The Hodge Theorem

The Hodge theorem is the central result of Hodge theory, asserting that every de Rham cohomology class on a compact Riemannian manifold has a unique harmonic representative. The proof combines elliptic PDE theory with functional analysis.

Statement

Theorem10.1Hodge theorem

Let $(M^n, g)$ be a compact oriented Riemannian manifold without boundary. Then:

The space of harmonic $k$ -forms $\mathcal{H}^k = \ker\Delta$ is finite-dimensional.
There is an $L^2$ -orthogonal decomposition $\Omega^k(M) = \mathcal{H}^k \oplus \Delta\Omega^k(M) = \mathcal{H}^k \oplus d\Omega^{k-1} \oplus \delta\Omega^{k+1}$ .
The inclusion $\mathcal{H}^k \hookrightarrow Z^k(M) \twoheadrightarrow H^k_{\mathrm{dR}}(M)$ is an isomorphism.

Proof

The proof has three main ingredients: elliptic regularity, Fredholm theory, and the orthogonal decomposition.

Step 1: Elliptic regularity. The Hodge Laplacian $\Delta = d\delta + \delta d$ is a second-order elliptic differential operator (its principal symbol is $|\xi|^2 \mathrm{Id}$ ). By elliptic regularity theory: if $\Delta\omega = f$ with $f$ smooth, then $\omega$ is smooth. More generally, $\omega \in H^s$ and $\Delta\omega \in H^k$ implies $\omega \in H^{k+2}$ (where $H^s$ are Sobolev spaces).

Step 2: Weak formulation and Sobolev spaces. Extend $\Delta$ to a bounded operator $\Delta: H^{s+2}\Omega^k \to H^s\Omega^k$ on Sobolev spaces of $k$ -forms. By the Rellich lemma, the inclusion $H^{s+2} \hookrightarrow H^s$ is compact on a compact manifold. This makes $\Delta$ a Fredholm operator.

Step 3: Fredholm alternative. Since $\Delta$ is self-adjoint and non-negative ( $\langle \Delta\omega, \omega \rangle = |d\omega|^2 + |\delta\omega|^2 \geq 0$ ), the Fredholm alternative gives:

L^2\Omega^k = \ker\Delta \oplus \overline{\mathrm{im}(\Delta)}.

Moreover, $\ker\Delta$ is finite-dimensional and $\mathrm{im}(\Delta)$ is closed. By elliptic regularity, the smooth forms in $\ker\Delta$ and $\mathrm{im}(\Delta)$ are the same as the Sobolev/distributional ones, giving the decomposition at the smooth level.

Step 4: Refining the decomposition. Since $\Delta = d\delta + \delta d$ :

\mathrm{im}(\Delta) \subset \mathrm{im}(d) + \mathrm{im}(\delta).

For the reverse: if $\omega \perp \ker\Delta$ and $\omega = d\alpha + \delta\beta$ , then $\omega = \Delta\eta$ for some $\eta$ (by the Fredholm property). Expanding gives $\mathrm{im}(\Delta) = \mathrm{im}(d) \oplus \mathrm{im}(\delta)$ .

Step 5: Hodge isomorphism. Every closed form $\omega$ decomposes as $\omega = h + d\alpha + \delta\beta$ . Since $d\omega = 0$ : $d\delta\beta = 0 \Rightarrow |\delta\beta|^2 = \langle d\delta\beta, \beta \rangle = 0$ . So $\omega = h + d\alpha$ , and $[\omega] = [h]$ in cohomology. Uniqueness: if $h = d\alpha$ is harmonic and exact, then $|h|^2 = \langle d\alpha, h \rangle = \langle \alpha, \delta h \rangle = 0$ . $\blacksquare$

■

Applications

RemarkFinite-dimensionality of cohomology

The Hodge theorem gives an analytical proof that the de Rham cohomology of a compact manifold is finite-dimensional (since $\ker\Delta$ is finite-dimensional by elliptic theory). This is a highly non-trivial fact from the purely topological perspective.

ExampleThe Green's operator

The Hodge theorem constructs the Green's operator $G: \Omega^k \to \Omega^k$ satisfying $\Delta G = G\Delta = \mathrm{Id} - H$ , where $H$ is the harmonic projection. For any $\omega$ : $\omega = H\omega + dG\delta\omega + \delta G d\omega$ . This gives explicit formulas for solving the Poisson equation $\Delta\eta = f$ (solvable if and only if $f \perp \mathcal{H}^k$ , in which case $\eta = Gf$ ).