64c3814
==== TABLE OF CONTENTS ====
Cuntz-comparison-and-Jiang-Su-absorption//absorption.tex
Cuntz-comparison-and-Jiang-Su-absorption//assembly.tex
Cuntz-comparison-and-Jiang-Su-absorption//central.tex
Cuntz-comparison-and-Jiang-Su-absorption//comparison.tex
Cuntz-comparison-and-Jiang-Su-absorption//divisibility.tex
Cuntz-comparison-and-Jiang-Su-absorption//extended-traces.tex
Cuntz-comparison-and-Jiang-Su-absorption//introduction.tex
Cuntz-comparison-and-Jiang-Su-absorption//localization.tex
Cuntz-comparison-and-Jiang-Su-absorption//norm.tex
Cuntz-comparison-and-Jiang-Su-absorption//paper.tex
Cuntz-comparison-and-Jiang-Su-absorption//preliminaries.tex
Cuntz-comparison-and-Jiang-Su-absorption//references.tex
Cuntz-comparison-and-Jiang-Su-absorption//rounding.tex
Cuntz-comparison-and-Jiang-Su-absorption//shrinking.tex
Cuntz-comparison-and-Jiang-Su-absorption//support-tools.tex
Cuntz-comparison-and-Jiang-Su-absorption//tracial.tex
===========================
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//absorption.tex ====
\section{Dimension-drop relations and Jiang--Su absorption}
\label{abs:section}
Proposition~\ref{prop:central-filler} supplies a matrix cone and an
element filling its defect. We first turn their relations into unital
maps from individual prime dimension-drop algebras. These maps need
not be compatible: nuclear approximation and bounded completely
positive lifts will instead produce a single unital map
$\Z\to F_\omega(A)$. Finally, half-flip unitaries and an approximate
intertwining give $A\cong A\otimes\Z$.
This route from central dimension-drop maps to absorption is
established. For separable, possibly nonunital algebras,
Nawata \cite[Proposition~5.1]{Nawata2013} states the criterion using
the sequence quotient. It fits the strongly self-absorbing framework
of Toms--Winter \cite{TW2007}; a formulation for arbitrary strongly
self-absorbing algebras is given in
\cite[Theorem~2.2]{PTV2025}. We give the lifting and intertwining
steps for the fixed ultrafilter quotient used here.
\subsection{The unit in the annihilator quotient}
Put
\[
\Ann(A)=\{x\in A_\omega:xa=ax=0\text{ for every }a\in A\},
\qquad
F_\omega(A)=(A_\omega\cap A')/\Ann(A).
\]
The annihilator is a closed two-sided ideal of $A_\omega\cap A'$.
It is contained in that relative commutant, since all its products
with $A$ vanish.
If $(e_n)$ is a positive contractive sequential approximate unit of
$A$, then $e=[(e_n)]\in A_\omega\cap A'$, and its class is the unit
of $F_\omega(A)$. Indeed $ea=ae=a$ for $a\in A$, while for
$x\in A_\omega\cap A'$ the differences $ex-x$ and $xe-x$
annihilate $A$ on both sides. For example
$(ex-x)a=(ea-a)x=0$, and the other products follow in the same way.
Apply Proposition~\ref{prop:central-filler}. The selfadjoint element
$S^*S+h-e$ annihilates $A$ on the right by
\eqref{cen:filler-relations}, and therefore on the left by taking
adjoints. Writing $s$ for the quotient image of $S$, and using the
same letters $\alpha,h$ for the quotient images of the map and its
weight, gives the exact relations
\begin{equation}\label{abs:quotient-relations}
s^*s=1-h,\qquad \alpha(e_{11})s=s,
\qquad h=\alpha(1).
\end{equation}
In particular $s$ is a contraction in the quotient. The argument uses
the quotient unit represented by $(e_n)$, without requiring a unit in
$A$.
\subsection{An explicit dimension-drop model}
For positive integers $r,s$, write
\[
I(r,s)=\{f\in C([0,1],M_r\otimes M_s):
f(0)\in M_r\otimes1_s,\ f(1)\in1_r\otimes M_s\}.
\]
Relations~\eqref{abs:quotient-relations} are the order-zero form of the dimension-drop
presentation in \cite[Proposition~2.1]{Sato2010}, following
\cite[Proposition~5.1, (iii)$\Rightarrow$(iv)]{RW2010}. We give an explicit model and verify
that every representation of the relations factors through it.
\begin{proposition}\label{prop:dimensiondrop}
Let $B$ be a unital $C^*$-algebra, let $p\ge2$, and suppose that a
\cpc order-zero map $\alpha:M_p\to B$ and $s\in B$ satisfy
\[
h=\alpha(1),\qquad s^*s=1-h,\qquad \alpha(e_{11})s=s.
\]
There is a unital homomorphism
$I(p,p+1)\to B$ whose range contains $\alpha(M_p)$ and $s$.
Consequently $F_\omega(A)$ receives a unital homomorphism from
$I(p,p+1)$ for every $p\ge2$.
\end{proposition}
\begin{proof}
\emph{The model and its generators.}
Decompose the first
matrix leg of $M_{p+1}\otimes M_p$ into blocks of sizes $p$ and $1$,
and put
\begin{align}
A_a(t)&=\begin{pmatrix}a\otimes1_p&0\\0&ta\end{pmatrix},
&&a\in M_p,\label{abs:model-map}\\
T(t)&=\sqrt{1-t}\ e_{1,p+1}^{(p+1)}\otimes1_p.
\label{abs:model-element}
\end{align}
The upper diagonal block has size $p^2$ and the lower one size $p$.
At each $t$ the map $a\mapsto A_a(t)$ is completely positive,
contractive, and order zero. Direct matrix multiplication gives
\begin{equation}\label{abs:model-relations}
T^*T=1-A_1,\qquad A_{e_{11}}T=T,
\end{equation}
and
\begin{equation}\label{abs:scalar-identity}
T^*T+\sum_{i=1}^p A_{e_{i1}}TT^*A_{e_{1i}}
=(1-t)1_{p(p+1)}.
\end{equation}
For the last identity, $T^*T$ is $(1-t)$ times the bottom block
projection, and the $i$th summand is $(1-t)$ times the $i$th top
block projection.
Let
\[
j(a)=\operatorname{diag}(a\otimes1_p,a)
\]
and define the boundary algebra
\[
B_p=\{f\in C([0,1],M_{p+1}\otimes M_p):
f(0)\in M_{p+1}\otimes1_p,\ f(1)\in j(M_p)\}.
\]
We claim that $A_a$ and $T$ generate $B_p$.
First $A_1+T^*T=1$, and \eqref{abs:scalar-identity} gives the
coordinate scalar function $t1$. Thus their generated algebra
contains $C([0,1])1$.
At $t=1$ the generators have image $j(M_p)$.
At $0\le t<1$, the nonzero link $T(t)$, its adjoint, and the top
matrix action yield all matrix units of $M_{p+1}\otimes1_p$:
the links $A_{e_{i1}}(t)T(t)$ give the $i$th top-to-bottom units,
and their products give all top-to-top units and the bottom
projection. At $t=0$ these are exactly the prescribed fiber.
When $0<t<1$, compression of $A_a(t)$ to the bottom block and
division by the scalar $t$ gives the full bottom copy of $M_p$.
Multiplying by the preceding matrix units gives all of
$M_{p+1}\otimes M_p$ in that fiber.
Fiber generation and the scalar functions prove the claim in norm.
More explicitly, for $f\in B_p$ and $\eps>0$, at each $t$ choose
an element $g_t$ of the generated algebra whose value at $t$ agrees
with $f(t)$, or is within $\eps/2$ of it. Continuity gives an open
neighborhood on which $\|g_t-f\|<\eps$. A finite subcover and a
subordinate scalar partition of unity produce an element of the
generated algebra within $\eps$ of $f$ on the whole interval.
The representation $j$ of $M_p$ has multiplicity $p+1$, so some
unitary $V\in M_{p+1}\otimes M_p$ conjugates $j(M_p)$ onto
$1_{p+1}\otimes M_p$. The finite-dimensional unitary group is
path connected; choose a unitary path $V(t)$ from $1$ to $V$.
Conjugation by this path changes only the endpoint presentation:
the boundary algebra becomes the one with fibers
$M_{p+1}\otimes1_p$ at $0$ and $1_{p+1}\otimes M_p$ at $1$.
Reversing the interval and interchanging the tensor legs identifies
it with $I(p,p+1)$.
\emph{Arbitrary representations of the relations.}
We have identified the algebra generated by the model. To obtain the
asserted map into $B$, we now show that every pair $(\alpha,s)$ in the
statement gives a representation of that algebra. Represent $B$ faithfully and
unitally on a Hilbert space $H$. The order-zero structure theorem
gives a supporting representation $\rho:M_p\to\mathcal B(H)$,
with unit the support projection of $h=\alpha(1)$, such that
$\alpha(a)=h\rho(a)=\rho(a)h$.
In the polar decomposition
\[
s=u(1-h)^{1/2},
\]
the initial space of $u$ is
$H_s=1_{[0,1)}(h)H$. The equation $\alpha(e_{11})s=s$ implies
that $\alpha(e_{11})$ is the identity on the closed range of $s$.
Since $0\le\alpha(e_{11})\le h\le1$, both $h$ and
$\rho(e_{11})$ are the identity on this range. Thus $u$ is an
isometry from $H_s$ into
$1_{\{1\}}(h)\rho(e_{11})H$.
For $1\le i\le p$ the subspaces
$\rho(e_{i1})uH_s$ are mutually orthogonal. Their direct sum
$H_t$ reduces $\rho$ and is identified with $\C^p\otimes H_s$
by the unitary sending $e_i\otimes\zeta$ to
$\rho(e_{i1})u\zeta$. On $H_t$ the weight is $1$ and the
representation is $a\otimes1_{H_s}$. Since $H_t$ is contained
in the $h=1$ space, it is orthogonal to $H_s$.
The orthogonal complement
\[
H_r=1_{\{1\}}(h)H\ominus H_t
\]
reduces both generators: $s$ and $s^*$ vanish there, and $\alpha$
acts there as the unital matrix representation $\rho|_{H_r}$.
This summand is therefore obtained by evaluating $B_p$ at $1$.
On $H_t\oplus H_s\cong\C^{p+1}\otimes H_s$ the generators
are precisely
\begin{equation}\label{abs:arbitrary-representation}
\alpha(a)=
\begin{pmatrix}a\otimes1_{H_s}&0\\0&h_s\rho_s(a)\end{pmatrix},
\qquad
s=e_{1,p+1}^{(p+1)}\otimes(1-h_s)^{1/2},
\end{equation}
where $h_s$ and $\rho_s$ denote the source restrictions, with
$\rho_s=0$ on $\ker h_s$.
Split $H_s$ as
$H_0\oplus H_{\mathrm{mid}}$ according to the spectral sets
$\{0\}$ and $(0,1)$. On $\C^{p+1}\otimes H_0$,
\eqref{abs:arbitrary-representation} is the amplification of
evaluation at $0$: its bottom matrix action is zero and its link
has coefficient $1$.
On $H_{\mathrm{mid}}$ the representation $\rho_s$ is unital and
commutes with $h_s$. Continuous functional calculus therefore gives
a unital representation
\[
C([0,1])\otimes M_p\longrightarrow\mathcal B(H_{\mathrm{mid}}),
\qquad f\otimes a\longmapsto f(h_s)\rho_s(a).
\]
Amplifying by $M_{p+1}$ and restricting to $B_p$ sends
\eqref{abs:model-map}--\eqref{abs:model-element} to
\eqref{abs:arbitrary-representation} on this summand.
Together with the two endpoint representations this gives a unital
representation of $B_p$ on all of $H$, sending its generators to
$\alpha(a)$ and $s$. Since these generators generate $B_p$, the
image lies in the represented copy of $B$. The identification of
$B_p$ with $I(p,p+1)$ proves the assertion.
\end{proof}
\subsection{From prime blocks to a central copy of \texorpdfstring{$\Z$}{Z}}
We use the standard structural properties of the Jiang--Su algebra.
It is separable, unital, and nuclear, and is an injective unital
inductive limit of prime dimension-drop blocks
\cite[Proposition~2.5 and its proof]{JS1999}.
It is also strongly self-absorbing
\cite[Theorems~7.6 and~8.7]{JS1999}; see the framework in
\cite{TW2007} and the later construction in \cite{RW2010}.
We next give the passage from the
individual block maps to a homomorphism of the limit, since the block
maps obtained above need not be compatible.
\begin{lemma}\label{abs:prime-block-map}
If $r$ and $s$ are relatively prime, there is a unital homomorphism
$I(r,s)\to I(p,p+1)$ for some $p\ge2$.
\end{lemma}
\begin{proof}
The congruences $p\equiv0\pmod r$ and $p\equiv-1\pmod s$
have a simultaneous solution, and adding a sufficiently large multiple
of $rs$ makes it at least $2$. Write $p=mr$, $p+1=ns$, and use
the unital amplifications
\[
M_r\longrightarrow M_p,\quad a\longmapsto a\otimes1_m,
\qquad
M_s\longrightarrow M_{p+1},\quad b\longmapsto b\otimes1_n.
\]
Their tensor product, applied pointwise, sends the $0$ boundary
$M_r\otimes1_s$ into $M_p\otimes1_{p+1}$ and the $1$ boundary
$1_r\otimes M_s$ into $1_p\otimes M_{p+1}$.
\end{proof}
The next lemma lets us approximately retract a finite subset of $\Z$
into a sufficiently late stage. Composing the retraction with any
homomorphism from that stage gives a completely positive map defined
on all of $\Z$. Approximation on the prescribed elements and their
products makes this map approximately multiplicative.
\begin{lemma}\label{abs:approximate-retraction}
Let $D\subset L$ be a unital inclusion with $D$ nuclear. Given a
finite subset $G\subset D$ and $\eps>0$, there is a \cpc map
$R:L\to D$ with $\|R(d)-d\|<\eps$ for $d\in G$.
\end{lemma}
\begin{proof}
Choose a nuclear approximation
$D\xrightarrow{\theta}M_k\xrightarrow{\eta}D$, with both maps
completely positive and contractive, whose composition is within
$\eps$ of the identity on $G$. The matrix-valued extension theorem
of Arveson \cite[Theorem~1.2.3]{Arveson1969} extends $\theta$ to a \cpc map
$\widetilde\theta:L\to M_k$. Then
$R=\eta\widetilde\theta$ has the stated properties. Applying the
approximation also to $1_D$ makes $R(1_L)$ as close to $1_D$ as
desired; an exactly unital retraction is unnecessary.
\end{proof}
\begin{proposition}\label{prop:Z-map}
Let $A$ be a nonzero separable nuclear $C^*$-algebra satisfying
\textup{(C)}, and fix any free ultrafilter $\omega$ on $\mathbb N$.
Then there is a unital homomorphism $\Z\to F_\omega(A)$.
\end{proposition}
\begin{proof}
Realize $\Z$ as the closure of an increasing sequence of its
injectively included prime dimension-drop stages $D_l$.
By Proposition~\ref{prop:dimensiondrop} and
Lemma~\ref{abs:prime-block-map}, every $D_l$ has a unital
homomorphism $\chi_l:D_l\to F_\omega(A)$.
Let $\mathcal Z_0$ be a countable dense $*$-subalgebra over
$\mathbb Q+i\mathbb Q$ containing $1$, and exhaust it by finite
sets. For any finite set of contractions $F\subset\mathcal Z_0$
and any $\eps>0$, approximate $F\cup F^2\cup\{1\}$ by one
stage, and apply Lemma~\ref{abs:approximate-retraction} to those
stage approximants. With the errors chosen sufficiently small,
the resulting \cpc map $R:\Z\to D_l$, viewed in $\Z$, satisfies
\[
\|R(x)-x\|<\eps
\quad(x\in F\cup F^2\cup\{1\}).
\]
Consequently, for $x,y\in F$,
\[
\|R(xy)-R(x)R(y)\|
\le\|R(xy)-xy\|+\|xy-R(x)R(y)\|<3\eps.
\]
Here the norm calculation takes place in $\Z$, with $D_l$ included
isometrically. Composing with $\chi_l$ produces \cpc maps
$\beta_n:\Z\to F_\omega(A)$ such that
\begin{equation}\label{abs:quotient-approximation}
\|\beta_n(xy)-\beta_n(x)\beta_n(y)\|\longrightarrow0,
\qquad \|\beta_n(1)-1\|\longrightarrow0
\end{equation}
for every $x,y\in\mathcal Z_0$. Including successively larger
scaled tests handles elements outside the unit ball.
We now turn these approximate quotient relations into exact ones.
Put $C_A=A_\omega\cap A'$, and let $q:C_A\to F_\omega(A)$
be the quotient map. By the nuclear completely positive lifting
theorem of Choi--Effros
\cite[Theorem~3.10 and Corollary~3.11]{CE1976}, each $\beta_n$ has a \cpc
lift $L_n:\Z\to C_A$. Regarding $L_n$ as a map into $A_\omega$
and applying the same theorem to
$\ell^\infty(\N,A)\to A_\omega$ gives a \cpc lift
\[
\widetilde L_n:\Z\longrightarrow\ell^\infty(\N,A),
\qquad \widetilde L_n(x)=(L_{n,k}(x))_{k\ge1}.
\]
Each coordinate map $L_{n,k}:\Z\to A$ is completely positive
and contractive, so all its values on each fixed test have a bound
independent of $n$ and $k$.
The relevant quotient norm errors are multiplied errors.
If $d\in C_A$ and $a\in A$, then
\begin{equation}\label{abs:multiplied-quotient-norm}
\|da\|,\ \|ad\|\le\|q(d)\|\,\|a\|.
\end{equation}
Indeed, replacing $d$ by $d+j$ with $j\in\Ann(A)$ does not
change either product, and taking the infimum over such $j$ gives
the inequalities. Apply this to
\[
d=L_n(xy)-L_n(x)L_n(y),\qquad d=L_n(1)-e,
\]
where $e$ represents the unit as above. Equations
\eqref{abs:quotient-approximation} and
\eqref{abs:multiplied-quotient-norm} give errors tending to zero
in $A_\omega$ after multiplication by each fixed $a\in A$.
Also $[L_n(x),a]=0$ exactly for every $n,x,a$.
Choose a countable dense set $\{a_j\}$ in $A$, and diagonalize
over $\mathcal Z_0$, its product tests, and these $a_j$.
More explicitly, choose increasing row indices $m_l$ so that the
first $l$ multiplied defects in row $m_l$ have ultrapower norm less
than $2^{-l-1}$. Their coordinate errors, together with the first
$l$ commutator errors, are less than $2^{-l}$ on some set
$E_l\in\omega$; replacing these sets by their successive finite
intersections with $\{k\ge l\}$ makes them decreasing. Let $l(k)$
be the largest $l\le k$ with $k\in E_l$, or zero when no such level
exists. Put $\gamma_k=L_{m_{l(k)},k}$ if $l(k)>0$, and $\gamma_k=0$
otherwise. Then $l(k)\to_\omega\infty$, and
\[
\Gamma(x)=[(\gamma_k(x))_k]
\]
defines a \cpc map $\Gamma:\Z\to A_\omega\cap A'$. The maps
$\gamma_k$ are linear and completely positive at every coordinate; no
separate lifting of individual elements is involved. The diagonal
conditions say
\[
(\Gamma(xy)-\Gamma(x)\Gamma(y))a=0,\qquad
(\Gamma(1)-e)a=0
\quad(x,y\in\mathcal Z_0,\ a\in A).
\]
All the displayed defects are central relative to $A$, so they
annihilate it on both sides. Passing through $q$ therefore gives
multiplicativity and unitality on $\mathcal Z_0$. Continuity extends
these identities to $\Z$ and proves the proposition.
\end{proof}
\subsection{Connected half-flip unitaries and their lifts}
\begin{lemma}\label{abs:connected-unitaries}
The unitary groups of $\Z$ and $\Z\otimes\Z$ are path connected.
If $q:D\to D/J$ is a unital quotient map, every unitary in the
identity component of $\mathcal U(D/J)$ lifts to a unitary in $D$.
\end{lemma}
\begin{proof}
We first check connectedness for a prime block $I(r,s)$, with
$\gcd(r,s)=1$. A unitary is a continuous path in $U(rs)$ whose
endpoints belong to
\[
H_0=U(r)\otimes1_s,\qquad H_1=1_r\otimes U(s).
\]
Both endpoint groups are connected. Attach a path in $H_0$ from
$1$ to the initial value and a path in $H_1$ from the final value
to $1$. Attaching these paths gradually, while reparametrizing the
original path, is a homotopy through paths satisfying the same
endpoint conditions. The result is a based loop in $U(rs)$.
Changing the attached $H_0$ path by an endpoint loop changes its
determinant winding by an arbitrary multiple of $s$; changing the
$H_1$ path changes it by an arbitrary multiple of $r$. Bezout's
identity therefore permits choices for which the total winding is
zero. The determinant identifies $\pi_1(U(rs))$ with $\mathbb Z$,
so this loop contracts with its basepoint fixed. That contraction
retains endpoint value $1$, and hence remains in the prime block.
Thus every unitary of $I(r,s)$ is connected to $1$.
For a unitary in the injective unital inductive limit $\Z$, choose
a sufficiently close element of a stage. Its two unitary defects
are small, so its polar part is a unitary in the same stage and is
still close to the original unitary. Two unitaries of distance less
than $1$ are joined by the logarithm path for their quotient.
Connectedness of the stage unitary group proves connectedness of
$\mathcal U(\Z)$. Strong self-absorption gives an isomorphism
$\Z\otimes\Z\cong\Z$, proving the same assertion for the tensor
square.
Finally, a unitary $v$ connected to $1$ in $D/J$ is a finite product
of exponentials of selfadjoints. To see this, subdivide a continuous
unitary path finely enough that consecutive values differ by less
than $1$, and take a selfadjoint logarithm of each consecutive
quotient. Lift these finitely many selfadjoints through $q$ and
multiply their exponentials in $D$. The resulting unitary lifts $v$.
\end{proof}
\begin{lemma}\label{abs:finite-unitary-criterion}
Let $A$ be a separable $C^*$-algebra, and suppose there is a unital
homomorphism $\varphi:\Z\to F_\omega(A)$.
For finite sets $F\subset A$, $G\subset A\otimes\Z$, and
$\eps>0$, there is a unitary $u\in A^{\sim}\otimes\Z$ such that
\begin{equation}\label{abs:unitary-criterion}
\|[u,a\otimes1]\|<\eps\quad(a\in F),\qquad
\dist(u^*bu,A\otimes1)<\eps\quad(b\in G).
\end{equation}
\end{lemma}
\begin{proof}
Write $D=A^{\sim}\otimes\Z$ and $A_1=A\otimes1$, and form the
unital algebra and its annihilator quotient
\[
C_D=D_\omega\cap A_1',\qquad
Q=C_D/\Ann_{D_\omega}(A_1).
\]
First-factor sequences give a homomorphism
$F_\omega(A)\to Q$: centrality and annihilation are both preserved
by tensoring with $1$. This homomorphism is unital, since the
approximate-unit representative $e\otimes1$ differs from $1_{D_\omega}$
by an annihilator of $A_1$. Its composition with $\varphi$ is a
unital map $\varphi_1:\Z\to Q$ and commutes with the constant
second-factor map $\varphi_2(z)=1\otimes z+\Ann(A_1)$.
Because $\Z$ is nuclear, these two maps give a unital homomorphism
\[
\Theta:\Z\otimes\Z\longrightarrow Q,
\qquad \Theta(z\otimes z')=\varphi_1(z)\varphi_2(z').
\]
The approximately inner half-flip of the strongly self-absorbing
algebra $\Z$ \cite[Proposition~1.5]{TW2007} says that, for a finite set $H\subset\Z$
and $\delta>0$, there is a unitary $V\in\Z\otimes\Z$ with
\begin{equation}\label{abs:half-flip}
\|V^*(1\otimes z)V-z\otimes1\|<\delta\quad(z\in H).
\end{equation}
By Lemma~\ref{abs:connected-unitaries}, $V$ is connected to $1$.
Thus $\Theta(V)$ is connected to $1$ in $Q$, and the same lemma
lifts it to a unitary $W\in C_D$.
This unitary has a representing sequence of actual unitaries in $D$.
Indeed choose bounded representatives $x_k$. Since $W$ is unitary,
both $\|x_k^*x_k-1\|$ and $\|x_kx_k^*-1\|$ tend to zero along
$\omega$. On a good set both are less than $1/2$, and the polar
parts $u_k=x_k(x_k^*x_k)^{-1/2}$ are unitaries with
$\|u_k-x_k\|\to_\omega0$. Put $u_k=1$ on the remaining
coordinates. Then $W=[(u_k)]$, and $W\in C_D$ gives
\begin{equation}\label{abs:lift-centrality}
\|[u_k,a\otimes1]\|\longrightarrow_\omega0\quad(a\in A).
\end{equation}
For each $z\in H$ choose a lift $c_z\in A_\omega\cap A'$ of
$\varphi(z)$; a single \cpc lift of $\varphi$ is available by
Choi--Effros lifting, so these lifts may satisfy
$\|c_z\|\le\|z\|$. In $C_D$ the difference
\[
d_z=W^*(1\otimes z)W-c_z\otimes1
\]
has quotient norm less than $\delta$ by \eqref{abs:half-flip}.
The analogue of \eqref{abs:multiplied-quotient-norm} gives
\[
\|(a\otimes1)d_z\|\le\delta\|a\|.
\]
Since $W$ commutes with $a\otimes1$ in the ultrapower, this is
the concrete norm estimate
\begin{equation}\label{abs:half-flip-multiplied}
\|W^*(a\otimes z)W-(ac_z)\otimes1\|\le\delta\|a\|.
\end{equation}
Here $(ac_z)\otimes1$ is represented by a first-factor sequence
$(ac_{z,k})\otimes1$, with $ac_{z,k}\in A$.
Thus no estimate in an annihilator quotient is being used as an
unmultiplied estimate in a norm ultrapower.
For finitely many elementary tensors, choose one coordinate where
\eqref{abs:lift-centrality} and
\eqref{abs:half-flip-multiplied} all have the desired accuracy.
This gives \eqref{abs:unitary-criterion} for those tensors.
For a general finite $G$, approximate each of its elements by a
finite sum of elementary tensors, include all the corresponding
second factors in $H$, and choose $\delta$ small enough for the
finite sums of the bounds $\delta\|a\|$. Unitary conjugation
preserves the initial approximation errors, so the same conclusion
holds for $G$.
\end{proof}
\subsection{The final intertwining}
The construction of the central map used nuclearity and condition
\textup{(C)}. The implication from that map to absorption requires
only separability, as the following proposition makes explicit.
\begin{proposition}\label{prop:absorption}
If a separable $C^*$-algebra $A$ admits a unital homomorphism
$\Z\to F_\omega(A)$, then $A\cong A\otimes\Z$.
\end{proposition}
\begin{proof}
The zero algebra is immediate, so assume $A\ne0$.
Set $B=A\otimes\Z$ and let $\iota(a)=a\otimes1$.
The algebra $A^{\sim}\otimes\Z$ acts by multipliers on its ideal
$B$, so all the unitaries in Lemma~\ref{abs:finite-unitary-criterion}
act as automorphisms of $B$ by conjugation.
Choose sequences $(a_j)$ and $(b_j)$ dense in $A$ and $B$,
respectively, and put $\eps_n=2^{-n}$.
We construct products of multiplier unitaries $U_n$ and embeddings
$\iota_n=\Ad(U_n)\circ\iota$ as follows. Start with $U_0=1$.
Before step $n$, let $F_n$ be a finite subset of $A$ containing
$a_1,\ldots,a_n$ and all elements $c_{k,j}$ chosen at earlier
steps $k<n$, $1\le j\le k$. Apply
Lemma~\ref{abs:finite-unitary-criterion} with these domain tests and
the finite target set
\[
\{U_{n-1}^*b_jU_{n-1}:1\le j\le n\}.
\]
Obtain $u_n$ with errors less than $\eps_n$, and put
$U_n=U_{n-1}u_n$.
The commutator estimate gives
\begin{equation}\label{abs:successive-embeddings}
\|\iota_n(a)-\iota_{n-1}(a)\|<\eps_n
\quad(a\in F_n),
\end{equation}
and the range estimate gives elements $c_{n,j}\in A$ with
\begin{equation}\label{abs:stored-preimages}
\|\iota_n(c_{n,j})-b_j\|<\eps_n
\quad(1\le j\le n).
\end{equation}
Store these finitely many preimages in every subsequent $F_m$.
They are legitimate finite norm tests: each has finite norm, indeed
$\|c_{n,j}\|<\|b_j\|+\eps_n$, because $\iota_n$ is isometric.
For each dense domain element $a_j$, equation
\eqref{abs:successive-embeddings} holds at every step $n\ge j$.
Summability of $(\eps_n)$ gives convergence of $\iota_n(a_j)$.
As all $\iota_n$ are isometric, approximation by the dense set
extends this to point-norm convergence on $A$.
The limit $\Phi:A\to B$ is a homomorphism by continuity of
multiplication and adjoints, and it is isometric since
$\|\Phi(a)\|=\lim_n\|\iota_n(a)\|=\|a\|$.
Fix $j$ and take any $n\ge j$. The stored preimage $c_{n,j}$
belongs to every $F_m$ with $m>n$, so
\[
\|\Phi(c_{n,j})-b_j\|
\le\eps_n+\sum_{m>n}\eps_m=2^{1-n}.
\]
Letting $n\to\infty$ shows that every $b_j$ is in the closure
of $\Phi(A)$. The range of an isometric homomorphism is closed,
and the $b_j$ are dense, so $\Phi(A)=B$. This proves the desired
isomorphism.
\end{proof}
==== END Cuntz-comparison-and-Jiang-Su-absorption//absorption.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//assembly.tex ====
\section{Assembly of the absorption theorems}
\label{sec:assembly}
The preceding construction has used separability, nuclearity, and
condition (C), rather than the canonical-map hypothesis. We now apply
it to the comparison and Cuntz-semigroup criteria in the introduction.
\begin{proof}[Proof of Theorem~\ref{thm:condition-C}]
Fix the free ultrafilter in the theorem.
Proposition~\ref{prop:Z-map}, through the support, norm-cone,
tracial-cone, comparison, and packing constructions above, supplies
a unital $*$-homomorphism $\Z\to F_\omega(A)$.
Proposition~\ref{prop:absorption} then gives
$A\cong A\otimes_{\min}\Z$.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{ext:thm:main}]
The zero algebra, if admitted by the convention for simplicity, is
immediate. Nuclearity implies exactness, so
Corollary~\ref{ext:cor:condition-C} shows that $A$ satisfies
condition \textup{(C)}. Theorem~\ref{thm:condition-C} then gives
$A\cong A\otimes_{\min}\Z$.
\end{proof}
\begin{proof}[Proof of Corollary~\ref{cor:canonical}]
Again the zero algebra is immediate. Lemma~\ref{lem:C} derives
(C) from the canonical induced map $\Cu(\iota)$, using that
$\Cu(\iota)([b])=[b\otimes1_\Z]$. Theorem~\ref{thm:condition-C}
then proves the assertion.
\end{proof}
\begin{proof}[Proof of Corollary~\ref{cor:abstract-cu}]
The forward implication follows from functoriality of $\Cu$ under
$*$-isomorphisms. For the converse, let
$\Phi:\Cu(A)\to\Cu(A\otimes_{\min}\Z)$ be an abstract isomorphism.
The case $A=0$ is immediate, so put $D=A\otimes_{\min}\Z$ and suppose
$A\ne0$. Then $D$ is nonzero, separable, and nuclear.
Strong self-absorption of $\Z$ supplies an isomorphism
$\varphi:\Z\to\Z\otimes\Z$ approximately unitarily equivalent to
$z\mapsto z\otimes1_\Z$
\cite[Definition~1.3(iv) and Example~1.14(iv)]{TW2007}.
