When does a SES of vector bundles split?

Question

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?

Answer 1

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.

Answer 2

I can't say I'm aware of a proof that doesn't use a metric, but (in relation to Question 2), this is probably because the proof by means of a Riemannian or Hermitian metric is actually very general. applying to all paracompact locally compact Hausdorff spaces (e.g., compact Hausdorff spaces, CW complexes).

If $X$ is just a locally compact Hausdorff space, then you can construct a continuous partition of unity subordinate to any finite cover by means of Urysohn's Lemma. In particular, if $X$ is paracompact (every open cover admits a locally finite refinement), e.g., if $X$ is compact or if $X$ is a CW complex, then you can construct a continuous partition of unity subordinate to any open cover at all. So, over a paracompact locally compact Hausdorff space, you can can always construct Riemannian or Hermitian metrics as needed and hence construct a splitting.

Now, even if you're interested in locally compact Hausdorff spaces, without requiring paracompactness, it's worth noting that in many contexts, e.g., topological $K$-theory, one only considers vector bundles trivial at infinity (i.e., $E \to X$ such that for some compact $K \subset X$, $E|_{X \setminus K}$ is trivial), in which case you can always pass to a finite trivialising subcover, which therefore admits a continuous partition of unity. Hence, a short exact sequence of vector bundles trivial at infinity, over a locally compact Hausdorff space, should split.

Answer 3

I believe you can just use the Serre-Swan theorem, which gives an equivalence of categories between finite rank vector bundles over a smooth manifold and finitely generated projective modules over the ring of smooth functions. Since any short exact sequence of projective modules splits, you have the desired conclusion.

As to the generality of this method, there is a paper by Morye which discusses various ringed spaces where Serre-Swan type theorems hold.