Proof. It is sufficient to verify the claim for each of the formulae above. The action of the upper triangular matrices commutes with the action of any linear transformation that preserves the norm. The Fourier transform intertwines the action of $L\in \operatorname {GL}(B_v)$ with the action of $|\operatorname {det} L|_v^{-1t}L^{-1}$. Hence, it commutes with orthogonal transformations.

Proof. The actions of $B^\times$ and $\mathbf {G}(F)$ preserve the norm form, hence they factor through the orthogonal group.

Proof. This is established by Weil [Reference WeilWei64] for the space of Schwartz or Schwartz–Bruhat functions. The same proof works for $L^2(B_\infty )$.

Proof. It is enough to show that if $M,M'$ are Schwartz solutions to (18) with quantum numbers $m\neq m'$, then $\langle M, M' \rangle =0$. Because the operator $L_\omega$ is symmetric, we have $\langle {\omega m} M,M' \rangle = \langle L[M],M' \rangle =\langle M, L[M'] \rangle =\langle M, {\omega m'} M' \rangle$. We deduce that $\langle M,M' \rangle =0$ in the usual fashion.

Proof. We establish first the direct sum decomposition. Recall that we need the Laplacian in (18) to be defined consistently as having Fourier symbol $-4\pi ^2 \operatorname {Nr}$. Choose a coordinate system $x_1,\ldots,x_4$ for $B_\infty$ such that $\operatorname {Nr}(x)=\sum _{i=1}^4 \epsilon _i x_i^2$ with $\epsilon _i\in \{\pm 1\}$. The Laplacian for our fixed character can be written in this coordinate system as

\[ \frac{1}{4}\sum_{i=1}^{4} \epsilon_i \frac{\partial^2}{\partial x_i^2}. \]

The $1/4$ factor appears because the Fourier transform is defined with respect to the bilinear form $\sum _{i=1}^4 2\epsilon _i x_i y_i$. The space of solutions to the one-dimensional quantum harmonic oscillator with angular frequency $4\pi$, $n\in \mathbb {Z}_{\geq 0}$,

\[ -\tfrac{1}{4}f''(x)+4\pi^2 x^2 f(x)=(2n+1)\pi f(x) \]

is one-dimensional and the $L^2$-normalized solution is

\[ f_n(x)\colon=\frac{1}{\sqrt{2^{n-1} n!}}\exp\big({-}2\pi x^2\big) H_n(2\sqrt{\pi} x), \]

where $H_n$ are the physicist's Hermite polynomials. Moreover, these solutions form an orthonormal basis of the Hilbert space $L^2(\mathbb {R})$. Define for every $\underline {n}=(n_1,n_2,n_3,n_4)\in \mathbb {Z}_{\geq 0}^4$ the function $M_{\underline {n}}\colon B_\infty \to \mathbb {C}$ by

\[ M_{\underline{n}}(x) = \prod_{j=1}^4 f_{n_j}(x_j) = \prod_{j=1}^4 \frac{1}{\sqrt{2^{n_j-1}n_j!}} \exp\big({-}2\pi x_j^2\big) H_{n_j}(2\sqrt{\pi} {x_j}), \]

where $x_1,\ldots,x_4$ are the normal form coordinates for the quadratic form $\operatorname {Nr}(x)$. Because $L^2(B_\infty )\simeq L^2(\mathbb {R})^{\bigotimes 4}$, we deduce that the functions $M_{\underline {n}}$ form an orthonormal basis of $L^2(B_\infty )$. These are Schwartz functions, and a separation of variables computation shows that $M_{\underline {n}}$ solves (18) withFootnote ⁵ $2m=\sum _{j=1}^4 \epsilon _j (2n_j+1)$. This and Lemma 3.4 establish that $L^2(B_\infty )=\bigoplus _{m\in \mathbb {Z}} V_{m,2\pi }$ as claimed.

We need to prove that if $M\in V_{m,2\pi }$ then $\rho (k_\theta ).M=e^{i m \theta }M$ for all $\theta \in [0,2\pi )$. By continuity of the Weil representation it is enough to establish this for Schwartz functions. Because Schwartz functions are smooth vectors for the Weil representation it is sufficient to show $({{d/d} \theta } )\rho (k_\theta ).M=i m \big (\rho (k_\theta ).M\big )$. Because the group $\mathbf {SO}_2(\mathbb {R})$ is abelian it is enough to verify this ordinary differential equation (ODE) at $\theta =0$. The formula $k_\theta =\exp ( \theta w)$ for $w=\left (\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}\right )$ implies that the ODE at $\theta =0$ is equivalent to

(19)\begin{equation} {d} \rho(w). M= im M, \end{equation}

where ${d} \rho$ is the Lie algebra representation of $\mathfrak {sl}_2(\mathbb {R})$ on $\mathcal {S}(B_\infty )$ differentiated from the Weil representation of $\mathbf {SL}_2(\mathbb {R})$. Using the definition of the Weil action for upper diagonal unipotent matrices one easily computes that $\big ({d}\rho \left (\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}\right ).M\big )(x)=2\pi i \operatorname {Nr} (x) M(x)$. The formula $\left (\begin{smallmatrix} 0 & 0 \\ -1 & 0 \end{smallmatrix}\right )=w^{-1} \left (\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}\right ) w$, then implies that ${d}\rho \left (\begin{smallmatrix} 0 & 0 \\ -1 & 0 \end{smallmatrix}\right ).M=({1/2\pi i})\Delta M$, and $({d} \rho (w).M)(x)=({1/2\pi i} )\Delta M(x)+2\pi i \operatorname {Nr}(x) M(x)$. Thus, (19) is equivalent to (18).

Proof. We use the notation $a\colon= \operatorname {diag}\big (\sqrt { {2\pi / \omega }},\sqrt {{\omega/2\pi }}\big )$ and write $\rho (g).M=\rho (g a^{-1}) \rho (a).M$. The Iwasawa decomposition of $g a^{-1}$ is

\[ g a^{-1}=\begin{pmatrix} 1 & \frac{bd \frac{2 \pi}{\omega}+ac \frac{\omega}{2 \pi}}{D^2} \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{D} & 0 \\ 0 & D \end{pmatrix} \begin{pmatrix} \sqrt{\frac{2\pi}{\omega}}\frac{d}{D} & -\sqrt{\frac{\omega}{2\pi}}\frac{c}{D}\\ \sqrt{\frac{\omega}{2\pi}}\frac{c}{D} & \sqrt{\frac{2\pi}{\omega}}\frac{d}{D} \end{pmatrix}. \]

The function

\[ (\rho(a).M)(x) =\frac{2 \pi}{\omega}M\bigg(\sqrt{\frac{2 \pi}{\omega}}x\bigg) \]

is a solution of (18) with angular frequency $2\pi$. Hence, we can apply Lemma 3.5 to $\rho (g a^{-1}) \rho (a).M$ and the Iwasawa decomposition of $g a^{-1}$.

Proof. The space $V_{m,\omega }$ is invariant under $\operatorname {O}(B_\infty,\operatorname {Nr})$ because (18) commutes with orthogonal transformations. Also, the decay condition is invariant under orthogonal transformations. Invariance under the Weil action follows from Corollary 3.6.

Proof. All maximal orders in $B_p$ are conjugate to each other by an element of $B_p^\times$. Because the Weil action commutes with conjugation, it is enough to prove the claim for a specific maximal order. Moreover, the group $\mathbf {SL}_2(\mathbb {Z}_p)$ is generated by the subgroup $\mathcal {P}<\mathbf {SL}_2(\mathbb {Z}_p)$ of upper triangular integral matrices and the involution $w$. The fact that $\mathbb {1}_\mathcal {R}$ is invariant under $\mathcal {P}$ follows because we have assumed $\psi _v$ is unramified. If $B_p$ is split, fix an isomorphism $B_p\simeq \operatorname {Mat}_{2\times 2}(\mathbb {Q}_p)$ and we need only show that $\mathbb {1}_{\operatorname {Mat}_{2\times 2}(\mathbb {Z}_p)}$ is invariant under the Fourier transform. This follows from the fact that $\mathbb {1}_{\mathbb {Z}_p}$ is invariant under the Fourier transform on $\mathbb {Q}_p$ with an unramified character.

