Math 726: L-functions and modular forms Fall 2011

Week 3, lecture 9: Examples of Modular Forms

Instructor: Henri Darmon Notes written by: Luiz Kazuo Takei

Eisenstein Series

Let z ∈ H and k ∈ ℤ_{>2} even.

We define

G_{k}(z) := \begin{matrix}∑ & ′ \\ (m,n)∈ℤ^{2} \end{matrix}(mz + n)^{−k},

where the primed sum means we sum over all the pairs (m, n) except (0, 0).

Fact 1. It is easy to see that this sum converges absolutely and uniformly on compact subsets of H. So G_{k}(z) is holomorphic on H.

Let ( \begin{matrix}a & b \\ c & d \end{matrix}) ∈ SL_{2}(ℤ). Since

(\begin{matrix}a & b \\ c & d \end{matrix}) ℤ^{2} = ℤ^{2},

we obtain that

\begin{aligned}G_{k} (\frac{az+b}{cz+d}) & = ^{∑′} [m (\frac{az+b}{cz+d}) + n]^{−k} = (cz + d)^{k} ^{∑′} [m(az + b) + n(cz + d)]^{−k} \\ & = (cz + d)^{k} ^{∑′} [(am + nc)z + (bn + dn)]^{−k} = (cz + d)^{k}G_{k}(z). \end{aligned}

We also define

E_{k}(z) := \frac{1}{2}\begin{matrix}∑ & ′ \\ gcd(m,n)=1 \end{matrix}(mz + n)^{−k} = \frac{1}{ζ(2k)}G_{k}(z) = 1 + ⋯

(the sum is multiplied by 1/2 in order to ensure that the first coefficient in the Fourier expansion is 1).

Theorem 1. The Fourier expansion of E_{k} is given by

E_{k}(z) = 1 + \frac{(2πi)^{k}}{(k − 1)!ζ(k)} ∑_{n=1}^{∞} \frac{n^{k−1}q^{n}}{1 − q^{n}} = 1 + \frac{(2πi)^{k}}{(k − 1)!ζ(k)} ∑_{n=1}^{∞} σ_{k−1}(n)q^{n},

where σ_{k−1}(n) = ∑_{d>0 , d|n} d^{k−1} and q = e^{2πiz}.

Proof. We start with the Euler’s product expansion for sin(πz):

sin(πz) = πz ∏_{n=1}^{∞} (1 − \frac{z^{2}}{n^{2}} ) .

Taking the logarithmic derivative on both sides results

\frac{π cos(πz)}{sin(πz)} = \frac{1}{z} + ∑_{n=1}^{∞} (\frac{1}{z + n} + \frac{1}{z − n}) .

The left-hand side can be written as

πi\frac{e^{iπz} + e^{−iπz}}{e^{iπz} − e^{−iπz}} = πi\frac{e^{2πiz} + 1}{e^{2πiz} − 1} = πi\frac{q + 1}{q − 1} = −πi (1 + \frac{2q}{1 − q} ) .

This implies that

\frac{1}{z} + ∑_{n=1}^{∞} (\frac{1}{z + n} + \frac{1}{z − n}) = −πi − 2πi ∑_{n=1}^{∞} q^{n}.

Applying (\frac{d}{dz})^{k−1} to both sides yields

(−1)^{k−1}(k − 1)! [\frac{1}{z^{k}} + ∑_{n=1}^{∞} (\frac{1}{(z + n)^{k}} + \frac{1}{(z − n)^{k}} )] = −(2πi)^{k} ∑_{n=1}^{∞} n^{k−1}q^{n}.

Hence, since k is even,

∑_{n∈ℤ} \frac{1}{(z + n)^{k}} = \frac{(2πi)^{k}}{(k − 1)!} ∑_{n=1}^{∞} n^{k−1}q^{n},

which in turn implies that

2 ∑_{m=1}^{∞} ∑_{n∈ℤ} \frac{1}{(mz + n)^{k}} = 2\frac{(2πi)^{k}}{(k − 1)!} ∑_{m=1}^{∞} ∑_{n=1}^{∞} n^{k−1}q^{mn} = 2\frac{(2πi)^{k}}{(k − 1)!} ∑_{n=1}^{∞} σ_{k−1}(n)q^{n}.

The terms with m = 0 contribute with

∑_{n∈ℤ−\{0\}} \frac{1}{n^{k}} = 2ζ(k).

Therefore

G_{k}(z) = 2ζ(k) + 2\frac{(2πi)^{k}}{(k − 1)!} ∑_{n=1}^{∞} σ_{k−1}(n)q^{n}

and

E_{k}(z) = 1 + \frac{(2πi)^{k}}{(k − 1)!ζ(k)} ∑_{n=1}^{∞} σ_{k−1}(n)q^{n}.

Exercise 1. Use the functional equation of the Riemann zeta function ζ to show that

\frac{(2πi)^{k}}{(k − 1)!ζ(k)} = \frac{2}{ζ(1 − k)}

when k is even.

Example 2. We can now compute the first few Eisenstein series:

E_{4}(q) = 1 + 240(q + 9q^{2} + 28q^{3} + 73q^{4} + ⋯)

E_{6}(q) = 1 − 504(q + 33q^{2} + 244q^{3} + ⋯)

E_{8}(q) = 1 + 480(q + 129q^{2} + ⋯)

E_{6}^{2}(q) = 1 − 1008q + ⋯

E_{4}(q)E_{8}(q) = 1 + 720q + ⋯

Ramanujan Δ-function

The Ramanujan Δ-function is the non-zero cusp form of weight 12 defined by

Δ(q) := \frac{E_{4}(q)E_{8}(q) − E_{6}^{2}(q)}{1728} = q + · · · .

Proposition 1. (1) If k ≤ 2 or k is odd, then

dim_{ℂ} S_{k}(SL_{2}(ℤ)) = 0.

(2) Multiplication by Δ induces an isomorphism of ℂ-vector space

M_{k}(SL_{2}(ℤ)) ˜ ⟶S_{k+12}(SL_{2}(ℤ)).

(3) If k ≥ 4, then

dim_{ℂ} M_{k}(SL_{2}(ℤ)) = 1 + dim_{ℂ} S_{k}(SL_{2}(ℤ)).

Proof. Suppose k = 0. Then M_{0}(SL_{2}(ℤ)) is the space of SL_{2}(ℤ)-invariant functions on H^{∗} with rapid decay. These functions are bounded holomorphic functions on the compact Riemann surface SL_{2}(ℤ)\H^{∗}. Hence

S_{0}(SL_{2}(ℤ)) = {0} and M_{0}(SL_{2}(ℤ)) = ℂ .

(The proof will be continued in the next class.)