# Quadratic forms in 8 prime variables

Ben Green  Mathematical Institute
Radcliffe Observatory Quarter
Woodstock Road
Oxford OX2 6GG
England<EMAIL_ADDRESS>
###### Abstract.

We give an asymptotic for the number of prime solutions to
$Q(x_{1},\dots,x_{8})=N$, subject to a mild non-degeneracy condition on the
homogeneous quadratic form $Q$.

The argument initially proceeds via the circle method, but this does not
suffice by itself. To obtain a nontrivial bound on certain averages of
exponential sums, we interpret these sums as matrix coefficients for the Weil
representation of the symplectic group
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$. Averages of such matrix
coefficients are then bounded using an amplification argument and a
convergence result for convolutions of measures, which reduces matters to
understanding the action of certain 12-dimensional subgroups in the Weil
representation. Sufficient understanding can be gained by using the basic
represention theory of $\operatorname{SL}_{2}(k)$, $k$ a finite field.

The author is supported by a Simons Investigator grant and is grateful to the
Simons Foundation for their continued support.

###### Contents

  1. 1 Introduction
  2. 2 Outline of the argument
  3. 3 The circle method
  4. 4 The major arcs
  5. 5 Minor arcs: the integral over $\mathfrak{m}_{1}$
  6. 6 The integral over $\mathfrak{m}_{2}$ – first reductions
  7. 7 Exponential sums as matrix coefficients on $\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$
  8. 8 Averages of matrix coefficients
  9. 9 Uniform distribution of convolution powers
  10. 10 Identifying $\Gamma_{q}$
  11. 11 Quasirandomness of $\rho|_{\Gamma_{q}}$.
  12. A Facts about $\operatorname{SL}_{2}(k)$
  13. B Weil representation of $\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$
  14. C Almost-invariant measures

##  1\. Introduction

Let $Q:\mathbf{Z}^{8}\rightarrow\mathbf{Z}$ be a homogeneous quadratic form
with integer coefficients, and let $N\in\mathbf{Z}$. We will study solutions
to $Q(x_{1},\dots,x_{8})=N$ with the $x_{i}$ prime.

Throughout the paper, it is convenient to split
$\mathbf{Z}^{8}=\mathbf{Z}^{4}\times\mathbf{Z}^{4}$ and to write

$Q(x,y)=x^{T}ax+x^{T}by+y^{T}cy$

where $a,b,c\in\operatorname{Mat}_{4}(\mathbf{Z})$ with $a,c$ symmetric.
Equivalently,

$Q(x,y)=(x^{T},y^{T})\begin{pmatrix}a&b/2\\\
b^{T}/2&c\end{pmatrix}\binom{x}{y}.$

Assume that $b$ is invertible. An important role will be played by the 4-by-4
matrix

$\Delta:=4b^{-1}ab^{-T}c-I.$ (1.1)

This formally resembles the discriminant (of a form in two variables), only it
is matrix-valued.

We turn now to the main theorem of the paper. Here, $\Lambda$ is the von
Mangoldt function. For $q$ a positive integer,
$\Lambda_{\mathbf{Z}/q\mathbf{Z}}:\mathbf{Z}/q\mathbf{Z}\rightarrow\mathbf{C}$
takes the value $q/\phi(q)$ when $(x,q)=1$, and $0$ otherwise. As the notation
suggests, this is the natural local variant of the von Mangoldt function. The
normalisation is chosen so that the average value of
$\Lambda_{\mathbf{Z}/q\mathbf{Z}}$ is 1. We abuse notation by writing
$\Lambda_{\mathbf{Z}/p\mathbf{Z}}$ instead of
$\Lambda_{\mathbf{Z}/p^{n}\mathbf{Z}}$ (the domain will always be clear from
context). Finally, we write $\Lambda^{\otimes
4}(x)=\Lambda(x_{1})\Lambda(x_{2})\Lambda(x_{3})\Lambda(x_{4})$ for
$x\in\mathbf{Z}^{4}$, and $\Lambda_{\mathbf{Z}/q\mathbf{Z}}^{\otimes 4}$ is
defined analogously.

######  Theorem 1.1.

Suppose that $Q$ is a quadratic form for which $\det ab\neq 0$ and $\Delta$
has four distinct eigenvalues which lie in
$\overline{\mathbf{Q}}\setminus\\{-1,0\\}$. Then we have the expected local-
global estimate for the number of solutions to $Q(x,y)=N$ in primes, namely
for any real $A$

$\sum_{\begin{subarray}{c}x,y\in[X]^{4}\\\
Q(x,y)=N\end{subarray}}\Lambda^{\otimes
4}(x)\Lambda^{\otimes}(y)=\mathfrak{S}(N)X^{6}+O_{A,Q}(X^{6}\log^{-A}X).$

Here

$\mathfrak{S}(N)=\beta_{\infty}\prod_{p}\beta_{p}(N)$ (1.2)

where $\beta_{p}(N)=\lim_{n\rightarrow\infty}\beta_{p,n}(N)$ is the $p$-adic
density of solutions, where

$\beta_{p,n}(N):=p^{-7n}\sum_{\begin{subarray}{c}x,y\in(\mathbf{Z}/p^{n}\mathbf{Z})^{4}\\\
Q(x,y)\equiv 0(\mbox{\scriptsize{mod}}\,p^{n})\end{subarray}}\Lambda^{\otimes
4}_{\mathbf{Z}/p\mathbf{Z}}(x)\Lambda^{\otimes
4}_{\mathbf{Z}/p\mathbf{Z}}(y),$ (1.3)

and

$\beta_{\infty}:=\lim_{\delta\rightarrow
0}\frac{1}{2\delta}\mu_{\mathbf{R}^{8}}\\{x,y\in[0,1]^{4}:|Q(x,y)-\frac{N}{X^{2}}|\leqslant\delta\\}$

is an archimedean measure of the density of (positive, real) solutions.
Included in the statements is the fact that the limit in the definition of the
$p$-adic density (1.3) exists.

Throughout the paper, we will say that a form $Q$ is _generic_ if it satisfies
the conditions of this theorem, that is to say if $\det ab\neq 0$ and if
$\Delta$ has four distinct eigenvalues in $\overline{\mathbf{Q}}$, and that
neither $0$ nor $-1$ is one of these eigenvalues. The word generic is
appropriate, since this condition holds for a Zariski-dense set of $(a,b,c)$
in the 36-dimensional parameter space where $a,c$ are symmetric. To see this,
first note that the condition that $\Delta$ has distinct eigenvalues is
(Zariski-)closed, by considering the resultant of the characteristic
polynomial $\rho_{\Delta}(\lambda):=\det(\Delta-\lambda I)$ and its derivative
$\rho^{\prime}_{\Delta}(\lambda)$. The conditions that $\det ab=0$, and that
$\Delta$ has an eigenvalue $0$ or $-1$, are evidently closed conditions.
Finally, these conditions are nontrivial (i.e. not always satisfied) as one
can see by taking $a$ to be a diagonal matrix with distinct rational
eigenvalues (not $0$ or $\frac{1}{4}$) and $b=c=I$.

Note that $\Delta$ is not canonically associated to $Q$, being dependent on
the splitting of variables into two sets of four. However, one may observe
that if $a$ is invertible then

$\begin{pmatrix}a&b/2\\\ b^{T}/2&c\end{pmatrix}=\begin{pmatrix}1&0\\\
A^{T}&1\end{pmatrix}\begin{pmatrix}B&0\\\
0&C\end{pmatrix}\begin{pmatrix}1&A\\\ 0&1\end{pmatrix}$

with $A=\frac{1}{2}a^{-1}b$, $B=a$, $C=\frac{1}{4}b^{T}a^{-1}b\Delta$, and so

$\det Q:=\det\begin{pmatrix}a&b/2\\\ b^{T}/2&c\end{pmatrix}=2^{-16}(\det
b)^{2}\det\Delta.$ (1.4)

(This is also true without the assumption that $a$ is invertible by a
continuity argument.) Therefore the requirement that $0$ is not an eigenvalue
of $\Delta$ is essentially an invariant of $Q$, more-or-less equivalent to
$\det Q\neq 0$ (called the “regular” case in previous works such as [29]). We
do not expect any variant of our methods to handle the irregular case $\det
Q=0$, which essentially corresponds to forms in 7 or fewer variables.

_Previous results._ Jianya Liu [12] handled generic quadratic forms in 10
prime variables. Lilu Zhao [29] subsequently handled all regular quadratic
forms in 9 variables. These works use fairly classical forms of the Hardy-
Littlewood circle method. For reasons we will go into later, 9 variables
appears to be the limit of what any such method can give, and so far as I am
aware results in 8 or fewer variables are known only for particular types of
form with some degenerate and/or diagonal behaviour. For diagonal forms, 5
variables suffice by work of Hua [9]. If one is content with almost-primes
instead of primes, the number of variables can be reduced to $3$: see [3] and
subsequent works.

_Future work._ In our main theorem we imposed conditions on $Q$, namely that
$\det ab\neq 0$ and the matrix $\Delta$ has four distinct eigenvalues in
$\overline{\mathbf{Q}}\setminus\\{0,-1\\}$. Whilst these are not especially
severe restrictions, it nonetheless seems to be of interest to weaken them as
far as possible, and we intend to address this in future work.

There are at least three paths to pursue in this direction. First, there are
35 essentially different ways to split $8$ variables into two groups of $4$,
which one would expect to lead to $\Delta$s with different properties. Second,
many of the arguments of the paper can be modified to work in more degenerate
situations. Finally, in some highly degenerate situations Theorem 1.1 can be
established by classical methods such as those in [29]. One would expect this
to be the case when $\det b=0$ for all splittings of the 8 variables (the case
of low “off-diagonal rank”). We anticipate this to be a somewhat tedious
endeavour, with all the main ideas already present in this paper and [29].

On a different matter, allowing linear terms in $Q$ (that is, nonhomogeneous
quadratics) is probably possible but seems to require a fairly significant
modification of the method, and we will not attempt this here.

_Notation._ Most of our notation is fairly standard. We write $e(t)=e^{2\pi
it}$, and for $q$ a positive integer we write $e_{q}(x):=e^{2\pi ix/q}$. If
$S$ is a finite set and $F:S\rightarrow\mathbf{C}$ a function, we write
$\mathbb{E}_{x\in S}F(x)$ to mean the average of $F$ over $S$. We write
$\mathbf{T}=\mathbf{R}/\mathbf{Z}$, and we write $\|x\|_{\mathbf{T}}$ for the
distance from $x$ to the nearest integer. We write $[X]$ to denote the
discrete interval $\\{1,\dots,X\\}$, and $[0,X]$ for the continuous interval
$\\{x:0\leqslant x\leqslant X\\}\subset\mathbf{R}$.

If $G$ is a finite group then we write $\ell^{2}(G)$ for the vector space of
all functions $f:G\rightarrow\mathbf{C}$ together with the inner product
$\langle f_{1},f_{2}\rangle:=\mathbb{E}_{x\in G}f_{1}(x)\overline{f_{2}(x)}$
and the associated norm $\|f\|_{2}:=\sqrt{\langle f,f\rangle}$. Later on in
the paper we will also define norms of probability measures, and we caution
the reader that there we will use a different normalisation.

If $V$ is a Hermitian inner product space (such as $\ell^{2}(G)$) then we
write $\operatorname{U}(V)$ for the group of unitary transformations of $V$.

For $p$ a prime, we will freqently encounter the group
$\mathbf{Z}/p\mathbf{Z}$. When this arises as a group or a ring, we will write
it $\mathbf{Z}/p\mathbf{Z}$, but when it is important that it is a field, we
will write $\mathbf{F}_{p}$. This may seems slightly eccentric, but it does
not seem stylistically correct to talk about homomorphisms from
$\mathbf{Z}/q\mathbf{Z}$ to $\mathbf{F}_{p}$ (when $p|q$) and nor does it seem
right to discuss the field $(\mathbf{Z}/p\mathbf{Z})(\theta)$ or the algebraic
closure $\overline{\mathbf{Z}/p\mathbf{Z}}$. At times the distinction is
somewhat arbitrary.

Throughout the paper we abuse notation in certain standard ways which should
not cause any confusion. For example, we also write $Q(x,y)$ for the quadratic
form $x^{T}ax+x^{T}by+y^{T}cy$ over $\mathbf{Z}/q\mathbf{Z}$, by which we mean
that $x,y\in(\mathbf{Z}/q\mathbf{Z})^{4}$ and $a,b$ and $c$ are to be
considered $(\operatorname{mod}\,q)$. Slightly more subtly, we also consider
$\Delta$ (defined in (1.1) as an element of
$\operatorname{Mat}_{4}(\mathbf{Q})$) as an element of
$\operatorname{Mat}_{4}(\mathbf{Z}/p\mathbf{Z})$, which makes sense provided
$p\nmid\det b$. Similarly, we consider a certain 8-by-8 symplectic matrix $g$
(defined in (2.3) below) as an element of
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$ for squarefree $q$, which
again will make sense provided no prime factor of $q$ divides $\det b$.

Finally, we regard the quadratic form $Q$ as fixed throughout the paper and
will not explicitly indicate dependence on $Q$ in asymptotic notation such as
$\ll$ or $O()$.

_Acknowledgements._ It is a pleasure to thank Emmanuel Breuillard, Charlotte
Chan, Tom Fisher and Balázs Szendrői for helpful correspondence related to
this work and earlier versions of it, and Roger Baker and James Maynard for
discussions which introduced me to the problem in around 2014. The author is a
Simons Investigator and is very grateful to the Simons Foundation for their
continued support.

##  2\. Outline of the argument

The initial steps of the argument proceed in the classical fashion using the
circle method, which we set up in Section 3. We introduce the exponential sum

$S(\theta):=\sum_{x,y\in[X]^{4}}\Lambda^{\otimes 4}(x)\Lambda^{\otimes
4}(y)e(\theta Q(x,y)),$

where $\Lambda^{\otimes 4}(x)$ is shorthand for
$\Lambda(x_{1})\Lambda(x_{2})\Lambda(x_{3})\Lambda(x_{4})$, and of course
$\Lambda$ denotes the von Mangoldt function. Then by orthogonality we have

$\sum_{\begin{subarray}{c}x,y\in[X]^{4}\\\
Q(x,y)=N\end{subarray}}\Lambda^{\otimes 4}(x)\Lambda^{\otimes
4}(y)=\int_{\mathbf{T}}S(\theta)e(-\theta N)d\theta.$

We divide $\mathbf{T}$ into the major arcs $\mathfrak{M}$ (roughly, the set of
$\theta$ within distance $\sim X^{-2}$ from a rational $\frac{a}{q}$ with
$q\ll\log^{O(1)}X$) and the minor arcs $\mathfrak{m}$. The major arcs give the
main term in the asymptotic, and the analysis of them is entirely classical.
We give this analysis in Section 4, referring to [29] for the details when
possible.

For the minor arcs, the fact that we are discussing primes is essentially
irrelevant and the same arguments work with $\Lambda^{\otimes
4}(x)\Lambda^{\otimes 4}(y)$ replaced by $F_{1}(x)F_{2}(y)$ for any reasonably
bounded functions $F_{1},F_{2}$. We in fact divide the minor arcs into two
sets $\mathfrak{m}_{1}$ and $\mathfrak{m}_{2}$, with $\mathfrak{m}_{1}$ being
points not too close to a rational and $\mathfrak{m}_{2}$ being the points
very close to a rational (but with moderately large denominator). The precise
definitions are given at the start of Section 3. The treatment of the integral
over $\mathfrak{m}_{1}$ uses diophantine approximation arguments standard in
the area, and is given in Section 5.

The treatment of the minor arcs $\mathfrak{m}_{2}$ is the heart of the paper.
One may reduce to considering actual rational points $\frac{r}{q}$, with
$q>\log^{C}X$ moderately large, and one is then led naturally led to look at
exponential sums of the form

$T_{f_{1},f_{2}}(r):=q^{2}\mathbb{E}_{x,y\in(\mathbf{Z}/q\mathbf{Z})^{4}}f_{1}(x)f_{2}(y)e_{q}(rQ(x,y)),$
(2.1)

where here $r\in(\mathbf{Z}/q\mathbf{Z})^{*}$. There is a “trivial” upper
bound of $1$ for such sums when $\|f_{1}\|_{2},\|f_{2}\|_{2}\leqslant 1$,
which turns out to be (just) not good enough for the purposes of bounding the
integral over $\mathfrak{m}_{2}$. However, any improvement of it by a factor
$q^{-\delta}$ would suffice.

Unfortunately, there is no such improvement: the trivial bound is best
possible. However, by a less wasteful reduction we can arrange things so that
we consider instead the averages

$\frac{1}{\phi(q)}\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}|T_{f_{1},f_{2}}(r)|.$
(2.2)

Again, a saving of $q^{-\delta}$ over the trivial bound of $1$ would be
enough.

We incorporate some additional tricks which allow us to restrict attention to
the case $q$ squarefree and without very small prime factors, two features
which are vital in our later arguments. The details of these reductions are
given in Section 6.

The remainder of the paper is devoted to establishing a nontrivial bound of
the required strength for averages (2.2). To make progress on this problem, we
interpret the exponential sums $T_{f_{1},f_{2}}(r)$ as matrix coefficients
$\langle f_{1},\rho(g^{(r)})f_{2}\rangle$, where here
$\rho:\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})\rightarrow\operatorname{U}(\ell^{2}((\mathbf{Z}/q\mathbf{Z})^{4}))$
is a certain unitary representation of the symplectic group
$\operatorname{Sp}_{8}$ over $\mathbf{Z}/q\mathbf{Z}$ called the Weil
representation. After a brief introduction to the symplectic group and the
Weil representation, we give this interpretation in Section 7. Whilst the
theory of the Weil representation is well-known over $\mathbf{R}$ and somewhat
well-known over finite fields, we do not know of a good source for the theory
we need over $\mathbf{Z}/q\mathbf{Z}$, so we must develop some of this
ourselves. This is fairly straightforward given the finite field statements,
and is done in Appendix B.

The elements $g^{(r)}\in\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$ are
what we call “dilates” of a single element $g=g^{(1)}$ given by the formula

$g:=\begin{pmatrix}-2b^{-T}c&b^{-T}\\\ 4ab^{-T}c-b&-2ab^{-T}\end{pmatrix}.$
(2.3)

The _dilate_ of $\big{(}\begin{smallmatrix}A&B\\\
C&D\end{smallmatrix}\big{)}\in\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$
by $r\in(\mathbf{Z}/q\mathbf{Z})^{*}$ is
$\big{(}\begin{smallmatrix}A&r^{-1}B\\\ rC&D\end{smallmatrix}\big{)}$; this is
in fact an action of $(\mathbf{Z}/q\mathbf{Z})^{*}$ by automorphisms, as may
be easily checked.

One is therefore led to the question of bounding an average of matrix
coefficients $|\langle f_{1},\rho(g^{(r)})f_{2}\rangle|$, where $r$ ranges
over $(\mathbf{Z}/q\mathbf{Z})^{*}$.

In Section 8 we supply a general tool for bounding averages of matrix
coefficients, in principle applicable to any unitary representation $\psi$ of
any finite group $G$. This allows one to bound an average

$\int_{G}|\langle f_{1},\psi(x)f_{2}\rangle|d\mu(x),$

where $\mu$ is a probability measure on $G$, when two conditions are
satisfied:

  1. (1)

(convergence to uniform measure) Some symmetrised convolution power
$\mu^{\circ}\ast\mu\ast\mu^{\circ}\ast\mu\cdots$ of bounded order should be
close to the uniform measure on a subgroup $H\leqslant G$;

  2. (2)

(quasirandomness) $\psi|_{H}$ has no low-dimensional irreducible components.

We wish to apply this tool with
$G=\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$, $\psi=\rho$ being the Weil
representation, and $\mu$ being the uniform measure on the $\phi(q)$ points
$g^{(r)}$, $r\in(\mathbf{Z}/q\mathbf{Z})^{*}$. To do this we need to establish
the convergence and quasirandomness properties.

The task (1) of showing that (symmetrised) convolution powers of $\mu$
converge to a uniform measure on a subgroup suggests the literature on the
affine sieve, expanders and general measure convolutions in groups of Lie
type, in particular the work of Varjú [27] which provides results in the
appropriate generality. Some variant of this can probably be made to work in
our context. However, our particular measure $\mu$ has a rather algebraic
definition, being parametrised by (very simple) rational functions and we are
able to offer an alternative approach using the Lang-Weil estimate. This is
inspired by a blog post of Tao [25], giving an alternative proof (inspired by
model-theoretic work of Pillay and Starchenko [15] and unpublished notes of
Hrushovski) of his own algebraic regularity lemma [24]. This may be of
independent interest, though we only develop it in the specific setting of
interest to us here. This allows one to demonstrate rapid convergence of
(symmetrised) powers of $\mu$ to the uniform measure on the group $\Gamma_{q}$
generated by the elements $g^{-(r)}g^{(s)}$, without knowing _a priori_ what
this group is. The arguments may be found in Section 9.