If $B_p$ is a division algebra, we need only show invariance under the element $\left (\begin{smallmatrix} 1 & 0 \\ p & 1 \end{smallmatrix}\right ) =-w \left (\begin{smallmatrix} 1 & -p\\ 0 & 1 \end{smallmatrix}\right ) w$. This element and the upper triangular integral matrices generate $U_0(p)$. Because of the duality of the Fourier transform, this is equivalent to showing that $\rho \big (\!\left (\begin{smallmatrix} 1 & -p\\ 0 & 1 \end{smallmatrix}\right )\!\big ).\mathcal {F}\mathbb {1}_{\mathcal {R}}=\mathcal {F}\mathbb {1}_{\mathcal {R}}$.

Let $E=F(\sqrt {a})/F$ be the unique unramified quadratic extension and write $B_p$ as the cyclic algebra $\big (\frac {a,p}{\mathbb {Q}_p}\bigr )$ with the standard generators $i,j,k$ and $i^2=a$, $j^2=p$ and $ij=-ji=k$. As usual, we identify $E$ with the sub-ring $\mathbb {Q}_p+i\mathbb {Q}_p< B_p$. Denote by $\mathcal {O}_E$ the maximal order of $E$. Then, the unique maximal order of $B_p$ is $\mathcal {R}\simeq \mathcal {O}_E+j\mathcal {O}_E$. The Fourier transform on $B_p$ descends to the standard Fourier transform on $E$ with an unramified character. Identifying $B_p\simeq E\times E$ via $a+jb\mapsto (a,b)$, we can write the Fourier self-dual measure on $B_p$ in these coordinates as $p^{-1}m_E\times m_E$. The $p^{-1}$ factor normalizes the measure to be self-dual.

The Fourier transform on $E$ satisfies $\mathcal {F} \mathbb {1}_{\mathcal {O}_E}=\mathbb {1}_{\mathcal {O}_E}$. An explicit computation with the Fourier self-dual measure implies

\[ \mathcal{F} \mathbb{1}_{\mathcal{O}_E+j\mathcal{O}_E}=p^{-1}\mathbb{1}_{\mathcal{O}_E+j^{-1}\mathcal{O}_E}. \]

Hence, $\operatorname {Nr} x\in p^{-1}\mathbb {Z}_p$ for all $x\in \operatorname {supp} \mathcal {F} \mathbb {1}_{\mathcal {R}}$, from which we deduce $\big (\rho \big (\!\!\left (\begin{smallmatrix} 1 & -p \\ 0 & 1 \end{smallmatrix}\right )\!\!\big ). \mathcal {F} \mathbb {1}_{\mathcal {R}}\big )(x)=\psi _v(-p\operatorname {Nr} x) \big (\!\mathcal {F} \mathbb {1}_{\mathcal {R}}\big )(x)=\big (\!\mathcal {F} \mathbb {1}_{\mathcal {R}}\big )(x)$ and the claim follows.

Proof. Let $j$ be a uniformizer in $\mathcal {R}$. The field norm on $\mathcal {R}/j\mathcal {R}\simeq \mathbb {F}_{p^2}$ coincides with the reduction $\bmod p$ of the reduced norm in $\mathcal {R}$. This can be seen by taking a subfield $E\subset B_p$, such that $E$ is an unramified quadratic extension of $F$ that splits $B_p$. In these coordinates, we can write

(20)\begin{equation} B_p = E+jE, \quad \mathcal{R}=\mathcal{O}_E+j\mathcal{O}_E, \quad \widehat{\mathcal{R}}=\mathcal{O}_E+j^{-1}\mathcal{O}_E. \end{equation}

Now, the restriction of the reduced norm in $B_p$ to $E$ coincides with the field norm on $E$ and $\mathcal {R}/j\mathcal {R}=\mathcal {O}_E/p\mathcal {O}_E\simeq \mathbb {F}_{p^2}$.

Observe that $\widehat {\mathcal {R}}= j^{-1} \mathcal {R}$ and $\widehat {\mathcal {R}}/\mathcal {R}\simeq \mathcal {R}/j\mathcal {R}\simeq \mathbb {F}_{p^2}$. The last isomorphism is a field isomorphism and, thus, it commutes with taking norms. The first isomorphism is via the map $x\mapsto j x$ and $\operatorname {Nr} (j x)=-p\operatorname {Nr} x$. This establishes the claimed formula for norms. To summarize, the map $\jmath _v$ may thus be given explicitly as

(21)\begin{equation} \jmath_v(a+j^{-1}b)=b \mod p\mathcal{O}_E, \end{equation}

for $a+j^{-1}b \in \widehat {\mathcal {R}}$ with $a,b \in \mathcal {O}_E$.

Proof. As in the previous lemma, we put coordinates on $B_p$ corresponding to the cyclic algebra $\big (\frac {a,p}{\mathbb {Q}_p}\big )$ where $E=F(\sqrt {a})/F$ is the unique unramified quadratic extension, see (20). We shall make use of the explicit map $\jmath _v$ given by (21). It is an isomorphism of abelian groups $\widehat {\mathcal {R}}/\mathcal {R}\to \mathbb {F}_{p^2}$ and $-p \operatorname {Nr} x \bmod p\mathbb {Z}_p \equiv \operatorname {Nr} \jmath _v(x)$ for all $x\in \widehat {\mathcal {R}}$, where the norm on the right-hand side is the field norm $\mathbb {F}_{p^2}\to \mathbb {F}_p$.

For each $\alpha \in \mathbb {F}_{p^2}$, fix a representative $x_\alpha \in \jmath _v^{-1}(\alpha )$. Then,

(22)\begin{equation} \widehat{\mathcal{R}}=\bigsqcup_{\alpha\in \mathbb{F}_{p^2}} \big(x_\alpha+\mathcal{R}\big). \end{equation}

Explicitly, for each $\alpha \in \mathbb {F}_{p^2}$, we take $x_\alpha =j^{-1}\check {\alpha }$, where $\check {\alpha }\in \mathcal {O}_E$ satisfies $\check {\alpha } \bmod p\mathcal {O}_E=\alpha$. The duality between $\mathcal {R}$ and $\widehat {\mathcal {R}}$ implies that $\operatorname {Nr}(x_\alpha +\mathcal {R})\subset \operatorname {Nr} x_\alpha + \mathbb {Z}_p$.

Recall that $\mathbb {1}_{\mathcal {R}}$ is $\rho (U_0(p))$-invariant. Hence, we need only calculate the action of each representative of $\mathbf {SL}_2(\mathbb {Z}_p)/U_0(p)$ on $\mathbb {1}_{\mathcal {R}}$. A set of representatives is given by $w$, $\left (\begin{smallmatrix} 1 & 0 \\ t & 1\end{smallmatrix}\right )$, $0\leq t< p$. The action of $\rho (w)$ is the Fourier transform (multiplied by $\gamma =-1$) and we have already seen in the proof of Lemma 3.11 that

\[ \rho(w).\mathbb{1}_{\mathcal{R}}=-p^{-1}\mathbb{1}_{\widehat{\mathcal{R}}}. \]

Write $x=a+j^{-1}b$ for $x\in \widehat {\mathcal {R}}$. Now, we compute the action of $\left (\begin{smallmatrix} 1 & 0 \\ t & 1\end{smallmatrix}\right )=-w\left (\begin{smallmatrix} 1 & -t \\ 0 & 1 \end{smallmatrix}\right ) w$ usingFootnote ⁶ (22)

\begin{align*} \bigg(\rho\bigg(\!\!\begin{pmatrix}1 & 0 \\ t & 1\end{pmatrix}\!\!\bigg).\mathbb{1}_{\mathcal{R}}\bigg)(x)&=-p^{-1}\rho(w). \big(\psi(-t\operatorname{Nr} y)\mathbb{1}_{\widehat{\mathcal{R}}}(y)\big)(x)\\ &=-p^{-1}\rho(w).\bigg(\sum_{\alpha\in\mathbb{F}_{p^2}} \psi(-t\operatorname{Nr} x_\alpha) \mathbb{1}_{x_\alpha+\mathcal{R}}\bigg)(x)\\ &=p^{-2}\mathbb{1}_{\widehat{\mathcal{R}}}(x) \sum_{\alpha\in\mathbb{F}_{p^2}} \psi(-t\operatorname{Nr} x_\alpha+\langle x, x_\alpha \rangle). \end{align*}

