Hodge decomposition theorem

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.