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On $m$-forms in an oriented Riemannian manifold of dimension $2m$, the Hodge star operator is orthogonal and satisfies ${\ast }^2=(-1{)}^{m^2}$. HenceIt is self-adjoint when $m$ is even and skew-adjoint when $m$ is odd.

On an oriented Riemannian four-manifold, a two-form $\alpha$ is anti-self-dual when $\ast \alpha =-\alpha$.

On an oriented Riemannian four-manifold, a two-form is anti-self-dual when $\ast \alpha =-\alpha$. Here $\ast$ is the Hodge star operator.

For a Hermitian manifold with the pointwise Hermitian inner product linear in the first variable, define the conjugate-linear star by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\omega }^n/n!$. It is the usual real Hodge star operator extended complex-linearly and then composed with complex conjugation. It has the displayed bidegree and satisfies ${\ast }^2=(-1{)}^{p+q}$ on $(p,q)$-forms.

The dual electromagnetic field tensor is obtained by applying the Hodge star operator to the electromagnetic field tensor. For metric signature $(+---)$ and ${ϵ}^{0123}=+1$, define ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. With $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, it has components ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$. The homogeneous Maxwell equations become ${\partial }_{\mu }{\tilde{F}}^{\mu \nu }=0$.

Let $M$ be a compact oriented Riemannian manifold of dimension four, without boundary. An exact two-form $\beta =d\alpha$ that is an anti-self-dual differential form must vanish. Indeed the Generalized Stokes theorem and the Hodge star operator identity $\ast \beta =-\beta$ giveThe boundaryless hypothesis is essential. On a Euclidean four-ball, the nonzero constant two-form $d{x}^1\wedge d{x}^2-d{x}^3\wedge d{x}^4$ is both an exact differential form and an anti-self-dual differential form; the omitted boundary integral need not vanish there.

For the underlying Riemannian metric $g$ and complex structure $J$, the fundamental form is $\omega (u,v)=g(Ju,v)$. It is a real (1, 1)-form. In complex dimension $n$, the compatible orientation has Riemannian volume form ${\omega }^n/n!$, and its Hodge star operator satisfies

On a connected compact oriented Riemannian manifold without boundary, a top-degree form is $f{\mathrm{vol}}_g$. Its exterior derivative is automatically zero, and the Hodge codifferential vanishes exactly when $df=0$. Thus it is harmonic exactly when $f$ is constant. Equivalently the Hodge star operator identifies this space with harmonic functions. Without connectedness the dimension is the number of connected components.

On an oriented boundaryless Riemannian manifold, apply Stokes theorem to the compactly supported form $f\ast \alpha$. The graded Leibniz rule gives $\int df\wedge \ast \alpha =-\int fd\ast \alpha$. The Hodge star operator identifies these top forms with the scalar integrands displayed above. This proves that the codifferential on one-forms is $\delta =-\ast d\ast$, the formal adjoint of the exterior derivative. Compact support of $f$ suffices.

On an oriented Euclidean metric four-space, the Hodge star operator on two-forms is an orthogonal involution. Its two eigenspaces each have dimension three. A self-dual two-form has eigenvalue $+1$, while an anti-self-dual two-form has eigenvalue $-1$. The splitting uses the metric-normalized Riemannian volume form; an arbitrarily rescaled volume in the defining wedge identity would rescale the operator itself.

On an oriented Riemannian manifold, the codifferential convention $\delta =(-1{)}^{n(p+1)+1}\ast d\ast$ and ${\ast }^2=(-1{)}^{p(n-p)}$ give $d\ast \alpha =(-1{)}^p\ast \delta \alpha$ and $\delta \ast \alpha =(-1{)}^{p+1}\ast d\alpha$. Applying these twice proves the displayed identity. Thus the Hodge star operator preserves harmonic differential forms even without compactness.

Use the complex-linear Hodge star operator on the direct sum of all form degrees. Its square on degree $r$ is $(-1{)}^{r(n-r)}$. In positive odd dimension this is always one; in positive even dimension its sign is positive on even-degree forms and negative on odd-degree forms. Complementary basis wedges provide both roots of each sign, so the displayed eigenvalues all occur. On a single degree the Hodge star operator is an endomorphism only in middle degree; on middle degree $m$ in dimension $2m$ its possible eigenvalues are $\pm 1$ for even $m$ and $\pm i$ for odd $m$.

For the complex-linear Hodge star operator and complex orientation on a Hermitian manifold of complex dimension $n$, every $(n,0)$-form has the displayed eigenvalue. In an adapted orthonormal real coframe, each factor $d{x}_j+id{y}_j$ has two-dimensional star eigenvalue $-i$. Regrouping the $n$ factors introduces $(-1{)}^{n(n-1)/2}$, giving $(-1{)}^{n(n+1)/2}{i}^n$. This is a pointwise metric identity; the metric need not be Kähler.

