Hodge Laplacian ($\Delta =d\delta +\delta d$)

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 $\Delta$.