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Let $E\to M$ have rank $r$. A vector bundle covariant derivative is a real-linear operatorPairing its one-form slot with a vector field $X$ gives ${\nabla }_{X}s$, with ${\nabla }_{fX}s=f{\nabla }_{X}s$ and ${\nabla }_X(fs)=X(f)s+f{\nabla }_{X}s$.

In the horizontal description, a connection on a vector bundle $A$ is a smooth splitting $T_{a}E=V_{a}E\oplus H_{a}E$, where $V_{a}E=\ker (d\pi {)}_a$, and the horizontal choice is compatible with fibre addition and scalar multiplication. This linearity is essential; an arbitrary nonlinear horizontal splitting is not a vector-bundle connection. Identify a vertical vector at $a\in E_x$ with the element of $E_x$ obtained by differentiating the curve $a+tv$. Let $K_A:T_{a}E\to E_x$ be vertical projection followed by this identification. The induced vector bundle covariant derivative isIn a local frame of a vector bundle $e=(e_1,\dots ,e_r)$, a linear horizontal splitting has the formfor a matrix-valued connection one-form $\Gamma$. Thus for $s=ev$, ${\nabla }^{A}s=e(dv+\Gamma v)$, which directly verifies the Leibniz rule.

Conversely, the Leibniz rule makes $\nabla$ local: if a section vanishes near a point, multiply it by a smooth cutoff function supported in that neighborhood and equal to one near the point to see that its derivative vanishes there too. Local sections can therefore be differentiated by any extension agreeing near the point. Given $\nabla$, define its local connection one-form by $\nabla e_b={\sum }_a{e}_a{\Gamma }_b^a$. The Leibniz rule gives $\nabla (ev)=e(dv+\Gamma v)$. Under $e^{\prime }=eg$, $v^{\prime }=g^{-1}v$, one hasThe tangent fibre coordinate transforms by $w^{\prime }=g^{-1}w-g^{-1}(dg)(\xi )g^{-1}v$, so $w=-\Gamma (\xi )v$ transforms precisely to $w^{\prime }=-{\Gamma }^{\prime }(\xi )v^{\prime }$. Consequently these local horizontal subspaces of a vector bundle connection glue to a smooth linear splitting inducing the given $\nabla$. Its local formula also shows uniqueness. This proves the horizontal connection associated to a covariant derivative construction.

A horizontal lift $\tilde{\gamma }$ of $\gamma$ is a smooth curve in $E$ with $\pi \circ \tilde{\gamma }=\gamma$ and $\dot{\tilde{\gamma }}(t)\in H_{\tilde{\gamma }(t)}E$. In a local frame write $\tilde{\gamma }(t)=e(\gamma (t))v(t)$. Horizontality is exactly the linear ordinary differential equationWe use existence and uniqueness for an ODE with smooth coefficients, and the continuation theorem that a solution extends across a finite interval when it remains bounded inside the coordinate domain. For a linear system $\dot{v}=-C(t)v$ with continuous coefficients bounded by $L$ on a compact time interval, the Gronwall inequality bounds $|v(t)|$ by its initial norm times $e^{L|t-t_0|}$, preventing finite-time fibre escape. Cover a compact subinterval of $I$ by finitely many bundle charts and partition it so each piece lies in one chart. Solve successively, changing fibre frames at the partition points. Coordinate invariance of the horizontal equation and uniqueness make the pieces agree. Exhausting $I$ by compact intervals containing $0$ produces a unique horizontal lift on all of $I$ for every prescribed $a_0$. This is global continuation of linear parallel transport.

The lift equation is linear in its initial value. Hence the endpoint map $P_{\gamma }:E_{\gamma (0)}\to E_{\gamma (1)}$ is linear. Solving the same equation backwards, or transporting along $t\mapsto \gamma (1-t)$, gives its inverse, soThis is parallel transport. For two parallel lifts $s_1(t),s_2(t)$, the assumed metric connection identity extends along the embedded curve and becomesHere $D_t$ is the covariant derivative along a curve. The permitted extension of sections and the tangent vector along the embedded curve justifies applying the bundle identity; equivalently it follows in each local frame. Thereforeso parallel transport is an isometry. This is the principle that parallel transport preserves a fibre metric.

