# bielliptic surfaces

Definition: A surface is called $S$ bielliptic if $S \cong (E \times F) /G$, where $E,F$ are elliptic curves and $G$ is a finite group of translations of $E$ acting on $F$ such that $F /G \cong \mathbb{P}^1$.

How can $F /G \cong \mathbb{P}^1$? Can you give an example with an geometric intuition?

Thanks

Peter

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Take an elliptic curve $F$ and the natural involution $\iota \colon x \to -x$. Then the group $G:=\langle \iota \rangle$ has $4$ fixed points, which are precisely the $4$ points of order $2$ on $F$, so the quotient $F \to F/G$ realizes $F$ as a $2$-sheeted cover of $\mathbf{P}^1$, branched in $4$ points. Moreover the $j$-invariant of $F$ can be recoverd in terms of the cross ratio of these points. This is very classical and can be found in any introductory book on Riemann surfaces (Miranda's one, for instance). –  Francesco Polizzi Jan 9 '12 at 17:01
(just a retag...) –  Artie Prendergast-Smith Jan 9 '12 at 17:22