Consequently $\id_A\otimes\varphi:D\to D\otimes\Z$ is an
isomorphism approximately unitarily equivalent to the first-factor
map $j_D(d)=d\otimes1_\Z$. This also holds when $A$ is nonunital:
if $u_n\in\Z\otimes\Z$ are the implementing unitaries, then
$1_{M(A)}\otimes u_n\in M(A\otimes\Z\otimes\Z)$ implement the
equivalence, first on elementary tensors and then on all of $D$ by
density.
Approximately unitarily equivalent homomorphisms induce the same
map on $\Cu$. Indeed, after stabilization, norm approximation by
unitary conjugates gives comparison of every positive cut of either
image with the other image, by
Lemma~\ref{lem:pre:cut-placement}(1); approximate units place the
comparison witnesses in the algebra when the conjugating unitaries
are multipliers. Taking the suprema of cuts in
Equation~\eqref{eq:pre:cuts} gives equality of the two Cuntz classes.
Thus $\Cu(j_D)=\Cu(\id_A\otimes\varphi)$ is an isomorphism.
Lemma~\ref{lem:C}, applied to $D$, gives condition \textup{(C)} on
$\Cu(D)$. Both almost unperforation and the full almost-divisibility
sandwich are intrinsic ordered-semigroup properties, so $\Phi$
transports them to $\Cu(A)$. Theorem~\ref{thm:condition-C} now gives
$A\cong A\otimes_{\min}\Z$.
\end{proof}
==== END Cuntz-comparison-and-Jiang-Su-absorption//assembly.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//central.tex ====
\section{Simultaneous domination and a central defect filler}
\label{cen:section}
The slots of Proposition~\ref{prop:slots} provide a central matrix
cone $\alpha$ with weight $h$, a positive contraction $P=s_1$,
and orthogonal equivalent slots $s_j$. Put $E=A_\omega$ and
\[
\Phi(a)=(1-h)a,\qquad \Psi(a)=P^2a\quad(a\in A).
\]
We will extend the comparison relations to obtain
$\Phi(a)\in\Ideal_E(\Psi(a))$ for every $a\in A_+$.
The individual comparison witnesses depend on $a$. To fill the
defect centrally, for each prescribed finite $F\subset A$ and
$\delta>0$ we need one finite list $v_1,\ldots,v_N\in E$ such that
\[
\left\|\Phi(a)-\sum_{j=1}^N v_j^*\Psi(a)v_j\right\|<\delta
\qquad(a\in F).
\]
The first part of this section proves this implication for weighted
inclusions. Nuclearity allows us to pass from pointwise ideal
containment to an inclusion of support-representation kernels; the
two tensor-ideal lemmas establish the needed bridge.
The slots then pack the common sum into $z=\sum_j s_jv_j$.
These elements need not be uniformly bounded or central.
Approximate-unit tests will give bounded compressions with
commutators tending to zero, and hence a bounded central element
$S$ satisfying
\[
\alpha(e_{11})S=S,\qquad S^*S a=(1-h)a\quad(a\in A).
\]
This defect-filling argument is related to Matui--Sato's use of
excision in small central sequences and property \textup{(SI)}
\cite{MatuiSato2012}.
\subsection{Tensor ideals and support representations}
\begin{lemma}\label{cen:elementary-tensor}
Let $D$ and $C$ be $C^*$-algebras. Every nonzero closed two-sided ideal
of $D\otimes C$ contains a nonzero tensor $a\otimes b$ with
$a\in D_+$ and $b\in C_+$.
\end{lemma}
\begin{proof}
Let $0\ne x\in J_+$. There is a pure state $\lambda$ of the
unitization $D^{\sim}$ such that
$b=(\lambda\otimes\id_C)(x)\ne0$.
Indeed, product positive functionals detect nonzero positive elements of
a spatial tensor product. Fixing the second functional gives a nonzero
positive affine function on the state space of $D^{\sim}$, which is
nonzero at some extreme point. The resulting pure state does not
annihilate $D$. Its GNS representation is therefore nondegenerate on
$D$; write $\xi$ for its unit cyclic vector.
Lemma~\ref{norm:excision} gives,
on any prescribed finite set of coefficients in $D^{\sim}$, positive
contractions $g\in D^{\sim}$ with
\[
\lambda(g)=1,\qquad
\|gcg-\lambda(c)g^2\|\longrightarrow0.
\]
Here $\pi_\lambda(g)\xi=\xi$, since $g$ is a positive contraction.
Choose once and for all a positive contraction $u\in D$ such that
$\|\pi_\lambda(u)\xi-\xi\|<1/8$. Approximation of $x$ by finite
algebraic sums and excision of their first coefficients show that
\[
\dist\bigl((ug^2u)\otimes b,J\bigr)\longrightarrow0.
\]
The element used to approximate it is
$(ug\otimes1)x(gu\otimes1)$, which belongs to $J$; all products with
$1$ here are multiplier products. Moreover,
\[
\|ug^2u\|\ \ge\
\|\pi_\lambda(g)\pi_\lambda(u)\xi\|^2
\ > (7/8)^2>1/2.
\]
Thus the first factor retains a fixed norm lower bound as the excision
accuracy improves.
We use the following elementary cut observation. If $a,b$ are positive
and $\|a\otimes b+J\|<t^2$, then
\begin{equation}\label{cen:product-cut}
(a-t)_+\otimes(b-t)_+\in J.
\end{equation}
To see this, represent $(D\otimes C)/J$ faithfully and nondegenerately.
The commuting multiplier operators representing $a\otimes1$ and
$1\otimes b$ have product of norm less than $t^2$. Their joint spectrum
has no point with both coordinates larger than $t$. Joint continuous
functional calculus proves \eqref{cen:product-cut}.
Now fix $0<t<\min\{1/2,\|b\|\}$ and choose $g$ so that the preceding
distance is less than $t^2$. Both cuts in
$(ug^2u-t)_+\otimes(b-t)_+$ are nonzero, and their tensor product
belongs to $J$.
\end{proof}
\begin{lemma}\label{cen:nuclear-tensor-ideals}
If $D$ is nuclear and $C$ is arbitrary, every closed two-sided ideal
$J\subset D\otimes C$ is generated by the positive elementary tensors
that it contains.
\end{lemma}
\begin{proof}
Let $J_0$ be the ideal generated by those tensors. If $J_0\ne J$,
primitive ideal separation gives a nondegenerate irreducible
representation $\pi$ with $\pi(J_0)=0$ and $\pi(J)\ne0$.
Let $\pi_D$ and $\pi_C$ be its commuting leg representations, and put
$I=\ker\pi_D$, $L=\ker\pi_C$. Nuclearity of $D$ and of its quotient
$D/I$ \cite[Corollary~4]{CE1977}, together with the quotient identity for the maximal tensor
product, gives
\begin{equation}\label{cen:tensor-quotient}
q:D\otimes C\longrightarrow (D/I)\otimes(C/L),\qquad
\ker q=\overline{I\otimes C+D\otimes L}.
\end{equation}
For completeness, the maximal quotient identity follows by the
universal property of commuting representations: a representation
annihilating the indicated two leg ideals is exactly a pair factoring
through $D/I$ and $C/L$. The nuclearity of the first leg identifies
both maximal products in this statement with spatial products.
The representation induced by $\pi$ on the target of $q$ is faithful.
Here is the needed verification. Write
$M=\pi_D(D)''$ and $N=\pi_C(C)''$. The center of $M$ commutes with
both legs, hence with the irreducible representation $\pi$; it consists
of scalars. Thus $M$ is a factor, and the same holds for $N$.
If nonzero positive elements $x\in M$, $y\in N$ had $xy=0$, their
support projections $e\in M$, $f\in N\subset M'$ would satisfy
$ef=0$. Since $f$ commutes with $M$, it annihilates every $vev^*$,
$v\in\mathcal U(M)$. The join of these projections is the central
support of $e$, which is $1$ in the factor $M$. This would give $f=0$,
a contradiction. The leg representations of $D/I$ and $C/L$ are
faithful, so the induced tensor representation kills no nonzero positive
elementary tensor. Lemma~\ref{cen:elementary-tensor} applied to its
kernel proves faithfulness.
The nonzero ideal $q(J)$ therefore contains a nonzero positive tensor
$\bar a\otimes\bar b$. Choose positive lifts $a\in D$, $b\in C$.
Since $a\otimes b\in J+\ker q$, compression by
$(1-e)\otimes(1-f)$, where $e$ and $f$ run through positive contractive
approximate units of $I$ and $L$, gives
\[
\dist(a_{e}\otimes b_f,J)\longrightarrow0,
\quad a_e=(1-e)a(1-e),\quad b_f=(1-f)b(1-f).
\]
Indeed these compressions send each term of a finite sum from
$I\otimes C+D\otimes L$ to zero in norm, and they preserve $J$.
The quotient images of $a_e,b_f$ remain $\bar a,\bar b$. Fix
$0<t<\min\{\|\bar a\|,\|\bar b\|\}$. Once the displayed distance
is less than $t^2$, \eqref{cen:product-cut} yields
$(a_e-t)_+\otimes(b_f-t)_+\in J_0$ with nonzero image under $q$.
Faithfulness of the induced representation contradicts $\pi(J_0)=0$.
\end{proof}
\subsection{The simultaneous approximation}
Gabe's approximate domination theorem
\cite[Definition~3.1 and Theorem~3.3]{Gabe2020} applies to an exact
$C^*$-algebra $D$, an arbitrary $C^*$-algebra $E$, and nuclear
completely positive maps $\Psi,\Phi:D\to E$, with $\Psi$ of order
zero. It identifies the condition
$\Phi(a)\in\Ideal_E(\Psi(a))$ for every $a\in D_+$ with the
following finite-set approximation: for every finite $F\subset D$
and $\delta>0$, there are $N\in\N$ and $v_1,\ldots,v_N\in E$
such that
\begin{equation}\label{cen:domination-equation}
\left\|\Phi(a)-\sum_{j=1}^Nv_j^*\Psi(a)v_j\right\|<\delta
\qquad(a\in F).
\end{equation}
Nuclearity is required of the maps; no nuclearity or separability
assumption is imposed on $E$. We prove the weighted-inclusion case
used here, so that the passage from ideal containment to one common
finite list of compression coefficients is explicit.
\begin{lemma}[Simultaneous domination for weighted inclusions]
\label{cen:weighted-domination}
Let $D\subseteq E$ be a nuclear $C^*$-subalgebra of a $C^*$-algebra
$E$, and let $v,w$ be positive contractions in $\widetilde E\cap D'$.
Put $\Phi(a)=va$ and $\Psi(a)=wa$ for $a\in D$; these products
belong to $E$. Suppose that
\[
\Phi(a)\in\Ideal_E(\Psi(a))\qquad(a\in D_+).
\]
Then for every finite set $F\subset D$ and every $\delta>0$, there
are an integer $N\ge1$ and $v_1,\ldots,v_N\in E$ such that
\[
\left\|va-\sum_{j=1}^N v_j^*wa v_j\right\|<\delta
\qquad(a\in F).
\]
\end{lemma}
\begin{proof}
Let $\mathcal Q$ be the cone of maps
$a\mapsto\sum_{j=1}^N b_j^*wa b_j$, where $b_j\in E$.
It is convex: sums concatenate the lists, and multiplication by a
nonnegative scalar is absorbed into the $b_j$. It suffices first to
place $\Phi$ in its point-weak closure.
Fix $a_1,\ldots,a_m\in D$ and $\ell_1,\ldots,\ell_m\in E^*$.
We may represent all these functionals in one nondegenerate cyclic
representation $\sigma:E\to\mathcal B(H)$, with cyclic vector $\xi$,
in the form
\begin{equation}\label{cen:common-functional-model}
\ell_i(b)=\langle\sigma(b)c_i\xi,\xi\rangle,
\qquad c_i\in\sigma(E)'.
\end{equation}
To verify this, decompose the real and imaginary parts of each
$\ell_i$ into differences of positive functionals. A positive multiple
of their sum dominates all the positive summands. In its GNS
representation a dominated positive functional gives a bounded positive
form on vectors $\sigma(b)\xi$; its representing operator commutes
with $\sigma(E)$, as is seen by moving $\sigma(x)$ between the two
arguments of the form. Linear combinations give the $c_i$ in
\eqref{cen:common-functional-model}. Extend $\sigma$ unitally to
$E^{\sim}$ when necessary.
Put $C=C^*(1,c_1,\ldots,c_m)\subset\sigma(E)'$. The commuting
representations define
\[
\pi:D\otimes C\longrightarrow\mathcal B(H),\qquad
\pi(a\otimes c)=\sigma(a)c;
\]
continuity for the spatial tensor norm follows from nuclearity of $D$.
For $q=v,w$, let $p_q$ be the support projection of $\sigma(q)$ and
let $\pi_q$ be the restriction to $p_qH$. This is a reducing subspace
for $\pi$: each $q$ commutes with $D$, and each element of $C$ commutes
with $E^{\sim}$.
We claim
\begin{equation}\label{cen:support-kernel}
\ker\pi_w\subset\ker\pi_v.
\end{equation}
Suppose $a\otimes c\in\ker\pi_w$. Commutation with the weight
and positivity give
\[
\sigma(wa^*a)c^*c=0.
\]
Since $c^*c$ commutes with all of $\sigma(E)$, this identity persists
after multiplying $\sigma(wa^*a)$ on either side by elements of
$\sigma(E)$ and taking closed linear spans. The ideal hypothesis
$va^*a\in\Ideal_E(wa^*a)$ therefore gives
$\sigma(va^*a)c^*c=0$. This says
$\pi(a\otimes c)\sigma(v)^{1/2}=0$, and, by density of the range of
$\sigma(v)^{1/2}$ in $p_vH$, gives $a\otimes c\in\ker\pi_v$.
Lemma~\ref{cen:nuclear-tensor-ideals} says that the elementary tensors
in $\ker\pi_w$ generate that kernel, and proves
\eqref{cen:support-kernel}.
Let $K=\ker\pi_w$. The vectors
\[
\eta_b=\sigma(w)^{1/2}\sigma(b)\xi,\qquad b\in E,
\]
are dense in $p_wH$: cyclicity first gives density before application
of $\sigma(w)^{1/2}$, and its range is dense in its support space.
Finite sums of their vector functionals have weak-* closure equal to
the positive functional cone of $(D\otimes C)/K$. Here is the
separation argument. A selfadjoint element of this quotient that is
nonnegative on all these functionals has a positive representing
operator on $p_wH$, by the density of the $\eta_b$. Faithfulness of
the quotient representation makes the element positive. Thus the
polar of the cone of these vector functionals is exactly the positive
cone of the quotient. The bipolar form of Hahn--Banach separation
identifies its weak-* closure with all positive functionals on that
quotient. This argument also applies to a nonunital quotient.
The positive functional
\[
\gamma_v(x)=
\langle\pi(x)\sigma(v)^{1/2}\xi,
\sigma(v)^{1/2}\xi\rangle
\]
factors through this quotient by \eqref{cen:support-kernel}. Hence,
simultaneously on the finitely many tensors $a_i\otimes c_i$, it is
approximated by finite sums of vector functionals for $\eta_{b_j}$.
On these tensors the identities are
\begin{align*}
\gamma_v(a_i\otimes c_i)&=\ell_i(va_i),\\
\sum_j\langle\pi(a_i\otimes c_i)\eta_{b_j},\eta_{b_j}\rangle
&=\ell_i\left(\sum_jb_j^*wa_i b_j\right).
\end{align*}
They prove the required point-weak closure, with the same finite list
of coefficients for all the tests. Finally, for a fixed finite list
$a_1,\ldots,a_m$, the set
\[
\{(T(a_1),\ldots,T(a_m)):T\in\mathcal Q\}\subset E^m
\]
is convex. Its weak and norm closures coincide by Hahn--Banach
separation. This proves \eqref{cen:domination-equation}.
\end{proof}
\subsection{From slots to a bounded central element}
\begin{proposition}\label{prop:central-filler}
Let $A$ be a separable nuclear $C^*$-algebra satisfying
\textup{(C)}.
For each $p\ge2$ there are a \cpc order-zero map
$\alpha:M_p\to A_\omega\cap A'$, with $h=\alpha(1)$, and an element
$S\in A_\omega\cap A'$ such that
\begin{equation}\label{cen:filler-relations}
\alpha(e_{11})S=S,\qquad S^*S a=(1-h)a\quad(a\in A).
\end{equation}
One can choose $\|S\|\le2$. In the second equation the expression
$(1-h)a$ is computed in the unitization of $A_\omega$ and belongs to
$A_\omega$.
\end{proposition}
\begin{proof}
Fix the map and slots from Proposition~\ref{prop:slots}. Thus
$P=s_1\ge0$, all $s_j$ are central contractions,
\begin{equation}\label{cen:slot-recall}
s_i^*s_j=\delta_{ij}P^2,\qquad
\alpha(e_{11})s_j=s_j,
\end{equation}
and, for every element $a$ of the countable dense family
$\mathcal A$ of positive contractions fixed in
Section~\ref{sec:comparison}, closed under positive rational cuts, and
every rational $\eps>0$, there is a bounded element $r_{a,\eps}$
with
\[
((1-h)a-\eps)_+=r_{a,\eps}^*(P^2a)r_{a,\eps}.
\]
Set $E=A_\omega$, $v=1-h$, $w=P^2$, and define
$\Phi(a)=va$, $\Psi(a)=wa$. Both maps are completely positive
and contractive, and $\Psi$ has order zero. Since $A$ is nuclear,
composing finite-dimensional completely positive approximations to
$\id_A$ with either map proves that both maps are nuclear.
Letting $\eps\downarrow0$ in the last comparison gives
$\Phi(d)\in\Ideal_E(\Psi(d))$ for $d\in\mathcal A$. We verify
the extension to an arbitrary positive contraction $a\in A$. Choose
$d_n\in\mathcal A$ and positive rational $\delta_n\downarrow0$
with $\|d_n-a\|<\delta_n$. The cuts
$b_n=(d_n-\delta_n)_+$ belong to $\mathcal A\cap\Ideal_A(a)$,
by positive-element perturbation, and $b_n\to a$ in norm.
Centrality of $w$ implies
\begin{equation}\label{cen:ideal-extension}
w\Ideal_A(a)\subset\Ideal_E(wa):
\quad w(xay)=x(wa)y\quad(x,y\in A).
\end{equation}
The same assertion follows for the closed ideal by approximation with
finite sums. Consequently $\Psi(b_n)\in\Ideal_E(\Psi(a))$, and
\[
\Phi(b_n)\in\Ideal_E(\Psi(b_n))
\subset\Ideal_E(\Psi(a)).
\]
Taking the norm limit proves the ideal hypothesis for $a$; scaling
proves it for every positive element.
Apply Lemma~\ref{cen:weighted-domination}. Any resulting sum packs
into a single element
$z=\sum_{j=1}^N s_jv_j\in E$, because, for $a\in A$,
\begin{equation}\label{cen:packed-sum}
z^*az=\sum_{i,j}v_i^*s_i^*as_jv_j
=\sum_jv_j^*(P^2a)v_j.
\end{equation}
There is no uniform bound here on $z$ or on the individual $v_j$,
and the coefficients need not commute with $A$. A compression by
one approximate-unit element will supply both the bound needed for
diagonalization and, through simultaneous tests, control of the
commutators.
Fix a finite set $F$ of contractions in $A$, and let $\eta,\delta>0$.
Choose a positive contraction $t\in A$ from an approximate unit so
that, for all $a\in F$,
\[
\|t^2a-a\|<\eta,\qquad
\|tat-a\|<\eta,\qquad
\|ta^*at-a^*a\|<\eta.
\]
Approximate $\Phi$ to within $\delta$, using one sum in
\eqref{cen:packed-sum}, on
\[
\{t^2\}\cup\{tat,ta^*at:a\in F\}.
\]
Put $S'=tz$. The $t^2$ test gives
\begin{equation}\label{cen:bounded-compression}
\|S'\|^2=\|z^*t^2z\|\le\|vt^2\|+\delta\le1+\delta.
\end{equation}
Also $\alpha(e_{11})S'=S'$ exactly, since $\alpha(e_{11})$
commutes with $t$ and fixes all the $s_j$. Writing
$\kappa=\delta+\eta$, the three tests give
\begin{align}
\|S'^*S'a-va\|&<\kappa,\label{cen:first-test}\\
\|S'^*aS'-va\|&<\kappa,\label{cen:second-test}\\
\|S'^*a^*aS'-va^*a\|&<\kappa.\label{cen:third-test}
\end{align}
For the first line, multiply
$\|S'^*S'-vt^2\|<\delta$ by $a$; the other two lines follow
directly from their respective tests. Taking the adjoint of
\eqref{cen:second-test} controls $S'^*a^*S'$ as well.
Now expand the full commutator square:
\begin{align*}
(aS'-S'a)^*(aS'-S'a)
={}&S'^*a^*aS'-S'^*a^*S'a\\
&-a^*S'^*aS'+a^*S'^*S'a.
\end{align*}
Its four terms are within $\kappa$ of, respectively,
$va^*a$, $-va^*a$, $-va^*a$, and $va^*a$, because $v$ commutes
with $A$. Therefore
\begin{equation}\label{cen:commutator-bound}
\|[a,S']\|\le2\sqrt\kappa\qquad(a\in F).
\end{equation}
Take increasing finite sets dense in the unit ball of $A$ and choose
$\eta,\delta\downarrow0$, always with $\delta<1$. Each resulting
$S'$ has norm at most $\sqrt2$ by
\eqref{cen:bounded-compression}, and may be represented by a sequence
bounded by $2$. Fix representatives of the finitely many matrix
entries of $\alpha$ throughout. The countable diagonalization of
Lemma~\ref{lem:diagonal}, applied to these fixed representatives and
the norm tests \eqref{cen:first-test},
\eqref{cen:commutator-bound}, and
$\alpha(e_{11})S'=S'$, gives a single element $S$ of norm at most $2$.
It commutes with every element of $A$, retains the fixed support
equation, and satisfies \eqref{cen:filler-relations} on a dense set,
hence on all of $A$ by continuity. Only the uniformly bounded $S'$
are diagonalized at this last step; the compression coefficients and
their unbounded sums are not retained as variables.
\end{proof}
==== END Cuntz-comparison-and-Jiang-Su-absorption//central.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//comparison.tex ====
\section{Adaptive comparison and an unlimited supply of slots}
\label{sec:comparison}
Our goal is to compare a cut of the defect of a central matrix cone
with one of its diagonal corners, with a witness bound independent of
matrix size. Norm visibility supplies finite coordinate ideal
multiplicities $L_n$, which need not be bounded. We fix the norm
maps and these integers first, and then choose a tracial cone with
relative defect error $\eta_n/L_n$, where $\eta_n\to0$.
The ideal comparison bounds the controlling rank by a multiple of
$L_n$ times the target rank; passing to a diagonal corner adds only
the fixed matrix-size factor. The reciprocal $L_n$ in the trace
error cancels this multiplicity. We must therefore obtain an exact
coordinate Cuntz inequality before using it. A further spectral
margin bounds the witnesses independently of matrix size, allowing
the final construction of countably many orthogonal equivalent
central slots.
Fix a countable norm-dense family $\mathcal A$ in the positive
contractions of $A$, closed under positive rational spectral cuts.
Such a family is obtained by starting with a countable dense family
and adjoining its iterated rational cuts. Enumerate the tests
\[
(a_r,\eps_r)\in\mathcal A\times(\mathbb Q\cap(0,1)),
\qquad r\ge1.
\]
We shall use the fixed scalar functions
\begin{equation}\label{cmp:cutoff-functions}
f(t)=\min(1,2t),\qquad g(t)=\max(0,2t-1)
\quad(0\le t\le1).
\end{equation}
Thus $g$ is nondecreasing, vanishes on $[0,1/2]$, and $g(1)=1$;
also $f(t)\ge t$ and $fg=g$. Functional calculus on an order-zero
map has the meaning fixed in the preliminaries. The function $f$
enlarges its weight, while $g$ selects a part on which $f$ is one;
the enlarged cone will consequently fix the slots cut out by $g$.
\begin{proposition}[Coordinate ideal visibility]\label{prop:idealvisibility}
For every fixed $k\ge2$ there is a \cpc order-zero map
$\nu:M_k\to A_\omega\cap A'$, represented by \cpc maps
$\nu_n:M_k\to A$, with weights $h_n^\nu=\nu_n(1)$, such that the
following holds. For every test $(a,\eps)$, put
\[
b=(a-\eps/8)_+,\qquad
y_n^\nu=a^{1/2}g(h_n^\nu)^2a^{1/2}.
\]
There are integers $L_{a,\eps,n}\ge1$ and an $\omega$-large set
$E_{a,\eps}$ on which
\begin{equation}\label{cmp:I}
[b]\le L_{a,\eps,n}[(y_n^\nu-\eps/20)_+]
\quad\text{in }\Cu(A).
\end{equation}
The integers need not be bounded as $n$ varies.
\end{proposition}
\begin{proof}
First consider finitely many tests. For one of them apply
Proposition~\ref{prop:norm} with
\[
x=(a-\eps/16)_+,
\qquad \gamma=\eps/64.
\]
It gives a norm-central order-zero map with positive contractive
weight representatives $t_n$ and a constant $0<\kappa\le1$ such
that, on a common $\omega$-large set,
\begin{equation}\label{cmp:norm-trigger}
\|\sigma(x)\|\ge\gamma
\quad\Longrightarrow\quad
\|\sigma(xt_n)\|\ge\kappa
\qquad\text{for every irreducible }\sigma.
\end{equation}
If no representation satisfies the trigger, one can take the zero
map and any $0<\kappa\le1$; the assertions below then remain
vacuous on the relevant representations. Choose a continuous
positive contraction $\chi$ on $[0,1]$, zero at zero and equal to
one on $[\kappa/2,1]$, and reparametrize the order-zero map by $\chi$.
Do this for each of the finitely many tests, choosing commuting
maps by successive reindexing. Here reindexing preserves the
representation quantifier in \eqref{cmp:norm-trigger}. More
explicitly, at coordinate $n$, previously chosen matrix-unit lifts
form a finite list of elements of $A$. Choose an index of the next
map at which its norm commutators with those elements and the first
$n$ dense tests are small, its order-zero defects are small, and
its own uniform implication \eqref{cmp:norm-trigger} holds.
Each condition holds on an $\omega$-large set, and a finite
intersection is nonempty. The implication is simultaneous over
all representations at the selected index, so it is retained
without choosing representations during reindexing. Join the
reparametrized maps using Lemma~\ref{lem:join}.
Let $\nu$ denote this finite join for the moment, and choose \cpc
coordinate lifts with weights $h_n^\nu$. We claim that, on an
$\omega$-large set, simultaneously in all irreducible representations,
\begin{equation}\label{cmp:visibility-implication}
\|\sigma(a)\|>3\eps/32
\quad\Longrightarrow\quad
\|\sigma(a^{1/2}g(h_n^\nu)^2a^{1/2})\|>\eps/20.
\end{equation}
Suppose instead that its failure set belongs to $\omega$, and
choose a failing representation $\sigma_n$ at each such coordinate.
Intersect with the good set for \eqref{cmp:norm-trigger}. Since
\[
\|\sigma_n((a-\eps/16)_+)\|
>3\eps/32-\eps/16=\eps/32>\gamma,
\]
that trigger applies, even if the represented norms tend to the
boundary $3\eps/32$.
Take the Hilbert space ultraproduct of these representations, and
write $\bar a,\bar t,\bar H$ for the resulting positive operators
representing $a$, the old unmodified weight for this test, and the
joined weight. Norm centrality gives $[\bar a,\bar t]=0$, and
\eqref{cmp:norm-trigger} gives
\[
\| (\bar a-\eps/16)_+\bar t\|\ge\kappa.
\]
Consequently the joint spectral part where $\bar a>\eps/16$ and
$\bar t>\kappa/2$ is nonzero. Indeed the product is zero where
$\bar a\le\eps/16$ and has norm at most $\kappa/2$ where
$\bar t\le\kappa/2$. On the indicated nonzero part the
reparametrized weight $\chi(\bar t)$ is one. The joined weight
dominates this weight and commutes with it and with $\bar a$;
being a contraction, it too is one there. It follows that
\[
\|\bar a^{1/2}g(\bar H)^2\bar a^{1/2}\|
\ge\eps/16>\eps/20.
\]
This contradicts the upper bound at the failing coordinates.
Arbitrary positive weight lifts of the joined map give the same
limiting operators, so no particular lifting convention was needed
for this contradiction. This proves
\eqref{cmp:visibility-implication} for each test in the finite group.
For each $l$ perform this construction for the first $l$ tests.
Take \cpc coordinate lifts of the resulting maps. Choose a decreasing
sequence of $\omega$-large sets on which the first $l$ norm
centrality tests and matrix-unit order-zero identities hold to
accuracy tending to zero, and all the first $l$ implications
\eqref{cmp:visibility-implication} hold exactly. Include $\{n\ge l\}$
in the $l$th set, and use its largest admissible stage at coordinate
$n$. Lemma~\ref{lem:diagonal} applies, since all these maps and
weights are contractions. The diagonal defines a single
norm-central order-zero $\nu$, with \cpc lifts, and preserves each
of the implications on an $\omega$-large set.
At a coordinate where \eqref{cmp:visibility-implication} holds,
every irreducible representation annihilating
$(y_n^\nu-\eps/20)_+$ annihilates $(a-3\eps/32)_+$. Indeed
annihilation of the former is precisely the condition
$\|\sigma(y_n^\nu)\|\le\eps/20$, which implies
$\|\sigma(a)\|\le3\eps/32$. Primitive ideal separation gives
\[
(a-3\eps/32)_+\in\Ideal_A((y_n^\nu-\eps/20)_+).
\]
The additional positive cut $\eps/8-3\eps/32=\eps/32$ now gives a
finite integer $L_{a,\eps,n}$ in \eqref{cmp:I}, by the compact-cut
ideal comparison from the preliminaries. Choose the integer to be
at least one and put it equal to one outside the good set. This
makes all constants specified at all coordinates and proves the
proposition.
\end{proof}
\begin{proposition}[An adaptive rank gap]\label{prop:rankgap}
For every fixed $k\ge2$ there is a \cpc order-zero map
$\zeta^k:M_k\to A_\omega\cap A'$ with weight $H$ such that, for every
test $(a,\eps)$, if
\begin{equation}\label{cmp:u-y}
u^k(a)=(1-f(H))a,
\qquad y^k(a)=g(\zeta^k)(e_{11})^2a,
\end{equation}
there is $r_{a,\eps}^k\in A_\omega$ satisfying
\begin{equation}\label{cmp:bounded-comparison}
(u^k(a)-\eps)_+
=(r_{a,\eps}^k)^*y^k(a)r_{a,\eps}^k,
\qquad \|r_{a,\eps}^k\|^2\le32/\eps.
\end{equation}
In particular the witness bound is independent of $k$.
\end{proposition}
\begin{proof}
Apply Proposition~\ref{prop:idealvisibility} for this $k$.
\emph{Fix its coordinate maps $\nu_n$, their weights, and every
integer $L_{a,\eps,n}$ before making any subsequent choices.}
They will not be reindexed in this proof.
For a test $(a,\eps)$ retain $b=(a-\eps/8)_+$ and put
$d=(a-\eps/4)_+$. Then $d\in\Ideal_A(b)$ and
$[d]\le[b]$, so Proposition~\ref{prop:localized-trace} applies to
$(d,b)$ with relative rank constant one. Let $E_r$ be the good
sets in \eqref{cmp:I} for the enumerated tests. Form decreasing
$\omega$-large sets
\[
G_l=\{n\ge l\}\cap\bigcap_{r\le l}E_r,
\]
and let $l(n)$ be the largest $l\le n$ with $n\in G_l$, or zero
if there is none. Thus $l(n)\to_\omega\infty$, and all ideal
bounds for the first $l(n)$ tests are valid at coordinate $n$.
