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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.

A holomorphic line bundle is a complex rank-one bundle locally trivialized over holomorphic coordinate opens, with nowhere-zero holomorphic transition functions. A local holomorphic section has a holomorphic coefficient in a holomorphic local frame. Let $h$ be a smooth positive Hermitian metric on its fibres. A Chern connection is a connection compatible with $h$ whose $(0,1)$ part is the bundle's Dolbeault operator.

Fix the Hermitian-product convention linear in the first variable and a nonvanishing holomorphic local frame $e$. Put $h_e=h(e,e)>0$ and write $\nabla e=Ae$. Compatibility with the holomorphic structure forces $A^{0,1}=0$. Metric compatibility givesso taking its $(1,0)$ part determines $A$ uniquely:Define the connection by this formula. To check it exists globally, change frame to $e^{\prime }=ge$ for a nonzero holomorphic function $g$. Then $h_{e^{\prime }}=|g{|}^2{h}_e$ andwhich is exactly the connection transformation law obtained from $\nabla (ge)=dge+g\nabla e$. Hence the local connections glue. The formula and its conjugate give metric compatibility, and $A^{0,1}=0$ gives the required holomorphic structure. Since either condition already forced this unique formula in every frame, existence and uniqueness are proved.

For a line bundle, the curvature form of a connection is $F={\nabla }^2=dA$, as the scalar one-form satisfies $A\wedge A=0$. The local formula for the Chern connection on a line bundle therefore givesThis formula uses a nowhere-zero local holomorphic section; arbitrary sections with zeros are not frames there. It has type $(1,1)$, is $d$-closed and satisfies ${\bar{F}}_h=-F_h$, so $i{F}_h$ is a real $(1,1)$-form. The transformation calculation makes its frame independence explicit.

For two metrics, their ratio $r=h_1(e,e)/h_2(e,e)$ is a global positive smooth function, since the common frame factors cancel. The curvature difference of two Chern connections isEach curvature is $\bar{\partial }$-closed, and the difference has a global $(1,0)$ primitive for $\bar{\partial }$. ThusThis is equality in Dolbeault cohomology, rather than merely equality in de Rham cohomology; the displayed primitive proves precisely that distinction.

Finally let $s$ be a nowhere-zero smooth section of $\hat{L}$. In this global smooth trivialization write ${\bar{\partial }}_{\hat{L}}s=a\otimes s$ with a global $(0,1)$-form $a$. The holomorphic structure has ${\bar{\partial }}_{\hat{L}}^2=0$. Its Leibniz rule givessince a scalar one-form wedges with itself to zero. The hypothesis $H^{0,1}(X)=0$ makes $a=\bar{\partial }f$ for a global smooth complex function $f$. ThenThe exponential never vanishes. Therefore $e^{-f}s$ is a nowhere-zero holomorphic section, proving smoothly trivial holomorphic line bundles when H01 vanishes constructively, without a compactness assumption.

Fix the input-index conventionExtend the connection on a vector bundle by the graded Leibniz rule, $D(\alpha \otimes s)=d\alpha \otimes s+(-1{)}^{\mathrm{deg}\alpha }\alpha \wedge Ds$. Applying it twice givessoThis is the Cartan curvature equation with input indices. In the more usual coefficient-column convention $A={\theta }^t$, and $(A\wedge A{)}^t=-\theta \wedge \theta$ because the entries have degree one. Thus $F=dA+A\wedge A={\Theta }^t$; the sign difference is entirely conventional.

The Chern connection of a Hermitian vector bundle with holomorphic structure is characterized by metric compatibility and $D^{0,1}={\bar{\partial }}_E$. To construct it, use a holomorphic local frame and let $h_{ij}=⟨e_i,e_j⟩$, with the Hermitian inner product linear in the first variable. The second defining property makes $\theta$ type $(1,0)$. The first givesand therefore the only possible matrix isConversely, this formula has type $(1,0)$ and satisfies both type components of the metric identity, because $h={\overline{h}}^t$. Under a holomorphic frame change $e^{\prime }=ge$, $h^{\prime }=gh{\overline{g}}^t$ and direct differentiation givesThis is precisely the transformation obtained by differentiating $e^{\prime }=ge$. The locally constructed connections glue, proving existence, and the forced formula proves uniqueness of the Chern connection.

