Timeline for Must an algebraic variety with trivial tangent bundle be an abelian variety?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Nov 16, 2018 at 10:32 | comment | added | Watson | Related: mathoverflow.net/q/73846 | |
| Sep 28, 2016 at 18:55 | comment | added | S. Carnahan♦ | @oxeimon canonical is the top exterior power. | |
| Sep 28, 2016 at 14:22 | comment | added | Will Chen | I'm clearly missing something, but doesn't trivial canonical (cotangent?) bundle imply trivial tangent bundle? Isn't the dual of a free sheaf itself free? | |
| Sep 28, 2016 at 14:07 | history | edited | user39380 | CC BY-SA 3.0 |
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| Aug 12, 2016 at 22:06 | comment | added | Georges Elencwajg | mqx: you should erase your parenthetical second sentence since it is false, as explained by @Will's crystal-clear argument ( which moreover you seem to acknowledge in your third sentence!) | |
| Apr 28, 2015 at 0:29 | vote | accept | CommunityBot | ||
| Apr 23, 2015 at 0:36 | history | edited | user39380 | CC BY-SA 3.0 |
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| Apr 22, 2015 at 3:40 | comment | added | Will Sawin | Does the Hopf manifold really have trivial tangent bundle? The sections $zd/dz, wd/dz, zd/dw, wd/dw$ on $\mathbb C^2$ are scale-invariant and so descend to the Hopf surface. But the structure sheaf of the Hopf surface has a 1-dimensional space of sections, so a trivial rank two vector bundle should have a $2$-dimensional space of sections. But there is no linear relation among these. | |
| Apr 22, 2015 at 1:02 | history | edited | user39380 | CC BY-SA 3.0 |
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| Apr 21, 2015 at 9:31 | comment | added | Allen Knutson | ...since any affine algebraic group gives an example. | |
| Apr 21, 2015 at 8:18 | answer | added | Francesco Polizzi | timeline score: 28 | |
| Apr 20, 2015 at 22:29 | comment | added | ACL | And no if you don't assume some properness condition. | |
| Apr 20, 2015 at 22:24 | comment | added | Vesselin Dimitrov | Yes, assuming the variety is a complex projective manifold. (Prove that the assumption implies that the Albanese map $X \to \mathbb{Alb}(X)$ is etale.) No in positive characteristic. For examples, see Mehta and Srinivas, "Varieties in positive characteristic with trivial tangent bundle," Compositio Math., 1987. | |
| Apr 20, 2015 at 22:23 | comment | added | Gunnar Þór Magnússon | There's probably a more low-tech way to do this, but such a manifold is a Kahler manifold with trivial canonical bundle and thus has a finite cover that splits into a product of tori, hyperkahler and Calabi-Yau manifolds. Since it has a trivial canonical bundle there's only a torus factor in the cover, which is an abelian variety because it has a finite quotient that is projective. So up to finite quotients, yes. | |
| Apr 20, 2015 at 22:08 | history | asked | user39380 | CC BY-SA 3.0 |