Choose positive numbers $\eta_n\to0$, for example $\eta_n=1/n$.
At coordinate $n$ use the finite-coordinate conclusion of
Proposition~\ref{prop:localized-trace} for the first $l(n)$ pairs
$(d_r,b_r)$, with the individually prescribed accuracies
\[
\eta_n/L_{a_r,\eps_r,n}.
\]
Include the first $n$ members of a fixed dense family in $A$, and
all matrix-unit values $\nu_n(e_{ij})$, in the commutator tests.
Ask for norm commutators and norm order-zero defects less than
$1/n$. Denote the resulting \cpc map by $\theta_n$, and its weight
by $h_n^\theta$. Where $l(n)=0$ one may use the zero map. Every
finite problem here has already specified its finite integers;
its requested trace accuracy is allowed to be arbitrarily small.
Consequently the maps represent a norm-central order-zero map
$\theta$ commuting with $\nu$, and, for each fixed test, on an
$\omega$-large set,
\begin{equation}\label{cmp:adaptive-trace}
\tau\bigl(d^{1/2}(1-h_n^\theta)d^{1/2}\bigr)
\le\frac{\eta_n}{L_{a,\eps,n}}d_\tau(b)
\qquad(d_\tau(b)<\infty).
\end{equation}
At zero rank this asserts exact zero; alternatively it follows
directly from $[d]\le[b]$.
Join $\theta$ and $\nu$ to obtain $\zeta^k$. Use the following
specific positive contractive representatives of its weight:
\begin{equation}\label{cmp:positive-weight-lift}
H_n=1-(1-h_n^\theta)^{1/2}(1-h_n^\nu)
(1-h_n^\theta)^{1/2}.
\end{equation}
They belong to $A$: equivalently, they are
$h_n^\theta+(1-h_n^\theta)^{1/2}h_n^\nu(1-h_n^\theta)^{1/2}$.
They lie between $h_n^\theta$ and $1$. Since the two weights
commute in the ultrapower, they represent the joined weight $H$.
For $u^k(a)$ use the positive contractive representatives
\begin{equation}\label{cmp:source-lift}
u_n=(1-f(H_n))^{1/2}a(1-f(H_n))^{1/2}.
\end{equation}
For $y^k(a)$ choose positive contractive representatives $y_n$;
for example, if $B_n$ are positive contractive representatives of
$g(\zeta^k)(e_{11})$, one can take $y_n=B_naB_n$.
All products in \eqref{cmp:u-y} are positive because the matrix
map is norm-central.
We first convert the norm visibility into an exact coordinate
Cuntz inequality. In $A_\omega$ the old weight $h^\nu$ and $H$
commute, $h^\nu\le H$, and both commute with $a$. Monotonicity of
$g$ in this commutative functional calculus gives
\begin{equation}\label{cmp:total-weight-domination}
y^\nu=ag(h^\nu)^2\le ag(H)^2.
\end{equation}
Write $\varrho$ for the supporting homomorphism of $\zeta^k$.
Since $a$ commutes with the range of $\zeta^k$, it also commutes
with the supporting matrix units: these are the strong limits of
$\zeta^k(e_{ij})(H+t)^{-1}$ as $t\downarrow0$ in the bidual.
The right side is the orthogonal sum of
\[
z_i=ag(H)^2\varrho(e_{ii})\quad(1\le i\le k),
\qquad z_1=y^k(a).
\]
The products are in $A_\omega$. All $z_i$ are Cuntz equivalent,
as are their corresponding positive cuts. Indeed
$a^{1/2}g(H)\varrho(e_{ij})\in A_\omega$ has its two squared
moduli equal to $z_j,z_i$; taking polar cuts gives the same fact
for $(z_j-t)_+$ and $(z_i-t)_+$. The two positives in
\eqref{cmp:total-weight-domination} commute, so scalar functional
calculus, splitting a cut of an orthogonal sum, and enlarging its
individual cuts give
\begin{equation}\label{cmp:ultrapower-cut}
[(y^\nu-\eps/22)_+]
\le k[(y^k(a)-\eps/30)_+]
\quad\text{in }\Cu(A_\omega).
\end{equation}
Here the cut margins yield an exact inequality before any
multiplication by a varying integer. Put
$\Delta=\eps/20-\eps/22>0$.
Choose one rectangular comparison witness for
\eqref{cmp:ultrapower-cut} in $M_{k,1}(A_\omega)$ with norm error
less than $\Delta/2$. Such rectangular witnesses suffice because
the source is in one corner and the target is the diagonal of
$k$ elements of $A_\omega$. Choose bounded coordinate lifts $X_n$
of this witness. Since functional calculus commutes with the
quotient, on an $\omega$-large set,
\[
\left\|(y_n^\nu-\eps/22)_+
-X_n^*\operatorname{diag}\bigl((y_n-\eps/30)_+,
\ldots,(y_n-\eps/30)_+\bigr)X_n
\right\|<\Delta.
\]
Norm comparison at that coordinate gives
\[
[(y_n^\nu-\eps/20)_+]
\le k[(y_n-\eps/30)_+]
\le k[(y_n-\eps/32)_+].
\]
Intersect with the good set for \eqref{cmp:I}. Only now multiply
the exact Cuntz inequality by $L_n=L_{a,\eps,n}$ to obtain
\begin{equation}\label{cmp:exact-coordinate-bound}
[b]\le kL_n[(y_n-\eps/32)_+].
\end{equation}
The bound on $X_n$ may depend on the fixed $k$ and the test. No
norm error is multiplied by $L_n$ in this argument.
We next prove the coordinate rank gap
\begin{equation}\label{cmp:R}
d_\tau((u_n-\eps)_+)
\le\tfrac12 d_\tau((y_n-\eps/32)_+)
\qquad\text{for every }\tau
\end{equation}
on an $\omega$-large set. Put
$c_n=(1-f(H_n))^{1/2}$ and $v_n=c_ndc_n$.
Since $\|a-d\|\le\eps/4$, one has
$\|u_n-v_n\|\le\eps/4$. Comparing $u_n$ with
$(v_n-\eps/2)_+$ gives
\[
[(u_n-\eps)_+]\le[(v_n-\eps/2)_+],
\qquad
d_\tau((u_n-\eps)_+)\le(2/\eps)\tau(v_n).
\]
For the first inequality the norm error is at most
$\eps/4+\eps/2<\eps$; the second is the elementary spectral
trace bound. From the scalar inequality $f(t)\ge t$ and
\eqref{cmp:positive-weight-lift},
\[
f(H_n)\ge H_n\ge h_n^\theta.
\]
This uses functional calculus on $H_n$ itself and positive
domination of $H_n$; it makes no operator-monotonicity assertion
about $f$. Traciality and \eqref{cmp:adaptive-trace} consequently
give, whenever $d_\tau(b)<\infty$,
\begin{equation}\label{cmp:source-rank-estimate}
d_\tau((u_n-\eps)_+)
\le\frac{2}{\eps}
\tau\bigl(d^{1/2}(1-f(H_n))d^{1/2}\bigr)
\le\frac{2\eta_n}{\eps L_n}d_\tau(b).
\end{equation}
This estimate handles every possible trace value. Set
$R_n=d_\tau((y_n-\eps/32)_+)$. If $R_n<\infty$, the exact
inequality \eqref{cmp:exact-coordinate-bound} gives
$d_\tau(b)\le kL_nR_n<\infty$, and therefore
\[
d_\tau((u_n-\eps)_+)
\le (2k\eta_n/\eps)R_n.
\]
For the fixed $k$ and test, eventually $2k\eta_n/\eps\le1/2$.
If $R_n=0$, the same argument first forces $d_\tau(b)=0$ and then
gives exactly zero on the left. If $R_n=\infty$, inequality
\eqref{cmp:R} is automatic in $[0,\infty]$. All good sets used
in these alternatives were chosen uniformly over the trace
families; thus their intersection proves \eqref{cmp:R}
simultaneously for every lower semicontinuous extended trace.
Apply Lemma~\ref{lem:G} in $A$ at each coordinate of this common
good set. It gives
\[
(u_n-\eps)_+\precsim (y_n-s)_+,
\qquad s=\eps/32.
\]
Choose $x_n\in A$ with
\[
\|x_n^*(y_n-s)_+x_n-(u_n-\eps)_+\|<1/n.
\]
These witnesses need not themselves be bounded. Define the
continuous scalar function
\[
q_s(t)=
\begin{cases}\bigl((t-s)_+/t\bigr)^{1/2},&t>0,\\0,&t=0,
\end{cases}
\]
and set $r_n=q_s(y_n)x_n$ on the good set and $r_n=0$ elsewhere.
Then
\[
r_n^*y_nr_n=x_n^*(y_n-s)_+x_n,
\qquad
\|r_n\|^2\le s^{-1}\|x_n^*(y_n-s)_+x_n\|
\le(32/\eps)(1+1/n).
\]
The norm estimate follows from $q_s(t)^2\le s^{-1}(t-s)_+$.
Thus these representatives have the fixed bound
$\|r_n\|^2\le64/\eps$, and their ultrapower element has norm
squared at most $32/\eps$. The compression errors tend to zero
along $\omega$, proving \eqref{cmp:bounded-comparison}. Neither
bound depends on $k$ or on any $L_n$.
\end{proof}
\begin{proposition}[Infinitely many central slots]\label{prop:slots}
For every $p\ge2$ there are a \cpc order-zero map
$\alpha:M_p\to A_\omega\cap A'$, with weight $h=\alpha(1)$,
and contractions $s_j\in A_\omega\cap A'$ for $j\ge1$ such that,
putting $P=s_1$, one has
\begin{equation}\label{cmp:slot-relations}
P\ge0,\qquad s_i^*s_j=\delta_{ij}P^2,
\qquad \alpha(e_{11})s_j=s_j.
\end{equation}
For every test $(a,\eps)$ there is $r_{a,\eps}\in A_\omega$ with
\begin{equation}\label{cmp:slot-comparison}
((1-h)a-\eps)_+
=r_{a,\eps}^*(P^2a)r_{a,\eps},
\qquad \|r_{a,\eps}\|^2\le64/\eps.
\end{equation}
No centrality is asserted or needed for these comparison witnesses.
\end{proposition}
\begin{proof}
For fixed $m\ge1$, apply Proposition~\ref{prop:rankgap} with $k=pm$
and identify $M_{pm}=M_p\otimes M_m$. Define
\[
\alpha_m(z)=f(\zeta^{pm})(z\otimes1_m),\qquad
s_j^{(m)}=g(\zeta^{pm})(e_{11}^{(p)}\otimes e_{j1}^{(m)})
\quad(1\le j\le m),
\qquad P_m=s_1^{(m)}.
\]
These are norm-central contractions, and $P_m$ is positive.
The matrix-unit relations for the supporting homomorphism give
\[
(s_i^{(m)})^*s_j^{(m)}=\delta_{ij}P_m^2.
\]
The scalar identity $fg=g$ gives
$\alpha_m(e_{11})s_j^{(m)}=s_j^{(m)}$.
The weight of $\alpha_m$ is $f(H^{pm})$. Under the indicated
identification the first diagonal slot of $M_{pm}$ is
$e_{11}^{(p)}\otimes e_{11}^{(m)}$, so
Proposition~\ref{prop:rankgap} supplies, for every test,
\[
((1-\alpha_m(1))a-\eps)_+
=(r_{a,\eps}^{(m)})^*(P_m^2a)r_{a,\eps}^{(m)}.
\]
The witness norm bounds are independent of $m$.
We spell out the diagonalization because the number of slots grows.
For every $m$ choose \cpc coordinate lifts $\alpha_{m,n}$, contractive
lifts $s_{j,m,n}$ for $1\le j\le m$, and positive contractive lifts
$P_{m,n}=s_{1,m,n}$. Choose lifts of each comparison witness bounded
by $\sqrt{64/\eps}$; the sharper ultrapower bound in
Proposition~\ref{prop:rankgap} leaves room for this choice. Put
$h_{m,n}=\alpha_{m,n}(1)$. At stage $l$ use $m=l$ and require:
\begin{enumerate}
\item the matrix-unit order-zero identities of $\alpha_{l,n}$,
and its commutators with the first $l$ dense tests of $A$,
to have error less than $1/l$;
\item the centrality and the two algebraic relations in
\eqref{cmp:slot-relations} for all slot indices at most $l$
to have error less than $1/l$;
\item for the first $l$ comparison tests, the equations with
positive coordinate expressions
\[
\bigl((1-h_{l,n})^{1/2}a(1-h_{l,n})^{1/2}-\eps\bigr)_+
\quad\text{and}\quad
r_{a,\eps,l,n}^*P_{l,n}aP_{l,n}r_{a,\eps,l,n}
\]
to have error less than $1/l$.
\end{enumerate}
For fixed $l$ all these finitely many estimates hold on an
$\omega$-large set, because their equations are exact in the
ultrapower. The positive coordinate expressions represent the
desired products since the weights and slots commute with $A$ in
the ultrapower.
Intersect the first $l$ such good sets with $\{n\ge l\}$ to obtain
decreasing $E_l\in\omega$, and take $m(n)$ to be the largest
$l\le n$ with $n\in E_l$ (zero if none). At coordinate $n$ use
the row indexed by $m(n)$. For a slot or witness whose index is
larger than $m(n)$, use zero at that coordinate. Every fixed
variable is eventually present along $\omega$, since
$m(n)\to_\omega\infty$. All maps and slots are contractions,
and the witness for a fixed test has its individual bound
$\sqrt{64/\eps}$ independently of the row. These are precisely
the boundedness hypotheses of Lemma~\ref{lem:diagonal}.
The selected maps therefore define a \cpc norm-central order-zero
$\alpha$, and the selected slots and witnesses define elements
with \eqref{cmp:slot-relations}--\eqref{cmp:slot-comparison}.
No uniform trace estimate as $m$ grows was needed: each fixed
matrix size had already produced exact equations with the stated
individual norm bounds.
\end{proof}
==== END Cuntz-comparison-and-Jiang-Su-absorption//comparison.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//divisibility.tex ====
\section{Almost divisibility with a relative error}
\label{ext:sec:divisibility}
For a given class $x$ and integer $N\ge1$, we seek a single class $u$
with $\varphi(x)/(N+1)<\varphi(u)<\varphi(x)/N$ whenever
$0<\varphi(x)<\infty$. We choose $s$ between $1/(N+1)$ and $1/N$
and approximate $s\varphi(x)$ from below. The error is measured by one
fixed cut, so the same construction also controls functionals taking the
value infinity on $x$.
\begin{theorem}
\label{ext:thm:divisibility}
Let $A$ be a simple exact non-elementary $C^*$-algebra with strict
comparison as in Equation~\eqref{ext:eq:comparison}.
For every $x\in\Cu(A)$ and integer $N\ge1$, there exists
$u\in\Cu(A)$ such that $Nu\le x\le(N+1)u$.
\end{theorem}
\begin{proof}
The assertion is immediate for $x=0$. Otherwise write $x=[a]$,
where $a\in B_+$ has norm one, and put $e=(a-1/2)_+$.
If $\Fin(A)$ is empty, all functionals are zero or infinite on
every nonzero class. Equation~\eqref{ext:eq:comparison} then gives
$Nx\le x$, while $x\le(N+1)x$ is automatic. Take $u=x$.
We henceforth assume $\Fin(A)\ne\varnothing$.
\medskip\noindent
\textit{Choose supports with summable relative rank.}
Choose a real number $s$ and integers $M_j\ge1$, $j\ge1$, such that
\begin{equation}
\label{ext:eq:budget}
\frac1{N+1}<s<\frac1N,\qquad
E:=\sum_{j=1}^{\infty}\frac1{M_j}<s-\frac1{N+1}.
\end{equation}
Lemma~\ref{ext:lem:small} supplies nonzero $h_j\in B_+$ with
$M_j[h_j]\le[e]$.
To recover the full rank of $a$ from functions vanishing near zero,
choose increasing cutoffs $F_j$ converging to one on $(0,1]$ and
transfer their positive increments. Concretely, set
\[
F_0=0,\qquad F_j(t)=\min\{1,\max\{0,2^jt-1\}\},\qquad
r_j=s(F_j-F_{j-1})\quad(j\ge1).
\]
The functions $F_j$ increase, and $r_j\ge0$ vanishes on
$[0,2^{-j}]$. Put $p_j=r_j(a)$.
Lemma~\ref{ext:lem:transfer} gives $c_j\in\her(h_j)_+$ satisfying
\begin{equation}
\label{ext:eq:transferred}
\tau(c_j)=\tau(p_j)\quad\text{for all extended traces }\tau,
\qquad M_j[c_j]\le[e].
\end{equation}
\medskip\noindent
\textit{Round and sum inside the Cuntz semigroup.}
Choose integers $L_j>\|c_j\|$ and put
\begin{equation}
\label{ext:eq:u}
w_j=\sum_{i=1}^{L_j}[(c_j-i)_+],\qquad
u=\sup_{k\ge1}\sum_{j=1}^k w_j.
\end{equation}
For completeness, this supremum is represented by a positive
element of $B$: choose representatives of $w_j$, rescale them
to norm at most $2^{-j}$, and put them in mutually orthogonal
stable blocks. Their sum converges in norm. Each finite partial
sum is Cuntz below it. Conversely, comparison of all partial sums
with one target passes to the norm limit, proving the least-upper-bound
property. Therefore
\begin{equation}
\label{ext:eq:u-values}
\varphi(u)=\sum_{j=1}^{\infty}\varphi(w_j)
\quad(\varphi\in\F(\Cu(A))).
\end{equation}
Fix $\varphi\in\Fin(A)$ and choose its representing trace
using Equation~\eqref{ext:eq:trace-representation}.
Equations~\eqref{ext:eq:rounding} and~\eqref{ext:eq:transferred} give
\begin{equation}
\label{ext:eq:piece-error}
\varphi(w_j)\le\tau(p_j)
\le\varphi(w_j)+\frac{\varphi([e])}{M_j}.
\end{equation}
The finite value $\varphi([e])$ is the only trace-dependent
constant in the error estimate.
\medskip\noindent
\textit{Recover the rank of $a$.}
Telescoping gives
\[
\sum_{j=1}^k\tau(p_j)=s\tau(F_k(a)).
\]
We claim that
\begin{equation}
\label{ext:eq:recover}
\tau(F_k(a))\uparrow d_\tau(a)=\varphi(x).
\end{equation}
The values increase and are at most $d_\tau(a)$ by
Equation~\eqref{eq:pre:trace-rank} and hereditary containment.
Given $0<\epsilon<1$, for sufficiently large $k$ we have
$F_k(t)=1$ on $[\epsilon,1]$. Thus, simultaneously for all
integers $m\ge1$,
\[
(a-\epsilon)_+^{1/m}\le F_k(a).
\]
Taking traces and then letting $m\to\infty$ gives
$d_\tau((a-\epsilon)_+)\le\tau(F_k(a))$.
The cut-supremum formula in Equation~\eqref{ext:eq:cu-facts} now proves
Equation~\eqref{ext:eq:recover}.
Sum Equation~\eqref{ext:eq:piece-error}, then use
Equations~\eqref{ext:eq:u-values} and~\eqref{ext:eq:recover}. In extended
nonnegative arithmetic, for every $\varphi\in\Fin(A)$,
\begin{equation}
\label{ext:eq:total-error}
\varphi(u)\le s\varphi(x)
\le\varphi(u)+E\varphi([e]).
\end{equation}
If $\varphi(x)<\infty$, Equation~\eqref{ext:eq:controller} implies
\[
(s-E)\varphi(x)\le\varphi(u)\le s\varphi(x).
\]
Since $\varphi(x)>0$, Equation~\eqref{ext:eq:budget} yields
\begin{equation}
\label{ext:eq:strict-interval}
\frac{\varphi(x)}{N+1}<\varphi(u)<\frac{\varphi(x)}N.
\end{equation}
If instead $\varphi(x)=\infty$, then
$E\varphi([e])<\infty$ and Equation~\eqref{ext:eq:total-error} forces
$\varphi(u)=\infty$. No subtraction of infinite values is used.
Since $\Fin(A)$ is nonempty, these conclusions also prove $u\ne0$.
\medskip\noindent
\textit{Apply comparison with each target.}
Zero functionals vanish at both $x$ and $u$, and the
everywhere-infinite functionals are infinite at both.
For functionals in $\Fin(A)$, the preceding estimates give
\[
\varphi(Nu)\le\varphi(x),\qquad
\varphi(x)\le\varphi((N+1)u).
\]
The first inequality is strict whenever $0<\varphi(x)<\infty$.
For the second, if $0<\varphi((N+1)u)<\infty$, then the functional
belongs to $\Fin(A)$ and the infinite-value case excludes
$\varphi(x)=\infty$. Equation~\eqref{ext:eq:strict-interval} makes
that inequality strict as well. Applying
Equation~\eqref{ext:eq:comparison} twice proves
$Nu\le x\le(N+1)u$.
\end{proof}
\begin{corollary}[Condition (C) from comparison]
\label{ext:cor:condition-C}
Let $A$ be a simple exact non-elementary $C^*$-algebra with strict
comparison in the sense of Equation~\eqref{ext:eq:comparison}.
Then $\Cu(A)$ satisfies condition \textup{(C)}.
\end{corollary}
\begin{proof}
Theorem~\ref{ext:thm:divisibility} gives the full almost-divisibility
sandwich, and Lemma~\ref{lem:comparison-convention} gives almost
unperforation.
\end{proof}
\begin{remark}
The same sequence $(M_j)$ works for every functional. The values
$\varphi([e])$ need not be uniformly bounded over any trace family:
their pointwise finiteness is enough to handle $\varphi(x)=\infty$,
and their bound by $\varphi(x)$ handles finite values.
This is why no trace normalization is needed.
\end{remark}
\subsection{A unital uniform-trace illustration}
The simple route into (C) is complete. We record a separate consequence
of trace transfer and integer rounding when traces can be normalized.
It illustrates the approximation mechanism without an exactness,
nuclearity, or comparison hypothesis, and is not needed in the
absorption construction.
The spectral discretization in \cite[Section~5]{DadarlatToms2010} provides
related rank-approximation methods; here small support permits integer
spectral cuts with one support-rank of error.
\begin{proposition}[Uniform approximation from below]
\label{prop:unital-approx}
Let $A$ be a unital simple infinite-dimensional $C^*$-algebra with
nonempty tracial state space $T(A)$. For every $p\in A_+$ and
$\eta>0$, there are a finite integer $L$ and $w\in M_L(A)_+$
such that
\[
0\le\tau(p)-d_\tau(w)\le\eta
\qquad(\tau\in T(A)),
\]
where matrix traces are unnormalized. No exactness, nuclearity, or
comparison assumption is required.
\end{proposition}
\begin{proof}
Choose an integer $N$ with $1/N<\eta$. There are $N$ mutually
orthogonal nonzero positive elements in $A$: repeatedly split a
positive element at two distinct positive spectral values. If a
nonzero hereditary corner had no such element, it would contain a
minimal projection, and simplicity would identify $A$ with compact
operators; unitality would then make $A$ finite dimensional.
For any two nonzero positives $e,f$, simplicity supplies
$0\ne v\in eAf$. The class $[v^*v]=[vv^*]$ is nonzero and
below both $[e]$ and $[f]$. Iterating over the orthogonal family
gives $0\ne h\in A_+$ with
\[
d_\tau(h)\le 1/N\qquad(\tau\in T(A)).
\]
Fullness gives finitely many $s_i\in A$ with
$\sum_i s_i^*hs_i=1$. To see the exact equality, choose an
invertible finite sum $S=\sum_i r_iht_i$ approximating $1$.
For the row $R=(r_ih^{1/2})_i$ and column
$D=(h^{1/2}t_i)_i$, the identity $S=RD$ gives
$S^*S\le\|R\|^2D^*D$, so $D^*D$ is invertible.
Set $s_i=t_i(D^*D)^{-1/2}$.
The element
\[
c=\sum_i h^{1/2}s_i p s_i^*h^{1/2}\in\her_A(h)_+
\]
therefore satisfies $\tau(c)=\tau(p)$ and
$d_\tau(c)\le1/N$ for every $\tau\in T(A)$.
Choose an integer $L>\|c\|$ and let
$w=\bigoplus_{j=1}^L(c-j)_+$. Lemma~\ref{ext:lem:rounding},
applied in the stabilization, gives
\[
d_\tau(w)\le\tau(c)\le d_\tau(w)+d_\tau(c)
\le d_\tau(w)+1/N.
\]
This proves the required simultaneous inequalities.
\end{proof}
==== END Cuntz-comparison-and-Jiang-Su-absorption//divisibility.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//extended-traces.tex ====
\section{Simple algebras and extended traces}
\label{ext:sec:preliminaries}
The next three sections prove full almost divisibility from
extended-functional comparison. Two consequences of simplicity supply
the error estimates: there are arbitrarily small nonzero Cuntz classes,
and one fixed positive cut has finite positive value at every nonzero
functional that is finite somewhere. The first provides small hereditary
supports; the second measures the errors on all of them. No nuclear
approximation is used.
We work in $B=A\otimes\K$ and identify $\Cu(B)$ with $\Cu(A)$.
Matrix algebras over $B$ are placed in stable corners. These placements
preserve Cuntz classes: if $V$ is the corresponding multiplier isometry,
then $b$ and $VbV^*$ are the two squared moduli of $Vb^{1/2}$.
The trace and rank conventions are those of
Section~\ref{sec:preliminaries}.
\begin{lemma}[Small nonzero classes]
\label{ext:lem:small}
If $A$ is simple and non-elementary, then $B=A\otimes\K$ is simple
and has no minimal projections. For every nonzero $e\in B_+$ and
integer $M\ge1$, there is a nonzero $h\in B_+$ with
\[
M[h]\le[e].
\]
\end{lemma}
\begin{proof}
Matrix-unit multipliers extract a nonzero matrix entry from any
nonzero ideal of $B$ and put it into every matrix position.
Simplicity of $A$ then implies simplicity of $B$.
If $p\in B$ were minimal, $Bp$ would be a Hilbert space with
inner product $x^*y=\langle x,y\rangle p$. Left multiplication
is faithful by simplicity, and the elements $xy^*$ act as rank-one
operators. Since $p$ generates $B$ as an ideal, they span a dense
subspace of $B$, giving $B\cong\K(Bp)$. The multiplier corner
$1\otimes e_{11}$ would consequently identify $A$ with compact
operators on a Hilbert space, a contradiction.
There are $M$ nonzero mutually orthogonal positives
$e_1,\ldots,e_M$ in $\her(e)$. Indeed, split a nonzero positive
element at two distinct nonzero spectral points. If no such points
exist, the element is a multiple of a projection, whose corner is
not one-dimensional. A shifted nonscalar selfadjoint in that corner
has two distinct positive spectral points and can be split instead.
Recursion gives the required family while preserving orthogonality.
For nonzero $f,g\in B_+$, simplicity gives $fBg\ne0$.
If $0\ne v\in fBg$, then $[v^*v]=[vv^*]$ is a nonzero class
below both $[f]$ and $[g]$. Iterating this observation gives
$h\ne0$ with $[h]\le[e_i]$ for every $i$. Therefore
$M[h]\le\sum_i[e_i]\le[e]$.
\end{proof}
\begin{lemma}[Functional trichotomy]
\label{ext:lem:trichotomy}
Let $A$ be simple and nonzero. Every functional on $\Cu(A)$ is
either zero, infinite on every nonzero class, or positive on every
nonzero class and finite on every class $[(c-\delta)_+]$ with
$c\in(A\otimes\K)_+$ and $\delta>0$.
The third type consists exactly of the functionals finite and
positive at some class.
\end{lemma}
\begin{proof}
Every nonzero $b\in B_+$ is full. Lemma~\ref{lem:pre:cut-placement}
therefore bounds each positive cut of any $c\in B_+$ by finitely
many copies of $[b]$. If $\varphi([b])=0$, preservation of
increasing suprema gives $\varphi([c])=0$ for every $c$.
If $0<\varphi([b])<\infty$, the same cut comparison gives
finiteness on every cut, and the first observation gives positivity
on every nonzero class. These alternatives exhaust all functionals.
\end{proof}
Write $\Fin(A)$ for the third type in
Lemma~\ref{ext:lem:trichotomy}. This notation does not assert finiteness
on all classes. If $\|a\|=1$ and $e=(a-1/2)_+$, then
\begin{equation}
\label{ext:eq:controller}
0<\varphi([e])<\infty,\qquad
\varphi([e])\le\varphi([a])
\quad(\varphi\in\Fin(A)).
\end{equation}
==== END Cuntz-comparison-and-Jiang-Su-absorption//extended-traces.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//introduction.tex ====
\begin{abstract}
We prove that strict comparison in the extended-functional sense implies
Jiang--Su absorption for separable simple nuclear non-elementary
$C^*$-algebras. This resolves the corresponding implication of the
Toms--Winter regularity problem, including nonunital algebras and allowing
unbounded traces. More generally, every separable nuclear $C^*$-algebra
whose Cuntz semigroup is almost unperforated and fully almost divisible
absorbs the Jiang--Su algebra.
\end{abstract}
\section{Introduction}
\label{sec:introduction}
Comparison asks whether numerical size determines order among positive
operators. Divisibility asks whether those operators can be split into
pieces of nearly equal size. For nuclear $C^*$-algebras, these questions
are tied to tensorial absorption of the Jiang--Su algebra $\Z$:
\[
A\cong A\otimes_{\min}\Z.
\]
Jiang and Su constructed $\Z$ as a simple infinite-dimensional unital
algebra with the Elliott invariant of the complex numbers
\cite{JS1999}. Its absorption is a central regularity property in the
classification program. The purpose of this paper is to pass from
comparison and divisibility in the Cuntz semigroup to this tensorial
regularity, including algebras without a unit and algebras with ideals.
\subsection{Comparison and the simple theorem}
All algebras in this paper are complex. Let $\K$ be the compact
operators on a separable infinite-dimensional Hilbert space. For
$a,b\in(A\otimes\K)_+$, write $a\precsim b$ if
$v_j^*bv_j\to a$ in norm for some sequence $(v_j)$ in
$A\otimes\K$. Mutual comparison defines the Cuntz classes $[a]$.
Their ordered semigroup $\Cu(A)$ has addition given by orthogonal
sum and suprema of increasing sequences \cite{CEI2008}. A
\emph{functional} on $\Cu(A)$ is a map to $[0,\infty]$ preserving
zero, addition, order, and increasing suprema. Write
$\F(\Cu(A))$ for the cone of all these functionals.
We use the following precise extended-functional convention for
\emph{strict comparison}: for every $x,y\in\Cu(A)$,
\begin{equation}
\label{ext:eq:comparison}
\left.
\begin{aligned}
\varphi(x)&\le\varphi(y)
&&\text{for every }\varphi\in\F(\Cu(A)),\\
\varphi(x)&<\varphi(y)
&&\text{whenever }0<\varphi(y)<\infty
\end{aligned}
\right\}
\quad\Longrightarrow\quad x\le y.
\end{equation}
The second line concerns the same cone of functionals. Arithmetic at
zero and infinity is extended nonnegative arithmetic; in particular,
strict inequality is required precisely at positive finite target
values. An algebra is \emph{non-elementary} if it is not isomorphic to
the compact operators on any Hilbert space.
\begin{theorem}[Simple comparison and absorption]
\label{ext:thm:main}
Let $A$ be a separable simple nuclear non-elementary $C^*$-algebra.
If $A$ has strict comparison in the sense of
Equation~\eqref{ext:eq:comparison}, then $A\cong A\otimes_{\min}\Z$.
\end{theorem}
R\o rdam proved that $\Z$-absorption forces almost unperforation of
the finite-matrix Cuntz semigroup and, in the simple exact unital
setting, the corresponding trace comparison theorem
\cite[Theorem~4.5 and Corollary~4.6]{Rordam2004}. The reverse
implication is the strict-comparison-to-$\Z$-stability part of the
Toms--Winter regularity problem; see
\cite[Conjecture~5 and Problem~XVIII]{Problems2025}.
