# Unifying the geometric decompositions of full and trimmed polynomial spaces
in finite element exterior calculus

Toby Isaac

###### Abstract

Arnold, Falk, & Winther, in _Finite element exterior calculus, homological
techniques, and applications_ (2006), show how to geometrically decompose the
full and trimmed polynomial spaces on simplicial elements into direct sums of
trace-free subspaces and in _Geometric decompositions and local bases for
finite element differential forms_ (2009) the same authors give direct
constructions of extension operators for the same spaces. The two families –
full and trimmed – are treated separately, using differently defined
isomorphisms between each and the other’s trace-free subspaces and mutually
incompatible extension operators. This work describes a single operator
$\mathring{\star}_{T}$ that unifies the two isomorphisms and also defines a
weighted-$L^{2}$ norm appropriate for defining well-conditioned basis
functions and dual-basis functionals for geometric decomposition. This work
also describes a single extension operator $\dot{E}_{\sigma,T}$ that
implements geometric decompositions of all differential forms as well as for
the full and trimmed polynomial spaces separately.

##  1 Introduction

This introduction, describing the contribution of the subsequent sections, is
intended for those familiar with finite elements but not with exterior
calculus.

###  1.1 Trace-free scalar-valued finite elements

Let $\mathcal{T}_{h}$ be a triangulation of a domain
$\Omega\subset\mathbb{R}^{2}$, and consider the standard finite element space
$\mathcal{P}_{3}(\mathcal{T}_{h})$ of continuous, real-valued functions that
are polynomials with degree at most 3 when restricted to each triangle,

$\mathcal{P}_{3}(\mathcal{T}_{h}):=\\{f\in
C^{0}(\Omega):f|_{T}\in\mathcal{P}_{3}(T),T\in\mathcal{T}_{h}\\}.$

A standard finite element basis for $\mathcal{P}_{3}(\mathcal{T}_{h})$
associates one basis function to each vertex, two to each edge, and one to
each triangle, with each basis function supported only on the triangles
surrounding its associated mesh object.

Let $T$ be a triangle in $\mathcal{T}_{h}$ with vertices at
$\\{v_{0},v_{1},v_{2}\\}$, and let $\lambda:T\to\mathbb{R}^{3}$ be its
barycentric coordinates, the unique affine map satisfying
$\lambda_{i}(v_{j})=\delta_{ij}$ and $\sum_{i}\lambda_{i}=1$. The basis
function $\phi_{T}$ associated with $T$, to be fully continuous and supported
only on $T$, must be zero on the boundary $T$: we say that $\phi_{T}$ is
_trace-free_. Because $\phi_{T}$ is in $\mathcal{P}_{3}(T)$ and zero on each
of the three edges surrounding $T$, $\phi_{T}$ must be a multiple of the
function

$\lambda_{T}:=\lambda_{0}\lambda_{1}\lambda_{2}.$

This is the _bubble function_ associated with $T$.

Now let $E$ be an edge in the mesh with vertices $\\{v_{0},v_{1}\\}$, and let
$\lambda$ be its barycentric coordinates. The two basis functions
$\\{\phi_{E,1},\phi_{E,2}\\}$ associated with $E$, to be continuous and
supported only on the triangles on either side of $E$, must be zero at $v_{0}$
and $v_{1}$. But while the basis function $\phi_{T}$ was determined up to a
constant, there is more freedom in what $\phi_{E,1}$ and $\phi_{E,2}$ can be:
they must vanish on the non-shared edges of the surrounding triangles, but
when restricted to $E$, may satisfy

$\phi_{E,1}|_{E}=\lambda_{E}p_{1},\quad\phi_{E,2}|_{E}=\lambda_{E}p_{2},\quad\quad[\lambda_{E}:=\lambda_{0}\lambda_{1}]$

for any basis $\\{p_{1},p_{2}\\}$ of $\mathcal{P}_{1}(E)$.

The pattern that this example demonstrates is that, for scalar-valued
polynomials on an $n$-dimensional simplex $T$, there is an isomorphism between
$\mathcal{P}_{r}(T)$, the polynomials on $T$ up to degree $r$, and the trace-
free polynomials up to degree $r+n+1$,

$\mathring{\mathcal{P}}_{r+n+1}(T):=\\{p\in\mathcal{P}_{r+n+1}(T):p(x)|_{\partial
T}=0\\}.$

This isomorphism can be realized by an operator
$h_{T}:\mathcal{P}_{r}(T)\to\mathring{\mathcal{P}}_{r+n+1}(T)$ defined by
pointwise multiplication with the bubble function,

$h_{T}(p):=\lambda_{T}p.$

We point out that because this operator is defined pointwise it can be applied
to any smoothly continuous function on $\overline{T}$, not just a polynomial,
to construct a trace-free function, and it transforms any basis of
$\mathcal{P}_{r}(T)$ into a basis for $\mathring{\mathcal{P}}_{r+n+1}(T)$.

###  1.2 Scalar-valued finite element extension operators

To complete the definition of the edge-centered basis function $\phi_{E,1}$
from the above example, we must define its values $\phi_{E,1}^{T}$ in each
triangle $T$ adjacent to $E$: we call this _extending_ $\phi_{E,1}$ into $T$.
For the basis function to be a part of $\mathcal{P}_{3}(\mathcal{T}_{h})$, the
extension $\phi_{E,1}^{T}$ must be in $\mathcal{P}_{3}(T)$, and it must be
_consistent_ : restricting back to $E$ should match the definition on the
edge,

$\phi_{E,1}^{T}|_{E}=\phi_{E,1},$

and restricting it to any $\tilde{E}\neq E$ should be identically zero,

$\phi_{E,1}^{T}|_{\tilde{E}}=0,\quad\tilde{E}\neq E.$

Suppose $E$ is the edge bounded by $\\{v_{0},v_{1}\\}$ and $T$ is the triangle
bounded by $\\{v_{0},v_{1},v_{2}\\}$. Because
$\phi_{E,1}|_{E}=\lambda_{0}\lambda_{1}p_{1}$ for some
$p_{1}\in\mathcal{P}_{1}(E)$, one way we can define a consistent extension
$\phi_{E,1}^{T}$ is by projecting each point in $T$ to a point in $E$ and
evaluating $p_{1}$,

$\phi_{E,1}^{T}(\lambda_{0},\lambda_{1},\lambda_{2}):=\lambda_{1}\lambda_{2}p_{1}(\lambda_{0}+(1/2)\lambda_{2},\lambda_{1}+(1/2)\lambda_{2}).$

Note that this procedure works not just for extending
$\mathring{\mathcal{P}}_{3}(E)$ into $\mathcal{P}_{3}(T)$: it extends
$\mathring{\mathcal{P}}_{r}(E)$ into $\mathcal{P}_{r}(T)$ for any $r$, and its
extension of $\lambda_{0}\lambda_{1}g$ is a consistent extension for any $g$
that is smooth on $\overline{E}$.

###  1.3 Finite element exterior calculus

Finite element exterior calculus is a framework that unifies the theory behind
$H^{1}$-conforming finite elements (spanned by fully-continuous basis
functions like $\mathcal{P}_{3}(\mathcal{T}_{h})$ in the preceding section)
with that of $H(\mathrm{div})$-conforming elements (which are vector-valued
and spanned by basis functions whose normal components only must be continuous
on the facets between cells) and $H(\mathrm{curl})$-conforming elements (whose
tangential components only must be continuous on the facets and edges between
cells).

In finite element exterior calculus, for each $k\in\mathbb{N}_{0}$ there is a
space $\Lambda^{k}(\Omega):=C^{\infty}(\Omega;\mathrm{Alt}^{k}\mathbb{R}^{n})$
of _differential $k$-forms_, smooth functions that take values in
$\mathrm{Alt}^{k}\mathbb{R}^{n}$, the vector space of _algebraic $k$-forms_,
the alternating $k$-linear maps that map $\bigotimes_{i=1}^{k}\mathbb{R}^{n}$
to $\mathbb{R}$. For each $k$ there is a differential operator
$\mathrm{d}:\Lambda^{k}(\Omega)\to\Lambda^{k+1}(\Omega)$ that generalizes
$\mathrm{grad}$ ($k=0$), $\mathrm{curl}$ ($k=1$), and $\mathrm{div}$
($k=n-1$), and an associated Sobolev space $H\Lambda^{k}(\Omega)$ that is the
closure of $\Lambda^{k}(\Omega)$ under the norm

$\|v\|_{H\Lambda^{k}}^{2}:=\int_{\Omega}|v|^{2}+|\mathrm{d}v|^{2}\ dx.$

To construct $H\Lambda^{k}(\Omega)$-conforming finite element spaces, one must
construct piecewise smooth functions that are _trace-continuous_ at the
boundaries between cells, where trace continuity is the generalization of the
normal continuity of $H(\mathrm{div})$-conforming spaces and tangential
continuity of $H(\mathrm{curl})$-conforming spaces.

Given a simplex $T$, let $\mathring{\Lambda}^{k}(\overline{T})$ be the trace-
free $k$-forms that are smoothly continuous on $\overline{T}$. (This is
distinct from $\mathring{\Lambda}^{k}(T)$ as defined in other works, which is
the subspace of $\Lambda^{k}(T)$ of functions compactly supported in the
interior of $T$.) Given a function space
$X(T)\subset\Lambda^{k}(\overline{T})$, identifying its trace-free subspace
$\mathring{X}(T):=X(T)\cap\mathring{\Lambda}^{k}(\overline{T})$ is important
for finite element construction. These are the basis functions that can be
extended by zero outside of $T$ while maintaining trace continuity, so these
are the basis functions associated with $T$ in a basis for the whole mesh
$\mathcal{T}_{h}$. Likewise, identifying $\mathring{X}(E)$ for a lower-
dimensional simplex $E$ is important because the basis functions associated
with $E$ are extensions of $\mathring{X}(E)$ into the cells surrounding $E$.
Constructing $X(T)$ as the direct sum of $\mathring{X}(T)$ and extensions of
$\mathring{X}(E)$ for $E$ in the boundary of $T$ is a _geometric
decomposition_ of $X(T)$.

There are two primary families of finite elements for the simplex $T$ in
finite element exterior calculus, chosen for their homological properties: the
full polynomial spaces $\mathcal{P}_{r}\Lambda^{k}(T)$ and the trimmed spaces
$\mathcal{P}_{r}^{-}\Lambda^{k}(T)$, which are defined in section 2. These two
families are intertwined, because each is isomorphic to the family of trace-
free subspaces of the other:

$\displaystyle\mathcal{P}_{r}\Lambda^{k}(T)$
$\displaystyle\cong\mathring{\mathcal{P}}_{r+k+1}^{-}\Lambda^{n-k}(T);$ (1)
$\displaystyle\mathcal{P}_{r}^{-}\Lambda^{k}(T)$
$\displaystyle\cong\mathring{\mathcal{P}}_{r+k}\Lambda^{n-k}(T).$ (2)

The work of [2] is a standard reference to which we direct the reader for more
details on finite element exterior calculus as well as proofs of isomorphisms
(1) and (2) [2, theorems 4.16 and 4.22]. In that work, and in subsequent work
on the topic of geometric decompositions [3, 4], the two isomorphisms are
treated separately, in that two distinct linear operators are defined,
$h_{T}^{k}:\mathcal{P}_{r}\Lambda^{k}(T)\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathring{\mathcal{P}}_{r+k+1}^{-}\Lambda^{n-k}(T)$
and
$h_{T}^{k,-}:\mathcal{P}_{r}^{-}\Lambda^{k}(T)\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathring{\mathcal{P}}_{r+k}\Lambda^{n-k}(T)$
(though they are anonymous in [2]), and they are proved to be bijections.
Furthermore, the operators $h_{T}^{k}$ and $h_{T}^{k,-}$ are defined with
respect to particular choices of basis functions, such that is not clear
whether $h_{T}^{k,-}$ acts pointwise or not.

Subsequent works by the same authors and others [3, 4] give direct
constructions of the trace-free subspaces that are based on Bernstein
polynomials. Bernstein polynomials have desirable symmetry properties and low-
complexity evaluation algorithms but are known to be ill-conditioned for
finite element operations relative to other bases. Giving these tradeoffs, it
is potentially useful to have a uniform method for constructing trace-free
$k$-forms that is not tied to any basis.

In [3], the authors define two extension operators from a boundary simplex $f$
into a simplex $g$,

$\displaystyle E_{f,g}^{r,k}:\mathcal{P}_{r}\Lambda^{k}(f)$
$\displaystyle\to\mathcal{P}_{r}\Lambda^{k}(g),$ (3) $\displaystyle
E_{f,g}^{r,k,-}:\mathcal{P}_{r}^{-}\Lambda^{k}(f)$
$\displaystyle\to\mathcal{P}_{r}^{-}\Lambda^{k}(g),$ (4)

that have the required consistency for constructing local, trace-continuous
basis functions for the two spaces. These two extension operators are not only
defined differently, but are truly distinct in that neither is an extension
operator for the other: though
$\mathcal{P}_{r}^{-}\Lambda^{k}(f)\subset\mathcal{P}_{r}\Lambda^{k}(f)$,
$E_{f,g}^{r,k}[\mathcal{P}_{r}^{-}\Lambda^{k}(f)]\not\subset\mathcal{P}_{r}^{-}\Lambda^{k}(g)$,
and though
$\mathcal{P}_{r}\Lambda^{k}(f)\subset\mathcal{P}_{r+1}^{-}\Lambda^{k}(f)$,
$E_{f,g}^{r+1,k,-}[\mathcal{P}_{r}\Lambda^{k}(f)]\not\subset\mathcal{P}_{r}\Lambda^{k}(g)$.

###  1.4 Unifying the geometric decompositions of
$\mathcal{P}_{r}^{-}\Lambda^{k}(T)$ and $\mathcal{P}_{r}\Lambda^{k}(T)$

#### A unified construction of trace-free $k$-forms

This work defines an operator $\mathring{\star}_{T}$ from differential
$k$-forms to trace-free $(n-k)$-forms that acts pointwise and induces the
isomorphisms (1) and (2) and the more general isomorphism for smooth functions
on the closure of $T$:

$\displaystyle\mathring{\star}_{T}:\Lambda^{k}(\overline{T})\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathring{\Lambda}^{n-k}(\overline{T});$
$\displaystyle\mathring{\star}_{T}:\mathcal{P}_{r}\Lambda^{k}(T)\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathring{\mathcal{P}}_{r+k+1}^{-}\Lambda^{n-k}(T);$
$\displaystyle\mathring{\star}_{T}:\mathcal{P}_{r}^{-}\Lambda^{k}(T)\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathring{\mathcal{P}}_{r+k}\Lambda^{n-k}(T).$

The definition of $\mathring{\star}_{T}$ and proof of these claims are given
in the section 3.3. Here we translate $\mathring{\star}_{T}$ into operators
which construct normal-free functions (for $H(\mathrm{div})$-conforming finite
elements) and tangent-free functions (for $H(\mathrm{curl})$-conforming finite
elements).

##### Normal-free.

Let $\\{E_{i}\\}_{i=1}^{n_{e}}$ be the edges around $T$ ($n_{e}=3$ if $n=2$,
$n_{e}=6$ if $n=3$), let $e_{i}=v_{i,1}-v_{i,0}$, where
$\\{v_{i,0},v_{i,1}\\}$ are the vertices of edge $E_{i}$, and let $|T|$ be the
$n$-dimensional volume of $T$. Then for any $u\in
C^{\infty}(\overline{T};\mathbb{R}^{n})$,

$\mathring{\star}_{T}u=\frac{1}{n!|T|}\sum_{i=1}^{n_{e}}\langle
u,e_{i}\rangle\lambda_{E_{i}}e_{i}$

is a function with normal-free boundary trace. In this translation of
$\mathring{\star}_{T}$, the input vector-valued function is treated as a
function in $\Lambda^{1}(\overline{T})$, while the output is treated as a
function in $\mathring{\Lambda}^{n-1}(\overline{T})$. This operator transforms
any basis for a Nédélec edge element of the first (second) kind into a basis
for the normal-free subspace of a Nédélec face element of the second (first)
kind.

