# 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,

1. 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?

2. 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?

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

4. 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.

1. On differential forms, take complex conjugate to turn $\partial$ into $\bar\partial$, and holomorphic functions into conjugate holomorphic.
2. 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. $\bar\partial f=0$.
3. 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.
4. 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.
• This is all true. It might be worth stating the obvious: people care more about $\bar{\partial}$ because its kernel is the holomorphic functions. Since we care about holomorphic functions in complex analysis, the $\bar{\partial}$ complex is of more importance. Mar 29, 2018 at 16:12
• Thanks for your guidence. So I understood that the two operators are not fundamentally different and they are just conjugate to each other. This makes me comfortable. For the last question I found that it was unrelevant to operators as the wedge product is of type (n,n-1). Mar 29, 2018 at 16:20
• @AndrewMcHugh I think "passes through" means $\bar\partial(f\cdot g)=f\cdot\bar\partial g$, which holds when $\bar\partial f=0$, i.e., when $f$ is holomorphic. Apr 4, 2018 at 0:48

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.

1. $\partial$ 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.
2. 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.
3. 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.
• I'm not sure how "choice-independent" this example actually is. If $E$ is instead an "antiholomorphic" vector bundle (the transition maps are antiholomorphic rather than holomorphic), then we have the reverse situation. We can define an operator $\partial_E$ intrinsically, but not an operator $\bar\partial_E$. Aug 31, 2020 at 3:16