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The smooth exponential sequence isLocal logarithms make it exact as a sequence of sheaves. A line bundle has a transition class in $H^1(M,{\mathcal{C}}^{\infty }({\mathbb{C}}^{\times }))$; its image under the connecting map defines the First Chern class. For a holomorphic line bundle the holomorphic exponential sequence gives the same class by naturality.

Here is the explicit Čech-de Rham curvature descent. On a good cover, write $e_j=e_i{g}_{ij}$ and choose logarithms $g_{ij}=\exp (2\pi i{f}_{ij})$. The integer cocyclerepresents $c_1(E)$. By the proved frame formula,Put $B_i=-A_i/(2\pi i)$ and $F=i\Theta /(2\pi )=d{B}_i$. Then $\delta B=-df$ and $\delta f=c$. In the Čech-de Rham double complex, with total differential $D_{\mathrm{tot}}=\delta +(-1{)}^{p}d$ on Čech degree $p$,Thus the global closed differential form and the integer cocycle represent the same class under de Rham theorem:This proves integrality, including the sign and normalization. It identifies the image of the integral class; vector-bundle curvature alone cannot recover torsion classes lost in passage to complex coefficients.

Two connections on the same line bundle differ by a global scalar one-form $a$. Their vector-bundle curvatures satisfy ${\Theta }_1-\Theta =da$, so the normalized representatives differ by an exact differential form. Hence the class is independent of the connection.

Use the integral normalization of the Fubini-Study form. On the chart $Z_j\ne 0$, define the integrally normalized Fubini-Study form byThe potentials on overlaps differ by the logarithm of the squared modulus of a nowhere-zero holomorphic function, whose $\partial \bar{\partial }$ is zero. Thus the forms patch and are closed. The dual of the tautological metric on $\mathcal{O}(1)$ has local squared norm $h=(1+\sum |z_k{|}^2{)}^{-1}$. Its Chern connection has $A=\partial \log h$ and vector-bundle curvature $\Theta =\bar{\partial }\partial \log h=\partial \bar{\partial }\log (1+\sum |z_k{|}^2)$. Therefore $[{\omega }_{\mathrm{int}}]=c_1(\mathcal{O}(1))$, proving the requested integrality. With the unnormalized convention ${\omega }_{FS}=i\partial \bar{\partial }\log (1+\sum |z_k{|}^2)$, this is ${\omega }_{\mathrm{int}}={\omega }_{FS}/(2\pi )$.

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 Hermitian metric on cotangent vectors is the dual metric, and its exterior-power metric is defined on decomposable vectors by the determinant of their pairwise inner products. Tensor this with the metric on $E$. In a unitary coframe and unitary frame of $E$, the forms obtained by wedging distinct holomorphic and antiholomorphic coframe elements and tensoring with a frame vector form an orthonormal basis. Let $dV={\omega }^n/n!$, and use inner products linear in the first variable.

The bundle-valued conjugate-linear Hodge star is characterized by contraction of the $E,E^{\ast }$ coefficients inIt is conjugate-linear, maps type $(p,q)$ to type $(n-p,n-q)$, and uses the metric identification of $E$ with its dual. For total degree $k=p+q$, the ordinary star-square rule giveswhere $E^{\ast \ast }$ is identified with $E$. The conjugate-linear convention is essential for the bidegrees specified here.

If $\psi$ has degree $k-2$ and $\eta$ degree $k$, the even degree of the real form $\omega$ givesSince ${\ast }_E\Lambda \eta =L{\ast }_E\eta$ for $\Lambda \eta =(-1{)}^k{\ast }_{E^{\ast }}L{\ast }_E\eta$, it follows that $(L\psi ,\eta )=(\psi ,\Lambda \eta )$ pointwise. Thus $\Lambda$ is the adjoint Lefschetz operator.

For holomorphic $E$, ${\bar{\partial }}_E$ differentiates coefficients in holomorphic local frames and uses the graded Leibniz rule on forms. Its bundle-valued Dolbeault adjoint is the formal differential operatorIt lowers the antiholomorphic degree by one. To verify this on compact $M$, define $⟨\psi ,\eta {⟩}_{L^2}={\int }_M(\psi ,\eta )dV$. For $\psi \in A^{p,q-1}(E)$ and $\eta \in A^{p,q}(E)$, Stokes applied to the scalar $(n,n-1)$-form $\psi \wedge {\ast }_E\eta$ yieldsThe star-square identity converts the right side into $⟨\psi ,{\bar{\partial }}_E^{\ast }\eta {⟩}_{L^2}$. Compactness and absence of boundary remove boundary terms. This establishes the asserted global adjoint relation on smooth sections.

