# Discrete Contraction and Inner Product

Let $$\alpha$$ and $$\beta$$ be two 1-forms. Their inner product can be expressed as

$(\alpha,\beta)=\langle\alpha^{\sharp},\beta\rangle=\langle\beta^{\sharp},\alpha\rangle=(\alpha^{\sharp},\beta^{\sharp})$

where $$\langle\cdot,\cdot\rangle$$ represents the pairing between a vector field and a one form. In this context one can think of vector fields as row matrices, and forms as row matrices.

The sharp operator is a mapping from forms to vector fields, and the flat operator is its inverse.

\begin{split}\begin{aligned} \flat: & \quad & TM\to T^{*}M\\ \sharp: & \quad & T^{*}M\to TM\end{aligned}\end{split}

Contraction with a one form

The contraction with a one form is simply given by

$\langle\alpha^{\sharp},\beta\rangle=(\alpha,\beta)=\star(\alpha\wedge\star\beta)$

Contraction with a two form

Suppose the two form can be expressed as the wedge product of two one forms such as $$\beta\wedge\gamma$$. Then

\begin{split}\begin{aligned} \langle\alpha^{\sharp},\beta\wedge\gamma\rangle & = & \langle\alpha^{\sharp},\beta\rangle\wedge\gamma-\langle\alpha^{\sharp},\gamma\rangle\wedge\beta\\ & = & \star(\alpha\wedge\star\beta)\wedge\gamma-\star(\alpha\wedge\star\gamma)\wedge\beta\end{aligned}\end{split}

# The Hodge Star Operator and the Dual Mesh

On an $$n$$-dimensional Riemanian manifold the continuous Hodge Star operator establishes an isomorphism between $$k$$-forms and $$(n-k)$$-forms.

\begin{split}\begin{aligned} \star^{k}: & \quad & \Lambda^{k}\to\Lambda^{n-k}\\ \star^{n-k}: & \quad & \Lambda^{n-k}\to\Lambda^{k}\end{aligned}\end{split}

It satisfies the following identity

$\star^{n-k}\star^{k}=\left(-1\right)^{k(n-k)}\mathbf{id}^{k}$

where $$\mathbf{id}^{k}$$ is the identity map between $$k$$-forms (it maps the form to itself)

\begin{split}\begin{aligned} \mathbf{id}^{k}: & \quad & \Lambda^{k}\to\Lambda^{k}\\ & & \alpha\mapsto\alpha\end{aligned}\end{split}

The hodge star operators is its own inverse up to a sign. Let us consider only the positive case for now (an analogous argument will hold for the negative case).

We want to study under what conditions the discrete Hodge star operator satisfies the discrete counterpart of the above equation

$$$\mathbf{H}\mathbf{\widetilde{\mathbf{H}}}=\mathbf{\widetilde{\mathbf{H}}H}=\mathbf{Id}\label{eq:discrete}$$$

The discrete Hodge star operator acting on primal forms is given by:

$H^{ij}=\langle\tilde{\sigma}^{i},\star\phi^{j}\rangle\equiv\int_{\tilde{\sigma}^{i}}\star\phi^{j}$

The discrete Hodge star acting on dual forms, on the other hand, is given by

$\widetilde{H}_{ij}=\langle\sigma_{i},\star\tilde{\phi}_{j}\rangle\equiv\int_{\sigma_{i}}\star\tilde{\phi}_{j}$

Let us assume that eq. (\ref{eq:discrete} ) holds. Then in terms of the basis functions it becomes

$\langle\sigma_{i},\star\tilde{\phi}_{j}\rangle\langle\tilde{\sigma}^{j},\star\phi^{k}\rangle=\delta_{i}^{k}$

The basis functions by their own definition must satisfy

$\langle\sigma_{i},\phi^{j}\rangle=\delta_{i}^{j}=\langle\tilde{\sigma}^{j},\tilde{\phi}_{i}\rangle$

Additionally, we desire the dual basis functions to be linear combinations of the hodge-star of the primal ones, therefore

$$$\tilde{\phi}_{k}=A_{kn}\star\phi^{n}\label{eq:linearity}$$$

where the matrix $$\mathbf{A}$$ represents the transform between the two basis sets. The hodge star is required in order for the dual basis functions to correspond to the correct simplex dimensions. For example if we take the basis functions for 0-forms in 2d corresponding to vertices, the dual basis functions must correspond to the corresponding elements on the dual mesh wich are faces, and therefore must be 2-forms.

The following sequence of equations are equivalent

\begin{split}\begin{aligned} \langle\sigma_{i},\star\tilde{\phi}_{j}\rangle\langle\tilde{\sigma}^{j},\star\phi^{k}\rangle & =\delta_{i}^{k}\\ A_{jn}\langle\sigma_{i},\star\star\phi^{n}\rangle\langle\tilde{\sigma}^{j},\star\phi^{k}\rangle & =\delta_{i}^{k}\\ A_{jn}\delta_{i}^{n}\langle\tilde{\sigma}^{j},\star\phi^{k}\rangle & =\delta_{i}^{k}\\ A_{ji}\langle\tilde{\sigma}^{j},\star\phi^{k}\rangle & =\delta_{i}^{k}\\ A_{ji}H^{jk} & =\delta_{i}^{k}\\ A_{ki}H^{kj} & =\delta_{i}^{j}\quad(k\leftrightarrow j)\end{aligned}\end{split}

and for eq. (\ref{eq:linearity} )

\begin{split}\begin{aligned} \langle\tilde{\sigma}^{j},\tilde{\phi}_{i}\rangle & =\delta_{i}^{j}\\ A_{ik}\langle\tilde{\sigma}^{j},\star\phi^{k}\rangle & =\delta_{i}^{j}\\ A_{ik}H^{jk} & =\delta_{i}^{j}\end{aligned}\end{split}

In matrix form

$\mathbf{A}^{T}\mathbf{H}=\mathbf{Id}=\mathbf{A}\mathbf{H}^{T}$

and since $$\mathbf{H}$$ is of full rank, this implies that

$\mathbf{A}^{T}=\mathbf{H}^{-1}$

We have shown that eq. (\ref{eq:discrete} ) (the Hodge star and its dual are exact inverses) and eq. (\ref{eq:linearity} ) (dual basis functions are linear combinations of primal ones) are completely consistent with each other (can it be shown that one implies the other?).

Therefore, the primal basis functions $$\phi$$ and the dual simplices $$\tilde{\sigma}$$ completely determine the dual basis functions, and they are implicitly given by

$\tilde{\phi}_{i}H^{ij}=\star\phi^{j}$

or

$\tilde{\phi}_{i}\,\langle\tilde{\sigma}^{i},\star\phi^{j}\rangle=\star\phi^{j}$