Weierstrass Elliptic Function

Basic construction

Weierstrass p\mathscr{p} function is the fundamental, and probably the simplest function on the complex torus C/L\mathbb{C}/L, where LL is a lattice. It turns out that the Weierstrass p\mathscr{p}

Def. A lattice LL in C\mathbb{C} is a set of complex numbers determined by linear independent w1,w2Cw_1, w_2\in\mathbb{C}, such that


For a given lattice LL, define the complex torus to be the quotient space C/L\mathbb{C}/L. Notice that a meromorphic function on complex torus f:C/LCf:\mathbb{C}/L\rightarrow \mathbb{C}​ can be lifted to a meromorphic function


such that f~(z)=f~(z+L)\tilde{f}(z)=\tilde{f}(z+L). Such function f~\tilde{f} is called a elliptic function. Define the Weierstrass function p(z)=p(z,L)\mathscr{p}(z)=\mathscr{p}(z,L) by

(z):=1z2+wL0(1(zw)21w2)\wp(z):=\frac{1}{z^2}+\sum_{w\in L-0}\left(\frac{1}{(z-w)^2}-\frac{1}{w^2}\right)

The following propositions indicate the existence of p\mathscr{p}.

Prop. The series

wL01ws\sum_{w\in L-0} \frac{1}{|w|^s}

converges for s>2s>2.

Proof. Suppose that w2/w1=τw_2/w_1=\tau has imaginary part Im(τ)>0\mathrm{Im}(\tau)>0. Let τ=u+iv\tau=u+iv. Then

wL01ws=w1sa,bZ,(a,b)01a+bτs=w1sa,bZ,(a,b)0(1a2+b2u2+2b2uv+b2v2)s2w1sa,bZ,(a,b)0(1a2+b2τ2)s2\begin{aligned} \sum_{w\in L-0}\frac{1}{|w|^s} &=|w_1|^s\mathop{\sum\sum}\limits_{a,b\in\mathbb{Z}, (a,b)\neq 0}\frac{1}{|a+b\tau|^s}\\ &=|w_1|^s\mathop{\sum\sum}\limits_{a,b\in\mathbb{Z}, (a,b)\neq 0}\left(\frac{1}{a^2+b^2u^2+2b^2uv+b^2v^2}\right)^{\frac{s}{2}}\\ &\leq|w_1|^s\mathop{\sum\sum}\limits_{a,b\in\mathbb{Z}, (a,b)\neq 0}\left(\frac{1}{a^2+b^2|\tau|^2}\right)^{\frac{s}{2}}\\ \end{aligned}

Since the summand converges, then the original series converges. \square

Prop. The Weierstrass function p(z)\mathscr{p}(z) converges absolutely and uniformly in any compact set KC/LK\subset\mathbb{C}/L.

Proof. Since zz lies in a compact set, the for sufficiently large aa and bb, we can assume that

zR,w2R.|z|\leq R, \quad |w|\geq 2R.

for some sufficiently large R>0R>0. Since

wL0(1(zw)21w2)=wL02wzz2(zw)2w2,\sum_{w\in L-0}\left(\frac{1}{(z-w)^2}-\frac{1}{w^2}\right)=\sum_{w\in L-0}\frac{2wz-z^2}{(z-w)^2w^2},


2wzz2(zw)2w22Rz+z2(w/2)2w210Rw3\begin{aligned} \left|\frac{2wz-z^2}{(z-w)^2w^2}\right| &\leq \frac{2R|z|+|z|^2}{(|w|/2)^2|w|^2}\\ &\leq \frac{10R}{|w|^3} \end{aligned}

Thus the sequence converges uniformly, as by the previous property

wL010Rw3<.\sum_{w\in L-0}\frac{10R}{|w|^3}<\infty.

This completes the proof. \square

The Weierstrass p\mathscr{p} function is almost the simplest elliptic function. It has the following properties

Prop. Suppose that p\mathscr{p} is the Weierstrass function for any lattice LL. Then

  1. p\mathscr{p} is elliptic;
  2. p\mathscr{p} is even, whereas p\mathscr{p}' is even;
  3. The set of all poles of p\mathscr{p} is LL. Moreover, p\mathscr{p} has order 22 at each pole.

