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I am looking for an example of a compact complex surface with $h^{1,0} < h^{0,1}$. The bound that $h^{1,0} \leq h^{0,1}$ is known. In the Kähler case, $h^{p,q}=h^{q,p}$, so the example cannot be (for example) a projective variety or a complex torus. Does anyone know of such an example? Thanks.

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up vote 5 down vote accepted

The standard example of such a thing is the Hopf surface.

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For a compact Kähler manifold, $h^{p,q} = h^{q, p}$, so the odd Betti numbers are even. For a compact complex surface, the only potentially non-zero odd Betti numbers are $b_1$ and $b_3$; note that by Poincare duality, they are equal.

So if $X$ is a compact complex surface, and $X$ is Kähler, then $b_1$ is even. Surprisingly, the converse is also true. That is

Let $X$ be a compact complex surface. Then $X$ is Kähler if and only if $b_1$ is even.

In particular, in complex dimension two, the existence of a Kähler metric is a purely topological question. The above was originally a conjecture of Kodaira and was first shown by using the Enriques-Kodaira classification, with the final case of $K3$ surfaces done by Siu in $1983$. In $1999$, Buchdahl and Lamari independently gave direct proofs which did not rely on the classification.

As $b_1 = h^{1,0} + h^{0,1}$ for compact complex surfaces (see Barth, Hulek, Peters, & Van de Ven Compact Complex Surfaces (second edition), Chapter IV, Theorem $2.7$ (i)), and $h^{p,q} = h^{q,p}$ for compact Kähler manifolds, we can restate the above result as follows

Let $X$ be a compact complex surface. Then $X$ is Kähler if and only if $h^{1,0} = h^{0,1}$.

For any compact complex surface, one can show that the map $H^{1,0}_{\bar{\partial}}(X) \to H^{0,1}_{\bar{\partial}}(X)$, $\alpha \mapsto [\bar{\alpha}]$ is well-defined and injective, so $h^{1,0} \leq h^{0,1}$. Combining with the above result, we have

Let $X$ be a compact complex surface. Then $h^{1,0} < h^{0,1}$ if and only if $X$ is non-Kähler.

So surfaces of class VII (in particular Hopf surfaces), and Kodaira surfaces satisfy the condition $h^{1, 0} < h^{0,1}$.

Note, in the non-Kähler case, we actually have $h^{1,0} = h^{0,1} - 1$ (see Chapter IV, Theorem $2.7$ (iii) of the aforementioned book).

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