# On Basis Constructions in
Finite Element Exterior Calculus

Martin W. Licht  UCSD Department of Mathematics, 9500 Gilman Drive MC0112, La
Jolla, CA 92093-0112, USA<EMAIL_ADDRESS>
###### Abstract.

We give a systematic and self-contained account of the construction of
geometrically decomposed bases and degrees of freedom in finite element
exterior calculus. In particular, we elaborate upon a previously overlooked
basis for one of the families of finite element spaces, which is of interest
for implementations. Moreover, we give details for the construction of
isomorphisms and duality pairings between finite element spaces. These
structural results show, for example, how to transfer linear dependencies
between canonical spanning sets, or give a new derivation of the degrees of
freedom.

###### Key words and phrases:

barycentric differential form, canonical spanning sets, degrees of freedom,
finite element exterior calculus, geometrically decomposed bases

###### 2010 Mathematics Subject Classification:

65N30

This research was supported by the European Research Council through the
FP7-IDEAS-ERC Starting Grant scheme, project 278011 STUCCOFIELDS. This
research was supported in part by NSF DMS/RTG Award 1345013 and DMS/CM Award
1262982. Parts of this article are based on the author’s PhD thesis.

##  1\. Introduction

Exterior calculus is a canonical approach towards mathematical
electromagnetism. Utilizing exterior calculus in numerical analysis hence
seems to be a natural choice, and a multitude of finite element methods have
been formulated in what is now known as _Finite Element Exterior Calculus_
(FEEC [4]).

A particular achievement of FEEC has been the identification of spaces of
polynomial differential forms invariant under affine transformations, and
subsequently the construction of _finite element de Rham complexes_. Research
in numerical analysis has elaborated upon bases, degrees of freedom, and their
geometric decompositions for those finite element spaces (see [2, 3, 11, 12,
16, 17], for example).

The primary purpose of the present publication is to give a new self-contained
account of the construction of bases and degrees of freedom in finite element
exterior calculus. Additionally, we develop several new techniques which
complete previous results in the literature. One new outcome of our research
are concise derivations of geometrically decomposed bases for spaces of finite
element differential forms. A new result with particular relevance to
practical implementations is a new and simple basis for one of the families of
spaces in FEEC; to the author’s best knowledge, that basis is used implicitly
in the seminal work of Arnold, Falk, and Winther [2] but has not attracted
much attention since then. Furthermore, the present work elaborates on
isomorphisms and duality pairings which have been used implicitly in the
initial work of Arnold, Falk, and Winther, and which have been made explicit
recently by Christiansen and Rapetti [8]. This leads to a straightforward
construction of geometrically decomposed degrees of freedom for the finite
element spaces.

Our construction of geometrically decomposed bases for finite element spaces
builds upon previous publications in the literature. For the purpose of
comparison, we recall the approaches in two of the major references. We take
the exposition [2, Chapter 4] of Arnold, Falk, and Winther as a starting
point. First, their approach devises a basis for
${\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$. Then they determine a geometrically
decomposed basis of the dual space ${\mathcal{P}}_{r}\Lambda^{k}(T)^{\ast}$,
and subsequently a geometrically decomposed basis of the dual space
${\mathcal{P}}_{r}^{-}\Lambda^{k}(T)^{\ast}$. After outlining bases for spaces
with vanishing trace, they find geometrically decomposed bases for
${\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$ and, implicitly, for
${\mathcal{P}}_{r}\Lambda^{k}(T)$. In [3], we are given bases for the spaces
with vanishing trace, $\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ and
$\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$. The latter publication studies
extension operators and geometrically decomposed bases, in explicit form also
for ${\mathcal{P}}_{r}\Lambda^{k}(T)$, though the basis in [3] is generally
different from the one in [2].

We arrange this material in a different manner. In particular, we directly
construct geometrically decomposed bases of the spaces
${\mathcal{P}}_{r}\Lambda^{k}(T)$ which contain bases of
$\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$. It does not seem to be widely
known that explicit geometrically decomposed bases for the
${\mathcal{P}}_{r}\Lambda^{k}$-family can already be derived with the methods
in [2] and that these bases are different from the ones in [3]. For that
reason, the author hopes that the present exposition helps a larger audience
to harness these results. By contrast, our geometrically decomposed bases for
${\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$ and
$\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ coincide with those in [2]
and [3]. We emphasize that we develop bases for the finite element spaces
without referring to any degrees of freedom in the first place.

Having found bases for the finite element spaces, we study isomorphisms
between finite element spaces. Whenever $T$ is an $n$-dimensional simplex and
$k$ and $r$ are non-negative integers, we have isomorphisms

(1)
$\displaystyle{\mathcal{P}}_{r}\Lambda^{k}(T)\simeq\mathring{{\mathcal{P}}}^{-}_{r+n-k+1}\Lambda^{n-k}(T),\quad{\mathcal{P}}^{-}_{r+1}\Lambda^{n-k}(T)\simeq\mathring{{\mathcal{P}}}_{r+n-k+1}\Lambda^{k}(T).$

In fact, more is true: to each of these isomorphic pairs corresponds a duality
pairing. These isomorphic relations and duality pairings are used in the
seminal publication by Arnold, Falk, and Winther [2].

A novel result in our exposition is that both isomorphisms preserve the
canonical spanning sets. This allows to set into correspondence the linear
dependencies of the canonical spanning sets between finite element spaces; in
turn, any basis for one of the spaces gives a basis for the corresponding
“partner space”. Subsequently, we carefully analyze the duality pairings;
they, too, can be stated in terms of the spanning sets alone, and we prove a
new definiteness condition. These ideas enable a new presentation of how to
construct the degrees of freedom for the finite element spaces. At this point,
our findings draw major inspiration from recent work by Christiansen and
Rapetti [8], who have derived similar results for the first isomorphism in (1)
and devise an algebraic resolution of the corresponding finite element spaces.
Their results for the second isomorphic pair, however, are less extensive.
Using different methods, this article reproduces and refines the results in
[8] on the first isomorphic pair and gives an analogous result for the second
isomorphic pair.

Finding geometrically decomposed bases and degrees of freedom is easy for
finite element spaces of scalar functions. It is a difficult challenge,
however, for vector-valued finite element spaces. The reason is that the
latter’s canonical spanning sets are linearly independent. Bases and degrees
of freedom in FEEC are discussed in [2, 3]. We take inspiration from the
framework of _finite element systems_ [8] and earlier algebraic investigations
[11, 16, 17]. Other aspects of vector-valued finite element methods include
condition numbers and sparsity properties [18, 5, 6], or fast evaluation of
discrete operators [13, 14, 1]. Finite element differential forms are also
studied over quadrilaterals and general polytopes [10, 7]. This article
prepares further algebraic and combinatorial studies of finite element spaces.

The remainder of this work is structured as follows. In Section 2 we review
combinatorial results, exterior calculus, and polynomial differential forms.
Section 3 summarizes some auxiliary lemmas. In Section 4 we introduce spaces
of polynomial differential forms and construct geometrically decomposed bases.
Subsequently, we study the isomorphy relations in Section 5 and the duality
pairings in Section 6. We supplement applications to finite element spaces
over triangulations in Section 7.

##  2\. Preliminaries

We introduce or review notions regarding combinatorics and differential forms
over simplices. All vector spaces in this publication are over the complex
numbers unless noted otherwise; we write $\overline{z}\in{\mathbb{C}}$ for the
complex conjugate of $z\in{\mathbb{C}}$.

###  2.1. Combinatorics

We let $[m:n]=\\{m,\dots,n\\}$ for $m,n\in{\mathbb{Z}}$ with $m\leq n$. For
$m,n\in{\mathbb{Z}}$ with $m\neq n$ we let ${\epsilon}(m,n)=1$ if $m<n$ and
${\epsilon}(m,n)=-1$ if $m>n$.

For any mapping $\alpha\colon[m:n]\rightarrow{\mathbb{N}}_{0}$ we write
$|\alpha|:=\sum_{i=m}^{n}\alpha(i)$. Given $r,m,n\in{\mathbb{N}}_{0}$, we let
$A(r,m:n)$ be the set of all mappings
$\alpha\colon[m:n]\rightarrow{\mathbb{N}}_{0}$ for which $|\alpha|=r$. So
$A(r,m:n)$ is the set of _multiindices_ over the set $[m:n]$. We abbreviate
$A(r,n):=A(r,0:n)$. Whenever $\alpha\in A(r,m:n)$, we write

(2) $\displaystyle[\alpha]:=\left\\{\;i\in[m:n]\mid\alpha(i)>0\;\right\\},$

and we write $\lfloor\alpha\rfloor$ for the minimal element of $[\alpha]$
provided that $[\alpha]$ is not empty, and $\lfloor\alpha\rfloor=\infty$
otherwise. The sum $\alpha+\beta$ of $\alpha,\beta\in A(r,m:n)$ is defined in
the obvious manner. When $\alpha\in A(r,m:n)$ and $p\in[m:n]$, then $\alpha+p$
denotes the unique member of $A(r+1,m:n)$ with $(\alpha+p)(p)=\alpha(p)+1$ and
coincides with $\alpha$ otherwise; similarly, when $p\in[\alpha]$, then
$\alpha-p$ denotes the unique member of $A(r-1,m:n)$ with
$(\alpha-p)(p)=\alpha(p)-1$ and coincides with $\alpha$ otherwise.

For $a,b,m,n\in{\mathbb{N}}_{0}$, we let $\Sigma(a:b,m:n)$ be the set of
strictly ascending mappings from $[a:b]$ to $[m:n]$. We call those mappings
also _alternator indices_. We write $\Sigma(a:b,m:n):=\\{\emptyset\\}$
whenever $a>b$. For any $\sigma\in\Sigma(a:b,m:n)$ we let

(3) $\displaystyle[\sigma]:=\left\\{\sigma(i)\mid i\in[a:b]\right\\},$

and we write $\lfloor\sigma\rfloor$ for the minimal element of $[\sigma]$
provided that $[\sigma]$ is not empty, and $\lfloor\sigma\rfloor=\infty$
otherwise. Furthermore, if $q\in[m:n]\setminus[\sigma]$, then we write
$\sigma+q$ for the unique element of $\Sigma(a:b+1,m:n)$ with image
$[\sigma]\cup\\{q\\}$. In that case, we also write ${\epsilon}(q,\sigma)$ for
the signum of the permutation that orders the sequence
$q,\sigma(a),\dots,\sigma(b)$ in ascending order, and we write
${\epsilon}(\sigma,q)$ for the signum of the permutation that orders the
sequence $\sigma(a),\dots,\sigma(b),q$ in ascending order. Similarly, if
$p\in[\sigma]$, then we write $\sigma-p$ for the unique element of
$\Sigma(a:b-1,m:n)$ with image $[\sigma]\setminus\\{p\\}$. Note that
${\epsilon}(\sigma,q)=(-1)^{b-a+1}{\epsilon}(q,\sigma)$.

We use the abbreviations $\Sigma(k,n)=\Sigma(1:k,0:n)$ and
$\Sigma_{0}(k,n)=\Sigma(0:k,0:n)$. If $n$ is understood and $k,l\in[0:n]$,
then for any $\sigma\in\Sigma(k,n)$ we define $\sigma^{c}\in\Sigma_{0}(n-k,n)$
by the condition $[\sigma]\cup[\sigma^{c}]=[0:n]$, and for any
$\rho\in\Sigma_{0}(l,n)$ we define $\rho^{c}\in\Sigma(n-l,n)$ by the condition
$[\rho]\cup[\rho^{c}]=[0:n]$. In particular, $\sigma^{cc}=\sigma$ and
$\rho^{cc}=\rho$. Note that $\sigma^{c}$ and $\rho^{c}$ depend on $n$, which
we suppress in the notation.

When $\sigma\in\Sigma(k,n)$ and $\rho\in\Sigma_{0}(l,n)$ with
$[\sigma]\cap[\rho]=\emptyset$, then ${\epsilon}(\sigma,\rho)$ denotes the
signum of the permutation ordering the sequence
$\sigma(1),\dots,\sigma(k),\rho(0),\dots,\rho(l)$ in ascending order, and we
let $\sigma+\rho\in\Sigma(0:k+l,0:n)$ be the unique strictly ascending mapping
from $[0:k+l]$ to $[0:n]$ whose image is the set $[\sigma]\cup[\rho]$.

###  2.2. Simplices

Let $n\in{\mathbb{N}}_{0}$. An $n$-dimensional simplex $T$ is the convex
closure of pairwise distinct points $v^{T}_{0},\dots,v^{T}_{n}$ in Euclidean
space, called the _vertices_ of $T$, such that the vertices are an affinely
independent set. Note that the dimension of the ambient Euclidean space must
be at least $n$ but otherwise does not matter.

An _ordered simplex_ is a simplex with an ordering of its set of vertices (see
[9]) We henceforth assume that all simplices in this article are ordered.

We call $F\subseteq T$ a _subsimplex_ of $T$ if the set of vertices of $F$ is
a subset of the set of vertices of $T$. We write $\imath(F,T)\colon
F\rightarrow T$ for the set inclusion of $F$ into $T$.

Suppose that $F$ is an $m$-dimensional subsimplex of $T$ with ordered vertices
$v^{F}_{0},\dots,v^{F}_{m}$. With a mild abuse of notation, we let
$\imath(F,T)\in\Sigma_{0}(m,n)$ denote the unique mapping that satisfies
$v^{T}_{\imath(F,T)(i)}=v^{F}_{i}$.

###  2.3. Barycentric Coordinates and Differential Forms

Let $T$ be a simplex of dimension $n$. Following the notation of [2], we write
$\Lambda^{k}(T)$ for the space of _differential $k$-forms_ over $T$ with
smooth bounded coefficients of all orders, where $k\in{\mathbb{Z}}$. Recall
that these mappings take values in the $k$-th exterior power of the dual of
the tangential space of the simplex $T$. In the case $k=0$, the space
$\Lambda^{0}(T)=C^{\infty}(T)$ is just the space of smooth functions over $T$
with uniformly bounded derivatives. Furthermore, $\Lambda^{k}(T)$ is the
trivial vector space unless $0\leq k\leq n$.

We recall the _exterior product_ $\omega\wedge\eta\in\Lambda^{k+l}(T)$ for
$\omega\in\Lambda^{k}(T)$ and $\eta\in\Lambda^{l}(T)$ and that it satisfies
$\omega\wedge\eta=(-1)^{kl}\eta\wedge\omega$. We let
${\mathsf{d}}\colon\Lambda^{k}(T)\rightarrow\Lambda^{k+1}(T)$ denote the
_exterior derivative_. It satisfies
${\mathsf{d}}\left(\omega\wedge\eta\right)={\mathsf{d}}\omega\wedge\eta+(-1)^{k}\omega\wedge{\mathsf{d}}\eta$
for $\omega\in\Lambda^{k}(T)$ and $\eta\in\Lambda^{l}(T)$. We also recall that
the integral $\int_{T}\omega$ of a differential $n$-form over $T$ is well-
defined. We refer to [2] and [15] for more background. In this article, we
focus on a special class of differential forms, namely the _barycentric
differential forms_.

The _barycentric coordinates_
$\lambda^{T}_{0},\dots,\lambda^{T}_{n}\in\Lambda^{0}(T)$ are the unique affine
functions over $T$ that satisfy the _Lagrange property_

(4) $\displaystyle\lambda^{T}_{i}(v_{j})=\delta_{ij},\quad i,j\in[0:n].$

The barycentric coordinate functions of $T$ are linearly independent and
constitute a partition of unity:

(5) $\displaystyle 1=\lambda^{T}_{0}+\dots+\lambda^{T}_{n}.$

We write
${\mathsf{d}}\lambda_{0}^{T},{\mathsf{d}}\lambda_{1}^{T},\dots,{\mathsf{d}}\lambda_{n}^{T}\in\Lambda^{1}(T)$
for the exterior derivatives of the barycentric coordinates. The exterior
derivatives are differential $1$-forms and constitute a partition of zero:

(6) $\displaystyle
0={\mathsf{d}}\lambda^{T}_{0}+\dots+{\mathsf{d}}\lambda^{T}_{n}.$

It can be shown that this is the only linear independence between the exterior
derivatives of the barycentric coordinate functions.

We consider several classes of differential forms over $T$ that are expressed
in terms of the barycentric polynomials and their exterior derivatives. When
$r\in{\mathbb{N}}_{0}$ and $\alpha\in A(r,n)$, then the corresponding
_barycentric polynomial_ over $T$ is

(7) $\displaystyle\lambda^{T}_{\alpha}$
$\displaystyle:=\prod_{i=0}^{n}(\lambda_{i}^{T})^{\alpha(i)}.$

When $a,b\in{\mathbb{N}}_{0}$ and $\sigma\in\Sigma(a:b,0:n)$, the
corresponding _barycentric alternator_ is

(8)
$\displaystyle{\mathsf{d}}\lambda^{T}_{\sigma}:={\mathsf{d}}\lambda^{T}_{\sigma(a)}\wedge\dots\wedge{\mathsf{d}}\lambda^{T}_{\sigma(b)}.$

Here, we treat the special case $\sigma=\emptyset$ by defining
${\mathsf{d}}\lambda^{T}_{\emptyset}=1$. Finally, whenever
$a,b\in{\mathbb{N}}_{0}$ and $\rho\in\Sigma(a:b,0:n)$, then the corresponding
_Whitney form_ is

(9)
$\displaystyle\phi^{T}_{\rho}:=\sum_{p\in[\rho]}{\epsilon}(p,\rho-p)\lambda^{T}_{p}{\mathsf{d}}\lambda^{T}_{\rho-p}.$

In the special case that $\rho_{T}\colon[0:n]\rightarrow[0:n]$ is the single
member of $\Sigma_{0}(n,n)$, then we write $\phi_{T}:=\phi_{\rho_{T}}$ for the
associated Whitney form.

