Use the convention that a Hermitian metric $h$ is complex-linear in its first argument and conjugate-linear in its second. It is a smoothly varying positive-definite Hermitian form on each fibre. A connection on a vector bundle is a complex-linear operator $D:\Gamma (E)\to {\Omega }^1(X,E)$ satisfying $D(as)=da\otimes s+aDs$ for smooth complex functions $a$. Metric compatibility means, for every real vector field $V$,This is the metric-compatible connection condition with the sesquilinear convention fixed.

Apply the smooth Gram-Schmidt process to a local frame to obtain an $h$-orthonormal smooth frame $s_1,\dots ,s_r$. Positivity guarantees that all normalization denominators are nonzero and depend smoothly on the base point. Write $D{s}_j={\sum }_i{A}_{ij}{s}_i$. Differentiating $h(s_i,s_j)={\delta }_{ij}$ and using compatibility givesfor every real $V$. Thus $\boxed{A(V{)}^{†}=-A(V)}$: the connection matrix is skew-Hermitian. This smooth unitary frame for a Hermitian connection is generally not holomorphic; the requested local-frame assertion requires only a smooth frame.

The holomorphic dual vector bundle $E^{\ast }$ is obtained by dualizing fibres and using transition matrices $g_{ij}^{-T}$ when $E$ has transitions $g_{ij}$. Their entries are holomorphic because matrix inversion is holomorphic on $G{L}_r(\mathbb{C})$. Their cocycle property follows from preservation of the fibrewise evaluation pairing, defining the natural holomorphic bundle with fibre $(E_x{)}^{\ast }$.

The conjugate vector bundle $\overline{E}$ has the same underlying real fibres but opposite scalar action: $\lambda \cdot \overline{v}=\overline{\overline{\lambda }v}$. Its smooth transition matrices are $\overline{g_{ij}}$; they need not be holomorphic on $X$. The conjugate bundle is used here as a smooth complex bundle, whereas $E^{\ast }$ is holomorphic.

Define the smooth dual tensor $H$ byIt is complex-linear in each tensor factor, because conjugating the second bundle converts the conjugate-linearity of $h$ into linearity. Thus $H\in \Gamma ((E\otimes \overline{E}{)}^{\ast })$.

The conjugate connection is defined on real vector fields by ${\overline{D}}_V(\overline{s})=\overline{D_{V}s}$ and extended complex-linearly on the conjugate bundle. The tensor product connection isIts dual connection $D_0$ is uniquely characterized byThe Leibniz rule makes this a genuine connection on $(E\otimes \overline{E}{)}^{\ast }$. Evaluate it on the decomposable tensor $u=s\otimes \overline{t}$:Decomposable tensors span every fibre. Hence this tensor-valued one-form vanishes precisely when the compatibility identity holds for every $V,s,t$:This is metric compatibility as parallelism of a Hermitian tensor.

Write $L\alpha =\omega \wedge \alpha$, the Lefschetz operator of a Kähler manifold. The metric and volume form define the $L^2$ inner product on smooth complex forms, and $\Lambda =L^{\ast }$ is its formal adjoint. Similarly, ${\bar{\partial }}^{\ast }$ is the formal adjoint of the Dolbeault operator, characterized by $⟨\bar{\partial }\beta ,\alpha ⟩=⟨\beta ,{\bar{\partial }}^{\ast }\alpha ⟩$. The Dolbeault Laplacian isOn the compact manifold without boundary, integration by parts givesIf the Laplacian vanishes, both terms are zero. Conversely, if both operators annihilate $\alpha$, the defining formula annihilates it. Thus harmonicity is equivalent to being both $\bar{\partial }$-closed and ${\bar{\partial }}^{\ast }$-closed.

Because $\omega$ is closed and has type $(1,1)$, $\bar{\partial }\omega =0$ and $[L,\bar{\partial }]=0$. The supplied identity from the Kähler identities gives $[L,{\bar{\partial }}^{\ast }]=-i\partial$. With ordinary commutators for the even-degree operator $L$,The last identity follows from $d^2=0$. This also proves that the Lefschetz operator preserves harmonic forms.

For the cohomology map one can work directly with forms: $\bar{\partial }({\omega }^k\wedge \alpha )={\omega }^k\wedge \bar{\partial }\alpha$, since $\omega$ has even degree. It takes closed forms to closed forms and exact forms to exact forms. Therefore the $k$th power of $L$ inducesThe bidegree is $(p+k,q+k)$, including the zero groups outside the dimension range. No isomorphism claim is needed here.

Finally put $\theta =-i\partial f$, so $\bar{\partial }\theta =i\partial \bar{\partial }f={\omega }^{\prime }-\omega$. LetBoth Kähler forms are $\bar{\partial }$-closed, so $\bar{\partial }{T}_k=0$. For a closed representative $\alpha$,This primitive has type $(p+k,q+k-1)$; the degree-one sign produces no additional term because $T_k\wedge \alpha$ is closed. HenceThis is the dependence of Lefschetz maps on the Dolbeault class.