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The metric on covectors is induced by the inverse matrix $g^{-1}$, and on $p$-forms by the determinant pairingThe Riemannian volume form is the unique positive top form taking value one on every oriented orthonormal frame. The Hodge star operator is uniquely determined byNondegeneracy of the wedge pairing proves existence and uniqueness pointwise, and the smooth metric dependence makes $\ast$ a well-defined smooth bundle map.

On compactly supported forms, Stokes theorem and the graded Leibniz rule givewhere $m=\dim M$; this is the formal $L^2$ adjoint of $d$. The Hodge Laplace-Beltrami operator is

For $g_f=e^{2f}g$, the covector metric scales by $e^{-2f}$, the $p$-form metric by $e^{-2pf}$, and the volume form by $e^{mf}$. ThereforeIf $f$ is constant, the two star factors in the codifferential contribute $e^{-2f}$, so ${\delta }_f=e^{-2f}\delta$. Since $d$ is metric-independent,

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The metric and orientation determine the Riemannian volume form: in a positively oriented coordinate chart $(x^1,\dots ,x^n)$,Writing $|g|=\det (g_{ij})$ and $(g^{ij})=(g_{ij}{)}^{-1}$, the Dirichlet energy on a Riemannian manifold with source $f$ is

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An orientation selects the positive ordered bases in each tangent space. On an oriented $n$-dimensional Riemannian manifold, the Riemannian volume form ${\omega }_g$ is the unique smooth $n$-form satisfyingfor every positively oriented orthonormal frame. In positively oriented local coordinates,

The metric induces an inner product on the bundle ${\Lambda }^p{T}^{\ast }M$ of $p$-forms. The Hodge star operator is the unique linear mapsuch thatfor all $p$-forms $\alpha ,\beta$. With the codifferential $\delta$, the Laplace-Beltrami operator on differential forms is

The Hodge decomposition theorem says that on a compact oriented Riemannian manifold,an $L^2$-orthogonal direct sum, where ${\mathcal{H}}^p(M)=\ker \Delta$ is the finite-dimensional space of harmonic $p$-forms. Every de Rham cohomology class has exactly one harmonic representative.

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Yes to both questions. Since ${\omega }_g$ has top degree, $d{\omega }_g=0$. Moreover $\ast {\omega }_g=1$, so the formula for the codifferential givesThereforeand the Riemannian volume form is a harmonic differential form.

The Levi-Civita connection preserves both the Riemannian metric and its chosen orientation. At any point, extend a positively oriented orthonormal basis to a local frame whose covariant derivatives vanish at that point. Differentiating ${\omega }_g(e_1,\dots ,e_n)=1$ there gives $\nabla {\omega }_g=0$. Hence ${\omega }_g$ is a parallel differential form.

Solved by gpt-5.6-sol high.