A discrete Hodge decomposition expresses a discrete 1-form \(\eta \in \mathbb{R}^E\) as \[ \eta = d_0 \alpha + \star_1^{-1} d_1^T \beta + \gamma \] for some primal 0-form potential \(\alpha \in \mathbb{R}^V\), dual 2-form potential \(\beta \in \mathbb{R}^F\), and harmonic component \(\gamma \in \mathbb{R}^E\) with \(d_1 \gamma = 0\) and \(d_0^T \star_1 \gamma = 0\). If \(\star_1\) is a positive definite matrix, it is easy to prove the existence of such a decomposition (for instance, see the exercises in Section 8.1 of Keenan's DDG notes). But in the case where \(\star_1\) is indefinite (which is quite common), the situation becomes trickier. Today, I will show that the decomposition remains valid under the weaker assumptions that \(\star_1\) is symmetric, invertible, and \[\ker d_0^T \star_1 d_0 = \ker d_0,\quad \ker d_1 \star_1^{-1} d_1^T = \ker d_1^T.\]

Let \(\star_0 \in \mathbb{R}^{V \times V}, \star_1 \in \mathbb{R}^{E \times E}\) and \(\star_2 \in \mathbb{R}^{F \times F}\) be symmetric, nondegenerate (\ie{} invertible), but possibly-indefinite. Given a subspace \(U \subset \mathbb{R}^E\), we define the \(\star_1\)-orthogonal complement as \[ U^{\perp_{\star_1}} := \left\{v \in \mathbb{R}^E \;\middle|\; U^T \star_1 v = 0\right\}. \] Note that because \(\star_1\) is nondegenerate, \(\dim U^{\perp_{\star_1}} = \dim \ker U^T = |E| - \dim U\).

As usual, we define the \(\star\)-adjoints \(d_k^\dagger = \star_k^{-1}d_k^T\star_{k+1}\), and nondegeneracy of \(\star_1\) suffices to guarantee that \(d_0^\dagger d_1^\dagger = 0\). And we verify that as usual, \(\textrm{im}\, d_0 \in \left(\textrm{im}\, d_1^\dagger\right)^{\perp_{\star_1}}\): \[ \left(d_1^\dagger\right)^T \star_1 d_0 = \left(\star_1^{-1}d_1^T\star_2\right)^T\star_1 d_0 = \star_2 d_1 \star_1^{-1} \star_1 d_0 = \star_2 d_1 d_0 = 0. \]

We continue and observe that the \(\star_1\)-orthogonal complement of this space is the space of harmonic forms, \(\mathcal{H}^1 = \ker d_0^\dagger \cap \ker d_1\) as usual: \[ \left(\textrm{im}\, d_0 \cup \textrm{im}\, d_1^\dagger\right)^{\perp_{\star_1}} = \left\{v \in \mathbb{R}^E \;\middle|\;d_0^T \star_1 v = 0 = \left(d_1^\dagger\right)^T\star_1 v\right\} = \ker d_0^\dagger \cap \ker d_1 = \mathcal{H}^1. \]

However we are not finished, since in the indefinite case we may not have \(U + U^{\perp_{\star_1}} = \mathbb{R}^E\): the intersection \(U \cap U^{\perp_{\star_1}}\) may contain nontrivial null vectors \(v\) with \(v^T \star_1 v = 0\). Indeed, by dimension counting we see that \[ \dim\left(U + U^{\perp_{\star_1}}\right) = \dim U + \dim U^{\perp_{\star_1}} - \dim \left(U \cap U^{\perp_{\star_1}}\right) = |E| - \dim \left(U \cap U^{\perp_{\star_1}}\right). \]

Thus, our discrete Hodge decomposition covers the entire space of discrete 1-forms if and only if the intersection of \(\textrm{im}\, d_0 \cup \textrm{im}\, d_1^\dagger\) with its \(\star_1\)-orthogonal complement equals \(\{0\}\).

Expanding the definitions, we want \(v = 0\) to be the only vector of the form \[ v = d_0 \alpha + d_1^\dagger \beta,\quad\text{with}\;d_0^\dagger v = 0,\;\text{and}\;d_1 v = 0. \]

This condition will be satisfied if and only if \[ d_0^\dagger d_0 \alpha = 0 \iff d_0 \alpha = 0,\quad d_1 d_1^\dagger \beta = 0 \iff d_1^\dagger \beta = 0. \]

Expanding out the definitions of the \(d_k^\dagger\) operators and using the nondegeneracy to eliminate some \(\star_k\) operators, we arrive at the final condition as desired \[ \ker d_0^T \star_1 d_0 = \ker d_0,\quad \ker d_1 \star_1^{-1} d_1^T = \ker d_1^T. \]

Note that this condition holds automatically if \(\star_1\) is positive definite (since \(\alpha^T d_0^T \star_1 d_0 \alpha = 0 \iff d_0 \alpha = 0\)), but it can hold even when \(\star_1\) is not (such as the cotan weights of a non-Delaunay triangle mesh).