An elliptic function is a meromorphic function f : C → C on C such that f (z + ω1

1. Weierstrass ℘-function

Let {ω₁, ω₂} be an R-basis for C. Assume that τ = Im(ω2/ω₁) > 0. The lattice generated by {ω1, ω2} is denoted by L = Zω1⊕ Zω2.

An elliptic function is a meromorphic function f : C → C on C such that f (z + ω1) = f (z), f (z + ω) = f (z)

for all z ∈ C. The numbers ω1 and ω2 are called periods of f.

Lemma 1.1. A holomorphic elliptic function be a constant function.

The Weierstrass ℘-function associated with the lattice L is function defined by the infinite series

℘(z) = 1

z² + X

ω∈L\{0}

 1

(z − ω)² − 1 ω²

 .

Theorem 1.1. The function ℘(z) is an elliptic function of periods {ω1, ω2}. Its poles are given by z = ω for ω ∈ L. It has the following properties:

(i) the principal part of ℘(z) at z = 0 is 1/z², (ii) limz→0 ℘(z) −_z¹2
= 0,

(iii) ℘(−z) = ℘(z), (iv) ℘⁰(−z) = −℘⁰(z).

Proof. (iii) follows from the series itself. Differentiating ℘(−z) = ℘(z) with respect to z, we obtain (iv). The infinite series

℘(z) − 1

z² =X

ω6=0

 1

(z − ω)² − 1 ω²



converges for z 6∈ L. For every R > 0, the series converges uniformly in |z| ≤ R. (i) and (ii)

follows from here. 

Theorem 1.2. The elliptic function ℘(z) satisfies the differential equation (1.1) (℘⁰(z))² = 4(℘(z))³− g₂℘(z) − g₃,

where

g2= 60 X

ω∈L\{0}

1

ω⁴, g3 = 140 X

ω∈L\{0}

1 ω⁶. Proof. Consider the Laurent expansion of ℘(z):

℘(z) = 1 z² + 1

20g2z²+ 1

28g3z⁴+ O(z⁶).

Then we check

℘⁰(z) = −2 z³ + 1

10g2z +1

7g3z³+ O(z⁵) (℘(z))³ = 1

z⁶ + 3 20g2

1 z² + 3

28g3+ O(z²) (℘⁰(z))² = 4

z⁶ −2 5g₂ 1

z² −4

7g₃+ O(z²).

These imply that

(℘⁰(z))²− 4(℘(z))³+ g2(℘(z)) + g3= O(z²).

1

2

In other words, (℘(z)⁰)² − 4(℘(z))³+ g2(℘(z)) + g3 is a holomorphic elliptic function. By Lemma 1.1, it is a constant. Letting z → 0, we find that the constant is zero. Hence ℘(z) satisfies the following differential equation

(℘⁰(z))²= 4(℘(z))³− g₂(℘(z)) − g3. It also follows from the definition of ℘(z) that

g2= 60 X

ω∈L\{0}

1

ω⁴, g3 = 140 X

ω∈L\{0}

1 ω⁶.

 Let us consider the following path integral

z = Z ∞

w

dt

p4t³− g₂t − g₃.

Here the path of integration may be any curve which does not pass through a zero of 4t³− g₂t − g3. Then we get

(1.2)  dw

dz

2

= 4w³− g₂w − g3.

Proposition 1.1. Suppose that w = w(z) satisfies (1.1). Then w = ℘(z) for some constant α.

Proof. Let u = u(z) be a holomorphic coordinate change on C such that w = ℘(u(z)). Then dw

dz = ℘⁰(z)du dz. Since w satisfies (1.1), we see that

 du dz

2

= 1.

Hence u⁰(z) = ±1 implies that u(z) = ±z + α for some constant α. In other words, w =

℘(±z + α). Making w → ∞, we see that z → 0. This implies that ℘(α) = ∞. Hence α is a pole of ℘(z). We know that α ∈ L. Therefore

w = ℘(±z + α) = ℘(±z) = ℘(z)

by the fact that ℘(z) is an even doubly periodic function with period {1, τ }.  It follows from this proposition that

z = Z ∞

℘(z)

dt p4t³− g₂t − g3

. Theorem 1.3. Let g₂, g₃ be as above. Then

g³₂− 27g²₃ 6= 0.

Proof. Let e₁, e₂, e₃ be roots of

(1.3) 4t³− g₂t − g₃ = 0.

Then e₁+ e₂+ e₃= 0, e₁e₂+ e₁e₃+ e₂e₃ = −g₂, and e₁e₂e₃= g₃. g³₂− 27g²₃ = 16(e₁− e₂)²(e₂− e₃)²(e₃− e₁)².

Let D = g₂³− 27g²₃. To show that D 6= 0, we only need to show that e1, e2, e3 are distinct.

3

Since ℘⁰(z) is an odd function and use the period of ℘⁰(z), we know

℘⁰ω₁ 2



= −℘⁰

−ω₁ 2



= −℘⁰

−ω₁ 2 + ω

= −℘⁰ω₁ 2

 .

This implies ℘⁰(ω₁/2) = 0. Using the same method, we can show that ℘⁰(ω₂/2) = 0 and

℘⁰((ω1+ ω2)/2) = 0. Let e1= ℘(ω1/2), e2 = ℘(ω2/2) and e3= ℘((ω1+ ω2)/2). Since ω1/2, ω₂/2 and (ω₁+ ω₂)/2 are all different, e₁, e₂, e₃ are distinct. Then Equation (1.1) implies that e₁, e₂, e₃ are exactly the three roots of the polynomial (1.3). We complete the proof of

our assertion.