# Selberg’s Central Limit Theorem for $L$-functions of level aspect-II

Madhuparna Das  Department of Mathematics, University of Exeter, Stocker Rd,
Exeter EX4 4PY, UK<EMAIL_ADDRESS>
(Date: March 2021)

###### 2010 Mathematics Subject Classification:

11M06

##  1\. Introduction

Selberg [1, 2] observed that the logarithm of the Riemann zeta function as
well $L$-functions (which belong to the Selberg’s class) behaves like random
variable with mean 0 and variance $\frac{1}{2}\log\log|t|$. In 2017, Radziwill
and Soundararajan [3] gave an elegant proof of Selberg’s central limit theorem
for the Riemann zeta function by using basic properties of the Riemann zeta
function. Later P. Hsu and P. Wong [4] used the technique developed in [3] to
prove this result for Dirichlet $L$-functions. We have proved this result
following the method developed in [3] for $GL(2)$ $L$-functions [5] associated
with primitive holomorphic cusp form.

In this paper, we prove the Selberg’s Central limit theorem for the
$L$-function associated with the Hecke-Maass cusp form. Let $f$ be a Hecke-
Maass cusp form of type $\nu=(\nu_{1},\nu_{2})$ for $SL(3,\mathbb{Z})$. Let
$\lambda(m_{1},m_{2})$ (with $\lambda(1,1)=1$) be the normalized Fourier
coefficients of $f$. Then the $L$-function associated with $f$ is given by the
Dirichlet series

$\displaystyle L(f,s)=\sum_{n=1}^{\infty}\frac{\lambda(1,n)}{n^{s}}$

in domain $\Re(s)>1$.

Since the $L$-functions we have defined above also belong to the Selberg class
so we can prove Selberg’s influential theorem for this family of
$L$-functions.

######  Theorem 1.

Let $V$ be a fixed real number. Then for all large $T$,

$\displaystyle\frac{1}{T}meas\left\\{T\leq t\leq
2T:\log|L(f,\frac{1}{2}+it)|\geq V\sqrt{\frac{1}{2}\log\log
T}\right\\}\sim\frac{1}{\sqrt{2\pi}}\int_{V}^{\infty}{e^{-u^{2}/2}du},$

where $f$ is a Hecke-Maass cusp form.

After the derivation of the central limit theorem Selberg [2] mentioned his
orthogonality conjecture, primitive $L$-functions in the Selberg class are
statistically independent. Although, Selberg did not give any precise
description of independence. In the paper by P. H. Hsu and P. Wong [4], they
proved the same result for the multiple Dirichlet $L$-functions. The second
goal of this paper is to prove, the statistical independence multiple
$L$-functions of degree 2 (and degree 3).

######  Theorem 2.

Let $(g_{j})_{j=1}^{n}$ be a sequence of distinct primitive holomorphic cusp
forms. Then for all large $T$ and $t\in[T,2T]$ the random vector

$\displaystyle\left(\log|L(g_{1},\frac{1}{2}+it)|,\ldots,\log|L(g_{j},\frac{1}{2}+it)|\right)$

is approximately $n$-variate normal distribution with mean vector $0_{n}$ and
co-variance matrix $\frac{1}{2}(\log\log T)I_{n}$.
Consequently, the random variables $\log|L(g_{j},\frac{1}{2}+it)|$’s (for
$j=1,\ldots,n$) are independent and each $\log|L(g_{j},\frac{1}{2}+it)|_{g\in
J}$ forms a Gaussian process for any totally ordered set $J$ of distinct
primitive holomorphic cusp forms111In this case $L(g,s)$ is $GL(2)$
$L$-functions associated with the primitive holomorphic cusp form $g$. [6].

Since in this paper we are working with the $GL(3)$ $L$-functions associated
with the Hecke-Maass cusp form $f$ in addition to theorem 2 we can also think
of the independence of $L$-functions associated with the Hecke-Maass cusp form
$f$.

######  Theorem 3.

Let $(f_{j})_{j=1}^{n}$ be a sequence of distinct Hecke-Maass cusp forms. Then
for all large $T$ and $t\in[T,2T]$ the random vector

$\displaystyle\left(\log|L(f_{1},\frac{1}{2}+it)|,\ldots,\log|L(f_{j},\frac{1}{2}+it)|\right)$

is approximately $n$-variate normal distribution with mean vector $0_{n}$ and
co-variance matrix $\frac{1}{2}(\log\log T)I_{n}$.
Consequently, the random variables $\log|L(f_{j},\frac{1}{2}+it)|$’s (for
$j=1,\ldots,n$) are independent and each $\log|L(f_{j},\frac{1}{2}+it)|_{f\in
J}$ forms a Gaussian process for any totally ordered set $J$ of distinct
Hecke-Maass cusp forms.

First, we give a sketch of the proof of Theorem 1 then we get into the
details. In a later section, we prove the independence of $L$-functions (for
degree 2 and 3).

##  2\. The Setup

The proof of Theorem 1 is almost similar to the proof given in [5]. Since we
are dealing with $GL(3)$ $L$-functions here we need some modifications. We
prove the first theorem in four steps. In the first step, we take away the
problem from the critical line because of the tail distribution of the
$L$-function.

######  Proposition 1.

Let $T$ be large and suppose $t\leq t\leq 2T$. Then for any
$\sigma>\frac{1}{2}$ we have

$\displaystyle\int_{t-1}^{t+1}\left|\log|L(f,\frac{1}{2}+iy)|-\log|L(f,\sigma+iy)|\right|dy\ll\left(\sigma-\frac{1}{2}\right)\log
T.$

Although, we are using $GL(3)$ $L$-functions associated with the Hecke-Maass
cusp form $f$ some basic results we are going to use in the proof is slimier
for all automorphic $L$-functions in families. So, we can fix the parameters
as given in [3].

$\displaystyle W=(\log\log\log T)^{4},\;X=T^{1/(\log\log\log
T)^{2}},\;Y=T^{(1/\log\log T)^{2}},\;\sigma_{0}=\frac{1}{2}+\frac{W}{\log T},$

where $T>0$ is sufficiently large so that $W\geq 3$.

In the next step, we introduce the auxiliary series

$\displaystyle\mathcal{P}(f,s)=\mathcal{P}(f,s;X)=\sum_{2\leq n\leq
X}\frac{\Lambda_{f}(n)}{n^{s}\log n}$

where $f$ is a Hecke-Maass cusp form. We compute moments of this auxiliary
series to determine the distribution of $\mathcal{P}(f,s)$.

######  Proposition 2.

As $t$ varies in $T\leq t\leq 2T$, the distribution of
$\Re(\mathcal{P}(\sigma_{0}+it))$ is approximately normal with mean 0 and
variance $\sim\frac{1}{2}\log\log T$.