In the sequel, we call the differential forms (7), (8), (9), and their sums
and exterior products, _barycentric differential forms_ over $T$.

######  Remark 2.1.

Whenever a fixed simplex $T$ is understood and there is no danger of
ambiguity, we may simplify the notation by writing

$\displaystyle\lambda_{i}\equiv\lambda_{i}^{T},\quad\lambda^{\alpha}\equiv\lambda_{T}^{\alpha},\quad{\mathsf{d}}\lambda_{\sigma}\equiv{\mathsf{d}}\lambda_{\sigma}^{T},\quad\phi_{\rho}\equiv\phi_{\rho}^{T}.$

###  2.4. Traces

Let $T$ be an $n$-dimensional simplex and let $F\subseteq T$ be a subsimplex
of $T$ of dimension $m$. The inclusion $\imath(F,T)\colon F\rightarrow T$
introduced above naturally induces a mapping
$\operatorname{tr}_{T,F}\colon\Lambda^{k}(T)\rightarrow\Lambda^{k}(F)$ by
taking the pullback. We call $\operatorname{tr}_{T,F}$ the _trace_ from $T$
onto $F$. It is well-known that
${\mathsf{d}}\operatorname{tr}_{T,F}\omega=\operatorname{tr}_{T,F}{\mathsf{d}}\omega$
for all $\omega\in\Lambda^{k}(T)$, that is, the exterior derivative commutes
with taking traces. In the case of $0$-forms, the trace is just the natural
restriction operator of functions. The trace does not depend on the order of
the simplices.

Taking into account the ordering of the simplices, however, we obtain explicit
formulas for the traces of barycentric differential forms. Write
$[\imath(F,T)]$ for the set of indices of those vertices of $T$ that are also
vertices of $F$, which is compatible with prior definition of $[\imath(F,T)]$.
We write $\imath(F,T)^{\dagger}\colon[\imath(F,T)]\rightarrow[0:m]$ for the
inverse of the mapping $\imath(F,T)\colon[0:m]\rightarrow[\imath(F,T)]$

Consider $i\in[0:n]$. If $i\notin[\imath(F,T)]$, then $v_{i}^{T}$ is a vertex
of $T$ that is not a vertex of $F$, and in that case we have
$\operatorname{tr}_{T,F}\lambda_{i}^{T}=0$. If instead $i\in[\imath(F,T)]$,
then there exists $j\in[0:m]$ such that $i=\imath(F,T)(j)$, and in that case
we have $\operatorname{tr}_{T,F}\lambda_{i}^{T}=\lambda_{j}^{F}$ or,
equivalently,
$\operatorname{tr}_{T,F}\lambda_{i}^{T}=\lambda_{\imath(F,T)^{\dagger}i}^{F}$.
Analogous observations follow for the exterior derivatives of the barycentric
coordinates.

Let $\alpha\in A(r,0:n)$ be a multiindex. If
$[\alpha]\nsubseteq[\imath(F,T)]$, then we have
$\operatorname{tr}_{T,F}\lambda_{T}^{\alpha}=0$. If instead
$[\alpha]\subseteq[\imath(F,T)]$, then there exists $\hat{\alpha}\in A(r,m:n)$
with $\hat{\alpha}=\alpha\circ\imath(F,T)$, and we then have

(10)
$\displaystyle\operatorname{tr}_{T,F}\lambda_{T}^{\alpha}=\lambda_{F}^{\hat{\alpha}}$

Let $\sigma\in\Sigma(a:b,0:n)$ be a basic alternator. If
$[\sigma]\nsubseteq[\imath(F,T)]$, then we have
$\operatorname{tr}_{T,F}{\mathsf{d}}\lambda^{T}_{\sigma}=0$. If instead
$[\sigma]\subseteq[\imath(F,T)]$, then there exists
$\hat{\sigma}\in\Sigma(a:b,0:n)$ with $\imath(F,T)\circ\hat{\sigma}=\sigma$,
or equivalently, $\hat{\sigma}=\imath(F,T)^{\dagger}\circ\sigma$, and we then
have

(11)
$\displaystyle\operatorname{tr}_{T,F}{\mathsf{d}}\lambda_{T}^{\sigma}={\mathsf{d}}\lambda_{F}^{\hat{\sigma}},\quad\operatorname{tr}_{T,F}\phi_{\sigma}^{T}=\phi_{\hat{\sigma}}^{F}.$

##  3\. Auxiliary Lemmas

In this section we provide some auxiliary lemmas on barycentric differential
forms over an $n$-dimensional simplex $T$.

Suppose that $k\in[1:n]$ and $\sigma\in\Sigma(k,n)$. For any $p\in[\sigma]$ we
have

(12)
$\displaystyle{\mathsf{d}}\lambda_{\sigma}={\epsilon}(p,\sigma-p){\mathsf{d}}\lambda_{p}\wedge{\mathsf{d}}\lambda_{\sigma-p}.$

This follows from definitions and properties of the alternating product. The
analogous result for the Whitney forms is folklore and slightly more technical
to derive.

######  Lemma 3.1.

Let $k\in[0:n]$. If $\rho\in\Sigma_{0}(k,n)$ and $q\in[0:n]$ with
$q\notin[\rho]$, then

(13)
$\displaystyle{\epsilon}(q,\rho)\phi_{\rho+q}=\lambda_{q}{\mathsf{d}}\lambda_{\rho}-{\mathsf{d}}\lambda_{q}\wedge\phi_{\rho}.$

###### Proof.

If $\rho$ and $q$ are as in the statement of the lemma, then we find

$\displaystyle\lambda_{q}{\mathsf{d}}\lambda_{\rho}-{\epsilon}(q,\rho)\phi_{\rho+q}$
$\displaystyle=\lambda_{q}{\mathsf{d}}\lambda_{\rho}-{\epsilon}(q,\rho)\sum_{l\in[\rho+q]}{\epsilon}(l,\rho+q-l)\lambda_{l}{\mathsf{d}}\lambda_{\rho+q-l}$
$\displaystyle=-{\epsilon}(q,\rho)\sum_{l\in[\rho]}{\epsilon}(l,\rho+q-l)\lambda_{l}{\mathsf{d}}\lambda_{\rho+q-l}.$

Using definitions and (12), we get

$\displaystyle\sum_{l\in[\rho]}{\epsilon}(l,\rho+q-l)\lambda_{l}{\mathsf{d}}\lambda_{\rho+q-l}={\mathsf{d}}\lambda_{q}\wedge\sum_{l\in[\rho]}{\epsilon}(l,\rho+q-l){\epsilon}(q,\rho-l)\lambda_{l}{\mathsf{d}}\lambda_{\rho-l}.$

For any $l\in[\rho]$ one finds that

$\displaystyle{\epsilon}(l,\rho+q-l){\epsilon}(q,\rho-l)={\epsilon}(l,q){\epsilon}(l,\rho-l){\epsilon}(q,l){\epsilon}(q,\rho)=-{\epsilon}(l,\rho-l){\epsilon}(q,\rho).$

The desired statement now follows from the definition of $\phi_{\rho}$. ∎

Whenever $k\in[0:n]$ and $\rho\in\Sigma_{0}(k,n)$, then it follows from
definitions that we can express the differential of the corresponding Whitney
form by

(14) $\displaystyle{\mathsf{d}}\phi_{\rho}=(k+1){\mathsf{d}}\lambda_{\rho}.$

The following result, which has appeared as Proposition 3.4 in [8], and also
as Equation (6.6) in [3], can be seen as a converse to that.

######  Lemma 3.2.

Let $k\in[0:n]$ and $\rho\in\Sigma_{0}(k,n)$. Then

(15)
$\displaystyle{\mathsf{d}}\lambda_{\rho}=\sum_{q\in[\rho^{c}]}{\epsilon}(q,\rho)\phi_{\rho+q}.$

###### Proof.

We use Lemma 3.1 and see

$\displaystyle\sum_{q\in[\rho^{c}]}{\epsilon}(q,\rho)\phi_{\rho+q}$
$\displaystyle=\sum_{q\in[\rho^{c}]}\lambda_{q}{\mathsf{d}}\lambda_{\rho}-\sum_{q\in[\rho^{c}]}{\mathsf{d}}\lambda_{q}\wedge\phi_{\rho}.$

Using (9), (6), (12), and (5), we see that the last expression equals

$\displaystyle\sum_{q\in[\rho^{c}]}\lambda_{q}{\mathsf{d}}\lambda_{\rho}+\sum_{p\in[\rho]}{\mathsf{d}}\lambda_{p}\wedge\phi_{\rho}=\sum_{q\in[\rho^{c}]}\lambda_{q}{\mathsf{d}}\lambda_{\rho}+\sum_{p\in[\rho]}{\mathsf{d}}\lambda_{p}\wedge{\epsilon}(p,\rho-p)\lambda_{p}{\mathsf{d}}\lambda_{\rho-p}$
$\displaystyle\qquad=\sum_{q\in[\rho^{c}]}\lambda_{q}{\mathsf{d}}\lambda_{\rho}+\sum_{p\in[\rho]}\lambda_{p}{\mathsf{d}}\lambda_{\rho}=\sum_{i=0}^{n}\lambda_{i}{\mathsf{d}}\lambda_{\rho}={\mathsf{d}}\lambda_{\rho},$

which had to be shown. ∎

The following identity describes an elementary linear dependence between
Whitney forms of higher order; see also [3, Equation (6.5)] and [8,
Proposition 3.3].

######  Lemma 3.3.

Let $k\in[0:n]$ and $\rho\in\Sigma_{0}(k,n)$. Then

(16)
$\displaystyle\sum_{p\in[\rho]}{\epsilon}(p,\rho-p)\lambda_{p}\phi_{\rho-p}=0.$

###### Proof.

Using (9), we expand the left-hand side of (16) to see

$\displaystyle\sum_{p\in[\rho]}{\epsilon}(p,\rho-p)\lambda_{p}\phi_{\rho-p}$
$\displaystyle=\sum_{p\in[\rho]}{\epsilon}(p,\rho-p)\lambda_{p}\sum_{s\in[\rho-p]}\lambda_{s}{\epsilon}(s,\rho-
p-s){\mathsf{d}}\lambda_{\rho-p-s}$
$\displaystyle=\sum_{\begin{subarray}{c}p,s\in[\rho]\\\ p\neq
s\end{subarray}}{\epsilon}(p,\rho-p){\epsilon}(s,\rho-
p-s)\lambda_{p}\lambda_{s}{\mathsf{d}}\lambda_{\rho-p-s}.$

We have ${\epsilon}(s,\rho-p-s)={\epsilon}(s,\rho-s){\epsilon}(s,p)$ for
$s,p\in[\rho]$ with $s\neq p$. The desired statement now follows by reasoning
with antisymmetry of the summands. ∎

##  4\. Finite Element Spaces

In this section we introduce two families of barycentric differential forms
over a simplex and find geometrically decomposed bases. Throughout this
section, we let $T$ be a simplex of dimension $n$, let $r\in{\mathbb{N}}_{0}$,
and let $k\in[0:n]$.

In the sequel, we are particularly interested in the following two spaces of
barycentric differential forms:

(17) $\displaystyle{\mathcal{P}}_{r}\Lambda^{k}(T)$
$\displaystyle:=\operatorname*{span}\left\\{\;\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}\mathrel{}\middle|\mathrel{}\alpha\in
A(r,n),\;\sigma\in\Sigma(k,n)\;\right\\},$ (18)
$\displaystyle{\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$
$\displaystyle:=\operatorname*{span}\left\\{\;\lambda_{T}^{\alpha}\phi^{T}_{\rho}\mathrel{}\middle|\mathrel{}\alpha\in
A(r-1,n),\;\rho\in\Sigma_{0}(k,n)\;\right\\}.$

We also consider subspaces of differential forms with vanishing traces:

(19) $\displaystyle\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$
$\displaystyle:=\left\\{\;\omega\in{\mathcal{P}}_{r}\Lambda^{k}(T)\mathrel{}\middle|\mathrel{}\forall
F\subsetneq T:\operatorname{tr}_{T,F}\omega=0\;\right\\},$ (20)
$\displaystyle\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)$
$\displaystyle:=\left\\{\;\omega\in{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)\mathrel{}\middle|\mathrel{}\forall
F\subsetneq T:\operatorname{tr}_{T,F}\omega=0\;\right\\}.$

It is evident that these spaces are nested, as follows from definitions and
Lemma 3.2,

$\displaystyle{\mathcal{P}}_{r}\Lambda^{k}(T)\subseteq{\mathcal{P}}_{r+1}^{-}\Lambda^{k}(T)\subseteq{\mathcal{P}}_{r+1}\Lambda^{k}(T),$
$\displaystyle\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)\subseteq\mathring{{\mathcal{P}}}_{r+1}^{-}\Lambda^{k}(T)\subseteq\mathring{{\mathcal{P}}}_{r+1}\Lambda^{k}(T),$

and that they are closed under taking traces: if $F\subseteq T$ is a
subsimplex, then

$\displaystyle\operatorname{tr}_{T,F}{\mathcal{P}}_{r}\Lambda^{k}(T)={\mathcal{P}}_{r}\Lambda^{k}(F),\quad\operatorname{tr}_{T,F}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)={\mathcal{P}}_{r}^{-}\Lambda^{k}(F).$

We remark that our definitions (17) and (18) are different from but equivalent
to the corresponding definitions in [2], as is easily checked.

###  4.1. Basis construction for ${\mathcal{P}}_{r}\Lambda^{k}(T)$ and
$\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$

In this subsection we study spanning sets and bases for the spaces
${\mathcal{P}}_{r}\Lambda^{k}(T)$ and
$\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$. We introduce the sets of
barycentric differential forms

(21) $\displaystyle{\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(T)$
$\displaystyle:=\left\\{\;\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}\mathrel{}\middle|\mathrel{}\alpha\in
A(r,n),\;\sigma\in\Sigma(k,n)\;\right\\},$ (24)
$\displaystyle{\mathcal{S}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$
$\displaystyle:=\left\\{\;\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}\mathrel{}\middle|\mathrel{}\begin{array}[]{l}\alpha\in
A(r,n),\;\sigma\in\Sigma(k,n),\\\
{}[\alpha]\cup[\sigma]=[0:n]\end{array}\;\right\\}.$

Furthermore, under the restriction that $r\geq 1$, we consider the sets of
barycentric differential forms

(25) $\displaystyle{\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)$
$\displaystyle:=\left\\{\;\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}\mathrel{}\middle|\mathrel{}\alpha\in
A(r,n),\;\sigma\in\Sigma(k,n),\;\lfloor\alpha\rfloor\notin[\sigma]\;\right\\},$
(28) $\displaystyle{\mathcal{B}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$
$\displaystyle:=\left\\{\;\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}\mathrel{}\middle|\mathrel{}\begin{array}[]{l}\alpha\in
A(r,n),\;\sigma\in\Sigma(k,n),\\\
\lfloor\alpha\rfloor\notin[\sigma],\;[\alpha]\cup[\sigma]=[0:n]\end{array}\;\right\\}.$

We call ${\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(T)$ the _canonical spanning
set_ of ${\mathcal{P}}_{r}\Lambda^{k}(T)$, and we call
${\mathcal{S}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$ the _canonical
spanning set_ of $\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$; these names are
justified below. Evidently,

$\displaystyle{\mathcal{B}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)\subseteq{\mathcal{S}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T),\quad{\mathcal{S}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)\subseteq{\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(T),$
$\displaystyle{\mathcal{B}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)\subseteq{\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T),\quad{\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)\subseteq{\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(T).$

Suppose that $F\subseteq T$ is a subsimplex. From definitions it is clear that

$\displaystyle\operatorname{tr}_{T,F}{\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(T)={\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(F),\quad\operatorname{tr}_{T,F}{\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)={\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(F).$

In fact, the trace of any member of
${\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(T)$ onto $F$ is either zero or a
member of ${\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(F)$, and any member of
${\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(F)$ has exactly one preimage under
the trace in ${\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(T)$. If
$\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}\in{\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)$
with $[\alpha]\cup[\sigma]\subseteq[\imath(F,T)]$, then

$\displaystyle\operatorname{tr}_{T,F}\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}=\lambda_{F}^{\hat{\alpha}}{\mathsf{d}}\lambda^{F}_{\hat{\sigma}}\in{\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(F),$

where $\hat{\alpha}=\alpha\circ\imath(F,T)$ and
$\hat{\sigma}=\imath(F,T)^{\dagger}\circ\sigma$. In turn, if
$\lambda^{\alpha}_{F}{\mathsf{d}}\lambda^{F}_{\sigma}\in{\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(F)$,
then

$\displaystyle\lambda_{T}^{\tilde{\alpha}}{\mathsf{d}}\lambda^{T}_{\tilde{\sigma}}\in{\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T),\quad\operatorname{tr}_{T,F}\lambda_{T}^{\tilde{\alpha}}{\mathsf{d}}\lambda^{T}_{\tilde{\sigma}}=\lambda_{F}^{\alpha}{\mathsf{d}}\lambda^{F}_{\sigma},$

where $\tilde{\alpha}=\alpha\circ\imath(F,T)^{\dagger}$ over $[\imath(F,T)]$
and zero otherwise, and where $\tilde{\sigma}=\imath(F,T)\circ\sigma$.