The Dolbeault Laplacian ${\Delta }_{\bar{\partial }}=B{B}^{\ast }+B^{\ast }B$, with $B={\bar{\partial }}_E$, is formally self-adjoint. Consequently the adjoint of a commutator satisfiesFor the $(1,0)$ part $D^{\prime }$ of the Chern connection, the explicit corresponding adjoint iswhere $D_{E^{\ast }}^{\prime }$ is the $(1,0)$ part of the dual Chern connection. This is the same integration-by-parts formula in holomorphic degree.

Now assume the metric is Kähler. Since $\bar{\partial }\omega =0$, $B$ commutes with $L$. Taking adjoints provesThe scalar Kähler identities needed areThey extend to the bundle-valued Kähler identitiesHere is a local justification of the extension. At any chosen point, take holomorphic normal coordinates for the Kähler metric and a holomorphic local frame of $E$ with $H=I$ and $\partial H=0$ at that point. This is a Chern connection in a normal holomorphic frame. Such a frame is obtained first by normalizing the metric matrix at the point and then prescribing the holomorphic local frame change's first derivatives to cancel $A=H^{-1}\partial H$. The first-order operators and their formal adjoints at the point are then the scalar operators acting componentwise. The scalar identities therefore give the two displayed bundle identities there. Since the point was arbitrary, they hold globally.

Let $A=D^{\prime }$. Its square and $B^2$ vanish, whilebecause the Chern curvature has pure type $(1,1)$. Use the two commutators already proved to calculateTaking adjoints, including conjugation of the scalar $-i$, gives the Lefschetz-Dolbeault curvature commutatorIf $\Theta =0$, commutation follows. Conversely, vanishing of the commutator on all $E$-valued forms implies $\Theta \wedge s=0$ for every smooth section of a vector bundle $s$. At any point, a bump-supported local section can have any prescribed fiber value. Thus $\Theta$ annihilates every vector in every fiber, forcing $\Theta =0$ identically. This proves the flatness criterion from Lefschetz commutation; the test on degree-zero forms is why the assertion is made on the entire form space.

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 ${\mathcal{A}}^2(\mathrm{End}(E))$.

The Chern connection decomposes on vector-bundle-valued differential forms as $D=D^{\prime }+D^{\prime \prime }$, where $D^{\prime }={\nabla }_h^{\prime }$ raises holomorphic degree and $D^{\prime \prime }={\nabla }_h^{\prime \prime }={\bar{\partial }}_E$ raises antiholomorphic degree. In a holomorphic local frame, $D^{\prime }=\partial +A\wedge$ and $D^{\prime \prime }=\bar{\partial }$. The type of the curvature form of a connection givesThe Kähler metric and the Hermitian metric define the $L^2$ inner product and the formal adjoints $D^{\prime \ast },D^{\prime \prime \ast }$. The Dolbeault Laplacians areLet $L=\omega \wedge$ be the Lefschetz operator of a Kähler manifold and $\Lambda =L^{\ast }$ its adjoint Lefschetz operator. With the ordinary commutator convention $[A,B]=AB-BA$, the printed Kähler identities giveSubstitution into the two Dolbeault Laplacians, followed by expansion, yieldsThe middle line follows by cancelling the terms with $D^{\prime }\Lambda D^{\prime \prime }$ and $D^{\prime \prime }\Lambda D^{\prime }$; the remaining terms collect the anticommutator of the two differentials. Hence $\boxed{{\Delta }^{\prime \prime }={\Delta }^{\prime }+[i{F}_h,\Lambda ]}$, where $i{F}_h$ denotes its wedge action. This is the Bochner-Kodaira-Nakano identity with the paper's sign convention.

A positive holomorphic line bundle $F$ admits a Hermitian metric whose Chern connection curvature satisfies that $i{F}_F$ is a positive real (1, 1)-form. In a local holomorphic local frame with squared length $e^{-\phi }$, the local formula for the Chern connection on a line bundle gives $F_F=\partial \bar{\partial }\phi$, so positivity means $i\partial \bar{\partial }\phi$ is positive definite. Its closedness makes $\omega =i{F}_F$ a Kähler form; use this form to define the operators below.

Let $n={\dim }_{\mathbb{C}}X>0$, choose a Hermitian metric on $E$, and equip $G_m=E\otimes F^{-m}$ with the tensor-product metric. The curvature of a tensor product connection givesOn $G_m$-valued zero-forms, the Lefschetz commutator is $[L,\Lambda ]=-n\mathrm{id}$. Thus the Bochner-Kodaira-Nakano identity givesThis last operator is a fixed smooth self-adjoint bundle endomorphism. Compactness supplies a finite $C$ such that $⟨R_{E}v,v⟩\ge -C\Vert v{\Vert }^2$ at every point. For a holomorphic section $s$ of $G_m$, $D^{\prime \prime }s=0$ and $D^{\prime \prime \ast }s=0$ by degree, so integrating the identity givesChoose an integer $m_0$ with $m_{0}n>C$. Then $\boxed{H^0(X,E\otimes F^{-m})=0(m\ge m_0)}$. The threshold depends on the fixed bundle $E$, as the curvature bound makes explicit. Positive complex dimension is necessary: on a zero-dimensional manifold positivity is vacuous and a nonzero fibre has nonzero sections for every twist.