To complete the proof we need to connect the $L$-functions with the auxiliary
series $\mathcal{P}(f,s)$. We use the mollification technique for that. Define
the Dirichlet polynomial

$\displaystyle M(f,s)=\sum_{n}\frac{\mu(n)a(n)\lambda(1,n)}{n^{s}}.$

where $a(n)$ is defined as

$\displaystyle a(n)=\begin{cases}1&\text{if $n$ is composed only primes below
$X$ and has at most $100\log\log T$ primes below $Y$,}\\\ &\text{and at most
$100\log\log\log T$ primes between $Y$ and $X$.}\\\
0&\text{otherwise}.\end{cases}$

Observe the definition of $a(n)$, it takes the value 0 except when $n\leq
Y^{100\log\log T}X^{100\log\log\log T}<T^{\epsilon}$. It is evident that
$M(f,s)$ is a short Dirichlet polynomial. Next we show that $M(f,s)$ can be
approximated by $e^{-\mathcal{P}(f,s)}$.

######  Proposition 3.

For $T\leq t\leq 2T$

$\displaystyle M(\sigma_{0}+it)=(1+o(1))\exp(-\mathcal{P}(\sigma_{0}+it))$

except perhaps on a subset of measure $o(T)$.

It is easy to see that we need to prove $M(f,s)$ and $L(f,s)$ are inverse to
each other in order to complete the proof of Theorem 1.

######  Proposition 4.

For $T\leq t\leq 2T$,

$\displaystyle\int_{T}^{2T}\left|1-L(f,\sigma_{0}+it)M(f,\sigma_{0}+it)\right|^{2}dt=o(1)$

So that for sufficiently large $t$ we have

$\displaystyle L(f,\sigma_{0}+it)M(f,\sigma_{0}+it)=1+o(1),$

except perhaps on a set of measure $o(T)$.

We are now ready to prove the first theorem. We prove the propositions in
later sections.

Proof of Theorem 1: From Proposition 4, we can say that
$L(f,\sigma_{0}+it)\approx M(f,\sigma_{0}+it)^{-1}$. By Proposition 3 we have
$M(f,\sigma-0+it)^{-1}\approx\exp{\mathcal{P}(f,\sigma_{0}+it)}$. With the
help of Proposition 2, we can conclude that $\log|L(f,\sigma_{0}+it)|$ has
normal distribution with mean 0 and variance $\frac{1}{2}\log\log T$. Finally,
by Proposition 1 we deduce that $\log|L(f,\sigma_{0}+it)|$ and
$\log|L(f,\frac{1}{2}+it)|$ have the same distribution for $\sigma=\sigma_{0}$
(as we have set $\sigma_{0}$ before), under the estimate
$W^{2}=o(\sqrt{\log\log T})$, which completes the proof. ∎

##  3\. Proof of Proposition 1

Recall the complete $L$-function associated with the Hecke-Maass cusp form $f$

$\displaystyle\Lambda(f,s)=G(f,s)L(f,s)$

where $G(f,s)$ is given by

$\displaystyle
G(f,s)=\prod_{i=1}^{3}\pi^{-s/2}\Gamma\left(\frac{s-\alpha_{i}}{2}\right).$

Since $f$ is a Hecke-Maass cusp form of type $(\nu_{1},\nu_{2})$. Then we have
the Langland’s parameters $(\alpha_{1},\alpha_{2},\alpha_{3})$ associated with
$f$ are defined as

$\displaystyle\alpha_{1}=-\nu_{1}-2\nu_{2}+1$
$\displaystyle\alpha_{2}=-\nu_{1}+\nu_{2}$
$\displaystyle\alpha_{3}=2\nu_{1}+\nu_{2}-1$

First we prove that for sufficiently large $t$ if $y\in[t-1,t+1]$ then

(1)
$\displaystyle\left|\log\frac{G(f,\sigma+iy)}{G(\frac{1}{2}+iy)}\right|\ll\left(\sigma-\frac{1}{2}\right)\log
t.$

The Ramanujan-Selberg conjecture predicts that $\Re(\alpha_{i})=0$. Applying
Stirling formula on (1) we get

$\displaystyle\left|\log\frac{G(f,\sigma+iy)}{G(\frac{1}{2}+iy)}\right|$
$\displaystyle\sim$
$\displaystyle\left|\log\frac{\prod_{i=1}^{3}\pi^{-(\sigma+it)/2}\Gamma\left(\frac{\sigma+it-\alpha_{i}}{2}\right)}{\prod_{i=1}^{3}\pi^{-(\frac{1}{2}+it)/2}\Gamma\left(\frac{\frac{1}{2}+it-\alpha_{i}}{2}\right)}\right|$
$\displaystyle\ll$ $\displaystyle\left|\log t^{(\sigma-\frac{1}{2})}\right|$
$\displaystyle\ll$ $\displaystyle\left(\sigma-\frac{1}{2}\right)\log t$

Since $t$ is large enough we can ignore the $\pi$ term in the logarithm. Now
recall the definition of complete $L$-function. To prove Proposition 1 it is
enough to prove that

$\displaystyle\int_{t-1}^{t+1}\left|\log\left|\frac{\Lambda(f,\frac{1}{2}+iy)}{\Lambda(f,\sigma+iy)}\right|\right|dy\ll\left(\sigma-\frac{1}{2}\right)\log
T.$

We prove this by counting zeros of $L(f,s)$ and the rest of the proof is
similar as given in [5]. From the Hadamard’s factorization formula (see e.g.
Theorem 5.6 of [7]), there exist $a=a(f)$ and $b=b(f)$ (where
$b(f)=-\sum_{\rho}\Re(\frac{1}{\rho})$) such that

$\displaystyle(s(1-s))\Lambda(f,s)=e^{a+bs}\prod_{\rho\neq
0,1}\left(1-\frac{s}{\rho}\right)$

where $\rho$ ranges over all zeros $\Lambda(f,s)$ different from $0,1$.

