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A complex surface is a complex manifold of complex dimension two, hence real dimension four. Compact complex surfaces need not admit a Kähler metric; a Hopf surface is an example. Every holomorphic one-form on a compact complex surface is closed, and the Betti and Hodge bounds for compact complex surfaces still apply without Kähler assumptions.

A primary Hopf surface is a quotient of ${\mathbb{C}}^2\setminus \{0\}$ by an infinite cyclic group generated by a suitable contracting holomorphic map; reversing the generator gives the expanding dilation description. The scalar dilation $z\mapsto 2z$ gives a compact surface diffeomorphic to $S^3\times S^1$ via polar coordinates and log radius. Hence $b_1=1$. The Betti and Hodge bounds for compact complex surfaces force $h^{1,0}=0$ and $h^{0,1}\ge 1$, furnishing a strict example of $h^{1,0}<h^{0,1}$.

On the polydisc, put $b_i(z)=f_i(z)-f_i(a)$ and defineEach integral is along the straight coordinate segment, which remains in the polydisc. Holomorphic dependence on parameters and differentiation under the integral show that $F$ is a holomorphic function. For a fixed $j$, terms with $k<j$ do not depend on $z_j$, and the $k=j$ term contributes $b_j(z_1,\dots ,z_j,a_{j+1},\dots ,a_n)$. If $k>j$, use the assumed symmetry ${\partial }_j{b}_k={\partial }_k{b}_j$ to obtainThese differences telescope. ThereforeThis is the coordinate-integral primitive on a polydisc.

For a closed holomorphic one-form $\beta ={\sum }_i{c}_{i}d{z}_i$, closedness is exactly ${\partial }_i{c}_j={\partial }_j{c}_i$. The preceding construction plus ${\sum }_i{c}_i(a)(z_i-a_i)$ gives a local holomorphic function $G$ with $\partial G=\beta$. The kernel of $\partial$ on the structure sheaf of a complex manifold consists of locally constant functions: holomorphicity already gives $\bar{\partial }G=0$, and $\partial G=0$ then implies $dG=0$. Thus the holomorphic Poincaré lemma gives the short exact sequence of sheavesThe first object is the complex constant sheaf, and surjectivity is local, on stalks. It does not say that every global closed holomorphic one-form has a global primitive.

Now assume compactness. The constancy assertion is for a connected complex manifold; otherwise it holds on each connected component. A smooth complex-valued $u$ with $\partial \bar{\partial }u=0$ is a pluriharmonic function. Restricting to each complex coordinate line makes it a harmonic function of one complex variable. Choose $x_0$ where $|u|$ has its global maximum and a coordinate polydisc about $x_0$. The allowed harmonic maximum principle makes the first coordinate-line restriction constant. Every point on that line still has the same maximal modulus and the same value, so the second coordinate may then vary without changing $u$. Continue through all coordinates to conclude that $u$ equals $u(x_0)$ on the whole polydisc. The set where it has this value is closed and, by the same argument at each of its points, open. Connectedness therefore gives $u$ constant. In particular global holomorphic functions on a compact connected complex manifold are constant.

For the top-degree pairing, write locally $\theta =gd{z}_1\wedge \cdots \wedge d{z}_n$. With the complex orientation,It is everywhere nonnegative and is positive on an open set if $\theta$ is not identically zero. HenceThe phase factor is necessary: the unnormalized integral need not be a positive real number. This is the integral pairing of complex top forms.

If $\psi$ is a holomorphic differential form of degree $n-1$, then $d\psi =\partial \psi$ has type $(n,0)$ and $d\overline{d\psi }=0$. The graded Leibniz rule and Stokes theorem giveThe positive pairing just proved forces $d\psi =0$. Thus every holomorphic $(n-1)$-form on a compact complex $n$-manifold is closed, without a Kähler metric assumption.

On a compact complex surface, this proves closedness of all holomorphic one-forms. Sending such a form to its de Rham cohomology class defines the required inclusion. Indeed, if $\alpha =df$, its type $(1,0)$ gives $\bar{\partial }f=0$, so $f$ is a global holomorphic function and is constant. Therefore $\alpha =0$.

To prove the zero intersection, let $\alpha ,\beta$ be holomorphic one-forms and suppose $[\alpha ]=[\overline{\beta }]$. There is a smooth complex-valued $f$ with $\alpha -\overline{\beta }=df$. Its type components are $\partial f=\alpha$ and $\bar{\partial }f=-\overline{\beta }$. Closedness of $\beta$ gives $\partial \overline{\beta }=0$, hence $\partial \bar{\partial }f=0$. Compact pluriharmonic rigidity now makes $f$ constant and $\alpha =\beta =0$. The two spaces are therefore linearly independent in de Rham cohomology, and conjugation preserves their dimensions:For the other bound, take the long exact sequence in sheaf cohomology of the short exact sequence above. The map $H^0(M,\underline{\mathbb{C}})\to H^0(M,{\mathcal{O}}_M)$ is an isomorphism, componentwise. Since every global holomorphic one-form is closed, the relevant portion isIdentify the middle term with first de Rham cohomology, and the last with $H_{\bar{\partial }}^{0,1}(M)$ by the Dolbeault theorem. Taking dimensions givesThese are the Betti and Hodge bounds for compact complex surfaces.

A strict example is the scalar Hopf surface $X=({\mathbb{C}}^2\setminus \{0\})/⟨z\mapsto 2z⟩$. Polar coordinates identify it smoothly with $S^3\times S^1$: the dilation fixes the angular coordinate and shifts $\log |z|$ by $\log 2$. Its first Betti number is one, since its fundamental group is $\mathbb{Z}$. The proved lower bound then forces $h^{1,0}=0$, and the upper bound forces $h^{0,1}\ge 1$. Thus $h^{1,0}<h^{0,1}$ on this compact complex surface; no unsupported cohomology computation is needed.