MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?



share|cite|improve this question
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...) – user5117 Jan 9 '12 at 17:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.