Dolbeault Cohomology is invariant under homeomorphisms If $X$ and $Y$ are two complex manifolds, which are homeomorphic but not necessarily diffeomorphic, must their Dolbeault cohomology groups be isomorphic? Here the Dolbeault cohomology groups $H^{p,q}(X)$ are defined as the quotient of the vector space of $(p,q)$-forms with $\bar\partial\alpha = 0$ by the space of $\bar\partial$-exact $(p,q)$-forms.
I tried looking at various manifolds which admit distinct complex structures, e.g. the torus. However, in all of the cases where I can compute the Dolbeault cohomology, it is determined by the Betti numbers, so is the same for homeomorphic manifolds. 
 A: Consider the Hodge numbers
$$h^{p,q}(X) = \dim(H^{p,q}(X)).$$
Then we can ask if the $h^{p,q}$ are topological invariants; if they are not, then clearly there exist homeomorphic complex manifolds $X$ and $Y$ with
$$H^{p,q}(X) \not\cong H^{p,q}(Y).$$
This is an old question, first asked in 1954 by Hirzebruch in the following form:

Are the $h^{p,q}$ and the Chern characteristic numbers of an algebraic variety $V_n$ topological invariants of $V_n$? If not, determine all those linear combinations of the $h^{p,q}$ and the Chern characteristic numbers which are topological invariants.

The answer to the first question is no. Some specific counterexamples are as follows. In 1986, Gang Xiao constructed two complex surfaces $S$ and $S'$ which are homeomorphic but not diffeomorphic and have different Hodge numbers. This shows that Hodge numbers are not topological invariants. But even further, one can show that $S \times S$ and $S' \times S'$ are diffeomorphic via an orientation-preserving diffeomorphism. $S \times S$ and $S' \times S'$ will still have different Hodge numbers, so this example shows that Hodge numbers are not even an invariant of oriented diffeomorphism type. See the following two references for more detail.

Xiao, Gang. An example of hyperelliptic surfaces with positive index. Northeast. Math. J. 2 (1986), no. 3, 255–257.
Campana, Frédéric. Une remarque sur les nombres de Hodge des variétés projectives complexes (Unpublished). Available here.

So the Hodge numbers are not topological invariants. But one can ask the second part of Hirzebruch's question: which linear combinations of Hodge numbers are topological invariants? This question was fully answered in quite a strong form recently by Kotschick and Schreieder.

Theorem. (Kotschick-Schreieder) The mod $m$ reduction of a $\Bbb Z$-linear combination of Hodge numbers of smooth complex projective varieties is
  
  
*
  
*An oriented homeomorphism invariant or an oriented diffeomorphism invariant if and only if it is congruent mod $m$ to a linear combination of the signature, the even-degree Betti numbers and the halves of the odd-degree Betti numbers.
  
*An unoriented homeomorphism invariant in any dimension, or an unoriented diffeomorphism invariant in complex dimensions $n \neq 2$, if and only if it is congruent mod $m$ to a linear combination of the even-degree Betti numbers and the halves of the odd-degree Betti numbers.

Their paper also classifies which linear combinations of both Hodge and Chern numbers are topological invariants. See the following for this interesting work:

Kotschick, Dieter and Schreieder, Stefan. The Hodge ring of Kähler manifolds. arXiv:1202.2676v2 [math.AG].

Now after all that, you may ask if there are some Hodge numbers which are invariant under some equivalence relation for some class of manifolds. There are some well-known results here.


*

*Hodge numbers are homeomorphism invariants of complex curves and surfaces.

*The Hodge numbers $h^{p,0}$ are birational invariants of smooth projective varieties. Not all Hodge numbers are birational invariants, as one can see by considering the blowup of a smooth projective variety.

*Hodge numbers are an invariant of Kähler manifolds under deformation equivalence. Recall that two complex manifolds $X$ and $Y$ are deformation equivalent if there exists a proper, holomorphic submersion $\pi: E \longrightarrow B$ such that $X = \pi^{-1}(b)$ and $Y = \pi^{-1}(b')$ for some $b, b' \in B$. More generally, if $\pi: E \longrightarrow B$ is a proper, holomorphic submersion and $b_0 \in B$ is such that $\pi^{-1}(b_0)$ is Kähler, then there is a neighborhood $U$ of $b_0$ such that the Hodge numbers of all the $\pi^{-1}(b)$ are equal for all $b \in U$.
