Timeline for Algebraic vs. homological equivalence for curves on a smooth complex projective surface
Current License: CC BY-SA 3.0
13 events
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| Jan 17, 2017 at 7:32 | vote | accept | dorebell | ||
| Jan 16, 2017 at 3:42 | comment | added | R. van Dobben de Bruyn | The argument presented in Lazarsfeld uses CM regularity in a different way (and of course, it is needed for the construction of the Hilbert and Quot schemes, but I guess you could take their existence for granted). I guess strictly speaking the key point is not CM regularity itself, but rather Mumford's observation that the regularity of a sheaf only depends on suitable numerical data plus some mild boundedness condition. | |
| Jan 16, 2017 at 2:35 | comment | added | dorebell | Thanks for the followup! Is Castelnuovo-Mumford regularity the key point in both references, or just Mumford's? (e.g. is Fujita vanishing sufficient for Lazarsfeld's argument?) | |
| Jan 15, 2017 at 23:49 | comment | added | R. van Dobben de Bruyn | And it's true that $H^1(X,\mathcal O_X)/H^1(X,\mathbb Z)$ is a very analytic object. To see a concrete example of how to make it into an algebraic object, you can look at Weierstrass $\wp$-functions on $1$-dimensional complex tori. See for example chapter VI of Silverman or section 1.4 of Diamond–Shurman for an algebraic treatise (I guess any book on Riemann surfaces should give you a more analytic treatise of the same material). | |
| Jan 15, 2017 at 23:26 | comment | added | R. van Dobben de Bruyn | I realised there was a gap in the algebraic proof I presented. I found references where they fix it, but instead of copy-pasting the argument, I decided to leave it as a reference. The argument is not very difficult, but it does involve the notion of Castelnuovo–Mumford regularity (which it develops on the way). This is a very important notion that you should learn about at some point anyway. | |
| Jan 15, 2017 at 23:22 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
Gave reference for missing argument.
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| Oct 13, 2016 at 9:55 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
Fixed some typos.
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| Oct 13, 2016 at 9:50 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
Added explanation about algebraicity of the Picard variety.
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| Oct 13, 2016 at 8:14 | comment | added | dorebell | Wow - great answer! I'm not so familiar with Picard and Hilbert schemes (although I want to be!), so I'm not so sure how to think about the argument that two divisors in the same component of the Picard scheme are algebraically equivalent. It seems like $H^1(X, \mathcal{O}_X)/H^1(X, \mathbb{Z})$ is a very analytic object (integer cohomology), so it's weird to me to think of it as a variety. Of course, it's a complex torus, so we can connect any two points with some complex line, but this may not be algebraic (and also is only genus 0!). Is there an elementary way to see what's going on? | |
| Oct 13, 2016 at 3:11 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
Remark on higher codimension.
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| Oct 12, 2016 at 10:29 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
Remark on Riemann–Roch.
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| Oct 12, 2016 at 10:17 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
Two more small corrections.
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| Oct 12, 2016 at 9:42 | history | answered | R. van Dobben de Bruyn | CC BY-SA 3.0 |