Fix the convention that the Hermitian metric is conjugate-linear in its first argument. In a holomorphic local frame $e=(e_1,\dots ,e_r)$, let $H_{ij}=h(e_i,e_j)$, so $h(ev,ew)=v^{†}Hw$. Write $De=e\theta$, or equivalently $D(ev)=e(dv+\theta v)$.

Compatibility with the holomorphic vector bundle structure means $D^{0,1}={\bar{\partial }}_E$. Since the frame is holomorphic, this forces ${\theta }^{0,1}=0$. Metric compatibility meansor, in the chosen frame, $dH={\theta }^{†}H+H\theta$. Conjugate transpose here also conjugates the one-form factors. Taking the $(1,0)$ component gives $\partial H=H\theta$, and henceThis proves uniqueness and suggests existence. The proposed $\theta$ has type $(1,0)$, and its conjugate transpose obeys ${\theta }^{†}H=\bar{\partial }H$, since $H=H^{†}$. Therefore $dH={\theta }^{†}H+H\theta$, proving metric compatibility as well.

Under a change of holomorphic local frame $e^{\prime }=eG$, we have $H^{\prime }=G^{†}HG$ and $\partial G^{†}=0$. A direct computation yieldsBecause $G$ is holomorphic, $\partial G=dG$; this is exactly the transformation rule for a connection one-form. The locally defined connections therefore glue to a global connection on a vector bundle. The Chern connection exists uniquely, and its local formula for the Chern connection on a vector bundle is

In a unitary frame the metric matrix is $I$. If $A$ is the connection one-form, metric compatibility gives $0=dI=A^{\ast }+A$, sothe matrix is Skew-Hermitian. In a holomorphic local frame, every frame vector is annihilated by ${\bar{\partial }}_E$, and compatibility with the holomorphic structure gives

In a holomorphic local frame $e$, let $H=(h(e_i,e_j))$ and write $De=eA$. Holomorphic compatibility forces $A^{0,1}=0$, while metric compatibility forces the local formula for the Chern connection on a vector bundleThis proves uniqueness. The formula transforms by the connection one-form law under a holomorphic frame change, so the local definitions glue and satisfy both conditions. This proves existence of the unique Chern connection.

A holomorphic line bundle is a complex line bundle with holomorphic transition functions, and a holomorphic section is one whose coefficient in every holomorphic local frame is holomorphic. Given a Hermitian metric on a holomorphic vector bundle $h$, its Chern connection is the connection $\nabla$ satisfying

Let $e$ be a nonvanishing holomorphic local frame, put $H=h(e,e)>0$, and write $\nabla e=Ae$. The first condition forces $A^{0,1}=0$, while metric compatibility forcesThis determines $\nabla$ uniquely and also constructs it. If $e^{\prime }=ge$ for a nowhere-zero holomorphic function $g$, then $A^{\prime }=A+g^{-1}\partial g$, exactly the connection one-form transformation law, so the local constructions glue.

For a line bundle, $A\wedge A=0$, and the curvature form of a connection isIt has type $(1,1)$. Under $e^{\prime }=ge$, the extra term $g^{-1}\partial g$ is closed, so the curvature is unchanged and therefore global. This is the local formula for the Chern connection on a line bundle.

Any other Hermitian metric has the form $\hat{h}=e^{u}h$ for a global smooth real function $u$. Its local squared norm is $\hat{H}=e^{u}H$, whence

Connections ${\nabla }_1$ and ${\nabla }_2$ induce the tensor product connectionIts connection form in a product frame is $A_1+A_2$, so the curvature of a tensor product connection is $F_1+F_2$. For Chern connections, equip $L_1\otimes L_2$ with the product metricThe tensor product connection has the correct $(0,1)$ part and preserves this metric, so uniqueness identifies it with the Chern connection of $h$.