Math 678, Lecture 2

Notes by Bryden Cais

Sept. 6, 2002

1 Elliptic Functions

Modular functions naturally arise from the theory of elliptic functions. An elliptic function is a meromorphic function f on ℂ satisfying for all z ∈ ℂ.

f(z + ω_{1}) = f(z + ω_{2}) = f(z)

for ω_{1}, ω_{2} ∈ ℂ with ℑ(ω_{1}/ω_{2}) > 0 (so that the ratio is not real). Thus if we let Λ = {mω_{1} + nω_{2} : (m, n) ∈ ℤ^{2}} be the lattice in ℂ spanned by ω_{1}, ω_{2}, we see that an elliptic function is just a meromorphic function on ℂ/Λ.

Exercise 1. Discuss the case when ω_{1}/ω_{2} ∈ ℝ.

We will construct an important example of a doubly periodic function by the process of “averaging.” The idea is o pick some suitable function g that decays rapidly and to consider the sum

f(z) = ∑_{λ∈Λ} g(z + λ).

Let us first derive some general properties of elliptic functions.

2 General Properties of Elliptic Functions

Let F be a fundamental parallelogram for the elliptic function f (that is, F is the parallelogram with sides ω_{1}, ω_{2}). It is not difficult to show that

\frac{1}{2πi} ∫_{F} \frac{f′(z)}{f(z)} dz = 0

sine the integrals along opposite sides of F cancel. This implies that any elliptic function has the same number of zeroes as poles, counted according to multiplicity. (Another way to see this is to realize that ℂ/Λ as a Riemann surface of genus 1). Similarly, we see that no elliptic function can be holomorphic in F, for then it would be holomorphic on ℂ and thus, by Louiville’s theorem, a constant. Finally, observe that

∫_{F} f(z) dz = 0,

so that the sum of the residues of any elliptic function is zero. Thus, no elliptic function can have a single, simple pole. Moreover, if f has only a single pole in F, we see that the residue there must be 0.

3 The Weierstrass ℘ Function

With these generalities in mind, consider the function g(z) = z^{−k}. We then ask when ∑ g(z + λ) converges absolutely. Observe that in a circle of radius R there are approximately R^{2} lattice points. More precisely, it is not hard to see that the number of lattice points inside a circle of radius R is

\frac{π}{A(F)}R^{2} + O(R),

where A(F) denotes the area of F. In particular, the number of lattice points in the annulus 2^{j} ≤ |λ| ≤ 2^{j+1} is bounded by C2^{2j} for some constant C. Therefore, we see that

∑_{λ∈Λ} \frac{1}{(z + λ)^{k}}

converges absolutely if and only if

∑_{j=0}^{∞} ∑_{2^{j}≤|λ|≤2^{j+1}} \frac{1}{λ^{k}}

converges. By our remarks above, this sum is bounded above by

C ∑_{j=1}^{∞} \frac{2^{2j}}{2^{jk}} = C ∑_{j=1}^{∞} \frac{1}{2^{j(k−2)}} ,

for some constant C. Clearly, this converges for any k > 2.

As an example, we see that the sum

f(z) = ∑_{λ∈Λ} \frac{1}{(z + λ)^{3}}

converges absolutely so that f(z) is an elliptic function. Moreover, the sum

℘(z) = \frac{1}{z^{2}} + ∑_{λ∈Λ′} \{\frac{1}{(z + λ)^{2}} − \frac{1}{λ^{2}} \}

converges absolutely (as long as we do not rearrange anything inside the curly brackets) since each term behaves like λ^{−3}. Here, Λ′ = Λ \ {0}. Observe that ℘(z) is clearly an even function of z and that ℘′(z) = −2f(z). This shows that for any λ ∈ Λ, one has

℘(z + λ) = ℘(z) + C_{λ}

for some constant C_{λ} depending only on λ. Setting z = −λ/2 and using the fact that ℘(z) is an even function shows that C_{λ} = 0 for any λ, and hence that ℘(z) is an elliptic function.

