One form of de Franchis theorem for algebraic curves is the following: let $X$ be an algebraic curve (defined over $\mathbb{C}$ say) with genus $g > 1$. Then there are only finitely many (isomorphism classes of) curves $Y$ with genus $g' > 1$ such that there is a non-constant map $f : X \rightarrow Y$.

My question is the following: suppose that $X, X^\prime$ are two curves of the same genus $g > 1$ admitting a dominant map to the same curve $Y$ of genus $h > 1$. What can be concluded about $X,X^\prime$? Can there be infinitely many isomorphism classes of such $X$, say?

  • 2
    I don't understand your question. What is fixed? $X$, $Y$, ...? – abx Dec 4 at 5:20
up vote 3 down vote accepted

Let $Y$ be an arbitrary fixed curve of genus $h>1$ and $\mathscr L$ a fixed ample line bundle. Further let $m\in\mathbb N$ fixed such that $\mathscr L^{\otimes m}$ is very ample and let $D_\lambda\subseteq Y$ be general members of the linear system corresponding to $\mathscr L^{\otimes m}$. Finally, let $X_\lambda\to Y$ be the cyclic branch cover corresponding to the divisor $D_\lambda$.

By construction $X_\lambda$ is a smooth curve of genus $g>1$ for an appropriate $g$ which only depends on the choice of $\mathscr L$ and $m$, but not on $\lambda$. Another part of de Franchis's theorem says that for fixed $X$ and $Y$ as in the question there are only finitely many different maps $X\to Y$ (one could think of this as a generalization of the finiteness of the automorphism groups of curves of genus at least two). Notice further that the maps we defined cannot be the same even if the $X_\lambda$ are isomorphic as their branch divisors are different. These observations imply that any given $X_\lambda$ could only be isomorphic to finitely many others and hence there are infinitely many different isomorphism classes among them. Note that in addition to the original requirements these curves admit maps of the same degree to this fixed target curve.

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