The derivative of $h^{-1}$ gives $\partial \theta =\theta \wedge \theta$. Consequently the Chern curvature isThe second assertion is ${\bar{\partial }}^2\theta =0$ in a holomorphic local frame; holomorphic transition functions make it intrinsic. Thus the curvature Dolbeault class is $\Psi =[\Theta ]\in H^1(M,{\Omega }_M^1\otimes \mathrm{End}E)$ by the Dolbeault theorem. Here ${\Omega }^1(\mathrm{End}E)$ denotes this tensor product. We use the identification with bundle-valued Dolbeault forms in which the antiholomorphic exterior factor is placed first, so $\bar{\partial }\theta$ has the stated positive descent sign.

For the stated transitions $e^{(\beta )}=g_{\alpha \beta }{e}^{(\alpha )}$, differentiating the frame relation yieldsTransport the alpha matrix to the beta frame before subtracting it. The difference is the holomorphic sectionOn a triple intersection, $g_{\alpha \gamma }=g_{\beta \gamma }{g}_{\alpha \beta }$, soThis is exactly the Čech cocycle condition after transporting all representatives to the gamma frame.

To identify its class, let $D_{\alpha }$ be the local holomorphic connection making the alpha frame parallel. Then $t_{\alpha }=D-D_{\alpha }$ is a genuine local section of $A^{1,0}(\mathrm{End}E)$, represented by ${\theta }_{\alpha }$, with $\bar{\partial }{t}_{\alpha }=\Theta$. Its Čech coboundary, using $t_{\beta }-t_{\alpha }$, is the intrinsic section represented by ${\sigma }_{\alpha \beta }$ in the beta frame. This local-primitive construction is the Čech–Dolbeault identification, and proves that $({\sigma }_{\alpha \beta })$ maps to $\Psi$. For an arbitrary trivializing cover use its natural map to sheaf cohomology; a Stein refinement gives the usual Čech computation. The resulting object is the Čech cocycle for the curvature Dolbeault class.

If the metric changes, the difference $B=D^{\prime }-D$ is a global smooth $(1,0)$ form with values in $\mathrm{End}E$. The local curvature formulas give ${\Theta }^{\prime }-\Theta =\bar{\partial }B$, so $\Psi$ is metric independent. If holomorphic frames change by $e_{\alpha }^{\prime }=u_{\alpha }{e}_{\alpha }$, the corresponding local tensors $t_{\alpha }^{\prime }-t_{\alpha }$, expressed in the old frame, are $B_{\alpha }=u_{\alpha }^{-1}\partial u_{\alpha }$. They are holomorphic, and their differences change the cocycle by $\delta B$. Hence $\Psi$ depends only on the holomorphic bundle, not on its metric or trivializations.

Apply exterior product and endomorphism composition to curvature to defineThese are closed for the Dolbeault operator by the product rule. Their independence can also be verified without commuting matrices. If ${\Theta }_1-{\Theta }_0=\bar{\partial }B$, thenThis follows by the noncommutative telescoping identity and the even total degree of curvature. The matrix products are interpreted in the fixed frame convention; powers of the degree-two curvature transpose to the corresponding usual curvature powers. The trace contraction is invariant under frame changes, so its class is well defined. This proves metric independence of curvature Dolbeault powers; for $k>\dim M$ the forms and groups in question vanish.

