I'm asked to show that the Weierstrass's elliptic function $\wp: \mathbb{C}/\Gamma \rightarrow \mathbb{C}P^1$ has exactly 4 branch points. My problem is that I don't see why there are 4 branch points and not just 3.

I looked at the zeroes of the derivative $\wp '$. Since $\wp '$ is doubly periodic and odd this implies that $\frac{w_1}{2}$, $\frac{w_2}{2}$ and $\frac{w_1+w_2}{2}$ are zeros of $\wp '$, where $w_1,w_2$ span $\Gamma$. But I know that an elliptic function has the same number of poles as it has zeros (where the order of the poles / zeroes matters). Since $\wp '$ has only one pole (of order 3) I know that the three zeros are all of order 1 and in particular there can't be a fourth zero.

Where is my mistake?

  • $\begingroup$ You mean the inverse of $\wp$ (the elliptic integral) is a function $\mathbb{C}P^1 \to \mathbb{C}/\Gamma$ with $k=4$ branch points ? $\endgroup$ – reuns May 27 '17 at 20:06
  • $\begingroup$ $\wp$ has a pole of order $2$ at $z=0$, so locally around $z=0$ it is like $1/z^2$ whose inverse $1/\sqrt{z}$ has a branch point at $z=\infty$. The 3 other branch points are the one you found where $\wp'(z) =0$ $\endgroup$ – reuns May 27 '17 at 20:12
  • $\begingroup$ Sorry, I didn't see this conversation. I personally was taught that branch points live in the domain of the map, not in the codomain. It's true at the branch point lies above the point at infinity of $\mathbb{CP}^1$, but we would still say that the branch point itself is the point $z = 0$ on $\mathbb C / \Gamma$. $\endgroup$ – Kenny Wong May 27 '17 at 20:18
  • $\begingroup$ @KennyWong $\wp$ is meromorphic (no branch point) $\mathbb{C} / \Gamma $ to the Riemann sphere. $\endgroup$ – reuns May 27 '17 at 20:20
  • $\begingroup$ @user1952009 It's an unfortunate overlap of terminology. Here, "branch point" means a branch point of the branched covering of the sphere, not branch point as a singular point like for $\log z$ or $\sqrt{z}$. $\endgroup$ – Daniel Fischer May 27 '17 at 20:22

The final branch point is at $z = 0$, and the reason for this is a bit subtle.

The image of $z = 0$ under $\wp$ is the "point at infinity" on $\mathbb {CP}^1$. In fact, $z = 0$ is the unique point on $\mathbb C / \Gamma$ whose image under $\wp$ is the point at infinity on $\mathbb {CP}^1$. Since the map $\mathbb C/\Gamma \to \mathbb {CP}^1$ is of degree two, this observation alone is enough for us to conclude that $z = 0$ is a branch point.

For an alternative perspective, recall that our map $\mathbb C / \Gamma \to \mathbb {CP}^1$ is given by $z \mapsto [1 : \wp(z)] $ for $z \neq 0$. But to define the map at $z = 0$, we need to rewrite the map as $z \mapsto [\wp(z)^{-1} : 1]$. Now $\wp(z)$ has a double pole at $z = 0$, so $\wp(z)^{-1}$ has a double zero at $z = 0$, hence $z = 0$ is a branch point of our covering with ramification index equal to two.

  • $\begingroup$ $\wp$ is meromorphic, can you explain why we say zeros of order $2$ are branch points ? $\endgroup$ – reuns May 27 '17 at 20:16
  • $\begingroup$ But why doesn't this contradict the fact that $\wp '$ has only one pole of order 3, so it can only have the 3 zeroes. What confuses me even more is the fact that since $\wp '$ is odd, we have $\wp '(0)=-\wp '(0)$, so $0$ is a zero of $\wp '$. $\endgroup$ – user450093 May 27 '17 at 20:20
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    $\begingroup$ @user450093 It's like $z \mapsto z^2$, as a covering of the sphere, it has two branch points, $0$ and $\infty$, although its derivative has only one zero. $\endgroup$ – Daniel Fischer May 27 '17 at 20:24
  • $\begingroup$ @user450093 as a function $\mathbb{C}\to \mathbb{C}$ : $\wp'(z)$ is odd and has a pole at $z=0$. as a function $\mathbb{C}\to \mathbb{C} \cup \{\infty\}$ it has a zero at $z=0$ because $\frac{1}{\wp(z)}$ has a zero of order $2$ at $z=0$ $\endgroup$ – reuns May 27 '17 at 20:27
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    $\begingroup$ @user450093 Actually the "derivative of the function" does vanish at the point $z = 0$. But given that we must write our map using a coordinate chart on $\mathbb {CP}^1$ that includes the point at infinity, the function whose derivative we care about is $\wp(z)^{-1}$, not $\wp(z)$. The statement that $\wp(z)^{-1}$ has vanishing derivative at $z = 0$ follows immediately from the fact that $\wp(z)$ has a double pole at $z = 0$. $\endgroup$ – Kenny Wong May 27 '17 at 20:37

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