Hodge diamonds of complex threefolds There is no closed complex curve or surface with $h^{1, 0}-h^{0, 1}=1$.
Now consider threefolds. Can this condition be satisfied?
Is Serre duality in fact the only restriction on the Hodge diamond?
 A: Let $R$ be a commutative ring with identity. The three-dimensional Heisenberg group over $R$ is
$$\mathbb{H}(3, R) = \left\{\begin{bmatrix} 1 & z^1 & z^3\\ 0 & 1 & z^2\\ 0 & 0 & 1\end{bmatrix} : z^1, z^2, z^3 \in R\right\}.$$
The Iwasawa manifold $\mathbb{I}_3$ is the quotient of $\mathbb{H}(3, \mathbb{C})$ by the discrete subgroup $\mathbb{H}(3, \mathbb{Z}[i])$ acting on the left, i.e. $\mathbb{I}_3 := \mathbb{H}(3, \mathbb{Z}[i])\setminus\mathbb{H}(3, \mathbb{C})$. It is a compact complex threefold with $h^{1,0}(\mathbb{I}_3) = 3$ and $h^{0,1}(\mathbb{I}_3) = 2$. The holomorphic one-forms $dz^1$, $dz^2$, and $dz^3 - z^1dz^2$ on $\mathbb{H}(3, \mathbb{C})$ descend to $\mathbb{I}_3$ and form a basis for $H^{1,0}_{\bar{\partial}}(\mathbb{I}_3)$, while the $(0,1)$-forms $d\bar{z}^1$ and $d\bar{z}^2$ descend to a basis of $H^{0,1}_{\bar{\partial}}(\mathbb{I}_3)$.
As for your second question, consider the following potential Hodge diamond (which satisfies the symmetries imposed by Serre duality):
$$\begin{array}{ccccccc}&&&1&&&\\&&0&&0&&\\&0&&0&&0&\\0&&0&&0&&0\\&0&&0&&0&\\&&0&&0&&\\&&&1&&&\end{array}$$
There does not exist a compact complex threefold with this Hodge diamond. In particular, Serre duality is not the only restriction on the Hodge diamond of a compact complex threefold.
To see this, first note that on a compact complex manifold, it follows from the Frölicher spectral sequence that $b_k \leq \sum_{p+q=k}h^{p,q}$. So if $X$ were to have the Hodge diamond above, it would satisfy $b_0(X) = b_6(X) = 1$ and $b_i(X) = 0$ otherwise; that is, $X$ is a six-dimensional rational homology sphere. It follows that $c_1(X) \in H^2(X; \mathbb{Z})$ and $c_2(X) \in H^4(X; \mathbb{Z})$ are torsion, so
$$\chi(X, \mathcal{O}) = \int_X\operatorname{Td}(X) = \int_X\frac{1}{24}c_1(X)c_2(X) = 0,$$
but $\chi(X, \mathcal{O}) = h^{0,0}(X) - h^{0,1}(X) + h^{0,2}(X) - h^{0,3}(X) = 1 \neq 0$. Therefore, no such $X$ exists.
This fact can be generalised in two ways.

*

*The above argument can be used to show that if a compact complex threefold $X$ has $h^{1,1}(X) = 0$, then $1 + h^{0,2}(X) = h^{0,1}(X) + h^{0,3}(X)$ which gives a restriction on the Hodge diamond which does not follow from Serre duality.


*Consider the potential Hodge diamond with $h^{0,0} = h^{n,n} = 1$ and all other numbers zero (the case $n = 3$ gives the Hodge diamond above). This cannot be realised unless $n = 1$. As before, we see that if $X$ has the given Hodge diamond, then $X$ is a $2n$-dimensional rational homology sphere. In this paper, Aleksandar Milivojevic and I showed that a rational homology sphere which admits an almost complex structure must have dimension $2$ or $6$, see Theorem 2.2, so the Hodge diamond cannot be realised for $n \neq 1, 3$. For $n = 1$, the Hodge diamond is realised by $\mathbb{CP}^1$, while the argument above shows that it cannot be realised for $n = 3$.
