lemmatheorem lemma propositiontheorem proposition corollarytheorem corollary
claimtheorem claim definitiontheorem definition notationtheorem notation
remarktheorem remark remarkstheorem remarks exampletheorem example
examplestheorem examples questiontheorem question

#  Images of Order Zero Maps

Kristin Courtney  Department of Mathematics and Computer Science, University
of Southern Denmark, Campusvej 55, 5230 Odense M, Denmark
<EMAIL_ADDRESS>and  Wilhelm Winter  Mathematical Institute, WWU
Münster, Einsteinstr. 62, 48149 Münster, Germany<EMAIL_ADDRESS>Dedicated to the memory of Eberhard Kirchberg

###### Abstract.

We give sufficient conditions allowing one to build a
$\mathrm{C}^{*}$-algebraic structure on a self-adjoint linear subspace of a
$\mathrm{C}^{*}$-algebra in such a way that the subspace is naturally
identified with the resulting $\mathrm{C}^{*}$-algebra via a completely
positive order zero map. This leads to necessary and sufficient conditions to
realize a self-adjoint linear subspace of a $\mathrm{C}^{*}$-algebra as the
image of an order zero map from a $\mathrm{C}^{*}$-algebra and to determine
whether it is completely order isomorphic to a $\mathrm{C}^{*}$-algebra via an
order zero map.

###### Key words and phrases:

Completely positive approximation, nuclearity, inductive limits, order zero
maps, completely positive maps, NF systems, NF algebras

###### 2010 Mathematics Subject Classification:

46L05

This research was supported by the Deutsche Forschungsgemeinschaft (DFG,
German Research Foundation) under Germany’s Excellence Strategy – EXC 2044 –
390685587, Mathematics Münster – Dynamics – Geometry – Structure, the Deutsche
Forschungsgemeinschaft (DFG, German Research Foundation) –- Project-ID
427320536 –- SFB 1442, and ERC Advanced Grant 834267 – AMAREC

###### Contents

  1. 1 Order zero order embeddings
  2. 2 Order Units
  3. 3 A $\mathrm{C}^{*}$-structure
  4. 4 A c.p.c. order zero map
  5. 5 Unitizations
  6. 6 Closed subspaces and order units

## Introduction

Completely positive order zero maps between $\mathrm{C}^{*}$-algebras are
those which preserve the ∗-linear and matrix order structures as well as
orthogonality. They were introduced in this form in [20], following ideas of
[18] and [13] (a collaboration of the second named author with Kirchberg) in
the case of finite-dimensional domains, and the pioneering work [22] of Wolff
in a both more special and more general context. Order zero maps play a vital
role in the structure and classification theory of nuclear
$\mathrm{C}^{*}$-algebras and in approximation properties related to
$\mathrm{C}^{*}$-dynamical systems, and they are a highly effective tool when
passing between $\mathrm{C}^{*}$-algebraic and von Neumann algebraic
regularity properties; see [6, 7, 8, 15, 16, 17, 19, 21] for examples of
applications of order zero maps.

In [9], we take inspiration from the work of Blackadar and Kirchberg (cf. [2,
3, 4, 5]) and develop notions of inductive systems with completely positive
asymptotically order zero maps which allow for inductive limit
characterizations of nuclearity and of finite nuclear dimension for all
separable $\mathrm{C}^{*}$-algebras. The proofs and constructions in the first
article were heavily informed by the systematic study in the present paper,
and the second directly depends on it. However, applications of the results
here are not limited to inductive limit constructions. One notable upshot is a
characterization of when a self-adjoint linear subspace of a
$\mathrm{C}^{*}$-algebra is completely order isomorphic to a
$\mathrm{C}^{*}$-algebra _via an order zero map_ (Introduction). This answers,
in a special case, the question of how to detect when a self-adjoint linear
subspace of a $\mathrm{C}^{*}$-algebra is completely order isomorphic to a
$\mathrm{C}^{*}$-algebra. The order zero condition is natural in this context
because it occurs in abundance (in particular for nuclear
$\mathrm{C}^{*}$-algebras; c. f. [9]), and since it yields an easy
characterization of complete order embeddings which is not available for
general completely positive maps (c. f. Proposition 1.1 and Section 1 below).

As a consequence of the structure theorem for order zero maps, when the domain
of a completely positive map is a unital $\mathrm{C}^{*}$-algebra $A$, the map
is order zero exactly when the following identity holds for all elements
$a,b\in A$:

(a) $\displaystyle\theta(1)\theta(ab)=\theta(a)\theta(b).$

From this one directly obtains two necessary conditions for a space to be of
the form $\theta(A)$, i.e., an image of a c.p. order zero map from a unital
$\mathrm{C}^{*}$-algebra: First, $\theta(1_{A})$ commutes with the image,
$\theta(1_{A})\in\theta(A)^{\prime}$, and second
$\theta(A)^{2}=\theta(1_{A})\theta(A)$.

Our main result in this article (Theorem A) draws inspiration from the
characterization (a) to provide sufficient conditions allowing one to realize
a self-adjoint linear subspace of a $\mathrm{C}^{*}$-algebra as a
(pre-)$\mathrm{C}^{*}$-algebra (albeit likely not a
sub-$\mathrm{C}^{*}$-algebra of the ambient $\mathrm{C}^{*}$-algebra). One of
these conditions, an adaptation of the notion of a local order unit from the
theory of abstract operator systems (as described in [14, Chapter 13]),
requires a brief introduction. It is necessary since we do not want to
restrict ourselves to unital domain $\mathrm{C}^{*}$-algebras.

######  Definition .

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace. We call an element $e\in B_{+}^{1}$ a _local order unit_ for $X$ if,
for any $x\in X_{\mathrm{s.a.}}$, there exists an $R>0$ so that $Re\geq x$. We
call $e$ an _order unit_ for $X$ if there exists an $R>0$ so that $R\|x\|e\geq
x$ for every $x\in X_{\mathrm{s.a.}}$. For an order unit, the minimal111See
Remark 2(ii) for a proof that the scaling factor is in fact a minimum. such
$R$ is called the scaling factor of $e$.

The unit of a $\mathrm{C}^{*}$-algebra is an order unit with scaling factor
$1$ for any self-adjoint linear subspace. The image of the unit under a
completely positive map into another $\mathrm{C}^{*}$-algebra is still a local
order unit for the image of the domain, and when the map is bounded below it
remains an order unit, with scaling factor dependent on the norm of the
inverse of the map.

Now we are ready to state our first main result, which appears as Theorem 3.1
below.

######  Theorem A.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace. Suppose there exists a local order unit $e\in B_{+}^{1}\cap
X^{\prime}$ for $X$ so that $X^{2}\subset eX$. Then there is an associative
bilinear map
${\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\times X\longrightarrow X$ satisfying

(b) $\displaystyle
xy=e(x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y)$

for all $x,y\in X$, so that
$(X,{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})$
is a ∗-algebra, with unit $e$ when $e\in X$, and there is a norm
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\longrightarrow[0,\infty)$ with respect to which
$(X,{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})$
is a pre-$\mathrm{C}^{*}$-algebra.

It follows that the completion of the ∗-algebra
$(X,\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,)$
with respect to
${\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}}$
is a $\mathrm{C}^{*}$-algebra, which we call the _$\mathrm{C}^{*}$ -algebra
associated to $(X,e)$_ and denote with
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$.
(We will make sure that suppressing the $e$ from the notation results in no
ambiguity). From the structure theorem for order zero maps ([20, Theorem 3.3])
we know that an example of a self-adjoint linear subspace of a
$\mathrm{C}^{*}$-algebra $B$ satisfying the assumptions of Theorem A is the
image of a c.p.c. order zero map $\theta\colon A\longrightarrow B$ from a
unital $\mathrm{C}^{*}$-algebra (with $\theta(1_{A})=e$). Another example
comes from [9, Section 2]. In [9, Proposition 2.6], the role of $B$ was played
by $F_{\infty}$, $X$ was $\overset{\leavevmode\hbox to28.85pt{\vbox
to2.8pt{\pgfpicture\makeatletter\hbox{\hskip 28.65276pt\lower-0.2pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{}{{}}{} {}{}{}{{
{\pgfsys@beginscope
\pgfsys@setdash{}{0.0pt}\pgfsys@roundcap\pgfsys@roundjoin{} {}{}{} {}
\pgfsys@moveto{-2.07999pt}{2.39998pt}\pgfsys@curveto{-1.69998pt}{0.95998pt}{-0.85318pt}{0.28pt}{0.0pt}{0.0pt}\pgfsys@lineto{-0.4pt}{0.0pt}\pgfsys@stroke\pgfsys@endscope}}
}{}{}{{}}\pgfsys@moveto{-28.45276pt}{0.0pt}\pgfsys@lineto{-0.4pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
}{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-0.2pt}{0.0pt}\pgfsys@invoke{
}\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}{{}}}} \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}{(F_{n},\rho_{n})}_{n}$,
and the element $[(\rho_{n+1,n}(1_{F_{n}}))_{n}]\in(F_{\infty})_{+}^{1}$ was
suggestively denoted by $e$. In this setting, [9, Lemma 2.5(i) and (ii)]
established that $e\in X^{\prime}$ is an order unit for $X$ with scaling
factor $1$. (In [9, Section 2], the norm
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
was simply the norm on $F_{\infty}$. We will shortly see why a scaling factor
$1$ guarantees that the two norms coincide.)

If $X$ is already complete with respect to
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$,
then
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
and $X$ are equal as ∗-vector spaces. This happens for instance when $X$ is
the image of an order zero map $\theta\colon A\longrightarrow B$ from a unital
$\mathrm{C}^{*}$-algebra with $e=\theta(1_{A})$ (see Introduction) and also
when $X$ is closed in $B$. (Indeed, we shall see that $X$ is closed in $B$ if
and only if it is closed with respect to
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
and $e$ is an order unit for $X$, cf. Corollary 6.1.) Though many of our
results hold outside the setting where $X$ is complete with respect to
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$,
our main applications and examples satisfy this assumption, and so we mostly
adopt it for the current exposition.

As mentioned above, the image of a c.p.c. order zero map from a unital
$\mathrm{C}^{*}$-algebra satisfies the assumptions of Theorem A. A key feature
of our study is the converse.

######  Theorem B.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace. Suppose there exists an $e\in B$ as in _Theorem A_, and assume $X$
is complete with respect to the norm
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$.
Then the map
$\Phi\coloneqq\mathrm{id}_{X}\colon\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\longrightarrow
B$ is c.p.c. order zero where
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
denotes the associated $\mathrm{C}^{*}$-algebra as above.

When $X$ is closed in $B$, then $e$ is an order unit and
$\Phi\coloneqq\mathrm{id}_{X}\colon\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\\\
\longrightarrow B$ is a bounded below c.p.c. map with c.p. inverse
$\Phi^{-1}\colon
X\longrightarrow\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
such that $\|\Phi^{-1}\|$ is equal to the scaling factor of $e$. When $e$ has
scaling factor $1$, $\Phi$ is a complete isometry.

This in particular tells us that not only are $X$ and
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
equal as ∗-vector spaces, they (and their matrix amplifications) also have the
same positive cones, i.e., $x\in X$ is positive in
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
if and only if it is positive in $B$, and the same holds for matrix
amplifications.

Now combining Theorem B with the characterization (a) of order zero maps out
of unital $\mathrm{C}^{*}$-algebras, we get a characterization of images of
order zero maps as well as a characterization of when a self-adjoint linear
subspace of a $\mathrm{C}^{*}$-algebra is completely order isomorphic to a
unital $\mathrm{C}^{*}$-algebra via an order zero map; cf. Proposition 4.3 and
Corollary 6.3 below.

######  Corollary B.

Let $X$ be a self-adjoint linear subspace of a $\mathrm{C}^{*}$-algebra $B$
and $e\in B_{+}^{1}$. Then the following are equivalent.

  1. (i)

There exists a unital $\mathrm{C}^{*}$-algebra $A$ and a c.p.c. order zero map
$\theta\colon A\longrightarrow B$ with $\theta(A)=X$ and $\theta(1_{A})=e$.

  2. (ii)

The element $e$ lies in $X\cap X^{\prime}$ and is a local order unit for $X$
such that $X^{2}=eX$, and $X$ is complete with respect to
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$.

######  Corollary B.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace. Then $X$ is completely order isomorphic to a unital
$\mathrm{C}^{*}$-algebra via a c.p. order zero map if and only if $X$ is
closed in $B$ and there exists an order unit $e\in B_{+}^{1}\cap X\cap
X^{\prime}$ for $X$ with scaling factor $1$ so that $X^{2}=eX$.

Since the construction of
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
heavily utilizes the local order unit $e$, it is natural to ask how much it
depends on the choice of $e$. Below we summarize some of our results in this
regard (cf. Proposition 5.2, Proposition 5.3, and Proposition 6.3).

######  Proposition .

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a closed self-adjoint
linear subspace. Suppose there exists a local order unit $e\in B_{+}^{1}\cap
X^{\prime}$ for $X$ so that $X^{2}\subset eX$.

  1. (i)

If $e\in X$ and there exists $h\in B_{+}^{1}$ which satisfies $hx=ex$ for all
$x\in X$, then $h\in X^{\prime}$ is an order unit for $X$ with the same
scaling factor as $e$, such that $X^{2}\subset hX$. Moreover, the associative
bilinear maps and norms induced by $e$ and $h$ on $X$ agree, and the induced
$\mathrm{C}^{*}$-algebras are equal.

  2. (ii)

If there exists another local order unit $h\in B_{+}^{1}\cap X\cap X^{\prime}$
so that $X^{2}\subset hX$, then the induced $\mathrm{C}^{*}$-algebras are
unitally ∗-isomorphic.

  3. (iii)

If $e\in X$ and has scaling factor $1$, then the abstract operator system
$(X,\\{\mathrm{M}_{r}(X)\cap\mathrm{M}_{r}(B)_{+}\\}_{r},e)$ is completely
order isomorphic to
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$,
and so
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
is canonically isomorphic to the enveloping $\mathrm{C}^{*}$-algebra for this
system.

In several of the above results, the local order unit $e$ for $X$ need not
actually lie in $X$. In this situation, one can replace $X$ with
$\textrm{span}\\{X,e\\}$ and obtain a unitization of
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
and of the map $\Phi$ from Theorem B.

######  Theorem C.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace. Suppose there exists an $e\in B$ as in _Theorem A_, and assume $X$
is complete with respect to the norm
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$.
Then we have the following:

  1. (i)

The associative bilinear map from _Theorem A_ extends to an associative
bilinear map on $Z\coloneqq\textrm{span}\\{X,e\\}$ which also satisfies (b),
and the norm
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
from _Theorem A_ extends to a $\mathrm{C}^{*}$-norm on $Z$.

  2. (ii)

The inclusion $X\subset Z$ induces an inclusion of
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
in
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$
as a two-sided closed ideal. When $e\in X$ or
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
is essential in
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$,
we may identify the smallest unitization
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)^{\sim}$
with
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$;
otherwise
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
is already unital, and
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$
is a forced unitization,
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)^{\dagger}\cong\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$.

  3. (iii)

The map $\Phi$ from _Theorem B_ extends to a c.p.c. order zero map
$\Phi^{\dagger}\coloneqq\mathrm{id}_{Z}\colon\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)\longrightarrow
B$. In case we identify
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)^{\sim}$
with
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$
as above, $\Phi^{\dagger}$ is bounded below (resp. a complete order
isomorphism) if and only if $\Phi$ is.

The article is organized as follows. In Section 1, we recall relevant
definitions and results regarding completely positive maps, order zero maps,
and complete order embeddings. We also give our characterization of order zero
complete order embeddings and show by examples that it fails without the order
zero hypothesis. In Section 2, we investigate properties of (commuting) order
units which will feature heavily throughout later sections. Section 3 is
dedicated to establishing Theorem A, and Section 4 to the c.p.c. order zero
map $\Phi$ from Theorem B. In Section 5, we consider unitizations of our
construction, establishing most of Theorem C. We also address here the
dependence of our constructions on the choice of order unit. Finally, in
Section 6 we incorporate the further assumption that our subspace is closed
and strengthen results from the previous sections accordingly, completing our
proof of Theorem B.

