Timeline for Is the moduli space of stable vector bundles over a smooth projective curve fano?
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8 events
| when toggle format | what | by | license | comment | |
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| Sep 21, 2013 at 23:54 | comment | added | Sándor Kovács | @Chen:(This assumes that we're in characteristic zero. Otherwise the statement is not true.) If $X$ is unirational, then $K_X$ cannot be trivial or even torsion. If the Picard number is one, then any such divisor is either ample or negative ample. Again if $X$ is unirational, then $K_X$ cannot be ample. So $-K_X$ is ample, and hence $X$ is Fano. (If the Picard group is $\mathbb Z$, then you don't even need to worry about torsion.) | |
| Sep 21, 2013 at 14:48 | vote | accept | Chen | ||
| Sep 21, 2013 at 14:48 | |||||
| Sep 21, 2013 at 11:17 | comment | added | Chen | @ulrich: Could you please give me a reference. | |
| Sep 21, 2013 at 9:41 | comment | added | naf | @Chen: Unirationality and Picard number 1 does. | |
| Sep 21, 2013 at 9:15 | comment | added | Chen | @ulrich: Does unirational imply Fano? | |
| Sep 21, 2013 at 7:12 | comment | added | naf | @Jim Bryan: It is easy to see that the moduli space (with fixed determinant bundle) is unirational. | |
| Sep 21, 2013 at 3:53 | comment | added | Jim Bryan | I'm confused. Why do you say that Picard number 1 implies it is Fano? A K3 surface can have $\mathbb{Z}$ for a Picard group. | |
| Sep 21, 2013 at 2:43 | history | answered | user37622 | CC BY-SA 3.0 |