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So I was asked to prove that every Elliptic function of order $2$ whose pole set is contained in the lattice $\Lambda$ is of the form $a+b\wp$, where $\wp$ is the Weierstrass-p function:

$$\wp(z)=\frac{1}{z^2}+\sum_{\omega\in\Lambda - \{0\}}\big [\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\big ].$$

After screwing around with Laurent series for a couple of hours I finally gave up and just looked up the answer:

Given our assumptions on $f$, subtract off a scalar multiple of $\wp$ to obtain an elliptic function of order $1$, which must be constant (by a previous theorem).

Talk about an unsatisfying answer to a pretty powerful classification result! I certainly can't see why it ought to be true. Can anyone construct $a+b\wp$ directly from the assumptions on $f$?

My attempt was to note that $Res (f,\omega)=\lim_{z\rightarrow\omega}\frac{d}{dz}(z-\omega)^2f(z)$ exists and is the same for all $\omega\in\Lambda$. And thus $\frac{d}{dz}(z-\omega)^2f(z)$ is analytic in a neighborhood of $\omega$, which can be integrated to obtain the function $(z-\omega)^2f(z)$, analytic in a nbhd of $\omega$. I can't however figure out where to go from here.

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  • $\begingroup$ I would be glad to help but I don't understand where do you have problems. That holomorphic bounded function on $\mathbb{C}$ can only be a constant is a very basic and powerful fact which is probably the main tool in the whole elliptic functions business. $\endgroup$ – Start wearing purple Jul 11 '13 at 10:47
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By assumption on $f$, for every $a\in\Lambda$ the Laurent series of $f$ about $a$ has the form $$f(z)=\sum_{n=-2}^\infty c_n (z-a )^n $$ where $c_{-2}\ne 0$. The coefficients $c_n$ are independent of which lattice point we look at, because of the periodicity. Consider $g(z)=f(z)-c_{-2}\wp(z)$ where $\wp$ is built on the same lattice $\Lambda$. For every $a\in\Lambda$ the Laurent series of $g$ about $a$ has the form $$g(z)=\sum_{n=-1}^\infty c_n (z-a )^n $$ because the power $-2$ was subtracted off. There are no poles outside of the lattice, since neither $f$ nor $\wp$ have such poles. If $c_{-1}\ne 0$, then $g$ has a single pole of order $1$ in its fundamental region, which makes it an elliptic function of order $1$. A theorem says there is no such thing as an elliptic function of order $1$.

Thus, $c_{-1}=0$. This makes $g$ an entire function. Being doubly-periodic, it must be bounded. A bounded entire function is constant. Conclusion: $f(z)=c_0+c_{-2}\wp(z)$.

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