On certain multiple Dirichlet series

TL;DR

This paper investigates the analytic properties of a specific multiple Dirichlet series linked to binary cubic forms, contributing to the understanding of their mathematical structure.

Contribution

It introduces new insights into the analytic behavior of multiple Dirichlet series related to binary cubic forms.

Findings

Analysis of the series' convergence properties

Identification of functional equations

Insights into the distribution of values

Abstract

In this paper we study the analytic properties of a multiple Dirichlet series associated to the prehomogeneous vector space of binary cubic forms.

Equations

$$f(x,y)=a{x}^3+b{x}^{2}y+cx{y}^2+d{y}^3.$$

$$r_1(f)$$

$$r_2(f)$$

$$r_3(f)$$

$$r_4(f)$$

$$\left\{(a,b,c,d)\in {\mathbb{Z}}^4:a>0,0\le b\le 3a-1,b^2-3ac<0\right\}.$$

$$\sum _{f\in {\Gamma }_{\infty }\setminus X^{+}}\frac{1}{|r_1(f){|}^{s_1}|r_2(f){|}^{s_2}|r_3(f){|}^{s_3}|r_4(f){|}^{s_4}}.$$

$$Z(s_1,s_2)=\sum _{\genfrac{}{}{0ex}{}{f\in ({\Gamma }_{\infty }\setminus X^{+})/\sim }{r_2(f)\text{ odd, sq. free}}}\frac{1}{|r_1(f){|}^{s_1}|r_2(f){|}^{s_2}}.$$

$$\sum _{m_1,\cdots ,m_n\in \mathbb{N}}\frac{a(m_1,\cdots ,m_n)}{m_1^{s_1}{m}_2^{s_2}\cdots m_n^{s_n}}.$$

$$r_1(T)$$

$$r_2(T)$$

$$r_3(T)$$

$$T=\left[\begin{array}{ccc}2p& s& t\\ s& 2q& u\\ t& u& 2r\end{array}\right]\in X_3^{+}(\mathbb{Z}).$$

$$Z(s_1,s_2,s_3)=\sum _{T\in P\backslash X_3^{+}(\mathbb{Z})}\frac{1}{|r_1(T){|}^{s_1}|r_2(T){|}^{s_2}|r_3(T){|}^{s_3}}.$$

$$Z(s_1,s_2)=\sum _{\begin{array}{c}a,b,c\in \mathbb{Z}\\ a>0\\ 0\le b\le 3a-1\\ b^2-3ac<0,sq.free,odd\end{array}}\frac{1}{a^{s_1}(-b^2+3ac{)}^{s_2}}.$$

$$Z(s_1,s_2)=\sum _{\begin{array}{c}m,n\in \mathbb{N}\\ nsq.free,odd\end{array}}\frac{C(3m,-n)}{m^{s_1}{n}^{s_2}},$$

$$C(m,n)=\#\{x\mathrm{mod}m:x^2\equiv n\mathrm{mod}m\}.$$

$$Z(s_1,s_2)=\sum _{\begin{array}{c}m,n\in \mathbb{N}\\ nsq.free,odd\end{array}}\frac{a(m,n)}{m^{s_1}{n}^{s_2}},$$

$$\#\{x\mathrm{mod}3m:x^2\equiv -n\mathrm{mod}3m\}.$$

$$C(2^{\alpha },n)=\{\begin{array}{ll}1& \alpha =1,\\ 2& \alpha =2,n\equiv 1\mathrm{mod}4,\\ 4& \alpha \ge 3,n\equiv 1\mathrm{mod}8,\\ 0& \text{mbox}otherwise.\end{array}$$

$$C(p^{\alpha },p^r{n}_0)=\{\begin{array}{ll}{p}^{\lfloor \frac{\alpha }{2}\rfloor }& r\ge \alpha ,\\ p^{\frac{r}{2}}C(p^{\alpha -r},n_0)& r<\alpha ,r\text{mbox}even,\\ 0& \text{mbox}otherwise.\end{array}$$

$$C(m,n)=\prod _{i}C(p_i^{{\alpha }_i},n),$$

$$x$$

$$=p^k\left(a_k+a_{k+1}p+\cdots +a_{\alpha -1}{p}^{\alpha -k-1}\right)$$

$$x^2$$

$$y^2\equiv n_0\mathrm{mod}{p}^{\alpha -r}.$$

$$Z(s_1,s_2)=\sum _{\begin{array}{c}n=1\\ nsq.free,odd\end{array}}^{\infty }\frac{1}{n^{s_2}}\sum _{m=1}^{\infty }\frac{C(3m,-n)}{m^{s_1}}.$$

