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For the $(1,0)$ component of a Chern connection, $(D_E^{\prime }{)}^{\ast }=-{\ast }_{E^{\ast }}{D}_{E^{\ast }}^{\prime }{\ast }_E$, where the middle operator is the corresponding component of the dual Chern connection. This is the holomorphic-degree counterpart of the bundle-valued Dolbeault adjoint.

The Hermitian metric on cotangent vectors is the dual metric, and its exterior-power metric is defined on decomposable vectors by the determinant of their pairwise inner products. Tensor this with the metric on $E$. In a unitary coframe and unitary frame of $E$, the forms obtained by wedging distinct holomorphic and antiholomorphic coframe elements and tensoring with a frame vector form an orthonormal basis. Let $dV={\omega }^n/n!$, and use inner products linear in the first variable.

The bundle-valued conjugate-linear Hodge star is characterized by contraction of the $E,E^{\ast }$ coefficients inIt is conjugate-linear, maps type $(p,q)$ to type $(n-p,n-q)$, and uses the metric identification of $E$ with its dual. For total degree $k=p+q$, the ordinary star-square rule giveswhere $E^{\ast \ast }$ is identified with $E$. The conjugate-linear convention is essential for the bidegrees specified here.

If $\psi$ has degree $k-2$ and $\eta$ degree $k$, the even degree of the real form $\omega$ givesSince ${\ast }_E\Lambda \eta =L{\ast }_E\eta$ for $\Lambda \eta =(-1{)}^k{\ast }_{E^{\ast }}L{\ast }_E\eta$, it follows that $(L\psi ,\eta )=(\psi ,\Lambda \eta )$ pointwise. Thus $\Lambda$ is the adjoint Lefschetz operator.

For holomorphic $E$, ${\bar{\partial }}_E$ differentiates coefficients in holomorphic local frames and uses the graded Leibniz rule on forms. Its bundle-valued Dolbeault adjoint is the formal differential operatorIt lowers the antiholomorphic degree by one. To verify this on compact $M$, define $⟨\psi ,\eta {⟩}_{L^2}={\int }_M(\psi ,\eta )dV$. For $\psi \in A^{p,q-1}(E)$ and $\eta \in A^{p,q}(E)$, Stokes applied to the scalar $(n,n-1)$-form $\psi \wedge {\ast }_E\eta$ yieldsThe star-square identity converts the right side into $⟨\psi ,{\bar{\partial }}_E^{\ast }\eta {⟩}_{L^2}$. Compactness and absence of boundary remove boundary terms. This establishes the asserted global adjoint relation on smooth sections.

The Dolbeault Laplacian ${\Delta }_{\bar{\partial }}=B{B}^{\ast }+B^{\ast }B$, with $B={\bar{\partial }}_E$, is formally self-adjoint. Consequently the adjoint of a commutator satisfiesFor the $(1,0)$ part $D^{\prime }$ of the Chern connection, the explicit corresponding adjoint iswhere $D_{E^{\ast }}^{\prime }$ is the $(1,0)$ part of the dual Chern connection. This is the same integration-by-parts formula in holomorphic degree.

Now assume the metric is Kähler. Since $\bar{\partial }\omega =0$, $B$ commutes with $L$. Taking adjoints provesThe scalar Kähler identities needed areThey extend to the bundle-valued Kähler identitiesHere is a local justification of the extension. At any chosen point, take holomorphic normal coordinates for the Kähler metric and a holomorphic local frame of $E$ with $H=I$ and $\partial H=0$ at that point. This is a Chern connection in a normal holomorphic frame. Such a frame is obtained first by normalizing the metric matrix at the point and then prescribing the holomorphic local frame change's first derivatives to cancel $A=H^{-1}\partial H$. The first-order operators and their formal adjoints at the point are then the scalar operators acting componentwise. The scalar identities therefore give the two displayed bundle identities there. Since the point was arbitrary, they hold globally.

Let $A=D^{\prime }$. Its square and $B^2$ vanish, whilebecause the Chern curvature has pure type $(1,1)$. Use the two commutators already proved to calculateTaking adjoints, including conjugation of the scalar $-i$, gives the Lefschetz-Dolbeault curvature commutatorIf $\Theta =0$, commutation follows. Conversely, vanishing of the commutator on all $E$-valued forms implies $\Theta \wedge s=0$ for every smooth section of a vector bundle $s$. At any point, a bump-supported local section can have any prescribed fiber value. Thus $\Theta$ annihilates every vector in every fiber, forcing $\Theta =0$ identically. This proves the flatness criterion from Lefschetz commutation; the test on degree-zero forms is why the assertion is made on the entire form space.