We call ${\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(T)$ the canonical spanning
set of ${\mathcal{P}}_{r}\Lambda^{k}(T)$ because

$\displaystyle{\mathcal{P}}_{r}\Lambda^{k}(T)=\operatorname*{span}{\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(T)$

by definition. However, ${\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(T)$ is
generally not linearly independent, and thus does not form a basis. However,
its subset ${\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)$ does.

######  Lemma 4.1.

Let $r\geq 1$. The set ${\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)$ is a
basis of ${\mathcal{P}}_{r}\Lambda^{k}(T)$.

###### Proof.

The claim holds in the case $k=0$, so let us assume that $k>0$. First we show
that ${\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)$ spans
${\mathcal{P}}_{r}\Lambda^{k}(T)$. For any $\alpha\in A(r,n)$ and
$\sigma\in\Sigma(k,n)$ with $\lfloor\alpha\rfloor\in[\sigma]$ we find

$\displaystyle\lambda^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}$
$\displaystyle={\epsilon}({\lfloor\alpha\rfloor},\sigma-{\lfloor\alpha\rfloor})\lambda^{\alpha}{\mathsf{d}}\lambda^{T}_{\lfloor\alpha\rfloor}\wedge{\mathsf{d}}\lambda^{T}_{\sigma-{\lfloor\alpha\rfloor}}$
$\displaystyle=-{\epsilon}({\lfloor\alpha\rfloor},\sigma-{\lfloor\alpha\rfloor})\sum_{q\in[\sigma^{c}]}\lambda^{\alpha}{\mathsf{d}}\lambda^{T}_{q}\wedge{\mathsf{d}}\lambda^{T}_{\sigma-{\lfloor\alpha\rfloor}}$
$\displaystyle=-{\epsilon}({\lfloor\alpha\rfloor},\sigma-{\lfloor\alpha\rfloor})\sum_{q\in[\sigma^{c}]}{\epsilon}(q,\sigma-{\lfloor\alpha\rfloor})\lambda^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma-{\lfloor\alpha\rfloor}+q}.$

Hence ${\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)$ is a spanning set. It
remains to show that ${\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)$ is
linearly independent. Let $\omega\in{\mathcal{P}}_{r}\Lambda^{k}(T)$. Then
there exist coefficients $\omega_{\alpha\sigma}\in{\mathbb{C}}$ such that

$\displaystyle\omega=\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\end{subarray}}\sum_{\begin{subarray}{c}\sigma\in\Sigma(k,n)\\\
\lfloor\alpha\rfloor\notin[\sigma]\end{subarray}}\omega_{\alpha\sigma}\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}.$

Suppose that $\omega=0$ while not all coefficients vanish. Consider the
constant $k$-forms

$\displaystyle V_{\alpha}:=\sum_{\begin{subarray}{c}\sigma\in\Sigma(k,n)\\\
\lfloor\alpha\rfloor\notin[\sigma]\end{subarray}}\omega_{\alpha\sigma}{\mathsf{d}}\lambda^{T}_{\sigma},\quad\alpha\in
A(r,n).$

For each $\alpha\in A(r,n)$ we have $V_{\alpha}=0$ if and only if for all
$\sigma\in\Sigma(k,n)$ with $\lfloor\alpha\rfloor\notin[\sigma]$ we have
$\omega_{\alpha\sigma}=0$. Since we assume that not all coefficients vanish,
there exists $\alpha\in A(r,n)$ with $V_{\alpha}\neq 0$. Letting
${\mathcal{V}}_{\alpha}$ be the constant $k$-vector field dual to
$V_{\alpha}$,

$\displaystyle 0$ $\displaystyle=\omega({\mathcal{V}}_{\alpha})=\sum_{\beta\in
A(r,n)}\lambda_{T}^{\beta}V_{\beta}({\mathcal{V}}_{\alpha})=\lambda^{\alpha}_{T}+\sum_{\begin{subarray}{c}\beta\in
A(r,n)\\\
\beta\neq\alpha\end{subarray}}\lambda_{T}^{\beta}V_{\beta}({\mathcal{V}}_{\alpha}).$

But this contradicts the linear independence of the $\lambda_{T}^{\alpha}$.
Hence all coefficients must vanish. This shows linear independence, and thus
completes the proof. ∎

The following result shows that the subset
${\mathcal{B}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)\subseteq{\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)$
is a basis of subspace
$\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)\subseteq{\mathcal{P}}_{r}\Lambda^{k}(T)$.
Moreover, it justifies why we call
${\mathcal{S}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)\subseteq{\mathcal{S}}{\mathcal{P}}_{r}\Lambda^{k}(T)$
a canonical spanning set.

######  Theorem 4.2.

Let $r\geq 1$. The set
${\mathcal{B}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$ is a basis for
$\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$, and
${\mathcal{S}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$ is a spanning set
for that space.

###### Proof.

Let $\omega\in\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$. Then
$\omega\in{\mathcal{P}}_{r}\Lambda^{k}(T)$, and thus there exist unique
coefficients $\omega_{\alpha\sigma}\in{\mathbb{C}}$ such that

$\displaystyle\omega=\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\end{subarray}}\sum_{\begin{subarray}{c}\sigma\in\Sigma(k,n)\\\
\lfloor\alpha\rfloor\notin[\sigma]\end{subarray}}\omega_{\alpha\sigma}\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}.$

When $F$ is a lower-dimensional subsimplex of $T$, then
$0=\operatorname{tr}_{T,F}\omega$ leads to

$\displaystyle 0=\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
\lfloor\alpha\rfloor\notin[\sigma]\end{subarray}}\omega_{\alpha\sigma}\operatorname{tr}_{T,F}\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}=\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\ \sigma\in\Sigma(k,n)\\\ \lfloor\alpha\rfloor\notin[\sigma]\\\
[\alpha]\cup[\sigma]\subseteq[\imath(F,T)]\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha\circ\imath(F,T)}_{F}{\mathsf{d}}\lambda^{F}_{\imath(F,T)^{\dagger}\circ\sigma}.$

Since the last sum runs over linearly independent differential forms, we thus
find that $\omega_{\alpha\sigma}=0$ for all
$[\alpha]\cup[\sigma]\subseteq[\imath(F,T)]$. Since $F$ was assumed to be an
arbitrary proper subsimplex of $T$, we get that $\omega_{\alpha\sigma}=0$ when
$[\alpha]\cup[\sigma]\neq[0:n]$. So
${\mathcal{B}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$ is a spanning set of
$\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$. It is linearly independent,
being a subset of ${\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)$. Hence
${\mathcal{S}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$ is a spanning set,
as claimed. ∎

We define an extension operator that facilitates a geometric decomposition.
Whenever $F$ is a subsimplex of $T$, we consider the operator

$\displaystyle\operatorname{ext}^{r,k}_{F,T}:\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(F)\rightarrow{\mathcal{P}}_{r}\Lambda^{k}(T),$

which is defined by setting

$\displaystyle\operatorname{ext}^{r,k}_{F,T}\lambda_{F}^{\alpha}{\mathsf{d}}\lambda^{F}_{\sigma}=\lambda_{T}^{\tilde{\alpha}}{\mathsf{d}}\lambda^{T}_{\tilde{\sigma}},\quad\lambda_{F}^{\alpha}{\mathsf{d}}\lambda^{F}_{\sigma}\in{\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(F),$

where $\tilde{\alpha}=\alpha\circ\imath(F,T)^{\dagger}$ over $[\imath(F,T)]$
and zero otherwise, and where $\tilde{\sigma}=\imath(F,T)\circ\sigma$.

We see that whenever $f\subseteq F$ is a subsimplex of $F$, then

$\displaystyle\operatorname{tr}_{T,F}\operatorname{ext}^{r,k}_{f,T}=\operatorname{ext}^{r,k,-}_{f,F},$

and that whenever $G\subset T$ is a subsimplex of $T$ with $F\cap
G=\emptyset$, then

$\displaystyle\operatorname{tr}_{T,G}\operatorname{ext}^{r,k}_{F,T}=0.$

######  Remark 4.3.

We give a brief overview of the literature. Our basis
${\mathcal{B}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$ of
$\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$ appears in [2] together with the
same extension operators. Our basis
${\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)$ of
${\mathcal{P}}_{r}\Lambda^{k}(T)$, however, it is not explicitly described
there even though it emerges naturally with their tools.

We remark that ${\mathcal{B}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$ can
also be written as

$\displaystyle{\mathcal{B}}\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$
$\displaystyle=\left\\{\;\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}\mathrel{}\middle|\mathrel{}\begin{array}[]{l}\alpha\in
A(r,n),\;\sigma\in\Sigma(k,n),\\\
\lfloor\alpha\rfloor=\min([0:n]\setminus[\sigma]),\;[\alpha]\cup[\sigma]=[0:n]\end{array}\;\right\\}$
$\displaystyle=\left\\{\;\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}\mathrel{}\middle|\mathrel{}\begin{array}[]{l}\alpha\in
A(r,n),\;\sigma\in\Sigma(k,n),\\\
\lfloor\alpha\rfloor\geq\min([0:n]\setminus[\sigma]),\;[\alpha]\cup[\sigma]=[0:n]\end{array}\;\right\\}.$

To see this, suppose that $\alpha\in A(r,n)$ and $\sigma\in\Sigma(k,n)$ with
$[\alpha]\cup[\sigma]=[0:n]$. We have equivalence of
$\lfloor\alpha\rfloor\notin[\sigma]$ and
$\lfloor\alpha\rfloor\in[0:n]\setminus[\sigma]$. And by
$[\alpha]\cup[\sigma]=[0:n]$, we have
$\lfloor\alpha\rfloor\in[0:n]\setminus[\sigma]$ if and only if
$\lfloor\alpha\rfloor=\min([0:n]\setminus[\sigma])$ if and only if
$\lfloor\alpha\rfloor\geq\min([0:n]\setminus[\sigma])$. In particular, we
recover the basis description in Theorem 6.1 of [3]. The same basis of
$\mathring{{\mathcal{P}}}\Lambda^{k}(T)$ is used implicitly in Theorem 4.22 of
[2]. Furthermore, our extension operator is used in the seminal publication by
Arnold, Falk, and Winther [2, p.56]. It differs from the extension operator in
[3].

The basis ${\mathcal{B}}_{r}\Lambda^{k}(T)$ above allows for a geometric
decomposition and can be therefore be used in the construction of basis for an
entire finite element space. We outline another basis, which will be of
technical interest in the next subsection. Consider the set of barycentric
differential forms

(29) $\displaystyle{\mathcal{B}}_{0}{\mathcal{P}}_{r}\Lambda^{k}(T)$
$\displaystyle:=\left\\{\;\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}\mathrel{}\middle|\mathrel{}\alpha\in
A(r,n),\;\sigma\in\Sigma(k,n),\;\lfloor\sigma\rfloor>0\;\right\\}.$

This is a basis of ${\mathcal{B}}_{0}{\mathcal{P}}_{r}\Lambda^{k}(T)$, which
can be seen, e.g., via the transformation to a reference simplex. The basis
${\mathcal{B}}_{0}{\mathcal{P}}_{r}\Lambda^{0}(T)$ is easy to describe, but it
is not amenable for a geometric decomposition of the space
${\mathcal{P}}_{r}\Lambda^{k}(T)$. For example, restricting the elements of
${\mathcal{B}}_{0}{\mathcal{P}}_{r}\Lambda^{k}(T)$ to faces $F$ of $T$
produces the basis ${\mathcal{B}}_{0}{\mathcal{P}}_{r}\Lambda^{k}(F)$
generally only under the condition that $F$ contains the $0$-th vertex.

######  Remark 4.4.

Our basis ${\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)$ is generally
different from ${\mathcal{B}}_{0}{\mathcal{P}}_{r}\Lambda^{k}(T)$. However,
they adhere to the following same idea. If for each multiindex $\alpha\in
A(r,n)$ we pick an index $j_{\alpha}\in[0:n]$, then a basis of
${\mathcal{P}}_{r}\Lambda^{k}(T)$ is given by the set

$\displaystyle\left\\{\;\lambda_{T}^{\alpha}{\mathsf{d}}\lambda^{T}_{\sigma}\mathrel{}\middle|\mathrel{}\alpha\in
A(r,n),\;\sigma\in\Sigma(k,n),\;j_{\alpha}\notin[\sigma]\;\right\\}.$

In the case ${\mathcal{B}}_{0}{\mathcal{P}}_{r}\Lambda^{k}(T)$ one always
picks $j_{\alpha}=0$ for every $\alpha\in A(r,n)$. In the case
${\mathcal{B}}{\mathcal{P}}_{r}\Lambda^{k}(T)$ one always picks
$j_{\alpha}=\lfloor\alpha\rfloor$ for every $\alpha\in A(r,n)$. The basis
${\mathcal{B}}_{0}{\mathcal{P}}_{r}\Lambda^{k}(T)$ coincides with the basis
given in Theorem 6.1 of [3] and also used implicitly in Theorem 4.16 of [2].

###  4.2. Basis construction for ${\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$ and
$\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$

This subsection follows a similar paths as the previous one. We study spanning
sets and bases for the spaces ${\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$ and
$\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)$. We introduce the sets of
barycentric differential forms

(30) $\displaystyle{\mathcal{S}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$
$\displaystyle:=\left\\{\;\lambda_{T}^{\alpha}\phi^{T}_{\rho}\mathrel{}\middle|\mathrel{}\alpha\in
A(r-1,n),\;\rho\in\Sigma_{0}(k,n)\;\right\\},$ (33)
$\displaystyle{\mathcal{S}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$
$\displaystyle:=\left\\{\;\lambda_{T}^{\alpha}\phi_{\rho}^{T}\mathrel{}\middle|\mathrel{}\begin{array}[]{l}\alpha\in
A(r-1,n),\;\rho\in\Sigma_{0}(k,n),\\\
{}[\alpha]\cup[\rho]=[0:n]\end{array}\;\right\\}.$

Furthermore, under the restriction that $r\geq 1$, we consider the sets of
barycentric differential forms

(36) $\displaystyle{\mathcal{B}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$
$\displaystyle:=\left\\{\;\lambda_{T}^{\alpha}\phi_{\rho}^{T}\mathrel{}\middle|\mathrel{}\begin{array}[]{l}\alpha\in
A(r-1,n),\;\rho\in\Sigma_{0}(k,n),\\\
\lfloor\alpha\rfloor\geq\lfloor\rho\rfloor\end{array}\;\right\\},$ (39)
$\displaystyle{\mathcal{B}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$
$\displaystyle:=\left\\{\;\lambda_{T}^{\alpha}\phi_{\rho}^{T}\mathrel{}\middle|\mathrel{}\begin{array}[]{l}\alpha\in
A(r-1,n),\;\rho\in\Sigma_{0}(k,n),\\\
\lfloor\rho\rfloor=0,\;[\alpha]\cup[\rho]=[0:n]\end{array}\;\right\\}.$

We call ${\mathcal{S}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$ the _canonical
spanning set_ of ${\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$, and we call
${\mathcal{S}}\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)$ the _canonical
spanning set_ of $\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)$; again these
names will be justified shortly. It is evident that

$\displaystyle{\mathcal{B}}\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)\subseteq{\mathcal{S}}\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T),\quad{\mathcal{S}}\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)\subseteq{\mathcal{S}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T),$
$\displaystyle{\mathcal{B}}\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)\subseteq{\mathcal{B}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T),\quad{\mathcal{B}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)\subseteq{\mathcal{S}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T).$

Suppose that $F\subseteq T$ is a subsimplex. From definitions it is clear that

$\displaystyle\operatorname{tr}_{T,F}{\mathcal{S}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)={\mathcal{S}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(F),\quad\operatorname{tr}_{T,F}{\mathcal{B}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)={\mathcal{B}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(F).$

In fact, the trace of any member of
${\mathcal{S}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$ onto $F$ is either zero or
a member of ${\mathcal{S}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(F)$, and any member
of ${\mathcal{S}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(F)$ has exactly one preimage
under the trace in ${\mathcal{S}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$. If
$\lambda_{T}^{\alpha}\phi^{T}_{\rho}\in{\mathcal{B}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$
with $[\alpha]\cup[\rho]\subseteq[\imath(F,T)]$, then

$\displaystyle\operatorname{tr}_{T,F}\lambda_{T}^{\alpha}\phi^{T}_{\rho}=\lambda_{F}^{\hat{\alpha}}\phi^{F}_{\hat{\rho}}\in{\mathcal{B}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(F),$

where $\hat{\alpha}=\alpha\circ\imath(F,T)$ and
$\hat{\rho}=\imath(F,T)^{\dagger}\circ\rho$. In turn, if
$\lambda_{F}^{\alpha}\phi^{F}_{\rho}\in{\mathcal{B}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(F)$,
then

$\displaystyle\lambda_{T}^{\tilde{\alpha}}\phi^{T}_{\tilde{\rho}}\in{\mathcal{B}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(T),\quad\operatorname{tr}_{T,F}\lambda_{T}^{\tilde{\alpha}}\phi^{T}_{\tilde{\rho}}=\lambda^{\alpha}_{F}\phi^{F}_{\rho},$

where $\tilde{\alpha}=\alpha\circ\imath(F,T)^{\dagger}$ over $[\imath(F,T)]$
and zero otherwise, and where $\tilde{\rho}=\imath(F,T)\circ\rho$.

