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A complex manifold such that any Building all holomorphic vector bundle can be builtbundles from the tangent bundle

Let $M$$V$ be a closedsmooth projective complex manifoldvariety such that the canonical bundle is not trivial. We can construct some vector bundles over $M$$V$ by starting with the tangent bundle and applying tensor products and Homs and taking subbundles, quotient bundles and extensions (so for instance we can construct the trivial line bundle and the symmetric powers of the tangent bundleincluding direct sums). Are there manifolds for whichDo we get all isomorphism classes of holomorphic vector bundles this way?

A complex manifold such that any holomorphic vector bundle can be built from the tangent bundle

Let $M$ be a closed complex manifold. We can construct some vector bundles over $M$ by starting with the tangent bundle and applying tensor products and Homs and taking subbundles, quotient bundles and extensions (so for instance we can construct the trivial line bundle and the symmetric powers of the tangent bundle). Are there manifolds for which we get all isomorphism classes of holomorphic vector bundles this way?

Building all holomorphic vector bundles from the tangent bundle

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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user145520

Let $M$ be a closed complex manifold. We can construct some vector bundles over $M$ by starting with the tangent bundle and applying Schur functors, tensor products, direct sums and Homs and taking subbundles, quotient bundles and extensions (so for instance we can construct the trivial line bundle and the symmetric powers of the tangent bundle). Are there manifolds for which we get all isomorphism classes of holomorphic vector bundles this way?

$M$ can not be a curve for example because the Picard group has to be generated by the canonical class.

Let $M$ be a closed complex manifold. We can construct some vector bundles over $M$ by starting with the tangent bundle and applying Schur functors, tensor products, direct sums and Homs. Are there manifolds for which we get all isomorphism classes of holomorphic vector bundles this way?

$M$ can not be a curve for example because the Picard group has to be generated by the canonical class.

Let $M$ be a closed complex manifold. We can construct some vector bundles over $M$ by starting with the tangent bundle and applying tensor products and Homs and taking subbundles, quotient bundles and extensions (so for instance we can construct the trivial line bundle and the symmetric powers of the tangent bundle). Are there manifolds for which we get all isomorphism classes of holomorphic vector bundles this way?

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Let $M$ be a closed complex manifold. We can construct some vector bundles over $M$ by starting with the tangent bundle and applying duals, Schur functors, tensor products and, direct sums and Homs. Are there manifolds for which we get all isomorphism classes of holomorphic vector bundles this way?

$M$ can not be a curve for example because the Picard group has to be generated by the canonical class.

Let $M$ be a closed complex manifold. We can construct some vector bundles over $M$ by starting with the tangent bundle and applying duals, Schur functors, tensor products and direct sums. Are there manifolds for which we get all isomorphism classes of holomorphic vector bundles this way?

$M$ can not be a curve for example because the Picard group has to be generated by the canonical class.

Let $M$ be a closed complex manifold. We can construct some vector bundles over $M$ by starting with the tangent bundle and applying Schur functors, tensor products, direct sums and Homs. Are there manifolds for which we get all isomorphism classes of holomorphic vector bundles this way?

$M$ can not be a curve for example because the Picard group has to be generated by the canonical class.

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