2. Preliminary considerations

In this section we are putting in some ground work on which the following sections will rely.

Recall that $\pi$ was a cuspidal automorphic representation. Since we are assuming that $\pi_v$ is unramified for $v\neq p$ , the (arithmetic)-conductor of $\pi$ is $p^{n_{\pi}}$ for $n_{\pi}\in \mathbb{N}\cup\{0\}$ . When $n_{\pi}=0$ we have $d=1$ and our theorem reduces to the local bound in the spectral aspect, so that without loss of generality we can assume $n_{\pi}\geq 1$ throughout. By Flath’s factorisation theorem we can fix an isomorphism $\pi \cong \bigotimes \pi_v$ . Note that also the central character of $\pi$ factors as $\omega_{\pi}=\bigotimes_v \omega_{\pi_v}$ where $\omega_{\pi_v}$ is the central character of $\pi_v$ . For $v\neq p$ we can fix a spherical (i.e. $K_v$ -invariant vector) $\phi_v^{\circ}\in \pi_v^{K_v}$ . This vector is unique up to scaling. Recall that $\phi_i$ , $i=1,\ldots,d$ forms an orthonormal basis of $V=\pi^{K(p^{m_{\pi}})}$ . Thus there is $\phi_p^{(i)}$ so that we can identify

\begin{equation*} \phi_i = \phi_p^{(i)} \otimes \bigotimes_{v\neq p} \phi_v^{\circ}.\end{equation*}

Since the spherical functions $\phi_p^{\circ}$ are well understood much of our work boils down to understanding properties of an orthogonal basis

\begin{equation*} \text{span}\{ \phi_p^{(1)},\ldots,\phi_p^{(d)} \} = \pi_p^{K_p(m_{\pi})}.\end{equation*}

This a purely local problem, which we investigate in the following subsection.

2·1. Local considerations

We now focus on properties of the local representation $\pi_p$ . We start by recalling the classification of local representations. But before we do so we need some more notation. Given a (quasi)-character $\chi\colon \mathbb{Q}_p^{\times}\to \mathbb{C}^{\times}$ we write $a(\chi)$ for the (exponent)-conductor. Further write

\begin{align*}I_0(p)=\left[\begin{matrix} \mathbb{Z}_p^{\times} &\mathbb{Z}_p \\ p\mathbb{Z}_p & \mathbb{Z}_p^{\times}\end{matrix}\right]\subset K_p\end{align*}

for an Iwahori subgroup. Let $K_p'=N_{G(\mathbb{Q}_p)}(I_0(p))$ be the normaliser of $I_0(p)$ in $G(\mathbb{Q}_p)$ . We also need the filtration

\begin{equation*} K_p'(m) = 1+\left[\begin{matrix}p\mathbb{Z}_p & \mathbb{Z}_p \\ p\mathbb{Z}_p & p\mathbb{Z}_p\end{matrix}\right]^m\end{equation*}

of $K_p'$ by normal subgroups. Finally given two quasi characters $\chi_1,\chi_2\colon \mathbb{Q}_p^{\times}\to\mathbb{C}^{\times}$ we form the (normalised) induced representation on $\operatorname{Ind}_B^{G(\mathbb{Q}_p)}(\chi_1\otimes \chi_2)$ as usual. If this representation is irreducible, then we denote the so obtained representation by $\chi_1\boxplus\chi_2$ . We write St for the Steinberg representation which we may identify with the unique irreducible subspace of $\operatorname{Ind}_B^{G(\mathbb{Q}_p)}(\lvert{\cdot}\rvert^{\frac{1}{2}}\otimes \lvert{\cdot}\rvert^{-\frac{1}{2}})$ . We are now ready to recite the following well-known classification.

Lemma 2·1. The representation $\pi_p$ falls into one of the following three cases:

1. (i) Case 1 (Principal series): there are (quasi)-characters $\chi_i\colon \mathbb{Q}_p^{\times}\to \mathbb{C}^{\times}$ such that $\chi_1\chi_2=\omega_{\pi_p}$ , $a(\chi_1)+a(\chi_2)=n_{\pi}$ and $\pi_p = \chi_1\boxplus \chi_2$ .

2. (ii) Case 2 (special): there is a (quasi)-character $\chi\colon \mathbb{Q}_p^{\times}\to \mathbb{C}^{\times}$ with $n_{\pi}=2a(\chi)$ if $a(\chi)>0$ or $n_{\pi}=1$ otherwise, $\chi^2=\omega_{\pi_p}$ and $\pi_p=\chi\otimes \text{St}$ .

3. (iii) Case 3 (Supercuspidal): the representation $\pi_p$ is supercuspidal. In this case we can write $\pi_p=\chi \cdot \pi_p'$ for a (quasi)-character $\chi\colon \mathbb{Q}_p^{\times}\to\mathbb{C}^{\times}$ and some twist-minimal representation $\pi_p'$ of conductor $n_{\pi}'$ which is constructed in one of the following two ways:

   1. (a) Case 3.1 ( $n_{\pi}'$ even): there is an irreducible representation $\tau$ of $Z\cdot K_p$ with $\tau\vert_{Z} = \chi^{-2}\cdot \omega_{\pi_p}$ which factors through $K_p({n_{\pi}'}/{2})$ so that $\pi_p' = c-\operatorname{Ind}_{ZK_p}^{G(\mathbb{Q}_p)}\tau$ .

   2. (b) Case 3.2 ( $n_{\pi}'$ odd): there is an irreducible representation $\tau$ of $K_p'$ which is invariant by $K_p'(n_{\pi}'-1)$ with $\tau\vert_Z=\chi^{-2}\cdot \omega_{\pi_p}$ such that $\pi_p'=c-\operatorname{Ind}_{K_p'}^{G(\mathbb{Q}_p)}\tau$ .

With this classification at hand we continue to study the subspaces V in more detail.

Lemma 2·2. Suppose $\pi_p$ is twist minimal and let $m_{\pi}$ be as in (2), then $V=\pi_p^{K(m_{\pi})}$ is irreducible as $K_p$ -module and we have:

1. (i) the invariant $m_{\pi}$ is given by

   \begin{align*}m_{\pi}=\begin{cases} n_{\pi} &\text{ if } \pi_p \text{ is Case 1},\\ \lfloor \frac{n_{\pi}+1}{2}\rfloor & \text{ if }\pi_p \text{ is Case 2,3;}\end{cases}\end{align*}

2. (ii) the dimension of V is given by

   \begin{align*}d= p^{m_{\pi}}\cdot \begin{cases} (1+\frac{1}{p}) &\text{ if $\pi_p$ is Case 1,}\\1 &\text{ if $\pi_p$ is Case 2,} \\ (1-\frac{1}{p}) &\text{ if $\pi_p$ is Case 3$\cdot$1 and }\\ (1-\frac{1}{p^2}) &\text{ if $\pi_p$ is Case 3$\cdot$2;} \end{cases}\end{align*}

Proof. This is not new and we only have to ensemble the pieces appropriately. Let us proceed case by case.

First, if $\pi_p$ is in Case 1, then twist-minimality implies that $\chi_2$ (or similarly $\chi_1$ ) is unramified. Thus we have $n_{\pi}=a(\chi_1)$ and the results on d and $m_{\pi}$ follow from [ Reference Miyauchi and Yamauchi20 , proposition 4.3]. Irreducibility can be seen by direct computation.

Second, if $\pi_p$ is in Case 2 and twist minimal, then $\pi_p = \text{St}$ and $n_{\pi}=1$ . The results on d and $m_{\pi}$ follow again from [ Reference Miyauchi and Yamauchi20 , proposition 4.3]. In this case irreducibility follows from [ Reference Casselman8 , theorem 1].

Finally, if $\pi_p$ belongs to Case 3, then the full statement is given in [ Reference Loeffler and Weinstein17 , theorem 3·5]. (See also [ Reference Miyauchi and Yamauchi20 , lemma 4·5, corollary 4·7] for the computation of $m_{\pi}$ and d.)

2·2. A generating domain

We now switch to the global picture again and aim to produce a suitable set $\mathcal{F}\subset G(\mathbb{A})$ which reduces our problem to studying

\begin{equation*} \mathcal{S}(\Phi,\mathcal{F}) = \sup_{g\in\mathcal{F}} \lvert{\Phi(g)}\rvert.\end{equation*}

Let $\mathcal{F}$ be the standard fundamental domain for $\operatorname{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ , which we identify with a subset of $\operatorname{GL}_2(\mathbb{R})$ by identifying $z=x+iy\in\mathbb{H}$ with $n(x)a(y)\in B(\mathbb{R})\subset\operatorname{GL}_2(\mathbb{R})$ . Here

\begin{equation*} n(x) =\left(\begin{matrix} 1&x\\0&1\end{matrix}\right) \text{ and } a(y) =\left(\begin{matrix} y&0\\0&1\end{matrix}\right).\end{equation*}

We further view $\mathcal{F}$ as a subset of $G(\mathbb{A})$ by identifying it with its image under the usual embedding $G(\mathbb{R}) \to G(\mathbb{A})$ . The same series of identifications allows us to write $\Phi(z)$ for $z\in \mathbb{H}$ .

Lemma 2·3. We have

\begin{equation} \Vert \Phi\Vert_{\infty} = \mathcal{S}(\Phi,\mathcal{F}).\nonumber\end{equation}

Proof. First we take $g\in G(\mathbb{A})$ and observe that by strong approximation we can write

\begin{equation} g=\gamma zb k \text{ with }\gamma\in G(\mathbb{Q}),\, z\in Z(\mathbb{R}),\, b\in\mathcal{F} \text{ and }k\in K.\nonumber\end{equation}

We directly obtain $\lvert{\phi_i(g)}\rvert=\lvert{\phi_i(bk)}\rvert$ by automorphy and the action of Z via a unitary character. However, we now observe that if $\phi_1,\ldots,\phi_d$ forms an orthonormal basis of V, then so does $\pi(k)\phi_1,\ldots, \pi(k)\phi_d$ . Let us write $\Phi^{(k)}$ for the average constructed from the latter basis. Recall that $\Phi$ was independent of the choice of the underlying orthonormal basis, so that we have $\Phi^{(k)}=\Phi$ . We conclude that

\begin{equation} \Phi(g) \leq \mathcal{S}(\Phi^{(k)}, \mathcal{F})=\mathcal{S}(\Phi, \mathcal{F}).\nonumber\end{equation}

3. The Whittaker bound

We will now start the process of deriving a first bound for $\Phi$ which will be valid (high) up in the cusp. This is done by estimating $\Phi$ using the Whittaker expansions of the $\phi_i$ ’s. Throughout we will be working with an arbitrary orthogonal basis $\phi_1,\ldots,\phi_d$ and consider only $g\in \mathcal{F}$ .

3·1. Reduction to a local problem

Let $\phi=\phi_i$ for some $i=1,\ldots, d$ . The global Whittaker period is given by

\begin{equation} W_{\phi}(g) = \int_{\mathbb{Q}\backslash \mathbb{A}} \phi(n(x)g)\psi_{\mathbb{A}}(x)^{-1}dx,\nonumber\end{equation}

where $\psi_{\mathbb{A}}$ is the standard character of $\mathbb{Q}\backslash\mathbb{A}$ which has a factorisation $\psi_{\mathbb{A}} = \bigotimes_{v}\psi_v$ for $\psi_{\infty}(x_{\infty}) = e(x_{\infty})$ and $\psi_l$ unramified for all primes l. Note that $W_{\phi}(\!\cdot)$ is right $K(p^{m_{\pi}})$ -invariant and transforms with respect to $\psi_{\mathbb{A}}$ when acted on by $N(\mathbb{A})$ from the left. Thus a standard trick shows that $W_{\phi}(a(q)g_{\infty})=0$ unless $0\neq q\in \frac{1}{p^{m_{\pi}}}\mathbb{Z}$ . Indeed, for any x with $n(x)\in K(p^{m_{\pi}})$ , one computes

\begin{equation} W(a(q)g_{\infty})=W(a(q)g_{\infty}n(x)) = W(n(xq)a(q)g_{\infty}) = \psi_{\mathbb{A}}(xq)W(a(q)g_{\infty}).\nonumber\end{equation}

we conclude that, if $W(a(q)g_{\infty})\neq 0$ , then we have $\psi_{\mathbb{A}}(xq)=1$ for all such x. This gives precisely the condition $q\in {1}/{p^{m_{\pi}}}\mathbb{Z}$ .