We call ${\mathcal{S}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$ the canonical
spanning set of ${\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$ because by definition it
is a spanning set for the higher-order Whitney forms,

$\displaystyle{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)=\operatorname*{span}{\mathcal{S}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T).$

However, ${\mathcal{S}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$ is generally not
linearly independent, and thus does not form a basis. Analogously to the
previous subsection, we show that its subset
${\mathcal{B}}{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$ is a basis, and we also
show that ${\mathcal{S}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ is a
spanning set and that
${\mathcal{B}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ is a basis of
$\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)$

######  Lemma 4.5.

The set ${\mathcal{B}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$ is a basis of
${\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$. The set
${\mathcal{B}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ is a basis of
$\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)$, and the set
${\mathcal{S}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ is a spanning
set for that space.

###### Proof.

We first show that ${\mathcal{B}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$ spans
${\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$. If $r=1$, then
${\mathcal{B}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(T)={\mathcal{S}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$,
so it remains to consider the case $r\geq 2$. Let $\alpha\in A(r-1,n)$ and
$\rho\in\Sigma_{0}(k,n)$, let $p:=\lfloor\alpha\rfloor$, and assume
$p<\lfloor\rho\rfloor$. There exists $\beta\in A(r-2,n)$ with
$\lambda_{T}^{\alpha}=\lambda_{T}^{\beta}\lambda^{T}_{p}$. Using Lemma 3.3, we
find that

$\displaystyle\lambda_{T}^{\alpha}\phi^{T}_{\rho}=\lambda_{T}^{\beta}\lambda^{T}_{p}\phi^{T}_{\rho}=\lambda_{T}^{\beta}\sum_{j=0}^{k}(-1)^{j}\lambda^{T}_{\rho(j)}\phi^{T}_{\rho+p-\rho(j)}.$

Hence all members of ${\mathcal{S}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$ are
linear combinations of members of
${\mathcal{B}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$.

Next we show that
${\mathcal{B}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ is linearly
independent. Let $\omega\in{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$ be in the span
of ${\mathcal{B}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$. Thus we can
write

$\displaystyle\omega=\sum_{\begin{subarray}{c}\alpha\in
A(r-1,n)\end{subarray}}\sum_{\begin{subarray}{c}\rho\in\Sigma_{0}(k,n)\\\
\lfloor\rho\rfloor=0\end{subarray}}\omega_{\alpha\rho}\lambda_{T}^{\alpha}\phi^{T}_{\rho}$

where $\omega_{\alpha\rho}\in{\mathbb{C}}$ for each $(\alpha,\rho)\in
A(r-1,n)\times\Sigma_{0}(k,n)$. Hence $\omega=\omega_{0}+\omega_{+}$, where

$\displaystyle\omega_{0}$ $\displaystyle:=\sum_{\begin{subarray}{c}\alpha\in
A(r-1,n)\end{subarray}}\sum_{\begin{subarray}{c}\rho\in\Sigma_{0}(k,n)\\\
\lfloor\rho\rfloor=0\end{subarray}}\omega_{\alpha\rho}\lambda_{T}^{\alpha}\lambda_{0}^{T}{\mathsf{d}}\lambda^{T}_{\rho-0},$
$\displaystyle\omega_{+}$ $\displaystyle:=\sum_{\begin{subarray}{c}\alpha\in
A(r-1,n)\end{subarray}}\sum_{\begin{subarray}{c}\rho\in\Sigma_{0}(k,n)\\\
\lfloor\rho\rfloor=0\end{subarray}}\sum_{\begin{subarray}{c}p\in[\rho]\\\
p\neq
0\end{subarray}}\omega_{\alpha\rho}{\epsilon}(p,\rho-p)\lambda_{T}^{\alpha}\lambda_{p}^{T}{\mathsf{d}}\lambda^{T}_{\rho-p}$
$\displaystyle=-\sum_{\begin{subarray}{c}\alpha\in
A(r-1,n)\end{subarray}}\sum_{\begin{subarray}{c}\rho\in\Sigma_{0}(k,n)\\\
\lfloor\rho\rfloor=0\end{subarray}}\sum_{\begin{subarray}{c}p\in[\rho]\\\
p\neq
0\end{subarray}}\sum_{i=1}^{n}\omega_{\alpha\rho}{\epsilon}(p,\rho-p)\lambda_{T}^{\alpha+p}{\mathsf{d}}\lambda_{i}^{T}\wedge{\mathsf{d}}\lambda^{T}_{\rho-p-0}$
$\displaystyle=\sum_{\begin{subarray}{c}\alpha\in
A(r-1,n)\end{subarray}}\sum_{\begin{subarray}{c}\rho\in\Sigma_{0}(k,n)\\\
\lfloor\rho\rfloor=0\end{subarray}}\sum_{\begin{subarray}{c}p\in[\rho]\\\
p\neq 0\end{subarray}}\sum_{\begin{subarray}{c}i\in[1:n]\\\
i\notin[\rho-p]\end{subarray}}\omega_{\alpha\rho}{\epsilon}(p,\rho-p){\epsilon}(i,\rho-p)\lambda_{T}^{\alpha+p}{\mathsf{d}}\lambda^{T}_{\rho-p-0+i}.$

These differential forms are expressed in terms of
${\mathcal{B}}_{0}{\mathcal{P}}_{r}\Lambda^{k}(T)$. Suppose $\omega=0$.

We use induction to prove that all $\omega_{\alpha\rho}$ vanish. First, it is
evident that $\omega_{\alpha\rho}=0$ for $\alpha(0)=r-1$. Now let us assume
that $s\in[1:r-1]$ such that $\omega_{\alpha\rho}=0$ for all
$\alpha(0)\in[s:r-1]$. Since the terms $\lambda_{T}^{\alpha}\lambda_{0}^{T}$
with $\alpha(s)=s-1$ in the definition of $\omega_{0}$ always have a higher
exponent in index $0$ than the terms $\lambda_{T}^{\alpha}\lambda_{p}^{T}$ in
the definition of $\omega_{+}$, we conclude that $\omega_{\alpha\rho}=0$ for
$\alpha(s)=s-1$. Repeating this argument yields $\omega_{\alpha\rho}=0$ for
all coefficients. Thus
${\mathcal{B}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ is linearly
independent.

It remains to show that
${\mathcal{B}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ spans
$\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ and that
${\mathcal{B}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$ is linearly independent. We
use induction over the dimension of $T$ for both claims.

First, the two claims hold if $\dim T=k$ because
${\mathcal{P}}^{-}_{r}\Lambda^{k}(T)={\mathcal{P}}_{r}\Lambda^{k}(T)$ and
${\mathcal{B}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)={\mathcal{B}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$
in that case.

Suppose that the two claims hold for simplices of dimension at most $m\geq k$
and that $\dim T=m+1$. Let $\omega\in{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$, so
there exist coefficients $\omega_{\alpha\rho}\in{\mathbb{C}}$ with

$\displaystyle\omega=\sum_{\begin{subarray}{c}\alpha\in
A(r-1,n)\end{subarray}}\sum_{\begin{subarray}{c}\rho\in\Sigma_{0}(k,n)\\\
\lfloor\alpha\rfloor\geq\lfloor\rho\rfloor\end{subarray}}\omega_{\alpha\rho}\lambda_{T}^{\alpha}\phi^{T}_{\rho}.$

We prove that if $\omega\in\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)$,
then $\omega_{\alpha\rho}=0$ for all $\alpha\in A(r-1,n)$ and
$\sigma\in\Sigma_{0}(k,n)$ with $[\alpha]\cup[\rho]=[0:n]$. Let us assume that
$\omega\in\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)$, and let $F$ be any
proper face of $T$. Then $0=\operatorname{tr}_{T,F}\omega$ leads to

$\displaystyle 0=\sum_{\begin{subarray}{c}\alpha\in A(r-1,n)\\\
\rho\in\Sigma_{0}(k,n)\\\
\lfloor\alpha\rfloor\geq\lfloor\rho\rfloor\end{subarray}}\omega_{\alpha\rho}\operatorname{tr}_{T,F}\lambda_{T}^{\alpha}\phi^{T}_{\rho}=\sum_{\begin{subarray}{c}\alpha\in
A(r-1,n)\\\ \rho\in\Sigma_{0}(k,n)\\\
\lfloor\alpha\rfloor\geq\lfloor\rho\rfloor\\\
[\alpha]\cup[\rho]\subseteq[\imath(F,T)]\end{subarray}}\omega_{\alpha\rho}\lambda^{\alpha\circ\imath(F,T)}_{F}\phi^{F}_{\imath(F,T)^{\dagger}\circ\rho}.$

By the induction assumption, this expresses $0=\operatorname{tr}_{T,F}\omega$
in terms of basis of ${\mathcal{P}}_{r}^{-}\Lambda^{k}(F)$. Hence
$\omega_{\alpha\rho}=0$ when $[\alpha]\cup[\rho]\subseteq[\imath(F,T)]$. Since
$F$ was assumed to be an abitrary proper face of $T$, we get that
$\omega_{\alpha\rho}=0$ when $[\alpha]\cup[\rho]\neq[0:n]$. So
${\mathcal{B}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ spans
$\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)$. Thus
${\mathcal{B}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ is a basis of
$\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)$, and
${\mathcal{S}}\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ is a spanning
set. Since $\omega=0$ implies
$\omega\in\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)$, we now also see
that ${\mathcal{B}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$ is linearly
independent and thus a basis of ${\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$. This
completes the induction step, and the desired claim follows. ∎

Similar as before, we can define an extension operator that facilitates a
geometric decomposition. Whenever $F$ is a subsimplex of $T$, we consider the
operator

$\displaystyle\operatorname{ext}^{r,k,-}_{F,T}:\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(F)\rightarrow{\mathcal{P}}^{-}_{r}\Lambda^{k}(T),$

which is defined by setting

$\displaystyle\operatorname{ext}^{r,k,-}_{F,T}\lambda_{F}^{\alpha}\phi^{F}_{\rho}=\lambda_{T}^{\tilde{\alpha}}\phi^{T}_{\tilde{\rho}},\quad\lambda_{F}^{\alpha}\phi^{F}_{\rho}\in{\mathcal{B}}{\mathcal{P}}^{-}_{r}\Lambda^{k}(F),$

where $\tilde{\alpha}=\alpha\circ\imath(F,T)^{\dagger}$ over $[\imath(F,T)]$
and zero otherwise, and where $\tilde{\rho}=\imath(F,T)\circ\rho$.

Similar as before, we note that whenever $f\subseteq F$ is a subsimplex of
$F$, then

$\displaystyle\operatorname{tr}_{T,F}\operatorname{ext}^{r,k,-}_{f,T}=\operatorname{ext}^{r,k,-}_{f,F},$

and that whenever $G\subset T$ is a subsimplex of $T$ with $F\cap
G=\emptyset$, then

$\displaystyle\operatorname{tr}_{T,G}\operatorname{ext}^{r,k,-}_{F,T}=0.$

######  Remark 4.6.

The bases for ${\mathcal{P}}^{-}_{r}\Lambda^{k}(T)$ and
$\mathring{{\mathcal{P}}}^{-}_{r}\Lambda^{k}(T)$ are identical to the bases
presented or implied in Section 4 of [2] (see Theorems 4.4 and 4.16 there) or
in [3], which are all the same. Our extension operator coincides with the
extension operator for the higher-order Whitney forms in [2].

##  5\. Linear dependencies

We have previously encountered canonical spanning sets for the spaces of
polynomial differential forms over a simplex $T$. The goal of this section is
to improve our understanding of the linear dependencies of those spanning
sets. As a byproduct, we improve our understanding of the isomorphisms

$\displaystyle{\mathcal{P}}_{r}\Lambda^{k}(T)\simeq\mathring{{\mathcal{P}}}_{r+k}^{-}\Lambda^{n-k}(T),\quad{\mathcal{P}}_{r+1}^{-}\Lambda^{n-k}(T)\simeq\mathring{{\mathcal{P}}}_{r+n-k+1}\Lambda^{k}(T),$

between the finite element spaces, which have been used earlier in [2].

######  Lemma 5.1.

Let $r\in{\mathbb{N}}_{0}$ and $k\in[0:n]$. Let
$\omega_{\alpha\sigma}\in{\mathbb{C}}$ for $\sigma\in\Sigma(k,n)$ and
$\alpha\in A(r,n)$. Then

(40) $\displaystyle\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}{\mathsf{d}}\lambda_{\sigma}=0\quad\Longleftrightarrow\quad\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}{\epsilon}(\sigma,\sigma^{c})\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma}\phi_{\sigma^{c}}=0,$

each of which is the case if and only if

(41)
$\displaystyle\omega_{\alpha\sigma}-\sum_{p\in[\sigma]}{\epsilon}(p,\sigma-p)\omega_{\alpha,\sigma-p+0}=0$

holds for $\alpha\in A(r,n)$ and $\sigma\in\Sigma(k,n)$ with
$0\notin[\sigma]$.

###### Proof.

The statement is trivial if $k=0$, so assume that $1\leq k\leq n$. Define

$\displaystyle S_{L}$ $\displaystyle:=\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}{\mathsf{d}}\lambda_{\sigma}=\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\ \sigma\in\Sigma(k,n)\\\
0\notin[\sigma]\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}{\mathsf{d}}\lambda_{\sigma}+\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\ \sigma\in\Sigma(k,n)\\\
0\in[\sigma]\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}{\mathsf{d}}\lambda_{\sigma}.$

For $\sigma\in\Sigma(k,n)$ with $0\in[\sigma]$ we observe

$\displaystyle{\mathsf{d}}\lambda_{\sigma}={\mathsf{d}}\lambda_{0}\wedge{\mathsf{d}}\lambda_{\sigma-0}$
$\displaystyle=-\sum_{q\in[\sigma^{c}]}{\mathsf{d}}\lambda_{q}\wedge{\mathsf{d}}\lambda_{\sigma-0}=\sum_{q\in[\sigma^{c}]}{\epsilon}(q,\sigma){\mathsf{d}}\lambda_{\sigma-0+q},$

Direct application of this observation gives

$\displaystyle S_{L}$ $\displaystyle=\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\ \sigma\in\Sigma(k,n)\\\
0\notin[\sigma]\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}{\mathsf{d}}\lambda_{\sigma}+\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\ \sigma\in\Sigma(k,n)\\\
0\in[\sigma]\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}\sum_{q\in[\sigma^{c}]}{\epsilon}(q,\sigma){\mathsf{d}}\lambda_{\sigma-0+q}$
$\displaystyle=\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
0\notin[\sigma]\end{subarray}}\left(\omega_{\alpha\sigma}+\sum_{p\in[\sigma]}{\epsilon}(p,\sigma-p+0)\omega_{\alpha,\sigma-p+0}\right)\lambda^{\alpha}{\mathsf{d}}\lambda_{\sigma}$
$\displaystyle=\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
0\notin[\sigma]\end{subarray}}\left(\omega_{\alpha\sigma}-\sum_{p\in[\sigma]}{\epsilon}(p,\sigma-p)\omega_{\alpha,\sigma-p+0}\right)\lambda^{\alpha}{\mathsf{d}}\lambda_{\sigma}.$

This is an expression in a basis of ${\mathcal{P}}_{r}\Lambda^{k}(T)$. On the
other hand, define $S_{R}$ by

$\displaystyle S_{R}$ $\displaystyle:=\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}{\epsilon}(\sigma,\sigma^{c})\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma}\phi_{\sigma^{c}}$
$\displaystyle=\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
0\notin[\sigma]\end{subarray}}{\epsilon}(\sigma,\sigma^{c})\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma}\phi_{\sigma^{c}}+\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\ \sigma\in\Sigma(k,n)\\\
0\in[\sigma]\end{subarray}}{\epsilon}(\sigma,\sigma^{c})\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma}\phi_{\sigma^{c}}.$

Using Lemma 3.3, for $\sigma\in\Sigma(k,n)$ with $0\in[\sigma]$ we observe

$\displaystyle\lambda_{\sigma}\phi_{\sigma^{c}}=\lambda_{\sigma-0}\lambda_{0}\phi_{\sigma^{c}}=\lambda_{\sigma-0}\sum_{q\in[\sigma^{c}]}{\epsilon}(q,\sigma)\lambda_{q}\phi_{\sigma^{c}-q+0}.$

Using previous observations, we calculate that $S_{R}$ equals

$\displaystyle\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
0\notin[\sigma]\end{subarray}}\left({\epsilon}(\sigma,\sigma^{c})\omega_{\alpha\sigma}+\sum_{p\in[\sigma]}{\epsilon}(\sigma-p+0,\sigma^{c}+p-0){\epsilon}(p,\sigma^{c}-0)\omega_{\alpha,\sigma-p+0}\right)\lambda^{\alpha}\lambda_{\sigma}\phi_{\sigma^{c}}.$

This is an expression in terms of a basis of
$\mathring{{\mathcal{P}}}_{r+k}^{-}\Lambda^{k}(T)$. Note that

$\displaystyle{\epsilon}(\sigma-p+0,\sigma^{c}+p-0){\epsilon}(p,\sigma^{c}-0)$
$\displaystyle=(-1)^{k+1}{\epsilon}(\sigma-p,\sigma^{c}+p){\epsilon}(p,\sigma^{c})$
$\displaystyle=-{\epsilon}(\sigma,\sigma^{c}){\epsilon}(p,\sigma-p)$

for $\sigma\in\Sigma(k,n)$, $p\in[\sigma]$ and $0\notin[\sigma]$. Thus
$S_{L}=0$ if and only if $S_{R}=0$, which is the case if and only if (41)
holds. This completes the proof. ∎

######  Lemma 5.2.

Let $r\in{\mathbb{N}}_{0}$ and $k\in[0:n]$. Let
$\omega_{\alpha\sigma}\in{\mathbb{C}}$ for $\sigma\in\Sigma(k,n)$ and
$\alpha\in A(r,n)$. Then

(42) $\displaystyle\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}{\epsilon}(\sigma,\sigma^{c})\omega_{\alpha\sigma}\lambda^{\alpha}\phi_{\sigma^{c}}=0\quad\Longleftrightarrow\quad\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma^{c}}{\mathsf{d}}\lambda_{\sigma}=0,$

each of which is the case if and only if

(43)
$\displaystyle\omega_{\alpha\sigma}-\sum_{\begin{subarray}{c}q\in[\sigma]\cap[\alpha]\end{subarray}}{\epsilon}(\lfloor\sigma^{c}\rfloor,\sigma-q){\epsilon}(q,\sigma-q)\omega_{\alpha+\lfloor\sigma^{c}\rfloor-q,\sigma+\lfloor\sigma^{c}\rfloor-q}=0$

holds for $\alpha\in A(r,n)$ and $\sigma\in\Sigma(k,n)$ with
$\lfloor\alpha\rfloor\geq\lfloor\sigma^{c}\rfloor$.