##  1\. Order zero order embeddings

For a $\mathrm{C}^{*}$-algebra $A$, we write $A_{+}$ for the cone of positive
elements, $A_{\mathrm{s.a.}}$ for its set of self-adjoint elements, and
$A^{1}$, $A^{1}_{+}$, and $A^{1}_{\mathrm{s.a.}}$ for the respective closed
unit balls. We denote the minimal unitization of $A$ by $A^{\sim}$ with the
convention that $A=A^{\sim}$ when $A$ is unital, and we write $A^{\dagger}$
for the smallest forced unitization of $A$, which agrees with $A^{\sim}$ when
$A$ is non-unital and contains $1_{A}$ as a central co-rank 1 projection when
$A$ is unital. The multiplier algebra of $A$ (regarded as a subalgebra of the
double dual $A^{**}$) is denoted by $\mathcal{M}(A)$. For a self-adjoint
linear subspace $X$ of a larger $\mathrm{C}^{*}$-algebra $M$ containing $A$
and an element $e\in A$, we write $X^{\prime}\cap A\subset M$ for the relative
commutant of $X$ in $M$, and we define

$\displaystyle X^{2}$ $\displaystyle\coloneqq\\{xy\mid x,y\in X\\}\ \text{
and}$ $\displaystyle eX$ $\displaystyle\coloneqq\\{ex\mid x\in X\\}.$

In the literature $X^{2}$ sometimes denotes $\textrm{span}\\{xy\mid x,y\in
X\\}$; throughout this article, the two will coincide because of assumptions
placed on $X^{2}$. For $r\geq 1$, we write $\mathrm{M}_{r}(X)$ for $r\times r$
matrices over $X$. We will fix throughout an approximate inverse
$\\{h_{j}\\}_{j}\in C_{0}((0,1])_{+}\subset C([0,1])_{+}$ for
$\mathrm{id}_{[0,1]}$ by defining for each $j\geq 1$

(1.3) $\displaystyle
h_{j}(t)=\left\\{\begin{array}[]{lr}j^{2}t&:t\in[0,\frac{1}{j}]\\\
\frac{1}{t}&:t\in[\frac{1}{j},1]\end{array}\right..$

Next we highlight a few relevant properties and observations for completely
positive maps and completely positive order zero maps. For a full introduction
to completely positive and completely bounded maps and see [14].

######  Definition 1.1.

Let $A$ and $B$ be $\mathrm{C}^{*}$-algebras with self-adjoint linear
subspaces $X\subset A$, $Y\subset B$, and $\theta\colon X\longrightarrow Y$ a
linear map. We say $\theta$ is _positive_ if $\theta(x)\in Y\cap B_{+}$ for
all $x\in X\cap A_{+}$. We say it is _completely positive_ (c.p.) if this
holds for all matrix amplifications
$\theta^{(r)}\colon\mathrm{M}_{r}(X)\longrightarrow\mathrm{M}_{r}(Y)$. We say
$\theta$ is _completely contractive_ if $\sup_{r\geq 1}\|\theta^{(r)}\|\leq
1$. We say a map is c.p.c. when it is completely positive and (completely)
contractive.

We say $\theta$ is a _complete order embedding_ if $\theta$ is c.p. and
completely isometric with c.p. inverse
$\theta^{-1}\colon\theta(A)\longrightarrow A$. A surjective complete order
embedding is a _complete order isomorphism_. We say $\theta$ is _bounded
below_ if there exists $R>0$ so that $\|\theta(x)\|\geq R\|x\|$ for all $x\in
X$.

######  Remarks .

(i) Under certain assumptions, some of the properties of a complete order
embedding are implied by the others, which is why this term tends to have
varying definitions in the literature. Many of these require that the map is
unital, but even in the nonunital setting, if the domain of an isometric c.p.
map is a (not-necessarily unital) $\mathrm{C}^{*}$-algebra, then the map is
automatically a complete order embedding if it is completely isometric or
order zero (see [9, Remark 1.7(ii)] and [9, Proposition 1.8], respectively).

(ii) Complete order isomorphisms between $\mathrm{C}^{*}$-algebras are
automatically ∗-homomorphisms (see e.g., [1, Theorem II.6.9.17]). It follows
that a self-adjoint linear subspace $X$ of a $\mathrm{C}^{*}$-algebra $B$
which is completely order isomorphic to a $\mathrm{C}^{*}$-algebra $A$
captures the isomorphism class of $A$ in that another $\mathrm{C}^{*}$-algebra
$C$ is ∗-isomorphic to $A$ if and only if it is completely order isomorphic to
$X$.

(iii) Bounded below maps are exactly the invertible bounded linear maps with
bounded linear inverses (defined on the image). A linear map between Banach
spaces is bounded below if and only if it is injective with closed range. It
follows that all matrix amplifications of a bounded below map $\theta\colon
X\longrightarrow Y$ between closed self-adjoint linear subspaces of
$\mathrm{C}^{*}$-algebras are bounded below. However, the lower bound need not
be uniform, i.e., $\theta^{(r)}$ has a bounded linear inverse for each $r\geq
1$, but we are not guaranteed that $\theta^{-1}$ is completely bounded. Even
an isometric c.p. map between closed self-adjoint linear subspaces of
$\mathrm{C}^{*}$-algebras with c.p. inverse can fail to be completely
isometric as the following example shows.

######  Example .

We construct a map $\varphi\colon\mathrm{M}_{3}\longrightarrow A$, where $A$
is a homogeneous $\mathrm{C}^{*}$-algebra, such that $\varphi$ is c.p. and
isometric with c.p. inverse (defined on its image) but is _not_ completely
isometric.

We denote by $\mathrm{Gr}(2,\mathbb{C}^{3})$ the Grassmannian manifold of all
$2$\- dimensional subspaces of $\mathbb{C}^{3}$, which is a compact Hausdorff
space. Set

$B\coloneqq C(\mathrm{Gr}(2,\mathbb{C}^{3}),\mathrm{M}_{3}),$

and define $\theta\colon\mathrm{M}_{3}\longrightarrow B$ by
$\theta(a)(V)\coloneqq p_{V}ap_{V}$ where $p_{V}\in\mathrm{M}_{3}$ is the
projection onto $V$. Define

$\displaystyle\varphi\coloneqq\theta\oplus\textstyle{\frac{1}{2}}\mathrm{id}_{\mathrm{M}_{3}}\colon\mathrm{M}_{3}$
$\displaystyle\longrightarrow B\oplus\mathrm{M}_{3}=:A.$

First, we claim that $\varphi$ is c.p. and isometric, which comes down to
checking that $\theta$ is c.p. and isometric. Since

$\theta^{(r)}\colon\mathrm{M}_{r}(\mathrm{M}_{3})\longrightarrow\mathrm{M}_{r}\big{(}C(\mathrm{Gr}(2,\mathbb{C}^{3})),\mathrm{M}_{3}\big{)}\cong
C\big{(}\mathrm{Gr}(2,\mathbb{C}^{3}),\mathrm{M}_{r}(\mathrm{M}_{3})\big{)},$

to show that $\theta$ is c.p., it suffices to show that
$\text{ev}_{V}\circ\theta$ is c.p. for every
$V\in\mathrm{Gr}(2,\mathbb{C}^{3})$. But $\text{ev}_{V}\circ\theta$ is simply
the compression by a projection, which gives a c.p. map. To see that $\theta$
is isometric, set $a\in\mathrm{M}_{3}\backslash\\{0\\}$ and
$0\neq\xi\in\mathbb{C}^{3}$ such that $\|a\xi\|=\|a\|\|\xi\|$, and choose
$V\in\mathrm{Gr}(2,\mathbb{C}^{3})$ such that $\xi$ and $a\xi\in V$. Then
$\|\theta(a)(V)\|=\|p_{V}ap_{V}\|=\|a\|$, and hence
$\|\theta(a)\|\geq\|a\|\geq\|\theta(a)\|$.

We claim that $\varphi$ is not 2-isometric. Since the second summand,
$\frac{1}{2}\mathrm{id}_{\mathrm{M}_{3}}$ is strictly contractive, we only
need to demonstrate that $\theta^{(2)}$ is not isometric. Consider

$\displaystyle s=\textstyle{\frac{1}{\sqrt{2}}}(e_{12}\otimes
e_{13}+e_{22}\otimes e_{23})\ \in\mathrm{M}_{2}\otimes\mathrm{M}_{3}.$

Then $\|s\|=1$, $s$ is rank 1, and

(1.4) $\displaystyle s^{*}s$ $\displaystyle=e_{22}\otimes e_{33}\ \text{ and
}$ (1.5) $\displaystyle ss^{*}$
$\displaystyle=\textstyle{\frac{1}{2}}(e_{11}\otimes e_{11}+e_{12}\otimes
e_{12}+e_{21}\otimes e_{21}+e_{22}\otimes e_{22}).$

Suppose there exists a $V\in\mathrm{Gr}(2,\mathbb{C}^{3})$ so that

$\|\theta^{(2)}(s)(V)\|=\|(1_{\mathrm{M}_{2}}\otimes
p_{V})s(1_{\mathrm{M}_{2}}\otimes p_{V})\|=\|s\|=1.$

Then there exists $\xi\in V\oplus V$ with $\|\xi\|=1$ such that
$\|(1_{\mathrm{M}_{2}}\otimes p_{V})s\xi\|=\|s\xi\|=\|s\|\|\xi\|=1$, and hence
$s\xi\in V\oplus V$. Since $s$ is rank 1, that means $V\oplus V$ is an
invariant subspace for $s$, and hence $(1_{\mathrm{M}_{2}}\otimes
p_{V})s=s(1_{\mathrm{M}_{2}}\otimes p_{V})$. It follows that
$ss^{*}\geq(1_{\mathrm{M}_{2}}\otimes
p_{V})ss^{*}=ss^{*}(1_{\mathrm{M}_{2}}\otimes p_{V})\neq 0$ and likewise with
$s^{*}s$. Moreover, from $[ss^{s},1_{\mathrm{M}_{2}}\otimes
p_{V}]=[s^{*}s,1_{\mathrm{M}_{2},p_{V}}]=0$, (1.4), and (1.5) it follows that
$[p_{V},e_{11}]=[p_{V},e_{22}]=[p_{V},e_{33}]=0$. Since
$(1_{\mathrm{M}_{2}}\otimes p_{V})s^{*}s=e_{22}\otimes p_{V}e_{33}$ is
positive and nonzero, we see immediately that $p_{V}e_{33}\neq 0$. Combined
with the fact that $[p_{V},e_{33}]=0$, we have $p_{V}\geq e_{33}$. Since
$(1_{\mathrm{M}_{2}}\otimes p_{V})ss^{*}=\textstyle{\frac{1}{2}}(e_{11}\otimes
p_{V}e_{11}+e_{12}\otimes p_{V}e_{12}+e_{21}\otimes p_{V}e_{21}+e_{22}\otimes
p_{V}e_{22})$ is positive and nonzero, we have $p_{V}e_{11}\neq 0$ or
$p_{V}e_{22}\neq 0$. If $p_{V}e_{22}=0$, then $(1_{M_{2}}\otimes
p_{V})ss^{*}=\frac{1}{2}(e_{11}\otimes e_{11})$, contradicting
$ss^{*}\geq(1_{M_{2}}\otimes p_{V})ss^{*}$. Likewise we must also have
$p_{V}e_{11}\neq 0$, and so $p_{V}\geq e_{11},e_{22}$ as well. But this means
$\dim(V)\geq 3$, a contradiction.

Hence we must have $\|\theta^{(2)}(s)(V)\|<1$ for all
$V\in\mathrm{Gr}(2,\mathbb{C}^{3})$. Since $\mathrm{Gr}(2,\mathbb{C}^{3})$ is
compact, that means $\|\theta^{(2)}(s)\|<1=\|s\|$, and therefore $\theta$ is
not 2-isometric, let alone completely isometric.

This demonstrates that, in the non-unital setting, even if the domain is a
$\mathrm{C}^{*}$-algebra, an isometric and c.p. map with c.p. inverse may fail
to be a complete order embedding, whence assumptions are necessary.

######  Definition 1.2.

A c.p. map $\theta\colon A\longrightarrow B$ between $\mathrm{C}^{*}$-algebras
is _order zero_ if for any $a,b\in A_{+}$,

$ab=0\ \Longrightarrow\ \theta(a)\theta(b)=0.$

While by Stinespring’s Theorem, every c.p. map can be written as a compression
of a ∗-homomorphism, order zero maps are characterized in [20, Theorem 3.3] as
compressions of ∗-homomorphisms by commuting positive elements.

######  Theorem 1.1.

_(Structure theorem for order zero maps)_ Let $A$ and $B$ be
$\mathrm{C}^{*}$-algebras, $\theta\colon A\longrightarrow B$ a c.p. order zero
map, $C=\mathrm{C}^{*}(\theta(A))$, $\\{u_{\lambda}\\}_{\lambda}$ an
increasing approximate identity for $A$, and
$h=\emph{s.o.-}\lim_{\lambda}\theta(u_{\lambda})\in C^{**}$. Then $0\leq
h\in\mathcal{M}(C)\cap C^{\prime}\subset C^{**}$ with $\|h\|=\|\theta\|$, and
there exists a ∗-homomorphism $\pi_{\theta}\colon
A\longrightarrow\mathcal{M}(C)$ so that for all $a\in A$,

$\theta(a)=h\pi_{\theta}(a)=h^{1/2}\pi_{\theta}(a)h^{1/2}.$

######  Remarks .

(i) ([20, Proposition 3.2]) When $A$ is unital, we have $h=\theta(1_{A})$, and
in general the map $\theta^{\sim}\colon A^{\sim}\longrightarrow\mathcal{M}(C)$
given by $\theta^{\sim}(a+\lambda 1_{A^{\sim}})=\theta(a)+\lambda h$ is the
unique c.p. order zero extension of $\theta$. In any case, $h\geq\theta(a)$
for all $a\in A_{+}^{1}$. Note that for $\theta^{\sim}$ to be order zero, one
can replace $h$ with $1_{\mathcal{M}(C)}$ only if $\theta$ is in fact
multiplicative, since u.c.p. order zero maps are automatically
∗-homomorphisms.

(ii) As pointed out in the introduction, the structure theorem gives a useful
characterization of when a c.p. map with unital domain is order zero: Assume
$A$ and $B$ are $\mathrm{C}^{*}$-algebras with $A$ unital. Then a c.p. map
$\theta\colon A\longrightarrow B$ is order zero if and only if (a) holds,
i.e., $\theta(a)\theta(b)=\theta(1_{A})\theta(ab)$, for all $a,b\in A$.

(iii) Using the orthogonal decomposition of self-adjoint elements, one can
show that for $\theta\colon A\longrightarrow B$ c.p. order zero, for any $a\in
A$ with $\theta(a)\geq 0$ there exists some $b\in A_{+}$ with
$\theta(a)=\theta(b)$. It follows that for all $r\geq 1$

$\theta^{(r)}(\mathrm{M}_{r}(A)_{+})=\mathrm{M}_{r}(\theta(A))\cap\mathrm{M}_{r}(B)_{+}.$

When $\theta$ is moreover assumed to be injective, for any $r\geq 1$ and $a\in
M_{r}(A)$ we have $\theta^{(r)}(a)\geq 0$ if and only if $a\geq 0$.

######  Proposition 1.1.

Suppose $\theta\colon A\longrightarrow B$ is a c.p. order zero map. Then the
following are equivalent.

  1. (i)

$\theta$ is isometric on positive elements of $A$.

  2. (ii)

$\theta$ is isometric.

  3. (iii)

$\theta$ is a complete order embedding.

Moreover, if $A$ is nonzero and if $\theta$ is a complete order embedding
which extends to a c.p.c. order zero map $\theta^{\dagger}\colon
A^{\dagger}\longrightarrow B$ on the (forced) unitization $A^{\dagger}$ of
$A$, then $\theta^{\dagger}$ is also a complete order embedding.

