Why is there no $\partial_E$?

Question

On Hybrechts's book, there exists a natural linear operator $\overline{\partial}_E$:

But why is there no ${\partial}_E$? Why doesn't ${\partial}_E:=\partial\otimes id_E$ make sense?

Answer 1

If you try to define a $\partial_E$ like this, then you'll compute different derivatives for different choices of trivialisation.

The key point is that the transition matrix $\psi_{ij}$ has holomorphic entries, since the vector bundle is a holomorphic bundle. This means that $\bar\partial \psi_{ij} = 0$ for each $i$ and $j$.

Huybrecht's uses this fact to show that $$\sum_{i,j} \bar\partial (\alpha_i \psi_{ij})\otimes s'_j = \sum_{i,j} \bar\partial (\alpha_i ) \psi_{ij} \otimes s'_j, $$ and from this, he deduces that his definition of $\bar\partial_E$ is independent of the choice of trivialisation.

However, it's not the case that $\partial \psi_{ij} = 0$, so this argument doesn't work for $\partial$.