Theorem~\ref{ext:thm:main} gives a positive resolution of this
implication with the extended-functional convention above. Its
nonunital formulation includes stably projectionless algebras and
allows unbounded traces.
The comparison argument has a stronger intermediate scope.
Theorem~\ref{ext:thm:divisibility} proves, for every simple exact
non-elementary $A$ satisfying Equation~\eqref{ext:eq:comparison},
that each $x\in\Cu(A)$ and integer $N\ge1$ admit a single class $u$
with $Nu\le x\le(N+1)u$. Neither separability nor nuclearity enters
that theorem. Nuclearity is used in constructing central maps from
this order-theoretic information.
\subsection{The nonsimple absorption mechanism}
The first order property we require is \emph{almost unperforation}:
\begin{equation}
(l+1)x\le ly\quad\Longrightarrow\quad x\le y
\qquad(x,y\in\Cu(A),\ l\ge1).
\label{eq:pre:au}
\end{equation}
The second is \emph{full almost divisibility}: for every
$x\in\Cu(A)$ and integer $N\ge1$, there is $u\in\Cu(A)$ such that
\begin{equation}
Nu\le x\le(N+1)u.
\label{eq:pre:div}
\end{equation}
Here $l$ is also an integer. We call the conjunction of these two
properties \textup{(C)}. The same class $u$ must satisfy both
inequalities in Equation~\eqref{eq:pre:div}.
The comparison convention in Equation~\eqref{ext:eq:comparison} is
equivalent to almost unperforation of $\Cu(A)$
\cite[Proposition~6.2]{ERS2011}. Lemma~\ref{lem:comparison-convention}
explains this equivalence, including the zero and infinite values.
Thus the extended-functional formulation agrees with the usual
Cuntz-semigroup comparison condition.
To state the central conclusion, fix a free ultrafilter $\omega$ on
$\mathbb N$. The norm ultrapower and its central annihilator quotient
are
\[
A_\omega=\ell^\infty(\mathbb N,A)/
\{(a_n):\lim_{n\to\omega}\|a_n\|=0\},\qquad
F_\omega(A)=(A_\omega\cap A')/\Ann(A),
\]
where $A$ is the constant copy and
$\Ann(A)=\{z\in A_\omega:zA=Az=0\}$. For separable $A$,
a sequential approximate unit represents the unit of $F_\omega(A)$.
The zero algebra absorbs $\Z$ immediately; the central-map statement
below is formulated for nonzero algebras.
\begin{theorem}[Absorption from condition \textup{(C)}]
\label{thm:condition-C}
Let $A$ be a nonzero separable nuclear $C^*$-algebra satisfying
\textup{(C)}. For every fixed free ultrafilter $\omega$ on
$\mathbb N$, there is a unital $*$-homomorphism
$\Z\to F_\omega(A)$. Consequently $A\cong A\otimes_{\min}\Z$.
\end{theorem}
\begin{corollary}[The canonical Cuntz-semigroup map]
\label{cor:canonical}
Let $A$ be a separable nuclear $C^*$-algebra, and let
$\iota:A\to A\otimes_{\min}\Z$ be $\iota(a)=a\otimes1_\Z$.
If $\Cu(\iota):\Cu(A)\to\Cu(A\otimes_{\min}\Z)$ is an
isomorphism, then $A\cong A\otimes_{\min}\Z$.
\end{corollary}
The isomorphism in this hypothesis is the map induced by $\iota$.
Corollary~\ref{cor:canonical} resolves positively implication
(iii)$\Rightarrow$(ii) in the nonsimple Toms--Winter problem
\cite[Section~22, Problem~LXXV]{Problems2025}. It applies to every
separable nuclear $A$ with the indicated canonical map; an additional
restriction on its elementary ideal subquotients is unnecessary.
Condition \textup{(C)} depends only on the ordered-semigroup structure.
Consequently, the same absorption theorem also removes the requirement
that the Cuntz-semigroup isomorphism be the canonical one.
\begin{corollary}[An abstract Cuntz-semigroup criterion]
\label{cor:abstract-cu}
Let $A$ be a separable nuclear $C^*$-algebra. Then
\[
A\cong A\otimes_{\min}\Z
\quad\Longleftrightarrow\quad
\Cu(A)\cong\Cu(A\otimes_{\min}\Z)
\text{ in the category }\mathbf{Cu}.
\]
The isomorphism on the right is abstract: it is not required to be
induced by the first-factor embedding, or by any $*$-homomorphism.
\end{corollary}
Here the category $\mathbf{Cu}$ retains zero, addition, order, increasing
sequential suprema, and the way-below relation of the Cuntz semigroup.
Corollary~\ref{cor:abstract-cu} answers positively the abstract
Cuntz-semigroup question in
\cite[Section~9, Problem~XXVI]{Problems2025}, without assuming
unitality, simplicity, or stable finiteness. Separability and nuclearity
remain hypotheses.
\subsection{History and the main obstacles}
The comparison of positive elements and their dimension functions goes
back to Cuntz and Blackadar--Handelman
\cite{Cuntz1978,BlackadarHandelman1982}. The completed Cuntz semigroup
of Coward--Elliott--Ivanescu \cite{CEI2008} makes increasing suprema
part of the invariant. Elliott--Robert--Santiago's correspondence
\cite{ERS2011} identifies its functionals with lower semicontinuous
extended quasitraces. For exact algebras, Haagerup's additivity theorem
\cite{Haagerup2014}, in the extended form of Blanchard--Kirchberg
recalled in \cite[Remark~4.3]{ERS2011}, makes these quasitraces traces.
The associated rank is
$d_\tau(a)=\lim_n\tau(a^{1/n})$: the functional corresponding to
$\tau$ sends $[a]$ to $d_\tau(a)$.
The reverse comparison implication has a substantial history of
partial results. Matui and Sato proved it for unital separable simple
nuclear infinite-dimensional algebras with finitely many extremal
traces, using excision in small central sequences and property
\textup{(SI)} \cite[Theorem~1.1]{MatuiSato2012}.
Sato, Kirchberg--R\o rdam, and Toms--White--Winter extended the
tracial-boundary approach to nonempty compact extreme boundaries of
finite covering dimension \cite[Corollary~1.2]{Sato2012},
\cite[Corollary~7.9]{KirchbergRordam2014},
\cite[Corollary~4.7]{TWW2015}.
Castillejos, Evington, Tikuisis, and White isolated the remaining
uniform central divisibility in trace: for simple separable unital
nuclear algebras with nonempty tracial state space, strict comparison
together with uniform property~$\Gamma$ is equivalent to
$\Z$-stability \cite[Theorem~5.6]{CETW2022}. Uniform property~$\Gamma$
requires approximately central partitions that split the trace values
of finitely many elements evenly, uniformly over all traces.
The present argument constructs
central maps from condition \textup{(C)} without imposing a
tracial-boundary or uniform-$\Gamma$ hypothesis.
The Toms--Winter problem also links these properties to finite nuclear
dimension, a finite-colour form of completely positive approximation.
Winter's absorption theorem and the converse of Castillejos, Evington,
Tikuisis, White, and Winter establish the equivalence of finite nuclear
dimension and $\Z$-stability in the simple separable unital nuclear
non-elementary setting \cite[Corollary~6.3]{Winter2012}
\cite[Theorem~A]{CETWW2021}. The comparison implication is the
remaining direction addressed here.
The distinction between comparison and divisibility is already
visible in the range of possible rank functions of positive elements.
Thiel proved that strict comparison yields almost divisibility and
the canonical Cuntz-semigroup isomorphism for separable simple unital
non-elementary algebras of stable rank one
\cite[Corollary~8.12]{Thiel2020}. Theorem~\ref{ext:thm:divisibility}
uses exactness without a stable-rank-one hypothesis.
The fractional spectral discretization in Dadarlat--Toms
\cite[Theorem~5.1 and its proof]{DadarlatToms2010} follows earlier
rank-realization ideas of Brown--Perera--Toms \cite[Section~5]{BPT2008}.
Our first construction moves trace values
into a hereditary subalgebra of small rank before taking integer
spectral cuts. The resulting error is at most one support rank,
regardless of the norm of the transferred element or the number of
spectral levels. Proposition~\ref{prop:unital-approx} gives the unital
uniform-trace version of this elementary construction.
For nonunital algebras, there may be no normalized trace space on
which all errors can be bounded. The substitute is one fixed nonzero
spectral cut of the element being divided. Simplicity makes its rank
positive and finite for every functional that is finite and positive
somewhere, even when the original element has infinite rank.
We measure all rounding errors relative to this same cut and make them
summable. Zero functionals and functionals infinite on every nonzero
class are handled separately. This keeps the full extended trace cone
available without requiring a common normalization; the precise
three-way distinction is proved in Lemma~\ref{ext:lem:trichotomy}.
Comparison and divisibility are also studied together through the
notion of pureness. For the completed semigroup, the usual almost
divisibility condition asks for $Nu\le x$ and $x'\le(N+1)u$ whenever
$x'$ is compactly contained in $x$; together with almost unperforation
this defines pureness \cite{APT2018}. Full almost divisibility as
used here implies that condition.
For separable simple unital non-elementary algebras of locally
finite nuclear dimension, Winter proved that pureness implies
Jiang--Su absorption \cite[Corollary~6.2]{Winter2012}. Under the same
local nuclear-dimension hypothesis, he also obtained an abstract
criterion using $W(A)$, the Cuntz semigroup of positive elements in
finite matrices over $A$ \cite[Corollary~6.4]{Winter2012}.
Corollary~\ref{cor:abstract-cu} instead uses the completed invariant
$\Cu(A)$ and has no local nuclear-dimension assumption.
The tensor-product approach to regularity of abstract Cuntz semigroups
was developed by Antoine, Perera, and
Thiel \cite{APT2018}; more recent work of Antoine, Perera, Thiel,
and Vilalta characterizes pureness by controlled comparison and
functional divisibility \cite[Theorem~C]{APTV2024}.
Regularity of $\Cu(A)$ and regularity of the central
sequence algebra are distinct inputs. In particular,
Perera, Thiel, and Vilalta characterize Jiang--Su absorption by
pureness of the central sequence algebra for every separable $A$
\cite[Theorem~3.6]{PTV2025}. Theorem~\ref{thm:condition-C}
constructs the required central structure from the stated order
properties of $\Cu(A)$ itself.
For the central construction, a \emph{matrix cone} means a completely
positive contractive order-zero map from $M_k$: it preserves
orthogonality of positive elements. Its \emph{weight} is the image of
$1_{M_k}$. Norm-central cones take values in $A_\omega\cap A'$,
with coordinate maps whose centrality and order-zero errors tend to
zero along $\omega$.
Robert and Tikuisis developed a nonsimple route from regularity to
central structure using finite nuclear dimension
\cite[Theorem~1.1]{RobertTikuisis2017}. For separable algebras of
finite nuclear dimension, they characterized absorption by the
existence of two full orthogonal elements in the central sequence
algebra \cite[Theorem~1.2]{RobertTikuisis2017}.
Our hypothesis is instead the order condition \textup{(C)}, with no
bound on nuclear dimension.
The nonsimple problem has a different obstruction. A positive element
can be large in every representation of interest without a single
uniform integer controlling all the associated ideal comparisons.
At the same time, traces may be finite only after hereditary
localization. The proof treats these two forms of largeness
separately. Pure-state compression and support shrinking construct
norm-visible central matrix cones. Convex order-zero nuclear
approximations, originating in the work of Hirshberg, Kirchberg,
and White \cite{HKW2012} and refined by Brown, Carri\'on,
and White \cite{BCW2019}, supply cones large for bounded traces.
We localize the latter estimates and then remove a relative trace
ideal to obtain norm centrality.
The matrix-cone calculus uses Winter and Zacharias's structure
theorem for order-zero maps \cite{WZ2009}. The final packing step is
an instance of Gabe's approximate domination theorem
\cite{Gabe2020}; its weighted-inclusion form is proved here.
Choi--Effros lifting and Arveson extension allow us to retain
completely positive maps during diagonalization
\cite{CE1976,Arveson1969}. The absorption step uses the structural
properties of $\Z$ established by Jiang--Su, Toms--Winter, and
R\o rdam--Winter \cite{JS1999,TW2007,RW2010}. Nawata's nonunital
central dimension-drop criterion \cite[Proposition~5.1]{Nawata2013}
is implemented here by an explicit lifting and intertwining argument.
\subsection{Proof strategy and organization}
Figure~\ref{fig:proof-flow} separates the two routes into condition
\textup{(C)} from the common central-sequence argument. The full
proof of Theorem~\ref{thm:condition-C} is contained in this paper.
Its concrete target, for each $p\ge2$, is a matrix cone
$\alpha:M_p\to F_\omega(A)$ and an element $s$ with
\[
s^*s=1-\alpha(1),\qquad \alpha(e_{11})s=s.
\]
Thus $s$ fills the cone's defect inside one diagonal corner.
These relations give a unital map from the prime dimension-drop
algebra $I(p,p+1)$, whose continuous functions on $[0,1]$ take values in
$M_p\otimes M_{p+1}$ and have endpoint values in
$M_p\otimes1$ and $1\otimes M_{p+1}$, respectively.
Such maps supply the central copy of $\Z$.
\begin{figure}[H]
\centering
\setlength{\tabcolsep}{7pt}
\begin{tabular}{c@{\qquad}c}
\fbox{\parbox{.39\linewidth}{\centering Simple exact non-elementary algebra:\\
extended-functional comparison}} &
\fbox{\parbox{.39\linewidth}{\centering Separable nuclear algebra:\\
abstract $\Cu(A)\cong\Cu(A\otimes\Z)$}}\\[5pt]
rounding and comparison arithmetic $\downarrow$ & $\downarrow$ transfer of tensor regularity\\[5pt]
\multicolumn{2}{c}{\fbox{\parbox{.67\linewidth}{\centering
Condition (C): almost unperforation and full almost divisibility}}}\\[7pt]
\multicolumn{2}{c}{\textit{Under separability and nuclearity:}}\\[5pt]
\fbox{\parbox{.39\linewidth}{\centering Norm-visible cones\\
fix the coordinate multiplicities}} &
\fbox{\parbox{.39\linewidth}{\centering Localized tracial cones\\
choose relative errors afterwards}}\\[5pt]
\multicolumn{2}{c}{$\downarrow$ bounded comparison witnesses and central slots}\\[5pt]
\multicolumn{2}{c}{$\downarrow$ central defect filler and dimension-drop maps}\\[5pt]
\multicolumn{2}{c}{$\Z\longrightarrow F_\omega(A)\quad\Longrightarrow\quad
A\cong A\otimes\Z$}
\end{tabular}
\caption{The order of the constructions. The simple comparison route
supplies condition (C) already under exactness. Both absorption routes
then use the same separable nuclear construction. The norm-cone
multiplicities are fixed before the localized trace accuracies.}
\label{fig:proof-flow}
\end{figure}
Section~\ref{sec:preliminaries} introduces the common Cuntz-semigroup
and trace conventions, together with the support calculus for spectral
cuts. Sections~\ref{ext:sec:preliminaries}--\ref{ext:sec:divisibility}
then prove the simple exact almost-divisibility theorem. Trace values
are transferred to successively smaller hereditary supports, rounded
at integer levels, and summed in one norm-convergent stable block sum.
A fixed nonzero cut controls the summable errors even when the original
class has infinite functional value. Comparison supplies the two sides
of the divisibility sandwich, completing the simple route to
condition \textup{(C)}.
Section~\ref{sec:support-tools} begins the common absorption argument.
It derives condition \textup{(C)} from the canonical Cuntz-semigroup
map and develops its consequences for rank comparison and divisible
matrix cones. It also records the ultrapower and bounded
diagonalization tools used in the central constructions.
Sections~\ref{shr:section} and~\ref{norm:section} construct the
norm-visible cones. Two-support shrinking preserves the
representations detecting a prescribed positive cut while making
cross terms small. Pure-state word models can then be combined into
approximately central matrix cones. Their norm visibility supplies
finite coordinate ideal multiplicities.
Sections~\ref{sec:tracial} and~\ref{sec:localization} construct the
tracial cones. Two support compressions keep their errors independent
of the number of nuclear approximation summands; hereditary
localization and a relative trace ideal turn tracial centrality into
norm centrality.
Section~\ref{sec:comparison} combines these constructions in that
order. Its trace accuracies depend on the already fixed ideal
multiplicities. Spectral margins first produce exact coordinate
Cuntz inequalities and then bound the comparison witnesses
independently of matrix size. This permits a slowly growing supply
of orthogonal equivalent central slots.
Section~\ref{cen:section} packs finite simultaneous compressions
into those slots; approximate-unit tests make the packed element
bounded and central. Section~\ref{abs:section} turns the resulting
relations into dimension-drop maps, constructs a single unital
central copy of $\Z$, and performs the nonunital intertwining.
Section~\ref{sec:assembly} assembles the four stated results.
==== END Cuntz-comparison-and-Jiang-Su-absorption//introduction.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//localization.tex ====
\section{Joining matrix cones and localizing traces}
\label{sec:localization}
Our goal is to retain the tracial mass from the previous section while
obtaining norm-central maps. The trace-kernel ideal and relative
commutant lifting methods have predecessors in Kirchberg--R\o rdam
\cite[Section~4]{KirchbergRordam2014}. Here the relative ideal is
specified by hereditary rank tests, including the traces which vanish
on the controlling positive element.
Throughout this section, traces on $A$ are the lower semicontinuous
extended traces of the preliminaries. In particular, they need not be
finite on a dense ideal. If $D\subseteq A$ is hereditary, write
$T_{\leq1}(D)$ for the bounded positive tracial functionals on $D$ of
norm at most one, and put
\[
\|z\|_{2,T_{\leq1}(D)}
=\sup_{\sigma\in T_{\leq1}(D)}\sigma(z^*z)^{1/2}.
\]
The supremum is zero if there are no nonzero such functionals. Expressions
involving $1-h$ are evaluated in a unitization; a sandwich by an element
of $D$ still belongs to $D$.
\begin{lemma}[Joining two matrix cones]\label{lem:join}
Let $B$ be a $C^*$-algebra, and let $\phi_1,\phi_2:M_k\to B$ be
\cpc order-zero maps with commuting ranges. Set $h_i=\phi_i(1)$.
There is a \cpc order-zero map $\phi:M_k\to
C^*(\phi_1(M_k),\phi_2(M_k))$ such that
\begin{equation}\label{loc:join-weight}
\phi(1)=1-(1-h_1)(1-h_2)=h_1+h_2-h_1h_2.
\end{equation}
Its weight dominates both $h_1$ and $h_2$. If both original maps commute
with a subset of an ambient algebra, the joined map also commutes with
that subset.
\end{lemma}
\begin{proof}
Let $\mathcal C_k=C_0((0,1],M_k)$. Its unitization is the algebra of continuous
$M_k$-valued functions on $[0,1]$ whose value at zero is scalar. Realize
$\widetilde{\mathcal C}_k\otimes\widetilde{\mathcal C}_k$ as the algebra of continuous
$M_k\otimes M_k$-valued functions $F$ on the square satisfying
\[
F(x,0)\in M_k\otimes1,
\qquad F(0,y)\in1\otimes M_k.
\]
These conditions also make $F(0,0)$ scalar. Here is a direct verification
that these conditions characterize the tensor product. The tensor
product contains every scalar continuous function on the square, and
its evaluation at each point is the entire indicated fiber: use scalar
functions vanishing at zero to realize matrix entries on either
nonzero coordinate. Given a continuous function satisfying the
conditions, choose, at each point, an element of the tensor product
having the same value there. Continuity gives a neighborhood on which
the error is less than a prescribed tolerance. A finite scalar
partition of unity subordinate to these neighborhoods gives an
approximation in the tensor product. Closedness proves the assertion.
Let $S$ be the unitary that interchanges the two factors of
$\C^k\otimes\C^k$, and put $Q=(1-S)/2$. Since $S=S^*=S^{-1}$,
$Q$ is a projection, and
\[
U(t)=\exp(i\pi tQ),\qquad 0\le t\le1,
\]
is a unitary path from $1$ to $S$. For $z\in M_k$ define
\[
F_z(x,y)=
\begin{cases}
(x+y-xy)\,U\bigl(y/(x+y)\bigr)(z\otimes1)
U\bigl(y/(x+y)\bigr)^*, &(x,y)\ne(0,0),\\
0,&(x,y)=(0,0).
\end{cases}
\]
At $y=0$ this is $xz\otimes1$, and at $x=0$ it is $y1\otimes z$.
At the origin continuity follows from
$\|F_z(x,y)\|\le(x+y)\|z\|$; no limit of the unitary itself is needed.
Thus $F_z$ belongs to the tensor product. Pointwise it is a positive
scalar contraction times a $*$-homomorphism, so $z\mapsto F_z$ is
completely positive, contractive, and order zero. Its weight is
$(x+y-xy)1$.
The cone homomorphisms associated to $\phi_1$ and $\phi_2$ extend to
commuting unital homomorphisms from their unitizations. Nuclearity of
the cones lets their product be evaluated on the spatial tensor
product. The functions $F_z$ vanish at $(0,0)$. The ideal of functions
vanishing there is generated by $\mathcal C_k\otimes1$ and $1\otimes\mathcal C_k$:
quotienting by those two ideals leaves only the scalar evaluation at
the origin. Its image consequently lies in
$C^*(\phi_1(M_k),\phi_2(M_k))$, even when $B$ is nonunital. Evaluation
of $F_z$ gives the asserted map and its weight. Finally
$\phi(1)-h_1=(1-h_1)h_2\ge0$, and similarly for $h_2$.
The last assertion follows from containment of the range in the
generated algebra.
\end{proof}
\begin{lemma}[Filling bounded traces]\label{loc:bounded-fill}
Let $D\subseteq A$ be hereditary, $k\ge2$, $F\subset D$ finite, and
$\mathcal P\subset D$ a finite set of positive contractions. For every
$\rho>0$ there is a \cpc map $\mu:M_k\to D$ such that, for all matrix
units $e_{ij},e_{rs}$ and all $x\in F$,
\begin{align}
\|[\mu(e_{ij}),x]\|_{2,T_{\leq1}(D)}&<\rho,\label{loc:fill-comm}\\
\|\mu(e_{ij})\mu(e_{rs})-\mu(1)\mu(e_{ij}e_{rs})
\|_{2,T_{\leq1}(D)}&<\rho,\label{loc:fill-oz}\\
\sigma\bigl(a^{1/2}(1-\mu(1))a^{1/2}\bigr)&<\rho
\quad(a\in\mathcal P,\ \sigma\in T_{\leq1}(D)).
\label{loc:fill-trace}
\end{align}
\end{lemma}
\begin{proof}
The sequences that vanish in the uniform $2$-seminorm define a closed
two-sided ideal $J_2$ of $D_\omega$. Indeed the uniform $2$-seminorm is
at most the norm, and
$\|uz\|_2,\|zu\|_2\le\|u\|\|z\|_2$ for every bounded trace; the
triangle inequality and the same inequalities establish ideality and
closedness. Write $B=D_\omega/J_2$. Every choice
$\sigma_n\in T_{\leq1}(D)$ gives a bounded trace
\[
\sigma_\omega((z_n))=\lim_{n\to\omega}\sigma_n(z_n)
\]
on $D_\omega$, which factors through $B$ by Cauchy--Schwarz. These
choices are allowed to depend on $n$.
Fix $0<\delta<1$. Proposition~\ref{prop:bounded-trace}, applied at
coordinate $n$ with its auxiliary error tending to zero, has the
following consequence. Given finitely many positive contractions in
$B$ and finitely many elements with which to commute, there is a
\cpc order-zero map $\psi:M_k\to B$ commuting with those elements and
satisfying
\begin{equation}\label{loc:fraction}
\sigma_\omega(c\psi(1))
\ge\frac{\delta}{3}\sigma_\omega(c)-C_k\delta^{3/2}
\end{equation}
for all the chosen targets and all the traces just described.
To see the quantifiers, choose positive contractive lifts of the
finitely many targets and bounded lifts of the tests, and apply that
proposition separately at every coordinate, including real and
imaginary parts of the tests, rescaled to contractions, with the
requested errors tightened by their fixed norms. The coordinate errors tend to zero
uniformly over $T_{\leq1}(D)$. The coordinate maps are order zero,
and their commutator sequences belong to $J_2$. This proves the
assertion. The lifts exist by completely positive lifting for the
finite-dimensional domain when a map, rather than a single element,
is being lifted.
Start with joined weight $H_0=0$. Having constructed the preceding
joined map, let $r=1-H_j$ and prescribe the targets
$r^{1/2}ar^{1/2}$ for $a\in\mathcal P$. Prescribe commutation with
the preceding map and with the fixed tests. Apply
\eqref{loc:fraction} and join the new map to the preceding one using
Lemma~\ref{lem:join}. The new weight commutes with $r$, and hence, for
any trace $\sigma_\omega$, the defect
\[
D_j(a)=\sigma_\omega\bigl(a^{1/2}(1-H_j)a^{1/2}\bigr)
\]
satisfies
\[
D_{j+1}(a)\le(1-\delta/3)D_j(a)+C_k\delta^{3/2}.
\]
This follows by subtracting
$\sigma_\omega(r^{1/2}ar^{1/2}\psi(1))$ from $D_j(a)$;
traciality and commutation of $\psi(1)$ with $r$ justify the subtraction
without any commutation assumption on $a$. Since $D_0(a)\le1$, induction
gives
\begin{equation}\label{loc:geometric}
D_N(a)\le(1-\delta/3)^N+3C_k\sqrt\delta.
\end{equation}
Choose $\delta$ so that the second term is less than $\rho/4$, and
then a finite $N$ so that the first term is less than $\rho/4$.
Lift the resulting \cpc map from $B$ to coordinate \cpc maps into $D$.
Its exact order-zero and commutation identities in $B$ give
\eqref{loc:fill-comm}--\eqref{loc:fill-oz} on an $\omega$-large set.
Its trace estimates also give \eqref{loc:fill-trace} on an
$\omega$-large set. Otherwise choose, at every failing coordinate, a
violating tracial contraction and one of the finitely many targets;
pass to an $\omega$-large set where the target is fixed. Their
ultralimit would contradict the strict bound $\rho/2$ from
\eqref{loc:geometric}. Intersect these finitely many good sets and
choose a coordinate. This proves all the required uniform estimates.
\end{proof}
We now seek norm-central order-zero maps whose trace deficits against
a positive contraction $a$ are arbitrarily small relative to the rank
of a fixed $b\in A_+$. Here $a\in\Ideal_A(b)$ and
$d_\tau(a)\le C d_\tau(b)$ for every $\tau$ and one finite $C>0$; the precise
statement is Proposition~\ref{prop:localized-trace} below. For positive
contractive representatives $(h_n)$ of the weight, the target is
\[
\tau\bigl(a^{1/2}(1-h_n)a^{1/2}\bigr)\le\eta d_\tau(b)
\]
on a common $\omega$-large set for each $\eta>0$, uniformly over traces
with $d_\tau(b)<\infty$. To remove the norm defects without losing this
estimate, we measure an error by the rank of each positive cut of its modulus.
Taking cuts makes this smallness condition stable under norm-null
changes of representatives; traces with $d_\tau(b)=0$ will require
exact vanishing.
\begin{definition}\label{loc:relative-ideal}
For $b\in A_+$ define $J_b\subseteq A_\omega$ as follows. An element
$z=(z_n)$ belongs to $J_b$ if, for every $s>0$ and $\eta>0$, there is
$E\in\omega$ such that
\begin{equation}\label{loc:J-predicate}
d_\tau((|z_n|-s)_+)\le\eta d_\tau(b)
\quad\text{for every }n\in E\text{ and every }\tau
\text{ with }d_\tau(b)<\infty.
\end{equation}
For $d_\tau(b)=0$ this means exact zero on the left. The set $E$ is
common to the entire indicated trace family. It is enough to test
positive rational $s,\eta$.
\end{definition}
\begin{lemma}[The relative trace ideal]\label{loc:sigma-ideal}
Definition~\ref{loc:relative-ideal} is independent of the bounded
representative. It defines a closed two-sided ideal of $A_\omega$.
Moreover, for every separable $C^*$-subalgebra $C_0\subseteq A_\omega$
there is a positive contraction $e\in J_b\cap C_0'$ such that
$ey=y$ for every $y\in J_b\cap C_0$.
\end{lemma}
\begin{proof}
We record the rank facts used in verifying ideality. For $u,v,x\in A$,
\begin{equation}\label{loc:rank-calculus}
d_\tau(|u^*|)=d_\tau(|u|),\quad
d_\tau(|xu|),d_\tau(|ux|)\le d_\tau(|u|),\quad
d_\tau(|u+v|)\le d_\tau(|u|)+d_\tau(|v|).
\end{equation}
The first follows from Cuntz equivalence of the two squared moduli.
For the second use $u^*x^*xu\le\|x\|^2u^*u$, and then the first
identity for right multiplication. For the last use
$(u+v)^*(u+v)\le2(u^*u+v^*v)$ and the Cuntz inequality for a sum of
positive elements.
For $t>0$ the polar cut
\[
u^{[t]}=u\,q_t(|u|),\qquad
q_t(\lambda)=
\begin{cases}(\lambda-t)_+/\lambda,&\lambda>0,\\0,&\lambda=0,
\end{cases}
\]
belongs to $A$, has modulus $(|u|-t)_+$, and satisfies
$\|u-u^{[t]}\|\le t$. Consequently a fixed positive cut of a sum or
product can be Cuntz compared to the modulus of a sum or product of
sufficiently small polar cuts. In detail, choose $t$ so small that
the norm difference of the two expressions makes the difference of
their moduli less than the desired cut. This is possible uniformly
on the bounded sets at issue, by uniform continuity of
$z\mapsto|z|$. Norm comparison and \eqref{loc:rank-calculus} then bound
that cut rank by the sum of the ranks of the input polar cuts, or by
one such rank for multiplication and adjoints. Apply
\eqref{loc:J-predicate} with error $\eta/2$ for each of the two
summands, and with error $\eta$ for a product or adjoint. Finite
intersection of the good sets proves closure under addition,
adjoints, and left and right multiplication by bounded sequences.
Scalar multiplication follows in the same way.
Here are the perturbation details, including the zero-rank case.
If $(z_n)$ and $(z'_n)$ represent the same element, then, for a fixed
$s>0$, on an $\omega$-large set
$\||z'_n|-|z_n|\|<s/2$. Therefore
\begin{equation}\label{loc:representative-cut}
[ (|z'_n|-s)_+ ]\le[ (|z_n|-s/2)_+ ],
\end{equation}
by comparing $|z'_n|$ with $(|z_n|-s/2)_+$ in norm, at error strictly
less than $s$. This proves representative independence. If
$z^{(l)}\in J_b$ converges in the norm of $A_\omega$ to $z$, fix a cut
$s>0$ and choose $l$ so that bounded lifts of $z^{(l)}$ and $z$ are
close enough on an $\omega$-large set to give
\eqref{loc:representative-cut} with $z^{(l)}$ on the right. Apply its
predicate at cut $s/2$ and arbitrary error $\eta$. This proves norm
closedness. When $d_\tau(b)=0$, every rank comparison in these
arguments has an exactly zero right side on the common good set.
No assertion about the uncut rank of a norm-small perturbation is
being used.
Put $I_0=J_b\cap C_0$. Choose dense sequences $(c_j)$ in the unit ball
of $C_0$ and $(y_j)$ in the unit ball of $I_0$. A quasicentral
approximate unit for $I_0$ in $C_0$ supplies positive contractions
$e^{(l)}\in I_0$ such that, for $j\le l$,
\[
\|[e^{(l)},c_j]\|<1/l,\qquad
\|e^{(l)}y_j-y_j\|<1/l.