$\displaystyle\log\left|\frac{\Lambda(f,1/2+iy)}{\Lambda(f,\sigma+iy)}\right|$
(2) $\displaystyle=$
$\displaystyle\sum_{\rho}\log\left|\frac{(f,1/2+iy)-\rho}{(f,\sigma+iy)-\rho}\right|$

Suppose $\rho=\beta+i\gamma$ is a non-trivial zero of $L(f,s)$. If
$|t-\gamma|\geq 2$ then we check that

$\displaystyle\int_{t-1}^{t+1}\left|\log\left|\frac{(f,1/2+iy)-\rho}{(f,\sigma+iy)-\rho}\right|\right|dy\ll\frac{(\sigma-1/2)}{(t-\gamma)^{2}}$

Take the range $|t-\gamma|\leq 2$ we use

$\displaystyle\int_{t-1}^{t+1}\left|\log\left|\frac{(f,1/2+iy)-\rho}{(f,\sigma+iy)-\rho}\right|\right|dy$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{-\infty}^{\infty}\left|\log\left|\frac{(\beta-\frac{1}{2})^{2}+x^{2}}{(\beta-\sigma)^{2}+x^{2}}\right|\right|dx$
$\displaystyle=$ $\displaystyle\pi\left(\sigma-\frac{1}{2}\right)$

Thus in either case

$\displaystyle\int_{t-1}^{t+1}\left|\log\left|\frac{(f,1/2+iy)-\rho}{(f,\sigma+iy)-\rho}\right|\right|dy\ll\frac{(\sigma-\frac{1}{2})}{1+(t-\gamma)^{2}}$

Insert this in (3) and note that there are $\ll\log(t+k^{\prime})$ zeros with
$k^{\prime}\leq|t-\gamma|<k^{\prime}+1$ and the proposition follows. ∎

##  4\. Proof of Proposition 2

This proposition study the statistical behaviour of the auxiliary series
$\mathcal{P}(f,s)$. We can restrict the sum to primes, because the higher-
order terms (for $p^{k}$ for $k\geq 3$) contribute $O(1)$. For the terms
$p^{2}$ as argued in [3] they contribute $O(T/R^{2})$ for where contribution
of the prime square larger than $R$ (say).

The coefficient $\Lambda_{f}(s)$ of the auxiliary series satisfies that
$\Lambda_{f}(p)=\lambda(1,p)\log p$, since we are working with $GL(3)$
$L$-functions. So, we have the auxiliary series where the sum is restricted to
the primes given by

(3)
$\displaystyle\mathcal{P}_{0}(f,\sigma_{0}+it)=\mathcal{P}_{0}(f,\sigma_{0}+it;X)=\sum_{p\leq
X}\frac{\lambda(1,p)}{p^{\sigma_{0}+it}}$

We study the moments of (3) in order to prove the proposition.

######  Lemma 1.

Suppose that $k$ and $l$ are non-negative integers with $X^{k+l}\ll T$. Then
if $k\neq l$

$\displaystyle\int_{T}^{2T}\mathcal{P}_{0}(\sigma_{0}+it)^{k}\overline{\mathcal{P}_{0}(\sigma_{0}+it)}^{l}dt\ll
T$

If $k=l$, we have

$\displaystyle\int_{T}^{2T}\left|\mathcal{P}_{0}(\sigma_{0}+it)\right|^{2k}dt=k!T(\log\log
T)^{k}+O_{k}(T(\log\log T)^{k-1+\epsilon})$

###### Proof.

Write $\mathcal{P}_{0}(s)^{k}=\sum_{n}\frac{a(n)\lambda(1,n)}{n^{s}}$ where

$\displaystyle
a_{k}(n)=\begin{cases}\frac{k!}{\alpha_{1}!\cdots\alpha_{r}!}&{\text{ if
}}n=\prod_{j=1}^{r}{p_{j}}^{\alpha_{j}},p_{1}<\ldots<p_{r}<X,\sum_{j=1}^{r}\alpha_{j}=k.\\\
0&{\text{ otherwise}}.\end{cases}$

Therefore,

$\displaystyle\int_{T}^{2T}\mathcal{P}_{0}(\sigma_{0}+it)^{k}\overline{\mathcal{P}_{0}(\sigma_{0}+it)}^{l}dt$
$\displaystyle=$
$\displaystyle\int_{T}^{2T}\sum_{n}\frac{a_{k}(n)\lambda_{f}(n)}{n^{s}}\sum_{n}\frac{a_{l}(n){\bar{\lambda}}_{f}(n)}{n^{\bar{s}}}dt$
$\displaystyle=$
$\displaystyle\sum_{n}a_{k}(n)a_{l}(n)\lambda(1,n)\bar{\lambda}(1,n)\int_{T}^{2T}\frac{1}{n^{2\sigma_{0}}}dt$
$\displaystyle=$ $\displaystyle
T\sum_{n}\frac{a_{k}(n)a_{l}(n)\lambda(1,n)\bar{\lambda}(1,n)}{n^{2\sigma_{0}}}+O\left(\sum_{m\neq
n}\frac{a_{k}(n)a_{l}(m)\lambda(1,n)\bar{\lambda}(1,m)}{(mn)^{\sigma_{0}}}\frac{1}{\log|m/n|}\right)$

Before approximating the above equation let us recall that
$\bar{\lambda}(1,n)=\lambda(n,1)$. The Ramanujan conjecture gives
$\lambda(m,n)\ll(mn)^{\epsilon}$.

Contribution of the diagonal term is 0 because for $k\neq l$
$a_{k}(n)a_{l}(n)$ is 0 by definition. For the off diagonal term we have

$\displaystyle\sum_{m\neq n}a_{k}(n)a_{l}(m)\lambda(1,n)\lambda(m,1)\ll
X^{k+l+\epsilon}\ll T.$

We conclude the first assertion of this lemma. For the second assertion we
have the diagonal term
$\sum_{n}\frac{a_{k}(n)^{2}\lambda(1,n)^{2}}{n^{2\sigma_{0}}}$. It can be seen
easily that for $k=l$ the off-diagonal terms contribution (for $n$ not being
square free) is $O_{k}((\log\log T)^{k-2(\epsilon+1)})$. The diagonal term for
$n$ being square free contribute [8, 9]

$\displaystyle k!\sum_{\begin{subarray}{c}p_{1},\ldots,p_{k}\leq X\\\
{\text{all $p_{j}$'s are
distinct}},(p_{j},q)=1\end{subarray}}\frac{\lambda(1,p)^{2}}{(p_{1}\cdots
p_{k})^{2\sigma_{0}}}=k!\left(\sum_{p\leq
X}\frac{\lambda(1,p)^{2}}{p^{2\sigma_{0}}}\right)^{k}=k!\left(\log\log
T\right)^{k}+O_{k}\left((\log\log T)^{k-1}\right).$

Recall the definition of $X$ we conclude the proof. ∎

Proof of Proposition 2: For any odd $k$ from lemma 1 we have

$\displaystyle\int_{T}^{2T}\left(\Re\left(\mathcal{P}_{0}(\sigma_{0}+it)\right)\right)^{k}dt=$
$\displaystyle\int_{T}^{2T}\frac{1}{2^{k}}\left(\mathcal{P}_{0}(\sigma_{0}+it)+\overline{\mathcal{P}_{0}(\sigma_{0}+it)}\right)^{k}dt$
$\displaystyle=$
$\displaystyle\frac{1}{2^{k}}\sum_{l=0}^{k}\binom{l}{k}\int_{T}^{2T}\left(\mathcal{P}_{0}(\sigma_{0}+it)\right)^{l}\overline{\left(\mathcal{P}_{0}(\sigma_{0}+it)\right)^{k-l}}dt\ll
T.$

Since $k,l$ are non-negative integers, $l=k-l$, does not hold for any odd $k$.
Then for even $k$ we have $l=k-l=k/2$, using the second assertion of Lemma 1
we have

$\displaystyle\frac{1}{T}\int_{T}^{2T}\left(\Re\left(\mathcal{P}_{0}(\sigma_{0}+it)\right)\right)^{k}dt=2^{-k}\binom{k}{k/2}\left(\frac{k}{2}\right)!(\log\log
T)^{\frac{k}{2}}+O_{k}\left((\log\log T)^{\frac{k}{2}-1+\epsilon}\right).$

The above equation matches with the moments of the Gaussian random variable
with mean 0 and variance $\sim\frac{1}{2}\log\log T$, completing the proof. ∎

##  5\. Proof of Proposition 3

We decompose $\mathcal{P}(f,s)$ and $M(f,s)$ to prove this proposition. Since
we will be mostly dealing with $\Lambda_{f}(n)$ here so, there is not much
difference in this proof from the proof given in [5].

We start by decomposing $\mathcal{P}(f,s)$ as $\mathcal{P}_{1}(f,s)$ and
$\mathcal{P}_{2}(f,s)$ as

$\displaystyle\mathcal{P}_{1}(f,s)=\sum_{2\leq n\leq
Y}\frac{\Lambda_{f}(n)}{n^{s}\log n}$
$\displaystyle\mathcal{P}_{2}(f,s)=\sum_{Y<n\leq
X}\frac{\Lambda_{f}(n)}{n^{s}\log n}$

In this proposition we are connecting the auxiliary series $\mathcal{P}(f,s)$
with the mollifier $M(f,s)$. For that we set

$\displaystyle\mathcal{M}_{1}(f,s)=\sum_{0\leq k\leq 100\log\log
T}\frac{(-1)^{k}}{k!}\mathcal{P}_{1}(s)^{k}$
$\displaystyle\mathcal{M}_{2}(f,s)=\sum_{0\leq k\leq 100\log\log\log
T}\frac{(-1)^{k}}{k!}\mathcal{P}_{2}(s)^{k}$

Evidently, we set $\mathcal{M}(f,s)$ to find the connection between
$\mathcal{P}(f,s)$ and $M(f,s)$. First we connect $\mathcal{M}(f,s)$ with
$\mathcal{P}(f,s)$.

######  Lemma 2.

For $T\leq t\leq 2T$ we have

(4) $\displaystyle|\mathcal{P}_{1}(\sigma_{0}+it)|\leq\log\log T$
$\displaystyle|\mathcal{P}_{2}(\sigma_{0}+it)|\leq\log\log\log T.$

except perhaps for a set of measure $\frac{T}{\log\log\log T}$.

(5)
$\displaystyle\mathcal{M}_{1}(\sigma_{0}+it)=\exp(-\mathcal{P}_{1}(\sigma_{0}+it))(1+O(\log
T)^{-99})$
$\displaystyle\mathcal{M}_{2}(\sigma_{0}+it)=\exp(-\mathcal{P}_{2}(\sigma_{0}+it))(1+O(\log\log
T)^{-99})$

###### Proof.

Let us start with the integration

$\displaystyle\int_{T}^{2T}|\mathcal{P}_{1}(\sigma_{0}+it)|^{2}dt$
$\displaystyle\ll\sum_{2\leq n_{1}=n_{2}\leq
Y}\frac{\Lambda_{f}{(n_{1})}\Lambda_{f}{(n_{2})}}{(n_{1}n_{2})^{\sigma_{0}}\log
n_{1}\log n_{2}}+\sum_{2\leq n_{1}\neq n_{2}\leq
Y}\frac{\Lambda_{f}{(n_{1})}\Lambda_{f}{(n_{2})}}{(n_{1}n_{2})^{\sigma_{0}}\log
n_{1}\log n_{2}}\sqrt{n_{1}n_{2}}$ $\displaystyle\ll T\log\log T$

Similarly, we have

$\displaystyle\int_{T}^{2T}|\mathcal{P}_{2}(\sigma_{0}+it)|^{2}dt\ll
T\log\log\log T$

and (4) holds.

If $|z|\leq k$ then using Stirling’s formula it is straight forward to check
that

$\displaystyle\left|e^{z}-\sum_{0\leq K\leq 100k}\frac{z^{k}}{k!}\right|\leq
e^{-99k}.$

Take $z=-\mathcal{P}_{1}(\sigma_{0}+it)$ and $k=\log\log T$ and (5) holds. ∎

Now it is time to connect $\mathcal{M}(f,s)$ with $M(f,s)$ to complete proof
of this proposition.

As we have decomposed $\mathcal{P}(f,s)$ similarly, we decompose $M(f,s)$. But
we need to decompose $a(n)$ first coming from the definition of $M(f,s)$.

$\displaystyle a_{1}(n)=\begin{cases}1&\text{if $n$ has at most $100\log\log
T$ prime factors with all $p\leq Y$ }\\\ 0&\text{otherwise }.\end{cases}$
$\displaystyle a_{2}(n)=\begin{cases}1&\text{if $n$ has at most
$100\log\log\log T$ prime factors with all $Y<p\leq X$ }\\\ 0&\text{otherwise
}.\end{cases}$

Therefore we have,

$\displaystyle M(f,s)=M_{1}(s)M_{2}(s)$ $\displaystyle
M_{1}(f,s)=\sum_{n}\frac{\mu(n)a_{1}(n)\lambda(1,n)}{n^{s}}$ $\displaystyle
M_{2}(f,s)=\sum_{n}\frac{\mu(n)a_{2}(n)\lambda(1,n)}{n^{s}}$

Next lemma connect the mollifier with $\mathcal{M}(f,s)$.

######  Lemma 3.

We have

$\displaystyle\int_{T}^{2T}|\mathcal{M}_{1}(\sigma_{0}+it)-M_{1}(\sigma_{0}+it)|^{2}dt\ll
T(\log T)^{-60}$
$\displaystyle\int_{T}^{2T}|\mathcal{M}_{2}(\sigma_{0}+it)-M_{2}(\sigma_{0}+it)|^{2}dt\ll
T(\log\log T)^{-60}$

###### Proof.

Define

$\displaystyle\mathcal{M}_{1}(f,s)=\sum_{n}\frac{b(n)\lambda(1,n)}{n^{s}}.$

where $b(n)$ satisfies the following properties.

  1. (1)

$|b(n)|\leq 1$ for all $n$.

  2. (2)

$b(n)=0$ unless $n\leq Y^{100\log\log T}$ has only prime factors below $Y$.

  3. (3)

$b(n)=\mu(n)a_{1}(n)$ unless $\Omega(n)>100\log\log T$ or, $p\leq Y$ s.t
$p^{k}|n$ with $p^{k}>Y$.

