtained from
$Q(x_{1},\ldots,x_{r},0,0,x_{r+3},\ldots,x_{n})$ by adding a form which
vanishes mod $p^{2}$. Since the extra conditions on $Q$ in the definition of
the $\delta_{i}$ for $i=1,2$ are conditions on the coefficients mod $p^{2}$,
these are not affected by this change. ∎

######  Lemma 5.3.

For $i,j\in\\{0,1,2\\}$ and $n\geqslant i+j$ we have

$\displaystyle\delta_{i}(k;0,j,n)$ $\displaystyle=\sum_{l\geqslant
0}\sum_{m=0}^{2}\pi_{i}(l,m,n-j)\delta_{j}(k;l,m,n).$

Moreover, if $n=i+j$ then

$\delta_{i}(k;0,j,n)=\left\\{\begin{array}[]{ll}1&\text{ if }k=0;\\\ 0&\text{
if }k\geqslant 1.\end{array}\right.$

The condition $n\geqslant i+j$ ensures that $\pi_{i}(l,m,n-j)$ is defined. It
can only be non-zero if $2l+m\leqslant n-j$, in which case
$\delta_{j}(k;l,m,n)$ is defined. In particular the sum over $l$ is finite.

###### Proof.

Let $Q$ be a $p$-adic integral quadratic form of dimension $n$ whose reduction
mod $p$ belongs to the class $[0,j,n]$. By an equivalence over
${\mathbb{Z}}_{p}$ we may suppose that the reduction of $Q$ mod $p$ is an
anisotropic form in the last $j\in\\{0,1,2\\}$ variables. We multiply each of
the last $j$ variables by $p$, and divide through by $p$ to give a new
$p$-adic integral quadratic form, whose reduction mod $p$ involves only the
first $n-j$ variables. If $i=1$ or $2$ then the additional conditions in
Definition 5.1 give the additional conditions in Definition 4.1. The reduction
mod $p$ now has class $[l,m,n]$ with probability $\pi_{i}(l,m,n-j)$, and in
this case the form is $k$-isotropic with probability $\delta_{j}(k;l,m,n)$. In
checking this last statement, notice that the extra conditions in Definition
5.1 when $j=1$ or $2$ are satisfied relative to the last $j$ variables rather
than the first $j$ variables. This change clearly does not matter. Summing
over all possibilities for $l$ and $m$ gives the result.

For the final part we show that if $n=i+j$ then the forms considered in the
definition of $\delta_{i}(k;0,j,n)$ are anisotropic over ${\mathbb{Q}}_{p}$.
For example, if $i=j=2$ and $n=4$ then the reduction of
$Q(x_{1},\ldots,x_{4})$ mod $p$ is an anisotropic form in $x_{3}$ and $x_{4}$,
and the reduction of $\frac{1}{p}Q(x_{1},x_{2},px_{3},px_{4})$ mod $p$ is an
anisotropic form in $x_{1}$ and $x_{2}$. Supposing that
$Q(a_{1},\ldots,a_{4})=0$ for some $a_{1},\ldots,a_{4}\in{\mathbb{Z}}_{p}$ not
all divisible by $p$ these conditions quickly lead to a contradiction. The
other cases are similar. ∎

######  Proposition 5.4.

The relations in Lemmas 5.2 and 5.3 are sufficient to determine all the
$\delta_{i}(k;l,m,n)$ and to show that they are rational functions in $p$. The
same is therefore true of $\rho_{p}(k,n)=\delta_{0}(k;0,0,n)$.

###### Proof.

Combining the two lemmas shows that

$\delta_{i}(k;0,j,n)=\frac{1}{p^{\binom{n+1-i-j}{2}}}\delta_{j}(k;0,i,n)+\ldots$

where the terms omitted involve either a smaller value of $n$ or a larger
value of $i+j$. Assuming all such previous values have been computed, we can
uniquely solve for $\delta_{i}(k;0,j,n)$ and $\delta_{j}(k;0,i,n)$ provided
that $n>i+j$. It is clear from Definition 5.1 that we must have $n\geqslant
i+j$ and the remaining case where $n=i+j$ is covered by the last part of Lemma
5.3. Finally we use Lemma 5.2 to compute the $\delta_{i}(k;l,m,n)$ with $l>0$.

Since we saw in Section 4 that the $\pi_{i}(l,m,n)$ are rational functions in
$p$, it follows that the $\delta_{i}(k;l,m,n)$ are also rational functions in
$p$. ∎

Proposition 5.4 together with the results of the next section are all we shall
need for the proof of Theorem 1.2. It is nonetheless still interesting to find
explicit closed formulae for the $\delta_{i}(k;l,m,n)$. We do this now,
leaving some of the details to Appendix A.

