How is a complex elliptic curve a torus?

Question

I have been learning about elliptic curves. In particular, I have been trying to understand the proof that an elliptic curve (let's assume it's nonsingular, if it matters), say $E:y^2=x^3+ax+b=:f(x)$ over $\mathbb C$ is always isomorphic as a topological group to $\mathbb C/\Lambda$ for some lattice $\Lambda$ (that of course depends on the coefficients $a,b$).

I think that I can follow the argument based on showing that the Weierstrass $\wp$-function works as such an isomorphism. However, I cannot reconcile this with my mental picture about what an elliptic curve looks like. Just topologically, I don't see how $E(\mathbb C)$ could possibly be homeomorphic to a torus.

Let's assume that $f$ has distinct roots $x_1,x_2,x_3$. Then, for every $x$ in the complex plane except for these three points, there are exactly two corresponding values of $y$ on the elliptic curve due to the fundamental theorem of algebra. Therefore, my naïve mental picture for $E(\mathbb C)$ is that of two infinite planes joined at $3$ points corresponding to the roots of $f$. Even accounting for the points at infinity, my mental image ends up being something like a sphere "squashed down" such that the northern and southern hemispheres touch at three points.

What I described doesn't seem like it could be possibly homeomorphic to a torus, although I have to admit I cannot prove this rigorously.

What's gone wrong in my logic? Thanks in advance!

Answer 1

Let's look at $y^2 = x$ first. This doesn't look like two planes glued together at a point, it looks like one plane folded in on itself rotationally (the ray of intersection is an artifact of being immersed in 3-dimensions).

For an elliptic curve, the picture is we have a torus and then there's this skewer through the centre of it ($\phi$) and when we do the same kind of rotational fold, we get 4, so-called "rammification points" where the skewer intersects the torus.

There are some pictures in the notes here around pages 25 to 27.

Answer 2

One way to think of the elliptic curve $$ E\!: y = x^3 + ax + b = (x-e_1)(x-e_2)(x-e_3) $$ is as $$ y = f(x) := \sqrt{(x-e_1)(x-e_2)(x-e_3)}. $$ The Riemann surface of $\,y = f(x)\,$ (just as $\,y = \sqrt{x}\,$) is two-sheeted or two valued over the extended complex plane (Riemann sphere). This can be understood with the use of branch cuts. For the function $\,f(x)\,$ there is a cut from, say, $e_1$ to $e_2$ and another from $e_3$ to $\infty$. With the cuts, the sphere is homeomorphic to an annulus, a cylinder, or to a torus sliced in half. Gluing this half torus with a copy of itself forms a complete torus. Of course, the four points $\,e_1,e_2,e_3,\infty$ are ramification points just as for the two points $0,\infty$ for $\sqrt{x}$. This may be hard to visualize, but if you can understand the the Riemann surface for $\sqrt{x}$ first, then it is not so hard.