# Proposition about Weierstrass's elliptic function, or $\wp$ function.

I cannot prove the proposition about Elliptic functions.

Proposition;

Every Elliptic function $$f$$ of order 2 whose pole set is contained in the lattice $$\Lambda$$ is written as $$f=a+b \wp$$, where $$\wp$$ is Weierstrass's elliptic function, i.e.

$$\begin{equation} \wp (z)=\dfrac{1}{z^2} + \sum_{\omega \in \Lambda-\{0\}} \bigg( \dfrac{1}{(z-\omega)^2} - \dfrac{1}{\omega^2} \Bigg) \end{equation}$$

In this page Direct construction of an arbitrary elliptic function of order $2$ with pole set contained in its lattice. , it is said that for every $$a \in \Lambda$$, the Laurent series of $$f$$ about $$a$$ is written as $$\begin{equation} f(z)=\sum_{n=-2}^{\infty} c_n (z-a)^n \quad (c_{-2}\neq 0). \end{equation}$$

And let $$g:=f-c_{-2} \wp$$.

The Laurent series of $$g(z)$$ about $$a$$ is written as $$\begin{equation} g(z)=\sum_{n=-1}^{\infty} c_{n} (z-a)^n. \end{equation}$$

But I cannot understand why $$c_{-2}$$ disappears.

I have $$\begin{equation} g(z)= \sum_{n=-2}^{\infty} c_n (z-a)^n -c_{-2} \left( \dfrac{1}{z^2} + \sum_{\omega \in \Lambda-\{0\}} \bigg( \dfrac{1}{(z-\omega)^2} - \dfrac{1}{\omega^2} \Bigg) \right). \end{equation}$$

I don't know what should I do in order to see the Laurent series of $$g(z)$$ about $$a$$. How should I deformate $$g(z)$$?

I would like you to give me some ideas.

• Every element of $\Lambda$ is a period for both $\wp$ and $f$, so it suffices to check that this holds at $a=0$. – Jyrki Lahtonen Jan 26 at 5:28
• It is $g(z)=\sum_{n=-1}^{\infty} d_{n} (z-a)^n$ not $c_n$. – reuns Jan 26 at 5:58

## 1 Answer

Suppose $$f(z)=\sum\limits_{n=-2}^{\infty} c_n(z-a)^n$$ is $$f$$'s Laurent series at $$z=a$$.

We know the first term in $$\wp(z)$$'s Laurent expansion around $$z=a$$ is $$(z-a)^{-2}$$, so therefore $$f(z)$$ and $$c_{-2}\wp(z)$$ share the same first term of their Laurent expansions, which means if we subtract one from the other that term cancels out, and you're only left with powers $$(z-a)^n$$ for $$n\ge-1$$.