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On a complex manifold, holomorphic coordinates give the decomposition $d=\partial +\bar{\partial }$, withFor $\alpha ={\sum }_{I,J}{a}_{I\bar{J}}d{z}^I\wedge d{\bar{z}}^J$, the conjugate Dolbeault operator $\partial$ and the Dolbeault operator $\bar{\partial }$ are explicitlyThese are the type components of the intrinsic exterior derivative, so are independent of coordinates. The identity $d^2=0$, separated by type, gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$. A holomorphic p-form on a complex manifold is a $(p,0)$-form whose coefficients are holomorphic functions; equivalently it is a $(p,0)$-form killed by the Dolbeault operator.

A rank-$r$ holomorphic vector bundle has local trivializations $E{|}_{U_i}\cong U_i\times {\mathbb{C}}^r$ whose changes are $(x,v)\mapsto (x,g_{ij}(x)v)$, with $g_{ij}:U_i\cap U_j\to {\mathrm{GL}}_r(\mathbb{C})$ holomorphic. These transition functions of a vector bundle satisfy $g_{ii}=I$ and $g_{ij}{g}_{jk}=g_{ik}$. Equivalently, for local frames we use the convention $e_j=e_i{g}_{ij}$. A holomorphic section is a section having holomorphic coordinate functions in every such holomorphic local frame; the condition is unchanged by the holomorphic invertible transitions.

The holomorphic cotangent bundle has local frames $d{z}^1,\dots ,d{z}^n$. Under another holomorphic coordinate system $w$, the chain rule givesThe Jacobian matrix is holomorphic and invertible, with holomorphic inverse supplied by the inverse coordinate change. These are precisely holomorphic bundle transitions. A section written as ${\sum }_b{f}_b(z)d{z}^b$ is holomorphic exactly when every $f_b$ is holomorphic, sothe space of holomorphic one-forms. In particular, being a holomorphic section is not the same as being a $d$-closed one-form on an arbitrary complex manifold.

The complex tautological line bundle isOn $U_i=\{Z_i\ne 0\}$ use the nowhere-zero frame $s_i([Z])=Z/Z_i$. Its components are holomorphic affine coordinates and the constant one in position $i$. On overlapsso the frame changes are nowhere-zero holomorphic functions. This proves that $\mathcal{O}(-1)$ is a holomorphic line bundle, and its inclusion into the trivial bundle is a holomorphic bundle map.

A global holomorphic section composed with that inclusion has $n+1$ component holomorphic functions on compact connected Complex projective space. Each is constant: its absolute value attains a maximum; in a coordinate ball the one-variable maximum modulus principle along complex lines makes it locally constant, and connectedness propagates that constant globally. Thus the section is a fixed vector $v\in {\mathbb{C}}^{n+1}$ lying in every fibre, namely every line $\mathbb{C}Z$. For $n\ge 1$ the intersection of all these lines is zero. HenceThe positive-dimension convention is necessary: ${\mathbb{CP}}^0$ is one point and this section space is $\mathbb{C}$.

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 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.

The complex structure splits the complexified cotangent bundle into its $(1,0)$ and $(0,1)$ parts. Thus the space of smooth differential forms of type (p, q) isHere $\Gamma$ denotes smooth sections of a vector bundle; exterior multiplication identifies the displayed bundle with a subbundle of the complexified exterior algebra. In holomorphic coordinates, such a differential form of type (p, q) has the expressionwhere the coefficients are smooth and both multi-indices are increasing. Negative bidegrees, or bidegrees exceeding the complex dimension, give the zero space.

On a complex manifold, the exterior derivative has just two type components, $d=\partial +\bar{\partial }$. The conjugate Dolbeault operator and the Dolbeault operator are, respectively,Keeping the differentiating factor at the front fixes the signs. These definitions are intrinsic because holomorphic coordinate changes preserve type. The two operators are the type projections of the exterior derivative, raising $p$ and $q$, respectively.

For a holomorphic map $f:M\to N$, its differential is a complex-linear map, so the pullback of a differential form sends $(1,0)$ covectors to $(1,0)$ covectors and $(0,1)$ covectors to $(0,1)$ covectors. In coordinates $w$ on $N$,The pullback of a differential form commutes with the wedge product of differential forms, so it preserves the counts of these factors. ConsequentlyIt also commutes separately with the Dolbeault operator and the conjugate Dolbeault operator, because it commutes with $d$ and preserves type.

If $M$ has complex dimension $n$, its complexified cotangent bundle splits into the $(1,0)$ and $(0,1)$ cotangent bundles. The space of global smooth differential forms of type (p, q) isIn holomorphic coordinates, its elements have the form $\alpha ={\sum }_{I,J}{a}_{I,J}d{z}^I\wedge d{\bar{z}}^J$, where the coefficients are complex-valued smooth functions, $|I|=p$, and $|J|=q$.

The exterior derivative decomposes as $d=\partial +\bar{\partial }$. The conjugate Dolbeault operator $\partial$ and Dolbeault operator $\bar{\partial }$ are its type components:They have bidegrees $(1,0)$ and $(0,1)$ respectively. Keeping the displayed wedge order avoids hiding a sign; moving $d{\bar{z}}_k$ past $d{z}^I$ introduces $(-1{)}^p$. Decomposing $d^2=0$ by type gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$.

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: