The first Liouville theorem states
Any elliptic function without poles is constant.
Let $\wp$ be the Weierstrass elliptic function of a lattice $L$ and let $\sigma$ be the Weierstrass sigma function of a lattice $L$. Now I'm not able to follow the following reasoning which uses the first Liouville theorem:
- Since $\left(\frac{\wp ''(z)}{2}\right)^2$ and $\wp'(z)^2(\wp (2z)+2\wp (z))$ are both elliptic functions without poles outside of $L$ and their Laurent series are equal up to $O(z^2)$, the first Liouville theorem yields $$\left(\frac{\wp''(z)}{2}\right)^2-\wp'(z)^2 (\wp (2z)+2\wp (z))=0.$$
How exactly is the fact that "their Laurent series are equal up to $O(z^2)$" used?
- Let $f(z)=\frac{\sigma (2z)}{\sigma (z)^4}$. Then $\frac{f(z)}{\wp'(z)}$ is a constant, comparing the Laurent series at $z=0$ gives $f(z)=2z^{-3}+O(z^{-1})$ and $\wp'(z)=-2z^{-3}+O(z^{-1})$, thus $\frac{f(z)}{\wp '(z)}=-1$.
How exactly is the fact that "$f(z)=2z^{-3}+O(z^{-1})$ and $\wp'(z)=-2z^{-3}+O(z^{-1})$" used?
The difference $$ h(z) = \left(\frac{\wp''(z)}{2}\right)^2-\wp'(z)^2 (\wp (2z)+2\wp (z)) $$ is an elliptic function without poles outside $L$. At $z=0$ the functions $\left(\frac{\wp''(z)}{2}\right)^2$ and $\wp'(z)^2 (\wp (2z)+2\wp (z))$ have the same Laurent series up to the $O(z^2)$ terms. Therefore $h$ has a removable singularity at $z= 0$ with $h(0) = 0$, and the same is true at all lattice points.
So $h$ is an elliptic function without poles, and therefore constant. Since $h(0) = 0$, $h$ is identically zero.
A similar argument is used for the second part.