###### Proof.

Clearly (iii) implies (i), and that (i) implies (ii) is a direct consequence
of Theorem 1.1, since for any $a\in A$

(1.6) $\displaystyle\|\theta(a)\|^{2}$
$\displaystyle=\|\theta(a)^{*}\theta(a)\|$
$\displaystyle=\|h^{2}\pi_{\theta}(a^{*}a)\|$
$\displaystyle=\|h^{2}\pi_{\theta}(|a|^{2})\|$
$\displaystyle=\|h^{2}\pi_{\theta}(|a|)^{2}\|$
$\displaystyle=\|\theta(|a|)^{2}\|$ $\displaystyle=\|\theta(|a|)\|^{2}.$

That (ii) implies (iii) was shown in [9, Proposition 1.8].

Now suppose $\theta$ is a complete order embedding that extends to a c.p.c.
order zero map on the (forced) unitization $A^{\dagger}$ of $A$. Note that
$\|\theta^{\dagger}(1_{A^{\dagger}})\|=1$ since $\theta^{\dagger}$ was assumed
to be c.p.c. and $\theta^{\dagger}(1_{A^{\dagger}})\geq\theta(a)$ for any
$a\in A_{+}$ with $\|a\|=\|\theta(a)\|=1$. (We want $A\neq\\{0\\}$ here.) By
the previous implications, we only need to show that $\theta^{\dagger}$ is
isometric on any fixed $a+\lambda 1_{A^{\dagger}}\in A^{\dagger}_{+}$. Note
that in this case $a=a^{*}$, and we may assume $a\neq 0$ and $\lambda\geq 0$.

For any $0<\varepsilon<\|a\|$ we define $g_{\varepsilon}\in C_{0}((0,\|a\|])$
to be linear on $(0,\varepsilon]$ and $1$ on $(\varepsilon,\|a\|]$. Set
$u_{\varepsilon}\coloneqq g_{\varepsilon}(|a|)$ and define
$b_{\varepsilon}=(a_{+}-\varepsilon)_{+}-(a_{-}-\varepsilon)_{+}$ to be the
difference of the $\varepsilon$-cutdowns of $a_{+}$ and $a_{-}$, and note that
$b_{\varepsilon}+\lambda 1_{A^{\dagger}}\geq 0$ by functional calculus. Then
we have $u_{\varepsilon}\in A_{+}$ with $\|u_{\varepsilon}\|=1$ and
$b_{\varepsilon}\in A$ with $b_{\varepsilon}=b_{\varepsilon}^{*}$,
$\|a-b_{\varepsilon}\|<\varepsilon$, $0\leq b_{\varepsilon}+\lambda
u_{\varepsilon}$, and
$u_{\varepsilon}b_{\varepsilon}=b_{\varepsilon}u_{\varepsilon}=b_{\varepsilon}$.
Since $0\in\sigma(b_{\varepsilon})$, for $\chi_{\\{1\\}}(u_{\varepsilon})\in
A^{**}$, we can identify
$\mathrm{C}^{*}(b_{\varepsilon})^{\sim}\cong\mathrm{C}^{*}(b_{\varepsilon},\chi_{\\{1\\}}(u_{\varepsilon}))\cong\mathrm{C}^{*}(b_{\varepsilon},1_{A^{\dagger}})$,
which gives

$\displaystyle\|b_{\varepsilon}+\lambda 1_{A^{\dagger}}\|$
$\displaystyle=\|(b_{\varepsilon}+\lambda\chi_{\\{1\\}}(u_{\varepsilon}))\|$
$\displaystyle=\|(b_{\varepsilon}+\lambda
u_{\varepsilon})\chi_{\\{1\\}}(u_{\varepsilon})\|$
$\displaystyle\leq\|b_{\varepsilon}+\lambda u_{\varepsilon}\|.$

Since $\theta$ is isometric on $A$, we have

$\displaystyle\|a+\lambda 1_{A^{\dagger}}\|$
$\displaystyle\leq\varepsilon+\|b_{\varepsilon}+\lambda 1_{A^{\dagger}}\|$
$\displaystyle\leq\varepsilon+\|b_{\varepsilon}+\lambda u_{\varepsilon}\|$
$\displaystyle=\varepsilon+\|\theta(b_{\varepsilon}+\lambda
u_{\varepsilon})\|$
$\displaystyle\leq\varepsilon+\|\theta^{\dagger}(b_{\varepsilon}+\lambda
1_{A^{\dagger}})\|$ $\displaystyle\leq
2\varepsilon+\|\theta^{\dagger}(a+\lambda 1_{A^{\dagger}})\|.$

It follows that $\theta^{\dagger}$ is isometric on $A^{\dagger}_{+}$ and hence
also on $A^{\dagger}$. ∎

##  2\. Order Units

Drawing inspiration from order units for abstract operator systems, we study
local order units for a (not-necessarily-unital) self-adjoint linear subspace
$X$ of a $\mathrm{C}^{*}$-algebra $B$. Using the ambient
$\mathrm{C}^{*}$-algebraic structure, we can place more sophisticated
assumptions on our local order unit, in particular that $e\in X^{\prime}$, and
then take full advantage of the functional calculus in $B$ to draw conclusions
about $X$ and the pair $(X,e)$.

######  Definition 2.1.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace. We say $e\in B^{1}_{+}$ is a _local order unit_ for $X$ if for any
self-adjoint element $x\in X$, there exists an $R>0$ so that $Re\geq x$. We
call $e$ a _commuting local order unit_ for $X$ if $e$ is a local order unit
for $X$ and $e\in X^{\prime}$.

######  Remarks .

(i) We can view a self-adjoint linear subspace $X\subset B$ of a
$\mathrm{C}^{*}$-algebra as a matrix ordered ∗-vector space as in [14, Chapter
13] by endowing it with the matrix ordering
$\\{\mathrm{M}_{r}(X)\cap\mathrm{M}_{r}(B)_{+}\\}_{r}$ it inherits from $B$.
In this setting a local order unit for a self-adjoint linear subspace of a
$\mathrm{C}^{*}$-algebra $B$ is a local order unit in the sense of Choi and
Effros (see [14, Chapter 13]) for the matrix ordered ∗-vector space
$(X,\\{\mathrm{M}_{r}(X)\cap\mathrm{M}_{r}(B)_{+}\\}_{r})$. (Note that if $e$
is a local order unit for $X\subset B$, then it is automatically a matrix
local order unit, i.e., $e^{(r)}$ is a local order unit for
$\mathrm{M}_{r}(X)$ for all $r\geq 1$. It is also Archimedean in the sense of
[14, Chapter 13] since $\mathrm{M}_{r}(B)_{+}$ is closed.)

(ii) A novelty of our definition is that we do not require that $e\in X$, in
which case it is possible that $X\cap B_{+}=\\{0\\}$ with $X\neq\\{0\\}$, and
we are only guaranteed that $X$ contains nontrivial self-adjoint elements.
(This problem will disappear when we incorporate the further assumption that
$X^{2}\subseteq eX$, see Section 3.)

######  Examples .

(i) Let $X$ be a self-adjoint linear subspace of a unital
$\mathrm{C}^{*}$-algebra $B$ with $1_{B}\in X$. Then any local order unit of
$X$ must be invertible in $B$ by virtue of dominating $1_{B}$ up to some
scalar factor, and so the collection of local order units for $X$ is exactly
$\text{GL}(B)\cap B^{1}_{+}$.

(ii) The subspace $C_{c}((0,1])$ of the non-unital $\mathrm{C}^{*}$-algebra
$C_{0}((0,1])$ has local order unit $\mathrm{id}_{(0,1]}$.

(iii) If $\theta\colon A\longrightarrow B$ is a c.p.c. map from a unital
$\mathrm{C}^{*}$-algebra, then $\theta(1_{A})$ is a local order unit for
$\theta(A)$.

(iv) It follows from Theorem 1.1 that if $\theta\colon A\longrightarrow B$ is
a c.p.c. order zero map from a unital $\mathrm{C}^{*}$-algebra, then
$\theta(1_{A})$ is a commuting local order unit for the self-adjoint linear
subspace $\theta(A)\subset B$. Even if $A$ is non-unital, we can, in some
cases (see for example Corollary 5.1), leverage the structure theorem for c.p.
order zero maps (as in Remark 1(i)) to find a c.p.c. order zero extension
$\theta^{\sim}\colon A^{\sim}\longrightarrow B$ with
$\theta^{\sim}(1_{A^{\sim}})=h$ (with $h$ as in Remark 1(i)). In this case
$\theta^{\sim}(1_{A^{\sim}})\in B^{1}_{+}\cap\theta(A)^{\prime}$ is a
commuting local order unit that is typically _not_ in $\theta(A)$.

The following two propositions give technical facts for commuting local order
units which will be used throughout the paper.

######  Proposition 2.1.

Let $B$ be a $\mathrm{C}^{*}$-algebra.

  1. (i)

For $y,z\in B$ such that $z\geq 0$ and $yz=zy$,

(2.1) $\displaystyle z\geq|y|\ \Longleftrightarrow\ z|y|\geq y^{*}y.$

  2. (ii)

If $X\subset B$ is a self-adjoint linear subspace, then the following are
equivalent for $e\in B^{1}_{+}\cap X^{\prime}$:

    1. (a)

$e$ is a local order unit for $X$.

    2. (b)

For each $x\in X$, there exists an $R>0$ so that $Re\geq|x|$.

  3. (iii)

When either of the equivalent conditions in (ii) hold, the
$\mathrm{C}^{*}$-algebra $\mathrm{C}^{*}(X)$ generated by $X$ in $B$ is
contained in the hereditary $\mathrm{C}^{*}$-algebra $\overline{eBe}$
generated by $e$. In particular, for any $a\in\mathrm{C}^{*}(X)$,
$\lim_{j}\|h_{j}(e)ea-a\|=0$, where $\\{h_{j}\\}_{j}$ are as in (1.3).

###### Proof.

(i) For $s,t\in[0,\infty)$, we have $s\geq t\longleftrightarrow st\geq t^{2}$.
Hence (2.1) follows from the fact that there is an isomorphism
$\mathrm{C}^{*}(z,y^{*}y)\cong C_{0}(\Omega)$ for some locally compact
$\Omega\subset\sigma(z)\times\sigma(y^{*}y)$ which sends $z$ and $|y|$ to the
functions $((s,t)\mapsto s)$ and $((s,t)\mapsto t)$, respectively.

(ii) (b) $\Longrightarrow$ (a) since $|x|\geq x$ for all self-adjoint $x\in
X$. To see that (a) $\Longrightarrow$ (b), fix $x\in X$. Then there exists
$R>0$ so that $Re\geq x^{*}+x$ and $Re\geq i(x-x^{*})$. Since $e$ and
$x+x^{*}$ commute, we also obtain $R^{2}e^{2}\geq(x+x^{*})^{2}$ and likewise
$R^{2}e^{2}\geq(i(x-x^{*}))^{2}$. Then

$\displaystyle R^{2}e^{2}$
$\displaystyle\geq\textstyle{\frac{1}{2}}((x+x^{*})^{2}-(x-x^{*})^{2})=x^{*}x+xx^{*}\geq
x^{*}x,$

and hence $Re\geq|x|$.

(iii) To see that $\mathrm{C}^{*}(X)\subset\overline{eBe}$, it suffices to
show that $x\in\overline{eBe}$ for any self-adjoint $x\in X$. For $x=x^{*}\in
X$ there exists $R>0$ so that $Re\pm x\geq 0$, and so $Re\geq|x|\geq 0$. Hence
$\mathrm{Re}+|x|\in\overline{eBe}$ and so also $Re+x\in\overline{eBe}$ since
$Re+|x|\geq Re+x\geq 0$. It follows that $x=Re+x-Re\in\overline{eBe}$, and so
$\mathrm{C}^{*}(X)\subset\overline{eBe}$. Since $\\{h_{j}(e)e\\}_{j}$ forms an
(increasing) approximate identity of positive contractions for
$\overline{eBe}$, the final claim follows. ∎

######  Proposition 2.2.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace with local order unit $e\in B_{+}^{1}\cap X^{\prime}$. Then the
following hold for all $x\in\textrm{span}\\{X,e\\}$.

  1. (i)

If $e^{n}x=0$ for some $n\geq 1$, then $x=0$. (Nondegeneracy)

  2. (ii)

If $ex=(ex)^{*}$, then $x=x^{*}$.

  3. (iii)

If $ex\geq 0$, then $x\geq 0$.

Moreover, these claims hold for all matrix amplifications $\mathrm{M}_{r}(X)$
with order units $e^{(r)}$.

###### Proof.

Note that if $e$ is a commuting local order unit for $X$, then it will also be
a commuting local order unit for $\textrm{span}\\{X,e\\}$. Hence, for
notational simplicity, we reduce to the assumption $X=\textrm{span}\\{X,e\\}$.
Fix $x\in X$.

For (i), we first handle the case where $x=x^{*}$. Since $ex=xe$, we may
identify $\mathrm{C}^{*}(e,x)$ with $C_{0}(\Omega)$ for some locally compact
Hausdorff space $\Omega\subset\sigma(e)\times\sigma(x)$. If $e^{n}x=0$ for
some $n\geq 1$, then for each pair $(s,t)\in\Omega$ either $s=0$ or $t=0$.
Since $e$ is a local order unit, by Proposition 2.1(ii) there exists an $R>0$
such that $Rs\geq|t|$ for all $(s,t)\in\Omega$. Together these imply that
$\sigma(x)=\\{0\\}$, and so $x=0$.

Now, let $x\in X$ be any element. If $e^{n}x=0$ for some $n\geq 1$, then
$e^{n}x^{*}=(e^{n}x)^{*}=0$, and so $e^{n}(i(x^{*}-x))=0$ where $i(x^{*}-x)$
is self-adjoint. It follows from the part of the proof above that
$i(x^{*}-x)=0$, which means $x$ is self-adjoint. But in that case, our
assumption that $e^{n}x=0$ implies $x=0$.

From (i) we immediately get (ii). Indeed, if $x\in X$ and
$ex=(ex)^{*}=ex^{*}$, then $e(x-x^{*})=0$, and so $x=x^{*}$.

For (iii), assume $ex\geq 0$ for some $x\in X$. Then by (ii), we know
$x=x^{*}$, and so we can again identify $\mathrm{C}^{*}(e,x)$ with
$C_{0}(\Omega)$ as in the proof of (i). Suppose
$t_{0}\in\sigma(x)\cap(-\infty,0)$, and let $s_{0}\in\sigma(e)$ such that
$(s_{0},t_{0})\in\Omega$. Since $e$ is a local order unit, there exists $R>0$
such that $Rs_{0}\geq|t_{0}|>0$. But this implies $s_{0}>0$ and hence
$s_{0}t_{0}<0$, a contradiction.

Finally, since $e$ is automatically a matrix local order unit for $X$ (as
mentioned in Section 2), it follows that for any $r\geq 1$,
$e^{(r)}\in\mathrm{M}_{r}(B)_{+}^{1}\cap\mathrm{M}_{r}(X)^{\prime}$ is a local
order unit for $\mathrm{M}_{r}(X)$, and so claims (i)-(iii) hold for all
matrix amplifications. ∎

######  Remark .

Suppose $A,B$, and $C$ are $\mathrm{C}^{*}$-algebras with $A$ and $B$ unital.
If there exist bounded below c.p. order zero maps $\psi\colon A\longrightarrow
B$ and $\varphi\colon C\longrightarrow B$ with $\psi(A)=\varphi(C)$ and
$\psi(1_{A})=\varphi(1_{C})$, then $\theta\coloneqq\varphi^{-1}\circ\psi\colon
A\longrightarrow C$ is a unital ∗-isomorphism – regardless of the norms of the
maps. Indeed, first note that $\theta$ is a well-defined, bijective, bounded
∗-linear map, so it suffices to show that $\theta(ab)=\theta(a)\theta(b)$ for
fixed $a,b\in A$. The structure theorem for order zero maps tells us that

(2.2)
$\displaystyle\psi(1_{A})\psi(ab)=\psi(a)\psi(b)=\varphi(\theta(a))\varphi(\theta(b))=\varphi(1_{C})\varphi(\theta(a)\theta(b)).$

In Example 2(iv), we saw that $X\coloneqq\psi(A)=\varphi(C)$ and
$e\coloneqq\psi(1_{A})=\varphi(1_{C})$ satisfy the assumptions of Proposition
2.2, and so applying nondegeneracy (Proposition 2.2(i)) to (2.2) tells us that
$\psi(ab)\\\ =\varphi(\theta(a)\theta(b))$. So we have

$\displaystyle\theta(ab)=\varphi^{-1}(\psi(ab))=\varphi^{-1}(\varphi(\theta(a)\theta(b)))=\theta(a)\theta(b).$

This fact can also be deduced readily from other results in the literature
(e.g. Corollary D with Theorem 2.2 from [10] (via [11])).

