# Convergence to the Plancherel measure of Hecke Eigenvalues

Peter Sarnak Nina Zubrilina

###### Abstract

We give improved uniform estimates for the rate of convergence to Plancherel
measure of Hecke eigenvalues of holomorphic forms of weight $2$ and level $N$.
These are applied to determine the sharp cutoff for the non-backtracking
random walk on arithmetic Ramanujan graphs and to Serre’s problem of bounding
the multiplicities of modular forms whose coefficients lie in number fields of
degree $d\geq 1$.

##  1 Introduction

It is well known that the distribution of Hecke eigenvalues of modular forms
at primes $p_{1},\ldots,p_{r}$ converges to the product of the corresponding
$p$-adic Plancherel measures as one varies over certain families ([Sar87],
[Ser97], [CDF97]). Our aim in this paper is to establish uniform rates on this
convergence and to apply these to problems of sharp cutoff for random walks on
Ramanujan graphs ([NS21]) and to the factorization of the Jacobian of the
modular curve $X_{0}(N)$ as in [Ser97].

The Eichler-Selberg trace formula expresses the trace of the Hecke operator
$T_{n}$ on the space $S(N):=S_{2}(N)$ of holomorphic cusp forms of weight $2$
for $\Gamma_{0}(N)$ in terms of class numbers of binary quadratic forms. Using
this, one can show ([Ser97], $\text{Prop. }4$) that as $f$ runs over a Hecke
basis $H(N)$ of such eigenforms with eigenvalues

$T_{n}f=:\lambda_{f}(n)\sqrt{n}\cdot f,\ \ (n,N)=1,$

we will have

$\displaystyle\frac{1}{\absolutevalue{H(N)}}\absolutevalue{\sum_{f\in
H(N)}\lambda_{f}(n)-\frac{\delta_{N}(n,\square)}{12}\psi(N)}\ll_{n}(n/N)^{1/2}\cdot
d(N),$ (1)

where

$\displaystyle\delta_{N}(n,\square):=\begin{cases}1&\text{ if }n\text{ is a
square modulo }N,\\\ 0&\text{ otherwise,}\end{cases}$

$d(N):=\sum_{d|N}1\ll_{\varepsilon}N^{\varepsilon}$ is the divisor function,
and $\psi(N):=N\prod_{p|N}(1+1/p)$ is the Dedekind psi function. Murty and
Sinha ([MS09]) give explicit and effective bounds in (1).

To extend the range of $n$ for which the left-hand side of (1) goes to $0$
with $N$, we introduce and remove $1/L(1,\mathrm{Sym}^{2}f)$ weights into the
sum. This allows us to use the Petersson trace formula, replacing class
numbers with Kloosterman sums, which enjoy sharp bounds coming from the
arithmetic geometry of curves ([Wei48]). This technique applied to a similar
problem is outlined in [Sar02] and is used on other problems in [Iwa84] and
[ILS00]. It allows us to double the exponent in the range of $n$, which is
crucial for certain applications.

In what follows, our aim is to establish sharp estimates, and to simplify the
analysis, we assume that $N$ is prime. In much of what we do, this assumption
can be removed. We address this further in the final section 6.

######  Theorem 1.

Let $\varepsilon>0$ and let $m,n$ be integers coprime to $N$. Then:

$\frac{1}{\absolutevalue{H(N)}}\sum_{f\in
H(N)}\lambda_{f}(m)\lambda_{f}(n)=\sum_{\begin{subarray}{c}d|(m,n)\\\
d^{2}k^{2}=mn\end{subarray}}\frac{1}{k}+\mathrm{O}_{\varepsilon}\left(\frac{(mn)^{1/8}}{N^{1/2}}(mnN)^{\varepsilon}\right).$

With this theorem, we can formulate and prove a corresponding uniform
convergence to the Plancherel measure. Let $r\geq 1$, and for
$\ell_{1},\ldots,\ell_{r}\geq 0$, let $\mathcal{P}^{\ell_{1},\ldots,\ell_{r}}$
denote the set of polynomials in $x_{1},\ldots,x_{r}$ of degrees at most
$\ell_{1},\ldots,\ell_{r}$, respectively, that is,

$\mathcal{P}^{\ell_{1},\ldots,\ell_{r}}:=\Big{\\{}\sum_{j_{1}=0}^{\ell_{1}}\cdots\sum_{j_{r}=0}^{\ell_{r}}a_{j_{1},\ldots,j_{r}}x_{1}^{j_{1}}\cdots
x_{r}^{j_{r}}\ \large|\ a_{j_{1},\ldots,j_{r}}\in\mathbb{C}\Big{\\}}.$

For $p\nmid N$, let $\theta_{f}(p)\in[0,\pi]$ be such that

$\lambda_{f}(p)=2\cos(\theta_{f}(p))$

(such a $\theta_{f}$ exists because of self-adjointness of $T_{p}$ and thanks
to the Ramanujan bound $\absolutevalue{\lambda_{f}(p)}\leq 2$ due to Eichler
[Eic54]). Let $\mu_{p}$ be the $p$-adic Plancherel measure:

$\displaystyle
d\mu_{p}:=\frac{2}{\pi}\cdot\dfrac{(p+1)\sin^{2}\theta}{(p^{1/2}+p^{-1/2})^{2}-4\cos^{2}\theta}d\theta.$
(2)

We have the following uniform convergence result:

######  Theorem 2.

Let $r\geq 1$ and $\eta>0$. Then uniformly for $p_{1}^{\ell_{1}}\cdots
p_{r}^{\ell_{r}}<N^{2-\eta}$ and $P\in\mathcal{P}^{\ell_{1},\ldots,\ell_{r}}$,

$\frac{1}{\absolutevalue{H(N)}}\sum_{f\in
H(N)}\absolutevalue{P\left(\cos\theta_{f}(p_{1}),\ldots,\cos\theta_{f}(p_{r})\right)}^{2}=(1+o(1))\int_{0}^{\pi}\cdots\int_{0}^{\pi}\absolutevalue{P}^{2}d\mu_{p_{1}}\dots
d\mu_{p_{r}}$

as $N\to\infty$.

This result with an exponent of $N$ larger than $1$ (which corresponds to $mn$
going up to $N^{2-\varepsilon}$ in Theorem 1) is what is needed to settle the
cutoff window for the non-backtracking random walks on Ramanujan graphs
([NS21]). In fact, it yields the conjectured asymptotics for the variance for
these walks (see end of section 3).

Another application of Theorem 2 is to multiplicities of $f$’s in a Hecke
basis with given $\lambda_{f}(p)$’s for $p\in\\{p_{1},\ldots,p_{r}\\}$. Let
$s(N):=\absolutevalue{H(N)}=\mathrm{dim}\ S(N)$, so for $N$ prime,
$s(N)=\lfloor\frac{N+1}{12}\rfloor-1$ when $N\equiv 1\pmod{12}$ and
$s(N)=\lfloor\frac{N+1}{12}\rfloor$ otherwise. For $\varphi\in S(N)$, let

$M_{N}(y,\varphi):=\\#\absolutevalue{\left\\{f\in
H(N):\lambda_{f}(p)=\lambda_{\varphi}(p)\text{ for }p\leq
y,(p,N)=1\right\\}}.$

For a fixed $y$, Theorem 2 implies that uniformly in $\varphi$,

$\displaystyle M_{N}(y,\varphi)\ll\frac{s(N)}{(\log N)^{r}},$ (3)

where $r=\pi(y)$ is the number of primes up to $y$. If $y$ is allowed to
increase with $N$, then one can exploit that for $f$ in the set defining
$M_{N}(y,\varphi)$, we also have $\lambda_{f}(m)=\lambda_{\varphi}(m)$ for all
$y$-smooth numbers $m$ (which are numbers all of whose prime factors are at
most $y$). This allows one to improve (3) vastly.

Such an argument using the large sieve for Dirichlet characters is due to
Linnik ([Lin41]). In the modular form setting, Duke and Kowalski ([DK00])
establish that the number of non-monomial newforms of square-free level up to
$N$ that have prescribed eigenvalues $\lambda_{\varphi}(p)$ at primes $p\leq
y=(\log N)^{\beta}$ satisfies

$M_{\leq N}(y,\varphi)^{\\#}\ll_{\beta}N^{10/\beta+\varepsilon},$

which is non-trivial for $\beta>5$. Lau and Wu ([LW08]) show that for $y=C\log
N$ with $C$ a large constant, there is $c>0$ s.t.

$M_{N}(y,\varphi)\ll\exp\left(\dfrac{-c\log N}{\log\log N}\right)s(N).$

Our interest is in smaller $y$’s, namely $y=(\log N)^{\beta}$ with
$0<\beta<1$.

######  Theorem 3.

Fix $\beta\in(0,1)$. Then for $y=(\log N)^{\beta}$ and uniformly in $\varphi$,

$M_{N}(y,\varphi)\leq\exp\left(-\frac{1-\beta}{\beta}(\log
N)^{\beta}+o\left((\log N)^{\beta}\right)\right)s(N)$

as $N\to\infty$.