If $t\neq 0$, then the sum above becomes

\begin{align*} \sum_{\alpha\in\mathbb{F}_{p^2}} \psi\bigg(\frac{\operatorname{Nr} x- \operatorname{Nr}(x-tx_\alpha)}{t} \bigg)&= \psi\bigg(\frac{\operatorname{Nr} x}{t}\bigg)\sum_{\alpha\in\mathbb{F}_{p^2}}\psi\bigg(\frac{\operatorname{Nr}(b-t\alpha)}{tp}\bigg) =\psi\bigg(\frac{\operatorname{Nr} x}{t}\bigg)\sum_{\alpha_0\in\mathbb{F}_{p^2}}\psi\bigg(\frac{\operatorname{Nr} \alpha_0}{tp}\bigg)\\ &=\psi\bigg(\frac{\operatorname{Nr} x}{t}\bigg)\bigg((p+1)\sum_{\beta\in\mathbb{F}_p^\times} \psi\bigg(\frac{\beta}{p}\bigg)+1\bigg) =-p\psi\bigg(\frac{\operatorname{Nr} x}{t}\bigg). \end{align*}

We have used the fact that every element of $\mathbb {F}_p^\times$ is the norm of exactly $p+1$ elements in $\mathbb {F}_{p^2}^\times$. This establishes the claim for $0< t< p$. For $t=0$, the sum becomes

\[ \sum_{\alpha\in\mathbb{F}_{p^2}} \psi\bigg(-\frac{\operatorname{Tr}(\alpha b)}{p}\bigg)= \begin{cases} p^2, & b\equiv 0 \bmod p\mathcal{O}_E,\\ 0, & \textrm{otherwise.} \end{cases} \]

In addition, $\mathbb {1}_{\widehat {\mathcal {R}}}(x)\delta _{b\equiv 0 \bmod p\mathcal {O}_E}(x)=\mathbb {1}_{\mathcal {R}}(x)$. Of course, the case of $t=0$ is actually trivial to compute because it corresponds to the identity representative.

Proof. Fix a maximal order $\mathcal {R}\subset B_p$. We first claim that for every Schwartz–Bruhat function $M\colon B_p\to \mathbb {C}$ there is some diagonal matrix $a\in \mathbf {SL}_2(\mathbb {Q}_p)$ such that $\rho (a).M$ is a linear combination of translates of $\mathbb {1}_{\mathcal {R}}$. Equivalently $\rho (a).M(x+\mathcal {R})=\rho (a).M(x)$. Because $M$ is Schwartz–Bruhat, there is a small neighborhood of the origin $\mathcal {V}\subset B_p$, such that $M(x+\mathcal {V})=M(x)$. There is $k\geq 1$ such that $p^k \mathcal {R}\subset \mathcal {V}$. The function $x\mapsto M(p^{k}x)$ is invariant under translations by $\mathcal {R}$. Set $a=\operatorname {diag}(p^k,p^{-k})$, then $\rho (a).M$ is a linear combination of translates of $\mathbb {1}_\mathcal {R}$ as claimed.

Fix $b\in B_p$ and consider the group

\[ \mathcal{A}(b)=\big\{\!\operatorname{diag}(u, u^{-1})\colon u\in\mathbb{Z}_p^\times \textrm{ and } bu-b\in\mathcal{R} \big\}. \]

Then, $\mathcal {A}(b)$ is an open subgroup of the diagonal group in $\mathbf {SL}_2(\mathbb {Z}_p)$ and $\rho (\mathcal {A}(b)).\mathbb {1}_{b+\mathcal {R}}=\mathbb {1}_{b+\mathcal {R}}$. Taking a finite intersection of such subgroups we find an open subgroup $\mathcal {A}_0$ of the diagonal group of $\mathbf {SL}_2(\mathbb {Z}_p)$, such that $\rho (\mathcal {A}_0 a).M=\rho (a).M$. Hence, $M$ is invariant under $\rho (\mathcal {A}_0)$.

In a similar fashion, let $k_b\geq 0$ such that $p^{k_b} b\in \mathcal {R}$ and define $\mathcal {N}(b)=\big (\begin{smallmatrix}1 & p^{k_b} \mathbb {Z}_p \\ 0 & 1 \end{smallmatrix}\big )$. Then, $\mathcal {N}(b)$ is an open subgroup of the upper-triangular unipotent group of $\mathbf {SL}_2(\mathbb {Z}_p)$ and $\rho (\mathcal {N}_b).\mathbb {1}_{b+\mathcal {R}}=\mathbb {1}_{b+\mathcal {R}}$. Taking a finite intersection of such subgroups we can find an open subgroup $\mathcal {N}_1'$ of the integral upper-triangular unipotent subgroup, such that $\rho (\mathcal {N}_1'a).M=\rho (a).M$. Set $\mathcal {N}_1=a^{-1} \mathcal {N}_1' a\cap \mathbf {SL}_2(\mathbb {Z}_p)$. Then, $\rho (\mathcal {N}_1).M=M$ and $\mathcal {N}_1$ is an open subgroup of the upper unipotent integral group. Replacing $M$ by $\rho (w).M$, we can find $\mathcal {N}_2$ such that $\rho (w^{-1} \mathcal {N_2} w).M=M$, and $w^{-1}\mathcal {N}_2 w$ is an open subgroup of the lower-triangular integral unipotent group.

Set now $U_M$ to be the group generated by $\mathcal {A}_0,\mathcal {N}_1,w^{-1}\mathcal {N}_2w$. Then, $U_M$ is an open subgroup of $\mathbf {SL}_2(\mathbb {Z}_p)$ and satisfies the requirements of the claim.

Proof. The action of $B_p^\times$ on the vertices of the Bruhat–Tits tree is transitive on pairs of vertices of a fixed distance,Footnote ⁷ thus it acts transitively by conjugation on the set of Eichler orders of a fixed level $p^n$. Because the conjugation action commutes with the Weil representation, it is enough to consider a single Eichler order of the form

\[ \mathcal{R}=\begin{pmatrix} \mathbb{Z}_p & \mathbb{Z}_p \\ p^n \mathbb{Z}_p & \mathbb{Z}_p \end{pmatrix}. \]

Indeed, invariance of $\mathbb {1}_{\mathcal {R}}$ under upper-triangular integral matrices is immediate and it is enough to check invariance under the element $\left (\begin{smallmatrix} 1 & 0 \\ p^{n} & 1 \end{smallmatrix}\right ) =-w \left (\begin{smallmatrix} 1 & -p^{n}\\ 0 & 1 \end{smallmatrix}\right ) w$. Equivalently, we need to show $\rho \big (\!\left (\begin{smallmatrix} 1 & -p^{n}\\ 0 & 1 \end{smallmatrix}\right ) \!\big ).\mathcal {F}\mathbb {1}_{\mathcal {R}}=\mathcal {F}\mathbb {1}_{\mathcal {R}}$. We can compute the Fourier transform of $\mathbb {1}_{\mathcal {R}}$ explicitly and arrive at

\[ {\mathcal{F}}\mathbb{1}_{\begin{pmatrix} \mathbb{Z}_p & \mathbb{Z}_p \\ p^n \mathbb{Z}_p & \mathbb{Z}_p \end{pmatrix}}=p^{-n} \mathbb{1}_{\begin{pmatrix} \mathbb{Z}_p & p^{-n} \mathbb{Z}_p \\ \mathbb{Z}_p & \mathbb{Z}_p \end{pmatrix}}=p^{-n}\mathbb{1}_{\widehat{\mathcal{R}}}. \]

In particular, for all $x\in \operatorname {supp} \mathcal {F} \mathbb {1}_{\mathcal {R}}$, we have $\operatorname {Nr} x=\operatorname {det} x\in p^{-n}\mathbb {Z}_p$ and

\[ \bigg(\rho\bigg(\!\!\begin{pmatrix} 1 & -p^{n} \\ 0 & 1 \end{pmatrix}\!\!\bigg). \mathcal{F} \mathbb{1}_{\mathcal{R}}\bigg)(x)=\psi_v(-p^{n}\operatorname{Nr} x) \big(\mathcal{F} \mathbb{1}_{\mathcal{R}}\big)(x)=\big(\mathcal{F} \mathbb{1}_{\mathcal{R}}\big)(x) \]

as necessary.

