dg.differential geometry - Hodge Index theorem for Complex Manifolds

Complementing Andrea's posting: the answer to the question as it is stated is no. Indeed, the proof of the Hodge index formula for Kaehler manifolds uses the strong Lefschetz decomposition, which does not exist for arbitrary complex manifolds.

A counter-example in the analytic case is given by the Hopf surface $H$, which is the quotient of $mathbf{C}^2$ minus the origin by the group generated by $(x,y)mapsto (2x,2y)$. Indeed, it is not too difficult to show that $H$ does not admit a holomorphic 1-form, i.e. $h^{1,0}(H)=0$. It is a non-trivial theorem (Barth, Peters, van de Ven, Compact complex surfaces, p. 117) that the Hodge to de Rham spectral sequence of any complex surface degenerates at $E_1$. Using this and the Serre duality we can compute all other Hodge numbers $h^{p,q}(H)$, which turn out to be 1 for $(p,q)=(0,0),(0,1),(2,1),(2,2)$ and zero otherwise. Plugging this into the right hand side of the index formula, we get 4. On the other hand, $H$ is diffeomorphic to $S^1times S^3$ and so $H^2(H,mathbf{R})=0$.