Notice that for the unit of a $\mathrm{C}^{*}$-algebra, the $R>0$ by which we
scale $1_{A}$ in order to dominate some $a\in A_{\mathrm{s.a.}}$ can always be
taken to be $\|a\|$. This uniformity motivates the following.

######  Definition 2.2.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace. We say an element $e\in B_{+}^{1}$ is an _order unit_ for $X$ if
there exists a $K>0$ so that $K\|x\|e\geq x$ for every $x\in
X_{\mathrm{s.a.}}$. We call the minimal such $K$ the scaling factor of $e$.

######  Remarks .

(i) If $X\neq\\{0\\}$, the scaling factor for a commuting order unit must be
at least 1. Indeed, suppose $x\in X_{\text{s.a.}}$ with $\|x\|=1$ and
$K\|x\|e=Ke\geq x$ for some $K>0$. Let $\Omega\subset\sigma(e)\times\sigma(x)$
be a locally compact Hausdorff space so that $\mathrm{C}^{*}(e,x)\cong
C_{0}(\Omega)$. Then $K\|x\|e=Ke\geq x$ implies that $Ks\geq|t|$ for all
$(s,t)\in\Omega$. Since the spectral radius of $x$ is $1$, we must have some
$s\in\sigma(e)\subset[0,1]$ with $Ks\geq 1$. Hence $K\geq 1$.

(ii) Note also that the scaling factor is indeed a minimum since the set
$\\{K\geq 1\mid K\|x\|e\geq x,\ \forall\ x\in X_{\mathrm{s.a.}}\\}$ is an
interval and $B_{+}$ is closed.

(iii) Just as a bounded map is not automatically completely bounded, if $e$ is
an order unit for $X$, the same scaling factor may not suffice for all matrix
amplifications (unless the scaling factor is $1$ and $X$ is closed; cf.
Corollary 6.2).

######  Examples .

(i) We saw in Example 2(i) that the local order units for a _unital_ self-
adjoint linear subspace $X$ of a unital $\mathrm{C}^{*}$-algebra $B$ are just
$\mathrm{GL}(B)\cap B^{1}_{+}$. These are also the order units, and the
scaling factor for $e\in\mathrm{GL}(B)\cap B_{+}^{1}$ is the minimum value
$R>0$ so that $Re\geq 1_{B}$, i.e., $\|e^{-1}\|$. In particular, that means
that if $1_{B}\in X$, then $1_{B}$ is the unique matrix (local) order unit in
$X$ with scaling factor $R=1$.

(ii) If $\theta\colon A\longrightarrow B$ is a bounded below c.p. map from a
unital $\mathrm{C}^{*}$-algebra $A$, then the image of the unit becomes an
order unit for $\theta(A)$ with scaling factor at most $\|\theta^{-1}\|$ since

(2.3)
$\displaystyle\|\theta^{-1}\|\|\theta(a)\|\theta(1_{A})\geq\|a\|\theta(1_{A})\geq\theta(a)$

for all $\theta(a)\in\theta(A)$.

When $\theta$ is order zero, $\|\theta^{-1}\|$ is the scaling factor of
$\theta(1_{A})$ as the following proposition shows.

######  Proposition 2.3.

Let $\theta\colon A\longrightarrow B$ be an injective c.p.c. order zero map.

  1. (i)

When $A$ is unital, we define $R_{0}\coloneqq 0$ and

$R_{b}\coloneqq\min\\{R\geq 1\mid R\|\theta(b)\|\theta(1_{A})\geq\theta(b)\\}$

for $b\in A_{\mathrm{s.a.}}\backslash\\{0\\}$. Then
$R_{|a|}\|\theta(a)\|=\|a\|$ for each $a\in A\backslash\\{0\\}$. If
$\|\theta(a)\|\theta(1_{A})\geq\theta(a)$ for all $a\in A_{+}$, then $\theta$
is a complete order embedding.

  2. (ii)

If $\theta$ is bounded below and extends to a c.p.c. order zero map
$\theta^{\sim}\colon A^{\sim}\longrightarrow B$, then $\theta^{\sim}$ is also
bounded below, and $\theta^{\sim}(1_{A^{\sim}})$ is an order unit for
$\theta(A)$ with scaling factor $\|\theta^{-1}\|$. When $\theta$ is isometric,
this is also its scaling factor for $\theta^{\sim}(A^{\sim})$. In particular,
the scaling factor of $\theta^{\sim}(1_{A^{\sim}})$ is $1$ if and only if
$\theta^{\sim}$ is completely isometric.

######  Remark .

When $A$ is unital, then $\theta^{\sim}=\theta$, and Proposition 2.3(ii) tells
us $\theta(1_{A})$ is an order unit for $\theta(A)$ with order scale
$\|\theta^{-1}\|$. When $A$ is not unital, applying this same argument to
$A^{\sim}$ tells us $\theta^{\sim}(1_{A^{\sim}})$ is an order unit for
$\theta^{\sim}(A^{\sim})$ with scaling factor $\|(\theta^{\sim})^{-1}\|$. Also
note that the first claim in (ii) is about the scaling factor of
$\theta^{\sim}(1_{A^{\sim}})$ for $\theta(A)$, not the scaling factor of
$\theta^{\sim}(1_{A^{\sim}})$ for $\theta^{\sim}(A^{\sim})$, which, as we have
just explained, is $\|(\theta^{\sim})^{-1}\|$.

###### Proof.

(i) Assume $A$ is unital, and let $b\in A_{\text{s.a.}}\backslash\\{0\\}$.
Since $\theta$ is positive,
$\textstyle{\frac{\|b\|}{\|\theta(b)\|}}\|\theta(b)\|\theta(1_{A})\geq\theta(b)$,
and so the interval $\\{R\geq 1\mid
R\|\theta(b)\|\theta(1_{A})\geq\theta(b)\\}$ is nonempty. Since $B_{+}$ is
closed and $\theta$ is continuous, this interval has a minimum, and so $R_{b}$
is well-defined and moreover $\textstyle{\frac{\|b\|}{\|\theta(b)\|}}\geq
R_{b}$.

Let $c\in A_{+}$. Since $\theta$ is injective c.p.c. order zero and
$R_{c}\|\theta(c)\|\theta(1_{A})-\theta(c)\geq 0$, Remark 1(iii) tells us that
$\theta^{-1}\big{(}R_{c}\|\theta(c)\|\theta(1_{A})-\theta(c)\big{)}=R_{c}\|\theta(c)\|1_{A}-c\geq
0$. Hence we must also have $R_{c}\|\theta(c)\|\geq\|c\|$, and so
$R_{c}\|\theta(c)\|=\|c\|$. Then it follows from (1.6) that
$R_{|a|}\|\theta(a)\|=\|a\|$ for any $a\in A$.

Now suppose $R_{b}=1$ for all $b\in A_{+}\backslash\\{0\\}$. Then it follows
that $\theta$ is isometric on $A_{+}$, and hence Proposition 1.1 tells us
$\theta$ is a complete order embedding.

(ii) Now assume $\theta$ is bounded below and $A$ is non-unital. We first
observe that $\theta^{\sim}$ is also bounded below, or equivalently injective
with closed image. Indeed, as Banach spaces,222We use here the well-known
Banach space fact that a linear subspace of a Banach space $\mathcal{X}$ is
closed if and only if its linear span with any finite-dimensional subspace of
$\mathcal{X}$ is closed. $\theta(A)$ is closed if and only if
$\theta^{\sim}(A^{\sim})=\textrm{span}\\{\theta(A),\theta^{\sim}(1_{A^{\sim}})\\}$
is, and so we only need to verify that $\theta^{\sim}$ is injective. To that
end, let $a\in A$ and $\lambda\in\mathbb{C}$ with $\theta^{\sim}(a+\lambda
1_{A^{\sim}})=\theta(a)+\lambda\theta^{\sim}(1_{A^{\sim}})=0$. Then either
$\theta(a)=0$ and $\lambda=0$, whence $a+\lambda 1_{A^{\sim}}=0$ since
$\theta$ is injective, or $u\coloneqq-a/\lambda\in A$ maps to
$\theta^{\sim}(1_{A^{\sim}})$. But then
$\theta^{\sim}(1_{A^{\sim}})\theta^{\sim}(b)=\theta^{\sim}(u)\theta^{\sim}(b)=\theta^{\sim}(1_{A^{\sim}})\theta^{\sim}(ub)$
for any $b\in A$. By nondegeneracy (Proposition 2.2(i)), it follows that
$\theta(b)=\theta(ub)$ and similarly $\theta(b)=\theta(bu)$ for all $b\in A$.
It follows from injectivity that $u$ is the unit of $A$, contradicting our
assumption. Hence $\theta^{\sim}$ is also bounded below.

We then know from Example 2(ii) that $\theta^{\sim}(1_{A^{\sim}})$ is an order
unit for $\theta^{\sim}(A^{\sim})$ and hence also for $\theta(A)$. It remains
to check that its scaling factor for $\theta(A)$ is $\|\theta^{-1}\|$.

Let $R$ denote the scaling factor of $\theta^{\sim}(1_{A^{\sim}})$ for
$\theta(A)$. Then $R\leq\|\theta^{-1}\|$ by (2.3). Note that if $b\in
A_{\mathrm{s.a.}}$ then $R_{b}\leq R_{|b|}$ since

$R_{|b|}\|\theta(b)\|\theta^{\sim}(1_{A^{\sim}})\overset{\eqref{eq: nom on
pos}}{=}R_{|b|}\|\theta(|b|)\|\theta^{\sim}(1_{A^{\sim}})\geq\theta(|b|)\geq\theta(b).$

Putting all this together, we have using (i)

$\displaystyle\|\theta^{-1}\|$ $\displaystyle\geq R$
$\displaystyle=\textstyle{\sup_{a\in A_{\mathrm{s.a.}}}R_{a}}$
$\displaystyle=\textstyle{\sup_{a\in A_{+}}R_{a}}$
$\displaystyle=\textstyle{\sup_{a\in A}R_{|a|}}$
$\displaystyle=\textstyle{\sup_{a\in
A\backslash\\{0\\}}\frac{\|a\|}{\|\theta(a)\|}}$
$\displaystyle=\|\theta^{-1}\|.$

It follows that $R=1$ if and only if $\theta$ is isometric if and only if
$\theta$ is completely isometric if and only if $\theta^{\sim}$ is completely
isometric by Proposition 1.1. Applying the result now to $\theta^{\sim}$ tells
us the scaling factor of $\theta^{\sim}(1_{A^{\sim}})$ for
$\theta^{\sim}(A^{\sim})$ is also $1$. ∎

######  Example .

Using the same arguments as in [9, Lemma 3.4], one shows that a system of c.p.
approximations
$A\xlongrightarrow{\psi_{n}}F_{n}\xlongrightarrow{\varphi_{n}}A$ of a
separable nuclear $\mathrm{C}^{*}$-algebra $A$ with $\psi_{n}$ and
$\varphi_{n}\circ\psi_{n}$ all c.p.c. and with
$\limsup_{n}\|\varphi_{n}\|<\infty$ induces a bounded below c.p.c. map
$\Psi\colon A\longrightarrow\prod F_{n}/\bigoplus F_{n}$, which is order zero
when the downwards maps $(\psi_{n})_{n}$ are approximately order zero. When
$A$ is unital, $\Psi(1_{A})$ is an order unit with scaling factor
$R\leq\limsup_{n}\|\varphi_{n}\|$. If we assume
$A\xlongrightarrow{\psi_{n}}F_{n}\xlongrightarrow{\varphi_{n}}A$ is a c.p.c.
approximation, then we automatically get $R=1$.

It is perhaps enlightening to see when a local order unit fails to be an order
unit.

######  Example .

(i) Fix $h\coloneqq((n+1)^{-2})_{n}\in C_{0}(\mathbb{N})\subset
C_{b}(\mathbb{N})$. Define $\theta\colon C_{b}(\mathbb{N})\longrightarrow
C_{b}(\mathbb{N})$ by $x\longmapsto hx$. Then this map is c.p.c. order zero
and injective, and hence $\theta(1)=h\in C_{b}(\mathbb{N})_{+}^{1}$ is a local
order unit for $\theta(C_{b}(\mathbb{N}))$. For each $n\in\mathbb{N}$, set
$x_{n}\coloneqq(n+1)\delta_{n}\in C_{b}(\mathbb{N})$. Then
$\|\theta(x_{n})\|=(n+1)^{-1}$ for each $n\in\mathbb{N}$, and for any $R>0$,

$\displaystyle\big{(}R\|\theta(x_{n})\|h-\theta(x_{n})\big{)}(n)=\textstyle{\frac{R-(n+1)^{2}}{(n+1)^{3}}}.$

Hence there is no $R>0$ so that $R\|\theta(x_{n})\|h-\theta(x_{n})\geq 0$ for
all $n\in\mathbb{N}$, and $h$ is not an order unit for
$\theta(C_{b}(\mathbb{N}))$.

Note that $\theta$ does not have closed image (and hence is not bounded
below).

(ii) Similarly, consider the $\mathrm{C}^{*}$-algebra $C([0,1])$, the element
$h\coloneqq\mathrm{id}_{[0,1]}\in C([0,1])$, and the c.p. map $\theta\colon
C([0,1])\longrightarrow C_{0}((0,1])$ given by $\theta(f)=(hf)|_{(0,1]}$
(which we write as just $hf$). Here again, $h=\theta(1)$ is a local order unit
but not an order unit. For each $j\geq 1$, define $h_{j}\in C([0,1])$ as in
(1.3). Then for each $j\geq 1$, we have $\|\theta(h_{j})\|=1$, and so there is
no single value $R>0$ so that for every $j\geq 1$ and $t\in(0,1]$,

$R\|\theta(h_{j})\|h(t)-\theta(h_{j})(t)=t(R-h_{j}(t))\geq 0.$

Again, notice that $\theta\big{(}C([0,1])\big{)}$ is not closed in
$C_{0}((0,1])$.

##  3\. A $\mathrm{C}^{*}$-structure

In this section, we prove our main theorem:

######  Theorem 3.1.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace. Suppose there exists a local order unit $e\in B_{+}^{1}\cap
X^{\prime}$ for $X$ so that $X^{2}\subset eX$. Then there is an associative
bilinear map
${\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\times X\longrightarrow X$ satisfying

(3.1) $\displaystyle
xy=e(x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y)$

for all $x,y\in X$, so that
$(X,{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})$
is a ∗-algebra. When $e\in X$, then $e$ is the unit for
$(X,{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})$.

Moreover, the map
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\longrightarrow[0,\infty)$ defined for each $x\in X$ by

(3.2)
$\displaystyle\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}=\lim_{j}\|h_{j}(e)x\|$

with $\\{h_{j}\\}_{j}\subset C_{0}((0,1])_{+}$ as given in (1.3), is a
pre-$\mathrm{C}^{*}$-norm on
$(X,{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})$.

We begin by establishing further descriptions of
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$.
(Note that the following Proposition does not require the $X^{2}\subset eX$
assumption.)

