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Good evening!

One of Liouville's theorems about elliptic functions states that there is no such function that has only one pole of order 1. This result is very well known and easily proven using the residue formula.

My question is the following: Is there some topological reason/ intuition for this fact?

For example, one can see the fact that $\int\limits_{\partial B_1(0)} f(z) d z \neq 0$ for some functions that are holomorphic in $G = \mathbb{C}\setminus\{0\}$ as a result of $G$ having non-vanishing first homology group (or plainly having one $0$-dimensional hole).

So my intuition would be that an elliptic function essentially lives on a torus (after identifying opposing sides of the fundamental lattice), whose first homology group is isomorphic to $\mathbb{Z}\times\mathbb{Z}$... however, I do not really see how to go from there.

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If you know that an elliptic function has the same number of zeroes and poles, then there is indeed a simple topological proof. If $f$ is an elliptic function with just one simple pole, then it also has one zero, as does $f-a$ for each $a\in\mathbb{C}$. So, $f$ takes every value in $\mathbb{C}\cup\{\infty\}$ exactly once (in the fundamental region). That is, $f$ induces a bijection $\mathbb{C}/\Lambda\to\mathbb{C}\cup\{\infty\}$ from the torus corresponding to the period lattice to the Riemann sphere. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism, so this would give a homeomorphism between a torus and a sphere. Now it is a simple theorem in topology that a torus and a sphere are not homeomorphic (e.g., by comparing their first homology groups, or their fundamental groups), so no such $f$ can exist.

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  • $\begingroup$ Thank you very much, that was exactly what I was looking for! $\endgroup$ Aug 15 at 11:47

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