Finally take a smooth unitary frame. Metric compatibility gives ${\theta }^{†}=-\theta$, and the Cartan curvature equation with input indices gives ${\Theta }^{†}=-\Theta$. ThusThe second expression is therefore a real differential form. It is also $d$-closed: differentiating Cartan's equation gives $d\Theta =\theta \wedge \Theta -\Theta \wedge \theta$, and cyclicity of trace cancels the resulting commutator in $d\mathrm{tr}({\Theta }^k)$. On a compact Kähler manifold, the Hodge decomposition theorem for compact Kähler manifolds identifies $H^k(M,{\Omega }_M^k)$ with the $(k,k)$ component of complex de Rham cohomology. HenceIt is $k!$ times the degree-$2k$ component of the Chern character. Reality does not assert integrality of these individual power-sum classes.

The connection difference as an endomorphism-valued one-form follows because the graded Leibniz rule terms cancel:Thus $a=D_1-D$ is a global endomorphism-valued one-form. To define the endomorphism bundle connection, requireThe right side is $C^{\infty }$-linear in $s$, so it is indeed an endomorphism-valued one-form, and replacing $\varphi$ by $f\varphi$ gives $df\otimes \varphi +f\tilde{D}\varphi$. In a frame with connection matrix $A$, its extension to an endomorphism-valued $k$-form $b$ isThe endomorphism-valued exterior product combines the wedge of differential forms with composition in $\mathrm{End}E$. In particular,Expanding $d(A+a)+(A+a)\wedge (A+a)$ proves the curvature difference formula:

The Chern connection is characterized by metric compatibility and $D^{0,1}={\bar{\partial }}_E$. Take Hermitian inner products linear in the first variable. In a holomorphic local frame write $⟨eu,ev⟩=v^{†}Hu$, with $H$ a positive Hermitian matrix. The second condition forces $A^{0,1}=0$, and metric compatibility isIts $(1,0)$ component determines $A$ uniquely:Conversely this formula and its conjugate transpose satisfy the metric equation. Under a holomorphic local frame change, $H^{\prime }=g^{†}Hg$ and $\partial g^{†}=0$, so the displayed formula transforms by $A^{\prime }=g^{-1}Ag+g^{-1}\partial g$. Hence the local matrices patch, proving existence and uniqueness.

Differentiating $H^{-1}$ givesTherefore the $(2,0)$ vector-bundle curvature is zero. There is no $(0,2)$ component, andThe last equality is ${\bar{\partial }}^{2}A=0$ in holomorphic local frames, so is intrinsic. This is Dolbeault closedness of Chern curvature. The induced endomorphism bundle connection is itself the Chern connection for the Hilbert-Schmidt metric on $\mathrm{End}E$: the tensor product of a metric-compatible connection and its dual preserves the induced metric, and its $(0,1)$ part is the induced holomorphic Dolbeault operator. These two properties and uniqueness suffice; no further matrix calculation is needed.

For two metrics, both Chern connections have the same $(0,1)$ part, so $a\in A^{1,0}(\mathrm{End}E)$. The $(1,1)$ part of the curvature difference formula is just ${\bar{\partial }}_{\mathrm{End}E}a$; the other two terms have type $(2,0)$, which must cancel because both vector-bundle curvatures have type $(1,1)$. ThusBy Dolbeault theorem, this proves that the curvature Dolbeault classis independent of the metric.

Endomorphism composition and wedge product of differential forms define the powers in $A^{k,k}(\mathrm{End}E)$, rather than in a tensor power of the endomorphism bundle. The Dolbeault operator is a derivation, so $\bar{\partial }({\Theta }^k)=0$. The noncommutative telescoping identity gives the stronger explicit independence statementAll powers of vector-bundle curvature have even total degree, so the derivation introduces no extra signs before $a$. Consequentlyis metric independent. For $k>n$, the form space and the resulting class are zero. This proves metric independence of curvature Dolbeault powers without assuming that endomorphism-valued forms commute.

Let ${\mathcal{A}}^{p,q}(X)$ be the space of smooth complex differential forms of type $(p,q)$. The Dolbeault operator satisfies ${\bar{\partial }}^2=0$, and the Dolbeault cohomology isFor $q=0$ the denominator is zero; forms outside the dimension range are zero.

