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For a smooth complex vector bundle $E$, a connection is a complex-linear map $D:\Gamma (E)\to A^1(M;E)$ satisfying $D(fs)=df\otimes s+fDs$ for smooth complex functions $f$. Local trivializations supply flat local connections. A locally finite smooth partition of unity $({\rho }_i)$ subordinate to them gives $D={\sum }_i{\rho }_i{D}_i$; local finiteness makes the sum smooth and ${\sum }_i{\rho }_i=1$ gives the Leibniz rule. Thus connections on a vector bundle always exist on the usual paracompact smooth manifold.

Extend $D$ to bundle-valued forms by $D(\alpha \otimes s)=d\alpha \otimes s+(-1{)}^{\mathrm{deg}\alpha }\alpha \wedge Ds$. Squaring shows that $D^2(fs)=f{D}^{2}s$, so the curvature form is the endomorphism-valued two-form $\Theta =D^2$. With a row frame $e=(e_1,\dots ,e_r)$, write $De=eA$; for coefficient columns $s=ev$, $Ds=e(dv+Av)$. Applying $D$ again and using the graded Leibniz rule givesThe cross terms $-A\wedge dv$ and $A\wedge dv$ cancel. Hence the Cartan curvature matrix equation in this convention isFor a frame change $e^{\prime }=eg$, the connection matrix and curvature transform as $A^{\prime }=g^{-1}Ag+g^{-1}dg$ and ${\Theta }^{\prime }=g^{-1}\Theta g$. Trace is invariant under conjugation, so the local forms $\mathrm{Tr}\Theta$ patch into the global trace of vector-bundle curvature.

The induced determinant connection on $\det E={\Lambda }^{r}E$ iswhere the one-form factor of each $D{s}_j$ is placed in front of the section factors. In the local determinant frame $e_1\wedge \cdots \wedge e_r$, only the diagonal terms survive, so its connection form is $\mathrm{Tr}A$. Its curvature is $d\mathrm{Tr}A$. Sinceby pairing off-diagonal terms and using $A_{ii}\wedge A_{ii}=0$, we obtainIt is closed locally by $d^2=0$, and thus closed globally.

For two connections, $B=D_1-D_0$ is a global endomorphism-valued one-form: the derivative terms cancel in their Leibniz rules. In a local frame $A_1=A_0+B$, soTaking traces cancels the mixed terms, because both factors are one-forms, and cancels $\mathrm{Tr}(B\wedge B)$. Therefore the trace curvature transgression formula isThe right-hand side is globally exact, proving independence of the de Rham cohomology class without needing a sheaf-cohomology identification.

Finally let $e$ be a holomorphic frame for the holomorphic line bundle, with $h=\Vert e{\Vert }^2>0$. Define $De=e\partial \log h$, extending by the Leibniz rule. If $e^{\prime }=eg$ for a nowhere-zero holomorphic function $g$, then $h^{\prime }=|g{|}^{2}h$ andThis is exactly the line-bundle frame-change law for a connection, so the local definitions glue globally. Its $(0,1)$ part is the Dolbeault operator. Metric compatibility follows from $d\log h=\partial \log h+\bar{\partial }\log h$, or $dh=h(A+\overline{A})$ for $A=\partial \log h$. These conditions uniquely force this $A$, identifying the Chern connection. Since a scalar one-form wedges with itself to zero,This agrees with the local formula for the Chern connection on a line bundle; keeping the order of the two Dolbeault differentials fixes the sign.

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.

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.

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.

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.