2. Background on a pair of quadrics

Exponential sums for a pair of quadrics will feature prominently in this work. Let $Q_1(\underline {x}), Q_2(\underline {x})$ be a pair of quadratic forms in $n$ variables with integer coefficients and consider the variety defined by

\[ V: Q_1(\underline{x})=Q_2(\underline{x})=0, \]

$\underline {x}\in \overline {\mathbb {Q}}^n$. Let $\mathrm {Sing}_K(V)$ to be the (projective) singular locus of $V$ over field $K$. When $Q_1$ and $Q_2$ intersect properly, namely, if $V$ is of projective dimension $n-3$, then we can express the singular locus of $V$ as follows:

(2.1)\begin{equation} \mathrm{Sing}_K(V):=\bigg\{\underline{x}\in \mathbb{P}^{n-1}_{\overline{K}} \,\bigg{|}\, \underline{x}\in V, \ \mathrm{Rank} \begin{pmatrix} \nabla Q_1(\underline{x})\\ \nabla Q_2(\underline{x}) \end{pmatrix} <2 \bigg\}. \end{equation}

We say that the intersection variety of $Q_1(\underline {x})$ and $Q_2(\underline {x})$, $V$, is non-singular if $\dim \mathrm {Sing}_K (V)=-1$ and singular otherwise. It should be noted that (2.1) only truly encapsulates the set of singular points when $Q_1,Q_2$ have a proper intersection over $K$ (that is, the forms $Q_1(\underline {x})$, $Q_2(\underline {x})$ share no common factor over K). However, $\mathrm {Sing}_K(V)$ is still a well-defined set with a well-defined dimension, even when $Q_1$ and $Q_2$ intersect improperly, and so we will also use this definition in this case.

We will now begin by noting a slight generalisation of [Reference Marmon and VisheMV19, Lemma 4.1] in the context of two quadrics, which will be vital in various stages of this paper.

Lemma 2.1 Let $Q_1,Q_2$ be a pair of quadratic forms defining a complete intersection $X=V(Q_1,Q_2)$. Let $\Pi$ be a collection of primes such that $\#\Pi =r\geq 0$ and define $\Pi _a:=\{p\in \Pi \,|\, p>a\}$ for every $a\in \mathbb {N}$. Then there exists a constant $c'=c'(n)$ and a set of primitive linearly independent vectors

\[ \underline{e}_1,\ldots,\underline{e}_{n}\in\mathbb{Z}^n \]

satisfying the following property for any integer $0\leq \eta \leq n-1$, any subset $\emptyset \neq I\subset \{1,2\}$ and any $\upsilon \in \{\infty \}\cup \Pi _{2c'}$: The subspace $\Lambda _\eta \subset \mathbb {P}^{n-1}_{\mathbb {F}_\upsilon }$ spanned by the images of $\underline {e}_1,\ldots,\underline {e}_{n-\eta }$ is such that

(2.2)\begin{equation} \dim(X_I\cap\Lambda_\eta)_\upsilon=\max\{-1,\dim(X_I)_\upsilon-\eta\} \end{equation}

and

(2.3)\begin{equation} \dim \mathrm{Sing}((X_I\cap \Lambda_\eta)_\upsilon)=\max\{-1,\dim \mathrm{Sing}((X_I)_\upsilon)-\eta\}. \end{equation}

Here given $\emptyset \neq I\subseteq \{1,2\}$, let $X_I$ denote the complete intersection variety defined by the forms $\{F_i:i\in I\}$. Moreover, the basis vectors $\underline {e}_i$ can be chosen so that

(2.4)\begin{equation} L/2\leq |\underline{e}_i|\leq L \end{equation}

for every $i=1,\ldots, n$ and

(2.5)\begin{equation} L^n\ll \det(\underline{e}_1,\ldots,\underline{e}_n)\ll L^n \end{equation}

for some constant $L=O_n(r+1).$

Proof. Note that the statement of this lemma is identical to that of [Reference Marmon and VisheMV19, Lemma 4.1] except that in the latter there is an additional assumption that the closed subscheme $X_I\subset \mathbb {P}^{n-1}_{\mathbb {Z}}$ defined by $F_i=0$ for all $i\in I$ satisfies

(2.6)\begin{equation} \dim(X_I)_\upsilon=n-1-|I|. \end{equation}

This is equivalent to the case when $X_1$ and $X_2$ intersect properly. Therefore, it is enough to consider different cases where we have an improper intersection. In each of these particular cases, somewhat softer argument works.

In the trivial case when $Q_1=Q_2=0$, the singular locus would be of dimension $n-1$ and, therefore, any basis $\underline {e}_1,\ldots,\underline {e}_n$ will work.

When $Q_2=\lambda Q_1$, where $\lambda \in K$ and $Q_1$ a non-zero quadratic form, our singular locus would consist of the hypersurface $Q_1=0$ of dimension $n-2$. Here, we may apply [Reference Marmon and VisheMV19, Lemma 4.1] only to the hypersurface $X_1$ to find a basis $\underline {e}_1,\ldots,\underline {e}_n$ which is chosen such that (2.2) and (2.3) hold for $I=\{1\}$. This choice will clearly work for all $I\subset \{1,2\}$.

In the remaining case when $Q_1=L_1L_2,Q_2=L_1L_3$, where $L_i=\underline {v}_i\cdot \underline {x}$ and $L_2$ is not a scalar multiple of $L_3$. In this case, it is easy to check that the singular locus of $X_1\cap X_2$ to is the hyperplane $L_1=0$ (of dimension $n-2$). Here, we may apply [Reference Marmon and VisheMV19, Lemma 4.1] to the single variety defined by the cubic form $L_1L_2L_3=0$. The basis $\Lambda$ that we get from this process will work here as well.

Since $Q_1$ and $Q_2$ are quadratic forms, we may also define $M_1$, $M_2$ to be their respective associated coefficient matrices defined as follows: if

\[ Q_i(\underline{x}):=\sum_{j=1}^n\sum_{k=j}^n b_{j,k}^{(i)} x_j x_k, \]

then

(2.7) \begin{equation} (M_i)_{j,k}:= \begin{cases} b_{j,k}^{(i)} & \text{ if } j=k\\ \frac{1}{2}b_{j,k}^{(i)} & \text{ if } j< k \\ \frac{1}{2}b_{k,j}^{(i)} & \text{ if } j> k. \end{cases} \end{equation}

We clearly have that $M_1,M_2\in M_n(\mathbb {Z}/2)$, the set of $n\times n$ matrices with coefficients of the form $a/2$, $a\in \mathbb {Z}$, since $b_{k,j}\in \mathbb {Z}$. For the rest of this section (and, in fact, the rest of the paper by Remark 3.1), we will assume without loss of generality that $M_1,M_2\in M_n(\mathbb {Z})$. This is because even if $M_1,M_2\not \in M_n(\mathbb {Z})$, we certainly have $2M_1, 2M_2\in M_n(\mathbb {Z})$, and so we may work with $2Q_1$, $2Q_2$ and relabel instead.

We are now ready to prove the following generalisation of [Reference Heath-Brown and PierceHP17, Proposition 2.1]. This will be particularly helpful for us when we are working with exponential sums of the form

\[ \sideset{}{^*}\sum_{{\underline{a}}}^q \sum_{\underline{x}}^q e_{q}(a_1Q_1(\underline{x})+a_2Q_2(\underline{x})+\underline{c}\cdot\underline{x}), \]

where $q$ is square-full, in § 5.3. Here, as is standard, the $*$ next to the sum denotes that the sum is over $({\underline {a}},q)=1$, and $e_q(x):=\exp (2\pi i x/q)$.

Proposition 2.2 Let $\nu$ either denote a finite prime $\nu \gg _n 1$ or the infinite prime, let $\mathbb {F}_\nu$ either denote the corresponding finite field or $\mathbb {Q}$, and let

(2.8)\begin{equation} s_{\nu}(Q_1,Q_2):=\dim\mathrm{Sing}_{\mathbb{F}_\nu}(V), \end{equation}

where $V$ is defined as above. Let ${\underline {a}}\in \mathbb {F}_\nu ^2\backslash (0,0)$ and $a_1M_1+a_2M_2$ be the matrix associated to the quadratic form $a_1Q_1+a_2Q_2$. Then

(2.9)\begin{equation} \mathrm{Rank}(a_1M_1+a_2M_2)\geq n-s_{\nu}(Q_1,Q_2)-2 \end{equation}

for any such ${\underline {a}}$. Furthermore, there exists a set $\Gamma =\{\gamma _1,\ldots,\gamma _k\}\subset \overline {\mathbb {F}}_\nu$ such that $1\leq i\leq k\leq n$ and

(2.10)\begin{equation} \mathrm{Rank}(a_1M_1+a_2M_2)\geq n-s_{\nu}(Q_1,Q_2)-1, \end{equation}

unless $a_2=0$ or $a_1= \gamma a_2$ for some $\gamma \in \Gamma$.

Proof. Let $M_1$ and $M_2$ denote the integer matrices defining the forms $Q_1$ and $Q_2$, respectively. We first note that for $s_{\nu }(Q_1,Q_2)=-1$, we recover (2.9) from [Reference Heath-Brown and PierceHP17, Proposition 2.1]. In this case, may also use [Reference Heath-Brown and PierceHP17, Proposition 2.1] to simultaneously diagonalise $M_1$, $M_2$, letting us instead work with

\[ Q_i'(\underline{x}):=\sum_{j=1}^n\lambda_{i,j}x_j^2, \quad M_i':= \mathrm{Diag}(\underline{\lambda_{i}}). \]

In particular, we have

\[ s_{\nu}(Q_1',Q_2')=s_{\nu}(Q_1,Q_2)=-1, \quad \mathrm{Rank}(a_1M_1'+a_2M_2')=\mathrm{Rank}(a_1M_1+a_2M_2), \]

for every ${\underline {a}}\in \mathbb {F}^2_{\nu }\backslash (0,0)$. Next, we note that $\mathrm {Rank}(a_1M_1'+a_2M_2')< n$ if and only if there is some $j\in \{1,\ldots, n\}$ such that $a_1\lambda _{1,j}+a_2\lambda _{2,j}=0$, which imposes the desired restriction on $(a_1,a_2)$ provided that $(\lambda _{1,j},\lambda _{2,j})\neq (0,0)$. However, if $(\lambda _{1,j},\lambda _{2,j})=(0,0)$, then it is easy to see from the definition of $Q_i'(\underline {x})$ that

\[ \nabla Q_1'(m\underline{e}_j)=\nabla Q_2'(m\underline{e}_j)=\underline{0} \]

for every $m\in \overline {\mathbb {F}}_{\nu }$ (provided $\nu > 2$), where $\underline {e}_j$ is the $j$th vector in the standard basis. This implies that $m\underline {e}_j\in \mathrm {Sing}(Q_1',Q_2')$, and so $s_{\nu }(Q_1',Q_2')\geq 0$, giving us a contradiction.

If $s_{\nu }(Q_1,Q_2)\neq -1$, we invoke Lemma 2.1. As long as $\nu \gg _n 1$, we obtain a basis $\underline {e}_1,\ldots,\underline {e}_n$ of $\mathbb {F}_\nu ^n$ such that the system of quadrics $\tilde {Q}_1,\tilde {Q}_2$ corresponding to the restriction of $Q_1$ and $Q_2$ onto the subspace $\Lambda _{n-s_{\nu }-1}$ obeys (2.2)–(2.3). This clearly defines a system of non-singular quadratic forms defined over $n-s_{\nu }-1$, whose complete intersection is non-singular over $\overline {\mathbb {F}}_\nu$ as well. Now let $\tilde {M}_1$ and $\tilde {M}_2$ denote the integer matrices defining the forms $\tilde {Q}_1$, and $\tilde {Q}_2$ respectively. The lemma now follows from noticing that

\[ \mathrm{Rank}(a_1M_1+a_2M_2)\geq \mathrm{Rank}(a_1\tilde{M}_1+a_2\tilde{M}_2), \]

for any pair $(a_1,a_2)\in \mathbb {F}_\nu ^2\setminus (0,0)$ and, further, using our analysis of the non-singular case above.

One of the key bounds for exponential sums in this work will be provided by Weyl differencing. Typically, these bounds use a ‘Birch-type’ singular locus $\sigma _K'$ as defined in (2.12) instead of the singular locus (2.1) used here. A relation between the two has been studied in [Reference Browning and Heath-BrownBH17]. A minor modification of [Reference MyersonMye18, Lemma 1.1] readily provides us with the following result.

Lemma 2.3 Let $F,G$ be non-constant forms of any degree, $K$ be a field, and let

(2.11)\begin{align} \sigma_K(F)&:=\dim \big\{ \underline{x}\in \mathbb{P}^{n-1}_{\overline{K}} \, : \, F(\underline{x})=0,\, \nabla F(\underline{x})=\underline{0}\big\}, \end{align}

(2.12)\begin{align} \sigma_K'(F,G)&:=\dim \bigg\{ \underline{x}\in \mathbb{P}^{n-1}_{\overline{K}} \, : \, \mathrm{Rank} \begin{pmatrix} \nabla F(\underline{x})\\ \nabla G(\underline{x}) \end{pmatrix} <2 \bigg\}, \end{align}

(2.13)\begin{align} \sigma_K(F,G)&:=\dim \mathrm{Sing}_K(F,G). \end{align}

Then, we have

\[ \sigma_K(a_1F+a_2G)\leq \sigma_K'(F,G)\leq \sigma_K(F,G)+1, \]

for any $(a_1,a_2)\in K\backslash \{(0,0)\}$.

Our main exponential sum bound for square-full moduli $q$ will be in terms of the size of the null set

(2.14)\begin{equation} \mathrm{Null}_{q}(M):=\{\underline{x}\in (\mathbb{Z}/q\mathbb{Z})^n:M\underline{x}=\underline{\mathrm{0}}\}, \end{equation}

for some matrix $M\in M_n(\mathbb {Z})$. The following three lemmas will be related to this set.

Lemma 2.4 For every $u,v\in \mathbb {N}$, and every $M\in M_n(\mathbb {Z})$, we have

\[ \#\mathrm{Null}_{uv}(M)\leq \#\mathrm{Null}_u(M)\#\mathrm{Null}_v(M), \]

with equality if $(u,v)=1$.

Proof. It is easy to prove that $\#\mathrm {Null}_{q}(M)$ is a multiplicative function, so we will not prove that

(2.15)\begin{equation} \#\mathrm{Null}_{uv}(M)= \#\mathrm{Null}_u(M)\#\mathrm{Null}_v(M), \end{equation}

when $(u,v)=1$. We will be brief when showing the inequality, as this is a standard Hensel lemma type of argument. If $\underline {x}\in \mathrm {Null}_{uv}(M)$, then we must have $M\underline {x}\equiv \underline {0} \bmod u$. Hence, if we write $\underline {x}:=\underline {y}+u\underline {z}$, where $\underline {y}\in (\mathbb {Z}/u\mathbb {Z})^n$, $\underline {z}\in (\mathbb {Z}/v\mathbb {Z})^n$, then $\underline {y}$ must be in $\mathrm {Null}_u(M)$.

Now, fix $\underline {y}$ and assume that there is some $\underline {z}_1,\underline {z}_2$ (not necessarily distinct) such that $\underline {y}+u\underline {z}_i\in \mathrm {Null}_{uv}(M)$. Then

\[ M(\underline{y}+u\underline{z}_i)\equiv \underline{0} \mod uv, \]

and so

\[ M(\underline{y}+u\underline{z}_2)-M(\underline{y}+u\underline{z}_1)=uM(\underline{z}_2-\underline{z}_1)\equiv \underline{0} \mod uv. \]

Therefore, upon letting $\underline {z}_2:=\underline {z}_1+\underline {z}'$ we must have

\[ M\underline{z}'\equiv \underline{0} \mod v. \]

Hence, there can only be at most $\#\mathrm {Null}_v(M)$ possible values for $\underline {z}'$ and so there can only be at most $\#\mathrm {Null}_v(M)$ values for $\underline {z}$ such that $\underline {y}+u\underline {z}\in \mathrm {Null}_{uv}(M)$ for any given $\underline {y}$. We also have that $\underline {y}$ must be in $\mathrm {Null}_{u}(M)$. This gives us

\[ \#\mathrm{Null}_{uv}(M)\leq \#\mathrm{Null}_{u}(M)\#\mathrm{Null}_{v}(M), \]

as required.

In both §§ 5 and 6, we will need to bound $\#\mathrm {Null}_{p}(M)$ for matrices of the form $M({\underline {a}}):=a_1M_1+a_2M_2$, where $M_1$ and $M_2$ are symmetric matrices associated to some quadratic forms $Q_1(\underline {x})$, $Q_2(\underline {x})$. In Proposition 2.2, we noted that for most values of ${\underline {a}}$, $\mathrm {Rank}_{p}(M({\underline {a}}))\geq n-s_{p}-1$, but there were potentially a few lines of ${\underline {a}}$ where $\mathrm {Rank}_{p}(M({\underline {a}}))= n-s_{p}-2$. Naturally, a lower bound on the size of the rank of a matrix leads to an upper bound on the dimension of the nullspace of a matrix (due to the rank-nullity theorem), and so using $\mathrm {Rank}_{p}(M({\underline {a}}))\geq n-s_{p}-2$ in order to bound $\#\mathrm {Null}_{p}(M({\underline {a}}))$ for every ${\underline {a}}$ would be wasteful. This will lead us to considering averages of $\#\mathrm {Null}_{p}(M({\underline {a}}))$, where ${\underline {a}}$ is allowed to vary (this is the topic of the next lemma).

