(8.0.6) $\eta^{\dagger}Y_{i}\eta\equiv\lambda_{i}N\bmod{\mathrm{nrd}(\eta)},$

and therefore

(8.0.7)
$\gamma^{\prime}_{i}\eta\equiv-\upsilon_{i}\lambda_{i}ZN\bmod{\mathrm{nrd}(\eta)},\textnormal{
or equivalently,
}\gamma^{\prime}_{i}\in-\upsilon_{i}\lambda_{i}Z\eta^{\dagger}N_{0}+\mathcal{O}_{D}\eta^{\dagger}.$

We will bound the measure of the set of $Z$ satisfying such a congruence on
average in Lemma 8.3 below. It strengthens the exponent of the bound in
Theorem 9.1 for the sum over $\gamma\neq 0$ from $3n+2+\epsilon$ to
$3n+\epsilon.$

Given $\eta\in\mathcal{O}_{D}-\varpi\mathcal{O}_{D}$, we define the map

(8.0.8)
$\Theta:\mathcal{O}_{F}^{n}\times\mathcal{O}_{D}\to(\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger})^{n},\quad(\lambda,Z)\mapsto\lambda
Z\eta^{\dagger}N_{0}.$

For $M_{0}\in\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger}$, let

(8.0.9)
$\mathcal{W}(M_{0},\eta):=\sup_{\begin{subarray}{c}\mu\in\mathcal{O}_{F}^{n}-\varpi\mathcal{O}_{F}^{n}\end{subarray}}\int_{\mathcal{O}_{D}}\mathbf{1}_{\mu
M_{0}\in\Theta(\mathcal{O}_{F}^{n},Z)}\,dZ.$

We now have developed enough notation that we can state the main theorem of
this section:

######  Theorem 8.2.

Assume $v\nmid 2$. Let $\gamma\in D^{n}$. Write
$\delta=\varpi^{v(\delta)}\eta.$ Then $I_{0}(\delta,\gamma)=0$ unless
$\gamma\in\varpi^{v(\delta)}\mathcal{O}_{D}^{n}$ and there exists an
$M_{0}\in\mathcal{O}_{D}$ and a
$\mu\in\mathcal{O}_{F}^{n}-\varpi\mathcal{O}_{F}^{n}$ such that

(8.0.10) $\displaystyle\gamma\varpi^{-v(\delta)}\eta\in\mu
M_{0}\eta+\mathrm{nrd}(\eta)\mathcal{O}_{D}^{n}.$

Assume $\gamma\in\varpi^{v(\delta)}\mathcal{O}_{D}^{n}$ and write
$\gamma^{\prime}:=\gamma\varpi^{-v(\delta)}.$ If (8.0.10) holds then

$\displaystyle\lvert
I_{0}(\delta,\gamma)\rvert\leq\frac{\mathcal{W}(M_{0},\eta)\,|\mathrm{nrd}(\eta)|^{3n/2}\left\lVert\delta\right\rVert^{n}}{\max(\left\lVert(\gamma^{\prime}-\gamma^{\prime\dagger})\eta\right\rVert,|\mathrm{nrd}(\eta)|)^{n/2}\max(\left\lVert\gamma^{\prime}\right\rVert,\left\lVert\delta\right\rVert|\mathrm{nrd}(\eta)|)^{n}}.$

The proof will occupy the remainder of this section.

######  Remark.

We point out that Theorem 8.2 implies that the integral $I_{0}(\delta,\gamma)$
vanishes unless the image of $\gamma^{\prime}\in\mathcal{O}_{D}^{n}$ in
$(\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger})^{n}$ is in the image of the
diagonal embedding
$\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger}\to(\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger})^{n}$,
at least up to
$(\mathcal{O}_{F}/\mathrm{nrd}(\eta)\mathcal{O}_{F})^{n}$-scaling. Thus the
exponential sums are supported on a relatively small set of residue classes
modulo $\mathrm{nrd}(\eta).$

In our eventual application of Theorem 8.2 to the proof of Theorem 9.1 we will
require the average estimate for $\mathcal{W}(M_{0},\eta)$ given by the
following lemma:

######  Lemma 8.3.

We have

$\sum_{M_{0}\in\mathcal{O}_{F}^{\times}\backslash(\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger})}\mathcal{W}(M_{0},\eta)\leq
v(\mathrm{nrd}(\eta))+1.$

###### Proof.

Given $M_{0}\in\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger}$ and
$\mu\in\mathcal{O}_{F}^{n}-\varpi\mathcal{O}_{F}^{n}$, we have

$\mu
M_{0}\in\Theta(\mathcal{O}_{F}^{n},Z)\Leftrightarrow\mu_{1}M_{0},\dots,\mu_{n}M_{0}\in\mathcal{O}_{F}Z\eta^{\dagger}N_{0}+\mathcal{O}_{D}\eta^{\dagger}\Leftrightarrow
M_{0}\in\mathcal{O}_{F}Z\eta^{\dagger}N_{0}+\mathcal{O}_{D}\eta^{\dagger}.$

The condition on the right is independent of $\mu$. Therefore, by (8.0.9), we
have

$\mathcal{W}(M_{0},\eta)=\int_{\mathcal{O}_{D}}\mathbf{1}_{M_{0}\in\mathcal{O}_{F}Z\eta^{\dagger}N_{0}+\mathcal{O}_{D}\eta^{\dagger}}\,dZ=\int_{\mathcal{O}_{D}}\mathbf{1}_{[M_{0}]\in[\mathcal{O}_{F}Z\eta^{\dagger}N_{0}+\mathcal{O}_{D}\eta^{\dagger}]}\,dZ,$

where $[A]$ denotes the image of $A$ in the quotient set
$\mathcal{O}_{F}^{\times}\backslash(\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger})$.
Here, if $A$ is an element (resp. subset) of either $\mathcal{O}_{D}$ or
$\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger}$, then we interpret $[A]$ as an
element (resp. subset) of
$\mathcal{O}_{F}^{\times}\backslash(\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger})$.
By Fubini’s theorem, then,

(8.0.11)
$\sum_{M_{0}\in\mathcal{O}_{F}^{\times}\backslash(\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger})}\mathcal{W}(M_{0},\eta)=\int_{\mathcal{O}_{D}}\sum_{M_{0}\in\mathcal{O}_{F}^{\times}\backslash(\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger})}\mathbf{1}_{M_{0}\in[\mathcal{O}_{F}Z\eta^{\dagger}N_{0}+\mathcal{O}_{D}\eta^{\dagger}]}\,dZ.$

Given $Z$, the inner sum over $M_{0}$ is simply
$\\#[\mathcal{O}_{F}Z\eta^{\dagger}N_{0}+\mathcal{O}_{D}\eta^{\dagger}]$.
However,

$[\mathcal{O}_{F}Z\eta^{\dagger}N_{0}+\mathcal{O}_{D}\eta^{\dagger}]=\bigcup_{0\leq
k\leq
v(\mathrm{nrd}(\eta))}[\varpi^{k}Z\eta^{\dagger}N_{0}+\mathcal{O}_{D}\eta^{\dagger}],$

which is a union of at most $v(\mathrm{nrd}(\eta))+1$ distinct elements of
$\mathcal{O}_{F}^{\times}\backslash(\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger})$.
So the right-hand side of (8.0.11) is less than or equal to
$v(\mathrm{nrd}(\eta))+1$. ∎

###  8.1. Beginning of the proof of Theorem 8.2

In the rest of the section, we assume $v\nmid 2$. We may assume
$I_{0}(\delta,\gamma)\neq 0$, or there is nothing to prove. Then by (8.0.2),
there exists an element $Z\in\mathcal{O}_{D}$ with $I_{0}(Z,\delta,\gamma)\neq
0$. Now $\gamma\in\varpi^{v(\delta)}\mathcal{O}_{D}^{n}$ by (8.0.4). In the
congruence (8.0.7), replace each $\lambda_{i}$ with a lift
$\lambda_{i}\in\mathcal{O}_{F}\cap F^{\times}$ and let
$\gcd(\upsilon_{1}\lambda_{1},\dots,\upsilon_{n}\lambda_{n})\in F^{\times}$ be
any element such that
$\gcd(\upsilon_{1}\lambda_{1},\dots,\upsilon_{n}\lambda_{n})\mathcal{O}_{F}=\upsilon_{1}\lambda_{1}\mathcal{O}_{F}+\dots+\upsilon_{n}\lambda_{n}\mathcal{O}_{F}.$
Then condition (8.0.10) holds with

$M_{0}:=\gcd(\upsilon_{1}\lambda_{1},\dots,\upsilon_{n}\lambda_{n})Z\eta^{\dagger}N_{0}\in\mathcal{O}_{D},\quad\mu_{i}:=-\upsilon_{i}\lambda_{i}/\gcd(\upsilon_{1}\lambda_{1},\dots,\upsilon_{n}\lambda_{n})\in\mathcal{O}_{F},$

where the factor of
$\gcd(\upsilon_{1}\lambda_{1},\dots,\upsilon_{n}\lambda_{n})\in\mathcal{O}_{F}\cap
F^{\times}$ ensures that
$\mu=(\mu_{1},\dots,\mu_{n})\in\mathcal{O}_{F}^{n}-\varpi\mathcal{O}_{F}^{n}$.
We emphasize that $M_{0}$ and $\mu$ may depend on $(\delta,\gamma)$ and $Z$.
However, all that matters is that $M_{0}$ does not depend on $i$.

######  Lemma 8.4.

Let $(\mu,M_{0})$ be any pair satisfying (8.0.10). We have

$\lvert
I_{0}(\delta,\gamma)\rvert\leq\mathcal{W}(M_{0},\eta)\,\sup\limits_{Z\in\mathcal{O}_{D}}|I_{0}(Z,\delta,\gamma)|.$

###### Proof.

By assumption we have $\gamma^{\prime}\in\mu
M_{0}+\mathcal{O}_{D}^{n}\eta^{\dagger}$. If $I_{0}(Z,\delta,\gamma)\neq 0$,
then (8.0.7) implies $\gamma^{\prime}\in\Theta(\mathcal{O}_{F}^{n},Z)$ or
equivalently $\mu M_{0}\in\Theta(\mathcal{O}_{F}^{n},Z)$. Therefore, by
(8.0.9) we have $\int_{\mathcal{O}_{D}}\mathbf{1}_{I_{0}(Z,\delta,\gamma)\neq
0}\,dZ\leq\mathcal{W}(M_{0},\eta)$. The lemma now follows from (8.0.2). ∎

By Lemma 8.4, to prove Theorem 8.2 it suffices to prove that for any
$Z\in\mathcal{O}_{D}$

(8.1.1)
$\displaystyle|I_{0}(Z,\delta,\gamma)|\leq\frac{|\mathrm{nrd}(\eta)|^{3n/2}\left\lVert\delta\right\rVert^{n}}{\max(\left\lVert(\gamma^{\prime}-\gamma^{\prime\dagger})\eta\right\rVert,|\mathrm{nrd}(\eta)|)^{n/2}\max(\left\lVert\gamma^{\prime}\right\rVert,\left\lVert\delta\right\rVert|\mathrm{nrd}(\eta)|)^{n}}.$

By definition $\eta=\delta\varpi^{-v(\delta)}$ and
$\gamma^{\prime}=\gamma\varpi^{-v(\delta)},$ so we can rewrite (8.0.3) as

$I_{0}(Z,\delta,\gamma)=\int_{\mathcal{O}_{D}^{n}}\mathbf{1}_{\mathrm{nrd}(\eta)\mathcal{O}_{D}}(\upsilon_{i}Z\eta^{\dagger}Y_{i}\eta+\gamma^{\prime}_{i}\eta)\,\psi\left(\frac{\mathrm{trd}(Z\eta^{\dagger}P(Y)+\gamma^{\prime}\cdot
Y)}{\mathrm{nrd}(\eta)\varpi^{v(\delta)}}\right)\,dY.$