\]
Choose positive contractive lifts $e^{(l)}_n$ and bounded lifts of the
$c_j,y_j$. For each $l$, there is an $\omega$-large set $G_l$ on
which these finitely many norm estimates hold with $2/l$ in place of
$1/l$ and, simultaneously, for $1\le j\le l$,
\begin{equation}\label{loc:sigma-stage}
d_\tau((e^{(l)}_n-1/j)_+)\le l^{-1}d_\tau(b)
\quad\text{for all }\tau\text{ with }d_\tau(b)<\infty.
\end{equation}
Each displayed inequality is one uniform coordinate predicate, so
only finitely many predicates are intersected at this stage. In
particular, this construction does not enumerate traces.
Replace $G_l$ by the decreasing sets
$E_l=\{n\ge l\}\cap\bigcap_{r\le l}G_r$.
For $n\in E_1$ let $l(n)$ be the largest $l\le n$ with $n\in E_l$,
and set $e_n=e^{(l(n))}_n$; put $e_n=0$ elsewhere. Then
$l(n)\to_\omega\infty$. The norm tests imply that $e=(e_n)$
commutes with $C_0$ and absorbs $I_0$. Given $s,\eta>0$, choose
$j$ with $1/j<s$ and then require $l(n)\ge j$ and $1/l(n)\le\eta$.
Equation~\eqref{loc:sigma-stage} implies
\eqref{loc:J-predicate}. For zero-rank traces it continues to assert
exact zero at each selected coordinate. Thus $e\in J_b$, completing
the proof.
\end{proof}
\begin{proposition}[Localized norm-central trace filling]
\label{prop:localized-trace}
Let $k\ge2$, $b\in A_+$, and let $a\in\Ideal_A(b)$ be a positive
contraction. Suppose that, for some finite $C>0$,
\begin{equation}\label{loc:relative-rank}
d_\tau(a)\le C d_\tau(b)\qquad\text{for every }\tau.
\end{equation}
There is a \cpc order-zero map $\phi:M_k\to A_\omega\cap A'$ with
weight $h$ such that every positive contractive representative
$(h_n)$ of $h$ has the following property: for every $\eta>0$, on a
common $\omega$-large set of coordinates,
\begin{equation}\label{loc:TL}
\tau\bigl(a^{1/2}(1-h_n)a^{1/2}\bigr)
\le\eta d_\tau(b)
\qquad(d_\tau(b)<\infty).
\end{equation}
In particular the left side is exactly zero when $d_\tau(b)=0$.
The finite-coordinate form permits arbitrary relative trace accuracies
together with prescribed finite norm tests. Given a
finite set $F\subset A$, finitely many pairs $(a_l,b_l)$ satisfying
\eqref{loc:relative-rank} with finite constants $C_l$, positive
accuracies $\eta_l$, and $\rho>0$, there is a \cpc map
$\theta:M_k\to A$ whose matrix-unit order-zero defects and
commutators with $F$ have norm less than $\rho$, and whose weight
satisfies
\begin{equation}\label{loc:TL-coordinate}
\tau\bigl(a_l^{1/2}(1-\theta(1))a_l^{1/2}\bigr)
\le\eta_l d_\tau(b_l)
\qquad(d_\tau(b_l)<\infty)
\end{equation}
for every $l$, simultaneously over all the indicated traces.
\end{proposition}
\begin{proof}
The relative rank bound will control both the approximate-unit tail
in the hereditary construction and the trace mass lost when we remove
the ideal $J_b$.
Write $I=\Ideal_A(b)$ and choose a sequential positive contractive
quasicentral approximate unit $v_i$ for $I$ in $A$. Fix $t_i>0$
tending to zero, and set
\[
p_i=(v_i-2t_i)_+,\qquad q_i=\chi_i(v_i),
\]
where $\chi_i$ is zero on $[0,t_i]$, increases linearly to one on
$[t_i,2t_i]$, and is one thereafter. We may assume $2t_i<1$.
Then $p_iq_i=p_i$, $\|p_i-v_i\|\le2t_i$, and $(p_i)$ is still a
quasicentral approximate unit. Moreover
$[q_i]\le[(v_i-t_i/2)_+]$. Compact containment and $v_i\in I$ give
an integer $L_i\ge1$ with
\begin{equation}\label{loc:hereditary-bound}
[q_i]\le L_i[b].
\end{equation}
No uniform bound on these integers is required.
Let $D_i=\her_A(q_i)$. If $d_\tau(b)<\infty$, the restriction of
$\tau$ to $D_i$ is a bounded positive tracial functional with
\begin{equation}\label{loc:trace-norm}
\|\tau|_{D_i}\|=d_\tau(q_i)\le L_i d_\tau(b).
\end{equation}
Indeed every positive contraction of $D_i$ is supported by $p_{q_i}$,
so its trace is at most $d_\tau(q_i)$; an approximate unit obtained
from $q_i$ shows equality. Equivalently, use $q_i^{1/m}$ and lower
semicontinuity. Finiteness of the bound turns the additive tracial
map on positives into a bounded positive linear functional on $D_i$.
For $d_\tau(b)=0$ that functional is zero. For a positive finite
$d_\tau(b)$, dividing it by $L_i d_\tau(b)$ gives an element of
$T_{\leq1}(D_i)$.
Choose a dense sequence $(x_j)$ in the unit ball of $A$, and apply
Lemma~\ref{loc:bounded-fill} inside $D_i$, with tests
\[
p_i,\quad q_i x_j q_i\ (j\le i)
\quad\text{and their real and imaginary parts},
\]
and positive target $p_iap_i$. Choose its accuracy
$\rho_i>0$ after $L_i$ has been fixed, so that
\begin{equation}\label{loc:accuracy}
L_i\rho_i\longrightarrow0,
\qquad L_i\rho_i^2\longrightarrow0;
\end{equation}
for example $\rho_i=1/(i(1+L_i))$ works. Denote the resulting maps
by $\mu_i:M_k\to D_i$, and put
$\theta_i(z)=p_i\mu_i(z)p_i$. These are \cpc maps into $A$.
Their trace defects have the required relative bound already before
correction. Put
$\lambda_i=\|a^{1/2}(1-p_i^2)a^{1/2}\|$, so $\lambda_i\to0$.
The positive element defining $\lambda_i$ is supported by $p_a$;
using \eqref{loc:relative-rank}, its trace is at most
$C\lambda_i d_\tau(b)$. The other part of the defect is, by
traciality,
\[
\tau\bigl((1-\mu_i(1))^{1/2}p_iap_i
(1-\mu_i(1))^{1/2}\bigr)
\le L_i\rho_i d_\tau(b).
\]
Thus
\begin{equation}\label{loc:uncorrected-trace}
\tau\bigl(a^{1/2}(1-\theta_i(1))a^{1/2}\bigr)
\le(C\lambda_i+L_i\rho_i)d_\tau(b)
\qquad(d_\tau(b)<\infty).
\end{equation}
At zero rank both contributions are exactly zero.
Let $\mu,\theta$ be the \cpc maps represented by the coordinate
maps. Their relevant relations first hold modulo $J_b$. In fact,
for each tested $2$-norm defect $z_i$, normalization in
\eqref{loc:trace-norm} and functional calculus give
\[
d_\tau((|z_i|-s)_+)
\le s^{-2}\tau(z_i^*z_i)
\le s^{-2}L_i\rho_i^2d_\tau(b).
\]
The estimate holds for every fixed $s>0$; at zero rank it holds because
all the coordinates at issue belong to $I$ and the restricted trace
is zero. Equation~\eqref{loc:accuracy} proves that these defect
sequences belong to $J_b$. Consequently, modulo $J_b$, the map
$\mu$ is order zero and commutes with $p=(p_i)$ and each
$(q_i x_jq_i)$. Also
$[p_i,q_ix_jq_i]=q_i[p_i,x_j]q_i\to0$ in norm. It follows that
$\theta=p\mu(\cdot)p$ is order zero modulo $J_b$ and commutes there
with $(q_ix_jq_i)$. It commutes there with the constant $x_j$ as
well: using $q_ip_i=p_i$, the difference between the two commutators
has norm at most $2\|[p_i,x_j]\|$. For example
$p_i\mu_i(z)p_i x_j(1-q_i)
=p_i\mu_i(z)[p_i,x_j](1-q_i)$ for $\|z\|\le1$; the other term is
its analogous left-sided expression. Density gives the assertion
for all of $A$.
Apply Lemma~\ref{loc:sigma-ideal} to
$C_0=C^*(A,\theta(M_k))$, and let $e$ be its absorbing positive
contraction. Define
\[
\phi(z)=(1-e)\theta(z).
\]
Since $e$ commutes with the range, this is c.p.c. Every order-zero
polynomial defect of $\theta$ belongs to $J_b\cap C_0$ and is
annihilated by $1-e$. Multiplication by $(1-e)^2$ therefore proves
the exact order-zero identities for $\phi$. Likewise
$[\phi(z),x]=(1-e)[\theta(z),x]=0$ for $x\in A$. Thus
$\phi:M_k\to A_\omega\cap A'$ has the required norm relations.
To check the trace loss, the removed weight
$j=e\theta(1)$ is a positive contraction in $J_b$. Choose positive
contractive representatives $j_i$ and $h_i$ of $j$ and $h=\phi(1)$.
Then $\theta_i(1)-h_i-j_i\to_\omega0$ in norm. For $s>0$,
functional calculus gives $j_i\le s1+(j_i-s)_+$, whence
\begin{equation}\label{loc:correction-cost}
\tau(a^{1/2}j_i a^{1/2})
\le s d_\tau(a)+d_\tau((j_i-s)_+).
\end{equation}
Here the second term bounds the trace of a compression of the
positive contraction $(j_i-s)_+$, and its rank is unchanged by a
positive rescaling if necessary. Given a desired relative error,
first choose $s$ so that $sC$ is small, then use
\eqref{loc:J-predicate} at that fixed cut with a sufficiently small
error. Norm-null differences of the lifts contribute at most their
norm times $d_\tau(a)\le C d_\tau(b)$. Combining these observations
with \eqref{loc:uncorrected-trace} proves \eqref{loc:TL}, uniformly
for positive finite $d_\tau(b)$. If $d_\tau(b)=0$, then
$d_\tau(a)=0$ and every positive contraction $h_i$ has exactly zero
defect against $a$. This also proves the assertion for every choice
of positive contractive representative of $h$.
For completeness, the passage to several pairs permits arbitrary
finite coordinate tests. Construct the map for the first pair.
For the next pair use the preceding construction and reindex its
coordinate lifts as follows. At coordinate $n$ the matrix-unit
values of all already fixed maps are a finite list of ordinary
elements of $A$. Norm centrality of the new map allows an index at
which its commutators with that list and with the first $n$ fixed
dense tests are less than $1/n$. Intersect this set of indices with
the sets imposing its order-zero errors and its uniform trace
predicate for its fixed pair, to any desired accuracy tending to
zero. A finite intersection is $\omega$-large and hence nonempty.
The selected sequence still has \eqref{loc:TL} and now commutes with
the earlier maps. Repeat for the finite list of pairs and join.
The joined weight $H$ dominates every original weight $h^{(l)}$.
This domination also preserves the uniform trace estimates at
coordinates. Indeed choose positive lifts $D^{(l)}_n$ of
$H-h^{(l)}$; for any positive contractive lifts $H_n,h^{(l)}_n$,
\[
H_n\ge h^{(l)}_n-\delta_{l,n}1,
\qquad \delta_{l,n}\to_\omega0.
\]
Their trace-defect difference against $a_l$ is therefore at most
$C_l\delta_{l,n}d_\tau(b_l)$ when that rank is finite; at zero rank
both defects vanish. Lift the joined \cpc map and select one
coordinate in the finite intersection imposing all desired norm
relations, all prescribed commutators, and all relative trace
bounds with strict slack. This gives
\eqref{loc:TL-coordinate}. The finite set $F$, and each $\eta_l$,
are arbitrary at this final step.
\end{proof}
==== END Cuntz-comparison-and-Jiang-Su-absorption//localization.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//norm.tex ====
\section{Norm-visible central matrix cones}\label{norm:section}
The construction in this section detects a prescribed positive element
uniformly in the irreducible representations at each coordinate. Its
building blocks are finite word spaces at pure states. Lemma~\ref{lem:shrinking}
lets us separate their hereditary supports while retaining all the
representations that must be detected.
\begin{proposition}\label{prop:norm}
Suppose that \(A\) is separable and satisfies \textup{(C)}. For every positive contraction \(x\in A\), every
\(\gamma>0\), and every integer \(k\ge2\), there are \cpc maps
\(\nu_n:M_k\to A\) such that
\[
\nu:M_k\longrightarrow \Aomega\cap A',
\qquad \nu(z)=[(\nu_n(z))_n],
\]
is order zero, and there is a number \(\kappa>0\) for which
\begin{equation}\label{norm:coordinate-visibility}
\begin{gathered}
\text{for every }n\in\N\text{ and every irreducible representation }\sigma
\text{ of }A,\\
\|\sigma(x)\|\ge\gamma
\ \Longrightarrow\
\|\sigma(x\nu_n(1))\|\ge\kappa .
\end{gathered}
\end{equation}
One may take \(\kappa=\gamma^2/32\). In particular, the estimate holds on
one \(\omega\)-large set of coordinates simultaneously for every
irreducible representation, including representations chosen separately
at each coordinate.
\end{proposition}
We first establish the uniformity and compression facts used in the proof.
All support projections below are taken in the bidual, or in the universal
representation, and each representation is extended normally to the
bidual when support estimates are used.
\subsection{Finite covers and pure-state compression}
For \(x\in A_+\) and \(\gamma>0\), write
\[
\mathcal K(x,\gamma)
=\{\sigma:\sigma\text{ is an irreducible representation of }A,\
\|\sigma(x)\|\ge\gamma\}.
\]
There is no compactness assumption on the primitive ideal space.
\begin{lemma}\label{norm:finite-cover}
Let \(x\in A_+\), \(\gamma>0\), and let
\((s_\alpha)_{\alpha\in\Lambda}\) be positive elements of \(A\).
Suppose that \(\mathcal K(x,\gamma)\ne\varnothing\) and that, for
every \(\sigma\in\mathcal K(x,\gamma)\), some
\(\sigma(s_\alpha)\) is nonzero. Then there are
\(\alpha_1,\ldots,\alpha_m\) and \(t>0\) such that
\[
\sigma\in\mathcal K(x,\gamma)
\quad\Longrightarrow\quad
\max_{1\le i\le m}\|\sigma(s_{\alpha_i})\|>t.
\]
Thus the nonvanishing loci of the cuts
\((s_{\alpha_i}-t)_+\) still cover \(\mathcal K(x,\gamma)\).
For a finite covering family of positive contractions, continuous
functional calculus can replace its members by positive contractions
with the same supports and with maximum represented norm equal to one
at every \(\sigma\in\mathcal K(x,\gamma)\).
\end{lemma}
\begin{proof}
Let \(I\) be the closed ideal generated by all the \(s_\alpha\), and let
\(q:A\to A/I\) be the quotient map. We claim that
\(\|q(x)\|<\gamma\). Otherwise, a pure state of the unitization of
\(A/I\) can be chosen to take the value \(\|q(x)\|\) on \(q(x)\).
For completeness, the states taking that value form a nonempty compact
face of the state space: nonemptiness follows by extending the
norm-attaining state of the commutative algebra generated by \(q(x)\)
and the unit. An extreme point of that face is a pure state.
Its value on \(q(x)\) is positive, so its irreducible representation
restricts nontrivially to \(A/I\). Pulling that representation back to
\(A\) gives an element of \(\mathcal K(x,\gamma)\) annihilating every
\(s_\alpha\), a contradiction.
The strict quotient-norm inequality permits a finite sum
\[
y=\sum_{j=1}^r u_j s_{\alpha(j)}v_j,\qquad
u_j,v_j\in\widetilde A,\qquad \|x-y\|<\gamma .
\]
Indeed, these finite sums are dense in \(I\), and the distance from
\(x\) to \(I\) is strictly less than \(\gamma\). Choose \(t>0\) so
small that
\[
y_t=\sum_{j=1}^r u_j(s_{\alpha(j)}-t)_+v_j
\quad\text{satisfies}\quad \|x-y_t\|<\gamma .
\]
If a representation annihilates every cut appearing in this sum, then
\(\|\sigma(x)\|\le\|x-y_t\|<\gamma\). This proves the first assertions.
For the last assertion apply what was just proved to the given finite
family and put \(f_t(u)=\min\{u/t,1\}\) for \(u\ge0\).
The function \(f_t\) vanishes only at zero, so it preserves supports.
Whenever \(\|\sigma(s_i)\|>t\), functional calculus gives
\(\|\sigma(f_t(s_i))\|=1\).
\end{proof}
The following compression is the pure-state excision principle of
Akemann, Anderson, and Pedersen \cite[Proposition~2.2]{AAP1986}.
We include a proof using Kadison's transitivity theorem and the
pure-state kernel identity \cite[Theorem~1 and Corollary~2]{Kadison1957}.
\begin{lemma}[Pure-state compression]\label{norm:excision}
Let \(\lambda\) be a pure state of a unital \(C^*\)-algebra \(D\), with
irreducible GNS representation \((\pi,\mathcal H,\xi)\).
Given a finite subset \(\mathcal E\subset D\) and \(\epsilon>0\), there
is a positive contraction \(q\in D\) such that
\[
\pi(q)\xi=\xi,\qquad
\|qyq-\lambda(y)q^2\|<\epsilon\quad(y\in\mathcal E).
\]
\end{lemma}
\begin{proof}
Set
\[
J=\{z\in D:\pi(z)\xi=0\},\qquad H=J\cap J^*.
\]
The space \(J\) is a closed left ideal, and \(H\) is a hereditary
\(C^*\)-subalgebra. For the latter assertion, it is a closed
selfadjoint subalgebra, and if \(0\le a\le b\) with \(b\in H_+\), then
\(\lambda(a)\le\lambda(b)=0\), so \(\pi(a)\xi=0\).
We show that
\begin{equation}\label{norm:state-kernel}
\ker\lambda=J+J^*.
\end{equation}
The inclusion from right to left is immediate. For the converse,
let \(\lambda(y)=0\) and put \(\zeta=\pi(y)\xi\), which is orthogonal
to \(\xi\). If \(\zeta=0\), then \(y\in J\). Otherwise, on
\(K=\operatorname{span}\{\xi,\zeta\}\), define \(R\xi=\zeta\) and
\(R\zeta=0\). The selfadjoint form of Kadison's Transitivity Theorem
\cite{Kadison1957} supplies selfadjoint \(h,g\in D\) whose restrictions
to \(K\) are respectively \(\operatorname{Re}R\) and
\(\operatorname{Im}R\). Thus \(v=h+ig\) satisfies
\[
\pi(v)\xi=\zeta,\qquad \pi(v^*)\xi=0.
\]
Consequently \(y-v\in J\) and \(v\in J^*\), proving
\eqref{norm:state-kernel}.
Let \((e_\alpha)\) be a positive contractive approximate unit for \(H\)
and put \(q_\alpha=1-e_\alpha\). Each \(q_\alpha\) fixes \(\xi\).
For \(z\in J\), the element \(z^*z\) belongs to \(H\), and hence
\[
\|zq_\alpha\|^2
=\|q_\alpha z^*zq_\alpha\|\longrightarrow0.
\]
Its adjoint gives \(\|q_\alpha z^*\|\to0\). Applying
\eqref{norm:state-kernel} to \(y-\lambda(y)1\) now gives
\[
\|q_\alpha(y-\lambda(y)1)q_\alpha\|\longrightarrow0
\qquad(y\in D).
\]
Choose one index sufficiently late for all the finitely many tests.
\end{proof}
We will apply excision to make finitely many word representatives
almost orthonormal on one hereditary support. Their Gram matrix will
control the norm of the row formed from these representatives. We
also estimate how that row intertwines the action of each prescribed
element below the outermost word level; a taper will later remove
that boundary level.
To construct this matrix form of the compression, fix a finite set
\(F\subset A\) of selfadjoint contractions, \(d\ge1\), and
\(0<\eta<1\). Suppose \(\sigma\in\mathcal K(x,\gamma)\), where \(x\)
is a positive contraction. Choose a unit vector \(\xi\) with
\[
\lambda(x):=\langle\sigma(x)\xi,\xi\rangle>\gamma/2.
\]
Every nonzero vector in an irreducible representation is cyclic, so
this is a pure state of \(\widetilde A\). Let \(\mathcal H_l\) be the
span of the vectors \(\sigma(w)\xi\), where \(w\) is a word of length
at most \(l\) in \(F\), with the empty word equal to \(1\). These
finite-dimensional spaces form an increasing filtration
\[
\C\xi=\mathcal H_0\subseteq\mathcal H_1
\subseteq\cdots\subseteq\mathcal H_d.
\]
Choose an orthonormal basis \(\xi_1,\ldots,\xi_m\) adapted to this
filtration, with \(\xi_1=\xi\). Write
\(\xi_j=\sigma(b_j)\xi\), where \(b_j\in\widetilde A\) is a finite
linear combination of the appropriate words; take \(b_1=1\).
Let \(l_j\in\{0,\ldots,d\}\) be the level of \(\xi_j\). There is
exactly one level-zero vector.
For \(a\in F\), let \(M_a\in M_m\) be the matrix of its compression
to \(\mathcal H_d\) in this basis. It is a selfadjoint contraction.
Moreover,
\begin{equation}\label{norm:banded-model}
(M_a)_{ij}=0\quad\text{if }|l_i-l_j|>1.
\end{equation}
Indeed, \(a\mathcal H_l\subseteq\mathcal H_{l+1}\) for \(l<d\);
orthogonality to the filtration proves one half of the assertion,
and selfadjointness proves the other. For \(l_j<d\), put
\[
r_j(a)=ab_j-\sum_{l=1}^m b_l(M_a)_{lj}.
\]
Since \(a\xi_j\in\mathcal H_d\), these residuals satisfy
\(\sigma(r_j(a))\xi=0\).
\begin{lemma}\label{norm:prototype}
For this finite word model there is a positive contractive seed
\(s\in A\) with \(\sigma(s)\ne0\), whose support \(P=p_s\) satisfies
\begin{align}
PxP&\ge(\gamma/4)P, \label{norm:positive-compression}\\
\big\|(Pb_i^*b_jP)_{i,j=1}^m-I_m\otimes P\big\|
&\le\eta^2/4, \label{norm:local-gram}\\
\big\|(Pr_i(a)^*r_j(a)P)_{l_i,l_j<d}\big\|
&\le\eta^2/4\qquad(a\in F). \label{norm:local-residual}
\end{align}
Every positive element in \(\her(s)\) has a support on which the
same three estimates hold.
\end{lemma}
\begin{proof}
Apply Lemma~\ref{norm:excision} in \(\widetilde A\) to \(x\), all
\(b_i^*b_j\), and all \(r_i(a)^*r_j(a)\). Their state values are
respectively \(\lambda(x)>\gamma/2\), \(\delta_{ij}\), and zero.
Choose the excision error at most \(\gamma/16\) for \(x\), and at
most \(\eta^2/(16m)\) for each of the matrix entries.
Put
\[
P_0=1_{(1/2,1]}(q),\qquad v=q^{-1}P_0
\quad\text{in }\widetilde A^{**}.
\]
Here \(\|v\|\le2\), and
\[
P_0(y-\lambda(y)1)P_0
=v\big(qyq-\lambda(y)q^2\big)v.
\]
The estimate for \(x\) therefore gives
\(P_0xP_0\ge(\lambda(x)-\gamma/4)P_0\ge(\gamma/4)P_0\).
For the matrices, each entry has error at most
\(\eta^2/(4m)\). An \(r\)-by-\(r\) operator matrix whose entries
have norm at most \(\delta\) has norm at most \(r\delta\), by the
Cauchy--Schwarz inequality for its row action. Since each matrix
here has size at most \(m\), both matrix errors are at most
\(\eta^2/4\).
Choose a continuous function \(f:[0,1]\to[0,1]\) that is zero on
\([0,1/2]\) and has \(f(1)=1\), and set
\(s=f(q)xf(q)\). This is a positive contraction in \(A\), because
\(A\) is an ideal in its unitization. Also
\(\lambda(s)=\lambda(x)>0\), so \(\sigma(s)\ne0\), and \(p_s\le P_0\).
Compressing the preceding inequalities to \(p_s\) proves the
claims. Further hereditary restriction only decreases the support
projection, so preserves all the estimates.
\end{proof}
\subsection{Separating the models}
The next observation records precisely why iterating
Lemma~\ref{lem:shrinking} does not lose the representation cover.
\begin{lemma}\label{norm:separated-cover}
Assume that $A$ satisfies \textup{(C)}. Suppose that positive
contractions \(s_1,\ldots,s_m\in A\) have
nonvanishing loci covering \(\mathcal K(x,\gamma)\).
For each pair \(i\ne j\), prescribe a finite set
\(\mathcal E_{ij}\subset\widetilde A\), and let \(\delta>0\).
There are positive contractions \(t_i\in\her(s_i)\), with supports
\(Q_i=p_{t_i}\), such that
\begin{align}
\max_i\|\sigma(t_i)\|&=1
&&(\sigma\in\mathcal K(x,\gamma)), \label{norm:retained-cover}\\
\|Q_i z Q_j\|&\le\delta
&&(i\ne j,\ z\in\mathcal E_{ij}). \label{norm:cross-compressions}
\end{align}
\end{lemma}
\begin{proof}
If \(\mathcal K(x,\gamma)\) is empty, take every \(t_i=0\).
Otherwise proceed as follows.
Let \(c<1\) be the constant in Lemma~\ref{lem:shrinking}.
Before each shrinking operation, apply the last assertion of
Lemma~\ref{norm:finite-cover} to the current finite family.
This leaves every support unchanged and ensures that each
representation in \(\mathcal K(x,\gamma)\) gives norm one to at
least one of the current seeds.
For one prescribed triple \(i,j,z\), let
\(\beta=\|p_{s_i}zp_{s_j}\|\), using the current seeds. If
\(\beta=0\), nothing is needed. Otherwise apply
Lemma~\ref{lem:shrinking} to the current seeds \(s_i,s_j\) and to
\(z/\beta\). Their replacements have supports on which the
compression of \(z\) has norm at most \(c\beta\).
For any \(\sigma\in\mathcal K(x,\gamma)\), choose a current seed of
represented norm one. If its index is outside \(\{i,j\}\), it
survives unchanged. If its index is in \(\{i,j\}\), the
representation clause in Lemma~\ref{lem:shrinking} ensures that
at least one replacement is nonzero. Thus the new family still
covers \(\mathcal K(x,\gamma)\), allowing the next
reparametrization.
All new supports are subprojections of their previous supports.
Consequently no previously obtained compression bound can increase.
For a fixed coefficient \(z\), finitely many repetitions suffice:
after \(r\) repetitions its bound is at most \(c^r\|z\|\).
There are only finitely many prescribed coefficients and pairs.
Treating them successively therefore gives
\eqref{norm:cross-compressions}. A final support-preserving
reparametrization gives \eqref{norm:retained-cover}.
\end{proof}
\subsection{The row map and the taper}
\begin{lemma}\label{norm:finite-stage}
Assume that $A$ satisfies \textup{(C)}. Let \(x\in A\) be a positive
contraction, \(\gamma>0\), \(k\ge2\),
\(F\subset A\) a finite set of selfadjoint contractions,
\(d\ge1\), and \(0<\eta<1\).
There is a \cpc map \(V:M_k\to A\) such that
\begin{align}
\|V(z)V(w)\|&\le\eta^2
&&(z,w\in(M_k)_+,\ \|z\|,\|w\|\le1,\ zw=0), \label{norm:order-zero-error}\\
\|[a,V(z)]\|&\le2\eta+2/d
&&(a\in F,\ \|z\|\le1), \label{norm:central-error}\\
\|\sigma(xV(1))\|&\ge\gamma^2/32
&&(\sigma\in\mathcal K(x,\gamma)). \label{norm:finite-visibility}
\end{align}
\end{lemma}
\begin{proof}
If \(\mathcal K(x,\gamma)\) is empty, take \(V=0\). Otherwise
construct the preceding word model and the seed in
Lemma~\ref{norm:prototype} for each representation in that set.
The seeds cover the set. Lemma~\ref{norm:finite-cover} therefore
retains finitely many models, indexed by \(i=1,\ldots,m\).
Write \(n_i\) for their basis sizes,
\(b_j^{(i)}\) for their basis representatives, \(l_j^{(i)}\) for
their levels, and \(M_a^{(i)}\) and \(r_j^{(i)}(a)\) for their
model matrices and residuals. Set \(N=\sum_i n_i\).
Apply Lemma~\ref{norm:separated-cover} with
\(\delta=\eta^2/(2N)\) to all coefficients
\[
(b_j^{(i)})^*b_l^{(h)}
\quad\text{and}\quad
(r_j^{(i)}(a))^*r_l^{(h)}(a)
\qquad(i\ne h,\ a\in F),
\]
where the residual indices have levels less than \(d\).
Let \(s_i\) denote the resulting seeds and \(P_i=p_{s_i}\).
All the estimates of Lemma~\ref{norm:prototype} remain valid.
The number of models and their sizes have already been fixed
before this cross-compression tolerance is chosen.
By Lemma~\ref{lem:placement}, choose \cpc order-zero maps
\(\phi_i:M_k\to\her(s_i)\) such that
\begin{equation}\label{norm:cone-norm-one}
\|\sigma(s_i)\|=1\quad\Longrightarrow\quad
\|\sigma(\phi_i(1))\|=1.
\end{equation}
In fact it suffices to apply that Lemma to the fixed cut
\((s_i-1/2)_+\). In the universal representation on
\(\mathcal H\), consider
\[
\mathcal L=\bigoplus_{i=1}^m\bigoplus_{j=1}^{n_i}P_i\mathcal H
\quad\text{and}\quad
B:\mathcal L\to\mathcal H,\qquad
B(\zeta_{ij})=\sum_{i,j}b_j^{(i)}\zeta_{ij}.
\]
The row \(B\) is only an operator on these support spaces.
Its Gram matrix has \((ij,hl)\)-entry
\(P_i(b_j^{(i)})^*b_l^{(h)}P_h\).
The diagonal blocks differ from the corresponding identity blocks
by at most \(\eta^2/4\). The off-diagonal matrix has \(N\) rows
and columns with entries bounded by \(\delta\), so has norm at most
\(N\delta=\eta^2/2\). In particular,
\begin{equation}\label{norm:row-gram}
\|B^*B-1_{\mathcal L}\|\le\eta^2,\qquad
\|B\|\le(1+\eta^2)^{1/2}.
\end{equation}
Let
\[
M_a=\bigoplus_i(M_a^{(i)}\otimes1_{P_i\mathcal H})
\quad\text{on }\mathcal L,
\]
and let \(Q\) be the projection onto the slots of level less than
\(d\). The columns of \((aB-BM_a)Q\) are
\(r_j^{(i)}(a)P_i\). Its squared norm is the norm of their Gram
matrix. The local residual estimates and the same off-diagonal
estimate therefore give
\begin{equation}\label{norm:row-residual}
\|(aB-BM_a)Q\|\le\eta\qquad(a\in F).
\end{equation}
For \(z\in M_k\), define operators on \(\mathcal L\) by
\[
D_z=\bigoplus_{i,j}\phi_i(z)|_{P_i\mathcal H},\qquad
T=\bigoplus_{i,j}(1-l_j^{(i)}/d)1_{P_i\mathcal H},\qquad
C_z=TD_z .
\]
The map \(z\mapsto C_z\) is \cpc and order zero: it is the direct
sum of the order-zero maps \(\phi_i\), each multiplied by a
nonnegative scalar at most one. It is supported on \(Q\), since
the taper \(T\) vanishes at level \(d\).