$$Z_{n,p}(s)=\{\begin{array}{ll}\sum _{k=0}^{\infty }\frac{C(p^k,-n)}{p^{ks}}& p\ne 3,\\ \sum _{k=0}^{\infty }\frac{C(3^{k+1},-n)}{3^{ks}}& p=3.\end{array}$$

$$Z_{n,p}(s)$$

$$=\frac{1+p^{-s}\left(\frac{-n}{p}\right)}{1-p^{-s}}\cdot \frac{1-p^{-s}\left(\frac{-n}{p}\right)}{1-p^{-s}\left(\frac{-n}{p}\right)}=\frac{1-p^{-2s}}{1-p^{-s}}\cdot \frac{1}{1-p^{-s}\left(\frac{-n}{p}\right)}.$$

$$Z_{n,2}(s)=\sum _{k=0}^{\infty }\frac{C(2^k,-n)}{2^{ks}}=1+\frac{1}{2^s}+\sum _{k=2}^{\infty }\frac{C(2^k,-n)}{2^{ks}}=1+\frac{1}{2^s}=\frac{1-2^{-2s}}{1-2^{-s}}.$$

$$Z_{n,2}(s)=\sum _{k=0}^{\infty }\frac{C(2^k,-n)}{2^{ks}}=1+\frac{1}{2^s}+\frac{2}{2^{2s}}+\sum _{k=3}^{\infty }\frac{C(2^k,-n)}{2^{ks}}=1+\frac{1}{2^s}+\frac{2}{2^{2s}}.$$

$$Z_{n,2}(s)=\sum _{k=0}^{\infty }\frac{C(2^k,-n)}{2^{ks}}=1+\frac{1}{2^s}+\sum _{k=2}^{\infty }\frac{C(2^k,-n)}{2^{ks}}=1+\frac{1}{2^s}=\frac{1-2^{-2s}}{1-2^{-s}}.$$

$$Z_{n,2}(s)$$

Full text

1. 1 Introduction

2. 2 Proof of the main theorem

3. 2.1 Reduction of the Multiple Dirichlet Series

4. 2.1.1 Local Euler Factors

5. 2.2 Domain of Meromorphy

6. 2.3 Functional Equation

7. 2.4 Conclusion

1. Introduction

Let $X$ be the prehomogeneous vector space of binary cubic forms, and denote the standard element of $X$ by

[TABLE]

Let $X^{+}$ be the space of binary cubic forms with $a>0$ and $b^{2}-3ac<0$. Geometrically, this would mean that the cubic function $y=ax^{3}+bx^{2}+cx+d$ is everywhere increasing.

Let $\Gamma_{\infty}=\left<\begin{pmatrix}1&1\\ 0&1\end{pmatrix}\right>$ be the upper triangular unipotent elements of $\operatorname{SL}_{2}(\mathbb{Z})$. Then we have a left action of $\Gamma_{\infty}$ on $X^{+}$, given by $\gamma\circ f(x,y)=f((x,y)\circ\gamma^{T})$. There are four relative invariants of $X^{+}$ under the action of $\Gamma_{\infty}$, namely

[TABLE]

Note that the last one of these invariants $r_{4}$ is the discriminant of $f(x,y)$ and these four relative invariants are linearly independent. See Section 3.1 of [3] for more information on the relative invariants.

A coset representative for $\Gamma_{\infty}\setminus X^{+}$ is

[TABLE]

In order to understand the distribution of the relative invariants of binary cubic forms, we form the four variable multiple Dirichlet series

[TABLE]

At present, determining the fine analytic properties of the this zeta function appears out of reach. For this reason we consider a simpler two variable zeta function. Define a further equivalence relation on $\Gamma_{\infty}\backslash X^{+}$ by letting $f_{1}\sim f_{2}$ if $r_{1}(f_{1})=r_{1}(f_{2})$ and $r_{2}(f_{1})=r_{2}(f_{2})$. We then set

[TABLE]

Theorem 1.1**.**

The multiple Dirichlet series $Z(s_{1},s_{2})$ can be meromorphically continued to the whole of $\mathbb{C}^{2}$.

We note that, as opposed to the original Shintani zeta functions in [15] where one considers over all equivalence classes of integral binary cubic forms, our zeta function involves only a portion of this set. We make some comments about the possible poles of the zeta function in section 2.2.

