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A well-known theorem of Grauert and Fischer states that a smooth proper family of complex manifolds is a locally trivial fibration as soon as all the fibers are isomorphic. It is also easy to obtain an algebro-geometric variant of this; cf. Lemma 1.3 in Buium's Differential Algebra and Diophantine Geometry.

For a complete (generically smooth) family of curves $X/B$, isotriviality (take this to mean that all smooth fibers are isomorphic) is thus a priori equivalent to having a quasi-finite dominant map $B' \to B$ over which $X \times_B B'$ is a product family. Now, for families of positive genus, it is often taken for granted that $B' \to B$ may be taken to be surjective (that is, finite). Why is this, or where is it proved? This of course is false for $\mathbb{P}^1$-bundles: if there were a finite cover $f: C \to \mathbb{P}^1$ under which $\mathbb{P}(\mathcal{O}_{\mathbb{P}^1} \oplus \mathcal{O}_{\mathbb{P}^1}(1)) \to \mathbb{P}^1$ pulled back to the product bundle $C \times \mathbb{P}^1$, then the line bundles $f^* \mathcal{O}_{\mathbb{P}^1}$ and $f^*\mathcal{O}_{\mathbb{P}^1}(1)$ would be isomorphic, which they are not: one is ample, the other is not.

For elliptic surfaces $E/B$ we can see it as follows. First, following a finite base change attaining stable reduction, we may assume that the family is smooth. Consider then the line bundle $\omega := 0^* \Omega_{E/B}^1$ on $B$. We know that $\omega^{\otimes 12}$ is trivial: it has the nowhere vanishing global section $\Delta$. It follows that there is a finite etale covering $p :B' \to B$ of degree dividing $12$ with $p^*\omega \cong \mathcal{O}_{B'}$ trivial. Then it is easy to see that $E \times_B B' \to B'$ splits (cf. section 3.2 of Ulmer's survey Elliptic curves over function fields).

The question: If $X/B$ is a family of abelian varieties or of curves of genus $> 0$, with all members isomorphic (a smooth isotrivial morphism), is there a finite etale covering $B' \to B$ such that $X \times_B B'$ is a product family? Does the same statement hold for smooth isotrivial families of projective varieties of non-negative Kodaira dimension?

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This is a question where the answer is different in the analytic and algebraic settings. There is an analytic smooth isotrivial family of elliptic curves over $P^1$, the Hodge surface. –  Ben Wieland Jul 29 at 15:40

3 Answers 3

up vote 8 down vote accepted

Let me work out the case of an isotrivial family $X\rightarrow B$ of curves -- the same argument applies to abelian varieties. The point is that the moduli space $\mathscr{M}_g(n)$ of curves $C$ of genus $g$ with a level $n$ structure (that is, a symplectic isomorphism $(\mathbb{Z}/n)^{2g}\stackrel{\sim}{\rightarrow }H^1(C,\mathbb{Z}/n)$) is a fine moduli space for $g\geq 2$, $n\geq 3$. Let $B'\rightarrow B$ be a finite étale cover such that the local system $H^1(X_b,\mathbb{Z}/n)_{b\in B}$ becomes trivial on $B'$. Since the curves $X_b$ are all isomorphic, the classifying map $B'\rightarrow \mathscr{M}_g(n)$ must be constant; since $\mathscr{M}_g(n)$ is a fine moduli space, this implies that the pull back of the fibration $X\rightarrow B$ to $B'$ is trivial.

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Thank you! It seems that the same should hold for smooth isotrivial families of varieties of non-negative Kodaira dimension? –  Vesselin Dimitrov Jul 24 at 20:53
    
It seems likely, but I am not completely sure: you might get into trouble with automorphisms acting trivially in cohomology, for instance. –  abx Jul 25 at 6:54
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At least in characteristic 0, I think it should be fine for all non-uniruled varieties. Of course everything is fine when the automorphism group is finite (or can be "naturally" reduced to a finite group). That leaves varieties whose automorphism group contains a positive-dimensional (connected) linear group. But all such groups are (geometrically) rational. Thus the orbits are unirational. So the variety is uniruled. –  Jason Starr Jul 25 at 12:44

abx and Jason Starr already answered your question. Allow me to just add one remark.

