Weierstrass Elliptic Function
[This is just a note to keep me remember things]
I have been reading some old "recipes" about constructing an elliptic function. As far as I know, there are two such classic developments, one is due to the Jacobi, and the other is due to Weierstrass.
Weierstrass defined the Elliptic Function \(f: \Omega \rightarrow \mathbb{C}\) as a doubly periodic meromorphic function.
Doubly Periodic means that there exist two non-zero complex numbers $\omega_1$ and $\omega_2$, satisfying $f(z+\omega_i)=f(z)$, for $i=1,2$ and $\frac{\omega_1}{\omega_2} \not \in \mathbb{R}$.
These two numbers are called the periods of $f$, and whenever any other period $f$ is of the form $m \omega_1 + n \omega_2$ with $m,n\in \mathbb{Z}$, we say that $f$ has fundamental pair of periods.
It can be proved that for any non-constant elliptic curve, the fundamental pair always exists.
We will study the behaviour of $f$ on the parallelogram with vertices $0, \omega_1,\omega_2, \omega_1+\omega_2$
For a fundamental pairs $(\omega_1,\omega_2)$, the contour integral of $f$ on each of these parallelogram is zero since the parallel sides cancel each other. Thus by Cauchy Residue Theorem
\[0=\int_{P} f(z) \, dz = 2\pi i \sum_{z_k} Res(z_k)\]
where $P$ is the parallelogram and $z_k$ are the poles of $f$, thus:
The sum of the residues of elliptic curve on its poles is zero.
For any elliptic function $f$, we can prove that $\frac{f^{\prime}}{f}$ is also an elliptic function with the same period with $f$. Therefore
\[\int_{P} \frac{f^{\prime}(z)}{f(z)} \, dz =0\]
the above integral counts the difference of the number of singular points and the numbers of roots of $f$ (each with multiplicity ), since $f$ is meromorphic, its singular points are its poles thus:
In the Parallelogram the number of zeros of $f$ equals to the number of poles of $f$
The number of poles of an elliptic function is called the order
From these informations, the construction of an elliptic function of order two, might start by guessing that its Laurent series has the following terms
\[\frac{A}{(z-\omega)^2}+\frac{B}{z-\omega}\]
for any period $\omega$, that is for any $\omega=m \omega_1 + n \omega_2$ where $(\omega_1,\omega_2)$ is fundamental pair.
However the double series
\[\sum_{m} \sum_{n} \frac{1}{(m \omega_1 + n \omega_2 -z)^{\alpha}}\]
is absolutely convergent only for $\alpha>2$ and divergent for $\alpha \leq 2$. With some elaborations from the above fact, we can prove that the function
\[f(z)=\sum_{m,n \in \mathbb{Z}}\frac{1}{(z-m\omega_1-n\omega_2)^3}\]
is an elliptic function with periods $\omega_1$ and $\omega_2$ and with poles of order $3$.
For convenience we write $A=\{m\omega_1+n\omega_2 | m,n\in\mathbb{Z}\}$, we call this $A$ a period lattice, thus
\[f(z)=\sum_{\omega \in A} \frac{1}{(z-\omega)^3}\]
is an elliptic function with periods $\omega_1$ and $\omega_2$ and with poles of order $3$.
For convenience we write $A=\{m\omega_1+n\omega_2 | m,n\in\mathbb{Z}\}$, we call this $A$ a period lattice, thus
\[f(z)=\sum_{\omega \in A} \frac{1}{(z-\omega)^3}\]
We can check that $f(z+\omega_1)$ is just the derangement of $f(z)$ and by the absolute convergence of the series it converges to the same value of $f(z)$.
We might have temporarily given up on the construction of Elliptic Function of order two and constructing of order three instead, but we can still proceed by integrating the above function, hoping that one of the pole will vanish. We want to integrate $f(w)$ from $w=0$ to $w=z$, but the term $\frac{1}{w^3}$ (when $m=n=0$) should be removed first
\[\int_0^z \sum_{\omega \in A/\{0\}} \frac{1}{(w-\omega)^3} \, dw=\sum_{\omega \in A/\{0\}} \frac{-2}{(w-\omega)^2}+ \frac{2}{\omega^2}\]
We remove the constant $-2$ since it does not effect the analyticity and add the removed term $\frac{1}{w^3}$ which is now also being integrated to $\frac{1}{z^2}$, we get
\[\wp(z)=\frac{1}{z^2} + \sum_{\omega \in A/\{0\}} \left(\frac{1}{(z-\omega)^2} - \frac{1}{\omega^2}\right)\]
\[\int_0^z \sum_{\omega \in A/\{0\}} \frac{1}{(w-\omega)^3} \, dw=\sum_{\omega \in A/\{0\}} \frac{-2}{(w-\omega)^2}+ \frac{2}{\omega^2}\]
We remove the constant $-2$ since it does not effect the analyticity and add the removed term $\frac{1}{w^3}$ which is now also being integrated to $\frac{1}{z^2}$, we get
\[\wp(z)=\frac{1}{z^2} + \sum_{\omega \in A/\{0\}} \left(\frac{1}{(z-\omega)^2} - \frac{1}{\omega^2}\right)\]
This function is called Weierstrass Elliptic Function, It is analytic except on any points in $A$ where it has pole of order two, it is also an even function.