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Yesterday I saw the definition of Dolbeaut cohomology: let $(M,J)$ be a complex manifold, write the exterior derivative as ${\rm d} = \partial +\overline{\partial}$, where $\partial: \Omega^{\ell,m}(M) \to \Omega^{\ell+1,m}(M)$ and $\overline{\partial}: \Omega^{\ell,m}(M) \to \Omega^{\ell,m+1}(M)$, look at the complex $$\cdots \xrightarrow{\hspace{.4cm}\overline{\partial}\hspace{.4cm}} \Omega^{\ell,m-1}(M)\xrightarrow{\hspace{.4cm}\overline{\partial}\hspace{.4cm}} \Omega^{\ell,m}(M) \xrightarrow{\hspace{.4cm}\overline{\partial}\hspace{.4cm}} \Omega^{\ell,m+1}(M)\xrightarrow{\hspace{.4cm}\overline{\partial}\hspace{.4cm}} \cdots$$and put $$H^{\ell,m}_{\rm Dolbeaut}(M) \doteq \frac{\ker(\overline{\partial}: \Omega^{\ell,m}(M) \to \Omega^{\ell,m+1}(M))}{{\rm Im}(\overline{\partial}: \Omega^{\ell,m-1}(M) \to \Omega^{\ell,m}(M))}.$$Great.

Question: Why did we took the above complex instead of $$\cdots \xrightarrow{\hspace{.4cm}\partial\hspace{.4cm}} \Omega^{\ell-1,m}(M)\xrightarrow{\hspace{.4cm}\partial\hspace{.4cm}} \Omega^{\ell,m}(M) \xrightarrow{\hspace{.4cm}\partial\hspace{.4cm}} \Omega^{\ell+1,m}(M)\xrightarrow{\hspace{.4cm}\partial\hspace{.4cm}} \cdots?$$

I get that $f: M \to \Bbb C$ is $J$-holomorphic if and only if $\overline{\partial}f = 0$ so that $\overline{\partial}$ sort of measures how stuff in general would be away from being holomorphic. But I'd like a more solid explanation.

Ivo Terek
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    Note that $\bar\partial$ gives rise to well-defined cohomology for holomorphic vector bundles. If you want to switch to anti-holomorphic bundles on anti-complex manifolds, go ahead :) – Ted Shifrin May 17 '18 at 00:51
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    Yes, people do consider that complex and its cohomology (when it is well-defined). To my knowledge, the reason why Dolbeaut's complex is mentioned more often is exactly your last paragraph and the above comment. – Student May 17 '18 at 00:51
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    From a strictly pragmatical standpoint (from someone who knows nothing about Dolbeaut cohomology), when one computes things using (co)homology or for (co)homology, one frequently needs to get their hands on representatives of classes. Knowing that a representative will be a $J$-holomorphic mapping is an advantage, regardless of what it is measuring. – Aloizio Macedo May 17 '18 at 00:55

2 Answers2

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Both $\partial$ and $\bar{\partial}$ are well defined locally. However, when we are looking at a global section of a holomorphic vector bundle, then $\partial$ is in general not well defined! (When both are well-defined, indeed I have seen people who consider both complexes and cohomologies.)

The point is that to describe a section, we usually pass to local trivialization, and patch them together with transition functions. When the bundle is holomorphic, by definition there is an atlas such that the transition functions are holomorphic. So by product rule and that $\bar{\partial}f=0$ for any holomorphic $f$, $\bar{\partial}$ can be extended from local to global.

Usually people care about holomorphic bundles more (I wonder why). If you are dealing with anti-holomorphic bundles, then indeed $\partial$ is well-defined.

Student
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The differential $\partial$ depends on the trivialisation of your vector bundle but not $\overline \partial$, see lemma 2.33 page 58 in Voisin's book.

Nicolas Hemelsoet
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