&&(a/\overline{a})^{1/2}\end{array}\right)\left(\begin{array}[]{ccc}(a\overline{a})^{1/2}&&\\\
&1&\\\ &&(a\overline{a})^{-1/2}\end{array}\right)$

where $z_{\infty}$ is in the center $Z_{G^{\prime}}({\mathbb{R}})$ and
$a_{\infty}^{1}$ has determinant $1$.

######  Lemma 8.3.

Let notation be as before. Then

(8.10)
$W_{n,\infty}(g_{\infty})\cdot\prod_{p<\infty}W_{n,p}(\textbf{1}_{p})=e^{-ik\alpha}e^{-2\pi
na\overline{a}+2\pi nit}\cdot\left({a}{\overline{a}}\right)^{k/2}a_{n}.$

where $a_{n}$ denote the $n$-th Fourier coefficient of the classical form
$\phi^{\prime}$ (cf. (4.39)).

###### Proof.

By (4.36), we have (since $\varphi^{\prime}$ is invariant by
$Z_{G^{\prime}}({\mathbb{R}})$)

$\displaystyle W_{n}(g_{\infty};\varphi^{\prime})=$
$\displaystyle\psi_{n,\infty}(n_{\infty})\int_{N^{\prime}(\mathbb{Z})\backslash
N^{\prime}(\mathbb{R})}\varphi^{\prime}(u_{\infty}a^{1}_{\infty}k_{\infty})\overline{\psi_{n,\infty}(u_{\infty})}du_{\infty}$
$\displaystyle=$
$\displaystyle\psi_{n,\infty}(n_{\infty})\int_{N^{\prime}(\mathbb{Z})\backslash
N^{\prime}(\mathbb{R})}\frac{\phi^{\prime}(\iota^{-1}(u_{\infty}a^{1}_{\infty}k_{\infty}).i)}{j(\iota^{-1}(u_{\infty}a^{1}_{\infty}k_{\infty}),i)^{k}}\overline{\psi_{n,\infty}(u_{\infty})}du_{\infty}$
$\displaystyle=$
$\displaystyle\frac{\psi_{n,\infty}(n_{\infty})}{j(\iota^{-1}(a^{1}_{\infty}),i)^{k}j(\iota^{-1}(k_{\infty}),i)^{k}}\int_{N^{\prime}(\mathbb{Z})\backslash
N^{\prime}(\mathbb{R})}\phi^{\prime}(\iota^{-1}(u_{\infty}a^{1}_{\infty}).i)\overline{\psi_{n,\infty}(u_{\infty})}du_{\infty}$
$\displaystyle=$
$\displaystyle\frac{\psi_{n,\infty}(n_{\infty})}{j(\iota^{-1}(a^{1}_{\infty}),i)^{k}j(\iota^{-1}(k_{\infty}),i)^{k}}\int_{0}^{1}\phi^{\prime}(a\overline{a}i+t)e^{-2\pi
nit}dt$ $\displaystyle=$
$\displaystyle\frac{\psi_{n,\infty}(n_{\infty})}{(a\overline{a})^{-k/2}e^{ik\alpha}}\int_{0}^{1}\phi^{\prime}(a\overline{a}i+t)e^{-2\pi
nit}dt.$

Then (8.10) follows from (8.9), (4.39) and the equality
$\psi_{n,\infty}(n_{\infty})=e^{2\pi nit}.$ ∎

####  8.2.2. The archimedean test tunction

Let $x_{\infty},$ $y_{\infty}\in G^{\prime}(\mathbb{R})$ written in their
Iwasawa forms:

$\displaystyle x_{\infty}=$
$\displaystyle\left(\begin{array}[]{ccc}e^{i\alpha_{1}}&&\\\ &1&\\\
&&e^{i\alpha_{1}}\end{array}\right)\left(\begin{array}[]{ccc}a_{1}&&\\\ &1&\\\
&&{a}_{1}^{-1}\end{array}\right)\left(\begin{array}[]{ccc}\cos\alpha&&\Delta\sin\alpha\\\
&1&\\\ -\Delta^{-1}\sin\alpha&&\cos\alpha\end{array}\right);$ $\displaystyle
y_{\infty}=$ $\displaystyle\left(\begin{array}[]{ccc}e^{i\alpha_{2}}&&\\\
&1&\\\
&&e^{i\alpha_{2}}\end{array}\right)\left(\begin{array}[]{ccc}a_{2}&&-{a}_{2}^{-1}\Delta
t\\\ &1&\\\
&&{a}_{2}^{-1}\end{array}\right)\left(\begin{array}[]{ccc}\cos\beta&&\Delta\sin\beta\\\
&1&\\\ -\Delta^{-1}\sin\beta&&\cos\beta\end{array}\right),$

where $\alpha_{1},\alpha_{2},\alpha,\beta\in[-\pi,\pi);$
$a_{1},a_{2}\in\mathbb{R}^{\times}$ and $t\in\mathbb{R}.$

######  Lemma 8.4.

With the notations above, we have

$f_{\infty}(x_{\infty}^{-1}\gamma(1)y_{\infty})=2^{k}{\big{[}(a_{1}^{-1}a_{2}+a_{1}a_{2}^{-1})-a_{1}^{-1}a_{2}^{-1}i(t+\Delta^{-1}/2)\big{]}^{-k}}\cdot
e^{ik(\alpha-\beta)}.$

###### Proof.

It follows from the definition that
$f_{\infty}(x_{\infty}^{-1}\gamma(1)y_{\infty})$ does not depend on
$\alpha_{i},\ i=1,2$ so we may assume that $\alpha_{1}=\alpha_{2}=0$.

Let $t^{\prime}=t-\Delta^{-1}/2.$ Computing the matrices one obtains

$\displaystyle x_{\infty}=$
$\displaystyle\left(\begin{array}[]{ccc}a_{1}\cos\alpha&&a_{1}\Delta\sin\alpha\\\
&1&\\\
-{a}_{1}^{-1}\Delta^{-1}\sin\alpha&&{a}_{1}^{-1}\cos\alpha\end{array}\right),$
$\displaystyle y_{\infty}=$
$\displaystyle\left(\begin{array}[]{ccc}a_{2}\cos\beta+{a}_{2}^{-1}t\sin\beta&&a_{2}\Delta\sin\beta-{a}_{2}^{-1}\Delta
t\cos\beta\\\ &1&\\\
-{a}_{2}^{-1}\Delta^{-1}\sin\beta&&{a}_{2}^{-1}\cos\beta\end{array}\right).$

We set

$a=a_{1}\cos\alpha,\ b=a_{1}\Delta\sin\alpha,\
c=-{a}_{1}^{-1}\Delta^{-1}\sin\alpha,\ d=a^{-1}_{1}\cos\alpha;$

and

$a^{\prime}=a_{2}\cos\beta+{a}_{2}^{-1}t\sin\beta,\
b^{\prime}=a_{2}\Delta\sin\beta-{a}_{2}^{-1}\Delta t\cos\beta,$
$c^{\prime}=-{a}_{2}^{-1}\Delta^{-1}\sin\beta,\
d^{\prime}={a}_{2}^{-1}\cos\beta.$

With these notations we have that $x_{\infty}^{-1}\gamma(1)y_{\infty}$ is
equal to

$\displaystyle\left(\begin{array}[]{ccc}\overline{d}&&\overline{b}\\\ &1&\\\
\overline{c}&&\overline{a}\end{array}\right)\gamma(1)\left(\begin{array}[]{ccc}a^{\prime}&&b^{\prime}\\\
&1&\\\ c^{\prime}&&d^{\prime}\end{array}\right)=$
$\displaystyle\left(\begin{array}[]{ccc}{d}&&-{b}\\\ &1&\\\
-{c}&&{a}\end{array}\right)\gamma(1)\left(\begin{array}[]{ccc}a^{\prime}&&b^{\prime}\\\
&1&\\\ c^{\prime}&&d^{\prime}\end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccc}{d}(a^{\prime}-\frac{c^{\prime}}{2})-{b}c^{\prime}&{d}&{d}(b^{\prime}-\frac{d^{\prime}}{2})-{b}d^{\prime}\\\
-c^{\prime}&1&-d^{\prime}\\\
-{c}(a^{\prime}-\frac{c^{\prime}}{2})+{a}c^{\prime}&-{c}&-{c}(b^{\prime}-\frac{d^{\prime}}{2})+{a}d^{\prime}\end{array}\right).$

Then by definition (4.19) we have (recall that $k_{1}=-k,\ k_{2}=k/2$)

(8.11)
$f_{\infty}(x_{\infty}^{-1}\gamma(1)y_{\infty})=2^{k}\cdot(\overline{A}-\overline{B})^{-k},$

where

$A={d}(a^{\prime}-\frac{c^{\prime}}{2})-{b}c^{\prime}-{c}(b^{\prime}-\frac{d^{\prime}}{2})+{a}d^{\prime}$

and

$B=\big{[}{d}(b^{\prime}-\frac{d^{\prime}}{2})-{b}d^{\prime}\big{]}\cdot|D_{E}|^{-1/2}+\big{[}-{c}(a^{\prime}-\frac{c^{\prime}}{2})+{a}c^{\prime}\big{]}\cdot|D_{E}|^{1/2}.$

Substituting expressions of $a,b,c,d$ and
$a^{\prime},b^{\prime},c^{\prime},d^{\prime}$ we then have

(8.12)
$A=(a_{1}^{-1}a_{2}+a_{1}a_{2}^{-1})\cos(\alpha-\beta)-a_{1}^{-1}a_{2}^{-1}t^{\prime}\sin(\alpha-\beta).$

Then

(8.13)
$B=-i\Bigl{[}(a_{1}^{-1}a_{2}+a_{1}a_{2}^{-1})\sin(\alpha-\beta)+a_{1}^{-1}a_{2}^{-1}t^{\prime}\cos(\alpha-\beta)\Bigr{]}.$

so that

$A-B=\big{[}(a_{1}^{-1}a_{2}+a_{1}a_{2}^{-1})+a_{1}^{-1}a_{2}^{-1}it^{\prime}\big{]}\cdot
e^{i(\alpha-\beta)}.$

Hence the Lemma follows from (8.11), (8.12), (8.13) and this last identity. ∎

####  8.2.3. Orbital Integrals

In this subsection, we will combine Lemma 8.3 and Lemma 8.4 to compute the
archimedean unipotent orbital integral. We set

(8.14)
$|W_{n,f}(\textbf{1})|^{2}:=\prod_{p<\infty}|W_{n,p}(\textbf{1}_{p})|^{2}$

######  Proposition 8.5.

Let notation be as before. If $|W_{n,f}(\textbf{1})|^{2}=0$ we have

$\mathcal{O}_{\infty}(f;n)=0.$

Otherwise we have

$\displaystyle\frac{\mathcal{O}_{\infty}(f;n)}{|W_{n,f}(\textbf{1})|^{2}}$
$\displaystyle=\frac{2^{3}\pi^{2}}{2^{4(k-1)}}\frac{\Gamma(k-1)^{2}}{\Gamma(k)}\frac{|a_{n}|^{2}}{(4\pi
n)^{k-1}}e^{-\pi nD_{E}^{-1/2}}$ (8.15)
$\displaystyle=\frac{2^{4}\pi^{5}}{3.2^{4(k-1)}}\frac{1}{(k-1)^{2}}{\prod_{p|N^{\prime}}(1+\frac{1}{p})}\frac{|\lambda_{\pi^{\prime}}(n)|^{2}}{L(\pi^{\prime},\operatorname{Ad},1)}e^{-\pi
nD_{E}^{-1/2}}.$

In particular we have

(8.16)
${\mathcal{O}_{\infty}(f;n)}\leq\frac{(kN^{\prime}n)^{o(1)}}{2^{4k}k^{2}}e^{-\pi
nD_{E}^{-1/2}}\prod_{p<\infty}|W_{n,p}(\textbf{1}_{p})|^{2}.$

###### Proof.

Suppose $|W_{n,f}(\textbf{1})|^{2}\not=0$. We recall that

$\frac{\mathcal{O}_{\infty}(f;n)}{|W_{n,f}(\textbf{1})|^{2}}=\int_{N^{\prime}({\mathbb{R}})\backslash{G}^{\prime}({\mathbb{R}})}\int_{{G}^{\prime}({\mathbb{R}})}\frac{\overline{W_{n,\infty}}(x){W_{n,\infty}(y)}}{|W_{n,f}(\textbf{1})|^{2}}f_{\infty}\left(x^{-1}\gamma(1)y\right)dxdy.$

By Lemma 8.3 and Lemma 8.4 we have

$\displaystyle\frac{\mathcal{O}_{\infty}(f;n)}{|W_{n,f}(\textbf{1})|^{2}}=$
$\displaystyle
2^{k}|a_{n}|^{2}\int_{\mathbb{R}}\int_{\mathbb{R}_{+}}\int_{\mathbb{R}_{+}}\int_{0}^{2\pi}\int_{0}^{2\pi}e^{-2\pi
n(a_{1}^{2}+a_{2}^{2})-2\pi nit}$
$\displaystyle\frac{a_{1}^{k}a_{2}^{k}}{\big{[}(a_{1}^{-1}a_{2}+a_{1}a_{2}^{-1})-a_{1}^{-1}a_{2}^{-1}i(t+\Delta^{-1}/2)\big{]}^{k}}\frac{d\alpha
d\beta}{4\pi^{2}}\frac{da_{1}}{a_{1}^{3}}\frac{da_{2}}{a_{2}^{3}}dt.$

Since $k=|k_{1}|\geq 8$, the integral $\mathcal{O}_{\infty}(f;n)$ converges
absolutely. After passing to polar coordinates,
$a_{1}^{2}+a_{2}^{2}=r^{2}(\cos^{2}\theta+\sin^{2}\theta)$, we obtain

$\displaystyle\frac{\mathcal{O}_{\infty}(f;n)}{|W_{n,f}(\textbf{1})|^{2}}=$
$\displaystyle
2^{k}|a_{n}|^{2}\cdot\int_{\mathbb{R}_{+}}\int_{\mathbb{R}_{+}}\int_{\mathbb{R}}\frac{e^{-2\pi
n(a_{1}^{2}+a_{2}^{2})-2\pi
nit}\cdot(a_{1}a_{2})^{2k}}{\big{[}a_{1}^{2}+a_{2}^{2}-i(t+\Delta^{-1}/2)\big{]}^{k}}dt\frac{da_{1}}{a_{1}^{3}}\frac{da_{2}}{a_{2}^{3}}$
$\displaystyle=$ $\displaystyle 2^{k}|a_{n}|^{2}e^{\pi
in\Delta^{-1}}\iint_{\mathbb{R}^{2}_{+}}\int_{\mathbb{R}}\frac{e^{-2\pi
n(a_{1}^{2}+a_{2}^{2})-2\pi
nit}\cdot(a_{1}a_{2})^{2k}}{(a_{1}^{2}+a_{2}^{2}-it)^{k}}dt\frac{da_{1}}{a_{1}^{3}}\frac{da_{2}}{a_{2}^{3}}$
$\displaystyle=$ $\displaystyle 2^{k-2k+3}|a_{n}|^{2}e^{\pi
in\Delta^{-1}}\int_{0}^{\infty}\int_{0}^{\frac{\pi}{2}}\int_{\mathbb{R}}\frac{e^{-2\pi
nr^{2}-2\pi
nit}\cdot(r^{2}2\cos\theta\sin\theta)^{2k-3}}{(r^{2}-it)^{k}}dtrdrd\theta$
$\displaystyle=$ $\displaystyle\frac{2}{2^{k-1}}|a_{n}|^{2}{e^{\pi
in\Delta^{-1}}}\int_{0}^{{\pi}}(\sin\theta)^{2k-3}d\theta\cdot\int_{0}^{\infty}\int_{\mathbb{R}}\frac{e^{-2\pi
nr-2\pi nit}\cdot r^{2k-3}}{(r-it)^{k}}dtdr.$

after making the changes of variable $2\theta\leftrightarrow\theta$,
$r^{2}\leftrightarrow r$.

We have

$\int_{0}^{{\pi}}\sin^{2k-{3}}\theta
d\theta=\pi^{1/2}\frac{\Gamma(k-{1})}{\Gamma(k-1/2)}.$

Appealing to Cauchy integral formula we then obtain

$\displaystyle\frac{1}{e^{\pi
in\Delta^{-1}}}\frac{\mathcal{O}_{\infty}(f;n)}{|W_{n,f}(\textbf{1})|^{2}}=$
$\displaystyle\frac{\pi^{1/2}}{2^{k-2}}|a_{n}|^{2}\frac{\Gamma(k-{1})}{\Gamma(k-1/2)}\frac{(-1)^{k}}{i}\int_{0}^{\infty}\int_{i\mathbb{R}}\frac{e^{-2\pi
nr-2\pi nz}\cdot r^{2k-{3}}}{(z-r)^{k}}dzdr$ $\displaystyle=$
$\displaystyle\frac{2\pi^{1/2}}{2^{k-1}}|a_{n}|^{2}\frac{\Gamma(k-{1})}{\Gamma(k-1/2)}\cdot\frac{2\pi(-1)^{k-1}}{(k-1)!}\int_{0}^{\infty}\frac{r^{2k-{3}}}{e^{2\pi
nr}}\Bigg{[}\frac{d^{k-1}e^{-2\pi nz}}{dz^{k-1}}\Bigg{]}_{z=r}dr$
$\displaystyle=$
$\displaystyle\frac{2\pi^{1/2}}{2^{k-1}}|a_{n}|^{2}\frac{\Gamma(k-{1})}{\Gamma(k-1/2)}\frac{2\pi(-1)^{k-1}}{\Gamma(k)}(-2\pi
n)^{k-1}\cdot\int_{0}^{\infty}e^{-4\pi nr}\cdot r^{2k-{2}}\frac{dr}{r}$
$\displaystyle=$
$\displaystyle\frac{2\pi^{1/2}}{2^{k-1}}|a_{n}|^{2}\frac{\Gamma(k-{1})}{\Gamma(k-1/2)}\frac{2\pi}{\Gamma(k)}\frac{(2\pi
n)^{k-1}}{(4\pi n)^{2(k-1)}}\cdot\Gamma(2(k-1))$ $\displaystyle=$
$\displaystyle\frac{2\pi^{1/2}}{2^{k-1}}|a_{n}|^{2}\frac{\Gamma(k-{1})}{\Gamma(k-1/2)}\frac{2\pi}{\Gamma(k)}\frac{(2\pi
n)^{k-1}}{(4\pi n)^{2(k-1)}}\pi^{1/2}2^{1-2(k-1)}\Gamma(k-1)\Gamma(k-1/2)$
$\displaystyle=$
$\displaystyle\frac{2^{3}\pi^{2}}{2^{4(k-1)}}\frac{\Gamma(k-1)^{2}}{\Gamma(k)}\frac{|a_{n}|^{2}}{(4\pi
n)^{k-1}}$

on using the duplication formula

$\Gamma(2(k-1))=\pi^{1/2}2^{1-2(k-1)}\Gamma(k-1)\Gamma(k-1/2).$

Then formula (8.15) follows from (4.40) and the bound (8.16) results from
(4.41) and (4.44). ∎

######  Remark 8.2.

The reason for this normalization by the factor

$|W_{n,f}(\textbf{1})|^{2}:=\prod_{p<\infty}|W_{n,p}(\textbf{1}_{p})|^{2}$

is that as we will see in the forthcoming lemma, for any $n$, the local
orbital integrals ${\mathcal{O}}_{p}(f;n)$ appearing in Lemma 8.2 are equal to
$|W_{n,p}(\textbf{1}_{p})|^{2}$ for almost every $p$.

###  8.3. Computation of $\mathcal{O}_{p}(f;n)$ when $\pi^{\prime}_{p}$ is
unramified

In this section and the next ones, we compute the local orbital integrals over
nonarchimedean places.

Let $p$ be a rational prime we denote by $\nu$ the usual $p$-adic valution.

In this subsection, we will assume $p\nmid N^{\prime}.$ Thus
$\xi_{p}^{\prime}$ is a spherical vector.

The next three lemmatas consider this situation when

  * •

$p$ splits in $E$.

  * •

$p$ is inert in $E$ and $\pi_{p}$ is unramified (ie. $p\not|N$)

  * •

$p$ is inert in $E$ and $\pi_{p}$ is ramified (ie. $p|N$).

  * •

$p$ is ramified in $E$.

For the sequel we denote by $\nu$ the usual valuation on ${\mathbb{Q}}_{p}$.

######  Lemma 8.6.

Let notation be as before. Let $p$ be a prime split in $E$ and coprime to $N$.
Under the isomorphism

$G^{\prime}(\mathbb{Q}_{p})\simeq\operatorname{GL}(2,\mathbb{Q}_{p}),$

$\pi_{p}^{\prime}$ is isomorphic to a principal series representation

$\pi_{p}^{\prime}\simeq\operatorname{Ind}(\chi\otimes\overline{\chi})$

for some unramified unitary character $\chi$. We have

(8.17) ${\mathcal{O}_{p}(f;n)}={|W_{n,p}(\textbf{1}_{p})|^{2}}\frac{e^{2\pi
nir_{p}(-\frac{1}{2\Delta})}}{{p^{\nu(n)}}}\cdot\sum_{l=0}^{\nu(n)}(l+1)\cdot\Bigg{|}\frac{\chi(p)^{\nu(n)-l+1}-\overline{\chi}(p)^{\nu(n)-l+1}}{\chi(p)^{\nu(n)+1}-\overline{\chi}(p)^{\nu(n)+1}}\Bigg{|}^{2}.$

In particular, (8.17) equals zero if $\nu(n)<0$ and equals $1$ if $\nu(n)=0.$

###### Proof.

Given $g_{p}\in G^{\prime}({\mathbb{Q}}_{p})$ with the Iwasawa decomposition
$g_{p}=u_{p}a_{p}\kappa_{p}$. Then

(8.18)
$W_{n,p}(g_{p})=|n|_{p}^{-1/2}{\psi}_{n,p}(u_{p})\overline{\chi}(n)W_{1,p}\left(\begin{pmatrix}n&&\\\
&1&\\\ &&1\end{pmatrix}a_{p}\right).$

Write $a_{p}=\operatorname{diag}(p^{j_{1}},1,p^{j_{2}}),$ where
$(j_{1},j_{2})\in\mathbb{Z}^{2}.$ Then it follows from (8.18) that
$W_{n,p}(g_{p})=0$ unless $j_{1}\geq j_{2}-\nu(n)$.

Let $x_{p}\in N^{\prime}({\mathbb{Q}}_{p})\backslash
G^{\prime}({\mathbb{Q}}_{p})$ and $y_{p}\in G^{\prime}({\mathbb{Q}}_{p}).$ We
can write them into their Iwasawa coordinates:

$\displaystyle x_{p}=\begin{pmatrix}p^{i_{1}}&&\\\ &1&\\\
&&p^{i_{2}}\end{pmatrix}\kappa_{1},\quad y_{p}=\begin{pmatrix}1&&\Delta t\\\
&1&\\\ &&1\end{pmatrix}\begin{pmatrix}p^{j_{1}}&&\\\ &1&\\\
&&p^{j_{2}}\end{pmatrix}\kappa_{2},$

where $(i_{1},i_{2})\in\mathbb{Z}^{2},$ $(j_{1},j_{2})\in\mathbb{Z}^{2};$
$t\in\mathbb{Q}_{p}$ and $\kappa_{1},\kappa_{2}\in
G^{\prime}(\mathbb{Z}_{p}).$

Since $p\nmid N,$ $\iota(\kappa_{1}),\iota(\kappa_{2})\in
K_{p}=\operatorname{GL}(3,\mathbb{Z}_{p})$ via the natural embedding (2.5) and
in this case we have

$f_{p}=\textbf{1}_{\operatorname{GL}(3,\mathbb{Z}_{p})}.$

Hence $f_{p}(x_{p}^{-1}\gamma(1)y_{p})\neq 0$ if and only if

$\displaystyle\begin{pmatrix}p^{-i_{1}}&&\\\ &1&\\\
&&p^{-i_{2}}\end{pmatrix}\left(\begin{array}[]{ccc}1&1&-1/2\\\ &1&-1\\\
&&1\end{array}\right)\begin{pmatrix}p^{j_{1}}&&p^{j_{2}}\Delta t\\\ &1&\\\
&&p^{j_{2}}\end{pmatrix}\in\operatorname{GL}(3,\mathbb{Z}_{p}).$

which, by a direct computation, can be further reduced to

$\displaystyle\left(\begin{array}[]{ccc}p^{j_{1}-i_{1}}&p^{-i_{1}}&p^{j_{2}-i_{1}}(\Delta
t-\frac{1}{2})\\\ &1&-p^{j_{2}}\\\
&&p^{j_{2}-i_{2}}\end{array}\right)\in\operatorname{GL}(3,\mathbb{Z}_{p}).$

Hence $j_{1}=i_{1}\leq 0$ and $j_{2}=i_{2}\geq 0.$ By the support of Whittaker
functions we have

$\overline{W_{n,p}}(x_{p}){W_{n,p}(y_{p})}=0$

unless $i_{1}\geq i_{2}-\nu(n)$ and $j_{1}\geq j_{2}-\nu(n).$ Hence
$\mathcal{O}_{p}(f;n)=0$ unless

$0\geq i_{1}\geq i_{2}-\nu(n)\geq-\nu(n).$

In particular we have $\nu(n)\geq 0$.