##### Tangent-free.

Let $\\{F_{i}\\}_{i=0}^{n}$ be the facets around $T$, let $\nu_{i}$ be the
unit normal vector for facet $i$, and let $\tilde{\nu}_{i}$ be $\nu_{i}$
scaled by $(n-1)!|F_{i}|$, the determinant of a matrix formed from any set of
$n-1$ edge vectors around $F_{i}$ (in 2D, $\tilde{\nu}_{i}$ is a 90 degree
rotation of $v_{i,1}-v_{i,0}$; in 3D, $\tilde{\nu}_{i}$ is the cross product
$(v_{i,1}-v_{i,0})\times(v_{i,2}-v_{i,0})$). Then for any $u\in
C^{\infty}(\overline{T};\mathbb{R}^{n})$,

$\mathring{\star}_{T}u=\frac{1}{n!|T|}\sum_{i=0}^{n}\langle
u,\tilde{\nu}_{i}\rangle\lambda_{F_{i}}\tilde{\nu}_{i}$

is a function with tangent-free boundary trace. This translation of
$\mathring{\star}_{T}$ is the opposite of the last: the input is treated as a
function in $\Lambda^{n-1}(\overline{T})$ and the output in
$\mathring{\Lambda}^{1}(\overline{T})$. This operator transforms any basis for
a Nédélec face element of the first (second) kind into a basis for the
tangent-free subspace of a Nédélec edge element of the second (first) kind.

#### A unified extension of trace-free $k$-forms

This work also defines an extension operator
$\dot{E}_{f,g}:\mathring{\Lambda}^{k}(\overline{f})\to\Lambda^{k}(\overline{g})$
that acts pointwise and is a consistent extension operator for full and
trimmed polynomials and indeed all differential $k$-forms: for every simplex
$g$ in the triangulation $\mathcal{T}_{h}$,

$\displaystyle\Lambda^{k}(\overline{g})$
$\displaystyle=\bigoplus_{f\in\Delta(g)}\dot{E}_{f,g}[\mathring{\Lambda}^{k}(\overline{f})],$
$\displaystyle\mathcal{P}_{r}\Lambda^{k}(\overline{g})$
$\displaystyle=\bigoplus_{f\in\Delta(g)}\dot{E}_{f,g}[\mathring{\mathcal{P}}_{r}\Lambda^{k}(\overline{f})],$
$\displaystyle\mathcal{P}_{r}^{-}\Lambda^{k}(\overline{g})$
$\displaystyle=\bigoplus_{f\in\Delta(g)}\dot{E}_{f,g}[\mathring{\mathcal{P}}_{r}^{-}\Lambda^{k}(\overline{f})].$

We do not translate $\dot{E}_{f,g}$ into $H(\mathrm{curl})$ or
$H(\mathrm{div})$ notation here, but do note that it is based on a
decomposition that generalizes the decomposition present in the hierarchical
bases for $\mathcal{P}_{r}\Lambda^{k}(T)$ of [1]. This work shows that,
although trimmed polynomials are not explicitly represented in those bases,
their extensions remain trimmed polynomials of the same order.

##  2 Preliminaries and notation

##### Integer maps, permutations, and multi-indices.

We let $[a$ be the set $\\{a,a+1,\dots,b\\}$ if $b\geq a$ and $[a=\emptyset$
otherwise. For a map $\rho:[a\to S$ and $i\in[a$ we define
$\rho\setminus\rho(i):[a\to S$ by

$((\rho\setminus\rho(i))(a),\dots,(\rho\setminus\rho(i))(b-1))=(\rho(a),\dots,\hat{\rho}(i),\dots\rho(b)),$

where $\hat{\rho}(i)$ is the omitted range index. Given any map $\rho:M\to S$
we let $[\\![\rho]\\!]$ be the range set $\\{\rho(i)\\}_{i\in M}$.

Given any two subsets $M$ and $S$ of $\mathbb{N}_{0}$ we let $\Sigma(M,S)$ be
the set of increasing maps from $M$ to $S$. We adopt the convention that
$\Sigma(\emptyset,S)$ contains only the empty map $\emptyset\to\emptyset$. If
$M$ is a subset of $S$ and $\rho\in\Sigma(M,S)$, we define its complement to
be $\rho^{*}\in\Sigma(S\setminus M,S)$ such that
$[\\![\rho]\\!]\cup[\\![\rho^{*}]\\!]=S$. The two complementary maps together
define a permutation: we let $\mathop{\mathrm{sign}}(\rho)\in\\{-1,1\\}$ be
$1$ if that permutation is even and $-1$ if it is odd.

When treating a map into $\alpha:S\to\mathbb{N}_{0}$ as a multi-index, we use
the standard notation $|\alpha|=\sum_{i\in S}\alpha(i).$

##### Exterior algebra.

We let $\mathrm{Alt}^{k}V$ denote the alternating $k$-linear forms, or
_algebraic $k$-forms_, on the vector space $V$. By “alternating” we mean that
$\omega\in\mathrm{Alt}^{k}V$ satisfies

$\omega(v_{1},\dots,v_{i},\dots,v_{j},\dots,v_{k})=-\omega(v_{1},\dots,v_{j},\dots,v_{i},\dots,v_{k}),\quad
i,j\in[1.$

Given $\omega\in\mathrm{Alt}^{j}V$ and $\mu\in\mathrm{Alt}^{k}V$, the _wedge
product_ $\omega\wedge\mu\in\mathrm{Alt}^{j+k}V$ is defined by

$(\omega\wedge\mu)(v_{1},\dots,v_{j+k}):=\sum_{\sigma\in\Sigma([1,[1)}\mathop{\mathrm{sign}}(\sigma)\omega(v_{\sigma(1)},\dots,v_{\sigma(j)})\mu(v_{\sigma^{*}(j+1)},\dots,v_{\sigma^{*}(j+k)}).$

The wedge product is bilinear and anti-commuting,
$\mu\wedge\omega=(-1)^{jk}\omega\wedge\mu$.

Given a linear map $J:V\to W$, the _pullback_ is
$J^{*}:\mathrm{Alt}^{k}W\to\mathrm{Alt}^{k}V$ defined by

$(J^{*}\omega)(v_{1},\dots,v_{k}):=\omega(Jv_{1},\dots,Jv_{k}),\quad\omega\in\mathrm{Alt}^{k}W,$

which distributes over the wedge product
$J^{*}(\omega\wedge\mu)=(J^{*}\omega)\wedge(J^{*}\mu)$.

Given $\omega\in\mathrm{Alt}^{k}V$ and $v\in V$, the _interior product_
$\omega\lrcorner v\in\mathrm{Alt}^{k-1}V$ is defined by

$(\omega\lrcorner v)(v_{1},\dots,v_{k-1}):=\omega(v,v_{1},\dots,v_{k-1}).$

The interior product follows the product rule

$(\omega\wedge\mu)\lrcorner v=(\omega\lrcorner
v)\wedge\mu+(-1)^{j}\omega\wedge(\mu\lrcorner
v),\quad\omega\in\mathrm{Alt}^{j}V,\mu\in\mathrm{Alt}^{k}V.$

The _inner product_ on $\mathrm{Alt}^{k}V$ when $V$ has its own inner product
is defined using any orthonormal basis $\\{q_{k}\\}_{k=1}^{n}$ of $V$ as

$\langle\omega,\mu\rangle:=\sum_{\rho\in\Sigma([1,[1))}\omega(q_{\rho(1)},\dots,q_{\rho(k)})\mu(q_{\rho(1)},\dots,q_{\rho(k)}),\quad\omega,\mu\in\mathrm{Alt}^{k}V.$

If the space not only has an inner product but a sign convention that the
ordering $(q_{1},\dots,q_{k})$ is positively oriented, then the _volume form_
$\mathrm{vol}\in\mathrm{Alt}^{n}V$ is defined by

$\mathrm{vol}(q_{1},\dots,q_{n}):=1.$

The _Hodge star operator_
$\star:\mathrm{Alt}^{k}V\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathrm{Alt}^{n-k}V$
is an isometry defined by

$\omega\wedge\mu:=\langle\star\omega,\mu\rangle\mathrm{vol},\quad\omega\in\mathrm{Alt}^{k}V,\mu\in\mathrm{Alt}^{n-k}V.$

##### Exterior calculus.

Exterior calculus can be defined for general manifolds: in this work we only
need to consider subsets $\Omega$ of $\mathbb{R}^{n}$ which have the same
tangent space $T_{x}\Omega$ for each $x\in\Omega$, either because they are
$n$-dimensional and $T_{x}\Omega=\mathbb{R}^{n}$ or because they are
$d$-dimensional subsets of $d$-dimensional hyperplanes.

The set of smooth, $\mathrm{Alt}^{k}T_{x}\Omega$-valued function on
$\Omega\subset\mathbb{R}^{n}$ are _differential $k$-forms_ and are denoted
$\Lambda^{k}(\Omega)$. The _exterior derivative_ on $k$-forms,
$\mathrm{d}:\Lambda^{k}(\Omega)\to\Lambda^{k+1}(\Omega)$, is defined by

$(\mathrm{d}\omega)_{x}(v_{1},\dots,v_{k+1}):=\sum_{j=1}^{k+1}(-1)^{j+1}\nabla_{x}(\omega_{x}(v_{1},\dots,\hat{v}_{j},\dots,v_{k+1}))\cdot
v_{j},$

where again $\hat{v}_{j}$ is the omitted argument. The exterior derivative
applied to each of the coordinate functions $x_{1},\dots,x_{n}$ is a constant
$1$-form, so we consider $\mathrm{d}x_{i}$ an algebraic 1-form for each $i$.
The set $\\{\mathrm{d}x_{i}\\}_{i=1}^{n}$ forms a basis of algebraic $k$-forms
that are dual to the canonical basis $\\{e_{i}\\}_{i=1}^{n}$ of
$\mathbb{R}^{n}$,

$\mathrm{d}x_{i}(e_{j})=\delta_{ij},\quad i,j\in[1.$

Given a sequence $\rho:[a\to S$, not necessarily increasing, we let
$x_{\rho}:=\prod_{i=a}^{b}x_{\rho(i)}$ and

$(\mathrm{d}x)_{\rho}:=\mathrm{d}x_{\rho(a)}\wedge\dots\wedge\mathrm{d}x_{\rho(b)},$

with $x_{\rho}:=(\mathrm{d}x)_{\rho}:=1$ if $[a=\emptyset$. With this notation
the canonical orthonormal basis of $\mathrm{Alt}^{k}\mathbb{R}^{n}$ in terms
of the coordinate functions is written as
$\\{(\mathrm{d}x)_{\rho}\\}_{\rho\in\Sigma([1,[1)}$, which we call the set of
_coordinate $k$-forms_.

If $\phi:T\to U$ is a smooth map, then
$\phi^{*}:\Lambda^{k}(U)\to\Lambda^{k}(T)$ is the pullback for differential
$k$-forms, defined by

$(\phi^{*}\omega)_{x}:=(D\phi)_{x}^{*}\omega_{\phi(x)},\quad\omega\in\Lambda^{k}(U),$

where $(D\phi)_{x}$ is the Jacobian of $\phi$ at $x$.

We let $\kappa_{x}:\Lambda^{k}(\Omega)\to\Lambda^{k-1}(\Omega)$ be the _Koszul
operator_ centered at $x$, which is defined by

$(\kappa_{x}\omega)_{y}:=\omega_{y}\lrcorner(y-x).$

An important fact for this work is that the Koszul operator commutes with the
pullback of an affine map up to a constant interior product: given affine
$\phi:T\to\hat{T}$,

$\kappa_{x}(\phi^{*}\omega)=\phi^{*}(\kappa_{\hat{x}}\omega-\omega\lrcorner(\phi(x)-\hat{x})).$
(5)

##### Oriented simplices.

In all that follows $T$ is a positively oriented $n$-simplex in
$\mathbb{R}^{n}$, $|T|$ is its volume and $\\{\lambda_{i}\\}_{i=0}^{n}$ are
its barycentric coordinates. We define $\lambda_{\rho}$ and
$(\mathrm{d}\lambda)_{\rho}$ similarly to $x_{\rho}$ and
$(\mathrm{d}x)_{\rho}$, and we refer to $(\mathrm{d}\lambda)_{\rho}$ as a
_barycentric $k$-form_. Given maps $\rho:S\to[0$ and
$\alpha:S\to\mathbb{N}_{0}$ with the same domain $S$, a barycentric monomial
is a polynomial in $\mathcal{P}_{|\alpha|}(T)$ defined by

$(\lambda_{\rho})^{\alpha}:=\Pi_{i\in S}\lambda_{\rho(i)}^{\alpha(i)}.$

A positively oriented simplex has the property [2, Section 4.1] that for any
permutation $\pi:[0\to[0$ and any $i\in[0$

$(\mathrm{d}\lambda)_{\pi\setminus\pi(i)}=(-1)^{i}\frac{\mathop{\mathrm{sign}}\pi}{n!|T|}\mathrm{vol}.$

##### Boundary simplices.

For each $\rho\in\Sigma([0,[0)$, the $d$-simplex with vertices
$\\{v_{\rho(0)},\dots,v_{\rho(d)}\\}$ in the boundary of $T$ is $f_{\rho}$. If
$i_{\rho}:f_{\rho}\hookrightarrow T$ is the inclusion map, then the trace
operator
$\mathrm{Tr}_{\rho}:\Lambda^{k}(\overline{T})\to\Lambda^{k}(\overline{f_{\rho}})$
is its pullback, $\mathrm{Tr}_{\rho}:=i_{\rho}^{*}$. In section 4, when we
need the trace of $\omega\in\Lambda^{k}(\overline{f_{\rho}})$ on another
boundary simplex $f_{\sigma}$, we denote it $\mathrm{Tr}_{\rho,\sigma}\omega$.

The trace operator lets us more formally define the space
$\mathring{\Lambda}^{k}(\overline{T})$ from the introduction as

$\mathring{\Lambda}^{k}(\overline{T}):=\\{\omega\in\Lambda^{k}(\overline{T}):\mathrm{Tr}_{\rho}u=0,f_{\rho}\neq
T\\}.$

##### Whitney forms and trimmed polynomials.

The Whitney form $\phi_{\rho}\in\mathcal{P}_{1}\Lambda^{d}(T)$ associated with
$f_{\rho}$ is

$\phi_{\rho}:=\sum_{i=0}^{d}(-1)^{i}\lambda_{\rho(i)}(\mathrm{d}\lambda)_{\rho\setminus\rho(i)},$

and the trimmed polynomials are spanned by the product of Whitney forms and
scalar polynomials,

$\mathcal{P}_{r}^{-}\Lambda^{k}(T)=\mathrm{span}\\{a_{\rho}\phi_{\rho}:a_{\rho}\in\mathcal{P}_{r-1}(T),\rho\in\Sigma([0,[0)\\}.$

An important fact about the trimmed polynomials in this work is that
$\mathcal{P}_{r}^{-}\Lambda^{k}(T)$ is the largest subspace of
$\mathcal{P}_{r}\Lambda^{k}(T)$ such that
$\kappa_{x}p\in\mathcal{P}_{r}\Lambda^{k-1}(T)$ for every
$p\in\mathcal{P}_{r}^{-}\Lambda^{k}(T)$ and every $x$.

