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What are concrete examples of fibre bundles with fibre genus $\geq 2$? I am trying to find examples I can use to work throught the following construction:

Suppose $X \to C$ is a fibre bundle with $C$ a curve and for which the fibre $F$ is a curve of genus $\geq 2$. Then $\mathrm{Aut}(F)$ is a finite group and we can look at $\bar{F} := F/\mathrm{Aut}(F)$. We get a natural morphism $X \to \bar{F} \times C$. We can compose this with Belyi morphisms for $\bar{F}$ and $C$ to obtain a cover of $\mathbb{P}^1 \times \mathbb{P}^1$.

Hence with 'concrete' I mean that by following the above construction I can actually see what the resulting cover of $\mathbb{P}^1 \times \mathbb{P}^1$ looks like.

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  • $\begingroup$ I assume that by "Belyi morphism" you mean a finite morphism to $\mathbb{P}^1$ that is an unbranched over away from three points of $\mathbb{P}^1$. Why do you believe that either of $\overline{F}$ or $C$ admits a Belyi morphism to $\mathbb{P}^1$? $\endgroup$
    – Jason Starr
    Commented Dec 23, 2013 at 15:38
  • $\begingroup$ Belyi's theorem says: A complete nonsingular algebraic curve over a field of characteristic zero can be defined over $\bar{\mathbb{Q}}$ if and only if it can cover $\mathbb{P}^1$ with ramification over three points. So by picking my curves to satisfy those conditions, I can be certain to find a Belyi morphism. $\endgroup$
    – Peter Lombaers
    Commented Dec 24, 2013 at 10:30
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    $\begingroup$ I know what Belyi's theorem says. However, nowhere in your statement do you specify that your curves are defined over $\mathbb{Q}$. $\endgroup$
    – Jason Starr
    Commented Dec 24, 2013 at 13:21

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All such fiber bundles are constructed in the following way : choose an étale covering
$\tilde{C}\rightarrow C $ with Galois group $G\subset \mathrm{Aut}(F)$, and take $S=(\tilde{C}\times F )/G$, with $G$ acting on each factor in the obvious way. The fiber bundle structure is given by the first projection $S\rightarrow C$. This is quite concrete.

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