We apply this to a question of Serre ([Ser97]). Assume that all $f=\sum_{n\geq
1}a(n)e(nz)\in H(N)$ are normalized so $a(1)=1$. The Fourier coefficients
$a(n)$ are algebraic integers in a totally real field of degree $d(f)$. For
$d\geq 1$, let $s(N)_{d}$ denote the number of $f$’s for which $d(f)=d$. Serre
shows that for $d$ fixed, $s(N)_{d}=o(s(N))$ (see also [MS09], Theorem $5$),
and asks for stronger upper bounds. Theorem 3 implies such a bound.

######  Theorem 4.

Fix $d\geq 1$ and $\beta<\frac{2}{d+2}$. Then as $N\to\infty$,

$s(N)_{d}\leq\exp(-c(d,\beta)(\log N)^{\beta}+o((\log N)^{\beta}))s(N),$

where $c(d,\beta):=(1-\beta)/\beta-d/2>0.$

This falls short of Serre’s conjecture, which asserts that Theorem 4 holds for
$\beta=1$ and $c=c(d)>0$ (i.e., $s(N)_{d}\ll s(N)^{\alpha}$ for some
$\alpha<1$).

Acknowledgments: We thank L. Alpöge, G. Harcos, E. Nestoridi, and Z. Rudnick
for discussions related to the questions addressed in this paper.

##  2 Weight Removal in the Petersson Formula

Throughout this section we assume $N$ is prime. Let $H(N)$ denote a
simultaneous eigenbasis of Hecke operators $T_{k},(k,N)=1$, acting on the
space $S(N)$ of dimension $s(N)$ of weight $2$ level $N$ cusp forms for
$\Gamma_{0}(N)$, and for $f\in H(N)$, let $a_{f}(n)$ and $\lambda_{f}(n)$ be
such that

$f(z)=\sum_{n\geq 1}a_{f}(n)e(nz)=\sum_{n\geq 1}\sqrt{n}\
\lambda_{f}(n)e(nz),$

where $e(z):=e^{2\pi iz}$. Assume $f$ are normalized so $a_{f}(1)=1$.

Our starting point for this section is the Petersson trace formula estimated
via the Weil bound on Kloosterman sums, as presented in [ILS00], Corollary
$2.2$ or [IK04], Corollary $14.24$:

######  Petersson Formula.

With $H(N)$ as above, $(mn,N)=1$, and $\varepsilon>0$,

$\displaystyle\sideset{}{{}^{h}}{\sum}_{f\in
H(N)}\lambda_{f}(m)\lambda_{f}(n)=\delta(m,n)+\mathrm{O}_{\varepsilon}\left(\frac{(mn)^{1/4}}{N}(mnN)^{\varepsilon}\right).$
(4)

Here the $h$ superscript signifies adding "harmonic" weights: $\sum_{f\in
H(N)}^{h}\alpha_{f}=\frac{1}{4\pi}\sum_{f\in
H(N)}\frac{\alpha_{f}}{\norm{f}^{2}}$, where $\norm{\cdot}$ denotes the
Petersson norm.

We derive Theorem 1 from the Petersson formula by removing the harmonic
weights. The Petersson norm is related to the special value of the symmetric
square $L$-function at $1$ ([ILS00], Lemma $2.5$) via:

$4\pi\norm{f}^{2}=\frac{s(N)}{\zeta(2)}L(\mathrm{Sym}^{2}f,1)$

(where $L(\mathrm{Sym}^{2}f,s)=\zeta(2s)(1-N^{-2s})\sum_{n\geq
1}\lambda_{f}(n^{2})n^{-s}:=\sum_{n=1}^{\infty}\lambda_{\mathrm{sym}^{2}f}(n)n^{-s}$),
so

$\displaystyle\frac{1}{s(N)}\sum_{f\in H(N)}\lambda_{f}(m)\lambda_{f}(n)$
$\displaystyle=\sideset{}{{}^{h}}{\sum}_{f\in
H(N)}\lambda_{f}(m)\lambda_{f}(n)\cdot\frac{4\pi\norm{f}^{2}}{s(N)}=\sideset{}{{}^{h}}{\sum}_{f\in
H(N)}\lambda_{f}(m)\lambda_{f}(n)\frac{L(\mathrm{Sym}^{2}f,1)}{\zeta(2)},$ (5)

and to prove Theorem 1, we need to derive a suitable approximation for
$L(\mathrm{Sym}^{2}f,s)$.

Let $\Psi(x)\geq 0$ be a smooth function supported on $[-1,1]$ with
$\Psi(0)=1$. The Mellin transform

$\tilde{\Psi}(s)=\int_{0}^{\infty}\Psi(x)x^{s}\frac{dx}{x}$

is an analytic function of $s=\sigma+it$ for $\sigma>-1$, except for a simple
pole at $0$ with residue $1$, and decreases rapidly as
$\absolutevalue{t}\to\infty$ for $-1\leq\sigma\leq 2$.

For a parameter $x>0$, let

$\displaystyle\textbf{A}:$ $\displaystyle=\frac{1}{2\pi
i}\int_{\sigma=2}L(\mathrm{Sym}^{2}f,s+1)x^{s}\tilde{\Psi}(s)ds$
$\displaystyle=\frac{1}{2\pi i}\int_{\sigma=2}\sum_{\nu\geq
1}\frac{\lambda_{\mathrm{sym}^{2}f}(\nu)}{\nu^{s+1}}x^{s}\tilde{\Psi}(s)ds$
$\displaystyle=\sum_{\nu\geq
1}\frac{\lambda_{\mathrm{sym}^{2}f}(\nu)}{\nu}\cdot\frac{1}{2\pi
i}\int_{\sigma=2}\left(\frac{\nu}{x}\right)^{-s}\tilde{\Psi}(s)ds$
$\displaystyle=\sum_{\nu\leq
x}\frac{\lambda_{\mathrm{sym}^{2}f}(\nu)}{\nu}\Psi\left(\frac{\nu}{x}\right)$

by the Mellin inversion theorem. Shifting the integral defining A to
$\real(s)=-1/2$ picks up the simple pole of $\tilde{\Psi}$ at $s=0$, so by the
residue theorem,

$\displaystyle L(\mathrm{Sym}^{2}f,1)=\sum_{\nu\leq
x}\frac{\lambda_{\mathrm{sym}^{2}f}(\nu)}{\nu}\Psi\left(\frac{\nu}{x}\right)+R(f,x),$
(6)

where

$R(f,x):=-\frac{1}{2\pi}\int_{-\infty}^{\infty}L(\mathrm{Sym}^{2}f,1/2+it)x^{-\frac{1}{2}+it}\tilde{\Psi}(-1/2+it)dt.$

Now, by Cauchy-Schwarz,

$\displaystyle\absolutevalue{R(f,x)}^{2}$
$\displaystyle=\frac{1}{4\pi^{2}x}\absolutevalue{\int_{-\infty}^{\infty}L(\mathrm{Sym}^{2}f,1/2+it)x^{it}\tilde{\Psi}(-1/2+it)dt}^{2}$
$\displaystyle\ll
x^{-1}\int_{-\infty}^{\infty}\absolutevalue{L(\mathrm{Sym}^{2}f,1/2+it)}^{2}\cdot\big{|}\tilde{\Psi}(-1/2+it)\big{|}dt,$

so

$\sideset{}{{}^{h}}{\sum}_{f\in H(N)}\absolutevalue{R(f,x)}^{2}\ll
x^{-1}\int_{-\infty}^{\infty}\big{|}\tilde{\Psi}(-1/2+it)\big{|}\sideset{}{{}^{h}}{\sum}_{f\in
H(N)}\absolutevalue{L(1/2+it,\mathrm{Sym}^{2}f)}^{2}dt.$

According to the Lindelöf on average result due to Iwaniec and Michel for this
family of $L$-functions ([IM01]),

$\sideset{}{{}^{h}}{\sum}_{f\in
H(N)}\absolutevalue{L(1/2+it,\mathrm{Sym}^{2}f)}^{2}\ll_{\varepsilon}N^{\varepsilon}(\absolutevalue{t}+1)^{8},$

so

$\displaystyle\sideset{}{{}^{h}}{\sum}_{f\in
H(N)}\absolutevalue{R(f,x)}^{2}\ll_{\varepsilon}x^{-1}N^{\varepsilon}.$ (7)

Substituting (7) and (6) into (5),

$\displaystyle\frac{1}{s(N)}\sum_{f\in H(N)}\lambda_{f}(m)\lambda_{f}(n)$
$\displaystyle=\frac{1}{\zeta(2)}\sideset{}{{}^{h}}{\sum}_{f\in
H(N)}\lambda_{f}(m)\lambda_{f}(n)\left(\sum_{\nu\leq
x}\frac{\lambda_{\mathrm{sym}^{2}f}(\nu)}{\nu}\Psi\left(\frac{\nu}{x}\right)+R(f,x)\right)$
$\displaystyle=\frac{1}{\zeta(2)}(\textbf{I}+\textbf{II}),$ (8)

where

$\textbf{I}=\sum_{\nu\leq
x}\frac{\lambda_{\mathrm{sym}^{2}f}(\nu)}{\nu}\Psi\left(\frac{\nu}{x}\right)\sideset{}{{}^{h}}{\sum}_{f\in
H(N)}\lambda_{f}(m)\lambda_{f}(n)$

and

$\absolutevalue{\textbf{II}}\leq\sideset{}{{}^{h}}{\sum}_{f\in
H(N)}\absolutevalue{\lambda_{f}(m)\lambda_{f}(n)}\absolutevalue{R(f,x)}\ll_{\varepsilon}{N^{\varepsilon}}\left(\sideset{}{{}^{h}}{\sum}_{\
f\in H(N)}(mn)^{\varepsilon}\right)^{1/2}\left(\sideset{}{{}^{h}}{\sum}_{\
f\in
H(N)}\absolutevalue{R(f,x)}^{2}\right)^{1/2}\ll_{\varepsilon}\frac{(mnN)^{\varepsilon}}{\sqrt{x}},$

where we used Cauchy-Schwartz and (7).