The remaining task (2) is to establish the quasirandomness property for
$\rho|_{\Gamma_{q}}$. It is easy to see that
$\Gamma_{q}\cong\prod_{p|q}\Gamma_{p}$, and so by using the basic theory of
tensor product representations it turns out to be enough to understand the
case $q=p$ prime. First, we identify $\Gamma_{p}$ explicitly. I was initially
under the impression that the elements $g^{-(r)}g^{(s)}$ might generically
generate the whole of $\operatorname{Sp}_{8}(\mathbf{Z}/p\mathbf{Z})$ (which
has size $\sim p^{36}$), on the grounds that there is no immediately evident
reason why they should not, and for the analogous situation in
$\operatorname{Sp}_{2}(\mathbf{Z}/p\mathbf{Z})$ this is true. However, it
turns out that this is not the case, and that $\Gamma_{p}$ is (generically) a
group of size $\sim p^{12}$, a conjugate (in
$\operatorname{GL}_{8}(\mathbf{Z}/p\mathbf{Z})$) of
$\operatorname{SL}_{2}(\mathbf{F}_{p}[\Delta])$, where here $\Delta$ is the
matrix discriminant given in (1.1). Establishing this takes some work,
involving calculations in $\operatorname{SL}_{2}$ together with applications
of lemmas of Goursat and Ribet on subgroups of direct products. A number of
facts about $\operatorname{SL}_{2}$ of a finite field are required here, and
these are collated in Appendix A. These tasks are accomplished in Section 10.

With $\Gamma_{p}$ identified explicitly, we turn to the quasirandomness
property itself. It is essentially automatic from the representation theory of
$\operatorname{SL}_{2}(\mathbf{F}_{p^{n}})$ that if $\rho|_{\Gamma_{p}}$ has
an irreducible component of small degree, then this component must be the
trivial representation: that is, $\Gamma_{p}$, acting via the Weil
representation on $\ell^{2}((\mathbf{Z}/p\mathbf{Z})^{4})$, would have a
nontrivial fixed vector. The final task of the paper, then, is to rule this
out. We do this in Section 11 using rather direct and explicit (that is, not
using any representation theory) arguments.

##  3\. The circle method

In this section we describe the basic setup of the circle method. As is
typical in problems of this type we will be aiming for error terms in our main
theorem of $O_{A}(X^{6}\log^{-A}X)$, for an arbitrary positive real number
$A$. Fix such an $A$, without loss of generality $A\geqslant 10$, and set

$M:=\log^{C_{1}}X,\quad M^{\prime}:=\log^{C_{2}}X,$ (3.1)

where $C_{1}:=10A$ and $C_{2}:=10^{5}A\delta^{-1}$, where $\delta$ is the
exponent appearing in Proposition 6.2 below (these choices are by no means
optimal, but this is inconsequential).

Set

$K:=8\max_{ij}|b_{ij}|,$ (3.2)

where the $b_{ij}$ are the entries of the 4-by-4 matrix $b$. Thus $K$ is a
constant depending only on the quadratic form $Q$. For $q\in\mathbf{N}$ and
for $r\in(\mathbf{Z}/q\mathbf{Z})^{*}$, denote

$I_{r,q}:=\\{\theta\in\mathbf{T}:|\theta-\frac{r}{q}|\leqslant\frac{M}{X^{2}}\\}$
(3.3)

and

$\tilde{I}_{r,q}:=\\{\theta\in\mathbf{T}:|\theta-\frac{r}{q}|\leqslant\frac{1}{KqX}\\}.$
(3.4)

Define the _major arcs_

$\mathfrak{M}:=\bigcup_{\begin{subarray}{c}q\leqslant M^{\prime}\\\
r\in(\mathbf{Z}/q\mathbf{Z})^{*}\end{subarray}}I_{r,q},$ (3.5)

and set

$\mathfrak{m}_{1}:=\bigcup_{\begin{subarray}{c}q\leqslant KX\\\
r\in(\mathbf{Z}/q\mathbf{Z})^{*}\end{subarray}}(\tilde{I}_{r,q}\setminus
I_{r,q}),\quad\mathfrak{m}_{2}:=\bigcup_{\begin{subarray}{c}M^{\prime}<q\leqslant
KX\\\ r\in(\mathbf{Z}/q\mathbf{Z})^{*}\end{subarray}}I_{r,q}.$ (3.6)

######  Lemma 3.1.

We have $\mathfrak{M}\cup\mathfrak{m}_{1}\cup\mathfrak{m}_{2}=\mathbf{T}$.

###### Proof.

By Dirichlet’s theorem on diophantine approximation,

$\mathbf{T}=\bigcup_{\begin{subarray}{c}q\leqslant KX\\\
r\in(\mathbf{Z}/q\mathbf{Z})^{*}\end{subarray}}\tilde{I}_{r,q}.$

The result then follows immediately. ∎

Set

$S(\theta):=\sum_{x,y\in[X]^{4}}\Lambda^{\otimes 4}(x)\Lambda^{\otimes
4}(y)e(\theta Q(x,y)),$ (3.7)

where, recall, $\Lambda^{\otimes 4}(x)$ is a convenient shorthand for
$\prod_{i=1}^{4}\Lambda(x_{i})$. Then by orthogonality we have

$\sum_{\begin{subarray}{c}x,y\in[X]^{4}\\\
Q(x,y)=N\end{subarray}}\Lambda^{\otimes 4}(x)\Lambda^{\otimes
4}(y)=\int_{\mathbf{T}}S(\theta)e(-\theta N)d\theta.$ (3.8)

We evaluate this by considering the contributions to the integral from
$\mathfrak{M},\mathfrak{m}_{1},\mathfrak{m}_{2}$ separately. The major arcs
$\mathfrak{M}$ give the main term in the asymptotic, as the following result
shows.

######  Proposition 3.2 (Major arcs).

Suppose that $Q$ is regular, that is to say
$\big{(}\begin{smallmatrix}a&b/2\\\ b^{T}/2&c\end{smallmatrix}\big{)}$ is
nonsingular. Then we have

$\int_{\mathfrak{M}}S(\theta)e(-\theta
N)d\theta=\mathfrak{S}(N)X^{6}+O_{A}(X^{6}\log^{-A}X),$

where the singular series $\mathfrak{S}(N)$ is as described in Theorem 1.1.

We will prove this in the next section using classical methods, referring to
[29] for most of the details.

Now we turn to the minor arcs $\mathfrak{m}_{1}$ and $\mathfrak{m}_{2}$. Here,
as previously remarked, the fact that we are dealing with primes and the von
Mangoldt function is essentially irrelevant. For any functions
$F_{1},F_{2}:[X]\rightarrow\mathbf{C}$ we introduce the sums

$S_{F_{1},F_{2}}(\theta):=\sum_{x,y\in[X]^{4}}F_{1}(x)F_{2}(y)e(\theta
Q(x,y)).$ (3.9)

######  Proposition 3.3 (Minor arcs $\mathfrak{m}_{1}$).

Suppose that $\det b\neq 0$. Then we have

$\int_{\mathfrak{m}_{1}}|S_{F_{1},F_{2}}(\theta)|d\theta\ll_{A}X^{6}\log^{-A-8}X,$

uniformly for all $1$-bounded functions $F_{1},F_{2}$.

We will prove this in Section 5, using diophantine approximation arguments
typical of the circle method.

######  Proposition 3.4 (Minor arcs $\mathfrak{m}_{2}$).

Suppose that $Q$ is generic (that is, $\Delta$ has four distinct eigenvalues
in $\overline{\mathbf{Q}}\setminus\\{0,-1\\}$). Then we have

$\int_{\mathfrak{m}_{2}}|S_{F_{1},F_{2}}(\theta)|d\theta\ll_{A}X^{6}\log^{-A-8}X,$

uniformly for all $1$-bounded functions $F_{1},F_{2}$.

This proof of this, which is a substantial undertaking, contains the new ideas
of the paper and occupies the remaining sections.

Let us conclude this section by remarking that Propositions 3.2, 3.3 and 3.4
easily combine to establish Theorem 1.1. Indeed, by (3.8) and Proposition 3.2
we have

$\displaystyle\big{|}\sum_{\begin{subarray}{c}x,y\in[X]^{4}\\\
Q(x,y)=N\end{subarray}}\Lambda^{\otimes 4}(x)$ $\displaystyle\Lambda^{\otimes
4}(y)-\mathfrak{S}(N)X^{6}\big{|}$ $\displaystyle\leqslant
O_{A}(X^{6}\log^{-A}X)+|\int_{\mathbf{T}\setminus\mathfrak{M}}S(\theta)e(-\theta
N)d\theta|.$

By the triangle inequality and Lemma 3.1, the second term on the right is
bounded above by

$\int_{\mathfrak{m}_{1}}|S(\theta)|d\theta+\int_{\mathfrak{m}_{2}}|S(\theta)|d\theta.$

By Propositions 3.3 and 3.4 (taking $F_{1}=F_{2}=(\log X)^{-4}\Lambda^{\otimes
4}$), both of these terms are bounded by $O_{A}(X^{6}\log^{-A}X)$.

##  4\. The major arcs

In this section we establish Proposition 3.2. The argument is very classical
and in fact large portions of it may be simply quoted from [29]. For this part
of the argument, similar results hold with as few as 5 variables. Define

$C(q,r):=\sum_{\begin{subarray}{c}x,y\in(\mathbf{Z}/q\mathbf{Z})^{4}\\\
(x,q)=(y,q)=1\end{subarray}}e_{q}(rQ(x,y)),$ (4.1)
$B_{Q,N}(q):=\frac{1}{\phi(q)^{8}}\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}C(q,r)e_{q}(-rN),$
(4.2) $\mathfrak{B}_{Q,N}:=\sum_{q=1}^{\infty}B_{Q,N}(q),$ (4.3)
$I(t):=\int_{u,v\in[0,X]^{4}}e(tQ(u,v))dudv$ (4.4)

and

$\mathfrak{J}_{Q,N}(X):=\int^{\infty}_{-\infty}I(t)e(-Nt).$ (4.5)

These are the same definitions as those in [29, Section 3], with some
notational substitutions (Zhao’s $f,t,\beta,\mathfrak{S}(f,t)$ become our
$Q,N,t,\mathfrak{B}_{Q,N}$ respectively). Also, we notate our quadratic forms
using two variables $x,y$. Definitions like these will be familiar to anyone
with knowledge of the circle method. The following is [29, Lemma 3.6].

######  Lemma 4.1 (Major arcs).

We have

$\int_{\mathfrak{M}}S(\theta)e(-\theta
N)d\theta=\mathfrak{B}_{Q,N}\mathfrak{J}_{Q,N}(X)+O_{A}(X^{6}\log^{-A}X).$
(4.6)

_Remarks._ Our choice of $C_{1}$ and $C_{2}$ in the definition (3.1) of
$M,M^{\prime}$ ensures that our major arcs are amply wide enough that the
error term in [29, Lemma 3.6] is $O_{A}(X^{6}\log^{-A}X)$. The only
consequence of taking the major arcs this wide is that the choice of exponent
in the error term of the Siegel-Walfisz theorem towards the end of [29,
Section 3] (which is, in any case, not made explicit there) must be larger.

There is one further inconsequential difference between our setup and that in
[29]. In [29] the major arc about $\frac{r}{q}$ has width $\frac{M}{qX^{2}}$,
whereas we have taken the width to be $\frac{M}{X^{2}}$. The only other tiny
change required is in (3.17), (3.19) of [29] where the integrals should be
taken over our slightly longer range $|\beta|\leqslant M/X^{2}$ (which
actually helps slightly).

To reconcile this with Proposition 3.2 we must express $\mathfrak{B}_{Q.N}$
and $\mathfrak{J}_{Q,N}(X)$ in terms of the local densities
$\beta_{p},\beta_{\infty}$, whose definitions are given in the statement of
Theorem 1.1. This is again a standard endeavour, but it is not done in Zhao’s
paper so we give brief details now.

Recall the definition (1.3) of $\beta_{p,n}(N)$. By orthogonality, we have

$\beta_{p,n}(N)=p^{-8n}\sum_{r\in\mathbf{Z}/p^{n}\mathbf{Z}}\sum_{x,y\in(\mathbf{Z}/p^{n}\mathbf{Z})^{4}}\Lambda_{\mathbf{Z}/p\mathbf{Z}}^{\otimes
4}(x)\Lambda_{\mathbf{Z}/p\mathbf{Z}}^{\otimes 4}(y)e_{p^{n}}(r(Q(x,y)-N)).$

In the sum over $r$, write $r=p^{n-j}r^{\prime}$ with $(r^{\prime},p)=1$. One
may check that the contribution from a particular $j$ is $B_{Q,N}(p^{j})$, and
so

$\beta_{p,n}(N)=\sum_{j=0}^{n}B_{Q,N}(p^{j}).$

Taking the limit as $n\rightarrow\infty$ gives

$\beta_{p}(N)=\sum_{j=0}^{\infty}B_{Q,N}(p^{j}).$

Finally, since $B_{Q,N}(q)$ is a multiplicative function of $q$ (see [29,
Lemma 3.1]) we have

$\prod_{p}\beta_{p}(N)=\sum_{q=1}^{\infty}B_{Q,N}(q)=\mathfrak{B}_{Q,N}.$
(4.7)

There are, of course, convergence issues to be dealt with here, but these are
fully fleshed out in [29, Lemma 3.4].

To handle the archimedean factor $\beta_{\infty}$, we proceed is as follows
(we leave detailed analytic justifications to the reader). For
$\varepsilon>0$, set

$f_{\varepsilon}(w):=\frac{1}{2\varepsilon}\mu_{\mathbf{R}^{4}}\\{u,v\in\mathbf{R}^{4}:u,v\in[X]^{4},w-\varepsilon\leqslant
Q(u,v)\leqslant w+\varepsilon\\}.$

Fourier inversion then gives

$f_{\varepsilon}(N)=\int^{\infty}_{-\infty}\hat{f}_{\varepsilon}(t)e(Nt)dt.$

However,

$I(t)=\lim_{\varepsilon\rightarrow 0}\hat{f}_{\varepsilon}(-t).$

Taking limits as $\varepsilon\rightarrow 0$ (and substituting
$x:=\frac{u}{X}$, $y:=\frac{v}{X}$, $\delta:=\frac{\varepsilon}{X^{2}}$ and
using the homogeneity of $Q$) gives

$\mathfrak{J}_{Q,N}(X)=\beta_{\infty}X^{6}.$ (4.8)

Substituting (4.7) and (4.8) into Lemma 4.1 gives Proposition 3.2.

##  5\. Minor arcs: the integral over $\mathfrak{m}_{1}$

In this section we prove Proposition 3.3. The reader may wish to recall the
definitions of $I_{r,q}$, $\tilde{I}_{r,q}$ and $\mathfrak{m}_{1}$, which are
(3.3), (3.4) and (3.6) respectively.

###### Proof of Proposition 3.3.

Observe that $\tilde{I}_{r,q}\setminus I_{r,q}$ is empty if $q>X/KM$, and
therefore

$\mu(\mathfrak{m}_{1})\leqslant\sum_{q\leqslant
X/KM}\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}\frac{2}{KqX}<\frac{1}{M}.$ (5.1)

Write

$S_{F_{1},F_{2}}(\theta):=\sum_{x,y\in[X]^{4}}F^{\prime}_{1}(x)F^{\prime}_{2}(y)e(\theta
x^{T}by)$ (5.2)

where $F^{\prime}_{1}(x):=F_{1}(x)e(\theta x^{T}ax)$,
$F^{\prime}_{2}(y):=F_{2}(y)e(\theta y^{T}cy)$. By Cauchy-Schwarz,

$|S_{F_{1},F_{2}}(\theta)|\leqslant
X^{2}\big{(}\sum_{y,y^{\prime}\in[X]^{4}}F^{\prime}_{2}(y)\overline{F^{\prime}_{2}(y^{\prime})}\sum_{x\in[X]^{4}}e(\theta
x^{T}b(y-y^{\prime}))\big{)}^{1/2}.$

By the standard estimate

$|\sum_{x\in[X]}e(\xi x)|\ll\min(X,\|\xi\|_{\mathbf{T}}^{-1})$

and since $\|F_{2}\|_{\infty}\leqslant 1$ it follows that

$|S_{F_{1},F_{2}}(\theta)|\ll
X^{2}\big{(}\sum_{y,y^{\prime}\in[X]^{4}}\prod_{j=1}^{4}\min(X,\|\theta(b(y-y^{\prime}))_{j}\|_{\mathbf{T}}^{-1})\big{)}^{1/2}.$
(5.3)

Now the image of $[X]^{4}\times[X]^{4}$ under the map $(y,y^{\prime})\mapsto
b^{T}(y-y^{\prime})$ is contained in the box
$[-\frac{1}{2}KX,\frac{1}{2}KX]^{4}$ and the fibres are of size at most
$X^{4}$ (since $b$ is nonsingular; recall also from the definition (3.2) that
$K=8\max_{ij}|b_{ij}|$). It follows from (5.3) that

$|S_{F_{1},F_{2}}(\theta)|\ll
X^{4}\big{(}\sum_{\begin{subarray}{c}h\in\mathbf{Z}\\\ |h|\leqslant
KX/2\end{subarray}}\min(X,\|\theta h\|_{\mathbf{T}}^{-1})\big{)}^{2}.$ (5.4)

Suppose now that $\theta\in\tilde{I}_{r,q}\setminus I_{r,q}$, thus

$\theta=\frac{r}{q}+\eta\quad\mbox{where}\quad\frac{M}{X^{2}}<|\eta|\leqslant\frac{1}{KqX}.$

Foliating into progression modulo $q$ we have

$\displaystyle\sum_{|h|\leqslant KX/2}\min(X,$ $\displaystyle\|\theta
h\|_{\mathbf{T}}^{-1})$ $\displaystyle=\sum_{b(\mbox{\scriptsize
mod}\,q)}\sum_{\begin{subarray}{c}|h|\leqslant KX/2\\\ h\equiv
b(\mbox{\scriptsize
mod}\,q)\end{subarray}}\min(X,\|(\frac{r}{q}+\eta)h)\|_{\mathbf{T}}^{-1}).$
(5.5)

We evaluate the contributions from $b=0$ and $b\neq 0$ separately. If $b\neq
0$, $h\equiv b(\operatorname{mod}\,q)$ and $|h|\leqslant KX/2$ then

$\|(\frac{r}{q}+\eta)h\|_{\mathbf{T}}\geqslant\|\frac{rb}{q}\|_{\mathbf{T}}-\frac{KX}{2}|\eta|\geqslant\|\frac{rb}{q}\|_{\mathbf{T}}-\frac{1}{2q}.$

Thus, if $b\neq 0$,

$\sum_{\begin{subarray}{c}|h|\leqslant KX/2\\\ h\equiv b(\mbox{\scriptsize
mod}\,q)\end{subarray}}\min(X,\|(\frac{r}{q}+\eta)h)\|_{\mathbf{T}}^{-1})\ll\frac{X}{q}\big{(}\|\frac{rb}{q}\|_{\mathbf{T}}-\frac{1}{2q}\big{)}^{-1}.$
(5.6)

(Recall here that $q\leqslant KX$, so the number of terms in the sum over $h$
is indeed $\ll X/q$.) Now as $b$ ranges over
$(\mathbf{Z}/q\mathbf{Z})\setminus\\{0\\}$, so does $rb$. Thus

$\sum_{\begin{subarray}{c}b(\mbox{\scriptsize mod}\,q)\\\ b\neq
0\end{subarray}}\big{(}\|\frac{rb}{q}\|_{\mathbf{T}}-\frac{1}{2q}\big{)}^{-1}=\sum_{\begin{subarray}{c}s(\mbox{\scriptsize
mod}\,q)\\\ s\neq
0\end{subarray}}\big{(}\|\frac{s}{q}\|_{\mathbf{T}}-\frac{1}{2q}\big{)}^{-1}\ll
q\log q.$

Substituting into (5.6), we see that the contribution to the right-hand side
of (5.5) from the terms with $b\neq 0$ is $\ll X\log q=O(X\log X)$.

