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The dual connection is induced by a connection on a vector bundle through the evaluation pairing. A connection on $E$ induces one on $E^{\ast }$ byIts connection matrices are the negatives of the transposes of those for $E$.

With a Hermitian form linear in its first slot, $H(s\otimes \bar{t})=h(s,t)$ is a complex-linear dual tensor. The original connection, its conjugate connection, the tensor product connection and the dual connection induce $D_0$. Its derivative is $(D_{0,V}H)(s\otimes \bar{t})=Vh(s,t)-h(D_{V}s,t)-h(s,D_{V}t)$. Consequently $D_{0}H=0$ is exactly metric compatibility.

The Riemannian volume form of an oriented $n$-dimensional Riemannian manifold is the unique positive $n$-form taking value one on every positively oriented orthonormal frame. In positive coordinates,The metric induces an inner product on exterior powers of the cotangent bundle, with wedges of distinct orthonormal covectors forming an orthonormal basis. The Hodge star is the pointwise linear map $\ast :{\Lambda }^p{T}^{\ast }M\to {\Lambda }^{n-p}{T}^{\ast }M$ specified byIt is an isometry and satisfies $\ast \ast \alpha =(-1{)}^{p(n-p)}\alpha$. With the positive-sign convention, the Hodge Laplacian on differential forms isHere $d$ is the exterior derivative and $\delta$ its formal adjoint; $\delta$ on functions is zero. On functions this is the Laplace-Beltrami operatorThe convention $\mathrm{div}\mathrm{grad}$ gives the negative of this operator. All subsequent Laplacians use the nonnegative convention above.

On a compact oriented manifold without boundary, the Hodge decomposition theorem gives an orthogonal decomposition of smooth real forms,The three component forms are uniquely determined, although the potentials in the last two summands need not be unique. The space ${\mathcal{H}}^p$ is finite-dimensional, and each de Rham cohomology class has a unique harmonic representative. Integration by parts gives $⟨\Delta \alpha ,\alpha {⟩}_{L^2}=\Vert d\alpha {\Vert }_{L^2}^2+\Vert \delta \alpha {\Vert }_{L^2}^2$, so harmonic forms are exactly the closed and coclosed forms. Set ${\Omega }^k=0$ outside $0\le k\le n$. Compactness without boundary is a hypothesis of this version of the theorem; orientation alone does not supply it.

A connection on $TM$ induces a dual connection on $T^{\ast }M$ and then connections on its tensor and exterior powers by the Leibniz rule. Explicitly, for a $p$-form $\alpha$,To prove the parallelism of the Riemannian volume form, use a local positively oriented orthonormal frame $E_i$, its dual coframe ${\theta }^i$, and write ${\nabla }_X{E}_j={\sum }_i{a}_{ij}(X)E_i$. Metric compatibility makes the matrix $a(X)$ skew-symmetric. Since ${\omega }_g={\theta }^1\wedge \cdots \wedge {\theta }^n$, differentiating the wedge yieldsOff-diagonal terms vanish because they repeat a coframe factor. Thus $\nabla {\omega }_g=0$; equivalently parallel transport preserves both the metric and the continuously chosen orientation.

Now let $M$ be compact, oriented and without boundary. For representing functionals on harmonic forms by wedge pairing, choose an $L^2$-orthonormal basis $h_1,\dots ,h_b$ of ${\mathcal{H}}^p$ and defineIf $\phi ={\sum }_a{c}_a{h}_a$, linearity givesThis also works when $b=0$, using the empty sum. Since the Hodge star commutes with the Hodge Laplacian, ${\psi }_0$ is harmonic of degree $n-p$.

For the full ambiguity of harmonic wedge-pairing representatives, write another representative as ${\psi }_0+\chi$. Because the star is an isometry,The forms $\ast \phi$ range over all of ${\mathcal{H}}^{n-p}$. Hence $\chi$ gives zero functional exactly when its harmonic projection is zero. Hodge decomposition proves that the complete answer isBoth summands in this ambiguity are permitted; it is not only an exact-form ambiguity. The exact and coexact component forms are orthogonal to every harmonic form, so every displayed choice works. Conversely a nonzero harmonic component would pair nontrivially with some $\ast \phi$, and therefore cannot occur. The harmonic representative $\ast h_f$ is unique; the arbitrary smooth representative generally is not. The formula also covers the endpoint degrees and any exceptional case where the ambiguity space is zero.

For a concrete coexact ambiguity, take the torus $(\mathbb{R}/2\pi \mathbb{Z}{)}^2$ with metric $d{x}^2+d{y}^2$. The form $\chi =\delta (\sin ydx\wedge dy)=\cos ydx$ annihilates every harmonic one-form by orthogonality, yet $d\chi =\sin ydx\wedge dy$ is nonzero. Thus an annihilating representative need not even be closed, let alone exact.

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.

In coordinates, expanding $d\alpha$ and the Lie bracket of vector fields showsThe terms differentiating components of $X,Y$ cancel against the bracket, leaving $({\partial }_i{\alpha }_j-{\partial }_j{\alpha }_i)X^i{Y}^j$. This is the exterior derivative of a one-form evaluated on vector fields.

A connection on a vector bundle is a linear map $\nabla :\Gamma (E)\to {\Omega }^1(E)$ satisfying $\nabla (fs)=df\otimes s+f\nabla s$; evaluation on $X$ defines ${\nabla }_{X}s$. Its curvature form of a connection is $d_A^2$, locally $F_A=dA+A\wedge A$. Direct expansion, using the displayed exterior-derivative identity, givesThe derivative terms on a scalar multiplying $s$ cancel, and the expression is also linear over smooth functions in $X,Y$. Thus it is an alternating tensor with values in $\mathrm{End}(E)$, namely an element of ${\Omega }^2(\mathrm{End}(E))$.

The dual connection and tensor product connection induce the endomorphism bundle connection on $E^{\ast }\otimes E$:Expanding twice cancels the cross terms and givesHere $\varphi ,s$ are local smooth sections; derivatives are not defined for isolated fiber elements without extensions. Subtract the corresponding ${\tilde{\nabla }}_{[X,Y]}$ identity to get the curvature of an endomorphism bundle connection:This vanishes for all $\varphi$ exactly when each $F_A(X,Y)$ is central in the full matrix algebra, hence scalar. For rank $r>0$, $\omega =r^{-1}\mathrm{tr}{F}_A$ is a smooth two-form andConversely scalar curvature commutes with every endomorphism. This is the scalar-curvature criterion for a flat endomorphism connection; it permits a nonflat connection on $E$ itself.

Regard an $E^{\vee }$-valued $r$-form $\lambda$ as a row vector and an $\mathrm{End}(E)$-valued $r$-form $\mu$ as a matrix. The dual connection and endomorphism connection areFor the curvature two-form,because substituting $F=dA+A\wedge A$ makes all terms cancel. This is the Bianchi identity.

For an endomorphism-valued $r$-form $\mu$ and an $E$-valued form $\sigma$, direct expansion gives the compatible Leibniz ruleThe cancellation of the two middle $A$-terms proves the identity.

The induced dual connection is uniquely defined bywhich immediately gives the required pairing identity. Ifthen

A connection on $TB$ is symmetric, or torsion-free, whenequivalently ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$ in coordinates. With $\mathrm{Alt}B(X,Y)=B(X,Y)-B(Y,X)$,