Put
\begin{equation}\label{norm:row-map}
\lambda_0=(1+\eta^2)^{-1},\qquad
V(z)=\lambda_0 BC_zB^*
=\lambda_0\sum_{i,j}(1-l_j^{(i)}/d)
b_j^{(i)}\phi_i(z)(b_j^{(i)})^* .
\end{equation}
The last formula shows that \(V(z)\in A\), including when \(A\)
is nonunital. Row compression gives complete positivity.
Furthermore \(0\le C_1\le1_{\mathcal L}\), so
\(\|V(1)\|\le\lambda_0\|B\|^2\le1\); hence \(V\) is contractive.
For orthogonal positive contractions \(z,w\in M_k\), we have
\(C_zC_w=0\). Thus
\[
V(z)V(w)
=\lambda_0^2 BC_z(B^*B-1_{\mathcal L})C_wB^*,
\]
and \eqref{norm:row-gram} gives
\[
\|V(z)V(w)\|
\le\lambda_0^2(1+\eta^2)\eta^2
\le\eta^2 .
\]
This proves \eqref{norm:order-zero-error}.
We next bound the commutator. Let \(E_l\) be the projection onto
all slots of level \(l\), summed over the prototypes. By
\eqref{norm:banded-model}, the selfadjoint contraction \(M_a\)
has only its level diagonal and its two adjacent-level diagonals.
The operators
\[
M_{a,+}=\sum_{l=0}^{d-1}E_{l+1}M_aE_l,\qquad
M_{a,-}=M_{a,+}^*
\]
each have norm at most one: their separate summands have
orthogonal initial spaces and orthogonal final spaces. The
level diagonal commutes with \(T\), and
\[
[M_a,T]=d^{-1}(M_{a,+}-M_{a,-}),\qquad
\|[M_a,T]\|\le2/d .
\]
Within each prototype the entries of \(M_a\) are scalar and the
operator \(\phi_i(z)\) is the same in every slot. Thus
\([M_a,D_z]=0\), and
\begin{equation}\label{norm:taper-bound}
\|[M_a,C_z]\|\le2/d\qquad(\|z\|\le1).
\end{equation}
With \(R_a=aB-BM_a\), selfadjointness gives
\(R_a^*=B^*a-M_aB^*\), and a direct expansion yields
\[
[a,V(z)]
=\lambda_0\big(R_aC_zB^*+
B[M_a,C_z]B^*-BC_zR_a^*\big).
\]
Because \(C_z=QC_zQ\), \eqref{norm:row-residual} bounds both
outside error terms. Consequently
\[
\|[a,V(z)]\|
\le\lambda_0\big(2\eta\|B\|+(2/d)\|B\|^2\big)
\le2\eta+2/d .
\]
This proves \eqref{norm:central-error}; the estimate is uniform
over the unit ball of \(M_k\).
Finally fix \(\sigma\in\mathcal K(x,\gamma)\). The retained cover
gives an index \(i\) with \(\|\sigma(s_i)\|=1\), so that
\(h_i=\sigma(\phi_i(1))\) has norm one by
\eqref{norm:cone-norm-one}. Denote the represented support by
\(P_i^\sigma\), and put \(X=\sigma(x)\), \(H=\sigma(V(1))\).
The unique level-zero slot of the \(i\)-th model has representative
\(b_1^{(i)}=1\) and taper coefficient one. Every summand of
\eqref{norm:row-map} at \(z=1\) is positive, so
\(H\ge\lambda_0 h_i\). Also
\[
P_i^\sigma X^2P_i^\sigma
\ge(P_i^\sigma XP_i^\sigma)^2
\ge(\gamma/4)^2P_i^\sigma,
\]
where the first inequality follows by expanding
\(P_i^\sigma X(1-P_i^\sigma)XP_i^\sigma\ge0\), and the second
uses \eqref{norm:positive-compression}.
Since \(h_i=P_i^\sigma h_iP_i^\sigma\),
\[
\|Xh_iX\|
=\|h_i^{1/2}X^2h_i^{1/2}\|
\ge(\gamma/4)^2\|h_i\|
=(\gamma/4)^2 .
\]
Positive domination and \(\|X\|\le1\) now imply
\[
\|XH\|\ge\|XHX\|
\ge\lambda_0\|Xh_iX\|
\ge\gamma^2/32,
\]
because \(\eta<1\) gives \(\lambda_0\ge1/2\).
This proves \eqref{norm:finite-visibility}.
\end{proof}
\begin{proof}[Proof of Proposition~\ref{prop:norm}]
Choose increasing finite sets \(F_n\) of selfadjoint contractions
whose union is dense in the selfadjoint unit ball of \(A\).
Apply Lemma~\ref{norm:finite-stage} with
\(F=F_n\), \(d=n\), and \(\eta=(n+1)^{-1}\), and denote its maps
by \(\nu_n\). Complete positivity and contractivity pass to the
ultrapower. The error in \eqref{norm:order-zero-error} tends to
zero, so the resulting map is order zero. The estimate in
\eqref{norm:central-error}, density, and
\(\|\nu_n(z)\|\le\|z\|\) imply that its range commutes with \(A\).
All maps have norm at most one, which supplies the bound required
to form their representing sequences. The final estimate in
Lemma~\ref{norm:finite-stage} holds for every \(n\) and every
\(\sigma\in\mathcal K(x,\gamma)\), with the same constant
\(\gamma^2/32\). This proves the stated simultaneous coordinate
quantifiers.
\end{proof}
\begin{remark}\label{norm:scope}
Proposition~\ref{prop:norm} controls irreducible representations of
the coordinate algebra \(A\). It supplies the representation
estimates needed for later ideal comparisons; it makes no
assertion of fullness in \(\Aomega\cap A'\) or in its annihilator
quotient.
\end{remark}
==== END Cuntz-comparison-and-Jiang-Su-absorption//norm.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//paper.tex ====
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\title{Cuntz comparison and Jiang--Su absorption}
\author{OpenAI}
\date{September 23, 2026}
\begin{document}
\maketitle
\input{sections/introduction}
{\small\setlength{\baselineskip}{11pt}\tableofcontents}
\clearpage
\input{sections/preliminaries}
\input{sections/extended-traces}
\input{sections/rounding}
\input{sections/divisibility}
\input{sections/support-tools}
\input{sections/shrinking}
\input{sections/norm}
\input{sections/tracial}
\input{sections/localization}
\input{sections/comparison}
\input{sections/central}
\input{sections/absorption}
\input{sections/assembly}
\input{references}
\end{document}
==== END Cuntz-comparison-and-Jiang-Su-absorption//paper.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//preliminaries.tex ====
\section{Cuntz comparison and support calculus}
\label{sec:preliminaries}
We collect the comparison and trace facts used by both routes into
condition (C). Cut placement lets us move positive elements into prescribed
hereditary subalgebras. The trace correspondence then makes numerical
estimates available to the Cuntz semigroup, with zero and infinite values
retained.
All tensor products are spatial unless a maximal tensor product is
explicitly indicated. Unitization is denoted by a tilde. We permit the
adjoined unit in formulas, but maps and constructed elements take
values in the stated algebra.
We write \(\K\) for the compact operators on a separable
infinite-dimensional Hilbert space.
For \(b\in D_+\), set
\[
\her_D(b)=\overline{bDb},\qquad
\Ideal_D(b)=\overline{\operatorname{span}}(DbD),\qquad
p_b=1_{(0,\infty)}(b)\in D^{**}.
\]
The ambient algebra is omitted when clear. Spectral projections are
also used in representations. A comparison or support relation for a
matrix over \(D\) is understood in the corresponding matrix
amplification.
\subsection{Order zero and the Cuntz semigroup}
A completely positive contractive map is abbreviated c.p.c. It has
\emph{order zero} if it preserves orthogonality of positive elements.
For a c.p.c.\ order-zero map \(\theta:E\to D\) with unital domain, the
order-zero structure theorem~\cite{WZ2009} gives
\[
\theta(z)=h\rho(z)=\rho(z)h,\qquad
h=\theta(1),\qquad \rho(1)=p_h,
\]
where \(\rho:E\to D^{**}\) is a supporting homomorphism. Its elements
act as multipliers on \(\her(h)\). If \(f:[0,1]\to[0,1]\) is continuous
and \(f(0)=0\), then \(f(h)\rho(z)\) belongs to \(D\) and defines another
c.p.c.\ order-zero map, denoted \(f(\theta)\). A \emph{matrix cone}
means a c.p.c.\ order-zero map from \(M_k\), for some \(k\).
For \(a,b\in(D\otimes\K)_+\), the notation \(a\precsim b\) means that
\(r_j^*br_j\to a\) in norm for suitable \(r_j\in D\otimes\K\).
The Cuntz semigroup \(\Cu(D)\) consists of equivalence classes for the
associated mutual comparison relation, with addition given by
orthogonal sum. We use its usual sequential-supremum structure
\cite{CEI2008}; in particular,
\begin{equation}
\label{eq:pre:cuts}
[(a-\epsilon)_+]\ll[a],
\qquad [a]=\sup_{\epsilon>0}[(a-\epsilon)_+].
\end{equation}
Here \(x\ll y\) means that whenever an increasing sequence has
supremum at least \(y\), one of its terms is at least \(x\).
We will use $[a+c]\le[a]+[c]$ and the identities
\begin{equation}
\label{ext:eq:cu-facts}
[z^*z]=[zz^*],\qquad
c\in\her(b)_+\Longrightarrow[c]\le[b],\qquad
[c]=\sup_{n\ge1}[(c-1/n)_+].
\end{equation}
For mutually orthogonal positive elements, the class of the sum is the
sum of the classes: the row of their square roots compares the sum with
its diagonal matrix. Subequivalence to a fixed positive element is closed
under norm limits in the left variable, by choosing a sufficiently good
comparison witness for each approximant.
The semigroup \(W(D\otimes\K)\), defined using finite matrices over the
stable algebra, is \(\Cu(D)\).
We record the elementary comparison facts which will control the
locations and norms of witnesses.
\begin{lemma}[Cuts, placement, and witness bounds]
\label{lem:pre:cut-placement}
Let \(D\) be a \(C^*\)-algebra.
\begin{enumerate}
\item If \(a,b\in D_+\) and \(\|a-r^*br\|<\epsilon\), then there is
\(v\in D\) with
\[
v^*v=(a-\epsilon)_+,\qquad vv^*\in\her(b).
\]
Consequently, if \(a\precsim b\), the hereditary algebra of every
positive cut of \(a\) can be placed in \(\her(b)\) by a polar
isometry.
\item If \(a\in\Ideal(b)_+\), then for every \(\epsilon>0\) there is
a finite integer \(L\) such that
\[
[(a-\epsilon)_+]\le L[b].
\]
If \([a]\le L[b]\), every positive cut of \(a\) still compares to
\(L[(b-\delta)_+]\) for a sufficiently small \(\delta>0\).
\item Comparisons between elements of a hereditary subalgebra can
be witnessed in that subalgebra. Comparisons between specified
matrix corners can be witnessed by the appropriate rectangular
matrices.
\item If \(u\precsim(y-\beta)_+\), where \(0\le\alpha<\beta\), then for
each \(\eta>0\) there is \(r\in D\) such that
\[
\|u-r^*(y-\alpha)_+r\|<\eta,\qquad
\|r\|^2\le\frac{\|u\|+\eta}{\beta-\alpha}.
\]
\end{enumerate}
\end{lemma}
\begin{proof}
For the first assertion, choose
\(\|a-r^*br\|<\delta<\epsilon\), and a continuous \(g\ge0\), vanishing
on \([0,\delta]\), which is one on \([\epsilon,\|a\|]\).
For \(z=b^{1/2}rg(a)\),
\[
(\delta-\|a-r^*br\|)\,g(a)^2
\le z^*z
\le (\|a\|+\|a-r^*br\|)\,g(a)^2.
\]
Thus \(z^*z\) and \(g(a)^2\) generate the same hereditary algebra,
which contains \((a-\epsilon)_+\). If \(w\) is the polar isometry of
\(z\), conjugation by \(w\) maps \(\her(z^*z)\) into \(\her(b)\).
Moreover \(wx\in D\) for \(x\in\her(z^*z)\), by approximation using
\(w|z|=z\). Taking \(v=w(a-\epsilon)_+^{1/2}\) proves the assertion.
Its matrix version gives the claimed hereditary placement.
For the second assertion, approximate $a^{1/2}$ by a finite sum
$q=\sum_i x_i b y_i$ so closely that $\|a-q^*q\|<\epsilon/2$.
Write $q=XY$, where $X=(x_i b^{1/2})_i$ is a row and
$Y=(b^{1/2}y_i)_i$ is a column. Then
$q^*q\le\|X\|^2Y^*Y$, while $YY^*$ belongs to the hereditary
algebra of $H=\operatorname{diag}(b,\ldots,b)$.
Positive domination and equality of the classes of the two squared
moduli give
\[
q^*q\precsim Y^*Y\sim YY^*\precsim H.
\]
Choose a comparison witness with error less than $\epsilon/2$ and
apply the first assertion to obtain the claimed cut comparison.
Alternatively, for a fixed comparison witness against
\(\operatorname{diag}(b,\ldots,b)\), replacing \(b\) by
\((b-\delta)_+\) changes the compression by at most
\(\delta\|r\|^2\). This also proves the assertion about an extra target
cut.
Let \(C\subseteq D\) be hereditary and let \((e_i)\) be an approximate
unit of \(C\). Given \(a,b\in C_+\) and a witness \(r\), first choose
an index for which the errors in \(e_i b e_i\) and \(e_i a e_i\) are
small. Then \(e_i r e_i\in C\), and its compression is as close to
\(a\) as desired. This proves the third assertion. The same
compression in matrix algebras gives rectangular witnesses. More
generally, an element whose two squared moduli belong to the
specified hereditary corners lies in their rectangular block, as
is seen by multiplying by approximate units on both sides.
For the last assertion choose \(x\) with
\(\|u-x^*(y-\beta)_+x\|<\eta\), and put
\[
f(t)=
\left(\frac{(t-\beta)_+}{(t-\alpha)_+}\right)^{1/2},
\qquad r=f(y)x,
\]
where the quotient is zero when its denominator vanishes. This is a
continuous function vanishing near zero, and
\[
f(y)(y-\alpha)_+f(y)=(y-\beta)_+,\qquad
f(y)^2\le(\beta-\alpha)^{-1}(y-\beta)_+.
\]
The compression and the norm bound follow. If the original
comparison is in the stabilization, compression to the given corner
first gives \(x\in D\).
\end{proof}
\subsection{Extended traces and rank}
We allow all lower semicontinuous extended traces
\(\tau:A_+\to[0,\infty]\), without dense finiteness. Thus \(\tau\) is
additive and positively homogeneous on positives, is lower
semicontinuous in norm, vanishes at zero, and satisfies
\(\tau(x^*x)=\tau(xx^*)\). For \(b\ge0\), write
\[
d_\tau(b)=\lim_{j\to\infty}\tau(b^{1/j}),
\]
first rescaling to a contraction when necessary. The rank is unchanged
by positive scalar rescaling. Traces
are extended to matrices and to stabilization using the unnormalized
matrix trace: on a positive infinite matrix, the extended trace is the
sum of the traces of its diagonal entries. Additivity and lower
semicontinuity justify these sums, and traciality follows by summing the
squared moduli of the matrix entries.
For a positive contraction \(b\) and \(s>0\),
\begin{equation}
\label{eq:pre:trace-rank}
\tau(b)\le d_\tau(b),\qquad
d_\tau((b-s)_+)\le s^{-1}\tau(b).
\end{equation}
These inequalities follow from functional calculus. Rank respects
Cuntz comparison. For a bounded trace we use its normal extension
to the bidual, so \(d_\tau(b)=\tau(p_b)\).
Every such trace defines a functional $[b]\mapsto d_\tau(b)$ on
$\Cu(A)$. Conversely, if $A$ is exact, every
$\varphi\in\F(\Cu(A))$ has the form
\begin{equation}
\label{ext:eq:trace-representation}
\varphi([b])=d_\tau(b),\qquad b\in(A\otimes\K)_+,
\end{equation}
for a lower semicontinuous extended trace $\tau$ on the stabilization.
Elliott--Robert--Santiago's correspondence first gives an extended
quasitrace; Haagerup's theorem, in its extended form for exact algebras,
makes it additive, and stabilization preserves the trace cone
\cite[Proposition~4.2 and Remarks~4.3 and~4.5]{ERS2011}.
Neither boundedness nor dense finiteness is imposed.
Nuclear algebras are exact. Indeed, for an ideal $J\lhd D$, the quotient
norm on $A\odot(D/J)$ induced by
$(A\otimes_{\min}D)/(A\otimes_{\min}J)$ dominates the spatial norm and
is a $C^*$-cross-norm, as seen by elementary-tensor lifts. Here $\odot$
denotes the algebraic tensor product. Uniqueness of the tensor norm for
nuclear $A$ identifies this quotient with $A\otimes_{\min}(D/J)$,
giving the exact tensor-product sequence.
\subsection{The comparison convention}
The extended-functional convention in
Equation~\eqref{ext:eq:comparison} is equivalent to almost
unperforation. The non-strict inequality on the whole functional cone
records ideal containment; the strict inequalities at finite positive
target values then determine comparison.
\begin{lemma}
\label{lem:comparison-convention}
For any $C^*$-algebra $A$, strict comparison in the sense of
Equation~\eqref{ext:eq:comparison} is equivalent to almost unperforation
of $\Cu(A)$.
\end{lemma}
\begin{proof}
Suppose first that Equation~\eqref{ext:eq:comparison} holds and that
$(l+1)x\le ly$. Applying a functional gives
$\varphi(x)\le\varphi(y)$, with strict inequality whenever
$0<\varphi(y)<\infty$. If $\varphi(y)=0$, then $\varphi(x)=0$;
if $\varphi(y)=\infty$, the non-strict inequality is automatic.
Thus Equation~\eqref{ext:eq:comparison} gives $x\le y$.
Conversely, suppose $\Cu(A)$ is almost unperforated, and let
$x=[a]$ and $y=[b]$ satisfy the two functional inequalities in
Equation~\eqref{ext:eq:comparison}. In $B=A\otimes\K$, set
$I=\Ideal_B(b)$. The map that is zero on $I_+$ and infinite outside
$I_+$ is a lower semicontinuous extended trace: a sum of positives
belongs to an ideal exactly when both summands do, $z^*z$ belongs to it
exactly when $zz^*$ does, and the ideal is norm closed. Its rank
functional vanishes at $y$, so the first inequality forces $a\in I$.
Lemma~\ref{lem:pre:cut-placement}(2) now gives
$x\le\sup_{n\ge1}ny$. For every functional normalized by
$\varphi(y)=1$, the second inequality gives $\varphi(x)<1$.
The comparison criterion of
\cite[Proposition~6.2]{ERS2011} therefore yields $x\le y$.
\end{proof}
==== END Cuntz-comparison-and-Jiang-Su-absorption//preliminaries.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//references.tex ====
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\emph{Nuclear \(C^*\)-algebras: 99 problems},
\href{https://arxiv.org/abs/2506.10902v2}{arXiv:2506.10902v2}
(2025; version 2, 2026).
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H.~Thiel,
\emph{Ranks of operators in simple $C^*$-algebras with stable rank one},
Comm. Math. Phys. \textbf{377} (2020), 37--76.
\href{https://arxiv.org/abs/1711.04721v3}{arXiv:1711.04721v3}.
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A.~S.~Toms, S.~White, and W.~Winter,
\emph{$\mathcal Z$-stability and finite-dimensional tracial boundaries},
Int. Math. Res. Not. \textbf{2015} (2015), no.~10, 2702--2727.
\href{https://doi.org/10.1093/imrn/rnu001}{doi:10.1093/imrn/rnu001}.
\bibitem{TW2007}
A.~S.~Toms and W.~Winter,
\emph{Strongly self-absorbing \(C^*\)-algebras},
Trans. Amer. Math. Soc. \textbf{359} (2007), no.~8, 3999--4029.
\href{https://arxiv.org/abs/math/0502211v2}{arXiv:math/0502211v2}.
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W.~Winter, \emph{Nuclear dimension and $\mathcal Z$-stability of pure $C^*$-algebras},
Invent. Math. \textbf{187} (2012), no.~2, 259--342.
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\end{thebibliography}
==== END Cuntz-comparison-and-Jiang-Su-absorption//references.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//rounding.tex ====
\section{Trace transfer and spectral rounding}
\label{ext:sec:rounding}
We approximate trace values by ranks in two steps. First, a positive
function of an element that vanishes near zero is transferred into a
small hereditary support without changing any trace value. Integer
spectral cuts then approximate that value with an error bounded by the
rank of the support. The transferred element may have large norm; the
error will be independent of that norm.
\begin{lemma}[Trace transfer]
\label{ext:lem:transfer}
Let $B$ be a simple stable $C^*$-algebra, let $a\in B_+$ be a
contraction, and let $0\ne h\in B_+$. Suppose that
$r:[0,1]\to[0,\infty)$ is continuous and vanishes on $[0,\delta]$
for some $0<\delta<1$. There is $c\in\her_B(h)_+$ such that
\[
\tau(c)=\tau(r(a))
\]
for every lower-semicontinuous extended trace $\tau$ on $B$.
\end{lemma}
\begin{proof}
Fullness of $h$ and Lemma~\ref{lem:pre:cut-placement} give
$[(a-\delta/4)_+]\le m[h]$ for some finite $m\ge1$.
Apply that lemma in $M_m(B)$, with cut parameter
$\epsilon=\delta/4$ and comparison error less than $\delta/4$,
to the source $\operatorname{diag}((a-\delta/4)_+,0,\ldots,0)$
and target $H=\operatorname{diag}(h,\ldots,h)$.
The resulting placement matrix has only its first column nonzero,
because all other diagonal entries of its initial squared modulus
are zero. Thus there is a column $v=(v_i)_{i=1}^m$ such that
\[
v^*v=(a-\delta/2)_+,\qquad
vv^*\in M_m(\her_B(h)).
\]
Define
\[
g(t)=\left(\frac{r(t)}{(t-\delta/2)_+}\right)^{1/2},
\]
setting it equal to zero where the denominator vanishes.
It is continuous and vanishes near zero because $r$ vanishes on
$[0,\delta]$. For $z_i=v_i g(a)$, we have
\[
\sum_{i=1}^m z_i^*z_i=r(a),\qquad
c:=\sum_{i=1}^m z_i z_i^*\in\her_B(h).
\]
For the containment, use
$z_i z_i^*\le\|g\|^2v_i v_i^*$ and heredity.
Additivity and traciality give the asserted identity for every
extended trace, with the same $c$.
\end{proof}
\begin{lemma}[Integer spectral rounding]
\label{ext:lem:rounding}
Let $B$ be a stable $C^*$-algebra, $c\in B_+$, and let $L$ be an
integer greater than $\|c\|$. Put
\[
w=\sum_{i=1}^L[(c-i)_+]\in\Cu(B).
\]
For every lower-semicontinuous extended trace $\tau$, writing
$\varphi=d_\tau$, one has
\begin{equation}
\label{ext:eq:rounding}
\varphi(w)\le\tau(c)\le\varphi(w)+d_\tau(c).
\end{equation}
\end{lemma}
\begin{proof}
For $q(t)=\min(t,1)$ on $[0,\infty)$, scalar functional calculus
gives the exact identity
\begin{equation}
\label{ext:eq:staircase}
c=\sum_{i=1}^L q((c-(i-1))_+).
\end{equation}
Set $R=\max(1,\|c\|)$. For every integer $m\ge1$,
\[
\big((c-i)_+/R\big)^{1/m}\le q((c-(i-1))_+).
\]
Indeed, below the spectral threshold $i$ the left side is zero,
and above it the right side is one. Consequently,
\[
d_\tau((c-i)_+)
\le\tau(q((c-(i-1))_+))
\le d_\tau((c-(i-1))_+).
\]
The last inequality uses Equation~\eqref{eq:pre:trace-rank} and
hereditary containment. Summing and using $(c-L)_+=0$ proves
Equation~\eqref{ext:eq:rounding}. Only the first rank $d_\tau(c)$
remains as error; the number $L$ of spectral levels does not occur
in the error bound.
\end{proof}
We now combine the two lemmas. If $M[h]\le[e]$ and $c$ is obtained
by trace transfer into $\her(h)$, then $M[c]\le[e]$.
For a functional with $\varphi([e])<\infty$, rounding therefore
approximates $\tau(r(a))$ to within $\varphi([e])/M$.
A single cut-supported function does not recover the full rank of an
arbitrary element. In the next section we write that rank as a sum of
such trace values and choose the support ranks so that the errors are
summable. Neither the norm of $c$ nor the number of its spectral levels
has to be bounded uniformly.
==== END Cuntz-comparison-and-Jiang-Su-absorption//rounding.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//shrinking.tex ====
\section{Shrinking two hereditary supports}
\label{shr:section}
We will shrink two hereditary supports so that a prescribed
cross-compression decreases by a fixed factor, while at least one new
support remains nonzero in each irreducible representation where an
original element has norm at least $1/2$. Repeated use of this operation
will separate the supports
in the norm-cone construction of Section~\ref{norm:section}.
Only the two Cuntz-semigroup properties in \textup{(C)} are needed;
nuclearity is not used here. All
Cuntz comparisons in hereditary subalgebras are understood in the ambient
stabilization, with their witnesses compressed into the indicated corners
as in the preliminaries. If $\sigma$ is a representation, its matrix
amplifications will also be denoted by $\sigma$.
\begin{lemma}[Two-support shrinking]\label{lem:shrinking}
There is a numerical constant $0<c<1$ with the following property for
every $C^*$-algebra $A$ satisfying \textup{(C)}. If $a,b\in A$ are positive
contractions and $z\in\widetilde A$ satisfies
$\|p_a z p_b\|\le1$, there are positive contractions
$a'\in\her(a)$ and $b'\in\her(b)$ such that
\begin{equation}\label{shr:conclusion}
\|p_{a'}z p_{b'}\|\le c
\end{equation}
and, for every irreducible representation $\sigma$ of $A$,
\begin{equation}\label{shr:visibility}
\max\{\|\sigma(a)\|,\|\sigma(b)\|\}\ge\tfrac12
\quad\Longrightarrow\quad
\sigma(a')\ne0\ \text{or}\ \sigma(b')\ne0.
\end{equation}
The support norms are computed in the bidual, or equivalently in the
universal representation.
\end{lemma}
We first prove a transfer statement for two hereditary supports. It
replaces finitely many rectangular elements by one contraction which
has norm one in every representation where a prescribed element reaches
a fixed norm threshold.
\begin{lemma}[Transfer between two supports]\label{lem:transfer}
Suppose that $A$ satisfies \textup{(C)}. Let $B\subseteq A$ be hereditary,
let $D\in M_l(B)_+$ and $E\in B_+$, and suppose that
$w_1,\ldots,w_m\in M_{l,1}(B)$ satisfy
\[
w_iw_i^*\in\her(D),\qquad w_i^*w_i\in\her(E).
\]
For each $\delta>0$ there is a contraction $v\in M_{l,1}(B)$ such that
\[
vv^*\in\her(D),\qquad v^*v\in\her(E),
\]
and, in every representation $\sigma$ of $B$,
\[
\max_{1\le i\le m}\|\sigma(w_i)\|\ge\delta
\quad\Longrightarrow\quad \|\sigma(v)\|=1.
\]
\end{lemma}
\begin{proof}
Set $T_0=\operatorname{diag}(w_1^*w_1,\ldots,w_m^*w_m)$. The hereditary
conditions and $[w_i^*w_i]=[w_iw_i^*]$ give
\[
[T_0]\le m[E],\qquad [T_0]\le m[D].
\]
By divisibility in \textup{(C)}, choose a positive element $r$ in a
stabilization with
\[
(m+1)[r]\le[T_0]\le(m+2)[r].
\]
Almost unperforation, applied with $m$, gives $[r]\le[E]$ and
$[r]\le[D]$. Choose $\eta>0$ sufficiently small that
\begin{equation}\label{shr:transfer-cut}
[(T_0-\delta^2/2)_+]\le(m+2)[(r-2\eta)_+].
\end{equation}
This is possible because the class on the left is compactly contained in
$[T_0]$, and the classes $(m+2)[(r-2\eta)_+]$ increase to $(m+2)[r]$.
The cut-placement result in the preliminaries supplies partial isometries
$u_E,u_D$ which identify $\her((r-\eta)_+)$ with hereditary subalgebras of
$\her(E)$ and $\her(D)$, respectively. Their initial projection is the
support of $(r-\eta)_+$. Choose a continuous function $j$ with
$0\le j\le1$, $j=0$ on $[0,\eta]$, and $j=1$ on $[2\eta,\infty)$.
Then
\[
v=u_D j(r)u_E^*
\]
belongs to $M_{l,1}(B)$: factor it as
$(u_Dj(r)^{1/2})(u_Ej(r)^{1/2})^*$. Each factor belongs to the ambient
stabilized algebra by cut placement, and the product's left and right
supports lie in the prescribed finite matrix corners. It is a contraction, and
\[
v^*v=u_Ej(r)^2u_E^*\in\her(E),\qquad
vv^*=u_Dj(r)^2u_D^*\in\her(D).
\]
If $\|\sigma(w_i)\|\ge\delta$ for some $i$, then
$\sigma((T_0-\delta^2/2)_+)\ne0$. By
\eqref{shr:transfer-cut}, the source copy
$u_E(r-2\eta)_+u_E^*$ is nonzero in this representation: otherwise the
comparison would make the represented left side zero. The operator
$\sigma(v^*v)$ is the identity on the support of that nonzero copy, since
$j$ is one on its spectrum. Thus $\|\sigma(v)\|=1$.
\end{proof}
\begin{proof}[Proof of Lemma~\ref{lem:shrinking}]
Suppose that no such constant exists. Fix numbers $c_n\in(0,1)$ increasing
to $1$. For each $n$, choose an algebra satisfying \textup{(C)} and data
$a,b,z$ for which the conclusion fails with $c=c_n$. The algebras are
allowed to vary; we suppress $n$ until the limiting argument. The
construction below always gives \eqref{shr:conclusion}, so failure will
provide a representation in which both constructed positives vanish.
\medskip
\noindent\textbf{Step 1: a matrix corner and two prospective supports.}
Apply Lemma~\ref{lem:placement}, with seven slots and a cut of $a$ below
$1/2$, to obtain a c.p.c.\ order-zero map $M_7\to\her(a)$ with weight $d$
satisfying
\begin{equation}\label{shr:d-visibility}
\|\sigma(a)\|\ge\tfrac12\quad\Longrightarrow\quad\|\sigma(d)\|=1.
\end{equation}
Its supporting matrix units identify
\[
\her(d)=M_7(B),\qquad d=\operatorname{diag}(P),
\]
where $B$ is the hereditary algebra generated by its first diagonal
weight and $P\in B$ is a strictly positive contraction. Concretely, in
any representation, the support space of $d$ is identified with
$\mathcal H^{\oplus7}$, where $B$ acts nondegenerately on $\mathcal H$.
Choose $Q=\min(1,2b)$ by functional calculus. Then $Q\in\her(b)$ is a
positive contraction and
\begin{equation}\label{shr:Q-visibility}
\|\sigma(b)\|\ge\tfrac12\quad\Longrightarrow\quad\|\sigma(Q)\|=1.
\end{equation}
We shall choose contractions
$v_0\in M_{2,1}(B)$ and $v_1\in M_{3,1}(B)$. For any such choice set,
in multiplier notation,
\begin{equation}\label{shr:U}
U=\begin{pmatrix}
\frac12\,1\\[2pt] \frac12 v_0\\[2pt] \frac12 v_1\\[2pt]
\bigl(\frac34\,1-\frac14v_0^*v_0-\frac14v_1^*v_1\bigr)^{1/2}
\end{pmatrix},\qquad
L=UPU^*\in M_7(B),\qquad t=\tfrac12.