Now we look at the contribution to the right-hand side of (5.5) from $b=0$.
Making the substitution $h=kq$, this is

$\sum_{|k|\leqslant KX/2q}\min(X,\|\eta kq\|_{\mathbf{T}}^{-1}).$ (5.7)

We have

$|\eta kq|\leqslant\frac{1}{KqX}\cdot\frac{KX}{2q}\cdot q<\frac{1}{2},$

so $\|\eta kq\|_{\mathbf{T}}=|\eta kq|$. Therefore (5.7) is

$\sum_{|k|\leqslant KX/2q}\min(X,|\eta kq|^{-1})\leqslant
X+\sum_{0<|k|<KX/2q}|\eta kq|^{-1}\ll X+\frac{1}{\eta q}\log X.$

Substituting these bounds for $b\neq 0$ and $b=0$ into (5.5), we obtain

$\sum_{|h|\leqslant KX/2}\min(X,\|\theta
h\|_{\mathbf{T}}^{-1})\ll(X+\frac{1}{\eta q})\log X.$

Substituting into (5.4) gives, for
$\theta=\frac{r}{q}+\eta\in\tilde{I}_{r,q}\setminus I_{r,q}$,

$|S_{F_{1},F_{2}}(\theta)|\ll
X^{6}\log^{2}X+\frac{X^{4}\log^{2}X}{\eta^{2}q^{2}}.$

To complete the proof of Proposition 3.3, we need to integrate this estimate
over $\theta\in\mathfrak{m}_{1}$, that is to say over all
$\tilde{I}_{r,q}\setminus I_{r,q}$ with $q\leqslant KX$ and
$r\in(\mathbf{Z}/q\mathbf{Z})^{*}$. The contribution from the first term
$X^{6}\log^{2}X$ is at most $X^{6}M^{-1}\log^{2}X$ by (5.1). The contribution
from the second term is

$\displaystyle\ll X^{4}\log^{2}X$ $\displaystyle\sum_{q\leqslant
KX}\frac{1}{q^{2}}\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}\int_{M/X^{2}}^{1/KqX}\eta^{-2}d\eta$
$\displaystyle\leqslant X^{4}\log^{2}X\sum_{q\leqslant
KX}\frac{1}{q}\frac{X^{2}}{M}\ll X^{6}M^{-1}\log^{3}X.$

Recalling that $M=\log^{C_{1}}X$ with $C_{1}=10A$, this completes the proof. ∎

##  6\. The integral over $\mathfrak{m}_{2}$ – first reductions

We now begin the lengthy task of establishing Proposition 3.4. Once again, the
reader may wish to begin by recalling the pertinent definitions, which are
those of $I_{r,q}$ (see (3.3)), $\mathfrak{m}_{2}$ (see (3.6)) and
$S_{F_{1},F_{2}}(\theta)$ (given in (3.9)).

At the heart of our analysis will be certain complete exponential sums
$T_{f_{1},f_{2}}$. Let $q$ be a positive integer, and suppose that
$f_{1},f_{2}:(\mathbf{Z}/q\mathbf{Z})^{4}\rightarrow\mathbf{C}$. Define

$T_{f_{1},f_{2}}(r):=q^{2}\mathbb{E}_{x,y\in(\mathbf{Z}/q\mathbf{Z})^{4}}f_{1}(x)f_{2}(y)e_{q}(rQ(x,y)).$
(6.1)

_Remark._ Of course, $T$ is also depends on $q$, but we omit explicit mention
of this from the notation. There should not be any danger of confusion. For
fixed $r$ and general $f_{1},f_{2}$ we have the following bound.

######  Lemma 6.1.

Suppose that $f_{1},f_{2}:(\mathbf{Z}/q\mathbf{Z})^{4}\rightarrow\mathbf{C}$.
Suppose that $\det b\neq 0$. Then for any $r\in(\mathbf{Z}/q\mathbf{Z})^{*}$
we have

$|T_{f_{1},f_{2}}(r)|\leqslant(q,\det b)^{2}\|f_{1}\|_{2}\|f_{2}\|_{2}.$ (6.2)

###### Proof.

Modifying $f_{1}(x)$ to $f_{1}(x)e_{q}(rx^{T}ax)$ and $f_{2}(y)$ to
$f_{2}(y)e_{q}(ry^{T}cy)$, it suffices to show that

$\mathbb{E}_{x,y\in(\mathbf{Z}/q\mathbf{Z})^{4}}f_{1}(x)f_{2}(y)e_{q}(rx^{T}by)\leqslant(q,\det
b)^{2}q^{-2}\|f_{1}\|_{2}\|f_{2}\|_{2}.$

By Cauchy-Schwarz, it suffices to show that

$\mathbb{E}_{y,y^{\prime}\in(\mathbf{Z}/q\mathbf{Z})^{4}}f_{2}(y)\overline{f_{2}(y^{\prime})}\mathbb{E}_{x\in(\mathbf{Z}/q\mathbf{Z})^{4}}e_{q}(rx^{T}b(y-y^{\prime}))\leqslant(q,\det
b)^{4}q^{-4}\|f_{2}\|_{2}^{2}.$

By orthogonality, and since $(r,q)=1$, the left-hand side is

$\displaystyle\mathbb{E}_{y,y^{\prime}\in(\mathbf{Z}/q\mathbf{Z})^{4}}f_{2}(y)$
$\displaystyle\overline{f_{2}(y^{\prime})}1_{b(y-y^{\prime})\equiv
0(\mbox{\scriptsize mod}\,q)}$
$\displaystyle=q^{-4}\sum_{\begin{subarray}{c}h\in(\mathbf{Z}/q\mathbf{Z})^{4}\\\
bh\equiv 0(\mbox{\scriptsize
mod}\,q)\end{subarray}}\mathbb{E}_{y\in(\mathbf{Z}/q\mathbf{Z})^{4}}f_{2}(y)\overline{f_{2}(y+h)}.$

By Cauchy-Schwarz this is at most

$q^{-4}\\#\\{h\in(\mathbf{Z}/q\mathbf{Z})^{4}:bh\equiv
0(\operatorname{mod}\,q)\\}\|f_{2}\|_{2}^{2},$

and so it is enough to show that

$\\#\\{h\in(\mathbf{Z}/q\mathbf{Z})^{4}:bh\equiv
0(\operatorname{mod}\,q)\\}\leqslant(q,\det b)^{4}.$ (6.3)

Now if $bh\equiv 0(\operatorname{mod}\,q)$ then, multiplying on the left by
$\operatorname{adj}(b)$, we have $(\det b)h\equiv 0(\operatorname{mod}\,q)$,
i.e. if $h=(h_{1},h_{2},h_{3},h_{4})$ then $q|(\det b)h_{i}$ . The number of
choices of each $h_{i}$ is therefore $(q,\det b)$ and so (6.3) follows. This
concludes the proof of (6.2). ∎

The bound in Lemma 6.1 is best possible, at least when $(q,\det b)=1$. To see
this, let $\psi:(\mathbf{Z}/q\mathbf{Z})^{4}\rightarrow\mathbf{C}$ be any
function with $\|\psi\|_{2}=1$, and take

$f_{1}(x):=q^{2}\mathbb{E}_{y\in(\mathbf{Z}/q\mathbf{Z})^{4}}\overline{\psi}(y)e_{q}(-rQ(x,y)),\quad
f_{2}(x):=\psi(x).$

Then one may check using the orthogonality relations that

$|T_{f_{1},f_{2}}(r)|=\|f_{1}\|_{2}=\|f_{2}\|_{2}=1.$ (6.4)

A more conceptual explanation of this is as follows. First note that

$T_{f_{1},f_{2}}(r)=\overline{\langle\overline{f}_{1},\Phi f_{2}\rangle},$
(6.5)

where the map
$\Phi:\ell^{2}((\mathbf{Z}/q\mathbf{Z})^{4})\rightarrow\ell^{2}((\mathbf{Z}/q\mathbf{Z})^{4})$
is given by

$\Phi
f(x):=q^{2}\mathbb{E}_{y\in(\mathbf{Z}/q\mathbf{Z})^{4}}f(y)e_{q}(rQ(x,y)).$

One may then make the key observation that $\Phi$ is unitary (being a
composition of invertible dilations, quadratic modulations and Fourier
transform). Then we have $f_{1}:=\overline{\Phi\psi}$, $f_{2}=\psi$, and the
relations (6.4) are clear from (6.5) and the unitary nature of $\Phi$.

This also allows a very short (albeit ultimately equivalent) proof of Lemma
6.1 in the case $(q,\det b)=1$. Indeed, by Cauchy-Schwarz and unitarity we
have

$|T_{f_{1},f_{2}}(r)|=|\langle\overline{f}_{1},\Phi
f_{2}\rangle|\leqslant\|f_{1}\|_{2}\|\Phi
f_{2}\|_{2}=\|f_{1}\|_{2}\|f_{2}\|_{2}.$

One may, using arguments similar to those below, use the bound obtained in
Lemma 6.1 to show that (roughly speaking)

$\int_{\theta\in\bigcup_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}I_{r,q}}|S_{F_{1},F_{2}}(\theta)|d\theta\lessapprox
X^{6}q^{-1}.$

Unfortunately the sum over $q$ does not converge and so this (just) fails to
give the desired estimate Proposition 3.4. It is this, and the sharpness of
Lemma 6.1, which ultimately explain the failure of the classical circle method
to handle the problem of quadratic forms in 8 prime variables.

To get around this issue we introduce the following improvement on (6.1) when
an average over $r$ is included (at least when $q$ is squarefree and has no
small prime factors, and $Q$ is generic).

######  Proposition 6.2.

There is an absolute constant $\delta>0$ with the following property. Suppose
that $Q$ is generic. Then there is $p_{0}(Q)$ such that if $q$ is squarefree
and with all prime factors greater than $p_{0}(Q)$, then we have

$\frac{1}{\phi(q)}\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}|T_{f_{1},f_{2}}(r)|\ll
q^{-\delta}\|f_{1}\|_{2}\|f_{2}\|_{2}$

for any $f_{1},f_{2}\in\ell^{2}((\mathbf{Z}/q\mathbf{Z})^{4})$.

The proof of this proposition occupies most of the rest of the paper. The
remainder of this section is devoted to deriving Proposition 3.4 from it.

First we observe that Lemma 6.1 and Proposition 6.2 have a fairly
straightforward application to the sums $S_{F_{1},F_{2}}(\theta)$ for
$\theta=\frac{r}{q}$, which we record now.

######  Corollary 6.3.

Suppose that $F_{1},F_{2}:[X]^{4}\rightarrow\mathbf{C}$ are $1$-bounded.
Suppose that $\det b\neq 0$. Then we have the pointwise bound

$|S_{F_{1},F_{2}}(\frac{r}{q})|\ll X^{8}q^{-2}.$ (6.6)

Suppose additionally that $q$ is squarefree and has no prime factors of size
$\leqslant p_{0}(Q)$, and that $Q$ is generic. Suppose that $q\leqslant KX$.
Then

$\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}|S_{F_{1},F_{2}}(\frac{r}{q})|\ll
X^{8}q^{-1-\delta}.$ (6.7)

###### Proof.

Extend $F_{1},F_{2}$ to functions on $\mathbf{Z}$ by defining them to be $0$
outside of $[X]$. Fix $x_{0},y_{0}\in q\mathbf{Z}^{4}$. Then

$\displaystyle\sum_{\begin{subarray}{c}x\in x_{0}+[q]^{4}\\\ y\in
y_{0}+[q]^{4}\end{subarray}}$ $\displaystyle F_{1}(x)F_{2}(y)e_{q}(rQ(x,y))$
$\displaystyle=\sum_{x^{\prime},y^{\prime}\in[q]^{4}}F_{1}(x_{0}+x^{\prime})F_{2}(y_{0}+y^{\prime})e_{q}(rQ(x_{0}+x^{\prime},y_{0}+y^{\prime})).$

Since $x_{0},y_{0}$ are both multiples of $q$ we have
$Q(x_{0}+x^{\prime},y_{0}+y^{\prime})\equiv
Q(x^{\prime},y^{\prime})(\operatorname{mod}\,q)$ and therefore, recalling the
definition (6.1),

$\sum_{\begin{subarray}{c}x\in x_{0}+[q]^{4}\\\ y\in
y_{0}+[q]^{4}\end{subarray}}F_{1}(x)F_{2}(y)e_{q}(rQ(x,y))=q^{6}T_{f_{1},f_{2}}(r)$
(6.8)

where

$f_{1}(x^{\prime}):=F_{1}(x_{0}+x^{\prime})\quad\mbox{and}\quad
f_{2}(y^{\prime}):=F_{2}(y_{0}+y^{\prime}).$

By Lemma 6.1 we therefore have

$\big{|}\sum_{\begin{subarray}{c}x\in x_{0}+[q]^{4}\\\ y\in
y_{0}+[q]^{4}\end{subarray}}F_{1}(x)F_{2}(y)e_{q}(rQ(x,y))\big{|}\leqslant
q^{6}(\det b)^{2}.$

Covering the range $[X]^{4}\times[X]^{4}$ by $\leqslant(\frac{X}{q}+1)^{8}\ll
X^{8}q^{-8}$ boxes of the form $(x_{0}+[q]^{4})\times(y_{0}+[q]^{4})$ gives
(6.6).

To obtain (6.7), we instead apply Proposition 6.2 to (6.8), obtaining

$\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}|\sum_{x\in x_{0}+[q]^{4},y\in
y_{0}+[q]^{4}}F_{1}(x)F_{2}(y)e_{q}(rQ(x,y))|\ll q^{6-\delta}\phi(q)\leqslant
q^{7-\delta}.$

Dividing into $O(X^{8}q^{-8})$ boxes as before gives (6.7). ∎

###### Proof of Proposition 3.4, assuming Proposition 6.2.

Recall that

$\mathfrak{m}_{2}=\bigcup_{\begin{subarray}{c}M^{\prime}\leqslant q\leqslant
KX\\\ r\in(\mathbf{Z}/q\mathbf{Z})^{*}\end{subarray}}I_{r,q},$

with $I_{r,q}$ as defined in (3.3), $M^{\prime}=\log^{C_{2}}X$ (with $C_{2}$
as described in (3.1)), and $K=8\max_{ij}|b_{ij}|$ being a constant associated
to the form $Q$.

Therefore the bound we are trying to prove is

$\sum_{M^{\prime}<q\leqslant
KX}\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}\int_{|\eta|\leqslant
M/X^{2}}|S_{F_{1},F_{2}}(\frac{r}{q}+\eta)|d\eta\ll_{A}X^{6}\log^{-A-8}X.$
(6.9)

We begin by using some Fourier analysis to handle the inner integral over
$\eta$. Recall the definition (3.9) of $S_{F_{1},F_{2}}$, that is to say

$S_{F_{1},F_{2}}(\theta):=\sum_{x,y\in[X]^{4}}F_{1}(x)F_{2}(y)e(\theta
Q(x,y)).$

Let $w:\mathbf{R}^{2}\rightarrow\mathbf{R}$ be some fixed smooth compactly-
supported function with $w(u,v)=1$ for $(u,v)\in[0,1]^{2}$ and set, for any
real parameter $\lambda$,

$W_{\lambda}(u,v):=w(u,v)e(\lambda Q(u,v)).$ (6.10)

Then

$S_{F_{1},F_{2}}(\frac{r}{q}+\eta)=\mathbb{E}_{x,y\in[X]^{4}}F_{1}(x)F_{2}(y)e_{q}(rQ(x,y))W_{\eta
X^{2}}(\frac{x}{X},\frac{y}{X}).$ (6.11)

Now by integration by parts and Leibniz’s rule we have

$|\hat{W}_{\lambda}(\xi,\xi^{\prime})|\ll|\xi|^{-2}|\xi^{\prime}|^{-2}\|\frac{\partial^{4}W_{\eta}}{\partial^{2}u\partial^{2}v}\|_{1}\ll\max(1,|\lambda|^{4})|\xi|^{-2}|\xi^{\prime}|^{-2}.$

Since we also have the trivial bound

$|\hat{W}_{\lambda}(\xi,\xi^{\prime})|\leqslant\|W_{\lambda}\|_{1}=\|w\|_{1}\ll
1$

it follows that

$\|\hat{W}_{\lambda}\|_{1}\ll\max(1,|\lambda|)^{4}.$ (6.12)

By Fourier inversion

$W_{\lambda}(u,v)=\int_{\mathbf{R}}\hat{W}_{\lambda}(\xi,\xi^{\prime})e(\xi
u+\xi^{\prime}v)d\xi d\xi^{\prime};$

substituting into (6.11) gives

$S_{F_{1},F_{2}}(\frac{r}{q}+\eta)=\int\hat{W}_{\eta
X^{2}}(\xi,\xi^{\prime})S_{F_{1,\xi},F_{2,\xi^{\prime}}}(\frac{r}{q})d\xi
d\xi^{\prime},$

where

$F_{1,\xi}(x):=F_{1}(x)e(\xi x/X),\quad
F_{2,\xi^{\prime}}(y):=F_{2}(y)e(\xi^{\prime}y/X).$

Therefore

$\displaystyle\sum_{M^{\prime}<q\leqslant KX}$
$\displaystyle\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}|S_{F_{1},F_{2}}(\frac{r}{q}+\eta)|\leqslant$
$\displaystyle\int_{\mathbf{R}}|\hat{W}_{\eta
X^{2}}(\xi,\xi^{\prime})|\sum_{M^{\prime}<q\leqslant
KX}\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}|S_{F_{1,\xi},F_{2,\xi^{\prime}}}(\frac{r}{q})|d\xi
d\xi^{\prime}.$ (6.13)

We claim the estimate

$\sum_{M^{\prime}<q\leqslant
KX}\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}|S_{F^{\prime}_{1},F^{\prime}_{2}}(\frac{r}{q})|\ll
X^{8}(M^{\prime})^{-\delta/100},$ (6.14)

uniformly for all $1$-bounded $F^{\prime}_{1},F^{\prime}_{2}$, where $\delta$
is the exponent appearing in Proposition 6.2. Assuming this claim, (6.12) and
(6.13) then imply that

$\sum_{M^{\prime}<q\leqslant
KX}\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}|S_{F_{1},F_{2}}(\frac{r}{q}+\eta)|\ll
X^{8}(M^{\prime})^{-\delta/100}\max(1,|\eta|X^{2})^{4}.$

Then, integrating over $|\eta|\leqslant M/X^{2}$ we obtain

$\int_{\mathfrak{m}_{2}}|S_{F_{1},F_{2}}(\theta)|d\theta\ll
X^{6}M^{5}(M^{\prime})^{-\delta/100}\ll_{A}X^{6}(\log X)^{-A-8}.$

For the last step, we recall that we chose $M=\log^{C_{1}}X$,
$M^{\prime}=\log^{C_{2}}X$ with the particular choice of $C_{1},C_{2}$
specified at the start of Section 3. This completes the proof of Proposition
3.4, assuming the claim (6.14).

Now we must establish (6.14). The pointwise bound (6.6) is not good enough,
but we do have the improved average bound (6.7), albeit only for squarefree
$q$ with no small prime factors. Most $q$ do not have this form, and so we
need the following lemma to allow us to reduce matters to the consideration of
to those that do.

######  Lemma 6.4.

Suppose that $q=q_{0}q_{1}$ with $(q_{0},q_{1})=1$. Then

$\sup_{F_{1},F_{2}}\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}|S_{F_{1},F_{2}}(\frac{r}{q})|\ll
q_{1}^{9}\sup_{F_{1},F_{2}}\sum_{r_{0}\in(\mathbf{Z}/q_{0}\mathbf{Z})^{*}}|S_{F_{1},F_{2}}(\frac{r_{0}}{q_{0}})|,$

where in both cases the $\sup$ is over $1$-bounded functions
$F_{1},F_{2}:[X]^{4}\rightarrow\mathbf{C}$.

###### Proof.

By the Chinese remainder theorem we have

$\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}|S_{F_{1},F_{2}}(\frac{r}{q})|=\sum_{\begin{subarray}{c}r_{0}\in(\mathbf{Z}/q_{0}\mathbf{Z})^{*}\\\
r_{1}\in(\mathbf{Z}/q_{1}\mathbf{Z})^{*}\end{subarray}}|S_{F_{1},F_{2}}(\frac{r_{0}}{q_{0}}+\frac{r_{1}}{q_{1}})|.$
(6.15)

Foliating into residue classes modulo $q_{1}$, we have

$\displaystyle S_{F_{1},F_{2}}($
$\displaystyle\frac{r_{0}}{q_{0}}+\frac{r_{1}}{q_{1}})$
$\displaystyle=\sum_{u,v\in(\mathbf{Z}/q_{1}\mathbf{Z})^{4}}e_{q_{1}}(r_{1}Q(u,v))\sum_{x,y\in[X]^{4}}F_{1,u}(x)F_{2,v}(y)e_{q_{0}}(r_{0}Q(x,y))$

where $F_{1,u}(x):=F_{1}(x)1_{x\equiv u(\mbox{\scriptsize mod}\,q_{1})}$,
$F_{2,v}(y):=F_{2}(y)1_{y\equiv v(\mbox{\scriptsize mod}\,q_{1})}$ and so by
(6.15)

$\displaystyle\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}$
$\displaystyle|S_{F_{1},F_{2}}(\frac{r}{q})|$ $\displaystyle\leqslant
q_{1}^{9}\sup_{u,v}\sum_{r_{0}\in(\mathbf{Z}/q_{0}\mathbf{Z})^{*}}\big{|}\sum_{x,y\in[X]^{4}}F_{1,u}(x)F_{2,v}(y)e_{q_{0}}(r_{0}Q(x,y))\big{|}$
$\displaystyle=q_{1}^{9}\sup_{u,v}\sum_{r_{0}\in(\mathbf{Z}/q_{0}\mathbf{Z})^{*}}|S_{F_{1,u},F_{2,v}}(\frac{r_{0}}{q_{0}})|.$

The lemma follows. ∎

Let us turn to the actual proof of (6.14). Let $p_{0}=p_{0}(Q)$ be the
threshold appearing in Proposition 6.2. For any $q$, write $q_{0}$ for the
product of all primes $p>p_{0}$ which divide $q$ precisely once, and set
$q_{1}:=q/q_{0}$, thus $q_{1}$ is the product of all prime powers $p^{j}\|q$
with $p\leqslant p_{0}$ or $j\geqslant 2$. Note that $q_{0},q_{1}$ are
coprime. By Lemma 6.4 and (6.6), (6.7) we have

$\sum_{r\in(\mathbf{Z}/q\mathbf{Z})^{*}}|S_{F_{1},F_{2}}(\frac{r}{q})|\ll
X^{8}\min(q^{-1},q_{1}^{9}q_{0}^{-1-\delta})\leqslant
X^{8}\min(q^{-1},q_{1}^{11}q^{-1-\delta}).$

It therefore suffices to prove that

$\sum_{q>M^{\prime}}\min(q^{-1},q_{1}^{11}q^{-1-\delta})\ll(M^{\prime})^{-\delta/100}.$
(6.16)

The contribution from $q$ with $q_{1}<q^{\delta/22}$ is acceptable (using the
second term in the $\min$).

