Let $E,F,G$ be holomorphic vector bundle, if we have the short exact sequence of holomorphic vector bundle:

$$0\to E\to F\to G \to 0$$

We have the splitting as smooth vector bundle however not the holomorphic vector bundle.

I don't know where the following arguement goes wrong, consider the local holomorphic frame on $E$ as $e_1,...,e_r$ and on $G$ as $f_1,...,f_k$.

As the short exact sequence is fiberwise split, we have the pointwise defined basis on each fiber of $F$ as $\{e_1,...,e_r,f_1,...,f_k\}$.

Therefore the transition matrix for $F$ should be diagonal in this setting ?? However I see in some place the transition function for $F$ is upper triangular matrix not the diagonal matrix I don't understand where goes wrong with the arguement above?(is it because the pointwise defined $\{e_1,...,e_r,f_1,...,f_k\}$ basis needs not to be holomorphic frame)

I know to prove the splitting for smooth setting is to use the fact that any complex vector bundle admits the hermitian metric,therefore the metric on $F$ will induce metric on the subbundle $E$(by restriction), and then we have the normal subbundle under this Hermitian metric on $F$ (this normal complement needs not to holomorphic bundle in general, since Hermitian metric as smooth bundle morphism, not holomorphic bundle morphism )

However I don't know where goes wrong in the arguement in the box, and what's the transition function for $F$ given $E$ and $G$?

  • 2
    $\begingroup$ Your idea for an argument proves that the exact sequence is locally split. $\endgroup$ Commented Sep 1, 2022 at 7:01
  • $\begingroup$ I don't quite understand , Can you elaborate more? There is equivalence between locally free sheaf and category of vector bundle, do you mean I just prove the spliting at the stalk level? @Mariano Suárez-Álvarez $\endgroup$
    – yi li
    Commented Sep 1, 2022 at 7:04
  • $\begingroup$ Is $\{e_1,...,e_r,f_1,...,f_k\} $ local holomorphic frame ? $\endgroup$
    – yi li
    Commented Sep 1, 2022 at 7:08

1 Answer 1


First note that since $E$ is subbundle of $F$ , we have the local trivialization $$\Phi_F:\pi_E^{-1}(U)\to U\times\Bbb{C}^{r+s}\\\Phi_E:\pi^{-1}_F(U)\to U\times \Bbb{C}^r(\times \{0\})\subset U\times \Bbb{C}^{r+s}$$

Observe that it different coordinate we always has first $r$ component span the fiber of $E$, therefore the transition for $F$ under this trivialization is upper triangular and the first upper left most block correspond to the transition for $E$.Since the sequence is exact, (by definition it means fiberwise exact just like the stalk wise exact of sequence of sheaves.)

Therefore the fiber of $G$ isomorphic to vector space $F_x/E_x$. Moreover the holomorphic bundle morphism $\psi:F\to G$ will maps holomorphic local frame on $F$ denote it $(s_1,..,s_{r+s})$to holomorphic local span set on $G$ (denote it $\psi\circ s_1,...,\psi\circ s_{r+s}$). Since $F\to G$ will collaps the first $s_1,..,s_r$ to zero, we see $\psi\circ s_{r+1},...,\psi\circ s_{r+s}$ is holomorphic frame for the bundle $G$. Assume the transition on $F$ is

$$\tilde{s}_k = \rho_{k}^js_j$$ with $k\ge r+1$ then apply $\psi$ on both side we see $$\psi\circ s_k = \rho^j_k \psi\circ s_j$$ where we use the fact $\psi$ as holomorphic bundle morphism is complex linear on each fiber. however $\psi\circ s_j = 0$ for $j\le r$ therefore the transition function for $G$ is just the lower right most block.

As a direct COR we see the determinant bundle has transition function the determinant of the transition matrix for the vector bundle therefore

$$\det F \cong \det E \otimes \det G$$

As we have seen, the transition matrix needs not to be diagonal, the problem is you may try to pull back the holomorphic frame on $G$ to $F$, but it's not guarantee that it's holomorphic on $F$. While if we push forward the holomorphic frame on $F$ to $G$ it's still holomorphic on $G$.(As we did above)

There is a question why for the smooth setting we can "pull back" from $G$ to $F$ hence getting the diagonal cocyle?

For the completeness let me make the proof that smooth short exact sequence split more clear.

Since we have the Hermitian metric which can be treated as smooth bundle morphism, therefore $E^\perp$ is just the kernel of the smooth bundle morphism, it's still smooth vector bundle.Therefore we have two smooth bundle morphism :

$$E\oplus E^\perp \to G\\E\oplus E^\perp \to E^\perp$$ since both are surjective bundle morphism(the first one by the sequence is exact, the second one is the one associated to direct sum), To prove $G\cong E^\perp$ (hence the sequence split) only needs to prove they have the same kernel.Both of them are $E$ , therefore by the universal property they are isomorphic $G \cong E^\perp$

  • $\begingroup$ Although using the metric seems auxiliary in the proof of smooth setting, but it's not clear from the direct argument above can get the diagonal spliting. $\endgroup$
    – yi li
    Commented Sep 1, 2022 at 14:06
  • $\begingroup$ You can always pullback an holomorphic frame to a holomorphic frame locally. That is not the problem. The problem is that splitting locally does not imply splitting globally. $\endgroup$ Commented Sep 1, 2022 at 16:57
  • $\begingroup$ I got it , thank you , To make it more clear do you mean under the local frame $s_{r+1},...s_{r+s}$ spans a locally a subbundle denote it $G'$ which is a holomorphic bijective bundle morphism $G' \to G|_U$ ,hence it's biholomorphic bundle morphism, therefore we can locally pull back it correct? @Mariano Suárez-Álvarez $\endgroup$
    – yi li
    Commented Sep 2, 2022 at 1:06
  • $\begingroup$ To be honest, I did not understamd what you wrote. $\endgroup$ Commented Sep 2, 2022 at 4:36
  • $\begingroup$ I suggest that ypou try tyo construct an actual section $G\to F$ of the morphism $F\to G$ on the whole base and not only locally. $\endgroup$ Commented Sep 2, 2022 at 4:42

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