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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.

A connection on a vector bundle is a complex-linear map from smooth sections to vector-bundle-valued one-forms satisfyingChoose a locally finite trivializing cover, its componentwise flat connections on a vector bundle $D_i$, and a subordinate smooth partition of unity ${\rho }_i$. The formula $D={\sum }_i{\rho }_i{D}_i$ is globally meaningful: each weighted term extends by zero outside its chart. Its Leibniz rule follows from ${\sum }_i{\rho }_i=1$, proving every smooth complex vector bundle admits a connection.

Extend the connection on a vector bundle to bundle-valued differential forms byThe curvature form of a connection is ${\Theta }_D=D^2$. Applying this twice to $fs$ shows that the $df$ terms cancel, so $D^2$ is tensorial and defines an $\mathrm{End}(E)$-valued two-form. With column coordinates for sections and connection one-form $A$, the local expression is $D=d+A\wedge$. Direct expansion on an arbitrary bundle-valued form givesThe terms containing a derivative of the argument cancel by the graded Leibniz rule. Thus Cartan curvature matrix equation isHere multiplication includes matrix multiplication and the exterior product of form entries. For a frame change $e^{\prime }=eg$, the connection one-form and curvature form of a connection transform asThe trace is consequently frame independent. Alsothe diagonal terms vanish and the off-diagonal terms cancel in pairs. Locally $\mathrm{Tr}\Theta =d\mathrm{Tr}A$, and hence $d\mathrm{Tr}\Theta =0$. The local primitives need not agree, but the two-form does. We obtain a global closed two-form $\mathrm{Tr}{\Theta }_D$.

The determinant connection on $\det E={\bigwedge }^{r}E$ is defined intrinsically bywhere the differential-form coefficient of $D{s}_j$ is placed first. In a local frame, only the diagonal components contribute to the derivative of $e_1\wedge \cdots \wedge e_r$, so its connection one-form is $\mathrm{Tr}A$. A line-bundle connection has curvature $d\mathrm{Tr}A$, because a scalar one-form wedges with itself to zero. ThereforeTo see independence of the de Rham cohomology class, write $D_1-D_0=a$, a globally defined endomorphism-valued one-form. The curvature difference formula and the same trace cancellations give the trace curvature transgressionTheir difference is globally an exact differential form, so their de Rham cohomology classes coincide.

For the Hermitian metric on a holomorphic vector bundle, use the displayed ordering of the local matrices and put $B=(\partial h)h^{-1}$. The identity $\bar{\partial }{h}^{-1}=-h^{-1}(\bar{\partial }h)h^{-1}$ givesThe plus sign comes from applying the graded Leibniz rule to the one-form $\partial h$. By the derivative formula for a determinant,The Hermitian positive-definite matrix $h$ has positive real determinant, so this logarithm is an ordinary smooth real function. The trace of Chern curvature in these conventions is thereforeIt is locally an exact differential form and in particular closed.

A holomorphic local frame change multiplies $\det h$ by $|\det g{|}^2$. A nowhere-zero holomorphic function has a local holomorphic logarithm, so $\bar{\partial }\partial \log |\det g{|}^2=0$. Thus the preceding expression for $\alpha$ is independent of the frame and glues globally. For two Hermitian metrics on a holomorphic vector bundle, the functionis globally defined because the frame-change factors cancel. Their two-forms differ by $d(\partial f)$. Hence the class of $\alpha$ is independent of the metric. The two-form is generally imaginary-valued; the metric independence statement is in complex de Rham cohomology, or equivalently for the real form $i\alpha$.

The trace of the curvature form of a connection is a global closed differential form of degree two. Under a frame change $e^{\prime }=eg$, the connection matrix and curvature change by $A^{\prime }=g^{-1}Ag+g^{-1}dg$ and ${\Theta }^{\prime }=g^{-1}\Theta g$. Thus trace is independent of frame. The Bianchi identity gives $d\Theta =\Theta \wedge A-A\wedge \Theta$. Taking the trace cancels the two terms: one-forms commute with two-forms in the graded sense, and renaming the matrix indices gives equality of their traces. Locally $\mathrm{tr}\Theta =d\mathrm{tr}A$, since the trace of $A\wedge A$ vanishes by pairing its off-diagonal terms. It is the curvature of the determinant connection.