If $q_{1}\geqslant q^{\delta/22}$ then suppose $q_{1}=\prod_{p}p^{v_{p}}$ with
$v_{p}\geqslant 2$ for $p\geqslant p_{0}$. Set $q_{2}:=\prod_{p}p^{\lfloor
v_{p}/2\rfloor}$. Then $q_{2}^{2}|q$. Moreover, if $v_{p}\geqslant 2$ then
$p^{\lfloor v_{p}/2\rfloor}\geqslant p^{v_{p}/3}$, so $q_{2}\geqslant
cq^{\delta/66}$ (with $c>0$ depending only on $p_{0}$). Thus the contribution
of these $q$ to (6.16) can be bounded by

$\sum_{d>c(M^{\prime})^{\delta/66}}\sum_{\begin{subarray}{c}q\ll
d^{66/\delta}\\\
d^{2}|q\end{subarray}}q^{-1}\ll\sum_{d>c(M^{\prime})^{\delta/66}}\frac{\log
d}{d^{2}}\ll(M^{\prime})^{-\delta/100}.$

This concludes the proof.∎

##  7\. Exponential sums as matrix coefficients on
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$

The remainder of the paper is occupied with the proof of Proposition 6.2, the
statement of which the reader may wish to recall at this point.

We remarked after the proof of Lemma 6.1 that we can write

$T_{f_{1},f_{2}}(r)=\overline{\langle\overline{f}_{1},\Phi_{r}f_{2}\rangle},$
(7.1)

where

$\Phi_{r}f=q^{2}\mathbb{E}_{y}f(y)e_{q}(rQ(x,y))$

is unitary. The crucial observation which drives our whole argument is that
the subgroup of $\operatorname{U}(\ell^{2}((\mathbf{Z}/q\mathbf{Z})^{4}))$
(the group of all unitary operators on
$\ell^{2}((\mathbf{Z}/q\mathbf{Z})^{4})$) generated by operators $\Phi$ of
this type (over all $Q$) is rather small. Indeed, as we shall shortly see, it
has size $q^{36+o(1)}$. This means that the specific operators $\Phi_{r}$
(with $Q$ fixed but $r$ allowed to vary over $(\mathbf{Z}/q\mathbf{Z})^{*}$)
already occupy a reasonable portion of this group.

This group turns out to be the symplectic group
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$. Let us recall what these
groups are, in a very concrete way. Let $R=\mathbf{Z}/q\mathbf{Z}$ with $q$
odd. Then $\operatorname{Sp}_{8}(R)$ is a group of $8\times 8$ matrices over
$R$, which we will write in $2\times 2$ block form with each block being a
$4\times 4$ matrix.

######  Definition 7.1.

We define the symplectic group $\operatorname{Sp}_{8}(R)$ to be the group
consisting of all $8\times 8$ block matrices
$g=\big{(}\begin{smallmatrix}A&B\\\ C&D\end{smallmatrix}\big{)}$ with entries
in $R$ and $A^{T}C=C^{T}A$, $B^{T}D=D^{T}B$ and $A^{T}D-C^{T}B=I$.

Define

$J:=\begin{pmatrix}0&I\\\ -I&0\end{pmatrix}.$ (7.2)

It is an simple exercise to check that $g\in\operatorname{Sp}_{8}(R)$ if and
only if $g^{T}Jg=J$. In fact, this is the more usual definition of the
sympletic group, but it suits us to be more explicit.

Note that if $g=\big{(}\begin{smallmatrix}A&B\\\
C&D\end{smallmatrix}\big{)}\in\operatorname{Sp}_{8}(R)$ then $g$ is left-
invertible with left-inverse $\big{(}\begin{smallmatrix}D^{T}&-B^{T}\\\
-C^{T}&A^{T}\end{smallmatrix}\big{)}$. This is then, of course, also a right-
inverse for $g$, and this gives us the additional relations

$AB^{T}=BA^{T},\;CD^{T}=DC^{T}\quad\mbox{and}\quad AD^{T}-BC^{T}=I$

for any symplectic matrix.

We will also need the fact that $(\mathbf{Z}/q\mathbf{Z})^{*}$ acts on
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$ by “dilation” automorphisms.
If $g=\big{(}\begin{smallmatrix}A&B\\\ C&D\end{smallmatrix}\big{)}$ and if
$r\in(\mathbf{Z}/q\mathbf{Z})^{*}$ then we define

$g^{(r)}=\begin{pmatrix}A&r^{-1}B\\\ rC&D\end{pmatrix}.$ (7.3)

It is then easy to see that this gives an action of
$(\mathbf{Z}/q\mathbf{Z})^{*}$ on
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$ by automorphisms.

Finally, we note for future reference (see, for example, [16]) that

$|\operatorname{Sp}_{8}(\mathbf{F}_{p})|=p^{16}(p^{8}-1)(p^{6}-1)(p^{4}-1)(p^{2}-1)=(1+o(1))p^{36}.$
(7.4)

Now we come to the key link between $\operatorname{Sp}_{8}$ and operators such
as $\Phi_{r}$ in (7.1), which stems from [28] and is thus known as the Weil
representation (or, depending on the context, the Segal-Shale-Weil
representation or the oscillator representation).

######  Proposition 7.2 (Weil representation).

Let $q$ be squarefree and odd. Then there is a unitary representation

$\rho:\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})\rightarrow\operatorname{U}(\ell^{2}((\mathbf{Z}/q\mathbf{Z})^{4}))$

and a function
$\xi:\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})\rightarrow\\{z\in\mathbf{C}:|z|=1\\}$
satisfying the following:

  * •

(Dilations) If $g=s(E)$ where $s(E):=\big{(}\begin{smallmatrix}E&0\\\
0&E^{-T}\end{smallmatrix}\big{)}$ with $E$ invertible then

$\rho(g)f(x)=\xi(s(E))f(E^{-1}x);$

  * •

(Fourier transform) If $g=J$ then

$\rho(g)f(x)=\xi(J)q^{2}\mathbb{E}_{y\in(\mathbf{Z}/q\mathbf{Z})^{4}}f(y)e(x^{T}y);$

  * •

(Quadratic modulations) If $g=l(W)$ where
$l(W):=\big{(}\begin{smallmatrix}I&0\\\ W&I\end{smallmatrix}\big{)}$ with $W$
symmetric then

$\rho(g)f(x)=\xi(l(W))e_{q}(-\frac{1}{2}x^{T}Wx)f(x).$

_Remarks._ _1._ There is nothing special about $\operatorname{Sp}_{8}$ here;
similar results hold for $\operatorname{Sp}_{2m}$ for any positive integer
$m$. The Weil representation is well-known over $\mathbf{R}$ (where one needs
to pass to the double cover of the symplectic group), but in finite situations
it seems to me that it is only at all widely discussed over finite fields. In
this case, the construction is given in detail in (for example) the paper [16]
of Neuhauser. The analogue of this in the lower-dimensional setting of
$\operatorname{Sp}_{2}(\mathbf{Z}/p\mathbf{Z})=\operatorname{SL}_{2}(\mathbf{Z}/p\mathbf{Z})$
already contains the key ideas, and a very nice description of this may be
found in the notes of Charlotte Chan [5], which I found helpful in preparing
this material. It is not difficult to derive the case $q$ squarefree from the
prime case, and we do this in Appendix B.

_2._ The phase $\xi$ can be given explicitly if desired. When $q=p$ is an odd
prime, we can take $\xi(l(W))=\xi(J)=1$ and $\xi(s(E))=\left(\frac{\det
E}{p}\right)$, and the general squarefree case can then be deduced from the
arguments in Appendix B. For details of these calculations (which are somewhat
involved) see [16]. In this paper, we will not need explicit values of $\xi$,
and the mere existence is a much easier statement to prove, this being [16,
Theorem 4.3].

_3._ If desired one can also add in the translations $f(x)\mapsto f(x-v)$ and
the linear modulations $f(x)\mapsto e(-t^{T}x)f(x)$, getting an action by the
“Jacobi group”
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})\ltimes\operatorname{H}_{8}(\mathbf{Z}/q\mathbf{Z})$,
where $\operatorname{H}_{8}$ is the Heisenberg group on
$(\mathbf{Z}/q\mathbf{Z})^{8}\times\mathbf{Z}/q\mathbf{Z}$.

_4._ It is not really correct to call $\rho$ “the” Weil representation. In the
case $q=p$ a prime, further representations $\tilde{\rho}$ of the same
dimension can be obtained by twisting with the dilation $\sigma(g)=g^{(r)}$,
that is to say $\tilde{\rho}(g):=\rho(g^{(r)})$. When $r$ is not a square in
$\mathbf{F}_{p}^{*}$, the dilation is an outer automorphism and it is known
that $\tilde{\rho}\not\cong\rho$. (In the literature this would be described
in terms of different central characters on the Heisenberg group giving
different Weil representations, see [16, Section 7] or [23, Proposition 4]).
Thus there are two Weil representations of
$\operatorname{Sp}_{8}(\mathbf{Z}/p\mathbf{Z})$, and the one we are
considering is a concrete realisation of one of them.

When $q=p_{1}\cdots p_{n}$, one may obtain $2^{n}$ non-isomorphic
representations of $\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$ by taking
tensor products. The representation whose existence we assert in Proposition
7.2 is one of these. However, it turns out _not_ to be simply the tensor
product of the $\rho$s associated to each $p_{i}$: we must first apply some
twists. The details are given in Appendix B.

_5._ Even when $q=p$, the Weil representation is not irreducible. It splits
into its actions on odd and even functions, which _are_ irreducible
representations of degrees $\frac{1}{2}(p^{4}\pm 1)$. We will not need this
fact here.

Now that we have defined the Weil representation, we can interpret the
exponential sums $T_{f_{1},f_{2}}(r)$ as matrix coefficients. We begin with an
important definition which will be relevant for the rest of the paper.

######  Definition 7.3 (Symplectic element).

Suppose that $Q(x,y)=x^{T}ax+x^{T}by+y^{T}cy$ is a quadratic form and that
$\det b\neq 0$. Then we associate to $Q$ the element
$g=g(Q)\in\operatorname{Sp}_{8}(\mathbf{Q})$ defined by

$g:=\begin{pmatrix}-2b^{-T}c&b^{-T}\\\ 4ab^{-T}c-b&-2ab^{-T}\end{pmatrix}.$

We call this the _symplectic element_ associated to $Q$.

As mentioned in the introduction, we will abuse notation by regarding $g$ as
an element of $\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$ for squarefree
odd $q$, coprime to $\det b$. Here is the promised interpretation of
exponential sums as matrix coefficients.

######  Proposition 7.4.

Suppose that $q$ is odd, squarefree and coprime to $\det b$. Then for any
$f_{1},f_{2}\in\ell^{2}((\mathbf{Z}/q\mathbf{Z})^{4})$ we have

$|T_{f_{1},f_{2}}(r)|=|\langle\overline{f}_{1},\rho(g^{(r)})f_{2}\rangle|,$

where $g^{(r)}$ is the dilate of $g$ (regarded as an element of
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$) by $r$ as defined in (7.3)
and $\rho$ is the Weil representation described in Proposition 7.2.

###### Proof.

It suffices to establish the case $r=1$, since then the formula for general
$r$ follows by applying that case with $Q$ replaced by $rQ$ (or, to be
pedantic, $\overline{r}Q$ for some $r\in\mathbf{Z}$ projecting to
$r(\operatorname{mod}\,q)$).

To handle the case $r=1$, note that
$q^{2}\mathbb{E}_{y\in(\mathbf{Z}/q\mathbf{Z})^{4}}f(y)e(Q(x,y))$ may be built
up as a composition of four (unitary) operations, as follows:

  1. (1)

A quadratic modulation $f(y)\mapsto f(y)e_{q}(y^{T}cy)$;

  2. (2)

Fourier transform $f\mapsto
q^{2}\mathbb{E}_{y\in(\mathbf{Z}/q\mathbf{Z})^{4}}f(y)e_{q}(x^{T}y)$;

  3. (3)

Dilation $f(x)\mapsto f(b^{T}x)$;

  4. (4)

Quadratic modulation $f(x)\mapsto f(x)e_{q}(x^{T}ax)$.

In the Weil representation these four operations correspond, up to scalar
multiplication by unit complex numbers, respectively, to the following
elements of $\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$:
$l(-2c),J,s(b^{-T})$ and $l(-2a)$. Therefore by Proposition 7.2 (since $\rho$
is a homomorphism!) we have

$q^{2}\mathbb{E}_{y\in(\mathbf{Z}/q\mathbf{Z})^{4}}f(y)e(Q(x,y))=z\rho(g)f(x)$

for some unit complex number $z=z(Q)$ where

$g=l(-2a)\cdot s(b^{-T})\cdot J\cdot l(-2c)$

is the product of the four elements just written down. A short computation
confirms that $g$ is the symplectic element of $Q$ as defined in Definition
7.3.

Finally, we have

$\displaystyle T_{f_{1},f_{2}}(1)$
$\displaystyle=\mathbb{E}_{x\in(\mathbf{Z}/q\mathbf{Z})^{4}}f_{1}(x)\mathbb{E}_{y\in(\mathbf{Z}/q\mathbf{Z})^{4}}f_{2}(y)e(Q(x,y))$
$\displaystyle=z\mathbb{E}_{x\in(\mathbf{Z}/q\mathbf{Z})^{4}}f_{1}(x)(\rho(g)f_{2})(x)$
$\displaystyle=z\overline{\langle\overline{f}_{1},\rho(g)f_{2}\rangle}.$

This completes the proof.∎

The following definition will play a key role in what follows.

######  Definition 7.5.

Fix $Q(x,y)=x^{T}ax+x^{T}by+y^{T}cy$, a quadratic form over $\mathbf{Z}$ with
$\det b\neq 0$. Let $g$ be the symplectic element of $Q$ (see Definition 7.3).
Then for every odd squarefree $q$ coprime to $\det b$ we associate a
probability measure $\mu_{q}$ on
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$, which puts weight
$\frac{1}{\phi(q)}$ on each of the points $g^{(r)}$,
$r\in(\mathbf{Z}/q\mathbf{Z})^{*}$.

We are now in a position to rephrase Proposition 6.2 in terms of matrix
coefficients.

######  Proposition 7.6.

Suppose that $Q$ is generic. Then there is $p_{0}(Q)$ such that the following
is true. Suppose that $q$ is squarefree with all prime factors greater than
$p_{0}(Q)$. Then

$\int|\langle f_{1},\rho(x)f_{2}\rangle|d\mu_{q}(x)\ll
q^{-\delta}\|f_{1}\|_{2}\|f_{2}\|_{2}$

for all $f_{1},f_{2}\in\ell^{2}((\mathbf{Z}/q\mathbf{Z})^{4})$.

_Remarks._ For a discussion of the integral notation, see the start of the
next section. By taking $p_{0}(Q)$ larger than any prime factor of $\det b$,
we can ensure that the conditions of Proposition 7.4 are satisfied. For
notational simplicity we switched $\overline{f}_{1}$ to $f_{1}$, which makes
no difference since these functions have the same $\ell^{2}$-norm.

##  8\. Averages of matrix coefficients

In this section we give a general bound for averages of matrix coefficients.
Whilst we do not know of a reference for quite this result, the first part of
the argument is related to earlier work, particularly Bourgain [1] and
Skhredov [20, Section 4]. The second idea, of using quasirandomness (no small-
dimensional representations apart from the trivial one), is also by no means
new. It is exploited in related ways in many works starting with Sarnak–Xue
[18] and continuing with, for instance, Bourgain–Gamburd [2] and Gowers [8].

_Probability measures._ We begin by recalling some basic notions about
probability measures on (finite) groups. Let $G$ be a finite group. A
probability measure $\mu$ on $G$ is simply a function $\mu:G\rightarrow[0,1]$
with $\sum_{x\in G}\mu(x)=1$. The opposite measure $\mu^{\circ}$ is defined by
$\mu^{\circ}(x)=\mu(x^{-1})$. If $\mu^{\circ}=\mu$ then we say that $\mu$ is
_symmetric_. If $\mu_{1},\mu_{2}$ are two probability measures then their
convolution $\mu_{1}\ast\mu_{2}$ is defined by
$\mu_{1}\ast\mu_{2}(x)=\sum_{g_{1}g_{2}=x}\mu_{1}(g_{1})\mu_{2}(g_{2})$. This
is also a probability measure. If $\mu$ is a probability measure and $2m$ a
positive even integer, we write $\mu^{(2m)}$ for the $2m$-fold _symmetrised_
convolution power $\mu^{\circ}\ast\mu\ast\mu^{\circ}\ast\cdots\ast\mu$. This
is slightly non-standard, but very convenient as these are the only types of
convolution power we will consider in this paper. At one place in Section 9 we
will use a similar notation with an odd power, thus
$\mu^{(2m-1)}(x)=\mu^{\circ}\ast\mu\ast\mu^{\circ}\ast\cdots\ast\mu^{\circ}$.
This, of course, is not necessarily a symmetric measure. We have
$\mu^{(2m)}=\mu^{(2m-1)}\ast\mu$. If $H\leqslant G$ is a subgroup then we
write $\mu_{H}$ for the uniform probability measure on $H$, that is to say
$\mu_{H}(x)=|H|^{-1}1_{x\in H}$.

If $\mu$ is a probability measure on a finite group $G$ then we write
$\|\mu\|=\big{(}\sum_{x}\mu(x)^{2}\big{)}^{1/2}$. Note that this is normalised
differently to the $\ell^{2}$-norm of functions which has appeared in previous
sections: to reduce the potential for confusion, we omit any subscript from
the norm. We extend this notion to differences of measures in the obvious way,
thus $\|\mu-\nu\|=\big{(}\sum_{x}(\mu(x)-\nu(x))^{2}\big{)}^{1/2}$.

If $F:G\rightarrow\mathbf{C}$, we will adopt the fairly standard convention in
this context of writing $\int F(x)d\mu(x)$ instead of $\sum_{x}F(x)\mu(x)$.

We will need the following consequence of Schur’s lemma which is standard but
cannot be reliably found in every textbook.

######  Lemma 8.1.

Let $\psi:G\rightarrow\operatorname{U}(V)$ be an irreducible representation of
a finite group $G$. Suppose that $v,w\in V$. Then we have

$\int|\langle
v,\psi(x)w\rangle|^{2}d\mu_{G}(x)=\frac{1}{\dim\psi}\|v\|^{2}\|w\|^{2}.$

###### Proof.

See [14, Proposition 4.3.5]. Note that here $\dim\psi$ is defined to be $\dim
V$.∎

######  Proposition 8.2.

Let $G$ be a finite group, and let $\rho:G\rightarrow\operatorname{U}(V)$ be a
finite-dimensional unitary representation of $G$. Let $\mu$ be a probability
measure on $G$. Let $H\leqslant G$ be the group generated by
$\operatorname{Supp}(\mu^{(2)})=\operatorname{Supp}(\mu^{\circ}\ast\mu)$.
Suppose that

  * •

(Almost uniform distribution of convolution powers) For some real number
$K\geqslant 1$ and for some power of two $m$ we have

$\mu^{(m)}(x)\leqslant K\mu_{H}(x)$ (8.1)

pointwise;

  * •

(Quasirandomness) If $\rho|_{H}=\bigoplus_{i}\psi_{i}$ as a sum of irreducible
representations (of $H$) then $\dim\psi_{i}\geqslant D$ for all $i$.

Then we have the bound

$\int|\langle v,\rho(x)w\rangle|d\mu(x)\leqslant K^{1/m}D^{-1/2m}$ (8.2)

for all $v,w\in V$ with $\|v\|=\|w\|=1$.

_Remark._ Note that the trivial bound is $1$ (by the unitary nature of $\rho$
and Cauchy-Schwarz). If $K\approx 1$ and $D$ is somewhat large, (8.2) is
therefore an appreciable improvement of the trivial bound.