######  Definition 5.5.

For $i,j\in\\{0,1,2\\}$ and $n\geqslant i+j$ we define

$\displaystyle\phi(i,j,n)$
$\displaystyle=\big{(}(j-1)p^{d}+(i-1)\big{)}\cdot\prod_{r=1}^{d}\frac{p^{2r-1}-1}{p^{2r}-1},$
$\displaystyle\psi(i,j,n)$
$\displaystyle=\frac{\big{(}(j-1)p^{d}+(i-1)\big{)}\big{(}(j-1)p^{d+2}-(i-1)\big{)}-\delta_{i1}p+\delta_{j1}p^{2d+1}}{(p+1)(p^{2d+1}-1)},$

where $d=\lfloor\frac{n+1-i-j}{2}\rfloor$ and $\delta_{ij}$ is the Kronecker
delta.

######  Proposition 5.6.

Let $i,j\in\\{0,1,2\\}$ and $n\geqslant i+j$. Then

$\phi(i,j,n)=\sum_{l\geqslant 0}\sum_{m=0}^{2}\pi_{i}(l,m,n-j)\phi(j,m,n-2l),$

and if $n$ is even then

$\psi(i,j,n)=\sum_{l\geqslant 0}\sum_{m=0}^{2}\pi_{i}(l,m,n-j)\psi(j,m,n-2l).$

###### Proof.

We prove this in Appendix A by adapting methods of Kovaleva [6]. ∎

Theorem 1.2 is the special case $i=j=0$ of the following result.

######  Theorem 5.7.

For any $i,j\in\\{0,1,2\\}$ and $n\geqslant i+j$ we have

$\delta_{i}(k;0,j,n)=\left\\{\begin{array}[]{ll}\vspace{1ex}0&\text{ if
}n\leqslant 2k-1;\\\
\vspace{1ex}\frac{1}{4}\big{(}-\phi(i,j,n)+\psi(i,j,n)\big{)}&\text{ if
}n=2k;\\\ \vspace{1ex}\frac{1}{2}\big{(}1-\phi(i,j,n)\big{)}&\text{ if
}n=2k+1;\\\
\vspace{1ex}1-\frac{1}{4}\big{(}\phi(i,j,n)+\psi(i,j,n)\big{)}&\text{ if
}n=2k+2;\\\ 1&\text{ if }n\geqslant 2k+3.\end{array}\right.$

###### Proof.

By Proposition 5.6 these are solutions to the recurrence relations in
Proposition 5.4. These particular linear combinations of $1$, $\phi$ and
$\psi$ also satisfy the initial conditions, that is, we checked they give the
correct answers when $n=i+j$. ∎

As explained in the introduction, the following corollary is interesting since
it generalises a phenomenon studied in [1].

######  Corollary 5.8.

The probabilities $\delta_{i}(k;l,j,n)$ and $\delta_{j}(k;l,i,n)$ are rational
functions in $p$ that are exchanged when we replace $p$ by $1/p$. In
particular $\rho_{p}(k,n)=\delta_{0}(k;0,0,n)$ is unchanged when we replace
$p$ by $1/p$.

###### Proof.

By Lemma 5.2 it suffices to prove the case $l=0$. The symmetries claimed then
follow from Definition 5.5 and Theorem 5.7. ∎

##  6 Second method: Using a theorem of Kovaleva

In this section we deduce Theorem 1.2 from a result of Kovaleva [6]. First we
recall the classification of quadratic forms over ${\mathbb{Q}}_{p}$ up to
equivalence.

######  Definition 6.1.

Let $a$, $b\in{\mathbb{Q}}_{p}^{*}$. The Norm-Residue symbol, denoted
$\binom{a,b}{p}$ or more simply as $(a,b)$, is set to be $1$ when the form
$ax^{2}+by^{2}-z^{2}$ vanishes for some $x,y,z\in{\mathbb{Q}}_{p}$ not all
zero, and $-1$ otherwise.

######  Definition/Lemma 6.2.

Let $Q\in{\mathbb{Q}}_{p}[x_{1},\ldots,x_{n}]$ be a quadratic form of rank $n$
which is equivalent to a diagonal form
$Q^{\prime}(x_{1},\ldots,x_{n})=\sum_{i=1}^{n}a_{i}x_{i}^{2}$. The Hasse-
Minkowski invariant of the form $Q$ is defined as
$c(Q)=\prod_{i<j}(a_{i},a_{j})$. This is independent of the choice of diagonal
form.

