# An extension of line bundles splits locally

Consider an extension $0\rightarrow L \overset{\alpha}{\rightarrow} E \overset{\beta}{\rightarrow} L' \rightarrow 0$ of bundles and bundle homomorphisms, where $L$ and $L'$ are line bundles. (Let's assume that all bundles and maps are holomorphic.)

Does it split locally? i.e. is there an open cover $\{U_i\}$ and maps $\sigma_i: L'\vert_{U_i} \rightarrow E\vert_{U_i}$ such that $\beta\circ \sigma_i = \text{id}$?

You could use a common trivialization for $L$, $L'$, and $E$, and then you get an extension of trivial bundles, say $0\rightarrow U\times \mathbb{C} \rightarrow U\times \mathbb{C}^2 \rightarrow U\times \mathbb{C} \rightarrow 0$. So a related question is: do extensions of trivial bundles split locally (or even globally)?

In the smooth case you can use a partition of unity to construct a Hermitian metric, and then you get splitting globally. I was hoping to get this local splitting for vector bundles in any setting, i.e. smooth, holomorphic, algebraic, etc.

I'll just shortly elaborate about the algebraic proof. In fact line bundles are in one-to-one correspondence with locally free sheaves of rank 1. They are locally given by projective modules (locally free = projective), and then the proof is just a simple algebraic fact that a surjection onto a projective modules splits, which is just the definition of projective.

• We'll have a sequence of sheaves $0\rightarrow \mathcal{F} \rightarrow \mathcal{E} \rightarrow \mathcal{F}'\rightarrow 0$, and there's an open cover $\{U_i\}$ such that $\mathcal{F}\vert_{U_i} \simeq \mathcal{O}_X\vert_{U_i}$ and $\mathcal{F}'\vert_{U_i}\simeq \mathcal{O}_X\vert_{U_i}$ as sheaves. The $\mathcal{F}'(V)$'s, for $V\subseteq U_i$, is free, so the stalk $\mathcal{F}'_p$ is projective for all $p\in U_i$, hence the extension splits at each stalk. That means it splits over all of $U_i$? Jun 27, 2014 at 13:29
• Don't argue with stalks, argue with restrictions. (Recall the definition of "locally free"!) Jun 27, 2014 at 21:04
• For affine schemes you may argue using projective modules. More generally you can argue like this locally: A locally trivial sheaf $E$ has an open cover $U_i$ where $E_{U_i}$ is trivial, hence projective. Hence local splittings exist. Jul 8, 2021 at 18:39

FWIW, an extension $0 \to F \to E \to Q \to 0$ of holomorphic vector bundles splits locally (holomorphically, but not naturally) by your "common trivialization" argument: Pick a local holomorphic frame $\{e_{1},\dots,e_{n}\}$ for $E$ whose first $m$ elements span the fibre of $F$ at some point $p$. Extend to a local frame $\{f_{1},\dots,f_{m}\}$ for $F$, and note that there exist $mn$ (local) holomorphic functions $a_{jk}$ such that $f_{j} = \sum_{k} a_{jk} e_{k}$.

The elements $\{e_{m+1},\dots,e_{n}\}$ induce a frame of $Q$ provided the sum $$\operatorname{span}(f_{1},\dots,f_{m}) + \operatorname{span}(e_{m+1},\dots,e_{n})$$ is direct. This is true in some neighborhood of $p$ by continuity of holomorphic functions.

Given a complex manifold $$X$$ and two locally free sheaves (=vector bundles) $$\mathcal E, \mathcal G$$ on $$X$$, the extensions $$0\to \mathcal E\to \mathcal F \to \mathcal G \to 0$$ are classified by the complex vector space $$H^1(X,\mathcal {Hom}(\mathcal G ,\mathcal E ))$$.
The trivial extension $$0\to \mathcal E\to \mathcal E\oplus \mathcal G\to \mathcal G \to 0$$ corresponds to the zero element of that vector space.
In particular over a Stein manifold all extensions are trivial (Theorem B of Cartan-Serre) and even more in particular all extensions are trivial over a manifold isomorphic to a ball, for example the domain of a chart.
Conclusion
Yes, over holomorphic manifolds all exact sequences of vector bundles split locally.
And the same proof, mutatis mutandis, shows that this also holds for differential manifolds, complex analytic spaces, algebraic varieties and schemes.
Bibliography
The classification of extensions mentioned above is Proposition 2 of Atiyah's seminal paper Complex analytic connexions in fibre bundles.

Question: "In the smooth case you can use a partition of unity to construct a Hermitian metric, and then you get splitting globally. I was hoping to get this local splitting for vector bundles in any setting, i.e. smooth, holomorphic, algebraic, etc."

Answer: If $$(X, \mathcal{O}_X)$$ is any ringed topological space and

$$\phi: E \rightarrow F \rightarrow 0$$

is a surjection of locally free sheaves on $$X$$ where $$rk(E)=n, rk(F)=m$$ and $$m\leq n$$ let $$U_i$$ be an open cover where $$E,F$$ trivialize and let

$$\phi_{U_i}:E_{U_i}\cong \mathcal{O}_{U_i}\{e_1,..,e_n\} \rightarrow F_{U_i} \cong \mathcal{O}_{U_i}\{f_1,..,f_m\}$$

be the induced map. Choose $$\omega_j\in E(U_i)$$ with $$\phi_{U_i}(\omega_j)=f_j$$ and define for $$V \subseteq U_i$$

$$\psi_{U_i}(V): F(V) \rightarrow F(V)$$

by

$$\psi_{U_i}(V)(\sum_j a_j f_j)=\sum_j (\omega_j)_{V}e_j.$$

It follows $$\psi_{U_i}: F_{U_i}\rightarrow E_{U_i}$$ is a local splitting of $$\phi_{U_i}$$. There is an equality

$$\phi_{U_i} \circ \psi_{u_i}=Id_{F_{U_i}}.$$

Hence you can construct such local splitting for any ringed topological space $$(X, \mathcal{O}_X)$$.