Set $c(n)=(b(n)-\mu(n)a_{1}(n))\lambda(1,n)$, then we have

$\displaystyle\int_{T}^{2T}$
$\displaystyle|\mathcal{M}_{1}(\sigma_{0}+it)-M_{1}(\sigma_{0}+it)|^{2}dt$
$\displaystyle\ll\sum_{n_{1},n_{2}}\frac{|c(n_{1})c(n_{2})|}{(n_{1}n_{2})}\min\left(T,\frac{1}{|\log(n_{1}/n_{2})|}\right)$
$\displaystyle\ll T(\log T)^{-60}$

We note that our $a_{1}(n),a_{2}(n),b(n)$ are exactly the same as Radziwill
and Soundarajan’s method. The only difference is that now we have a further
twisting by $|\lambda(1,n)|$. If we use the Ramanujan conjecture we can obtain
the above bound. ∎

##  6\. Proof of Proposition 4

First, we evaluate the explicit formula and the approximate functional
equation for $L(f,s)$ associated with the Hecke-Maass cusp form $f$ then we
prove the proposition. Recall the functional equation of $L$-functions

$\displaystyle\Lambda(f,s)=G(f,s)L(f,s)=G(\bar{f},1-s)L(\bar{f},1-s)=\bar{\Lambda}(f,s)$

Here $\bar{f}$ is the dual form have the Langland’s parameter
$(-\alpha_{3},-\alpha_{2},-\alpha_{1})$.

######  Lemma 4.

For $c>0$ define,

$\displaystyle I(f,s)$ $\displaystyle\coloneqq\frac{1}{2\pi
i}\int_{(c)}\Lambda(f,z+s,)\Lambda(\bar{f},z+\bar{s})e^{z^{2}}\frac{dz}{z},$
$\displaystyle I(\bar{f},1-s)$ $\displaystyle\coloneqq\frac{1}{2\pi
i}\int_{(c)}\Lambda(\bar{f},-z+(1-s))\Lambda(f,-z+(1-\bar{s}))e^{z^{2}}\frac{dz}{z}.$

where $(c)$ denotes the line from $c-i\infty$ to $c+i\infty$. Then we have

$\displaystyle|L(f,s)|^{2}=\frac{1}{|G(f,s)|^{2}}\left(I(f,s)+I(\bar{f},1-s)\right).$

The proof of this lemma is similar to the proof of Lemma 4 given in [5]. The
only modification we need is the change of the sign of the $L$-function and
instead of a cusp form we have the Hecke-Maass cusp form here. The behaviour
of the dual of $f$ is important to prove this lemma but the proof technique
and the mechanism are the same. Theorem 5.11 of [7] is a standard reference
for the proof of this lemma.

######  Lemma 5.

Let $h,k$ be non-negative integers with $h,k\leq T$. Then for any
$\frac{1}{2}<\sigma\leq 1$,

$\displaystyle\int_{T}^{2T}\left(\frac{h}{k}\right)^{it}|L(f,\sigma+it)|^{2}=$
$\displaystyle\int_{T}^{2T}\left(\zeta(2\sigma)\left(\frac{(h,k)^{2}}{hk}\right)^{1-\sigma}+\left(\frac{qt}{2\pi}\right)^{1-2\sigma}\zeta(2-2\sigma)\left(\frac{(h,k)}{hk}\right)^{\sigma}\right)dt$
$\displaystyle+O(T^{1-\sigma+\epsilon}\min\\{h,k\\}).$

###### Proof.

Denote $s=\sigma+it$ with $T\leq t\leq 2T$ and $\frac{1}{2}\leq\sigma\leq 1$.
if $z$ is a complex number with real part $c=1-\sigma+\frac{1}{\log T}$ then

$\displaystyle\frac{G(f,z+s)G(\bar{f},z+\bar{s})}{|G(f.s)|^{2}}=\left(\frac{qt}{2\pi}\right)^{z}\left(1+O\left(\frac{|z|}{T}\right)\right).$

Therefore,

$\displaystyle\frac{I(f,s)}{|G(f,s)|^{2}}=\frac{1}{2\pi
i}\int_{\left(1-\sigma+\frac{1}{\log
T}\right)}\left(\frac{tq}{2\pi}L(z+s,f)L(z+\bar{s},\bar{f})e^{z^{2}}\frac{dz}{z}\right)+O(T^{1-\sigma+\epsilon})$

We are in the region of absolute convergence so we can interchange the
summations and integrals.

$\displaystyle\int_{T}^{2T}\left(\frac{h}{k}\right)^{it}\frac{I(f,s)}{|G(f,s)|^{2}}dt$
$\displaystyle=$ $\displaystyle\frac{1}{2\pi
i}\int_{\left(1-\sigma+\frac{1}{\log
T}\right)}\frac{e^{z^{2}}}{z}\sum_{m,n=1}^{\infty}\frac{\lambda(1,n)\bar{\lambda}(1,m)}{(mn)^{z+\sigma}}\left(\int_{T}^{2T}\left(\frac{h}{k}\right)^{it}\left(\frac{tq}{2\pi}\right)^{z}dt\right)dz+O(T^{1-\sigma+\epsilon})$

The diagonal term can be counted by taking $hm=kn$ and the off-diagonal term
can be counted by taking $hm\neq kn$. To calculate the diagonal term we
parametrized $m$ and $n$ as

$\displaystyle m=\frac{Nk}{(h,k)},\quad n=\frac{Nh}{(h,k)}.$

First we calculate the sum

(6)
$\displaystyle\sum_{m,n=1}^{\infty}\frac{\lambda(1,n)\bar{\lambda}(1,m)}{(mn)^{z+\sigma}}$

We know that $\bar{\lambda}(1,m)=\lambda(m,1)$. We can write
$\lambda(1,n)\lambda(m,1)$ as

$\displaystyle\lambda(1,n)\lambda(m,1)=\sum_{d|(m,n)}\lambda\left(\frac{m}{d},\frac{n}{d}\right)$

Applying Ramanujan conjecture, by interchanging the variables in (6) we obtain

$\displaystyle\left(\frac{(h,k)^{2}}{hk}\right)^{z+\sigma-1}\sum_{N=1}^{\infty}\left(\frac{1}{N}\right)^{2(z+\sigma-1)}$
$\displaystyle=$
$\displaystyle\left(\frac{(h,k)^{2}}{hk}\right)^{z+\sigma-1}\zeta(2(z+\sigma-1)).$

So the diagonal term become

$\displaystyle\frac{1}{2\pi i}\int_{(1-\sigma+1/\log
T)}\frac{e^{z^{2}}}{z}\zeta{(2(z+\sigma-1))}\left(\frac{(h,k)^{2}}{hk}\right)^{z+\sigma-1}\left(\int_{T}^{2T}\left(\frac{tq}{2\pi}\right)^{z}dt\right)dz.$

For the off-diagonal terms the inner integral over $t$ may be bounded by $\ll
T^{1-\sigma}\min(T,1/|\log(hm/kn)|)$, and therefore these contribute

(7) $\displaystyle\ll T^{1-\sigma}\sum_{\begin{subarray}{c}m,n=1\\\ hm\neq
kn\end{subarray}}^{\infty}\frac{1}{(mn)^{1+1/\log
T}}\min\left(T,\frac{1}{|\log(hm/kn)|}\right)\ll\min(h,k)T^{1-\sigma+\epsilon}.$

Combining (6) and (7) we obtain

$\displaystyle\int_{T}^{2T}\left(\frac{h}{k}\right)^{it}\frac{I(f,s)}{|G(f,s)|^{2}}dt=\frac{1}{2\pi
i}\int_{(1-\sigma+1/\log
T)}\frac{e^{z^{2}}}{z}\zeta{(2(z+\sigma-1))}\left(\frac{(h,k)^{2}}{hk}\right)^{z+\sigma-1}\left(\int_{T}^{2T}\left(\frac{tq}{2\pi}\right)^{z}dt\right)dz$
$\displaystyle+O(\min(h,k)T^{1-\sigma+\epsilon}).$

By a similar argument we have

$\displaystyle\int_{T}^{2T}\left(\frac{h}{k}\right)^{it}\frac{I(f,1-s)}{|G(f,s)|^{2}}dt=\frac{1}{2\pi
i}\int_{(\sigma+1/\log
T)}\frac{e^{z^{2}}}{z}\zeta{(2(z-\sigma))}\left(\frac{(h,k)^{2}}{hk}\right)^{z-\sigma}\left(\int_{T}^{2T}\left(\frac{tq}{2\pi}\right)^{z}dt\right)dz$
$\displaystyle+O(\min(h,k)T^{1-\sigma+\epsilon}).$

With a suitable change of variable, we can obtain the main term

$\displaystyle\frac{1}{2\pi i}\int_{(1+1/\log
T)}\zeta(2z)\left(\frac{(h,k)^{2}}{hk}\right)^{z}\left(\int_{T}^{2T}\left(\frac{tq}{2\pi}\right)^{z-\sigma}dt\right)\left(\frac{e^{(z+\sigma-1)^{2}}}{z+\sigma-1}+\frac{e^{(z-\sigma)^{2}}}{z-\sigma}\right)dz$

by moving the line of integration to the left we obtain the main term of the
lemma as the residues of the poles at $z=\sigma$ and $z=1-\sigma$, which
completes the proof. ∎

The approximate functional equation (see Theorem 5.13 of [7]) gives,

$\displaystyle
L(f,s)=\sum_{n}\frac{\lambda(1,n)}{n^{s}}V\left(\frac{n}{X}\right)+\sum_{n}\frac{\bar{\lambda}(1,n)}{n^{1-s}}V(nX)$

where the function $V$ is given by

$\displaystyle V(y)=\frac{1}{2\pi
i}\int_{(3)}y^{-u}\frac{G(f,s+u)}{G(f,s)}\frac{du}{u}$

For small $y$ we can shift the contour to the left up to
$\sigma_{0}=\frac{1}{2}+\varepsilon$. We get the asymptotic expansion

$\displaystyle V(y)=1+O(y^{1/2+\varepsilon}).$

Now are ready to prove the proposition.

Proof of Proposition 4: With the help of the approximate functional equation
of the $L$-function we can write

$\displaystyle\int_{T}^{2T}L(f,\sigma_{0}+it)M(f,\sigma_{0}+it)dt$
$\displaystyle=$ $\displaystyle\sum_{n\leq
T}\frac{\lambda(1,n)}{n^{\sigma}}\sum_{m}\frac{a(m)\mu(m)}{(m)^{\sigma}}\int_{T}^{2T}(mn)^{-it}+O(T^{1/2+\epsilon})dt$
$\displaystyle=$ $\displaystyle T+O(T^{1/2+\epsilon}).$

Expanding the equation given in Proposition 4 we get,

$\displaystyle\int_{T}^{2T}|1-L(f,\sigma_{0}+it)M(f,\sigma_{0}+it)|^{2}dt$
$\displaystyle=$
$\displaystyle\int_{T}^{2T}|L(f,\sigma_{0}+it)M(f,\sigma_{0}+it)|^{2}dt-T+O(T^{1/2+\epsilon}).$

Evaluating the above integral is our next step. By Lemma 5, we have

(8) $\displaystyle\int_{T}^{2T}|1-L(f,\sigma_{0}+it)M(f,\sigma_{0}+it)|^{2}dt$
$\displaystyle=$
$\displaystyle\sum_{h,k}\frac{\mu(h)\mu(k)a(h)a(k)}{(hk)^{\sigma_{0}}}\int_{T}^{2T}\left(\frac{h}{k}\right)^{it}|L(f,\sigma_{0}+it)|^{2}dt-T+O(T^{1/2+\epsilon}).$

To conclude the proof of proposition 4 we have to show that the first term of
(8) is $\sim T$. Recalling the definition of $a(n)$, we know that $a(n)\leq 1$
for any $n\leq T^{\epsilon}$ and 0 otherwise.

Now the integral in (8) equals to $-T+O(T^{1/2+\epsilon})+o(T)$

$\displaystyle+\sum_{h,k}\frac{\mu(h)\mu(k)a(h)a(k)}{(hk)^{\sigma_{0}}}\int_{T}^{2T}\zeta(2\sigma_{0})\left(\frac{(h,k)^{2}}{hk}\right)^{1-\sigma_{0}}dt$
$\displaystyle+$
$\displaystyle\sum_{h,k}\frac{\mu(h)\mu(k)a(h)a(k)}{(hk)^{\sigma_{0}}}\int_{T}^{2T}\left(\frac{qt}{2\pi}\right)^{1-2\sigma_{0}}\zeta(2-2\sigma_{0})\left(\frac{(h,k)^{2}}{hk}\right)^{\sigma_{0}}dt$

The first term is the main term which is $\sim T$, as argued in [3]. The
contribution of the second term is $o(T)$, which completes the proof. ∎

##  7\. Independence of the automorphic $L$-functions

In this section, we prove Theorem 2 and 3 by adapting P. Hsu’s et al’s
technique. These two theorems state about the independence of $GL(2)$ and
$GL(3)$ $L$-functions respectively. We prove Theorem 2 here and the result
follows similarly for Theorem 3 with minor modifications.