##  3 Trace-free operators

###  3.1 Previous trace-free operators and bases

We can now define the trace-free operators $h_{T}^{k}$ and $h_{T}^{k,-}$ given
in [2] that induce isomorphisms (1) and (2):

$\displaystyle h_{T}^{k}:\sum_{\begin{subarray}{c}\rho\in\Sigma([0,[0)\\\
\rho(0)=0\end{subarray}}a_{\rho}(\mathrm{d}\lambda)_{\rho^{*}}\mapsto\sum_{\begin{subarray}{c}\rho\in\Sigma([0,[0)\\\
\rho(0)=0\end{subarray}}a_{\rho}\lambda_{\rho^{*}}\phi_{\rho}$
$\displaystyle[a_{\rho}\in\mathcal{P}_{r}(T)];$ (6) $\displaystyle
h_{T}^{k,-}:\sum_{\rho\in\Sigma([0,[0)}a_{\rho}\phi_{\rho}\mapsto\sum_{\rho\in\Sigma([0,[0)}a_{\rho}\lambda_{\rho}(\mathrm{d}\lambda)_{\rho*}$
$\displaystyle[a_{\rho}=a_{\rho}(\lambda_{\rho(0)},\lambda_{\rho(0)+1},\dots,\lambda_{n})\in\mathcal{P}_{r-1}(T)].$
(7)

The set
$\\{(\mathrm{d}\lambda)_{\rho^{*}}\\}_{\rho\in\Sigma([0,[0),\rho(0)=0}$ is a
basis for $\mathrm{Alt}^{k}\mathbb{R}^{n}$, so $h_{T}^{k}$ can be computed
pointwise, but the set $\\{(\phi_{\rho})_{x}\\}_{\rho\in\Sigma([0,[0)}$ has
$\binom{n+1}{k}$ non-zero elements for $x$ in the interior of $T$ while
$|\mathrm{Alt}^{k}\mathbb{R}^{n}|=\binom{n}{k}$, so the values of the
coefficients $\\{a_{\rho}\\}_{\rho\in\Sigma([0,[0)}$ at a point are
overdetermined in eq. 7, making it appear that $h_{T}^{k,-}$ is not a
pointwise operator. We show that this appearance is deceptive because both
$h_{T}^{k}$ and $h_{T}^{k,-}$ are equivalent to the pointwise operator
$\mathring{\star}_{T}$ we define in section 3.3.

These maps directly lead to the constructive bases given for the trace-free
subspace in [3, 4] in the following way. In (6), we can express
$a_{\rho}\in\mathcal{P}_{r}(T)$ as a polynomial of $n+1$ arguments in the
barycentric coordinates, $a_{\rho}=a_{\rho}(\lambda_{0},\dots,\lambda_{n})$,
so that both (6) and (7) use polynomials of barycentric coordinates. If one
expands each $a_{\rho}$ in terms of a monomial basis of the barycentric
coordinates (which is equivalent up to a multiplicative constant to the
Bernstein polynomial basis), the right hand side of (6) defines a basis for
$\mathring{\mathcal{P}}_{r+k+1}^{-}\Lambda^{n-k}(T)$ where each basis function
is the product of a barycentric monomial and a Whitney form, and the right
hand side of (7) defines a basis for
$\mathring{\mathcal{P}}_{r+k}\Lambda^{n-k}(T)$ which each basis function is
the product of a barycentric monomial and an $(n-k)$-form
$(\mathrm{d}\lambda)_{\rho^{*}}$ for some $\rho\in\Sigma([0,[0)$.

###  3.2 The $\star_{T}$ operator

The duality between $\mathrm{Alt}^{k}\mathbb{R}^{n}$ and
$\mathrm{Alt}^{n-k}\mathbb{R}^{n}$ is clearly relevant to the isomorphisms (1)
and (2), but the Hodge star operator
$\star:\mathrm{Alt}^{k}\mathbb{R}^{n}\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathrm{Alt}^{n-k}\mathbb{R}^{n}$
does not appear to play any role. While this is surprising at first, it
becomes clear that the Hodge star cannot be used because it is not affine
invariant: given a bijection
$\phi:T\stackrel{{\scriptstyle\sim\,}}{{\to}}\hat{T}$ between simplices, in
general $\star\neq\phi^{*}\circ\star\circ\phi^{-*}$.

For each simplex $T$, we define a bijection between
$\mathrm{Alt}^{k}\mathbb{R}^{n}$ and $\mathrm{Alt}^{n-k}\mathbb{R}^{n}$ based
instead on the barycentric coordinates of $T$:

$\star_{T}:\omega\mapsto\frac{n!|T|}{\sqrt{n+1}}\sum_{\rho\in\Sigma([0,[0)}\star(\omega\wedge(\mathrm{d}\lambda)_{\rho})(\mathrm{d}\lambda)_{\rho}.$
(8)

######  Lemma 3.1.

$\star_{T}$ is a bijection,
$\star_{T}:\mathrm{Alt}^{k}\mathbb{R}^{n}\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathrm{Alt}^{n-k}\mathbb{R}^{n}$.

###### Proof.

We can define a symmetric bilinear form on $\mathrm{Alt}^{k}\mathbb{R}^{n}$,

$\langle\omega,\mu\rangle_{\star_{T}}:=\star(\omega\wedge\star_{T}\mu)=\frac{n!|T|}{\sqrt{n+1}}\sum_{\rho\in\Sigma([0,[0)}\star(\omega\wedge(\mathrm{d}\lambda)_{\rho})\
{\star(\mu\wedge(\mathrm{d}\lambda)_{\rho})}.$

If $\star_{T}\omega=0$, then $\langle\omega,\omega\rangle_{\star_{T}}=0$, and
so $\star(\omega\wedge(\mathrm{d}\lambda)_{\rho})=0$ for all
$\rho\in\Sigma([0,[0)$. These $(n-k)$-forms span
$\mathrm{Alt}^{n-k}\mathbb{R}^{n}$, and
$\mathrm{Alt}^{n-k}\mathbb{R}^{n}\cong(\mathrm{Alt}^{n}\mathbb{R}^{n})^{*}$,
so $\omega=0$. ∎

Because $\star_{T}$ is defined by a simplex instead of being universal, it can
have a useful form of affine invariance.

######  Lemma 3.2.

Given an affine bijection
$\phi:T\stackrel{{\scriptstyle\sim\,}}{{\to}}\hat{T}$,
$\star_{T}=(D\phi)^{*}\circ\star_{\hat{T}}\circ(D\phi)^{-*}$.

###### Proof.

If $\hat{\lambda}$ are the barycentric coordinates of $\hat{T}$, then
$(\mathrm{d}\hat{\lambda})_{\rho}=(D\phi)^{-*}(\mathrm{d}\lambda)_{\rho}$ and
$(D\phi)^{-*}\mathrm{vol}=(|T|/|\hat{T}|)\mathrm{vol}$. Hence

$\displaystyle((D\phi)^{*}\circ\star_{\hat{T}}\circ(D\phi)^{-*})(\omega)$
$\displaystyle=(D\phi)^{*}\left(\frac{n!|\hat{T}|}{\sqrt{n+1}}\sum_{\rho\in\Sigma([0,[0)}\star((D\phi)^{-*}\omega\wedge(\mathrm{d}\hat{\lambda})_{\rho})\
(\mathrm{d}\hat{\lambda})_{\rho}\right)$
$\displaystyle=(D\phi)^{*}\left(\frac{n!|\hat{T}|}{\sqrt{n+1}}\sum_{\rho\in\Sigma([0,[0)}\star((D\phi)^{-*}\omega\wedge(D\phi)^{-*}(\mathrm{d}\lambda)_{\rho})\
(D\phi)^{-*}(\mathrm{d}\lambda)_{\rho}\right)$
$\displaystyle=\frac{n!|\hat{T}|}{\sqrt{n+1}}\sum_{\rho\in\Sigma([0,[0)}\star(D\phi)^{-*}(\omega\wedge(\mathrm{d}\lambda)_{\rho})\
(\mathrm{d}\lambda)_{\rho}$
$\displaystyle=\frac{n!|T|}{\sqrt{n+1}}\sum_{\rho\in\Sigma([0,[0)}\star(\omega\wedge(\mathrm{d}\lambda)_{\rho})\
(\mathrm{d}\lambda)_{\rho}$ $\displaystyle=\mathring{\star}_{T}\omega.$

∎

######  Lemma 3.3.

If $T_{\mathrm{eq}}$ is an equilateral simplex with edge length $\sqrt{2}$,
then $\star_{T_{\mathrm{eq}}}=\star$.

###### Proof.

If we take the convention that the canonical basis of $\mathbb{R}^{n+1}$ is
numbered $e_{0},\dots,e_{n}$, then for every simplex the barycentric $k$-forms
are pullbacks of the coordinate $k$-forms,

$(\mathrm{d}\lambda)_{\rho}=(D\lambda)^{*}(\mathrm{d}x)_{\rho},\quad\rho\in\Sigma([0,[0).$

In the case of $T_{\mathrm{eq}}$, the Jacobian $D\lambda$ is isometric:
$\|(D\lambda)v\|_{2}=\|v_{2}\|$ for all $v\in\mathbb{R}^{n}$. This is clear
when we think of $\lambda$ as mapping $T$ to the standard barycentric simplex,
which is the simplex in the positive orthant of $\mathbb{R}^{n+1}$ connecting
the $e_{i}$ unit vectors, which is also equilateral and has the same edge
length.

Because $(D\lambda)$ is isometric, the
$\binom{n+1}{k+1}\times\binom{n+1}{k+1}$ matrix of inner products between
barycentric $k$-forms,

$M_{\rho,\sigma}:=\langle(\mathrm{d}\lambda)_{\rho},(\mathrm{d}\lambda)_{\sigma}\rangle=\langle(D\lambda)^{*}(\mathrm{d}x)_{\rho},(D\lambda)^{*}(\mathrm{d}x)_{\sigma}\rangle,\quad\rho,\sigma\in\Sigma([0,[0),$

is an orthogonal projection matrix for every $k$. This implies that the set
$\\{(\mathrm{d}\lambda)_{\rho}\\}_{\rho\in\Sigma([0,[0)}$ is a normalized
tight frame [6, Theorem 2.5], that is

$\omega=\sum_{\rho\in\Sigma([0,[0)}\langle\omega,(\mathrm{d}\lambda)_{\rho}\rangle(\mathrm{d}\lambda)_{\rho},\quad\omega\in\mathrm{Alt}^{k}\mathbb{R}^{n}.$

To compute $\star_{T_{\mathrm{eq}}}$ for
$\omega\in\mathrm{Alt}^{k}\mathbb{R}^{n}$, we apply the above fact to the
$(n-k)$-forms, and note that $|T_{\mathrm{eq}}|=\sqrt{n+1}/{n!}$ so that the
leading constant in (8) is cancelled, to see

$\star_{T_{\mathrm{eq}}}\omega=\sum_{\rho\in\Sigma([0,[0)}\star(\omega\wedge(\mathrm{d}\lambda)_{\rho})(\mathrm{d}\lambda)_{\rho}=\sum_{\rho\in\Sigma([0,[0)}\langle\star\omega,(\mathrm{d}\lambda)_{\rho}\rangle(\mathrm{d}\lambda)_{\rho}=\star\omega.$

∎

These lemmas imply that, like $\star$ itself,
$\star_{T}\circ\star_{T}=(-1)^{k(n-k)}$.

######  Corollary 3.4.

$\star_{T}\circ\star_{T}=(-1)^{k(n-k)}.$

###### Proof.

Let $\phi:T\stackrel{{\scriptstyle\sim\,}}{{\to}}T_{\mathrm{eq}}$ be an affine
bijection. Then

$\star_{T}\circ\star_{T}=(D\phi)^{*}\circ\star_{T_{\mathrm{eq}}}\circ(D\phi)^{-*}\circ(D\phi)^{*}\circ\star_{T_{\mathrm{eq}}}\circ(D\phi)^{-*}=(D\phi)^{*}\circ\star\circ\star\circ(D\phi)^{-*}=(-1)^{k(n-k)}.$

∎

###  3.3 The $\mathring{\star}_{T}$ operator

Having defined an isomorphism $\star_{T}$ between algebraic $k$\- and
$(n-k)$-form with the desired variance, we are now prepared to define an
similar isomorphism between differential $k$\- and trace-free $(n-k)$-forms. A
similar construction to (8) defines a linear operator
$\mathring{\star}_{T}:\Lambda^{k}(\overline{T})\to\mathring{\Lambda}^{n-k}(\overline{T})$
by multiplying the forms summed in (8) by their complementary bubble
functions. We define

$\mathring{\star}_{T}:\omega\mapsto
n!|T|\sum_{\rho\in\Sigma([0,[0)}\star(\omega\wedge(\mathrm{d}\lambda)_{\rho})\lambda_{\rho^{*}}(\mathrm{d}\lambda)_{\rho}.$
(9)

######  Lemma 3.5.

$\mathring{\star}_{T}$ is affine invariant: given affine bijection
$\phi:T\stackrel{{\scriptstyle\sim\,}}{{\to}}\hat{T}$,
$\mathring{\star}_{T}=\phi^{*}\circ\mathring{\star}_{\hat{T}}\circ\phi^{-*}.$

###### Proof.

The proof is essentially the same as for lemma 3.2. ∎

######  Lemma 3.6.

$\mathring{\star}_{T}:\Lambda^{k}(\overline{T})\to\mathring{\Lambda}^{n-k}(\overline{T})$
is an injection.

###### Proof.

To show that the range of $\mathring{\star}_{T}$ is in
$\mathring{\Lambda}^{n-k}(\overline{T})$ it is sufficient to show that
$\lambda_{\rho^{*}}(\mathrm{d}\lambda)_{\rho}\in\mathring{\Lambda}^{n-k}(\overline{T})$
for each $\rho\in\Sigma([0,[0)$. This is so because
$\mathrm{Tr}_{\sigma}\lambda_{\rho^{*}}=0$ if
$[\\![\rho^{*}]\\!]\not\subseteq[\\![\sigma]\\!]$ and
$\mathrm{Tr}_{\sigma}(\mathrm{d}\lambda)_{\rho}=0$ if
$[\\![\rho]\\!]\not\subseteq[\\![\sigma]\\!]$. Therefore
$\mathrm{Tr}_{\sigma}\lambda_{\rho^{*}}(\mathrm{d}\lambda)_{\rho}=0$ for any
$[\\![\sigma]\\!]\not\supseteq[\\![\rho]\\!]\cup[\\![\rho^{*}]\\!]$, that is
for any $f_{\sigma}\neq T$.

To show injectivity, as in lemma 3.1, we can define a symmetric bilinear form

$\langle\omega,\mu\rangle_{\mathring{\star}_{T}}:=\int_{T}\omega\wedge\mathring{\star}_{T}\mu=n!|T|\int_{T}\mathrm{vol}\sum_{\rho\in\Sigma([0,[0)}\lambda_{\rho^{*}}\
{\star(\omega\wedge(\mathrm{d}\lambda)_{\rho})}\
{\star(\mu\wedge(\mathrm{d}\lambda)_{\rho})}.$ (10)

If $\mathring{\star}_{T}\omega=0$, then
$\langle\omega,\omega\rangle_{\mathring{\star}_{T}}=0$, and for every $x$ in
$T$

$\sum_{\rho\in\Sigma([0,[0)}\lambda_{\rho^{*}}(\star(\omega_{x}\wedge(\mathrm{d}\lambda)_{\rho}))^{2}=0.$

For every $\rho\in\Sigma([0,[0)$ and every $x$ in the interior of $T$ where
$\lambda_{\rho^{*}}>0$ this implies that
$\star(\omega_{x}\wedge(\mathrm{d}\lambda)_{\rho})=0$ and so $\omega_{x}=0$.
By continuity, we conclude $\omega=0$. ∎

The operator $\mathring{\star}_{T}$ has an important property, analogous to
corollary 3.4, that applying $\mathring{\star}_{T}$ twice is like multiplying
by $(-1)^{k(n-k)}\lambda_{T}$.

