# Dirichlet L-functions

As we know that there are three important properties about the Riemann $\zeta$-function: infinite product expression, analytic continuation and the functional equation.

One natural generalization of the Riemann $\zeta$-function is the Dirichlet L-function. And indeed these L-functions share the three properties mentioned above.

First let’s introduce the Dirichlet character. Given a positive integer $N$, then consider the group homomorphisms from the multiplicative group $(\mathbb{Z}/N\mathbb{Z})^{\times}$ to the multiplicative group $\mathbb{C}^{\times}$. We extend this map to all of $\mathbb{Z}/N\mathbb{Z}$ by setting the values at those missing points to be $0$. We call one such map a Dirichlet character of conductor $N$. If $n$ is a divisor of $N$, then we have a natural map from $(\mathbb{Z}/N\mathbb{Z})^{\times}$ to $(\mathbb{Z}/n\mathbb{Z})^{\times}$, i.e. the restriction map. So all the characters of conductor $n$ can be extended to characters of conductor $N$. But the inverse doesn’t hold, that is to say, there is some character $\chi$ of conductor $N$ which is not the extension of any character of conductor $n$ for $n(we call this character the primitive character of $\mathbb{Z}/N\mathbb{Z}$, this is a consequence of the Euler formula $N=\sum_{n|N}\psi(n)$). This simple fact can be deduced from the structure theorem of finite abelian groups, and we omit the proof.

Now we define the Gauss sum of a character $\Gamma(\chi,m)=\sum_{n\in\mathbb{Z}/N\mathbb{Z}}\chi(n)e^{2\pi imn/N}$.

Note we can make an analogy between the group $(\mathbb{Z}/N\mathbb{Z})^{\times}$ and the group $\mathbb{R}_{>0}$. Recall the $\Gamma$ function, $\Gamma(s)=\int_{0}^{\infty}e^{-t}t^sdt/t$. Note that $\mathbb{R}_{>0}\rightarrow\mathbb{C},t\mapsto t^s$ is a character for the multiplicative group $\mathbb{R}_{>0}$. It is easy to show that all the continuous characters of $\mathbb{R}_{>0}$ are of the form $t\mapsto t^s$ for some complex number $s$. This is just like $\chi$ on $(\mathbb{Z}/N\mathbb{Z})^{\times}$. Moreover, the exponential function is a character of the additive group $\mathbb{R}$, just like $e^{2\pi imn}$ on $\mathbb{Z}/N\mathbb{Z}$. So, in some sense, we can also define $\Gamma(s,m)=\int_0^{\infty}t^se^{-mt}dt/t$. An easy calculation shows that $\Gamma(s,m)=m^{-s}\Gamma(s,1)$. We will show that there is a similar relation for the $\Gamma$ functions on $\mathbb{Z}/N\mathbb{Z}$:

$\Gamma(\chi,m)=\overline(\chi(m))\Gamma(\chi,1)$

There are two cases, the first is that $gcd(m,N)=1$.  Then we have that $\Gamma(\chi,m)=\sum_n\chi(mn)\chi(m)^{-1}e^{2\pi imn/N}=\chi(m)^{-1}\sum_n\chi(n)e^{2\pi in/N}$ $=\chi(m)^{-1}\Gamma(\chi,1)=\overline{\chi(m)}\Gamma(\chi,1)$. So, the first case is easily done. As for the second case, we suppose that $gcd(m,N)=d>1, m=dm',N=dN'$. We can find $c$ such that $c=1(N'),gcd(c,dN')=1$ with $\chi(c)$(if not, then for all $c$ with $c=1(N'), gcd(c,dN')=1$, we have that $\chi(c)=1$. This means that $\chi$ can be defined on $\mathbb{Z}/N'\mathbb{Z}$. in other words, $\chi$ is an extension of some character of $\mathbb{Z}/N'\mathbb{Z}$, which contradicts the fact that $\chi$ is not any extension). Then note that $\Gamma(\chi,m)=\sum_{r(\text{mod}N')}\sum_{n=r(N)}\chi(n)e^{2\pi inm'/N'}=\sum_{r(N')}(\sum_{n=r(N)}\chi(n))e^{2\pi irm'/N'}$. Then $\sum_{n=r(N)}\chi(n)=\sum_{n=r(N)}\chi(nc)\chi(c)^{-1}=\chi(c)^{-1}\sum_{nc=r(N)}\chi(nc)=\chi(c)^{-1}\sum_{n=r(N)}\chi(n)$. Since $\chi(c)\neq1$, we have that this sum is zero, and so $\Gamma(\chi,m)=0=\overline{\chi(m)}\Gamma(\chi,1)$. So, the result is proved.

