Modular Forms, PS 7

Bryden Cais

October 28, 2002

Exercise 1. Recall we showed in class that

Θ(s)E(z, s) = Θ(s)y^{s} + Θ(s − 1/2)y^{1−s} + ∑_{r≠0} \frac{W_{s}(rz)}{\sqrt{|r|}} ( ∑_{ab=|r|} (\frac{a}{b})^{s−1/2}) ,

where

Θ(s) = π^{−s}Γ(s)ζ(2s).

Observe that Θ(s) has simple poles corresponding to a pole of the Gamma function at s = 0 and a pole of ζ(2s) as s = 1/2. Since ζ(2s) has simple zeroes at s = −n for any positive integer n, we see that Θ(s) is holomorphic everywhere except at s = 0, 1/2. The residue at s = 1/2 is readily seen to be

R_{1/2}(Θ(s)) = π^{−1/2}Γ(1/2)\frac{1}{2} = \frac{1}{2},

while at 0 we have

R_{0}(Θ(s)) = ζ(0) = −\frac{1}{2}.

We have seen that W_{s}(z) is a holomorphic function of s, for any value of z, and that as ℑ(z) tends to infinity, W_{s}(z) decays exponentially. It follows that any pole of Θ(s)E(z, s) comes from the term Θ(s)y^{s} +Θ(s−1/2)y^{1−s}. Clearly this term has (potential) simple poles at s = 1/2, 0, 1 and nowhere else. By our residue calculations above, the residue of Θ(s)y^{s} + Θ(s − 1/2)y^{1−s} at s = 1/2 is

\frac{1}{2}y^{1/2} − \frac{1}{2}y^{1−1/2} = 0,

so that, in fact, E(z, s) is holomorphic at s = 1/2. A short computation shows that the residue at 0 is

−\frac{1}{2},

while at 1 it is

R_{1}(Θ(s − 1/2)) = \frac{1}{2}.

Since Θ(s)E(z, s) has a simple pole of residue −\frac{1}{2} at s = 0, we see that E(z, s) is holomorphic at s = 0. Finally, since Θ(z, s) is holomorphic at s = 1, we see that E(z, s) has a simple pole at s = 1 and that

R_{1}(Θ(s)E(z, s)) = \frac{1}{2}.

Observe that W_{s} = W_{1−s} since the Bessel function K_{v} is an even function of v and W_{s} = K_{s−1/2}. Moreover, it is clear that the divisor sum

∑_{ab=|r|} (\frac{a}{b})^{s−1/2}

is invariant under s ↦ 1 − s. We thus have

Θ(1 − s)E(z, 1 − s) = Θ(1 − s)y^{1−s} + Θ(1/2 − s)y^{s} + ∑_{r≠0} \frac{W_{1−s}(rz)}{\sqrt{|r|}} ( ∑_{ab=|r|} (\frac{a}{b})^{1/2−s})

= Θ(1 − s)y^{1−s} + Θ(1/2 − s)y^{s} + ∑_{r≠0} \frac{W_{s}(rz)}{\sqrt{|r|}} ( ∑_{ab=|r|} (\frac{b}{a})^{s−1/2}) .

Since Θ(s) = π^{−s}Γ(s)ζ(2s), we have Θ(1/2 − s) = Θ(s). It follows that

Θ(1 − s)E(z, 1 − s) = Θ(s − 1/2)y^{1−s} + Θ(s)y^{s} + ∑_{r≠0} \frac{W_{s}(rz)}{\sqrt{|r|}} ( ∑_{ab=|r|} (\frac{b}{a})^{s−1/2}) = Θ(s)E(z, s).

Exercise 2. Let f be a normalized Hecke Eigenform with fourier expansion

∑_{n=1}^{∞} a_{f}(n)n^{\frac{k−1}{2}} q^{n}.

For a prime p, write a_{f}(p) = α_{p} + β_{p} with α_{p}β_{p} = 1. Recall that we have the relations

a_{f}(m)a_{f}(n) = ∑_{d|(m,n)} a_{f} (\frac{mn}{d^{2}})

and

a_{f}(p^{n}) = α_{p}^{n} + α_{p}^{n−1}β_{p} + ⋯ + β_{p}^{n} = \frac{α_{p}^{n+1} − β_{p}^{n+1}}{α_{p} − β_{p}} ,

with appropriate interpretation if α_{p} = β_{p}. We can therefore write

ζ(2s) ∑_{n=1}^{∞} \frac{a_{f}(n)ā_{f}(n)}{n^{s}} = ∏_{p} \frac{1}{1 − p^{−2s}} ∑_{n=0}^{∞} \frac{a_{f}(p^{n})ā_{f}(p^{n})}{p^{ns}}

= ∏_{p} \frac{(1 − p^{−2s})^{−1}}{(α_{p} − β_{p})(ᾱ_{p} − β̄_{p})} ∑_{n=0}^{∞} \frac{(α_{p}^{n+1} − β_{p}^{n+1})(ᾱ_{p}^{n+1} − β̄_{p}^{n+1})}{p^{ns}}

= ∏_{p} \frac{(1 − p^{−2s})^{−1}}{(α_{p} − β_{p})(ᾱ_{p} − β̄_{p})} (\frac{α_{p}ᾱ_{p}}{1 − \frac{α_{p}ᾱ_{p}}{p^{s}}} − \frac{α_{p}β̄_{p}}{1 − \frac{α_{p}β̄_{p}}{p^{s}}} − \frac{β_{p}ᾱ_{p}}{1 − \frac{β_{p}ᾱ_{p}}{p^{s}}} + \frac{β_{p}β̄_{p}}{1 − \frac{β_{p}β̄_{p}}{p^{s}}} )

