x\right\rvert}_{1}},\cdots,\log{\mathinner{\\!\left\lvert
x\right\rvert}_{n}}),$

where $\mathinner{\\!\left\lvert x\right\rvert}_{i}$ is the absolute value
with respect to the $i$-th archimedean place. For all $1\leq i\leq n-1$, we
see from the proof of the Dirichlet unit theorem that $S(\beta_{i})$ is a
basis for the full lattice in $W$. We write the coordinates of $W$ with
respect to these generators as $y=(y_{1},\cdots,y_{n-1})$ by which we mean
that a unit $u\in{\mathcal{O}}^{*}_{K}$ can be written as

$u=\pm\beta_{1}^{y_{1}}\cdots\beta_{n-1}^{y_{n-1}}.$

######  Definition 4.2.

Define the _interpolation function_ $\theta^{\pm}$ from ${\mathbb{R}}^{2n-1}$
to ${\mathbb{C}}$ as

$\theta^{\pm}(x_{1},\cdots,x_{n},y_{1},\cdots,y_{n-1})=\\\
{\mathsf{p}}^{\frac{{\mathsf{k}}}{2}}\prod_{i=1}^{n}\left(2\mathinner{\\!\left\lvert\frac{x_{i}^{2}}{\pm
4\beta_{1,i}^{y_{1}}\cdots\beta_{n-1,i}^{y_{n-1}}\varepsilon_{i}}-1\right\rvert}^{1/2}g_{\nu_{i},1}(x_{i},\pm
4\beta_{1,i}^{y_{1}}\cdots\beta_{n-1,i}^{y_{n-1}}\varepsilon_{i})+g_{\nu_{i},2}(x_{i},\pm
4\beta_{1,i}^{y_{1}}\cdots\beta_{n-1,i}^{y_{n-1}}\varepsilon_{i})\right),$

where $\beta_{t,i}=\sigma_{i}(\beta_{t})$ and the $\varepsilon_{i}$ are the
real embedding of the fixed $\varepsilon$.

The domain of this function is the product of the domains of each local
component, which is the $(r,N)$ plane with both the parabola $N=r^{2}$ and the
$N$-axis removed. Moreover, it is compactly supported because
$f_{\nu_{1}},\cdots,f_{\nu_{n}}$ are.

######  Notation.

To simplify notation when we write our equations, we will follow the following
conventions:

  * •

we write $\theta^{\pm}(x,y)$ instead of
$\theta^{\pm}(x_{1},\cdots,x_{n},y_{1},\cdots,y_{n-1})$.

  * •

we write $\beta$ to denote a _generic unit_ in the ${\mathbb{Z}}^{n-1}$. In
other words, the units of ${\mathcal{O}}_{K}$ are $\pm\beta$ as $\beta$
varies.

  * •

we continue to write $u$ to denote a unit in ${\mathcal{O}}_{K}^{*}$ with the
understanding that if we are evaluating the function $\theta^{\pm}(x,y)$ we
are referring to its $\beta$-component under the isomorphism induced by the
Dirichlet unit theorem.

  * •

we identify $\tau$ and $T(\tau)$.

The regular elliptic part of the trace formula can then be written as

$\displaystyle R(f)$ $\displaystyle=\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\in
L(u)}\frac{L(1,\chi_{\gamma}){\mathcal{O}}(f_{\operatorname{fin}},\gamma(\tau,u))}{\operatorname{\mathsf{N}}_{K}\left(S_{\gamma}\right)}\theta^{\pm}(\tau,u)$
$\displaystyle=\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}\sum_{\pm}\sum_{\beta}\sum_{\tau\in
L(u)}\frac{L(1,\chi_{\gamma}){\mathcal{O}}(f_{\operatorname{fin}},\gamma(\tau,u))}{\operatorname{\mathsf{N}}_{K}\left(S_{\gamma}\right)}\theta^{\pm}(\tau_{1},\cdots,\tau_{n},y_{1},\ldots,y_{n-1}).$

Note that $\tau\in{\mathcal{O}}_{K}$ can be viewed as an element of
${\mathbb{R}}^{n}$, so it makes sense to set $x=\tau$ as an input for
$\theta^{\pm}$. On the other hand, when we write $y=u$, we mean that
$y=(y_{1},\cdots,y_{n-1})$ corresponding to the unit
$u=\pm\beta_{1}^{y_{1}}\cdots\beta_{n-1}^{y_{n-1}}$.

We wish to perform Poisson summation on the variables $(\tau,u)$; but we need
to address the following issues:

  1. (a)

the function $\theta^{\pm}$ has singularities.

  2. (b)

the coefficient
$\frac{L(1,\chi_{\gamma}){\mathcal{O}}(f_{\operatorname{fin}},\gamma(\tau,u))}{\operatorname{\mathsf{N}}_{K}\left(S_{\gamma}\right)}$
in front of $\theta^{\pm}$ is not yet a smooth function with compact support
when evaluated at the points of a lattice.

  3. (c)

we do not have a complete lattice; the lack of completeness is due to missing
$\tau$’s.

This will be done in §6 and §7.

We conclude this section with the following result

######  Theorem 4.3.

With notation and assumptions introduced so far, the regular elliptic part of
$G(K)$ can be written as

$R(f)=\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\in
L(u)}\theta^{\pm}(\tau,u)L(1,\chi_{\gamma})\left(\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})}\prod_{{\mathfrak{p}}\mid(S_{\gamma}/{\mathfrak{d}})}\left(1-\frac{\chi_{\gamma}({\mathfrak{p}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{p}})}\right)\right).$

###### Proof.

This follows from substituting Theorem 3.19 into our previous expansion of the
regular elliptic part. ∎

In the next part, we will carry out a detailed study of the sum over
incomplete $L-$functions and the sum over the finite orbital integrals via the
Approximate Functional Equation.

##  5\. The Approximate Functional Equation

Let $z$ be a complex number and consider a non-square integer $\delta\equiv
0,1\pmod{4}$. In this setting, the Zagier zeta function is defined as below

$L(z,\delta)\mathrel{\mathop{\mathchar
58\relax}}=\sum_{d^{2}\mid\delta}^{\prime}\frac{1}{d^{2z-1}}L\left(z,\binom{\frac{\delta}{d^{2}}}{\cdot}_{K}\right).$

Here, we use the notation $\binom{\frac{\delta}{d^{2}}}{\cdot}_{K}$ to denote
the Kronecker symbol. The sum is taken only over those $d$ such that
$\delta/d^{2}\equiv 0,1\pmod{4}$. This congruence condition is indicated by
the ’ on top of the summation sign.

The Zagier zeta function has the following two pivotal roles in [Alt15]:

  1. (a)

it recovers the values of the finite orbital integrals when evaluated at
$z=1$.

  2. (b)

an appropriate completion of the Zagier zeta function satisfies a functional
equation to which it is possible to apply the approximate functional equation.

This allows us to fix the issues preventing us from performing Poisson
summation.

######  Remark 5.1.

For a general number field $F$, Arthur conjectures in [Art18, §3] that the
generalization of the Zagier zeta function for GL$(n,F)$ is obtained via the
work of Z. Yun, see [Yun13]. This is verified by the second named author in
[Esp23a] in the case $n=2$. We will make use of this construction below. For
more details about the Zagier zeta function over a general number field, we
refer the reader to [Esp23a].

###  5.1. Multiplicative formula of Langlands

In this section, we define the Zagier zeta function for a totally real number
field and prove a functional equation. As before, let $\chi_{\gamma}$ be the
quadratic sign character associated to the extension generated by the regular
elliptic element $\gamma$.

######  Definition 5.2.

Let $K$ be a totally real number field555The definition does not require that
$K$ is totally real and holds true for all number fields. and let $z$ be a
complex parameter with $\Re(z)>1$. Set

${\mathcal{O}}(z,\gamma)\mathrel{\mathop{\mathchar
58\relax}}=\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{z}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{(1-2z)}\prod_{{\mathfrak{q}}\mid{\mathfrak{d}}^{\prime}}\left(1-\frac{\chi_{\gamma}({\mathfrak{q}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{q}})^{z}}\right),$

where ${\mathfrak{d}}^{\prime}\mathrel{\mathop{\mathchar
58\relax}}=\frac{S_{\gamma}}{{\mathfrak{d}}}$ for any ideal ${\mathfrak{d}}$
such that ${\mathfrak{d}}\mid S_{\gamma}$. Writing $L(z,\chi_{\gamma})$ for
the Hecke $L$-function of the character $\chi_{\gamma}$, define

$L(z,\gamma)\mathrel{\mathop{\mathchar
58\relax}}=L(z,\chi_{\gamma})\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{(1-2z)}\prod_{{\mathfrak{q}}\mid{\mathfrak{d}}^{\prime}}\left(1-\frac{\chi_{\gamma}({\mathfrak{q}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{q}})^{z}}\right).$

This is the _(generalized) multiplicative formula of Langlands_ for $G(K)$.
Next define its completion as

${\Lambda}(z,\gamma)\mathrel{\mathop{\mathchar
58\relax}}=\Lambda(z,\chi_{\gamma}){\mathcal{O}}(z,\gamma).$

This is the _(generalized) completed multiplicative formula of Langlands_ for
$G(K)$.

######  Notation.

Set

$L_{\nu}(z,\chi_{\gamma})=L_{\mathbb{R}}(z)\mathrel{\mathop{\mathchar
58\relax}}=\pi^{-z/2}\Gamma\left(\dfrac{z}{2}\right).$

For a totally real field $K$, the completed Dedekind zeta function is defined
as

$\xi_{K}(z)\mathrel{\mathop{\mathchar 58\relax}}=\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{z/2}L_{{\mathbb{R}}}(z)^{n}\zeta_{K}(z).$

Also, the $L$-factors at the archimedean places for $\chi_{\gamma}$ are given
by

$L_{\nu}(z,\chi_{\gamma})=L_{{\mathbb{R}}}(z+\delta_{\nu,\gamma})=\pi^{-z/2}\Gamma((z+\delta_{\nu,\gamma})/2)$

with $\delta_{\nu,\gamma}=1$ if $\chi_{\gamma}$ is ramified at $\nu$ and 0
otherwise. Thus

$\Lambda(z,\chi_{\gamma})=\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{z/2}\operatorname{\mathsf{N}}(\Delta_{\gamma})^{z/2}L(z,\chi_{\gamma})\prod_{\nu\mid\infty}L_{\mathbb{R}}(z+\delta_{\nu,\gamma}).$

We first show that $\Lambda(z,\gamma)$ behaves nicely. More precisely,

######  Theorem 5.3.

The completed multiplicative formula of Langlands ${\Lambda}(z,\gamma)$,
admits an analytic continuation to an entire function and satisfies the
functional equation

${\Lambda}(1-z,\gamma)={\Lambda}(z,\gamma).$

###### Proof.

Writing $\mu(\cdot)$ to denote the Möbius function of an ideal of
${\mathcal{O}}_{K}$, we have

(9)
$\prod_{{\mathfrak{q}}\mid{\mathfrak{d}}^{\prime}}\left(1-\frac{\chi_{\gamma}({\mathfrak{q}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{q}})^{s}}\right)=\sum_{{\mathfrak{r}}\mid{\mathfrak{d}}^{\prime}}\frac{\chi_{\gamma}({\mathfrak{r}})\mu({\mathfrak{r}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{r}})^{s}}.$

On the left hand side, the primes are not repeated in the product. So, we only
obtain terms with different prime factors whose sign depends on the number of
such factors - this is controlled by the Möbius function on the right.

Substituting (9) into definition of the completed multiplicative formula
yields,

$\displaystyle{\Lambda}(z,\gamma)$
$\displaystyle=\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{z}\Lambda(z,\chi_{\gamma})\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}\sum_{{\mathfrak{r}}\mid{\mathfrak{d}}^{\prime}}\frac{\chi_{\gamma}({\mathfrak{r}})\mu({\mathfrak{r}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{r}})^{z}}$
$\displaystyle=\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{z}\Lambda(z,\chi_{\gamma})\sum_{{\mathfrak{d}}\mid
S_{\gamma},{\mathfrak{r}}\mid{\mathfrak{d}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}\frac{\chi_{\gamma}({\mathfrak{r}})\mu({\mathfrak{r}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{r}})^{z}}$
$\displaystyle=\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{z}\Lambda(z,\chi_{\gamma})\sum_{{\mathfrak{r}}\mid
S_{\gamma},{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}\frac{\chi_{\gamma}({\mathfrak{r}})\mu({\mathfrak{r}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{r}})^{z}}\qquad\text{
where }{\mathfrak{r}}^{\prime}=\frac{S_{\gamma}}{{\mathfrak{r}}}$
$\displaystyle=\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{z}\Lambda(z,\chi_{\gamma})\sum_{{\mathfrak{r}}\mid
S_{\gamma}}\frac{\chi_{\gamma}({\mathfrak{r}})\mu({\mathfrak{r}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{r}})^{z}}\sum_{{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}$
$\displaystyle=\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{z}\Lambda(z,\chi_{\gamma})\sum_{{\mathfrak{r}}\mid
S_{\gamma}}\frac{\chi_{\gamma}({\mathfrak{r}})\mu({\mathfrak{r}})}{\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{z}}\operatorname{\mathsf{N}}_{K}({\mathfrak{r}}^{\prime})^{z}\sum_{{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}$
$\displaystyle=\Lambda(z,\chi_{\gamma})\sum_{{\mathfrak{r}}\mid
S_{\gamma}}\frac{\chi_{\gamma}({\mathfrak{r}})\mu({\mathfrak{r}})}{\sqrt{\operatorname{\mathsf{N}}_{K}(S_{\gamma})}}\sqrt{\operatorname{\mathsf{N}}_{K}(S_{\gamma})}\operatorname{\mathsf{N}}_{K}({\mathfrak{r}}^{\prime})^{z}\sum_{{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}$
$\displaystyle=\Lambda(z,\chi_{\gamma})\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{1/2}\sum_{{\mathfrak{r}}\mid
S_{\gamma}}\frac{\chi_{\gamma}({\mathfrak{r}})\mu({\mathfrak{r}})}{\sqrt{\operatorname{\mathsf{N}}_{K}({\mathfrak{r}}{\mathfrak{r}}^{\prime})}}\operatorname{\mathsf{N}}_{K}({\mathfrak{r}}^{\prime})^{z}\sum_{{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}$
$\displaystyle=\Lambda(z,\chi_{\gamma})\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{1/2}\sum_{{\mathfrak{r}}\mid
S_{\gamma}}\frac{\chi_{\gamma}({\mathfrak{r}})\mu({\mathfrak{r}})}{\sqrt{\operatorname{\mathsf{N}}_{K}({\mathfrak{r}})}}\operatorname{\mathsf{N}}_{K}({\mathfrak{r}}^{\prime})^{z-1/2}\sum_{{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}.$

Shifting our focus to the term

$\operatorname{\mathsf{N}}_{K}({\mathfrak{r}}^{\prime})^{z-1/2}\sum_{{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}},$

we see it remains unchanged when we replace $z$ by $1-z$. Indeed

$\displaystyle\operatorname{\mathsf{N}}_{K}({\mathfrak{r}}^{\prime})^{(1-z)-1/2}\sum_{{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2(1-z)-1}}$
$\displaystyle=\operatorname{\mathsf{N}}_{K}({\mathfrak{r}}^{\prime})^{1/2-z}\sum_{{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{1-2z}}$
$\displaystyle=\operatorname{\mathsf{N}}_{K}({\mathfrak{r}}^{\prime})^{1/2-z}\sum_{{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}\left(\frac{{\mathfrak{r}}^{\prime}}{{\mathfrak{d}}}\right)^{1-2z}}$
$\displaystyle=\operatorname{\mathsf{N}}_{K}({\mathfrak{r}}^{\prime})^{1/2-z-(1-2z)}\sum_{{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}$
$\displaystyle=\operatorname{\mathsf{N}}_{K}({\mathfrak{r}}^{\prime})^{z-1/2}\sum_{{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}.$

Hence, when we substitute $1-z$ for $z$ in the whole expression we obtain

$\displaystyle{\Lambda}(1-z,\gamma)$
$\displaystyle=\Lambda(1-z,\chi_{\gamma})\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{1/2}\sum_{{\mathfrak{r}}\mid
S_{\gamma}}\frac{\chi({\mathfrak{r}})\mu({\mathfrak{r}})}{\sqrt{\operatorname{\mathsf{N}}_{K}({\mathfrak{r}})}}\operatorname{\mathsf{N}}_{K}({\mathfrak{r}}^{\prime})^{(1-z)-1/2}\sum_{{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2(1-z)-1}}$
$\displaystyle=\Lambda(z,\chi_{\gamma})\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{1/2}\sum_{{\mathfrak{r}}\mid
S_{\gamma}}\frac{\chi({\mathfrak{r}})\mu({\mathfrak{r}})}{\sqrt{\operatorname{\mathsf{N}}_{K}({\mathfrak{r}})}}\operatorname{\mathsf{N}}_{K}({\mathfrak{r}}^{\prime})^{z-1/2}\sum_{{\mathfrak{d}}\mid{\mathfrak{r}}^{\prime}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}$
$\displaystyle={\Lambda}(z,{\gamma}).$

This manipulation required using the functional equation for the primitive
quadratic Hecke character. ∎

######  Remark 5.4.

Theorem 5.3 is an adaptation of [SY13, Equations (4), (5), and Lemma 2.1],
where they work over $\mathbb{Z}$.

The following result connects this construction with our work on the regular
elliptic part of the trace formula.

######  Proposition 5.5.

With notation as introduced above,

${\mathcal{O}}(f_{\operatorname{fin}},\gamma)={\mathsf{p}}^{-{\mathsf{k}}/2}{\mathcal{O}}(1,\gamma).$

###### Proof.

This is a reformulation of Proposition 3.19. ∎

We now define certain quadratic characters that arise naturally in our
calculations.

