Why only $\bar\partial$ but not $\partial$ in Dolbeault cohomology
- On differential forms, take complex conjugate to turn $\partial$ into $\bar\partial$, and holomorphic functions into conjugate holomorphic.
- All of the proofs about differential forms then go through the complex conjugation effortlessly, including the Poincare lemma. We use $\bar\partial$ because we like holomorphic functions, i.e. $\bar\partial f=0$.
- As for vector bundles, the operator $\partial_E$ is defined only for conjugate holomorphic vector bundles, i.e. with conjugate holomorphic transition maps, because the $\partial$ operator passes through conjugate holomorphic functions: $\partial (fg)=f \partial g$ for all $g$ precisely for $f$ conjugate holomorphic, so in a conjugate holomorphic local trivialization.
- In local holomorphic coordinates $z^1,\dots,z^n$, count numbers of $dz^1, \dots, dz^n$ in the wedge products; you already have $n$ of them, so you get zero if you wedge in another.
If $X$ is a complex manifold and $E\to X$ is a holomorphic vector bundle, only $\bar\partial_E$ can be defined naturally, i.e., it depends only on the complex structures of $X$ and $E$. The $\partial_E$ operator cannot be defined intrinsically.
If $E$ has a flat connection, you can also define the $\partial$-operator. This is the case when $E$ is trivial, and then the $\partial$ operator you mention is the one we all know. The flat connection is $d$.
In general, the operators $\bar\partial$ and $\partial$ have different roles, even if $\partial$ is well-defined.
To answer your questions:
$\partial$ is not defined, therefore it cannot be used to define some cohomology. If you have a holomorphic vector bundle over $X$, then the $\bar\partial$ complex give you some cohomology which is isomorphic to the Cech cohomology, and this a very deep result.
Sure, the $\bar\partial$-Poincare lemma is a local statement, and it can be conjugated to get a statement about the $\partial$ operator. No problem here.
Because $\partial_E$ cannot be defined naturally. You need some other conditions (like flatness or a Hermitian metric) in order to define the $\partial$ operator.
"Again why don't we define $\partial_E$?" Because it doesn't exist.
- You already have some metric since you can define the star operator. The short answer is "for bidegree reasons"