# The sup-norm problem beyond the newform

Edgar Assing

(Date: 1st November 2021)

###### Abstract.

In this note we take up the classical sup-norm problem for automorphic forms
and view it from a new angle. Given a twist minimal automorphic representation
$\pi$ we consider a special small $GL_{2}(\mathbb{Z}_{p})$-type $V$ in $\pi$
and proof global sup-norm bounds for an average over an orthonormal basis of
$V$. We achieve a non-trivial saving when the dimension of $V$ grows.

##  1\. Introduction

It is a classical problem in analysis and mathematical physics, more precisely
Quantum Chaos, to bound the $L^{\infty}$-norm of certain eigenfunctions on
manifolds. In the most basic situation one considers a Riemann surfaces $X$ of
finite volume and eigenfunctions $\phi$ of the Laplace-Beltrami operator
$\Delta_{X}$. A sup-norm bound in the spectral aspect is then an estimates of
the form

$\frac{\|\phi\|_{\infty}}{\|\phi\|_{2}}\ll_{X}(1+\lvert{t_{\phi}}\rvert)^{\frac{1}{2}-\delta+\epsilon},$
(1)

where $\lambda_{\phi}=\frac{1}{4}+t_{\phi}^{2}$ is the Laplace-Beltrami
eigenvalue of $\phi$. The local bound corresponds to $\delta=0$ and is known
in great generality. The sup-norm problem asks for improved bounds featuring
some $\delta>0$. The sup-norm problem has only been solved for very special
surfaces $X$ and is hopeless in general. Indeed there is a well known
obstruction to the sup-norm problem coming from large eigenspaces
$V_{\lambda}$ given by the inequality

$\dim_{\mathbb{C}}V_{\lambda}\ll_{X}\sup_{\phi\in
V_{\lambda}}\frac{\|\phi\|_{\infty}}{\|\phi\|_{2}}.$

This observation is enough to establish the well known fact, that the local
bound (i.e. (1) with $\delta=0$) can not be improved for the sphere $X=S^{2}$.
So far we have only described the most basic version of the sup-norm problem
which, is already very interesting on its own. In addition it admits many
variations which have been studied throughout the years. An example for such a
variation is the so called level aspect where the base manifold changes in
some convenient family $X_{1},X_{2},\ldots$ and one keeps track of this change
in the sup-norm bound (1) using a suitable parameter called the level. Another
generalisation that should be mentioned allows $X$ to be a manifold of higher
dimension and rank.

Essentially any progress that has been made towards the sup-norm problem as
introduced above relies on the arithmeticity of $X$. The basic idea introduced
in the monumental paper [11] is to introduce additional symmetries (in the
form of Hecke operators), which are used to build a spectral projector that is
sharper than the one constructed with only the Laplace-Beltrami operator at
hand. Morally this might be thought of as forcing a multiplicity one situation
even if the Laplace-Beltrami eigenspaces can not be rigorously controlled. The
result of this method is a bound as in (1) with $\delta=\frac{1}{12}$ for
compact quotients $X=\Gamma\backslash\mathbb{H}$ constructed from maximal
orders in quaternion algebras.

Since its appearance the method from [11] has been tweaked, modified and
generalised, see for example [5, 20, 18, 1, 6] and the references within. Much
work is concerned with congruence quotients
$X=\Gamma_{0}(N)\backslash\mathbb{H}$ on which so called Hecke-Maaß newforms
are considered. Since these newforms enjoy a nice multiplicity one property
they are natural candidates for the sup-norm problem. In this note we are
going beyond the case of newforms and consider situations where the dimension
of the underlying $p$-adic representation grows. In other words, we solve the
sup-norm problem in the dimension aspect. This aspect is a new facet of the
sup-norm problem which seems extremely interesting and is not yet well
studied. While our result is the first in $p$-adic setting it is only preceded
by [4] where an archimedean version of this aspect is discussed.

To explain our result and its connection to the work of Blomer, Harcos, Maga
and Milićević it will be most convenient to leave the classical world of
Hecke-Maaß-newforms behind and work in the language of automorphic forms and
automorphic representations.

The sup-norm problem we will consider is connected to small
$\operatorname{GL}_{2}(\mathbb{Z}_{p})$-types in cuspidal automorphic
representations $\pi$, where $p$ is any prime. Comparing this to the recent
work [4] we are replacing the archimedean place $\infty$ by a finite place $p$
and the minimal $U(2)$-type of some automorphic representation by a suitably
chosen $\operatorname{GL}_{2}(\mathbb{Z}_{p})$-type. Note that in order to
afford interesting $K$-types at the archimedean place it is necessary to work
over fields admitting complex places or in higher rank. In the $p$-adic world
we already meet interesting cases when working with automorphic forms for
$\operatorname{GL}_{2}$ over $\mathbb{Q}$.

###  1.1. Set-up and main result

Before we continue our discussion we need to fix some notation. Let
$G(R)=\operatorname{GL}_{2}(R)$ for some ring $R$ and let $\mathbb{A}$ be the
adele ring over $\mathbb{Q}$. We will be working with cuspidal automorphic
representations $\pi$ of $G(\mathbb{A})$ with central character
$\omega_{\pi}$. Abusing notation we will write $\pi\subset
L_{0}^{2}(G(\mathbb{Q})\backslash G(\mathbb{A}),\omega_{\pi})$ assuming that
$\pi$ acts on an irreducible subspace of cuspidal automorphic forms by right
translation. Given a compact subgroup $H$ we write $\pi^{H}$ for the space of
$H$-invariant elements in $\pi$.

Set $K_{\infty}=SO(2)$ and $K_{l}=\operatorname{GL}_{2}(\mathbb{Z}_{l})$ for
primes $l$. Combining these we get the compact subgroup
$K=\prod_{v}K_{v}\subset G(\mathbb{A})$. Given a prime $p$ and $m>0$ we
consider the smaller compact subgroup

$K(p^{m})=K_{\infty}\times K_{p}(m)\times\prod_{l\neq p}K_{l}\text{ for
}K_{p}(p^{m})=1+p^{m}\cdot\operatorname{Mat}_{2\times
2}(\mathbb{Z}_{p})\subset K_{p}.$

Note that $K(p^{m})$ is normal and of finite index in $K$.

Throughout we restrict ourselves to the situation where $\pi$ is unramified
(i.e. spherical) away from $p$. In particular it is spherical at $\infty$ and
one associates the spectral parameter $t_{\pi}$. Set
$T=1+\lvert{t_{\pi}}\rvert$. Further, we have

$m_{\pi}=\min\\{m\in\mathbb{N}\colon\pi^{K(p^{m})}\neq\\{0\\}\\}<\infty.$

We set $V=\pi^{K(p^{m_{\pi}})}$ and observe that $\pi|_{K}$ endows $V$ with
the structure of a $K$-module. It turns out that, if $\pi$ is twist minimal,
$V$ is irreducible (see Lemma 2.2 below) by construction it is a $K$-type of
$\pi$. Set $d=\dim_{\mathbb{C}}V$, choose an orthonormal basis
$\phi_{1},\ldots,\phi_{d}$ for $V$ and define

$\Phi(g)=\left(\sum_{i=1}^{d}\lvert{\phi_{i}(g)}\rvert^{2}\right)^{\frac{1}{2}}.$

We are concerned with the sup-norm of $\Phi(g)$ and obtain the following
theorem which is a close analogue to [4, Theorem 1].

######  Theorem 1.1.

Suppose $\pi$ is twist minimal. In the notation above we have

$\|\Phi\|_{\infty}\ll T^{\frac{1}{2}+\epsilon}d^{\frac{11}{12}+\epsilon}.$

If the (arithmetic)-conductor of $\pi$ is a perfect square (i.e. the exponent-
conductor of the $p$-component $\pi_{p}$ of $\pi$ is even) or the
$p$-component $\pi_{p}$ of $\pi$ is not supercuspidal, then we have the better
bound

$\|\Phi\|_{\infty}\ll T^{\frac{1}{2}+\epsilon}d^{\frac{5}{6}+\epsilon}.$ (2)

While in the spectral aspect (i.e. the $T$-aspect in our statement) we only
recover the local bound, the key feature of our theorem is the sub-local
exponent in the dimension aspect $d$. Given the obstruction to the sup-norm
problem coming from growing eigenspaces the aspect under consideration may
seem counter intuitive. However, we are letting the dimension of the
eigenspace vary in a controlled manner and manage to show that one can still
achieve a considerable power saving in $d$ on average over any orthonormal
basis.