We will present the proof of this theorem in the next section. Many mathematicians have studied multiple Dirichlet series of the form

[TABLE]

See, e.g., [4], [5], [6], [7], and [8] to name a few publications, and our proof of the theorem follows the (by now standard) techniques used in these works.

M. Sato and Shintani [14] first defined zeta functions associated to prehomogeneous vector spaces with single relative invariants, and proved some properties including analytic continuation and functional equations. Then F. Satō undertook the study of multiple Dirichlet series associated to prehomogeneous vector spaces. See [11], [12] and [13] for more details.

Our work in this paper is directly motivated by the thesis of Li-Mei Lim [10] who considered multiple Dirichlet series associated to the prehomogeneous vector space of ternary quadratic forms in an attempt to generalize the result of Chinta and Offen [2] on the orthogonal period of a GL3 Eisenstein series. Namely, let $X_{3}$ be the space of ternary quadratic forms, and let $X_{3}^{+}$ be the space of positive definite ternary quadratic forms. Also, let $P$ be the minimal parabolic subgroup in $\operatorname{SL}_{3}(\mathbb{Z})$. The action of $P$ on $X_{3}^{+}$ has three relative invariants, namely,

[TABLE]

for

[TABLE]

Using these relative invariants, one can form a multiple Dirichlet series:

[TABLE]

Lim proved that the $k^{-s_{1}}$ coefficient in the series $Z(s_{1},s_{2},s_{3})$ is both equal to a sum of special values of the minimal parabolic $\operatorname{GL}_{3}$ Eisenstein series, and a linear combination of double Dirichlet series which arise as Fourier coefficients of the Eisenstein series on a double cover of $\operatorname{GL}_{3}$.

We would like to think of our work as only the beginning of an investigation into an interesting realm of problems. For example, it would be desirable to give interpretations of the multiple Dirichlet series we consider here in terms of canonical objects of the theory of automorphic forms, e.g., Eisenstein series on higher rank groups.

The authors wish to thank Gautam Chinta for useful conversations, and for his hospitality during the first author’s visit to CCNY in 2018. We also benefited from conversations with Li-Mei Lim. We also wish to thank the referee and Evan O’Dorney for pointing out many inaccuracies in earlier drafts of this paper. The second author is partially supported by a Collaboration Grant from Simons Foundation.

2. Proof of the main theorem

Let $f(x)=ax^{3}+bx^{2}y+cxy^{2}+dy^{3}\in X^{+}$. Then $|r_{1}(f)|=a$ and $|r_{2}(f)|=3ac-b^{2}$. Then

[TABLE]

The following is the broad outline of the proof of Theorem 1.1:

1. (1)

We reduce the multiple Dirichlet series into a finite linear combination of standard type double zeta functions; 2. (2)

we find a fairly large domain of meromorphy; 3. (3)

we find a functional equation, and use the functional equation to extend the domain of meromorphy; 4. (4)

and finally, we use a convexity argument to get the final result.

2.1. Reduction of the Multiple Dirichlet Series

Lemma 2.1**.**

We formally have

[TABLE]

where

[TABLE]

Proof.

Let $m=a$ and $n=-b^{2}+3ac$. Write

[TABLE]

for integers $a(m,n)$. Then, since $b^{2}=3ac-n\equiv-n\mod 3m$, the coefficient $a(m,n)$ is

[TABLE]

∎

The following proposition shows the explicit formulae for $C(m,n)$.

Proposition 2.2**.**

The following properties hold.

1. i)

For any fixed n, $C(m,n)$ is a weekly multiplicative function in $m$. In particular, $C(1,n)=1$ for all $n$. 2. ii)

If any prime $p\neq 2$ and $p\nmid n$, then $C(p^{\alpha},n)=1+\left(\dfrac{n}{p}\right)$ for $\alpha>0$. 3. iii)

If $p=2$ and $n$ is odd, then for $\alpha>0$,

[TABLE] 4. iv)

If $n=p^{r}n_{0}$ with $p\nmid n_{0}$, then for $\alpha>0$,

[TABLE]

Proof.

