[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*}
[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]
[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*}
[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]
[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]