Note that the sup-norm bound given in the theorem holds globally. Thus, unlike
the one in [4][Theorem 1], no restriction to a compact domain is necessary
here. As usual when proving global sup-norm bound the argument consists of two
steps. First, a bound via the Whittaker expansion takes care of the regions
close to the cusps. This part of the argument is fairly standard but requires
some new computations of ramified Whittaker vectors. Second, a bound obtained
from the amplified pre-trace inequality is used to handle the bulk. It is here
were we exploit that we are only treating the average function $\Phi$. Indeed,
this allows us to identify the test function on the geometric side as a
character of a finite group which is nice to analyse.

To end this section let us briefly discuss the numerology of the exponents in
the $d$-aspect. For simplicity we restrict this discussion to the cases in
which our result gives the strong bound (2). Let us start by talking about the
local- (not to say trivial-) bound (in the bulk). To obtain this we can follow
Marshall’s strategy (see [16]) which leads to the following. Let $F$ be any
cuspidal automorphic form so that the translates $\phi(\cdot k)$, $k\in K$,
generate an irreducible $K$-module $W_{F}$. Then choosing certain $K$-matrix-
coefficients as test functions in the pre-trace inequality yields

$\frac{\|F|_{\Omega}\|_{\infty}}{\|F\|_{2}}\ll\dim_{\mathbb{C}}(W_{F})^{\frac{1}{2}}.$
(3)

Applying this to $\Phi$ upon noting that $\|\Phi\|_{2}=d^{\frac{1}{2}}$
suggests the local bound

$\|\Phi\|_{\infty}\ll d^{1+\epsilon}.$

(The same bound can also be obtained from the Whittaker expansion coupled with
a suitable generating domain.) Thus amplification allows us to improve the
exponent from the local bound by $\frac{1}{6}$, which should be an familiar
exponent. More suggestively we can write our main result as

$\frac{\|\Phi\|_{\infty}}{\|\Phi\|_{2}}\ll
d^{\frac{1}{2}-\frac{1}{6}+\epsilon}.$

One could say that Theorem 1.1 implies $\|\phi_{i}\|_{\infty}\ll
d^{\frac{1}{3}}$ on average. Note that if $p^{m_{\pi}}$ agrees with the
arithmetic conductor $p^{n_{\pi}}$ of $\pi$, then this result is not very
interesting. Indeed, in this case we can generate the elements
$\phi_{1},\ldots,\phi_{d}$ in $V$ directly from the newform $\phi_{\circ}$ in
$\pi$. By now there are very good bounds for this newform (and thus also for
the $\phi_{i}$’s) known in the literature. See [20] if $m_{\pi}=n_{\pi}=1$ or
[8, 12] in general. However, in the remaining cases (since $\pi$ is assumed to
be twist minimal these correspond to the situation where $\pi$ is
supercuspidal at $p$) our result provides new information in the sup-norm
problem. Indeed one can still generate $V$ from a translate of the newfom
$\phi_{\circ}$. (This is precisely the strategy used in [16, 18] to derive
local bounds for the newform of arbitrary level using (3).) Translated into
the level-aspect our result now essentially says that the sup-norm of the
$\phi_{i}$’s is bounded by $p^{\frac{1}{3}\lceil\frac{n_{\pi}}{2}\rceil}$ on
average. To the best of our knowledge this can not be derived from any known
sup-norm results on the newform $\phi_{\circ}$.

Finally we want to compare our result to the guiding archimedean example [4,
Theorem 1]. Recall that we need to replace the $K$-module $V$ by some
irreducible $U(2)$ representation $W$. This representation $W$ will occur as
minimal $U(2)$-type in some cuspidal automorphic $\pi$ of
$G(\mathbb{A}_{\mathbb{Q}(i)})$. Note that if $\dim_{\mathbb{C}}W\asymp l$ we
can think of $\pi$ (or rather $\pi_{\infty}$) having spectral density $\asymp
l^{2}$. This explains the local bounds

$\frac{\|\Phi\|_{\infty}}{\|\Phi\|_{2}}\ll l^{1+\epsilon}\text{ or
}\|\Phi\|_{\infty}\ll l^{\frac{3}{2}+\epsilon},$

where $\Phi$ is constructed as an average over some suitable basis of $W$
similar to our construction above. As result of an amplification process the
authors of [4] arrive at

$\frac{\|\Phi|_{\Omega}\|_{\infty}}{\|\Phi\|_{2}}\ll(l^{2})^{\frac{1}{2}-\frac{1}{12}+\epsilon}.$

Our notation suggests that in the result from [4] the number $l^{2}$ playes
the role of our $d$. This can be explained via the spectral density of
$\pi_{\infty}$ and respectively $\pi_{p}$. Indeed while in the archimedean
situation the spectral density is roughly $l^{2}$ in our case the spectral
density is linearly related to $d$. Thus in both cases the square root of the
spectral density seems to determine the trivial bound. (This is only
reasonable because we are considering minimal or close to minimal $K$-types in
both cases.) Note that the quality of the saving $\frac{1}{6}$ in the $p$-adic
versus $\frac{1}{12}$ in the archimedean case comes from slightly different
behaviour of the spectral transform.

Finally, let us remark that if the exponent conductor of $\pi_{p}$ is odd and
$\pi_{p}$ is supercuspidal, then our bounds for the spectral transform, which
in this case are linked to certain badly behaved characters of
$\operatorname{GL}_{2}$ over finite rings, are comparable to those used in
[4]. This explains that in this case we have matching numerology and obtain
only a saving of $\frac{1}{12}$ in the final exponent. Translated to the level
aspect our result states that on average the $\phi_{i}$’s are bounded by
$p^{\frac{5(n_{\pi}+1)}{24}}$. Bounds of this quality are known for newforms
only in the compact setting, see [10].

######  Remark 1.2.

Questions of these type should be even more interesting when considered in
higher rank. The reason is that in higher rank the analogously defined small
$K$-types can not be generated from translates of the newform. For example if
one considers a depth-zero supercuspidal representation $\pi_{p}$ of
$\operatorname{GL}_{3}(\mathbb{Q}_{p})$, then it has (arithmetic)-conductor
$p^{3}$ and the space $\pi_{p}^{K_{p}^{(3)}(1)}$, where $K_{p}^{(3)}(1)$ is
the principal congruence subgroup modulo $p$ in
$\operatorname{GL}_{3}(\mathbb{Z}_{p})$, is non-zero. However, it seems
impossible to find a translate of the newform that generates
$\pi_{p}^{K_{p}^{(3)}(1)}$. Indeed this would mean finding
$g\in\operatorname{GL}_{3}(\mathbb{Q}_{p})$ with

$K_{p}^{(3)}(1)\subset
g^{-1}\left[\begin{matrix}\mathbb{Z}_{p}&\mathbb{Z}_{p}&\mathbb{Z}_{p}\\\
\mathbb{Z}_{p}&\mathbb{Z}_{p}&\mathbb{Z}_{p}\\\
p\mathbb{Z}_{p}&p\mathbb{Z}_{p}&1+p\mathbb{Z}_{p}\end{matrix}\right]g.$

However, the question treated in this paper still makes sense and trying to
answer it is work in progress.

Acknowledgments: We would like to thank Prof. Dr. V. Blomer for fruitful
discussions and useful comments on an earlier draft of this manuscript.

##  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 in
$\omega_{\pi}=\bigotimes_{v}\omega_{\pi_{v}}$ 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

$\phi_{i}=\phi_{p}^{(i)}\cdot\prod_{v\neq p}\phi_{v}^{\circ}.$

Since the spherical functions $\phi_{p}^{\circ}$ are well understood much of
our work boils down to understanding properties of a orthonormal basis

$\text{span}\\{\phi_{p}^{(1)},\ldots,\phi_{p}^{(d)}\\}=\pi_{p}^{K_{p}(m_{\pi})}.$

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

$I_{0}(p)=\left[\begin{matrix}\mathbb{Z}_{p}^{\times}&\mathbb{Z}_{p}\\\
p\mathbb{Z}_{p}&\mathbb{Z}_{p}^{\times}\end{matrix}\right]\subset K_{p}$

for a Iwahori subgroup and put
$K_{p}^{\prime}=N_{G(\mathbb{Q}_{p})}(I_{0}(p))$. We also need the filtration

$K_{p}^{\prime}(m)=1+\left[\begin{matrix}p\mathbb{Z}_{p}&\mathbb{Z}_{p}\\\
p\mathbb{Z}_{p}&p\mathbb{Z}_{p}\end{matrix}\right]^{n}$

of $K_{p}^{\prime}$ 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:

  * •

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
$\pi_{p}=\chi_{1}\boxplus\chi_{2}$.