1. i)

This is true by the Chinese Remainder Theorem. In particular, if $m=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{k}^{\alpha_{k}}$ for all $p_{i}$’s prime, then

[TABLE]

hence, we only have to analyze $C(p^{\alpha},n)$ for any prime $p$, $\alpha$, and $n$. 2. ii)

By Hensel’s Lemma, if $p$ is an odd prime and $p\nmid n$, $x^{2}\equiv n\mod p^{\alpha}$ is solvable if and only if $x^{2}\equiv n\mod p$ is solvable and in such cases, $x^{2}\equiv n\mod p^{\alpha}$ has exactly two solutions. 3. iii)

The cases of $\alpha=1$ and $\alpha=2$ are trivial.

Now, let’s look at the cases of $\alpha\geq 3$.

Remark*.*

The equation $x^{2}\equiv 1\mod 2^{\alpha}$ for $\alpha\geq 3$ has exactly four solutions, which are $1$, $-1$, $1+2^{\alpha-1}$, and $-1+2^{\alpha-1}$.

For $x_{1}$ and $x_{2}$ odd, if $x_{1}^{2}\equiv n\mod 2^{\alpha}$ and $x_{2}^{2}\equiv n\mod 2^{\alpha}$, then $x_{1}^{2}\equiv x_{2}^{2}\mod 2^{\alpha}$, hence $(x_{1}x_{2}^{-1})^{2}\equiv 1\mod 2^{\alpha}$. By the above remark, $x_{1}x_{2}^{-1}$ will be one of the four. Hence, if there is a solution to $x^{2}\equiv n\mod 2^{\alpha}$, then there will be exactly four of them.

Now, let’s show that $x^{2}\equiv n\mod 2^{\alpha}$ is solvable if and only if $x^{2}\equiv n\mod 8$ is solvable, which is when $n\equiv 1\mod 8$.

Recall the generalized version of Hensel’s Lemma. For $p$ a prime and $f(x)\in\mathbb{Z}[x]$, if $p^{2N+1}|f(a)$ and $p^{N}|f^{\prime}(a)$ but $p^{N+1}\nmid f^{\prime}(a)$, then for all $M>N$, there is unique $x_{M}\mod p^{M}$ such that $f(x_{M})\equiv 0\mod p^{M}$, and $x_{M}\equiv a\mod p^{N+1}$.

Take $f(x)=x^{2}-n$ for an odd $n$, $p=2$, $N=1$. Then if $2^{3}|a^{2}-n$, $2|2a$, and $2^{2}\nmid 2a$, i.e. $a$ odd and $a^{2}-n\equiv 0\mod 8$, then for all $M>1$, there exists unique $x_{M}$ such that $x_{M}^{2}-n\equiv 0\mod 2^{M}$. That is to say, $x^{2}\equiv n\mod 2^{\alpha}$ is solvable if and only if $x^{2}\equiv n\mod 8$ is solvable, i.e., if $n\equiv 1\mod 8$. 4. iv)

We first show that $C(p^{\alpha},0)=p^{\left\lfloor\frac{\alpha}{2}\right\rfloor}$. Write $x$ in base $p$. Then

[TABLE]

So we need $2k\geq\alpha$, i.e. $k\geq\left\lceil\frac{\alpha}{2}\right\rceil$. Then $x=a_{\left\lceil\frac{\alpha}{2}\right\rceil}p^{\left\lceil\frac{\alpha}{2}\right\rceil}+\cdots+a_{\alpha-1}p^{\alpha-1}$ and any of these will work. Since the number of choices for each $a_{i}$ is $p$, $C(p^{\alpha},0)=p^{(\alpha-1)-\left\lceil\frac{\alpha}{2}\right\rceil+1}=p^{\alpha-\left\lceil\frac{\alpha}{2}\right\rceil}=p^{\left\lfloor\frac{\alpha}{2}\right\rfloor}$.

Now, let $x^{2}\equiv p^{r}n_{0}\mod p^{\alpha}$ with $p\nmid n_{0}$. If $r\geq\alpha$, then $x^{2}\equiv 0\mod p^{\alpha}$, so $C(p^{\alpha},p^{r}n_{0})=p^{\left\lceil\frac{\alpha}{2}\right\rceil}$. For $r<\alpha$, $x^{2}\equiv p^{r}n_{0}\mod p^{\alpha}$ shows that $r$ is even. Write $r=2\ell$, then $x=p^{\ell}y$ for some $y$. Then $x^{2}=p^{2\ell}y^{2}=p^{r}y^{2}\equiv p^{r}n_{0}\mod p^{\alpha}$, hence

[TABLE]

We know the number of solutions to (2) for $y$ is $C(p^{\alpha-r},n_{0})$. Now we have to lift $y$’s up to mod $p^{\alpha}$. The number of solutions of $y$ for $y^{2}\equiv n_{0}\mod p^{\alpha}$ is $p^{r}C(p^{\alpha-r},n_{0})$. Since $x=p^{\ell}y$ and multiplication by $p^{\ell}$ is a $p^{\ell}$-to-1 function mod $p^{\alpha}$, there are $p^{\ell}C(p^{\alpha-r},n_{0})$ many solutions of $x$ to $x^{2}\equiv p^{r}n_{0}\mod p^{\alpha}$.