###### Proof.

See [3, pp. 56–58]. ∎

######  Theorem 6.3.

A quadratic form $Q\in{\mathbb{Q}}_{p}[x_{1},\ldots,x_{n}]$ of rank $n$ is
uniquely determined up to ${\mathbb{Q}}_{p}$-equivalence by its determinant
$d(Q)\in{\mathbb{Q}}_{p}^{*}/({\mathbb{Q}}_{p}^{*})^{2}$ and its Hasse-
Minkowski invariant $c(Q)\in\\{\pm 1\\}$.

###### Proof.

See [3, p. 61]. ∎

The next lemma explains why we only need to consider forms of full rank over
${\mathbb{Q}}_{p}$.

######  Lemma 6.4.

A $p$-adic integral quadratic form, with coefficients chosen independently
from ${\mathbb{Z}}_{p}$ according to Haar measure, is singular with
probability zero.

###### Proof.

We write the form as $Q(x_{1},x_{2},\ldots,x_{n})=a_{11}x_{1}^{2}+x_{1}\cdot
f(x_{2},\ldots,x_{n})+g(x_{2},\ldots,x_{n})$ for some linear form
$f\in{\mathbb{Z}}_{p}[x_{2},\ldots,x_{n}]$ and quadratic form
$g\in{\mathbb{Z}}_{p}[x_{2},\ldots,x_{n}]$.

If $n=1$, the form is singular when $a_{11}$ is zero, which happens with
probability zero. Inductively, for $n>1$, we can assume the form $g$ to be
non-singular. For each linear form $f$ and non-singular form $g$, there is
only one value of $a_{11}$ that makes $Q$ singular, corresponding to the
determinant of the coefficient matrix being zero. This value is attained by
$a_{11}$ with probability zero, hence the form is singular with probability
zero by induction. ∎

We now take $p$ an odd prime. The following theorem, due to Kovaleva, gives
for each triple $(n,d,c)$ the probability that a random $p$-adic integral
quadratic form $Q$ in $n$ variables has determinant $d(Q)=d$ and Hasse-
Minkowski invariant $c(Q)=c$. Since $p$ is an odd prime, the quotient
${\mathbb{Q}}_{p}^{*}/({\mathbb{Q}}_{p}^{*})^{2}$ has order $4$, with coset
representatives $\\{1,u,p,up\\}$ where $u$ is a quadratic non-residue modulo
$p$.

######  Theorem 6.5 (Kovaleva).

Let $p$ be an odd prime, and let
$Q\in{\mathbb{Z}}_{p}[x_{1},x_{2},\ldots,x_{n}]$ be a random $p$-adic integral
quadratic form in $n$ variables. Let $\varepsilon$ and $s$ denote the Legendre
symbols $(\frac{-1}{p})$ resp. $(\frac{d}{p})$ and let $u$ be a quadratic non-
residue modulo $p$. Then the probability $\mathbb{P}_{n}(d(Q)=d,c(Q)=c)$ that
$Q$ has determinant $d\in{\mathbb{Q}}_{p}^{*}/({\mathbb{Q}}_{p}^{*})^{2}$ and
Hasse-Minkowski invariant $c\in\\{\pm 1\\}$ is given by

$\displaystyle\mathbb{P}_{2k+1}(d(Q)=d,c(Q)=c)$
$\displaystyle=\left\\{\begin{aligned}
&\frac{1}{4}\cdot\frac{p}{p+1}+\frac{1}{4}\cdot c\cdot
p^{k+1}\cdot\prod_{i=1}^{k+1}\frac{p^{2i-1}-1}{p^{2i}-1}&&\text{if}\hskip
6.0ptd\in\\{1,u\\};\\\ &\frac{1}{4}\cdot\frac{1}{p+1}+\frac{1}{4}\cdot
c\cdot\varepsilon^{k}\cdot\prod_{i=1}^{k+1}\frac{p^{2i-1}-1}{p^{2i}-1}&&\text{if}\hskip
6.0ptd\in\\{p,up\\};\\\ \end{aligned}\right.$
$\displaystyle\mathbb{P}_{2k}(d(Q)=d,c(Q)=c)$
$\displaystyle=\left\\{\begin{aligned}
&\frac{1}{4}\cdot(p^{k}+s\varepsilon^{k})\cdot\left(\frac{(p^{k+2}-s\varepsilon^{k})}{(p+1)(p^{2k+1}-1)}+c\cdot\prod_{i=1}^{k}\frac{p^{2i-1}-1}{p^{2i}-1}\right)&&\text{if}\hskip
6.0ptd\in\\{1,u\\};\\\
&\frac{1}{4}\cdot\frac{p}{p+1}\cdot\frac{p^{2k}-1}{p^{2k+1}-1}&&\text{if}\hskip
6.0ptd\in\\{p,up\\}.\\\ \end{aligned}\right.$

###### Proof.

See [6, Theorem 1.3]. ∎

We deduce Theorem 1.2 for $p$ odd using the following lemma. Recall that we
wrote $\rho_{p}(k,n)$ for the probability that a random $p$-adic integral
quadratic form in $n$ variables is $k$-isotropic.