  * •

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 $\pi_{p}=\chi\otimes\text{St}$.

  * •

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

    * –

Case 3.1 ($n_{\pi}^{\prime}$ even): There is an irreducible representation
$\tau$ of $Z\cdot K_{p}$ with $\tau|_{Z}=\chi^{-2}\cdot\omega_{\pi_{p}}$ which
factors through $K_{p}(\frac{n_{\pi}^{\prime}}{2})$ so that
$\pi_{p}^{\prime}=c-\operatorname{Ind}_{ZK_{p}}^{G(\mathbb{Q}_{p})}\tau$.

    * –

Case 3.2 ($n_{\pi}^{\prime}$ odd): There is an irreducible representation
$\tau$ of $K_{p}^{\prime}$ which factors through
$K_{p}^{\prime}(n_{\pi}^{\prime}-1)$ with
$\tau|_{Z}=\chi^{-2}\cdot\omega_{\pi_{p}}$ such that
$\pi_{p}^{\prime}=c-\operatorname{Ind}_{K_{p}^{\prime}}^{G(\mathbb{Q}_{p})}\tau$.

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

######  Lemma 2.2.

Suppose $\pi_{p}$ is twist minimal, then $V=\pi_{p}^{K(m_{\pi})}$ is
irreducible as $K_{p}$-module and we have:

  1. (1)

The invariant $m_{\pi}$ is given by

$m_{\pi}=\begin{cases}n_{\pi}&\text{ if }\pi_{p}\text{ is Case~{}1},\\\
\lfloor\frac{n_{\pi}+1}{2}\rfloor&\text{ if }\pi_{p}\text{ is
Case~{}2,3.}\end{cases}$

  2. (2)

The dimension of $V$ is given by

$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~{}3.1 and }\\\ (1-\frac{1}{p^{2}})&\text{ if $\pi_{p}$ is
Case~{}3.2.}\end{cases}$

###### 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 [17,
Proposition 4.3]. Irreducibility can be seen by direct computation.

Second, if $\pi_{p}$ is in Case 1 and twist minimal, then $\pi_{p}=\text{St}$
and $n_{\pi}=1$. The results on $d$ and $m_{\pi}$ follow again from [17,
Proposition 4.3]. In this case irreducibility follows from [7, Theorem 1].

Finally, if $\pi_{p}$ belongs to Case 3, then the full statement is given in
[15, Theorem 3.5]. (See also [17, 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

$\mathcal{S}(\Phi,\mathcal{F})=\sup_{g\in\mathcal{F}}\lvert{\Phi(g)}\rvert.$

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

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

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.

Suppose

$\mathcal{S}(\Phi,\mathcal{F})\leq A$

holds for $\Phi^{2}=\sum_{i=1}^{d}\lvert{\phi_{i}}\rvert^{2}$ constructed from
an arbitrary orthonormal basis $\phi_{1},\ldots,\phi_{d}$ of $V$, then

$\|\Phi\|_{\infty}\leq A.$

###### Proof.

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

$g=\gamma zbk\text{ with }\gamma\in G(\mathbb{Q}),\,z\in
Z(\mathbb{R}),\,b\in\mathcal{F}\text{ and }k\in K.$

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. We thus have

$\Phi(g)\leq\mathcal{S}(\Phi^{(k)},\mathcal{F})\leq A$

by assumption. ∎

##  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

$W_{\phi}(g)=\int_{\mathbb{Q}\backslash\mathbb{A}}\phi(n(x)g)\psi_{\mathbb{A}}(x)^{-1}dx.$

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 $0\neq
q\in\frac{1}{p^{m_{\pi}}}\mathbb{Z}$. This leads to the Whittaker expansion

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

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

$\mathcal{W}(\pi,\psi_{\mathbb{A}})=\bigotimes_{v}\mathcal{W}(\pi_{v},\psi_{v}).$

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

$W_{v}(n(x)a(y))=\frac{\lvert{y}\rvert^{\frac{1}{2}}K_{it_{\pi}}(2\pi\lvert{y}\rvert)}{2\lvert{\Gamma(\frac{1}{2}+it_{\pi})\Gamma(\frac{1}{2}-it_{\pi})}\rvert^{\frac{1}{2}}}e(x),$

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

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

where $dy$ is th 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

$W_{v}(a(y))=\lvert{y}\rvert_{v}^{\frac{1}{2}}\lambda_{\pi}(l^{v_{l}(y)}).$

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

$L^{\\{\infty,p\\}}(s,\pi)=\prod_{v\neq
p,\infty}L_{v}(s,\pi_{v})=\sum_{(n,p)=1}\lambda_{\pi}(n)n^{-s}$

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 the image of
$\mathbb{C}\phi_{p}^{(i)}$ in the Whittaker model
$\mathcal{W}(\pi_{p},\psi_{p})$ such that

$\langle
W_{p}^{(i)},W_{p}^{(i)}\rangle_{\mathcal{W}(\pi_{p},\psi_{p})}=\int_{\mathbb{Q}_{p}^{\times}}\lvert{W_{p}^{(i)}(a(y))}\rvert^{2}\frac{dy}{\lvert{y}\rvert}=1,$

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

With these choices made we observe that

$\frac{W_{\phi_{i}}(g)}{\|\phi_{i}\|_{2}^{\frac{1}{2}}}=C_{\pi}^{(i)}\cdot\prod_{v}W_{v}(g_{v}),$

where $C_{\pi}^{(i)}\in\mathbb{C}^{\times}$ satisfies

$\lvert{C_{\pi}^{(i)}}\rvert^{2}=\lim_{s\to
1}\frac{\zeta^{\\{p,\infty\\}}(1)\zeta^{\\{p,\infty\\}}(2)}{L^{\\{p,\infty\\}}(s,\pi\otimes\check{\pi})}\ll_{\epsilon}p^{\epsilon
n_{\pi}}\cdot(1+t_{\pi})^{\epsilon}.$

Note that we choose the global measure on
$Z(\mathbb{A})G(\mathbb{Q})\backslash G(\mathbb{A})$ to be the Tamagawa
measure.

Combining everything we end up with

$\frac{\phi_{i}(n(x_{\infty})a(y_{\infty}))}{\|\phi_{i}\|_{2}^{\frac{1}{2}}}=C_{\pi}^{(i)}\sum_{k\in\mathbb{N}_{0}}\sum_{\begin{subarray}{c}0\neq
n\in\mathbb{Z},\\\
(n,p)=1\end{subarray}}\operatorname{sgn}(n)^{\rho}\frac{\lambda_{\pi}\left(n\right)}{\sqrt{\lvert{n}\rvert}}W_{p}^{(i)}(a(np^{k-m_{\pi}}))W_{\infty}\left(\frac{n}{p^{m_{\pi}-k}}y\right)e\left(\frac{n}{p^{m_{\pi}-k}}x\right),$

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

$v\mapsto W_{v}\in\mathcal{W}(\pi_{p},\psi_{p}).$

Define

$\mathcal{S}_{\pi_{p}}(g_{p})=\sum_{i=1}^{d}\frac{\lvert{W_{v_{i}}(g_{p})}\rvert^{2}}{\langle
W_{v_{i}},W_{v_{i}}\rangle_{\mathcal{W}(\pi_{p},\psi_{p})}}.$

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

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

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

###### Proof.

To safe space we put

$a(x)=\operatorname{sgn}(n)^{\rho}\frac{\lambda_{\pi}\left(n\right)}{\sqrt{\lvert{n}\rvert}}W_{\infty}\left(\frac{n}{p^{m_{\pi}-k}}y\right)e\left(\frac{n}{p^{m_{\pi}-k}}x\right)\text{
and }b_{i}(x)=W_{p}^{(i)}(a(np^{k-m_{\pi}}))$

