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On a polydisc, $H_{BC}^{p,q}=0$ for $p,q>0$. A Poincaré lemma primitive for a closed pure-type form can be modified by exact terms until only bidegrees $(p-1,q)$ and $(p,q-1)$ remain. The Dolbeault-Poincaré lemma and conjugate Dolbeault-Poincaré lemma then make the original form $\partial \bar{\partial }$-exact.

On a polydisc, a $\partial$-closed differential form of type (p, q) with $p>0$ is $\partial$-exact. Apply the Dolbeault-Poincaré lemma to its complex conjugate, whose antiholomorphic degree is $p$.

For a $\bar{\partial }$-closed positive-antiholomorphic-degree differential form of type (p, q) on a polydisc, write $\phi =d{\bar{z}}_j\wedge \alpha +\beta$ with no $d{\bar{z}}_j$ in $\alpha ,\beta$. Apply the Cauchy-Green operator in coordinate $j$ to the coefficients of $\alpha$. Subtracting its Dolbeault operator derivative removes every $d{\bar{z}}_j$ term. Closedness then makes the remaining coefficients holomorphic in $z_j$. Eliminating successive coordinates preserves holomorphicity in earlier coordinates and produces a local primitive after finitely many steps.

If the right-hand side and initial data are jointly holomorphic, the local solution is jointly holomorphic in its independent variable and parameters. On a smaller closed polydisc, bound $|A|$ by $M$ and $|{\partial }_{w}A|$ by $L$. Choose $r$ so that $rM$ stays inside the dependent-variable disc and $rL<1$. The integral contraction mapping has jointly holomorphic iterates converging uniformly, so locally uniform convergence of holomorphic functions proves the assertion. Shifting the dependent variable turns initial values into parameters.

Choose holomorphic coordinates centered at the point $p$ and a polydisc $U\subset {\mathbb{C}}^n$ about the origin. The local blowup of a complex manifold at a point is the incidence manifoldAway from $z=0$ the direction is forced to be $[\ell ]=[z]$, giving a biholomorphism with $U\setminus \{0\}$. Glue this punctured part to $X\setminus \{p\}$. The charts below make the result a complex manifold. The projection is proper near $p$, since it is the restriction of projection from a closed subset of $U\times {\mathbb{CP}}^{n-1}$; globally the modification is proper and is an isomorphism off $p$. Its exceptional divisor is

On ${\ell }_i\ne 0$, let $t=z_i$ and $u_j={\ell }_j/{\ell }_i$ for $j\ne i$. ThenThese are coordinates on the open domain where $t$ and all $t{u}_j$ lie in the original polydisc; at $t=0$ all finite $u_j$ are allowed. Their inverse reads the direction ratios and $z_i$ from the incidence point. In this chart $E$ is the nonsingular hypersurface $t=0$. On overlap with chart $k\ne i$, the holomorphic coordinate changes areThey are holomorphic with holomorphic inverses where $u_k\ne 0$, and the $n$ charts cover every point of $E$. Intrinsically, changing the original coordinates by $F$ lifts via $[\ell ]\mapsto [A(z)\ell ]$, where $F(z)=A(z)z$, $A(0)=d{F}_0$, and $A$ is holomorphic and invertible near zero. One can take $A(z)={\int }_0^{1}d{F}_{sz}ds$ on a sufficiently small polydisc. Away from the origin the lift is uniquely forced; this proves coordinate-independence of the gluing near $E$.

The canonical bundle is $K_X={\Lambda }^n{\Omega }_X^1$. In the $i$th blowup chart the top-form Jacobian iswhere the $u_j$ occur in increasing order. Every term with a second $dt$ vanishes in the wedge product of differential forms. Thus the differential gives a map ${\sigma }^{\ast }{K}_X\to K_{\tilde{X}}$ vanishing to order $n-1$ along $E$ and nowhere else. To express this through the supplied meromorphic-section hypothesis, take a nonzero meromorphic section $s$ of $K_X$ and put $D=\mathrm{div}(s)$. Locally $s=f(z)d{z}_1\wedge \cdots \wedge d{z}_n$, soHere ${\sigma }^{\ast }D$ is the total divisor pullback, including any additional exceptional multiplicity from $f$; it is not merely the strict transform. A line bundle with a nonzero meromorphic section is isomorphic to the line bundle of that section's divisor, as its local coefficient ratios are its frame transitions. Pullback of divisor line bundles and additivity of analytic divisor line bundles therefore give the canonical bundle formula for a point blowup:For $n=1$ the blowup is already an isomorphism, consistently with the zero exponent.

