Duality in Finite Element Exterior Calculus and Hodge Duality on the Sphere

Abstract

Finite element exterior calculus refers to the development of finite element methods for differential forms, generalizing several earlier finite element spaces of scalar fields and vector fields to arbitrary dimension n, arbitrary polynomial degree r, and arbitrary differential form degree k. The study of finite element exterior calculus began with the ${\mathcal {P}}_r\varLambda ^k$ and ${\mathcal {P}}_r^-\varLambda ^k$ families of finite element spaces on simplicial triangulations. In their development of these spaces, Arnold, Falk, and Winther rely on a duality relationship between ${\mathcal {P}}_r\varLambda ^k$ and $\mathring{{\mathcal {P}}}_{r+k+1}^-\varLambda ^{n-k}$ and between ${\mathcal {P}}_r^-\varLambda ^k$ and $\mathring{{\mathcal {P}}}_{r+k}\varLambda ^{n-k}$. In this article, we show that this duality relationship is, in essence, Hodge duality of differential forms on the standard n-sphere, disguised by a change of coordinates. We remove the disguise, giving explicit correspondences between the ${\mathcal {P}}_r\varLambda ^k$, ${\mathcal {P}}_r^-\varLambda ^k$, $\mathring{{\mathcal {P}}}_r\varLambda ^k$ and $\mathring{{\mathcal {P}}}_r^-\varLambda ^k$ spaces and spaces of differential forms on the sphere. As a direct corollary, we obtain new pointwise duality isomorphisms between ${\mathcal {P}}_r \varLambda ^k$ and $\mathring{{\mathcal {P}}}_{r+k+1}^-\varLambda ^{n-k}$ and between ${\mathcal {P}}_r^-\varLambda ^k$ and $\mathring{{\mathcal {P}}}_{r+k} \varLambda ^{n-k}$. These isomorphisms can be implemented via a simple computation, which we illustrate with examples.

Appendices

Appendix A. Equivalence of Definitions of ${\mathcal {P}}_r^-\varLambda ^k(T^n)$

The definition of ${\mathcal {P}}_r^-\varLambda ^k(T^n)$ that we gave in Definition 2.9 is different from the definition given by Arnold, Walk, and Winther in [4]. In this appendix, we show that the two definitions are equivalent.

The definition in [4, Subsection 3.2] involves choosing an arbitrary point z in the simplex $T^n$ and defining a vector field on $T^n$ such that the vector based at $x\in T^n$ points away from z with magnitude $\left|{x-z}\right|$. Setting z to be the center of the simplex $\left( \frac{1}{n+1},\dotsc ,\frac{1}{n+1}\right)$, one can check that this vector field is given by the formula

$$\begin{aligned} V_\kappa =\sum _{i=1}^{n+1}\left( x_i-\tfrac{1}{n+1}\right) \frac{\partial {}}{\partial {x_i}} \end{aligned}$$

for $x\in T^n$. Using this vector field, we can define ${\mathcal {P}}_r^-\varLambda ^k(T^n)$.

Definition A.1

For $r\ge 1$, Arnold, Falk, and Winther [4] define ${\mathcal {P}}_r^-\varLambda ^k(T^n)$ to be

$$\begin{aligned} {\mathcal {P}}_{r-1}\varLambda ^k(T^n)+\kappa {\mathcal {P}}_{r-1}\varLambda ^{k+1}(T^n), \end{aligned}$$

where $\kappa$ denotes the interior product with the vector field $V_\kappa$.

Proposition A.2

The definitions of ${\mathcal {P}}_r^-\varLambda ^k(T^n)$ given by Definitions 2.9 and A.1 are equivalent.

Proof

Let $t=x_1+\cdots +x_{n+1}$. With the equation

$$\begin{aligned} V_\kappa :=\sum _{i=1}^{n+1}\left( x_i-\tfrac{t}{n+1}\right) \frac{\partial {}}{\partial {x_i}}=X-\tfrac{t}{n+1}\nabla t, \end{aligned}$$

we can extend the definition of $V_\kappa$ to all of ${\mathbb {R}}^{n+1}$. Geometrically, $V_\kappa$ is the projection of the radial vector field X to the simplicies $t=\text {const}$. We can extend the definition of $\kappa$ to $\kappa :\varLambda ^{k+1}({\mathbb {R}}^{n+1}) \rightarrow \varLambda ^k({\mathbb {R}}^{n+1})$ to be the interior product with the vector field $V_\kappa$, and so we have

$$\begin{aligned} \kappa =i_X-\tfrac{t}{n+1}i_{\nabla t}. \end{aligned}$$

Assume now that a satisfies Definition 2.9, so a can be extended to a differential form $\hat{a}=i_X\hat{b}$, where $\hat{b}\in {\mathcal {P}}_{r-1}\varLambda ^{k+1}({\mathbb {R}}^{n+1})$. Then,

$$\begin{aligned} \hat{a}=\kappa \hat{b}+\tfrac{t}{n+1}i_{\nabla t}\hat{b} \end{aligned}$$