######  Proposition 3.1.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace with local order unit $e\in B_{+}^{1}\cap X^{\prime}$. Then the map
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\longrightarrow[0,\infty)$ defined in (3.2) is a norm on $X$. Moreover, for
each $x\in X$ we have

(3.3)
$\displaystyle\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
$\displaystyle=\lim_{j}\|h_{j}(e)x\|$ $\displaystyle=\sup_{j}\|h_{j}(e)x\|$
$\displaystyle=\lim_{j}\|h_{j}(e)^{2}ex\|$
$\displaystyle=\sup_{j}\|h_{j}(e)^{2}ex\|.$

###### Proof.

Fix $x\in X$. First we show that the limit in (3.2) exists. Since
$(h_{j})_{j}$ is increasing, for all $j\geq 1$ we have (using that $x$ and $e$
commute)

$\|h_{j}(e)x\|^{2}=\|x^{*}h_{j}(e)^{2}x\|\leq\|x^{*}h_{j+1}(e)^{2}x\|=\|h_{j+1}(e)x\|^{2}.$

So to show the limits exist, it suffices to show that the sequence is bounded.

Since $e$ is a local order unit for $X$, we can choose for any self-adjoint
$y\in X$ an $R>0$ such that $y\leq Re$. Since $e\in X^{\prime}$, we have
$\|h_{j}(e)y\|\leq\|Rh_{j}(e)e\|\leq R$ for all $j\geq 1$. Applying this to
$(x+x^{*})/2$ and $(x-x^{*})/(2\mathrm{i})$, we conclude that
$(\|h_{j}(e)x\|)_{j}$ is bounded and moreover

(3.4)
$\displaystyle\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}=\lim_{j}\|h_{j}(e)x\|=\sup_{j}\|h_{j}(e)x\|<\infty.$

It is straightforward to check that
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
is a seminorm. To see that it is a norm, suppose $x\in X$ and
$0=\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}=\sup_{j}\|h_{j}(e)x\|$.
Because $0\leq e\leq h_{j}(e)$ for each $j\geq 1$ and $e$ commutes with
$x^{*}x$, we have

$\|e|x|\|^{2}=\|e^{2}|x|^{2}\|\leq\|ex^{*}x\|\leq\|h_{j}(e)x^{*}x\|\leq\|x^{*}\|\|h_{j}(e)x\|=0$

for all $j\geq 1$. So $e|x|=0$, $e$ commutes with $|x|$, and, by Proposition
2.1(ii), there exists $R>0$ with $Re\geq|x|$. From this, Proposition 2.1(i)
tells us $x^{*}x\leq Re|x|=0$, and hence $x=0$.

All that remains is to show the additional claim: for each fixed $x\in
X\backslash\\{0\\}$ we have

(3.5)
$\displaystyle\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}=\lim_{j}\|h_{j}(e)^{2}ex\|=\sup_{j}\|h_{j}(e)^{2}ex\|.$

Since $0\leq h_{j}(e)^{4}e^{2}\leq h_{j+1}(e)^{4}e^{2}$ and $0\leq
h_{j}(e)^{4}e^{2}\leq h_{j}(e)^{2}$ for all $j\geq 1$, we have

$\displaystyle\|h_{j}(e)^{2}ex\|^{2}$
$\displaystyle=\|x^{*}h_{j}(e)^{4}e^{2}x\|\leq\|x^{*}h_{j+1}(e)^{4}e^{2}x\|=\|h_{j+1}(e)^{2}ex\|^{2}\
\text{ and}$ $\displaystyle\|h_{j}(e)^{2}ex\|^{2}$
$\displaystyle=\|x^{*}h_{j}(e)^{4}e^{2}x\|\leq\|x^{*}h_{j}(e)^{2}x\|=\|h_{j}(e)x\|^{2}\overset{\eqref{mon
and
bdd}}{\leq}\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}^{2}$

for all $j\geq 1$. This implies the sequence
$\big{(}\|h_{j}(e)^{2}ex\|\big{)}_{j}$ is bounded and nondecreasing, and so
the limit in (3.5) exists and furthermore

$\displaystyle\lim_{j}\|h_{j}(e)^{2}ex\|=\sup_{j}\|h_{j}(e)^{2}ex\|\leq\sup_{j}\|h_{j}(e)x\|=\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}.$

It remains to show that this is an equality. Since $e\in B_{+}\cap
X^{\prime}$, we have $\|h_{j}(e)|x|\|=\|h_{j}(e)x\|$ and likewise
$\|h_{j}(e)^{2}e|x|\|=\|h_{j}(e)^{2}ex\|$ for all $j\geq 1$, and so we instead
show that

$\sup_{j}\|h_{j}(e)|x|\|\leq\sup_{j}\|h_{j}(e)^{2}e|x|\|.$

We fix $k\geq 1$ and aim to find $j\geq 1$ so that
$\|h_{j}(e)^{2}e|x|\|\geq\|h_{k}(e)|x|\|$. Since $e$ commutes with $|x|$, we
identify $\mathrm{C}^{*}(e,|x|)=C_{0}(\Omega)$ as before for some locally
compact Hausdorff space $\Omega\subset\sigma(e)\times\sigma(|x|)$. First, we
claim that there exists $\varepsilon>0$ such that

(3.6)
$\displaystyle\|h_{k}(e)|x|\|=\sup_{(s,t)\in\Omega,s\geq\varepsilon}|h_{k}(s)t|.$

Since $(s,t)\longmapsto h_{k}(s)t$ is a continuous function on
$[0,1]\times[0,\infty)$, there exists a $\delta>0$ such that
$|h_{k}(s)t|<\|h_{k}(e)|x|\|/2$ whenever $|(s,t)|<\delta$ (since $x\neq 0$, we
can conclude from Proposition 2.2(ii) along with Proposition 2.1(i) that
$\|h_{k}(e)|x|\|>0$). By Proposition 2.1(ii), there exists $R>0$ so that
$Rs\geq t$ for all $(s,t)\in\Omega$. For
$s<\varepsilon\coloneqq\delta(R^{2}+1)^{-1/2}$, we have

$|(s,t)|^{2}=s^{2}+t^{2}\leq s^{2}+R^{2}s^{2}<\delta^{2},$

which establishes the claim.

Now choose $j\geq k$ such that $1/j<\varepsilon$, and consider
$(s,t)\in\Omega$ with $s\geq\varepsilon$. Then $s>1/j$ and so
$h_{j}(s)^{2}st=\textstyle{\frac{t}{s}}$. If $s>1/k$, then

$h_{k}(s)t=\textstyle{\frac{t}{s}}=h_{j}(s)^{2}st,$

and if $s\leq 1/k$, then

$h_{k}(s)t=k^{2}st\leq\textstyle{\frac{t}{s}}=h_{j}(s)^{2}st.$

Hence for all $(s,t)\in\Omega$ with $s\geq\varepsilon$, we have

$h_{j}(s)^{2}st=\textstyle{\frac{t}{s}}\geq h_{k}(s)t,$

which gives us

$\displaystyle\|h_{j}(e)^{2}e|x|\|\geq\sup_{(s,t)\in\Omega,s\geq\varepsilon}|h_{j}(s)^{2}st|\geq\sup_{(s,t)\in\Omega,s\geq\varepsilon}|h_{k}(s)t|\overset{\eqref{eq:
cts}}{=}\|h_{k}(e)|x|\|.$

This establishes (3.5). ∎

We are now ready to prove Theorem 3.1. Since we are establishing an
alternative product on a subspace of a $\mathrm{C}^{*}$-algebra, we emphasize
here that multiplication in $B$ will always be denoted by the usual
concatenation (i.e., $xy$). For example for
$e(x\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,y)$,
denotes the $B$-product of $e$ and
$x\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,y$.

###### Proof of Theorem 3.1.

The fact that $X^{2}\subset eX$ along with the nondegeneracy of $e$ (i.e.,
Proposition 2.2(i)) guarantee that for each $(x,y)\in X\times X$, there exists
a unique element $z$ in $X$ such that $xy=ez$. We denote this element by
$x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y$,
and conclude that the map
${\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\times X\longrightarrow X$ given by $(x,y)\longmapsto
x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y$ is
well-defined.

The arguments for associativity and bilinearity use nondegeneracy of $e$ in
the same way, so we just give one argument for associativity. Let $x,y,z\in X$
and $\lambda\in\mathbb{C}$. Then we have

$\displaystyle
e^{2}(x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(y{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}z))$
$\displaystyle=ex(y{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}z)$
$\displaystyle=xe(y{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}z)$
$\displaystyle=x(yz)$ $\displaystyle=(xy)z$
$\displaystyle=e(x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y)z$
$\displaystyle=e^{2}((x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y){\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}z).$

By nondegeneracy of $e$ (or actually $e^{2}$ in this case), it follows that
$x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(y{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}z)=(x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y){\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}z$.

To see that
$(X,{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})$
is a ∗-algebra, we compute for $x,y\in X$

$e(x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y)^{*}=(e(x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y))^{*}=(xy)^{*}=y^{*}x^{*}=e(y^{*}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x^{*}).$

Then it follows from nondegeneracy that
$(x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y)^{*}=y^{*}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x^{*}$.

When $e\in X$, nondegeneracy also gives

$0=ex-
ex=e(e{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x)-ex=e(e{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x-x)\Longrightarrow
e{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x=x,$

and similarly
$x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}e=x$.
Hence $e$ becomes the unit for
$(X,{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})$.

From Proposition 3.1, we know that
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\longrightarrow[0,\infty)$ defines a norm on $X$. It remains to show that it
is a pre-$\mathrm{C}^{*}$-norm on
$(X,{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})$.
We begin by checking that
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
is a algebra norm with respect to
${\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\times X\longrightarrow X$ (i.e., it is submultiplicative). Let $x,y\in X$.
With (3.3) we have that

$\displaystyle\|x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
$\displaystyle=\lim_{j}\|h_{j}(e)^{2}e(x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y)\|$
$\displaystyle=\lim_{j}\|h_{j}(e)^{2}xy\|$
$\displaystyle\leq(\lim_{j}\|h_{j}(e)x\|)(\lim_{j}\|h_{j}(e)y\|)$
$\displaystyle=\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\|y\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}.$

Likewise we check the $\mathrm{C}^{*}$-identity for each $x\in X$ again using
(3.3)

$\displaystyle\|x^{*}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
$\displaystyle=\lim_{j}\|h_{j}(e)^{2}e(x^{*}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x)\|$
$\displaystyle=\lim_{j}\|h_{j}(e)^{2}x^{*}x\|$
$\displaystyle=\lim_{j}\|h_{j}(e)x\|^{2}$
$\displaystyle=\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}^{2}.$

Hence
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
is a pre-$\mathrm{C}^{*}$-norm on
$(X,\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,)$.
∎

######  Definition 3.1.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace with local order unit $e\in B_{+}^{1}\cap X^{\prime}$ with
$X^{2}\subset eX$. Let
${\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\times X\longrightarrow X$ be the product and
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\longrightarrow[0,\infty)$ the norm from Theorem 3.1. We denote the induced
$\mathrm{C}^{*}$-algebra
$\overline{(X,{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})}^{\|\cdot\|_{{}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}}}$
by
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$,
and we call this the _$\mathrm{C}^{*}$ -algebra associated to $(X,e)$_.

Note that when $X$ is already complete with respect to
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$,
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
and $X$ are equal as ∗-vector spaces. This is often the case in our
applications.

######  Examples .

(i) Let $B$ be a $\mathrm{C}^{*}$-algebra and $e\in B_{+}$ with $\|e\|\leq 1$.
Then $X=\textrm{span}\\{e\\}$ is readily seen to satisfy the criteria of
Theorem 3.1, and in this case
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\cong\mathbb{C}$.

(ii) Let $A$ and $B$ be $\mathrm{C}^{*}$-algebras and $\theta\colon
A\longrightarrow B$ a c.p.c. map. Then $X=\theta(A)$ is a self-adjoint linear
subspace of $B$, and we can define an associative bilinear map
${\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\times X\longrightarrow X$ given by
$\theta(a){\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\theta(b)=\theta(ab)$.
When $A$ is unital and $\theta$ is moreover order zero,
$e\coloneqq\theta(1_{A})\in B_{+}^{1}\cap X\cap X^{\prime}$ is a local order
unit for $X$ and $eX=X^{2}$ by the structure theorem for order zero maps. In
particular, we have
$e(\theta(a){\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\theta(b))=\theta(1_{A})\theta(ab)=\theta(a)\theta(b)$
for all $\theta(a),\theta(b)\in X$. Hence $(\theta(A),\theta(1_{A}))$ also
satisfy the criteria of Theorem 3.1.

######  Proposition 3.2.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace with local order unit $e\in B_{+}^{1}\cap X^{\prime}$ with
$X^{2}\subset eX$. Then we have the following for all $x,y\in X$ with respect
to the product  from Theorem 3.1.

  1. (i)

$xy=0$ if and only if
$x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y=0$,

  2. (ii)

$xy=yx$ if and only if
$x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y=y{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x$,
and

  3. (iii)

$x^{*}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x\in
B_{+}$.

###### Proof.

Set $x,y\in X$. By nondegeneracy (Proposition 2.2(i)) we have that

$0=xy=e(x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y)\iff
x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y=0.$

If $xy=yx$, then

$e(x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y)=xy=yx=e(y{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x)\
\Longrightarrow\
x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y=y{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x,$

and likewise
$x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y=y{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x$
implies $xy=yx$. Finally,
$e(x^{*}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x)=x^{*}x\geq
0$ implies
$x^{*}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x\geq
0$ by Proposition 2.2(iv). ∎

######  Remark .

In particular, this tells us that, even when $e\notin X$, we are still
guaranteed to have $X\cap B_{+}\neq\\{0\\}$ when $X\neq\\{0\\}$, which
resolves the concern raised in Remark 2(ii).

We close by confirming that our construction is compatible with matrix
amplifications.

######  Corollary 3.1.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace with local order unit $e\in B_{+}^{1}\cap X^{\prime}$ with
$X^{2}\subset eX$. Then for every $r\geq 1$,
$e^{(r)}\in\mathrm{M}_{r}(B)_{+}^{1}\cap\mathrm{M}_{r}(X)^{\prime}$ is a local
order unit for $\mathrm{M}_{r}(X)$ with $\mathrm{M}_{r}(X)^{2}\subset
e^{(r)}\mathrm{M}_{r}(X)$, where $e^{(r)}\coloneqq 1_{\mathrm{M}_{r}}\otimes
e$. The $\mathrm{C}^{*}$-algebra associated to $(\mathrm{M}_{r}(X),e^{(r)})$
is
$\mathrm{M}_{r}\big{(}\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\big{)}$.

###### Proof.

Fix $r\geq 1$. By Remark 2(i) $e$ is automatically a matrix local order unit.
Moreover, $e^{(r)}\in\mathrm{M}_{r}(X)^{\prime}$, and we have the inclusions

$\mathrm{M}_{r}(X)^{2}\subset\textrm{span}\\{\mathrm{M}_{r}(X^{2})\\}\subset\mathrm{M}_{r}(eX)=e^{(r)}\mathrm{M}_{r}(X).$

Applying Theorem 3.1 gives us an associative bilinear map
$\star\colon\mathrm{M}_{r}(X)\times\mathrm{M}_{r}(X)\longrightarrow\mathrm{M}_{r}(X)$
satisfying

$\displaystyle xy=e^{(r)}(x{\star}y)$

for all $x,y\in\mathrm{M}_{r}(X)$, and a pre-$\mathrm{C}^{*}$-norm
$\|\cdot\|_{\star}$ on $(\mathrm{M}_{r}(X),{\star})$ defined for each
$x\in\mathrm{M}_{r}(X)$ by

$\displaystyle\|x\|_{\star}\coloneqq\lim_{j}\|(1_{\mathrm{M}_{r}}\otimes
h_{j}(e))x\|=\lim_{j}\|h_{j}(1_{\mathrm{M}_{r}}\otimes
e)x\|=\lim_{j}\|h_{j}(e^{(r)})x\|,$

where $\\{h_{j}\\}_{j}$ is defined as in (1.3).