To estimate I, we use Hecke relations

$\lambda_{f}(m)\lambda_{f}(n)=\sum_{d|(m,n)}\lambda_{f}\left(\frac{mn}{d^{2}}\right)$

and the formula

$\lambda_{\mathrm{Sym}^{2}f}(\nu)=\sum_{\begin{subarray}{c}t^{2}k=\nu\\\
(t,N)=1\end{subarray}}\lambda_{f}(k^{2}).$

From this,

$\textbf{I}=\sum_{t^{2}k\leq
x}\frac{\Psi\left(\frac{t^{2}k}{x}\right)}{t^{2}k}\sum_{d|(m,n)}\sum_{f\in
H(N)}^{h}\lambda_{f}\left(\frac{mn}{d^{2}}\right)\lambda_{f}\left(k^{2}\right),$

so by the Petersson formula,

I $\displaystyle=\sum_{t^{2}k\leq
x}\frac{\Psi\left(\frac{t^{2}k}{x}\right)}{t^{2}k}{t^{2}k}\sum_{\begin{subarray}{c}d|(m,n)\\\
mn=d^{2}k^{2}\end{subarray}}1+\mathrm{O}_{\varepsilon}\left(\sum_{t^{2}k\leq
x}\frac{(mn)^{\varepsilon}}{t^{2}k}\frac{(mnk^{2})^{1/4}}{N}(Nmn)^{\varepsilon}\right)$
$\displaystyle=\left(\sum_{\begin{subarray}{c}d|(m,n)\\\
mn=d^{2}k^{2}\end{subarray}}\frac{1}{k}\right)\sum_{t^{2}k\leq
x/k}\frac{\Psi\left(\frac{t^{2}k}{x}\right)}{t^{2}}+\mathrm{O}_{\varepsilon}\left(\frac{x^{1/2}(mn)^{1/4}}{N}(mnN)^{\varepsilon}\right)$
$\displaystyle=\zeta(2)\left(\sum_{\begin{subarray}{c}d|(m,n)\\\
mn=d^{2}k^{2}\end{subarray}}\frac{1}{k}\right)+\mathrm{O}_{\varepsilon}\left(\frac{x^{1/2}(mn)^{1/4}}{N}(mnN)^{\varepsilon}\right).$

Combining estimates of I and II with (8),

$\frac{1}{s(N)}\sum_{f\in
H(N)}\lambda_{f}(m)\lambda_{f}(n)=\sum_{\begin{subarray}{c}d|(m,n)\\\
mn=d^{2}k^{2}\end{subarray}}\frac{1}{k}+\mathrm{O}_{\varepsilon}((mnN)^{\varepsilon}/x^{1/2})+\mathrm{O}_{\varepsilon}\left(\frac{x^{1/2}(mn)^{1/4}}{N}(mnN)^{\varepsilon}\right).$

Choosing $x=\frac{N}{(mn)^{1/4}}$, we finish the proof of Theorem 1.

##  3 Convergence to the Plancherel Product Measure

In this section we address Theorem 2. Fix an integer $r>0$, let
$\ell_{1},\ldots,\ell_{r}>0$, and let $p_{1},\ldots,p_{r}$ be distinct primes
with $(p_{j},N)=1$.

Consider a polynomial $P(x_{1},\ldots,x_{r})\in\mathbb{C}[x_{1},\ldots,x_{r}]$
of degree at most $\ell_{i}$ in $x_{i}$ for $1\leq i\leq r$. For $n\geq 0$,
let

$U_{n}(\cos\theta):=\frac{\sin((n+1)\theta)}{\sin\theta}$

be the $n$th Chebyshev polynomial of the second kind. $U_{n}$ is a degree $n$
polynomial in $\cos\theta$ with real coefficients, so we can find
$a_{t_{1},\ldots,t_{r}}\in\mathbb{C}$ such that

$\displaystyle
P(x_{1},\ldots,x_{r})=\sum_{t_{1}=0}^{\ell_{1}}\cdots\sum_{t_{r}=0}^{\ell_{r}}a_{t_{1},\ldots,t_{r}}U_{t_{1}}(x_{1})\cdots
U_{t_{r}}(x_{r}).$ (9)

Moreover, Hecke relations imply that for $(p,N)=1$,
$U_{n}(\lambda(p))=\lambda(p^{n})$. From this,

$\displaystyle\absolutevalue{P(\theta_{f}(p_{1}),\ldots,\theta_{f}(p_{r}))}^{2}$
$\displaystyle=\sum_{t_{1},s_{1}=0}^{\ell_{1}}\cdots\sum_{t_{r},s_{r}=0}^{\ell_{r}}a_{t_{1},\ldots,t_{r}}\overline{a_{s_{1},\ldots,s_{r}}}U_{t_{1}}(\theta_{f}(p_{1}))\overline{U_{s_{1}}(\theta_{f}(p_{1}))}\cdots
U_{t_{r}}(\theta_{f}(p_{r}))\overline{U_{s_{r}}(\theta_{f}(p_{r}))}=$
$\displaystyle=\sum_{t_{i},s_{i}}a_{t_{1},\ldots,t_{r}}\overline{a_{s_{1},\ldots,s_{r}}}\lambda_{f}(p_{1}^{t_{1}})\overline{\lambda_{f}(p_{1}^{s_{1}})}\cdots\lambda_{f}(p_{r}^{t_{r}})\overline{\lambda_{f}(p_{r}^{s_{r}})}$
$\displaystyle=\sum_{t_{i},s_{i}}a_{t_{1},\ldots,t_{r}}\overline{a_{s_{1},\ldots,s_{r}}}\lambda_{f}(p_{1}^{t_{1}}\cdots
p_{r}^{t_{r}})\overline{\lambda_{f}(p_{1}^{s_{1}}\cdots p_{r}^{s_{r}})},$

so by Theorem 1,

$\displaystyle\frac{1}{\absolutevalue{H(N)}}\sum_{f\in
H(N)}\absolutevalue{P(\theta_{f}(p_{1}),\ldots,\theta_{f}(p_{r}))}^{2}$
$\displaystyle=\sum_{t_{i},s_{i}}a_{t_{1},\ldots,t_{r}}\overline{a_{s_{1},\ldots,s_{r}}}\cdot\frac{1}{\absolutevalue{H(N)}}\sum_{f\in
H(N)}\lambda_{f}(p_{1}^{t_{1}}\cdots
p_{r}^{t_{r}})\overline{\lambda_{f}(p_{1}^{s_{1}}\cdots p_{r}^{s_{r}})}$
$\displaystyle=\textbf{I}+\textbf{II},$ (10)

where

$\textbf{I}=\sum_{\begin{subarray}{c}s_{i},t_{i}\leq\ell_{i}\\\
m=p_{1}^{t_{1}}\cdots p_{r}^{t_{r}}\\\ n=p_{1}^{t_{1}}\cdots
p_{r}^{t_{r}}\end{subarray}}a_{t_{1},\ldots,t_{r}}\overline{a_{s_{1},\ldots,s_{r}}}\sum_{\begin{subarray}{c}d|(m,n)\\\
mn=d^{2}k^{2}\end{subarray}}\frac{1}{k},$

and

II
$\displaystyle\ll_{\varepsilon}\frac{1}{\sqrt{N}}\sum_{\begin{subarray}{c}s_{i},t_{i}\leq\ell_{i}\\\
m=p_{1}^{t_{1}}\cdots p_{r}^{t_{r}}\\\ n=p_{1}^{t_{1}}\cdots
p_{r}^{t_{r}}\end{subarray}}\absolutevalue{a_{t_{1},\ldots,t_{r}}\overline{a_{s_{1},\ldots,s_{r}}}}(mn)^{1/8}(mnN)^{\varepsilon}$
$\displaystyle\ll\frac{1}{\sqrt{N}}\left(\sum_{0\leq
t_{i}\leq\ell_{i}}\absolutevalue{a_{t_{1},\ldots,t_{r}}}p_{1}^{t_{1}/8}\cdots
p_{r}^{t_{r}/8}\right)^{2}(p_{1}^{2\ell_{1}}\cdots
p_{r}^{2\ell_{r}}N)^{\varepsilon}=(*).$

