# The Uniformization Theorem and Elliptic Curves

In the theory of elliptic curves, I have read that the elliptic curves is topologically equivalent to a torus, given by $\mathbb{C}$/ $\Lambda$, where $\Lambda$ is a lattice.

The proof appears to use the Uniformization Theorem, which states that every simply connected Riemann surface is topologically equivalent to either the open unit disk, the complex plane, or the Riemann sphere. I believe I understand how two spheres with two branch cuts each, joined at the branch cuts yield a torus. Could someone explain a general sketch of the proof (and correct any misunderstandings)?

• You're looking for a sketch of the proof of the uniformization theorem...? It's a very difficult theorem and the proof I know involves a lot of heavy machinery like sheaves and connections, and I'm not sure how well one could sketch the proof here. In addition, it's much stronger than asserting topological equivalence: it asserts conformal equivalence. – user98602 Jan 11 '14 at 0:43
• Read chapter VI in Silverman's "The Arithmetic of Elliptic Curves". The proof that an elliptic curve over $\;\Bbb C\;$ is a (complex) torus is pretty quick and direct there, and no big weapons like the UT is required. If I just could find my thesis I could try to see what in the world I wrote there, but the proof was also nothing too messy...perhaps later. – DonAntonio Jan 11 '14 at 1:26
• Or perhaps even better: read chapter 9 in Husemöller's "Elliptic Curves" ! – DonAntonio Jan 11 '14 at 1:33

Theorem: Let $A$ be an abelian variety over $\mathbb{C}$ (that is, a connected projective algebraic group; in particular if its dimension is 1 then it's an elliptic curve). Then $A$ is analytically isomorphic to a complex torus.
Sketch of proof: $A$ is in particular a complex compact Lie group, and we have the exponential map $\mbox{exp}:T_0A\simeq\mathbb{C}^n\to A$. It is surjective and is a homomorphism, since $A$ is commutative. Since it is a local diffeomorphism, its kernel is discrete, and so is a lattice in $\mathbb{C}^n$. Therefore $A\simeq\mathbb{C}^n/\ker(\mbox{exp})$. $\Box$
• The above proof doesn't use the projective nature of abelian varieties, but you're right, I should have clarified this. All that is used is the fact that $A$ is a connected compact complex Lie group. – rfauffar Jan 11 '14 at 12:14