This observation leads to the Whittaker expansion

\begin{equation} \phi(g_{\infty}) = \sum_{n\in \mathbb{Z}\setminus \{0\}} W_{\phi}\left(a(\frac{n}{p^{m_{\pi}}})g_{\infty}\right).\nonumber\end{equation}

We need to exploit the factorisation of the Whittaker function $W_{\phi}$ . To do so we first observe that we have the factorisation of Whittaker models

\begin{equation} \mathcal{W}(\pi,\psi_{\mathbb{A}}) = \bigotimes_v \mathcal{W}(\pi_v,\psi_v).\nonumber\end{equation}

Using the factorisation of $\phi$ will now determine distinguished elements in the local Whittaker models as follows. Starting at $v=\infty$ we set

\begin{equation} W_v(n(x)a(y)) = \frac{\lvert{y}\rvert^{\frac{1}{2}}K_{it_{\pi}}(2\pi\lvert{y}\rvert)}{2\lvert{\Gamma(\frac{1}{2}+it_{\pi})\Gamma(\frac{1}{2}-it_{\pi})}\rvert^{\frac{1}{2}}}e(x),\nonumber\end{equation}

where $t_{\pi}$ is the spectral parameter of $\pi_{\infty}$ . Of course $W_{v}$ is the spherical Whittaker function and is normalised so that

\begin{equation} \int_{\mathbb{R}^{\times}} \lvert{W_{v}(a(y))}\rvert^2\frac{dy}{\lvert{y}\rvert} = 1, \nonumber\end{equation}

where dy is the normal Lebesgue measure.

We turn towards the finite places $v\neq p$ given by some prime $l\neq p$ . The spherical Whittaker function in $\mathcal{W}(\pi_v,\psi_v)$ is then given by

\begin{equation} W_v(a(y)) = \lvert{y}\rvert_v^{\frac{1}{2}}\lambda_{\pi}(l^{v_l(y)}).\nonumber\end{equation}

Here $\lambda_{\pi}(n)$ is defined by

\begin{align*}L^{\{\infty,p\}}(s,\pi)= \prod_{v\neq p,\infty} L_v(s,\pi_v)=\sum_{(n,p)=1}\lambda_{\pi}(n)n^{-s}\end{align*}

in analytic normalisation. We have set things up so that $W_v(1)=1$ .

Finally we turn towards $v=p$ . Here we write $W_p^{(i)}$ for an element in the image of $\mathbb{C} \phi_p^{(i)}$ in the Whittaker model $\mathcal{W}(\pi_p,\psi_p)$ such that

\begin{equation} \langle W_p^{(i)},W_p^{(i)}\rangle_{\mathcal{W}(\pi_p,\psi_p)} = \int_{\mathbb{Q}_p^{\times}}\lvert{W_p^{(i)}(a(y))}\rvert^2 \frac{dy}{\lvert{y}\rvert}=1,\nonumber\end{equation}

here dy is the Haar measure of $\mathbb{Q}_p$ normalised so that $\operatorname{Vol}(\mathbb{Z}_p,dy)=1$ .

With these choices made there are constants $C_{\pi}^{(i)}\in \mathbb{C}^{\times}$ so that

\begin{equation} \frac{W_{\phi_i}(g)}{\Vert \phi_i \Vert_2} = C_{\pi}^{(i)} \cdot \prod_{v}W_v(g_v). \nonumber\end{equation}

As shown in [ Reference Michel and Venkatesh19 , (4·16)] (see also [ Reference Lapid and Mao16 , section 4]) the absolute values of these constants satisfy

\begin{equation} \lvert{C_{\pi}^{(i)}}\rvert^{2} = \lim_{s\to 1} \frac{\zeta^{\{p,\infty\}}(1)\zeta^{\{p,\infty\}}(2)}{L^{\{p,\infty\}}(s,\pi\otimes\check{\pi})}.\nonumber\end{equation}

Note that we choose the global measure on $Z(\mathbb{A})G(\mathbb{Q})\backslash G(\mathbb{A})$ to be the Tamagawa measure. In particular, the absolute value is independent of i and using [ Reference Hoffstein and Lockhart12 ] we get

\begin{equation} \lvert{C_{\pi}^{(i)}}\rvert^{2} \ll_{\epsilon} p^{\epsilon n_{\pi}}\cdot (1+t_{\pi})^{\epsilon}. \nonumber\end{equation}

Combining everything we end up with

\begin{multline} \frac{\phi_i(n(x)a(y))}{\Vert \phi_i\Vert_2} = C_{\pi}^{(i)} \sum_{k\in \mathbb{N}_0}\sum_{\substack{ 0 \neq n\in\mathbb{Z},\\ (n,p)=1}}\operatorname{sgn}(n)^{\rho}\frac{\lambda_{\pi}\left(n\right)}{\sqrt{\lvert{n}\rvert}}W_p^{(i)}(a(np^{k-m_{\pi}})) \\ \cdot W_{\infty}\left(a\left(\frac{n}{p^{m_{\pi}-k}}y\right)\right)e\left(\frac{n}{p^{m_{\pi}-k}}x\right), \nonumber\end{multline}

for $x\in \mathbb{R}$ and $y\in \mathbb{R}^+$ . Here $\rho\in\{0,1\}$ depends on whether $\phi_1,\ldots,\phi_d$ are even or odd.

Let $v_1,\ldots, v_d$ be an orthogonal basis of $\pi_p^{K_p(m_{\pi})}$ . We fix a Whittaker functional and thus an embedding

\begin{align*}v\longmapsto W_v\in\mathcal{W}(\pi_p,\psi_p).\end{align*}

Define

\begin{equation} \mathcal{S}_{\pi_p}(g_p) = \sum_{i=1}^d \frac{\lvert{W_{v_i}(g_p)}\rvert^2}{\langle W_{v_i},W_{v_i}\rangle_{\mathcal{W}(\pi_p,\psi_p)}}.\nonumber\end{equation}

Note that $\mathcal{S}_{\pi_p}$ is well defined as it is independent of the choice of Whittaker functional and the choice of basis $v_1,\ldots,v_d$ .

Lemma 3·1. For any orthonormal basis $\phi_1,\ldots,\phi_d$ we have

\begin{equation} \Phi(g)\leq (dT)^{\epsilon}\sum_{k\in \mathbb{N}_0}\sum_{\substack{ 0 \neq n\in\mathbb{Z},\\ (n,p)=1}}\frac{\lvert{\lambda_{\pi}\left(n\right)}\rvert}{\sqrt{\lvert{n}\rvert}} \cdot \left\lvert{W_{\infty}\left(\frac{n}{p^{m_{\pi}-k}}y\right)}\right\rvert \cdot\mathcal{S}_{\pi_p}(a(np^{k-m_{\pi}}))^{\frac{1}{2}},\nonumber\end{equation}

where $g=n(x)a(y)\in \mathcal{F}$ .

Proof. To simplify notation we define

\begin{equation} a(t) = \operatorname{sgn}(n)^{\rho}\frac{\lambda_{\pi}\left(n\right)}{\sqrt{\lvert{n}\rvert}}W_{\infty}\left(\frac{n}{p^{m_{\pi}-k}}y\right)e\left(\frac{n}{p^{m_{\pi}-k}}x\right) \text{ and }b_i(t) = W_p^{(i)}(a(np^{k-m_{\pi}})) \nonumber\end{equation}

if $t=np^{k-m_{\pi}}$ for $k\in\mathbb{N}_0$ and $(n,p)=1$ , and $a(t)=0=b_i(t)$ otherwise. The Whittaker expansion now neatly reads

\begin{equation} \phi_i(n(x)a(y)) = C_{\pi}^{(i)}\sum_{t\in\mathbb{Q}^{\times}}a(t)b_i(t).\nonumber\end{equation}

With this at hand we estimate

\begin{align} \Phi(g) &= \left(\sum_{i=1}^d\left\vert C_{\pi}^{(i)}\sum_{t\in\mathbb{Q}^{\times}}a(t)b_i(t) \right\vert^2\right)^{\frac{1}{2}} \nonumber \\ &\leq\max_i\lvert{C_{\pi}^{(i)}}\rvert\cdot\left(\sum_{t_1\in\mathbb{Q}^{\times}}\sum_{t_2\in\mathbb{Q}^{\times}}a(t_1)\overline{a(t_2)}\sum_{i=1}^d b_i(t_1)\overline{b_i(t_2)}\right)^{\frac{1}{2}} \nonumber \\ & \ll (dT)^{\epsilon} \left(\sum_{t_1\in\mathbb{Q}^{\times}}\sum_{t_2\in\mathbb{Q}^{\times}}\lvert{a(t_1)a(t_2)}\rvert\left(\sum_{i=1}^d\lvert{b_i(t_1)}\rvert^2\right)^{\frac{1}{2}}\left(\sum_{i=1}^d\lvert{b_i(t_2)}\rvert^2\right)^{\frac{1}{2}}\right)^{\frac{1}{2}} \nonumber\\ &=(dT)^{\epsilon}\sum_{t\in\mathbb{Q}^{\times}}\lvert{a(t)}\rvert\left(\sum_{i=1}^d\lvert{b_i(t)}\rvert^2\right)^{\frac{1}{2}}.\nonumber\end{align}

The claim follows by inserting the definitions of a(t) and $b_i(t)$ .

Before we can estimate this expression we need to investigate the size of the local average $\mathcal{S}_{\pi_p}(a(y))$ . This is the content of the following subsection.

3·2. Computing the local averages

The computation of $\mathcal{S}_{\pi_p}(a(y))$ involves a case study and each case will be treated using different techniques. Finally, combining all possible cases, will lead to the bound

(5) \begin{equation} \mathcal{S}_{\pi_p}(a(p^{-m_{\pi}}y)) \ll d^{1+\epsilon}\cdot \lvert{y}\rvert_p. \end{equation}

See Lemma 3·3, 3·6 and 3·7 below.

3·2·1. The Steinberg representation

Let $V=\operatorname{Ind}_B^G(\lvert{\cdot}\rvert^{\frac{1}{2}}\otimes \lvert{\cdot}\rvert^{-\frac{1}{2}})$ . Then we can identify $\pi=\text{St}$ with the unique irreducible generic subspace of V. Let $V^{\vee}=\operatorname{Ind}_B^G(\lvert{\cdot}\rvert^{-\frac{1}{2}}\otimes \lvert{\cdot}\rvert^{\frac{1}{2}})$ . This is the dual space of V and the invariant bilinear pairing is given by

\begin{equation} \langle f,f^{\vee}\rangle = \int_K f(k)f^{\vee}(k)dk.\nonumber\end{equation}

Further $\tilde{\pi}=\text{St}$ can be identified as the unique irreducible generic sub-quotient of $V^{\vee}$ .

Next we choose a basis $v_0,\ldots,v_p$ of $V^{K_p(1)}$ . (In an analogous way one constructs the dual basis $v_0^{\vee},\ldots,v_p^{\vee}$ in $(V^{\vee})^{K_p(1)}$ .) This is done as follows: we first construct

\begin{equation} v_p(g) = \operatorname{Vol}(B(\mathbb{Z}_p)K_p(1),dk)^{-\frac{1}{2}}\cdot \begin{cases} \displaystyle\lvert{\frac{a}{d}}\rvert &\text{ if }g=\left(\begin{matrix} a&b\\0&d\end{matrix}\right)k\in B(\mathbb{Q}_p)K_p(1),\\ 0&\text{ else.} \end{cases}\nonumber\end{equation}

Further $\gamma_i=wn(i)$ for $i=0,\ldots,p-1$ . For consistency of the indices we put $\gamma_p=1$ so that we can identify

\begin{equation} B(\mathbb{Z}_p)\backslash K_p/K_p(1) = \{\gamma_0,\ldots, \gamma_p \} \nonumber\end{equation}

via the Bruhat decomposition of $G(\mathbb{F}_p)$ . (Note that $\gamma_0=w$ .) Finally define $v_i(g) = v_p(g\cdot \gamma_i^{-1})$ . This is the desired basis.

