Why do the simplest elliptic functions need to have order $2$?

Question

Not sure if this has been asked before. I understand the proof that the sum of the residues of an elliptic function is zero. However I fail to see that an elliptic function of order one cannot have an irreducible pole of residue zero. I don't know if I'm missing something simple or not. I can't seem to find the answer anywhere.

Answer 1

You seem to be asking why an elliptic function of order $1$ cannot exist.

1. One (perfectly fine) argument is given by Daniel Fischer in the comments. Namely, let $z_0$ be a pole of an elliptic function $f\colon \mathbb{C}\rightarrow\mathbb{C}$. Consider the Laurent expansion $\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$ of $f$ around $z_0$. In one common definition, the order of $f$ at $z_0$ is given by $$\operatorname{ord}_{z_0}(f)=-\operatorname{min}\{n\ \vert \ a_n\neq 0\}.$$ Thus, if $z_0$ is a pole of order $1$, we have $$\operatorname{res}_{z_0}(f)=a_{-1}\neq 0.$$ Therefore, since the sum of the residues of an elliptic function vanishes, $f$ cannot have just one simple pole.

2. An alternative argument goes as follows:
   Assume for a contradiction that there existed an elliptic function $f\colon \mathbb{C}\rightarrow \mathbb{C}$ of order $1$ with respect to a lattice $\Lambda$. By defintion of an elliptic function, $f$ would extend to a meromorphic function $\hat{f}\colon \mathbb{C}/\Lambda\rightarrow \mathbb{C}$. Here, $\mathbb{C}/\Lambda$ denotes the complex torus with respect to the lattice $\Lambda$. By Riemann's Removable Singularity Theorem, $\hat{f}$ would further extend to a holomorphic mapping $F\colon \mathbb{C}/\Lambda\rightarrow \mathbb{CP}^1$. Here, $\mathbb{CP}^1$ denotes the Riemann sphere. This proper non-constant holomorphic mapping would have degree $1$ by assumption. Thus, it would be an isomorphism of Riemann surfaces. In particular, it would be a homeomorphism. This contradicts the fact that the topological torus $S^1\times S^1$ and the $2$-sphere $S^2$ are not homeomorphic.