Applying Cauchy-Schwartz and summing the geometric series,

$(*)\leq\frac{(p_{1}^{2\ell_{1}}\cdots
p_{r}^{2\ell_{r}}N)^{\varepsilon}}{\sqrt{N}}\sum_{0\leq
t_{i}\leq\ell_{i}}p_{1}^{t_{1}/4}\cdots p_{r}^{t_{r}/4}\sum_{0\leq
t_{i}\leq\ell_{i}}\absolutevalue{a_{t_{1},\ldots,t_{r}}}^{2}\leq(p_{1}^{2\ell_{1}}\cdots
p_{r}^{2\ell_{r}}N)^{\varepsilon}\left(\frac{p_{1}^{\ell_{1}}\cdots
p_{r}^{\ell_{r}}}{N^{2}}\right)^{\frac{1}{4}}\sum_{0\leq
t_{i}\leq\ell_{i}}\absolutevalue{a_{t_{1},\ldots,t_{r}}}^{2}.$

Let $\mu_{\infty}(\theta):=\frac{2}{\pi}\sin^{2}\theta d\theta$ be the Sato-
Tate measure on $[0,\pi]$. From the definition of $d\mu_{p}$, it follows that

$d\mu_{\infty}\cdots d\mu_{\infty}\ll_{r}d\mu_{p_{1}}\cdots d\mu_{p_{r}}.$

Hence, from (9) and the orthonormality of $U_{n}$ with respect to
$d\mu_{\infty}$, we have

$\sum_{0\leq
t_{i}\leq\ell_{i}}\absolutevalue{a_{t_{1},\ldots,t_{r}}}^{2}\ll\norm{P}^{2}_{\mu_{p_{1}},\ldots,\mu_{p_{r}}}.$

We conclude that

$\textbf{II}\ll_{\varepsilon}(p_{1}^{\ell_{1}}\cdots
p_{r}^{\ell_{r}}N)^{2\varepsilon}\left(\frac{p_{1}^{\ell_{1}}\cdots
p_{r}^{\ell_{r}}}{N^{2}}\right)^{\frac{1}{4}}\norm{P}^{2}_{\mu_{p_{1}},\ldots,\mu_{p_{r}}}.$

It remains to evaluate I. To interpret I as the integral against the
Plancherel measure, we need the following observation:

######  Proposition 3.1.

Let $m,n\geq 0$, and let $d\mu_{p}$ be the $p$-adic Plancherel measure. Then:

$\displaystyle\int_{0}^{\pi}U_{n}(\theta)U_{m}(\theta)d\mu_{p}=\left[\begin{array}[]{ll}\dfrac{p}{(p-1)}\left(\dfrac{1}{p^{\frac{\absolutevalue{m-n}}{2}}}-\dfrac{1}{p^{\frac{m+n}{2}+1}}\right)&\text{if
}m\equiv n\pmod{2}\\\ 0&\text{otherwise.}\end{array}\right.$ (13)

We leave the proof until the end of the section. From (9),

$\displaystyle\int_{[0,\pi]^{n}}\absolutevalue{P(\theta_{1},\ldots,\theta_{r})}^{2}d\mu_{p_{1}}\cdots
d\mu_{p_{r}}$
$\displaystyle=\sum_{t_{i},s_{i}\leq\ell_{i}}a_{t_{1}}\overline{a_{s_{1}}}\cdots
a_{t_{r}}\overline{a_{s_{r}}}\int_{[0,\pi]^{n}}U_{t_{1}}(\theta_{1})U_{s_{1}}(\theta_{1})\cdots
U_{t_{r}}(\theta_{r})U_{s_{r}}(\theta_{r})d\mu_{p_{1}}\cdots d\mu_{p_{r}}$
$\displaystyle=\sum_{t_{i},s_{i}\leq\ell_{i}}a_{t_{1}}\overline{a_{s_{1}}}\cdots
a_{t_{r}}\overline{a_{s_{r}}}\cdot\prod_{i}\int_{0}^{\pi}U_{t_{i}}(\theta)U_{s_{i}}(\theta)d\mu_{p_{i}}.$
(14)

We substitute the inner product (13) into (3). Observe that

$\displaystyle\frac{p}{(p-1)}\left(\frac{1}{p^{\frac{\absolutevalue{m-n}}{2}}}-\frac{1}{p^{\frac{m+n}{2}+1}}\right)$
$\displaystyle=\frac{1}{p^{\frac{\absolutevalue{m-n}}{2}}}+\frac{1}{p^{\frac{\absolutevalue{m-n}}{2}}+1}+\cdots+\frac{1}{p^{\frac{m+n}{2}}}$
$\displaystyle=\sum_{p^{\alpha}|(p^{m},p^{n})}\frac{1}{p^{\frac{m+n}{2}-\alpha}}$
$\displaystyle=\sum_{\begin{subarray}{c}d|(p^{m},p^{n})\\\
d^{2}k^{2}=p^{m}p^{n}\end{subarray}}\frac{1}{k},$

so (13) implies

$\displaystyle\norm{P}^{2}_{\mu_{p_{1}},\ldots,\mu_{p_{r}}}=(\ref{prodmeasureinner})$
$\displaystyle=\sum_{\begin{subarray}{c}s_{i},t_{i}\leq\ell_{i}\\\
s_{i}\overset{\mathrm{mod}2}{\equiv}t_{i}\end{subarray}}a_{t_{1}}\overline{a_{s_{1}}}\cdots
a_{t_{r}}\overline{a_{s_{r}}}\cdot\prod_{i}\sum_{\begin{subarray}{c}d|(p_{i}^{t_{i}},p_{i}^{s_{i}})\\\
d^{2}k^{2}=p_{i}^{t_{i}+s_{i}}\end{subarray}}\frac{1}{k}=$
$\displaystyle\sum_{\begin{subarray}{c}s_{i},t_{i}\leq\ell_{i}\\\
s_{i}\overset{\mathrm{mod}2}{\equiv}t_{i}\end{subarray}}a_{t_{1}}\overline{a_{s_{1}}}\cdots
a_{t_{r}}\overline{a_{s_{r}}}\cdot\sum_{\begin{subarray}{c}d|(p_{1}^{t_{1}}\cdots
p_{r}^{t_{r}},p_{1}^{s_{1}}\cdots p_{r}^{s_{r}})\\\
d^{2}k^{2}=p_{1}^{t_{1}+s_{1}}\cdots
p_{r}^{t_{r}+s_{r}}\end{subarray}}\frac{1}{k}=\textbf{I}.$

Combining I and II,

$\displaystyle\frac{1}{\absolutevalue{H(N)}}\sum_{f\in
H(N)}\absolutevalue{P(\theta_{f}(p_{1}),\ldots,\theta_{f}(p_{r}))}^{2}$
$\displaystyle=\norm{P}^{2}_{\mu_{1},\ldots,\mu_{r}}\left(1+\mathrm{O}\left((p_{1}^{\ell_{1}}\cdots
p_{r}^{\ell_{r}}N)^{2\varepsilon}\left(\frac{p_{1}^{\ell_{1}}\cdots
p_{r}^{\ell_{r}}}{N^{2}}\right)^{\frac{1}{4}}\right)\right).$

Suppose now that the conditions of Theorem 2 are met, i.e.,

$p_{1}^{\ell_{1}}\cdots p_{r}^{\ell_{r}}<N^{2-\eta}$

for some $\eta>0$. Then (3) is equal to

$\norm{P}^{2}_{\mu_{1},\ldots,\mu_{r}}(1+\mathrm{O}(N^{6\varepsilon}\cdot
N^{-\eta/4}))=\norm{P}^{2}_{\mu_{1},\ldots,\mu_{r}}(1+o(1))$

for small enough $\varepsilon$. This concludes the proof of Theorem 2.

###### Proof of Proposition 3.1.

From the trigonometric identity for the product of sines,

$\displaystyle\int_{0}^{\pi}U_{n}(\theta)U_{m}(\theta)d\mu_{p}$
$\displaystyle=\frac{p+1}{2\pi}\int_{0}^{\pi}\frac{\sin((n+1)\theta)\sin((m+1)\theta)}{\left(\frac{p^{1/2}+p^{-1/2}}{2}\right)^{2}-\cos^{2}\theta}d\theta$
$\displaystyle=\frac{p+1}{4\pi}\int_{0}^{\pi}\frac{\cos((m-n)\theta)-\cos((m+n+2)\theta)}{\frac{(p-1)^{2}}{4p}+\sin^{2}\theta}d\theta$
$\displaystyle=\mathcal{I}(\absolutevalue{m-n})-\mathcal{I}(m+n+2),$

where
$\mathcal{I}(k):=\frac{p+1}{4\pi}\int_{0}^{\pi}\frac{\cos(k\theta)}{\frac{(p-1)^{2}}{4p}+\sin^{2}\theta}d\theta$.