Since $\operatorname{vol}(G^{\prime}(\mathbb{Z}_{p}))=1.$ Therefore, we have

$\displaystyle{\mathcal{O}_{p}(f;n)}=$
$\displaystyle\int_{N^{\prime}(\mathbb{Q}_{p})\backslash
G^{\prime}(\mathbb{Q}_{p})}\int_{G^{\prime}(\mathbb{Q}_{p})}{f_{p}(x_{p}^{-1}\gamma(1)y_{p})\overline{W_{n,p}}(x_{p}){W_{n,p}(y_{p})}}dx_{p}dy_{p}$
$\displaystyle=$
$\displaystyle|n|_{p}^{-1}\sum_{i_{1}=-\nu(n)}^{0}\sum_{i_{2}=0}^{i_{1}+\nu(n)}p^{2i_{1}-2i_{2}}I_{p}(i_{1},i_{2})\Big{|}W_{1,p}\begin{pmatrix}np^{i_{1}}&&\\\
&1&\\\ &&p^{i_{2}}\end{pmatrix}\Big{|}^{2},$

where

$I_{p}(i_{1},i_{2}):=\int_{\mathbb{Q}_{p}}\textbf{1}_{\mathbb{Z}_{p}}(p^{i_{2}-i_{1}}(\Delta
t-1/{2}))\theta_{n,p}(t)dt.$

Since $\Delta\in{\mathbb{Z}_{p}}^{\times}$ and $i_{2}-i_{1}\leq\nu(n),$ we
have

$I_{p}(i_{1},i_{2})=e^{2\pi nir_{p}(-\Delta^{-1}/2)}p^{i_{2}-i_{1}}.$

Hence we have

$\displaystyle{\mathcal{O}_{p}(f;n)}=$ $\displaystyle\frac{e^{2\pi
nir_{p}(-\Delta^{-1}/2)}}{|n|_{p}}\cdot\sum_{l=0}^{\nu(n)}(\nu(n)-l+1)p^{l-\nu(n)}\Bigg{|}W_{1,p}\left(\iota\begin{pmatrix}p^{l}&\\\
&1\end{pmatrix}\right)\Bigg{|}^{2}.$

Substituting Casselman-Shalika formula (cf. [CS80]) into the above computation
we then obtain

$\displaystyle\frac{\mathcal{O}_{p}(f;n)}{e^{2\pi
nir_{p}(-\frac{1}{2\Delta})}}=$
$\displaystyle\frac{|W_{1,p}(\textbf{1}_{p})|^{2}}{|n|_{p}}\cdot\sum_{l=0}^{\nu(n)}(\nu(n)-l+1)p^{l-\nu(n)}\cdot\Bigg{|}\frac{p^{-\frac{l}{2}}(\chi(p)^{l+1}-\overline{\chi}(p)^{l+1})}{\chi(p)-\overline{\chi}(p)}\Bigg{|}^{2}$
$\displaystyle=$
$\displaystyle|W_{n,p}(\textbf{1}_{p})|^{2}\cdot\sum_{l=0}^{\nu(n)}(l+1)p^{-\nu(n)}\cdot\Bigg{|}\frac{\chi(p)^{\nu(n)-l+1}-\overline{\chi}(p)^{\nu(n)-l+1}}{\chi(p)^{\nu(n)+1}-\overline{\chi}(p)^{\nu(n)+1}}\Bigg{|}^{2}.$

Then (8.17) follows. ∎

######  Lemma 8.7.

Let notation be as before. Let $p$ be a rational prime inert in $E$ and such
that $p\nmid N^{\prime}$.

For

$f_{p}=\mathrm{1}_{G({\mathbb{Z}_{p}})},$

we have

(8.19)
${\mathcal{O}_{p}(f;n)}={|W_{n,p}(\textbf{1}_{p})|^{2}}\sum_{-\nu(n)/2\leq
i\leq
0}\Bigg{|}\frac{\chi(p)^{\nu(n)+2i+1}-\overline{\chi}(p)^{\nu(n)+2i+1}}{\chi(p)^{\nu(n)+1}-\overline{\chi}(p)^{\nu(n)+1}}\Bigg{|}^{2}.$

In particular, (8.19) equal $0$ if $\nu(n)<0$ and equals $1$ if $\nu(n)=0.$

For

$f_{p}=\mathrm{1}_{G({\mathbb{Z}_{p}})A_{r}G({\mathbb{Z}_{p}})},$

for $r\geq 1$ (cf. (5.64)), we have

(8.20)
${\mathcal{O}_{p}(f;n)}\ll(r+\nu(n))^{4}p^{r}{|W_{n,p}(\textbf{1}_{p})|^{2}},$

where the implied constant is absolute.

###### Proof.

By our assumptions

$K_{p}^{\prime}(N)=G^{\prime}(\mathbb{Z}_{p})=U(L\otimes_{\mathbb{Q}}\mathbb{Q}_{p}).$

Let $g_{p}\in G^{\prime}(\mathbb{Q}_{p}).$ Write $g_{p}=u_{p}a_{p}k_{p}$ be
the Iwasawa decomposition of $g_{p}.$ Then one has

(8.21)
$W_{n,p}(g_{p})=W_{n,p}(u_{p}a_{p}k_{p})={\psi}_{n,p}(u_{p})W_{n,p}(a_{p}).$

For $a_{p}=\operatorname{diag}(p^{i},1,p^{-i}),$ where $i\in\mathbb{Z}$, we
have, by Casselman-Shalika formula, that

(8.22) $W_{n,p}\begin{pmatrix}p^{i}&&\\\ &1&\\\
&&p^{-i}\end{pmatrix}=\frac{\textbf{1}_{\mathcal{O}_{E_{p}}}(np^{2i})}{p^{i}}\cdot\frac{\chi(p)^{\nu(n)+2i+1}-\overline{\chi}(p)^{\nu(n)+2i+1}}{\chi(p)^{\nu(n)+1}-\overline{\chi}(p)^{\nu(n)+1}}\cdot
W_{n,p}(\textbf{1}_{p}).$

Let $x_{p}\in
N^{\prime}(\mathbb{Q}_{p})\backslash{G}^{\prime}(\mathbb{Q}_{p})$ and
$y_{p}\in{G}^{\prime}(\mathbb{Q}_{p}).$ We can write them into their Iwasawa
coordinates:

$\displaystyle x_{p}=\begin{pmatrix}p^{i}&&\\\ &1&\\\
&&p^{-i}\end{pmatrix}\kappa_{1},\quad y_{p}=\begin{pmatrix}1&&\Delta t\\\
&1&\\\ &&1\end{pmatrix}\begin{pmatrix}p^{j}&&\\\ &1&\\\
&&p^{-j}\end{pmatrix}\kappa_{2},$

for $i,j\in\mathbb{Z}^{2}$; $t\in\mathbb{Q}_{p}$ and $\kappa_{1},\kappa_{2}\in
K^{\prime}_{p}(N).$

By definition, $f_{p}=\textbf{1}_{K_{p}}.$ Noting that $p\nmid N,$ then
$\kappa_{1},\kappa_{2}\in K_{p}$ via the natural embedding (2.5). Hence
$f_{p}(x_{p}^{-1}\gamma(1)y_{p})\neq 0$ if and only if

$\displaystyle\begin{pmatrix}p^{-i}&&\\\ &1&\\\
&&p^{i}\end{pmatrix}\left(\begin{array}[]{ccc}1&1&-1/2\\\ &1&-1\\\
&&1\end{array}\right)\begin{pmatrix}1&&\Delta t\\\ &1&\\\
&&1\end{pmatrix}\begin{pmatrix}p^{j}&&\\\ &1&\\\ &&p^{-j}\end{pmatrix}\in
G(\mathbb{Z}_{p}),$

which could be further reduced to

$\displaystyle\left(\begin{array}[]{ccc}p^{j-i}&p^{-i}&p^{-j-i}(\Delta
t-\frac{1}{2})\\\ &1&-p^{-j}\\\ &&p^{-j+i}\end{array}\right)\in
G(\mathbb{Z}_{p}).$

Hence $j=i\leq 0.$ By the support of Whittaker functions we have

$\overline{W_{n,p}}(x_{p}){W_{n,p}(y_{p})}=0$

unless

$\nu(n)+2i\geq 0,\ \nu(n)+2j\geq 0.$

In particular $\nu(n)\geq 0$.

It follows that $\mathcal{O}_{p}(f;n)=0$ unless

$i=j\geq-\nu(n)/2.$

Therefore, we have

$\displaystyle\mathcal{O}_{p}(f;n)=$
$\displaystyle\int_{N^{\prime}(\mathbb{Q}_{p})\backslash{G}^{\prime}(\mathbb{Q}_{p})}\int_{{G}^{\prime}(\mathbb{Q}_{p})}f_{p}(x_{p}^{-1}\gamma(1)y_{p})\overline{W_{n,p}}(x_{p}){W_{n,p}(y_{p})}dx_{p}dy_{p}$
$\displaystyle=$ $\displaystyle\sum_{-\nu(n)/2\leq i\leq
0}{p^{4i}}\int_{{\mathbb{Q}}_{p}}\textbf{1}_{\mathcal{O}_{E_{p}}}(p^{-2i}(\Delta
t-1/{2}))\theta_{n,p}(t)dt\cdot\Bigg{|}W_{n,p}\begin{pmatrix}p^{i}&&\\\ &1&\\\
&&p^{-i}\end{pmatrix}\Bigg{|}^{2}.$

The condition $p^{-2i}(\Delta t-1/{2})\in{\mathcal{O}}_{E_{p}}$ is equivalent
to $t\in p^{2i}{\mathbb{Z}_{p}}$ and since $-2i\leq\nu(n)$, we have

$\int_{{\mathbb{Q}}_{p}}\textbf{1}_{\mathcal{O}_{E_{p}}}(p^{-2i}(\Delta
t-1/{2}))\theta_{n,p}(t)dt=p^{-2i}.$

This in conjunction with formula (8.22) implies that

$\displaystyle\frac{\mathcal{O}_{p}(f;n)}{|W_{n,p}(\textbf{1}_{p})|^{2}}=\sum_{-\nu(n)/2\leq
i\leq
0}\Bigg{|}\frac{\chi(p)^{\nu(n)+2i+1}-\overline{\chi}(p)^{\nu(n)+2i+1}}{\chi(p)^{\nu(n)+1}-\overline{\chi}(p)^{\nu(n)+1}}\Bigg{|}^{2},$

proving (8.19).

Now let’s take
$f_{p}=\mathrm{1}_{G({\mathbb{Z}_{p}})A_{r}G({\mathbb{Z}_{p}})},$ for $r\geq
1.$ Then $f_{p}(x_{p}^{-1}\gamma(1)y_{p})\neq 0$ if and only if

$\displaystyle\begin{pmatrix}p^{-i}&&\\\ &1&\\\
&&p^{i}\end{pmatrix}\left(\begin{array}[]{ccc}1&1&-1/2\\\ &1&-1\\\
&&1\end{array}\right)\begin{pmatrix}1&&\Delta t\\\ &1&\\\
&&1\end{pmatrix}\begin{pmatrix}p^{j}&&\\\ &1&\\\ &&p^{-j}\end{pmatrix}\in
G(\mathbb{Z}_{p})A_{r}G(\mathbb{Z}_{p}),$

which could be further reduced to $t\in\mathcal{E}(i,j;r).$ Here
$\mathcal{E}(i,j;r)$ is defined by

$\displaystyle\Bigg{\\{}t\in\mathbb{Q}_{p}:\
\left(\begin{array}[]{ccc}p^{j-i}&p^{-i}&p^{-j-i}(\Delta t-\frac{1}{2})\\\
&1&-p^{-j}\\\ &&p^{-j+i}\end{array}\right)\in
G(\mathbb{Z}_{p})\left(\begin{array}[]{ccc}p^{r}&\\\ &1&\\\
&&p^{-r}\end{array}\right)G(\mathbb{Z}_{p})\Bigg{\\}}.$

Note that $\mathcal{E}(i,j;r)$ is empty unless $i\leq r,$ $j\leq r,$ and
$\Delta t-1/2\in p^{i+j-r}{\mathbb{Z}_{p}}.$

Considering the support of Whittaker functions as above, it follows that

$\displaystyle\mathcal{O}_{p}(f;n)=$
$\displaystyle\int_{N^{\prime}(\mathbb{Q}_{p})\backslash{G}^{\prime}(\mathbb{Q}_{p})}\int_{{G}^{\prime}(\mathbb{Q}_{p})}f_{p}(x_{p}^{-1}\gamma(1)y_{p})\overline{W_{n,p}}(x_{p}){W_{n,p}(y_{p})}dx_{p}dy_{p}$

is equal to

$\displaystyle\sum_{\begin{subarray}{c}-\frac{\nu(n)}{2}\leq i\leq r\\\
-\frac{\nu(n)}{2}\leq j\leq
r\end{subarray}}p^{2(i+j)}\int_{{\mathbb{Q}}_{p}}\textbf{1}_{\mathcal{E}(i,j;r)}(t)\theta_{n,p}(t)dtW_{n,p}\begin{pmatrix}p^{i}&&\\\
&1&\\\ &&p^{-i}\end{pmatrix}\overline{W_{n,p}\begin{pmatrix}p^{j}&&\\\ &1&\\\
&&p^{-j}\end{pmatrix}}.$

Appealing to the triangle inequality, $|\mathcal{O}_{p}(f;n)|$ is

$\displaystyle\leq$
$\displaystyle\sum_{\begin{subarray}{c}-\frac{\nu(n)}{2}\leq i\leq r\\\
-\frac{\nu(n)}{2}\leq j\leq
r\end{subarray}}p^{2(i+j)}\int_{{\mathbb{Q}}_{p}}\textbf{1}_{\mathcal{E}(i,j;r)}(t)dt\Bigg{|}W_{n,p}\begin{pmatrix}p^{i}&&\\\
&1&\\\ &&p^{-i}\end{pmatrix}\overline{W_{n,p}\begin{pmatrix}p^{j}&&\\\ &1&\\\
&&p^{-j}\end{pmatrix}}\Bigg{|}$ $\displaystyle\leq$
$\displaystyle\sum_{\begin{subarray}{c}-\frac{\nu(n)}{2}\leq i\leq r\\\
-\frac{\nu(n)}{2}\leq j\leq
r\end{subarray}}p^{r}\Bigg{|}\frac{\chi(p)^{\nu(n)+2i+1}-\overline{\chi}(p)^{\nu(n)+2i+1}}{\chi(p)^{\nu(n)+1}-\overline{\chi}(p)^{\nu(n)+1}}\cdot\frac{\chi(p)^{\nu(n)+2j+1}-\overline{\chi}(p)^{\nu(n)+2j+1}}{\chi(p)^{\nu(n)+1}-\overline{\chi}(p)^{\nu(n)+1}}\Bigg{|}{|W_{n,p}(\textbf{1}_{p})|^{2}}.$

Hence (8.20) follows. ∎

If $p\mid N$ then $p$ is inert in $E$ and $\pi_{p}\simeq\mathrm{St}_{p}$ is
the Steinberg representation. From (4.23) we have

$G(\mathbb{Z}_{p})=U(\Gamma\otimes_{\mathbb{Q}}\mathbb{Q}_{p})\hbox{and
}G^{\prime}(\mathbb{Z}_{p})=U(L\otimes_{\mathbb{Q}}\mathbb{Q}_{p}).$

Explicitly,

$\displaystyle G(\mathbb{Z}_{p})$ $\displaystyle=\\{g\in G(\mathbb{Q}_{p}):\
g.\Gamma\otimes_{\mathbb{Q}}\mathbb{Q}_{p}=\Gamma\otimes_{\mathbb{Q}}\mathbb{Q}_{p}\\}=G(\mathbb{Q}_{p})\cap\operatorname{GL}(3,\mathcal{O}_{E_{p}}),$
$\displaystyle G^{\prime}(\mathbb{Z}_{p})$ $\displaystyle=\\{g\in
G^{\prime}(\mathbb{Q}_{p}):\
g.L\otimes_{\mathbb{Q}}\mathbb{Q}_{p}=L\otimes_{\mathbb{Q}}\mathbb{Q}_{p}\\}=G^{\prime}(\mathbb{Q})\cap\operatorname{GL}(2,\mathcal{O}_{E_{p}}).$

######  Lemma 8.8.

Let notation be as before. Suppose that $p\mid N$ (in particular $p$ is inert)
and $(p,n)=1$. Then

(8.23)
$\mathcal{O}_{p}(f;n)=|W_{n,p}(\textbf{1}_{p})|^{2}\frac{\mu(I_{p}^{\prime})^{2}}{\mu(I_{p})}.$

###### Proof.

Let $x_{p}\in
N^{\prime}(\mathbb{Q}_{p})\backslash{G}^{\prime}(\mathbb{Q}_{p})$ and
$y_{p}\in{G}^{\prime}(\mathbb{Q}_{p})$. By Lemma 5.10 and Iwasawa
decomposition, we can write

$x_{p}=\begin{pmatrix}p^{i}&&\\\ &1&\\\
&&p^{-i}\end{pmatrix}\mu_{1}\kappa_{1}^{\prime},\ y_{p}=\begin{pmatrix}1&&t\\\
&1&\\\ &&1\end{pmatrix}\begin{pmatrix}p^{j}&&\\\ &1&\\\
&&p^{-j}\end{pmatrix}\mu_{2}\kappa_{2}^{\prime},$

where $\kappa_{1}^{\prime},\kappa_{2}^{\prime}\in I_{p}^{\prime};$
$i,j\in\mathbb{Z};$ and $\mu_{1},$ $\mu_{2}$ are either trivial or of form
$\begin{pmatrix}\delta&&1\\\ &1&\\\ 1&&\end{pmatrix}$ as in Lemma 5.10. By
definition, $f_{p}(x_{p}^{-1}\gamma(1)y_{p})\neq 0$ is equivalent to

$\displaystyle x_{p}^{-1}\gamma(1)y_{p}=\mu_{1}^{-1}\begin{pmatrix}p^{-i}&&\\\
&1&\\\ &&p^{i}\end{pmatrix}\begin{pmatrix}1&&\Delta t\\\ &1&\\\
&&1\end{pmatrix}\begin{pmatrix}p^{j}&&\\\ &1&\\\
&&p^{-j}\end{pmatrix}\mu_{2}\in I_{p},$

which could be further simplified to

(8.24) $\mu_{1}^{-1}\left(\begin{array}[]{ccc}p^{j-i}&p^{-i}&p^{-j-i}(\Delta
t-\frac{1}{2})\\\ &1&-p^{-j}\\\ &&p^{-j+i}\end{array}\right)\mu_{2}\in
I_{p}\subseteq G(\mathbb{Z}_{p}).$

Hence $j=i\leq 0.$ However, when $i<0,$ from the properties of Whittaker
functions (since $(n,p)=1$) we have
$W_{n,p}(x_{p})=W_{n,p}(\operatorname{diag}(p^{i},1,p^{-i}))=0.$ Then
$\mathcal{O}_{p}(f;n)=0$ is this case. So $i=j=0$ and (8.24) becomes

(8.25) $\mu_{1}^{-1}\left(\begin{array}[]{ccc}1&1&\Delta t-\frac{1}{2}\\\
&1&-1\\\ &&1\end{array}\right)\mu_{2}\in I_{p}.$

If $\mu_{1}\neq\textbf{1}_{p}$ and $\mu_{2}\neq\textbf{1}_{p},$ we may write
$\mu_{1}=\begin{pmatrix}\delta_{1}&&1\\\ &1&\\\ 1&&\end{pmatrix},$
$\mu_{2}=\begin{pmatrix}\delta_{2}&&1\\\ &1&\\\ 1&&\end{pmatrix},$ with
$\delta_{j}+\overline{\delta}_{j}=0,$ $j=1,2.$ Since

$\displaystyle\begin{pmatrix}\delta_{1}&&1\\\ &1&\\\
1&&\end{pmatrix}^{-1}\left(\begin{array}[]{ccc}1&1&\Delta t-\frac{1}{2}\\\
&1&-1\\\ &&1\end{array}\right)\begin{pmatrix}\delta_{2}&&1\\\ &1&\\\
1&&\end{pmatrix}=$ $\displaystyle\left(\begin{array}[]{ccc}1&0&1\\\ -1&1&-1\\\
\Delta t-\frac{1}{2}-\delta_{1}+\delta_{2}&1&1\end{array}\right)$

does not belong to $I_{p},$ (8.25) cannot hold. A contradiction! Hence our
assumption fails, namely, at least one $\mu_{1}$ and $\mu_{2}$ is equal to
$\textbf{1}_{p}.$

Let $b\in\mathcal{O}_{E_{p}}^{\times}.$ From the following computations

$\displaystyle\left(\begin{array}[]{ccc}1&1&\Delta t-\frac{1}{2}\\\ &1&-1\\\
&&1\end{array}\right)\begin{pmatrix}\delta_{2}&&1\\\ &1&\\\
1&&\end{pmatrix}=\left(\begin{array}[]{ccc}\Delta
t-\frac{1}{2}+\delta_{2}&1&1\\\ -1&1&\\\ 1&&\end{array}\right)\not\in I_{p},$
$\displaystyle\begin{pmatrix}\delta_{1}&&1\\\ &1&\\\
1&&\end{pmatrix}^{-1}\left(\begin{array}[]{ccc}1&1&\Delta t-\frac{1}{2}\\\
&1&-1\\\ &&1\end{array}\right)=\left(\begin{array}[]{ccc}&&1\\\ &1&-1\\\
1&1&\Delta t-\frac{1}{2}-\delta_{1}\end{array}\right)\not\in I_{p},$

we see that in order for (8.25) to hold, one must have
$\mu_{1}=\mu_{2}=\textbf{1}_{p}.$

Therefore,

$\displaystyle\frac{\mathcal{O}_{p}(f;n)}{|W_{n,p}(\textbf{1}_{p})|^{2}}=$
$\displaystyle\int_{N^{\prime}(\mathbb{Q}_{p})\backslash
G^{\prime}(\mathbb{Q}_{p})}\int_{G^{\prime}(\mathbb{Q}_{p})}\frac{f_{p}(x_{p}^{-1}\gamma(1)y_{p})\overline{W_{n,p}}(x_{p}){W_{n,p}(y_{p})}}{|W_{n,p}(\textbf{1}_{p})|^{2}}dx_{p}dy_{p}$
$\displaystyle=$
$\displaystyle\frac{\mu(I_{p}^{\prime})^{2}}{\mu(I_{p})}\int_{{\mathbb{Q}}_{p}}\textbf{1}_{\mathcal{O}_{E,p}}(\Delta
t-\frac{1}{2}){\psi}_{p}(nt)dt=\frac{\mu(I_{p}^{\prime})^{2}}{\mu(I_{p})}.$

Thus (8.23) follows. ∎

######  Lemma 8.9.

Let notation be as before. Suppose that $p\mid(n,N)$. Then we have

(8.26)
$\frac{\mathcal{O}_{p}(f;n)}{|W_{n,p}(\textbf{1}_{p})|^{2}}=\frac{\mu(I_{p}^{\prime})^{2}}{\mu(I_{p})}\cdot\Bigg{\\{}1+2p\sum_{-\nu(n)/2\leq
i\leq-1}\Bigg{|}\frac{\chi(p)^{\nu(n)+2i+1}-\overline{\chi}(p)^{\nu(n)+2i+1}}{\chi(p)^{\nu(n)+1}-\overline{\chi}(p)^{\nu(n)+1}}\Bigg{|}^{2}\Bigg{\\}}.$

In particular (8.26) equals $0$ if $\nu(n)\leq 0$.