######  Definition 5.6.

Let ${\mathfrak{d}}$ be an ideal dividing $S_{\gamma}$. Define an _imprimitive
character_ $\chi_{{\mathfrak{d}}}$, obtained from $\chi_{\gamma}$, by setting
$\chi({\mathfrak{q}})=0$ for any
${\mathfrak{q}}\mid\frac{S_{\gamma}}{{\mathfrak{d}}}$. Associated to this
imprimitive quadratic Hecke character is an $L$-function

$L(z,\chi_{{\mathfrak{d}}})=L(z,\chi_{\gamma})\prod_{{\mathfrak{q}}\mid\frac{S_{\gamma}}{{\mathfrak{d}}}}\left(1-\frac{\chi_{\gamma}({\mathfrak{q}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{q}})^{z}}\right),$

where $L(z,\chi_{\gamma})$ is the Hecke $L$-function associated to the
quadratic character $\chi_{\gamma}$.

Notice that

(10) $L(z,\gamma)=L(z,\chi_{\gamma})\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}\prod_{{\mathfrak{p}}\mid(S_{\gamma}/{\mathfrak{d}})}\left(1-\frac{\chi_{\gamma}({\mathfrak{p}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{p}})^{z}}\right)=\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}L(z,\chi_{{\mathfrak{d}}}).$

Recall from the expression of $R(f)$ obtained in Theorem 4.3 the presence of
the factor

$L(1,\chi_{\gamma})\left(\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})}\prod_{{\mathfrak{p}}\mid(S_{\gamma}/{\mathfrak{d}})}\left(1-\frac{\chi_{\gamma}({\mathfrak{p}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{p}})}\right)\right),$

which is precisely $L(1,\gamma)$. We now Theorem 4.3 as follows.

######  Proposition 5.7.

The regular elliptic part of the trace formula is

$R(f)=\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\in
L(u)}\theta^{\pm}(\tau,u)L(1,\gamma).$

######  Remark 5.8.

For $K=\mathbb{Q}$ the above decomposition becomes

$\sum_{\pm}\sum_{\tau\in L(\pm 1)}\theta^{\pm}(\tau)L(1,\tau^{2}\pm
4p^{\mathsf{k}}),$

which corresponds to [Alt15, (unnumbered) equation at the start of §4]. An
important deviation from the work of Altuğ is that in order to define
$L(z,\tau^{2}\pm 4p^{\mathsf{k}})$, he uses a sum which has some congruence
condition as opposed to the condition $d\mid S_{\gamma}$. We will discuss this
at length in §7.1.

When $K={\mathbb{Q}}$, the sum over $\pm$ comes from the Dirichlet unit
theorem (but there is no $\beta$-sum, i.e., sum over fundamental units). In
the general case, we have to deal with the extra $\beta$-sum. The factor
${\mathsf{p}}^{-{\mathsf{k}}/2}$ that we have and Altuğ does not is just a
manifestation of the slight difference between our interpolation functions.

Another fundamental difference is that $\chi_{d}$ coincides with the Kronecker
symbol which is well-understood and can be easily computed in $\mathbb{Q}$.
For a general number field, calculating the symbols is non-trivial which we
explore in §7.2.

###  5.2. Statement and Proof of Approximate Functional Equation

#### 5.2.1.

The approximate functional equation requires the use of a smooth cut-off
function. While it is not necessary to specify this function precisely, it is
possible for us to do so as in [Alt15]. For example, consider the smooth
function

(11) $F(x)=\frac{1}{2K_{0}(2)}\int_{x}^{\infty}e^{-y-1/y}\frac{dy}{y}\textrm{
where }x>0,$

where $K_{0}$ is the zero-th Bessel function of the second kind. We know from
[IK04, p. 257] that $0<F(x)<\frac{e^{-x}}{2K_{0}(2)}$.

Define its Mellin transform,

$\widetilde{F}(z)\mathrel{\mathop{\mathchar
58\relax}}=\int_{0}^{\infty}F(x)x^{z}\frac{dx}{x}.$

Moreover, the following bound will be required.

######  Lemma 5.9.

The Mellin transform $\widetilde{F}(z)$ is meromorphic with a unique simple
pole at $z=0$ with residue $1$. Furthermore, it is odd and is uniformly
bounded for $z=\sigma+it$, with the uniform bound

$\widetilde{F}(z)\ll\mathinner{\\!\left\lvert
z\right\rvert}^{\mathinner{\\!\left\lvert\sigma\right\rvert}-1}e^{-\frac{\pi\mathinner{\\!\left\lvert
t\right\rvert}}{2}}.$

###### Proof.

This follows from the properties of $F_{1}$ in [IK04, pp. 257–258]. In
particular we note that the residue of $\widetilde{F}_{1}(z)$ at $z=0$ is $1$,
and the bounds [IK04, (10.57)] also hold for $\widetilde{F}$. ∎

#### 5.2.2.

We are now in a position to state and prove the _Approximate Functional
Equation_. The key idea is to apply contour integration to the right integral.

As in §5.1, we need to work with the $L$-factor at the infinite places which
are defined in terms of local Gamma factors (see, [Roh11, Lecture 2] or [IK04,
Equation (5.3)]).

######  Notation.

Let $z$ be a complex number and define

$L_{\infty}(z,\chi_{\gamma})\mathrel{\mathop{\mathchar
58\relax}}=\prod_{\nu\mid\infty}L_{\mathbb{R}}(z+\delta_{\nu,\gamma})$

where $\delta_{\nu,\gamma}=1$ if $\chi_{\gamma}$ is ramified at $\nu$.

######  Theorem 5.10 (Approximate Functional Equation).

Let $\gamma$ be a regular elliptic element and let $\chi_{\gamma}$ denote the
quadratic sign character associated the quadratic extension $L/K$ generated by
$\gamma$. Let $F$ be a smooth function as defined above. For any
$y\in{\mathbb{R}}$, further define666In this definition (and later on in the
proof), note that $u$ is a choice of variable and is not to be confused with
the unit $u$, which does not appear in this section.

(12) $H_{\gamma}(z,y)\mathrel{\mathop{\mathchar 58\relax}}=\frac{1}{2\pi
i}\int_{\Re(u)=1}y^{-u}\widetilde{F}(u)\frac{L_{\infty}(1-z+u,\chi_{\gamma})}{L_{\infty}(z-u,\chi_{\gamma})}du.$

Let $A$ be any real positive number. Then

$\displaystyle L(z,{\gamma})$ $\displaystyle=\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{d}}\right)^{2z-1}}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}{({\mathfrak{a}})}}{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{a}}\right)^{z}}F\left(\frac{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{d}}\right)^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{A}\right)$
$\displaystyle+\left(\operatorname{\mathsf{N}}_{K}\left(S_{\gamma}\right)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})^{1/2}\right)^{1-2z}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{d}}\right)^{1-2z}}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}\left({\mathfrak{a}}\right)}{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{a}}\right)^{1-z}}H_{\gamma}\left(z,\frac{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{d}}\right)^{2}\operatorname{\mathsf{N}}_{K}\left({\mathfrak{a}}\right)A}{\operatorname{\mathsf{N}}_{K}\left(S_{\gamma}\right)^{2}\mathinner{\\!\left\lvert
D_{K}\right\rvert}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})}\right),$

where the sum over ${\mathfrak{a}}$ runs over all ideals of
${\mathcal{O}}_{K}$.

###### Proof.

Fix $z\in{\mathbb{C}}$ and a positive real number $\sigma$ such that
$\sigma+\Re(z)>1$ and $\Re(z)-\sigma<0$. Define the integral

(13) $I\mathrel{\mathop{\mathchar 58\relax}}=\frac{1}{2\pi
i}\int_{\Re(u)=\sigma}{L}(z+u,\gamma)\widetilde{F}(u)A^{u}du.$

By the choice of $\sigma$, this integral is absolutely convergent. By equation
(10),

$\displaystyle{L}(z+u,\gamma)$ $\displaystyle=\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z+2u-1}}L(z+u,\chi_{{\mathfrak{d}}})$
$\displaystyle=\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z+2u-1}}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{z+u}}.$

The last equality follows since we are in the half-plane of absolute
convergence of the Dirichlet series for each $\chi_{{\mathfrak{d}}}$.
Substituting this in (13) gives

$\displaystyle I$ $\displaystyle=\frac{1}{2\pi
i}\int_{\Re(u)=\sigma}{L}(z+u,\gamma)\widetilde{F}(u)A^{u}du$
$\displaystyle=\frac{1}{2\pi
i}\int_{\Re(u)=\sigma}\left(\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z+2u-1}}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{z+u}}\right)\widetilde{F}(u)A^{u}du$
$\displaystyle=\frac{1}{2\pi i}\int_{\Re(u)=\sigma}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\sum_{{\mathfrak{a}}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z+2u-1}}\frac{\chi_{{\mathfrak{d}}}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{z+u}}\widetilde{F}(u)A^{u}du$
$\displaystyle=\frac{1}{2\pi i}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\sum_{{\mathfrak{a}}}\int_{\Re(u)=\sigma}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z+2u-1}}\frac{\chi_{{\mathfrak{d}}}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{z+u}}\widetilde{F}(u)A^{u}du$
$\displaystyle=\frac{1}{2\pi i}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}{({\mathfrak{a}})}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{z}}\int_{\Re(u)=\sigma}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2u}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{u}}\widetilde{F}(u)A^{u}du$
$\displaystyle=\frac{1}{2\pi i}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}{({\mathfrak{a}})}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{z}}\int_{\Re(u)=\sigma}\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{-2u}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{-u}\widetilde{F}(u)A^{u}du.$

It is possible to interchange the order of integration as the integrand is
absolutely convergent. Using the Mellin inversion formula for the last
integral gives

$\frac{1}{2\pi
i}\int_{\Re(u)=\sigma}\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{-2u}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{-u}\widetilde{F}(u)A^{u}du=F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{A}\right).$

Hence,

$I=\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z-1}}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}{({\mathfrak{a}})}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{z}}F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{A}\right).$

Shifting the contour from $\Re(u)=\sigma$ to $\Re(u)=\sigma^{\prime}$ for any
$\sigma^{\prime}<0$, we get a residue from the pole of $\widetilde{F}(u)$ at
$u=0$. This pole is simple, so the residue is $\lim_{u\rightarrow
0}u\widetilde{F}(u)$. We deduce the equality

$I=L(z,\gamma)+\frac{1}{2\pi
i}\int_{\Re(u)=\sigma^{\prime}}{L}(z+u,\gamma)\widetilde{F}(u)A^{u}du,$

or equivalently, writing $I^{\prime}$ to denote the integral running over
$\sigma^{\prime}$ instead of $\sigma$, we have that

${L}(z,\gamma)=I-I^{\prime}.$

Next we evaluate $I^{\prime}$. Making the change of variables $u\rightarrow-u$
and using that $\widetilde{F}$ is odd

$\displaystyle I^{\prime}$ $\displaystyle=-\frac{1}{2\pi
i}\int_{\Re(u)=\sigma^{\prime}}{L}(z-u,\gamma)\widetilde{F}(-u)A^{-u}du$
$\displaystyle=-\frac{1}{2\pi
i}\int_{\Re(u)=\sigma^{\prime}}{L}(z-u,\gamma)\widetilde{F}(u)A^{-u}du.$

Evaluating ${L}(z,\gamma)$ at $1-(z-u)$, a point in the half plane of absolute
convergence, we obtain the Dirichlet series

${L}(1-z+u,\gamma)=\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{1-2z+2u}}L(1-z+u,\chi_{{\mathfrak{d}}}).$

Substituting this into $I^{\prime}$ and using the functional equation of the
completed Zagier zeta function,

$\displaystyle-I^{\prime}$ $\displaystyle=\frac{1}{2\pi
i}\int_{\Re(u)=\sigma^{\prime}}{L}(z-u,\gamma)\widetilde{F}(u)A^{-u}du.$
$\displaystyle=\frac{1}{2\pi
i}\int_{\Re(u)=\sigma^{\prime}}\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{1-2(z-u)}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{(1-2(z-u))/2}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})^{(1-2(z-u))/2}$
$\displaystyle\hskip
216.81pt\times\frac{L_{\infty}(1-z+u,\chi_{\gamma})}{L_{\infty}(z-u,\chi_{\gamma})}{L}(1-z+u,\gamma)\widetilde{F}(u)A^{-u}du$
$\displaystyle=\frac{1}{2\pi
i}\int_{\Re(u)=\sigma^{\prime}}\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{1-2(z-u)}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{(1-2(z-u))/2}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})^{(1-2(z-u))/2}\frac{L_{\infty}(1-z+u,\chi_{\gamma})}{L_{\infty}(z-u,\chi_{\gamma})}$
$\displaystyle\hskip
216.81pt\times\widetilde{F}(u)A^{-u}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{1-2z+2u}}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{1-z+u}}du.$
$\displaystyle=\frac{1}{2\pi
i}\int_{\Re(u)=\sigma^{\prime}}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\sum_{{\mathfrak{a}}}\frac{\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{1-2(z-u)}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{(1-2(z-u))/2}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})^{(1-2(z-u))/2}\widetilde{F}(u)A^{-u}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{1-2z+2u}}$
$\displaystyle\hskip
216.81pt\times\frac{L_{\infty}(1-z+u,\chi_{\gamma})}{L_{\infty}(z-u,\chi_{\gamma})}\frac{\chi_{{\mathfrak{d}}}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{1-z+u}}du.$

As the Dirichlet series converges absolutely and uniformly in a half-plane
away from the abscissa of convergence, we may interchange the order of
integration. Shifting the integral to $\sigma=1$ since there are no poles in
between,

$\displaystyle-I^{\prime}$ $\displaystyle=\frac{1}{2\pi
i}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\sum_{{\mathfrak{a}}}\int_{\Re(u)=\sigma^{\prime}}\frac{\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{1-2(z-u)}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{(1-2(z-u))/2}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})^{(1-2(z-u))/2}\widetilde{F}(u)A^{-u}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{1-2z+2u}}$
$\displaystyle\hskip
216.81pt\times\frac{L_{\infty}(1-z+u,\chi_{\gamma})}{L_{\infty}(z-u,\chi_{\gamma})}\frac{\chi_{{\mathfrak{d}}}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{1-z+u}}du$
$\displaystyle=\frac{\left(\operatorname{\mathsf{N}}_{K}\left(S_{\gamma}\right)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})^{1/2}\right)^{1-2z}}{2\pi
i}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{d}}\right)^{1-2z}}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{1-z}}$
$\displaystyle\hskip
144.54pt\times\int_{\Re(u)=\sigma^{\prime}}\frac{\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{2u}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{u}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})^{u}\widetilde{F}(u)A^{-u}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2u}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{u}}\frac{L_{\infty}(1-z+u,\chi_{\gamma})}{L_{\infty}(z-u,\chi_{\gamma})}du.$

Collecting terms inside the last integral, it is possible to write it as

$\int_{\Re(u)=1}\left(\frac{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{d}}\right)^{2}\operatorname{\mathsf{N}}_{K}\left({\mathfrak{a}}\right)A}{\operatorname{\mathsf{N}}_{K}\left(S_{\gamma}\right)^{2}\mathinner{\\!\left\lvert
D_{K}\right\rvert}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})}\right)^{-u}\frac{L_{\infty}\left(1-z+u,\chi_{\gamma}\right)}{L_{\infty}\left(z-u,\chi_{\gamma}\right)}\widetilde{F}\left(u\right)du.$

This motivates the definition in the theorem

$H_{\gamma}(z,y)\mathrel{\mathop{\mathchar 58\relax}}=\frac{1}{2\pi
i}\int_{\Re(u)=1}y^{-u}\widetilde{F}(u)\frac{L_{\infty}(1-z+u,\chi_{\gamma})}{L_{\infty}(z-u,\chi_{\gamma})}du.$

Therefore, we have proven

$-I^{\prime}=\left(\operatorname{\mathsf{N}}_{K}\left(S_{\gamma}\right)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})^{1/2}\right)^{1-2z}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{d}}\right)^{1-2z}}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}\left({\mathfrak{a}}\right)}{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{a}}\right)^{1-z}}H_{\gamma}\left(z,\frac{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{d}}\right)^{2}\operatorname{\mathsf{N}}_{K}\left({\mathfrak{a}}\right)A}{\operatorname{\mathsf{N}}_{K}\left(S_{\gamma}\right)^{2}\mathinner{\\!\left\lvert
D_{K}\right\rvert}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})}\right).$

Hence, putting together our evaluations of $I$ and $I^{\prime}$ we get

$\displaystyle{L}(z,\gamma)$ $\displaystyle=\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{d}}\right)^{2z-1}}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}{({\mathfrak{a}})}}{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{a}}\right)^{z}}F\left(\frac{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{d}}\right)^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{A}\right)$
$\displaystyle+\left(\operatorname{\mathsf{N}}_{K}\left(S_{\gamma}\right)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})^{1/2}\right)^{1-2z}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{d}}\right)^{1-2z}}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}\left({\mathfrak{a}}\right)}{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{a}}\right)^{1-z}}H_{\gamma}\left(z,\frac{\operatorname{\mathsf{N}}_{K}\left({\mathfrak{d}}\right)^{2}\operatorname{\mathsf{N}}_{K}\left({\mathfrak{a}}\right)A}{\operatorname{\mathsf{N}}_{K}\left(S_{\gamma}\right)^{2}\mathinner{\\!\left\lvert
D_{K}\right\rvert}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})}\right).$

∎

###  5.3. The function $H_{(x,y)}$

The following result connects the archimedean behaviour of the quadratic
character $\chi_{\gamma}$ with the regions we defined in §3.7.

