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. )

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    $\begingroup$ I am confused about the Grauert-Fischer theorem, and I don't have a reference at hand (not to mention my german is not in very good shape). Certainly there exist proper maps $\pi \colon X \to S$, with both $X$ and $S$ compact, which are isotrivial on an open subset $U \subset S$, but not locally trivial. It seems to me that the restriction $\pi^{-1}(U) \to U$ is still proper, but does not satisfy the conclusion of the G-F theorem. :-? $\endgroup$ – Andrea Ferretti Feb 26 '11 at 15:19
  • $\begingroup$ I don't know the answer; but are there any know examples where the fibers are all isomorphic, but the family is not locally trivial? $\endgroup$ – Angelo Feb 26 '11 at 15:40
  • $\begingroup$ Dear Andrea, you write that there "certainly" exist counterexamples to Fischer-Grauert's theorem (I suppose that by isotrivial you mean isomorphic fibers ?). What makes you think so ? You can find that theorem stated in English in Barth, Peters and Van de Ven's classic book Compact Complex Surfaces in Chapter I,Theorem 10.1 on page 29. In a way, I'm happy about your doubts: it shows how unbelievable Fischer-Grauert's theorem is! $\endgroup$ – Georges Elencwajg Feb 26 '11 at 16:23
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    $\begingroup$ Dear Angelo: no, there are no such examples because they are excluded by Fischer-Grauert's theorem! It seems this theorem doesn't have the celebrity it deserves, maybe because of the already mentioned relative obscurity of the journal where it was published. And also because of the language barrier: if articles are to be read, they must be written in Modern Latin, i.e. English! $\endgroup$ – Georges Elencwajg Feb 26 '11 at 16:32
  • $\begingroup$ So you are claiming that every isotrivial family of compact complex manifolds is actually locally trivial? I have to admit I do not have a proof of non-local-triviality for any isotrivial family (over $\mathbb{C}$) I can think of (and foor good reasons, apparently!) but then I wonder why the terminology... I wish I was always just told "here is a fiber bundle with fiber $F$". $\endgroup$ – Andrea Ferretti Feb 26 '11 at 16:47

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.

  • $\begingroup$ This is an interestingly unexpected (by me) technique, Sándor. $\endgroup$ – Georges Elencwajg Feb 27 '11 at 0:28
  • $\begingroup$ It's ad hoc. I tried to grasp the fact that the projection maps the complement of two lines and a single point to the complement of two points. If the map were algebraic one could argue many ways that this can't happen, but you seem to be hung up on holomorphic things. (wink-wink) :) $\endgroup$ – Sándor Kovács Feb 27 '11 at 0:56
  • $\begingroup$ To return the compliment, you seem to know a suspicious lot of holomorphic notions -like hyperbolic manifolds mentioned in your comment above- for a tough-boiled algebraic geometer :) $\endgroup$ – Georges Elencwajg Feb 27 '11 at 12:52
  • $\begingroup$ Georges, you are too kind to call this "returning the compliment". It's more like a gracious way to respond to a silly joke. On the other hand, I believe that a true tough-boiled algebraic geometer ought to know a little complex geometry and a little topology if for nothing else but motivation. $\endgroup$ – Sándor Kovács Feb 27 '11 at 19:09

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.

  • $\begingroup$ Dear Jason, thank you for your interesting answer. I'd like to insist that we are in a purely holomorphic setting, so I interpret $\mathbb A^1$ and $\mathbb A^2$ as $\mathbb C^1$ and $\mathbb C^2$ respectively and geometric fibers as fibers, all isomorphic to $\mathbb C^2 \setminus \{{0,1 \}}$ . Your example is then very convincing but I confess that I cannot write down a proof that it is not locally trivial. $\endgroup$ – Georges Elencwajg Feb 26 '11 at 21:45
  • $\begingroup$ Consider the following variation on your construction. Start with $\mathbb C^2$ with coordinates $x,y$ ; delete both the $x$-axis and the diagonal $x=y$ and also the point $(0,1)$ . Call $X$ the remaining open subset $X \subset \mathbb C^2$ and project it down to the $x$-axis. Do you think the submersion $ X \to \mathbb C$ thus obtained is locally trivial holomorphically? $\endgroup$ – Georges Elencwajg Feb 26 '11 at 21:47
  • $\begingroup$ Georges, I don't think that this example will be locally trivial. I included a proof below that shows that if you do the same with adding the points at infinity on the fibers then it is not locally trivial. I think that a similar but more laborious proof would show that this example is also not locally trivial. $\endgroup$ – Sándor Kovács Feb 26 '11 at 23:13
  • $\begingroup$ Thanks again for your interest, Sándor. I also thought that my modification of Jason's example wasn't locally trivial but couldn't (and still can't) prove it. I have the feeling that Hartogs's theorem might be used in this context initiated by Jason. Anyway I am happy that your subtle argument below confirms that it is not trivial to show that all these submersions are not locally trivial! $\endgroup$ – Georges Elencwajg Feb 27 '11 at 0:22
  • $\begingroup$ Georges, I think you are right. By Hartog's theorem you can extend the projection map into the single point and then argue that the rest can't map to where it is supposed to. $\endgroup$ – Sándor Kovács Feb 27 '11 at 1:55

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.


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

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 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 ${\rm Isom}_S(X,F\times S)\to S$).

  • $\begingroup$ Dear Sándor, although my question in principle concerns the holomorphic category, I thank you for your interest in it and for your remarks on their algebraic equivalent and the role of (multi)sections there. $\endgroup$ – Georges Elencwajg Feb 26 '11 at 22:45

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