Thus, $a=\kappa b+c$, where b and c are the restrictions of $\hat{b}$ and $\tfrac{t}{n+1}i_{\nabla t}\hat{b}$ to $T^n$, respectively. By definition, $b\in {\mathcal {P}}_{r-1}\varLambda ^{k+1}(T^n)$. Meanwhile, since $t=1$ on $T^n$, we know that c is also the restriction to $T^n$ of $\tfrac{1}{n+1}i_{\nabla t}\hat{b}$, which is in ${\mathcal {P}}_{r-1}\varLambda ^k({\mathbb {R}}^{n+1})$ because $\nabla t$ has polynomial degree zero. Thus, $c\in {\mathcal {P}}_{r-1}\varLambda ^k(T^n)$. We conclude that $a\in \kappa {\mathcal {P}}_{r-1}\varLambda ^{k+1}(T^n)+{\mathcal {P}}_{r-1}\varLambda ^k(T^n)$, so a satisfies Definition A.1.

Conversely, assume that a satisfies Definition A.1. We consider the two summands as separate cases.

If $a\in {\mathcal {P}}_{r-1}\varLambda ^k(T^n)$, then let $\hat{a}'\in {\mathcal {P}}_{r-1}\varLambda ^k({\mathbb {R}}^{n+1})$ be an arbitrary extension of a, and let

$$\begin{aligned} \hat{a}&:=i_X(\mathrm {d}t\wedge \hat{a}') \end{aligned}$$

(11)

$$\begin{aligned}&:=i_X(\mathrm {d}t)\hat{a}'-\mathrm {d}t\wedge i_X\hat{a}' \end{aligned}$$

(12)

$$\begin{aligned}&:=t\hat{a}'-\mathrm {d}t\wedge i_X\hat{a}'. \end{aligned}$$

(13)

Using Eq. (13), since the restriction of t to $T^n$ is 1 and the restriction of $\mathrm{d}t$ to $T^n$ is zero, the restriction of $\hat{a}$ to $T^n$ is the same as the restriction of $\hat{a}'$, namely a. Meanwhile, using Eq. (11), we have that $\hat{a}\in i_X{\mathcal {P}}_{r-1}\varLambda ^{k+1}({\mathbb {R}}^{n+1})$ because $\mathrm {d}t\in {\mathcal {P}}_0\varLambda ^1({\mathbb {R}}^{n+1})$, so $\mathrm {d}t\wedge \hat{a}'\in {\mathcal {P}}_{r-1}\varLambda ^{k+1}({\mathbb {R}}^{n+1})$. Thus, a satisfies Definition 2.9.

If $a=\kappa b$ for some $b\in {\mathcal {P}}_{r-1}\varLambda ^{k+1}(T^n)$, then let $\hat{b}'\in {\mathcal {P}}_{r-1}\varLambda ^{k+1}({\mathbb {R}}^{n+1})$ be an arbitrary extension of b. Set

$$\begin{aligned} \hat{b}&:=\tfrac{1}{n+1}i_{\nabla t}(\mathrm {d}t\wedge \hat{b}') \end{aligned}$$

(14)

$$\begin{aligned}&:=\tfrac{1}{n+1}i_{\nabla t}(\mathrm {d}t)\hat{b}'-\tfrac{1}{n+1}\mathrm {d}t\wedge i_{\nabla t}\hat{b}' \end{aligned}$$

(15)

$$\begin{aligned}&:=\hat{b}'-\tfrac{1}{n+1}\mathrm {d}t\wedge i_{\nabla t}\hat{b}'. \end{aligned}$$

(16)

As before, observe from Eq. (16) that the restriction of $\hat{b}$ to $T^n$ is the same as the restriction of $\hat{b}'$, namely b. Because $V_\kappa$ is tangent to $T^n$, we can then conclude that the restriction of $\kappa \hat{b}$ to $T^n$ is $\kappa b=a$. Next, observe from Eq. (14) that $i_{\nabla t}\hat{b}=0$. Consequently, we can set

$$\begin{aligned} \hat{a}:=i_X\hat{b}=\kappa \hat{b}+\tfrac{t}{n+1}i_{\nabla t}\hat{b}=\kappa \hat{b}, \end{aligned}$$

from which we see that $\hat{a}$ is an extension of a. Finally, observe that $\hat{b}\in {\mathcal {P}}_{r-1}\varLambda ^{k+1}({\mathbb {R}}^{n+1})$ because $\mathrm{d}t$ and $\nabla t$ both have polynomial degree zero. Thus, a satisfies Definition 2.9, as desired. $\square$

Appendix B. Vector Space Identities

In this appendix, we provide proofs of two vector space identities involving the Hodge star. I believe that these identities are well-known, but I have not been able to find published references for them.

Let V be a vector space equipped with an inner product and an orientation. This structure defines an inner product $\langle \cdot ,\cdot \rangle$ on the exterior algebra $\bigwedge ^*V^*$ and a Hodge star map $*:\bigwedge ^*V^*\rightarrow \bigwedge ^*V^*$.