######  Lemma 3.7.

$\mathring{\star}_{T}\circ\mathring{\star}_{T}=(-1)^{k(n-k)}\lambda_{T}$.

###### Proof.

Let $\omega\in\Lambda^{k}(\overline{T})$ be given and first expand
$\mathring{\star}_{T}\mathring{\star}_{T}\omega:$

$\mathring{\star}_{T}\mathring{\star}_{T}\omega=(n!|T|)^{2}\sum_{\begin{subarray}{c}\rho\in\Sigma([0,[0)\\\
\sigma\in\Sigma([0,[0)\end{subarray}}\lambda_{\rho^{*}}\lambda_{\sigma^{*}}\
{\star(\omega\wedge(\mathrm{d}\lambda)_{\rho})}\
{\star((\mathrm{d}\lambda)_{\rho}\wedge(\mathrm{d}\lambda)_{\sigma})}\
(\mathrm{d}\lambda)_{\sigma}.$

If $[\\![\rho]\\!]\cap[\\![\sigma]\\!]\neq\emptyset$, then
$(\mathrm{d}\lambda)_{\rho}\wedge(\mathrm{d}\lambda)_{\sigma}=0$. So for each
nonzero summand there is a map $\tau\in\Sigma([0,[0)$ and a map
$\hat{\rho}\in\Sigma([0,[0)$ such that $\rho=\tau\circ\hat{\rho}$ and
$\sigma=\tau\circ\hat{\rho}^{*}$. We use this fact to reorganize the sum,
noting that the union of the sets $[\\![(\tau\circ\rho)^{*}]\\!]$ and
$[\\![(\tau\circ\rho^{*})^{*}]\\!]$ is $[0$ and their intersection is
$[\\![\tau^{*}]\\!]$:

$\displaystyle\mathring{\star}_{T}\mathring{\star}_{T}\omega$
$\displaystyle=(n!|T|)^{2}\sum_{\begin{subarray}{c}\tau\in\Sigma([0,[0)\\\
\hat{\rho}\in\Sigma([0,[0)\end{subarray}}\lambda_{(\tau\circ\hat{\rho})^{*}}\lambda_{(\tau\circ\hat{\rho}^{*})^{*}}\
{\star(\omega\wedge(\mathrm{d}\lambda)_{\tau\circ\hat{\rho}})}\
{\star((\mathrm{d}\lambda)_{\tau\circ\hat{\rho}}\wedge(\mathrm{d}\lambda)_{\tau\circ\hat{\rho}^{*}})}\
(\mathrm{d}\lambda)_{\tau\circ\hat{\rho}^{*}}.$
$\displaystyle=\lambda_{T}\sum_{\tau\in\Sigma([0,[0)}\lambda_{\tau^{*}}\left\\{(n!|T|)^{2}\sum_{\hat{\rho}\in\Sigma([0,[0)}{\star(\omega\wedge(\mathrm{d}\lambda)_{\tau\circ\hat{\rho}})}\
{\star((\mathrm{d}\lambda)_{\tau\circ\hat{\rho}}\wedge(\mathrm{d}\lambda)_{\tau\circ\hat{\rho}^{*}})}\
(\mathrm{d}\lambda)_{\tau\circ\hat{\rho}^{*}}\right\\}.$

We show that the term in braces is $(-1)^{k(n-k)}\omega$ for each $\tau$.

Let us again number the canonical basis of $\mathbb{R}^{n}$ starting from
zero, $e_{0},\dots,e_{n-1}$, and let $T_{\mathrm{unit}}$ be the unit right
simplex in the positive orthant,
$T_{\mathrm{unit}}:=\\{x\in\mathbb{R}^{n}:x_{i}\geq
0,\sum_{i=0}^{n-1}x_{i}\leq 1\\}$. For each $\tau\in\Sigma([0,[0)$, define the
affine bijection
$\phi_{\tau}:T\stackrel{{\scriptstyle\sim\,}}{{\to}}T_{\mathrm{unit}}$ such
that the $\tau(i)$th vertex maps to $e_{i}$ for $i\in[0$ and the remaining
vertex maps to the origin. We note that because the volume of
$T_{\mathrm{unit}}$ is $1/n!$, the volume pullback is
$(D\phi_{\tau})^{*}\mathrm{vol}=\pm(1/(n!|T|))\mathrm{vol}$. We also note that
under $\phi_{\tau}$ the barycentric $k$-forms are pullbacks of coordinate
$k$-forms if their indices are subsets of $\tau$:
$(\mathrm{d}\lambda)_{\tau\circ\hat{\rho}}=(D\phi_{\tau})^{*}(\mathrm{d}x)_{\hat{\rho}}$
for each $\hat{\rho}\in\Sigma([0,[0)$. Last, we note that the Hodge star maps
a coordinate $k$-forms $(\mathrm{d}x)_{\hat{\rho}}$ to the coordinate
$(n-k)$-forms with complementary indices,
$\star(\mathrm{d}x)_{\hat{\rho}}=\pm(\mathrm{d}x)_{\hat{\rho}^{*}}$. Thus for
each $\tau\in\Sigma([0,[0)$, the term in braces above becomes

$\displaystyle(n!|T|)^{2}\sum_{\hat{\rho}\in\Sigma([0,[0)}{\star(\omega\wedge(\mathrm{d}\lambda)_{\tau\circ\hat{\rho}})}\
{\star((\mathrm{d}\lambda)_{\tau\circ\hat{\rho}}\wedge(\mathrm{d}\lambda)_{\tau\circ\hat{\rho}^{*}})}\
(\mathrm{d}\lambda)_{\tau\circ\hat{\rho}^{*}}$ $\displaystyle={}$
$\displaystyle(-1)^{k(n-k)}(n!|T|)^{2}\sum_{\hat{\rho}\in\Sigma([0,[0)}{\star(\omega\wedge(\mathrm{d}\lambda)_{\tau\circ\hat{\rho}})}\
{\star((\mathrm{d}\lambda)_{\tau\circ\hat{\rho}^{*}}\wedge(\mathrm{d}\lambda)_{\tau\circ\hat{\rho}})}\
(\mathrm{d}\lambda)_{\tau\circ\hat{\rho}^{*}}$ $\displaystyle={}$
$\displaystyle(-1)^{k(n-k)}(D\phi_{\tau})^{*}\sum_{\hat{\rho}\in\Sigma([0,[0)}{\star((D\phi_{\tau})^{-*}\omega\wedge(\mathrm{d}x)_{\hat{\rho}})}\
{\star((\mathrm{d}x)_{\hat{\rho}^{*}}\wedge(\mathrm{d}x)_{\hat{\rho}})}\
(\mathrm{d}x)_{\hat{\rho}^{*}}$ $\displaystyle={}$
$\displaystyle(-1)^{k(n-k)}(D\phi_{\tau})^{*}\sum_{\hat{\rho}\in\Sigma([0,[0)}\langle(D\phi_{\tau})^{-*}\omega,(\mathrm{d}x)_{\hat{\rho}^{*}}\rangle\langle(\mathrm{d}x)_{\hat{\rho}^{*}},(\mathrm{d}x)_{\hat{\rho}^{*}}\rangle(\mathrm{d}x)_{\hat{\rho}^{*}}$
$\displaystyle={}$
$\displaystyle(-1)^{k(n-k)}(D\phi_{\tau})^{*}(D\phi_{\tau})^{-*}\omega$
$\displaystyle={}$ $\displaystyle(-1)^{k(n-k)}\omega.$

Summing these contributions over all $\tau$, we get

$\mathring{\star}_{T}\mathring{\star}_{T}\omega=(-1)^{k(n-k)}\lambda_{T}\omega\sum_{\tau\in\Sigma([0,[0)}\lambda_{\tau^{*}}=(-1)^{k(n-k)}\lambda_{T}\omega.$

∎

We are now ready to prove the main results of this section, that
$\mathring{\star}_{T}$ implements the isomorphisms (1) and (2).

######  Theorem 3.8.

$\mathring{\star}_{T}:\mathcal{P}_{r}\Lambda^{k}(T)\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathring{\mathcal{P}}_{r+k+1}^{-}\Lambda^{n-k}(T).$

###### Proof.

Applying $\mathring{\star}_{T}$ to $(\mathrm{d}\lambda)_{\sigma}$ for some
$\sigma\in\Sigma([0,[0)$, each term in (9) where
$[\\![\rho]\\!]\cap[\\![\sigma]\\!]\neq\emptyset$ vanishes. The only nonzero
summands are for $\rho$ where
$[\\![\rho]\\!]=[\\![\sigma^{*}]\\!]\setminus\\{\sigma^{*}(j)\\}$ for some
$j$. Hence

$\displaystyle\mathring{\star}_{T}(\mathrm{d}\lambda)_{\sigma}$
$\displaystyle=n!|T|\sum_{j=0}^{n-k-1}\star((\mathrm{d}\lambda)_{\sigma}\wedge(\mathrm{d}\lambda)_{\sigma^{*}\setminus\sigma^{*}(j)})\lambda_{\sigma}\lambda_{\sigma^{*}(j)}(\mathrm{d}\lambda)_{\sigma^{*}\setminus\sigma^{*}(j)}$
$\displaystyle=(-1)^{k}(\mathop{\mathrm{sign}}\sigma)\lambda_{\sigma}\sum_{j=0}^{n-k-1}(-1)^{j}\lambda_{\sigma^{*}(j)}(\mathrm{d}\lambda)_{\sigma^{*}\setminus\sigma^{*}(j)}$
$\displaystyle=(-1)^{k}(\mathop{\mathrm{sign}}\sigma)\lambda_{\sigma}\phi_{\sigma^{*}}.$

As a result we conclude
$\mathring{\star}_{T}(\mathrm{d}\lambda)_{\sigma}\in\mathring{\mathcal{P}}_{k+1}^{-}\Lambda^{n-k}(T)$.
Each polynomial in $\mathcal{P}_{r}\Lambda^{k}(T)$ has a representation of the
form

$\sum_{\sigma\in\Sigma([0,[0)}a_{\sigma}(\mathrm{d}\lambda)_{\sigma},\quad
a_{\sigma}\in\mathcal{P}_{r}(T),$

so $\mathring{\star}_{T}$ maps $\mathcal{P}_{r}\Lambda^{k}(T)$ into
$\mathring{\mathcal{P}}_{r+k+1}^{-}\Lambda^{n-k}(T)$. By lemma 3.6,
$\mathring{\star}_{T}$ is injective, and it has already been established that
the spaces have the same dimension, so the operator is an isomorphism. ∎

######  Theorem 3.9.

$\mathring{\star}_{T}:\mathcal{P}_{r}^{-}\Lambda^{k}(T)\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathring{\mathcal{P}}_{r+k}\Lambda^{n-k}(T).$

###### Proof.

Let $\sigma\in\Sigma([0,[0)$ be given. From the proof of theorem 3.8, we have

$\phi_{\sigma}=(-1)^{n-k}\frac{\mathop{\mathrm{sign}}\sigma^{*}}{\lambda_{\sigma^{*}}}\mathring{\star}_{T}(\mathrm{d}\lambda)_{\sigma^{*}}.$

Applying $\mathring{\star}_{T}$ to both sides, we have

$\displaystyle\mathring{\star}_{T}\phi_{\sigma}$
$\displaystyle=(-1)^{n-k}\frac{\mathop{\mathrm{sign}}\sigma^{*}}{\lambda_{\sigma^{*}}}\mathring{\star}_{T}\mathring{\star}_{T}(\mathrm{d}\lambda)_{\sigma^{*}}$
$\displaystyle=(-1)^{(k+1)(n-k)}\frac{(\mathop{\mathrm{sign}}\sigma^{*})\lambda_{T}}{\lambda_{\sigma^{*}}}(\mathrm{d}\lambda)_{\sigma^{*}}$
$\displaystyle=(\mathop{\mathrm{sign}}\sigma)\lambda_{\sigma}(\mathrm{d}\lambda)_{\sigma^{*}},$

which is in $\mathcal{P}_{k+1}\Lambda^{n-k}(T)$. Therefore, as every function
in $\mathcal{P}_{r}^{-}\Lambda^{k}(T)$ has a representation of the form

$\sum_{\sigma\in\Sigma([0,[0)}a_{\sigma}\phi_{\sigma},$

where $a_{\sigma}\in\mathcal{P}_{r-1}(T)$, $\mathring{\star}_{T}$ maps
$\mathcal{P}_{r}^{-}\Lambda^{k}(T)$ into
$\mathring{\mathcal{P}}_{r+k}\Lambda^{n-k}(T)$ injectively, which concludes
the proof. ∎

We note that the proofs of theorems 3.8 and 3.9 show that
$\mathring{\star}_{T}$ differs from $h_{T}^{k}$ and $h_{T}^{k,-}$ only by sign
conventions.

To complete the claim that $\mathring{\star}_{T}$ induces an isomorphism
between $\Lambda^{k}(\overline{T})$ and
$\mathring{\Lambda}^{n-k}(\overline{T})$, we need to prove that
$\mathring{\star}_{T}:\Lambda^{k}(\overline{T})\to\mathring{\Lambda}^{n-k}(\overline{T})$
is a surjection, for which we require one additional lemma.

Let us define the space of $k$-forms that are not only trace-free but vanish
at the boundary,

$\mathring{C}^{\infty}(\overline{T};\mathrm{Alt}^{n-k}\mathbb{R}^{n}):=\\{\omega\in\Lambda^{n-k}(\overline{T}):\omega|_{\partial
T}=0\\}.$

It turns out that $\mathring{\star}_{T}$ maps
$\mathring{\Lambda}^{k}(\overline{T})$ into this space.

######  Lemma 3.10.

$\mathring{\star}_{T}:\mathring{\Lambda}^{k}(\overline{T})\to\mathring{C}^{\infty}(\overline{T};\mathrm{Alt}^{n-k}\mathbb{R}^{n})$.

###### Proof.

Let $x\in f_{\sigma}\neq T$ be given. If
$\omega\in\mathring{\Lambda}^{k}(\overline{T})$, then
$(\mathrm{Tr}_{\sigma}\omega)_{x}=(i_{\sigma}^{*}\omega)_{x}=(Di_{\sigma})^{*}\omega_{x}=0$.
We use the fact that the nullspace of $(Di_{\sigma})^{*}$ is spanned by
barycentric $k$-forms whose indices are not contained in $\sigma$,
$\\{(\mathrm{d}\lambda)_{\rho}\in\Sigma([0,[0):[\\![\rho]\\!]\not\subseteq[\\![\sigma]\\!]\\}$,
to say that $\omega_{x}$ is a linear combination of these $k$-forms. For such
$(\mathrm{d}\lambda)_{\rho}$, the proof of theorem 3.8 shows that

$(\mathring{\star}_{T}(\mathrm{d}\lambda)_{\rho})_{x}=(-1)^{k}(\mathop{\mathrm{sign}}\rho)(\lambda_{\rho}(\mathrm{d}\lambda)_{\rho^{*}})_{x}=0,$

a conclusion we reach because $(\lambda_{\rho})_{x}=0$ for $x\in f_{\sigma}$
and $[\\![\rho]\\!]\not\subseteq[\\![\sigma]\\!]$. Therefore
$(\mathring{\star}_{T}\omega)_{x}=0$ as well. ∎

We are now able to present our final theorem of this section.

