THE ANALYTIC CONTINUATION OF DIRICHLET L-SERIES

CRISTINA TOROPU

Definition 1. Let G be a finite abelian group.A character of Gis a homomorphism of G in C^{∗}

χ : G → C^{∗}such thatχ(ab) = χ(a)χ(b)andχ(1) = 1

\overline{χ}(a) = \overline{χ(a)} for all a ∈ G Definition 2. Let G = (\frac{Z}{mZ} )^{∗} for all m ≥ 1

A function χ^{∗} : Z → C^{∗}

χ^{∗}(a) = \{ \begin{matrix}χ(\overline{a}) & if (a, m) = 1 \\ 0 & if (a, m) > 1 \end{matrix}

is called Dirichlet character modulo m.

Let d | m and let χ be a character modulo d.Let’s define

χ′(a) = \{ \begin{matrix}χ(a) & if (a, m) = 1 \\ 0 & if (a, m) > 1 \end{matrix}

We say that χ induces χ′.

Definition 3. Let χ be a character modulo m.If there exists d | m and a character χ′ modulo d such that χ′ induces χ,then χ is called nonprimitive.Otherwise is called primitive. Proposition 1. A χ modulo m is a primitive character if ∀d | m ∃x ≡ 1 (mod m) such that χ(x) ≠ 1

Proof. It is equivalent to prove that χ is nonprimitive if ∃d | m for all x ≡ 1 mod m such that χ(x) = 1. Since χ is nonprimitive there exists a character χ′ mod d such that χ(x) = χ′(x) = 1 □

Dirichlet L-functions

If χ is a Dirichlet character,primitive or not,we define a L-function

L(s, χ) = ∑_{n=1}^{∞} \frac{χ(n)}{n^{s}}

By comparison with Riemann zeta-function it is absolutely convergent for Re(s) > 1. It has an Euler product. For re(s) > 1 we prove that

(1)  L(s, χ) = ∏_{p} (1 − \frac{χ(p)}{p^{s}})^{−1}

(1 − \frac{χ(p)}{p^{s}})^{−1} = ∑_{k=o}^{∞} \frac{χ(p)^{k}}{p^{ks}}

For any positive integer n there exist exactly one way to write

n^{−s} = (∏ p_{i}^{k_{i}})^{−s}

On expanding

∏_{p} ∑_{k=0}^{∞} \frac{χ(p)^{k}}{p^{ks}}

the coefficient of n^{−s} is ∏_{i} χ(p_{i}^{k_{i}}) = χ(n) Let χ be a primitive character modulo N

The Gauss sum τ(χ) is defined by

(2)  τ(χ) = ∑_{n mod N} χ(n)e^{\frac{2πin}{N}}  n = 1, …, N − 1

We prove that :

(3)  ∑_{n mod N} χ(n)e^{\frac{2πinm}{N}} = \overline{χ(m)}τ(χ)

We have 2 cases: If (m, N) = 1 |χ(m)| = 1 Making the change of variables n → nm

(4)  τ(χ) = ∑_{n mod N} χ(nm)e^{\frac{2πinm}{N}} = χ(m) ∑_{n mod N} χ(n)e^{\frac{2πinm}{N}}

Multiplying this by χ^{−1}(m) = \overline{χ(m)} we obtain (3) If (m, n) > 1 let d = (m, n) ⇒ m = dM and N = dN_{1} So

∑_{n mod N} χ(n)e^{\frac{2πinm}{N_{1}}} = ∑_{r mod N_{1}} \{ ∑ ; \begin{matrix}n mod N \\ n ≡ r mod N_{1} \end{matrix} χ(n)\}e^{\frac{2πirM}{N_{1}}}

