Hodge star operator ($\ast$)

On an oriented Riemannian $n$-manifold, the Hodge star is determined byIt maps $p$-forms to $(n-p)$-forms and satisfies ${\ast }^2=(-1{)}^{p(n-p)}$ on $p$-forms.

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.

For inner products linear in the first argument, define ${\ast }_E$ by contraction in $\psi \wedge {\ast }_E\eta =(\psi ,\eta )dV$. It is conjugate-linear and takes $(p,q)$-forms with values in $E$ to $(n-p,n-q)$-forms with values in $E^{\ast }$. With induced dual metrics, ${\ast }_{E^{\ast }}{\ast }_E=(-1{)}^{p+q}$.

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.

The Hodge star operator is self-adjoint on real two-forms in four Euclidean dimensions, so its opposite eigenspaces are orthogonal. Since $H\wedge G=⟨H,\ast G⟩{\mathrm{vol}}_g$, a self-dual two-form and an anti-self-dual two-form also have zero wedge product of differential forms. Applied to Lie-algebra components with an invariant pairing, this identity eliminates mixed terms from the Yang-Mills action.

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.

If a metric in dimension $n$ is rescaled by $\tilde{g}=e^{2\omega }g$, then its Hodge star on $p$-forms satisfies $\tilde{\ast }=e^{(n-2p)\omega }\ast$. The exponent vanishes for middle-degree forms with $n=2p$, so their self-dual and anti-self-dual subspaces depend only on the conformal class.

The Hodge star extends complex-linearly to complex differential forms. With the Hermitian pointwise inner product, it is characterized byOn a Hermitian manifold it maps forms of type $(p,q)$ to forms of type $(n-q,n-p)$.

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.

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

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

For the orientation $d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d{x}^4$, define ${\eta }_{ij}^a={ϵ}_{aij}$, ${\eta }_{i4}^a={\delta }_{ai}$, and extend antisymmetrically. The three forms $\frac{1}{2}{\eta }_{\mu \nu }^{a}d{x}^{\mu }\wedge d{x}^{\nu }$ are self-dual. Reversing the sign of the components involving index four gives the anti-self-dual symbols $\bar{\eta }$. They identify the two three-dimensional eigenspaces of the Hodge star on differential two-forms and give compact formulas for the BPST instanton.

For $w=x^1+i{x}^2$, $z=x^3+i{x}^4$ and the standard oriented Euclidean metric, the real and imaginary parts of $dw\wedge dz$, together with $i(dw\wedge d\bar{w}+dz\wedge d\bar{z})$, span the self-dual two-forms. The printed complex four-form $dw\wedge dz\wedge d\bar{w}\wedge d\bar{z}$ is four times the metric volume form and gives the same orientation.

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.

If an exact two-form $\beta =d\alpha$ on a compact oriented Riemannian four-manifold is anti-self-dual, thenso $\beta =0$.

The codifferential is the formal $L^2$ adjoint of the exterior derivative. On $p$-forms in dimension $n$, one convention is

On a closed oriented Riemannian manifold, let $\alpha$ have degree $r-1$ and $\beta$ degree $r$. The graded Leibniz rule and Stokes theorem give $\int d\alpha \wedge \ast \beta =(-1{)}^r\int \alpha \wedge d\ast \beta$. The codifferential convention $\delta \beta =(-1{)}^{n(r+1)+1}\ast d\ast \beta$, combined with the square of the Hodge star, gives $\ast \delta \beta =(-1{)}^{r}d\ast \beta$. This proves that the codifferential is the formal adjoint of the exterior derivative. A boundary requires the corresponding boundary term or specified boundary conditions.

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.

A differential form is coclosed when its codifferential vanishes. On a closed Riemannian manifold, a form is a harmonic differential form exactly when it is both a closed differential form and coclosed: the identity $⟨\Delta \alpha ,\alpha ⟩=\Vert d\alpha {\Vert }^2+\Vert \delta \alpha {\Vert }^2$ proves the converse as well as necessity.

The Hodge Laplacian acts on differential forms by $\Delta =d\delta +\delta d$. On a compact manifold,

It is one of the Laplace operators in differential geometry.

This Poisson equation uses the Hodge Laplacian on differential forms. On a closed Riemannian manifold, the obstruction to solving it is the component of the source along harmonic differential forms. The Green operator of the Hodge Laplacian supplies a solution when that component vanishes.

The Poisson equation for differential forms on a closed oriented Riemannian manifold is solvable exactly when its source is $L^2$-orthogonal to every harmonic differential form of the same degree. Necessity follows from the formal adjoint identity $⟨\Delta \alpha ,h⟩=⟨\alpha ,\Delta h⟩=0$. For sufficiency, the Green operator of the Hodge Laplacian gives $\alpha =G\beta$, since $\Delta G=I-P_{\mathcal{H}}$.

When the Poisson equation for differential forms has one solution ${\alpha }_0$, all solutions are its translates by the kernel of a linear map of the Hodge Laplacian, namely the harmonic differential forms. The Hodge decomposition theorem identifies this translation vector space canonically with de Rham cohomology. There is no preferred affine origin until an extra normalization is chosen; orthogonality to harmonic forms picks the Green-operator solution.

