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Let $S\overset{\pi}{\to} E$ be a ruled surface over an elliptic curve over complex field. Clearly, there are rational curves and elliptic curves on $S$. Is there any higher genus curves on $S$. Are all the elliptic curves isomorphic. The reason for the second question is that, all the smooth sections are isomorphic elliptic curves, except this sections is there any other elliptic curves?If so, how do I find them?

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2 Answers

up vote 5 down vote accepted

Any surface has lots of curves of high genus. Just take a generic hypersurface section of high degree.

Any other elliptic curve will be isogenous to $E$ with $\pi$ inducing the isogeny. I think you can embed any isogenous curve in $S$.

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Thanks for the answer. But I still have questions. Is there a way to find a concrete curve? Why the elliptic curves are all isogenous to $E$? –  Fei YE Dec 18 '09 at 19:24
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Concrete curve of high genus? How is your surface given in the first place? If you have your surface in $P^n$, just take your favorite homogeneous polynomial in $n+1$ variables not vanishing on the surface and consider the curve inside the surface where this polynomial vanishes. One can't be more concrete than that. If $E' \subset S$ is an elliptic curve, then $E'$ is not contained in a fiber, since the fibers are rational, so $\pi : E' \to E$ is non-constant, but a non-constant map between elliptic curves must be an isogeny (modulo translations but you can choose the origin in $E'$) –  Felipe Voloch Dec 18 '09 at 20:00
    
Ruled surfaces come from rank two vector bundles on E, and these are all explicitly known. So the surface is `given', in a sense. –  t3suji Dec 18 '09 at 20:20
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To find higher genus curves without using a specific embedding $S \subset \mathbb{P}^n$, it could help to think first about the case when your surface is actually a product $S=\mathbb{P}^1 \times E$. Let $C$ be a curve which admits two branched covers, $f\colon C \to E$ and $g \colon C \to \mathbb{P}^1$. Then the product $f \times g \colon C \to S$ maps into the surface $S$. If the branch points of $f$ and $g$ are different then $f \times g$ will even be an embedding.

In general, let $V \to E$ be your rank-two vector bundle, so $S=\mathbb{P}(V)$. Given a banched cover $f \colon C \to E$, you pull back $V$ to a bundle $V' \to C$. Now every time you have a line sub-bundle $L$ of $V'\to C$ you get a section of $\mathbb{P}(V')$ which plays the role of $g$ in the first paragraph. It can be combined with $f \colon C \to E$ to give a map $C \to S$. Depending on how much you know about $E$ and $V$, hopefully this should help you find plenty of explicit curves in $S$.

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Thanks. The explain is really helpful! –  Fei YE Dec 19 '09 at 1:08
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