######  Lemma 6.6.

We have $\rho_{p}(k,n)=0$ for $n\leqslant 2k-1$ and $\rho_{p}(k,n)=1$ for
$n\geqslant 2k+3$. If $p$ is odd then

$\displaystyle\rho_{p}(k,2k)$
$\displaystyle=\mathbb{P}_{2k}(d(Q)=(-1)^{k},c(Q)=1);$
$\displaystyle\rho_{p}(k,2k+1)$
$\displaystyle=\sum_{a\in{\mathbb{Q}}_{p}^{*}/({\mathbb{Q}}_{p}^{*})^{2}}\mathbb{P}_{2k+1}(d(Q)=(-1)^{k}a,c(Q)=(-1,a)^{k});$
$\displaystyle\rho_{p}(k,2k+2)$
$\displaystyle=1-\mathbb{P}_{2k+2}(d(Q)=(-1)^{k-1},c(Q)=-1).$

###### Proof.

We first note that if $n\leqslant 2k-1$ then every $k$-isotropic form of
dimension $n$ is singular, and so $\rho_{p}(k,n)=0$ by Lemma 6.4. We now
suppose that $n\geqslant 2k$. By Lemma 2.6 every regular quadratic form of
dimension $n$ over ${\mathbb{Q}}_{p}$ is equivalent to one of the form

$Q(x_{1},\ldots,x_{n})=\sum_{i=1}^{l}x_{2i-1}x_{2i}+f(x_{2l+1},\ldots,x_{2l+m}),$
(10)

where $l$ is the number of hyperbolic planes in the decomposition,
$f(x_{2l+1},\ldots,x_{2l+m})$ is an anisotropic form over ${\mathbb{Q}}_{p}$
of rank $m$, and $n=2l+m$. By Theorem 2.9(ii) we have $m\leqslant 4$. It
follows that if $n\geqslant 2k+3$ then $k\leqslant l$ and so
$\rho_{p}(k,n)=1$.

We now take $p$ an odd prime. If $n=2k$ then for the form in (10) to be
$k$-isotropic we need $l=k$ and $m=0$. There is only one such form up to
${\mathbb{Q}}_{p}$-equivalence. It has determinant $d(Q)=(-1)^{k}$ and Hasse-
Minkowski invariant $c(Q)=1$. This gives the formula for $\rho_{p}(k,2k)$ as
stated. If $n=2k+1$ then for $k$-isotropy we need $l=k$ and $m=1$. The
anisotropic form in (10) is $f(x_{n})=ax_{n}^{2}$ for some
$a\in\\{1,u,p,up\\}$. This gives four ${\mathbb{Q}}_{p}$-equivalence classes
of forms, with invariants $d(Q)=(-1)^{k}a$ and $c(Q)=(-1,a)^{k}$. Finally we
take $n=2k+2$. For $Q$ not to be $k$-isotropic we need $l=k-1$ and $m=4$. The
rank $4$ anisotropic form $f$ has determinant $d(f)=1$ and Hasse-Minkowski
invariant $c(f)=-1$ (see [3, p. 59]). It follows that $d(Q)=(-1)^{k-1}$ and
$c(Q)=-1$, giving the result as stated. ∎

Theorem 1.2 for $p$ odd now follows from Theorem 6.5 and Lemma 6.6. The
interesting thing to note is that the Legendre symbols $\varepsilon$ and $s$
cancel, giving answers that are rational functions in $p$. Indeed when $n=2k$
we have $s\varepsilon^{k}=1$. When $n=2k+1$ the contributions for
$a\in\\{1,u\\}$ have $c=1$ and the contributions for $a\in\\{p,up\\}$ have
$c=\varepsilon^{k}$. When $n=2k+2$ we employ the corresponding formula in
Theorem 6.5 with $k$ replaced by $k+1$ and $s$ replaced by
$\varepsilon^{k-1}$.

We saw in Proposition 5.4 that $\rho_{p}(k,n)$ is given by a rational function
in $p$, where the same rational function works for all primes $p$ including
the prime $p=2$. Since we proved Theorem 1.2 in the last paragraph for
infinitely many primes (in fact for all odd primes), the theorem is therefore
true for all primes.

##  Appendix A Solving the recurrence relations for the first method

In this appendix we prove Proposition 5.6. This is not needed for the proof of
our main theorems as stated in the introduction, but is needed to compute all
the $\delta_{i}(k;l,m,n)$ (see Theorem 5.7) and hence to see that they satisfy
some interesting symmetries (see Corollary 5.8). The proof is based on that of
Theorem 6.5, but we could not see a way to directly cite Kovaleva’s work
without reworking all the details.

