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For a smooth complex vector bundle $E$, a connection is a complex-linear map $D:\Gamma (E)\to A^1(M;E)$ satisfying $D(fs)=df\otimes s+fDs$ for smooth complex functions $f$. Local trivializations supply flat local connections. A locally finite smooth partition of unity $({\rho }_i)$ subordinate to them gives $D={\sum }_i{\rho }_i{D}_i$; local finiteness makes the sum smooth and ${\sum }_i{\rho }_i=1$ gives the Leibniz rule. Thus connections on a vector bundle always exist on the usual paracompact smooth manifold.

Extend $D$ to bundle-valued forms by $D(\alpha \otimes s)=d\alpha \otimes s+(-1{)}^{\mathrm{deg}\alpha }\alpha \wedge Ds$. Squaring shows that $D^2(fs)=f{D}^{2}s$, so the curvature form is the endomorphism-valued two-form $\Theta =D^2$. With a row frame $e=(e_1,\dots ,e_r)$, write $De=eA$; for coefficient columns $s=ev$, $Ds=e(dv+Av)$. Applying $D$ again and using the graded Leibniz rule givesThe cross terms $-A\wedge dv$ and $A\wedge dv$ cancel. Hence the Cartan curvature matrix equation in this convention isFor a frame change $e^{\prime }=eg$, the connection matrix and curvature transform as $A^{\prime }=g^{-1}Ag+g^{-1}dg$ and ${\Theta }^{\prime }=g^{-1}\Theta g$. Trace is invariant under conjugation, so the local forms $\mathrm{Tr}\Theta$ patch into the global trace of vector-bundle curvature.

The induced determinant connection on $\det E={\Lambda }^{r}E$ iswhere the one-form factor of each $D{s}_j$ is placed in front of the section factors. In the local determinant frame $e_1\wedge \cdots \wedge e_r$, only the diagonal terms survive, so its connection form is $\mathrm{Tr}A$. Its curvature is $d\mathrm{Tr}A$. Sinceby pairing off-diagonal terms and using $A_{ii}\wedge A_{ii}=0$, we obtainIt is closed locally by $d^2=0$, and thus closed globally.

For two connections, $B=D_1-D_0$ is a global endomorphism-valued one-form: the derivative terms cancel in their Leibniz rules. In a local frame $A_1=A_0+B$, soTaking traces cancels the mixed terms, because both factors are one-forms, and cancels $\mathrm{Tr}(B\wedge B)$. Therefore the trace curvature transgression formula isThe right-hand side is globally exact, proving independence of the de Rham cohomology class without needing a sheaf-cohomology identification.

Finally let $e$ be a holomorphic frame for the holomorphic line bundle, with $h=\Vert e{\Vert }^2>0$. Define $De=e\partial \log h$, extending by the Leibniz rule. If $e^{\prime }=eg$ for a nowhere-zero holomorphic function $g$, then $h^{\prime }=|g{|}^{2}h$ andThis is exactly the line-bundle frame-change law for a connection, so the local definitions glue globally. Its $(0,1)$ part is the Dolbeault operator. Metric compatibility follows from $d\log h=\partial \log h+\bar{\partial }\log h$, or $dh=h(A+\overline{A})$ for $A=\partial \log h$. These conditions uniquely force this $A$, identifying the Chern connection. Since a scalar one-form wedges with itself to zero,This agrees with the local formula for the Chern connection on a line bundle; keeping the order of the two Dolbeault differentials fixes the sign.

Let another local frame be $e^{\prime }=eA$ with $A:U\to G{L}_r(\mathbb{R})$ smooth. Since the coefficient columns satisfy $u=A{u}^{\prime }$, the Leibniz rule gives the change of frame of a vector-bundle connection:The curvature form of a connection is linear over smooth functions in its section argument, so its matrix transforms homogeneously:ConsequentlyThe entries of $A$ are functions, of degree zero, so ordinary cyclic invariance of trace applies without a graded sign. The local forms therefore glue to a globally defined two-form $\alpha =\mathrm{tr}\Theta$, the trace of vector-bundle curvature.

Apply the Bianchi identity proved in part (b):For the first term,The middle interchange has sign $(-1{)}^{2\cdot 1}=1$, and the next equality simply renames $i,j$. ThusOne can also check closedness locally: the off-diagonal terms in $\mathrm{tr}(\theta \wedge \theta )={\sum }_{i,j}{\theta }_{ij}\wedge {\theta }_{ji}$ cancel in pairs, while the diagonal terms vanish. Hence $\alpha =d(\mathrm{tr}\theta )$ in each frame, and $d^2=0$ gives the same conclusion.