######  Theorem 5.11.

Let $\gamma\in\operatorname{GL}(2,K)$ be a regular elliptic matrix and $\nu$
be a real place of $K$. Then, $\chi_{\gamma}$ is unramified at $\nu$ if and
only if $(r_{\nu},N_{\nu})\in S_{\operatorname{spl}}$.

######  Notation.

In this proof we write $\chi$ in place of $\chi_{\gamma}$ for ease of
notation.

###### Proof.

Let $\nu$ denote a real place of $K$ and $K_{\nu}={\mathbb{R}}$ be the
localization. Let

$\rho_{\gamma}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{I}_{K}\longrightarrow\\{1,-1\\}$

be the idèlic Artin map associated to the quadratic extension $K_{\gamma}/K$.
This may also be viewed as an idèlic Hecke character associated to $\chi$.
Thus, for the real place $\nu$ we have

$\chi_{\nu}(r)=\rho_{\gamma}(1,\ldots,r,\ldots,1),$

where $(1,\ldots,r,\ldots,1)$ is the idèle with $r$ in the $\nu$-th entry and
$1$ everywhere else. We write $c_{\nu}$ to denote the element
$(1,\ldots,-1,\ldots,1)$ where $-1$ is the $\nu$-th entry.

Recall that

$\displaystyle\chi\textrm{ is unramified}$
$\displaystyle\Longleftrightarrow\chi_{\nu}(-1)=1\textrm{ by definition}$
$\displaystyle\Longleftrightarrow\rho_{\gamma}(c_{\nu})=1$
$\displaystyle\Longleftrightarrow
c_{\nu}\in\ker(\rho_{\gamma})=K^{*}\operatorname{\mathsf{N}}_{K_{\gamma}/K}(K_{\gamma}^{*})\textrm{
by Artin's reciprocity law.}$

Thus, there exists an idèle $b\in\mathbb{I}_{K_{\gamma}}$ and an $\alpha\in
K^{*}$ such that

$c_{\nu}=\alpha\operatorname{\mathsf{N}}_{K_{\gamma}/K}(b).$

Comparing the $\omega$-th entry, and using the definition of the idèlic norm,
we get

$(c_{\nu})_{\omega}=\alpha\prod_{\omega^{\prime}\mid\omega}\operatorname{\mathsf{N}}_{K_{\gamma,\omega^{\prime}}/K_{\omega}}(b_{\omega^{\prime}}).$

where $\omega^{\prime}$ varies over the places of $K_{\gamma}$ above $\omega$.
This happens only when
$-1\in\operatorname{\mathsf{N}}_{K_{\gamma,\omega^{\prime}}/K_{\omega}}(b_{\omega^{\prime}})$.
However, by definition of the local norm in the archimedean case, this is not
possible if $\nu$ ramifies (i.e. if $K_{\gamma,\omega^{\prime}}=\mathbb{C}$).
Thus, $\nu$ splits.

Thus, we have proven that $\chi$ is unramified at $\nu$ if and only if
$K_{\gamma,\omega^{\prime}}=\mathbb{R}$. We know that this extension is
precisely $K_{\gamma,\nu}(\sqrt{\tau_{\nu}^{2}-4\delta_{\nu}})$. This
extension is $\mathbb{R}$ if and only if $\tau_{i}^{2}-4\delta_{i}>0$.

Using the $(r_{\nu},N_{\nu})$-coordinates, $\chi$ is unramified at $\nu$ if
and only if $r_{\nu}^{2}-N_{\nu}>0$. So, $(r_{\nu},N_{\nu})$ lies in
$S_{\operatorname{spl}}$. ∎

######  Proposition 5.12.

Let $\gamma_{1},\gamma_{2}$ be regular elliptic matrices. If
$\gamma_{1}\operatorname{\overset{\infty}{\;\sim\;}}\gamma_{2}$ then
$H_{\gamma_{1}}=H_{\gamma_{2}}$.

###### Proof.

By definition

$H_{\gamma}(z,y)\mathrel{\mathop{\mathchar 58\relax}}=\frac{1}{2\pi
i}\int_{\Re(u)=1}y^{-u}\widetilde{F}(u)\frac{L_{\infty}(1-z+u,\chi_{\gamma})}{L_{\infty}(z-u,\chi_{\gamma})}du.$

The dependence of $\gamma$ is on the factors at infinity. But, these factors
are determined by the ramification of $\gamma$ at infinity. Theorem 5.11
precisely states that the ramification behaviour of $\chi_{\gamma}$ remains
invariant if $\gamma_{1}\operatorname{\overset{\infty}{\;\sim\;}}\gamma_{2}$.
∎

######  Definition 5.13.

Let $(x,y)\in\mathbb{R}^{n}\times\mathbb{R}^{n-1}$ belong to the region
$S_{J}$ for the multi-index $J=(j_{1},\ldots,j_{n})$ where each
$j_{t}\in\\{{\operatorname{spl}},{\operatorname{ell}}\\}$. Define
$H_{(x,y)}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{C}\times\mathbb{R}\longrightarrow\mathbb{C}$ by

$H_{(x,y)}(z,r)\mathrel{\mathop{\mathchar 58\relax}}=H_{\gamma}(z,r)$

for any regular elliptic matrix with $\gamma\in\operatorname{GL}(2,K)$ with
$\operatorname{CH}(\gamma)\in S_{J}$.

######  Remark 5.14.

  1. (i)

The above definition is well-defined because all maximal tori of $G_{\infty}$
span all possible splitting behaviour and Proposition 5.3 asserts that the
function $H_{\gamma}$ is invariant under splitting behaviour.

  2. (ii)

Definition 5.13 allows us to extend the definition of
$H_{\gamma}=H_{(\tau,u)}$ to _all_ points $(x,y)\in{\mathbb{R}}^{2n-1}$. [Here
we are using the fact that $\gamma$ is characterized by $(\tau,u)$ where $u$
is once again a unit in ${\mathcal{O}}_{K}$.]

  3. (iii)

Notice that the ratio

$\frac{L_{\infty}(1-z+u,\chi_{\gamma})}{L_{\infty}(z-u,\chi_{\gamma})}$

is independent of $\gamma$ and only depends on the behaviour at infinity
(because the $L$-factors only depend on the factors at infinity). For ease of
notation, when we want to emphasize this independence we will suppress the
subscript $\gamma$ and write

$\frac{L_{\infty}(1-z+u,\chi)}{L_{\infty}(z-u,\chi)}.$

In particular, we will use this notation in Corollary 10.2.

###  5.4. Estimates on $H_{\gamma}(1,x)$

The difference for each $H_{\gamma}(1,x)$ as defined in Theorem 5.10 comes
from the $L_{\infty}$ factors. We need to calculate the bound when $z=1$.

######  Lemma 5.15.

For $x\geq 1$,

$H_{\gamma}(1,x)\ll\frac{1}{x}e^{-(\sqrt[n+1]{x})}$

where the implied constant is absolute.

######  Notation.

In this proof we write $u$ to denote a variable and it should not be confused
with a unit of ${\mathcal{O}}_{K}$.

###### Proof.

We begin the proof by recalling Stirling’s formula for the Gamma function,

$\Gamma(u)=\sqrt{2\pi}u^{u-1/2}e^{-u}\left(1+O\left(\mathinner{\\!\left\lvert
u\right\rvert}^{-1/2}\right)\right).$

Therefore,

$\displaystyle\Gamma\left((u+\delta_{\nu,\gamma})/2\right)$
$\displaystyle=\sqrt{2\pi}\left((u+\delta_{\nu,\gamma})/2\right)^{(u+\delta_{\nu,\gamma}-1)/2}e^{(-u-\delta_{\nu,\gamma})/2}\left(1+O\left(\mathinner{\\!\left\lvert(u+\delta_{\nu,\gamma})/2\right\rvert}^{-1/2}\right)\right)$
$\displaystyle\Gamma\left((1-u+\delta_{\nu,\gamma})/2\right)$
$\displaystyle=\sqrt{2\pi}\left((1-u+\delta_{\nu,\gamma})/2\right)^{(-u+\delta_{\nu,\gamma})/2}e^{(-1+u-\delta_{\nu,\gamma})/2}\left(1+O\left(\mathinner{\\!\left\lvert(1-u+\delta_{\nu,\gamma})/2\right\rvert}^{-1/2}\right)\right).$

Combining the above two expressions,

$\displaystyle\frac{\Gamma\left((u+\delta_{\nu,\gamma})/2\right)}{\Gamma\left((1-u+\delta_{\nu,\gamma})/2\right)}$
$\displaystyle=e^{-u+1/2}\left(\frac{u+\delta_{\nu,\gamma}}{2}\right)^{u-1/2}\left(\frac{u+\delta_{\nu,\gamma}}{2}\right)^{-u/2+\delta_{\nu,\gamma}/2}\left(\frac{1-u+\delta_{\nu,\gamma}}{2}\right)^{u-\delta_{\nu,\gamma}/2}\left(1+O\left({\mathinner{\\!\left\lvert
u\right\rvert}^{-1/2}}\right)\right)$
$\displaystyle=\left(\frac{u+\delta_{\nu,\gamma}}{2e}\right)^{u-1/2}\left(\frac{1-u+\delta_{\nu,\gamma}}{u+\delta_{\nu,\gamma}}\right)^{u/2-\delta_{\nu,\gamma}/2}\left(1+O\left({\mathinner{\\!\left\lvert
u\right\rvert}^{-1/2}}\right)\right)$
$\displaystyle=\left(\frac{u}{2e}\right)^{u-1/2}\left(\frac{u+\delta_{\nu,\gamma}}{u}\right)^{u-1/2}\left(\frac{1-u+\delta_{\nu,\gamma}}{u+\delta_{\nu,\gamma}}\right)^{(u-\delta_{\nu,\gamma})/2}\left(\frac{u}{u}\right)^{(u-\delta_{\nu,\gamma})/2}\left(1+O\left({\mathinner{\\!\left\lvert
u\right\rvert}^{-1/2}}\right)\right)$
$\displaystyle=\left(\frac{u}{2e}\right)^{u-1/2}\left(\frac{u+\delta_{\nu,\gamma}}{u}\right)^{u-1/2}\left(\frac{u}{u+\delta_{\nu,\gamma}}\right)^{(u-\delta_{\nu,\gamma})/2}\left(\frac{1-u+\delta_{\nu,\gamma}}{u}\right)^{(u-\delta_{\nu,\gamma})/2}\left(1+O\left({\mathinner{\\!\left\lvert
u\right\rvert}^{-1/2}}\right)\right)$
$\displaystyle=\left(\frac{u}{2e}\right)^{u-1/2}\left(1+\frac{\delta_{\nu,\gamma}}{u}\right)^{u/2+(\delta_{\nu,\gamma}-1)/2}\left(\frac{1+\delta_{\nu,\gamma}}{u}-1\right)^{(u-\delta_{\nu,\gamma})/2}\left(1+O\left({\mathinner{\\!\left\lvert
u\right\rvert}^{-1/2}}\right)\right).$

For
$\mathinner{\\!\left\lvert\delta_{\nu,\gamma}\right\rvert}<\mathinner{\\!\left\lvert
u\right\rvert}$ we have that $\mathinner{\\!\left\lvert
1+\delta_{\nu,\gamma}/u\right\rvert}\leq 2$ and
$\mathinner{\\!\left\lvert(1+\delta_{\nu,\gamma})/u-1\right\rvert}\leq 1$
which gives

$\frac{\Gamma\left((u+\delta_{\nu,\gamma})/2\right)}{\Gamma\left((1-u+\delta_{\nu,\gamma})/2\right)}\ll\left(\frac{u}{\sqrt{2}e}\right)^{u-1/2}$

where the implied constant is absolute. Hence we have

$\frac{L_{\infty}(u,\chi_{\gamma})}{L_{\infty}(1-u,\chi_{\gamma})}=\prod_{j=1}^{n}\frac{\Gamma\left((u+\delta_{\nu,\gamma})/2\right)}{\Gamma\left((1-u+\delta_{\nu,\gamma})/2\right)}\ll\left(\frac{u}{\sqrt{2}e}\right)^{n(u-1/2)}.$

Substituting into the definition for $H_{\gamma}(1,x)$ and using the bound on
$\widetilde{F}(x)$ given in Lemma 5.9,

$\displaystyle H_{\gamma}(1,x)$
$\displaystyle\ll\int_{\Re(u)=1}\mathinner{\\!\left\lvert\frac{u}{\sqrt{2}e}\right\rvert}^{n(\Re(u)-1/2)}x^{-\Re(u)}\mathinner{\\!\left\lvert
u\right\rvert}^{\Re(u)-1}e^{-\mathinner{\\!\left\lvert\Im(u)\right\rvert}\pi/2}$
$\displaystyle\ll\int_{\Re(u)=1}\mathinner{\\!\left\lvert
u\right\rvert}^{(n+1)\Re(u)-(n+2)/2}(e^{n+1}x)^{-\Re(u)}e^{-\pi\mathinner{\\!\left\lvert\Im(u)\right\rvert}/2}du.$

Shifting the contour to $\Re(u)=\max\\{1,\sqrt[n+1]{x}\\}$ gives

$\displaystyle H_{\gamma}(1,x)$
$\displaystyle\ll\frac{1}{x^{\frac{n+2}{2(n+1)}}}e^{-(\sqrt[n+1]{x})}$
$\displaystyle\ll\frac{1}{x}e^{-(\sqrt[n+1]{x})}.\qed$

######  Corollary 5.16.

For $\Re(x)\geq 1$,

$H_{(x,y)}(1,r)\ll\frac{1}{r}e^{-(\sqrt[n+1]{r})}$

where the implied constant is absolute.

###  5.5. Manipulation using the Approximate Functional Equation

At this point, we have the following expression for the regular elliptic part
of the trace formula

$R(f)={\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\in
L(u)}\theta^{\pm}(\tau,u)L(1,\gamma).$

Using the approximate functional equation (see Theorem 5.10) at $z=1$, we find

$\displaystyle L(1,\gamma)$ $\displaystyle=\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}{({\mathfrak{a}})}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{A}\right)$
$\displaystyle+\frac{1}{\operatorname{\mathsf{N}}_{K}(S_{\gamma})\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})^{1/2}}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{-1}}\sum_{{\mathfrak{a}}}\chi_{{\mathfrak{d}}}({\mathfrak{a}})H_{\gamma}\left(1,\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})A}{\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{2}\mathinner{\\!\left\lvert
D_{K}\right\rvert}\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})}\right).$

Recall that $\partial=\tau^{2}-4\det(\gamma)$, therefore

$\operatorname{\mathsf{N}}_{K}(\partial)=\frac{\mathinner{\\!\left\lvert
D_{L}\right\rvert}\operatorname{\mathsf{N}}_{K}(S_{\gamma}^{2})}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{2}}=\operatorname{\mathsf{N}}_{K}(\Delta_{\gamma})\operatorname{\mathsf{N}}_{K}(S_{\gamma})^{2}.$

Combining these expressions, we obtain that

$\begin{split}R(f)&={\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{u\in{\mathcal{O}}^{*}_{K}}\sum_{\tau\in
L(u)}\theta^{\pm}(\tau,u)\left(\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}{({\mathfrak{a}})}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{A}\right)\right.\\\
&+\left.\frac{1}{\operatorname{\mathsf{N}}_{K}(\partial)^{1/2}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{-1}}\sum_{{\mathfrak{a}}}\chi_{{\mathfrak{d}}}({\mathfrak{a}})H_{\gamma}\left(1,\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})A}{\operatorname{\mathsf{N}}_{K}(\partial)\mathinner{\\!\left\lvert
D_{K}\right\rvert}}\right)\right).\end{split}$

######  Remark 5.17.

At this stage two sums have appeared: one runs over all ideals
${\mathfrak{a}}$ (this sum arises from the Dirichlet series of the imprimitive
character). The other runs over ${\mathfrak{d}}\mid S_{\gamma}$ (this arises
from the definition of the generalized Multiplicative Formula of Langlands).
The sum over ${\mathfrak{a}}$ has no dependence on any of the other variables
involved.

##  6\. Smoothing the Singularities

Throughout this section we fix a totally positive system of fundamental units
of $K$ denoted by $\\{\beta_{1},\ldots,\beta_{n-1}\\}$. Let $\gamma$ be a
contributing regular elliptic matrix that corresponds to the unit $u$ and the
trace $\tau$. Recall from (2) that $\varepsilon=\rho^{{\mathsf{k}}^{\prime}}$
is a fixed generator for the principal ideal
$\left(\det(\gamma)\right)={\mathfrak{p}}^{{\mathsf{k}}}$ with
$\operatorname{\mathsf{N}}_{K}({\mathfrak{p}})={\mathsf{p}}$.

######  Notation.

We write $\varepsilon_{i}=\sigma_{i}(\varepsilon)$ and
$\beta_{t,i}=\sigma_{i}(\beta_{t})$ for $1\leq t\leq n-1$, where $\sigma_{i}$
is the $i$-th real embedding of $K$.

