Timeline for On sufficient conditions on an analytic map to be algebraic(=regular)
Current License: CC BY-SA 3.0
21 events
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| Aug 29, 2011 at 17:22 | history | bounty ended | Hugo Chapdelaine | ||
| Aug 22, 2011 at 17:02 | history | bounty started | Hugo Chapdelaine | ||
| Aug 17, 2011 at 0:00 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Aug 16, 2011 at 23:57 | comment | added | Hugo Chapdelaine | @Samuele, just to make sure I understand your argument, are you using implicitly the fact that the group of analytic automorphisms of $\mathbf{C}$ is equal to $z\mapsto az+b$ for $a\neq 0$ so that you know that over a small open set of the base you can normalize in a "coherent way" so that the tangent vector $0$ maps to $0$? | |
| Aug 16, 2011 at 22:00 | answer | added | anon | timeline score: 4 | |
| Aug 16, 2011 at 20:33 | comment | added | Samuele | @Hugo: map a fiber of $L$ into $C\times\mathbb{C}$ through the alg.var.isomorphism, then project down to $C$; it must be constant (genus is high), so any isomorphism between the two bundles sends fiber to fiber (possibly on a different point). Now you can produce a bundle isomorphism by composing with the appropriate automorphism of the basis and adding on each fiber the right vector to fix the origin. | |
| Aug 16, 2011 at 18:01 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Aug 16, 2011 at 17:48 | comment | added | Hugo Chapdelaine | so I just reedited my question. | |
| Aug 16, 2011 at 17:47 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Aug 16, 2011 at 17:36 | comment | added | Hugo Chapdelaine | I see, so missed this subtle point, thanks @ulrich for the explanation | |
| Aug 16, 2011 at 17:26 | comment | added | naf | @Hugo: There will always be an algebraic structure on $X$ which will make $f$ algebraic (as you say, by Grauert-Remmert and GAGA) but there could be algebraic structures on $X$ which are not the restriction of the algebraic structure from $\bar{X}$ (if $\bar{X}$ is fixed). | |
| Aug 16, 2011 at 17:10 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Aug 16, 2011 at 17:00 | comment | added | Sam Gunningham | More examples of non-isomormphic algebraic varieties which are analytically isomorphic are given here: mathoverflow.net/questions/68421/… | |
| Aug 16, 2011 at 16:56 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
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| Aug 16, 2011 at 16:40 | comment | added | Hugo Chapdelaine | So @Samuele, how do you prove your last statement: "But if $L$ and $C\times\mathbf{C}$ are algebraically isomorphic as varieties, then they are isomorphic as algebraic line bundles". | |
| Aug 16, 2011 at 16:37 | comment | added | Hugo Chapdelaine | Hi @Samuele, so your example seems to work. So there must be something wrong in my reasoning... | |
| Aug 16, 2011 at 16:18 | comment | added | Samuele | On the example of a complex manifold carrying two different algebraic structures, take an affine curve $C$ of high genus and an algebraically non trivial line bundle $L$ on it. As $C$ is a Stein space, every line bundle on it is analytically trivial. But if $L$ and $C\times\mathbb{C}$ are algebraically isomorphic as varieties, then they are isomorphic as algebraic line bundles and that's impossible. | |
| Aug 16, 2011 at 15:59 | comment | added | Hugo Chapdelaine | So what is wrong in the following sketch: By Grauert-Remmert one may extend f to a holomorphic function $\bar{f}:\bar{X}\rightarrow \bar{Y}$ where the bars denote compactifications. Now look at the analytic coherent (algebra)-sheaf $\bar{f}_*O_{\bar{X}}^{an}$. By GAGA it comes from an (unique) algebraic (algebra)-coherent sheaf on $\bar{Y}$. Therefore $X$ and $f$ are algebraic. Isn't? | |
| Aug 16, 2011 at 15:52 | comment | added | Hugo Chapdelaine | @Ulrich, Could you give me such an example? | |
| Aug 16, 2011 at 15:41 | comment | added | naf | 1) is not quite correct since there are non-isomorphic algebraic varieties which are isomorphic as complex manifolds. | |
| Aug 16, 2011 at 15:30 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |