When is a holomorphic submersion with isomorphic fibers locally trivial?

Question

A justly celebrated theorem by Ehresmann states that a proper smooth submersion $\pi: X\to S$ between smooth manifolds is locally trivial in the sense that every point $s\in S$ downstairs has a neighbourhood $ U$ such that $\pi ^{-1} (U) $ is $S$-diffeomorphic (=fiber preserving diffeomorphism) to $U\times F$ for some manifold $F$, called the typical fiber. Of course the holomorphic analogon is completely false: deformation theory might be described as the study of this failure!

To give a concrete example, consider the family $X \subset S\times \mathbb P^2 (\mathbb C) $ of elliptic curves $y^2z=x(x-z)(x-\lambda z)$ with $\lambda \in S=\mathbb C \setminus \{0,1\} $ and the corresponding proper holomorphic submersion $\pi: X \to S: (\lambda , [x:y:z]) \mapsto \lambda $. This $\pi$ is certainly not locally trivial downstairs because its fibers are not mutually isomorphic.

However, in the proper case, non-isomorphy of fibers is the only obstruction to being locally trivial. Indeed, Fischer and Grauert proved that a proper holomorphic submersion having all its fibers isomorphic is locally trivial downstairs. I wonder what can be salvaged of their theorem in the non-proper case:

Question: Is there a class $\mathcal C$ of non-compact complex manifolds such that the following holds. If a holomorphic submersion $\pi: X\to S$ between complex manifolds has all its fibers $\pi^{-1}(s)\;(s\in S)$ isomorphic to the same $F \in \mathcal C$, then $\pi$ is locally trivial downstairs with typical fiber $F$.

Bibliography Fischer and Grauert unfortunately published their result in a rather confidential journal: W. Fischer, H. Grauert, Lokal-triviale Familien kompakter komplexer Mannigfaltigkeiten, Nachr. Akad. Wiss. G¨ottingen Math.-Phys. Kl. II (1965), 89–94.

(Please note that the Fischer here is Grauert's doctoral student Wolfgang Fischer, not the complex geometer Gerd Fischer. )

Answer 1

Here is a variant of Jason's example with a proof that it is not even topologically locally trivial. Let $T$ be a (complex) manifold that admits a morphism $\phi$ onto $\mathbb P^1=\mathbb P^1_{\mathbb C}$ (or $S^2$ if you prefer) and there exists a point $a\in T$ with $b=\phi(a)\in \mathbb P^1$ such that $\{a\}=\phi^{-1}(b)$. This is satisfied for example if $\phi={\rm id}_{\mathbb P^1}$. Let $\Gamma\subset T\times \mathbb P^1$ be the graph of $\phi$ and let and $c\in \mathbb P^1, c\neq b$.

Now let $X=T\times \mathbb P^1\setminus \bigg( \Gamma\cup \big(T\times \{b\}\big)\cup \{(a,c)\}\bigg)$ with the natural projection $\pi:X\to T$. Then every fiber of $\pi$ is isomorphic to $\mathbb P^1\setminus \{0,\infty\}\simeq \mathbb C^*\sim S^2\setminus \{P,Q\}$ (for two points $P,Q\in S^2$).

Claim $\pi$ is not topologically locally trivial near $a\in T$.

Proof Suppose $a$ has a neighbourhood $U\subseteq T$ such that $Y:=\pi^{-1}U\simeq U\times S^2\setminus \{P,Q\}$. Then there exists a projection $p:Y\to S^2\setminus \{P,Q\}$. Consider a circle in $S^2\setminus \{P,Q\}$ that's non-trivial in $H_1(S^2\setminus \{P,Q\}, \mathbb Z)$. Since $p$ is an isomorphism between $\pi^{-1}(a)$ and $S^2\setminus \{P,Q\}$, the same circle lives in $\pi^{-1}(a)$ as well. Then the homology class of the circle can be represented by a "small" circle around the point $(a,c)$ (this point is not in $X$!). Next take a "small" ball inside $T\times \mathbb P^1$ with center at $(a,c)$ that contains the previous "small" circle. By the construction of $X$, the intersection of this ball with $X$ is the entire ball except its center $(a,c)$. Therefore the homology class of that "small" circle in $X$ is trivial. However, this is a contradiction, because it was chosen in a way that its image via $p_*$ would be a nontrivial homology class. $\qquad {\rm Q.E.D.}$

Remark I suppose a similar proof works to show that Jason's example is also not locally trivial.

Answer 2

If you drop the properness hypothesis, there are easy counterexamples. For instance, begin with A^1 with coordinate t, and with A^2 with coordinates x and y. Consider the projection p_1: A^1 x A^2 --> A^1, (t,(x,y)) --> t. Now remove the closed subset C = Z(x(x-t),y) as well as the disjoint closed set which is the singleton {(0,(0,1))}. Let X be the open complement. Then p_1:X --> A^1 is flat, all geometric fibers are isomorphic -- each being the complement of two points in A^2, yet the morphism is not locally trivial.

Regarding Sandor's suggestion, something like this is written up in my joint paper with Johan de Jong on "almost properness" of GIT stacks.

Answer 3

Taking your question to the realm of schemes I think that assuming something like that ${\rm Aut} F$ has a natural scheme structure gives you something that could be considered the algebraic equivalent of this statement.

Here is the argument. You can decide what needs to be assumed in addition to the above to get your goal.

Let $\pi:X\to S$ be a smooth morphism and $F$ a (smooth) variety such that $F\simeq X_s$ for all $s\in S$. Consider the relative ${\rm Isom}$ scheme $$ {\rm Isom}_S(X,F\times S)\to S. $$ This is a problem point as it might not exist. Or rather, the ${\mathscr Isom}$ functor is not necessarily representable. See this answer for a sketch of why this functor is representable for projective families and Torsten Ekedahl's answer to the same question for an easy example when it is not.

Anyway, if $I:={\rm Isom}_S(X,F\times S)$ exists, then consider the base change of your family to $I$, $$ X_I=X\times_S I\to I, $$ and consider the relative ${\rm Isom}$ scheme for the new family: $$ {\rm Isom}_I(X_I,F\times I) \to I. $$ Since the fibers of $X$ are isomorphic to $F$, the fibers of this scheme are isomorphic to ${\rm Aut} F$.

From the definition of the ${\mathscr Isom}$ functor it is clear that $$ {\rm Isom}_I(X_I,F\times I)\simeq {\rm Isom}_S(X,F\times S)\times_SI = I\times_S I, $$ so it admits a natural section over $I$. In other words, $X_I\simeq F\times I$.

So this proof seems to show that if ${\rm Isom}_S(X,F\times S)$ exists, then there is a base-change that trivializes $\pi:X\to S$. As a next step I would try to take multisections of ${\rm Isom}_S(X,F\times S)\to S$ to get a finite cover. You would probably want for each $s\in S$ a multisection that is unramified over $s$ to get an étale local trivialization. If you are happy with a local trivialization in the Euclidean topology (assuming you're working over $\mathbb C$) then this should do it. For the issue of local trivialization in the Zariski topology, see below.

Finally, this proof definitely shows that if $F$ is projective with a finite automorphism group, then $\pi:X\to S$ is étale locally trivial: the projectivity of $F$ implies the existence of ${\rm Isom}_S(X,F\times S)$ and the finiteness of ${\rm Aut} F$ implies that ${\rm Isom}_S(X,F\times S)\to S$ is a finite étale morphism.

To see the last statement, recall that the $I:={\rm Isom}_S(X,F\times S)$ functor/scheme is defined as an open subfunctor/subscheme of ${\rm Hilb}_S(X\times_S(F\times S))\simeq {\rm Hilb}_S(X\times F)=:H$. From the latter description it is clear that $H$ is flat over $S$. As $I$ is open in $H$, its closure in $H$, say $\overline I\subseteq H$ consist of a union of irreducible components of $H$ and hence it is also flat over $S$. Furthermore, $H$ is projective over $S$ and this implies that then $\overline I$ is finite and flat over $S$. Then it follows that the length of the fibers of $\overline I$ is constant. But the same is true for $I$ because its fibers over $S$ are isomorphic to the same automorphism group. Finally, as $\overline I$ is finite over $S$, $\overline I\setminus I$ cannot be dominant onto $S$, which implies that the general fiber of $\overline I$ and $I$ coincide. So the constant length of the fibers of both $\overline I$ and $I$ are the same and hence they are the same. In other words, $\overline I=I$ and hence $I$ is flat over $S$. As it has isomorphic fibers, it has to be unramified and hence étale.

Addendum

Regarding the discussion of families that are analytically but not algebraically locally trivial, an important difference is whether they have a section or not.

Assume that $\dim S=1$. Then if $\pi:X\to S$ is a smooth projective family without a section, then it cannot be algebraically locally trivial, since the closure of a section of the trivial part over a Zariski open set would give a section of the family. So, this way it is easy to find tons of examples.

On the other hand, a (quasi-)projective family over a curve will always have multisections, so it will have a section after a base change. So, for algebraically locally trivializing an isotrivial family the best hope is to do it after a finite base change. The above proof shows that if the appropriate $\mathscr Isom$ functor is representable by a quasi-projective scheme then this is doable. In particular, if $\pi:X\to S$ is projective, it can be trivialized with a finite base change (take a multisection of ${\rm Isom}_S(X,F\times S)\to S$).