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

$\phi_{i}(n(x)a(y))=C_{\pi}^{(i)}\sum_{x\in\mathbb{Q}^{\times}}a(x)b_{i}(x).$

With this at hand we estimate

$\displaystyle\Phi(g)$
$\displaystyle=\left(\sum_{i=1}^{d}\left|C_{\pi}^{(i)}\sum_{x\in\mathbb{Q}^{\times}}a(x)b_{i}(x)\right|^{2}\right)^{\frac{1}{2}}$
$\displaystyle=\max_{i}\lvert{C_{\pi}^{(i)}}\rvert\cdot\left(\sum_{x_{1}\in\mathbb{Q}^{\times}}\sum_{x_{2}\in\mathbb{Q}^{\times}}a(x_{1})\overline{a(x_{2})}\sum_{i=1}^{d}b_{i}(x_{1})\overline{b_{i}(x_{2})}\right)^{\frac{1}{2}}$
$\displaystyle\ll(dT)^{\epsilon}\left(\sum_{x_{1}\in\mathbb{Q}^{\times}}\sum_{x_{2}\in\mathbb{Q}^{\times}}\lvert{a(x_{1})a(x_{2})}\rvert\left(\sum_{i=1}^{d}\lvert{b_{i}(x_{1})}\rvert^{2}\right)^{\frac{1}{2}}\left(\sum_{i=1}^{d}\lvert{b_{i}(x_{2})}\rvert^{2}\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}$
$\displaystyle=(dT)^{\epsilon}\sum_{x\in\mathbb{Q}^{\times}}\lvert{a(x)}\rvert\left(\sum_{i=1}^{d}\lvert{b_{i}(x)}\rvert^{2}\right)^{\frac{1}{2}}.$

The claim follows by inserting the definitions of $a(x)$ and $b_{i}(x)$. ∎

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 $\mathcal{S}_{\pi_{p}}(a(y))$ will involve a case study and
each case we will feature its own approach to the problem. Finally, combining
all possible cases, will lead to the bound

$\mathcal{S}_{\pi_{p}}(p^{-m_{\pi}}y)\ll
d^{1+\epsilon}\cdot\lvert{y}\rvert_{p}.$ (4)

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

$\langle f,f^{\vee}\rangle=\int_{K}f(k)f^{\vee}(k)dk.$

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

$v_{p}(g)=\operatorname{Vol}(B(\mathbb{Z}_{p})K_{p}(1),dk)^{-\frac{1}{2}}\cdot\begin{cases}\lvert{\frac{a}{d}}\rvert&\text{
if }g=\left(\begin{matrix}a&b\\\ 0&d\end{matrix}\right)k\in
B(\mathbb{Q}_{p})K_{p}(1),\\\ 0&\text{ else.}\end{cases}$

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

$B(\mathbb{Z}_{p})\backslash
K_{p}/K_{p}(1)=\\{\gamma_{0},\ldots,\gamma_{p}\\}$

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

$W_{v}(g)=\Lambda(g.v)\text{ or }W_{v^{\vee}}(g)=\Lambda^{\vee}(g.v^{\vee}).$

We will first consider the related average

$\mathcal{S}_{V}(y)=\sum_{i=0}^{p}W_{v_{i}}(a(y))W_{v_{i}^{\vee}}(a(y)).$

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
$V^{\vee}$.

By [14, Lemma 4.4 and Remark 4.6] we get

$\displaystyle W_{v_{i}}(a(y))W_{v_{i}^{\vee}}(a(y))$
$\displaystyle=\int_{\mathbb{Q}_{p}}^{st}\langle
n(x)a(y).v_{i},a(y).v_{i}^{\vee}\rangle\psi_{p}(x)^{-1}dx$
$\displaystyle=\lvert{y}\rvert_{p}\int_{\mathbb{Q}_{p}}^{st}\langle
n(x)v_{i},v_{i}^{\vee}\rangle\psi_{p}(xy)^{-1}dx.$

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

$\langle
n(x).v_{i},v_{i}^{\vee}\rangle=\frac{\sharp[B(\mathbb{Z}_{p})/K_{p}(1)\cap
B(\mathbb{Z}_{p})]^{\frac{1}{2}}}{\operatorname{Vol}(K_{p}(1),dk)^{\frac{1}{2}}}\cdot\int_{K_{p}(1)}v_{0}(\gamma_{i}^{-1}kn\gamma_{i})dk.$

The case $i=0$ is somehow special and will be treated later. For now let us
assume $i>0$. To take advantage of the support of $v_{0}$ we have to
investigate

$kn=\gamma_{i}b\gamma_{i}^{-1}\tilde{k}\in wBw\cdot K_{p}(1).$

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

$\langle\pi(n(x))v_{i},v_{i}^{\vee}\rangle=\delta_{n\in N(p\mathbb{Z}_{p})}.$

With this at hand it is easy to compute

$\displaystyle\int_{\mathbb{Q}_{p}}^{st}\langle\pi(n(x))v_{i},v_{i}^{\vee}\rangle\psi_{p}(yx)^{-1}dx=\int_{p\mathbb{Z}_{p}}\psi(yx)^{-1}dx=p^{-1}\delta_{y\in
p^{-1}Z_{p}},$

for $1\leq i\leq p$.

We turn towards $i=0$, so that $\gamma_{0}=1$. Further we consider
$y\in\mathbb{Z}_{p}$. Recall that every $k\in K_{p}(1)$ can be written as
$k=t_{k}\overline{n}_{k}n_{k}\in
B(\mathbb{Z}_{p})N(p\mathbb{Z}_{p})^{t}N(p\mathbb{Z}_{p})$ by using the
Iwahori-factorisation. We obtain

$\displaystyle\int_{\mathbb{Q}_{p}}^{st}\langle\pi_{p}(n(x))v_{0},v_{0}^{\vee}\rangle\psi_{p}(yp^{-1}x)^{-1}dx$
$\displaystyle\quad=\frac{\sharp[B(\mathbb{Z}_{p})/K_{p}(1)\cap
B(\mathbb{Z}_{p})]^{\frac{1}{2}}}{\operatorname{Vol}(K_{p}(1),dk)^{\frac{1}{2}}}\cdot\int_{\mathbb{Q}_{p}}^{st}\int_{K_{p}(1)}v_{0}(kn)dk\psi_{p}(yp^{-1}x)^{-1}dx$
$\displaystyle\quad=\frac{\sharp[B(\mathbb{Z}_{p})/K_{p}(1)\cap
B(\mathbb{Z}_{p})]^{\frac{1}{2}}}{\operatorname{Vol}(K_{p}(1),dk)^{\frac{1}{2}}}\cdot\int_{K_{p}(1)}\int_{\mathbb{Q}_{p}}^{st}v_{0}(\overline{n}_{k}n)\psi_{p}(yp^{-1}x)^{-1}dxdk.$

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

$\int_{K_{p}(1)}f(k)dk=\frac{\operatorname{Vol}(K_{p}(1),dk)}{\operatorname{Vol}(B(\mathbb{Z}_{p})\cap
K_{p}(1),db)}\int_{B(\mathbb{Z}_{p})\cap
K_{p}(1)}p\int_{p\mathbb{Z}_{p}}f(b\overline{n}(pu))dudb.$

If we write $\tilde{v}_{0}$ to be the re-normalisation of $v_{0}$ with
$\tilde{v}_{0}(1)=1$, then we have

$\displaystyle\int_{\mathbb{Q}_{p}}^{st}\langle\pi_{p}(n(x))v_{0},v_{0}^{\vee}\rangle\psi_{p}(yp^{-1}x)^{-1}dx$
$\displaystyle=p\int_{\mathbb{Q}_{p}}^{st}\int_{p\mathbb{Z}_{p}}\tilde{v}_{0}(n(z)^{t}n(x))\psi_{p}(yp^{-1}x)^{-1}dzdx$
$\displaystyle=p\int_{\mathbb{Q}_{p}}^{st}\int_{p\mathbb{Z}_{p}}\tilde{v}_{0}\left(\left(\begin{matrix}1&x-z^{-1}\\\
z&zx\end{matrix}\right)\right)\psi_{p}(yp^{-1}(x-z^{-1}))^{-1}dzdx.$

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

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

Inserting this and using the transformation behaviour of $\tilde{v}_{0}$ one
obtains

$\int_{\mathbb{Q}_{p}}^{st}\langle\pi_{p}(n(x))v_{0},v_{0}^{\vee}\rangle\psi_{p}(yp^{-1}x)^{-1}dx\\\
=p\int_{p\mathbb{Z}_{p}}^{st}\psi_{p}(yz^{-1}p^{-1})\frac{dz}{\lvert{z}\rvert_{p}^{2}}\int_{\mathbb{Q}_{p}}^{st}\tilde{v}_{0}\left(\left(\begin{matrix}1&0\\\
x^{-1}&1\end{matrix}\right)\right)\psi_{p}(-xyp^{-1})\frac{dx}{\lvert{x}\rvert_{p}^{2}}.$
(5)