4 The Laurent Expansion of ℘(z)

Let us now compute the laurent expansion of ℘(z). One readily computes

\frac{1}{(z + λ)^{2}} = \frac{1}{λ^{2}} \frac{1}{(1 + (z/λ))^{2}}

= \frac{1}{λ^{2}} (1 − 2\frac{z}{λ} + 3\frac{z^{2}}{λ^{2}} − ⋯) ,

so that

℘(z) = \frac{1}{z^{2}} + ∑_{λ∈Λ′} ∑_{n=1}^{∞}(−1)^{n}(n + 1)\frac{z^{n}}{λ^{n+2}}

= \frac{1}{z^{2}} + ∑_{n=1}^{∞}(−1)^{n}(n + 1)G_{n+2}(Λ)z^{n},

where

G_{n}(Λ) := ∑_{λ∈Λ′} \frac{1}{λ^{n}} ,

and the change in order of summation above is justified by absolute convergence. Now observe that if λ ∈ Λ′ then also −λ ∈ Λ′ so that G_{n}(Λ) = 0 if n is odd. Thus we have

℘(z) = \frac{1}{z^{2}} + ∑_{n=1}^{∞}(2n + 1)G_{2n+2}(Λ)z^{2n}.

5 The Differential Equation Satisfied by ℘(z)

With the laurent expansion of ℘(z) we can now compute

(℘′(z))^{2} = \frac{4}{z^{6}} + \frac{c_{1}}{z^{2}} + c_{2}z^{2} + ⋯
(℘(z))^{3} = \frac{1}{z^{6}} + \frac{b_{1}}{z^{2}} + b_{2}z^{2} + ⋯,

so that there exists a constant d such that

(℘′(z))^{2} − 4(℘(z))^{3} − d℘(z) = z^{2} + ⋯

But the left and side is obviously an elliptic function. On the other hand, the only possible pole is at z = 0 and we see that the left hand side is analytic there. Thus it is a holomorphic elliptic function and therefore a constant. This will turn out to have profound consequences.

6 Eisenstein Series

The series G_{n}(Λ) are known as Eisenstein Series. It is clear that they converge absolutely for n ≥ 3 and that G_{n}(Λ) = 0 if n is odd. Observe that G_{n}Λ depends only on the lattice λ and not any chosen basis for λ. Let

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

Then it is clear that γ takes a basis of Λ to itself where we have

ω_{1} ↦ aω_{1} + bω_{2}
ω_{2} ↦ cω_{1} + dω_{2}.

This gives an action of SL_{2}(ℤ) on the G_{n}(Λ) under which G_{n}(Λ) is invariant. Observe that we can require, without loss of generality, that z = ω_{1}/ω_{2} ∈ ℍ. Then the action of SL_{2}(ℤ) on Λ induces the usual fractional linear action on ℍ, that is,

γz = \frac{az + b}{cz + d}.

Write

G_{k}(Λ) = ∑_{(m,n)∈ℤ^{2}\\{0\}} \frac{1}{(mω_{1} + nω_{2})^{k}}

= \frac{1}{ω_{2}^{k}} ∑_{(m,n)∈ℤ^{2}\\{0\}} \frac{1}{(m(ω_{1}/ω_{2}) + n)^{k}} .

Therefore we define

G_{k}(z) = ∑_{(m,n)∈ℤ^{2}\\{0\}} \frac{1}{(mz + n)^{k}} .

By our discussion above, we readily see that

G_{k}(γz) = (cz + d)^{k}G_{k}(z)

for γ ∈ SL_{2}(ℤ) as above.

7 Modular Forms

Definition 1. A Modular Form of weight k for Γ = SL_{2}(ℤ) is a holomorphic function f on ℍ such that

1. f(γz) = (cz + d)^{k}f(z) for any γ = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(ℤ)

2. f(z) is holomorphic at the cusps of ℍ/Γ.

Item 2 requires explanation. Observe that any modular form is periodic with period 1 (by item 1) and therefore has a fourier expansion, or, equivalently, a laurent expansion in the variable q = e(z). Write

f(z) = ∑_{n=M}^{∞} a_{n}q^{n},

where M is possibly negative. Observe that q : ℍ ⟶ D \ {0} where D is the unit disc. We say that f is holomorphic at the cusps of ℍ/Γ if the function ∑_{n=M}^{∞} a_{n}q^{n} is holomorphic at q = 0 (corresponding to z = ∞ ∈ ℍ).

Definition 2. A cusp form for SL_{2}(ℤ) is a modular form that is 0 at the cusps; i.e. a modular form f(z) = ∑_{n=0}^{∞} a_{n}q^{n} with a_{0} = 0.