 Solution

The Chern connection of a Hermitian metric on a holomorphic vector bundle is the unique connection on a vector bundle compatible with that metric and whose $(0,1)$ part is the bundle's Dolbeault operator ${\bar{\partial }}_E$. Fix a holomorphic local frame and column coefficients for sections. Write the metric as $h(s,t)={\bar{s}}^{T}Ht$, conjugate-linear in the first argument, and write the connection as $Ds=ds+As$. The condition $D^{0,1}={\bar{\partial }}_E$ forces $A$ to have type $(1,0)$. Metric compatibility requireswhere the dagger conjugates the differential-form coefficients as well as transposing the matrix. Taking the $(1,0)$ part gives $\boxed{A=H^{-1}\partial H}$. Its conjugate-transpose supplies the $(0,1)$ metric equation because $H$ is Hermitian. This proves uniqueness and local existence.

Under a holomorphic change of frame $e^{\prime }=eg$, the metric matrix becomes $H^{\prime }=g^{†}Hg$. The local formula for the Chern connection on a vector bundle then givesThis is precisely the transformation rule for a connection on a vector bundle, so the local connections glue and establish global existence. No Kähler hypothesis is needed for this part.

Extend $D$ to vector-bundle-valued differential forms by the graded Leibniz rule. The curvature form of a connection is the tensorial square $F_h=D^2$, acting by exterior multiplication. In the chosen frame,Indeed $\partial A+A\wedge A=0$ by differentiating $H^{-1}H=I$. It follows that $F_h$ has type $(1,1)$; the gauge change is $F_h^{\prime }=g^{-1}{F}_{h}g$. Thus it is a global smooth two-form with values in $\mathrm{End}(E)$, namely an element of ${\mathcal{A}}^2(\mathrm{End}(E))$.