Proof. Because all local Eichler orders of fixed level are conjugate, it is enough to verify the claim for

\[ \mathcal{R}_1=\begin{pmatrix} \mathbb{Z}_p & \mathbb{Z}_p \\ \mathbb{Z}_p & \mathbb{Z}_p \end{pmatrix}, \quad \mathcal{R}_2=\begin{pmatrix} \mathbb{Z}_p & p^{-n} \mathbb{Z}_p \\ p^n \mathbb{Z}_p & \mathbb{Z}_p \end{pmatrix}. \]

In this case, we have

\[ \mathcal{R}=\begin{pmatrix} \mathbb{Z}_p & \mathbb{Z}_p \\ p^n \mathbb{Z}_p & \mathbb{Z}_p \end{pmatrix}, \quad \widehat{\mathcal{R}}=\begin{pmatrix} \mathbb{Z}_p & p^{-n} \mathbb{Z}_p \\ \mathbb{Z}_p & \mathbb{Z}_p \end{pmatrix}. \]

We define the abelian homomorphism $\jmath _v\colon \widehat {\mathcal {R}}\to \big ({\mathbb{Z}}/ {p^n}\mathbb{Z} \big )^{2}$ explicitly as

\[ \begin{pmatrix} a & b/p^n \\ c & d \end{pmatrix}\mapsto (c,b) \bmod p^n\mathbb{Z}_p. \]

A direct computation shows that this homomorphism has kernel $\mathcal {R}$ and that it satisfies the claimed properties.

This isomorphism is unique up to post-composition with a linear automorphism of the first and second coordinate of $\big ({\mathbb{Z}}/ {p^n}\mathbb{Z}\big )^{2}$, i.e. a transformation of the form $(b,c)\mapsto (b u_1, c u_2)$ for $u_1,u_2\in ({\mathbb{Z}}/ {p^n}\mathbb{Z})^{\times }$. The requirement that the quadratic form $-p^n\operatorname {Nr}$ descends to the product form $(b,c)\mapsto b\cdot c$ forces $u_2=u_1^{-1}$.

Proof. Again, as all Eichler orders are conjugate, the claim reduces to the case of

\[ \mathcal{R}=\begin{pmatrix} \mathbb{Z}_p & \mathbb{Z}_p \\ p^n \mathbb{Z}_p & \mathbb{Z}_p \end{pmatrix}, \quad \widehat{\mathcal{R}}=\begin{pmatrix} \mathbb{Z}_p & p^{-n} \mathbb{Z}_p \\ \mathbb{Z}_p & \mathbb{Z}_p \end{pmatrix}, \]

and

\[ \jmath_v\bigg(\!\!\begin{pmatrix} a & b/p^n \\ c & d \end{pmatrix}\!\!\bigg) =(c,b) \mod p^n\mathbb{Z}_p. \]

Because of Lemma 3.17, it is enough to compute $\rho (s).\mathbb {1}_{\mathcal {R}}$ for each of the representatives of ${\mathbf {SL}_2(\mathbb {Z}_p)/ U_0(p^n)}$. To find these representatives, we first write representatives for ${\mathbf {SL}_2(\mathbb {Z}_p)/ U_0(p)}$

\[ \mathbf{SL}_2(\mathbb{Z}_p)=w U_0(p)\sqcup \bigsqcup_{0< t \leq p} \begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix} U_0(p). \]

This decomposition follows from the fact that $U_0(p)$ is the stabilizer of an edge in the apartment of the diagonal torus in the Bruhat–Tits tree of $\mathbf {SL}_2(\mathbb {Q}_p)$ and the representatives above permute the $p+1$ neighbors of the vertex stabilized by $\mathbf {SL}_2(\mathbb {Z}_p)$.

Next, we find representatives for ${U_0(p)/ U_0(p^n)}$ using their definition as congruence subgroups

\[ U_0(p)=\bigsqcup_{0 < t_0 \leq p^{n-1}} \begin{pmatrix} 1 & 0 \\ p t_0 & 1 \end{pmatrix} U_0(p^n). \]

By combining, we arrive at

\[ \mathbf{SL}_2(\mathbb{Z}_p)=w \bigsqcup_{0 < t_0 \leq p^{n-1}} \begin{pmatrix} 1 & 0 \\ p t_0 & 1 \end{pmatrix} U_0(p^n) \sqcup \bigsqcup_{0 < t \leq p^n} \begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix} U_0(p^n). \]

We now compute explicitly the action of all representatives. We need to decompose $\widehat {\mathcal {R}}$ into $\mathcal {R}$-cosets

\[ \widehat{\mathcal{R}}=\bigsqcup_{0\leq \alpha,\beta< p^n} \begin{pmatrix} 0 & \alpha/p^n\\ \beta & 0 \end{pmatrix} +\mathcal{R}. \]

To simplify notation, we denote $x_{\alpha,\beta }\colon= \big (\begin{smallmatrix} 0 & \alpha /p^n\\ \beta & 0 \end{smallmatrix}\big )$, with $\jmath _v(x_{\alpha,\beta })=(\beta,\alpha )$. The duality between $\widehat {\mathcal {R}}$ and $\mathcal {R}$ implies $\operatorname {Nr}(x_{\alpha,\beta }+\mathcal {R})=\operatorname {Nr} x_{\alpha,\beta } + \mathbb {Z}_p=-\alpha \beta /p^n+\mathbb {Z}_p$. Write $\left (\begin{smallmatrix} 1 & 0 \\ t & 1 \end{smallmatrix}\right )=-w \left (\begin{smallmatrix} 1 & -t \\ 0 & 1 \end{smallmatrix}\right ) w$ and $x=\big (\begin{smallmatrix} a & b/p^n \\ c & d \end{smallmatrix}\big )$. Then,

\begin{align*} \rho\bigg(\!\!\begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix}\!\!\bigg).\mathbb{1}_{\mathcal{R}} (x) &=p^{-n}\rho(w) \rho\bigg(\!\!\begin{pmatrix} 1 & -t \\ 0 & 1 \end{pmatrix}\!\!\bigg). \mathbb{1}_{\widehat{\mathcal{R}}} (x) =p^{-n}\rho(w).\big( \psi_v(-t \operatorname{Nr} x) \mathbb{1}_{\widehat{\mathcal{R}}}(x)\big)\\ &= p^{-n}\rho(w).\bigg(\sum_{0\leq \alpha,\beta < p^n} \psi_v(t \alpha\beta/p^n) \mathbb{1}_{x_{\alpha,\beta}+ \mathcal{R}}(x) \bigg)\\ &= \mathbb{1}_{\widehat{\mathcal{R}}}(x) \cdot p^{-2n} \sum_{0\leq \alpha,\beta < p^n} \psi_v(t \alpha\beta/p^n+\langle x_{\alpha,\beta},x \rangle)\\ &= \mathbb{1}_{\widehat{\mathcal{R}}}(x) \cdot p^{-2n} \sum_{0\leq \alpha,\beta < p^n} \psi_v \bigg(\frac{-\alpha c- \beta b +t \alpha\beta}{p^n}\bigg). \end{align*}

Let $k=\operatorname {ord}_p t$. Then, summing first over $\alpha$ we deduce $p^k \mid c$ and summing first over $\beta$ we see that $p^k \mid b$. Using this input, we can sum first over $\alpha$ and then over $\beta$ to arrive at

\begin{align*} &\rho\bigg(\!\!\begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix}\!\!\bigg).\mathbb{1}_{\mathcal{R}} (x)\\ &\quad=\mathbb{1}_{\widehat{\mathcal{R}}}(x) p^{-(n-k)} \begin{cases} \psi_v\bigg(-\dfrac{(c/p^k)(b/p^k)}{(t/p^k)p^{n-k}}\bigg)=\psi_v\bigg(\dfrac{\nu_{p^k}(x)}{tp^{n-2k}}\bigg), & \jmath_v(x)\equiv (0,0) \bmod p^k, \\ 0, & \textrm{otherwise.} \end{cases} \end{align*}

We need only deal now with representatives of the form $w\left (\begin{smallmatrix} 1 & 0 \\ p t_0 & 1\end{smallmatrix}\right )=\left (\begin{smallmatrix} 1 & -p t_0 \\ 0 & 1 \end{smallmatrix}\right ) w$ which are easier to compute

\[ \rho\bigg(w\begin{pmatrix} 1 & 0 \\ p t_0 & 1\end{pmatrix}\!\!\bigg).\mathbb{1}_{\mathcal{R}} (x)=p^{-n}\psi_v(-p t_0 \operatorname{Nr} x) \mathbb{1}_{\widehat{\mathcal{R}}}(x).\]