Since $\sin^{2}(\theta)=\sin^{2}(\pi-\theta)$ and $\cos(k(\pi-x))=-\cos kx$
for odd $k$, $\mathcal{I}(k)=0$ for odd $k$, so the integral is $0$ when $m$
and $n$ have different parity. To prove the proposition, it suffices to show
that for all integers $T\geq 0$,

$\displaystyle\mathcal{I}(2T)=\frac{(p+1)}{4\pi}\int_{0}^{\pi}\frac{\cos(2T\theta)}{\frac{(p-1)^{2}}{4p}+\sin^{2}\theta}d\theta=\frac{p}{(p-1)p^{T}}.$
(16)

We prove this statement by induction. Let $\zeta:=e^{ix}$,
$c:=\frac{(p-1)^{2}}{4p}$, and let $\alpha:=2+4c=p+1/p$. Then:

$\displaystyle\int_{0}^{\pi}\frac{\cos(2T\theta)}{\frac{(p-1)^{2}}{4p}+\sin^{2}\theta}d\theta$
$\displaystyle=\frac{1}{2}\int_{0}^{2\pi}\frac{(\zeta^{2T}+\zeta^{-2T})/2}{c+((\zeta-\zeta^{-1})/2i)^{2}}dx=$
$\displaystyle=-\int_{0}^{2\pi}\frac{(\zeta^{2T}+\zeta^{-2T})}{-4c+(\zeta-\zeta^{-1})^{2}}dx$
$\displaystyle=-\int_{S^{1}}\frac{(\zeta^{4T}+1)}{\zeta^{2T-1}(\zeta^{4}-\alpha\zeta^{2}+1)}d\zeta$
$\displaystyle=-\int_{S^{1}}\frac{(\zeta^{4T}+1)}{\zeta^{2T-1}(\zeta^{2}-p)(\zeta^{2}-1/p)}d\zeta.$

We evaluate the integral using the residue theorem. For $T=0$, the poles are
at $\pm\sqrt{1/p}$, and both residues are equal to
$\frac{1/\sqrt{p}}{(p-1/p)\cdot 2/\sqrt{p}}=\frac{1}{p-1/p}$, so

$\mathcal{I}(0)=(p+1)\cdot 2\pi(2p/(p^{2}-1))/4\pi=p/(p-1).$

For $T=1$, the pole at $0$ has residue $-1$ and the poles at $\pm 1/\sqrt{p}$
have residues
$\frac{1/p^{2}+1}{(1/\sqrt{p})(p-1/p)(2/\sqrt{p})}=\frac{p^{2}+1}{2(p^{2}-1)}$,
so

$\mathcal{I}(2)=(p+1)\cdot 2\pi(-1+(p^{2}+1)/(p^{2}-1))/4\pi=1/(p-1).$

Assume now $T\geq 2$. The rational function
$-\dfrac{x^{4T}+1}{x^{2T-1}(x^{2}-p)(x^{2}-1/p)}$ has three poles inside the
unit circle: $0,\omega=1/\sqrt{p}$ and $-\omega$, and the two latter ones have
the same residue. Let $A(T)$, $B(T)$ be the residues at $0$ and $\omega$
respectively. Then:

$B(T)=-\dfrac{(\omega^{2})^{2T}+1}{2(\omega^{2})^{T}(\omega^{2}-p)}=\frac{1}{2(p-1/p)}(p^{T}+1/p^{T}),$

and $A(T)$ is the coefficient of $x^{2T-2}$ in $1/(1-\alpha
x^{2}+x^{4})=\sum_{r\geq 0}(x^{4}-\alpha x^{2})^{r}$.

Notice that both $A(T)$ and $B(T)$ satisfy the recurrence relation

$F(T+2)-\alpha F(T+1)+F(T)=0.$

It remains to notice $\dfrac{p}{(p-1)p^{T}}$ satisfies the same recurrence
relation, which proves (16). ∎

To end this section, we apply Theorem 2 to the question of sharp cutoff of
random walks on certain Ramanujan graphs. Let $\ell$ be a fixed prime and $p$
a large prime, $p\equiv 1\pmod{12}$ (the notation here is made to conform with
[CGL09]). The Brandt-Ihara-Pizer "super singular isogeny graphs," $G(p,\ell)$,
are $d:=\ell+1$ regular graphs on $n:=\frac{p-1}{12}+1$ vertices (see [CGL09],
page $4$, for a description). The non-trivial eigenvalues of $G(p,\ell)$ are
the numbers $2\sqrt{\ell}\cos(\theta_{f}(\ell))$ for $f\in H(N)$ ($N=p$ in our
notation). The $G(p,\ell)$’s are $d$-regular Ramanujan graphs on $n$ vertices.
The $L^{2}$-variance, $W_{2}(t)$, for the $t$-step non-backtracking random
walk on $G(p,\ell)$ is given by (see [NS21], page $13$)

$W_{2}(t)=\frac{\ell^{t}}{n}\sum_{f\in
H(N)}\absolutevalue{R_{t}(\cos\theta_{f}(\ell))}^{2},$

where $R_{t}$ is the $t^{\text{th}}$ orthogonal polynomial on $[0,\pi]$ with
respect to $d\mu_{\ell}$, normalized so that

$\int_{0}^{\pi}\absolutevalue{R_{t}(\theta)}^{2}d\mu_{\ell}(\theta)=\frac{\ell+1}{\ell}.$

Applying Theorem 2 with $r=1$, $p_{1}=\ell$, and $\ell_{1}=t$ yields that
uniformly for $t<(2-\eta)\log_{\ell}n$,

$W_{2}(t)\sim(\ell+1)\ell^{t-1}\text{ as }n\to\infty.$

Note that $N(t)$, the number of non-backtracking walks of length $t$, is
$(\ell+1)\ell^{t-1}$, so that

$W_{2}(t)\sim N(t)\text{ for }t<(2-\eta)\log_{\ell}n.$

This proves conjecture $1.8$ in [NS21] for graphs $G(p,\ell)$. For the
application to bounded window cutoff one needs $t$ to be as large as
$(1+\varepsilon)\log_{\ell}n$, which is provided by the key doubling of the
degree of $P$ in Theorem 2. In order to prove Conjecture $1.8$ in [NS21] for
the more general Ramanujan graphs constructed using modular forms, one would
need to identify the images of division algebras in $H(N)$ under the Jacquet-
Langlands correspondence and restrict the sums in Theorem $2$ to those forms.

##  4 Multiplicity of Eigenvalue Tuples

Recall that for a fixed prime level $N$ and $\varphi\in S(N)$ a weight $2$
holomorphic cusp form for $\Gamma_{0}(N)$, we let $M_{N}(y,\varphi)$ be the
multiplicity of the tuple of eigenvalues of $\varphi$ at primes up to $y$ in a
Hecke basis $H(N)$, i.e.

$M_{N}(y,\varphi):=\\#\absolutevalue{\left\\{f\in
H(N):\lambda_{f}(p)=\lambda_{\varphi}(p)\text{ for }p\leq
y,(p,N)=1\right\\}}.$

In this section we bound $M_{N}(y,\varphi)$ uniformly in $\varphi$ in the
range $y=(\log N)^{\beta}$ for a fixed $\beta\in(0,1)$. Specifically, we prove
Theorem 3 via the large sieve and smooth number estimates.

From now on, we assume $y=o(\log N)$. We let $p_{1},\ldots,p_{r}$ denote the
first $r$ prime numbers, where $r=\pi(y)$ is the number of primes up to $y$.

An integer $m$ is called $y$-smooth if all primes $p|m$ satisfy $p\leq y$. The
set of $y$-smooth numbers is denoted with $\mathcal{S}_{y}$, and the de Bruijn
function $\Psi(y,M)$ is the counting function for $y$-smooth numbers up to
$M$:

$\Psi(y,M):=\\#\absolutevalue{\\{m\in\mathcal{S}_{y},m\leq M\\}}.$

We use the large sieve inequality as in [IK04], Theorem $7.26$ (the inequality
is stated in [IK04] for weight $k>2$ but holds for $k=2$ as well – see comment
after the proof):

######  Large Sieve Inequality.

Let $\mathcal{F}$ be an orthonormal basis of $S(N)$,
$f(z):=\sum\rho_{f}(n)e(nz)$ for $f\in\mathcal{F}$. Then for any complex
numbers $\\{c_{n}\\}$ we have

$\displaystyle\sum_{f\in\mathcal{F}}\absolutevalue{\sum_{n\leq
M}\frac{c_{n}\rho_{f}(n)}{\sqrt{n}}}^{2}\ll(1+M/N)\norm{c}^{2},$ (17)

where $\norm{c}^{2}=\sum_{n\leq M}\absolutevalue{c_{n}}^{2}$ and the implied
constant is absolute.

