[Math] Why only $\bar\partial$ but not $\partial$ in Dolbeault cohomology

complex-geometrycv.complex-variablesseveral-complex-variables

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.

Best Answer

  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.
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