Invariance of the Discrete Wedge Product

Consider a manifold \(\mathcal{M}\) and its polyhedrization in the context of discrete exterior calculus, and let \(\sigma_{i}\) denote the simplices and \(\phi^{i}\) the corresponding basis functions, such that the pairing given by integration is natural

\[\langle\sigma_{i},\phi^{j}\rangle\equiv\int_{\sigma_{i}}\phi^{j}=\delta_{i}^{\phantom{i}j}\]

Given the projection \(\mathcal{P}\) and reconstruction operators \(\mathcal{R}\),

\[\begin{split}\begin{aligned} \mathcal{P}:\quad & \Lambda^{k}\to\bar{\Lambda}^{k}\\ & \alpha\mapsto\bar{\alpha}_{i}=\int_{\sigma_{i}}\alpha\end{aligned}\end{split}\]
\[\begin{split}\begin{aligned} \mathcal{R}:\quad & \bar{\Lambda}^{k}\to\Lambda^{k}\\ & \bar{\alpha}_{i}\mapsto\alpha=\bar{\alpha}_{i}\phi^{i}\end{aligned}\end{split}\]

the discrete wedge product is defined as a three tensor

\[\bar{\gamma}=\mathbf{W}(\bar{\alpha},\bar{\beta})=\mathcal{P}\left(\left(\mathcal{R}\bar{\alpha}\right)\wedge\left(\mathcal{R}\bar{\beta}\right)\right)\]
\[\bar{\gamma}_{i}=\mathbf{W}_{i}^{\phantom{i}jk}\bar{\alpha}_{j}\bar{\beta}_{k}\]

where

\[\mathbf{W}_{i}^{\phantom{i}jk}=\left[\mathbf{W}(\phi^{j},\phi^{k})\right]_{i}=\int_{\sigma_{i}}\phi^{j}\wedge\phi^{k}\]

Consider now a second manifold \({\cal M}^{\prime}\) and a diffeomorphism

\[\varphi:\quad{\cal M}\to{\cal M}^{'}\]

which induces a new polyderization with the corresponding simplices \(\sigma^{\prime}=\varphi_{*}\sigma\) and basis functions \(\phi^{\prime}=\varphi_{*}\phi\).

Given a simplex \(\sigma\) and a form \(\phi\), the following identities hold for push-forwards and pull-backs:

\[\begin{equation}\int_{\varphi_{*}\sigma}\varphi_{*}\phi=\int_{\sigma}\phi\label{eq:integration}\end{equation}\]
\[\begin{equation}\varphi_{*}(\alpha\wedge\beta)=(\varphi_{*}\alpha)\wedge(\varphi_{*}\beta)\label{eq:wedge}\end{equation}\]

The discrete wedge product on this new manifold will be elementwise equal to the wedge product on the original manifold. To see why this is the case:

\[\begin{split}\begin{aligned} \mathbf{W}_{\phantom{\prime}i}^{\prime\phantom{i}jk} & =\int_{\sigma_{i}^{\prime}}\phi^{\prime j}\wedge\phi^{\prime k}\\ & =\int_{\varphi_{*}\sigma_{i}}\left(\varphi_{*}\phi^{j}\right)\wedge\left(\varphi_{*}\phi^{k}\right)\\ & =\int_{\varphi_{*}\sigma_{i}}\varphi_{*}\left(\phi^{j}\wedge\phi^{k}\right)\quad(\mbox{eq.}\ref{eq:wedge})\\ & =\int_{\sigma_{i}}\phi^{j}\wedge\phi^{k}\quad(\mbox{eq.}\ref{eq:integration})\\ & =\mathbf{W}_{i}^{\phantom{i}jk}\end{aligned}\end{split}\]

So we have shown that the discrete wedge product is indeed invariant under diffeomorphisms.

\[\mathbf{W}_{\phantom{\prime}i}^{\prime\phantom{i}jk}=\mathbf{W}_{\phantom{}i}^{\phantom{i}jk}\]

This result is consistent with theory in differential geometry, where the wedge product is thought of as a pureley topological operator that is independent of the metric.