[proofplan] We compute the Chern connection in a local holomorphic frame and use metric compatibility to identify its connection form as $-\partial\varphi$. Taking the [exterior derivative](/theorems/1525) gives the curvature form $\Theta_h(L)=\partial\bar\partial\varphi$ under the stated convention. We then check that changing the frame by a nowhere-vanishing [holomorphic function](/page/Holomorphic%20Function) changes $\varphi$ only by a pluriharmonic term, so the local formula patches consistently. Finally, evaluating the resulting $(1,1)$-form on $(1,0)$ tangent vectors identifies positivity with the sign of the complex Hessian matrix. [/proofplan]

[step:Compute the Chern connection form in the holomorphic frame]Let $\nabla_h$ denote the Chern connection of the Hermitian holomorphic line bundle $(L,h)$. Let \begin{align*} \bar\partial_L:\Omega^0(U;L|_U)&\to\Omega^{0,1}(U;L|_U) \end{align*} denote the Dolbeault operator defining the holomorphic structure on $L|_U$. Since $e$ is a nowhere-vanishing local frame on $U$, there is a unique complex-valued $1$-form \begin{align*} A \in \Omega^1(U;\mathbb{C}) \end{align*} such that \begin{align*} \nabla_h e = A \otimes e. \end{align*} Because $e$ is holomorphic and the $(0,1)$-part of the Chern connection equals $\bar\partial_L$, we have $A^{0,1}=0$. Hence $A$ is a $(1,0)$-form. Metric compatibility gives \begin{align*} d(h(e,e)) = h(\nabla_h e,e)+h(e,\nabla_h e). \end{align*} Using $h(e,e)=e^{-\varphi}$ and $\nabla_h e=A\otimes e$, the $(1,0)$-part of this identity is \begin{align*} \partial(e^{-\varphi}) = A e^{-\varphi}. \end{align*} Since \begin{align*} \partial(e^{-\varphi}) = -e^{-\varphi}\partial\varphi, \end{align*} division by the positive function $e^{-\varphi}$ gives \begin{align*} A=-\partial\varphi. \end{align*}[/step]

[guided]We first translate the connection into a scalar $1$-form. Since $L|_U$ is generated by the single nowhere-vanishing section $e$, every covariant derivative of $e$ must be a scalar $1$-form times $e$. Thus there is a unique complex-valued $1$-form \begin{align*} A \in \Omega^1(U;\mathbb{C}) \end{align*} with \begin{align*} \nabla_h e = A \otimes e. \end{align*} Let \begin{align*} \bar\partial_L:\Omega^0(U;L|_U)&\to\Omega^{0,1}(U;L|_U) \end{align*} denote the Dolbeault operator defining the holomorphic structure on $L|_U$. The Chern connection is characterized by two properties: it is compatible with the Hermitian metric, and its $(0,1)$-part agrees with $\bar\partial_L$. Since $e$ is holomorphic, $\bar\partial_L e=0$. Therefore the $(0,1)$-part of $\nabla_h e$ vanishes, so $A^{0,1}=0$. Hence $A$ is of type $(1,0)$. Now use metric compatibility. The function $h(e,e):U\to\mathbb{R}_{>0}$ is $e^{-\varphi}$ by definition of $\varphi$, and compatibility says \begin{align*} d(h(e,e)) = h(\nabla_h e,e)+h(e,\nabla_h e). \end{align*} Substituting $\nabla_h e=A\otimes e$ gives \begin{align*} d(e^{-\varphi}) = A\,e^{-\varphi}+\overline{A}\,e^{-\varphi}. \end{align*} Taking the $(1,0)$-part removes the $\overline{A}$ term because $A$ is a $(1,0)$-form, so \begin{align*} \partial(e^{-\varphi}) = A e^{-\varphi}. \end{align*} The chain rule for the smooth function $t\mapsto e^{-t}$ gives \begin{align*} \partial(e^{-\varphi}) = -e^{-\varphi}\partial\varphi. \end{align*} Since $e^{-\varphi}>0$ on $U$, we may divide by $e^{-\varphi}$ and obtain \begin{align*} A=-\partial\varphi. \end{align*}[/guided]

