Given two nonzero complex numbers $\omega_1, \omega_2$, with nonreal ratio, we define the period module $$M= \omega_1 \mathbb Z+ \omega_2 \mathbb Z= \{n_1 \omega_1+ n_2 \omega_2:n_1,n_2 \in \mathbb Z \} $$
and the Weierstrass $\wp$ function $$\wp(z; \omega_1, \omega_2) \equiv \wp(z;M)= \frac{1}{z^2}+ \sum_{\omega \in M \setminus \{0 \}} \frac{1}{(z- \omega)^2}-\frac{1}{\omega^2} .$$
I want to solve the following exercise from Ahlfors' complex analysis text (page 274):
Show that any even elliptic function with periods $\omega_1$, $\omega_2$ can be expressed in the form
$$C \prod_{k=1}^n \frac{\wp(z)-\wp(a_k)}{\wp(z)-\wp(b_k)} \text{ ($C$=const.)} $$
provided that $0$ is neither a zero nor a pole. What is the corresponding form if the function either vanishes or becomes infinite at the origin?
My attempt:
Let $f$ be an even elliptic function with periods $\omega_1,\omega_2$, and suppose for the moment that $f$ has neither a zero nor a pole at the origin. If $f$ is constant, we have an empty product representation $$f(z)= C \prod_{k=1}^0 \left( \dots \right). $$
Suppose now that $f$ isn't constant. As an elliptic function $f$ has equal number of (congruent) zeros and poles. Denote its zeros by $a_1, \dots,a_n$ and its poles by $b_1, \dots ,b_n$ (multiple points being repeated), and define $$g(z)=f(z) \bigg/ \prod_{k=1}^n \frac{\wp(z)-\wp(a_k)}{\wp(z)-\wp(b_k)} $$
What I want to say is that any numerator $$\wp(z)-\wp(a_k) $$ has a simple zero at $a_k$, and any denominator $$\frac{1}{\wp(z)-\wp(b_k)}$$ has a simple pole at $b_k$. If that's true then $g$ is a holomorphic elliptic function, which reduces to a constant $C$.
If $f$ has a zero of order $2m$ at the origin we repeat the proof for $\tilde{f}= \wp^m f$ and we obtain the representation $$f(z)=C \wp(z)^{-m} \prod_{k=1}^n \frac{\wp(z)-\wp(a_k)}{\wp(z)-\wp(b_k)} \text{ ($C$=const.)} $$
If $f$ has a pole of order $2m$ at the origin we repeat the proof for $\tilde{f}= \wp^{-m} f$ and we obtain the representation $$f(z)=C \wp(z)^{+m} \prod_{k=1}^n \frac{\wp(z)-\wp(a_k)}{\wp(z)-\wp(b_k)} \text{ ($C$=const.)} .$$
My question:
Why are all values of $\wp$ (except $\infty$) taken "simply" (that is with non-vanishing derivative at the point)?
I tried considering the "fundamental parallelogram" with vertices at $a,a+\omega_1,a+\omega_2,a+\omega_1+\omega_2 $ where $a=-\frac{1}{2} \omega_1-\frac{1}{2} \omega_2$, and WLOG the uppermost and rightmost edges are included. It is known that in this parallelogram all complex values are taken twice. Since $\wp$ is even, if $a$ is an interior point, the value $\wp(a)$ is taken at least twice, at the points $\pm a$.
However, if $a$ lies on the part of the boundary of the parallelogram which is included, I can't use the evenness argument, and as far as I can tell $a$ might be a double value of $\wp$ (?)
Is my solution correct so far? and can you please help me with the question in boldface?
Thanks!
Best Answer
Your solution is basically correct, there are just some special cases that need to be handled. It is not always true that $\wp(z) - \wp(a_k)$ has a simple zero in $a_k$. If $a_k$ is a zero of $\wp'$, then $\wp(z)-\wp(a_k)$ has a double zero in $a_k$, and similar for the poles $b_k$ of course. If none of the zeros or poles of $f$ coincides with a zero of $\wp'$, then the construction goes through without any problems, and you have
$$f(z) = C\prod_{k=1}^n \frac{\wp(z)-\wp(a_k)}{\wp(z)-\wp(b_k)}$$
where the $a_k$ resp. $b_k$ are the zeros resp. poles of $f$ in the fundamental parallelogram for an even elliptic $f$ that has neither a pole nor a zero in $0$.
What if one (or more) of the $a_k$ resp. $b_k$ is a zero of $\wp'$?
In this question, we saw that $\wp'$ has the three distinct zeros
$$\rho_1 = \frac{\omega_1}{2},\; \rho_2 = \frac{\omega_1+\omega_2}{2},\; \rho_3 = \frac{\omega_2}{2},$$
and since the order of $\wp'$ is three, these are all simple zeros, and $\wp'$ has no other zeros (modulo the lattice $\Omega = \langle \omega_1,\omega_2\rangle$). The argument used the oddness and periodicity of $\wp'$, but of course $f'$ is also an odd elliptic function for the lattice $\Omega$, so the same argument yields
$$-f'(\rho_1) = f'(-\rho_1) = f'(-\rho_1+\omega_1) = f'(\rho_1),$$
hence $f'(\rho_1) = 0$, if $f$ doesn't have a pole in $\rho_1$,and similar for $\rho_2$, $\rho_3$. Thus if any of the $\rho_i$ is a zero of $f$, it is a zero of even order (if the order is greater than $2$, divide out one factor $\wp(z)-\wp(\rho_i)$ and repeat the argument), and you include the factor $\wp(z)-\wp(\rho_i)$ only half as often in the product. If one of the $\rho_i$ is a pole of $f$, the same argument for $1/f$ shows that the pole must have even order, and then you include the factor $\dfrac{1}{\wp(z)-\wp(\rho_i)}$ only half as often as the multiplicity of the pole would indicate.
Now, if $f$ has a zero or a pole in any of the $\rho_i$, it may happen that the halving of the factors $\wp(z) - \wp(\rho_i)$ produces a different number of factors in the numerator than in the denominator. But that means that $f$ then must have either a zero or a pole in $0$, so this cannot happen for an even elliptic function that has neither a zero nor a pole in $0$ (sorry, I'd rather have a more elegant proof of that fact, but this will have to do for now).