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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 ${\mathcal{A}}^2(\mathrm{End}(E))$.

The Chern connection decomposes on vector-bundle-valued differential forms as $D=D^{\prime }+D^{\prime \prime }$, where $D^{\prime }={\nabla }_h^{\prime }$ raises holomorphic degree and $D^{\prime \prime }={\nabla }_h^{\prime \prime }={\bar{\partial }}_E$ raises antiholomorphic degree. In a holomorphic local frame, $D^{\prime }=\partial +A\wedge$ and $D^{\prime \prime }=\bar{\partial }$. The type of the curvature form of a connection givesThe Kähler metric and the Hermitian metric define the $L^2$ inner product and the formal adjoints $D^{\prime \ast },D^{\prime \prime \ast }$. The Dolbeault Laplacians areLet $L=\omega \wedge$ be the Lefschetz operator of a Kähler manifold and $\Lambda =L^{\ast }$ its adjoint Lefschetz operator. With the ordinary commutator convention $[A,B]=AB-BA$, the printed Kähler identities giveSubstitution into the two Dolbeault Laplacians, followed by expansion, yieldsThe middle line follows by cancelling the terms with $D^{\prime }\Lambda D^{\prime \prime }$ and $D^{\prime \prime }\Lambda D^{\prime }$; the remaining terms collect the anticommutator of the two differentials. Hence $\boxed{{\Delta }^{\prime \prime }={\Delta }^{\prime }+[i{F}_h,\Lambda ]}$, where $i{F}_h$ denotes its wedge action. This is the Bochner-Kodaira-Nakano identity with the paper's sign convention.

The Kähler form and the Hermitian metric give the $L^2$ inner product on vector-bundle-valued differential forms, using volume ${\omega }^n/n!$. Define the formal adjoint ${\bar{\partial }}_E^{\ast }$ and the elliptic, self-adjoint, nonnegative Dolbeault LaplacianIts harmonic space isThe equality follows from $⟨{\Delta }_E^{\prime \prime }\alpha ,\alpha ⟩=\Vert {\bar{\partial }}_E\alpha {\Vert }^2+\Vert {\bar{\partial }}_E^{\ast }\alpha {\Vert }^2$. On compact $X$, the bundle-valued Dolbeault Hodge decomposition states that this space is finite dimensional and thatThe sum is orthogonal for the $L^2$ inner product and all summands here consist of smooth forms. Every Dolbeault cohomology class has a unique harmonic representative, giving $H_{\bar{\partial }}^{p,q}(X,E)\cong {\mathcal{H}}^{p,q}(X,E)$ and, by the Dolbeault theorem, the corresponding sheaf cohomology isomorphism. This is a decomposition for ${\bar{\partial }}_E$, whose square is zero; it does not require the full Chern connection to be flat.