######  Definition 6.1.

Let $x=(x_{1},\cdots,x_{n})\in{\mathbb{R}}^{n}$ and
$y=(y_{1},\cdots,y_{n-1})\in{\mathbb{R}}^{n-1}$. Define the discriminant
function $\mathcal{P}^{\pm}\mathrel{\mathop{\mathchar
58\relax}}{\mathbb{R}}^{n}\times{\mathbb{R}}^{n-1}\rightarrow{\mathbb{R}}$ by

$\displaystyle\mathcal{P}^{\pm}(x,y)$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{2^{n}}\prod_{i=1}^{n}\left(\frac{\mathinner{\\!\left\lvert
x_{i}^{2}\pm
4\beta_{1,i}^{y_{1}}\cdots\beta_{n-1,i}^{y_{n-1}}\varepsilon_{i}\right\rvert}^{-1/2}}{\mathinner{\\!\left\lvert
4\beta_{1,i}^{y_{1}}\cdots\beta_{n-1,i}^{y_{n-1}}\varepsilon_{i}\right\rvert}^{-1/2}}\right).$

In the following proposition, we evaluate the above defined function at
$x=\tau$ and $y=u$. Our notation here will be similar to the one adapted for
$\theta^{\pm}$. We explain it once again for convenience. Since
$\tau\in{\mathcal{O}}_{K}$, it can be viewed as an element of
${\mathbb{R}}^{n}$. On the other hand, when we write $y=u$, we mean that
$y=(y_{1},\cdots,y_{n-1})$ corresponding to the unit
$u=\pm\beta_{1}^{y_{1}}\cdots\beta_{n-1}^{y_{n-1}}$. The relevance of the
functions $\mathcal{P}^{\pm}$ is clear from the following result.

######  Proposition 6.2.

With notation as above,

$\mathcal{P}^{\pm}(\tau,u)=\frac{{\mathsf{p}}^{{\mathsf{k}}/2}}{\sqrt{\operatorname{\mathsf{N}}_{K}(\partial)}}.$

###### Proof.

By hypothesis we are evaluating the functions $\mathcal{P}^{\pm}$ at $x=\tau$
and $y=u$. Since there exists a regular elliptic matrix corresponding to
$(\tau,u)$, for each $i$ we can write

$\frac{\mathinner{\\!\left\lvert x_{i}^{2}\pm
4\beta_{1,i}^{y_{1}}\cdots\beta_{n-1,i}^{y_{n-1}}\varepsilon_{i}\right\rvert}^{-1/2}}{\mathinner{\\!\left\lvert
4\beta_{1,i}^{y_{1}}\cdots\beta_{n-1,i}^{y_{n-1}}\varepsilon_{i}\right\rvert}^{-1/2}}=\mathinner{\\!\left\lvert\frac{\tau_{i}}{4u_{i}\varepsilon_{i}}\pm
1\right\rvert}^{-1/2}=2D(\sigma_{i}(\lambda_{1}),\sigma_{i}(\lambda_{2})),$

where $\lambda_{1},\lambda_{2}$ are the eigenvalues of $\gamma(\tau,u)$. By
Proposition 4.1

$\mathcal{P}^{\pm}(\tau,u)=\frac{{\mathsf{p}}^{{\mathsf{k}}/2}}{\sqrt{\operatorname{\mathsf{N}}_{K}(\partial)}}.\qed$

Recall that our interpolation function $\theta^{\pm}$ has singularities. In
what follows we define smoothed-out versions of this interpolation function
which then appear more naturally in the expression for the regular elliptic
part.

######  Definition 6.3.

Let $F$ be the smooth function defined in (11). For a parameter $0<\alpha<1$,
and for $(x,y)\in\mathbb{R}^{n}\times\mathbb{R}^{n-1}$ in the union of the
open regions $S_{J}$ define the functions

$\displaystyle\Phi_{(\pm,\alpha)}(x,y)$
$\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\Phi_{\pm}(x,y)\mathrel{\mathop{\mathchar
58\relax}}=\theta^{\pm}(x,y)F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}}){\mathsf{p}}^{-{\mathsf{k}}\alpha}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\mathcal{P}^{\pm}(x,y)^{2\alpha}\right),$
$\displaystyle\Psi_{(\pm,\alpha)}(x,y)$
$\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\Psi_{\pm}(x,y)\mathrel{\mathop{\mathchar
58\relax}}=\theta^{\pm}(x,y)\mathcal{P}^{\pm}(x,y)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{-1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}H_{(x,y)}\left(1,\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}}){\mathsf{p}}^{-{\mathsf{k}}(1-\alpha)}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\mathcal{P}^{\pm}(x,y)^{2(1-\alpha)}\right).$

Suppose that $x=\tau$ and $y=u$. Then

(14)
$\begin{split}\Phi_{\pm}(\tau,u)&=\theta^{\pm}(\tau,u)F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}}){\mathsf{p}}^{-{\mathsf{k}}\alpha}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\mathcal{P}^{\pm}(\tau,u)^{2\alpha}\right)\\\
&=\theta^{\pm}(\tau,u)F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}}){\mathsf{p}}^{-{\mathsf{k}}\alpha}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\left(\frac{{\mathsf{p}}^{{\mathsf{k}}/2}}{\operatorname{\mathsf{N}}_{K}(\partial)^{1/2}}\right)^{2\alpha}\right)\quad\textrm{by
Proposition}~{}\ref{Prop: restatement}\\\
&=\theta^{\pm}(\tau,u)F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\right).\end{split}$

Recall the definition of the function $H_{\gamma}(z,y)$ introduced in (12). A
similar calculation as the one above yields

(15)
$\begin{split}\Psi_{\pm}(\tau,u)&=\theta^{\pm}(\tau,u)\mathcal{P}^{\pm}(\tau,u)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{-1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}H_{\gamma}\left(1,\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}}){\mathsf{p}}^{-{\mathsf{k}}(1-\alpha)}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\mathcal{P}^{\pm}(\tau,u)^{2(1-\alpha)}\right)\\\
&=\theta^{\pm}(\tau,u)\left(\frac{{\mathsf{p}}^{{\mathsf{k}}/2}}{\operatorname{\mathsf{N}}_{K}(\partial)^{1/2}}\right)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{-1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}H_{\gamma}\left(1,\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}}){\mathsf{p}}^{-{\mathsf{k}}(1-\alpha)}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\left(\frac{{\mathsf{p}}^{{\mathsf{k}}/2}}{\operatorname{\mathsf{N}}_{K}(\partial)^{1/2}}\right)^{2(1-\alpha)}\right)\\\
&=\frac{\theta^{\pm}(\tau,u)}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}\operatorname{\mathsf{N}}_{K}(\partial)^{1/2}}H_{\gamma}\left(1,\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{1-\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\right),\end{split}$

where $\gamma=\gamma(\tau,u)$ is the corresponding regular elliptic matrix.

######  Proposition 6.4.

Let $\alpha>0$ and let $\phi$ be a Schwartz function777For our purposes,
$\phi$ will either be the function $F$ or the function $H_{\gamma}$. on
$\mathbb{R}$. Then

$\theta^{\pm}(x,y)\phi(\mathcal{P}^{\pm}(x,y)^{2\alpha}),\qquad\theta^{\pm}(x,y)\mathcal{P}^{\pm}(x,y)\phi(\mathcal{P}^{\pm}(\tau,u)^{2(1-\alpha)})$

are smooth in $(x,y)$ and compactly supported.

###### Proof.

The proof is essentially that of [GKM+18, Theorem 4.1], also [Alt15,
Proposition 4.1], in the setting of $\operatorname{GL}(n,\mathbb{Q})$ and
$\operatorname{GL}(2,\mathbb{Q})$ respectively. Following the discussion of
[GKM+18, §4], the singularities of the archimedean orbital integral
$\theta^{\pm}$ in $\mathbb{R}^{2n-1}$ are precisely the zero locus of the
discriminant, which is given by the vanishing of $\mathcal{P}^{\pm}(\tau,u)$,
and are at most $\|(x,y)\|^{-\beta}$ for some $\beta\geq 0$. Then for any
singular point $a$ of $\theta^{\pm}$ and $\alpha>0$ we have
$\mathcal{P}^{\pm}(x,y)^{2\alpha}\to\infty$ as $v=(x,y)\to a$, and

$\phi(\mathcal{P}^{\pm}(x,y)^{2\alpha})\ll_{M}\mathcal{P}^{\pm}(x,y)^{-2\alpha
M}$

for any $M>0.$ In particular, taking $M>(\beta+1)/\alpha$ and arguing as in
the proof of [GKM+18, Theorem 4.1], we see that the product
$\theta^{\pm}(x,y)\phi(\mathcal{P}^{\pm}(\tau,u)^{2\alpha})$ and its partial
derivatives vanish as $(x,y)$ approaches $a$. The argument for the second
product
$\theta^{\pm}(x,y)\mathcal{P}^{\pm}(x,y)\phi(\mathcal{P}^{\pm}(x,y)^{2(1-\alpha)})$
follows in a similar manner.

Finally, at the archimedean places, we had imposed the condition on the test
functions that their orbital integrals must have compact support. This ensures
that the functions of interest are compactly supported. ∎

######  Remark 6.5.

The proof shows these functions are now defined in
$\mathbb{R}^{n}\times\mathbb{R}^{n-1}$ and not only in the regular set (i.e.,
union of the open regions $S_{J}$).

Upon choosing a precise value of $A$ in Theorem 5.10, it is possible to ensure
that $\Phi_{\pm}$ and $\Psi_{\pm}$ emerge in our manipulation of the regular
elliptic part.

######  Proposition 6.6.

For a regular elliptic class $\gamma$ and a parameter $0<\alpha<1$,

$\displaystyle L(1,\gamma)$ $\displaystyle=\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}{({\mathfrak{a}})}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\right)$
$\displaystyle+\frac{1}{\sqrt{\operatorname{\mathsf{N}}_{K}(\partial)}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{-1}}\sum_{{\mathfrak{a}}}\chi_{{\mathfrak{d}}}({\mathfrak{a}})H_{\gamma}\left(1,\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{1-\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\right),$

###### Proof.

This follows from the Approximate Functional Equation upon choosing

$A=\operatorname{\mathsf{N}}_{K}(\partial)^{\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}.\qed$

Using Proposition 6.6 and the definitions of $\Psi_{\pm}$ and $\Phi_{\pm}$, we
get

(16) $\begin{split}R(f)&={\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}\sum_{u\in{\mathcal{O}}_{K}^{*}}}\sum_{\tau\in
L(u)}\theta^{\pm}(\tau,u)\left(\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}{({\mathfrak{a}})}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\right)\right)\\\
&+\frac{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}{\operatorname{\mathsf{N}}_{K}(\partial)^{1/2}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}}\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\in
L(u)}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{-1}}\sum_{{\mathfrak{a}}}\chi_{{\mathfrak{d}}}({\mathfrak{a}})H_{\gamma}\left(1,\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{1-\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\right)\\\ &={\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\in
L(u)}\left(\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}{({\mathfrak{a}})}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}\Phi_{\pm}(\tau,u)+\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\sum_{{\mathfrak{a}}}\chi_{{\mathfrak{d}}}({\mathfrak{a}})\Psi_{\pm}(\tau,u)\right)\\\
&={\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\in
L(u)}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\left(\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})}\sum_{{\mathfrak{a}}}\frac{\chi_{{\mathfrak{d}}}{({\mathfrak{a}})}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}\Phi_{\pm}(\tau,u)+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\sum_{{\mathfrak{a}}}\chi_{{\mathfrak{d}}}({\mathfrak{a}})\Psi_{\pm}(\tau,u)\right)\\\
&={\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\in
L(u)}\sum_{{\mathfrak{d}}\mid
S_{\gamma}}\sum_{{\mathfrak{a}}}\chi_{{\mathfrak{d}}}({\mathfrak{a}})\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right).\end{split}$

##  7\. The Problem of Completion

In this section, we perform two main tasks. First, we replace the sum over
ideals ${\mathfrak{d}}\mid S_{\gamma}$ arising in the expression of the
regular elliptic part by a congruence condition. Second, we introduce a
modified Hilbert symbol which correctly detects the character
$\chi_{{\mathfrak{d}}}(-)$ in our case and clarifies the dependence on $\tau$.
These two steps will then allow us to ‘complete the lattice’ and set the stage
for Poisson summation.

###  7.1. The Congruence Conditions

So far we have dealt with the condition ${\mathfrak{d}}\mid S_{\gamma}$.
However, to perform Poisson summation over the variables $u,\tau$ we need to
be explicit regarding its role. We will now aim to replace this divisibility
condition by a congruence condition as was done by Altuğ. The main result of
this section is stated below.

Let ${\mathfrak{p}}_{1},\cdots,{\mathfrak{p}}_{n}$ be the primes above $2$ in
${\mathcal{O}}_{K}$, that is,

$2{\mathcal{O}}_{K}={\mathfrak{p}}_{1}\cdots{\mathfrak{p}}_{n}.$

We assume $\pi_{{\mathfrak{p}}_{i}}$ is a uniformizer.

######  Theorem 7.1.

Let $\gamma$ be a contributing conjugacy class. The following two conditions
are equivalent

  1. (i)

${\mathfrak{d}}\mid S_{\gamma}$,

  2. (ii)

${\mathfrak{d}}^{2}\mid\tau^{2}-4\det(\gamma)$ and for each $i=1,\cdots,n$ the
following congruence condition holds in
${\mathcal{O}}_{K_{{\mathfrak{p}}_{i}}}$,

$\frac{\tau^{2}-4\det(\gamma)}{\pi_{{\mathfrak{p}}_{i}}^{2\operatorname{val}_{{\mathfrak{p}}_{i}}({\mathfrak{d}})}}\equiv
0,1\pmod{\pi_{{\mathfrak{p}}_{i}}^{2}}.$

###### Proof.

This is a special case of [Esp23a, Proposition 9]. ∎

Using this result we have

$R(f)={\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\in
L(u)}\sum_{{\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon}^{{}^{\prime}}\sum_{{\mathfrak{a}}}\chi_{{\mathfrak{d}}}({\mathfrak{a}})\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right),$

where following Altuğ we use ′ to denote the congruence conditions given by
the previous theorem.

###  7.2. The Normalized Character

The primary goal of this section is to clarify the dependence of
$\chi_{{\mathfrak{d}}}$ on $\tau$.

######  Remark 7.2.

Over ${\mathbb{Q}}$, the ideals can be replaced by rational elements and the
Kronecker symbol satisfies

$\chi_{d}(a)=\binom{\frac{\tau^{2}-4p^{{\mathsf{k}}}}{d^{2}}}{a}_{K},$

making the following step unnecessary (see [Alt15, (4)]). However, we want to
provide a uniform proof and obtain an expression similar to the one above for
$\chi_{{\mathfrak{d}}}$.

#### 7.2.1.

We begin by recalling the definition of the Hilbert Symbol in our context.

######  Definition 7.3.

Let $K$ be a number field and let ${\mathfrak{q}}$ be a prime of $K$. The
_Hilbert Symbol_ of $K_{{\mathfrak{q}}}$ is defined in
$K_{{\mathfrak{q}}}^{*}\times K_{{\mathfrak{q}}}^{*}$ given by

$\binom{a,b}{{\mathfrak{q}}}_{H}=\left\\{\begin{array}[]{ll}1&\text{ if
}z^{2}=ax^{2}+by^{2}\text{ has solutions in }{\mathcal{O}}_{K_{q}}\\\
-1&\text{ otherwise}.\end{array}\right.$

Fix a uniformizer $\pi_{{\mathfrak{q}}}$ and define the _restricted Hilbert
Symbol_ as the function in ${\mathcal{O}}_{K_{{\mathfrak{q}}}}^{*}$ given by

$\binom{a}{{\mathfrak{q}}}_{rH}=\left\\{\begin{array}[]{ll}\binom{a,\pi_{{\mathfrak{q}}}}{{\mathfrak{q}}}_{\tiny{H}}&\text{
if }a\in{\mathcal{O}}_{K_{{\mathfrak{q}}}}^{*}\\\ 0&\text{
otherwise}\end{array}\right.$

and we extend it to $K_{{\mathfrak{q}}}$ as $0$ for all $a$ without
$\mathinner{\\!\left\lvert a\right\rvert}_{{\mathfrak{q}}}\neq 1$.

######  Remark 7.4.

For primes ${\mathfrak{q}}\nmid 2$, the restricted Hilbert Symbol coincides
with the quadratic character of ${\mathcal{O}}_{K_{{\mathfrak{q}}}}^{*}$,
i.e., the character whose value is $\pm 1$ according as $a$ is a square or
not. It is independent of the choice of the uniformizer
$\pi_{{\mathfrak{q}}}$, see [Lem10, chapter 4]. For the primes above $2$ this
is not the case but the above definition will allow us make explicit the
character $\chi_{{\mathfrak{d}}}$.

We record the following fact of the Hilbert Symbol that will be useful for us.
We include the proof of (ii) as we were not able to find it written in the
literature.

######  Lemma 7.5.

Let ${\mathfrak{q}}$ be a prime of $K$ and $a_{1},a_{2}$ elements of
${\mathcal{O}}_{K_{{\mathfrak{q}}}}^{*}$. Then

  1. (i)

If ${\mathfrak{q}}\nmid 2$ and $a_{1}\equiv a_{2}\pmod{\pi_{{\mathfrak{q}}}}$,
then

$\binom{a_{1},\pi_{{\mathfrak{q}}}}{{\mathfrak{q}}}_{H}=\binom{a_{2},\pi_{{\mathfrak{q}}}}{{\mathfrak{q}}}_{H}.$

  2. (ii)

If ${\mathfrak{q}}\mid 2$, has ramification degree $e_{{\mathfrak{q}}}$, and
$a_{1}\equiv a_{2}\pmod{\pi_{{\mathfrak{q}}}^{2e_{{\mathfrak{q}}}+1}}$, then

$\binom{a_{1},\pi_{{\mathfrak{q}}}}{{\mathfrak{q}}}_{H}=\binom{a_{2},\pi_{{\mathfrak{q}}}}{{\mathfrak{q}}}_{H}.$

###### Proof.

Both these assertions are applications of Hensel’s Lemma. We show (ii) and
leave (i) as an exercise for the reader. If ${\mathfrak{q}}$ ramifies, then
Hensel’s lemma asserts that $u\in{\mathcal{O}}_{K_{{\mathfrak{q}}}}^{*}$ is a
square if and only if $u$ is a square modulo
$\pi^{2e_{{\mathfrak{q}}}+1}_{{\mathfrak{q}}}$.