The identities we seek to prove are ones between rational functions in $p$. So
it suffices to prove them for any infinite set of primes. There is therefore
no loss of generality in assuming (as we now do) that $p$ is odd. This allows
us to identify quadratic forms and symmetric matrices in the usual way.

######  Definition A.1.

Let $\lambda(r,n)$ be the probability that a randomly chosen $n\times n$
symmetric matrix over ${\mathbb{F}}_{p}$ has rank $r$. This is computed in [6,
Corollary 4.2]. In the notation of Section 4 we have

$\lambda(r,n)=\left\\{\begin{array}[]{ll}\pi_{0}(l,0,n)+\pi_{0}(l-1,2,n)&\text{
if }r=2l;\\\ \pi_{0}(l,1,n)&\text{ if }r=2l+1.\end{array}\right.$ (11)

######  Lemma A.2.

For any $x\in{\mathbb{R}}$ we have

$\sum_{\begin{subarray}{c}r=0\\\ n-r\text{ even
}\end{subarray}}^{n}\lambda(r,n)\frac{x-p^{r}}{p^{n+1}-p^{r}}\,\,+\sum_{\begin{subarray}{c}r=0\\\
n-r\text{ odd
}\end{subarray}}^{n}\lambda(r,n)=\left\\{\begin{array}[]{ll}\vspace{1ex}\displaystyle\frac{x-1}{p^{n+1}-1}&\text{
if $n$ is even;}\\\ \displaystyle\frac{x}{p^{n+1}}&\text{ if $n$ is
odd}.\end{array}\right.$

###### Proof.

Since each side is linear in $x$, it suffices to prove the identity for just
two values of $x$. If $x=p^{n+1}$ then this is just the fact that
$\sum_{r=0}^{n}\lambda(r,n)=1$. If $n=2k$ and $x=1$ then the left hand side is

$\sum_{s=1}^{k}\left(\lambda(2s,n)\frac{1-p^{2s}}{p^{n+1}-p^{2s}}+\lambda(2s-1,n)\right)$

whereas if $n=2k+1$ and $x=0$ then the left hand side is

$\sum_{s=1}^{k}\left(\lambda(2s,n)-\lambda(2s+1,n)\frac{p^{2s}}{p^{n+1}-p^{2s}}\right).$

It may be checked using (11) that in each of these last two sums all the
summands are zero. ∎

######  Lemma A.3.

Let $\pi_{n}=\prod_{i=1}^{n}(1-p^{-i})$. Then for any $m,n\geqslant 0$ we have

$\sum_{r=0}^{\min(m,n)}\frac{\pi_{m}\pi_{n}}{p^{(m-r)(n-r)}\pi_{r}\pi_{m-r}\pi_{n-r}}=1.$

###### Proof.

This is [6, Corollary 2.5]. The $r$th summand is the probability that an
$m\times n$ matrix over ${\mathbb{F}}_{p}$ has rank $r$. This may be computed
by considering the action of
$\operatorname{GL}_{m}({\mathbb{F}}_{p})\times\operatorname{GL}_{n}({\mathbb{F}}_{p})$
via $(A,B):X\mapsto AXB^{T}$ and applying the orbit-stabiliser theorem. ∎

We define

$A(n)=\prod_{i=1}^{\lfloor\frac{n+1}{2}\rfloor}\frac{p^{2i}-p}{p^{2i}-1},\qquad
B(n)=\prod_{i=1}^{\lfloor\frac{n+1}{2}\rfloor}\frac{p^{2i-1}-1}{p^{2i}-1}.$
(12)

######  Lemma A.4.

For any $x,y,z\in{\mathbb{R}}$ we have

$\displaystyle\sum_{\begin{subarray}{c}r=0\\\ n-r\text{ even
}\end{subarray}}^{n}$
$\displaystyle\lambda(r,n)B(n-r)(x+yp^{n-r})\,\,+\sum_{\begin{subarray}{c}r=0\\\
n-r\text{ odd }\end{subarray}}^{n}\lambda(r,n)B(n-r)(x+zp^{n+1-r})$
$\displaystyle=\left\\{\begin{array}[]{ll}\vspace{1ex}\left(x+y+\left(1-\displaystyle\frac{1}{p^{n}}\right)z\right)A(n)&\text{
if $n$ is even;}\\\
\left(x+\left(1-\displaystyle\frac{1}{p^{n+1}}\right)y+z\right)A(n)&\text{ if
$n$ is odd}.\end{array}\right.$