[step:Take the curvature of the local connection form]We use the curvature convention in which the local curvature endomorphism of a connection form is $dA+A\wedge A$. Since $L$ has rank one, $A$ is scalar-valued, so $A\wedge A=0$ and the curvature of the local connection form $A$ is the scalar $2$-form \begin{align*} \Theta_h(L)|_U = dA. \end{align*} Using $A=-\partial\varphi$, we obtain \begin{align*} \Theta_h(L)|_U = d(-\partial\varphi) = -(\partial+\bar\partial)\partial\varphi = -\partial^2\varphi-\bar\partial\partial\varphi. \end{align*} Since $\partial^2=0$ and $\bar\partial\partial=-\partial\bar\partial$ on smooth functions, \begin{align*} \Theta_h(L)|_U=\partial\bar\partial\varphi. \end{align*} Multiplying by $i$ gives \begin{align*} i\Theta_h(L)|_U=i\partial\bar\partial\varphi. \end{align*}[/step]

[guided]The previous step reduced the geometry of the connection to the scalar identity \begin{align*} A=-\partial\varphi. \end{align*} We use the convention that the local curvature endomorphism of a connection form is $dA+A\wedge A$. For a line bundle, the connection form is scalar-valued, so the wedge product term $A\wedge A$ vanishes. Thus the curvature is \begin{align*} \Theta_h(L)|_U=dA. \end{align*} Substituting the connection form gives \begin{align*} \Theta_h(L)|_U = d(-\partial\varphi). \end{align*} We decompose the exterior derivative as $d=\partial+\bar\partial$ on complex forms: \begin{align*} d(-\partial\varphi) = -(\partial+\bar\partial)\partial\varphi = -\partial^2\varphi-\bar\partial\partial\varphi. \end{align*} For smooth functions, $\partial^2=0$ and the anticommutation relation is $\bar\partial\partial=-\partial\bar\partial$. Therefore \begin{align*} -\partial^2\varphi-\bar\partial\partial\varphi = 0+\partial\bar\partial\varphi. \end{align*} Hence \begin{align*} \Theta_h(L)|_U=\partial\bar\partial\varphi. \end{align*} Multiplication by $i$ gives the stated curvature formula: \begin{align*} i\Theta_h(L)|_U=i\partial\bar\partial\varphi. \end{align*}[/guided]

[step:Check that the local formula is independent of the holomorphic frame]Let \begin{align*} e': U \to L|_U \end{align*} be another nowhere-vanishing holomorphic frame. Then there is a nowhere-vanishing holomorphic function \begin{align*} g: U \to \mathbb{C}^\times \end{align*} such that $e'=g e$. Let \begin{align*} \varphi':U\to\mathbb{R} \end{align*} be defined by $|e'|_h^2=e^{-\varphi'}$. Then \begin{align*} e^{-\varphi'} = h(ge,ge) = |g|^2 e^{-\varphi}, \end{align*} so \begin{align*} \varphi'=\varphi-\log |g|^2. \end{align*} Since $g$ is holomorphic and nowhere zero, locally $\log |g|^2$ is the real part of a holomorphic logarithm multiplied by $2$, and therefore \begin{align*} \partial\bar\partial \log |g|^2=0. \end{align*} Consequently \begin{align*} \partial\bar\partial\varphi' = \partial\bar\partial\varphi. \end{align*} Thus the local expression $i\partial\bar\partial\varphi$ is independent of the chosen holomorphic frame.[/step]