Let

(8.1.2) $\displaystyle
b:=\min(v(Z\eta^{\dagger}),v(\delta)+v(\mathrm{nrd}(\eta))).$

By shifting $Y_{i}\mapsto Y_{i}+\mathrm{nrd}(\eta)\varpi^{v(\delta)-b}M_{i}$
for $M_{i}\in\mathcal{O}_{D}$, we find that $I_{0}(Z,\delta,\gamma)=0$ unless

(8.1.3) $\gamma^{\prime}\in\varpi^{b}\mathcal{O}_{D}^{n}.$

Therefore, we can write

(8.1.4) $\displaystyle
I_{0}(Z,\delta,\gamma)=\int_{\mathcal{O}_{D}^{n}}\mathbf{1}_{\mathrm{nrd}(\eta)\mathcal{O}_{D}}(\upsilon_{i}Z\eta^{\dagger}Y_{i}\eta+\gamma^{\prime}_{i}\eta)\,\psi\left(\frac{\mathrm{trd}(WP(Y)+\rho\cdot
Y)}{\mathrm{nrd}(\eta)\varpi^{v(\delta)-b}}\right)\,dY,$

where

(8.1.5) $\displaystyle\rho$
$\displaystyle:=\gamma^{\prime}/\varpi^{b}\in\mathcal{O}_{D}^{n},\qquad
W:=(Z\eta^{\dagger})/\varpi^{b}\in\mathcal{O}_{D}.$

From the quadratic form $\mathrm{trd}(WP(Y))$ in (8.1.4), we proceed to factor
out generalized Gauss sums along various directions, in an appropriate
iterative manner. This is like a “partial diagonalization” process, but we
will phrase it in the most conceptual way possible. We begin with the most
significant direction of cancellation: translation by scalars. The invertible
affine map $Y_{i}\mapsto Y_{i}+k_{i}$, for $k_{i}\in\mathcal{O}_{F}$, leaves
$Z\eta^{\dagger}Y_{i}\eta\bmod{\mathrm{nrd}(\eta)}$ invariant. Therefore,
changing variables $Y\mapsto Y+k$ in (8.1.4) gives

(8.1.6)
$I_{0}(Z,\delta,\gamma)=\int_{\mathcal{O}_{D}^{n}}\mathbf{1}_{\mathrm{nrd}(\eta)\mathcal{O}_{D}}(\upsilon_{i}Z\eta^{\dagger}Y_{i}\eta+\gamma^{\prime}_{i}\eta)\,\psi\left(\frac{\mathrm{trd}(WP(Y)+\rho\cdot
Y)}{\mathrm{nrd}(\eta)\varpi^{v(\delta)-b}}\right)J_{1}(Y,Z)\,dY,$

where

$J_{1}(Y,Z):=\int_{\mathcal{O}_{F}^{n}}\psi\left(\frac{\mathrm{trd}(W(P(Y+k)-P(Y))+\rho\cdot
k)}{\mathrm{nrd}(\eta)\varpi^{v(\delta)-b}}\right)\,dk.$

Here $P(Y+k)-P(Y)=\sum_{1\leq i\leq
n}(\upsilon_{i}k_{i}^{2}+2\upsilon_{i}k_{i}Y_{i})$, because $k_{i}$ commutes
with $Y_{i}$.

Let

(8.1.7) $\displaystyle a$
$\displaystyle:=\begin{cases}\mathrm{trd}(W)=\mathrm{trd}(Z\eta^{\dagger})\varpi^{-b}&\textrm{
if }\mathbf{1}_{\mathrm{trd}(W)\mid\mathrm{nrd}(\eta)\varpi^{v(\delta)-b}},\\\
\mathrm{nrd}(\eta)\varpi^{v(\delta)-b}&\textrm{ if
}\mathbf{1}_{\mathrm{trd}(W)\nmid\mathrm{nrd}(\eta)\varpi^{v(\delta)-b}},\end{cases}$

so
$v(a)=\min(v(\mathrm{trd}(Z\eta^{\dagger})),v(\delta)+v(\mathrm{nrd}(\eta)))-b\geq
0,$ i.e., $a\in\mathcal{O}_{F}\cap F^{\times}.$ Since $v\nmid 2$, we may use
(2.4.2) and (2.4.3) to evaluate $J_{1}$ and obtain444If $v\mid 2$, this
approach would require extra (doable) casework. But we note that a less
explicit version of the current method, based on changes of variables of the
form $Y\mapsto Y+k+R$ restricted to $\mathrm{trd}(WR)=0$, would likely work at
once for all $v$. However, we find the present iterative approach much more
enlightening.

(8.1.8) $\begin{split}J_{1}(Y,Z)&=\prod_{1\leq i\leq
n}\mathcal{G}(\upsilon_{i}a,\mathrm{nrd}(\eta)\varpi^{v(\delta)-b},\mathrm{trd}(2\upsilon_{i}WY_{i}+\rho_{i}))\\\
&=\mathbf{1}_{a\mathcal{O}_{F}}(\mathrm{trd}(2\upsilon_{i}WY_{i}+\rho_{i}))\,\psi\left(-\sum_{1\leq
i\leq
n}\frac{\mathrm{trd}(2\upsilon_{i}WY_{i}+\rho_{i})^{2}}{4\upsilon_{i}a\,\mathrm{nrd}(\eta)\varpi^{v(\delta)-b}}\right)G_{1}(W),\end{split}$

where $G_{1}(W):=\prod_{1\leq i\leq
n}\mathcal{G}(\upsilon_{i}a,\mathrm{nrd}(\eta)\varpi^{v(\delta)-b},0)$. Here
$G_{1}(W)$ depends (implicitly) on $\delta$ and $Z$, but not on $Y$. Moreover,
by (2.4.1), we have

(8.1.9) $\lvert G_{1}(W)\rvert\leq\prod_{1\leq i\leq
n}\frac{\lvert\mathrm{nrd}(\eta)\varpi^{v(\delta)-b}\rvert^{1/2}}{\lvert\upsilon_{i}a\rvert^{1/2}}=\frac{q^{v(a)n/2}\lvert\mathrm{nrd}(\eta)\rvert^{n/2}}{(q^{v(\delta)-b})^{n/2}}.$

Substituting the formula (8.1.8) for $J_{1}$ into (8.1.6), we get

(8.1.10)
$I_{0}(Z,\delta,\gamma)=\int_{\mathcal{O}_{D}^{n}}S_{1}(Y,Z)\psi\left(\frac{\mathrm{trd}(WP(Y)+\rho\cdot
Y)}{\mathrm{nrd}(\eta)\varpi^{v(\delta)-b}}-\sum_{1\leq i\leq
n}\frac{\mathrm{trd}(2\upsilon_{i}WY_{i}+\rho_{i})^{2}}{4\upsilon_{i}a\,\mathrm{nrd}(\eta)\varpi^{v(\delta)-b}}\right)\,dY,$

where

(8.1.11)
$S_{1}(Y,Z):=\mathbf{1}_{\mathrm{nrd}(\eta)\mathcal{O}_{D}}(\upsilon_{i}Z\eta^{\dagger}Y_{i}\eta+\gamma^{\prime}_{i}\eta)\,\mathbf{1}_{a\mathcal{O}_{F}}(\mathrm{trd}(2\upsilon_{i}WY_{i}+\rho_{i}))\,G_{1}(W).$

Thus

(8.1.12) $\displaystyle\begin{split}\lvert
I_{0}(Z,\delta,\gamma)\rvert&\leq\int_{\mathcal{O}_{D}^{n}}\lvert
S_{1}(Y,Z)\rvert\,dY\\\
&=|G_{1}(W)|\int_{\mathcal{O}_{D}^{n}}\mathbf{1}_{\mathrm{nrd}(\eta)\mathcal{O}_{D}}(\upsilon_{i}Z\eta^{\dagger}Y_{i}\eta+\gamma^{\prime}_{i}\eta)\,\mathbf{1}_{a\mathcal{O}_{F}}(\mathrm{trd}(2\upsilon_{i}WY_{i}+\rho_{i}))\,dY.\end{split}$

######  Lemma 8.5.

One has

$\int_{\mathcal{O}_{D}}\mathbf{1}_{\mathrm{nrd}(\eta)\mathcal{O}_{D}}(\upsilon_{i}Z\eta^{\dagger}Y_{i}\eta+\gamma^{\prime}_{i}\eta)\,dY_{i}\leq
q^{h}|\mathrm{nrd}(\eta)|,$

where $h:=\min(v(ZN),v(\mathrm{nrd}(\eta)))$.

###### Proof.

By Lemma 8.1, the map
$Y_{i}\mapsto\upsilon_{i}Z\eta^{\dagger}Y_{i}\eta\bmod{\mathrm{nrd}(\eta)}$
has image $\mathcal{O}_{F}\upsilon_{i}ZN\bmod{\mathrm{nrd}(\eta)}$, which is
of order $|\mathrm{nrd}(\eta)|^{-1}q^{-h}$ since
$\gcd(\upsilon_{i}ZN,\mathrm{nrd}(\eta))=\varpi^{h}$. Thus the kernel of the
map has order $|\mathrm{nrd}(\eta)|^{-3}q^{h}.$ If $\gamma^{\prime}_{i}\eta$
is in the image, the integral is thus $|\mathrm{nrd}(\eta)|q^{h},$ otherwise
it is zero. ∎

######  Lemma 8.6.

We have $h=\min(b,v(\mathrm{nrd}(\eta)))$.

###### Proof.

Both sides are zero if $v(\mathrm{nrd}(\eta))=0$. Now suppose
$v(\mathrm{nrd}(\eta))>0$. Then

$I_{2}=\left(\begin{smallmatrix}1-\mathrm{nrd}(\eta)&0\\\
0&\mathrm{nrd}(\eta)\end{smallmatrix}\right)+\left(\begin{smallmatrix}\mathrm{nrd}(\eta)&0\\\
0&1-\mathrm{nrd}(\eta)\end{smallmatrix}\right)\in\mathcal{O}_{D}^{\times}\eta\mathcal{O}_{D}^{\times}+\mathcal{O}_{D}^{\times}\eta\mathcal{O}_{D}^{\times},$

by the Cartan decomposition. Therefore, $I_{2}$ lies in the _two-sided ideal_
$\mathcal{O}_{D}\eta\mathcal{O}_{D}$ of $\mathcal{O}_{D}$. Since $N$ generates
$\eta^{\dagger}\frac{\mathcal{O}_{D}}{\mathrm{nrd}(\eta)\mathcal{O}_{D}}\eta$
by definition, we conclude that

$\displaystyle h$ $\displaystyle=\min(v(ZN),v(\mathrm{nrd}(\eta)))=\min_{Y\in
Z\eta^{\dagger}\mathcal{O}_{D}\eta}(v(Y),v(\mathrm{nrd}(\eta)))$
$\displaystyle=\min_{Y\in
Z\eta^{\dagger}\mathcal{O}_{D}\eta\mathcal{O}_{D}}(v(Y),v(\mathrm{nrd}(\eta)))=\min(v(Z\eta^{\dagger}),v(\mathrm{nrd}(\eta)))=\min(b,v(\mathrm{nrd}(\eta))).$

∎

######  Lemma 8.7.

One has

$\int_{\mathcal{O}_{D}}\mathbf{1}_{a\mathcal{O}_{F}}(\mathrm{trd}(2\upsilon_{i}WY_{i}+\rho_{i}))\,dY_{i}=\lvert
a\rvert.$

