Math 726: L-functions and modular forms Fall 2011

Week 2, lectures 5 and 6: The functional equation for Dirichlet L-functions. Remarks on Artin L-functions attached to higher-dimensional representations.

Instructor: Henri Darmon Notes written by: Juan Ignacio Restrepo

Recall

Last lecture we explained L-functions attached to one dimensional representations of the absolute Galois group of ℚ, G_{ℚ} := Gal(\overline{ℚ}/ℚ). By Class Field Theory, we showed that L(ρ, s) = L(χ, s) for some Dirichlet character χ : (ℤ/qℤ)^{×} ⟶ ℂ^{×} for some integer q, called the conductor of χ. Throughout these lectures, we will assume that q is prime. As an exercise, think about the case of general conductor q.

Functional Equation of L(χ, s)

For the case of an even characater, i.e. χ(−1) = 1, we introduced

Λ(χ, s) = π^{−s/2}q^{s/2}Γ (\frac{s}{2}) L(χ, s) = ∫_{0}^{∞} ω(t, χ)t^{s/2} \frac{dt}{t} = M(ω(t, χ), \frac{s}{2}),

where ω(t, χ) = ∑_{n=1}^{∞} χ(n)e^{−πn^{2}t/q} = \frac{1}{2} ∑_{n∈ℤ}χ(n)e^{−πn^{2}t/q} = \frac{1}{2}θ(t, χ).

As with the Riemann ζ function, using the Poisson Summation Formula we will derive a functional equation for θ(t, χ) of the form

θ (\frac{1}{t}, χ) = ∗\sqrt{t}θ(t, \overline{χ}),

where ∗ is a factor we will determine soon.

Last time we derived the equation

∑_{n∈ℤ}e^{−π(a+qn)^{2}/qt} = \sqrt{\frac{t}{q}} ∑_{n∈ℤ}e^{−πn^{2}t/q}ψ_{a}(n),  (1)

where ψ_{a} : ℤ/qℤ ⟶ ℂ^{×} is the additive character given by ψ_{a}(n) = e^{2πian/q}. Now, to relate the right hand side to θ(t, χ) it is necessary to replace the additive character ψ_{a} by by the multiplicative character χ, or, more precisely, to express the latter as a linear combination of the characters ψ_{a} for some values of a. In order to do this we will do \underline{Fourier Analysis} on the finite group ℤ/qℤ.

Let L^{2}(ℤ/qℤ, ℂ) = {ℂ-valued functions on ℤ/qℤ} be the Hilbert space equipped with the inner product

⟨f, g⟩ = \frac{1}{q} ∑_{a∈ℤ/qℤ}f(a)\overline{g(a)}.

The characters ψ_{0}, …, ψ_{q−1} are an orthonormal basis of L^{2}(ℤ/qℤ, ℂ) and in particular, for

all f ∈ L^{2}(ℤ/qℤ, ℂ) we have f = ∑_{j=0}^{q−1}⟨f, ψ_{j}⟩ψ_{j}. We are mainly interested in f = χ, in which

case we have the following properties:

(a) ⟨χ, ψ_{0}⟩ = 0 (because χ ≠ 1).

(b) ⟨χ, ψ_{1}⟩ = \frac{1}{q} ∑_{j=0}^{q−1} χ(j)e^{−2πij/q} =: \frac{τ(χ)}{q}. τ(χ) is called the \underline{Gauss sum} attached to χ. Some

properties of the Gauss sums, which are left as an exercise are

• τ(\overline{χ}) = χ(−1)\overline{τ(χ)}.

• τ(χ)\overline{τ(χ)} = q, i.e., ∥τ(χ)∥ = \sqrt{q}.

(c) If a ∈ (ℤ/qℤ)^{×}, then ∃a′ ∈ (ℤ/qℤ)^{×} such that aa′ ≡ 1 (mod q). As j runs from 0 to q − 1, so does a′j so

⟨χ, ψ_{a}⟩ = \frac{1}{q} ∑_{j=0}^{q−1} χ(j)e^{−2πiaj/q} = \frac{1}{q} ∑_{j=0}^{q−1} χ(a′j)e^{−2πiaa′j/q}

= \frac{1}{q} ∑_{j=0}^{q−1} χ(a′j)e^{−2πij/q} = \overline{χ}(a) · \frac{1}{q} ∑_{j=0}^{q−1} χ(j)e^{−2πij/q}

= \overline{χ}(a)\frac{τ(χ)}{q}.

