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Use the standard boundaryless-manifold convention, so $M$ is a closed manifold. Its orientation and Riemannian metric determine a unique positive Riemannian volume form ${\mu }_g$ taking value one on every positively oriented orthonormal frame. In an oriented atlas it isFor the coordinate invariance of the Riemannian volume form, on an overlap put $J=\partial y/\partial x$. Then $g_x=J^T{g}_{y}J$ and $d{y}^1\wedge \cdots \wedge d{y}^n=(\det J)d{x}^1\wedge \cdots \wedge d{x}^n$. The oriented transition has $\det J>0$, so the square root of the metric determinant and the coordinate volume transform together. The local forms therefore agree and define ${\mu }_g$ globally.

For real $r$-forms the Hodge star operator is defined byThe pointwise wedge pairing is nondegenerate, so this uniquely defines a smooth map $\ast :{\Omega }^r(M)\to {\Omega }^{n-r}(M)$. On an oriented orthonormal coframe, it sends a basis wedge to the signed complementary wedge. Applying it twice gives

For Hodge star eigenvalues on graded differential forms, extend $\ast$ complex-linearly to the direct sum of all form degrees. This interpretation matters: on a fixed degree it is an endomorphism only in middle degree. In positive odd dimension, $r(n-r)$ is even for every $r$, so ${\ast }^2=I$ and only $\pm 1$ can occur. Both occur, with eigenforms $1+{\mu }_g$ and $1-{\mu }_g$.

In positive even dimension, ${\ast }^2$ is $+I$ on even-degree forms and $-I$ on odd-degree forms. Hence ${\ast }^4=I$ and the only possible eigenvalues are $\pm 1,\pm i$. The first two occur as above. For the latter, choose a smooth local orthonormal frame and multiply its first dual differential one-form by a nonzero smooth bump function supported in the chart. This gives a form $\eta$ for which $\eta$ and $\ast \eta$ are independent. For $\lambda =\pm i$,Thus all four occur. Consequently, for $n\ge 1$,On degree $m$ in dimension $2m$, the Hodge star has eigenvalues $\pm 1$ for even $m$ and $\pm i$ for odd $m$; signed complementary middle-degree wedges show both occur for $m\ge 1$. In zero dimensions it is multiplication by the signed unit volume at each point, so only the signs actually present occur.

For Hodge integration by parts in arbitrary degree, let $\alpha \in {\Omega }^{r-1}(M)$ and $\beta \in {\Omega }^r(M)$. The graded Leibniz rule and Stokes theorem on the closed manifold giveUsing the given codifferential and the square of the Hodge star,Thereforewhich proves that $\delta$ is the formal adjoint of $d$. The complex version inserts conjugation in the usual Hermitian inner product.

The Hodge decomposition theorem gives an $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^k(M)=\ker \Delta$ is finite dimensional and consists of the harmonic differential forms. Each de Rham cohomology class has a unique harmonic representative. Equivalently, with harmonic projection $P_{\mathcal{H}}$ there is a smooth Green operator of the Hodge Laplacian $G$ satisfying $\Delta G=G\Delta =I-P_{\mathcal{H}}$.

The solvability condition for the Hodge Poisson equation isNecessity follows from $⟨\Delta \alpha ,h⟩=⟨\alpha ,\Delta h⟩=0$. Conversely the orthogonality condition gives $P_{\mathcal{H}}\beta =0$, so ${\alpha }_0=G\beta$ solves the Poisson equation for differential forms.

The nonnegativity of the Hodge Laplacian identityidentifies its kernel with closed and coclosed forms. Thus the affine space of solutions of the Hodge Poisson equation isFor completeness, if a closed form is decomposed as $\omega =h+d\xi +\delta \zeta$, then $d\delta \zeta =0$, and $\Vert \delta \zeta {\Vert }^2=⟨\zeta ,d\delta \zeta ⟩=0$. Hence its cohomology class is represented by $h$. An exact harmonic form $h=d\xi$ must vanish, since $\Vert h{\Vert }^2=⟨\delta h,\xi ⟩=0$. This proves the canonical isomorphism ${\mathcal{H}}^k(M)\cong H_{\mathrm{dR}}^k(M)$, so the solution set is an affine space whose translation vector space is isomorphic to the requested de Rham cohomology.

The Levi-Civita connection of a Riemannian manifold $(M,g)$ is the unique connection on a vector bundle on $TM$ that is torsion-free and compatible with the Riemannian metric. Any such connection must satisfy the Koszul formulaNondegeneracy of $g$ determines ${\nabla }_{X}Y$ uniquely from the right-hand side. Conversely, define $\nabla$ by this formula. Direct substitution shows that it is $C^{\infty }$-linear in $X$, satisfies the Leibniz rule in $Y$, preserves $g$, and obeys ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$. It is therefore a torsion-free metric connection. This proves the existence and uniqueness of the Levi-Civita connection.

For the metric vector bundle $E$, choose the stated orthonormal local frame $(e_1,\dots ,e_m)$ and write $d_A{e}_j=A^i{}_j{e}_i$. Since $⟨e_i,e_j⟩={\delta }_{ij}$, metric compatibility givesHence the connection matrix in an orthonormal frame is skew-symmetric:

On an oriented Riemannian $d$-manifold, the Hodge star operator is defined byIt is an orthogonal map $\ast :{\Lambda }^r{T}_x^{\ast }M\to {\Lambda }^{d-r}{T}_x^{\ast }M$ and satisfies ${\ast }^2=(-1{)}^{r(d-r)}$. In dimension $d=2n$ on middle-degree forms, the Adjoint of the Hodge star on middle-degree forms isThus it is self-adjoint when $n$ is even, but it is not always self-adjoint. For $n=1$ on the oriented Euclidean plane,so its matrix in the orthonormal basis $(dx,dy)$ is skew-adjoint.

The Laplace-Beltrami operator on differential forms is the Hodge Laplacianwhere the codifferential $\delta$ is the formal $L^2$ adjoint of the exterior derivative. On a compact manifold without boundary, if $\Delta \omega =\lambda \omega$ for a nonzero differential form $\omega$, then integration by parts givesTherefore the nonnegativity of the Hodge Laplacian yields