MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 13 down vote accepted

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).

share|cite|improve this answer
How does this relate to versions that involve the rank of the Neron-Severini group? – Aston Smythe Jan 29 '10 at 14:33

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 $H^2(H,\mathbf{R})=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.