###### Proof.

By the proof of Lemma 8.8 (we use the same notation here),
$f_{p}(x_{p}^{-1}\gamma(1)y_{p})\neq 0$ is equivalent to (8.24), i.e.,

$\mu_{1}^{-1}\left(\begin{array}[]{ccc}p^{j-i}&p^{-i}&p^{-j-i}(\Delta
t-\frac{1}{2})\\\ &1&-p^{-j}\\\ &&p^{-j+i}\end{array}\right)\mu_{2}\in
I_{p}\subseteq G(\mathbb{Z}_{p}).$

If $\mu_{1}\neq\textbf{1}_{p}$ and $\mu_{2}\neq\textbf{1}_{p},$ we may write
$\mu_{1}=\begin{pmatrix}\delta_{1}&&1\\\ &1&\\\ 1&&\end{pmatrix},$
$\mu_{2}=\begin{pmatrix}\delta_{2}&&1\\\ &1&\\\ 1&&\end{pmatrix},$ with
$\delta_{j}+\overline{\delta}_{j}=0,$ $j=1,2.$ Then the condition (8.24)
becomes

$\left(\begin{array}[]{ccc}1&&\\\ p^{-i}&1&\\\ p^{-2i}(\Delta
t-\frac{1}{2})+\delta_{2}-\delta_{1}&p^{-i}&1\end{array}\right)\in
I_{p}\subseteq G(\mathbb{Z}_{p}),$

which is equivalent to

$i\leq-1,\ p^{-2i}(\Delta t-\frac{1}{2})\in\mathcal{O}_{E_{p}},\hbox{ and
}\delta_{1}-\delta_{2}\equiv p^{-2i}(\Delta t-\frac{1}{2})\pmod{N_{1}}.$

On the other hand, we have

$\displaystyle\left(\begin{array}[]{ccc}1&p^{-i}&p^{-2i}(\Delta
t-\frac{1}{2})\\\ &1&-p^{-i}\\\
&&1\end{array}\right)\begin{pmatrix}\delta_{2}&&1\\\ &1&\\\
1&&\end{pmatrix}=\left(\begin{array}[]{ccc}\frac{\Delta
t-{1}/{2}}{p^{2i}}+\delta_{2}&p^{-i}&1\\\ -p^{-i}&1&\\\
1&&\end{array}\right)\not\in I_{p},$
$\displaystyle\begin{pmatrix}\delta_{1}&&1\\\ &1&\\\
1&&\end{pmatrix}^{-1}\left(\begin{array}[]{ccc}1&p^{-i}&p^{-2i}(\Delta
t-\frac{1}{2})\\\ &1&-p^{-i}\\\
&&1\end{array}\right)=\left(\begin{array}[]{ccc}&&1\\\ &1&-p^{-i}\\\
1&p^{-i}&\frac{\Delta t-{1}/{2}}{p^{2i}}-\delta_{1}\end{array}\right)\not\in
I_{p},$

therefore to make (8.25) hold, one must have either
$\mu_{1}=\mu_{2}=\textbf{1}_{p}$ or

$\mu_{1}=\begin{pmatrix}\delta_{1}&&1\\\ &1&\\\ 1&&\end{pmatrix},\
\mu_{2}=\begin{pmatrix}\delta_{2}&&1\\\ &1&\\\ 1&&\end{pmatrix},$

for some $\delta_{j}\in\mathcal{O}_{E_{p}}/p\mathcal{O}_{E_{p}},$
$\delta_{j}+\overline{\delta}_{j}=0,$ $j=1,2.$

By (8.22) we obtain

$\displaystyle\frac{\mathcal{O}_{p}(f;n)}{\mu(I_{p}^{\prime})^{2}/\mu(I_{p})}=$
$\displaystyle\iint
f_{p}(x_{p}^{-1}\gamma(1)y_{p})\overline{W_{n,p}}(x_{p}){W_{n,p}(y_{p})}dx_{p}dy_{p}$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}i\geq-\nu(n)/2\\\
i\leq
0\end{subarray}}\int_{{\mathbb{Q}}_{p}}{p^{4i}\textbf{1}_{\mathcal{O}_{E_{p}}}(p^{-2i}(\Delta
t-1/{2}))\theta_{n,p}(t)}dt\Bigg{|}W_{n,p}\begin{pmatrix}p^{i}&&\\\ &1&\\\
&&p^{-i}\end{pmatrix}\Bigg{|}^{2}$ $\displaystyle\quad+\sum_{-\nu(n)/2\leq
i\leq-1}\int_{{\mathbb{Q}}_{p}}{p^{4i}}\cdot\Bigg{|}W_{n,p}\begin{pmatrix}p^{i}&&\\\
&1&\\\ &&p^{-i}\end{pmatrix}\Bigg{|}^{2}$
$\displaystyle\qquad\sum_{\begin{subarray}{c}\delta_{1},\delta_{2}\in\mathcal{O}_{E_{p}}/N\mathcal{O}_{E_{p}}\\\
\delta_{1}+\overline{\delta}_{1}=0,\ \delta_{2}+\overline{\delta}_{2}=0\\\
\delta_{1}-\delta_{2}\equiv p^{-2i}(\Delta
t-\frac{1}{2})\pmod{N}\end{subarray}}{\textbf{1}_{\mathcal{O}_{E_{p}}}(p^{-2i}(\Delta
t-1/{2}))\theta_{n,p}(t)}dt$ $\displaystyle=$
$\displaystyle|W_{n,p}(\textbf{1}_{p})|^{2}\Bigg{\\{}\sum_{-\nu(n)/2\leq i\leq
0}\Bigg{|}\frac{\chi(p)^{\nu(n)+2i+1}-\overline{\chi}(p)^{\nu(n)+2i+1}}{\chi(p)^{\nu(n)+1}-\overline{\chi}(p)^{\nu(n)+1}}\Bigg{|}^{2}$
$\displaystyle\qquad+\sum_{-\nu(n)/2\leq
i\leq-1}p\Bigg{|}\frac{\chi(p)^{\nu(n)+2i+1}-\overline{\chi}(p)^{\nu(n)+2i+1}}{\chi(p)^{\nu(n)+1}-\overline{\chi}(p)^{\nu(n)+1}}\Bigg{|}^{2}\Bigg{\\}}.$

Then Lemma 8.9 follows. ∎

######  Lemma 8.10.

Let notation be as before. Let $p$ be a prime ramified in $E.$ We have

(8.27)
$\frac{\mathcal{O}_{p}(f;n)}{|W_{n,p}(\textbf{1}_{p})|^{2}}=\sum_{-\nu(n)+1\leq
i\leq-2\nu(D_{E})}\Bigg{|}\frac{\chi(p)^{\nu(n)+2i+1}-\overline{\chi}(p)^{\nu(n)+2i+1}}{\chi(p)^{\nu(n)+1}-\overline{\chi}(p)^{\nu(n)+1}}\Bigg{|}^{2}.$

In particular, $\mathcal{O}_{p}(f;n)=0$ unless $\nu(n)\geq 2\nu(D_{E})+1.$

###### Proof.

Let $p=\mathfrak{p}^{2}$ be ramified in $E$ and $\nu_{\mathfrak{p}}$ be the
corresponding valuation ($\nu_{\mathfrak{p}}=2\nu$ on ${\mathbb{Q}}_{p}$).
Then $p\mid D_{E}.$ Let $\varpi$ be a uniformizer in $\mathfrak{p}.$ Writing
$g_{p}=u_{p}a_{p}k_{p}$ for the Iwasawa decomposition of $g_{p}\in
G^{\prime}(\mathbb{Q}_{p})$, we have again

(8.28)
$W_{n,p}(g_{p})=W_{n,p}(u_{p}a_{p}k_{p})={\psi}_{n,p}(u_{p})W_{n,p}(a_{p}).$

Let $a_{p}=\operatorname{diag}(\varpi^{j},1,\varpi^{-j}),$ where
$j\in\mathbb{Z}.$ From the support properties of the Whittaker function, we
have $W_{n,p}(g_{p})=0$ unless $2j+\nu(n)\geq 0.$

Let $x_{p}\in
N^{\prime}(\mathbb{Q}_{p})\backslash{G}^{\prime}(\mathbb{Q}_{p})$ and
$y_{p}\in{G}^{\prime}(\mathbb{Q}_{p}).$ We can write them into their Iwasawa
coordinates:

$\displaystyle x_{p}=\begin{pmatrix}\varpi^{i}&&\\\ &1&\\\
&&\overline{\varpi}^{-i}\end{pmatrix}\kappa_{1},\quad
y_{p}=\begin{pmatrix}1&&\Delta t\\\ &1&\\\
&&1\end{pmatrix}\begin{pmatrix}\varpi^{j}&&\\\ &1&\\\
&&\overline{\varpi}^{-j}\end{pmatrix}\kappa_{2},$

where $i,j\in\mathbb{Z}^{2}$, $t\in{\mathbb{Q}}_{p}$ and
$\kappa_{1},\kappa_{2}\in K^{\prime}_{p}(N).$

By definition, $f_{p}=\textbf{1}_{K_{p}}.$ Since $p\nmid N,$ we have
$\kappa_{1},\kappa_{2}\in K_{p}$ and $f_{p}(x_{p}^{-1}\gamma(1)y_{p})\neq 0$
if and only if

$\displaystyle\begin{pmatrix}\varpi^{-i}&&\\\ &1&\\\
&&\overline{\varpi}^{i}\end{pmatrix}\left(\begin{array}[]{ccc}1&1&-1/2\\\
&1&-1\\\ &&1\end{array}\right)\begin{pmatrix}1&&\Delta t\\\ &1&\\\
&&1\end{pmatrix}\begin{pmatrix}\varpi^{j}&&\\\ &1&\\\
&&\overline{\varpi}^{-j}\end{pmatrix}\in G(\mathbb{Z}_{p}),$

which could be further reduced to

$\displaystyle\left(\begin{array}[]{ccc}\varpi^{j-i}&\varpi^{-i}&\varpi^{-j}\overline{\varpi}^{-i}(\Delta
t-\frac{1}{2})\\\ &1&-\overline{\varpi}^{-j}\\\
&&\overline{\varpi}^{-j+i}\end{array}\right)\in G(\mathbb{Z}_{p}).$

Hence $j=i\leq-2\nu(D_{E}).$ By the support of Whittaker functions we have
$\overline{W_{n,p}}(x_{p}){W_{n,p}(y_{p})}=0$ unless
$2i+\nu_{\mathfrak{p}}(n)\geq 0.$ Hence $\mathcal{O}_{p}(f;n)=0$ unless

$-\nu(n)\leq i=j\leq-2\nu(D_{E}).$

Therefore, we have

$\displaystyle\mathcal{O}_{p}(f;n)=$
$\displaystyle\int_{N^{\prime}(\mathbb{Q}_{p})\backslash{G}^{\prime}(\mathbb{Q}_{p})}\int_{{G}^{\prime}(\mathbb{Q}_{p})}f_{p}(x_{p}^{-1}\gamma(1)y_{p})\overline{W_{n,p}}(x_{p}){W_{n,p}(y_{p})}dx_{p}dy_{p}$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}i\geq-\nu(n)\\\
i\leq-2\nu(D_{E})\end{subarray}}{p^{2i}}\int_{E_{\mathfrak{p}}}\textbf{1}_{\mathcal{O}_{E_{p}}}(p^{-i}(\Delta
t-1/{2}))\theta_{n,p}(t)dt\cdot\Bigg{|}W_{n,p}\begin{pmatrix}\varpi^{i}&&\\\
&1&\\\ &&\overline{\varpi}^{-i}\end{pmatrix}\Bigg{|}^{2}$ $\displaystyle=$
$\displaystyle|W_{n,p}(\textbf{1}_{p})|^{2}\cdot\sum_{-\nu(n)+\nu(D_{E})/2\leq
i\leq-2\nu(D_{E})}\Bigg{|}\frac{\chi(p)^{\nu(n)+2i+1}-\overline{\chi}(p)^{\nu(n)+2i+1}}{\chi(p)^{\nu(n)+1}-\overline{\chi}(p)^{\nu(n)+1}}\Bigg{|}^{2}.$

We then conclude (8.27). ∎

###  8.4. Computation of $\mathcal{O}_{p}(f;n)$ when $\pi^{\prime}_{p}$ is
ramified

In this subsection we deal with the case $p=N^{\prime}$. In particular $p$ is
split and $\pi_{p}^{\prime}\simeq\mathrm{St}^{\prime}_{p}$ is the Steinberg
representation. We continue to denote by $\nu$ the usual valuation on
${\mathbb{Q}}_{p}$.

######  Lemma 8.11.

Suppose that $p=N^{\prime}$.

For $\nu(n)<0$, we have

$\mathcal{O}_{p}(f^{\mathfrak{n}_{p}};n)=0.$

For $\nu(n)=0$ we have

(8.29)
$\frac{\mathcal{O}_{p}(f^{\mathfrak{n}_{p}};n)}{|W_{n,p}(\textbf{1}_{p})|^{2}}=\frac{(p-2)\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}.$

For $\nu(n)\geq 1$ we have

(8.30)
$\frac{\mathcal{O}_{p}(f^{\mathfrak{n}_{p}};n)}{|W_{n,p}(\textbf{1}_{p})|^{2}}\ll\frac{\nu(n)^{2}}{p\mu(K_{p})},$

where the implied constant is absolute. We recall that $I^{\prime}_{p}(1)$ is
defined in (5.98) (when we identify it with a subgroup of
$\operatorname{GL}_{2}({\mathbb{Z}_{p}})$).

###### Proof.

By definition of the test function $f_{p}$, the orbital integral

$\displaystyle\mathcal{O}_{p}(f;n)=$
$\displaystyle\int_{N^{\prime}(\mathbb{Q}_{p})\backslash
G^{\prime}(\mathbb{Q}_{p})}\int_{G^{\prime}(\mathbb{Q}_{p})}f_{p}(\mathfrak{n}_{p}^{-1}x_{p}^{-1}\gamma(1)y_{p}\mathfrak{n}_{p})\overline{W_{n,p}}(x_{p}){W_{n,p}(y_{p})}dx_{p}dy_{p},$

is zero unless

(8.31) $\mathfrak{n}_{p}^{-1}\left(\begin{array}[]{ccc}a&b&\\\ c&d&\\\
&&1\end{array}\right)^{-1}\left(\begin{array}[]{ccc}1&-\frac{1}{2}&1\\\ &1&\\\
&-1&1\end{array}\right)\left(\begin{array}[]{ccc}a^{\prime}&b^{\prime}&\\\
c^{\prime}&d^{\prime}&\\\ &&1\end{array}\right)\mathfrak{n}_{p}\in K_{p}.$

By the Iwasawa decomposition for $G({\mathbb{Q}}_{p})$ and the Iwahori
decomposition for $K^{\prime}_{p}$, we write

(8.32) $\left(\begin{array}[]{ccc}a^{\prime}&b^{\prime}&\\\
c^{\prime}&d^{\prime}&\\\
&&1\end{array}\right)\in\left(\begin{array}[]{ccc}p^{j^{\prime}}&&\\\
&p^{k^{\prime}}&\\\
&&1\end{array}\right)\left(\begin{array}[]{ccc}1&b^{\prime}&\\\ &1&\\\
&&1\end{array}\right)\left(\begin{array}[]{ccc}\delta&&\\\ &1&\\\
&&1\end{array}\right)I_{p}^{\prime}(1)$

or

(8.33) $\left(\begin{array}[]{ccc}a^{\prime}&b^{\prime}&\\\
c^{\prime}&d^{\prime}&\\\
&&1\end{array}\right)\in\left(\begin{array}[]{ccc}p^{j^{\prime}}&b^{\prime}p^{k^{\prime}}&\\\
&p^{k^{\prime}}&\\\ &&1\end{array}\right)\begin{pmatrix}\mu_{2}&1&\\\ 1&&\\\
&&1\end{pmatrix}\left(\begin{array}[]{ccc}\delta&&\\\ &1&\\\
&&1\end{array}\right)I_{p}^{\prime}(1)$

  1. 1.

Suppose $x_{p}$ and $y_{p}$ are both of the form in (8.32), namely, suppose

$\displaystyle x_{p}=$ $\displaystyle\begin{pmatrix}p^{j}&&\\\ &p^{k}&\\\
&&1\end{pmatrix}\begin{pmatrix}\delta&&\\\ &1&\\\ &&1\end{pmatrix}\gamma_{1},$
$\displaystyle y_{p}=$ $\displaystyle\begin{pmatrix}p^{j^{\prime}}&&\\\
&p^{k^{\prime}}&\\\ &&1\end{pmatrix}\begin{pmatrix}1&b^{\prime}&\\\ &1&\\\
&&1\end{pmatrix}\begin{pmatrix}\delta&&\\\ &1&\\\ &&1\end{pmatrix}\gamma_{2},$

where $\gamma_{1},\gamma_{2}\in I_{p}^{\prime}(1).$ Denote by
$\mathcal{O}_{p}^{(1)}(f;n)$ the contribution of $x_{p},y_{p}$ in the above
forms. Then (8.31) is equivalent to

$\displaystyle\mathfrak{n}_{p}^{-1}\left(\begin{array}[]{ccc}\tau p^{j}&&\\\
&p^{k}&\\\ &&1\end{array}\right)^{-1}\left(\begin{array}[]{ccc}1&-1/2&1\\\
&1&\\\ &-1&1\end{array}\right)\left(\begin{array}[]{ccc}p^{j^{\prime}}&&\\\
&p^{k^{\prime}}&\\\
&&1\end{array}\right)\left(\begin{array}[]{ccc}\delta&b^{\prime}&\\\ &1&\\\
&&1\end{array}\right)\mathfrak{n}_{p}\in K_{p}$

A direct calculation shows that

$\displaystyle\left(\begin{array}[]{ccc}p^{j^{\prime}-j}&p^{j^{\prime}-j}b^{\prime}-2^{-1}p^{k^{\prime}-j}+\tau
p^{k-1}&\delta p^{j^{\prime}-j-1}+p^{-j}-\tau p^{-1}\\\ &p^{k^{\prime}-k}&\\\
&-p^{k^{\prime}}&1\end{array}\right)\in K_{p}.$

Similarly, taking the inverse, we have

$\displaystyle\left(\begin{array}[]{ccc}p^{-j^{\prime}}&-b^{\prime}p^{-k^{\prime}}&-\delta
p^{-1}\\\ &p^{-k^{\prime}}&\\\
&&1\end{array}\right)\left(\begin{array}[]{ccc}1&-1/2&-1\\\ &1&\\\
&1&1\end{array}\right)\left(\begin{array}[]{ccc}p^{j}&&\tau p^{j-1}\\\
&p^{k}&\\\ &&1\end{array}\right)\in K_{p}.$

Expanding the matrices we then obtain

$\displaystyle\left(\begin{array}[]{ccc}p^{j-j^{\prime}}&-2^{-1}p^{k-j^{\prime}}-b^{\prime}p^{k-k^{\prime}}-\delta
p^{k-1}&\tau p^{k-j^{\prime}-1}-p^{-j^{\prime}}-\delta p^{-1}\\\
0&p^{k-k^{\prime}}&0\\\ 0&p^{k}&1\end{array}\right)\in K_{p}.$

So $j-j^{\prime}=k-k^{\prime}$ and $k=k^{\prime}\geq 0$, therefore
$j=j^{\prime}.$ Hence the above constraints reduces to the following:

$\displaystyle\left(\begin{array}[]{ccc}1&b^{\prime}-2^{-1}p^{k-j}&(\delta-\tau)p^{-1}+p^{-j}\\\
0&1&0\\\ 0&-p^{k}&1\end{array}\right)\in K_{p}.$
$\displaystyle\left(\begin{array}[]{ccc}1&-2^{-1}p^{k-j}-b^{\prime}-\delta
p^{k-1}&\tau p^{k-j-1}-p^{-j}-\delta p^{-1}\\\ 0&1&0\\\
0&p^{k}&1\end{array}\right)\in K_{p}.$

Since $\tau p^{k-j-1}-p^{-j}-\delta^{-1}p^{-1}\in\mathbb{Z}_{p}$ and $k\geq
0,$ then $0\leq j\leq 1.$ Also, $p^{k-j}+\delta p^{k-1}\in\mathbb{Z}_{p}.$ So
$j\neq 0,$ forcing that $j=1.$ From

$\tau p^{k-j-1}-p^{-j}-\delta^{-1}p^{-1}\in\mathbb{Z}_{p}$

we then see that $k\geq 1.$

Also, by the support properties of Whittaker functions, we have

$\nu(n)+1\geq k\geq 1.$

If $\nu(n)=0$ we have thus $k=1$ and

$\displaystyle\mathcal{O}_{p}^{(1)}(f;n)=$
$\displaystyle\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\sum_{\begin{subarray}{c}\delta,\tau\\\
(\delta-\tau)p^{-1}+p^{-1}\in\mathbb{Z}_{p}\end{subarray}}\bigg{|}W_{n,p}\left(\textbf{1}_{p}\right)\bigg{|}^{2}$
$\displaystyle=$
$\displaystyle\frac{(p-2)\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}|W_{n,p}(\textbf{1}_{p})|^{2}.$

Assume now that $\nu(n)\geq 1$. If $k\geq 2,$ then there are only one choice
for the pair $(\delta,\tau)$ and

$\displaystyle\mathcal{O}_{p}^{(1)}(f;n)=$
$\displaystyle\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\sum_{1\leq
k\leq\nu(n)}p^{1-k}\bigg{|}W_{n,p}\begin{pmatrix}p&&\\\ &1&\\\
&&p^{k}\end{pmatrix}\bigg{|}^{2}\ll\frac{\mu(I_{p}^{\prime}(1))^{2}\nu(n)}{\mu(K_{p})}|W_{n,p}(\textbf{1}_{p})|^{2}.$

  2. 2.

Suppose $x_{p}$ and $y_{p}$ are both of the form in (8.33), namely, suppose

$\displaystyle x_{p}=$ $\displaystyle\begin{pmatrix}p^{j}&&\\\ &p^{k}&\\\
&&1\end{pmatrix}\begin{pmatrix}\mu_{1}&1&\\\ 1&&\\\
&&1\end{pmatrix}\begin{pmatrix}\tau^{-1}&&\\\ &1&\\\
&&1\end{pmatrix}\gamma_{1},$ $\displaystyle y_{p}=$
$\displaystyle\begin{pmatrix}p^{j^{\prime}}&&\\\ &p^{k^{\prime}}&\\\
&&1\end{pmatrix}\begin{pmatrix}1&b^{\prime}&\\\ &1&\\\
&&1\end{pmatrix}\begin{pmatrix}\mu_{2}&1&\\\ 1&&\\\
&&1\end{pmatrix}\begin{pmatrix}\delta^{-1}&&\\\ &1&\\\
&&1\end{pmatrix}\gamma_{2},$

where $\gamma_{1},\gamma_{2}\in I_{p}^{\prime}(1).$ Then (8.31) is equivalent
to

$\displaystyle\begin{pmatrix}p^{-j}&&\\\ &p^{-k}&-\tau p^{-1}\\\
&&1\end{pmatrix}\begin{pmatrix}1&-1/2&1\\\ &1&\\\
&-1&1\end{pmatrix}\begin{pmatrix}p^{j^{\prime}}&bp^{j^{\prime}}&\delta^{-1}bp^{j^{\prime}-1}\\\
&p^{k^{\prime}}&\delta^{-1}p^{k^{\prime}-1}\\\ &&1\end{pmatrix}\in K_{p},$

where $b=b^{\prime}+\mu_{2}-\mu_{1}p^{j-k+k^{\prime}-j^{\prime}}.$ Expanding
the left hand side we obtain

$\displaystyle\begin{pmatrix}p^{j^{\prime}-j}&bp^{j^{\prime}-j}-2^{-1}p^{k^{\prime}-j}&z\\\
0&p^{k^{\prime}-k}+\tau p^{k^{\prime}-1}&\delta^{-1}p^{k^{\prime}-1-k}-\tau
p^{-1}+\tau\delta^{-1}p^{k^{\prime}-2}\\\
0&-p^{k^{\prime}}&1-\delta^{-1}p^{k^{\prime}-1}\end{pmatrix}\in K_{p},$

where
$z=\delta^{-1}bp^{j^{\prime}-j-1}-2^{-1}\delta^{-1}p^{k^{\prime}-1-j}+p^{-j}.$

Taking the inverse of the above matrix we then obtain

$\displaystyle\begin{pmatrix}p^{j-j^{\prime}}&-2^{-1}p^{k-j^{\prime}}-bp^{k-k^{\prime}}&z^{\prime}\\\
0&p^{k-k^{\prime}}-\delta p^{k-1}&\tau^{-1}p^{k-k^{\prime}-1}-\delta
p^{-1}-\delta\tau^{-1}p^{k-2}\\\ 0&p^{k}&1+\tau^{-1}p^{k-1}\end{pmatrix}\in
K_{p},$

where

$z^{\prime}=-2^{-1}\tau^{-1}p^{k-j^{\prime}-1}-p^{-j^{\prime}}-b\tau^{-1}p^{k-k^{\prime}-1}.$

In particular we have

$k\geq 1,\ k^{\prime}\geq 1,\ j-j^{\prime}=k-k^{\prime}.$

If $k>k^{\prime}$ then $k\geq 2,$ which contradicts the condition

$\tau^{-1}p^{k-k^{\prime}-1}-\delta
p^{-1}-\delta\tau^{-1}p^{k-2}\in\mathbb{Z}_{p}.$

Hence $k\leq k^{\prime}.$ Likewise, $k^{\prime}\leq k.$ So we must have
$k=k^{\prime}$ and therefore $j=j^{\prime}$. Eventually, the above constraints
reduces to

$\displaystyle\begin{pmatrix}1&b-2^{-1}p^{k-j}+\mu_{1}p^{k-1}&z_{1}\\\
0&1+\tau p^{k-1}&\delta^{-1}p^{-1}-\tau p^{-1}+\tau\delta^{-1}p^{k-2}\\\
0&-p^{k}&1-\delta^{-1}p^{k-1}\end{pmatrix}\in K_{p},$

where $z_{1}=\delta^{-1}bp^{-1}-2^{-1}\delta^{-1}p^{-1-j}+p^{-j}.$ From

$bp^{j^{\prime}-j}-2^{-1}p^{k^{\prime}-j}\in\mathbb{Z}_{p},\ \
-2^{-1}p^{k-j^{\prime}}-bp^{k-k^{\prime}}\in\mathbb{Z}_{p}$

one has $k\geq j$ and $b\in\mathbb{Z}_{p}.$ On the other hand, from the
support of Whittaker functions we have necessarily that $\nu(n)+j\geq k\geq
1.$

We have the following cases

     * (a)

Suppose $\nu(n)=0.$ Then $k=j\geq 1.$ Since $b\in\mathbb{Z}_{p},$ then from

$z^{\prime}=-2^{-1}\tau^{-1}p^{k-j^{\prime}-1}-p^{-j^{\prime}}-b\tau^{-1}p^{k-k^{\prime}-1}\in\mathbb{Z}_{p}$

one concludes that $-j\geq-1,$ i.e., $j\leq 1.$ Hence $j=k=1.$ Then it follows
from

(8.34) $\delta^{-1}p^{k^{\prime}-1-k}-\tau
p^{-1}+\tau\delta^{-1}p^{k^{\prime}-2},\ \tau^{-1}p^{k-k^{\prime}-1}-\delta
p^{-1}-\delta\tau^{-1}p^{k-2}\in\mathbb{Z}_{p}$

that $\tau\equiv-\delta\pmod{p}$ and $1+\tau+\tau^{2}\equiv 0\pmod{p}.$ From
$\tau z^{\prime}+\delta z\in\mathbb{Z}_{p}$ we have $1+2\tau\equiv 0\pmod{p},$
which in conjunction with $1+\tau+\tau^{2}\equiv 0\pmod{p}$ implies $p\mid 3.$
However, since we have assumed that $p\geq 5$ we encounter a contradiction if
$\nu(n)=0.$

     * (b)

Therefore $\nu(n)\geq 1.$ Note that from (8.34) $\delta$ and $\tau$ should
satisfy either $\delta\tau\equiv 1\pmod{p}$ or $\tau^{2}+\tau+1\equiv
0\pmod{p}$ and $\tau+\delta\equiv 0\pmod{p}.$ Hence, in conjunction with

$\displaystyle
W_{n,p}(x_{p})=\theta_{n,p}(p^{j-k}\mu_{1})W_{n,p}\begin{pmatrix}p^{j-k}&&\\\
&1&\\\ &&1\end{pmatrix}$ $\displaystyle
W_{n,p}(y_{p})=\theta_{n,p}(p^{j-k}\mu_{2})\theta_{n,p}(p^{j-k}b^{\prime})W_{n,p}\begin{pmatrix}p^{j-k}&&\\\
&1&\\\ &&1\end{pmatrix},$

we conclude that

$\displaystyle\mathcal{O}_{p}^{(2)}(f;n)\leq$ $\displaystyle
p^{3}\mu(I_{p}^{\prime}(1))^{2}\sum_{k=1}^{\nu(n)+1}\sum_{j=k-\nu(n)}^{1}p^{j-k}\bigg{|}W_{n,p}\begin{pmatrix}p^{j-k}&&\\\
&1&\\\
&&1\end{pmatrix}\bigg{|}^{2}\ll\frac{\nu(n)^{2}|W_{n,p}(\textbf{1}_{p})|^{2}}{p\mu(K_{p})}.$

  3. 3.

