# 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:

$$$\int_{\varphi_{*}\sigma}\varphi_{*}\phi=\int_{\sigma}\phi\label{eq:integration}$$$
$$$\varphi_{*}(\alpha\wedge\beta)=(\varphi_{*}\alpha)\wedge(\varphi_{*}\beta)\label{eq:wedge}$$$

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.