[proofplan] Work locally on the [open set](/page/Open%20Set) $U$ and express the Chern connection in the holomorphic frame $e$. The condition $(\nabla^E)^{0,1}=\bar{\partial}_E$ forces the connection matrix to have only type $(1,0)$, while metric compatibility determines that matrix by differentiating the entries $H_{ij}=h(e_i,e_j)$. This gives $\partial H=H\theta$, hence $\theta=H^{-1}\partial H$. Finally, the curvature is computed from the local formula $d\theta+\theta\wedge\theta$, and the identity $\partial(H^{-1}\partial H)=-\theta\wedge\theta$ leaves only the $\bar{\partial}$ term. [/proofplan]

[step:Use the holomorphic frame to restrict the connection matrix to type $(1,0)$] Let $\omega \in \Omega^1(U;\operatorname{Mat}_r(\mathbb{C}))$ denote the full connection matrix of $\nabla^E$ in the row-frame convention: \begin{align*} \nabla^E e = e\omega. \end{align*} Thus, for each $j \in \{1,\dots,r\}$, \begin{align*} \nabla^E e_j = \sum_{i=1}^r e_i\omega_{ij}. \end{align*} Since $e_1,\dots,e_r$ are holomorphic local sections and $(\nabla^E)^{0,1}=\bar{\partial}_E$, we have \begin{align*} (\nabla^E)^{0,1}e_j = \bar{\partial}_E e_j = 0 \end{align*} for every $j$. Therefore the $(0,1)$ part of every entry $\omega_{ij}$ is zero. Hence $\omega$ has type $(1,0)$. We write \begin{align*} \theta := \omega \in \Omega^{1,0}(U;\operatorname{Mat}_r(\mathbb{C})). \end{align*} [/step]

[step:Apply metric compatibility to derive $\partial H=H\theta$]Fix indices $i,j \in \{1,\dots,r\}$. Since $\nabla^E h=0$, metric compatibility gives \begin{align*} d(h(e_i,e_j)) = h(\nabla^E e_i,e_j) + h(e_i,\nabla^E e_j). \end{align*} Using $\nabla^E e_i=\sum_{k=1}^r e_k\theta_{ki}$ and $\nabla^E e_j=\sum_{\ell=1}^r e_\ell\theta_{\ell j}$, together with conjugate-linearity of $h$ in the first argument and linearity in the second, this becomes \begin{align*} dH_{ij} = \sum_{k=1}^r \overline{\theta_{ki}}H_{kj} + \sum_{\ell=1}^r H_{i\ell}\theta_{\ell j}. \end{align*} The forms $\theta_{\ell j}$ have type $(1,0)$, so the forms $\overline{\theta_{ki}}$ have type $(0,1)$. Taking the $(1,0)$ part gives \begin{align*} \partial H_{ij} = \sum_{\ell=1}^r H_{i\ell}\theta_{\ell j}. \end{align*} Since this holds for every pair $(i,j)$, the matrix identity is \begin{align*} \partial H = H\theta. \end{align*}[/step]