Now we can define the Dirichlet $L$-function for the primitive character $\chi$, $L(s,\chi)=\sum_{n\in\mathbb{N}-0}\chi(n)n^{-s}$. Using the multiplicity of $\chi$, we have that $L(s,\chi)=\prod_{p,\text{prime}}(1-\frac{\chi(p)}{p^s})$. As for $\chi$, there are two possibilities for $\chi(-1)=1,-1$. We define that $\epsilon(\chi)=0$ if $\chi(-1)=1$ and $\epsilon(\chi)=1$ if $\chi(-1)=-1$. And we define the complete Dirichlet $L$-function to be $\hat{L}(s,\chi)=N^{s/2}\pi^{-(s+\epsilon(\chi))/2}\Gamma(\frac{s+\epsilon(\chi)}{2})L(s,\chi)$. Then we have the corresponding result for this $L$-function:

$L(s,\chi)$ can be extended to the whole complex plane for any non-trivial character on $\mathbb{Z}/N\mathbb{Z}$. And if $\chi$ is a primitive character, then we have the functional equation $\hat{L}(s,\chi)=\frac{\Gamma(\chi,1)}{i^{\epsilon(\chi)}\sqrt{N}}\hat{L}(1-s,\overline{\chi})$.

Note that we have used the same letter $\Gamma$ to refer to both the $\Gamma$-function on $\mathbb{R}_{>0}$ and those on $\mathbb{Z}/N\mathbb{Z}$. When the variables of the function are $(s,m)$, we refer to the first kind, and when the variables are $(\chi,m)$, we refer to the second. Not too confusing.

First of all, let’s look at what $\hat{L}(s,\chi)$ is(we first consider the case $\chi(-1)=1$). $\hat{L}(s,\chi)=(\frac{N}{\pi})^{s/2}\sum_n\int e^{-t}t^{s/2}\frac{\chi(n)}{n^s}dt/t=\sum_n\chi(n)\int e^{-t}(\frac{Nt}{\pi n^2})^{s/2}dt/t$ $=\int t^{s/2}dt/t \sum_{n>0}\chi(n)e^{-t\pi n^2/N}=\int t^{s/2}dt/t f(t,\chi)$ where we denote $f(t,\chi)=\sum_{n>0}\chi(n)e^{-t\pi n^2/N}=\frac{1}{2}\sum_{n\in\mathbb{Z}}\chi(n)e^{-t\pi n^2/N}$(for the lastidentity we have to pose the condition that $\chi(-1)=1$). The form of $f(t,\chi)$ reminds us of the Poisson summation formula. If there is no term $\chi(n)$, then we can exactly apply this formula. Recall that there is a formula $\Gamma(\chi,m)=\overline{\chi}(m)\Gamma(\chi,1)$. In other words, $\Gamma(\overline{\chi},m)=\chi(m)\Gamma(\overline{\chi},1)$. Thus we have that $\Gamma(\overline{\chi},1)f(t,\chi)=1/2\sum_n e^{-t\pi n^2/N}(\Gamma(\overline{\chi},1)\chi(n))=1/2\sum_ne^{-t\pi n^2/N}\Gamma(\overline{\chi},n)$ $=1/2\sum_{k(N)}\overline{\chi}(k)\sum_ne^{-t\pi n^2/N+2\pi in/N}$. If we define $F(x)=e^{-t\pi x^2/N+2\pi ix/N}$, then we got that $\Gamma(\overline{\chi},1)f(t,\chi)=1/2\sum_{k(N)}\overline{\chi}(k)\sum_nF(n)$. So, now we can use the Poisson summation formula to $F(x)$, and we get $\sum_nF(n)=\sum_n\tilde{F}(n)=\sqrt{N/t}\sum_ne^{-\pi(k-nN)^2/Nt}$. So, there is $\Gamma(\overline{\chi},1)f(t,\chi)=1/2\sqrt{N/t}\sum_{k(N)}\overline{\chi}(k)\sum_{n}e^{-\pi(k-nN)^2/Nt}$. Note that $\chi(k)=\chi(k-nN)$ for any character. Moreover, the whole sum $k-nN(k=0,1,...,N-1;n\in\mathbb{Z})$ runs through the whole $\mathbb{Z}$ once and only once, so the last term becomes $1/2\sqrt{N/t}\sum_{m\in\mathbb{Z}}\overline{\chi}(m)e^{-\pi m^2/Nt}$. As a result, we get that $\Gamma(\overline{\chi},1)f(t,\chi)=\sqrt{N/t}f(1/t,\overline{\chi})$. Or, equivalently, $\Gamma(\overline{\chi},1)f(1/t,\chi)=\sqrt{Nt}f(t,\overline{\chi})$. So the next step is to express $\hat{L}(s,\chi)$ in a form more symmetric. In the $\theta$ function, we split the integral into two parts, here we want to try something formal. In fact, $\Gamma(\overline{\chi},1)\hat{L}(s,\chi)=\int t^{s/2}dt/t f(t,\chi)\Gamma(\overline{\chi},1)=\int t^{s/2}dt/t \sqrt{N/t}f(1/t,\overline{\chi})$. After a change of variables $t\mapsto 1/t$, we get that $\Gamma(\overline{\chi},1)\hat{L}(s,\chi)=\sqrt{N}\int t^{(1-s)/2}dt/tf(r,\overline{\chi})=\sqrt{N}\hat{L}(1-s,\overline{\chi})$(we have to say that this process is purely formal, it doesn’t consider any convergence. The legal process should still split the integral into two parts, $\hat{L}(s,\chi)=\int_0^1t^{s/2}dt/tf(t,\chi)+\int_1^{\infty}t^{s/2}dt/tf(t,\chi)=\int_1^{\infty}dt/t(f(1/t,\chi)t^{-s/2}+f(t,\chi)t^{s/2})$, and then use the following identity to get show the symmetry of $\hat{L}(s,\chi)$). So we have really finished our proof except the last identity