For $x=(x_{kl})_{k,l=1}^{r}$ and $y=(y_{kl})_{k,l=1}^{r}\in\mathrm{M}_{r}(X)$,

$\displaystyle e^{(r)}(x{\star}y)$ $\displaystyle=xy$
$\displaystyle=\big{(}\textstyle{\sum}_{m=1}^{r}x_{km}y_{ml}\big{)}_{k,l=1}^{r}$
$\displaystyle=\big{(}\textstyle{\sum}_{m=1}^{r}e(x_{km}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y_{ml})\big{)}_{k,l=1}^{r}$
$\displaystyle=e^{(r)}\big{(}\textstyle{\sum}_{m=1}^{r}(x_{km}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y_{ml})\big{)}_{k,l=1}^{r},$

and so nondegeneracy (Proposition 2.2(i)) tells us that

$x{\star}y=\big{(}\textstyle{\sum}_{m=1}^{n}(x_{km}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y_{ml})\big{)}_{k.l=1}^{r},$

whence we may write  for both products.

Next, since for each $j\geq 1$ and
$x=(x_{kl})_{k,l=1}^{r}\in\mathrm{M}_{r}(X)$

$\max_{1\leq k,l\leq
r}\|h_{j}(e)x_{kl}\|\leq\|(h_{j}(e)x_{kl})_{k,l=1}^{r}\|\leq r\max_{1\leq
k,l\leq r}\|h_{j}(e)x_{kl}\|,$

it follows that

$\displaystyle\max_{1\leq k,l\leq
r}\|x_{kl}\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\leq\|x\|_{\star}\leq
r\max_{1\leq k,l\leq
r}\|x_{kl}\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}.$

Hence we have

$\overline{\mathrm{M}_{r}(X)}^{\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}}=\mathrm{M}_{r}(\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)).\qed$

With this, we may write
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\big{(}\mathrm{M}_{r}(X)\big{)}=\mathrm{M}_{r}\big{(}\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\big{)}$
for all $r\geq 1$ without ambiguity.

##  4\. A c.p.c. order zero map

We now turn our attention to the map
$\mathrm{id}_{X}:(X,\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})\longrightarrow(X,\|\cdot\|)$
between the normed linear spaces
$(X,\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})$
and $(X,\|\cdot\|)$ (where
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
is as defined in (3.2) and $\|\cdot\|$ is the norm inherited from the
$\mathrm{C}^{*}$-algebra $B$ as before), which we want to extend to a c.p.c.
order zero map
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\longrightarrow
B$. The map $\mathrm{id}_{X}$ is already a linear ∗-preserving bijection, but
in order to extend it to
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$,
we need to show that it is bounded. In fact, it is completely contractive as
the following proposition shows.

######  Proposition 4.1.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace with local order unit $e\in B_{+}^{1}\cap X^{\prime}$, and let
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
be as defined in (3.2). Then for each $r\geq 1$ and $x\in\mathrm{M}_{r}(X)$,

$\|x\|\leq\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}.$

Moreover, if we assume that $X$ is closed in $\|\cdot\|$, then there exists an
$R\geq 1$ such that for all $x\in X$,

$\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\leq
R\|x\|.$

In particular, if $X=\overline{X}^{\|\cdot\|}$, then
$X=\overline{X}^{\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}}$.

###### Proof.

Let $\\{h_{j}\\}_{j}$ be as defined (1.3), and fix $r\geq 1$ and
$x\in\mathrm{M}_{r}(X)$. Since $e^{(r)}$ is still a local order unit for
$\mathrm{M}_{r}(X)$, $\\{h_{j}(e^{(r)})e^{(r)}\\}_{j}\subset X^{\prime}$ forms
an increasing approximate identity for $X$ by Proposition 2.1(iii). Combining
this with Corollary 3.1 and (3.3), we have

(4.1) $\displaystyle\|x\|$ $\displaystyle=\lim_{j}\|h_{j}(e^{(r)})e^{(r)}x\|$
$\displaystyle=\sup_{j}\|h_{j}(e^{(r)})e^{(r)}x\|$
$\displaystyle\leq\sup_{j}\|e^{(r)}\|\|h_{j}(e^{(r)})x\|$
$\displaystyle\leq\|e^{(r)}\|\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
$\displaystyle\leq\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}.$

If $X$ is closed in $\|\cdot\|$, then $X$ is a Banach space with respect to
$\|\cdot\|$, and the maps $H_{j}\colon X\longrightarrow B$ given for each
$j\geq 1$ for all $x\in X$ by

(4.2) $\displaystyle H_{j}(x)=h_{j}(e)x$

are bounded linear maps. Since
$\sup_{j}\|H_{j}(x)\|=\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}<\infty$
for each $x\in X$ (by Proposition 3.1), the Banach–Steinhaus theorem (or
Uniform Boundedness Principle) guarantees that
$0\leq\sup_{j}\|H_{j}\|=:R<\infty$. Thus we have for all $x\in X$

(4.3)
$\displaystyle\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
$\displaystyle=\lim_{j}\|h_{j}(e)x\|$ $\displaystyle=\sup_{j}\|h_{j}(e)x\|$
$\displaystyle=\sup_{j}\|H_{j}(x)\|$ $\displaystyle\leq\sup_{j}\|H_{j}\|\|x\|$
$\displaystyle=R\|x\|.$

Moreover, (4.1) tells us $R\geq 1$. ∎

######  Theorem 4.1.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace with local order unit $e\in B_{+}^{1}\cap X^{\prime}$ with
$X^{2}\subset eX$. Let
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
be the associated $\mathrm{C}^{*}$-algebra as in Definition 3.1. Then the map
$\mathrm{id}_{X}:(X,\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})\longrightarrow(X,\|\cdot\|)$
extends to a c.p.c. order zero map
$\Phi\colon\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\longrightarrow
B$ with
$\Phi(\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X))\subset\overline{X}^{\|\cdot\|}$.

If $\Phi$ is injective, then $\Phi^{(r)}(x)\geq 0$ if and only if $x\geq 0$
for any $r\geq 1$ and
$x\in\mathrm{M}_{r}(\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X))$.
Moreover $X$ is closed in $\|\cdot\|$ if and only if $\Phi=\mathrm{id}_{X}$
and $\Phi$ is bounded below.

###### Proof.

It follows from Proposition 4.1 that the map
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\supset
X\xlongrightarrow{\mathrm{id}_{X}}X\subset B$ is completely contractive and
hence extends to a completely contractive map
$\Phi\colon\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\longrightarrow
B$ with
$\Phi(\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X))\subset\overline{X}^{\|\cdot\|}$.
To see that $\Phi$ is completely positive, fix $r\geq 1$. By Corollary 3.1, we
may apply Proposition 3.2(iii) and conclude that for each
$x\in\mathrm{M}_{r}(X)$,

$\Phi^{(r)}(x^{*}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x)=x^{*}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x\in\mathrm{M}_{r}(B)_{+}.$

Since
$(\mathrm{M}_{r}(X),{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})$
is a dense ∗-subalgebra of
$\mathrm{M}_{r}(\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X))=\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(\mathrm{M}_{r}(X))$,
it follows that
$\\{x^{*}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x\
:\ x\in\mathrm{M}_{r}(X)\\}$ is dense in
$\mathrm{M}_{r}(\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X))_{+}$,
and so we have

$\displaystyle\Phi^{(r)}(\mathrm{M}_{r}(\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X))_{+})$
$\displaystyle\subset\overline{\Phi^{(r)}(\\{x^{*}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x\
:\ x\in\mathrm{M}_{r}(X)\\})}^{\|\cdot\|}$
$\displaystyle=\overline{\\{x^{*}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}x\
:\ x\in\mathrm{M}_{r}(X)\\}}^{\|\cdot\|}\subset\mathrm{M}_{r}(B)_{+}.$

To see that $\Phi$ is order zero, suppose
$a,b\in\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)_{+}$
with
$a{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}b=0$,
and let $(x_{n})_{n},(y_{n})_{n}\subset X$ such that
$\lim_{n}\|x_{n}-a\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}=0$
and
$\lim_{n}\|y_{n}-b\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}=0$.
Then
$\lim_{n}\|x_{n}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y_{n}\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}=0$
and so

$\displaystyle 0$
$\displaystyle=\lim_{n}\|e\Phi(x_{n}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y_{n})\|$
$\displaystyle=\lim_{n}\|e(x_{n}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y_{n})\|$
$\displaystyle=\lim_{n}\|x_{n}y_{n}\|$
$\displaystyle=\lim_{n}\|\Phi(x_{n})\Phi(y_{n})\|$
$\displaystyle=\|\Phi(a)\Phi(b)\|,$

using continuity of $\Phi$ for the first and last equalities.

Remark 1(iii) tells us that if $\Phi$ is injective, then for any $r\geq 1$ and
$x\in\mathrm{M}_{r}(\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X))$,
we have $\Phi^{(r)}(x)\geq 0$ if and only if $x\geq 0$.

If $X$ is closed in $\|\cdot\|$, then Proposition 4.1 tells us that $X$ is
already complete with respect to
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$.
Hence
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)=X$
as sets, and $\Phi=\mathrm{id}_{X}$. In particular, $\Phi$ is injective with
closed range and hence bounded below. On the other hand, if $\Phi$ is bounded
below, then its image is closed in $B$, and so
$\Phi(\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X))=\overline{X}^{\|\cdot\|}$.
If moreover $\Phi=\mathrm{id}_{X}$, then it follows that $X$ is closed in $B$.
∎

Recall from Example 3(ii) that for a given a c.p.c. order zero map
$\theta\colon A\longrightarrow B$ from a unital $\mathrm{C}^{*}$-algebra,
$\theta(A)$ and $\theta(1_{A})$ satisfy the assumptions of Theorem 3.1. In
this case, it turns out that $\theta(A)$ is automatically closed with respect
to the induced
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
norm. This is essentially Section 2 in our  -language.

######  Proposition 4.2.

Suppose $\theta\colon A\longrightarrow B$ is a c.p.c. order zero map between
$\mathrm{C}^{*}$-algebras and there exists a local order unit $e\in
B_{+}^{1}\cap\theta(A)^{\prime}$ for $\theta(A)$ such that
$\theta(a)\theta(b)=e\theta(ab)$ for all $a,b\in A$. Then
$\theta(ab)=\theta(a){\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\theta(b)$
for all $a,b\in A$, and $\eta\coloneqq\mathrm{id}_{\theta(A)}\circ\theta\colon
A\longrightarrow\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(\theta(A))$
is a surjective ∗-homomorphism which is a ∗-isomorphism when $\theta$ is
injective. In particular, $\theta(A)$ is closed with respect to
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$.

###### Proof.

Since $(\theta(A),e)$ satisfy the assumptions of Theorem 3.1 (see Example
3(ii)) we have a product  on $\theta(A)$ satisfying
$e\theta(ab)=\theta(a)\theta(b)\\\
=e(\theta(a){\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\theta(b))$
for all $a,b\in A$. It then follows from nondegeneracy (Proposition 2.2(i))
that
$\theta(ab)=\theta(a){\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\theta(b)$
for all $a,b\in A$.

Since the map
$\mathrm{id}_{\theta(A)}:(\theta(A),\|\cdot\|)\longrightarrow(\theta(A),\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})$
between the normed ∗-linear spaces is linear and ∗-preserving (though a priori
not necessarily bounded), the map $\eta\colon
A\longrightarrow\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(\theta(A))$
is also linear and ∗-preserving with $\eta(A)=\theta(A)$ dense in
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(\theta(A))$.
Moreover, for each $a,b\in A$, we have
$\eta(ab)=\theta(ab)=\theta(a){\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\theta(b)=\eta(a){\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\eta(b)$.
Hence $\eta$ is a surjective ∗-homomorphism, which implies that its image,
$\theta(A)$, is closed in
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(\theta(A))$,
i.e.
$\theta(A)=\overline{\theta(A)}^{\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}}$.
When $\theta$ is moreover injective, so is $\eta$. ∎

######  Remark .

It follows from the (proof of the) structure theorem for c.p.c. order zero
maps ([20, Theorem 3.3]) that the kernel of any c.p.c. order zero map
$\theta\colon A\longrightarrow B$ from a unital $\mathrm{C}^{*}$-algebra is an
ideal. In this case,
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(\theta(A))$
is isomorphic to $A/\ker({\theta})$, which is in turn isomorphic to
$\pi_{\theta}(A)$ where $\pi_{\theta}\colon
A\longrightarrow\mathcal{M}(\mathrm{C}^{*}(\theta(A)))$ is the ∗-homomorphism
guaranteed by the structure theorem.

We are now equipped to say moreover when a self-adjoint linear subspace of a
$\mathrm{C}^{*}$-algebra is actually the image of a c.p.c. order zero map from
a unital $\mathrm{C}^{*}$-algebra.

######  Proposition 4.3.

Let $B$ be a $\mathrm{C}^{*}$-algebra with $e\in B_{+}^{1}$ and $X\subset B$ a
self-adjoint linear subspace. Then the following are equivalent.

  1. (i)

There exists a unital $\mathrm{C}^{*}$-algebra $A$ and c.p.c. order zero map
$\theta\colon A\longrightarrow B$ with $\theta(A)=X$ and $\theta(1_{A})=e$.

  2. (ii)

$e\in X\cap X^{\prime}$ is a local order unit for $X$, $X^{2}=eX$, and
$X=\overline{X}^{\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}}$.

###### Proof.

Example 3(ii) and Proposition 4.2 give (i) $\Longrightarrow$ (ii).

On the other hand, if $e\in X\cap X^{\prime}$ is a local order unit such that
$X^{2}=eX$ and
$X=\overline{X}^{\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}}$,
then Theorem 3.1 gives us a unital $\mathrm{C}^{*}$-algebra
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$,
and Theorem 4.1 gives us
$\Phi=\mathrm{id}_{X}\colon\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\longrightarrow
B$, the desired c.p.c. order zero map. ∎

##  5\. Unitizations

In this section, we describe a unitization of our construction and consider
the dependence of the construction on the given local order unit. This will
also yield the proof of Theorem C (by combining Proposition 5.1 with Remarks
5).

######  Proposition 5.1.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace, and suppose there exists a local order unit $e\in B_{+}^{1}\cap
X^{\prime}$ for $X$ so that $X^{2}\subset eX$. Set
$Z\coloneqq\textrm{span}\\{X,e\\}$.

Then $e\in B_{+}^{1}\cap Z^{\prime}$ is a local order unit for $Z$ so that
$Z^{2}=eZ$, and the induced product and norm from _Theorem 3.1_ agree with
those on $X$.

Moreover, the inclusion $X\subset Z$ induces an inclusion of
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
in
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$
as a two-sided closed ideal, which has co-dimension 1 unless $e\in X$, and the
map $\mathrm{id}_{Z}$ extends to a c.p.c. order zero map
$\Phi^{\dagger}\colon\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)\longrightarrow
B$ which agrees with $\mathrm{id}_{Z}$ on $Z$.

###### Proof.

If $e\in X$, then there is nothing to show, so we assume $e\notin X$. We want
to apply Theorem 3.1 to $Z$. One readily verifies that $e\in Z^{\prime}\cap
B_{+}^{1}$ is a local order unit for $Z$. For $\alpha,\beta\in\mathbb{C}$ and
$x,y\in X$, let $z\in X$ so that $xy=ez$. Then

$\displaystyle(\alpha e+x)(\beta e+y)$ $\displaystyle=\alpha\beta e^{2}+\alpha
ey+\beta ex+xy$ $\displaystyle=\alpha\beta e^{2}+\alpha ey+\beta ex+ez\in eZ.$

Hence $Z^{2}=eZ$. Applying Theorem 3.1 to both $(X,e)$ and $(Z,e)$ gives us
associative bilinear maps
${\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\times X\longrightarrow X$ and $\star\colon Z\times Z\longrightarrow Z$ such
that for all $x,y\in X$ and $w,z\in Z$,

$\displaystyle
xy=e(x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y)\hskip
14.22636pt\text{ and }\hskip 14.22636ptwz=e(w\star z).$

Nondegeneracy (Proposition 2.2(i)) for $(Z,e)$ tells us that for all $x,y\in
X\subset Z$, we have

$e(x\star
y)=xy=e(x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y)\
\Longrightarrow x\star
y=x{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y.$

Then $\star\colon Z\times Z\longrightarrow Z$ extends
${\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\times X\longrightarrow X$, and we may write  for the $\star$ multiplication
on $Z$ without confusion.