###### Proof.

Set

$\eta:=\int|\langle v,\rho(x)w\rangle|d\mu(x).$

For each $x\in G$, let $\xi(x)=e(\arg\langle v,\rho(x)w\rangle)$, so $\xi(x)$
is a unit complex number and

$\langle v,\int\xi(x)\rho(x)wd\mu(x)\rangle=\int\overline{\xi(x)}\langle
v,\rho(x)w\rangle d\mu(x)=\eta.$

By Cauchy-Schwarz,

$\big{\|}\int\xi(x)\rho(x)wd\mu(x)\big{\|}\geqslant\eta.$

Squaring and expanding out gives

$\int\xi(x)\overline{\xi(x^{\prime})}\langle\rho(x)w,\rho(x^{\prime})w\rangle
d\mu(x)d\mu(x^{\prime})\geqslant\eta^{2},$

thus

$\int|\langle\rho(x)w,\rho(x^{\prime})w\rangle|d\mu(x)d\mu(x^{\prime})\geqslant\eta^{2}.$

Since $\rho$ is a unitary representation, this implies

$\int|\langle
w,\rho(x^{-1}x^{\prime})w\rangle|d\mu(x)d\mu(x^{\prime})\geqslant\eta^{2},$

or in other words

$\int|\langle w,\rho(x)w\rangle|d\mu^{(2)}(x)\geqslant\eta^{2}.$

We may now apply the same argument again repeatedly, noting that
$\mu^{(2)},\mu^{(4)},\dots$ are symmetric, to obtain

$\int|\langle w,\rho(x)w\rangle|d\mu^{(m)}(x)\geqslant\eta^{m}$

for any power of two $m$. By the almost uniform distribution assumption (8.1),
this implies (with $m$ as in (8.1)) that

$\int|\langle w,\rho(x)w\rangle|d\mu_{H}(x)\geqslant K^{-1}\eta^{m}.$ (8.3)

Now decompose $V=\bigoplus_{i=1}^{n}V_{i}$ as a sum of orthogonal
$\rho(H)$-invariant subspaces, irreducible for $\rho|_{H}$. Let $w_{i}$ be the
projection of $w$ to $V_{i}$, so $w=\sum_{i=1}^{n}w_{i}$ and

$\sum_{i=1}^{n}\|w_{i}\|^{2}=1.$ (8.4)

By Lemma 8.1 we have for $i=1,\dots,n$

$\int|\langle w_{i},\rho(x)w_{i}\rangle|^{2}d\mu_{H}(x)=\frac{1}{\dim
V_{i}}\|w_{i}\|^{4},$

so by Cauchy-Schwarz and the quasirandomness assumption

$\int|\langle w_{i},\rho(x)w_{i}\rangle|d\mu_{H}(x)\leqslant
D^{-1/2}\|w_{i}\|^{2}.$ (8.5)

Since (by orthogonality)

$\langle w,\rho(x)w\rangle=\sum_{i=1}^{n}\langle w_{i},\rho(x)w_{i}\rangle$

for all $x\in H$, it follows from (8.4) and (8.5) that

$\int|\langle w,\rho(x)w\rangle|d\mu_{H}(x)\leqslant D^{-1/2}.$

Comparing this with (8.3) gives the claimed bound. ∎

We now outline the rest of the paper. Recall that we have reduced the proof of
our main theorem to the task of proving Proposition 7.6. We now have a tool,
Proposition 8.2, to use on this problem. However, we must verify the two
requirements, the uniform distribution property (8.1) and the quasirandomness
property, in our setting.

The formal statements are Propositions 8.4 and 8.5 below. First, we give a
definition which will play an important role in the rest of the paper.

######  Definition 8.3.

Suppose that $q$ is odd, squarefree and coprime to $\det b$. Let
$\Gamma_{q}\leqslant\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$ be the
group generated by the elements $g^{-(r)}g^{(s)}$,
$r,s\in(\mathbf{Z}/q\mathbf{Z})^{*}$, or equivalently by the support of
$\operatorname{Supp}(\mu_{q}^{(2)})$.

######  Proposition 8.4.

Let $Q$ be a quadratic form. Then there is some $p_{0}(Q)$ such that the
following is true. Suppose that $q$ is squarefree and has all prime factors
greater than $p_{0}(Q)$. Let $\mu_{q}$ be the measure described in Definition
7.5. Then there is power of two $m=O(1)$ such that
$\mu_{q}^{(m)}\ll\mu_{\Gamma_{q}}$ pointwise, with the implied constant being
absolute.

We will prove this in Section 9. It does not require any genericity assumption
on the form $Q$.

######  Proposition 8.5.

Let $Q$ be a generic quadratic form. Then there is some $p_{0}(Q)$ such that
the following is true. Let $\rho$ be the Weil representation on
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$ (as given in Proposition 7.2).
Suppose that $q$ is squarefree and has all prime factors greater than
$p_{0}(Q)$. Then $\rho|_{\Gamma_{q}}$ splits into irreducible
subrepresentations of dimensions $\geqslant q^{1-o(1)}$.

We will prove this in Sections 10 and 11.

Propositions 8.4 and 8.5 are precisely what is needed in order to apply
Proposition 8.2, and the conclusion is precisely Proposition 7.6. Therefore we
have, as the remaining outstanding tasks, the proofs of these two
propositions.

##  9\. Uniform distribution of convolution powers

In this section we establish Proposition 8.4. Our proof of this statement has
a hint of model theory about it, though we will not use that language. As
remarked in the introduction, it is somewhat related to Tao’s argument in
[25]. Here is a rough plan of the proof.

  1. (1)

(Step 1) Consider first the case $q=p$ a sufficiently large prime. We argue
that the sequence $\|\mu_{p}^{(2^{j})}\|$, $j=0,1,2,\dots$ (which is non-
increasing by Young’s inequality) stabilises at some time $t=O(1)$, in the
sense that $\|\mu_{p}^{(2^{t+1})}\|=(1+O(p^{-1/2}))\|\mu_{p}^{(2^{t})}\|$.
This uses the Lang-Weil bound from algebraic geometry.

  2. (2)

(Step 2) By standard arguments from additive combinatorics (recalled in
Appendix C), this implies that $\|\mu_{p}^{(2^{t})}-\mu_{H}\|\ll
p^{-c}\|\mu_{H}\|$, for some subgroup
$H\leqslant\operatorname{Sp}_{8}(\mathbf{Z}/p\mathbf{Z})$.

  3. (3)

(Step 3) By some group-theoretic arguments, $H$ must in fact be $\Gamma_{p}$.

  4. (4)

(Step 4) Taking a few further convolution powers, we upgrade the estimate to a
much stronger bound $\|\mu^{(s)}-\mu_{\Gamma_{p}}\|_{\infty}\ll p^{-100}$.

  5. (5)

(Step 5) We deduce the general squarefree case of Proposition 8.4.

_Lang-Weil estimate._ We keep algebro-geometric terminology to an absolute
minimum. A good down-to-earth account of what we need may be found in [26,
Chapter 9]. Let $M$ be a real parameter. Then for the purposes of this paper,
by a variety of complexity $\leqslant M$ defined over $\mathbf{F}_{p}$ we mean
a set of points of the form

$V:=\\{x\in\overline{\mathbf{F}}_{p}^{n}:P_{1}(x)=\cdots=P_{m}(x)=0\\},$

where $P_{1},\dots,P_{m}\in\mathbf{F}_{p}[X_{1},\dots,X_{n}]$ are polynomials
all of degree $\leqslant M$, and $m,n\leqslant M$. Denote by
$V(\mathbf{F}_{p})$ the $\mathbf{F}_{p}$-points of $V$, that is to say the
points of $V$ all of whose coordinates lie in $\mathbf{F}_{p}$.

######  Proposition 9.1 (Lang-Weil).

We have

$|V(\mathbf{F}_{p})|=(c(V)+O_{M}(p^{-1/2}))p^{\dim V},$

for some integer $c(V)=O_{M}(1)$.

_Remark._ In fact, $c(V)$ is the number of top-dimensional components of $V$
which are definable over $\mathbf{F}_{p}$, but we shall not need this
description. Nor will we need to really know what dimension means, other than
that it is an integer in the range $0\leqslant\dim V\leqslant n$. All we need
is the fact that the quantities $|V(\mathbf{F}_{p})|$ are restricted to a
rather discretised set of values. This kind of application of Lang-Weil has
appeared in several model-theoretic works and is related to the concept of
_stability_.

The Lang-Weil estimate has the following consequence for convolution powers of
our measures $\mu_{p}$.

######  Lemma 9.2.

There are functions
$\alpha,\beta:\mathbf{N}\times\\{\textup{primes}\\}\rightarrow\mathbf{Z}_{\geqslant
0}$ and non-decreasing functions $\alpha_{*}\rightarrow\mathbf{Z}_{\geqslant
0}$ and $p_{0}:\mathbf{N}\rightarrow\mathbf{N}$ such that for all $j\geqslant
1$ and for all primes $p$ we have

$\|\mu_{p}^{(2^{j})}\|^{2}=(\alpha(j,p)+O_{j}(p^{-1/2}))p^{-\beta(j,p)},$
(9.1)

where $\alpha(j,p),\beta(j,p)$ are integers with
$0<\alpha(j,p)\leqslant\alpha_{*}(j)$. Moreover, if $p\geqslant p_{0}(j)$ then
we have

$0\leqslant\beta(j,p)\leqslant 36;$ (9.2) $\beta(j+1,p)\geqslant\beta(j,p)$
(9.3)

and

$\alpha(j+1,p)\leqslant\alpha(j,p)\quad\mbox{if}\quad\beta(j+1,p)=\beta(j,p).$
(9.4)

###### Proof.

The key point is to interpret $\|\mu_{p}^{(2^{j})}\|^{2}$ in terms of the
number of $\mathbf{F}_{p}$-points on a variety of bounded complexity. To this
end, we have

$\displaystyle\|\mu^{(2^{j})}_{p}\|^{2}=\sum_{x}\mu_{p}^{(2^{j})}(x)^{2}$
$\displaystyle=(p-1)^{-2^{j+1}}\\#\\{(r,r^{\prime})\in(\mathbf{F}_{p}^{*})^{2^{j}}\times(\mathbf{F}_{p}^{*})^{2^{j}}:g^{-(r_{1})}g^{(r_{2})}\cdots
g^{(r_{2^{j}})}=$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=g^{-(r^{\prime}_{1})}g^{(r^{\prime}_{2})}\cdots
g^{(r^{\prime}_{2^{j}})}\\}$
$\displaystyle=(p-1)^{-2^{j+1}}\\#\\{(r,r^{\prime},y,y^{\prime})\in\mathbf{F}_{p}^{2^{j}}\times\mathbf{F}_{p}^{2^{j}}\times\mathbf{F}_{p}\times\mathbf{F}_{p}:r_{1}\cdots
r_{2^{j}}y=$ $\displaystyle\qquad\qquad=r^{\prime}_{1}\cdots
r^{\prime}_{2^{j}}y^{\prime}=1,\;\;g^{-(r_{1})}g^{(r_{2})}\cdots
g^{(r_{2^{j}})}=g^{-(r^{\prime}_{1})}g^{(r^{\prime}_{2})}\cdots
g^{(r^{\prime}_{2^{j}})}\\}$
$\displaystyle=(p-1)^{-2^{j+1}}|V_{j}(\mathbf{F}_{p})|=(1+O_{j}(\frac{1}{p}))p^{-2^{j+1}}|V_{j}(\mathbf{F}_{p})|,$

where $V_{j}\subset\overline{\mathbf{F}}_{p}^{2^{j+1}+2}$ is some variety of
complexity $O_{j}(1)$, defined over $\mathbf{F}_{p}$. Note here that, although
(for instance)
$g^{(r_{1})}=\big{(}\begin{smallmatrix}-2b^{-T}c&r_{1}^{-1}b^{-T}\\\
r_{1}(4ab^{-T}c-b)&-2ab^{-T}\end{smallmatrix}\big{)}$ is not _a priori_ given
by polynomials, we can express

$g^{(r_{1})}=\begin{pmatrix}-2b^{-T}c&r_{2}\cdots r_{2^{j}}yb^{-T}\\\
r_{1}(4ab^{-T}c-b)&-2ab^{-T}\end{pmatrix},$

and this _is_ given by polynomials. (Alternatively, one could talk about
quasiprojective varieties, but the trick of introducing $y,y^{\prime}$ avoids
the need to do that.)

This immediately implies, by the Lang-Weil bound, the first statement (9.1)
(with $\beta(j,p)=2^{j+1}-\dim V_{j}$). We now proceed to derive the
additional statements (9.2), (9.3) and (9.4), which we do by combinatorial
means (with reference to (9.1)).

For (9.2), note that _any_ probability measure $\nu$ on a finite group $G$
satisfies $|G|^{-1}\leqslant\|\nu\|^{2}\leqslant 1$; the lower bound is
Cauchy-Schwarz, and the upper bound is the trivial bound (with equality only
if $\nu$ is concentrated at one point). Since (see (7.4))
$|\operatorname{Sp}_{8}(\mathbf{Z}/p\mathbf{Z})|=(1+o(1))p^{36}$, (9.2)
follows if $p$ is large enough.

For items (9.3) and (9.4) we use Young’s inequality (Lemma C.1), which implies
that $\|\mu_{p}^{(2^{j})}\|^{2}$ is a non-increasing function of $j$.
Therefore we have, by (9.1),

$(\alpha(j+1,p)+O_{j}(p^{-1/2}))p^{-\beta(j+1,p)}\leqslant(\alpha(j,p)+O_{j}(p^{-1/2}))p^{-\beta(j,p)}.$

Here, $\alpha(j+1,p),\alpha(j,p)$ are positive integers of size $O_{j}(1)$ and
so by taking $p$ sufficiently large in terms of $j$ both (9.3) and (9.4)
follow.

Finally, note that $\alpha(j,p)$ is bounded above by $O_{M_{j}}(1)$, where
$M_{j}=O_{j}(1)$ is an upper bound for the complexity of $V_{j}$. Thus
$\alpha(j,p)\leqslant\alpha_{*}(j)$ for some function $\alpha_{*}$, which we
may clearly assume to be non-decreasing (else replace it by $\sup_{i\leqslant
j}\alpha_{*}(i)$). For the same reasons, we may also assume that $p_{0}$ is a
non-decreasing function. ∎

_Remark._ There should probably be a more purely algebro-geometric way to see
the inequalities (9.3), (9.4) in terms of the number of connected components
and dimensions of the relevant $V$s, but this would certainly take much longer
to set up than the argument we gave above.

We now turn to the realisation of Step 1 of the outline.

######  Proposition 9.3.

Suppose that $p$ is sufficiently large. Then there is some $t=O(1)$ such that
$\|\mu_{p}^{(2^{t+1})}\|=(1+O(p^{-1/2}))\|\mu_{p}^{(2^{t})}\|$.

###### Proof.

We use Lemma 9.2 and the notation there. It clearly suffices to show that, for
some $t=O(1)$, we have

$\alpha(t+1,p)=\alpha(t,p)\quad\mbox{and}\quad\beta(t+1,p)=\beta(t,p).$ (9.5)

Define a sequence $T_{1},T_{2},\dots,T_{37}$ as follows. Set
$T_{1}:=\alpha_{*}(1)$, and then inductively define
$T_{i+1}:=T_{i}+\alpha_{*}(T_{i})$ for $i=1,2,3,\dots,36$. If $p\geqslant
p_{0}(T_{37})$ we have the bounds (9.2), (9.3) and (9.4), for all $j\leqslant
T_{37}$.

We claim that there is some $t\leqslant T_{37}$ such that (9.5) holds. Suppose
not. Then, by (9.3), (9.4) we have that for all $j\leqslant T_{37}$ either

  1. (1)

$\beta(j+1,p)>\beta(j,p)$ or

  2. (2)

$\beta(j+1,p)=\beta(j,p)$ and $\alpha(j+1,p)<\alpha(j,p)$.

By (9.2), there are at most $36$ values of $j$ for which (1) occurs; suppose
they are $T^{\prime}_{1},\dots,T^{\prime}_{m}$, $m\leqslant 36$.

For $j=1,2,\dots,T^{\prime}_{1}-1$ we must have (2), which means that
$\alpha(T^{\prime}_{1},p)\leqslant\alpha(1,p)-T^{\prime}_{1}\leqslant\alpha_{*}(1)-T^{\prime}_{1}+1=T_{1}-T^{\prime}_{1}+1$.
Since $\alpha(j,p)$ is always positive, this implies that
$T^{\prime}_{1}\leqslant T_{1}$.

Now for $j=T^{\prime}_{1}+1,\dots,T^{\prime}_{2}-1$ we must also have (2),
which means that

$\displaystyle\alpha(T^{\prime}_{2},p)$
$\displaystyle\leqslant\alpha(T^{\prime}_{1},p)-(T^{\prime}_{2}-T^{\prime}_{1}-1)$
$\displaystyle\leqslant\alpha_{*}(T^{\prime}_{1})-(T^{\prime}_{2}-T^{\prime}_{1}-1)\leqslant\alpha_{*}(T_{1})+T_{1}-T^{\prime}_{2}+1.$

Since $\alpha(j,p)$ is always positive, this implies that
$T^{\prime}_{2}\leqslant\alpha_{*}(T_{1})+T_{1}=T_{2}$. Continuing in this
manner we see inductively that $T^{\prime}_{m}\leqslant T_{m}\leqslant
T_{36}$. Continuing now with $j=T^{\prime}_{m}+i$, $i=1,2,\dots$, only (2) can
occur, and so

$\alpha(T^{\prime}_{m}+i,p)\leqslant\alpha(T^{\prime}_{m},p)-i\leqslant\alpha_{*}(T^{\prime}_{m})-i\leqslant\alpha_{*}(T_{36})-i.$

Since (yet again) $\alpha(j,p)$ is always positive, this can only continue as
far as $i=\alpha_{*}(T_{36})-1$ before we get a contradiction. Note that then
$j\leqslant T^{\prime}_{m}+\alpha_{*}(T_{36})\leqslant T_{37}$, so all the
appeals we made to (9.2), (9.3) and (9.4) were indeed valid.

This contradiction shows that we were wrong to assume that there is no
$t\leqslant T_{37}$ for which (9.5) holds. ∎

Step 2. The conclusion of Step 1 (Proposition 9.3) is that for some $t$,
$1\leqslant t\ll O(1)$, we have, for the symmetic measure $\nu:=\mu^{(2^{t})}$
the very strong “flattening”

$\|\nu\ast\nu\|=(1+O(p^{-1/2})\|\nu\|.$

It is well-known that any probability measure satisfying this kind of property
is close to uniform on a subgroup. The precise statement we need is Corollary
C.3 in Appendix C, from which we conclude that there is some subgroup $H$ such
that

$\|\mu_{p}^{(2^{t})}-\mu_{H}\|\ll p^{-c}\|\mu_{H}\|,$ (9.6)
$\mu_{p}^{(2^{t})}(H)\geqslant 1-O(p^{-c}),$ (9.7)

and

$|\operatorname{Supp}(\mu_{p}^{(2^{t})})|>(1-O(p^{-c}))|H|,$ (9.8)

where $\mu_{H}$ is the uniform measure on $H$. It follows from (9.7) that

$\displaystyle 1-O(p^{-c})\leqslant\mu_{p}^{(2^{t})}(H)$
$\displaystyle=(\mu_{p}^{(2^{t}-1)}\ast\mu_{p})(H)=$
$\displaystyle=\sum_{x}\mu_{p}^{(2^{t}-1)}(x)\mu_{p}(x^{-1}H)\leqslant\sup_{x}\mu_{p}(xH).$

Thus there is some coset $xH$ such that

$\mu_{p}(xH)\geqslant 1-O(p^{-c}).$ (9.9)

Step 3. In this step we use a group-theoretic argument, making use of some
slightly specific features of the problem, to upgrade the statement (9.9) to
$\mu(xH)=1$, or in other words (recalling Definitions 7.3 and 7.5) to show
that all $g^{(r)}$, $r\in(\mathbf{Z}/p\mathbf{Z})^{*}$, lie in $xH$.

Let $R:=\\{r\in(\mathbf{Z}/p\mathbf{Z})^{*}:g^{(r^{-1})}\in xH\\}$. Thus, by
(9.9),

$|R|\geqslant(1-O(p^{-c}))(p-1).$ (9.10)

Perform the following algorithm to generate distinct elements
$r_{1},r_{2},\dots$ of $R$ as long as possible. Write
$S_{j}:=\bigcap_{i\leqslant j}(xH)^{(r_{i})}$. Each $S_{j}$ is a coset (of
some subgroup of $\operatorname{Sp}_{8}(\mathbf{Z}/p\mathbf{Z})$) and, no
matter how we choose the $r_{i}$, we have the nesting

$S_{1}\supseteq S_{2}\supseteq\cdots$

If, at step $j$ of the construction, it is possible to choose $r_{j+1}\in R$
so that $S_{j+1}$ is a _proper_ subset of $S_{j}$ then do so; otherwise, stop.

Note that, as long as the algorithm continues, we have
$|S_{j+1}|\leqslant\frac{1}{2}|S_{j}|$ (since the $S_{j}$ are all cosets of
subgroups). Therefore, the algorithm stops in at most $O(\log p)$ steps.