Now there is an (up to scaling) unique $\psi_p$ -Whittaker functional $\Lambda\colon V\to \mathbb{C}$ (resp. a unique $\psi_p^{-1}$ -Whittaker functional $\Lambda^{\vee}\colon V^{\vee}\to\mathbb{C}$ ). As usual we set

\begin{equation} W_v(g) = \Lambda(g.v) \text{ or }W_{v^{\vee}}(g) = \Lambda^{\vee}(g.v^{\vee}).\nonumber\end{equation}

We will first consider the related average

\begin{equation} \mathcal{S}_V(y) = \sum_{i=0}^p W_{v_i}(a(y))W_{v_i^{\vee}}(a(y)). \nonumber\end{equation}

Note that also this is independent of the choice of the particular basis $v_0,\ldots,v_p$ as long as one considers the corresponding dual basis of $V^{\vee}$ .

We will write $\int^{st}$ for the stable integral as defined [ Reference Lapid and Mao16 , definition 2·1]. By [ Reference Lapid and Mao16 , lemma 4·4 and remark 4·6] we get

\begin{align} W_{v_i}(a(y))W_{v_i^{\vee}}(a(y)) &= \int_{\mathbb{Q}_p}^{st}\langle n(x)a(y).v_i,a(y).v_i^{\vee}\rangle\psi_p(x)^{-1}dx\nonumber \\ &= \lvert{y}\rvert_p\int_{\mathbb{Q}_p}^{st}\langle n(x)v_i,v_i^{\vee}\rangle\psi_p(xy)^{-1}dx.\nonumber\end{align}

Knowing the exact shape of the $v_i$ ’s we can compute these integrals. First, we observe that a simple change of variables yields

\begin{align} \langle n(x).v_i,v_i^{\vee} \rangle &= \int_K v_p(kn(x)\gamma_i^{-1})v_p^{\vee}(k\gamma_i^{-1})dk=\int_K v_p(k\gamma_in(x)\gamma_i^{-1})v_p^{\vee}(k)dk\nonumber\\ &= \frac{\sharp [B(\mathbb{Z}_p)/K_p(1)\cap B(\mathbb{Z}_p)]^{\frac{1}{2}}}{\operatorname{Vol}(K_p(1),dk)^{\frac{1}{2}}}\cdot \int_{K_p(1)} v_p(k\gamma_in(x)\gamma_i^{-1})dk.\nonumber\end{align}

The case $i=p$ is somehow special and will be treated later. For now let us assume $0\leq i<p$ . In this case we have

\begin{equation} \gamma_in(x)\gamma_i^{-1} = wn(i)n(x)n(\!-i)w^{-1}=wn(x)w^{-1}.\nonumber\end{equation}

To take advantage of the support of $v_p$ we have to investigate

\begin{equation} kwn(x)w^{-1}=b\tilde{k} \in B \cdot K_p(1).\nonumber\end{equation}

In view of the Iwahori-factorisation of k we find that $n(x)\in N(\mathbb{Q}_p)\cap K_p(1)$ is necessary for the integral to be non-zero. Thus one gets

\begin{equation} \langle \pi(n(x))v_i,v_i^{\vee} \rangle = \delta_{n\in N(p\mathbb{Z}_p)}.\nonumber\end{equation}

With this at hand it is easy to compute

\begin{align} \int_{\mathbb{Q}_p}^{st}\langle \pi(n(x))v_i,v_i^{\vee} \rangle \psi_p(yx)^{-1}dx = \int_{p\mathbb{Z}_p} \psi(yx)^{-1}dx = p^{-1}\delta_{y\in p^{-1}\mathbb{Z}_p},\nonumber\end{align}

for $0\leq i< p$ .

We turn towards $i=p$ , so that $\gamma_p=1$ . Further we replace y by $yp^{-1}$ and consider $y\in\mathbb{Z}_p$ . Recall that every $k\in K_p(1)$ can be written as $k=t_k\overline{n}_kn_k \in B(\mathbb{Z}_p)N(p\mathbb{Z}_p)^t N(p\mathbb{Z}_p)$ by using the Iwahori-factorisation. We obtain

\begin{align} &\int_{\mathbb{Q}_p}^{st}\langle \pi_p(n(x))v_p,v_p^{\vee}\rangle \psi_p(yp^{-1}x)^{-1}dx \nonumber \\ &\quad = \frac{\sharp [B(\mathbb{Z}_p)/K_p(1)\cap B(\mathbb{Z}_p)]^{\frac{1}{2}}}{\operatorname{Vol}(K_p(1),dk)^{\frac{1}{2}}}\cdot\int_{\mathbb{Q}_p}^{st}\int_{K_p(1)} v_p(kn(x))dk \psi_p(yp^{-1}x)^{-1}dx \nonumber \\ &\quad = \frac{\sharp [B(\mathbb{Z}_p)/K_p(1)\cap B(\mathbb{Z}_p)]^{\frac{1}{2}}}{\operatorname{Vol}(K_p(1),dk)^{\frac{1}{2}}}\cdot \int_{K_p(1)}\int_{\mathbb{Q}_p}^{st} v_p(\overline{n}_kn(x)) \psi_p(yp^{-1}x)^{-1}dxdk.\nonumber\end{align}

Note that the integrand only depends on $\overline{n}_k$ . Therefore we start by discussing a suitable measure on $K_p(1)$ . Indeed using the Iwahori factorisation we can write

\begin{equation} \int_{K_p(1)} f(k)dk = \frac{\operatorname{Vol}(K_p(1),dk)}{\operatorname{Vol}(B(\mathbb{Z}_p)\cap K_p(1),db)}\int_{B(\mathbb{Z}_p)\cap K_p(1)}p\int_{p\mathbb{Z}_p} f(b\overline{n}(pu))dudb.\nonumber\end{equation}

If we write $\tilde{v}_p$ to be the re-normalisation of $v_p$ with $\tilde{v}_p(1)=1$ , then we have

\begin{align} \int_{\mathbb{Q}_p}^{st}\langle \pi_p(n(x))v_p,v_p^{\vee}\rangle \psi_p(yp^{-1}x)^{-1}dx &= p\int_{\mathbb{Q}_p}^{st}\int_{p\mathbb{Z}_p} \tilde{v}_p(n(z)^tn(x)) \psi_p(yp^{-1}x)^{-1}dzdx \nonumber \\ &{=} p\int_{\mathbb{Q}_p}^{st}\int_{p\mathbb{Z}_p} \!\!\tilde{v}_p\left(\left(\begin{matrix} 1&\!x-z^{-1} \\ z &\! zx\end{matrix}\right)\right) \psi_p(yp^{-1}(x{-}z^{-1}))^{-1}dzdx.\nonumber\end{align}

In the last step we simply made a change of variables in the x-integral. A simple matrix computation shows that

\begin{equation} \left(\begin{matrix} 1&x-z^{-1} \\ z & zx\end{matrix}\right) = n(\star)\left(\begin{matrix} (zx)^{-1} & 0 \\ z & zx \end{matrix}\right).\nonumber\end{equation}

Inserting this and using the transformation behaviour of $\tilde{v}_p$ one obtains

\begin{multline} \int_{\mathbb{Q}_p}^{st}\langle \pi_p(n(x))v_p,v_p^{\vee}\rangle \psi_p(yp^{-1}x)^{-1}dx \\ = p\int_{p\mathbb{Z}_p}\psi_p(yz^{-1}p^{-1})\frac{dz}{\lvert{z}\rvert_p^2}\int_{\mathbb{Q}_p}^{st}\tilde{v}_p\left(\left(\begin{matrix} 1& 0\\x^{-1} & 1 \end{matrix}\right)\right)\psi_p(\!-xyp^{-1})\frac{dx}{\lvert{x}\rvert_p^2}.\nonumber\end{multline}

Both integrals can now be computed quite easily. Starting from the first one we obtain

\begin{multline} p\int_{p\mathbb{Z}_p}\psi_p(yz^{-1}p^{-1})\frac{dz}{\lvert{z}\rvert_p^2} = \sum_{l=1}^{\infty}p^{l+1}\int_{\mathbb{Z}_p^{\times}}\psi_p(zyp^{-1-l})dz \nonumber \\ = \sum_{l=1}^{v_p(y)-1}p^{l+1}(1-p^{-1})-\delta_{v_p(y)\geq 1}p^{v_p(y)} = -p\delta_{v_p(y)\geq 1}.\nonumber\end{multline}

Turning to the other integral we find

\begin{align} \int_{\mathbb{Q}_p}^{st}\tilde{v}_p\left(\left(\begin{matrix} 1& 0\\x^{-1} & 1 \end{matrix}\right)\right)\psi_p(\!-xyp^{-1})\frac{dx}{\lvert{x}\rvert_p^2} &= \int_{\mathbb{Q}_p\setminus \mathbb{Z}_p}^{st}\psi_p(\!-xyp^{-1})\frac{dx}{\lvert{x}\rvert_p^2} \nonumber \\ &= \sum_{l=1}^{\infty}p^{-l}\int_{\mathbb{Z}_p^{\times}}\psi_p(\!-xyp^{-1-l})dx \nonumber\\ &= \sum_{l=1}^{v_p(y)-1}p^{-l}(1-p^{-1})-\delta_{v_p(y)\geq 1}p^{-v_p(y)-1} \nonumber \\ &= \delta_{v_p(y)\geq 2}(p^{-1}-p^{-v_p(y)})-\delta_{v_p(y)\geq 1}p^{-1-v_p(y)}.\nonumber\end{align}

In particular we have

\begin{equation} \int_{\mathbb{Q}_p}^{st}\langle \pi_p(n(x))v_0,v_0^{\vee}\rangle \psi_p(yp^{-1}x)^{-1}dx =\delta_{v_p(y)\geq 2}(p^{1-v_p(y)}-1)+\delta_{v_p(y)\geq 1}p^{-v_p(y)}. \nonumber\end{equation}

Note that this can be negative, but for non-unitary representations there is no expectation for these integrals to be non-negative.

Combining the computations above and swapping back to $y\in p^{-1}\mathbb{Z}_p$ leads us to the following result.

Lemma 3·2. In the notation above we have

\begin{equation} \mathcal{S}_V(y) = \lvert{y}\rvert_p\left[\delta_{v_p(y)\geq -1}+\delta_{v_p(y)\geq 0} p^{-1}\lvert{y}\rvert_p+\delta_{v_p(y)\geq 1}(\lvert{y}\rvert_p-1)\right].\nonumber\end{equation}

We will obtain the desired estimate by relating $\mathcal{S}_{\text{St}}(a(y))$ to $\mathcal{S}_V$ .

Lemma 3·3. For $\pi_p=\text{St}$ and $y\in \mathbb{Q}_p$ we have

\begin{equation} \mathcal{S}_{\pi_p}(a(y)) \ll \lvert{y}\rvert_p.\nonumber\end{equation}

Proof. Recall that the definitions of $\mathcal{S}_{\pi_p}$ and $\mathcal{S}_V$ are independent of the choice of the underlying basis. Thus we can choose an orthogonal basis $w_1,\ldots,w_p$ of $\pi_p^{K_p(1)}$ . Viewing $\pi$ as invariant subspace of V we can assume that the $w_i$ ’s are in V. We then have

\begin{equation} \mathcal{S}_{\pi}(a(y)) = \sum_{i=1}^p \frac{\lvert{W_{w_i}(a(y))}\rvert^2}{\langle W_{w_i},W_{w_i}\rangle} = \sum_{i=1}^p \frac{W_{w_i}(a(y))W_{w_i^{\vee}}(a(y))}{\langle w_i,w_i^{\vee}\rangle}. \nonumber\end{equation}

Finally if we choose $w_0^{\vee}\in (V^{\vee})^{K_p(1)}$ in the annihilator of the $w_1,\ldots,w_p$ and let $w_0\in V^{K_p(1)}$ be the dual element then after renormalising we have

\begin{equation} \mathcal{S}_V(y) = \mathcal{S}_{\pi}(a(y)) + W_{w_0}(a(y))W_{w_0^{\vee}}(a(y)).\nonumber\end{equation}

However, since there is a unique Whittaker functional on $V^{\vee}$ which descents to the unique Whittaker functional on $\pi$ when viewed as a sub-quotient we must have $W_{w_0^{\vee}}(a(y))=0$ . (Since the unique invariant subspace is non-generic.) Thus $\mathcal{S}_V(y) = \mathcal{S}_{\pi_p}(a(y))$ and the desired estimate follows directly from the previous lemma.

