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S Jul 4, 2020 at 11:08 history bounty ended CommunityBot
S Jul 4, 2020 at 11:08 history notice removed CommunityBot
Jun 26, 2020 at 12:29 comment added Libli If you assume that the canonical bundle is ample or anti-ample, then I think your question has an affirmative answer. As you noticed, the determinant of the tangent bundle is a sub-bundle of the $n$-th ($n = \dim V$) power of the tangent bundle. Up to a dual, you get an ample line bundle. Now, up to some twists by this ample line bundle, every vector bundle is a quotient of a trivial vector bundle (global generation up to twists).
Jun 26, 2020 at 12:15 comment added abx It's not a good idea to change completely a question when there are already comments and answers.
S Jun 26, 2020 at 9:33 history bounty started CommunityBot
S Jun 26, 2020 at 9:33 history notice added user145520 Improve details
Jun 24, 2020 at 15:37 history edited user145520 CC BY-SA 4.0
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Jun 24, 2020 at 15:25 comment added Jason Starr If you also allow direct sums, the rephrased question has a positive answer for all compact (connected) Riemann surfaces of genus $g\neq 1$.
Jun 24, 2020 at 15:23 history became hot network question
Jun 24, 2020 at 12:01 history edited user145520 CC BY-SA 4.0
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Jun 24, 2020 at 8:21 history edited user145520 CC BY-SA 4.0
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Jun 24, 2020 at 8:08 comment added Piotr Achinger Vector bundles on $\mathbb{P}^n$ have continuous moduli for $n>1$. Every projective variety admits a finite morphism $X\to \mathbb{P}^{\dim X}$, so the same should hold for $X$. It therefore seems that examples of the kind you want do not exist among projective varieties (and perhaps also among proper algebraic varieties).
Jun 24, 2020 at 8:05 answer added Sasha timeline score: 9
Jun 24, 2020 at 7:59 history edited user145520 CC BY-SA 4.0
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Jun 24, 2020 at 7:23 history asked user145520 CC BY-SA 4.0