\end{equation}
The expression under the square root is at least $\frac14\,1$, and
$U^*U=1$. Define
\begin{equation}\label{shr:ab-prime}
a'=(L-t)_+,\qquad
b'=Q\bigl[Q(c_n^2t-z^*Lz)Q\bigr]_+Q.
\end{equation}
These belong to $\her(a)$ and $\her(b)$, respectively. They are
contractions: $0\le L\le1$, and the selfadjoint operator inside the positive
part in the second expression is bounded above by $c_n^2t\,1$.
By Lemma~\ref{lem:support}, the compression of $z^*Lz$ to the support of
$b'$ is at most $c_n^2t$. Since $L\ge t p_{a'}$, this implies
\[
t p_{b'}z^*p_{a'}z p_{b'}
\le p_{b'}z^*Lz p_{b'}\le c_n^2t p_{b'},
\]
which proves \eqref{shr:conclusion} with $c_n$, for every choice of the two
columns.
\medskip
\noindent\textbf{Step 2: data chosen before either column.}
Put $Z=p_d zQ$ and write its rows in groups of sizes $1,2,3,1$ as
$Z_{\mathrm f},Z_0,Z_1,Z_{\mathrm{last}}$. These are operators from the
original representation space to the corresponding copies of
$\mathcal H$. The hypothesis on $z$ gives
\begin{equation}\label{shr:Z-bound}
Z^*Z\le Q^2,
\end{equation}
because $p_d\le p_a$ and $p_bQ=Q$. Thus $Z$ has norm at most one, even
though no uniform bound on $z$ itself is assumed.
The next choices address the only way the prospective supports can fail
to retain visibility. For any choice of the columns, a representation
in which $a'=b'=0$ would have $P\le t$ and, by the zero case of
Lemma~\ref{lem:support},
\begin{equation}\label{shr:alignment-chain}
c_n^2tQ^2\le Z^*UPU^*Z\le tZ^*Z\le tQ^2.
\end{equation}
As $c_n\to1$, these inequalities force the limiting range of $Z$ into
the range of $U$, with its $U$-coordinate in the $t$-spectral subspace
of $P$. Step~5 proves this alignment precisely. Since the first row
of $U$ is $1/2$, the first row of $Z$ then recovers that coordinate.
The following cutoffs implement this recovery within the hereditary
corner, before either column has been chosen.
Fix continuous functions on $[0,1]$ by
\[
g(x)=\min\{1,\max\{0,8x-2\}\},\qquad G(x)=\min\{1,4x\}.
\]
They vanish at zero, satisfy $Gg=g$, and $g(t)=G(t)=1$. Set
\begin{equation}\label{shr:XY}
X=2g(P)Z_{\mathrm f},\qquad h_0=XX^*,\qquad
Y_i=2\operatorname{diag}(G(P))Z_iX^*\quad(i=0,1).
\end{equation}
Here $h_0\in B_+$, $Y_0\in M_{2,1}(B)$, and
$Y_1\in M_{3,1}(B)$. For example, their entries are the appropriate
entries of sandwiches of $zQ^2z^*$ by $g(d)$ and $G(d)$; these sandwiches
belong to $\her(d)$. We have
\begin{equation}\label{shr:initial-bounds}
\|X\|\le2,\quad \|h_0\|\le4,\quad \|Y_i\|\le4,
\qquad G(P)h_0=h_0=h_0G(P).
\end{equation}
In particular, $G(P)$ acts as the identity on the support of $h_0$.
None of these data depends on $v_0$ or $v_1$.
For later use fix
\[
r=\tfrac18,\qquad
W_j=r^{-1}(Y_0)_j^*\quad(j=1,2),\qquad
f=f(h_0),\quad f(x)=\min\{1,\max\{0,4x-1\}\}\quad(x\ge0).
\]
Thus $f$ vanishes on $[0,1/4]$ and is one on $[1/2,\infty)$.
With $u=W_1f$ and $c_f=(1-f^2)^{1/2}$, put $V=u+c_f$ in
$\widetilde B$ and define
\begin{equation}\label{shr:T}
T=1-VV^*=f^2-uu^*-u c_f-c_f u^*\in B.
\end{equation}
This selfadjoint element records the prospective range complement of
$V$. At this stage $V$ need not be an isometry and $T$ need not be
positive. Step~7 will produce a reducing subspace in the failure limit
on which $V$ is an isometry and $T$ is a nonzero projection. We use
the positive spectral part of $T$ now to prepare the second column,
which will rule out that possibility.
The bounds above give
$\|W_j\|\le32$, $\|u\|\le32$, and $\|T\|\le1089$. Fix on $\R$ the
continuous functions
\[
\beta(x)=\min\{1,\max\{0,2x-1\}\},\qquad
\chi(x)=\min\{1,\max\{0,4x-1\}\}.
\]
They vanish at zero, $\beta$ vanishes on $(-\infty,1/2]$,
$\beta(1)=1$, and $\chi\beta=\beta$. Put
\begin{equation}\label{shr:BT-y}
b_T=\beta(T),\qquad B_T=\her_B(b_T),\qquad
y=\operatorname{diag}(b_T)Y_1b_T\in M_{3,1}(B_T),\qquad
S=\operatorname{diag}(b_T)\in M_3(B_T).
\end{equation}
We have $\|y\|\le4$, and $S$ is strictly positive in $M_3(B_T)$.
\medskip
\noindent\textbf{Step 3: the second column and bounded comparison witnesses.}
Apply Lemma~\ref{lem:placement} with two slots inside
$\her((yy^*-10^{-3})_+)$, using the narrower cut
$(yy^*-10^{-2})_+$. The plateau of the slot weights may be chosen equal
to one on the transported narrower element. To use the contraction
form of that Lemma explicitly, take
$b=(yy^*-10^{-3})_+/16$ and $\epsilon=9\cdot10^{-3}/16$;
positive rescaling preserves all the hereditary algebras and Cuntz
classes in question. We obtain equivalent
orthogonal positive contractions $k,k'\in M_3(B_T)$ with
\begin{equation}\label{shr:k-comparison}
k,k'\in\her((yy^*-10^{-3})_+),\qquad
[(yy^*-10^{-2})_+]\le3[(k-0.9)_+].
\end{equation}
If this hereditary algebra is zero, take $k=k'=0$; the same assertions
hold. Choose three matrices $x_j\in M_3(B_T)$ witnessing
\[
\left\|(yy^*-0.01)_+-\sum_{j=1}^3x_j^*(k-0.9)_+x_j\right\|<0.005.
\]
There are three matrices because the right side of the comparison is a
threefold direct sum; rectangular witnesses can be split into its three
blocks. Replace $x_j$ by
\[
q_j=\left(\frac{(k-0.9)_+}{(k-0.8)_+}\right)^{1/2}x_j,
\]
where the quotient is defined to be zero when its denominator is zero.
This is a continuous function of $k$. Then
\begin{equation}\label{shr:bounded-witnesses}
\left\|(yy^*-0.01)_+-\sum_{j=1}^3q_j^*(k-0.8)_+q_j\right\|<0.005,
\qquad \|q_j\|\le\sqrt{160}<13.
\end{equation}
Indeed the compression sum before replacement is positive and has norm
less than $16$, so each summand has norm less than $16$. Also
\[
\frac{(k-0.9)_+}{(k-0.8)_+}\le10(k-0.9)_+.
\]
These give $q_j^*q_j\le10x_j^*(k-0.9)_+x_j$ and the stated bound. The
positive gap between the two cuts is what makes these witnesses uniformly
bounded.
Choose $p'=\min(1,k/0.8)$ and put
\[
D'=(1-p')S(1-p')\in M_3(B_T)_+.
\]
We claim that
\begin{equation}\label{shr:Dprime-comparison}
3[b_T]=[S]\le[k]+[D']\le2[D'].
\end{equation}
For the first inequality, sandwich
$\frac12\,1\le(p')^2+(1-p')^2$ by $S^{1/2}$. The resulting two summands
have the classes of $p'Sp'$ and $(1-p')S(1-p')$, and
$p'Sp'\in\her(k)$. For the second inequality it suffices to check that
$k'\in\her(D')$, since $[k]=[k']$. Orthogonality gives
$(1-p')k'(1-p')=k'$. Since $S$ is strictly positive, there are positive
elements $c_j\to k'$ with $c_j\le C_jS$ for finite constants $C_j$:
one may take $c_j=e_jk'e_j$ with
$e_j=S(S+1/j)^{-1}$. Consequently
\[
(1-p')c_j(1-p')\le C_jD',\qquad
(1-p')c_j(1-p')\longrightarrow k',
\]
which proves the inclusion and \eqref{shr:Dprime-comparison}.
Almost unperforation now gives $[b_T]\le[D']$. Cut placement supplies a
column $v_1\in M_{3,1}(B_T)$ with
\begin{equation}\label{shr:v1}
v_1^*v_1=(b_T-1/4)_+,\qquad
v_1v_1^*\in\her(D'),\qquad \|v_1\|\le1.
\end{equation}
Since $(k-0.8)_+(1-p')=0$, its range condition implies
\begin{equation}\label{shr:v1-annihilation}
(k-0.8)_+v_1=0.
\end{equation}
We shall also retain the following fixed continuous support identities:
\begin{equation}\label{shr:fixed-supports}
(10^{-3}-yy^*)_+k=0,\qquad
\chi(T)(v_1)_j=(v_1)_j=(v_1)_j\chi(T)\quad(1\le j\le3).
\end{equation}
The first follows from \eqref{shr:k-comparison}; the second holds because
$\chi(T)b_T=b_T$ and the entries of $v_1$ belong to $B_T$. These equations,
unlike statements about varying support projections, pass directly to an
operator ultraproduct.
\medskip
\noindent\textbf{Step 4: finite word tests and the first column.}
Let $p$ be the continuous function of $Y_0Y_0^*$ which is zero up to
$r^2$, linear between $r^2$ and $(r+1/n)^2$, and one thereafter. Put
\begin{equation}\label{shr:DE}
s=\operatorname{diag}((h_0-1/4)_+)\in M_2(B),\qquad
E=(h_0-1/2)_+\in B,\qquad
D=s\bigl[s(\tfrac14-p)s\bigr]_+s\in M_2(B).
\end{equation}
Choose an ordered alphabet consisting of all entries of
\begin{equation}\label{shr:alphabet}
P,h_0,Y_0,Y_1,p,s,E,D,T,b_T,y,k,k',D',v_1,q_1,q_2,q_3
\end{equation}
and their adjoints. The number of letters is independent of $n$, as all
matrix sizes are fixed. Their norms are bounded independently of the
instance: the bounds already given apply, $\|s\|,\|E\|\le4$,
$\|D\|\le256$, and $p,k,k',b_T,D',v_1$ are contractions. Thus, for
example, $1089$ is a common bound on every letter. Neither $v_0$ nor $U$
is a letter.
Apply Lemma~\ref{lem:transfer} to the finite list of columns
\begin{equation}\label{shr:word-tests}
D\mathbf e_jxE,\qquad j=1,2,\quad
x\text{ a word of length at most }n,
\end{equation}
including the empty word $x=1$, with threshold $1/n$. Here
$\mathbf e_j$ is the $j$th scalar basis column. The left and right squared
moduli of these columns lie in $\her(D)$ and $\her(E)$, respectively.
The resulting contraction $v_0\in M_{2,1}(B)$ satisfies
\begin{equation}\label{shr:v0}
v_0v_0^*\in\her(D),\qquad v_0^*v_0\in\her(E),\qquad
v_0^*pv_0\le\tfrac14v_0^*v_0.
\end{equation}
The last assertion is Lemma~\ref{lem:support} applied to the pencil
defining $D$. In addition, we have the exact operator identities
\begin{equation}\label{shr:v0-fixed}
(1/2-h_0)_+v_0^*=0,\qquad
\operatorname{diag}(G(P))v_0=v_0.
\end{equation}
For the first use the right hereditary condition. For the second, the
support of $D$ is contained in that of $s$, and $G(P)$ fixes the support
of $h_0$ by \eqref{shr:initial-bounds}. The choice of $v_1$, all the
comparison witnesses, and the whole alphabet precedes the choice of
$v_0$, so the construction is not circular.
\medskip
\noindent\textbf{Step 5: failure forces alignment in a Hilbert ultraproduct.}
Make the choices above at coordinate $n$ and define $a',b'$ by
\eqref{shr:ab-prime}. Since they satisfy the required angle bound, the
assumed failure gives an irreducible representation $\sigma_n$ such that
\[
\max\{\|\sigma_n(a)\|,\|\sigma_n(b)\|\}\ge\tfrac12,
\qquad \sigma_n(a')=\sigma_n(b')=0.
\]
Use the same symbols for the represented operators. The first vanishing
gives $L\le t$, hence $P=U^*LU\le t$. By
\eqref{shr:d-visibility}, the $a$-trigger is impossible. Therefore
$\|Q\|=1$ by \eqref{shr:Q-visibility}. The two vanishings also give
the conditional inequalities \eqref{shr:alignment-chain} established
in Step~2.
Pass along the fixed free ultrafilter to Hilbert-space ultraproducts of
the representation spaces and of their first matrix-coordinate spaces.
For clarity, such an ultraproduct is the completion of bounded sequences
of vectors modulo those whose norms tend to zero along the ultrafilter,
with inner product the ultralimit of the coordinate inner products.
Uniformly bounded operator sequences act on these spaces, respect sums,
products and adjoints, and have norm equal to the ultralimit of their
norms. The last assertion follows by choosing a nearly norm-maximizing
unit vector at each coordinate. Fixed continuous functional calculus is
preserved, by uniform polynomial approximation on a common spectral
interval. All operators used here are uniformly bounded: in addition to
the alphabet this includes $Q,Z,X,U,v_0$, by
\eqref{shr:Z-bound}, \eqref{shr:initial-bounds}, and \eqref{shr:U}.
We keep the same notation for their limits.
Equality holds throughout \eqref{shr:alignment-chain}. In particular
$Z^*Z=Q^2$ and
$Z^*(t1-UPU^*)Z=0$. Since
\[
t1-UPU^*=t(1-UU^*)+U(t-P)U^*
\]
is a sum of positive operators, both positive summands annihilate $Z$.
For $\xi=U^*Z$, it follows that
\begin{equation}\label{shr:alignment}
Z=U\xi,\qquad P\xi=t\xi,\qquad \xi^*\xi=Q^2.
\end{equation}
The first row of $U$ is $1/2$, so $X=2g(P)Z_{\mathrm f}=\xi$.
Consequently
\begin{equation}\label{shr:limit-Y}
\|h_0\|=1,\qquad
Y_i=\operatorname{diag}(G(P))v_i h_0,\qquad
Y_0=v_0h_0.
\end{equation}
The last equality uses the second identity in \eqref{shr:v0-fixed}.
\medskip
\noindent\textbf{Step 6: the moving cutoff makes every word test vanish.}
Although the function defining $p$ changes with $n$, $p$ itself is a
bounded alphabet variable. At each coordinate it commutes with
$R_0=Y_0Y_0^*$ and satisfies
\[
(r^2-R_0)_+p=0,\qquad
\|(1-p)(R_0-r^2)_+\|\le(r+1/n)^2-r^2.
\]
Thus in the limit
\begin{equation}\label{shr:moving-cutoff}
[p,R_0]=0,\qquad (r^2-R_0)_+p=0,\qquad
(1-p)(R_0-r^2)_+=0.
\end{equation}
The first two equations show that the support of $p$ lies in the spectral
subspace of $R_0$ for $[r^2,\infty)$, hence in the closure of the range of
$Y_0$, and therefore of $v_0$ by \eqref{shr:limit-Y}. On this last closed
range the inequality $v_0^*pv_0\le\frac14v_0^*v_0$ says that the quadratic
form of $p$ is at most one quarter of the squared norm. This follows
first on vectors $v_0\eta$, then by density; on the orthogonal complement
$p$ is zero. Hence
\begin{equation}\label{shr:p-small}
0\le p\le\tfrac14\,1.
\end{equation}
The last equation in \eqref{shr:moving-cutoff}, and invertibility of
$1-p$, now give $R_0\le r^2$. The other support equation then gives
\begin{equation}\label{shr:R0}
\|Y_0\|\le r,\qquad R_0p=r^2p.
\end{equation}
Every fixed word $x$ in the limiting alphabet satisfies
\begin{equation}\label{shr:all-word-tests}
D\mathbf e_jxE=0\qquad(j=1,2).
\end{equation}
Indeed, if such an operator had norm $\delta>0$, its coordinate norms
would be at least $\delta/2$ on an ultrafilter-large set. On an
ultrafilter-large subset we also have $n\ge\operatorname{length}(x)$ and
$1/n\le\delta/2$. The transfer condition used in
\eqref{shr:word-tests} would give $\|v_0\|=1$ in the ultraproduct.
The first equation in \eqref{shr:v0-fixed} puts the range of $v_0^*$ in
the $h_0\ge1/2$ spectral subspace. Inverting $h_0$ on that subspace,
with inverse norm at most two, gives
\[
\|v_0h_0\|\ge\tfrac12\|v_0\|=\tfrac12.
\]
This contradicts $Y_0=v_0h_0$ and \eqref{shr:R0}, since $r=1/8$.
\medskip
\noindent\textbf{Step 7: a nonzero defect projection on the word space.}
Let $\mathcal H$ denote the limiting single-coordinate Hilbert space and
set
\[
\mathcal L=\overline{\operatorname{span}}
\{xE\eta:x\text{ a word in the alphabet},\ \eta\in\mathcal H\}.
\]
The alphabet includes adjoints, so $\mathcal L$ reduces every letter and
the $C^*$-algebra they generate. Restrict letters to this space, and
matrix variables to its amplifications. Equation
\eqref{shr:all-word-tests} gives $D=0$ on $\mathcal L^{\oplus2}$.
Moreover $E\mathcal H\subseteq\mathcal L$, so the restriction retains
$\|E\|=1/2$, hence $\|h_0\|=1$. We do not assert that $v_0$ reduces
$\mathcal L$. All further calculations involve only alphabet operators
and their continuous functions. The needed range information from
$v_0$ has already become the identity
\begin{equation}\label{shr:W-range}
(1/2-h_0)_+W_j=0\qquad(j=1,2),
\end{equation}
which follows on the full limiting space from
$W_j=r^{-1}h_0(v_0)_j^*$ and \eqref{shr:v0-fixed}, and thus restricts to
$\mathcal L$.
By the zero case of Lemma~\ref{lem:support}, $D=0$ implies
$s(\frac14-p)s\le0$. Equation \eqref{shr:p-small} gives the reverse
inequality, so $s(\frac14-p)s=0$. Taking a positive square root shows
$(\frac14-p)s=0$, or $ps=s/4$. Together with \eqref{shr:R0}, this yields
\begin{equation}\label{shr:Rs}
R_0s=r^2s.
\end{equation}
Let $e=1_{(1/4,\infty)}(h_0)$ on $\mathcal L$. The range of $s$ is dense
in $(e\mathcal L)^{\oplus2}$, so \eqref{shr:Rs} implies
\[
(W_i e)^*(W_j e)=\delta_{ij}e\qquad(1\le i,j\le2).
\]
Thus $W_1,W_2$ restricted to $e\mathcal L$ are isometries with orthogonal
ranges. By \eqref{shr:W-range}, their ranges lie in the
$h_0\ge1/2$ subspace, where $f=1$.
It follows that $u=W_1f$ satisfies $u^*u=f^2$ and $c_fu=0$, hence also
$u^*c_f=0$. Consequently
\[
V^*V=f^2+c_f^2=1.
\]
Its range-complement projection is exactly
\begin{equation}\label{shr:T-projection}
1-VV^*=f^2-uu^*-uc_f-c_fu^*=T.
\end{equation}
Both the initial and final supports of $u$ are contained in $e$, and
$f=ef$. As $c_f$ commutes with $e$, formula \eqref{shr:T-projection}
shows $T=eTe$. In particular
\begin{equation}\label{shr:ThT}
Th_0T\ge\tfrac14 T.
\end{equation}
This projection is nonzero. Indeed $W_2f\ne0$ because
$\|h_0\|=\|f\|=1$ and $W_2$ is isometric on $e\mathcal L$, whereas
\[
V^*W_2f=fW_1^*W_2f+c_fW_2f=0.
\]
Its nonzero range is therefore contained in the range of $T$.
\medskip
\noindent\textbf{Step 8: the second column gives the contradiction.}
Since $T$ is now a projection, $b_T=\beta(T)=T$ and $\chi(T)=T$.
The fixed identities \eqref{shr:fixed-supports} give
\[
T(v_1)_j=(v_1)_j=(v_1)_jT.
\]
They hold after restriction because $v_1$ was included in the alphabet.
As $G(P)$ fixes the support of $h_0$ and $T=eTe$, it also fixes $T$;
thus $\operatorname{diag}(G(P))v_1=v_1$. Using
\eqref{shr:BT-y}, \eqref{shr:v1}, and \eqref{shr:limit-Y}, we obtain
\begin{equation}\label{shr:final-y}
y=v_1Th_0T,\qquad v_1^*v_1=\tfrac34 T,\qquad
y^*y=\tfrac34(Th_0T)^2\ge\tfrac{3}{64}T.
\end{equation}
In particular, $\|yy^*\|\ge3/64$.
The three uniformly bounded witnesses in
\eqref{shr:bounded-witnesses} survive both the ultraproduct and the
restriction, giving
\[
\left\|(yy^*-0.01)_+-\sum_{j=1}^3q_j^*(k-0.8)_+q_j\right\|\le0.005.
\]
Hence $q=(k-0.8)_+$ is nonzero: otherwise
$\|yy^*\|\le0.015<3/64$, contradicting \eqref{shr:final-y}.
On the other hand, \eqref{shr:v1-annihilation} and
\eqref{shr:final-y} imply $qy=0$. The first fixed support identity in
\eqref{shr:fixed-supports} puts the range of $k$, and hence of $q$, in
the spectral subspace of $yy^*$ for $[10^{-3},\infty)$. Therefore
\[
0=qyy^*q\ge10^{-3}q^2,
\]
forcing $q=0$. This contradiction proves the existence of the constant
$c$ and completes the proof.
\end{proof}
==== END Cuntz-comparison-and-Jiang-Su-absorption//shrinking.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//support-tools.tex ====
\section{Comparison and support tools under condition (C)}
\label{sec:support-tools}
Corollary~\ref{ext:cor:condition-C} has supplied condition (C) from
simple extended-functional comparison. We first complete the other
route, from the canonical Cuntz-semigroup map. The remaining lemmas turn
(C) into the hereditary supports, matrix cones, and bounded witnesses
used by the common absorption construction. They apply without
simplicity; exactness is needed for the trace comparison, and
separability enters the central-sequence constructions.
\subsection{The consequences of the canonical Cu map}
\begin{lemma}[Cuntz regularity]
\label{lem:C}
Under the hypothesis of Corollary~\ref{cor:canonical}, the semigroup
$\Cu(A)$ satisfies condition \textup{(C)}, namely
Equations~\eqref{eq:pre:au}--\eqref{eq:pre:div}.
\end{lemma}
\begin{proof}
The algebra \(A\otimes\Z\) and its stabilization absorb \(\Z\).
R{\o}rdam's Theorem~4.5~\cite{Rordam2004} makes their \(W\)-semigroup
almost unperforated. Transport through \(\Cu(\iota)\) proves
\eqref{eq:pre:au}.
For an integer $N\ge1$, R\o rdam's divisibility lemma
\cite[Lemma~4.2]{Rordam2004} supplies $z\in\Z_+$ such that
\[
N[z]\le[1_{\Z}]\le(N+1)[z].
\]
For \(x=[b]\), tensor these comparisons with \(b\). Comparison is
preserved by tensoring with a fixed positive element
\cite[Lemma~4.1]{Rordam2004}: tensor the witnesses and use a bounded
approximate unit on the other factor.
Thus
\[
N[b\otimes z]\le[b\otimes1_{\Z}]
\le(N+1)[b\otimes z].
\]
The middle class is exactly \(\Cu(\iota)(x)\). Pulling back the
outside class proves \eqref{eq:pre:div}.
\end{proof}
\subsection{Comparison from a rank gap}
The later construction will compare a defect with a matrix-cone weight
by proving a uniform gap between their ranks. The following form allows
traces that vanish on an ideal or take infinite values. Exactness turns
the rank inequalities into functional inequalities, and almost
unperforation supplies comparison through
Lemma~\ref{lem:comparison-convention}.
\begin{lemma}[Comparison from a uniform rank gap]
\label{lem:G}
Suppose \(A\) is exact and satisfies \eqref{eq:pre:au}. If \(v,w\in A_+\)
satisfy
\[
d_\tau(v)\le\tfrac12d_\tau(w)
\]
for every lower semicontinuous extended trace on \(A\), then
\(v\precsim w\).
\end{lemma}
\begin{proof}
Let $\varphi\in\F(\Cu(A))$. By
Equation~\eqref{ext:eq:trace-representation}, it is represented by an
extended trace on $A\otimes\K$. Restricting that trace to the original
corner and applying the hypothesis gives
\[
\varphi([v])\le\tfrac12\varphi([w])\le\varphi([w]).
\]
The inequality is strict whenever $0<\varphi([w])<\infty$.
Since $\Cu(A)$ is almost unperforated,
Lemma~\ref{lem:comparison-convention} applies and gives $[v]\le[w]$.
\end{proof}
\subsection{Placing divisible matrix cones}
Almost divisibility provides equal Cuntz classes below a prescribed
positive element. Cutting these classes lets us place the corresponding
matrix cone inside its hereditary algebra. The two forms below retain,
respectively, norm visibility in representations and a prescribed
fraction of the rank for bounded traces.
\begin{lemma}[Cone placement]
\label{lem:placement}
Assume \eqref{eq:pre:au}--\eqref{eq:pre:div}.
\begin{enumerate}
\item Let \(B\) be a hereditary subalgebra of a matrix algebra over
\(A\otimes\K\), let \(b\in B_+\) be a contraction, and let \(N\ge1\)
and \(\epsilon>0\). There is a c.p.c.\ order-zero map
\(\psi:M_N\to\her_B(b)\) such that
\[
\|\sigma(\psi(1))\|=1
\quad\text{whenever}\quad
\sigma((b-\epsilon)_+)\ne0.
\]
Its \(N\) diagonal weights are orthogonal and equivalent in
\(\her_B(b)\). Each diagonal weight, cut at \(1/2\), dominates a
positive class \(u_0\) for which
\[
[(b-\epsilon)_+]\le(N+1)u_0.
\]
\item Let \(C\) be isomorphic to a finite direct sum of hereditary
subalgebras as above. For \(b\in C_+\) a contraction, \(k\ge1\), and
\(0<q\le1\), there is a c.p.c.\ order-zero map
\(\psi:M_k\to\her_C(b)\), with weight \(t=\psi(1)\), such that
\begin{equation}
\label{eq:pre:P}
d_\sigma(t)\le q\,d_\sigma(b),\qquad
d_\sigma((t-\tfrac12)_+)
\ge(q/3)d_\sigma((b-\tfrac12)_+)
\end{equation}
for every bounded positive trace \(\sigma\) on \(C\).
\end{enumerate}
\end{lemma}
\begin{proof}
Choose \(u=[r]\) with \(N[r]\le[b]\le(N+1)[r]\).
By \eqref{eq:pre:cuts}, a sufficiently small \(\eta>0\) gives
\[
[(b-\epsilon)_+]\le(N+1)[(r-2\eta)_+].
\]
Lemma~\ref{lem:pre:cut-placement} places the hereditary algebra of
\(\operatorname{diag}((r-\eta)_+,\ldots,(r-\eta)_+)\), with \(N\)
diagonal entries, inside \(\her_B(b)\). In this hereditary matrix
algebra take the cone \(e_{ij}\mapsto e_{ij}\otimes g(r)\), where
\(0\le g\le1\), \(g=0\) on \([0,\eta]\), and \(g=1\) on
\([2\eta,\infty)\). Transport it by the polar isometry.
Every diagonal acts as one on the corresponding transported
\((r-2\eta)_+\), proving both its cut comparison and the norm-one
assertion. The off-diagonal entries give equivalence of the
orthogonal weights.
For the second assertion do this in each summand, with
\(N=\lceil k/q\rceil\) and \(\epsilon=1/2\), and restrict to the first
\(k\) slots. The complete \(N\)-slot weight lies in \(\her(b)\),
so equivalence and additivity of rank give
\[
d_\sigma(t)\le(k/N)d_\sigma(b)\le q\,d_\sigma(b).
\]
The retained narrow classes and the cut comparison give
\[
d_\sigma((t-\tfrac12)_+)
\ge\frac{k}{N+1}d_\sigma((b-\tfrac12)_+)
\ge(q/3)d_\sigma((b-\tfrac12)_+).
\]
All comparisons take place in the indicated hereditary algebra,
by Lemma~\ref{lem:pre:cut-placement}. Thus arbitrary bounded traces
on that algebra may be used, whether or not they extend boundedly
to \(A\). Summing the finitely many summands proves the claim.
\end{proof}
\subsection{A support compression}
To make an error operator small on a hereditary support, we compress its
complement below a chosen threshold and take the positive part. The
following observation gives the resulting bound on the support, without
requiring the error operator to commute with the compressing element.
\begin{lemma}
\label{lem:support}
Let \(K=K^*\) be a bounded operator on a Hilbert space, let \(s\ge0\)
be bounded, and let \(\lambda\in\R\). On the support of
\[
b=s[s(\lambda-K)s]_+s
\]
the compression of \(K\) is at most \(\lambda\).
If \(b=0\), then \(s(\lambda-K)s\le0\).
\end{lemma}
\begin{proof}
Put \(H=s(\lambda-K)s\). For a vector \(\xi\) in the positive spectral
space of \(H\),
\[
\langle(\lambda-K)s\xi,s\xi\rangle
=\langle H\xi,\xi\rangle\ge0.
\]
The range closure of \(sH_+^{1/2}\) is the support space of \(b\).
It is contained in the closure of the images \(s\xi\) just used.
Continuity of the quadratic form proves the compression bound.
If \(b=0\), then \(H_+^{1/2}s=0\). Since \(H_+\) is supported on
\(p_s\), and the range of \(s\) is dense in its support space, this implies
\(H_+=0\), as required.
\end{proof}
\subsection{Ultrapowers and bounded diagonalization}
Fix a free ultrafilter \(\omega\) on \(\N\). We write
\[
D_\omega=\ell^\infty(\N,D)/
\{(x_n):\lim_{n\to\omega}\|x_n\|=0\},
\]
and identify \(D\) with its constant copy. For separable \(A\),
\[
\Ann(A)=\{x\in A_\omega\cap A':xa=ax=0
\text{ for every }a\in A\}.
\]
An approximate unit of \(A\) gives the unit in
\((A_\omega\cap A')/\Ann(A)\).
We use c.p.c.\ lifting for separable nuclear domains
\cite[Theorem~3.10 and Corollary~3.11]{CE1976}; in particular maps from a matrix algebra can be lifted
to c.p.c.\ coordinate maps.
\begin{lemma}[Bounded diagonalization]
\label{lem:diagonal}
Suppose a countable family of norm-continuous algebraic and
functional-calculus conditions, with fixed coefficients in a norm
ultrapower, is approximately satisfiable on every finite subfamily.
Suppose each unknown has a fixed individual norm bound in these
finite solutions. Then the conditions can be satisfied exactly in
the ultrapower. Additional countably many commutator conditions and
conditions after multiplication by a fixed countable subset of the
algebra may be imposed in the same way.
\end{lemma}
\begin{proof}
For each \(j\), choose an ultrapower solution of the first \(j\)
conditions to accuracy \(1/j\), and choose bounded coordinate
representatives. Slightly enlarging each fixed bound once, and
truncating representatives, gives bounds independent of \(j\).