The substitution

n → cn

permutes the residue classes n mod N such that

n ≡ rmodN_{1}

among themselves,so

∑ ; \begin{matrix}n mod N \\ n ≡ rmodN_{1} \end{matrix} χ(n) = ∑ ; \begin{matrix}n mod N \\ n ≡ rmodN_{1} \end{matrix} χ(cn) = χ(c) ∑ ; \begin{matrix}n mod N \\ n ≡ rmodN_{1} \end{matrix} χ(n)

Since χ is a primitive character ⇒ ∀d|N there exist c ≡ 1 mod d such that χ(c) ≠ 1 So

∑ ; \begin{matrix}n mod N \\ n ≡ rmodN_{1} \end{matrix} χ(n) = 0

,so (3) is proved We prove τ(χ) ≠ 0

(5)  |τ(χ)| = \sqrt{N}

Because:

\overline{χ(n)e^{\frac{2πinm}{N}}} = \overline{χ(n)}e^{\frac{−2πinm}{N}}

| ∑_{n mod N} χ(τ)e^{\frac{2πinm}{N}} |^{2} = ∑ ; \begin{matrix}n_{1}, n_{2} mod N \\ (n_{1}n_{2}, N) = 1 \end{matrix} χ(n_{1})\overline{χ(n_{2}}e^{\frac{2πi(n_{1}−n_{2})m}{N}}

This becomes by(3)

\{ \begin{matrix}|τ(χ)|^{2} & if (m, N) = 1 \\ 0 & if (m, N) > 1 \end{matrix}

Summing over all m modulo N

Φ(N)|τ(χ)|^{2} = ∑_{m mod N} ∑ ; \begin{matrix}n_{1}, n_{2}bmodN \\ (n_{1}n_{2}, N) = 1 \end{matrix} χ(n_{1})\overline{χ(n_{2})}e^{\frac{2πi(n_{1}−n_{2})m}{N}}

whereΦ(N) is the number of residue classes mod N prime to N Because

∑_{m mod N} e^{\frac{2πiam}{N}} = \{ \begin{matrix}N & if N|a \\ 0 & if N doesn’t divide a \end{matrix}

Only terms with n_{1} ≡ n_{2} mod N contribute

χ(n_{1})\overline{χ(n_{2})} = 1

So

Φ(N)|χ(τ)|^{2} = ∑ ; \begin{matrix}n_{1}, n_{2} mod N \\ (n_{1}n_{2}, N) = 1 \\ n_{1} ≡ n_{2}modN \end{matrix} N = Φ(N)N

⇒ |χ(τ)|^{2} = \sqrt{N}

Since

τ(χ) ≠ 0

we interpolate a primitive Dirichlet character between integers. We interchange

χ → \overline{χ}

in(3) and change n with m

(6)  ⇒ χ(n) = \frac{1}{τ(\overline{χ})} ∑_{m mod N} \overline{χ(m)}e^{\frac{2πimn}{N}}

Since

τ(\overline{χ}) = \overline{τ(χ)} · χ(−1)
|τ(χ)|^{2} = N ⇒ τ(χ)\overline{τ(χ)} = N
(7)  → χ(n) = \frac{χ(−1)τ(χ)}{N} ∑_{m mod N} \overline{χ(m)}e^{\frac{2πimn}{N}}

This is defined when n is an arbitrary real number. We prove more twisted Poisson summation:

(8)  ∑_{n=−∞}^{∞} χ(n)f(n) = \frac{χ(−1)τ(χ)}{N} ∑_{−∞}^{∞} \overline{χ(n)}f̂(\frac{n}{N} )

But

∑_{−∞}^{∞} χ(n)f(n) = ∑ f_{1}(n)

where

f_{1}(x) = \frac{χ(−1)τ(χ)}{N} ∑_{m mod N} \overline{χ(m)}e^{\frac{2πixm}{N}} f(x)

Applying Poisson summation formula we get:

f̂_{1}(x) = \frac{χ(−1)τ(χ)}{N} ∑_{m mod N} \overline{χ(m)}f̂(\frac{Nx + m}{N})