Suppose $x_{p}$ is of the form in (8.32) and $y_{p}$ is of the form in (8.33),
namely, suppose

$\displaystyle x_{p}=$ $\displaystyle\begin{pmatrix}p^{j}&&\\\ &p^{k}&\\\
&&1\end{pmatrix}\begin{pmatrix}\tau&&\\\ &1&\\\ &&1\end{pmatrix}\gamma_{1},$
$\displaystyle y_{p}=$ $\displaystyle\begin{pmatrix}p^{j^{\prime}}&&\\\
&p^{k^{\prime}}&\\\ &&1\end{pmatrix}\begin{pmatrix}1&b^{\prime}&\\\ &1&\\\
&&1\end{pmatrix}\begin{pmatrix}\mu_{2}&1&\\\ 1&&\\\
&&1\end{pmatrix}\begin{pmatrix}\delta&&\\\ &1&\\\ &&1\end{pmatrix}\gamma_{2},$

where $\gamma_{1},\gamma_{2}\in I_{p}^{\prime}(1).$ Denote by
$\mathcal{O}_{p}^{(3)}(f;n)$ the contribution of $x_{p},y_{p}$ in the above
forms. Then (8.31) is equivalent to

$\displaystyle\left(\begin{array}[]{ccc}\tau^{-1}p^{-j}&&-p^{-1}\\\
&p^{-k}&\\\ &&1\end{array}\right)\begin{pmatrix}1&-1/2&1\\\ &1&\\\
&-1&1\end{pmatrix}\begin{pmatrix}p^{j^{\prime}}&bp^{j^{\prime}}&\delta^{-1}bp^{j^{\prime}-1}\\\
&p^{k^{\prime}}&\delta^{-1}p^{k^{\prime}-1}\\\ &&1\end{pmatrix}\in K_{p}$

where $b=b^{\prime}+\mu_{2}.$ Expanding the left hand side, the above
constraint becomes

$\displaystyle\begin{pmatrix}p^{j^{\prime}-j}&bp^{j^{\prime}-j}-2^{-1}p^{k^{\prime}-j}+\tau
p^{k^{\prime}-1}&z_{2}\\\ 0&p^{k^{\prime}-k}&\delta^{-1}p^{k^{\prime}-k-1}\\\
0&-p^{k^{\prime}}&1-\delta^{-1}p^{k^{\prime}-1}\end{pmatrix}\in K_{p},$

where

$z_{2}:=\delta^{-1}bp^{j^{\prime}-j-1}-2^{-1}\delta^{-1}p^{k^{\prime}-j-1}+p^{-j}-\tau
p^{-1}+\tau\delta^{-1}p^{k^{\prime}-2}.$

Considering the inverse as before, we obtain

$\displaystyle\begin{pmatrix}p^{-j^{\prime}}&-bp^{-k^{\prime}}&\\\
&p^{-k^{\prime}}&-\delta p^{-1}\\\
&&1\end{pmatrix}\begin{pmatrix}p^{j}&-2^{-1}p^{k}&\tau^{-1}p^{j-1}-1\\\
&p^{k}&\\\ &p^{k}&1\end{pmatrix}\in K_{p},$

which amounts to the following condition:

$\displaystyle\begin{pmatrix}p^{j-j^{\prime}}&-2^{-1}p^{k-j^{\prime}}-bp^{k-k^{\prime}}&\tau^{-1}p^{j-j^{\prime}-1}-p^{-j^{\prime}}\\\
0&p^{k-k^{\prime}}-\delta p^{k-1}&-\delta p^{-1}\\\ 0&p^{k}&1\end{pmatrix}\in
K_{p}.$

Then we get a contradiction. So in this case, one has
$\mathcal{O}_{p}^{(3)}(f;n)=0.$

  4. 4.

Suppose $x_{p}$ is of the form in (8.33) and $y_{p}$ is of the form in (8.32),
namely, suppose

$\displaystyle x_{p}=$ $\displaystyle\begin{pmatrix}p^{j}&&\\\ &p^{k}&\\\
&&1\end{pmatrix}\begin{pmatrix}\mu_{1}&1&\\\ 1&&\\\
&&1\end{pmatrix}\begin{pmatrix}\tau&&\\\ &1&\\\ &&1\end{pmatrix}\gamma_{1},$
$\displaystyle y_{p}=$ $\displaystyle\begin{pmatrix}p^{j^{\prime}}&&\\\
&p^{k^{\prime}}&\\\ &&1\end{pmatrix}\begin{pmatrix}1&b^{\prime}&\\\ &1&\\\
&&1\end{pmatrix}\begin{pmatrix}\delta&&\\\ &1&\\\ &&1\end{pmatrix}\gamma_{2},$

where $\gamma_{1},\gamma_{2}\in I_{p}^{\prime}(1).$ Denote by
$\mathcal{O}_{p}^{(4)}(f;n)$ the contribution of $x_{p},y_{p}$ in the above
forms. Then (8.31) is equivalent to

$\displaystyle\begin{pmatrix}p^{-j}&&-\mu_{1}p^{-1}\\\ &p^{-k}&-\tau p^{-1}\\\
&&1\end{pmatrix}\left(\begin{array}[]{ccc}p^{j^{\prime}}&-2^{-1}p^{k^{\prime}}&1\\\
&p^{k^{\prime}}&\\\
&-p^{k^{\prime}}&1\end{array}\right)\left(\begin{array}[]{ccc}1&b^{\prime}&\delta^{-1}p^{-1}\\\
&1&\\\ &&1\end{array}\right)\in K_{p}.$

Expanding the left hand side this becomes

$\displaystyle\left(\begin{array}[]{ccc}p^{j^{\prime}-j}&p^{j^{\prime}-j}b^{\prime}-2^{-1}p^{k^{\prime}-j}+\mu_{1}p^{k^{\prime}-1}&\delta^{-1}p^{j^{\prime}-j-1}+p^{-j}-\mu_{1}p^{-1}\\\
0&p^{k^{\prime}-k}+\tau p^{k^{\prime}-1}&-\tau p^{-1}\\\
0&-p^{k^{\prime}}&1\end{array}\right)\in K_{p}.$

Taking the inverse we then obtain

$\displaystyle\begin{pmatrix}p^{-j^{\prime}}&-bp^{-k^{\prime}}&-\delta
p^{-1}\\\ &p^{-k^{\prime}}&\\\
&&1\end{pmatrix}\begin{pmatrix}p^{j}&-2^{-1}p^{k}&-1\\\ &p^{k}&\\\
&p^{k}&1\end{pmatrix}\begin{pmatrix}1&\mu_{1}&\mu_{1}\tau^{-1}p^{-1}\\\
&1&\tau^{-1}p^{-1}\\\ &&1\end{pmatrix}\in K_{p}$

By a calculation, this is equivalent to

$\displaystyle\begin{pmatrix}p^{j-j^{\prime}}&-2^{-1}p^{k-j^{\prime}}-bp^{k-k^{\prime}}-\delta
p^{k-1}&z_{2}^{\prime}\\\ 0&p^{k-k^{\prime}}&\tau^{-1}p^{k-k^{\prime}-1}\\\
0&p^{k}&1+\tau^{-1}p^{k-1}\end{pmatrix}\in K_{p},$

where

$z_{2}^{\prime}=\mu_{1}\tau^{-1}p^{j-j^{\prime}-1}-2^{-1}\tau^{-1}p^{k-j^{\prime}-1}-p^{-j^{\prime}}-\tau^{-1}bp^{k-k^{\prime}-1}-\delta
p^{-1}-\tau^{-1}\delta p^{k-2}.$

So $j=j^{\prime}$ and $k=k^{\prime}\geq 1.$ Thus,

$\displaystyle\left(\begin{array}[]{ccc}1&b^{\prime}-2^{-1}p^{k-j}+\mu_{1}p^{k-1}&\delta^{-1}p^{-1}+p^{-j}-\mu_{1}p^{-1}\\\
0&1+\tau p^{k-1}&-\tau p^{-1}\\\ 0&-p^{k}&1\end{array}\right)\in K_{p}.$

Then we get a contradiction and conclude that $\mathcal{O}_{p}^{(4)}(f;n)=0.$

In conclusion, (8.29) and (8.30) hold, giving Lemma 8.11. ∎

###  8.5. An upper bound for the global orbital integral
$\mathcal{O}_{\gamma(1)}(f;\varphi^{\prime})$

In this section, we combine results from previous sections to deduce an upper
bound for $\mathcal{O}_{\gamma(1)}(f,\varphi^{\prime}).$

Let $\nu(f)$ be the set of (inert) primes $p$ (coprime with $NN^{\prime}D$)
such that

$f_{p}=\mathrm{1}_{G({\mathbb{Z}_{p}})A_{r_{p}}G({\mathbb{Z}_{p}})}$

for some $r_{p}\geq 1.$

######  Proposition 8.12.

Let notation be as before. We have

(8.35) $\mathcal{O}_{\gamma(1)}(f,\varphi^{\prime})\leq\frac{(\ell
N^{\prime})^{o(1)}}{2^{4k}k^{2}}\frac{N}{{N^{\prime}}^{3}}\prod_{p\in\nu(f)}p^{r_{p}}.$

where the implicit constants depend on $E$ and $\pi^{\prime}$ (via
$L(\pi^{\prime},\operatorname{Ad},1)$).

###### Proof.

It follows from (8.7), (8.15), and Lemmas 8.6, 8.7, 8.8, 8.9, 8.10, 8.11 that

(8.36)
$\mathcal{O}_{\gamma(1)}(f,\varphi^{\prime})=\eta(\Delta)\frac{C_{k}C_{N,N^{\prime}}}{L(\pi^{\prime},\operatorname{Ad},1)}\sum_{n\geq
1}|\lambda_{\pi^{\prime}}(n)|^{2}e^{-\frac{\pi
n}{{|D_{E}|^{1/2}}}}\prod_{p<\infty}I_{p}(n,\varphi^{\prime}),$

where $\eta(\Delta)$ is a complex number of modulus $1$,

$C_{k}=\frac{2^{4}\pi^{5}}{3.2^{4(k-1)}}\frac{1}{(k-1)^{2}},$

(8.37) $\displaystyle C_{N,N^{\prime}}$ $\displaystyle=\prod_{p\mid
N}\frac{\mu(I_{p}^{\prime})^{2}}{\mu(K_{p})}\prod_{p\in
N^{\prime}}(1+\frac{1}{p})(p-2)\mu(I_{p}^{\prime}(1))^{2}$
$\displaystyle=\prod_{p\mid N}\frac{p^{2}-p+1}{p+1}\prod_{p\mid
N^{\prime}}\frac{(p-2)(p+1)}{p(p^{2}-1)^{2}}\ll\frac{N}{{N^{\prime}}^{3}}.$

and where $I_{p}(n,\varphi^{\prime})=1$ if $(n,p\ell)=1$ and in general
satisfies,

(8.38) $\begin{cases}I_{p}(n,\varphi^{\prime})\leq(n,p)^{1+o(1)},&\ \text{if
$p\notin\nu(f)$}\\\
I_{p}(n,\varphi^{\prime})\ll(r_{p}+1)^{4}(n,p)^{1+o(1)}p^{r_{p}},&\ \text{if
$p\in\nu(f)$}.\end{cases}$

Hence, (8.36), Deligne’s bound (4.41) and (4.44) yields

(8.39) $\mathcal{O}_{\gamma(1)}(f,\varphi^{\prime})\ll\frac{(\ell
N^{\prime})^{o(1)}}{2^{4k}k^{2}}\frac{N}{{N^{\prime}}^{3}}\cdot\prod_{p\in\nu(f)}p^{r_{p}}.$

∎

###  8.6. Computation of the remaining unipotent orbital integrals

In this subsection, we prove the claim (6.16) and reduce the computation of
$\mathcal{O}_{\gamma(x)}(f^{\mathfrak{n}},\varphi^{\prime})$ $x\in E^{1}$ to
$\mathcal{O}_{\gamma(1)}(f^{\mathfrak{n}},\varphi^{\prime})$. Our result is
the following:

######  Proposition 8.13.

Let notation be as before. Let $x\in E^{1}.$ Then

$\displaystyle\mathcal{O}_{\gamma(x)}(f^{\mathfrak{n}},\varphi^{\prime})=\begin{cases}\mathcal{O}_{\gamma(1)}(f^{\mathfrak{n}},\varphi^{\prime}),\
\text{if $x\in\mathcal{O}_{E}^{1};$}\\\ 0,\ \text{otherwise}.\end{cases}$

###### Proof.

Let $x\neq 1.$ By Lemma 6.3 we have

$\gamma(x)=g_{1}.x\gamma(1).g_{2},\ g_{1},g_{2}\in G^{\prime}({\mathbb{Q}})$

and by $G^{\prime}({\mathbb{Q}})$ invariance of $\varphi^{\prime}$, we have

(8.40)
$\begin{split}\mathcal{O}_{\gamma(x)}(f^{\mathfrak{n}},\varphi^{\prime})&=\int_{{H}_{\gamma(x)}({\mathbb{Q}})\backslash
G^{\prime}(\mathbb{A})^{2}}f^{\mathfrak{n}}(u^{-1}g_{1}.x\gamma(1).g_{2}v)\overline{\varphi}^{\prime}(u){\varphi}^{\prime}(v)dudv\\\
&=\mathcal{O}_{x\gamma(1)}(f^{\mathfrak{n}},\varphi^{\prime}).\end{split}$

Similar to the proof of Lemma 8.1 the orbital integral
$\mathcal{O}_{x\gamma(1)}(f^{\mathfrak{n}},\varphi^{\prime})$ is equal to

(8.41) $\int_{N^{\prime}(\mathbb{A})\backslash
G^{\prime}(\mathbb{A})}\int_{G^{\prime}(\mathbb{A})}f^{\mathfrak{n}}\left(y_{1}^{-1}x\gamma(1)y_{2}\right)\int_{[N^{\prime}]}\overline{\varphi}^{\prime}(vy_{1}){\varphi}^{\prime}(vy_{2})dvdy_{1}dy_{2},$

which is absolutely converging by Proposition 8.12.

Note that $f^{\mathfrak{n}}\left(y_{1}^{-1}x\gamma(1)y_{2}\right)=0$ unless

$\widetilde{\mathfrak{n}}^{-1}y_{1}^{-1}x\gamma(1)y_{2}\widetilde{\mathfrak{n}}\in\prod_{\begin{subarray}{c}p<\infty\\\
p\notin\nu(f)\end{subarray}}K_{p}(N)\prod_{p\in\nu(f)}G(\mathbb{Z}_{p})A_{p^{r_{p}}}G(\mathbb{Z}_{p}).$

Also,

$\displaystyle\left(\begin{array}[]{ccc}a&&b\\\ &1&\\\
c&&d\end{array}\right)\left(\begin{array}[]{ccc}1&1&-1/2\\\ &1&-1\\\
&&1\end{array}\right)\left(\begin{array}[]{ccc}a^{\prime}&&b^{\prime}\\\
&1&\\\
c^{\prime}&&d^{\prime}\end{array}\right)=\left(\begin{array}[]{ccc}*&&*\\\
&1&\\\ &&*\end{array}\right).$

Hence $f^{\mathfrak{n}}\left(y_{1}^{-1}x\gamma(1)y_{2}\right)=0$ unless
$\nu_{v}(x)\geq 0$ (i.e., $x\in\mathcal{O}_{E_{v}}$) for all places $v$ in $E$
such that $v$ is not above $N^{\prime}$ and $v\notin\nu(f).$

Let $p\in\nu(f).$ Then
$f_{p}^{\mathfrak{n}_{p}}\left(y_{1}^{-1}x\gamma(1)y_{2}\right)=0$ unless

(8.42)
$\widetilde{\mathfrak{n}}_{p}^{-1}y_{1,p}^{-1}\gamma(1)y_{2,p}\widetilde{\mathfrak{n}}_{p}\in
x^{-1}G(\mathbb{Z}_{p})A_{p^{r_{p}}}G(\mathbb{Z}_{p}).$

Taking the determinant on both sides of (8.42) we then derive that
$|x|_{p}=1,$ implying that $x\in\mathcal{O}_{E_{p}}^{\times}.$

Let $p\mid N^{\prime}.$ Then
$f_{p}^{\mathfrak{n}_{p}}\left(y_{1}^{-1}x\gamma(1)y_{2}\right)=0$ unless

(8.43)
$\widetilde{\mathfrak{n}}_{p}^{-1}y_{1,p}^{-1}\gamma(1)y_{2,p}\widetilde{\mathfrak{n}}_{p}\in
x^{-1}K_{p}.$

We then apply the manipulation as in the proof of Lemma 8.11 to show that, if
(8.43) holds for some $x\in E^{1},$ then necessarily
$x\in\mathcal{O}_{E_{p}}^{\times}.$

Since $p|N^{\prime}$ we may identify $G(\mathbb{Q}_{p})$ (resp.
$G^{\prime}(\mathbb{Q}_{p})$) with $\operatorname{GL}(3,\mathbb{Q}_{p})$
(resp. $\operatorname{GL}(2,\mathbb{Q}_{p})$); suppose for the contrary that
$\nu(x)\neq 0.$ In the notations of the proof of Lemma 8.11 we have four cases
to consider.

In Case 1, (8.43) implies that

$\displaystyle\left(\begin{array}[]{ccc}p^{j-j^{\prime}}&-2^{-1}p^{k-j^{\prime}}-b^{\prime}p^{k-k^{\prime}}-\delta
p^{k-1}&\tau p^{k-j^{\prime}-1}-p^{-j^{\prime}}-\delta p^{-1}\\\
0&p^{k-k^{\prime}}&0\\\ 0&p^{k}&1\end{array}\right)\in x^{-1}K_{p},$

which contradicts the assumption that $\nu(x)<0.$ Similarly, taking the
inverse, we see that $\nu(x)>0$ cannot happen as well.

In the Case 3, (8.43) implies that

(8.44)
$x\begin{pmatrix}p^{j^{\prime}-j}&bp^{j^{\prime}-j}-2^{-1}p^{k^{\prime}-j}+\tau
p^{k^{\prime}-1}&z_{2}\\\ &p^{k^{\prime}-k}&\delta^{-1}p^{k^{\prime}-k-1}\\\
&-p^{k^{\prime}}&1-\delta^{-1}p^{k^{\prime}-1}\end{pmatrix}\in K_{p},$

where

$z_{2}:=\delta^{-1}bp^{j^{\prime}-j-1}-2^{-1}\delta^{-1}p^{k^{\prime}-j-1}+p^{-j}-\tau
p^{-1}+\tau\delta^{-1}p^{k^{\prime}-2}.$

Considering the inverse as before, we then obtain

(8.45)
$x^{-1}\begin{pmatrix}p^{j-j^{\prime}}&-2^{-1}p^{k-j^{\prime}}-bp^{k-k^{\prime}}&\tau^{-1}p^{j-j^{\prime}-1}-p^{-j^{\prime}}\\\
0&p^{k-k^{\prime}}-\delta p^{k-1}&-\delta p^{-1}\\\ 0&p^{k}&1\end{pmatrix}\in
K_{p}.$

By (8.45) one has $\nu(x)\leq-1$ and $j-j^{\prime}=\nu(x).$ Computing the
determinant of (8.44) we then have

$\displaystyle j-j^{\prime}+k-k^{\prime}=3\nu(x),\ i.e.,\
k-k^{\prime}=2\nu(x).$

Therefore

$j=1,j^{\prime}=2,k=-1,k^{\prime}=1,\hbox{ and }\nu(x)=-1.$

Applying inversion in (8.45) we then get $\nu(x)-1\geq 0$ by considering its
$(2,3)$-th entry and therefore $\nu(x)\geq 1$, a contradiction! In conclusion,
the Case 3 does not contribute to the orbital integral, just as the situation
in Lemma 8.11. Similarly the contribution of Case 4 is also zero.

Finally we consider the Case 2; where (8.43) implies that

$\displaystyle\begin{pmatrix}p^{j^{\prime}-j}&bp^{j^{\prime}-j}-2^{-1}p^{k^{\prime}-j}+\mu_{1}p^{k^{\prime}-1}&z\\\
&p^{k^{\prime}-k}+\tau p^{k^{\prime}-1}&\delta^{-1}p^{k^{\prime}-1-k}-\tau
p^{-1}+\tau\delta^{-1}p^{k^{\prime}-2}\\\
&-p^{k^{\prime}}&1-\delta^{-1}p^{k^{\prime}-1}\end{pmatrix}\in x^{-1}K_{p},$

where

$z=\delta^{-1}bp^{j^{\prime}-j-1}-2^{-1}\delta^{-1}p^{-1-j}+p^{-j}-\mu_{1}p^{-1}+\mu_{1}\delta^{-1}p^{k^{\prime}-2}.$

Suppose $\nu(x)\leq-1.$ Then $k^{\prime}=1=k$ and $j^{\prime}-j=-\nu(x)$ by
analyzing the diagonals. However, taking determinant we then obtain

$j^{\prime}-j+k^{\prime}-k=-3\nu(x),$

a contradiction! Suppose $\nu(x)\geq 1.$ Then taking the inverse of the above
matrix we then obtain

$\displaystyle\begin{pmatrix}p^{j-j^{\prime}}&\mu_{1}p^{j-j^{\prime}}-2^{-1}p^{k-j^{\prime}}-bp^{k-k^{\prime}}&z^{\prime}\\\
&p^{k-k^{\prime}}-\delta p^{k-1}&\tau^{-1}p^{k-k^{\prime}-1}-\delta
p^{-1}-\delta\tau^{-1}p^{k-2}\\\ &p^{k}&1+\tau^{-1}p^{k-1}\end{pmatrix}\in
xK_{p},$

where

$z^{\prime}=\mu_{1}\tau^{-1}p^{j-j^{\prime}-1}-2^{-1}\tau^{-1}p^{k-j^{\prime}-1}-p^{-j^{\prime}}-b\tau^{-1}p^{k-k^{\prime}-1}.$

So $k=k^{\prime}=1$ and $j-j^{\prime}=1.$ Again, we encounter an contradiction
by taking determinant. Hence, $\nu(x)=0,$ i.e.,
$x\in\mathcal{O}_{E_{p}}^{\times}.$

In summary, we have shown that $\nu(x)\geq 0$ in all finite places. So
$x\in\mathcal{O}_{E}.$ Since $x\overline{x}=1$ we have
$x\in\mathcal{O}_{E}^{1}.$

Since the test function $f$ is $Z_{G}(\mathcal{O}_{E}^{1})$-invariant, then
Proposition 8.13 follows. ∎

As a consequence of Proposition 8.12 and 8.13 we have

######  Corollary 8.14.

The sum of the unipotent orbital integrals satisfies the bound

$\displaystyle\sum_{x\in
E^{1}}\mathcal{O}_{\gamma(x)}(f,\varphi^{\prime})\ll\frac{(\ell
N^{\prime})^{o(1)}}{2^{4k}k^{2}}\frac{N}{{N^{\prime}}^{3}}\prod_{p\in\nu(f)}p^{r_{p}}$

where the implicit constant depend on $E$ and $\pi^{\prime}$.

##  9\. The Regular Orbital Integrals

In this section and the next we handle the contribution to (6.17) of the
regular orbital integrals, ie. the
${\mathcal{O}}_{\gamma(x)}(f,\varphi^{\prime})$ when $x\in
E^{\times}\\!-E^{1}$.

In the case of $\operatorname{GL}(2)\times\operatorname{GL}(1)$ this was
handled in [RR05, §2.6]; here the geometric structure of $U(2,1)\times U(1,1)$
is much more complicated.

In this section, we provide bounds for some local integrals which we combine
together in Section 10 to prove Theorem 10.1.

###  9.1. The Stabilizer $H_{\gamma(x)}$

We start by computing the stabilizer $H_{\gamma(x)}$ associated to $\gamma(x)$
when $x\in E^{\times}\\!-E^{1}.$ Recall that for any ${\mathbb{Q}}$-algebra
$R$

$\displaystyle H_{{\gamma(x)}}(R):=\big{\\{}(g_{1},g_{2})\in G^{\prime}(R):\
g_{1}^{-1}\gamma(x)g_{2}=\gamma(x)\big{\\}}.$

To save space we will represent matrices in $G^{\prime}$ either as $2\times 2$
or a $3\times 3$ matrices (in the standard bases $\\{e_{-1},e_{1}\\}$ or
$\\{e_{-1},e_{0},e_{1}\\}$), that is $\hbox{ either }\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\hbox{ or }\begin{pmatrix}a&&b\\\ &1&\\\ c&&d\end{pmatrix}.$
Of course whenever matrices from $G$ are involved we will use the $3\times 3$
notation.