If $\binom{a_{1},\pi_{{\mathfrak{q}}}}{{\mathfrak{q}}}_{H}=1$ then there exist
$X_{0},Y_{0},Z_{0}$ (not all equal to 0) satisfying

$Z_{0}^{2}=a_{1}X_{0}^{2}+\pi_{{\mathfrak{q}}}Y_{0}^{2}.$

Using an infinite descent argument, we assume without loss of generality that
$Z_{0},X_{0}$ are in ${\mathcal{O}}_{K_{{\mathfrak{q}}}}^{*}$. In particular,

$a_{2}X_{0}^{2}+\pi_{{\mathfrak{q}}}Y_{0}^{2}\equiv
a_{1}X_{0}^{2}+\pi_{{\mathfrak{q}}}Y_{0}^{2}\equiv
Z_{0}^{2}\pmod{\pi_{{\mathfrak{q}}}^{2e_{{\mathfrak{q}}}+1}},$

where the first congruence follows from the assumptions. In view of Hensel’s
Lemma, we deduce that $a_{2}X_{0}^{2}+\pi_{{\mathfrak{q}}}Y_{0}^{2}$ is a
square; i.e., there exists $Z_{1}\in{\mathcal{O}}_{K_{{\mathfrak{q}}}}^{*}$
with

$Z_{1}^{2}=a_{2}X_{0}^{2}+\pi_{{\mathfrak{q}}}Y_{0}^{2}.$

Thus, the Hilbert symbol on the right is also $1$. Both sides play symmetric
roles so we deduce that both sides are either $\pm 1$. This concludes the
proof. ∎

#### 7.2.2.

Now, we define a symbol which allows us to recover the characters
$\chi_{{\mathfrak{d}}}$. It also clarifies the role of $\tau$ and $u$.

######  Definition 7.6.

Let ${\mathfrak{a}},\mathfrak{b}$ be any ideals of ${\mathcal{O}}_{K}$ and let
$\alpha$ be an element of $K$. For each prime ideal ${\mathfrak{q}}$ of $K$,
let $\pi_{{\mathfrak{q}}}$ be a uniformizer of $K_{{\mathfrak{q}}}$. Define
the _modified Hilbert symbol_ , with respect to the $\pi_{{\mathfrak{q}}}$, as

$\binom{\alpha,{\mathfrak{a}}}{\mathfrak{b}}=\prod_{{\mathfrak{q}}\mid\mathfrak{b}}\binom{\alpha\pi_{{\mathfrak{q}}}^{-2\operatorname{val}_{{\mathfrak{q}}}({\mathfrak{a}})}}{{\mathfrak{q}}}^{\operatorname{val}_{{\mathfrak{q}}}(\mathfrak{b})}_{rH}.$

######  Remark 7.7.

The modified Hilbert symbol doesn’t depend on the choice of the uniformizer
because from the definition of the Artin symbol

$\binom{\alpha,\pi_{{\mathfrak{q}}}}{{\mathfrak{q}}}\sqrt{\alpha}=\left(\pi_{{\mathfrak{q}}},K_{{\mathfrak{q}}}(\gamma)/K_{{\mathfrak{q}}}\right)_{A}\sqrt{\alpha}.$

and the Artin Symbol doesn’t depend on uniformizers.

Now we prove a crucial property of the modified symbol: it recovers the
quadratic Hecke character at the primes.

######  Lemma 7.8.

For $u\in{\mathcal{O}}_{K}^{*}$ and $\tau\in L(u)$, let $\gamma$ be the
regular elliptic class corresponding to the pair $(\tau,u)$. Let
${\mathfrak{q}}$ be any prime of ${\mathcal{O}}_{K}$. Then,

$\chi_{\gamma}({\mathfrak{q}})=\binom{\tau^{2}-4u\varepsilon,S_{\gamma}}{{\mathfrak{q}}}=\binom{(\tau^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2\operatorname{val}_{{\mathfrak{q}}}(S_{\gamma})}}{{\mathfrak{q}}}_{rH}.$

###### Proof.

First, consider the case that ${\mathfrak{q}}\nmid 2$. Since

$(\tau^{2}-4u\det(\gamma))=(\tau^{2}-4u\varepsilon)=S_{\gamma}^{2}\Delta_{\gamma},$

we deduce

$\operatorname{val}_{{\mathfrak{q}}}(\tau^{2}-4u\varepsilon)-2\operatorname{val}_{{\mathfrak{q}}}(S_{\gamma})=\operatorname{val}_{{\mathfrak{q}}}(\Delta_{\gamma}).$

The upper entry of the modified Hilbert symbol has ${\mathfrak{q}}$-valuation
equal to that of $\Delta_{\gamma}$.

Note that if ${\mathfrak{q}}$ is a ramified prime, then both sides of the
equality (we wish to prove) are $0$. If ${\mathfrak{q}}$ is not ramified, then
$\alpha\mathrel{\mathop{\mathchar
58\relax}}=(\tau^{2}-4u\varepsilon){\pi_{{\mathfrak{q}}}^{-2\operatorname{val}_{{\mathfrak{q}}}(S_{\gamma})}}$
is an element of ${\mathcal{O}}^{*}_{K_{{\mathfrak{q}}}}$. By Hensel’s Lemma
(applied to the polynomial $X^{2}-\alpha$) we know that $\alpha$ is a square
in ${\mathcal{O}}_{K_{\mathfrak{q}}}$ if and only if $\alpha$ is a square
modulo ${\mathfrak{q}}$. Recall that $\alpha$ is a square in
${\mathcal{O}}_{K_{\mathfrak{q}}}$ precisely when the extension
$K(\sqrt{\alpha})/K$ is either degree $1$ or $2$; i.e., ${\mathfrak{q}}$ is
either split or inert. On the other hand, the condition $\alpha$ is a square
(resp. non-square) modulo ${\mathfrak{q}}$ is equivalent to the symbol being
$1$ (resp. $-1$) by definition. This concludes the proof for odd primes.

Now, suppose that ${\mathfrak{q}}\mid 2$. If ${\mathfrak{q}}$ is ramified, the
proof is the same as before. So, we only consider the case that
${\mathfrak{q}}\mid 2$ and is unramified. It follows from the definition of
the Artin symbol that

$\binom{\alpha,\pi_{{\mathfrak{q}}}}{{\mathfrak{q}}}\sqrt{\alpha}=\left(\pi_{{\mathfrak{q}}},K_{{\mathfrak{q}}}(\gamma)/K_{{\mathfrak{q}}}\right)_{A}\sqrt{\alpha}.$

Since $\pi_{{\mathfrak{q}}}$ is a uniformizer, the Artin symbol
$\left(\pi_{{\mathfrak{q}}},K_{{\mathfrak{q}}}(\gamma)/K_{{\mathfrak{q}}}\right)_{A}$
is the Frobenius element of ${\mathfrak{q}}$. This takes the value $1$ or $-1$
according as ${\mathfrak{q}}$ is split or inert. This is precisely the
definition of $\chi_{\gamma}({\mathfrak{q}})$. Hence,

$\binom{\alpha,\pi_{{\mathfrak{q}}}}{{\mathfrak{q}}}=\chi_{\gamma}({\mathfrak{q}}).$

This completes the proof. ∎

######  Theorem 7.9.

Let $u$ be an element of ${\mathcal{O}}^{*}_{K}$ and let $\tau\in L(u)$.
Denote by $\gamma$ the regular elliptic class that corresponds to the pair
$(\tau,u)$. Let ${\mathfrak{d}}$ be an ideal of ${\mathcal{O}}_{K}$ satisfying
${\mathfrak{d}}\mid S_{\gamma}$ and let ${\mathfrak{a}}$ be any ideal of
${\mathcal{O}}_{K}$. Write

${\mathfrak{a}}=\prod{\mathfrak{q}}^{v_{{\mathfrak{q}}}},\;{\mathfrak{d}}=\prod{\mathfrak{q}}^{r_{{\mathfrak{q}}}},$

where $v_{{\mathfrak{q}}}$ or $r_{{\mathfrak{q}}}$ may be equal to 0. Then,

$\chi_{{\mathfrak{d}}}({\mathfrak{a}})=\binom{\tau^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}.$

###### Proof.

By definition

$\binom{\tau^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}=\prod_{{\mathfrak{q}}\mid{\mathfrak{a}}}\binom{(\tau^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r_{{\mathfrak{q}}}}}{{\mathfrak{q}}}^{v_{{\mathfrak{q}}}}_{rH}.$

As before, set ${\mathfrak{d}}^{\prime}=S_{\gamma}/{\mathfrak{d}}$. Recall
that the definition of $\chi_{{\mathfrak{d}}}$ is

$\chi_{{\mathfrak{d}}}=\left\\{\begin{array}[]{ll}\chi_{\gamma}({\mathfrak{a}})&\mbox{if
}({\mathfrak{a}},{\mathfrak{d}}^{\prime})=1\\\ 0&\mbox{if
}({\mathfrak{a}},{\mathfrak{d}}^{\prime})\neq 1.\end{array}\right.$

Hence, we divide the proof into two cases depending on
$({\mathfrak{a}},{\mathfrak{d}}^{\prime})$.

_Case 1:_ If $({\mathfrak{a}},{\mathfrak{d}}^{\prime})\neq 1$ then
$\chi_{{\mathfrak{d}}}({\mathfrak{a}})=0$ by definition and for some
${\mathfrak{q}}$ both $r_{{\mathfrak{q}}}$ and $v_{{\mathfrak{q}}}$ are non-
zero. Hence

$\operatorname{val}_{{\mathfrak{q}}}((\tau^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r_{\mathfrak{q}}})=\operatorname{val}_{{\mathfrak{q}}}({\mathfrak{d}}^{\prime}\Delta_{\gamma})\geq
1.$

Hence, $(\tau^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r_{\mathfrak{q}}}$ is
not a unit in ${\mathcal{O}}_{K_{{\mathfrak{q}}}}$; and since
$v_{{\mathfrak{q}}}\geq 1$

$\binom{(\tau^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r_{{\mathfrak{q}}}}}{{\mathfrak{q}}}^{v_{{\mathfrak{q}}}}_{rH}=0.$

By the definition of the modified Hilbert symbol, we have

$\binom{\tau^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}=0.$

_Case 2:_ Let us now suppose $({\mathfrak{a}},{\mathfrak{d}}^{\prime})=1$ and
pick a prime ${\mathfrak{q}}\mid{\mathfrak{a}}$. When ${\mathfrak{q}}$ is
ramified, i.e., ${\mathfrak{q}}\mid\Delta_{\gamma}$,

$\binom{\tau^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}=0.$

This is because

$\operatorname{val}_{{\mathfrak{q}}}((\tau^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r_{\mathfrak{q}}})=\operatorname{val}_{{\mathfrak{q}}}({\mathfrak{d}}^{\prime}\Delta_{\gamma})\geq
1.$

On the other hand, since a ramified prime divides ${\mathfrak{a}}$,

$\chi_{{\mathfrak{d}}}({\mathfrak{a}})=\chi_{\gamma}({\mathfrak{a}})=0.$

Thus we are left to consider the case in which all the primes that divide
${\mathfrak{a}}$ are not ramified. In this situation,
$(\tau^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r_{\mathfrak{q}}}$ is a unit
for each ${\mathfrak{q}}\mid{\mathfrak{a}}$. We have

$\operatorname{val}_{{\mathfrak{q}}}(S_{\gamma})=\operatorname{val}_{{\mathfrak{q}}}({\mathfrak{d}})+\operatorname{val}_{{\mathfrak{q}}}({\mathfrak{d}}^{\prime})=\operatorname{val}_{{\mathfrak{q}}}({\mathfrak{d}})=r_{\mathfrak{q}}.$

Hence,

$\displaystyle\binom{\tau^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}$
$\displaystyle=\prod_{{\mathfrak{q}}\mid{\mathfrak{a}}}\binom{(\tau^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r_{{\mathfrak{q}}}}}{{\mathfrak{q}}}_{rH}^{v_{{\mathfrak{q}}}}$
$\displaystyle=\prod_{{\mathfrak{q}}\mid{\mathfrak{a}}}\binom{(\tau^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2\operatorname{val}_{{\mathfrak{q}}}(S_{\gamma})}}{{\mathfrak{q}}}_{rH}^{v_{{\mathfrak{q}}}}$
$\displaystyle=\prod_{{\mathfrak{q}}\mid{\mathfrak{a}}}\chi_{\gamma}({\mathfrak{q}})^{v_{{\mathfrak{q}}}}\quad\text{by
Lemma}~{}\ref{lemma: to check}$
$\displaystyle=\chi_{\gamma}({\mathfrak{a}}).\qed$

Theorem 7.9 together with the congruence conditions from §7.1 will allow us to
clarify how every term of the trace formula depends on $\tau$. We now prove
that the modified Hilbert symbol is periodic.

######  Lemma 7.10.

Let $\tau_{1}\equiv\tau_{2}\pmod{4{\mathfrak{a}}{\mathfrak{d}}^{2}}$ in
${\mathcal{O}}_{K}$. Then

$\binom{\tau_{1}^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}=\binom{\tau_{2}^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}.$

###### Proof.

Let ${\mathfrak{q}}\mid{\mathfrak{a}}$ and set
$r=\operatorname{val}_{{\mathfrak{q}}}({\mathfrak{d}})$. Let
$\pi_{{\mathfrak{q}}}$ be a uniformizer of $K_{\mathfrak{q}}$. By definition
of the modified Hilbert symbol, it is enough to prove

$\binom{(\tau_{1}^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r}}{{\mathfrak{q}}}=\binom{(\tau_{2}^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r}}{{\mathfrak{q}}}.$

Notice that $\pi_{{\mathfrak{q}}}^{2r}\mid\tau_{1}^{2}-\tau_{2}^{2}$. Hence
$(\tau_{1}^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r}$ is an integer if and
only if $(\tau_{2}^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r}$ is an
integer. Since ${\mathfrak{a}}\mid\tau_{1}-\tau_{2}$, we get an extra factor
$\pi_{{\mathfrak{q}}}$ than those than come from ${\mathfrak{d}}^{2}$. Whence

$\pi_{{\mathfrak{q}}}^{2r+1}\mid\tau_{1}^{2}-\tau_{2}^{2}=(\tau_{1}^{2}-4u\varepsilon)-(\tau_{2}^{2}-4u\varepsilon).$

So, in case both ‘numerators’ of the symbols are integers, we have

$(\tau_{1}^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r}\equiv(\tau_{1}^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r}\pmod{\pi_{{\mathfrak{q}}}}.$

Lemma 7.5(i) asserts that if ${\mathfrak{q}}\nmid 2$, then

$\binom{(\tau_{1}^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r}}{{\mathfrak{q}}}=\binom{(\tau_{2}^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r}}{{\mathfrak{q}}}.$

On the other hand, if ${\mathfrak{q}}\mid 2$, then

$\operatorname{val}_{{\mathfrak{q}}}(4{\mathfrak{a}}{\mathfrak{d}}^{2})=2e_{{\mathfrak{q}}}+\operatorname{val}_{{\mathfrak{q}}}({\mathfrak{a}})+r\geq
2e_{{\mathfrak{q}}}+1,$

where $e_{{\mathfrak{q}}}$ is the ramification of ${\mathfrak{q}}$ above $2$.
Notice equality can occur if $r=0$ and
$\operatorname{val}_{{\mathfrak{q}}}({\mathfrak{a}})=1$. Hence

$\pi_{{\mathfrak{q}}}^{2e_{{\mathfrak{q}}}+1}\mid(\tau_{1}^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r}-(\tau_{2}^{2}-4u\varepsilon)\pi_{{\mathfrak{q}}}^{-2r}.$

By Lemma 7.5(ii) we conclude the proof. ∎

Next, we prove the periodicity of the congruence conditions.

######  Lemma 7.11.

Let ${\mathfrak{a}}$ and ${\mathfrak{d}}$ be ideals in ${\mathcal{O}}_{K}$. If
$\tau_{1}\equiv\tau_{2}\pmod{4{\mathfrak{a}}{\mathfrak{d}}^{2}}$ then

${\mathfrak{d}}^{2}\mid\tau_{1}^{2}-4u\varepsilon\Longleftrightarrow{\mathfrak{d}}^{2}\mid\tau_{2}^{2}-4u\varepsilon.$

Moreover, if ${\mathfrak{d}}^{2}\mid\tau_{i}^{2}-4u\varepsilon$, then

$\frac{\tau_{1}^{2}-4u\varepsilon}{{\mathfrak{d}}^{2}}\equiv\frac{\tau_{2}^{2}-4u\varepsilon}{{\mathfrak{d}}^{2}}\pmod{4}.$

###### Proof.

Recall that $4{\mathfrak{d}}^{2}\mid\tau_{1}^{2}-\tau_{2}^{2}$. Therefore

$4{\mathfrak{d}}^{2}\mid(\tau_{1}^{2}-4u\varepsilon)-(\tau_{2}^{2}-4u\varepsilon).$

The result now follows. ∎

###  7.3. Completion of the Lattice

Using the modified Hilbert symbol, we can rewrite (16) as

$R(f)=\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\in
L(u)}\sum_{{\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon}^{{}^{\prime}}\sum_{{\mathfrak{a}}}\binom{\tau^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right).$

Notice that the expression

$\binom{\tau^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right)$

makes sense for any value of $\tau,u,{\mathfrak{d}},{\mathfrak{a}}$ regardless
of whether there is a regular elliptic matrix $\gamma$ (corresponding to
$\tau$ and $u$) whose quadratic character produces the divisors
${\mathfrak{d}}\mid S_{\gamma}$. In particular, we can consider

$\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\notin
L(u)}\sum_{{\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon}^{{}^{\prime}}\sum_{{\mathfrak{a}}}\binom{\tau^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right).$

Introducing the congruence condition in place of the condition
${\mathfrak{d}}\mid S_{\gamma}$ allowed us to replace the ${\mathfrak{d}}$-sum
in (16) with a sum whose condition depends on $\tau$ and $u$ (but not on
whether there is associated to such ${\mathfrak{d}}$ a quadratic extension
which determines $S_{\gamma}$).

