[ANT] Dirichlet characters and Dirichlet L-functions

본문은 아래 링크의 후속편이다.

https://whitemask.tistory.com/251

Characters of Groups

 

원활한 이해를 위해 아래 링크의 지식이 필요할 수도 있다.

https://whitemask.tistory.com/247

[ANT] Estimating abscissa of convergence of Dirichlet series

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Definition. A Dirichlet character $\chi$ modulo $q\in\mathbb{N}$ is a character of finite abelian group $(\mathbb{Z}/q\mathbb{Z})^{\times}$, extended to a function on $\mathbb{Z}$ by

$$\chi(n)=\begin{cases} \chi(n+q\mathbb{Z}) & (n, q)=1 \\ 0 & \text{otherwise} \end{cases} $$

Since a Dirichlet character $\chi$ has domain $\mathbb{Z}$, we may consider it as an arithmetic function. A character assuming only real values are called real characters, and others are called complex characters. A non-trivial real character is called the quadratic character.

Note that we make convention that $ (\mathbb{Z}/1\mathbb{Z})^{\times}=\{1\}$. Thus the only Dirichlet character modulo $1$ is the function $u(n)=1$.

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Proposition 1. Dirichlet characters modulo $q$ are periodic modulo $q$ and completely multiplicative. There are $\varphi(q)$ of Dirichlet characters modulo $q$, which forms a group under pointwise multiplication, where $\varphi$ is the Euler totient function. The identity is the trivial character $\chi_0(n)=\textbf{1}_{(n, q)=1}(n)$, and inverse of $\chi$ is $\overline{\chi}$. The orthogonalities holds, and more generally

$$\frac{1}{\varphi(q)}\sum_{n=1}^q \chi(n)= \begin{cases} 1 & \chi=\chi_0\\ 0 & \text{otherwise} \end{cases}$$

and for $(a, q)=1$,

$$\frac{1}{\varphi(q)}\sum_{\chi^{(\bmod q)}} \overline{\chi}(a)\chi(n)=\begin{cases} 1 & n\equiv a\pmod q \\ 0 & \text{otherwise} \end{cases}$$

proof. First suppose $(n, q)=1$ then $\chi(n+q)=\chi(n)$ by definition and if $(n, q)\ne 1$ then $\chi(n+q)=\chi(n)=0$ hence Dirichlet characters modulo $q$ are periodic modulo $q$.

Since the Dirichlet character $\chi$ modulo $q$ is the extension of the character of $(\mathbb{Z}/q\mathbb{Z})^{\times}$, $\chi(ab)=\chi(a)\chi(b)$ for $a, b\in(\mathbb{Z}/q\mathbb{Z})^{\times}$ where $(a, q)=1$ and $(b, q)=1$. Note that $(ab, q)=1$ also. Now suppose  $m, n\in\mathbb{N}$ are relatively prime with $q$, that is, $(m, q)=1$ and $(n, q)=1$. Then $(mn, q)=1$ and we have $a, b\in(\mathbb{Z}/q\mathbb{Z})^{\times}$ such that $\chi(m)=\chi(a)$ and $\chi(n)=\chi(b)$. So $\chi(mn)=\chi(ab)=\chi(a)\chi(b)=\chi(m)\chi(n)$. If one of $m, n$ is not relatively prime with $q$, then $\chi(mn)=0=\chi(m)\chi(n)$ so $\chi$ is completely multiplicative.

Now note that $(\mathbb{Z}/q\mathbb{Z})^{\times}$ is finite abelian group. For finite abelian group $G$, we know that $G\cong \widehat{G}$ so there are $\varphi(q)$ of Dirichlet characters modulo $q$ and this forms a group under the pointwise multiplication. It is clear that the identity is $\chi_0(n)=\textbf{1}_{(n, q)=1}(n)$. We also know that for finite group $G$, $|\chi(g)|=1$ for all $g\in G$ hence the inverse of a character $\chi$ is $\overline{\chi}$.

For the orthogonalities, note that $|(\mathbb{Z}/q\mathbb{Z})^{\times}|=\varphi(q)$ then the first orthogonality relation is immediate. The second orthogonality relation is obvious since $\overline{\chi}(a)\chi(n)$ is $1$ whenever $n\equiv a\pmod q$ and $0$ otherwise. 증명 끝.

We now turn to the problem of explicitly determining the Dirichlet series. We will deal with this problem in the next article.

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Definition. The Dirichlet L-function associated to a Dirichlet character $\chi$ modulo $q$, denoted by $L(s, \chi)$, is a Dirichlet series $\mathcal{D}_\chi(s)$.