When the algorithm finishes, we have $r_{1},\dots,r_{m}\in R$, $m=O(\log p)$,
and a coset $S:=S_{m}=\bigcap_{i=1}^{m}(xH)^{(r_{i})}$ (of some subgroup).
Note that, since $r_{i}\in R$, we have $g^{(r_{i}^{-1})}\in xH$ and so
$g\in(xH)^{(r_{i})}$, and therefore $g\in S$.

Now set

$R^{\prime}:=r_{1}^{-1}R\cap\cdots\cap r_{m}^{-1}R,$

and suppose that $r\in R^{\prime}$. Since the algorithm we described stopped
at the $m$th stage, we have

$S\cap(xH)^{(r_{i}r)}=S\qquad\mbox{for $i=1,\dots,m$},$

since otherwise we could take $r_{m+1}:=r_{i}r$ (which would be an element of
$R$ by the definition of $R^{\prime}$). It follows that

$S\cap S^{(r)}=S\cap\bigcap_{i=1}^{m}(xH)^{(r_{i}r)}=S.$

That is, if $r\in R^{\prime}$ then $S=S^{(r)}$. It follows that $S$ is
invariant under the entire subgroup of $(\mathbf{Z}/p\mathbf{Z})^{*}$
generated by $R^{\prime}$. However,

$|R^{\prime}|\geqslant 1-m|(\mathbf{Z}/p\mathbf{Z})^{*}\setminus
R|>\frac{1}{2}(p-1),$

by (9.10) and the fact that $m=O(\log p)$. Therefore the group generated by
$R^{\prime}$ is the whole of $(\mathbf{Z}/p\mathbf{Z})^{*}$, and so we have
that $S=S^{(r)}$ for all $r\in(\mathbf{Z}/p\mathbf{Z})^{*}$.

We showed earlier that $g\in S$. It now follows that $g^{(rr_{1})}\in S$ for
all $r\in(\mathbf{Z}/p\mathbf{Z})^{*}$. In particular,
$g^{(rr_{1})}\in(xH)^{(r_{1})}$, which implies that $g^{(r)}\in xH$ for all
$r\in(\mathbf{Z}/p\mathbf{Z})^{*}$, which is what we wanted to show.

Consequently, all the elements $g^{-(r)}g^{(s)}$ lie in $H$. By definition
(Definition 8.3), we have $\Gamma_{p}\leqslant H$.

It follows that $\operatorname{Supp}(\mu_{p}^{(2^{t})})\subseteq H$. However,
we showed in (9.8) that
$|\operatorname{Supp}(\mu_{p}^{(2^{t})})|>(1-O(p^{-c}))|H|>\frac{1}{2}|H|$,
and therefore the group generated by $\operatorname{Supp}(\mu_{p}^{(2^{t})})$
is all of $H$. However, the group generated by
$\operatorname{Supp}(\mu_{p}^{(2^{t})})$ is the group generated by
$\operatorname{Supp}(\mu_{p}^{(2)})$ which, as we remarked earlier, is
precisely $\Gamma_{p}$. Finally, we may conclude that $H=\Gamma_{p}$.
Therefore (9.6) may be rewritten as

$\|\mu_{p}^{(2^{t})}-\mu_{\Gamma_{p}}\|\ll p^{-c}\|\mu_{\Gamma_{p}}\|.$ (9.11)

Step 4. In this step of the argument we upgrade (9.11) to a highly uniform
estimate by taking a few extra convolution powers. By Cauchy-Schwarz we have

$\displaystyle\|\mu_{p}^{(2^{t+1})}-\mu_{\Gamma_{p}}\|_{\infty}$
$\displaystyle=\|(\mu_{p}^{(2^{t})}-\mu_{\Gamma_{p}})\ast(\mu_{p}^{(2^{t})}-\mu_{\Gamma_{p}})\|_{\infty}$
$\displaystyle\leqslant\|\mu_{p}^{(2^{t})}-\mu_{\Gamma_{p}}\|^{2}$
$\displaystyle\ll p^{-2c}\|\mu_{\Gamma_{p}}\|^{2}=p^{-2c}|\Gamma_{p}|^{-1}.$
(9.12)

However, if $\nu$ is some probability measure on a finite group $\Gamma$ of
size $N$ and if

$\|\nu-\mu_{\Gamma}\|_{\infty}\leqslant\frac{\varepsilon}{N}$

then

$\displaystyle|\nu\ast\nu(x)-\frac{1}{N}|$
$\displaystyle=|(\nu-\mu_{\Gamma})\ast(\nu-\mu_{\Gamma})(x)|$
$\displaystyle\leqslant\sum_{y}|\nu(y)-\frac{1}{N}||\nu(y^{-1}x)-\frac{1}{N}|\leqslant\frac{\varepsilon^{2}}{N},$

that is to say

$\|\nu^{(2)}-\mu_{\Gamma}\|_{\infty}\leqslant\frac{\varepsilon^{2}}{N}.$

Applying this $s$ times to (9.12) gives

$\|\mu_{p}^{(2^{t+1+s})}-\mu_{\Gamma_{p}}\|_{\infty}\leqslant(Cp^{-2c})^{2^{s}}|\Gamma_{p}|^{-1},$

and so, taking a suitably large $s$, there is some power of two
$m=2^{t+1+s}=O(1)$ such that

$\|\mu_{p}^{(m)}-\mu_{\Gamma_{p}}\|_{\infty}<p^{-38},$ (9.13)

provided (as always) $p$ is sufficiently large. Since we are free to choose
any sufficiently large $s=O(1)$, we can make the choice so that $m$ is
independent of $p$.

Note that (9.13) is a much stronger version of Proposition 8.4 in the case
$q=p$ a sufficiently large prime.

Since $|\Gamma_{p}|\leqslant|\operatorname{Sp}_{8}(\mathbf{Z}/p\mathbf{Z})|\ll
p^{36}$ it follows from (9.13) that

$\mu_{p}^{(m)}\leqslant(1+O(p^{-2}))|\Gamma_{p}|^{-1}$ (9.14)

pointwise. This one-sided estimate is the only one we will need subsequently.

Step 5. Finally, we turn to the deduction of Proposition 8.4 itself. That is,
we pass from the case $q=p$ a prime to the general case. Suppose then that $q$
is squarefree, and that all its prime factors are sufficiently large (larger
than $p_{0}(T_{37})$, the quantity appearing in Step 1, is enough). We have a
natural homomorphism

$\pi:\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})\rightarrow\prod_{p|q}\operatorname{Sp}_{8}(\mathbf{Z}/p\mathbf{Z}).$

By the Chinese remainder theorem, the measure $\mu_{q}$ pushes forward under
$\pi$ to the product $\times_{p|q}\mu_{p}$: we have
$\pi(g^{(r)})=(g^{(r(\mbox{\scriptsize mod}\,p))})_{p|q}$, and the tuple
$(r(\operatorname{mod}\,p))_{p|q}$ takes all values in
$\prod_{p|q}(\mathbf{Z}/p\mathbf{Z})^{*}$ as $r$ ranges over
$(\mathbf{Z}/q\mathbf{Z})^{*}$.

Recall that $\Gamma_{q}$ is the group generated by the $g^{-(r)}g^{(s)}$,
$r,s\in(\mathbf{Z}/q\mathbf{Z})^{*}$. These groups also behave nicely under
projection, as the following lemma shows.

######  Lemma 9.4.

Suppose that $q$ is squarefree, and let
$\pi:\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})\rightarrow\prod_{p|q}\operatorname{Sp}_{8}(\mathbf{Z}/p\mathbf{Z})$
be the natural isomorphism. Then $\pi(\Gamma_{q})=\prod_{p|q}\Gamma_{p}$.

###### Proof.

It is easy to see that $\pi(\Gamma_{q})\subseteq\prod_{p|q}\Gamma_{p}$. To see
that the two are in fact equal, suppose we have elements
$\gamma_{p}\in\Gamma_{p}$. For some $N$, we may write
$\gamma_{p}=g^{-(r_{1,p})}g^{(s_{1,p})}\cdots g^{-(r_{N,p})}g^{(s_{N,p})}$ for
elements $r_{i,p},s_{i,p}\in(\mathbf{Z}/p\mathbf{Z})^{*}$, that is to say as a
word consisting of a product of $N$ of the generators. Note that we can use
the same $N$ for each $p$ by padding with exponents $r_{i,p}=s_{i,p}$, if
necessary, each of which contributes the identity, and we have also taken
advantage of the fact that $(g^{-(r)}g^{(s)})^{-1}=g^{-(s)}g^{(r)}$, which
means we do not need to worry about including inverses separately. By the
Chinese remainder theorem there are
$r_{i},s_{i}\in(\mathbf{Z}/q\mathbf{Z})^{*}$ such that
$r_{i}(\operatorname{mod}\,p)=r_{i,p}$, and similarly for the $s_{i}$. Setting
$\gamma:=g^{-(r_{1})}g^{(s_{1})}\cdots
g^{(-r_{N})}g^{(s_{N})}\in\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$, we
see that $\pi(\gamma)=(\gamma_{p})_{p|q}$, as desired. ∎

_Remark._ We caution that this lemma is a rather specific result. If, for
example, we defined $\tilde{\Gamma}_{q}$ to be the group generated by the
$g^{(r)}$, $r\in(\mathbf{Z}/q\mathbf{Z})^{*}$, the same argument would not
work (consider, for example, the question of how to find an element whose
projection to $\operatorname{Sp}_{8}(\mathbf{Z}/p\mathbf{Z})$ is $g$, and
whose projection to $\operatorname{Sp}_{8}(\mathbf{Z}/p^{\prime}\mathbf{Z})$
is $g^{-1}$).

With these facts in hand, we may now complete the proof. Let $m$, a power of
two, be as in (9.14). Then

$\|\mu_{q}^{(m)}\|_{\infty}=\prod_{p|q}\|\mu_{p}^{(m)}\|_{\infty}\leqslant\prod_{p|q}(1+O(p^{-2}))|\Gamma_{p}|^{-1}\ll\prod_{p|q}|\Gamma_{p}|^{-1}=|\Gamma_{q}|^{-1}.$

This bound, coupled with the fact that
$\operatorname{Supp}(\mu^{(m)}_{q})\subset\Gamma_{q}$, implies Proposition
8.4.

##  10\. Identifying $\Gamma_{q}$

We turn now to the task of proving Proposition 8.5. The first stage is to
actually identify the group $\Gamma_{q}$ that is to say the subgroup of
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$ generated by the
$g^{-(r)}g^{(s)}$, $r,s\in(\mathbf{Z}/q\mathbf{Z})^{*}$, in an explicit
algebraic form. Recall that the symplectic element $g$ is a particular element
of $\operatorname{Sp}_{8}(\mathbf{Q})$ associated to the quadratic form $Q$
(see Definition 7.3). In view of Lemma 9.4, it is enough to consider the prime
case $\Gamma_{p}$.

To understand this group, it seems best to think of
$\operatorname{Sp}_{8}(\mathbf{Z}/p\mathbf{Z})$ as embedded in the full group
$\operatorname{GL}_{8}(\mathbf{Z}/p\mathbf{Z})$ (in the obvious way). It then
turns out that $\Gamma_{p}$ is a conjugate (in
$\operatorname{GL}_{8}(\mathbf{Z}/p\mathbf{Z})$) of the group
$\operatorname{SL}_{2}(\mathbf{F}_{p}[\Delta])\leqslant\operatorname{GL}_{8}(\mathbf{Z}/p\mathbf{Z})$,
where (recall) $\Delta=4b^{-1}ab^{-T}c-I$ is the matrix discriminant of our
form $Q$. We will discuss such groups at much greater length in due course.
For now, note that $\mathbf{F}_{p}[\Delta]$ is an algebra over
$\mathbf{F}_{p}$ of dimension at most $4$ (by the Cayley-Hamilton theorem).
Generically, this algebra will have dimension exactly $4$ and in this case
$\operatorname{SL}_{2}(\mathbf{F}_{p}[\Delta])$ (and hence $\Gamma_{p}$) will
be a group of of size $\sim p^{12}$.

The precise structure of $\operatorname{SL}_{2}(\mathbf{F}_{p}[\Delta])$ will
depend on how the characteristic polynomial $\rho_{\Delta}$ for $\Delta$
splits over $\mathbf{F}_{p}$, but it will be a direct product of some groups
$\operatorname{SL}_{2}(\mathbf{F}_{p^{n}})$ for various $n$.

Suppose henceforth that $Q$ is generic, and that $p$ is large enough that
$a,b,\Delta$ are invertible over $\mathbf{F}_{p}$. In order to examine the
elements $g^{-(r)}g^{(s)}$ (which, by definition, generate $\Gamma_{p}$), one
eventually hits upon the idea of looking at what might be called a “block DUL
factorisation” of $g$. Namely, there is an invertible $A$ and symmetric $B,C$
(all 4-by-4 matrices) such that

$g=\begin{pmatrix}A&0\\\ 0&A^{-T}\end{pmatrix}\begin{pmatrix}I&B\\\
0&I\end{pmatrix}\begin{pmatrix}I&0\\\ C&I\end{pmatrix}.$ (10.1)

To see that such a factorisation exists is simple: writing
$g=\big{(}\begin{smallmatrix}P&Q\\\ R&S\end{smallmatrix}\big{)}$ where
$P=-2b^{-T}c$, $Q=b^{-T}$, $R=4ab^{-T}c-b$, $S=-2ab^{-T}$, we can take

$A=S^{-T},\;B=S^{T}Q,\;C=S^{-1}R.$ (10.2)

Note that $S$ is well-defined and invertible (over $\mathbf{Q}$ and over
$\mathbf{F}_{p}$) since both $a$ and $b$ are invertible.

The matrices $B$ and $C$ here will be somewhat important in their own right.
We calculate

$B=-2b^{-1}ab^{-T},\quad C=-2c+\frac{1}{2}b^{T}a^{-1}b,\quad BC=\Delta.$
(10.3)

We note that the appearance of $\Delta$ here is the reason for its definition.
Since $\Delta$ is assumed invertible over $\mathbf{F}_{p}$, both $B$ and $C$
are invertible over $\mathbf{F}_{p}$.

The purpose of looking at the DUL factorisation (10.1) is that it renders the
action of dilation easy to understand. Indeed,

$g^{-(r)}g^{(s)}=\begin{pmatrix}I&0\\\
-rC&I\end{pmatrix}\begin{pmatrix}I&-r^{-1}B\\\
0&I\end{pmatrix}\begin{pmatrix}I&s^{-1}B\\\
0&I\end{pmatrix}\begin{pmatrix}I&0\\\ sC&I\end{pmatrix}.$ (10.4)

Now set

$\tau:=\begin{pmatrix}I&0\\\
0&B\end{pmatrix}\in\operatorname{GL}_{8}(\mathbf{Q}).$ (10.5)

This is invertible over $\mathbf{F}_{p}$ and so may considered as an element
of $\operatorname{GL}_{8}(\mathbf{Z}/p\mathbf{Z})$. Now observe that for
$\lambda\in\mathbf{F}_{p}^{*}$ we have

$\begin{pmatrix}I&\lambda B\\\
0&I\end{pmatrix}=\tau^{-1}\begin{pmatrix}I&\lambda I\\\
0&I\end{pmatrix}\tau,\quad\begin{pmatrix}I&0\\\ \lambda
C&I\end{pmatrix}=\tau^{-1}\begin{pmatrix}I&0\\\
\lambda\Delta&I\end{pmatrix}\tau.$ (10.6)

It follows from this and (10.4) that $g^{-(r)}g^{(s)}$ takes values in the
subgroup
$\tau^{-1}\operatorname{SL}_{2}(\mathbf{F}_{p}[\Delta])\tau\leqslant\operatorname{Sp}_{8}(\mathbf{Z}/p\mathbf{Z})$.

_Remark._ It is important to note that neither $\tau$ nor
$\operatorname{SL}_{2}(\mathbf{F}_{p}[\Delta])$ are contained in
$\operatorname{Sp}_{8}(\mathbf{Z}/p\mathbf{Z})$ in general, although of course
the conjugate $\tau^{-1}\operatorname{SL}_{2}(\mathbf{F}_{p}[\Delta])\tau$ is.
It turns out that (assuming $Q$ is generic, and for sufficiently large $p$)
this is the group $\Gamma_{p}$. This is the first key result of the section.

######  Proposition 10.1.

Suppose that $Q$ is generic and that $p$ is sufficiently large in terms of
$Q$. Then
$\Gamma_{p}=\tau^{-1}\operatorname{SL}_{2}(\mathbf{F}_{p}[\Delta])\tau$.

Before turning to the proof, we assemble some lemmas. We will need the
following two facts about polynomials.

######  Lemma 10.2.

Let $\eta>0$. Suppose that $F\in\overline{\mathbf{F}}_{p}[X,Y]$ has total
degree $D$ and that $p>2D/\eta$. Then

  1. (1)

If $F(x,y)=0$ for at least a proportion $\eta$ of all pairs
$(x,y)\in\mathbf{F}_{p}\times\mathbf{F}_{p}$, then $F$ is identically zero;

  2. (2)

If $F(x,y)$ takes values in some subfield $k_{0}<\overline{\mathbf{F}}_{p}$,
for at least a proportion $\eta$ of all pairs
$(x,y)\in\mathbf{F}_{p}\times\mathbf{F}_{p}$, then all the coefficients of $F$
lie in $k_{0}$.

###### Proof.

(i) This is an immediate consequence of the Schwartz-Zippel lemma, which
states that if $F$ is not identically zero then the number of solutions to
$F(x,y)=0$ with $x,y\in\mathbf{F}_{p}$ is at most $Dp$, which is less than
$\eta p^{2}$ under the assumptions of the lemma.

(ii) We begin with a 1-variable version. If $f(x)$ is a polynomial of degree
$D$ which takes values in $k_{0}$ for $D+1$ different values of
$x\in\mathbf{F}_{p}$ then it follows from Lagrange interpolation that all the
coefficients of $f$ lie in $k_{0}$.

Turning to the 2-variable statement we actually want, write
$F(X)=\sum_{i=0}^{D}f_{i}(X)Y^{i}$, where $\deg f_{i}\leqslant D$. Let
$S\subset\mathbf{F}_{p}\times\mathbf{F}_{p}$ be the set of pairs $(x,y)$ for
which $F(x,y)\in k_{0}$. For each $x$, let $S_{x}:=\\{y:(x,y)\in S\\}$. Then
there is a set $A\subset\mathbf{F}_{p}$, $|A|\geqslant\eta p/2$, such that
$|S_{x}|\geqslant\eta p/2$ for all $x\in A$. If $x\in A$, the 1-variable
result implies (since $p>2D/\eta$) that all the $f_{i}(x)$ lie in $k_{0}$. A
second application of the 1-variable result then implies that all the
coefficients of each $f_{i}$ lie in $k_{0}$.

We remark that the proof technique for (ii) can also be used for (i) (in fact
this is essentially the usual proof of Schwartz-Zippel by induction). ∎

We will also need a couple of lemmas about subgroups of direct products. Both
may be found in [19, Chapter 1]. The first result is well-known.

######  Lemma 10.3 (Goursat’s lemma).

Let $G_{1},G_{2}$ be groups. Consider the direct product $G_{1}\times G_{2}$
and let $\pi_{i}:G_{1}\times G_{2}\rightarrow G_{i}$ be the two projection
maps. Let $H\leqslant G_{1}\times G_{2}$ be a subgroup, and suppose that
$\pi_{i}(H)=G_{i}$ for $i=1,2$. Then there are normal subgroups $N_{i}\lhd
G_{i}$ and an isomorphism $\phi:G_{1}/N_{1}\rightarrow G_{2}/N_{2}$ such that
$H$ has the form $\\{(g_{1},g_{2})\in G_{1}\times
G_{2}:\phi(\overline{g}_{1})=\overline{g}_{2}\\}$, where $\overline{g}_{i}$ is
the image of $g_{i}$ in $G_{i}/N_{i}$.

###### Proof.

See, for example, [19, Proposition 1.6]. ∎

The second result is somewhat less well-known and is called Ribet’s lemma by
Serre [19].

######  Lemma 10.4 (Ribet’s Lemma).

Let $G_{1},\dots,G_{n}$ be perfect groups, that is to say equal to their own
commutator subgroups. Let $H\leqslant G_{1}\times\cdots\times G_{n}$ be such
that the projection $\pi_{ij}$ of $H$ to $G_{i}\times G_{j}$ is surjective for
every pair $(i,j)$. Then $H$ is the whole of $G_{1}\times\cdots\times G_{n}$.

###### Proof.

See [19, Proposition 1.8]. ∎

The next lemma, which looks a little _ad hoc_ , is in some sense the scalar
version of Proposition 10.1, and is the heart of the proof of it.

