Hodge decomposition

In the discrete 2d case, there is an orthogonal decomposition of an edge flow on a graph into three components:

• a gradient flow that is globally acyclic,
• a harmonic flow that is locally acyclic but globally cyclic, and
• a curl flow that is locally cyclic.

More generally...

Suppose $M$ is a Riemannian manifold and suppose $\delta =d^*$ is the codifferential on $M$. We say that a differential form $\omega$ is

• co-closed if $\delta\omega=0$;
• co-exact if $\omega=\delta\alpha$ for some form $\alpha$;
• harmonic if $\Delta\omega=0$, where $\Delta$ is the Laplace-de Rham operator.

Hodge decomposition theorem. Any $k$-form $\omega$ on $M$ can be split into three components: $$\phi = d\alpha +\delta \beta + \gamma \,$$ with $\gamma$ is harmonic.

Why? The proof is based on these two facts:

• Exact forms are orthogonal to co-closed forms, and also
• Co-exact forms are orthogonal to closed forms.

Therefore, the orthogonal complement to the set of all exact and co-exact forms consists of forms that are both closed and co-closed, i.e., harmonic.

The orthogonality is defined with respect to the $L^2$-inner product on the cochain complex $\Omega^k(M)$ or $C^k(M)$: $$<\alpha,\beta>=\int_M \alpha \wedge *\beta.$$

As a single formula: $$\Omega ^j \cong d\Omega ^{j-1}(M) \oplus H_{dR}^j(M)\oplus d^*\Omega ^{j+1}(M),$$ where $H_{dR}^j(M)$ is the de Rham cohomology.