For a complex manifold $M$, one can consider (A) its de Rham cohomology, or (B) its Dolbeault cohomology. I'm looking for some motivation as to why one would bother introducing Dolbeault cohomology. To be more specific, here are some straight questions.


1. What can Dolbeault tell us that de Rham can't?
2. Does there exist some simple relationship between these two cohomologies?
3. When are they equal?
4. Do things become simpler for the Kahler case?
5. What happens for the projective spaces?
6. Why does nobody talk about the holomorphic cohomology?



de Rham vs Dolbeault Cohomology
Let $\Omega^{p,q}(M)$ be the $C^{\infty}$ $(p,q)$-forms. One always has a double complex with $\Omega^{p,q}(M)$ in position $(p,q)$. The cohomology in the $q$ direction is Dolbeault cohomology, the cohomology of the total complex is deRham cohomology. (In each case, essentially by definition.) Whenever you have a double complex, you get a spectral sequence. On the first page of the spectral sequence is Dolbeault cohomology; the spectral sequence converges to deRham cohomology. This is sometimes called the Hodge-de Rham spectral sequence, and sometimes called the Frolicher spectral sequence.


**If $M$ is compact Kahler (in particular projective)** then the spectral sequence collapses at the first page; all maps between Dolbeault groups are zero. So $H^k(M) \cong \bigoplus\_{p+q=k} H^{p,q}(M)$ in this case.


**If $M$ is Stein (in particular, affine)** then the only nonzero Dolbeault groups are the $H^{p,0}(M)$, corresponding to holmorphic $p$-forms. The next page takes the cohomology of the complex of holomorphic $p$-forms; I'll term this ``holomorphic deRham". After that, there are no further maps, so holomorphic deRham equals deRham.


In general, the spectral sequence [can be arbitrarily nondegenerate](http://arxiv.org/abs/0709.0481).


