11
$\begingroup$

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 $h^{p,q}(M)$?

$\endgroup$

2 Answers 2

13
$\begingroup$

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).

$\endgroup$
1
  • 1
    $\begingroup$ I think that if $M$ is Kähler, then the Hodge-Frölicher spectral sequence rather degenerates in $E_1^\bullet$, not at the $E_2$ page. $\endgroup$
    – diverietti
    Mar 22, 2019 at 21:41
15
$\begingroup$

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: $$ \chi_{\text{top}}(X)=\sum_{k=0}^{2n}(-1)^kb_k=\sum_{p,q=0}^n(-1)^{p+q}h^{p,q}. $$

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – diverietti
    Nov 17, 2010 at 10:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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