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A complex bundle connection induces a connection on its conjugate vector bundle by the displayed rule on real tangent fields. The conjugate scalar action gives the correct Leibniz rule. In conjugate frames the matrix is the complex conjugate of the original connection matrix, with evaluation of one-forms on real tangent vectors understood.

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.