For the projective-plane construction, an invertible linear change of homogeneous coordinates puts $x=[1:0:0]$. Lines through $x$ are parametrized by $[a:b]\in {\mathbb{CP}}^1$, with equation $b{Z}_1-a{Z}_2=0$. Thus the required map on the punctured plane isIt is well-defined under rescaling and holomorphic on the covering sets $Z_1\ne 0$ and $Z_2\ne 0$, where its target coordinates are $Z_2/Z_1$ and $Z_1/Z_2$. These sets cover the complement of $x$.

The graph closure isOver the complement of $x$ it is the graph of $f$. In the affine chart $Z_0\ne 0$, its equation is exactly the two-dimensional local incidence model for the blowup of a complex manifold at a point, so its first projection identifies it with the required blowup globally. The second projection is a holomorphic map $\tilde{f}:S\to {\mathbb{CP}}^1$ extending $f\circ \sigma$. Explicitly, the two blowup charts giveThese formulas agree when $v=1/u$ and remain holomorphic at $t=0$. Hence the pencil extends over the exceptional divisor, where it records the projective tangent direction. This is the projective pencil resolved by a point blowup.

Choose $0<R<R_1$, $0<S<S_1$ and $0<T<T_1$. On this smaller closed polydisc, the holomorphic function $A$ and its $w$-derivative are bounded; write $|A|\le M$ and $|{\partial }_{w}A|\le L$. Choose $0<r<R$ with $rM<S/2$ and $rL<1$, replacing a zero bound by any positive upper bound if necessary, and choose $0<\epsilon <T$.

Consider functions $u(z,\alpha )$ continuous on $\overline{D(0,r)}\times \overline{D(0,\epsilon )}$, jointly holomorphic in the interior, with $u(0,\alpha )=0$ and $\Vert u{\Vert }_{\infty }\le S/2$. This is a closed subset of a Banach space in the supremum norm. The integral operatoris well defined on the straight segment and is jointly holomorphic. Indeed, it equals $z{\int }_0^{1}A(tz,u(tz,\alpha ),\alpha )dt$, an integral of jointly holomorphic functions locally uniformly bounded in both variables. Convexity of the $w$-disc and the derivative bound giveThus the Banach fixed-point theorem gives a fixed point $f(z,\alpha )$. More explicitly, the successive integral iterates starting from zero converge uniformly, and locally uniform convergence of holomorphic functions makes their limit jointly holomorphic. Differentiating the integral equation gives the required complex ordinary differential equation and initial value. This proves local existence, uniqueness and joint holomorphic dependence on $(z,\alpha )$. Uniqueness among arbitrary local solutions follows first on a small disc where their values stay within $|w|<S$, by the same contraction mapping estimate, and then throughout their common connected domain by the identity theorem.

For holomorphic dependence of ordinary differential equations on parameters in the initial value, fix an admissible initial value $w_{\ast }$ and put $u=f-w_0$. The equation becomesIts right-hand side is jointly holomorphic in $(z,u,w_0)$ near $(0,0,w_{\ast })$. The preceding construction supplies a common disc and a jointly holomorphic solution for $w_0$ near $w_{\ast }$. Consequently $f=u+w_0$ is jointly holomorphic in $z$ and its initial value $w_0$.

The complex structure on a complex manifold splits its complexified cotangent bundle into the $+i$ and $-i$ eigenspaces, locally spanned by $d{z}^j$ and $d{\bar{z}}^j$. A differential form of type (p, q) is a smooth section of ${\bigwedge }^p(T^{1,0}X{)}^{\ast }\otimes {\bigwedge }^q(T^{0,1}X{)}^{\ast }$, so in holomorphic coordinates it has the expressionHere the increasing multi-indices avoid duplicating alternating factors. The exterior derivative has just two type components: the conjugate Dolbeault operator $\partial$ and the Dolbeault operator $\bar{\partial }$. On such a local expression they areHolomorphic changes of coordinates preserve these types, so the definitions are intrinsic. Comparing type components of $d^2=0$ gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$.

A real (p, p)-form satisfies $\bar{\eta }=\eta$, where complex conjugation acts on both coefficients and differential factors. Conjugation exchanges the two Dolbeault operators, soThus $\bar{\partial }\eta =0$ forces $\partial \eta =0$ as well. Conversely, if $d\eta =0$, its distinct $(p+1,p)$ and $(p,p+1)$ components both vanish. ThereforeThe assertion remains valid in top degree, when the indicated type components are zero automatically.

