# Dolbeault cohomology

Let $X=S^{2n-1} \times S^1$. I have to compute $H^{(1,0)}_{\bar{\partial}} (X)$ and

$H^{(0,1)}_{\bar{\partial}}(X)$ . I don't know how to do this but if we use Kunnet formula we

have that $dim(H^{(0,1)}_{\bar{\partial}}(X))$ $+$

$dim(H^{(1,0)}_{\bar{\partial}}(X))=1$, so we

have that $H^{(1,0)}_{\bar{\partial}}(X)\simeq \mathbb{C}$

and $H^{(0,1)}_{\bar{\partial}} (X)\simeq 0$

or vice versa. I find that $X \simeq (\mathbb{C}^{n}-\{0\})/\mathbb{Z}$ via the

action $d:(z_1, \cdots, z_n) \mapsto(\lambda^dz_1, \cdots, \lambda^dz_n)$ with $n>1$,

$\lambda \in \mathbb{C}$ and $|\lambda|>1$. Futhermore I read that

$H=\bar{\partial}log(|z_1|^2+ \cdots +|z_n|^2)$ represents a non trivial class. But how can

I prove that $H$ is a closed but non exact form?

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## 1 Answer

You seem to have some basic misunderstandings here.

For non-Kähler manifolds, like the Hopf manifold that you consider, the Dolbeault cohomology is not in general topological, and it depends a lot on the complex structure that you have. Note that there are many different (i.e. not biholomorphic) complex structure on the smooth manifold $S^{2n-1}\times S^1$.

Also, the Künneth formula as you stated it is wrong, and you cannot just use topology to compute the Hodge numbers. But note that the Frölicher spectral sequence in particular implies that $h^{1,0}(X)+h^{0,1}(X)\geq b_1(X)$ for any compact complex manifold $X$.

Having said this, let's suppose that you look at the Hopf manifold $X$ that you define as your explicit $\mathbb{Z}$ quotient (this defines the complex structure uniquely, independent of the value of your parameter $\lambda$). Then its Hodge numbers were calculated a long time ago (first by A.Borel, I believe, again using spectral sequences for the more general class of Calabi-Eckmann manifolds). You can find the result conveniently packaged in this paper of Höfer, see page 232. In particular, we have $h^{0,1}(X)=1, h^{1,0}(X)=0$ for this particular Hopf manifold (here I am assuming $n\geq 2$).

The last question that you ask is a nice exercise for you!

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@YangMills But I have to find a method to show that $H$ is an exact but not closed form... –  Newlander May 15 '13 at 22:52