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A differential form of type (p, q) of type $(p,0)$ is holomorphic when its local coefficients in holomorphic coordinates are holomorphic functions, equivalently $\bar{\partial }\alpha =0$. It need not be $d$-closed on every complex manifold. Top-degree holomorphic forms are closed by bidegree. On a compact complex surface, holomorphic forms of all degrees are closed, even without a Kähler metric.

A complex K3 surface is a compact, simply connected complex surface with trivial canonical bundle. For projective algebraic surfaces, one can equivalently require a trivial canonical bundle and $H^1(X,{\mathcal{O}}_X)=0$. Nonprojective complex K3 surfaces are also included, which is essential for the full moduli space of Ricci-flat K3 metrics.

On a complex manifold, the exterior derivative splits by type as $d=\partial +\bar{\partial }$. The conjugate Dolbeault operator $\partial$ raises holomorphic degree by one and the Dolbeault operator $\bar{\partial }$ raises antiholomorphic degree by one. Their identities areUse the convention $d^c=i(\bar{\partial }-\partial )$ for the d c operator; then $d{d}^c=2i\partial \bar{\partial }$. The alternative convention with a factor $1/2$ changes this last factor only. Complex conjugation of differential-form type interchanges $\partial$ and $\bar{\partial }$ without reversing the wedge order. Thus, for a real form $\eta$,The resulting form is real, independently of the degree or type components of $\eta$.

The Dolbeault cohomology isIn degree $(0,0)$ there is no incoming image, and the kernel consists of global holomorphic functions. On a compact connected $X$, $|f|$ attains a maximum. In a coordinate polydisc around a maximizing point, the one-variable maximum modulus principle applied successively along the coordinate directions makes $f$ constant on a smaller polydisc. Alternatively the same argument shows the maximum set is open; it is closed and nonempty, so connectedness makes it all of $X$, and the local constancy then makes $f$ globally constant. Consequently

A holomorphic p-form on a complex manifold is a form of type $(p,0)$ with holomorphic coordinate coefficients, equivalently one killed by $\bar{\partial }$. For degree $n={\dim }_{\mathbb{C}}X$, it is $d$-closed: its $\bar{\partial }$ vanishes by holomorphicity and its $\partial$ has impossible bidegree $(n+1,0)$. Its conjugate is also closed.

The key positivity behind nonzero holomorphic top forms are not exact is explicit. If $\alpha =fd{z}_1\wedge \cdots \wedge d{z}_n$, complex orientation givesThis is nonnegative everywhere and positive on some open set when $\alpha \not\equiv 0$, so its integral on the compact manifold is strictly positive. If $\alpha =d\gamma$, closedness of $\bar{\alpha }$ and the Stokes theorem would giveHere Stokes is used on a compact oriented manifold without boundary, and complex-valued forms are handled by linearity. The contradiction proves a nonzero holomorphic top form is not $d$-exact. For dimension zero the same nonexactness follows directly from the absence of degree $-1$ forms.

Now take a compact complex surface, with no Kähler metric assumed. For a holomorphic one-form $\alpha$, the form $\beta =d\alpha =\partial \alpha$ has type $(2,0)$ and is holomorphic, since $\bar{\partial }\partial \alpha =-\partial \bar{\partial }\alpha =0$. It is exact. The top-form result forces $\beta =0$, proving closedness. This establishes holomorphic forms on a compact complex surface are closed by an argument which does not require a Kähler identity.

If a holomorphic one-form were exact, write $\alpha =df$ for a global smooth complex function. Comparing types gives $\bar{\partial }f=0$. The preceding compactness argument makes $f$ constant on every connected component, hence $\alpha =0$. Holomorphic two-forms are closed and nonexact unless zero by the top-form result. Holomorphic zero-forms are constant on each component, and a nonzero zero-form is never exact because there are no negative-degree forms. Degrees above two contain only zero forms. Therefore every holomorphic form on a compact complex surface is closed, and a nonzero one is never exact.

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.