For a holomorphic function $f$, $\bar{\partial }f=0$ and $\partial \bar{f}=0$. Writing its real part as $u=(f+\bar{f})/2$ givesHence the real part of a holomorphic function is a pluriharmonic function.

For the local exactness assertion, work on a smaller polydisc whose closure lies in the original one. The allowed one-variable integral is the Cauchy-Green operator $T_j$ in coordinate $z_j$. Applied coefficientwise, it satisfies ${\partial }_{{\bar{z}}_j}{T}_{j}a=a$ on a smaller disc and commutes with differentiation in all other coordinates; the smooth parameter dependence follows by applying the integral to a smooth compactly supported extension in the integration variable. This extension can be made with a cutoff equal to one on the smaller disc.

Begin with a Dolbeault operator closed form $\phi$ and decompose it uniquely as $\phi =d{\bar{z}}^1\wedge \alpha +\beta$, where neither $\alpha$ nor $\beta$ contains $d{\bar{z}}^1$. Set ${\psi }_1=T_1\alpha$. The $d{\bar{z}}^1$ component of $\bar{\partial }{\psi }_1$ is exactly $d{\bar{z}}^1\wedge \alpha$, sohas no $d{\bar{z}}^1$ factor. It is still $\bar{\partial }$-closed. The $d{\bar{z}}^1$ component of $\bar{\partial }{\phi }_1=0$ now shows that every coefficient of ${\phi }_1$ is holomorphic in $z_1$.

Apply the same construction in $z_2$ to remove its antiholomorphic differential factor. The operator $T_2$ preserves holomorphicity in $z_1$, since it commutes with ${\partial }_{{\bar{z}}_1}$. Consequently the second subtraction introduces no previously removed factor. Continue through all $n$ coordinates, shrinking the polydisc when necessary. The final remainder has positive antiholomorphic degree but contains none of the $d{\bar{z}}^j$, and hence is zero. This coordinate elimination for local Dolbeault primitives constructsIf $q>n$, the original form is already zero, so one can take $\psi =0$.

Finally let $\phi$ be a smooth $(0,1)$-form on the complex projective line. Choose $0<r<R$ and the two coordinate neighborhoods $U=\{|z|<R\}$ and $V=\{|z|>r\}\cup \{\infty \}$, the latter expressed in $w=1/z$. They cover the sphere. The one-variable construction, integrating over slightly larger closed discs in each coordinate chart, gives smooth functions $a$ on $U$ and $b$ on $V$ with $\bar{\partial }a=\phi {|}_U$ and $\bar{\partial }b=\phi {|}_V$.

Their difference $h=a-b$ is a holomorphic function on the annulus $r<|z|<R$. Its Laurent series splits asThe first series extends holomorphically throughout $U$, while the second extends throughout $V$, including infinity, where it has value zero. Thus $a-h_{+}$ and $b+h_{-}$ agree on the overlap and glue to a global smooth function $F$. Both corrections are holomorphic, soThis Laurent gluing of Dolbeault primitives on the projective line uses no Hodge theorem.

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.

We prove the Bott-Chern Poincaré lemma on a polydisc $B$. Let $p,q\ge 1$ and $d\alpha =0$, with $\alpha$ of type $(p,q)$. The allowed Dolbeault-Poincaré lemma first gives $\alpha =\bar{\partial }{b}_0$, where $b_0$ has type $(p,q-1)$. Since $\bar{\partial }\partial b_0=-\partial \alpha =0$, repeatedly applying the same lemma constructsZero spaces outside the allowed bidegrees cause no difficulty. The form $\partial b_{q-1}$ has type $(p+q,0)$, is holomorphic because its Dolbeault operator vanishes, and is $\partial$-closed. The holomorphic Poincaré lemma on a polydisc gives a holomorphic form $h$ of type $(p+q-1,0)$ with $\partial h=\partial b_{q-1}$. This holomorphic version follows from the radial homotopy formula: on positive-degree holomorphic forms the ordinary radial Poincaré lemma homotopy preserves holomorphic coefficients. If the displayed form is zero, take $h=0$.

Now $b_{q-1}-h$ is $\partial$-closed. The conjugate Dolbeault-Poincaré lemma, obtained by conjugating the allowed lemma, gives $b_{q-1}-h=\partial c_{q-1}$, since its holomorphic degree is positive. Moving backwards, if the last modified form is $\partial c_j$, thenApply the conjugate Dolbeault-Poincaré lemma again to obtain $b_{j-1}+\bar{\partial }{c}_j=\partial c_{j-1}$. The modification does not change its Dolbeault operator, so this process continues down to $b_0$. In the case $q=1$, subtract $h$ at this last step instead. Finally,Thus $\boxed{H_{BC}^{p,q}(B)=0(p,q\ge 1)}$.

