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The Dolbeault theorem identifies this group with first Čech cohomology of the trivial holomorphic line bundle. On the two standard affine charts, a holomorphic overlap function has a Laurent series. Its nonnegative powers extend to the chart at zero, and its negative powers extend to the chart at infinity. Every overlap cocycle is therefore a coboundary. Consequently the Dolbeault operator from functions to $(0,1)$-forms is surjective.

This is the cohomology of ${\mathcal{A}}^{p,\bullet }(X,E)$ with the bundle Dolbeault operator. The Dolbeault theorem identifies it with $H^q(X,{\Omega }_X^p\otimes \mathcal{O}(E))$. For compact Hermitian $X$, Dolbeault Hodge decomposition gives unique harmonic representatives.

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.

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.

The Dolbeault theorem identifieswhere ${\Omega }_X^p$ is the sheaf of holomorphic differential forms of degree $p$.

The local analytic ingredient is the Dolbeault-Poincaré lemma: a $\bar{\partial }$-closed smooth $(p,q)$-form with $q>0$ is locally $\bar{\partial }$-exact. Here is a local proof. On a polydisc, the one-variable Cauchy-Green operator in coordinate $z_j$ iswith a smooth cutoff supported in the coordinate disc and equal to one on a smaller disc. The fundamental-solution identity ${\partial }_{\bar{z}}(1/(\pi z))={\delta }_0$ gives ${\partial }_{{\bar{z}}_j}{T}_{j}g=g$ on that smaller disc. The operator is smooth in the parameters and commutes with derivatives in the other coordinates.

For a $(0,q)$-form, write $\alpha =d{\bar{z}}_m\wedge \beta +\gamma$, with neither $\beta$ nor $\gamma$ containing $d{\bar{z}}_m$. Subtract $\bar{\partial }(T_m\beta )$. The remainder contains no $d{\bar{z}}_m$, remains closed, and its coefficients are holomorphic in $z_m$. Repeat with $z_{m-1}$ and then the other coordinates, shrinking discs as needed. Each new integral preserves the holomorphic dependence already achieved. At the end the closed remainder has positive antiholomorphic degree but contains no antiholomorphic differential, so it is zero. Factoring out each holomorphic basis form $d{z}_I$ gives the same assertion for $(p,q)$-forms, with the fixed degree sign included in the primitive. At degree zero, the kernel of $\bar{\partial }$ consists precisely of holomorphic coefficients.

Thus the following is an exact Dolbeault resolution of holomorphic differential forms of sheaves:Each sheaf of smooth forms is a fine sheaf: multiplication by a smooth partition of unity supplies endomorphisms supported in its open cover. Fine sheaves on a paracompact manifold have zero positive sheaf cohomology; the Čech contraction sums a cochain against the partition of unity, giving $\delta h+h\delta =I$ in positive degrees. The general acyclic resolution theorem therefore computes ${\check{H}}^q(X,{\Omega }_X^p)$ as the cohomology of the global-section complex above. That complex is exactly the Dolbeault resolution of holomorphic differential forms, proving the claimed isomorphism. The sheaf-cohomology/Čech comparison and the acyclic-resolution principle are the stated general Čech properties used here.

Cover the product by $U_0={\mathbb{C}}_z\times {\mathbb{C}}_w$ and $U_{\infty }={\mathbb{C}}_{\zeta }\times {\mathbb{C}}_w$, with $\zeta =1/z$ on their intersection ${\mathbb{C}}^{\ast }\times \mathbb{C}$. The permitted vanishing and the Dolbeault theorem make this an acyclic cover for both $\mathcal{O}$ and ${\Omega }^2$. The acyclic cover theorem lets its two-term Čech complex compute the cohomology. In particular, all groups with $q\ge 2$ vanish.

For $p=0$, a global holomorphic function is constant on each compact complex projective line fibre, by the maximum modulus principle. Its remaining dependence on $w$ is entire. Hence $H_{\bar{\partial }}^{0,0}=\mathcal{O}(\mathbb{C})$. For the first Čech group, a function on the intersection has a Laurent seriesEach coefficient is entire in $w$ by its Cauchy integral formula. The nonnegative powers extend to $U_0$, and the negative powers extend to $U_{\infty }$ in coordinate $\zeta$. These two series converge locally uniformly with the parameter $w$, by the usual Laurent estimates on compact parameter sets. Thus every intersection function is a Čech coboundary and $H_{\bar{\partial }}^{0,1}=0$.

For $p=2$, write a two-form on the intersection as $F(z,w)dz\wedge dw$. The other chart hasForms extending from $U_0$ have coefficient powers $j\ge 0$; those extending from $U_{\infty }$ have powers $j\le -2$. A globally defined two-form would need to have both types of expansion, so it is zero. In the first Čech quotient, precisely the $z^{-1}$ term remains, and its coefficient is an arbitrary entire function of $w$. ConsequentlyThe second nonzero group is represented in Čech cohomology by $a(w)z^{-1}dz\wedge dw$. This explicit residue description is the Dolbeault cohomology of the projective line times the affine line; it also identifies that group naturally with the holomorphic one-forms on the affine factor.

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.

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.

The Kähler form and the Hermitian metric give the $L^2$ inner product on vector-bundle-valued differential forms, using volume ${\omega }^n/n!$. Define the formal adjoint ${\bar{\partial }}_E^{\ast }$ and the elliptic, self-adjoint, nonnegative Dolbeault LaplacianIts harmonic space isThe equality follows from $⟨{\Delta }_E^{\prime \prime }\alpha ,\alpha ⟩=\Vert {\bar{\partial }}_E\alpha {\Vert }^2+\Vert {\bar{\partial }}_E^{\ast }\alpha {\Vert }^2$. On compact $X$, the bundle-valued Dolbeault Hodge decomposition states that this space is finite dimensional and thatThe sum is orthogonal for the $L^2$ inner product and all summands here consist of smooth forms. Every Dolbeault cohomology class has a unique harmonic representative, giving $H_{\bar{\partial }}^{p,q}(X,E)\cong {\mathcal{H}}^{p,q}(X,E)$ and, by the Dolbeault theorem, the corresponding sheaf cohomology isomorphism. This is a decomposition for ${\bar{\partial }}_E$, whose square is zero; it does not require the full Chern connection to be flat.

Take the threshold $k_0\ge 1$ in the stated spectral gap for powers of a positive line bundle. For $q\ge 1$ and $k\ge k_0$, it givesA harmonic form must therefore be zero. The Dolbeault Hodge decomposition and the Dolbeault theorem identify its zero harmonic space with $H^q(X,\mathcal{O}(L^k))$. Hence $\boxed{H^q(X,L^k)=0(q\ge 1,k\ge k_0)}$. For $q>{\dim }_{\mathbb{C}}X$ the form spaces already vanish. The gap controls the lowest eigenvalue including zero, so it rules out harmonic forms rather than just controlling the positive spectrum.

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.

Evaluation is surjective as a sheaf morphism: near $p$, any prescribed value is supplied by a constant holomorphic function. We therefore have a short exact sequence of sheaveswhere ${\mathcal{I}}_p$ is the point ideal sheaf and ${\mathcal{O}}_X$ the sheaf of holomorphic functions. By part (c), the skyscraper sheaf is flasque, so $H^0(X,{\mathbb{C}}_p)=\mathbb{C}$ and $H^i(X,{\mathbb{C}}_p)=0$ for $i>0$.

On the connected compact elliptic curve, the maximum modulus principle makes every global holomorphic function constant. Consequently $H^0(X,{\mathcal{O}}_X)=\mathbb{C}$, and evaluation on global sections is an isomorphism. Also $H^1(X,{\mathcal{O}}_X)\cong \mathbb{C}$: by Serre duality for compact complex manifolds, its dual is $H^0(X,K_X)$, generated by the nowhere-zero form $dz$ descended from $\mathbb{C}$. Higher cohomology of ${\mathcal{O}}_X$ vanishes by the Dolbeault theorem in complex dimension one.

The long exact sequence in sheaf cohomology thus beginsIt follows that

The Dolbeault theorem identifies $H^2(M,{\mathcal{O}}_M)$ with $H_{\bar{\partial }}^{0,2}(M)$. It vanishes by hypothesis, and complex conjugation in the Hodge decomposition theorem for compact Kähler manifolds makes $H^{2,0}$ vanish too. Thus every real degree-two cohomology class has type $(1,1)$.

