Re: On Hodge and Betti numbers

From: Torsten Ekedahl (teke_at_math.su.se)
Date: 10/14/04 

Date: Thu, 14 Oct 2004 13:48:38 GMT

Marcel Vonk wrote:

> While giving a seminar today, I got the following question to which I
> don't know the answer.
>
> On a Kaehler manifold, one can quite easily prove that the p-th Betti
> number b^p equals the sum of the Hodge numbers b^{r,s}, where the sum
> is over r and s such that r+s=p. The proof uses Hodge's theorem
> (saying that in each cohomology class there is a unique harmonic form)
> and the fact that the De Rham Laplacian equals (two times) the
> Dolbeault Laplacian, so the harmonic forms with respect to the two
> exterior derivatives are the same.
>
> For a non-Kaehler complex manifold, this proof does not work, since
> the two Laplacians may differ.
>
> (a) Does this mean the statement is also not true in this case?
> (b) If so, does anyone know a counter-example?
>
> Best regards,
>
> Marcel Vonk

There is a spectral sequence whose E_1-term is the Hodge cohomology with the
map induced by the exterior differential as E_1-differential and converging
to de Rham cohomology. Hence the sum of the b^{r,s} equals b^p for all p
precisely when this spectral sequence degenerates. In particular if there
are non-closed holomorphic forms then the sum of the b^{r,s} is strictly
larger than the b^p.

An example of a compact complex manifold with non-closed is obtained as
follows:

Let H(C) be the group of upper triangular complex 3x3-matrices with 1's
along the diagonal and let H(Z[i]) be the subgroup of matrices with entries
in the Gaussian integers, Z+Zi. H(Z[i]) is a discrete subgroup and as
C/Z[i] is compact so is X=H(C)/H(Z[i]), which is a homogeneous variety
under left multiplication by H(C). The cotangent bundle is trivial spanned
by translation invariant forms and hence, by the compactness of X, the
holomorphic k-forms are exactly the translation invariant forms so that the
space of k-forms may be identified with \Lambda^k g^*, the space of k-forms
on the Lie algebra g of H(C). The exterior differential d from 1-forms to
2-forms, d: g^* ---> \Lambda^2 g^* is dual to the the commutator map
\Lambda^2 g ---> g on the Lie algebra g (by the well-known formula
expressing the exterior differential in terms of commutators of vector
fields). As g is a non-commutative Lie algebra there are non-closed
1-forms.

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