A sequence of sheaves is an exact sequence of sheaves precisely when its sequence of stalks at every point is exact. Here this means the first map is injective on stalks, the last is surjective on stalks, and the image equals the kernel at the middle stalk. In particular, a section of the last sheaf need only have local lifts; surjectivity on all global sections is not required.

The associated long exact sequence in sheaf cohomology of Čech cohomology, understood in the direct limit over open covers, isThe degree-zero groups are global sections. The connecting map is obtained by locally lifting a Čech cocycle to the middle sheaf and taking its Čech coboundary, which takes values in the first sheaf. Different lifts change it by a Čech coboundary.

Write $L\alpha =\omega \wedge \alpha$, the Lefschetz operator of a Kähler manifold. The metric and volume form define the $L^2$ inner product on smooth complex forms, and $\Lambda =L^{\ast }$ is its formal adjoint. Similarly, ${\bar{\partial }}^{\ast }$ is the formal adjoint of the Dolbeault operator, characterized by $⟨\bar{\partial }\beta ,\alpha ⟩=⟨\beta ,{\bar{\partial }}^{\ast }\alpha ⟩$. The Dolbeault Laplacian isOn the compact manifold without boundary, integration by parts givesIf the Laplacian vanishes, both terms are zero. Conversely, if both operators annihilate $\alpha$, the defining formula annihilates it. Thus harmonicity is equivalent to being both $\bar{\partial }$-closed and ${\bar{\partial }}^{\ast }$-closed.

Because $\omega$ is closed and has type $(1,1)$, $\bar{\partial }\omega =0$ and $[L,\bar{\partial }]=0$. The supplied identity from the Kähler identities gives $[L,{\bar{\partial }}^{\ast }]=-i\partial$. With ordinary commutators for the even-degree operator $L$,The last identity follows from $d^2=0$. This also proves that the Lefschetz operator preserves harmonic forms.

For the cohomology map one can work directly with forms: $\bar{\partial }({\omega }^k\wedge \alpha )={\omega }^k\wedge \bar{\partial }\alpha$, since $\omega$ has even degree. It takes closed forms to closed forms and exact forms to exact forms. Therefore the $k$th power of $L$ inducesThe bidegree is $(p+k,q+k)$, including the zero groups outside the dimension range. No isomorphism claim is needed here.

Finally put $\theta =-i\partial f$, so $\bar{\partial }\theta =i\partial \bar{\partial }f={\omega }^{\prime }-\omega$. LetBoth Kähler forms are $\bar{\partial }$-closed, so $\bar{\partial }{T}_k=0$. For a closed representative $\alpha$,This primitive has type $(p+k,q+k-1)$; the degree-one sign produces no additional term because $T_k\wedge \alpha$ is closed. HenceThis is the dependence of Lefschetz maps on the Dolbeault class.

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 printed assertion for every cohomological degree is false. The valid conclusion supplied by the stated Hartogs extension theorem is the degree-zero isomorphism. Indeed, around each removed point choose a small coordinate ball on which the holomorphic vector bundle $E$ is trivial. A holomorphic section on the punctured ball has finitely many holomorphic coefficient functions, each of which extends by Hartogs extension theorem when the complex dimension is at least two. Uniqueness of holomorphic extension makes these local extensions agree with the original section on overlaps, givingThis also identifies the direct image sheaf $j_{\ast }\mathcal{O}(E{|}_{X_0})$ with $\mathcal{O}(E)$ for the inclusion $j:X_0\hookrightarrow X$. It does not identify higher sheaf cohomology.

