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I read that an elliptic curve $E({\mathbb R})$ is isomorphic to ${\mathbb R}/{\mathbb Z}$ if $x^3 + ax + b$ has only one real root, but what is the exact map? Does it come from the Weierstrass function?

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Basically, yes.

If $\Lambda$ is a lattice in $\Bbb C$ the map $$z\mapsto (\wp(z),\wp'(z))$$ is a parametrisation of the complex points of the elliptic curve $$E:\qquad y^2=4x^3-g_2x-g_3$$ where $g_2$ and $g_3$ depend on $\Lambda$. It induces a group isomorphism $\Bbb C/\Lambda\to E(\Bbb C)$. When $\Lambda=\Bbb Z r+\Bbb Z is$ with $r$, $s$ real, then $g_2$ and $g_3$ are real, and $t\mapsto (\wp(t),\wp'(t))$ restricts to an isomorphism $\Bbb R/r\Bbb Z\to E(\Bbb R)$.

This encompasses all real elliptic curves with $E(\Bbb R)$ connected.

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  • $\begingroup$ what is $s$? typo in $\Lambda$? $\endgroup$ – Andres Mejia Aug 6 '18 at 18:54

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