Given a short exact sequence of smooth vector bundles,
$$0\to A \to B \to C \to 0$$
on a manifold $M$, it is an easy exercise that sequence splits. One approach is to pick a Riemannian metric on $B$ and show that $C$ is isomorphic to the orthogonal complement of $A$. This proof extends to complex line bundles by choosing a Hermitian metric.
If we leave the category of smooth bundles, a lack of bump functions means we can no longer assume a metric exists, although if we consider bundles equipped with a metric, the same proof should work.
Question 1: Is there a proof that does not make use of a metric?
Question 2: In what generality does it hold that short exact sequences of vector bundles split? I know that vector bundles correspond to projective modules, which says that we have splittings over affine schemes, but what about more generally?
Best Answer
Not totally sure what you're asking (particularly, are you asking about splitting in algebraic/holomorphic settings?), but the following comments may be relevant:
Because of the principle that "curvature decreases in holomorphic sub-bundles and increases in quotient bundles", it's "rare" for a short exact sequence of holomorphic vector bundles to split holomorphically. For example, the sequence $$ 0 \to \mathcal{O}(-1) \to \mathbf{C}^{n+1} \to T\mathbf{P}^{n}(-1) \to 0 $$ over the complex projective space $\mathbf{P}^{n}$ does not split holomorphically. In fact, the trivial bundle in the middle is split only by trivial sub-bundles.
Metrics don't help in the holomorphic category, because the components of a non-constant Hermitian metric are never holomorphic functions in local holomorphic coordinates.