Modular Forms, PS 5

Bryden Cais October 5, 2002

Exercise 1. Recall the functional equation for the Riemann zeta function:

π^{−s/2}Γ (\frac{s}{2}) ζ(s) = π^{(s−1)/2}Γ (\frac{1 − s}{2}) ζ(1 − s).

Supposing that k is any nonnegative even integer and replacing s by s − k + 1 above yields

π^{(k−s−1)/2}Γ (\frac{s − k + 1}{2}) ζ(s − k + 1) = π^{(s−k)/2}Γ (\frac{k − s}{2}) ζ(k − s).

Multiplying these two equalities together gives

π^{(k−2s−1)/2}Γ (\frac{s}{2}) Γ (\frac{s − k + 1}{2}) ζ(s − k + 1)ζ(s)

= π^{(2s−k−1)/2}Γ (\frac{k − s}{2}) Γ (\frac{1 − s}{2}) ζ(1 − s)ζ(k − s).  (1)

Since k is even, k/2 = l is an integer. Therefore, using the recurrence

Γ(1 + z) = zΓ(z)

we arrive at

Γ (\frac{s + 1}{2}) = (\frac{s + 1}{2} − l) (\frac{s + 1}{2} − l + 1) ⋯ (\frac{s + 1}{2} − 1) Γ (\frac{s + 1}{2} − l)
Γ (l − \frac{s}{2}) = (−\frac{s}{2} + l − 1) (−\frac{s}{2} + l − 2) ⋯ (−\frac{s}{2} + 1) Γ (1 − \frac{s}{2}) .

Now recall the duplication formula for the Γ function:

Γ(2x) = \frac{2^{2x−1}}{\sqrt{π}} Γ(x)Γ (x + \frac{1}{2}) .

We thus obtain

Γ (\frac{s}{2}) Γ (\frac{s − k + 1}{2}) = \frac{Γ (\frac{s}{2}) Γ (\frac{s+1}{2})}{(\frac{s+1}{2} − l) (\frac{s+1}{2} − l + 1) ⋯ (\frac{s+1}{2} − 1)}

= \frac{π^{1/2}Γ(s)}{2^{s−1}} \frac{1}{(\frac{s+1}{2} − l) (\frac{s+1}{2} − l + 1) ⋯ (\frac{s+1}{2} − 1)}

and

Γ (\frac{k − s}{2}) Γ (\frac{1 − s}{2}) = (−\frac{s}{2} + l − 1) (−\frac{s}{2} + l − 2) ⋯ (−\frac{s}{2} + 1) Γ (1 − \frac{s}{2}) Γ (\frac{1 − s}{2})

= (−\frac{s}{2} + l − 1) (−\frac{s}{2} + l − 2) ⋯ (−\frac{s}{2} + 1) \frac{π^{1/2}Γ(1 − s)}{2^{−s}}.

Substituting these expressions in to (1) yields

π^{(k−2s−1)/2} \frac{π^{1/2}Γ(s)}{2^{s−1}} \frac{1}{(\frac{s+1}{2} − l) (\frac{s+1}{2} − l + 1) ⋯ (\frac{s+1}{2} − 1)}ζ(s − k + 1)ζ(s)

= π^{(2s−k−1)/2} (−\frac{s}{2} + l − 1) (−\frac{s}{2} + l − 2) ⋯ (−\frac{s}{2} + 1) \frac{π^{1/2}Γ(1 − s)}{2^{−s}}ζ(1 − s)ζ(k − s)

Now observe that

(\frac{s + 1}{2} − l) (\frac{s + 1}{2} − l + 1) ⋯ (\frac{s + 1}{2} − 1) (−\frac{s}{2} + l − 1) (−\frac{s}{2} + l − 2) ⋯ (−\frac{s}{2} + 1) Γ(1 − s)

= (i)^{k} (\frac{k − 1 − s}{2}) (\frac{k − 2 − s}{2}) ⋯ (\frac{1 − s}{2}) Γ(1 − s)

= \frac{(i)^{k}}{2^{k−1}} Γ(k − s),

so that we have

π^{(k−2s−1)/2} \frac{Γ(s)}{2^{s}}ζ(s − k + 1)ζ(s) = (i)^{k}π^{(2s−k−1)/2} \frac{Γ(k − s)}{2^{k−s}}ζ(1 − s)ζ(k − s),

which upon simplification gives

(2π)^{−s}Γ(s)ζ(s − k + 1)ζ(s) = (i)^{k}(2π)^{s−k}Γ(k − s)ζ(1 − s)ζ(k − s),

as required.