######  Lemma 9.1.

Let notation be as above, for any ${\mathbb{Q}}$-algebra $R$,
$H_{\gamma(x)}(R)$ is equal to

$\displaystyle\Bigg{\\{}$
$\displaystyle\begin{pmatrix}1+\frac{y(x-1)(\overline{x}+1)}{2}&\frac{y(1+x)(1+\overline{x})}{4}\\\
y(1-x)(1-\overline{x})&1+\frac{y(x+1)(\overline{x}-1)}{2}\end{pmatrix}\times\begin{pmatrix}1+\frac{y(x-1)(\overline{x}+1)}{2}&y\\\
\frac{y(1+x)(1+\overline{x})(1-x)(1-\overline{x})}{4}&1+\frac{y(x+1)(\overline{x}-1)}{2}\end{pmatrix}$
$\displaystyle\qquad\in G^{\prime}(R)\times G^{\prime}(R):y\in E^{\times}(R)\
\text{and}\ y+\overline{y}+y\overline{y}(x\overline{x}-1)=0\Bigg{\\}}.$

###### Proof.

Let

$g=\begin{pmatrix}a&b\\\
c&d\end{pmatrix},g^{\prime}=\begin{pmatrix}a^{\prime}&b^{\prime}\\\
c^{\prime}&d^{\prime}\end{pmatrix}\in G^{\prime}(R).$

Assume

(9.1)
$g\left(\begin{array}[]{ccc}\frac{x\overline{x}+3\overline{x}-x+1}{4}&\frac{1+x}{2}&-\frac{1}{2}\\\
\frac{(x+1)(\overline{x}-1)}{2}&x&-1\\\
-\frac{(1-x)(1-\overline{x})}{2}&1-x&1\end{array}\right)=\left(\begin{array}[]{ccc}\frac{x\overline{x}+3\overline{x}-x+1}{4}&\frac{1+x}{2}&-\frac{1}{2}\\\
\frac{(x+1)(\overline{x}-1)}{2}&x&-1\\\
-\frac{(1-x)(1-\overline{x})}{2}&1-x&1\end{array}\right)g^{\prime}.$

Expanding both sides of equation (9.1) we then obtain an equality between the
matrix

$\left(\begin{array}[]{ccc}a^{\prime}(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-\frac{c^{\prime}}{2}&\frac{1+x}{2}&b^{\prime}(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-\frac{d^{\prime}}{2}\\\
\frac{a^{\prime}(x+1)(\overline{x}-1)}{2}-c^{\prime}&x&\frac{b^{\prime}(x+1)(\overline{x}-1)}{2}-d^{\prime}\\\
-\frac{a^{\prime}(1-x)(1-\overline{x})}{2}+c^{\prime}&1-x&-\frac{b^{\prime}(1-x)(1-\overline{x})}{2}+d^{\prime}\end{array}\right)$

and the matrix

$\left(\begin{array}[]{ccc}a(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-\frac{b(1-x)(1-\overline{x})}{2}&\frac{a(1+x)}{2})+b(1-x)&-\frac{a}{2}+b\\\
\frac{(x+1)(\overline{x}-1)}{2}&x&-1\\\
c(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-\frac{d(1-x)(1-\overline{x})}{2}&\frac{c(1+x)}{2}+d(1-x)&-\frac{c}{2}+d\end{array}\right),$

which is equivalent to the system of equations

(9.2)
$\begin{cases}-\frac{a}{2}+b=b^{\prime}(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-\frac{d^{\prime}}{2}\\\
\frac{a(1+x)}{2})+b(1-x)=\frac{1+x}{2}\\\
a(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-\frac{b(1-x)(1-\overline{x})}{2}=a^{\prime}(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-\frac{c^{\prime}}{2}\\\
-\frac{c}{2}+d=-\frac{b^{\prime}(1-x)(1-\overline{x})}{2}+d^{\prime}\\\
\frac{c(1+x)}{2}+d(1-x)=1-x\\\
c(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-\frac{d(1-x)(1-\overline{x})}{2}=-\frac{a^{\prime}(1-x)(1-\overline{x})}{2}+c^{\prime}\\\
-1=\frac{b^{\prime}(x+1)(\overline{x}-1)}{2}-d^{\prime}\\\
\frac{(x+1)(\overline{x}-1)}{2}=\frac{a^{\prime}(x+1)(\overline{x}-1)}{2}-c^{\prime}\end{cases}$

Now we need to solve (9.2) to find the group $H_{\gamma(x)}(R).$ Note that

$\displaystyle(-\frac{1}{2},1)\frac{(1-x)(1+\overline{x})}{2}+(\frac{1+x}{2},1-x)\overline{x}=(\frac{x\overline{x}+3\overline{x}-x-1}{4},-\frac{(1-x)(1-\overline{x})}{2}).$

Hence we obtain from (9.2) the following equations on
$(a^{\prime},b^{\prime},c^{\prime},d^{\prime}):$

(9.3)
$\begin{cases}(\frac{b^{\prime}(x\overline{x}+3\overline{x}-x-1)}{4}-\frac{d^{\prime}(x-1)(1+\overline{x})}{4}+\frac{(1+x)\overline{x}}{2}=\frac{a^{\prime}(x\overline{x}+3\overline{x}-x-1)}{4}-\frac{c^{\prime}}{2}\\\
(-\frac{b^{\prime}(1-x)(1-\overline{x})}{2}+d^{\prime})\frac{(x-1)(1+\overline{x})}{2}+\overline{x}(1-x)=-\frac{a^{\prime}(1-x)(1-\overline{x})}{2}+c^{\prime}\\\
-1=\frac{b^{\prime}(x+1)(\overline{x}-1)}{2}-d^{\prime}\\\
\frac{(x+1)(\overline{x}-1)}{2}=\frac{a^{\prime}(x+1)(\overline{x}-1)}{2}-c^{\prime}.\end{cases}$

A computation shows that the companion matrix is singular. By Gauss
elimination, we can further simplify the system of equations (9.3) to get

(9.4)
$\begin{cases}c^{\prime}=\frac{b^{\prime}(x+1)(\overline{x}-1)(x-1)(\overline{x}+1)}{4}\\\
d^{\prime}=\frac{b^{\prime}(x+1)(\overline{x}-1)}{2}+1\\\
c^{\prime}=\frac{(a^{\prime}-1)(x+1)(\overline{x}-1)}{2}.\end{cases}$

Denote by $y=\frac{(x-1)(\overline{x}+1)}{2}.$ Since
$\begin{pmatrix}a^{\prime}&b^{\prime}\\\ c^{\prime}&d^{\prime}\end{pmatrix}\in
G^{\prime}(R),$ we have

$\displaystyle\begin{pmatrix}\overline{b}^{\prime}y+1&\overline{b}^{\prime}\\\
\overline{b}^{\prime}y\overline{y}&\overline{b}^{\prime}\overline{y}+1\end{pmatrix}\begin{pmatrix}b^{\prime}y+1&b^{\prime}\\\
b^{\prime}y\overline{y}&b^{\prime}\overline{y}+1\end{pmatrix}=\operatorname{Id},$

which turns out to be equivalent to
$b^{\prime}\overline{b}^{\prime}(y+\overline{y})+b^{\prime}+\overline{b}^{\prime}=0,$
i.e.,

(9.5)
$b^{\prime}\overline{b}^{\prime}(x\overline{x}-1)+b^{\prime}+\overline{b}^{\prime}=0.$

Substituting (9.4) into (9.2) we then obtain

$\displaystyle\begin{cases}-\frac{a}{2}+b=b^{\prime}(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-b^{\prime}(\frac{(1-x)(1-\overline{x})}{4}-\frac{1-\overline{x}}{2})-\frac{1}{2}=\frac{b^{\prime}(1+\overline{x})}{2}-\frac{1}{2}\\\
\frac{a(1+x)}{2}+b(1-x)=\frac{1+x}{2}\\\
a(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-\frac{b(1-x)(1-\overline{x})}{2}=a^{\prime}(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-(a^{\prime}-1)(\frac{(1-x)(1-\overline{x})}{4}-\frac{1-\overline{x}}{2})\\\
-\frac{c}{2}+d=-\frac{b^{\prime}(1-x)(1-\overline{x})}{2}+d^{\prime}=b^{\prime}(\frac{(1-x)(1-\overline{x})}{2}+\overline{x}-1)+1-\frac{b^{\prime}(1-x)(1-\overline{x})}{2}\\\
\frac{c(1+x)}{2}+d(1-x)=1-x\\\
c(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-\frac{d(1-x)(1-\overline{x})}{2}=-\frac{a^{\prime}(1-x)(1-\overline{x})}{2}+(a^{\prime}-1)(\frac{(1-x)(1-\overline{x})}{2}-1+\overline{x}).\end{cases}$

Using Gauss elimination method we then see the above system becomes

(9.6) $\begin{cases}a=1-b^{\prime}(1-x)\frac{1+\overline{x}}{2}\\\
b=\frac{b^{\prime}(1+\overline{x})}{2}-\frac{b^{\prime}(1-x)}{2}(1-\frac{\overline{x}}{2})=\frac{b^{\prime}(1+x)(1+\overline{x})}{4}\\\
c=b^{\prime}(1-x)(1-\overline{x})\\\
d=1-b^{\prime}(1-\overline{x})+\frac{b^{\prime}(1-x)(1-\overline{x})}{2}\\\
a(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-\frac{b(1-x)(1-\overline{x})}{2}=\frac{a^{\prime}(1+\overline{x})}{2})+(\frac{(1-x)(1-\overline{x})}{4}-\frac{(1-\overline{x})}{2})\\\
c(\overline{x}+\frac{(1-x)(1-\overline{x})}{4})-\frac{d(1-x)(1-\overline{x})}{2}=-a^{\prime}(1-\overline{x})-(\frac{(1-x)(1-\overline{x})}{2}-1+\overline{x}).\end{cases}$

A calculation shows that the last two equations in (9.6) are redundant. That
is, (9.6) is equivalent to

(9.7) $\begin{cases}a=1-\frac{b^{\prime}(1-x)(1+\overline{x})}{2}\\\
b=\frac{b^{\prime}(1+\overline{x})}{2}-\frac{b^{\prime}(1-x)(1+\overline{x})}{4}=\frac{b^{\prime}(1+x)(1+\overline{x})}{4}\\\
c=b^{\prime}(1-x)(1-\overline{x})\\\
d=1-\frac{b^{\prime}(1+x)(1-\overline{x})}{2}.\end{cases}$

Since $\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\in G^{\prime}(\mathbb{Q}),$ we
then have

(9.8) $\begin{pmatrix}\overline{d}&\overline{b}\\\
\overline{c}&\overline{a}\end{pmatrix}\begin{pmatrix}a&b\\\
c&d\end{pmatrix}=\operatorname{Id}.$

Substituting (9.7) into (9.8) we then conclude together with (9.5) that

(9.9)
$\begin{cases}\big{[}b^{\prime}+\overline{b}^{\prime}+b^{\prime}\overline{b}^{\prime}(x\overline{x}-1)\big{]}\frac{(1-x)(1+\overline{x})}{2}=0\\\
b^{\prime}+\overline{b}^{\prime}+b^{\prime}\overline{b}^{\prime}(x\overline{x}-1)=0.\end{cases}$

Then it follows from (9.9) that
$b^{\prime}+\overline{b}^{\prime}+b^{\prime}\overline{b}^{\prime}(x\overline{x}-x-\overline{x})=0.$
Hence Lemma 9.1 follows. ∎

We will now use that $x\notin E^{1}$, that is $x.\overline{x}-1\not=0$. It
turns out that the conjugate stabilizer of $\gamma(x)J$ (which is a conjugate
to that of $\gamma(x)$ since $J\in G^{\prime}({\mathbb{Q}})$) has a nice
parametrization and we will use it instead:

######  Corollary 9.2.

Notation be as before, for $x\in E^{\times}\\!-E^{1}$ set

$s(x):=\frac{(x-1)(\overline{x}+1)}{2}\in E^{\times}.$

For $R$ any $\mathbb{Q}$-algebra, the stabilizer of $\gamma(x)J$ is given by

$H_{\gamma(x)J}(R)=\big{\\{}(h_{1}(w),h_{2}(w))\in G^{\prime}(R)\times
G^{\prime}(R):\ w\in E^{\times}(R),\ w\overline{w}=1\big{\\}},$

where

(9.10)
$\begin{cases}h_{1}(w):=&\begin{pmatrix}1-\frac{(1-w)s(x)}{s(x)+s(\overline{x})}&-\frac{(1-w)s(x)s(\overline{x})}{(s(x)+s(\overline{x}))x\overline{x}}\\\
-\frac{(1-w)x\overline{x}}{s(x)+s(\overline{x})}&1-\frac{(1-w)s(\overline{x})}{s(x)+s(\overline{x})}\end{pmatrix}\\\
\\\
h_{2}(w):=&\begin{pmatrix}1-\frac{(1-w)s(\overline{x})}{s(x)+s(\overline{x})}&-\frac{(1-w)s(x)s(\overline{x})}{s(x)+s(\overline{x})}\\\
-\frac{(1-w)}{s(x)+s(\overline{x})}&1-\frac{(1-w)s(x)}{s(x)+s(\overline{x})}\end{pmatrix}.\end{cases}$

In particular, $h_{1}(w)$ and $h_{2}(w)$ are determined uniquely by their
determinant:

$\det h_{1}(w)=\det h_{2}(w)=w.$

###### Proof.

Since $x\not\in E^{1},$ $x\overline{x}-1\neq 0.$ Now we consider the equation

$y+\overline{y}+y\overline{y}(x\overline{x}-1)=0,$

whose locus is a conic (since $x\overline{x}-1\neq 0$) and by the intersection
method, its $R$-rational points are parametrized by $\mathbb{P}^{1}(R)$.
Explicitely, write $y=a+at\sqrt{-D},$ where $a,t\in R.$ Then

$a(1+t^{2})(x\overline{x}-1)+2=0.$

and we obtain

$\displaystyle
y=-\frac{2(1+t\sqrt{-D})}{(1+t^{2}D)(x\overline{x}-1)}=-\frac{2}{(1-t\sqrt{-D})(x\overline{x}-1)},\quad
t\in\mathbb{P}^{1}(R).$

Then the parametrization (9.10) follows from Lemma 9.1 and the fractional
linear transform replacing $t$ with $w$ such that
$(1-w)/2=(1-t\sqrt{-D})^{-1}.$ We have again used the fact that $J\in
G^{\prime}({\mathbb{Q}}).$ ∎

######  Remark 9.1.

For $x_{0}\in E^{1},$ the above computation of $H_{\gamma(x)J}$ is consistent
with what we obtained before for $H_{\gamma(x_{0})}$ after taking limit
$x\mapsto x_{0}$ in (9.10).

We will henceforth write $H_{x}$ for the image of $H_{\gamma(x)J}\subset
G^{\prime}\times G^{\prime}$ under the projection to the first component,
i.e.,

$H_{x}(R)=\\{h_{1}(u),\ u\in E^{\times}(R),\ u\overline{u}=1\\}$

for $h_{1}(u)$ defined in (9.10).

As a consequence of Corollary 9.2, we obtain

######  Corollary 9.3.

Let $x\in E^{\times}\\!-E^{1}.$ Then $H_{x}$ is isomorphic to $U(1),$ the
unitary group of rank $1.$ In particular, $H_{x}(\mathbb{Q})\backslash
H_{x}(\mathbb{A})$ has volume $2$ for the Tamagawa measure.

###### Proof.

A direct computation shows that

$h_{1}(u_{1})h_{1}(u_{2})=h_{1}(u_{1}u_{2}).$

This gives the group law of $H_{x}(R),$ where $R$ is a $\mathbb{Q}$-algebra.
In particular, we have a natural isomorphism with the unitary group

$H_{x}\simeq U(1)=E^{1}.$

∎

###  9.2. Reduction to factorable integrals

To simplify notations we will write

${\mathcal{O}}_{x}(f^{\mathfrak{n}},\varphi^{\prime}):={\mathcal{O}}_{\gamma(x)}(f^{\mathfrak{n}},\varphi^{\prime})$

Observe that since $J\in G^{\prime}({\mathbb{Q}})$ and $\varphi^{\prime}$ is
$G^{\prime}({\mathbb{Q}})$-invariant we have by suitable changes of variables
that for $x\in E^{\times}\\!-E^{1}$

$\displaystyle{\mathcal{O}}_{x}(f^{\mathfrak{n}},\varphi^{\prime})={\mathcal{O}}_{\gamma(x)}(f^{\mathfrak{n}},\varphi^{\prime})={\mathcal{O}}_{\gamma(x)J}(f^{\mathfrak{n}},\varphi^{\prime})$
$\displaystyle=\int_{H_{\gamma(x)J}(\mathbb{Q})\backslash
G^{\prime}(\mathbb{A})\times
G^{\prime}(\mathbb{A})}f^{\mathfrak{n}}(y_{1}^{-1}\gamma(x)Jy_{2})\overline{\varphi}^{\prime}(y_{1}){\varphi}^{\prime}(y_{2})dy_{1}dy_{2}$
$\displaystyle=\int_{H_{x}(\mathbb{A})\backslash G^{\prime}(\mathbb{A})\times
G^{\prime}(\mathbb{A})}f^{\mathfrak{n}}(y_{1}^{-1}\gamma(x)Jy_{2})\int_{[H_{\gamma(x)J}]}\overline{\varphi}^{\prime}(h_{1}y_{1}){\varphi}^{\prime}(h_{2}y_{2})dudy_{1}dy_{2}$

where

$[H_{\gamma(x)J}]:=H_{\gamma(x)J}(\mathbb{Q})\backslash
H_{\gamma(x)J}(\mathbb{A}),$

and

$(h_{1},h_{2})=(h_{1}(u),h_{2}(u))\in[H_{\gamma(x)J}].$

For any $(y_{1},y_{2})$, the $u$-integral can be evaluated as an (infinite)
sum of Waldspurger’s period integrals (see [Wal85]). However, we will be much
more coarse and will content ourselves with the upperbound

(9.11)
$\int_{[H_{\gamma(x)J}]}\varphi^{\prime}(h_{1}y_{1})\overline{\varphi}^{\prime}(h_{2}y_{2})du\ll\|\varphi^{\prime}\|_{\infty}^{2}.$

Therefore we obtain

(9.12)
$|\mathcal{O}_{x}(f^{\mathfrak{n}},\varphi^{\prime})|\ll\|\varphi^{\prime}\|_{\infty}^{2}{\mathcal{I}}(f^{\mathfrak{n}},x)$

where

${\mathcal{I}}(f^{\mathfrak{n}},x)=\int_{H_{\gamma(x)J}(\mathbb{A})\backslash
G^{\prime}(\mathbb{A})^{2}}|f^{\mathfrak{n}}(u^{-1}\gamma(x)Jv)|dudv,$

say.

The main benefit of this rather crude treatment is that the resulting integral
is factorable: we have

${\mathcal{I}}(f^{\mathfrak{n}},x)=\prod_{v}{\mathcal{I}}_{v}(f^{\mathfrak{n}}_{v},x)$

where

(9.13)
${\mathcal{I}}_{v}(f^{\mathfrak{n}}_{v},x):=\int_{H_{\gamma(x)J}({\mathbb{Q}}_{v})\backslash(G^{\prime}\times
G^{\prime})({\mathbb{Q}}_{v})}|f_{v}^{\mathfrak{n}}(u^{-1}\gamma(x)Jv)|dudv;$

indeed as we will verify in the next subsections below, for any given $x\in
E-E_{1}$, we have

${\mathcal{I}}_{p}(f^{\mathfrak{n}}_{p},x)=1$

for all but finitely many $p$.

In the subsequent subsections we analyse these local integrals
${\mathcal{I}}_{v}(f^{\mathfrak{n}}_{v},x)$ give critera for vanishing and
provide upper bounds.

In the sequel to simplify notations we will write
${\mathcal{I}}(x),{\mathcal{I}}_{v}(x)$ in place of
${\mathcal{I}}(f^{\mathfrak{n}},x)$ and
${\mathcal{I}}_{v}(f^{\mathfrak{n}}_{v},x)$.

###  9.3. A vanishing criterion for the non-archimedan local integrals

Let $p$ be a prime, ${\mathfrak{p}}$ the place above $p$ we have fixed (we
take $\varpi_{p}=p$ if $p$ is inert) and let $\nu$ be the associated valuation
in the local field $E_{\mathfrak{p}}$.

######  Proposition 9.4.

Let $x\in E^{\times}\\!-E^{1}$ a non-zero regular element. The following holds

  * •

If $p$ is ramified, then ${\mathcal{I}}_{p}(x)=0$ unless

$x\in{\mathcal{O}}_{E,p}.$

  * •

If $p$ is inert, then ${\mathcal{I}}_{p}(x)=0$ unless

$x\in p^{-\nu(\ell)}{\mathcal{O}}_{E,p}\ \hbox{ and }x\overline{x}\equiv
1\,(\mathrm{mod}\,{(N,p)}).$

  * •

If $p$ is split then ${\mathcal{I}}_{p}(x)=0$ unless

$\nu(N^{\prime}x),\nu(N^{\prime}\overline{x})\geq 0.$

From this we deduce immediately the a global vanishing criterion:

######  Theorem 9.5.

Notations being as above, let

(9.14) $\mathfrak{X}(N,N^{\prime},\ell)=\big{\\{}x\in E^{\times}\\!-E^{1},\
x\in(\ell N^{\prime})^{-1}\mathcal{O}_{E}:\ x\overline{x}\equiv 1\
(\mathrm{mod}\ N)\big{\\}}.$

For $x\in E^{\times}\\!-E^{1}$ and not contained in
$\mathfrak{X}(N,N^{\prime},\ell)$, we have

${\mathcal{I}}(f^{\mathfrak{n}},x)=0$

and by (9.12) we have

(9.15) $\mathcal{O}_{\gamma(x)}(f^{\mathfrak{n}},\varphi^{\prime})=0$

######  Remark 9.2.

This criterion is similar to the phenomenon encountered in the case of the
forms of $\operatorname{GL}(2)\times\operatorname{GL}(1)$ in [RR05, §2.6] and
[FW09].

We will prove this through case by case analysis: see Propositions 9.6, 9.7,
9.8, In fact we will also provide upper bounds for ${\mathcal{I}}_{p}(x)$. The
non-split case is discussed in §9.4 and the following subsection and the split
case in §9.5.

For any ${\mathbb{Q}}$-algebra $R$ let

$SG^{\prime}(R):=SU(W)=\\{g\in G^{\prime}(R),\ \det g=1\\}.$

We start with the observation that a fundamental domain for the
quotient$H_{\gamma(x)J}({\mathbb{Q}}_{p})\backslash(G^{\prime}\times
G^{\prime})({\mathbb{Q}}_{p})$ is the subgroup

$SG^{\prime}({\mathbb{Q}}_{p})\times G^{\prime}({\mathbb{Q}}_{p}).$

Indeed any $(u,v)\in G^{\prime}\times G^{\prime}({\mathbb{Q}}_{p})$ can be
written

$(u,v)=(g_{1}(w)u^{\prime},g_{2}(w)v^{\prime})$

with

$w=\det(u)=\det(g_{1}(w)),\ u^{\prime}=g_{1}(w)^{-1}u\in
SG^{\prime}({\mathbb{Q}}_{p}),\ v^{\prime}=g_{2}(w)^{-1}v.$

Moreover for $u_{1},u^{\prime}_{1}\in SG^{\prime}({\mathbb{Q}}_{p})$ and
$v,v^{\prime}\in G^{\prime}({\mathbb{Q}}_{p})$ we have

$(u^{\prime}_{1},v^{\prime})=(g_{1}(w)u_{1},g_{2}(w)v)\Longrightarrow w=1,\
g_{1}(w)=g_{2}(w)=\operatorname{Id}_{3},\ u_{1}=u^{\prime}_{1},\
v^{\prime}=v.$

We have therefore

${\mathcal{I}}_{p}(x):=\int_{SG^{\prime}({\mathbb{Q}}_{p})\times
G^{\prime}({\mathbb{Q}}_{p})}|f_{v}^{\mathfrak{n}}(u^{-1}\gamma(x)Jv)|dudv.$

###  9.4. The non-split case

In this section we evaluate the local integral ${\mathcal{I}}_{p}(x)$ when $p$
is non-split: this implies that $\overline{\varpi}_{p}=\pm\varpi_{p}$, that
${\mathfrak{n}}_{p}=\operatorname{Id}_{3}$ and that
$f^{{\mathfrak{n}}}_{p}=f_{p}$ is a scalar times the characteristic function
of either $K_{p}(N)$ or of

$G({\mathbb{Z}_{p}})A_{r}G({\mathbb{Z}_{p}})$

for $A_{r},\ r\geq 0$ the diagonal matrix defined in (5.64) (if $p$ is inert
and coprime to $N$).

We use the Iwasawa decomposition of for $u,v\in
SG^{\prime}({\mathbb{Q}}_{p})\times G^{\prime}(\mathbb{Q}_{p})$.

We have $\det u=1$ and from the description of $G^{\prime}({\mathbb{Q}}_{p})$
$\det v$ has valuation $0$ so that (here we represent $u$ and $v$ as $2\times
2$ matrices)

(9.16) $\displaystyle u=\begin{pmatrix}\varpi_{p}^{i}&\\\
&\overline{\varpi}_{p}^{-i}\end{pmatrix}\begin{pmatrix}1&b\\\
&1\end{pmatrix}k_{1}=u^{\prime}k_{1}\kappa_{1}^{i},\ \ \
v=\begin{pmatrix}\varpi_{p}^{j}&\\\
&\overline{\varpi}_{p}^{-j}\end{pmatrix}\begin{pmatrix}1&b^{\prime}\\\
&1\end{pmatrix}k_{2}=v^{\prime}k_{2},$

with $i,j\in\mathbb{Z},$ $b,b^{\prime}\in E^{0}_{p}$, $k_{1},k_{2}\in
G^{\prime}(\mathbb{Z}_{p})$, $\det k_{1}=1$ and
$\kappa_{1}=\begin{pmatrix}\overline{\varpi}_{p}&\\\
&\varpi_{p}\end{pmatrix}$.

By definition, the function $f_{p}$ is bi-$K_{p}(N)$-invariant so that

$f_{p}(u^{-1}\gamma(x)Jv)=f_{p}(k_{1}^{-1}{u^{\prime}}^{-1}\gamma(x)Jv^{\prime}k_{2}).$

We have

$\displaystyle\gamma(x)J=\left(\begin{array}[]{ccc}-\frac{1}{2}&\frac{1+x}{2}&\frac{x\overline{x}+3\overline{x}-x+1}{4}\\\
-1&x&\frac{(x+1)(\overline{x}-1)}{2}\\\
1&1-x&-\frac{(1-x)(1-\overline{x})}{2}\end{array}\right)$

and

(9.17)
${u^{\prime}}^{-1}\gamma(x)Jv^{\prime}=\begin{pmatrix}-\frac{\varpi_{p}^{j-i}}{2}-\overline{\varpi}_{p}^{i}\varpi_{p}^{j}b&e&z\\\
-\varpi_{p}^{j}&x&f\\\
\varpi_{p}^{i+j}&\overline{\varpi}_{p}^{i}(1-x)&g\end{pmatrix},$

where

$\displaystyle
e=\frac{\varpi_{p}^{-i}(1+x)}{2}-\overline{\varpi}_{p}^{i}b(1-x),$ (9.18)
$\displaystyle
f=-\varpi_{p}^{j}b^{\prime}+\overline{\varpi}_{p}^{-j}\frac{(x+1)(\overline{x}-1)}{2},$
$\displaystyle
g=\overline{\varpi}_{p}^{i}\varpi_{p}^{j}b^{\prime}-\overline{\varpi}_{p}^{i-j}\frac{(x-1)(\overline{x}-1)}{2},$

and

$\displaystyle
z=-\frac{1}{2}\varpi_{p}^{j-i}b^{\prime}+\varpi_{p}^{-i}\overline{\varpi}_{p}^{-j}\frac{x\overline{x}+3\overline{x}-x+1}{4}-\overline{\varpi}_{p}^{i}\varpi_{p}^{j}bb^{\prime}+\overline{\varpi}_{p}^{i-j}b\frac{(x-1)(\overline{x}-1)}{2}.$
(9.19)
$\displaystyle=\frac{1-x\overline{x}}{1-x}\varpi_{p}^{-i}\overline{\varpi}_{p}^{-j}+\frac{f}{1-x}\varpi_{p}^{-i}-\frac{1-\overline{x}}{1-x}e\overline{\varpi}_{p}^{-j}+\frac{ef}{1-x}$

####  9.4.1. The non-split case $f_{p}=1_{K_{p}(N)}$

######  Proposition 9.6.

We assume here that $p$ is non-split and that $f_{p}=1_{K_{p}(N)}$. The
following hold:

  * •

If $\nu(x)<0$ we have ${\mathcal{I}}_{p}(x)=0$.