###### Proof.

If $r=0$, then the two sums in (42) are already stated in terms of bases of
${\mathcal{P}}_{r+1}^{-}\Lambda^{n-k}(T)$ and
$\mathring{{\mathcal{P}}}_{r+n-k+1}\Lambda^{k}(T)$, and (43) just reduces to
all coefficients vanishing. So it remains to study the case $r\geq 1$. In the
special case $k=0$, the statement is trivial. So let us assume $k>0$. We
define $S_{L}$ by setting

$\displaystyle S_{L}:=\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}{\epsilon}(\sigma,\sigma^{c})\omega_{\alpha\sigma}\lambda^{\alpha}\phi_{\sigma^{c}}.$

Using Lemma 3.3, for each $\sigma\in\Sigma(k,n)$ and $\alpha\in A(r,n)$ with
$\lfloor\alpha\rfloor<\lfloor\sigma^{c}\rfloor$ we have

$\displaystyle\lambda^{\alpha}\phi_{\sigma^{c}}=\lambda^{\alpha-\lfloor\alpha\rfloor}\lambda_{\lfloor\alpha\rfloor}\phi_{\sigma^{c}}=\sum_{\begin{subarray}{c}q\in[\sigma^{c}]\end{subarray}}{\epsilon}(q,\sigma^{c}-q)\lambda^{\alpha-\lfloor\alpha\rfloor+q}\phi_{\sigma^{c}+\lfloor\alpha\rfloor-q}.$

Therefore we can rewrite $S_{L}$ as

$\displaystyle\quad\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
\lfloor\alpha\rfloor\geq\lfloor\sigma^{c}\rfloor\end{subarray}}{\epsilon}(\sigma,\sigma^{c})\omega_{\alpha\sigma}\lambda^{\alpha}\phi_{\sigma^{c}}+\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\ \sigma\in\Sigma(k,n)\\\
\lfloor\alpha\rfloor<\lfloor\sigma^{c}\rfloor\end{subarray}}{\epsilon}(\sigma,\sigma^{c})\omega_{\alpha\sigma}\lambda^{\alpha}\phi_{\sigma^{c}}$
$\displaystyle=\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
\lfloor\alpha\rfloor\geq\lfloor\sigma^{c}\rfloor\end{subarray}}{\epsilon}(\sigma,\sigma^{c})\omega_{\alpha\sigma}\lambda^{\alpha}\phi_{\sigma^{c}}+\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\ \sigma\in\Sigma(k,n)\\\
\lfloor\alpha\rfloor<\lfloor\sigma^{c}\rfloor\\\
q\in[\sigma^{c}]\end{subarray}}{\epsilon}(\sigma,\sigma^{c}){\epsilon}(q,\sigma^{c}-q)\omega_{\alpha\sigma}\lambda^{\alpha-\lfloor\alpha\rfloor+q}\phi_{\sigma^{c}+\lfloor\alpha\rfloor-q}.$

Let $\sigma\in\Sigma(k,n)$, $\alpha\in A(r,n)$ and $q\in[\sigma^{c}]$ with
$\lfloor\alpha\rfloor<\lfloor\sigma^{c}\rfloor$. We set
$\beta=\alpha-\lfloor\alpha\rfloor+q$ and
$\rho=\sigma-\lfloor\alpha\rfloor+q$. Then
$\lfloor\beta\rfloor\geq\lfloor\alpha\rfloor=\lfloor\rho^{c}\rfloor$, thus
$\beta+\lfloor\rho^{c}\rfloor-q=\alpha$ and
$\rho+\lfloor\rho^{c}\rfloor-q=\sigma$. Hence $q\in[\rho]$ and $q\in[\beta]$.
Based on these observations,

$\displaystyle\quad\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
\lfloor\alpha\rfloor<\lfloor\sigma^{c}\rfloor\end{subarray}}\sum_{\begin{subarray}{c}q\in[\sigma^{c}]\end{subarray}}{\epsilon}(\sigma,\sigma^{c}){\epsilon}(q,\sigma^{c}-q)\omega_{\alpha\sigma}\lambda^{\alpha-\lfloor\alpha\rfloor+q}\phi_{\sigma^{c}+\lfloor\alpha\rfloor-q}$
$\displaystyle=\sum_{\begin{subarray}{c}\beta\in A(r,n)\\\
\rho\in\Sigma(k,n)\\\ \lfloor\beta\rfloor\geq\lfloor\rho^{c}\rfloor\\\
q\in[\rho]\cap[\beta]\end{subarray}}{\epsilon}(\rho+\lfloor\rho^{c}\rfloor-q,\rho^{c}-\lfloor\rho^{c}\rfloor+q){\epsilon}(q,\rho^{c}-\lfloor\rho^{c}\rfloor)\omega_{\beta+\lfloor\rho^{c}\rfloor-q,\rho+\lfloor\rho^{c}\rfloor-q}\lambda^{\beta}\phi_{\rho^{c}}.$

For $\rho\in\Sigma(k,n)$, $\beta\in A(r,n)$ and $q\in[\rho]\cap[\beta]$ such
that $\lfloor\beta\rfloor\geq\lfloor\rho^{c}\rfloor$, we make the
combinatorial observation

$\displaystyle{\epsilon}(\rho+\lfloor\rho^{c}\rfloor-q,\rho^{c}-\lfloor\rho^{c}\rfloor+q){\epsilon}(q,\rho^{c}-\lfloor\rho^{c}\rfloor)$
$\displaystyle\quad=-{\epsilon}(\rho,\rho^{c}){\epsilon}(\rho-q,q){\epsilon}(\lfloor\rho^{c}\rfloor,\rho^{c}-\lfloor\rho^{c}\rfloor){\epsilon}(\rho-q,\lfloor\rho^{c}\rfloor)$
$\displaystyle\quad=-{\epsilon}(\rho,\rho^{c}){\epsilon}(\rho-q,q){\epsilon}(\rho-q,\lfloor\rho^{c}\rfloor).$

It can thus be seen that $S_{L}$ equals

$\displaystyle\sum_{\begin{subarray}{c}\beta\in A(r,n)\\\
\rho\in\Sigma(k,n)\\\
\lfloor\beta\rfloor\geq\lfloor\rho^{c}\rfloor\end{subarray}}{\epsilon}(\rho,\rho^{c})\left(\omega_{\beta\rho}-\sum_{\begin{subarray}{c}q\in[\rho]\cap[\beta]\end{subarray}}{\epsilon}(\rho-q,q){\epsilon}(\rho-q,\lfloor\rho^{c}\rfloor)\omega_{\beta+\lfloor\rho^{c}\rfloor-q,\rho+\lfloor\rho^{c}\rfloor-q}\right)\lambda^{\beta}\phi_{\rho^{c}}.$

This an expression in terms of a basis of
${\mathcal{P}}_{r+1}^{-}\Lambda^{n-k}(T)$. Now, define $S_{R}$ by

$\displaystyle S_{R}:=\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma^{c}}{\mathsf{d}}\lambda_{\sigma}.$

For any $\alpha\in A(r,n)$ and $\sigma\in\Sigma(k,n)$ with
$\lfloor\alpha\rfloor<\lfloor\sigma^{c}\rfloor$ we see

$\displaystyle\lambda^{\alpha}\lambda_{\sigma^{c}}{\mathsf{d}}\lambda_{\sigma}$
$\displaystyle=\lambda^{\alpha}\lambda_{\sigma^{c}}{\epsilon}(\lfloor\alpha\rfloor,\sigma-\lfloor\alpha\rfloor){\mathsf{d}}\lambda_{\lfloor\alpha\rfloor}\wedge{\mathsf{d}}\lambda_{\sigma-\lfloor\alpha\rfloor}$
$\displaystyle=\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma^{c}}{\epsilon}(\lfloor\alpha\rfloor,\sigma-\lfloor\alpha\rfloor){\epsilon}(q,\sigma-\lfloor\alpha\rfloor){\mathsf{d}}\lambda_{\sigma-\lfloor\alpha\rfloor+q}.$

Hence

$\displaystyle\quad-\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
\lfloor\alpha\rfloor<\lfloor\sigma^{c}\rfloor\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma^{c}}{\mathsf{d}}\lambda_{\sigma}$
$\displaystyle=\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
\lfloor\alpha\rfloor<\lfloor\sigma^{c}\rfloor\end{subarray}}\sum_{\begin{subarray}{c}q\in[\sigma^{c}]\\\
q\neq\lfloor\alpha\rfloor\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma^{c}}{\epsilon}(\lfloor\alpha\rfloor,\sigma-\lfloor\alpha\rfloor){\epsilon}(q,\sigma-\lfloor\alpha\rfloor){\mathsf{d}}\lambda_{\sigma-\lfloor\alpha\rfloor+q}.$

Arguing similarly as above, we see that the last expression is identical to

$\displaystyle\sum_{\begin{subarray}{c}\beta\in A(r,n)\\\
\rho\in\Sigma(k,n)\\\ \lfloor\beta\rfloor\geq\lfloor\rho^{c}\rfloor\\\
q\in[\rho]\cap[\beta]\end{subarray}}\omega_{\beta+\lfloor\rho^{c}\rfloor-q,\rho+\lfloor\rho^{c}\rfloor-q}\lambda^{\beta+\lfloor\rho^{c}\rfloor-q}\lambda_{\rho^{c}-\lfloor\rho^{c}\rfloor+q}{\epsilon}(\lfloor\rho^{c}\rfloor,\rho-q){\epsilon}(q,\rho-q){\mathsf{d}}\lambda_{\rho}.$

Note that we can simplify
$\lambda^{\beta+\lfloor\rho^{c}\rfloor-q}\lambda_{\rho^{c}-\lfloor\rho^{c}\rfloor+q}=\lambda^{\beta}\lambda_{\rho^{c}}$
for each $\beta\in A(r,n)$, $\rho\in\Sigma(k,n)$, and $q\in[\rho]\cap[\beta]$
with $\lfloor\beta\rfloor\geq\lfloor\rho^{c}\rfloor$ in this sum.
Consequently, we see that $S_{R}$ equals

$\displaystyle\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
\lfloor\alpha\rfloor\geq\lfloor\sigma^{c}\rfloor\end{subarray}}\left(\omega_{\alpha\sigma}-\sum_{\begin{subarray}{c}q\in[\sigma]\cap[\alpha]\end{subarray}}{\epsilon}(\lfloor\sigma^{c}\rfloor,\sigma-q){\epsilon}(q,\sigma-q)\omega_{\alpha+\lfloor\sigma^{c}\rfloor-q,\sigma+\lfloor\sigma^{c}\rfloor-q}\right)\lambda^{\alpha}\lambda_{\sigma^{c}}{\mathsf{d}}\lambda_{\sigma}.$

This is an expression in terms of a basis of
$\mathring{{\mathcal{P}}}_{r+n-k+1}\Lambda^{k}(T)$. Thus $S_{L}=0$ if and only
if $S_{R}=0$, which is the case if and only if (43) holds. The proof is
complete. ∎

The point of these results is that we have a correspondence between the linear
dependencies of the canonical spanning sets of
${\mathcal{P}}_{r}\Lambda^{k}(T)$ and
$\mathring{{\mathcal{P}}}^{-}_{r+k+1}\Lambda^{n-k}(T)$, and a correspondence
between the linear dependencies of the canonical spanning sets of
$\mathring{{\mathcal{P}}}_{r+n-k+1}\Lambda^{k}(T)$ and
${\mathcal{P}}^{-}_{r+1}\Lambda^{n-k}(T)$.

An immediate application is the well-definedness of the following
isomorphisms. There exists a linear isomorphism from
${\mathcal{P}}_{r}\Lambda^{k}(T)$ to
$\mathring{{\mathcal{P}}}^{-}_{r+k+1}\Lambda^{n-k}(T)$ that in terms of
coefficients can be written as

(44) $\displaystyle\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}{\mathsf{d}}\lambda_{\sigma}\mapsto\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma}\phi_{\sigma^{c}}$

and we have a linear isomorphism from ${\mathcal{P}}_{r+1}\Lambda^{n-k}(T)$ to
$\mathring{{\mathcal{P}}}_{r+n-k+1}\Lambda^{k}(T)$, that in terms of
coefficients can be written as

(45) $\displaystyle\sum_{\begin{subarray}{c}\alpha\in A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}\phi_{\sigma^{c}}\mapsto\sum_{\begin{subarray}{c}\alpha\in
A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma^{c}}{\mathsf{d}}\lambda_{\sigma}$

That these mappings are indeed well-defined follows immediately from Lemma 5.1
and Lemma 5.2. We refer to Remark 5.5 below an example.

We give two more auxiliary results, Lemma 5.3 and Lemma 5.4, which are stated
and proven below. They give conditions on the coefficients that are equivalent
to the ones encountered in the previous two lemmas, but which seem more
“natural” than the latter. This not only rounds up the theory, but will also
be instrumental in the next section.

######  Lemma 5.3.

Let $r\in{\mathbb{N}}_{0}$ and $k\in[0:n]$. Let $\omega_{\alpha\sigma}$ be a
family of complex numbers indexed over $\alpha\in A(r,n)$ and
$\sigma\in\Sigma(k,n)$. We have that

(46)
$\displaystyle\omega_{\alpha\sigma}-\sum_{p\in[\sigma]}{\epsilon}(p,\sigma-p)\omega_{\alpha,\sigma-p+0}=0$

holds for all $\alpha\in A(r,n)$ and $\sigma\in\Sigma(k,n)$ with
$0\notin[\sigma]$ if and only

(47)
$\displaystyle\sum_{p\in[\theta]}{\epsilon}(p,\theta-p)\omega_{\alpha,\theta-p}=0$

holds for all $\alpha\in A(r,n)$ and $\theta\in\Sigma(k+1,n)$.

###### Proof.

The lemma is trivial in the special case $k=0$. So let us assume that $1\leq
k\leq n$. Clearly, the second claim implies the first. So let us suppose the
first claim holds. Then the second claim holds for all $\theta$ with
$0\in[\theta]$. If instead $0\notin[\theta]$, then we find

$\displaystyle\sum_{p\in[\theta]}{\epsilon}(p,\theta-p)\omega_{\alpha,\theta-p}$
$\displaystyle=\sum_{p\in[\theta]}\sum_{s\in[\theta-p]}{\epsilon}(p,\theta-p){\epsilon}(s,\theta-
p-s)\omega_{\alpha,\theta-p-s+0}$
$\displaystyle=\sum_{p\in[\theta]}\sum_{s\in[\theta-p]}{\epsilon}(p,s){\epsilon}(p,\theta-p){\epsilon}(s,\theta-
p-s)\omega_{\alpha,\theta-p-s+0}.$

This sum vanishes as follows by antisymmetry. The lemma follows. ∎

######  Lemma 5.4.

Let $r\in{\mathbb{N}}_{0}$ and $k\in[0:n]$. Let $\omega_{\alpha\sigma}$ be a
family of complex numbers indexed over $\alpha\in A(r,n)$ and
$\sigma\in\Sigma(k,n)$. We have that

$\displaystyle\omega_{\alpha\sigma}-\sum_{\begin{subarray}{c}q\in[\sigma]\cap[\alpha]\end{subarray}}{\epsilon}(\lfloor\sigma^{c}\rfloor,\sigma-q){\epsilon}(q,\sigma-q)\omega_{\alpha+\lfloor\sigma^{c}\rfloor-q,\sigma+\lfloor\sigma^{c}\rfloor-q}=0$

holds for all $\alpha\in A(r,n)$ and $\sigma\in\Sigma(k,n)$ with
$\lfloor\alpha\rfloor\geq\lfloor\sigma\rfloor$ if and only

$\displaystyle\sum_{p\in[\theta]\cap[\beta]}{\epsilon}(\theta-p,p)\omega_{\beta-p,\theta-p}=0$

holds for all $\beta\in A(r+1,n)$ and $\theta\in\Sigma(k+1,n)$.

