[proofplan] The assertion is local on $X$, so we compute in an arbitrary local holomorphic frame of $E$. In such a frame, the Chern connection on $E$ has a matrix of one-forms $\theta$, and its curvature matrix is $d\theta + \theta \wedge \theta$. The induced connection on $\det E$ acts on the wedge frame by differentiating one factor at a time, so its connection one-form is $\operatorname{tr}(\theta)$. Taking curvature and using the cancellation $\operatorname{tr}(\theta \wedge \theta)=0$ for a matrix of one-forms gives $\Theta_{\det E}=\operatorname{tr}(\Theta_E)$. [/proofplan]

[step:Compute the induced connection form on the determinant frame]Let $U \subset X$ be an [open set](/page/Open%20Set) on which $E$ admits a holomorphic frame $e_1,\dots,e_r$. Let $A^1(U)$ denote the space of smooth complex-valued one-forms on $U$. Define one-forms $\theta_{ij} \in A^1(U)$, for $1 \le i,j \le r$, by \begin{align*}\nabla^E e_j = \sum_{i=1}^r e_i \otimes \theta_{ij}.\end{align*} Let $\theta=(\theta_{ij})$ denote the corresponding matrix-valued one-form. The determinant line bundle $\det E|_U$ has local holomorphic frame \begin{align*}s := e_1 \wedge \cdots \wedge e_r.\end{align*} The connection induced by $\nabla^E$ on $\det E$ satisfies the Leibniz rule on exterior products, hence \begin{align*}\nabla^{\det E}s = \sum_{j=1}^r e_1 \wedge \cdots \wedge e_{j-1} \wedge \nabla^E e_j \wedge e_{j+1} \wedge \cdots \wedge e_r.\end{align*} Substituting the definition of $\theta_{ij}$, every term with $i \ne j$ contains two equal frame vectors in the wedge product and therefore vanishes. The surviving term in the $j$th summand is $s \otimes \theta_{jj}$. Thus \begin{align*}\nabla^{\det E}s = s \otimes \sum_{j=1}^r \theta_{jj}.\end{align*} Therefore the connection one-form of $\nabla^{\det E}$ in the frame $s$ is \begin{align*}\operatorname{tr}(\theta) := \sum_{j=1}^r \theta_{jj}.\end{align*}[/step]

[guided]We work locally because curvature forms of connections are determined by their values in local frames, and equality of local representatives on every such open set gives equality globally. Choose an open set $U \subset X$ with a holomorphic frame $e_1,\dots,e_r$ for $E|_U$. Let $A^1(U)$ denote the space of smooth complex-valued one-forms on $U$. The Chern connection $\nabla^E$ can be written in this frame by declaring one-forms $\theta_{ij} \in A^1(U)$, for $1 \le i,j \le r$, through the formula \begin{align*}\nabla^E e_j = \sum_{i=1}^r e_i \otimes \theta_{ij}.\end{align*} The matrix $\theta=(\theta_{ij})$ is the local connection matrix of $\nabla^E$. Before computing on the determinant line, we must identify which connection we are computing. Let $\widetilde{\nabla}$ denote the connection on $\Lambda^r E$ induced from $\nabla^E$ by differentiating one wedge factor at a time. We claim that $\widetilde{\nabla}$ is the same connection as the Chern connection $\nabla^{\det E}$ appearing in the theorem. The reason is that the defining properties of the Chern connection are preserved by taking the determinant exterior power. First, $\widetilde{\nabla}$ is compatible with the Hermitian metric on $\Lambda^r E$ induced from the metric $h_E$ on $E$: for decomposable local sections $u_1\wedge\cdots\wedge u_r$ and $v_1\wedge\cdots\wedge v_r$, the induced metric is the determinant of the matrix whose $(i,j)$ entry is $h_E(u_i,v_j)$. Differentiating this determinant and using the metric-compatibility of $\nabla^E$ differentiates one $u_i$ or one $v_j$ at a time, which is exactly the metric-compatibility identity for $\widetilde{\nabla}$. Second, the $(0,1)$-part of $\widetilde{\nabla}$ is the holomorphic structure on $\det E$: since $\nabla^E$ is the Chern connection, $(\nabla^E)^{0,1}=\bar\partial_E$, and the induced operator on $u_1\wedge\cdots\wedge u_r$ is the exterior derivation induced by $\bar\partial_E$, namely $\bar\partial_{\det E}$. Therefore $\widetilde{\nabla}$ satisfies the two defining properties of the Chern connection on the holomorphic Hermitian line bundle $\det E$. By uniqueness of the Chern connection, $\widetilde{\nabla}=\nabla^{\det E}$. Now pass to the determinant line. Its local frame over $U$ is \begin{align*}s := e_1 \wedge \cdots \wedge e_r.\end{align*} The induced connection on an exterior product differentiates one factor at a time. Therefore \begin{align*}\nabla^{\det E}s = \sum_{j=1}^r e_1 \wedge \cdots \wedge e_{j-1} \wedge \nabla^E e_j \wedge e_{j+1} \wedge \cdots \wedge e_r.\end{align*} Substitute the connection formula for $\nabla^E e_j$. In the $j$th summand, the term involving $e_i \otimes \theta_{ij}$ contributes \begin{align*}e_1 \wedge \cdots \wedge e_{j-1} \wedge e_i \wedge e_{j+1} \wedge \cdots \wedge e_r \otimes \theta_{ij}.\end{align*} If $i \ne j$, then $e_i$ already appears elsewhere in the wedge product, so the alternating property of the exterior product makes this term zero. If $i=j$, the wedge product is exactly $s$. Hence only the diagonal terms survive, and we get \begin{align*}\nabla^{\det E}s = s \otimes \sum_{j=1}^r \theta_{jj}.\end{align*} This proves that the local connection one-form of the determinant connection is the trace one-form \begin{align*}\operatorname{tr}(\theta) := \sum_{j=1}^r \theta_{jj}.\end{align*}[/guided]