######  Theorem 3.11.

$\mathring{\star}_{T}:\Lambda^{k}(\overline{T})\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathring{\Lambda}^{k}(\overline{T}).$

###### Proof.

Lemma 3.6 already shows that the map is an injection: it remains to show that
it is a surjection.

Let $\omega\in\Lambda^{k}(\overline{T})$ be given. By lemma 3.7,
$\mathring{\star}_{T}\mathring{\star}_{T}\omega=(-1)^{k(n-k)}\lambda_{T}\omega$,
so $\mathring{\star}_{T}\circ\mathring{\star}_{T}$ is a bijection,
$\mathring{\star}_{T}\circ\mathring{\star}_{T}:\Lambda^{k}(\overline{T})\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathring{C}^{\infty}(\overline{T};\mathrm{Alt}^{k}\mathbb{R}^{n})$.
Given $\mu\in\mathring{\Lambda}^{n-k}(\overline{T})$, by lemma 3.10 we can
define an element
$\omega:=(\mathring{\star}_{T}\circ\mathring{\star}_{T})^{-1}\mathring{\star}_{T}\mu\in\Lambda^{k}(\overline{T})$.
$\mathring{\star}_{T}\omega$ is in $\mathring{\Lambda}^{n-k}(\overline{T})$,
and $\mathring{\star}_{T}\mathring{\star}_{T}\omega=\mathring{\star}_{T}\mu$.
By the injectivity of $\mathring{\star}_{T}$,
$\mathring{\star}_{T}\omega=\mu$. ∎

###  3.4 Optimal dual basis functionals via the $\mathring{\star}_{T}$ inner
product

Because the operator $\mathring{\star}_{T}$ is defined pointwise, it can be
used in practice to adapt an unrestricted basis with desirable qualities –
some combination of numerical stability and computational efficiency – into a
basis for trace-free subspaces.

Implementations of the finite element method may also define the basis
functions indirectly, opting instead to define a unisolvent set of functionals
and constructing the basis functions by inverting the generalized vandermonde
matrix of a numerically stable basis for the primal space. The canonical basis
dual basis functionals for the trace free subspace
$\mathring{\mathcal{P}}_{r}\Lambda^{k}(T)$ given in [2, Sections 4.5] are

$\phi_{i}(\omega)=\int_{T}\omega\wedge\eta_{i},$

where $\\{\eta_{i}\\}$ is a basis of
$\mathcal{P}_{r-(n-k)}^{-}\Lambda^{k}(T)$; for
$\mathring{\mathcal{P}}_{r}^{-}\Lambda^{k}(T)$, they are the basis functionals
given in [2, Sections 4.6],

$\phi_{i}(\omega)=\int_{T}\omega\wedge\eta_{i},$

where $\\{\eta_{i}\\}$ is a basis of $\mathcal{P}_{r-(n-k)-1}\Lambda^{k}(T)$.

That work does not suggest which dual basis differential forms
$\\{\eta_{i}\\}$ to use in either case. The inner product
$\langle\cdot,\cdot\rangle_{\mathring{\star}_{T}}$ defined in (10) in the
proof of lemma~3.6 is useful for this purpose. If a basis $\\{\eta_{i}\\}$ is
chosen that is orthonormal with respect to this inner product, then the
corresponding basis functions can be computed directly as
$\\{\mathring{\star}_{T}\eta_{i}\\}$. For scalar polynomials, such orthonormal
polynomials can be evaluated explicitly using Dubiner-type combinations of
Gauss-Jacobi polynomials, such as in the basis of [5]. We do not investigate
the construction of an explicit orthonormal basis for
$\langle\cdot,\cdot\rangle_{\mathring{\star}_{T}}$ further in this work.

##  4 Extension operators

###  4.1 Consistent families of extension operators

In [3], the authors define a _consistent family of extension operators_ for
geometric decomposition. Letting $X$ stand in for $\mathcal{P}_{r}\Lambda^{k}$
or $\mathcal{P}_{r}^{-}\Lambda^{k}$, a consistent family of extension
operators is a set of operators $\\{E_{f,g}:X(f)\to
X(g),f,g\in\mathcal{T}_{h}\\}$ for any pair of simplices $f$ and $g$ in a mesh
$\mathcal{T}_{h}$, for which trace and extension operators commute in the
following sense: if $f$ and $g$ are boundary simplices of simplex $h$, and
$f\cap g$ is the intersection of their boundaries, then

$\mathrm{Tr}_{h,g}E_{f,h}\omega=E_{f\cap g,g}\mathrm{Tr}_{f,f\cap
g}\omega\quad\omega\in X(f).$ (11)

This property guarantees that the extension of a trace-free $k$-form
$\omega\in X(f_{\sigma})$ into neighboring cells has trace continuity at the
boundaries between the cells, and so it can serve as a local basis function in
an $H\Lambda^{k}(\Omega$)-conforming finite element space.

There are many ways to define consistent families of extension operators, but
in this work we only define extension operators that are affine invariant and
respect all of the symmetries of the simplex. This allows us to give our
definitions with respect to the single-element mesh made up of the simplex $T$
and its boundary simplices, and the results can be mapped to a general mesh
$\mathcal{T}_{h}$ with properties preserved. We also note that, for the
purpose of geometric decomposition, it is not necessary to design an extension
operator whose domain is all of $X(f)$, because the only $k$-forms that are
extended are in $\mathring{X}(f)$.

Using these facts, in this section we design a family of extension operators
$\\{\dot{E}_{\rho,\sigma}:\mathring{X}(f_{\rho})\to
X(f_{\sigma}),[\\![\rho]\\!],[\\![\sigma]\\!]\subseteq[0\\}$ such that for any
$f_{\tau}$ such that $[\\![\rho]\\!],[\\![\sigma]\\!]\subset[\\![\tau]\\!]$,

$\mathrm{Tr}_{\tau,\sigma}\dot{E}_{\rho,\tau}\omega=\dot{E}_{\rho\cap\sigma,\sigma}\mathrm{Tr}_{\rho,\rho\cap\sigma}\omega,\quad\omega\in
X(f_{\rho}).$

This is sufficient to show that one has the geometric decomposition

$X(T)=\bigoplus_{\sigma}\dot{E}_{\sigma,T}[\mathring{X}(f_{\sigma})],$

where we can take not only $X(f)=\mathcal{P}_{r}\Lambda^{k}(f)$ or
$X(f)=\mathcal{P}_{r}^{-}\Lambda^{k}(f)$, but even
$X=\Lambda^{k}(\overline{f})$.

###  4.2 Additional notation

We identify $k$-forms whose domain if $f_{\sigma}$ with a superscript, such as
$\lambda^{(\sigma)}$, $\mathrm{d}\lambda^{(\sigma)}$, $\phi_{\rho}^{(\tau)}$.

##### Centroid projectors.

For each $d$-simplex $f_{\sigma}$ in the boundary of $T$, the _centroid
projector_ $P_{T,\sigma}:T\to f_{\sigma}$ is an affine map defined in terms of
its actions on the vertices $\\{v_{i}\\}_{i=0}^{n}$ of $T$: if $i\in\sigma$,
then $P_{T,\sigma}(v_{i})=v_{i}$, otherwise $v_{i}$ is mapped to the centroid
of $f_{\sigma}$, $P_{T,\sigma}(v_{i})=\frac{1}{d}\sum_{i=0}^{d}v_{\sigma(i)}$.
Given two boundary simplices $f_{\sigma}$ and $f_{\rho}$, with
$[\\![\rho]\\!]\subset[\\![\sigma]\\!]$, then we define the centroid projector
$P_{\sigma,\rho}:f_{\sigma}\to f_{\rho}$ analogously. Centroid projectors
compose with each other, in that if
$[\\![\rho]\\!]\subseteq[\\![\sigma]\\!]\subseteq[\\![\tau]\\!]$, then

$P_{\sigma,\rho}\circ P_{\tau,\sigma}=P_{\tau,\rho}.$ (12)

The centroid project $P_{\sigma,\rho}$ is also a left inverse of the inclusion
map $i_{\rho,\sigma}:f_{\rho}\hookrightarrow f_{\sigma}$, which implies that
$\mathrm{Tr}_{\sigma,\rho}$ is a left inverse for $P_{\sigma,\rho}^{*}$.
Combining this fact with (12), we get

$P_{\sigma,\rho}^{*}=\mathrm{Tr}_{\tau,\sigma}\circ P_{\tau,\rho}^{*}.$ (13)

##### Centroid Koszul operators.

For each simplex $f_{\sigma}$, we let $\kappa^{(\sigma)}$ be the Koszul
operator centered at the centroid of $f_{\sigma}$. Because the centroid
projector $P_{\sigma,\rho}$ maps the centroid of $f_{\sigma}$ to the centroid
of $f_{\rho}$, (5) implies

$\kappa^{(\sigma)}P_{\sigma,\rho}^{*}=P_{\sigma,\rho}^{*}\kappa^{(\rho)}.$

###  4.3 Previous extension operators

In [3], the authors define two extension operators,
$E_{\sigma,T}^{r,k}:\mathcal{P}_{r}\Lambda^{k}(f_{\sigma})\to\mathcal{P}_{r}\Lambda^{k}(T)$
and
$E_{\sigma,T}^{r,k,-}:\mathcal{P}_{r}^{-}\Lambda^{k}(f_{\sigma})\to\mathcal{P}_{r}^{-}\Lambda^{k}(T)$,
and show that they are consistent extension operators that can be used in
geometric decompositions of their target spaces. Those extension operators are
defined as follows:

$\displaystyle
E_{\sigma,T}^{r,k}:(\lambda_{\sigma}^{(\sigma)})^{\alpha}(\mathrm{d}\lambda^{\sigma})_{\tau}$
$\displaystyle\mapsto(\lambda_{\sigma})^{\alpha}P_{T,\sigma,\alpha}^{*}(\mathrm{d}\lambda^{\sigma})_{\tau},$
$\displaystyle|\alpha|=r,\tau\in\Sigma([0,{[\\![\sigma]\\!]});$ (14)
$\displaystyle
E_{\sigma,T}^{r,k,-}:(\lambda_{\sigma}^{(\sigma)})^{\alpha}\phi_{\tau}^{(\sigma)}$
$\displaystyle\mapsto(\lambda_{\sigma})^{\alpha}\phi_{\tau},$
$\displaystyle|\alpha|=r-1,\tau\in\Sigma([0,{[\\![\sigma]\\!]}).$ (15)

The projector $P_{T,\sigma,\alpha}:T\to f_{\sigma}$ used in (14) differs from
the centroid projector in that the vertices that are not in $f_{\sigma}$ map
not to the centroid of $f_{\sigma}$, but to $x_{\alpha}$, a weighted average
of the vertices of $f_{\sigma}$ with weights determined by the multi-index
$\alpha$,

$x_{\alpha}=\frac{1}{|\alpha|}\sum_{i=0}^{d}\alpha(i)v_{\sigma(i)}.$

As was the case with the trace-free maps in section 3.1, these extension
operators are defined in terms of their action on the products of scalar
polynomials and barycentric $k$-forms (for full polynomials) or Whitney forms
(for trimmed). Unlike that case, though, these operators are mutually
incompatible, in that neither maps the other into the correct space.

We demonstrate their incompatibility when extending 1-forms from the triangle
$\sigma=(v_{1},v_{2},v_{3})$ to the tetrahedron $T=(v_{0},v_{1},v_{2},v_{3})$
by finding specific examples where a 1-form extends to the wrong space.

Example showing
$E_{\sigma,T}^{k,r,-}\mathring{\mathcal{P}}_{r-1}\Lambda^{k}(f_{\sigma})\not\to\mathcal{P}_{r-1}\Lambda^{k}(T)$.
In this case $k=1$ and $r=3$. Let
$\omega=\lambda_{1}^{(\sigma)}\lambda_{2}^{(\sigma)}\mathrm{d}\lambda_{3}^{(\sigma)}\in\mathring{\mathcal{P}}_{2}\Lambda^{1}(f_{\sigma})$.

We can expand $\omega$ in a Whitney form basis as
$\omega=\lambda_{1}^{(\sigma)}\lambda_{2}^{(\sigma)}(\phi_{23}^{(\sigma)}+\phi_{13}^{(\sigma)})$,
so applying $E_{\sigma,T}^{3,1,-}$ by the definition in (15) yields

$E_{\sigma,T}^{3,1,-}\omega=E_{\sigma,T}^{3,1,-}(\lambda_{1}^{(\sigma)}\lambda_{2}^{(\sigma)}(\phi_{23}^{(\sigma)}+\phi_{13}^{(\sigma)}))=\lambda_{1}\lambda_{2}(\phi_{23}+\phi_{13}).$

On the tetrahedron $\phi_{23}+\phi_{13}\neq\mathrm{d}\lambda_{3}$, so
$E_{\sigma,T}^{3,1,-}\omega\not\in\mathcal{P}_{2}\Lambda^{1}(T).$

Example showing
$E_{\sigma,T}^{k,r}\mathring{\mathcal{P}}_{r}^{-}\Lambda^{k}(f_{\sigma})\not\to\mathcal{P}_{r}^{-}\Lambda^{k}(T)$.
Let
$\omega=\lambda_{1}^{(\sigma)}\lambda_{2}^{(\sigma)}\phi_{23}^{(\sigma)}\in\mathring{\mathcal{P}}_{3}^{-}\Lambda^{1}(f_{\sigma})$
(in this case again $k=1$ and $r=3$). Expanded in a barycentric 1-form basis,
$\omega$ is
$\lambda_{1}^{(\sigma)}(\lambda_{2}^{(\sigma)})^{2}\mathrm{d}\lambda_{3}^{(\sigma)}-\lambda_{1}^{(\sigma)}\lambda_{2}^{(\sigma)}\lambda_{3}^{(\sigma)}\mathrm{d}\lambda_{2}^{(\sigma)}.$
According to (14), $E_{\sigma,T}^{3,1}$ extends the two 1-forms in this
representation of $\omega$ by the pullbacks of two different projections,
$P_{T,\sigma,(1,2,0)}$ and $P_{T,\sigma,(1,1,1)}$, determined by their
barycentric monomials. The first projection $P_{T,\sigma,(1,2,0)}$ can be
described in terms of barycentric coordinates as

$(\lambda_{1}^{(\sigma)},\lambda_{2}^{(\sigma)},\lambda_{3}^{(\sigma)})\leftarrow(\lambda_{1}+(1/3)\lambda_{0},\lambda_{2}+(2/3)\lambda_{0},\lambda_{3}),$

so
$P_{T,\sigma,(1,2,0)}^{*}\mathrm{d}\lambda_{3}^{(\sigma)}=\mathrm{d}\lambda_{3}.$
The second projection $P_{T,\sigma,(1,1,1)}$ is

$(\lambda_{1}^{(\sigma)},\lambda_{2}^{(\sigma)},\lambda_{3}^{(\sigma)})\leftarrow(\lambda_{1}+(1/3)\lambda_{0},\lambda_{2}+(1/3)\lambda_{0},\lambda_{3}+(1/3)\lambda_{0}),$

so
$P_{T,\sigma,(1,1,1)}^{*}\mathrm{d}\lambda_{2}^{(\sigma)}=\mathrm{d}\lambda_{2}+(1/3)\mathrm{d}\lambda_{0}.$
All together, this shows that the extension of $\omega$ is

$E_{T,\sigma}^{3,1}\omega=\lambda_{1}\lambda_{2}^{2}\mathrm{d}\lambda_{3}-\lambda_{1}\lambda_{2}\lambda_{3}(\mathrm{d}\lambda_{2}+(1/3)\mathrm{d}\lambda_{0}).$

The easiest way to show that this is not in
$\mathcal{P}_{3}^{-}\Lambda^{1}(T)$ is to apply the Koszul operator because
$\kappa_{x}p\in\mathcal{P}_{3}\Lambda^{0}(T)$ for every
$p\in\mathcal{P}_{3}^{-}\Lambda^{1}(T)$ and every $x$. By [2, Theorem 3.1] we
have
$\kappa_{x}(\lambda^{\alpha}\mathrm{d}\lambda_{i})=\lambda^{\alpha}(\lambda_{i}-\lambda_{i}(x))$
for each multi-index $\alpha$ and each $i\in[0$. Choosing $x=v_{1}$ where
$\lambda_{0}(v_{1})=\lambda_{2}(v_{i})=\lambda_{3}(v_{1})=0$ gives us

$\kappa_{v_{1}}E_{T,\sigma}^{3,1}\omega=-(1/3)\lambda_{0}\lambda_{1}\lambda_{2}\lambda_{3}\not\in\mathcal{P}_{3}\Lambda^{0}(T).$

###  4.4 The bubble decomposition of $\mathring{\Lambda}^{k}(\overline{T})$

Even though each $\omega\in\mathring{\Lambda}^{k}(\overline{T})$ is trace-
free, $\omega_{x}$ for $x\in\partial T$ is not necessarily zero because there
can be components of $\omega_{x}$ that are perpendicular to the trace operator
$\mathrm{Tr}$ at $x$. In this section we define a way to decompose a trace-
free $k$-form based on these perpendicular traces called the _bubble
decomposition_. This decomposition is the foundation of the unified extension
operator we define below. We present the bubble decomposition for the
$n$-simplex $T$: the definition extends naturally to
$\mathring{\Lambda}^{k}(\overline{f_{\sigma}})$ for each boundary simplex
$f_{\sigma}$.