Let $\omega$ be a Kähler form. Choose a rational cohomology class $\eta$ sufficiently close to $[\omega ]$. More concretely, using harmonic differential forms as representatives for a fixed Kähler metric, the harmonic form representing $\eta$ is close to $\omega$ in every smooth norm: harmonic differential forms constitute a finite-dimensional vector space. It is a real closed $(1,1)$ form, and is positive if sufficiently close to $\omega$, by compactness. This is the openness of the Kähler cone. Multiplying by a positive integer gives an integral Kähler class $\kappa =N\eta$.

The holomorphic exponential sequence contains the cohomology segmentChoose an integral lift of $\kappa$. Exactness gives a holomorphic line bundle $\mathcal{L}$ having that First Chern class. We must ensure that it has a positive metric, rather than merely a positive representative of its class.

Choose any Hermitian metric $h_0$ and let ${\theta }_0$ be its normalized curvature form of a Hermitian holomorphic line bundle. If $\theta$ is the positive form representing $\kappa$, then $\theta -{\theta }_0$ is an exact real $(1,1)$ form. The ddbar lemma gives a real smooth function $u$ withReplace $h_0$ by $h=e^{-u}{h}_0$. Its normalized curvature is $\theta$, so $\mathcal{L}$ is a positive holomorphic line bundle. The Kodaira embedding theorem now applies: sufficiently many sections of a sufficiently high tensor power define the desired embedding. Therefore

For the Calabi-Yau threefold in the stated convention, the Hodge decomposition theorem for compact Kähler manifolds givesIn particular, the Dolbeault theorem yields $H^1(M,{\mathcal{O}}_M)=0$. By Serre duality for compact complex manifolds in dimension three,The canonical bundle is trivial, so the group on the right equals $H^1(M,{\mathcal{O}}_M)=0$. Thus $H^2(M,{\mathcal{O}}_M)=0$, and part (c) gives

Use the usual connected-manifold convention for this assertion. Suppose, for contradiction, that $H^2(M,{\mathcal{O}}_M)=0$. The Dolbeault theorem gives $H^{0,2}=0$, and the Hodge decomposition theorem for compact Kähler manifolds and its conjugation symmetry give $H^{2,0}=0$. Hence every real class in $H^2(M,\mathbb{R})$ has a real harmonic differential form of type $(1,1)$ as its representative.

The intersection form is $Q([u],[v])={\int }_{M}u\wedge v$. Its value on the Kähler class is positive: $Q([\omega ],[\omega ])={\int }_M{\omega }^2>0$. Every real harmonic $(1,1)$ form splits uniquely as $c\omega +u_0$, with $u_0$ primitive and $c$ constant. Indeed, $\Lambda u$ is a harmonic function, hence constant by connectedness, and $\Lambda \omega =2$; subtracting $c\omega$ kills $\Lambda u$, and part (b) makes $u_0$ primitive.

Part (c) gives $\ast u_0=-u_0$, while $\omega \wedge u_0=0$. ThusThis proves the Hodge index theorem for compact Kähler surfaces in this case: $Q$ has signature $(1,b_2-1)$.

On the alleged two-dimensional totally isotropic subspace $V$, the linear functional $v\mapsto Q(v,[\omega ])$ has a nonzero vector in its kernel. Its harmonic differential form representative is primitive, so the preceding negative-definiteness statement gives $Q(v,v)<0$, contradicting the vanishing of $Q$ on $V$. ThereforeConnectedness is essential if one's definition of a manifold allows disconnected spaces. For $M=Y\bigsqcup Y$, with $Y={\mathbb{CP}}^1\times {\mathbb{CP}}^1$, one has $H^2(M,{\mathcal{O}}_M)=0$. Take on each component the pullback of the degree-two class from its first factor. Each class has square zero, and their mutual pairing is zero because their supports lie on different components. Their span is then a two-dimensional totally isotropic subspace, disproving the unqualified disconnected version.

Let ${\Omega }^p$ be the sheaf of holomorphic $p$-forms and ${\mathcal{A}}^{p,q}$ the sheaf of smooth $(p,q)$-forms. The Dolbeault-Poincaré lemma makesan exact resolution by fine sheaves, which are acyclic for global sections. Its global cochain complex computes Dolbeault cohomology, proving the Dolbeault theorem

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.

For a positive Hermitian holomorphic line bundle on compact $X$, using its curvature as the Kähler form, a linear lower bound in $k$ for the full Dolbeault Laplacian on positive antiholomorphic degrees rules out harmonic forms for sufficiently large $k$. The Dolbeault theorem and Dolbeault Hodge decomposition then imply $H^{q>0}(X,L^k)=0$.