Quotient of Elliptic Curve $E$ homeomorphic to $\mathbb{P}^1_{\mathbb{C}}$

Question

Let $E$ be a complex elliptic curve and let $-1$ be the involution on the group structure over $E$. I would like to prove that $E/{-1}$ is homeomorphic to $\mathbb{P}^1$. I am following two main ideas:

1. I tried to explicitely draw the quotient as a polygon with some weird identification on the sides. I'm wondering if I can use something like "cut and paste" techniques to get a fully topological argumet.

2. I would like to prove that the quotient map $E \rightarrow E/{-1}$ is exactly the map induced by some linear system. For example, taking a divisor $D$ such that $\ell(D)=2$, we get a morphism $E \rightarrow \mathbb{P}^1$. But how can I construct exactly the one I want and how can I conclude?

Every hint is appreciated.

Answer 1

I think you can view this topologically as a successive quotient $\mathbb{R}^2 \to \mathbb{R}^2/\mathbb{Z}^2 =E \to E/-1$. These actions are free and the second is orientation preserving (since $\det(-I_2) = 1$), so $E/-1$ is a compact oriented 2-manifold, which are classified by their genus, $g(-) = \textrm{rank}_{\mathbb{Z}}(H_1(-; \mathbb{Z}))$. Hence we want to argue that this quotient has trivial first homology.

There are a couple ways to do this. One is to note that the closure of the fundamental domain for the action of $\mathbb{Z}^2 \times \mathbb{Z}/2\mathbb{Z}$ is the right triangle obtained from taking the convex hull of $(1,0), (0,1) \in \mathbb{R}^2$. We can then use this to give the quotient a CW/$\Delta$ complex structure and this allows one to compute the homology.

Another is to view $E/-1$ as a Riemann Surface, where $E \to E/-1$ is a holomorphic quotient map, ramified over the $2$-torsion points of $E$. There are exactly four of those points (each with ramification index $2$), so by Riemann-Hurwitz,

$$2g(E) - 2 = 0 = 2*(2g(E/-1) - 2) - 4 = 4g(E/-1)$$ which implies $g(E/-1) = 0$ as required.

Lastly, being an algebraic geometer, my favorite way to see this is using a linear system. Fix an origin point $O \in E$ and consider the linear system $|2O|$. By Riemann-Roch, this is a one-dimensional linear system of degree 2, and it is basepoint-free so it defines a map $E \to \mathbb{P}(H^0(E, \mathcal{O}(2O)))$, which is described as follows.

(We essentially use the coordinate-free description of this map described here, keeping in mind that $\mathbb{P}^1$ is canonically self-dual. )

If $P \in E$, by basepoint freeness there is a surjective evaluation map $H^0(E, \mathcal{O}(2O)) \to \mathbb{C}$ so the kernel is one-dimensional, generated by some nonzero section $s_P$, so we send $P \mapsto [s_P] \in \mathbb{P}(H^0(E, \mathcal{O}(2O))) = \mathbb{P}^1$.

But then, $P, Q$ map to the same $[s] \in \mathbb{P}(H^0(E,\mathcal{O}(2O)))$ if and only if $P + Q = \operatorname{Div}(s) \sim 2O$, which is the same as saying $P = -Q$ in the group structure of $E$. Hence, the map $E \to \mathbb{P}^1$ is surjective an has fibers precisely the orbits of this $\mathbb{Z}/2\mathbb{Z}$ action on $E$, so it's the required quotient map giving $\mathbb{P}^1 \cong E/-1$.