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<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.

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