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For the Chern connection of a Hermitian holomorphic vector bundle on a Kähler manifold, the two Dolbeault Laplacians differ by $[iF\wedge ,\Lambda ]$, where $F$ is its curvature form of a connection and $\Lambda$ is the adjoint Lefschetz operator. The curvature term controls vanishing estimates.

In complex dimension $n$, the Lefschetz operator of a Kähler manifold and adjoint Lefschetz operator satisfy $[\Lambda ,L]=(n-k)\mathrm{id}$ on degree $k$. In middle degree, this yields $\Vert L\alpha {\Vert }^2=\Vert \Lambda \alpha {\Vert }^2$.

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.

Let $\omega$ be the Kähler form associated with the metric. The Lefschetz operator of a Kähler manifold and adjoint Lefschetz operator areHere the adjoint is taken with respect to the pointwise Hermitian inner product on complexified forms, or equivalently the global $L^2$ inner product. It lowers degree by two and bidegree by $(1,1)$. In terms of the Hodge star operator, $\Lambda ={\ast }^{-1}L\ast$.

Since $d\omega =0$, $dL=Ld$, so $L$ takes closed differential forms to closed differential forms and exact differential forms to exact differential forms. This already gives its action on de Rham cohomology. For both operators together, use the Kähler identities: $L$ commutes with the Hodge Laplacian, and taking adjoints shows that $\Lambda$ does too. Thus both preserve harmonic differential forms.

When $M$ is compact, the Hodge decomposition theorem identifies $H^k(M,\mathbb{C})$ with its space of harmonic $k$-forms. If $h_{\alpha }$ is the unique harmonic differential form representing $[\alpha ]$, defineThe resulting forms are harmonic and therefore closed, and uniqueness of $h_{\alpha }$ makes these definitions independent of the initial representative. The cohomological definition of $\Lambda$ uses harmonic differential forms as representatives: $\Lambda$ need not take an arbitrary closed form to a closed form.

Represent the middle-degree class by its unique harmonic differential form $h$. The Lefschetz commutator on degree $k$ is $[\Lambda ,L]=(n-k)\mathrm{id}$, so at $k=n$ the operators commute. Since the adjoint Lefschetz operator is the adjoint of the Lefschetz operator of a Kähler manifold,Both $Lh$ and $\Lambda h$ are harmonic. A harmonic form represents zero in de Rham cohomology exactly when the form itself is zero. Consequently $L[\alpha ]=0$ if and only if $Lh=0$, which by the norm identity is equivalent to $\Lambda h=0$, hence to $\Lambda [\alpha ]=0$. ThereforeThis identifies the two middle-degree descriptions of a primitive differential form on a Kähler manifold at the cohomology level.