###### Proof.

By the definitions of $b$ in (8.1.2) and $W$ in (8.1.5), we have

$\min(v(\upsilon_{i}W),v(\mathrm{nrd}(\eta))+v(\delta)-b)=0.$

By (8.1.7) we have $v(a)\leq v(\mathrm{nrd}(\eta))+v(\delta)-b$, whence
$\min(v(\upsilon_{i}W),v(a))=0$. Thus $\min(v(2\upsilon_{i}W),v(a))=0$ since
$v\nmid 2$. Therefore the linear congruence
$\mathrm{trd}(2\upsilon_{i}WY_{i}+\rho_{i})\equiv 0\bmod{a}$ has exactly
$\lvert a\rvert^{-3}$ solutions $Y_{i}\bmod{a}$. Multiplying by $\lvert
a\rvert^{4}$ gives the desired equality. ∎

By Lemmas 8.5, 8.6, and 8.7, substituting (8.1.9) into (8.1.12) gives

(8.1.13) $\lvert
I_{0}(Z,\delta,\gamma)\rvert\leq(q^{-t})^{n}\cdot\frac{q^{v(a)n/2}\lvert\mathrm{nrd}(\eta)\rvert^{n/2}}{q^{(v(\delta)-b)n/2}},$

where

(8.1.14) $\displaystyle q^{-t}:=\min(q^{h}|\mathrm{nrd}(\eta)|,\lvert
a\rvert)=\min(q^{b}|\mathrm{nrd}(\eta)|,\lvert a\rvert).$

Equivalently,

(8.1.15)
$\begin{split}t=\max(v(\mathrm{nrd}(\eta))-b,v(a))=v(\mathrm{nrd}(\eta))+v(a)-\min(v(\mathrm{nrd}(\eta)),v(a)+b).\end{split}$

This completes our study of cancellation in the exponential sum using changes
of variables of the form $Y\mapsto Y+k.$

Let us now consider changes of variables of the form $Y\mapsto Y+kR$ in
(8.1.10), for some $R\in\mathcal{O}_{D}$. Such a change of variables will
preserve $S_{1}(Y,Z)$ if we have $Z\eta^{\dagger}R\eta\equiv
0\bmod{\mathrm{nrd}(\eta)}$ and $\mathrm{trd}(2WR)\equiv 0\bmod{a}$. The
analogue of the argument above will then introduce Gauss sums over
$k_{i}\in\mathcal{O}_{F}$ with modulus
$\mathrm{nrd}(\eta)\varpi^{v(\delta)-b}$ and leading quadratic coefficient

$\upsilon_{i}\mathrm{trd}(WR^{2})-\mathrm{trd}(2\upsilon_{i}WR)^{2}/(4\upsilon_{i}a)=\upsilon_{i}\,(\mathrm{trd}(WR^{2})-\mathrm{trd}(WR)^{2}/a),$

for each $i$. Thus for any $R\in\mathcal{O}_{D}$ such that
$Z\eta^{\dagger}R\eta\equiv 0\bmod{\mathrm{nrd}(\eta)}$ and
$\mathrm{trd}(2WR)\equiv 0\bmod{a}$, we have

(8.1.16) $\displaystyle\lvert
I_{0}(Z,\delta,\gamma)\rvert\leq\int_{\mathcal{O}_{D}^{n}}\frac{|\mathrm{nrd}(\eta)|^{n/2}|S_{1}(Y,Z)|}{(q^{v(\delta)-b}|\mathrm{trd}(WR^{2})-\mathrm{trd}(WR)^{2}/a|)^{n/2}}\,dY$

by (2.4.1) applied to the $n$ Gauss sums just described. Applying our previous
estimates on $|S_{1}(Y,Z)|$ using Gauss sum estimates and Lemmas 8.5 and 8.7
as in (8.1.13) we obtain

(8.1.17) $\displaystyle\lvert
I_{0}(Z,\delta,\gamma)\rvert\leq\frac{q^{v(a)n/2}|\mathrm{nrd}(\eta)|^{n}}{q^{(t+v(\delta)-b)n}|\mathrm{trd}(WR^{2})-\mathrm{trd}(WR)^{2}/a)|^{n/2}}.$

Let us now try to maximize $|\mathrm{trd}(WR^{2})-\mathrm{trd}(WR)^{2}/a|$.

######  Lemma 8.8.

Assume $v(a)<v(\delta)+v(\mathrm{nrd}(\eta))-b$. Then there exists
$R\in\mathcal{O}_{D}$, with $Z\eta^{\dagger}R\eta\equiv
0\bmod{\mathrm{nrd}(\eta)}$ and $\mathrm{trd}(WR)=0$, such that
$v(\mathrm{trd}(WR^{2})-\mathrm{trd}(WR)^{2}/a)=t$.

###### Proof.

Throughout this proof, we let

$m:=\min(v(\mathrm{nrd}(\eta)),v(a)+b).$

By the Cartan decomposition, there exists an element
$\eta_{0}\in\mathcal{O}_{D}\cap D^{\times}$ such that
$v(\mathrm{nrd}(\eta_{0}))=m$ and
$\eta\equiv\eta_{0}\bmod{\varpi^{m}}$.555Indeed, if
$\eta=\left(\begin{smallmatrix}\mathrm{nrd}(\eta)&0\\\
0&1\end{smallmatrix}\right)$, then we could take
$\eta_{0}=\left(\begin{smallmatrix}\varpi^{m}&0\\\
0&1\end{smallmatrix}\right)$. This ensures that

(8.1.18)
$Z\eta_{0}^{\dagger}+\eta_{0}Z^{\dagger}=\mathrm{trd}(Z\eta_{0}^{\dagger})\equiv\mathrm{trd}(Z\eta^{\dagger})=\varpi^{b}\mathrm{trd}(W)\equiv
0\bmod{\varpi^{m}},$

by the definition of $W$ in (8.1.5) and the inequality $v(a)\leq
v(\mathrm{trd}(W)).$

Let
$R_{0}:=\eta\eta_{0}^{-1}\in(\eta_{0}+\mathcal{O}_{D}\mathrm{nrd}(\eta_{0}))\eta_{0}^{-1}=1+\mathcal{O}_{D}\eta_{0}^{\dagger}$.
Then $\eta=R_{0}\eta_{0}$, and $R_{0}\in\mathcal{O}_{D}$, with

(8.1.19)
$v(\mathrm{nrd}(R_{0}))=v(\mathrm{nrd}(\eta))-v(\mathrm{nrd}(\eta_{0}))=v(\mathrm{nrd}(\eta))-m.$

By our assumption $v(a)<v(\delta)+v(\mathrm{nrd}(\eta))-b$, we have

(8.1.20) $v(\mathrm{trd}(W))=v(a)<v(\delta)+v(\mathrm{nrd}(\eta))-b.$

In particular, $W\neq 0$. So $WR_{0}\neq 0$, since $R_{0}\in D^{\times}$ by
(8.1.19). Let $W^{\prime}:=WR_{0}/\varpi^{v(WR_{0})}$.

Next, we construct $S\in\mathcal{O}_{D}$ with $\mathrm{trd}(WR_{0}S)=0$ and
$\mathrm{nrd}(S)\in\mathcal{O}_{F}^{\times}$. For any $W^{\prime}$ we have a
hyperplane $\mathrm{trd}(W^{\prime}S)=0$ in
$\mathbb{P}^{3}_{\mathcal{O}_{F}/\varpi\mathcal{O}_{F}}$, while
$\mathrm{nrd}(S)=0$ defines an irreducible quadric surface. Because no conic
passes through every point of
$\mathbb{P}^{2}(\mathcal{O}_{F}/\varpi\mathcal{O}_{F})$,666Indeed,
$\\#\mathbb{P}^{2}(\mathcal{O}_{F}/\varpi\mathcal{O}_{F})=q^{2}+q+1$, while a
(possibly degenerate) conic has at most $2q+1$ points valued in
$\mathcal{O}_{F}/\varpi\mathcal{O}_{F}$. we conclude that exists
$S_{0}\in\mathcal{O}_{D}$ such that
$\mathrm{trd}(W^{\prime}S_{0})\in\varpi\mathcal{O}_{F}$ and
$\mathrm{nrd}(S_{0})\in\mathcal{O}_{F}^{\times}$. By Hensel’s lemma, there
exists $S\in S_{0}+\varpi\mathcal{O}_{D}$ with $\mathrm{trd}(W^{\prime}S)=0$.
Then $\mathrm{nrd}(S)\equiv\mathrm{nrd}(S_{0})\not\equiv 0\bmod{\varpi}$, so
$S\in\mathcal{O}_{D}^{\times}$.

Finally, let $R:=R_{0}S\in\mathcal{O}_{D}$. Then by (8.1.5), we have
$\mathrm{trd}(Z\eta^{\dagger}R)=\varpi^{b}\mathrm{trd}(WR)=0$, so

$(Z\eta^{\dagger}R)\eta=-(R^{\dagger}\eta
Z^{\dagger})\eta=-(S^{\dagger}R_{0}^{\dagger})(R_{0}\eta_{0})Z^{\dagger}\eta=-S^{\dagger}\mathrm{nrd}(R_{0})(\eta_{0}Z^{\dagger})\eta.$

By (8.1.18), (8.1.19), and the congruence
$\eta\equiv\eta_{0}\bmod{\varpi^{m}}$, we conclude that

$Z\eta^{\dagger}R\eta\equiv
S^{\dagger}\mathrm{nrd}(R_{0})(Z\eta_{0}^{\dagger})\eta_{0}=S^{\dagger}\mathrm{nrd}(R_{0})Z\mathrm{nrd}(\eta_{0})\equiv
0\bmod{\mathrm{nrd}(\eta)},$

since $\mathrm{nrd}(R_{0})\mathrm{nrd}(\eta_{0})=\mathrm{nrd}(\eta)$. Finally,
by Cayley–Hamilton,

$\mathrm{trd}(WR^{2})-\mathrm{trd}(WR)^{2}/a=\mathrm{trd}(R)\mathrm{trd}(WR)-\mathrm{nrd}(R)\mathrm{trd}(W)-0=-\mathrm{nrd}(R)\mathrm{trd}(W),$

since $\mathrm{trd}(WR)=0$. Since $S\in\mathcal{O}_{D}^{\times}$, we have
$v(\mathrm{nrd}(R))=v(\mathrm{nrd}(R_{0}))$, so

$v(\mathrm{nrd}(R))+v(\mathrm{trd}(W))=v(\mathrm{nrd}(\eta))-m+v(a)=t,$

by (8.1.19), (8.1.20), and (8.1.15). Thus
$v(\mathrm{trd}(WR^{2})-\mathrm{trd}(WR)^{2}/a)=t$ and we are done. ∎

###  8.2. Proof of the inequality (8.1.1)

Consider first the case $v(a)<v(\mathrm{nrd}(\eta))+v(\delta)-b$, then by
(8.1.17) and Lemma 8.8, we have

(8.2.1) $\displaystyle\lvert
I_{0}(Z,\delta,\gamma)\rvert\leq\frac{q^{v(a)n/2}|\mathrm{nrd}(\eta)|^{n}}{q^{(t/2+v(\delta)-b)n}}=q^{(v(\mathrm{nrd}(\eta))+v(a)-t+2b)n/2}|\mathrm{nrd}(\eta)|^{3n/2}\left\lVert\delta\right\rVert^{n}.$

The same bound applies to the case $v(a)\geq
v(\mathrm{nrd}(\eta))+v(\delta)-b$. Indeed, in this case we have
$t=v(a)=v(\delta)+v(\mathrm{nrd}(\eta))-b$ by (8.1.7) and (8.1.15). Therefore,
by (8.1.13)

$\displaystyle\lvert
I_{0}(Z,\delta,\gamma)\rvert\leq(q^{-t})^{n}\cdot\frac{q^{v(a)n/2}\lvert\mathrm{nrd}(\eta)\rvert^{n/2}}{q^{(v(\delta)-b)n/2}}=\frac{q^{v(a)n/2}|\mathrm{nrd}(\eta)|^{n}}{q^{(t/2+v(\delta)-b)n}}.$

We claim

$\displaystyle
q^{(v(\mathrm{nrd}(\eta))+v(a)-t+2b)n/2}\leq\max(\left\lVert(\gamma^{\prime}-\gamma^{\prime\dagger})\eta\right\rVert,|\mathrm{nrd}(\eta)|)^{-n/2}\max(\left\lVert\gamma^{\prime}\right\rVert,\left\lVert\delta\right\rVert|\mathrm{nrd}(\eta)|)^{-n}$

and thus (8.1.1) whence Theorem 8.2 follows from (8.2.1). By (8.1.2) and
(8.1.3), we have

$b\leq\min(v(\delta)+v(\mathrm{nrd}(\eta)),v(\gamma^{\prime})).$

On the other hand, by Lemma 8.9 below if $I_{0}(Z,\delta,\gamma)\neq 0$ then

$\min(v((\gamma^{\prime}-\gamma^{\prime\dagger})\,\eta),v(\mathrm{nrd}(\eta)))\geq\min(v(\mathrm{nrd}(\eta)),v(a)+b)=v(\mathrm{nrd}(\eta))+v(a)-t,$

where the final equality holds by (8.1.15). This justifies the claim.