Proposition 2. If $\chi$ is trivial, then $\sigma_a(\chi)=\sigma_c(\chi)=1$ and

$$L(s, \chi)=\zeta(s)\prod_{p | q}\left(1-\frac{1}{p^s}\right)$$

for $\text{Re}(s)=\sigma>1$. If $\chi$ is non-trivial, then $\sigma_a(\chi)=1$ and $\sigma_c(\chi)=0$.

proof. Since $\chi_0$, the trivial homomorphism is non-negative so $\sigma_a(\chi_0)=\sigma_c(\chi_0)$. Now we use the fact that

$$\sigma_c(a)=\inf\{\sigma:A(x)=O(x^\sigma)\}$$

where $A(x)=\sum_{n\le x} a(n)$ for arithmetic function $a$. Note that we have

$$\varphi(q)\left[\frac{x}{q}\right]\le\sum_{n\le x}\chi_0(n)\le x$$

for $x\ge q$ and fixed $\chi_0$, by the periodicity of the Dirichlet character and its orthogonality. But since

$$x=O\left(\varphi(q)\left[\frac{x}{q}\right]\right)$$

from $[x]=x+O(1)$. If there is $\sigma_0<1$ such that $\sum_{n\le x}\chi_0(n)=O(x^{\sigma_0})$ then we have $x\le C_1\cdot\varphi(q)\left[\frac{x}{q}\right]$ and $\sum_{n\le x}\chi_0(n)\le C_2\cdot x^{\sigma_0}$ for all large $x$. Then we have

$$\frac{1}{C_1}x\le\varphi(q)\left[\frac{x}{q}\right]\le\sum_{n\le x}\chi_0(n)\le C_2\cdot x^{\sigma_0}$$

so $x^{1-\sigma_0}\le C_1C_2$ for all large $x$. But since $1-\sigma_0>0$, this is contradiction. Thus if $\sum_{n\le x}\chi_0(n)=O(x^{\sigma_0})$ then $\sigma_0\ge 1$ so we may conclude that

$$\sigma_c(a)=\inf\{\sigma:A(x)=O(x^\sigma)\}=1$$

Now for $\sigma>1$, we apply the Euler product for Dirichlet series to get

$$L(s, \chi_0)=\prod_{p}\left(1-\frac{\chi_0(p)}{p^s}\right)^{-1}=\prod_{p\nmid q}\left(1-\frac{1}{p^s}\right)^{-1}=\zeta(s)\prod_{p | q}\left(1-\frac{1}{p^s}\right)$$

since we have

$$\zeta(s)=\prod_{p}\left(1-\frac{1}{p^s}\right)^{-1}$$

Now if $\chi$ is non-trivial, we know that $|\chi|=1$ so $|\chi(g)|=\chi_0(g)$ so $\sigma_a(\chi)=\sigma_a(\chi_0)=1$. To calculate $\sigma_c(\chi)$, note that the partial sum $\sum_{n\le x} \chi(n)$ is bounded from its periodicity and that

 

$$\sum_{n=1}^q \chi(n)=0$$

 

by the orthogonality. Since $D_a(s)=\sum a(n)/n^s$ is locally uniform hence holomorphic for $\sigma>\sigma_0$ if $D_a(s_0)=\sum a(n)/n^{s_0}$ is bounded, we have $\sigma_c(\chi)\le 0$. While the partial sum $\sum_{n\le x} \chi(n)$ is bounded, it does not converge as $x\to \infty$ and it just oscillates. By the same argument, we have $\sigma_c(\chi)=0$. 증명 끝.

 

 

Corollary. The Dirichlet L-functions are holomorphic for $\sigma>0$. Only for $L(s, \chi_0)$ where $\chi_0$ is trivial, there is a simple pole at $s=1$ with residue $\varphi(q)/q$.

 

proof. It is obvious for non-trivial case. For trivial case, $L(s, \chi_0)$ is holomorphic if $\sigma>1$. And the meromorphic continuation of zeta function

 

$$\zeta(s)=\frac{s}{s-1}-s\int_1^\infty\frac{\{u\}}{u^{s+1}}du$$

 

for $\sigma>0$ gives a meromorphic continuation of $L(s, \chi_0)$ to $\sigma>0$, proving the corollary. For the last statement, since the residue of meromophic continuation of zeta function at $s=1$ is $1$, we have

 

$$\text{Res}(L(s, \chi), s=1)=1\cdot\prod_{p|q}\left(1-\frac{1}{p}\right)=\frac{\varphi(q)}{q}$$

 

by the definition of Euler totient function $\varphi$.