Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For a non-Kahler complex manifold $M$, we still have the decomposition of differential forms into differential forms of type $(p,q)$ and we can write $d=\partial+\bar\partial$ and we can define cohomology classes $H^{p,q}_{\bar\partial}(M)$.

In general is there any relation between $h^r(M)$ and $h^{p,q}(M)$?

share|improve this question
add comment

2 Answers

up vote 9 down vote accepted

For any compact complex manifold there is a spectral sequence with $E_1$ term $H^{p,q}(M)$ which converges to $H^{p+q}(M)$. If $M$ were Kahler, then this spectral sequence would degenerate at the $E_2$ page, giving the familiar Hodge decomposition on cohomology.

In general, there is still a filtration on the cohomology, and the associated graded pieces will be sub-quotients of the $H^{p,q}(M)$.

So there is such a relationship between Hodge numbers and Betti numbers, but being able to write it down depends on being able to calculate the differentials in the spectral sequence.

Voisin's book, Hodge Theory and Complex Algebraic Geometry has more information. There is also an interesting exercise in which you can calculate the precise relationship in the case when $M$ is a complex surface (not necessarily Kahler).

share|improve this answer
add comment

Let $X$ be a compact complex manifold of complex dimension $n$. The Hodge-Frölicher spectral sequence starts with $$ E_1^{p,q}=H^{p,q}(X,\mathbb C) $$ and the limit term $E^{p,q}_\infty$ is the graded module associated to a filtration of the de Rham cohomology group $H^{p+q}_{\text{dR}}(X,\mathbb C)$. In particular $$ b_k=\sum_{p+q=k}\dim E_\infty^{p,q}\le\sum_{p+q=k}\dim E_1^{p,q}=\sum_{p+q=k}h^{p,q}. $$ The equality is equivalent to the degeneration of the spectral sequence at the $E_1^\bullet$ level.

On the other hand, an elementary lemma on bounded complexes of finite dimensional vector spaces applied to $E^\bullet_r$, tells you that you always have equality for the (topological) Euler characteristic: $$ \chi_{\text{top}}(X)=\sum_{k=0}^{2n}(-1)^kb_k=\sum_{p,q=0}^n(-1)^{p+q}h^{p,q}. $$

share|improve this answer
    
Complement: the elementary lemma I speak about, just says that the Euler characteristic of a bounded complex of finite dimensional vector spaces equals the Euler characteristic of its cohomology module. –  diverietti Nov 17 '10 at 10:27
add comment

Your Answer

 
discard

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.