Assume that $\nu(x)\geq 0$ (ie. $x\in{\mathcal{O}}_{E,p}$)

  * •

if $p\nmid 2D_{E}N$ and $\nu(x\overline{x}-1)=0$, we have

${\mathcal{I}}_{p}(x)=1.$

  * •

if $p\mid N,$ and $\nu(x\overline{x}-1)\leq 0$ we have
$f_{p}(u^{-1}\gamma(x)Jv)=0$.

  * •

In general, we have the bound

(9.20) ${\mathcal{I}}_{p}(x)\ll
e_{p}(x)(N,p)N(\varpi_{p})^{3\nu(x\overline{x}-1)};$

here we have set

$e_{p}(x)=(1+\nu(x-1))^{2}(1+\nu(x\overline{x}-1)),$
$N(\varpi_{p})=|{\mathcal{O}}_{E_{\mathfrak{p}}}/\varpi_{p}{\mathcal{O}}_{E_{\mathfrak{p}}}|=p^{f_{p}},\
f_{p}=\begin{cases}2&\hbox{ if $p$ is inert}\\\ 1&\hbox{ if $p$ is
ramified}\end{cases}$

and the implicit constant is absolute.

###### Proof.

As noted previously $f_{p}(u^{-1}\gamma(x)Jv)$ is non zero if and only if

(9.21) ${u^{\prime}}^{-1}\gamma(x)Jv^{\prime}\in
k_{1}K_{p}(N)k_{2}^{-1}\subset G({\mathbb{Z}_{p}}).$

This implies that

(9.22) $\begin{cases}\nu(x)\geq 0,\ \nu(z)\geq 0\\\ j\geq 0\\\ i+\nu(1-x)\geq
0\\\
e,f,g,z,-\frac{\varpi_{p}^{j-i}}{2}-\overline{\varpi}_{p}^{i}\varpi_{p}^{j}b\in\mathcal{O}_{E_{p}}\\\
\end{cases}$

In particular we have $f_{p}(u^{-1}\gamma(x)Jv)=0$ unless $\nu(x)\geq 0$. The
proves the first part of Proposition 9.6.

Since $j\geq 0,$ then it follows from the equality

$\varpi_{p}^{j}e-[-\frac{1}{2}\varpi_{p}^{j-i}-\overline{\varpi}_{p}^{i}\varpi_{p}^{j}b](1-x)=\varpi_{p}^{j-i}$

that $j\geq i.$

Note that

$\overline{\varpi}_{p}^{i}f+g=\overline{\varpi}_{p}^{i-j}(\overline{x}-1).$

Hence

(9.23) $i-j+\nu(x-1)\geq\min\\{0,i\\}.$

From the condition $e,f,z\in{\mathcal{O}}_{E,p}$, the equality (9.19) and the
fact that $\nu(1-x)=\nu(1-\overline{x})$, we have

(9.24) $-i-j-\nu(x-1)+\nu(x\overline{x}-1)\geq\min\\{-i-\nu(x-1),-j\\}.$

Suppose now that $\nu(x\overline{x}-1)=0$; this implies that $\nu(x-1)=0$ and
therefore by (9.22) $i\geq 0$ and eventually $i\geq j$ by (9.23), therefore
$i=j\geq 0.$ By (9.24) we have $-2i\geq-i$ and therefore

$i=j=0$

and from (9.17) we conclude that

$b,b^{\prime}\in\mathcal{O}_{E_{p}}.$

If $p$ does not divide $2N$ we have $K_{p}(N)=G({\mathbb{Z}_{p}})$ so that if
$\nu(x\overline{x}-1)=0$ we have

$|f_{p}(u^{-1}\gamma(x)Jv)|=1$

precisely, if

$i=j=0,\ k_{1}\in SK_{p},\ k_{2}\in K_{p},\
b,b^{\prime}\in\mathcal{O}_{E_{p}}$

and otherwise it is zero. It follows that

$\displaystyle\int_{H_{\gamma(x)J}({\mathbb{Q}}_{p})\backslash(G^{\prime}\times
G^{\prime})(\mathbb{Q}_{p})}\big{|}f_{p}(u^{-1}\gamma(x)Jv)\big{|}dudv=\mu(\mathcal{O}_{E_{p}})^{2}=1.$

This proves the generic part of Proposition 9.6.

We return to the general case: given two integers $i,j$ such that $u,v$ have
Iwasawa decomposition given by (9.16) and such that
$f_{p}(u^{-1}\gamma(x)Jv)\not=0$. From the previous discussion we see that

$b=\frac{\varpi_{p}^{-i}\overline{\varpi}_{p}^{-i}(1+x)}{2(1-x)}-\frac{e}{\overline{\varpi}_{p}^{i}(1-x)}\in
B_{i},$
$b^{\prime}=\varpi_{p}^{-j}\overline{\varpi}_{p}^{-j}\frac{(x+1)(\overline{x}-1)}{2}-\varpi_{p}^{-j}f\in
B^{\prime}_{j}$

where

$B_{i}:=\frac{\varpi_{p}^{-i}\overline{\varpi}_{p}^{-i}(1+x)}{2(1-x)}-\frac{1}{\overline{\varpi}_{p}^{i}(1-x)}\mathcal{O}_{E_{p}},$
$B^{\prime}_{j}:=\varpi_{p}^{-j}\overline{\varpi}-p^{-j}\frac{(x+1)(\overline{x}-1)}{2}-\varpi_{p}^{-j}\mathcal{O}_{E_{p}}$

We have

(9.25)
$\int_{B_{i}}db\int_{B_{j}^{\prime}}db^{\prime}\ll\mathrm{Nr}(\varpi_{p})^{i+\nu(1-x)+j}.$

Suppose $i\leq 0.$ Then by (9.23) we have $j\leq\nu(x-1)$ and

$-\nu(x-1)\leq i\leq 0\leq j\leq\nu(x-1).$

Then by (9.25)

(9.26) $\mathop{\sum\sum}\limits_{-\nu(x-1)\leq i\leq 0\leq
j\leq\nu(x-1)}\int_{B_{i}}db\int_{B_{j}^{\prime}}db^{\prime}\leq(1+\nu(x-1))^{2}N(\varpi_{p})^{2\nu(x-1)}.$

Suppose $i\geq 1.$ Then similar arguments as before show that $j-\nu(x-1)\leq
i\leq j$ and $1\leq j\leq\nu(x\overline{x}-1).$ So in this range we have

(9.27)
$\mathop{\sum\sum}\limits_{i,j\cdots}\int_{B_{i}}db\int_{B_{j}^{\prime}}db^{\prime}\leq
e_{p}(x)N(\varpi_{p})^{2\nu(x\overline{x}-1)+\nu(x-1)}.$

Then (9.20) follows from (9.26) and (9.27). When $p=2,$ a similar argument
also applies with some worse (but absolute) implied constant. The proves the
last part of Proposition 9.6.

Suppose now that $p\mid N.$ Then $K_{p}(N)=K_{p}(p)\subset K_{p}$ is the
Iwahori subgroup and its intersection with $G^{\prime}(\mathbb{Z}_{p})$ is
$K_{p}^{\prime}(N)=K^{\prime}_{p}(p)$ the Iwahori subgroup of
$G^{\prime}(\mathbb{Z}_{p}).$ It follows that the function on
$G^{\prime}({\mathbb{Z}_{p}})\times G^{\prime}({\mathbb{Z}_{p}})$

$(u,v)\mapsto f_{p}(u^{-1}\gamma(x)Jv)$

is bi-$K^{\prime}_{p}(p)$ invariant.

Because of this we will evaluate the integral ${\mathcal{I}}_{p}(x)$ using the
Iwasawa decompositions of $u$ and $v$ (9.16) and in particular decompose the
two integrals over the $k_{1}\in SG^{\prime}({\mathbb{Z}_{p}})$ and $k_{2}\in
G^{\prime}({\mathbb{Z}_{p}})$ variable into a sum of $(p+1)^{2}$-integrals
supported along $SK^{\prime}_{p}(p)\times K^{\prime}_{p}(p)$-cosets using
Lemma 5.10.

Given $u,v$ such that $f_{p}(u^{-1}\gamma(x)v)\neq 0$ and whose Iwasawa
decompositions are given by (9.16) and let

$\displaystyle g:=\begin{pmatrix}1&&-b\\\ &1&\\\
&&1\end{pmatrix}\begin{pmatrix}\varpi_{p}^{-i}&\\\ &1&\\\
&&\overline{\varpi}_{p}^{i}\end{pmatrix}\gamma(x)J\begin{pmatrix}\varpi_{p}^{j}&\\\
&1&\\\
&&\overline{\varpi}_{p}^{-j}\end{pmatrix}\begin{pmatrix}1&&b^{\prime}\\\
&1&\\\ &&1\end{pmatrix},$

Since $f_{p}$ is supported on $K_{p}(p)$, for the integrand in one of these to
be non-zero one of the following hold :

$g\in K_{p}(p),\hbox{ or }J\begin{pmatrix}1&&-\delta\\\ &1&\\\
&&1\end{pmatrix}g\in K_{p}(p)\hbox{ or }g\begin{pmatrix}1&&\delta^{\prime}\\\
&1&\\\ &&1\end{pmatrix}J\in K_{p}(p)$ $\hbox{ or
}J\begin{pmatrix}1&&-\delta\\\ &1&\\\
&&1\end{pmatrix}g\begin{pmatrix}1&&\delta^{\prime}\\\ &1&\\\
&&1\end{pmatrix}J\in K_{p}(p)\hbox{ for some
}\delta,\delta^{\prime}\in{\mathbb{Z}_{p}}/p{\mathbb{Z}_{p}}.$

We will show in each case that $\nu(x\overline{x}-1)\geq 1.$

If we assume instead that $\nu(x\overline{x}-1)=0$, we have

$\nu(x-1)=\nu(\overline{x}+1)=\nu(1-\overline{x})=0.$

We will obtain a contradiction for each of the $1+2p+p^{2}$ possible locations
of $g$:

  1. (i)

Assume $g\in K_{p}(p).$ Then by (9.17) we have $i\geq 1,$ $j\geq 1$ and
$e,f,z\in\mathcal{O}_{E_{p}}.$ But this contradicts the algebraic relation
(9.19), which forces that $z\not\in\mathcal{O}_{E_{p}}$ a contradiction and
therefore $\nu(x\overline{x}-1)\geq 1$.

  2. (ii)

Suppose $J\begin{pmatrix}1&&-\delta\\\ &1&\\\ &&1\end{pmatrix}g\in K_{p}(p).$
This implies that

(9.28)
$\begin{pmatrix}-\frac{\varpi_{p}^{j-i}}{2}-\overline{\varpi}_{p}^{i}\varpi_{p}^{j}b&e&z\\\
-\varpi_{p}^{j}&x&f\\\
\varpi_{p}^{i+j}&\overline{\varpi}_{p}^{i}(1-x)&\overline{\varpi}_{p}^{i}\varpi_{p}^{j}b^{\prime}-\overline{\varpi}_{p}^{i-j}\frac{(x-1)(\overline{x}-1)}{2}\end{pmatrix}\in
JK_{p}(p).$

here we have made the change of variable

$b\mapsto b+\delta.$

Hence $i\geq 0,$ $j\geq 1,$ $\nu(e)\geq 1,$ $\nu(f)\geq 0$ and $\nu(z)\geq 0.$
However, this contradicts (9.19) as well.

Notice then, that for $\nu(x\overline{x}-1)\geq 1$, (9.28) leads to

$-\frac{\varpi_{p}^{j-i}}{2}-\overline{\varpi}_{p}^{i}\varpi_{p}^{j}b\in\varpi_{p}\mathcal{O}_{E_{p}}$

and $\varpi_{p}^{i+j}\in\mathcal{O}_{E_{p}}^{\times}.$ So there is only one
possible choice for $b\mod{\varpi}_{p}.$

  3. (iii)

Suppose that $g\begin{pmatrix}1&&\delta^{\prime}\\\ &1&\\\
&&1\end{pmatrix}J\in K_{p}(p).$ This is similar to the preceding case (ii).
Again we must have $\nu(x\overline{x}-1)\geq 1$ and there is only one choice
for $b^{\prime}\mod{\varpi}_{p}.$

  4. (iv)

Suppose finally that $J\begin{pmatrix}1&&-\delta\\\ &1&\\\
&&1\end{pmatrix}g\begin{pmatrix}1&\delta^{\prime}\\\ &1&\\\
&&1\end{pmatrix}J\in K_{p}(p).$ We then have again

$\begin{pmatrix}-\frac{\varpi_{p}^{j-i}}{2}-\overline{\varpi}_{p}^{i}\varpi_{p}^{j}b&e&z\\\
-\varpi_{p}^{j}&x&f\\\
\varpi_{p}^{i+j}&\overline{\varpi}_{p}^{i}(1-x)&\overline{\varpi}_{p}^{i}\varpi_{p}^{j}b^{\prime}-\overline{\varpi}_{p}^{i-j}\frac{(x-1)(\overline{x}-1)}{2}\end{pmatrix}\in
JK_{p}(p);$

where this time we have made the change of variables

$b\mapsto b+\delta,\ b^{\prime}\mapsto b^{\prime}+\delta^{\prime}.$

We have therefore $i\geq 0,$ $j\geq 0,$ $\nu(e)\geq 1,$ $\nu(f)\geq 1$ and
$\nu(z)\geq 1,$ which contradicts (9.19) and necessarily
$\nu(x\overline{x}-1)\geq 1$. Since $\nu(e)\geq 1$ and $\nu(f)\geq 1,$ this
implies that

$\displaystyle b\in
C_{i}:=\frac{\varpi_{p}^{-i}\overline{\varpi}_{p}^{-i}(1+x)}{2(1-x)}-\frac{1}{\overline{\varpi}_{p}^{i}(1-x)}\varpi_{p}\mathcal{O}_{E_{p}},$
$\displaystyle b^{\prime}\in
C^{\prime}_{j}:=\varpi_{p}^{-j}\overline{\varpi_{p}}-p^{-j}\frac{(x+1)(\overline{x}-1)}{2}-\varpi_{p}^{1-j}\mathcal{O}_{E_{p}}.$

Therefore,

$\int_{C_{i}}db\int_{C^{\prime}_{j}}db^{\prime}\ll
N(\varpi_{p})^{i+\nu(x-1)+j-2}.$

Notice the extra saving $p^{-2}$ (compared with (9.25)): it comes to
compensate the contribution of the $p^{2}$ choices of $\delta$ and
$\delta^{\prime}$ in ${\mathbb{Z}_{p}}/p{\mathbb{Z}_{p}}$.

In all case we have $\nu(x\overline{x}-1)\geq 1$. Integrating over these
$SK^{\prime}_{p}(p)\times K^{\prime}_{p}(p)$ cosets, we conclude that when
$p\mid N,$ one has

$\displaystyle{\mathcal{I}}_{p}(x)\ll\frac{e_{p}(x)N(\varpi_{p})^{3\nu(x\overline{x}-1)}\mu(SK_{p}^{\prime}(p))\mu(K_{p}^{\prime}(p))}{\mu(K_{p}(N))}\ll\frac{(1/p)^{2}}{1/p^{3}}e_{p}(x)N(\varpi_{p})^{3\nu(x\overline{x}-1)}.$

This completes the proof of the Proposition when $p|N$. ∎

####  9.4.2. The inert Hecke case

We now evaluate ${\mathcal{I}}_{p}(x)$ for $p$ inert in the last remaining
case.

######  Proposition 9.7.

Let $p$ be an inert prime, $r\geq 0$ and suppose that

$f_{p}=1_{G({\mathbb{Z}_{p}})A_{p^{r}}G({\mathbb{Z}_{p}})}.$

We have

${\mathcal{I}}_{p}(x)=0$

unless $\nu(x)\geq-r$ in which case we have

(9.29)
${\mathcal{I}}_{p}(x)\ll(r+|\nu(1-x\overline{x})|+|\nu(1-x)|)(p^{3r+2\nu(1-x)}+p^{7r+2\nu(1-x\overline{x})}).$

Here the implied constant is absolute.

###### Proof.

We use again the notations on §9.4. Let $u,v$ such that

$f_{p}(u^{-1}\gamma(x)Jv)=f_{p}({u^{\prime}}^{-1}\gamma(x)Jv^{\prime})\not=0,$

which, by (9.17), amounts to

(9.30)
${u^{\prime}}^{-1}\gamma(x)Jv^{\prime}=\begin{pmatrix}-\frac{p^{j-i}}{2}-p^{j+i}b&e&z\\\
-p^{j}&x&f\\\ p^{i+j}&p^{i}(1-x)&g\end{pmatrix}\in
G({\mathbb{Z}_{p}})A_{r}G({\mathbb{Z}_{p}}).$

where

$\displaystyle e=\frac{p^{-i}(1+x)}{2}-p^{i}b(1-x),$ (9.31) $\displaystyle
f=-p^{j}b^{\prime}+p^{-j}\frac{(x+1)(\overline{x}-1)}{2},$ $\displaystyle
g=p^{i+j}b^{\prime}-p^{i-j}\frac{(x-1)(\overline{x}-1)}{2}.$

and

$z=\frac{1-x\overline{x}}{1-x}p^{-i-j}+\frac{f}{1-x}p^{-i}-\frac{1-\overline{x}}{1-x}ep^{-j}+\frac{ef}{1-x}.$

Analyzing the entries in (9.30) we then derive that $z,e,f\in
p^{-r}{\mathbb{Z}_{p}}.$ Hence,

(9.32) $\begin{cases}\nu(x)\geq-r,\ \nu(g)\geq-r,\\\ j\geq-r,\
i\geq-r-\nu(1-x),\\\
f=-p^{j}b^{\prime}+p^{-j}\frac{(x+1)(\overline{x}-1)}{2}\in
p^{-r}{\mathbb{Z}_{p}},\\\ e=\frac{p^{-i}(1+x)}{2}-p^{i}b(1-x)\in
p^{-r}{\mathbb{Z}_{p}},\\\
z=\frac{1-x\overline{x}}{1-x}p^{-i-j}+\frac{f}{1-x}p^{-i}-\frac{1-\overline{x}}{1-x}ep^{-j}+\frac{ef}{1-x}\in
p^{-r}{\mathbb{Z}_{p}},\end{cases}$

where the last constraint yields that

(9.33)
$-i-j+\nu(1-x\overline{x})\geq\min\big{\\{}-r-i,-r-j+\nu(1-x),-2r,-r+\nu(1-x)\big{\\}}.$

Since $p^{i}f+g=p^{i-j}(\overline{x}-1),$ then

(9.34) $i-j+\nu(1-x)=\min\\{i+\nu(f),\nu(g)\\}\geq\min\\{i-r,-r\\}.$

Since $p^{j}e-(1-x)a=p^{j-i},$ then

(9.35) $j-i=\min\\{j+\nu(e),\nu(1-x)+\nu(a)\\}\geq\min\\{j-r,\nu(1-x)-r\\}.$

We now separate the cases to derive the ranges of $i$ and $j$ as follows.

  1. (1)

Suppose $i\leq 0.$ Then by (9.34) we obtain that $i-j+\nu(1-x)\geq i-r,$
implying that $j\leq r+\nu(1-x).$ Thus in this case we have $-r\leq i\leq 0$
and $-r\leq j\leq r+\nu(1-x).$

  2. (2)

Suppose that $i\geq 1.$ Then (9.34) gives

(9.36) $i-j+\nu(1-x)\geq-r.$

     * •

Suppose further that $j\leq\nu(1-x).$ Then it follows from (9.35) that
$j-i\geq j-r,$ i.e., $i\leq r.$ Hence, in this case we have $1\leq i\leq r$
and $-r\leq j\leq\nu(1-x).$

     * •

Suppose that $j>\nu(1-x).$ Then by (9.35) we have $j-i\geq\nu(1-x)-r$, i.e.,
$-r-i\geq-2r-j+\nu(1-x).$ Substituting this into (9.33) to obtain

$\displaystyle-i-j+\nu(1-x\overline{x})\geq$
$\displaystyle\min\big{\\{}-2r-j+\nu(1-x),-2r,-r+\nu(1-x)\big{\\}}$
$\displaystyle=$
$\displaystyle\min\big{\\{}-2r-j+\nu(1-x),-2r\big{\\}}=-2r-j+\nu(1-x).$

Here we note the fact that $-2r\leq-r+\nu(1-x)$ and $j>\nu(1-x).$ Therefore,
we have $1\leq i\leq 2r+\nu(1-x\overline{x})-\nu(1-x).$ By (9.36),

$\nu(1-x)<j\leq i+r+\nu(1-x)\leq 3r+\nu(1-x\overline{x}).$

Denote by $\mathcal{S}$ the support of $(i,j)\in\mathbb{Z}^{2}$ determined by
(9.32). From the above discussion we see that $\\#\mathcal{S}\ll
r+|\nu(1-x\overline{x})|+|\nu(1-x)|$ and for $(i,j)\in\mathcal{S},$

$i+j\leq\max\\{r+\nu(1-x),5r+2\nu(1-x\overline{x})-\nu(1-x)\\}.$

Note that (9.32) also implies that $b$ (resp. $b^{\prime}$) ranges over a
translate of $p^{-i-r-\nu(1-x)}{\mathbb{Z}_{p}}$ (resp.
$p^{-j-r}{\mathbb{Z}_{p}}$). Therefore,

${\mathcal{I}}_{p}(x)\ll\sum_{(i,j)\in\mathcal{S}}\int_{p^{-i-r-\nu(1-x)}{\mathbb{Z}_{p}}}db\int_{p^{-j-r}{\mathbb{Z}_{p}}}db^{\prime}\ll\sum_{(i,j)\in\mathcal{S}}p^{i+j}\cdot
p^{\nu(1-x)+2r}.$

Hence, (9.29) follows. ∎

###  9.5. The split case

Let $p$ be a prime split in $E$, the corresponding ideal of $K$ decomposes as
$\mathfrak{p}{\overline{\mathfrak{p}}}=(p)$ and we have
$E\otimes_{\mathbb{Q}}{\mathbb{Q}}_{p}\simeq E_{\mathfrak{p}}\times
E_{\overline{\mathfrak{p}}}\simeq{\mathbb{Q}}_{p}\times{\mathbb{Q}}_{p}$. Let
$\varpi$ in $E_{\mathfrak{p}}\simeq{\mathbb{Q}}_{p}$ be an uniformizer.

In the split case, we have
$G^{\prime}(\mathbb{Q}_{p})\simeq\operatorname{GL}(2,\mathbb{Q}_{p})$ and by
Corollary 9.2 one has

$H_{x}(\mathbb{Q}_{p})\simeq\operatorname{GL}(1,\mathbb{Q}_{p}).$

Given $u,v\in G^{\prime}(\mathbb{Q}_{p})$, we write them in Iwasawa
coordinates:

(9.37) $u=\begin{pmatrix}p^{i_{1}}&\\\
&p^{i_{2}}\end{pmatrix}\begin{pmatrix}1&b\\\ &1\end{pmatrix}k_{1},\ \
v=\begin{pmatrix}p^{j_{1}}&\\\
&p^{j_{2}}\end{pmatrix}\begin{pmatrix}1&b^{\prime}\\\ &1\end{pmatrix}k_{2},$

where $i_{1},i_{2},j_{1},j_{2}\in\mathbb{Z},$ $b,b^{\prime}\in\mathbb{Q}_{p}$
and $k_{1},k_{2}\in G^{\prime}(\mathbb{Z}_{p})$. In particular if $\nu(\det
u)=0$ we have $i_{2}=-i_{1}$.

Applying this decomposition to the variable $u$ in the integral
${\mathcal{I}}_{p}(x)$ we have

$\displaystyle{\mathcal{I}}_{p}(x)=\sum_{i\in\mathbb{Z}}\int_{SG^{\prime}({\mathbb{Z}_{p}})}\int_{\mathbb{Q}_{p}}\int_{G^{\prime}(\mathbb{Q}_{p})}\Big{|}f^{\mathfrak{n}}_{p}\left(k_{1}^{-1}\begin{pmatrix}p^{-i}&&-p^{i}b\\\
&1&\\\ &&p^{i}\end{pmatrix}\gamma(x)Jv\right)\Big{|}dbdk_{1}dv.$

######  Proposition 9.8.

Let $p$ be a prime, split in $E$ and $x\in E^{\times}\\!-E^{1}.$

  1. (1)

If $p\nmid N^{\prime}$ we have

${\mathcal{I}}_{p}(x)=0$

unless $\nu(x),\ \nu(\overline{x})\geq 0$; in that case we have the bound

(9.38)
${\mathcal{I}}_{p}(x)\ll(1+\nu(P(x,\overline{x}))p^{3\nu(x\overline{x}-1)},$

where $P(X,Y)\in{\mathbb{Z}}[X,Y]$ is a polynomial independent of $p$ whose
degree and coefficients are absolutely bounded; moreover the implicit constant
is absolute.

– In addition, if $p\neq 2$ and

$\nu(x\overline{x}-1)=\nu(1-x)=\nu(1-\overline{x})=0$

we have ${\mathcal{I}}_{p}(x)=1.$

  2. (2)

If $p|N^{\prime}$, we have

(9.39) ${\mathcal{I}}_{p}(x)=0$

unless $x\in p^{-1}\mathcal{O}_{E_{p}},$ i.e., $\nu(x)\geq-1$ and
$\nu(\overline{x})\geq-1.$

– Moreover if

$\nu(x),\nu(\overline{x})\geq 0,$

one has

(9.40) ${\mathcal{I}}_{p}(x)\ll(1+\nu(P(x,\overline{x})))^{2}\\\
\big{(}p^{\nu(x\overline{x}(1-x)^{2}(1-\overline{x})^{2})}+p^{\nu(1-x\overline{x})-1}+p^{2\nu(1-x\overline{x})-2}+p^{2\nu(1-x\overline{x})-1}\big{)}$

where $P(X,Y)\in{\mathbb{Z}}[X,Y]$ is a polynomial independent of $p$ whose
degree and coefficients are absolutely bounded; moreover the implicit constant
is absolute.