Lemma 2.5 Let $Q_1,Q_2$ be quadratic forms in $n$ variables, $q\in \mathbb {N}$, and $d$ be a square-free divisor of $q$. Furthermore, let $M_1, M_2$ be integer matrices defining $Q_1$ and $Q_2$, respectively, and let $s_p=s_p(Q_1,Q_2)$ be as defined in (2.8) for $K=\mathbb {F}_p$, $p$ a prime. If $d=\prod _{i=1}^r p_i$ for some primes $p_i$, then

\[ \sideset{}{^*}\sum_{{\underline{a}} \bmod{q}} \#\mathrm{Null}_d(a_1M_1+a_2M_2)\ll_n q^2 \prod_{i=1}^r p_i^{s_{p_i}+1}. \]

Proof. For the duration of this proof only, we will use the notation

\[ \mathrm{D}(d,q):=\sideset{}{^*}\sum_{{\underline{a}} \bmod{q}} \#\mathrm{Null}_d(a_1M_1+a_2M_2). \]

We first note that upon setting ${\underline {a}}=\underline {b}+d\underline {c}$,

(2.16)\begin{align} \mathrm{D}(d,q)&\leq \sum_{\substack{{\underline{a}} \bmod{q}\\ (a_1,a_2,d)=1}} \#\mathrm{Null}_d(a_1M_1+a_2M_2)\nonumber\\ &= \sum_{\substack{\underline{b} \bmod{d}\\ (b_1,b_2,d)=1}} \#\mathrm{Null}_d(b_1M_1+b_2M_2) \sum_{\underline{c} \bmod{q/d}} 1\nonumber\\ &= \bigg(\frac{q}{d}\bigg)^2\sideset{}{^*}\sum_{\underline{b} \bmod{d}} \#\mathrm{Null}_d(b_1M_1+b_2M_2)\nonumber\\ &= \bigg(\frac{q}{d}\bigg)^2 \mathrm{D}(d,d). \end{align}

For convenience, define

(2.17)\begin{equation} T(d):=\mathrm{D}(d,d). \end{equation}

Using the Chinese remainder theorem, it is easy to see that $T(d)$ is a multiplicative function. In particular, we have

(2.18)\begin{equation} T(d)=\prod_{\substack{i=1\\ p_i\mid d \textrm{ where } p_i \textrm{ prime }}}^r T(p_i). \end{equation}

It is therefore sufficient to consider

(2.19)\begin{equation} T(p)=\sideset{}{^*}\sum_{{\underline{a}} \bmod{p}} \#\{\underline{x}\bmod{p}:(a_1M_1+a_2M_2)\underline{x}\equiv \underline{\mathrm{0}}\bmod{p}\}, \end{equation}

where $p$ is a prime. When $p\ll _n 1$, the right-hand side is trivially $O(p^{2})$. It is therefore enough to consider the case $p\gg _n 1$, where the implied constant is chosen as in the statement in Proposition 2.2. Proposition 2.2 now implies that except for $O_n(p)$ different exceptional pairs $(a_1,a_2)$, $\mathrm {Rank}(a_1M_1+a_2M_2)\geq n-s_p-1$. Moreover, for the exceptional pairs we still have $\mathrm {Rank}(a_1M_1+a_2M_2)= n-s_p-2$. Finally, we note that if $M$ is an integer matrix rank $k$ over $\mathbb {F}_p$, it is easy to see that

\[ \#\{\underline{x}\in\mathbb{F}_p^n:M\underline{x}=\underline{\mathrm{0}}\}\ll p^{n-k}. \]

Applying these results to (2.19) gives us

\begin{align*} T(p)&\ll \sideset{}{^*}\sum_{\substack{{\underline{a}} \bmod{p}\\ \mathrm{Rank}(a_1M_1+a_2M_2)\geq n-s_p-1}} p^{s_p+1} + \sideset{}{^*}\sum_{\substack{{\underline{a}} \bmod{p}\\ \mathrm{Rank}(a_1M_1+a_2M_2)= n-s_p-2}} p^{s_p+2}\\ &\ll p^2\times p^{s_p+1}+p\times p^{s_p+2}\\ &\ll p^{2+s_p+1}, \end{align*}

and so

\[ T(d)\ll\prod_{i=1}^r p_i^{2+s_{p_i}+1} = d^2 \prod_{i=1}^r p_i^{s_{p_i}+1} \]

by (2.18). Hence, by (2.16)–(2.17), we have

\[ D(d,q)\leq \bigg{(}\frac{q}{d}\bigg{)}^2 T(d)\ll q^2 \prod_{i=1}^r p_i^{s_{p_i}+1}, \]

as required.

During the process of bounding quadratic exponential sums, we will need to bound the size of the set

(2.20)\begin{equation} N_{\underline{b},q}(M):=\bigg\{\underline{x}\in (\mathbb{Z}/q\mathbb{Z})^n \, : \, M\underline{x} \equiv \frac{q}{2}\,\underline{b}\ ({\rm mod}\ {q})\bigg\}. \end{equation}

The next lemma will help us to do this by letting us relate $N_{\underline {b},q}(M)$ to $\mathrm {Null}_q(M)$.

Lemma 2.6 Let $q\in \mathbb {N}$ be even, $M\in M_n(\mathbb {Z}/q\mathbb {Z})$ and let $N_{\underline {b},q}(M)$ be defined as in (2.20). Then for every $\underline {b}\in \{0,1\}^n$, either $N_{\underline {b},q}(M)=\emptyset$ or there exists some $\underline {y}_{\underline {b}}\in (\mathbb {Z}/q\mathbb {Z})^n$ such that

\[ N_{\underline{b},q}(M)= \underline{y}_{\underline{b}} + \mathrm{Null}_q(M). \]

We will not prove this here as the argument used in the classical proof of Hensel's lemma can be trivially adapted to prove this lemma.

3. Initial setup

In this section we will start with some initial considerations which will help us to properly set up the circle method and state our main results which will be used to prove Theorem 1.2. As stated previously, the Hardy Littlewood circle method transforms the task of answering Theorem 1.2 to proving an asymptotic formula:

(3.1)\begin{equation} \int_0^1 \int_0^1 K(\alpha_1,\alpha_2) \,d\alpha_1\,d\alpha_2= C_{X}P^{n-6}+o(P^{n-6}). \end{equation}

Here $K(\underline {\alpha })$ is the exponential sum as defined in (1.5) and $C_X$ denotes a product of local densities.

We will start by splitting the box $[0,1]^2$ into a set of major arcs and minor arcs as follows. For any pair $(\alpha _1,\alpha _2)$, we can use a two-dimensional version of Dirichlet's approximation theorem to find a simultaneous approximation $(a_1/q,a_2/q)$. In particular, upon taking $Q=\lfloor P^{3/2} \rfloor$, there exists $\underline {a}=(a_1,a_2)\in \mathbb {Z}^2$ and $q\in \mathbb {N}$ such that $(a_1,a_2,q)=1$, $q\leq Q$, and

(3.2)\begin{equation} \bigg{|}\alpha_1-\frac{a_1}{q}\bigg{|}\leq \frac{1}{qQ^{1/2}}, \quad \bigg{|}\alpha_2-\frac{a_2}{q}\bigg{|}\leq \frac{1}{qQ^{1/2}}. \end{equation}

We can therefore write

(3.3)\begin{equation} \alpha_1=\frac{a_1}{q}+z_1, \quad \alpha_2=\frac{a_2}{q}+z_2, \end{equation}

for some $|\underline {z}|:=\max \{|z_1|,|z_2|\}\leq 1/qQ^{1/2}$. The choice $Q=\lfloor P^{3/2} \rfloor$ arises from our final optimisation of various bounds. We explain this in detail in § 9.3.1.

Now let $0<\Delta <1$ be some small parameter also to be chosen later, and define

\[ \mathfrak{M}_{q,\underline{a}}(\Delta):=\bigg{\{}(\alpha_1,\alpha_2) \mod 1 \,:\, \bigg{|}\alpha_i-\frac{a_i}{q}\bigg{|}\leq P^{-3+\Delta},\ i=1,2\bigg{\}}. \]

We then define the set of major arcs to be

(3.4)\begin{equation} \mathfrak{M}=\mathfrak{M}(\Delta):=\bigcup_{q\leq P^\Delta}\bigcup_{\substack{\underline{a}\bmod{q}\\ (\underline{a},q)=1}} \mathfrak{M}_{q,\underline{a}}(\Delta). \end{equation}

This union of sets is disjoint if $P^{-2\Delta }\geq 2 P^{-3+\Delta }$, namely when $\Delta <1$ and when $P$ is sufficiently large. Moreover, it is easy to check that $P^{-3+\Delta }<1/qQ^{1/2}$ for any $q\leq Q$, provided that $Q< P^{3-\Delta }$. This is certainly true for our final choice $Q=P^{3/2}$ since we assumed $\Delta <1$, and so we have that each set $\mathfrak {M}_{q,\underline {a}}$ is contained in the corresponding range from (3.2). Therefore, the major arcs give the following contribution to the integral in (3.1):

(3.5)\begin{equation} S_{\mathfrak{M}}:=\sum_{1\leq q\leq P^{\Delta}} \,\,\sideset{}{^*}\sum_{\underline{a}\bmod{q}}\int_{|\underline{z}|\leq P^{-3+\Delta}} K({\underline{a}}/q+\underline{z}) \,d\underline{z}. \end{equation}

We then define the minor arcs to be $\mathfrak {m}=[0,1]^2\backslash \mathfrak {M}$. By the construction of $\mathfrak {M}$, the individual minor arcs must therefore either have

(3.6)\begin{equation} P^\Delta< q\leq Q \textrm{ and } |\underline{z}|<(qQ^{1/2})^{-1} \quad\textrm{or} \quad 1\leq q\leq P^\Delta \textrm{ and } P^{-3+\Delta}<|\underline{z}|<(qQ^{1/2})^{-1}. \end{equation}

Hence, we can bound the minor arcs contribution, upon further bringing the average over ${\underline {a}}$ inside the integral in (3.1), by

(3.7)\begin{equation} S_{\mathfrak{m}}=\sum_{1\leq q\leq P^{\Delta}}\int_{P^{-3+\Delta} \leq |\underline{z}|\leq 1/qQ^{1/2}} K(q,\underline{z}) \,d\underline{z} + \sum_{P^{\Delta}\leq q\leq Q }\int_{|\underline{z}|\leq 1/qQ^{1/2}} K(q,\underline{z}) \,d\underline{z}. \end{equation}

Here

(3.8)\begin{equation} K(q,\underline{z}):=\sideset{}{^*}\sum_{{\underline{a}}\bmod{q}}|K({\underline{a}}/q+\underline{z})|. \end{equation}

Our techniques for dealing with the major arcs contribution are standard. Let

(3.9)\begin{equation} \begin{aligned} \mathfrak{S}(R) & :=\sum_{q=1}^R q^{-n}\sideset{}{^*}\sum_{\underline{a}\bmod{q}} \sum_{\underline{x} \bmod{q}} e_q(a_1F(\underline{x})+a_2G(\underline{x})),\\ \mathfrak{J}(R) & :=\int_{|\underline{z}|< R}\int_{\mathbb{R}^n}\omega(\underline{x})e(z_1F(\underline{x})+z_2G(\underline{x}))\, d\underline{x}\,d\underline{z}, \end{aligned} \end{equation}

and let

(3.10)\begin{equation} \mathfrak{S}:=\lim_{R\rightarrow\infty} \mathfrak{S}(R),\quad \mathfrak{J}=\lim_{R\rightarrow \infty} \mathfrak{J}(R), \end{equation}

denote the singular series and the corresponding singular integral, provided the limits exist. Our main major arcs estimate is the following lemma.

Lemma 3.2 Assume that $n-\sigma (F,G)\geq 34$, where $\sigma (F,G):=\sigma (X_{F,G})$ as defined in (1.1), and assume that $\mathfrak {S}$ is absolutely convergent, satisfying

\[ \mathfrak{S}(R)=\mathfrak{S}+O_{\phi}(R^{-\phi}), \]

for some $\phi >0$. Then provided that we have $\Delta \in (0,1/7)$,

\[ S_{\mathfrak{M}}=\mathfrak{S}\mathfrak{J}P^{n-6}+O_{\phi}(P^{n-6-\delta}). \]

The proof of this lemma, along with the proof of convergence of the singular series will be established in § 10.

The majority of our effort will be spent in bounding the minor arcs contribution. In order to state the proposition we aim to prove for the minor arcs, we need to further specify our choice of weight function and the point which it will centred on. Let $\underline {x}_0$ be a fixed point satisfying $|\underline {x}_0|<1$ and

(3.11)\begin{equation} \mathrm{Rank} \begin{pmatrix} \nabla F(\underline{x}_0)\\ \nabla G(\underline{x}_0) \end{pmatrix}=2. \end{equation}

Without loss of generality, we may assume that

(3.12)\begin{equation} |\nabla F(\underline{x}_0)\cdot \nabla G(\underline{x}_0)|\leq C'\|\nabla F(\underline{x}_0)\|\|\nabla G(\underline{x}_0)\|, \end{equation}

for some $0< C'<1$, possibly depending on $\underline {x}_0$. Here and throughout, by $\|\underline {x}\|$ we denote the $\ell ^2$ norm of the vector $\underline {x}$. Note that this norm is equivalent to the sup-norm on $\mathbb {R}^n$. We will also slightly expand our definition of the test function $\omega$ to assume it to be supported in a box $\underline {x}_0+(-\rho,\rho )^n$, for a small parameter $\rho >0$ to be chosen in due course. Moreover, we ask that $\omega \in \mathcal {W}_n$, where $\mathcal {W}_n$ is defined to be the set of infinitely differentiable functions $\hat {\omega }: \mathbb {R}^n\rightarrow \mathbb {R}_{\geq 0}$ with compact support contained within $[-S_n,S_n]^n$ for some fixed $S_n$, and with the following bound to be true on its derivatives:

(3.13)\begin{equation} \max\bigg{\{}\bigg{|}\frac{\partial^{j_1+\cdots + j_n}}{\partial x_1^{j_1}\cdots \partial x_n^{j_n}} \hat{\omega}(\underline{x})\bigg{|} \mid \underline{x}\in\mathbb{R}, j_1+\cdots +j_n=j\bigg{\}}\ll_{j,n} 1 \end{equation}

for every $j\geq 0$. A satisfactory bound for the minor arcs will be produced by the following proposition, which we aim to prove.

Proposition 3.3 Let $F,G$ be a system of two cubic forms with a smooth intersection satisfying $n\geq 39$, and let $\omega \in \mathbb {C}^\infty _c(\underline {x}_0+(-\rho,\rho )^n)$ satisfy (3.13), where $\underline {x}_0$ satisfies (3.12). Then there exists some $\delta =\delta (\Delta )>0$ and some $\rho _0>0$, such that for any $0<\Delta < 1/7$ and for any $0<\rho <\rho _0$, we have

\[ S_{\mathfrak{m}}=O_{n,\rho,\Delta, \|F\|,\|G\|}(P^{n-6-\delta}). \]

Here, given a polynomial $F$, let $\|F\|$ denote the maximum of all its coefficients.

A major part of the rest of this work will be dedicated to proving Proposition 3.3, which will ultimately be achieved in § 9. Before we move on, it will be desirable to obtain a consequence of our choice of $\omega$ and $\underline {x}_0$, akin to the conditions [Reference Marmon and VisheMV19, (2.15) and (2.16)]. This will be our aim in Lemma 3.4 below, which will be useful in setting up a two-dimensional van der Corput differencing argument in § 4 and, in particular, in the proof of Lemma 4.3. In order to state Lemma 3.4, we will choose an orthonormal basis for the two-dimensional vector space spanned by $\{\nabla F(\underline {x}_0),\nabla G(\underline {x}_0)\}$:

(3.14)\begin{equation} \underline{e}_1':=\frac{\nabla F(\underline{x}_0)}{\|\nabla F(\underline{x}_0)\|},\quad \underline{e}_2':=\frac{\nabla G(\underline{x}_0)-\gamma\underline{e}_1'}{\gamma_1}, \end{equation}

where $\gamma =\nabla G(\underline {x}_0)\cdot \underline {e}_1'$, and $\gamma _1=\|\nabla G(\underline {x}_0)-\gamma \underline {e}_1'\|$ is a non-zero constant by (3.12).

Lemma 3.4 Let $F$ and $G$ be cubic forms and $\omega$ be a compactly supported function supported in $\underline {x}_0+(-\rho,\rho )^n$ satisfying (3.13), where $\underline {x}_0$ satisfies (3.12). Then there exist constants $M_1,M_2>0$ and there exists some $0<\rho _0\leq 1$ such that if $\rho \leq \rho _0$, then

(3.15)\begin{gather} \min_{\underline{x}\in \mathrm{Supp}(P\omega)}|\nabla F(\underline{x})\cdot \underline{e}_1'|\geq M_1P^2, \quad \min_{\underline{x}\in \mathrm{Supp}(P\omega)}|\nabla G(\underline{x})\cdot \underline{e}_2'|\geq M_1P^2, \end{gather}

(3.16)\begin{gather} \max_{\underline{x}\in\mathrm{Supp} (P\omega)}\{|\nabla F(\underline{x})\cdot \underline{e}_2'| \}\leq \rho M_2P^2, \quad \max_{\underline{x}\in\mathrm{Supp} (P\omega)}\{|\nabla G(\underline{x})\cdot \underline{e}_1'|\}\leq M_2P^2. \end{gather}

Furthermore, $M_1$ and $M_2$ depend only on $F$, $G$ and our choice of $\underline {x}_0$ (in particular, $M_1$ and $M_2$ do not depend on $\rho$).