[guided]We must verify that the expression obtained from a weight is not an artifact of the chosen generator of the line bundle. Let \begin{align*} e':U\to L|_U \end{align*} be another nowhere-vanishing holomorphic frame. Since both $e$ and $e'$ generate the same one-dimensional fiber at each point of $U$, there is a unique nowhere-vanishing holomorphic function \begin{align*} g:U\to\mathbb{C}^{\times} \end{align*} such that \begin{align*} e'=g e. \end{align*} Define \begin{align*} \varphi':U\to\mathbb{R} \end{align*} by $|e'|_h^2=e^{-\varphi'}$. The Hermitian metric is sesquilinear, so \begin{align*} e^{-\varphi'} = h(e',e') = h(ge,ge) = |g|^2 h(e,e) = |g|^2 e^{-\varphi}. \end{align*} Taking logarithms gives \begin{align*} \varphi'=\varphi-\log |g|^2. \end{align*} Now we compute the extra term. Since $g$ is holomorphic and nowhere zero, every point of $U$ has a smaller open neighbourhood $V\subset U$ on which there is a holomorphic function \begin{align*} \ell:V\to\mathbb{C} \end{align*} with $g=e^\ell$ on $V$. On $V$, \begin{align*} \log |g|^2=\ell+\overline{\ell}. \end{align*} Because $\ell$ is holomorphic, $\bar\partial\ell=0$; because $\overline{\ell}$ is antiholomorphic, $\partial\overline{\ell}=0$. Therefore \begin{align*} \partial\bar\partial\log |g|^2 = \partial\bar\partial(\ell+\overline{\ell}) = \partial(0)+\partial\bar\partial\overline{\ell} = 0. \end{align*} The identity is local and hence holds on all of $U$. Applying $\partial\bar\partial$ to $\varphi'=\varphi-\log |g|^2$ gives \begin{align*} \partial\bar\partial\varphi' = \partial\bar\partial\varphi-\partial\bar\partial\log |g|^2 = \partial\bar\partial\varphi. \end{align*} Thus the curvature form computed from the weight is independent of the holomorphic frame.[/guided]

[step:Express the curvature form in holomorphic coordinates] Let $(V,z)$ be a holomorphic coordinate chart with $V\subset U$ and coordinate map \begin{align*} z:V&\to z(V)\subset\mathbb{C}^n,\\ x&\mapsto (z_1(x),\dots,z_n(x)). \end{align*} On $V$, the operators $\partial$ and $\bar\partial$ give \begin{align*} \partial\varphi = \sum_{j=1}^{n}\frac{\partial\varphi}{\partial z_j}\,dz_j. \end{align*} Therefore \begin{align*} \partial\bar\partial\varphi = \sum_{j,k=1}^{n} \frac{\partial^2\varphi}{\partial z_j\partial \bar z_k} \,dz_j\wedge d\bar z_k, \end{align*} with the sign convention already fixed by the identity $\Theta_h(L)=\partial\bar\partial\varphi$. Hence \begin{align*} i\Theta_h(L) = i\sum_{j,k=1}^{n} \frac{\partial^2\varphi}{\partial z_j\partial \bar z_k} \,dz_j\wedge d\bar z_k. \end{align*} [/step]

[step:Identify positivity with the sign of the complex Hessian] Fix $x\in V$. Let \begin{align*} \xi \in T_x^{1,0}X \end{align*} be written in the coordinate frame as \begin{align*} \xi=\sum_{j=1}^{n}\xi_j\frac{\partial}{\partial z_j}\bigg|_x, \end{align*} where $\xi_1,\dots,\xi_n\in\mathbb{C}$. By convention, the Hermitian form associated to the real $(1,1)$-form \begin{align*} i\Theta_h(L) = i\sum_{j,k=1}^{n} a_{j\bar k}\,dz_j\wedge d\bar z_k \end{align*} is the coefficient form \begin{align*} H_x(\xi,\xi) = \sum_{j,k=1}^{n}a_{j\bar k}(x)\,\xi_j\,\overline{\xi_k}, \end{align*} not the literal alternating-form evaluation on the pair $(\xi,\overline{\xi})$. For the present curvature form this gives \begin{align*} H_x(\xi,\xi) = \sum_{j,k=1}^{n} \frac{\partial^2\varphi}{\partial z_j\partial \bar z_k}(x) \,\xi_j\,\overline{\xi_k}. \end{align*} Thus $i\Theta_h(L)$ is positive definite, positive semidefinite, or negative definite at $x$ exactly when the Hermitian matrix \begin{align*} \left( \frac{\partial^2\varphi}{\partial z_j\partial \bar z_k}(x) \right)_{j,k=1}^{n} \end{align*} has the corresponding sign. Since the previous step proved that this matrix representation changes only by the usual coordinate change law for Hermitian forms and is independent of the holomorphic frame, the criterion holds in every holomorphic coordinate chart and every local holomorphic frame. This proves the stated equivalence for positivity, semipositivity, and negativity of $(L,h)$. [/step]