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I'm working my way through Silverman and Tate's Undergraduate Introduction to Elliptic Curves. I haven't yet been able to study complex analysis, so it comes as no surprise that I'm having a tough time with that portion of the book right now.

Let $\omega_1, \omega_2 \in \mathbb{C}$ be two complex numbers which are $\mathbb{R}$-linearly independent and let: $$L = \mathbb{Z}\omega_1 + \mathbb{Z}\omega_2 = \{n_1\omega_1 + n_2\omega_2 : n_1, n_2 \in \mathbb{Z}\}$$ Let $\wp(u) = \frac{1}{u^2} + \sum\limits_{\omega \in L, \omega \neq 0} \left(\frac{1}{(u-\omega)^2} - \frac{1}{\omega^2}\right)$ Show that $\wp$ is a doubly periodic function, that is, show that $$\wp(u + \omega) = \wp(u)$$

If you are able, please give me a shove in the right direction. Thank you!

Dear Answerers: Thank you, I have been able to figure it out. Yes, convergence was quite tricky and I was trying to make this particular question much more difficult than it actually was. Thank you!

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3 Answers 3

You don't need complex analysis for this one. Just write down the series definition for $\wp(u+\omega)$. If you understand what is summed over, it should be pretty clear that you are summing "the same terms" as before.

Once you have this idea, making a rigorous proof is an easy exercise in analysis.

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Hopefully the OP has shown that the series is absolutely convergent, uniformly convergent on compact sets, etc. –  Dylan Moreland Feb 2 '12 at 22:30
    
And that is the hard part, from what I remember... –  Zhen Lin Feb 2 '12 at 23:40

The derivative is obviously periodic, so the original function must be also.

For completeness, you should also show that the function is absolutely convergent (so the order in which you sum over the lattice doesn't matter) and uniformly convergent on compact subsets (so it defines a meromorphic function). If you haven't had any complex analysis, you may be able to omit the last bit without any trouble down the line.

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This affirmation is not true in general. For example, $1$ is doubly periodic, but $z$ is not. –  glebovg Sep 8 at 0:19
    
@glebovg Indeed. I'm sure not sure what I was thinking here. –  Potato Sep 8 at 4:06
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However, in this case it is true. The periodicity of $\wp'(z)$ implies that $\wp(z + \omega_k) - \wp(z) = A = \text{const}$, but $\wp(z)$ is even, so upon setting $z = -\omega_k/2$ we deduce that $A = 0$. –  glebovg Sep 8 at 23:24

It is easy to show that $\wp'(z)$ is doubly periodic because replacing $z$ by $z + \omega_k$ in $\wp'(z)$ does not change the series because it is doubly infinite. Now, the periodicity of $\wp'(z)$ implies that $\wp'(z + \omega_k) - \wp'(z) = 0$, that is, $\wp(z + \omega_k) - \wp(z) = A = \text{const}$, but $\wp(z)$ is even, so upon setting $z = -\omega_k/2$ we deduce that $A = 0$.

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