###### Proof.

We start with the case $p\nmid N^{\prime}$. We have
$f_{p}^{\mathfrak{n}}=f_{p}$ and

$f_{p}(u^{-1}\gamma(x)Jv)=0$

if and only if $u^{-1}\gamma(x)Jv\in G({\mathbb{Z}_{p}})$. Since the
determinant of $v$ has valuation $0$, its Iwasawa coordinates (9.37) satisfy
$j_{1}+j_{2}=0$ . Hence ${\mathcal{I}}_{p}(x)$ is bounded from above by

$\displaystyle\sum_{i\in\mathbb{Z}}\sum_{j\in\mathbb{Z}}\int_{\mathbb{Q}_{p}}\int_{\mathbb{Q}_{p}}|f_{p}\left(\begin{pmatrix}p^{i}&&p^{i}b\\\
&1&\\\
&&p^{-i}\end{pmatrix}^{-1}\gamma(x)J\begin{pmatrix}p^{j}&&p^{j}b^{\prime}\\\
&1&\\\ &&p^{-j}\end{pmatrix}\right)|dbdb^{\prime}.$

The proof then follow the same lines as for the nonsplit case considered in
Proposition 9.6. We also note that at a split place $p$, we always have
$G^{\prime}(\mathbb{Z}_{p})\subset G(\mathbb{Z}_{p})$ if $p\nmid N^{\prime}$.

We now consider the case of a split prime $p\mid N^{\prime}.$ Let
$b,b^{\prime}\in\mathbb{Q}_{p}$ and $i,j\in\mathbb{Z}.$ We have

(9.41) $\begin{pmatrix}p^{i}&&p^{i}b\\\ &1&\\\
&&p^{-i}\end{pmatrix}^{-1}\gamma(x)J\begin{pmatrix}p^{j}&&p^{j}b^{\prime}\\\
&1&\\\ &&p^{-j}\end{pmatrix}=\begin{pmatrix}a&e&z\\\ -p^{j}&x&f\\\
p^{i+j}&p^{i}(1-x)&g\end{pmatrix},$

where

(9.42) $\begin{cases}a=-\frac{1}{2}p^{j-i}-p^{i+j}b\\\
e=\frac{p^{-i}(1+x)}{2}-p^{i}b(1-x)\\\
f=-p^{j}b^{\prime}+p^{-j}.\frac{(x+1)(\overline{x}-1)}{2}\\\
g=p^{i+j}b^{\prime}-\frac{(x-1)(\overline{x}-1)}{2}p^{i-j}\\\
z=-\frac{1}{2}p^{j-i}b^{\prime}+p^{-i-j}y-p^{i+j}bb^{\prime}+p^{i-j}b\frac{(x-1)(\overline{x}-1)}{2}.\end{cases}$

where

$y=\frac{x\overline{x}+3\overline{x}-x+1}{4}.$

Then one has an explicit algebraic relation

(9.43)
$z=\frac{1-x\overline{x}}{1-x}p^{-i-j}+\frac{f}{1-x}p^{-i}-\frac{1-\overline{x}}{1-x}ep^{-j}-\frac{ef}{1-x}.$

Taking inverse of (9.41) to obtain

(9.44) $\begin{pmatrix}p^{j}&&p^{j}b^{\prime}\\\ &1&\\\
&&p^{-j}\end{pmatrix}^{-1}J\gamma(x)^{-1}\begin{pmatrix}p^{i}&&p^{i}b\\\
&1&\\\
&&p^{-i}\end{pmatrix}=\begin{pmatrix}a^{\prime}&e^{\prime}&z^{\prime}\\\
p^{i}(1-\overline{x})&\overline{x}&f^{\prime}\\\
p^{i+j}&-p^{j}&g^{\prime}\end{pmatrix},$

where

(9.45)
$\begin{cases}a^{\prime}=-\frac{(1-x)(1-\overline{x})}{2}p^{i-j}-p^{i+j}b^{\prime}\\\
e^{\prime}=\frac{(x-1)(\overline{x}+1)}{2}p^{-j}+p^{j}b^{\prime}\\\
f^{\prime}=(1-\overline{x})p^{i}b+\frac{1+\overline{x}}{2}p^{-i}\\\
g^{\prime}=p^{i+j}b-\frac{1}{2}p^{j-i}\\\
z^{\prime}=-p^{i-j}b\frac{(x-1)(\overline{x}-1)}{2}+p^{-i-j}\overline{y}-p^{i+j}bb^{\prime}+\frac{1}{2}p^{j-i}b^{\prime}\end{cases}$

and one notes the algebraic relation

(9.46)
$z^{\prime}=\frac{1-x\overline{x}}{1-\overline{x}}p^{-i-j}+\frac{e^{\prime}}{1-\overline{x}}p^{-i}-\frac{1-x}{1-\overline{x}}f^{\prime}p^{-j}-\frac{e^{\prime}f^{\prime}}{1-\overline{x}}.$

We also note that

(9.47) $\displaystyle a+g^{\prime}=-p^{j-i},\
a^{\prime}+g=-{(1-x)(1-\overline{x})}p^{i-j},$ $\displaystyle
p^{i}.f+g=p^{i-j}(\overline{x}-1),\
p^{j}f^{\prime}-(1-\overline{x})g^{\prime}=p^{j-i}$

By definition (9.13), we have

$\displaystyle{\mathcal{I}}_{p}(x)=\int_{SG^{\prime}(\mathbb{Q}_{p})\times
G^{\prime}(\mathbb{Q}_{p})}\big{|}f_{p}(\widetilde{\mathfrak{n}}_{p}^{-1}u^{-1}\gamma(x)Jv\widetilde{\mathfrak{n}}_{p})\big{|}dudv.$

where we recall that $f_{p}=\textbf{1}_{K_{p}}/\mu(K_{p}),\
K_{p}=G({\mathbb{Z}_{p}})$ and

$\widetilde{\mathfrak{n}}_{p}=w^{\prime}\mathfrak{n}_{p}w^{\prime}=\begin{pmatrix}1&p^{-1}&\\\
&1&\\\ &&1\end{pmatrix}$

We will apply the Iwasawa decomposition as in the beginning of the proof of
Proposition 9.8.

Due to the conjugation by $\widetilde{\mathfrak{n}}_{p},$ the function

$(u,v)\mapsto
f_{p}(\widetilde{\mathfrak{n}}_{p}^{-1}u^{-1}\gamma(x)Jv\widetilde{\mathfrak{n}}_{p})$

is bi $I^{\prime}_{p}(1)$-invariant (since
$\widetilde{\mathfrak{n}}_{p}I^{\prime}_{p}(1)\widetilde{\mathfrak{n}}_{p}^{-1}\subset
K_{p}$) so we shall further decompose $G^{\prime}(\mathbb{Z}_{p})$ into a
union of disjoint cosets of $I_{p}^{\prime}(1)$: we start with the Iwahori
decomposition for $G^{\prime}$ (see (5.95))

$G^{\prime}({\mathbb{Z}_{p}})=I_{p}^{\prime}\sqcup
I_{p}^{\prime}JI_{p}^{\prime}=I_{p}^{\prime}\sqcup\bigsqcup_{\delta\in({\mathbb{Z}}/p{\mathbb{Z}})^{\times}}I_{p}^{\prime}J\begin{pmatrix}\delta&&\\\
&1&\\\ &&1\end{pmatrix}I_{p}^{\prime}(1)$

We then check four cases as in Lemma 9.6.

### Case I

We start with the most complicated case:

$\displaystyle u=$ $\displaystyle\begin{pmatrix}p^{i}&&p^{i}b\\\ &1&\\\
&&p^{-i}\end{pmatrix}\begin{pmatrix}1&&\mu\\\ &1&\\\
&&1\end{pmatrix}J\begin{pmatrix}\tau^{-1}&&\\\ &1&\\\ &&1\end{pmatrix}k_{1},$
$\displaystyle v=$ $\displaystyle\begin{pmatrix}p^{j}&&p^{j}b^{\prime}\\\
&1&\\\ &&p^{-j}\end{pmatrix}\begin{pmatrix}1&&\mu^{\prime}\\\ &1&\\\
&&1\end{pmatrix}J\begin{pmatrix}\delta&&\\\ &1&\\\ &&1\end{pmatrix}k_{2},$

with $\mu,\mu^{\prime}\in{\mathbb{Z}_{p}}$ run over representatives of
${\mathbb{Z}_{p}}/p{\mathbb{Z}_{p}}$ and
$\tau,\delta\in{\mathbb{Z}^{\times}_{p}}$ run over representatives of
$({\mathbb{Z}_{p}}/p{\mathbb{Z}_{p}})^{\times},$ and $k_{1},k_{2}\in
I_{p}^{\prime}(1).$

Let ${\mathcal{I}}_{p}(x;1)$ be the contribution to ${\mathcal{I}}_{p}(x)$ of
all the $u,v$ whose Iwasawa decomposition is of the above form. We first look
for some necessary condition for ${\mathcal{I}}_{p}(x;1)$ to be non zero.

For $\delta\in{\mathbb{Z}^{\times}_{p}},$ we denote by

$\mathfrak{u}_{\delta}=\begin{pmatrix}1&&\\\ &1&\\\ &\delta
p^{-1}&1\end{pmatrix}=J\begin{pmatrix}\delta&&\\\ &1&\\\
&&1\end{pmatrix}\widetilde{\mathfrak{n}}_{p}J.$

Then, since $J\in K_{p}$,
$f_{p}(\widetilde{\mathfrak{n}}_{p}^{-1}u^{-1}\gamma(x)Jv\widetilde{\mathfrak{n}}_{p})\neq
0$ if and only if

(9.48) $\mathfrak{u}_{\tau}^{-1}\begin{pmatrix}p^{i}&&p^{i}(b+\mu)\\\ &1&\\\
&&p^{-i}\end{pmatrix}^{-1}\gamma(x)J\begin{pmatrix}p^{j}&&p^{j}(b^{\prime}+\mu^{\prime})\\\
&1&\\\ &&p^{-j}\end{pmatrix}\mathfrak{u}_{\delta}\in K_{p}$

or equivalently

(9.49) $\begin{pmatrix}a&e+\frac{\delta}{p}z&z\\\
-p^{j}&x+\frac{\delta}{p}f&f\\\ p^{i+j}+\tau
p^{j-1}&p^{i}(1-x)+\frac{\delta}{p}g-\frac{\tau}{p}(x+\delta p^{-1}f)&g-\tau
p^{-1}f\end{pmatrix}\in K_{p},$

and taking the inverse of (9.49) we also obtain the condition

(9.50)
$\begin{pmatrix}a^{\prime}&e^{\prime}+\frac{\tau}{p}z^{\prime}&z^{\prime}\\\
p^{i}(1-\overline{x})&\overline{x}+\frac{\tau}{p}f^{\prime}&f^{\prime}\\\
p^{i+j}-\delta
p^{i-1}(1-\overline{x})&-p^{j}+\frac{\tau}{p}g^{\prime}-\frac{\delta}{p}(\overline{x}+\frac{\tau}{p}f^{\prime})&g^{\prime}-\frac{\delta}{p}f^{\prime}\end{pmatrix}\in
K_{p}.$

Here $a,e,f,g,z$ are defined in (9.42) and
$a^{\prime},e^{\prime},f^{\prime},g^{\prime},z^{\prime}$ as defined in (9.45).

These conditions imply that

$\nu(f),\ \nu(f^{\prime}),\ \nu(x+\delta p^{-1}f),\ \nu(\overline{x}+\tau
p^{-1}f^{\prime})\geq 0$

which in turn imply that

$\nu(x),\ \nu(\overline{x})\geq-1.$

This proves Proposition 9.8 in Case I.

Note also that (9.49) and (9.50) imply that

(9.51) $\displaystyle i^{\prime}:=i+\nu(1-\overline{x})\geq 0,\ j\geq 0,$
$\displaystyle\nu(z),\ \nu(z^{\prime})\geq 0,\ \nu(e),\
\nu(e^{\prime})\geq-1.$

We have

(9.52)
${\mathcal{I}}_{p}(x;1)\leq\mu(I_{p}^{\prime}(1))^{2}p^{2}\sum_{i\geq-\nu(1-\overline{x})}\sum_{j\geq
0}\int_{\mathbb{Q}_{p}}\int_{\mathbb{Q}_{p}}\sum_{\tau}\sum_{\delta}\textbf{1}_{K_{p}}(\cdots)dbdb^{\prime},$

where the “$\cdots$” in the parenthesis indicates the left hand side of (9.49)
with

$\mu=\mu^{\prime}=0.$

Indeed, up to changing variables in the $b$ and $b^{\prime}$ integrals, we may
assume this is the case.

Define

$\displaystyle{\mathcal{I}}_{p}^{j=0}(x;1):=\mu(I_{p}^{\prime}(1))^{2}p^{2}\sum_{i\geq-\nu(1-\overline{x})}\sum_{j=0}\int_{\mathbb{Q}_{p}}\int_{\mathbb{Q}_{p}}\sum_{\tau}\sum_{\delta}\textbf{1}_{K_{p}}(\cdots)dbdb^{\prime},$
$\displaystyle{\mathcal{I}}_{p}^{j>0}(x;1):=\mu(I_{p}^{\prime}(1))^{2}p^{2}\sum_{i\geq-\nu(1-\overline{x})}\sum_{j>0}\int_{\mathbb{Q}_{p}}\int_{\mathbb{Q}_{p}}\sum_{\tau}\sum_{\delta}\textbf{1}_{K_{p}}(\cdots)dbdb^{\prime}.$

We now prove Proposition 9.8 in case I and when assuming that

$\nu(x),\ \nu(\overline{x})\geq 0$

and therefore

$\nu(1\pm x),\ \nu(1\pm\overline{x})\geq 0.$

By (9.49) and (9.50), we have

(9.53) $\displaystyle\nu(f)\geq 1,\ \nu(f^{\prime})\geq 1$ $\displaystyle
g=p^{i+j}b^{\prime}-\frac{(x-1)(\overline{x}-1)}{2}p^{i-j}\in{\mathbb{Z}_{p}},\
g^{\prime}=-\frac{p^{j-i}}{2}-p^{i+j}b\in\mathbb{Z}_{p}.$

Also since (see (9.47))

$p^{j-i}=a+g^{\prime}\in{\mathbb{Z}_{p}}\hbox{ and
}-(x-1)(\overline{x}-1)p^{i-j}=a^{\prime}+g\in{\mathbb{Z}_{p}}$

we have

(9.54) $0\leq j-i\leq\ \nu(1-x)+\nu(1-\overline{x}).$

From $\nu(e^{\prime})\geq-1$ and $\nu(f)\geq 1$ we then have

(9.55) $j\leq 1+\nu(x(\overline{x}-1)).$

and likewise

$j\leq 1+\nu(\overline{x}({x}-1))$

so that

(9.56) $j\leq 1+\frac{\nu(x\overline{x}({x}-1)(\overline{x}-1))}{2}.$

#### Localization of $b,b^{\prime}$

Another general observation in Case I is that, since

(9.57) $f=-p^{j}b^{\prime}+p^{-j}.\frac{(x+1)(\overline{x}-1)}{2}\in
p{\mathbb{Z}_{p}},\
f^{\prime}=(1-\overline{x})p^{i}b+\frac{1+\overline{x}}{2}p^{-i}\in
p{\mathbb{Z}_{p}},$

$b$ and $b^{\prime}$ are contained, respectively, in translates of
$p^{-i-\nu(1-\overline{x})+1}{\mathbb{Z}_{p}}$ and $p^{-j+1}{\mathbb{Z}_{p}}$
(depending only on $x$) whose volumes are $p^{i+\nu(1-\overline{x})-1}$ and
$p^{j-1}$.

Similarly since

(9.58)
$g=p^{i+j}b^{\prime}-\frac{(x-1)(\overline{x}-1)}{2}p^{i-j}\in{\mathbb{Z}_{p}},\
g^{\prime}=-\frac{p^{j-i}}{2}-p^{i+j}b\in\mathbb{Z}_{p},$

$b$ and $b^{\prime}$ are contained in translates of
$p^{-(i+j)}{\mathbb{Z}_{p}}$ (depending only on $x$) whose volumes are
$p^{i+j}$.

We will use either of these informations depending on the values of

$\min(i+\nu(1-\overline{x})-1,i+j)\hbox{ and }\min(j-1,i+j).$

For this we need to split the discussion into further cases:

#### The case $j=0$

Since $p^{i}+\tau p^{-1}\in\mathbb{Z}_{p}$ we have $i=-1$ and
$\tau\equiv-1\pmod{p}$. Since $i+j=-1$ we have that $b$ and $b^{\prime}$ each
belong to some translates of $p\mathbb{Z}_{p}$. Hence we have

(9.59)
${\mathcal{I}}_{p}^{0}(x;1)\leq\frac{\mu(I_{p}^{\prime}(1))^{2}p^{2}(p-1)}{\mu(K_{p})}\int_{p\mathbb{Z}_{p}}\int_{p\mathbb{Z}_{p}}dbdb^{\prime}\leq\frac{1}{p^{2}(p-1)}.$

#### The case $j\geq 1$: $\nu(x)=\nu(\overline{x}-1)=0$

Suppose first that $\nu(x)=\nu(\overline{x}-1)=0.$ From (9.51), (9.55) and
(9.54), we have

$0\leq i\leq j=1.$

Finally, since

$p^{i+1}-\delta p^{i-1}(1-\overline{x})\in{\mathbb{Z}_{p}}$

we conclude that

$i=j=1.$

In particular by (9.57), $b$ and $b^{\prime}$ are contained in single
translates of $\mathbb{Z}_{p}$.

We need to split the discussion into two further cases:

– Suppose that, given $x,e,f,z$, there exists at most one $\delta\
(\mathrm{mod}\ p)$ satisfying (9.49) and (9.50) we then have (under the
assumption that $\nu(x)=\nu(\overline{x}-1)=0$)

(9.60)
${\mathcal{I}}_{p}^{>0}(x;1)\leq\frac{\mu(I_{p}^{\prime}(1))^{2}(p-1)}{\mu(K_{p})}\int_{\mathbb{Z}_{p}}\int_{\mathbb{Z}_{p}}dbdb^{\prime}\leq\frac{1}{p^{2}(p-1)}.$

– Suppose instead that, given $x,e,f,z$, there exists
$\delta_{1}\not\equiv\delta_{2}\ (\mathrm{mod}\ p)$ satisfying (9.49) and
(9.50); we then have $(\delta_{1}-\delta_{2})p^{-1}z\in\mathbb{Z}_{p}$; this
implies that $\nu(e)\geq 0$ and

$(1-x\overline{x})p^{-2}+{f}p^{-1}-(1-\overline{x})ep^{-1}-{ef}\in(1-x)p{\mathbb{Z}_{p}}.$

Since the last three terms belong to $p^{-1}{\mathbb{Z}_{p}}$ we must have

$\nu(x\overline{x}-1)\geq 1.$

In all cases, we conclude that

(9.61)
${\mathcal{I}}_{p}^{>0}(x;1)\leq\frac{\mu(I_{p}^{\prime}(1))^{2}(p-1)^{2}}{\mu(K_{p})}\int_{\mathbb{Z}_{p}}\int_{\mathbb{Z}_{p}}dbdb^{\prime}\leq\frac{1}{p^{2}}\leq\frac{p^{\nu(x\overline{x}-1)}}{p^{2}(p-1)}.$

#### The case $j\geq 1$: $\nu(x)+\nu(\overline{x}-1)\geq 1$

Suppose now that $\nu(\overline{x}-1)+\nu(x)\geq 1.$ From (9.51), (9.54) and
(9.56) we find that ${\mathcal{I}}_{p}^{>0}(x;1)$ is bounded by

$\displaystyle\frac{\mu(I_{p}^{\prime}(1))^{2}p^{2}(p-1)^{2}}{\mu(K_{p})}\sum_{j=1}^{\frac{\nu(x\overline{x}(x-1)(\overline{x}-1))}{2}+1}\sum_{i=-\nu({x}-1)}^{j}\int_{p^{-(i+\nu(1-\overline{x})-1)}{\mathbb{Z}_{p}}}\int_{p^{-(j-1)}{\mathbb{Z}_{p}}}dbdb^{\prime}$
$\displaystyle\ll\sum_{j=1}^{\frac{\nu(x\overline{x}(x-1)(\overline{x}-1))}{2}+1}\sum_{i=-\nu({x}-1)}^{j}p^{i+j+\nu(1-\overline{x})-2}$
$\displaystyle\ll(1+\nu(x\overline{x}(x-1)(\overline{x}-1)))^{2}p^{\nu(x\overline{x}(x-1)(\overline{x}-1))+\nu(1-\overline{x})}$

Since $\nu(1-x)\geq 0$ we can replace the exponent in $p$ above by the more
symmetric expression

$\nu(x\overline{x})+2\nu((x-1)(\overline{x}-1)),$

so that combining this with (9.59), (9.60), (9.61)

(9.62)
${\mathcal{I}}_{p}(x;1)\ll(1+\nu(x\overline{x}(x-1)(\overline{x}-1)))^{2}p^{\nu(x\overline{x})+2\nu((x-1)(\overline{x}-1))}+\frac{p^{\nu(1-x\overline{x})}}{p^{3}}.$

### Case II

We assume now that the Iwasawa decomposition of $u$ and $v$ are of the form

$\displaystyle u=\begin{pmatrix}p^{i}&&p^{i}b\\\ &1&\\\
&&p^{-i}\end{pmatrix}\begin{pmatrix}\tau^{-1}&&\\\ &1&\\\
&&1\end{pmatrix}k_{1},\ v=\begin{pmatrix}p^{j}&&p^{j}b^{\prime}\\\ &1&\\\
&&p^{-j}\end{pmatrix}\begin{pmatrix}\delta&&\\\ &1&\\\ &&1\end{pmatrix}k_{2},$

where $\tau,\delta\in(\mathbb{Z}/p\mathbb{Z})^{\times},$ and $k_{1},k_{2}\in
I_{p}^{\prime}(1).$

Let ${\mathcal{I}}_{p}(x;2)$ be the contribution of $u,v$ of the above forms
to $\mathcal{I}_{p}(x).$ We first look for some necessary condition for
${\mathcal{I}}_{p}(x;2)$ to be non zero.

Computing $u^{-1}\gamma(x)Jv$ we see that
$f_{p}^{\mathfrak{n}_{p}}(u^{-1}\gamma(x)Jv)\neq 0$ if and only if

(9.63) $\begin{pmatrix}a+\tau p^{j-1}&e+\delta p^{-1}a-\tau p^{-1}(x-\delta
p^{j-1})&z-\tau p^{-1}f\\\ -p^{j}&x-\delta p^{j-1}&f\\\
p^{i+j}&p^{i}(1-x)+\delta p^{i+j-1}&g\end{pmatrix}\in K_{p}.$

Taking inverse, we obtain

(9.64) $\begin{pmatrix}a^{\prime}-\delta
p^{i-1}(1-\overline{x})&e^{\prime}+\frac{\tau}{p}a^{\prime}-\frac{\delta}{p}(\overline{x}+\tau
p^{i-1}(1-\overline{x}))&z^{\prime}-\frac{\delta}{p}f^{\prime}\\\
p^{i}(1-\overline{x})&\overline{x}+\tau p^{i-1}(1-\overline{x})&f^{\prime}\\\
p^{i+j}&-p^{j}+\tau p^{i+j-1}&g^{\prime}\end{pmatrix}\in K_{p}.$

Here, as in Case I, $a,e,f,g,z$ are defined in (9.42) and
$a^{\prime},e^{\prime},f^{\prime},g^{\prime},z^{\prime}$ as defined in (9.45).

These conditions imply in particular that

(9.65) $i+\nu(1-\overline{x})\geq 0,\ j\geq 0,\ i+j\geq 0$

which together with $\nu(x-\delta p^{j-1}),\ \nu(\overline{x}-\tau
p^{i}(1-\overline{x})p^{-1})\geq 0$ implies that

$\nu(x),\ \nu(\overline{x})\geq-1.$

This proves Proposition 9.8 in Case II.

We also note that since

$f,f^{\prime},g,g^{\prime},\ z-\tau p^{-1}f,\ z^{\prime}-\delta
p^{-1}f^{\prime}\in{\mathbb{Z}_{p}}$

we have

$\nu(z),\nu(z^{\prime})\geq-1$

and

(9.66)
$\frac{1-x\overline{x}}{1-x}p^{-i-j}+\frac{f}{1-x}p^{-i}-\frac{1-\overline{x}}{1-x}ep^{-j}-\frac{ef}{1-x}=z\in
p^{-1}{\mathbb{Z}_{p}}.$

We considered Case II when $p|N^{\prime}$ and assume that

$\nu(x),\ \nu(\overline{x})\geq 0.$

In particular $\nu(1\pm x),\ \nu(1\pm\overline{x})\geq 0.$

A few general remarks:

  * •

Since $\nu(x)\geq 0$ and $\nu(x-\delta p^{j-1})\geq 0$ we have $j\geq 1$ and
therefore since $a+\tau p^{j-1}\in{\mathbb{Z}_{p}}$ we have
$a\in{\mathbb{Z}_{p}}$.