∑_{n=−∞}^{∞} χ(n)f(n) = \frac{χ(−1)τ(χ)}{N} ∑_{m mod N} ∑_{n=−∞}^{∞} \overline{χ(m)}f̂(\frac{Nn + m}{N}) = \frac{χ(−1)τ(χ)}{N} ∑_{m mod N} ∑_{n=−∞}^{∞} \overline{χ(n)}f̂(\frac{n}{N} )

This is happend because

χ(m) = χ (Nm + m)

and Nn + m runs uniquely through Z when m runs through a set of residue classes mod N and n runs through Z We compute Fourier transform of the Gaussian distribution.Let t have positive real part and let

f_{t} = e^{−πtx^{2}}

f_{t}(x) is of faster –than–polynomial decay as x → ±∞

f̂_{t}(x) = \frac{1}{\sqrt{t}}f_{\frac{1}{t}}

Both sides are analytical functions of t defined whenre(t) > 0.By completing square:

f̂_{t}(x) = ∫_{−∞}^{∞} e^{−π(ty^{2}−2πixy)}dy = e^{− \frac{πx^{2}}{t}} ∫_{−∞}^{∞} e^{−π(\sqrt{t}y− \frac{ix}{\sqrt{t}})^{2}}dy

By using Cauchy’s theorem to shift the path of integration with respect to y vertically by a constant distance depending on x.

y → y + \frac{ix}{t}

⇒ f̂_{t}(x) = e^{\frac{−πx^{2}}{t}} ∫_{−∞}^{∞} e^{−πty^{2}}dy = \frac{c}{\sqrt{t}}e^{\frac{−πx^{2}}{t}}

where

c = ∫_{−∞}^{∞} e^{−πy^{2}}dy

⇒ ˆ t̂_{t} = \frac{c}{\sqrt{t}}f̂_{\frac{1}{t}} = c^{2}f_{t}

We use Fourier inversion formula

ˆf̂_{t}(x) = f(−x)

if f is any continuous function such that f and f̂ are in L^{1}(R)

Applying this to even function ⇒ ˆ f̂_{t} = f_{t} and ⇒ c^{2} = 1 ⇒ c = 1 and c > 0 Let’s define some theta functions.

Let χ be a primitive character mod N. First assume χ(−1) = 1 θ_{χ} = \frac{1}{2} ∑_{n=−∞}^{∞} χ(n)e^{−πn^{2}t} = \frac{1}{2}χ(0) + ∑_{n=1}^{∞} χ(n)e^{−πn^{2}t} when re(s) > 0 (χ(0) = 0 unless N = 1 because χ(n) is primitive) Using f̂_{t} = \frac{1}{\sqrt{t}}f_{\frac{1}{t}} and twisted Poisson formula we obtain the functional equation:

∑_{n=−∞}^{∞} χ(n)f(n) = \frac{χ(−1)}{N}τ(χ) ∑_{n=−∞}^{∞} \overline{χ(n)}f̂(\frac{n}{N})

θ_{χ}(t) = \frac{τ(χ)}{N\sqrt{t}}θ\overline{χ}(\frac{1}{N^{2}t}) if χ(−1) = 1 Now assume χ(−1) = −1 Let g_{t}(x) = xe^{−πtx^{2}} = −\frac{1}{2πt}(\frac{df_{t}}{dx} )(x)

ĝ_{t}(x) = \frac{i}{t^{\frac{3}{2}}} g_{\frac{1}{t}}

ĝ_{t}(x) = ∫_{−∞}^{∞} g_{t}(y)e^{2πixy}dy = −\frac{1}{2πt} ∫_{−∞}^{∞} (\frac{df_{t}}{dy} )(y)e^{2πixy}dy