######  Definition 7.12.

Define

$\displaystyle\Sigma(\square)$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\notin
L(u)}\sum_{{\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon}^{{}^{\prime}}\sum_{{\mathfrak{a}}}\binom{\tau^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right).$

For this definition makes sense, we need to prove the following theorem.

######  Theorem 7.13.

The series defining $\Sigma(\square)$ is an absolutely convergent sum.

###### Proof.

The absolute value of the character is bounded by $1$. Thus, what we must
prove is that

$\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\notin
L(u)}\sum_{{\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon}^{{}^{\prime}}\sum_{{\mathfrak{a}}}\frac{\mathinner{\\!\left\lvert\Phi_{\pm}(\tau,u)\right\rvert}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}=\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\notin
L(u)}\mathinner{\\!\left\lvert\theta^{\pm}(\tau,u)\right\rvert}\sum_{{\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon}^{{}^{\prime}}\sum_{{\mathfrak{a}}}\frac{\mathinner{\\!\left\lvert
F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\right)\right\rvert}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}$

and

$\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\notin
L(u)}\sum_{{\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon}^{{}^{\prime}}\sum_{{\mathfrak{a}}}\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\mathinner{\\!\left\lvert\Psi_{\pm}(\tau,u)\right\rvert}=\\\
\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\notin
L(u)}\mathinner{\\!\left\lvert\frac{\theta^{\pm}(\tau,u)}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}\operatorname{\mathsf{N}}_{K}(\partial)^{1/2}}\right\rvert}\sum_{{\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon}^{{}^{\prime}}\sum_{{\mathfrak{a}}}\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\mathinner{\\!\left\lvert
H_{\gamma}\left(1,\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{1-\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\right)\right\rvert},$

are both convergent. If we prove the inner sums over ${\mathfrak{d}}$ and
${\mathfrak{a}}$ are absolutely convergent we are done, because the outer sums
over $u$ and $\tau$ are finite (since $\theta^{\pm}$ are of compact support).
Thus they contribute a finite nonzero terms.

Let us study the interior sums separately. We know that for
${\mathfrak{d}},{\mathfrak{a}}$ large enough

$\mathinner{\\!\left\lvert
F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\right)\right\rvert}\ll\exp\left(-\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\right)\ll\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\right)^{-1}.$

Thus

$\sum_{{\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon}^{{}^{\prime}}\sum_{{\mathfrak{a}}}\frac{\mathinner{\\!\left\lvert
F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\right)\right\rvert}}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}\ll\sum_{{\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon}^{{}^{\prime}}\sum_{{\mathfrak{a}}}\dfrac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{3}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{2}}<\infty.$

Using Lemma 5.15, we have

$\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\mathinner{\\!\left\lvert
H_{\gamma}\left(1,\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{1-\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\right)\right\rvert}\ll\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})}{\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{1-\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}}\exp\left(-\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}{\operatorname{\mathsf{N}}_{K}(\partial)^{1-\alpha}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\right)^{1/n+1}\right)$

For large enough
$\operatorname{\mathsf{N}}_{K}({\mathfrak{a}}),\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})$
the above expression can be bounded by

$\ll\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})\left(\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})\right)^{1/n+1}},$

where the implied constant only depends on $\tau,u$, and by the finiteness of
their contribution, it can be taken to be absolute. This completes the proof.
∎

######  Remark 7.14.

The proof works exactly the same way if we instead use $\tau\in L(u)$, even
though we already know that that sum converges absolutely by the approximate
functional equation (since it comes from the Zagier Zeta Function which is
entire). We cannot use the approximate functional equation directly for the
cases of $\Sigma(\square)$ because the character in that case returns the zeta
function, and thus has poles itself.

######  Definition 7.15.

We define the _completed regular elliptic part_ of the trace formula as

(17) $\overline{R}(f)={\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{\tau\in{\mathcal{O}}_{K}}\sum_{{\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon}^{{}^{\prime}}\sum_{{\mathfrak{a}}}\binom{\tau^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right)=R(f)+\Sigma(\square).$

######  Remark 7.16.

The name comes from the fact that we have added those
$\tau\in{\mathcal{O}}_{K}$ that were missing from the sum. We have completed
the lattice of integers.

In the following result we show that the order of summation in the completed
regular elliptic part can be swapped.

######  Proposition 7.17.

The order of summation in the completed regular elliptic part can be swapped.
More precisely,

$\overline{R}(f)={\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\sum_{u\in{\mathcal{O}}_{K}^{*}}\sum_{{\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon}^{{}^{\prime}}\binom{\tau^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right).$

###### Proof.

We have defined $\overline{R}(f)$ as the sum of two absolutely convergent
series. Thus we can arrange the series in any alternative order without
altering the convergence properties or values. ∎

######  Remark 7.18.

In the expression defining $\overline{R}(f)$, we see that $\tau$ is a free
variable and ${\mathfrak{d}}$ runs over the possible ideals such that
${\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon$ and satisfies certain
congruence conditions. After changing the order of summation, ${\mathfrak{d}}$
is a free variable and the innermost sum is now over the $\tau$ such that
$\tau^{2}-4u\varepsilon$ has ${\mathfrak{d}}^{2}$ as divisor, with the
appropriate congruence conditions. Both sums look the same but it is pertinent
to note that they have different free and dependent variables.

##  8\. Poisson Summation

###  8.1. Explicitly writing the lattice

Recall from §4.2 that we fixed the fundamental units
$\beta_{1},\cdots,\beta_{n-1}$ in ${\mathcal{O}}_{K}^{*}$ (which are embedded
into $W$ under the function $S$) satisfying (Hyp-pos). Any unit
$u\in{\mathcal{O}}_{K}^{*}$ can be written as

$u=\pm\beta_{1}^{y_{1}}\cdots\beta_{n-1}^{y_{n-1}}.$

In particular, $u=u(\pm,y)$ where
$y=(y_{1},\ldots,y_{n-1})\in{\mathbb{Z}}^{n-1}$.

######  Notation.

In what follows, we write $u$ in place of $u(\pm,y)$ for a unit in
${\mathcal{O}}_{K}^{*}$.

The expression obtained in Proposition 7.17 can be rewritten as

(18) $\overline{R}(f)={\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{\pm}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\sum_{y\in\mathbb{Z}^{n-1}}\sum_{{\mathfrak{d}}^{2}\mid\tau^{2}-4u\varepsilon}^{{}^{\prime}}\binom{\tau^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right).$

Given ideals ${\mathfrak{d}}$ and ${\mathfrak{a}}$ in ${\mathcal{O}}_{K}$, the
periodicity of the congruence conditions proved in Lemma 7.11 implies that for
a given congruence class modulo $4{\mathfrak{a}}{\mathfrak{d}}^{2}$, either
every term of the arithmetic progression appears in the sum or no term
appears. Therefore, the expression (18) can be rewritten as

${\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{\pm}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\sum_{y\in\mathbb{Z}^{n-1}}\sum_{\mu\in{\mathcal{O}}_{K}/4{\mathfrak{a}}{\mathfrak{d}}^{2}}^{{}^{\prime}}\sum_{\tau\equiv\mu}\binom{\tau^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right).$

By Lemma 7.11 we know that the symbol is periodic so the above expression can
be further rewritten as

${\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{\pm}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\sum_{y\in\mathbb{Z}^{n-1}}\sum_{\mu\in{\mathcal{O}}_{K}/4{\mathfrak{a}}{\mathfrak{d}}^{2}}^{{}^{\prime}}\binom{\mu^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\sum_{\tau\equiv\mu}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right).$

Within the lattice $\mathbb{Z}^{n-1}$ there is the sub-lattice
$(2\mathbb{Z})^{n-1}$ and the quotient is $\mathbb{F}_{2}^{n-1}$, where
$\mathbb{F}_{2}$ is the field of two elements. Let $Y$ be a variable that runs
over this quotient. We can rewrite the above expression as

${\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{\pm}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{y\equiv
Y}\sum_{\mu\in{\mathcal{O}}_{K}/4{\mathfrak{a}}{\mathfrak{d}}^{2}}^{{}^{\prime}}\binom{\mu^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\sum_{\tau\equiv\mu}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right).$

Fix the sign $\pm$ (at the start) and a vector
$Y=(Y_{1},\cdots,Y_{n-1})\in\mathbb{F}_{2}^{n-1}$. If $y\in{\mathbb{Z}}^{n-1}$
satisfies $y\equiv Y$, there exists another unit $v\in{\mathcal{O}}_{K}^{*}$
such that $u=u(y)=u(Y)v^{2}$ with

$u(Y)=\beta_{1}^{Y_{1}}\cdots\beta_{n-1}^{Y_{n-1}}$

once the sign is fixed and each $Y_{i}\in\\{0,1\\}$. This unit $v$ is uniquely
determined by $Y$ and $y$, if we impose the extra condition that it lies in
$W$, which is to say, that the two units have the same sign. The condition
$y\equiv Y$ implies the existence of
$y^{\prime}=(y^{\prime}_{1},\cdots,y^{\prime}_{n-1})\in{\mathbb{Z}}^{n-1}$
such that for each $1\leq t\leq n-1$,

$y_{t}=Y_{t}+2y^{\prime}_{t}.$

This vector $y^{\prime}\in{\mathbb{Z}}^{n-1}$ is in one-to-one correspondence
with $v=\beta_{1}^{y^{\prime}_{1}}\cdots\beta_{n-1}^{y^{\prime}_{n-1}}$. In
other words, the unit $v$ can be associated with an $(n-1)$-dimensional vector
given by

$\frac{1}{2}\left(y_{1}-Y_{1},\cdots,y_{n-1}-Y_{n-1}\right).$

We can then rewrite the completed regular elliptic part as

${\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{\pm}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{y\equiv
Y}\sum_{\mu\in{\mathcal{O}}_{K}/4{\mathfrak{a}}{\mathfrak{d}}^{2}}^{{}^{\prime}}\binom{\mu^{2}-4{u(Y)}v^{2}\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\sum_{\tau\equiv\mu}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right).$

Since

$\binom{\mu^{2}-4{u(Y)}v^{2}\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}=\binom{v^{2}(\mu
v^{-1})^{2}-4{u(Y)}v^{2}\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}=\binom{(\mu
v^{-1})^{2}-4{u(Y)}\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}},$

the completed regular elliptic part is

${\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{\pm}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{y\equiv
Y}\sum_{\mu\in{\mathcal{O}}_{K}/4{\mathfrak{a}}{\mathfrak{d}}^{2}}^{{}^{\prime}}\binom{(\mu
v^{-1})^{2}-4{u(Y)}\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\sum_{\tau\equiv\mu}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right).$

Since $v$ is a unit, multiplication by it only permutes the classes modulo
$4{\mathfrak{a}}{\mathfrak{d}}^{2}$. Relabeling $\mu v^{-1}$ as $\mu$, we may
remove $v^{-1}$ from the symbol at the expense of changing the inner sum to
$\tau\equiv\mu v$. Therefore,

$\overline{R}(f)={\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{\pm}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{y\equiv
Y}\sum_{\mu\in{\mathcal{O}}_{K}/4{\mathfrak{a}}{\mathfrak{d}}^{2}}^{{}^{\prime}}\binom{\mu^{2}-4{u(Y)}\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\sum_{\tau\equiv\mu
v}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right).$

We can record the main theorem of this section.

######  Theorem 8.1.

The completed regular elliptic part can be rewritten as

$\overline{R}(f)={\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}}\sum_{\pm}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{\mu\in{\mathcal{O}}_{K}/4{\mathfrak{a}}{\mathfrak{d}}^{2}}^{{}^{\prime}}\binom{\mu^{2}-4{u(Y)}\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\sum_{y\equiv
Y}\sum_{\tau\equiv\mu
v}\left(\frac{\Phi_{\pm}(\tau,u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}(\tau,u)\right).$

In particular, the symbol does not detect the sum $y\equiv Y$.

######  Remark 8.2.

When $K={\mathbb{Q}}$, the unit group is a finite set - there are two units,
namely, the two roots of unity $\pm 1$. Hence, the sums over $Y$ and $y$ are
both empty. The ideals correspond to the integers in this case. For each
ideal, we always pick the positive representative, and we pick $p$ to be the
representative of ${\mathfrak{p}}$. Therefore, Theorem 8.1 asserts that the
completed regular elliptic part of the trace formula can be written as

$\sum_{\pm}\sum_{d}\sum_{a}\sum_{\mu\mod{4ad^{2}}}\binom{(\mu^{2}\pm
4p^{k})/d^{2}}{a}\sum_{\tau\equiv\mu}^{{}^{\prime}}\left(\frac{\Phi_{u}(\tau)}{da}+d\Psi_{u}(\tau)\right).$

This is exactly (changing our $a$ for his $l$, our $d$ for his $f$, our $\mu$
for his $a$ and our $\tau$ for his $m$) what appears in the proof of [Alt15,
Theorem 4.2], before performing Poisson summation.

We must study the indexing set of the two inner sums appearing in the
expression in Theorem 8.1 and rewrite them as a lattice. The following lemma
will ensure that the indexing sets form an honest lattice inside
${\mathbb{R}}^{2n-1}$ which will then allow us to perform the Poisson
summation in a future step.

######  Lemma 8.3.

Fix a set of totally positive fundamental units $\beta_{1},\cdots,\beta_{n-1}$
of ${\mathcal{O}}_{K}^{*}$, a choice of sign, and a vector
$Y=(Y_{1},\cdots,Y_{n-1})\in\mathbb{F}_{2}^{n-1}$. Let $u=u(\pm,Y)$ be the
unit corresponding to this choice. Fix an integral basis
$\omega_{1},\ldots,\omega_{n}$ for the ideal
$4{\mathfrak{a}}{\mathfrak{d}}^{2}$. Finally, fix a representative $\mu$ of
some class of ${\mathcal{O}}_{K}/4{\mathfrak{a}}{\mathfrak{d}}^{2}$.

For each embedding $\sigma_{i}\mathrel{\mathop{\mathchar
58\relax}}K\hookrightarrow\mathbb{R}$, set $A_{i}=\sigma_{i}(u\mu)$ and
$\omega_{j,i}=\sigma_{i}(\omega_{j})$. Then the pairs $(\tau,u(y))$ that
appear in Theorem 8.1 are those with coordinates

$\left(A_{1}\beta_{1,1}^{y_{1}}\cdots\beta_{n-1,1}^{y_{n}-1}+x_{1}\omega_{1,1}+\cdots+x_{n}\omega_{n,1},\ldots,A_{n}\beta_{1,n}^{y_{1}}\cdots\beta_{n-1,n}^{y_{n}-1}+x_{1}\omega_{1,n}+\cdots+x_{n}\omega_{n,n},Y_{1}+2y_{1},\cdots,Y_{n-1}+2y_{n-1}\right),$

for some $x=(x_{1},\cdots,x_{n})\in{\mathbb{Z}}^{n}$ and
$(y_{1},\ldots,y_{n-1})\in{\mathbb{Z}}^{n-1}$.

###### Proof.

The condition $y\equiv Y$ implies the existence of
$y^{\prime}=(y^{\prime}_{1},\cdots,y^{\prime}_{n-1})\in{\mathbb{Z}}^{n-1}$
such that for each $1\leq t\leq n-1$,

$y_{t}=Y_{t}+2y^{\prime}_{t}.$

This vector $y^{\prime}\in{\mathbb{Z}}^{n-1}$ is in one-to-one correspondence
with $v=\beta_{1}^{y^{\prime}_{1}}\cdots\beta_{n-1}^{y^{\prime}_{n-1}}$.
Therefore, for $\mu\in{\mathcal{O}}_{K}/4{\mathfrak{a}}{\mathfrak{d}}^{2}$,
the equivalence $\tau\equiv\mu v$ can be rewritten (component-wise) as

$\tau_{i}=\sigma_{i}(\beta_{1}^{y^{\prime}_{1}}\cdot\ldots\cdot\beta_{n-1}^{y^{\prime}_{n-1}}\mu)+z_{i},$

for some integer $z_{i}$, where $1\leq i\leq n$. The result now follows (by
relabelling $y_{t}^{\prime}$ by $y_{t}$). ∎

In view of the above discussion, we make the following definition.

######  Definition 8.4.

Keep the notation introduced above except now suppose that
$(x,y)\in{\mathbb{R}}^{n}\times{\mathbb{R}}^{n-1}$. Define

$\displaystyle C_{i}(y)$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\beta_{1,i}^{y_{1}}\cdots\beta_{n-1,i}^{y_{n-1}}\textrm{ and }$
$\displaystyle A(x,y)$ $\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\left(A_{1}C_{1}(y)+x_{1}\omega_{1,1}+\cdots+x_{n}\omega_{n,1},\ldots,A_{n}C_{n}(y)+\cdots+x_{n}\omega_{n,n},Y_{1}+2y_{1},\cdots,Y_{n-1}+2y_{n-1}\right).$

######  Proposition 8.5 (Change of Variables).

Keep the notation introduced above and let
$(x,y)\in{\mathbb{R}}^{n}\times{\mathbb{R}}^{n-1}$. Then

  1. (i)

$A(x,y)$ is an invertible function.