[step:Compare the two local curvature forms] Let $\Omega=(\Omega_{ij})$ denote the curvature matrix of $\nabla^E$ in the frame $e_1,\dots,e_r$. With the convention $\Theta_E=(\nabla^E)^2$, its local matrix is \begin{align*}\Omega = d\theta + \theta \wedge \theta.\end{align*} Thus \begin{align*}\operatorname{tr}(\Theta_E)|_U = \operatorname{tr}(\Omega) = d\operatorname{tr}(\theta) + \operatorname{tr}(\theta \wedge \theta).\end{align*} We compute the last term directly: \begin{align*}\operatorname{tr}(\theta \wedge \theta) = \sum_{i=1}^r \sum_{k=1}^r \theta_{ik} \wedge \theta_{ki}.\end{align*} The diagonal summands satisfy $\theta_{ii}\wedge\theta_{ii}=0$ because $\theta_{ii}$ is a one-form. For $i \ne k$, the two summands indexed by $(i,k)$ and $(k,i)$ cancel: \begin{align*}\theta_{ik}\wedge\theta_{ki}+\theta_{ki}\wedge\theta_{ik}=0.\end{align*} Therefore \begin{align*}\operatorname{tr}(\theta \wedge \theta)=0.\end{align*} Hence \begin{align*}\operatorname{tr}(\Theta_E)|_U = d\operatorname{tr}(\theta).\end{align*} [/step]

[step:Identify this form with the curvature of the determinant connection] Since $\det E$ is a line bundle and its local connection one-form in the frame $s$ is $\operatorname{tr}(\theta)$, the local curvature form of $\nabla^{\det E}$ is \begin{align*}\Theta_{\det E}|_U = d\operatorname{tr}(\theta).\end{align*} The previous step gives \begin{align*}\Theta_{\det E}|_U = \operatorname{tr}(\Theta_E)|_U.\end{align*} Because $U$ was an arbitrary local holomorphic frame domain, these local equalities patch to a global equality of $A^{1,1}(X)$-forms under the scalar-multiplication identification $A^{1,1}(X;\operatorname{End}(\det E))\cong A^{1,1}(X)$ stated in the theorem. Therefore \begin{align*}\Theta_{\det E}=\operatorname{tr}(\Theta_E).\end{align*} This proves the theorem. [/step]