###### Proof.

It suffices to prove this identity for three linearly independent choices of
$(x,y,z)$. If $(x,y,z)=(1,-1,0)$ and $n$ is even or $(x,y,z)=(1,0,-1)$ and $n$
is odd then the terms with $r=2s$ and $r=2s+1$ cancel, giving the result in
these cases. The proof is completed by the next lemma which proves the cases
$(x,y,z)=(0,1,0)$ and $(x,y,z)=(0,0,1)$. ∎

######  Lemma A.5.

We have

$\sum_{\begin{subarray}{c}l=0\\\ l\text{ even
}\end{subarray}}^{n}\lambda(n-l,n)B(l)p^{l}=\left\\{\begin{array}[]{ll}A(n)&\text{
if $n$ is even; }\\\ \left(1-\frac{1}{p^{n+1}}\right)A(n)&\text{ if $n$ is
odd, }\end{array}\right.$

and

$\sum_{\begin{subarray}{c}l=0\\\ l\text{ odd
}\end{subarray}}^{n}\lambda(n-l,n)B(l)p^{l+1}=\left\\{\begin{array}[]{ll}\left(1-\frac{1}{p^{n}}\right)A(n)&\text{
if $n$ is even; }\\\ A(n)&\text{ if $n$ is odd. }\end{array}\right.$

###### Proof.

As before let $\pi_{n}=\prod_{i=1}^{n}(1-p^{-i})$. We put
$\beta_{2n}=\prod_{i=1}^{n}(1-p^{-2i})$. If $n=2k$ or $2k+1$ then

$\displaystyle\sum_{\begin{subarray}{c}l=0\\\ l\text{ even
}\end{subarray}}^{n}\lambda(n-l,n)B(l)p^{l}$
$\displaystyle=\sum_{\begin{subarray}{c}l=0\\\ l\text{ even
}\end{subarray}}^{2k}\frac{\pi_{n}}{p^{\binom{l+1}{2}}\pi_{l}\beta_{2k-l}}\cdot\frac{\pi_{l}}{p^{l/2}\beta_{l}^{2}}\cdot
p^{l}$
$\displaystyle=\frac{\pi_{n}}{\beta_{2k}^{2}}\sum_{\begin{subarray}{c}l=0\\\
l\text{ even
}\end{subarray}}^{2k}\frac{\beta_{2k}^{2}}{p^{l^{2}/2}\beta_{2k-l}\beta_{l}^{2}}$
$\displaystyle=\frac{\pi_{n}}{\beta_{2k}^{2}}\sum_{s=0}^{k}\frac{\beta_{2k}^{2}}{p^{2(k-s)^{2}}\beta_{2s}\beta_{2(k-s)}^{2}}.$

The last sum here is $1$, as is seen by taking $(m,n)=(k,k)$ in Lemma A.3 and
replacing $p$ by $p^{2}$. This leaves us with $\pi_{n}/\beta_{2k}^{2}$ which,
upon splitting into the cases $n$ even and $n$ odd, agrees with the answer in
the statement of the lemma.

If $n=2k+1$ or $2k+2$ then

$\displaystyle\sum_{\begin{subarray}{c}l=0\\\ l\text{ odd
}\end{subarray}}^{n}\lambda(n-l,n)B(l)p^{l+1}$
$\displaystyle=\sum_{\begin{subarray}{c}l=1\\\ l\text{ odd
}\end{subarray}}^{2k+1}\frac{\pi_{n}}{p^{\binom{l+1}{2}}\pi_{l}\beta_{2k+1-l}}\cdot\frac{\pi_{l}}{p^{(l+1)/2}\beta_{l-1}\beta_{l+1}}\cdot
p^{l+1}$
$\displaystyle=\frac{\pi_{n}}{\beta_{2k}\beta_{2k+2}}\sum_{\begin{subarray}{c}l=1\\\
l\text{ odd
}\end{subarray}}^{2k+1}\frac{\beta_{2k}\beta_{2k+2}}{p^{(l^{2}-1)/2}\beta_{2k+1-l}\beta_{l-1}\beta_{l+1}}$
$\displaystyle=\frac{\pi_{n}}{\beta_{2k}\beta_{2k+2}}\sum_{s=0}^{k}\frac{\beta_{2k}\beta_{2k+2}}{p^{2(k+1-s)(k-s)}\beta_{2s}\beta_{2(k-s)}\beta_{2(k+1-s)}}.$

The last sum here is $1$, as is seen by taking $(m,n)=(k,k+1)$ in Lemma A.3
and replacing $p$ by $p^{2}$. This leaves us with
$\pi_{n}/(\beta_{2k}\beta_{2k+2})$ which, upon splitting into the cases $n$
even and $n$ odd, agrees with the answer in the statement of the lemma. ∎

######  Lemma A.6.

Let $\pi_{i}(l,m,n)$ be as defined in Section 4.