3·2·2. Twist minimal principal series

Turning to this case we assume that $\pi_p=\chi \cdot \lvert{\cdot}\rvert^{\rho}\boxplus \lvert{\cdot}\rvert^{-\rho}$ where $a(\chi)=n_{\pi}>0$ . Without loss of generality we can assume that $\chi(p)=1$ . (If we assume that $\pi$ is unitary then it is tempered so that $\rho\in i\mathbb{R}$ .) Now we can choose a basis in the induced picture essentially as above, but we need to find a suitable decomposition of $B(\mathbb{Z}_p)\backslash K_p/K_p(n_{\pi})$ (since the Bruhat decomposition does not hold in $G(\mathbb{Z}_p/p^m\mathbb{Z}_p)$ if $m>1$ ). First we start by defining

\begin{equation} v_0(g) = \operatorname{Vol}(B(\mathbb{Z}_p)K_p(m_{\pi}),dk)^{-\frac{1}{2}}\cdot \begin{cases} \chi(a)\lvert{\frac{a}{d}}\rvert^{\frac{1}{2}+\rho} &\text{ if }g=\left(\begin{matrix} a&b\\0&d\end{matrix}\right)k\in B(\mathbb{Q}_p)K_p(m_{\pi}),\\ 0&\text{ else.}\end{cases}\nonumber\end{equation}

From this element we can construct a basis of $\pi_p^{K_p(m_{\pi})}$ as in the Steinberg case. Indeed, we fix a system of representatives $\{\gamma_j\}$ for $B(\mathbb{Z}_p)\backslash K_p/K_p(m_{\pi})$ and set $v_j=\pi_p(\gamma_j^{-1})v_0$ .

In order to explicate this basis we need to compute a suitable coset decomposition for $B(\mathbb{Z}_p)\backslash K_p/K_p(m_{\pi})$ . This is the content of the following lemma.

Lemma 3·4. We have

\begin{equation} K_p = \bigsqcup_{a\in \mathbb{Z}_p/p^{m_{\pi}}\mathbb{Z}_p} B(\mathbb{Z}_p) \gamma_{0,a} K_p(m_{\pi}) \sqcup \bigsqcup_{i=1}^{m_{\pi}}\bigsqcup_{a\in (\mathbb{Z}_p/p^{m_{\pi}-i}\mathbb{Z}_p)^{\times}} B(\mathbb{Z}_p)\gamma_{i,a} K_p(m_{\pi}),\nonumber\end{equation}

for

\begin{equation} \gamma_{i,a} = \begin{cases} \left(\begin{matrix} 0& -1 \\ 1 & a\end{matrix} \right) & \text{ if }i=0,\\ \left(\begin{matrix} 1 & 0 \\ ap^i &1 \end{matrix} \right) &\text{ if } 1\leq i\leq m_{\pi}-1, \\ 1_2 &\text{ if }i=m_{\pi}. \end{cases} \nonumber\end{equation}

Proof. Take $g=\left(\begin{matrix} a&b\\c & d\end{matrix}\right)\in K_p$ and set $i=v_p(c)$ . We treat several cases distinguished by the value of i.

First, if $i=m_{\pi}$ , then we have

\begin{equation} g=\left(\begin{matrix} a & b\\ c & d\end{matrix}\right) = \left( \begin{matrix} a&b\\0&d \end{matrix}\right) \left(\begin{matrix} 1-\frac{bc}{ad} & 0 \\ \frac{c}{d} & 1 \end{matrix}\right) \in B(\mathbb{Z}_p)K_p(m_{\pi}).\nonumber\end{equation}

Second, for $1\leq i<m_{\pi}$ we have

\begin{equation} g=\left( \begin{matrix} a-\frac{bc}{d} & b \\ 0&d\end{matrix}\right) \left( \begin{matrix} 1 & 0 \\ \frac{c}{d} & 1 \end{matrix}\right).\nonumber\end{equation}

By right multiplication with elements in $K_p(m_{\pi})$ we can view ${c}/{d} \in p^i\mathbb{Z}_p^{\times}/p^{m_{\pi}}\mathbb{Z}_p$ .

The critical contribution is given by the matrices with $i=0$ . We can write

\begin{equation} g= \left(\begin{matrix} \frac{ad}{c}-b & a+(b-\frac{ad}{c})p^{m_{\pi}}\\ 0 & c \end{matrix}\right)\left(\begin{matrix}0 & -1 \\ 1 & \frac{d}{c} \end{matrix}\right)\left(\begin{matrix} 1+\frac{d}{c}p^{m_{\pi}} & \frac{d^2}{c^2}p^{m_{\pi}} \\ -p^{m_{\pi}} & 1-\frac{d}{c}p^{m_{\pi}} \end{matrix}\right). \nonumber\end{equation}

Given $v \in \pi^{K_p(n_{\pi})}$ we can compute the Jacquet Integral as follows. Without loss of generality assume $v_p(y)\geq -n_{\pi}$ , since otherwise the Whittaker function $W_v$ vanishes for trivial reasons. We compute

\begin{align} W_v(a(y)) &= \int_{\mathbb{Q}_p} v(wn(x)a(y))\psi_p(x)^{-1}dx \nonumber \\ &= \lvert{y}\rvert_p^{\frac{1}{2}-\rho}\int_{\mathbb{Q}_p}v(wn(x))\psi_p(xy)^{-1}dx\nonumber \\ &= \lvert{y}\rvert_p^{\frac{1}{2}-\rho} p^{-n_{\pi}}\sum_{a\in \mathbb{Z}_p/p^{n_{\pi}}\mathbb{Z}_p} \psi_p(ay)^{-1}v(wn(a)) \nonumber \\ &\qquad + \lvert{y}\rvert_p^{\frac{1}{2}-\rho}\int_{\mathbb{Q}_p\setminus \mathbb{Z}_p}\lvert{x}\rvert^{-2\rho} \chi(x)^{-1}\psi(xy)^{-1}v(n(x^{-1})^t)\frac{dx}{\lvert{x}\rvert_p}.\nonumber\end{align}

Note that $wn(a) = \gamma_{0,a}$ . Now we will have a closer look at the remaining integral:

\begin{align} &\int_{\mathbb{Q}_p\setminus \mathbb{Z}_p}\lvert{x}\rvert^{-2\rho} \chi(x)^{-1}\psi(xy)^{-1}v(n(x^{-1})^t)\frac{dx}{\lvert{x}\rvert_p} \nonumber \\ &\qquad = \int_{p\mathbb{Z}_p}\lvert{x}\rvert^{2\rho} \chi(x)\psi(x^{-1}y)^{-1}v(n(x)^t)\frac{dx}{\lvert{x}\rvert_p} \nonumber\\ &\qquad = v(1)\int_{p^{m_{\pi}}\mathbb{Z}_p}\chi(x)\lvert{x}\rvert^{2\rho}\psi(x^{-1}y)^{-1}\frac{dx}{\lvert{x}\rvert_p} \nonumber \\ &\qquad\qquad +\sum_{i=1}^{m_{\pi}-1}\sum_{b\in (\mathbb{Z}_p/p^{m_{\pi}-i}\mathbb{Z}_p)^{\times}}p^{-2\rho i}\chi(b)^{-1}v(n(bp^i)^t)\cdot \int_{1+p^{m_{\pi}-i}\mathbb{Z}_p}\chi(x)^{-1}\psi(\!-byxp^{-i})dx.\nonumber\end{align}

Note that $n(bp^i)^t=\gamma_{i,b}$ .

Since the v(1)-contribution is easily computed we arrive at the following lemma.

Lemma 3·5. For $v_p(y)\geq -m_{\pi}$ we have

\begin{align} W_v(a(y)) &= \lvert{y}\rvert_p^{\frac{1}{2}-\rho}p^{-m_{\pi}}\sum_{a\in \mathbb{Z}_p/p^{m_{\pi}}\mathbb{Z}_p}\psi_p(ay)^{-1}v(\gamma_{0,a}) \nonumber \\ &\qquad + \lvert{y}\rvert_p^{\frac{1}{2}-\rho}\sum_{i=1}^{m_{\pi}-1}\sum_{b\in (\mathbb{Z}_p/p^{m_{\pi}-i}\mathbb{Z}_p)^{\times}}p^{-2\rho i}\chi(b)^{-1}v(\gamma_{i,b})G_{m_{\pi}-i}(\!-byp^{-i},\chi^{-1}) \nonumber\\ &\qquad +\lvert{y}\rvert_p^{\frac{1}{2}-\rho}\cdot v(\gamma_{m_{\pi},0})\cdot \int_{p^{m_{\pi}}\mathbb{Z}_p}\chi(x)\lvert{x}\rvert^{2\rho}\psi(x^{-1}y)^{-1}\frac{dx}{\lvert{x}\rvert_p}, \nonumber\end{align}

for

\begin{equation} G_l(y,\chi) = \int_{1+p^l\mathbb{Z}_p}\chi(x)\psi(yx)dx,\nonumber\end{equation}

This supplies us with the necessary ingredients to show the required estimate for $\mathcal{S}_{\pi_p}$ .

Lemma 3·6. For $\pi_p=\chi\lvert{\cdot}\rvert^{\rho}\boxplus \lvert{\cdot}\rvert^{-\rho}$ unitary and $y\in \mathbb{Q}_p$ we have

\begin{equation} \mathcal{S}_{\pi_p}(a(p^{-m_{\pi}}y)) \ll d^{1+\epsilon}\lvert{y}\rvert_p. \nonumber\end{equation}

Proof. Note that since all $v_j$ ’s are translates of $v_0$ their Whittaker-norm all coincides. So it suffices to compute one of these norms and it is easy to see that

\begin{equation} \langle W_{v_0},W_{v_0}\rangle = \int_{\mathbb{Q}_p^{\times}}\lvert{W_{v_0}(a(y))}\rvert^2\frac{dy}{\lvert{y}\rvert} = (1+p^{-1}). \nonumber\end{equation}

Next we observe that one can choose representatives so that $v_j=\pi_p(\gamma_{i,a}^{-1})v_0$ for some $i=i(j)$ and $a=a(j)$ . In particular, we can sort the terms of the sum $\mathcal{S}_{\pi_p}(a(y))$ according to this i. We get

\begin{equation} \mathcal{S}_{\pi_p}(a(y)) = \sum_{i=0}^{m_{\pi}} \mathcal{S}_i(y) \text{ for } \mathcal{S}_i(y) = (1+p^{-1})^{-1}\sum_{a\in \mathbb{Z}_p/p^{m_{\pi}-i}\mathbb{Z}_p} \lvert{W_{\pi_p(\gamma_{i,a}^{-1})v_0}(a(y))}\rvert^2.\nonumber\end{equation}

Applying the previous Lemma with $v=\pi_p(\gamma_{i,a}^{-1})v_0$ and taking support properties of $v_0$ into account provides us with nice formulae for the $W_{\pi_p(\gamma_{i,a}^{-1})v_0}(a(y))$ .

As soon as we can show that $\mathcal{S}_i(y)\ll \lvert{y}\rvert_p$ for all i we are done. We start with $i=0$ . Here we have the explicit formula

\begin{equation} \mathcal{S}_i(y) = (1+p^{-1})^{-1}\sum_{a\in \mathbb{Z}_p/p^{m_{\pi}}\mathbb{Z}_p} p^{-2m_{\pi}}\lvert{y}\rvert_p v_0(1)^2 = \frac{(1+p^{-1})^{-1}p^{-m_{\pi}}\lvert{y}\rvert_p}{ \operatorname{Vol}(B(\mathbb{Z}_p)K_p(m_{\pi}),dk)} = \lvert{y}\rvert_p\nonumber\end{equation}

for $y\in p^{-m_{\pi}}\mathbb{Z}_p$ .

