Timeline for Cup Product with Ample Line Bundles
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 22, 2019 at 6:05 | comment | added | Yosemite Stan | @abx Oops! I guess I should say that the real (1,1) Hodge classes are dual to the real (n-1,n-1) Hodge classes and that the ample cone is open in the space of real Hodge classes. I think that corrects the argument... | |
| Nov 22, 2019 at 1:39 | comment | added | Winnie_XP | @abx But isn't the tensor with enough power of an ample line bundle makes arbitrary line bundle ample ? I know this is a result for Noetherian schemes, but I suppose that it also holds for cpt projective algebraic complex manifolds ? | |
| Nov 21, 2019 at 9:45 | comment | added | abx | @Yosemite Stan: no, in general the ample cone does not span $H^{1,1}$, and is not open there. Think of a general surface of degree $\geq 4$ in $\Bbb{P}^3$: the ample cone is a half-line, while $H^{1,1}$ is quite large. | |
| Nov 21, 2019 at 7:49 | vote | accept | Winnie_XP | ||
| Nov 21, 2019 at 7:43 | comment | added | Winnie_XP | @YosemiteStan Thanks ! I just realized i was asking a stupid question. Thanks for the reply ! | |
| Nov 21, 2019 at 7:35 | comment | added | Yosemite Stan | @Winnie_XP Yes it should be true in this case. $H^{n-1,n-1}(X)$ and $H^{1,1}(X)$ are dual under cup product. The ample cone in $H^{1,1}(X)$ is open (and spans $H^{1,1}(X)$). So if a class $\alpha\in H^{n-1,n-1}(X)$ vanishes after cupping with arbitrary ample, then it vanishes when cupping with an arbitrary (1,1)-class. By duality it is 0. | |
| Nov 21, 2019 at 5:53 | history | answered | abx | CC BY-SA 4.0 |