[proofplan] We express $c_k(E,h)$ as the value of the $k$th determinant coefficient, an invariant polynomial on $\mathfrak{gl}_r(\mathbb C)$, applied to the curvature of the Chern connection. The closedness follows from the Chern-Weil differential identity together with the Bianchi identity for the curvature. To prove independence of the Hermitian metric, we join two metrics by a smooth path, differentiate the corresponding Chern-Weil forms, and use the Chern-Weil transgression formula to show that the derivative is exact. Finally, the standard Chern-Weil theorem identifies these metric-independent de Rham classes, with the stated normalization, with the Chern classes of $E$. [/proofplan]

[step:Represent the determinant coefficients as invariant polynomials] Let $r=\operatorname{rank}_{\mathbb C} E$. Let $\mathfrak{gl}_r(\mathbb C)$ denote the complex Lie algebra of all $r \times r$ complex matrices, let $GL_r(\mathbb C)$ denote the group of invertible $r \times r$ complex matrices, and let $I_r \in GL_r(\mathbb C)$ denote the identity matrix. For each $0 \leq k \leq r$, define the homogeneous polynomial \begin{align*} P_k: \mathfrak{gl}_r(\mathbb C) &\to \mathbb C \end{align*} by requiring \begin{align*} \det\left(I_r + \frac{i}{2\pi}A\right) = \sum_{k=0}^{r} P_k(A) \end{align*} for every $A \in \mathfrak{gl}_r(\mathbb C)$, where $P_k(A)$ is homogeneous of degree $k$ in the matrix entries of $A$. Because the determinant is invariant under conjugation, each $P_k$ satisfies \begin{align*} P_k(gAg^{-1}) = P_k(A) \end{align*} for every $g \in GL_r(\mathbb C)$ and every $A \in \mathfrak{gl}_r(\mathbb C)$. Let $\widetilde P_k$ denote the symmetric $k$-linear polarization of $P_k$. If $\nabla^h$ is the Chern connection and $\Theta_h \in \Omega^2(X;\operatorname{End}(E))$ is its curvature form, then \begin{align*} c_k(E,h) = \widetilde P_k(\Theta_h,\dots,\Theta_h). \end{align*} This equality is exactly the degree-$2k$ component of \begin{align*} \det\left(I_E + \frac{i}{2\pi}\Theta_h\right). \end{align*} [/step]

[step:Use the Bianchi identity to prove the Chern-Weil forms are closed]Fix $k \in \{0,\dots,r\}$. The case $k=0$ gives $c_0(E,h)=1$, so $d c_0(E,h)=0$. Assume $k \geq 1$. Let \begin{align*} d_{\nabla^h}: \Omega^m(X;\operatorname{End}(E)) \to \Omega^{m+1}(X;\operatorname{End}(E)) \end{align*} denote the exterior covariant derivative induced by $\nabla^h$ on $\operatorname{End}(E)$-valued forms. Since $P_k$ is invariant under conjugation, the Chern-Weil differential identity gives \begin{align*} d\,\widetilde P_k(\Theta_h,\dots,\Theta_h)=k\,\widetilde P_k(d_{\nabla^h}\Theta_h,\Theta_h,\dots,\Theta_h). \end{align*} By the Bianchi identity for the curvature of the connection $\nabla^h$, one has \begin{align*} d_{\nabla^h}\Theta_h = 0. \end{align*} Therefore \begin{align*} d c_k(E,h)=d\,\widetilde P_k(\Theta_h,\dots,\Theta_h)=0. \end{align*} Thus $c_k(E,h)$ is closed.[/step]