Here is an explicit higher-degree obstruction. Take $X={\mathbb{P}}^2$, remove $[0:0:1]$, and take the trivial line bundle. Cover $X_0$ by $U=\{x\ne 0\}$ and $V=\{y\ne 0\}$. On $U$ use coordinates $t=y/x$, $s=z/x$; on $V$ use $u=x/y=1/t$, $v=z/y=s/t$. Both charts are ${\mathbb{C}}^2$, and the overlap is ${\mathbb{C}}^{\ast }\times \mathbb{C}$. For every integer $m\ge 2$, the holomorphic functiondefines a Čech cocycle. It cannot be a Čech coboundary $b(1/t,s/t)-a(t,s)$ with $a,b$ entire on their charts. Taking the coefficient of $s^m$, the term from $a$ has only nonnegative powers of $t$, whereas the term from $b$ has the form $t^{-m}{b}_m(1/t)$ and only powers at most $-m$. Neither can provide the coefficient $t^{-1}$. Uniqueness of Laurent series proves the contradiction. The same argument applied to each fibre degree proves that all the classes $[g_m]$, $m\ge 2$, are linearly independent.

These classes remain nonzero in Čech cohomology of $X_0$, rather than merely of this cover. The low-degree Mayer-Vietoris sequence for sheaf cohomology injects the quotient of overlap sections by the two chart-section groups into $H^1(X_0,\mathcal{O})$. Concretely, a partition of unity gives smooth $f_U,f_V$ with $f_V-f_U=g$; their common Dolbeault operator defines a global $(0,1)$-form. If that form were $\bar{\partial }$-exact, subtracting its global smooth primitive would make $f_U,f_V$ holomorphic and split $g$. The nonsplitting just proved therefore gives the same obstruction in Dolbeault cohomology, and the Dolbeault theorem identifies it with sheaf cohomology. Thus $H^1(X_0,\mathcal{O})$ is infinite dimensional. In contrast, $H^1({\mathbb{P}}^2,\mathcal{O})$ is finite dimensional by compact Dolbeault Hodge decomposition for the Fubini-Study form. Hence the two degree-one groups cannot be isomorphic, disproving the printed all-degree request already in complex dimension two.

In complex dimension one, even the degree-zero conclusion fails. Take $X={\mathbb{P}}^1$, remove its point at infinity and use the trivial line bundle. Holomorphic functions on compact connected ${\mathbb{P}}^1$ are constant by the maximum modulus principle, whereas $X_0\cong \mathbb{C}$ has nonconstant entire functions such as $z$. Therefore the restriction map is not onto even in degree zero.

The Chern connection of a Hermitian metric on a holomorphic vector bundle is the unique connection on a vector bundle compatible with that metric and whose $(0,1)$ part is the bundle's Dolbeault operator ${\bar{\partial }}_E$. Fix a holomorphic local frame and column coefficients for sections. Write the metric as $h(s,t)={\bar{s}}^{T}Ht$, conjugate-linear in the first argument, and write the connection as $Ds=ds+As$. The condition $D^{0,1}={\bar{\partial }}_E$ forces $A$ to have type $(1,0)$. Metric compatibility requireswhere the dagger conjugates the differential-form coefficients as well as transposing the matrix. Taking the $(1,0)$ part gives $\boxed{A=H^{-1}\partial H}$. Its conjugate-transpose supplies the $(0,1)$ metric equation because $H$ is Hermitian. This proves uniqueness and local existence.

Under a holomorphic change of frame $e^{\prime }=eg$, the metric matrix becomes $H^{\prime }=g^{†}Hg$. The local formula for the Chern connection on a vector bundle then givesThis is precisely the transformation rule for a connection on a vector bundle, so the local connections glue and establish global existence. No Kähler hypothesis is needed for this part.

Extend $D$ to vector-bundle-valued differential forms by the graded Leibniz rule. The curvature form of a connection is the tensorial square $F_h=D^2$, acting by exterior multiplication. In the chosen frame,Indeed $\partial A+A\wedge A=0$ by differentiating $H^{-1}H=I$. It follows that $F_h$ has type $(1,1)$; the gauge change is $F_h^{\prime }=g^{-1}{F}_{h}g$. Thus it is a global smooth two-form with values in $\mathrm{End}(E)$, namely an element of ${\mathcal{A}}^2(\mathrm{End}(E))$.