We turn towards $1\leq i \leq m_{\pi}-1$ . In this range we get

\begin{align} \mathcal{S}_i(y) &= (1+p^{-1})^{-1}v_0(1)^2\lvert{y}\rvert_p \sum_{a\in (\mathbb{Z}_p/p^{m_{\pi}-i}\mathbb{Z}_p)^{\times}}\lvert{G_{m_{\pi}-i}(\!-ayp^{-i},\chi^{-1})}\rvert^2.\nonumber\end{align}

Thus we need to bound the integrals $G_l(z,\chi)$ , which are somehow incomplete Gauß sums in the sense that one sums only over a specific congruence class. These sums were essentially computed in the proof of [ Reference Assing2 , lemma 5·8]. Indeed one extracts

\begin{equation} G_l(zp^{-k},\chi) = \begin{cases} \epsilon(\frac{1}{2},\chi^{-1})\chi^{-1}(z)p^{-\frac{k}{2}} &\text{ if }l\leq \lfloor \frac{a(\chi)}{2}\rfloor, k=a(\chi) \text{ and }z\in b(\chi)+p^l\mathbb{Z}_p,\\ \psi_p(zp^k)p^{-l} &\text{ if }l\geq \lceil \frac{a(\chi)}{2}\rceil, k=a(\chi) \text{ and }z\in -b(\chi)+p^{a(\chi)-l}\mathbb{Z}_p,\\ 0&\text{ else}, \end{cases} \nonumber\end{equation}

for $z\in \mathbb{Z}_p^{\times}$ , $k\in \mathbb{Z}$ and $b(\chi)\in \mathbb{Z}_p^{\times}$ is determined by $\chi$ . With this at hand we can easily evaluate $\mathcal{S}_i$ . For $i\leq \lfloor {m_{\pi}}/{2}\rfloor$ we have

\begin{align} \mathcal{S}_i(y) &= \delta_{v_p(y) =i-m_{\pi}}(1+p^{-1})^{-1}v_0(1)^2\lvert{y}\rvert_p \sum_{b\in (\mathbb{Z}_p/p^{m_{\pi}-i}\mathbb{Z}_p)^{\times}}p^{2i-2m_{\pi}}\delta_{-byp^{-v_p(y)}\in -b(\chi)+p^i\mathbb{Z}_p} \nonumber\\ &= \delta_{v_p(y) =i-m_{\pi}}\lvert{y}\rvert_p.\nonumber\end{align}

Similarly for $i\geq \lceil {m_{\pi}}/{2}\rceil$ we have

\begin{align} \mathcal{S}_i(y) &= \delta_{v_p(y) =i-m_{\pi}}(1+p^{-1})^{-1}v_0(1)^2\lvert{y}\rvert_p \sum_{b\in (\mathbb{Z}_p/p^{m_{\pi}-i}\mathbb{Z}_p)^{\times}}p^{-m_{\pi}}\cdot \delta_{-byp^{-v_p(y)}\in b(\chi)+p^{m_{\pi}-i}\mathbb{Z}_p} \nonumber\\ &= \delta_{v_p(y) =i-m_{\pi}}\lvert{y}\rvert_p.\nonumber\end{align}

Finally consider $i=m_{\pi}$ . We have

\begin{align} \mathcal{S}_{m_{\pi}}(y) &= (1+p^{-1})v_0(1)^2\lvert{y}\rvert_p \lvert{\int_{p^{m_{\pi}}\mathbb{Z}}\chi(x)\lvert{x}\rvert^{2\rho}\psi(x^{-1}y)^{-1}\frac{dx}{\lvert{x}\rvert_p}}\rvert^2 \nonumber \\ &= p^{m_{\pi}}\lvert{y}\rvert_p\cdot \lvert{\int_{p^{m_{\pi}}\mathbb{Z}_p}\chi(x)\lvert{x}\rvert^{2\rho}\psi(x^{-1}y)^{-1}\frac{dx}{\lvert{x}\rvert_p}}\rvert^2.\nonumber\end{align}

Therefore it suffices to compute the remaining integral. By some basic Gauß sum evaluations one gets

\begin{equation} \int_{p^{m_{\pi}}\mathbb{Z}_p}\chi(x)\lvert{x}\rvert^{2\rho}\psi(x^{-1}y)^{-1}\frac{dx}{\lvert{x}\rvert_p} = \delta_{y\in \mathbb{Z}_p}\epsilon(\frac{1}{2},\chi)\chi(y)p^{-\frac{m_{\pi}}{2}-2\rho[v_p(y)+m_{\pi}]}. \nonumber\end{equation}

Inserting this above concludes the proof since it implies $\mathcal{S}_{m_{\pi}}(y) =\delta_{v_p(y)>0}\lvert{y}\rvert_p$ .

3·2·3. Supercuspidal representations

Let $\mathfrak{X}_k$ be the set of character $\chi\colon \mathbb{Q}_p^{\times} \to\mathbb{C}^{\times}$ with $a(\chi)\leq k$ and $\chi(p)=1$ . Note that

\begin{equation} \sharp \mathfrak{X}_k = p^{k-1}(p-1).\nonumber\end{equation}

For $\chi\in \mathfrak{X}_k$ and $m\in \mathbb{Z}$ we will consider the functions $\xi_{\chi}^{(m)} \in \mathcal{C}_c^{\infty}(\mathbb{Q}_p^{\times})$ given by

\begin{equation} \xi_{\chi}^{(m)}(y) = \mathbb{1}_{p^{-m}\mathbb{Z}_p^{\times}}(y)\chi(y).\nonumber\end{equation}

Given any representation $\pi_p$ we write $\mathcal{K}_{\psi_p}(\pi_p)$ for the corresponding $\psi_p$ -Kirillov model. Note that this model contains the Schwartz functions so that we have $\xi_{\chi}^{(m)}\in \mathcal{K}_{\psi_p}(\pi_p)$ . Note that by construction of the Kirillov model we have

\begin{equation} W_f(a(y))=f(y) \text{ for } f\in\mathcal{K}_{\psi_p}(\pi_p) \text{ and }y\in \mathbb{Q}_p^{\times}.\nonumber\end{equation}

Thus we compute

\begin{multline} \langle W_{\xi_{\chi_1}^{(m_1)}},W_{\xi_{\chi_2}^{(m_2)}}\rangle = \int_{\mathbb{Q}_p^{\times}}\xi_{\chi_1}^{(m_1)}(y)\overline{\xi_{\chi_2}^{(m_2)}(y)}d^{\times}y = \delta_{m_1=m_2} \int_{p^{-m_1}\mathbb{Z}_p^{\times}}\chi_1(y)\chi_2^{-1}(y)d^{\times}y\\ = \delta_{\substack{m_1=m_2,\\ \chi_1=\chi_2}}.\nonumber\end{multline}

This suffices to compute $S_{\pi_p}(a(y))$ for supercuspdial representations $\pi_p$ .

Lemma 3·7. Suppose $\pi_p$ is a twist minimal supercuspidal representation with (exponent)-conductor $n_{\pi}$ . Then the following is true:

1. (i) If $n_{\pi}=2m_{\pi}$ , then

   \begin{equation} S_{\pi_p}(a(y)) = \frac{p^{m_{\pi}}}{\zeta_p(1)}\cdot \delta_{y\in p^{-m_{\pi}}\mathbb{Z}_p^{\times}};\nonumber \end{equation}

2. (ii) If $n_{\pi} = 2m_{\pi}-1$ , then

   \begin{equation} S_{\pi_p}(a(y)) = \frac{p^{m_{\pi}}}{\zeta_p(1)}\cdot \delta_{y\in p^{-m_{\pi}}\mathbb{Z}_p^{\times}}+\frac{p^{m_{\pi}-1}}{\zeta_p(1)}\cdot \delta_{y\in p^{1-m_{\pi}}\mathbb{Z}_p^{\times}};\nonumber \end{equation}

In general we have the bound

\begin{align*}S_{\pi_p}(a(p^{-m_{\pi}}y)) \ll d\cdot\lvert{y}\rvert_p\cdot \delta_{y\in \mathbb{Z}_p}.\end{align*}

Proof. We start with the case $n_{\pi} = 2m_{\pi}$ . By [ Reference Miyauchi and Yamauchi20 , lemma 4·4] we find that a basis for $\pi^{K_p(m_{\pi})}$ in the Kirillov model is given by

\begin{equation} \{ \xi_{\chi}^{(m_{\pi})}\colon \chi\in \mathfrak{X}_{m_{\pi}}\}.\nonumber\end{equation}

Note that we already took advantage of twist-minimality using that $n_{\chi\pi} =n_{\pi}$ for all $\chi\in \mathfrak{X}_{m_{\pi}}$ . Our computations above show that this basis is orthonormal (with respect to the Whittaker inner product). Thus we have

\begin{equation} \mathcal{S}_{\pi_p}(y) = \sum_{\chi\in \mathfrak{X}_{m_{\pi}}} \lvert{\xi_{\chi}^{(m_{\pi})}(y)}\rvert^2 = \delta_{y\in p^{-m_{\pi}}\mathbb{Z}_p^{\times}}\cdot \sharp \mathfrak{X}_{m_{\pi}}.\nonumber\end{equation}

We turn towards the second case where $n_{\pi}$ is odd. Then we get the orthonormal basis

\begin{equation} \{\xi_{\chi}^{m_{\pi}}\colon \chi \in \mathfrak{X}_{m_{\pi}}\} \cup \{\xi_{\chi}^{m_{\pi}-1}\colon \chi \in \mathfrak{X}_{m_{\pi}-1}\}.\nonumber\end{equation}

It is again easy to compute the desired quantity:

\begin{equation} \mathcal{S}_{\pi_p}(y) = \delta_{y\in p^{-m_{\pi}}\mathbb{Z}_p^{\times}}\cdot \sharp \mathfrak{X}_{m_{\pi}}+\delta_{y\in p^{1-m_{\pi}}\mathbb{Z}_p^{\times}}\cdot \sharp \mathfrak{X}_{m_{\pi}-1}.\nonumber\end{equation}

The result follows directly.

3·3. Conclusion

We can now give a decent bound for $\Phi(z)$ using the Whittaker expansion. We will use the bound (5) and follow the standard procedure.

Lemma 3·8. We have

\begin{equation} \Phi(z) \ll \left(\frac{dT}{y}\right)^{\epsilon}\left( d^{\frac{1}{2}}T^{\frac{1}{6}}+\frac{dT^{\frac{1}{2}}}{y^{\frac{1}{2}}}\right).\nonumber\end{equation}

Proof. Inserting (5) into Lemma 3·1 yields

\begin{equation} \Phi(g)\ll d^{\frac{1}{2}+\epsilon}T^{\epsilon}\sum_{0 \neq n\in\mathbb{Z}}\frac{\lvert{\lambda_{\pi}\left(n/(n,p^{\infty})\right)}\rvert}{\sqrt{\lvert{n}\rvert}} \lvert{W_{\infty}\left(n\frac{y}{p^{m_{\pi}}}\right)}\rvert.\nonumber\end{equation}

Estimating the remaining n-sum as for example in [ Reference Templier23 ] or [ Reference Saha21 ] yields the desired result.

4. A bound via the pre-trace formula

The next bound will be derived from the pre-trace inequality. We start by discussing the local test functions. At the archimedean place we closely follow [ Reference Saha21 , section 3.5] and fix $f_{\infty}$ so that it satisfies:

1. (1) $f_{\infty}(g) = 0$ unless $g\in G(\mathbb{R})^+$ and $u(g)\leq 1$ ;

2. (2) $\hat{f}_{\infty}(\sigma)>0$ for all irreducible spherical unitary principal series representations $\sigma$ of $G(\mathbb{R})$ ;

3. (3) $\hat{f}(\pi_{\infty})\gg 1$ ;

4. (4) $\lvert{f_{\infty}(g)}\rvert\leq T$ and if $u(g)\geq T^{-2}$ , then $\lvert{f_{\infty}(g)}\rvert\leq T^{\frac{1}{2}}u(g)^{-\frac{1}{4}}$ .

