= Solution
On the <polydisc>, put $b_i(z)=f_i(z)-f_i(a)$ and define
$$
F(z)=\sum_{k=1}^n\int_{a_k}^{z_k}b_k(z_1,\ldots,z_{k-1},t,a_{k+1},\ldots,a_n)\,dt.
$$
Each 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 $b_j(z_1,\ldots,z_j,a_{j+1},\ldots,a_n)$. If $k>j$, use the assumed symmetry $\partial_jb_k=\partial_kb_j$ to obtain
$$
\partial_j\int_{a_k}^{z_k}b_k(\ldots,t,\ldots)dt=b_j(z_1,\ldots,z_k,a_{k+1},\ldots,a_n)-b_j(z_1,\ldots,z_{k-1},a_k,\ldots,a_n).
$$
These differences telescope. Therefore
$$
\boxed{\frac{\partial F}{\partial z_j}=f_j(z)-f_j(a),\qquad F(a)=0.}
$$
This is the <coordinate-integral primitive on a polydisc>.
For a closed <holomorphic one-form> $\beta=\sum_i c_i dz_i$, closedness is exactly $\partial_ic_j=\partial_jc_i$. 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 <sheaves>
$$
\boxed{0\longrightarrow\underline{\mathbb C}\longrightarrow\mathcal O_M\xrightarrow{\partial}\widetilde\Omega_M^1\longrightarrow0.}
$$
The 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 \b[$u$ constant]. In particular global <holomorphic functions> on a compact connected <complex manifold> are constant.
For the top-degree pairing, write locally $\theta=g\,dz_1\wedge\cdots\wedge dz_n$. With the complex orientation,
$$
i^{n^2}\theta\wedge\overline\theta=2^n|g|^2\,dx_1\wedge dy_1\wedge\cdots\wedge dx_n\wedge dy_n.
$$
It is everywhere nonnegative and is positive on an open set if $\theta$ is not identically zero. Hence
$$
\boxed{i^{n^2}\int_M\theta\wedge\overline\theta>0,\quad\text{so}\quad\int_M\theta\wedge\overline\theta\ne0.}
$$
The 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> give
$$
\int_Md\psi\wedge\overline{d\psi}=\int_Md(\psi\wedge\overline{d\psi})=0.
$$
The positive pairing just proved forces $d\psi=0$. Thus \b[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:
$$
\boxed{H^0(M,\Omega_M^1)\cap\overline{H^0(M,\Omega_M^1)}=0,\qquad2h^{1,0}\leq b_1.}
$$
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 is
$$
0\longrightarrow H^0(M,\Omega_M^1)\longrightarrow H^1(M,\underline{\mathbb C})\longrightarrow H^1(M,\mathcal O_M).
$$
Identify the middle term with first <de Rham cohomology>, and the last with $H^{0,1}_{\bar\partial}(M)$ by the <Dolbeault theorem>. Taking dimensions gives
$$
\boxed{2h^{1,0}\leq b_1\leq h^{1,0}+h^{0,1},\qquad h^{1,0}\leq h^{0,1}.}
$$
These are the <Betti and Hodge bounds for compact complex surfaces>.
A strict example is the scalar <Hopf surface> $X=(\mathbb C^2\setminus\{0\})/\langle z\mapsto2z\rangle$. Polar coordinates identify it smoothly with $S^3\times S^1$: the dilation fixes the angular coordinate and shifts $\log|z|$ by $\log2$. 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 $h^{0,1}\geq1$. Thus \b[$h^{1,0}<h^{0,1}$ on this compact complex surface]; no unsupported cohomology computation is needed.
Back to article page