∎

Next, $Z(s_{1},s_{2})$ can be rewritten as

[TABLE]

Let $\displaystyle Z_{n}(s)=\sum_{m=1}^{\infty}\dfrac{C(3m,-n)}{m^{s}}$.

Proposition 2.3**.**

$\displaystyle Z_{n}(s)=\prod_{p\mbox{ }\operatorname{prime}}Z_{n,p}(s)$, for

[TABLE]

Proof.

Trivial due to the multiplicativity of $C(m,n)$ for fixed $n$.∎

It is worth noting that the above general results are presented for the reference. We will use only the cases in which $n$ is square-free and odd.

2.1.1. Local Euler Factors

We explicitly compute the local Euler factors via local computations for various $p$’s and $n$’s.

CASE 1: $p\nmid 6n$

[TABLE]

CASE 2: $p=2$, $n$ odd

1. i)

$n\equiv 1\mod 8$, i.e., $-n\equiv 7\mod 8$:

[TABLE] 2. ii)

$n\equiv 3\mod 8$, i.e., $-n\equiv 5\mod 8$:

[TABLE] 3. iii)

$n\equiv 5\mod 8$, i.e., $-n\equiv 3\mod 8$:

[TABLE] 4. iv)

$n\equiv 7\mod 8$, i.e., $-n\equiv 1\mod 8$:

[TABLE]

CASE 3: $p\neq 3$, $p|n$, i.e., $n=p^{r}n_{0}$

1. i)

$r$ is odd:

[TABLE]

In particular, if $r=1$, then $Z_{n,p}(s)=\dfrac{1-p^{-2s}}{1-p^{-s}}$. 2. ii)

$r$ is even:

[TABLE]

CASE 4: $p=3$

1. i)

$3\nmid n$:

[TABLE]

In particular, if $-n\equiv 2\mod 3$, i.e. $n\equiv 1\mod 3$, then $Z_{n,3}(s)=0$. 2. ii)

$n=3^{r}n_{0}$, $3\nmid n_{0}$, $r$ odd:

[TABLE]

In particular, if $r=1$, then $Z_{n,3}(s)=1$. 3. iii)

$n=3^{r}n_{0}$, $3\nmid n_{0}$, $r$ even:

[TABLE]

We collect these results in the following proposition:

Proposition 2.4**.**

Define a Dirichlet character $\eta_{-n}(p)$ by setting $\eta_{-n}(p)=\left(\dfrac{-n}{p}\right)$, and denote the Dirichlet $L$-function associated to $\eta_{-n}$ by $L(s,\eta_{-n})$.

1. i)

If $n\equiv 1\mod 3$, then $Z_{n}(s)=0$. 2. ii)

If $n=q_{1}^{r_{1}}q_{2}^{r_{2}}\cdots q_{k}^{r_{k}}$ for $q_{i}\neq 2,3$ primes, $r_{i}\geq 1$ odd, then

1. (a)

$n\equiv 5\mod 12$:

[TABLE] 2. (b)

$n\equiv 23\mod 24$:

[TABLE] 3. (c)

$n\equiv 11\mod 24$:

[TABLE] 3. iii)

Define

[TABLE]

Then, if $n$ is square-free,

[TABLE]

So from the above computations, letting $n$ be square-free, we conclude that

[TABLE]

Here, $L_{2,3}$ is a Dirichlet $L$-function with the Euler factors at $2,3$ removed, and $a_{n}(s_{1})$ is a function defined to correct those removed Euler factors, which depends on the residue of $n$ modulo 24.

Let us assume that $n$ is coprime to $6$. Let $\delta_{j}(n)$ to be equal to $1$ if $n\equiv j\mod 24$, otherwise equal to [math]. Then

[TABLE]

Then if $\gcd(n,24)=1$, we have

[TABLE]

Hence

[TABLE]

This shows that $Z(s_{1},s_{2})$ is a finite linear combination of standard type multiple Dirichlet series of the form

[TABLE]

and this will be analyzed in the following section.