Proof. A key in the proof here will be the following bound, which is an easy consequence of the mean value theorem: given any $\underline {x}\in \mathrm {Supp}(P\omega )$, we have

(3.17)\begin{equation} \|\nabla F(\underline{x})-\nabla F(P\underline{x}_0)\|\ll_{\|F\|} \rho P^2 \quad \textrm{and}\quad \|\nabla G(\underline{x})-\nabla G(P\underline{x}_0)\|\ll_{\|G\|} \rho P^2. \end{equation}

Let us first prove that the conditions for $\nabla F(\underline {x})$ in (3.15)–(3.16) are met. The key here is conditions (3.11) and (3.12). Clearly, using (3.17) we have

\begin{align*} \nabla F(\underline{x})\cdot \underline{e}_1'&=(\nabla F(\underline{x})-\nabla F(P\underline{x}_0))\cdot \underline{e}_1'+\nabla F(P\underline{x}_0)\cdot \underline{e}_1'\\ &=(\nabla F(\underline{x})-\nabla F(P\underline{x}_0))\cdot \underline{e}_1'+P^2\nabla F(\underline{x}_0)\cdot \nabla F(\underline{x}_0)/\|\nabla F(\underline{x}_0)\|\\ &=(\nabla F(\underline{x})-\nabla F(P\underline{x}_0))\cdot \underline{e}_1'+P^2\|\nabla F(\underline{x}_0)\|\\ &\geq (1-O(\rho))P^2\|\nabla F(\underline{x}_0)\|\\ &\geq M_{F,1} P^2 \end{align*}

for some $M_{F,1}>0$ which is independent of $\rho$, provided that $\rho$ is chosen to be small enough. Similarly, we may also assure that

(3.18)\begin{equation} |\nabla G(\underline{x})\cdot \nabla G(\underline{x}_0)| \geq (1-O(\rho))P^2\|\nabla G(\underline{x}_0)\|^2. \end{equation}

In both of these equations, the implied constants only depend on $\| F\|$, $\| G\|$ and $n$. This will be a feature of all implied constants appearing in this proof. On the other hand, since $\nabla F(\underline {x}_0)=\|\nabla F(\underline {x}_0)\|\,\underline {e}_1'$ is orthogonal to $\underline {e}_2'$, we have

(3.19)\begin{equation} |\nabla F(\underline{x})\cdot \underline{e}_2'|=|(\nabla F(\underline{x})-P^2\nabla F(\underline{x}_0))\cdot \underline{e}_2'|\leq \|(\nabla F(\underline{x})-\nabla F(P\underline{x}_0))\|\ll_{\|F\|} \rho P^2 \end{equation}

by (3.17). In other words, there is some $M_{F,2}>0$ independent of $\rho$ such that

\[ |\nabla F(\underline{x})\cdot \underline{e}_2'|\leq M_{F,2} \,\rho P^2. \]

To deal with the inequalities concerning $G$, we use (3.12), which hands us a constant $0< C'<1$ satisfying

(3.20)\begin{equation} \gamma\|\nabla F(\underline{x}_0)\|=|\nabla F(\underline{x}_0)\cdot \nabla G(\underline{x}_0)|\leq C' \|\nabla F(\underline{x}_0)\|\|\nabla G(\underline{x}_0)\|. \end{equation}

Therefore, for any $\underline {x}\in \mathrm {Supp}(P\omega )$, by (3.17) and (3.20), we have that

\begin{align*} |\nabla F(\underline{x}_0)\cdot \nabla G(\underline{x})|&\leq |\nabla F(\underline{x}_0)\cdot \nabla G(P\underline{x}_0)|+ |\nabla F(\underline{x}_0)\cdot (\nabla G(\underline{x})-\nabla G(P\underline{x}_0))|\\ &\leq C' P^2\|\nabla G(\underline{x}_0)\|\|\nabla F(\underline{x}_0)\|+O_{\|G\|}(\rho) P^2\|\nabla F(\underline{x}_0)\|. \end{align*}

Hence (since $\|\nabla G(\underline {x}_0)\|>0$ is a constant), provided that the support $\rho$ is sufficiently small, we may choose some $0< C''<1$ independent of $\rho$ such that

(3.21)\begin{equation} |\nabla F(\underline{x}_0)\cdot \nabla G(\underline{x})|\leq C''P^2\|\nabla F(\underline{x}_0)\|\|\nabla G(\underline{x}_0)\|. \end{equation}

Thus, for any $\underline {x}\in \mathrm {Supp}(P\omega )$,

\begin{align*} |\nabla G(\underline{x})\cdot(\nabla G(\underline{x}_0)-\gamma \underline{e}_1')|&=|\nabla G(\underline{x})\cdot \nabla G(\underline{x}_0)- \gamma\|\nabla F(\underline{x}_0)\|^{-1}\nabla G(\underline{x})\cdot \nabla F(\underline{x}_0)|\\ &\geq (1-O(\rho)-C'C'')P^2\|\nabla G(\underline{x}_0)\|^2, \end{align*}

where we have used (3.20) to bound $\gamma$ by $C'\|\nabla G(\underline {x}_0)\|$, as well as (3.21) and (3.18). Hence, provided that the support $\rho$ is chosen to be sufficiently small, there is some $M_{G,1}>0$ such that

\[ |\nabla G(\underline{x})\cdot \underline{e}_2'|=\gamma_1^{-1}|\nabla G(\underline{x})\cdot(\nabla G(\underline{x}_0)-\gamma \underline{e}_1')|\geq M_{G,1} P^2. \]

Hence, upon taking

\[ M_1:=\min\{M_{F,1}, M_{G,1}\}, \]

we conclude that (3.15) is true. Finally, (3.21) also hands us

(3.22)\begin{equation} |\nabla G(\underline{x})\cdot \underline{e}_1'|=\|\nabla F(\underline{x}_0)\|^{-1}|\nabla F(\underline{x}_0)\cdot \nabla G(\underline{x})|\leq C'' P^2\| G (\underline{x}_0)\|, \end{equation}

for any $\underline {x}\in \mathrm {Supp}(P\omega )$. Therefore, upon setting $M_{2,G}:=C''\| G (\underline {x}_0)\|$, and taking

\[ M_2:=\max\{M_{F,2}, M_{G,2}\}, \]

we are now able to verify (3.16). Furthermore, there is some $\rho _0>1$, such that $M_1$ and $M_2$ are independent of $\rho$ provided that $\rho \leq \rho _0$. This concludes the proof of the lemma.

4. Van der Corput differencing

In this section, we will use van der Corput differencing to bound $K({\underline {a}}/q+\underline {z})$ by a quadratic exponential sum. We will introduce the topic by beginning with the simpler pointwise van der Corput differencing used in [Reference Browning and Heath-BrownBH09] before attempting to generalise the differencing arguments from [Reference HanselmannHan12] and [Reference VisheVis23] to attain a bound which also takes advantage of averaging over the both $\underline {z}$ integrals. In both cases, we will innovate on the standard differencing approach in order to introduce a path to attaining Kloosterman refinement.

4.1 Pointwise van der Corput

For convenience, we will set

(4.1)\begin{equation} \hat{F}_{\underline{a},q,\underline{z}}(\underline{x}):=(a_1/q+z_1)F(\underline{x})+(a_2/q+z_2)G(\underline{x}), \end{equation}

where $F$ and $G$ are cubic forms. Since $\underline {x}$ is summed over all of $\mathbb {Z}^n$, we can replace $\underline {x}$ with $\underline {x}+\underline {h}$, for any $\underline {h}\in \mathbb {Z}^n$, giving

(4.2)\begin{equation} K(q,\underline{z})=\sideset{}{^*}\sum_{{\underline{a}}}\bigg{|}\sum_{\underline{x}\in\mathbb{Z}^n} \omega((\underline{x}+\underline{h})/P)e(\hat{F}_{\underline{a},q,\underline{z}}(\underline{x}+\underline{h}))\bigg{|}, \end{equation}

where $K(q,\underline {z})$ is as defined in (3.8). Let $\mathcal {H}\subset \mathbb {Z}^n$ be a set of lattice points (which we may choose freely). In the case of pointwise van der Corput differencing, we can just take $\mathcal {H}$ to be the set of lattice points ${\underline {h}}$ such that $|{\underline {h}}|< H$, for some $1\leq H\ll P$ which we may choose freely. However, we will not specify this in the arguments that follow since we will need a different choice of $\mathcal {H}$ when we come to averaged van der Corput differencing later. Applying the Cauchy–Schwarz inequality to (4.2) gives the following

\begin{align*} \#\mathcal{H}K(q,\underline{z})&= \sideset{}{^*}\sum_{{\underline{a}}}\bigg{|}\sum_{\underline{h}\in\mathcal{H}}\sum_{\underline{x}\in\mathbb{Z}^n} \omega((\underline{x}+\underline{h})/P)e(\hat{F}_{\underline{a},q,\underline{z}}(\underline{x}+\underline{h}))\bigg{|}\\ &\leq \sideset{}{^*}\sum_{{\underline{a}}}\sum_{\underline{x}\in\mathbb{Z}^n} \bigg{|}\sum_{\underline{h}\in\mathcal{H}}\omega((\underline{x}+\underline{h})/P)e(\hat{F}_{\underline{a},q,\underline{z}}(\underline{x}+\underline{h}))\bigg{|}\\ &\leq \bigg{(} \sideset{}{^*}\sum_{{\underline{a}}}\sum_{|\underline{x}|<2P} 1 \bigg{)}^{1/2} \bigg{(}\sideset{}{^*}\sum_{{\underline{a}}}\sum_{\underline{x}\in\mathbb{Z}^n}\bigg{|}\sum_{\underline{h}\in\mathcal{H}} \omega((\underline{x}+\underline{h})/P)e(\hat{F}_{\underline{a},q,\underline{z}}(\underline{x}+\underline{h}))\bigg{|}^2\bigg{)}^{1/2}\\ &\ll q P^{n/2} \bigg{(}\sideset{}{^*}\sum_{{\underline{a}}}\sum_{\underline{x}\in\mathbb{Z}^n} \sum_{\underline{h}_1,\underline{h}_2\in\mathcal{H}} \omega((\underline{x}+\underline{h}_1)/P)\overline{\omega((\underline{x}+\underline{h}_2)/P)}\\ &\quad \times e(\hat{F}_{\underline{a},q,\underline{z}}(\underline{x}+\underline{h}_1)) \overline{e(\hat{F}_{\underline{a},q,\underline{z}}(\underline{x}+\underline{h}_2))}\bigg{)}^{1/2}. \end{align*}

The key difference between this and the standard van der Corput differencing process is the introduction of the ${\underline {a}}$ sum in the Cauchy–Schwarz step. In particular, this enables us to bring the ${\underline {a}}$ sum inside of the bracket in the final step which, in turn, gives us a path to Kloosterman refinement. We still need to write $K(q,\underline {z})$ in terms of a quadratic exponential sum however, so we will come back to Kloosterman refinement later.

Set $\underline {y}:=\underline {x}+\underline {h}_2$, $\underline {h}=\underline {h}_1-\underline {h}_2$ and recall that we defined $\omega$ to be a real weight function. Therefore, after setting

(4.3)\begin{equation} N(\underline{h}):=\#\{\underline{h}_2-\underline{h}_1=\underline{h}:\underline{h}_1,\underline{h}_2\in\mathcal{H}\},\quad \textrm{and}\quad \omega_{\underline{h}}(\underline{x}):=\omega(\underline{x}+P^{-1}\underline{h})\omega(\underline{x}), \end{equation}

we get

\[ |K(q,\underline{z})|^2\ll \#\mathcal{H}^{-2}q^2P^{n}\sideset{}{^*}\sum_{{\underline{a}}}\sum_{\underline{y}\in\mathbb{Z}^n} \sum_{\underline{h}\in\mathcal{H}} N(\underline{h})\omega_{\underline{h}}(\underline{y}/P) e(\hat{F}_{\underline{a},q,\underline{z}}(\underline{y}+\underline{h}) -\hat{F}_{\underline{a},q,\underline{z}}(\underline{y})). \]

Recall that $\hat {F}_{\underline {a},q,\underline {z}}(\underline {x})=(a_1/q+z_1)F(\underline {x})+(a_2/q+z_2)G(\underline {x})$. Therefore if we set $F_{{\underline {h}}}$ and $G_{{\underline {h}}}$ be the differenced polynomials

\[ F_{\underline{h}}(\underline{y}):=F(\underline{y}+\underline{h})-F(\underline{y}),\quad G_{\underline{h}}(\underline{y}):=G(\underline{y}+\underline{h})-G(\underline{y}), \]

we have

\[ \hat{F}_{\underline{a},q,\underline{z}}(\underline{y}+\underline{h}) -\hat{F}_{\underline{a},q,\underline{z}}(\underline{y})=(a_1/q+z_1)F_{{\underline{h}}}(\underline{y}) +(a_2/q+z_2)G_{{\underline{h}}}(\underline{y}). \]

Hence,

(4.4)\begin{equation} |K(q,\underline{z})|^2\ll \#\mathcal{H}^{-2}P^{n}q^2 \sum_{\underline{h}\in \mathcal{H}} N(\underline{h})T_{\underline{h}}(q,\underline{z}), \end{equation}

where

(4.5)\begin{equation} T_{\underline{h}}(q,\underline{z}):= \sideset{}{^*}\sum_{{\underline{a}}\bmod{q}}\sum_{\underline{y}\in\mathbb{Z}^n} \omega_{\underline{h}}(\underline{y}/P)e((a_1/q+z_1)F_{{\underline{h}}}(\underline{y})+(a_2/q+z_2)G_{{\underline{h}}}(\underline{y})) \end{equation}

denote the corresponding exponential sum for the system of quadratic polynomials $F_{{\underline {h}}}$ and $G_{{\underline {h}}}$. Note that the top form of $F_{{\underline {h}}}$, $F_{{\underline {h}}}^{(0)}$, is precisely (1.8). Finally, by noting that $N({\underline {h}})\leq \#\mathcal {H}=H^n$, we arrive at the following.

Lemma 4.1 For any $1\leq H\ll P$, for any fixed choice of $\underline {z}\in [0,1]^2$, we have

\[ |K(q,\underline{z})|\ll H^{-n/2}P^{n/2}q\bigg{(}\sum_{\underline{h}\ll H}|T_{\underline{h}}(q,\underline{z})|\bigg{)}^{1/2}. \]

This bound will be useful to us when $t:=|\underline {z}|$ is small, say of size $P^{-3-\Delta }$, since it is wasteful to use averaged van der Corput differencing in this case. We will now set up averaged van der Corput differencing, which will be a key in proving Proposition 3.3.

4.2 Averaged van der Corput

Throughout this section, $\underline {x}_0$ will denote a fixed point satisfying $|\underline {x}_0|<1$ in $\underline {x}_0\in \mathrm {Supp}(\omega )$, where $\mathrm {Supp}(\omega )$ is contained in the set $\underline {x}_0+(-\rho,\rho )^n$. Likewise, $F$ and $G$ will be cubic polynomials whose leading forms satisfy (3.15) and (3.16) for a fixed orthonormal set of vectors $\underline {e}_1',\underline {e}_2'$ (see (3.14)). Let

(4.6)\begin{align} \{\underline{e}_1',\ldots,\underline{e}_n'\}, \end{align}

denote an extended orthonormal basis of $\mathbb {R}^n$. We will begin our effort to bound the sum

(4.7)\begin{equation} \sum_{P^{\Delta}\leq q\leq Q }\int_{P^{-3-\Delta}\leq|\underline{z}|\leq 1/qQ^{1/2}} K(q,\underline{z})\,d\underline{z}, \end{equation}

where $K(q,\underline {z})=\sideset {}{^*}\sum _{{\underline {a}}\bmod {q}} |K({\underline {a}}/q+\underline {z})|$ is as defined in (3.8). As in the previous section, let $1\leq H\ll P$ be a parameter to be chosen later. Typically, $H$ will be chosen as a small power of $P$, so it is safe to further assume $H\log P\ll P$. In addition, let $\varepsilon >0$ be an arbitrarily small absolute constant to be chosen at the end. Note that the implied constants will be allowed to depend on the choice of $\varepsilon$ after it is introduced into our bounds. As is standard (see, for example, [Reference VisheVis23]), we start by splitting the integral over $\underline {z}$ above as a sum over $O(P^\varepsilon )$ dyadic intervals of the form $[t,2t]$ where $P^{-3+\Delta }\leq t\leq 1/(qQ^{1/2})$. For convenience, given $t\in \mathbb {R}_{>0}^2$, we will set

\[ I(q,t):=\int_{t\leq |\underline{z}|\leq 2t} K(q,\underline{z})\, d\underline{z}. \]

Analogous to [Reference HanselmannHan12] and [Reference Marmon and VisheMV19, Section 3], for a fixed value of $P^{-3-\Delta }< t<1/qQ^{1/2}$ we choose two sets $T_1$, $T_2$, each of cardinality $O(1+tHP^2)$ such that

(4.8)\begin{align} \{\underline{z}: t\leq |\underline{z}|\leq 2t\}&\subseteq \bigcup_{\underline{\tau} \in T_1\times T_2} {[}\tau_1-(HP^2)^{-1},\tau_1+(HP^2)^{-1}{]}\times {[}\tau_2-(HP^2)^{-1},\tau_2+(HP^2)^{-1}{]}\nonumber\\ &\subseteq \{\underline{z}: t\leq|\underline{z}|\leq 2(t+(HP^2)^{-1})\}. \end{align}

Thus, an application of Cauchy–Schwarz further gives

(4.9)\begin{align} I(q,t)&\ll ((HP^2)^{-1}+t)\bigg(\int_{t\leq|\underline{z}|\leq 2(HP^2)^{-1}+t)}|K(q,\underline{z})|^2\,d\underline{z}\bigg)^{1/2}\nonumber\\ &\ll((HP^2)^{-1}+t) \bigg(\sum_{\underline{\tau}\in \underline{T}} \mathcal{M}_q(\underline{\tau},H)\bigg)^{1/2}, \end{align}

where

(4.10)\begin{align} \mathcal{M}_q(\underline{\tau},H)&:=\int_{\underline{\tau}-(HP^2)^{-1}}^{\underline{\tau}+(HP^2)^{-1}} |K(q,\underline{z})|^2 \,d\underline{z}\nonumber\\ &\ll \int_{\mathbb{R}^2} \exp(-H^2P^4[(\tau_1-z_1)^2+(\tau_2-z_2)^2])|K(q,\underline{z})|^2 \,d\underline{z}. \end{align}

Here we have used $\underline {T}:=T_1\times T_2$, and $\int _{\underline {\tau }-(HP^2)^{-1}}^{\underline {\tau }+(HP^2)^{-1}}$ to denote the integral

\[ \int_{(\tau_1-(HP^2)^{-1}, \tau_1+(HP^2)^{-1})\times (\tau_2-(HP^2)^{-1}, \tau_2+(HP^2)^{-1})} \]

in order to simplify the notation. After an inspection of the right-hand side of (4.8), it is easy to see that

\[ \int_{P^{-3-\Delta}\leq |\underline{z}|\leq 1/qQ^{1/2}} K(q,\underline{z})\, d\underline{z}\ll \sum_{t}((HP^2)^{-1}+t)\bigg(\sum_{\underline{\tau}\in \underline{T}} \mathcal{M}_q(\underline{\tau},H)\bigg)^{1/2}, \]

where the sum over $t$ runs over $O_\varepsilon (P^\varepsilon )$ choices satisfying

(4.11)\begin{equation} P^{-3-\Delta}\leq t\leq 1/(qQ). \end{equation}

Note that the choice of the parameter $H$ will ultimately depend on $t$. For now, we will assume $t$ to be fixed.

