When is a holomorphic submersion with isomorphic fibers locally trivial? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:40:36Z http://mathoverflow.net/feeds/question/56737 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56737/when-is-a-holomorphic-submersion-with-isomorphic-fibers-locally-trivial When is a holomorphic submersion with isomorphic fibers locally trivial? Georges Elencwajg 2011-02-26T14:00:23Z 2011-02-27T01:15:46Z <p>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!</p> <p>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.</p> <p>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:</p> <p><strong>Question:</strong> 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 $\pi^{-1}(s), s\in \mathcal C$ isomorphic to the same $F \in \mathcal C$, then $\pi$ is locally trivial downstairs with typical fiber $F$.</p> <p><strong>Bibliography</strong> 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.</p> <p>(Please note that the Fischer here is Grauert's doctoral student <em>Wolfgang</em> Fischer, not the complex geometer Gerd Fischer. )</p> http://mathoverflow.net/questions/56737/when-is-a-holomorphic-submersion-with-isomorphic-fibers-locally-trivial/56750#56750 Answer by Sándor Kovács for When is a holomorphic submersion with isomorphic fibers locally trivial? Sándor Kovács 2011-02-26T17:11:45Z 2011-02-26T21:51:04Z <p>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.</p> <p>Here is the argument. You can decide what needs to be assumed in addition to the above to get your goal.</p> <p>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 <a href="http://mathoverflow.net/questions/55042/automorphism-group-of-a-scheme/55050#55050" rel="nofollow">this answer</a> 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.</p> <p>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 ${\rm Aut} F$.</p> <p>From the definition of the ${\mathscr Isom}$ <em>functor</em> 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, $X_I\simeq F\times I$. </p> <p>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.</p> <p>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. </p> <p><strong>Addendum</strong></p> <p>Regarding the discussion of families that are analytically but not algebraically locally trivial, the important difference is whether they have a section or not.</p> <p>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.</p> <p>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 ${\rm Isom}_S(X,F\times S)\to S$). </p> http://mathoverflow.net/questions/56737/when-is-a-holomorphic-submersion-with-isomorphic-fibers-locally-trivial/56761#56761 Answer by Jason Starr for When is a holomorphic submersion with isomorphic fibers locally trivial? Jason Starr 2011-02-26T19:32:42Z 2011-02-26T19:32:42Z <p>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. </p> <p>Regarding Sandor's suggestion, something like this is written up in my joint paper with Johan de Jong on "almost properness" of GIT stacks.</p> http://mathoverflow.net/questions/56737/when-is-a-holomorphic-submersion-with-isomorphic-fibers-locally-trivial/56773#56773 Answer by Sándor Kovács for When is a holomorphic submersion with isomorphic fibers locally trivial? Sándor Kovács 2011-02-26T23:03:33Z 2011-02-27T01:15:46Z <p>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 <code>$\{a\}=\phi^{-1}(b)$</code>. 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 $c\in \mathbb P^1, c\neq b$.</p> <p>Now let <code>$X=T\times \mathbb P^1\setminus \bigg( \Gamma\cup \big(T\times \{b\}\big)\cup \{(a,c)\}\bigg)$</code> with the natural projection $\pi:X\to T$. Then every fiber of $\pi$ is isomorphic to <code>$\mathbb P^1\setminus \{0,\infty\}\simeq \mathbb C^*\sim S^2\setminus \{P,Q\}$</code> (for two points $P,Q\in S^2$).</p> <p><strong>Claim</strong> $\pi$ is not topologically locally trivial near $a\in T$.</p> <p><strong>Proof</strong> Suppose $a$ has a neighbourhood $U\subseteq T$ such that <code>$Y:=\pi^{-1}U\simeq U\times S^2\setminus \{P,Q\}$</code>. Then there exists a projection <code>$p:Y\to S^2\setminus \{P,Q\}$</code>. Consider a circle in <code>$S^2\setminus \{P,Q\}$</code> that's non-trivial in <code>$H_1(S^2\setminus \{P,Q\}, \mathbb Z)$</code>. Since $p$ is an isomorphism between $\pi^{-1}(a)$ and <code>$S^2\setminus \{P,Q\}$</code>, 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. $\qquad {\rm Q.E.D.}$</p> <p><strong>Remark</strong> I suppose a similar proof works to show that Jason's example is also not locally trivial.</p>