From the previous properties we deduce that χ(n) = \frac{τ(χ)}{q} ∑_{a=0}^{q−1} \overline{χ}(a)ψ_{a}(n). We multiply (1)

by \frac{τ(χ)}{q} · \overline{χ}(a) and we sum over a = 0, …, q − 1 to obtain

\frac{τ(χ)}{q} ∑_{a=0}^{q−1} \overline{χ}(a) ∑_{n∈ℤ}e^{−π(a+qn)^{2}/qt} = \sqrt{\frac{t}{q}} ∑_{n∈ℤ}e^{−πn^{2}t/q}χ(n).

The LHS can be written as

\frac{τ(χ)}{q} ∑_{n∈ℤ} ∑_{a=0}^{q−1} \overline{χ}(a + qn)e^{−π(a+qn)^{2}/qt},

because \overline{χ} is q-periodic, but as a runs from 0 to q − 1 and n runs over all integers, a + qn runs over all integers exactly once, so the LHS simplifies to

\frac{τ(χ)}{q} ∑_{n∈ℤ} \overline{χ}(n)e^{−πn^{2}/qt}

and thus we obtain

\frac{τ(χ)}{q}θ (\frac{1}{t}, \overline{χ}) = \sqrt{\frac{t}{q}} θ(t, χ),

which yields

θ (\frac{1}{t}, \overline{χ}) = (\frac{\sqrt{q}}{τ(χ)}) · \sqrt{t}θ(t, χ)

It is now easy to write down the functional equation for Λ(χ, s) and show that it extends analytically to the whole complex plane. We will apply exactly the same techniques we applied for the Riemann ζ-function; namely, we will use its expression as a Mellin transform, we will split the integral into two parts, one of them with nice convergence everywhere and in the other one we apply the substitution t ↦ 1/t, where we can use the functional equation we just derived for θ(t, χ) as follows:

2Λ(χ, s) = ∫_{0}^{∞} θ(t, χ)t^{s/2} \frac{dt}{t} = ∫_{0}^{1} θ(t, χ)t^{s/2} \frac{dt}{t} + ∫_{1}^{∞} θ(t, χ)t^{s/2} \frac{dt}{t}

= ∫_{1}^{∞} θ (\frac{1}{t}, χ) t^{−s/2} \frac{dt}{t} + ∫_{1}^{∞} θ(t, χ)t^{s/2} \frac{dt}{t}

= \frac{\sqrt{q}}{τ(\overline{χ})} ∫_{1}^{∞} \sqrt{t}θ(t, \overline{χ})t^{−s/2} \frac{dt}{t} + ∫_{1}^{∞} θ(t, χ)t^{s/2} \frac{dt}{t}

= \frac{τ(χ)}{\sqrt{q}} ∫_{1}^{∞} θ(t, \overline{χ})t^{\frac{1−s}{2}} \frac{dt}{t} + ∫_{1}^{∞} θ(t, χ)t^{s/2} \frac{dt}{t},

which clearly is entire, so Λ(χ, s) extends analytically to the whole complex plane. Furthermore, we can see the relation

Λ(χ, s) = (\frac{τ(χ)}{\sqrt{q}}) · Λ(\overline{χ}, 1 − s).

\underline{Remarks}:

• Λ(χ, s) is everywhere holomorphic (no poles at s = 0 and/or s = 1).

• The functional equations relates Λ(χ, s) to Λ(\overline{χ}, 1 − s) and involves a ”root number” τ(χ)/\sqrt{q}.

• Since L(χ, s) admits an Euler product expansion for ℜ[s] > 1, it does not vanish in said half-plane. The analyticity of Λ(χ, s) and the poles of Γ(s) at the nonpositive integers yield zeroes of L(χ, s) at the nonpositive even integers. These are the socalled trivial zeroes of the L-function. The functional equation shows that there are no more zeroes in the half-plane ℜ[s] < 0 and it is conjectured that all the zeroes in the critical strip (i.e., 0 ≤ ℜ[s] ≤ 1) lie on the line ℜ[s] = \frac{1}{2}. This is the Generalized Riemann Hypothesis.

