Consider the Weierstrass Elliptic function, $$\rho(z) = \frac{1}{z^{2}} + \sum\bigg(\frac{1}{(z-m-nw)^{2}} - \frac{1}{(m+nw)^{2}}\bigg), $$ where $m,n \neq 0,0 $. How can one show that it is meromorphic on the complex plane with double poles at the lattice points $m+nw?$ I know that the lattice points would be clearly points of singularities, but how to show that they are poles and of the second order also how can I prove that there won't be any other types of singularities i.e., just poles or meromorphic?

  • $\begingroup$ @daOnlyBG thanks for the edit. $\endgroup$ – Deven Feb 24 '15 at 15:22
  • $\begingroup$ you're welcome. $\endgroup$ – daOnlyBG Feb 24 '15 at 15:23

The pole is right there in the formula $\frac1{(z-m-nw)^2}$. If you subtract just that term, then you are in the same situation as if $z$ were not near a lattice point. The other contributions are $O(m^{-3},n^{-3})$ for large $m$ and $n$. so they give a finite sum.
You can differentiate the function term by term, and what is left still gives a finite sum for the same reason, so there are no other singularities.
$$\frac1{(z-m-nw)^2}-\frac1{(m+nw)^2}=\frac{2z(m+nw)-z^2}{(z-m-nw)^2(m+nw)^2}$$ When $m$ and $n$ are large, this fraction is $O(m,n)/O(m^4,n^4)$ or $O(m^{-3},n^{-3})$.
I'm not sure what's missing about $\frac1{(z-m-nw)^2}$ being a pole of order two at $m+nw$. Perhaps I should call it $\frac1{(z-m'-n'w)^2}$ so I don't confuse a particular pole with the great mass of other contributions to the sum.

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  • $\begingroup$ +1 IIRC the point is that any tail of the sum (sorted according to the absolute value of the lattice point) converges absolutely and uniformly on any compact subset $K$ of the complex plane (tail= the terms with poles outside of $K$). Therefore term-by-term differentiation is justified, and the claim follows. $\endgroup$ – Jyrki Lahtonen Feb 24 '15 at 15:40
  • $\begingroup$ @Michael can you please explain it in more detail like how is the order of the pole 2 and how the order contributions are $O(m^{-3},n^{-3})$ $\endgroup$ – Deven Feb 24 '15 at 16:27
  • $\begingroup$ @JyrkiLahtonen Can you please explain why there would be no other singularities, I could not understand the explanation by Michael. $\endgroup$ – Deven Feb 25 '15 at 10:41

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