Exercise 2. Observe that the above functional equation applies for k = 2: that is, if we set

Λ(s) = (2π)^{−s}Γ(s)ζ(s)ζ(s − 1)

then we have

Λ(s) = −Λ(2 − s).

Let us apply Heck’s argument for proving the converse theorem to see what object gives rise to Λ(s). Observe that Λ(s) has simple poles at s = 0, 1, 2 (the poles at 0, 2 are clear since ζ has a simple pole at s = 1, and Γ has a simple pole at s = 0, while the pole at 1 arises since ζ(0) ≠ 0). Now let c ∈ ℝ be large. Then

\frac{1}{2πi} ∫_{c−i∞}^{c−i∞} Λ(s)y^{−s} ds = \frac{1}{2πi} ∫_{c−i∞}^{c−i∞} ∑_{n=1}^{∞} ∑_{m=1}^{∞} \frac{m}{mn^{s}} Γ(s)(2πy)^{−s} ds

= \frac{1}{2πi} ∫_{c−i∞}^{c−i∞} ∑_{n=1}^{∞} \frac{σ_{1}(n)}{n^{s}}Γ(s)(2πy)^{−s} ds

= \frac{1}{2πi} ∑_{n=1}^{∞} σ_{1}(n) ∫_{c−i∞}^{c−i∞} Γ(s)(2πny)^{−s} ds,

where the change in order of summation is justified since we have chosen c large enough that the sum converges absolutely and since for any s with large imaginary part, Γ(s) decays exponentially as ℑ(s) → ∞ so that the integral also converges absolutely. Employing the fact that

\frac{1}{2πi} ∫_{c−i∞}^{c−i∞} Γ(s)x^{−s} ds = e^{−x},

we therefore have

f(iy) − a(0) := \frac{1}{2πi} ∫_{c−i∞}^{c−i∞} Λ(s)y^{−s} ds = ∑_{n=1}^{∞} σ_{1}(n)e^{−2πny}.

We now evaluate the LHS by considering

∫_{C} Λ(s)y^{−s} ds,

where C is the rectangle with vertical sides passing through c and 2−c, and horizontal sides passing through ±iN for large N. Indeed, we will consider this integral as N → ∞: it is not hard to see that the integrals along the horizontal sides tend rapidly to 0—this follows from the behavior of Γ(s) for s with large imaginary part. Therefore, we have

\frac{1}{2πi} ∫_{c−i∞}^{c−i∞} Λ(s)y^{−s} ds − \frac{1}{2πi} ∫_{2−c−i∞}^{2−c−i∞} Λ(s)y^{−s} ds = ∑_{ρ} Res_{ρ}(Λ),  (2)

where the sum is over all poles ρ of Λ inside C and Res_{ρ}(Λ) denotes the residue of Λ at ρ. As mentioned above, the only poles inside C are simple poles at s = 0, 1, 2. The determination of the residues is routine:

Res_{Λ}(0) = ζ(0)ζ(−1) = \frac{1}{24}
Res_{Λ}(1) = (2πy)^{−1}ζ(0) = \frac{1}{4πy}
Res_{Λ}(2) = (2πy)^{−2}ζ(2) = \frac{1}{24y^{2}} .

We now use the functional equation Λ(s) = −Λ(2 − s) to rewrite the integral in the RHS of (2) as

\frac{1}{2πi} ∫_{c−i∞}^{c−i∞} Λ(s)y^{−s} ds + \frac{1}{2πi} ∫_{c−i∞}^{c−i∞} Λ(2 − s)y^{s−2} ds = \frac{1}{24} + \frac{1}{24y^{2}} + \frac{1}{4πy} ,  (3)

so that by our definition of f(iy) we have

(f(iy) − a(0)) + \frac{1}{y^{2}} (f(i/y) − a(0)) = \frac{1}{24} + \frac{1}{24y^{2}} + \frac{1}{4πy} .  (4)

Therefore, with a(0) = 1/24, or equivalently, with

F (iy) := 1 − 24 ∑_{n=1}^{∞} σ_{1}(n)e^{−2πny},

we have that

F (iy) = \frac{i^{2}}{y^{2}} F (i/y) + \frac{12}{2πi(iy)},

and hence by analytic continuation that

z^{2}F (z) = F (−1/z) + \frac{12z}{2πi}

for all z ∈ ℍ. This is the transformation formula for E_{2}. Integrating, exponentiating, and taking 24^{th} powers shows that Δ (as given by the product q ∏(1 − q^{n})^{24}) is a cusp form of weight 12.