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I know that over $\mathbb{C}\mathbb{P}^1$, every vector bundle decomposes as a direct sum of line bundles. When else does this happen? My question is, what assumptions do I need to put on a complex manifold (especially a surface) and/or what assumptions do I need to put on the vector bundles so that I guarantee the vector bundle decomposes into a direct sum of line bundles? I think some things are know for vector bundles over curves, but what about surfaces?

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  • $\begingroup$ Hmm.... No. Even over curves (projective) vector bundles seldom split, projective line being the unique exception. Beyond that very little can be said in general. $\endgroup$ – Mohan Oct 14 '15 at 16:16
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    $\begingroup$ The only positive dimensional projective variety on which all vector bundles split into a direct sum of line bundles is $\mathbb P^1$. See here $\endgroup$ – Georges Elencwajg Oct 14 '15 at 16:44
  • $\begingroup$ So it sound like the conditions to put on your space $X$ is that $X = \mathbb{C}\mathbb{P}^1$ : ( $\endgroup$ – user46348 Oct 19 '15 at 14:16
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This is rare even in the topological setting; it's analogous to asking when the representations of a group decompose into a direct sum of $1$-dimensional representations. A necessary topological condition for a (complex, to fix ideas) vector bundle $V$ on a space $X$ to split as a direct sum $L_1 \oplus \dots \oplus L_n$ of line bundles is that its total Chern class $c(V)$ splits as a product

$$c(V) = c(L_1) \dots c(L_n) = (1 + c_1(L_1)) \dots (1 + c_1(L_n)).$$

This usually won't happen. One obstruction is that this condition implies that the Chern classes of $V$ must lie in the subalgebra of $H^{\bullet}(X, \mathbb{Z})$ generated by $H^2(X, \mathbb{Z})$.

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