######  Theorem 4.

Let $g_{1}$ and $g_{2}$ be distinct primitive holomorphic cusp forms. Let $V$
be a fixed real number. As $T\to\infty$, for $a_{1},a_{2}\in\mathbb{R}$, we
have,

$\displaystyle\frac{1}{T}meas\\{t\in[T,2T]:\log|\mathcal{L}_{a_{1},a_{2}}(\frac{1}{2}+it)|\geq
V\sqrt{\frac{{a_{1}}^{2}+{a_{2}}^{2}}{2}\log\log
T}\\}\sim\frac{1}{\sqrt{2\pi}}\int_{V}^{\infty}e^{-\frac{x^{2}}{2}}dx,$

where

$\displaystyle\mathcal{L}_{a_{1},a_{2}}(s)=\mathcal{L}(g_{1},g_{2},s;a_{1},a_{2})\coloneqq|L(g_{1},s)|^{a_{1}}|L(g_{2},s)|^{a_{2}}.$

Observe that Theorem 4, can be proved with a similar argument of Theorem 1. So
we define the auxiliary series,

$\displaystyle\mathcal{P}_{a_{1},a_{2},0}(s)=\mathcal{P}_{0}(g_{1},g_{2},s;X)\coloneqq\sum_{p\leq
X}\frac{a_{1}\lambda_{g_{1}}(p)+a_{2}\lambda_{g_{2}}(p)}{p^{s}},$

where $\lambda_{g_{1}},\lambda_{g_{2}}$ are the Fourier coefficients of the
primitive holomorphic cusp forms $g_{1},g_{2}$ respectively. Similarly to the
proof of Theorem 1, we need moment calculation for
$\mathcal{P}_{a_{1},a_{2},0}(s)$.

######  Lemma 6.

Let $g_{1},g_{2}$ be distinct primitive holomorphic cusp forms. Suppose, $k,l$
are non-negative integers with $X^{k+l}\ll T$. Then for any real numbers
$a_{1},a_{2}$, we have,

$\displaystyle\int_{T}^{2T}\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)^{k}\overline{\mathcal{P}_{a_{1},a_{2},0}(s)^{l}}dt\ll
T,$

for $k\neq l$, and

$\displaystyle\int_{T}^{2T}|\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)|^{2k}dt=k!T((a_{1}^{2}+a_{2}^{2})\log\log
T)^{k}+O_{k}(T(\log\log T)^{k-1+\epsilon})$

###### Proof.

Set $\psi(p)=\psi_{a_{1},a_{2}}(p)\coloneqq
a_{1}\lambda_{g_{1}}(p)+a_{2}\lambda_{g_{2}}(p)$ and
$\Psi_{k}(n)\coloneqq\prod_{j=1}^{r}\psi(p_{j})^{\alpha_{j}}$, where
$n=p_{1}^{\alpha_{1}}\cdots p_{r}^{\alpha_{r}}$ with
$\alpha_{1}+\cdots+\alpha_{r}=k$, we can write

$\displaystyle\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)^{k}=\sum_{n}\frac{a_{k}(n)\Psi_{k}(n)}{n^{\sigma_{0}+it}},$

where $a_{k}(n)$ is same as defined before in the proof of Proposition 4.
Hence, with a similar argument from Lemma 1 we have,

$\displaystyle\int_{T}^{2T}\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)^{k}\overline{\mathcal{P}_{a_{1},a_{2},0}(s)^{l}}dt$
$\displaystyle=$ $\displaystyle
T\sum_{n}\frac{a_{k}(n)a_{l}(n)\Psi_{k}(n)\overline{\Psi_{l}(n)}}{n^{2\sigma_{0}}}+O\left(\sum_{n\neq
m}\frac{a_{k}(n)a_{l}(m)|\Psi_{k}(n)\overline{\Psi_{l}(m)}|}{(nm)^{\sigma_{0}-\frac{1}{2}}}\right).$

From the definition of $\Psi_{k}(n)$ that for
$n=\prod_{j=1}^{r}p_{j}^{\alpha_{j}}$ we have

$\displaystyle|\Psi_{k}(n)|=\prod_{j=1}^{r}|\psi(p_{j})|^{\alpha_{j}}\leq\prod_{j=1}^{r}(|a_{1}\lambda_{g_{1}}(p_{j})|+|a_{2}\lambda_{g_{2}}(p_{j})|)=X(|a_{1}|+|a_{2}|)^{k}.$

Thus, the big-$O$ term is at most

$\displaystyle\sum_{n\neq m}a_{k}(n)a_{l}(m)X^{2}(|a_{1}|+|a_{2}|)^{k+l}\ll
X^{(k+1)+(l+1)}\ll T.$

By the definition $a_{k}(n)a_{l}(n)=0$ if $k\neq l$, we conclude the first
assertion of the lemma. It remains to prove the lemma for $k=l$.

For $n=\prod_{j=1}^{r}p_{j}^{\alpha_{j}}$, we write

$\displaystyle\Psi_{k}(n)\overline{\Psi_{k}(n)}=\prod_{j=1}^{r}\left(a_{1}^{2}+a_{2}^{2}+a_{1}a_{2}\lambda_{g}(p_{j})+a_{1}a_{2}\bar{\lambda}_{g}(p_{j})\right)^{\alpha_{j}},$

where $\lambda_{g}\coloneqq\lambda_{g_{1}}\bar{\lambda}_{g_{2}}$. From the
above equation and the fact that $\sum_{j}\alpha_{j}=k$, for the non-square
free $n$ we have

$\displaystyle\left|\sum_{n\text{ non square
free}}\frac{a_{k}(n)a_{k}(n)\Psi_{k}(n)\overline{\Psi_{k}(n)}}{n^{2\sigma_{0}}}\right|\leq(|a_{1}|+|a_{2}|)^{2k}\sum_{n}\frac{a_{k}(n)^{2}}{n^{2\sigma_{0}}}$

Consider the each term $\prod_{j=1}^{r}a_{1}a_{2}\lambda_{f}(p_{j})$. From the
fundamental theorem of arithmetic each $p_{j}$’s are distinct primes. Since,
$\lambda_{g}(n)$ is a real valued function and gcd of each pair
$(p_{j},p_{i})$ equals to 1 for $i\neq j$ we can write
$\prod_{j=1}^{r}|(a_{1}a_{2}\lambda_{g}(p_{j}))|^{\alpha_{j}}=|a_{1}a_{2}|^{k}\prod_{j=1}^{r}|\lambda_{g}(p_{j})|^{\alpha_{j}}=|a_{1}a_{2}|^{k}(p_{1}^{\alpha_{1}}\cdots
p_{j}^{\alpha_{j}})^{\delta}=|a_{1}a_{2}|^{k}n^{\delta}$, where $\delta$ is
the appropriate constant satisfying $|\lambda_{g}(n)|\leq n^{\delta}$.