###### Proof.

The lemma is trivial in the special case $k=0$. So let us assume that $1\leq
k\leq n$. The first condition has several equivalent formulations:

$\displaystyle\quad\quad\quad\omega_{\alpha\sigma}-\sum_{\begin{subarray}{c}q\in[\sigma]\cap[\alpha]\end{subarray}}{\epsilon}(\lfloor\sigma^{c}\rfloor,\sigma-q){\epsilon}(q,\sigma-q)\omega_{\alpha+\lfloor\sigma^{c}\rfloor-q,\sigma+\lfloor\sigma^{c}\rfloor-q}=0$
$\displaystyle\Longleftrightarrow\quad\omega_{\alpha\sigma}-\sum_{\begin{subarray}{c}q\in[\sigma]\cap[\alpha]\end{subarray}}{\epsilon}(\lfloor\sigma^{c}\rfloor,q){\epsilon}(\lfloor\sigma^{c}\rfloor,\sigma){\epsilon}(q,\sigma-q)\omega_{\alpha+\lfloor\sigma^{c}\rfloor-q,\sigma+\lfloor\sigma^{c}\rfloor-q}=0$
$\displaystyle\Longleftrightarrow\quad{\epsilon}(\lfloor\sigma^{c}\rfloor,\sigma)\omega_{\alpha\sigma}+\sum_{\begin{subarray}{c}q\in[\sigma]\cap[\alpha]\end{subarray}}{\epsilon}(q,\lfloor\sigma^{c}\rfloor){\epsilon}(q,\sigma-q)\omega_{\alpha+\lfloor\sigma^{c}\rfloor-q,\sigma+\lfloor\sigma^{c}\rfloor-q}=0$
$\displaystyle\Longleftrightarrow\quad{\epsilon}(\lfloor\sigma^{c}\rfloor,\sigma)\omega_{\alpha\sigma}+\sum_{\begin{subarray}{c}q\in[\sigma]\cap[\alpha]\end{subarray}}{\epsilon}(q,\sigma+\lfloor\sigma^{c}\rfloor-q)\omega_{\alpha+\lfloor\sigma^{c}\rfloor-q,\sigma+\lfloor\sigma^{c}\rfloor-q}=0$
$\displaystyle\Longleftrightarrow\quad\sum_{\begin{subarray}{c}q\in[\sigma+\lfloor\sigma^{c}\rfloor]\cap[\alpha+\lfloor\sigma^{c}\rfloor]\end{subarray}}{\epsilon}(q,\sigma+\lfloor\sigma^{c}\rfloor-q)\omega_{\alpha+\lfloor\sigma^{c}\rfloor-q,\sigma+\lfloor\sigma^{c}\rfloor-q}=0.$

It is now obvious that the second condition implies the first condition.

Let us assume in turn that the first condition holds, and derive the second
condition. From the first condition we conclude that the second condition
already holds for $\beta\in A(r+1,n)$ and $\theta\in\Sigma(k+1,n)$ for which
there exists $\sigma\in\Sigma(k,n)$ and $\alpha\in A(r,n)$ such that
$\theta=\sigma+\lfloor\sigma^{c}\rfloor$ and
$\beta=\alpha+\lfloor\sigma^{c}\rfloor$.

But since $0\in[\sigma]\cup[\sigma^{c}]$, we know that
$\theta=\sigma+\lfloor\sigma^{c}\rfloor$ if and only if $0\in[\theta]$ and
$\lfloor\sigma^{c}\rfloor=0$. So it remains to show the second condition for
the case $0\notin[\theta]\cap[\beta]$. For such $\theta$ and $\beta$, we find

$\displaystyle\quad\sum_{p\in[\theta]\cap[\beta]}{\epsilon}(\theta-p,p)\omega_{\beta-p,\theta-p}$
$\displaystyle=-\sum_{p\in[\theta]\cap[\beta]}\sum_{s\in[\theta]\cap[\beta]\setminus\\{p\\}}{\epsilon}(\theta-p,p){\epsilon}(s,\theta-p+0-s)\omega_{\beta-p+0-s,\theta-p+0-s},$

using the first condition. But with the combinatorial observation

$\displaystyle{\epsilon}(\theta-p,p){\epsilon}(s,\theta-p+0-s)$
$\displaystyle={\epsilon}(\theta+0-p,p){\epsilon}(s,\theta-p+0-s)$
$\displaystyle=-{\epsilon}(\theta+0-p,p){\epsilon}(s,p){\epsilon}(s,\theta+0-s)$

we conclude that the sum vanishes if and only if

$\displaystyle 0=\sum_{\begin{subarray}{c}s,p\in[\theta]\cap[\beta]\\\ p\neq
s\end{subarray}}{\epsilon}(\theta+0-p,p){\epsilon}(s,p){\epsilon}(s,\theta+0-s)\omega_{\beta-p+0-s,\theta-p+0-s}.$

This holds because the terms in the sum cancel. The statement is proven. ∎

######  Remark 5.5.

The results of this section show the correspondence of linear independencies
between finite element spaces: a basis for one space is induced by one and
only one basis for the other space.

Note that the first identity in Lemma 5.1 is already contained Proposition 3.7
of [8]. The latter reference, however, does not state further details about
the conditions on the coefficients. Our analogous result in Lemma 5.2 is a
natural analogue of their result and has not appeared previously in the
literature.

The isomorphism (44) is identical to the isomorphism used in Theorem 4.16 of
[2], and the isomorphism (45) is identical to the isomorphism used in Theorem
4.22 of [2]. In that reference, the isomorphisms are only stated in terms of
basis forms. We emphasize that the isomorphisms can be stated naturally in
terms of the canonical spanning sets.

##  6\. Duality Pairings

We have seen in the last section that there exist isomorphisms

$\displaystyle{\mathcal{P}}_{r}\Lambda^{k}(T)\simeq\mathring{{\mathcal{P}}}^{-}_{r+k+1}\Lambda^{n-k}(T),\quad{\mathcal{P}}^{-}_{r+1}\Lambda^{n-k}(T)\simeq\mathring{{\mathcal{P}}}_{r+n-k+1}\Lambda^{k}(T).$

In this section, we extend those results and introduce a non-degenerate
bilinear pairings between the spaces ${\mathcal{P}}_{r}\Lambda^{k}(T)$ and
$\mathring{{\mathcal{P}}}^{-}_{r+k+1}\Lambda^{n-k}(T)$, and between the spaces
$\mathring{{\mathcal{P}}}_{r+n-k+1}\Lambda^{k}(T)$ and
${\mathcal{P}}^{-}_{r+1}\Lambda^{n-k}(T)$. We begin with a technical auxiliary
result.

######  Lemma 6.1.

Let $r\in{\mathbb{N}}_{0}$ and $k\in[0:n]$. Let $\sigma,\rho\in\Sigma(k,n)$.
Then

(51)
$\displaystyle{\mathsf{d}}\lambda_{\sigma}\wedge\phi_{\rho^{c}}=\left\\{\begin{array}[]{rcl}0&\text{if}&|[\sigma]\cap[\rho^{c}]|>1,\\\
(-1)^{k}{\epsilon}(\sigma,\sigma^{c})\sum_{q\in[\sigma^{c}]}\lambda_{q}\phi_{T}&\text{if}&[\sigma]\cap[\rho^{c}]=\emptyset,\\\
(-1)^{k+1}{\epsilon}(\rho,\rho^{c}){\epsilon}(p,\sigma-p){\epsilon}(q,\sigma-p)\lambda_{p}\phi_{T}&\text{if}&|[\sigma]\cap[\rho^{c}]|=1,\end{array}\right.$

where in the last case $q\in[\sigma^{c}]$ and $p\in[\sigma]$ are the unique
solutions of $\rho=\sigma-p+q$.

###### Proof.

Let $\sigma,\rho\in\Sigma(k,n)$, so $\rho^{c}\in\Sigma_{0}(n-k,n)$. Exactly
one of the cases on the right-hand side of (51) is true.

Firstly, suppose that $|[\sigma]\cap[\rho^{c}]|>1$. Then it is easy to verify
that

$\displaystyle{\mathsf{d}}\lambda_{\sigma}\wedge\phi_{\rho^{c}}=0.$

This can be seen by expanding the Whitney form $\phi_{\rho^{c}}$ according to
(9) and using the properties of the alternating product.

Secondly, suppose that $[\sigma]\cap[\rho^{c}]=\emptyset$. This is equivalent
to $|[\sigma]\cap[\rho^{c}]|=0$ and, in particular, to $\sigma=\rho$. We see,
using (9), (12) and Lemma 3.2, that

$\displaystyle{\mathsf{d}}\lambda_{\sigma}\wedge\phi_{\sigma^{c}}$
$\displaystyle={\mathsf{d}}\lambda_{\sigma}\wedge\sum_{q\in[\sigma^{c}]}\lambda_{q}{\epsilon}(q,\sigma^{c}-q){\mathsf{d}}\lambda_{\sigma^{c}-q}$
$\displaystyle=\sum_{q\in[\sigma^{c}]}\lambda_{q}{\epsilon}(q,\sigma^{c}-q){\epsilon}(\sigma,\sigma^{c}-q){\mathsf{d}}\lambda_{\sigma+\sigma^{c}-q}$
$\displaystyle=\sum_{q\in[\sigma^{c}]}\lambda_{q}{\epsilon}(q,\sigma^{c}-q){\epsilon}(\sigma,\sigma^{c}-q){\epsilon}(q,\sigma+\sigma^{c}-q)\phi_{T}.$

From the combinatorial observation that

$\displaystyle{\epsilon}(q,\sigma^{c}-q){\epsilon}(\sigma,\sigma^{c}-q){\epsilon}(q,\sigma+\sigma^{c}-q)=(-1)^{k}{\epsilon}(\sigma,\sigma^{c}),$

we conclude the desired expression for
${\mathsf{d}}\lambda_{\sigma}\wedge\phi_{\sigma^{c}}$ in the second case.

Lastly, suppose that $|[\sigma]\cap[\rho^{c}]|=1$. There exists a unique
$p\in[\sigma]\cap[\rho^{c}]$. Then there exists a unique
$q\in[\sigma^{c}]\cap[\rho]$ such that $\rho=\sigma-p+q$ and
$\rho^{c}=\sigma^{c}-q+p$. We see that the right-hand side of (51) is well-
defined. We have $[\sigma]\cap[\rho^{c}]=\\{p\\}$ and
$[\sigma^{c}]\cap[\rho]=\\{q\\}$. We find, similar as above, that

$\displaystyle{\mathsf{d}}\lambda_{\sigma}\wedge\phi_{\rho^{c}}$
$\displaystyle={\mathsf{d}}\lambda_{\sigma}\wedge\phi_{\sigma^{c}-q+p}$
$\displaystyle={\epsilon}(p,\sigma^{c}-q)\lambda_{p}{\mathsf{d}}\lambda_{\sigma}\wedge{\mathsf{d}}\lambda_{\sigma^{c}-q}$
$\displaystyle={\epsilon}(p,\sigma^{c}-q){\epsilon}(\sigma,\sigma^{c}-q)\lambda_{p}{\mathsf{d}}\lambda_{\sigma+\sigma^{c}-q}$
$\displaystyle={\epsilon}(p,\sigma^{c}-q){\epsilon}(\sigma,\sigma^{c}-q){\epsilon}(q,\sigma+\sigma^{c}-q)\lambda_{p}\phi_{T}$
$\displaystyle=(-1)^{k}{\epsilon}(p,\sigma^{c}-q){\epsilon}(\sigma,\sigma^{c}){\epsilon}(q,\sigma^{c}-q)\lambda_{p}\phi_{T}.$

With the combinatorial observation

$\displaystyle{\epsilon}(\sigma-p+q,\sigma^{c}-q+p)$
$\displaystyle\quad={\epsilon}(\sigma,\sigma^{c}){\epsilon}(\sigma-p,p){\epsilon}(q,\sigma^{c}-q)(-1){\epsilon}(\sigma-p,q){\epsilon}(p,\sigma^{c}-q),$

we derive

$\displaystyle(-1)^{k}{\epsilon}(p,\sigma^{c}-q){\epsilon}(\sigma,\sigma^{c}){\epsilon}(q,\sigma^{c}-q)$
$\displaystyle\quad=(-1)^{k+1}{\epsilon}(\sigma-p+q,\sigma^{c}-q+p){\epsilon}(\sigma-p,p){\epsilon}(\sigma-p,q)$
$\displaystyle\quad=(-1)^{k+1}{\epsilon}(\sigma-p+q,\sigma^{c}-q+p){\epsilon}(p,\sigma-p){\epsilon}(q,\sigma-p).$

This, together with $\rho=\sigma-p+q$, leads to the identity

$\displaystyle{\mathsf{d}}\lambda_{\sigma}\wedge{\epsilon}(\rho,\rho^{c})\phi_{\rho^{c}}$
$\displaystyle=(-1)^{k+1}{\epsilon}(\sigma-p+q,\sigma^{c}-q+p){\epsilon}(p,\sigma-p){\epsilon}(q,\sigma-p)\lambda_{p}\phi_{T}$
$\displaystyle=(-1)^{k+1}{\epsilon}(\rho,\rho^{c}){\epsilon}(p,\sigma-p){\epsilon}(q,\sigma-p)\lambda_{p}\phi_{T}.$

The proof is complete. ∎

######  Lemma 6.2.

Let $r\in{\mathbb{N}}_{0}$ and $k\in[0:n]$. Let $\sigma,\rho\in\Sigma(k,n)$.
Then

(52)
$\displaystyle{\mathsf{d}}\lambda_{\sigma}\wedge{\epsilon}(\rho,\rho^{c})\lambda_{\rho}\phi_{\rho^{c}}={\mathsf{d}}\lambda_{\rho}\wedge{\epsilon}(\sigma,\sigma^{c})\lambda_{\sigma}\phi_{\sigma^{c}}.$

Moreover, we have

(53)
$\displaystyle{\mathsf{d}}\lambda_{\sigma}\wedge{\epsilon}(\sigma,\sigma^{c})\lambda_{\sigma}\phi_{\sigma^{c}}=(-1)^{k}\lambda_{\sigma}\sum_{q\in[\sigma^{c}]}\lambda_{q}\phi_{T}.$

and, if $\rho=\sigma-p+q$ for $p\in[\sigma]$ and $q\in[\sigma^{c}]$, then we
have

(54)
$\displaystyle{\mathsf{d}}\lambda_{\sigma}\wedge{\epsilon}(\rho,\rho^{c})\lambda_{\rho}\phi_{\rho^{c}}=(-1)^{k+1}{\epsilon}(p,\sigma-p){\epsilon}(q,\sigma-p)\lambda_{\rho}\lambda_{p}\phi_{T}.$

###### Proof.

We use Lemma 6.1 above. Firstly, if $[\sigma]\cap[\rho]=\emptyset$, then we
obtain (52) by

$\displaystyle{\mathsf{d}}\lambda_{\sigma}\wedge\lambda_{\rho}\phi_{\rho^{c}}={\mathsf{d}}\lambda_{\rho}\wedge\lambda_{\sigma}\phi_{\sigma^{c}}=0.$

Secondly, if $\sigma=\rho$, then (52) holds trivially and (53) is an easy
observation.

