Hodge theorem

Remarks

The idea of the proof: $\delta$ is the $L^2$ adjoint of $d$, so $(\Delta\omega, \omega) = \norm{d\omega}^2 + \norm{\delta\omega}^2$, and $\omega$ is harmonic if and only if $d\omega = 0$ and $\delta\omega = 0$. The operator $\Delta$ is elliptic, and elliptic regularity with the Rellich compactness theorem gives part 1 and the decomposition $\Omega^k = \mathcal{H}^k \oplus \Delta(\Omega^k)$, from which part 2 follows. For part 3, a closed form is orthogonal to $\delta\Omega^{k+1}$, so it is its harmonic part plus an exact form.

Example: on the flat torus $\R^n / \Z^n$ the harmonic $k$-forms are the forms with constant coefficients in $dx^{i_1} \wedge \dots \wedge dx^{i_k}$. So $\dim H^k_{dR} = \binom{n}{k}$.

Compactness is needed: on $\R$, every function $a x + b$ is harmonic, and the space of harmonic $0$-forms is $2$-dimensional while $H^0_{dR}(\R)$ is $1$-dimensional.

Going deeper

For a student

Since $\star$ maps harmonic $k$-forms to harmonic $(n-k)$-forms, the theorem gives Poincaré duality $b_k = b_{n-k}$ for the Betti numbers. Combined with Weitzenböck identities, it turns curvature conditions into vanishing of cohomology, as in Bochner's theorem. On compact Kähler manifolds a refinement gives the Hodge decomposition of complex cohomology into $(p, q)$-types.