For a holomorphic vector bundle $E$, the sheaf of vector-bundle-valued differential forms iswhere sections are smooth and restrictions are the usual ones. In a holomorphic local frame of $E$, the Dolbeault operator ${\bar{\partial }}_E$ acts coefficientwise and satisfies ${\bar{\partial }}_E^2=0$. Its Dolbeault cohomology with values in a holomorphic vector bundle is the cohomology of the global section complex ${\mathcal{A}}^{p,\bullet }(X,E)$.

The bundle-valued Dolbeault theorem identifies this with sheaf cohomology:Here ${\Omega }_X^p$ is the sheaf of holomorphic differential forms, and $\mathcal{O}(E)$ is the sheaf of holomorphic sections of a vector bundle. Equivalently, the complexis a fine sheaf resolution, by the local Dolbeault-Poincaré lemma and smooth partitions of unity. In particular $p=0$ computes $H^q(X,\mathcal{O}(E))$; for $p>0$ the holomorphic differential-form factor must be retained.

Choose a positive Hermitian metric on $L$. Its curvature form $\omega =i{F}_h$ is a Kähler form on the compact complex manifold of dimension one. Part (c) gives a threshold $k_0$ with $H^1(X,\mathcal{O}(L^k))=0$ for every $k\ge k_0$. We use this to prescribe a finite principal part of a meromorphic section.

Fix a smooth cutoff $\chi$ supported inside the given coordinate chart and equal to one near $x_0$. Let $P(z)={\sum }_{j=-r}^{-1}{a}_j{z}^j$. On the punctured manifold define $t_k=\chi P(z){\zeta }^k$ inside the chart, extended by zero outside. Its Dolbeault operatoris smooth globally: it vanishes near the pole and near the boundary of the chart. It is ${\bar{\partial }}_{L^k}$-closed, either by the square-zero identity or because there are no $(0,2)$-forms in complex dimension one. The vanishing of $H^1(X,\mathcal{O}(L^k))$, together with the Dolbeault theorem, therefore gives a global smooth section $u_k$ with ${\bar{\partial }}_{L^k}{u}_k={\eta }_k$.

The section $s_k=t_k-u_k$ is holomorphic on $X\setminus \{x_0\}$. Near $x_0$, ${\eta }_k=0$, so $u_k=f_k(z){\zeta }^k$ with $f_k$ holomorphic through zero. Its Taylor series then giveswhere the second series converges near zero and its coefficients are those of $-f_k$. The threshold is determined by the fixed bundle $L$ and $X$, and is independent of the prescribed coefficients. In fact the same threshold works for every finite pole order $r$, since the Dolbeault cohomology obstruction is the same $H^1(X,L^k)$ for every such cutoff construction.

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:

Complexifying the cotangent bundle of an almost complex manifold gives the $\pm i$ eigenspace decomposition $T^{\ast }X\otimes \mathbb{C}=T^{1,0\ast }X\oplus T^{0,1\ast }X$. DefineFor a form $\alpha$ of type $(p,q)$, the Dolbeault operator is the type-$(p,q+1)$ projection of $d\alpha$:

For a holomorphic line bundle on a complex curve, a meromorphic section modulo sections holomorphic through a point is its principal part. In a local coordinate and holomorphic local frame it is represented by finitely many negative powers. A vanishing $H^1(X,\mathcal{O}(E))$ permits prescribed principal parts, by a cutoff followed by solving a global Dolbeault operator equation.

On a complex manifold, ${\Omega }_X^p$ is the sheaf of holomorphic degree-$p$ forms. It is the kernel of the Dolbeault operator on ${\mathcal{A}}^{p,0}$. The Dolbeault theorem computes its sheaf cohomology using smooth forms.