Proof. Only by observing the expression of p\mathscr{p}. \square

Elliptic functions

Given a lattice LCL\subset\mathbb{C}. Define the set of all elliptic function by

EL:={fM(C):f(z)=f(z+w),wL}.\mathcal{E}_L:=\{f\in\mathcal{M}(\mathbb{C}):f(z)=f(z+w),\forall w\in L\}.

It can be verified that EL\mathcal{E}_L is a field. Since p\mathscr{p} is an even function, we can sieve out those even elliptic functions

EL+:={fEL:f(z)=f(z),zC}.\mathcal{E}_L^+:=\{f\in\mathcal{E}_L:f(z)=f(-z),\forall z\in\mathbb{C}\}.

Prop. Let fEL+f\in\mathcal{E}_L^+ such that the poles of ff lie in LL. Then fC[p]f\in\mathbb{C}[\mathscr{p}].

Proof. Consider the Laurent expansion of ff at 00. Since ff is even, then the the coefficients are only nonzero at even terms. Namely, ff can be written as


where gg is a holomorphic function. By subtracting and comparing the coefficients,


we can lower the highest order term. Since nn is finite, then doing the above procedure recursively give a holomorphic function in C\mathbb{C}, and thus ff can be written as a polynomial of variable p\mathscr{p}. \square

Prop. EL+C(p)\mathcal{E}_L^+\cong\mathbb{C}(\mathscr{p})​.

Proof. Let z1,,zmz_1,\ldots,z_m be the poles of fEL+f\in\mathcal{E}_L^+, where ziLz_i\notin L is the pole of order nin_i. Then ff has finitely many poles in the fundamental domain. Consider


Then gg has only poles in LL. By the above proposition, gC[p]g\in\mathbb{C}[\mathscr{p}]. Namely, there exists a polynomial PC[z]P\in\mathbb{C}[z], such that g(z)=P(p)g(z)=P(\mathscr{p}). Therefore,


and EL+C[p]\mathcal{E}_L^+\subset\mathbb{C}[\mathscr{p}]. The other inclusion is obvious, which establishes the proposition. \square

Prop. ELC(p,p)\mathcal{E}_L\cong\mathbb{C}(\mathscr{p},\mathscr{p}').

Proof. Using the fact that p\mathscr{p}' is an odd function, and every function ff can be written as the summand of its odd part and even part. The even part has been discussed, whereas the odd part can be written as

even part\wp'\cdot\text{even part}

This completes the proof. \square

Algebraic properties

The Weierstrass p\mathscr{p} function also satisfies some interesting algebraic equation.

Prop. The Laurent expansion of p\mathscr{p} is


where Gk(L)=wL01wkG_k(L)=\sum\limits_{w\in L-0}\frac{1}{w^k}.

Proof. Let p~(z)=p(z)1z2\tilde{\mathscr{p}}(z)=\mathscr{p}(z)-\frac{1}{z^2}. Then

~(z)=wL02(zw)3\tilde{\wp}(z)=\sum_{w\in L-0}\frac{-2}{(z-w)^3}

and the nn-th derivative of p~\tilde{p} is

p~(n)(z)=wL0(1)n(n+1)!(zw)n+2.\tilde{p}^{(n)}(z)=\sum_{w\in L-0}\frac{(-1)^n(n+1)!}{(z-w)^{n+2}}.


~(2k)(z)=(2k+1)!wL01w2k+2=(2k+1)!G2k+2\tilde{\wp}^{(2k)}(z)=(2k+1)!\sum_{w\in L-0}\frac{1}{w^{2k+2}}=(2k+1)!G_{2k+2}

Hence the Laurent expansion of p\mathscr{p} can be written as


by using the expansion of p~\tilde{p}. \square

Prop. p\mathscr{p} satisfies the algebraic differential equation:


where g2=60G4g_2=60G_4 and g3=140G6g_3=140G_6. Therefore,


and [M(C):M(C/L)]=2[\mathcal{M}(\mathbb{C}):\mathcal{M}(\mathbb{C/L})]=2.

Mapping behavior of p\mathscr{p}