  1. (i).

For $i,m\in\\{0,1,2\\}$ and $n\geqslant i$ we have

$\pi_{i}(l,m,n)=\frac{1}{2}\left(1+\delta_{m1}+\frac{(i-1)(m-1)}{p^{r/2}}\right)\lambda(r,n-i)$

where $r=2l+m-i$ and $\delta_{m1}$ is the Kronecker delta.

  2. (ii).

Suppose that $f(i,j,n)=\sum_{t=0}^{2}(j-1)^{t}f_{t}(i,n-i-j)$. Then for
$i,j\in\\{0,1,2\\}$ and $n^{\prime}=n-i-j\geqslant 0$ we have

$\displaystyle\sum_{l\geqslant 0}$
$\displaystyle\sum_{m=0}^{2}\pi_{i}(l,m,n-j)f(j,m,n-2l)=\sum_{r=0}^{n^{\prime}}\lambda(r,n^{\prime})f_{0}(j,n^{\prime}-r)$
$\displaystyle+(i-1)\sum_{\begin{subarray}{c}r=0\\\ r\text{ even
}\end{subarray}}^{n^{\prime}}\lambda(r,n^{\prime})p^{-r/2}f_{1}(j,n^{\prime}-r)+\sum_{\begin{subarray}{c}r=0\\\
r+i\text{ even
}\end{subarray}}^{n^{\prime}}\lambda(r,n^{\prime})f_{2}(j,n^{\prime}-r).$

###### Proof.

(i) This follows from the formulae for the $\pi_{i}(l,m,n)$ in Section 4.
Notice that the term involving $p^{r/2}$ only contributes when $i$ and $m$ are
even, in which case $r/2$ is an integer.
(ii) Replacing $l$ by $(r+i-m)/2$ and using (i) the left hand side becomes

$\displaystyle\sum_{\begin{subarray}{c}r=0\\\ r+i\text{ even
}\end{subarray}}^{n^{\prime}}$
$\displaystyle\left(\frac{1}{2}\left(1-\frac{i-1}{p^{r/2}}\right)\lambda(r,n^{\prime})f(j,0,n-i-r)\right.$
$\displaystyle\left.+\,\,\frac{1}{2}\left(1+\frac{i-1}{p^{r/2}}\right)\lambda(r,n^{\prime})f(j,2,n+2-i-r)\right)$
$\displaystyle+\sum_{\begin{subarray}{c}r=0\\\ r+i\text{ odd
}\end{subarray}}^{n^{\prime}}\lambda(r,n^{\prime})f(j,1,n+1-i-r).$

Writing $f$ in terms of $f_{0}$, $f_{1}$, $f_{2}$ this becomes

$\displaystyle\sum_{\begin{subarray}{c}r=0\\\ r+i\text{ even
}\end{subarray}}^{n^{\prime}}$
$\displaystyle\left(\frac{1}{2}\left(1-\frac{i-1}{p^{r/2}}\right)\lambda(r,n^{\prime})\big{[}f_{0}(j,n^{\prime}-r)-f_{1}(j,n^{\prime}-r)+f_{2}(j,n^{\prime}-r)\big{]}\right.$
$\displaystyle\left.+\,\,\frac{1}{2}\left(1+\frac{i-1}{p^{r/2}}\right)\lambda(r,n^{\prime})\big{[}f_{0}(j,n^{\prime}-r)+f_{1}(j,n^{\prime}-r)+f_{2}(j,n^{\prime}-r)\big{]}\right)$
$\displaystyle+\sum_{\begin{subarray}{c}r=0\\\ r+i\text{ odd
}\end{subarray}}^{n^{\prime}}\lambda(r,n^{\prime})f_{0}(j,n^{\prime}-r).$

This simplifies to the expression in the statement of the lemma. Notice that
the sum involving $f_{1}$ only contributes for $i\in\\{0,2\\}$ and so the
condition “$r+i$ even” simplifies to “$r$ even”. ∎

The functions $\phi(i,j,n)$ and $\psi(i,j,n)$ were defined in Definition 5.5.
The aim of this appendix is to prove Proposition 5.6 which for convenience we
now restate.

######  Proposition A.7.

Let $i,j\in\\{0,1,2\\}$ and $n\geqslant i+j$. Then

$\phi(i,j,n)=\sum_{l\geqslant 0}\sum_{m=0}^{2}\pi_{i}(l,m,n-j)\phi(j,m,n-2l),$
(13)

and if $n$ is even then

$\psi(i,j,n)=\sum_{l\geqslant 0}\sum_{m=0}^{2}\pi_{i}(l,m,n-j)\psi(j,m,n-2l).$
(14)

The condition $n\geqslant i+j$ ensures that $\pi_{i}(l,m,n-j)$ is defined. It
can only be non-zero if $2l+m\leqslant n-j$, equivalently $n-2l\geqslant j+m$,
in which case $\phi(j,m,n-2l)$ and $\psi(j,m,n-2l)$ are defined.

