I am reading Apostle's Modular Functions and Dirichlet Series in Number Theory. Theorem 1.7 says that a non-constant elliptic function has at least 2 simple poles or a double pole in each period parallelogram. And the book went on saying that these are the only 2 possibilities, and that they led to two different theories proposed by Weierstrass and Jacobi.

I am wondering why cannot we have a triple pole or just any poles of orders higher than 1.

  • $\begingroup$ by double periodicity $\int_{\partial P} f(z)dz = 0 = 2 i \pi \sum_{a \in P} Res(f(z),a)$ where $P$ is a period parallelogram chosen such that $f(z)$ has no zero on $\partial P$, and $a$ are the poles of $f(z)$ $\endgroup$ – reuns Nov 1 '16 at 3:22
  • $\begingroup$ I understand that point. But that does not eliminate the possibility of having higher orders of poles. The argument principle pertains only to the residues, with only concerns the first negative power in Laurent expansion... $\endgroup$ – Sekots Reivan Nov 1 '16 at 3:24
  • $\begingroup$ The weierstrass function has only one pole of order $2$ by parallelogram, and its derivatives have only one pole of order $2+k$ $\endgroup$ – reuns Nov 1 '16 at 3:25
  • $\begingroup$ And you should try another book : diamond shurman "a first course in modular forms" pdf $\endgroup$ – reuns Nov 1 '16 at 3:47
  • $\begingroup$ Thanks! I have that as well. $\endgroup$ – Sekots Reivan Nov 1 '16 at 3:48

You can have poles of higher orders. For example, $\wp^2$ has poles of order $4$ in every point of the period lattice. Also the derivative $\wp'$ has a pole of order $3$.

A theorem of Abel says that you can find an elliptic function with any prescribed zeros and poles $\alpha_i$ and orders $m_i$ (positive $m_i$ being the order of a zero and negative $m_i$ the order of a pole) if and only if $\sum m_i = 0$ and $\sum m_i \alpha_i \in L$ lies in the lattice.

This excludes the possibility of a single simple pole $\alpha$ already: the first sum forces the existence of exactly one simple zero, and the second equation forces that zero to appear at $\alpha$.

| cite | improve this answer | |
  • $\begingroup$ well $\wp'(z)$ has a pole of order $3$ due to direct differentiation $\endgroup$ – reuns Nov 1 '16 at 3:25
  • $\begingroup$ @user1952009 Yes you're right $\endgroup$ – user384359 Nov 1 '16 at 3:34
  • $\begingroup$ and $\sum m_i = 0$ is the argument principle : $\int_{\partial P} \frac{f'(z)}{f(z)}dz=0$, while $\sum m_i a_i \in \Lambda$ is $2i \pi \sum m_i a_i = \int_{\partial P} z \frac{f'(z)}{f(z)}dz = \int_{\partial P} \log f(z) dz = 2i \pi \lambda$ where $\lambda \in \Lambda$ (since $\log f(z+w_1) = \log f(z)+ 2i k_1 \pi$ and $\log f(z+w_2) = \log f(z)+ 2i k_2 \pi$) $\endgroup$ – reuns Nov 1 '16 at 3:36
  • $\begingroup$ @user1952009 Abel's theorem really refers to the converse, existence of elliptic functions to given divisors, which is not so elementary $\endgroup$ – user384359 Nov 1 '16 at 5:02
  • $\begingroup$ Yes of course. Anyway answering to such an elementary question with a complicated theorem wasn't a good idea. $\endgroup$ – reuns Nov 1 '16 at 5:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.