First of all I want to say that algebraic geometry is not "my field of research" so I apologize if the notation is not standard.

$S$ is a smooth complex projective surface with a fibration $f$ over $\mathbb P^1(\mathbb C)$. Moreover suppose that the following properties hold for $f$:

  1. $f$ has singular fibers (a finite number).

  2. Each singular fiber is a curve with a single knot. (In this case I think that $f$ is called a stable fibration.)

Under the above hypotheses I have to prove (or disprove) that the fibration $f$ can't be isotrivial. Here isotrivial means that any two smooth fibers are isomorphic as algebraic varieties.

Many thanks.

  • 2
    $\begingroup$ A quick proof can be given if you take the existence of moduli spaces for granted. The fibration gives a map $\mathbb{P}^1 \to M, x \mapsto [f^{-1}(x)]$ where $M$ is the moduli space and $[C]$ is the class of a curve $C$. This map is non-constant since singular fibers cannot map to the same point as a non-singular fiber and your fibration has both. OTOH the map would be constant if the fibration were isotrivial. But I am sure this statement has to be proved on the process of constructing moduli spaces, so I am cheating a bit. $\endgroup$ Oct 28, 2014 at 20:18
  • $\begingroup$ My argument above doesn't work for genus zero, where $M$ is a point. I only saw this after reading Roberto's answer that has a genus zero example. $\endgroup$ Oct 29, 2014 at 9:02
  • 2
    $\begingroup$ I think your argument works. The point is that Roberto's example is not stable (the rational tail meets the other component of the singular fibre in just one point), hence it does not provide directly a map to the moduli space (this is precisely the reason why (semi)-stable reduction is introduced). $\endgroup$ Oct 29, 2014 at 9:15
  • 2
    $\begingroup$ In other words, since the fibration is stable, it provides a map to the stable moduli space of curves $\overline{\mathcal{M}}_g$. The image of this map meets both the boundary of the moduli space (because of the presence of the nodal, irreducible fibres) and its interior $\mathcal{M}_g$, hence it cannot be constant. This is precisely your argument, of course. $\endgroup$ Oct 29, 2014 at 9:21

2 Answers 2


There are three important invariants for any relatively minimal fibration $f:\,S \to B$ from a smooth complex projective surface $S$ to a smooth curve $B$: the self-intersection $\omega_{S/B}^2$, the degree of $f_*\omega_{S/B}$, and the singular index $e_f$ of $f$, where $\omega_{S/B}:=\omega_S\otimes\omega_B^{\vee}$ is the relative canonical sheaf of $f$. Here "relatively minimal" means that there is no exceptional curves (i.e., a curve isomorphic to $\mathbb P^1$ and have self-intersection equals $-1$) contained in fibers of $f$. If $f$ is stable (or semi-stable), $e_f$ is actually equal to the number of nodes contained in the singular fibers of $f$. These three invariants are all non-negative and satisfy the Noether formula: $$12\deg f_*\omega_{S/B}=\omega_{S/B}^2+e_f,$$ and the slope inequality $$\omega_{S/B}^2 \geq \frac{4(g-1)}{g} \deg f_*\omega_{S/B},~\text{equivalently,}~ (8g+4)\deg f_*\omega_{S/B}\geq g e_f,$$ where $g$ is genus of a general fiber of $f$.

An important description of these invariants for semi-stable (or stable) fibrations is as follows. Let $j:B \to \overline M_g$ be the induced morphism from $B$ to compactification of the moduli space $M_g$ of smooth curves of genus $g$. Then there exist divisors $\lambda, \kappa, \delta$ on $\overline M_g$ such that $$\deg f_*\omega_{S/B}=\deg j^*\lambda, \quad \omega_{S/B}^2=\deg j^*\kappa,\quad e_f=\deg j^*\delta.$$

Now by your assumption, $f$ is stable and we have $e_f>0$. On the other hand, if $f$ is isotrivial, then the image of $j$ is a point, hence the pulling-back of any divisor is trivial and so $e_f=\deg j^*\delta=0$, which is a contradiction.

We remark that the condition "$e_f >0$" implies in particular that $g>0$, since the moduli space of genus zero is a single point.

PS: from the arguments, the conclusion is also true for any base $B$, not only for $B=\mathbb P^1$.


If the genus of the fibre is not 0, by Theorem 2.1 in Serrano's paper "Fibrations on algebraic surfaces" any isotrivial fibration is birational to $(A \times B)/G \rightarrow B/G$ where $G$ is a finite group acting on both curves $A$ and $B$ and considered acting diagonally on the product $A \times B$. The singularities of $A \times B/G $ are cyclic quotient singularities, and solving them one gets a smooth surface with an isotrivial fibration birational to the starting fibration such that all singular fibres are normal crossing divisors (union of smooth curves intersecting each other at most transversally in a point) whose dual graph is a tree. So your starting surface is obtained by this surface by a finite sequence of blow ups and blow downs, where you are not allowed to blow down curves transversal to the fibration. The proofs of what I say are in Serrano's paper and the references therein.

If I do understand correctly in your assumptions you want all fibres to be irreducible (you say "curve") possibly not reduced, so you have a fibre which is irreducible and with an ordinary double point. So, performing first the blow ups (that can always be done), the dual graph of your fibre remains a tree (with possibly more vertices), and then you are contracting all curves but one. Then it is rather easy to show that you can't obtain a curve with a single ordinary double point, because you will never contract a curve intersecting another curve in two distinct points, the graph is a tree, at the end you will get either a smooth curve or a curve with a singular point which is a cusp (or worse).

If the genus of the fibre is zero every book on algebraic surface tells you that the fibration is birational to a bundle, and then the same argument applies.

If on the contrary you allow reducible fibres, then here is a simple counterexample. Pick any curve $C$ and blow up $C \times {\mathbb P}^1$ in a point. Then the composition of the blow up with the projection on the second factor gives a fibration on ${\mathbb P}^1$ with exactly one singular fibre, having two components respectively isomorphic to $C$ and to ${\mathbb P}^1$ intersecting transversally into a point. That's a trivial example for every algebraic geometer, I wrote it only because you said that algebraic geometry is not your field of research.

Edit: As pointed out by Francesco in the comments, the main point of the above simple counterexample is indeed not really the reducibility of the special fibre but its instability. Again I agree with Francesco that the "moduli" argument works, with the advantage that works in greater generality, as you only need to know that your fibres are stable, while the argument above requires you to know much more in detail which fibres you have. My argument is then weaker although more direct and elementary, as it does not need any deformation theory. Moreover, as pointed out in another answer, the genus of the base clearly plays no role at all.


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