######  Lemma 8.9.

Assume $I_{0}(Z,\delta,\gamma)\neq 0$. Then the following congruence holds:

$(\gamma^{\prime}_{i}-\gamma_{i}^{\prime\dagger})\,\eta\equiv
0\bmod{\varpi^{\min(v(\mathrm{nrd}(\eta)),b+v(a))}\mathcal{O}_{D}}.$

###### Proof.

By (8.1.10), we have $S_{1}(Y,Z)\neq 0$ for some $Y$. Then by (8.1.11) , we
have $\upsilon_{i}Z\eta^{\dagger}Y_{i}\eta+\gamma^{\prime}_{i}\eta\equiv
0\bmod{\mathrm{nrd}(\eta)\mathcal{O}_{D}}$,
$\mathrm{trd}(\upsilon_{i}Z\eta^{\dagger})=\upsilon_{i}\varpi^{b}\mathrm{trd}(W)\equiv
0\bmod{\varpi^{b}a\mathcal{O}_{F}}$ (because $v(a)\leq v(\mathrm{trd}(W))$),
and
$\mathrm{trd}(2\upsilon_{i}Z\eta^{\dagger}Y_{i}+\gamma^{\prime}_{i})=\varpi^{b}\mathrm{trd}(2\upsilon_{i}WY_{i}+\rho_{i})\equiv
0\bmod{\varpi^{b}a\mathcal{O}_{F}}$. Thus the conditions of Lemma 8.10 below
hold with parameters
“$(\delta,A,M,Z,K):=(\eta,Y_{i},\gamma^{\prime}_{i},\upsilon_{i}Z,\varpi^{\min(v(\mathrm{nrd}(\eta)),b+v(a))})$”,
and the desired congruence follows by Lemma 8.10. ∎

######  Lemma 8.10.

Let $A,M,Z\in\mathcal{O}_{D}$ and $K\in\mathcal{O}_{F}$ with
$\mathrm{nrd}(\delta)\in K\mathcal{O}_{F}$. Suppose
$Z\delta^{\dagger}A\delta+M\delta\equiv 0\bmod{K\mathcal{O}_{D}}$,
$\mathrm{trd}(Z\delta^{\dagger})\equiv 0\bmod{K\mathcal{O}_{F}}$, and
$\mathrm{trd}(2Z\delta^{\dagger}A+M)\equiv 0\bmod{K\mathcal{O}_{F}}$. Then
$(M-M^{\dagger})\delta\equiv 0\bmod{K\mathcal{O}_{D}}$.

###### Proof.

Since $Z\delta^{\dagger}+\delta
Z^{\dagger}=\mathrm{trd}(Z\delta^{\dagger})I_{2}\equiv
0\bmod{K\mathcal{O}_{D}}$, we have $Z\delta^{\dagger}\equiv-\delta
Z^{\dagger}\bmod{K\mathcal{O}_{D}}$. On replacing $Z\delta^{\dagger}$ with
$-\delta Z^{\dagger}$ in the assumption, we get $M\delta\equiv\delta
Z^{\dagger}A\delta\bmod{K\mathcal{O}_{D}}$ and
$\mathrm{trd}(M)\equiv\mathrm{trd}(2\delta
Z^{\dagger}A)\bmod{K\mathcal{O}_{F}}$. Right-multiplying the latter by
$\delta$ produces the congruence

$(M+M^{\dagger})\delta\equiv(2\delta
Z^{\dagger}A+2A^{\dagger}Z\delta^{\dagger})\delta\equiv 2\delta
Z^{\dagger}A\delta\equiv 2M\delta\bmod{K\mathcal{O}_{D}}.$

Subtracting $2M\delta$ from both sides gives the result. ∎

∎

##  9\. Applying the geometry of numbers

Let $F$ be a number field. Assume $P(\gamma)=\sum_{1\leq i\leq
n}\upsilon_{i}\gamma_{i}^{2}$, where $\upsilon_{i}\in\mathcal{O}_{F}^{\times}$
are fixed. Recall the $\gamma\neq 0$ contribution $E_{1}(X)$ to (4.0.6),
defined as follows:

$E_{1}(X):=c_{\Phi,X}X^{4(n-1)}\sum_{\gamma\in D^{n}-\\{0\\}}\sum_{\delta\in
D^{\times}}\left(I_{1}\left(\frac{\delta}{X},\frac{\gamma}{X}\right)-I_{0}\left(\frac{\delta}{X},\frac{\gamma}{X}\right)\right).$

This section is devoted to proving the following result:

######  Theorem 9.1.

Assume $F=\mathbb{Q}$ and $D_{\infty},D_{2}$ nonsplit. Fix compactly supported
smooth functions $f=\otimes_{v}f_{v},$ $\Phi=\otimes_{v}\Phi_{v}$ where
$f_{v}=\mathbf{1}_{\mathcal{O}_{D_{v}}^{n}}$ and
$\Phi_{v}=\mathbf{1}_{\mathcal{O}_{D_{v}}^{2}}$ whenever $D_{v}$ is split.
Then $E_{1}(X)\ll_{\epsilon}X^{3n+\epsilon}$ for all $\epsilon>0$.

The idea is to use good exponential sum vanishing results and estimates where
possible, and to control the likelihood of bad events on average over $\gamma$
using Lemma 9.3 based on the geometry of numbers. The assumption
$F=\mathbb{Q}$ is only to avoid the difficulty of treating multiple infinite
places.

To ease the notation, in this section for $M\in\mathcal{O}_{D}^{r}$ we let
$\left\lVert M\right\rVert:=\left\lVert M\right\rVert_{\infty}$ be its box
norm at the archimedean place.

###  9.1. Proof of Theorem 9.1

Throughout this proof we say a place $v$ is split (resp. nonsplit) if $D_{v}$
is split (resp. nonsplit). By the symmetry property (4.0.7) relating $I_{1}$
to $I_{0}$ and the fact that at split places $v$ one has
$\mathbf{1}_{\mathcal{O}_{D_{v}}^{n}}=\mathbf{1}_{\mathcal{O}_{D_{v}}^{n}}\circ\dagger$
and
$\mathbf{1}_{\mathcal{O}_{D_{v}}^{2}}=\mathbf{1}_{\mathcal{O}_{D_{v}}^{2}}^{\mathrm{sw}}\circ\dagger,$
it suffices to show

(9.1.1) $\displaystyle\sum_{\gamma\in D^{n}-\\{0\\}}\sum_{\delta\in
D^{\times}}\left|I_{0}\left(\frac{\delta}{X},\frac{\gamma}{X}\right)\right|\ll_{\epsilon}X^{4-n+\epsilon}.$

Recall that since $f$ and $\Phi$ are pure tensors,

$\displaystyle
I_{0}\left(\frac{\delta}{X},\frac{\gamma}{X}\right)=I_{0,\infty}\left(\frac{\delta}{X},\frac{\gamma}{X}\right)I_{0}^{\infty}\left(\delta,\gamma\right)=I_{0,\infty}\left(\frac{\delta}{X},\frac{\gamma}{X}\right)\prod_{v\neq\infty}I_{0,v}(\delta,\gamma).$

######  Proposition 9.2.

There exists $C\in\mathcal{O}_{F}\cap F^{\times}$ with $|C|_{v}=1$ at all
split places $v$ such that $I_{0}^{\infty}\left(\delta,\gamma\right)=0$ unless
$C\delta\in\mathcal{O}_{D}\cap D^{\times}$. Furthermore, there is a positive
integer $J$ such that

$\displaystyle C\delta$ $\displaystyle=J\eta,\,\textnormal{ where
$\eta\in\mathcal{O}_{D}\cap D^{\times}$ is primitive,}$ $\displaystyle
C^{2}\gamma$ $\displaystyle=J\gamma^{\prime},\textnormal{ where
}\gamma^{\prime}\in\mathcal{O}_{D}^{n}-\\{0\\},$

and

(9.1.2) $\displaystyle C\gamma^{\prime}\eta\in\mu
M_{0}\eta+\mathrm{nrd}(\eta)\mathcal{O}_{D}^{n}$

for some $M_{0}\in\mathcal{O}_{D}$ and primitive
$\mu\in\mathcal{O}^{n}_{F}-\\{0\\}$. Furthermore,
$\left|I_{0}^{\infty}\left(\delta,\gamma\right)\right|$ is dominated by

$\displaystyle\prod_{\begin{subarray}{c}v\neq\infty\\\ \textrm{ nonsplit
}\end{subarray}}\frac{\lvert\mathrm{nrd}(\delta)\rvert_{v}^{n}}{\max(\left\lVert\gamma\right\rVert_{v},\lvert\mathrm{nrd}(\delta)\rvert_{v})^{n}}\prod_{\begin{subarray}{c}v\textrm{
split
}\end{subarray}}\frac{\mathcal{W}_{v}(M_{0},\eta)\,|\mathrm{nrd}(\eta)|^{3n/2}_{v}\left\lVert\delta\right\rVert_{v}^{n}}{\max(\left\lVert(\gamma^{\prime}-\gamma^{\prime\dagger})\eta\right\rVert_{v},|\mathrm{nrd}(\eta)|_{v})^{n/2}\max(\left\lVert\gamma^{\prime}\right\rVert_{v},\left\lVert\delta\right\rVert_{v}|\mathrm{nrd}(\eta)|_{v})^{n}},$

where $\mathcal{W}_{v}(M_{0},\eta)$ is defined in (8.0.9).

###### Proof.

If we ignore (9.1.2), the existence of $C$ and $J$ follows from Proposition
7.5(1) at nonsplit primes and Theorem 8.2 together with (8.0.1) at split
primes.

For (9.1.2), as we can always take $\mu$ to be primitive by scaling $M_{0}$
and $\mu$ appropriately, it suffices to construct any pair
$(\mu,M_{0})\in(\mathcal{O}^{n}_{F}-\\{0\\},\mathcal{O}_{D})$ such that
(9.1.2) holds. Observe that the constraint (9.1.2) over a place $v$ is
trivially satisfied when $|\mathrm{nrd}(\eta)|_{v}=1,$ i.e. no restrictions on
$\mu$ and $M_{0}$ need to be imposed. Therefore, it suffices to verify the
constraint is satisfied locally at a place $v$. For split $v,$ it follows from
Theorem 8.2. For nonsplit $v,$ as $\eta$ is primitive, we have
$v(\mathrm{nrd}(\eta))\in\\{0,1\\}$ by Lemma 7.6. Then (9.1.2) is satisfied
for an appropriate $\mu$ and $M_{0}$ as long as $v(C)\geq 1,$ which we can
always assume to be true.