A Riemannian metric induces a dual inner product on each cotangent space. On decomposable covectors define $⟨{\xi }_1\wedge \cdots \wedge {\xi }_p,{\zeta }_1\wedge \cdots \wedge {\zeta }_p⟩=\det (⟨{\xi }_i,{\zeta }_j⟩)$ and extend bilinearly. Increasing wedges of an orthonormal coframe form an orthonormal basis. This positive-definite inner product is the one used to define the Hodge star operator.

With zero cosmological constant and the component Hodge star operator convention $(\star j{)}_{abc}={ϵ}_{abcd}{j}^d$, the Einstein field equations and Killing equation give $d\star dm=-16\pi \star (T_{ab}{m}^b-\frac{1}{2}T{m}_a)$. If $m$ is tangent to a spacelike hypersurface, the trace term pulls back to zero. For $\partial \Sigma =S_{\infty }-{\bigcup }_i{S}_i$, Stokes theorem yieldsThe inner Komar charges cannot be dropped for a vacuum Kerr black hole, whose angular momentum is nonzero although its stress-energy tensor vanishes. A matter-only volume formula requires a regular filling with no inner boundaries.

A Komar charge is an oriented surface integral constructed from a Killing vector field and the metric tensor. Let $k$ be the metric-dual differential one-form of a stationary Killing vector field. In four dimensions, with geometrized units $G=c=1$ and the standard future/outward orientation, its mass normalization iswhere $\star$ is the Hodge star operator and $S$ is a closed spacelike two-surface. The normalization of the Killing vector field fixes the time scale: for Schwarzschild spacetime, unit time translation at infinity gives the black-hole mass parameter.

The Killing equation relates the divergence of $dk$ to a contraction of the Ricci tensor. Consequently, homologous enclosing surfaces give the same mass in a vacuum region with zero cosmological constant. Matter or a nonzero cosmological constant introduces a volume contribution; an arbitrary finite-radius surface integral need not equal the asymptotic mass. Axial symmetry instead gives Komar angular momentum with a different normalization.

For $f=1-s/r^2+q/r^4$ and the spacetime orientation $-dt\wedge dr\wedge d{\Omega }_3$, the Hodge star operator gives ${\int }_{S^3}\star d{k}^{♭}=2{\pi }^2{r}^3{f}^{\prime }\to 4{\pi }^{2}s$. A prefactor $1/(16\pi )$ produces $\pi s/4$, whereas the dimension-adjusted ADM mass normalization produces $3\pi s/8$. The normalization and orientation must be specified rather than imported from four dimensions.

Komar mass is a gravitational surface charge associated with a stationary Killing vector field that is timelike at infinity. In four spacetime dimensions with $G=c=1$, let $k$ be the one-form dual to the Killing field, normalized to unit time translation at infinity. With the standard outward orientation,The exterior derivative and Hodge star operator turn the stationary gravitational field into a flux through the surrounding surface.^[ref]

In a vacuum region with zero cosmological constant, the field equations make this flux independent of smooth surface deformations that cross no source or inner boundary. A nonzero charge can therefore surround a vacuum black hole. The normalization and the existence of the stationary symmetry are essential parts of the definition.

For the magnetically charged dilaton black hole, the metric-dual one-form of $k={\partial }_t$ is $k^{♭}=-Vdt$. With orientation $dt\wedge dr\wedge d\theta \wedge d\varphi$, its Hodge star operator gives $\star d{k}^{♭}=-V^{\prime }{R}^2\sin \theta d\theta \wedge d\varphi$. Therefore the Komar mass is $R^2{V}^{\prime }/2=r_{+}/2-P^2/r$. Its radial dependence is allowed because the Ricci tensor is nonzero; the divergence of $d{k}^{♭}$ is sourced by $R_{ab}{k}^b$. At the horizon $M_K(r_{+})=(r_{+}-r_{-})/2=\kappa A_H/(4\pi )$.

The metric signature records the signs of the diagonalized nondegenerate metric tensor, equivalently its numbers of negative and positive directions. For a four-dimensional Lorentzian manifold, both $(-,+,+,+)$ and its overall negative convention are common. The numerical signs of contractions and component Hodge star operator identities must be consistent with that choice and the orientation.

Use ${ϵ}_{123}=1$ and orientation $d\tau \wedge d{x}^1\wedge d{x}^2\wedge d{x}^3$. A $\tau$-independent gauge curvature has mixed component ${\mathcal{F}}_{i\tau }=D_i\Phi$. If $B_i=D_i\Phi$, its two-form is ${\sum }_{\mathrm{cyclic}}{B}_i(d{x}^j\wedge d{x}^k-d\tau \wedge d{x}^i)$. The Hodge star operator interchanges the two displayed basis forms, so $\ast \mathcal{F}=-\mathcal{F}$. With orientation $d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d\tau$ the same lift is self-dual; changing the sign of its $d\tau$ component instead recovers anti-self-duality. Orientation must therefore accompany any stated sign in this reduction.