(The final property is not really necessary because we are ignoring the spectral aspect for now.) Note that $\hat{f}$ is the spherical transform (also Selberg/Harish–Chandra transform) of f and u(g) is the point-pair invariant on group level.

At the place $v=p$ we define multiple test functions:

\begin{equation} f_p^{(i)}(g) = \mathbb{1}_{ZK_p}(g) \frac{\overline{\langle \pi(g)\phi_p^{(i)},\phi_p^{(i)}\rangle_{\pi_p}}}{\langle \phi_p^{(i)},\phi_p^{(i)}\rangle_{\pi_p}}.\nonumber\end{equation}

The operators are self-adjoint since $f_p^{(i)}(g^{-1})=\overline{f_p^{(i)}(g)}$ . To see non-negativity we will show the convolution identity

\begin{equation} f_p^{(i)} = \dim_{\mathbb{C}}\pi^{K_{p,1}(m_{\pi})} \cdot (f_p^{(i)}\star f_p^{(i)}).\nonumber\end{equation}

Indeed we compute

\begin{align} [f_p^{(i)}\star f_p^{(i)}](h) &=\int_{Z\backslash G} f_p^{(i)}(g^{-1})f_p^{(i)}(gh)dg \nonumber \\ &= \frac{\mathbb{1}_{ZK}(h)}{{\langle \phi_p^{(i)},\phi_p^{(i)}\rangle_{\pi_p}}^2} \int_{K_p}\langle \pi_p(g)\phi_p^{(i)},\phi_p^{(i)}\rangle_{\pi_p}\overline{\langle \pi_p(gh)\phi_p^{(i)},\phi_p^{(i)}\rangle_{\pi_p}}dk\nonumber \\ &= \dim_{\mathbb{C}}(\pi^{K_{p}(m_{\pi})})^{-1}\frac{\mathbb{1}_{ZK}(h)}{\langle \phi_p^{(i)},\phi_p^{(i)}\rangle_{\pi_p}}\overline{\langle \pi_p(h)\phi_p^{(i)},\phi_p^{(i)}\rangle_{\pi_p}}\nonumber\end{align}

and the claimed identity follows directly.

It remains to show the final claim. First observe that the image of $\pi_p(f_p^{(i)})\phi_p^{(i)}$ is obviously $K_p(m_{\pi})$ -invariant. Thus it suffices to show that

\begin{align*}\langle\pi_p(f_p^{(i)})\phi_p^{(i)},w\rangle_{\pi_p} = \dim_{\mathbb{C}}(\pi_p^{K_p(m_{\pi})})^{-1}\cdot\langle\phi_p^{(i)},w\rangle_{\pi_p}\end{align*}

for any $w\in \pi_p^{K_p(m_{\pi})}$ But this follows again from the orthogonality relations since $\pi_p^{K_p(m_{\pi})}$ is irreducible (as $K_p$ -module) and

\begin{equation} \langle\pi_p(f_p^{(i)})\phi_p^{(i)},w\rangle_{\pi_p} = \langle \int_{Z\backslash G} f_p^{(i)}(g)\pi_p(g)\phi_p^{(i)}dg,w\rangle =\int_{Z\backslash G}f_p^{(i)}(g)\langle \pi_p(g)\phi_p^{(i)},w\rangle_{\pi_p}dg.\nonumber\end{equation}

Finally we define the unramified part of the test function $f_{ur}$ by setting

\begin{multline} f_{ur} = \left( \sum_{l\in S} c_l \kappa_l\right)\star \left( \sum_{l\in S} c_l \kappa_l \right)^*+\left( \sum_{l\in S} c_{l^2} \kappa_{l^2}\right)\star \left( \sum_{l\in S} c_{l^2} \kappa_{l^2} \right)^*, \\ \text{ for } c_r=\begin{cases} \frac{\lvert{\lambda_{\pi}(r)}\rvert}{\lambda_{\pi}(r)} &\text{ if $r=l$ or $r=l^2$ for $l\in S$},\\ 0&\text{ else},\end{cases} \nonumber\end{multline}

for a set of primes S (to be determined) and normalised rth Hecke-operators $\kappa_r$ . This implements the usual amplification procedure. Finally we define the global test functions

\begin{equation} f^{(i)} = f_{\infty}\otimes f_p^{(i)}\otimes f_{ur} \text{ and } f=\sum_i f^{(i)}.\nonumber\end{equation}

We introduce

\begin{equation} M(l, g) = \{A\in M_2(\mathbb{Z}) \colon \det(A)=l,\, A\equiv g \text{ mod }p^{m_{\pi}} \} \text{ for }g\in \operatorname{GL}_2(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z}).\nonumber\end{equation}

Further let $\sigma$ denote the irreducible representation of $\operatorname{GL}_2(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})$ through which the irreducible $K_p$ -module $\pi^{K_p(m_{\pi})}$ factors. This is a representation of a finite group and we write $\chi_{\sigma}$ for its character. Finally we define the coefficients $y_r$ by linearising the convolutions of Hecke-operators in the definition of $f_{ur}$ . More precisely we write

\begin{equation} f_{ur} = \sum_{r} y_r \kappa_r.\nonumber\end{equation}

This can be compared to the analogous expression in [ Reference Saha21 , section 7].

The following pre-trace inequality provides the transition to the counting problem.

Lemma 4·2. For $z\in \mathcal{F}$ we have

\begin{equation} \frac{(\sharp S)^2}{\dim_{\mathbb{C}}\pi_p^{K_p(m_{\pi})}} \cdot \Phi(z)^2 \ll \sum_{r} \frac{\lvert{y_r}\rvert}{\sqrt{r}} \sum_{g\in \operatorname{GL}_2(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})} \lvert{\chi_{\sigma}(g)}\rvert \sum_{A\in M(l,g)} \lvert{f_{\infty}(u(Az,z))}\rvert.\nonumber\end{equation}

Proof. We start by considering the spectral expansion of the automorphic kernel $k_{f^{(i)}}$ associated to the self-adjoint operators $R(f^{(i)})$ and dropping all terms except $\phi_i$ . The latter is possible by positivity. We obtain

\begin{equation} \frac{(\sharp S)^2}{\dim_{\mathbb{C}}\pi_p^{K_p(m_{\pi})}} \cdot \phi_i(g_{\infty})^2 \leq k_{f^{(i)}}(g,g) = \sum_{r}y_r \sum_{\gamma\in Z(\mathbb{Q})\backslash G(\mathbb{Q})} f_p^{(i)}(\gamma)\kappa_r(\gamma)f_{\infty}(g_{\infty}^{-1}\gamma g_{\infty}).\nonumber\end{equation}

We now sum this inequality over i to obtain

\begin{equation} \frac{(\sharp S)^2}{\dim_{\mathbb{C}}\pi_p^{K_p(m_{\pi})}}\Phi(g_{\infty})^2 \leq \sum_r y_r \sum_{\gamma\in Z(\mathbb{Q})\backslash G(\mathbb{Q})} \left(\sum_{i=1}^d f_p^{(i)}(\gamma)\right) \kappa_r(\gamma)f_{\infty}(g_{\infty}^{-1}\gamma g_{\infty}).\nonumber\end{equation}

Recall that the test functions $f_p^{(i)}$ are supported in $ZK_p$ . Thus, after choosing a suitable representative for $\gamma$ modulo $Z(\mathbb{Q})$ , we can assume that $\gamma\in K_p$ and write $\overline{\gamma}$ for the image of $\gamma$ in $\operatorname{GL}_2(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})$ . For such $\gamma$ we get

\begin{equation} \sum_{i=1}^d f_p^{(i)}(\gamma) = \chi_{\sigma}(\overline{\gamma}).\nonumber\end{equation}

The rest of the argument is standard and can for example be found in [ Reference Saha21 ].

By the choice of $f_{\infty}$ we can already eliminate the archimedean influence from the right-hand side. (Note that we are not aiming to amplify in the T-aspect.)

Corollary 4·3. For $z\in \mathcal{F}$ we have

\begin{equation} \frac{(\sharp S)^2}{\dim_{\mathbb{C}}\pi_p^{K_p(m_{\pi})}} \cdot \Phi(z)^2 \ll T \sum_{g\in \operatorname{GL}_2(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})} \lvert{\chi_{\sigma}(g)}\rvert \sum_{r} \frac{\lvert{y_r}\rvert}{\sqrt{r}} \cdot \sharp M_z(r,g),\nonumber\end{equation}

for

\begin{equation} M_z(r, g) = \{A\in M_2(\mathbb{Z}) \colon \det(A)=r,\, A\equiv g \text{ mod }p^{m_{\pi}} \text{ and }u(Az,z)\leq 1 \}.\nonumber\end{equation}

This last corollary tells us that we need to control the character $\chi_{\sigma}$ and solve a counting problem estimating $M_z(r,g)$ .

To estimate the character we need to define certain level sets.

\begin{equation} K_{m,\lambda} = \{g\in G(\mathbb{Z}/p^m\mathbb{Z})\colon g\equiv 1 \text{ mod }p^{\lambda} \} \text{ for } 0\leq \lambda \leq m.\nonumber\end{equation}

Note that $K_{m,m} = \{1\}$ and $K_{m,0} = G(\mathbb{Z}/p^m\mathbb{Z})$ .

Before we continue let us make a remark concerning notation. We will write $A\neg B = A\cap B^c$ for the set-theoretic difference. We are not using $A\setminus B$ , since it may be confused with the quotient when A and B are groups.

Lemma 4·4. Suppose $p>3$ . Let $\pi$ belong to Case 1, 2 or 3·1. Then, for $0\leq \lambda < m_{\pi}$ and $g\in Z\cdot K_{m_{\pi},\lambda}\neg Z\cdot K_{m_{\pi},\lambda+1}$ we have

(6) \begin{equation} \chi_{\sigma}(g) \ll p^{\lambda}.\end{equation}

If $\pi$ belongs to Case 3·2 and $\lambda$ and g are as above, then we have the slightly weaker bound

\begin{equation} \chi_{\sigma}(g) \ll p^{\frac{m_{\pi}+\lambda}{2}} .\nonumber\end{equation}

Furthermore, let $h\in G(\mathbb{Z}/p^m\mathbb{Z})$ be a diagonal matrix such that $G(\mathbb{Z}/p^m\mathbb{Z})=Z\cdot \operatorname{SL}_2(\mathbb{Z}/p^m\mathbb{Z})\sqcup hZ\cdot \operatorname{SL}_2(\mathbb{Z}/p^m\mathbb{Z})$ . Then the same estimates hold for $hZ \cdot K_{m_{\pi},\lambda}\neg hZ\cdot K_{m_{\pi},\lambda+1}$ , where $\det(h)$ is not a square modulo $p^{m_{\pi}}$ .

The representations of $\operatorname{GL}_2$ over finite rings such as $\mathbb{Z}/p^m\mathbb{Z}$ and their characters are well studied but explicit estimates for the characters as needed here seem to be hard to find. We choose to use the character tables for $\operatorname{SL}_2(\mathbb{Z}/p^m\mathbb{Z})$ computed by Kutzko [ Reference Kutzko15 ]. This makes it necessary to pass from $\operatorname{SL}_2$ to $\operatorname{GL}_2$ using Mackey Theory. Note that the character values in question were calculated in [ Reference Leigh, Cliff and Wen3 ]. However, they remain hard to extract and we hope our approach is more transparent.