Finally, the bound of $\left|I_{0}^{\infty}\left(\delta,\gamma\right)\right|$
follows from Proposition 7.5(2) and Theorem 8.2. ∎

Since $\Phi$ is compactly supported, if
$I_{0}\left(\frac{\delta}{X},\frac{\gamma}{X}\right)\neq 0,$ then
$\left\lVert\delta\right\rVert\ll X$ and

$\displaystyle
I_{0,\infty}\left(\frac{\delta}{X},\frac{\gamma}{X}\right)\ll_{A,\epsilon}(1+\left\lVert\gamma\right\rVert/\left\lVert\delta\right\rVert)^{-A}(1+\left\lVert
X\gamma/\mathrm{nrd}(\delta)\right\rVert)^{\epsilon-n}$

for any $A\geq 0$ and $0<\epsilon<1/2$ by Proposition 7.4. Assume first that
$\delta$ and $\gamma$ satisfy $\left\lVert\gamma\right\rVert\geq
X^{\epsilon}\left\lVert\delta\right\rVert.$ Then
$I_{0,\infty}\left(\frac{\delta}{X},\frac{\gamma}{X}\right)\ll_{\epsilon,A}X^{-A}.$
On the other hand, by Proposition 9.2, $I_{0}^{\infty}(\delta,\gamma)$ is
supported in a compact subset of $D_{\mathbb{A}_{F}^{\infty}}\times
D_{\mathbb{A}_{F}^{\infty}}^{n}$ and one trivially has
$I_{0}^{\infty}(\delta,\gamma)\ll 1.$ Hence the contribution to $E_{1}(X)$ of
$\delta$ and $\gamma$ satisfying $\left\lVert\gamma\right\rVert\geq
X^{\epsilon}\left\lVert\delta\right\rVert$ is $\ll_{\epsilon,A}X^{-A}$ for any
$A\in\mathbb{R}.$

We henceforth assume $\left\lVert\gamma\right\rVert\leq
X^{\epsilon}\left\lVert\delta\right\rVert,$ so

(9.1.3) $\displaystyle
I_{0,\infty}\left(\frac{\delta}{X},\frac{\gamma}{X}\right)$
$\displaystyle\ll_{\epsilon}(1+\left\lVert
X\gamma/\mathrm{nrd}(\delta)\right\rVert)^{\epsilon-n}.$

Let us rewrite the sum over $(\delta,\gamma)$ using the constraint (9.1.2).
Let $m=\mathrm{nrd}(\eta)\in\mathcal{O}_{F}\cap F^{\times}.$ For
$H,K\in\mathcal{O}_{F}\cap F^{\times},$ define the $\mathcal{O}_{F}$-lattice

$\begin{split}\Lambda(H,K,m,\eta,M_{0}):=\left\\{M\in
H\mathcal{O}_{D}:\begin{array}[]{ll}(M-M^{\dagger})\eta\in
K\mathcal{O}_{D},\\\
M\eta\in\mathcal{O}_{F}M_{0}\eta+m\mathcal{O}_{D}\end{array}\right\\}.\end{split}$

Note that for a given $\eta,$ $\Lambda(H,K,m,\eta,M_{0})$ depends only on the
class of $M_{0}$ in the quotient set

(9.1.4)
$(\mathcal{O}_{F}/m\mathcal{O}_{F})^{\times}\backslash(\mathcal{O}_{D}/\mathcal{O}_{D}\eta^{\dagger})\cong\prod_{\textrm{finite
}v\mid
m}(\mathcal{O}_{F_{v}}^{\times}\backslash(\mathcal{O}_{D_{v}}/\mathcal{O}_{D_{v}}\eta^{\dagger})).$

Take $H:=\gcd(\gamma^{\prime},Jm).$ Then we have
$C\gamma^{\prime}\in\Lambda(H,1,m,\eta,M_{0})^{n}$ by definition. Let
$K=K(\gamma^{\prime},\eta,M_{0})$ be the largest divisor of $m$ such that

$C\gamma^{\prime}\in\Lambda(H,K,m,\eta,M_{0})^{n}.$

Let $S=S(\eta)\subseteq\mathcal{O}_{D}$ be a complete set of representatives
for the quotient set (9.1.4). Then by replacing $M_{0}$ with the
representative for its image in (9.1.4), we may assume $M_{0}\in S$, while
keeping $\mu$ primitive.

If $v$ is split, we have

$\begin{split}\frac{\mathcal{W}_{v}(M_{0},\eta)\,|\mathrm{nrd}(\eta)|^{3n/2}_{v}\left\lVert\delta\right\rVert_{v}^{n}}{\max(\left\lVert(\gamma^{\prime}-\gamma^{\prime\dagger})\eta\right\rVert_{v},|\mathrm{nrd}(\eta)|_{v})^{n/2}\max(\left\lVert\gamma^{\prime}\right\rVert_{v},\left\lVert\delta\right\rVert_{v}|\mathrm{nrd}(\eta)|_{v})^{n}}=\frac{\mathcal{W}_{v}(M_{0},\eta)\,|J^{2}m^{3}|_{v}^{n/2}}{|KH^{2}|^{n/2}_{v}}.\end{split}$

When $v$ is finite nonsplit,

$\frac{\lvert\mathrm{nrd}(\delta)\rvert_{v}^{n}}{\max(\left\lVert\gamma\right\rVert_{v},\lvert\mathrm{nrd}(\delta)\rvert_{v})^{n}}=\frac{\lvert
J^{2}m/C^{2}\rvert_{v}^{n}}{\max(\left\lVert
J\gamma^{\prime}/C^{2}\right\rVert_{v},\lvert
J^{2}m/C^{2}\rvert_{v})^{n}}\asymp_{v}\frac{|J^{2}m^{3}|_{v}^{n/2}}{|KH^{2}|^{n/2}_{v}},$

since $0\leq v(K)\leq v(m)=v(\mathrm{nrd}(\eta))\leq 1$. By Proposition 9.2
and the product formula, we have

(9.1.5)
$I_{0}^{\infty}(\delta,\gamma)\ll\frac{K^{n/2}H^{n}}{J^{n}m^{3n/2}}\mathcal{W}(M_{0},\eta).$

where $\mathcal{W}(M_{0},\eta):=\prod_{\textnormal{split }v\mid
Jm}\mathcal{W}_{v}(M_{0},\eta)$. Say

$\displaystyle\left\lVert\gamma^{\prime}\right\rVert\in[R,2R),$

where $R$ is _dyadic_ , i.e. $R\in\\{2^{t}:t\in\mathbb{Z}\\}$. Then by (9.1.3)
we have

(9.1.6)
$\begin{split}I_{0,\infty}\left(\frac{\delta}{X},\frac{\gamma}{X}\right)&\ll_{\epsilon}(1+\left\lVert
X\gamma/\mathrm{nrd}(\delta)\right\rVert)^{\epsilon-n}\\\
&\asymp(1+(XJR)/(J^{2}m))^{\epsilon-n}=(1+XR/(Jm))^{\epsilon-n}.\end{split}$

Since $\left\lVert\delta\right\rVert\ll X$, we have
$\left\lVert\eta\right\rVert\ll X/J$ and by definition
$m\asymp\left\lVert\eta\right\rVert^{2}$ as $D_{\infty}$ is nonsplit. Observe
that

$1\leq H\ll\left\lVert\gamma^{\prime}\right\rVert\asymp
R\asymp\left\lVert\gamma/J\right\rVert\leq
X^{\epsilon}\left\lVert\delta/J\right\rVert\asymp
X^{\epsilon}\left\lVert\eta\right\rVert.$

In summary, $H\ll R\ll X^{\epsilon}\left\lVert\eta\right\rVert$. Then
combining (9.1.5) and (9.1.6), we have

(9.1.7) $\begin{split}&\sum_{\gamma\in D^{n}-\\{0\\}}\sum_{\delta\in
D^{\times}}\left|I_{0}\left(\frac{\delta}{X},\frac{\gamma}{X}\right)\right|\\\
&\ll_{\epsilon}\sum_{J\geq 1}\sum_{\left\lVert\eta\right\rVert\ll
X/J}\sum_{\begin{subarray}{c}H\mid Jm\\\ K\mid
m\end{subarray}}\sum_{\begin{subarray}{c}\textnormal{dyadic }R\\\ H\ll R\ll
X^{\epsilon}\left\lVert\eta\right\rVert\end{subarray}}\sum_{M_{0}\in
S}\sum_{\begin{subarray}{c}\gamma^{\prime}\in C^{-1}\Lambda^{n}\\\
0<\left\lVert\gamma^{\prime}\right\rVert\ll
R\end{subarray}}\frac{\mathcal{W}(M_{0},\eta)K^{n/2}H^{n}}{(1+XR/(Jm))^{n-\epsilon}J^{n}m^{3n/2}},\\\
&\ll_{\epsilon}X^{\epsilon}\sum_{J\geq 1}\sum_{\left\lVert\eta\right\rVert\ll
X/J}\sum_{\begin{subarray}{c}H\mid Jm\\\ K\mid
m\end{subarray}}\sum_{\begin{subarray}{c}\textnormal{dyadic }R\\\ H\ll R\ll
X^{\epsilon}\left\lVert\eta\right\rVert\end{subarray}}\sup_{M_{0}\in\mathcal{O}_{D}}\sum_{\begin{subarray}{c}\gamma^{\prime}\in
C^{-1}\Lambda^{n}\\\ 0<\left\lVert\gamma^{\prime}\right\rVert\ll
R\end{subarray}}\frac{K^{n/2}H^{n}}{(1+XR/(Jm))^{n-\epsilon}J^{n}m^{3n/2}},\end{split}$

where $\Lambda:=\Lambda(H,K,m,\eta,M_{0})$, and where in the final step we sum
$\mathcal{W}(M_{0},\eta)$ over $M_{0}\in S$ via Lemma 8.3 at split places
$v\mid m$ and via the bound
$\\#(\mathcal{O}_{F_{v}}^{\times}\backslash(\mathcal{O}_{D_{v}}/\mathcal{O}_{D_{v}}\eta^{\dagger}))\leq\\#(\mathcal{O}_{D_{v}}/m\mathcal{O}_{D_{v}})=\lvert
m\rvert_{v}^{-4}\leq q_{v}^{4}$ at nonsplit places $v\mid m$.

We postpone the proof of the following key lemma:

######  Lemma 9.3.

Assume $F=\mathbb{Q}$ and $D$ nonsplit. Let $H,K,m\geq 1$ be integers. Suppose
$K\mid m\mid\mathrm{nrd}(\eta)$, where $\eta\in\mathcal{O}_{D}\cap D^{\times}$
is primitive. Let $M_{0}\in\mathcal{O}_{D}$. Then for all $R>0$, we have

$\begin{split}&\\#\\{M\in\Lambda(H,K,m,\eta,M_{0}):\left\lVert
M\right\rVert\leq R\\}\\\ &\ll
1+\frac{R}{H}+\frac{(R/H)^{2}}{(K^{\prime})^{1/2}}+\frac{(R/H)^{3}}{(K^{\prime}m^{\prime})^{1/2}}+\frac{(R/H)^{4}}{K^{\prime}m^{\prime}},\end{split}$

where $K^{\prime}:=K/\gcd(K,H)$ and $m^{\prime}:=m/\gcd(m,H)$.

