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I was wondering in the world of non-Kähler compact complex manifold, what is the relationship between $h^1(\mathcal{O}_X)$, $h^0(\Omega_X)$, and first betti number?

The only thing I know is that for surfaces, we have $h^1(\mathcal{O}_X) \geq \frac{b_1}{2} \geq h^0(\Omega_X)$, the reason being that every holomorphic 1-form on a compact surface must be closed. Does this inequality hold true for general compact complex manifold? I'm mainly interested in the relation between $h^1(O_X)$ and $h^0(\Omega_X)$.

Another somewhat related question is: does there exist a compact complex manifold with non-closed holomorphic 1-forms?

Thanks to an anonymous friend's comment. On page 444 of Griffiths and Harris' book Principles of Algebraic Geometry, they gave an example (Iwasawa manifold) of a non-closed holomorphic 1-form.

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    $\begingroup$ Can you explain where the inequality you state for surfaces comes from? I don't see how to get this from the spectral sequence and closedness of global $1$-forms alone. If Hodge symmetry fails, couldn't you get things like $h^1(\mathcal O_X) = 0$ and $h^0(\Omega^1_X) = 1$ while the spectral sequence degenerates? $\endgroup$ Oct 18 '16 at 14:57
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There are lots of non-closed holomorphic 1-forms on $SL(2,\mathbb{C})/\Gamma$ for any cocompact lattice $\Gamma \subset SL(2,\mathbb{C})$. See Étienne Ghys, Déformations des structures complexes sur les espaces homogènes de SL(2, C), J. Reine Angew. Math. 468 (1995), 113–138. MR 1361788 (96m:32017).

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  • $\begingroup$ Split the right invariant Maurer-Cartan 1-form into its components. $\endgroup$
    – Ben McKay
    Oct 18 '16 at 7:06
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There is a spectral sequence $$ E^{pq}_1 = H^q(X, \Omega^p_X) \quad\Rightarrow\quad H^{p+q}(X, \Omega^\bullet_X) \cong H^{p+q}(X, \mathbb{C})$$ (which degenerates if $X$ is Kahler). In particular, $h^1(\mathcal{O}_X) + h^0(\Omega^1_X) \geq b_1$, and there is a short exact sequence $$ 0 \to H^0(X, \Omega^1_X)^{d=0}\to H^1(X, \mathbb{C}) \to \ker\left(\partial_2: H^1(X, \mathcal{O}_X)^{d=0} \to H^0(X, \Omega^2_X)\right)\to 0. $$ Here $\partial_2$ is the $E_2$-differential, and $(-)^{d=0}$ means kernel of the map induced by $d$, i.e., the kernel of the $E_1$-differential.

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  • $\begingroup$ Is there an inequality between $h^{0,1}$ and $h^{1,0}$? $\endgroup$
    – S. Li
    Oct 18 '16 at 14:28
  • $\begingroup$ No such inequality in general. $\endgroup$
    – YangMills
    Oct 19 '16 at 5:39
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Let $h^{10} = \dim H^0(X, \Omega^1)$, $z^{10} = \dim H^0(X, Z^1)$ where $Z^1$ is the closed holomorphic $1$-forms, $h^{01} = \dim H^1(X, \mathcal{O})$ and $b_1 = \dim H^1(X, \mathbb{C})$. We claim that we have the following inequalities:

$$h^{10} \geq z^{10}$$ $$h^{01}+z^{10} \geq b_1$$ $$b_1 \geq 2 z^{10}.$$

These imply $h^{01} \geq z^{10}$ (add the second and third and rearrange), but they don't imply $h^{01} \geq h^{10}$ and I don't see why we should expect $h^{01} \geq h^{10}$ if $h^{01} > z^{10}$, that is to say, if there are non-closed holomorphic $1$-forms.

The first inequality is obvious, and the second is proved by abx. For the third, map $H^0(X, Z^1) \oplus \overline{H^0(X, Z^1)} \to H^1_{DR}(X)$ by sending $(\alpha, \beta)$ to the closed $1$-form $\alpha+\overline{\beta}$. We need to show this map is injective -- that is to say, that $\alpha+\overline{\beta}$ is not exact unless $\alpha=\beta = 0$. Suppose, to the contrary, that $\alpha+\overline{\beta} = df$, so $\alpha = \partial f$ and $\beta = \overline{\partial} f$. Since we assumed $\alpha$ is holomorphic, this means $0 = \overline{\partial} \alpha = \overline{\partial} \partial f$ and thus $\nabla^2 f = 0$. But, on a compact manifold, the only harmonic functions are locally constant, so $f$ is locally constant and $\alpha = \beta = 0$.

