Question

The Riemann-Roch theorem is a result about Riemann Surfaces that was extended to the Hirzebruch–Riemann–Roch theorem, a result about compact complex manifolds. The Hodge Index theorem is a result about Riemann surfaces (I'm just worried about the complex case) that is proved using Riemann-Roch. Has the Hirzebruch–Riemann–Roch theorem been used to extend the Hodge Index theorem to a result about compact complex manifolds.

Answer 1

The Hodge index theorem IS a result on compact Kahler manifolds of complex dimension 2n.

It states that the signature of the intersection form on $H^{2n}(X, \mathbb{R})$ equals $\sum (-1)^a h^{a, b}(X)$, where $h^{a, b}$ are the Hodge numbers.

See Voisin, Hodge theory and complex algebraic geometry I, theorem 6.33

By the way, the result you probably have in mind is on complex surfaces, which are NOT Riemann surfaces (these are complex curves).

Answer 2

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 $S^1\times S^3$ and so $H^2(H,\mathbf{R})=0$.