Let $k\in[0$ be given. For each $d\geq n-k$ and each $\sigma\in\Sigma([0,[0)$,
let $\omega_{\sigma}$ be a $(k-(n-d))$-form in
$\Lambda^{k-(n-d)}(\overline{f_{\sigma}})$. Define
$\mathring{E}_{\sigma,T}:\Lambda^{k-(n-d)}(\overline{f_{\sigma}})\to\mathring{\Lambda}^{k}(\overline{T})$
by

$\mathring{E}_{\sigma,T}:\omega_{\sigma}\mapsto
P_{T,\sigma}^{*}\omega_{\sigma}\wedge(\lambda_{\sigma}(\mathrm{d}\lambda)_{\sigma^{*}}).$
(16)

The range of $\mathring{E}_{\sigma,T}$ is trace-free because the form
$\lambda_{\sigma}(\mathrm{d}\lambda)_{\sigma^{*}}$ appearing in the right hand
side of (16) is trace-free and because trace operators distribute over the
wedge product. We first show that $\mathring{E}_{\sigma,T}$ is an injection.

######  Lemma 4.1.

$\mathring{E}_{\sigma,T}$ defined by (16) is injective.

###### Proof.

Let $\hat{\Sigma}=\Sigma([1,[\\![\sigma\setminus\sigma(0)]\\!])$. We note that
$\\{(\mathrm{d}\lambda^{(\sigma)})_{\hat{\rho}}\\}_{\hat{\rho}\in\hat{\Sigma}}$
is a basis for $\mathrm{Alt}^{k-(n-d)}T_{x}f_{\sigma}$, so it is sufficient to
show that
$\mathring{E}_{\sigma,T}\alpha(\mathrm{d}\lambda^{(\sigma)})_{\hat{\rho}}\neq
0$ for every nonzero $\alpha\in C^{\infty}(\overline{f_{\sigma}})$ and every
$\hat{\rho}\in\hat{\Sigma}$.

We note that the pullback of a barycentric $1$-form by $(DP_{T,\sigma})^{*}$
is

$(DP_{T,\sigma})^{*}\mathrm{d}\lambda_{i}^{(\sigma)}=\mathrm{d}\lambda_{i}+\frac{1}{d+1}\sum_{j\in[\\![\sigma^{*}]\\!]}\mathrm{d}\lambda_{j},\quad
i\in[\\![\sigma]\\!].$

Because $\mathrm{d}\lambda_{j}\wedge\mathrm{d}\lambda_{\sigma^{*}}=0$ if and
only if $j\in[\\![\sigma^{*}]\\!]$, this implies
$(DP_{T,\sigma})^{*}\mathrm{d}\lambda_{i}^{(\sigma)}\wedge\mathrm{d}\lambda_{\sigma^{*}}=\mathrm{d}\lambda_{i}\wedge\mathrm{d}\lambda_{\sigma^{*}}$
for each $i\in[\\![\sigma]\\!]$, and in general

$(DP_{T,\sigma})^{*}(\mathrm{d}\lambda^{(\sigma)})_{\hat{\rho}}\wedge\mathrm{d}\lambda_{\sigma^{*}}=(\mathrm{d}\lambda)_{\hat{\rho}}\wedge\mathrm{d}\lambda_{\sigma^{*}},\quad[\\![\hat{\rho}]\\!]\subseteq[\\![\sigma]\\!].$

Therefore

$\displaystyle\mathring{E}_{\sigma,T}\alpha(\mathrm{d}\lambda^{(\sigma)})_{\hat{\rho}}$
$\displaystyle=P_{T,\sigma}^{*}(\alpha(\mathrm{d}\lambda^{(\sigma)})_{\hat{\rho}})\wedge(\lambda_{\sigma}(\mathrm{d}\lambda)_{\sigma^{*}})$
$\displaystyle=(\lambda_{\sigma}P_{T,\sigma}^{*}\alpha)\
((DP_{T,\sigma})^{*}(\mathrm{d}\lambda^{(\sigma)})_{\hat{\rho}}\wedge(\mathrm{d}\lambda)_{\sigma^{*}})$
$\displaystyle=(\lambda_{\sigma}P_{T,\sigma}^{*}\alpha)\
((\mathrm{d}\lambda)_{\hat{\rho}}\wedge(\mathrm{d}\lambda)_{\sigma^{*}}).$

Neither the scalar term nor the algebraic $k$-form term is identically zero,
so
$\mathring{E}_{\sigma,T}\alpha(\mathrm{d}\lambda^{(\sigma)})_{\hat{\rho}}\neq
0$. ∎

The following lemma shows how $\mathring{E}_{\sigma,T}$ can represent the
components of $\omega_{x}$ perpendicular to $\mathrm{Tr}_{\sigma}$ for $x\in
f_{\sigma}$ and $\omega\in\mathring{\Lambda}^{k}(\overline{T})$.

######  Lemma 4.2.

Let $\omega\in\mathring{\Lambda}^{k}(\overline{T})$ be given and let a
$d$-simplex $f_{\sigma}$ be given such that $d\geq n-k$. Suppose that $\omega$
vanishes at the boundaries of $f_{\sigma}$, that is $\omega|_{f_{\tau}}=0$ for
every $\tau$ such that $[\\![\tau]\\!]\subsetneq[\\![\sigma]\\!]$. Then there
is a unique $\omega_{\sigma}\in\Lambda^{k-(n-d)}(\overline{f_{\sigma}})$ such
that

$\omega|_{f_{\sigma}}=\mathring{E}_{\sigma,T}(\omega_{\sigma})|_{f_{\sigma}}.$

###### Proof.

By theorem 3.11, there exists $\mu\in\Lambda^{n-k}(\overline{T})$ such that
$\omega=\mathring{\star}_{T}\mu$, so for $x\in f_{\sigma}$,

$\displaystyle\omega_{x}$
$\displaystyle=n!|T|\sum_{\rho\in\Sigma([0,[0)}\star(\mu_{x}\wedge(\mathrm{d}\lambda)_{\rho})\
(\lambda_{\rho^{*}})_{x}(\mathrm{d}\lambda)_{\rho}$
$\displaystyle=n!|T|\sum_{\begin{subarray}{c}\rho\in\Sigma([0,[0)\\\
[\\![\rho^{*}]\\!]\subseteq[\\![\sigma]\\!]\end{subarray}}\star(\mu_{x}\wedge(\mathrm{d}\lambda)_{\rho})\
(\lambda_{\rho^{*}})_{x}(\mathrm{d}\lambda)_{\rho},$

where we have eliminated terms from the sum that are zero because
$\lambda_{\rho^{*}}|_{f_{\sigma}}=0.$ But if $\rho\in\Sigma([0,[0)$ and
$[\\![\rho^{*}]\\!]\subseteq[\\![\sigma]\\!]$, then there is
$\hat{\rho}\in\Sigma([0,[\\![\sigma]\\!])$ such that
$(\mathrm{d}\lambda)_{\rho}=\pm(\mathrm{d}\lambda)_{\hat{\rho}}\wedge(\mathrm{d}\lambda)_{\sigma^{*}}.$
Therefore there exist some smooth functions $\\{\alpha_{\hat{\rho}}\\}\subset
C^{\infty}(\overline{f_{\sigma}})$ such that

$\omega|_{f_{\sigma}}=\sum_{\hat{\rho}\in\Sigma([0,[\\![\sigma]\\!])}\alpha_{\hat{\rho}}((\mathrm{d}\lambda)_{\hat{\rho}}\wedge(\mathrm{d}\lambda)_{\sigma^{*}}).$

By the same argument from the proof of lemma 4.1, we have
$(\mathrm{d}\lambda)_{\hat{\rho}}\wedge(\mathrm{d}\lambda)_{\sigma^{*}}=P_{T,\sigma}^{*}(\mathrm{d}\lambda^{(\sigma)})_{\hat{\rho}}\wedge(\mathrm{d}\lambda)_{\sigma^{*}}$,
so

$\omega|_{f_{\sigma}}=\sum_{\hat{\rho}\in\Sigma([0,[\\![\sigma]\\!])}\alpha_{\hat{\rho}}(P_{T,\sigma}^{*}(\mathrm{d}\lambda^{(\sigma)})_{\hat{\rho}}\wedge(\mathrm{d}\lambda)_{\sigma^{*}}).$

Finally, we stipulated that $\omega$ vanishes at the boundaries of
$f_{\sigma}$, so for each $\alpha_{\hat{\rho}}$ there is
$\tilde{\alpha}_{\hat{\rho}}\in C^{\infty}(\overline{f_{\sigma}})$ such that
$\alpha_{\hat{\rho}}=\lambda_{\sigma}\tilde{\alpha}_{\hat{\rho}}$, and so

$\displaystyle\omega|_{f_{\sigma}}$
$\displaystyle=P_{T,\sigma}^{*}\big{(}\sum_{\hat{\rho}\in\Sigma([0,[\\![\sigma]\\!])}\tilde{\alpha}_{\hat{\rho}}(\mathrm{d}\lambda^{(\sigma)})_{\hat{\rho}}\big{)}\wedge(\lambda_{\sigma}(\mathrm{d}\lambda)_{\sigma^{*}})$
$\displaystyle=\mathring{E}_{\sigma,T}\big{(}\sum_{\hat{\rho}\in\Sigma([0,[\\![\sigma]\\!])}\tilde{\alpha}_{\hat{\rho}}(\mathrm{d}\lambda^{(\sigma)})_{\hat{\rho}}\big{)}\big{|}_{f_{\sigma}}.$

The uniqueness follows from lemma 4.1. ∎

Having established a way to encode the perpendicular component of
$\omega|_{f_{\sigma}}$ for a single boundary simplex $f_{\sigma}$ using
$\mathring{E}_{\sigma,T}$, we now construct the full bubble decomposition of a
trace-free $k$-form.

We denote with $Z^{k}(T)$ the set of increasing maps $\sigma$ for which
$\mathring{E}_{\sigma,T}$ is defined,

$Z^{k}(T):=\bigcup_{d=n-k}^{n}\Sigma([0,[0),$

and use $\sigma\in Z^{k}(T)$ to index the _$k$ -bubble trace space_
$\mathbf{B}^{k}(T)$, the product of the spaces for which
$\mathring{E}_{\sigma,T}$ is defined,

$\mathbf{B}^{k}(T):=\bigotimes_{\sigma\in
Z^{k}(T)}\Lambda^{k-(n-\dim(f_{\sigma}))}(\overline{f_{\sigma}}).$

Finally, we define
$\mathring{E}_{T}:\mathbf{B}^{k}(T)\to\mathring{\Lambda}^{k}(\overline{T})$ to
be the sum of the extensions of each of the components of the product space:
given $W=\otimes_{\sigma\in Z^{k}(T)}\omega_{\sigma}$,

$\mathring{E}_{T}:W\mapsto\sum_{\sigma\in
Z^{k}(T)}\mathring{E}_{\sigma,T}\omega_{\sigma}.$ (17)

######  Theorem 4.3.

$\mathring{E}_{T}$ is a bijection,
$\mathring{E}_{T}:\mathbf{B}^{k}(T)\stackrel{{\scriptstyle\sim\,}}{{\to}}\Lambda^{k}(\overline{T}).$

###### Proof.

Lemma 4.1 showed that $\mathring{E}_{\sigma,T}$ is injective for each
individual boundary simplex $f_{\sigma}$, but we must show that the sum of
contributions from different boundary simplices is still injective.

Suppose $\omega=\mathring{E}_{T}W=\sum_{\sigma\in
Z^{k}(T)}\mathring{E}_{\sigma,T}\omega_{\sigma}=0.$ Let $d$ be the smallest
dimension such that $\omega_{\sigma}=0$ for every $\sigma\in Z^{k}(T)$ such
that $\dim(f_{\sigma})<d$, and choose $\hat{\sigma}\in Z^{k}(T)$ such that
$\dim(f_{\hat{\sigma}})=d$. Given $\tau\in Z^{k}(T)$ such that
$\tau\neq\sigma$ and $\dim(f_{\tau})\geq d$, we have

$(\mathring{E}_{\tau,T}\omega_{\tau})|_{f_{\hat{\sigma}}}=(P_{\tau,T}^{*}\omega)\wedge(\lambda_{\tau}(\mathrm{d}\lambda)_{\tau^{*}})|_{f_{\hat{\sigma}}}=0,$
(18)

a conclusion we reach because
$[\\![\tau]\\!]\not\subseteq[\\![\hat{\sigma}]\\!]$, and so
$\lambda_{\tau}|_{f_{\hat{\sigma}}}=0.$ Therefore

$0=\omega|_{f_{\sigma}}=(\mathring{E}_{T}W)|_{f_{\sigma}}=(\mathring{E}_{\hat{\sigma},T}\omega_{\hat{\sigma}})|_{f_{\sigma}}.$

By the same argument as in lemma 4.1, this implies $\omega_{\hat{\sigma}}=0.$
Because $\hat{\sigma}$ was arbitrarily chosen, $\omega_{\sigma}=0$ for each
$d$-simplex $f_{\sigma}$, but this violates the way $d$ was chosen, so we must
conclude $W=0$.