By definition the norm
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
Z\longrightarrow[0,\infty)$ guaranteed by Theorem 3.1 restricts to the norm
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\longrightarrow[0,\infty)$ guaranteed by Theorem 3.1, and hence
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)=\overline{\textrm{span}\\{X,e\\}}^{\|\cdot\|_{{}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}}}=\textrm{span}\\{\overline{X}^{\|\cdot\|_{{}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}}},e\\}=\textrm{span}\\{\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X),e\\}$.333As
a consequence of the Banach space fact from footnote 2, it follows that for
any linear subspaces $\mathcal{Y}$ and $\mathcal{Z}$ of a Banach space
$\mathcal{X}$ with $\mathcal{Z}$ finite-dimensional,
$\overline{\textrm{span}\\{\mathcal{Y},\mathcal{Z}\\}}=\textrm{span}\\{\overline{\mathcal{X}},\mathcal{Z}\\}$.
Hence the inclusion $X\subset Z$ induces an isometric embedding
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\subset\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$.
Since $e$ is the unit of
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$
(by Theorem 3.1), it follows that
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
sits as a closed 2-sided ideal with co-dimension at most 1. Note that if
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)=\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$,
then
$(X,{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})\subset\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$
is a dense 2-sided ideal, which must then contain the unit $e$. So
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
has co-dimension 1 if and only if $e\notin X$.

It follows from Theorem 4.1 that $\mathrm{id}_{Z}$ extends to a c.p.c. order
zero map
$\Phi^{\dagger}\colon\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)\longrightarrow
B$, and
$X\subset\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$
is closed with respect to
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
if and only if $Z$ is.444See footnote 2. ∎

######  Remarks .

(i) In this generality, there is nothing ruling out the possibility that
$e\notin X$ but
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
is nonetheless unital, in which case
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$
would end up being the forced unitization
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)^{\dagger}$
of
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$.
(Indeed a unital $\mathrm{C}^{*}$-algebra sitting inside its forced
unitization is already an example of this.) However, note that
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\triangleleft\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$
is an essential ideal if and only if
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
is not unital. Hence, when we really want to deal with
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)^{\sim}$,
we use
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$
when
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\triangleleft\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$
is essential and
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)^{\sim}$
otherwise. In this case, we write $\Phi^{\sim}$ for $\Phi^{\dagger}$.

(ii) As a Banach space, $Z$ is closed with respect to
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
if and only if $X$ is.555See footnote 2. Likewise, when
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)=\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)^{\sim}$,
$\Phi^{\sim}$ is bounded below if and only if $\Phi$ is. Indeed, since
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)=\textrm{span}\\{\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X),e\\}$
and $\Phi$ is linear, it follows that
$\Phi^{\sim}\big{(}\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)\big{)}=\textrm{span}\\{\Phi\big{(}\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\big{)},e\\}$,
and so $\Phi$ has closed image if and only if $\Phi^{\sim}$ does. Clearly
$\Phi$ is injective if $\Phi^{\sim}$ is, and the other direction we checked in
Proposition 2.3(ii).

Now, Theorem C follows from Proposition 5.1 and Section 5.

As a corollary to Proposition 5.1, we get sufficient conditions for when a
c.p.c. order zero map $A\longrightarrow B$ from a non-unital
$\mathrm{C}^{*}$-algebra extends to a c.p.c. order zero map
$A^{\sim}\longrightarrow B$.

######  Corollary 5.1.

Let $A$ be a non-unital $\mathrm{C}^{*}$-algebra and $\theta\colon
A\longrightarrow B$ a c.p.c. order zero map. If there exists a local order
unit $e\in B_{+}^{1}\cap\theta(A)^{\prime}$ for $\theta(A)$ such that
$e\theta(ab)=\theta(a)\theta(b)$ for all $a,b\in A$, then $\theta^{\sim}\colon
A^{\sim}\longrightarrow B$ given by $\theta^{\sim}(a+\lambda
1_{A^{\sim}})=\theta(a)+\lambda e$ is a c.p.c. order zero extension of
$\theta$.

###### Proof.

Set $X=\theta(A)$ and $Z=\textrm{span}\\{X,e\\}$. Proposition 4.2 tells us
$X=\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
as sets, and so by Remark 5(ii)
$Z=\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$
as well, and the c.p.c. order zero map
$\Phi^{\dagger}\colon\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)\longrightarrow
B$ from Proposition 5.1 is just $\mathrm{id}_{Z}$.

Let $\eta\coloneqq\mathrm{id}_{X}\circ\theta\colon
A\longrightarrow\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
be the ∗-homomorphism from Proposition 4.2. Then $\eta$ extends to a unital
∗-homomorphism $\eta^{\sim}\colon
A^{\sim}\longrightarrow\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$
given by $\eta^{\sim}(a+\lambda 1_{A^{\sim}})=\eta(a)+\lambda
e=\theta(a)+\lambda e$. Hence
$\theta^{\sim}\coloneqq\Phi^{\dagger}\circ\eta^{\sim}$ is a c.p.c. order zero
map, and it is given by $\theta^{\sim}(a+\lambda
1_{A^{\sim}})=\theta(a)+\lambda e$, as desired. ∎

For the remainder of this section, we consider how our construction of
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
depends on the distinguished commuting local order unit $e$. The following is
essentially what is employed in [9, Proposition 3.5].

######  Proposition 5.2.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace with local order unit $e\in B_{+}^{1}\cap X^{\prime}$ such that
$X^{2}\subset eX$. Suppose $h\in B_{+}^{1}$ satisfies $hx=ex$ for all $x\in
X$. Then $h\in X^{\prime}$ and $h$ is a local order unit for $X$ such that
$X^{2}\subset hX$. If $e$ is moreover an order unit then $h$ is also an order
unit with the same scaling factor as $e$. Moreover, the products and norms
induced by $e$ and $h$ on $X$ agree, and the induced $\mathrm{C}^{*}$-algebras
are equal.

###### Proof.

For $x,y\in X$, there exists $z\in X$ so that $xy=ez=hz$, and so it follows
that $X^{2}\subset hX$. Moreover,

$\|hx-xh\|=\|ex-xh\|=\|xe-xh\|=\|ex^{*}-hx^{*}\|=0,$

and so $h\in X^{\prime}$.

In order to check that $h$ is a local order unit, first note that for all
$a\in\mathrm{C}^{*}(X)\subset B$ (the $\mathrm{C}^{*}$-algebra generated by
$X$ in $B$) we also have $ea=ha$. In particular, $e|x|=h|x|$ for all $x\in X$.
Let $x\in X$ be self-adjoint. Then Proposition 2.1(ii) guarantees an $R>0$ so
that $Re\geq|x|$, which by Proposition 2.1(i) implies $Re|x|\geq x^{*}x$.
Since $Re|x|=Rh|x|$, it follows that $Rh|x|\geq x^{*}x$, which by Proposition
2.1(i) implies $Rh\geq|x|$. Then Proposition 2.1(ii) tells us $h$ is a local
order unit for $X$.

If $e$ is moreover an order unit with scaling factor $R$, then the same
argument shows that $h$ is an order unit with scaling factor at most as large
as $R$. A symmetric argument now shows that $R$ is the scaling factor of $h$.

Finally, it follows from nondegeneracy and the fact that $h_{j}(e)x=h_{j}(h)x$
for all $x\in X$ and $j\geq 1$ that the $\mathrm{C}^{*}$-algebras associated
to $(X,e)$ and $(X,h)$, as guaranteed by Theorem 3.1, are equal. ∎

The previous lemma applies in particular to the case where
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
is unital but does not contain $e$, as in the following corollary.

######  Corollary 5.2.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace with order unit $e\in B_{+}^{1}\cap X^{\prime}$ such that
$X^{2}\subset eX$, and let
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
denote the associated $\mathrm{C}^{*}$-algebra. If
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
is unital but $e\notin X$, then the unit of
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$,
denoted by $h$, is also an order unit for $X$ with the same scaling factor as
$e$, and the $\mathrm{C}^{*}$-algebra associated to $(X,h)$ is exactly
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$.

###### Proof.

Let $x\in X$. Since
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
is a $\mathrm{C}^{*}$-algebra, there exist
$y,z\in\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
so that
$x=y\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,z$.
Then since $\Phi$ is order zero, we have

$hx=\Phi(h)\Phi(y\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,z)=\Phi(y)\Phi(z)=yz=e(y\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,z)=ex.$

By Proposition 5.2 we are done. ∎

######  Proposition 5.3.

Let $X\subset B$ be a self-adjoint linear subspace of a
$\mathrm{C}^{*}$-algebra $B$ with local order units $e,h\in B_{+}^{1}\cap
X\cap X^{\prime}$ with $hX=X^{2}=eX$. Let
$\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,\colon
X\times X\longrightarrow X$,
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$,
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
denote the multiplication, norm, and $\mathrm{C}^{*}$-algebra induced by $e$
and $\star\colon X\times X\longrightarrow X$, $\|\cdot\|_{\star}$, and
$\mathrm{C}^{*}_{\star}(X)$ the multiplication, norm, and
$\mathrm{C}^{*}$-algebra induced by $h$. If $X$ is complete with respect to
both $\|\cdot\|_{\star}$ and
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$,
then
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
and $\mathrm{C}^{*}_{\star}(X)$ are unitally ∗-isomorphic.

###### Proof.

First, we remark that our assumptions guarantee that
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)=X=\mathrm{C}^{*}_{\star}(X)$
as ∗-vector spaces, and $e$, resp. $h$, is the unit of
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$,
resp. $\mathrm{C}^{*}_{\star}(X)$. We also note that $e\in
Z(\mathrm{C}^{*}_{\star}(X))$ and $h\in
Z(\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X))$
by Proposition 3.2(ii). By Theorem 4.1, for all $x\in X$

$x\in\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)_{+}\
\Longleftrightarrow\ x\in B_{+}\ \Longleftrightarrow\
x\in\mathrm{C}^{*}_{\star}(X)_{+}.$

So
$Rh-e\in\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)_{+}$
and $Re-h\in\mathrm{C}^{*}_{\star}(X)_{+}$ for some $R>0$, which means
$h\in\mathrm{GL}(\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X))$
and $e\in\mathrm{GL}(\mathrm{C}^{*}_{\star}(X))$. Let $h^{-1}$ denote the
element of $X$ that acts as the inverse of $h$ in
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$,
i.e.,
$h\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,h^{-1}=h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,h=e$,
and let $e^{-1}$ likewise denote the inverse of $e$ in
$\mathrm{C}^{*}_{\star}(X)$. Then for all $x,y\in X$, we have

$\displaystyle h(x\star y)$ $\displaystyle=xy$
$\displaystyle=e(x\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,y)$
$\displaystyle=e(h\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,x\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,y)$
$\displaystyle=e(h\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,(h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,x\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,y))$
$\displaystyle=h(h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,x\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,y).$

Nondegeneracy of $h$ implies that

(5.1) $\displaystyle x\star
y=h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,x\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,y$

for all $x,y\in X$, and likewise, we have

(5.2) $\displaystyle
x\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,y=e^{-1}\star
x\star y$

for all $x,y\in X$. In particular, we have for all $x,y\in X$

(5.3) $\displaystyle h$ $\displaystyle=e\star
e^{-1}=h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,e\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,e^{-1}=h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,e^{-1},\
\mathrm{and}$ (5.4) $\displaystyle e$
$\displaystyle=h\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,h^{-1}=e^{-1}\star
h\star h^{-1}=e^{-1}\star h^{-1}.$

Now define
$\varphi\colon\mathrm{C}^{*}_{\star}(X)\longrightarrow\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
by
$\varphi(x)=h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,x$
for all $x\in X$. Then $\varphi$ is a unital ∗-homomorphism. Indeed,
$\varphi(h)=h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,h=e$,
and for $x,y\in X$, we have

$\displaystyle\varphi(x\star
y)=h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,(x\star
y)\overset{\eqref{prod1}}{=}h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,(h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,x\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,y)=(h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,x)\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,(h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,y).$

Likewise,
$\psi\colon\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\longrightarrow\mathrm{C}^{*}_{\star}(X)$
given by $\psi(x)=e^{-1}\star x$ is a unital ∗-homomorphism. Moreover, we have
for any $x\in X$,

$\displaystyle\psi(\varphi(x))$
$\displaystyle=e^{-1}\star(h^{-1}\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,x)\overset{\eqref{prod2}}{=}e^{-1}\star(e^{-1}\star
h^{-1}\star x)\overset{\eqref{inverses2}}{=}e^{-1}\star(e\star x)=x.$

Likewise using (5.3), we have $\varphi\circ\psi(x)=x$. So $\psi=\varphi^{-1}$.
∎

######  Example .

Let $A,B,$ and $C$ be $\mathrm{C}^{*}$-algebras with $A$ and $C$ unital.
Suppose we have c.p. order zero maps $\varphi\colon A\longrightarrow B$ and
$\psi\colon C\longrightarrow B$ with $\varphi(A)=\psi(C)$. Then
$X=\varphi(A)=\psi(C)$ and $e=\varphi(1_{A})$ and $h=\psi(1_{C})$ satisfy the
assumptions of Proposition 5.3, and so, with the notation above,
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(\varphi(A))\cong\mathrm{C}^{*}_{\star}(\psi(C))$.
Note that it now follows from Proposition 4.2 that this
$\mathrm{C}^{*}$-algebra is a quotient of both $A$ and $C$.

##  6\. Closed subspaces and order units

The theory is better behaved when our self-adjoint linear subspace $X\subset
B$ is assumed to be closed and also when the commuting local order unit for
$X$ is assumed to be an order unit. It turns out these two notions are deeply
intertwined.

######  Proposition 6.1.

Suppose $X\subset B$ is a self-adjoint linear subspace of a
$\mathrm{C}^{*}$-algebra with order unit $e\in B_{+}^{1}\cap X^{\prime}$ with
scaling factor $R$. Writing $\overline{X}$ for the closure of $X$ with respect
to $\|\cdot\|$, we have the following.

  1. (i)

$e$ is an order unit for $\overline{X}$ with scaling factor $R$.

  2. (ii)

$R\|ex\|\geq\|x\|$ for each self-adjoint $x\in X$. In particular,
multiplication by $e$ is bounded below on $\overline{X}$, and
$\overline{eX}=e\overline{X}$.

  3. (iii)

If $X^{2}\subset\overline{eX}$, then $(\overline{X},e)$ satisfies the
assumptions of Theorem 3.1, where the induced norm and product agree with
those on $X$, and we have
$\overline{X}=\overline{X}^{\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}}$
and
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)=\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(\overline{X})$.

###### Proof.

(i) Suppose $x\in\overline{X}$ is self-adjoint and $(x_{n})_{n}$ is a sequence
of self-adjoint elements in $X$ with $\lim_{n}\|x_{n}-x\|=0$. Then
$R\|x_{n}\|e+x_{n}\geq 0$ for all $n\in\mathbb{N}$, which implies
$R\|x\|e+x\geq 0$ since $B_{+}$ is closed. Since $X\subset\overline{X}$, the
scaling factor for $e$ with respect to $\overline{X}$ is at least $R$, and so
it is $R$.

(ii) Fix $x\in X$, and assume for now that $x$ is self-adjoint. Since $e\in
X^{\prime}$, $R\|x\|e\geq x$ implies $R\|x\|e|x|\geq x^{2}$ by Proposition
2.1(i), and so

$R\|x\|\|ex\|=R\|x\|\|e|x|\|\geq\|x^{2}\|=\|x\|^{2}.$

Now, for general $x\in X$, we have

(6.1) $\displaystyle 2\|x\|$ $\displaystyle\leq\|x+x^{*}\|+\|i(x-x^{*})\|$
$\displaystyle\leq R\|e(x+x^{*})\|+R\|e(i(x-x^{*}))\|$ $\displaystyle\leq
4R\|ex\|.$

It follows that multiplication by $e$ is bounded below on $\overline{X}$, and
hence $e\overline{X}=\overline{eX}$.

