Timeline for homologically trivial $1$-cycles and surfaces
Current License: CC BY-SA 3.0
8 events
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Mar 13, 2018 at 5:36 | comment | added | naf | For a projective example, simply take any cycle which is homologically equivalent to zero but not algebraically equivalent to zero. (On a surface, any 1-cycle which is homologically trivial is algebraically trivial.) | |
Mar 12, 2018 at 15:49 | comment | added | Jason Starr | I suspect that this fails already for Hironaka's example, as described in Appendix B of Hartshorne's "Algebraic geometry". | |
Mar 12, 2018 at 14:55 | history | edited | YCor |
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Mar 12, 2018 at 14:39 | comment | added | pi_1 | Indeed, my question is whether one can choose a class on $\widetilde S$ which is homologically trivial. | |
Mar 12, 2018 at 14:19 | comment | added | byu | If you take any surface containing the curves in the support of $\gamma$, and take a resolution of it, you get a class in $Pic(\widetilde S)$ that pushes forward to $\gamma$ like you say. Is your question whether we can take this to be homologically trivial? | |
Mar 12, 2018 at 12:38 | comment | added | pi_1 | Thank you. No, I do mean "curves". | |
Mar 12, 2018 at 12:26 | comment | added | Qfwfq | You mean $\gamma$ in $\mathrm{CH}^1(X)$ (codimension $1$)? | |
Mar 12, 2018 at 11:53 | history | asked | pi_1 | CC BY-SA 3.0 |