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I have read that a short exact sequence of differentiable vector bundles is always split. I was interpreting this as the splitting being fiber wise, i.e that the fibers of the middle component of the short exact sequence are isomorphic as a $\mathbb{C}$-module to the direct sum of the fibers of the outside components.

However, I read that short exact sequences of holomorphic vector bundles are not necessarily split. This makes me believe that my fiber wise interpretation is wrong since the fibers are always vector spaces over $\mathbb{C}$ and thus being split would not depend on the vector bundle being holomorphic or not.

What is wrong with my original interpretation of what it means for a sequence of vector bundles to be split?

user7090
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If you have an exact sequence $0\rightarrow E\rightarrow F\rightarrow G\rightarrow 0$ of vector bundles over the manifold $M$. Take a differentiable metric $b$ on $F$, and the orthogonal of $i(E)$ is isomorphic to $G$. You don't have necessarily holomorphic metric on complex vector bundles.

There is an old paper of Atiyah (I believe) which constructs a class which is the obstruction to the existence of such a splitting for complex bundles.

Tsemo Aristide
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  • Interesting. I thought it might be possible to approach this problem without metrics. – user7090 Jul 08 '17 at 19:11
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    With sufficient energy, you can remove all mentions to the metric and use the fact that the manifold is paracompact and that there are partitions of unity. That is is an essentially equivalent route, only longer and more complicated. You can get complements to $E$ in $F$ over balls of $M$ and use a partition of unity to glue them into a global complement, with sufficient ingenuity. – Mariano Suárez-Álvarez Jul 08 '17 at 19:15
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    You certainly can define hermitian metrics on any complex vector bundle. The issue is that you do not stay in the holomorphic category when you use a hermitian metric. – Ted Shifrin Jul 08 '17 at 19:18
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    @Tsemo, what you do not necessarily have is holomorphic hermitian metrics on vector bundles on complex manifolds. Over a paracompact manifold (space, even), every complex vector bundle has a hermitian metric. – Mariano Suárez-Álvarez Jul 08 '17 at 19:19
  • When we write $\mathcal{F} \cong \mathcal{E} \oplus \mathcal{G}$ I am assuming now that $\cong$ is referring to a biholomorphism of the vector bundles viewed as complex manifolds and maybe this is a stronger condition than all the fibers being isomorphic as $\mathbb{C}$-modules + commuting condition (which from what I understand is what it means to have an isomorphism in the category of vector bundles). I'm having a hard time understanding in what category these isomorphisms are taking place. – user7090 Jul 08 '17 at 19:49
  • @Janziek The relevant category is that of smooth/holomorphic vector bundles over a manifold $M$. If you only look at each fiber separately, you lose the whole point in vector bundles! Indeed, the fibers of two vector bundles of the same rank are always isomorphic to one another. – Amitai Yuval Jul 08 '17 at 19:56
  • Oh wow after reading this I completely see my error in thought. I did not read the definition of a morphism carefully enough to recognize that we first must have a morphism on the vector bundles (as complex manifolds) and this induces the map on the fibers. Sort of like in sheaf theory that we must have a map of the sheaves first before we look at the map on the residue fields! – user7090 Jul 08 '17 at 20:01
  • @Janziek Yes, it is the same. In fact, vector bundles are, in a way, a particular kind of sheaves. – Amitai Yuval Jul 09 '17 at 14:49