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.

The orthogonal projection ${\pi }_S$ is a smooth bundle map, since the Hermitian metric and the subbundle vary smoothly. The mapis independent of the representative, bijective on each fibre, and smooth with smooth inverse supplied by the quotient map on $S^{\perp }$. Hence it is a natural isomorphism of smooth vector bundles. It need not be holomorphic. The quotient Hermitian metric isIt is well-defined and positive definite because $S^{\perp }$ represents each quotient class uniquely.

On sections of $S$, set $D^{\prime }={\pi }_S{D}_E$. The Leibniz rule makes $D^{\prime }$ a connection on a vector bundle. For $s,t\in \Gamma (S)$, the orthogonal terms vanish in the metric pairing, soThus $D^{\prime }$ has metric compatibility. Because $S$ is a holomorphic subbundle, the $(0,1)$ part of $D_E$ preserves $S$ and restricts to ${\bar{\partial }}_S$. Therefore $(D^{\prime }{)}^{0,1}={\bar{\partial }}_S$. The uniqueness of the Chern connection in part (b) gives the projected Chern connection formulaIt follows that $A(s)=(I-{\pi }_S)D_{E}s$ is a quotient-valued one-form. The printed target $T_M\otimes Q$ is missing a dual: the cotangent bundle is the correct factor,Complex-valued forms use the complexified cotangent bundle here. In fact the $(0,1)$ part cancels, so $A$ is the second fundamental form of a holomorphic subbundle, an element of ${\mathcal{A}}^{1,0}(\mathrm{Hom}(S,Q))$. Finally the two connection Leibniz rules giveThis proves its tensoriality and all the asserted quotient and projection properties.

Use the integral normalization of the Chern connection curvature. For a holomorphic local frame $e$, put $h_e=h(e,e)>0$. The fundamental form of a Hermitian holomorphic line bundle isThis follows from the local formula for the Chern connection on a line bundle, $F_h=\bar{\partial }\partial \log h_e$. It is a closed real (1, 1)-form representing the First Chern class. If $h_e=e^{-\phi }$, the formula becomes $(i/(2\pi ))\partial \bar{\partial }\phi$.

The Hermitian metric is positive when this is a positive real (1, 1)-form: for every nonzero $v\in T^{1,0}M$,Equivalently, the matrix $({\partial }^2\phi /\partial z_j\partial {\bar{z}}_k)$ is a positive-definite matrix in every frame. Such a metric makes $\mathcal{L}$ a positive holomorphic line bundle. Omitting the normalization factor $2\pi$ changes none of these positivity conditions.

In a holomorphic local frame $e$, let $H=(h(e_i,e_j))$ and write $De=eA$. Holomorphic compatibility forces $A^{0,1}=0$, while metric compatibility forces the local formula for the Chern connection on a vector bundleThis proves uniqueness. The formula transforms by the connection one-form law under a holomorphic frame change, so the local definitions glue and satisfy both conditions. This proves existence of the unique Chern connection.

Because the two Chern connections have the same $(0,1)$ part, $a=D_1-D_2$ has type $(1,0)$. The curvature difference formula has only $(2,0)$ and $(1,1)$ parts. Both Chern curvatures have type $(1,1)$, so the total $(2,0)$ part vanishes. The remaining part is obtained from $D_2^{0,1}={\bar{\partial }}_{\mathrm{End}E}$, proving the curvature difference of two Chern connections

A holomorphic line bundle is a complex line bundle with holomorphic transition functions, and a holomorphic section is one whose coefficient in every holomorphic local frame is holomorphic. Given a Hermitian metric on a holomorphic vector bundle $h$, its Chern connection is the connection $\nabla$ satisfying

Let $e$ be a nonvanishing holomorphic local frame, put $H=h(e,e)>0$, and write $\nabla e=Ae$. The first condition forces $A^{0,1}=0$, while metric compatibility forcesThis determines $\nabla$ uniquely and also constructs it. If $e^{\prime }=ge$ for a nowhere-zero holomorphic function $g$, then $A^{\prime }=A+g^{-1}\partial g$, exactly the connection one-form transformation law, so the local constructions glue.