######  Lemma 10.5.

Suppose that $p\geqslant 5$ and that
$\theta\in\overline{\mathbf{F}}_{p}\setminus\\{0,-1\\}$. Then the matrices

$M_{\theta}(r,s)=\begin{pmatrix}1&0\\\
-r\theta&1\end{pmatrix}\begin{pmatrix}1&-r^{-1}\\\
0&1\end{pmatrix}\begin{pmatrix}1&s^{-1}\\\
0&1\end{pmatrix}\begin{pmatrix}1&0\\\ s\theta&1\end{pmatrix},$

as $r,s$ range over $\mathbf{F}_{p}^{*}$, generate
$\operatorname{SL}_{2}(\mathbf{F}_{p}(\theta))$.

###### Proof.

Set $k:=\mathbf{F}_{p}(\theta)$. We use the fact that any proper subgroup of
$\operatorname{SL}_{2}(k)$ has a subgroup of index $C=O(1)$ which is conjugate
to a subgroup of one of

  1. (1)

the group of upper triangular matrices or

  2. (2)

$\operatorname{SL}_{2}(k_{0})$ for some proper subfield $k_{0}<k$.

See Appendix A for further comments on this fact.

Suppose, then, that the $M_{\theta}(r,s)$ generate a proper subgroup
$\Gamma<\operatorname{SL}_{2}(\mathbf{F}_{p}(\theta))$. Let
$\Gamma^{\prime}\leqslant\Gamma$, $[\Gamma:\Gamma^{\prime}]\leqslant C$, be a
subgroup conjugate to a group of type (1) or (2) above. By the pigeonhole
principle there is some coset $\Gamma^{\prime}x$ containing the elements
$g^{-(r)}g$, $r\in R$, for some set $R\subset\mathbf{F}_{p}^{*}$ of size at
least $\frac{1}{C}(p-1)$. Then $\Gamma^{\prime}$ contains the elements
$g^{-(r)}g^{(s)}=g^{-(r)}g^{-1}(g^{-(s)}g^{-1})^{-1}$ for all $r,s\in R$, that
is to say for more than $\eta p^{2}$ pairs $(r,s)$, where
$\eta=(2C^{2})^{-1}$.

Now we may explicitly compute that if $P=\big{(}\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\big{)}\in\operatorname{SL}_{2}(k)$ (here $a,b,c$ are
local to the proof of Lemma 10.5, not the coefficients of $Q$) then the
bottom-left entry of $PM_{\theta}(r,s)P^{-1}$ is

$\frac{1}{rs}\big{(}d^{2}\theta(1+\theta)(s^{2}r-r^{2}s)+c^{2}(s-r)+cd\theta(r^{2}-s^{2})\big{)}.$

Consider the bracketed expression as a polynomial in $r$ and $s$. If this is
to be zero for at least $\eta p^{2}$ pairs
$(r,s)\in\mathbf{F}_{p}^{*}\times\mathbf{F}_{p}^{*}$ then the first part of
Lemma 10.2 implies that (if $p$ is sufficiently large) all the coefficients of
this polynomial vanish, and so $c^{2}=cd\theta=d^{2}\theta(1+\theta)=0$. We
cannot have $c=d=0$ (since $P$ is invertible) and so, since $\theta\neq 0,-1$,
possibility (1) is excluded.

Turning to the subfield case (2), we proceed similarly, but now using the
second part of Lemma 10.2. This implies that all of
$c^{2}=cd\theta=d^{2}\theta(1+\theta)$ lie in $k_{0}$. If $cd\neq 0$, it
follows that

$1+\frac{1}{\theta}=\frac{c^{2}\cdot d^{2}\theta(1+\theta)}{(cd\theta)^{2}}\in
k_{0},$

and so $\theta\in k_{0}$. This, however, is impossible since
$k=\mathbf{F}_{p}(\theta)$ and $k_{0}$ is assumed to be a _proper_ subfield of
$k$.

If $c=0$, we consider additionally the top-left entry of $PM(r,s)P^{-1}$,
which (when $c=0$) is

$\frac{1}{r}\big{(}ad(1+\theta)r-ad\theta
s+bd(rs-r^{2})\theta(1+\theta)\big{)}.$

Note that $ad=\det P=1$. Therefore, by the second part of Lemma 10.2,
$\theta\in k_{0}$. This is again a contradiction.

Finally if $d=0$, the top-left entry of $PM(r,s)P^{-1}$ is

$\frac{1}{rs}\big{(}ac(s-r)-bc(1+\theta)rs+bc\theta r^{2}\big{)}.$

Note that $bc=\det P=-1$, and so again we get $\theta\in k_{0}$. ∎

###### Proof of Proposition 10.1.

We must show that the elements $\gamma(r,s):=\tau g^{-(r)}g^{(s)}\tau^{-1}$
generate $\operatorname{SL}_{2}(\mathbf{F}_{p}[\Delta])$. From (10.4), (10.6)
we have

$\gamma(r,s)=\begin{pmatrix}I&0\\\
-r\Delta&I\end{pmatrix}\begin{pmatrix}I&-r^{-1}I\\\
0&I\end{pmatrix}\begin{pmatrix}I&s^{-1}I\\\
0&I\end{pmatrix}\begin{pmatrix}I&0\\\ s\Delta&I\end{pmatrix}.$ (10.7)

This should be compared with the definition of $M_{\theta}(r,s)$ in Lemma
10.5.

Now we are assuming that $Q$ is generic which, by definition, means that the
characteristic polynomial $\rho_{\Delta}(\lambda)=\det(\Delta-\lambda I)$ has
four distinct roots in $\overline{\mathbf{Q}}\setminus\\{0,-1\\}$.
Consequently, it will also be the minimum polynomial of $\Delta$ over
$\mathbf{Q}$.

Let $\overline{\rho}_{\Delta}\in\mathbf{F}_{p}[X]$ be the reduction of
$\rho_{\Delta}$ modulo $p$. If $p$ is sufficiently large,
$\overline{\rho}_{\Delta}$ will have four distinct roots in
$\overline{\mathbf{F}}_{p}\setminus\\{-1,0\\}$. (The resultant
$\operatorname{Res}(\rho_{\Delta},\rho^{\prime}_{\Delta})\in\mathbf{Q}[X]$ is
not the zero polynomial, by assumption, and so it is also not the zero
polynomial when reduced mod $p$, for $p$ sufficiently large; also
$\rho_{\Delta}(0),\rho_{\Delta}(-1)\neq 0$ in $\mathbf{Q}$, and hence in
$\mathbf{F}_{p}$ for $p$ sufficiently large.) Since it has distinct roots,
$\overline{\rho}_{\Delta}$ will be the minimal polynomial of $\Delta$ over
$\mathbf{F}_{p}$.

Henceforth, suppose that $p\geqslant p_{0}(Q)$. Suppose that the factorisation
of $\overline{\rho}_{\Delta}$ into irreducibles polynomials over
$\mathbf{F}_{p}$ is $f_{1}\cdots f_{n}$. Then, since
$\overline{\rho}_{\Delta}$ has distinct roots in $\overline{\mathbf{F}}_{p}$,
the $f_{i}$ will be coprime. For each $i$, let
$\alpha_{i}\in\overline{\mathbf{F}}_{p}$ be a root of $f_{i}$ and consider the
map

$\Phi:\mathbf{F}_{p}[\Delta]\rightarrow\prod_{i=1}^{n}\mathbf{F}_{p}(\alpha_{i})$

given by

$\Phi(F(\Delta))=(F(\alpha_{1}),\dots,F(\alpha_{n}))$

for any $F\in\mathbf{F}_{p}[X]$. This is a well-defined ring homomorphism: if
$F_{1}(\Delta)=F_{2}(\Delta)$ then $\overline{\rho}_{\Delta}|F_{1}-F_{2}$
(since $\overline{\rho}_{\Delta}$ is the minimal polyomial of $\Delta$) and
hence $f_{i}|F_{1}-F_{2}$ for each $i$, whence
$F_{1}(\alpha_{i})=F_{2}(\alpha_{i})$. We claim that $\Phi$ is injective. If
$\Phi(F(\Delta))=0$ then for $i=1,\dots,n$ we have $F(\alpha_{i})=0$ which
implies $f_{i}|F$. Since the $f_{i}$ are coprime, $\overline{\rho}_{\Delta}|F$
and so $F(\Delta)=0$. Both the domain and range of $\Phi$ have size $p^{4}$
and so it is in fact a ring isomorphism.

Therefore there are five possible isomorphism types for the ring
$\mathbf{F}_{p}[\Delta]$, namely $\mathbf{F}_{p^{4}}$ (if
$\overline{\rho}_{\Delta}$ is irreducible over $\mathbf{F}_{p}$),
$\mathbf{F}_{p^{3}}\times\mathbf{F}_{p}$,
$\mathbf{F}_{p^{2}}\times\mathbf{F}_{p^{2}}$,
$\mathbf{F}_{p^{2}}\times\mathbf{F}_{p}\times\mathbf{F}_{p}$, or
$\mathbf{F}_{p}\times\mathbf{F}_{p}\times\mathbf{F}_{p}\times\mathbf{F}_{p}$
(if $\overline{\rho}_{\Delta}$ splits completely over $\mathbf{F}_{p}$.) Note
that by standard algebraic number theory we can expect all of these
possibilities to occur as $p$ varies over primes.

The map $\Phi$ induces a group isomorphism

$\Phi:\operatorname{SL}_{2}(\mathbf{F}_{p}[\Delta])\rightarrow\prod_{i=1}^{n}\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{i})).$

In view of (10.7), we have

$\Phi(\gamma(r,s))=(M_{\alpha_{i}}(r,s))_{i=1}^{n},$

where the $M_{\alpha_{i}}(r,s)$ are as defined in Lemma10.5. It follows from
this and Lemma 10.5 that, if
$\Gamma=\langle\gamma(r,s):r,s\in\mathbf{F}_{p}^{*}\rangle=\tau\Gamma_{p}\tau^{-1}$
is the group generated by the $\gamma(r,s)$, then the projection of
$\Phi(\Gamma)$ on to each factor
$\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{i}))$ is surjective. When $n=1$,
this is the end of the proof, but we must work a little harder in the other
cases.

Let us begin by looking at $H$, the projection of $\Phi(\Gamma)$ to the
product
$\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{1}))\times\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{2}))$
of two of the factors (without loss of generality, the first two). Write
$\pi_{i}$, $i=1,2$ for projection onto each factor. As we have remarked,
$\pi_{i}(H)=\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{i}))$. This allows us
to apply Goursat’s lemma (Lemma 10.3). We conclude that there are
$N_{i}\lhd\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{i}))$, and an
isomorphism
$\phi:\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{1}))/N_{1}\rightarrow\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{2}))/N_{2}$
such that
$H=\\{(g_{1},g_{2})\in\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{1}))\times\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{2})):\phi(\overline{g}_{1})=\overline{g}_{2}\\}$,
where $\overline{g}_{i}$ denotes reduction mod $N_{i}$.

Now the $\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{i}))$ are almost simple:
each $N_{i}$ must be either trivial, $\\{\pm I\\}$ or
$\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{i}))$. See Appendix A. Moreover,
the fact that $\phi$ is an isomorphism, and the fact that
$\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{i}))$ is not isomorphic to
$\operatorname{PSL}_{2}(\mathbf{F}_{p}(\alpha_{j}))$ (consider cardinalities),
means that up to relabelling there are only three essentially different cases,
which we consider separately below.

_Case 1._ $N_{1}=\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{1}))$. Then
$N_{2}=\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{2}))$, and $H$ is the
whole of the product
$\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{1}))\times\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{2}))$.

_Case 2._ $N_{1}=N_{2}=\\{I\\}$. Then
$H=\\{(x,\phi(x)):x\in\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{1}))\\}$,
where
$\phi:\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{1}))\rightarrow\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{2}))$
is some isomorphism. Note that $H$ contains the elements
$(M_{\alpha_{1}}(r,s),M_{\alpha_{2}}(r,s))$, $r,s\in\mathbf{F}_{p}^{*}$, so in
this scenario we must have

$\phi(M_{\alpha_{1}}(r,s))=M_{\alpha_{2}}(r,s)$ (10.8)

for all $r,s$. By looking at cardinalities, the fields
$\mathbf{F}_{p}(\alpha_{1})$ and $\mathbf{F}_{p}(\alpha_{2})$ must be
isomorphic, so to ease notation we may suppose that
$\alpha_{2}\in\mathbf{F}_{p}(\alpha_{1})$.

Now it is known (see Appendix A) that the automorphism group of
$\operatorname{SL}_{2}(k)$ is generated by conjugation by elements of
$\operatorname{GL}_{2}(k)$ and field automorphisms. Therefore for some
$P\in\operatorname{GL}_{2}(\mathbf{F}_{p}(\alpha_{1}))$ and for some field
automorphism $\sigma$ of $\mathbf{F}_{p}(\alpha_{1})$ we have

$\phi(M_{\alpha_{1}}(r,s))=PM_{\sigma(\alpha_{1})}(r,s)P^{-1}.$

Comparing with (10.8) gives

$PM_{\sigma(\alpha_{1})}(r,s)=M_{\alpha_{2}}(r,s)P$ (10.9)

for all $r,s$. Writing $P=\big{(}\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\big{)}$ and comparing top left entries gives, writing
$\theta=\sigma(\alpha_{1})$ and $\theta^{\prime}=\alpha_{2}$,

$a(1+\theta-\frac{s}{r}\theta)+b(s-r)\theta(1+\theta)=a(1+\theta^{\prime}-\frac{s}{r}\theta^{\prime})+(\frac{1}{s}-\frac{1}{r})c$
(10.10)

for all $r,s\in\mathbf{F}_{p}^{*}$. By Lemma 10.2 (and since $\theta\neq
0,-1$) we have $b=c=0$, thus $a\neq 0$ and $\theta=\theta^{\prime}$. That is,
$\sigma(\alpha_{1})=\alpha_{2}$, and so $\alpha_{1},\alpha_{2}$ have the same
minimal polynomial over $\mathbf{F}_{p}$. This is a contradiction, since we
assumed that the minimal polynomials of $\alpha_{1},\alpha_{2}$ over
$\mathbf{F}_{p}$ (the $f_{i}$, that is to say the factors of the minimal
polynomial of $\Delta$ over $\mathbf{F}_{p}$) are coprime.

_Case 3._ $N_{i}=N_{j}=\\{\pm I\\}$. We reprise the argument from Case 2, only
now we must allow a sign error. Included in the classification of
automorphisms of $\operatorname{PSL}_{2}(\mathbf{F}_{p}(\alpha_{1}))$ (see
Appendix A) is the fact that such automorphisms lift to automorphisms of
$\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{1}))$. Therefore
$H=\\{(x,\varepsilon(x)\phi(x)):x\in\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{1})),\varepsilon(x)\in\pm
I\\}$ for some isomorphism
$\phi:\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{1}))\rightarrow\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{2}))$.
We may now proceed as before but with an additional sign error, thus (10.9)
becomes

$PM_{\sigma(\alpha_{1})}(r,s)=\varepsilon_{r,s}M_{\alpha_{2}}(r,s)P$ (10.11)

for all $r,s\in\mathbf{F}_{p}^{*}$ and for some choice of signs
$\varepsilon_{r,s}\in\\{\pm 1\\}$. If $\varepsilon_{r,s}=1$ for at least half
of all pairs $(r,s)\in\mathbf{F}_{p}^{*}\times\mathbf{F}_{p}^{*}$ then we are
done, exactly as before (taking $\eta=\frac{1}{4}$ in Lemma 10.2). If
$\varepsilon_{r,s}=-1$ for at least half of all pairs
$(r,s)\in\mathbf{F}_{p}^{*}\times\mathbf{F}_{p}^{*}$ then (10.10) is modified
to

$a(1+\theta-\frac{s}{r}\theta)+b(s-r)\theta(1+\theta)=-a(1+\theta^{\prime}-\frac{s}{r}\theta^{\prime})-(\frac{1}{s}-\frac{1}{r})c,$
(10.12)

for half of all pairs $(r,s)\in\mathbf{F}_{p}^{*}\times\mathbf{F}_{p}^{*}$.
From this we conclude that $b=c=0$, hence $a\neq 0$ and so both
$\theta=-\theta^{\prime}$ and $1+\theta=-(1+\theta^{\prime})$. This is
impossible.

Since only Case 1 in the above analysis did not lead to a contradiction (and
since we can replace $\\{1,2\\}$ by any pair $\\{i,j\\}$), we have now shown
that the projection of $\Phi(\Gamma)$ to the product
$\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{i}))\times\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{j}))$
of any pair of factors is surjective. Proposition 10.1 now follows from Lemma
10.4, together with the fact (see Appendix A) that all the factors
$\operatorname{SL}_{2}(\mathbf{F}_{p}(\alpha_{i}))$ are perfect. ∎

##  11\. Quasirandomness of $\rho|_{\Gamma_{q}}$.

We turn now to the proof of Proposition 8.5 itself. Let us begin by recalling
the statement.

######  Proposition 11.1 (Proposition 8.5).

Let $Q$ be a generic quadratic form. Then there is some $p_{0}(Q)$ such that
the following is true. Let $\rho$ be the Weil representation on
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$ (as given in Proposition 7.2).
Suppose that $q$ is squarefree and has all prime factors greater than
$p_{0}(Q)$. Then $\rho|_{\Gamma_{q}}$ splits into irreducible
subrepresentations of dimensions $\geqslant q^{1-o(1)}$.

We begin by reducing to the prime case. Write $q=p_{1}\cdots p_{n}$. The
representation $\rho$ is constructed in Appendix B as a tensor product
$\otimes_{i=1}^{n}\tilde{\rho}_{i}$, where
$\tilde{\rho}_{i}:\operatorname{Sp}_{8}(\mathbf{Z}/p_{i}\mathbf{Z})\rightarrow\operatorname{U}(\ell^{2}((\mathbf{Z}/p_{i}\mathbf{Z})^{4})$
is a twisted version of the mod $p_{i}$ Weil representation $\rho_{i}$, given
by $\tilde{\rho}_{i}(g)=\rho_{i}(g^{\sigma_{i}})$ where
$\big{(}\begin{smallmatrix}M_{1}&M_{2}\\\
M_{3}&M_{4}\end{smallmatrix}\big{)}^{\sigma_{i}}=\big{(}\begin{smallmatrix}M_{1}&\lambda_{i}M_{2}\\\
\lambda_{i}^{-1}M_{3}&M_{4}\end{smallmatrix}\big{)}$ and
$\lambda_{i}=\prod_{j\neq i}p_{i}$. This tensor product may be realised on
$\ell^{2}((\mathbf{Z}/q\mathbf{Z})^{4})$ by
$\rho(g)f=\prod_{i=1}^{n}\tilde{\rho}_{i}(g)f_{i}(x)$ for “pure tensors”
$f(x)=\prod_{i=1}^{n}f_{i}(x)$, where $f_{i}$ factors through the projection
$\pi_{i}:(\mathbf{Z}/q\mathbf{Z})^{4}\rightarrow(\mathbf{Z}/p_{i}\mathbf{Z})^{4}$
(see Appendix B for a discussion of the notation here). Consider the
restriction to $\Gamma_{q}\cong\prod_{i=1}^{n}\Gamma_{p_{i}}$. The
decomposition of $\rho|_{\Gamma_{q}}$ into irreducibles is then given by
decomposing each $\ell^{2}((\mathbf{Z}/p_{i}\mathbf{Z})^{4})$ into irreducible
$\tilde{\rho}_{i}|_{\Gamma_{p_{i}}}$-invariant subspaces $V_{i}$ and taking
tensor products. (Here we use the fact that if $V_{i}$ is an irreducible
$G_{i}$ representation then $\bigotimes_{i=1}^{n}V_{i}$ is an irreducible
$\times_{i=1}^{n}G_{i}$-representation, which is a standard fact of
representation theory. See for instance [10, Theorem 19.18].)

Now each $\Gamma_{p_{i}}$ is a direct product of groups
$\operatorname{SL}_{2}(\mathbf{F}_{p_{i}^{j}})$, and therefore by Appendix A
any irreducible representation of $\Gamma_{p_{i}}$ is either trivial, or has
dimension at least $\frac{1}{2}(p_{i}-1)$. That is, if $\dim V_{i}\neq 1$ then
$\dim V_{i}\geqslant\frac{1}{2}(p_{i}-1)$. Consequently, if for all $i$ the
representation $\tilde{\rho}_{i}|_{\Gamma_{p_{i}}}$ has no invariant vector
(that is, 1-dimensional invariant subspace) then
$\dim\rho\geqslant\prod_{p|q}\frac{1}{2}(p-1)=q^{1-o(1)}$, as desired.

This reduces the task of proving Proposition 8.5 to the following, which is
the final task for the main part of the paper.

######  Proposition 11.2.

Suppose that $Q$ is generic and that $p$ is sufficiently large in terms of
$Q$. Let
$\rho:\operatorname{Sp}_{8}(\mathbf{Z}/p\mathbf{Z})\rightarrow\operatorname{U}(\ell^{2}((\mathbf{Z}/p\mathbf{Z})^{4}))$
be the mod $p$ Weil representation. Let $r\in(\mathbf{Z}/p\mathbf{Z})^{*}$,
and let $\tilde{\rho}$ be the twist of $\rho$ by dilation by $r$, that is to
say $\tilde{\rho}(x)=\rho(x^{(r)})$. Then $\tilde{\rho}|_{\Gamma_{p}}$ has no
nontrivial invariant vector.