Proof. The claim for $v=\infty$ follows immediately from Corollary 3.6. Fix now $v<\infty$. We may cover the set $\mathcal {K}$ by the product of its projections onto $\mathbf {SL}_2(F)$ and $\operatorname {O}(B_v,\operatorname {Nr})$, which is compact. Hence, we may assume without loss of generality that $\mathcal {K}=\mathcal {K}_0 \times \mathcal {K}_1$ is a compact product set. Note that it is enough to show that there is some $\mathcal {C}_0\subset B_v$ such that $|\rho (s).{M}|\ll \mathbb {1}_{\mathcal {C}_0}$ for $s\in \mathcal {K}_0$. In particular, the compact set $\mathcal {C}=\bigcup _{L\in \mathcal {K}_1} L^{-1}\mathcal {C}_0$ will satisfy the claimed properties. Using the Iwasawa decomposition, we can find a compact subset $C_P$ of the group of upper triangular matrices such that $\mathcal {K}_0\subset C_P\mathbf {SL}_2(\mathbb {Z}_p)$. Recall from Lemma 3.15 that the $\rho \big (\mathbf {SL}_2(\mathbb {Z}_p)\big )$-orbit of $M$ is finite and a uniform bound on $\rho (\mathcal {K_0}).M$ will follow from a uniform bound on $\rho (C_P).M'$ for every $M'$ in $\rho \big (\mathbf {SL}_2(\mathbb {Z}_p)\big ).M$. A uniform bound on $|\rho (C_P).M'|$ follows immediately from the formulae for the Weil action of diagonal and upper unipotent matrices.

Proof. Because the theta series converges absolutely, we may exchange summation and integration in the definition of $W_{\Theta _M(l,r;\bullet )}$ and write

\begin{align*} W_{\Theta_M(l,r;\bullet)}(s,\alpha)&= \sum_{\xi\in B }\int_{[\mathbf{N}(\mathbb{A})]} \bigg(\rho\bigg(\!\!\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}s\bigg)M \bigg) (l^{-1}\xi r) \psi(-\alpha n) \,{d} n\\ &= \sum_{\xi\in B }\int_{[\mathbf{N}(\mathbb{A})]} \big(\rho\big(s\big)M \big) (l^{-1}\xi r) \psi(\operatorname{Nr} \xi\cdot n-\alpha n) \,{d} n\\ &=\sum_{\substack{\xi\in B \\ \operatorname{Nr} \xi=\alpha}} \big(\rho(s)M\big)(l^{-1} \xi r). \end{align*}

Proof. This follows from combining the local contributions as calculated in Propositions 3.13 and 3.21 and the remarks following these propositions.

Proof. This follows from the discussion above, Lemma 3.17, and the local-to-global principle for lattices that implies

\[ \mathcal{R}=\bigcap_{v<\infty} \mathcal{R}_v, \]

where the intersection is taken in $B$.

Proof. Fix $M=M_\infty \cdot \prod _{v<\infty }M_v \in \Omega$ such that $M_v=\mathbb {1}_{\mathcal {R}_v}$ for all finite $v$. Then, Proposition 4.7 and the isomorphism (24) imply

\[ \frac{1}{\operatorname{covol}(\Lambda)}\int_{{\Lambda}\backslash{\mathbf{SL}_2(\mathbb{R})}} |\vartheta_{M_\infty}(l_\infty, r_\infty;s_\infty)|^2 \,{d} s_\infty= \int_{[\mathbf{SL}_2(\mathbb{A})]} |\Theta_M(l_\infty K_{\mathcal{R}}, r_\infty K_{\mathcal{R}};s)|^2 \,{d} s. \]

We proceed to bound the adelic integral by expanding the domain of integration to a Siegel set.

Denote by $\mathbf {N}<\mathbf {SL}_2$ the algebraic subgroup of upper triangular matrices. We have $\mathbf {N}\simeq \mathbb {G}_a$ and a fundamental domain for the action of $\mathbf {N}(\mathbb {Q})$ on $\mathbf {N}(\mathbb {A})$ is

\[ \mathcal{N}=\begin{pmatrix} 1 & [0,1) \\ 0 & 1 \end{pmatrix} \cdot \prod_{v<\infty} \begin{pmatrix} 1 & \mathbb{Z}_v \\ 0 & 1 \end{pmatrix}. \]

Set $A^>\colon= \big \{\!\operatorname {diag}(y^{1/2},y^{-1/2}) \colon y>\sqrt {3}/2\big \}\subset \mathbf {SL}_2(\mathbb {R})$. A Siegel set for the action of $\mathbf {SL}_2(\mathbb {Q})$ on $\mathbf {SL}_2(\mathbb {A})$ is given by

\[ \mathcal{S}=\mathcal{N}\cdot A^> \cdot \mathbf{SO}_2(\mathbb{R}) \mathbf{SL}_2(\hat{\mathbb{Z}}). \]

Denote $l=(l_\infty,e,e,\ldots )$ and similarly $r=(r_\infty,e,e,\ldots )$. Because the Siegel set contains a fundamental domain for the action of the lattice, we can write

\begin{align*} &\int_{[\mathbf{SL}_2(\mathbb{A})]} |\Theta_M(l_\infty K_{\mathcal{R}}, r_\infty K_{\mathcal{R}};s)|^2 \,{d} s\\ &\quad \leq \int_{\mathcal{S}} |\Theta_M(l_\infty K_{\mathcal{R}}, r_\infty K_{\mathcal{R}};s)|^2 \,{d} s\\ &\quad =\varrho_{\mathbf{SL}_2}^{-1}\int_{[\mathbf{N}(\mathbb{A})]} \int _{\sqrt{3}/2}^\infty \int_{\mathbf{SL}_2(\hat{\mathbb{Z}})} \int_{\mathbf{SO}_2(\mathbb{R})} \bigg|\sum_{\xi\in B} (\rho(n)\rho(\operatorname{diag}(y^{1/2},y^{-1/2}) k k_f).M)(l^{-1} \xi r) \bigg|^2 \,{d} n \frac{{d} y}{y^2} \,{d} k_f\,{d} k\\ &\quad=\varrho_{\mathbf{SL}_2}^{-1}\sum_{\alpha\in\mathbb{Q}} \int _{\sqrt{3}/2}^\infty \int_{\mathbf{SL}_2(\hat{\mathbb{Z}})} \int_{\mathbf{SO}_2(\mathbb{R})} \bigg|\sum_{\substack{\xi\in B \\ \operatorname{Nr} \xi=\alpha}} (\rho(\operatorname{diag}(y^{1/2},y^{-1/2}) k)\rho(k_f).M)(l^{-1} \xi r) \bigg|^2 \,{d} k_f \frac{{d} y}{y^2} \,{d} k\\ &\quad \leq\varrho_{\mathbf{SL}_2}^{-1}\sum_{\alpha\in\mathbb{Q}} \int _{\sqrt{3}/2}^\infty \int_{\mathbf{SL}_2(\hat{\mathbb{Z}})} \int_{\mathbf{SO}_2(\mathbb{R})} \bigg|\sum_{\substack{\xi\in B \\ \operatorname{Nr} \xi=\alpha}} \big|(\rho(\operatorname{diag}(y^{1/2},y^{-1/2}) k)\rho(k_f).M)(l^{-1} \xi r)\big| \bigg|^2 \,{d} k_f \frac{{d} y}{y^2}\, {d} k. \end{align*}

The last equality follows from the computation of the Fourier coefficients in the unipotent variable in Proposition 3.25 and the orthogonality of characters. We normalize the measure on $\mathbf {SL}_2(\mathbb {A}_f)$ so that $\mathbf {SL}_2(\widehat {\mathbb {Z}})$ has volume $1$. Then, the global measure normalization constant $\varrho _{\mathbf {SL}_2}$ is equal to the volume of ${\mathbf {SL}_2(\mathbb {Z})\backslash \mathbb {H}}$ with respect to the standard hyperbolic measure ${{d} x\, {d} y/y^2}$, i.e. $\varrho _{\mathbf {SL}_2}=\operatorname {covol}(\mathbf {SL}_2(\mathbb {Z}))= {\pi /3}$.

