TITLE: de Rham vs Dolbeault Cohomology
QUESTION [27 upvotes]: 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.

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

REPLY [23 votes]: Although the main questions have been answered quite well, I would like to say a few words
about the first question "why would one bother...". 
De Rham and Dolbeault cohomology are measuring different things. The first, which measures the failure of the necessary condition $d\alpha=0$ to guarantee a potential $\alpha=d\beta$, is a topological invariant.
The second gives the obstruction to solving the similar problem for the Cauchy-Riemann operator; it measures the holomorphic complexity  and it has a priori nothing to do with the topology. That they turn out to be the "same" in good (e.g. compact Kähler) cases
is sort of a miracle. Even in such cases, these spaces are not literally the same,
and this can be exploited in interesting ways.
There is an isomorphism which can be represented by a matrix called a period matrix,
which is sensitive to the complex structure. To see how this works in the simplest interesting case, let $X= \mathbb{C}/(\mathbb{Z}+\mathbb{Z}\tau)$ be an elliptic curve. As a topological space it is just a torus, and independent of the choice of $\tau$,  but as a Riemann surface it is sensitive to this choice. The natural basis of the first Dolbeault cohomology $H^{10}(X)\oplus H^{01}(X)$is
$\lbrace dz,d\bar z\rbrace$. This maps to $\lbrace (1,\tau), (1,\bar \tau)\rbrace$ under the basis of $H^1_{DR}(X)$ dual to loops given by projecting  $[0,1],[0,\tau]\subset \mathbb{C}$  to $X$. Thus  $X$ can be recovered from its period matrix.