###### Proof.

We will show that the conclusion holds under the assumption that $\Delta$ has
distinct eigenvalues and is invertible over $\mathbf{F}_{p}$. This includes
all sufficiently large primes $p$. Indeed $Q$ is generic, so by definition
$\Delta$ has distinct eigenvalues and is invertible over $\mathbf{Q}$.
Therefore the same is true over $\mathbf{F}_{p}$, $p$ sufficiently large, for
the reasons detailed at the start of the proof of Proposition 10.1.

Suppose from now on that $\Delta$ has distinct eigenvalues and is invertible
over $\mathbf{F}_{p}$. Since $\Gamma_{p}$ is invariant under the dilation
$\gamma\mapsto\gamma^{(r)}$, it suffices to consider the case
$\tilde{\rho}=\rho$. Suppose, then, that

$\rho(\gamma)f=f$ (11.1)

for all $\gamma\in\Gamma_{p}$. Our aim is to show that $f$ is identically
zero. To examine the condition (11.1) we will look at the following particular
elements $\gamma$, where $B,C\in\operatorname{Mat}_{4}(\mathbf{F}_{p})$ are
the specific symmetric matrices described in (10.3):

  1. (1)

the upper triangular elements $u(MB)=\big{(}\begin{smallmatrix}I&MB\\\
0&I\end{smallmatrix}\big{)}$, where $M\in\mathbf{F}_{p}[\Delta]$;

  2. (2)

the lower triangular elements $l(CM)=\big{(}\begin{smallmatrix}I&0\\\
CM&I\end{smallmatrix}\big{)}$, where $M\in\mathbf{F}_{p}[\Delta]$;

  3. (3)

the diagonal elements $s(\lambda I)=\big{(}\begin{smallmatrix}\lambda I&0\\\
0&\lambda^{-1}I\end{smallmatrix}\big{)}$, $\lambda\in\mathbf{F}_{p}^{*}$.

Now that we know from Proposition 10.1 that
$\Gamma_{p}=\tau^{-1}\operatorname{SL}_{2}(\mathbf{F}_{p}[\Delta])\tau$ (where
$\tau$ is defined in (10.5)), so one may easily check using (10.6) that all of
these elements do lie in $\Gamma_{p}$.

Now we already have formulae for the actions of the elements in (2) and (3),
directly from Proposition 7.2. Namely,

$\rho(l(CM))f(x)=\xi_{M}e_{p}(-\frac{1}{2}x^{T}CMx)f(x)$ (11.2)

for some unit complex number $\xi_{M}=\xi(l(CM))$ (we do not care exactly what
this is) and

$\rho(s(\lambda I))f(x)=\xi^{\prime}_{\lambda}f(\lambda^{-1}x)$ (11.3)

for some unit complex number $\xi^{\prime}_{\lambda}$. If (11.1) holds, it
follows from (11.2) that for each $M$

$\operatorname{Supp}(f)\subset\\{x:x^{T}CMx=t_{M}\\}$

for some parameters $t_{M}$ (in fact satisfying
$e_{p}(-\frac{1}{2}t_{M})=\xi_{M}$). Now (11.3) tells us that
$\operatorname{Supp}(f(x))=\operatorname{Supp}(f(\lambda^{-1}x))$ for all
$\lambda\in\mathbf{F}_{p}^{*}$, therefore (taking $\lambda\neq\pm 1$) we see
that in fact all the $t_{M}$ must be zero, that is

$\operatorname{Supp}(f)\subset\\{x:x^{T}CMx=0\\}\quad\mbox{for all
$M\in\mathbf{F}_{p}[\Delta]$}.$ (11.4)

To get a formula for the upper triangular action, we note the identity

$u(W)=J^{-1}l(-W)J$

for any matrix $W$, where as usual $J=\big{(}\begin{smallmatrix}0&I\\\
-I&0\end{smallmatrix}\big{)}$. Thus if $f$ is invariant under $\rho(u(MB))$
then $\rho(J)f$ is invariant under $\rho(l(-MB))$. As above, this implies that
$\operatorname{Supp}(\rho(J)f)\subset\\{x:x^{T}MBx=t^{\prime}_{M}\\}$. Since
$J^{-1}s(\lambda I)J=s(\lambda^{-1}I)$, we see that $\rho(J)f$ is invariant
(up to multiplication by a scalar) by dilation $x\mapsto\lambda^{-1}x$.
Therefore all of the $t^{\prime}_{M}$ are in fact zero. Moreover, from
Proposition 7.2 we know that $\rho(J)f$ is the (normalised) Fourier transform
of $f$ times a scalar, and so we come to the conclusion that

$\operatorname{Supp}(\hat{f})\subset\\{x:x^{T}MBx=0\\}\quad\mbox{for all
$M\in\mathbf{F}_{p}[\Delta]$}.$ (11.5)

Conditions (11.4) and (11.5) seem highly incompatible and should, for example,
violate the uncertainty principle under reasonable assumptions. However, a
proof seems not to be completely straightforward (and indeed the statement
fails in sufficiently degenerate situations, for example if we were to allow
$\Delta=0$).

Let us begin the argument. Consider the bilinear form $\phi(x,y)=x^{T}Cy$ on
$\overline{\mathbf{F}}_{p}^{4}\times\overline{\mathbf{F}}_{p}^{4}$. Since
$BC=\Delta$ and $\Delta$ is invertible over $\mathbf{F}_{p}$, $C$ is
invertible over $\mathbf{F}_{p}$ and so this is a non-degenerate form. We note
that $\Delta$ is self-adjoint with respect to $\phi$:

$\phi(x,\Delta y)=x^{T}CBCy,$ $\phi(\Delta x,y)=x^{T}\Delta^{T}Cy=x^{T}CBCy$

(note that $B,C$ are both symmetric). By the usual argument, this means that
eigenvectors of $\Delta$ with distinct eigenvalues are orthogonal with respect
to $\phi$. Indeed, if $\Delta v_{1}=\lambda_{1}v_{1}$ and $\Delta
v_{2}=\lambda_{2}v_{2}$ then

$\lambda_{2}\phi(v_{1},v_{2})=\phi(v_{1},\Delta v_{2})=\phi(\Delta
v_{1},v_{2})=\lambda_{1}\phi(v_{1},v_{2}).$

Now since $\Delta$ has distinct eigenvalues, it is diagonalisable over
$\overline{\mathbf{F}}_{p}$. Suppose that $v_{i}$ are eigenvectors with
(distinct) eigenvalues $\lambda_{i}$, $i=1,2,3,4$. These are a basis for
$\overline{\mathbf{F}}_{p}^{4}$. Write $x\in\overline{\mathbf{F}}_{p}^{4}$ as
$\sum_{i=1}^{4}x_{i}v_{i}$. Then

$x^{T}C\Delta^{j}x=\phi(x,\Delta^{j}x)=\sum_{i=1}^{4}x_{i}^{2}\lambda_{i}^{j}\phi(v_{i},v_{i}).$
(11.6)

Note that $\phi(v_{i},v_{i})\neq 0$ (if it was, $v_{i}$ would be orthogonal
with respect to $\phi$ to all of $v_{1},v_{2},v_{3},v_{4}$ and hence to all of
$\overline{\mathbf{F}}_{p}^{4}$, contrary to the fact that $\phi$ is non-
degenerate).

Therefore the matrix with $(i,j)$-entry $\lambda_{i}^{j}\phi(v_{i},v_{i})$
($1\leqslant i\leqslant 4,0\leqslant j\leqslant 3$) is non-singular, its
determinant being a non-zero multiple $\prod_{i=1}^{4}\phi(v_{i},v_{i})$ of a
certain Vandermonde determinant. It follows from (11.6) that if
$x^{T}C\Delta^{j}x=0$ for $j=0,1,2,3$ then $x=0$, and so any $f$ satisfying
(11.4) is supported only at zero.

Noting that $x^{T}\Delta^{j+1}Bx=(Bx)^{T}C\Delta^{j}(Bx)$, we can also
conclude that if $x^{T}\Delta^{j+1}Bx=0$ for $j=0,1,2,3$ then $x=0$, and so
any $f$ satisfying (11.5) has $\hat{f}$ supported only at zero.

These two facts about $f$ are completely incompatible, unless $f$ is
identically zero: if $f$ is supported at zero, $\hat{f}$ is in fact constant
on $(\mathbf{Z}/p\mathbf{Z})^{4}$.

This completes the proof of Proposition 11.2, and hence that of Proposition
8.5. ∎

All of the main results in the paper are now established.

_Remark._ It is in fact possible to show (under the assumptions on $\Delta$ in
force throughout this section) that $\rho|_{\Gamma_{p}}$ is isomorphic to a
Weil representation of $\operatorname{SL}_{2}(\mathbf{F}_{p}[\Delta])$ on
$\ell^{2}(\mathbf{F}_{p}[\Delta])$, by giving an explicit intertwining map.
This fact can be used to give an alternative proof of Proposition 11.2 which,
while more natural than the _ad hoc_ arguments presented here, requires quite
a bit more setting up. We intend to give a full account in future work.

##  Appendix A Facts about $\operatorname{SL}_{2}(k)$

We collect various well-known facts about $\operatorname{SL}_{2}(k)$, $k$ a
finite field, which we used in the main text. For our purposes, “rough”
versions of these facts (passing to subgroups of index $O(1)$, etc) would be
quite sufficient but we use the precise versions when sufficiently clean
results are relatively easily-available.

######  Proposition A.1.

Let $k$ be a finite field of odd characteristic. Then

  1. (1)

The smallest nontrivial complex representation of $\operatorname{PSL}_{2}(k)$
has dimension at least $\frac{1}{2}(|k|-1)$.

  2. (2)

If $k$ has order at least $5$ then $\operatorname{SL}_{2}(k)$ is perfect.

  3. (3)

Any proper subgroup of $\operatorname{SL}_{2}(k)$ has a subgroup of index at
most $C$ which is conjugate to a subgroup of one of (i) the group of upper
triangular matrices or (ii) $\operatorname{SL}_{2}(k_{0})$ for some proper
subfield $k_{0}<k$.

  4. (4)

Every automorphism of $\operatorname{PSL}_{2}(k)$ or
$\operatorname{SL}_{2}(k)$ is a composition of a conjugation by elements of
$\operatorname{GL}_{2}(k)$, and a power of the Frobenius automorphism of $k$.

###### Proof.

(1) goes back well over a century, to Jordan and Schur. For a nice and easy-
to-access discussion, see Prasad’s notes [17].

(2) See Lang [11, Chapter XIII, Theorem 8.3].

(3) The rough statement given here, which suffices for our purposes is [26,
Theorem 5.2.7]. As one would expect, a detailed classification of maximal
subgroups of $\operatorname{SL}_{2}(k)$ has been known for more than a
century. It is somewhat complicated; the details, as well as references to the
original papers, may be found in [13].

(4) This is certainly well-known. A standard reference is [21, 3.2]. See also
[22]. The MathOverflow post [30] is helpful in navigating these papers.

∎

##  Appendix B Weil representation of
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$

In the section we construct the representation of
$\rho:\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})\rightarrow\operatorname{U}(\ell^{2}((\mathbf{Z}/q\mathbf{Z})^{4})$
which we have been using throughout the paper, and whose properties are
detailed in Proposition 7.2. Suppose that $q=p_{1}\cdots p_{n}$ is squarefree
and odd. We assume the existence, for each $i$, of the Weil representations
$\rho_{i}:\operatorname{Sp}_{8}(\mathbf{Z}/p_{i}\mathbf{Z})\rightarrow\operatorname{U}(\ell^{2}((\mathbf{Z}/p_{i}\mathbf{Z})^{4}))$,
the construction of which is given in detail in [16] and shown to satisfy the
properties of Proposition 7.2 (and in fact that paper gives details of the
multiplier $\xi$, whose precise properties are unimportant in this paper).
Since
$\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})\cong\prod_{i=1}^{n}\operatorname{Sp}_{8}(\mathbf{Z}/p_{i}\mathbf{Z})$,
one thinks of looking at the (exterior) tensor product
$\rho=\otimes_{i=1}^{n}\rho_{i}$. However, this turns out to need a small
modification.

For each $i$, denote by
$\sigma_{i}:\operatorname{Sp}_{8}(\mathbf{Z}/p_{i}\mathbf{Z})\rightarrow\operatorname{Sp}_{8}(\mathbf{Z}/p_{i}\mathbf{Z})$
the automorphism defined by $\big{(}\begin{smallmatrix}M_{1}&M_{2}\\\
M_{3}&M_{4}\end{smallmatrix}\big{)}^{\sigma_{i}}:=\big{(}\begin{smallmatrix}M_{1}&\lambda_{i}M_{2}\\\
\lambda_{i}^{-1}M_{3}&M_{4}\end{smallmatrix}\big{)}$, where
$\lambda_{i}:=\prod_{i\neq j}p_{i}$. Note that this is, in fact, one of the
dilates we considered earlier (see (7.3)), but with parameter
$r=\lambda_{i}^{-1}$. It may not be an inner automorphism (this depends on
whether or not $\lambda_{i}$ is a square mod $p_{i}$). Each $\rho_{i}$ may be
twisted by $\sigma_{i}$ to give a representation
$\tilde{\rho}_{i}:\operatorname{Sp}_{8}(\mathbf{Z}/p_{i}\mathbf{Z})\rightarrow\operatorname{U}(\ell^{2}((\mathbf{Z}/p_{i}\mathbf{Z})^{4}))$,
defined by $\tilde{\rho}_{i}(g)f:=\rho(g^{\sigma_{i}})f$. This will be
isomorphic to $\rho_{i}$ if $\lambda_{i}$ is a square mod $p_{i}$, but not
otherwise. However, we will not need this last fact.

We now construct $\rho$ as the tensor product
$\bigotimes_{i=1}^{n}\tilde{\rho}_{i}$, which we shall shortly show how to
realise concretely in $\ell^{2}((\mathbf{Z}/q\mathbf{Z})^{4})$. From here on,
we will abuse notation by omitting explicit notation for projection maps from
$(\mathbf{Z}/q\mathbf{Z})^{4}\rightarrow(\mathbf{Z}/p_{i}\mathbf{Z})^{4}$,
from $\operatorname{Sp}_{8}(\mathbf{Z}/q\mathbf{Z})$ to
$\operatorname{Sp}_{8}(\mathbf{Z}/p_{i}\mathbf{Z})$, or from
$\operatorname{Mat}_{4}(\mathbf{Z}/q\mathbf{Z})$ to
$\operatorname{Mat}_{4}(\mathbf{Z}/p_{i}\mathbf{Z})$ when the domain is clear
from context. Thus, for example, for functions
$f_{i}\in\ell^{2}((\mathbf{Z}/p_{i}\mathbf{Z})^{4})$ we define the “pure
tensor” $f(x):=\prod_{i=1}^{n}f_{i}(x)$, but it would be more correct, though
cumbersome, to write $\prod_{i=1}^{n}f_{i}(\pi_{i}(x))$ where
$\pi_{i}:(\mathbf{Z}/q\mathbf{Z})^{4}\rightarrow(\mathbf{Z}/p_{i}\mathbf{Z})^{4}$
is the natural projection.

For a pure tensor $f$ as above define

$\rho(g)f(x)=\prod_{i=1}^{n}\tilde{\rho}_{i}(g)f_{i}(x)=\prod_{i=1}^{n}\rho_{i}(g^{\sigma_{i}})f_{i}(x).$
(B.1)

This is well-defined by the universal property of tensor products.

We now turn to the verification of the properties stated in Proposition 7.2.
Recall that the properties to be established are as follows (for some unit
complex numbers $\xi(\cdot)$):

$\rho(s(E))f(x)=\xi(s(E))f(E^{-1}x)$ (B.2)

for $s(E):=\big{(}\begin{smallmatrix}E&0\\\ 0&E^{-T}\end{smallmatrix}\big{)}$
with $E$ invertible;

$\rho(J)f(x)=\xi(J)q^{2}\mathbb{E}_{y\in(\mathbf{Z}/q\mathbf{Z})^{4}}f(y)e(x^{T}y);$
(B.3)

$\rho(l(W))f(x)=\xi(l(W))e_{q}(-\frac{1}{2}x^{T}Wx)f(x)$ (B.4)

for $l(W):=\big{(}\begin{smallmatrix}I&0\\\ W&I\end{smallmatrix}\big{)}$ with
$W$ symmetric.

We will also, of course, be using the corresponding properties for the prime
case $\rho_{i}$. To avoid confusion, we write $\xi_{i}(\cdot)$ for the
corresponding unit complex numbers. It is enough to check (B.2), (B.3) and
(B.4) for pure tensors $f$.

_Proof of ( B.2)._ We have

$\displaystyle\rho(s(E))f(x)$
$\displaystyle=\prod_{i=1}^{n}\rho_{i}(s(E)^{\sigma_{i}})f_{i}(x)=\prod_{i=1}^{n}\rho_{i}(s(E))f_{i}(x)$
$\displaystyle=\prod_{i=1}^{n}\xi_{i}(s(E))f_{i}(E^{-1}x)=\xi(s(E))f(E^{-1}x),$

where $\xi(s(E)):=\prod_{i=1}^{n}\xi_{i}(s(E))$. This establishes (B.2).

For the remaining two parts, we will need the relation

$e_{q}(\lambda t)=\prod_{i=1}^{n}e_{p}(t),$ (B.5)

where $\lambda=\sum_{i=1}^{n}\lambda_{i}$.

_Proof of ( B.3)._ We have

$\displaystyle\rho(J)f(x)$
$\displaystyle=\prod_{i=1}^{n}\rho_{i}(J^{\sigma_{i}})f_{i}(x)=\prod_{i=1}^{n}\rho_{i}(s(\lambda_{i}I)J)f_{i}(x)$
$\displaystyle=q^{2}\xi(J)\prod_{i=1}^{n}\mathbb{E}_{y_{i}\in(\mathbf{Z}/p_{i}\mathbf{Z})^{4}}f_{i}(y_{i})e_{p_{i}}(\lambda_{i}^{-1}x^{T}y_{i}),$

where $\xi(J):=\prod_{i=1}^{n}\xi_{i}(s(\lambda_{i}I))\xi_{i}(J)$. Applying
(B.5) gives

$\prod_{i=1}^{n}e_{p_{i}}(\lambda_{i}^{-1}x^{T}y_{i})=e_{q}(x^{T}y),$

where $y\in(\mathbf{Z}/q\mathbf{Z})^{4}$ reduces to $y_{i}$ in
$(\mathbf{Z}/p_{i}\mathbf{Z})^{4}$, and the claim follows.

_Proof of ( B.4)._ We have

$\displaystyle\rho(l(W))f(x)$
$\displaystyle=\prod_{i=1}^{n}\rho_{i}(l(W)^{\sigma_{i}})f_{i}(x)=\prod_{i=1}^{n}\rho_{i}(l(\lambda_{i}^{-1}W))f_{i}(x)$
$\displaystyle=\xi(l(W))\prod_{i=1}^{n}e_{p_{i}}(-\frac{1}{2}x^{T}\lambda_{i}^{-1}Wx)f_{i}(x),$

where $\xi(l(W)):=\prod_{i=1}^{n}\xi_{i}(l(\lambda_{i}^{-1}W))$. The claim
again follows using (B.5).

##  Appendix C Almost-invariant measures

We invite the reader to recall the notation concerning probability measures on
finite groups $G$ as described at the start of Section 8. In particular,
$\|\cdot\|$ means the $2$-norm with respect to the counting measure on $G$,
$\|\mu\|=\big{(}\sum_{x\in G}\mu(x)^{2}\big{)}^{1/2}$. We extend the notation
$\mu_{H}$ for the uniform measure on a subgroup $H$ to arbitrary sets: thus if
$A\subset G$ is a finite set then we write $\mu_{A}$ for the uniform measure
on $A$, that is to say the measure which puts weight $|A|^{-1}$ on each point
of $A$.

We have the following instance of Young’s inequality.

######  Lemma C.1 (Young’s inequality).

Let $\mu,\nu$ be two probability measures on $G$. Then we have
$\|\mu\ast\nu\|\leqslant\|\mu\|$.

###### Proof.

We have

$\mu\ast\nu(x)=\sum_{y\in
G}\big{(}\mu(y)^{2}\nu(y^{-1}x)\big{)}^{1/2}\nu(y^{-1}x)^{1/2}.$

By Cauchy-Schwarz and the fact that $\nu$ is a probability measure,

$\mu\ast\nu(x)^{2}\leqslant\sum_{y\in G}\mu(y)^{2}\nu(y^{-1}x).$

Finally, summing over $x\in G$ gives the result. ∎

In the main text we required a statement about almost equality here in the
case that $\mu=\nu$ and both are symmetric. The actual statement we quoted in
the main text is Corollary C.3 below, but the heart of it is Lemma C.2. This
result should be thought of as “well-known”, but it is hard to give a precise