  2. (ii)

The absolute value of the determinant of the Jacobian matrix of $A(x,y)$ is
$2^{3n-1}N({\mathfrak{a}})N({\mathfrak{d}})^{2}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}$.

###### Proof.

  1. (i)

One can write explicitly the inverse of this function. This is left as an
exercise for the reader.

  2. (ii)

By computing partial derivatives the reader can check that the Jacobian matrix
has the form

$\begin{pmatrix}\omega_{1,1}&\cdots&\omega_{n,1}&*&\cdots&*\\\
\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\\
\omega_{n,1}&\cdots&\omega_{n,n}&*&\cdots&*\\\ 0&\cdots&0&2&\cdots&0\\\
0&\cdots&0&\vdots&\ddots&\vdots\\\ 0&\cdots&0&0&\cdots&2\\\ \end{pmatrix}.$

Thus, the determinant of the above matrix is

$2^{n-1}\det\begin{pmatrix}\omega_{1,1}&\cdots&\omega_{n,1}\\\
\vdots&\ddots&\vdots\\\ \omega_{n,1}&\cdots&\omega_{n,n}\\\ \end{pmatrix}$

By definition, the determinant of the $n\times n$ matrix is
$N(4{\mathfrak{a}}{\mathfrak{d}}^{2})\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}=4^{n}N({\mathfrak{a}})N({\mathfrak{d}})^{2}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}$. ∎

######  Notation.

The transformation $A$ depends on ${\mathfrak{a}}$, ${\mathfrak{d}}$, $\pm$,
$Y$, and $\mu$ but we omit this from the notation.

We now rewrite the inner two sums of the completed regular elliptic part.

######  Proposition 8.6.

The inner two sums of the expression in Theorem 8.1 can be rewritten as

$\sum_{y\in\mathbb{Z}^{n-1}}\sum_{x\in\mathbb{Z}^{n}}\left(\frac{\Phi_{\pm}\left(A(x,y)\right)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}\left(A(x,y)\right)\right).$

###  8.2. Performing Poisson Summation.

For $(t,s)\in\mathbb{R}^{n}\times\mathbb{R}^{n-1}$, let $G(t,s)$ be a smooth
function with compact support. Set $\widehat{G}$ to denote the _Fourier
transform_ of $G$, i.e.,

$\widehat{G}(\xi,\eta)\mathrel{\mathop{\mathchar
58\relax}}=\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n-1}}G(t,s)\exp(-t\cdot\xi-s\cdot\eta)dsdt,$

where $(\xi,\eta)\in{\mathbb{Z}}^{n}\times{\mathbb{Z}}^{n-1}$. Then, Poisson
summation is the following assertion

(19)
$\sum_{(t,s)\in{\mathbb{Z}}^{n}\times{\mathbb{Z}}^{n-1}}G(t,s)=\sum_{(\xi,\eta)\in{\mathbb{Z}}^{n}\times{\mathbb{Z}}^{n-1}}\widehat{G}(\xi,\eta).$

The following important result essentially follows from unraveling the
definitions.

######  Proposition 8.7.

Let $A$ be the transformation from Definition 8.4. Let $G(x,y)=H(A(x,y))$ for
some smooth function $H$ with compact support, then

$\widehat{G}(\xi,\eta)=\frac{1}{2^{3n-1}N({\mathfrak{a}})N({\mathfrak{d}})^{2}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}}\int\int H(t,s)\exp(-(\xi,\eta)\cdot
A^{-1}(t,s))dsdt.$

###### Proof.

This follows from definition of Fourier transform and the change of variable
formula using Proposition 8.5. ∎

######  Definition 8.8.

For the functions $\Phi_{\pm}$ and $\Psi_{\pm}$ introduced in (14) and (15),
define the _scaled Fourier transforms_

$\displaystyle\Phi_{\pm}^{1}(\xi,\eta)$
$\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\int\int\Phi_{\pm}(t,s)\exp(-(\xi,\eta)\cdot A^{-1}(t,s))dsdt,$
$\displaystyle\Psi_{\pm}^{1}(\xi,\eta)$
$\displaystyle\mathrel{\mathop{\mathchar
58\relax}}=\int\int\Psi_{\pm}(t,s)\exp(-(\xi,\eta)\cdot A^{-1}(t,s))dsdt.$

######  Remark 8.9.

We worked with these scaled Fourier transforms because it allowed us to keep
track of the constants after Poisson summation more easily.

Applying Poisson summation, we get the following result.

######  Theorem 8.10.

The completed regular elliptic part is

$\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{\mu\in{\mathcal{O}}_{K}/4{\mathfrak{a}}{\mathfrak{d}}^{2}}^{{}^{\prime}}\binom{\mu^{2}-4{u(Y)}\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\sum_{\xi\in\mathbb{Z}^{n}}\sum_{\eta\in\mathbb{Z}^{n-1}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}}\left(\frac{\Phi_{\pm}^{1}(\xi,\eta)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}^{1}(\xi,\eta)\right).$

###### Proof.

This is a direct consequence of Propositions 8.6 and 8.7. ∎

Our analysis will now deviate from [Alt15]. If we were to proceed in grouping
terms as in _op. cit._ , we would have to analyze the exponential term that
appears inside the integrals. Our approach will avoid this step and we will
focus only on the sum obtained when $(\xi,\eta)=0$.

######  Notation.

We write $\overline{\Sigma_{0}}(f)$ to denote the expression obtained in
Theorem 8.10 with $\xi=\eta=0$.

Note that the term

$\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}}\left(\frac{\Phi_{\pm}^{1}(0,0)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})}+\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})\Psi_{\pm}^{1}(0,0)\right)$

is independent of the value of $Y$ and $\mu$. We can therefore reorder and
simplify the expression.

######  Proposition 8.11.

With notation as above

$\overline{\Sigma_{0}}(f)=\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\left(\frac{\Phi_{\pm}^{1}(0,0)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{3}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{2}}+\frac{\Psi_{\pm}^{1}(0,0)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})}\right)\left(\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{\mu\in{\mathcal{O}}_{K}/4{\mathfrak{a}}{\mathfrak{d}}^{2}}^{{}^{\prime}}\binom{\mu^{2}-4{u(Y)}\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}\right).$

To further simplify notation, we introduce the following definition.

######  Definition 8.12.

Given ideals ${\mathfrak{a}},{\mathfrak{d}}$ of ${\mathcal{O}}_{K}$ and a unit
$u\in{\mathcal{O}}_{K}^{*}$, define the _Kloosterman-type sums_ as

$K_{{\mathfrak{a}},{\mathfrak{d}}}(u)\mathrel{\mathop{\mathchar
58\relax}}=\sum_{\mu\in{\mathcal{O}}_{K}/4{\mathfrak{a}}{\mathfrak{d}}^{2}}^{{}^{\prime}}\binom{\mu^{2}-4u\varepsilon,{\mathfrak{d}}}{{\mathfrak{a}}}.$

######  Remark 8.13.

The definition of the Kloosterman-type sum does not require any hypothesis on
$K$.

Proposition 8.11 can now be rewritten as follows.

######  Theorem 8.14.

With notation as before,

$\overline{\Sigma_{0}}(f)=\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\left(\frac{\Phi_{\pm}^{1}(0,0)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{3}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{2}}+\frac{\Psi_{\pm}^{1}(0,0)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})}\right)K_{{\mathfrak{a}},{\mathfrak{d}}}(\pm
u(Y)).$

Our goal in the following section is to evaluate these sums precisely. We will
see that the sum over $Y$ is in fact straightforward. It is the sum over $\mu$
which presents challenges.

##  9\. Evaluation of Kloosterman-type Sums

The evaluation of the Kloosterman-type sums is central to the paper, but the
calculations are long and technical. For ease of following the argument, we
omit the calculations from this section and have deferred them to Appendix A.
Here, we only state the main result.

Throughout this section we invoke the following hypothesis.

(Hyp-split) Let $K$ be a number field such that 2 splits completely in $K$.

Let ${\mathfrak{p}}_{1},\cdots,{\mathfrak{p}}_{n}$ be the primes above $2$ in
${\mathcal{O}}_{K}$, that is,

$2{\mathcal{O}}_{K}={\mathfrak{p}}_{1}\cdots{\mathfrak{p}}_{n}.$

By assumption, $K_{{\mathfrak{p}}_{i}}\cong{\mathbb{Q}}_{2}$. We assume
$\pi_{{\mathfrak{p}}_{i}}$ corresponds to $2$ under the above isomorphism.
This fixes the uniformizer uniquely as there is only one field isomorphism
$\mathbb{Q}\longrightarrow\mathbb{Q}$.

######  Remark 9.1.

The splitting condition at the prime 2 is a technical hypothesis we need to
impose to evaluate the Kloosterman-type sums. We believe that the main result
of this section should be true in general. We note that in this section we do
not require the hypothesis that $K$ is totally real.

######  Definition 9.2.

Let $u\in{\mathcal{O}}_{K}^{*}$ be a unit and let ${\mathfrak{d}}$ and
${\mathfrak{a}}$ be arbitrary ideals of ${\mathcal{O}}_{K}$. For a complex
parameter $z=\sigma+it$ with $\sigma>1$, define

${\mathbb{D}}(z,u)\mathrel{\mathop{\mathchar
58\relax}}=\sum_{{\mathfrak{d}}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2z+1}}\sum_{{\mathfrak{a}}}\frac{K_{{\mathfrak{a}},{\mathfrak{d}}}(u)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{z+1}}.$

The main result of this section is the following theorem. We provide a proof
of this result in Appendix A.

######  Theorem 9.3.

Let $K$ be a number field satisfying (Hyp-split) and let
$({\mathfrak{p}},{\mathsf{k}})$ be the pair of prime ideal and corresponding
positive integer satisfying (Hyp-div) fixed in §3.1. Write
$\mathsf{p}=\operatorname{\mathsf{N}}_{K}({\mathfrak{p}})$. Let
$u\in{\mathcal{O}}_{K}^{*}$ be a unit. For a complex parameter $z$ with
$\Re(z)>1$,

${\mathbb{D}}(z,u)=4^{n}\frac{\zeta_{K}(2z)}{\zeta_{K}(z+1)}\cdot\frac{1-1/{\mathsf{p}}^{z({\mathsf{k}}+1)}}{1-1/{\mathsf{p}}^{z}}.$

In particular, ${\mathbb{D}}(z,u)$ is independent of $u$. Moreover,
${\mathbb{D}}(z,u)$ admits an analytic continuation to a meromorphic function
in the whole complex plane with poles at $z=0$ and $z=\frac{1}{2}$.

##  10\. Isolation of the Contribution of the Trivial and Special
Representations

In this section of the paper, the goal is to isolate the contribution of the
special representation and finish the proof of the main theorem. Recall from
Theorem 8.14 that the dominant term is,

(20)
$\overline{\Sigma_{0}}(f)=\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\left(\frac{\Phi_{\pm}^{1}(0,0)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{3}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{2}}+\frac{\Psi_{\pm}^{1}(0,0)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})}\right)K_{{\mathfrak{a}},{\mathfrak{d}}}(\pm
u(Y)).$

######  Notation.

In Theorem 9.3 we proved that ${\mathbb{D}}(z,u(Y))$ is independent of $u(Y)$.
To avoid confusion, we will henceforth drop the dependence on $u$ from the
notation.

###  10.1. Some algebraic number theory and a discussion

Let $\gamma\in\operatorname{GL}(2,K)$ be a regular elliptic matrix with trace
$\tau$ and determinant $\delta$. For

$(\gamma_{1},\cdots,\gamma_{n})\in G_{\infty},$

where $\gamma_{i}$ is the embedding of $\gamma$ in the $i$-th real embedding
of $K$, define

$C(\gamma)\mathrel{\mathop{\mathchar
58\relax}}=C(\operatorname{CH}((\gamma_{1},\cdots,\gamma_{n}))),$

where $C$ is the indicator function defined in Definition 3.32.

######  Lemma 10.1.

The following equality holds

$\lim_{z\to
0}\frac{L_{\infty}(z,\chi_{\gamma})}{L_{\infty}(1-z,\chi_{\gamma})}\frac{\zeta_{K}(2z)}{\zeta_{K}(z+1)}=-C(\gamma)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}2^{n-1}$

where $L_{\infty}(z,\chi_{\gamma})=\prod_{\nu}L_{\nu}(z,\chi_{\gamma})$ and
$\nu$ varies over all real places.

######  Notation.

In this proof we write $\chi$ in place of $\chi_{\gamma}$ for ease of
notation.

###### Proof.

Recall that for a totally real field $K$, the completed Dedekind zeta function
is defined as

$\xi_{K}(z)\mathrel{\mathop{\mathchar 58\relax}}=\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{z/2}L_{{\mathbb{R}}}(z)^{n}\zeta_{K}(z).$

Then

$\frac{\xi_{K}(2z)}{\xi_{K}(z+1)}=\frac{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{z}L_{\mathbb{R}}(2z)^{n}\zeta_{K}(2z)}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{(z+1)/2}L_{\mathbb{R}}(z+1)^{n}{\zeta_{K}(z+1)}}=\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{(z-1)/2}\frac{L_{\mathbb{R}}(2z)^{n}\zeta_{K}(2z)}{L_{\mathbb{R}}(z+1)^{n}\zeta_{K}(z+1)}$

whose limit at $z=0$ is equal to $-1/2$. Rearranging terms we write

$\frac{\zeta_{K}(2z)}{\zeta_{K}(z+1)}=\frac{\xi_{K}(2z)}{\xi_{K}(z+1)}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{(1-z)/2}\frac{L_{\mathbb{R}}(z+1)^{n}}{L_{\mathbb{R}}(2z)^{n}}$

We can therefore obtain

$\frac{L_{\infty}(z,\chi)}{L_{\infty}(1-z,\chi)}\frac{\zeta_{K}(2z)}{\zeta_{K}(z+1)}=\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{(1-z)/2}\prod_{\nu}\frac{L_{\mathbb{R}}(z+\delta_{\nu})L_{\mathbb{R}}(z+1)}{L_{\mathbb{R}}(1-z+\delta_{\nu})L_{\mathbb{R}}(2z)}\frac{\xi_{K}(2z)}{\xi_{K}(z+1)},$

where as before, $\delta_{\nu}=\delta_{\nu,\gamma}=1$ if $\chi$ is ramified at
$\nu$ and 0 otherwise. As $z\to 0$, we have

$\lim_{z\to
0}\frac{L_{\infty}(z,\chi)}{L_{\infty}(1-z,\chi)}\frac{\zeta_{K}(2z)}{\zeta_{K}(z+1)}=-\frac{1}{2}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}\lim_{z\to
0}\left(\prod_{\nu}\frac{L_{\mathbb{R}}(z+\delta_{\nu})L_{\mathbb{R}}(z)}{L_{\mathbb{R}}(1-z+\delta_{\nu})L_{\mathbb{R}}(2z)}\right),$

and it remains to compute the limit explicitly. We have

$\displaystyle\lim_{z\to
0}\frac{L_{\mathbb{R}}(z+\delta_{\nu})L_{\mathbb{R}}(1+z)}{L_{\mathbb{R}}(2z)L_{\mathbb{R}}(1-z+\delta_{\nu})}$
$\displaystyle=\lim_{z\to
0}\pi^{z}\frac{\Gamma(\frac{z+\delta_{\nu}}{2})\Gamma(\frac{z+1}{2})}{\Gamma(\frac{1-z+\delta_{\nu}}{2})\Gamma(z)}$
$\displaystyle=\lim_{z\to
0}2^{1-z}\pi^{z+1/2}\frac{\Gamma(\frac{z+\delta_{\nu}}{2})}{\Gamma(\frac{1-z+\delta_{\nu}}{2})\Gamma(\frac{z}{2})}=\begin{cases}2,&\textrm{
if }\delta=0\\\ 0,&\textrm{ if }\delta=1\end{cases}$

Altogether then

$\lim_{z\to
0}\frac{L_{\infty}(z,\chi)}{L_{\infty}(1-z,\chi)}\frac{\zeta_{K}(2z)}{\zeta_{K}(z+1)}=-\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}2^{n-1}$

when $\chi$ is unramified at every real place and $0$ otherwise. By the
Theorem 5.11, this dichotomy is equivalent to $C(\gamma)=1$ or $0$,
respectively. ∎

From the above and proposition (5.12) we get immediately

######  Corollary 10.2.

The following limit holds for any $(x,y)\in\mathbb{R}^{2n-1}$

$\lim_{z\to
0}\frac{L_{\infty}(z,\chi)}{L_{\infty}(1-z,\chi)}\frac{\zeta_{K}(2z)}{\zeta_{K}(z+1)}=-C(x,y)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}2^{n-1}.$

######  Remark 10.3.

Recall that the $L$-factors at infinity only depend on the regions, i.e. the
behaviours at infinity, and thus we can disregard the dependence on any
particular $\gamma$.