(iii) If $X^{2}\subset\overline{eX}$, then by (ii), we have
$(\overline{X})^{2}\subset\overline{X^{2}}\subset\overline{eX}=e\overline{X}$,
and so $(\overline{X},e)$ satisfies the assumptions of Theorem 3.1. Then we
have an associative bilinear map
$\overline{X}\times\overline{X}\longrightarrow\overline{X}$ satisfying (3.1),
which nondegeneracy tells us agrees with
${\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\times X\longrightarrow X$ on $X\times X$. We also have a
$\mathrm{C}^{*}$-norm on $\overline{X}$ given by (3.2), which agrees by
definition with
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\colon
X\longrightarrow[0,\infty)$ on $X$. Hence, we can denote the induced
multiplication and norm on $\overline{X}$ by  and
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$,
and we have a natural inclusion
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\subset\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(\overline{X})$
coming from
$(X,{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})\subset(\overline{X},{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,})$.
Since $\overline{X}$ is closed, it then follows from Proposition 4.1 that
$\|\cdot\|$ and
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$
are equivalent norms on $X$, and so
$\overline{X}=\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(\overline{X})=\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$.
∎

######  Remark .

The weaker assumption $X^{2}\subset\overline{eX}$ may seem out of place, but
it will be useful in upcoming work and only requires a minimal adjustment
here.

Now we are equipped to explicitly describe the relationship between closed
self-adjoint linear subspaces and order units.

######  Corollary 6.1.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace with local order unit $e\in B_{+}^{1}\cap X^{\prime}$ such that
$X^{2}\subset eX$. Then $X$ is closed with respect to $\|\cdot\|$ if and only
if $e$ is an order unit and $X$ is closed with respect to
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$.
In particular, if $A$ is a unital $\mathrm{C}^{*}$-algebra and $\theta\colon
A\longrightarrow B$ a c.p.c. order zero map, then $\theta(A)$ is closed if and
only if $\theta(1_{A})$ is an order unit for $\theta(A)$.

###### Proof.

Set $Z=\textrm{span}\\{X,e\\}$. If $X$ is closed with respect to $\|\cdot\|$,
then by footnote 2 so is $Z$. Then it follows from Proposition 4.1 that $Z$ is
also closed with respect to
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$,
and from Theorem 4.1 that
$\Phi^{\dagger}\colon\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)\longrightarrow
Z$ is bounded below with
$Z=\Phi^{\dagger}(\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z))$.
Then as we saw in Example 2(ii), $e$ is an order unit for $Z$ in $B$ and hence
also for $X$. Conversely, if $e$ is an order unit for $X$ and
$X=\overline{X}^{\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}}$,
then since $X^{2}\subset eX\subset\overline{eX}$, it follows from Proposition
6.1(iii) that
$\overline{X}^{\|\cdot\|}=\overline{X}^{\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}}=X$.

When $X=\theta(A)$, Proposition 4.2 says that $X$ is automatically closed with
respect to
$\|\cdot\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$.
Hence $X$ is closed if and only if $e$ is an order unit. ∎

Applying Proposition 2.3 we can now characterize the scaling factor for our
commuting order unit $e$ as $\|\Phi^{-1}\|$.

######  Corollary 6.2.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a closed self-adjoint
linear subspace with local order unit $e\in B_{+}^{1}\cap X^{\prime}$ such
that $X^{2}\subset eX$. Then
$\Phi\colon\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\longrightarrow
B$ is bounded below and $e$ is an order unit with scaling factor
$\|\Phi^{-1}\|=\sup_{j}\|H_{j}\|$, where $H_{j}\in\mathcal{L}(X,B)$ are
defined as in (4.2). When the scaling factor of $e$ is $1$, $\Phi$ is a
complete order embedding and $\|ex\|=\|x\|$ for all $x\in X$.

###### Proof.

We know from Corollary 6.1 that $e$ is an order unit for $X$. Set
$Z=\text{span}\\{X,e\\}$. If
$e\in\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
or
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
is non-unital, then by Proposition 5.1,
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)^{\sim}=\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(Z)$.
Since $X$ is closed in $\|\cdot\|$, so is $Z$,666See footnote 2. and so
$\mathrm{id}_{Z}$ is a bounded below c.p. order zero extension of $\Phi$ to
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)^{\sim}$
(by Theorem 4.1). Then Proposition 2.3(ii) tells us that the scaling factor of
$e$ for $X$ is $\|\Phi^{-1}\|$. If
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
is unital with
$e\notin\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$,
then Proposition 2.3(ii) tells us
$\Phi(1_{\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)})$
is an order unit with scaling factor $\|\Phi^{-1}\|$, and Corollary 5.2 tells
us this is then also the scaling factor for $e$. Furthermore, it follows from
(4.3) that for all $x\in X$ with $\|x\|=1$ and all $j\geq 1$

$\|H_{j}(x)\|\leq\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}=\sup_{j}\|h_{j}(e)x\|=\sup_{j}\|H_{j}(x)\|\leq\sup_{j}\|H_{j}\|<\infty,$

(where $H_{j}$ are as defined in (4.2)), and so for all $x\in X$

$\|H_{j}(x)\|\leq\|\Phi^{-1}(x)\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}\leq\sup_{j}\|H_{j}\|.$

Hence $\|H_{j}\|\leq\|\Phi^{-1}\|\leq\sup_{j}\|H_{j}\|$ for all $j\geq 1$, and
so $\|\Phi^{-1}\|=\sup_{j}\|H_{j}\|$.

In particular, if the scaling factor of $e$ is $1$, then $\Phi$ is an isometry
(since $\Phi$ was already contractive). Since $\Phi$ is c.p. order zero,
Proposition 1.1 tells us it is a complete order embedding. Moreover for any
$x\in X\backslash\\{0\\}$,

$\displaystyle\|x\|^{4}$ $\displaystyle=\|x^{*}xx^{*}x\|$
$\displaystyle\leq\|x^{*}\Phi(x\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,x^{*})x\|$
$\displaystyle\leq\|x^{*}\Phi^{\sim}(\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}^{2}e)x\|$
$\displaystyle\leq\|x\|\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}^{2}\|ex\|.$

Since
$\|x\|=\|x\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}$,
it follows that $\|ex\|=\|x\|$ for all $x\in X$. ∎

Now combining Theorem 4.1, Corollary 6.1, and Corollary 6.2, we can finish the
proof of Theorem B:

######  Corollary 6.3.

Let $B$ be a $\mathrm{C}^{*}$-algebra and $X\subset B$ a self-adjoint linear
subspace. Then $X$ is completely order isomorphic to a unital
$\mathrm{C}^{*}$-algebra via a c.p. order zero map if and only if $X$ is
closed in $B$ and there exists an order unit $e\in B_{+}^{1}\cap X\cap
X^{\prime}$ for $X$ with scaling factor $1$ so that $X^{2}=eX$.

###### Proof.

Suppose $A$ is a unital $\mathrm{C}^{*}$-algebra and $\theta\colon
A\longrightarrow B$ is a completely isometric c.p. order zero map with
$\theta(A)=X$. Set $e\coloneqq\theta(1_{A})$. Then $e\in B_{+}^{1}\cap X\cap
X^{\prime}$ is an order unit for $X$ with scaling factor $1$ such that
$X^{2}=eX$ by the structure theorem for order zero maps (see Section 3) and
Proposition 2.3(ii).

On the other hand, if $X$ is closed in $B$ and there exists an order unit
$e\in B_{+}^{1}\cap X\cap X^{\prime}$ for $X$ with scaling factor $1$ so that
$X^{2}=eX$, then by Theorem 3.1, Theorem 4.1, and Corollary 6.2,
$\Phi\coloneqq\mathrm{id}_{X}\colon\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)\longrightarrow
B$ is a complete order embedding from the unital $\mathrm{C}^{*}$-algebra
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
with image $X$. ∎

When $X$ is closed, we can detect an approximate identity from
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
by how it behaves in $B$ (this essentially underlies [9, Corollary 2.11]).

######  Proposition 6.2.

Let $X\subset B$ be a closed self-adjoint linear subspace with order unit
$e\in B_{+}^{1}\cap X^{\prime}$ such that $X^{2}\subset eX$, and let
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
be the associated $\mathrm{C}^{*}$-algebra and
$C\subset\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
a sub-$\mathrm{C}^{*}$-algebra. An increasing net
$(u_{\lambda})_{\Lambda}\subset C^{1}_{+}$ is an approximate unit for $C$ if
and only if $\lim_{\lambda}\|u_{\lambda}y-ey\|=0$ for all $y\in C$.

###### Proof.

Since $X$ is closed,
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)=X$
as vector spaces, and so $C$ is also a subspace of $B$. From Corollary 6.2, we
know the scaling factor for $e$ is $\|\Phi^{-1}\|$, and with this (6.1) tells
us $2\|\Phi^{-1}\|\|ex\|\geq\|x\|$ for any $x\in X$. Now fix $y\in C$ and
$\lambda\in\Lambda$. Then
$\|u_{\lambda}y-ey\|=\|e(u_{\lambda}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y-y)\|$,
and

$\displaystyle\frac{\|u_{\lambda}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y-y\|_{{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}}}{2\|\Phi^{-1}\|^{2}}$
$\displaystyle\leq\frac{\|u_{\lambda}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y-y\|}{2\|\Phi^{-1}\|}$
$\displaystyle\leq\|e(u_{\lambda}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y-y)\|$
$\displaystyle\leq\|u_{\lambda}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y-y\|$
$\displaystyle\leq\|u_{\lambda}{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}y-y\|_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,},$

where the last inequality follows from Proposition 4.1. The claim follows. ∎

Finally, when our local order unit $e$ lies in $X$, the triple
$(X,\\{\mathrm{M}_{r}(X)\cap\mathrm{M}_{r}(B)_{+}\\}_{r},e)$ forms an abstract
operator system (see [14, Chapter 13]), to which we can associate the
enveloping $\mathrm{C}^{*}$-algebra, denoted $\mathrm{C}^{*}_{\text{min}}(X)$
or $\mathrm{C}^{*}_{\text{env}}(X)$ (as in [12]). When $X$ is moreover closed,
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
is $\mathrm{C}^{*}_{\text{min}}(X)$.

######  Proposition 6.3.

Let $B$ be a $\mathrm{C}^{*}$-algebra.

  1. (i)

Suppose $A$ is a unital $\mathrm{C}^{*}$-algebra and $\theta\colon
A\longrightarrow B$ is isometric and c.p.c. order zero. Then

$\mathrm{C}^{*}_{\emph{min}}(\theta(A))\cong
A\cong\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(\theta(A)).$

  2. (ii)

If $X\subset B$ is a closed self-adjoint linear subspace with order unit $e\in
X\cap X^{\prime}\cap B_{+}^{1}$ with scaling factor $R=1$ such that
$X^{2}=eX$, then
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$
is the enveloping $\mathrm{C}^{*}$-algebra of
$(X,\\{\mathrm{M}_{r}(X)\cap\mathrm{M}_{r}(B)_{+}\\},e)$.

###### Proof.

(i) Since $\theta$ is an isomorphic order zero map, it is a complete order
embedding by Proposition 1.1, and so the first isomorphism follows from the
universal property of the minimal $\mathrm{C}^{*}$-algebra associated to the
abstract operator system
$(\theta(A),\\{\mathrm{M}_{r}(X)\cap\mathrm{M}_{r}(B)_{+}\\}_{r},\theta(1_{A}))$.
The second isomorphism follows from Proposition 4.2.

Now (ii) follows immediately from (i) and the fact that
$\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(\Phi(\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)))\\\
=\mathrm{C}^{*}_{\,\begin{picture}(-1.0,1.0)(-1.0,-2.5)\circle*{3.0}\end{picture}\,\,\,}(X)$.
∎

## References

  * [1] B. Blackadar.  Operator algebras, volume 122 of Encyclopaedia of Mathematical Sciences.  Springer-Verlag, Berlin, 2006.  Theory of $\mathrm{C}^{*}$-algebras and von Neumann algebras, Operator Algebras and Non-commutative Geometry, III.
  * [2] B. Blackadar and E. Kirchberg.  Generalized inductive limits of finite-dimensional $\mathrm{C}^{*}$-algebras.  Math. Ann., 307(3):343–380, 1997.
  * [3] B. Blackadar and E. Kirchberg.  Generalized inductive limits and quasidiagonality.  In $\mathrm{C}^{*}$-algebras (Münster, 1999), pages 23–41. Springer, Berlin, 2000.
  * [4] B. Blackadar and E. Kirchberg.  Inner quasidiagonality and strong NF algebras.  Pacific J. Math., 198(2):307–329, 2001.
  * [5] B. Blackadar and E. Kirchberg.  Irreducible representations of inner quasidiagonal $\mathrm{C}^{*}$-algebras.  Canad. Math. Bull., 54(3):385–395, 2011.
  * [6] J. Bosa, J. Gabe, A. Sims, and S. White.  The nuclear dimension of $\mathcal{O}_{\infty}$-stable $\mathrm{C}^{*}$-algebras.  Adv. Math., 401:Paper No. 108250, 51, 2022.
  * [7] J. Castillejos, S. Evington, A. Tikuisis, S. White, and W. Winter.  Nuclear dimension of simple $\mathrm{C}^{*}$-algebras.  Invent. Math., 224(1):245–290, 2021.
  * [8] E. Christensen, A. Sinclair, Roger R. Smith, S. White, and W. Winter.  Perturbations of nuclear $\mathrm{C}^{*}$-algebras.  Acta Math., 208(1):93–150, 2012.
  * [9] K. Courtney and W. Winter.  Nuclearity and $\mathrm{CPC}^{*}$-systems.  arXiv:arXiv:2304.01332.
  * [10] E Gardella and H Thiel.  Weighted homomorphisms between $\mathrm{C}^{*}$-algebras.  arXiv:2203.16702.
  * [11] Eusebio Gardella and Hannes Thiel.  Zero-product balanced algebras.  Linear Algebra Appl., 670:121–153, 2023.
  * [12] M. Hamana.  Injective envelopes of operator systems.  Publ. Res. Inst. Math. Sci., 15(3):773–785, 1979.
  * [13] E. Kirchberg and W. Winter.  Covering dimension and quasidiagonality.  Internat. J. Math., 15(1):63–85, 2004.
  * [14] V. Paulsen.  Completely bounded maps and operator algebras, volume 78 of Cambridge Studies in Advanced Mathematics.  Cambridge University Press, Cambridge, 2002.
  * [15] Y. Sato.  2-positive almost order zero maps and decomposition rank.  J. Operator Theory, 85(2):505–526, 2021.
  * [16] Y. Sato, S. White, and W. Winter.  Nuclear dimension and $\mathcal{Z}$-stability.  Invent. Math., 202(2):893–921, 2015.
  * [17] A. Tikuisis, S. White, and W. Winter.  Quasidiagonality of nuclear $\mathrm{C}^{*}$-algebras.  Ann. of Math. (2), 185(1):229–284, 2017.
  * [18] W. Winter.  Covering dimension for nuclear $\mathrm{C}^{*}$-algebras. II.  Trans. Amer. Math. Soc., 361(8):4143–4167, 2009.
  * [19] W. Winter.  Nuclear dimension and $\mathcal{Z}$-stability of pure $\mathrm{C}^{*}$-algebras.  Invent. Math., 187(2):259–342, 2012.
  * [20] W. Winter and J. Zacharias.  Completely positive maps of order zero.  Münster J. Math., 2:311–324, 2009.
  * [21] W. Winter and J. Zacharias.  The nuclear dimension of $\mathrm{C}^{*}$-algebras.  Adv. Math., 224(2):461–498, 2010.
  * [22] M. Wolff.  Disjointness preserving operators on $\mathrm{C}^{*}$-algebras.  Arch. Math. (Basel), 62(3):248–253, 1994.