###### Proof.

We have $\phi(i,j,n)=(j-1)A(n-i-j)+(i-1)B(n-i-j)$ where $A$ and $B$ were
defined in (12). It follows that
$\phi(i,j,n)=\sum_{t=0}^{2}(j-1)^{t}\phi_{t}(i,n-i-j)$ where

$\phi_{0}(i,n)=(i-1)B(n),\quad\phi_{1}(i,n)=A(n),\quad\phi_{2}(i,n)=0.$

By Lemma A.6(ii) the right hand side of (13) is

$(j-1)\sum_{r=0}^{n^{\prime}}\lambda(r,n^{\prime})B(n^{\prime}-r)+(i-1)\sum_{\begin{subarray}{c}r=0\\\
r\text{ even
}\end{subarray}}^{n^{\prime}}\lambda(r,n^{\prime})p^{-r/2}A(n^{\prime}-r)$

where $n^{\prime}=n-i-j$. By Lemma A.4 the first sum is $A(n^{\prime})$ and
the second sum is $B(n^{\prime})$. This gives
$(j-1)A(n^{\prime})+(i-1)B(n^{\prime})=\phi(i,j,n)$, which is the left hand
side of (13) as required.

We have $\psi(i,j,n)=\sum_{t=0}^{2}(j-1)^{t}\psi_{t}(i,n-i-j)$ where

$\psi_{0}(i,n)=\frac{p^{2d+1}-p^{\delta_{i1}}}{(p+1)(p^{2d+1}-1)},\quad\psi_{1}(i,n)=\frac{(i-1)p^{d}(p^{2}-1)}{(p+1)(p^{2d+1}-1)},\quad\psi_{2}(i,n)=\frac{p^{2d+1}(p-1)}{(p+1)(p^{2d+1}-1)},$

and $d=\lfloor\frac{n+1}{2}\rfloor$.

Now suppose that $n$ is even. If $j\in\\{0,2\\}$ then by Lemma A.6(ii) the
right hand side of (14) is

$\frac{1}{p+1}\left(\sum_{\begin{subarray}{c}r=0\\\ n^{\prime}-r\text{ even
}\end{subarray}}^{n^{\prime}}\lambda(r,n^{\prime})\frac{x-p^{r}}{p^{n^{\prime}+1}-p^{r}}\,\,+\sum_{\begin{subarray}{c}r=0\\\
n^{\prime}-r\text{ odd
}\end{subarray}}^{n^{\prime}}\lambda(r,n^{\prime})\right)$

where $n^{\prime}=n-i-j$ and
$x=p^{n^{\prime}+2}+(i-1)(j-1)p^{n^{\prime}/2}(p^{2}-1)$. By Lemma A.2 this is
equal to $(x-1)/((p+1)(p^{n^{\prime}+1}-1))$ if $i\in\\{0,2\\}$ and $p/(p+1)$
if $i=1$. This is equal to $\psi(i,j,n)$ as required.

If $j=1$ then by Lemma A.6(ii) the right hand side of (14) is

$\frac{p}{p+1}\left(\sum_{\begin{subarray}{c}r=0\\\ n^{\prime}-r\text{ even
}\end{subarray}}^{n^{\prime}}\lambda(r,n^{\prime})\frac{p^{n^{\prime}}-p^{r}}{p^{n^{\prime}+1}-p^{r}}\,\,+\sum_{\begin{subarray}{c}r=0\\\
n^{\prime}-r\text{ odd
}\end{subarray}}^{n^{\prime}}\lambda(r,n^{\prime})\right).$

By Lemma A.2 this is equal to
$(p^{n^{\prime}+1}-p)/((p+1)(p^{n^{\prime}+1}-1))$ if $i=1$ and $1/(p+1)$ if
$i\in\\{0,2\\}$. This is equal to $\psi(i,j,n)$ as required. ∎

## References

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  * [2] Bhargava, M., Cremona, J. E., Fisher, T. A., Jones, N. G., and Keating, J. P. (2016). What is the probability that a random integral quadratic form in n variables has an integral zero?. International Mathematics Research Notices, Issue 12, pp. 3828–3848.
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Lycka Drakengren, Trinity College, Cambridge CB3 9DH, UK

E-mail address<EMAIL_ADDRESS>
Tom Fisher, University of Cambridge, DPMMS, Centre for Mathematical Sciences,
Wilberforce Road, Cambridge CB3 0WB, UK

E-mail address<EMAIL_ADDRESS>