• Dirichlet proved that L(χ, 1) ≠ 0, which is a key ingredient in his proof of the theorem with his name on primes in arithmetic sequences.

Now we’ll deal with odd characters, i.e., χ(−1) = −1. Using exactly the same ideas as for even characters is useless because θ(t, χ) is 0, unless we modify the definition of ω(t, χ), so we will multiply by another function, or quasi-character, to make it an even character, namely, instead of χ(n) we will write nχ(n). More concretely, the idea we will use is defining

ω(t, χ) = ∑_{n=1}^{∞} nχ(n)e^{−πn^{2}t/q} = \frac{1}{2} ∑_{n∈ℤ}nχ(n)e^{−πn^{2}t/q} = \frac{1}{2}θ(t, χ).

Proposition 1. M (ω, \frac{s}{2}) = q^{s/2}π^{−s/2}Γ (\frac{s}{2}) L(χ, s − 1).

Proof. It’s just a simple computation.

M (ω, \frac{s}{2}) = ∫_{0}^{∞} ω(t, χ)t^{s/2} \frac{dt}{t} = ∫_{0}^{∞} ∑_{n=1}^{∞} nχ(n)e^{−πn^{2}t/q}t^{s/2} \frac{dt}{t}

= ∑_{n=1}^{∞} nχ(n) (∫_{0}^{∞} e^{−πn^{2}t/q}t^{s/2} \frac{dt}{t})

= ∑_{n=1}^{∞} nχ(n) ∫_{0}^{∞} e^{−u}q^{s/2}π^{−s/2}n^{−s}u^{s/2} \frac{du}{u}

= (∑_{n=1}^{∞} χ(n)n^{1−s}) Γ (\frac{s}{2}) π^{−s/2}q^{s/2}

Corollary 1. M (ω, \frac{s + 1}{2}) = q^{\frac{s+1}{2}} π^{− \frac{s+1}{2}} Γ (\frac{s + 1}{2}) L(χ, s).

Since we are using a different θ(t, χ), we can’t expect the same functional equation to work. Instead, we will derive another one. Recall the identity obtained from the Poisson Summation Formula,

∑_{n∈ℤ}e^{−π(x+n)^{2}/t} = \sqrt{t} ∑_{n∈ℤ}e^{−πn^{2}t}e^{2πinx}.

Since we have nice convergence properties on both sides of the equality, we are allowed to take derivatives with respect to x and exchange the limit processes of summation and differentiation, yielding the equation

∑_{n∈ℤ} \frac{−2π(x + n)}{t}e^{−π(x+n)^{2}/t} = \sqrt{t} ∑_{n∈ℤ}2πine^{−πn^{2}t}e^{2πinx},

whence

i ∑_{n∈ℤ}(x + n)e^{−π(x+n)^{2}/t} = t^{3/2} ∑_{n∈ℤ}ne^{−πn^{2}t}e^{2πinx}.

Set x = \frac{a}{q}, a ∈ ℤ and replace t by \frac{t}{q} to obtain

i ∑_{n∈ℤ}(a + qn)e^{−π(a+qn)^{2}/tq} = \frac{t^{3/2}}{\sqrt{q}} ∑_{n∈ℤ}ne^{−πn^{2}t/q}ψ_{a}(n).

After multiplying by \frac{τ(χ)}{q} · \overline{χ}(a) on both sides and summing over a = 0, …, q − 1 we find that

i · \frac{τ(χ)}{q} ∑_{n∈ℤ}n\overline{χ}(n)e^{−πn^{2}/tq} = \frac{t^{3/2}}{\sqrt{q}} ∑_{n∈ℤ}nχ(n)e^{−πn^{2}tq}

and it follows that

θ (\frac{1}{t}, \overline{χ}) = (\frac{\sqrt{q}}{iτ(χ)}) · t^{3/2}θ(t, χ).

Now we get the following corollary.

Corollary 2. Let χ be an odd Dirichlet character. The function

Λ(χ, s) = q^{\frac{s+1}{2}} π^{− \frac{s+1}{2}} Γ (\frac{s + 1}{2}) L(χ, s)

satisfies the functional equation

Λ(χ, s) = (\frac{iτ(χ)}{\sqrt{q}}) Λ(\overline{χ}, 1 − s)

and is everywhere holomorphic.