For a real one-form $\alpha$, the Hodge Laplacian differs from the rough Laplacian by the action of Ricci curvature: $(\mathrm{Ric}\cdot \alpha )(X)=\alpha ({\mathrm{Ric}}^{♯}X)$, where $g({\mathrm{Ric}}^{♯}X,Y)=\mathrm{Ric}(X,Y)$. Equivalently, with the nonnegative scalar Laplacian convention,

On a closed Riemannian manifold with nonnegative Ricci curvature, integration of the Bochner-Weitzenbock formula for one-forms for a harmonic one-form givesBoth terms are nonnegative, so $\nabla \alpha =0$. Riemannian volume density makes the argument valid even without orientation.

On a compact connected Riemannian manifold without boundary with nonnegative Ricci curvature, the Bochner formula for one-forms makes every harmonic one-form parallel. Evaluation at any point is then injective: a parallel form that is zero there stays zero along every path. Hence $\dim {\mathcal{H}}^1(M)\le \dim M$. Flat tori attain equality. Positive-definite Ricci curvature at even one point instead forces ${\mathcal{H}}^1=0$, because the Bochner integral makes its Ricci contraction vanish everywhere.

On a closed connected Riemannian manifold with nonnegative Ricci curvature, harmonic one-forms are parallel under nonnegative Ricci curvature. Evaluation at one point injects their space into the cotangent space. The Hodge decomposition theorem identifies its dimension with the first Betti number, giving $b_1(M)\le \dim M$. For example, a genus-two closed orientable surface times a circle has dimension three and first Betti number five, so no Riemannian metric on that product has nonnegative Ricci curvature.

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.

On a compact oriented Riemannian manifold without boundary, the codifferential is the $L^2$ adjoint operator of the exterior derivative. ConsequentlyEvery eigenvalue of the Hodge Laplacian is therefore nonnegative.

On a compact oriented Riemannian manifold, the Green operator $G$ is the inverse of the Hodge Laplacian on the orthogonal complement of harmonic forms and is zero on harmonic forms. If $H$ is harmonic projection, thenIt commutes with every differential operator that commutes with $\Delta$.

A differential form is harmonic when it lies in the kernel of the Hodge Laplacian $\Delta =d\delta +\delta d$. On a compact manifold, this is equivalent to $d\alpha =0$ and $\delta \alpha =0$.

Its relation to cohomology is part of Hodge theory.

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.

A harmonic one-form is a differential one-form $\alpha$ satisfying $\Delta \alpha =0$ for the Hodge Laplacian. On a closed manifold, the Hodge decomposition theorem identifies these forms with degree-one de Rham cohomology. The Bochner-Weitzenbock formula for one-forms implies that harmonic one-forms are parallel under nonnegative Ricci curvature.

On a closed oriented Riemannian manifold, a smooth isometric action of a connected Lie group fixes every harmonic differential form. Each group element is connected to the identity through a smooth path, so its action is smoothly homotopic to the identity and fixes de Rham cohomology. Isometries preserve the Hodge Laplacian; uniqueness of the harmonic representative, from the Hodge decomposition theorem, then makes each form itself fixed.

On a compact oriented Riemannian manifold,as an $L^2$-orthogonal direct sum. Every de Rham cohomology class consequently has a unique harmonic representative.

This orthogonal-splitting theorem is one result of Hodge theory.

On a compact oriented Riemannian manifold without boundary, every real linear functional on ${\mathcal{H}}^p$ has a unique $L^2$ representing harmonic form $h_f$: $f(\phi )=⟨\phi ,h_f{⟩}_{L^2}$. If $h_a$ is an orthonormal harmonic basis, then $h_f={\sum }_{a}f(h_a)h_a$. The Hodge star gives a harmonic $(n-p)$-form ${\psi }_0=\ast h_f$ satisfying $f(\phi )={\int }_M\phi \wedge {\psi }_0$. The ambiguity of harmonic wedge-pairing representatives describes all smooth representatives with this same functional.

On a compact oriented Riemannian manifold without boundary, a smooth $(n-p)$-form $\chi$ annihilates all harmonic $p$-forms under wedge integration exactly when its harmonic projection vanishes. Indeed ${\int }_M\phi \wedge \chi =⟨\ast \phi ,\chi {⟩}_{L^2}$, and the Hodge star maps ${\mathcal{H}}^p$ isomorphically onto ${\mathcal{H}}^{n-p}$. By the Hodge decomposition theorem, this annihilator isThus both exact and coexact additions are allowed; a harmonic representative is unique. The potentials of these additions need not be unique, and out-of-range degree spaces are 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.

On a compact oriented Riemannian manifold of dimension four without boundary, every exact three-form has a self-dual two-form primitive. Write $\beta =d\omega$ and use the Hodge decomposition theorem to split $\omega =h+d\xi +\delta \psi$. Then $\beta =d\delta \psi$. For $a=\delta \psi$, the codifferential identity $\delta =-\ast d\ast$ and the fact ${\ast }^2=1$ on two-forms give $\ast a=-d\ast \psi$, so $d\ast a=0$. Thus $\sigma =a+\ast a$ satisfies $\ast \sigma =\sigma$ and $d\sigma =\beta$. It is twice the self-dual projection of $a$, rather than the projection itself.