In the last line we have inserted the absolute value into the sum using the triangle inequality, and we continue to evaluate the integral along $\mathbf {SL}_2(\hat {\mathbb {Z}})$. From Lemmata 3.11 and 3.17, we know that the integrand is invariant under the finite index subgroup $U_\mathcal {R}<\mathbf {SL}_2(\hat {\mathbb {Z}})$. We decompose the integral into $\big [\mathbf {SL}_2(\hat {\mathbb {Z}})\colon U_{\mathcal {R}}\big ]$-integrals along the different cosets of $U_{\mathcal {R}}$ in $\mathbf {SL}_2(\hat {\mathbb {Z}})$. We have an exact expression for the integrand on each coset due to Proposition 4.5. The phases of the form $\psi \big ( {t \cdot \nu _a(x)/qD_B/a}\big )$ that appear in each element in the $\rho \big (\mathbf {SL}_2(\widehat {\mathbb {Z}})\big )$-orbit are discarded because of the innermost absolute value. Hence, each $U_{\mathcal {R}}$-coset reduces to a sum over elements in a lattice $\widehat {\mathcal {R}}^{(a)}$ for $a\mid q D_B$. From Remark 4.6, we deduce for any $a \mid q D_B$ that the weight of the sum over $\mathbb {1}_{\widehat {R}^{(a)}}$ is $\rho (a \mid q D_B) \varrho _{\mathbf {SL}_2}^{-1} \big [\mathbf {SL}_2(\widehat {\mathbb {Z}})\colon U_{\mathcal {R}}\big ]^{-1}$. Because $\Lambda =U_{\mathcal {R}}\cap \mathbf {SL}_2(\mathbb {A}_f)$ and $\mathbf {SL}_2(\mathbb {Z})=\mathbf {SL}_2(\widehat {\mathbb {Z}})\cap \mathbf {SL}_2(\mathbb {A}_f)$, the index satisfies $\varrho _{\mathbf {SL}_2}\big [\mathbf {SL}_2(\widehat {\mathbb {Z}})\colon U_{\mathcal {R}}\big ]=\operatorname {covol}(\Lambda )$. Because $\Lambda =\Gamma _0(qD_B)$, we see that the weight factor is equal to

\[ \operatorname{covol}(\Lambda)^{-1}\rho(a \mid q D_B)=\frac{\prod_{p\mid \gcd(q D_B/a,a)}(1-p^{-1})}{(\pi/3) q D_B\prod_{p\mid q D_B}(1+p^{-1})}\leq (q D_B)^{-1}.\]

Proof. This is simple to verify if $B$ is split at $p$ by conjugating $\mathcal {R}_p$ to $\big (\begin{smallmatrix} \mathbb {Z}_p & \mathbb {Z}_p \\ p^n \mathbb {Z}_p & \mathbb {Z}_p \end{smallmatrix}\big )$. If $B$ is ramified at $p$, then $\mathcal {R}_p^\times =\mathcal {O}_{B_p}^\times$ is the unit group of the ring of algebraic integers in $B_p$. The algebra $B_p$ contains an unramified quadratic extension $E/\mathbb {Q}_p$, hence $\operatorname {Nr} \mathcal {R}_p^\times$ contains $\operatorname {Nr} \mathcal {O}_E^\times =\mathbb {Z}_p^\times$.

Proof. The integral decomposes into a product of local integrals $\varrho _{\mathbf {G}}^{-1}\prod _p \int _{\mathbf {G}(\mathbb {Q}_p)} \mathbb {1}_{p^{-k_p}\mathcal {R}_p}(l_p^{-1}\xi ) \,{d} l_p$. All elements of $p^{-k_p}\mathcal {R}_p$ have norms in $p^{-2 k_p} \mathbb {Z}_p$. Hence, the local integral vanishes if $\operatorname {Nr}(l_p^{-1}\xi )=\operatorname {Nr}(\xi )\ \not \in p^{-2 k_p} \mathbb {Z}_p$. Because $\operatorname {Nr} \xi \in \mathbb {Q}$, the non-vanishing conditions at all primes $p$ imply that $\int _{\mathbf {G}(\mathbb {A}_f)}\cdots$ vanishes if $\operatorname {Nr} \xi \not \in N^{-1}\mathbb {Z}$.

Fix now $p$ and assume $\operatorname {Nr}\xi \in p^{-2k_p} \mathbb {Z}_p$. Then, the local integral is equal to $\int _{\mathbf {G}(\mathbb {Q}_p)} \mathbb {1}_{\mathcal {R_p}}(l_p^{-1}p^{k_p}\xi )\, {d} l_p$. The integrand is right-invariant under $K_p$. Denote by $\mathcal {R}_p(\alpha )$ the set of elements in $\mathcal {R}_p$ of norm $\alpha \in \mathbb {Q}_p^{\times }$. Of course, $\mathcal {R}_p(\alpha )=\emptyset$ if $\alpha \not \in \mathbb {Z}_p$. The set $\mathcal {R}_p(\alpha )$ is left-invariant under multiplication by $K_p \colon- \mathcal {R}_p(1)$, and

\[ \int_{\mathbf{G}(\mathbb{Q}_p)} \mathbb{1}_{\mathcal{R_p}}(l_p^{-1}p^{k_p}\xi) \,{d} l_p = m_{\mathbf{G}(\mathbb{Q}_p)}(K_p) \cdot\#\Big({K_p ^{\backslash ^{\mathcal{R}_p\big(p^{2k_p}\operatorname{Nr}\xi\big)}}} \Big). \]

We have $m_{\mathbf {G}(\mathbb {Q}_p)}(K_p)=1$ if $p\not \mid q$ and $m_{\mathbf {G}(\mathbb {Q}_p)}(K_p)=(p+1)^{-1}p^{-n+1}$ if $p^n\parallel q$ with $n>0$.

We now estimate $\#\big ({K_p \backslash \mathcal {R}_p(\alpha )} \big )$. Define

\[ \mathcal{R}_p(\alpha)^{\dagger}=\big\{x\in\mathcal{R}_p\colon |\operatorname{Nr} x|_p=|\alpha|_p \big\}, \]

evidently $\mathcal {R}_p(\alpha )\subset \mathcal {R}_p(\alpha )^{\dagger}$. The set $\mathcal {R}_p(\alpha )^{\dagger}$ is invariant under left multiplication by $\mathcal {R}_p^\times$. Because $\operatorname {Nr} \mathcal {R}_p^\times =\mathbb {Z}_p^\times$, each coset of ${\mathcal {R}_p^\times \backslash \mathcal {R}_p(\alpha )^{\dagger} }$ contains exactly one coset of ${K_p \backslash \mathcal {R}_p(\alpha )}$. Thus, $\#\big ({K_p \backslash \mathcal {R}_p(\alpha )} \big )=\#\big ({\mathcal {R}_p^\times \backslash \mathcal {R}_p(\alpha )^{\dagger} } \big )$.

If $B_p$ ramifies, then the fact that $\operatorname {ord}_p(\operatorname {Nr} \bullet )$ is a valuation on $B_p$ implies that $\mathcal {R}(\alpha )^{\dagger}$ is a single coset of $\mathcal {R}_p^\times$ if $\alpha \in \mathbb {Z}_p$. In the split case, we can assume $B_p=\operatorname {Mat}_{2\times 2}(\mathbb {Q}_p)$ and $\mathcal {R}_p=\big (\begin{smallmatrix}\mathbb {Z}_p & \mathbb {Z}_p \\ p^n \mathbb {Z}_p & \mathbb {Z}_p \end{smallmatrix}\big )$, where $p^n\parallel q$. Let $\widetilde {\mathcal {R}}_p^\times$ be the image of $\mathcal {R}_p^\times$ in $\mathbf {PGL}_2(\mathbb {Q}_p)$. Then, the map ${\mathcal {R}_p^\times \backslash \mathcal {R}_p(\alpha )^{\dagger} }\to {\widetilde {\mathcal {R}}_p^\times \backslash \mathbf {PGL}_2(\mathbb {Q}_p)}$ is injective because $Z_{\mathbf {GL}_2}(\mathbb {Q}_p)\cap \{g\in \mathbf {GL}_2(\mathbb {Q}_p)\colon |\operatorname {det} g|_p=1\}\subset \mathcal {R}_p^\times$. Hence, it is enough to find an upper bound for the number of $\widetilde {\mathcal {R}}_p^\times$ cosets in the image of $\mathcal {R}_p(\alpha )^{\dagger}$ in $\mathbf {PGL}_2(\mathbb {Q}_p)$.