2.2. Domain of Meromorphy

In this section, we adapt the methods of Diaconu, Goldfeld and Hoffstein from [4].

Let $w,s_{1},s_{2},\cdots,s_{m}\in\mathbb{C}$ with $\operatorname{Re}(w)>1,\operatorname{Re}(s_{i})>1$ for $i=1,\cdots,m$. Consider the absolutely convergent multiple Dirichlet series

[TABLE]

where $\chi(d)$ is some Dirichlet character and

[TABLE]

To see where the poles and residues of $Z(w,s_{1},\cdots,s_{m})$ are, let us define an adjusted multiple Dirichlet series:

[TABLE]

We set

[TABLE]

Proposition 2.5**.**

For $\sigma>0$, $Z^{\pm}$ can be continued meromorphically to the domain $\operatorname{Re}(w)>1+m\left(\frac{1}{2}+\sigma\right)$, and $\operatorname{Re}(s_{i})>-\sigma$ for $i=1,\cdots,m$. In this region, the only poles are at $s_{i}=1$ for $i=1,\cdots,m$.

Before proving the statement, it is worth noting that we do not find the exact poles, but the above Proposition 2.5 can be used to precisely determine the poles of $\tilde{Z}_{\chi}(s_{1},s_{2})$. Since $Z(s_{1},s_{2})$ is a linear combination of $\tilde{Z}_{\chi}(s_{1},s_{2})$, the set of poles of $Z(s_{1},s_{2})$ is a subset of that of $Z^{*}(s_{1},s_{2})$.

Proof.

We can rearrange the series and see the following:

$\displaystyle Z_{1}^{\pm}(w,s_{1},\cdots,s_{m})=\sum_{\begin{subarray}{c}\pm d>0\\ d\equiv 1\mod 4\\ d:\operatorname{sq.free}\\ 3\nmid d\end{subarray}}\dfrac{\chi(d)L(s_{1},\chi_{d})\cdots L(s_{m},\chi_{d})}{|d|^{w}}\\ \phantom{Z_{1}^{\pm}(s_{1},\cdots,s_{m},w)}=\sum_{\begin{subarray}{c}\pm d>0\\ d\equiv 1\mod 4\\ d:\operatorname{sq.free}\\ 3\nmid d\end{subarray}}\dfrac{\chi(d)}{|d|^{w}}\prod_{j=1}^{m}\sum_{n_{j}=1}^{\infty}\dfrac{\chi_{d}(n_{j})}{n_{j}^{s_{j}}}\\ \phantom{Z_{1}^{\pm}(s_{1},\cdots,s_{m},w)}=\sum_{n_{1},\cdots,n_{m}}\dfrac{1}{n_{1}^{s_{1}}\cdots n_{m}^{s_{m}}}\sum_{\begin{subarray}{c}\pm d>0\\ d\equiv 1\mod 4\\ d:\operatorname{sq.free}\\ 3\nmid d\end{subarray}}\dfrac{\chi(d)\chi_{d}(n_{1}n_{2}\cdots n_{m})}{|d|^{w}}.$

For any fixed $m$-tuple of positive integers, $\{n_{1},\cdots,n_{m}\}$, we can write

[TABLE]

with the following properties:

• •

$n$: square-free,

• •

If $p|N$, then $p|n$,

• •

$\gcd(M,n)=1$,

• •

$n$, $N$, $M$ odd.

Hence, the inner summation becomes:

$\displaystyle\sum_{\begin{subarray}{c}\pm d>0\\ d\equiv 1\mod 4\\ d:\operatorname{sq.free}\\ 3\nmid d\end{subarray}}\dfrac{\chi(d)\chi_{d}(n_{1}n_{2}\cdots n_{m})}{|d|^{w}}=\sum_{\begin{subarray}{c}\pm d>0\\ d\equiv 1\mod 4\\ d:\operatorname{sq.free}\\ 3\nmid d\\ (d,M)=1\end{subarray}}\dfrac{\chi(d)\chi_{d}(2)^{c}\chi_{d}(n)}{|d|^{w}}=\sum_{\begin{subarray}{c}\pm d>0\\ d\equiv 1\mod 4\\ d\operatorname{sq.free}\\ 3\nmid d\\ (d,M)=1\end{subarray}}\dfrac{\chi(d)\chi_{2}(d)^{c}\chi_{n}(d)}{|d|^{w}}\\ \phantom{\sum_{\begin{subarray}{c}\pm d>0\\ d\equiv 1\mod 4\\ d:\operatorname{sq.free}\\ 3\nmid d\end{subarray}}\dfrac{\chi(d)\chi_{d}(n_{1}n_{2}\cdots n_{m})}{|d|^{w}}}=\dfrac{1}{2}\sum_{\begin{subarray}{c}\pm d>0\\ d\operatorname{sq.free}\\ 3\nmid d\\ (d,M)=1\\ 2\nmid d\end{subarray}}\dfrac{\chi_{2}(d)^{c}\chi_{n}(d)\chi(d)}{|d|^{w}}+\dfrac{1}{2}\sum_{\begin{subarray}{c}\pm d>0\\ d\operatorname{sq.free}\\ 3\nmid d\\ (d,M)=1\\ 2\nmid d\end{subarray}}\dfrac{\chi_{2}(d)^{c}\chi_{-1}(d)\chi_{n}(d)\chi(d)}{|d|^{w}}.$ Before we continue we state and prove a lemma. Let $\psi$ be a Dirichlet character of conductor $n$. Define

[TABLE]

Lemma 2.6**.**

For a Dirichlet character $\psi$ mod $n$, $L(2w,\psi^{2})L_{b}(w,\psi)$ is holomorphic on $\operatorname{Re}(w)>0$ except when $\psi$ is trivial mod $n$, in which case, there is a unique pole at $w=1$ with residue

[TABLE]

Lemma 2.7**.**

For a Dirichlet character $\psi$ mod $n$, $L(2w,\psi^{2})L_{b}(w,\psi)$ is holomorphic on $\operatorname{Re}(w)>0$ except when $\psi$ is trivial mod $n$, in which case, there is a unique pole at $w=1$ with residue

[TABLE]

Proof.

For a primitive $\psi_{1}\mod\tilde{n}$, which is extended to $\psi\mod n$, we have

[TABLE]

Since $\displaystyle L(2w,\psi_{1}^{2})=L(2w,\psi^{2})\prod_{p|n}\dfrac{1}{1-\psi_{1}^{2}(p)p^{-2w}}$,

[TABLE]

has a pole at $w=1$ if and only if $\psi_{1}$ is trivial. Since $\psi_{1}$ is primitive, in order to have a pole, $\tilde{n}=1$. Hence, the residue of $L(2w,\psi^{2})L_{b}(w,\psi)$ at $w=1$ is

[TABLE]

∎

We now return to the proof of Proposition 2.5.

For

[TABLE]

$L(2w,\chi^{2})Z(w)$ is holomorphic on $\operatorname{Re}(w)>0$ and meromorphic with a pole at $w=1$ if and only if either $\chi_{2}^{c}\chi_{n}\chi$ or $\chi_{2}^{c}\chi_{-1}\chi_{n}\chi$ is trivial. In particular, if $\chi$ is not quadratic, then there is no pole for $L(2w,\chi^{2})Z(w)$. Note that $\chi_{2}^{c}\chi_{n}\chi$ and $\chi_{2}^{c}\chi_{-1}\chi_{n}\chi$ cannot be both trivial. ∎

Proposition 2.8**.**

For a Dirichlet character $\chi$, we have

[TABLE]

provided that the primitive part of $\chi$ is trivial. Otherwise, there is no pole.

In order to prove above proposition, we will need the following lemma:

Lemma 2.9**.**

For a character $\psi(n)\operatorname{mod}N$, $\displaystyle\sum_{\begin{subarray}{c}n\operatorname{sq.free}\\ (n,m)=1\end{subarray}}\dfrac{\psi(n)}{n^{w}}=\dfrac{L(w,\psi)}{L_{m}(2w,\psi^{2})}$ for $m|\operatorname{cond}\psi$.

Proof.

We have

[TABLE]

∎

Proof of Proposition 2.8.