Adding level structure is one way to "trivialize" isotrivial families of varieties by a finite etale base change. It's a powerful tool because it corresponds to the existence of a finite etale atlas of the corresponding moduli stack. As far as I know, this is possible for families of principally polarized abelian varieties (in all characteristics coprime to the "level"). Thus, consequently, by passing to the Jacobian (resp. Kuga-Satake resp. intermediate Jacobian) and applying infinitesimal Torelli, this is also possible for families of curves of genus at least two (resp. polarized K3 surfaces resp. cubic threefolds). On the other hand, I don't know whether there is a notion of "level structure" for (polarized) families of Calabi-Yau threefolds (for instance). Is there such a notion?

If $X \to B$ is an isotrivial family of (smooth projective) varieties over an algebraically closed field $k$ and $F$ is a fiber of $X \to B$, the family trivializes over the Isom-scheme Isom$(X, B \times_k F) \to B$ (for tautological reasons). Thus, to answer your question (as Jason Starr says), an isotrivial family trivializes over a finite etale base change if the Isom-scheme above is finite etale. Thus, if $k$ is of characteristic zero, it suffices for your purposes that the above Isom-scheme is finite over $B$. As Isom-schemes are usually affine (at least in the cases you are interested in), it suffices for the Isom-scheme to be proper. This translates into a statement about lifting $K$-automorphisms on the generic fibre of $X\to B$ uniquely to $R$-automorphisms (via the valuative criterion) with $K$ the function field of a dvr $R$. The latter statement was proven by Matsusaka and Mumford for "one-parameter" polarized families of varieties of non-negative Kodaira dimension; see Theorem 2 in http://dash.harvard.edu/handle/1/3450065 . This answers your question positively for (polarized) families when $B$ is one-dimensional.

By the way, this is where you need polarizations in general. Else the Isom-schemes are clearly not finite (in the case of families of abelian varieties this difficulty can be circumvented in many ways: adding level structure is just one way). So to be really precise, it's not the above Isom-scheme that you should use, but rather the Isom-scheme taking into account polarizations.

To finish, let me mention a a fairly general statement which answers your question for families of varieties with ample canonical bundle. In fact, in Lemma 7.3 of the paper (by Kovacs and Kebekus)

http://home.mathematik.uni-freiburg.de/kebekus/research/papers/08-KMcK3.pdf

it is shown that if the fibres of the family $X \to B$ are canonically polarized (i.e., have ample canonical bundle), and $B$ is a curve, the Isom-scheme is finite etale. Their result is slightly more general and even allows $B$ to be higher-dimensional under the condition that $B$ fibres smoothly over a $\dim B -1$-dimensional variety. It might be possible to extend the arguments in the proof of Lemma 7.3 to your case.

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Thank you very much! –  Vesselin Dimitrov Jul 29 at 13:41
    
You are welcome. –  Ari Jul 30 at 7:11

This is false for abelian varieties, as eluded by Ari. The moduli of polarized abelian varieties is a DM stack, as they have only finitely many automorphisms preserving the polarization, but some abelian varieties have infinite order automorphisms, which must permute the polarizations. In particular, the product of a non-CM elliptic curve with itself has automorphism group $GL_2(\mathbb Z)$. An arbitrary automorphism can be realized as monodromy over the base a nodal curve by taking the trivial family over the normalization and gluing by the automorphism.

But you may consider a nodal curve pathological. A normal variety has profinite étale fundamental group and cannot have such exotic automorphisms as monodromy. An abelian variety over a normal base must support a polarization.

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Also, since analytic fundamental groups are discrete, this allows the creation of an analytic bundles of abelian varieties over a smooth complete base (say, an elliptic curve) that is isotrivial but not trivial on a finite cover. These is a very different example than the Hodge surface. –  Ben Wieland Jul 31 at 23:52

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