Timeline for When is a holomorphic submersion with isomorphic fibers locally trivial?
Current License: CC BY-SA 2.5
9 events
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| Feb 27, 2011 at 19:20 | comment | added | Sándor Kovács | Georges, actually I think that the real reason is topological: you can't map a sphere minus $g$ points to a sphere minus $g+1$ points, especially that in this case it would have to be one-to-one outside one point. It was just easier above to say that a sphere minus two points is hyperbolic. However, I think to prove that it is, one might actually argue topologically. In other words, the same construction should work with cutting out more lines from C2 and making up with cutting isolated points out of the fibers where those lines intersect. For that the hyperbolicity argument would not work | |
| Feb 27, 2011 at 12:48 | comment | added | Georges Elencwajg | Great, Sándor, I had this feeling that the concept "hyperbolic" might be relevant, but I was too ignorant to know whether this made sense. | |
| Feb 27, 2011 at 7:17 | comment | added | Sándor Kovács | ...what I meant was to extend the projection on the total space and then restrict to the fiber and get that $\mathbb C\setminus\{0\}$ would have to map onto $\mathbb C\setminus\{0,1\}$ which cannot happen as the latter is hyperbolic. | |
| Feb 27, 2011 at 1:55 | comment | added | Sándor Kovács | 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. | |
| Feb 27, 2011 at 0:22 | comment | added | Georges Elencwajg | 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! | |
| Feb 26, 2011 at 23:13 | comment | added | Sándor Kovács | 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. | |
| Feb 26, 2011 at 21:47 | comment | added | Georges Elencwajg | 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? | |
| Feb 26, 2011 at 21:45 | comment | added | Georges Elencwajg | 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. | |
| Feb 26, 2011 at 19:32 | history | answered | Jason Starr | CC BY-SA 2.5 |