For a line bundle, $A\wedge A=0$, and the curvature form of a connection isIt has type $(1,1)$. Under $e^{\prime }=ge$, the extra term $g^{-1}\partial g$ is closed, so the curvature is unchanged and therefore global. This is the local formula for the Chern connection on a line bundle.

Any other Hermitian metric has the form $\hat{h}=e^{u}h$ for a global smooth real function $u$. Its local squared norm is $\hat{H}=e^{u}H$, whence

Connections ${\nabla }_1$ and ${\nabla }_2$ induce the tensor product connectionIts connection form in a product frame is $A_1+A_2$, so the curvature of a tensor product connection is $F_1+F_2$. For Chern connections, equip $L_1\otimes L_2$ with the product metricThe tensor product connection has the correct $(0,1)$ part and preserves this metric, so uniqueness identifies it with the Chern connection of $h$.

A real (1, 1)-form is a form $\phi \in {\Omega }^{1,1}(X)$ satisfying $\bar{\phi }=\phi$. In holomorphic coordinates it has the formIt is a positive real (1, 1)-form whenfor every nonzero tangent vector $\xi$ of type $(1,0)$, equivalently when the Hermitian matrix $(h_{j\bar{k}})$ is positive definite.

A holomorphic local trivialization of a holomorphic line bundle $L\to X$ is equivalently a nowhere-zero holomorphic local frame $e$. A connection $A$ is unitary when it preserves the fiberwise Hermitian inner product:The Chern connection is the unique unitary connection whose $(0,1)$ part is the bundle's Dolbeault partial connection ${\bar{\partial }}_L$.

In a holomorphic frame, put $h=h(e,e)$. The local formula for the Chern connection on a line bundle isThe curvature therefore has type $(1,1)$. Since a unitary connection has imaginary curvature, $\overline{F(A)}=-F(A)$, and henceThus $iF(A)$ is a real $(1,1)$-form.

For connections $A$ on $L$ and $\hat{A}$ on $\hat{L}$, the tensor product connection is defined on decomposable local sections byThe curvature of a tensor product connection on line bundles is additive:Consequentlywhich is positive whenever both summands are positive.

For the final assertion, simultaneously diagonalize the positive Hermitian matrices of $\phi$ and $\psi$ by congruence at the chosen point. In the resulting coframe,A direct wedge-product calculation givesIf $\xi$ and $\eta$ are linearly independent, at least one of these $2\times 2$ minors is nonzero. Every coefficient is positive, so the sum is strictly positive. This is wedge positivity for two positive (1, 1)-forms.

A Hermitian metric on a holomorphic vector bundle $E\to X$ is a smoothly varying family of positive-definite Hermitian forms $h_x$ on its fibers. A Chern connection is a connection on a vector bundle $\nabla$ that is compatible with $h$ and whose $(0,1)$ part is the bundle's Dolbeault partial connection, ${\nabla }^{0,1}={\bar{\partial }}_E$.

Choose a local holomorphic frame $e$ and write $h=h(e,e)>0$. The connection form in this frame isIf $e^{\prime }=ge$ for a nowhere-zero holomorphic function $g$, then $h^{\prime }=|g{|}^{2}h$ andwhich is exactly the connection-form transformation law. The local formulas therefore define a global connection.

Identify $\mathcal{O}(-1)$ with the tautological bundle over Complex projective space. The standard Hermitian inner product of ${\mathbb{C}}^{n+1}$ restricts to each tautological line. On the affine chart $U_i=\{Z_i\ne 0\}$, put $w_j=Z_j/Z_i$ for $j\ne i$ and use the holomorphic frameThenThe curvature form of a connection isConsequently Chern-Weil theory gives the closed representativewhere ${\omega }_{FS}$ is the Fubini-Study form.

The curvature of the Chern connection on the complex line bundle $TX$ has type $(1,1)$, and $i{F}_{\nabla }$ is real. Since $X$ has complex dimension one, every real two-form is a unique smooth multiple of its nonvanishing area form, sofor some real smooth function $\lambda$. By Chern-Weil theory,The hyperplane class has degree $d$ on a degree-$d$ plane curve, so

A holomorphic line bundle is positive when it admits a Hermitian metric whose Chern connection has positive curvature form $i{F}_{\nabla }$, equivalently when $i{F}_{\nabla }$ is a positive real (1, 1)-form.

The positive-curvature definition uses a positive real (1, 1)-form.

If $S$ is a holomorphic subbundle of a Hermitian vector bundle $E$, its induced Chern connection is $D_S={\pi }_S{D}_E$ on sections of $S$. Orthogonal projection preserves metric compatibility, and holomorphicity of $S$ preserves the required $(0,1)$ operator. Uniqueness of the Chern connection therefore identifies the projected connection.