We are therefore first led to find a bound for $|K(q,\underline {z})|^2$ using van der Corput differencing. Recall that results (4.4) and (4.5) from § 4.1 hold for any subset of integer vectors $\mathcal {H}$ satisfying $|{\underline {h}}|\ll P$ for every ${\underline {h}}\in \mathcal {H}$. Therefore, by (4.4), (4.9) and (4.10), we have shown the following.

Lemma 4.2 For any $1\leq H\leq P$, $\mathcal {H}\subset \mathbb {Z}^n$ and $t$ satisfying (4.11) we have

(4.12)\begin{align} I(q,t)&\ll ((HP^2)^{-1}+t)\#\mathcal{H}^{-1}P^{n/2}q\nonumber\\ &\quad \times\bigg{(} \sum_{\underline{\tau}\in \underline{T}}\sum_{\underline{h}\in \mathcal{H}} N(\underline{h})\int_{\mathbb{R}^2} \exp(-H^2P^4[(\tau_1-z_1)^2+(\tau_2-z_2)^2])T_{\underline{h}}(q,\underline{z})\, d\underline{z}\bigg{)}^{1/2}. \end{align}

Since we intend to develop a two-dimensional version of averaged van der Corput differencing, we intend to choose $\mathcal {H}$ to be a set of size $O(P^2H^{n-2})$ and then use averaging over $z_1$ and $z_2$ to show that for all but $O((H\log (P))^n)$ of $\underline {h}\in \mathcal {H}$, the value of the averaged integral $\mathcal {M}_q(\underline {\tau },H)$ defined in (4.10) is negligible. This will enable us to ‘win’ an extra factor of $P/H$ in our final estimate for (4.7) when compared with pointwise van der Corput differencing.

Our choice of $\mathcal {H}$ will be informed by the following lemma.

Lemma 4.3 For any $\underline {h}\in \mathbb {R}^n$, any $1\leq H\leq P$, any fixed $\underline {\tau }$ and any $N>0$,

\[ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp(-H^2P^4[(\tau_1-z_1)^2+(\tau_2-z_2)^2])T_{\underline{h}}(q,\underline{z})\, d\underline{z}\ll_NP^{-N}, \]

provided that $\underline {h}=\sum _{i=1}^n h_i'\underline {e}_i'$ satisfies

(4.13)\begin{equation} H\mathcal{L}\ll \sup\{|h_1'|,|h_2'|\}\ll P,\quad|h_i'|< H \text{ for } i\in\{3,\ldots,n\}, \end{equation}

where $\mathcal {L}=\log (P)$, $\{\underline {e}_1',\ldots,\underline {e}_n'\}$ denote the basis chosen in (4.6) and the implied constants only depend on $n,\|F\|$ and $\|G\|$.

Proof. We start by rewriting

\begin{align*} &\int_{\mathbb{R}^2}\exp(-H^2P^4[(\tau_1-z_1)^2+(\tau_2-z_2)^2])T_{\underline{h}}(q,\underline{z})\, d\underline{z}\\ &\quad =\sum_{\underline{y}\in\mathbb{Z}^n}\sum_{{\underline{a}}}{}^*\omega_{\underline{h}} (\underline{y}/P)e_q(a_1F_{{\underline{h}}}(\underline{y})+a_2G_{{\underline{h}}}(\underline{y}))J(\underline{h},\underline{y}), \end{align*}

where

(4.14)\begin{equation} J(\underline{h},\underline{y})=\int_{\mathbb{R}^2}\exp(-H^2P^4[(\tau_1-z_1)^2 +(\tau_2-z_2)^2])e(z_1F_{\underline{h}}(\underline{y})+z_2G_{\underline{h}}(\underline{y}))\, d\underline{z}, \end{equation}

and $e_q(x):=e^{2\pi i x/q}$. We may separate the two integrals over $\underline {z}$ and integrate them to get

\[ J(\underline{h},\underline{y})= \frac{\pi}{H^2P^4} \exp\bigg{(}-\frac{\pi^2}{H^2P^4}{(}|F_{\underline{h}}(\underline{y})|^2 +|G_{\underline{h}}(\underline{y})|^2{)}\bigg{)}e(-\tau_1F_{\underline{h}}(\underline{y})-\tau_2G_{\underline{h}}(\underline{y})). \]

We note that if either $|F_{\underline {h}}(\underline {y})|$ or $|G_{\underline {h}}(\underline {y})|$ are $\gg HP^2\mathcal {L}$, then trivially bounding everything in $J$ from above gives

\begin{align*} \sum_{\underline{y}\in\mathbb{Z}^n}\sideset{}{^*}\sum_{{\underline{a}}\bmod{q}}\omega_{\underline{h}}(\underline{y}/P) e_q(a_1F_{{\underline{h}}}(\underline{y})+a_2G_{{\underline{h}}}(\underline{y}))J(\underline{h},\underline{y})&\ll P^nq^2\frac{1}{H^2P^4}\exp(-m\mathcal{L}^2)\\ &\ll_N P^{-N}, \end{align*}

for some constant $m>0$. Therefore, it is sufficient to show that there exist constants $0< c_1, c_2<1$ such that for every

(4.15)\begin{equation} \underline{h}=\sum_{i=1}^n h_i\underline{e}_i=\sum_{i=1}^n h_i'\underline{e}_i' \end{equation}

with ${\underline {h}}\in \mathbb {R}^n$,

(4.16)\begin{equation} |h_1'|< c_1P,\ |h_2'|< c_2P,\ |h_i'|< H \quad\text{ for } i\in\{3,\ldots,n\}, \text{ and } H\mathcal{L}\ll \sup\{|h_1'|,|h_2'|\}, \end{equation}

we have

(4.17)\begin{equation} |F_{\underline{h}}(\underline{y})|\gg HP^2\mathcal{L}\quad \text{or}\quad |G_{\underline{h}}(\underline{y})|\gg HP^2\mathcal{L}. \end{equation}

We will rewrite $F_{\underline {h}}$ as follows:

\[ F_{\underline{h}}(\underline{y})=\nabla F(\underline{y})\cdot \underline{h}+\underline{h}^t\mathcal{H}_F(\underline{y})\underline{h}+F_{\underline{h}}^{(2)} \]

where $F_{\underline {h}}^{(2)}$ is the constant part of $F_{\underline {h}}$ and $\mathcal {H}_F(\underline {y})$ is the Hessian of $F$ evaluated at $\underline {y}$. Now for $\underline {h}$ satisfying (4.16), we have

(4.18)\begin{align} F_{\underline{h}}(\underline{y})&=\nabla F(\underline{y})\cdot \underline{h}+\bigg{(}\sum h_i'\underline{e}_i'\bigg{)}^t\mathcal{H}_F(\underline{y})\bigg{(}\sum h_i'\underline{e}_i'\bigg{)}+F_{\underline{h}}^{(2)}\nonumber\\ &=\nabla F(\underline{y})\cdot \underline{h}+F_{\underline{h}}^{(2)}+O(|h_1'|^2P)+O(|h_2'|^2P)+O(HP^2), \end{align}

where $F_{{\underline {h}}}^{(2)}$ is a cubic polynomial in ${\underline {h}}$, and the implied constants depend only on $\|F\|$, $\|G\|$ and $n$. Note that

\[ F_{\underline{h}}^{(2)}= O(|h_1'|^3)+O(|h_2'|^3)+O(H^3), \]

and so we may simplify (4.18) to

(4.19)\begin{equation} F_{\underline{h}}(\underline{y})=\nabla F(\underline{y})\cdot \underline{h} +O(|h_1'|^2P)+O(|h_2'|^2P)+O(HP^2), \end{equation}

since $H, |h_1'|, |h_2'|< P$. We also write ${\underline {h}}=h_1'\underline {e}_1'+\cdots +h_n'\underline {e}_n'$ and invoke (3.15) and (3.16) to further get that for all $\underline {y}\in \mathrm {Supp}(P\omega )$ we have

\[ |\nabla F(\underline{y})\cdot \underline{h}|\geq |h_1'| M_1P^2+O(\rho |h_2'| P^2)+O(HP^2), \]

and so we get

(4.20)\begin{equation} |F_{\underline{h}}(\underline{y})|\geq M_1|h_1'| P^2+O(\rho |h_2'| P^2)+O(|h_1'|^2P)+O(|h_2'|^2P)+O(HP^2), \end{equation}

by (4.19). For now, let us focus on the case $|h_2'|\ll \rho ^{-1/2}| h_1'|$. In this case, we must have that $h_1'$ satisfies $H\mathcal {L}\ll |h_1'|$. Furthermore, upon choosing $c_1\leq \rho ^2$ and by (4.16), we have

\begin{gather*} \rho |h_2'| P^2\ll \rho^{1/2} |h_1'|P^2,\quad |h_1'|^2P\leq c_1 |h_1'|P^2 \leq \rho^2 |h_1'|P^2,\\ |h_2'|^2P\ll \rho^{-1}|h_1'|^2P\leq \rho^{-1}c_1|h_1'| P^2 \leq \rho |h_1'|P^2. \end{gather*}

Hence, we may simplify (4.20) to obtain

\[ |F_{\underline{h}}(\underline{y})|\geq M_1|h_1'| P^2+O(\rho^{1/2} |h_1'| P^2)\gg |h_1'|P^2\gg HP^2\mathcal{L}, \]

provided that $\rho$ is chosen to be sufficiently small with respect to $M_1$.

It now remains to study the case $|h_1'|\ll \rho ^{1/2}|h_2'|$. In this case, we instead have that $|h_2'|\gg H\mathcal {L}$. We now apply the same process used to obtain (4.19) to $G_{\underline {h}}(\underline {y})$ to obtain

(4.21)\begin{equation} G_{\underline{h}}(\underline{y})=\nabla G(\underline{y})\cdot \underline{h}+O(|h_1'|^2P)+O(|h_2'|^2P)+O(HP^2), \end{equation}

where the implied constants again depend only on $n$, $\|F\|$ and $\|G\|$. Note again that

\[ \nabla G(\underline{y})\cdot {\underline{h}}=h_1'\nabla G(\underline{y})\cdot \underline{e}_1'+h_2'\nabla G(\underline{y})\cdot \underline{e}_2'+O(HP^2). \]

Combining this with (4.21), and applying (3.15)–(3.16) gives

(4.22)\begin{equation} |G_{\underline{h}}(\underline{y})|\geq M_1|h_2'| P^2+O(|h_1'| P^2)+O(|h_1'|^2P)+O(|h_2'|^2P)+O(HP^2). \end{equation}

We now aim to simplify (4.22). Using the assumption that $|h_1'|\ll \rho ^{1/2}|h_2'|$, the fact that $|h_2'|$ must obey (4.16) in this case, and setting $c_2\leq \rho$ we have

\begin{gather*}|h_1'| P^2 \ll \rho^{1/2} |h_2'|P^2,\quad |h_1'|^2P\ll \rho |h_2'|^2P\leq \rho c_2|h_2'| P^2 \leq \rho^2 |h_2'|P^2,\\ |h_2'|^2P \leq c_2 |h_2'|P^2 \leq \rho |h_1'|P^2,\quad HP^2\ll |h_1'|P^2 \mathcal{L}^{-1}\ll \rho |h_1'|P^2. \end{gather*}

Hence,

\[ |G_{\underline{h}}(\underline{y})|\geq M_1|h_2'|P^2+O(\rho^{1/2} |h_2'| P^2)\gg |h_2'|P^2\gg HP^2\mathcal{L}, \]

as long as $\rho$ is chosen small enough depending only on $M_1$ and $M_2$.

The lemma above leads to the following natural choice for $\mathcal {H}$:

(4.23)\begin{equation} \mathcal{H}:=\{\underline{h}\in\mathbb{Z}^n:0\leq h_1'< c_1P,\ 0\leq h_2'< c_2P,\ 0\leq h_i'< H \text{ for } i\in\{3,\ldots,n\}\}, \end{equation}

where $c_1$ and $c_2$ are the implied constants arising in (4.13). Essentially, $\mathcal {H}$ is chosen to be the collection of lattice points inside of a fixed $n$-dimensional cuboid, $B_P$, centred at the origin, with volume $\mathrm {Vol}(B_P)=c_1c_2P^2H^{n-2}$. The sides of the cuboid are in the direction of the basis vectors $\{\underline {e}_1',\ldots,\underline {e}_n'\}$. We now claim that

(4.24)\begin{equation} P^2H^{n-2}\ll\#\mathcal{H}\ll P^2H^{n-2}. \end{equation}

This follows very easily from the following asymptotic formula for a general cuboid $B$ with side lengths $l_1,\ldots,l_n$. It is easy to see that

\[ \#\{\mathbb{Z}^n\cap B\}=\mathrm{Vol}(B)+\sum_{i=1}^n O\bigg(\prod_{j\neq i} l_j\bigg). \]

The error comes from estimating the $(n-1)$-dimensional boundary of $B$. In our case $l_1=c_1P$,$l_2=c_2P$, $l_i=H$ for $i\geq 3$, which leads to (4.24). Note that $\mathcal {H}$ is chosen as in (4.23) so that we can use the bound Lemma 4.3. In particular, we can now show the following.

Lemma 4.4 Let $1\leq H\leq P$ and let

\[ \tilde{\mathcal{H}}:=\{\underline{h}\in\mathbb{Z}^n: |{\underline{h}}|\ll H\mathcal{L}\}. \]

Then for any $1\leq H\leq P$, any $1\leq N$, and any $t>0$ such that (4.11) holds, we have

\[ I(q,t)\ll H^{-n/2+1}\log(P)P^{n/2-1}q((HP^2)^{-1}+t)^2\bigg{(}\sum_{\underline{h}\in \tilde{\mathcal{H}}}\max_{\underline{z}}|T_{\underline{h}}(q,\underline{z})|\bigg{)}^{1/2}+O_N(P^{-N}), \]

where the maximum over $\underline {z}$ is taken over the set

(4.25)\begin{equation} t\leq |\underline{z}|\leq 2(t+(HP^2)^{-1}\mathcal{L}). \end{equation}

Proof. Let $\mathcal {H}$ be as in (4.23). Then we use the decomposition $\mathcal {H}=(\tilde {\mathcal {H}}\cap \mathcal {H})\bigcup \mathcal {H}\backslash \tilde {\mathcal {H}}$. By construction,

\[ \mathcal{H}\backslash\tilde{\mathcal{H}}=\{\underline{h}\in\mathbb{Z}^n: |{\underline{h}}_1'|< c_1P,|h_2'|< c_2P, |h_i'|< H, \text{ for } i\in\{3,\ldots,n\}; H\mathcal{L}\ll \max\{|h_1'|,|h_2'|\}\}. \]

Furthermore, note that for any fixed ${\underline {h}}$, $N({\underline {h}})$ as defined in (4.3) satisfies the bound

(4.26)\begin{equation} N({\underline{h}})\ll \#\mathcal{H}\ll P^2H^{n-2}. \end{equation}

Therefore, by Lemma 4.3, and a bound $\#\underline {T}\ll (1+tHP^2)^2\ll P^6$, which arises from using crude bounds $t\leq 1$ and $1\leq H\leq P$

\[ \#\mathcal{H}^{-1}\bigg(\sum_{\underline{\tau}\in\underline{T}}\sum_{\underline{h}\in \mathcal{H}\setminus\tilde{\mathcal{H}}} N(\underline{h})\int_{\mathbb{R}^2} \exp(-H^2P^4[(\tau_1-z_1)^2+(\tau_2-z_2)^2])T_{\underline{h}}(q,\underline{z}) \,d\underline{z}\bigg)^{1/2} \ll P^{-N}. \]

Further combining with the bounds $q\leq Q\leq P^{3/2}$, we may bound the contribution from the sum over ${\underline {h}}\in \mathcal {H}\setminus \tilde {\mathcal {H}}$ in (4.12) as follows:

\begin{align*} &\ll((HP^2)^{-1}+t)P^{n/2}q\#\mathcal{H}^{-1}\\ &\quad\times\bigg{(}\sum_{\underline{\tau}\in \underline{T}}\sum_{\underline{h}\in \mathcal{H}\setminus \tilde{\mathcal{H}}} N(\underline{h})\int_{\mathbb{R}^2} \exp(-H^2P^4[(\tau_1-z_1)^2+(\tau_2-z_2)^2])T_{\underline{h}}(q,\underline{z})\,d\underline{z}\bigg{)}^{1/2}\\ &\ll_N P^{n/2+3/2-N}\ll_{n,N} P^{-N}, \end{align*}

as $N$ is allowed to be arbitrarily large. Therefore, combining this with Lemma 4.2, we get

(4.27)\begin{align} I(q,t)&\ll ((HP^2)^{-1}+t)\#\mathcal{H}^{-1/2}P^{n/2}q\nonumber\\ &\quad \times\bigg{(}\sum_{\underline{\tau}\in \underline{T}}\sum_{\underline{h}\in \tilde{\mathcal{H}}}\int_{\mathbb{R}^2} \exp(-H^2P^4[(\tau_1-z_1)^2+(\tau_2-z_2)^2])T_{\underline{h}}(q,\underline{z})\, d\underline{z}\bigg{)}^{1/2}\nonumber\\ &\quad +O_{n,N}(P^{-N}). \end{align}

Further note that for a fixed $\tau$ and for any $z$ satisfying $|z-\tau |\geq HP^2\mathcal {L}$ we have the following decay of the function in the integrand:

(4.28)\begin{equation} \exp(-H^2P^4(\tau-z)^2)\ll \frac{\exp(-\mathcal{L}^2/2)}{|z-\tau|^2+1}\ll_N \frac{P^{-N}}{|z-\tau|^2+1}. \end{equation}

Thus, in the same vein as before, using the bound (4.28) in (4.27) we may obtain

\[ I(q,t)\ll ((HP^2)^{-1}+t)\#\mathcal{H}^{-1/2}P^{n/2}q\bigg{(}\sum_{\underline{\tau}\in \underline{T}}\sum_{\underline{h}\in \tilde{\mathcal{H}}}\int_{\underline{\tau}-(HP^2)^{-1}\mathcal{L}}^{\underline{\tau}+(HP^2)^{-1}\mathcal{L}}|T_{\underline{h}}(q,\underline{z})|\, d\underline{z}\bigg{)}^{1/2}+O_{n,N}(P^{-N}). \]

The lemma now follows after using (4.24) to estimate $\#\mathcal {H}$, using the estimate $\#\underline {T}=O((1+tHP^2)^2)$, and (4.8) which allows us to take the maximum over all possible $\underline {z}$ appearing in the expression.