Proof. Exercise. (Hint: Follow the walkthrough given for even characters.)

Corollary 3. If χ is an odd Dirichlet character, L(χ, s) is an entire function with zeroes at all negative odd integers. Moreover, it vanishes nowhere else outside of the critical strip 0 ≤ ℜ[s] ≤ 1.

\underline{Question:} What about more general continuous ρ : G_{ℚ} ⟶ GL_{d}(ℂ) ≅ Aut(V_{ρ}), V_{ρ} ≅ ℂ^{d},

L(ρ, s) = ∏_{p} det(1 − ρ(σ_{p})p^{−s} ⟲ V_{ρ}^{I_{p}})^{−1}

= ∑_{n=1}^{∞} a_{n}n^{−s} a_{p} = tr(σ_{p} ⟲ V_{ρ}),

where σ_{p} is the Frobenius element at p? What kind of patterns do the coefficients in this Dirichlet series satisfy?

\underline{Artin-Conjecture:} Let ρ be any continuous non trivial irreducible representation of G_{ℚ}. Then L(ρ, s) extends to an analytic function on all of ℂ. Moreover, there is a prediction for what the functional equation looks like!! (Generalization of Class Field Theory to nonabelian representations.)

Status of the conjecture:

(a) It is known for d = 2, det ρ(−1) = −1. This was finished around 2006 putting together the work of Hecke, Langlands-Tunnell, Serre-Deligne, Wiles, Taylor, Khare-Wintenberger. It is highly nontrivial.

(b) Very little is known for d > 2.

A much easier result is available.

Theorem 1 (Artin). The function L(ρ, s) extends to a meromorphic function of s ∈ ℂ.

Sketch of Proof. Observe that if ρ is a general representation, we can write ρ = ⨁_{i=1}^{t} m_{i}ρ_{i},

where m_{i} ∈ ℤ^{>0} and the ρ_{i} are irreducible. L(ρ, s) = ∏_{i=1}^{t} L(ρ_{i}, s)^{m_{i}}.

\underline{Key remark:} Class Field Theory over a field K allows us to analyse the L-function L_{K}(ρ, s) attached to ρ : G_{K} := Gal(\overline{K}/K) ⟶ ℂ^{×},

L_{K}(ρ, s) = ∏_{℘◁O_{K}}(1 − ρ(F rob_{℘})N(℘)^{−s})^{−1} (= 0 if ρ(I_{℘}) ≠ 1).

Class Field Theory for K shows that L_{K}(ρ, s) = L_{K}(χ, s) where χ is a ”Hecke character” of K.

Theorem 2 (Hecke(Tate’s Thesis)). L_{K}(ρ, s) extends to a holomorphic function of s when ρ ≠ 1 (with eventually a pole at s = 1).

We have the examples ζ_{K}(s) = L(Ind_{K}^{ℚ}1_{K}, s) (assignment) and L_{K}(χ, s) = L(Ind_{K}^{ℚ}χ, s), where Ind_{K}^{ℚ} is a one dimensional representation of G_{ℚ}. Then, Artin’s conjecture is true for all ρ which are induced from abelian characters of K (K varying). More generally, if there exist K_{1}, …, K_{t} number fields and χ_{1}, …, χ_{t} Hecke characters of these fields such

that ρ = ⨁_{i=1}^{t} m_{i}Ind_{K_{i}}^{ℚ}χ_{i} with m_{i} ≥ 1 then L(ρ, s) satisfies Artin’s conjecture.

Theorem 3 (Brauer’s Theorem). Let ρ be any representation of a finitely generated group G. Then, there exist

• subgroups H_{1}, …, H_{t} ≤ G

• charachters χ_{j} : H_{j} ⟶ ℂ^{×}

• (not necessarily positive) integers m_{1}, …, m_{t}

such that

ρ = ∑_{i=1}^{t} m_{i}Ind_{H_{i}}^{G}χ_{i}

When applied to ρ : Gal(K/ℚ) ⟶ GL_{d}(ℂ) we get K_{i} = K^{H_{i}} and χ_{i} such that

L(ρ, s) = ∏_{i=1}^{t} L_{K_{i}}(χ_{i}, s)^{m_{i}}.