The group $\mathbf {PGL}_2(\mathbb {Z}_p)$ is the stabilizer of a vertex $v_0$ in the Bruhat–Tits tree of $\mathbf {PGL}_2(\mathbb {Q}_p)$ and $\operatorname {diag} (p^{-n}, 1) \mathbf {PGL}_2(\mathbb {Z}_p) \operatorname {diag}(p^n,1)$ is a stabilizer of a vertex $v_n$, with $\operatorname {dist}(v_0,v_n)=n$. Hence, $\widetilde {\mathcal {R}}_p^\times$, which is the intersection of the two, is the stabilizer of the geodesic path of length $n$ connecting $v_0$ and $v_n$ in the tree. Because $\mathbf {PGL}_2(\mathbb {Q}_p)$ acts strongly transitively on its Bruhat–Tits tree, it acts transitively on the set of geodesic paths of length $n$. Hence, the map $g_p\mapsto (g_p^{-1}.v_0, g_p^{-1}.v_n)$ is a bijection between ${\widetilde {\mathcal {R}}_p^{\times }\backslash \mathbf {PGL}_2(\mathbb {Q}_p)}$ and the set of oriented geodesic paths of length $n$ in the tree. We need to find an upper bound on the number of paths that correspond to the image of $\mathcal {R}_p(\alpha )^{\dagger}$.

Denote $\operatorname {ord}_p(\alpha )=m$. If $g_p\in \mathcal {R}_p(\alpha )^{\dagger}$, then the existence of the Smith normal form for $g_p\in M_2(\mathbb {Z}_p)$ implies that $g_p\in \mathbf {GL}_2(\mathbb {Z}_p) \operatorname {diag}(p^{m_1}, p^{m_2})\mathbf {GL}_2(\mathbb {Z}_p)$ for some $m_2\geq m_1\geq 0$, with $m_2+m_1=m$. Then, $\operatorname {dist}(g_p^{-1}.v_0,v_0)=m_2-m_1\leq m$. Hence, the number of possibilities for the first vertex of the $n$-path is at most the number of vertices in a ball of radius $m$, that is $1+ ((p+1)/(p-1))(p^m-1)\ll p^m$. Because the length of the path is $n$, the number of possibilities for the final vertex, after the first vertex has been fixed, is at most $\lfloor (p+1)p^{n-1}\rfloor =m_{\mathbf {G}(\mathbb {Q}_p)}(K_p)^{-1}$. We conclude that if $B_p$ is split, then

\[ m_{\mathbf{G}(\mathbb{Q}_p)}(K_p) \cdot\#\Big({K_p ^{\backslash ^{\mathcal{R}_p(p^{2k_p}\operatorname{Nr}\xi)}}} \Big)\ll p^{2k_p+\operatorname{ord}_p(\operatorname{Nr}\xi)}\ll |N \operatorname{Nr} \xi|_p^{-1}. \]

Multiplying the contributions from all primes $p$ we arrive at the claimed bound.

Proof. The theta series $\Theta _M$ can be rewritten as a sum over $B^\times$ due to the vanishing condition for norm-zero elements. A priori, we do not even know that the left-hand side is integrable. Thus, we proceed with the following computation as formal operations which hold for positive-valued functions. We will then use the positive-valued case to show absolute convergence which will justify these operations in general.

Unfold first the integral along the $l$ variable to rewrite the left-hand side above as

\[ \int_{[\mathbf{G}(\mathbb{A})]}\varphi'(r) \sum_{\xi\in {\mathbf{G}(\mathbb{Q})\backslash B^\times}} \int_{\mathbf{G}(\mathbb{A})} M(l^{-1}\xi r) \varphi(l) \,{d} l \,{d} r. \]

Two elements in $B^\times$ are in the same left $\mathbf {G}(\mathbb {Q})$-orbit exactly when they have the same norm. Thus, the equality in question holds if we can establish absolute integrability. To show absolute integrability, we first bound

\[ \int_{[\mathbf{G}(\mathbb{A})]} \int_{[\mathbf{G}(\mathbb{A})]} \sum_{\xi\in B}\big| M(l^{-1}\xi r) \varphi(l) \varphi'(r)\big| \,{d} l \,{d} r \leq \|\varphi\|_\infty \|\varphi'\|_\infty \sum_{\alpha\in \mathbb{Q}^\times} \int_{[\mathbf{G}(\mathbb{A})]} \int_{\mathbf{G}(\mathbb{A})} \big| M(l^{-1}\xi_\alpha r) \big| \,{d} l \,{d} r . \]

By expanding the function $M$ into finite summands of standard test functions, we reduce to the case that $M=\prod _v M_v$. Furthermore, because we are only interested in upper bounds, we can further reduce to the case that in any finite place $v=p$ the function $M_v$ is a multiple of the characteristic function of $p^{-k_p} \mathcal {R}_p$, where $k_p=0$ for almost all $p$. Taking into account this reduction, the function $M$ is right- and left-invariant under $K_{\mathcal {R}}$ and we can apply Lemma 5.3 above. First we deduce that the integral over $l$ vanishes unless $\alpha \in N^{-1}\mathbb {Z}$ for some fixed integer $N$ depending only on $M$. In addition, by using the bound from Lemma 5.3, we can write

(25) \begin{align} &\int_{[\mathbf{G}(\mathbb{A})]} \int_{[\mathbf{G}(\mathbb{A})]} \sum_{\xi\in B}\big| M(l^{-1}\xi r) \varphi(l) \varphi'(r)\big| \,{d} l \,{d} r\nonumber\\ &\quad \ll_{\,\mathbf{G},\varphi,\varphi', M,\varepsilon} Nq^{\varepsilon} \sum_{0\neq \alpha\in N^{-1}\mathbb{Z}} |\alpha| \int_{{\Lambda \backslash \mathbf{G}(\mathbb{R})}} \int_{\mathbf{G}(\mathbb{R})} \big|M_\infty(l^{-1}\xi_\alpha r)\big|\,{d} l \,{d} r\nonumber\\ &\quad = Nq^{\varepsilon}\sum_{0\neq \alpha\in N^{-1}\mathbb{Z}} |\alpha| \int_{{\Lambda \backslash \mathbf{G}(\mathbb{R})}} \int_{\mathbf{G}(\mathbb{R})} \big|M_\infty(l^{-1}\xi_\alpha)\big| \,{d} l \,{d} r\nonumber\\ &\quad \ll_{\,\mathbf{G},\Lambda} Nq^{\varepsilon}\sum_{0\neq \alpha\in N^{-1}\mathbb{Z}} |\alpha| \int_{\mathbf{G}(\mathbb{R})} \big|M_\infty(l^{-1}\xi_\alpha)\big|\,{d} l\nonumber\\ &\quad \ll Nq^{\varepsilon}\sum_{0\neq \alpha\in N^{-1}\mathbb{Z}} |\alpha| \int_{\mathbf{G}(\mathbb{R})} \big(1+\big\|l^{-1}\xi_\alpha \big\|\big)^{-4-\delta} \,{d} l. \end{align}

In the second line, we have made a change of variable $l\mapsto \xi _\alpha r \xi _\alpha ^{-1}l$. Note that we can take here $\xi _\alpha$ to be any real matrix of determinant $\alpha$, choose $\xi _\alpha =\operatorname {diag} \big (\sqrt {|\alpha |}, \operatorname {sign}(\alpha )\sqrt {|\alpha |}\big )$. The last integral in (25) can be computed using the formula for the Haar measure in Cartan coordinates

\begin{align*} &\int_{\mathbf{G}(\mathbb{R})} \big(1+\sqrt{|\alpha|}\big\|l^{-1}\operatorname{diag}(1,\operatorname{sign}(\alpha)) \big\|\big)^{-4-\delta} \, {d} l\\ &\quad \ll \int_0^{\infty}\big(1+\sqrt{2|\alpha|\cosh(t))}\big)^{-4-\delta}\sinh(t) \,{d} t \\ &\quad\ll |\alpha|^{-2-\delta/2}\int_0^\infty \frac{\sinh(t)}{\cosh(t)^{2+\delta/2}} \,{d} t\\ &\quad =|\alpha|^{-2-\delta/2}\bigg[\frac{-1}{\cosh(t)^{1+\delta/2}(1+\delta/2)}\bigg]_0^\infty\ll |\alpha|^{-2-\delta/2}. \end{align*}