Since $H$ is arbitrary, we may relabel $H\mathcal {L}$ as $H$ at the expense of a factor of size at most $O_\varepsilon (P^\varepsilon )$ we can now conclude as follows.

Lemma 4.5 For any $1\leq H\ll P$, any $0<\varepsilon <1$, any $\underline {t}$ satisfying (4.11) and any $N\geq 1$ we have

\[ I(q,t)\ll_{\varepsilon,n,N} H^{-n/2+1}P^{n/2-1+\varepsilon}q((HP^2)^{-1}+t)^2\bigg{(}\max_{|\underline{z}|}\sum_{|{\underline{h}}|\ll H}|T_{\underline{h}}(q,\underline{z})|\bigg{)}^{1/2}+P^{-N}, \]

where the maximum over $\underline {z}$ is taken over the set

(4.29)\begin{equation} t\leq |\underline{z}|\leq 2(t+P^\varepsilon(HP^2)^{-1}). \end{equation}

5. Quadratic exponential sums: initial consideration

The differencing technique used in § 4 leads us to consider quadratic exponential sums $T_{\underline {h}}(q,\underline {z})$ (see (4.5)) for a family of differenced quadratic forms $F_{{\underline {h}}}$ and $G_{\underline {h}}$. Throughout this section, let $q$ denote an arbitrary but fixed integer. Our main goal here is to estimate quadratic sums corresponding to a general system of quadratic polynomials $f,g$ defined as

(5.1)\begin{equation} T(q,\underline{z}):=\sideset{}{^*}\sum_{{\underline{a}}}^q\sum_{\underline{y}\in\mathbb{Z}^n} \omega(\underline{y}/P)e((a_1/q+z_1)f(\underline{y})+(a_2/q+z_2)g(\underline{y})). \end{equation}

Here $f$ and $g$ denote a system of quadratic polynomials with integer coefficients and $\omega$ denotes a compactly supported function on $\mathbb {R}^n$. Let us denote leading quadratic parts of $f$ and $g$ by $f^{(0)}$ and $g^{(0)}$, respectively. We further assume that the quadratic forms $f^{(0)}$ and $g^{(0)}$ are defined by integer matrices $M_1$ and $M_2$, respectively. We will later apply the estimates in this section by setting $f=F_{\underline {h}}$ and $g=G_{\underline {h}}$.

Given a (finite or infinite) prime $p$, by $s_p$ we denote

(5.2)\begin{equation} s_p:=s_p(f^{(0)},g^{(0)}), \end{equation}

where, further, given a set of forms $F_1,F_2$, $s_p(F_1,F_2)$ denotes the dimension of singular locus of the projective complete intersection variety defined by the simultaneous zero locus of the forms $F_1,F_2$. That is,

\[ s_p(F_1,F_2):=\dim \big\{\underline{x}\in \mathbb{P}_{\overline{\mathbb{F}}_p}^n \, : \, F_1(\underline{x})=F_2(\underline{x})=0,\mathrm{Rank}_p (\nabla F_1(\underline{x}), F_2(\underline{x}))<2\big\}. \]

When $n\geq 2$, given an integer $q$, we define $D(q)$ by

(5.3)\begin{equation} D_{f,g}(q)=D(q):=\prod_{\substack{p\mid q\\ p \textrm{ prime }}} p^{s_p(f^{(0)},g^{(0)})+1}. \end{equation}

On the other hand, when $n=1$, we define $D(q)$ as

(5.4)\begin{equation} D(q):=(q,\mathrm{Cont}(f^{(0)}),\mathrm{Cont}(g^{(0)})), \end{equation}

where, given a polynomial $f$, $\mathrm {Cont}(f)$ is the gcd of all its coefficients.

As is standard, we begin by applying Poisson summation to $T(q,\underline {z})$. This will allow us separate the sum over ${\underline {a}}$ and the integral over $\underline {z}$, into an exponential sum and an exponential integral respectively. In particular, applying Poisson summation gives us the following.

Lemma 5.1 We have

\[ T(q,\underline{z})=q^{-n} \sum_{\underline{m}\in\mathbb{Z}} S(q;\underline{m})I(\underline{z};q^{-1}\underline{m}), \]

where

(5.5)\begin{equation} S(q; \underline{m},f,g)=S(q; \underline{m}):=\sideset{}{^*}\sum_{{\underline{a}}}^q\sum_{\underline{u} \bmod{q}} e_q(a_1f(\underline{u})+a_2g(\underline{u})+\underline{m}\cdot\underline{u}), \end{equation}

and

(5.6)\begin{equation} I(\underline{\gamma};\underline{k}):=\int_{\mathbb{R}^n}\omega(\underline{x}/P) e(\gamma_1f(\underline{x})+\gamma_2g(\underline{x})-\underline{k}\cdot \underline{x}) \,d\underline{x}. \end{equation}

Proof. The proof of Lemma 5.1 is standard and can be obtained by slightly modifying [Reference Browning and Heath-BrownBH09, Lemma 8]: let $\underline {x}=\underline {u}+q\underline {v}$. Then

\begin{align*} T(q,\underline{z})&=\sideset{}{^*}\sum_{{\underline{a}}}^q\sum_{\underline{u} \bmod{q}}\sum_{\underline{v}\in\mathbb{Z}^n} \omega((\underline{u}+q\underline{v})/P)e([a_1/q+z_1]f(\underline{u}+q\underline{v})+[a_2/q+z_2]g(\underline{u}+q\underline{v}))\\ &=\sideset{}{^*}\sum_{{\underline{a}}}^q\sum_{\underline{u} \bmod{q}} e_q(a_1f(\underline{u})+a_2g(\underline{u}))\sum_{\underline{v}\in\mathbb{Z}^n}\omega((\underline{u}+q\underline{v})/P)e(z_1f(\underline{u}+q\underline{v})+z_2g(\underline{u}+q\underline{v})). \end{align*}

We now apply Poisson summation on the second sum (and use the substitution $\underline {x}=\underline {u}+q\underline {v}$) to get

\begin{align*} T(q,\underline{z})&=\sideset{}{^*}\sum_{{\underline{a}}}^q\sum_{\underline{u} \bmod{q}} e_q(a_1f(\underline{u})+a_2g(\underline{u})) \\ &\quad\times\sum_{\underline{m}\in\mathbb{Z}^n}\int_{\mathbb{R}^n} \omega((\underline{u}+q\underline{v})/P)e(z_1f(\underline{u}+q\underline{v})+z_2g(\underline{u}+q\underline{v})-\underline{m}\cdot \underline{v})\, d\underline{v}\\ &=q^{-n}\sum_{\underline{m}\in\mathbb{Z}^n}\sideset{}{^*}\sum_{{\underline{a}}}^q\sum_{\underline{u} \bmod{q}} e_q(a_1f(\underline{u})+a_2g(\underline{u})+\underline{m}\cdot \underline{u})\\ &\quad \times\int_{\mathbb{R}^n} \omega(\underline{x}/P)e(z_1f(\underline{x})+z_2g(\underline{x})-q^{-1}\underline{m}\cdot \underline{x}),\, d\underline{x} \end{align*}

as required.

As a result, we trivially have the following pointwise bound

(5.7)\begin{equation} |T(q,\underline{z})|\leq q^{-n}\sum_{\underline{m}\in\mathbb{Z}} |S(q;\underline{m})|\cdot |I(\underline{z};q^{-1}\underline{m})|. \end{equation}

The treatment of the exponential integral is standard. In particular, upon letting $\|f\|$ denote the supremum of absolute values of its coefficients and defining

(5.8)\begin{equation} \|f\|_P:=\|P^{-\deg(f)}f(Px_1,\ldots,Px_n)\|, \end{equation}

we can use the following lemma to bound $I(\underline {z};q^{-1}\underline {m})$.

Lemma 5.2 Let $f,g$ be quadratic polynomials such that $\max \{\|f\|_P,\|g\|_P\}\ll H$. Furthermore, let $V:=1+q P^{\varepsilon -1}\max \{1,HP^2|\underline {z}|\}^{1/2}$, $\varepsilon >0$, and $N\in \mathbb {N}$. Then

\[ I(\underline{z};q^{-1}\underline{m})\ll_N P^{-N}+ \mathrm{meas}(\{\underline{y}\in P\,\mathrm{Supp}(\omega_{\underline{h}}):\: |\nabla \hat{f}_{\underline{z}}(\underline{y})-\underline{m}|\leq V\}), \]

where

\[ \hat{f}_{\underline{z}}(\underline{x}):=qP^{-1}z_1f(\underline{x})+qP^{-1}z_2g(\underline{x}). \]

Furthermore, if $|\underline {m}|\geq q P^{\epsilon -1}\max \{1,HP^2|\underline {z}|\}$, then we have

\[ I(\underline{z};q^{-1}\underline{m})\ll_N P^{-N}|\underline{m}|^{-N}. \]

The proof of this is almost identical to the proofs of [Reference Browning, Dietmann and Heath-BrownBDH15, Lemmas 6.5 and 6.6], and so we will not provide details here. In particular, the only thing in the proofs that needs to be tweaked in order to verify Lemma 5.2 is that $\Theta$ in [Reference Browning, Dietmann and Heath-BrownBDH15, equation (6.11)] must be replaced with

\[ \Theta':=1+|z_1|HP^2+|z_2|HP^2. \]

We also note that we use $|\nabla \hat {f}_{\underline {z}}(\underline {y})-\underline {m}|\leq V$ instead of $Pq^{-1}|\nabla \hat {f}_{\underline {z}}(\underline {y})-\underline {m}|\leq Pq^{-1}V$ since we are using slightly different notation.

The latter bound enables us to handle the tail of the sum over $\underline {m}$. Let $\hat {V}:=q P^{\epsilon -1}\max \{1,HP^2|\underline {z}|\}$. By trivially bounding $|S(q;\underline {m})|$ by $q^n$, and setting $N\geq n+2$, it is easy to show that

\[ q^{-n}\sum_{|\underline{m}|\gg \hat{V}} |S(q;\underline{m})|\cdot |I(\underline{z};q^{-1}\underline{m})|\ll 1, \]

by the second half of Lemma 5.2. Hence,

\[ \implies |T(q,\underline{z})|\ll 1+q^{-n}\sum_{|\underline{m}|\ll \hat{V}} |S(q;\underline{m})|\cdot |I(\underline{z};q^{-1}\underline{m})|. \]

Now by the first half of Lemma 5.2 (setting $N\geq n+4$), we have

\begin{align*} |T_{\underline{h}}(q,\underline{z})|&\ll 1+q^{-n}\sum_{|\underline{m}|\ll \hat{V}} |S(q;\underline{m})|\cdot \mathrm{meas}(\{\underline{y}\in P\,\mathrm{Supp}(\omega)\,:\, |\nabla \hat{f}_{\underline{z}}(\underline{y})-\underline{m}|\leq V\}\\ &= 1+q^{-n}\sum_{|\underline{m}|\ll \hat{V}} |S(q;\underline{m})|\int_{\underline{y}\in P\,\mathrm{Supp}(\omega)} \mathrm{Char}(\underline{m},\underline{y})\, d\underline{y}, \end{align*}

where

\[ \mathrm{Char}(\underline{m},\underline{y})=\begin{cases} 1 & \text{if }\; |\nabla \hat{f}_{\underline{z}}(\underline{y})-\underline{m}|\leq V,\\ 0 & \text{else,} \end{cases} \]

\begin{align*} \implies |T_{\underline{h}}(q,\underline{z})|&\ll 1+q^{-n}\int_{\underline{y}\in P\,\mathrm{Supp}(\omega)}\sum_{\substack{|\underline{m}|\ll \hat{V}\\|\nabla \hat{f}_{\underline{z}}(\underline{y})-\underline{m}|\leq V}}|S(q;\underline{m})| \,d\underline{y}\\ &\ll 1+q^{-n}\int_{\underline{y}\in P\,\mathrm{Supp}(\omega)}\sum_{|\underline{m}-\underline{m}_0(\underline{y})|\leq V}|S(q;\underline{m})| \,d\underline{y}. \end{align*}

where $\underline {m}_0(\underline {y}):=\nabla \hat {f}_{\underline {z}}(\underline {y})$. Hence, we have the following.

Proposition 5.3 Let $|\underline {z}|=\max \{|z_1|,|z_2|\}$. Then for any $q\in \mathbb {N}$,

\[ |T(q,\underline{z})|\ll 1+q^{-n}P^n\sup_{\underline{y}\in P\,\mathrm{Supp}(\omega)}\bigg{\{}\sum_{|\underline{m}-\underline{m}_0(\underline{y})|\leq V}|S(q;\underline{m})| \bigg{\}}, \]

for some $\underline {m}_0(\underline {y})$, where

(5.9)\begin{gather} V:=1+q P^{-1+\varepsilon}\max\{1,HP^2|\underline{z}|\}^{1/2}. \end{gather}

Our attention now turns to finding a suitable bound for $|S(q;\underline {m})|$. As is standard when dealing with exponential sum bounds, we will take advantage of the multiplicative property of $S(q;\underline {m})$ and decompose $q$ into its square-free, square and cube-full components so that we can use better bounds in the former two cases (in particular, we will make use of the ${\underline {a}}$ sum to improve our bounds in the former cases). Indeed, we may use a lemma of Hooley [Reference HooleyHoo78, Lemma 3.2] to get the following result.

Lemma 5.4 Let $\underline {a}\in \mathbb {Z}^2$ such that $(q,\underline {a})=1$, $q=rs$ where $(r,s)=1$ and $\underline {m}\in \mathbb {Z}^n$. Then

(5.10)\begin{equation} S(rs; \underline{m})=S(r; \bar{s}\underline{m})S(s;\bar{r}\underline{m}), \end{equation}

where $r\bar {r}+s\bar {s}=1$.

The above lemma is proved using a very standard argument akin to [Reference Browning and Heath-BrownBH09, Lemma 10] and [Reference Marmon and VisheMV19, Lemma 4.5], and therefore we will skip its proof here. Our treatment of bounds for the quadratic exponential sums will vary depending on whether $q$ is square-free, a square or cube-full. Since the exponential sums satisfy the mutliplicativity relation (5.10), it is natural to set $q=b_1b_2q_3$ where

(5.11)\begin{equation} b_1:=\prod_{p||q}p,\quad b_2:=\prod_{p^2||q}p^2,\quad q_3:=\prod_{\substack{p^e||q\\ e>2}}p^e. \end{equation}

Then by Lemma 5.4, we have that

(5.12)\begin{equation} S(q; \underline{m})=S(b_1; c_1\underline{m})S(b_2; c_2\underline{m})S(q_3; c_3\underline{m}), \end{equation}

for some constants $c_1,c_2,c_3$ such that $(b_1,c_1)=(b_2,c_2)=(q_3,c_3)=1$. Finding suitable bounds for the size of these three exponential sums will be the topic of the rest of this section.

5.1 Square-free exponential sums

In this section, we will briefly consider the quadratic exponential sums $S(b_1;\underline {m})$ when $q=b_1$ is square-free. This case is extensively studied in [Reference Marmon and VisheMV19, Section 5], where bounds are obtained for exponential sums for a general system of polynomials $f$ and $g$. Using the multiplicativity of the exponential sum in (5.10), it is enough to consider the sums $S(p,\underline {m})$ where $p$ is a prime. We may rewrite

(5.13)\begin{equation} S(p,\underline{m})=\Sigma_1-\Sigma_4, \end{equation}

where

(5.14)\begin{equation} \Sigma_1:=\sum_{a_1=1}^p\sum_{a_2=1}^p \sum_{\underline{u} \bmod{q}} e_p(a_1f(\underline{u})+a_2g(\underline{u})+\underline{m}\cdot\underline{u})\quad\textrm{ and }\quad\Sigma_4:=\sum_{\underline{u} \bmod{q}} e_p(\underline{m}\cdot\underline{u}). \end{equation}

Here the notation $\Sigma _1$ and $\Sigma _4$ is used to correspond to the corresponding sums in [Reference Marmon and VisheMV19, Section 5]. Note that the argument in [Reference Marmon and VisheMV19, Section 5] does not depend on the degree of the forms $f$ and $g$. In fact, our exponential sums are more ‘natural’ than those which appear in [Reference Marmon and VisheMV19] and, as a result, only sums $\Sigma _1$ and $\Sigma _4$ appear in our analysis. We may now use the results in [Reference Marmon and VisheMV19, Section 5] directly here as they do indeed bound the sums $\Sigma _1$ and $\Sigma _4$ as well, but only in the case where $f^{(0)}$ and $g^{(0)}$ intersect properly over $\overline {\mathbb {F}}_p$. When $n\geq 2$, we may use [Reference Marmon and VisheMV19, Prop. 5.2, Lemma 5.4] to get the following.

Proposition 5.5 Let $f,g\in \mathbb {Z}[x_1,\ldots,x_n]$ be quadratic polynomials such that $s_{\infty }(f^{(0)},g^{(0)})=-1$. Let $b_1$ be a square-free number. If $n>1$, then there exists some $\Phi _{f,g}=\Phi \in \mathbb {Z}[x_1,\ldots,x_n]$ such that

\begin{align*} S(b_1,\underline{m})\ll_n b_1^{1+n/2+\varepsilon}D(b_1)(b_1,\Phi(\underline{m}))^{1/2} \end{align*}

for every $\underline {m}\in \mathbb {Z}^n$. Furthermore, $\Phi$ has the following properties:

1. (1) $\Phi$ is homogeneous;

2. (2) $\deg (\Phi )\ll _n 1$;

3. (3) $\log \|\Phi \|\ll _n \log \|f\| + \log \|g\|$;

4. (4) $\mathrm {Cont}(\Phi )=1$.

Note that all the implied constants here only depend on $n$ and are independent of $\|f\|$ and $\|g\|$.