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

$p\int_{p\mathbb{Z}_{p}}^{st}\psi_{p}(yz^{-1}p^{-1})\frac{dz}{\lvert{z}\rvert_{p}^{2}}=\sum_{l=1}^{\infty}p^{l}\int_{\mathbb{Z}_{p}^{\times}}\psi_{p}(zyp^{-1-l})dz\\\
=\sum_{l=1}^{v_{p}(y)-1}p^{l}(1-p^{-1})-\delta_{v_{p}(y)\geq
1}p^{v_{p}(y)-1}=-\delta_{v_{p}(y)\geq 1}.$ (6)

Turning to the other integral we find

$\displaystyle\int_{\mathbb{Q}_{p}}^{st}\tilde{v}_{0}\left(\left(\begin{matrix}1&0\\\
x^{-1}&1\end{matrix}\right)\right)\psi_{p}(-xyp^{-1})\frac{dx}{\lvert{x}\rvert_{p}^{2}}$
$\displaystyle=\int_{\mathbb{Q}_{p}\setminus\mathbb{Z}_{p}}^{st}\psi_{p}(-xyp^{-1})\frac{dx}{\lvert{x}\rvert_{p}^{2}}$
$\displaystyle=\sum_{l=1}^{\infty}p^{-l}\int_{\mathbb{Z}_{p}^{\times}}\psi_{p}(-xyp^{-1-l})dx$
$\displaystyle=\sum_{l=1}^{v_{p}(y)-1}p^{-l}(1-p^{-1})-\delta_{v_{p}(y)\geq
1}p^{-v_{p}(y)-1}$ $\displaystyle=\delta_{v_{p}(y)\geq
2}(p^{-1}-p^{-2-v_{p}(y)})-\delta_{v_{p}(y)\geq 1}p^{-1-v_{p}(y)}.$

In particular we have

$\int_{\mathbb{Q}_{p}}^{st}\langle\pi_{p}(n(x))v_{0},v_{0}^{\vee}\rangle\psi_{p}(yp^{-1}x)^{-1}dx=\delta_{v_{p}(y)\geq
2}(p^{-2-v_{p}(y)}-p^{-1})+\delta_{v_{p}(y)\geq 1}p^{-1-v_{p}(y)}.$

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 leads us to the following result.

######  Lemma 3.2.

In the notation above we have

$\mathcal{S}_{V}(y)=\lvert{y}\rvert_{p}(\delta_{v_{p}(y)\geq-1}+\delta_{v_{p}(y)\geq
0}p^{-2}\lvert{y}\rvert_{p}+\delta_{v_{p}(y)\geq
1}(p^{-3}\lvert{y}\rvert_{p}-p^{-1}).$

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

$\mathcal{S}_{\pi_{p}}(a(y))\ll d\lvert{y}\rvert_{p}.$

###### 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

$\mathcal{S}_{\pi}(a(y))=\sum_{i=1}^{p}\frac{\lvert{W_{w_{i}}(a(y))}\rvert^{2}}{\langle
W_{w_{i}},W_{w_{i}}\rangle}=\sum_{i=1}^{p}\frac{W_{w_{i}}(a(y))W_{w_{i}^{\vee}}(a(y))}{\langle
w_{i},w_{i}^{\vee}\rangle}.$

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

$\mathcal{S}_{V}(y)=\mathcal{S}_{\pi}(a(y))+W_{w_{0}}(a(y))W_{w_{0}^{\vee}}(a(y)).$

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

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

To obtain the other basis elements 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

$K_{p}=\bigsqcup_{a\in\mathbb{Z}_{p}/p^{m_{\pi}}\mathbb{Z}_{p}}B(\mathbb{Z}_{p})\gamma_{0,a}K_{p}(m_{\pi})\sqcup\bigsqcup_{i=1}^{m_{\pi}}\bigsqcup_{a\in(\mathbb{Z}_{p}/p^{m_{\pi}-i}\mathbb{Z}_{p})^{\times}}B(\mathbb{Z}_{p})\gamma_{i,a}K_{p}(m_{\pi}).$

for

$\gamma_{i,a}=\begin{cases}\left(\begin{matrix}0&-1\\\
1&a\end{matrix}\right)&\text{ if }i=0,\\\ \left(\begin{matrix}1&0\\\
ap^{i}&1\end{matrix}\right)&\text{ if }1\leq i\leq m_{\pi}-1,\\\ 1_{2}&\text{
if }i=m_{\pi}.\end{cases}$

###### 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

$g=\left(\begin{matrix}a&b\\\
0&d\end{matrix}\right)=\left(\begin{matrix}a&b\\\
0&c\end{matrix}\right)\left(\begin{matrix}1-\frac{bc}{ad}&0\\\
\frac{c}{d}&1\end{matrix}\right)\in B(\mathbb{Z}_{p})K_{p}(m_{\pi}).$

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

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

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

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

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

∎

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

$\displaystyle W_{v}(a(y))$
$\displaystyle=\int_{\mathbb{Q}_{p}}v(wn(x)a(y))\psi_{p}(x)^{-1}dx$
$\displaystyle=\lvert{y}\rvert_{p}^{\frac{1}{2}-\rho}\int_{\mathbb{Q}_{p}}v(wn(x))\psi_{p}(xy)^{-1}dx$
$\displaystyle=\lvert{y}\rvert_{p}^{\frac{1}{2}-\rho}p^{-n_{\pi}}\sum_{a\in\mathbb{Z}_{p}/p^{n_{\pi}}\mathbb{Z}_{p}}\psi_{p}(ay)^{-1}v(wn(a))$
$\displaystyle\qquad+\lvert{y}\rvert_{p}^{\frac{1}{2}-\rho}\int_{\mathbb{Q}_{p}\setminus\mathbb{Z}_{p}}\lvert{x}\rvert^{-2\rho}\chi(x)^{-1}\psi(xy)^{-1}v(n(x^{-1})^{t})\frac{dx}{\lvert{x}\rvert_{p}}.$

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

$\displaystyle\int_{\mathbb{Q}_{p}\setminus\mathbb{Z}_{p}}\lvert{x}\rvert^{-2\rho}\chi(x)^{-1}\psi(xy)^{-1}v(n(x^{-1})^{t})\frac{dx}{\lvert{x}\rvert_{p}}$
$\displaystyle\qquad=\int_{p\mathbb{Z}_{p}}\lvert{x}\rvert^{2\rho}\chi(x)\psi(x^{-1}y)^{-1}v(n(x)^{t})\frac{dx}{\lvert{x}\rvert_{p}}$
$\displaystyle\qquad=v(1)\int_{p^{m_{\pi}}\mathbb{Z}_{p}}\chi(x)\lvert{x}\rvert^{2\rho}\psi(x^{-1}y)^{-1}\frac{dx}{\lvert{x}\rvert_{p}}$
$\displaystyle\qquad\qquad+\sum_{i=1}^{m_{\pi}-1}\sum_{b\in(\mathbb{Z}_{p}/p^{m_{\pi}-i}\mathbb{Z}_{p})^{\times}}p^{-2\rho
i}\chi(b)\cdot\int_{1+p^{m_{\pi}-i}\mathbb{Z}_{p}}\chi(x)^{-1}\psi(-byxp^{-i})dx.$

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

$\displaystyle W_{v}(a(y))$
$\displaystyle=\lvert{y}\rvert_{p}^{\frac{1}{2}-\rho}p^{-n_{\pi}}\sum_{a\in\mathbb{Z}_{p}/p^{m_{\pi}}\mathbb{Z}_{p}}\psi_{p}(ay)^{-1}v(\gamma_{0,a})$
$\displaystyle\qquad+\lvert{y}\rvert_{p}^{\frac{1}{2}-\rho}\sum_{i=1}^{m_{\pi}-1}\sum_{b\in(\mathbb{Z}_{p}/p^{m_{\pi}-i}\mathbb{Z}_{p})^{\times}}p^{-2\rho
i}\chi(b)v(\gamma_{i,b})G_{m_{\pi}-i}(-byp^{-i},\chi^{-1})$
$\displaystyle\qquad+\lvert{y}\rvert_{p}^{\frac{1}{2}-\rho}\cdot
v(\gamma_{m_{\pi},0})\cdot\int_{p^{m_{\pi}}\mathbb{Z}_{p}}\chi(x)\lvert{x}\rvert^{2\rho}\psi(x^{-1}y)^{-1}\frac{dx}{\lvert{x}\rvert_{p}}.$

for

$G_{l}(y,\chi)=\int_{1+p^{l}\mathbb{Z}_{p}}\chi(x)\psi(yx)dx.$

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

$\mathcal{S}_{\pi_{p}}(a(p^{-m_{\pi}}y))\ll
d^{1+\epsilon}\lvert{y}\rvert_{p}.$

###### 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

$\langle W_{v_{i}},W_{v_{i}}\rangle=(1+p^{-1}).$

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

$\mathcal{S}_{\pi_{p}}(a(y))=\sum_{i=0}^{m_{\pi}}\mathcal{S}_{i}(y)\text{ for
}\mathcal{S}_{i}(y)=(1+p^{-1})^{-1}\sum_{a\in\mathbb{Z}_{p}/p^{m_{\pi}-i}\mathbb{Z}_{p}}\lvert{W_{\pi_{p}(\gamma_{i,a}^{-1})v_{0}}}\rvert^{2}.$

