Let $X$ and $Y$ be two smooth projective varieties over $\mathbb{C}$ such that $X(\mathbb{C})$ is homeomorphic to $Y(\mathbb{C})$. Is it true that $\dim\_{\mathbb{C}} H^k(X,\mathcal{O}\_X)=\dim\_{\mathbb{C}} H^k(Y,\mathcal{O}\_Y)$ for $k\ge 2$? (Note that for $k=1$ it follows by Hodge theory).



Is $h^{0,k}$ a topological invariant?
The answer is *no*. 


There are counterexamples already for surfaces, due to X. Gang and F. Campana (unpublished). The link to Campana's article is [here](http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/hodgenumbers.pdf), and the relevant result is 



> 
> **Proposition 0.1** Il existe des surfaces projectives complexes $S$, $S\_0$ simplement connexes et de type général homéomorphes, mais ayant des nombres de Hodge différents.
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You can look at this paper and at the references cited therein for further details. 