Assuming Lemma 9.3, we have

$\sup_{M_{0}\in\mathcal{O}_{D}}\sum_{\begin{subarray}{c}\gamma^{\prime}\in
C^{-1}\Lambda^{n}\\\ 0<\left\lVert\gamma^{\prime}\right\rVert\ll
R\end{subarray}}1\ll\left(1+\frac{R}{H}+\frac{(R/H)^{2}}{(K^{\prime})^{1/2}}+\frac{(R/H)^{3}}{(K^{\prime}m^{\prime})^{1/2}}+\frac{(R/H)^{4}}{K^{\prime}m^{\prime}}\right)^{n},$

Since $R\ll X^{\epsilon}\left\lVert\eta\right\rVert\asymp
X^{\epsilon}m^{1/2},$

$\frac{(R/H)^{4}}{K^{\prime}m^{\prime}}\ll
X^{\epsilon}\frac{(R/H)^{3}}{(K^{\prime}m^{\prime})^{1/2}}\ll
X^{2\epsilon}\frac{(R/H)^{2}}{(K^{\prime})^{1/2}},$

by comparing consecutive terms using the bound $R/X^{\epsilon}\ll m^{1/2}\leq
H(m^{\prime})^{1/2}$. Since $1\ll\frac{R}{H}$ as well, the penultimate display
now implies the simpler bound

(9.1.8)
$\sup_{M_{0}\in\mathcal{O}_{D}}\sum_{\begin{subarray}{c}\gamma^{\prime}\in
C^{-1}\Lambda^{n}\\\ 0<\left\lVert\gamma^{\prime}\right\rVert\ll
R\end{subarray}}1\ll
X^{2n\epsilon}\left(\frac{R}{H}+\frac{R^{2}}{H^{2}(K^{\prime})^{1/2}}\right)^{n}.$

A direct application of (9.1.8), for each $\Lambda$, gives (for any $R$ with
$1\ll R\ll X^{\epsilon}\left\lVert\eta\right\rVert$)

$\begin{split}\sum_{K\mid m}\sum_{\begin{subarray}{c}1\leq H\ll R\\\ H\mid
Jm\end{subarray}}\sup_{M_{0}\in\mathcal{O}_{D}}\sum_{\begin{subarray}{c}\gamma^{\prime}\in
C^{-1}\Lambda^{n}\\\ 0<\left\lVert\gamma^{\prime}\right\rVert\ll
R\end{subarray}}K^{n/2}H^{n}&\ll X^{2n\epsilon}\sum_{K\mid
m}\sum_{\begin{subarray}{c}1\leq H\ll R\\\ H\mid
Jm\end{subarray}}K^{n/2}H^{n}\left(\frac{R^{n}}{H^{n}}+\frac{R^{2n}}{H^{2n}(K^{\prime})^{n/2}}\right)\\\
&=X^{2n\epsilon}\sum_{K\mid m}\sum_{\begin{subarray}{c}1\leq H\ll R\\\ H\mid
Jm\end{subarray}}\left(K^{n/2}R^{n}+\frac{\gcd(K,H)^{n/2}R^{2n}}{H^{n}}\right)\\\
&\ll_{\epsilon}(Xm)^{2n\epsilon}\sum_{\begin{subarray}{c}1\leq H\ll R\\\ H\mid
Jm\end{subarray}}\left(m^{n/2}R^{n}+\frac{\gcd(m,H)^{n/2}R^{2n}}{H^{n}}\right),\end{split}$

by the divisor bound on $m$. Now, by the divisor bound on $Jm$, this is
dominated by

$\begin{split}&(XJm)^{4n\epsilon}\max_{\begin{subarray}{c}1\leq H\ll R\\\
H\mid
Jm\end{subarray}}\left(m^{n/2}R^{n}+\frac{\gcd(m,H)^{n/2}R^{2n}}{H^{n}}\right)\\\
&\leq(XJm)^{4n\epsilon}(m^{n/2}R^{n}+R^{2n})\\\
&\ll_{\epsilon}(XJm)^{5n\epsilon}m^{n/2}R^{n},\end{split}$

since $R\ll_{\epsilon}X^{\epsilon}m^{1/2}$. Substituting this into our
previous bound (LABEL:INEQ:first-E(X)-bound), we get

$\displaystyle X^{O(\epsilon)}\sum_{J\geq
1}\sum_{\left\lVert\eta\right\rVert\ll
X/J}\sum_{\begin{subarray}{c}\textnormal{dyadic }R\\\ 1\ll R\ll
X^{\epsilon}\left\lVert\eta\right\rVert\end{subarray}}\frac{m^{n/2}R^{n}}{(1+XR/(Jm))^{n-\epsilon}J^{n}m^{3n/2}}$
$\displaystyle\ll X^{-n+O(\epsilon)}\sum_{J\geq
1}\sum_{\left\lVert\eta\right\rVert\ll
X/J}\sum_{\begin{subarray}{c}\textnormal{dyadic }R\\\ 1\ll R\ll
X^{\epsilon}\left\lVert\eta\right\rVert\end{subarray}}1$ $\displaystyle\ll
X^{-n+O(\epsilon)}\sum_{J\geq 1}\sum_{\left\lVert\eta\right\rVert\ll X/J}1.$

However,

$\sum_{J\geq 1}\sum_{\left\lVert\eta\right\rVert\ll X/J}1\ll\sum_{J\geq
1}(X/J)^{4}\ll X^{4},$

so (9.1.1) holds as desired. ∎

We are left with proving Lemma 9.3 which concerns the
$\mathcal{O}_{F}$-lattices

(9.1.9) $\begin{split}\Lambda(H,K,m,\eta,M_{0}):=\left\\{M\in
H\mathcal{O}_{D}:\begin{array}[]{ll}(M-M^{\dagger})\eta\in
K\mathcal{O}_{D},\\\
M\eta\in\mathcal{O}_{F}M_{0}\eta+m\mathcal{O}_{D}\end{array}\right\\}.\end{split}$

for $H,K,m\in\mathcal{O}_{F}\cap F^{\times}$, primitive
$\eta\in\mathcal{O}_{D}\cap D^{\times}$ such that $K\mid
m\mid\mathrm{nrd}(\eta)$, and $M_{0}\in\mathcal{O}_{D}.$ We need to show for
$R\geq 0$

(9.1.10) $\begin{split}&\\#\\{M\in\Lambda(H,K,m,\eta,M_{0}):\left\lVert
M\right\rVert\leq R\\}\\\ &\ll
1+\frac{R}{H}+\frac{(R/H)^{2}}{(K^{\prime})^{1/2}}+\frac{(R/H)^{3}}{(K^{\prime}m^{\prime})^{1/2}}+\frac{(R/H)^{4}}{K^{\prime}m^{\prime}},\end{split}$

where $K^{\prime}:=K/\gcd(K,H)$ and $m^{\prime}:=m/\gcd(m,H)$.

We remark that in the proof of Theorem 9.1 we apply the lemma with
$m=\mathrm{nrd}(\eta).$ However, allowing general $m\mid\mathrm{nrd}(\eta)$
turns out to be useful for the statement and proof of the lemma. The proof of
Lemma 9.3 proceeds by techniques from Geometry of Numbers. While the messier
case of $H\neq 1$ plays an important role in the proof of Theorem 9.1, we show
that the general case follows from the case $H=1$ in Lemma 9.5 below, so that
one can actually neglect the awkward condition $M\in H\mathcal{O}_{D}$ in
their first reading.

###  9.2. Proof of Lemma 9.3

Since $\eta$ is primitive, we have $0\leq v(\mathrm{nrd}(\eta))\leq 1$ if
$D_{v}$ is nonsplit, by Lemma 7.6. Since $K\mid m\mid\mathrm{nrd}(\eta)$, to
prove Lemma 9.3 it suffices to prove the lemma in the special case where
$v(K)=v(m)=0$ if $v$ is a finite place where $D_{v}$ is nonsplit. In this
case, we will need the following preliminary observations.

######  Proposition 9.4.

Assume $F=\mathbb{Q}$. Let $\eta\in\mathcal{O}_{D}\cap D^{\times}$ be
primitive and let $K\geq 1$ be such that $K\mid\mathrm{nrd}(\eta)$, where $D$
is split at all primes dividing $K$. Then the following statements hold.

  1. (1)

Suppose that $A\in\mathcal{O}_{D}$ and $A\eta\equiv 0\bmod{K\mathcal{O}_{D}}$.
Then $K\mid\mathrm{nrd}(A)$.

  2. (2)

One has

$\\#\\{\theta\in\mathcal{O}_{D}/K\mathcal{O}_{D}:\eta\theta\equiv
0\bmod{K\mathcal{O}_{D}}\\}=K^{2}.$

  3. (3)

There exists $\theta\in\mathcal{O}_{D}$ such that $\eta\theta\equiv
0\bmod{K\mathcal{O}_{D}}$ and $0<\left\lVert\theta\right\rVert\ll K^{1/2}$.

  4. (4)

There exists $\eta^{\prime}\in\eta\mathcal{O}_{D}+K\mathcal{O}_{D}$ such that
$0<\left\lVert\eta^{\prime}\right\rVert\ll K^{1/2}$.

###### Proof.

For (1) and (2), we observe that, by factoring $K$ into prime powers, and
using the assumption that $D$ splits at all prime divisors of $K$, an
application of the Chinese Remainder Theorem renders it sufficient to prove
the following local statements, in both of which we assume that $\eta\in
M_{2}(\mathcal{O}_{F_{v}})\cap\mathrm{GL}_{2}(F_{v})$ is primitive, and
$K\in\mathcal{O}_{F_{v}}$ is such that $K\mid\mathrm{nrd}(\eta)$:

  1. (1’)

Suppose that $A\in M_{2}(\mathcal{O}_{F_{v}})$ is such that $A\eta\equiv
0\bmod{K}$. Then $K\mid\mathrm{nrd}(A)$.

  2. (2’)

The congruence

$\eta\theta\equiv 0\bmod{K}\text{,}$

to be solved in $\theta\in M_{2}(\mathcal{O}_{F_{v}})$, has exactly $K^{2}$
solutions modulo $K$.

To prove (1’), by a Cartan decomposition argument we may assume that
$\eta=\left(\begin{smallmatrix}\varpi^{k}&0\\\ 0&1\end{smallmatrix}\right)$
for some integer $k\geq v(K)$. If $A=\left(\begin{smallmatrix}s&u\\\
t&v\end{smallmatrix}\right)$, then the condition $A\eta\equiv 0\bmod{K}$
becomes $K\mid u,v$, whence $K\mid sv-tu=\mathrm{nrd}(A)$. To prove (2’),
again by a Cartan decomposition argument we may assume that
$\eta=\left(\begin{smallmatrix}\varpi^{k}&0\\\ 0&1\end{smallmatrix}\right)$,
whence if $\theta=\left(\begin{smallmatrix}a&b\\\
c&d\end{smallmatrix}\right)$, the congruence to be solved becomes $c\equiv
d\equiv 0\bmod{K}$, after which the result is clear. This finishes the proofs
of (1’) and (2’), and hence of (1) and (2).

To prove (3), consider the lattice

$L=\\{\theta\in\mathcal{O}_{D}:\eta\theta\equiv 0\bmod{K}\\}\subseteq
D\text{.}$

Note that $K\mathcal{O}_{D}\subseteq L\subseteq\mathcal{O}_{D}$, and by (2) we
have $[L:K\mathcal{O}_{D}]=K^{2}$. Therefore,

$\mathrm{covol}(L)\asymp[\mathcal{O}_{D}:L]=\frac{[\mathcal{O}_{D}:K\mathcal{O}_{D}]}{[L:K\mathcal{O}_{D}]}=\frac{K^{4}}{K^{2}}=K^{2}\text{.}$

Hence, by Minkowski’s Theorem, $L$ contains a nonzero vector of magnitude at
most $O(\mathrm{covol}(L)^{1/4})=O(K^{1/2})$, which translates into the
desired result.