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

$\mathcal{S}_{i}(y)=(1+p^{-1})^{-1}\sum_{a\in\mathbb{Z}_{p}/p^{m_{\pi}}\mathbb{Z}_{p}}p^{-2m_{\pi}}\lvert{y}\rvert_{p}v_{0}(1)^{2}=\frac{(1+p^{-1})^{-1}p^{-m_{\pi}}}{\operatorname{Vol}(B(\mathbb{Z}_{p})K_{p}(m_{\pi}),dk)}=\lvert{y}\rvert_{p}$

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

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

$\displaystyle\mathcal{S}_{i}(y)$
$\displaystyle=(1+p^{-1})^{-1}v_{0}(1)^{2}\lvert{y}\rvert_{p}\sum_{a\in(\mathbb{Z}_{p}/p^{m_{\pi}-i}\mathbb{Z}_{p})^{\times}}\lvert{G_{m_{\pi}-i}(-byp^{-i},\chi^{-1})}\rvert^{2}.$

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 [2,
Lemma 5.8]. Indeed one extracts

$G_{l}(zp^{k},\chi)=\begin{cases}\epsilon(\frac{1}{2},\chi^{-1})\chi^{-1}(y)p^{-\frac{k}{2}}&\text{
if }l\leq\lfloor\frac{a(\chi)}{2}\rfloor,k=-a(\chi)\text{ and }z\in
b(\chi)+p^{l}\mathbb{Z}_{p},\\\ \psi_{p}(zp^{k})p^{-l}&\text{ if
}l\geq\lceil\frac{a(\chi)}{2}\rceil,k=-a(\chi)\text{ and
}z\in-b(\chi)+p^{a(\chi)-l}\mathbb{Z}_{p},\\\ 0&\text{ else.}\end{cases}$

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
$i\leq\lfloor\frac{m_{\pi}}{2}\rfloor$ we have

$\displaystyle\mathcal{S}_{i}(y)$
$\displaystyle=\delta_{v_{p}(y)=i-m_{\pi}}(1+p^{-1})^{-1}v_{0}(1)^{2}\lvert{y}\rvert_{p}\sum_{b\in(\mathbb{Z}_{p}/p^{m_{\pi}-i}\mathbb{Z}_{p})^{\times}}p^{2i-2m_{\pi}}\delta_{-byp^{v_{p}(y)}\in-b(\chi)+p^{i}\mathbb{Z}_{p}}$
$\displaystyle=\lvert{y}\rvert_{p}.$

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

$\displaystyle\mathcal{S}_{i}(y)$
$\displaystyle=\delta_{v_{p}(y)=i-m_{\pi}}(1+p^{-1})^{-1}v_{0}(1)^{2}\lvert{y}\rvert_{p}\sum_{b\in(\mathbb{Z}_{p}/p^{m_{\pi}-i}\mathbb{Z}_{p})^{\times}}p^{-m_{\pi}}\cdot\delta_{-byp^{v_{p}(y)}\in
b(\chi)+p^{m_{\pi}-i}\mathbb{Z}_{p}}$ $\displaystyle=\lvert{y}\rvert_{p}.$

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

$\displaystyle\mathcal{S}_{m_{\pi}}(y)$
$\displaystyle=(1+p^{-1})v_{0}(1)^{2}\lvert{y}\rvert_{p}\lvert{\int_{p^{m_{\pi}}\mathbb{Z}}\lvert{x}\rvert^{2\rho}\psi(x^{-1}y)^{-1}\frac{dx}{\lvert{x}\rvert_{p}}}\rvert^{2}$
$\displaystyle=p^{m_{\pi}}\lvert{y}\rvert_{p}\cdot\lvert{\int_{p^{m_{\pi}}\mathbb{Z}_{p}}\chi(x)\lvert{x}\rvert^{2\rho}\psi(x^{-1}y)^{-1}\frac{dx}{\lvert{x}\rvert_{p}}}\rvert^{2}.$

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

$\int_{p^{m_{\pi}}\mathbb{Z}_{p}}\chi(x)\lvert{x}\rvert^{2\rho}\psi(x^{-1}y)^{-1}\frac{dx}{\lvert{x}\rvert_{p}}=\delta_{y\in\mathbb{Z}_{p}}\epsilon(\frac{1}{2},\chi)\chi(y)p^{-\frac{m_{\pi}}{2}-2\rho[v_{p}(y)+m_{\pi}]}.$

Inserting this above concludes the proof since it implies
$\mathcal{S}_{m_{\pi}}(y)=\lvert{y}\rvert_{p}$. ∎

####  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

$\sharp\mathfrak{X}_{k}=p^{k-1}(p-1).$

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

$\xi_{\chi}^{(m)}(y)=\mathbbm{1}_{p^{-m}\mathbb{Z}_{p}^{\times}}(y)\chi(y).$

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

$W_{f}(a(y))=f(y)\text{ for }f\in\mathcal{K}_{\psi_{p}}(\pi_{p})\text{ and
}y\in\mathbb{Q}_{p}^{\times}.$

Thus we compute

$\langle
W_{\xi_{\chi_{1}}^{(m_{1})}},W_{\xi_{\chi_{2}}^{(m_{2})}}\rangle=\int_{\mathbb{Q}_{p}^{\times}}\xi_{\chi_{1}}^{(m_{1})}(y)\overline{\xi_{\chi_{2}}^{(m_{2})}(y)}d^{\times}y=\delta_{m_{1}=m_{2}}\int_{p^{-m_{1}}\mathbb{Z}_{p}^{\times}}\chi_{1}(y)\chi_{2}^{-1}(y)d^{\times}y\\\
=\delta_{\begin{subarray}{c}m_{1}=m_{2},\\\ \chi_{1}=\chi_{2}\end{subarray}}.$
(7)

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 we have

  * •

If $n_{\pi}=2m_{\pi}$, then we have

$S_{\pi_{p}}(a(y))=\frac{p^{m_{\pi}}}{\zeta_{p}(1)}\cdot\delta_{y\in
p^{-m_{\pi}}\mathbb{Z}_{p}^{\times}}.$

  * •

If $n_{\pi}=2m_{\pi}-1$, then we have

$S_{\pi_{p}}(a(y))=\frac{p^{m_{\pi}}}{\zeta_{p}(1)}\cdot\delta_{y\in
p^{-m_{\pi}}\mathbb{Z}_{p}^{\times}}+\frac{p^{m_{\pi}-1}}{\zeta_{p}(1)}\cdot\delta_{y\in
p^{1-m_{\pi}}\mathbb{Z}_{p}^{\times}}.$

In general we have the bound

$S_{\pi_{p}}(a(p^{-m_{\pi}}y))\ll
d\cdot\lvert{y}\rvert_{p}\cdot\delta_{y\in\mathbb{Z}_{p}}.$

###### Proof.

We start with the case $n_{\pi}=2m_{\pi}$. By [17, Lemma 4.4] we find that a
basis for $\pi^{K_{p}(m_{\pi})}$ in the Kirillov model is given by

$\\{\xi_{\chi}^{(m_{\pi})}\colon\chi\in\mathfrak{X}_{m_{\pi}}\\}.$

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

$\mathcal{S}_{\pi_{p}}(y)=\sum_{\chi\in\mathfrak{X}_{m_{\pi}}}\lvert{\xi_{\chi}^{(m_{\pi})}(y)}\rvert^{2}=\delta_{y\in
p^{-m_{\pi}}\mathbb{Z}_{p}^{\times}}\cdot\sharp\mathfrak{X}_{m_{\pi}}.$

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

$\\{\xi_{\chi}^{m_{\pi}}\colon\chi\in\mathfrak{X}_{m_{\pi}}\\}\cup\\{\xi_{\chi}^{m_{\pi}-1}\colon\chi\in\mathfrak{X}_{m_{\pi}-1}\\}.$

It is again easy to compute the desired quantity:

$\mathcal{S}_{\pi_{p}}(y)=\delta_{y\in
p^{-m_{\pi}}\mathbb{Z}_{p}^{\times}}\cdot\sharp\mathfrak{X}_{m_{\pi}}+\delta_{y\in
p^{1-m_{\pi}}\mathbb{Z}_{p}^{\times}}\cdot\sharp\mathfrak{X}_{m_{\pi}-1}.$

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 (4) and follow standard procedure.

