I am currently trying to prove various facts about $\wp'$, considered as a meromorphic map from $\mathbb{C}/\Lambda\to\mathbb{C}$, where $$\wp'(z) = -2\sum_{w\in\Lambda}\frac{1}{(z-w)^3}.$$ In particular, I am interested in the degree of this map and its ramification points. Here is what I have so far:

  • There is a triple pole at $z=0$, so $0$ is a ramification point of index 3, and this is the only pole (since the series converges for all values of $0\neq w\in\Lambda$), so $$\deg(\wp')=3$$ (by the definition of the degree).
  • By Riemann-Hurwitz, $$\sum_{r\in\Omega}(v_{\wp'}(r)-1)=b(\wp')=6,$$ where $\Omega$ is the set of ramification points and $v_{\wp'}(r)$ is the valency/ramification index.
  • Since the index of a ramification point is bounded by the degree, any $r\in\Omega$ has $$2\leqslant v_{\wp'}(r)\leqslant3.$$

Now putting these facts together we see that we have only three possibilities for describing $\Omega\setminus\{0\}$ (since there are only three ways to partition $4$ as a sum of $\{1,2\}$, up to reordering):

  1. four points of index $2$;
  2. two points of index $2$ and one of index $3$;
  3. two points of index $3$.

I have three questions (though answering the one of them will probably answer some of the others):

(a) Is all of the above correct? All of the literature I've read about ramification points etc. deals with holomorphic maps between Riemann surfaces, but here we have meromorphic maps. Just kind of ignoring this fact seems to make sense, since projective space is homogenous in the sense that it doesn't matter which point we pick to be infinity, and so having poles etc. just translates to having solutions to $\wp'(z)=\infty$ (or so I tell myself...).

Basically, I'm pretty certain that meromorphic maps are just holomorphic maps to $\mathbb{C}\mathbb{P}^1$. Is this true?

(b) How can we proceed from here?

I tried looking at the zeros of $\wp'''$, and this gave me all of the half-lattice points, which led to me claiming that these three points were all of ramification index 3, which contradicts Riemann-Hurwitz. So, finally

(c) Is this sum definition the local form of $\wp'$?. It doesn't look like it, but then again we're dealing with meromorphic maps, and not holomorphic ones. Either way, can we still look at derivatives of this form to find ramification points, or do we need the local form explicitly. The chain rule seems to say to me that it is ok to use the form we are given.

(c) In the references from which I am working, we are told that a holomorphic map $F:X\to Y$ between Riemann surfaces has a locally holomorphic form $f:\mathbb{D}\to\mathbb{D}$, where $\mathbb{D}$ is the unit disc in $\mathbb{C}$ taken to be at the origin, for ease of notation (really, in fact, this is the definition of what it means for a function to be holomorphic between Riemann surfaces.) We are then told that a ramification point occurs when $F'=0$, and it seems to me, by the chain rule, since $f=\varphi_U^{-1}\circ F\circ\varphi_V$, that this happens if and only if $f'=0$ at that point under the coordinates given by $\varphi_U$. Is this the case? Since here we are looking at derivatives of $\wp$, which is not holomorphic from $\mathbb{D}\to\mathbb{D}$, so is not a 'local form'.


First of all, a small correction: you have the wrong expression for $\wp'$. Since $$ \wp(z) = \frac{1}{z^2} + \sum_{0 \neq \omega \in \Lambda} \Big(\frac{1}{(z - \omega)^2} - \frac{1}{\omega^2}\Big) $$ it follows that $$ \wp'(z) = -2\sum_{\omega \in \Lambda}\frac{1}{(z - \omega)^3} $$ (i.e. you have the wrong exponent)

A) You are correct; a meromorphic function on a Riemann surface is just a holomorphic map to $\mathbb{CP}^1$.

B) If you want to understand the ramification of $\wp'$ you need to look at where $\wp'' = 0$. In order to do so, one can look at the following differential equation satisfied by $\wp$: $$ (\wp')^2 = 4\wp^3 - g_2\wp - g_3 $$ where $g_2, g_3$ are constants determined by the lattice $\Lambda$. If we now differentiate this expression with respect to $z$, we get $$ 2\wp'\wp'' = 12\wp^2\wp' - g_2\wp' $$ and so away from the points where $\wp' = 0$ (which we can avoid since $\wp$ is simply ramified as it is a degree 2 map), we find that $\wp''$ satisfies $$ \wp'' = 12\wp^2 - g_2 $$ i.e. when $\wp(z)^2 = \frac{g_2}{12}$. However, there are four values of $z$ for which this is true, since $\wp$ is a degree 2 map from $\mathbb{C}/\Lambda \to \mathbb{CP}^1$. So there are four other ramification points, which all must be simple ramification.

We can check this by showing that $\wp''' \neq 0$ at each of these points. Differentiating again we find that $\wp''' = 24\wp\wp'$. Since we know that $\wp \neq 0$ at each of these points, we would have to have that $\wp' = 0$ which is also not possible, as $\wp$ is simply ramified at each of the half-lattice points.

C) If I understand your question correctly, you are asking if the sum that you wrote is a valid expression for $\wp'$. Other than the issue about the exponent it is. This follows due to the fact that we can differentiate holomorphic functions term by term.

You are correct about the chain rule argument, that despite the fact that we have not written $\wp'$ as a function $\mathbb{D} \to \mathbb{D}$, that we can determine the ramification by looking at where the function $\wp'$ is zero instead of writing it in local coordinates around the desired points. So this method works for exactly the reasons you describe.

Hopefully this clears it up!

  • $\begingroup$ Fixed the mistake about the exponent, just a typo, but thanks! As for the third question, I'm just about to do an edit that should hopefully clarify what I'm asking, but apart from that this is very very great, thanks! $\endgroup$ – Tim Mar 31 '15 at 11:18
  • $\begingroup$ absolutely brilliant answer, and thanks for answering my edited question as well! I can't award the bounty yet, but will accept your answer for the moment :) $\endgroup$ – Tim Mar 31 '15 at 12:42
  • $\begingroup$ There is an exception: if g_2=0, then there are only two additional ramification points, and all have index 3 (Simon's argument does show this!) $\endgroup$ – quim Oct 2 '15 at 9:45

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.