$\Gamma(\overline{\chi},1)\Gamma(\chi,1)=\chi(-1)N$.

The proof is a bit tricky. We note that $\overline{\chi}(n)\Gamma(\chi,1)=\Gamma(\chi,n), \chi(n)\Gamma(\overline{\chi},1)=\Gamma(\overline{\chi},n)$. So, there is $|\chi(n)|^2\Gamma(\overline{\chi},1)\Gamma(\chi,1)=\Gamma{\overline{\chi},n}\Gamma(\chi,n)$.Summing over $\mathbb{Z}/N\mathbb{Z}$, we get that $\phi(N)\Gamma(\overline{\chi},1)\Gamma(\chi,1)=\sum_{a,b}\overline{\chi}(a)\chi(b)\sum_ne^{2\pi i(a+b)n/N}$(where $\phi$ is the Euler function). Note that, if $a+b\neq 0$, then $\sum_ne^{2\pi i(a+b)n/N}=0$, and if $a+b=0$, we have that $\sum_ne^0=N$. So, at last we get that $\phi(N)\Gamma(\overline{\chi},1)\Gamma(\chi,1)=\sum_aN\overline{\chi}(a)\chi(-a)=\phi(N)N\chi(-1)$. And this is $\Gamma(\overline{\chi},1)\Gamma(\chi,1)=\chi(-1)N$. Note also that $\chi(-1)=1$(our first assumption), thus, plugging all this into the above equation, we have that $\hat{L}(s,\chi)=\frac{\Gamma(\chi,1)}{\sqrt{N}}\hat{L}(1-s,\overline{\chi})$.