######  Lemma 3.8.

We have

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

###### Proof.

Inserting (4) into Lemma 3.1 yields

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

Estimating the remaining $n$-sum as for example in [20] or [18] yields the
desired result. ∎

##  4\. A bound via the pre-trace formula

A next bound will be derived from the pre-trace inequality. We start by
discussing the local test functions. At the archimedean place we fix
$f_{\infty}$ as usual. It will satisfy

  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.)

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

$f_{p}^{(i)}(g)=\mathbbm{1}_{ZK}(g)\frac{\overline{\langle\pi(g)\phi_{p}^{(i)},\phi_{p}^{(i)}\rangle_{\pi_{p}}}}{\langle\phi_{p}^{(i)},\phi_{p}^{(i)}\rangle_{\pi_{p}}}.$

######  Lemma 4.1.

For every irreducible admissible unitary representation $\sigma$ of
$G(\mathbb{Q}_{p})$ the operator $\sigma(f_{p}^{(i)})$ is non-negative and
self-adjoint. Furthermore we have
$\pi_{p}(f_{p}^{(i)})\phi_{p}^{(i)}=\dim_{\mathbb{C}}(\pi_{p}^{K_{p}(m_{\pi})})^{-1}\cdot\phi_{p}^{(i)}$.

###### Proof.

The proof is standard and relies on Schur’s orthogonality relations for
irreducible representations of $K_{p}$. To apply this it will be important to
keep in mind that $\pi^{K_{p,1}(m_{\pi})}$ is irreducible.

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

$f_{p}^{(i)}=\dim_{\mathbb{C}}\pi^{K_{p,1}(m_{\pi})}\cdot(f_{p}^{(i)}\star
f_{p}^{(i)}).$

Indeed we compute

$\displaystyle[f_{p}^{(i)}\star f_{p}^{(i)}](h)$
$\displaystyle=\int_{Z/G}f_{p}^{(i)}(g^{-1})f_{p}^{(i)}(gh)dg$
$\displaystyle=\frac{\mathbbm{1}_{ZK}(h)}{\langle\phi_{p}^{(i)},\phi_{p}^{(i)}\rangle_{\pi_{p}}^{2}}\int_{K_{p}}\langle\pi_{p}(g)\phi_{p}^{(i)},\phi_{p}^{(i)}\rangle_{\pi_{p}}\overline{\langle\pi_{p}(gh)\phi_{p}^{(i)},\phi_{p}^{(i)}\rangle_{\pi_{p}}}dk$
$\displaystyle=\dim_{\mathbb{C}}(\pi^{K_{p,1}(m_{\pi})})^{-1}\frac{\mathbbm{1}_{ZK}(h)}{\langle\phi_{p}^{(i)},\phi_{p}^{(i)}\rangle_{\pi_{p}}}\overline{\langle\pi_{p}(gh)\phi_{p}^{(i)},\phi_{p}^{(i)}\rangle_{\pi_{p}}}$

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

$\langle\pi_{p}(f_{p}^{(i)})\phi_{p}^{(i)},w\rangle_{\pi_{p}}=\dim_{\mathbb{C}}(\pi_{p}^{K_{p}(m_{\pi})})^{-1}\cdot\langle\phi_{p}^{(i)},w\rangle_{\pi_{p}}$

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

$\langle\pi_{p}(f_{p}^{(i)})\phi_{p}^{(i)},w\rangle_{\pi_{p}}=\langle\int_{Z\backslash
G}f_{p}^{(i)}(g)\pi_{p}(g)\phi_{p}^{(i)}dg,w\rangle=\int_{Z\backslash
G}f_{p}^{(i)}(g)\langle\pi_{p}(g)\phi_{p}^{(i)},w\rangle_{\pi_{p}}dg.$

∎

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

$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}$ (8)

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

$f^{(i)}=f_{\infty}\otimes f_{p}\otimes f_{ur}\text{ and }f=\sum_{i}f^{(i)}.$

We introduce

$M(l,g)=\\{A\in M_{2}(\mathbb{Z})\colon\det(A)=l,\,A\equiv g\text{ mod
}p^{m_{\pi}}\\}\text{ for
}g\in\operatorname{GL}_{2}(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z}).$

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

$f_{ur}=\sum_{r}y_{r}\kappa_{r}.$

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

######  Lemma 4.2.

For $z\in\mathcal{F}$ we have

$\frac{(\sharp
S)^{2}}{\dim_{\mathbb{C}}\pi_{p}^{K_{p}(m_{\pi})}}\cdot\Phi(z)^{2}\ll\sum_{r}\frac{\lvert{y_{r}}\rvert}{\sqrt{r}}\sum_{g\in\operatorname{GL}_{2}(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})}\lvert{\chi_{\sigma}(g)}\rvert\sum_{A\in
M(l,g)}\lvert{f_{\infty}(u(Az,z))}\rvert.$

###### 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

$\frac{(\sharp
S)^{2}}{\dim_{\mathbb{C}}\pi_{p}^{K_{p}(m_{\pi})}}\cdot\phi_{i}(g_{\infty})^{2}\leq
k_{f^{(i)}}(g,g)=\sum_{r}y_{r}\sum_{\gamma\in Z(\mathbb{Q})\backslash
G(\mathbb{Q})}f_{p}^{(i)}(\gamma)\kappa_{r}(\gamma)f_{\infty}(g_{\infty}^{-1}\gamma
g_{\infty}).$

We now sum this inequality over $i$ to obtain

$\frac{(\sharp
S)^{2}}{\dim_{\mathbb{C}}\pi_{p}^{K_{p}(m_{\pi})}}\Phi(g_{\infty})^{2}\leq\sum_{r}y_{r}\sum_{\gamma\in
Z(\mathbb{Q})\backslash
G(\mathbb{Q})}\left(\sum_{i=1}^{d}f_{p}^{(i)}(\gamma)\right)\kappa_{r}(\gamma)f_{\infty}(g_{\infty}^{-1}\gamma
g_{\infty}).$

Now we write $\overline{\gamma}$ for the image of $\gamma$ in
$\operatorname{GL}_{2}(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})$. Note that this is
well defined as long as $\gamma\in K_{p}$. For such $\gamma$ we get

$\sum_{i=1}^{d}f_{p}^{(i)}(\gamma)=\chi_{\sigma}(\overline{\gamma}).$

The rest of the argument is standard and can for example be found in [18]. ∎

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

$\frac{(\sharp
S)^{2}}{\dim_{\mathbb{C}}\pi_{p}^{K_{p}(m_{\pi})}}\cdot\Phi(z)^{2}\ll
T\sum_{g\in\operatorname{GL}_{2}(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})}\lvert{\chi_{\sigma}(g)}\rvert\sum_{r}\frac{y_{r}}{\sqrt{r}}\cdot\sharp
M_{z}(r,g).$

for

$M_{z}(r,g)=\\{A\in M_{2}(\mathbb{Z})\colon\det(A)=r,\,A\equiv g\text{ mod
}p^{m_{\pi}}\text{ and }u(Az,z)\leq 1\\}.$

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.

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

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

######  Lemma 4.4.

Let $\pi$ belong to Case 1, 2 or 3.1. Then, for $0\leq\lambda<m_{\pi}$ and
$g\in Z\cdot K_{m_{\pi},\lambda}\setminus Z\cdot K_{m_{\pi},\lambda+1})$ we
have

$\chi_{\sigma}(g)\ll p^{\lambda}.$ (9)

If $\pi$ belongs to Case 3.2 and $\lambda$ and $g$ are as above, then we have
the slightly weaker bound

$\chi_{\sigma}(g)\ll p^{\frac{m_{\pi}+\lambda}{2}}.$

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 in his
PhD thesis. 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 [3]. However, they remain hard to extract and we
hope our approach is more transparent.