This vanishing fails on general complex manifolds. On the complex projective line, let $\omega$ be its Fubini-Study form. It is a $d$-closed $(1,1)$-form with ${\int }_{{\mathbb{P}}^1}\omega >0$. If it were $\partial \bar{\partial }f$, it would be the exact differential form $d(\bar{\partial }f)$, whose integral is zero by Stokes theorem. Therefore $\boxed{H_{BC}^{1,1}({\mathbb{P}}^1)\ne 0}$.

The assertion requires $p+q>0$. As printed, it also includes $(p,q)=(0,0)$, where the nonzero constant function $1$ is closed but cannot be the exterior derivative of a negative-degree form. The zero constant is harmless; below we prove the intended positive-degree result.

Set $k=p+q>0$, let ${\rho }_t(z)=tz$, and letThe polydisc is preserved by these dilations. The radial homotopy operator isHere ${\iota }_R$ is the interior product of a differential form. For a smooth $k$-form, its pulled-back contraction is $O(t^k)$, so the integral and its coefficient derivatives converge at zero. The identity for a pullback of a differential form along the radial flow and Cartan's magic formula giveIntegrating yields $dH\alpha +Hd\alpha =\alpha -{\rho }_0^{\ast }\alpha$. The final pullback vanishes in positive degree; since $d\alpha =0$, we have $dH\alpha =\alpha$.

Each ${\rho }_t$ is a holomorphic map, so its pullback of a differential form preserves bidegree. Contracting with the $(1,0)$ part of $R$ lowers $p$ by one, and contracting with its $(0,1)$ part lowers $q$ by one. Therefore the primitive has precisely the permitted types:This is a type-preserving refinement of the Poincaré lemma; when one index is zero, the corresponding negative-degree summand is simply absent.

Complex conjugation interchanges the Dolbeault operator and conjugate Dolbeault operator. Thus $\partial \alpha =0$ implies $\bar{\partial }\bar{\alpha }=0$, and $\bar{\alpha }$ has type $(q,p)$. Its antiholomorphic degree is $p\ge 1$, so the allowed Dolbeault-Poincaré lemma on the polydisc gives a form $\gamma \in {\mathcal{A}}^{q,p-1}$ satisfying $\bar{\partial }\gamma =\bar{\alpha }$.

Conjugate this identity and put $\beta =\bar{\gamma }$. This proves the conjugate Dolbeault-Poincaré lemma:

A pure-type $d$-closed form is both $\partial$-closed and $\bar{\partial }$-closed, because these two derivatives have different types. Let $k=p+q$. The smooth Poincaré lemma on the polydisc gives $\eta \in {\mathcal{A}}_{\mathbb{C}}^{k-1}$ with $d\eta =\alpha$. We will modify $\eta$ by exact terms without changing $d\eta$.

Write $\eta ={\sum }_r{\eta }^{r,k-1-r}$. Starting with the smallest holomorphic degree, eliminate every component with $r<p-1$. Once the lower components are zero, the $(r,k-r)$ component of $d\eta$ says $\bar{\partial }{\eta }^{r,k-1-r}=0$, since $r<p$. Its antiholomorphic degree is positive, so the Dolbeault-Poincaré lemma supplies ${\nu }^{r,k-2-r}$ with $\bar{\partial }\nu ={\eta }^{r,k-1-r}$. Replace $\eta$ by $\eta -d\nu$: this removes the component at $r$ and changes only the next holomorphic degree.

Next, work downwards from the largest holomorphic degree and eliminate the components with $r>p$. Once the higher components are zero, the corresponding component of $d\eta$ gives $\partial {\eta }^{r,k-1-r}=0$. By part (b), write this as $\partial {\nu }^{r-1,k-1-r}$ and again subtract $d\nu$. This affects only the next lower holomorphic degree. The two finite procedures leaveThe components of $d\eta =\alpha$ outside type $(p,q)$ now give $\bar{\partial }u=0$ and $\partial v=0$, while its $(p,q)$ component gives $\alpha =\partial u+\bar{\partial }v$.

Since $q\ge 1$, the Dolbeault-Poincaré lemma gives $u=\bar{\partial }a$ with $a$ of type $(p-1,q-1)$. Since $p\ge 1$, part (b) gives $v=\partial b$ of the same type. The anticommutation identity therefore yieldsThis proves the Bott-Chern Poincaré lemma using precisely the permitted primitives: