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Given nonzero complex constants $\omega_1,\omega_2$, with nonreal ratio, we define $$\wp(z;\omega_1,\omega_2)=\frac{1}{z^2}+ \sum_\omega \frac{1}{(z-\omega)^2}-\frac{1}{\omega^2} $$ where the sum is taken over all nonzero linear combinations $\omega=n_1 \omega_1+n_2 \omega_2$ with integer coefficients.

It is known that $\wp$ is of order 2, which means that for any $c \in \hat{\mathbb C}$ the equation $\wp(z)=c$ has two non-congruent solutions (two points are called congruent if their difference is linear combination of $\omega_1,\omega_2$ with integer coefficients).

In addition, it is known that $\wp$ is even, and that the poles on the "lattice" $ \omega_1 \mathbb Z+\omega_2 \mathbb Z$ are all of order 2.

My question is: Could there be a point $z_0$ such that $\wp'(z_0)=0$? This would imply that the value $\wp(z_0)$ is taken twice at $z_0$.

I tried using the evenness of the function to show that there isn't such point. However starting with $z_0=\frac{1}{2} \omega_1+\frac{1}{2} \omega_2$ this approach fails.

Thanks.

user1337
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1 Answers1

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In fact, since $\wp'$ is of order $3$, it has three zeros. Since $\wp'$ is odd and periodic, the zeros are

$$\frac{\omega_1}{2},\, \frac{\omega_2}{2},\, \frac{\omega_1+\omega_2}{2}.$$

By the oddness of $\wp'$, we have

$$\wp'\left(-\frac{\omega_1}{2} \right) = -\wp'\left(\frac{\omega_1}{2} \right).$$

On the other hand,

$$\wp'\left(\frac{\omega_1}{2} \right) = \wp'\left(-\frac{\omega_1}{2} +\omega_1\right) = \wp'\left(-\frac{\omega_1}{2} \right)$$

by the periodicity. Together, that implies

$$\wp'\left(\frac{\omega_1}{2} \right) = 0.$$

The argument for the other two points is analogous. Since $\wp'$ has only three zeros (in a fundamental mesh), there are no others.

Daniel Fischer
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  • Thanks Daniel. I understand that $\wp$ is of order three, but how did you find the zeros so easily? – user1337 Nov 16 '13 at 21:15
  • Odd and periodic, let me expand in the answer. – Daniel Fischer Nov 16 '13 at 21:16
  • Thanks again. While your answer is absolutely correct, it saddens me, because it means that my approach on an earlier question was wrong. – user1337 Nov 16 '13 at 21:26
  • Sorry to hear that. What approach to which question, if I may ask? – Daniel Fischer Nov 16 '13 at 21:27
  • In this question I was trying to express a general even elliptic function $f$ in terms of $\wp$. Now that some values may be double, I think I'm doing something wrong. – user1337 Nov 16 '13 at 21:30
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    It's in principle the right approach. There are just special cases to take care of. When you want to say that any numerator $\wp(z) - \wp(a_k)$ has a simple zero in $a_k$, that need not always be the case, as we saw here. But if it's not a simple zero, i.e. $\wp'(a_k) = 0$, then what can $a_k$ only be? And what does the argument above tell you about $f'(a_k)$? – Daniel Fischer Nov 16 '13 at 21:36
  • Thank you, I will think about it. – user1337 Nov 16 '13 at 21:51