# Doubly periodic meromorphic function with prescribed poles and zeros

The field of the meromorphic functions on a complex torus $$\mathbb{C} \mathbin{/} \Lambda$$ is $$\mathbb{C}(\wp, \wp')$$, where $$\wp$$ is the weierstrass p-function to the lattice $$\Lambda$$. Furthermore, for such a function $$f$$ and its finite set $$U$$ of poles and zeros holds: $$\sum_{ u \in U } \operatorname{ord}_u(f) = 0$$ and $$\sum_{ u \in U } u \cdot \operatorname{ord}_u(f) \in \Lambda$$, where $$\operatorname{ord}_u(f)$$ is the order of the pole (if negative) resp. the zero (if positive) of $$f$$ at $$u$$.

If now some points $$U$$ and their orders are given, and fulfill constraints above, I believe (because of the Riemann–Roch theorem) that a corresponding meromorphic function exists and is unique (up to a multiplicative constant), but I cannot figure out how to construct it from $$\wp$$ and $$\wp'$$.

Are my claims correct? And if yes, how to construct the meromorphic function in question (with closed-form formula, or recursively)?

• I changed $ord_u(f)$ to $\operatorname{ord}_u(f),$ coded as \operatorname{ord}_u(f). The effect is not only to prevent italicization. Note that the space to the right of $\operatorname{ord}$ is different in the two expressions $\operatorname{ord}f$ (coded as \operatorname{ord}f) and $\operatorname{ord}(f)$ (coded as \operatorname{ord}(f)). The spacing becomes context-dependent and in some cases the formatting of sub- and superscripts is affected. In genuine LaTeX (as opposed to MathJax, which is used here), you can put \newcommand{\ord}{\operatorname{ord}} above the$\,\ldots \qquad$ Sep 8, 2020 at 17:07
• $\ldots\,$above the \begin{document} and then in the body of the document just write \ord each time. $\qquad$ Sep 8, 2020 at 17:08
• @MichaelHardy That is ok, thank you for the latex hints.
– Loic
Sep 8, 2020 at 17:09
• At least one standard proof that the field of elliptic functions for a given lattice is generated by $\wp$ and $\wp'$ does give what amounts to a recursive procedure to express the thing in terms of them... Sep 8, 2020 at 17:41

We can just ignore the extra constraints, since it is enough to prove every meromorphic function on $$\mathbb{C}/\Lambda$$ is rational in $$\wp$$ and $$\wp'$$. Given a meromorphic function $$f$$ on $$\mathbb{C}/\Lambda$$, there is indeed a standard proof that $$f \in \mathbb{C}(\wp,\wp')$$. It goes something like this. Write $$f$$ as a sum of an even function and an odd function: $$f(z) = \frac{f(z) + f(-z)}{2} + \frac{f(z) - f(-z)}{2}$$ Using this trick we may assume that $$f$$ is odd, or that $$f$$ is even. In fact we can assume $$f$$ is an even function, since if $$f$$ is an odd elliptic function then $$\wp' \cdot f$$ is an even elliptic function. Thus it is enough to show that if $$f$$ is an even elliptic function then $$f \in \mathbb{C}(\wp)$$.

For even elliptic functions $$f$$, the identity $$\operatorname{ord}_w f = \operatorname{ord}_{-w} f$$ holds for all $$w \in \mathbb{C}$$. Furthermore, if $$2 w \in \Lambda$$, then $$\operatorname{ord}_w f$$ is even, because the $$i$$-th derivative satisfies $$f^{(i)}(-w) = f^{(i)}(w) = (-1)^i f^{(i)}(-w)$$ for all odd values of $$i$$ (the first equality follows because $$2 w \in \Lambda$$, and the last equality is achieved by repeatedly applying the chain rule). Therefore $$\operatorname{div}(f) = \sum_{w \in H} n_w ((w) + (-w))$$ for some set of integers $$n_w$$, where $$H$$ is half of a fundamental parallelogram for $$\Lambda$$, and the sum has only finitely many nonzero terms.

Consider the function $$g(z) = \prod_{w \in H\setminus \{0\}} (\wp(z) - \wp(w))^{n_w}.$$ We have $$\operatorname{div}(\wp(z)-\wp(w)) = (w) + (-w) - 2(0)$$, so $$\operatorname{div}(g)$$ and $$\operatorname{div}(f)$$ are identical except possibly at $$(0)$$. Since every principal divisor has degree zero, $$\operatorname{div}(g)$$ and $$\operatorname{div}(f)$$ must in fact also be identical at $$(0)$$. Hence $$f/g$$ is an elliptic function with no poles, so it is constant. But then $$f \in \mathbb{C}(\wp)$$ since $$g \in \mathbb{C}(\wp)$$.

• That is a proof that meromorphic functions are rational functions of $\wp$ and $\wp'$, but I seems to me that the existence of a suitable meromorphic function is assumed, whereas its existence should be proved. Also notice the zeros of a function and those of its even resp. odd part are related in a non-obvious way (the addition of functions does no preserve the zeros). In my question, the zeros and poles are prescribed, how can they be dispatched such that they correspond to the even resp. odd part of a meromorphic function?
– Loic
Sep 11, 2020 at 9:58
• Sorry I'm not able to write this up right here as it's kind of long, but I think you want Lemma 2.5.2 in my course notes math.uwaterloo.ca/~djao/co789.2007/co789.pdf which tells you how to transform a divisor into any equivalent divisor that sums to the same value under the elliptic curve group law. Using Lemma 2.5.2, one can (by induction) show that every degree zero divisor which sums to zero under the elliptic curve group law is principal (this is done essentially in Theorem 3.1.7 -- and yes the proof is constructive -- it gives you the rational function).
– djao
Sep 11, 2020 at 23:09
• Do note that your condition $\sum_{u\in U} u\cdot \operatorname{ord}_u(f)\in \Lambda$ is exactly the condition that a divisor sums to zero under the elliptic curve group law, so what I wrote above applies.
– djao
Sep 11, 2020 at 23:16
• I think that the induction you mention in the comments is a valid answer. I had similar thoughts (including considering initially the very same proof as you, until I realized that it proves something else). But I wanted a second opinion about the new stuff I just had learned, and I am more confident now that I understood those concepts correctly. Thank you.
– Loic
Sep 12, 2020 at 12:05
• I studied the example 2.4.5 of your course notes. For the case 5., I suggest to mention the fact the no rational function can have single pole (by Theorem 2.2.5 a), such that, in the linear combination in question, the pole at $(-s_1)$ is cancelled (by construction), the pole at $(0)$ is not cancelled (because it occurs only in one of the summands), and thus the pole at $(s_1)$ is not cancelled (because there cannot be only one simple pole).
– Loic
Oct 12, 2020 at 10:46