Proof. To begin, since $s_p(f^{(0)},g^{(0)})=-1$, we may use a $\mathbb {Q}$ version of the dual variety explicitly described in [Reference VisheVis23, Lemma 4.2] to see that the polynomial defining the dual variety of the intersection variety of $f^{(0)},g^{(0)}$ satisfies the four conditions for $\Phi$ in the statement of this proposition. Hence, we may let $\Phi$ be this polynomial. This allows us to improve the first assertion of [Reference Marmon and VisheMV19, Proposition 5.2]: indeed, if we let $\delta _p(\underline {v}):=s_p(f^{(0)},g^{(0)},L_{\underline {v}})$, where $L_{\underline {v}}$ is the hyperplane defined by $\underline {v}$, then $\delta _p(\underline {v})\leq s_p(f^{(0)},g^{(0)})$ whenever $p\nmid \Phi (\underline {v})$. We automatically get this since

\[ s_p(f^{(0)},g^{(0)},L_{\underline{v}})\leq s_p(f^{(0)},g^{(0)}) \]

for every $\underline {v}$ not on the dual variety of $f^{(0)},g^{(0)}$ (over $\overline {\mathbb {F}}_p$), and we must have $p\mid \Phi (\underline {v})$ when $\underline {v}$ is on the dual variety by our choice of $\Phi$.

Note that if $f^{(0)},g^{(0)}$ intersect properly, then the bounds in [Reference Marmon and VisheMV19, Lemmas 5.1 and 5.4] hold for all polynomials $f,g$ regardless of any condition on their degrees. The key difference to the situation here is that, in the case when we have improper intersection, we define the singular locus differently to [Reference Marmon and VisheMV19]. This is due to both of our polynomials varying here, whilst one of the corresponding polynomials in [Reference Marmon and VisheMV19] is fixed. In particular, in this case our singular locus can either be $n-2$ or $n-1$ as discussed in the proof of Lemma 2.1, while in [Reference Marmon and VisheMV19] in the case of improper intersection of top forms, the singular locus is defined to be $n-1$ uniformly. Our proof here will follow that of [Reference Marmon and VisheMV19, Lemma 5.4].

In the case when quadratic polynomials $f^{(0)},g^{(0)}$ intersect properly over $\overline {\mathbb {F}}_p$, [Reference Marmon and VisheMV19, Lemma 5.4] (and our improvement to [Reference Marmon and VisheMV19, Proposition 5.2]) goes through handing us

(5.15)\begin{equation} S(p,\underline{m})\ll_n p^{1+n/2+\varepsilon}p^{(s_p(f^{(0)},g^{(0)})+1)/2}(p,\Phi(\underline{m}))^{1/2} =p^{1+n/2+\varepsilon}D(p)^{1/2}(p,\Phi(\underline{m}))^{1/2}. \end{equation}

It now remains to consider the case when $f^{(0)}$ and $g^{(0)}$ intersect improperly in greater detail. In each of the cases of improper intersection of $f^{(0)}$ and $g^{(0)}$, the singular locus $s_{p}(f^{(0)},g^{(0)})\geq n-2$. We therefore note that the trivial bound

\[ \Sigma_4\ll p^n\ll p^{1+s_p(f^{(0)},g^{(0)})+1}=pD(p) \]

suffices for every $n\geq 2$. We now turn our attention to $\Sigma _1$. We will first show that

\[ |\Sigma_1|\ll p^2D(p) \]

in the case that $f^{(0)}$ and $g^{(0)}$ intersect improperly over $\overline {\mathbb {F}}_p$. In the case where $n>1$, there are two cases to consider: $s_{p}(f^{(0)},g^{(0)})= n-1$ and $s_{p}(f^{(0)},g^{(0)})= n-2$ (see the proof of Lemma 2.1). In the former case, we may again use the trivial bound:

\[ |\Sigma_1|\ll p^{2+n}=p^{2+s_p(f^{(0)},g^{(0)})+1}=p^2D(p). \]

When $s_{p}(f^{(0)},g^{(0)})= n-2$, we instead note that

\begin{align*} |\Sigma_1|&=\bigg{|}p^2\sum_{\substack{\underline{x}\bmod{p}\\f(\underline{x})\equiv g(\underline{x})\equiv 0 \bmod{p}}} e_p(\underline{m}\cdot\underline{x})\bigg{|}\nonumber\\ &\leq p^2 \#\{\underline{x}\in \mathbb{F}_p^n \, : \, f(\underline{x})=g(\underline{x})=0\}\\ &\ll p^{2+n-1}=p^{2+s_{p}(f^{(0)},g^{(0)})}=p^2D(p). \end{align*}

Here, we could bound $\#\{\underline {x}\in \mathbb {F}_p^n \, : \, f(\underline {x})=g(\underline {x})=0\}$ by $O(p^{n-1})$ due to the fact that

\[ s_{p}(f^{(0)},g^{(0)})= n-2 \implies f\not\equiv 0 \text{ or } g\not\equiv 0. \]

Hence, we have shown that when $f^{(0)}$, $g^{(0)}$ intersect improperly, we have

\[ |\Sigma_1|\ll p^2D(p)\leq p^{1+n/2}D(p), \]

provided that $n\geq 2$, as required. Therefore, we may conclude that for a general $p$ (irrespective of whether or not the intersection is proper)

\[ S(p,\underline{m})\leq C(n) p^{1+n/2+\varepsilon}D(p)(p,\Phi(\underline{m}))^{1/2}, \]

where $C$ is some constant. Finally, by Lemma 5.4, we have

\begin{align*} S(b_1,\underline{m})&=\prod_{p\mid b_1} S(p,c_p\underline{m})\\ &\leq C(n)^{d(b_1)} b_1^{1+n/2+\varepsilon}D(b_1)\prod_{p\mid b_1}(p,\Phi(c_p\underline{m}))^{1/2}\\ &=C(n)^{d(b_1)} b_1^{1+n/2+\varepsilon}D(b_1)(b_1,\Phi(\underline{m}))^{1/2}, \end{align*}

where $d(b_1):=\#\{p\mid b_1\}$ is the divisor function of $b_1$. We could replace $(p,\Phi (c_p\underline {m}))$ with $(b_1,\Phi (\underline {m}))$ because $\Phi$ is homogeneous and $(p,c_p)=1$. All that is left to do is show that $C(n)^{d(b_1)}$ does not contribute more than $O(P^{\varepsilon })$. To see this, we note that $d(b_1)\ll \log (b_1)/\log \log (b_1)$. Hence, there is some constant $d$ such that

\[ C(n)^{d(b_1)}\leq C(n)^{d\log(b_1)/\log\log(b_1)}\ll b_1^{d\log(C(n))/\log\log(b_1)}\ll b_1^{\varepsilon} \]

provided that $b_1\gg _{\varepsilon } 1$. We automatically have $d(b_1)\ll 1$ if $b_1\not \gg 1$, so we get $c^{d(b_1)}\ll 1\ll b_1^{\varepsilon }$ in that case. Hence, we may conclude that Proposition 5.5 is true. We will bound the $C(n)$ term in future lemmas by $b_1^{\epsilon }$ without further comment.

We also must consider when $n=1$. In this case, it is sufficient for us to use a weaker bound than [Reference Marmon and VisheMV19, Lemma 5.5]. We will show the following.

Proposition 5.6 Let $f,g\in \mathbb {Z}[x]$ be quadratic polynomials and let $b_1$ be a square-free integer. Then

\[ S(b_1,m)\ll b_1^{2+\varepsilon}D(b_1). \]

Proof. The proof of Proposition 5.6 is almost trivial. We start by applying Lemma 5.4 so that we may consider $S(p; cm)$ for some $p\nmid c$. We note that

\[ |\Sigma_1|= p^2 \#\{x \ {\rm mod}\ p \, :\, f(x)\equiv g(x) \equiv 0 \ {\rm mod}\ p\}\ll p^2(p,\mathrm{Cont}(f),\mathrm{Cont}(g)), \]

and we trivially have $|\Sigma _4|\leq p$. Hence, by (5.4) and noting that $(p,\mathrm {Cont}(f),\mathrm {Cont}(g))\leq (p,\mathrm {Cont}(f^{(0)}),\mathrm {Cont}(g^{(0)}))$:

\[ |S(p;cm)|\leq |\Sigma_1|+|\Sigma_4|\ll p^2D(p), \]

and so

\[ |S(b_1;m)|\ll b_1^{2+\varepsilon} D(b_1) \]

for any $m\in \mathbb {Z}$.

5.2 Square-full bound

In this section, we will derive the bound which will be used when $q$ is square-full. When $q$ is square-full, we give up on saving $q$ over the ${\underline {a}}$ sum, and instead start with the bound

(5.16)\begin{equation} |S(q;\underline{m})|\leq \sideset{}{^*}\sum_{{\underline{a}}}^q |S({\underline{a}},q;\underline{m})|, \end{equation}

where $f,g$ are quadratic polynomials, and

\[ S(\underline{a}, q; \underline{m}):=\sum_{\underline{x} \bmod{q}} e_{q}(a_1f(\underline{x})+a_2g(\underline{x})+\underline{m}\cdot\underline{x}). \]

For a fixed value of ${\underline {a}}$, the exponential sum $S(\underline {a}, q; \underline {m})$ is a standard quadratic exponential sum with leading quadratic part defined by the matrix

(5.17)\begin{equation} M({\underline{a}}):=M:=a_1M_1+a_2M_2. \end{equation}

We will assume further that $2\,|\, (\mathrm {Cont}(f^{(0)}),\mathrm {Cont}(g^{(0)}))$ so that $M({\underline {a}})\in M_n(\mathbb {Z})$ for every ${\underline {a}}$.

Remark 5.7 In the broader context of the argument that we are building, the reason why we may assume that $2\,|\, (\mathrm {Cont}(f^{(0)}),\mathrm {Cont}(g^{(0)}))$ is due to Remark 3.1: if the coefficients of our original cubic forms in § 3 are divisible by $2$, then the coefficients of the differenced quadratic polynomials coming from § 4 must also be divisible by $2$.

A standard squaring argument as obtained in [Reference VisheVis23, Lemma 2.5], for example, readily hands us a bound

(5.18)\begin{equation} |S(\underline{a}, q; \underline{m})|\ll q^{n/2}\#\mathrm{Null}_q(M)^{1/2}, \end{equation}

where $\#\mathrm {Null}_q(M)$ denotes the number of solutions of the equation $M\underline {x}\equiv \underline {\mathrm {0}}\bmod {q}$ as defined in (2.14). To estimate this, we will resort to using a Smith normal form of the matrix $M$. The Smith normal form of $M$ hands us invertible integer matrices $S$ and $T$ be with determinant $\pm 1$ such that

(5.19)\begin{equation} SMT=\mathrm{Smith}(M)=\begin{pmatrix} \lambda_1 & 0 & 0 & \cdots & 0\\ 0 & \lambda_2 & 0 & \cdots & 0 \\ 0 & 0 & \ddots & & \vdots\\ \vdots & \vdots & & \ddots\\ 0 & 0 & \cdots & & \lambda_n \end{pmatrix}\in M_n(\mathbb{Z}), \end{equation}

where $\lambda _1\mid \lambda _2\mid \cdots \mid \lambda _n$. Since the forms $f^{(0)}$ and $g^{(0)}$ are assumed to be arbitrary for now, it is easy to conclude that

(5.20)\begin{equation} |S(\underline{a}, q; \underline{m})|\ll q^{n/2}\prod_{i=1}^n \lambda_{q,i}^{1/2}, \end{equation}

where

(5.21)\begin{equation} \lambda_{q,i}:=(q,\lambda_i). \end{equation}

Remark 5.8 Recall that we aim to finally substitute $f=F_{\underline {h}}$ and $g=G_{\underline {h}}$. Note that the extra factor appearing on the right-hand side of (5.20) is a generalisation of the factor $D(b_1)^{1/2}$ appearing in Proposition 5.5. This is a drawback of van der Corput differencing that although one starts with a nice pair of forms $F$ and $G$, one ends up with exponential sums of differenced polynomials $F_{\underline {h}}$ and $G_{\underline {h}}$, which can be highly singular modulo $q$. If $q=p^\ell$ for some prime $p$, if the singular locus $s_p$ as defined in (5.2) is large, then this gives restrictions on the vector ${\underline {h}}\bmod {p}$. When $\ell$ is small, the extra factors appearing can be compensated from the corresponding bounds on the ${\underline {h}}$ sum. However, in the case at hand, when $q=p^\ell$ for a large $\ell$, we cannot rule out the possibility that for many ${\underline {h}}$, there may exist a large $q$ such that the factor $\prod _{i=1}^n \lambda _{q,i}^{1/2}$ is as large as $q^{n/2}$. This complication arises partly due to the simplicity of the quadratic exponential sums appearing. However, later we would need to average the sums over various $|\underline {m}-\underline {m}_0|\leq V$. We will aim to salvage some of this loss by gaining a congruence condition on $\underline {m}$ instead and saving from the sum over $\underline {m}$. This idea partly has already featured in Vishe's work [Reference VisheVis23, Lemma 6.4]. However, in [Reference VisheVis23], the authors are dealing with fixed $f$ and $g$, which is not the case here.

Our main goal here is to prove the following result.

Proposition 5.9 Let ${\underline {a}}\in \mathbb {Z}^2$ and $q\in N$ be such that $({\underline {a}},q)=1$, let $\underline {m}\in \mathbb {Z}^n$ and let $f,g$ be quadratic polynomials such that $2\,|\, (\mathrm {Cont}(f^{(0)}),\mathrm {Cont}(g^{(0)}))$. Furthermore, let

(5.22)\begin{equation} (a_1f+a_2g)(\underline{x})=\underline{x}^t M\underline{x}+\underline{\mathfrak{b}}\cdot\underline{x}+\mathfrak{c}. \end{equation}

Then

\[ |S(\underline{a}, q; \underline{m})|\leq 2^{n/2}q^{n/2}\#\mathrm{Null}_q(M)^{1/2}\Delta_{q}(\underline{m}+\underline{\mathfrak{b}}), \]

where

(5.23)\begin{equation} \Delta_q(\underline{m}):=\Delta_{T,q}(\underline{m}):=\begin{cases} 1 & \text{if } \lambda_{q,i} \mid (T^t\underline{m})_i\textrm{ for } 1\leq i \leq n,\\ 0 & \text{else.} \end{cases} \end{equation}

Here, $T$ is the matrix appearing in the Smith normal form of $M$ in (5.19), the $\lambda _{q,i}$ are defined in (5.21) and given a vector $\underline {v}$, $(\underline {v})_i$ denotes its $i$th component.

Proof. To estimate $|S({\underline {a}},q;\underline {m})|$, we begin by working with its square:

\begin{align*} |S(\underline{a},q;\underline{m})|^2&= \sum_{\underline{x},\underline{y} \bmod{q}} e_{q}((a_1f+a_2g)(\underline{x})+\underline{m}\cdot\underline{x})\overline{e_{q}((a_1f+a_2g)(\underline{y})+\underline{m}\cdot\underline{y})}\\ &= \sum_{\underline{x},\underline{y} \bmod{q}} e_{q}(\underline{x}^t M\underline{x}-\underline{y}^t M\underline{y}+(\underline{m}+\underline{\mathfrak{b}})\cdot(\underline{x}-\underline{y})). \end{align*}

We will now change order of summation by setting $\underline {x}=\underline {y}+\underline {z}$. Then

\begin{align*} |S(\underline{a},q;\underline{m})|^2&= \sum_{\underline{y},\underline{z} \bmod{q}} e_{q}(\underline{z}^t M\underline{z}+(\underline{m}+\underline{\mathfrak{b}})\cdot\underline{z} +2\underline{y}^t M \underline{z})\\ &=\sum_{\underline{z} \bmod{q}} e_{q}(\underline{z}^t M\underline{z}+\underline{m}'\cdot\underline{z})\sum_{\underline{y} \bmod{q}} e_{q}(\underline{y}\cdot 2M\underline{z}), \end{align*}

where $\underline {m}'=\underline {m}+\underline {\mathfrak {b}}$. Therefore,

(5.24)\begin{equation} |S(\underline{a},q;\underline{m})|^2=q^{n}\sum_{\underline{z} \bmod{q}} e_{q}(\underline{z}^t M \underline{z}+\underline{m}'\cdot\underline{z})\delta_{2M}(\underline{z}), \end{equation}

where

(5.25)\begin{equation} \delta_{M}(\underline{z}):=\begin{cases} 1 & \text{if } M\underline{z}\equiv 0\mod q,\\ 0 & \text{otherwise.} \end{cases} \end{equation}

The ‘2’ appearing in $\delta _{2M}(\underline {z})$ gives rise to some minor technical difficulties in the case when $q$ is even. Therefore, we will start by first considering the case when $q$ is odd.

5.2.1 Case: $q$ odd

In this case, $\delta _{2M}(\underline {z})=1$ if and only if $M\underline {z}\equiv \underline {0} \mod {q}$, and so we may replace $\delta _{2M}(\underline {z})$ in (5.24) by $\delta _M(\underline {z})$. Furthermore, we note that $M\underline {z}\equiv \underline {0} \mod {q}$ implies that $\underline {z}^t M\underline {z}\equiv \underline {0} \mod {q}$. Hence, (5.24) simplifies as

(5.26)\begin{equation} |S(\underline{a},q;\underline{m})|^2=q^{n}\sum_{\underline{z} \bmod{q}} e_{q}(\underline{m}'\cdot\underline{z})\delta_{M}(\underline{z}). \end{equation}

Now, $M$ has a Smith normal form $\mathrm {Smith}(M):=SMT$, for some matrices $S,T\in SL_n(\mathbb {Z})$. In particular, matrices $S$ and $T$ are invertible modulo $q$, for any $q\in \mathbb {N}$.

