I cannot prove the proposition about Elliptic functions.


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.

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

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 $n\ge-1$.


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