Recall that p is assumed to be odd, write $\omega_{\sigma}$ for the central character of $\sigma$ and let $\tilde{\sigma}$ be an irreducible component of $\sigma\vert_{\operatorname{SL}_2(\mathbb{Z}/p^m\mathbb{Z})}$ . Further, fix h such that $\operatorname{GL}_2(\mathbb{Z}/p^m\mathbb{Z}) = Z\operatorname{SL}_2(\mathbb{Z}/p^m\mathbb{Z})\cup h\cdot Z\operatorname{SL}_2(\mathbb{Z}/p^m\mathbb{Z})$ and write $\tilde{\sigma}^h(g) = \tilde{\sigma}(hgh^{-1})$ . If $\tilde{\sigma}\not\cong\tilde{\sigma}^h$ , then

(7) \begin{equation} \sigma = \operatorname{Ind}_{Z\cdot \operatorname{SL}_2(\mathbb{Z}/p^m\mathbb{Z})}^{\operatorname{GL}_2(\mathbb{Z}/p^m\mathbb{Z})}(\omega_{\sigma}\cdot \tilde{\sigma}) .\end{equation}

In this case $\sigma\vert_{\operatorname{SL}_2(\mathbb{Z}/p^m\mathbb{Z})}=\tilde{\sigma}+\tilde{\sigma}^h$ and we have $\chi_{\sigma}(zs)=w_{\sigma}(z)[\chi_{\tilde{\sigma}}(s)+\chi_{\tilde{\sigma}}^h(s)]$ for $z\in Z$ and $s\in \operatorname{SL}_2(\mathbb{Z}/p^m\mathbb{Z})$ . Otherwise, if $\tilde{\sigma}\cong\tilde{\sigma}^h$ , then

(8) \begin{equation} \sigma\oplus \sigma' = \operatorname{Ind}_{Z\cdot \operatorname{SL}_2(\mathbb{Z}/p^m\mathbb{Z})}^{\operatorname{GL}_2(\mathbb{Z}/p^m\mathbb{Z})}(\omega_{\sigma}\cdot \tilde{\sigma}),\end{equation}

where $\sigma'$ is another irreducible representation of $\operatorname{GL}_2(\mathbb{Z}/p^m_{\pi}\mathbb{Z})$ . Here we have $\sigma\vert_{\operatorname{SL}_2(\mathbb{Z}/p^m\mathbb{Z})}=\tilde{\sigma}$ and obtain $\chi_{\sigma}(zs) = \omega_{\sigma}(z)\chi_{\tilde{\sigma}}(s)$ .

The subgroups $K_{m,\lambda}$ are all normal in $K_{m,0}$ , so that the sets $Z\cdot K_{m_{\pi},\lambda}\neg K_{m_{\pi},\lambda+1}$ and $hZ \cdot K_{m_{\pi},\lambda}\neg hZ\cdot K_{m_{\pi},\lambda+1}$ can be decomposed into (disjoint) $\operatorname{GL}_2(\mathbb{Z}/p^{m_{\pi}})$ conjugacy classes. We focus on estimating $\chi_{\sigma}$ for

\begin{align*}g=z\cdot s\in Z\cdot K_{m_{\pi},\lambda}\neg K_{m_{\pi},\lambda+1} \subseteq Z\cdot \operatorname{SL}_2(\mathbb{Z}/p^m\mathbb{Z}).\end{align*}

The case $g\in hZ\cdot \operatorname{SL}_2(\mathbb{Z}/p^mZ)$ is similar.

Let us recall that non-trivial irreducible representations of $\operatorname{SL}_2(\mathbb{Z}/p^m\mathbb{Z})$ with $m>1$ have dimensions $p^m(1-p^{-1})$ , $p^m(1+p^{-1})$ or ${1}/{2}p^m(1-p^{-2})$ . If $m=1$ , we are dealing with the representation theory of $\operatorname{SL}_2(\mathbb{F}_p)$ which was for example studied by Schur in [ Reference Schur22 ]. It turns out that, besides the normal representations of dimensions $p+1$ and $p-1$ , we have certain special cases. Indeed, we have the Steinberg representation $\rho_{\mathrm{St}}$ of dimension p and two irreducible representations $\rho_+$ and $\rho_+^h$ (resp. $\rho_-$ and $\rho_-^h$ ) of dimension ${1}/{2}(p+1)$ (resp. ${1}/{2}(p-1)$ ).

We now treat each case appearing in Lemma 2·1 separately.

1. (a) Suppose $\pi$ belongs to Case 1. We recall from Lemma 2·2 that in this case $\sigma$ has dimension $p^{m_{\pi}}(1+{1}/{p})$ . We need to look into two sub cases:

   1. (i) Suppose $m_{\pi}=1$ . In this case two situations can occur. If $\sigma$ stays irreducible when restricted to $\operatorname{SL}_2(\mathbb{Z}/p\mathbb{Z})$ , then we are in case (8) and $\vert\chi_{\sigma}(zs)\vert =\vert \chi_{\tilde{\sigma}}(s)\vert$ with an irreducible representation $\tilde{\sigma}$ of $\operatorname{SL}_2(\mathbb{Z}/p\mathbb{Z})$ of dimension $p+1$ . On the other hand, if $\sigma\vert_{\operatorname{SL}_2(\mathbb{Z}/p\mathbb{Z})} = \rho_+\oplus\rho_-$ , then we have $\vert\chi_{\sigma}(zs)\vert =\vert \chi_{\rho_+}(s)+\chi_{\rho_+^h}(s)\vert$ . In both cases one concludes by looking up the corresponding character values, for example in [ Reference Kutzko15 , table V]. (Alternatively one can compute the character of $\sigma$ directly by observing that it can be realised as an irreducible parabolically induced representation of $\operatorname{GL}(\mathbb{F}_p)$ .)

   2. (ii) Suppose $m_{\pi}>1$ . By looking at the possible dimensions of non-trivial irreducible representations of $\operatorname{SL}_2(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})$ we see that $\tilde{\sigma}$ must have dimension $p^{m_{\pi}}(1+{1}/{p})$ as well. Thus $\sigma$ remains irreducible when restricted to $\operatorname{SL}_2(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})$ and we are in the situation of (8). We have $\vert \chi_{\sigma}(zs)\vert =\vert \chi_{\tilde{\sigma}}(s)\vert$ and the relevant character values for $\chi_{\tilde{\sigma}}$ can be found in [ Reference Kutzko15 , table III].

2. (b) Suppose $\pi$ belongs to Case 2. Here we automatically have $m_{\pi}=1$ and applying Lemma 2·2 yields that the dimension of $\sigma$ is p. We conclude that we are in the situation of (8) and $\tilde{\sigma}=\rho_{}$ is the Steinberg representation. We have $\vert \chi_{\sigma}(zs)\vert = \vert\chi_{\tilde{\sigma}}(s)\vert = 1$ as can be seen from [ Reference Kutzko15 , table V]. (Alternatively one could note that $\sigma$ itself is the Steinberg representation of $\operatorname{GL}_2(\mathbb{F}_p)$ , whose character is also well known.)

3. (c) Suppose $\pi$ belongs to Case 3·1. An application of Lemma 2·2 yields that $\sigma$ has dimension $p^{m_{\pi}}(1-{1}/{p})$ . As in Case 1 we need to distinguish two sub cases:

   1. (i) Suppose $m_{\pi}=1$ . We can have $\sigma\vert_{\operatorname{SL}_2(\mathbb{Z}/p\mathbb{Z})}=\rho_-\oplus\rho_-^h$ . In this case we estimate $\vert\chi_{\sigma}(zs)\vert =\vert \chi_{\rho_+}(s)+\chi_{\rho_+^h}(s)\vert$ . Otherwise, $\sigma\vert_{\operatorname{SL}_2(\mathbb{Z}/p\mathbb{Z})} = \tilde{\sigma}$ for an irreducible representation $\tilde{\sigma}$ of $\operatorname{SL}_2(\mathbb{Z}/p\mathbb{Z})$ of dimension $p-1$ . Here we simply have $\vert\chi_{\sigma}(zs)\vert =\vert \chi_{\tilde{\sigma}}(s)\vert$ . The desired estimate is produced by looking up the relevant character values. These can be found in [ Reference Kutzko15 , table V] for example. (Alternatively one can observe that $\sigma$ is a cuspidal representation of $\operatorname{GL}_2(\mathbb{F}_p)$ . It turns out that values of characters of cuspidal representations are even known for $\operatorname{GL}_n(\mathbb{F}_p)$ , see for example [ Reference Green10 ].)

   2. (ii) Suppose $m_{\pi}>1$ . In this case we again must be in the situation of (8). In particular, $\tilde{\sigma}$ is a representation of dimension $p^{m_{\pi}}(1-{1}/{p})$ . The relevant character values can be found in [ Reference Kutzko15 , table III].

4. (d) Suppose $\pi$ belongs to Case 3.2. Note that in this case we automatically have $m_{\pi}>1$ . Furthermore, Lemma 2·2 tells us that the dimension of $\sigma$ is $p^{m_{\pi}}(1-{1}/{p^2})$ . Thus, taking the dimensions of (irreducible) representations of $\operatorname{SL}_2(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})$ into account, we see that $\sigma\vert_{\operatorname{SL}_2(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})}$ must be reducible, so that we are in the situation of (7). We estimate $\vert\chi_{\sigma}(zs)\vert \leq 2\max(\vert\chi_{\tilde{\sigma}}(s)\vert,\vert\chi_{\tilde{\sigma}^h}(s)\vert)$ . Both representations $\tilde{\sigma}$ and $\tilde{\sigma}^h$ must have dimension ${1}/{2}p^{m_{\pi}}(1-{1}/{p^2})$ and the corresponding character values can be found in [ Reference Kutzko15 , table IV].Footnote 1

This exhausts all the cases and the proof is complete.

Before continuing we will discuss our choice of S. But first recall that $d=\dim_{\mathbb{C}}\pi_p^{K_p(m_{\pi})} \asymp p^{m_{\pi}}$ . Further note that $y_r=0$ unless $r=1,l_1,l_1l_2,l_1^2l_2^2$ for $l_1,l_2\in S$ . Put $\Lambda = d^{\frac{1}{3}}$ or $\Lambda= d^{\frac{1}{6}}$ . This is a slight spoiler but for experts in amplification it should be no surprise that this is the optimal size of the amplifier in this setting. Let

\begin{align*}S =\{ l \text{ prime }\colon l\asymp \Lambda \}.\end{align*}

By the prime number theorem (assuming d is sufficiently large, which is no problem) we have $\sharp S \sim \Lambda/\log(\Lambda)$ , but for us the following crude bound suffices

\begin{equation} \sharp S \gg_{\epsilon} \Lambda^{1-\epsilon}. \nonumber\end{equation}

Before we are ready to prove our key estimate we need to establish some counting results. Let

\begin{equation} M_z^{(\lambda)}(r) = \{ A\in \left[\begin{matrix} \mathbb{Z} &p^{\lambda}\mathbb{Z} \\ p^{\lambda} \mathbb{Z} & \mathbb{Z} \end{matrix}\right]\colon \det(A)=r \text{ and } u(Az,z)\leq 1 \}.\nonumber\end{equation}

The case $\lambda=0$ is easily handled using existing results. For example taking $N=\delta=1$ in [ Reference Templier23 , proposition 6·1]. We follow standard procedure and write

\begin{equation} M_z^{(\lambda)}(r)=M_{z,\star}^{(\lambda)}(r)\sqcup M_{z,p}^{(\lambda)}(r)\sqcup M_{z,u}^{(\lambda)}(r).\nonumber\end{equation}

Here the subscript $\star$ indicated that we are dealing with generic matrices $A=\left(\begin{matrix} a& b\\ c&d\end{matrix}\right)$ with $c\neq 0$ and $(a+d)^2\neq 4r$ . On the other hand u stands for unipotent so that $A\in M_{z,u}^{(\lambda)}(r)$ if and only if $c=0$ . Finally, $A\in M_{z,p}^{(\lambda)}(r)$ if $c\neq 0$ and $(a+d)^2= 4r$ . These are the parabolic matrices.

Counting the contribution of generic matrices is a standard lattice point counting argument. Note that, since we can take z in the classical fundamental domain for $\operatorname{SL}_2(\mathbb{Z})$ , this argument can be simplified. Indeed, in contrast to [ Reference Harcos and Templier11 , 23] we do not need to invoke Atkin–Lehner operators here.

