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Let $M$ be a complex manifold, $A^{p,q}(M)$ be $C^{\infty}$ $(p,q)$ form.

Dolbeault cohomology $H^{p,q}_{\bar{\partial}}(M)$ is defined as the cohomology with boundary map $\bar{\partial}$, but why not define $H^{p,q}_{\partial}(M)$, with boundary map $\partial$?

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up vote 6 down vote accepted

It might be that part of the reason is you don't really get any new information by doing so. There is a natural conjugate linear isomorphism $H^{p,q}_{\overline{\partial}}\to H^{q,p}_\partial$ given by conjugation: $[\omega]\mapsto [\overline{\omega}]$. Thus once you know $H^{p,q}_{\overline{\partial}}$ you know $H^{q,p}_{\partial}$ as well.

Also, $\overline{\partial}$ is in some ways more useful because its kernel consists of holomorphic objects, whereas the kernel of $\partial$ consists of anti-holomorphic objects.

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In addition to the two points above, one should mention that, if $M$ is compact Kahler, then one has a natural isomorphism $H^{p,q}_{\partial}(M) \cong H^{p,q}_{\overline{\partial}}(M)$. –  David Speyer Jan 30 '12 at 15:46

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