First, we note that

\[ \delta_M=\delta_{SM}. \]

Therefore, on using the substitution $\underline {z}\mapsto T^{-1}\underline {z}$, (5.24) becomes

(5.27)\begin{equation} |S(\underline{a},q;\underline{m})|^2=q^{n}\sum_{\underline{z} \bmod{q}} e_{q}(\underline{m}'\cdot T\underline{z})\delta_{SMT}(\underline{z}), \end{equation}

since $\delta _{SM}(T\underline {z})=\delta _{SMT}(\underline {z})$ by (5.25). We will now work towards determining which $\underline {z}$ make $\delta _{SMT}(\underline {z})$ non-zero. By definition, $\delta _{SMT}(\underline {z})\neq 0$ if and only if

\[ SMT\underline{z}\equiv \underline{0} \mod q, \]

or, equivalently,

\[ \underline{z}\in \mathrm{Null}_q(SMT):=\{\underline{x}\in {(}\mathbb{Z}/q\mathbb{Z}{)}^n \mid SMT\underline{x} \equiv \underline{0} \ {\rm mod}\ q\}. \]

Hence, we may simplify (5.27) as follows:

(5.28)\begin{align} |S(\underline{a},q;\underline{m})|^2&=q^{n}\sum_{\underline{z}\in \mathrm{Null}_q(SMT)} e_{q}(\underline{m}'\cdot T\underline{z})\nonumber\\ &=q^{n}\sum_{\underline{z}\in \mathrm{Null}_q(SMT)} e_{q}(\underline{z}\cdot T^t\underline{m}'), \end{align}

where $T^t$ is the transpose of $T$. This is true because

\[ \underline{m}'\cdot T\underline{z}=(T \underline{z})^t\underline{m}'=\underline{z}^t T^t\underline{m}'=\underline{z}\cdot T^t\underline{m}'. \]

We now turn our attention to the structure of the $\mathrm {Null}_q(SMT)$. Since $S$ and $T$ are defined to be the unique matrices (up to units) such that $SMT=\mathrm {Smith}(M)$, it is quite easy to determine precisely when $\underline {z}\in \mathrm {Null}_q(SMT)$. Therefore, $SMT\underline {z}\equiv \underline {0} \mod q$ if and only if

(5.29)\begin{equation} \frac{q}{\lambda_{q,i}}\bigg| z_i \end{equation}

for every $i\in \{1,\ldots, n\}$. Therefore,

(5.30)\begin{equation} \#\mathrm{Null}_q(SMT)=\prod_{i=1}^n \lambda_{q,i}. \end{equation}

Hence, by (5.21) and (5.28)–(5.29), we have the following:

(5.31)\begin{equation} |S(\underline{a},q;\underline{m})|^2=q^{n}\prod_{i=1}^n\sum_{({q}/{\lambda_{q,i}})\mid z_i}e_q(z_i(T^t\underline{m}')_i)=q^{n}\prod_{i=1}^n\sum_{x_i=1}^{\lambda_{q,i}}e_{\lambda_{q,i}}(x_i(T^t\underline{m}')_i) =q^{n}\prod_{i=1}^n \lambda_{q,i} \delta_{q,i}(\underline{m}'), \end{equation}

where

(5.32)\begin{equation} \delta_{q,i}(\underline{u}):=\begin{cases} 1 & \text{if } \lambda_{q,i} \mid (T^t\underline{u})_i,\\ 0 & \text{otherwise,} \end{cases} \end{equation}

and $(\underline {v})_i$ is the $i$th component of vector $\underline {v}$. Therefore, by (5.30) and (5.31):

\[ |S(\underline{a},q;\underline{m})|^2=q^{n}\#\mathrm{Null}_q(SMT)\prod_{i=1}^n\delta_{q,i}(\underline{m}'). \]

Finally, it is easy to check that

\[ \#\mathrm{Null}_q(SMT)=\#\mathrm{Null}_q(M), \]

since $S$ and $T$ are both invertible over $\mathbb {Z}/q\mathbb {Z}$ and, therefore, in this case we establish

\[ |S(\underline{a}, q; \underline{m})|= q^{n/2}\#\mathrm{Null}_q(M)^{1/2}\Delta_{q}(\underline{m}+\underline{\mathfrak{b}}), \]

which clearly suffices.

5.2.2 Case: $q$ even

We now turn to the case where $q$ is even. In this case, the above argument needs to be modified due to not being able to directly replace the condition $\delta _{2M}(\underline {z})$ with $\delta _M(\underline {z})$ in (5.24). Instead, we note that $\delta _{2M}(\underline {z})\neq 0$ if and only if $M\underline {z}\equiv \underline {0} \mod q/2$. In particular, there must be some $\underline {c}\in \{0,1\}^n$ such that

\[ M\underline{z} \equiv \frac{q}{2}\underline{c} \mod{q}. \]

Therefore, if we let

\[ N_{\underline{c},q}(M):=\bigg\{\underline{x}\ {\rm mod}\ q \, : \, M\underline{x} \equiv \frac{q}{2}\underline{c}\ {\rm mod}\ {q}\bigg\}, \]

then $\delta _{2M(\underline {z})}\neq 0$ if and only if $\underline {z}\in N_{\underline {c},q}$ for some $\underline {c}$. Hence, we may rewrite (5.24) as follows:

(5.33)\begin{equation} |S(\underline{a}, q; \underline{m})|^2= q^n\sum_{\underline{c}\in \{0,1\}^n} \sum_{\underline{z}\in N_{\underline{c},q}(M)} e_q(\underline{z}^t M \underline{z} + \underline{m}'\cdot \underline{z}). \end{equation}

We now wish to write $N_{\underline {c},q}$ in terms of $\mathrm {Null}_q(M)$ as this will enable us to use the arguments discussed in the odd case. To do this, we invoke Lemma 2.6 to see that either $N_{\underline {c},q}=\emptyset$ or there exists some $\underline {y}_{\underline {c}}\in (\mathbb {Z}/q\mathbb {Z})^n$ such that

\[ N_{\underline{c},q}= \underline{y}_{\underline{c}}+ \mathrm{Null}_q(M). \]

Hence,

(5.34)\begin{align} |S(\underline{a}, q; \underline{m})|^2&= q^n\sum_{\substack{\underline{c}\in \{0,1\}^n\\ N_{\underline{c},q}(M)\neq \emptyset}} \sum_{\underline{z}\in \underline{y}_{\underline{c}} + \mathrm{Null}_{q}(M)} e_q(\underline{z}^t M \underline{z} + \underline{m}'\cdot \underline{z})\nonumber\\ &=q^n\sum_{\substack{\underline{c}\in \{0,1\}^n\\ N_{\underline{c},q}(M)\neq \emptyset}} \sum_{\underline{z}\in \mathrm{Null}_{q}(M)} e_q([\underline{y}_{\underline{c}}+\underline{z}]^t M [\underline{y}_{\underline{c}}+\underline{z}] + \underline{m}'\cdot [\underline{y}_{\underline{c}}+\underline{z}])\nonumber\\ &= q^n\sum_{\substack{\underline{c}\in \{0,1\}^n\\ N_{\underline{c},q}(M)\neq \emptyset}} e_q(\underline{y}_{\underline{c}}^t M \underline{y}_{\underline{c}} + \underline{m}'\cdot \underline{y}_{\underline{c}}) \sum_{\underline{z}\in \mathrm{Null}_{q}(M)} e_q((\underline{z}+2\underline{y}_{\underline{c}})^t M \underline{z} + \underline{m}'\cdot \underline{z})\nonumber\\ &\leq q^n\sum_{\underline{c}\in \{0,1\}^n} \bigg{|} \sum_{\underline{z}\in \mathrm{Null}_{q}(M)} e_q((\underline{z}+2\underline{y}_{\underline{c}})^t M \underline{z} + \underline{m}'\cdot \underline{z})\bigg{|}. \end{align}

Finally, we note that $M\underline {z}\equiv \underline {0} \mod q$ since $\underline {z}\in \mathrm {Null}_q(M)$, and so by (5.34), we have the following:

\begin{align*} |S(\underline{a}, q; \underline{m})|^2 &\leq q^n\sum_{\underline{c}\in \{0,1\}^n} \bigg{|} \sum_{\underline{z}\in \mathrm{Null}_{q}(M)} e_q(\underline{m}'\cdot \underline{z})\bigg{|}\\ &=2^n q^n \bigg{|} \sum_{\underline{z} \bmod{q}} e_q(\underline{m}'\cdot \underline{z}) \delta_{M}(\underline{z})\bigg{|}. \end{align*}

This is precisely (5.26) with an extra factor of $2^n$ and some absolute value signs around the sum (which are irrelevant). We may therefore repeat the arguments in the $q$ odd case which follow from (5.26) to establish Proposition 5.9.

5.2.3 Special case: $n=1$

We will now briefly consider the case when $n=1$, as we will need to deal with this case separately later. The arguments used above are still valid in this case, but the bound that we get is simpler due to the matrix, $M$, becoming an integer. In particular, Proposition 5.9 becomes the following.

Proposition 5.10 Let ${\underline {a}}\in \mathbb {Z}^2$ and $q\in N$ be such that $({\underline {a}},q)=1$, let $m\in \mathbb {Z}$ and let $f,g\in \mathbb {Z}[x]$ be quadratic polynomials such that $2\,|\, (\mathrm {Cont}(f^{(0)}),\mathrm {Cont}(g^{(0)}))$. Let

(5.35)\begin{equation} (a_1f+a_2g)(x)=Mx^2+bx+c. \end{equation}

Then

\[ |S(\underline{a}, q; m)|\leq 2^{1/2}q^{1/2}(q,M)^{1/2}\Delta_{q}'(m+b), \]

where

(5.36)\begin{equation} \Delta_{q}'(m):=\begin{cases} 1 & \text{if } (q,M) \mid m,\\ 0 & \text{otherwise.} \end{cases} \end{equation}

We will use Propositions 5.9 and 5.10 directly in our future treatment of the cube-full part of $S(q_3;\underline {m})$ (see (5.12)) in order to get additional saving over the $\underline {m}$ sum. For the perfect square part, $b_2$, however, we will derive a slightly weaker bound from this which will be used to get saving over the ${\underline {h}}$ sum later on in the argument.

5.3 Cube-free square exponential sums

In this section, we will assume that $q=b_2$ or, equivalently, that $q$ is a cube-free square. In this case, we will give up on the potential saving we could attain via the $\underline {m}$ sum from the $\Delta _q(\underline {m}')$ term in Proposition 5.9, and bound $\#\mathrm {Null}_q(M({\underline {a}}))^{1/2}$ in terms of the singular locus of $f^{(0)},g^{(0)}$, where $M({\underline {a}})$ is defined as in (5.22). In this special case, we will need to obtain a pointwise saving over the ${\underline {a}}$ sum in order for our bound to be useful. We will start with the case when $n\geq 2$. Upon letting $b_2=c^2$, by Proposition 5.9, Lemmas 2.4–2.5 and (5.16) we have

(5.37)\begin{align} |S(b_2,\underline{m})|&\ll \sideset{}{^*}\sum_{{\underline{a}}}^{b_2}|S(\underline{a}, b_2; \underline{m})|\ll b_2^{n/2}\sideset{}{^*}\sum_{{\underline{a}}}^{b_2}\#\mathrm{Null}_{c^2}(M({\underline{a}}))^{1/2}\ll b_2^{n/2}\sideset{}{^*}\sum_{{\underline{a}}}^{b_2}\#\mathrm{Null}_{c}(M({\underline{a}}))\nonumber\\ &\ll b_2^{2+n/2}c^{s_p+1}=b_2^{2+n/2}\prod_{\substack{p^2\mid q\nonumber\\ p \textrm{ prime }}} p^{s_p+1} =b_2^{2+n/2}D(b_2). \end{align}

When $n=1$, we have $M({\underline {a}})=a_1d_f+a_2d_g$ for some constants $d_f,d_g$. By Proposition 5.10,

(5.38) \begin{align} |S(p^2,\underline{m})|&\ll p\sideset{}{^*}\sum_{{\underline{a}}\bmod{p^2}}(p^2,M({\underline{a}}))^{1/2}\ll p\sideset{}{^*}\sum_{{\underline{a}}\bmod{p^2}}(p,a_1d_f+a_2d_g)\nonumber\\ &= p \bigg{(} \sideset{}{^*}\sum_{\substack{{\underline{a}}\bmod{p^2}\\ \underline{p} | a_1d_f+a_2d_g}} p + \sideset{}{^*}\sum_{\substack{{\underline{a}}\bmod{p^2}\\ \underline{p} \nmid a_1d_f+a_2d_g}} 1 \bigg{)}\nonumber\\ &\leq \begin{cases} 2p^{5} & \text{if } (d_f, d_g, p)=1, \\ p^6 &\text{otherwise. }\end{cases} \end{align}

Hence, upon recalling (5.4), we may bound (5.38) by

\[ |S(p^2,\underline{m})|\ll p^{5}D(p). \]

We may then use the multiplicativity relation in Lemma 5.4 to get

\[ |S(b_2,\underline{m})|\ll b_2^{2+1/2+\varepsilon}D(b_2). \]

Combining this with (5.37) gives us the following.

Proposition 5.11 Let $b_2\in \mathbb {N}$ be a cube-free square. Then

\[ S(b_2,\underline{m})\ll b_2^{2+n/2+\varepsilon}D(b_2). \]

6. Quadratic exponential sums: finalisation

In this section, we will combine all of the bounds we have found in § 5 to reach our final estimate for $T(q,\underline {z})$. Recall that Proposition 5.3 hands us

(6.1)\begin{equation} |T(q,\underline{z})|\ll 1+q^{-n}P^n\sup_{\underline{y}\in P\,\mathrm{Supp}(\omega)}\bigg{\{}\sum_{|\underline{m}-\underline{m}_0(\underline{y})|\leq V}|S(q;\underline{m})|\bigg{\}}. \end{equation}

In the last section, we focused on getting bounds for individual exponential sums $|S(q;\underline {m})|$. We begin by considering averages of exponential sums. Throughout, let $\underline {m}_0$ be an arbitrary but fixed vector in $\mathbb {Z}^n$ and let ${\underline {\mathfrak {b}}}({\underline {a}})={\underline {\mathfrak {b}}}$ be defined as in (5.22). For $n\geq 2$: by Lemma 5.4 and Propositions 5.5, 5.9, and 5.11, there are some constants $c_1,c_2,c_3$ such that $(b_1,c_1)=(b_2,c_2)=(q_3,c_3)=1$, and

(6.2)\begin{align} \sum_{|\underline{m}-\underline{m}_0|\leq V}|S(q;\underline{m})|&\leq \sum_{|\underline{m}-\underline{m}_0|\leq V}|S(b_1;c_1\underline{m})|\cdot|S(b_2;c_2\underline{m})|\cdot|S(q_3;c_3\underline{m})|\nonumber\\ &\ll q^{n/2+\varepsilon} b_1b_2^2 D(b_1b_2)\sum_{|\underline{m}-\underline{m}_0(\underline{m}_0)|\leq V} (\Phi(c_1\underline{m}),b_1)^{1/2}\nonumber\\ &\quad \times\sideset{}{^*}\sum_{{\underline{a}}}^{q_3}\#\mathrm{Null}_{q_3}(a_1M_1+a_2M_2)^{1/2}\Delta_{T,q_3}(c_3\underline{m}+{\underline{\mathfrak{b}}}) \nonumber\\ &:= q^{n/2+\varepsilon} b_1b_2^2 D(b_1b_2)\sideset{}{^*}\sum_{{\underline{a}}}^{q_3}\#\mathrm{Null}_{q_3}(M({\underline{a}}))^{1/2}B(b_1,q_3, V; \underline{m}_0), \end{align}

where $M({\underline {a}})$ be as in (5.17) and

(6.3)\begin{equation} B(b_1,q_3, V; \underline{m}_0):= \sum_{|\underline{m}-\underline{m}_0|\leq V} (\Phi(\underline{m}),b_1)^{1/2}\cdot \Delta_{T,q_3}(\underline{m}+{\underline{\mathfrak{b}}}'), \end{equation}

where ${\underline {\mathfrak {b}}}'\equiv c_3^{-1}{\underline {\mathfrak {b}}} \mod q_3$. We used $(\Phi (\underline {m}),b_1)$ instead of $(\Phi (c_1\underline {m}),b_1)$ in the definition of $B(b_1,q_3,V;\underline {m}_0)$ because $\Phi$ is homogeneous and $(b_1,c_1)=1$. Likewise, by inspecting the definition of $\Delta$, we can use $\Delta _{T,q_3}(\underline {m}+{\underline {\mathfrak {b}}}')$ in the definition of $B(b_1,q_3,V;\underline {m}_0)$ instead of $\Delta _{T,q_3}(c_3\underline {m}+{\underline {\mathfrak {b}}})$ since we can ‘divide through’ by $c_3$, as $(c_3,q_3)=1$ (in particular, $(c_3,\lambda )=1$ for any divisor, $\lambda$, of $q_3$).

The first and most difficult task for this section is to bound $B(b_1,q_3, V; \underline {m}_0)$. This will be quite a delicate task since we need to save over the $\underline {m}$ sum in two different ways, simultaneously. The following lemma will provide our main estimate for this sum.