  * •

This together with $e+\delta p^{-1}a-\tau p^{-1}(x-\delta
p^{j-1})\in{\mathbb{Z}_{p}}$ implies that

$\nu(e)\geq-1.$

  * •

Since $f,f^{\prime},g,g^{\prime}\in{\mathbb{Z}_{p}}$, we see that
$b,b^{\prime}$ are contained in translates of respectively

$p^{-\min(i+\nu(1-\overline{x}),i+j)}{\mathbb{Z}_{p}}\hbox{ and
}p^{-\min(j,i+j)}{\mathbb{Z}_{p}}$

which have volumes

$p^{\min(i+\nu(1-\overline{x}),i+j)}{\mathbb{Z}_{p}}\hbox{ and
}p^{\min(j,i+j)}{\mathbb{Z}_{p}}.$

#### The case $\nu(\overline{x}-1)=0$

Since $\overline{x}+\tau p^{i-1}(1-\overline{x})\in{\mathbb{Z}_{p}}$ we have
$i\geq 1$ and since

(9.67) $p^{i}.f+p^{i-j}(1-\overline{x})=-g\in{\mathbb{Z}_{p}},\
p^{j}f^{\prime}-(1-\overline{x})g^{\prime}=p^{j-i}\in{\mathbb{Z}_{p}}$

we have

$j=i\geq 1.$

Since

$x-\delta p^{j-1},a,e+\delta p^{-1}a-\tau p^{-1}(x-\delta
p^{j-1})\in{\mathbb{Z}_{p}}$

we have $\nu(e)\geq-1$

We now look at the variable $z$: under our current assumptions (9.66) becomes

$\frac{1-x\overline{x}}{1-x}p^{-2i}+\frac{f}{1-x}p^{-i}-\frac{1-\overline{x}}{1-x}ep^{-i}-\frac{ef}{1-x}=z\in
p^{-1}{\mathbb{Z}_{p}}.$

The valuation of the first term is $\geq$ of the minimum of the valuations of
the three other terms and of $z$:

$\nu(1-x\overline{x})-2i\geq\min(-i+\nu(f),\nu(1-\overline{x})+\nu(e)-i,\nu(e)+\nu(f),\nu(z)+\nu(1-x))$

which yields (since $\nu(1-\overline{x}),\nu(f)\geq 0$ and $\nu(e),\
\nu(z)\geq-1$)

$\nu(1-x\overline{x})-2i\geq-i-1$

so that

$1\leq i=j\leq\nu(1-x\overline{x})+1.$

Notice also that if $\nu(f)=0$ or $\nu(f^{\prime})=0$ (which is the generic
case) the relations

$z-\tau p^{-1}f\in\mathbb{Z}_{p},\ z^{\prime}-\delta p^{-1}f^{\prime}$

uniquely determine $\delta\,(\mathrm{mod}\,{p})$ and
$\tau\,(\mathrm{mod}\,{p})$. On the other hand, if either $f\in
p{\mathbb{Z}_{p}}$ or $f^{\prime}\in p{\mathbb{Z}_{p}}$, then $b^{\prime}$ or
$b$ belong to fixed translates of $p^{-i+1}{\mathbb{Z}_{p}}$ (whose volume is
smaller by a factor $p$): Therefore, we have

$\displaystyle{\mathcal{I}}_{p}(x;2)\ll$ $\displaystyle\
\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\sum_{1\leq i\leq
1+\nu(1-x\overline{x})}\iint_{b\in p^{-i+1}{\mathbb{Z}_{p}},b^{\prime}\in
p^{-i+1}{\mathbb{Z}_{p}}}\mathop{\sum\sum}\limits_{\delta,\tau\,(\mathrm{mod}\,{p})}dbdb^{\prime}$
$\displaystyle\ +2\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\sum_{1\leq
i\leq 1+\nu(1-x\overline{x})}\iint_{b\in p^{-i}{\mathbb{Z}_{p}},b^{\prime}\in
p^{-i+1}{\mathbb{Z}_{p}}}\sum_{\tau\,(\mathrm{mod}\,{p})}dbdb^{\prime}$
$\displaystyle+\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\sum_{1\leq i\leq
1+\nu(1-x\overline{x})}\iint_{b\in p^{-i}{\mathbb{Z}_{p}},b^{\prime}\in
p^{-i}{\mathbb{Z}_{p}}}dbdb^{\prime}$ (9.68) $\displaystyle\ll$
$\displaystyle\frac{1}{p^{4}}(1+\nu(1-x\overline{x}))p^{2\nu(1-x\overline{x})+2}\ll(1+\nu(1-x\overline{x}))\frac{p^{2\nu(1-x\overline{x})}}{p^{2}}$

if $\nu(1-\overline{x})=0$.

#### The case $\nu(1-\overline{x})\geq 1$

Since $\overline{x}+\tau p^{i-1}(1-\overline{x})\in{\mathbb{Z}_{p}}$, $j\geq
1$, $f^{\prime},g^{\prime}\in{\mathbb{Z}_{p}}$ and
$p^{j-i}=p^{j}f^{\prime}-(1-\overline{x})g^{\prime}$ we have

$i\geq 1-\nu(1-\overline{x}),\ j\geq i+1.$

By (9.66) we have (since $\nu(f)\geq 0$, $\nu(e),\nu(z),\nu(1-x)\geq 0$)

$\nu(1-x\overline{x})-i-j\geq\min(-i,-j+\nu(1-\overline{x})-1,-1)$

or equivalently

(9.69) $i+j\leq\nu(1-x\overline{x})+\max(i,j+1-\nu(1-\overline{x}),1).$

Also since $p^{i}.f+p^{i-j}(1-\overline{x})=-g\in{\mathbb{Z}_{p}}$ we have
$i-j+\nu(1-\overline{x})\geq\min(i,0)$ or equivalently

(9.70) $j\leq\max(i,0)+\nu(1-\overline{x}).$

If $i\leq 0,$ this gives $j\leq\nu(1-\overline{x})$ and by (9.69)

(9.71) $i+j\leq\nu(1-x\overline{x})+1.$

We have therefore

$1-\nu(1-\overline{x})\leq i\leq 0,\ 1\leq j\leq\nu(1-x\overline{x})+1-i.$

The contribution of this configuration is bounded by

$\displaystyle\ll$ $\displaystyle\
\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\mathop{\sum\sum}\limits_{\stackrel{{\scriptstyle{\scriptstyle
1-\nu(1-\overline{x})\leq i\leq 0}}}{{{\scriptstyle 1\leq
j\leq\nu(1-x\overline{x})+1-i}}}}\iint_{\stackrel{{\scriptstyle{\scriptstyle
b\in p^{-i-\nu(1-\overline{x})+1}{\mathbb{Z}_{p}}}}}{{{\scriptstyle
b^{\prime}\in
p^{-j+1}{\mathbb{Z}_{p}}}}}}\mathop{\sum\sum}\limits_{\delta,\tau\,(\mathrm{mod}\,{p})}1dbdb^{\prime}$
$\displaystyle\
+\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\mathop{\sum\sum}\limits_{\stackrel{{\scriptstyle{\scriptstyle
1-\nu(1-\overline{x})\leq i\leq 0}}}{{{\scriptstyle 1\leq
j\leq\nu(1-x\overline{x})+1-i}}}}\iint_{\stackrel{{\scriptstyle{\scriptstyle
b\in p^{-i-\nu(1-\overline{x})+1}{\mathbb{Z}_{p}}}}}{{{\scriptstyle
b^{\prime}\in
p^{-j}{\mathbb{Z}_{p}}}}}}\sum_{\tau\,(\mathrm{mod}\,{p})}1dbdb^{\prime}$
$\displaystyle\
+\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\mathop{\sum\sum}\limits_{\stackrel{{\scriptstyle{\scriptstyle
1-\nu(1-\overline{x})\leq i\leq 0}}}{{{\scriptstyle 1\leq
j\leq\nu(1-x\overline{x})+1-i}}}}\iint_{\stackrel{{\scriptstyle{\scriptstyle
b\in p^{-i-\nu(1-\overline{x})}{\mathbb{Z}_{p}}}}}{{{\scriptstyle
b^{\prime}\in
p^{-j+1}{\mathbb{Z}_{p}}}}}}\sum_{\delta\,(\mathrm{mod}\,{p})}1dbdb^{\prime}$
$\displaystyle\
+\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\mathop{\sum\sum}\limits_{\stackrel{{\scriptstyle{\scriptstyle
1-\nu(1-\overline{x})\leq i\leq 0}}}{{{\scriptstyle 1\leq
j\leq\nu(1-x\overline{x})+1-i}}}}\iint_{\stackrel{{\scriptstyle{\scriptstyle
b\in p^{-i-\nu(1-\overline{x})}{\mathbb{Z}_{p}}}}}{{{\scriptstyle
b^{\prime}\in p^{-j}{\mathbb{Z}_{p}}}}}}dbdb^{\prime}$

and using (9.71) this is bounded by

(9.72)
$\ll(1+\nu(1-x\overline{x}))(1+\nu(1-\overline{x}))\frac{p^{\nu(1-x\overline{x})+\nu(1-\overline{x})}}{p^{3}}.$

If $i\geq 1$ then by (9.70) we have $j\leq\nu(1-\overline{x})+i$ and by (9.69)

$j\leq\nu(1-x\overline{x})+1\hbox{ and }i\leq j-1\leq\nu(1-x\overline{x}).$

$\displaystyle\ll$ $\displaystyle\
\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\mathop{\sum\sum}\limits_{\stackrel{{\scriptstyle{\scriptstyle
1\leq i\leq\nu(1-x\overline{x})}}}{{{\scriptstyle i+1\leq
j\leq\nu(1-x\overline{x})+1}}}}\iint_{\stackrel{{\scriptstyle{\scriptstyle
b\in p^{-i-\nu(1-\overline{x})+1}{\mathbb{Z}_{p}}}}}{{{\scriptstyle
b^{\prime}\in
p^{-j+1}{\mathbb{Z}_{p}}}}}}\mathop{\sum\sum}\limits_{\delta,\tau\,(\mathrm{mod}\,{p})}1dbdb^{\prime}$
$\displaystyle\
+\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\mathop{\sum\sum}\limits_{\stackrel{{\scriptstyle{\scriptstyle
1\leq i\leq\nu(1-x\overline{x})}}}{{{\scriptstyle i+1\leq
j\leq\nu(1-x\overline{x})+1}}}}\iint_{\stackrel{{\scriptstyle{\scriptstyle
b\in p^{-i-\nu(1-\overline{x})+1}{\mathbb{Z}_{p}}}}}{{{\scriptstyle
b^{\prime}\in
p^{-j}{\mathbb{Z}_{p}}}}}}\sum_{\tau\,(\mathrm{mod}\,{p})}1dbdb^{\prime}$
$\displaystyle\
+\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\mathop{\sum\sum}\limits_{\stackrel{{\scriptstyle{\scriptstyle
1\leq i\leq\nu(1-x\overline{x})}}}{{{\scriptstyle i+1\leq
j\leq\nu(1-x\overline{x})+1}}}}\iint_{\stackrel{{\scriptstyle{\scriptstyle
b\in p^{-i-\nu(1-\overline{x})}{\mathbb{Z}_{p}}}}}{{{\scriptstyle
b^{\prime}\in
p^{-j+1}{\mathbb{Z}_{p}}}}}}\sum_{\delta\,(\mathrm{mod}\,{p})}1dbdb^{\prime}$
$\displaystyle\
+\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\mathop{\sum\sum}\limits_{\stackrel{{\scriptstyle{\scriptstyle
1\leq i\leq\nu(1-x\overline{x})}}}{{{\scriptstyle i+1\leq
j\leq\nu(1-x\overline{x})+1}}}}\iint_{\stackrel{{\scriptstyle{\scriptstyle
b\in p^{-i-\nu(1-\overline{x})}{\mathbb{Z}_{p}}}}}{{{\scriptstyle
b^{\prime}\in p^{-j}{\mathbb{Z}_{p}}}}}}dbdb^{\prime}$

which is bounded by

(9.73)
$\ll(1+\nu(1-x\overline{x}))^{2}\frac{p^{2\nu(1-x\overline{x})+\nu(1-\overline{x})}}{p^{3}}.$

Combining (9.68) with the bounds (9.72) and (9.73) for
$\nu(1-\overline{x})\geq 1$ we obtain

(9.74)
${\mathcal{I}}_{p}(x;2)\ll(1+\nu(1-x\overline{x}))(1+\nu(1-x\overline{x})+\nu(1-\overline{x}))\frac{p^{2\nu(1-x\overline{x})+\max(1,\nu(1-\overline{x}))}}{p^{3}}.$

### Case III

We assume now that the Iwasawa decomposition of $u$ and $v$ are of the form

$\displaystyle u=$ $\displaystyle\begin{pmatrix}p^{i}&&p^{i}b\\\ &1&\\\
&&p^{-i}\end{pmatrix}\begin{pmatrix}1&&\mu\\\ &1&\\\
&&1\end{pmatrix}J\begin{pmatrix}\tau&&\\\ &1&\\\ &&1\end{pmatrix}\gamma_{1},$
$\displaystyle v=$ $\displaystyle\begin{pmatrix}p^{j}&&p^{j}b^{\prime}\\\
&1&\\\ &&p^{-j}\end{pmatrix}\begin{pmatrix}\delta&&\\\ &1&\\\
&&1\end{pmatrix}\gamma_{2},$

where $\tau,\delta\in(\mathbb{Z}/p\mathbb{Z})^{\times},$
$\mu\in\mathbb{Z}/p\mathbb{Z},$ and $\gamma_{1},\gamma_{2}\in
I_{p}^{\prime}(1).$

Let ${\mathcal{I}}_{p}(x;3)$ be the contribution of $u,v$ of the above forms
to $\mathcal{I}_{p}(x).$ We first look for some necessary condition for
${\mathcal{I}}_{p}(x;3)$ to be non zero.

We have $f_{p}^{\mathfrak{n}_{p}}(u^{-1}\gamma(x)Jv)\neq 0$ if and only if

(9.75) $\begin{pmatrix}a&e+\delta p^{-1}a&z\\\ -p^{j}&x-\delta p^{j-1}&f\\\
p^{i+j}-\tau p^{j-1}&p^{i}(1-x)+\delta p^{i+j-1}-\frac{\tau}{p}(x-\delta
p^{j-1})&g-\tau p^{-1}f\end{pmatrix}\in K_{p}.$

Taking inverse of (9.75) we then obtain that

(9.76) $\begin{pmatrix}a^{\prime}-\delta
p^{i-1}(1-\overline{x})&e^{\prime}-\frac{\delta}{p}\overline{x}+\frac{\tau}{p}(z^{\prime}-\delta
p^{-1}f^{\prime})&z^{\prime}-\delta p^{-1}f^{\prime}\\\
p^{i}(1-\overline{x})&\overline{x}+\tau p^{-1}f^{\prime}&f^{\prime}\\\
p^{i+j}&-p^{j}+\tau p^{-1}g^{\prime}&g^{\prime}\end{pmatrix}\in K_{p}.$

These conditions imply in particular that

(9.77) $i+j\geq 0,\ j\geq 1,i+\nu(1-\overline{x})\geq 0$

and therefore $\nu(x)\geq 0$. Also $f^{\prime},\overline{x}+\tau
p^{-1}f^{\prime}\in{\mathbb{Z}_{p}}$ implies that $\nu(\overline{x})\geq-1$.
This proves (9.8) (in a stronger form) in Case III.

We assume now that $x,\overline{x}\in{\mathbb{Z}_{p}}$. We have

$f^{\prime},g^{\prime}\in p{\mathbb{Z}_{p}}$

So $j\geq 1,$ $\nu(f^{\prime})\geq 1$ and $\nu(g^{\prime})\geq 1.$
Consequently, since

$p^{j-i}=p^{j}f^{\prime}-(1-\overline{x})g^{\prime}\in p{\mathbb{Z}_{p}}$

(see (9.47)) we have

(9.78) $i\leq j-1$

and $a\in p{\mathbb{Z}_{p}}$ which implies that $e\in{\mathbb{Z}_{p}}.$

Since $z\in{\mathbb{Z}_{p}}$ by (9.43) we have

$\frac{1-x\overline{x}}{1-x}p^{-i-j}+\frac{f}{1-x}p^{-i}-\frac{1-\overline{x}}{1-x}ep^{-j}-\frac{ef}{1-x}=z\in{\mathbb{Z}_{p}}$

and since $f,e\in{\mathbb{Z}_{p}}$ we have

$\nu(1-x\overline{x})-i-j\geq\min\\{\nu(1-x),-i,-j+\nu(1-\overline{x}),0\\}=\min\\{-i,-j+\nu(1-\overline{x}),0\\}.$

or

(9.79) $i+j\leq\nu(1-x\overline{x})+\max\\{i,j-\nu(1-\overline{x}),0\\}.$

Also, we have $\nu(g)\geq-1$ so from the relation
$p^{i}f+g=p^{i-j}(1-\overline{x})$ we conclude that

(9.80) $j-i-\nu(1-\overline{x})\leq\max\\{1,-i\\}$

and substracting $i$ from (9.79) we obtain

(9.81) $1\leq j\leq\nu(1-x\overline{x})+\max\\{1,-i\\}.$

#### Localisation of $b$ and $b^{\prime}$

Finally we observe that since $f\in{\mathbb{Z}_{p}},\ f^{\prime}\in
p{\mathbb{Z}_{p}}$, we see from the expression of $f$ and $f^{\prime}$ in
(9.42) and (9.45) that $b$ and $b^{\prime}$ are contained in translates of

$p^{1-i-\nu(1-\overline{x})}{\mathbb{Z}_{p}}\hbox{ and
}p^{-j}{\mathbb{Z}_{p}}\hbox{ respectively}$

which have volumes

$p^{i+\nu(1-\overline{x})-1}\hbox{ and }p^{j}.$

Moreover we notice that if $\nu(f)=0$ then since $g-\tau
p^{-1}f\in{\mathbb{Z}_{p}}$ the congruence class $\tau$ is uniquely determined
by $g$ and $f$ (which depend on $x$ and $b^{\prime}$) while for $\nu(f)\geq 1$
there is no constraint on $\tau$ but $b^{\prime}$ varies over a translate of
$p^{-j+1}{\mathbb{Z}_{p}}$.

#### The case $\nu(1-\overline{x})=0$

We have $1\leq j\leq\max\\{i+1,0\\}\leq j$ so $j\leq i+1$ and

$j=i+1.$

Therefore, the contribution from this case to ${\mathcal{I}}_{p}(x;3)$ is
bounded by

$\displaystyle\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}\sum_{\mu\in\mathbb{Z}/p\mathbb{Z}}\sum_{0\leq
i\leq\nu(1-x\overline{x})}\sum_{\tau}\sum_{\delta}\int_{p^{1-i}\mathbb{Z}_{p}}db\int_{p^{-i-1}\mathbb{Z}_{p}}db^{\prime}$
$\displaystyle\ll$
$\displaystyle{p\mu(I_{p}^{\prime}(1))^{2}(1+\nu(1-x\overline{x}))p^{2\nu(1-x\overline{x})}}\ll(1+\nu(1-x\overline{x}))\frac{p^{2\nu(1-x\overline{x})}}{p^{3}}.$

#### The case $\nu(1-\overline{x})\geq 1$

We have $j-\nu(1-\overline{x})\leq j-1$ and (since $i\leq j-1$) we have

$i+j\leq\nu(1-x\overline{x})+j-1\Longleftrightarrow
i\leq\nu(1-x\overline{x})-1.$

– If $i\geq 0$ we have by (9.81)

$j\leq\nu(1-x\overline{x})+1$

and by (9.78)

$i+j\leq 2\nu(1-x\overline{x})+1.$

– If $i\leq-1$ we have by (9.80)

$-i,j\leq\nu(1-\overline{x})$

and

$i+j\leq\nu(1-x\overline{x})$

We conclude that the contribution from the case $\nu(1-\overline{x})\geq 1$ to
${\mathcal{I}}_{p}(x;3)$ is bounded by (see the paragraph on the localisation
of $b$ and $b^{\prime}$)

$\displaystyle\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}$
$\displaystyle\sum_{\mu\in\mathbb{Z}/p\mathbb{Z}}\mathop{\sum\sum}\limits_{\stackrel{{\scriptstyle{\scriptstyle-\nu(1-\overline{x})\leq
i\leq j-1\leq\nu(1-x\overline{x})}}}{{{\scriptstyle i+j\leq
2\nu(1-x\overline{x})+1}}}}\int_{p^{1-i-\nu(1-\overline{x})}\mathbb{Z}_{p}}\int_{p^{-j+1}\mathbb{Z}_{p}}\sum_{\tau}\sum_{\delta}1dbdb^{\prime}$
$\displaystyle+\frac{\mu(I_{p}^{\prime}(1))^{2}}{\mu(K_{p})}$
$\displaystyle\sum_{\mu\in\mathbb{Z}/p\mathbb{Z}}\mathop{\sum\sum}\limits_{\stackrel{{\scriptstyle{\scriptstyle-\nu(1-\overline{x})\leq
i\leq j-1\leq\nu(1-x\overline{x})}}}{{{\scriptstyle i+j\leq
2\nu(1-x\overline{x})+1}}}}\int_{p^{1-i-\nu(1-\overline{x})}\mathbb{Z}_{p}}\int_{p^{-j}\mathbb{Z}_{p}}\sum_{\delta}1dbdb^{\prime}$
$\displaystyle\ll(1+\nu(1-\overline{x})+\nu(1-x\overline{x}))^{2}\frac{p^{2\nu(1-x\overline{x})+\nu(1-\overline{x})}}{p^{2}}.$

Putting the above discussion together we then obtain

(9.82)
${\mathcal{I}}_{p}(x;3)\ll(1+\nu(1-\overline{x})+\nu(1-x\overline{x}))^{2}\frac{p^{2\nu(1-x\overline{x})+\nu(1-\overline{x})}}{p^{2}}.$

### Case IV

Consider $u$ and $v$ in their Iwasawa forms:

$\displaystyle u=$ $\displaystyle\begin{pmatrix}p^{i}&&p^{i}b\\\ &1&\\\
&&p^{-i}\end{pmatrix}\begin{pmatrix}\tau&&\\\ &1&\\\
&&1\end{pmatrix}\gamma_{1},$ $\displaystyle v=$
$\displaystyle\begin{pmatrix}p^{j}&&p^{j}b^{\prime}\\\ &1&\\\
&&p^{-j}\end{pmatrix}\begin{pmatrix}1&&\mu\\\ &1&\\\
&&1\end{pmatrix}J\begin{pmatrix}\delta&&\\\ &1&\\\
&&1\end{pmatrix}\gamma_{2},$

where $\tau,\delta\in(\mathbb{Z}/p\mathbb{Z})^{\times},$
$\mu\in\mathbb{Z}/p\mathbb{Z},$ and $\gamma_{1},\gamma_{2}\in
I_{p}^{\prime}(1).$

Let ${\mathcal{I}}_{p}(x;4)$ be the contribution of $u,v$ of the above forms
to ${\mathcal{I}}_{p}(x).$ We first look for some necessary condition for
${\mathcal{I}}_{p}(x;4)$ to be non zero.

Then $f_{p}^{\mathfrak{n}_{p}}(u^{-1}\gamma(x)Jv)\neq 0$ if and only if

(9.83) $\begin{pmatrix}a+\tau
p^{j-1}&e+\frac{\delta}{p}z-\frac{\tau}{p}(x+\frac{\delta}{p}f)&z-\frac{\tau}{p}f\\\
-p^{j}&x+\frac{\delta}{p}f&f\\\
p^{i+j}&p^{i}(1-x)+\frac{\delta}{p}g&g\end{pmatrix}\in K_{p}.$

Taking inverse of (9.83) we then obtain that

(9.84)
$\begin{pmatrix}a^{\prime}&e^{\prime}+\frac{\tau}{p}a^{\prime}&z^{\prime}\\\
p^{i}(1-\overline{x})&\overline{x}+\tau p^{i-1}(1-\overline{x})&f^{\prime}\\\
p^{i+j}-\frac{\delta}{p}p^{i}(1-\overline{x})&-p^{j}+p^{i+j}\frac{\tau}{p}-\frac{\delta}{p}(\overline{x}+\frac{\tau}{p}p^{i}(1-\overline{x}))&g^{\prime}-\frac{\delta}{p}f^{\prime}\end{pmatrix}\in
K_{p}.$

These conditions imply in particular that

$j\geq 0,i+j\geq 0,\ i+\nu(1-\overline{x})\geq 1.$

Moreover since $f,x+\delta f/p\in{\mathbb{Z}_{p}}$ we have $\nu(x)\geq-1$.
Since

$p^{i}(1-\overline{x})\in p{\mathbb{Z}_{p}},\overline{x}+\tau
p^{i}(1-\overline{x})/p\in{\mathbb{Z}_{p}}$

we have $\nu(\overline{x})\geq 0$. This proves Proposition 9.8 in case IV (in
a stronger form).

We assume now that $x,\overline{x}\in{\mathbb{Z}_{p}}$. We have

$f\in p{\mathbb{Z}_{p}},z\in{\mathbb{Z}_{p}}.$

Since $z\in{\mathbb{Z}_{p}}$ by (9.43) we have

$\frac{1-x\overline{x}}{1-x}p^{-i-j}+\frac{f}{1-x}p^{-i}-\frac{1-\overline{x}}{1-x}ep^{-j}-\frac{ef}{1-x}=z\in{\mathbb{Z}_{p}}$

and since $f\in p{\mathbb{Z}_{p}},\ e\in p^{-1}{\mathbb{Z}_{p}}$ we have

$\nu(1-x\overline{x})-i-j\geq\min\\{\nu(1-x),-i+1,-j+\nu(1-\overline{x})-1,0\\}=\min\\{-i+1,-j+\nu(1-\overline{x})-1,0\\}.$

or

$i+j\leq\nu(1-x\overline{x})+\max\\{i-1,j-\nu(1-\overline{x})+1,0\\}.$

Also from $g\in{\mathbb{Z}_{p}}$ and $p^{i}f+g=p^{i-j}(1-\overline{x})$ we
have

$j-\nu(1-\overline{x})-i\leq\max(-i-1,0)$

which implies that

$j-\nu(1-\overline{x})\leq\max(-1,i)$

and

$i+j\leq\nu(1-x\overline{x})+\max\\{i+1,0\\}.$

We also have $f^{\prime}\in{\mathbb{Z}_{p}},g^{\prime}\in
p^{-1}{\mathbb{Z}_{p}}$ and since

$p^{j-i}=p^{j}f^{\prime}-(1-\overline{x})g^{\prime}$

we have

$j-i\geq\min(j,\nu(1-\overline{x})-1)$

and therefore

$1-\nu(1-\overline{x})\leq i\leq\max(0,j+1-\nu(1-\overline{x})).$

### The case $i\geq 0$

In that case we have

$j-\nu(1-\overline{x})\leq i,\ j\leq\nu(1-x\overline{x})+1$

and therefore

$0\leq i\leq\nu(1-x\overline{x})+2$

so that

$i+j\leq 2\nu(1-x\overline{x})+3$

### The case $i<0$

We have

$1-\nu(1-\overline{x})\leq i\leq 0\leq j\leq\nu(1-\overline{x})-1$

and

$i+j\leq\nu(1-x\overline{x})\leq 2\nu(1-x\overline{x})+3.$

#### Localisation of $b$ and $b^{\prime}$

we observe that since $f\in p{\mathbb{Z}_{p}},\
f^{\prime}\in{\mathbb{Z}_{p}}$, we see from the expression of $f$ and
$f^{\prime}$ in (9.42) and (9.45) that $b$ and $b^{\prime}$ are contained in
translates of

$p^{-i-\nu(1-\overline{x})}{\mathbb{Z}_{p}}\hbox{ and