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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.