Timeline for compute the Kähler moduli of an elliptic curve
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 14, 2010 at 21:36 | vote | accept | Dan | ||
| Sep 14, 2010 at 18:34 | answer | added | Eric Zaslow | timeline score: 5 | |
| Sep 14, 2010 at 14:47 | comment | added | Emerton | There is a missing wedge in the above: I should have written $-i dz \wedge d\overline{z}.$ | |
| Sep 14, 2010 at 14:46 | comment | added | Emerton | Kevin Buzzard, yes, the coherent sheaf of (1,0) forms is called the sheaf of Kahler differentials. Kevin Lin, you are right, the Kahler moduli here is just one-dimensional. Gunnar, Yes, and I think the correct sign is $-i \alpha \wedge \overline{\alpha}$. (Think of $\alpha$ as being $dz = dx + i dy$; then $-i dz d\overline{z}$ is a positive multiple of $dx\wedge dy$.) | |
| Sep 14, 2010 at 11:20 | comment | added | Gunnar Þór Magnússon | Can't we make use of Kevin's comment? Take his (1,0) form \alpha and set \omega = i * \alpha \wedge \overline{\alpha}. Then \omega is a real (1,1)-form, and Kahler if it is non-degenarate (and positive, but take -\omega if it's negative). | |
| Sep 14, 2010 at 9:51 | comment | added | Kevin Buzzard | @Emerton: sounds like I misunderstood the question. I'd delete my comment were it not for the fact that it would make your comment look meaningless :-) Yes, I described a global holomorphic 1-form. I thought these were called Kaehler 1-forms by some people and assumed this was what the questioner was asking about. | |
| Sep 14, 2010 at 8:25 | comment | added | Kevin H. Lin | @Dan: $H^{1,1}$ is 1 dimensional. You just have to write down the Kähler form. Then the Kähler moduli is just a 1 dimensional space given by scalings of the Kähler form. | |
| Sep 14, 2010 at 7:01 | comment | added | Emerton | Kevin, you have described an element of $H^0(\Omega^1)$. Isn't the Kahler form a 2-form, giving the hyperplane class in $H^2$ (so it should be a (1,1) form on the elliptic curve, not a (1,0) form). | |
| Sep 14, 2010 at 6:38 | comment | added | Kevin Buzzard | Put it into the more usual form of $y^2=f(x)$ with $f$ cubic (see Cassels book on elliptic curves for how to do this; Cassels gives lots of transformations for getting general genus 1 curves into this form, including the type you have here) and then it's just $dy/x$ (with the new coordinates). | |
| Sep 14, 2010 at 6:31 | history | edited | Harry Gindi | CC BY-SA 2.5 |
edited title
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| Sep 14, 2010 at 5:42 | history | asked | Dan | CC BY-SA 2.5 |