Finally, to prove (4), consider the lattice

$R=\eta\mathcal{O}_{D}+K\mathcal{O}_{D}\text{.}$

Note that $K\mathcal{O}_{D}\subseteq R\subseteq\mathcal{O}_{D}$, and

(9.2.1)
$\mathrm{covol}(R)\asymp[\mathcal{O}_{D}:R]=\frac{[\mathcal{O}_{D}:K\mathcal{O}_{D}]}{[R:K\mathcal{O}_{D}]}=\frac{K^{4}}{[R:K\mathcal{O}_{D}]}\text{.}$

Consider the $\mathbb{Z}$-module homomorphism

$\begin{array}[]{ccc}\varphi:&\mathcal{O}_{D}/K\mathcal{O}_{D}&\longrightarrow\mathcal{O}_{D}/K\mathcal{O}_{D}\\\
&\theta+K\mathcal{O}_{D}&\longmapsto\eta\theta+K\mathcal{O}_{D}\end{array}$

and observe that $[R:K\mathcal{O}_{D}]=|\mathrm{im}(\varphi)|$. Note that
$|\ker(\varphi)|=K^{2}$ by (2). By the First Isomorphism Theorem it follows
that

$[R:K\mathcal{O}_{D}]=\frac{|\mathcal{O}_{D}/K\mathcal{O}_{D}|}{|\ker(\varphi)|}=\frac{K^{4}}{K^{2}}=K^{2}\text{.}$

Inserting this into (9.2.1), it follows that

$\mathrm{covol}(R)\asymp\frac{K^{4}}{K^{2}}=K^{2}\text{.}$

Now Minkowski’s Theorem implies that $R$ contains a nonzero vector of
magnitude at most $O(\mathrm{covol}(R)^{1/4})=O(K^{1/2})$, as desired. ∎

The following lemma will help us deal with the awkward condition $M\in
H\mathcal{O}_{D}$ in (9.1.9).

######  Lemma 9.5.

Let $F,D,H,K,m,\eta,M_{0},K^{\prime},m^{\prime}$ be as in Lemma 9.3. Assume
that $v(K)=v(m)=0$ if $D_{v}$ is nonsplit. Then there exist integers
$K^{\star},m^{\star}\geq 1$, a vector $Y\in\mathcal{O}_{D}$, and a primitive
vector $\sigma\in\mathcal{O}_{D}\cap D^{\times}$, such that the following
hold:

  1. (1)

$K^{\star}\mid m^{\star}\mid\mathrm{nrd}(\sigma)$;

  2. (2)

$K^{\star}\asymp K^{\prime}$ and $m^{\star}\asymp m^{\prime}$;

  3. (3)

$\left\lVert\sigma\right\rVert\asymp(m^{\star})^{1/2}$ and
$\lvert\mathrm{nrd}(\sigma)\rvert\asymp m^{\star}$;

  4. (4)

there is an injection

$\displaystyle\phi:\Lambda(H,K,m,\eta,M_{0})$
$\displaystyle\longrightarrow\Lambda(1,K^{\star},m^{\star},\sigma,Y)$
$\displaystyle M$ $\displaystyle\longmapsto M/H.$

###### Proof.

For convenience, we define the map

$\Psi:H\mathcal{O}_{D}\to\mathcal{O}_{D}/m\mathcal{O}_{D},\quad M\mapsto
M\eta+m\mathcal{O}_{D}.$

For each $A\in\mathcal{O}_{D}$, we also define the map

$T_{A}:\mathcal{O}_{F}/m\mathcal{O}_{F}\to\mathcal{O}_{D}/m\mathcal{O}_{D},\quad\lambda\mapsto\lambda
A\eta+m\mathcal{O}_{D}.$

Observe that for any $M\in H\mathcal{O}_{D}$, we have

$M\eta\in\mathcal{O}_{F}A\eta+m\mathcal{O}_{D}\Leftrightarrow\Psi(M)\in\mathrm{im}(T_{A}).$

Both $\Psi$ and $T_{A}$ are $\mathcal{O}_{F}$-module homomorphisms. We will
use these maps to gradually alter $M_{0}$ into a friendlier form. Consider the
ideal

$I:=T_{M_{0}}^{-1}(\mathrm{im}(\Psi))=\\{x\in\mathcal{O}_{F}/m\mathcal{O}_{F}:T_{M_{0}}(x)\in\mathrm{im}(\Psi)\\}\leq\mathcal{O}_{F}/m\mathcal{O}_{F}.$

All ideals of $\mathcal{O}_{F}/m\mathcal{O}_{F}$ are principal, so $I$ is
generated by some $\lambda_{0}\in\mathcal{O}_{F}$. Let
$M_{1}:=\lambda_{0}M_{0}\in\mathcal{O}_{D}$. Then

$\mathrm{im}(T_{M_{1}})=T_{M_{0}}(I)=\mathrm{im}(\Psi)\cap\mathrm{im}(T_{M_{0}}).$

So for any $M\in H\mathcal{O}_{D}$, we have

$\Psi(M)\in\mathrm{im}(T_{M_{0}})\Leftrightarrow\Psi(M)\in\mathrm{im}(\Psi)\cap\mathrm{im}(T_{M_{0}})\Leftrightarrow\Psi(M)\in\mathrm{im}(T_{M_{1}}).$

Thus $\Lambda(H,K,m,\eta,M_{0})=\Lambda(H,K,m,\eta,M_{1})$. Yet
$T_{M_{1}}^{-1}(\mathrm{im}(\Psi))=\mathcal{O}_{F}/m\mathcal{O}_{F}$, since
$\mathrm{im}(T_{M_{1}})\leq\mathrm{im}(\Psi)$. Replacing $M_{0}$ with $M_{1}$
if necessary, we may thus assume that $I=\mathcal{O}_{F}/m\mathcal{O}_{F}$.

Since $1\in I$, there exists $M_{2}\in H\mathcal{O}_{D}$ such that
$T_{M_{0}}(1)=\Psi(M_{2})$, i.e.

$M_{0}\eta\equiv M_{2}\eta\bmod{m\mathcal{O}_{D}}.$

But then $\mathrm{im}(T_{M_{0}})=\mathrm{im}(T_{M_{2}})$, so
$\Lambda(H,K,m,\eta,M_{0})=\Lambda(H,K,m,\eta,M_{2})$. Replacing $M_{0}$ with
$M_{2}$ if necessary, we may therefore assume that $M_{0}\in
H\mathcal{O}_{D}$. Let $M_{3}:=M_{0}/H\in\mathcal{O}_{D}$.

Since $\operatorname{lcm}(H,K)=HK^{\prime}$ and
$\operatorname{lcm}(H,m)=Hm^{\prime}$, we have

$\Lambda(H,K,m,\eta,M_{0})=\Lambda(H,HK^{\prime},Hm^{\prime},\eta,M_{0})$

by (9.1.9), because $H\mid M_{0}$. Now, dividing the conditions in (9.1.9)
through by $H$, we see that the formula $M\mapsto M/H$ maps
$\Lambda(H,K,m,\eta,M_{0})$ into
$\Lambda(1,K^{\prime},m^{\prime},\eta,M_{3})$.

Clearly $K^{\prime}\mid m^{\prime}\mid\mathrm{nrd}(\eta)$, since $K\mid
m\mid\mathrm{nrd}(\eta)$. By Proposition 9.4(4) with
“$(\eta,K):=(\eta,m^{\prime})$”, there exists a nonzero vector
$\eta^{\prime}\in\eta\mathcal{O}_{D}+m^{\prime}\mathcal{O}_{D}$ such that
$\left\lVert\eta^{\prime}\right\rVert\ll(m^{\prime})^{1/2}$. Since $D$ is
nonsplit, we have $\eta^{\prime}\in D^{\times}$, i.e.
$\mathrm{nrd}(\eta^{\prime})\neq 0$. Say
$\eta^{\prime}\in\eta\alpha+m^{\prime}\mathcal{O}_{D}$, where
$\alpha\in\mathcal{O}_{D}$. Then multiplying the conditions in (9.1.9) by
$\alpha$ on the right, we find that

$\Lambda(1,K^{\prime},m^{\prime},\eta,M_{3})\subseteq\Lambda(1,K^{\prime},m^{\prime},\eta^{\prime},M_{3}).$

Moreover, $\mathrm{nrd}(\eta^{\prime})\equiv\mathrm{nrd}(\eta\alpha)\equiv
0\bmod{m^{\prime}}$, since $m^{\prime}\mid\mathrm{nrd}(\eta)$. In particular,
$\lvert\mathrm{nrd}(\eta^{\prime})\rvert\asymp m^{\prime}$ and
$\left\lVert\eta^{\prime}\right\rVert\asymp(m^{\prime})^{1/2}$, since
$\left\lVert\eta^{\prime}\right\rVert\ll(m^{\prime})^{1/2}$.

We are almost done, but $\eta^{\prime}\in\mathcal{O}_{D}\cap D^{\times}$ need
not be primitive. Write $\eta^{\prime}=z\sigma$, where $z\geq 1$ and
$\sigma\in\mathcal{O}_{D}\cap D^{\times}$ is primitive. Then
$\gcd(m^{\prime},z)\mid\eta\alpha$, since
$\eta^{\prime}\in\eta\alpha+m^{\prime}\mathcal{O}_{D}$. So by Proposition
9.4(1) with
“$(\eta,K,A):=(\eta^{\dagger},\gcd(m^{\prime},z),\alpha^{\dagger})$”, we have
$\gcd(m^{\prime},z)\mid\mathrm{nrd}(\alpha)$. Thus

$\begin{split}\mathrm{nrd}(\eta^{\prime})=(\eta^{\prime})(\eta^{\prime})^{\dagger}&\in(\eta\alpha+m^{\prime}\mathcal{O}_{D})((\eta\alpha)^{\dagger}+m^{\prime}\mathcal{O}_{D})\\\
&\subseteq\mathrm{nrd}(\eta\alpha)+\gcd(m^{\prime},z)m^{\prime}\mathcal{O}_{D}=\gcd(m^{\prime},z)m^{\prime}\mathcal{O}_{D},\end{split}$

since $m^{\prime}\mid\mathrm{nrd}(\eta)$. But
$\mathrm{nrd}(\eta^{\prime})=z^{2}\mathrm{nrd}(\sigma)$, so it follows that

$\gcd(m^{\prime},z)m^{\prime}\mid\gcd(z^{2},\gcd(m^{\prime},z)m^{\prime})\mathrm{nrd}(\sigma)=\gcd(m^{\prime},z)^{2}\mathrm{nrd}(\sigma),$

whence $m^{\prime}\mid\gcd(m^{\prime},z)\mathrm{nrd}(\sigma)\mid
z\,\mathrm{nrd}(\sigma)$. Yet
$\left\lVert\sigma\right\rVert=\left\lVert\eta^{\prime}\right\rVert/z\asymp(m^{\prime})^{1/2}/z$,
so we conclude that

$m^{\prime}\leq z\,\lvert\mathrm{nrd}(\sigma)\rvert\ll
z\left\lVert\sigma\right\rVert^{2}\asymp z(m^{\prime}/z^{2})=m^{\prime}/z.$

Thus $z\ll 1$. In other words, $\eta^{\prime}$ is either primitive, or nearly
primitive.

