Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\partial$?

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.