We apply this with $M=N$ and

$\displaystyle c_{n}:=\begin{cases}\overline{\lambda_{\varphi}(n)}&\text{ if
}n\in\mathcal{S}_{y}\\\ 0&\text{ otherwise.}\end{cases}$

Using that $\rho_{f}(n)=\sqrt{n}\cdot\lambda_{f}(n)\rho_{f}(1)$ and the
definition of $\mathcal{S}_{y}$, we have

$M_{N}(y,\varphi)\absolutevalue{\rho^{2}_{\varphi}(1)}\Big{|}\sum_{\begin{subarray}{c}n\leq
N\\\
n\in\mathcal{S}_{y}\end{subarray}}\absolutevalue{\lambda_{\varphi}(n)}^{2}\Big{|}^{2}\leq\sum_{\begin{subarray}{c}f\in\mathcal{F}:\\\
\lambda_{f}(m)=\lambda_{\varphi}(m)\\\ \text{for }m\in\mathcal{S}_{y},m\leq
N\end{subarray}}\Big{|}\rho_{\varphi}(1)\sum_{\begin{subarray}{c}n\leq N\\\
n\in\mathcal{S}_{y}\end{subarray}}\absolutevalue{\lambda_{\varphi}(n)}^{2}\Big{|}^{2}\leq\sum_{f\in\mathcal{F}}\Big{|}\sum_{\begin{subarray}{c}n\leq
N\\\
n\in\mathcal{S}_{y}\end{subarray}}\rho_{f}(1)\lambda_{f}(n)\overline{\lambda_{\varphi}(n)}\Big{|}^{2}\overset{\text{(\ref{sieve})
}}{\ll}\sum_{\begin{subarray}{c}n\leq N\\\
n\in\mathcal{S}_{y}\end{subarray}}\absolutevalue{\lambda_{\varphi}(n)}^{2},$

which, combined with the Hoffstein-Lockhart estimate
$\absolutevalue{\rho^{2}_{\varphi}(1)}\gg N^{-1}(\log N)^{-2}$ ([HL94]) yields

$\displaystyle M_{N}(y,\varphi)\ll N(\log
N)^{2}/\sum_{\begin{subarray}{c}n\leq N\\\
n\in\mathcal{S}_{y}\end{subarray}}\absolutevalue{\lambda_{\varphi}(n)}^{2}.$
(18)

To prove Theorem 3, we bound $\sum_{\begin{subarray}{c}n\leq N\\\
n\in\mathcal{S}_{y}\end{subarray}}\absolutevalue{\lambda_{\varphi}(n)}^{2}$
away from $0$ uniformly in $\varphi$. We use the following fact:

######  Proposition 4.1.

Let $k,x\in\mathbb{R}$. Then:

$\displaystyle\max\left\\{\absolutevalue{\sin kx/\sin
x},\absolutevalue{\sin((k+1)x)/\sin x}\right\\}\geq 1/2.$ (19)

(where the functions are extended continuously to the $x$ with $\sin x=0$).

###### Proof.

If $\sin x=0$, (19) is true since $\max\\{k,k+1\\}\geq 1/2$, so assume $\sin
x\neq 0$. Let $\frac{\sin^{2}kx}{\sin^{2}x}=\varepsilon^{2}$, where
$0\leq\varepsilon<1$ (since if $\varepsilon\geq 1$, (19) clearly holds). Then:

$\absolutevalue{\cos
kx}=\sqrt{1-\varepsilon^{2}\sin^{2}x}=\sqrt{1-\varepsilon^{2}+\varepsilon^{2}\cos^{2}x},$

so by the trigonometric identity for sine of a sum,

$\displaystyle\absolutevalue{\frac{\sin(kx+x)}{\sin x}}$
$\displaystyle=\absolutevalue{\frac{\sin kx}{\sin x}\cos x+\cos
kx}\geq\absolutevalue{\cos kx}-\varepsilon\absolutevalue{\cos
x}=\sqrt{1-\varepsilon^{2}+\varepsilon^{2}\absolutevalue{\cos
x}^{2}}-\varepsilon\absolutevalue{\cos x}.$

It remains to minimize
$f_{\varepsilon}(t):=\sqrt{1-\varepsilon^{2}+\varepsilon^{2}t^{2}}-\varepsilon
t$ for $t\in[0,1]$. The function $f_{\varepsilon}(t)$ has a non-vanishing
derivative in this interval when $\varepsilon<1$, so

$\absolutevalue{\frac{\sin((k+1)x)}{\sin
x}}\geq\min\\{f_{\varepsilon}(0),f_{\varepsilon}(1)\\}=\min\\{\sqrt{1-\varepsilon^{2}},1-\varepsilon\\}=1-\varepsilon,$

and so

$\max\left\\{\absolutevalue{\frac{\sin kx}{\sin
x}},\absolutevalue{\frac{\sin((k+1)x)}{\sin
x}}\right\\}\geq\max\\{\varepsilon,1-\varepsilon\\}\geq 1/2.$

∎

Using the Hecke relation

$\lambda_{\varphi}(p^{k})=\frac{\sin((k+1)\theta_{\varphi}(p))}{\sin(\theta_{\varphi}(p))},$

(19) implies that

$\max\\{\absolutevalue{\lambda_{\varphi}(p^{k})},\absolutevalue{\lambda_{\varphi}(p^{k+1})}\\}\geq
1/2$

for all $\varphi$ and $k\geq 0$. Since $\lambda_{\varphi}(n)$ is
multiplicative, for any $r$-tuple $(\alpha_{1},\ldots,\alpha_{r})$ of non-
negative integers, there are
$(\delta_{1},\ldots,\delta_{r})\subseteq\\{0,1\\}^{r}$ such that

$\displaystyle\absolutevalue{\lambda_{\varphi}(p_{1}^{2\alpha_{1}+\delta_{1}}\cdots
p_{r}^{2\alpha_{r}+\delta_{r}})}^{2}\geq 4^{-r}.$ (20)

The set $\mathcal{S}_{y}$ of all $y$-smooth numbers is a disjoint union of
sets

$\mathcal{E}_{\alpha_{1},\ldots,\alpha_{r}}:=\\{p_{1}^{2\alpha_{1}+\delta_{1}}\cdots
p_{r}^{2\alpha_{r}+\delta_{r}}|\delta_{i}\in\\{0,1\\}\\}$

of size $2^{r}$, and (20) implies that each
$\mathcal{E}_{\alpha_{1},\ldots,\alpha_{r}}$ contains an element $t$ with
$\absolutevalue{\lambda_{\varphi}(t)}^{2}\geq 4^{-r}$. Moreover, for every
$s\in S_{y}\cap[0,N/(p_{1}\cdots p_{r})]$, the set $\mathcal{E}_{\alpha}$
containing $s$ is fully contained in $\mathcal{S}_{y}\cap[0,N]$. Hence, there
are at least $\Psi[y,N/(p_{1}\cdots p_{r})]/2^{r}$ sets $\mathcal{E}_{\alpha}$
fully contained in $\mathcal{S}_{y}\cap[0,N]$, so

$\displaystyle\sum_{\begin{subarray}{c}n\leq N\\\
n\in\mathcal{S}_{y}\end{subarray}}\absolutevalue{\lambda_{\varphi}(n)}^{2}\gg
8^{-r}\Psi\left(y,\frac{N}{\prod_{p<y}p}\right)=8^{-r}\Psi\left(y,\frac{N}{N^{\frac{(1+o(1))y}{\log
N}}}\right),$ (21)

where we used the prime number theorem in the second step. We use a result of
Hildebrand and Tenenbaum on the size of $\Psi(y,X)$ in the range $y=o(\log
X)$:

######  Theorem ([HT86], Corollary 1).

Let $y=o(\log X)$ such that $y\to\infty$ as $X\to\infty$. Let

$\alpha(X,y):=(1+o(1))\frac{y}{\log X\log y},$

and

$\zeta(\alpha,y):=\prod_{p\leq y}(1-p^{-\alpha})^{-1}.$

Then:

$\displaystyle\Psi(X,y)=(1+o(1))X^{\alpha}\zeta(\alpha,y)\sqrt{\log y/(2\pi
y)}.$ (22)

From (22),

$\displaystyle\log\Psi(X,y)=\alpha\log X+\log\zeta(\alpha,y)+\mathrm{O}(\log
y)=\log\zeta(\alpha,y)+\mathrm{O}(y/\log y),$ (23)

and using the Taylor series expansion,

$\displaystyle\log\zeta(\alpha,y)$ $\displaystyle=-\sum_{p\leq
y}\log(1-e^{-\alpha\log p})$ $\displaystyle=-\sum_{p\leq y}\log(\alpha\log
p(1+\mathrm{O}(\alpha\log p)))$ $\displaystyle=-\pi(y)\log\alpha-\sum_{p\leq
y}\log\log p+O(\alpha)\sum_{p\leq y}\log p.$ (24)