[guided]We now prove closedness carefully. The object being differentiated is not an arbitrary scalar expression: it is obtained by applying an invariant polynomial to an endomorphism-valued curvature form. This is exactly the setting of the Chern-Weil differential identity. Let \begin{align*} d_{\nabla^h}: \Omega^m(X;\operatorname{End}(E)) \to \Omega^{m+1}(X;\operatorname{End}(E)) \end{align*} be the exterior covariant derivative induced by the Chern connection $\nabla^h$. The curvature of $\nabla^h$ is the form \begin{align*} \Theta_h \in \Omega^2(X;\operatorname{End}(E)). \end{align*} For $k=0$, the definition gives $c_0(E,h)=1$, hence $d c_0(E,h)=0$. Now assume $k \geq 1$. The determinant coefficient $P_k$ is invariant under conjugation because \begin{align*} \det\left(I_r + \frac{i}{2\pi}gAg^{-1}\right) = \det\left(g\left(I_r + \frac{i}{2\pi}A\right)g^{-1}\right) = \det\left(I_r + \frac{i}{2\pi}A\right) \end{align*} for every $g \in GL_r(\mathbb C)$ and $A \in \mathfrak{gl}_r(\mathbb C)$. This invariance is the precise algebraic hypothesis needed by the Chern-Weil differential identity. Applying that identity to the connection $\nabla^h$ and the invariant polynomial $P_k$ gives \begin{align*} d\,\widetilde P_k(\Theta_h,\dots,\Theta_h) = k\,\widetilde P_k(d_{\nabla^h}\Theta_h,\Theta_h,\dots,\Theta_h). \end{align*} The remaining input is the Bianchi identity. For the curvature of any connection, and hence for the curvature of the Chern connection $\nabla^h$, the Bianchi identity states that \begin{align*} d_{\nabla^h}\Theta_h = 0. \end{align*} Substituting this into the preceding Chern-Weil identity yields \begin{align*} d c_k(E,h) = d\,\widetilde P_k(\Theta_h,\dots,\Theta_h) = k\,\widetilde P_k(0,\Theta_h,\dots,\Theta_h) = 0. \end{align*} Therefore $c_k(E,h)$ is a closed complex-valued $2k$-form on $X$.[/guided]

[step:Connect two Hermitian metrics by a smooth path] Let $h_0$ and $h_1$ be smooth Hermitian metrics on $E$. Define a smooth path of Hermitian metrics \begin{align*} h_t: E \times_X E &\to \mathbb C \end{align*} for $t \in [0,1]$ by \begin{align*} h_t = (1-t)h_0 + t h_1. \end{align*} For each $t \in [0,1]$, the form $h_t$ is Hermitian and positive definite on every fiber because it is a convex combination of two positive definite Hermitian forms. Let \begin{align*} \nabla_t: \Omega^0(X;E) &\to \Omega^1(X;E) \end{align*} be the Chern connection of $(E,h_t)$, and let \begin{align*} \Theta_t \in \Omega^2(X;\operatorname{End}(E)) \end{align*} be its curvature. Define the connection variation \begin{align*} A_t \in \Omega^1(X;\operatorname{End}(E)) \end{align*} by \begin{align*} A_t = \frac{\partial \nabla_t}{\partial t}. \end{align*} [/step]

[step:Apply the transgression formula to the path of Chern connections]Fix $k \geq 1$. The Chern-Weil transgression formula for the smooth path of connections $(\nabla_t)_{t \in [0,1]}$ states that \begin{align*} \frac{\partial}{\partial t}\widetilde P_k(\Theta_t,\dots,\Theta_t) = d\left(k\,\widetilde P_k(A_t,\Theta_t,\dots,\Theta_t)\right), \end{align*} where $A_t=\partial \nabla_t/\partial t$ occupies the first argument and $\Theta_t$ occupies the remaining $k-1$ arguments. The hypotheses of this formula are satisfied because $(\nabla_t)_{t \in [0,1]}$ is a smooth path of connections on the fixed smooth complex vector bundle $E$, and $P_k$ is invariant under conjugation. Thus \begin{align*} \frac{\partial}{\partial t}c_k(E,h_t) = d\,T_{k,t}, \end{align*} where the transgression form \begin{align*} T_{k,t} \in \Omega^{2k-1}(X;\mathbb C) \end{align*} is defined by \begin{align*} T_{k,t}=k\,\widetilde P_k(A_t,\Theta_t,\dots,\Theta_t). \end{align*} For $k=0$, the form $c_0(E,h_t)=1$ is independent of $t$.[/step]