Let $K^{\star}:=K^{\prime}/\gcd(z^{2},K^{\prime})$ and
$m^{\star}:=m^{\prime}/\gcd(z^{2},m^{\prime})$. Since $z\ll 1$, we have

$K^{\star}\asymp
K^{\prime},\quad\lvert\mathrm{nrd}(\sigma)\rvert\asymp\lvert\mathrm{nrd}(\eta^{\prime})\rvert\asymp
m^{\prime}\asymp
m^{\star},\quad\left\lVert\sigma\right\rVert\asymp\left\lVert\eta^{\prime}\right\rVert\asymp(m^{\prime})^{1/2}\asymp(m^{\star})^{1/2}.$

Also, since $K^{\prime}\mid
m^{\prime}\mid\mathrm{nrd}(\eta^{\prime})=z^{2}\mathrm{nrd}(\sigma)$, we have
$K^{\prime}\mid m^{\prime}\mid\gcd(z^{2},m^{\prime})\mathrm{nrd}(\sigma)$,
whence $K^{\star}\mid m^{\star}\mid\mathrm{nrd}(\sigma)$. Moreover, since
$\eta^{\prime}=z\sigma$, it is clear that

$\Lambda(1,K^{\prime},m^{\prime},\eta^{\prime},M_{3})\subseteq\Lambda(1,K^{\star},m^{\star},\sigma,M_{3}),$

by (9.1.9). Let $Y:=M_{3}\in\mathcal{O}_{D}$, and take $\phi$ to be the
composition

$\Lambda(H,K,m,\eta,M_{0})\to\Lambda(1,K^{\prime},m^{\prime},\eta,M_{3})\to\Lambda(1,K^{\prime},m^{\prime},\eta^{\prime},M_{3})\to\Lambda(1,K^{\star},m^{\star},\sigma,M_{3}),$

where the first arrow is $M\mapsto M/H$ and the second and third arrows are
inclusion. This finishes the job, with conditions (1)–(4) all satisfied. ∎

###### Proof of Lemma 9.3.

In the notation of Lemma 9.5, we have

$\\#\\{M\in\Lambda(H,K,m,\eta,M_{0}):\left\lVert M\right\rVert\leq
R\\}\leq\\#\\{M\in\Lambda(1,K^{\star},m^{\star},\sigma,Y):\left\lVert
M\right\rVert\leq R/H\\}.$

If (LABEL:INEQ:main-geom-numbers) holds for
$\Lambda(1,K^{\star},m^{\star},\sigma,Y)$, then the right-hand side above is

$\ll
1+\frac{R}{H}+\frac{(R/H)^{2}}{(K^{\star})^{1/2}}+\frac{(R/H)^{3}}{(K^{\star}m^{\star})^{1/2}}+\frac{(R/H)^{4}}{K^{\star}m^{\star}}.$

This would imply (LABEL:INEQ:main-geom-numbers) for
$\Lambda(H,K,m,\eta,M_{0})$, because $K^{\star}\asymp K^{\prime}$ and
$m^{\star}\asymp m^{\prime}$.

Therefore, we are reduced to proving (LABEL:INEQ:main-geom-numbers) for
$\Lambda(1,K^{\star},m^{\star},\sigma,Y)$. In other words, by Lemma 9.5, it
suffices to prove Lemma 9.3 under the conditions

(9.2.2) $H=1,\quad\left\lVert\eta\right\rVert\asymp
m^{1/2},\quad\lvert\mathrm{nrd}(\eta)\rvert\asymp m,$

which we henceforth assume. Then $(K^{\prime},m^{\prime})=(K,m)$, since $H=1$.

For the rest of the proof, we abbreviate $\Lambda(H,K,m,\eta,M_{0})$ to
$\Lambda$. Trivially,

(9.2.3) $m\mathcal{O}_{D}\subseteq\Lambda\subseteq\mathcal{O}_{D},$

by (9.1.9) and the condition $K\mid m$. In particular, $\Lambda$ is a full
lattice in $D_{\infty}\cong\mathbb{R}^{4}$. We claim that its successive
minima $\lambda_{i}>0$ (for $1\leq i\leq 4$) satisfy

  1. (1)

$\lambda_{1}\gg 1$;

  2. (2)

$\lambda_{2}\gg K^{1/2}$;

  3. (3)

$\lambda_{4}\ll(Km)^{1/2}$;

  4. (4)

$\lambda_{1}\lambda_{2}\lambda_{3}\lambda_{4}\asymp\mathrm{covol}(\Lambda)\gg
Km$.

Assuming the claim, the inequality (LABEL:INEQ:main-geom-numbers) follows from
the classical fact, recorded in [BLBS23, Lemma 3.6 in the published version;
Lemma 3.5 in arXiv:2006.02356v1] and based on [Sch68, Lemma 2], that if
$\Lambda\subseteq\mathbb{R}^{n}$ is a full lattice and
$\lambda_{1}\leq\cdots\leq\lambda_{n}$ denote its successive minima, and $S$
denotes the ball of radius $R$ centered at the origin, then

(9.2.4)
$|S\cap\Lambda|\ll_{n}\sum_{j=0}^{n}\frac{R^{j}}{\lambda_{1}\cdots\lambda_{j}}.$

Indeed, $\lambda_{1}\lambda_{2}\lambda_{3}\gg(Km)^{1/2}$ by (3) and (4). Thus
we are reduced to proving the claim.

The proof of (1) is immediate from the containment
$\Lambda\subseteq\mathcal{O}_{D}$. For (2) we begin by observing that if
$U\neq 0$ and $U\eta\equiv 0\bmod{K}$, then $\left\lVert U\right\rVert\gg
K^{1/2}$: this follows by Proposition 9.4(1), which implies that under these
assumptions $K\mid\mathrm{nrd}(U)\ll\left\lVert U\right\rVert^{2}$, where
$\mathrm{nrd}(U)\neq 0$ because $D$ is nonsplit. Now if $M\in\Lambda$ is not a
scalar multiple of the identity, then $M-M^{\dagger}\neq 0$ and hence, by the
previous observation, $\left\lVert M-M^{\dagger}\right\rVert\gg K^{1/2}$.
Since $\left\lVert M-M^{\dagger}\right\rVert\ll\left\lVert M\right\rVert$, it
follows that for $M\in\Lambda$ not a scalar multiple of the identity we have
$\left\lVert M\right\rVert\gg K^{1/2}$, whence $\lambda_{2}\gg K^{1/2}$.

We moreover have
$\lambda_{1}\lambda_{2}\lambda_{3}\lambda_{4}\asymp\mathrm{covol}(\Lambda)$,
by Minkowski’s second theorem, as is conveniently recorded in [BLBS23, (3.5)],
say. By (9.2.3), we may write

(9.2.5)
$\mathrm{covol}(\Lambda)\asymp[\mathcal{O}_{D}:\Lambda]=\frac{[\mathcal{O}_{D}:m\mathcal{O}_{D}]}{[\Lambda:m\mathcal{O}_{D}]}=\frac{m^{4}}{[\Lambda:m\mathcal{O}_{D}]}=\frac{m^{4}}{[\tau\Lambda:\mathrm{nrd}(\eta)\mathcal{O}_{D}]},$

where $\tau:=\mathrm{nrd}(\eta)/m\in\mathbb{Z}$. By the final condition in
(9.1.9), elements of

$\tau\Lambda/\mathrm{nrd}(\eta)\mathcal{O}_{D}=\Lambda(\tau,\tau K,\tau
m,\eta,\tau M_{0})/\mathrm{nrd}(\eta)\mathcal{O}_{D}$

can be written in the form $M=aM_{0}+A\eta^{\dagger}$, where
$a\in\tau\mathcal{O}_{F}/\mathrm{nrd}(\eta)\mathcal{O}_{F}$ and
$A\in\mathcal{O}_{D}/\mathcal{O}_{D}\eta$. The condition
$(M-M^{\dagger})\eta\equiv 0\bmod{\tau K}$ implies $(M-M^{\dagger})\eta\equiv
0\bmod{K}$, which translates into

(9.2.6) $a(M_{0}-M_{0}^{\dagger})\eta-\eta A^{\dagger}\eta\equiv 0\bmod{K}.$

Note that $\eta A^{\dagger}\eta\bmod{K}$ is well-defined for
$A\in\mathcal{O}_{D}/\mathcal{O}_{D}\eta$, since
$K\mid\mathrm{nrd}(\eta)=\eta\eta^{\dagger}$. Since
$\mathcal{O}_{D_{v}}^{\times}\eta\mathcal{O}_{D_{v}}^{\times}=\mathcal{O}_{D_{v}}^{\times}\eta^{\dagger}\mathcal{O}_{D_{v}}^{\times}$
for all $v\mid K$, we find that by Lemma 8.1, the kernel of $A\mapsto\eta
A^{\dagger}\eta\bmod{K}$ has index $K$ in $\mathcal{O}_{D}$ (or in
$\mathcal{O}_{D}/\mathcal{O}_{D}\eta$). This implies that once $a$ is
specified, the congruence (9.2.6) has either $0$ or
$\frac{|\mathcal{O}_{D}/\mathcal{O}_{D}\eta|}{K}$ solutions
$A\in\mathcal{O}_{D}/\mathcal{O}_{D}\eta$. On summing over
$a\in\tau\mathcal{O}_{F}/\mathrm{nrd}(\eta)\mathcal{O}_{F}$, we get

$[\tau\Lambda:\mathrm{nrd}(\eta)\mathcal{O}_{D}]\leq\sum_{a\in\tau\mathcal{O}_{F}/\mathrm{nrd}(\eta)\mathcal{O}_{F}}\frac{|\mathcal{O}_{D}/\mathcal{O}_{D}\eta|}{K}\ll\frac{m^{3}}{K},$

since $|\tau\mathcal{O}_{F}/\mathrm{nrd}(\eta)\mathcal{O}_{F}|=m$ and
$|\mathcal{O}_{D}/\mathcal{O}_{D}\eta|\asymp\mathrm{nrd}(\eta)^{2}=O(m^{2})$.
Hence by (9.2.5) we obtain $\mathrm{covol}(\Lambda)\gg mK$, which yields (4).

Finally, we establish (3), for which it suffices to show that $\Lambda$
contains a full sublattice spanned by elements of magnitude $O((Km)^{1/2})$.
For this we observe that there exists $\theta\in\mathcal{O}_{D}$ such that
$0<\left\lVert\theta\right\rVert\ll K^{1/2}$ and $\eta\theta\equiv 0\bmod{K}$,
by Proposition 9.4(3). We necessarily have $\theta\in D^{\times}$, because $D$
is nonsplit. Now consider the lattice

$\Lambda^{\prime}:=\\{B\theta^{\dagger}\eta^{\dagger}:B\in\mathcal{O}_{D}\\}\qquad\text{(a
full lattice in $D_{\infty}$, because $\theta,\eta\in D^{\times}$).}$

Clearly $\Lambda^{\prime}\subseteq\Lambda$, by (9.1.9), since
$K\mid\mathrm{nrd}(\eta)$ and $m\mid\mathrm{nrd}(\eta)$. Moreover, if
$e_{1},e_{2},e_{3},e_{4}$ denotes an integral basis of $\mathcal{O}_{D}$, with
each $e_{i}$ having magnitude $O(1)$, then
$e_{1}\theta^{\dagger}\eta^{\dagger},e_{2}\theta^{\dagger}\eta^{\dagger},e_{3}\theta^{\dagger}\eta^{\dagger},e_{4}\theta^{\dagger}\eta^{\dagger}$
is an integral basis of $\Lambda^{\prime}$ formed by vectors of magnitude
$O(\left\lVert\theta\right\rVert\left\lVert\eta\right\rVert)=O((Km)^{1/2})$. ∎

######  Remark.

In key ranges of parameters, we expect Lemma 9.3 to be optimal for some
$M_{0}$, such as the scalar $1$, but not necessarily for all $M_{0}$. For
example, one may usually expect to have $1\notin\Lambda$, in which case our
lower bound for $\lambda_{1}$ could be improved. It is an interesting question
to try to improve our lattice bounds, at least for typical choices of $M_{0}$.
Indeed, this is one plausible route to decreasing the $n$ required in Theorem
1.1.

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