# A Torus and the Weierstrass P function?

Let $\wp$ be the Weierstrass function.

From what I understand, $\wp$ maps the torus to $CP^1 \times CP^1$ in the following way:

$a \mapsto (\wp(a),\wp'(a)) = (z,w)$

Furthermore, the image of this map lies on the zero set of the polynomial $P(z,w) = 4(z-e_1)(z-e_2)(z-e_3) - w^2.$

What I don't get is the description of the inverse to this map, which is supposedly the integral of the differential form $\frac{dz}{w}$ from $\infty$ to a point $Q$ along a path $c$.

1. I don't understand what $\infty$ means here.
2. I don't understand why this is would be the inverse.

Heuristically, I see that

\begin{equation} \int_\infty^Q \frac{dz}{w} = \int_0^z \frac{\wp'(u)}{\wp'(u)} du = \int_0^a du = a. \end{equation}

But unfortunately, the computation above makes little sense. I suppose I am attempting pull-back by setting $z=\wp$ and $w=\wp'$. But isn't this a map into the complex plane, and not the Torus?

Also, why is infinity the branch point?

More generally, let $w^2 = p(z)$ with degree of $p$ odd. Why is infinity one of the branch points, and why isn't it a branch point when the degree of $p$ is even?

• Click the "x hours ago" link after "edited", to see the $\LaTeX$. For comments and unedited stuff, you can right click on the equations and then click "Show source". – J. M. is a poor mathematician Oct 30 '10 at 23:09
It is better to think of mapping to $\mathbb C P^2$ rather than a product of $\mathbb C P^1$s. The point $\infty$ is then the unique point at infinity on the cubic $y^2 = 4(x-e_1)(x-e_2)(x-e_3).$
Also, the integral depends on the path, not just the endpoint; the ambiguity in the value of the integral if you just specify the endpoints is given by integrating over closed loops, i.e. over representatives of the 1st homology. These integrals span a lattice $\Lambda$ in $\mathbb C$ (the original lattice with respect to which you defined $\wp$), and so the integrals really take values in $\mathbb C/\Lambda$, not in $\mathbb C$.
• I'm not sure if I understand the map into $\mathbb{C}P^2$. If I'm using homogeneous coordinates on $\mathbb{C}P^2$, is it kind of like $(z,w) \mapsto [z:w:1]$? What is the point at infinity on the cubic? – Braindead Oct 30 '10 at 14:18
• So I think... I kind of understand. So this map is embedding the torus into $\mathbb{C}P^2$ as a projective variety of the polynomial $P(z,w,u) = 4(z - e_1u)(z-e_2u)(z-e_3u) - w^2 u$, and the point at infinity on the cubic is the point $[0:1:0]$, is that correct? – Braindead Oct 30 '10 at 16:28
• @Braindead: Yes, that's correct. If you consider the orders of the poles of $\wp$ and $\wp'$ at the lattice points, you'll see that they map the lattice points to this point at infinity. – Matt E Oct 30 '10 at 20:35