If a function's zeroes are doubly periodic, must that function be elliptic?

Asked Jun 01 '14 at 17:48

Active Jun 01 '14 at 17:48

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Proposition: Let a meromorphic function $f(z)$ be such that $f(z)=0\Leftrightarrow z \in\Lambda=\{n\omega_1+m \omega_2, (n,m)\in \mathbb{Z}^2\}$. Except in the following exception, must $f(z)$ be elliptic with period lattice $\Lambda$?

Exception: $f(z)$ is the product of some elliptic function $e(z)_{\Lambda}$ with another function $g(z)$ such that $g(z)$ has no zeroes outside $\Lambda$. Example: $f(z)=\wp(z,\Lambda)e^z$.

Note: there may be a couple more exceptions, but my question is really whether the proposition is true with not too many sets of counterexamples. However, I'd accept a counterexample that is clearly nontrivial.

asked Jun 01 '14 at 17:48

Meow

6,353

Presumably by "function" you mean "meromorphic function"? It seems easy to come up with counterexamples if you drop this condition. – Qiaochu Yuan Jun 01 '14 at 17:51
@QiaochuYuan Of course. – Meow Jun 01 '14 at 17:51
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The exceptions cover all the set: $f(z)$ is a product of elliptic function $e(z)=1$ and a function with no zeroes outside $\Lambda$. It seems that something needs to be made more precise. – Start wearing purple Jun 28 '14 at 10:45
You want your exceptions to be $\wp(z)e^{g(z)}$ for entire functions $g$. – anon May 14 '15 at 20:06

If a function's zeroes are doubly periodic, must that function be elliptic?

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