Using partial summation,

$\absolutevalue{\sum_{p\leq y}\log\log p-\pi(Y)\log\log
Y}\ll\int_{2}^{Y}\frac{dt}{\log^{2}t}=\left(\mathrm{li}(t)-\frac{t}{\log
t}\right)\Big{|}_{2}^{Y}\ll\frac{Y}{\log^{2}Y},$

and by the prime number theorem, $\sum_{p\leq y}\log p=(1+o(1))y$. Hence,

$\displaystyle(\ref{tag})$ $\displaystyle=\pi(y)\left(-\log y+\log\log
X+\log\log y+\mathrm{O}(1)-\log\log
y\right)+\mathrm{O}(y/\log^{2}y)+\mathrm{O}(y^{2}/(\log y\log X))$
$\displaystyle=(1+o(1))\frac{y}{\log y}\log\left(\frac{\log X}{y}\right),$

and so (23) gives

$\displaystyle\log\Psi(X,y)=(1+o(1))\frac{y}{\log y}\log\left(\frac{\log
X}{y}\right).$ (25)

Finally, we combine the results above. Let $X=N^{1-(1+o(1))\frac{y}{\log N}}$,
so

$\log X=\log N\left(1-(1+o(1))\frac{y}{\log N}\right)=\log N(1+o(1))$

when $y=o(\log N)$. Clearly, for such $y$ we also have $y=o(\log X)$, so it
follows from Theorem Theorem that (25) holds for such $y$ and $X$. Finally,
combining (25) with (18) and (21) and using that $r=\pi(y)=o(y)$ (i.e.
$\log(8^{r})=o(y)$), we see that

$\log(M_{N}(y,\varphi)/N)\leq\mathrm{O}(\log\log N)+o(y)+(1+o(1))\frac{y}{\log
y}\log\left(\frac{\log N}{y}\right)$

as long as $y=o(\log N)$ and $y\to\infty$ as $N\to\infty$. In particular, when
$y=(\log N)^{\beta}$ for $0<\beta<1$,

$\log\left(\frac{\log N}{y}\right)=\frac{1-\beta}{\beta}\log y,$

so

$\log(M_{N}(y,\varphi)/N)\leq(1+o(1))\frac{1-\beta}{\beta}y.$

Since $s(N)\asymp N$, this proves Theorem 3.

##  5 Number of Forms with Degree $d$ Hecke Fields

For a prime level $N$, let $H(N)_{d}\subseteq H(N)$ denote Hecke forms whose
Hecke eigenvalues span a number field of degree exactly $d$. We bound the size
of $H(N)_{d}$ using the multiplicity bound from the previous section.

Specifically, let $y>0$, $r=\pi(y)$, and for $f\in H(N)_{d}$, and let
$a_{f}(p)=\lambda_{f}(p)\sqrt{p}$ be the $p^{\text{th}}$ Hecke operator
eigenvalue of $f$. To prove Theorem 4, we combine the multiplicity bound with
an upper bound on the set

$T_{N}(y)_{d}:=\\{(a_{f}(p_{1}),\ldots,a_{f}(p_{r}))|f\in H(N)_{d}\\}$

of possible tuples of eigenvalues of a Hecke form at the first $r$ primes. We
get this bound by exploiting that $a_{f}(p)$ is a totally real algebraic
integers whose conjugates are bounded by $2\sqrt{p}$ in size.

######  Proposition 5.1.

$\displaystyle\\#\absolutevalue{T_{N}(y)_{d}}\leq\exp\left(yd/2+o_{d}(y)\right).$
(26)

######  Lemma 5.2.

For $f\in H(N)_{d}$, let
$K_{f,r}:=\mathbb{Q}(a_{f}(p_{1}),\ldots,a_{f}(p_{r}))$. Then:

$\\#\absolutevalue{\\{K_{f,r}|f\in H(N)_{d}\\}}\ll_{d}y^{\kappa}$

where $\kappa=\kappa(d)$ is a constant depending on $d$.

###### Proof.

Let $K=K_{f,r}$ for some $f\in H(N)_{d}$. Let
$K_{i}:=\mathbb{Q}(a_{f}(p_{i}))$ be of degree $d_{i}\leq d$ with discriminant
$\Delta_{i}$, and let $P_{i}(x)=\prod(x-\beta_{j})$ be the minimal polynomial
of $a_{f}(p_{i})$. Then

$\absolutevalue{\Delta_{i}}=\frac{\absolutevalue{\mathrm{disc}(P_{i})}}{[\mathcal{O}_{K_{i}}:\mathbb{Z}[a_{f}(p_{i})]]^{2}}\leq\prod_{i\neq
j}\absolutevalue{(\beta_{i}-\beta_{j})}\leq(4\sqrt{p_{i}})^{d_{i}(d_{i}-1)}\ll_{d}y^{d^{2}/2}.$

Since $K$ has degree at most $d$, it can be expressed as a composition of at
most $\log_{2}d$ fields $K_{i}$, so the discriminant $\Delta$ of $K$ satisfies

$\absolutevalue{\Delta}\ll_{d}y^{k}$

for some constant $k$ depending only on $d$. This implies a bound of the same
form on the number of possibilities for $K$ by the Theorem of Schmidt
([Sch95]). ∎

######  Lemma 5.3.

Let $K$ be a totally real number field of degree $\leq d$. Then for $M>1$, the
number of $\alpha\in\mathcal{O}_{K}$ such that all the Galois conjugates of
$\alpha$ are bounded by $M$ is at most $C(d)M^{d}$ for some constant $C(d)$
which does not depend on $K$.

###### Proof.

Consider that standard embedding $\iota:K\hookrightarrow\mathbb{R}^{d}$. For
$\alpha\in\mathcal{O}_{K}$, the coordinates of $\iota(\alpha)$ are the Galois
conjugates of $\alpha$; their product is a non-zero integer, so the non-zero
vectors in the lattice formed by the image of $\mathcal{O}_{K}$ under $\iota$
have length $\geq 1$. From this, sphere packing bounds imply immediately that
the number of lattice points in the box $[-M,M]^{d}$ is bounded by
$O_{d}(1)M^{d}$ (this can be seen, for example, by placing (disjoint) balls of
diameter $1$ at each lattice point in the box and comparing volumes). ∎

###### Proof of Proposition 5.1.

From Lemma 5.3, we see that for a fixed degree $K$ number field, the number of
possible tuples $(a_{f}(p_{1}),\ldots,a_{f}(p_{r}))$ with $a_{f}(p_{i})\in K$
is at most

$\prod_{p\leq y}C(d)(2\sqrt{p})^{d}=(2C(d))^{r}\exp((d/2)\sum_{p\leq y}\log
p)=\exp(dy/2+o_{d}(y)),$

where the last step uses the prime number theorem. On the other hand, from
Lemma 5.2, the number of choices for $K$ is $\exp(O_{d}(\log
y))=\exp(o_{d}(y))$, so multiplying the two proves the proposition statement.
∎

Combining this Proposition with Theorem 3,

$\log
s(N)_{d}/s(N)\leq-\left(\frac{1-\beta}{\beta}-\frac{d}{2}\right)y+o_{d}(y),$

which concludes the proof of Theorem 4.

Note that for the coefficient of $y$ to be negative, we have to choose $\beta$
small, which is why we dealt with multiplicity bounds in Theorem $3$ only for
$0\leq\beta\leq 1$.

##  6 Composite Level

The discussion up to this point was restricted to weight $k=2$ and level $N$
being prime. Theorems 1 and 2 can be extended without much change to allow
varying weight and general $N$, as long as the relatively prime conditions
$(mn,N)=1$ and $(p_{j},N)=1$ are maintained. Indeed, the starting point, which
is an application of the Petersson formula ([ILS00], Corollary $2.2$ or
[IK04], Corollary $14.24$), gives the desired uniformity in the "harmonic"
weighted form. To remove the weights, one has to take care with new and old
forms and the Atkin-Lehner involutions in relating $\norm{f}^{2}_{2}$ and
$L(1,\mathrm{Sym}^{2}f)$.

On the other hand, in the proof of the multiplicity bounds in Theorem 3, we
used $y$-smooth numbers and the assumption that $(p,N)=1$ for $p\leq y<\log
N$. As we show in Theorem 5 below, similar bounds can be proved for $N$’s that
do not have an abnormal number of small prime factors. For "super-smooth"
numbers, such as $N=\prod_{p\leq t}p$, we cannot make use of the approach to
the Plancherel measure of the Hecke eigenvalues for small primes, and our
bounds in Theorem 3 and 4 don’t apply.

In what follows, we restrict ourselves to the $s^{*}(N)$-dimensional space
$S^{*}(N)$ of weight $2$ level $N$ newforms, which admits a simultaneous
eigenbasis $H^{*}(N)$ with respect to Hecke operators $T_{n}$ with $(n,N)=1$
(we assume these forms are normalized to have constant Fourier coefficient
$1$).