Now we prove that $\mathring{E}_{T}$ is surjective. Suppose
$\omega\in\Lambda^{k}(\overline{T})$ vanishes at $\partial T$. Then there
exists $\tilde{\omega}\in\Lambda^{k}(\overline{T})$ such that
$\omega=\lambda_{T}\tilde{\omega}=\mathring{E}_{T,T}\tilde{\omega}.$

Now suppose that we have shown that the range of $\mathring{E}_{T}$ includes
all $k$-forms that vanish at all $f_{\sigma}$ with $\dim(f_{\sigma})\leq d$,
and let $\omega\in\Lambda^{k}$ be a $k$-form that vanishes at each
$f_{\sigma}$ with $\dim(f_{\sigma})<d.$ For each $\hat{\sigma}$ such that
$\dim(f_{\hat{\sigma}})=d$ there is, by lemma 4.2,
$\hat{\omega}_{\hat{\sigma}}\in\Lambda^{k-(n-d)}(\overline{f_{\hat{\sigma}}})$
such that
$(\mathring{E}_{\hat{\sigma},T}\hat{\omega}_{\hat{\sigma}})|_{f_{\hat{\sigma}}}=\omega|_{f_{\hat{\sigma}}}.$
Let

$\tilde{\omega}=\omega-\sum_{\hat{\sigma}\in\Sigma([0,[0)}\mathring{E}_{\hat{\sigma},T}\hat{\omega}_{\hat{\sigma}}.$

By (18), $\tilde{\omega}$ vanishes at all $d$ simplices, so there exists
$\tilde{W}=\otimes_{\sigma\in Z^{k}(T)}\tilde{\omega}_{\sigma}$ such that
$\tilde{\omega}=\mathring{E}_{T}\tilde{W}$. Hence defining
$W=\otimes_{\sigma\in Z^{k}(T)}\omega_{\sigma}$ by

$\omega_{\sigma}=\begin{cases}\tilde{\omega}_{\sigma},&\dim(f_{\sigma})\neq
d,\\\
\tilde{\omega}_{\sigma}-\hat{\omega}_{\sigma},&\dim(f_{\sigma})=d,\end{cases}$

we have constructed $W$ such that $\mathring{E}_{T}W=\omega$. By induction, we
conclude $\mathring{E}_{T}$ is surjective. ∎

The proof of the bijection
$\mathring{E}_{T}:\mathbf{B}^{k}(T)\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathring{\Lambda}^{k}(\overline{T})$
shows that $\mathring{E}_{\sigma,T}$ operators define a decomposition of
$\mathring{\Lambda}(\overline{T})$,

$\mathring{\Lambda}^{k}(\overline{T})=\bigoplus_{\sigma\in
Z^{k}(T)}\mathring{E}_{\sigma,T}[\Lambda^{k-(n-d)}(\overline{f}_{\sigma})].$
(19)

We define the inverse of $\mathring{E}_{T}$ to be the operator
$\dot{B}_{T}^{k}:\mathring{\Lambda}^{k}(\overline{T})\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathbf{B}^{k}(T)$,
and we let
$\dot{B}_{T,\sigma}^{k}:\mathring{\Lambda}^{k}(\overline{T})\stackrel{{\scriptstyle\sim\,}}{{\to}}\Lambda^{k-(n-d)}(\overline{f_{\sigma}})$
be the projection of $\dot{B}_{T,\sigma}^{k}$ onto the $\sigma$ component. We
have two remarks about this operator.

First, the proof of theorem 4.3 shows a constructive procedure for evaluating
$(B_{T,\sigma}^{k}\omega)_{x}$. One must evaluate $\omega$ at $P_{T,\rho}(x)$
for each simplex $f_{\rho}$ such that
$[\\![\rho]\\!]\subseteq[\\![\sigma]\\!]$, and solve the system

$\omega_{P_{T,\rho}(x)}=\lambda_{\rho}(B_{T,\rho}^{k}\omega)_{P_{T,\rho}(x)}+\sum_{\begin{subarray}{c}\hat{\rho}\in
Z^{k}(T)\\\
[\\![\hat{\rho}]\\!]\subset[\\![\rho]\\!]\end{subarray}}(\mathring{E}_{\hat{\rho},T}B_{T,\hat{\rho}}^{k}\omega)_{P_{T,\rho}(x)},\quad[\\![\rho]\\!]\subseteq[\\![\sigma]\\!].$

This is a triangular system that can be solved from the smallest dimension to
the largest.

The second remark is that, while we have shown that this bubble decomposition
exists for all of $\mathring{\Lambda}^{k}(\overline{T})$, it cannot be used to
define a basis for every subspace $X(T)\subset\Lambda^{k}(\overline{T})$. On
the one hand, it can be used to define a basis for the trace-free full
polynomials $\mathring{\mathcal{P}}_{r}\Lambda^{k}(T),$

$\mathring{\mathcal{P}}_{r}\Lambda^{k}(T)=\bigoplus_{\sigma\in
Z^{k}(T)}\mathring{E}_{\sigma,T}[\mathcal{P}_{r-\dim(f_{\sigma})-1}\Lambda^{k-(n-\dim(f_{\sigma}))}(f_{\sigma})].$
(20)

On the other hand, the components of a trace-free trimmed $k$-form
$\omega\in\mathring{\mathcal{P}}_{r}^{-}\Lambda^{k}(\overline{T})$ will not
necessarily be in the same space. As a simple example, consider
$\lambda_{0}\phi_{12}\in\mathring{\mathcal{P}}_{2}^{-}\Lambda^{1}(T)$ for the
triangle $T=\\{v_{0},v_{1},v_{2}\\}$. Its bubble decomposition is

$\lambda_{0}\phi_{12}=\lambda_{0}\lambda_{2}\mathrm{d}\lambda_{1}-\lambda_{0}\lambda_{1}\mathrm{d}\lambda_{2}=\underbrace{P_{T,(0,2)}^{*}(1)\wedge(\lambda_{0}\lambda_{2}\mathrm{d}\lambda_{1})}_{\mathring{E}_{(0,2),T}(1)}+\underbrace{P_{T,(0,1)}^{*}(-1)\wedge(\lambda_{0}\lambda_{1}\mathrm{d}\lambda_{2})}_{\mathring{E}_{(0,1),T}(-1)},$

and each of its bubble components is in $\mathcal{P}_{2}\Lambda^{1}(T)$ but
not $\mathcal{P}_{2}^{-}\Lambda^{1}(T)$.

###  4.5 The $\dot{E}_{\sigma,T}$ operator

We can generalize the bubble decomposition from the previous section to
boundary simplices. Given $\sigma$ and $\xi$ such that
$[\\![\sigma]\\!]\subseteq[\\![\xi]\\!]$, we can define
$\mathring{E}_{\sigma,\xi}:\Lambda^{k-(\dim(f_{\xi})-\dim(f_{\sigma}))}(\overline{f_{\sigma}})\to\mathring{\Lambda}^{k}(\overline{f_{\xi}})$
by

$\mathring{E}_{\sigma,\xi}:\omega_{\sigma}\mapsto
P_{\xi,\sigma}^{*}\omega_{\sigma}\wedge(\lambda_{\sigma}^{(\xi)}(\mathrm{d}\lambda^{(\xi)})_{\xi\setminus\sigma}),$
(21)

and we have a similar decomposition result to (19),

$\mathring{\Lambda}^{k}(\overline{f_{\xi}})=\bigoplus_{\sigma\in
Z^{k}(f_{\xi})}\mathring{E}_{\sigma,\xi}[\Lambda^{k-(\dim(f_{\xi})-\dim(f_{\sigma}))}(\overline{f}_{\sigma})].$

We use the bubble decomposition to define the extension operator
$\dot{E}_{\tau,\xi}:\mathring{\Lambda}^{k}(\overline{f_{\sigma}})\to\Lambda^{k}(\overline{f_{\xi}})$
when $[\\![\tau]\\!]\subseteq[\\![\xi]\\!]$ by its action on the bubble
components:

$\dot{E}_{\tau,\xi}:\mathring{E}_{\sigma,\tau}\omega_{\sigma}\mapsto
P_{\xi,\sigma}^{*}\omega_{\sigma}\wedge(\lambda_{\sigma}^{(\xi)}\wedge(\mathrm{d}\lambda^{(\xi)})_{\tau\setminus\sigma}).$
(22)

Note the difference between (21) and (22) is only in the barycentric form
wedged with the pullback: for
$\dot{E}_{\tau,\xi}\mathring{E}_{\sigma,\tau}\omega_{\sigma}$ it is
$(\mathrm{d}\lambda^{\xi})_{\tau\setminus\sigma}$, so that the product
$\lambda_{\sigma}^{(\xi)}(\mathrm{d}\lambda^{(\xi)})_{\tau\setminus\sigma}$ is
_not_ trace-free on $f_{\xi}$, but has nonzero trace on any boundary simplex
of $f_{\xi}$ that contains $f_{\tau}$.

######  Theorem 4.4.

Let a family of extension operators be defined by

$\\{\dot{E}_{\tau,\xi}:\mathring{\Lambda}^{k}(\overline{f_{\tau}})\to\Lambda^{k}(\overline{f_{\xi}}),[\\![\tau]\\!]\subseteq[\\![\xi]\\!]\\},$

where $\dot{E}_{\tau,\xi}$ is defined (22). The family is a consistent family,
that is for simplices $f_{\rho}$, $f_{\tau}$ and $f_{\xi}$ such that
$[\\![\rho]\\!],[\\![\tau]\\!]\subseteq[\\![\xi]\\!]$, we have

$\mathrm{Tr}_{\xi,\rho}\dot{E}_{\tau,\xi}\omega=\dot{E}_{\rho\cap\tau,\rho}\mathrm{Tr}_{\tau,\rho\cap\tau}\omega,\quad\omega\in\mathring{\Lambda}^{k}(\overline{f_{\sigma}}).$
(23)

###### Proof.

Let $\xi$ be a $d$-simplex and let $\tau$ be a boundary simplex,
$[\\![\tau]\\!]\subseteq[\\![\xi]\\!]$. Let $f_{\sigma}$ be any boundary
simplex of $f_{\tau}$ such that $\dim(f_{\sigma})\geq\dim(f_{\tau})-k$, and
let
$\omega_{\sigma}\in\Lambda^{k-(\dim(f_{\tau})-\dim(f_{\sigma}))}(\overline{f_{\sigma}})$
be given. $\mathring{E}_{\sigma,\tau}\omega_{\sigma}$ is a $k$-form in
$\mathring{\Lambda}^{k}(\overline{f_{\sigma}})$.

First let a simplex $f_{\rho}$ be given such that
$[\\![\tau]\\!]\not\subseteq[\\![\rho]\\!]$. Because
$\mathring{E}_{\sigma,\tau}\omega_{\sigma}$ is trace-free,
$\mathrm{Tr}_{\tau,\rho\cap\tau}\mathring{E}_{\sigma,\tau}\omega_{\sigma}=0$,
and so

$\dot{E}_{\rho\cap\tau,\rho}\mathrm{Tr}_{\tau,\rho\cap\tau}\mathring{E}_{\sigma,\tau}\omega_{\sigma}=0.$

By the definition in (22), we see that
$\mathrm{Tr}_{\xi,\rho}\dot{E}_{\tau,\xi}\mathring{E}_{\sigma,\tau}\omega_{\sigma}=0$
as well, because
$\mathrm{Tr}_{\xi,\rho}\lambda_{\sigma}^{(\xi)}(\mathrm{d}\lambda^{(\xi)})_{\tau\setminus\sigma}=0.$

Now let a simplex $f_{\rho}$ be given such that
$[\\![\tau]\\!]\subseteq[\\![\rho]\\!]$, which implies
$\mathrm{Tr}_{\tau,\rho\cap\tau}\mathring{E}_{\sigma,\tau}\omega_{\sigma}=\omega_{\sigma}.$
We compute the left hand side of (23):

$\displaystyle\mathrm{Tr}_{\xi,\rho}\dot{E}_{\tau,\xi}(\mathring{E}_{\sigma,\tau}\omega_{\sigma})$
$\displaystyle=\mathrm{Tr}_{\xi,\rho}(P_{\xi,\sigma}^{*}\omega_{\sigma}\wedge(\lambda^{(\xi)}_{\sigma}(\mathrm{d}\lambda^{(\xi)})_{\tau\setminus\sigma}))$
$\displaystyle=(\mathrm{Tr}_{\xi,\rho}P_{\xi,\sigma}^{*}\omega_{\sigma})\wedge(\mathrm{Tr}_{\xi,\rho}\lambda^{(\xi)}_{\sigma}(\mathrm{d}\lambda^{(\xi)})_{\tau\setminus\sigma})$
$\displaystyle=(P_{\rho,\sigma}^{*}\omega_{\sigma})\wedge(\lambda^{(\rho)}_{\sigma}(\mathrm{d}\lambda^{(\rho)})_{\tau\setminus\sigma})$
$\displaystyle=\dot{E}_{\tau,\rho}\mathring{E}_{\sigma,\tau}\omega_{\sigma}=\dot{E}_{\rho\cap\tau,\rho}\mathrm{Tr}_{\tau,\rho\cap\tau}(\mathring{E}_{\sigma,\tau}\omega_{\sigma}),$

where we have used (13) between the second and third right hand sides. Because
every $\omega\in\mathring{\Lambda}^{k}(\overline{f_{\tau}})$ has a bubble
decomposition, this completes the proof. ∎

###  4.6 Geometric decompositions

With $\dot{E}_{\sigma,\tau}$ defining a consistent family of extension
operators, we can now show that it defines a geometric decomposition of all of
$\Lambda^{k}(\overline{T})$.

######  Theorem 4.5.

With $\dot{E}_{\sigma,T}$ defined as in (22) taking $f_{\xi}=T$,

$\Lambda^{k}(\overline{T})=\bigoplus_{\begin{subarray}{c}\sigma\\\
\dim(f_{\sigma})\geq
k\end{subarray}}\dot{E}_{\sigma,T}[\mathring{\Lambda}^{k}(\overline{f_{\sigma}})].$

###### Proof.

The proof proceeds along the same lines as the proof of the bubble
decompostion in theorem 4.3. Implicit in the fact that $\dot{E}_{\sigma,\tau}$
defines a consistent family of extension operators is that

$\mathrm{Tr}_{\sigma}\dot{E}_{\sigma,T}\omega=\omega,\quad\omega\in\mathring{\Lambda}^{k}(\overline{f_{\sigma}}).$

This means that if $\omega\in\mathring{\Lambda}^{k}(\overline{T})$ then
$\omega=\dot{E}_{T,T}\omega$. From this base case one can inductively assume
that every $\omega$ has a geometric decomposition in the
$\dot{E}_{\sigma,\tau}$ family if it is trace-free on boundary simplices with
dimension less than or equal to $d$. Then taking $\omega$ that is trace-free
on boundary simplicies with dimension less than $d$ and define

$\tilde{\omega}=\omega-\sum_{\sigma\in\Sigma([0,[0)}\dot{E}_{\sigma,T}\mathrm{Tr}_{\sigma}\omega,$

which by the inductive assumption has a geometric decomposition in the
$\dot{E}_{\sigma,\tau}$ family, and so $\omega$ has one as well. By the
inductive assumption, this completes the proof. ∎

We now show that the $\dot{E}_{\sigma,\tau}$ extension operator defines a
geometric decomposition of full polynomial $k$-forms.

######  Theorem 4.6.

With $\dot{E}_{\sigma,T}$ defined as in (22) taking $f_{\xi}=T$,

$\mathcal{P}_{r}\Lambda^{k}(T)=\bigoplus_{\begin{subarray}{c}\sigma\\\
\dim(f_{\sigma})\geq
k\end{subarray}}\dot{E}_{\sigma,T}[\mathring{\mathcal{P}}_{r}\Lambda^{k}(f_{\sigma})].$

###### Proof.

Let $f_{\tau}$ such that $\dim(f_{\tau})\geq k$ be given. By (20),
$\mathcal{P}_{r}\Lambda^{k}(f_{\tau})$ is spanned by bubble functions of the
form $\mathring{E}_{\sigma,\tau}\omega_{\sigma}$ for
$\omega_{\sigma}\in\mathcal{P}_{r-\dim(f_{\sigma})-1}\Lambda^{k-(\dim(f_{\tau})-\dim(f_{\sigma}))}(f_{\sigma}).$
By inspection of (22) we see that this means
$\dot{E}_{\sigma,T}\mathring{E}_{\sigma,\tau}\omega_{\sigma}\in\mathcal{P}_{r}\Lambda^{k}(T).$
The injectivity of $\dot{E}_{\sigma,T}$, the fact that it defines a consistent
family of extension operators, and a counting argument are sufficient to
complete the proof. ∎

Our final theorem shows that $\dot{E}_{\sigma,\tau}$ defines a geometric
decomposition of trimmed polynomial $k$-forms as well.