Lemma 4·5. For $\lambda>1$ and $y\geq {\sqrt{3}}/{2}$ we have

\begin{equation} \sum_{r\asymp K} [\sharp M_{z,\star}^{(\lambda)}(r)+M_{z,p}^{(\lambda)}(r)] \ll \frac{K^{\frac{3}{2}}}{p^\lambda}+\frac{K^2}{p^{2\lambda}}\nonumber\end{equation}

and

\begin{equation} \sum_{\substack{r\asymp K, \\ r=\square}} \sharp M_{z,\star}^{(\lambda)}(r) \ll \frac{K^{1+\epsilon}}{p^\lambda}+\frac{K^{\frac{3}{2}+\epsilon}}{p^{2\lambda}}.\nonumber\end{equation}

We closely follow the argument in [ Reference Templier23 ].

Proof. Write $A=\left(\begin{matrix} a&b \\ c & d\end{matrix}\right)$ . Since $c\neq0$ and $c\ll \sqrt{r}y^{-1}$ we have $\ll {K^{\frac{1}{2}}}/{p^{\lambda}y}$ choices for c. Similarly, using the bound $\lvert{a+d}\rvert\ll\sqrt{K}$ , we have $\ll K^{\frac{1}{2}}$ possibilities to choose $a+d$ . Finally, we have the bound

(9) \begin{equation} \lvert{-cz^2+(a-d)z+b}\rvert^2 \leq 4Ky^2.\end{equation}

Write $b=p^{\lambda}b'$ and consider the lattice $L=\langle z,p^{\lambda}\rangle$ . Note that L has cocolume $\asymp p^{\lambda}y$ and first successive minima $\gg 1$ (since $y\gg 1$ ). Let B be the ball of radius $2\sqrt{K}y$ around $-cz^2$ . Then we have

\begin{equation} \sharp\{ (b',a-d)\} \ll \sharp(B\cap L) \ll 1+\sqrt{K}y+\frac{Ky}{p^{\lambda}}.\nonumber\end{equation}

We have counted the number of possibilities for the admissible quadruples $(c,b',a+d,a-d)$ . Since each of those quadruples uniquely determines a matrix A we have established

\begin{align} \sharp \{ A\in M_{z,\star}^{(\lambda)}(r)\colon r\asymp K \} &\ll \frac{K^{\frac{1}{2}}}{p^{\lambda}y}\cdot K^{\frac{1}{2}}\cdot (1+\sqrt{K}y+\frac{Ky}{p^{\lambda}}) \nonumber \\ &\ll \frac{K}{p^{\lambda}}+\frac{K^{\frac{3}{2}}}{p^\lambda}+\frac{K^2}{p^{2\lambda}}.\nonumber\end{align}

The case when we are only considering square matrices only needs a minor modification. Indeed, instead of counting $a+d$ trivially as earlier we observe that

\begin{equation} (a-d)^2+4bc=(a+d)^2-4(\sqrt{r})^2. \nonumber\end{equation}

If the matrix is parabolic the right-hand side would be 0. Thus we now consider only generic matrices. For those we can fix the left-hand side first, so that we determine $(a+d,\sqrt{r})$ essentially as solutions to a generalised Pell equation. There are at most $\ll K^{\epsilon}$ possibilities.

Lemma 4·6. We have

\begin{equation} \sharp M_{z,u}^{(\lambda)}(r) \ll r^{\epsilon}(1+\sqrt{r}p^{-\lambda}y). \nonumber\end{equation}

\begin{align} &\sum_{l\in S}M_{z,u}^{(\lambda)}(l) \ll \Lambda+\frac{\Lambda^{\frac{3}{2}} y}{p^{\lambda}}, \nonumber \\ &\sum_{l_1,l_2\in S}M_{z,u}^{(\lambda)}(l_1l_2) \ll \Lambda^2+\frac{\Lambda^3y}{p^{\lambda}} \text{ and }\nonumber\\ &\sum_{l_1,l_2\in S}M_{z,u}^{(\lambda)}(l_1^2l_2^2) \ll \Lambda^2+\frac{\Lambda^{4}y}{p^{\lambda}}.\nonumber\end{align}

The other bounds are derived elementary using only the fact that S contains only primes. (In contrast to [ Reference Harcos and Templier11 , lemma 2·4] we do not need a lattice counting argument, because we have an additional congruence condition on b that we can use.) We will only show the final estimate, since the others are derived similarly.

There are $\ll \Lambda^2$ possible choices for $r=l_1^2l_2^2\asymp \Lambda^4$ . Having fixed the determinant of this form we find that there are only $\ll 1$ choices for (a,d) with $ad=l_1^2l_2^2$ . Finally we observe that we can choose b in $\ll 1+{\Lambda^2y}/{p^{\lambda}}$ ways, since $p^\lambda\mid b$ and $\lvert{b}\rvert\ll \Lambda^2 y$ . Putting these estimates together completes the proof.

Remark 4·7. Suppose $y\geq {\sqrt{3}}/{2}$ . As in [ Reference Iwaniec and Sarnak14 , (A.7)] we have the bound

(10) \begin{equation} \lvert{c}\rvert\leq \frac{\sqrt{8r}}{y} \leq \sqrt{\frac{2^5r}{3}}. \end{equation}

Thus if $p^{\lambda}\geq 3,5\cdot\sqrt{r}$ , we must have $c=0$ . This is because $p^{\lambda}\mid c$ . In this case we obtain

\begin{equation} M_z^{(\lambda)}(r)=M_{z,u}^{(\lambda)}(r). \nonumber\end{equation}

Finally we need to consider the parabolic contribution.

Lemma 4·8. For $\lambda\geq 1$ and $y\geq {\sqrt{3}}/{2}$ we have

\begin{equation} \sum_{\substack{r\asymp K, \\ r=\square}} \sharp M_{z,p}^{(\lambda)}(r) \ll \frac{K^{\frac{3}{2}}}{p^{\lambda}}.\nonumber\end{equation}

Proof. This follows along the lines of [ Reference Blomer, Harcos, Maga and Milićević4 , lemma 14]. We provide some details for the convenience of the reader. Let $A=\left(\begin{matrix} a & b \\ c & d \end{matrix}\right)$ be a matrix contributing the the count $M_{z,p}^{(\lambda)}(r)$ for some $r\asymp K$ with $r=\square$ . We start by recalling the bounds

\begin{equation} cy \ll \sqrt{K} \text{ and }\vert 2cx-(a-d)\vert \ll \sqrt{K}.\nonumber\end{equation}

The first one is (10) and the second one follows from (9) when looking at the imaginary part.

Now, since A is parabolic, we must have

\begin{equation} (a+d)^2= 4(ad-bc)=4r.\nonumber\end{equation}

The latter equality can be rewritten as

\begin{equation} (a-d)^2 +4bc= 0.\nonumber\end{equation}

This implies that, if c is fixed, then $a-d$ has a fixed divisor of size at least $\sqrt{\vert c\vert}$ . We conclude that there are $\ll 1+{\sqrt{K}}/{\sqrt{\vert c\vert}}$ possible choices for $a-d$ . By summing over all possibilities for $c\neq 0$ and $p^\lambda\mid c$ we find that

\begin{equation} \sharp\{ c,a-d \}\ll \frac{K}{p^\lambda}.\nonumber\end{equation}

Finally, note that $\vert a+d\vert = 2\sqrt{r} \asymp \sqrt{K}$ , so that there are $\ll \sqrt{K}$ possible choices for $a+d$ . Since $a-d$ and c determine b we are done after gathering all the contributions.

We can now prove the main estimate of this section.

Lemma 4·9. Assume $p>3$ . Suppose $\pi$ belongs to Case 1, 2 or 3·1. Then, for ${\sqrt{3}}/{2}\ll y\ll \sqrt{d}$ we have

\begin{equation} \Phi(x+iy) \ll \sqrt{T}d^{\frac{5}{6}}.\nonumber\end{equation}

If $\pi$ belongs to Case 3.2, then we have the weaker bound

\begin{equation} \Phi(x+iy) \ll \sqrt{T}d^{\frac{11}{12}} \text{ for } \frac{\sqrt{3}}{2}\ll y\ll d^{\frac{1}{4}}.\nonumber\end{equation}

Proof. We start with Cases 1, 2 or 3·1. Our starting point is Corollary 4·3. Breaking the g-sum up into pieces on which we can estimate the character using Lemma 4·4 we get

\begin{equation} \frac{(\sharp S)^2}{\dim_{\mathbb{C}}\pi_p^{K_p(m_{\pi})}} \cdot \Phi(z)^2 \ll T\sum_{0\leq \lambda \leq m_{\pi}}p^{\lambda} \sum_{r} \frac{\lvert{y_r}\rvert}{\sqrt{r}} \cdot \sharp M_z^{(\lambda)}(r).\nonumber\end{equation}

We first consider the contribution of $\lambda=0$ . In this case the counting problem is independent of p and relatively easy. Indeed we have

(11) \begin{equation} \sum_{r} \frac{y_r}{\sqrt{r}}\sharp \{A\in M_2(\mathbb{Z}) \colon \det(A)=r \text{ and }u(Az,z)\leq 1 \} \ll \Lambda^4+\Lambda^{\frac{5}{2}}y.\end{equation}

This is for example [ Reference Templier23 , proposition 6·1] with $N=\delta=1$ and $z\in\mathcal{F}$ so that $y\gg 1$ .

Next we assume $\lambda>0$ . We summarise the results from Lemma 4·5, 4·8 and 4·6 in Table I below. Note that for the contribution of $r=1$ we have used Remark 4·7.

Table 1. Summary of counting results

All together this gives a contribution of

\begin{equation} \sum_{r}\frac{\lvert{y}\rvert}{\sqrt{r}}\cdot \sharp M_z^{(\lambda)}(r) \ll \Lambda+\frac{\Lambda^4}{p^\lambda}+\frac{\Lambda^2y}{p^\lambda}.\nonumber\end{equation}

Inserting these estimates in our (amplified) pre-trace inequality we get

\begin{equation} \frac{(\sharp S)^2}{\dim_{\mathbb{C}}\pi_p^{K_p(m_{\pi})}} \cdot \Phi(z)^2 \ll T\left(\Lambda^4+\Lambda^{\frac{5}{2}}y+\sum_{1\leq \lambda \leq m_{\pi}}p^{\lambda}\left[\Lambda+\frac{\Lambda^4}{p^\lambda}+\frac{\Lambda^2y}{p^\lambda}\right]\right).\nonumber\end{equation}

This simplifies to

\begin{equation} \Phi(z)^2 \ll \frac{Td^{1+\epsilon}}{\Lambda^{2-\epsilon}}\left(\Lambda^{\frac{5}{2}}y+\Lambda^4+\Lambda d \right). \nonumber\end{equation}

Inserting $\Lambda=d^{\frac{1}{3}}$ yields

\begin{align*}\Phi(z)^2 \ll Td^{1+\epsilon}(d^{\frac{2}{3}}+d^{\frac{1}{6}}y)\end{align*}

which directly implies the result for the non-exceptional cases 1, 2 and 3·1.

Finally if $\pi$ belongs to Case 3·2, then the same analysis with the weaker character estimates yields

\begin{align} \frac{(\sharp S)^2}{\dim_{\mathbb{C}}\pi_p^{K_p(m_{\pi})}} \cdot \Phi(z)^2 &\ll T\left(d^{\frac{1}{2}}\Lambda^4+d^{\frac{1}{2}}\Lambda^{\frac{5}{2}}y+\sum_{1\leq \lambda \leq m_{\pi}}p^{\frac{m_{\pi}+\lambda}{2}}\left[\Lambda+\frac{\Lambda^4}{p^\lambda}+\frac{\Lambda^2y}{p^\lambda}\right]\right) \nonumber \\ &\ll T\left(d^{\frac{1}{2}}\Lambda^4+d^{\frac{1}{2}}\Lambda^{\frac{5}{2}}y + \Lambda d^{1+\epsilon}\right). \nonumber\end{align}

This prompts the choice $\Lambda=d^{\frac{1}{6}}$ and we find

\begin{align*}\Phi(z)^2 \ll Td^{1+\epsilon}(d^{\frac{5}{6}}+d^{\frac{7}{12}}y)\end{align*}

This completes the proof.