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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!

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1 Answer 1

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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).

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  • $\begingroup$ Wow, that exercise is much deeper than I thought to be. Thanks! $\endgroup$
    – user1337
    Commented Nov 17, 2013 at 13:54
  • $\begingroup$ @Daniel Fischer What do you mean with "if the order is greater than 2, divide out one factor $\wp(z)-\wp(\rho_i)$ and repeat the argument" ? $\endgroup$
    – user95747
    Commented Jul 18, 2016 at 15:52
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    $\begingroup$ @user95747 If $f$ has a zero of order $> 2$ at $\rho_i$, consider the elliptic function $g(z) = \dfrac{f(z)}{\wp(z) - \wp(\rho_i)}$. That is also an even elliptic function (with period group containing $M$), and the order of the zero of $g$ at $\rho_i$ is two smaller than the order of the zero of $f$ there. The argument to show that the order of the zero of $f$ at $\rho_i$ was at least $2$ then also works for $g$. Inductively it follows that the order of the zero of $f$ at $\rho_i$ is even. It would perhaps have been better to call the order of the zero of $f$ there $k$, and then consider $\endgroup$ Commented Jul 18, 2016 at 16:19
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    $\begingroup$ $h(z) = \dfrac{f(z)}{\bigl(\wp(z) - \wp(\rho_i)\bigr)^{\lfloor k/2\rfloor}}$. Then either $h(\rho_i) \neq 0$ (when $k$ is even), or $h$ has a simple zero at $\rho_i$ (when $k$ is odd), since $2\lfloor k/2\rfloor \leqslant k \leqslant 2\lfloor k/2\rfloor + 1$. But $h$ is an even elliptic function, and the argument shows that $h$ cannot have a simple zero at $\rho_i$, that gets us to the result in one step rather than $k/2$ steps. $\endgroup$ Commented Jul 18, 2016 at 16:19
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    $\begingroup$ @user686624 Are you referring to the last paragraph of the answer? I don't say that $f$ having a zero or pole at $\rho_i$ always means that $f$ has a zero or pole at $0$. I'm saying that follows if the number of occurrences of $\wp(z)$ in the numerator ($n$) is different from the number of occurrences in the denominator ($d$). Since $\wp$ has a pole of order $2$ at $0$, the product has a pole of order $2(n-d)$ at $0$ if $n > d$, and it has a zero of order $2(d-n)$ at $0$ if $d > n$. If $d = n$, then the product has a nonzero finite value at $0$. $\endgroup$ Commented Dec 22, 2019 at 14:26

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