Therefore as argued in [4] the non-square free term contributes a quantity of
order $O((\log\log T)^{k-1})$. For $n$ square free we can express

$\displaystyle(a_{1}+a_{2})^{2k}k!\sum_{\begin{subarray}{c}p_{1},\ldots,p_{k}\leq
X\\\ \text{all $p_{j}$'s are distinct and each pair}\\\
\text{$(p_{j},N_{1})=1$ and
$(p_{j},N_{2})=1$}\end{subarray}}\frac{\lambda_{g}(p)^{2}}{(p_{1}\cdots
p_{k})^{2\sigma_{0}}}$ $\displaystyle=$
$\displaystyle(a_{1}+a_{2})^{2k}k!\left(\sum_{p\leq
X}\frac{\lambda_{g}(p)^{2}}{p^{2\sigma_{0}}}\right)^{k}$ $\displaystyle=$
$\displaystyle k!((a_{1}+a_{2})\log\log T)^{2k},$

where $N_{1},N_{2}$ are the conductor for $L(g_{1},s)$ and $L(g_{2},s)$
respectively. Hence, the second assertion of the lemma follows. ∎

Now we are ready to prove Theorem 4.

Proof of Theorem 4: Set $Z_{j}=\log|L(g_{j},\sigma_{0}+it)|$ for $j=1,2$.
Observe that $\log|\mathcal{L}_{a_{1},a_{2}(s)}|=a_{1}Z_{1}+a_{2}Z_{2}$. Like
an similar argument of Proposition 1 we have

(9)
$\displaystyle\int_{t-1}^{t+1}\left|\log|\mathcal{L}_{a_{1},a_{2}(\frac{1}{2}+iy)}|-|\mathcal{L}_{a_{1},a_{2}(\sigma_{0}+iy)}|\right|dy\ll\left(\sigma_{0}-\frac{1}{2}\right)\log
T.$

By a similar argument of the proof of Proposition 1 the above equation is
true. However, we can study the function
$\log|\mathcal{L}_{a_{1},a_{2}}(\sigma_{0}+it)|$ away from the critical line
$\Re(s)=\frac{1}{2}$.

Define $\mathcal{P}_{a_{1},a_{2}}(s)\coloneqq
a_{1}\mathcal{P}(g_{1},s)+a_{2}\mathcal{P}(g_{2},s)$ where
$\mathcal{P}(g_{j},s)$’s are defined as the auxiliary series
$\mathcal{P}(f,s)$. The difference is we are dealing with holomorphic cusp
forms instead of Hecke-Maass cusp forms. Similarly, the contribution of the
higher order terms of the primes is at most $O(1)$. For the terms involving
$p^{2}$ we can consider $\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)$, as
defined. Now by Lemma 6 for $X^{k}\ll T$ and odd $k$,

$\displaystyle\int_{T}^{2T}(\Re(\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)))^{k}dt=\frac{1}{2^{k}}\sum_{l=0}^{k}\binom{k}{l}\int_{T}^{2T}\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)^{l}\overline{\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)^{k-l}}dt\ll
T.$

Also, by Lemma 6, for even $k$,

$\displaystyle\int_{T}^{2T}(\Re(\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)))^{k}dt=2^{-k}\binom{k}{k/2}\left(\frac{k}{2}\right)!\left(\frac{a_{1}^{2}+a_{2}^{2}}{2}\log\log
T\right)^{\frac{k}{2}}+O_{k}((\log\log T)^{\frac{k}{2}-1+\epsilon}).$

The proves that the real part of $\mathcal{P}_{a_{1},a_{2}}(\sigma_{0}+it)$
has an approximately normal distribution with mean 0 and variance
$\frac{a_{1}^{2}+a_{2}^{2}}{2}\log\log T$.

Now it remains to connect $\mathcal{L}_{a_{1},a_{2}}(\sigma_{0}+it)$ with
$\mathcal{P}_{a_{1},a_{2}}(\sigma_{0}+it)$. From the similar argument of
Proposition 3 and Proposition 4,
$|L(g_{j},\sigma_{0}+it)|=(1+o(1))\exp(\Re(\mathcal{P}_{a_{1},a_{2}}(\sigma_{0}+it)))$,
except for a possible set of measure $o(T)$, for each $j$. Since,
$\mathcal{L}_{a_{1},a_{2}}(s)=|L(g_{1},s)|^{a_{1}}|L(g_{2},s)|^{a_{2}}$ and
$\mathcal{P}_{a_{1},a_{2}}(s)=a_{1}\mathcal{P}(g_{1},s)+a_{2}\mathcal{P}(g_{2},s)$,
we have

$\displaystyle\mathcal{L}_{a_{1},a_{2}}(\sigma_{0}+it)=(1+o(1))\exp(\Re(\mathcal{P}_{a_{1},a_{2}}(\sigma_{0}+it))).$

It shows $\log|\mathcal{L}_{a_{1},a_{2}}(\sigma_{0}+it)|$ has Gaussian
distribution with mean 0 and variance $\frac{a_{1}^{2}+a_{2}^{2}}{2}\log\log
T$, combining with (9) we conclude the proof. ∎

Proof of Theorem 2: For the sake of the calculation we are proving this
theorem for two distinct holomorphic cusp form $g_{1}$ and $g_{2}$ instead of
$(g_{j})_{j=1}^{n}$. Let
$\mathbf{X}\coloneqq(X_{j})_{j=1}^{2}=\log|L(g_{j},\frac{1}{2}+it)|_{j=1}^{2}$.

We know that $(X_{j})_{j=1}^{n}$ is an $n$-variate normal distribution iff
linear combination of $X_{j}$’s is a normal distribution. Furthermore, if
$(X_{k},X_{l})$ is a bi-variate normal distribution the $X_{k}$ and $X_{l}$
are independent iff they are uncorrelated.

Combining this fact with Theorem 4, we have $\mathbf{X}$ is an approximate bi-
variate normal distribution. Finally, from Theorem 4 with $a_{1}=a_{2}=1$, we
deduce that $X_{1},X_{2}$ are uncorrelated because

$\displaystyle\mathrm{Var}(X_{1})+\mathrm{Var}(X_{2})=\mathrm{Var}(X_{1}+X_{2}).$

Hence, $X_{1}$ and $X_{2}$ are independent, completes the proof. ∎

Proof of Theorem 3, is similar to the proof of Theorem 2. We just need to take
care of the coefficients of the Hecke-Maass cusp forms.

##  8\. Concluding Remarks

In this paper we have proved the Selberg’s central limit theorem for $GL(3)$
$L$-function associated with Hecke-Mass cusp form. We have used the method
developed in [3]. Since we have to deal with the coefficient of the Hecke-Mass
cusp form we need modifications from the proof of [5]. To prove Theorem 1 we
have assumed the Ramanujan conjecture which gives the upper-bound for the
normalized Fourier coefficient for the Hecke-Mass cusp form.

Moreover, we have shown the independence of the automorphic $L$-functions of
degree 2 and 3. The proof of this result can be derived from the proof of
Theorem 1. One can extend this result for the $L$-functions associated with
distinct cusp forms (or Maass forms) at distinct point on the critical line.

We have proved the theorems for level aspect. It is possible to prove these
results for $q$-aspect as well.

## Acknowledgement

I would like to express my gratitude to my primary supervisor, Prof. Gihan
Marasingha, who guided me throughout this project.

## References

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