1.1 B.1. The Interior Product, the Exterior Product, and the Hodge Star

Let $X\in V$, and let $\nu \in V^*$ be the dual vector corresponding to X with respect to the inner product. It is a standard result that the adjoint of $i_X$ is $\nu \wedge$ in the sense that

$$\begin{aligned} \langle i_X\hat{\alpha },\hat{\beta }\rangle =\langle \hat{\alpha },\nu \wedge \hat{\beta }\rangle , \end{aligned}$$

where $\hat{\alpha }\in \bigwedge ^kV^*$ and $\hat{\beta }\in \bigwedge ^{k-1}V^*$.

This adjoint relationship has the following consequence for the Hodge star on $\bigwedge ^*V^*$.

Proposition B.1

Let V be an oriented inner product space. If $X\in V$ and $\nu \in V^*$ are dual to one another with respect to the inner product, then

$$\begin{aligned} i_X(*\hat{\alpha })=*(\hat{\alpha }\wedge \nu ). \end{aligned}$$

for all $\hat{\alpha }\in \bigwedge ^kV^*$.

Proof

Let $\dim V=n+1$. For all $\hat{\beta }\in \bigwedge ^{n-k}V^*$, we have

$$\begin{aligned} \langle i_X(*\hat{\alpha }),\hat{\beta }\rangle \mathrm{vol}= & {} \langle *\hat{\alpha },\nu \wedge \hat{\beta }\rangle \mathrm{vol}=\langle \hat{\alpha },*^{-1}(\nu \wedge \hat{\beta })\rangle \mathrm{vol}\\= & {} \hat{\alpha }\wedge \nu \wedge \hat{\beta } =\langle \hat{\alpha }\wedge \nu ,*^{-1}\hat{\beta }\rangle \mathrm{vol}=\langle *(\hat{\alpha }\wedge \nu ),\hat{\beta }\rangle \mathrm{vol}. \end{aligned}$$

$\square$

1.2 B.2 The Hodge Star on a Hyperplane

In this subsection, we consider a hyperplane H that is orthogonal to a unit covector $\nu \in V^*$. The inner product on V induces an inner product on H, and the orientation on V along with the choice of unit conormal $\nu$ induces an orientation on H. Thus, in addition to a Hodge star operator $\bigwedge ^*V^*\rightarrow \bigwedge ^*V^*$, we also have a Hodge star operator $\bigwedge ^*H^*\rightarrow \bigwedge ^*H^*$. We denote these by $*_V$ and $*_H$, respectively.

We show that $*_H$ can be computed from $*_V$ and $\nu$ as follows.

Proposition B.2

With notation as above, let $\alpha \in \bigwedge ^kH^*$ be the restriction to H of $\hat{\alpha }\in \bigwedge ^kV^*$. Then, $*_H\alpha$ is the restriction to H of

$$\begin{aligned} *_V(\nu \wedge \hat{\alpha }). \end{aligned}$$

Proof

Let $n=\dim H$, let $\hat{\beta }\in \bigwedge ^{n-k}V^*$ denote $*_V(\nu \wedge \hat{\alpha })$, and let $\beta \in \bigwedge ^{n-k}H^*$ be the restriction of $\hat{\beta }$ to H. We prove that $*_H\alpha =\beta$ by verifying it on a basis for $\hat{\alpha }$.

Let $\nu ,e_1,\dotsc ,e_n$ be an oriented orthonormal basis for $V^*$, so $e_1,\dotsc ,e_n$ is an oriented orthonormal basis for $H^*$, and we have that $\mathrm{vol}_H=e_1\wedge \cdots e_n$ and $\mathrm{vol}_V=\nu \wedge \mathrm{vol}_H$. For $I=\{i_1<i_2<\cdots <i_k\} \subseteq \{1,\dotsc ,n\}$, let $e_I$ denote $e_{i_1}\wedge \cdots \wedge e_{i_k}$.

If $\hat{\alpha }=\nu \wedge e_I$, then $\alpha =0$ because the restriction of $\nu$ to H is zero. Also, $\nu \wedge \hat{\alpha }=0$, so $\hat{\beta }$ and $\beta$ are zero as well, as desired.

Now let $\hat{\alpha }=e_I$. By the definition of $*_V$ with respect to the oriented basis $\nu ,e_1,\dotsc ,e_n$, we see that $\hat{\beta }=*_V(\nu \wedge e_I)$ has the form $\pm e_J$, where $J=\{1,\dotsc ,n\}{\setminus } I$ and the sign is chosen so that $(\nu \wedge e_I)\wedge (\pm e_J)=\mathrm{vol}_V$. Since $\nu \wedge (e_I\wedge \pm e_J)=\mathrm{vol}_V$, we conclude that $e_I\wedge \pm e_J=\mathrm{vol}_H$. Thus, by the definition of $*_H$ with respect to the oriented basis $e_1,\dotsc ,e_n$, we have $*_He_I=\pm e_J$, that is, $*_H\alpha =\beta$. $\square$