Lastly, consider the case $|[\sigma]\cap[\rho]|=1$. In that case, there exist
$p\in[\sigma]$ and $q\in[\sigma^{c}]$ such that $\rho=\sigma-p+q$. Then

(55)
$\displaystyle{\mathsf{d}}\lambda_{\sigma}\wedge{\epsilon}(\rho,\rho^{c})\lambda_{\rho}\phi_{\rho^{c}}=(-1)^{k+1}{\epsilon}(p,\sigma-p){\epsilon}(q,\sigma-p)\lambda_{\rho}\lambda_{p}\phi_{T}$

on the one hand, proving (54), while

(56)
$\displaystyle\begin{split}{\mathsf{d}}\lambda_{\rho}\wedge{\epsilon}(\sigma,\sigma^{c})\lambda_{\sigma}\phi_{\sigma^{c}}&=(-1)^{k+1}{\epsilon}(q,\rho-q){\epsilon}(p,\rho-q)\lambda_{\sigma}\lambda_{q}\phi_{T}\\\
&=(-1)^{k+1}{\epsilon}(p,\sigma-p){\epsilon}(q,\sigma-p)\lambda_{\rho}\lambda_{p}\phi_{T}\end{split}$

on the other hand. The identity (52) follows. The proof is complete. ∎

Without much further ado, we give our first main result in this section:

######  Theorem 6.3.

Let $r\in{\mathbb{N}}_{0}$ and $k\in[0:n]$. Let $\omega_{\alpha\sigma}$ be a
family of complex numbers indexed over $\alpha\in A(r,n)$ and
$\sigma\in\Sigma(k,n)$. Then we have

(57) $\displaystyle\begin{split}&\sum_{\alpha,\beta\in
A(r,n)}\sum_{\sigma,\rho\in\Sigma(k,n)}\int_{T}\omega_{\alpha\sigma}\lambda^{\alpha}{\mathsf{d}}\lambda_{\sigma}\wedge{\epsilon}(\rho,\rho^{c})\overline{\omega_{\beta\rho}}\lambda^{\beta}\lambda_{\rho}\phi_{\rho^{c}}\\\
&\quad=(-1)^{k}\sum_{\theta\in\Sigma(k+1,n)}\int_{T}\lambda_{\theta}\Big{|}\sum_{\alpha\in
A(r,n)}\sum_{p\in[\theta]}{\epsilon}(p,\theta-p)\lambda^{\alpha}\omega_{\alpha,\theta-p}\Big{|}^{2}.\end{split}$

In particular, this term is zero if and only one of the equivalent conditions
of Lemma 5.1 and Lemma 5.3 is satisfied.

###### Proof.

For the proof, we introduce some additional notation. Let us write

$\displaystyle S(\theta,\alpha,\omega)$
$\displaystyle:=\sum_{p\in[\theta]}{\epsilon}(p,\theta-p)\omega_{\alpha,\theta-p},\quad\theta\in\Sigma(k+1,n),\quad\alpha\in
A(r,n).$

We write $S(\omega)$ for the left-hand side of (57), and we moreover write

$\displaystyle S_{d}(\omega)$
$\displaystyle:=\sum_{\begin{subarray}{c}\alpha,\beta\in A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}\int_{T}\lambda^{\alpha+\beta}\omega_{\alpha\sigma}\overline{\omega_{\beta\sigma}}\lambda_{\sigma}{\mathsf{d}}\lambda_{\sigma}\wedge{\epsilon}(\sigma,\sigma^{c})\phi_{\sigma^{c}}$
$\displaystyle S_{o}(\omega)$
$\displaystyle:=\sum_{\begin{subarray}{c}\alpha,\beta\in A(r,n)\\\
\sigma,\rho\in\Sigma(k,n)\\\
\sigma\neq\rho\end{subarray}}\int_{T}\lambda^{\alpha+\beta}\omega_{\alpha\sigma}\overline{\omega_{\beta\rho}}{\mathsf{d}}\lambda_{\sigma}\wedge{\epsilon}(\rho,\rho^{c})\lambda_{\rho}\phi_{\rho^{c}}.$

So $S(\omega)=S_{d}(\omega)+S_{o}(\omega)$ splits into a _diagonal part_
$S_{d}(\omega)$ and an _off-diagonal part_ $S_{o}(\omega)$. We apply our
previous observations and find that $S(\omega)$ equals

$\displaystyle\sum_{\begin{subarray}{c}\alpha,\beta\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
q\in[\sigma^{c}]\end{subarray}}\int_{T}\lambda^{\alpha+\beta}(-1)^{k}\lambda_{\sigma+q}\omega_{\alpha\sigma}\left(\overline{\omega_{\beta\sigma}}-\sum_{p\in[\sigma]}{\epsilon}(p,\sigma-p){\epsilon}(q,\sigma-p)\overline{\omega_{\beta,\sigma-p+q}}\right)\phi_{T}.$

With the combinatorial observation

$\displaystyle{\epsilon}(p,\sigma-p){\epsilon}(q,\sigma-p)$
$\displaystyle={\epsilon}(p,\sigma+q-p){\epsilon}(p,q){\epsilon}(q,\sigma){\epsilon}(q,p)$
$\displaystyle=-{\epsilon}(p,\sigma+q-p){\epsilon}(\sigma,q),$

we simplify this sum further to

$\displaystyle\sum_{\begin{subarray}{c}\alpha,\beta\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
q\in[\sigma^{c}]\end{subarray}}\int_{T}\lambda^{\alpha+\beta}(-1)^{k}\lambda_{\sigma+q}\omega_{\alpha\sigma}{\epsilon}(q,\sigma)\left(\sum_{p\in[\sigma+q]}{\epsilon}(p,\sigma-p+q)\overline{\omega_{\beta,\sigma+q-p}}\right)\phi_{T}$
$\displaystyle=\sum_{\begin{subarray}{c}\alpha,\beta\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
q\in[\sigma^{c}]\end{subarray}}\int_{T}\lambda^{\alpha+\beta}(-1)^{k}\lambda_{\sigma+q}\omega_{\alpha\sigma}{\epsilon}(q,\sigma)\overline{S(\sigma+q,\beta,\omega)}\phi_{T}.$

This leads to

$\displaystyle S(\omega)$ $\displaystyle=(-1)^{k}\sum_{\alpha,\beta\in
A(r,n)}\int_{T}\lambda^{\alpha+\beta}\sum_{\theta\in\Sigma(k+1,n)}\lambda_{\theta}\sum_{p\in[\theta]}\omega_{\alpha,\theta-p}{\epsilon}(p,\theta-p)\overline{S(\theta,\beta,\omega)}\phi_{T}$
$\displaystyle=(-1)^{k}\sum_{\theta\in\Sigma(k+1,n)}\int_{T}\lambda_{\theta}\sum_{\alpha,\beta\in
A(r,n)}\lambda^{\alpha+\beta}S(\theta,\alpha,\omega)\overline{S(\theta,\beta,\omega)}\phi_{T}$
$\displaystyle=(-1)^{k}\sum_{\theta\in\Sigma(k+1,n)}\int_{T}\lambda_{\theta}\Big{|}\sum_{\alpha\in
A(r,n)}\lambda^{\alpha}S(\theta,\alpha,\omega)\Big{|}^{2}\phi_{T}.$

The integrand is non-negative. Hence the integral vanishes if and only if for
all $\theta\in\Sigma(k+1,n)$ we have

$\displaystyle 0=\sum_{\alpha\in
A(r,n)}\lambda^{\alpha}S(\theta,\alpha,\omega).$

Since the $\lambda^{\alpha}$ are linearly independent for $\alpha\in A(r,n)$,
this holds if and only if one of the equivalent conditions of Lemma 5.1 and
Lemma 5.3 is satisfied. ∎

######  Theorem 6.4.

Let $r\in{\mathbb{N}}_{0}$ and $k\in[0:n]$. Let $\omega_{\alpha\sigma}$ be a
family of complex numbers indexed over $\alpha\in A(r,n)$ and
$\sigma\in\Sigma(k,n)$. Then we have

(58) $\displaystyle\begin{split}&\sum_{\alpha,\beta\in
A(r,n)}\sum_{\sigma,\rho\in\Sigma(k,n)}\int_{T}\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma^{c}}{\mathsf{d}}\lambda_{\sigma}\wedge{\epsilon}(\rho,\rho^{c})\omega_{\beta\rho}\lambda^{\beta}\phi_{\rho^{c}}\\\
&\quad=(-1)^{k}\sum_{\theta\in\Sigma(k+1,n)}\int_{T}\lambda_{\theta^{c}}\Big{|}\sum_{\alpha\in
A(r,n)}\sum_{p\in[\theta]}{\epsilon}(p,\theta-p)\lambda^{\alpha}\lambda_{p}\omega_{\alpha,\theta-p}\Big{|}^{2}.\end{split}$

In particular, this term is zero if and only one of the equivalent conditions
of Lemma 5.2 and Lemma 5.4 is satisfied.

###### Proof.

Let us write $S(\omega)$ for the left-hand side in the equality (58). We can
split that sum into two parts. On the one hand, for the _diagonal part_ ,

$\displaystyle S_{d}(\omega)$
$\displaystyle:=\sum_{\begin{subarray}{c}\alpha,\beta\in A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}\int_{T}\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma^{c}}{\mathsf{d}}\lambda_{\sigma}\wedge{\epsilon}(\sigma,\sigma^{c})\overline{\omega_{\beta\sigma}}\lambda^{\beta}\phi_{\sigma^{c}}$
$\displaystyle=\sum_{\begin{subarray}{c}\alpha,\beta\in A(r,n)\\\
\sigma\in\Sigma(k,n)\end{subarray}}\int_{T}\omega_{\alpha\sigma}\overline{\omega_{\beta\sigma}}\lambda^{\alpha+\beta}\lambda_{\sigma^{c}}(-1)^{k}\sum_{q\in[\sigma]}\lambda_{q}\phi_{T},$

while on the other hand, for the _off-diagonal part_ ,

$\displaystyle S_{o}(\omega)$
$\displaystyle:=\sum_{\begin{subarray}{c}\alpha,\beta\in A(r,n)\\\
\sigma,\rho\in\Sigma(k,n)\\\
\sigma\neq\rho\end{subarray}}\int_{T}\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma^{c}}{\mathsf{d}}\lambda_{\sigma}\wedge{\epsilon}(\rho,\rho^{c})\overline{\omega_{\beta\rho}}\lambda^{\beta}\phi_{\rho^{c}}$
$\displaystyle=\sum_{\begin{subarray}{c}\sigma\in\Sigma(k,n)\\\
\alpha,\beta\in A(r,n)\\\ p\in[\sigma]\\\
q\in[\sigma^{c}]\end{subarray}}\int_{T}\omega_{\alpha\sigma}\overline{\omega_{\beta,\sigma-p+q}}\lambda^{\alpha+\beta}\lambda_{\sigma^{c}}(-1)^{k+1}{\epsilon}(p,\sigma-p){\epsilon}(q,\sigma-p)\lambda_{p}\phi_{T}.$

Since $S(\omega)=S_{d}(\omega)+S_{o}(\omega)$, we combine that
$(-1)^{k}S(\omega)$ equals

$\displaystyle\quad\sum_{\begin{subarray}{c}\alpha,\beta\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
q\in[\sigma^{c}]\end{subarray}}\int_{T}\lambda^{\alpha+\beta}\omega_{\alpha\sigma}\lambda_{\sigma^{c}}\left(\overline{\omega_{\beta\rho}}\lambda_{q}-\sum_{p\in[\sigma]}{\epsilon}(p,\sigma-p){\epsilon}(q,\sigma-p)\overline{\omega_{\beta,\sigma-p+q}}\lambda_{p}\right)\phi_{T}$
$\displaystyle=\sum_{\begin{subarray}{c}\alpha,\beta\in A(r,n)\\\
\sigma\in\Sigma(k,n)\\\
q\in[\sigma^{c}]\end{subarray}}\int_{T}\lambda^{\alpha+\beta}\omega_{\alpha\sigma}\lambda_{\sigma^{c}}{\epsilon}(q,\sigma)\left(\sum_{p\in[\sigma+q]}{\epsilon}(p,\sigma-p+q)\overline{\omega_{\beta,\sigma-p+q}}\lambda_{p}\right)\phi_{T}$
$\displaystyle=\sum_{\begin{subarray}{c}\alpha,\beta\in A(r,n)\\\
\theta\in\Sigma(k+1,n)\\\
p\in[\theta^{c}]\end{subarray}}\int_{T}\lambda^{\alpha+\beta}{\epsilon}(p,\theta-p)\omega_{\alpha,\theta-p}\lambda_{\theta^{c}}\lambda_{p}\left(\sum_{p\in[\theta]}{\epsilon}(p,\theta-p)\overline{\omega_{\beta,\theta-p}}\lambda_{p}\right)\phi_{T}$
$\displaystyle=\sum_{\theta\in\Sigma(k+1,n)}\int_{T}\lambda_{\theta^{c}}\Big{|}\sum_{\alpha\in
A(r,n)}\sum_{p\in[\theta]}{\epsilon}(p,\theta-p)\omega_{\alpha,\theta-p}\lambda_{\alpha}\lambda_{p}\Big{|}^{2}\phi_{T}$
$\displaystyle=\sum_{\theta\in\Sigma(k+1,n)}\int_{T}\lambda_{\theta^{c}}\Big{|}\sum_{\beta\in
A(r+1,n)}\sum_{p\in[\theta]}{\epsilon}(\theta-p,p)\omega_{\beta-p,\theta-p}\lambda_{\beta}\Big{|}^{2}\phi_{T}.$

The integrand is non-negative. Moreover, we see that it vanishes if and only
if the conditions of Lemma 5.2 and Lemma 5.4. This completes the proof. ∎

######  Remark 6.5.

A careful inspection of the foregoing proofs shows that the statements of
Theorems 6.3 and 6.4 remain true even with the integral sign removed.

We apply the former two theorems in our study of duality pairings between
spaces of finite element differential forms. Let us write

$\displaystyle{\mathcal{P}}(r,k,n):={\mathbb{C}}^{A(r,n)\times\Sigma(k,n)}$

for the abstract complex vector space generated by the set
${A(r,n)\times\Sigma(k,n)}$. The members of that vector space represent the
coefficients in linear combinations of the canonical spanning sets.

We have a bilinear form over ${\mathcal{P}}(r,k,n)$ which for
$\omega,\eta\in{\mathcal{P}}(r,k,n)$ is given by

$\displaystyle(\omega,\eta)\mapsto\sum_{\alpha,\beta\in
A(r,n)}\sum_{\sigma,\rho\in\Sigma(k,n)}\int_{T}\omega_{\alpha\sigma}\lambda^{\alpha}{\mathsf{d}}\lambda_{\sigma}\wedge{\epsilon}(\rho,\rho^{c})\overline{\eta_{\beta\rho}}\lambda^{\beta}\lambda_{\rho}\phi_{\rho^{c}},$

and another bilinear form over ${\mathcal{P}}(r,k,n)$ which for
$\omega,\eta\in{\mathcal{P}}(r,k,n)$ is given by

$\displaystyle(\omega,\eta)\mapsto\sum_{\alpha,\beta\in
A(r,n)}\sum_{\sigma,\rho\in\Sigma(k,n)}\int_{T}\omega_{\alpha\sigma}\lambda^{\alpha}\lambda_{\sigma^{c}}{\mathsf{d}}\lambda_{\sigma}\wedge{\epsilon}(\rho,\rho^{c})\overline{\eta_{\beta\rho}}\lambda^{\beta}\phi_{\rho^{c}}.$

Theorems 6.3 and 6.4 have the following implications. We see that these
bilinear forms are symmetric and semi-definite. In fact, they are positive
semidefinite for $k$ even and negative semidefinite for $k$ odd.

The degeneracy space of the first bilinear form is exactly the linear subspace
of ${\mathcal{P}}(r,k,n)$ spanned by those coefficient vectors that satisfy
the conditions of Lemma 5.1 and Lemma 5.3. In particular, it follows that the
bilinear form

(59)
$\displaystyle(\omega,\eta)\mapsto\int_{T}\omega\wedge\eta,\quad\omega\in{\mathcal{P}}_{r}\Lambda^{k}(T),\quad\eta\in\mathring{{\mathcal{P}}}^{-}_{r+k+1}\Lambda^{n-k}(T),$

is non-degenerate.

The degeneracy space of the first bilinear form is exactly the linear subspace
of ${\mathcal{P}}(r,k,n)$ spanned by those coefficient vectors that satisfy
the conditions of Lemma 5.2 and Lemma 5.4. In particular, it follows that the
bilinear form

(60)
$\displaystyle(\omega,\eta)\mapsto\int_{T}\omega\wedge\eta,\quad\omega\in\mathring{{\mathcal{P}}}_{r+n-k+1}\Lambda^{k}(T),\quad\eta\in{\mathcal{P}}^{-}_{r+1}\Lambda^{n-k}(T),$

is non-degenerate.

######  Remark 6.6.

Theorem 6.3 refines Proposition 3.7 in [8], while Theorem 6.4 states the
natural but hitherto unpublished analogue for the second isomorphism relation.
Our first duality pairing is also used in Lemma 4.11 of [2], whereas our
second duality pairing is utilized in Lemma 4.7 of [2].

##  7\. Geometric Decompositions and Degrees of Freedom

In this section we supplement the results of the preceding sections with the
larger context and consider finite element spaces over triangulations.
Throughout this section, we let ${\mathcal{T}}$ be a collection of simplices
satisfying the following conditions: (i) for every $T\in{\mathcal{T}}$ and
every subsimplex $F\subseteq T$ we have $F\in{\mathcal{T}}$, (ii) for every
two $T,T^{\prime}\in{\mathcal{T}}$ we either have $T\cap T^{\prime}=\emptyset$
or $T\cap T^{\prime}\in{\mathcal{T}}$, (iii) we have $\dim T\leq n$ for every
$T\in{\mathcal{T}}$.