= ∏_{p} \frac{(1 − p^{−2s})^{−1}}{(α_{p} − β_{p})(ᾱ_{p} − β̄_{p})} (\frac{1}{β_{p}β̄_{p} − \frac{1}{p^{s}}} − \frac{1}{β_{p}ᾱ_{p} − \frac{1}{p^{s}}} − \frac{1}{α_{p}β̄_{p} − \frac{1}{p^{s}}} + \frac{1}{α_{p}ᾱ_{p} − \frac{1}{p^{s}}} )

= ∏_{p} \frac{(1 − p^{−2s})^{−1}}{(α_{p} − β_{p})(ᾱ_{p} − β̄_{p})} (\frac{α_{p}ᾱ_{p} + β_{p}β̄_{p} − \frac{2}{p^{s}}}{1 − \frac{1}{p^{s}} (α_{p}ᾱ_{p} + β_{p}β̄_{p}) + \frac{1}{p^{2s}}} − \frac{α_{p}β̄_{p} + ᾱ_{p}β_{p} − \frac{2}{p^{s}}}{1 − \frac{1}{p^{s}} (α_{p}β̄_{p} + β_{p}ᾱ_{p}) + \frac{1}{p^{2s}}} )

= ∏_{p} (\frac{1}{(1 + \frac{1}{p^{2s}} )^{2} − \frac{1}{p^{s}} (1 + \frac{1}{p^{2s}} ) (α_{p} + β_{p})(ᾱ_{p} + β̄_{p}) + \frac{1}{p^{2s}} (α_{p}^{2} + β_{p}^{2} + ᾱ_{p}^{2} + β̄_{p}^{2})})

Since we have normalized f so that its fourier coefficients are precisely the eigenvalues of the T_{n}, and since these eigenvalues are real, it follows that

α_{p} + β_{p} = ᾱ_{p} + β̄_{p}

and

α_{p}^{2} + β_{p}^{2} = (α_{p} + β_{p})^{2} − 2 = (ᾱ_{p} + β̄_{p})^{2} − 2 = ᾱ_{p}^{2} + β̄_{p}^{2}.

Therefore, we have

ζ(2s) ∑_{n=1}^{∞} \frac{a_{f}(n)ā_{f}(n)}{n^{s}} = ∏_{p} (\frac{1}{(1 + \frac{1}{p^{2s}} )^{2} − \frac{1}{p^{s}} (1 + \frac{1}{p^{2s}} ) (α_{p} + β_{p})^{2} + \frac{2}{p^{2s}} (α_{p}^{2} + β_{p}^{2})})

= ∏_{p} (\frac{1}{(1 + \frac{1}{p^{2s}} )^{2} − (1 + \frac{1}{p^{2s}} ) \frac{α_{p}^{2}+β_{p}^{2}}{p^{s}} − \frac{2}{p^{s}} (1 + \frac{1}{p^{2s}} ) + \frac{2}{p^{s}} \frac{α_{p}^{2}+β_{p}^{2}}{p^{s}}} )

= ∏_{p} (\frac{1}{(1 − \frac{α_{p}^{2}+β_{p}^{2}}{p^{s}} + \frac{1}{p^{2s}} ) (1 − \frac{2}{p^{s}} + \frac{1}{p^{2s}} )})

= ∏_{p} (1 − \frac{α_{p}^{2}}{p^{s}} )^{−1} (1 − \frac{β_{p}^{2}}{p^{s}} )^{−1} (1 − \frac{1}{p^{s}} )^{−2} ,

as required.

Exercise 3. Let 𝔉 be the standard fundamental domain for SL_{2}(ℤ)/ℍ and let {γ_{i}} for 1 ≤ i ≤ n be a complete set of coset representatives for SL_{2}(ℤ)/Γ where Γ ⊂ SL_{2}(ℤ) is a subgroup of finite index. Then we claim that no two interior points of

D = ⋃_{i=1}^{n} γ_{i}𝔉

are Γ equivalent. For suppose to the contrary that z, gz ∈ D for some g ∈ Γ. If z, gz ∈ γ_{i}𝔉 for some i then γ_{i}^{−1}z, γ_{i}^{−1}gz ∈ 𝔉 are two SL_{2}(ℤ) equivalent interior (see below) points of 𝔉, so they must be the same point, implying that z = gz. Otherwise, z ∈ γ_{i}𝔉 and gz ∈ γ_{j}𝔉 for some i ≠ j. Then γ_{i}^{−1}z, γ_{j}^{−1}gz ∈ 𝔉. Since we have supposed z to be an interior point of γ_{i}𝔉, we see that γ_{i}^{−1}z, γ_{j}^{−1}gz are interior points of 𝔉 with

γ_{j}^{−1}gγ_{i}(γ_{i}^{−1}z) = γ_{j}^{−1}gz.

Since any interior point of 𝔉 has trivial stabilizer in SL_{2}(ℤ), it follows that γ_{i}^{−1}z = γ_{j}^{−1}gz and

gγ_{i} = γ_{j}.

But since γ_{i}, γ_{j} are distinct coset representatives of Γ in SL_{2}(ℤ), they cannot be Γ equivalent. This is a contradiction, and no two interior points of D are Γ equivalent.