Integrating by parts this becomes equal to \frac{ix}{t} ∫_{−∞}^{∞} f_{t}(y)e^{2πixy}dy = \frac{ix}{t} f̂_{t} = \frac{ix}{t} \frac{1}{\sqrt{t}}f_{1}t = \frac{i}{t^{3}2}g_{\frac{1}{t}} If χ(−1) = −1 we define θ_{χ}(t) = \frac{1}{2} ∑_{n=−∞}^{∞} nχ(n)e^{−πn^{2}t} = ∑_{n=1}^{∞} nχ(n)e^{−πn^{2}t} Applying twisted Poisson summation to g_{t} ∑_{n=−∞}^{∞} χ(n)f(n) = \frac{χ(−1)τ(χ)}{N} ∑_{n=−∞}^{∞} \overline{χ(n)}f̂(\frac{n}{N})

θ_{χ}(t) = \frac{−iτ(χ)}{N^{2}t^{\frac{3}{2}}}θ\overline{χ}(\frac{1}{N^{2}t}) if χ(−1) = −1 Theorem 1. Let χ be a primitive Dirichlet character with conductor N and let ϵ = 0or1 be chosen so that χ(−1) = (−1)^{ϵ}.Let

Λ(s, χ) = π^{− \frac{(s+ϵ)}{2}} Γ (\frac{s + ϵ}{2}) L(s, χ)

ThenΛ(s, χ) has a meromorphic continuation to all s:indeed ifχ ≠ 1 it is entire,while if χ = 1 it is analytic for all s except s = 1 or s = 0 where it has simple poles.We have the functional equation

Λ(s, χ) = (−1)^{ϵ}τ(χ)N^{−s}Λ(1 − s, \overline{χ})

Proof. If χ is the trivial character,χ primitive ⇒ N = 1 ⇒ L(s, χ) is just Riemann zeta function. Letχ ≠ 1.Because χ(0) = 0,the series θ_{χ}(t) is a sum of terms of n^{ϵ}χ(n)e^{−πn^{2}t} with |n| ≥ 1 ,each of which is of very decay as t → ∞ Combining θ_{χ} for χ(−1) = −1

θ_{χ}(t) = \frac{(−1)^{ϵ}τ(χ)}{N^{1+ϵ}t^{ϵ+ \frac{1}{2}}}θ\overline{χ} (\frac{1}{N^{2}t})

→ θ_{χ(t)} is also of very decay as t → 0 →Mellin transform

∫_{0}^{∞} θ_{χ}(t)t^{\frac{s+ϵ}{2}} \frac{dt}{t}

is convergent for all s.If re(s) > 1 we have the identity:

∫_{0}^{∞} e^{−πtn^{2}}t^{\frac{s+ϵ}{2}} \frac{dt}{t} = π^{− \frac{s+ϵ}{2}} Γ (\frac{s + ϵ}{2}) n^{−s−ϵ}

So∫_{0}^{∞} θ_{χ}(t)t^{\frac{s+ϵ}{2}} \frac{dt}{t} = Λ(s, χ)

Because Mellin transform is convergent for all s,and clearly defines an analytic function,this gives the analytic continuation of Λ(s, χ) We are in the case χ(−1) = −1 because χ ≠ 1 So substituting θ_{χ}(t) = \frac{−iτ(χ)}{N^{2}t^{\frac{3}{2}}}θ\overline{χ}(\frac{1}{N^{2}t}) in Mellin transform and making the change

of variables t → \frac{1}{N^{2}t}

∫_{0}^{∞} θ_{χ}(t)t^{\frac{s+ϵ}{2}} \frac{dt}{t} = (−1)^{ϵ}τ(χ)N^{−s} ∫_{0}^{∞} θ\overline{χ}(t)t^{\frac{1−s+ϵ}{2}} \frac{dt}{t} = (−1)^{ϵ}τ(χ)N^{−s}Λ(1 − s, \overline{χ})

So

Λ(s, χ) = (−i)^{ϵ}τ(χ)N^{−s}Λ(1 − s, \overline{χ})
□