We formulate our results within an abstract framework. We assume to be given
$X^{k}(T)\subseteq\Lambda^{k}(T)$ for each cell $T\in{\mathcal{T}}$ such that
for every $F,T\in{\mathcal{T}}$ with $F\subseteq T$ we have the surjectivity
condition $\operatorname{tr}_{T,F}X^{k}(T)=X^{k}(F)$. We write
$\mathring{X}^{k}(T)$ for the subspace of forms with vanishing boundary
traces:

(61) $\displaystyle\mathring{X}^{k}(T)=\left\\{\;\omega\in
X^{k}(T)\mathrel{}\middle|\mathrel{}\forall F\in{\mathcal{T}},F\subsetneq
T:\operatorname{tr}_{T,F}\omega=0\;\right\\}.$

Let us abbreviate
$X^{k}_{-1}({\mathcal{T}}):=\bigoplus_{T\in{\mathcal{T}},\dim T=n}X^{k}(T)$
for the direct sum of vector spaces associated to the $n$-simplices. We say
that $\omega\in X^{k}_{-1}({\mathcal{T}})$ is _single-valued_ if for all
$n$-dimensional simplices $T,T^{\prime}\in{\mathcal{T}}$ with non-empty
intersection $F=T\cap T^{\prime}$ we have
$\operatorname{tr}_{T,F}\omega_{T}=\operatorname{tr}_{T^{\prime},F}\omega_{T^{\prime}}$.
The single-valued members of $X^{k}_{-1}({\mathcal{T}})$ constitute a vector
space on their own that we denote by $X^{k}({\mathcal{T}})$.

The definition suggests a natural way to define the trace of any $\omega\in
X^{k}({\mathcal{T}})$ onto any simplex $F\in{\mathcal{T}}$. We introduce the
_global trace operators_

(62)
$\displaystyle\operatorname{Tr}_{{\mathcal{T}},F}:X^{k}({\mathcal{T}})\rightarrow
X^{k}(F).$

######  Remark 7.1.

Let $r\in{\mathbb{N}}$ and $k\in[0:n]$. We consider two prototypical instances
of our abstract framework. On the one hand, we have the full spaces of
barycentric polynomial differential forms, where
$X^{k}(T)={\mathcal{P}}_{r}\Lambda^{k}(T)$ and
$\mathring{X}^{k}(T)=\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(T)$ for each
$T\in{\mathcal{T}}$. Here,
${\mathcal{P}}_{r}\Lambda^{k}({\mathcal{T}})=X({\mathcal{T}})$ is common
notation. On the other hand, we have the spaces of higher order Whitney forms,
where $X^{k}(T)={\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$ and
$\mathring{X}^{k}(T)=\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(T)$ for each
$T\in{\mathcal{T}}$. In this case,
${\mathcal{P}}_{r}^{-}\Lambda^{k}({\mathcal{T}})=X({\mathcal{T}})$ is common
notation. From these examples we see that $X^{k}({\mathcal{T}})$ captures the
idea of a _conforming_ finite element space.

Our abstract framework relies on extension operators. For all
$F,T\in{\mathcal{T}}$ with $F\subseteq T$ we assume to have a linear mapping

$\displaystyle\operatorname{ext}_{F,T}:\mathring{X}^{k}(F)\rightarrow
X^{k}(T).$

We assume that these are generalized inverses of the trace operators,

(63)
$\displaystyle\operatorname{tr}_{T,F}\operatorname{ext}_{F,T}\omega=\omega,\quad\omega\in\mathring{X}^{k}(F),$

and satisfy the two conditions

(64)
$\displaystyle\operatorname{ext}_{F,G}\omega=\operatorname{tr}_{T,G}\operatorname{ext}_{F,T}\omega,\quad\omega\in\mathring{X}^{k}(F),\quad
F\subseteq G\subseteq T,\quad F,G,T\in{\mathcal{T}},$ (65)
$\displaystyle\operatorname{tr}_{T,G}\operatorname{ext}_{F,T}\omega=0,\quad\omega\in\mathring{X}^{k}(F),\quad
F,G\subseteq T,\quad F\nleq G,\quad F,G,T\in{\mathcal{T}}.$

The identity (64) formalizes that extensions to different simplices have the
same trace on common subsimplices, while the identity (65) formalizes that the
extension is local in the sense that the extension has zero trace on all
simplices of ${\mathcal{T}}$ that do not contain the original simplex.

Under these assumptions, we easily verify that the _global extension
operators_

(66)
$\displaystyle\operatorname{Ext}_{F,{\mathcal{T}}}:\mathring{X}^{k}(F)\rightarrow
X^{k}({\mathcal{T}}),\quad\omega_{F}\mapsto\sum_{\begin{subarray}{c}F,T\in{\mathcal{T}}\\\
F\subseteq T,\;\dim T=n\end{subarray}}\operatorname{ext}_{F,T}\omega_{F},$

are well-defined. We can state this section’s main result.

######  Theorem 7.2.

Suppose that $\omega\in X^{k}({\mathcal{T}})$. Then there exist unique
$\omega_{F}\in\mathring{X}^{k}(F)$ for every $F\in{\mathcal{T}}$ such that

(67)
$\displaystyle\omega=\sum_{F\in{\mathcal{T}}}\operatorname{Ext}_{F,{\mathcal{T}}}\omega_{F}.$

###### Proof.

Let $\omega\in X^{k}({\mathcal{T}})$. We prove the statement of the theorem by
a recursion argument. We let
$\omega_{V}:=\operatorname{Tr}_{{\mathcal{T}},V}\omega\in\mathring{X}^{k}(V)$
for every vertex $V\in{\mathcal{T}}$ of the simplicial complex. Set

$\displaystyle\omega^{(0)}:=\sum_{V\in{\mathcal{T}},\;\dim
T=0}\operatorname{Ext}_{V,{\mathcal{T}}}\omega_{V}.$

Then $\operatorname{Tr}_{{\mathcal{T}},V}(\omega-\omega^{(0)})=0$ for every
$0$-dimensional $V\in{\mathcal{T}}$.

Now assume that for some $m\in[0:n-1]$ the following holds: for every
$F\in{\mathcal{T}}$ of dimension at most $m$ there exists
$\omega_{F}\in\mathring{X}^{k}(F)$ such that, letting

$\displaystyle\omega^{(m)}:=\sum_{F\in{\mathcal{T}},\;\dim F\leq
m}\operatorname{Ext}_{F,{\mathcal{T}}}\omega_{F},$

we have $\operatorname{Tr}_{{\mathcal{T}},F}(\omega-\omega^{(m)})=0$ for every
$F\in{\mathcal{T}}$ of dimension at most $m$. Now, for every
$F\in{\mathcal{T}}$ of dimension $m+1$ we set
$\omega_{F}:=\operatorname{Tr}_{{\mathcal{T}},F}\omega\in\mathring{X}^{k}(F)$
for every $F\in{\mathcal{T}}$ of dimension at most $m+1$. Letting

$\displaystyle\omega^{(m+1)}:=\sum_{F\in{\mathcal{T}},\;\dim F\leq
m+1}\operatorname{Ext}_{F,{\mathcal{T}}}\omega_{F},$

it follows that $\operatorname{Tr}_{{\mathcal{T}},F}(\omega-\omega^{(m)})=0$
for every $F\in{\mathcal{T}}$ of dimension at most $m+1$.

Iterating this construction produces $\omega_{F}\in\mathring{X}^{k}(F)$ for
every $F\in{\mathcal{T}}$ such that

$\displaystyle\omega-\sum_{F\in{\mathcal{T}}}\operatorname{Ext}_{F,{\mathcal{T}}}\omega_{F}$

has vanishing trace on every $F\in{\mathcal{T}}$. Thus (67) follows,
completing the proof. ∎

######  Remark 7.3.

The two families of extension operators defined previously,

(68)
$\displaystyle\operatorname{ext}^{k,r}_{F,T}:\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(F)\rightarrow{\mathcal{P}}_{r}\Lambda^{k}(T),\quad\operatorname{ext}^{k,r,-}_{F,T}:\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(F)\rightarrow{\mathcal{P}}_{r}^{-}\Lambda^{k}(T),$

satisfy the required conditions of this section, and thus lead to geometric
decompositions of ${\mathcal{P}}_{r}\Lambda^{k}({\mathcal{T}})$ and
${\mathcal{P}}_{r}^{-}\Lambda^{k}({\mathcal{T}})$, respectively.

######  Remark 7.4.

Any basis of $\mathring{X}^{k}(F)$ induces a basis of
$\operatorname{Ext}_{F,{\mathcal{T}}}\mathring{X}^{k}(F)$. In the light of the
geometric decomposition (67), we see that choosing a basis for the space of
vanishing trace for each simplex leads to a basis for the entire finite
element space.

The extension operators are defined on the spaces with vanishing traces but
using the geometric decomposition, we can extend them to the full space on
each cell. For each $T\in{\mathcal{T}}$ we define the operator

$\displaystyle\operatorname{ext}_{F,T}\colon X^{k}(F)\rightarrow
X^{k}(T),\quad\sum_{f\in{\mathcal{T}},\;f\subseteq
F}\operatorname{ext}_{f,F}\omega_{f}\mapsto\sum_{f\in{\mathcal{T}},\;f\subseteq
F}\operatorname{ext}_{f,T}\omega_{f},$

where the argument in is expressed in terms of the geometric decomposition
with $\omega_{f}\in\mathring{X}^{k}(f)$ for each subsimplex of $F$. The
operator $\operatorname{ext}_{F,T}\colon X^{k}(F)\rightarrow X^{k}(T)$ extends
the operator
$\operatorname{ext}_{F,T}\colon\mathring{X}^{k}(F)\rightarrow\mathring{X}^{k}(T)$,
as is easily seen.

######  Remark 7.5.

The identities (63), (64) and (65) are satisfied by this definition of
extension operator in the general case $\omega\in X^{k}(F)$. Moreover, for
$\omega\in X^{k}(F)$ and $F,G,T\in{\mathcal{T}}$ with $F,G\subseteq T$ and
$F\cap G\neq\emptyset$ one can verify

(69) $\displaystyle\operatorname{ext}_{F\cap G,G}\operatorname{tr}_{F,F\cap
G}\omega=\operatorname{tr}_{T,G}\operatorname{ext}_{F,T}\omega,$

and in particular, these extension operators are _consistent_ in the
terminology of [3, Section 4]. In that manner, we obtain operators

(70)
$\displaystyle\operatorname{ext}^{k,r}_{F,T}:{\mathcal{P}}_{r}\Lambda^{k}(F)\rightarrow{\mathcal{P}}_{r}\Lambda^{k}(T),\quad\operatorname{ext}^{k,r,-}_{F,T}:{\mathcal{P}}_{r}^{-}\Lambda^{k}(F)\rightarrow{\mathcal{P}}_{r}^{-}\Lambda^{k}(T).$

The mapping $\operatorname{ext}^{k,r}_{F,T}$ has not appeared before in the
literature, whereas $\operatorname{ext}^{k,r,-}_{F,T}$ appears in [3].

We finish this section with an outline of the degrees of freedom. For each
$F\in{\mathcal{T}}$ of dimension $\dim F=m$ we define the spaces of
functionals

$\displaystyle W_{r,k}(F)$
$\displaystyle:=\left\\{\phi\in{\mathcal{P}}_{r}\Lambda^{k}({\mathcal{T}})^{\ast}\mathrel{}\middle|\mathrel{}\exists\eta\in{\mathcal{P}}_{r+k-m}^{-}\Lambda^{m-k}(F):\phi(\cdot)=\int_{F}\eta\wedge\operatorname{Tr}\cdot\right\\},$
$\displaystyle W_{r,k}^{-}(F)$
$\displaystyle:=\left\\{\phi\in{\mathcal{P}}_{r}^{-}\Lambda^{k}({\mathcal{T}})^{\ast}\mathrel{}\middle|\mathrel{}\exists\eta\in{\mathcal{P}}_{r+k-m-1}\Lambda^{m-k}(F):\phi(\cdot)=\int_{F}\eta\wedge\operatorname{Tr}\cdot\right\\}.$

We have isomorphisms
$W_{r,k}(F)\simeq\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(F)^{\ast}$ and
$W_{r,k}^{-}(F)\simeq\mathring{{\mathcal{P}}}_{r}^{-}\Lambda^{k}(F)^{\ast}$,
as is evident considering the pairings (59) and (60). We define the spaces

$\displaystyle
W_{r,k}({\mathcal{T}}):=\sum_{F\in{\mathcal{T}}}W_{r,k}(F),\quad
W_{r,k}^{-}({\mathcal{T}}):=\sum_{F\in{\mathcal{T}}}W_{r,k}^{-}(F).$

These are not only spaces of functionals over conforming finite element spaces
but in fact the entire dual spaces, as expressed in the following two
theorems.

######  Theorem 7.6.

For $r\in{\mathbb{N}}_{0}$ and $k\in[0:n]$ we have

(71)
$\displaystyle{\mathcal{P}}_{r}\Lambda^{k}({\mathcal{T}})^{\ast}=W_{r,k}({\mathcal{T}}),\quad{\mathcal{P}}_{r}^{-}\Lambda^{k}({\mathcal{T}})^{\ast}=W_{r,k}^{-}({\mathcal{T}}).$

###### Proof.

We state the proof for the first identity; the proof for the second identity
is completely analogous. Recall that
$W_{r,k}({\mathcal{T}})\subseteq{\mathcal{P}}_{r}\Lambda^{k}({\mathcal{T}})^{\ast}$.
Let $\omega\in{\mathcal{P}}_{r}\Lambda^{k}({\mathcal{T}})$ such that
$\eta(\omega)=0$ for all $\eta\in W_{r,k}({\mathcal{T}})$. We show that
$\omega=0$.

By Theorem 7.2 we have
$\omega=\sum_{F\in{\mathcal{T}}}\operatorname{Ext}_{F,{\mathcal{T}}}\omega_{F}$
with unique $\omega_{F}\in\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(F)$ for each
$F\in{\mathcal{T}}$. Suppose that for some $m\in{\mathbb{N}}_{0}$ we have
$\omega_{F}=0$ for each $F\in{\mathcal{T}}$ with $\dim F<m$ and consider some
$F\in{\mathcal{T}}$ with $\dim F=m$. By assumption,
$\operatorname{Tr}_{F}\omega=\omega_{F}$, and since $\eta(\omega)=0$ for all
$\eta\in W_{r,k}(F)$, we have $\omega_{F}=0$. By induction, we find
$\omega=0$. Hence $W_{r,k}({\mathcal{T}})$ spans
${\mathcal{P}}_{r}\Lambda^{k}({\mathcal{T}})^{\ast}$. ∎

######  Theorem 7.7.

For $r\in{\mathbb{N}}_{0}$ and $k\in[0:n]$ we have direct sums

(72)
$\displaystyle{\mathcal{P}}_{r}\Lambda^{k}({\mathcal{T}})^{\ast}=\sum_{F\in{\mathcal{T}}}W_{r,k}(F),\quad{\mathcal{P}}_{r}^{-}\Lambda^{k}({\mathcal{T}})^{\ast}=\sum_{F\in{\mathcal{T}}}W_{r,k}^{-}(F).$

###### Proof.

Suppose that we have $\eta_{F}\in W_{r,k}(F)$ for each $F\in{\mathcal{T}}$,
not all zero. Write ${\mathcal{T}}^{m}$ for the set of $m$-dimensional
simplices of ${\mathcal{T}}$ and abbreviate
$\eta_{m}:=\sum_{l=k}^{m}\sum_{F\in{\mathcal{T}}^{l}}\eta_{F}$. We use
induction to find $\omega\in{\mathcal{P}}_{r}\Lambda^{k}({\mathcal{T}})$ such
that $\eta(\omega)>0$.

First, consider the smallest $m\in{\mathbb{N}}_{0}$ for which there exists an
$m$-dimensional $F\in{\mathcal{T}}$ with $\eta_{F}$ nonzero. For each
$m$-dimensional $F\in{\mathcal{T}}$ we choose
$\omega_{F}\in\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(F)$ such that
$\eta_{F}(\omega_{F})>0$ if $\eta_{F}$ is nonzero and let $\omega_{F}=0$
otherwise. It follows that
$\omega_{k}:=\sum_{F\in{\mathcal{T}}^{m}}\operatorname{Ext}_{F,{\mathcal{T}}}\omega_{F}$
satisfies $\eta_{k}(\omega_{k})>0$.

For the induction step, suppose that for some $m\in{\mathbb{N}}$ we have
$\omega_{m-1}\in{\mathcal{P}}_{r}\Lambda^{k}({\mathcal{T}})$ such that
$\eta_{m-1}(\omega_{m-1})>0$. For every $m$-dimensional $F\in{\mathcal{T}}$ we
then choose $\omega_{F}\in\mathring{{\mathcal{P}}}_{r}\Lambda^{k}(F)$
satisfying
$\eta_{F}(\operatorname{Ext}_{F,{\mathcal{T}}}\omega_{F})>\eta_{F}(\omega_{m-1})$
if $\eta_{F}$ is nonzero and $\omega_{F}=0$ otherwise. It follows that
$\omega_{m}:=\omega_{m-1}+\sum_{F\in{\mathcal{T}}^{m}}\operatorname{Ext}_{F,{\mathcal{T}}}\omega_{F}$
satisfies $\eta_{m}(\omega_{m})>0$. Repeating this, we get
$\omega\in{\mathcal{P}}_{r}\Lambda^{k}({\mathcal{T}})$ such that
$\eta(\omega)>0$, finishing the proof. ∎

### Acknowledgements

The author acknowledges helpful discussions with Douglas N. Arnold and Snorre
H. Christiansen.

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