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This is actually a comment on Piotr Achinger's answer, but a bit too long for a comment. I don't think you need the spectral sequence, nor the differential $\partial_2$. From the exact sequence $0\rightarrow \mathbb{C}\rightarrow \mathcal{O}_X \rightarrow (\Omega ^1_X)_{d=0}\rightarrow 0\ $ you get an exact sequence $$0\rightarrow H^0(X,\Omega ^1_X)_{d=0}\rightarrow H^1(X,\mathbb{C})\rightarrow H^1(X,\mathcal{O}_X)\xrightarrow{d} H^1(X,(\Omega ^1_X)_{d=0})$$from which the inequality $b_1\leq h^{1,0}+h^{0,1}$ follows immediately.

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  • $\begingroup$ Is there an inequality between $h^{0,1}$ and $h^{1,0}$? $\endgroup$
    – S. Li
    Oct 18 '16 at 14:28
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It feels awkward to answer the question posed by myself, but I was told it's okay to do so. There are few points I want to say here.

First of all, regarding Remy's question. On a compact complex surface $X$, every holomorphic 1-form is closed, hence $h^0(\Omega_X) \leq h^1(O_X)$ holds true in this case (by David Speyer's answer). So suppose $\eta$ is a holomorphic 1-form on $X$, then $\int_X d\eta \wedge \bar{d\eta}=0$ by Stokes. Then one expands $d\eta \wedge \bar{d\eta}$ in local coordinate to see that this would imply $d\eta=0$. More generally on a compact complex n-fold, every holomorphic (n-1)-form is closed by the same argument.

Secondly, my main question is actually whether the inequality $h^0(\Omega_X) \leq h^1(O_X)$ holds in general. Now I knew a counterexample to this inequality which is Iwasawa manifold. One can just consult this and that. Probably I should describe Iwasawa manifold here in case someone forgot the definition. Let $G=\{\begin{bmatrix} 1 &a & b \\ 0 &1 & c \\ 0 &0 & 0 \end{bmatrix}|a, b, c \in \mathbb{C}\}$ be the Heisenberg group of complex coefficient and let $\Gamma$ be the subgroup with entries in $\mathbb{Z}[i]$, then we form a quotient $X=\Gamma \backslash G$. By projecting to $(a,c)$ coordinates, we exhibit $X$ as a holomorphic fibration with fiber $E$ and base $E \times E$, where $E$ is the beautiful elliptic with j-invariant 1728. One compute directly that $\Gamma^{ab}=\mathbb{Z}[i]^2$, hence $h^1(X,\mathbb{C})=4$. One sees (arguably not so obvious) that $da$, $dc$, and $db-adc$ form a basis of $H^0(X,\Omega^1_X)$. Applying Leray spectral sequence to the projection $X \to E \times E$, one can actually show that $h^1(O_X)=2$ (the point being that the differential $d_2: H^0(R^1\pi_*(O_X)=O_{E \times E}) \to H^2(O_{E\times E})$ is nonzero, and it sends 1 to $\bar{da} \wedge \bar{dc}$ if one normalizes everything appropriately).

Thirdly, Hopf surface would also give an example where the inequality $h^{01} \geq z^{10}$ in Davide Speyer's answer is not an equality ($z^{10}=h^{10}=0$ in that case).

Lastly, I was also wondering if there's an example where the Hodge-deRham doesn't degenerate at $E_2$ page. I was informed by my friend yesterday that such examples (nilmanifold) do exist, but I know nothing about it so perhaps I should stop here.

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    $\begingroup$ Examples of non-degeneration of Hodge-deRham complex arxiv.org/abs/0709.0481 $\endgroup$ Oct 19 '16 at 18:20
  • $\begingroup$ @DavidSpeyer Thanks for providing me this reference. And thanks again for answering my question. $\endgroup$
    – S. Li
    Oct 19 '16 at 21:20

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