Why only $\bar\partial$ but not $\partial$ in Dolbeault cohomology While I learn about $\partial$ and $\bar{\partial}$ operators, I had some questions about the reason why people prefer $\bar\partial$ over $\partial$. Specifically, 


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*When defining Dolbeault cohomology, one uses $\bar{\partial}$ but not $\partial$. I wonder whether there happens any problem if one define a cohomology by $\partial$. Or is this because it isn't interesting?

*Any textbook on complex variables say only about $\bar{\partial}$-Poincare lemma. Is there a version with $\partial$? If not, where the difference between two operators fundamentally comes?

*For a holomorphic vector bundle $E$, we define the operator $\bar{\partial}_E$ only. Again why don't we define $\partial_E$?

*Let $(E,h)$ be an hermitian holomorphic vector bundle on a compact hermitian manifold $(X,g)$. When we show that the operator $\bar\partial_E^*:=-\bar*_{E^*}\circ\bar\partial_{E^*}\circ\bar*_E$ on $A^{p,q}(X,E)$ is adjoint to $\bar\partial_E$, one uses $$\int_X\bar\partial(\alpha\wedge\bar*_E\beta)=\int_X d(\alpha\wedge\bar*_E\beta)$$ for $\alpha\in A^{p,q}(X,E)$ and $\beta\in A^{p,q+1}(X,E)$(c.f. "complex geometry" by Huybrechts, p.170). Here how we know $$\int_X\partial(\alpha\wedge\bar*_E\beta)=0$$? Again, the two operators appear to have different rules.
 A: 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:


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*$\partial$ 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"

A: *

*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.

