Based on the sketch given in Canonical divisor of an abelian variety and The genus of a curve with a group structure, I tried to work out the proof for an abelian variety $X$ of dimension 1 with multiplication map $m$ has genus 1. Denote $e$ the neutral element. Assume $X = \mathbb{V}(f).$
- Consider the tangent bundle $TX$ to an abelian variety of dimension 1. Let $$L_g: X \to X: h \to m(g,h).$$ $T_p(X) = \mathbb{V}(dF|_p(x-p))$ according to "An invitation to Algebraic Geometry". Then we can define $$\phi: G \times T_e G \to TG: (g,p) \to (g, {L_g}_\star(p-e)+g).$$ How can I show that this is an isomorphism?
- If the tangent bundle is trivial, then the cotangent bundle is trivial as well. Now the differential 1-forms are the sections of this bundle, so it are the constants. Then $$deg(K) = 0=2g-2$$ for a canonical divisor $K$; this equation is a corollary from Riemann-Roch, so the genus is 1. I am not sure how I can declare the step from differential 1-forms to holomorphic 1-forms, since these are the one used in Riemann-Roch. I am not familiar with sheaves, so that's why they are avoided.