###  10.2. Rewriting the dominant term

We do this calculation in two steps. First, we simplify the
$\Phi^{1}_{\pm}(0,0)$-part:

(21)
$\begin{split}&\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\left(\frac{\Phi_{\pm}^{1}(0,0)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{3}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{2}}\right)K_{{\mathfrak{a}},{\mathfrak{d}}}(\pm
u(Y))\\\
&=\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{{\mathfrak{d}}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{3}}\sum_{{\mathfrak{a}}}\int\int\theta^{\pm}(x,y)F\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}}){\mathsf{p}}^{-{\mathsf{k}}\alpha}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\mathcal{P}^{\pm}(x,y)^{2\alpha}\right)\left(\frac{K_{{\mathfrak{a}},{\mathfrak{d}}}(\pm
u(Y))}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{2}}\right)dydx\\\
&=\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{{\mathfrak{d}}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{3}}\sum_{{\mathfrak{a}}}\frac{K_{{\mathfrak{a}},{\mathfrak{d}}}(\pm
u(Y))}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{2}}\left(\int\int\frac{\theta^{\pm}(x,y)}{2\pi
i}\int_{\Re{(z)}=1}\widetilde{F}(z)\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}}){\mathsf{p}}^{-{\mathsf{k}}\alpha}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\mathcal{P}^{\pm}(x,y)^{2\alpha}\right)^{-z}dzdydx\right)\\\
&=\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\left(\int\int\frac{\theta^{\pm}(x,y)}{2\pi
i}\int_{\Re{(z)}=1}\widetilde{F}(z)\left(\frac{{\mathsf{p}}^{-{\mathsf{k}}\alpha}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\mathcal{P}^{\pm}(x,y)^{2\alpha}\right)^{-z}\sum_{{\mathfrak{d}}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2(z+1)+1}}\sum_{{\mathfrak{a}}}\frac{K_{{\mathfrak{a}},{\mathfrak{d}}}(\pm
u(Y))}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{(z+1)+1}}dzdydx\right)\\\
&=\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\left(\int\int\frac{\theta^{\pm}(x,y)}{2\pi
i}\int_{\Re{(z)}=1}\widetilde{F}(z)\left(\frac{{\mathsf{p}}^{-{\mathsf{k}}\alpha}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\mathcal{P}^{\pm}(x,y)^{2\alpha}\right)^{-z}{\mathbb{D}}(z+1)dzdydx\right)\\\
&={\mathsf{p}}^{-{\mathsf{k}}/2}\sum_{\pm}\left(\int\int\frac{\theta^{\pm}(x,y)}{2\pi
i}\int_{\Re{(z)}=1}\widetilde{F}(z)\left(\frac{{\mathsf{p}}^{-{\mathsf{k}}\alpha}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\mathcal{P}^{\pm}(x,y)^{2\alpha}\right)^{-z}\frac{\zeta_{K}(2z+2)}{\zeta_{K}(z+2)}\cdot\frac{1-1/{\mathsf{p}}^{(z+1)({\mathsf{k}}+1)}}{1-1/{\mathsf{p}}^{z+1}}dzdydx\right).\end{split}$

The last line is obtained by applying Theorem 9.3 which brings out a $4^{n}$,
and by evaluating the sum over $Y\in\mathbb{F}_{2}^{n-1}$. These two
contributions cancel the $2^{3n-1}$ in the denominator. Note that after
evaluating the Kloosterman-type sums, nothing in the integrals depends on $Y$.
Similarly for the second part of the sum, we have

$\begin{split}&\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{{\mathfrak{a}}}\sum_{{\mathfrak{d}}}\left(\frac{\Psi_{\pm}^{1}(0,0)}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})}\right)K_{{\mathfrak{a}},{\mathfrak{d}}}(\pm
u(Y))\\\
&=\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{{\mathfrak{d}}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{3}}\sum_{{\mathfrak{a}}}\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})K_{{\mathfrak{a}},{\mathfrak{d}}}(\pm
u(Y))}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{2}}\Psi_{\pm}^{1}(0,0)\right)\\\
&=\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{{\mathfrak{d}}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{3}}\sum_{{\mathfrak{a}}}\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})K_{{\mathfrak{a}},{\mathfrak{d}}}(\pm
u(Y))}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{2}}\right)\\\
&\quad\times\int\int\theta^{\pm}(x,y)\mathcal{P}^{\pm}(x,y)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{-1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}H_{\gamma}\left(1,\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}}){\mathsf{p}}^{-{\mathsf{k}}(1-\alpha)}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\mathcal{P}^{\pm}(x,y)^{2(1-\alpha)}\right)dydx\\\
&=\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\sum_{{\mathfrak{d}}}\frac{1}{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{3}}\sum_{{\mathfrak{a}}}\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})K_{{\mathfrak{a}},{\mathfrak{d}}}(\pm
u(Y))}{\operatorname{\mathsf{N}}_{K}({\mathfrak{a}})^{2}}\right)\\\
&\times\int\int\frac{\theta^{\pm}(x,y)}{2\pi
i}\mathcal{P}^{\pm}(x,y)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{-1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}\int_{\Re(z)=1}\widetilde{F}(z)\left(\frac{\operatorname{\mathsf{N}}_{K}({\mathfrak{d}})^{2}\operatorname{\mathsf{N}}_{K}({\mathfrak{a}}){\mathsf{p}}^{-{\mathsf{k}}(1-\alpha)}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\mathcal{P}^{\pm}(x,y)^{2(1-\alpha)}\right)^{-z}\frac{L_{\infty}(z,\chi)}{L_{\infty}(1-z,\chi)}dzdydx\\\
&=\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{2^{3n-1}}\sum_{\pm}\sum_{Y\in\mathbb{F}_{2}^{n-1}}\int\int\frac{\theta^{\pm}(x,y)}{2\pi
i}\mathcal{P}^{\pm}(x,y)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{-1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}\int_{\Re(z)=1}\widetilde{F}(z)\left(\frac{{\mathsf{p}}^{-{\mathsf{k}}(1-\alpha)}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\mathcal{P}^{\pm}(x,y)^{2(1-\alpha)}\right)^{-z}\frac{L_{\infty}(z,\chi)}{L_{\infty}(1-z,\chi)}{\mathbb{D}}(z)dzdydx.\end{split}$

Once again, we apply Theorem 9.3 to the above expression and evaluating the
sum over $Y\in\mathbb{F}_{2}^{n-1}$. This cancels the $2^{3n-1}$ in the
denominator and the above expression simplifies to

(22) $\begin{split}{{\mathsf{p}}^{-{\mathsf{k}}}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{-1/2}}&\sum_{\pm}\int\int\frac{\theta^{\pm}(x,y)}{2\pi
i}\mathcal{P}^{\pm}(x,y)\\\
&\times\int_{\Re(z)=1}\widetilde{F}(z)\left(\frac{{\mathsf{p}}^{-{\mathsf{k}}(1-\alpha)}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\mathcal{P}^{\pm}(x,y)^{2(1-\alpha)}\right)^{-z}\frac{L_{\infty}(z,\chi)}{L_{\infty}(1-z,\chi)}\frac{\zeta_{K}(2z)}{\zeta_{K}(z+1)}\cdot\frac{1-1/{\mathsf{p}}^{(z)({\mathsf{k}}+1)}}{1-1/{\mathsf{p}}^{z}}dzdydx.\end{split}$

Note that (20) is the sum of (LABEL:eq:contour1new) and (22). However, we
write the summands separately as we work with each piece one-at-a-time. The
trivial representation should come from (LABEL:eq:contour1new), whereas the
special representation should come from (22) after contour shifting. In both
cases, they will arise from the residue at $z=0$.

###  10.3. Calculating the residue

While our main interest is the residue at $z=0$, we shall follow [Alt15] to
shift past the line $\Re(z)=-1/2$. More precisely, we shift the contour to the
line $\mathcal{C}_{1}$ defined by $\Re(z)=\sigma$ for $-1<\sigma<-1/2$.

######  Lemma 10.4.

The contribution (LABEL:eq:contour1new) is equal to the sum of

(23)
$\displaystyle{\mathsf{p}}^{-{\mathsf{k}}/2}\frac{1-{\mathsf{p}}^{-({\mathsf{k}}+1)}}{1-{{\mathsf{p}}^{-1}}}\sum_{\pm}\int\int\theta^{\pm}(x,y)dydx,$
(24)
$\displaystyle\dfrac{{\mathsf{p}}^{-{\mathsf{k}}/2}\kappa}{\zeta_{K}(3/2)}\frac{1-{\mathsf{p}}^{-({\mathsf{k}}+1)/2}}{1-{\mathsf{p}}^{-1/2}}\widetilde{F}\left(-\frac{1}{2}\right)\sum_{\pm}\left(\int\int{\theta^{\pm}(x,y)}\left(\frac{{\mathsf{p}}^{-{\mathsf{k}}\alpha}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\mathcal{P}^{\pm}(x,y)^{2\alpha}\right)^{1/2}dydx\right),\textrm{
and }$ (25)
$\displaystyle\frac{{\mathsf{p}}^{-{\mathsf{k}}/2}}{4^{n}}\sum_{\pm}\left(\int\int\frac{\theta^{\pm}(x,y)}{2\pi
i}\int_{\mathcal{C}_{1}}\widetilde{F}(z)\left(\frac{{\mathsf{p}}^{-{\mathsf{k}}\alpha}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\mathcal{P}^{\pm}(x,y)^{2\alpha}\right)^{-z}{\mathbb{D}}(z+1)dzdydx\right).$

###### Proof.

The constant in front of (LABEL:eq:contour1new) is
${\mathsf{p}}^{-{\mathsf{k}}/2}$.

To shift the contour of integration to the line $\mathcal{C}_{1}$, we describe
the poles in the region that is traversed. The integral over $(x,y)$ can be
taken over the regular set, so that $\mathcal{P}^{\pm}(x,y)$ is finite and
non-zero. The Mellin transform $\widetilde{F}(z)$ has a simple pole at $z=0$
with residue equal to $1$. So, the total residue at $z=0$ is equal to (23).

The auxiliary Dirichlet series ${\mathbb{D}}(z)$ has a simple pole at $z=1/2$
with residue equal to

(26) $\lim_{z\rightarrow 1/2}(z-1/2){\mathbb{D}}(z)=\lim_{z\rightarrow
1/2}(z-1/2)4^{n}\frac{\zeta_{K}(2z)}{\zeta_{K}(z+1)}\cdot\frac{1-1/{\mathsf{p}}^{z({\mathsf{k}}+1)}}{1-1/{\mathsf{p}}^{z}}=\frac{4^{n}\kappa}{\zeta_{K}(3/2)}\frac{1-{\mathsf{p}}^{-({\mathsf{k}}+1)/2}}{1-{\mathsf{p}}^{-1/2}}.$

By a change of variables, we have the same for ${\mathbb{D}}(z+1)$ at
$z=-1/2$. However, notice that we do not have ${\mathbb{D}}(z+1)$ but rather
$4^{-n}{\mathbb{D}}(z+1)$ (i.e. we do not include the power of $4$ when
substituting the residue). Thus the total residue there is equal to (24).
Notice that the residue at $z=-1/2$ of the quotient of zeta functions is

$\displaystyle\lim_{z\longrightarrow-1/2}(z+1/2)\dfrac{\zeta_{K}(2z+2)}{z+2}=\dfrac{\kappa}{\zeta_{K}(3/2)},$

which explains the constant in front of equation (24).

The two residues together yield the main term; the remainder (25) is described
by the integral over $\mathcal{C}_{1}$. The $4^{n}$ appears in the denominator
because we rewrote the expression using ${\mathbb{D}}(z+1)$; we put back the
factor $4^{n}$ needed. ∎

######  Lemma 10.5.

The contribution of (22) is equal to the sum of the following three
expressions

(27)
$\displaystyle-2^{n-1}({\mathsf{k}}+1){\mathsf{p}}^{-{\mathsf{k}}}\sum_{\pm}\int\int\theta^{\pm}(x,y)\mathcal{P}^{\pm}(x,y)C(x,y)dxdy,$
(28)
$\displaystyle{\mathsf{p}}^{-{\mathsf{k}}/2}\frac{\kappa}{\zeta_{K}(3/2)}\frac{1-{\mathsf{p}}^{-({\mathsf{k}}+1)/2}}{1-{\mathsf{p}}^{-1/2}}\widetilde{F}\left(\dfrac{1}{2}\right)\sum_{\pm}\int\int\theta^{\pm}(x,y)\mathcal{P}^{\pm}(x,y)\left(\frac{{\mathsf{p}}^{-{\mathsf{k}}\alpha}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\mathcal{P}^{\pm}(x,y)^{2\alpha}\right)^{1/2}dydx,\textrm{
and }$ (29)
$\begin{split}\dfrac{{\mathsf{p}}^{-{\mathsf{k}}}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{-1/2}}{4^{n}}\sum_{\pm}\int\int\frac{\theta^{\pm}(x,y)}{2\pi
i}&\mathcal{P}^{\pm}(x,y)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}{\mathsf{p}}^{-{\mathsf{k}}/2}\\\
&\times\int_{\mathcal{C}_{2}}\widetilde{F}(z)\left(\frac{{\mathsf{p}}^{-{\mathsf{k}}(1-\alpha)}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\mathcal{P}^{\pm}(x,y)^{2(1-\alpha)}\right)^{-z}\frac{L_{\infty}(z,\chi)}{L_{\infty}(1-z,\chi)}{\mathbb{D}}(z)dzdydx.\end{split}$

###### Proof.

Shifting the the left, we see that at $z=1/2$ the Dirichlet series
${\mathbb{D}}(z+1)$ has a pole with the same residue (26), so the contribution
there is equal to

(30) ${\mathsf{p}}^{-{\mathsf{k}}}\mathinner{\\!\left\lvert
D_{k}\right\rvert}^{-1/2}\frac{\kappa}{\zeta_{K}(3/2)}\frac{1-{\mathsf{p}}^{-({\mathsf{k}}+1)/2}}{1-{\mathsf{p}}^{-1/2}}\widetilde{F}\left(\frac{1}{2}\right)\sum_{\pm}\int\int\theta^{\pm}(x,y)\mathcal{P}^{\pm}(x,y)\left(\frac{{\mathsf{p}}^{-{\mathsf{k}}(1-\alpha)}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\mathcal{P}^{\pm}(x,y)^{2(1-\alpha)}\right)^{-1/2}dydx.$

Notice that

${\mathsf{p}}^{-{\mathsf{k}}}\mathcal{P}^{\pm}(x,y)\left(\frac{{\mathsf{p}}^{-{\mathsf{k}}(1-\alpha)}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1-\alpha}}\mathcal{P}^{\pm}(x,y)^{2(1-\alpha)}\right)^{-1/2}={\mathsf{p}}^{-{\mathsf{k}}/2}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}\dfrac{{\mathsf{p}}^{-{\mathsf{k}}\alpha/2}\mathcal{P}^{\pm}(x,y)^{\alpha}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha/2}}.$

Thus the residue (30) can be rewritten as

${\mathsf{p}}^{-{\mathsf{k}}/2}\frac{\kappa}{\zeta_{K}(3/2)}\frac{1-{\mathsf{p}}^{-({\mathsf{k}}+1)/2}}{1-{\mathsf{p}}^{-1/2}}\widetilde{F}\left(\frac{1}{2}\right)\sum_{\pm}\int\int\theta^{\pm}(x,y)\left(\frac{{\mathsf{p}}^{-{\mathsf{k}}\alpha}}{\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{\alpha}}\mathcal{P}^{\pm}(x,y)^{2\alpha}\right)^{1/2}dydx.$

This is (28) above. We now move on to $z=0$ and note that

$\frac{1-{\mathsf{p}}^{-z({\mathsf{k}}+1)}}{1-{\mathsf{p}}^{-z}}=\sum_{n=0}^{{\mathsf{k}}}{\mathsf{p}}^{-nz}.$

At $z=0$ the right hand side888We must use the right hand side of the
geometric series as otherwise the expression is undefined.  contributes
${\mathsf{k}}+1$. Recall from Lemma 10.1 that

$\lim_{z\to
0}\frac{L_{\infty}(z,\chi)}{L_{\infty}(1-z,\chi)}\frac{\zeta_{K}(2z)}{\zeta_{K}(z+1)}=-C(x,y)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}2^{n-1}$

and that the residue at $z=0$ of $\widetilde{F}(z)$ is $1$. Thus the residue
at $z=0$ is

(31) ${\mathsf{p}}^{-{\mathsf{k}}}\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{-1/2}\sum_{\pm}\int\int\theta^{\pm}(x,y)\mathcal{P}^{\pm}(x,y)\cdot(-C(x,y)\mathinner{\\!\left\lvert
D_{K}\right\rvert}^{1/2}2^{n-1})\cdot({\mathsf{k}}+1)dydx,$

which becomes

(32)
$-2^{n-1}({\mathsf{k}}+1){\mathsf{p}}^{-{\mathsf{k}}}\sum_{\pm}\int\int\theta^{\pm}(x,y)\mathcal{P}^{\pm}(x,y)\cdot
C(x,y)dydx,$

Thus the total residue is given by (27) as desired. The two residues together
yield the main term, while the remainder (29) is described by the integral
over $\mathcal{C}_{2}$. Once more, the $4^{n}$ appears precisely because we
have put ${\mathbb{D}}(z)$ again. ∎

Recall that $\widetilde{F}$ is an odd function which allows us to conclude
that (24) + (28) = 0. We obtain the following result.

######  Proposition 10.6.

The dual sum (20) is equal to $\eqref{eq:triv res new}+\eqref{eq:spec
residue2new}+\eqref{eq:C(1)new}+\eqref{C2new}$.

##  11\. Computation of the trace of the trivial and the special
representations

In this section, we compute the traces of the representations we desire to
cancel; namely the trivial representation and the special representations. The
core of the argument lies the computing the trace at the archimedean places.

###  11.1. The main statement

Given a pair of integers $(i,j)$, in [Lan04, p. 631], Langlands introduces a
representation of $\operatorname{GL}(2,{\mathbb{R}})$ induced from
$T_{\operatorname{spl}}$ given by

$\chi_{i,j}\begin{pmatrix}\alpha&\\\
&\beta\end{pmatrix}=\mbox{sgn}{(\alpha)}^{i}\mbox{sgn}{(\beta)}^{j}.$

######  Notation.

Only the parities of $i$ and $j$ matter, so we simplify notation and refer to
this as $\chi_{\pm,\pm}$. We can unitarily induce this representation to