For a positive integer $N$, the number of distinct prime divisors of $N$ is at
most

$\displaystyle(1+o(1))\frac{\log N}{\log\log N}=:\bf{P}.$ (27)

Let $y=y(N)=o(\log N)$ be a parameter going to infinity with $N$, and let
$r:=\pi(y)\sim y/\log y$. Let $q_{1},\ldots,q_{r}$ be the first $r$ primes
which don’t divide $N$. Since $q_{k}$ is at most the
$(k+\mathbf{P})^{\text{th}}$ prime and $k\leq r=o(\bf{P}),$ we can conclude
via the prime number theorem that

$\displaystyle q_{k}\leq(1+o(1))(\bf{P}\log\bf{P}).$ (28)

In the spirit of section 4, we want to give a lower bound for the function

$\Phi(q_{1},\ldots,q_{r},X):=\\#\absolutevalue{\\{(\alpha_{1},\ldots,\alpha_{r}):q_{1}^{\alpha_{1}}\cdots
q_{r}^{\alpha_{r}}\leq X\\}}$

for $X$ (to be chosen later) satisfying

$\displaystyle\log X=(1+o(1))\log N.$ (29)

By (28),

$\displaystyle\Phi(q_{1},\ldots,q_{r},X)$
$\displaystyle=\\#\absolutevalue{\\{(\alpha_{1},\ldots,\alpha_{r}):q_{1}^{\alpha_{1}}\cdots
q_{r}^{\alpha_{r}}\leq X\\}}\geq$
$\displaystyle\geq\\#\absolutevalue{\\{(\alpha_{1},\ldots,\alpha_{r}):\big{(}(1+o(1))\mathbf{P}\log\mathbf{P}\big{)}^{\alpha_{1}+\cdots+\alpha_{r}}\leq
X\\}}=$
$\displaystyle=\\#\absolutevalue{\\{(\alpha_{1},\ldots,\alpha_{r}):\alpha_{1}+\cdots+\alpha_{r}\leq\log
X/\log\big((1+o(1))(\mathbf{P}\log\mathbf{P})\big{missing})\\}}\overset{\tiny{\text{
by }}(\ref{sizeofp}),(\ref{xislogn})}{=}$
$\displaystyle=\\#\absolutevalue{\\{(\alpha_{1},\ldots,\alpha_{r}):\alpha_{1}+\cdots+\alpha_{r}\leq(1+o(1))\log
X/\log\log X\\}}.$ (30)

The number of non-negative integer solutions to $x_{1}+\ldots+x_{A}\leq B$ is

${A+B\choose A}\geq\left(\frac{B}{A}\right)^{A},$

so (6) implies

$\displaystyle\log\Phi(q_{1},\ldots,q_{r},X)\geq r\log\frac{(1+o(1))\log
X}{r\log\log X}.$ (31)

For $\varphi\in S^{*}(N)$ a weight $2$ holomorphic cusp newform for
$\Gamma_{0}(N)$, we let $M^{*}_{N}(q_{1},\ldots,q_{r},\varphi)$ be the
multiplicity of the tuple of eigenvalues of $\varphi$ at primes $q_{i}$, i.e.

$M^{*}_{N}(q_{1},\ldots,q_{r},\varphi):=\\#\absolutevalue{\left\\{f\in
H^{*}(N):\lambda_{f}(q_{i})=\lambda_{\varphi}(q_{i})\text{ for all }i\leq
r\right\\}}.$

We bound $M^{*}_{N}(q_{1},\ldots,q_{r},\varphi)$ for a fixed $\varphi$ via the
large sieve inequality identically to section 4. Taking

$\displaystyle c_{n}:=\begin{cases}\overline{\lambda_{\varphi}(n)}&\text{ if
}n=q_{1}^{\alpha_{1}}\cdots q_{r}^{\alpha_{r}}\leq X\\\ 0&\text{
otherwise,}\end{cases}$

we get

$\displaystyle M^{*}_{N}(q_{1},\ldots,q_{r},\varphi)\ll N(\log
N)^{2}/\sum_{\begin{subarray}{c}n=q_{1}^{\alpha_{1}}\cdots
q_{r}^{\alpha_{r}}\\\ n\leq
N\end{subarray}}\absolutevalue{\lambda_{\varphi}(n)}^{2}.$ (32)

Recreating the proof in section 4, we can see that

$\displaystyle\sum_{\begin{subarray}{c}n\leq N\\\
n\in\mathcal{S}_{q_{1},\ldots,q_{r}}\end{subarray}}\absolutevalue{\lambda_{\varphi}(n)}^{2}\gg
8^{-r}\Phi\left(q_{1},\ldots,q_{r},\frac{N}{q_{1}\cdots q_{r}}\right)\geq
8^{-r}\Phi\left(q_{1},\ldots,q_{r},\frac{N}{\mathbf{P}^{2r}}\right).$ (33)

Let $X:=N/\mathbf{P}^{2r}$. Recall that $r=\pi(o(\log N))=o(\log N/\log\log
N),$ so

$\log X=\log N-2r(1+o(1))\log\log N=\log N(1+o(1)),$

which means this choice of $X$ satisfies (29) and hence also satisfies (31).
Let $0<\beta<1$ and let

$y:=(\log N)^{\beta},$

so

$r=(1+o(1))(\log X)^{\beta}/\beta\log\log X.$

Then (31) becomes

$\log\Phi(q_{1},\ldots,q_{r},X)\geq(1+o(1))\frac{(\log
X)^{\beta}}{\beta\log\log X}\log\frac{\beta\log X}{(\log
X)^{\beta}}=(1+o(1))\frac{1-\beta}{\beta}(\log X)^{\beta},$

i.e.,

$\log\Phi(q_{1},\ldots,q_{r},N)\geq(1+o(1))\frac{1-\beta}{\beta}(\log
N)^{\beta}.$

Finally, by (32) and (33),

$\displaystyle\log M^{*}_{N}(q_{1},\ldots,q_{r},\varphi)/N\leq\log\log N+r\log
8-(1+o(1))\frac{1-\beta}{\beta}(\log
N)^{\beta}=-(1+o(1))\frac{1-\beta}{\beta}(\log N)^{\beta}$ (34)

(note that this bound is identical to the one in section 4 which is sharp).

We apply (34) to extend Theorem 4 to more general $N$’s. For $T\geq 1$ fixed
and for some $y=y(N)$ with $\log\log N\ll y\ll\log N$, we say that a large $N$
is $T$-super-smooth if

$\frac{\pi(y^{T};N)}{\pi(y)}=o(1),$

where $\pi(z;N)=\\#\absolutevalue{\\{p\leq z:(p,N)=1\\}}$ is the number of
primes up to $z$ that don’t divide $N$. Clearly, very few numbers are
$T$-super-smooth for all $T$.

Let $H^{*}(N)_{d}:=\\{f\in H^{*}(N):d(f)=d\\}$ be the set of Hecke newforms
whose Fourier coefficients span a number field of degree $d$,
$s^{*}(N)_{d}=\absolutevalue{H^{*}(N)_{d}}$. The following theorem extends
Theorem 4 to non-super-smooth numbers.

######  Theorem 5.

Let $0\leq\beta\leq 1$, $y=(\log N)^{\beta}$, $T\geq 1$, and $d\geq 1$. Then
for $N$ not $T$-super-smooth,

$s^{*}(N)_{d}\leq\exp\left(-\left(\frac{1-\beta}{\beta}-\frac{dT}{2}\right)y+o_{T,d}(y)\right)s^{*}(N)$

as $N\to\infty$.

###### Proof.

The proof emulates that of section 5. For $f\in H^{*}(N)_{d}$, let
$a_{f}(q_{i})=\lambda_{f}(q_{i})\sqrt{q_{i}}$ be the eigenvalue of $f$ for the
Hecke operator $T_{q_{i}}$, and let

$T^{*}_{N}(q_{1},\ldots,q_{r})_{d}:=\\{(a_{f}(q_{1}),a_{f}(q_{2}),\ldots,a_{f}(q_{r}))|f\in
H^{*}(N)_{d}\\}$

denote the set of possible eigenvalue tuples of a form in $H^{*}(N)_{d}$ at
the first $r$ primes not dividing $N$. Repeating verbatim the proof of Lemma
5.2, there are $\ll_{d}y^{T\kappa(d)}$ possible number fields of the form
$\mathbb{Q}(a_{f}(q_{1}),\ldots,a_{f}(q_{r}))$. Tautologically, for $N$ as in
the statement of the theorem, the first $r$ primes $q_{1},\ldots,q_{r}$ not
dividing $N$ satisfy $q_{i}\leq y^{T}$, so using Lemma 5.3, we get

$\\#\absolutevalue{T^{*}_{N}(q_{1},\ldots,q_{r})}\leq
y^{T\kappa(d)}\prod_{i\leq r}C(d)y^{Td/2}\leq\exp((d/2)Tr\log
y+o_{T,d}(y))\leq\exp((d/2)Ty+o_{d,T}(y)).$

Combined with (34), this gives

$\log
s(N)_{d}/N\leq-\left(\frac{1-\beta}{\beta}-\frac{dT}{2}\right)y+o_{T,d}(y).$

It remains to note that $s^{*}(N)\asymp\varphi(N)$, the Euler’s totient
function, and $\log\varphi(N)=\log N+O(\log\log\log N),$ which completes the
proof.

∎

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