Using Lemma 2.9, we get the follwoing:

[TABLE]

Hence, we have the following identity:

[TABLE]

Since $\displaystyle\dfrac{L(2s_{2},\chi^{2})}{L_{m}(2s_{2},\chi^{2})}=\prod_{p|m}\left(1-\dfrac{\chi^{2}(p)}{p^{2s_{2}}}\right)^{-1}$, we get

[TABLE]

Now, if $m$ is a square, then

[TABLE]

and if $m$ is not a square,

[TABLE]

since $\operatorname{cond}\chi(\cdot)\left(\dfrac{\cdot}{m}\right)=\underbrace{\operatorname{cond}\chi(\cdot)}_{|24}\underbrace{\operatorname{cond}\left(\dfrac{\cdot}{m}\right)}_{|m}\neq 1$. Therefore,

[TABLE]

Here, the inner product can be analyzed as follows:

[TABLE]

Recall that the product $\displaystyle\prod_{n}(1+a_{n})$ converges absolutley if $\displaystyle\sum_{n}|a_{n}|<\infty$. For $\operatorname{Re}s_{1}>0$, since

[TABLE]

and

[TABLE]

we conclude

[TABLE]

converges absolutely. Hence, the only pole of $\displaystyle\mathop{Res}_{s_{2}=1}L(2s_{2},\chi^{2})\tilde{Z}_{\chi}(s_{1},s_{2})$ comes from

[TABLE]

and the pole is at $s_{1}=\frac{1}{2}$.

Finally,

[TABLE]

This concludes the proof of Proposition 2.8. ∎

2.3. Functional Equation

Recall the fuctional equation of $L(s,\chi)$ for $\chi$ a primitive Dirichlet character mod $k$. Define

[TABLE]

where

[TABLE]

For $\displaystyle\tau(\chi)=\sum_{n=1}^{\infty}\chi(n)\exp(2\pi in/k)$, we have the following functional equation:

[TABLE]

It is a well-known (see, e.g., Exercise 21 on page 45 of [1]) that for $\chi$ a real primitive character modulo $k$ where $k>2$,

[TABLE]

Now,

[TABLE]

In this case, if $m$ is even, then due to the presence of $\chi_{4}(m)$, the whole term disappears: take only odd $m$’s. Now, letting $n=2^{c}N$ for $N$ odd,

[TABLE]

This gives the conductor of $\left(\dfrac{n}{m}\right)=\begin{cases}8N&\mbox{ if }c\,\,\operatorname{odd}\\ 4N&\mbox{ if }c\,\,\operatorname{even}\end{cases}$, so that the conductor of $\chi_{4}(m)\left(\dfrac{n}{m}\right)\mathbb{1}_{3}(m)$ is $\begin{cases}3\cdot 8N&c\,\,\operatorname{odd}\\ 3\cdot 4N&c\,\,\operatorname{even}\end{cases}$.

Now we apply the above functional equation to the function

[TABLE]

for $\psi_{n}$ primitive Dirichlet character mod $3\cdot 4n$.

From (3), since

[TABLE]

$a_{n}=1$, and we can define

[TABLE]

Finally, the functional equation for $\Lambda$ is:

[TABLE]

for $\tau(\psi_{n})=\displaystyle\sum_{\ell=1}^{12n}\psi_{n}(\ell)\exp(2\pi i\ell/12n)$.

Since $\psi_{n}$ is real, $\overline{\psi_{n}}=\psi_{n}$. Hence,

[TABLE]

Since $\tau(\psi_{n})=i\cdot(12n)^{1/2}$ from (4), we have the following functional equation:

[TABLE]

Using the definition of $\Lambda$ in (5) and the functional equation in (6), we get

[TABLE]

Hence,

[TABLE]

Going back to $\tilde{Z}_{\chi}(s_{1},s_{2})$,

[TABLE]

2.4. Conclusion

From the sections 4.2 and 4.3, we get the domain of meromorphy of

• •

$\operatorname{Re}s_{1}>0,\operatorname{Re}s_{2}\gg 0$ and

• •

$\operatorname{Re}s_{1}>\frac{3}{2},\operatorname{Re}s_{1}+\operatorname{Re}s_{2}>\frac{3}{2}$.

-2$$2$$4$$M$$\sigma_{1}$$\sigma_{2}

Next, we transform this domain using the transformation

[TABLE]

to obtain the following figure:

-2$$2$$4$$M$$\sigma_{1}$$\sigma_{2}

Now, we can use the properties of tube domains [9, Theorem 2.5.10] to conclude that our $\tilde{Z}_{\chi}(s_{1},s_{2})$ can be extended to a function on the convex hull, which is the whole $\mathbb{C}^{2}$.

The fact that $Z(s_{1},s_{2})$ is a finite linear combination of $\tilde{Z}_{\chi}(s_{1},s_{2})$ concludes the proof of Theorem 1.1.