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In a conference on elliptic curves (an introduction to the subject), the speaker said that an elliptic curve (I.e. an equation of the form $y^2=x^3+ax+b $ where the RHS has distinct roots) is, in the complex space, a torus/Riemann surface of genus 1.

What is meant by that? Are we talking about a 2-dimensional manifold the 4D space?

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  • $\begingroup$ One can also use Riemann-Roch for see that the Abel-Jacobi map is an isomorphism. $\endgroup$ – user171326 Jul 24 '17 at 2:26
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If $\Lambda = \mathbb{Z}+\tau \mathbb{Z}, \tau \not\in \mathbb{R}$ is a lattice then $\mathbb{C}/\Lambda$ is a complex torus. We have the Weierstrass function of the lattice $$\wp(z) = \frac{1}{z^2}+\sum_{\lambda \in \Lambda^*} \frac{1}{(z-\lambda)^2} -\frac{1}{\lambda^2}$$ which is meromorphic and $\Lambda$ periodic, with a double pole on $\mathbb{C}/\Lambda$. From the Laurent expansion $\wp(z) = z^{-2}+0z^0+g_2 z^2+ g_3 z^4+\mathcal{O}(z^6)$ and because an entire doubly periodic function is constant, we find a non-linear relation : $$\wp'(z)^2 = 4 \wp(z)^3- g_2 \wp(z)-g_3$$ In other words, we have found an isomorphism (of Riemann surface and group) $$\varphi : \mathbb{C}/\Lambda \to E, \qquad \varphi(z) = (\wp(z),\wp'(z))$$ where $E$ is the complex elliptic curve $$E = \{ (x,y) \in \mathbb{C}^2, y^2 = 4x^3-g_2 x-g_3\}$$ Finally, applying change of variables to $x,y $ we find any complex elliptic curve is isomorphic to such a complex torus by the mean of its Weierstrass function.

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  • $\begingroup$ Your answer is quite advanced for me but it givese the idea. Thank you! $\endgroup$ – Friedrich Jul 23 '17 at 19:40
  • $\begingroup$ @Friedrich For me the complex analysis point of view is easier than the algebraic geometry's one. There are different approaches. Often, books on elliptic curves start by showing the abstract statement that any Riemann surface of genus $1$ is a complex torus, by looking at the invariant differential and the function fields. $\endgroup$ – reuns Jul 23 '17 at 19:46
  • $\begingroup$ It's important to be careful with the point at infinity in $E$ though. $\endgroup$ – Mr. Chip Nov 7 '18 at 6:35

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