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This question comes from Rational Points on Elliptic Curves (Silverman & Tate) exercise $2.3$ (a):

$2.3$ (a)

Show that the series $$\wp(u) = \frac{1}{u^2} + \sum_{\substack{\omega \in L \\ \omega \neq 0}}\left(\frac{1}{(u-\omega)^2}-\frac{1}{\omega^2}\right) $$ defining the Weierstrass $\wp$-function is absolutely and uniformly convergent on any compact subset of the complex $u$-plane that does not contain any of the points of $L$. Conclude that $\wp$ is a meromorphic function with a double pole at each point of $L$ and no other poles.

If the compact subspace we are summing contains no points in $L$, then how can we complete the sum for $\omega \in L$? Does this just mean that the function turns into $\wp(u) = \dfrac 1 {u^2}$?

How can we use this to show convergence?

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2 Answers 2

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The summation doesn't change: It runs over all $\omega \in L$. In the definition of absolute convergence over a compact subset $K$ that avoids $L$, it's the possible $u$ values that are being required to come from $K$.

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  • $\begingroup$ Okay, I understand. How can we use this to show convergence? $\endgroup$ Commented Feb 20, 2023 at 22:51
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    $\begingroup$ That seems to be a new addition to your question but see if you can fill in from the sketch on slide 4 here: people.math.harvard.edu/~smarks/mod-forms-tutorial/… $\endgroup$
    – Arkady
    Commented Feb 20, 2023 at 23:02
  • $\begingroup$ I also agree that what you are asking the two of us is a different question alltogether. $\endgroup$
    – F. Conrad
    Commented Feb 20, 2023 at 23:24
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It means that $\wp(u)$ has poles precisely on $L$. The poles are second order in this case and you can infer that $$ \wp(u)=(z-\omega)^{-2}h(z) $$ for some holomorphic non-vanishing function $h(z)$ around each $\omega \in L$. So we only have $$ \wp(u)=\frac{1}{u^2}h(z) $$ around $0$ as opposed to "$\wp(u)=\frac{1}{u^2}$".

In fact, it has poles nowhere else other than on the lattice points, since it converges uniformly on compact subsets and each partial sum is a holomorphic function on $\mathbb{C} \setminus L$.

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  • $\begingroup$ Okay, I understand. How can we use this to show convergence? $\endgroup$ Commented Feb 20, 2023 at 22:51
  • $\begingroup$ The proof I know uses a comparison test with the Eisenstein series and you dont use the fact above at all. $\endgroup$
    – F. Conrad
    Commented Feb 20, 2023 at 23:02

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