[proofplan] We compute the Chern connection in the global holomorphic frame $e$. The Hermitian metric is represented in this frame by the positive smooth function $g := h(e,e)=e^{-\varphi}$, so the Chern connection one-form is $g^{-1}\partial g$. Differentiating this connection form by $\bar{\partial}$ gives the curvature, and the anticommutation identity $\bar{\partial}\partial=-\partial\bar{\partial}$ on smooth functions converts the result into $\partial\bar{\partial}\varphi$. [/proofplan]

[step:Represent the Hermitian metric by its scalar weight in the holomorphic frame] Define the smooth positive function \begin{align*} g: X &\to \mathbb{R}_{>0} \\ x &\mapsto h_x(e(x),e(x)). \end{align*} By hypothesis, \begin{align*} g = e^{-\varphi}. \end{align*} Since $e$ is a global holomorphic frame for the line bundle $L$, every smooth local section $s$ of $L$ over an [open set](/page/Open%20Set) $U \subseteq X$ has a unique representation $s=f e|_U$ with $f:U\to\mathbb{C}$ smooth. [/step]

[step:Compute the Chern connection form from metric compatibility]Let $\nabla_h$ denote the Chern connection of $(L,h)$. In the frame $e$, define the connection one-form \begin{align*} A_e \in \Omega^{1,0}(X) \end{align*} by \begin{align*} \nabla_h e = A_e \otimes e. \end{align*} Here $\Omega^{1,0}(X)$ denotes the space of smooth complex-valued one-forms of type $(1,0)$ on $X$. Because $e$ is holomorphic, the $(0,1)$-part of the Chern connection is the holomorphic structure, so $A_e$ has type $(1,0)$. Metric compatibility gives the identity \begin{align*} \partial h(e,e) &= h(\nabla_h^{1,0}e,e)+h(e,\nabla_h^{0,1}e). \end{align*} Because $e$ is holomorphic and the $(0,1)$-part of the Chern connection is the holomorphic structure, $\nabla_h^{0,1}e=\bar\partial e=0$. Therefore \begin{align*} \partial h(e,e) &= h(\nabla_h^{1,0}e,e) \\ &= h(A_e\otimes e,e) \\ &= A_e\,h(e,e) \\ &= A_e\,g. \end{align*} Since $g>0$, division by $g$ is valid, and hence \begin{align*} A_e = g^{-1}\partial g. \end{align*} Using $g=e^{-\varphi}$, the chain rule gives \begin{align*} \partial g = \partial(e^{-\varphi}) = -e^{-\varphi}\partial\varphi = -g\,\partial\varphi. \end{align*} Therefore \begin{align*} A_e = -\partial\varphi. \end{align*}[/step]

[guided]The goal of this step is to translate the metric into the connection form. Since $L$ is a line bundle and $e$ is a nowhere-vanishing holomorphic frame, the connection is completely determined by the one-form $A_e$ defined by \begin{align*} \nabla_h e = A_e \otimes e. \end{align*} The Chern connection is characterized by two properties: its $(0,1)$-part agrees with the holomorphic structure, and it is compatible with the Hermitian metric. The first property implies that $A_e$ has type $(1,0)$, because $e$ is holomorphic. Now apply metric compatibility to the scalar function $h(e,e)=g$. Taking the $(1,0)$-part gives the full identity \begin{align*} \partial h(e,e) &= h(\nabla_h^{1,0}e,e)+h(e,\nabla_h^{0,1}e). \end{align*} The second term vanishes for a specific reason: $e$ is holomorphic, and the defining property of the Chern connection says that $\nabla_h^{0,1}$ is the holomorphic structure operator $\bar\partial$. Hence \begin{align*} \nabla_h^{0,1}e=\bar\partial e=0. \end{align*} Therefore \begin{align*} \partial h(e,e) &= h(\nabla_h^{1,0}e,e) \\ &= h(A_e\otimes e,e) \\ &= A_e\,h(e,e) \\ &= A_e\,g. \end{align*} The function $g$ is strictly positive because it is the squared norm of the frame vector $e(x)\neq 0$, so we may divide by $g$ and obtain \begin{align*} A_e = g^{-1}\partial g. \end{align*} Substituting the given weight $g=e^{-\varphi}$ and using the ordinary chain rule for smooth functions, \begin{align*} \partial g = \partial(e^{-\varphi}) = -e^{-\varphi}\partial\varphi = -g\,\partial\varphi. \end{align*} Thus the connection one-form in the frame $e$ is \begin{align*} A_e = -\partial\varphi. \end{align*}[/guided]

[step:Differentiate the connection form to obtain the curvature]For the Chern connection in the holomorphic frame $e$, the curvature is the $(1,1)$-part of the exterior covariant square of the connection. Since $L$ has rank one and $A_e$ has type $(1,0)$, the scalar wedge term $A_e\wedge A_e$ vanishes by antisymmetry of the wedge product. Therefore the curvature form in the frame $e$ is represented by \begin{align*} \Theta_h = \bar{\partial}A_e. \end{align*} Using the computed connection form $A_e=-\partial\varphi$, we obtain \begin{align*} \Theta_h = \bar{\partial}(-\partial\varphi) = -\bar{\partial}\partial\varphi. \end{align*} For a smooth function $\varphi:X\to\mathbb{R}$, the Dolbeault operators anticommute on functions: \begin{align*} \bar{\partial}\partial\varphi = -\partial\bar{\partial}\varphi. \end{align*} Therefore \begin{align*} \Theta_h = \partial\bar{\partial}\varphi. \end{align*}[/step]

[guided]The curvature is obtained from the exterior covariant square of the Chern connection. In a holomorphic frame, the connection form has type $(1,0)$, so the $(1,1)$-part of the curvature is obtained from $\bar{\partial}A_e$ together with the wedge product term. Here the bundle has rank one, so the connection form is scalar-valued. The scalar wedge term is \begin{align*} A_e\wedge A_e = 0, \end{align*} because the wedge product of a one-form with itself is zero. Hence the curvature form in the frame $e$ is \begin{align*} \Theta_h = \bar{\partial}A_e. \end{align*} From the previous step, \begin{align*} A_e=-\partial\varphi. \end{align*} Substituting this into the curvature formula gives \begin{align*} \Theta_h = \bar{\partial}(-\partial\varphi) = -\bar{\partial}\partial\varphi. \end{align*} The remaining point is the sign. On smooth functions, the Dolbeault operators satisfy the anticommutation identity \begin{align*} \bar{\partial}\partial\varphi = -\partial\bar{\partial}\varphi. \end{align*} This identity is the complex-coordinate form of $d^2=0$ together with the decomposition $d=\partial+\bar{\partial}$. Hence \begin{align*} \Theta_h = -\bar{\partial}\partial\varphi = \partial\bar{\partial}\varphi. \end{align*}[/guided]

[step:Multiply by $i$ to obtain the stated real curvature convention] Multiplying the identity $\Theta_h=\partial\bar{\partial}\varphi$ by the scalar $i\in\mathbb{C}$ gives \begin{align*} i\Theta_h = i\partial\bar{\partial}\varphi. \end{align*} This is exactly the asserted consequence. [/step]