[guided]The purpose of introducing the path $h_t$ is to turn metric independence into a differential statement in the parameter $t$. If we can show that the $t$-derivative of $c_k(E,h_t)$ is an exact form, then the de Rham cohomology class cannot change as $t$ varies. For each $t \in [0,1]$, let \begin{align*} \nabla_t: \Omega^0(X;E) \to \Omega^1(X;E) \end{align*} be the Chern connection of $(E,h_t)$, and let \begin{align*} \Theta_t \in \Omega^2(X;\operatorname{End}(E)) \end{align*} be its curvature. The derivative of the connection in the parameter $t$ is an endomorphism-valued one-form \begin{align*} A_t \in \Omega^1(X;\operatorname{End}(E)), \end{align*} defined by \begin{align*} A_t = \frac{\partial \nabla_t}{\partial t}. \end{align*} This is a tensorial object because the difference of two connections on the same vector bundle is an $\operatorname{End}(E)$-valued one-form, and the derivative of a smooth path of connections is obtained as the limit of such differences. We apply the Chern-Weil transgression formula to the smooth path $(\nabla_t)_{t \in [0,1]}$ and the invariant polynomial $P_k$. The formula requires two hypotheses: first, that $(\nabla_t)$ is a smooth path of connections on a fixed complex vector bundle, and second, that $P_k$ is invariant under conjugation. The first holds because $h_t$ is a smooth path of Hermitian metrics and the Chern connection depends smoothly on the metric. The second was proved from determinant invariance. Therefore \begin{align*} \frac{\partial}{\partial t}\widetilde P_k(\Theta_t,\dots,\Theta_t) = d\left(k\,\widetilde P_k(A_t,\Theta_t,\dots,\Theta_t)\right). \end{align*} Since \begin{align*} c_k(E,h_t)=\widetilde P_k(\Theta_t,\dots,\Theta_t), \end{align*} we obtain \begin{align*} \frac{\partial}{\partial t}c_k(E,h_t) = d\,T_{k,t}, \end{align*} where \begin{align*} T_{k,t}:=k\,\widetilde P_k(A_t,\Theta_t,\dots,\Theta_t) \end{align*} is a complex-valued differential form of degree $2k-1$. This is the exactness statement needed for metric independence. For $k=0$, no transgression is needed, since $c_0(E,h_t)=1$ for all $t$.[/guided]

[step:Integrate the exact derivative to prove metric independence] Assume $k \geq 1$. Let $\mathcal L^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $[0,1]$. Integrating the identity \begin{align*} \frac{\partial}{\partial t}c_k(E,h_t)=d\,T_{k,t} \end{align*} over $t \in [0,1]$ with respect to $\mathcal L^1$ gives \begin{align*} c_k(E,h_1)-c_k(E,h_0) = \int_0^1 d\,T_{k,t}\,d\mathcal L^1(t). \end{align*} Because exterior differentiation acts only on the $X$-variables and the path is smooth in $t$, differentiation commutes with integration over $[0,1]$. Hence \begin{align*} c_k(E,h_1)-c_k(E,h_0) = d\left(\int_0^1 T_{k,t}\,d\mathcal L^1(t)\right). \end{align*} The form \begin{align*} \int_0^1 T_{k,t}\,d\mathcal L^1(t) \in \Omega^{2k-1}(X;\mathbb C) \end{align*} is smooth because $T_{k,t}$ depends smoothly on $t$. Therefore $c_k(E,h_1)-c_k(E,h_0)$ is exact, so \begin{align*} [c_k(E,h_0)] = [c_k(E,h_1)] \in H_{\mathrm{dR}}^{2k}(X;\mathbb C). \end{align*} For $k=0$, both classes are represented by $1$, so the same conclusion holds. [/step]

[step:Identify the common class with the Chern class] By the standard Chern-Weil theorem for complex vector bundles, applied with the invariant polynomial determined by the normalization \begin{align*} \det\left(I_E+\frac{i}{2\pi}\Theta\right), \end{align*} the de Rham cohomology class represented by the $2k$-degree component of this Chern-Weil form for any connection compatible with the complex vector bundle structure is precisely the de Rham image of the topological Chern class $c_k(E)$ in $H_{\mathrm{dR}}^{2k}(X;\mathbb C)$. Applying this theorem to the Chern connection $\nabla^h$ gives \begin{align*} [c_k(E,h)] = c_k(E) \in H_{\mathrm{dR}}^{2k}(X;\mathbb C), \end{align*} where the right-hand side denotes the de Rham image of the topological Chern class. Together with closedness and metric independence proved above, this proves the theorem. [/step]