Let $V$ be a smooth projective complex variety such that the canonical bundle is not trivial. We can construct some vector bundles over $V$ by starting with the tangent bundle and applying tensor products and Homs and taking subbundles, quotient bundles and extensions (including direct sums). Do we get all isomorphism classes of holomorphic vector bundles this way?
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5$\begingroup$ 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). $\endgroup$– Piotr AchingerCommented Jun 24, 2020 at 8:08

$\begingroup$ If you also allow direct sums, the rephrased question has a positive answer for all compact (connected) Riemann surfaces of genus $g\neq 1$. $\endgroup$– Jason StarrCommented Jun 24, 2020 at 15:25

1$\begingroup$ It's not a good idea to change completely a question when there are already comments and answers. $\endgroup$– abxCommented Jun 26, 2020 at 12:15

$\begingroup$ If you assume that the canonical bundle is ample or antiample, then I think your question has an affirmative answer. As you noticed, the determinant of the tangent bundle is a subbundle 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). $\endgroup$– LibliCommented Jun 26, 2020 at 12:29
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I think it is a result of Verbitsky (probably proved in Coherent sheaves on general K3 surfaces and tori) that if $M$ is a very general (nonalgebraic) K3 surface then the condition you want is satisfied.