###### Proof.

Note that if $m_{\pi}=1$, then $\sigma$ is the character of
$\operatorname{GL}_{2}(\mathbb{Z}/p\mathbb{Z})$. If we are in Case 1 or 2,
then the representation is constructed by parabolic induction and the
character is easily computed. Otherwise we must be in Case 3.1 in which case
$\sigma$ is cuspidal. In this case the character values are well known, see
for example [9]. (Alternatively one can use Mackey Theory to reduce to the
case of characters for $\operatorname{SL}_{2}(\mathbb{Z}/p\mathbb{Z})$ and use
the corresponding character table given in [19, p. 128].)

We will now assume $m_{\pi}>1$. Since Case 2 can not occur we treat Case 1 and
3.1, leaving Case 3.2 for later. Our approach is based on reduction to the
case of characters for $\operatorname{SL}_{2}(\mathbb{Z}/p^{m}\mathbb{Z})$
using Mackey-Theory. Recall that we are assuming $p$ to be odd. Let
$\omega_{\sigma}$ be the central character of $\sigma$ and $\tilde{\sigma}$ be
an irreducible component of $\sigma|_{\operatorname{SL}_{2}(\mathbb{Z})}$. 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

$\sigma=\operatorname{Ind}_{Z\cdot\operatorname{SL}_{2}(\mathbb{Z}/p^{m}\mathbb{Z})}^{\operatorname{GL}_{2}(\mathbb{Z}/p^{m}\mathbb{Z})}(\omega_{\sigma}\cdot\tilde{\sigma}).$

Otherwise, if $\tilde{\sigma}\cong\tilde{\sigma}^{h}$, then

$\sigma\oplus\sigma^{\prime}=\operatorname{Ind}_{Z\cdot\operatorname{SL}_{2}(\mathbb{Z}/p^{m}\mathbb{Z})}^{\operatorname{GL}_{2}(\mathbb{Z}/p^{m}\mathbb{Z})}(\omega_{\sigma}\cdot\tilde{\sigma}).$
(10)

Now we observe that the dimensions of irreducible representations of
$\operatorname{SL}_{2}(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})$ are given by
$p^{m_{\pi}}(1+p^{-1})$, $p^{m_{\pi}}(1-p^{-1})$ and
$\frac{1}{2}p^{m_{\pi}}(1-p^{-2})$. Thus by recalling that $\sigma$ has
dimension $p^{m_{\pi}}(1\pm p^{-1})$ (in Cases 1 and 3.1) we find that we must
be in the situation described in (10). In particular the character
$\chi_{\sigma}$ is essentially given by $\chi_{\tilde{\sigma}}$. We conclude
by referring to [13, Table III] where the character values of characters of
dimensions $p^{m_{\pi}}(1\pm p^{-1})$ are listed.

We now turn towards Case 3.2. Note that this case is exceptional in the sense
that the restriction of $\sigma$ to
$\operatorname{SL}_{2}(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})$ is reducible. As a
consequence the character $\chi_{\sigma}$ can only be described by a
combination of two character of
$\operatorname{SL}_{2}(\mathbb{Z}/p^{m_{\pi}}\mathbb{Z})$. The corresponding
character values are listed in [13, Table IV] and the claimed bound is derived
directly by ignoring any possible cancellation between the two characters
$\chi_{\tilde{\sigma}}$ and $\chi_{\tilde{\sigma}^{h}}$. (Even though such
cancellation can be observed in the $m_{\pi}=1$ situation this phenomenon does
not seem to generalise.)

∎

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 $r=1,l_{1},l_{1}l_{2},l_{1}^{2}l_{2}^{2}$ for
$l_{1},l_{2}\in S$. Put will $\Lambda=d^{\frac{1}{3}}$ 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

$S_{0}=\\{l\text{ prime }\colon l\asymp\Lambda\\}.$

By the prime number theorem (assuming $d$ is sufficiently large, which is no
problem) we have $\sharp S_{0}\sim\Lambda/\log(\Lambda)$. We define now
$S\subset S_{0}$ by requiring that
$l_{1},l_{1}l_{2},l_{1}^{2}l_{2}^{2}\not\equiv 1\text{ mod }p$ for all
$l_{1},l_{2}\in S$. This can be achieved by removing a moderate number of
primes from $S_{0}$ and one sees that one still has

$\sharp S\gg_{\epsilon}\Lambda^{1-\epsilon}.$

We are now ready to prove the following key estimate.

######  Lemma 4.5.

Suppose $\pi$ belongs to Case 1, 2 or 3.1. Then, for $\frac{\sqrt{3}}{2}\ll
y\ll\sqrt{d}$ we have

$\Phi(x+iy)\ll\sqrt{T}d^{\frac{5}{6}}.$

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

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

###### Proof.

We start with Cases 1, 2 or 3.1. Our starting point is Corollary 4.3. We break
up the geometric side into pieces according to the estimates on the character.
In particular we define the sets

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

We obtain

$\frac{(\sharp
S)^{2}}{\dim_{\mathbb{C}}\pi_{p}^{K_{p}(m_{\pi})}}\cdot\Phi(z)^{2}\ll
T\sum_{0\leq\lambda\leq
m_{\pi}}p^{\lambda}\sum_{r}\frac{y_{r}}{\sqrt{r}}\cdot\sharp
M_{z}^{(\lambda)}(r).$

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

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

This is for example [20, Proposition 6.1] with $N=\delta=1$ and
$z\in\mathcal{F}$ so that $y\gg 1$.

Next we assume $\lambda>0$. Here we observe that if $A\in M_{z}^{(\lambda)}$,
then $r=\det(A)\equiv 1\text{ mod }p$. But by construction of $S$ and the
support of $y_{r}$ this happens only for $r=1$. Thus we get

$\sum_{r}\frac{y_{r}}{\sqrt{r}}\cdot\sharp
M_{z}^{(\lambda)}(r)=y_{1}M_{z}^{\lambda}(1).$

Given any matrix $\left(\begin{matrix}a&b\\\ c&d\end{matrix}\right)\in
M_{z}^{\lambda}(1)$ we have the usual bound

$c\leq 2^{\frac{3}{2}}y^{-1}\leq\frac{2^{\frac{5}{2}}}{\sqrt{3}}.$

Thus since also $p^{\lambda}\mid c$ we must have $c=0$. (The cases $p=2,3$ and
$\lambda=1$ can be treated trivially.) However, upper triangular matrices with
$u(zA,z)\leq 1$ are easily counted. We obtain

$\sharp M_{z}^{(\lambda)}(r)\ll 1+\sqrt{y}p^{-\lambda}.$

Taking these counting results into account yields

$\frac{(\sharp
S)^{2}}{\dim_{\mathbb{C}}\pi_{p}^{K_{p}(m_{\pi})}}\cdot\Phi(z)^{2}\ll
T\left(\Lambda^{4}+\Lambda^{\frac{5}{2}}y+\sum_{1\leq\lambda\leq
m_{\pi}}\Lambda(p^{\lambda}+\sqrt{y})\right).$

This simplifies to

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

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

$\Phi(z)^{2}\ll Td^{1+\epsilon}(d^{\frac{2}{3}}+d^{\frac{1}{6}}y)$

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

$\displaystyle\frac{(\sharp
S)^{2}}{\dim_{\mathbb{C}}\pi_{p}^{K_{p}(m_{\pi})}}\cdot\Phi(z)^{2}$
$\displaystyle\ll
T\left(d^{\frac{1}{2}}\Lambda^{4}+d^{\frac{1}{2}}\Lambda^{\frac{5}{2}}y+\sum_{1\leq\lambda\leq
m_{\pi}}\Lambda(p^{\frac{m_{\pi}+\lambda}{2}}+p^{\frac{m_{\pi}-\lambda}{2}}\sqrt{y})\right)$
$\displaystyle\ll
T\left(d^{\frac{1}{2}}\Lambda^{4}+d^{\frac{1}{2}}\Lambda^{\frac{5}{2}}y+\Lambda
d^{1+\epsilon}+\Lambda d^{\frac{1}{2}+\epsilon}\sqrt{y}\right).$

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

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

This completes the proof. ∎

Proof of Theorem 1.1: We are now ready to complete the proof of our main
theorem. First of all note that it is enough to bound $\Phi(z)$ for
$z\in\mathcal{F}$. Next, if $y\ll\sqrt{d}$ (resp. $y\ll d^{\frac{1}{4}}$),
then we are done by the previous lemma. For larger $y$ the Whittaker expansion
(see Lemma 3.8) gives even better result. ∎

## References

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