[guided]The point of this step is that metric compatibility is the condition that determines the unknown $(1,0)$ connection matrix. We already know from the holomorphic-frame condition that no $(0,1)$ component remains, so the only unknown object is \begin{align*} \theta \in \Omega^{1,0}(U;\operatorname{Mat}_r(\mathbb{C})). \end{align*} For fixed indices $i,j \in \{1,\dots,r\}$, the metric matrix entry is the smooth function \begin{align*} H_{ij}: U \to \mathbb{C}, \qquad H_{ij}=h(e_i,e_j). \end{align*} Metric compatibility of $\nabla^E$ with $h$ means that differentiating this scalar function is the same as differentiating each vector argument using the connection: \begin{align*} dH_{ij} = d(h(e_i,e_j)) = h(\nabla^E e_i,e_j) + h(e_i,\nabla^E e_j). \end{align*} Now insert the local connection formula. The row-frame convention $\nabla^E e=e\theta$ means precisely that \begin{align*} \nabla^E e_i = \sum_{k=1}^r e_k\theta_{ki} \end{align*} and \begin{align*} \nabla^E e_j = \sum_{\ell=1}^r e_\ell\theta_{\ell j}. \end{align*} Because the Hermitian metric is conjugate-linear in the first argument and linear in the second, the scalar one-forms factor out as \begin{align*} h(\nabla^E e_i,e_j) = \sum_{k=1}^r \overline{\theta_{ki}}H_{kj} \end{align*} and \begin{align*} h(e_i,\nabla^E e_j) = \sum_{\ell=1}^r H_{i\ell}\theta_{\ell j}. \end{align*} Therefore \begin{align*} dH_{ij} = \sum_{k=1}^r \overline{\theta_{ki}}H_{kj} + \sum_{\ell=1}^r H_{i\ell}\theta_{\ell j}. \end{align*} Why does the desired formula involve $\partial H$ rather than all of $dH$? The reason is type decomposition. Each $\theta_{\ell j}$ is a $(1,0)$ form, hence each $\overline{\theta_{ki}}$ is a $(0,1)$ form. Taking the $(1,0)$ component of the preceding identity discards the first sum and keeps the second: \begin{align*} \partial H_{ij} = \sum_{\ell=1}^r H_{i\ell}\theta_{\ell j}. \end{align*} This is exactly the $(i,j)$ entry of the matrix product $H\theta$, where multiplication means ordinary matrix multiplication together with wedge product of forms. Hence \begin{align*} \partial H = H\theta. \end{align*}[/guided]

[step:Solve the matrix equation for the connection matrix] For every point of $U$, the matrix $H$ is Hermitian positive definite, hence invertible. Multiplying the identity $\partial H=H\theta$ on the left by $H^{-1}$ gives \begin{align*} \theta = H^{-1}\partial H. \end{align*} This proves the local formula for the connection matrix. [/step]

[step:Compute the curvature matrix and cancel the $(2,0)$ part]By definition of the curvature matrix in the same row frame, \begin{align*} (\nabla^E)^2 e = e\Theta_E. \end{align*} We compute this expression directly from $\nabla^E e=e\theta$. The connection extends to $E$-valued forms by the exterior covariant derivative, so \begin{align*} (\nabla^E)^2 e=\nabla^E(e\theta)=(\nabla^E e)\wedge\theta+e\,d\theta=e(\theta\wedge\theta+d\theta). \end{align*} Comparing with $(\nabla^E)^2e=e\Theta_E$ in the local frame $e$ gives \begin{align*} \Theta_E = d\theta + \theta\wedge\theta. \end{align*} Since $\theta$ has type $(1,0)$, \begin{align*} d\theta = \partial\theta + \bar{\partial}\theta. \end{align*} It remains to compute the $(2,0)$ term. Let $I_r \in \operatorname{Mat}_r(\mathbb{C})$ denote the $r \times r$ identity matrix. From the identity $HH^{-1}=I_r$, applying $\partial$ gives \begin{align*} (\partial H)H^{-1} + H\partial(H^{-1}) = 0. \end{align*} Multiplying on the left by $H^{-1}$ yields \begin{align*} \partial(H^{-1}) = -H^{-1}(\partial H)H^{-1}. \end{align*} Using $\theta=H^{-1}\partial H$ and $\partial^2H=0$, we obtain \begin{align*} \partial\theta = \partial(H^{-1}\partial H) = \partial(H^{-1})\wedge\partial H. \end{align*} Substituting the formula for $\partial(H^{-1})$ gives \begin{align*} \partial\theta = -H^{-1}(\partial H)H^{-1}\wedge\partial H = -\theta\wedge\theta. \end{align*} Therefore \begin{align*} \Theta_E = \partial\theta+\bar{\partial}\theta+\theta\wedge\theta = \bar{\partial}\theta. \end{align*} Finally substituting $\theta=H^{-1}\partial H$ gives \begin{align*} \Theta_E = \bar{\partial}(H^{-1}\partial H). \end{align*}[/step]