######  Theorem 4.7.

With $\dot{E}_{\sigma,T}$ defined as in (22) taking $f_{\xi}=T$,

$\mathcal{P}_{r}^{-}\Lambda^{k}(T)=\bigoplus_{\begin{subarray}{c}\sigma\\\
\dim(f_{\sigma})\geq
k\end{subarray}}\dot{E}_{\sigma,T}[\mathring{\mathcal{P}}_{r}^{-}\Lambda^{k}(f_{\sigma})].$

###### Proof.

It is sufficient to show that
$\dot{E}_{\tau,T}\omega\in\mathcal{P}_{r}^{-}\Lambda^{k}(T)$ for each
$f_{\tau}$ such that $\dim(f_{\tau})\geq k$ and each
$\omega\in\mathring{\mathcal{P}}_{r}^{-}\Lambda^{k}(f_{\tau})$. By the proof
of theorem 4.6, it is already known that
$\dot{E}_{\tau,T}\omega\in\mathcal{P}_{r}\Lambda^{k}(T)$. We shall apply the
Koszul operator centered at the centroid of $f_{\tau}$ to
$\dot{E}_{\tau,T}\omega$: if we can show that
$\kappa^{(\tau)}\dot{E}_{\tau,T}\omega\in\mathcal{P}_{r}\Lambda^{k-1}(T)$,
that is sufficient to prove that
$\dot{E}_{\tau,T}\omega\in\mathcal{P}_{r}^{-}\Lambda^{k}(T)$.

Let the bubble decomposition of $\omega$ be

$\omega=\sum_{\sigma\in
Z^{k}(f_{\tau})}\mathring{E}_{\sigma,\tau}\omega_{\sigma}.$

We apply $\kappa^{(\tau)}$ to the extension of a single bubble component, but
shift it to be centered at $\kappa^{(\sigma)}$, which introduces a term
involving the interior product with some constant vector $b_{\sigma,\tau}$,
which we denote $\delta_{\sigma,1}$:

$\kappa^{(\tau)}\dot{E}_{\tau,T}\mathring{E}_{\sigma,\tau}\omega_{\sigma}=\kappa^{(\sigma)}\dot{E}_{\tau,T}\mathring{E}_{\sigma,\tau}\omega_{\sigma}+\underbrace{(\dot{E}_{\tau,T}\mathring{E}_{\sigma,\tau}\omega_{\sigma})\lrcorner
b_{\sigma,\tau}}_{\delta_{\sigma,1}}.$

Because the vector $b_{\sigma,\tau}$ is constant, the $(k-1)$-form
$\delta_{\sigma,1}$ is in $\mathcal{P}_{r}\Lambda^{k-1}(T)$.

We now address the term
$\kappa^{(\sigma)}\dot{E}_{\tau,T}\mathring{E}_{\sigma,\tau}\omega_{\sigma}$.
We wish to show that in this case $\kappa^{(\sigma)}$ and $\dot{E}_{\tau,T}$
commute. Using the product rule for the Koszul operator, we have

$\displaystyle\kappa^{(\sigma)}\dot{E}_{\tau,T}\mathring{E}_{\sigma,\tau}\omega_{\sigma}$
$\displaystyle=\kappa^{(\sigma)}[P_{T,\sigma}^{*}\omega_{\sigma}\wedge(\lambda_{\sigma}(\mathrm{d}\lambda)_{\tau\setminus\sigma})]$
(24)
$\displaystyle=(\kappa^{(\sigma)}P_{T,\sigma}^{*}\omega_{\sigma})\wedge(\lambda_{\sigma}(\mathrm{d}\lambda)_{\tau\setminus\sigma})+(-1)^{\tilde{k}}P_{T,\sigma}^{*}\omega_{\sigma}\wedge(\lambda_{\sigma}\kappa^{(\sigma)}(\mathrm{d}\lambda)_{\tau\setminus\sigma}),$
(25)

where $\tilde{k}=k-(\dim(f_{\tau})-\dim(f_{\sigma}))$. The centroid of
$f_{\sigma}$ is fixed by $P_{T,\sigma}$, so $\kappa^{(\sigma)}$ commutes with
$P_{T,\sigma}^{*}$ and the first term becomes

$\displaystyle(\kappa^{(\sigma)}P_{T,\sigma}^{*}\omega_{\sigma})\wedge(\lambda_{\sigma}(\mathrm{d}\lambda)_{\tau\setminus\sigma})$
$\displaystyle=P_{T,\sigma}^{*}(\kappa^{(\sigma)}\omega_{\sigma})\wedge(\lambda_{\sigma}(\mathrm{d}\lambda)_{\tau\setminus\sigma})$
$\displaystyle=\dot{E}_{\tau,T}\mathring{E}_{\sigma,\tau}(\kappa^{(\sigma)}\omega_{\sigma}).$

Because $\lambda_{j}$ vanishes at the centroid of $f_{\sigma}$ if
$j\in[\\![\tau\setminus\sigma]\\!]$, we have
$\kappa^{(\sigma)}\mathrm{d}\lambda_{j}=\lambda_{j}$ for each such $j$. Let
$m+1$ be the number of indices in $[\\![\tau\setminus\sigma]\\!]$ and let
$\rho\in\Sigma([0,[0)$ be an increasing map such that
$[\\![\rho]\\!]=[\\![\tau\setminus\sigma]\\!]$. By the product rule

$\kappa^{(\sigma)}(\mathrm{d}\lambda)_{\tau\setminus\sigma}=\sum_{j=0}^{m}(-1)^{m}\lambda_{\rho(j)}(\mathrm{d}\lambda)_{\rho\setminus\rho(j)}.$

Putting this into the second term in (25), we get

$\displaystyle(-1)^{\tilde{k}}P_{T,\sigma}^{*}\omega_{\sigma}\wedge(\lambda_{\sigma}\kappa^{(\sigma)}(\mathrm{d}\lambda)_{\tau\setminus\sigma})$
$\displaystyle=\sum_{j=0}^{m}(-1)^{j+\tilde{k}}P_{T,\sigma}^{*}\omega_{\sigma}\wedge(\lambda_{\sigma\cup\rho(j)}(\mathrm{d}\lambda)_{\rho\setminus\rho(j)})$
$\displaystyle=\sum_{j=0}^{m}P_{T,\sigma\cup\rho(j)}^{*}P_{\sigma\cup\rho(j),\sigma}^{*}((-1)^{j+\tilde{k}}\omega_{\sigma})\wedge(\lambda_{\sigma\cup\rho(j)}(\mathrm{d}\lambda)_{\rho\setminus\rho(j)})$
$\displaystyle=\dot{E}_{\tau,T}\big{[}\sum_{j=0}^{m}\mathring{E}_{\sigma\cup\rho(j),\tau}(P_{\sigma\cup\rho(j),\sigma}^{*}((-1)^{j+\tilde{k}}\omega_{\sigma}))\big{]}.$

Combining both terms, we have

$\kappa^{(\sigma)}\dot{E}_{\tau,T}\mathring{E}_{\sigma,\tau}\omega_{\sigma}=\dot{E}_{\tau,T}\big{[}\mathring{E}_{\sigma,\tau}(\kappa^{(\sigma)}\omega_{\sigma})+\sum_{j=0}^{m}\mathring{E}_{\sigma\cup\rho(j),\tau}(P_{\sigma\cup\rho(j),\sigma}^{*}(-1)^{j+\tilde{k}}\omega_{\sigma})\big{]}.$

We now want to show that the term in brackets is
$\kappa^{(\sigma)}\mathring{E}_{\sigma,\tau}\omega_{\sigma}.$ We essentially
follow the steps above in reverse. For the first term,

$\displaystyle\mathring{E}_{\sigma,\tau}(\kappa^{(\sigma)}\omega_{\sigma})$
$\displaystyle=P_{\tau,\sigma}^{*}(\kappa^{(\sigma)}\omega_{\sigma})\wedge(\lambda^{(\tau)}_{\sigma}(\mathrm{d}\lambda^{(\tau)})_{\tau\setminus\sigma})$
$\displaystyle=(\kappa^{(\sigma)}P_{\tau,\sigma}^{*}\omega_{\sigma})\wedge(\lambda^{(\tau)}_{\sigma}(\mathrm{d}\lambda^{(\tau)})_{\tau\setminus\sigma}).$

For the second term,

$\displaystyle\sum_{j=0}^{m}\mathring{E}_{\sigma\cup\rho(j),\tau}(P_{\sigma\cup\rho(j),\sigma}^{*}((-1)^{j+\tilde{k}}\omega_{\sigma}))$
$\displaystyle=\sum_{j=0}^{m}P_{\tau,\sigma\cup\rho(j)}^{*}P_{\sigma\cup\rho(j),\sigma}^{*}((-1)^{j+\tilde{k}}\omega_{\sigma})\wedge(\lambda_{\sigma\cup\rho(j)}^{(\tau)}(\mathrm{d}\lambda^{(\tau)})_{\rho\setminus\rho(j)})$
$\displaystyle=\sum_{j=0}^{m}(-1)^{j+\tilde{k}}P_{\tau,\sigma}^{*}\omega_{\sigma}\wedge(\lambda_{\sigma\cup\rho(j)}^{(\tau)}(\mathrm{d}\lambda^{(\tau)})_{\rho\setminus\rho(j)})$
$\displaystyle=(-1)^{\tilde{k}}P_{\tau,\sigma}^{*}\omega_{\sigma}\wedge(\lambda_{\sigma}^{(\tau)}\kappa^{(\sigma)}(\mathrm{d}\lambda^{(\tau)})_{\tau\setminus\sigma}).$

Combining these back together we get the desired result,

$(\kappa^{(\sigma)}P_{\tau,\sigma}^{*}\omega_{\sigma})\wedge(\lambda^{(\tau)}_{\sigma}(\mathrm{d}\lambda^{(\tau)})_{\tau\setminus\sigma})+(-1)^{\tilde{k}}P_{\tau,\sigma}^{*}\omega_{\sigma}\wedge(\lambda_{\sigma}^{(\tau)}\kappa^{(\sigma)}(\mathrm{d}\lambda^{(\tau)})_{\tau\setminus\sigma})=\kappa^{(\sigma)}\mathring{E}_{\sigma,\tau}\omega_{\sigma},$

that is,

$\kappa^{(\sigma)}\dot{E}_{\tau,T}\mathring{E}_{\sigma,\tau}\omega_{\sigma}=\dot{E}_{\tau,T}\kappa^{(\sigma)}\mathring{E}_{\sigma,\tau}\omega_{\sigma}.$

We now shift the center of the Koszul operator back to $\kappa^{(\tau)}$,
which again introduces an interior product with the constant vector which we
denote $\delta_{\sigma,2}$:

$\dot{E}_{\tau,T}\kappa^{(\sigma)}\mathring{E}_{\sigma,\tau}\omega_{\sigma}=\dot{E}_{\tau,T}(\kappa^{(\tau)}\mathring{E}_{\sigma,\tau}\omega_{\sigma}-\underbrace{(\mathring{E}_{\sigma,\tau}\omega_{\sigma})\lrcorner
b_{\sigma,\tau}}_{\delta_{\sigma,2}}).$

As with the $\delta_{\sigma,1}$, we know that
$\delta_{\sigma,2}\in\mathcal{P}_{r}\Lambda^{k-1}(f_{\sigma})$.

Why have we gone through the steps of shifting the Koszul operator between the
centroids of $f_{\sigma}$ and $f_{\tau}$? Because although
$\kappa^{(\sigma)}\mathring{E}_{\sigma,\tau}\omega_{\sigma}$ is trace-free (as
demonstrated in the steps that it took to show that $\kappa^{(\sigma)}$ and
$\dot{E}_{\tau,T}$ commuted),
$\kappa^{(\tau)}\mathring{E}_{\sigma,\tau}\omega_{\sigma}$ is not necessarily
trace-free. We cannot apply $\dot{E}_{\tau,T}$ to
$\kappa^{(\tau)}\mathring{E}_{\sigma,\tau}\omega_{\sigma}$ or
$\delta_{\sigma,2}$ separately, because the domain of $\dot{E}_{\tau,T}$ is
trace-free $k$-forms.

Nevertheless, we can now combine into $\kappa^{(\tau)}\dot{E}_{\tau,T}\omega$
the contributions from the different terms in the bubble decomposition:

$\displaystyle\kappa^{(\tau)}\dot{E}_{\tau,T}\omega$
$\displaystyle=\sum_{\sigma\in
Z^{k}(f_{\tau})}\dot{E}_{\tau,T}(\kappa^{(\tau)}\mathring{E}_{\sigma,T}\omega_{\sigma}+\delta_{\sigma,2})+\delta_{\sigma,1}$
$\displaystyle=\dot{E}_{\tau,T}\left[\kappa^{(\tau)}\omega+\sum_{\sigma\in
Z^{k}(f_{\tau})}\delta_{\sigma,2}\right]+\sum_{\sigma\in
Z^{k}(f_{\tau})}\delta_{\sigma,1}.$

By the assumption that
$\omega\in\mathring{\mathcal{P}}_{r}^{-}\Lambda^{k}(f_{\sigma})$, we know that
$\kappa^{(\tau)}\omega\in\mathcal{P}_{r}\Lambda^{k-1}(f_{\sigma})$. Therefore
the term in brackets is in
$\mathring{\mathcal{P}}_{r}\Lambda^{k-1}(f_{\sigma})$. By theorem 4.6, that
implies that the first term, and thus the whole expression, is in
$\mathcal{P}_{r}\Lambda^{k-1}(T)$. ∎

The main results of this work combine to unify the geometric decompositions of
full polynomials, trimmed polynomials, and all $k$-forms through the action of
just two operators acting on the unrestricted polynomial spaces.

######  Corollary 4.8.

Generalizing the operator $\mathring{\star}_{T}$ to
$\mathring{\star}_{\sigma}:\Lambda^{k}(\overline{f_{\sigma}})\stackrel{{\scriptstyle\sim\,}}{{\to}}\mathring{\Lambda}^{\dim(f_{\sigma})-k}(\overline{f_{\sigma}})$,
we have

$\displaystyle\Lambda^{k}(\overline{T})$
$\displaystyle=\bigoplus_{\begin{subarray}{c}\sigma\\\ \dim(f_{\sigma})\geq
k\end{subarray}}\dot{E}_{\sigma,T}\
\mathring{\star}_{\sigma}[\Lambda^{\dim(f_{\sigma})-k}(\overline{f_{\sigma}})];$
(26) $\displaystyle\mathcal{P}_{r}\Lambda^{k}(T)$
$\displaystyle=\bigoplus_{\begin{subarray}{c}\sigma\\\ \dim(f_{\sigma})\geq
k\end{subarray}}\dot{E}_{\sigma,T}\
\mathring{\star}_{\sigma}[\mathcal{P}_{r-\dim(f_{\sigma})-k}^{-}\Lambda^{\dim(f_{\sigma})-k}(f_{\sigma})];$
(27) $\displaystyle\mathcal{P}_{r}^{-}\Lambda^{k}(T)$
$\displaystyle=\bigoplus_{\begin{subarray}{c}\sigma\\\ \dim(f_{\sigma})\geq
k\end{subarray}}\dot{E}_{\sigma,T}\
\mathring{\star}_{\sigma}[\mathcal{P}_{r-\dim(f_{\sigma})-k-1}\Lambda^{\dim(f_{\sigma})-k}(f_{\sigma})].$
(28)

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