The Hodge Decomposition

The Hodge decomposition theorem provides an orthogonal splitting of the space of differential forms into exact, co-exact, and harmonic components, revealing the deep interplay between analysis and topology.

The Hodge Decomposition Theorem

Theorem10.1Hodge decomposition

Let $(M^n, g)$ be a compact oriented Riemannian manifold without boundary. The space of smooth $k$ -forms admits an $L^2$ -orthogonal decomposition:

\Omega^k(M) = \mathcal{H}^k(M) \oplus d\Omega^{k-1}(M) \oplus \delta\Omega^{k+1}(M),

where $\mathcal{H}^k(M) = \{\omega \in \Omega^k : \Delta\omega = 0\}$ is the space of harmonic $k$ -forms. The three summands are pairwise $L^2$ -orthogonal:

$\mathcal{H}^k \perp d\Omega^{k-1}$ : harmonic forms are orthogonal to exact forms.
$\mathcal{H}^k \perp \delta\Omega^{k+1}$ : harmonic forms are orthogonal to co-exact forms.
$d\Omega^{k-1} \perp \delta\Omega^{k+1}$ : exact and co-exact forms are orthogonal.

Consequences

Theorem10.2Hodge isomorphism

The natural map $\mathcal{H}^k(M) \to H^k_{\mathrm{dR}}(M)$ sending a harmonic form to its cohomology class is an isomorphism:

\mathcal{H}^k(M) \cong H^k_{\mathrm{dR}}(M).

In particular, every cohomology class has a unique harmonic representative, and $b_k = \dim \mathcal{H}^k(M)$ .

Proof

Surjectivity: Given a closed $k$ -form $\omega$ , the Hodge decomposition writes $\omega = h + d\alpha + \delta\beta$ . Since $d\omega = 0$ : $d\delta\beta = 0$ , which gives $\langle d\delta\beta, \beta \rangle = \langle \delta\beta, \delta\beta \rangle = 0$ , so $\delta\beta = 0$ . Hence $\omega = h + d\alpha$ , and $[\omega] = [h]$ in $H^k_{\mathrm{dR}}$ .

Injectivity: If $h \in \mathcal{H}^k$ is exact, $h = d\alpha$ . Then $0 = \langle h, h \rangle = \langle d\alpha, h \rangle = \langle \alpha, \delta h \rangle = 0$ , and also $\langle h, h \rangle = 0$ , so $h = 0$ . $\blacksquare$

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The Hodge Star and Poincare Duality

Theorem10.3Hodge duality

The Hodge star operator $*: \mathcal{H}^k(M) \to \mathcal{H}^{n-k}(M)$ is an isomorphism. Combined with the Hodge isomorphism, this gives an analytical proof of Poincare duality: $H^k(M) \cong H^{n-k}(M)$ .

Proof

If $\Delta\omega = 0$ , then $d\omega = 0$ and $\delta\omega = 0$ . We check $\Delta(*\omega) = 0$ : since $d*\omega = \pm *\delta\omega = 0$ and $\delta*\omega = \pm *d\omega = 0$ . So $*$ maps harmonic forms to harmonic forms. Since $**$ is (up to sign) the identity, $*$ is an isomorphism. $\blacksquare$

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ExampleHodge numbers of complex manifolds

On a compact Kahler manifold $X$ of complex dimension $n$ , the Hodge decomposition refines to:

H^k(X; \mathbb{C}) = \bigoplus_{p+q=k} H^{p,q}(X),

where $H^{p,q} \cong H^q(X, \Omega^p_X)$ (Dolbeault cohomology). The dimensions $h^{p,q} = \dim H^{p,q}$ are the Hodge numbers, satisfying $h^{p,q} = h^{q,p}$ (complex conjugation) and $h^{p,q} = h^{n-p,n-q}$ (Serre duality).

RemarkDependence on the metric

The Hodge decomposition depends on the Riemannian metric $g$ (since $\delta$ and $\Delta$ do), but the harmonic space $\mathcal{H}^k$ is isomorphic to $H^k_{\mathrm{dR}}(M)$ regardless of $g$ . Changing the metric changes which specific forms are harmonic, but not the dimension of the harmonic space.