[guided]The curvature of a connection in a local frame is obtained by applying the connection twice, but the sign and order depend on the frame convention. Here we are using the row-frame convention $\nabla^E e=e\theta$. Since $e$ is a row vector of $E$-valued $0$-forms, the exterior covariant derivative gives \begin{align*} (\nabla^E)^2e=\nabla^E(e\theta)=(\nabla^E e)\wedge\theta+e\,d\theta. \end{align*} Substituting $\nabla^E e=e\theta$ yields \begin{align*} (\nabla^E)^2e=e(\theta\wedge\theta+d\theta). \end{align*} The curvature matrix is defined by $(\nabla^E)^2e=e\Theta_E$, so comparison of coefficients in the local frame gives \begin{align*} \Theta_E=d\theta+\theta\wedge\theta. \end{align*} Here $\theta\wedge\theta$ means matrix multiplication with wedge product in the entries: \begin{align*} (\theta\wedge\theta)_{ij}=\sum_{k=1}^r \theta_{ik}\wedge\theta_{kj}. \end{align*} Since $\theta$ is a matrix of $(1,0)$ forms, its [exterior derivative](/theorems/1525) splits by type: \begin{align*} d\theta=\partial\theta+\bar{\partial}\theta. \end{align*} Thus \begin{align*} \Theta_E = \partial\theta+\bar{\partial}\theta+\theta\wedge\theta. \end{align*} The theorem claims that only the $\bar{\partial}\theta$ term remains. So the key calculation is the cancellation \begin{align*} \partial\theta+\theta\wedge\theta=0. \end{align*} We prove this cancellation from $\theta=H^{-1}\partial H$. Let $I_r \in \operatorname{Mat}_r(\mathbb{C})$ denote the $r \times r$ identity matrix. First differentiate the identity \begin{align*} HH^{-1}=I_r. \end{align*} Since $H$ and $H^{-1}$ are matrix-valued functions, they have form degree $0$, so the product rule gives \begin{align*} (\partial H)H^{-1}+H\partial(H^{-1})=0. \end{align*} Multiplying on the left by $H^{-1}$ gives \begin{align*} \partial(H^{-1})=-H^{-1}(\partial H)H^{-1}. \end{align*} Now differentiate the formula $\theta=H^{-1}\partial H$: \begin{align*} \partial\theta=\partial(H^{-1}\partial H). \end{align*} The product rule gives \begin{align*} \partial\theta=\partial(H^{-1})\wedge\partial H+H^{-1}\partial^2H. \end{align*} The operator $\partial$ satisfies $\partial^2=0$ on smooth forms, so the second term vanishes: \begin{align*} \partial\theta=\partial(H^{-1})\wedge\partial H. \end{align*} Substituting the formula for $\partial(H^{-1})$ gives \begin{align*} \partial\theta=-H^{-1}(\partial H)H^{-1}\wedge\partial H. \end{align*} Because $\theta=H^{-1}\partial H$, the right-hand side is exactly \begin{align*} -\theta\wedge\theta. \end{align*} Therefore \begin{align*} \partial\theta+\theta\wedge\theta=0. \end{align*} Returning to the curvature formula, we get \begin{align*} \Theta_E=d\theta+\theta\wedge\theta=\partial\theta+\bar{\partial}\theta+\theta\wedge\theta=\bar{\partial}\theta. \end{align*} Finally, replacing $\theta$ by $H^{-1}\partial H$ gives \begin{align*} \Theta_E=\bar{\partial}(H^{-1}\partial H). \end{align*}[/guided]

[step:Conclude the connection and curvature formulas] The connection matrix in the holomorphic frame $e$ is \begin{align*} \theta=H^{-1}\partial H, \end{align*} so $\nabla^E e=e\theta$. The curvature matrix in the same convention is \begin{align*} \Theta_E=\bar{\partial}(H^{-1}\partial H). \end{align*} These are precisely the asserted local formulas for the Chern connection and its curvature. [/step]