There is an \(\omega\)-large set on which these first \(j\)
coordinate conditions hold to accuracy \(2/j\). Replace these sets
by decreasing finite intersections, also intersecting the \(j\)-th
set with \(\{n\ge j\}\). They have empty intersection. At coordinate
\(n\), use the solution of the largest applicable level \(j\le n\),
with an arbitrary bounded choice when no level applies.
The chosen level tends to infinity along \(\omega\); each fixed
condition therefore has error tending to zero. The bounds ensure
that every unknown represents an element of the norm ultrapower.
Positivity or complete positivity can be preserved by choosing
positive or completely positive lifts at each level.
\end{proof}
We will also use the following reindexing observation. If a central
sequence is to commute with finitely many already chosen coordinate
values, then, at each output coordinate, those values are ordinary
elements of \(A\). Centrality supplies the required good set of old
indices. It can be intersected with finitely many other good sets
before choosing the new index. Any uniform coordinate predicate
retained in this selection continues to hold. In the nonunital
quotient, algebraic and unit defects are tested after multiplication
by a dense subset of \(A\); boundedness then extends the resulting
relations to all of \(A\).
==== END Cuntz-comparison-and-Jiang-Su-absorption//support-tools.tex ====
==== BEGIN Cuntz-comparison-and-Jiang-Su-absorption//tracial.tex ====
\section{Matrix cones large for bounded traces}
\label{sec:tracial}
From this section onward, unless stated otherwise, $A$ is a separable
nuclear $C^*$-algebra satisfying \textup{(C)}.
The matrix cones in this section are approximately central in tracial
$2$-norm. The construction uses a convex nuclear approximation, and its
estimates are independent of the number of maps in that approximation.
Throughout, $D\subseteq A$ is hereditary and $T_{\leq1}(D)$ denotes the
positive bounded traces on $D$ of norm at most one. We extend these traces
normally to $D^{**}$. On matrices we use the unnormalized extension
$\mathcal T=\Tr\otimes\tau$, and write
$\|x\|_{2,\tau}=\mathcal T(x^*x)^{1/2}$, also for rectangular matrices.
The traces need not be faithful. All estimates below hold simultaneously
for every $\tau\in T_{\leq1}(D)$.
\begin{proposition}[Bounded-trace matrix cones]\label{prop:bounded-trace}
Fix $k\geq2$ and put
\[
L=(20k)^2,\qquad C_k=1+4L.
\]
Let $F\subset D$ be a finite set of selfadjoint contractions, let
$\mathcal P\subseteq F$ consist of positive contractions, and let
$0<\delta<1$. For every $\varepsilon>0$ there is a \cpc order-zero map
$V:M_k\to D$ such that, for every $\tau\in T_{\leq1}(D)$,
\begin{align}
\sup_{\|x\|\leq1}\|[a,V(x)]\|_{2,\tau}
&<\varepsilon &&(a\in F),\label{tr:central-target}\\
\tau(aV(1))
&\geq \frac\delta3\tau(a)-C_k\delta^{3/2}-\varepsilon
&&(a\in\mathcal P).\label{tr:mass-target}
\end{align}
In particular, $k$ is fixed before the accuracy is chosen; no bound on the
number of nuclear approximation summands is required.
\end{proposition}
We first record the functional-calculus estimate used to regularize a
row whose operator norm need not be uniformly bounded.
\begin{lemma}[Lipschitz calculus in tracial $2$-norm]\label{tr:lipschitz}
Let $\mathcal M$ be a von Neumann algebra with a finite normal trace $T$,
let $H=H^*\in\mathcal M$, and let $B\in\mathcal M$. If $f$ is real-valued
and Lipschitz on an interval containing the spectrum of $H$, then
\[
\|[f(H),B]\|_{2,T}
\leq\operatorname{Lip}(f)\|[H,B]\|_{2,T}.
\]
Consequently, if $W$ is a finite row, $a=a^*$ and $m=m^*$ are compatible
contractions, and
\[
R=W(1+W^*W)^{-1/2},
\]
then $\|R\|\leq1$ and
\[
\|aR-Rm\|_{2,T}\leq\|aW-Wm\|_{2,T}.
\]
The constants do not depend on the size of the matrices or on $\|W\|$.
\end{lemma}
\begin{proof}
For a finite spectral resolution $H=\sum_j\alpha_j e_j$, the blocks
$e_iBe_j$ are orthogonal in $L^2(\mathcal M,T)$. Thus
\begin{align*}
\|[f(H),B]\|_{2,T}^2
&=\sum_{i,j}|f(\alpha_i)-f(\alpha_j)|^2\|e_iBe_j\|_{2,T}^2\\
&\leq\operatorname{Lip}(f)^2
\sum_{i,j}|\alpha_i-\alpha_j|^2\|e_iBe_j\|_{2,T}^2.
\end{align*}
Approximate an arbitrary $H$ in operator norm by finite spectral step
functions of $H$. The corresponding functions $f(H)$ also converge in
norm. Since $T(1)<\infty$, passage to $2$-norm proves the assertion.
This passage is made separately in each finite matrix algebra; the
resulting inequality has no factor involving its size or $T(1)$.
For the row assertion use
\[
H=\begin{pmatrix}0&W\\W^*&0\end{pmatrix},\qquad
B=\begin{pmatrix}a&0\\0&m\end{pmatrix},\qquad
f(t)=\frac{t}{\sqrt{1+t^2}}.
\]
This function has Lipschitz constant one, and
$f(H)=\left(\begin{smallmatrix}0&R\\R^*&0\end{smallmatrix}\right)$.
The squares of the two commutator norms in the preceding inequality are
respectively $2\|aR-Rm\|_{2,T}^2$ and
$2\|aW-Wm\|_{2,T}^2$.
\end{proof}
We shall also use the following description of a finite-dimensional
commutant. If $h\in D_+$ and
$\rho:E\to M(\her_D(h))$ is a unital homomorphism from a finite-dimensional
algebra, with unit $p_h$ in $D^{**}$, write
$E=\bigoplus_jM_{n_j}$ and let $v^{(j)}_{rs}$ be its represented matrix
units. Then
\begin{equation}\label{tr:commutant-corners}
\her_D(h)\cap\rho(E)'
\cong\bigoplus_j v^{(j)}_{11}\her_D(h)v^{(j)}_{11},\qquad
(b_j)_j\longmapsto\sum_{j,r}v^{(j)}_{r1}b_jv^{(j)}_{1r}.
\end{equation}
Indeed, commuting with the diagonal matrix units kills every unwanted
block, and commuting with the remaining matrix units identifies the
retained diagonal blocks by the displayed formula. Each corner on the
right is hereditary in $D$. Averaging over the compact unitary group
of $E$ gives a trace-preserving conditional expectation
\[
\mathbb E_\rho:p_hD^{**}p_h\longrightarrow
p_hD^{**}p_h\cap\rho(E)'.
\]
It preserves $\her_D(h)$ because the represented unitaries are
multipliers. It is contractive in tracial $2$-norm, by the triangle
inequality for the integral. If $b$ is in its range, then
$\tau(b\mathbb E_\rho(x))=\tau(bx)$ whenever these expressions are defined.
These observations allow us to apply Lemma~\ref{lem:placement} in the
commutant, simultaneously to all its bounded positive traces.
\begin{proof}[Proof of Proposition~\ref{prop:bounded-trace}]
We will combine small matrix cones chosen in the commutants of the
nuclear approximation maps. Their supports are chosen to make the
model and overlap errors summable, after which a regularized row gives
a contraction. A second support compression makes this row bounded
below on the retained support. The polar isometry of the compressed
row then transports an exact matrix cone to $D$ while retaining the
tracial mass needed for the stated estimate.
Fix $F,\mathcal P,k,\delta$. Our approximation parameter is
$0<u<1/4$. The notation $o_u(1)$ below always means a quantity tending
to zero uniformly over $T_{\leq1}(D)$ with this fixed data. It has no
dependence on the number of approximation summands.
\medskip\noindent
\textbf{The convex approximation and its weighted errors.}
The hereditary algebra $D$ is nuclear. The convex order-zero return
approximation is due to Hirshberg, Kirchberg, and White
\cite[Theorem~1.4]{HKW2012}; it is also part of the stronger
Brown--Carri\'on--White approximation theorem
\cite[Theorem~3.1]{BCW2019}. This supplies a finite-dimensional unital algebra
$E$, one \cpc map $\alpha:D\to E$, positive numbers
$\lambda_1,\ldots,\lambda_r$ with $\sum_l\lambda_l=1$, and \cpc
order-zero maps $\beta_l:E\to D$, such that
\[
\Phi=\sum_{l=1}^r\lambda_l\beta_l\alpha
\]
is as close to the identity as prescribed on a given finite set.
The integer $r$ may depend on $u$ without restriction. Put
\[
h_l=\beta_l(1),\quad p_l=p_{h_l},\quad
\beta_l(x)=h_l\rho_l(x),\quad
\mu_l=\rho_l\alpha,\quad\theta_l=h_l\mu_l.
\]
Here $\rho_l$ is the supporting homomorphism of the order-zero map
\cite{WZ2009}; it acts by multipliers of $\her_D(h_l)$.
In particular $h_l$ commutes with $\mu_l(D)$.
By approximating $F$, its squares, and a sufficiently late positive
contractive approximate-unit element $e\in D$, we arrange
\begin{align}
\left\|\sum_l\lambda_l(a-\theta_l(a))^2\right\|
&\leq u^4 &&(a\in F),\label{tr:mean-square}\\
\left\|a^{1/2}\left(1-\sum_l\lambda_lh_l\right)a^{1/2}\right\|
&\leq u^4 &&(a\in\mathcal P).\label{tr:unit-error}
\end{align}
For completeness, the Schwarz inequality gives
\[
0\leq\sum_l\lambda_l(a-\theta_l(a))^2
\leq a^2-a\Phi(a)-\Phi(a)a+\Phi(a^2),
\]
whose norm is at most $3v$ if both relevant approximation errors are
at most $v$. For the second estimate choose $e$ so that
$\|a^{1/2}(1-e)a^{1/2}\|\leq u^4/2$ for all targets, and approximate
$e$ within $u^4/2$. Since $\theta_l(e)\leq h_l$, this proves
\eqref{tr:unit-error}. All appearances of $1$ in this proof are allowed
in the bidual; no unit of $D$ is assumed.
Choose the following explicit continuous cutoff on $[0,1]$:
\[
g(t)=\begin{cases}
0,&t\leq1-2u,\\
(t-1+2u)/u,&1-2u<t<1-u,\\
1,&t\geq1-u.
\end{cases}
\]
Set $g_l=g(h_l)$ and $z_l=1_{(1-2u,1]}(h_l)$. Thus
$p_{g_l}\leq z_l\leq p_l$, and the scalar inequality
\begin{equation}\label{tr:g-loss}
1-g_l^4\leq u^{-1}(1-h_l)
\end{equation}
holds by functional calculus. With
\[
D_l=\sum_{a\in F}p_l(a-\mu_l(a))^2p_l,\qquad
B_u=2u^4+8u^2,
\]
we have, for every $a\in F$,
\begin{equation}\label{tr:model-square}
\sum_l\lambda_l\|(a-\mu_l(a))z_l\|_{2,\tau}^2\leq B_u,
\qquad
\sum_l\lambda_l\tau(z_lD_l)\leq |F|B_u.
\end{equation}
Indeed, the contribution of $a-\theta_l(a)$ is bounded by $u^4$
using \eqref{tr:mean-square} and the trace inequality
$\tau(z_lxz_l)\leq\tau(x)$ for $x\geq0$. On $z_l$,
$\|(\theta_l(a)-\mu_l(a))z_l\|\leq2u$.
Apply $\|x+y\|_2^2\leq2\|x\|_2^2+2\|y\|_2^2$ and
$\sum_l\lambda_l\tau(z_l)\leq1$.
Weighted Cauchy--Schwarz now gives, for $a\in\mathcal P$,
\begin{align}
\sum_l\lambda_l\tau(\mu_l(a)g_l^4)
&\geq\tau(a)-u^3-\sqrt{B_u},\label{tr:T1-lower}\\
\sum_l\lambda_l\tau(\mu_l(a)z_l)
&\leq\tau(a)+\sqrt{B_u}.\label{tr:T1-upper}
\end{align}
To verify the replacement of $\mu_l(a)$ by $a$, for example,
\[
\left|\sum_l\lambda_l\tau((\mu_l(a)-a)g_l^4)\right|
\leq
\left(\sum_l\lambda_l\|(\mu_l(a)-a)z_l\|_2^2\right)^{1/2}
\left(\sum_l\lambda_l\tau(g_l^8)\right)^{1/2}
\leq\sqrt{B_u}.
\]
The same argument applies to $z_l$. The lost $a$-mass in the first
inequality is at most $u^3$, by \eqref{tr:g-loss} and
\eqref{tr:unit-error}; in the second use $z_l\leq1$.
\medskip\noindent
\textbf{The first pencil: small supports and summable overlaps.}
For each $l$ put
\[
C_l=\her_D(h_l)\cap\rho_l(E)',\qquad
\mathbb E_l=\mathbb E_{\rho_l}.
\]
Construct maps $\psi_l:M_k\to C_l$ inductively, writing
$t_l=\psi_l(1)$. Once the earlier maps have been chosen, we penalize
both the model error and overlap with their weights by setting
\begin{align}
S_l&=\sum_{i<l}t_i^2,\qquad
K_l=\mathbb E_l\big(u^{-1}D_l+\delta^{-1/2}p_lS_lp_l\big),\nonumber\\
X_l&=g_l(p_l-K_l)g_l,\qquad
b_l=g_l(X_l)_+g_l.\label{tr:first-pencil}
\end{align}
These formulas define $b_l\in C_l$. To check membership before taking
positive parts, sandwich each term of $D_l$ by $g_l$. Terms involving
$a\in D$ then lie in $\her_D(h_l)$, and terms involving $\mu_l(a)$ lie
there by multiplier action. The same is true of $g_lS_lg_l$.
Averaging preserves this hereditary algebra and commutes with
multiplication by $g_l$. Hence $X_l$ belongs to it and to the
commutant, as does $b_l$.
The inequalities $X_l\leq p_l$ and $X_l\geq-(\|K_l\|+1)p_l$
imply $0\leq(X_l)_+\leq p_l$ by scalar functional calculus. Therefore
\begin{equation}\label{tr:first-pencil-bounds}
0\leq b_l\leq g_l^2\leq z_l,\qquad
b_l\geq g_l^4-g_l^2K_lg_l^2.
\end{equation}
The second inequality follows from $(X_l)_+\geq X_l$.
Lemma~\ref{lem:support}, applied to the pencil in
\eqref{tr:first-pencil}, gives
\begin{equation}\label{tr:first-support}
p_{b_l}K_lp_{b_l}\leq p_{b_l}.
\end{equation}
Use the tracial conclusion of Lemma~\ref{lem:placement} in $C_l$,
which has the form \eqref{tr:commutant-corners}, with
$q=\delta\lambda_l$ and positive contraction $b_l$.
It gives a \cpc order-zero map
$\psi_l:M_k\to\her_{C_l}(b_l)$ whose weight $t_l$ satisfies
\begin{equation}\label{tr:placement}
d_\sigma(t_l)\leq\delta\lambda_l d_\sigma(b_l),\qquad
d_\sigma((t_l-1/2)_+)
\geq\frac{\delta\lambda_l}{3}d_\sigma((b_l-1/2)_+)
\end{equation}
for every bounded positive trace $\sigma$ on $C_l$.
In particular, on restricting $\tau$ to $C_l$,
\begin{equation}\label{tr:support-budget}
\tau(p_{t_l})\leq\delta\lambda_l,\qquad
\sum_l\tau(p_{t_l})\leq\delta,\qquad \tau(S_l)\leq\delta.
\end{equation}
Here ranks are the traces of the corresponding support projections in
the bidual. As $p_{t_l}\leq p_{b_l}$, \eqref{tr:first-support} gives
\[
\sum_l\tau(t_l^2K_l)\leq\sum_l\tau(t_l^2)\leq\delta.
\]
We may remove $\mathbb E_l$ inside these traces because $t_l^2\in C_l$.
Both summands defining $K_l$ are positive. It follows that
\begin{align}
\sum_l\|(a-\mu_l(a))t_l\|_{2,\tau}^2
&\leq u\delta &&(a\in F),\label{tr:T2-model}\\
\sum_l\tau(t_l^2S_l)&\leq\delta^{3/2}.
\label{tr:T2-overlap}
\end{align}
These are unweighted sums: the individual support allowances
$\delta\lambda_l$ in \eqref{tr:support-budget} have absorbed the
convex coefficients. Their total is $\delta$, so increasing the number
of approximation summands does not enlarge either error bound.
We next obtain a lower bound for their high spectral cuts. Let
\[
s_l=1_{(1/2,1]}(t_l),\qquad e_l=1_{(1/2,1]}(b_l).
\]
Since $b_l$ is a contraction supported under $z_l$, scalar functional
calculus and \eqref{tr:first-pencil-bounds} give
\begin{equation}\label{tr:cut-lower}
e_l\geq2b_l-z_l\geq2g_l^4-2g_l^2K_lg_l^2-z_l.
\end{equation}
For $a\in\mathcal P$, the functional
\[
\sigma_{l,a}(c)=\tau(\mu_l(a)c)\qquad(c\in C_l)
\]
is a bounded positive trace. Indeed $\mu_l(a)$ is a positive
contraction commuting with $C_l$, so positivity and the trace identity
follow from those of $\tau$. Normality shows that
$d_{\sigma_{l,a}}(t_l)=\tau(\mu_l(a)p_{t_l})$, and gives the same
formula for every positive spectral cut. Thus
\eqref{tr:placement} applies to this single weighted trace and yields
$\tau(\mu_l(a)s_l)\geq(\delta\lambda_l/3)\tau(\mu_l(a)e_l)$.
Using \eqref{tr:cut-lower} and $0\leq\mu_l(a)\leq1$, we obtain
\begin{align}
\sum_l\tau(\mu_l(a)s_l)
&\geq \frac\delta3\sum_l\lambda_l
\tau(\mu_l(a)(2g_l^4-z_l))
-\frac{2\delta}{3}\sum_l\lambda_l\tau(g_l^2K_lg_l^2).
\label{tr:weighted-mass}
\end{align}
For the last term, remove the expectation against $g_l^4\in C_l$:
\begin{align*}
\sum_l\lambda_l\tau(g_l^2K_lg_l^2)
&=\sum_l\lambda_l\left(u^{-1}\tau(g_l^4D_l)
+\delta^{-1/2}\tau(g_l^4S_l)\right)\\
&\leq u^{-1}|F|B_u+\sqrt\delta.
\end{align*}
In the inequality use $g_l^4\leq z_l$, \eqref{tr:model-square}, and
$\tau(S_l)\leq\delta$. Combining this with
\eqref{tr:T1-lower}--\eqref{tr:T1-upper} proves
\begin{equation}\label{tr:T3}
\sum_l\tau(\mu_l(a)s_l)
\geq\frac\delta3\tau(a)-\delta^{3/2}-r_u
\qquad(a\in\mathcal P),
\end{equation}
where one may take the explicit error
\[
r_u=\frac\delta3\left(2u^3+3\sqrt{B_u}
+2|F|B_u/u\right)\longrightarrow0.
\]
\medskip\noindent
\textbf{Regularizing the combined row.}
Form matrices over $D$ and models in its bidual as follows:
\[
W=(t_1,\ldots,t_r),\quad D_0=\operatorname{diag}(t_l^2),\quad
m(a)=\operatorname{diag}(\mu_l(a)),\quad p=p_{D_0},
\]
\[
R=W(1+W^*W)^{-1/2},\qquad Q_0=R^*R,\qquad
G_0=D_0(1+D_0)^{-1}.
\]
Here $\mathcal T(p)\leq\delta$ by \eqref{tr:support-budget}.
Each $t_l$ commutes with $\mu_l(a)$, so \eqref{tr:T2-model} says
\[
\|aW-Wm(a)\|_{2,\tau}\leq\sqrt{u\delta}=:e_u.
\]
The diagonal of $W^*W-D_0$ is zero, and its $(i,j)$ entry off the
diagonal is $t_it_j$. Consequently
\begin{equation}\label{tr:off-diagonal-cost}
\|W^*W-D_0\|_{2,\tau}^2
=2\sum_l\sum_{i<l}\tau(t_l^2t_i^2)
=2\sum_l\tau(t_l^2S_l)\leq2\delta^{3/2}.
\end{equation}
Lemma~\ref{tr:lipschitz} gives the first of the following estimates;
the second follows by expanding the commutator of $R^*R$:
\begin{align}
\|aR-Rm(a)\|_{2,\tau}&\leq e_u,
&\|[Q_0,m(a)]\|_{2,\tau}&\leq2e_u.
\label{tr:regularized-central}
\end{align}
Finally, $Q_0=1-(1+W^*W)^{-1}$ and
$G_0=1-(1+D_0)^{-1}$. The resolvent identity and the contractivity of
both resolvents show that
\begin{equation}\label{tr:regularized-overlap}
\|Q_0-G_0\|_{2,\tau}^2\leq2\delta^{3/2}.
\end{equation}
This regularization has removed every occurrence of $\|W\|$ from the
estimates.
\medskip\noindent
\textbf{The second pencil in one matrix commutant.}
Write $B_0=\her_{M_r(D)}(D_0)$. Both $W^*W$ and $D_0$ belong to
$B_0$: for the former, use its entries $t_it_j$ and an approximate
unit obtained by functional calculus of $D_0$. Hence $Q_0,G_0\in B_0$.
Let
\[
\pi:M_k\longrightarrow M(B_0)
\]
be the diagonal of the supporting homomorphisms of $\psi_l$, whose
unit in the bidual is $p$. Its values commute with
$\operatorname{diag}(t_l)$ and with $m(a)$. To see the latter assertion,
recover each supporting homomorphism by bounded strong limits from
$t_l$ and $\psi_l(M_k)$, all of which commute with $\mu_l(a)$.
The multiplier assertion follows because its products with
$\operatorname{diag}(t_l)$ belong to $M_r(D)$; the same observation
shows that $m(a)$ acts by multipliers on $B_0$ and commutes with $p$.
The row $R$ is contractive, but its Gram operator $Q_0$ need not
commute with $\pi(M_k)$, so $R\pi(\cdot)R^*$ need not have order zero.
We use the small overlap cost in \eqref{tr:regularized-overlap} to find
a $\pi(M_k)$-invariant support on which $R$ is bounded below. The polar
isometry of the compressed row will then transport the matrix action
while the estimates below control the lost tracial mass.
Let $u_k,v_k\in M_k$ be the clock and shift unitaries. The map
\[
\mathbb E_\pi(X)=\frac1{k^2}\sum_{i,j=0}^{k-1}
\pi(u_k^iv_k^j)X\pi(u_k^iv_k^j)^*
\]
is the trace-preserving expectation onto the commutant of $\pi(M_k)$
in the corner with unit $p$. The usual matrix-unit calculation shows
that the average kills the nonconstant matrix coefficients, and fixes
the commutant. Since the identity conjugation is one of its $k^2$
positive summands,
\begin{equation}\label{tr:finite-average-bound}
X\leq k^2\mathbb E_\pi(X)\qquad(X\geq0).
\end{equation}
Put
\[
K=\mathbb E_\pi((Q_0-G_0)^2).
\]
Then $K\in B_0\cap\pi(M_k)'$, and
\begin{equation}\label{tr:second-cost}
\mathcal T(K)\leq2\delta^{3/2},\qquad
\|[m(a),K]\|_{2,\tau}\leq4e_u.
\end{equation}
For the commutator use $[m(a),G_0]=0$, $\|Q_0-G_0\|\leq1$,
\eqref{tr:regularized-central}, and the $2$-norm contractivity of the
expectation, which commutes with multiplication by $m(a)$.
Choose the fixed piecewise linear cutoff
\[
g'(t)=\begin{cases}
0,&t\leq1/10,\\
10t-1,&1/10<t<1/5,\\
1,&t\geq1/5,
\end{cases}
\qquad g'=g'(G_0),
\]
and, with $L=(20k)^2$, set
\begin{equation}\label{tr:second-pencil}
X'=g'(p-LK)g',\qquad c=g'(X')_+g'.
\end{equation}
As for the first pencil, $c\in B_0\cap\pi(M_k)'$ and
\begin{equation}\label{tr:second-pencil-bounds}
0\leq c\leq(g')^2\leq p,\qquad
c\geq(g')^4-L(g')^2K(g')^2,
\qquad p_c\leq1_{(1/10,\infty)}(G_0).
\end{equation}
Lemma~\ref{lem:support} gives $p_cKp_c\leq L^{-1}p_c$.
For $\xi\in p_c\mathcal H$, \eqref{tr:finite-average-bound} implies
\[
\|(Q_0-G_0)\xi\|^2
\leq k^2\langle K\xi,\xi\rangle
\leq\frac1{400}\|\xi\|^2.
\]
On the same space $G_0$ has compression at least $1/10$. Therefore
\begin{equation}\label{tr:polar-lower}
c^2\geq cQ_0c\geq\frac1{20}c^2.
\end{equation}
Furthermore $m(a)$ commutes with $g'$. Lemma~\ref{tr:lipschitz},
applied to the scalar function $t\mapsto t_+$ in
\eqref{tr:second-pencil}, yields the useful explicit bound
\begin{equation}\label{tr:c-commutator}
\|[m(a),c]\|_{2,\tau}
\leq L\|[m(a),K]\|_{2,\tau}\leq4Le_u.
\end{equation}
There remains a lower bound on the mass surviving this second pencil.
Set $s'=\operatorname{diag}(s_l)$. On $s_l$ one has $t_l>1/2$,
so $t_l^2/(1+t_l^2)>1/5$, and hence $g's'=s'$.
The scalar inequality $c^2\geq2c-p$, followed by
\eqref{tr:second-pencil-bounds}, gives
\[
s'c^2s'\geq2s'cs'-s'\geq s'-2Ls'Ks'.
\]
For $a\in\mathcal P$, $m(a)$ is a positive contraction commuting with
$s'$. Decomposing the trace along $s'$ and $p-s'$ shows that
$\mathcal T(m(a)c^2)\geq\mathcal T(m(a)s'c^2s')$;
the omitted diagonal contribution is nonnegative. Thus
\begin{align}
\mathcal T(m(a)c^2)
&\geq\mathcal T(m(a)s')-2L\mathcal T(K)\nonumber\\
&\geq\frac\delta3\tau(a)-(1+4L)\delta^{3/2}-r_u.
\label{tr:corrected-mass}
\end{align}
Here we used \eqref{tr:T3} and \eqref{tr:second-cost}. The argument
only uses the commutation of $m(a)$ with $s'$; it does not require
$c$ to commute with $s'$.
\medskip\noindent
\textbf{Polar correction and exact order zero.}
Put $d=cQ_0c$ and let $U$ be the polar partial isometry of the row $Rc$
in the bidual. Equation~\eqref{tr:polar-lower} implies
$p_d=p_c$, so $U^*U=p_c$. We claim that the row
\[
Y=Uc
\]
belongs to $M_{1,r}(D)$ and satisfies
\begin{equation}\label{tr:polar-intertwining}
\sup_{\tau\in T_{\leq1}(D)}\|aY-Ym(a)\|_{2,\tau}=o_u(1)
\qquad(a\in F).
\end{equation}
We give uniform estimates for this passage, since $r$ can grow with $u$.
For $\eta>0$ define the algebraic row
\[
Y_\eta=Rc(d+\eta)^{-1/2}c\in M_{1,r}(D).
\]
The domination $c^2\leq20d$ implies a factorization
$c=d^{1/2}x$ with $\|x\|\leq\sqrt{20}$. One elementary proof is to
define $x^*$ initially on the range of $d^{1/2}$ by
$x^*(d^{1/2}\xi)=c\xi$; the stated domination makes this well-defined
and bounded, and it extends by zero on the orthogonal complement.
Using $Rc=Ud^{1/2}$ now gives
\begin{equation}\label{tr:uniform-polar-norm}
\|Y-Y_\eta\|
\leq\sqrt{20}\sup_{t\geq0}
\sqrt t\left(1-\sqrt{\frac{t}{t+\eta}}\right)
\leq\sqrt{20\eta}.
\end{equation}
For the last scalar bound split at $t=\eta$: below it use $\sqrt t$,
and above it use
$1-\sqrt{t/(t+\eta)}\leq\eta/t$.
In particular the norm limit $Y$ lies in the rectangular algebra.
Both $Y$ and $Y_\eta$ have right support under $p$, so
\begin{equation}\label{tr:uniform-polar-two}
\|Y-Y_\eta\|_{2,\tau}\leq\sqrt{20\eta\delta}.
\end{equation}
This uses $\mathcal T(p)\leq\delta$, rather than a bound on the trace
of the full matrix unit.
At fixed $\eta$ the intertwining error for $Y_\eta$ tends to zero
uniformly with $u$. To spell this out, \eqref{tr:regularized-central}
and \eqref{tr:c-commutator} imply
\[
\|[m(a),d]\|_{2,\tau}\leq(8L+2)e_u.
\]
On $[0,1]$ the function $(t+\eta)^{-1/2}$ has norm at most
$\eta^{-1/2}$ and Lipschitz constant at most
$\tfrac12\eta^{-3/2}$. Expand $aY_\eta-Y_\eta m(a)$ through its four
factors and apply Lemma~\ref{tr:lipschitz}. This gives
\[
\|aY_\eta-Y_\eta m(a)\|_{2,\tau}
\leq\left((1+8L)\eta^{-1/2}
+(4L+1)\eta^{-3/2}\right)e_u.
\]
Together with \eqref{tr:uniform-polar-two}, this bounds the error for
$Y$ by the displayed right-hand side plus $2\sqrt{20\eta\delta}$.
First choose $\eta$ small and then $u$ small. This proves
\eqref{tr:polar-intertwining} without any restriction on $r$.
Define
\[
V(x)=Uc\pi(x)cU^*=Y\pi(x)Y^*\qquad(x\in M_k).
\]
These elements lie in $D$: the rows $Y_\eta\pi(x)$ belong to
$M_{1,r}(D)$ by multiplier action, and converge in norm to
$Y\pi(x)$ by \eqref{tr:uniform-polar-norm}.
The map is completely positive and contractive, since
$V(1)=Uc^2U^*\leq1$. It is exactly order zero. Indeed $c$ commutes
with $\pi(M_k)$, $U^*U=p_c$, and for orthogonal positive $x,y\in M_k$,
\[
V(x)V(y)=Uc^4\pi(xy)U^*=0.
\]
For a selfadjoint $a\in F$ set $\Delta_a=aY-Ym(a)$. Since $m(a)$
commutes with $\pi(M_k)$ and $\|Y\|\leq1$,
\[
[a,V(x)]=\Delta_a\pi(x)Y^*-Y\pi(x)\Delta_a^*,\qquad
\|[a,V(x)]\|_{2,\tau}\leq2\|x\|\|\Delta_a\|_{2,\tau}.
\]
This proves \eqref{tr:central-target} when $u$ is sufficiently small.
Finally $Y^*Y=c^2$ and $\|Y\|_{2,\tau}\leq\sqrt\delta$. Rectangular
traciality and Cauchy--Schwarz give
\[
\left|\tau(aV(1))-\mathcal T(m(a)c^2)\right|
=\left|\mathcal T(Y^*\Delta_a)\right|
\leq\sqrt\delta\,\|\Delta_a\|_{2,\tau}=o_u(1).
\]
Combining this with \eqref{tr:corrected-mass}, and choosing $u$ to make
$r_u+\sqrt\delta\sup_\tau\|\Delta_a\|_{2,\tau}<\varepsilon$ for the
finitely many targets, proves \eqref{tr:mass-target}. All choices were
made for the entire family $T_{\leq1}(D)$ at once, completing the proof.
\end{proof}
==== END Cuntz-comparison-and-Jiang-Su-absorption//tracial.tex ====