Question that seems pretty easy, but I can't formalize it:

Let $L \subset C$ be a lattice, and $f(z)$ be an elliptic function for $L$, that is a meromorphic function so that $f(z+w) = f(z)$ for all $\omega \in L$. Assume that $f$ is analytic except for double poles at each point of the lattice $L$. Show that $f = a\wp + b$ for some constants $a,b$.

What I tried: $\displaystyle f(z) = \prod_{\omega \in L} {\frac{g(z)}{(z-\omega)^2}}$ , $g(z)$ is analytic and therefore constant in the fundamental domain. Now what is left to do, is to take the product apart to partial fractions, and then I get almost what needed, except it's not one constant $a$ and $b = \sum_{\omega \in L} -\frac1{\omega^2}$.

Am I right? How do I proceed?

Thanks in advance.


Here is a hint to help orient you: Suppose $f$ is doubly periodic on the lattice $L_{\tau} = \mathbb{Z} \oplus \tau \mathbb{Z}$, where $\tau \in \mathbb{H}$, and $f$ is assumed analytic everywhere save for double poles on the lattice points of $L_{\tau}$. What can you say about the function $f/\wp$ on $L_{\tau}$?

Compare the position of the poles of both $f$ and $\wp$. If they occur at the same place, then $f/\wp$ has no poles (provided that the order of the poles is the same), and so is doubly periodic as well as analytic, hence constant of $\Lambda_{\tau}$.

Now generalize $f$ to an arbitrary lattice $\omega_1 \mathbb{Z} \oplus \omega_2 \mathbb{Z}$, as the one you post, and consider the function $g = (f - a)/\wp = f/\wp - a/\wp$, where $a$ is an arbitrary constant. What properties will $g$ have on the lattice?

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  • $\begingroup$ For the first question - I suppose the function $f/\wp$ wouldn't have poles (I don't fully understand why), and therefore analytic and constant. I don't really understand why the same property might hold for an arbitrary $a$. Can you please collaborate more on the topic? It's very new for me. $\endgroup$ – Pavel Mar 5 '11 at 12:36
  • $\begingroup$ Okay, let's see if I got it straight: $f/\wp$ is elliptic as a quotient of two elliptic functions, so is $a/\wp$ and their sum. Moreover, $g$ doesn't have any poles and therefore is constant (Liouville). Didn't understand the integration significance - is it integration on the boundary? $\endgroup$ – Pavel Mar 5 '11 at 13:34
  • $\begingroup$ That seems about right. The integration was just to get you thinking about periodicity. $\endgroup$ – user02138 Mar 5 '11 at 13:48

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