Lemma 6.1 Let $b_1,q_3, V\in \mathbb {N}$ and $\underline {m}_0\in \mathbb {Z}^n$. Furthermore, let $c$ and $q_3$ be defined as follows:

(6.4)\begin{equation} \hat{q}_3:=\prod_{\substack{p^e||q_3\\ 2\nmid e}} p, \quad q_3=c^2\hat{q}_3. \end{equation}

Then

\[ B(b_1,q_3, V; \underline{m}_0) \ll b_1^{\varepsilon}(b_1^{1/2}c^{n/2}+ V^{n-1}b_1^{1/2}c^{1/2}+ V^n) \#\mathrm{Null}_c(M({\underline{a}}))^{-1}. \]

Proof. We begin by noting that

\[ (T^t\underline{x})_i\equiv 0 \ {\rm mod}\ (q_3,\lambda_i) \;\implies\; (T^t\underline{x})_i\equiv 0 \ {\rm mod} \ (c,\lambda_i), \]

and so by the definition of $\Delta _{T, q_3}$ (5.23) we clearly have that

\[ \Delta_{T, q_3}(\underline{x})=1 \;\implies\; \Delta_{T, c}(\underline{x})=1, \]

for any $\underline {x}\in \mathbb {Z}^n$. Therefore, since we are looking for an upper bound of $B(b_1,q_3, V; \underline {m}_0)$, we may replace $\Delta _{T, q_3}(\underline {m}+{\underline {\mathfrak {b}}}')$ in (6.3) with $\Delta _{T, c}(\underline {m}+{\underline {\mathfrak {b}}}')$. Furthermore, since all elements of our sum are non-negative, we may extend the sum in (6.3) if we wish. In particular, the following bound must be true:

(6.5)\begin{equation} B(b_1,q_3, V; \underline{m}_0)\leq \sum_{|\underline{m}-\underline{m}_0|\leq \hat{V}} (\Phi(\underline{m}),b_1)^{1/2}\cdot \Delta_{T,c}(\underline{m}+{\underline{\mathfrak{b}}}'), \end{equation}

where

(6.6)\begin{equation} \hat{V}:=\max\{V,c\}. \end{equation}

We have extended the sum up to $\hat {V}$ so that we can consider complete sums modulo $c$, as this will make it easier to acquire saving from $\Delta _{T,c}$ later. To this end, let $\underline {m}:=\underline {m}_0+\underline {m}_1+c\underline {m}_2$, where $\underline {m}_1\in (\mathbb {Z}/c\mathbb {Z})^n$ and $|\underline {m}_2|\leq \hat {V}/c$. Applying this decomposition on the right-hand side of (6.5) gives

(6.7)\begin{align} B(b_1,q_3, V; \underline{m}_0)&\leq \sum_{\underline{m}_1 \bmod{c}}\,\sum_{|\underline{m}_2|\leq \hat{V}/c} (\Phi(\underline{m}_0+\underline{m}_1+c\underline{m}_2),b_1)^{1/2}\Delta_{T,c}(\underline{m}_0+\underline{m}_1 +c\underline{m}_2+{\underline{\mathfrak{b}}}')\nonumber\\ &=\sum_{\underline{m}_1 \bmod{c}}\Delta_{T,c}(\underline{m}_0+\underline{m}_1+{\underline{\mathfrak{b}}}')\sum_{\underline{m}_2\in U(\underline{m}_1)} (\Phi(\underline{m}_0+\underline{m}_1+c\underline{m}_2),b_1)^{1/2}. \end{align}

The upshot of reordering our sum in this way is that we have managed to separate $\Delta _{T,c}(\underline {m}_0+\underline {m}_1+{\underline {\mathfrak {b}}}')$ and $(\Phi (\underline {m}_0+\underline {m}_1+c\underline {m}_2),b_1)^{1/2}$. In particular, we can treat $\underline {m}_1$ as fixed for now, and since $\underline {m}_0$ and $c$ are also fixed, we may focus on acquiring saving in the $\underline {m}_2$ sum via $(\Phi _{c,\underline {m}_1}(\underline {m}_2),b_1)^{1/2}$, where

\[ \Phi_{c,\underline{m}_0,\underline{m}_1}(\underline{m}_2):=\Phi(\underline{m}_0+\underline{m}_1+c\underline{m}_2). \]

We observe that $(\Phi _{c,\underline {m}_0,\underline {m}_1}(\underline {m}_2),b_1)$ must be equal to some divisor of $b_1$, so we will decompose the $\underline {m}_2$ sum as follows:

(6.8)\begin{equation} \sum_{|\underline{m}_2|\leq \hat{V}/c} (\Phi(\underline{m}_0+\underline{m}_1+c\underline{m}_2),b_1)^{1/2}=\sum_{d\mid b_1} d^{1/2} \#\{|\underline{x}|\leq \hat{V}/c : \Phi(\underline{m}_0+\underline{m}_1+c\underline{x})\equiv 0 \mod d\}. \end{equation}

We now aim to use [Reference Browning and Heath-BrownBH09, Lemma 4] to bound the right-hand side. Since $\Phi$ is homogeneous with $\mathrm {Cont}(\Phi )=1$ by Proposition 5.5, and since $c$ and $d$ are co-prime, we have that

\[ \big(\mathrm{Cont}\big(\Phi_{c,\underline{m}_0,\underline{m}_1}\big),d\big)\leq \big(\mathrm{Cont}\big(\Phi_{c,\underline{m}_0,\underline{m}_1}^{(0)}\big),d\big)\leq(c^{\deg(\Phi)} \mathrm{Cont}(\Phi), d)=(c^{\deg(\Phi)},d)=1. \]

Hence, for every prime $p$ dividing $d$, $\Phi (\underline {m}_0+\underline {m}_1+c\underline {x})$ is a non-trivial polynomial and therefore the corresponding variety is of dimension $n-1$. Therefore, we may now use [Reference Browning and Heath-BrownBH09, Lemma 4] to conclude that

\[ \#\{|\underline{x}|\leq \hat{V}/c: \Phi(\underline{m}_0+\underline{m}_1+c\underline{x})\equiv 0 \mod d\}\ll 1+\bigg{(} \frac{V}{c}\bigg{)}^{n-1}+ \bigg{(} \frac{V}{c}\bigg{)}^n d^{-1}. \]

Substituting this back into (6.8) gives the following:

(6.9)\begin{align} \sum_{|\underline{m}_2|\leq \hat{V}/c} (\Phi(\underline{m}_0+\underline{m}_1+c\underline{m}_2),b_1)^{1/2}&\ll \sum_{d\mid b_1} d^{1/2}+\bigg{(} \frac{V}{c}\bigg{)}^{n-1}d^{1/2}+ \bigg{(} \frac{V}{c}\bigg{)}^n d^{-1/2}\nonumber\\ &\ll b_1^{\varepsilon}\bigg{(} b_1^{1/2}+ \bigg{(} \frac{V}{c}\bigg{)}^{n-1}b_1^{1/2}+ \bigg{(} \frac{V}{c}\bigg{)}^n \bigg{)}. \end{align}

This, in turn, will enable us to find a suitable bound for $B(b_1,q_2,V;\underline {m}_0)$. By (6.7) and (6.9), we have

(6.10)\begin{equation} B(b_1,q_3, V; \underline{m}_0)\ll b_1^{\varepsilon}\bigg{(} b_1^{1/2}+ \bigg{(} \frac{V}{c}\bigg{)}^{n-1}b_1^{1/2}+ \bigg{(} \frac{V}{c}\bigg{)}^n \bigg{)} \sum_{\underline{x}\bmod{c}}\Delta_{T,c}(\underline{x}+\underline{m}_0+{\underline{\mathfrak{b}}}'). \end{equation}

In order to find the bound we desire for $B(b_1,q_3, V; \underline {m}_0)$, we will need to turn our attention to the sum of type

\[ \sum_{\underline{x}\bmod{c}}\Delta_{T,c}(\underline{x}+\underline{l}), \]

for some fixed $\underline {l}\in \mathbb {Z}^n$. Our bound here will be independent of the choice of the vector $\underline {l}$. This sum is much easier to handle since we have a complete sum at hand. It is easy to check from the definition of $\Delta _{T,c}(\underline {x}+\underline {l})$ (and the fact that $\det (T^t)=1$) that

\begin{align*} \sum_{\underline{x}\bmod{c}}\Delta_{T,c}(\underline{x}+\underline{l})&=\#\{\underline{x} \bmod{c} \, : \, (T^t\underline{x})_i\equiv -(T^t\underline{l})_i \bmod{\lambda_{c,i}}, \; i\in\{1,\ldots, n\}\}\\ &\leq \#\{\underline{x} \bmod{c} \, : \, (T^t\underline{x})_i\equiv 0 \bmod{\lambda_{c,i}}, \; i\in\{1,\ldots, n\}\}\\ &= \#\{\underline{x} \bmod{c} \, : \, x_i\equiv 0 \bmod{\lambda_{c,i}}, \; i\in\{1,\ldots, n\}\}\\ &=\frac{c^n}{\prod_i \lambda_{c,i}}=c^n \#\mathrm{Null}_c(M({\underline{a}}))^{-1}. \end{align*}

Therefore, by (6.10), we have

(6.11)\begin{equation} B(b_1,q_3, V; \underline{m}_0)\leq b_1^{\varepsilon}(b_1^{1/2}c^n+ V^{n-1}b_1^{1/2}c+ V^n) \#\mathrm{Null}_c(M({\underline{a}}))^{-1}, \end{equation}

as required.

We are now ready to obtain a final bound for $\sum _{|\underline {m}-\underline {m}_0|\leq V} |S(q;\underline {m})|$. Before substituting (6.11) back into (6.2), we will perform some simplifications. First, we note that by (6.4) and Lemma 2.4, we have

(6.12)\begin{equation} \#\mathrm{Null}_{q_3}(M({\underline{a}}))^{1/2}\leq \#\mathrm{Null}_{c}(M({\underline{a}}))\#\mathrm{Null}_{\hat{q}_3}(M({\underline{a}}))^{1/2}. \end{equation}

Furthermore, by Lemma 2.5, we have

(6.13)\begin{equation} \sideset{}{^*}\sum_{{\underline{a}}}^{q_3}\#\mathrm{Null}_{\hat{q}_3}(M({\underline{a}}))^{1/2}\leq \sideset{}{^*}\sum_{{\underline{a}}}^{q_3}\#\mathrm{Null}_{\hat{q}_3}(M({\underline{a}})) \ll q_3^{2+\varepsilon} \prod_{p_i | \hat{q}_3} p_i^{s_{p_i}(f^{(0)},g^{(0)})+1} =q_3^{2+\varepsilon} D(\hat{q}_3). \end{equation}

Finally, by combining (6.11)–(6.13) with (6.2), we arrive at the following bound.

Lemma 6.2 For every $q\in \mathbb {N}$, if $n>1$, then

\[ \sum_{|\underline{m}-\underline{m}_0|\leq V}|S(q;\underline{m})|\ll q^{1+n/2+\varepsilon} b_2q_3 D(b_1b_2\hat{q}_3)(b_1^{1/2}c^{n}+ V^{n-1}b_1^{1/2}c+ V^n), \]

where $q_3=c^2\hat {q}_3$ as defined in the statement of Lemma 6.1.

Recall that our ultimate goal was to find a suitable bound for $|T(q,\underline {z})|$. Upon noting that the above treatment of $\sum _{|\underline {m}-\underline {m}_0|\leq V}|S(q;\underline {m})|$ works for any value of $\underline {y}\in P\,\mathrm {Supp}(\omega )$ we may now substitute the bound in Lemma 6.2 into (6.1) to get the following bound for $T(q,\underline {z})$:

\[ |T(q,\underline{z})|\ll 1+ P^n q^{1-n/2+\varepsilon} b_1^{1/2}b_2q_3 D(b_1b_2\hat{q}_3) (V^n b_1^{-1/2} + V^{n-1}c + c^n). \]

If $q$ is sufficiently small ($q< P^2$ say), then the right-hand term dominates over $1$ for every $n\geq 1$. Therefore, we finally reach the following bound for $|T(q,\underline {z})|$:

\[ |T(q,\underline{z})|\ll P^n q^{1-n/2+\varepsilon} b_1^{1/2}b_2q_3 D(b_1b_2\hat{q}_3)( V^n b_1^{-1/2} + V^{n-1}c + c^n), \]

where $q_3=c^2\hat {q}_3$ as defined in Lemma 6.1. Note that if we use a weaker bound $c\leq b_3^{1/3}q_4^{1/2}$ and $D(\hat {q}_3)\leq D(q_3)$, and use the quality $q_3=b_3q_4$, where $b_3$ is the fourth power-free cube part of $q$ and $q_i$ is the $i$th power-full part of $q$, the above bound becomes the following.

Proposition 6.3 For every $q=b_1b_2q_3< P^2$, $\underline {z}$ and every $\varepsilon >0$, if $n>1$, we have

\[ |T(q,\underline{z})|\ll P^n q^{1-n/2+\varepsilon} b_1^{1/2}b_2q_3 D(q)(V^n b_1^{-1/2} + V^{n-1}b_3^{1/3}q_4^{1/2} + b_3^{n/3}q_4^{n/2}), \]

where $n$ is the number of variables of $f, g$.

The bound for the $n=1$ case is much simpler to derive than in the $n>1$ case. By Lemma 5.4 and Propositions 5.6, 5.10 and 5.11, we have

(6.14)\begin{align} \sum_{|m-m_0|\leq V}|S(q;m)| &\ll q^{1/2+\varepsilon}b_1^{3/2} b_2^2D(b_1b_2) \sideset{}{^*}\sum_{{\underline{a}}}^{q_3}(q_3,M({\underline{a}}))^{1/2}\sum_{|m-m_0|\leq V} \Delta'_{q_3}(c_3m+\mathfrak{b})\nonumber\\ &\leq q^{1/2+\varepsilon}b_1^{3/2} b_2^2D(b_1b_2)\sideset{}{^*}\sum_{{\underline{a}}}^{q_3}(q_3,M({\underline{a}}))^{1/2}\bigg{(} 1 + \frac{V}{(q_3,M({\underline{a}}))} \bigg{)}\nonumber\\ &\leq q^{1/2+\varepsilon}b_1^{3/2} b_2^2D(b_1b_2)\sideset{}{^*}\sum_{{\underline{a}}}^{q_3}{(}(q_3,M({\underline{a}}))^{1/2} + V {)}. \end{align}

We trivially have $\sum _{{\underline {a}}} V\leq q_3^2 V$. As for the other part of the sum, upon recalling that $q_3=\hat {q}_3c^2$, we have

\[ \sideset{}{^*}\sum_{{\underline{a}}}^{q_3}(q_3,M({\underline{a}}))^{1/2}\leq c \sideset{}{^*}\sum_{{\underline{a}}}^{q_3}(\hat{q}_3,M({\underline{a}}))^{1/2}\leq c^5 \sideset{}{^*}\sum_{{\underline{a}}}^{\hat{q}_3}(\hat{q}_3,M({\underline{a}}))^{1/2}\ll q_3^2 c D(\hat{q}_3)\leq q_3^2 c D(q_3), \]

by the same argument as the proof of Proposition 5.11. Combining this with (6.14) gives the following result.

Lemma 6.4 Let $q=b_1b_2q_3\in \mathbb {N}$, $m\in \mathbb {Z}$ and $q_3:=c^2 \hat {q}_3$ be defined as in Lemma 6.1. Then for every $\varepsilon >0$,

\[ \sum_{|m-m_0|\leq V}|S(q;m)| \ll q^{2+\varepsilon} (b_2q_3)^{1/2} D(q) (V+c). \]

Finally, upon recalling that $q_3=b_3q_4$, $c\leq b_3^{1/3}q_4^{1/2}$, we may combine this lemma with (6.1) to get our final bound for $|T(q,\underline {z})|$ in the $n=1$ case.

Proposition 6.5 For every $q< P^2$, $\underline {z}$ and every $\varepsilon >0$, if $n=1$, we have

\[ |T(q,\underline{z})|\ll P q^{1+\varepsilon} (b_2q_3)^{1/2} D(q) (V + b_3^{1/3}q_4^{1/2}), \]

where $n$ is the number of variables of $f, g$.

7. Finalisation of the Poisson bound

In this section, we will adapt the arguments used in [Reference Browning and Heath-BrownBH09, Section 7] and [Reference Marmon and VisheMV19, Section 8] to our context in order to finalise our main bounds coming from Poisson summation. Throughout this section, we treat $\underline {z}$ as fixed. Lemmas 4.1 and 4.5 allow us to consider bounding the sum

\[ \sum_{|\underline{h}|\ll H}\, |T_{\underline{h}}(q,\underline{z})|, \]

where

\[ T_{\underline{h}}(q,\underline{z}):= \sideset{}{^*}\sum_{{\underline{a}}\bmod{q}}\sum_{\underline{x}\in\mathbb{Z}^n} \omega_{\underline{h}}(\underline{x}/P)e((a_1/q+z_1)F_{{\underline{h}}}(\underline{x})+(a_2/q+z_2)G_{{\underline{h}}}(\underline{x})), \]

is the quadratic exponential sum as defined in (4.5). We may therefore apply our bounds for quadratic exponential sums in Propositions 6.3 and 6.5 to estimate these.

Now that $\underline {h}$ is allowed to vary, we will define

(7.1)\begin{equation} s_p(\underline{h}):=s_p\big(F_{\underline{h}}^{(0)},G_{\underline{h}}^{(0)}\big), \end{equation}

where $F_{\underline {h}}^{(0)}$ and $G_{\underline {h}}^{(0)}$ denote the leading quadratic parts of $F_{\underline {h}}$ and $G_{\underline {h}}$, respectively. We recall that $q=b_1b_2q_3$, where $q_3$ is the cube-full part of $q$, and $b_1,b_2$ are the square-free and cube-free square parts of $q$. Since we are fixing $q$ for now, $b_1$, $b_2$ and $q_3$ are also fixed. Recall that we may write $b_i=b_{i,0}b_{i,1}\cdots b_{i,n}$, $q_3=q_{3,0}q_{3,1}\cdots q_{3,n}$ where $b_{i,j}$, and $q_{3,j}$ now depend on $\underline {h}$ and are defined to be

\[ b_{i,j}(\underline{h})=\prod_{\substack{p^i||b_i\\s_p(\underline{h})=j-1}} p^i,\quad q_{3,j}(\underline{h})=\prod_{\substack{p^e||q_3\\s_p(\underline{h})=j-1}} p^e. \]

We see that for any $q$ fixed, there are at most $O(q^{\varepsilon })=O(P^{\varepsilon })$ possible choices for

\[ \underline{c}=(b_{1,0},\ldots, b_{1,n},b_{2,0},\ldots, b_{2,n},q_{3,0},\ldots, q_{3,n}) \]

since there are only at most $O(q^{\varepsilon })$ partitions of $q$ into multiplicative factors. Therefore, using the triangle inequality, we have that

\[ \sum_{\underline{h}\ll H} |T_{\underline{h}}(q,z)|\leq P^{\varepsilon}\max_{\underline{c}}\bigg{\{}\sum_{\substack{\underline{h}\\ \underline{c}(\underline{h})=\underline{c}}} |T_{\underline{h}}(q,z)|\bigg{\}}=P^{\varepsilon}\sum_{\substack{\underline{h}\\ \underline{c}(\underline{h})=\underline{c}'}} |T_{\underline{h}}(q,z)| \]

for some particular $\underline {c}'$, and $\underline {c}(\underline {h}):=(b_{1,0}(\underline {h}),\ldots, q_{3,n}(\underline {h}))$. We can then decompose this sum further by grouping the $\underline {h}$ with $s_{\infty }(\underline {h})=s$:

(7.2)\begin{equation} \implies \sum_{\underline{h}\ll H} |T_{\underline{h}}(q,z)|\leq P^{\varepsilon}\sum_{s=-1}^{n-1}\sum_{\underline{h}\in\mathcal{H}_s} |T_{\underline{h}}(q,z)|, \end{equation}

where

(7.3)\begin{equation} \mathcal{H}_s:=\{\underline{h}\in\mathbb{Z}^n:\, \underline{h}\ll H,\quad \underline{c}'(\underline{h})=\underline{c}',\ s_{\infty}(\underline{h})=s\}. \end{equation}

Here, given $\nu$ either a prime or $\infty$, we define