Finally, we see that the expression in (25) is bounded from above by

\[ N^{2+\delta/2}q^{\varepsilon}\sum_{0\neq n \in \mathbb{Z}} \frac{1}{|n|^{1+\delta/2}}<\infty.\]

Proof. By Proposition 5.4, we know that $F(s)$ is well-defined pointwise and can be rewritten as

\[ F(s)=\sum_{\alpha \in \mathbb{Q}^\times} \int_{[\mathbf{G}(\mathbb{A})]} \varphi'(r) \int_{\mathbf{G}(\mathbb{A})} \big(\rho(s)M\big)(l^{-1}\xi_\alpha r) \varphi(l) \,{d} l \,{d} r. \]

To calculate $\int _{[\mathbf {SL}_2(\mathbb {A})]} |F(s)|^2\,{d} s$, we will bound the integral over $[\mathbf {SL}_2(\mathbb {A})]$ by an integral over a Siegel set $\mathcal {S}=\mathcal {N}\cdot A^> \cdot \mathbf {SO}_2(\mathbb {R}) \mathbf {SL}_2(\hat {\mathbb {Z}})$ as in the proof of Proposition 4.9. Because $M$ belongs to $\Omega$ and $M_\infty \in V_{m,2\pi }$, the function $M$ has a finite $\rho \big ( \mathbf {SL}_2(\hat {\mathbb {Z}})\big )$-orbit and $\rho \big (\mathbf {SO}_2(\mathbb {R})\big )$-isotypic. Hence, it is enough to bound $\int _{\mathcal {N}\cdot A^>} F_1(z) \overline {F_2(z)} \,{d} z$ where $F_1$, $F_2$ are defined in the same manner as $F$ but with $M$ replaced by test functions $M_1$, $M_2$ in the $\rho \big (\mathbf {SO}_2(\mathbb {R}) \mathbf {SL}_2(\hat {\mathbb {Z}})\big )$-orbit of $M$. Denote $a(y)=\operatorname {diag}(y^{1/2},y^{-1/2})\in \mathbf {SL}_2(\mathbb {R})$. Using the orthogonality relation of additive characters and the sup-norm bound on $\varphi$, $\varphi '$, we arrive at

(26)\begin{align} &\bigg|\int_{\mathcal{N}\cdot A^>} F_1(z) \overline{F_2(z)} \,{d} z\bigg|\nonumber\\ &\quad \ll_{\,\varphi,\varphi'} \sum_{\alpha\in\mathbb{Q}^\times} \int_{[\mathbf{G}(\mathbb{A})]} \int_{\mathbf{G}(\mathbb{A})} \int_{[\mathbf{G}(\mathbb{A})]} \int_{\mathbf{G}(\mathbb{A})}\nonumber\\ &\qquad \int_{\sqrt{3}/2}^{\infty} \bigg|\rho_\infty(a(y)).M_1(l_1^{-1} \xi_\alpha r_1) \overline{\rho_\infty(a(y)). M_2(l_2^{-1} \xi_\alpha r_2)}\bigg| \frac{{d} y}{y^2} \,{d} l_2 \,{d} r_2 \,{d} l_1 \,{d} r_1\nonumber\\ &\quad =\sum_{\alpha\in\mathbb{Q}^\times} \int_{\mathbf{G}(\mathbb{A})} \int_{\mathbf{G}(\mathbb{A})} \int_{\sqrt{3}/2}^{\infty} \bigg|\rho_\infty(a(y)).M_1(l_1^{-1} \xi_\alpha) \overline{\rho_\infty(a(y)).M_2(l_2^{-1} \xi_\alpha )}\bigg|\frac{{d} y}{y^2} \,{d} l_2 \,{d} l_1. \end{align}

As in the proof of Proposition 5.4, we apply Lemma 5.3 to the integrals in the $l_1$ and $l_2$ variables. The integral vanishes unless $\alpha \in N^{-1}\mathbb {Z}$ for some integer $N>0$ depending only on $M$. For every $\varepsilon >0$, we can bound (26) from above by

\begin{align*} &\varrho_{\mathbf{G}}^{-2}\sum_{0\neq \alpha\in N^{-1}\mathbb{Z}} |N\alpha|^{2}q^{\varepsilon} \int_{\mathbf{G}(\mathbb{R})} \int_{\mathbf{G}(\mathbb{R})} \int_{\sqrt{3}/2}^{\infty} \Big|\rho_\infty(a(y)).M_1(l_1^{-1} \xi_\alpha) \overline{\rho_\infty(a(y)).M_2(l_2^{-1} \xi_\alpha )}\Big| \frac{{d} y}{y^2} \,{d} l_2 \,{d} l_1\\ &\quad=\varrho_{\mathbf{G}}^{-2} \sum_{0\neq \alpha\in N^{-1}\mathbb{Z}} |N\alpha|^{2}q^{\varepsilon} \int_{\mathbf{G}(\mathbb{R})} \int_{\mathbf{G}(\mathbb{R})} \int_{\sqrt{3}/2}^{\infty} \Big|M_1(\sqrt{y} l_1^{-1} \xi_\alpha) \overline{M_2(\sqrt{y} l_2^{-1} \xi_\alpha )}\Big|\, {{d} y} \,{d} l_2 \,{d} l_1\\ &\quad\ll_{\mathbf{G},M} \sum_{0\neq \alpha\in N^{-1}\mathbb{Z}} |N\alpha|^{2}q^{\varepsilon} \int_{\mathbf{G}(\mathbb{R})} \int_{\mathbf{G}(\mathbb{R})} \int_{\sqrt{3}/2}^{\infty} \|\sqrt{y} l_1^{-1} \xi_\alpha\|^{-4-\delta}\|\sqrt{y} l_2^{-1} \xi_\alpha\|^{-4-\delta} \,{{d} y} \,{d} l_2 \,{d} l_1\\ &\quad \ll \sum_{0\neq \alpha\in N^{-1}\mathbb{Z}} |N\alpha|^{2}q^{\varepsilon} \int_{\mathbf{G}(\mathbb{R})} \int_{\mathbf{G}(\mathbb{R})} \|l_1^{-1} \xi_\alpha\|^{-4-\delta}\|l_2^{-1} \xi_\alpha\|^{-4-\delta} \,{d} l_2 \,{d} l_1 . \end{align*}

Take $\xi _\alpha =\operatorname {diag}(\sqrt {|\alpha |},\operatorname {sign}(\alpha )\sqrt {|\alpha |})$ and bound the last integral from above in the same manner as in the proof of Proposition 5.4 by a multiple of $|\alpha |^{-4-\delta }$. It follows that

\[ \int_{[\mathbf{SL}_2(\mathbb{A})]} |F(s)|^2 \,{d} s \ll_{\varphi,\varphi',G, M, \varepsilon} N^{4+\delta}q^{\varepsilon}\sum_{0\neq n \in \mathbb{Z}} \frac{1}{|n|^{2+\delta}}<\infty.\]

Proof. We only establish the second claim as it immediately implies the first. Proposition 5.6 implies $(\varphi \otimes \overline {\varphi '} )_M(s)\in L^2([\mathbf {SL}_2(\mathbb {A})])$. We then apply Propositions 5.4 to deduce (29).

We use the notation $u_n\colon= \left (\begin{smallmatrix} 1 & n \\ 0 & 1 \end{smallmatrix}\right )$. Fubini's theorem and the orthogonality of characters imply for all $\alpha,\beta \in \mathbb {Q}^\times$, $s \in \mathbf {SL}_2(\mathbb {A})$, and $x\in \mathbf {G}(\mathbb {A})$

\[ \int_{[\mathbf{N}(\mathbb{A})]} T^{\rho(u_n s)M}_\alpha \varphi(x) \psi(-\beta n)\,{d} n=\begin{cases} T^{\rho(s)M}_\alpha\varphi (x), & \alpha=\beta,\\ 0, & \alpha \neq \beta. \end{cases} \]

The claim follows from substituting this expression in the definition of the Whittaker function applied to (29).