It is known that non-isotrivial fibrations of genus $g>0$ curves over the projective line have a bunch of singular fibers. There are at least three of them.

It is not difficult to prove that an elliptic fibration $X \longrightarrow \mathbb{P}^1$ have at least three singular fibers and at least two of them are not multiple of a smooth curve, provided that it is not isotrivial.

I was wondering if the number of singular fibers that are not multiples of a smooth curve is at least three but I didn't find any proof of that.

Is that true? If an elliptic fibration with an arbitrary number of singular fibers but only two of them are not multiple of a smooth curve, then it is isotrivial?

The surface $X$ is projective and smooth over $\mathbb{C}$


1 Answer 1


Consider first an elliptic fibration with a section over $\mathbb{P}^1$. (In this case none of the singular fibers are multiples of smooth curves.) Assume that the minimal discriminant has degree $12n$. Then the number of components of singular fibers which do not intersect the zero section equals $12n-2a-m$, where $a$ is the number of additive fibers and $m$ the number of multiplicative fibers.

By the Shioda-Tate formula this number is at most $\rho(X)-2$, and by Lefschetz $(1,1)$ and the fact that $h^{1,1}=10n$, we get that $2a+m\geq 2n+2$. Hence to have $a+m=2$ we need that $n=1,m=0$ and $a=2$. In this case there is a straightforward argument to show that none of the singular fibers of type $I_\nu^*$, so the family is isotrivial.

You can then get the general case by considering the Jacobian of the elliptic fibration. If the original fibration has a singular fiber over a point $p$, and this fiber is not a multiple of a smooth fiber, then the fiber of $p$ in the associated Jacobian fibration is also singular. For more details on this final step, see the final chapter of the book by Dolgachev and Cossec on Enriques' surfaces.


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