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Let $k$ be an algebraically closed field, it is well known that $\mathbb P^1$ is simply connected, but how about smooth projective surfaces $X$ with a smooth morphism to $\Bbb P^1$?

Except the case $X$ is a product of curves or a projective bundle like Hirzebruch surfaces, could we classify all such $X \rightarrow \Bbb P^1$ ? What about higher dimensional cases?

Motivation: Shafarevich conjecture over function field.

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Let $k$ be an algebraically closed field.

Let $f:X\to \mathbb{P}^1$ be a smooth proper morphism with fibres of dimension one. Note that the fibres of $f$ are geometrically connected by Stein factorization and the fact that $\mathbb{P}^1$ is simply connected (Riemann-Hurwitz).

Theorem. The morphism $f$ is isotrivial.

Proof. Let $g$ be the genus of the fibres. Clearly, if $g=0$, then all geometric fibres are isomorphic. Thus, if $g=0$, then $f$ is isotrivial. Next, assume that $g=1$. Then, as the moduli space of elliptic curves is affine, the moduli map associated to the Jacobian of $f$ is constant, so that $Jac(f)$ has constant $j$-invariant. It follows that $f$ is isotrivial. Finally, it is a theorem of Moret-Bailly that any genus $g>1$ curve over $\mathbb{P}^1$ is isotrivial; see Lemme 5 in [1]. QED

[1] Laurent Moret-Bailly. Un théorème de pureté pour les familles de courbes lisses. C. R. Acad. Sc. Paris, t. 300, Serie I, n. 14, 1985.

Remark. If $k$ has characteristic zero, then Moret-Bailly's theorem is due to Parshin. Parshin's theorem follows from the "hyperbolicity" of the moduli stack of genus $g$ ($g>1$) curves. There are many notions of "hyperbolicity" lurking around there, and all of them imply the statement you want (in characteristic zero). (More generally, if $k$ is of characteristic zero, any smooth proper morphism $X\to \mathbb{P}^1$ whose fibres are smooth proper connected varieties with ample canonical bundle is isotrivial by work of Kovács, Migliorini, Viehweg-Zuo.)

Let $F$ be a fibre of $f$ over a closed point of $\mathbb{P}^1_k$.

If $g>1$, then the Isom-scheme $Isom(F\times \mathbb{P}^1, X)\to \mathbb{P}^1$ is finite etale and thus trivial. Thus, $f$ is trivial.

If $g=1$, then probably the family $X\to \mathbb{P}^1$ is trivial Will Sawin's argument below.

If $g=0$, as Daniel Loughran explains in the comments below, the (isotrivial) morphism $f$ has a section by Tsen's theorem, so that $X$ is a Hirzebruch surface.

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    $\begingroup$ If $g=1$ then the automorphism of the Jacobean, as a pointed elliptic curve, are finite so the same argument shows isotrivial implies trivial. So we have to classify $H^1(\mathbb P^1, E)$ for elliptic $E$. Use that it is algebraic to find a multi section, hence show it’s torsion, then use the vanishing of $ H^1(\mathbb P^, E[n])$ to conclude. Without algebraicity, the Hopf surface is a counterexample. $\endgroup$
    – Will Sawin
    Oct 15, 2018 at 23:27
  • $\begingroup$ Thanks, the tool of moduli space is wonderful. Last but not least, what about the case $g=0$? $\endgroup$
    – sawdada
    Oct 16, 2018 at 3:22
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    $\begingroup$ For $g=0$ there is a section by Tsen's theorem. So you get a $\mathbb{P}^1$-bundle over $\mathbb{P}^1$ with a section, which is thus isomorphic to a Hirzebruch surfaces (so not necessarily trivial). $\endgroup$ Oct 16, 2018 at 8:35

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