As for the case $\chi(-1)=-1$, we can note write directly $\hat{L}(s,\chi)$ is a symmetric form as above. But let’s try anyway. We have that $\hat{L}(s,\chi)=N^{s/2}\pi^{-(s+1)/2}\Gamma((s+1)/2)L(s,\chi)=\frac{1}{\sqrt{N}}\int t^{(s+1)/2}dt/t\sum_{n>0}n\chi(n)e^{-t\pi n^2/N}$. Thus we denote $g(t,\chi)=\sum_{n>0}n\chi(n)e^{-t\pi n^2/N}=1/2\sum_{n\in\mathbb{Z}}n\chi(n)e^{-t\pi n^2/N}$. Using the same technique as above, we have that $\Gamma(\overline{\chi},1)g(t,\chi)=1/2\sum_{k(N)}\overline{\chi}(k)\sum_nne^{-t\pi n^2/N+2\pi ikn/N}$. So we define $G(x)=ze^{-t\pi x^2/N+2\pi ikx/N}$(which depends on $k$, of course). And applying the Poisson summation formula, we get that $\sum_nG(n)=\sum_n\tilde{G}(n)$ $=\sum_n\int xe^{-t\pi x^2/N+2\pi ikx/N-2\pi inx}=\sum_n i\sqrt{N}e^{-\pi(k-nN)^2/Nt}(k-nN)1/\sqrt{t^3}$. Thus we get that $\Gamma(\overline{\chi},1)g(t,\chi)=\frac{i\sqrt{N}}{2\sqrt{t^3}}\sum_{k(N)}\sum_n\overline{\chi}(k-nN)(k-nN)e^{-\pi(k-nN)^2/Nt}$ $=\frac{i\sqrt{N}}{2\sqrt{t^3}}\sum_m\overline{\chi}(m)e^{-\pi m^2/Nt}=i\sqrt{N/t^3}g(1/t,\overline{\chi})$. So, we have that $\Gamma(\overline{\chi},1)\hat{L}(s,\chi)=1/\sqrt{N}\int t^{(s+1)/2}dt/t\Gamma(\overline{\chi},1)g(t,\chi)=\frac{i\sqrt{N}}{\sqrt{N}}\int t^{(s+1-3)/2}dt/tg(1/t,\overline{\chi})$. This time, we prove the symmetry of the $L$-function rigorously. $\Gamma(\chi,1)\hat{L}(1-s,\overline{\chi})=1/\sqrt{N}\int_0^1t^{(2-s)/2}\Gamma(\chi,1)g(t,\overline{\chi})dt/t+\int_1^{\infty}t^{(2-s)/2}\Gamma(\chi,1)g(t,\overline{\chi})dt/t=1/\sqrt{N}\int_1^{\infty}(t^{(s-2)/2}\Gamma(\chi,1)g(1/t,\overline{\chi})+t^{(2-s)/2}\Gamma(\chi,1)g(t,\overline{\chi}))dt/t$ $=1/\sqrt{N}\int_1^{\infty}(t^{(s-2)/2}i\sqrt{NT^3}g(t,\chi)+t^{(2-s)/2}\Gamma(\chi,1)g(t,\overline{\chi}))dt/t=i\int_1^{\infty}t^{(s+1)/2}g(t,\chi)dt/t+1/\sqrt{N}\int_0^1t^{(s-2)/2}\Gamma(\chi,1)g(1/t,\overline{\chi})dt/t$ $=i\int_1^{\infty}t^{(s+1)/2}g(t,\chi)dt/t+1/\sqrt{N}\int_0^1t^{(s-2)/2}i\sqrt{Nt^3}g(t,\chi)dt/t=i\int t^{(s+1)/2}g(t,\chi)dt/t=i\hat{L}(s,\chi)$ where we have used several times the following identities proved above $\Gamma(\overline{\chi},1)g(t,\chi)=i\sqrt{N/t^3}g(1/t,\overline{\chi})$ and its conjugate $\Gamma(\overline{\chi},1)g(1/t,\chi)=i\sqrt{Nt^3}g(t,\overline{\chi})$.

In the above proofs, we have used substantially the identity

$\overline{\chi}(n)\Gamma(\chi,1)=\Gamma(\chi,n)$ for a primitive character $\chi$. Note that, we really should understand this identity in this way $\Gamma(\chi,n)=\chi(n)^{-1}\Gamma(\chi,1)$, which is a perfect analogy of the real number case, $\Gamma(s,m)=m^{-s}\Gamma(s,1)$.

The fact that $L(s,\chi)$ can be analytically continued to the whole complex plane for non-trivial characters is just a consequence of the fact that $\sum_n\chi(n)=0$ and the Abel’s summation theorem. And we omit the proof here.