[proofplan] We compute locally in a holomorphic frame of $E$. The induced determinant frame is the wedge of the frame vectors, and the induced connection differentiates each factor in turn, so its connection one-form is the trace of the connection matrix of $E$. Taking curvature gives $d\operatorname{tr}A$, while the trace of the vector bundle curvature matrix is $\operatorname{tr}(dA+A\wedge A)=d\operatorname{tr}A$ because $\operatorname{tr}(A\wedge A)=0$ for a matrix of one-forms. The component formula is then just the local expression of this trace, with the bundle indices contracted by the Hermitian metric. [/proofplan]

[step:Write the Chern connection in a local holomorphic frame] Fix an [open set](/page/Open%20Set) $U\subset X$ on which $E$ admits a holomorphic frame $e_1,\dots,e_r$. Let \begin{align*} \nabla^E:\Gamma(U,E)\to \Gamma(U,T^*X\otimes E) \end{align*} be the Chern connection of $(E,h)$ restricted to $U$. Define the connection matrix $A=(A_{\alpha\beta})_{1\le \alpha,\beta\le r}$ by the identities \begin{align*} \nabla^E e_\beta=\sum_{\alpha=1}^{r} e_\alpha\otimes A_{\alpha\beta}. \end{align*} Here each $A_{\alpha\beta}$ is a complex-valued one-form on $U$. The curvature of $\nabla^E$ is \begin{align*} \Theta(E,h)=(\nabla^E)^2, \end{align*} and in the chosen frame its matrix is \begin{align*} F=dA+A\wedge A. \end{align*} Thus the trace of the curvature form on $U$ is \begin{align*} \operatorname{tr}\Theta(E,h)=\operatorname{tr}F. \end{align*} [/step]

[step:Compute the induced connection on the determinant frame]Let \begin{align*} s:U\to \det E \end{align*} be the local holomorphic frame of $\det E$ defined by \begin{align*} s=e_1\wedge e_2\wedge \cdots \wedge e_r. \end{align*} Let \begin{align*} \nabla^{\det E}:\Gamma(U,\det E)\to \Gamma(U,T^*X\otimes \det E) \end{align*} be the connection induced from $\nabla^E$ on $\det E$. By the Leibniz rule for the exterior product connection, \begin{align*} \nabla^{\det E}s=\sum_{\beta=1}^{r} e_1\wedge\cdots\wedge \nabla^E e_\beta\wedge\cdots\wedge e_r. \end{align*} Substituting the definition of $A_{\alpha\beta}$, every term with $\alpha\ne\beta$ contains two equal frame vectors in the wedge product and is therefore zero. The remaining terms are \begin{align*} \nabla^{\det E}s=\left(\sum_{\beta=1}^{r}A_{\beta\beta}\right)\otimes s. \end{align*} Therefore the connection one-form of $\nabla^{\det E}$ in the determinant frame $s$ is \begin{align*} \operatorname{tr}A=\sum_{\beta=1}^{r}A_{\beta\beta}. \end{align*}[/step]

[guided]The determinant bundle is the top exterior power of $E$, so the natural local frame is the wedge \begin{align*} s=e_1\wedge e_2\wedge \cdots \wedge e_r. \end{align*} The induced connection on an exterior product is defined by differentiating one factor at a time. Thus \begin{align*} \nabla^{\det E}s=\sum_{\beta=1}^{r} e_1\wedge\cdots\wedge \nabla^E e_\beta\wedge\cdots\wedge e_r. \end{align*} Now insert the connection matrix formula \begin{align*} \nabla^E e_\beta=\sum_{\alpha=1}^{r} e_\alpha\otimes A_{\alpha\beta}. \end{align*} For fixed $\beta$, the contribution with index $\alpha$ replaces the factor $e_\beta$ in the wedge by $e_\alpha$. If $\alpha\ne\beta$, then $e_\alpha$ already appears elsewhere in the wedge, so the alternating property of the exterior product makes that term vanish. If $\alpha=\beta$, the wedge remains exactly $s$. Therefore only the diagonal connection forms survive, and we get \begin{align*} \nabla^{\det E}s=\left(\sum_{\beta=1}^{r}A_{\beta\beta}\right)\otimes s. \end{align*} This is precisely the trace of the connection matrix: \begin{align*} \sum_{\beta=1}^{r}A_{\beta\beta}=\operatorname{tr}A. \end{align*} Hence the induced connection on $\det E$ has local connection one-form $\operatorname{tr}A$ in the frame $s$.[/guided]

[step:Take curvature and identify it with the trace of the vector bundle curvature] Since $\det E$ is a line bundle and its connection one-form in the frame $s$ is $\operatorname{tr}A$, its curvature form on $U$ is \begin{align*} \Theta(\det E,h_{\det E})=d(\operatorname{tr}A). \end{align*} On the other hand, \begin{align*} \operatorname{tr}F=\operatorname{tr}(dA+A\wedge A)=d(\operatorname{tr}A)+\operatorname{tr}(A\wedge A). \end{align*} It remains to compute the last term. By the definition of matrix multiplication with wedge product, \begin{align*} \operatorname{tr}(A\wedge A)=\sum_{\alpha=1}^{r}\sum_{\beta=1}^{r}A_{\alpha\beta}\wedge A_{\beta\alpha}. \end{align*} The diagonal summands vanish because a one-form wedges with itself to zero. For $\alpha\ne\beta$, the two summands indexed by $(\alpha,\beta)$ and $(\beta,\alpha)$ cancel since \begin{align*} A_{\beta\alpha}\wedge A_{\alpha\beta}=-A_{\alpha\beta}\wedge A_{\beta\alpha}. \end{align*} Hence \begin{align*} \operatorname{tr}(A\wedge A)=0. \end{align*} Therefore \begin{align*} \operatorname{tr}\Theta(E,h)=\operatorname{tr}F=d(\operatorname{tr}A)=\Theta(\det E,h_{\det E}) \end{align*} on $U$. [/step]

[step:Pass from the local computation to the global identity and read off components] The open set $U$ and holomorphic frame were arbitrary, and both sides are globally defined two-forms on $X$. Since the local representatives agree in every holomorphic trivialisation, the global identity follows: \begin{align*} \Theta(\det E,h_{\det E})=\operatorname{tr}\Theta(E,h). \end{align*} For the component formula, fix a point $x\in X$, local holomorphic coordinates $z_1,\dots,z_n$ near $x$, and an $h$-unitary frame $e_1,\dots,e_r$ of $E_x$. Let $e^1,\dots,e^r$ denote the [dual basis](/theorems/414) of $E_x^*$, defined by $e^\beta(e_\alpha)=1$ when $\alpha=\beta$ and $e^\beta(e_\alpha)=0$ when $\alpha\ne\beta$. Write the curvature components at $x$ as \begin{align*} \Theta(E,h)_x=\sum_{i,j,\alpha,\beta}R_{i\bar j\alpha\bar\beta}\,dz_i\wedge d\bar z_j\otimes e_\alpha\otimes e^\beta. \end{align*} Taking the trace over the bundle indices gives \begin{align*} \operatorname{tr}\Theta(E,h)_x=\sum_{i,j}\left(\sum_{\alpha=1}^{r}R_{i\bar j\alpha\bar\alpha}\right)dz_i\wedge d\bar z_j. \end{align*} Since this equals $\Theta(\det E,h_{\det E})_x$, the determinant curvature components satisfy \begin{align*} R^{\det E}_{i\bar j}=\sum_{\alpha=1}^{r}R_{i\bar j\alpha\bar\alpha}. \end{align*} This is the desired local coordinate formula. [/step]